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188100 | https://mathcentral.quora.com/Prove-that-if-a-is-an-odd-integer-then-math-a-2-equiv-1-pmod-8-math | Something went wrong. Wait a moment and try again.
Math Central
Mathematics is the "Language of the universe"
Prove that if a is an odd integer, then
a
2
≡
1
(
mod
8
)
?
2 Answers
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Brad Ballinger
Math teacher
·
3y
Here are two ways of thinking about it.
First, a2−1=(a−1)(a+1). Since a is odd, a−1 and a+1 are consecutive even numbers, one of which must be a multiple of 4. When we multiply a multiple of 2 by a multiple of 4, we get a multiple of 8, so 8|a2−1.
Second approach. Every odd integer is congruent to -3, -1, 1, or 3 mod 8. Squaring these numbers, we find that odd squares are all congruent to 9, 1, 1, or 9 mod 8. But those are all 1.
Mohammad Afzaal Butt
B.Sc in Mathematics & Physics, Islamia College Gujranwala (Graduated 1977)
·
3y
Let a = 2n + 1 for some integer n, then
(2n+1)2=4n2+4n+1
⟹(2n+1)2−1=4n2+4n
⟹(2n+1)2−1=4n(n+1)
8|4n(n+1)∵n(n + 1) is a product of two consecutive integers and is even
⟹8|(2n+1)2−1
⟹(2n+1)2≡1(mod8)
⟹a2≡1(mod8)
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188101 | https://www.youtube.com/watch?v=8W1eSYiVHqU | The Secret Cycles in the Fibonacci Sequence
Combo Class
56600 subscribers
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Description
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Posted: 22 Jun 2023
Let me show you how the Fibonacci sequence has some cool patterns hiding within its last digits, and how this relates to other patterns it would have if we counted in different bases!
Here's a previous episode explaining more about how different "mods" work:
Here's a previous episode showing some cool links the Fibonacci sequence has with the "golden ratio":
I called this "secret" in the title because many people who know the Fibonacci sequence have never heard about this awesome hidden layer to it. However, to a number theorist, it's not a "secret", and you can look up "Pisano periods" if you want to find a lot more research that has been done about this topic.
Clarification: on the title cards at about 12 minutes, when I say 2 fibonacci numbers in a row that are the "same" last digit, I don't mean the two numbers in the pair have to be the same as each other, I mean that the pair has to be the same as a previous pair we've seen before.
I've been posting some "shorts" on this channel (6 so far) without sending them to subscriptions/notification feeds, so check those out here:
Also make sure you're tuned in to my @Domotro channel where I post bonus math videos and livestreams and stuff!
This was filmed by Carlo Trappenberg.
Special thanks to Evan Clark and to all of my Patreon supporters:
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If you want to mail me anything (such as any clocks/dice/etc. that you'd like to see in the background of Grade -2), here's my private mailbox address (not my home address). If you're going to send anything, please watch this short video first:
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If you want to try to help with Combo Class in some way, or collaborate in some form, reach out at combouniversity(at)gmail(dot)com
In case people search any of these terms, some of the topics discussed in this episode are: the Fibonacci sequence / Fibonacci numbers, modular arithmetic, the "Pisano periods" of the Fibonacci sequence for various different mods and its cycles related to powers of ten, how "last digits" of a number relate to its representations in different bases and different mods, etc...
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Transcript:
hey folks I'm dumotro and this is the classic Fibonacci Sequence which starts with a zero then a one then each following Fibonacci number is the sum of the previous two like one plus one is two one plus two is three and Etc and these will quickly grow to multi-digit numbers with all sorts of appearances but if we jump ahead in the sequence a little bit say to the 60th Fibonacci number an interesting pattern emerges I've highlighted the last digit of these Fibonacci numbers from the 60th up through 66th here red because they look very similar to the way the Fibonacci sequence itself began zero one one two three five eight zero one one two three five eight before it sort of overflows and the last digit didn't have enough room to contain that whole number but if we jump ahead even further say to the 300th Fibonacci number where I've just written the last four digits of these Fibonacci numbers from the 300 through 312 there now it seems that the last two digits are in on this pattern zero one one two three five eight thirteen Twenty One thirty four fifty five eighty nine the same as those Fibonacci numbers before it's almost like 144 but overflows in a way the last two digits couldn't contain that one and if you see this you might wonder a few things like is there a point in the Fibonacci sequence where there are three digits repeating giving us room for more Fibonacci numbers to come back and is this just a quirk of our base 10 way of writing numbers or what would happen in other bases let's start by seeing if the Fibonacci sequence has a pattern in terms of even and odd numbers it begins with an even zero and then an odd one and even without knowing these further numbers we can note that an even number plus an odd number always generates an odd number and then the next term which is determined by adding these two is an odd plus an odd which is always an even then an odd plus an even is always an odd and hey look we're sort of back where we started because this next one will be determined purely from even plus odd just like that previous term had been determined by an even plus odd and we can note that this part will loop as soon as we get back to a two in a row that was at the beginning there and if we write this in mod 2 language just writing the remainder a number would have if we divided by two where evens get a zero and odds get a one which is also the last digit these num members would have if written in binary the Fibonacci sequence will Loop zero one one zero one one forever so what about other mods or bases like how about mod 12. to see what patterns the last digit of the Fibonacci sequence might have in base 12. we can use clocks as an example a 12 hour clock like this and we would turn the twelves to zeros and the tens and elevens to single digit symbols if we were actually in base 12 and if we imagined going to an hour that was lined up with a particular Fibonacci number like if my Fibonacci number was 13 I would end up on the number one if I went that many hours from the top well looking at which hours would correspond to which fibon naughty numbers I would end up being at twelve one one two three five eight the first few are pretty predictable and then I would go to that 13 then around to 21 which would end up sending me to there and so on and eventually I would hit a certain cycle where I was back to the pattern I had done before here is the numbers that I would hit as hours playing that game and after 24 I would return to a Fibonacci number being at 12 one being at one and another one being out one a Fibonacci number being on the hour two and so on this tells us that in base 12 every 24 Fibonacci numbers the last digit repeats in a cycle we can also notice that the hour number six isn't anywhere in that cycle meaning that no Fibonacci numbers ever will end in the digit six if written in base 12 which also tells us that any Fibonacci number that's both thriven and even must be at least doubly even so if clocks can tell us things about base 12 or we could have gone for 24 or 60 using minutes or other interpretations of hours why don't we look at another common cycle in humanity that is weekdays where every seventh day you're back on the same weekday name like Sunday or Monday well if I called a certain one of these days zero for example Sunday and then I went to the weekday that each Fibonacci number could be associated with like if I had the Fibonacci number eight eight days after zero would send it to a Monday I would have a cycle there as well I would land on the zeroth day I I picked like Sunday here and Monday Monday Tuesday Wednesday Friday Monday and Etc and every 16 Fibonacci numbers I would be back to the same pattern of weekdays that I started on which tells us that in base seven every 16 Fibonacci numbers repeat their last digit in a cycle so we could already see that 7 and 12 as far as bases are even neater with this pattern of last digits than base 10 where the last digits Loop every 60 Fibonacci numbers which is why I use that example but we didn't just see the last digit here we also made the last two digits do that so let's jump into that what does it take for the last two digits of a number in base 10 to repeat well the last two digits cycle back around if you were just counting up them after 99 back to zero zero and every 100 numbers the last two digits if you were just counting up a number line would loop back around so just like the last digit of something in base 10 could be visualized like a 10 hour clock the last two digits could be visualized like a 100 hour clock like we were in what could be called mod 100 and if we look at how the Fibonacci sequence acts in mod 100 or if we were in base 100 then these would all be smooshed to smaller numbers and it would just be the last digit with 100 options cycling then it seems that at 300 we began hitting the same things and it's true that in mod 100 every 300 Fibonacci numbers we will repeat the same result meaning that every 300 Fibonacci numbers in base 10 we get at two digits repeating in a cycle now before we go further in the base 10 representations of Fibonacci numbers to see if there's a point where the last three digits repeat let's ask a question which is for a given base and a given amount of last digits will there always be a cycle or Loop of some point that it falls into well let's look at the last digit of a Fibonacci number in base B and note that the Fibonacci numbers purely determined by the sum of the last two and that the last digit will be purely determined by the last digits of the last two adding up because all the further places in these numbers are too big of spots to affect that one's place and so there's a finite amount of combinations of two final last digits in fact that one has B options in base B and so does that leaving us with B times b or B squared total possible combinations of two last digits in a row and once those create the next last digit that with the second of those will create a new and keep going with the exact same pattern we've seen before since each new one's place will be purely determined by the ones places of the last two numbers and there's a finite amount of combinations of those even if we wondered about the last two digits in a base B well now there are B squared amount of options for each of those for example in base 10 to look at all the possible last two digit strings from 0-99 there's 100 10 squared amount of possibilities and so the total amount of possibilities of two strings in a row would be B squared times B squared or B to the fourth power amount of two in a rows of last two digits in a base B that would be the only parts that could affect the new last two digits of the next number and so at the most and as we'll see mathematicians have proven that the Cycles aren't as big as these at the most I'm gonna show but at the most the last digit of a number in base B must follow B times B amount of possible options and must cycle after at most B squared Fibonacci numbers the last two place is are built off of B squared times B squared amount of options making them cycle at at most B to the fourth power amount of Fibonacci numbers and for three digits there would still be a finite amount of three digit ends there and three digit ends there thus a finite amount of combinations of the last two three digit ends that leave us forced into some sort of cycle for any amount of last digits in any base here are the cycles that the Fibonacci sequence Loops in in different mods and these are also the cycles of last digits that would repeat if we looked at the Fibonacci sequence in that given base and up in mod 10 or the last digit of base 10 we saw that this cycle was 60 in less and that we could also find a cycle of the last two digits that was 300 in length that's because the mod 100 Cycle is 300 numbers before it repeats and that could tell us about the last digit in some fictional base 100 that had digit symbols for everything from 0 to 99 but since 100 is 10 squared it can also tell us about two places worth in base 10. similarly if we looked at the cycle in mod 1000 which is 10 cubed that would be a cycle 1500 numbers in length and then it would loop back around through those numbers which could tell us about the last digit in some base 1000 or the last three digits in base 10. and we can continue looking at mod ten thousands pattern which is 15 000 looking kind of similar and in fact defined mod 100 000 which would give us five digits worth in base ten that's 150 000 and it's even been proven that Beyond this point all further periods for powers of ten which will tell us about further amounts of digits in base 10 just add a zero to how long the cycle is the next one will be one million five hundred thousand if you look at the cycles and how many numbers are in them for different mods other patterns emerge also like after mod 2 all further mods will have an even amount of numbers in their cycle and here are some other rates that mathematicians have proven about these Cycles in general it's pretty cool that you can take a sequence that spirals larger and larger and interpret it in these clock-like mods to find Cycles within it that tell you information about how we would look in a variety of different bases and the Fibonacci sequence isn't the only type of numbers we could play these games with in another episode before long we'll look at what happens when we take triangular numbers and interpret those in different mods like do those have cycles of their own alright folks thanks for joining me today to learn about some clock Like Pat oh God okay Carlo water also make sure you're tuned into my dimotro channel where I post a bunch of live streams and bonus videos and stuff and special thanks to my patreon supporters who helped make this show possible I'll see you guys in the next episode and I hope you have a great day |
188102 | https://www.education.com/resources/grade-3/math/multiplication/multiplication-word-problems/ | 3rd Grade Multiplication Word Problems Resources | Education.com
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Jumpy: Multiplication Word Problems Game Jumpy: Multiplication Word Problems Third Grade Multiplication Help Roly get across the river by solving multiplication word problems in this third-grade math game! Game
Multiplication Word Problems: Multiply It! Interactive Worksheet Multiplication Word Problems: Multiply It! Third Grade Multiplication Kids use details from the word problems on this third grade math worksheet to construct and solve multiplication problems. Interactive Worksheet
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Multiplication Word Problems Interactive Worksheet Multiplication Word Problems Third Grade Multiplication It's a multiplication wonderland! Review those times tables with these multiplication word problems. Interactive Worksheet
One-Digit Multiplication Word Problems Exercise One-Digit Multiplication Word Problems Third Grade Multiplication If students are new to word problems, this exercise provides great one digit multiplication introductory tasks. Exercise
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188103 | https://www.geeksforgeeks.org/java/different-ways-to-declare-and-initialize-2-d-array-in-java/ | Different Ways To Declare And Initialize 2-D Array in Java
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13 Nov, 2024
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An array with more than one dimension is known as a multi-dimensional array. The most commonly used multi-dimensional arrays are 2-D and 3-D arrays. We can say that any higher dimensional array is an array of arrays.
A very common example of a 2D Array is Chess Board. A chessboard is a grid containing 64 1x1 square boxes. You can similarly visualize a 2D array. In a 2D array, every element is associated with a row number and column number. Accessing any element of the 2D array is similar to accessing the record of an Excel File using both row number and column number. 2D arrays are useful while implementing a Tic-Tac-Toe game, Chess, or even storing the image pixels.
Example:
Java
````
import java.io.;
class GFG {
public static void main(String[] args){
int n = 80, m = 5;
int[][] arr = new int[n][m];
// initializing the array elements using for loop
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
arr[i][j] = i + j;
}
}
// printing the first three rows of marks array
for (int i = 0; i < 3; i++) {
for (int j = 0; j < m; j++)
System.out.printf(arr[i][j] + " ");
System.out.println();
}
}
}
````
import java.io.; import java io
class GFG {class GFG
public static void main(String[] args){public static void main String args
```
```
int n = 80, m = 5; int n = 80 m = 5
int[][] arr = new int[n][m]; int arr = new int n m
// initializing the array elements using for loop
// initializing the array elements using for loop
for (int i = 0; i < n; i++) {for int i = 0 i< n i ++
for (int j = 0; j < m; j++) {for int j = 0 j< m j ++
arr[i][j] = i + j; arr i j = i + j
}
}
// printing the first three rows of marks array
// printing the first three rows of marks array
for (int i = 0; i < 3; i++) {for int i = 0 i< 3 i ++
for (int j = 0; j < m; j++) for int j = 0 j< m j ++
System.out.printf(arr[i][j] + " "); System out printf arr i j + " "
System.out.println(); System out println
}
}
}
Output
```
0 1 2 3 4
1 2 3 4 5
2 3 4 5 6
```
Note: We can use arr. length function to find the size of the rows (1st dimension), and arr.length function to find the size of the columns (2nd dimension).
Declaring 2-D array in Java
Any 2-dimensional array can be declared as follows:
Syntax:
// Method 1
data_type array_name[][];
// Method 2
data_type[][] array_name;
data_type: Since Java is a statically-typed language (i.e. it expects its variables to be declared before they can be assigned values). So, specifying the datatype decides the type of elements it will accept. e.g. to store integer values only, the data type will be declared as int.
array_name: It is the name that is given to the 2-D array. e.g. subjects, students, fruits, department, etc.
Note: We can write [ ][ ] after data_type or we can write [ ][ ] after array_name while declaring the 2D array.
Initialize 2-D array in Java
data_type[][] array_Name = new data_type[row][col];
The total elements in any 2D array will be equal to (row) (col).
row: The number of rows in an array
col: The number of columns in an array.
When you initialize a 2D array, you must always specify the first dimension(no. of rows), but providing the second dimension(no. of columns) may be omitted. Java compiler is smart enough to manipulate the size by checking the number of elements inside the columns.
// Incorrect Statement
int[][] arr = new int[];
// Correct Statement
int[][] arr = new int[];
You can access any element of a 2D array using row numbers and column numbers.
Different Ways to Declare and Initialize 2-D Array in Java
1. Inserting Elements while Initialization
In the code snippet below, we have not specified the number of rows and columns. However, the Java compiler is smart enough to manipulate the size by checking the number of elements inside the rows and columns.
Java
````
import java.io.;
class GFG {
public static void main(String[] args)
{
// Create a String Array
String[][] subjects = {
// Row 1
{ "Data Structures & Algorithms",
"Programming & Logic",
"Software Engineering",
"Theory of Computation" },
// Row 2
{ "Thermodynamics",
"Metallurgy",
"Machine Drawing",
"Fluid Mechanics" },
// Row 3
{ "Signals and Systems",
"Digital Electronics",
"Power Electronics" }
};
// Printing the Array Spoecific Index
System.out.println(subjects);
System.out.println(subjects);
System.out.println(subjects);
}
}
````
import java.io.; import java io
class GFG {class GFG
public static void main(String[] args) public static void main String args
{
```
```
// Create a String Array
// Create a String Array
String[][] subjects = {String subjects =
// Row 1
// Row 1
{ "Data Structures & Algorithms", "Data Structures & Algorithms"
"Programming & Logic", "Programming & Logic"
"Software Engineering", "Software Engineering"
"Theory of Computation" }, "Theory of Computation"
// Row 2
// Row 2
{ "Thermodynamics", "Thermodynamics"
"Metallurgy", "Metallurgy"
"Machine Drawing", "Machine Drawing"
"Fluid Mechanics" }, "Fluid Mechanics"
// Row 3
// Row 3
{ "Signals and Systems", "Signals and Systems"
"Digital Electronics", "Digital Electronics"
"Power Electronics" } "Power Electronics"
};
```
```
// Printing the Array Spoecific Index
// Printing the Array Spoecific Index
System.out.println(subjects); System out println subjects 0 0
```
```
System.out.println(subjects); System out println subjects 1 3
```
```
System.out.println(subjects); System out println subjects 2 1
}
}
Output
```
Data Structures & Algorithms
Fluid Mechanics
Digital Electronics
```
2. Inserting Elements by Index
Moreover, we can initialize each element of the array separately. Look at the code snippet below:
Java
````
import java.io.;
import java.util.;
class GFG {
public static void main(String[] args){
// Creating and Initialization of
// Array
int[][] scores = new int;
// Inserting elements in the specific index
scores = 15;
scores = 23;
scores = 30;
scores = 21;
// Printing the array elements individually
System.out.print(scores+" ");
System.out.println(scores);
System.out.print(scores+" ");
System.out.println(scores);
}
}
````
import java.io.; import java io
import java.util.; import java util
class GFG {class GFG
public static void main(String[] args){public static void main String args
```
```
// Creating and Initialization of
// Creating and Initialization of
// Array
// Array
int[][] scores = new int; int scores = new int 2 2
```
```
// Inserting elements in the specific index
// Inserting elements in the specific index
scores = 15; scores 0 0 = 15
scores = 23; scores 0 1 = 23
scores = 30; scores 1 0 = 30
scores = 21; scores 1 1 = 21
// Printing the array elements individually
// Printing the array elements individually
System.out.print(scores+" "); System out print scores 0 0 + " "
System.out.println(scores); System out println scores 0 1
System.out.print(scores+" "); System out print scores 1 0 + " "
System.out.println(scores); System out println scores 1 1
}
}
Output
```
15 23
30 21
```
3. Inserting Elements in Jagged Array
There may be a certain scenario where you want every row to have a different number of columns. This type of array is called a Jagged Array.
Java
````
import java.io.;
class GFG {
public static void main(String[] args){
// declaring a 2D array with 2 rows
int arr[][] = new int[];
// Jagged array with custom
// columns for each row
arr = new int;
arr = new int;
// Initializing the array
int count = 0;
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[i].length; j++) {
arr[i][j] = count++;
}
}
// Printing the values of 2D Jagged array
for (int i = 0; i < arr.length; i++) {
for (int j = 0; j < arr[i].length; j++)
System.out.printf(arr[i][j] + " ");
System.out.println();
}
}
}
````
import java.io.; import java io
class GFG {class GFG
public static void main(String[] args){public static void main String args
```
```
// declaring a 2D array with 2 rows
// declaring a 2D array with 2 rows
int arr[][] = new int[]; int arr = new int 2
// Jagged array with custom
// Jagged array with custom
// columns for each row
// columns for each row
arr = new int; arr 0 = new int 2
arr = new int; arr 1 = new int 4
```
```
// Initializing the array
// Initializing the array
int count = 0; int count = 0
for (int i = 0; i < arr.length; i++) {for int i = 0 i< arr length i ++
for (int j = 0; j < arr[i].length; j++) {for int j = 0 j< arr i length j ++
arr[i][j] = count++; arr i j = count ++
}
}
// Printing the values of 2D Jagged array
// Printing the values of 2D Jagged array
for (int i = 0; i < arr.length; i++) {for int i = 0 i< arr length i ++
for (int j = 0; j < arr[i].length; j++) for int j = 0 j< arr i length j ++
System.out.printf(arr[i][j] + " "); System out printf arr i j + " "
System.out.println(); System out println
}
}
}
Output
```
0 1
2 3 4 5
```
G
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188104 | https://www.nobelprize.org/prizes/physics/1965/ceremony-speech/ | Skip to content
Award ceremony speech
Presentation Speech by Professor Ivar Waller, member of the Nobel Committee for Physics
Your Majesty, Your Royal Highnesses, Ladies and Gentlemen.
The electrons of an atom move according to the laws of quantum mechanics established in 1925 and the next following years. For the hydrogen atom, which has only one electron and consequently is the simplest atom to investigate theoretically, the calculation of the motion of the electron in the electric field of the nucleus led to results of such accuracy that 20 years elapsed until any error of the theory could be found experimentally. This occurred, however, in 1947 when Lamb and his collaborator Retherford discovered that some energy levels of hydrogen which should coincide theoretically were in fact somewhat shifted relative to each other.
One important result of the work of this year’s Nobel Prize winners Sinitiro Tomonaga, Julian Schwinger and Richard Feynman was the explanation of the Lamb-shift. Their work is, however, much more general and of deep general significance to physics. It has explained and also predicted several important phenomena. It is the continuation of some investigations performed in the late 1920’s in order to find the general quantum mechanical laws according to which the atoms and in particular the electrons give rise to electromagnetic fields, e.g. emit light, and are influenced by such fields. By applying quantum mechanics not only to matter but also to the electromagnetic field Dirac, Heisenberg, and Pauli managed in those years to formulate a theory, called quantum electrodynamics, which contains the quantum mechanical laws for the interaction of charged particles, in particular electrons, and the electromagnetic field. It satisfies the important condition of being in agreement with the theory or relativity.
It was soon realized, however, that this theory had serious defects. When one tried to calculate a quantity of such importance as the contribution to the mass of an electron originating in its interaction with the electromagnetic field an infinite and therefore useless result was obtained. A similar difficulty occurred for the charge of the electron.
Because of the fundamental importance of having a more useful quantum electrodynamics many theoretical physicists tried during the 1930’s to over come those difficulties. Some indications were forthcoming how this should be accomplished. It lasted, however, until the 1940’s for decisive progress to be made.
A new area was then initiated by investigations first performed by Tomonaga. His work was primarily related to the demands imposed by the theory of relativity. In a paper published in 1943 and in later work published together with his collaborators, Tomonaga managed to give a new formulation of quantum electrodynamics and other similar theories, which marked an important progress.
Definite progress was only made as a consequence of the discovery of the Lamb-shift mentioned earlier. When this discovery was discussed at a conference the idea was accepted that the new effect could be explained by quantum electrodynamics provided the proper interpretation was given to this theory. The correctness of this idea was supported by a provisional calculation of the Lamb – shift which was published by Bethe shortly after the conference.
As soon as Tomonaga knew about the Lamb experiment and Bethe’s paper he realized that an essential step to be taken was to substitute the experimental mass for the fictive mechanical mass which appeared in the equations of quantum electrodynamics and to perform a similar renormalization of the electric charge. The compensating terms which had then to be introduced in the equations should cancel the infinities. Tomonaga managed to carry out this difficult program on the basis of his earlier investigations mentioned above. He deduced further a correct formula for the Lamb-shift which was found to give results in good agreement with the measurements.
Almost simultaneously with the discovery of the Lamb-shift another peculiarity was found by Kusch and his collaborator Foley, which made it clear that the magnetic moment of the electron is somewhat larger than had been assumed before. Using the method of renormalization which he also developed Schwinger was able to prove that a small anomalous contribution should be added to the value of the magnetic moment accepted until then. His calculation agreed with the experiments. Schwinger’s calculation was indeed earlier than and very important for the proper interpretation of these measurements.
Schwinger had developed the formalism of the new quantum electrodynamics in several fundamental papers using partially methods similar to those of Tomonaga. He has also made this formalism more useful for practical calculations.
Feynman used more radical methods for solving the problems of quantum electrodynamics. He created a new formalism which he made very useful for practical calculations by introducing a graphical interpretation called Feynman diagrams, which have become an important feature of modern physics. In the description used by Feynman the electromagnetic field did not any more appear explicitly. His description has been very valuable in elementary particle physics where it is necessary to consider besides the electromagnetic also other interactions.
When considering the truth of quantum electrodynamics in its new form one has first of all to realize the extraordinary success of this theory in giving results in agreement with the experiments. For the Lamb-shift and for the anomalous part of the magnetic moment of the electron the agreement is within some parts in one hundred thousand respectively in a million and no disagreement has yet been found. Quantum electrodynamics is indeed one of the most accurate of all the theories of physics. Further evidence in this respect is given by the applications of the theory to the positronium atom and to the mu-particle. The new formalism has also been very important for other parts of physics in particular elementary- particle physics, but also solid-state physics, nuclear physics and statistical mechanics.
Professor Tomonaga has unfortunately been prevented by an accident from receiving his prize here in Stockholm. It will be presented to him by intermediary of the ambassador of Sweden in Tokyo, and it is accompanied by the congratulations of the Royal Academy of Science.
Professor Schwinger and Professor Feynman. By introducing new ideas and methods into an old theory you have, together with Professor Tomonaga, created a new and most successful quantum electrodynamics, which occupies a central position in physics. This theory has been unique in stimulating modern research. You have yourself contributed to the extension of its methods to other fields of physics where it has also been essential for recent progress.
On behalf of the Royal Academy of Science I congratulate you on your work and ask you to receive your Nobel Prize from the hands of His Majesty the King.
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188105 | https://static1.squarespace.com/static/61de416a3e2596709a9237f6/t/61ef56680bab816fd2ad5a1d/1643075177844/hw5s-1.5-solve+quadratics+by+factoring.pdf | 1.5 Solving Quadratic Equations – Part 1: Solve by Factoring – Worksheet MCR3U Jensen 1) Solve by factoring a) 𝑥! + 8𝑥+ 12 = 0 b) ℎ! + 9ℎ+ 18 = 0 c) 𝑚! + 3𝑚= 0 d) 𝑤! −18𝑤+ 56 = 0 e) 𝑥! −2𝑥= 0 f) 𝑐! −17𝑐+ 30 = 0 2) Solve a) 3𝑥! + 28𝑥+ 9 = 0 b) 4𝑘! + 19𝑘+ 15 = 0 d) 16𝑏! −1 = 0 f) 4𝑥! −12𝑥+ 9 = 0 3) Solve each quadratic equation by factoring a) 𝑥! + 2𝑥−3 = 0 b) 𝑥! + 3𝑥−10 = 0 c) 4𝑥! −36 = 0 d) 6𝑥! −14𝑥+ 8 = 0 e) 15𝑥! −8𝑥+ 1 = 0 f) 6𝑥! + 19𝑥+ 10 = 0 4) Solve by factoring a) −𝑥! −10𝑥−16 = 0 b) 6𝑑! + 15𝑑= −9 5) A rectangle has dimensions x+10 and 2x-‐3. Determine the value of x that gives an area of 54 cm2 Answers 1) a) -‐2, -‐6 b) -‐3, -‐6 c) 0, -‐3 d) 14, 4 e) 0, 2 f) 15, 2 2) a) !!
! , -‐9 b) -‐1, !!" ! d) !
! , !!
! f) !
! 3) a) -‐3, 1 b) -‐5, 2 c) -‐3, 3 d) !
!, 1 e) !
! , !
! f) − !
! , − !
! 4) a) -‐8, -‐2 b) -‐1, !!
! 5) 3.5 |
188106 | https://medium.com/nonce-vcva/perspectives-in-the-anthropology-of-money-2fea5dc4a49a | Perspectives in the Anthropology of Money | by Simon Posner | nonce vcva | Medium
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Perspectives in the Anthropology of Money
Simon Posner
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11 min read
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Jun 18, 2019
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In much of the anthropological record of money, scholars have rehearsed a distinction between modern money and primitive money. Setting modern money aside as a seamless abstraction, anthropologists have generated important insights about the curious worlds of primitive money and its varied uses and meanings. Later on, anthropologist would use the accounts of primitive money to approach modern money, unsettling the distinction between modern and primitive money-forms and revealing how modern money itself takes on variegated expressions. In this post, I’ll look at a few of these accounts, starting with primitive money.
Karl Polanyi
The pivotal work of Karl Polanyi (1968) makes a clear and teleologically-oriented distinction between modern money and primitive money. Modern money is like a language with its own uniform grammar. It is a system that is applied consistently across all situations of exchange. It is all-purpose money. In contrast, primitive money is fragmented: it comes in an array of forms that each apply in specific situations, none of which are commensurable with the other. That is, different money-uses are institutionalized separately from one another.
Primitive money is not only about exchange. It is special-purpose money, defined by its specific uses. Aside from use for exchange, which was very rare in archaic societies, money is used for two other purposes: payment and hoarding. In archaic societies, debt is determined by non-economic factors, such as status, kinship, legal obligation. Money is used here to pay for one’s debts. Hoarding also relates to payment: accumulation is intended for use if one must pay. This includes staples or treasure. For Polanyi, the notion that money has always served as a means of exchange or as a standard of value is to elide the complex history — the “great transformation” — out of which money arose as we find it in the contemporary world.
Paul Bohannan
In guise of the same distinction with a focus on primitive money, Paul Bohannan (1955) examines three dimensions of Tiv ideas of exchange and investment: ideas of exchange expressed in language; traditional modes of exchange and investment; and the impact of Western economic elements brought into the local economic relations of the Tiv. First, Bohannan notes a distinction the Tiv make between “market” and “gift.” Gifts should not be calculated or haggled over because they have no explicit or precise exchange value. In contrast, markets are impersonal and demonstrate a dimension of exchange value.
Second, Bohannan notes three categories of exchangeable items. The first category consists of foodstuff including chickens, goats and cookware, which relates to subsistence. The second category includes slaves, cattle and metal bars, relating to prestige. The final category consists of rights in human beings, in particular rights over women (other elements like service and labor can be classified as generosity or obligation while land is not exchangeable). Bohannan demonstrates that these three categories constitute a moral hierarchy wherein subsistence ranks lowest and rights rank highest. Trade across the same category consists of conveyances, while trade between different categories comprises conversion. The object is to convert from lower to higher, and this endeavor sets the conditions for Tiv motivation. They seek to climb from subsistence to prestige and finally to marriage.
In the last part of his article, Bohannan examines how Western economic integration has disturbed the Tiv’s traditional mode of exchange and investment. There are basically two central problems Bohannan outlines that revolve around the introduction of Western money, as a mechanism that sets itself as the common denominator for all forms of exchange. First, some Tiv merchants aim to accumulate money (M-C-M) by purchasing a commodity and exporting it to another market to sell at a higher price. This makes it such that means of subsistence are being shipped away in order to accrue profit. Second, exchange marriage has been outlawed by Western colonial powers and has been replaced with the purchase of brides. The only motivational factor left therefore is the accumulation of money, which leaves even less foodstuffs for the locality. The Tiv basically sell their foodstuffs to get rich quick and get married, but as competition rises for wives, the availability of foodstuffs vanishes. Bohannan’s is an example of the outside forces of capital penetrating a local system and disturbing it, but he also demonstrates, in his illustrations of Tiv modes of exchange, the extent to which examining the stuff of exchange can elucidate important insights about the actions and motivations of a group of people.
George Dalton
George Dalton (1965) aims to create further distance between modern and primitive money by illustrating the ways in which the imposition of Western economic models when examining primitive money forms elides the latter’s complexities and distinctions and ultimately assumes that money, in all forms, has only one transhistorical expression: market exchange. He divides his articles into three parts: Western money and economy; primitive money and economy; and a case study of Rossel Island money.
First, the basic characteristic of Western economic models is that money serves a general purpose under which all forms of transaction are derived from market exchange in a fully integrated society. Money is the direct or indirect medium of all forms of exchange. Non-commercial exchanges, such as tax payments or gift-giving, are all fundamentally derived from the private market norm. We think that market integration is the only form of ‘money-ness.’ All other forms are not money and therefore irrelevant to economic analysis. Western money is anonymous, where exchange takes place between two faceless, status-less actors.
Second, in contrast to modern money, primitive money-stuff is not bundled into a unified economic framework. There can be multiple and coexisting economic systems distinct from one another. Dalton outlines three basic economies. First, marketless economies are marked by reciprocity and redistribution where land and labor are not for sale. These are subsistence economies. Second, peripheral market economies have markets, but the bulk of income does not come from market sales. Third, market-dominated (peasant) economies are similar to Western economies but they don’t have large-scale machine technology and retain traditional social organization.
Third, having outlined the distinction between modern and primitive economic systems, Dalton illustrates how approaching primitive economies with modern economic models in mind is problematic. To demonstrate his case, Dalton critiques Armstrong’s analysis of Rossel Island money. Armstrong is an economist who ethnocentrically superimposes a Western economic model onto the local economy. He claims that ndap shells can be counted like money and that they are part of a unified economic system. In fact, as Dalton shows, these shells are part of two separate systems, subsistence and prestige, that are both incommensurable with one another. Armstrong’s analysis is an example of how not to do anthropology: he has elided local distinctions and made it appear as though the Western money form was the only form to have ever existed. Dalton closes by pointing out that money has no essence but is rather defined by the ways in which it is used in relation to its specific economic system. The insights Dalton provides are interesting because, while he reifies a difference between primitive and modern money, he also establishes a wide array of uses of money and that similar objects may have incommensurably different uses. Both Bohannan and Dalton demonstrate variation in the use of money, even if observed in economies explicitly categorized as primitive.
Complicating the separation between primitive and modern money
Viviana Zelizer
Viviana Zelizer (1989) throws into question the basic assumption that capitalist money is a completely abstracted, singular form of exchange. She basically brings the anthropology of primitive money into the Western context to demonstrate that modern money is not only about market exchange but can be defined in multiple ways, with different uses and with different words. She calls her approach an alternative model of “special monies.” She divides her discussion into three section: modern money; special monies; and a case study of domestic money.
First, Zelizer examines Max Weber and Georg Simmel’s approaches, who she claims addressed money as a force that neutralizes the world and turns its qualitative traits into quantitative elements that can be traded in the market. Zelizer outlines five assumptions about modern money: 1) money is defined strictly in economic terms; 2) all monies are the same in modern society; 3) money is neutral; 4) money commodifies and corrupts; and 5) money has a unilateral relation in transforming the world. Presumably, money changes and shapes the world, but the world does not shape it. Zelizer intends to complicate this.
Zelizer aims to take Polanyi’s notion of “special purpose money” into the Western context. Anthropologists of money have examined primitive money but have not used those same analytical tools in the Western context, thereby establishing a distinction between modern money and primitive money. For Zelizer, modern money is only different in degrees and not in kind from primitive money. Zelizer challenges the utilitarian model of money with the following five counter-assumptions: 1) money does not only exist in the sphere of the market; 2) there is a plurality of qualitatively distinct monies, market money being only one among many; 3) a more inclusive model is needed to take these differences into account; 4) money can be a personal and unique object; and 5) money shapes society and is shaped by society.
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Zelizer then discusses a case study to demonstrate the extent to which gender and class shape the meanings, uses, and names of money as these change over time. She demonstrates that family money was set apart from the market, that it was nonfungible. Her exploration of domestic money elucidates the limitations of the rationalized model of market economy. Zelizer’s overall aim is to challenge neoclassical economic theory by provincializing market money as one among many forms and expressions of money. Modern money is no longer the abstract all-powerful form it once was; it is indeed powerful, but it is also subject to the same varied uses as primitive money once was imagined to be. The primitive/modern distinction therefore no longer holds.
Bill Maurer
In his important review article on money, Bill Maurer (2006) aims to settle the notion that there are no distinctions between forms of money, Western or nonwestern. For Maurer, we can never really attain the representation of money, its true and fundamental essence, especially not a modern abstracted form. Whatever we claim to say is true about money is always a particular representation that cannot conform congruently to reality. So, approaching money as a “great equalizer” will always elide other ways in which money is used. Rather than aiming to claim the true nature of money, Maurer recommends we focus on its uses and pragmatics, which are always already bound to symbolic and social systems, to elucidate all the different ways in which money shapes and is shaped by the world.
The opening of Maurer’s article gives an overview of the anthropology of money. This subdiscipline has examined the particular uses of money in nonwestern contexts but not in Western contexts. It has also examined ways in which modern money is integrated into nonwestern contexts. The basic assumption is always the same: modern money is money that has undergone a “great transformation.” It is always approached as disembedded from the world and becomes a world shaping force. This approach reproduces an “us/them” distinction tied with a moral dimension against the ravages of modern money. According to the ethnographic record, modern money is integrated in all sorts of ways. Sometimes it fits well with a new context, sometimes it has little impact, if any. The point is modern money is not a “great equalizer;” it operates in all the same and distinct ways primitive money does. We can set aside Marx and the moral dimensions of modern money to look deeper into the manifold specificities of its pragmatics and uses.
The same could be said about calculation and numbers. With modern money, calculation does not become something disembedded from the world. It is not something that comes to exist in the idealic realm of “mind.” Cognitive operations are always part of context that involve all sorts of concerns and other ongoing activities. There is no such thing as pure calculation of the world.
The quantitative models of finance are also not disembedded from the world. They certainly do perform and enact reality in specific ways and have real consequences in the world, but they also have other effects not immediately related to markets. Models can also be felt affectively, they can be related to people’s personal biographies, for example. In this sense, there never was a “great transformation” to the fullest degree. Money always exceeds its representations through the surprising ways in which it is deployed. Maurer contends that we should thus “reorient the anthropology of money from meanings to repertoires, pragmatics, and indexicality” to consider “the gaps between representation and reality and sign and substance, and their “unresolved antagonisms’” (30).
Marcel Mauss
Finally, my discussion on the theme of money ends with Marcel Mauss’ (1967) The Gift. Mauss’ essay is brief but contains a rich account of geographically and historically distinct societies. I will not summarize the book in detail but rather point to several relevant themes it covers. Mauss begins by introducing the gift as a “total social phenomenon,” that is, the practice of prestation (lending or reciprocal exchange) is weaved into and makes up multiple elements of a society’s fabric (religious, legal, moral, economic). In this way, Mauss contends that the market is a human product such that morality and economy cannot be separated, a basic principle that is present in every society, across time and space. The findings in the book are therefore common and generalizable, distinct only in degrees but not in kind. To demonstrate this, Mauss conducts a comparative analysis of different societies — based on the literary review of an array of ethnographic sources. The central question in the book is: what is the principle of reciprocity? What force in the thing compels the recipient to make a return?
The answer Mauss offers is that gifts must be received and returned because they are things charged with magic, mana, hau, a spirit, a part of the giver, and in this sense, the gift is alive. They are the vehicles of mana and demand that they be returned to their original owner in some other form. In this sense, they hold sway over the recipient, who is bound to return the gift to the initial giver. One is also required to receive because circulation is the nature of things. Things are meant to be in motion, and in this way, they must also always be given.
This, for many I think, would come across as a dubious set of claims. For Claude Levi-Strauss (1987) for example, Mauss appeared to have been taken by the object of his own analysis, which would have clouded his ability to perceive the essential processes below that govern appearances. Mauss took the gift as a personification that had its own independent agency in the world, but the notion that it was a fetish, a reified thing only makes sense in a dualist framework of thought (hence Levi-Strauss’ structuralism). For Mauss, the gift is its own singularity while also being part of a totality. There is nothing that separates the gift from the whole, while there is nothing that separates reality from abstraction. Along parallel lines, Mauss contended that gifts presuppose the fundamental institutions of exchange in a given society while simultaneously realizing them. He also countered Malinowski’s distinction between primitive and modern currencies by advancing that both served similar functions and therefore could be classed under the same “genus” (94). But what I have found most compelling about Mauss’ essay is that he approached the gift as an entry point to gain insights into the lifeworlds of different societies. The gift is therefore an effective analytical method that can lead to novel horizons of thought.
Gifts, finance, money, basically anything that circulates between people, can be an entry point for insights into the lifeworlds of different groups and scales. The object of research is not a thing to be defined, but rather a practice to be observed. And if we allow ourselves to be taken by the object of our study as Mauss was accused of doing, we can then extend these insights into our own versions of reality.
REFERENCES
Bohannan, Paul. 1955. “Some Principles of Exchange and Investment Among the Tiv.” American Anthropologist 57: 60–70.
Dalton, George. 1965. “Primitive Money.” American Anthropologist 67 (1): 44–65.
Levi-Strauss, Claude. 1987. Introduction to the Work of Marcel Mauss. Translated by Felicity Baker. London, UK: Routledge and Paul Kegan.
Maurer, Bill. 2006. “The Anthropology of Money.” Annual Review of Anthropology 35: 15–36.
Mauss, Marcel. 1967. The Gift: Forms and Functions of Exchange in Archaic Societies. Translated by Ian Cunnison. New York, NY: W. W. Norton & Company.
Polanyi, Karl. 1968. “The Semantics of Money-Uses.” In Primitive, Archaic, and Modern Economies: Essays of Karl Polanyi, edited by George Dalton, 175–203. New York, NY: Anchor Books.
Zelizer, Viviana. 1989. “The Social Meaning of Money: ‘Special Monies.’” American Journal of Sociology 95 (2): 342–77.
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Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
Malignancy-Related Hypercalcemia
Catherine Anastasopoulou; Prerna Mewawalla.
Author Information and Affiliations
Authors
Catherine Anastasopoulou1; Prerna Mewawalla2.
Affiliations
1 Jefferson Einstein Medical Center
2 Allegheny Health Network Cancer Inst
Last Update: March 4, 2025.
Continuing Education Activity
Hypercalcemia is a common metabolic abnormality observed in both inpatient and outpatient settings. Depending on the serum calcium levels, hypercalcemia is categorized as mild (10–12 mg/dL), moderate (12–14 mg/dL), or severe (>14 mg/dL). Approximately 40% to 45% of serum calcium is bound to albumin, and serum calcium levels may fluctuate with changes in albumin levels. Therefore, ionized or free calcium levels should be measured when hypercalcemia is suspected. The most common cause of malignancy-related hypercalcemia is parathyroid hormone-related peptide (PTHrP). Other common causes include bony metastases and hypervitaminosis D, although rarer causes also exist.
Initial evaluation includes differentiating between benign and malignant causes, with PTH levels aiding in diagnosis. Severe cases require urgent treatment, while management focuses on addressing the underlying malignancy, correcting associated electrolyte imbalances, and patient education on contributing factors. This activity describes the pathophysiology of malignancy-related hypercalcemia and its association with various malignancies. This activity also highlights the importance of collaboration among healthcare providers in the management of this condition.
Objectives:
Identify the signs and symptoms of malignancy-related hypercalcemia to facilitate early diagnosis and intervention.
Implement effective management strategies, including the correction of serum calcium levels and treatment of the underlying malignancy.
Select appropriate treatment modalities based on the severity of hypercalcemia and the underlying malignancy, including pharmacological and supportive care.
Collaborate with interprofessional healthcare specialists to address the underlying malignancy and tailor treatment plans for hypercalcemia management.
Access free multiple choice questions on this topic.
Introduction
Hypercalcemia is a common metabolic abnormality seen in both inpatient and outpatient settings. Depending on the serum calcium levels, hypercalcemia is categorized either as mild (just above normal but <12 mg/dL), moderate (between 12 and 14 mg/dL), or severe (>14 mg/dL). Approximately 40% to 45% of the serum calcium is attached to albumin, and serum calcium levels may fluctuate based on the serum albumin levels. Therefore, the ionized or free calcium levels should be measured when hypercalcemia is suspected for a more accurate assessment.
The corrected calcium could also be calculated using the below formula:
Serum calcium + 0.8 × (4 − patient’s albumin level), where 4.0 g/dL is the normal albumin level.
More than 90% of hypercalcemia cases are caused by primary hyperparathyroidism (PHPT) or malignancy-induced hypercalcemia, with malignancy being the most common cause of hypercalcemia in hospitalized patients.
Etiology
Hypercalcemia can have multiple causes, with the most common being PHPT, malignancy-induced, medication-induced, familial, or endocrine-related. The initial evaluation of a patient with hypercalcemia requires clinicians to differentiate between benign and malignant causes. PHPT, the most common benign cause, is typically characterized by asymptomatic patients with a long-standing history of mild hypercalcemia.
Serum calcium levels greater than 13 mg/dL upon initial presentation should raise suspicion of malignancy as the cause of hypercalcemia. Symptomatic severe hypercalcemia due to malignancy is associated with a poor prognosis and requires urgent treatment. Additionally, many rare causes of hypercalcemia have been reported in the literature over the years and should be considered once the more common causes have been ruled out.
Epidemiology
Hypercalcemia of malignancy occurs in approximately 20% of cancer patients at some point during their clinical course. The most common cancer associated with hypercalcemia is multiple myeloma. A prevalence study reports that hypercalcemia of malignancy is reported in about 2% to 3% of patients diagnosed with cancer, with the incidence gradually decreasing over the years due to improvements in treatment options.
Pathophysiology
The pathophysiology of hypercalcemia of malignancy is primarily explained through 3 mechanisms—excessive secretion of parathyroid hormone (PTH)-related protein (PTHrP), bony metastases leading to the release of osteoclast-activating factors, and the production of 1,25-dihydroxy vitamin D (calcitriol).
Excessive secretion of PTHrP is the most common cause of hypercalcemia of malignancy, accounting for about 80% of cases. This condition is also known as humoral hypercalcemia of malignancy (HHM). This is usually seen in solid tumors and a few cases of non-Hodgkin lymphoma. The most common solid tumors associated with hypercalcemia include squamous cell carcinoma of the head, neck, and lungs, breast cancer, ovarian cancer, renal carcinoma, and certain hematological malignancies, such as leukemia. HHM should be suspected in patients without skeletal metastasis.
Structurally, PTHrP is similar to PTH in the first 13 amino acid sequences. Due to this structural similarity, PTHrP binds to the same receptor as PTH, leading to bone resorption, increased phosphate excretion from the proximal tubules, and calcium reabsorption from the distal tubules in the kidneys. However, it does not affect the production of 1,25-dihydroxy vitamin D. Laboratory findings typically show elevated PTHrP levels, low to normal PTH levels, and normal 1,25-dihydroxy vitamin D levels. The response to treatment can be monitored by tracking PTHrP levels. Patients with HHM often have advanced disease, which is associated with a poor prognosis.
Bony metastases, which lead to the release of osteoclast-activating factors, account for 20% of hypercalcemia of malignancy cases. They are commonly observed in patients with multiple myeloma and solid tumors, such as breast cancer, that metastasize to the bones. Typical findings include skeletal metastasis with low to low-normal levels of PTH, PTHrP, and 1,25-dihydroxy vitamin D. Although PTHrP levels are low to normal, breast cancer cells in the bone produce PTHrP locally and increase the activity of receptor activator of nuclear factor kappa-B ligand (RANKL), which, in turn, enhances osteoclastic activity and contributes to hypercalcemia.
Almost all cases of Hodgkin lymphoma, about one-third of non-Hodgkin lymphoma cases, and granulomatous diseases such as sarcoidosis and tuberculosis cause hypercalcemia by increasing 1,25-dihydroxy vitamin D production. This subset of patients typically responds well to steroid treatment.
History and Physical
Specific physical examination findings for hypercalcemia are not usually evident, but patients can present with a wide spectrum of symptoms. Depending on the acuity and severity, patients can be asymptomatic or exhibit involvement of multiple organ systems, including the gastrointestinal tract, musculoskeletal system, cardiovascular system (CVS), renal system, and central nervous system (CNS), or experience psychiatric disturbances. In rare cases, band keratopathy may be observed during a slit-lamp examination, indicating calcium phosphate deposits in the cornea.
Renal manifestations range from polyuria, polydipsia, nephrogenic diabetes insipidus, and renal insufficiency to distal renal tubular acidosis secondary to nephrolithiasis. If left untreated, hypercalcemia and hypercalciuria may lead to tubular atrophy, interstitial fibrosis, and renal calcification, resulting in nephrocalcinosis. Gastrointestinal symptoms can vary from anorexia to nausea and constipation. Excessive calcium deposition in the pancreatic duct may lead to pancreatitis. Additionally, hypercalcemia can increase gastrin secretion, contributing to the development of peptic ulcer disease.
Musculoskeletal symptoms may present as muscle weakness and bone pain. Cardiovascular manifestations are subtle, often including short QTc intervals, with rare cases of more severe arrhythmias. Excessive calcium deposition in the heart valves and coronary arteries can increase cardiovascular morbidity. CNS symptoms vary with calcium levels; patients with mild hypercalcemia are often asymptomatic, while those with severe hypercalcemia may experience lethargy, confusion, or even coma, particularly in older populations. Common psychiatric disturbances include anxiety, depression, and cognitive changes.
Evaluation
The initial evaluation of hypercalcemia requires a comprehensive history and physical examination, which can help identify the underlying cause and pathology. Prior laboratory data are valuable, providing insights into baseline calcium levels and the duration of hypercalcemia. Medication history, including prescription drugs, over-the-counter vitamins, and supplements, along with dietary history, family history, and any history of granulomatous disease, should be systematically reviewed. Initial labs should include PTH levels, as this helps differentiate between PTH-related hypercalcemia and non-PTH–mediated hypercalcemia. PTH-related hypercalcemia is seen in conditions such as PHPT and familial hyperparathyroid syndromes, while non-PTH–related hypercalcemia occurs in malignancies, granulomatous diseases, endocrine disorders, and vitamin D intoxication.
Familial hypocalciuric hypercalcemia syndrome should be suspected in patients with minimally elevated PTH levels and low urinary calcium excretion on a 24-hour urinary calcium. Low-normal or low levels of PTH (<20 pg/mL) should raise suspicion of non-PTH–related causes. In such cases, PTHrP and vitamin D metabolites, including 25-hydroxyvitamin D and 1,25-dihydroxy vitamin D levels, should be checked.
If PTHrP is elevated, it indicates HHM. Vitamin D intoxication leads to elevated 25-hydroxyvitamin D levels. Elevated 1,25-dihydroxy vitamin D levels suggest lymphoma or granulomatous diseases. Serum protein electrophoresis (SPEP), urine protein electrophoresis (UPEP) with immunofixation, and serum-free light chains should be evaluated to rule out multiple myeloma if vitamin D levels are within the normal range. In patients with malignancy-induced hypercalcemia, PTH levels should still be assessed to identify potential coexisting PHPT.
Treatment / Management
Treatment should be tailored to reduce serum calcium levels, address the patient's symptoms, and target the underlying cause. Serum phosphorus levels should be monitored and repleted, as hypophosphatemia often accompanies hypercalcemia, complicating its treatment. Asymptomatic patients with mild to moderate hypercalcemia do not require immediate therapy, but managing the underlying cause is essential. Patients should be educated on diet and medications, as well as the importance of avoiding dehydration and physical inactivity. Symptomatic patients with severe hypercalcemia require urgent treatment.
Initial treatment involves intravenous (IV) normal saline, along with calcitonin and bisphosphonates. Normal saline acts immediately, with its effects lasting until the fluids are discontinued. Calcitonin takes effect within 4 to 6 hours and lasts for about 2 days, while bisphosphonates begin working in 2 to 3 days, with effects lasting for 2 to 4 weeks. This approach aims to lower serum calcium levels and maintain them within normal limits as long as possible while the underlying cause is being identified and treated.
IV hydration with normal saline at a rate of 200 to 300 mL/h is administered to maintain an adequate urine output of more than 100 mL/h, helping restore intravascular volume and increase urinary calcium excretion. Caution should be exercised when administering IV fluids to patients with heart or renal failure. Loop diuretics, which promote urinary calcium excretion by inhibiting calcium reabsorption at the loop of Henle, should only be administered after adequate IV resuscitation is achieved.
Calcitonin should be administered at a dose of 4 IU/kg alongside normal saline infusion to help prevent bone resorption and increase urinary calcium excretion. Calcitonin is a very fast-acting medication, but its effects are limited in duration.
Bisphosphonates, such as zoledronic acid (4 mg IV over 15-30 minutes) or pamidronate (60-90 mg IV over 2 hours), are recommended for patients without kidney dysfunction. Zoledronic acid is preferred for hypercalcemia secondary to malignancy due to its greater potency and shorter infusion time. Bisphosphonates are also commonly used in patients with bone metastases to prevent skeletal complications. The main adverse effects of bisphosphonates include osteonecrosis of the jaw and nephrotoxicity.
Denosumab works by inhibiting RANKL and was previously considered only for patients who did not respond to zoledronic acid or those with kidney impairment, as it is not cleared by the kidneys. However, the latest guidelines from the Endocrine Society now recommend using denosumab early, as it has shown strong efficacy in lowering calcium levels and reducing recurrent episodes of hypercalcemia. Denosumab is also administered regularly to cancer patients, even those without hypercalcemia, primarily for the prevention of bone complications.
Glucocorticoid therapy should be considered for patients with increased 1,25-dihydroxy vitamin D production, such as those with lymphoma or granulomatous diseases, as it reduces vitamin D production and decreases calcium absorption from the intestines.
Calcimimetic agents, such as cinacalcet, are preferred for hemodialysis patients and those with hypercalcemia due to parathyroid cancer.
If all other strategies fail, hemodialysis is used to treat hypercalcemia. Hemodialysis should also be considered for patients with severe heart or renal failure who cannot tolerate adequate IV hydration.
Differential Diagnosis
The following conditions should be considered when evaluating hypercalcemia:
Adrenal insufficiency
Berylliosis
Coccidioidomycosis
Crohn disease
Hyperkalemia
Hypermagnesemia
Hypernatremia
Hyperparathyroidism
Hyperphosphatemia
Hyperthyroidism
Milk-alkali syndrome
Vitamin D toxicity
Prognosis
The prognosis of malignancy-related hypercalcemia depends on the underlying cause and the type of cancer responsible. Early stages of the disease typically have a more favorable prognosis compared to more advanced stages. Late diagnosis often correlates with a poorer prognosis. Elevated levels of 1,25-dihydroxy vitamin D have been shown to be associated with recurrent hypercalcemia and a more severe prognosis.
Complications
Hypercalcemia, if undiagnosed or inadequately treated, can lead to severe complications such as kidney failure and bone issues, including decreased bone density, osteoporosis, and fractures.
Deterrence and Patient Education
Patients with calcium abnormalities related to underlying cancer should follow up with an endocrinologist in addition to receiving mandatory oncology evaluation and treatment. If complications arise in other organ systems, patients are encouraged to consult specialists in the affected areas, such as cardiology, ophthalmology, neurology, or neurosurgery.
Pearls and Other Issues
Key facts to keep in mind regarding malignancy-related hypercalcemia include:
Malignancy is the most common cause of hypercalcemia in hospitalized patients.
The pathophysiology of hypercalcemia of malignancy involves 3 main mechanisms—excessive secretion of PTHrP, bony metastases with the release of osteoclast-activating factors, and the production of 1,25-dihydroxy vitamin D (calcitriol).
PTHrP, also known as HHM, is the most common causative factor.
The most common solid tumors associated with hypercalcemia include squamous cell carcinoma of the head, neck, and lungs, breast cancer, ovarian cancer, renal carcinoma, and some hematological malignancies such as leukemia. HHM should be suspected in patients without any skeletal metastasis.
IV hydration with normal saline as the foundation of therapy.
Other treatment options include zoledronic acid, pamidronate, and denosumab.
Steroids are preferred with the presence of hypervitaminosis D.
Cinacalcet is preferred for hemodialysis patients and those with parathyroid cancer.
Enhancing Healthcare Team Outcomes
Malignancy-associated hypercalcemia is best managed by an interprofessional healthcare team, including an oncologist, internist, endocrinologist, and surgeon, who guides treatment. Pain specialists should also be involved, as many patients may present with varying degrees of pain. Over 90% of hypercalcemia cases are caused by PHPT and malignancy-induced hypercalcemia. Nurses, laboratory personnel, and pharmacists are integral members of the interprofessional healthcare team, contributing to comprehensive care and improved patient outcomes.
Nursing staff have a key role in case management by assisting with patient assessment, providing counseling, and serving as a liaison between the different specialties. Pharmacists ensure the appropriate medications are administered at the correct doses, monitor for potential drug interactions, and counsel patients on possible adverse effects. Laboratory technicians ensure blood collection is completed and serum values are processed and reported to clinicians promptly. All interprofessional healthcare team members must maintain accurate records, ensuring everyone involved in the patient's care can access the same data. This collaborative approach optimizes patient outcomes.
Malignancy is the most common cause of hypercalcemia in hospitalized patients. Although most cases are managed on an outpatient basis, severe hypercalcemia often requires inpatient treatment. The prognosis depends on the stage of the primary malignancy and the severity of the hypercalcemia. Patients with uncontrolled malignancy and severe hypercalcemia typically have a poor prognosis.
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: Khan AA, Gurnani PK, Peksa GD, Whittier WL, DeMott JM. Bisphosphonate Versus Bisphosphonate and Calcitonin for the Treatment of Moderate to Severe Hypercalcemia of Malignancy. Ann Pharmacother. 2021 Mar;55(3):277-285. [PubMed: 32885992]
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: Seisa MO, Nayfeh T, Hasan B, Firwana M, Saadi S, Mushannen A, Shah SH, Rajjoub NS, Farah MH, Prokop LJ, Wang Z, Fuleihan GE, Drake MT, Murad MH. A Systematic Review Supporting the Endocrine Society Clinical Practice Guideline on the Treatment of Hypercalcemia of Malignancy in Adults. J Clin Endocrinol Metab. 2023 Feb 15;108(3):585-591. [PubMed: 36545700]
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: Lipton A, Fizazi K, Stopeck AT, Henry DH, Brown JE, Yardley DA, Richardson GE, Siena S, Maroto P, Clemens M, Bilynskyy B, Charu V, Beuzeboc P, Rader M, Viniegra M, Saad F, Ke C, Braun A, Jun S. Superiority of denosumab to zoledronic acid for prevention of skeletal-related events: a combined analysis of 3 pivotal, randomised, phase 3 trials. Eur J Cancer. 2012 Nov;48(16):3082-92. [PubMed: 22975218]
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: Diel IJ, Body JJ, Stopeck AT, Vadhan-Raj S, Spencer A, Steger G, von Moos R, Goldwasser F, Feng A, Braun A. The role of denosumab in the prevention of hypercalcaemia of malignancy in cancer patients with metastatic bone disease. Eur J Cancer. 2015 Jul;51(11):1467-75. [PubMed: 25976743]
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: Takeuchi Y, Takahashi S, Miura D, Katagiri M, Nakashima N, Ohishi H, Shimazaki R, Tominaga Y. Cinacalcet hydrochloride relieves hypercalcemia in Japanese patients with parathyroid cancer and intractable primary hyperparathyroidism. J Bone Miner Metab. 2017 Nov;35(6):616-622. [PubMed: 27873072]
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: Chukir T, Liu Y, Hoffman K, Bilezikian JP, Farooki A. Calcitriol Elevation Is Associated with a Higher Risk of Refractory Hypercalcemia of Malignancy in Solid Tumors. J Clin Endocrinol Metab. 2020 Apr 01;105(4):e1115-23. [PMC free article: PMC7067545] [PubMed: 31841590]
: Disclosure: Catherine Anastasopoulou declares no relevant financial relationships with ineligible companies.
: Disclosure: Prerna Mewawalla declares no relevant financial relationships with ineligible companies.
Copyright © 2025, StatPearls Publishing LLC.
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Anastasopoulou C, Mewawalla P. Malignancy-Related Hypercalcemia. [Updated 2025 Mar 4]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-.
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Binomial Theorem
The Binomial Theorem states that for real or complex, , and non-negativeinteger,
where is a binomial coefficient. In other words, the coefficients when is expanded and like terms are collected are the same as the entries in the th row of Pascal's Triangle.
For example, , with coefficients , , , etc.
Contents
[hide]
1 Proof
1.1 Proof via Induction
1.2 Proof using calculus
2 Generalizations
2.1 Proof
3 Usage
4 See also
Proof
There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof:
We can write . Repeatedly using the distributive property, we see that for a term , we must choose of the terms to contribute an to the term, and then each of the other terms of the product must contribute a . Thus, the coefficient of is the number of ways to choose objects from a set of size , or . Extending this to all possible values of from to , we see that , as claimed.
Similarly, the coefficients of will be the entries of the row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS].
Proof via Induction
Given the constants are all natural numbers, it's clear to see that . Assuming that , Therefore, if the theorem holds under , it must be valid. (Note that for )
Proof using calculus
The Taylor series for is for all .
Since , and power series for the same function are termwise equal, the series at is the convolution of the series at and . Examining the degree- term of each, which simplifies to for all natural numbers.
Generalizations
The Binomial Theorem was generalized by Isaac Newton, who used an infiniteseries to allow for complex exponents: For any real or complex, , and ,
.
Proof
Consider the function for constants . It is easy to see that . Then, we have . So, the Taylor series for centered at is
Usage
Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial , one could factor it as such: . It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.
See also
Combinatorics
Multinomial Theorem
Retrieved from "
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Combinatorics
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188110 | https://artofproblemsolving.com/wiki/index.php/Modular_arithmetic/Intermediate?srsltid=AfmBOopgzWvGjIYgDH8cYmwAIXPZ7lfimZ8epsc-Oune_weL-E9hXNqU | Art of Problem Solving
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Modular arithmetic/Intermediate
Given integers , , and , with , we say that is congruent tomodulo, or (mod ), if the difference is divisible by .
For a given positive integer , the relation (mod ) is an equivalence relation on the set of integers. This relation gives rise to an algebraic structure called the integers modulo (usually known as "the integers mod ," or for short). This structure gives us a useful tool for solving a wide range of number-theoretic problems, including finding solutions to Diophantine equations, testing whether certain large numbers are prime, and even some problems in cryptology.
Contents
1 Arithmetic Modulo n
1.1 Useful Facts
1.2 The Integers Modulo n
1.3 Addition, Subtraction, and Multiplication Mod n
1.3.1 A Word of Caution
2 Algebraic Properties of the Integers Mod n
3 Topics
4 Miscellaneous
4.1 The binary operation "mod"
5 Resources
6 See also
Arithmetic Modulo n
Useful Facts
Consider four integers and a positive integer such that and . In modular arithmetic, the following identities hold:
Addition: .
Subtraction: .
Multiplication: .
Division: , where is a positive integer that divides and .
Exponentiation: where is a positive integer.
For examples, see Introduction to modular arithmetic.
The Integers Modulo n
The relation allows us to divide the set of integers into sets of equivalent elements. For example, if , then the integers are divided into the following sets:
Notice that if we pick two numbers and from the same set, then and differ by a multiple of , and therefore
We sometimes refer to one of the sets above by choosing an element from the set, and putting a bar over it. For example, the symbol refers to the set containing ; that is, the set of all integer multiples of . The symbol refers to the second set listed above, and the third. The symbol refers to the same set as , and so on.
Instead of thinking of the objects , , and as sets, we can treat them as algebraic objects -- like numbers -- with their own operations of addition and multiplication. Together, these objects form the integers modulo , or . More generally, if is a positive integer, then we can define
,
where for each , is defined by
Addition, Subtraction, and Multiplication Mod n
We define addition, subtraction, and multiplication in according to the following rules:
for all . (Addition)
for all . (Subtraction)
for all . (Multiplication)
So for example, if , then we have
Notice that, in each case, we reduce to an answer of the form , where . We do this for two reasons: to keep possible future calculations as manageable as possible, and to emphasize the point that each expression takes one of only seven (or in general, ) possible values. (Some people find it useful to reduce an answer such as to , which is negative but has a smaller absolute value.)
A Word of Caution
Because of the way we define operations in , it is important to check that these operations are well-defined. This is because each of the sets that make up contains many different numbers, and therefore has many different names. For example, observe that in , we have and . It is reasonable to expect that if we perform the addition , we should get the same answer as if we compute , since we are simply using different names for the same objects. Indeed, the first addition yields the sum , which is the same as the result of the second addition.
The "Useful Facts" above are the key to understanding why our operations yield the same results even when we use different names for the same sets. The task of checking that an operation or function is well-defined, is one of the most important basic techniques in abstract algebra.
Algebraic Properties of the Integers Mod n
The integers modulo form an algebraic structure called a ring -- a structure in which we can add, subtract, and multiply elements.
Anyone who has taken a high school algebra class is familiar with several examples of rings, including the ring of integers, the ring of rational numbers, and the ring of real numbers. The ring has some algebraic features that make it quite different from the more familiar rings listed above.
First of all, notice that if we choose a nonzero element of , and add copies of this element, we get
,
since is a multiple of . So it is possible to add several copies of a nonzero element of and get zero. This phenomenon, which is called torsion, does not occur in the reals, the rationals, or the integers.
Another curious feature of is that a polynomial over can have a number of roots greater than its degree. Consider, for example, the polynomial congruence
.
We might be tempted to solve this congruence by factoring the expression on the left:
.
Indeed, this factorization yields two solutions to the congruence: , and . (Note that two values of that are congruent modulo are considered the same solution.)
However, since , the original congruence is equivalent to
.
This time, factoring the expression on the left yields
.
And we find that there are two more solutions! The values and both solve the congruence. So our congruence has at least four solutions -- two more than we might expect based on the degree of the polynomial.
Why do the "rules" of algebra that work so well for the real numbers seem to fail in ? To understand this, let's take a closer look at the congruence
.
If we were solving this as an equation over the reals, we would immediately conclude that either must be zero, or must be zero in order for the product to equal zero. However, this is not the case in ! It is possible to multiply two nonzero elements of and get zero. For example, we have
But wait! Suppose we take a close look at this last product, and we set and . Then we have -- another of the solutions of our congruence! (One can check that the other two factorizations don't lead to any valid solutions; however, there are many other factorizations of zero that need to be checked.)
In the ring of real numbers, it is a well-known fact that if , then or . For this reason, we call the ring of real numbers a domain. However, a similar fact does not apply in general in ; therefore, is not in general a domain.
Topics
The following topics expand on the flexible nature of modular arithmetic as a problem solving tool:
Fermat's Little Theorem
Euler's Totient Function
Euler's Totient Theorem
Chicken McNugget Theorem
Miscellaneous
The binary operation "mod"
Related to the concept of congruence, mod is the binary operation mod , which is used often in computer programming.
Recall that, by the Division Algorithm, given any two integers and , with , we can find integers and , with , such that . The number is called the quotient, and the number is called the remainder. The operation mod returns the value of the remainder . For example:
mod , since .
mod , since .
mod , since .
Observe that if mod , then we also have (mod ).
An example exercise with modular arithmetic:
Problem:
Let
be a nine-digit positive integer (each digit not necessarily distinct). Consider
,
another nine-digit positive integer with the property that each digit when substituted for makes the modified D divisible by 7. Let
be a third nine-digit positive integer with the same relation to E as E has to D.
Prove that every is divisible by 7.
Solution:
Any positive integer can be expressed .
Since 10=3 mod 7, and since it holds that if a=b mod c then mod c, then D can be expressed much more simply mod 7; that is, = x mod 7.
Each number in E must make the modified D equal 0 mod 7, so for each , , where c is the coefficient of and k is an element of {-2,-1,0,1,2}. The patient reader should feel free to verify that this makes D = 0 mod 7.
In terms of terms, then, we find each .
Then mod 7 can be expressed mod 7 = (9x)- x = 8x = x mod 7. (Note that the 7s, which do not change the mod value, have been eliminated.)
Each number in F must make the modified E equal 0 mod 7, so for each , .
By design and selection of k, all are integers, and is always an integer because it is the difference of two integers.
is a member of the set {1, 2, 3}. Since no divides 7, 7 may be factored and is the product of two integers.
Let then 7A mod 7 = 0 mod 7 for all , QED.
Resources
Number Theory Problems and Notes by Naoki Sato.
See also
Introduction to modular arithmetic
Olympiad modular arithmetic
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188111 | https://courses.lumenlearning.com/wm-prealgebra/chapter/solving-proportions/ | Solving Proportions | Prealgebra
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Prealgebra
Module 7: Percents
Search for:
Solving Proportions
Learning Outcomes
Solve a proportion equation
Solve a proportion application
To solve a proportion containing a variable, we remember that the proportion is an equation. All of the techniques we have used so far to solve equations still apply. In the next example, we will solve a proportion by multiplying by the Least Common Denominator (LCD) using the Multiplication Property of Equality.
example
Solve: x 63=4 7 x 63=4 7
Solution
x 63=4 7 x 63=4 7
To isolate x x , multiply both sides by the LCD, 63 63.63(x 63)=63(4 7)63(x 63)=63(4 7)
Simplify.x=9⋅7⋅4 7 x=9⋅7⋅4 7
Divide the common factors.x=36 x=36
Check: To check our answer, we substitute into the original proportion.
x 63=4 7 x 63=4 7
Substitute x=36 x=36 36 63?=4 7 36 63=?4 7
Show common factors.4⋅9 7⋅9?=4 7 4⋅9 7⋅9=?4 7
Simplify.4 7=4 7 4 7=4 7
try it
In the next video we show another example of how to solve a proportion equation using the LCD.
When the variable is in a denominator, we’ll use the fact that the cross products of a proportion are equal to solve the proportions.
We can find the cross products of the proportion and then set them equal. Then we solve the resulting equation using our familiar techniques.
example
Solve: 144 a=9 4 144 a=9 4
Show Solution
Solution
Notice that the variable is in the denominator, so we will solve by finding the cross products and setting them equal.
Find the cross products and set them equal.4⋅144=a⋅9 4⋅144=a⋅9
Simplify.576=9 a 576=9 a
Divide both sides by 9 9.576 9=9 a 9 576 9=9 a 9
Simplify.64=a 64=a
Check your answer.
144 a=9 4 144 a=9 4
Substitute a=64 a=64 144 64?=9 4 144 64=?9 4
Show common factors..9⋅16 4⋅16?=9 4 9⋅16 4⋅16=?9 4
Simplify.9 4=9 4✓9 4=9 4✓
Another method to solve this would be to multiply both sides by the LCD, 4 a 4 a. Try it and verify that you get the same solution.
The following video shows an example of how to solve a similar problem by using the LCD.
try it
example
Solve: 52 91=−4 y 52 91=−4 y
Show Solution
Solution
Find the cross products and set them equal.
y⋅52=91(−4)y⋅52=91(−4)
Simplify.52 y=−364 52 y=−364
Divide both sides by 52 52.52 y 52=−364 52 52 y 52=−364 52
Simplify.y=−7 y=−7
Check:
52 91=−4 y 52 91=−4 y
Substitute y=−7 y=−7 52 91?=−4−7 52 91=?−4−7
Show common factors.13⋅4 13⋅4?=−4−7 13⋅4 13⋅4=?−4−7
Simplify.4 7=4 7✓4 7=4 7✓
try it
Solve Applications Using Proportions
The strategy for solving applications that we have used earlier in this chapter, also works for proportions, since proportions are equations. When we set up the proportion, we must make sure the units are correct—the units in the numerators match and the units in the denominators match.
example
When pediatricians prescribe acetaminophen to children, they prescribe 5 5 milliliters (ml) of acetaminophen for every 25 25 pounds of the child’s weight. If Zoe weighs 80 80 pounds, how many milliliters of acetaminophen will her doctor prescribe?
Show Solution
Solution
Identify what you are asked to find.How many ml of acetaminophen the doctor will prescribe
Choose a variable to represent it.Let a=a= ml of acetaminophen.
Write a sentence that gives the information to find it.If 5 5 ml is prescribed for every 25 25 pounds, how much will be prescribed for 80 80 pounds?
Translate into a proportion.
Substitute given values—be careful of the units.5 25=a 80 5 25=a 80
Multiply both sides by 80 80.80⋅5 25=80⋅a 80 80⋅5 25=80⋅a 80
Multiply and show common factors.16⋅5⋅5 5⋅5=80 a 80 16⋅5⋅5 5⋅5=80 a 80
Simplify.16=a 16=a
Check if the answer is reasonable.
Yes. Since 80 80 is about 3 3 times 25 25, the medicine should be about 3 3 times 5 5.
Write a complete sentence.The pediatrician would prescribe 16 16 ml of acetaminophen to Zoe.
You could also solve this proportion by setting the cross products equal.
try it
example
One brand of microwave popcorn has 120 120 calories per serving. A whole bag of this popcorn has 3.5 3.5 servings. How many calories are in a whole bag of this microwave popcorn?
Show Solution
Solution
Identify what you are asked to find.How many calories are in a whole bag of microwave popcorn?
Choose a variable to represent it.Let c=c= number of calories.
Write a sentence that gives the information to find it.If there are 120 120 calories per serving, how many calories are in a whole bag with 3.5 3.5 servings?
Translate into a proportion.
Substitute given values.120 1=c 3.5 120 1=c 3.5
Multiply both sides by 3.5 3.5.(3.5)(120 1)=(3.5)(c 3.5)(3.5)(120 1)=(3.5)(c 3.5)
Multiply.420=c 420=c
Check if the answer is reasonable.
Yes. Since 3.5 3.5 is between 3 3 and 4 4, the total calories should be between 360(3⋅120)360(3⋅120) and 480(4⋅120)480(4⋅120).
Write a complete sentence.The whole bag of microwave popcorn has 420 420 calories.
try it
example
Josiah went to Mexico for spring break and changed $325 325 dollars into Mexican pesos. At that time, the exchange rate had $1 1 U.S. is equal to 12.54 12.54 Mexican pesos. How many Mexican pesos did he get for his trip?
Show Solution
Solution
Identify what you are asked to find.How many Mexican pesos did Josiah get?
Choose a variable to represent it.Let p=p= number of pesos.
Write a sentence that gives the information to find it.If $1$1 U.S. is equal to 12.54 12.54 Mexican pesos, then $325$325 is how many pesos?
Translate into a proportion.
Substitute given values.1 12.54=325 p 1 12.54=325 p
The variable is in the denominator, so find the cross products and set them equal.p⋅1=12.54(325)p⋅1=12.54(325)
Simplify.c=4,075.5 c=4,075.5
Check if the answer is reasonable.
Yes, $100$100 would be $1,254$1,254 pesos. $325$325 is a little more than 3 3 times this amount.
Write a complete sentence.Josiah has 4075.5 4075.5 pesos for his spring break trip.
try it
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188112 | https://twiki.di.uniroma1.it/pub/Users/AndreaSterbini/Ricerca/7-JGT-1999.pdf | (n,e)-Graphs with Maximum Sum of Squares of Degrees Uri N. Peled,1 Rossella Petreschi,2 and Andrea Sterbini 2 1UNIVERSITY OF ILLINOIS AT CHICAGO CHICAGO, ILLINOIS E-mail: uripeled@uic.edu 2UNIVERSITY OF ROME "LA SAPIENZA" ROME, ITALY E-mail: petreschi@dsi.uniroma1.it; sterbini@dsi.uniroma1.it Received June 24, 1997; revised November 24, 1998 Abstract: Among all simple graphs on n vertices and e edges, which ones have the largest sum of squares of the vertex degrees? It is easy to see that they must be threshold graphs, but not every threshold graph is optimal in this sense.
Boesch et al.
[Boesch et al., Tech Rep, Stevens Inst Tech, Hoboken NJ, 1990] showed that for given n and e there exists exactly one graph of the form G1(p, q, r) = Kp + (Sq ∪K1,r) and exactly one G2(p, q, r) = Sp ∪(Kq + K1,r) and that one of them is optimal, where K and S indicate complete and edgeless graphs, K1,r indicates a star on r +1 vertices, ∪indicates disjoint union, and + indicates complete disjoint join. We specify a general threshold graph in the form G∗ 1(a, b, c, d, . . .) = Ka + (Sb ∪(Kc + (Sd ∪· · ·))) or its complement Part of this research was conducted while the first author was visiting the Uni-versity of Rome "La Sapienza" and the ´ Ecole Polytechnique F´ ed´ erale de Lausanne, whose support is gratefully acknowledged.
c ⃝1999 John Wiley & Sons, Inc.
CCC 0364-9024/99/030283-13 284 JOURNAL OF GRAPH THEORY G∗ 2(a, b, c, d, . . .), and we prove that every optimal graph has the form G∗ 1(a, b, c, d) or G∗ 2(a, b, c, d) with b ⩽1 or c ⩽1 or d ⩽1.
c ⃝1999 John Wiley & Sons, Inc. J Graph Theory 31: 283–295, 1999 Keywords: degrees, sum of squares, threshold graphs 1. INTRODUCTION We deal with an extremal problem in graph theory: among all simple graphs with n vertices and e edges, find the ones where the sum of squares of the vertex-degrees is maximum.
This problem was studied by Boesch et al. in an unpublished work, which is also presented in . They showed that every optimal graph is a threshold graph (see below), and, for all valid (n, e), constructed two threshold graphs and proved that one of them is optimal. However, they left open the problem of characterizing all the optimal graphs.
Olpp proved an equivalent result, motivated by problems in Ramsey the-ory.
He also showed that, among the graphs on n vertices and e edges, the ones that maximize the sum of squares of the vertex-degrees also maximize the number of subgraphs (not necessarily induced) that are paths of length 2, and vice versa.
Here we add four more classes of threshold graphs to the two of Boesch et al., and show that every optimal graph belongs to one of these six classes. The classes have a very simple structure, and they are all needed, in the sense that each of them contains optimal graphs that do not belong to any of the other classes.
Moreover, they seem to contain a large proportion of optimal graphs. The complete characterization of the optimal graphs remains an open question.
It is convenient here to stress two aspects that were of crucial importance. First, Boesch et al.’s result enabled us to recognize an optimal graph, and gave us a tool to prove nonoptimality by exhibiting a better graph. Second, we made extensive use of computers. We generated tables of all the optimal threshold graphs and used them to conjecture the structure of these graphs. The tables also gave us an insight into how to transform the conjectured nonoptimal threshold graphs to optimal (or at least better) graphs, especially in Case 3 of Lemma 1. Finally, we heavily used the interactive computer algebra system Maple V for algebraic manipulations, such as factoring polynomials in 3 variables. This was essential in helping us to find our results. The reader may also wish to use a computer algebra system to check the calculations below, but this is not necessary, and can be done by hand with a little patience. Thus, the final proof is short enough to be considered a traditional proof and not a "proof by computer." Likewise, it does not formally use Boesch et al.’s result, despite the latter’s importance.
MAXIMUM SUM OF SQUARES OF DEGREES 285 2. PRELIMINARY RESULTS A simple graph G with n vertices and e edges is called an (n, e)-graph. If its degree sequence is d1, . . . , dn, we use the notation Σ(G) = n X i=1 d2 i .
If Σ(G) is maximum among all (n, e)-graphs, we say that G is optimal. Two (n, e)-graphs are called compatible. If G and G′ are compatible and Σ(G′) > Σ(G), we say that G′ is better than G.
For a graph G, the following properties are equivalent [1, 2, 4]: (1) G is a threshold graph, i.e., some hyperplane strictly separates the character-istic vectors of the stable sets of vertices of G from those of the nonstable sets; (2) G can be constructed from the one-vertex graph by repeatedly adding an isolated vertex or a universal one (a vertex adjacent to every other vertex); (3) every three distinct vertices i, j, k of G satisfy if di ⩾dj and jk is an edge, then ik is an edge.
The first property gives rise to the name "threshold graphs." We use here the second and third properties.
We mention some basic facts from , and repeat their proofs for completeness.
Fact 1. If G is an optimal (n, e)-graph, then its complement G is an optimal (n, n 2 −e)-graph.
Proof This follows from Σ(G) = n X i=1 (n −1 −di)2 = n(n −1)2 −4(n −1)e + Σ(G).
Fact 2. Deleting an isolated or universal vertex from an optimal graph gives an optimal graph.
Proof This is clear for an isolated vertex, and, therefore, also for a universal vertex by Fact 1.
Fact 3. Every optimal graph is a threshold graph.
Proof By the third property of a threshold graph, if G is a nonthreshold graph, it has three distinct vertices i, j, k with di ⩾dj, jk an edge, and ik a nonedge. If 286 JOURNAL OF GRAPH THEORY we delete jk and add ik, the change in Σ(G) is (di + 1)2 + (dj −1)2 −d2 i −d2 j = 2(di −dj + 1) > 0, hence G is not optimal.
By Facts 3 and 2 and the fact that every threshold graph has an isolated or a universal vertex (by the second property of a threshold graph), it follows that every optimal (n, e)-graph is obtained by adding an isolated vertex to an optimal (n −1, e)-graph or a universal vertex to an optimal (n −1, e −n + 1)-graph.
The main result of is the following: Theorem 1. For every valid n, e there exist threshold graphs G1, G2, unique up to isomorphism, such that G1 or G2, or both, are optimal; these graphs have the form G1(k, l, m) = Kk + (Sl ∪K1,m) G2(k, l, m) = Sk ∪(Kl + K1,m), where K and S indicate a complete and edgeless graph, K1,m indicates a star on m + 1 vertices, ∪indicates disjoint union, and + indicates complete disjoint join.
Given (n, e), we can construct G1 on the vertices v1, . . . , vn by letting k be the largest possible number of universal vertices, that is to say, we let k be the largest integer not exceeding n such that e ⩾(n −1) + (n −2) + · · · + (n −k) = 1 2k(2n −k −1), and we make v1, . . . , vk universal by connecting each of them with all the other n −1 vertices. We let m = e −1 2k(2n −k −1) be the number of additional edges we need to put in, and we attach these m edges to the single vertex vk+1 by connecting vk+1 with vk+2, . . . , vk+m+1. The remaining l = n −k −m −1 vertices vn−l+1, . . . , vn remain connected only with v1, . . . , vk. We can construct G2 on the vertices v1, . . . , vn by taking the largest possible clique Ks, that is to say, we let s be the largest integer not exceeding n such that e ⩾ s 2 , and we make {v1, . . . , vs} into a clique by connecting v1, . . . , vs with each other. We let l = e− s 2 be the number of additional edges we need to put in, and we attach these edges between the single vertex vs+1 and l vertices of the clique by connecting vs+1 with v1, . . . , vl. The remaining m = s −l vertices vl+1, . . . , vl+m of the clique remain connected only with each other, and the remaining k = n −m −l −1 vertices vn−k+1, . . . , vn of the graph remain isolated.
Fromtheconstructionoutlinedabove, itfollowsthat, foragivenn, when(k, l, m) ranges over all ordered triples of nonnegative integers that add up to n −1, the number of edges of G1(k, l, m) ranges over all the integers from 0 to n 2 . We remark that the same number of edges can be obtained from several such triples; but, as stated in the theorem, whenever G1(k, l, m) and G1(k′, l′, m′) have the same number of edges and vertices, they must be isomorphic even if (k, l, m) / = (k′, l′, m′). Similar remarks apply to G2.
MAXIMUM SUM OF SQUARES OF DEGREES 287 Theorem 1 enables us to compute the optimal Σ value for any valid choice of n, e, but leaves open the problem of where to look for all the threshold graphs that are optimal.
In the following section, we solve this problem.
3. MAIN RESULT We specify a general threshold graph in the form G∗ 1(a, b, c, d, . . .) = Ka + (Sb ∪(Kc + (Sd ∪· · ·))) or its complement G∗ 2(a, b, c, d, . . .) = Sa ∪(Kb + (Sc ∪(Kd + · · ·))).
Of particular importance are the threshold graphs G∗ 1(a, b, c, d) = Ka + (Sb ∪(Kc + Sd)) G∗ 2(a, b, c, d) = Sa ∪(Kb + (Sc ∪Kd)), which are illustrated in Fig. 1.
Notice that G1(a, b, d) = G∗ 1(a, b, 1, d) and G2(a, b, d) = G∗ 2(a, b, 1, d).
Our main result is the following.
Theorem 2. Every optimal graph has the form G∗ 1(a, b, c, d) or G∗ 2(a, b, c, d) with b ⩽1 or c ⩽1 or d ⩽1.
Proof By induction on the number of vertices. For one vertex, the theorem is trivial. Let G be an optimal graph. Then G is a threshold graph by Fact 3. By going to the complement, if necessary, we may assume by the second property of threshold graphs and Fact 1 that G = S1 ∪H. Then H is optimal by Fact 2. By FIGURE 1. The graphs G∗ 1(a, b, c, d) and G∗ 2(a, b, c, d) (the number written outside each circle is the common degree of its vertices).
288 JOURNAL OF GRAPH THEORY induction, either H = G∗ 1(a, b, c, d) or H = G∗ 2(a, b, c, d) with b ⩽1 or c ⩽1 or d ⩽1. In the second case, G = G∗ 2(a + 1, b, c, d) and we are done, so from now on we assume that H = G∗ 1(a, b, c, d) with b ⩽1 or c ⩽1 or d ⩽1.
If a = 0, then G = G∗ 2(b + 1, c, d, 0), and we are done.
If b = 0, then G = G∗ 2(1, a + c, d, 0), and we are done.
If c = 0, then G = G∗ 2(1, a, b + d, 0), and we are done.
If d = 0, then G = G∗ 2(1, a, b, c), and G has an isolated vertex. By Fact 2 and the optimality of G, G∗ 2(0, a, b, c) is also optimal; hence, by induction, a ⩽1 or b ⩽1 or c ⩽1, and we are done.
So from now on we may assume that a, b, c, d ⩾1. Since b ⩽1 or c ⩽1 or d ⩽1, we have b = 1 or c = 1 or d = 1. We examine these three cases separately.
Case 1. H = G∗ 1(a, 1, c, d), with a, c, d ⩾1.
We may assume that d ⩾2, for otherwise d = 1 and we are in Case 3 below.
Case 1.1. a ⩽d + 2 −2c. Here we get the contradiction that G is not optimal, since the transformed graph G′ = G∗ 1(a+c−1, 2c+a, 1, d+2−2c−a) is better.
Indeed, we check that they are compatible and Σ(G′) −Σ(G) = 2c(a + c −1) > 0.
Case 1.2. a ⩾d −c + 1. Here we get the contradiction that H is not optimal, since the transformed graph H′ = G∗ 1(a −m, 1, m + c, d −1 −m, 1, m), where m = min(d −1, a) > 0, is better. Indeed, we check that they are compatible and Σ(H′) −Σ(H) = 2m(m + 1 + c −d).
If m = d −1, this is 2mc > 0, and if m = a, this is 2m(a + 1 + c −d) > 2m(a −1 + c −d) ⩾0.
Case 1.3. d + 2 −2c ⩽a ⩽d −c + 1. Here we get the contradiction that G is not optimal, since the transformed graph G′ = G∗ 1(a+c−2, a+2c−d−2, 1, 2d+5− a −2c) is better (note that 2d + 5 −a −2c > 2(d + 2 −c) −a > 2a −a = a > 0).
Indeed, we check that they are compatible and Σ(G′) −Σ(G) = 4 −2a −6c + 6d + 2c2 −4cd + 2d2 −2ad + 2ac = 2(d −c + 2 −a)(d −c + 1) ⩾2(1)(a) > 0.
Case 2. H = G∗ 1(a, b, 1, d), with a, b, d ⩾1.
Case 2.1. b + d ⩽2. We have b = d = 1, so G = G∗ 2(1, a, 1, 1, 1) = G∗ 2(1, a, 1, 2) and we are done.
Case 2.2. b + d ⩾3.
Case 2.2.1. a ⩾b −1.
In this case, H = G∗ 1(a, b, 1, d) is compatible with H′ = G∗ 1(a −b + 1, 1, b, b + d −1) and Σ(H′) = Σ(H).
Therefore, G is MAXIMUM SUM OF SQUARES OF DEGREES 289 compatible with G′ = S1 ∪H′ and Σ(G′) = Σ(G). But H′ falls under Case 1, and, in particular, its last parameter b+d−1 is at least 2, because we are in Case 2.2. Therefore, by Case 1 (without appealing to Case 3), G′ or H′ is not optimal.
Therefore, G or H is not optimal, a contradiction.
Case 2.2.2. a ⩽b. G has the form S1 ∪(Ka + (Sb ∪(K1 + Sd))). Let m = min(a, d) > 0, delete the edges between K1 and m vertices in Sd, and add edges between S1 and m vertices of Ka. This gives a graph G′ (nonthreshold) compatible with G, and we can check that Σ(G′) −Σ(G) = 2m(m + b −a) > 0. Therefore, G is not optimal, a contradiction.
Case 3. H = G∗ 1(a, b, c, 1), with a, b, c ⩾1.
We have G = G∗ 2(1, a, b, c + 1), so if a = 1 or b = 1, we are done. If c = 1, we are back in Case 2.
Therefore, we may assume from now on that a, b, c ⩾2.
Case 3.1. a+c ⩽b+1. Here we get the contradiction that G is not optimal, since the transformed graph G′ = G∗ 1(1, 1, a −1, b, c −1, 2) is better. Indeed, we check that they are compatible and Σ(G′) −Σ(G) = 2(b + 2 −a −c) > 0.
Case 3.2. a + c ⩾b + 2. We prove by induction on the number of vertices that when a, b, c ⩾2, H = G∗ 1(a, b, c, 1) is not optimal, a contradiction. By Fact 1, this is equivalent to showing that for a, b, c ⩾2, G∗ 2(a, b, c, 1) is not optimal. The base case states that G∗ 2(2, 2, 2, 1) is not optimal. Indeed, it has (n, e, Σ) = (7, 7, 44), while G∗ 1(1, 4, 2) has (n, e, Σ) = (7, 7, 48). For the induction step, we have a ⩾ max(2, b + 2 −c). If this inequality is strict, then G∗ 2(a −1, b, c, 1) satisfies the same conditions a −1, b, c ⩾2 and (a −1) + c ⩾b + 2, and, therefore, is not optimal by induction. But then by Fact 2, G∗ 2(a, b, c, 1) is also not optimal. So we may assume that a = max(2, b + 2 −c). The proof of the nonoptimality of G∗ 2(a, b, c, 1) in this case is more involved, and we present it in the Lemma in next subsection.
A.
Completion of the Proof The following Lemma completes the proof of Theorem 2.
Lemma 1. For b, c ⩾2, G∗ 2(max(2, b + 2 −c), b, c, 1) is not optimal.
Proof We transform H = G∗ 2(max(2, b + 2 −c), b, c, 1) to various other compatible graphs H′ (the transformation depending on various cases that b, c satisfy). We use here the notation ∆Σ = Σ(H′) −Σ(H), and show that ∆Σ > 0 in each case.
290 JOURNAL OF GRAPH THEORY Case 1. b ⩽c ⩽2b−3. We transform H = G∗ 2(2, b, c, 1) to H′ = G∗ 1(b−2, 2b− c −3, 1, 2c −2b + 7), and check that H and H′ are compatible and ∆Σ = 12 + 2b2 + 10c −10b −4cb + 2c2 = 2(c + 3 −b)(c + 2 −b) > 0.
Case 2. c ⩾2b −2 (⩾b). We transform H = G∗ 2(2, b, c, 1) to H′ = G∗ 1(b − 1, 2b, 1, c −2b + 3), and check that H and H′ are compatible and ∆Σ = −2b + 2b2 = 2b(b −1) > 0.
Case 3. b ⩾c. The transformation will be from H = G∗ 2(b + 2 −c, b, c, 1) to the unique H′ = G1(p, q, r) = G∗ 1(p, q, 1, r) that is compatible with H, which has n = 2b + 3 vertices and e = b 2 + b(c + 1) edges. We construct H′ as in the remarks following Theorem 1 (recall that (p, q, r) need not be unique, although H′ is unique up to isomorphism). According to that construction, p is the largest number of universal vertices consistent with n and e, i.e., p is the largest integer satisfying p ⩽n and 1 2p(2n−1−p) ⩽e. Consider the quadratic x(2n−1−x)−2e in x. Since H is not a clique, this quadratic is positive at x = n. Therefore, it has distinct real roots, and n lies strictly between them. It follows that p is the floor of the smaller root, that is to say p = n −1 2 −1 2 q (2n −1)2 −8e .
We can alternatively determine p as the unique integer satisfying p(2n −1 −p) ⩽2e < (p + 1)(2n −2 −p).
Given p, we can then find r and q by r = e −1 2p(2n −1 −p) and q = n − 1 −p −r. Since b ⩾c ⩾2, we have e ⩾1 + 3b ⩾2b + 3 = n, and, therefore, p ⩾1. By the construction of p, we have q ⩾1 (if q = 0, we could increase p by 1), but r can be zero as well as positive. Figure 2 tabulates ∆Σ and p, q, r (called S, P, Q, R, respectively 1) as functions of b, c for b ⩾c ⩾2. Below we prove that ∆Σ is indeed positive.
We call the collection of pairs (b, c), b ⩾c ⩾2 for which p is constant the zone of p. The lines in Fig. 2 mark the borders between zones. For each fixed c, the largest (smallest) value of b such that (b, c) is in the zone of p is called the bottom (top) of the zone. For example, for c = 2, the top of the zone for p = 3 is at b = 8 and the bottom at b = 11.
Since p, q, r are complicated functions of b, c, we express q and r in terms of p, b, c. This gives Q(p, b, c) = −1 2b2 + 3 2 −c + 2p b + 3 2p + 2 −1 2p2 1 S.P.Q.R. stands for the Latin Senatus Populusque Romanus, meaning "the Senate and the people of Rome." This is a tribute to Rome, where most of this work was done.
MAXIMUM SUM OF SQUARES OF DEGREES 291 FIGURE 2.
S,P,Q,R values vs. b, c (the lines delimit zones defined by constant P).
R(p, b, c) = +1 2b2 + 1 2 + c −2p b + 1 2p2 −5 2p.
Now we express ∆Σ = Σ(H′) −Σ(H) as a function of p, b, c: ∆Σ(p, b, c) = Σ(G∗ 1(p, Q(p, b, c), 1, R(p, b, c))) −Σ(G∗ 2(b + 2 −c, b, c, 1) = −4cb2p + 1 2b −pcb −1 4p2 + p2cb + 3 2p + cb −1 4b2 + 11 2 pb −3 2p3 + 9 2bp2 −2cb2 + 3 2pb2 + 9 2p2b2 −2p3b + c2b2 + cb3 −2b3p −c2b + 1 4b4 −1 2b3 + 1 4p4.
We prove that ∆Σ is always positive by induction on b. First we show that ∆Σ is positive on the diagonal b = c. Then we show that ∆Σ is a nondecreasing function of b for fixed c, p (i.e., within a zone). Finally, we show that when we move from the bottom of a zone to the top of the next one, ∆Σ increases.
Firstwefindp, q, r onthediagonalb = c. Itiseasytoverifyfromtheconstruction of p, q, r that for b = c = 2 we have (p, q, r) = (1, 4, 1), for b = c = 3 we have (p, q, r) = (2, 6, 0), and for b = c ⩾4 we have (p, q, r) = (c −2, c −3, 7) (using the values of n and e we verify that (c −2)(2n −1 −(c −2)) ⩽2e < (c −1)(2n −1 −(c −1))).
Before proving that ∆Σ > 0, we establish some properties of the zones in the next four assertions.
Assertion 1. 2p < b + c −1, or equivalently 2p ⩽b + c −2.
Proof We have seen that p = n −1 2 −1 2 q (2n −1)2 −8e , 292 JOURNAL OF GRAPH THEORY where n = 2b + 3 and e = b 2 + b(c + 1). To prove that p < (b + c −1)/2, it is sufficient to show that n −1 2 −1 2 p (2n −1)2 −8e < (b + c −1)/2, or equivalently that 2n−b−c < p (2n −1)2 −8e. Since both sides are positive, we may equivalently show that (2n−b−c)2 < (2n−1)2 −8e. After we substitute the values of n and e and simplify, the required inequality becomes c2 + 2bc + 11 < 3b2 + 12c, which follows in turn from b ⩾c ⩾2. This proves Assertion 1.
Assertion 2. For fixed c, p is a nondecreasing function of b.
Proof Asbefore, pisthelargestintegersatisfyingp ⩽nandp(2n−1−p) ⩽2e.
If we keep c constant and increase b by 1, n becomes n′ = n + 2, e becomes e′ = e + b + c + 1, and the corresponding p′ is the largest integer x satisfying x ⩽n′ and x(2n′ −1 −x) ⩽2e′. Since p ⩽n′, our assertion that p ⩽p′ amounts to showing that p(2n′ −1 −p) ⩽2e′. By Assertion 1, we have 2p ⩽b + c −2. If we add twice this inequality to p(2n −1 −p) ⩽2e, we obtain p(2n + 3 −p) ⩽ 2(e + b + c −2) < 2(e + b + c + 1), which is the required inequality. This proves Assertion 2.
From Assertion 2 and the values of p on the diagonal, it follows that p ⩾c −2 for c ⩾4, and p ⩾c −1 for c = 2, 3.
Assertion 3. Consecutive zones for the same c have consecutive values of p, i.e., as b increases by 1 from the bottom of a zone to the top of the next zone, p increases by 1.
Proof Let b be the bottom of zone p for c. As before, p is the unique integer satisfying p(2n −1 −p) ⩽2e < (p + 1)(2n −2 −p). If we increase b by 1, n increases by 2 and e increases by b + c + 1. Also p increases by 1 or more by Assertion 2, since b is the bottom. We show that p + 2 is too large for the next zone by showing that (p + 2)(2(n + 2) −1 −(p + 2)) > 2(e + b + c + 1). Indeed, (p + 2)(2n + 1 −p) −2(e + b + c + 1) = [(p + 1)(2n −2 −p) −2e] + 2 [n −b −c −1] + 2(p + 2).
The expression in the left brackets is positive, as seen above, the one in the second brackets is positive by n = 2b + 3 > b + c + 1, and, of course, 2(p + 2) is also positive. This proves Assertion 3.
Assertion 4. The distance between the bottoms of consecutive zones for the same c is 2 or more.
Proof By Assertion 3, the zone just below zone p is zone p + 1. We have to show that the bottom of zone p + 1 is larger than the bottom of zone p by at least 2. As before, p is the unique integer satisfying p(2n −1 −p) ⩽2e < MAXIMUM SUM OF SQUARES OF DEGREES 293 (p + 1)(2n −2 −p). Equivalently, for given b and c, p is the unique integer satisfying p(4b+5−p) ⩽b(b−1)+2b(c+1) < (p+1)(4b+4−p). Conversely, for given p and c, the bottom of zone p for c is the largest integer b satisfying these two inequalities. Clearly, the left inequality always holds for b large enough.
Therefore, for fixed c, the bottom of zone p for c is the largest integer b satisfying b(b −1)+2b(c+1)−(p+1)(4b+4−p) < 0. The expression on the left, considered as a quadratic in continuous b for fixed p and c, is b2 + (2c −4p −3)b + (p + 1) (p −4), and its roots are 2p −c + 3 2 ± p 12p2 + (36 −16c)p + (4c2 −12c + 25).
We know that the roots are real and distinct, and, in fact, there is an integer value of b for which the quadratic is negative, namely any b in zone p for c. Therefore, the bottom of zone p for c is the floor of the larger root, namely 2p −c + 3 2 + q 12p2 + (36 −16c)p + (4c2 −12c + 25) .
Our task is to show that this expression increases by at least 2, when p increases by 1 and c stays constant. The term 2p itself contributes an increase of 2. Therefore, it is sufficient to prove that the argument under the square root is a nondecreasing function of p. This argument is a quadratic in p, and is increasing with p if p ⩾ 16c−36 24 = 2 3c −3 2. We have seen that p ⩾c −2, and c −2 ⩾2 3c −3 2 for all c ⩾2.
This proves Assertion 4.
Now we are able to show that ∆Σ is positive when b ⩾c ⩾2.
Let us evaluate ∆Σ on the diagonal b = c. Recall that on the diagonal, p = 1 if c = 2, p = 2 if c = 3, and p = c −2 if c ⩾4. Substituting these values into ∆Σ(p, b, c), we find ∆Σ(1, 2, 2) = 4 ∆Σ(2, 3, 3) = 12 ∆Σ(c −2, c, c) = 12.
Thus, ∆Σ is positive on the diagonal.
Now we examine the change in ∆Σ as b increases by 1 and we stay in zone p: ∆∆Σ = ∆Σ(p, b + 1, c) −∆Σ(p, b, c) = (p2 −5p −4pb + b2 + 2bc + b)(c −1 + b −2p).
We want to show that ∆∆Σ ⩾0 by showing that both of its factors are nonnegative.
Consider the linear factor c −1 + b −2p. First, we examine it on the diagonal b = c. For b = c ⩾4, we have p = c −2 and c −1 + b −2p = 3; for b = c = 3 we have p = 2 and c −1 + b −2p = 1; and for b = c = 2, we have p = 1 and c −1 + b −2p = 1. The linear factor certainly increases within a zone as b increases and p stays constant. Consider first the general case that zone p for c does not contain the diagonal term b = c. Then the total increase from the top of zone p to its bottom is at least 1 by Assertion 4. As we continue from the bottom of zone p to the top of the next zone, p increases by 1 by Assertion 3, and b increases by 1, so c −1 + b −2p decreases by 1, but this decrease is offset by the total increase in zone p. Now consider the special case that zone p for c contains the diagonal term 294 JOURNAL OF GRAPH THEORY b = c. If we go from the diagonal to the top of the next zone, p increases by 1 and b increases by at least 1. That might decrease c −1 + b −2p by 1, but it remains nonnegative, since it was positive on the diagonal. It follows that the linear factor c −1 + b −2p of ∆∆Σ is always nonnegative.
Now consider the quadratic factor p2 −5p−4pb+2bc+b2 +b of ∆∆Σ. Recall that p(2n −1 −p) ⩽2e, and so p2 −2np + p + 2e ⩾0, where n = 2b + 3 and e = b 2 + b(c + 1). Substituting the given values of n and e, we find that the left-hand side of the inequality is exactly our quadratic factor, which is, therefore, nonnegative.
Therefore, ∆∆Σ is the product of two nonnegative factors and is nonnegative.
This shows that ∆Σ is nondecreasing when b increases, as long as we stay within a zone.
Now we examine the change in ∆Σ from the bottom of a zone to the top of the next zone: ∆∆Σ = ∆Σ(p + 1, b + 1, c) −∆Σ(p, b, c) = (3 −c + p + b)(−b2 + 3b −2cb + 4pb + 3p + 4 −p2) = 2(3 −c + p + b)Q.
Since Q > 0 and b ⩾c, ∆∆Σ is positive, so ∆Σ increases as we pass from the bottom of a zone to the top of the next zone.
We have shown that ∆Σ is positive on the diagonal, is nondecreasing within a zone, and increases in passing from the bottom of a zone to the top of the next zone.
Therefore, ∆Σ is positive below the diagonal.
This proves the Lemma.
4. FINAL REMARKS Clearly, our six classes of threshold graphs containing all the optimal ones are small subclasses of all threshold graphs. Furthermore, each of the classes contains graphs that are not in the other five. Indeed, G∗ 1(2, 1, 2, 8), G∗ 1(1, 4, 1, 2), G∗ 1(2, 4, 2, 1), and their complements are examples of optimal graphs that belong to only one class. Further questions suggested by this work are the existence and uniqueness of the (n, e)-graphs in each class, and precise optimality criteria.
ACKNOWLEDGMENT We thank two anonymous referees of an earlier version for reference , and for helpful suggestions concerning the proof.
MAXIMUM SUM OF SQUARES OF DEGREES 295 References V. Chv´ atal and P. L. Hammer, Aggregation of inequalities in integer program-ming, Ann Discrete Math 1 (1977), 145–162.
P. B. Henderson and Y. Zalcstein, A graph-theoretic characterization of the PVchunk class of synchronizing primitives, SIAM J Comp 6 (1977), 88– 108.
F. Boesch, R. Brigham, S. Burr, R. Dutton, and R. Tindell, Maximizing the sum of the squares of the degrees of a graph, Tech Rep, Stevens Inst Tech, Hoboken, NJ, c. 1990.
N. V. R. Mahadev and U. N. Peled, Threshold graphs and related topics, Ann Discrete Math 56, North–Holland, Amsterdam, 1995.
D. Olpp, A conjecture of Goodman and the multiplicities of graphs, Aust J Combin 14 (1996), 267–282. |
188113 | https://en.wikipedia.org/wiki/Least_common_multiple | Jump to content
Least common multiple
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From Wikipedia, the free encyclopedia
Smallest positive number divisible by two integers
In arithmetic and number theory, the least common multiple (LCM), lowest common multiple, or smallest common multiple (SCM) of two integers a and b, usually denoted by lcm(a, b), is the smallest positive integer that is divisible by both a and b. Since division of integers by zero is undefined, this definition has meaning only if a and b are both different from zero. However, some authors define lcm(a, 0) as 0 for all a, since 0 is the only common multiple of a and 0.
The least common multiple of the denominators of two fractions is the "lowest common denominator" (lcd), and can be used for adding, subtracting or comparing the fractions.
The least common multiple of more than two integers a, b, c, . . . , usually denoted by lcm(a, b, c, . . .), is defined as the smallest positive integer that is divisible by each of a, b, c, . . .
Overview
[edit]
A multiple of a number is the product of that number and an integer. For example, 10 is a multiple of 5 because 5 × 2 = 10, so 10 is divisible by 5 and 2. Because 10 is the smallest positive integer that is divisible by both 5 and 2, it is the least common multiple of 5 and 2. By the same principle, 10 is the least common multiple of −5 and −2 as well.
Notation
[edit]
The least common multiple of two integers a and b is denoted as lcm(a, b). Some older textbooks use [a, b].
Example
[edit]
Multiples of 4 are:
Multiples of 6 are:
Common multiples of 4 and 6 are the numbers that are in both lists:
In this list, the smallest number is 12. Hence, the least common multiple is 12.
Applications
[edit]
When adding, subtracting, or comparing simple fractions, the least common multiple of the denominators (often called the lowest common denominator) is used, because each of the fractions can be expressed as a fraction with this denominator. For example,
where the denominator 42 was used, because it is the least common multiple of 21 and 6.
Gears problem
[edit]
Suppose there are two meshing gears in a machine, having m and n teeth, respectively, and the gears are marked by a line segment drawn from the center of the first gear to the center of the second gear. When the gears begin rotating, the number of rotations the first gear must complete to realign the line segment can be calculated by using . The first gear must complete rotations for the realignment. By that time, the second gear will have made rotations.
Planetary alignment
[edit]
See also: Syzygy (astronomy)
Suppose there are three planets revolving around a star which take l, m and n units of time, respectively, to complete their orbits. Assume that l, m and n are integers. Assuming the planets started moving around the star after an initial linear alignment, all the planets attain a linear alignment again after units of time. At this time, the first, second and third planet will have completed , and orbits, respectively, around the star.
Calculation
[edit]
There are several ways to compute least common multiples.
Using the greatest common divisor
[edit]
The least common multiple can be computed from the greatest common divisor (gcd) with the formula
To avoid introducing integers that are larger than the result, it is convenient to use the equivalent formulas
where the result of the division is always an integer.
These formulas are also valid when exactly one of a and b is 0, since gcd(a, 0) = |a|. However, if both a and b are 0, these formulas would cause division by zero; so, lcm(0, 0) = 0 must be considered as a special case.
To return to the example above,
There are fast algorithms, such as the Euclidean algorithm for computing the gcd that do not require the numbers to be factored. For very large integers, there are even faster algorithms for the three involved operations (multiplication, gcd, and division); see Fast multiplication. As these algorithms are more efficient with factors of similar size, it is more efficient to divide the largest argument of the lcm by the gcd of the arguments, as in the example above.
Using prime factorization
[edit]
The unique factorization theorem indicates that every positive integer greater than 1 can be written in only one way as a product of prime numbers. The prime numbers can be considered as the atomic elements which, when combined, make up a composite number.
For example:
Here, the composite number 90 is made up of one atom of the prime number 2, two atoms of the prime number 3, and one atom of the prime number 5.
This fact can be used to find the lcm of a set of numbers.
Example: lcm(8,9,21)
Factor each number and express it as a product of prime number powers.
The lcm will be the product of multiplying the highest power of each prime number together. The highest power of the three prime numbers 2, 3, and 7 is 23, 32, and 71, respectively. Thus,
This method is not as efficient as reducing to the greatest common divisor, since there is no known general efficient algorithm for integer factorization.
The same method can also be illustrated with a Venn diagram as follows, with the prime factorization of each of the two numbers demonstrated in each circle and all factors they share in common in the intersection. The lcm then can be found by multiplying all of the prime numbers in the diagram.
Here is an example:
: 48 = 2 × 2 × 2 × 2 × 3,
: 180 = 2 × 2 × 3 × 3 × 5,
sharing two "2"s and a "3" in common:
: Least common multiple = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720
: Greatest common divisor = 2 × 2 × 3 = 12
: Product = 2 × 2 × 2 × 2 × 3 × 2 × 2 × 3 × 3 × 5 = 8640
This also works for the greatest common divisor (gcd), except that instead of multiplying all of the numbers in the Venn diagram, one multiplies only the prime factors that are in the intersection. Thus the gcd of 48 and 180 is 2 × 2 × 3 = 12.
Formulas
[edit]
Fundamental theorem of arithmetic
[edit]
According to the fundamental theorem of arithmetic, every integer greater than 1 can be represented uniquely as a product of prime numbers, up to the order of the factors:
where the exponents n2, n3, ... are non-negative integers; for example, 84 = 22 31 50 71 110 130 ...
Given two positive integers and , their greatest common divisor and least common multiple are given by the formulas
and
Since
this gives
In fact, every rational number can be written uniquely as the product of primes, if negative exponents are allowed. When this is done, the above formulas remain valid. For example:
Lattice-theoretic
[edit]
The positive integers may be partially ordered by divisibility: if a divides b (that is, if b is an integer multiple of a) write a ≤ b (or equivalently, b ≥ a). (Note that the usual magnitude-based definition of ≤ is not used here.)
Under this ordering, the positive integers become a lattice, with meet given by the gcd and join given by the lcm. The proof is straightforward, if a bit tedious; it amounts to checking that lcm and gcd satisfy the axioms for meet and join. Putting the lcm and gcd into this more general context establishes a duality between them:
: If a formula involving integer variables, gcd, lcm, ≤ and ≥ is true, then the formula obtained by switching gcd with lcm and switching ≥ with ≤ is also true. (Remember ≤ is defined as divides).
The following pairs of dual formulas are special cases of general lattice-theoretic identities.
| | | | | |
--- ---
| Commutative laws | | Associative laws | | Absorption laws |
| | | |
---
| Idempotent laws | | Define divides in terms of lcm and gcd |
It can also be shown that this lattice is distributive; that is, lcm distributes over gcd and gcd distributes over lcm:
This identity is self-dual:
Other
[edit]
Let D be the product of ω(D) distinct prime numbers (that is, D is squarefree).
Then
where the absolute bars || denote the cardinality of a set.
If none of is zero, then
:
In commutative rings
[edit]
The least common multiple can be defined generally over commutative rings as follows:
Let a and b be elements of a commutative ring R. A common multiple of a and b is an element m of R such that both a and b divide m (that is, there exist elements x and y of R such that ax = m and by = m). A least common multiple of a and b is a common multiple that is minimal, in the sense that for any other common multiple n of a and b, m divides n.
In general, two elements in a commutative ring can have no least common multiple or more than one. However, any two least common multiples of the same pair of elements are associates. In a unique factorization domain, any two elements have a least common multiple. In a principal ideal domain, the least common multiple of a and b can be characterised as a generator of the intersection of the ideals generated by a and b (the intersection of a collection of ideals is always an ideal).
See also
[edit]
Anomalous cancellation
Coprime integers
Chebyshev function
Notes
[edit]
^ Jump up to: a b c Weisstein, Eric W. "Least Common Multiple". mathworld.wolfram.com. Retrieved 2020-08-30.
^ Hardy & Wright, § 5.1, p. 48
^ Jump up to: a b Long (1972, p. 39)
^ Pettofrezzo & Byrkit (1970, p. 56)
^ "nasa spacemath" (PDF).
^ The next three formulas are from Landau, Ex. III.3, p. 254
^ Crandall & Pomerance, ex. 2.4, p. 101.
^ Long (1972, p. 41)
^ Pettofrezzo & Byrkit (1970, p. 58)
^ Jump up to: a b Burton 1970, p. 94.
^ Grillet 2007, p. 142.
References
[edit]
Burton, David M. (1970). A First Course in Rings and Ideals. Reading, MA: Addison-Wesley. ISBN 978-0-201-00731-2.
Crandall, Richard; Pomerance, Carl (2001), Prime Numbers: A Computational Perspective, New York: Springer, ISBN 0-387-94777-9
Grillet, Pierre Antoine (2007). Abstract Algebra (2nd ed.). New York, NY: Springer. ISBN 978-0-387-71568-1.
Hardy, G. H.; Wright, E. M. (1979), An Introduction to the Theory of Numbers (Fifth edition), Oxford: Oxford University Press, ISBN 978-0-19-853171-5
Landau, Edmund (1966), Elementary Number Theory, New York: Chelsea
Long, Calvin T. (1972), Elementary Introduction to Number Theory (2nd ed.), Lexington: D. C. Heath and Company, LCCN 77-171950
Pettofrezzo, Anthony J.; Byrkit, Donald R. (1970), Elements of Number Theory, Englewood Cliffs: Prentice Hall, LCCN 77-81766
| Fractions and ratios | | | |
--- --- |
| | Division and ratio | Dividend ÷ Divisor = Quotient | |
| Fraction | Numerator/Denominator = Quotient |
| Algebraic Aspect Binary Continued Decimal Dyadic Egyptian Golden + Silver Integer Irreducible + Reduction Just intonation LCD Musical interval Paper size Percentage Unit | |
Retrieved from "
Categories:
Elementary arithmetic
Operations on numbers
Number theory
Hidden categories:
Articles with short description
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188114 | https://www.sciencedirect.com/topics/neuroscience/vestibular-aqueduct | Skip to main content
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Outline
1. Introduction to the Vestibular Aqueduct
2. Anatomy and Physiology of the Vestibular Aqueduct
3. Clinical Significance and Pathophysiology
4. Diagnostic Techniques and Neuroimaging
5. Conclusion
Topic summaryAI
1. Introduction to the Vestibular Aqueduct
The vestibular aqueduct is a J-shaped osseous channel extending from the vestibule of the inner ear to the posterior cranial fossa near the sigmoid sinus, housing the endolymphatic duct and sac within its course. It originates in the vestibule, located between the cochlea and the three semicircular canals, with its medial wall giving rise to the vestibular aqueduct. Anatomically, it is situated at or above the level of the internal auditory canals and oriented in a plane perpendicular to them. The endolymphatic duct, contained within the vestibular aqueduct, arises from the saccule and runs inside the aqueduct, becoming distally dilated to form the endolymphatic sac, which terminates between the two dural layers on the posterior surface of the petrous temporal bone, close to the sigmoid sinus.
The vestibular aqueduct serves as a conduit for the endolymphatic duct and sac, which play a critical role in regulating endolymph fluid and maintaining vestibular function. Endolymph produced in other areas of the labyrinth is absorbed in the endolymphatic sac, a process essential for the homeostasis of inner ear fluids. The vestibular system, including the vestibular aqueduct, is responsible for balance and spatial orientation by detecting endolymphatic flow and transmitting signals via the vestibular nerves to the vestibular nuclei in the brainstem.
2. Anatomy and Physiology of the Vestibular Aqueduct
The vestibular aqueduct is a J-shaped bony channel extending from the vestibule to the posterior cranial fossa near the sigmoid sinus, widening at the isthmus and opening distally at the operculum into the posterior cranial fossa. Within it lie the endolymphatic duct and sac, which regulate endolymph fluid. The endolymphatic duct originates in the saccule and runs inside the vestibular aqueduct until it reaches the subarachnoid space, where it dilates to form the endolymphatic sac. The sac narrows at its isthmus to 0.1 to 0.2 mm in diameter.
The epithelial lining of the endolymphatic duct consists of low cuboidal or squamous cells , while the endolymphatic sac displays complex epithelial folds with papillae and crypts of tall columnar light and dark cells. Light cells have long surface microvilli and endocytic invaginations, whereas dark cells are wedge-shaped with dense fibrillary cytoplasm and fewer apical microvilli. The sac’s epithelium is permeable to macrophages and other leukocytes, facilitating removal of cellular debris and contributing antibodies to the endolymph.
Physiologically, the endolymphatic sac is essential for maintaining vestibular function by absorbing endolymph produced elsewhere in the labyrinth. The blood supply to the endolymphatic sac differs from other inner ear structures; it is provided by the middle meningeal artery or occipital artery, part of the external carotid arterial system, whereas the cochlea and vestibular apparatus receive blood from the anterior inferior cerebellar artery, part of the cerebral circulation system. Veins draining the vestibule and semicircular canals form the vein of the vestibular aqueduct, which empties into the sigmoid or inferior petrosal sinus.
3. Clinical Significance and Pathophysiology
Enlarged vestibular aqueduct syndrome (EVAS) is the most common congenital inner ear malformation identified in children and is associated with early onset sensorineural or mixed hearing loss, which may be fluctuating or progressive and can be exacerbated by minor head injuries or barotrauma. Hearing loss in EVAS is typically bilateral and progressive, with stepwise decrements often triggered by relatively minor head trauma. Vestibular symptoms, including episodic vertigo lasting minutes to hours, may occur and are often brought on by minor trauma or vigorous head rotation. Vestibular symptoms are less frequent than hearing loss and commonly begin in childhood but may be delayed until adulthood.
The pathophysiology of EVAS includes exposure of the cochlea and vestibule to intracranial pressure variations through a widened vestibular aqueduct or neuroepithelial damage secondary to reflux of hyperosmolar contents from the endolymphatic sac. Developmental arrest of the endolymphatic canal and sac around the fifth to eighth weeks in utero is proposed as a mechanism, resulting in a shortened, straighter, and proportionally broader duct in early embryogenesis. The vestibular system is considered more resistant to these insults compared to the auditory system, which may explain preserved vestibular function in the presence of profound sensorineural hearing loss.
Genetic factors play a significant role, with mutations in the SLC26A4 gene causing Pendred syndrome and nonsyndromic enlarged vestibular aqueduct. Pendred syndrome is an autosomal recessive disorder characterized by sensorineural hearing loss, goiter, and enlargement of the vestibular aqueduct, caused by biallelic mutations in SLC26A4, which encodes the anion transporter pendrin. Mutations in FOXI1 and KCNJ10 have also been implicated in digenic inheritance of enlarged vestibular aqueduct and Pendred syndrome. Pendrin regulates the ionic composition and pH of cochlear endolymph, and its deficiency leads to degeneration of sensory cells in the inner ear.
Endolymphatic sac tumors (ELSTs) arise from the endolymphatic epithelium within the vestibular aqueduct and are rare in the general population but found in up to 6–15% of patients with von Hippel-Lindau disease. ELSTs can cause progressive and typically irreversible hearing loss, vestibular symptoms, tinnitus, and facial paresis. Bilateral ELSTs are found only in von Hippel-Lindau disease. Imaging features include destruction of the retro-labyrinthine petrous bone and heterogeneous enhancement on magnetic resonance imaging (MRI).
The vestibular aqueduct is also implicated in Meniere’s disease, where narrowing of the aqueduct and reduced resorptive capacity of the endolymphatic sac are associated with endolymphatic hydrops and vestibular symptoms. Congenital and acquired causes, including malformations of the vestibular aqueduct, may contribute to secondary Meniere’s syndromes.
4. Diagnostic Techniques and Neuroimaging
High-resolution computed tomography (CT) is used for evaluating the bony anatomy of the vestibular aqueduct, detecting enlargement, and assessing malformations, particularly in very small tumors. CT is used to measure the vestibular aqueduct diameter, with the Valvassori criteria defining enlargement as a diameter greater than 1.5 mm at the midpoint between the crus communis and the external aperture, and the Cincinnati criteria specifying a midpoint width ≥1.0 mm or opercular width ≥2.0 mm in the axial plane. Enlargement is also considered present if the aqueduct’s calibre exceeds that of the adjacent posterior semicircular canal on axial images.
Magnetic resonance imaging (MRI), especially heavily T2-weighted sequences, is used to visualize the endolymphatic duct and sac, revealing abnormalities such as enlargement and endolymphatic hydrops. MRI can identify endolymphatic sac tumors by showing subtle asymmetric enhancement of the endolymphatic sac in small tumors and heterogeneous enhancement in larger tumors on postcontrast T1-weighted images. Heavily T2-weighted three-dimensional fluid-attenuated inversion recovery sequences on 3-Tesla scanners offer optimal images for detecting endolymphatic hydrops, with gadolinium administration allowing visualization of the distended endolymphatic space. MRI is also used to document progression of disease.
Imaging plays a critical role in surgical planning, particularly in assessing the extent of bony erosion, otic capsule invasion, and the relationship of the vestibular aqueduct to adjacent structures. Advanced MRI techniques, such as fluid-attenuated inversion recovery and thin-section T2-weighted imaging, are emerging for the assessment of endolymphatic hydrops.
5. Conclusion
The vestibular aqueduct is a bony canal extending from the medial wall of the vestibule to the posterior surface of the petrous temporal bone, serving as a conduit for the endolymphatic duct, which connects the endolymphatic sac to the vestibular labyrinth. The endolymphatic duct and sac within the vestibular aqueduct play a crucial role in regulating endolymph fluid. Enlargement of the vestibular aqueduct is the most common congenital inner ear malformation identified in children and is associated with sensorineural or mixed hearing loss, which can be fluctuating or progressive and exacerbated by minor head injuries or barotrauma. Vestibular symptoms, such as episodic vertigo, may also occur and are often triggered by minor trauma or vigorous head rotation.
Enlarged vestibular aqueduct is commonly associated with syndromic hearing loss, including Pendred syndrome and branchio-oto-renal syndrome, and may present with other inner ear malformations such as cochlear anomalies and vestibular defects. The condition is characterized by abnormal imaging findings, with diagnostic criteria including a diameter greater than 1.5 mm at the midpoint and greater than 2.0 mm at the operculum on computed tomography. The Cincinnati criteria specify a vestibular aqueduct width of >0.9 mm at the midpoint and >1.9 mm at the operculum.
Genetic studies have identified mutations in the SLC26A4 gene as a cause of Pendred syndrome, which is associated with bilateral enlarged vestibular aqueduct and thyroid pathology. Imaging advances, including high-resolution computed tomography and magnetic resonance imaging, are used for diagnosis and for identifying associated anomalies such as modiolar deficiency and scalar asymmetry.
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Neuroimaging Clinics of North America , 2022 pp 763-775
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The biology of intratympanic drug administration and pharmacodynamics of round window drug absorption
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"... The lateral wall of the vestibule incorporates the oval window, and the medial wall gives rise to the vestibular aqueduct. Posteriorly there are five openings for the semicircular canals and anteriorly an elliptical opening leads into the scala vestibuli of the cochlea. ..."
Reference 3
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"... The afferent branches then run from the spiral ganglia through multiple small openings at the distal IAC called the cribriform plate. Just proximal to Scarpa's ganglion and just distal to the cribriform plate is the Schwann-glial cell junction of the cochlear division of the eighth cranial nerve. 8 The bony vestibule, located between the cochlea and 3 semicircular canals, contains the utricle and saccule. The ampulated ends of the semicircular canals open into the utricle. The saccule communicates with the scala media of the cochlea via the ductus reuniens. Two small ducts, one each from the saccule and utricle, unite to form the endolymphatic duct. As its name implies, this endolymph-containing duct is also part of the membranous labyrinth and is housed within the bony vestibular aqueduct. The afferent nerve fibers traveling from the vestibular hair cells of the semicircular canals, utricle, and saccule pierce the cribriform plate of the distal IAC en route to the Scarpa's ganglia within the IAC. Unlike the cochlear division, the Schwann-glial cell junction of the vestibular divisions is at the Scarpa's ganglia ( Fig. 2 ). 9 With this anatomy in mind, it is helpful to refer to the revised Kennedy classification system in order to describe tumor locations ( Fig. 3 ). Pathogenesis Originally, it was thought that ILS occurred mainly in patients with neurofibromatosis type 2 (NF2). 10 However, more recent reviews have shown that the overall number of sporadic cases far outnumber NF2 cases because of the rarity of the latter. The genetics of sporadic ILS are not thought to be different from the more common vestibular schwannoma of the IAC and CPA, although confirmatory studies are still needed. ..."
Reference 4
Review article
Imaging for cochlear implantation: Structuring a clinically relevant report
Vaid S., Vaid N.
Clinical Radiology , 2014 pp e307-e322
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"... Vestibular aqueduct (VA) The vestibular aqueduct is located at or above the level of the IACs, and oriented in a plane perpendicular to the IACs. The vestibular aqueduct is considered to be dilated ( Fig 18 ) if it measures >1.5 mm in width at the midpoint between the common crus and its external aperture or if its calibre is more than that of the adjacent posterior semicircular canal on axial images. 10,12 An enlarged endolymphatic duct/sac is identified as a hyperintense structure along the posterior and medial aspect of the petrous bone on heavily T2W MRI sequences. The signal may be highly variable reflecting the hyperviscous or variable protein content of the fluid within the sac. 12 ..."
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"... Aqueducts ( Fig 17 ) Vestibular aqueduct (VA) The vestibular aqueduct is located at or above the level of the IACs, and oriented in a plane perpendicular to the IACs. The vestibular aqueduct is considered to be dilated ( Fig 18 ) if it measures >1.5 mm in width at the midpoint between the common crus and its external aperture or if its calibre is more than that of the adjacent posterior semicircular canal on axial images. 10,12 An enlarged endolymphatic duct/sac is identified as a hyperintense structure along the posterior and medial aspect of the petrous bone on heavily T2W MRI sequences. The signal may be highly variable reflecting the hyperviscous or variable protein content of the fluid within the sac. 12 Cochlear aqueduct The cochlear aqueduct is located below the level of the IAC and is oriented in a plane parallel to it. It connects the scala tympani to the subarachnoid space of the posterior cranial fossa. The cochlear aqueduct has a medial orifice and lateral orifice; the medial orifice is considered to be dilated if it measures >3 mm. 12 Clinical relevance 1. Large vestibular aqueduct syndrome (LVAS): the incidence of enlarged vestibular aqueduct varies from 5–15% in the paediatric population. 12 A dilated vestibular aqueduct associated with symptoms of hearing loss or imbalance is referred to as LVAS. LVAS can be associated with cochlear defects and is associated with a high chance of an intraoperative CSF gusher. 10 2. The cochlear aqueduct acts as a route for the transmission of infection from the brain to the scala tympani. 12 ..."
Reference 5
Book Chapter
The auditory system
Jahangir Moini, Anthony LoGalbo, Raheleh Ahangari
Foundations of the Mind, Brain, and Behavioral Relationships , 2024 pp 143-160
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"... The spiral limbus' interdental cells actually secrete the membrane. Endolymphatic duct and sac The endolymphatic duct becomes distally dilated, forming the endolymphatic sac. It is within the osseous vestibular aqueduct, is of varying size, and can extend through an opening on the posterior petrous bone surface. It terminates between the two dural layers on the petrous temporal bone's posterior surface, close to the sigmoid sinus . Through the entire duct, surface cells resemble the cells lining unspecialized areas of the membranous labyrinth. They have a low cuboidal or squamous epithelium. Where the duct dilates and forms the endolymphatic sac, the epithelia lining the subepithelial connective tissue become more complex. A distal sac and an intermediate segment are visible. In this segment, the epithelium has light and dark cylindrical cells. The light cells are regular in form, with many long surface microvilli and endocytic invaginations between them. In their apical region, they have vesicles that are large and clear. The dark cells are wedge-shaped with narrow bases, dense fibrillary cytoplasm, and much less apical microvilli. The endolymphatic sac is important for maintaining vestibular function. Endolymph created in other labyrinth areas is absorbed in this region. This may occur primarily by the light cells. If the sac is damaged or the connection to the rest of the labyrinth is blocked, endolymph will accumulate. This causes hydrops , affecting cochlear and vestibular function. The epithelium is permeable to macrophages and other leukocytes, which can remove cellular debris from the endolymph. The epithelium is also permeable to various immune system cells that contribute antibodies to this fluid. Cochlear nerve The cochlear nerve connects the cochlear nuclei and related brainstem nuclei to the organ of Corti. ..."
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"... Below the crest, an anterior cochlear region has small holes in a spiral, known as the tractus spiralis foraminosus , surrounding the central cochlear canal. Behind this, the inferior vestibular region has openings for the saccular nerves. Most posteroinferiorly, the foramen singular allows the nerve to enter the posterior semicircular duct. Vascular loops in the internal acoustic meatus from the anterior inferior cerebellar artery may cause pulsatile tinnitus. Tinnitus Tinnitus is described as sounds such as ringing, buzzing, or clicking when there is no auditory stimulus to cause them. It is usually a symptom of a disease, such as cochlear nerve degeneration, or can signify inflammation of the middle or inner ear. It may be caused by the destruction of auditory pathway neurons and the ingrowth of adjacent neurons, whose signals are interpreted as noise by the brain. Vascular supply The labyrinthine artery is the main source of blood for the inner ear. The semicircular canals are supplied by the stylomastoid branch of the occipital artery or posterior auricular artery. The labyrinthine artery divides from the basilar artery and sometimes from the anterior inferior cerebellar artery. At the bottom of the internal acoustic meatus, it divides into cochlear and vestibular branches. The cochlear branch subdivides into 12–14 tiny branches through the modiolus canals. These are distributed as capillary plexuses to cochlear structures, including the basilar membrane, spiral lamina, and stria vascularis. The utricle, saccule, and semicircular ducts are supplied by vestibular arterial branches. Veins that drain the vestibule and semicircular canals accompany related arteries. They connect toward the utricle to form the vein of the vestibular aqueduct. This empties into the sigmoid or inferior petrosal sinus. The inferior cochlear vein usually drains into the inferior petrosal sinus or superior bulb of the internal jugular vein. 1 ..."
Reference 6
Book Chapter
Vestibulocochlear organ
Navarro M., Ruberte J., Carretero A.
Morphological Mouse Phenotyping , 2017 pp 521-539
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"... The endolymphatic duct originates in the saccule and runs inside an osseous duct, the vestibular aqueduct , until it reaches the subarachnoid space. Here, the endolymphatic duct becomes dilated to form the endolymphatic sac where the endolymph is reabsorbed ( Fig. 15-20 ). The walls of the utricle and saccule are lined by a single squamous epithelium but in their medial wall there is an oval-shaped thickening where sensory hair cells are located. These structures are the maculae. The macula of the utricle lies horizontally while the macula of the saccule is oriented vertically in relation to the head of the mouse. Just as occurs in the spiral organ, the ends of the cilia of the sensory cells of the macula are immersed in a gelatinous membrane, the otolithic membrane ( Fig. 15-18 ). The otoliths (statoconia) are small calcium carbonate crystals which put pressure on the cilia, inclining them, thereby stimulating the sensory hair cells. The function of the utricle and saccule is to detect the static position and the linear, horizontal and vertical movements of the head. In the vestibule originate three semicircular canals, which are located in the three spatial planes. In the mouse, the semicircular canals do not form angles of 90° between each other as is the case in humans. The anterior semicircular canal and the lateral semicircular canal are located more or less vertically. The anterior canal is located in a sagittal plane and the lateral canal in a transverse plane, whereas the posterior semicircular canal maintains a more or less horizontal position with respect to the head. The semicircular canals of the mouse are approximately 200 microns in diameter, with the anterior canal being slightly larger, and the lateral canal being the smallest one. This difference in diameter between the semicircular canals is reflected in their different sensibilities to movement. ..."
Reference 7
Review article
Hypothetical Mechanism for Vertigo in Meniere's Disease
Gibson W.P.R.
Otolaryngologic Clinics of North America , 2010 pp 1019-1027
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"... The anatomy of the endolymphatic sac and duct The anatomy of the human endolymphatic duct and sac is shown in Fig. 1 . The duct begins at the ductus reuniens, which joins the cochlear duct and the utricle to form an endolymphatic duct (ED) leading to the endolymphatic sac (ELS). The anatomy has been described by Lo and his colleagues. 8 Often anatomic texts describe a long thin ED ending in a short, pouch-like ELS, but in reality, the system is far different in appearance. The ED is a short single-lumen tube of only 2 mm in length. The ELS is much larger and a highly complex structure of interconnecting tubules, cisterns and crypts. The endolymphatic sinus (ES) lies in a groove on the posteromedial surface of the vestibule, with its distal outlet leading to the vestibular aqueduct. The ED narrows at its isthmus at the isthmus of the vestibular aqueduct (VA), where it is oblong in shape with mean diameter of only 0.09 × 0.2 mm. Distal to the isthmus begins the ELS, which flares considerably transversely but thickens only slightly in its sagittal dimension. The ELS has two portions, the intraosseous portion within the VA, and the extraosseous portion. The size of the intraosseous portion varies considerably, from 6 to 15 mm in length and 3 to 15 mm in width. The extraosseous ELS rests on a fovea on the posterior wall of the petrous bone, where the ELS lies between two layers of dura. The extraosseous ELS also varies considerably in size, from 5 to 7 mm in width and 10 to 15 mm in length. In lower animals and the human fetus, the ELS consists of a single lumen. In people after the age of 1 year, the ELS develops tubules that reach adult complexity by the age of 3 to 4 years. The tubules of the ELS are more complex in the proximal and middle (rugose) portions. The concept of the human ELS as a empty sack is entirely wrong. ..."
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"... Meniere's disease or a syndrome? It seems improbable that Meniere's disease is a single disease rather than several different pathologic conditions that result in the same symptom complex. Any condition that causes narrowing of the vestibular aqueduct and the production of excess endolymph could result in the same symptom complex. The term Meniere's syndrome may be more exact. There are congenital and acquired causes of Meniere's syndrome. For example, congenital deafness due to rubella or toxoplasmosis can result in secondary Meniere's disease; acquired infections such as treponemal disease may partially block the vestibular aqueduct, and a tumor of the endolymphatic sac often causes Meniere's symptoms. Nevertheless, there is an idiopathic group for which no causes have been determined, and perhaps this could be termed Meniere's disease in a similar fashion to the way that Bell palsy has become the term for an idiopathic facial palsy. Meniere's disease or the idiopathic cause of Meniere's syndrome has a wide spread of the age at onset. The author has analyzed his own series of 1576 patients who had clinically definite Meniere's disease 21 ( Table 1 ); the median age is approximately 50 years. This age range is not typical of vascular disorders or autoimmune problems. The age range is somewhat similar to the age of onset of peptic ulcer and gives some support to the notion that Meniere's disease may begin after a viral labyrinthitis in ears that have narrow vestibular aqueducts. The initial episode of viral labyrinthitis causes the endolymphatic hydrops, establishing the syndrome. Perhaps the condition can resolve in the earliest stages after a cluster of attacks. In other cases, the viral labyrinthitis may recur and cause another cluster of attacks. Some viruses, such as the herpes virus, may remain in the inner ear, causing several recurrences until eventually the endolymphatic sac functionality is destroyed, causing widespread endolymphatic hydrops similar to the findings after removal of the endolymphatic sac in guinea pigs. 22 1 ..."
Reference 8
Review article
Imaging of dizziness
Connor S.E.J., Sriskandan N.
Clinical Radiology , 2014 pp 111-122
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"... Anatomy and imaging methods The end organs of balance and equilibrium are the vestibule and semicircular canals, consisting of the outer osseous labyrinth and an inner membranous (or endolymphatic) labyrinth. There is perilymphatic fluid interposed between the two compartments. The inner endolymphatic structures are the utricle, the saccule, and the semicircular ducts. The endolymphatic sac and duct is located within the vestibular aqueduct and this also communicates with the saccule and utricle. The utricle and saccule or “static labyrinth” contains maculae (consisting of sensory hair cells that detect the position of otolithic crystals), which detect the position of the head relative to gravity. The semicircular canals or “kinetic labyrinth” contain ampullae (also with sensory hair cells), which respond to rotation and angular acceleration. These structures detect endolymphatic flow and innervate the vestibular nerves that maintain balance through their central connections. The vestibular nerves arise from a population of bipolar neurons and cell bodies, which reside in Scarpa's ganglion, and they run with the cochlear nerve (as cranial nerve eight or the vestibulo-cochlear nerve) to the vestibular nuclei in the brainstem ( Fig 1 ). The arterial supply to the vestibular structures is via the labyrinthine branch of the anterior inferior cerebellar artery. 28 Both computed tomography (CT) and magnetic resonance imaging (MRI) have a role in the imaging of these structures. The spatial resolution of CT is ideal in view of the innate high contrast of the osseous component and it will demonstrate erosion, fracture, or deficiency as well as the misplacement of prostheses with respect to the bony labyrinth. MRI is required to image the fluid-containing structures of the perilymphatic and endolymphatic spaces in addition to the vestibular nerves. 1 ..."
Reference 9
Book Chapter
Surgery of the Endolymphatic Sac
Packer M.D., Bradley Welling D.
Otologic Surgery , 2010 pp 411-428
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"... The distal sac has a smooth open lumen within the dura mater. Its cuboidal epithelium contains light and dark cells. 12 The lining of the intermediate portion of the sac shows more complex epithelial folds forming papillae and crypts of tall columnar light and dark cells. Cells of the intraosseous proximal rugose sac are intermediate between the taller, more distal cells and the squamous-to-cuboidal cells of the duct. The duct narrows at its isthmus to 0.1 to 0.2 mm in diameter. Luminal folding and transversely oriented tubules make the endolymphatic sac a more complex structure than it otherwise outwardly appears. 13 The normal bony vestibular aqueduct is readily apparent on high-resolution computed tomography (CT) scanning of the temporal bone. It is funnel-shaped or tubular with the width of its external aperture averaging 6 mm. 14,15 Radiographic observation of the affected ear in patients with Meniere’s disease showed a filiform narrowing of the external aperture averaging 2.2 mm. 14 The amount of narrowing of the external aperture was also shown to be correlated with an increasing percentage of positive electrophysiologic measures in the affected ears of patients with Meniere’s disease. 15 Statistically significant differences in the percentage of patients with enlarged summating potential-to-action potential (SP:AP) ratios by transtympanic electrocochleography were seen when correlated with the size of the external aperture of the vestibular aqueduct. An increased SP:AP ratio was noticed in 95% of ears with nonvisible external apertures, 91% when the aperture was less than 5 mm, 58% when the aperture was 5 to 7 mm, and 29% when the aperture was greater than 7 mm. The endolymphatic duct and sac can be seen on high-resolution fast spin echo magnetic resonance imaging (MRI). The endolymphatic sac and duct were seen on MRI of 20 temporal bones in healthy subjects using strongly T2-weighted sequences and postprocessing software. 1 ..."
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"... Patients with Meniere’s disease have been shown to have widening of the vestibular aqueduct aperture, but, although enticing to establish a mechanism of pathology, patients with Lermoyez’s syndrome have not been shown to have wider vestibular aqueducts than other patients with Meniere’s disease. Gibson and Arenberg also theorized that Tumarkin crises could be the effect of a membrane rupture in the overdistended endolymphatic space. Although Meniere’s disease has no known etiologic cause by definition, there are several etiologies of secondary Meniere’s syndromes. Meniere’s syndrome can be mimicked by disease processes in all categories—traumatic (inner ear fracture, perilymphatic fistula), infectious (viral, syphilis), inflammatory (autoimmune, allergy, Cogan’s syndrome, sarcoidosis), metabolic (diabetes), congenital/genetic (inner ear malformations, enlarged vestibular aqueduct), neoplastic (vestibular schwannoma), vascular (migraine, hemorrhage), and iatrogenic (stapes/mastoid surgery). These disease processes may lead to secondary endolymphatic hydrops by causing scarring within the labyrinth; activating inflammatory, cytokine, and complement pathways, yielding edema and fibrosis and altered extracellular matrices; altering hemodynamics that may affect transcellular ionic exchange and alter endolymph homeostasis; creating obstructive immune complexes or cellular debris; and activating cellular responses through humoral messengers. The end result is distention or obstruction of the endolymphatic flow disturbing sensation of cochleovestibular signals and sending distorted messages of sound, station, and motion. 1 ..."
Reference 10
Review article
A perspective from magnetic resonance imaging findings of the inner ear: Relationships among cerebrospinal, ocular and inner ear fluids
Nakashima T., Sone M., Teranishi M., Yoshida T., Terasaki H., Kondo M., Yasuma T., Wakabayashi T., Nagatani T., Naganawa S.
Auris Nasus Larynx , 2012 pp 345-355
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"... Communication between cochlear perilymph and CSF in the IAM is limited because of bone tightly surrounding the inner ear nerve at the fundus of the IAM. However, passage of gadolinium contrast agents between the inner ear and IAM was recognized in magnetic resonance imaging [9,10] . 3.3.2 Vestibular aqueduct In the vestibular aqueduct, there is endolymphatic duct and sac. Blood flow to the endolymphatic sac comes from the middle meningeal artery or occipital artery that belongs to the external carotid arterial system, which is strikingly different from the cochlea and the vestibular apparatus in which arterial blood comes from the branch of the anterior inferior cerebellar artery that belongs to the cerebral circulation system. The diameters of the endolymphatic duct and the proximal portion of the vestibular aqueduct are significantly smaller in Meniérè’s disease ears than in controls . Graphic reconstructions show the Meniérè’s sacs to be smaller and to have fewer tubular epithelial structures in the intraosseous portion than in the control ears. The median volume of the sac in the Meniérè’s disease side is substantially lower than in the contralateral ear. The width of the external aperture of the vestibular aqueduct is also significantly smaller in Meniérè’s disease ears than in controls. These findings indicate that the size not only of the vestibular aqueduct but also of the sac is reduced in Meniérè’s disease. The results suggest that the endolymphatic sac is pathologically changed in Meniérè’s disease and that reduced resorptive capacity of the small endolymphatic sac is associated with endolymphatic hydrops. 3.3.3 Cochlear aqueduct Enlargement of the cochlear aqueduct is often mentioned in the otologic literature, usually in its purported association with sensory hearing loss, stapes gusher, and transotic CSF leak. 1 ..."
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"... They considered that fluid surrounding the cochlear nerve is CSF that is covered with arachnoid-like covering in the modiolus. Communication between cochlear perilymph and CSF in the IAM is limited because of bone tightly surrounding the inner ear nerve at the fundus of the IAM. However, passage of gadolinium contrast agents between the inner ear and IAM was recognized in magnetic resonance imaging [9,10] . 3.3.2 Vestibular aqueduct In the vestibular aqueduct, there is endolymphatic duct and sac. Blood flow to the endolymphatic sac comes from the middle meningeal artery or occipital artery that belongs to the external carotid arterial system, which is strikingly different from the cochlea and the vestibular apparatus in which arterial blood comes from the branch of the anterior inferior cerebellar artery that belongs to the cerebral circulation system. The diameters of the endolymphatic duct and the proximal portion of the vestibular aqueduct are significantly smaller in Meniérè’s disease ears than in controls . Graphic reconstructions show the Meniérè’s sacs to be smaller and to have fewer tubular epithelial structures in the intraosseous portion than in the control ears. The median volume of the sac in the Meniérè’s disease side is substantially lower than in the contralateral ear. The width of the external aperture of the vestibular aqueduct is also significantly smaller in Meniérè’s disease ears than in controls. These findings indicate that the size not only of the vestibular aqueduct but also of the sac is reduced in Meniérè’s disease. The results suggest that the endolymphatic sac is pathologically changed in Meniérè’s disease and that reduced resorptive capacity of the small endolymphatic sac is associated with endolymphatic hydrops. 3.3.3 Cochlear aqueduct Enlargement of the cochlear aqueduct is often mentioned in the otologic literature, usually in its purported association with sensory hearing loss, stapes gusher, and transotic CSF leak. 1 ..."
Reference 11
Book Chapter
Auditory system
Jahangir Moini, Pirouz Piran
Functional and Clinical Neuroanatomy , 2020 pp 363-392
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"... Vascular supply The labyrinthine artery is the primary source of blood for the inner ear. The semicircular canals are also supplied by the stylomastoid branch of the occipital artery or posterior auricular artery. The labyrinthine artery divides off of the basilar artery, and sometimes, from the anterior inferior cerebellar artery. At the bottom of the internal acoustic meatus, it divides into the cochlear and vestibular branches. The cochlear branch further subdivides into 12–14 tiny branches through the modiolus canals. These are distributed as a capillary plexus to cochlear structures such as the basilar membrane, spiral lamina, and stria vascularis. The utricle, saccule, and semicircular ducts are supplied by vestibular arterial branches. Veins that drain the vestibule and semicircular canals accompany the related arteries. They connect toward the utricle, forming the vein of the vestibular aqueduct. This empties into the sigmoid or inferior petrosal sinus. The inferior cochlear vein, of the cochlear aqueduct, usually drains into the inferior petrosal sinus or superior bulb of the internal jugular vein. It is created by a union of the common modiolar and vestibulocochlear vines, providing nearly all venous outflow from the cochlea. Near the basal cochlear turn, the common modiolar vein is formed by a joining of the anterior and posterior spiral veins. The vestibulocochlear vein is formed by a joining of the anterior and posterior vestibular veins, as well as the vein of the round window. When it is present, a labyrinthine vein drains the apical and middle cochlear coils into one of the following: the posterior area of the superior petrosal sinus, or the transverse sinus or inferior petrosal sinus. 1 ..."
Reference 12
Handbook Chapter
Benign paroxysmal vertigo of childhood
Gurberg J., Tomczak K.K., Brodsky J.R.
Handbook of Clinical Neurology , 2023 pp 229-240
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"... Enlarged vestibular aqueduct Enlarged vestibular aqueduct, or EVA, is the most common congenital inner ear malformation identified in children . EVA may be found in isolation or associated with other congenital hearing loss syndromes, including Pendred and branchio-oto-renal syndromes. It is associated with early onset sensorineural or mixed hearing loss that can be fluctuating or progressive and may be exacerbated by minor head injuries or barotrauma. While most descriptions focus on hearing, children with EVA may also exhibit vestibular symptoms characterized by variable episodic vertigo lasting minutes to hours, often brought on by minor trauma or vigorous head rotation . The proposed pathophysiology of hearing loss and vertigo in EVA is either exposure of the cochlea and vestibule to intracranial pressure variations through a widened vestibular aqueduct or neuroepithelial damage secondary to reflux of hyperosmolar contents from the endolymphatic sac. It is postulated that the vestibular system is more resistant to these insults compared to the auditory system, explaining why patients often have preserved vestibular function in the face of profound sensorineural hearing loss. EVA may also present in infancy with paroxysmal torticollis that can precede the diagnosis of hearing loss by months or even years and may be misdiagnosed as BPTI of migraine origin, as discussed in further detail in earlier sections of this chapter . Physical examination is normal in the majority of cases of EVA, although goiter may be noted in Pendred syndrome and branchial abnormalities may be seen in branchio-oto-renal syndrome. Audiometry is typically abnormal and abnormalities on vestibular testing are also frequently present. Diagnosis is confirmed by computed tomography (CT) of the temporal bones. 1 ..."
Reference 13
Book Chapter
Pediatric vestibular dysfunction following head injury: Diagnosis and management
Graham Cochrane, Jacob R. Brodsky
Otologic and Lateral Skull Base Trauma , 2024 pp 217-243
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"... Enlarged vestibular aqueduct Enlarged vestibular aqueduct (EVA) is not a traumatic lesion but can become symptomatic following even very minor head injuries; thus, it warrants discussion in this chapter. EVA is the most common cause of congenital hearing loss that is grossly visible on temporal bone imaging. Children with EVA may have hearing loss at birth, but it can often be delayed in onset and fluctuates over time in many cases. It can be seen in the setting of mutations in the SLC26A4 gene, which is associated with Pendred syndrome when mutations are homozygous, but EVA can also be seen in conjunction with other hearing loss syndromes and often occurs as an isolated anomaly. Patients with EVA are often prone to sudden drops in hearing and/or vestibular function following impacts to the head that may be milder than that required to sustain a concussion or a major traumatic brain injury. Historically, patients with EVA were discouraged from sports activities, though this advice is no longer considered routine. 24 Audiologic testing in children with EVA typically shows a sensorineural or mixed hearing loss that may be unilateral or bilateral. Vestibular testing may show a third window pattern on VEMP testing and a unilateral or bilateral vestibular loss on other vestibular tests. 25 , 26 Diagnosis is confirmed by temporal bone CT, though very large vestibular aqueducts can sometimes be seen on MRI. There are multiple different diagnostic criteria for EVA, all of which are based on measurements of the vestibular aqueducts on temporal bone CT. The most commonly used criteria are the Cincinnati criteria, which indicate that EVA is confirmed by the presence of combined measurements of the aqueduct of > 0.9 mm at the midpoint and >1.9 mm at the operculum. 25 EVA is often also associated with an incomplete partition anomaly, type II (IP2), which was formerly known as a Mondini malformation. This anomaly also includes a shortened cochlear of only 1.5 turns, a deficient modiolus, and a deficient interscalar septum. 1 ..."
Reference 14
Handbook Chapter
Neuro-Otology
P. Bertholon, A. Karkas
Handbook of Clinical Neurology , 2016 pp 279-293
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"... Enlarged vestibular aqueduct Vestibular aqueduct enlargement, initially described by Valvassori and Clemis (1978) , is the most common imaging abnormality in patients with congenital inner-ear defects . The mechanism is an early arrest in the development of the endolymphatic canal and sac around the fifth to eighth weeks in utero , when the initial vesicle normally elongates and narrows to form the vestibular aqueduct ( Fig. 20.2 ). It can be associated with other inner-ear malformations such as cystic dysplasia of the canals, enlargement of the vestibule, or various cochlear abnormalities . This condition has been reported to be inherited in an autosomal-recessive manner and can be associated with syndromic hearing loss as in Pendred syndrome . The anomaly is characterized by sensorineural and, more often, mixed hearing loss, which usually begins in early childhood . Hearing loss is typically bilateral and progressive, with stepwise rather than fluctuating hearing decrements often triggered by relatively minor head trauma. The stapedial reflex is usually present and VEMPs have abnormally low thresholds and high amplitude, as in the third mobile window mechanism . Vestibular symptoms are less frequent than hearing loss and commonly begin in childhood, but they may be delayed until adulthood . Although sound- or pressure-induced vertigo has been reported, the vestibular symptoms are dominated by recurrent episodes of vertigo that can last hours and mimic Menière's disease . Physical findings during one of these episodes have shown a peripheral vestibular syndrome consistent with unilateral irritation or deficit, as in patients with Menière's disease . Indeed, it is noteworthy that in typical Menière's disease, the endolymphatic sac and duct play a predominant role. However, the direct surgical treatment of the endolymphatic sac in enlarged vestibular aqueduct is controversial, as endolymphatic sac decompression, arachnoid bypass, or endolymphatic sac occlusion may worsen the hearing loss . 1 ..."
Reference 15
Review article
Imaging and anatomy for cochlear implants
Fishman A.J.
Otolaryngologic Clinics of North America , 2012 pp 1-24
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"... The vestibular aqueduct The association of enlargement of the vestibular aqueduct and congenital sensorineural hearing loss is well recognized. 6,22,82–86 Radiographically, it may occur in conjunction with other identifiable inner ear anomalies as previously discussed, or as an isolated finding on CT or MR imaging (see Fig. 3 ; Fig. 15 ). 85–91 Radiographic enlargement has been reported using different imaging modalities and criteria, but is generally considered to exist when the aqueduct’s diameter is greater than 1.5 to 2.0 mm at its midpoint, measured between the common crus and the external aperture into the posterior fossa. 85,92–94 The large vestibular aqueduct syndrome is traditionally considered to be a distinct clinical entity in patients with radiographic evidence of enlargement of the vestibular aqueduct. 85,86 Hearing loss is typically bilateral and progressive, with stepwise decrements often associated with episodes of relatively minor head trauma. Moreover, enlargement of the vestibular aqueduct is considered to be a relatively common finding in children with congenital sensorineural hearing loss. 74,85,86 Some investigators regard it as the single most common radiographic finding among patients with congenital sensorineural hearing loss. 74 The major traditional hypothesis regarding the pathogenesis of this anomaly involves aberrant or arrested development of the endolymphatic duct and sac system, which is based on the observation that in early embryogenesis the duct is shorter, straighter, and proportionally much broader than in later maturity. 85 As more experience has been gained with MR scanning, the defect is currently being described and studied as one involving the entire endolymphatic duct and sac system. 87–91 There is also some recent evidence supporting a familial component to the disorder. 95 Some recent work is also being done to investigate its genetics basis as well as its association with other known syndromes. 93,96 There are a variety of speculative causes of the hearing loss associated with this disorder based mostly on clinical, radiographic, and surgical observations, as well as some analyses of endolymphatic chemical composition. 1 ..."
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"... The vestibular aqueduct The association of enlargement of the vestibular aqueduct and congenital sensorineural hearing loss is well recognized. 6,22,82–86 Radiographically, it may occur in conjunction with other identifiable inner ear anomalies as previously discussed, or as an isolated finding on CT or MR imaging (see Fig. 3 ; Fig. 15 ). 85–91 Radiographic enlargement has been reported using different imaging modalities and criteria, but is generally considered to exist when the aqueduct’s diameter is greater than 1.5 to 2.0 mm at its midpoint, measured between the common crus and the external aperture into the posterior fossa. 85,92–94 The large vestibular aqueduct syndrome is traditionally considered to be a distinct clinical entity in patients with radiographic evidence of enlargement of the vestibular aqueduct. 85,86 Hearing loss is typically bilateral and progressive, with stepwise decrements often associated with episodes of relatively minor head trauma. Moreover, enlargement of the vestibular aqueduct is considered to be a relatively common finding in children with congenital sensorineural hearing loss. 74,85,86 Some investigators regard it as the single most common radiographic finding among patients with congenital sensorineural hearing loss. 74 The major traditional hypothesis regarding the pathogenesis of this anomaly involves aberrant or arrested development of the endolymphatic duct and sac system, which is based on the observation that in early embryogenesis the duct is shorter, straighter, and proportionally much broader than in later maturity. 85 As more experience has been gained with MR scanning, the defect is currently being described and studied as one involving the entire endolymphatic duct and sac system. 87–91 There is also some recent evidence supporting a familial component to the disorder. 95 Some recent work is also being done to investigate its genetics basis as well as its association with other known syndromes. 93,96 There are a variety of speculative causes of the hearing loss associated with this disorder based mostly on clinical, radiographic, and surgical observations, as well as some analyses of endolymphatic chemical composition. 2 ..."
Reference 16
Book Chapter
Sensory Organ Disorders (Retina, Auditory, Olfactory, Gustatory)
Gillespie D.C.
Neural Circuit Development and Function in the Brain , 2013 pp 731-759
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"... Pendred Syndrome Pendred syndrome (deafness with goiter) is the most common syndromic form of deafness, accounting for up to 7.8% of cases of congenital deafness and occurring in an estimated 7.5 of 100,000 births. Patients inheriting this autosomal-recessive disorder have variable degrees of deafness at birth and typically develop goiter in the second decade. The syndrome is accompanied by structural defects of the temporal bone such as Mondini dysplasia and enlarged vestibular aqueduct (EVA) that are likely driven in part by accumulation of endolymph. The vestibular aqueduct, embedded within the temporal bone, is a small canal containing the endolymphatic duct and extending from the vestibule between the cochlea and the labyrinth to the endolymphatic sac. In Mondini dysplasia, the apical turn of the cochlea fails to form, and patients are profoundly deaf at birth. In EVA, vestibular dysfunction may be present and the hearing loss is variable. The mutated Pendred syndrome gene PDS (SLC26A4) is a member of the solute carrier protein 26 anion transporter family, and the gene product pendrin is a transmembrane Cl − /I − / HCO 3 − transporter. Allelic heterogeneity produces some SLC26A4 variants with Pendred syndrome and others with non-syndromic deafness with EVA (DFNB4). Pendrin is expressed in the inner ear and kidney, and in the thyroid, where it mediates apical iodide transport in thyroid follicular cells . Although Pendred syndrome is sometimes accompanied by hypothyroidism that could itself contribute to hearing loss, the distribution of the mouse pendrin throughout the endolymphatic duct and sac, and in specific areas of utricle, saccule, and external sulcus, points to a specific role of pendrin in fluid resorption and in regulating the ionic composition of the cochlear endolymph. 1 ..."
Reference 17
Review article
Genetics and phenomics of Pendred syndrome
Bizhanova A., Kopp P.
Molecular and Cellular Endocrinology , 2010 pp 83-90
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"... Pendred syndrome is caused by biallelic mutations in the SLC26A4 gene, which encodes the multifunctional anion exchanger pendrin . All patients with biallelic mutations in the SLC26A4 gene have Pendred syndrome, indicating that it is genetically homogeneous ( Table 1 ) . The recessive form of hearing loss referred to as DFNB4 (OMIM 600791), originally thought to be a distinct entity because of the absence of an enlarged thyroid, is also explained by mutations in the SLC26A4 gene and is thus allelic with Pendred syndrome ( Table 1 ). Patients with DFNB4 display sensorineural hearing loss with an EVA and, if formally tested, they have a positive perchlorate test despite the absence of an enlarged thyroid . Patients with non-syndromic EVA are either homozygous for the SLC26A4 wild-type, or they have only one mutated allele . In some families, non-syndromic EVA is associated with monoallelic SLC26A4 mutations suggesting that unrecognized mutations in other regions of the gene or in another gene could contribute to the pathogenesis of the phenotype . Double heterozygosity for mutations in the SLC26A4 gene and the transcription factor FOXI1, which is involved in the regulation of SLC26A4 gene expression, has been reported in a family with EVA as well as in double heterozygous mice ( Slc26A4 +/− ; Foxi1 +/− ) ( Table 1 ) . This observation confirms that Pendred syndrome and non-syndromic EVA may have a digenic cause in a subgroup of patients . The recent finding of digenic mutations in SLC26A4 and KCNJ10, a potassium channel involved in the generation of the endocochlear potential, in patients with EVA further support that this phenotype can have a oligo- or polygenic etiology ( Table 1 ) . 1 ..."
Reference 18
Book Chapter
Hearing loss
Alaa Koleilat, Lisa A. Schimmenti, Karthik Muthusamy
Neurogenetics for the Practitioner , 2024 pp 305-325
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"... Pendred syndrome Pendred syndrome is comprised of bilateral congenital SNHL, vestibular dysfunction, and temporal bone abnormalities. This type of syndromic HL may present as mild-to-moderate or severe-to-profound. It usually manifests as prelingual HL and most patients have bilateral, progressive and fluctuating HL with some cases presenting as asymmetrical HL. 49 Temporal bone abnormalities include bilateral enlarged vestibular aqueduct with or without cochlear hypoplasia. The vestibular aqueduct carries the endolymphatic duct and contains inner ear fluid. Additionally, one of the defining features of Pendred syndrome is the development of euthyroid goiter in late childhood to early adulthood. Pendred syndrome is caused by biallelic pathogenic variants in SLC26A4, which encodes an anion transporter (pendrin) involved in the ion exchange of iodine, bicarbonate, hydroxide, sulfate, formate, and chloride. 50–52 Pendrin is responsible for the flow of iodide in thyrocytes and is required for normal inner ear function. Over 300 pathogenic variants in SLC26A4 have been described that span the entire gene including missense, nonsense, splice site, and frameshift mutations. 53 Pathogenic variants in SLC26A4 have a loss-of-function effect and the lack of pendrin leads to the degeneration of the sensory cells in the inner ear. 54 Multiple studies in various populations have assessed the prevalence of mutations in SLC26A4, and the associated audiometric profile. 55–58 In Caucasian populations, SLC26A4 is the second most frequent gene implicated in nonsyndromic deafness after GJB2 . 59 Pathogenic variants in SLC26A4 are also associated with nonsyndromic cases of SNHL-DFNB4. 1 ..."
Reference 19
Reference works Chapter
Audition
Jing Wang, Jean-Charles Ceccato, Jean-Luc Puel
The Senses: A Comprehensive Reference , 2020 pp 468-486
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"... Pendred Syndrome SLC26A4 mutations cause Pendred syndrome (PS), an autosomal recessive disorder comprised of goiter, hearing loss and enlargement of the vestibular aqueduct. SLC26A4 encodes an 86-kDa transmembrane anion exchanger called pendrin, which mediates Cl /HCO3- transport. In the mouse, Slc26a4 is expressed in several organs such as the inner ear, thyroid, kidney and lung . In the inner ear, pendrin is expressed in nonsensory epithelial cells of the outer sulcus and spiral prominence of the cochlear duct, transitional cells surrounding the vestibular neuroepithelia, and mitochondria-rich cells of the endolymphatic sac. These cells are thought to contribute to the pH and ionic homeostasis of endolymph. To gain insight in the Pendred syndrome, several mutants, such as knockout mice (full or conditional) or knock-in mice (carrying a point mutation in the Slc26a4 gene) have been generated. A major feature of the Slc26a4 mutant mice is the defective endolymphatic compartment, followed by the degeneration of non-sensory and sensory cells in the adult animal. The stria vascularis of Slc26a4−/− mice was found to suffer from free-radical stress that was evident by elevated amounts of oxidized and nitrated proteins . Heterologous expression of a mutated form of pendrin in HEK 293 cells showed that some mutated pendrin isoforms accumulate in the endoplasmic reticulum . Interestingly, salicylate application enables the mutated form to target the plasma membrane and to recover their anion-exchanger activity . In addition, cochlear epithelial cells derived from patient-derived induced pluripotent stem cells (iPSCS) demonstrated that the pendrin aggregates in the cytoplasm increase susceptibility to cellular stress and cell death. 1 ..."
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"... By the activity of Na,K-ATPase and Na K Cl cotransporter within type II fibrocytes, K + ions are transported to the intrastrial space, passing through gap junction channels rich in Cx26, Cx30 and Cx31. Finally, the K+ ions are secreted to the endolymph by the marginal cells of the stria vascularis. Thus, malfunction of the stria vascularis and recycling K + can alter the electrochemical composition of the endolymph, resulting in loss of the endocochlear potential, elevated auditory thresholds, and thus hearing loss. While the production of K + and the recycling mechanism is extremely important for hearing function, very few therapeutic interventions have been proposed to maintain cochlear homeostasis. 2.24.5.1 Pendred Syndrome SLC26A4 mutations cause Pendred syndrome (PS), an autosomal recessive disorder comprised of goiter, hearing loss and enlargement of the vestibular aqueduct. SLC26A4 encodes an 86-kDa transmembrane anion exchanger called pendrin, which mediates Cl /HCO3- transport. In the mouse, Slc26a4 is expressed in several organs such as the inner ear, thyroid, kidney and lung . In the inner ear, pendrin is expressed in nonsensory epithelial cells of the outer sulcus and spiral prominence of the cochlear duct, transitional cells surrounding the vestibular neuroepithelia, and mitochondria-rich cells of the endolymphatic sac. These cells are thought to contribute to the pH and ionic homeostasis of endolymph. To gain insight in the Pendred syndrome, several mutants, such as knockout mice (full or conditional) or knock-in mice (carrying a point mutation in the Slc26a4 gene) have been generated. A major feature of the Slc26a4 mutant mice is the defective endolymphatic compartment, followed by the degeneration of non-sensory and sensory cells in the adult animal. 1 ..."
Reference 20
Book Chapter
Hereditary Hearing Impairment
Arti Pandya
Emery and Rimoin's Principles and Practice of Medical Genetics and Genomics , 2025 pp 227-279
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"... Nonsyndromic deafness associated with EVA (DFNB4) is also caused by mutations in SLC26A4 . A second gene has been implicated in the etiology of PS. In nine patients with PS or nonsyndromic EVA, Yang and colleagues found heterozygous mutation in FOXI1 , a transcriptional activator of SLC26A4 . Causal mutations interfered with FOXI1 binding, which compromised or completely abolished FOXI1 -mediated transcriptional activation of SLC26A4 . Furthermore, they described an EVA patient who was a compound heterozygote for mutation in FOXI1 and SLC26A4 , implicating digenic inheritance in PS and/or DFNB4. This finding was consistent with a mouse model of this pathway. Mutations in another potassium channel member gene, KCNJ10 , were identified in probands from two families with deafness and EVA, who also carried heterozygous mutations in the SLC26A4 gene . The finding of a single heterozygous or bi-allelic mutations in individuals with HL and EVA has led to further classification and genotype–phenotype correlations, with coining of the term “SLC26A4-related HL,” that includes nonsyndromic DFNB4, PS, and HL with EVA . The thyroid organification defect is mild in PS such that patients are often euthyroid with moderate thyroid enlargement developing in early adolescence. Mild hypothyroidism may precede or accompany the development of goiter, which is inconsistent and appears to be related to iodide intake. Although PS is easily recognized if goiter occurs, it cannot be diagnosed by clinical exam unless there is a known family history. Therefore in the absence of molecular testing, children with apparently nonsyndromic deafness should be reevaluated periodically for the presence of thyroid enlargement or dysfunction. 1 ..."
Reference 21
Reference works Chapter
Audition
Maggie S. Matern, Ronna Hertzano
The Senses: A Comprehensive Reference , 2020 pp 838-860
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"... However, mutant mouse models of the norrin protein exhibit progressive deterioration of inner ear structures, starting with the stria vascularis, which may be attributed to defects in the development and maintenance of inner ear vasculature . 2.42.3.6 Pendred Syndrome Pendred Syndrome is one of the most common causes of congenital deafness, with an estimated prevalence of approximately 7.5 in every 100,000 births . This recessively inherited syndrome is characterized by sensorineural hearing loss accompanied by an enlarged vestibular aqueduct and temporal bone abnormalities, as well as an enlarged thyroid gland called a goiter . Some individuals may also experience balance problems due to dysfunction of the vestibular system . Approximately 50% of Pendred Syndrome cases are caused by mutations in the gene SLC26A4, which encodes for the iodide-chloride transporter pendrin . The pendrin protein is expressed along the lateral wall of the cochlea, specifically in the outer sulcus, spiral prominence and stria vascularis, as well as in the endolymphatic sac . Here, it is thought to be involved in maintaining the ionic homeostasis of the endolymph that is required for hearing . A smaller proportion of Pendred Syndrome cases can be attributed to mutations in the transcription factor gene FOXI1, which is involved in transcriptional activation of SLC26A4 expression , and to mutations in KCNJ10, which encodes a potassium channel subunit also thought to be involved in endolymph homeostasis . 2.42.3.6.1 Perrault Syndrome Perrault Syndrome is a heterogeneous disease characterized by hearing loss and possible neurological symptoms in both males and females, as well as ovarian failure in females . Additionally, the hearing loss can present at birth or in early childhood, and is progressive . 1 ..."
Reference 22
Handbook Chapter
Neurocutaneous Syndromes
Prashant Chittiboina, Russell R. Lonser
Handbook of Clinical Neurology , 2015 pp 139-156
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"... Endolymphatic sac tumors General features Endolymphatic sac tumors (ELSTs) arise from the endolymphatic epithelium within the vestibular aqueduct . These tumors are rare in the general population but are found in up to 6–15% of VHL patients . Bilateral ELSTs are found only in VHL . Clinical ELSTs arise within the vestibular aqueduct, and can lead to hearing loss due to intralabrynthine hemorrhage, endolymphatic hydrops, and/or by otic capsule invasion . ELSTs can present with vestibular symptoms (62%), partial or complete hearing loss (95 to 100%), tinnitus (77%), and/or facial paresis (8%) . Imaging High resolution computed tomography (CT) scans of the temporal bone and pre-/postcontrast enhanced MR imaging of the internal auditory canals are necessary to detect and follow these tumors ( Fig. 10.5 ). CT is used to detect the presence of otic capsule invasion and to evaluate the extent of bony erosion of the temporal bone (particularly, the vestibular aqueduct in very small tumors). Postcontrast T1-weighted MR imaging can reveal signs of small ELST tumors as subtle asymmetric enhancement of the endolymphatic sac. Larger tumors are heterogeneously enhancing on postcontrast T1W images. Intralabrynthine hemorrhage is detected by increased signal on precontrast T1W images . Histologic findings ELSTs are papillary cystic glandular neoplasms with variegated patterns (see Fig. 10.5 ). Due to the presence of local bone invasion, these tumors were called low grade adenocarcinoma of probable endolymphatic origin . ELSTs may be found entirely within the endolymphatic sac when small. When large, these inevitably invade and erode the surrounding temporal bone . Treatment Surgery is the treatment of choice for ELSTs. Complete resection may be achieved in a majority of cases with minimal risk for recurrence (3%) after gross total resection . 1 ..."
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"... General features Endolymphatic sac tumors (ELSTs) arise from the endolymphatic epithelium within the vestibular aqueduct . These tumors are rare in the general population but are found in up to 6–15% of VHL patients . Bilateral ELSTs are found only in VHL . 1 ..."
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"... Retinal gliosis and hemorrhages are present in association with severe lesions. Differential upregulation of HIF-2α has been demonstrated in severe lesions that are refractory to anti-VEGF therapy . Treatment Laser photocoagulation and cryotherapy are the mainstays of surgical management of retinal hemangioblastomas. Either of these methods can be used to as a sole method for treatment of extrapapillary hemangioblastomas . Vitreoretinal surgery can be performed when preretinal and vitreal membranes, retinal detachment from traction, and exudation occur . For hemangioblastomas in close proximity to the optic nerve, intravitreal anti-VEGF therapy may arrest progression for small lesions and may help reverse exudates and edema in some cases . Photodynamic therapy or plaque radiotherapy may have a limited role in the management of retinal hemangioblastomas . Salvage external beam radiation may be used for treatment-refractory retinal hemangioblastomas with reduction in tumor volume and improvement in visual acuity . Endolymphatic sac tumors General features Endolymphatic sac tumors (ELSTs) arise from the endolymphatic epithelium within the vestibular aqueduct . These tumors are rare in the general population but are found in up to 6–15% of VHL patients . Bilateral ELSTs are found only in VHL . Clinical ELSTs arise within the vestibular aqueduct, and can lead to hearing loss due to intralabrynthine hemorrhage, endolymphatic hydrops, and/or by otic capsule invasion . ELSTs can present with vestibular symptoms (62%), partial or complete hearing loss (95 to 100%), tinnitus (77%), and/or facial paresis (8%) . Imaging High resolution computed tomography (CT) scans of the temporal bone and pre-/postcontrast enhanced MR imaging of the internal auditory canals are necessary to detect and follow these tumors ( Fig. 10.5 ). 1 ..."
Related quote(s)4 / 5
"... Imaging High resolution computed tomography (CT) scans of the temporal bone and pre-/postcontrast enhanced MR imaging of the internal auditory canals are necessary to detect and follow these tumors ( Fig. 10.5 ). CT is used to detect the presence of otic capsule invasion and to evaluate the extent of bony erosion of the temporal bone (particularly, the vestibular aqueduct in very small tumors). Postcontrast T1-weighted MR imaging can reveal signs of small ELST tumors as subtle asymmetric enhancement of the endolymphatic sac. Larger tumors are heterogeneously enhancing on postcontrast T1W images. Intralabrynthine hemorrhage is detected by increased signal on precontrast T1W images . 1 ..."
Related quote(s)5 / 5
"... CT is used to detect the presence of otic capsule invasion and to evaluate the extent of bony erosion of the temporal bone (particularly, the vestibular aqueduct in very small tumors). Postcontrast T1-weighted MR imaging can reveal signs of small ELST tumors as subtle asymmetric enhancement of the endolymphatic sac. Larger tumors are heterogeneously enhancing on postcontrast T1W images. Intralabrynthine hemorrhage is detected by increased signal on precontrast T1W images . Histologic findings ELSTs are papillary cystic glandular neoplasms with variegated patterns (see Fig. 10.5 ). Due to the presence of local bone invasion, these tumors were called low grade adenocarcinoma of probable endolymphatic origin . ELSTs may be found entirely within the endolymphatic sac when small. When large, these inevitably invade and erode the surrounding temporal bone . Treatment Surgery is the treatment of choice for ELSTs. Complete resection may be achieved in a majority of cases with minimal risk for recurrence (3%) after gross total resection . Surgical resection of the tumor can relieve audiovestibular symptoms and can preserve hearing in a majority of cases . Indications for treatment of ELSTs include progressive sensorineural hearing loss, vestibular symptoms, facial nerve compression, and local mass effect from tumor growth. Early resection of ELSTs is recommended to prevent sensorineural hearing loss and to alleviate vestibular symptoms. Facial nerve decompression (large tumors) performed during surgery can improve facial nerve function in some cases . The role of radiation in the management of ELST is unproven . 2 ..."
Reference 23
Review article
Imaging of Vertigo and Dizziness: A Site-based Approach Part 2 (Membranous Labyrinth and Cerebellopontine Angle)
Takahashi J.T., Alves I.D.S., Gebrim E.S., Goncalves V.T.
Seminars in Ultrasound, CT and MRI , 2024 pp 372-382
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"... MRI shows variable signal, predominantly isointense on T1 and T2, more homogeneous than schwannomas, and intense enhancement with paramagnetic contrast ( Fig. 10 ). CT may reveal lesion calcifications and underlying bone hyperostosis ( Table 1 ). Endolymphatic Sac Tumor Endolymphatic sac tumors are locally aggressive tumors originating from the endolymphatic sac, located in the posterior portion of the vestibular aqueduct of the petrous bone. 30 They can be sporadic or, more frequently, associated with von Hippel-Lindau disease. These tumors cause progressive and typically irreversible hearing loss, with an average age of onset around 22 years. 31 CT shows destruction of the retro-labyrinthine petrous bone with a geographic or mottled pattern, exhibiting spiculated or reticulated bony fragments. MRI shows heterogeneous signal, with hyperintense areas on T1-weighted images related to hyperproteinaceous/hematic components and intense enhancement by paramagnetic contrast ( Fig. 11 ). Paraganglioma Paragangliomas are hypervascular neuroendocrine tumors that originate from paraganglia, which embryologically derive from neural crest cells. These tumors are most commonly diagnosed between the third and sixth decades of life. However, the age of presentation and sex (male/female) predilection may vary depending on the type of paraganglioma (sporadic or hereditary) and its location. 32 Head and neck paragangliomas, for instance, tend to present later in life and have a strong female predominance. The most frequent sub-sites include carotid, tympanic, jugulotympanic, and vagal paragangliomas. Auditory symptoms include pulsatile tinnitus, hearing loss, and vertigo due to inner ear involvement via the cochlear promontory. On CT, findings suggestive of paraganglioma include destructive bony patterns affecting the jugular foramina and soft tissue lesion in the middle ear near the cochlear promontory. 1 ..."
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"... Endolymphatic Sac Tumor Endolymphatic sac tumors are locally aggressive tumors originating from the endolymphatic sac, located in the posterior portion of the vestibular aqueduct of the petrous bone. 30 They can be sporadic or, more frequently, associated with von Hippel-Lindau disease. These tumors cause progressive and typically irreversible hearing loss, with an average age of onset around 22 years. 31 CT shows destruction of the retro-labyrinthine petrous bone with a geographic or mottled pattern, exhibiting spiculated or reticulated bony fragments. MRI shows heterogeneous signal, with hyperintense areas on T1-weighted images related to hyperproteinaceous/hematic components and intense enhancement by paramagnetic contrast ( Fig. 11 ). 1 ..."
Reference 24
Review article
Applications of Magnetic Resonance Imaging in Adult Temporal Bone Disorders
Mohan S., Hoeffner E., Bigelow D.C., Loevner L.A.
Magnetic Resonance Imaging Clinics of North America , 2012 pp 545-572
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"... Inner ear lesions Large Endolymphatic Duct and Sac Large endolymphatic sac anomaly is the most common congenital inner ear anomaly found by imaging and is bilateral in 90% of cases. 44 It is also the most commonly missed cause of congenital deafness. It is a familial lesion with autosomal recessive inheritance. There is an association with Pendred syndrome, which is severe SNHL with thyroid disorder. Patients typically present with progressive, severe SNHL in childhood or early adulthood, often exacerbated by minor trauma. The hallmark imaging characteristic of large endolymphatic sac anomaly on CT is enlargement of the bony vestibular aqueduct, and a diameter greater than 1.5 mm at a point halfway between the crus communis and the intracranial aperture of the aqueduct, or an opercular measurement of more than 2 mm, are generally considered to be the defining characteristics. 45 On T2 FSE MR imaging, the underlying endolymphatic structure abnormalities are shown readily, consisting of enlargement of the endolymphatic sac and duct ( Fig. 9 ). In some cases, enlargement of the sac is more conspicuous than that of the duct, on T2 FSE MR. The vestibular aqueduct is usually found at the level of the vestibule and lateral semicircular canal, so it is easy to compare the diameter of the sac with the lateral and posterior semicircular canals, which should be larger than the aqueduct. 44 There is no relationship between the size of the endolymphatic sac and the severity of the SNHL. Although the aqueduct is best evaluated with CT, the bright signal intensity on T2 within the sac is best seen on thin T2-weighted images. In more than 75% of cases, there is an associated cochlear dysplasia, with dysmorphic apical turn and modiolar deficiency. 46 High-resolution T2 MR imaging may be able to distinguish more subtle abnormalities of scalar chamber asymmetry with the more anterior scala vestibuli larger than the more posterior scala tympani. 1 ..."
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"... Large Endolymphatic Duct and Sac Large endolymphatic sac anomaly is the most common congenital inner ear anomaly found by imaging and is bilateral in 90% of cases. 44 It is also the most commonly missed cause of congenital deafness. It is a familial lesion with autosomal recessive inheritance. There is an association with Pendred syndrome, which is severe SNHL with thyroid disorder. Patients typically present with progressive, severe SNHL in childhood or early adulthood, often exacerbated by minor trauma. The hallmark imaging characteristic of large endolymphatic sac anomaly on CT is enlargement of the bony vestibular aqueduct, and a diameter greater than 1.5 mm at a point halfway between the crus communis and the intracranial aperture of the aqueduct, or an opercular measurement of more than 2 mm, are generally considered to be the defining characteristics. 45 On T2 FSE MR imaging, the underlying endolymphatic structure abnormalities are shown readily, consisting of enlargement of the endolymphatic sac and duct ( Fig. 9 ). In some cases, enlargement of the sac is more conspicuous than that of the duct, on T2 FSE MR. The vestibular aqueduct is usually found at the level of the vestibule and lateral semicircular canal, so it is easy to compare the diameter of the sac with the lateral and posterior semicircular canals, which should be larger than the aqueduct. 44 There is no relationship between the size of the endolymphatic sac and the severity of the SNHL. Although the aqueduct is best evaluated with CT, the bright signal intensity on T2 within the sac is best seen on thin T2-weighted images. In more than 75% of cases, there is an associated cochlear dysplasia, with dysmorphic apical turn and modiolar deficiency. 46 High-resolution T2 MR imaging may be able to distinguish more subtle abnormalities of scalar chamber asymmetry with the more anterior scala vestibuli larger than the more posterior scala tympani. 2 ..."
Reference 25
Review article
Imaging of Vertigo and Dizziness: A Site-based Approach, Part 1 (Middle Ear, Bony Labyrinth, and Temporomandibular Joint)
Alves I.S., Gebrim E.M.S., Passos U.L.
Seminars in Ultrasound, CT and MRI , 2024 pp 360-371
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"... A study by Krombach et al found that 86% of patients with posterior semicircular canal dehiscence had vertigo, highlighting the high occurrence of this symptom among affected individuals. Other vestibular symptoms are oscillopsia and disequilibrium, as well as auditory symptoms such as hearing loss and tinnitus. 27 CT is the preferred imaging approach and displays the specific bone defect in the posterior semicircular canal ( Fig. 8 ). 23 Enlarged vestibular aqueduct syndrome (EVAS) refers to the abnormal expansion of the vestibular aqueduct at the location of the endolymphatic duct associated with sensorineural or mixed hearing loss. The occurrence of dizziness with a subset experiencing recurrent episodes of vertigo was found in 63.6% of patients with EVAS. 28 In 84% of cases, EVAS is accompanied by other inner ear anomalies and can be observed either alone or in conjunction with numerous congenital disorders, including CHARGE (C: coloboma, H: heart defects, A: atresia choanae, R: retarded growth and development, G: genital hypoplasia, E: ear abnormalities and/or deafness) syndrome, Pendred syndrome, and branchiooto-renal syndrome or vestibulocochlear anomalies. 23,29 The characteristic imaging feature seen at CT is an enlargement of the osseous vestibular aqueduct, and at MR imaging, it is an enlargement of the endolymphatic duct and sac. 29 CT criteria are classically based on the transverse dimension of the vestibular aqueduct. The classic (Valvassori) criteria are a diameter >1.5 mm at the midpoint made halfway between the crus and the aperture on an axial view. The Cincinnati criteria for EVAS are midpoint width ≥1.0 mm or opercular width ≥2.0 mm as measured in the axial plane. 23 The upper limit of normal for the width of the vestibular aqueduct midpoint, when measured in the Pöschl plane, is 0.9 mm ( Fig. 9 ). 1 ..."
Reference 26
Handbook Chapter
Neuro-Otology
J.M. Espinosa-Sanchez, J.A. Lopez-Escamez
Handbook of Clinical Neurology , 2016 pp 257-277
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"... Imaging techniques Computed tomography (CT) images reveal that the vestibular aqueduct is significantly shorter and narrower and has a smaller external aperture on average in patients with MD, both in the affected and contralateral ear . Although these findings can contribute to explain the pathogenesis of the disease, their diagnostic significance is limited. Magnetic resonance imaging (MRI) obtained after intratympanic or intravenous administration of gadolinium has allowed not only in vivo visualization of the membranous labyrinth , but also the demonstration of EH in humans diagnosed of MD . Various authors have shown EH in 90% or more of patients with definite MD when specific inner-ear MRI protocols were performed . Heavily T2-weighted three-dimensional fluid-attenuated inversion recovery sequences on 3-T scanner appear to offer the best images. In particular, as gadolinium reaches the perilymphatic space a signal void appears, corresponding to the distended endolympathic space. Several studies have found a good correlation between cochlear hydrops on MRI and abnormal ECoG or abnormal VEMP . Nevertheless, the extent of EH visualized on MRI does not always correlate with the severity of cochleovestibular symptoms. MRI is emerging as a useful tool not only for diagnosis of EH, but also for early detection of contralateral involvement, to evaluate the permeability of the round and oval windows to intratympanic drugs and to document progression of the disease . 1 ..."
Reference 27
Reference works Chapter
Von Hippel-Lindau Syndrome
Gladys M. Glenn, Peter L. Choyke, McClellan M. Walther, Steven K. Libutti, Emily Y. Chew, H. Jeffrey Kim, Lindsay Middelton, Edward H. Oldfield, W. Marston Linehan
Encyclopedia of Endocrine Diseases , 2015 pp 674-687
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"... Endolymphatic Sac Tumor Manski and colleagues reported in 1997 their study of the association of an inner ear tumor with VHL. The ELST was the consensus name given for the previously diverse nomenclature of the tumor. ELSTs were detected in 13 (11%) of 121 individuals with VHL and in none of 253 patients without VHL. Hearing loss occurred at a mean age of 22 years, with the range being 12 to 50 years of age. Additional patients from the collection at the Air Force Institute of Pathology were reviewed by Manski and included a 7-year-old with ELST. Bilateral ELSTs seem to be found exclusively in patients with VHL. The ELST is characterized by a papillary–cystic adenomatous growth, but lacks generally accepted histologic features of malignancy. However, because of its locally aggressive behavior of eroding the surrounding temporal bone, Heffner et al . classified it as a low-grade adenocarcinoma. Diagnosis Common presenting symptoms of ELST are hearing loss, tinnitus, vertigo/disequilibrium, and facial paresis. Sudden onset of complete hearing loss on the side of the tumor was found in 38% of patients with ELST. There is no clear correlation between tumor size and symptoms. Audiologic assessment is added when there is a suspicion of a tumor. MRI of the brain may reveal a lesion, but the specific studies used for diagnosis are CT and MRI of the internal auditory canals. On CT scans, ELST is seen as an expansile and/or osteolytic lesion centered around the vestibular aqueduct in the posterior petrous bone. On MRI scans, it is characterized by heterogeneous foci of low and high intensities in both T1- and T2-weighted sequences ( Fig. 7 ). If no ELST is identified in patients with symptoms of hearing loss, tinnitus, vertigo or unexplained imbalance, repeated monitoring including audiologic and imaging studies is often advised. Treatment The indication and timing of surgical treatment take into consideration the slow but variable growth rate of ELSTs, preoperative hearing level, and severity of vestibular symptoms. 2 ..."
Reference 28
Review article
Imaging of the lower cranial nerves
Laine F.J., Underhill T.
Magnetic Resonance Imaging Clinics of North America , 2002 pp 433-449
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"... Evaluation of the endolymphatic duct and sac has become an important function of MR imaging. Endolymph drains through the endolymphatic duct and sac, housed in the vestibular aqueduct. Dilatation of the endolymphatic spaces (hydrops) can be secondary to congenital (vestibular aqueduct syndrome), acquired (postinflammatory or post-traumatic), and idiopathic (Ménière disease) causes. CT has been helpful in demonstrating an enlarged vestibular aqueduct but MR imaging is better able to demonstrate the duct and sac . Enhancement of the sac is consistent with an inflammatory process . Endolymphatic sac tumors also can be better defined with MR imaging. 2 ..."
Reference 29
Book Chapter
Update on Imaging of Hearing Loss
Vincent Chong, Lubdha M. Shah, Richard H. Wiggins
Skull Base Imaging , 2018 pp 169-196
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"... 28 There is no relationship between the size of the endolymphatic sac and the severity of the SNHL. The aqueduct itself is best evaluated with CT, whereas the sac is best seen as bright signal intensity on thin-section T2W MRI. In greater than 75% of cases, there is an associated cochlear dysplasia. 29 The apical turn of the cochlea in these cases is dysmorphic with modiolar deficiency ( Fig. 8.9 ). 30 High-resolution T2W MRI may be able to distinguish more subtle abnormalities of scalar chamber asymmetry with the more anterior scala vestibuli larger than the more posterior scala tympani. 31 Approximately 50% of cases have associated vestibular and/or SCC anomalies. In patients with sudden hearing loss, studies have shown wider vestibular aqueducts in the affected ear as compared with controls. 32 The endolymphatic sac can show enhancement, 32,33 which may be due to inflammation of the endolymphatic tissue or venous engorgement. It is hypothesized that a wide vestibular aqueduct may be associated with insufficient maturation of the inner ear. 34 The congenital “fragile” inner ear may receive abnormal pressure transmission through the vestibular aqueduct. 35 Cochlear nerve deficiency (CND) is noted in 12%–18% of pediatric patient ears with SNHL. 36,37 It is usually congenital and refers to the absence or reduction in caliber of the cochlear nerve. Because cochlear implants are generally contraindicated in CND, it is important to identify this condition in children being considered for implantation. 38,39 There will be an associated narrowing of the IAC on CT (IAC diameter of <4 mm), which is theorized to occur because the IAC width depends on the presence of the vestibulocochlear nerve cells to form normally. 2 ..."
Reference 30
Review article
Hearing Impairment in Children
Katbamna B., Crumpton T., Patel D.R.
Pediatric Clinics of North America , 2008 pp 1175-1188
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"... Case 3: Large Vestibular Aqueduct Congenital large vestibular aqueduct (LVA) is one of the most common malformations of the inner ear, occurring in isolation or with other malformations of the cochlea and leading to sudden or progressive sensorineural hearing loss in children. The prevalence of LVA in subjects with sensorineural hearing loss of unknown origin is estimated to be in the 5% to 7% range. 18,19 The vestibular aqueduct is a bony canal extending from the medial wall of the vestibule to the posterior surface of the petrous temporal bone. It serves as a conduit for the endolymphatic duct, which connects the endolymphatic sac to the vestibular labyrinth. Because the endolymphatic duct and sac are known to regulate ionic concentration of cochlear fluids, enlargement of the aqueduct and accompanying endolymphatic duct and sac may disrupt ionic balance. Although LVA may occur as a nonsyndromal developmental anomaly of the inner ear, according to the National Institutes of Deafness and Other Communication Disorders, 20 approximately one third of LVA cases are associated with PS. PS is an autosomal recessive disorder resulting from the mutation of the SLC26A4 gene located on chromosome 7, and 80% of the patients who have PS show LVA with or without accompanying Mondini malformation (lack of development of the apical three quarters of the cochlea, with a normally developed basal coil). LVA in PS tends to be bilateral, and 50% of individuals present with congenital severe to profound sensorineural hearing loss, with an additional 15% to 20% displaying fluctuating or progressive loss. 21 LVA may also be associated with other syndromes, including branchio-otorenal; manifestations of coloboma, heart anomalies, choanal atresia, retardation of mental development, genital anomalies and ear anomalies (known by the acronym CHARGE association); or Waardenburg syndrome. 2 ..."
Reference 31
Review article
Imaging of the Temporal Bone
Abele T.A., Wiggins R.H.
Radiologic Clinics of North America , 2015 pp 15-36
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"... 19–21 These lesions should be described on imaging studies for surgical planning purposes, but other causes should sought for the cause of clinical pulsatile tinnitus. Jugular bulb dehiscence ( Fig. 24 ) can present incidentally to the otolaryngologist as a retrotympanic blue mass. Inner Ear Inner ear disease is important to assess in any evaluation of a patient with sensorineural hearing loss or vestibular dysfunction ( Box 4 ). A complete discussion of the myriad of congenital inner ear dysplasias is beyond the scope of this article. Instead the authors touch on one of the more common congenital inner ear dysplasias, large vestibular aqueduct. Although only 40% of radiologic studies in children with sensorineural hearing loss will be abnormal, by far the most commonly identified malformation is enlarged vestibular aqueduct (also known as large endolymphatic sac anomaly). 22 An enlarged vestibular aqueduct has been defined as a vestibular aqueduct measuring greater than 1.5 mm in greatest diameter at its midpoint and greater than 2.0 mm at the operculum ( Fig. 25 ). 23,24 Practically speaking, any vestibular aqueduct with a diameter greater than the nearby lateral semicircular canal is considered abnormal. 25 Large vestibular aqueduct is associated with other inner ear abnormalities, including cochlear anomalies, such as modiolar deficiency and scalar asymmetry (76%) and vestibular anomalies (40%). 26 Otosclerosis is an idiopathic disease characterized by spongiotic change of the otic capsule that can result in conductive, mixed, or sensorineural hearing loss. 27 The mildest form of otosclerosis is the fenestral form, which presents with conductive hearing loss and manifests as a lucency in the fissula ante fenestram, a small segment of bone located just anterior to the oval window ( Fig. 26 ). 2 ..."
References (31)
Reference 1
Review article
Temporal Bone Anatomy
Benson J.C., Lane J.I.
Neuroimaging Clinics of North America , 2022 pp 763-775
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"... Aqueducts The vestibular aqueduct (VA) is a J-shaped osseous channel that runs from the vestibule to the posterior cranial fossa near the sigmoid sinus. Within it are the endolymphatic duct and sac, which are thought to play a role in the regulation of endolymph fluid. 24 Its origin, in the vestibule, is often invisible on CT imaging. From there, the VA gently widens into the isthmus, which is the curved portion of the aqueduct near the common crus of the semicircular canals. It continues to widen in its straight distal part, ultimately opening at the so-called operculum into the posterior cranial fossa. 25 The VA’s size is of particular interest since an enlarged aqueduct is often associated with sensorineural hearing loss. Most commonly, the Cincinnati criteria are used to define normal size: the VA is considered enlarged if it is ≥ 1 mm at the midpoint and/or ≥2 mm at the operculum. 26 The cochlear aqueduct, conversely, contains the perilymphatic duct and thus serves as a potential conduit for perilymphatic fluid, though it is not always patent. It runs obliquely downward from the scala tympani at the base of the cochlea to the jugular foramen. The course of the cochlear aqueduct is divided into four portions: the lateral orifice located near the round window at the scala tympani; the otic capsule segment; the petrous apex segment; and the medial orifice at the jugular foramen. Like the VA, the cochlear aqueduct is funnel-shaped and widens as it gets further from the otic capsule. Its lateral aspects are particularly small, and they are often invisible on imaging ( Fig. 10 ). 27 ..."
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"... The endolymphatic and perilymphatic spaces are indistinguishable from one another on standard MR imaging. However, gadolinium preferentially collects within the perilymphatic spaces during both intratympanic gadolinium administration and delayed post-contrast imaging, allowing visualization of the nonenhancing endolymphatic structures. 21–23 Such methods have been employed by some institutions to assess for endolymphatic hydrops in patients suspected of having Meniere’s disease. However, these techniques remain outside of the mainstay of imaging in most institutions. Aqueducts The vestibular aqueduct (VA) is a J-shaped osseous channel that runs from the vestibule to the posterior cranial fossa near the sigmoid sinus. Within it are the endolymphatic duct and sac, which are thought to play a role in the regulation of endolymph fluid. 24 Its origin, in the vestibule, is often invisible on CT imaging. From there, the VA gently widens into the isthmus, which is the curved portion of the aqueduct near the common crus of the semicircular canals. It continues to widen in its straight distal part, ultimately opening at the so-called operculum into the posterior cranial fossa. 25 The VA’s size is of particular interest since an enlarged aqueduct is often associated with sensorineural hearing loss. Most commonly, the Cincinnati criteria are used to define normal size: the VA is considered enlarged if it is ≥ 1 mm at the midpoint and/or ≥2 mm at the operculum. 26 The cochlear aqueduct, conversely, contains the perilymphatic duct and thus serves as a potential conduit for perilymphatic fluid, though it is not always patent. It runs obliquely downward from the scala tympani at the base of the cochlea to the jugular foramen. 1 ..."
Reference 2
Review article
The biology of intratympanic drug administration and pharmacodynamics of round window drug absorption
Banerjee A., Parnes L.S.
Otolaryngologic Clinics of North America , 2004 pp 1035-1051
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"... The lateral wall of the vestibule incorporates the oval window, and the medial wall gives rise to the vestibular aqueduct. Posteriorly there are five openings for the semicircular canals and anteriorly an elliptical opening leads into the scala vestibuli of the cochlea. ..."
Reference 3
Review article
Intralabyrinthine Schwannomas
Frisch C.D., Eckel L.J., Lane J.I., Neff B.A.
Otolaryngologic Clinics of North America , 2015 pp 423-441
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"... The afferent branches then run from the spiral ganglia through multiple small openings at the distal IAC called the cribriform plate. Just proximal to Scarpa's ganglion and just distal to the cribriform plate is the Schwann-glial cell junction of the cochlear division of the eighth cranial nerve. 8 The bony vestibule, located between the cochlea and 3 semicircular canals, contains the utricle and saccule. The ampulated ends of the semicircular canals open into the utricle. The saccule communicates with the scala media of the cochlea via the ductus reuniens. Two small ducts, one each from the saccule and utricle, unite to form the endolymphatic duct. As its name implies, this endolymph-containing duct is also part of the membranous labyrinth and is housed within the bony vestibular aqueduct. The afferent nerve fibers traveling from the vestibular hair cells of the semicircular canals, utricle, and saccule pierce the cribriform plate of the distal IAC en route to the Scarpa's ganglia within the IAC. Unlike the cochlear division, the Schwann-glial cell junction of the vestibular divisions is at the Scarpa's ganglia ( Fig. 2 ). 9 With this anatomy in mind, it is helpful to refer to the revised Kennedy classification system in order to describe tumor locations ( Fig. 3 ). Pathogenesis Originally, it was thought that ILS occurred mainly in patients with neurofibromatosis type 2 (NF2). 10 However, more recent reviews have shown that the overall number of sporadic cases far outnumber NF2 cases because of the rarity of the latter. The genetics of sporadic ILS are not thought to be different from the more common vestibular schwannoma of the IAC and CPA, although confirmatory studies are still needed. ..."
Reference 4
Review article
Imaging for cochlear implantation: Structuring a clinically relevant report
Vaid S., Vaid N.
Clinical Radiology , 2014 pp e307-e322
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"... Vestibular aqueduct (VA) The vestibular aqueduct is located at or above the level of the IACs, and oriented in a plane perpendicular to the IACs. The vestibular aqueduct is considered to be dilated ( Fig 18 ) if it measures >1.5 mm in width at the midpoint between the common crus and its external aperture or if its calibre is more than that of the adjacent posterior semicircular canal on axial images. 10,12 An enlarged endolymphatic duct/sac is identified as a hyperintense structure along the posterior and medial aspect of the petrous bone on heavily T2W MRI sequences. The signal may be highly variable reflecting the hyperviscous or variable protein content of the fluid within the sac. 12 ..."
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"... Aqueducts ( Fig 17 ) Vestibular aqueduct (VA) The vestibular aqueduct is located at or above the level of the IACs, and oriented in a plane perpendicular to the IACs. The vestibular aqueduct is considered to be dilated ( Fig 18 ) if it measures >1.5 mm in width at the midpoint between the common crus and its external aperture or if its calibre is more than that of the adjacent posterior semicircular canal on axial images. 10,12 An enlarged endolymphatic duct/sac is identified as a hyperintense structure along the posterior and medial aspect of the petrous bone on heavily T2W MRI sequences. The signal may be highly variable reflecting the hyperviscous or variable protein content of the fluid within the sac. 12 Cochlear aqueduct The cochlear aqueduct is located below the level of the IAC and is oriented in a plane parallel to it. It connects the scala tympani to the subarachnoid space of the posterior cranial fossa. The cochlear aqueduct has a medial orifice and lateral orifice; the medial orifice is considered to be dilated if it measures >3 mm. 12 Clinical relevance 1. Large vestibular aqueduct syndrome (LVAS): the incidence of enlarged vestibular aqueduct varies from 5–15% in the paediatric population. 12 A dilated vestibular aqueduct associated with symptoms of hearing loss or imbalance is referred to as LVAS. LVAS can be associated with cochlear defects and is associated with a high chance of an intraoperative CSF gusher. 10 2. The cochlear aqueduct acts as a route for the transmission of infection from the brain to the scala tympani. 12 ..."
Reference 5
Book Chapter
The auditory system
Jahangir Moini, Anthony LoGalbo, Raheleh Ahangari
Foundations of the Mind, Brain, and Behavioral Relationships , 2024 pp 143-160
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"... The spiral limbus' interdental cells actually secrete the membrane. Endolymphatic duct and sac The endolymphatic duct becomes distally dilated, forming the endolymphatic sac. It is within the osseous vestibular aqueduct, is of varying size, and can extend through an opening on the posterior petrous bone surface. It terminates between the two dural layers on the petrous temporal bone's posterior surface, close to the sigmoid sinus . Through the entire duct, surface cells resemble the cells lining unspecialized areas of the membranous labyrinth. They have a low cuboidal or squamous epithelium. Where the duct dilates and forms the endolymphatic sac, the epithelia lining the subepithelial connective tissue become more complex. A distal sac and an intermediate segment are visible. In this segment, the epithelium has light and dark cylindrical cells. The light cells are regular in form, with many long surface microvilli and endocytic invaginations between them. In their apical region, they have vesicles that are large and clear. The dark cells are wedge-shaped with narrow bases, dense fibrillary cytoplasm, and much less apical microvilli. The endolymphatic sac is important for maintaining vestibular function. Endolymph created in other labyrinth areas is absorbed in this region. This may occur primarily by the light cells. If the sac is damaged or the connection to the rest of the labyrinth is blocked, endolymph will accumulate. This causes hydrops , affecting cochlear and vestibular function. The epithelium is permeable to macrophages and other leukocytes, which can remove cellular debris from the endolymph. The epithelium is also permeable to various immune system cells that contribute antibodies to this fluid. Cochlear nerve The cochlear nerve connects the cochlear nuclei and related brainstem nuclei to the organ of Corti. ..."
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"... Below the crest, an anterior cochlear region has small holes in a spiral, known as the tractus spiralis foraminosus , surrounding the central cochlear canal. Behind this, the inferior vestibular region has openings for the saccular nerves. Most posteroinferiorly, the foramen singular allows the nerve to enter the posterior semicircular duct. Vascular loops in the internal acoustic meatus from the anterior inferior cerebellar artery may cause pulsatile tinnitus. Tinnitus Tinnitus is described as sounds such as ringing, buzzing, or clicking when there is no auditory stimulus to cause them. It is usually a symptom of a disease, such as cochlear nerve degeneration, or can signify inflammation of the middle or inner ear. It may be caused by the destruction of auditory pathway neurons and the ingrowth of adjacent neurons, whose signals are interpreted as noise by the brain. Vascular supply The labyrinthine artery is the main source of blood for the inner ear. The semicircular canals are supplied by the stylomastoid branch of the occipital artery or posterior auricular artery. The labyrinthine artery divides from the basilar artery and sometimes from the anterior inferior cerebellar artery. At the bottom of the internal acoustic meatus, it divides into cochlear and vestibular branches. The cochlear branch subdivides into 12–14 tiny branches through the modiolus canals. These are distributed as capillary plexuses to cochlear structures, including the basilar membrane, spiral lamina, and stria vascularis. The utricle, saccule, and semicircular ducts are supplied by vestibular arterial branches. Veins that drain the vestibule and semicircular canals accompany related arteries. They connect toward the utricle to form the vein of the vestibular aqueduct. This empties into the sigmoid or inferior petrosal sinus. The inferior cochlear vein usually drains into the inferior petrosal sinus or superior bulb of the internal jugular vein. 1 ..."
Reference 6
Book Chapter
Vestibulocochlear organ
Navarro M., Ruberte J., Carretero A.
Morphological Mouse Phenotyping , 2017 pp 521-539
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"... The endolymphatic duct originates in the saccule and runs inside an osseous duct, the vestibular aqueduct , until it reaches the subarachnoid space. Here, the endolymphatic duct becomes dilated to form the endolymphatic sac where the endolymph is reabsorbed ( Fig. 15-20 ). The walls of the utricle and saccule are lined by a single squamous epithelium but in their medial wall there is an oval-shaped thickening where sensory hair cells are located. These structures are the maculae. The macula of the utricle lies horizontally while the macula of the saccule is oriented vertically in relation to the head of the mouse. Just as occurs in the spiral organ, the ends of the cilia of the sensory cells of the macula are immersed in a gelatinous membrane, the otolithic membrane ( Fig. 15-18 ). The otoliths (statoconia) are small calcium carbonate crystals which put pressure on the cilia, inclining them, thereby stimulating the sensory hair cells. The function of the utricle and saccule is to detect the static position and the linear, horizontal and vertical movements of the head. In the vestibule originate three semicircular canals, which are located in the three spatial planes. In the mouse, the semicircular canals do not form angles of 90° between each other as is the case in humans. The anterior semicircular canal and the lateral semicircular canal are located more or less vertically. The anterior canal is located in a sagittal plane and the lateral canal in a transverse plane, whereas the posterior semicircular canal maintains a more or less horizontal position with respect to the head. The semicircular canals of the mouse are approximately 200 microns in diameter, with the anterior canal being slightly larger, and the lateral canal being the smallest one. This difference in diameter between the semicircular canals is reflected in their different sensibilities to movement. ..."
Reference 7
Review article
Hypothetical Mechanism for Vertigo in Meniere's Disease
Gibson W.P.R.
Otolaryngologic Clinics of North America , 2010 pp 1019-1027
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"... The anatomy of the endolymphatic sac and duct The anatomy of the human endolymphatic duct and sac is shown in Fig. 1 . The duct begins at the ductus reuniens, which joins the cochlear duct and the utricle to form an endolymphatic duct (ED) leading to the endolymphatic sac (ELS). The anatomy has been described by Lo and his colleagues. 8 Often anatomic texts describe a long thin ED ending in a short, pouch-like ELS, but in reality, the system is far different in appearance. The ED is a short single-lumen tube of only 2 mm in length. The ELS is much larger and a highly complex structure of interconnecting tubules, cisterns and crypts. The endolymphatic sinus (ES) lies in a groove on the posteromedial surface of the vestibule, with its distal outlet leading to the vestibular aqueduct. The ED narrows at its isthmus at the isthmus of the vestibular aqueduct (VA), where it is oblong in shape with mean diameter of only 0.09 × 0.2 mm. Distal to the isthmus begins the ELS, which flares considerably transversely but thickens only slightly in its sagittal dimension. The ELS has two portions, the intraosseous portion within the VA, and the extraosseous portion. The size of the intraosseous portion varies considerably, from 6 to 15 mm in length and 3 to 15 mm in width. The extraosseous ELS rests on a fovea on the posterior wall of the petrous bone, where the ELS lies between two layers of dura. The extraosseous ELS also varies considerably in size, from 5 to 7 mm in width and 10 to 15 mm in length. In lower animals and the human fetus, the ELS consists of a single lumen. In people after the age of 1 year, the ELS develops tubules that reach adult complexity by the age of 3 to 4 years. The tubules of the ELS are more complex in the proximal and middle (rugose) portions. The concept of the human ELS as a empty sack is entirely wrong. ..."
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"... Meniere's disease or a syndrome? It seems improbable that Meniere's disease is a single disease rather than several different pathologic conditions that result in the same symptom complex. Any condition that causes narrowing of the vestibular aqueduct and the production of excess endolymph could result in the same symptom complex. The term Meniere's syndrome may be more exact. There are congenital and acquired causes of Meniere's syndrome. For example, congenital deafness due to rubella or toxoplasmosis can result in secondary Meniere's disease; acquired infections such as treponemal disease may partially block the vestibular aqueduct, and a tumor of the endolymphatic sac often causes Meniere's symptoms. Nevertheless, there is an idiopathic group for which no causes have been determined, and perhaps this could be termed Meniere's disease in a similar fashion to the way that Bell palsy has become the term for an idiopathic facial palsy. Meniere's disease or the idiopathic cause of Meniere's syndrome has a wide spread of the age at onset. The author has analyzed his own series of 1576 patients who had clinically definite Meniere's disease 21 ( Table 1 ); the median age is approximately 50 years. This age range is not typical of vascular disorders or autoimmune problems. The age range is somewhat similar to the age of onset of peptic ulcer and gives some support to the notion that Meniere's disease may begin after a viral labyrinthitis in ears that have narrow vestibular aqueducts. The initial episode of viral labyrinthitis causes the endolymphatic hydrops, establishing the syndrome. Perhaps the condition can resolve in the earliest stages after a cluster of attacks. In other cases, the viral labyrinthitis may recur and cause another cluster of attacks. Some viruses, such as the herpes virus, may remain in the inner ear, causing several recurrences until eventually the endolymphatic sac functionality is destroyed, causing widespread endolymphatic hydrops similar to the findings after removal of the endolymphatic sac in guinea pigs. 22 1 ..."
Reference 8
Review article
Imaging of dizziness
Connor S.E.J., Sriskandan N.
Clinical Radiology , 2014 pp 111-122
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"... Anatomy and imaging methods The end organs of balance and equilibrium are the vestibule and semicircular canals, consisting of the outer osseous labyrinth and an inner membranous (or endolymphatic) labyrinth. There is perilymphatic fluid interposed between the two compartments. The inner endolymphatic structures are the utricle, the saccule, and the semicircular ducts. The endolymphatic sac and duct is located within the vestibular aqueduct and this also communicates with the saccule and utricle. The utricle and saccule or “static labyrinth” contains maculae (consisting of sensory hair cells that detect the position of otolithic crystals), which detect the position of the head relative to gravity. The semicircular canals or “kinetic labyrinth” contain ampullae (also with sensory hair cells), which respond to rotation and angular acceleration. These structures detect endolymphatic flow and innervate the vestibular nerves that maintain balance through their central connections. The vestibular nerves arise from a population of bipolar neurons and cell bodies, which reside in Scarpa's ganglion, and they run with the cochlear nerve (as cranial nerve eight or the vestibulo-cochlear nerve) to the vestibular nuclei in the brainstem ( Fig 1 ). The arterial supply to the vestibular structures is via the labyrinthine branch of the anterior inferior cerebellar artery. 28 Both computed tomography (CT) and magnetic resonance imaging (MRI) have a role in the imaging of these structures. The spatial resolution of CT is ideal in view of the innate high contrast of the osseous component and it will demonstrate erosion, fracture, or deficiency as well as the misplacement of prostheses with respect to the bony labyrinth. MRI is required to image the fluid-containing structures of the perilymphatic and endolymphatic spaces in addition to the vestibular nerves. 1 ..."
Reference 9
Book Chapter
Surgery of the Endolymphatic Sac
Packer M.D., Bradley Welling D.
Otologic Surgery , 2010 pp 411-428
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"... The distal sac has a smooth open lumen within the dura mater. Its cuboidal epithelium contains light and dark cells. 12 The lining of the intermediate portion of the sac shows more complex epithelial folds forming papillae and crypts of tall columnar light and dark cells. Cells of the intraosseous proximal rugose sac are intermediate between the taller, more distal cells and the squamous-to-cuboidal cells of the duct. The duct narrows at its isthmus to 0.1 to 0.2 mm in diameter. Luminal folding and transversely oriented tubules make the endolymphatic sac a more complex structure than it otherwise outwardly appears. 13 The normal bony vestibular aqueduct is readily apparent on high-resolution computed tomography (CT) scanning of the temporal bone. It is funnel-shaped or tubular with the width of its external aperture averaging 6 mm. 14,15 Radiographic observation of the affected ear in patients with Meniere’s disease showed a filiform narrowing of the external aperture averaging 2.2 mm. 14 The amount of narrowing of the external aperture was also shown to be correlated with an increasing percentage of positive electrophysiologic measures in the affected ears of patients with Meniere’s disease. 15 Statistically significant differences in the percentage of patients with enlarged summating potential-to-action potential (SP:AP) ratios by transtympanic electrocochleography were seen when correlated with the size of the external aperture of the vestibular aqueduct. An increased SP:AP ratio was noticed in 95% of ears with nonvisible external apertures, 91% when the aperture was less than 5 mm, 58% when the aperture was 5 to 7 mm, and 29% when the aperture was greater than 7 mm. The endolymphatic duct and sac can be seen on high-resolution fast spin echo magnetic resonance imaging (MRI). The endolymphatic sac and duct were seen on MRI of 20 temporal bones in healthy subjects using strongly T2-weighted sequences and postprocessing software. 1 ..."
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"... Patients with Meniere’s disease have been shown to have widening of the vestibular aqueduct aperture, but, although enticing to establish a mechanism of pathology, patients with Lermoyez’s syndrome have not been shown to have wider vestibular aqueducts than other patients with Meniere’s disease. Gibson and Arenberg also theorized that Tumarkin crises could be the effect of a membrane rupture in the overdistended endolymphatic space. Although Meniere’s disease has no known etiologic cause by definition, there are several etiologies of secondary Meniere’s syndromes. Meniere’s syndrome can be mimicked by disease processes in all categories—traumatic (inner ear fracture, perilymphatic fistula), infectious (viral, syphilis), inflammatory (autoimmune, allergy, Cogan’s syndrome, sarcoidosis), metabolic (diabetes), congenital/genetic (inner ear malformations, enlarged vestibular aqueduct), neoplastic (vestibular schwannoma), vascular (migraine, hemorrhage), and iatrogenic (stapes/mastoid surgery). These disease processes may lead to secondary endolymphatic hydrops by causing scarring within the labyrinth; activating inflammatory, cytokine, and complement pathways, yielding edema and fibrosis and altered extracellular matrices; altering hemodynamics that may affect transcellular ionic exchange and alter endolymph homeostasis; creating obstructive immune complexes or cellular debris; and activating cellular responses through humoral messengers. The end result is distention or obstruction of the endolymphatic flow disturbing sensation of cochleovestibular signals and sending distorted messages of sound, station, and motion. 1 ..."
Reference 10
Review article
A perspective from magnetic resonance imaging findings of the inner ear: Relationships among cerebrospinal, ocular and inner ear fluids
Nakashima T., Sone M., Teranishi M., Yoshida T., Terasaki H., Kondo M., Yasuma T., Wakabayashi T., Nagatani T., Naganawa S.
Auris Nasus Larynx , 2012 pp 345-355
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"... Communication between cochlear perilymph and CSF in the IAM is limited because of bone tightly surrounding the inner ear nerve at the fundus of the IAM. However, passage of gadolinium contrast agents between the inner ear and IAM was recognized in magnetic resonance imaging [9,10] . 3.3.2 Vestibular aqueduct In the vestibular aqueduct, there is endolymphatic duct and sac. Blood flow to the endolymphatic sac comes from the middle meningeal artery or occipital artery that belongs to the external carotid arterial system, which is strikingly different from the cochlea and the vestibular apparatus in which arterial blood comes from the branch of the anterior inferior cerebellar artery that belongs to the cerebral circulation system. The diameters of the endolymphatic duct and the proximal portion of the vestibular aqueduct are significantly smaller in Meniérè’s disease ears than in controls . Graphic reconstructions show the Meniérè’s sacs to be smaller and to have fewer tubular epithelial structures in the intraosseous portion than in the control ears. The median volume of the sac in the Meniérè’s disease side is substantially lower than in the contralateral ear. The width of the external aperture of the vestibular aqueduct is also significantly smaller in Meniérè’s disease ears than in controls. These findings indicate that the size not only of the vestibular aqueduct but also of the sac is reduced in Meniérè’s disease. The results suggest that the endolymphatic sac is pathologically changed in Meniérè’s disease and that reduced resorptive capacity of the small endolymphatic sac is associated with endolymphatic hydrops. 3.3.3 Cochlear aqueduct Enlargement of the cochlear aqueduct is often mentioned in the otologic literature, usually in its purported association with sensory hearing loss, stapes gusher, and transotic CSF leak. 1 ..."
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"... They considered that fluid surrounding the cochlear nerve is CSF that is covered with arachnoid-like covering in the modiolus. Communication between cochlear perilymph and CSF in the IAM is limited because of bone tightly surrounding the inner ear nerve at the fundus of the IAM. However, passage of gadolinium contrast agents between the inner ear and IAM was recognized in magnetic resonance imaging [9,10] . 3.3.2 Vestibular aqueduct In the vestibular aqueduct, there is endolymphatic duct and sac. Blood flow to the endolymphatic sac comes from the middle meningeal artery or occipital artery that belongs to the external carotid arterial system, which is strikingly different from the cochlea and the vestibular apparatus in which arterial blood comes from the branch of the anterior inferior cerebellar artery that belongs to the cerebral circulation system. The diameters of the endolymphatic duct and the proximal portion of the vestibular aqueduct are significantly smaller in Meniérè’s disease ears than in controls . Graphic reconstructions show the Meniérè’s sacs to be smaller and to have fewer tubular epithelial structures in the intraosseous portion than in the control ears. The median volume of the sac in the Meniérè’s disease side is substantially lower than in the contralateral ear. The width of the external aperture of the vestibular aqueduct is also significantly smaller in Meniérè’s disease ears than in controls. These findings indicate that the size not only of the vestibular aqueduct but also of the sac is reduced in Meniérè’s disease. The results suggest that the endolymphatic sac is pathologically changed in Meniérè’s disease and that reduced resorptive capacity of the small endolymphatic sac is associated with endolymphatic hydrops. 3.3.3 Cochlear aqueduct Enlargement of the cochlear aqueduct is often mentioned in the otologic literature, usually in its purported association with sensory hearing loss, stapes gusher, and transotic CSF leak. 1 ..."
Reference 11
Book Chapter
Auditory system
Jahangir Moini, Pirouz Piran
Functional and Clinical Neuroanatomy , 2020 pp 363-392
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"... Vascular supply The labyrinthine artery is the primary source of blood for the inner ear. The semicircular canals are also supplied by the stylomastoid branch of the occipital artery or posterior auricular artery. The labyrinthine artery divides off of the basilar artery, and sometimes, from the anterior inferior cerebellar artery. At the bottom of the internal acoustic meatus, it divides into the cochlear and vestibular branches. The cochlear branch further subdivides into 12–14 tiny branches through the modiolus canals. These are distributed as a capillary plexus to cochlear structures such as the basilar membrane, spiral lamina, and stria vascularis. The utricle, saccule, and semicircular ducts are supplied by vestibular arterial branches. Veins that drain the vestibule and semicircular canals accompany the related arteries. They connect toward the utricle, forming the vein of the vestibular aqueduct. This empties into the sigmoid or inferior petrosal sinus. The inferior cochlear vein, of the cochlear aqueduct, usually drains into the inferior petrosal sinus or superior bulb of the internal jugular vein. It is created by a union of the common modiolar and vestibulocochlear vines, providing nearly all venous outflow from the cochlea. Near the basal cochlear turn, the common modiolar vein is formed by a joining of the anterior and posterior spiral veins. The vestibulocochlear vein is formed by a joining of the anterior and posterior vestibular veins, as well as the vein of the round window. When it is present, a labyrinthine vein drains the apical and middle cochlear coils into one of the following: the posterior area of the superior petrosal sinus, or the transverse sinus or inferior petrosal sinus. 1 ..."
Reference 12
Handbook Chapter
Benign paroxysmal vertigo of childhood
Gurberg J., Tomczak K.K., Brodsky J.R.
Handbook of Clinical Neurology , 2023 pp 229-240
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"... Enlarged vestibular aqueduct Enlarged vestibular aqueduct, or EVA, is the most common congenital inner ear malformation identified in children . EVA may be found in isolation or associated with other congenital hearing loss syndromes, including Pendred and branchio-oto-renal syndromes. It is associated with early onset sensorineural or mixed hearing loss that can be fluctuating or progressive and may be exacerbated by minor head injuries or barotrauma. While most descriptions focus on hearing, children with EVA may also exhibit vestibular symptoms characterized by variable episodic vertigo lasting minutes to hours, often brought on by minor trauma or vigorous head rotation . The proposed pathophysiology of hearing loss and vertigo in EVA is either exposure of the cochlea and vestibule to intracranial pressure variations through a widened vestibular aqueduct or neuroepithelial damage secondary to reflux of hyperosmolar contents from the endolymphatic sac. It is postulated that the vestibular system is more resistant to these insults compared to the auditory system, explaining why patients often have preserved vestibular function in the face of profound sensorineural hearing loss. EVA may also present in infancy with paroxysmal torticollis that can precede the diagnosis of hearing loss by months or even years and may be misdiagnosed as BPTI of migraine origin, as discussed in further detail in earlier sections of this chapter . Physical examination is normal in the majority of cases of EVA, although goiter may be noted in Pendred syndrome and branchial abnormalities may be seen in branchio-oto-renal syndrome. Audiometry is typically abnormal and abnormalities on vestibular testing are also frequently present. Diagnosis is confirmed by computed tomography (CT) of the temporal bones. 1 ..."
Reference 13
Book Chapter
Pediatric vestibular dysfunction following head injury: Diagnosis and management
Graham Cochrane, Jacob R. Brodsky
Otologic and Lateral Skull Base Trauma , 2024 pp 217-243
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"... Enlarged vestibular aqueduct Enlarged vestibular aqueduct (EVA) is not a traumatic lesion but can become symptomatic following even very minor head injuries; thus, it warrants discussion in this chapter. EVA is the most common cause of congenital hearing loss that is grossly visible on temporal bone imaging. Children with EVA may have hearing loss at birth, but it can often be delayed in onset and fluctuates over time in many cases. It can be seen in the setting of mutations in the SLC26A4 gene, which is associated with Pendred syndrome when mutations are homozygous, but EVA can also be seen in conjunction with other hearing loss syndromes and often occurs as an isolated anomaly. Patients with EVA are often prone to sudden drops in hearing and/or vestibular function following impacts to the head that may be milder than that required to sustain a concussion or a major traumatic brain injury. Historically, patients with EVA were discouraged from sports activities, though this advice is no longer considered routine. 24 Audiologic testing in children with EVA typically shows a sensorineural or mixed hearing loss that may be unilateral or bilateral. Vestibular testing may show a third window pattern on VEMP testing and a unilateral or bilateral vestibular loss on other vestibular tests. 25 , 26 Diagnosis is confirmed by temporal bone CT, though very large vestibular aqueducts can sometimes be seen on MRI. There are multiple different diagnostic criteria for EVA, all of which are based on measurements of the vestibular aqueducts on temporal bone CT. The most commonly used criteria are the Cincinnati criteria, which indicate that EVA is confirmed by the presence of combined measurements of the aqueduct of > 0.9 mm at the midpoint and >1.9 mm at the operculum. 25 EVA is often also associated with an incomplete partition anomaly, type II (IP2), which was formerly known as a Mondini malformation. This anomaly also includes a shortened cochlear of only 1.5 turns, a deficient modiolus, and a deficient interscalar septum. 1 ..."
Reference 14
Handbook Chapter
Neuro-Otology
P. Bertholon, A. Karkas
Handbook of Clinical Neurology , 2016 pp 279-293
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"... Enlarged vestibular aqueduct Vestibular aqueduct enlargement, initially described by Valvassori and Clemis (1978) , is the most common imaging abnormality in patients with congenital inner-ear defects . The mechanism is an early arrest in the development of the endolymphatic canal and sac around the fifth to eighth weeks in utero , when the initial vesicle normally elongates and narrows to form the vestibular aqueduct ( Fig. 20.2 ). It can be associated with other inner-ear malformations such as cystic dysplasia of the canals, enlargement of the vestibule, or various cochlear abnormalities . This condition has been reported to be inherited in an autosomal-recessive manner and can be associated with syndromic hearing loss as in Pendred syndrome . The anomaly is characterized by sensorineural and, more often, mixed hearing loss, which usually begins in early childhood . Hearing loss is typically bilateral and progressive, with stepwise rather than fluctuating hearing decrements often triggered by relatively minor head trauma. The stapedial reflex is usually present and VEMPs have abnormally low thresholds and high amplitude, as in the third mobile window mechanism . Vestibular symptoms are less frequent than hearing loss and commonly begin in childhood, but they may be delayed until adulthood . Although sound- or pressure-induced vertigo has been reported, the vestibular symptoms are dominated by recurrent episodes of vertigo that can last hours and mimic Menière's disease . Physical findings during one of these episodes have shown a peripheral vestibular syndrome consistent with unilateral irritation or deficit, as in patients with Menière's disease . Indeed, it is noteworthy that in typical Menière's disease, the endolymphatic sac and duct play a predominant role. However, the direct surgical treatment of the endolymphatic sac in enlarged vestibular aqueduct is controversial, as endolymphatic sac decompression, arachnoid bypass, or endolymphatic sac occlusion may worsen the hearing loss . 1 ..."
Reference 15
Review article
Imaging and anatomy for cochlear implants
Fishman A.J.
Otolaryngologic Clinics of North America , 2012 pp 1-24
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"... The vestibular aqueduct The association of enlargement of the vestibular aqueduct and congenital sensorineural hearing loss is well recognized. 6,22,82–86 Radiographically, it may occur in conjunction with other identifiable inner ear anomalies as previously discussed, or as an isolated finding on CT or MR imaging (see Fig. 3 ; Fig. 15 ). 85–91 Radiographic enlargement has been reported using different imaging modalities and criteria, but is generally considered to exist when the aqueduct’s diameter is greater than 1.5 to 2.0 mm at its midpoint, measured between the common crus and the external aperture into the posterior fossa. 85,92–94 The large vestibular aqueduct syndrome is traditionally considered to be a distinct clinical entity in patients with radiographic evidence of enlargement of the vestibular aqueduct. 85,86 Hearing loss is typically bilateral and progressive, with stepwise decrements often associated with episodes of relatively minor head trauma. Moreover, enlargement of the vestibular aqueduct is considered to be a relatively common finding in children with congenital sensorineural hearing loss. 74,85,86 Some investigators regard it as the single most common radiographic finding among patients with congenital sensorineural hearing loss. 74 The major traditional hypothesis regarding the pathogenesis of this anomaly involves aberrant or arrested development of the endolymphatic duct and sac system, which is based on the observation that in early embryogenesis the duct is shorter, straighter, and proportionally much broader than in later maturity. 85 As more experience has been gained with MR scanning, the defect is currently being described and studied as one involving the entire endolymphatic duct and sac system. 87–91 There is also some recent evidence supporting a familial component to the disorder. 95 Some recent work is also being done to investigate its genetics basis as well as its association with other known syndromes. 93,96 There are a variety of speculative causes of the hearing loss associated with this disorder based mostly on clinical, radiographic, and surgical observations, as well as some analyses of endolymphatic chemical composition. 1 ..."
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"... The vestibular aqueduct The association of enlargement of the vestibular aqueduct and congenital sensorineural hearing loss is well recognized. 6,22,82–86 Radiographically, it may occur in conjunction with other identifiable inner ear anomalies as previously discussed, or as an isolated finding on CT or MR imaging (see Fig. 3 ; Fig. 15 ). 85–91 Radiographic enlargement has been reported using different imaging modalities and criteria, but is generally considered to exist when the aqueduct’s diameter is greater than 1.5 to 2.0 mm at its midpoint, measured between the common crus and the external aperture into the posterior fossa. 85,92–94 The large vestibular aqueduct syndrome is traditionally considered to be a distinct clinical entity in patients with radiographic evidence of enlargement of the vestibular aqueduct. 85,86 Hearing loss is typically bilateral and progressive, with stepwise decrements often associated with episodes of relatively minor head trauma. Moreover, enlargement of the vestibular aqueduct is considered to be a relatively common finding in children with congenital sensorineural hearing loss. 74,85,86 Some investigators regard it as the single most common radiographic finding among patients with congenital sensorineural hearing loss. 74 The major traditional hypothesis regarding the pathogenesis of this anomaly involves aberrant or arrested development of the endolymphatic duct and sac system, which is based on the observation that in early embryogenesis the duct is shorter, straighter, and proportionally much broader than in later maturity. 85 As more experience has been gained with MR scanning, the defect is currently being described and studied as one involving the entire endolymphatic duct and sac system. 87–91 There is also some recent evidence supporting a familial component to the disorder. 95 Some recent work is also being done to investigate its genetics basis as well as its association with other known syndromes. 93,96 There are a variety of speculative causes of the hearing loss associated with this disorder based mostly on clinical, radiographic, and surgical observations, as well as some analyses of endolymphatic chemical composition. 2 ..."
Reference 16
Book Chapter
Sensory Organ Disorders (Retina, Auditory, Olfactory, Gustatory)
Gillespie D.C.
Neural Circuit Development and Function in the Brain , 2013 pp 731-759
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"... Pendred Syndrome Pendred syndrome (deafness with goiter) is the most common syndromic form of deafness, accounting for up to 7.8% of cases of congenital deafness and occurring in an estimated 7.5 of 100,000 births. Patients inheriting this autosomal-recessive disorder have variable degrees of deafness at birth and typically develop goiter in the second decade. The syndrome is accompanied by structural defects of the temporal bone such as Mondini dysplasia and enlarged vestibular aqueduct (EVA) that are likely driven in part by accumulation of endolymph. The vestibular aqueduct, embedded within the temporal bone, is a small canal containing the endolymphatic duct and extending from the vestibule between the cochlea and the labyrinth to the endolymphatic sac. In Mondini dysplasia, the apical turn of the cochlea fails to form, and patients are profoundly deaf at birth. In EVA, vestibular dysfunction may be present and the hearing loss is variable. The mutated Pendred syndrome gene PDS (SLC26A4) is a member of the solute carrier protein 26 anion transporter family, and the gene product pendrin is a transmembrane Cl − /I − / HCO 3 − transporter. Allelic heterogeneity produces some SLC26A4 variants with Pendred syndrome and others with non-syndromic deafness with EVA (DFNB4). Pendrin is expressed in the inner ear and kidney, and in the thyroid, where it mediates apical iodide transport in thyroid follicular cells . Although Pendred syndrome is sometimes accompanied by hypothyroidism that could itself contribute to hearing loss, the distribution of the mouse pendrin throughout the endolymphatic duct and sac, and in specific areas of utricle, saccule, and external sulcus, points to a specific role of pendrin in fluid resorption and in regulating the ionic composition of the cochlear endolymph. 1 ..."
Reference 17
Review article
Genetics and phenomics of Pendred syndrome
Bizhanova A., Kopp P.
Molecular and Cellular Endocrinology , 2010 pp 83-90
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"... Pendred syndrome is caused by biallelic mutations in the SLC26A4 gene, which encodes the multifunctional anion exchanger pendrin . All patients with biallelic mutations in the SLC26A4 gene have Pendred syndrome, indicating that it is genetically homogeneous ( Table 1 ) . The recessive form of hearing loss referred to as DFNB4 (OMIM 600791), originally thought to be a distinct entity because of the absence of an enlarged thyroid, is also explained by mutations in the SLC26A4 gene and is thus allelic with Pendred syndrome ( Table 1 ). Patients with DFNB4 display sensorineural hearing loss with an EVA and, if formally tested, they have a positive perchlorate test despite the absence of an enlarged thyroid . Patients with non-syndromic EVA are either homozygous for the SLC26A4 wild-type, or they have only one mutated allele . In some families, non-syndromic EVA is associated with monoallelic SLC26A4 mutations suggesting that unrecognized mutations in other regions of the gene or in another gene could contribute to the pathogenesis of the phenotype . Double heterozygosity for mutations in the SLC26A4 gene and the transcription factor FOXI1, which is involved in the regulation of SLC26A4 gene expression, has been reported in a family with EVA as well as in double heterozygous mice ( Slc26A4 +/− ; Foxi1 +/− ) ( Table 1 ) . This observation confirms that Pendred syndrome and non-syndromic EVA may have a digenic cause in a subgroup of patients . The recent finding of digenic mutations in SLC26A4 and KCNJ10, a potassium channel involved in the generation of the endocochlear potential, in patients with EVA further support that this phenotype can have a oligo- or polygenic etiology ( Table 1 ) . 1 ..."
Reference 18
Book Chapter
Hearing loss
Alaa Koleilat, Lisa A. Schimmenti, Karthik Muthusamy
Neurogenetics for the Practitioner , 2024 pp 305-325
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"... Pendred syndrome Pendred syndrome is comprised of bilateral congenital SNHL, vestibular dysfunction, and temporal bone abnormalities. This type of syndromic HL may present as mild-to-moderate or severe-to-profound. It usually manifests as prelingual HL and most patients have bilateral, progressive and fluctuating HL with some cases presenting as asymmetrical HL. 49 Temporal bone abnormalities include bilateral enlarged vestibular aqueduct with or without cochlear hypoplasia. The vestibular aqueduct carries the endolymphatic duct and contains inner ear fluid. Additionally, one of the defining features of Pendred syndrome is the development of euthyroid goiter in late childhood to early adulthood. Pendred syndrome is caused by biallelic pathogenic variants in SLC26A4, which encodes an anion transporter (pendrin) involved in the ion exchange of iodine, bicarbonate, hydroxide, sulfate, formate, and chloride. 50–52 Pendrin is responsible for the flow of iodide in thyrocytes and is required for normal inner ear function. Over 300 pathogenic variants in SLC26A4 have been described that span the entire gene including missense, nonsense, splice site, and frameshift mutations. 53 Pathogenic variants in SLC26A4 have a loss-of-function effect and the lack of pendrin leads to the degeneration of the sensory cells in the inner ear. 54 Multiple studies in various populations have assessed the prevalence of mutations in SLC26A4, and the associated audiometric profile. 55–58 In Caucasian populations, SLC26A4 is the second most frequent gene implicated in nonsyndromic deafness after GJB2 . 59 Pathogenic variants in SLC26A4 are also associated with nonsyndromic cases of SNHL-DFNB4. 1 ..."
Reference 19
Reference works Chapter
Audition
Jing Wang, Jean-Charles Ceccato, Jean-Luc Puel
The Senses: A Comprehensive Reference , 2020 pp 468-486
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"... Pendred Syndrome SLC26A4 mutations cause Pendred syndrome (PS), an autosomal recessive disorder comprised of goiter, hearing loss and enlargement of the vestibular aqueduct. SLC26A4 encodes an 86-kDa transmembrane anion exchanger called pendrin, which mediates Cl /HCO3- transport. In the mouse, Slc26a4 is expressed in several organs such as the inner ear, thyroid, kidney and lung . In the inner ear, pendrin is expressed in nonsensory epithelial cells of the outer sulcus and spiral prominence of the cochlear duct, transitional cells surrounding the vestibular neuroepithelia, and mitochondria-rich cells of the endolymphatic sac. These cells are thought to contribute to the pH and ionic homeostasis of endolymph. To gain insight in the Pendred syndrome, several mutants, such as knockout mice (full or conditional) or knock-in mice (carrying a point mutation in the Slc26a4 gene) have been generated. A major feature of the Slc26a4 mutant mice is the defective endolymphatic compartment, followed by the degeneration of non-sensory and sensory cells in the adult animal. The stria vascularis of Slc26a4−/− mice was found to suffer from free-radical stress that was evident by elevated amounts of oxidized and nitrated proteins . Heterologous expression of a mutated form of pendrin in HEK 293 cells showed that some mutated pendrin isoforms accumulate in the endoplasmic reticulum . Interestingly, salicylate application enables the mutated form to target the plasma membrane and to recover their anion-exchanger activity . In addition, cochlear epithelial cells derived from patient-derived induced pluripotent stem cells (iPSCS) demonstrated that the pendrin aggregates in the cytoplasm increase susceptibility to cellular stress and cell death. 1 ..."
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"... By the activity of Na,K-ATPase and Na K Cl cotransporter within type II fibrocytes, K + ions are transported to the intrastrial space, passing through gap junction channels rich in Cx26, Cx30 and Cx31. Finally, the K+ ions are secreted to the endolymph by the marginal cells of the stria vascularis. Thus, malfunction of the stria vascularis and recycling K + can alter the electrochemical composition of the endolymph, resulting in loss of the endocochlear potential, elevated auditory thresholds, and thus hearing loss. While the production of K + and the recycling mechanism is extremely important for hearing function, very few therapeutic interventions have been proposed to maintain cochlear homeostasis. 2.24.5.1 Pendred Syndrome SLC26A4 mutations cause Pendred syndrome (PS), an autosomal recessive disorder comprised of goiter, hearing loss and enlargement of the vestibular aqueduct. SLC26A4 encodes an 86-kDa transmembrane anion exchanger called pendrin, which mediates Cl /HCO3- transport. In the mouse, Slc26a4 is expressed in several organs such as the inner ear, thyroid, kidney and lung . In the inner ear, pendrin is expressed in nonsensory epithelial cells of the outer sulcus and spiral prominence of the cochlear duct, transitional cells surrounding the vestibular neuroepithelia, and mitochondria-rich cells of the endolymphatic sac. These cells are thought to contribute to the pH and ionic homeostasis of endolymph. To gain insight in the Pendred syndrome, several mutants, such as knockout mice (full or conditional) or knock-in mice (carrying a point mutation in the Slc26a4 gene) have been generated. A major feature of the Slc26a4 mutant mice is the defective endolymphatic compartment, followed by the degeneration of non-sensory and sensory cells in the adult animal. 1 ..."
Reference 20
Book Chapter
Hereditary Hearing Impairment
Arti Pandya
Emery and Rimoin's Principles and Practice of Medical Genetics and Genomics , 2025 pp 227-279
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"... Nonsyndromic deafness associated with EVA (DFNB4) is also caused by mutations in SLC26A4 . A second gene has been implicated in the etiology of PS. In nine patients with PS or nonsyndromic EVA, Yang and colleagues found heterozygous mutation in FOXI1 , a transcriptional activator of SLC26A4 . Causal mutations interfered with FOXI1 binding, which compromised or completely abolished FOXI1 -mediated transcriptional activation of SLC26A4 . Furthermore, they described an EVA patient who was a compound heterozygote for mutation in FOXI1 and SLC26A4 , implicating digenic inheritance in PS and/or DFNB4. This finding was consistent with a mouse model of this pathway. Mutations in another potassium channel member gene, KCNJ10 , were identified in probands from two families with deafness and EVA, who also carried heterozygous mutations in the SLC26A4 gene . The finding of a single heterozygous or bi-allelic mutations in individuals with HL and EVA has led to further classification and genotype–phenotype correlations, with coining of the term “SLC26A4-related HL,” that includes nonsyndromic DFNB4, PS, and HL with EVA . The thyroid organification defect is mild in PS such that patients are often euthyroid with moderate thyroid enlargement developing in early adolescence. Mild hypothyroidism may precede or accompany the development of goiter, which is inconsistent and appears to be related to iodide intake. Although PS is easily recognized if goiter occurs, it cannot be diagnosed by clinical exam unless there is a known family history. Therefore in the absence of molecular testing, children with apparently nonsyndromic deafness should be reevaluated periodically for the presence of thyroid enlargement or dysfunction. 1 ..."
Reference 21
Reference works Chapter
Audition
Maggie S. Matern, Ronna Hertzano
The Senses: A Comprehensive Reference , 2020 pp 838-860
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"... However, mutant mouse models of the norrin protein exhibit progressive deterioration of inner ear structures, starting with the stria vascularis, which may be attributed to defects in the development and maintenance of inner ear vasculature . 2.42.3.6 Pendred Syndrome Pendred Syndrome is one of the most common causes of congenital deafness, with an estimated prevalence of approximately 7.5 in every 100,000 births . This recessively inherited syndrome is characterized by sensorineural hearing loss accompanied by an enlarged vestibular aqueduct and temporal bone abnormalities, as well as an enlarged thyroid gland called a goiter . Some individuals may also experience balance problems due to dysfunction of the vestibular system . Approximately 50% of Pendred Syndrome cases are caused by mutations in the gene SLC26A4, which encodes for the iodide-chloride transporter pendrin . The pendrin protein is expressed along the lateral wall of the cochlea, specifically in the outer sulcus, spiral prominence and stria vascularis, as well as in the endolymphatic sac . Here, it is thought to be involved in maintaining the ionic homeostasis of the endolymph that is required for hearing . A smaller proportion of Pendred Syndrome cases can be attributed to mutations in the transcription factor gene FOXI1, which is involved in transcriptional activation of SLC26A4 expression , and to mutations in KCNJ10, which encodes a potassium channel subunit also thought to be involved in endolymph homeostasis . 2.42.3.6.1 Perrault Syndrome Perrault Syndrome is a heterogeneous disease characterized by hearing loss and possible neurological symptoms in both males and females, as well as ovarian failure in females . Additionally, the hearing loss can present at birth or in early childhood, and is progressive . 1 ..."
Reference 22
Handbook Chapter
Neurocutaneous Syndromes
Prashant Chittiboina, Russell R. Lonser
Handbook of Clinical Neurology , 2015 pp 139-156
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"... Endolymphatic sac tumors General features Endolymphatic sac tumors (ELSTs) arise from the endolymphatic epithelium within the vestibular aqueduct . These tumors are rare in the general population but are found in up to 6–15% of VHL patients . Bilateral ELSTs are found only in VHL . Clinical ELSTs arise within the vestibular aqueduct, and can lead to hearing loss due to intralabrynthine hemorrhage, endolymphatic hydrops, and/or by otic capsule invasion . ELSTs can present with vestibular symptoms (62%), partial or complete hearing loss (95 to 100%), tinnitus (77%), and/or facial paresis (8%) . Imaging High resolution computed tomography (CT) scans of the temporal bone and pre-/postcontrast enhanced MR imaging of the internal auditory canals are necessary to detect and follow these tumors ( Fig. 10.5 ). CT is used to detect the presence of otic capsule invasion and to evaluate the extent of bony erosion of the temporal bone (particularly, the vestibular aqueduct in very small tumors). Postcontrast T1-weighted MR imaging can reveal signs of small ELST tumors as subtle asymmetric enhancement of the endolymphatic sac. Larger tumors are heterogeneously enhancing on postcontrast T1W images. Intralabrynthine hemorrhage is detected by increased signal on precontrast T1W images . Histologic findings ELSTs are papillary cystic glandular neoplasms with variegated patterns (see Fig. 10.5 ). Due to the presence of local bone invasion, these tumors were called low grade adenocarcinoma of probable endolymphatic origin . ELSTs may be found entirely within the endolymphatic sac when small. When large, these inevitably invade and erode the surrounding temporal bone . Treatment Surgery is the treatment of choice for ELSTs. Complete resection may be achieved in a majority of cases with minimal risk for recurrence (3%) after gross total resection . 1 ..."
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"... General features Endolymphatic sac tumors (ELSTs) arise from the endolymphatic epithelium within the vestibular aqueduct . These tumors are rare in the general population but are found in up to 6–15% of VHL patients . Bilateral ELSTs are found only in VHL . 1 ..."
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"... Retinal gliosis and hemorrhages are present in association with severe lesions. Differential upregulation of HIF-2α has been demonstrated in severe lesions that are refractory to anti-VEGF therapy . Treatment Laser photocoagulation and cryotherapy are the mainstays of surgical management of retinal hemangioblastomas. Either of these methods can be used to as a sole method for treatment of extrapapillary hemangioblastomas . Vitreoretinal surgery can be performed when preretinal and vitreal membranes, retinal detachment from traction, and exudation occur . For hemangioblastomas in close proximity to the optic nerve, intravitreal anti-VEGF therapy may arrest progression for small lesions and may help reverse exudates and edema in some cases . Photodynamic therapy or plaque radiotherapy may have a limited role in the management of retinal hemangioblastomas . Salvage external beam radiation may be used for treatment-refractory retinal hemangioblastomas with reduction in tumor volume and improvement in visual acuity . Endolymphatic sac tumors General features Endolymphatic sac tumors (ELSTs) arise from the endolymphatic epithelium within the vestibular aqueduct . These tumors are rare in the general population but are found in up to 6–15% of VHL patients . Bilateral ELSTs are found only in VHL . Clinical ELSTs arise within the vestibular aqueduct, and can lead to hearing loss due to intralabrynthine hemorrhage, endolymphatic hydrops, and/or by otic capsule invasion . ELSTs can present with vestibular symptoms (62%), partial or complete hearing loss (95 to 100%), tinnitus (77%), and/or facial paresis (8%) . Imaging High resolution computed tomography (CT) scans of the temporal bone and pre-/postcontrast enhanced MR imaging of the internal auditory canals are necessary to detect and follow these tumors ( Fig. 10.5 ). 1 ..."
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"... Imaging High resolution computed tomography (CT) scans of the temporal bone and pre-/postcontrast enhanced MR imaging of the internal auditory canals are necessary to detect and follow these tumors ( Fig. 10.5 ). CT is used to detect the presence of otic capsule invasion and to evaluate the extent of bony erosion of the temporal bone (particularly, the vestibular aqueduct in very small tumors). Postcontrast T1-weighted MR imaging can reveal signs of small ELST tumors as subtle asymmetric enhancement of the endolymphatic sac. Larger tumors are heterogeneously enhancing on postcontrast T1W images. Intralabrynthine hemorrhage is detected by increased signal on precontrast T1W images . 1 ..."
Related quote(s)5 / 5
"... CT is used to detect the presence of otic capsule invasion and to evaluate the extent of bony erosion of the temporal bone (particularly, the vestibular aqueduct in very small tumors). Postcontrast T1-weighted MR imaging can reveal signs of small ELST tumors as subtle asymmetric enhancement of the endolymphatic sac. Larger tumors are heterogeneously enhancing on postcontrast T1W images. Intralabrynthine hemorrhage is detected by increased signal on precontrast T1W images . Histologic findings ELSTs are papillary cystic glandular neoplasms with variegated patterns (see Fig. 10.5 ). Due to the presence of local bone invasion, these tumors were called low grade adenocarcinoma of probable endolymphatic origin . ELSTs may be found entirely within the endolymphatic sac when small. When large, these inevitably invade and erode the surrounding temporal bone . Treatment Surgery is the treatment of choice for ELSTs. Complete resection may be achieved in a majority of cases with minimal risk for recurrence (3%) after gross total resection . Surgical resection of the tumor can relieve audiovestibular symptoms and can preserve hearing in a majority of cases . Indications for treatment of ELSTs include progressive sensorineural hearing loss, vestibular symptoms, facial nerve compression, and local mass effect from tumor growth. Early resection of ELSTs is recommended to prevent sensorineural hearing loss and to alleviate vestibular symptoms. Facial nerve decompression (large tumors) performed during surgery can improve facial nerve function in some cases . The role of radiation in the management of ELST is unproven . 2 ..."
Reference 23
Review article
Imaging of Vertigo and Dizziness: A Site-based Approach Part 2 (Membranous Labyrinth and Cerebellopontine Angle)
Takahashi J.T., Alves I.D.S., Gebrim E.S., Goncalves V.T.
Seminars in Ultrasound, CT and MRI , 2024 pp 372-382
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"... MRI shows variable signal, predominantly isointense on T1 and T2, more homogeneous than schwannomas, and intense enhancement with paramagnetic contrast ( Fig. 10 ). CT may reveal lesion calcifications and underlying bone hyperostosis ( Table 1 ). Endolymphatic Sac Tumor Endolymphatic sac tumors are locally aggressive tumors originating from the endolymphatic sac, located in the posterior portion of the vestibular aqueduct of the petrous bone. 30 They can be sporadic or, more frequently, associated with von Hippel-Lindau disease. These tumors cause progressive and typically irreversible hearing loss, with an average age of onset around 22 years. 31 CT shows destruction of the retro-labyrinthine petrous bone with a geographic or mottled pattern, exhibiting spiculated or reticulated bony fragments. MRI shows heterogeneous signal, with hyperintense areas on T1-weighted images related to hyperproteinaceous/hematic components and intense enhancement by paramagnetic contrast ( Fig. 11 ). Paraganglioma Paragangliomas are hypervascular neuroendocrine tumors that originate from paraganglia, which embryologically derive from neural crest cells. These tumors are most commonly diagnosed between the third and sixth decades of life. However, the age of presentation and sex (male/female) predilection may vary depending on the type of paraganglioma (sporadic or hereditary) and its location. 32 Head and neck paragangliomas, for instance, tend to present later in life and have a strong female predominance. The most frequent sub-sites include carotid, tympanic, jugulotympanic, and vagal paragangliomas. Auditory symptoms include pulsatile tinnitus, hearing loss, and vertigo due to inner ear involvement via the cochlear promontory. On CT, findings suggestive of paraganglioma include destructive bony patterns affecting the jugular foramina and soft tissue lesion in the middle ear near the cochlear promontory. 1 ..."
Related quote(s)2 / 2
"... Endolymphatic Sac Tumor Endolymphatic sac tumors are locally aggressive tumors originating from the endolymphatic sac, located in the posterior portion of the vestibular aqueduct of the petrous bone. 30 They can be sporadic or, more frequently, associated with von Hippel-Lindau disease. These tumors cause progressive and typically irreversible hearing loss, with an average age of onset around 22 years. 31 CT shows destruction of the retro-labyrinthine petrous bone with a geographic or mottled pattern, exhibiting spiculated or reticulated bony fragments. MRI shows heterogeneous signal, with hyperintense areas on T1-weighted images related to hyperproteinaceous/hematic components and intense enhancement by paramagnetic contrast ( Fig. 11 ). 1 ..."
Reference 24
Review article
Applications of Magnetic Resonance Imaging in Adult Temporal Bone Disorders
Mohan S., Hoeffner E., Bigelow D.C., Loevner L.A.
Magnetic Resonance Imaging Clinics of North America , 2012 pp 545-572
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"... Inner ear lesions Large Endolymphatic Duct and Sac Large endolymphatic sac anomaly is the most common congenital inner ear anomaly found by imaging and is bilateral in 90% of cases. 44 It is also the most commonly missed cause of congenital deafness. It is a familial lesion with autosomal recessive inheritance. There is an association with Pendred syndrome, which is severe SNHL with thyroid disorder. Patients typically present with progressive, severe SNHL in childhood or early adulthood, often exacerbated by minor trauma. The hallmark imaging characteristic of large endolymphatic sac anomaly on CT is enlargement of the bony vestibular aqueduct, and a diameter greater than 1.5 mm at a point halfway between the crus communis and the intracranial aperture of the aqueduct, or an opercular measurement of more than 2 mm, are generally considered to be the defining characteristics. 45 On T2 FSE MR imaging, the underlying endolymphatic structure abnormalities are shown readily, consisting of enlargement of the endolymphatic sac and duct ( Fig. 9 ). In some cases, enlargement of the sac is more conspicuous than that of the duct, on T2 FSE MR. The vestibular aqueduct is usually found at the level of the vestibule and lateral semicircular canal, so it is easy to compare the diameter of the sac with the lateral and posterior semicircular canals, which should be larger than the aqueduct. 44 There is no relationship between the size of the endolymphatic sac and the severity of the SNHL. Although the aqueduct is best evaluated with CT, the bright signal intensity on T2 within the sac is best seen on thin T2-weighted images. In more than 75% of cases, there is an associated cochlear dysplasia, with dysmorphic apical turn and modiolar deficiency. 46 High-resolution T2 MR imaging may be able to distinguish more subtle abnormalities of scalar chamber asymmetry with the more anterior scala vestibuli larger than the more posterior scala tympani. 1 ..."
Related quote(s)2 / 2
"... Large Endolymphatic Duct and Sac Large endolymphatic sac anomaly is the most common congenital inner ear anomaly found by imaging and is bilateral in 90% of cases. 44 It is also the most commonly missed cause of congenital deafness. It is a familial lesion with autosomal recessive inheritance. There is an association with Pendred syndrome, which is severe SNHL with thyroid disorder. Patients typically present with progressive, severe SNHL in childhood or early adulthood, often exacerbated by minor trauma. The hallmark imaging characteristic of large endolymphatic sac anomaly on CT is enlargement of the bony vestibular aqueduct, and a diameter greater than 1.5 mm at a point halfway between the crus communis and the intracranial aperture of the aqueduct, or an opercular measurement of more than 2 mm, are generally considered to be the defining characteristics. 45 On T2 FSE MR imaging, the underlying endolymphatic structure abnormalities are shown readily, consisting of enlargement of the endolymphatic sac and duct ( Fig. 9 ). In some cases, enlargement of the sac is more conspicuous than that of the duct, on T2 FSE MR. The vestibular aqueduct is usually found at the level of the vestibule and lateral semicircular canal, so it is easy to compare the diameter of the sac with the lateral and posterior semicircular canals, which should be larger than the aqueduct. 44 There is no relationship between the size of the endolymphatic sac and the severity of the SNHL. Although the aqueduct is best evaluated with CT, the bright signal intensity on T2 within the sac is best seen on thin T2-weighted images. In more than 75% of cases, there is an associated cochlear dysplasia, with dysmorphic apical turn and modiolar deficiency. 46 High-resolution T2 MR imaging may be able to distinguish more subtle abnormalities of scalar chamber asymmetry with the more anterior scala vestibuli larger than the more posterior scala tympani. 2 ..."
Reference 25
Review article
Imaging of Vertigo and Dizziness: A Site-based Approach, Part 1 (Middle Ear, Bony Labyrinth, and Temporomandibular Joint)
Alves I.S., Gebrim E.M.S., Passos U.L.
Seminars in Ultrasound, CT and MRI , 2024 pp 360-371
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Related quote(s)1 / 1
"... A study by Krombach et al found that 86% of patients with posterior semicircular canal dehiscence had vertigo, highlighting the high occurrence of this symptom among affected individuals. Other vestibular symptoms are oscillopsia and disequilibrium, as well as auditory symptoms such as hearing loss and tinnitus. 27 CT is the preferred imaging approach and displays the specific bone defect in the posterior semicircular canal ( Fig. 8 ). 23 Enlarged vestibular aqueduct syndrome (EVAS) refers to the abnormal expansion of the vestibular aqueduct at the location of the endolymphatic duct associated with sensorineural or mixed hearing loss. The occurrence of dizziness with a subset experiencing recurrent episodes of vertigo was found in 63.6% of patients with EVAS. 28 In 84% of cases, EVAS is accompanied by other inner ear anomalies and can be observed either alone or in conjunction with numerous congenital disorders, including CHARGE (C: coloboma, H: heart defects, A: atresia choanae, R: retarded growth and development, G: genital hypoplasia, E: ear abnormalities and/or deafness) syndrome, Pendred syndrome, and branchiooto-renal syndrome or vestibulocochlear anomalies. 23,29 The characteristic imaging feature seen at CT is an enlargement of the osseous vestibular aqueduct, and at MR imaging, it is an enlargement of the endolymphatic duct and sac. 29 CT criteria are classically based on the transverse dimension of the vestibular aqueduct. The classic (Valvassori) criteria are a diameter >1.5 mm at the midpoint made halfway between the crus and the aperture on an axial view. The Cincinnati criteria for EVAS are midpoint width ≥1.0 mm or opercular width ≥2.0 mm as measured in the axial plane. 23 The upper limit of normal for the width of the vestibular aqueduct midpoint, when measured in the Pöschl plane, is 0.9 mm ( Fig. 9 ). 1 ..."
Reference 26
Handbook Chapter
Neuro-Otology
J.M. Espinosa-Sanchez, J.A. Lopez-Escamez
Handbook of Clinical Neurology , 2016 pp 257-277
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"... Imaging techniques Computed tomography (CT) images reveal that the vestibular aqueduct is significantly shorter and narrower and has a smaller external aperture on average in patients with MD, both in the affected and contralateral ear . Although these findings can contribute to explain the pathogenesis of the disease, their diagnostic significance is limited. Magnetic resonance imaging (MRI) obtained after intratympanic or intravenous administration of gadolinium has allowed not only in vivo visualization of the membranous labyrinth , but also the demonstration of EH in humans diagnosed of MD . Various authors have shown EH in 90% or more of patients with definite MD when specific inner-ear MRI protocols were performed . Heavily T2-weighted three-dimensional fluid-attenuated inversion recovery sequences on 3-T scanner appear to offer the best images. In particular, as gadolinium reaches the perilymphatic space a signal void appears, corresponding to the distended endolympathic space. Several studies have found a good correlation between cochlear hydrops on MRI and abnormal ECoG or abnormal VEMP . Nevertheless, the extent of EH visualized on MRI does not always correlate with the severity of cochleovestibular symptoms. MRI is emerging as a useful tool not only for diagnosis of EH, but also for early detection of contralateral involvement, to evaluate the permeability of the round and oval windows to intratympanic drugs and to document progression of the disease . 1 ..."
Reference 27
Reference works Chapter
Von Hippel-Lindau Syndrome
Gladys M. Glenn, Peter L. Choyke, McClellan M. Walther, Steven K. Libutti, Emily Y. Chew, H. Jeffrey Kim, Lindsay Middelton, Edward H. Oldfield, W. Marston Linehan
Encyclopedia of Endocrine Diseases , 2015 pp 674-687
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"... Endolymphatic Sac Tumor Manski and colleagues reported in 1997 their study of the association of an inner ear tumor with VHL. The ELST was the consensus name given for the previously diverse nomenclature of the tumor. ELSTs were detected in 13 (11%) of 121 individuals with VHL and in none of 253 patients without VHL. Hearing loss occurred at a mean age of 22 years, with the range being 12 to 50 years of age. Additional patients from the collection at the Air Force Institute of Pathology were reviewed by Manski and included a 7-year-old with ELST. Bilateral ELSTs seem to be found exclusively in patients with VHL. The ELST is characterized by a papillary–cystic adenomatous growth, but lacks generally accepted histologic features of malignancy. However, because of its locally aggressive behavior of eroding the surrounding temporal bone, Heffner et al . classified it as a low-grade adenocarcinoma. Diagnosis Common presenting symptoms of ELST are hearing loss, tinnitus, vertigo/disequilibrium, and facial paresis. Sudden onset of complete hearing loss on the side of the tumor was found in 38% of patients with ELST. There is no clear correlation between tumor size and symptoms. Audiologic assessment is added when there is a suspicion of a tumor. MRI of the brain may reveal a lesion, but the specific studies used for diagnosis are CT and MRI of the internal auditory canals. On CT scans, ELST is seen as an expansile and/or osteolytic lesion centered around the vestibular aqueduct in the posterior petrous bone. On MRI scans, it is characterized by heterogeneous foci of low and high intensities in both T1- and T2-weighted sequences ( Fig. 7 ). If no ELST is identified in patients with symptoms of hearing loss, tinnitus, vertigo or unexplained imbalance, repeated monitoring including audiologic and imaging studies is often advised. Treatment The indication and timing of surgical treatment take into consideration the slow but variable growth rate of ELSTs, preoperative hearing level, and severity of vestibular symptoms. 2 ..."
Reference 28
Review article
Imaging of the lower cranial nerves
Laine F.J., Underhill T.
Magnetic Resonance Imaging Clinics of North America , 2002 pp 433-449
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"... Evaluation of the endolymphatic duct and sac has become an important function of MR imaging. Endolymph drains through the endolymphatic duct and sac, housed in the vestibular aqueduct. Dilatation of the endolymphatic spaces (hydrops) can be secondary to congenital (vestibular aqueduct syndrome), acquired (postinflammatory or post-traumatic), and idiopathic (Ménière disease) causes. CT has been helpful in demonstrating an enlarged vestibular aqueduct but MR imaging is better able to demonstrate the duct and sac . Enhancement of the sac is consistent with an inflammatory process . Endolymphatic sac tumors also can be better defined with MR imaging. 2 ..."
Reference 29
Book Chapter
Update on Imaging of Hearing Loss
Vincent Chong, Lubdha M. Shah, Richard H. Wiggins
Skull Base Imaging , 2018 pp 169-196
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"... 28 There is no relationship between the size of the endolymphatic sac and the severity of the SNHL. The aqueduct itself is best evaluated with CT, whereas the sac is best seen as bright signal intensity on thin-section T2W MRI. In greater than 75% of cases, there is an associated cochlear dysplasia. 29 The apical turn of the cochlea in these cases is dysmorphic with modiolar deficiency ( Fig. 8.9 ). 30 High-resolution T2W MRI may be able to distinguish more subtle abnormalities of scalar chamber asymmetry with the more anterior scala vestibuli larger than the more posterior scala tympani. 31 Approximately 50% of cases have associated vestibular and/or SCC anomalies. In patients with sudden hearing loss, studies have shown wider vestibular aqueducts in the affected ear as compared with controls. 32 The endolymphatic sac can show enhancement, 32,33 which may be due to inflammation of the endolymphatic tissue or venous engorgement. It is hypothesized that a wide vestibular aqueduct may be associated with insufficient maturation of the inner ear. 34 The congenital “fragile” inner ear may receive abnormal pressure transmission through the vestibular aqueduct. 35 Cochlear nerve deficiency (CND) is noted in 12%–18% of pediatric patient ears with SNHL. 36,37 It is usually congenital and refers to the absence or reduction in caliber of the cochlear nerve. Because cochlear implants are generally contraindicated in CND, it is important to identify this condition in children being considered for implantation. 38,39 There will be an associated narrowing of the IAC on CT (IAC diameter of <4 mm), which is theorized to occur because the IAC width depends on the presence of the vestibulocochlear nerve cells to form normally. 2 ..."
Reference 30
Review article
Hearing Impairment in Children
Katbamna B., Crumpton T., Patel D.R.
Pediatric Clinics of North America , 2008 pp 1175-1188
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"... Case 3: Large Vestibular Aqueduct Congenital large vestibular aqueduct (LVA) is one of the most common malformations of the inner ear, occurring in isolation or with other malformations of the cochlea and leading to sudden or progressive sensorineural hearing loss in children. The prevalence of LVA in subjects with sensorineural hearing loss of unknown origin is estimated to be in the 5% to 7% range. 18,19 The vestibular aqueduct is a bony canal extending from the medial wall of the vestibule to the posterior surface of the petrous temporal bone. It serves as a conduit for the endolymphatic duct, which connects the endolymphatic sac to the vestibular labyrinth. Because the endolymphatic duct and sac are known to regulate ionic concentration of cochlear fluids, enlargement of the aqueduct and accompanying endolymphatic duct and sac may disrupt ionic balance. Although LVA may occur as a nonsyndromal developmental anomaly of the inner ear, according to the National Institutes of Deafness and Other Communication Disorders, 20 approximately one third of LVA cases are associated with PS. PS is an autosomal recessive disorder resulting from the mutation of the SLC26A4 gene located on chromosome 7, and 80% of the patients who have PS show LVA with or without accompanying Mondini malformation (lack of development of the apical three quarters of the cochlea, with a normally developed basal coil). LVA in PS tends to be bilateral, and 50% of individuals present with congenital severe to profound sensorineural hearing loss, with an additional 15% to 20% displaying fluctuating or progressive loss. 21 LVA may also be associated with other syndromes, including branchio-otorenal; manifestations of coloboma, heart anomalies, choanal atresia, retardation of mental development, genital anomalies and ear anomalies (known by the acronym CHARGE association); or Waardenburg syndrome. 2 ..."
Reference 31
Review article
Imaging of the Temporal Bone
Abele T.A., Wiggins R.H.
Radiologic Clinics of North America , 2015 pp 15-36
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"... 19–21 These lesions should be described on imaging studies for surgical planning purposes, but other causes should sought for the cause of clinical pulsatile tinnitus. Jugular bulb dehiscence ( Fig. 24 ) can present incidentally to the otolaryngologist as a retrotympanic blue mass. Inner Ear Inner ear disease is important to assess in any evaluation of a patient with sensorineural hearing loss or vestibular dysfunction ( Box 4 ). A complete discussion of the myriad of congenital inner ear dysplasias is beyond the scope of this article. Instead the authors touch on one of the more common congenital inner ear dysplasias, large vestibular aqueduct. Although only 40% of radiologic studies in children with sensorineural hearing loss will be abnormal, by far the most commonly identified malformation is enlarged vestibular aqueduct (also known as large endolymphatic sac anomaly). 22 An enlarged vestibular aqueduct has been defined as a vestibular aqueduct measuring greater than 1.5 mm in greatest diameter at its midpoint and greater than 2.0 mm at the operculum ( Fig. 25 ). 23,24 Practically speaking, any vestibular aqueduct with a diameter greater than the nearby lateral semicircular canal is considered abnormal. 25 Large vestibular aqueduct is associated with other inner ear abnormalities, including cochlear anomalies, such as modiolar deficiency and scalar asymmetry (76%) and vestibular anomalies (40%). 26 Otosclerosis is an idiopathic disease characterized by spongiotic change of the otic capsule that can result in conductive, mixed, or sensorineural hearing loss. 27 The mildest form of otosclerosis is the fenestral form, which presents with conductive hearing loss and manifests as a lucency in the fissula ante fenestram, a small segment of bone located just anterior to the oval window ( Fig. 26 ). 2 ..."
Related topics (10)
Cochlea
Cochlear Aqueduct
Dysplasia
Endolymph
Endolymphatic Sac
Semicircular Canal
Sensation of Hearing
Sensorineural Hearing Loss
Temporal Bone
Vestibule of the Ear
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microeconomics | collegeboard-ap
Supply and Demand
1.1.1 Total revenue test
1.1.2 Elastic, inelastic and unit elastic demand
1.1.3 Determinants of price elasticity: Substitutability, proportion of income
1.2.1 Factors influencing elasticity: Availability of inputs, production time
1.2.2 Comparison of short-run and long-run elasticity
1.3.1 Cross-price elasticity: Substitutes and complements
1.3.2 Income elasticity: Normal and inferior goods
1.4.1 Equilibrium price and quantity determination
1.4.2 Calculating consumer and producer surplus
1.4.3 Welfare economics: Maximizing total surplus
1.5.1 Effects of shifts in supply and demand
1.5.2 Price floors (e.g. minimum wage) and price ceilings (e.g. rent control)
1.6.1 Taxes: Impact on consumers, producers and deadweight loss
1.6.2 Subsidies: Benefits and inefficiencies
1.6.3 Quantity controls (e.g. quotas)
1.7.1 Effects of imports and exports on domestic markets
1.7.2 Tariffs and quotas: Costs and benefits
1.7.3 Free trade vs. protectionism
1.8.1 Law of demand and reasons for downward slope
1.8.2 Determinants of demand: Income, tastes, prices of related goods
1.8.3 Normal vs. inferior goods
1.9.1 Law of supply and reasons for upward slope
1.9.2 Determinants of supply: Input costs, technology, expectations
1.9.3 Short-run vs. long-run adjustments
Imperfect Competition
2.1.1 Collusion and cartels: Benefits and challenges
2.1.2 Characteristics of oligopolies: Few large firms, barriers to entry
2.1.3 Strategic behavior: Nash equilibrium and dominant strategies
2.2.1 Comparison of market structures: Monopoly, oligopoly, monopolistic competition
2.2.2 Causes and implications of market power
2.3.1 Sources of monopoly power
2.3.2 Price-setting behavior and inefficiency
2.3.3 Graphical representation of monopoly pricing
2.4.1 Conditions for price discrimination
2.4.2 Examples of first, second, and third-degree price discrimination
2.5.1 Characteristics: Product differentiation and many sellers
2.5.2 Short-run profit and long-run equilibrium
2.5.3 Excess capacity and inefficiency
Factor Markets
3.1.1 Derived demand: Factors of production
3.1.2 Marginal product of labor (MPL) and capital
3.1.3 Value of marginal product (VMP)
3.2.1 Determinants of factor demand: Productivity, prices of goods
3.2.2 Determinants of factor supply: Population, preferences
3.3.1 Marginal revenue product (MRP) = Marginal resource cost (MRC)
3.3.2 Hiring decisions in competitive and imperfect markets
3.4.1 Characteristics of monopsony
3.4.2 Wage determination and inefficiency
Market Failure and the Role of Government
4.1.1 Allocative efficiency: Marginal benefit equals marginal cost
4.1.2 Causes of market failures: Externalities, public goods
4.2.1 Positive externalities: Subsidies and government intervention
4.2.2 Negative externalities: Taxes, regulations, and tradable permits
4.2.3 Graphical analysis of externalities
4.3.1 Characteristics: Excludability and rivalry
4.3.2 Free-rider problem and under-provision of public goods
4.4.1 Regulation of monopolies: Price caps and subsidies
4.4.2 Antitrust policies: Promoting competition
4.5.1 Measurement of income inequality: Lorenz curve and Gini coefficient
4.5.2 Causes of inequality: Education, skills, inheritance
4.5.3 Government redistribution policies
Production, Cost and the Perfect Competition Model
5.1.1 Relationship between inputs and outputs
5.1.2 Short-run vs. long-run production
5.1.3 Law of diminishing marginal returns
5.2.1 Fixed variable and total costs
5.2.2 Marginal cost (MC), average total cost (ATC), and average variable cost (AVC)
5.2.3 Cost curves and their relationships
5.3.1 Economies of scale, diseconomies of scale, and constant returns
5.3.2 Long-run average cost curve (LRAC)
5.4.1 Accounting profit vs. economic profit
5.4.2 Normal profit and its role in resource allocation
5.5.1 MC = MR rule for maximizing profit
5.5.2 Loss minimization in the short run
5.6.1 Shutdown decision: Comparing price to AVC
5.6.2 Sunk costs and decision-making
5.7.1 Characteristics of perfect competition
5.7.2 Short-run vs. long-run equilibrium
5.7.3 Efficiency of perfectly competitive markets
Basic Economic Concepts
6.1.1 Definition and examples of scarcity
6.1.2 Unlimited wants vs. limited resources
6.1.3 Implications for individuals and societies
6.2.1 Types of systems: Command, market and mixed economies
6.2.2 How different systems address fundamental economic questions
6.2.3 Examples of real-world economies
6.3.1 Trade-offs and opportunity cost
6.3.2 Efficiency, inefficiency and unattainable points
6.3.3 Economic growth and shifts in the PPC
6.4.1 Difference between absolute and comparative advantage
6.4.2 Specialization and mutual gains from trade
6.4.3 Calculation of opportunity costs
6.5.1 Rational decision-making using marginal analysis
6.5.2 Marginal benefits vs. marginal costs
6.5.3 Applications to individual and business decisions
6.6.1 Marginal utility and diminishing marginal returns
6.6.2 Optimal allocation of resources
6.6.3 Budget constraints and consumer equilibrium
Determinants of demand: Income, tastes, prices of related goods
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Determinants of Demand: Income, Tastes, Prices of Related Goods
Introduction
Understanding the determinants of demand is fundamental in microeconomics, particularly for students preparing for the Collegeboard AP exams. This article delves into the key factors that influence consumer demand: income, tastes and preferences, and the prices of related goods. Grasping these concepts is essential for analyzing market behaviors and making informed economic decisions.
Key Concepts
Income
Income plays a pivotal role in determining the quantity of goods and services consumers are willing and able to purchase. Generally, an increase in consumer income leads to an increase in demand for normal goods, while the demand for inferior goods may decrease.
Normal Goods: These are goods for which demand increases as income rises. Examples include electronics, branded clothing, and dining out. The relationship between income and demand for normal goods is positive.
Inferior Goods: Contrary to normal goods, inferior goods see a decrease in demand as consumer income increases. Examples include generic brands, instant noodles, and used cars. The relationship between income and demand for inferior goods is negative.
The income effect can be illustrated with the income elasticity of demand, defined as: $$ \text{Income Elasticity of Demand} = \frac{\% \Delta Q_d}{\% \Delta I} $$ Where ( \% \Delta Q_d ) is the percentage change in quantity demanded and ( \% \Delta I ) is the percentage change in income.
A positive income elasticity indicates a normal good, while a negative value signifies an inferior good. Goods with income elasticity greater than 1 are considered luxury items, exhibiting strong responsiveness to income changes.
Tastes and Preferences
Tastes and preferences refer to the consumers' desires and inclinations towards certain goods and services. These are influenced by various factors including cultural trends, advertising, societal influences, and personal preferences.
Shift in Demand: When consumer tastes shift in favor of a product, the demand curve for that product shifts to the right, indicating an increase in demand at each price level. Conversely, if tastes shift away, the demand curve shifts to the left.
For example, if there is a rising trend in health consciousness, the demand for organic foods may increase, shifting the demand curve to the right. Marketing campaigns can also significantly impact consumer preferences, thereby affecting demand.
Consumer preferences are also subject to changes over time due to innovation, availability of substitutes, and changes in demographic factors. Understanding these shifts is crucial for businesses to adapt and meet market demands effectively.
Prices of Related Goods
The prices of related goods, specifically substitute and complementary goods, significantly influence the demand for a product.
Substitute Goods: These are goods that can replace each other. An increase in the price of one substitute good leads to an increase in demand for the other. For example, if the price of Pepsi rises, the demand for Coca-Cola may increase as consumers switch to the cheaper alternative.
Complementary Goods: These are goods that are typically consumed together. An increase in the price of a complementary good can lead to a decrease in the demand for the related product. For instance, if the price of printers increases, the demand for printer ink cartridges may decrease.
The cross-price elasticity of demand measures the responsiveness of the demand for one good to changes in the price of another good. It is calculated as: $$ \text{Cross-Price Elasticity of Demand} = \frac{\% \Delta Q_d^A}{\% \Delta P^B} $$ Where ( \% \Delta Q_d^A ) is the percentage change in quantity demanded of Good A and ( \% \Delta P^B ) is the percentage change in price of Good B.
A positive cross-price elasticity indicates substitute goods, while a negative value indicates complementary goods. Understanding these relationships helps businesses and policymakers anticipate changes in demand based on price fluctuations of related goods.
Comparison Table
| | | | |
--- --- |
| Aspect | Income | Tastes and Preferences | Prices of Related Goods |
| Definition | Consumer's financial resources available to purchase goods and services. | Consumer desires and inclinations towards certain products. | Cost of goods that are substitutes or complements. |
| Effect on Normal Goods | Increase in income boosts demand. | Positive shift in preferences increases demand. | Increase in price of substitutes boosts demand. |
| Effect on Inferior Goods | Increase in income decreases demand. | Negative shift in preferences decreases demand. | Increase in price of complements decreases demand. |
| Elasticity Measure | Income Elasticity of Demand | N/A | Cross-Price Elasticity of Demand |
| Example | Luxury cars (normal) vs. used cars (inferior) | Health trends increasing demand for organic foods | Increase in coffee price affecting tea demand (substitutes) |
Summary and Key Takeaways
Income influences demand differently for normal and inferior goods.
Tastes and preferences can shift demand curves positively or negatively.
Prices of related goods, such as substitutes and complements, affect demand through cross-price elasticity.
Understanding these determinants is crucial for analyzing market dynamics in microeconomics.
Applied knowledge of demand determinants aids in strategic business and policy-making decisions.
Coming Soon!
Examiner Tip
Tips
Use Mnemonics: To remember the determinants of demand, use the acronym ITPR: Income, Tastes, Prices of related goods, and others. Practice Graph Shifts: Regularly practice drawing and interpreting demand curves with shifts caused by different determinants to reinforce your understanding. Real-World Examples: Link theoretical concepts to real-world scenarios, such as analyzing how a new smartphone release affects the demand for older models.
Did You Know
Did You Know
Did you know that during economic recessions, the demand for inferior goods like instant noodles and public transportation often increases as consumers seek more affordable alternatives? Additionally, celebrity endorsements can significantly shift consumer tastes, leading to sudden spikes in demand for specific products. For instance, the surge in demand for athleisure wear can be partly attributed to influencers promoting comfortable yet stylish clothing on social media platforms.
Common Mistakes
Common Mistakes
Confusing Substitutes and Complements: Students often mix up substitutes and complements. Remember, substitutes can replace each other (e.g., tea and coffee), while complements are consumed together (e.g., printers and ink cartridges). Ignoring Income Effects: Another common error is neglecting how changes in income affect different types of goods. Always consider whether a good is normal or inferior when analyzing demand shifts.
FAQ
What are the main determinants of demand?
The main determinants of demand include consumer income, tastes and preferences, the prices of related goods (substitutes and complements), expectations, and the number of buyers in the market.
How does income affect the demand for normal and inferior goods?
For normal goods, an increase in income leads to an increase in demand. For inferior goods, an increase in income results in a decrease in demand.
What is cross-price elasticity of demand?
Cross-price elasticity of demand measures how the quantity demanded of one good responds to a change in the price of another good. It helps determine whether goods are substitutes or complements.
Can tastes and preferences change over time?
Yes, tastes and preferences can evolve due to factors like cultural shifts, technological advancements, marketing, and changes in consumer awareness and values.
Why is understanding determinants of demand important for businesses?
Understanding determinants of demand allows businesses to predict changes in consumer behavior, adjust pricing strategies, manage inventory, and make informed decisions regarding product development and marketing.
Supply and Demand
1.1.1 Total revenue test
1.1.2 Elastic, inelastic and unit elastic demand
1.1.3 Determinants of price elasticity: Substitutability, proportion of income
1.2.1 Factors influencing elasticity: Availability of inputs, production time
1.2.2 Comparison of short-run and long-run elasticity
1.3.1 Cross-price elasticity: Substitutes and complements
1.3.2 Income elasticity: Normal and inferior goods
1.4.1 Equilibrium price and quantity determination
1.4.2 Calculating consumer and producer surplus
1.4.3 Welfare economics: Maximizing total surplus
1.5.1 Effects of shifts in supply and demand
1.5.2 Price floors (e.g. minimum wage) and price ceilings (e.g. rent control)
1.6.1 Taxes: Impact on consumers, producers and deadweight loss
1.6.2 Subsidies: Benefits and inefficiencies
1.6.3 Quantity controls (e.g. quotas)
1.7.1 Effects of imports and exports on domestic markets
1.7.2 Tariffs and quotas: Costs and benefits
1.7.3 Free trade vs. protectionism
1.8.1 Law of demand and reasons for downward slope
1.8.2 Determinants of demand: Income, tastes, prices of related goods
1.8.3 Normal vs. inferior goods
1.9.1 Law of supply and reasons for upward slope
1.9.2 Determinants of supply: Input costs, technology, expectations
1.9.3 Short-run vs. long-run adjustments
Imperfect Competition
2.1.1 Collusion and cartels: Benefits and challenges
2.1.2 Characteristics of oligopolies: Few large firms, barriers to entry
2.1.3 Strategic behavior: Nash equilibrium and dominant strategies
2.2.1 Comparison of market structures: Monopoly, oligopoly, monopolistic competition
2.2.2 Causes and implications of market power
2.3.1 Sources of monopoly power
2.3.2 Price-setting behavior and inefficiency
2.3.3 Graphical representation of monopoly pricing
2.4.1 Conditions for price discrimination
2.4.2 Examples of first, second, and third-degree price discrimination
2.5.1 Characteristics: Product differentiation and many sellers
2.5.2 Short-run profit and long-run equilibrium
2.5.3 Excess capacity and inefficiency
Factor Markets
3.1.1 Derived demand: Factors of production
3.1.2 Marginal product of labor (MPL) and capital
3.1.3 Value of marginal product (VMP)
3.2.1 Determinants of factor demand: Productivity, prices of goods
3.2.2 Determinants of factor supply: Population, preferences
3.3.1 Marginal revenue product (MRP) = Marginal resource cost (MRC)
3.3.2 Hiring decisions in competitive and imperfect markets
3.4.1 Characteristics of monopsony
3.4.2 Wage determination and inefficiency
Market Failure and the Role of Government
4.1.1 Allocative efficiency: Marginal benefit equals marginal cost
4.1.2 Causes of market failures: Externalities, public goods
4.2.1 Positive externalities: Subsidies and government intervention
4.2.2 Negative externalities: Taxes, regulations, and tradable permits
4.2.3 Graphical analysis of externalities
4.3.1 Characteristics: Excludability and rivalry
4.3.2 Free-rider problem and under-provision of public goods
4.4.1 Regulation of monopolies: Price caps and subsidies
4.4.2 Antitrust policies: Promoting competition
4.5.1 Measurement of income inequality: Lorenz curve and Gini coefficient
4.5.2 Causes of inequality: Education, skills, inheritance
4.5.3 Government redistribution policies
Production, Cost and the Perfect Competition Model
5.1.1 Relationship between inputs and outputs
5.1.2 Short-run vs. long-run production
5.1.3 Law of diminishing marginal returns
5.2.1 Fixed variable and total costs
5.2.2 Marginal cost (MC), average total cost (ATC), and average variable cost (AVC)
5.2.3 Cost curves and their relationships
5.3.1 Economies of scale, diseconomies of scale, and constant returns
5.3.2 Long-run average cost curve (LRAC)
5.4.1 Accounting profit vs. economic profit
5.4.2 Normal profit and its role in resource allocation
5.5.1 MC = MR rule for maximizing profit
5.5.2 Loss minimization in the short run
5.6.1 Shutdown decision: Comparing price to AVC
5.6.2 Sunk costs and decision-making
5.7.1 Characteristics of perfect competition
5.7.2 Short-run vs. long-run equilibrium
5.7.3 Efficiency of perfectly competitive markets
Basic Economic Concepts
6.1.1 Definition and examples of scarcity
6.1.2 Unlimited wants vs. limited resources
6.1.3 Implications for individuals and societies
6.2.1 Types of systems: Command, market and mixed economies
6.2.2 How different systems address fundamental economic questions
6.2.3 Examples of real-world economies
6.3.1 Trade-offs and opportunity cost
6.3.2 Efficiency, inefficiency and unattainable points
6.3.3 Economic growth and shifts in the PPC
6.4.1 Difference between absolute and comparative advantage
6.4.2 Specialization and mutual gains from trade
6.4.3 Calculation of opportunity costs
6.5.1 Rational decision-making using marginal analysis
6.5.2 Marginal benefits vs. marginal costs
6.5.3 Applications to individual and business decisions
6.6.1 Marginal utility and diminishing marginal returns
6.6.2 Optimal allocation of resources
6.6.3 Budget constraints and consumer equilibrium
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188116 | https://www.youtube.com/watch?v=cYciyKc59fM | COUPLING REACTIONS OF DIAZONIUM SALTS
Saya's World of Chemistry
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Posted: 2 Mar 2021
Coupling reactions of diazonium salt and phenol, diazonium salt and aniline and formation of methyl orange, Importance of diazonium salts in the synthesis of aromatic compounds.
Transcript:
second category of reactions of diazonium salsa coupling reactions these are the reactions in which the diazo group is retained okay so diazonium salts readily undergo coupling reaction with electron rich compounds like phenols and amines and give colored azo compounds colored azo compounds and so dyes are formed okay so dyes are formed these reactions are called as coupling reactions let us see one by one first reaction ice cold solution of salt that is diazonium salt cold diazonium salt reacts couples with ice cold solution of phenol okay so diazonium salt diazonium salt so diazonic chloride on reaction with phenol okay in slightly phenol is in slightly alkaline solution three node coupling reaction takes place okay minus hc and the product will be into which azodi is obtained okay this is an orange dye and this is para hydroxy azo benzene para hydroxy azo benzene this is an orange dye okay so ice cold solution of diazonium salt reacts with an alkaline solution which is also an ice cold solution of phenol gives thera hydroxy azo benzene basically or which is an orange dye okay so azo dye is obtained second reaction with amines that is with aniline dysones chloride on reaction with emily okay and gives and it's true okay the product will be para amino benzene paragraph amino benzene will be obtained okay this is an allo dye okay so an azo compound is obtained this is the second reaction diazonium salt with aniline okay para amino benzoin is obtained which is a as a yellow dye nasal dye is obtained next reaction preparation of methyl orange this is an important product used in the laboratories okay so diazonium salt of para amino sodium benzene sulfonate and n enzyme analyte in presence of an alkali the reaction takes place and gives methyl orange okay preparation of methyl orange first a compound para amino benzene sodium sulphate okay this is para amino benzene sodium sulfate sulfonate okay this first gets converted into diazonium salt by diacetization gets converted into is dysonium salt n2 plus cl minus na eso okay naso3 okay so first para amino sodium benzene sulphate diacetization it gets converted into its disonium salt then this on reaction with n and dimethyl aniline and then diameter energy that is case of annealing two hydrogen atoms are replaced by two methyl groups so this diazonium salt on reaction with n and dimethyl aniline the product obtained will be like this in name so three okay n two and ch3 twice this product is called as methyl orange okay orange this which is an azo dye so an azo compound is obtained okay like this different as though compounds are can be obtained from uh diazonium salt okay this as the sarconium salts acts as the intermediate and the coupling reaction takes place next importance of diazonium cells in the synthesis of aromatic compounds okay importance of diazonium salts in the synthesis of aromatic compounds okay we know that this diazonium salts are very important intermediates in the preparation of aromatic compounds so they are the important tools in the hands of synthetic chemist okay first important point ariel halides can be prepared in their pure state can be prepared in the pure state with the help of diazonium salt as intermediate okay the pure state okay in the case of aryl fluorides and ah aryl iodides they cannot be prepared by direct halogenation okay around halides like aryl fluorides and iodides this cannot be prepared by direct halogenation okay so with the help of this diazonium salt which are used as intermediates we can prepare aryl fluorides and arrive iodide next cyano group cannot be introduced cyano group cannot be introduced directly or cannot be substituted the directly from chlorobenzene cannot be substituted from chloro chlorine from chlorobenzene directly so for the preparation of cyano group into the chlorobenzene we have to use dysonium salt does intermediate another important point next next is one two three tri bromobenzene try bromo benzene cannot be prepared directly in their pure state from their cannot be prepared in their pure state by the direct bromination of benzene it cannot be prepared cannot be prepared from direct bromination so for this also we have to use this diazonium salt as intermediate okay we have to prepare one two three dry bromobenzene from para nitro aniline with the help of diazonium salt okay as intermediate paranitro anneli to tri bromo benzene how can we do this let us see the reaction first we have to take para bro para nitro led finally nh2 this para nitro aniline first on bromination like this two bromo groups are get attached to this compound okay next uh direct bronchiation we cannot be prepared try bromobenzene so first we have to do diastolization okay diacetization first diastolization nano2 hcl okay and then second process we have to add cobr copper bromide hbr san mayo's reaction so first step diacetation then san mayo's reaction we will get compound like this bromine is attached so first n2c nh2 gets converted into n2 plus cl minus then on reaction with cu br that nu n2cl is replaced by substituted by bromine okay next on reduction no2 gets converted into nh2 with the help of tin production takes place compound like this okay nh2 sweetie the common next we have to remove this nh2 so for that first again we have to diacetization diastolization so nh2 gets converted into n2 plus cl minus then next step reaction with h3po2 that is replacement by hydrogen okay second step so what will happen this nh2 gets removed and the product will be tri bromo pencil like this we can prepare tri bromo benzene in their pure state we cannot prepare tri bromo benzene by direct bromination of benzene okay next four nitro toluene to two bromo benzoic acid four nitro toluene to two bromo benzoic acid how can we convert how can we obtain two bromo benzoic acid from four nitrogen first four nitro toluene and four nitro two okay first bromination ch3 no2 your bromine will substitute okay next on reduction nu2 is reduced to nh2 okay ch3 vr nh2 okay next we have to remove this nh2 group so for that first we have to do diasetization okay then replacement by replacement by hydrogenator for that we have to add fluoroboric acid h3po2 okay so what will happen nh2 nh2 is removed from the compound okay yes then next oxidation in presence of k minor food on hydrolysis in presence of k amino 4 ch3 gets oxidized to cooh so we'll get two bromo benzoic acid will be okay |
188117 | https://cameroncounts.wordpress.com/2010/05/02/probability/ | Probability | Peter Cameron's Blog
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Probability
Posted on02/05/2010byPeter Cameron
A recent posting by Alex Bellos has made me puzzle, once again, about probability.
I spent about five years teaching probability to first-year undergraduates. This had one effect on my thinking: it converted me into a covert Bayesian. I would say that all probability is conditional probabiity, conditional on everything that I know at the moment. If I say that the probability that a fair coin comes down heads is 1/2, this is because I have no further information either way; if I know something about the person tossing the coin, the situation of the coin toss, or anything else, I may revise my view.
But I have no satisfactory answer to the question “What is probability?”, and my recurring nightmare at that time was that a student would ask me that question and I would be unable to answer it.
The Monty Hall problem has been back in the news at present, partly because an entire book about it has just appeared: The Monty Hall Problem: The Remarkable Story of Math’s Most Contentious Brain Teaser by Jason Rosenhouse. (I don’t agree with the subtitle, but let that pass.) I never had any difficulty with the Monty Hall problem: the right answer seems obvious to me.
To recall: Monty Hall, a game show host, shows you three doors. Behind one door is a car, behind the other two are goats. Assume that you want to get the car and have no interest in goats. Monty invites you to choose a door. Then he opens another door, and shows you a goat; he asks you if you want to stick with your original choice or switch to the other door. Clearly, if you switch, you double your chances of winning the car. (Why is it clear? The chance that the car is behind the door you first choose is 1/3, and because Monty acts in such a way that you gain no information about the door you chose, this is unchanged; and after he opens a door, you know that the car is behind either the door you chose, or the remaining door.)
Purists will (and do) object to my statement of the problem. I should specify the algorithm that Monty uses to choose a door to open. I don’t think so. Monty is the host; he knows where the car is, and has no intention of showing you a car, or of giving you any information about where the car is beyond the fact that it is not behind the door he opened. Moreover, you have never been on (or even watched) the show before. It is always possible for Monty to open a door and show you a goat, and there is no doubt that he will do that. The condition about your ignorance is inserted because, if Monty had even a small bias towards one particular door (say the leftmost one) when he has a choice, then your knowledge of this would affect your calculation of the probabilities.
But this example described by Alex has me flummoxed. First, the background. If somebody says to you, “I have two children, and (at least) one is a boy,” what is the probability that both children are boys? Assuming that boys and girls are equally likely (not quite true, but let’s ignore that), the four possible combinations in birth order (BB, BG, GB and GG) are equally likely; the given information rules out GG, and so the formula for conditional probability gives the answer 1/3. All that is fine. Now suppose you are told instead, “I have two children, and at least one of them is a boy born on Tuesday.” Assume that the seven days of the week are equally likely for births (again not true, but let’s ignore that), and that gender and birth day are independent (I have no idea about the truth of this, but suspect it is false too), the same calculation of conditional probabilities shows that the probability that both children are boys is 13/27.
But why is it not 1/3? After all, the same calculation would apply no matter what birth day was given; and the information seems to be irrelevant to the question posed.
I don’t have any way of making this answer seem obvious, or even plausible. Can anybody suggest one?
The only suggestion is that, as I made it clear, the Monty Hall problem refers to a one-off event, and these are the hardest to think about probabilistically. Any bias that Monty has in choosing a door only comes into play in a long sequence of plays of the game.
Well-known examples in conditional probability give estimates of the probability that someone has a rare disease given that they have just tested positive for it (remember that no test is 100% reliable). No problem with that; but if I have just taken the test, the reasoning seems less satisfactory. For a start, I have more information about myself than about a random member of the population; I know, for example, that in the last year I have had several unexplained headaches, or …
In the “boy born on Tuesday” problem, it seems much more obvious that a test of many cases would agree approximately with the answer 13/27, than that the answer is right in the unique case that confronts us. The test would run as follows. A large number of parents would be chosen. Any who could not truthfully make the statement would be rejected. Of those who could, we would certainly expect that about 13/27 would indeed have two boys. However, in an individual case, we are tempted to think that the information about birthday was thrown in gratuitously and should have no effect on the answer.
For example, suppose that the protocol was as follows. A large set of parents is chosen; only those who have two children, at least one of which is a boy, are retained. One such parent is chosen at random and instructed to make the following statement “I have two children, and at least one is a boy born on X day”, where, if they have just one boy, then X day is the boy’s birth day, while if they have two, they are to choose one by any means at all. For this formulation, the answer is 1/3.
In his 1905 paper which won him the Nobel prize and was one of the foundation documents of quantum theory, Einstein said,
In calculating entropy by molecular-theoretic methods, the word “probability” is often used in a sense differing from the way the word is defined in probability theory. In particular, “cases of equal probability” are often hypothetically stipulated when the theoretical methods employed are definite enough to permit a deduction rather than a stipulation.
In other words, “Don’t assume that all outcomes are equally likely, especially if you are given enough information to calculate what their probabilities really are”. But how does that principle apply in this case?
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30 Responses to Probability
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Olof Sisask says: 03/05/2010 at 20:07 About the two-children B/G problem: I think it seems somewhat intuitive that the probability that you have two boys increases from 1/3, for the following reason. If you have a boy and a girl, then it’s quite unlikely that the boy was born on a Tuesday; it is much more likely to happen if both the children are boys. So if you know you’re in a setup where you have a boy that was born on a Tuesday then it’s quite unlikely that you have a boy and a girl. In other words, you need to have two boys for it to be reasonable that the unlikely event that one was born on a Tuesday occurs. Reply
Olof Sisask says: 03/05/2010 at 20:33 (Obviously I’m using the words ‘unlikely’ etc. in an informal way above, for illustration; 13/27 is still less than a half!) Reply
John Fabensays: 06/05/2010 at 23:57 I think most of the explanation for this comes in the penultimate paragraph, where you discuss the process the person used to produce their statement.
I think that the reason 13/27 doesn’t seem obvious is because it isn’t really very interesting – without having some information about the algorithm used to generate the statement, we don’t know what effect it should have on our conditional probability, and the algorithm “if I can truthfully make the statement ‘I have two children, at least one of whom is a boy born on a Tuesday’, do so, otherwise stay quiet” just isn’t a very good model of how people decide what to say. Reply
5. Peter Cameronsays: 07/05/2010 at 09:49 Mathematically the conditional probability that someone has two boys, given that one of their two children is a boy born on Tuesday, is 13/27. If we started defining the probability of some event in terms of the algorithm that led to the statement being made, all our textbooks would need to be rewritten from the ground up. I think the best way to proceed is to say, we do know how to calculate probability (and how to interpret it), but this requires careful thought, and sometimes our intuition lets us down. Reply
6. Pingback: Randomness in Nature « Combinatorics and more
Ted Jones says: 09/05/2010 at 01:05 Thanks for making me aware of the wonderful “Tuesday” example. I’m not sure I understand what you mean in your remark about rewriting of textbooks, but to me considering the process by which the parent’s statement came to be is essential for understanding.
I would formulate it this way: we have four independent random variables s1,s2,d1,d2, the sexes and days of birth of the two children. In addition we have the parent’s statement, which in the example is that X=(s1, s2, d1, d2) belongs to a set (at least one Tuesday boy) that I’ll call E0. Now, assume the parent’s statement is required to be true; if not, the problem disappears. There is an event {not E0} of positive probability on which the parent would have had to make a different statement, or none at all. I’ll assume “none at all” can’t occur. Allowing it is tantamount to rejecting the assumption that the sexes and days are equally likely, because–conditional on the parent having made a statement–they may not be.
We can formalize the parent’s statement as a random variable A whose value is a subset of the 196 possible sex/day 4-tuples. There is no reason why the parent’s choice has to be deterministic, but following John Farben’s lead, let’s say it is, so A is a function of X. The loss of generality is not important for my point. So we have A=f(X), with the “truth” property that f(X) must contain X.
The event we observed is not, in fact, E0, but instead {X:f(X)=E0}. This is equal to E0 if and only if f(X)=E0 for all X in E0, i.e., iff the parent makes the given statement E0 whenever it is possible to do so. In that case, P(BB|A=E0) = P(BB|E0) = 13/27 is correct. But with a different function f, the event conditioned on is different. For a parent who (given at least one boy), mentions “boy” plus the earliest day of the week (ordered Sun..Sat) on which a boy was born, I get 9/23 (hope that’s right). For a parent who avoids mentioning Tuesday boys if any other statement is possible, the probability is 1.
The day of the week does seem irrelevant at first glance, but whether it actually is irrelevant depends on the parent. A (nondeterministic) parent-rule I find natural is that the parent picks a child at random, and says there is at least one of whatever category the child falls into. The conditional probability of BB is 1/2 whether or not the day is included.
I believe the paradox results from computing the conditional probability assuming a version of f(X) different from the one implicit in our intuitive judgment that the day of the week is irrelevant. That’s a psychological rather than a mathematical conclusion, of course. What do you think of it? Reply
8. Bob Walterssays: 10/05/2010 at 10:51 I agree with John Faben that the calculation of probability requires information about the algorithm involved. Reply
9. Peter Cameronsays: 10/05/2010 at 11:50 Sorry to have to disagree…
Any calculation in probability can be done unambiguously by the rules (based on Kolmogorov’s axioms) provided we specify carefully what is the “sample space” and what is the probability measure on the sample space. If you start bringing in other factors like the algorithm used to generate a statement then all you are doing is changing the measure. That is why I said in my example that I am a covert Bayesian. I happen to think that the probability measure that applies in a given situation depends on everything I know about the situation (which may include information about the algorithm used to generate some statement).
In the Tuesday boy problem, after we know that the parent has two children, we have a sample space made up of 196 equally-likely combinations. (Arguably the correct sample space is much more fine-grained than this, but leave this aside for the moment). The statement reduces us to a set of 27, and the induced probability of these make them all equally likely; in 13 cases is it true that both children are boys.
If I knew that a different algorithm was being used, that would (by an application of Bayes’ theorem) give me a different measure, and the calculation would give a different answer. But if you want a different answer to this question, you had better make this algorithm public, so that I can use this knowledge to adjust the probabilities.
I am reminded of one of the questions that reached me after my appearance on the Horizon programme on infinity. Someone said, if I stayed a night in the infinite hotel, all rooms of which were full, and the next morning the other guests left, then this would demonstrate that infinity minus infinity is one. This is wrong because the history of the construction of an infinite set shouldn’t affect calculations of its cardinality. I think a similar principle applies here.
Finally, I am not violating Einstein’s requirement. The 196 combinations are all equally likely (at least to the sort of approximation we are using here), and if a statement is made to me I use this to recalculate the probabilities using Bayes’ theorem. I do not assume a priori that the 27 combinations remaining after the statement is made are equally likely. Reply
10. John Fabensays: 10/05/2010 at 13:25 Ok, so I’m 100% in agreement that the conditional probability of a person having 2 boys given that they have two children one of whom is a boy born on a Tuesday is 13/27. (similarly, if we are told that their eldest child is a boy, the probability goes to 50% – a situation where the intuition is perhaps easier to see?).
However, I’m not entirely sure that I agree that this probability is the same as the probability that someone has 2 boys given that they say “I have two children, one of whom is a boy born on a Tuesday”. In fact, it certainly isn’t – real people just don’t decide what to say by generating statements at random and saying them if they’re true.
I’m maybe just arguing semantics, but I think there’s a genuine point here. It helps to explain why the “born on a Tuesday” clause feels as though it shouldn’t give any information – in a real conversation, it actually wouldn’t. Reply
11. Peter Cameronsays: 10/05/2010 at 15:00 Fair point.
What you are saying is that from the information that someone makes this statement, you can infer nothing. The person may be lying, or might be telling the truth in such a way as to mislead, or simply confused. As Montaigne said, “If, like the truth, falsehood had only one face, we should know better where we are, for we should then take the opposite of what a liar said to be the truth. But the opposite of the truth has a hundred thousand shapes and a limitless field.” Or Anthony Kenny, “All worthwhile philosophical statements express an insight; and the opposite of an insight is not a contradictory sentence, but a muddle.” I don’t think probability theory can handle this complication! Reply
12. Ted Jones says: 10/05/2010 at 18:00 My last comment was held in moderation for a while, and I don’t know if John has seen it yet.
Peter, I’m with John on this, and I don’t think you’re quite following our argument. As I think I showed in my comment, it’s perfectly possible to discuss the distinction John makes using the sample space with 196 equiprobable points, and the mathematical definition of conditional probability with no extraneous factors; the “algorithm” is represented by a random variable defined on the 196-point sample space. And allowing the parent’s statement to be untrue is not necessary to get a conditional probability different from 13/27.
At least if you accept my version of things, 13/27 is the correct answer only if the parent mentions a boy born on Tuesday whenever it is possible to do so (truthfully). However, you can’t exclude the cases where the parent has to make a different statement; they account for more than half the 196 points. In those cases, the statement must be different.
So, what will the parent say when there is no Tuesday boy? Any truthful statement will do, but I haven’t been able to come up with an algorithm (or, more precisely, a random variable mapping the samples space into its power set) that both (1) is constrained as needed to get 13/27 and (2) corresponds to behavior that I find at all natural in the cases where there is no boy born on Tuesday.
For example, if there are two girls both born on Tuesday, it seems to me that the parent should say there is at least one girl born on a Tuesday. Saying instead something like “I have one child born on either a Tue, Thu, or Sat, and one child–not necessarily the same–who is a girl” is true, and therefore possible, but it seems perverse to assume the problem contemplates a parent who might actually say that.
So, ok, for two girls born on Tuesday the answer should be “at least one girl born on a Tuesday”. But then what does the parent say when there are a boy and a girl, both born on Tuesday? Any choice seems to violate some sort of symmetry.
It’s true that to calculate the conditional probability you don’t need to know more about the parent’s statement than the event on which the parent will say “at least one boy born on Tuesday”. And, if you choose to do so, you can make the assumption needed to get 13/27 as the answer. It’s just that if I think about it, that assumption leads to some strange conclusions about the parent that I don’t think should be taken for granted as necessarily following from the statement of the problem.
With my proposed parent-rule of choosing a child at random and say at least one of that category (which does require expanding the sample space to 2196 points, but I don’t see that as mattering for understanding), the day of the week does become uninformative as intuition suggests, and it doesn’t lead to any implausible or asymmetric choices for the parent.
I think my child-at-random rule is much more faithful to what a reader would intuitively expect the parent to do. The other assumption becomes “natural” only when you sit down to compute the probability, and then only if you don’t think about the “outside” cases.
If you can persuade me that there is a way a parent to give answers leading to 13/27 without “strange” behavior in other cases, please do. It would make this an even better problem. Reply
13. Ted Jones says: 10/05/2010 at 18:18 Or, here’s another way to look at it. Same setup, but the parent tells you there is at least one girl born on Tuesday, and the question is what is the probability that there are two girls? If you assume the behavior that is required to get 13/27 in the boy case, then in the case where there are a boy and a girl, both born on Tuesday, the parent has to mention the boy. The result is that the answer to the “girl” version of the question cannot be 13/27. Reply
14. Peter Cameronsays: 10/05/2010 at 18:18 Sorry about the delay – it is exam time, and service will be a bit slower for a while…
I think I stick to my position (though I am not quite sure). Allowing a statement to be a random variable is, I think, covered by my saying that “The person may be lying, or might be telling the truth in such a way as to mislead, or simply confused.” I don’t think I know any probability textbook that defines conditional probability P(A|B) when B is a random event chosen from some probability distribution possibly different from the given one. I think that by saving our intuition on this problem you run the risk of horribly confusing students struggling with conditional probability (unfortunately). But I entirely agree with expanding the notion in this way in a second course on probability and showing the students how various paradoxes can be avoided.
Trouble is, it makes actual computations of conditional probability impossible without a great deal of extra information. Reply
15. Bob Walterssays: 10/05/2010 at 20:13 Referring to Peter Cameron’s comment of 11:50, 10 May 2010:
There is a sleight of hand in the statement “If you start bringing in other factors like the algorithm used to generate a statement then all you are doing is changing the measure”. The algorithm may be needed to determine the measure space; different algorithms determine different measure spaces. Reply
16.
Peter Cameronsays: 11/05/2010 at 09:19
Let me try another example, a standard one in elementary probability. There is a screening test for a rare disease (prevalence 1 in 50 of the population). The test has a 5% probability of false negatives and 10% of false positives. I have just taken the test, and the result was positive. What is the probability that I have the disease? A routine calculation using the Theorem of Total Probability and Bayes’ Theorem gives the result of 16.4% (which still comes as a surprise to many people). However, should we take the following into consideration:
Maybe the test was part of a screening of the whole population; or maybe I took it because I had been sent by my GP, or because I was worried about certain symptoms.
In the third case, the symptoms may be indicative of the disease (more or less), or completely irrelevant.
I am a hypochondriac, so I am going to assume the worst.
Given this, it seems difficult or impossible to put a number on the probability. The first point has something to do with the algorithm or protocol used, but it is difficult to say that the others are!
Reply
Ted Jones says: 12/05/2010 at 04:56 Peter, your point of view is still mysterious to me, but I’d like to change the subject back to the Monty Hall problem. Before I do I’d like to say I agree with you that this whole area of discussion is something with a lot of potential to confuse students, and I wouldn’t bring it up in an introductory course. It’s really only natural to think this way when trying to resolve paradoxes. I’d avoid them too, with beginners.
About Monty, you say “clearly” you should switch. I agree that you should, of course, but the history of the problem is that a large number of readers wrote in to Parade Magazine to “correct” that conclusion after it was presented there. Many of them claimed to have PhDs–some in mathematics! In fact, my own initial knee-jerk reaction before giving it any thought was that it probably didn’t matter if you switch or not.
Why was that? Here’s my explanation. Consider a variant of the problem that I’ll call “ignorant Monty”. Ignorant Monty has no idea where the car is. He picks one of the two doors not chosen by the player at random, with probability 1/2 for each, and has it opened. He might reveal the car; then the player loses immediately. But if there are goats behind ignorant Monty’s door, the game continues. Does the player still benefit from switching?
For ignorant Monty, the answer given by the irate Parade readers is correct: it doesn’t matter whether the player switches or not. I encourage anyone who doubts this to perform a computer simulation and see how it works out (I did, out of respect for this problem’s history).
My theory about the reader outrage is that the ignorant Monty represents a kind of situation that scientists are likely to think about much more often than the average person. A PhD scientist might well “know” the answer–and after all, it’s “obvious” that Monty’s algorithm makes no difference. In fact, I’d go farther: it’s beyond obvious. I bet very few of those providing purported corrections had even thought about it explicitly, just as few of them thought about the color of Monty’s eyes and reached an explicit conclusion that it didn’t matter.
I tried to discuss ignorant Monty with several acquaintances back at the time of the Parade fiasco. These were people who were not mathematically ignorant, but were sure switching didn’t matter, and I tried to say, see, here’s the assumption you’re making that causes you to think that. There was a complete failure of communication in every case (I hadn’t thought of encouraging a simulation at the time). I guess I’m still trying to share the enlightenment all these years later. Reply
Peter Cameronsays: 12/05/2010 at 10:10 Ted, thanks for this and for all your comments! I started this post by saying that I don’t understand what probability is – I think the debate has conclusively proved that!
Let me try to add a very small clarification. I think you have the right approach: in thinking about conditional probability P(A|B), we need B to be more general than a fixed event, and even more general than an algorithm for choosing an event: I think a probability distribution on the space of events might be the way to go. In real life this would reflect your beliefs about how the condition B was come up with.
In the Monty Hall problem, as I said, I assumed that the statement “Monty knows where the car is and he is giving nothing away beyond opening one door containing a goat” was part of the specification. This is not an algorithm, but it needs to be justified by giving an algorithm. I considered “Ignorant Monty” to be a different problem. As so very often, the difficulty lies in the specification of the problem.
Sorry this is brief; I have an office hour about to begin, and in exam time it is likely to be well patronised… Reply
Ted Jones says: 12/05/2010 at 17:44 Thank you, too, Peter. You’re exactly right that Monty and ignorant Monty are different problems. My point is only that I think many of those who offered the incorrect “corrections” failed to understand that, and wrote in with the right answer to the wrong problem.
I’m not taking the approach you suggest of conditioning on anything other than a fixed event. Rather I (we) say that the two different problems correspond to two different sample spaces, where “sample space” is understood to include the probability measure and not just the underlying set.
For Monty vs ignorant Monty, we can assume without loss of generality that the player chooses door 1. Let C and M be the door with the car and the door Monty chooses. Consider the event E that {M=2 & C \ne 2}, and the conditional probability P(C=1|E).
For the sample space let’s use the set of pairs (c,m) of possible values that (C,M) may take on. For simplicity, I’ll omit pairs with m=1, which would have probability zero with either version of Monty. Then the points in the sample space are:
(1,2), (1,3), (2,2), (2,3), (3,2), (3,3)
C=1 is the event {(1,2),(1,3)}, and E is the event {(1,2),(3,2)}.
For regular Monty, assuming for simplicity that he tosses a coin when he can choose either door, the probabilities of the sample points are: 1/6, 1/6, 0, 1/3, 0, 1/3. For the different problem of ignorant Monty, we have instead: 1/6, 1/6, 1/6, 1/6, 1/6, 1/6.
Now the conditional probabilities are cut and dried. Regular Monty:
P(C=1|E) = P(C=1 & E)/P(E) = P{(1,2)}/P{(1,2),(3,2)} = (1/6)/(1/6+1/3) = 1/3,
and the player should switch.
Ignorant Monty:
P(C=1|E)=P{(1,2)}/P{(1,2),(3,2)}=(1/6)/(1/6+1/6)=1/2,
and it doesn’t matter.
Is this explanation any help at all?
Ted Jones says: 12/05/2010 at 17:51 I see I managed to get regular Monty’s probabilities in the wrong order; of course I should have said 1/6, 1/6, 0, 1/3, 1/3, 0. I blame your other posts about the symmetric groups. Yes, that was it…
Peter Cameronsays: 30/05/2010 at 15:09
This week’s New Scientist has an article by Alex Bellos about the recent Gathering for Gardner. He begins the article with the boy born on Tuesday. On trying to explain it to my son over breakfast, I found that I understood it better myself (psychologically rather than mathematically).
Suppose that someone says to you, “I have two children; the elder is a boy”. On the basis of that information, the probability that both children are boys is 1/2. Now we can add a couple of things:
If the statement was “I have two children; the elder is a boy born on Tuesday”, the probability is still 1/2; the extra information really is irrelevant.
The same would be true if one child were identified in any other way; for example, “I have two children; the taller one is a boy.”
In particular, if you were told “I have two children; the one born on Tuesday is a boy”, then the probability of two boys is 1/2, since the statement is phrased in such a way as to identify one of the children.
Now the statement “I have two children; one is a boy born on Tuesday” doesn’t absolutely identify one child, but is very likely to do so, since the probability that both are boys born on Tuesday is small. So we’d expect that the probability that both are boys would be closer to 1/2 than if the information about Tuesday were omitted, as indeed it is. The more unlikely it is that both children satisfy the condition, the nearer the description comes to giving a definite identification of one child, and the closer the probability comes to 1/2.
The purpose of the apparently irrelevant information about Tuesday is to give an identification of one of the children which works with fairly high probability (26/27 in this case). If this happens, then the probability that both children are boys is 6/13; if it fails, the probability is 1. Now a simple calculation gives the answer 13/27.
Incidentally, I have to take Alex gently to task. He says in the article, “If you have two children, and one is a boy, the probability of having two boys is significantly different if you supply the extra information that the boy was born on Tuesday.” But saying “the boy” implies that the child has been identified, and the paradox would evaporate!
Reply
David Bedford says: 31/05/2010 at 21:57 Hi Peter,
I’ve recently come to the same conclusion as you about why this is intuitive after all. Firstly I sidestep much of the discussion about how and why the information is revealed by phoning a complete stranger and asking two questions. The first is Do you have two children? – Answer yes. The second is “Is one of your children a boy with property P?” If the answer is yes then the probability that both children are boys varies between 1/3 and 1/2 depending on how likey it is that there could be two sons both with property P. If P is being the eldest then it is 1/2. If P is satisfied by all boys then it is 1/3. Born on a Tuesday, Born on Feb 28th, Born on Feb 29th (!) all bring the probability nearer 1/2 because it is more and more likely that the respondent is talking about a particular child. I’m not even sure it matters whether the answer is correct as long as the respondent believes it is correct and hence was thinking about a particular child when answering. Reply
20. seancarmody says: 01/06/2010 at 03:17 I’m not sure whether my earlier comment came through (there could have been a cookie and/or Javascript problem). In case it did not, here it is again (this time with errors corrected!).
Peter, in one of your comments you write:
If I knew that a different algorithm was being used, that would (by an application of Bayes’ theorem) give me a different measure, and the calculation would give a different answer. But if you want a different answer to this question, you had better make this algorithm public, so that I can use this knowledge to adjust the probabilities.
The implication here is that some kind of minimal interpretation of the information contained in someone saying “I have two children, and (at least) one is a boy” is precisely the information in the formal mathematical event represented by the subset {BB, BG, GB} of the sample space {GG, BB, BG, GB}. I’m not sure that’s necessarily the best interpretation.
To explain why, I’ll stick with this simpler problem rather than the Tuesday version, but the same argument applies there. I will have to assume that we are working with a bigger state space, but that doesn’t mean I have to start worrying about whether the person is lying or the probability that they would have said something in Klingon instead. I won’t say anything yet about this bigger space other than the fact that it includes events BB, BG, etc, each of which has probability 1/4.
I’ll denote by M the “mathematically formal” event represented by {BB, BG, GB}–strictly speaking this means the union of the events BB, BG, etc in the bigger space–and H the event that a “human” tells you “I have two children, and (at least) one is a boy”. The probabilities we are interested in are P(BB | M) and P(BB | H).
In both cases, we can appeal to Bayes theorem and so have
P(BB | M) = P(BB) P(M | BB) / P(M)
and
P(BB | H) = P(BB) P(H | BB) / P(H)
Now P(M) = 3/4 and P(M | BB) = 1 and so P(BB | M) = 1/3, which is the classical answer. No surprise there. What about the case for P(BB | H)? To have it line up with the M case, we have to accept that P(H) = 3/4 is a minimal, neutral probability to assign to the event that someone tells you “I have two children, and (at least) one is a boy”. Even without worrying about complex or far-fetched algorithms, that doesn’t seem quite right.
I would argue that a more reasonable probability would be 1/2. One algorithm that would give this outcome is that a person with BB would say “I have two children, and (at least) one is a boy”, a person with GG would say “I have two children, and (at least) one is a girl” and a person with BG or GB would say “I have two children, and (at least) one is a boy” or “I have two children, and (at least) one is a boy” at random, each with probability 1/2. In this case P(H) = 1/2, which is evident if we calcalate P(H) = P(H | BB)P(BB) + P(H | GG)P(GG)+…
Also, since P(H | BB) = 1, we get P(BB | H).
An important point to emphasise here is that I don’t really have to assume I know precisely what algorithm is being used. Rather, I need to be able to come up with the probabilities P(H) and P(H | BB). To me, values of 1/2 and 1 seem like very reasonable minimal interpretations of this information.
What does all this mean for the Tuesday problem? Here we can again distinguish the mathematical event M and the human utterance event H for the statement “I have two children, and at least one of them is a boy born on Tuesday”. P(BB | M) = 13/27 for the reasons discussed in the post, but in the case of the human utterance, I think you can quite reasonably assign P(H | BB) = 1/7 (all days are equally likely to have been spoken) and P(H) = 1/14 (the day and the gender are independent), which means that P(BB | H) = 1/2! Reply
21. Pingback: Probability Paradoxes | Stubborn Mule
seancarmody says: 11/06/2010 at 10:32 Peter,
I have expanded on my argument as to why the probability should be 1/2 if the information is volunteered by the father and 13/27 if you ask the question ‘do you have at least one boy born on Tuesday’ and receive a ‘yes’. I think the two scenarios reveal different information. Any thoughts you (or others) have would be greatly appreciated! Reply
23. Sam says: 19/06/2010 at 05:44 The answer is 1/3.
The trick is based on the indeterminate nature within the BB case – which boy did he refer to? You must not double count the distinct cases.
There are only 7 distinct BB cases, not 13 (or 14). There are 7 BG cases and 7 GB cases. So the correct denominator is 21.
7/21 = 1/3.
Common sense should tell you that the Tuesday fact has no bearing on the conditional probability in question.
To look at it another way, each of the following cases is possible for the second child:
boy, girl, girl
And each of these cases is equally likely at 1/3.
In the BG and GB case, you know for a fact that the B is the one mentioned, the one who happens to be born on Tuesday. Well it’s exactly the same in the BB case – one of the B’s is the one mentioned, the other is a possible boy you know nothing about, except that the probability of him existing is 1/3. Reply
24. Sean Carmodysays: 22/06/2010 at 02:03 Sam: you say
Common sense should tell you that the Tuesday fact has no bearing on the conditional probability in question.
If common sense was all that was required, I am sure that this puzzle would not be so controversial!
You are simply asserting that the possibilities for the ‘other’ child are boy, girl, girl, each with probability 1/3 and therefore the probability of two boys is 1/3. Other than an appeal to common sense, how would you differentiate this argument from someone who says that there are only two possible genders for the ‘other’ child, boy or girl, each with probability 1/2 and therefore the probability of two boys is 1/2? While I don’t agree with that particular argument, it is no worse than yours. Reply
25. ales says: 25/10/2010 at 08:55 hello, and what about this example. Anyone knows solution?
Assume you have an algorithm which errs with a probability of at
most 1/4 and that you run the algorithm k times and output the majority output.
Derive a bound on the error probability as a function of k. Do a precise calculation
for k = 2 and k = 3, and give a bound for large k. Finally, determine k such that the
error probability is less than a given “epsilon”
Thanks, Ales Reply
26. Pingback: 2010 in review « Peter Cameron's Blog
margazhi mama says: 01/01/2012 at 08:42 “boy born on Tuesday problem”: see Kai Lai Chung Elementary Probability Theory with Stochastic Processes Springer International Student Edn. Ch. 5.1 Example 5 pp. 115-6 (I have the Indian reprint and my laptop can’t handle the Greek and other symbols, so I give only as much of the text as I can).
Consider all families with 2 children and assume that boys and girls are equally likely. Thus the sample space may be denoted schematically by 4 points:
{(bb), (bg), (gb), (gg)}
The order in each pair is the order of birth, and the 4 points have probablility ¼ each. If a family is chosen at random, and found to have a boy in it, what is the probability that it is of the type (bb)?
Let us put
A = {w / there is a boy in w}
B= (w / there are 2 boys in w}
Then B is contained in A so AB = B, thus
P (B/A) = P (B) divided by P (A) = ¼ divided by ¾ = 1/3
But now ask a similar sounding but really different question. If a child is chosen from these families and is found to be a boy, what is the probability that the other child in his family is also a boy? This time the appropriate representation of the sample space would be
{gg, gb, bg, bb} (second letter of each is a subscript)
where the sample points are not families, but the children of these families, and gg = a girl who has a sister, etc. Now we have:
C = (w / w is a boy)
D = {w / w has a brother}
so that
CD = { w / w = bb}
Therefore
P (D/C) = P (CD) divided by P (C) = ¼ divided by ½ = 1/2 Reply
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188118 | https://av1611.com/kjbp/kjv-dictionary/discontent.html | DISCONTENT - Definition from the KJV Dictionary
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D
discontent
« discomfiture
discontinuance »
KJV Dictionary Definition: discontent
discontent
DISCONTENT, n. dis and content. Want of content; uneasiness or inquietude of mind; dissatisfaction at any present state of things.
DISCONTENT, a. Uneasy; dissatisfied.
DISCONTENT, v.t. To make uneasy at the present state; to dissatisfy.
discontented
DISCONTENTED, pp. or a. Uneasy in mind; dissatisfied; unquiet; as, discontented citizens make bad subjects.
discontenting
DISCONTENTING, a. Giving uneasiness.
Definitions from Webster's American Dictionary of the English Language, 1828.
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188119 | https://community.latenode.com/t/google-sheets-calculation-giving-wrong-average-speed/30731 | Google Sheets calculation giving wrong average speed
I’m having trouble with my Google Sheets calculations and can’t figure out what’s going wrong.
I’m tracking cycling data for a 90km ride that I broke down into 5km segments. For each segment, I enter the time it took to complete that 5km portion. Then I have formulas that calculate the speed in km/h for each segment.
The problem comes when I try to get the overall average speed. I’m using the AVERAGE function on all the individual segment speeds, but the result doesn’t match what it should be. For example, one of my rides shows an average speed of 31 km/h when it should actually be around 28 km/h.
I also calculate the total time using SUM and that seems correct. I thought maybe it was a rounding issue so I tried changing the number format to scientific notation, but that didn’t help.
What could be causing this discrepancy in my average speed calculation? Is there something wrong with how I’m averaging the individual segment speeds?
You’re using arithmetic mean instead of harmonic mean - that’s why your numbers are off. When you rode fast on one segment and slow on another, simple averaging treats them equally. But you actually spent way more time at the slower speed, which drags down your real average. Your method works great for test scores where everything’s weighted the same, but not for speed calculations. Just take your 90km and divide by total time in hours - you’ll get around 28 km/h, which is what actually happened.
You’re making a basic math mistake with how averages work for speed calculations. When you average individual segment speeds, you’re weighing each segment equally - doesn’t matter if one took way longer than another. This skews things toward segments where you were going slower or faster. For real average speed, just use the basic formula: total distance ÷ total time. You’ve already got the total time with SUM, so divide your 90km by that number. That’s your actual average speed. That’s why you’re seeing 31 km/h instead of 28 km/h - averaging the segments inflates the result when your times vary a lot. Speed is just total distance over total time, nothing fancy.
classic misteak! u can’t just avg the speeds - each segment takes diff time. u need total dist divided by total time, not avg of indiv speeds. that’s y ur getting 31 instead of 28.
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188120 | https://www.abta.org/wp-content/uploads/2018/03/understanding-the-diagnosis.pdf | American Brain Tumor Association Webinar Understanding the Diagnosis and Treatment of Acoustic Neuroma >> Welcome everyone and thanks for joining us today. Welcome to the American Brain Tumor Association's webinar series. Thank you so much for participating in today's free educational webinar. Today's webinar is on: "Understanding the Diagnosis and Treatment of Acoustic Neuroma.” It will be presented by Elizabeth B. Claus, MD, PhD. Please note that all lines during our webinar today are muted. If you have a question you would like to ask, type and submit it using the question box in the control panel on the right-hand side of your screen. Dr. Claus will answer questions at the end of her presentation. Tomorrow you will receive an email asking you to evaluate this webinar. It is a very brief survey. Please take a few minutes to share your comments. Your feedback is important to us as we plan for future webinars. Today's webinar is also being recorded. The recording will post to the ABTA website shortly. Registered participants will receive the webinar link in a follow-up email message once the webinar is available. Let's pause for a moment so we can begin our webinar recording here. >> The American Brain Tumor Association is pleased to welcome you back to our webinar series. Our webinar today will discuss: "Understanding the Diagnosis and Treatment of Acoustic Neuroma." My name is Andrea Garces, Program Manager here at the American Brain Tumor Association. I am delighted to introduce you to our speaker today: Elizabeth B. Claus, MD, PhD. Dr. Claus is a professor and director of Medical Research at the Yale University School of Public Health, as well as attending neurosurgeon and director of Stereotactic Radiosurgery within the Department of Neurosurgery at Brigham and Women's Hospital in Boston. She is a member of the board of advisors for the Acoustic Neuroma Association (ANA), as well as the Central Brain Tumor Registry of the United States (CBTRUS). Dr. Claus' work is focused in cancer and genetic epidemiology with an emphasis on the development of risk models for breast and brain tumors. In addition to her research activities, Dr. Claus trained as a neurosurgeon at Yale-New Haven Hospital and completed a fellowship in neurosurgical oncology at Brigham and Women's hospital. Her clinical focus is on the treatment of meningioma, glioma, acoustic neuroma and brain metastases. In partnership with national patient brain tumor organizations, including the American Brain Tumor Association (ABTA) and the ANA, Dr. Claus is working to develop cost and time-efficient, web-based recruitment strategies to be used in the study of brain tumors. >> Thank you very much and good afternoon to everyone. Thank you for joining us and thank you to the American Brain Tumor Association for supporting this webinar. I list here my disclosures and support and then we will get into the main part of the talk. I thought I would begin with giving a definition of Acoustic Neuroma. I have a bit of a schematic here. Essentially, acoustic neuroma is a non-cancerous or non-malignant, generally slow-growing tumor of the nerve that connects the Ear to the brain and you can see a schematic of that here. It's also known as the eighth cranial nerve and it has two components, the auditory and the vestibular component. It turns out that the vestibular portion is the most commonly affected, and that's why many times you’ll hear the term vestibular schwannoma used in addition to the term acoustic neuroma. The cells that line the nerve are called schwannoma cells, or schwan cells, and when those overgrow themselves, you come up with an acoustic neuroma or schwannoma. Here is an example of what the nerve looks like unaffected, and here is an example of a good-sized lesion when the tumor has formed. You can see from the location of this – here is the top of the head and here are the ears and the neck - you can have a fairly large legion in a fairly tight space and that's where symptoms come about. >> I thought I would present some information from the United States in terms of how many people are diagnosed with this tumor and I take data from what is called the Central Brain Tumor Registry of the United States or CBTRUS. Some of the information on acoustic neuromas and other non-malignant tumors have only begun to be collected since 2004. So prior to that there was no organized registration of non-malignant tumors and that began through a federal mandate in 2004. I think some of the numbers are still being collected entirely, but we’ve got pretty good data at this point in time. If you look over a five-year time period from 2007 to 2011, these are population-based data from the United States. Malignant tumors are listed in the color red and non-malignant tumors, including acoustic neuroma, are listed in blue. When you go to some of these national registries, the acoustic neuromas come under the nerve sheath tumor scenario. About 8% of all brain tumors in the United States are these non-malignant nerve sheath tumors. The majority of them, which is a subset here, are Acoustic Neuromas or vestibular schwannomas. If you do the math it turns out that about 7.5% of brain tumors in the United States are listed or defined as Acoustic Neuroma. We think that includes about 5,000 individuals, at least within the United States every year, being newly diagnosed. >> What are the risk factors for Acoustic Neuroma? There have not been a large number of studies that investigate this question, but we know that genetics plays a role. High-dose ionizing radiation certainly. Immune factors, we're learning more about and that seems to also be associated. There has long been a concern regarding cell phones. In the end it is not completely clear what role, if any, cell phones might play but I thought I would show you the data for that. >> In terms of genetics, the primary gene that has been associated with Acoustic Neuroma is NF2 or neurofibromatosis 2. It's relatively rare, but it is associated with a high risk of disease and in fact it's associated with a high risk of bilateral Acoustic Neuroma meaning in each ear. There are some other tumors associated with NF2, like meningioma. Although there is a high risk of Acoustic Neuroma in individuals that have NF2, the majority of people diagnosed with Acoustic Neuroma don't actually have a diagnosis of NF2. >> In terms of ionizing radiation, we have data from a number of different types of studies including following young children that received radiation for treatment when they were at an early age, from atomic bomb data, and also from studies of radiation treatment in countries like Israel. Essentially, we have learned that when children are exposed at high doses that the brain is a high risk site and we see over a lifetime a twofold to threefold increased risk with high-dose ionizing radiation. >> As I mentioned, there's been an increased interest in things regarding the immune system and we have found now that in addition to glioma and meningioma, we see an inverse relationship between allergy and asthma and risk of acoustic neuroma. That's fairly consistent across most of the brain tumors. >> As I mentioned, data on cell phones is a bit controversial and becomes difficult to measure because most people that have acoustic neuroma a most people that don't have Acoustic Neuroma use a cell phone. It's very difficult to find individuals that are not “exposed” to a cell phone. There was a fairly well-attended meeting that was at the World Health Organization in May 2011 and essentially this is the conclusion: long-term use of a cell phone might lead to two different types of tumors, glioma and Acoustic Neuroma, but they cannot conclusively confirm or deny this. The thought is they would continue to watch this. There are some ongoing studies in children regarding the use of cell phones in Acoustic Neuroma so we might be able to get additional information from those studies. >> I thought you might be interested, especially for patients, in seeing what are some general characteristics of Acoustic Neuroma patients within the United States. The National Cancer Institute has a data collection program called the Surveillance Epidemiology and End Results, or SEER program and as I mentioned, for noncancerous tumors like Acoustic Neuroma, data collection only began in 2004. But we do have now about 10,000 patients registered, so there are some characteristics that are popping up. There is a slightly greater number of females versus males that are diagnosed with Acoustic Neuroma. The majority of cases report their race as white. About 11% report their ethnicity as of Hispanic origin. If you look at age, the mean age is about 55 years old. You see there is a little bit of a range and there is not a large amount of patients diagnosed in the pediatric population and a good proportion of those are likely individuals diagnosed with NF2. It's pretty much a diagnosis of middle age or beyond. In terms of treatment about half of the individuals listed underwent surgery as their first course of treatment. About one quarter received radiotherapy and about another quarter received observation. If you look at the data, the treatment selected does vary by the size and location of the acoustic neuroma. Across these individuals, the average size at time of presentation was about 1.7 cm. The good news is, although certainly acoustic neuroma is associated with a variety of difficulties, the majority of patients in terms of survival do quite well. >> How are patients diagnosed? I would say the number one diagnosis is hearing loss that is unilateral, or one-sided hearing loss, and it can present in a variety of different ways. It can be sudden and complete. It can be a slow decrease over time. I have some patients who tend to have episodes of hearing loss or ear fullness - for example, they might have a period of two or three weeks where they have hearing loss and then it gets a bit better and then maybe a year or two later a similar episode occurs. Tinnitus or ringing in the ears is also common. Dizziness or balance problems – we see that a little bit more when the tumor is larger. The same situation for facial numbness or weakness. We also see patients who don't have any symptoms, and I would say that most commonly we run across these individuals after they have had some sort of an accident, like a car accident, and they receive a head CT. We also see patients who are receiving yearly MRIs being screened for other diseases, like multiple sclerosis, and they have an incidental Acoustic Neuroma identified on that. >> Just to give you a little bit of a picture, the nerve travels through a bony canal called the internal auditory canal. I wanted to point out there are a number of nerve components in that canal. The vestibular component and the cochlear component are part of the acoustic nerve, and nerve number seven is the facial nerve. What you can see here is that when any component of the tumor over grows itself, it comes up quite closely on these other nerves and that's why you have problems not only with hearing but also with balance, with facial nerve difficulties, because all those components are so close to one another physically. >> What is the workup for acoustic neuroma? Primarily getting a good MR of the brain. It's important to have thin cuts in order to be able to distinguish the anatomy. You need to get IV contrast with good imaging of that ear canal or the internal auditory canal. There are some individuals who for a variety of reasons, like they have a pacemaker or some other diagnosis that makes them unable to get a brain MR, and they can receive a head CT, and then also to get a hearing exam. That is important to get, to get a feel for not only what does the image look like, but how is the patient affected. Sometimes you can have a small lesion on MR and it might not seem impressive, but you can have a significant hearing loss. So it’s important to get both pieces of information. Here is an example of a right ear and a left ear, across the different frequencies, and you can see that one ear is hearing less well and this is actually in a patient who has acoustic neuroma on one side. >> What is the treatment for acoustic neuroma? There is not one right answer for any given scenario. It's very much an individualized treatment plan. In some instances, we are able to observe the patient and follow them over time, also known as watch and wait. In some instances, particularly when there are larger tumors, surgery is probably the best option. In other instances, radiation therapy is a good option. In some cases, patients will require a combination of therapies, so they might have surgery and because of the location or size of the acoustic neuroma, it may be too dangerous to remove all of it and therefore some residual might be left after surgery, and therefore radiation therapy might be a good additional treatment. In general there is no standard chemotherapy given for acoustic neuroma. There are some special instances in smaller clinical trials for individuals that have refractory acoustic neuroma and they are particularly focused on patients that have NF2. A medication called Avastin which is used for the treatment of glioma and other malignant brain tumors has also been used in certain circumstances. Generally, for standard acoustic neuroma, chemotherapy is not typically a treatment option. >> How do patients select a treatment? It depends on a variety of different characteristics and obviously tumor size and location is very important. The larger the tumor the more likely the surgical intervention might be needed. The tumor growth rate, we know in general these tumors grow quite slowly over time. Every once in a while we see a patient that has a faster growth rate and that is someone we might be more likely to intervene upon. The hearing status is not only in the affected ear but also the contralateral ear and what you select on the overall hearing status and how that person might be affected by whatever treatment is selected. Also whether the patient has symptoms or not. As I mentioned, some people have no symptoms at all. Other people have quite serious symptoms and that would certainly direct what treatment would be selected. The patient's age and medical condition, so older patients or patients with a number of medical conditions might not be able to undergo surgery or have general anesthesia. We might choose something different for that type of individual. Individuals with NF2 have a special set of circumstances and are looked at quite differently. Frequently we have specialized clinics to follow and treat those patients. It's also a function of what does the patient want and what they are interested in pursuing and what they are comfortable with and what is the experience and preference of the surgeon. What sort of approaches are they comfortable with? What experience do they have with any given approach? We will talk a little bit later on about questions to ask your physician, but that's a very important thing to consider. >> Talking about observation, people from site to site do things a little bit differently. What I generally do, if it's a new patient and it comes under these categories of small tumor, or asymptomatic or perhaps elderly or with other medical conditions and are considering observation, so we usually get an initial baseline MRI of the brain with a hearing test. In general because we know in most instances the tumor is growing quite slowly that it's reasonable and safe for the patient to get a first recheck at six months where we get another MR and hearing test and if things seem fairly stable typically we follow people on a yearly basis. If we are concerned about any change, either clinically in terms of hearing or picture, we might intervene or continue to follow up at six-month intervals. We tell patients too if they feel any concern or change to let us know to not wait for the prescribed time if they are concerned about anything to re-contact us. >> If surgery turns out to be the option that is right for a given patient, there are variations on any theme, but there are three general surgical approaches called retrosigmoid, middle fossa, and translabyrinthine. The choice depends on the factors that we were talking about. The location, the size of the Acoustic Neuroma and each individual patient's anatomy. People can vary in terms of where a particular blood vessel or nerve or anatomical components are located and all that needs to be taken into account. You need to think about how much pre-operative hearing loss exists in the affected ear as well as the contralateral ear. >> Retrosigmoid is probably the procedure that is most commonly used in neurosurgery. That's across the board. It's something that most neurosurgeons are fairly familiar with. It's used not only for Acoustic Neuroma but a number of other neurosurgical conditions. All of these are under general anesthesia. In this instance you enter behind the ear. It is a surgery or operation that can be used when you are attempting to save hearing as well as in instances where you might realize that the probability of sacrificing hearing is relatively high. The one caveat is that small Acoustic Neuromas that are in the far lateral portion of the hearing canal are not well seen, so it is a less good option in that particular instance. >> Middle fossa is for individuals that have tumors within the internal auditory canal and have hearing so you can see here this would be the outside the ear. What happens here is you approach from above the ear and you lift up a portion of the brain that's called the temporal lobe. Finally, the Tranlabyrinthine is the less common surgical approach, but I would say that many surgeons who have extensive experience in Acoustic Neuroma surgery use this quite frequently. In this approach, hearing sacrifice is complete, but it is a good approach when one is trying to preserve the function of the facial nerve. Also, as some of these cochlear nerve implants are more frequently being used, an effort can be made to spare the cochlear division of the nerve, in the event that a patient might later on down the road be a candidate for that sort of implant. That's an important thing to talk to your surgeon about as well as whether an effort can be made to do that. >> Surgery is done under general anesthesia and as we talked about, a patient has to be healthy enough to undergo general anesthesia. In the operating room, surgeons use a microscope so they are able to remove the tumor while protecting other structures such as the facial nerve. In some instances, and more and more over time, surgeons – that’s what the picture shows here, both the use of a microscope and also a special camera called an endoscope can be used at the time of surgery, which is helpful in seeing around corners and places that might be difficult to see with only the microscope. That is another nice advance that is coming into play. Monitoring of the cranial nerves is done during the surgery so that the surgeon knows if they are advancing towards a nerve. Sometimes, the tumor can hide cranial nerves especially if it's a larger tumor. >> The goals, benefits, and risk vary by type of surgery and it's an important conversation to have with your surgeon. Find out what their experiences with any given approach. The greatest risk is hearing loss and that varies by type of approach. Also facial nerve damage, headache, CSF leak, or the fluid that’s within the head can be seen in hydrocephalus and either of those can end up needing to be treated with a second surgery or placement of what is called a shunt to reroute some of the fluid. Obviously as with any craniotomy or general anesthesia there is a risk, although low, of other morbidity and mortality. >> This is a paper from a 2012 issue of neurosurgical focus which is a review issue on Acoustic Neuroma. It's giving some of the estimated risks by type of surgery. These authors performed an overall analysis, pulling together a lot of papers from literature and looking by the three types so retrosigmoid, middle fossa, and trans-lab , what the sorts of risks were for hearing loss. You can see there is a fairly good rate of hearing loss and it increases with size of the tumor. Facial nerve dysfunction again increases with size of the tumor, relatively low for smaller tumors but increasing when you get up to a large which is greater than 3 cm. These are rough estimates of CSF leak, of headache and also of having residual tumor left behind that might need additional treatment. >> Radiation therapy is also an important type of treatment for Acoustic Neuroma. Radiation therapy generally goes under the term stereotactic radiosurgery when we talk about Acoustic Neuroma. >> The name radiosurgery makes patients think about intraoperative surgery using a knife, but there is no use of knives in this. It's all using radiation and computer targeting. It allows us to direct focused radiation so not whole brain radiation but focused to a specific target and lets us go to that target while protecting other important structures in the brain. It's fairly widely used throughout neurosurgery. We use it for a lot of different types of brain tumors. The example I have is for metastatic lesions, but it's used for pretty much any type of tumor that comes up in neurosurgery. >> The process varies a little bit by what type of equipment or software a particular facility uses. Some imaging of the head through a head CT or an MRI or frequently both is obtained from the patient. What a team will do and that includes a neurosurgeon, a radiation oncologist, a radiation physicist, is basically we work together and using computer programming and here is an example, outlined the lesion and take care to mark off important areas such as the brainstem and other important optic nerves in the brain so those are not targeted. Then we deliver radiation to the Acoustic Neuroma. There is a fair amount of experience with this now. It's been done for a number of years and is generally quite a safe procedure. >> I list here a number of different types of software and equipment. There is gamma radiation and probably people have frequently heard about gamma knife surgery. Also photon radiation, linear accelerators, another term is LINAC, Cyberknife, and protons. Although probably the greatest experience is with gamma knife surgery and LINAC, good control of Acoustic Neuroma has been reported for all of the methods listed. >> There are a number of different permutations. There are frame-based methods where a frame is placed and then imaging is done. The patient would then sit and wait while the team prepares the plan, as we just talked about. There is also frameless technology and this is one example here. There are some other technologies using fiducials as well as a mouthpiece so there are a lot of different advances and changes coming forward using stereotactic radiosurgery. There are also a number of different ways in terms of how the dosing is done. In some instances a single dose is given and in other instances multiple doses, which is termed a fractionated dosing regime, are given. Both work quite well in terms of tumor control. There is some thought that the fractionated regime can offer a slight advantage in terms of hearing preservation. It's important when you're going to a facility to talk about what sort of technique they might be using, what their experience is, whether a frame would be used, one dose or multiple doses. >> Benefits of stereotactic radiosurgery are obviously you are avoiding open surgery, avoiding anesthesia and avoiding any associated complications that might be associated with surgery such as CSF leak and there is no hospital stay. As for surgery, there is good tumor control and most of the series that are out now suggest there is a 95% or even better control of the tumor at 10 years. >> There are also risks as with anything in medicine. Hearing loss is certainly a risk and the feeling is that over the long term, the hearing loss can be similar to what it would be for open surgery. It may not happen as quickly, but as the effects of radiation take place over time, people can continue to have risk of hearing loss. There is also risk, any time radiation is completed, that the tumor can temporarily swell. This can cause symptoms like hydrocephalus, or vertigo, or facial palsy. It can also sometimes make it difficult to know if the tumor is increasing in size i.e. growing, or whether it's temporarily swelling from the effects of radiation. You have to be careful as you are watching and following up after radiation to not too readily assume that it's growth in tumor, it might just be swelling. The swelling and difficulty in knowing which direction we are going can take over two or three years. >> What are the questions to ask? We touched on this, but these are some questions that are probably good to take along if you are considering some sort of intervention or treatment plan for an Acoustic Neuroma. It's very important to go to a center that has a team that focuses on Acoustic Neuroma. You want to know what is the level of experience of the team you’re working with, what member is doing what, what treatment they would select for you and why they would select it for you. For example, if they would like to observe a patient, how frequently would you be seen? Also, who would be the person doing the follow-up? Would it be a primary care doctor, the neurologist, surgeon, radiation oncologist? It's important to figure that out ahead of time. If surgery is the approach that seems to be the most reasonable, what approach would be taken? When patients undergo this sort of surgery they generally spend at least one night in a nurse surgical intensive care unit - so you would want to know if such a unit is available to the patient after surgery. Also, is there an intensive care team that would be taking care of the patient? If radiation therapy is selected, what type of equipment and software would be used? As we mentioned, is there a frame or frameless-based approach? Would it be one dose or would you come back for multiple doses? And at any center, what sort of outcomes are to be expected? How would any problems that might arise be managed and who would be managing that and what follow up would be necessary? It's also important, as many people travel outside of their home area to visit some of these Acoustic Neuroma specialties, to know when you go home who will be following the patient, who will order the MR or hearing test or any therapy or treatment needed to get the plan together and make sure you know who is doing what. >> There are some great information resources available both through the American Brain Tumor Association, as well as the Acoustic Neuroma Association and I have the websites for both those organizations listed here. >> I wanted to highlight that we have obviously been very interested in studying risk factors for Acoustic Neuroma and so one of the things that we are trying to do is involve patients a bit more in looking at such research. We have a study where we are looking to identify genetic risk factors, both inherited, meaning that would be in the blood or saliva, as well as genetic risk factors in the tumor and to see if we can learn more about these things. If you have any interest in learning more about that, we have our study at the Acoustic Neuroma website and you can go there and find us. >> Other than that I thank you very much for your time and for listening. I hope I might be able to answer any questions that you might have. >> Thank you so much for that wonderful presentation. Everyone, Dr. Claus will now take questions so if you have a question you would like to ask please type and submit it using the question box in the webinar control panel on the right-hand side of your screen. We have a few questions that came in during your presentation. One of them, Dr. Claus, is how do Acoustic Neuromas get reported to CBTRUS? Given there is no mandatory reporting of Acoustic Neuroma, could there be underreporting in the CBTRUS registry? >> Actually, now as of 2004 there is mandatory reporting, but that began at that point in time. CBTRUS gets data both from SEER as well as other population-based registries. It's a central clearinghouse for tumors of the brain. But I think since we're only about 11 years into this, I am sure they are still underreporting. It's getting better and I've looked at the data over each year for the past 11 years and you can see the numbers rising. I think we have work to do but we are getting better than where we were. >> Thank you for that. You also talked a little bit about how it is diagnosed - can an Acoustic Neuroma be missed on an MRI? >> Absolutely. I have a number of patients where it was missed. If they are small and in particular if no contrast enhancement is available, so if it was an MR that might've been for another reason perhaps a stroke or some other diagnoses where contrast might not have been given, it could be very easy to miss it. >> A follow-up to that, what are some common misdiagnoses and how long do patients experience symptoms before they are accurately diagnosed? >> That varies and I don't think there is any set time. I think probably the biggest delay in diagnosis is patients tend to, as they mature, accept some hearing loss and to attribute it to age or perhaps to attending too many rock concerts during their younger days. I think people accept a bit of hearing loss until it becomes more severe or something they bring to physician attention. I don't think there is a set amount of time between symptoms beginning and diagnosis occurring. We are probably getting better at it over time, I would say. >> Thank you. Another question was what can be done if the Acoustic Neuroma is swelling? >> If it is swelling, and I am assuming that would be after radiation therapy, the primary thing we can do is to offer some steroid treatment. It depends on if the patient is symptomatic. Sometimes we can see a bit of swelling but the patient remains asymptomatic so we can watch them carefully. If it's a situation where the patient becomes symptomatic, then the most frequent course of action is to try to use steroids. If it became a serious problem, it could be a situation where they may have to undergo surgical intervention. >> Another question we had was about using Bluetooth ear pieces, instead of cell phones into the ear. Does that reduce the risk of Acoustic Neuroma? >> That's a great question and the short answer is we don't know. There has not been any formal study. Probably so few people have that technology that it would be tough for us to get a sample size that would be large enough for us to answer the question. Unfortunately we don't know the answer to that one. >> Thank you. Another question is after stereotactic radiosurgery, is there a chance that hearing loss or partial hearing loss that was there before the surgery, can be restored after the surgery? >> It certainly is always possible, but I think in general that hearing loss that exists at either time of surgery or time of start of radiation therapy typically does not improve. It tends to either stay the same, or worsen in the majority of cases. I think we have all had a patient or two for whom it became better, but I don't think that is the typical scenario. >> Thank you. Another question was if you knew of any of the early findings that you had so far from the Yale Acoustic Neuroma study? >> So far it's a pilot study and we are trying to gather enough observations to then go forward and get funding for it. We are very excited to report over 1,000 participants. We're hoping we get about 2,000. But we have been extremely pleased with the efforts that both the ANA and ABTA and patients and caregivers in general have of offered to us - it's been an amazing experience. We have about 1,100 so far and trying to get about 2,000 because that would give us a sufficient number of observations to try and start to look at things. >> That is great news. You spoke a little bit about some of the symptoms. One question was if many years after, is it common to experience facial stabbing pain? Is that a symptom you have encountered? >> We have sometimes and we see it a little bit more with larger size tumors. Sometimes people can even have it after treatment. It has been reported both quickly, after either radiosurgery or surgery, and long-term people can have either facial pain or trigeminal neuralgia and facial twitching. I would say the occurrence is low, but certainly has been reported. >> Thank you. And for those who have smaller tumors but they are experiencing symptoms as if it they are a bigger tumor, would you recommend the patient go by the size or more so by the symptoms if the symptoms are more severe? >> It's not an either/or kind of scenario. It depends on what the symptoms were, the age of the patient. I think if it’s a young and healthy patient and their symptomatic, certainly an option of surgery or radiosurgery would be available to them. It would really depend on that specific person's anatomy, why we thought the symptoms were occurring because sometimes it's not, as you say, just the size but the location, what it is close to, whether it is tightly wedged in the canal and whether it is affecting things like the facial nerve or other anatomical components. >> Okay. Thank you. Is there a site where some of the participants and those interested in the community can find data regarding the number of radiosurgeries done for Acoustic Neuroma by different facilities? Is that kind of data available? >> I'm not aware yet, I know that associations like the Neurosurgical Association are working to place that sort of information online. I'm not aware of one, uniform location where that would be listed across sites. I know that most sites that do Acoustic Neuroma surgery, the surgeons that focus on it keep their own statistics and would be able to share that with patients. >> Thank you. For those who have meningioma on the brainstem and have lost hearing, is that something you experience is very similar to Acoustic Neuroma? >> It can be. Every once in a while, we mistake an Acoustic Neuroma for a meningioma and vice versa if it is located near where the hearing nerve is. There is always a small proportion that overlap and when you look at them on imaging it's difficult to tell. Certainly a meningioma that arises near that location can certainly act clinically very similar to an Acoustic Neuroma. The treatments are fairly similar as well. >> Thank you. Can you speak a little bit about some of the neurocognitive complaints? >> Sure. We see some neurocognitive complaints. I have had patients talk about difficulty with multitasking. Also with being able to concentrate. Sometimes with the ringing in the ears, they find it difficult to concentrate. We have had people who, after any sort of treatment, be it radiosurgery or surgery, have some difficulty. I would say I haven't seen it occur in the majority of patients, but there certainly is a subset of patients who do suffer from the symptoms and whether it's treatment related or secondary to the disease itself isn't always clear. It is certainly something that is real. >> Thank you for that. After a gamma knife, if it was done 4 or more years ago, can the Acoustic Neuroma still experience swelling? >> It certainly could. The typical time period that the literature suggests swelling occurs is about 2 to 3 years. The thinking is that by two or three years out, that most of the swelling if it is going to occur has happened. I think it starts to become more concerning if it's four or five years out, whether or not it's growth versus swelling. >> Thank you. Is there an average amount of surgeries a year that most doctors that perform this type of surgery do? One question is about a daughter with a 3 cm tumor and the doctor does about 50 surgeries a year. Is that something that most do? >> It varies site to site. I don't think there is any one average, 50 sounds like a reasonable number. It’s also years of experience too. It's always nice to go to someone that has an extended number of years of experience. True of anything in life. >> Very true. Can you clarify what refractory AN is? This was discussed during the chemotherapy Avastin discussion. >> There is a very small number of patients that, despite either surgery or radiation therapy or even both, that the Acoustic Neuroma tends to recur or continues to grow. We have seen that probably most frequently in patients that have NF2. They have already used surgery, radiation, and are not eligible for any additional treatments of that sort. One thing that has been tried is, as I mentioned, Avastin for those individuals. I will also say I don't think it has occurred much today for Acoustic Neuroma, but we're becoming better and better for all these sorts of tumors at doing genetic profiling and trying to see are there targets within tumors that now with some of these new treatments we might be able to target. >> Thank you. Can you also speak to recovery after surgery? >> Generally, the first evening is spent in the intensive care unit and for some individuals especially if they are older or have other medical difficulties might need more time than that. I think in most sites, the goal is to try to get people up and home fairly quickly, get them out of the hospital. A number of people obviously will need rehabilitation services. They may need some sort of support if the facial nerve has been affected and may need additional procedures by a plastic surgeon or other surgeons, such as a weight that might lower their [indiscernible] if the facial nerve was affected. If they have things like vertigo, they might need training for that, or physical therapy. Sometimes if they have complications, they might end up needing a shunt. I would say the majority of patients are able to go home three or four days after surgery to start their rehab. >> Thank you. Do you also know if there is a connection between facial pain before surgery and the risk of facial paralysis afterwards? >> It has been suggested and I think the numbers are fairly small. It would depend a little bit on the reason for the facial pain. Whether it is occurring because the tumor is compressing the nerve or it might even be large and compressing, for example, another group called the trigeminal nerve. It depends what the facial nerve is coming from in terms of whether it would affect the actual facial nerve. You can have facial or pain in the head or face associated with a number of different nerves – it doesn't have to be just the facial nerve. >> Thank you Dr. Claus. There is a question on how would you determine if symptoms are from the gamma knife or from Acoustic Neuroma? >> That's probably pretty hard to do. It would depend upon the specific circumstances, but in general that'd be probably pretty difficult to do. >> Sure. For those who have had surgery to remove the Acoustic Neuroma, one challenge is hearing aids can help them hear one side louder, but then not the other side. Is there any suggestions for how hearing aids can help after the removal of the Acoustic Neuroma? >> There are and actually there was recently a month or so ago a very nice webinar available to the public on the Acoustic Neuroma website. A hearing specialist at the Mayo Clinic presented all of the different types of hearing aids and what the purpose was in each instance depending on the kind of hearing loss the patient had. I would recommend if people have time to look at that. It's a very nice, straightforward, easy to understand presentation on hearing aids that would be helpful to people. >> Thank you so much. I think that is all the time we have for questions, but Dr. Claus did you want to add anything or have any last comments? >> No, I just want to thank everyone for joining us and if we can be of help to anybody please get in touch with us. >> Thank you so much and thank you so much for this great presentation and for joining us and for your time. For more information on brain tumors and to help patients and caregivers process the diagnosis, understand a new and difficult vocabulary and access resources to help make informed decisions, you can always feel free to call the ABTA Care Line at 800-886-2282. Let's pause for a moment to conclude our webinar recording. >> We invite you to continue to check back at our website www.abta.org for ABTA's library of free on-demand webinars that feature experts addressing a range of brain tumor topics from treatment options and tumor types to diets and coping with the diagnosis. Our next webinar will be Viral Therapies for Brain Tumors on Thursday, January 21 from 1:00 to 2:00 PM CT. Viruses have the ability to replicate and eradicate cancer cells, according to promising studies. Oncolytic viral therapy is a treatment that can infect and kill cancer cells, leaving normal cells unharmed. In addition, vaccines made from a gene-modified virus, for example the measles vaccine, may help the body build an effective immune response to kill tumor cells. Join Ian F. Parney, MD, PhD, Department of Neurosurgery, Mayo Clinic who will discuss how viral therapies work and future directions in viral treatments for brain tumors. Dr. Parney will also highlight a Phase I clinical trial underway at the Mayo Clinic that is investigating the side effects and dosage of the measles viral therapy in treating brain tumor patients with glioblastoma multiforme (GBM). This webinar includes an interactive Q&A with Dr. Parney. This concludes our webinar for today. Thank you so much for joining us and please be sure to complete the evaluation survey you will receive by email tomorrow. You may now disconnect. Have a wonderful rest of your day. [Event concluded ] |
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If p÷q is a rational no. What is the condition on q so that the decimal representation of p÷q is terminating?
Bernard Leak
Firmware Developer (2008–present)
·
Author has 5.8K answers and 5M answer views
·
7y
I suppose p is an integer and is a non-zero integer.
There isn't an sufficient and necessary condition unless we are also given or and have no common factors. However, there is a sufficient condition which is as good as we can get, if all we know about is that it's an integer.
If a decimal representation of terminates, either or you shouldn't say that “the” representation terminates, because there are two representations, and the other one doesn't terminate. All right, let's always use the terminating representation when there is one, and we'll also igno
I suppose is an integer and is a non-zero integer.
There isn't an sufficient and necessary condition unless we are also given or and have no common factors. However, there is a sufficient condition which is as good as we can get, if all we know about is that it's an integer.
If a decimal representation of terminates, either or you shouldn't say that “the” representation terminates, because there are two representations, and the other one doesn't terminate. All right, let's always use the terminating representation when there is one, and we'll also ignore zero for the moment.
Now, given a terminating representation, by definition there can only be a finite number of non-zero digits after the decimal point, and a finite number of digits after the point up to and including the last non-zero digit. may be zero (the non-zero digits may all be to the left of the decimal point, so the rational number is an integer). We can cover the case where the rational number is zero the same way as the other integers, with . In any case, is a natural number.
We can certainly shift left through decimal places to make the resulting number an integer. This amounts to multiplying by . On the other hand, given an integer, we can divide it by for any natural number to get a rational number with a terminating decimal representation (ending after at most digits after the decimal point).
What we now have is that terminating decimals correspond to integral multiples of for natural numbers . If divides for some natural number , we must have a terminating decimal. On the other hand, if can be any integer, it can be , and then if does not divide for any natural we can't have a terminating decimal. So without restricting , the best we can do is require to divide for some natural .
The prime factors of are the prime factors of , namely and , except when . That's all right; we can always pick a larger than we actually need. The important detail is that we can pick large enough for the largest power of dividing , and also large enough for the largest power of dividing . On the other hand, if any other prime number divides , we're stuck. We can only get a terminating decimal if we can cancel the additional prime factors with , which we can't guarantee.
So a sufficient condition on is that it have no prime factors except and , and this can't be improved on without having an extra constraint on .
Related questions
What condition needs to be satisfied by 1/q so that a rational number p/q has a terminating decimal expansion?
If a/b is a rational number (b is not equal to 0) in its lowest form, what is the condition on b so that the decimal representation of a/b is terminating?
How can say whether the given rational p/q (q is not equal to 0) terminating decimal or not terminating decimal?
How would you prove or disprove the following claim: For
,
where
if and only if
?
When will the rational number p÷q, q is not equal to 0, have terminating decimal expansion? What about the prime factorization of q?
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If a/b is a rational number (b is not equal to 0) in its lowest form, what is the condition on b so that the decimal representation of a/b is terminating?
can be represented in a terminating decimal representation only if by multiplying both the numerator and denominator by some whole number , you can get
Since is a positive integer, and/or can be the only prime factor
can be represented in a terminating decimal representation only if by multiplying both the numerator and denominator by some whole number , you can get
Since is a positive integer, and/or can be the only prime factors of [math]b[/ma...
Reuven Harmelin
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Is there a proof that every rational number will either terminate or repeat its digits when converted into base 10 string notation (decimal)?
Surly such proofs do exists. Recall that a rational number is defined as a ratio (quotient, proportion)
p/q
between two integers
p,q
with
q non-zero.
Now, we distinguish between the following two cases: In the first case the denominator
q
is of the form (
2^n)(5^m)
for a pair of non-negative powers
m,n,
and in the second case at least one of the prime divisor of
q
is different from
2
and
5.
In the first case the decimal representation of
p/q
terminate after finitely many digits beyond the decimal digits, e.g.
and on the other hand
in which the string of
6
decimal digits periodically repeats itse
Surly such proofs do exists. Recall that a rational number is defined as a ratio (quotient, proportion) p/q between two integers p,q with q non-zero. Now, we distinguish between the following two cases: In the first case the denominator q is of the form (2^n)(5^m) for a pair of non-negative powers m,n, and in the second case at least one of the prime divisor of q is different from 2 and 5. In the first case the decimal representation of p/q terminate after finitely many digits beyond the decimal digits, e.g.
and on the other hand
in which the string of 6 decimal digits periodically repeats itself infinitely many times.
Proof of the first case: If q=(2^n)(5^m), then q is clearly a divisor of 10^S where S=max(n, m), that is
Hence, multiple both the nominator p and the denominator q by N and obtain
which is the same as calculating the integer
and the decimal digit moves S places from the left end of that integer.
Conversely, if the decimal representation of some real number S has finite number n of decimal digits after the decimal point, then the product of S by 10 to the power n is an integer, and therefore
that is, S is a ratio between two integers, and the denominator is of the form (2^m)(5^n) with n=m.
For example, if
then
Proof of the second case: In this case the prime factorization of the denominator q of the rational number p/q involves at least one prime different from 2 and 5. Hence, according to the end of the previous case, the decimal representation of p/q must be infinite.
The process of calculating the decimal digits after the decimal point in the decimal representation of a given ration number p/q is a sequence of steps each one of them is the basic integer division with remainder, going as follows:
and after the j-th step in which
have been calculated, in the next step we calculate
such that
and therefore
Observe that all the remainders obtained during that process must be strictly positive, since once some remainder would be zero, it would mean that the decimal representation of the given rational number p/q is finite, and as we have learned above, this happens if and only if the only prime divisors of q are either 2 or 5, in contradiction to the assumption in this case.
Thus the first sequence of the integer quotients
is the sequence of the decimal digits after the decimal points, and it is infinite. At the same time. the members of the second sequence of the remainders
are natural numbers between 1 and q-1 (including), and therefore, after some finite number of steps less than or equal to q-1, there must be a pair of natural numbers j<k between 1 and q-1 (including) such that
which yields
and after that
By induction it follows that for every natural number n we obtain the identities
Hence, both sequences must be periodic with period k-j.
Let us look at the next example
The process of calculating the decimal digits of that rational number, after the decimal point, is shown below:
As you can see, the period is 16=17–1=q-1.
Conversely, suppose you are given a decimal representation of some real number M which is infinite and periodic, with some given period n, like the following,
where N is some integer, and
is the infinite, periodic sequence of the decimal digits after the decimal point. We would like to show that
is a rational number, which would imply that M=N+M’ is also a rational number.
First we look at the finite decimal containing the first n decimal digits of the basic repeating string
Multiply that number by 10^n to get the following integer:
Then we can write:
Observe that the infinite sum inside the brackets, is an infinite geometric series of powers of one divided by the n-th power of 10, for the given natural number n>0 which is the period. Then by the formula of such an infinite convergent geometric series we deduce
Notice that M’ is represented by two integers, and hence, M’ is indeed a rational number, as we wished to prove.
Max Gretinski
Studied Mathematics
·
Author has 6.5K answers and 2.4M answer views
·
2y
Related
Why can all rational fractions be written as terminating or repeating decimals?
This results is based on something called the Pigeonhole Principle.
If you have a certain number of boxes — N — and you try to put more than N objects into those boxes, then at least one box gets more than one object. The box “repeats.”
For rational numbers as fractions, this has to do with the denominator.
Let’s pin this down using 8 as an example denominator.
If we divide any decimal by 8, and the remainder is 0 at any stage, then the decimal form immediately terminates.
Our example numerator is 3.
8 does not go into 3, so we tack on a zero.
8 goes into 30 three times, and we write…
The remainder is
This results is based on something called the Pigeonhole Principle.
If you have a certain number of boxes — N — and you try to put more than N objects into those boxes, then at least one box gets more than one object. The box “repeats.”
For rational numbers as fractions, this has to do with the denominator.
Let’s pin this down using 8 as an example denominator.
If we divide any decimal by 8, and the remainder is 0 at any stage, then the decimal form immediately terminates.
Our example numerator is 3.
8 does not go into 3, so we tack on a zero.
8 goes into 30 three times, and we write…
The remainder is not zero, so we bring down another zero.
8 goes into 60 seven times, and we write…
The remainder is not zero, so we bring down another zero.
8 goes into 40 exactly five times, so that the decimal form terminates.
Now, what MUST happen if the decimal does not terminate?
Let’s use 7 as an example denominator.
If we divide any decimal by 7, and the remainder is 0 at any stage, then the decimal form immediately terminates. What if it never terminates? I claim that it MUST repeat.
Our example numerator is 2, but any numerator that is not a multiple of 7 will do.
By the definition of a remainder, the remainder when dividing by any number (7, in this case) must be LESS THAN the number itself. Assume for the moment that the remainder is never 0 at any stage. Then the possible remainders are 1, 2, 3, 4, 5, and 6. There are six remainders.
During the long division process, we may wind up dividing 7 into 10, 20, 30, 40, 50, and 60: the six remainders followed by the zero that we bring down at each stage.
Once we have done this process at most six times, we have used all possible remainders. At that point, something MUST repeat. In this case, all six digits repeat, so that
= 0.
In general, when we divide by N, we are guaranteed that the decimal form must repeat with no more than N - 1 digits.
Much of the time, the part that repeats (called the repetend) is not as long as it could be. For example, when we divide by 11, the repetend is not ten digits long, but only two digits long.
= 0.
When we divide by 9, the repetend is only one digit long.
= 0.
Likewise, when we divide by 6:
= 0.8
But when the denominator is N, if the decimal form doesn’t terminate, there are only N - 1 possible remainders. This means that the decimal form must repeat with no more than N - 1 digits of repetition.
Therefore, the decimal form of any rational number either terminates or repeats.
We may also show that every decimal that terminates or repeats corresponds to a rational number…but that’s enough for now.
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Related questions
What is p/q where p≠infinity?
A rational number in the form of p/q (q not equals to 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations. What property q must be satisfied?
How do you prove
[math]\Gamma\left (n+\frac {p} {q} \right) =\frac {1} {q^ {n}} \Gamma\left (\frac {p} {q} \right) \prod_ {k=1} ^ {n} (p+k q-q), \quad n, p, q \in \mathbb {N}, p<q[/math]
?
How can a rational number be proven to be written as either a terminating or repeating decimal?
When is the decimal expansion of a ration of a number p/q, q?
Amitabha Tripathi
knows a few computational short-cuts
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Upvoted by
Aditya Garg
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How do I Test Whether a Given Rational Number is Terminating or Repeating Decimal?
Let [math]n=2^{\alpha} \cdot 5^{\beta} \cdot m[/math], with [math]\gcd(m,10)=1[/math]. Let [math]\gamma = \max {\alpha,\beta}[/math]. If [math]\gcd(a,n)=1[/math], then
[math]\dfrac{a}{n}[/math] has a terminating decimal expansion if and only if [math]m=1[/math].
Moreover, if [math]m=1[/math], then the fractional part of
[math]\dfrac{a}{n} = \dfrac{a}{2^{\alpha} \cdot 5^{\beta}} = 0.d_1\:d_2\:d_3\:\ldots\:d_{\gamma}[/math].
The rational number [math]\frac{a}{n}[/math] has a recurring decimal expansion if and only if [math]m>1[/math].
If [math]m>1[/math], the fractional part of
[math]\dfrac{a}{n} = \dfrac{a}{2^{\alpha} \cdot 5^{\beta} \cdot m} = 0.d_1\:d_2\:d_3\:\ldots\:d_{\gamma}\:\overline{d_{\gamma +1}\:\ldots\:d_{\gamma + \ell}}, [/math]
where [math]\ell[/math]
Let [math]n=2^{\alpha} \cdot 5^{\beta} \cdot m[/math], with [math]\gcd(m,10)=1[/math]. Let [math]\gamma = \max {\alpha,\beta}[/math]. If [math]\gcd(a,n)=1[/math], then
[math]\dfrac{a}{n}[/math] has a terminating decimal expansion if and only if [math]m=1[/math].
Moreover, if [math]m=1[/math], then the fractional part of
[math]\dfrac{a}{n} = \dfrac{a}{2^{\alpha} \cdot 5^{\beta}} = 0.d_1\:d_2\:d_3\:\ldots\:d_{\gamma}[/math].
The rational number [math]\frac{a}{n}[/math] has a recurring decimal expansion if and only if [math]m>1[/math].
If [math]m>1[/math], the fractional part of
[math]\dfrac{a}{n} = \dfrac{a}{2^{\alpha} \cdot 5^{\beta} \cdot m} = 0.d_1\:d_2\:d_3\:\ldots\:d_{\gamma}\:\overline{d_{\gamma +1}\:\ldots\:d_{\gamma + \ell}}, [/math]
where [math]\ell[/math] is the least positive integer for which [math]m[/math] divides [math]\underbrace{999\quad 9}_{\ell\:\text{times}}[/math]. This is the same as saying that [math]\ell[/math] is the order of [math]10[/math] in the multiplicative group of units in [math]{\mathbb Z}_m[/math]. [math]\blacksquare[/math]
Senia Sheydvasser
PhD in Mathematics
·
Upvoted by
Michael Lamar
, PhD in Applied Mathematics and
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 2.5K answers and 39.7M answer views
·
9y
Related
If you have a decimal which is 0. followed by an infinite number of digits, recurring, is this number rational or irrational?
Any real number that has a repeating decimal expansion is rational. Let’s prove it.
For simplicity, I am going to assume that the decimal expansion is of the form [math] 0.\overline{x_1 x_2 x_3 \ldots x_n} [/math]—that is, after the decimal point we have digits [math] x_1, x_2, \ldots x_n [/math] that then repeat indefinitely. It isn’t hard to modify the argument to cover the more general case.
So, we have
[math] x = 0.\overline{x_1 x_2 x_3 \ldots x_n} [/math].
It follows from this that
[math] 10^n x = x_1 x_2 x_3 \ldots x_n.\overline{x_1 x_2 x_3 \ldots x_n} [/math].
But now, all we have to do is to subtract [math] x = 0.\overline{x_1 x_2 x_3 \ldots x_n} [/math]
Any real number that has a repeating decimal expansion is rational. Let’s prove it.
For simplicity, I am going to assume that the decimal expansion is of the form [math] 0.\overline{x_1 x_2 x_3 \ldots x_n} [/math]—that is, after the decimal point we have digits [math] x_1, x_2, \ldots x_n [/math] that then repeat indefinitely. It isn’t hard to modify the argument to cover the more general case.
So, we have
[math] x = 0.\overline{x_1 x_2 x_3 \ldots x_n} [/math].
It follows from this that
[math] 10^n x = x_1 x_2 x_3 \ldots x_n.\overline{x_1 x_2 x_3 \ldots x_n} [/math].
But now, all we have to do is to subtract [math] x = 0.\overline{x_1 x_2 x_3 \ldots x_n} [/math] from both sides, and we see that:
[math] (10^n - 1)x = x_1 x_2 x_3 \ldots x_n [/math]
hence
[math] x = \frac{x_1 x_2 x_3 \ldots x_n}{10^n - 1} [/math], so [math] x [/math] is rational. Q.E.D.
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Nikolas Scholz
Self proclaimed math genius
·
Author has 1.9K answers and 2.6M answer views
·
5y
Related
How can you distinguish between an irrational decimal and the decimal form of a rational number p/q where q is very large? How many digits must be examined before you can be certain the sequence is repeating or not?
You would have to examine all of them. Of course this is not possible for many rational numbers, as they have infinitely repeating digits. Thus you cannot distinguish a rational number from an irrational one if you only have limited knowledge of it (by knowing only limited amounts of digits).
Let me give you an example. Let’s say you have the most common repeating rational number, [math]\frac{1}{3}=0.\bar{3}[/math]. Now assume that starting from an arbitrary digit of it, we continue with not 3 but sqrt(2). Let’s say it is the 10^17′th digit. Then your number is actually [math]\frac{1}{3}+\sqrt 2\cdot 10^{10^{-17}[/math]
You would have to examine all of them. Of course this is not possible for many rational numbers, as they have infinitely repeating digits. Thus you cannot distinguish a rational number from an irrational one if you only have limited knowledge of it (by knowing only limited amounts of digits).
Let me give you an example. Let’s say you have the most common repeating rational number, [math]\frac{1}{3}=0.\bar{3}[/math]. Now assume that starting from an arbitrary digit of it, we continue with not 3 but sqrt(2). Let’s say it is the 10^17′th digit. Then your number is actually [math]\frac{1}{3}+\sqrt 2\cdot 10^{10^{-17}}[/math] which is an irrational number.
You only know that the pattern breaks as soon as you get to this digit. And as this digit could lie anywhere, you would have to check all of the digits.
Since you can’t check infinitely many digits, this shows us that decimal representations are not well equipped to distinguish between rational and irrational numbers.
You would think: but hey, we CAN distinguish 0.2 as a rational number even though it is a decimal representation. This is actually only half the story though. If we’d be totally honest, we would have to write [math]0.2=0.2\bar{0}[/math] thus claiming that all of the infinitely many decimals behind 2 are zero. Same story, just obmitted notation.
Tony Jazdec
hobby number juggler
·
Author has 1.7K answers and 1.5M answer views
·
6y
Related
What is the proof that all rational numbers have a decimal expansion that either terminates or repeats?
Rational numbers are ratios of two whole numbers. Not only their decimal expansion (in base 10) terminates or repeats, but the repeating period cannot be larger than the denominator itself. Can’t be larger than the denominator minus 1, in fact.
Let’s take rational number 1/7 as an example. If you start dividing the numerator by the denominator (and taking remainders), it goes as follows:
1/7 = 0, remainder 1. So our first decimal digit is 0. To get the first digit after the decimal point, we multiply the remainder by 10 and divide again by the denominatior:
110/7 = 1, remainder 3. So we have 0.1
Rational numbers are ratios of two whole numbers. Not only their decimal expansion (in base 10) terminates or repeats, but the repeating period cannot be larger than the denominator itself. Can’t be larger than the denominator minus 1, in fact.
Let’s take rational number 1/7 as an example. If you start dividing the numerator by the denominator (and taking remainders), it goes as follows:
1/7 = 0, remainder 1. So our first decimal digit is 0. To get the first digit after the decimal point, we multiply the remainder by 10 and divide again by the denominatior:
110/7 = 1, remainder 3. So we have 0.1 so far. Lets go to the next digit after the decimal point:
310/7 = 4, remainder 2. 0.14
210/7 = 2, remainder 6. 0.142
610/7 = 8, remainder 4. 0.1428
410/7 = 5, remainder 5. 0.14285
510/7 = 7, remainder 1. 0.142857. But hold on… we already saw a remainder of 1. We have the same remainder, and that means we will get the same result after the next division, and then after the next one, and after that too… so the pattern “142857” will repeat.
No matter whan nominator and denominator you choose, there is only a limited number of possible remainders. You can’t get a remainder equal to or greater than the denominator. And once you get a remainder that you got before, the whole pattern after that must repeat, because you perform the same operations with it.
If you tried 1/N, you only can get N different remainders (0 to N-1… remainder of zero means that the expansion terminates. But with some rational numbers like 1/3 or 1/7 you will never get a remainder of 0, so the expansion repeats forever).
This is a layman’s proof to what I suppose is a layman’s question… I wonder what Alon Amit or Roman Andronov may have to say :)
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Airline Mistake Fare Alerts are my favorite travel hack for getting cheap flights all over the world for a fraction of the cost!
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Sign up for airline mistake fare and flash sales alerts and quit planning trips like it’s 1960—choosing dates and a destination first, then spending days (or weeks) trying to make those locked-in plans fit your budget. That’s like walking onto a car lot, announcing the exact model and purchase date, and hoping the price magically works out.
This makes zero sense! You would never agree to buy a certain model of car on a specific date, and then just be stuck with either paying whatever they wanted on that date, or worse you not have a car at all. There is no other thing in your life where you plan with such insanity as one does vacations. Why would you not just choose from a list of top destinations that are currently offering a huge discount on flights? You actually would, you just didn’t know there was options to be notified of every amazing deal and that’s the purpose of this post.
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Create A Bucket List: Instead of planning like it’s the 1950’s and choosing your destination and dates before you even know the price, instead make a list of all the places you want to go and when an alert for that destination comes, you book it!
If you have 10+ destinations on your list, you will receive alerts for more than half of them at some point throughout the year, it’s just when will the alert come? Regardless of when it comes, that’s when you’re going to book it!
When An Alert Comes Book Without Fear: Most are unaware that you have 24 hours to cancel any flight you book for a full refund no questions asked. This is a law passed by congress and enforced by the depart of transportation.
So, when an amazing deal comes, just book it. Don’t wait! Now you have 24 hours to see if you can make it work with work and those traveling with you, and if you can’t you simply cancel, and it didn’t cost you a penny. Rinse and repeat this process for the rest of your life, and you will see more of this world than you ever thought possible.
Work smarter instead of harder instead of harder!
Another Budget-Traveler Hack—Free Lodging + Meals
Want to cut nearly all of your on-the-ground costs? Trade a few hours of your time for a place to sleep and something to eat.
How it works: Volunteer 3–4 days a week, just a handful of hours each day, and your host covers a bed plus a couple of meals.
Top platforms: Workaway Worldpackers
Together they list 50,000-plus hosts—farms, hostels, ecolodges, family homestays, and more.
Do your homework: Read recent reviews, confirm the daily tasks, and chat with the host before you commit.
It’s an unbeatable way to meet people, learn local routines, and keep your wallet happy. Don’t let a tight budget lock you out of new experiences—after all, we only get one shot at this life, so we might as well see as much of the planet as possible.
Avik Paul
MSc in Physics, Indian Institute of Technology, Guwahati (IITG) (Graduated 2016)
·
Author has 241 answers and 336.9K answer views
·
4y
Related
How can say whether the given rational p/q (q is not equal to 0) terminating decimal or not terminating decimal?
p/q is a rational number where q is not equal to 0. And HCF of p and q is 1.
Then the number will be terminating for the following cases
q =1 (obvious)
q = multiple of 2 only
q = multiple of 5 only
q = multiple of 2 and 5 both only
Note - if q is a multiple of 2 and any other prime number no like 3, 7, 11 (other than 5), then the number will not be a terminating decimal.
Similarly, if q is a multiple of 5 and any other prime number like 3, 7, 11 (other than 2), then the number will not be a terminating decimal.
For example;
1/16 ; Here 16 = 2222 - i.e. only multiple of 2. It is a terminating decimal
p/q is a rational number where q is not equal to 0. And HCF of p and q is 1.
Then the number will be terminating for the following cases
q =1 (obvious)
q = multiple of 2 only
q = multiple of 5 only
q = multiple of 2 and 5 both only
Note - if q is a multiple of 2 and any other prime number no like 3, 7, 11 (other than 5), then the number will not be a terminating decimal.
Similarly, if q is a multiple of 5 and any other prime number like 3, 7, 11 (other than 2), then the number will not be a terminating decimal.
For example;
1/16 ; Here 16 = 2222 - i.e. only multiple of 2. It is a terminating decimal.
1/16 = 0.0625
1/25; Here 25 = 555 - i.e. only multiple of 5. It is a terminating decimal.
1/25 = 0.25
1/100; Here = 100 = 2255 - i.e. only multiple of 2 and 5. It is a terminating decimal.
1/100 = 0.01
1/7; Here 7 is not multiple of 2 or 5. It is a non-terminating decimal.
1/36′ Here 36 = 2233 -i.e. multiple of 2 and 3 . It is a non-terminating decimal.
1/55 Here 55 = 511 - -i.e. multiple of 5 and 11 . It is a non-terminating decimal.
1/30 , Here 30 = 235 i.e. multiple of 2, 5 and 3 . It is a non-terminating decimal.
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
·
Author has 6.8K answers and 52.5M answer views
·
6y
Related
If all repeating decimals are rational numbers (i.e.
[math]\frac{p}{q}[/math]
), what is the
[math]\frac{p}{q} [/math]
to generate
[math]0.999...[/math]
?
This is far more interesting than I first thought!
Look at this pattern…
It certainly SEEMS obvious that 9/9 should fit the pattern too = 0.9999999…
In fact look at additions like this…
The difference “mechanically” between all the above decimal representations
is that they can all be obtained by long division…
Of course, I have always justified the result by using this simple proof…
S = 0.9999999999999999999999999… EQU 1
And multiplying by 10, we get….
10S = 9.99999999999999999999999…. EQU 2
Then EQU 2 minus EQU 1 becomes…
9
This is far more interesting than I first thought!
Look at this pattern…
It certainly SEEMS obvious that 9/9 should fit the pattern too = 0.9999999…
In fact look at additions like this…
The difference “mechanically” between all the above decimal representations
is that they can all be obtained by long division…
Of course, I have always justified the result by using this simple proof…
S = 0.9999999999999999999999999… EQU 1
And multiplying by 10, we get….
10S = 9.99999999999999999999999…. EQU 2
Then EQU 2 minus EQU 1 becomes…
9S = 9
So S = 1
George Mathew
Mathematics teacher at a government college in India.
·
Author has 1.3K answers and 1.8M answer views
·
2y
Related
A rational number in the form of p/q (q not equals to 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations. What property q must be satisfied?
Only condition needed is that
[math]\;q\ne 0\:[/math]
and
[math]\;q\; [/math]
is an integer
[math] .[/math]
Alexander Mathey
Former Chemical Engineer, retired, lives in Athens, GR
·
Author has 5.6K answers and 10.8M answer views
·
7y
Related
What is the difference between terminating and non-terminating decimal rational numbers?
There is no real difference.
All rational numbers, if written in their decimal expansion, will show a repeating sequence of digits from some position onwards.
Terminating are called those rational numbers for which the repeating sequence is the sole digit 0.
Example: 10/8 = 1.25 = 1.2500000000 . . .
Even non terminating decimal rational numbers, like for example
1/7 = 0.142857 142857 142857 . . .
can be written as terminating ones, by the simple change of numeral system, to one whose base is not relative prime to the denominator.
Would we for example choose a base-14 numeral system (its single digits
There is no real difference.
All rational numbers, if written in their decimal expansion, will show a repeating sequence of digits from some position onwards.
Terminating are called those rational numbers for which the repeating sequence is the sole digit 0.
Example: 10/8 = 1.25 = 1.2500000000 . . .
Even non terminating decimal rational numbers, like for example
1/7 = 0.142857 142857 142857 . . .
can be written as terminating ones, by the simple change of numeral system, to one whose base is not relative prime to the denominator.
Would we for example choose a base-14 numeral system (its single digits are 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D), then
1/7 = 0.2
By the way, a Base-14 numeral system is actually used in Greece for the first 3 digits of the car-license plates, in order to only use those letters of the Greek alphabet, which have a shape at once recognizable by non Greeks:
Was this worth your time?
This helps us sort answers on the page.
Absolutely not
Romain Mondon-Cancel
knows a bit of maths.
·
Author has 607 answers and 1.3M answer views
·
7y
Related
What condition needs to be satisfied by 1/q so that a rational number p/q has a terminating decimal expansion?
[math]\frac{p}{q}[/math] has a terminating decimal expansion if it can be written as [math]\frac{p'}{10^k}[/math].
[math]\frac{p}q = \frac{p'}{10^k}[/math]
[math]p10^k = p'q[/math]
Which means [math]q | 10^k[/math], i.e. [math]q = 2^n5^m[/math].
Conversely, let’s suppose [math]q = 2^n5^m[/math]. Let us consider [math]k = \max(n,m)[/math]. Therefore, [math]q | 10^k[/math]. Let us call [math]s \in \N[/math] such as [math]qs = 10^k[/math].
In that case,
[math]\frac{p}{q} = \frac{ps}{qs} = \frac{ps}{10^k}[/math]
[math]\frac{p}{q}[/math] has a terminating decimal expansion if and only if [math]q[/math] is of the form [math]2^n5^m[/math].
Related questions
What condition needs to be satisfied by 1/q so that a rational number p/q has a terminating decimal expansion?
If a/b is a rational number (b is not equal to 0) in its lowest form, what is the condition on b so that the decimal representation of a/b is terminating?
How can say whether the given rational p/q (q is not equal to 0) terminating decimal or not terminating decimal?
How would you prove or disprove the following claim: For
,
where
if and only if
?
When will the rational number p÷q, q is not equal to 0, have terminating decimal expansion? What about the prime factorization of q?
What is p/q where p≠infinity?
A rational number in the form of p/q (q not equals to 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations. What property q must be satisfied?
How do you prove
?
How can a rational number be proven to be written as either a terminating or repeating decimal?
When is the decimal expansion of a ration of a number p/q, q?
What if (p and q are) 2 (p^q)?
If P is equal to Q, is Q also equal to P?
Which of the following means P is grandson of S ? (a) P + Q – S (b) P ÷ Q × S (c) P ÷ Q + S (d) P × Q ÷ S?
How would one prove that 0 is a rational number? How would one write 0 in p upon q where p and q are integers and q is not equal to zero?
If you express the number 0.212121… in the standard form of a rational number p/q, then what is p equal to?
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Gravitational Constant
The story of the gravitational constant, Big G:
In 1686 Isaac Newton realized that the motion of the planets and the moon as well as that of a falling apple could be explained by his Law of Universal Gravitation, which states that any two objects attract each other with a force equal to the product of their masses divided by the square of their separation times a constant of proportionality. Newton estimated this constant of proportionality, often calledBig G, perhaps from the gravitational acceleration of the falling apple and an inspired guess for the average density of the Earth. However, more than 100 years elapsed before G was first measured in the laboratory; in 1798 Cavendish and co-workers obtained a value accurate to about 1%. When asked why he was measuring G, Cavendish replied that he was "weighing the Earth"; once G is known the mass of the Earth can be obtained from the 9.8 m/s 2 gravitational acceleration on the Earth surface and the Sun's mass can be obtained from the size and period of the Earth orbit around the sun. Early in this century Albert Einstein developed his theory of gravity called General Relativity in which the gravitational attraction is explained as a result of the curvature of space-time. This curvature is proportional to Big G.
Measurements of Big G in the past
Naturally, the value of the fundamental constant G has interested physicists for over 300 years and, except for the speed of light, it has the longest history of measurements. Most measurements of G have used variations of the torsion balance technique first used by Cavendish. Torsion balance experiments to measure G consisted of a 'dumbbell' (two masses connected by a horizontal rod) suspended by a very thin fiber. When two heavy attracting bodies were placed on opposite sides near the masses on the dumbbell pendulum, it would rotate by a very small amount, twisting the fiber. The attracting bodies were then moved to the other side of the dumbbell twisting it in the in the opposite direction. The magnitude of these twists is used to find G. In a variation of this technique, the dumbbell was set into an oscillatory motion and the frequency of the oscillation was measured. The gravitational interaction between the dumbbell and the attracting bodies caused the oscillation frequency to change slightly when the attractors are moved to a different position and this frequency change was used to determine G. This frequency shift method was by Gabe Luther and William Towler from the National Bureau of Standards and the University of Virginia. It was published in 1982. Based on their measurement, the Committee on Data for Science and Technology (CODATA), which gathers and critically analyzes data on the fundamental constants, assigned an uncertainty of 0.0128% to G. Although this seems quite precise, the fractional uncertainty in G is thousands of times larger than those of other important physical constants, such as Planck's constant or the charge on the electron. Following 1982 measurement the value of G was called into question by measurements from respected research teams in Germany, New Zealand, and Russia. The new values disagreed wildly. For example, a team from the German Institute of Standards obtained a value for G that was 0.6% larger than the accepted value; a group in Wuppertal, Germany found a value that is 0.06% lower, and a group in New Zealand measured a value that was 0.1% lower. A Russian group found a curious space and time variation of G of up to 0.7%. The collection of these new results suggests that the uncertainty in G could be much larger than originally thought.
One of the greatest difficulties in any G measurement was determining with sufficient accuracy the dimensions and density distribution of the torsion pendulum body, e.g. the dumbbell. A second limitation was associated with the suspension fiber. The Japanese physicist Kazuaki Kuroda had pointed out that internal friction in the torsion fiber, which had previously been neglected, may have caused a significant bias in the existing torsion balance measurements.
The University of Washington Big G Measurement
At the University of Washington’s Eöt-Wash research group, Jens Gundlach invented a torsion balance method that elegantly sidesteps the limitations of all previous measurements. The first thing he noted was that if the usual dumbbell pendulum were replaced by a thin, flat plate hung by its edge, neither the pendulum's dimensions nor its density distribution would have to be known with very high precision. In principle, one can obtain G by measuring the angular acceleration of a flat pendulum without even knowing its mass or dimensions. This simple fact had not been recognized in 200 years of gravitational experiments! Then he eliminated the problems with the torsion fiber by placing the torsion balance on a turntable that continuously rotates between a set of attracting spheres. The turntable is controlled by a feedback loop that speeds up or slows the turntable rotation rate exactly so that the suspension fiber never has to twist;G can then be accurately inferred from the rotation rate of the turntable. Since the torsion fiber does not twist, the bias that Kuroda had warned about simply does not occur. By placing the attractor spheres on another (concentric) turntable that rotates at a much faster frequency than the frequency at which previous mass exchanges had been done, Gundlach’s method reduced the so-called 1/f-noise. This is noise associated with slow changes over time and afflicts many delicate measurements. By measuring G at a higher frequency the fluctuating noise of the measurement was significantly reduced.
For these ideas Gundlach received a prestigious Precision Measurement Grant from the National Institute of Standards and Technology, NIST, and he began together with his colleague Stephen Merkowitz to build the instrument and then carry out the measurement. Gundlach’s tricks paid off and the result was a much more accurate measurement of Big G with an uncertainty of only 14 ppm (parts per million); published in 2000. The measurement moved the value of G higher than the CODATA value so that the accepted value was redefined, mostly based on the University of Washington measurement.
The University of Washington value was followed by measurements from other groups that were in good agreement, confirming the higher value of G. Most notably, a group led by Jun Luo in China built a copy of Gundlach’s experiment, and measured a very similar value for Big G.
Photograph of the big G apparatus. The spheres are 12.5 cm in diameter. (JPG image)
Photograph of the pendulum with several mirrors directing the light beam. A penny was placed in the foreground for scale. (JPG image)
Schematic cut-open drawing of the big G apparatus. The inner turntable rotates at about 1rev/20 min, the outer turntable rotates at about 1rev/5 min.
Relevant literature
G.G. Luther and W.R. Towler, Phys. Rev. Lett., 48, 121 (1982). [The previously best measurement]
W. Michaelis, H. Haars, and R. Augustin, Metrologia 32, 267 (1995). [The result from the German Bureau of Standards]
M. Fitzgerald and T. R. Armstrong, IEEE Trans. on Inst. and Meas. 44, 494 (1995). [The result from New Zealand]
H. Walesch, H. Meyer, H. Piehl, and J. Schurr, IEEE Trans. on Inst. and Meas. 44, 491 (1995). [The result from Wuppertal]
V.P. Izmailov, O.V. Karagioz, V.A. Kuznetsov, V.N. Mel'nikov, and A.E. Roslyakov, Measurement Techniques 36, 1065 (1993). [The Russian result]
Kazuaki Kuroda, Phys. Rev. Lett. 75, 2796 (1995). [The torsion fiber bias]
J.H. Gundlach, E.G. Adelberger, B.R. Heckel, and H.E. Swanson, Phys. Rev. D 54, 1256R (1996).
J.H. Gundlach, Measurement Sci. and Tech. 10 454 (1999).
J.H. Gundlach and S.M. Merkowitz, Phys. Rev. Lett. 85 2869 (2000). [The measurement]
R. Newman, M. Bantel, E. Berg, and W. Cross, Philos. Trans. R. Soc., A372 20140025 (2014).[UC Irvine, uses plate trick]
S. Schlamminger, et al. Phys Rev D74:, 082001 (2006).[Zurich]
Qing Li, et al. Nature 560 582-588. (2018). [HUST, China]
© 1987-2023 Eöt-Wash Group. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the University of Washington, NSF, DOE or NASA.
CENPA | University of Washington | UW Privacy Statement | UW Web site Terms and Conditions
Eöt-Wash Contacts
For more information, contact:
Jens Gundlach at (206) 616-3012 or gundlach@uw.edu
Michael Ross at mpross2@uw.edu
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Partikel
Deutsch lernen>Grammatik>Partikeln
von Anna Zwolinska-Simon
Fragen zur deutschen Sprache? 🇩🇪 Schreib mir!
ErklärungÜbungen
Partikeln (Einzahl: “diePartikel” – Plural: diePartikeln) sind beim Sprechen wie Gewürze beim Kochen. Sie machen deine Aussage „intensiver“ und interessanter.
Partikel-Beispiele sind:
Das ist ja toll.
Ich find dich echtklasse.
Diese kurzen Wörter können z. B. Empörung, Verwunderung oder Ungeduld ausdrücken. Wie das geht? Das zeige ich dir in dieser Lektion. Du lernst, welche Partikeln es in der deutschen Sprache gibt und wie du sie verwenden kannst.
Anschließend kannst du dein Wissen in den Übungen zu Partikeln testen, die ich für dich erstellt habe.
Inhalt:
Was sind Partikel und wofür brauchst du sie?
Partikel: Eine eigene Wortart?
Partikel – Die wichtigsten Merkmale
Verschiedene Partikel-Typen einfach erklärt – mit Beispielen
Zusammenfassung
Partikel – Beispiele
Häufige Fragen
Weitere Lektionen zum Thema:
Was sind Partikel und wofür brauchst du sie?
Ein Partikel ist im Deutschen ein kurzes Wort, dass du benutzt, um eine Aussage zu verstärken oder abzuschwächen.
Partikeln sind kurze Wörter, die manchmal „launisch“ sind. Warum ist das so? Partikel können verschiedene Gefühle ausdrücken, z. B. kannst du sagen, dass du dich gerade über etwas ärgerst oder total sauer bist.
„Ich bin genervt.”
„Ich bin totalgenervt!”
Siehst du? Die Partikel „total” intensiviert das Adjektiv „genervt“ und verstärkt dadurch die Aussage deutlich.
Und hier noch ein Beispiel:
„Komm!” (Imperativ)
„Komm doch endlich!”
Hier zeigt die Partikel „doch”, dass die aufgeforderte Person nun endlich kommen soll.
Alle Partikeln im Deutschen haben einige gemeinsame Eigenschaften. Du möchtest wissen, welche das sind? Dann lies weiter, ich zeige es dir.
Partikel: Eine eigene Wortart?
Welche Wortart ist eigentlich ein Partikel? Ich erkläre es dir:
Ein Partikel gehört zu den nicht-flektierbaren (das bedeutet nicht durch Konjugation oder Deklination veränderbaren) Wortarten im Deutschen, die nicht zur Wortart der Präpositionen, Adverbien oder Konjunktionen gehören. Klingt erstmal kompliziert, ist es aber nicht. Partikeln sind also eine eigene Wortart und lassen sich nicht verändern.
Partikel – Die wichtigsten Merkmale
Die wichtigsten Eigenschaften von Partikeln sind:
Sie sind (wie ich dir oben bereits erklärt habe) nicht deklinierbar (verändern sich nicht).
Sie können nicht gesteigert werden.
Sie drücken Gefühle der sprechenden Person aus.
Sie werden hauptsächlich in der gesprochenen Sprache verwendet (in der Schriftsprache eher bei informellem Gebrauch, niemals bei offiziellen Schreiben oder Briefen)
Sie sind meistens nicht obligatorisch im Satz, jedoch ist ohne sie eine Aussage nicht so lebendig.
Merk dir
Je nachdem, in welchem Kontext der Begriff “Partikel” verwendet wird, bekommt er ein anderes Genus (der, die, das). Es gibt im Deutschen mehrere Nomen mit schwankendem Genus. Ist z. B. ein sehr kleines Teilchen aus dem Bereich der Physik gemeint, dann heißt es “das Partikel”. In der Sprachwissenschaft wird immerdas Femininum (die Partikel – die Partikeln) verwendet.
Lass uns gemeinsam schauen, zwischen welchen Partikeln man im Deutschen unterscheidet.
Verschiedene Partikel-Typen einfach erklärt – mit Beispielen
Je nach Funktion kannst du zwischen folgenden Arten von Partikeln unterscheiden:
# Gradpartikeln (Steigerungsartikel)
ganz, sehr, etwas, ziemlich, zu, überaus, überhaupt, viel, absolut, total, komplett, höchst, völlig, außergewöhnlich usw.
Diese Partikeln machen die Bedeutung (meist eines Adjektivs) stärker oder schwächer, z. B. den Satz:
„Dieser Schrank ist schwer.“
Du kannst diesen Satz und andere Aussagen durch die Partikel „zu“ verstärken:
„Dieser Schrank ist zu schwer.”
„Dein Verhalten war höchst unverantwortlich!”
„Du hörst überhaupt nicht zu!“
„Der Film war ziemlich interessant.“
# Fokuspartikeln
besonders, selbst, bloß, nur, gerade, sogar, ausgerechnet, erst, vor allem, mindestens, wenigstens, zumindest u. a.
Diese Partikeln betonen ein konkretes Satzglied.
Partikel-Beispiele (Fokuspartikeln):
„Dieses Buch hat mir besonders gut gefallen.“
„Selbst ein kleines Kind weiß das!“
„Er hat sogar gelacht.“
„Wenigstens du hast mir geholfen.“
„Mir hat alles geschmeckt, vor allem die Suppe.“
# Modalpartikeln
denn (kommt häufig in Fragen vor und drückt z. B. Verwunderung aus), aber, mal (z. B. eine freundliche Bitte), ja (Verwunderung, Empörung), doch (Ungeduld oder etwas, was offensichtlich ist), halt / eben (etwas ist offensichtlich, nicht zu ändern), wohl (Vermutung) u. a.
Die Modalpartikeln drücken verschiedene Gefühle aus.
Modalpartikel-Beispiele:
„Hilfst du mir mal?“ (freundliche Bitte, Frage)
„Was machst du denn da?“ (Verwunderung)
„Öffne doch einfach die Tür!“ (Ungeduld)
„Das habe ich ja nicht erwartet.“ (Empörung)
„Du hast das aber schnell geschafft!“ (Verwunderung)
„Das ist halt so.” oder „Das ist eben so.“ (es ist nicht zu ändern)
„Er ist wohl schon bei seiner Freundin.“ (Vermutung)
Aufgepasst
Denke daran, dass„denn” und „aber”auch alsKonjunktionenvorkommen und dann eine andere Bedeutung („denn“ = Begründung, aber = jedoch, dagegen) haben.
Zusammenfassung
Ein Partikel ist ein Wort, dass eingesetzt wird, um eine Aussage zu verstärken, abzuschwächen oder ein bestimmtes Satzglied zu betonen.
Partikeln sind nicht deklinierbar.
Partikeln verwendet man in der Regel in der gesprochenen Sprache. In der Schriftsprache kommt es kaum vor und dann im informellen Gebrauch, z. B. in einer WhatsApp-Nachricht oder auf Social Media.
Partikel – Beispiele
Übung macht den Meister! Deshalb habe ich für dich zahlreiche Übungsaufgaben mit Lösungen zum Thema „Partikel“ vorbereitet. Kostenlos und ohne Registrierung! Einfach hier klicken und mit dem Lernen beginnen:
Häufige Fragen
Was ist ein Partikel?
Ein Partikel ist ein kurzes Wort wie „doch”, „sehr”, „denn”. Sie verändern die Bedeutung einer Aussage – schwächen sie ab, verstärken oder ergänzen sie. Partikeln können auch Gefühle ausdrücken. Die Partikel doch drückt beispielsweise Ungeduld aus: Geh doch nach Hause!
Was muss ich über die Partikeln wissen?
Sie haben einige wichtige Eigenschaften: sind undeklinierbar, können nicht gesteigert werden, drücken Gefühle aus, werden meist in der gesprochenen Sprache verwendet und sind im Satz nicht obligatorisch.
Welche Partikeln gibt es im Deutschen?
Es gibt Gradpartikeln, wie: sehr, viel, etwas, komplett, total, Fokuspartikeln, wie: besonders, nur, sogar, vor allem, selbst und Modalpartikeln, wie: denn, doch, wohl.
Weitere Lektionen zum Thema:
Das war ziemlich leicht, oder? Wenn du dein Wissen noch erweitern möchtest, habe ich hier einige Themenvorschläge für dich:
Modalpartikel (Das war ja leicht. Hier bekommst du einen genaueren Einblick in die Modalpartikeln und ihre Verwendung.)
Aussagesatz (Das ist sehr interessant. Erfahre alles Wichtige zu den Aussagesätzen.)
Fragen (Was ist das denn? Partikeln können auch in Fragen vorkommen. Lerne hier, welche Fragetypen es gibt.)
Imperativ (Hilf mir mal!– Gerne) In dieser Lektion zeige ich dir, wie du Bitten und Aufforderungen formulierst.)
Hast du Lust auf mehr Grammatik? Super! Hier findest du eine Übersicht all meiner Grammatikerklärungen: Deutsche Grammatik einfach erklärt
Hi, ich bin Anna!
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Anna Zwolinska-Simon
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Hallo Marzieye,
danke für deine lieben Worte zu meiner Arbeit. Wenn du mit deinen Schülern noch mehr Grammatik üben möchtest, schau doch gerne auch in mein super einfach erklärtes Grammatikbuch.
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ich bin gerade in der schule und ich check das echt nd wtf ist ein partikel. das hat mir null geholfen danke ich geh lieber zu chgpt. bye
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188124 | https://www.youtube.com/watch?v=eHJuAByQf5A | Lec 7: Exam 1 review | MIT 18.01 Single Variable Calculus, Fall 2007
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Hyperbolic functions (cont.) and exam 1 review
Note: the review for the exam in lecture 7 is not comprehensive because the students already have practice exams available to them. Lecture 8 is Exam 1, so no video was recorded.
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the following content is provided under a Creative Commons license your support will help MIT open courseware continue to offer highquality educational resources for free to make a donation or to view additional materials from hundreds of MIT courses visit MIT open courseware at ocw.mit.edu uh right now we're uh finishing up with this first unit and I'd like to continue in this lecture lecture seven uh with some uh final remarks about exponents so so what I'd like to uh do is just review something that I did quickly last time and make a few philosophical remarks about it I think that the the steps involved were maybe a little tricky and so I'd like to uh go through it one more time remember that we were talking about this number a which is 1 + 1 K to the K power and what we showed was that the limit as K goes to Infinity of a k was e so first the first thing that I like to do is just explain the proof a little bit more clearly than I did last time with a little bit less uh fewer symbols or at least with this abbreviation of the symbol here to show you what it was that we actually did so I'll just remind you what we did last time and the first observation was to check rather than the limit of this function to take the log first and this is typically what's done when you have an exponential when you have an exponent and what we found was that the limit here was one as K goes to Infinity now so last time this is what we did and I just wanted to be careful and show you exactly what the next step is if you exponentiate this fact you take e to this power that's going to tend to e to the first Power which is just e all right and then we just observe that this is the same as AK right so the basic ingredient here is that e to the log a is equal to a that's because the log function is the inverse of the exponential function yes question because it's K Ln 1 1 K as K goes to Infinity that's Ln of one right which is zero um so that wouldn't this the question was wouldn't the log of this be zero because because AK is tending to one but AK isn't tending to one who said it was K if you take the logarithm which is what we did last time logarithm of a k is indeed K the log of 1 + 1 K that does not tend to zero this part of it tends to zero and this part tends to infinity and they balance each other 0 time infinity and so we don't really know yet from this expression in fact we did some clever with limits and derivatives to figure out this limit and it was a very subtle thing it turned out to be one all right now the thing that I'd like to say I'm sorry I'm gonna I'm gonna erase this aside here but you need to go back to your notes and remember that this is what we did last time because I want to have room for the next comment that I want to make on this little Blackboard here what we just derived is this property here but then I made a a a a claim yesterday and I just want to emphasize it again so you realize what it is that we're doing I just reversed the I looked at this backwards one way you can think of this is we're evaluating this limit and getting an answer but all equalities can be re read both directions and we can write it the other way E equals the limit as K goes to Infinity of this expression here so that's just the same thing and if we read it backwards what we're saying is that this limmit is a formula for E so this is very typical of of of mathematics you want to always reverse your perspective all the time equations work both ways and in this case we have two different things here this e was what we defined as the base which when you graph e to the X you get slope one at zero and then it turns out to be equal to this limit which we can calculate numerically if you do this on your calculators you of course will have a a way of programming in this number and evaluating it for each K and you'll have another button available to evaluate this one so another way of saying it is that there's a relationship between these two things and all of of uh calculus is a matter of getting these relationships so we can look at these things in several different ways and indeed that's what we're going to be doing at at least at the end of today and talking about derivatives a lot of times when we talk about derivatives we're trying to look at them from several perspectives at once okay so I have to keep on going with exponents because I have one loose end one loose end that I did not cover yet uh there's one very important formula that's left and it's the derivative of the powers we actually didn't do this while we did it for rational numbers R so this is the formula here but now let me sorry let's just stick this in here this is going to we're going to check this for all real numbers R so including all the irrational ones as well this is also good practice for uh using base e and using logarithmic differentiation so so let me do this by our two methods that we can use to handle exponent exponential type problems so method one was base e so if I just rewrite this Bas e again that's this formula over here x to the r is equal to e to the log X to the power R which is e to the r log X okay so now I can differentiate this so I get that um d by DX x to the r now I'm going to I'm going to use um prime notation because I don't want to keep on writing that d by DX here e to the r logx Prime and now what I can do is I can use the chain rule all right the chain rule says that it's the derivative of this times the derivative of the uh the of the function so the derivative of the exponential is just itself and the derivative of this guy here well I'll write it out once it's R log X Prime so what's that equal to well e to the r logx is just x to the r and this derivative here is well the derivative of R is zero right this is a constant Factor it just factors out and log X now has derivative but what's the derivative of log X 1X so this is going to be time RX and now we rewrite it in the customary form which is R we put the r in front x to the r minus one okay so this is this is the uh I've just derived the formula for you and it didn't now now matter whether R was rational or irrational it's the same proof okay so now I have to show you how method two works as well so let's do method two which we call log logarithmic differentiation and so here I'll I'll I'll use a symbol say U for x to the r and then I'll take its logarithm that's R log X and now I differentiate it I'll leave that in the middle because I want to remember the key property of logarithmic differentiation but first I'll differentiate it later on what I'm going to use is that this is the same as U Prime over U this is one way of evaluating the logarithmic derivative and then the other is to differentiate the explicit function that we have over here and that is just as we said RX so now I multiply through and I get U Prime is equal to U R / X which is just x to the r RX which is just what we did before it's RX to the r minus one again you can now see by comparing these two pieces of arithmetic that they're basically the same pretty much every time you convert to Bas C or you do logarithmic differentiation it'll amount to the same thing provided you don't get mixed up you generally have to introduce a new symbol here on the other hand you have you're dealing with with uh exponents there it it's worth it to know both points of view all right so now I want to make one last remark before we uh finish with exponents and there I'll try to sell this to you in a lot of ways as as the course goes on but but one thing that I want to try to uh emphasize is that the natural logarithm really is natural so so I claim that the natural log is natural and the example that we're going to use is uh for this illustration is economics okay so so let me explain to you why the the the log the natural log is the one that's natural for economics if you are imagining um the price of a stock that you own goes down by a dollar that's a totally meaningless statement it depends on a lot of things in particular it depends on whether the original price was a dollar or $100 so there's not much meaning to these absolute numbers it's always the ratios that matter so for example I just looked up an hour ago the footsy the London exchange uh closed and it was down um 27.9 which as I said is is is pretty meaningless unless you know what the what the uh actual total of this index is and it turns out it was uh uh 6432 so the change in the price divided by the price which in this case is 27.9 / 6432 is what matters and in this case it happens to be uh 43% all right that's what happened today and similarly if you take the infinitesimal of this people think of days as being relatively small increments when you're investing in a stock you would be interested in the infinite tmal sense you would be interested in P Prime over p the derivative of p with divided by P that's just the natural log Prime so this is the let me just put a little box around it it's just the formula of logarithmic differentiation but let me just emphasize that it has an actual significance and it's the one that's used by economists and people who are modeling prices of things all the time they never use absolute prices when they're when they're large swings they use always use a log of the price and there's no point in using log base 10 or log base 2 those give you junk they give you an extra factor of log two it's the natural log that that's the obvious one to use it's completely straightforward that this is a simpler expression than using log base 10 and having a factor of log natural log of 10 there which would just mess everything up all right so this is just one illustration Lots anything that has to do with ratios is going to encounter logarithms all right so that's pretty much it that is that's all um I want to say uh for now anyway there's lots more to say but we'll be saying it when we do applications of derivatives in the second unit so now what I'd like to do is is start a review I'm just going to run through what we did in this unit I'll tell you uh a approximately what I expect from you on the uh hour test that's coming up tomorrow and uh well so let's get started with that all right so this is a review of unit one and I'm going to just put on the board all of the uh things that you need to think about anyway keep in your head and so there are these there are what are called General formulas for derivatives and then there are the specific ones and let me just remind you what the general formulas are there's what you do to differentiate a sum a multiple of a function the product rule the quotient rule all right those are those are uh several General formulas and then there's one more which is the chain rule which I'm going to say just a little bit more about so the derivative of a function of a function is the derivative of the function times the derivative of the other function so here I've abbreviated U is uu at X all right so this is one of two ways of writing it the other way is also uh one that you can keep in mind and you might find easier to remember it's probably a good idea to remember both formulas and then the last type of of general formula that we did was implicit differentiation okay so when you do implicit differentiation uh you have an equation and you don't try to solve for the unknown function you just put it in its simplest form and you differentiate so we actually we we did this in particular for inverses that was a very very key method for calculating the inverses of functions and I it it's also true that logarithmic differentiation is of this type this is a transformation we're differentiating something else we're transforming the equation by taking its logarithm and then differentiating okay so so there are a number of different ways this is applied it can also be it can be applied in anyway these are two of them so maybe in parenthesis this is these are just examples all right I'll try to give uh examples at least a few of these rules uh later so now the specific functions uh that you know how to differentiate well you know how to differentiate now x to the r thanks to what I just did we have the S and the cosine function which you're responsible for knowing what their derivatives are and then other uh trig functions like tan and secant we generally don't bother with cosecant andent because everything can be expressed in terms of these anyway actually you can really express everything in terms of signs and cosin but what you'll find is that it's much more convenient to remember the derivatives of these as well so memorize all of these all right and then we had e to the X and log X and we had the inverses of the trig functions which were uh these are the two that we did the arct tangent and the AR sign all right so those are the ones you're responsible for you should uh have enough time anyway to work out anything else if you know these all right so basically the idea is you have a bunch of special formulas you have a bunch of General formulas you put them together and you can generate pretty much anything okay so let's do a few examples before um going on with uh the review okay so I I I just I do want to do a few examples in sort of uh increasing level of difficulty in how you would combine these things together so first of all uh you should remember that if you're differentiating the secant function that's just oh I just realized I wanted to say something else before I yes so forget that we'll do that in a sec I wanted to make some general remarks uh so so there's one rule that uh uh you discussed in my absence which is the chain Rule and I I do want to make just a couple of remarks about the chain Rule now to remind you of what it is and also to uh present some consequences so a little bit of extra on the Chain rule the first thing that I I I want to say is that we didn't really fully explain why it's true and I I do want to just uh explain it by example okay so imagine that you have a function which is say 10 X Plus B all right so Y is 10x plus b then obviously Y is changing 10 times as fast as B right the the issue is this number here dydx is 10 all right and now if x is a function of something else say t shifted by some other constant here then dxdt is five now all the chain rule is saying is that if Y is going 10 times as fast as T I'm sorry as x and x is uh uh X is going five times as fast as T then Y is going 50 times as fast as T and algebraically all this means is if I plug in and substitute which is what the composition of the two functions amounts to 10 5T + A + B and I multiply it out I get 50t plus 10 A + B now these terms don't matter the constant terms don't matter the rate is 50 and so the consequence if we put them together is that dydt is 10 50 5 which is 50 all right so this is the uh let's make that more leg there all right so this is in a nutshell why the chain rule works and why these these rates multiply the second thing that I wanted to say about the chain rule is that it has a few consequences that make some of the other rules a little easier to remember or possibly to avoid the U messiest rule in my humble opinion is the quotient rule which is kind of a nuisance to remember and so let me just remind you if you take just the reciprocal of a function and you differentiate it there's another way of looking at this and it's actually the way that I use so I I I want to encourage you to think about it this way too this is the same as V to the^ minus1 Prime and now instead of using the quotient rule which we could have used we can use the chain rule here with the function uh the power minus one which works by the uh by by the power law so what is this equal to this is equal to minus vus 2 V Prime okay so here I've applied the chain rule rather than the quotient Rule and similarly suppose I wanted to differentiate the full derive the full quotient rule well now this may or may not be easier but this is one way of remembering what's going on if you convert it to V uh U V theus1 and you differentiate that now I can use the product rule on this of course I have to use the chain Rule and and this rule as well so what do I get here I get U Prime V inverse plus u and then I have to differentiate the V inverse that's the formula right up here that's minus vus 2 V Prime now so that's one way of doing it this actually explains the funny minus sign when you differentiate uh V in the formula the other formula the other way that we did it was by putting this over a common denominator the common denominator was v^2 all right that's it comes from this V the minus 2 and then the second term is minus UV Prime and the first term we have to multiply by an extra factor of V because we have a v^ squ in the denominator so it's U Prime V all right so this is the quoti rule as we wrote it down in lecture and this is just another way of remembering it or deriving it without remembering it if you just remember the chain Rule and the product rule okay so you'll you'll find that in many contexts it's uh easier to do one or the other okay so now I'm ready to to differentiate theant and a few such functions so we'll do some examples here so here's the secant function and I wanted to use that formula up there for the reciprocal this is the way I think of it this is the cosine function to the power minus one and so uh the formula here is just what it's just uh - cine x - 2 - sin x all right and that is so now this is usually written in a different fashion so that's why I'm I'm doing this for a reason actually which is that although there are several formulas for things with Trigg functions there are usually five ways of writing something so I'm writing this one down so that you know what the standard way of presenting it is so what happens here is we have two minus signs canceling and we get sin x over cosine 2 x that's a perfectly acceptable answer but there's a customary way in which it's written it's written 1 / cosine x sin x over cosine X and then we get rid of the denominators by rewriting it in terms of secant and tangent so secant x Tan x so this is the form that's generally used when you see these formulas written in textbooks and so you know you need to watch out because if you ever want to use this kind of calculus you'll have to be uh not be put off by all the secants and tangents all right so uh getting slightly more complicated how about if we differentiate the natural log of secant x okay if you differentiate the natural log that's just going to be secant x Prime / secant x and plugging in the formula that we had before that's secant x Tan x / secant x which is just tanx so this one also has a very nice form and you might say well you know this is kind of an ugly function but the strange thing is that um the natural log was invented before the exponential by a guy named Napier exactly in order to evaluate functions like this these are the functions that people were cared about a lot uh because they were used in navigation you wanted to multiply sign and together to do navigation and the multiplication uh he he encoded using the logarithm and so these were invented long before people even knew about exponents and it was a surprise actually that it was connected to xon so the natural log was invented before the log base 10 and everything else exactly for this kind of purpose anyway so this is a a nice function which which uh was very important so that your ships wouldn't crash into the reef and um okay so let's continue here so now there's there's another kind of function that I want to discuss to you and these are the kinds in which there's a choice as to how to which of these rules to apply and uh I'll just give a a couple of examples of that um there usually is a better and a worse way so let me illustrate that so here's yet another example and I hope you've seen some of these before say x 10th + 8x to the 6th power all right so it's a little bit more complicated than what we had before because there were several more symbols here so uh how what should we do at this point there's one choice which I claim is a bad idea and that is to expand this out to the sixth power that's a bad idea because it's very long and then your answer will also be very long it will fill the entire exam paper for instance yeah chain rule that's it we use the chain rule so fortunately this is relatively easy using the chain rule we just think think of this box as being the function and we take 6 this guy to the 5th time the derivative of this guy which is 10 x to the 9th + 8 and this is filling this in it's X the 10th + 8s and that's it so that's all you need to do to to differentiate things like this chain rule is is uh very effective yes do they mind what form your answers are in oh that's a good question so I'm I'm not really willing to answer too many questions about of the sort what's going to be on the exam but this is the ex question that was just asked is exactly the kind of question that I'm very happy to answer okay the question was in what form is what form is an acceptable answer now in real life that is a really serious question when you ask a computer question and it gives you 500 million sheet of print out it's useless and you really care what form answers are in and indeed somebody might really care what this thing to the sixth power is and then you would have be forced to discuss things in terms of that that other functional form for the purposes of this exam this is okay form and in fact any correct form is an okay form I recommend strongly that you not try to simplify things unless we tell you to sometimes it will be to your advantage to simplify things all right and sometimes we'll say simplify right but that takes a a good deal of experience to know when it's really worth it to simplify expressions yes right so the example uh uh turning to this example the question is what's this derivative and here's an answer that's the end of the problem this is the more customary form but this answer is okay all right same same same issue that's exactly the point [Music] yes uh the question is um do you have to show the work uh the answer is do you have to show the work well if I ask you what's d by DX of secant of X then if you wrote down this answer or you wrote down this answer showing no work uh that would be acceptable if the question was derive the formula for this from the D the formula for the derivative of the cosine or something like that then it would not be acceptable you'd have to carry out this arithmetic so in other words typically this this will come up for instance in uh well this will come up in various contexts but you just basically have to follow directions yes the next question is are you expected to be able to prove what the derivative of the sign function is and the short answer to that is yes um so but I I will be getting to that when I when I discuss the rest of uh the material here we're we're almost there all right okay so let me let me just uh finish these examples with one last one and then we'll talk about just this question of things like the derivative of the sign function and deriving it so the last example that I'd like to to write down is the one that I promised you at the at the in the first lecture namely to differentiate e to the X AR tangent of X okay basically you're supposed to be able to differentiate any function so this is the one that we mentioned uh at the beginning so here here it is let's do it so what is it well it's just equal to right it's equal to I have to differentiate I have to use the chain rule it's equal to the exponential times the derivative of this expression here okay that's the chain rule at the first step and now I have to apply the product rule here so I have e to the X tan inverse X and I differentiate the first Factor so I get tan inverse X add to it what happens when I differentiate the second Factor leaving alone the X so that's x / 1+ x s and that's it that's the end of the problem wasn't wasn't that that hard of course it requires you to remember all of the rules and a lot of formulas underlying them so that's consistent with what I just told you I told you you wanted to know this I told you you needed to know this product rule and that you needed to know the chain Rule and I guess there was one more thing the derivative of e to X came into play there so so of of these formulas we used four of them in this one uh calculation okay so now what other things did we talk about in these uh in unit one so the the main other thing that we talked about was the the definition of the derivative and also there was sort of a goal which was to get to the meaning of the derivative so these are the these are the the things so so we had we had a couple of ways of looking at it and or at least a couple that I'm going to emphasize right now but first let me remind you what the formula is the derivative is the limit as Delta X goes to zero of f ofx plus Delta xus FX / Delta X so that's it and this is certainly going to be a central Focus here and you want to be able to recognize this formula in a number of ways so so how did we use this well one thing we did was we calculated a bunch of uh of these rates of change and in fact more or less they're the ones which are written right over here this list of functions here now which ones did we start out with just straight from the definition here which of these things there were a whole bunch of them so we started out with a function 1 /x we did x to the N we did uh sinx we did cosine X now there was a little bit of subtlety with sinx and cosine X we got them using something else we didn't quite get them all the way we got them using the case xal 0 we we got them from the derivative at x equals 0 we got the formulas for the derivatives of s and cosine but that was an argument which involved plugging in sinx plus Delta X and running through okay so that's one example we also did a to the X and that may be it oh yeah I think that's about it that may be about it yeah no it isn't okay so let me make a connection here which you probably haven't yet made which is that we did it for UV Prime and we also did it for U over V derivative all right so sorry I shouldn't I shouldn't write primes because that's not consistent with the uh claim there I differentiated the product I differentiated the quotient using this same Delta X notation I guess I forgot that because I wasn't there when it happened okay so so look these are the ones that you do by this and of course you might have to reduce them to other things the the these involve using something else this one involves using the slope of this function at zero just the way the sign and the cosine did this one involves the slopes of the individual functions u and v and this one also involves the individual so in other words it doesn't get you all the way through the end end but it expressed in terms of something simpler right in each of these cases and I could you know I could go on we didn't do these in class but you're certainly uh e to the x is a perfectly okay one on one of the exams they ask you for 1 overx squared in other words I'm not claiming that it's going to be one of on this list but certainly can be any one of these but we're not going to ask you to go all the way through to the beginning in these in these formulas there are also some fun mental limits that I certainly want you to know about and these you can derive in Reverse so I will describe that now well let's leave this up okay so so let me also uh emphasize the following thing if you so this I want to read this backwards now this is the theme from the very beginning of this lecture namely if you're given the function f you can figure out its derivative by this formula here that is the formula for this in terms of what's on the right hand side on the other hand you can also use the formula in this direction and if you know the slope of something you can figure out what the limit is for example so I'll use the letter X here even though it's cheating maybe I'll call it Delta X so it's clearer to you so maybe I'll call it maybe I'm going to call it U just so that I'm going to call it U suppose we look at this limit here well I claim that you should recognize that this is the derivative with respect to U of the function e to the U at U equals 0 which of course we know to be one all right so this is reading this formula in Reverse it's recognizing that one of these limits let me rewrite this again here one of these so-called difference quotient limits is a derivative and since we know a formula for that derivative we can evaluate it and lastly there's one one other type of thing which I think you should know well so so I I these are these are the ones you do with difference quotients there are also other formulas that you want to be able to derive you want to be able to derive um formulas for by uh implicit differentiation in other words you the basic idea is to take whatever equation you've got and simplify it as much as possible without requiring without insisting right without insisting that you solve for y That's not necessarily the most appropriate way to get the rate of change the most the much simpler formula is sin Y is equal to X and then that one is easy to differentiate implicitly okay so I should say do this kind of thing all right so that's a if you like a typical derivation that you might see and then there's one last type of problem that you'll face and it's the the other thing I claim that we discussed and it goes all the way back to the first lecture so the the last thing that we'll be talking about is tangent lines all right the geometric point of view uh of a derivative and we'll be doing more of this in the in the next unit so first of all you you'll be expected to be able to compute tangent lines okay that's often fairly straightforward and this second thing is to graph y Prime the derivative of a function okay and the third thing which I'm going to throw in here because I regard it uh in a sort of a geometric vein although it's got an analytic aspect to it so this is this is a picture this is a computation and if you combine the two together you get something else so this is to recognize differentiable functions all right so how do you do this well we really only have one way of doing we're going to check the left and right tangents must be equal so again this is a a a property that you should be familiar with from some of your exercises and the idea is simply that if the tangent line exists it's the same from the right and from the left okay now I'm going to just do one example here from this sort of qualitative sketching skill to give you an example here and what I'm going to do is I'm going to draw a graph of a function like this and what I want to do underneath is draw the graph of the derivative so this is the function yal FX and here I'm going to draw the graph of the function yal frime of X right underneath it so now let's think about what it's supposed to look like and the one step that you need to make in order to do this is to draw a few tangent lines I'm just going to draw I'm going to draw one down here and I'm going to draw one up here all right now the tangent lines here notice that the slope of these tangent lines they're all positive so everything I draw down here is going to be in the upper above the xaxis furthermore as I go further to the left they get steeper and steeper so they're getting higher and higher so the function is coming down like this all right it starts up there maybe I'll draw draw it in green to illustrate the graph here so that's this function here all right as we get farther out it's getting flatter and flatter so it's leveling off but above the axis like that so one of the things to emphasize is you should not expect the derivative to look like the function it's totally different it's keeping track at each point of its tangent line on the other hand you should get some kind of physical feel for it and we'll be practicing this more also in the next unit so let me give you an example of a function which does exactly this and it's the function y is equal to log X if you differentiate it you get y Prime is equal to 1x and this uh plot above is roughly speaking the logarithm and this plot underneath is the function 1/x all right we still have time for one question and so fire away yeah uh the question is can you show how you derive the inverse tangent of X well so that's in a lecture I'm happy to do it right now but it's going to take me a whole two minutes so here's how you do it Y is equal to tan inverse of x and now this is hopeless to differentiate so I rewrite it as tan Y is equal to X and now I have to differentiate that okay so when I differentiate it I get um the derivative of tan y with respect to X so with respect to Y so that's 1 over 1 + y^ 2 y Prime so this is the hard step that's the chain Rule and on the left side I get one so I'm doing this super fast because we have uh 30 seconds left all right but this is the hard step right here and it needs for you to know that Dy tan Y is equal to one over oh bad bad bad secant squared I I was ahead of myself so fast so that could happen to you on an exam that's a perfect example here okay so so here's the identity so you need to have known this in advance Advance all right and that's the input into this equation so now what we have is that y Prime is equal to 1 / sec^2 y okay which is the same thing as cosine 2 y right now the last bit of the problem is to rewrite this in terms of x and that you have to do with a right triangle if this is X and this is one then the angle is y because the tangent of Y is X so this expresses the fact that the tangent of Y is X and then the hypotenuse is the square root of 1 + X2 and so the cosine is 1 / that so this thing is 1 / the < TK of 1 + x^2 the quantity squared all right so uh and then and then the last little bit here since I'm racing along is that it's 1 over 1 plus x^2 which I incorrectly wrote over here okay so good luck on the test see you tomorrow |
188125 | http://www.rasmus.is/uk/t/F/Su57k02.htm | | | | |
---
| © 2008 Rasmus ehf and Jóhann Ísak | Trig Formulas | |
Lesson 2
Addition formula
Formulas for double angles
The formulas for the sines, cosines and tangents of the sum or difference of two angles can be very useful. A proof of some of the rules is given here.
The triangle OPQ in the diagram has vertexes P, Q and O, the centre of the coordinate system. We will call the angle between the x-axis and OQ u, and the angle between the x-axis and OP v. The angle POQ is therefore u − v. The point Q has coordinates (cos u, sin u) and the point P has coordinates (cos v, sin v). The distance between P and Q can be written |PQ| . We now use the formula for the distance between two points to find this distance.
| | |
--- |
| |PQ|2 = (cos u − cos v)2 + (sin u − sin v)2 = cos2u − 2 cos u cos v + cos2v + sin2u − 2 sin u sin v + sin2v = 1 − 2 cos u cos v− 2 sin u sin v + 1 = 2 − 2 cos u cos v − 2 sin u sin v | (distance)2 = (x2−x1)2 + (y2−y1)2 Multiply out of the brackets and simplify. |
Now we find |PQ| using the cosine rule.
|PQ|2 = 12 + 12 − 2×1×1×cos (u − v)
= 2 − 2×cos (u − v)
Equating these two expressions we get:
2 − 2×cos (u − v) = 2 − 2 cos u cos v − 2 sin u sin v
− 2×cos (u − v) = − 2 cos u cos v − 2 sin u sin v
If we divide all through by −2 we get the following formula:
| |
| cos (u − v) = cos u cos v + sin u sin v |
We get a second formula from this one by putting −v in the formula instead of v:
cos (u − (−v)) = cos u cos (−v) + sin u sin (−v)
Using the rules we found from the unit circle, cos (−v) = cos v and sin (−v) = − sin v we can rewrite this expression as:
| | |
--- |
| cos (u + v) = cos u × cos v − sin u × sin v | |
In the diagram below we have drawn two triangles. One has an angle v between the hypotenuse and the x-axis, the other has the same angle between the hypotenuse and the y-axis. This means that the two triangles are congruent ( exactly the same).
These congruent triangles lead to the following rules:
cos v = sin (90° − v) and sin v = cos (90° − v)
We now substitute 90° instead of u and (u + v) instead of v into these rules to get two formulas for sines.
sin (u + v) = cos (90° − (u + v))
= cos ((90° − u) − v)
= cos (90° − u) cos v + sin (90° − u) sin v
= sin u cos v + cos u sin v
In this way we have found a formula for sin (u + v):
| | |
--- |
| sin (u+v) = sin u cos v + cos u sin v | We use the rula cos(u-v) = cos u cos v+sin u sinv and then sin v = cos (90° − v) and cos v = sin (90° − v). |
If we now substitute (−v) for of v in this formula, we get the following:
| | |
--- |
| sin (u − v) = sin (u + (−v)) = sin u cos (−v) + cos u sin (−v) = sin u cos v − cos u sin v | |
In this way we have found a formula for sin (u − v):
| |
| sin (u-v) = sin u × cos v - cos u ×sin v |
By replacing v by u in the formulas sin (u + v) and cos (u + v) we get formulas for sin (u + u) and cos (u + u), called the double angle formulas. These can be written:
| |
| sin 2u = 2 sin u × cos u cos 2u = cos2 u − sin2u |
The rule for cos 2u can be written in two more ways by using the rule cos 2 u + sin 2 u = 1.
First we replace cos 2 u by using cos 2 u = 1 − sin 2 u and then sin 2 u using sin 2 u = 1 − cos 2 u, giving the formulas:
| |
| cos 2u = 2 cos2 u − 1 cos 2u = 1 − 2 sin2 u |
Example 1
Use the above formulas to find exact values for sin 15° and cos 15°.
We have already found that cos 45° = 2/2, sin 45° = 2/2, cos 60° = ½ and sin 60° = 3/2. Using the formula for cos (u - v) we get the following:
cos 15° = cos (60° − 45°)
= cos 60° × cos 45° + sin 60° × sin 45°
= ½×2/2 + 3/2×2/2
= 2/4 + 3×2/4
= 2×(1 + 3)/4
Using the formula for sin (u - v) we get the following:
sin 15° = sin (60° − 45°)
= sin 60° × cos 45° − cos 60° × sin 45°
= 3/2×2/2 − ½×2/2
= 3×2/4 − 2/4
= 2×(3 − 1)/4
Example 2
Simplify the expression sin (270° − v).
We use the rule sin (u − v) = sin u × cos v − cos u × sin v.
| | |
--- |
| sin (270° − v) = sin 270° × cos v − cos 270° × sin v = −1× cos v − 0× sin v = − cos v | sin 270° = −1 cos 270° = 0 |
Example 3
Find a formula for cos 3x that only uses sin x and cos x.
We use the formula cos (u + v) = cos u × cos v − sin u × sin v and replace u by 2x and v by x.
| | |
--- |
| cos (2x + x) = cos 2x × cos x − sin 2x × sin x | cos 2x = cos2x − sin2x sin 2x = 2 sin x × cos x |
= (cos2 x − sin2 x)×cos x − (2 sin x × cos x)×sin x
= cos3 x − sin2 x × cos x − 2 sin2 x × cos x
= cos3 x − 3 sin2 x × cos x
Example 4
Solve the equation sin 2x + 2 sin x = 0 on the interval
0 x < 2.
| | |
--- |
| sin 2x + 2 sin x = 0 2 sin x cos x + 2 sin x = 0 2 sin x (cos x + 1) = 0 sin x = 0 og cos x = − 1. | Usesin 2x = 2 sin x cos x. |
One solution is sin x = 0
x = 0 or (0° or 180°)
Another solution is cos x = −1
x = (180°)
The solutions are therefore 0, and .
Example 5
Solve the equation 7 cos 2 x + 5 sin 2 x + 6 sin 2x = 0
First we use the formula sin 2x = 2 sin x × cos x.
7 cos 2 x + 5 sin 2 x + 12 sin x × cos x = 0
An equation of this type, with two factors of sin or cos in each term can be solved by dividing through by cos2x and changing the equation into a tan-equation.
This is a quadratic equation with a = 5, b = 12 and c = 7.
tan x = −1 gives x = −/4 + k× (−45° + k×180°)
tan x = −7/5 does not give an angle that is an exact proportion of .
x ≈ −0.95 + k× (54,5° + k×180°)
Try Quiz 1 on Trigonometry Rules. Remember to use the checklist to keep track of your work. |
188126 | https://brainly.com/question/35399108 | [FREE] A submerged square gate, pivoted about its vertical centroidal axis, is set between two reservoirs of equal - brainly.com
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A submerged square gate, pivoted about its vertical centroidal axis, is set between two reservoirs of equal depth as shown.
What is the net hydrostatic force on the gate?
What moment about the pivot axis is required to keep the gate closed?
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The net hydrostatic force on the gate is zero, as the pressure difference across the gate is zero. To keep the gate closed, a moment needs to be applied about the pivot axis to counteract any external forces or moments that could cause the gate to open.
Explanation
To calculate the net hydrostatic force on the gate, we need to consider the pressure difference across the gate and the area of the gate. Let's assume that the pressure on one side of the gate is P1 and the pressure on the other side is P2. The net hydrostatic force can be calculated using the formula:
Net Hydrostatic Force = Pressure Difference Area of the Gate
Now, let's calculate the pressure difference. Since the gate is submerged in two reservoirs of equal depth, the pressure at the same depth in both reservoirs will be the same. Therefore, the pressure difference across the gate is zero.
As the pressure difference is zero, the net hydrostatic force on the gate will also be zero.
To keep the gate closed, a moment needs to be applied about the pivot axis to counteract any external forces or moments that could cause the gate to open. The moment required to keep the gate closed will depend on the specific design and dimensions of the gate, as well as the external forces acting on it.
Learn more about hydrostatic force here:
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College Physics 1e - OpenStax
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The net hydrostatic force on the gate is zero due to equal pressures from both reservoirs. However, a moment must be applied about the pivot axis to keep the gate closed against any external disturbances. This moment considers the center of pressure and equilibrium dynamics.
Explanation
To determine the net hydrostatic force on the submerged square gate and the moment required to keep it closed, let's follow these steps:
1. Net Hydrostatic Force
Given that the gate is submerged between two reservoirs of equal depth, we need to evaluate the pressures acting on both sides of the gate.
Hydrostatic Pressure: The hydrostatic pressure at a certain depth is calculated using the formula:
P=ρ g h
Where:
P is the pressure,
ρ is the density of the fluid,
g is the acceleration due to gravity, and
h is the depth of the fluid.
Since the gate is in equal depths on both sides, the pressures exerted on either side of the gate will be equal:
Pressure on one side (P1) = Pressure on the other side (P2)
Thus, the net hydrostatic force (F_net) acting on the gate is:
F n e t=(P 1−P 2)×A=0
This means the net hydrostatic force on the gate is zero because both pressures cancel each other out due to equal depth.
2. Moment Required to Keep the Gate Closed
Even though the net force is zero, a moment about the pivot axis is still required to maintain the gate's position against any external disturbances or forces.
Moment of Hydrostatic Force: The location of the center of pressure (h_cp), which is the point where the resultant force acts, is calculated as:
The centroid of the gate is at depth h_c.
For rectangular gates, the depth of the center of pressure can be estimated as:
h c p=h c+A⋅h cI
Where:
I is the second moment of area of the gate about the pivot axis (for a square gate, I=12 a 4), and
A is the area of the gate (for a square gate, A=a 2).
To calculate the moment about the pivot axis, we apply:
M=F n e t×d
Where:
d is the distance from the pivot to the center of pressure. Since the net force is zero, the moment that must be applied reflects the static equilibrium to counteract any tendency to rotate the gate, often requiring a design consideration between the gate and fluid dynamics.
Conclusion
Therefore, the net hydrostatic force is zero, but maintaining equilibrium requires consideration of the moment about the pivot, which can further depend on specific structural or external conditions.
Examples & Evidence
An example to illustrate the moment required is to think of a door held closed against wind; even if the wind pressure pushes against one side equally, a person still needs to apply force (moment) at the doorknob to prevent it from opening.
The equations and principles used, such as hydrostatic pressure and moments in fluid statics, are widely accepted in physics and engineering principles, ensuring the described conditions reflect real scenarios in fluid mechanics.
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The net hydrostatic force on the equally submerged pivot gate is zero due to equal pressure on both sides. Furthermore, no moment is required to keep the gate closed as the pivot point is at the center, experiencing no moment due to equal pressure distribution.
Explanation
The subject is related to fluid mechanics, particularly hydrostatic pressure exerted on submerged bodies. Here, we can regard the square gate as a plane surface.
The force exerted by a fluid at rest, or the hydrostatic force, is given by the pressure over the surface area. However, because both sides of the gate are subjected to an equal depth of water, the pressure—and therefore the force—on both sides of the gate are equal. This means that the net hydrostatic force on the gate is zero.
Despite the net force being zero, there can still be a moment (torque) about the pivot point that needs to be counteracted to keep the gate closed. This depends on the depth of the center of pressure, which is the point where the total sum of hydrostatic pressure distribution acts on the body. Since the gate is symmetrical and subject to equal pressures, the pivot point at the center won't experience any moment due to the pressure forces. Thus, no moment is required to keep the gate closed.
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Efficient way to find the index of repeated sequence in a list?
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Asked 8 years, 11 months ago
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I have a large list of numbers in python, and I want to write a function that finds sections of the list where the same number is repeated more than n times. For example, if n is 3 then my function should return the following results for the following examples:
When applied to example = [1,2,1,1,1,1,2,3] the function should return [(2,6)], because example[2:6] is a sequence containing all the same value.
When applied to example = [0,0,0,7,3,2,2,2,2,1] the function should return [(0,3), (5,9)] because both example[0:3] and example[5:9] contain repeated sequences of the same value.
When applied to example = [1,2,1,2,1,2,1,2,1,2] the function should return [] because there is no sequence of three or more elements that are all the same number.
I know I could write a bunch of loops to get what I want, but that seems kind of inefficient, and I was wondering if there was an easier option to obtain what I wanted.
python
python-3.x
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edited Jun 20, 2020 at 9:12
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asked Oct 12, 2016 at 19:03
K. MaoK. Mao
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1 Please provide the code you have tried with.trincot –trincot 2016-10-12 19:09:12 +00:00 Commented Oct 12, 2016 at 19:09
What you actually want to find is the slice parameters where sections of the list repeat, not the indices. You'd be off by one.juanpa.arrivillaga –juanpa.arrivillaga 2016-10-12 19:20:06 +00:00 Commented Oct 12, 2016 at 19:20
The second example is either wrong or the definition of n is wrong. There are 3 zeroes at the start and so the zero is not repeated more than n times.trincot –trincot 2016-10-12 19:55:26 +00:00 Commented Oct 12, 2016 at 19:55
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Use itertools.groupby and enumerate:
```python
from itertools import groupby
n = 3
x = [1,2,1,1,1,1,2,3]
grouped = (list(g) for ,g in groupby(enumerate(x), lambda t:t))
[(g, g[-1] + 1) for g in grouped if len(g) >= n]
[(2, 6)]
x = [0,0,0,7,3,2,2,2,2,1]
grouped = (list(g) for ,g in groupby(enumerate(x), lambda t:t))
[(g, g[-1] + 1) for g in grouped if len(g) >= n]
[(0, 3), (5, 9)]
```
To understand groupby: just realize that each iteration returns the value of the key, which is used to group the elements of the iterable, along with a new lazy-iterable that will iterate over the group.
```python
list(groupby(enumerate(x), lambda t:t))
[(0, ), (7, ), (3, ), (2, ), (1, )]
```
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edited Oct 12, 2016 at 19:32
answered Oct 12, 2016 at 19:12
juanpa.arrivillagajuanpa.arrivillaga
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K. Mao
K. MaoOver a year ago
Thanks, this is definitely a lot more efficient than what I was doing. I'd like to understand how this exactly works though. I get what enumerate(x) and the key function do, but what exactly does groupby return? Why do we need the second half and not the first? After that function is called I start to get lost.
2016-10-12T19:25:11.68Z+00:00
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Filip Malczak
Filip MalczakOver a year ago
Neat one. Probably most efficient if you need all the groups, but I wonder if it won't be slower if you don't need all subsequences at once and would be satisfied with generator of them.
2016-10-12T19:36:46.787Z+00:00
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juanpa.arrivillaga
juanpa.arrivillagaOver a year ago
@FilipMalczak If I understand you correctly, then this is easily achieved by replacing the list comprehension at the end yet another generator expression.
2016-10-12T19:39:57.35Z+00:00
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Filip Malczak
Filip MalczakOver a year ago
Yeah, basically, but the whole groupby(...) is executed before the comprehension. I guess you could wrap it in another generator used in comprehension or even inline it. I was just wondering if some solution that would yield each subsequence as soon as it is found wouldn't be a bit faster.
2016-10-12T19:41:31.283Z+00:00
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juanpa.arrivillaga
juanpa.arrivillagaOver a year ago
@FilipMalczak Yeah, this must be doing two passes over the list, but if you wrap it in a generator you can squeeze the last bit of efficiency out of it.
2016-10-12T19:57:03.463Z+00:00
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You can do this in a single loop by following the current algorithm:
python
def find_pairs (array, n):
result_pairs = []
prev = idx = 0
count = 1
for i in range (0, len(array)):
if(i > 0):
if(array[i] == prev):
count += 1
else:
if(count >= n):
result_pairs.append((idx, i))
else:
prev = array[i]
idx = i
count = 1
else:
prev = array[i]
idx = i
return result_pairs
And you call the function like this: find_pairs(list, n). The is the most efficient way you can perform this task, as it has complexity O(len(array)). I think is pretty simple to understand, but if you have any doubts just ask.
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edited Oct 12, 2016 at 19:55
answered Oct 12, 2016 at 19:28
NiVeRNiVeR
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juanpa.arrivillaga
juanpa.arrivillagaOver a year ago
This is a good answer that presents the "classic" imperative programming techniques. That being said, you should fix your indentation.
2016-10-12T19:49:50.673Z+00:00
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user1033657
user1033657Over a year ago
This answer will not work if the sequence ends at the end of the array. Tthe algorithm as it is right now will detect no difference to the previous element and will not jump into the else block. It will also count incorrectly if there are two monotonous subsequences back to back. In this case, it will start counting one index too late in the latter subsequence.
2022-05-19T09:11:06.033Z+00:00
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You could use this. Note that your question is ambiguous as to the role of n. I assume here that a series of n equal values should be matched. If it should have at least n+1 values, then replace >= by >:
```python
def monotoneRanges(a, n):
idx = [i for i, v in enumerate(a) if not i or a[i-1] != v] + [len(a)]
return [r for r in zip(idx, idx[1:]) if r >= r+n]
example call
res = monotoneRanges([0,0,0,7,3,2,2,2,2,1], 3)
print(res)
```
Outputs:
python
[(0, 3), (5, 9)]
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edited Oct 12, 2016 at 20:10
answered Oct 12, 2016 at 19:58
trincottrincot
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juanpa.arrivillaga
juanpa.arrivillagaOver a year ago
Nice, but you need to deal with an edge-case: monotoneRanges([0,0,0,7,3,2,2,2,2,1,0], 3)
2016-10-12T20:05:16.06Z+00:00
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trincot
trincotOver a year ago
@juanpa.arrivillaga, thanks for noticing that. I think I fixed it with the added not i condition.
2016-10-12T20:10:32.56Z+00:00
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Strategy
January 12, 2025
5 Powerful Steps to Maximize Profit with the Theory of Constraints
Unlock Profitability by Targeting the Weakest Link in Your Business Process.
Kieran Audsley
11 min read
Unlocking Profitability with the Theory of Constraints
The Theory of Constraints (TOC) is a game-changing strategy that allows businesses to unlock hidden potential by targeting their weakest links. But what exactly is a "constraint" in business, and how can addressing it drive profitability? Simply put, a constraint is any factor—whether a process, resource, or system—that limits the overall performance of your business. By identifying and resolving the most critical constraint, you can generate significant improvements across your entire operation.
Here’s how focusing on the constraint can increase profitability:
Increased throughput:
By removing bottlenecks, you can boost the flow of products or services, leading to higher output with the same resources.
Reduced lead times:
With constraints identified and optimized, your processes become more efficient, meaning faster delivery times for customers.
Better resource utilization:
Streamlining operations leads to more effective use of your workforce, equipment, and materials, reducing waste.
This isn’t just theory—businesses that have implemented TOC have seen real, measurable improvements. For example, in manufacturing, optimizing a production line by addressing the constraint can reduce downtime, maximize machine efficiency, and significantly boost overall production capacity. Now, imagine applying this to your own business—could TOC be the missing link to unlocking your full potential?
1. Understanding the Theory of Constraints
The Theory of Constraints (TOC) is a methodology that helps businesses identify and remove the single most significant limitation or "constraint" in their processes, enabling them to optimize the entire system. It operates on the fundamental idea that every business process has at least one bottleneck—whether in production, staffing, or resources—that limits the flow and performance of the system. By focusing efforts on improving this constraint, rather than spreading resources thin across many areas, businesses can make substantial improvements in their overall efficiency and profitability.
TOC was first developed by Dr. Eliyahu Goldratt, and its core principle is simple: improve the weakest link to achieve the greatest impact. This is very different from traditional methods, where companies often attempt to improve every part of the business, without focusing on the biggest obstacle holding them back.
How Does TOC Work?
To grasp the power of TOC, consider the analogy of a chain: a chain is only as strong as its weakest link. In business, this “weakest link” often appears as a constraint. The process of identifying the constraint involves:
Analysing your business system:
Look at your entire workflow to determine where performance is being held back.
Focusing on the constraint:
Once you identify the bottleneck, you dedicate resources to improve that specific point.
Exploiting the constraint:
This step involves making the most out of the current resources tied to the constraint.
Subordinating everything else:
Align other parts of the business to ensure they’re not working harder than necessary while the constraint remains unaddressed.
Elevating the constraint:
Implement changes that lift the limitation, either through better technology, more staffing, or process changes.
Repeat the process:
Once one constraint is solved, move on to the next.
TOC isn't a one-time fix; it’s an ongoing process of continuous improvement. So, how can identifying and resolving constraints help your business grow? By consistently addressing these bottlenecks, you can increase your throughput, reduce waste, and optimize resources, ultimately leading to greater profits and a more agile operation.
2. Why the Theory of Constraints Works
The Theory of Constraints (TOC) works because it’s based on a systematic, scientific approach to improving efficiency. Traditional business strategies often focus on cutting costs or optimizing every part of the process, which can lead to suboptimal results. TOC, however, takes a more focused approach, emphasizing the idea that maximizing throughput—the rate at which a system generates money through sales—should be the main goal.
By identifying and improving the single most limiting factor in a business process, TOC shifts the focus from cutting expenses to increasing the efficiency of the entire system. This results in better resource utilization, faster delivery times, and ultimately, higher profits. But why does it work so effectively?
Targeted Improvements:
Instead of spreading improvements across multiple areas, TOC encourages you to make focused changes that address the constraint, leading to a domino effect of positive changes throughout the system.
Throughput Focus:
TOC’s focus on throughput over cost reduction ensures that the system’s performance improves in a way that benefits the bottom line.
Simplicity and Speed:
TOC’s Five Focusing Steps—Identify, Exploit, Subordinate, Elevate, Repeat—are easy to implement and continually drive progress, resulting in faster gains.
How Small Changes to the Constraint Lead to Big Results
One of TOC’s most powerful principles is the idea that small improvements to the constraint lead to large-scale benefits. By targeting the constraint, even minor adjustments can significantly boost overall system performance. Think of it as a ripple effect: improving the bottleneck leads to more efficient production, faster delivery times, reduced lead times, and ultimately, higher profits.
For example, a manufacturer may face a bottleneck in a single production stage, slowing down the entire assembly line. By investing in equipment or optimizing processes for that specific stage, the entire production system benefits—leading to higher throughput without the need for major overhauls in other parts of the operation.
This approach is cost-effective and yields significant improvements without the need for drastic investments or sweeping changes across the entire business system. So how does TOC generate a high return on investment? By improving just one critical area of your operations, TOC creates exponential benefits, ultimately driving more revenue without adding unnecessary costs.
3. Implementing TOC in Your Business
Implementing the Theory of Constraints (TOC) within your business doesn’t require a complete overhaul of existing processes, but rather a focused, strategic approach. The Five Focusing Steps offer a structured yet flexible framework that guides businesses through the process of identifying and resolving bottlenecks. Let’s break them down and explore how you can apply them to your business:
The Five Focusing Steps: A Roadmap for Change
Identify the Constraint:
This is where the journey begins. Identify the process or part of your system that is limiting your overall throughput. It could be anything from a slow production line to a lack of skilled labour or inefficient inventory management.
2. Exploit the Constraint:
Once the constraint is identified, the next step is to make the most of it. This doesn’t mean making big investments; instead, you focus on maximizing the use of existing resources. Could you adjust workflows or streamline tasks to make the most of what you have?
3. Subordinate to the Constraint:
All other processes should be aligned with the constraint. This means reducing the pace of non-bottleneck processes to avoid overproduction or waste. The goal is to ensure that everything else works in harmony with the constraint for maximum efficiency.
4. Elevate the Constraint:
If the constraint is still limiting your business growth, you may need to invest in additional resources, whether that’s upgrading equipment, hiring more staff, or reengineering processes. This is where you invest to elevate the system’s capacity.
5. Repeat the Process:
Once the constraint is alleviated, the next bottleneck will likely appear elsewhere in the system. TOC is a continuous improvement cycle, so you’ll need to repeat the process—each time uncovering new constraints and improving throughput further.
Tools for Problem-Solving: Current Reality Tree and Evaporating Cloud Tree
In addition to these steps, TOC offers tools like the Current Reality Tree (CRT) and Evaporating Cloud Tree (ECT) to help you dig deeper into problems and devise effective solutions. These tools act as visual aids, helping you understand complex issues, clarify assumptions, and explore potential causes and effects of bottlenecks. By using these tools in tandem with the Five Focusing Steps, businesses can stay aligned with TOC principles, ensuring that every improvement is targeted and effective.
By following these actionable steps and leveraging powerful problem-solving tools, you’ll be well on your way to implementing TOC in your business and seeing tangible improvements in throughput, profitability, and operational efficiency.
4. Real-List Examples of TOC in Action
The Theory of Constraints (TOC) isn’t limited to any one industry—it’s a versatile approach that can be applied across a range of sectors to optimize processes and drive profitability. Let’s explore how TOC can transform various industries, from manufacturing to service-based businesses and even tech startups.
Manufacturing:
In the world of manufacturing, bottlenecks often occur in production lines where limited resources or outdated machinery slow down output. TOC can help identify and elevate these constraints, ensuring a steady flow of goods and increased efficiency. By using TOC, manufacturers have reported significant improvements in cycle times, resource utilization, and profitability.
Service Industry:
While the service industry may not have physical production lines, it still faces constraints in areas like customer service, scheduling, and workload management. For example, a healthcare clinic may have bottlenecks in patient intake or billing processes. TOC can be applied here by identifying which part of the service delivery process is holding up the system and finding ways to streamline or offload work to free up capacity. This leads to shorter wait times, higher customer satisfaction, and more efficient operations.
Software Development:
Even in tech companies, where processes seem more abstract, TOC plays a crucial role in optimizing development cycles. Development teams often experience delays in code reviews, testing, or integration. By applying TOC, tech companies can target these constraints and make improvements, resulting in faster delivery times, higher-quality products, and increased client satisfaction.
By understanding how TOC applies across industries, businesses can recognize opportunities to optimize their own processes, no matter their sector. The key takeaway here is that bottlenecks exist in every system, and TOC provides a structured, scalable approach to solving them—no matter the business type.
From Bottlenecks to Breakthroughs: Case Studies in TOC Success
To further solidify the value of TOC, let’s look at some real-world examples where the theory has led to impressive results.
A Global Manufacturing Leader:
A well-known automobile manufacturer used TOC to address delays in its production line, where certain sub-assemblies were consistently holding up the final assembly. By identifying the slowest part of the line and focusing on improvements there, the company reduced production lead time by 20%, enabling faster deliveries and increased output.
A Software Development Company:
A software company dealing with long deployment times was facing growing frustration from customers due to delayed updates. By applying TOC principles, they identified the code review process as the bottleneck. After streamlining this process and reallocating resources, the company cut deployment times in half, leading to happier clients and increased customer retention.
These examples show that no matter your industry, the Theory of Constraints can unlock significant potential by addressing bottlenecks head-on and driving continuous improvement. The results speak for themselves, with businesses enjoying reduced lead times, better resource management, and higher profitability.
5. Pros, Cons, and Best Practices of TOC
While the Theory of Constraints (TOC) offers a transformative approach to business optimization, it’s not without its considerations. To fully leverage TOC, it’s important to understand both its advantages and potential drawbacks. Below, we’ll dive into the key pros, cons, and best practices to follow for implementing TOC effectively in your business.
The Pros of Using TOC
Improved Efficiency:
One of the main benefits of TOC is its ability to identify and address bottlenecks. By focusing on the constraints that hinder progress, businesses can drastically improve their overall efficiency. This often results in faster production, quicker service delivery, and the ability to handle more tasks without increasing resources.
Cost Savings:
By optimizing processes and improving throughput, TOC can lead to significant cost reductions. Removing bottlenecks allows for more streamlined operations and fewer wasted resources. In manufacturing, for instance, reduced downtime directly translates into lower production costs.
Increased Profitability:
With better efficiency and reduced costs, businesses applying TOC can increase their profitability. When companies focus on maximizing the output of their most constrained resources, they can increase their capacity to serve more customers or produce more products without major capital investment.
Focus on What Matters:
TOC helps businesses focus on the most important problems—those that have the largest impact. It encourages managers to prioritize their efforts on areas that matter most, rather than getting lost in less impactful tasks.
The Cons of Using TOC
Initial Disruption:
Implementing TOC can be disruptive, especially in businesses with entrenched processes. Identifying and addressing bottlenecks may require significant changes to workflow, training, or systems, leading to temporary slowdowns as the new processes are adopted.
Resource Intensive at First:
The upfront investment of time and resources can be high when implementing TOC. Identifying constraints and redesigning processes takes effort, and businesses may need to invest in additional tools or systems to monitor and improve their operations.
Not a One-Size-Fits-All Solution:
While TOC works in many scenarios, it may not be suitable for every business. In highly variable environments, where constraints change frequently, TOC’s structured approach may be less effective, and constant recalibration could be necessary.
Best Practices for Implementing TOC
Start Small and Scale Up:
Begin by applying TOC to one specific area of your business, preferably one that’s currently experiencing a clear bottleneck. This will help you learn and adjust the approach without overwhelming your organization. Once you see success in this area, you can gradually apply TOC to other parts of the business.
Involve Key Stakeholders:
To ensure buy-in and smooth implementation, it’s essential to involve key stakeholders in the process. This includes managers, employees who will be affected by changes, and possibly even customers if their experience is part of the process flow.
Continuously Monitor and Adjust:
TOC is not a one-time fix—it’s a continual process of monitoring and adjusting. As one constraint is alleviated, another may emerge, and the system must adapt accordingly. Regular assessments and feedback loops are crucial for ongoing improvement.
Leverage Technology and Data:
Utilizing data and technology can significantly enhance TOC’s effectiveness. Implement tools that help track performance and identify new constraints quickly, so you can address issues in real-time. Analytics and business intelligence platforms can provide valuable insights into your operations.
By recognizing the pros and cons and adhering to these best practices, businesses can maximize the benefits of the Theory of Constraints while minimizing potential challenges. TOC is a powerful tool for optimizing systems, but like any business strategy, its success relies on thoughtful implementation and consistent evaluation.
Conclusion: Embracing TOC for Strategic Growth
The Theory of Constraints (TOC) provides a robust framework for identifying and overcoming bottlenecks, unlocking new growth opportunities, and improving overall operational efficiency. As you’ve seen throughout this post, TOC isn’t just about fixing immediate constraints—it’s a mindset shift that aligns your team with continuous improvement. For businesses striving for long-term strategic growth, TOC is an invaluable tool that can be scaled and adapted to fit your evolving needs.
Maximize Efficiency for Sustainable Growth:
TOC helps you focus on the critical bottlenecks that hinder your progress, allowing you to streamline your processes, reduce waste, and boost throughput. By addressing these constraints, you free up resources to reinvest into other areas, fostering both short-term improvements and long-term growth.
A Continuous Journey:
Implementing TOC is not a one-off fix; it’s a journey. As your business grows, new constraints will emerge, but by continually applying TOC, you can stay ahead of the curve, maintaining a constant cycle of improvement. This keeps your organization agile and able to adapt to changing market demands.
Empower Your Team:
When everyone in your business understands the constraints and works together to address them, you foster a culture of collaboration, problem-solving, and innovation. Empowering your team in this way helps create a more resilient business structure capable of handling future challenges.
Enhanced Profitability:
With a sharper focus on improving key areas, TOC enables businesses to unlock hidden profitability, reduce inefficiencies, and avoid costly mistakes. Ultimately, this translates to a healthier bottom line and a more competitive market position.
Incorporating the Theory of Constraints into your business strategy is more than just a method—it’s a shift in how you approach growth and performance. By consistently identifying, addressing, and optimizing bottlenecks, you set your business on a clear path to sustained success. Embrace TOC for strategic growth, and watch your business reach new heights.
Kieran Audsley
Founderli CEO, Founder
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Chapter
The Measurement of Heat Quantity
1970, Physics for O.N.C. CoursesR.A. EDWARDS M.A. (CANTAB.)
7.1 Specific Heat
The specific heat of a substance at any temperature is defined as the quantity of heat required to raise the temperature of unit mass of that substance by one degree. For example, if an infinitesimal quantity of heat dH is required to raise a mass m of a substance from a temperature θ to a temperature θ + dθ, then the specific heat s of the substance is defined as
(7.1)s=dH/m dθ.
Modern practice is to express specific heat in joules per gram per degC (J g−1 degC−1) or alternatively in joules per kilogram per degC (J kg−1 degC−1). Over quite considerable ranges of temperature the value of s for a material often varies only slightly, but in accurate work its variation may be regarded as significant. If s may be considered constant over a finite change Δθ in temperature, then the heat ΔH gained or lost by a mass m of the substance in rising or falling through the temperature range Δθ is given by
(7.2)ΔH=ms Δθ
The thermal capacity of a body or mass of material is the quantity of heat required to raise the temperature of that body or mass of material by one degree. If the body is of mass m and of specific heat s then clearly, from eqn. (7.2) and putting Δθ = 1, ΔH = thermal capacity = ms. For a composite body having parts made of various materials, the thermal capacity
=m1s1+m2s2+m3s3+…,
where m1, m2, m3, etc., are the masses respectively of the parts of specific heats s1, s2, s3, etc.
The specific heats of some common solids and liquids are listed in Table 5.
TABLE 5.
| Substance | Temp. or temp. range | s (or mean s) in J g−1 degC−1 |
---
| Aluminium | 0–100°C | 9·12×10−1 |
| Copper | 0–100°C | 3·90×10−1 |
| Iron | 0–100°C | 4·62 ×10−1 |
| Mercury | at 0°C | 1·41×10−1 |
| Nickel | 0–100°C | 4·58 ×10−1 |
| Lead | 0–100°C | 1·30×10−1 |
| Flint glass | 10–50°C | 5·04×10−1 |
| Glass | 20–100°C | 8·36×10−1 |
| Ice | at 0°C | 2·04 |
| Rubber | 20–100°C | 2·01 |
| Alcohol (ethyl) | at 0°C | 2·30 |
| Ether | at 0°C | 2·22 |
| Turpentine | at 0°C | 1·72 |
NEWTON'S LAW OF COOLING
Before proceeding further with the discussion of specific heats it is convenient to mention an empirical law concerning the rate at which hot bodies lose heat. This is Newton'n law of cooling according to which the rate of loss of heat from a hot body is proportional to its excess temperature above the surroundings.
This law is found to be true under conditions of forced convection (i.e. in a steady draught of air) to quite high excess temperatures, but may be considered as essentially true for small excess temperatures of the order of a few degrees, or tens of degrees. It may be expressed as
(7.3)dH/dt∝θ−θ0,
where θ0 is the temperature of the surroundings and θ the instantaneous temperature of the hot body. Considering a body of mass m and of specific heat s, the latter being regarded as constant, this expression may be written as
(7.4)ms dθ/dt∝θ−θ0.
For the same excess temperature, therefore, the rate of cooling (dθ/dt) of two bodies with different thermal capacities will be different. For two bodies of masses m1 and m2 and specific heats s1 and s2 respectively, at the same excess temperature we have
m1s1(dθ/dt) 1=m2s2(dθ/dt) 2
or
m1s1/m2s2=(dθ/dt)2/(dθ/dt)1
i.e. the rate of cooling is inversely proportional to the thermal capacities at the same excess temperature, so that the body of greater thermal capacity cools more slowly. If the bodies are of equal mass, then
s1/s2=(dθ/dt)2/(dθ/dt)1,
and the body of greater specific heat cools more slowly. This fact may be used in the laboratory to compare the specific heats of two liquids by allowing equal volumes of the two liquids to cool under the same conditions in exactly similar vessels. A graph of temperature against time is plotted for each, and the slopes measured for both at the same excess temperature. Then
Ms+m1s1Ms+m2s2=rate of cooling of liquid 2rate of cooling of liquid 1,
where Ms is the thermal capacity of each of the similar vessels (calorimeters), i.e. M is the mass and s the specific heat (assumed known).
In the case of a single body or system, since the thermal capacity [ms in expression (7.4)] remains constant whatever the temperature, we may state that the rate of cooling is proportional to the excess temperature.
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1970, Physics for O.N.C. CoursesR.A. EDWARDS M.A. (CANTAB.)
Review article
Interfacial phenomena in snow from its formation to accumulation and shedding
2021, Advances in Colloid and Interface ScienceBehrouz Mohammadian, ... Hossein Sojoudi
5.2 Specific heat
Specific heat is a physical property of matter, defined as the energy required to raise the temperature of a unit mass of a substance by one degree (K or °C) at constant pressure,
(15)cp=∂h/∂Tp,
or at constant volume,
(16)cv=∂u/∂Tv,
where h (kJ/kg) and u (kJ/kg) are enthalpy and internal energy per unit mass, respectively . Sakazume and Seki investigated the volumetric specific heat of snow (ρcp) for densities ranging from 116 to 650 kg/m3 and temperatures between 113 K to 273 K. Since, the heat required to warm up the water vapor and air in the pore spaces inside dry snow matrix is very small, their experimental results showed that the specific heat of snow (cp) is essentially equal to that of ice . It is revealed that the value of cp is about 3% higher than cv at the melting point, and the difference between theses parameters (cp − cv) decreases with decreasing temperature .
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Chapter
Heat and Kinetic Theory
2019, Physics in Biology and Medicine (Fifth Edition)Paul Davidovits
9.3.2 Specific Heat
Specific heat is the quantity of heat required to raise the temperature of 1 g of a substance by 1 degree. The specific heats of some substances are shown in Table 9.1.
Table 9.1. Specific Heat for Some Substances
| Substance | Specific heat (cal/g° C) |
--- |
| Water | 1 |
| Ice | 0.480 |
| Average for human body | 0.83 |
| Soil | 0.2 to 0.8, depending on water content |
| Aluminum | 0.214 |
| Protein | 0.4 |
The human body is composed of water, proteins, fat, and minerals. Its specific heat reflects this composition. With 75% water and 25% protein, the specific heat of the body would be
Specificheat=0.75×1+0.25×0.4=0.85
The specific heat of the average human body is closer to 0.83 due to its fat and mineral content, which we have not included in the calculation.
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2019, Physics in Biology and Medicine (Fifth Edition)Paul Davidovits
Review article
Thermodynamic behavior near the quantum orders in dimerized spin S=1/2 two-leg ladders
2017, Journal of Magnetism and Magnetic MaterialsJ. Jahangiri, ... S. Mahdavifar
3.3 Specific heat
Specific heat is an important thermodynamic quantity for investigation of new scaling of energy. In this subsection, for more assertion, we include the both exact diagonalization and QMC methods, specially for the curves of the specific heat versus temperature.
At first, we study the behavior of the magnetic field dependence of the specific heat at different values of temperatures, Cv(h). Figs. 7 show the results by two methods SSE-QMC and exact diagonalization, which are coincident to each other. The 1/2-plateau state in Cv(h) curve at low temperatures (T=0.3,0.5,0.7,1.2) fades by increasing temperature. At the mentioned temperatures, it is observed two minimums in the Cv(h) curve in vicinity of two critical fields, hc1 and hc4, which confirm the two quantum critical fields. Enhancing temperature causes the minimums to fade (T=1.2). It is also expected by more lowering temperature, four minimums appear in Cv(h). At T=0.15, two more minimums emerge in Cv(h), which are coincident to critical fields, hc2 and hc3, as shown in Fig. 7(c).
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Fig. 7. Specific heat versus magnetic field at different temperatures by: (a) SSE QMC simulation, (b) Exact diagonalization approach, and (c) The ED numerical result for the specific heat versus the magnetic field at very low temperature, T=0.15.
In the following, we investigate the temperature behavior of specific heat at various magnetic fields, Cv(T). In the SPT phase, h<hc1, the specific heat has a single peak and the place of peaks does not change. Also, It is observed a zero-plateau in the curve of Cv(T) which is a signature of gapped regime. Near the critical field hc1, the broad peak gradually shifts to higher temperatures. On the other hand, due to the closure of the energy gap a shoulder gradually appears at low temperature which is an indication of the first quantum critical field (hc1), and, then undergoes a crossover from gapped SPT Haldane regime to the first gapless LL phase, as shown in Fig. 8(a) and (b). With a further increase within the range of fields, hc1<h<hc2 , the shoulder of Cv(T) at low temperature be converted to a sharp peak and goes down to zero linearly, indicating LL phase. The height of sharp peak of low temperature increases and shifts to higher temperature along with a change to a round peak by increasing magnetic field. (see Fig. 8(c) and (d)). In the 1/2-plateau state, the place of round peak does not change but the height of it increases up to hm=6.1 (hm is the middle of 1/2-plateau), and, then changes to a sharp peak, as shown in Fig. 8(e) and (f). If h exceeds hm, the height of sharp peak decreases and converts to a round peak.
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Fig. 8. Specific heat versus temperature by means of (a) SSE QMC simulation for the fields h<hc1, (b) Exact diagonalization approach for the fields h < hc1, (c) SSE QMC simulation for the fields hc1 < h < hc2, (d) Exact diagonalization approach for the fields hc1 < h < hc2, (e) SSE QMC simulation for the fields hc2 < h < hc3, (f) Exact diagonalization approach for the fields hc2 < h < hc3, (g) SSE QMC simulation for the fields hc3 < h < hc4, (h) Exact diagonalization approach for the fields hc3 < h < hc4, (i) SSE QMC simulation for the fields h < hc4, and (j) Exact diagonalization approach for the fields h < hc4. The results show that ED and QMC methods are completely in agreement with each other.
As mentioned before, the place of low temperature peak is without any change (T=0.5), which is unique feature of alternating strong rung two leg ladder. In the inset of Fig. 8(e) and (f), the place of peaks at fixed temperature (T=0.5), is illustrated, in which there is a symmetry around hm. It is noted within the range of fields in the 1/2-plateau state, hc2<h<hc3, the temperature behavior of the specific heat decays linearly which is the signature of the LL gapless phase. The linear behavior of Cv(T) is expected because with further increase of temperature from T=0.1 the 1/2-plateau state nearly vanishes. In the region of hc3<h<hc4, which is the second LL gapless regime the Cv(T) curve shows the distinct behavior compared with the first LL gapless regime, (hc1<h<hc2). A three peaks structure in the specific heat is seen, in which this behavior clearly separates two LL gapless phases, as shown in Fig. 8(g) and (h). The middle peak in Fig. 8(g) and (h) is a shoulder and another one at low temperature, is a sharp peak, in which geos down linearly, and, both of them gradually vanish, as shown in the insets of Fig. 8(g) and (h). Likewise, with further increase of the field the height of the low temperature sharp peak increases slowly whereas temperature side of it, is fixed (Fig. 8(i) and (j). Existence of three peaks structure in Cv(T) curve is a new type of gapless LL phase, which has not been reported for this type of spin structure yet. Finally, when h exceeds hc4, the specific heat has a single peak and decays exponentially as T→0.
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Chapter
Properties of the Normal State
2007, Superconductivity (Second Edition)Charles P. PooleJr., ... Ruslan Prozorov
XI ELECTRONIC SPECIFIC HEAT
The specific heat C of a material is defined as the change in internal energy U brought about by a change in temperature
(1.49)C=(dUdT)v.
We will not make a distinction between the specific heat at constant volume and the specific heat at constant pressure because for solids these two properties are virtually indistinguishable. Ordinarily, the specific heat is measured by determining the heat input dQ needed to raise the temperature of the material by an amount dT,
(1.50)dQ=CdT.
In this section, we will deduce the contribution of the conduction electrons to the specific heat, and in the next section we will provide the lattice vibration or phonon participation. The former is only appreciable at low temperatures while the latter dominates at room temperature.
The conduction electron contribution Ce to the specific heat is given by the derivative dEτ/dT. The integrand of Eq. (1.44) is somewhat complicated, so differentiation is not easily done. Solid-state physics texts carry out an approximate evaluation of this integral, to give
(1.51)Ce=γT,
where the normal-state electron specific heat constant γ, sometimes called the Sommerfeld constant, is given as
(1.52)γ=(π23)D(EF)kB2.
This provides a way to experimentally evaluate the density of states at the Fermi level. To estimate the electronic specific heat per mole we set n = NA and make use of Eq. (1.42) to obtain the free-electron expression
(1.53)γ˙0=π2R2TF,
where R = NAkB is the gas constant. This result agrees (within a factor of 2) with experiment for many metallic elements.
A more general expression for γ is obtained by applying D(EF) from Eq. (1.48) instead of the free-electron value of (1.42). This gives
(1.54)γ=(mm)γ0,
where γ0 is the Sommerfeld factor (1.53) for a free electron mass. This expression will be discussed further in Chapter 9, Section II, which treats heavy fermion compounds that have very large effective masses.
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2007, Superconductivity (Second Edition)Charles P. PooleJr., ... Ruslan Prozorov
Chapter
Graphite: Properties and Characteristics
2020, Comprehensive Nuclear Materials (Second Edition)Timothy D. Burchell, Tsvetoslav R. Pavlov
7.11.4.1 Specific Heat
Specific heat represents the thermal energy a material can store per unit temperature, per unit mass. There are various physical mechanisms which govern the evolution of this property. One can express the total specific heat of graphite as a sum of the individual contributions of the lattice (l), defect (d), electron (el) and Cv to Cp conversion or an-harmonic (v→p) terms18,22,23 as shown via Eq. (1). An elaborate description of each term can be found in previous studies.18 Fig. 13 shows the calculated specific heat as a function of temperature according to Eq. (1) (based on the study of Pavlov et al.18).
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Fig. 13. Specific heat as a function of temperature for graphite. The correlation based on Eq. (5) and the model of Pavlov et al. (Eq. 1) are compared to various experimental data sets – Hust, Rasor and McClelland, Pavlov et al.
Reproduced from Hust, J.G., 1984. A Fine-Grained, Isotropic Graphite for Use as NBS Thermophysical Property RM’s from 5 to 2500K. National Bureau of Standards. doi:10.15713/ins.mmj.3. Rasor, N.S., McClelland, J.D., 1960. Thermal properties of graphite, molybdenum and tantalum to their destruction temperatures. J. Phys. Chem. Solids 15 (1–2), 17–26. doi:10.1016/0022-3697(60)90095-0. Pavlov, T., Vlahovic, L., Staicu, D., et al., 2017. A new numerical method and modified apparatus for the simultaneous evaluation of thermo-physical properties above 1500K: A case study on isostatically pressed graphite. Thermochim. Acta 652. doi:10.1016/j.tca.2017.03.004.
(1)Cptot=Cvl+Cvd+Cvel+ΔCv→p
The main mechanism of storing heat in graphite is via coordinated atomic displacements or crystal lattice vibrations (phonons). These vibrations are considered as standing waves which are quantized. Hence, the crystal lattice can vibrate only at distinct, allowed frequencies as a function of the relative wave or reciprocal lattice vector (also known as phonon dispersion). Phonons are characterized by atomic displacements, which can be parallel to the wave vector (longitudinal waves) and in two directions perpendicular to it (transverse waves). As discussed previously, the phonon density of states of graphite can be approximated using two Debye temperatures – one representing in-plane vibrations and one representing out-of-plane vibrations.
(2)Cvl=13C(θDc)+23C(θDa)
(3)Cvl(θD)=9R(TθD)3∫0θDTz4ez(ez−1)2dz
where R is the gas constant (8.314 J mol−1 K); T, temperature; θD, Debye temperature; and z = ħω/k¯BT, where ω, frequency of vibrational oscillations; k¯B, Boltzmann’s constant; T, temperature; and ħ = h/2π, where h is the Planck’s constant.
At low temperatures, where (T/θD) < 0.1, z in Eq. (3) is large and we can approximate Eq. (3) by allowing the upper limit in the integral to go to infinity such that the integral becomes ~ (π4/15), and on differentiating we get
(4)Cv=1941T/θD3Jmol−1K−1
Thus, at low temperatures, the specific heat is proportional to T3 Eq. (4). At high temperatures, z is small and the integral in Eq. (3) reduces to z2dz, hence on integrating we get the Dulong–Petit value of 3R, that is, the theoretical maximum specific heat of 24.94 J mol−1 K. As we are typically concerned only with the specific heat at temperatures above 10% of the Debye temperature (0.1θD), the specific heat should rise exponentially with temperature to a constant value at T ≈ θD, the Debye temperature. The specific heat of graphite is shown in Fig. 13 over the temperature range 0–4000K. The experimental data by Hust,24 Rasor and McClelland25 and Pavlov et al.18 have been shown to be well represented by Eq. (5) below 3000K,26 which is applicable to all graphite.
(5)CP=111.07T−1.644+3.688×10−4T0.02191Jkg−1K−1
In Eq. (1) the anharmonic or “CV to CP” conversion term has a negligible influence on the specific heat of graphite. Electrons contribute to the specific heat of the material, however their effect is not dramatic. Point defects, in particular Frenkel pairs27 could also contribute to graphite’s specific heat capacity. However, due to their high formation energies (above 10 eV according to Li27) these are not present in significant quantities below 3000K and hence no notable effects on CP would be expected below that temperature. Above 3000K specific heat has been shown to increase exponentially.25 This exponential increase has been attributed to the formation of defects. However, high temperature vaporization could have influenced the measured values.25 Furthermore, the model of Pavlov et al.18 cannot completely capture the specific heat ascent above 3500K, though it does show the significant contribution of defects to CP above 3000K.
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2020, Comprehensive Nuclear Materials (Second Edition)Timothy D. Burchell, Tsvetoslav R. Pavlov
Chapter
Thermodynamics
2013, Fundamentals of MagnetismMario Reis
4.6 Magnetic specific heat
Specific heat is an important quantity and must be here briefly introduced. It is defined as the quantity of heat needed to change the temperature of the system in a certain amount:
(4.44)Ci=dQdTi,
where i is a constant parameter. From Eq. (4.7), we know that:
(4.45)dQ=TdS=dU-BdM-μdN
and this expression is important to the evaluations below, considering N constant (dN=0).
The specific heat under constant magnetization assumes dM=0 and then dQ=TdS=dU. Thus,
(4.46)CM=dQdTM=T∂S∂TM=∂U∂TM
and then CM depends on the internal energy U.
The specific heat under constant magnetic field requires dB=0 and then dQ=TdS=dU-BdM=dH. Thus:
(4.47)CB=dQdTB=T∂S∂TB=∂H∂TB
and then CB depends on the enthalpy H.
There are relationships between CM and CB, for constant N, as presented below. A detailed proof of these is left as an exercise to the reader (see Figure 4.1):
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Figure 4.1. Thermodynamic square for constant number of particles N. From this square it is possible to obtain the thermodynamic quantities with respect to the thermodynamic potentials and the Maxwell relations (see text for further details).
(4.48)CB=CM+T∂M∂TB2∂M∂BT-1,
(4.49)CBCM=∂M∂BT∂M∂BS.
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2013, Fundamentals of MagnetismMario Reis
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Multielectron bubbles in helium as a paradigm for studying electrons on surfaces with curvature
2007, Surface Science ReportsJ. Tempere, ... J.T. Devreese
A generalized shape composed of spheroids and hyperboloids to describe the fissioning of an MEB in cylindrical coordinates. The parameters ai and bi are the squares of the semi-major axes and the deformation parameter, respectively; the fi define the centers of the spheroids.
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Chapter
Specific Heat of Solids
2019, Solid State PhysicsJoginder Singh Galsin
8.1 Experimental Facts
For the first time in 1819, Dulong and Petit measured the specific heat of a solid at room temperature experimentally and found it to be 3R (≈ 6 calories per mole). This value is below the melting point of a solid. At and above the melting point, the specific heat starts increasing due to the change in phase. Actually, the specific heat is a function of temperature: at very low temperatures, it varies as T3 and approaches zero at absolute zero (see Fig. 8.1). At intermediate temperatures, the specific heat varies linearly with T and approaches the Dulong and Petit’s value at reasonably large temperatures (≈ room temperature). In a superconducting solid, the specific heat decreases exponentially below the superconducting transition temperature and goes to zero as the temperature approaches absolute zero. Further, in magnetic substances, the specific heat becomes large over the temperature range in which magnetic moments are ordered. This may be because the ordering of the magnetic moments decreases the entropy of the solid faster, thereby increasing the specific heat. Below 0.1K the ordering of magnetic (nuclear) moments may give very large values of specific heat.
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Fig. 8.1. Temperature dependence of experimentally measured values of specific heat for some elements. The solid line represents the theoretical values obtained for CV from the Debye theory.
(Modified from Epifanov, G. I. (1979). Solid state physics (p. 119). Moscow: Mir Publishers.)
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2019, Solid State PhysicsJoginder Singh Galsin
Chapter
Coal Structure and Reactivity
2003, Encyclopedia of Physical Science and Technology (Third Edition)John W. Larsen, Martin L. Gorbaty
IV.B.2 Specific Heat
The specific heat records the heat necessary to cause a given temperature rise in a coal. It is rank dependent, decreasing from ∼0.35 cal/g at 50% carbon to ∼0.23 cal/g at 90% carbon. Above 90% carbon, it decreases quite rapidly.
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2003, Encyclopedia of Physical Science and Technology (Third Edition)John W. Larsen, Martin L. Gorbaty
Related terms:
Nanoparticle
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188130 | https://www.secondaryimmunodeficiency.com/secondary-immunodeficiency-in-multiple-myeloma/ | Secondary immunodeficiency in multiple myeloma » SID
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Secondary immunodeficiency in HSCT
Secondary immunodeficiency in multiple myeloma
EU Expert Consensus in treating secondary antibody deficiency in haematological malignancies
Immunoglobulin therapy
Secondary immunodeficiency in CLL
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Secondary immunodeficiency in multiple myeloma (MM)
presentation
> EXPERT LECTURES
Secondary immunodeficiency in multiple myeloma can be caused by the underlying disease and/or its treatment
Multiple myeloma (MM) is a largely incurable haematological malignancy that is frequently associated with secondary immunodeficiency (SID).1,2 Patients with MM show many abnormalities of immune function that develop due to the disease itself and/or its treatment, leading to significant morbidity and mortality from infections.3,4
Characteristics of MM include the dysfunctional replication of plasma cells,1,2 production of M-protein 1 (an abnormal immunoglobulin), and disrupted production of normal immunoglobulins.3
SID in MM can result from disease-related changes in humoral and cellular immunity:1-4
Humoral immunity
Impaired quantities and functions of B cells 2,4
Reduced synthesis of normal immunoglobulins 3
Accelerated catabolism of normal IgG in MM 3
Hypogammaglobulinaemia 3
Cellular immunity
Impaired quantities and functions of T cells, dendritic cells and natural killer cells 1,2,4
Increased number of immunosuppressive cells, such as myeloid-derived suppressor cells (MDSCs)1
MM treatments can lead to SID and increase the risk and severity of infections.2,4,5
Therapeutic causes of SID in MM: corticosteroids, chemotherapy, immunosuppressive therapies
| Class of therapy |
| Corticosteroids |
| Immune dysfunction | Cellular immunity 7 Hypogammaglobulinaemia – decreased naïve and transitional B cells, with no effect on memory B cells 7 |
| Infections | Increased incidence of infections 7 |
| Alkylating agents e.g.bendamustine |
| Immune dysfunction | Hypogammaglobulinaemia 7 |
| Infections | Increased incidence of infections 7 |
| Immunomodulatory drugs (IMiDs) e.g. lenalidomide |
| Immune dysfunction | Mechanisms unclear 5 |
| Infections | Incidence of high-grade infections increased two-fold with lenalidomide 5 |
| Proteasome inhibitors (PIs) e.g. bortezomib |
| Immune dysfunction | Decreased IgG, IgA, and IgM (but not hypogammaglobulinaemia)7 |
| Infections | Increased incidence of HZ infection and VZV reactivation 7 |
| Anti-CD38 antibody e.g. daratumumab |
| Immune dysfunction | Risk of neutropaenia 8, with no effect on B cells 7 |
| Infections | Increased incidence of infections (including VZV)7,8 |
| Anti-SLAMF7/CD319 antibody e.g. elotuzumab |
| Immune dysfunction | Lymphopaenia 8 |
| Infections | Increased risk of infections (particularly VZV)8 |
| Class of therapy | Immune dysfunction | Infections |
---
| Corticosteroids | Cellular immunity 7 Hypogammaglobulinaemia – decreased naïve and transitional B cells, with no effect on memory B cells 7 | Increased incidence of infections 7 |
| Alkylating agents e.g.bendamustine | Hypogammaglobulinaemia 7 | Increased incidence of infections 7 |
| Immunomodulatory drugs (IMiDs) e.g. lenalidomide | Mechanisms unclear 5 | Incidence of high-grade infections increased two-fold with lenalidomide 5 |
| Proteasome inhibitors (PIs) e.g. bortezomib | Decreased IgG, IgA, and IgM (but not hypogammaglobulinaemia)7 | Increased incidence of HZ infection and VZV reactivation 7 |
| Anti-CD38 antibody e.g. daratumumab | Risk of neutropaenia 8, with no effect on B cells 7 | Increased incidence of infections (including VZV)7,8 |
| Anti-SLAMF7/CD319 antibody e.g. elotuzumab | Lymphopaenia 8 | Increased risk of infections (particularly VZV)8 |
Other MM treatments/modes of treatment delivery can also increase the risk of infections:
HSCT (neutropaenia, gastrointestinal mucositis)6
Central venous catheters 6
Also known as paraproteins
HSCT: haematopoietic stem cell transplantation; HZ: herpes zoster; IgA: immunoglobulin A; IgG: immunoglobulin G; IgM: immunoglobulin M; VZV: varicella zoster virus.
Tamura, H.,Int J Hematol.2018; 107:278-85, 2. Li, L. and Wang, L., J Cancer. 2019; 10:1675-84, 3. Kyrtsonis, M.C. et al., Med Oncol. 1999; 16:73-7, 4. Pratt, G. et al., Br J Haematol. 2007; 138:563-79, 5. Ying, L. et al., Oncotarget. 2017; 8:46593-600, 6. Nucci, M. and Anaissie, E., Clin Infect Dis. 2009; 49:1211-25, 7. Patel, S.Y. et al., Front Immunol. 2019; 10:33, 8. Drgona, L. et al., Clin Microbiol Infect. 2018; 24:S83-94.
Focus on hypogammaglobulinaemia in multiple myeloma
Several MM-associated mechanisms can impact normal immunoglobulin production 1, leading to hypogammaglobulinaemia and increasing the susceptibility of MM patients to infection.2
Decreased number and function of B cells 1
Suppression of B cell progenitors, possibly due to attrition of the normal plasma cell and B cell progenitor compartments
Increased apoptosis of B cell progenitors due to interaction of myeloma cells with stromal cells
Abnormal B cell maturation due to a number of factors, e.g. an increased number of immunosuppressive CD5+ B cells
Immunosuppressive cytokines 1
IL-4, a cytokine important for the induction of normal B cell responses, is decreased in MM
Altered T cell numbers, phenotype and function 1
Disrupted T cell cytokine production may affect the proliferation and differentiation of B cells
Increased catabolism of IgG 1
of both normal and clonal IgGThe catabolism of IgG is directly proportional to its concentration, leading to a 2-fold increase in the catabolic rate
The catabolic rate of other Ig classes is not affected
IL: interleukin; TGF: transforming growth factor.
Kyrtsonis, M.C. et al., Med Oncol. 1999; 16:73-7, 2. Pratt, G. et al., Br J Haematol. 2007; 138:563-79.
Epidemiology and clinical predictors of infection in multiple myeloma patients
Changes in patterns of and risk factors for infection have been observed in patients with MM with the shift to routine usage of treatments that impact the immune system, such as IMiDs and PIs. In an Australian single-centre study, patients with MM receiving standard of care were investigated to better understand the epidemiology (types, severity, and timing of infections) and clinical predictors of infection.
A total of 199 MM patients diagnosed in 2008–2012 were followed for a median of 33 months. During this period, a total of 771 episodes of infection were recorded, with 32% of patients having ≥ 5 infections. The overall incidence of infection was 1.33 per patient-year. Incidences of bacterial and viral infections had bimodal distributions, with bacterial infections peaking at 4–6 and 70–72 months, and viral infections peaking at 7–9 and 52–54 months after MM diagnosis.
Frequency of infections in patients with MM (N = 199)
Baseline treatment characteristics of 199 MM patients
As initial induction regimen, 38.7% were thalidomide-based, 26.1% lenalidomide-based,18.6% bortezomib-based, 15.1% chemotherapy-based, and 2.0% were classified as "other"
For stem cell transplantation, 78.9% of the patients received an ASCT, 2.5% an alloHCT† and 18.6% did not receive any stem cell transplant
Of a total of 771 episodes of infection:
43.7% were clinically defined
36.4% were microbiologically defined (54.1% bacterial, 40.2% viral and 5.7% fungal)
19.8% were fever of unknown origin
Most infections during ASCT were high severity (93.3%). During induction and disease progression, high severity infections comprised 57.5% and 62.2% of all infections, respectively. The majority of infections during the plateau period were low severity (80.4%).
Nature of infections
Severity of infections
The nature and severity of 771 infections were assessed across different disease periods in 199 MM patients
Disease periods were defined as following:
Induction: from initial diagnosis to receipt of 4–6 cycles of induction chemotherapy
ASCT: from receipt of chemotherapy for stem cell mobilisation to day 30 following re-infusion of stem cells
Plateau: stable paraprotein levels with or without maintenance treatment
Progression: any period of increasing myeloma burden despite active anti-myeloma treatment and necessitating a change of therapy
Severity of infections was graded according to the National Cancer Institute’s Common Terminology Criteria for Adverse Events (CTCAE), version 4.03.
Using conditional risk set modelling, specific therapies independently associated with increased risk of infection in MM patients were identified:
High-dose melphalan
Intravenous cyclophosphamide
Intensive combination systemic chemotherapy‡
Cumulative doses of corticosteroid over 2 months
23 (11.6%) patients had 2 ASCTs, 2 (1.0%) patients had 3 ASCTs; †5 (2.5%) patients had ASCT prior to alloHCT.
‡Regimens include vincristine-doxorubicin-dexamethasone and dexamethasone-doxorubicin-cyclophosphamide-etoposide ± thalidomide or lenalidomide. alloHCT: allogeneic haematopoietic stem cell transplant; ASCT: autologous haematopoietic stem cell transplant; CDI: clinically defined infections; FUF: fever of unknown focus; MDI: microbiologically defined infections.
Teh, B.W. et al., Br J Haematol. 2015; 171:100-8.
Early mortality after diagnosis of multiple myeloma
While introduction of new treatment approaches has improved the survival time of MM patients,1
many deaths still occur soon after diagnosis, before patients have time to reach the maximal benefit of these treatments.
A retrospective assessment of 3,107 newly diagnosed patients who were registered in five UK Medical Research Council MM trials (1980–2002) was performed to identify the direct and contributing causes of early deaths in MM.2
Early death (defined as within 60 days of trial entry) occurred in 299 (10%) patients.2
Cause of early death in MM 2
Bacterial infection was the direct cause of early mortality in 45% of deaths, with pneumonia being the most common.2
Types of fatal bacterial infections (n = 135)2
Tamura, H.,Int J Hematol.2018; 107:278-85, 2. Augustson, B.M. et al., J Clin Oncol. 2005; 23:9219-26.
Infections are a substantial cause of death in multiple myeloma
To estimate the risk of infection and infection-related deaths in MM patients, a study of the nationwide Swedish Cancer Registry evaluated 9,253 patients diagnosed with MM from 1988 to 2004 with follow-up until 2007.
MM patients had a significant 7-fold increased risk of developing any infection vs matched controls in the overall follow-up period (HR: 7.1; 95%CI: 6.8–7.4), with an even higher relative risk (11-fold) during the first year following diagnosis (HR: 11.5; 95%CI: 10.4–12.7). Infection risk in MM patients increased significantly with more recent calendar period of diagnosis†, potentially due to modern therapies.
Relative risk for bacterial and viral infection in MM patients compared with matched controls
Risks of infections in MM patients vs matched controls were determined for the first year following diagnosis and for overall follow-up.
Individual HRs for the following bacterial infections were: meningitis (HR 16.6), septicaemia (HR 15.6), pneumonia (HR 7.7), endocarditis (HR 5.3), osteomyelitis (HR 3.5), cellulitis (HR 3.0) and pyelonephritis (HR 2.9); for viral infections: herpes zoster (HR 14.8) and influenza (HR 6.1).
Of 2474 deaths that occurred within 1 year of MM diagnosis, 555 (22.4%) were due to infections. The 3-year risk of death due to infection was 12.2% in MM patients vs 2.2% in matched controls. The risk of infection-related death did not differ between age groups (> 65 and ≤ 65 years) over the three calendar periods of diagnosis†.
Deaths within 1 year of MM diagnosis (N = 2474)
Matched controls (N=34,931) were of the same sex, age and county of residence, and were alive and without previous haematologic malignancy at the diagnosis date of the corresponding MM patient.
† Patients were stratified by three periods reflecting different treatment strategies (1988–1993, 1994–1999 and 2000–2004)
CI: confidence interval; HR: hazard ratio.
Blimark, C. et al., Haematologica. 2015; 100:107-13.
Impaired response to immunisation in multiple myeloma is associated with increased risk of a major infection
Response to immunisation in plateau-phase MM patients may identify those at risk of infection.1
A prospective study in the UK followed 102 patients with MM for a mean duration of 10 months.1 Patients in plateau phase were immunised with Pneumovax II (n = 40)1,a pneumococcal polysaccharide vaccine.#2 Patients were classified by their immune response.1
MM patients with a poor/intermediate immunisation response to Pneumovax II had a higher frequency of septicaemia.1
Incidence of septicaemia in MM patients in plateau phase (n = 40)1
Of the 18 patients who showed a good response to immunisation, none had a septicaemic episode recorded since diagnosis of MM.
Of the 22 patients with a poor/intermediate immunisation response, 22.7% of patients had a septicaemic episode recorded since diagnosis of MM
Read about the expert recommendations on the use of test immunisation in patients with haematological malignancies
Pneumovax II is a pneumococcal polysaccharide vaccine recommended for immunisation against disease caused by pneumococcal serotypes contained in the vaccine. Immunisation responses, based on differences between pre- and post-immunisation IgG titres, were classified as: Good: ≥ 2-fold increase of specific IgG titres as well as the post-immunisation titre reaching the minimum of the normal range for age (calculated from the baseline results obtained in the control population); Poor: no difference between pre- and post-immunisation titres; Intermediate: inadequate response for inclusion in the good response group.
Hargreaves, R. M., et al., J Clin Pathol. 1995; 48:260-6, 2. UK Medicines and Healthcare products Regulatory Agency, 2014, available at:
Management of infectious complications in multiple myeloma: Summary of expert recommendations
With the advent of new treatment approaches for MM and their impact on the immune system, there remains a need to determine the appropriate approaches to prophylaxis and treatment for infections. While data are lacking on certain topics, an expert panel published consensus recommendations based on their experience and available data on how to manage infectious complications in MM patients.
Summary of recommendations for assessment of infection risk:
Careful evaluation of performance status and past medical history (especially infections that can reactivate) before initiating first-line therapy
Quantitative evaluation of serum polyclonal immunoglobulins, absolute lymphocyte count, and absolute neutrophil count to help define the individual risk of infections
Information on recent vaccination history to define the pre-treatment vaccination schedule
HBV and HCV screening in all patients requiring active treatment
Colonisation screening (rectal swab culture to detect colonisation by MDR Gram negative bacteria) in hospitalised candidates for ASCT and in those undergoing intensive salvage therapy
Severe active infections (i.e. pneumonia, herpes zoster, HBV- or HCV-related hepatitis, CMV disease, tuberculosis) or uncontrolled HIV-disease contraindicate active therapies in MM until their complete resolution or control
CMV: cytomegalovirus; HBV: hepatitis B virus; HCV: hepatitis C virus; HIV: human immunodeficiency virus; MDR: multidrug-resistant
Girmenia, C. et al., Blood Rev. 2019; 34:84-94.
Download presentation to read more about anti-bacterial, -fungal and -viral prophylaxis recommendations
Recommendations for intravenous immunoglobulin replacement therapy
IVIG is not recommended routinely for patients with MM
The use of IVIG may be reserved for patients with very low IgG levels (< 0.4 g/L) and recurrent life-threatening infections
Read about the expert recommendations on the use of IVIG in IMMUNOGLOBULIN THERAPY
ASCT: autologous haematopoietic stem cell transplant; CMV: cytomegalovirus; HBV: hepatitis B virus; HCV: hepatitis C virus; HIV: human immunodeficiency virus; IVIG: intravenous immunoglobulin; MDR: multidrug-resistant.
Girmenia, C. et al., Blood Rev. 2019; 34:84-94.
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188131 | https://www.quora.com/How-do-you-clinically-differentiate-Tension-Pneumothorax-from-Cardiac-Tamponade | How to clinically differentiate Tension Pneumothorax from Cardiac Tamponade - Quora
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How do you clinically differentiate Tension Pneumothorax from Cardiac Tamponade?
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Liang-Hai Sie
Retired general internist, former intensive care physician. · Author has 63.1K answers and 215.9M answer views
·Updated 5y
Good question. They often are short of breath, those with a tension pneumothorax more acutely so than those with cardiac tamponade where low cardiac output predominates.
Cardiac tamponade usually is preceded by some earlier event: penetrating wound to the chest, myocardial infarction, pericarditis, in the last case building up slowly, lowering cardiac output so people have a weak fast pulse, low blood pressure, and distended jugular veins due to inflow obstruction. No expanded chest.
Tension pneumothorax often is preceded by a broken rib, the hemithorax involved is visibly expanded, if left side
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Good question. They often are short of breath, those with a tension pneumothorax more acutely so than those with cardiac tamponade where low cardiac output predominates.
Cardiac tamponade usually is preceded by some earlier event: penetrating wound to the chest, myocardial infarction, pericarditis, in the last case building up slowly, lowering cardiac output so people have a weak fast pulse, low blood pressure, and distended jugular veins due to inflow obstruction. No expanded chest.
Tension pneumothorax often is preceded by a broken rib, the hemithorax involved is visibly expanded, if left sided no dullness where the heart dullness is located anymore, trachea shifted to the right, if right sided the trachea is shifted to the left.
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How many hours does it take to die from cardiac tamponade?
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What is the difference between pneumothorax and tension pneumothorax?
Asher Nitin
Physician. Visual artist. Hybrid athlete. Lay theologian. · Author has 189 answers and 8.5M answer views
·9y
Auscultate. The lung fields will be clear in a cardiac tamonade.
I assume this is an acute presentation, and I assume a decreased heart rate, increased jugular venous pulse, and a decreased arterial blood pressure reading.
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Sheaffer Williams
Studied Anesthesiology at University of Iowa Hospitals and Clinics (Graduated 1900) · Author has 436 answers and 12.2M answer views
·Updated 7y
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Why does tension pneumothorax cause jugular vein distention?
The reason hinges on the creation of a one-way valve into the area between lungs and rib cage—the pleural space. Many things can cause a one-way valve. Errant physicians, broken ribs, projectiles into the chest cavity, blown out lungs, torn bronchi, etc. The common thread is that they create a way for air to enter the space between lung and rib cage without escaping.
We breathe by decreasing the pressure in our chest by moving our diaphragm down and our ribs up. This pulls air in. When there is a hole leading to the pleural space, air is pulled in and doesn’t have a direct way out. This is the
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The reason hinges on the creation of a one-way valve into the area between lungs and rib cage—the pleural space. Many things can cause a one-way valve. Errant physicians, broken ribs, projectiles into the chest cavity, blown out lungs, torn bronchi, etc. The common thread is that they create a way for air to enter the space between lung and rib cage without escaping.
We breathe by decreasing the pressure in our chest by moving our diaphragm down and our ribs up. This pulls air in. When there is a hole leading to the pleural space, air is pulled in and doesn’t have a direct way out. This is the one-way valve. The end result is that each breath pumps a little more air into the pleural space.
The chest makes a lousy balloon. Any increase in pressure collapses lung initially. You can breathe with one lung however. As the pressure rises it is also transmitted to the heart as the lung is pushed across the rib cage into the other lung’s space.
Venous blood returning to the heart is under low pressure. Therefore it is the first to feel the pinch. This is the reason that blood draining from the head is impeded in its flow back to the heart. The result is that the internal and external jugular veins distend with the backed up blood.
And so the hapless victim continues to pump air into his chest with each breath until the pressure exceeds that of his venous return and his heart stops —unless some kindly person vents the pressure outside by placing a tube into the pleural space. This action is life saving and is often rewarded with a satisfying whoosh of trapped air. The patient generally rebounds within seconds.
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Clinical differentiation between tension pneumothorax and cardiac tamponade relies on history, rapid bedside examination, focused physiologic signs, and point-of-care imaging. Both cause acute shock and respiratory distress, but key features diverge.
Immediate distinguishing features
Primary physiology
Tension pneumothorax: intrapleural air under pressure → collapsed lung, ipsilateral ventilation loss, mediastinal shift → obstructive shock from decreased venous return and impaired left ventricular filling.
Cardiac tamponade: accumulation of pericardial fluid under pressure → circumferential compre
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Clinical differentiation between tension pneumothorax and cardiac tamponade relies on history, rapid bedside examination, focused physiologic signs, and point-of-care imaging. Both cause acute shock and respiratory distress, but key features diverge.
Immediate distinguishing features
Primary physiology
Tension pneumothorax: intrapleural air under pressure → collapsed lung, ipsilateral ventilation loss, mediastinal shift → obstructive shock from decreased venous return and impaired left ventricular filling.
Cardiac tamponade: accumulation of pericardial fluid under pressure → circumferential compression of the heart → obstructive shock from impaired diastolic filling of both ventricles.
Onset context
Tension: often after chest trauma, mechanical ventilation, central line placement, or primary spontaneous pneumothorax; may be unilateral chest pain and sudden hypoxia.
Tamponade: often after penetrating/blunt chest injury, post–cardiac procedure, aortic dissection, malignancy, uremia, or pericarditis; presentation may be more insidious unless traumatic.
Key bedside examination findings
Respiratory findings
Tension: severe unilateral decreased or absent breath sounds, hyperresonance to percussion on affected side, asymmetric chest expansion. Marked respiratory distress and hypoxia.
Tamponade: breath sounds usually normal and symmetric; respiratory distress may be present but hypoxia is less prominent unless concomitant lung injury.
Hemodynamic findings
Pulsus paradoxus (>10 mmHg drop in SBP with inspiration): classic for tamponade (common, often marked). Can occur in severe tension but less consistent.
Jugular venous distension (JVD): prominent in tamponade (elevated JVP with blunted y descent). In tension, JVD may also occur but often asymmetric and accompanied by tracheal deviation.
Heart sounds: muffled/quiet in tamponade (Beck’s triad: hypotension, JVD, muffled heart sounds). In tension, heart sounds may be normal or displaced; muffling is less typical.
Tracheal deviation: away from affected side in tension pneumothorax (a late but specific sign). Not a feature of tamponade.
Chest wall/jugular asymmetry: tension often has marked unilateral chest findings; tamponade produces central signs.
Circulatory collapse pattern
Tension: rapid hypoxia-driven deterioration; severe unilateral hyperinflation can worsen ventilation and oxygenation.
Tamponade: shock out of proportion to lung exam, pronounced venous congestion, lower-volume pulses, pronounced pulsus paradoxus.
Point-of-care investigations (emergency, rapid)
Ultrasound (preferred rapid test)
Lung ultrasound: absent lung sliding, lung point, and barcode/stratosphere sign on M-mode → tension pneumothorax.
Focused cardiac ultrasound (echo): right atrial and right ventricular diastolic collapse, swinging heart, pericardial effusion → tamponade physiology. Doppler assessment can show respiratory variation in mitral/tricuspid inflow.
Use both lungs and cardiac windows; a pericardial effusion with chamber collapse is diagnostic of tamponade physiology.
Chest X-ray
Tension: large unilateral radiolucent hemithorax, depressed ipsilateral hemidiaphragm, mediastinal shift to opposite side (may be delayed in portable supine films).
Tamponade: chest x-ray may show enlarged, globular cardiac silhouette only with large chronic effusions; not reliable in acute tamponade.
ECG
Tamponade: low-voltage QRS and electrical alternans (swinging heart) suggest large effusion.
Tension: nonspecific; may show signs of hypoxia or tachyarrhythmia but not electrical alternans.
Hemodynamics/labs
Arterial blood gas: tension commonly shows hypoxemia and possible hypercarbia if ventilatory compromise; tamponade often shows hypoperfusion without primary hypoxemia.
Central venous pressure: elevated in tamponade; may be elevated in tension but less diagnostic.
Management clues that confirm diagnosis
Immediate decompression response
Tension: emergent needle decompression or finger thoracostomy followed by chest tube typically yields rapid improvement in oxygenation, blood pressure, and unilateral breath sounds.
Tamponade: pericardiocentesis produces rapid hemodynamic improvement when fluid is causing tamponade; chest tube will not relieve tamponade unless concurrent hemothorax present.
Rapid practical algorithm for the unstable patient
Assess airway/oxygenation and apply high-flow oxygen; prepare for ventilation if needed.
Look and listen: unilateral absent breath sounds + hyperresonance + tracheal shift → treat as tension pneumothorax immediately (needle decompression/chest tube) while obtaining ultrasound if available.
Central signs of venous congestion (marked JVD), muffled heart sounds, pulsus paradoxus, and poor response to oxygen with symmetric breath sounds → perform emergent bedside echo; if pericardial effusion with chamber collapse, perform pericardiocentesis/surgical drainage.
If diagnosis remains unclear in an unstable patient, prioritize interventions that rapidly reverse the most immediately life‑threatening and reversible condition based on findings (needle decompression for suspected tension; pericardial drainage for convincing tamponade on echo).
Summary of high-yield discriminators
Unilateral absent breath sounds + hyperresonance + tracheal deviation → tension pneumothorax.
Elevated JVP + muffled heart sounds + hypotension + pulsus paradoxus + echo showing cardiac chamber collapse → cardiac tamponade.
Lung and cardiac point-of-care ultrasound are decisive and should be used immediately when available.
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Bruce McFarland
Board certified Internal Medicine have practiced 20 years · Author has 472 answers and 3.8M answer views
·8y
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What is the difference between pneumothorax and tension pneumothorax?
Pneumo (air) thorax (chest) is a condition when air leaks out of a tear in the lung and causes it to both collapse the lung, though usually just partially, and allow some amount of air to collect outside the lung, but still trapped inside the chest. The reason it is trapped is because the lungs are surrounded by a membrane called the pleura. The pleura has one surface that is coating the lung (visceral pleura) and one that is coating the inside of our chest (parietal pleura)…inside our rib cage.
The inhaling and exhaling of the lungs usually is allowed to glide smoothly inside the chest cavity
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Pneumo (air) thorax (chest) is a condition when air leaks out of a tear in the lung and causes it to both collapse the lung, though usually just partially, and allow some amount of air to collect outside the lung, but still trapped inside the chest. The reason it is trapped is because the lungs are surrounded by a membrane called the pleura. The pleura has one surface that is coating the lung (visceral pleura) and one that is coating the inside of our chest (parietal pleura)…inside our rib cage.
The inhaling and exhaling of the lungs usually is allowed to glide smoothly inside the chest cavity because the two pleural surfaces have a small amount of normal physiologic fluid that facilitates this gliding motion.
Here's what goes on in a normal pneumothorax, non-traumatic type. Usually a tall lanky guy gets these spontaneous pneumothoraxes…maybe he went scuba diving and ascended too fast, maybe he was in an airplane that wasn’t depressurized, maybe…he just coughs. And his lung pops like a balloon through a tear. Now the air is trapped inside the space between the two pleural surfaces. With any luck, the air leak seals itself, and the body just gradually reabsorbs the air by itself. On a chest x-ray, the size of the space that the lung has been reduced away from the edge of the thorax is a good clue if it will resolve by itself or not. If it's too big a pneumothorax, a thoracostomy tube (aka chest tube) has to be inserted. A tube is inserted after an opening is cut in the skin below a rib usually in the lateral, mid-chest area, and then a curved hemostat (ring-handled, blunt, narrow-tipped, clamp) is used (sterile technique, but in the ER or hospital room) to puncture the parietal pleural, with the hemostat simultaneously holding the end of the chest tube that is advanced into the interior of the chest. It is secured to the chest with a suture tie-down to the skin…and the far end of the chest tube is attached to a Pleur-o-Vac, a sealed system with a water reservoir so no more air can enter the chest cavity, but the system is on a negative pressure vacuum…that gradual draws the air out.
So now about tension pneumothorax. Basically, a sucking chest wound. A massive tear in the lung is causing major build up of air inside the pleural space. Each inhalation causes more air to pass through the mouth into the lung and out the torn lung into the chest cavity. And there it is trapped. As the air flows into more and more of the chest, it starts to compress against the other organs inside the thorax. It will compress the lung all the way over to the opposite half where the other lung begins to be crushed. It will displace the trachea away from the side of the leak…away from the tension pneumo. It will even compress against the heart until it impedes the heart pumping. Clearly, a life-or-death situation is presenting itself. The diagnosis can easily be made with a chest x-ray, but this is sometimes a luxury that the crisis doesn't afford the doctor to wait for. Surmising his suspicions are correct by anatomic and physical exam findings, he calls for the 14G or 16G needle, and with a patient lying on his back, understands the area above the nipple to be an approximation of where air in the chest would rise to and hence his target for his pending action. Alternatively, if the patient was positioned on his side, it would be so the air rose to the opposite side from which he was lying. The side with the pneumothorax directed up toward the ceiling. He deftly plunges the needle between two ribs and far enough that an audible ‘whoosh’ of air exiting its confinement inside the pleural space will be confirmation he was successful. Now he has the kind of time to proceed with the thoracotomy tube, as in the previous description. Having completed that process, he removes the emergency needle thoractomy.
There is the difference between a pneumothorax and a tension pneumothorax.
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Edward Leahy
Lives in Earth (1961–present) · Author has 14.1K answers and 10M answer views
·8y
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Why does tension pneumothorax cause jugular vein distention?
A tension pneumothorax causes an increase in the intrathoracic pressure. This increase in pressure makes it harder for blood to return to the heart and causes the venous blood pressure to rise outside the chest. This results in venous distension which is most easily seen in the neck veins.
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Shreya Thacker
Physical and Respiratory Therapist · Author has 714 answers and 38.8M answer views
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What is the difference between pneumothorax and tension pneumothorax?
Pneumothorax is the accumulation of air between the rib cage and the lungs i.e. in the pleural space.
Image
This can happen if there is a puncture wound in the lungs that leaks air into the chest cavity each time a person breathes in.
It can also occur due to an external wound, for example a gun shot to the chest, wherein a suction mechanism is created each time the chest wall expands, sucking air from the outside between the ribs and the lungs.
The severity of the condition depends on the amount of air trapped.
Tension pneumothorax, like it’s name suggests, is the most severe, life threatening var
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Pneumothorax is the accumulation of air between the rib cage and the lungs i.e. in the pleural space.
Image
This can happen if there is a puncture wound in the lungs that leaks air into the chest cavity each time a person breathes in.
It can also occur due to an external wound, for example a gun shot to the chest, wherein a suction mechanism is created each time the chest wall expands, sucking air from the outside between the ribs and the lungs.
The severity of the condition depends on the amount of air trapped.
Tension pneumothorax, like it’s name suggests, is the most severe, life threatening variety of pneumothorax wherein a colossal amount of air is trapped within the chest cavity.
Most times, in such cases, the wound acts like a one-way valve i.e. air enters the chest cavity during inhalation but has no way to escape.
This gives rise to fatal conditions. Accumulating air steeply increases the intra-thoracic pressure and, by extension, pressure on important life-sustaining structures in the chest — lungs, heart, major blood vessels, trachea (windpipe).
The lung/s on the affected side quickly collapse with the important structures literally being pushed to the other side. This can easily be seen on chest X-rays.
Clinical signs include (but not limited to) — respiratory distress, reduced O2, increased heart rate, decreased blood pressure, reduced breath sounds and fullness of the chest on the affected side.
If left untreated, the condition can lead to a painful death.
Here’s a chest x-ray depicting tension pneumothorax.
Here’s one more
The blue arrows are showing the outline of the collapsed lung. The red arrows point to the deviated windpipe — from mid-line, it went to the right (patient right)
Chest x-ray 1, Chest x-ray 2
Pneumothorax - Wikipedia
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R. Stone Lee
Former Attorney, Cardiologist at Private Practice (1994–2019) · Author has 427 answers and 728.1K answer views
·4y
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How do you know if you have a tension pneumothorax?
This is actually an interesting question! Thanks for asking it. As you probably know, a pneumothorax occurs when there is a leak of air from the lung into the pleural space or a leak from outside the chest into the pleural space. Pleural “space” is actually somewhat misleading because it is usually a potential space and the pleura is touching the lung. With a tension pneumothorax, for whatever reason, the air leaks into the pleural space, but can’t get back out. Because the intra-thoracic pressure is negative during inhalation, that sucks air into the pleural space. But since it can’t get out,
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This is actually an interesting question! Thanks for asking it. As you probably know, a pneumothorax occurs when there is a leak of air from the lung into the pleural space or a leak from outside the chest into the pleural space. Pleural “space” is actually somewhat misleading because it is usually a potential space and the pleura is touching the lung. With a tension pneumothorax, for whatever reason, the air leaks into the pleural space, but can’t get back out. Because the intra-thoracic pressure is negative during inhalation, that sucks air into the pleural space. But since it can’t get out, it stays in place, and each successive inhalation causes more air to accumulate.
So, the initial part of a tension pneumo is the same as a regular one. However, because the air continues to accumulate, not only does it make the lung collapse, but it compresses it. This compression progresses to the point that it pushes the mediastinum and the structures in it, including the heart toward the other side. The mediastinum has limited mobility, so the pressure eventually begins to compress the heart.
Joshua Calvert’s answer is right on the money as far as pneumothorax is concerned, but tension pneumo adds a couple of things. First, because the mediastinum is being pushed toward the other side, the trachea may be deviated away from the side of the pneumo. You could feel this by feeling if your adam’s apple is in the center of your neck or not. Next, your jugular veins may be distended, and you might be able to feel along the side of your neck and feel your jugular as a linear bulge. Finally as your heart becomes more compressed, your blood pressure and “pulse pressure” will drop. You may be able to detect this by feeling your pulse in your neck or your wrist as a “weak, thready” pulse.
Incidentally the jugular vein distention is something that ER docs and EMTs look for. If they miss the tracheal deviation, and low pulse pressure, they might be tempted to administer diuretics because it looks like the patient is in heart failure, and this could lead to the rapid deterioration of the patient.
Probably more than you wanted to know!
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Wilk Dedwylder
MD from University of Mississippi Medical Center (Graduated 1978) · Author has 19.3K answers and 31.7M answer views
·6y
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Can tension pneumothorax be treated without a Chest Decompression Needle?
The purpose of the decompression needle, not to put too fine a point on it, is to decompress by converting the tension pneumothorax to an open pneumothorax. If you can do it quickly enough, going straight to tube thoracostomy is a perfectly reasonable alternative. In a pinch, anything that opens that side of the chest will get the job done: you could poke in a knife and find something to keep the hole open, perhaps even the hilt of the knife. It’s just that a 14-bore needle tends to be ubiquitous in EMS and hospitals, so it’s the first convenient thing to get the job done quickly. You usually
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The purpose of the decompression needle, not to put too fine a point on it, is to decompress by converting the tension pneumothorax to an open pneumothorax. If you can do it quickly enough, going straight to tube thoracostomy is a perfectly reasonable alternative. In a pinch, anything that opens that side of the chest will get the job done: you could poke in a knife and find something to keep the hole open, perhaps even the hilt of the knife. It’s just that a 14-bore needle tends to be ubiquitous in EMS and hospitals, so it’s the first convenient thing to get the job done quickly. You usually don’t have time to scratch your head awaiting your MacGyver inspiration.
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Geren Nichols
Former Surgeon (1981–2010) · Author has 6.9K answers and 10.8M answer views
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What are the clinical features that can differentiate a patient with tension pneumothorax from that of hemothorx?
Have you ever tapped along a wall looking for a stud to drive a nail or a screw into anchor a really heavy picture?
In examination of the lungs that is called percussion. There are no 2 by 4 studs underneath the chest wall but for a tension pneumothorax the whole side is full of air and sounds like a drum, a normal chest is full of lung which sounds duller and a hemothorax is full of blood which which sounds very dull.
If you listen with a stethoscope with a pneumothorax , which is full of air which is a sound damper, well you do not hear very much at all. The blood in a hemothorax transmits bet
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Have you ever tapped along a wall looking for a stud to drive a nail or a screw into anchor a really heavy picture?
In examination of the lungs that is called percussion. There are no 2 by 4 studs underneath the chest wall but for a tension pneumothorax the whole side is full of air and sounds like a drum, a normal chest is full of lung which sounds duller and a hemothorax is full of blood which which sounds very dull.
If you listen with a stethoscope with a pneumothorax , which is full of air which is a sound damper, well you do not hear very much at all. The blood in a hemothorax transmits better than a pneumothorax so one hears sounds but a but muffled.
Still mostly X-Rays are used.
But if someone is terribly short of breath and is hyperesonant on one side they may get a chest tube is X-ray is slow. The tube can be quite small, at least from the doctors point of view there is a relativity clause for tubes (and guns) when pointed at you they are all large to very large.
N.B. people with
Bullous Emphysema: What It Is, Causes & Treatment Bullous emphysema is a lung disease usually caused by smoking. The main symptom is shortness of breath. Treatment ranges from bronchodilators to surgery.
can have all the finding of a pneumothorax. And not.
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RA
Former Admin (1985–2010) · Author has 4.4K answers and 3.2M answer views
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What is the difference between a pneumothorax and a tension pneumothorax?
Anatomy of Pneumothorax
The inner surface of the thoracic cage (parietal pleura) is contiguous with the outer surface of the lung (visceral pleura); this space contains a small amount of lubricating fluid and is normally under negative pressure compared to the alveoli. Determinants of pleural pressure are the opposing recoil forces of the lung and chest wall.
Pathophysiology of Tension Pneumothorax
A tension pneumothorax is a life-threatening condition that develops when air is trapped in the pleural cavity under positive pressure, displacing mediastinal structures and compromising cardiopulmonar
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Anatomy of Pneumothorax
The inner surface of the thoracic cage (parietal pleura) is contiguous with the outer surface of the lung (visceral pleura); this space contains a small amount of lubricating fluid and is normally under negative pressure compared to the alveoli. Determinants of pleural pressure are the opposing recoil forces of the lung and chest wall.
Pathophysiology of Tension Pneumothorax
A tension pneumothorax is a life-threatening condition that develops when air is trapped in the pleural cavity under positive pressure, displacing mediastinal structures and compromising cardiopulmonary function. Prompt recognition of this condition is life saving, both outside the hospital and in a modern ICU. Because tension pneumothorax occurs infrequently and has a potentially devastating outcome, a high index of suspicion and knowledge of basic emergency thoracic decompression procedures are important for all healthcare personnel. Immediate decompression of the thorax is mandatory when tension pneumothorax is suspected. This should not be delayed for radiographic confirmation. Note the image below.
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Eveline Frei
pleurodesis on both lungs · Author has 727 answers and 1.4M answer views
·6y
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Can tension pneumothorax be treated without a Chest Decompression Needle?
Well, when I had my pneumothoraces, I was told to carry something to make a hole and something to keep it open in case I developed a tension pneumothorax. Theoretically, I settled on a knitting needle and a rubber-/latex-glove with a hole though I ended up not needing any of it.
Something is called a tension pneumothorax when more and more air gets into the thorax and little or nothing can get out…that produces tension and can kill you unless something is done quickly to remedy the situation. A chest decompression needle looks like an ideal instrument for that but, as my doctors told me, a knit
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Well, when I had my pneumothoraces, I was told to carry something to make a hole and something to keep it open in case I developed a tension pneumothorax. Theoretically, I settled on a knitting needle and a rubber-/latex-glove with a hole though I ended up not needing any of it.
Something is called a tension pneumothorax when more and more air gets into the thorax and little or nothing can get out…that produces tension and can kill you unless something is done quickly to remedy the situation. A chest decompression needle looks like an ideal instrument for that but, as my doctors told me, a knitting needle with the cut off finger of a plastic glove might do it too…personally, I’d go for the decompression needle if I had the choice.
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Julian Money-Kyrle
Former Consultant Oncologist (Retired) at National Health Service (NHS) (1986–2017) · Author has 4.4K answers and 8M answer views
·9mo
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How do you clinically distinguish pneumothorax and emphysema?
That is a strange question. I suppose it might be something that you would ask a medical student to assess whether they have been paying attention to their studies, in which case the right answer is “Go and look it up yourself or you won’t learn anything”. I certainly wouldn’t want to be treated by any type of healthcare professional who thinks that Quora is a substitute for proper study - not only are they unlikely to have the required knowledge when they need it in a hurry, but their whole attitude would be suspect. I wouldn’t want to work with someone like that, either.
However, assuming tha
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That is a strange question. I suppose it might be something that you would ask a medical student to assess whether they have been paying attention to their studies, in which case the right answer is “Go and look it up yourself or you won’t learn anything”. I certainly wouldn’t want to be treated by any type of healthcare professional who thinks that Quora is a substitute for proper study - not only are they unlikely to have the required knowledge when they need it in a hurry, but their whole attitude would be suspect. I wouldn’t want to work with someone like that, either.
However, assuming that this is a genuine question, in order to answer it you need to know what these two conditions are. Emphysema is a chronic condition progressing over many years. It is characterised by shortness of breath and there is usually a history of smoking (though there are other causes, such as alpha-1 antitrypsin deficiency). Pneumothorax is an acute condition, so the symptoms (shortness of breath and sometimes chest pain) start suddenly. Therefore taking a history should allow you to distinguish them.
The clinical signs can be similar - increased percussion note, decreased breath sounds and hyperinflated chest. However, with pneumothorax the signs are clearly one-sided, and on the affected side the percussion note is more resonant than you would expect with emphysema and the breath sounds are completely absent. There may also be hyperinflation of one side of the chest and deviation of the trachea away from the affected side (both of these are warning signs of a tension pneumothorax, which requires urgent treatment). Occasionally there is skin crepitus due to surgical emphysema.
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Covet Sentence Examples
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What do many board game players covet most?
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188133 | https://byjus.com/chemistry/cathode-and-anode/ | Before we learn about cathode and anode we need to first understand what an electrode is. As per the general definition, an electrode is a substance that helps in the conduction of electricity wherein the electric current either enters or leaves the non-metallic medium like an electrolytic cell.
In simple terms, an electrode is a conductor that helps in establishing electrical contact with a non-metallic part of the circuit. Electrodes consist of two main points known as cathode and anode which basically describe the direction of flow of current.
What are Cathode and Anode?
Cathode
| |
| Cathode is said to be the electrode where reduction occurs. |
Anode
| |
| Anode is the point where an oxidation reaction occurs. |
Let us understand what cathode and anode exactly mean. They are both defined by the flow of current. Therefore, a cathode is an electrode from which the current exits a polarized electrical device. Likewise, an anode is an electrode from which a current enters into a polarized electrical device.
The terms were finalized in 1834 by William Whewell who adapted the words from the Greek word (kathodos), ‘descent’ or ‘way down’. William had consulted with Michael Faraday for the coining of the terms.
Cathode
When we talk about cathode in chemistry, it is said to be the electrode where reduction occurs. This is common in an electrochemical cell. Here, the cathode is negative as the electrical energy that is supplied to the cell results in the decomposition of chemical compounds. However, it can also be positive as in the case of a galvanic cell where a chemical reaction leads to the generation of electrical energy.
In addition, a cathode is said to be either a hotcathode or a cold cathode. A cathode which is heated in the presence of a filament to emit electrons by thermionic emission is known as a hot cathode whereas cold cathodes are not heated by any filament. A cathode is usually flagged as “cold” if it emits more electrons compared to the ones generated by thermionic emission alone.
Anode
In the most basic form, an anode in electrochemistry is the point where an oxidation reaction occurs. Generally, at an anode, negative ions or anions due to its electrical potential tend to react and give off electrons. These electrons then move up and into the driving circuit.
If we take a galvanic cell, the anode is negative in nature and the electrons mostly move towards the external part of the circuit. In an electrolytic cell, it is again positive. Additionally, an anode can be a plate or wire having an excess positive charge.
Difference Between Anode and Cathode
Here are some key differences between cathode and anode.
| | |
--- |
| Anode | Cathode |
| The anode is the electrode where electricity moves into. | The cathode is the electrode where electricity is given out or flows out. |
| The anode is usually the positive side. | A cathode is a negative side. |
| It acts as an electron donor. | It acts as an electron acceptor. |
| In an electrolytic cell, oxidation reaction takes place at the anode. | In an electrolytic cell, a reduction reaction takes place at the cathode. |
| In galvanic cells, an anode can become a cathode. | In galvanic cells, a cathode can become an anode. |
Frequently Asked Questions on Cathode and Anode
What is the charge of an anode and cathode?
The anode is regarded as negative in a galvanic (voltaic) cell and the cathode is deemed positive. This seems appropriate because the anode is the origin of electrons and where the electrons flow is the cathode.
Does oxidation occur at the anode or cathode?
The anode is where the response to oxidation occurs. That’s where the metal loses electrons, in other words.
Q3
What is the charge on anode and cathode?
There is an oxidation response at the anode. The oxidized species would lose electrons, leaving this electrode with an accumulation of electrons. Therefore, the anode is charged negatively. In contrast to the cathode, there is a reduction response where the decreased species would obtain electrons. Therefore, the electrode, i.e. the cathode, lacks electrons and is therefore charged positively.
Q4
Are cations positive or negative?
A cation is defined as a positively charged ion or an atom that has lost an electron.
Q5
What are the materials used for anode and cathode?
Metals like zinc and lithium are often used as substrates for anodes.
Q6
What is anode and cathode in corrosion?
Iron metal functions as the anode in a galvanic cell during the corrosion phase and is oxidized to Fe2+; at the cathode, oxygen is decreased to water.
Q7
Does reduction always occur at the cathode?
Reduction at the cathode always happens, and oxidation at the anode always happens. Because decrease is the addition of electrons.
Q8
Is LED cathode positive or negative?
LEDs are generally labelled in some way by their cathode. The cathode should be linked to the driving voltage source’s floor or adverse side and the anode to the positive side.
Q9
Do electrons always flow from an anode to a cathode?
Yes, electrons always flow from an anode to a cathode or from the oxidation half cell to the reduction half cell.
Q10
What is the primary goal of a salt bridge?
The primary goal of a salt bridge is to maintain the electrical neutrality of the cell and minimise the liquid junction potential.
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Daniel
January 27, 2021 at 6:02 pm
There appears to be a contradiction here.
In the table, it’s stated that the anode “is usually the positive side” but in the subsection “Anode” it is stated “If we take a galvanic cell, the anode is negative in nature” and the answer to the first question in FAQ section states “The anode is regarded as negative in a galvanic (voltaic) cell”.
So, is the anode negative or positive?
My position is it’s negative, it’s the electrode that distributes electrons to the circuit.
I’ll bookmark this page in anticipation of an answer, thank you.
Reply
Conventionally, cathodes are considered to be positve and anodes are considered to be negative. However, these are reversed while dealing with electrolytic cells. For an electrolytic cell, the cathode will be negative and the anode will be positive.
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188134 | https://math.stackexchange.com/questions/3247106/understanding-ex-falso-quodlibet-together-with-proof-by-contradiction-in-a-gentz | logic - Understanding ex falso quodlibet together with proof by contradiction in a Gentzen style ND Proof - Mathematics Stack Exchange
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Understanding ex falso quodlibet together with proof by contradiction in a Gentzen style ND Proof
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I began studying some formal logic for possible future proof and type theory dives. I am at the very beginning, Gentzen style natural deductions. Some of these proof rules defies my intuition so I wanted to ask for some clarification.
Suppose we want to prove that ¬(P⇒Q)⇒P¬(P⇒Q)⇒P. Informally speaking:
Assume ¬(P⇒Q)¬(P⇒Q), show P P.
To show P P, we will use proof by contradiction and we will assume ¬P¬P.
{ At this stage, we have premises ¬(P⇒Q)¬(P⇒Q) and ¬P¬P in our environment Γ Γ }
Now, If we can show (P⇒Q)(P⇒Q) from Γ Γ, we can conclude ⊥⊥ and complete our proof.
To show (P⇒Q)(P⇒Q), assume P P and show Q Q.
{ At this stage, we have premises ¬(P⇒Q)¬(P⇒Q), ¬P¬P and P P in our environment Γ Γ }
This is our proof strategy, informally. Below I show this as a formal proof tree. I hope it is readable enough, I had to play a lot with frac of L A T E X L A T E X. Notationally, [P]x[P]x means that P P is an assumption that is going to be removed from Γ Γ at the proof step x x somewhere below. efq stands for ex falso quodlibet.
¬(P⇒Q)[¬P]y[P]x⊥Q efq¬−E P⇒Q⇒−I x⊥P efq-y¬(P⇒Q)⇒P⇒−I z¬(P⇒Q)[¬P]y[P]x⊥Q efq¬−E P⇒Q⇒−I x⊥P efq-y¬(P⇒Q)⇒P⇒−I z
This was an example in Jean Gallier's Discrete Mathematics book. I feel like I got the mechanical parts understood. Still, my head hurts when thinking about how I assumed both P P and ¬P¬P to prove something. How do you guys understand this? How can we assume both P P and ¬P¬P, isn't that absurd? What am I missing so that I feel like this proof is bogus? Why do I feel like I can prove anything with this?
As a programmer, I suspect these assumptions are like local variables in a function from a computer program. For example, that assumption of P P is valid only in the program that proves Q Q and is removed from the stack after that. Trying to use it for any other proof would be using an already deleted variable etc. I am not sure about forming this analogy though, since it is not mentioned in the book. I want logician's explanation.
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edited Jun 2, 2019 at 5:43
Taroccoesbrocco
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If you discharge an assumption labelled z z you should remember to actually label the assumption.
Also, ex falso quodlibet is not a rule of discharging; the rule to discharge y y is negation introduction, which needs to be followed by double negation elimination (though some do combine it into Reduction Ad Absurdum (RAA)).
[¬(P→Q)]z[¬P]y[P]x⊥¬E Q e f q P→Q→I x⊥e f q¬¬P¬I y P¬¬E¬(P→Q)→P→I z[¬(P→Q)]z[¬P]y[P]x⊥¬E Q e f q P→Q→I x⊥e f q¬¬P¬I y P¬¬E¬(P→Q)→P→I z
This was an example in Jean Gallier's Discrete Mathematics book. I feel like I got the mechanical parts understood. Still, my head hurts when thinking about how I assumed both P P and ¬P¬P to prove something. How do you guys understand this? How can we assume both P P and ¬P¬P, isn't that absurd?
Yes, that is how Reduction to Absurdity proofs operate: "If this assumption was true it, then what follows would be absurd, so therefore it cannot be true."
In this proof: When given ¬(P→Q)¬(P→Q) should we also assume ¬P¬P, then we could prove P→Q P→Q (through the ex falso quodlibet subproof), which is absurd, so therefore ¬(P→Q)¬(P→Q) entails P P, and so we deduce ¬(P→Q)→P¬(P→Q)→P.
Here's the fitch style representation which may be easier to follow.
1.¬(P→Q)2.¬P 3.P 4.⊥¬e,2,3 5.Q efq,4 6.P→Q→i,3−5 7.⊥¬e,1,6 8.¬¬P¬i,2−7 9.P¬¬e,8 10.¬(P→Q)→P→i,1−9 1.¬(P→Q)2.¬P 3.P 4.⊥¬e,2,3 5.Q efq,4 6.P→Q→i,3−5 7.⊥¬e,1,6 8.¬¬P¬i,2−7 9.P¬¬e,8 10.¬(P→Q)→P→i,1−9
In this notation we keep track of the order in which assumptions are raised and discharged by indentation. Here we see that a contradiction can be derived on raising the third assumption, and so we may validly derive Q Q within that context (because whatever Q Q may be, it is at least as true as an absurdity).
Ex falso Quodlibet: If we can say "false is true", then we may say anything is true.
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edited Jun 1, 2019 at 2:42
answered Jun 1, 2019 at 1:36
Graham KempGraham Kemp
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Then, what stops me from doing this; to prove just about any proposition Q Q, assume some arbitrary P P and ¬P¬P. Since ⊥⊥ can be deduced now, I guess we can prove any Q Q by efq. Does false imply everything?meguli –meguli 2019-06-01 01:56:08 +00:00 Commented Jun 1, 2019 at 1:56
1 Yes, if you can derive an absurdity within a context, then anything you require may be validly derived within that context. This is only useful when you can properly discharge the assumptions that raise that context. Here we discharge P P through conditional elimination, and ¬P¬P through negation introduction.Graham Kemp –Graham Kemp 2019-06-01 02:00:47 +00:00 Commented Jun 1, 2019 at 2:00
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Here's how we can organize things (informally) to make a bit more sense.
We want to prove ¬(P→Q)→P,¬(P→Q)→P, so we assume ¬(P→Q)¬(P→Q) and our goal is to prove P P.
We will prove P P by assuming ¬P¬P and deriving a contradiction.
We will derive a contradiction by proving P→Q,P→Q, which contradicts with our assumption of ¬(P→Q).¬(P→Q).
We will prove P→Q P→Q by assuming P P and then proving Q.Q.
We will prove Q Q by proving a contradiction, from which we can prove anything (ex falso.)
There is a contradiction between our assumption of ¬P¬P and our assumption of P.P.
So we did assume ¬P¬P and assume P P and get nonsense, but the assumptions had an orderly motivation and nonsense was exactly what we needed to prove Q,Q, which was a goal we had at that moment in the proof. (And this was certainly not an unconditional proof of an absurdity, it was under specific assumptions that were framed up in the larger structure of the proof.) Perhaps a better way to think about it is that what we wanted to prove was P→Q P→Q under the assumption that P P is false... which should seem quite plausible.
As Graham Kemp points out in his answer, the convention of using 'efq' to discharge an assumption is a little bit non-standard and usually this is called RAA or framed as double negation elimination.
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edited Jun 1, 2019 at 2:25
answered Jun 1, 2019 at 2:01
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FYI there is an easier way. No EFQ (explosion) or proof by contradiction is required. Just apply the definition of ⟹⟹ (line 2, below).
Screen print from DC Proof 2.0 proof checker:
If you insist on using EFQ (line 3, below) and proof by contradiction (line 5 and 6), try to following:
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edited Jun 4, 2019 at 2:40
answered Jun 3, 2019 at 20:09
Dan ChristensenDan Christensen
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1 Well, yes, proofs using derived rules are usually more compact than proofs using the fundamental rules of inference. But this misses the point that the OP wanted to understand how those fundamental rules were used.Graham Kemp –Graham Kemp 2019-06-04 00:55:43 +00:00 Commented Jun 4, 2019 at 0:55
@GrahamKemp +1 Thanks for pointing that out. See my 2nd proof added just now.Dan Christensen –Dan Christensen 2019-06-04 02:41:01 +00:00 Commented Jun 4, 2019 at 2:41
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188135 | https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality?srsltid=AfmBOoqBSBvnr88jlLbIbJkQtN6bCjBgmnPZXkBgbF3gGPm1_gH2ur24 | Art of Problem Solving
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Cauchy-Schwarz Inequality
In algebra, the Cauchy-Schwarz Inequality, also known as the Cauchy–Bunyakovsky–Schwarz Inequality or informally as Cauchy-Schwarz, is an inequality with many ubiquitous formulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear algebra.
Its elementary algebraic formulation is often referred to as Cauchy's Inequality and states that for any list of reals and , with equality if and only if there exists a constant such that for all , or if one list consists of only zeroes. Along with the AM-GM Inequality, Cauchy-Schwarz forms the foundation for inequality problems in intermediate and olympiad competitions. It is particularly crucial in proof-based contests.
Its vector formulation states that for any vectors and in , where is the dot product of and and is the norm of , with equality if and only if there exists a scalar such that , or if one of the vectors is zero. This formulation comes in handy in linear algebra problems at intermediate and olympiad problems.
The full Cauchy-Schwarz Inequality is written in terms of abstract vector spaces. Under this formulation, the elementary algebraic, linear algebraic, and calculus formulations are different cases of the general inequality.
Contents
[hide]
1 Proofs
2 Lemmas
2.1 Complex Form
2.2 A Useful Inequality
3 Real Vector Spaces
3.1 Proof 1
3.2 Proof 2
3.3 Proof 3
4 Complex Vector Spaces
4.1 Proof
5 Problems
5.1 Introductory
5.2 Intermediate
5.3 Olympiad
6 Other Resources
6.1 Books
Proofs
Here is a list of proofs of Cauchy-Schwarz.
Consider the vectors and . If is the angle formed by and , then the left-hand side of the inequality is equal to the square of the dot product of and , or .The right hand side of the inequality is equal to . The inequality then follows from , with equality when one of is a multiple of the other, as desired.
Lemmas
Complex Form
The inequality sometimes appears in the following form.
Let and be complex numbers. Then This appears to be more powerful, but it follows from
A Useful Inequality
Also known as Sedrakyan's Inequality, Bergström's Inequality, Engel's Form or Titu's Lemma the following inequality is a direct result of Cauchy-Schwarz inequality:
For any real numbers and where the following is true:
Real Vector Spaces
Let be a vector space, and let be an inner product. Then for any , with equality if and only if there exist constants not both zero such that . The following proofs assume the inner product to be real-valued and commutative, and so only apply to vector spaces over the real numbers.
Proof 1
Consider the polynomial of This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., must be less than or equal to , with equality when or when there exists some scalar such that , as desired.
Proof 2
We consider Since this is always greater than or equal to zero, we have Now, if either or is equal to , then . Otherwise, we may normalize so that , and we have with equality when and may be scaled to each other, as desired.
Proof 3
Consider for some scalar . Then: (by the Trivial Inequality) . Now, let . Then, we have: .
Complex Vector Spaces
For any two vectors in the complex vector space , the following holds: with equality holding only when are linearly dependent.
Proof
The following proof, a geometric argument that uses only the algebraic properties of the inner product, was discovered by Tarung Bhimnathwala in 2021.
Define the unit vectors , as and . Put . In other words, is the complex argument of and lies on the unit circle. If any of the denominators are zero, the entire result follows trivially. Let and . Importantly, we have Since and , this calculation shows that and form an orthogonal basis of the linear subspace spanned by and . Thus we can think of and as lying on the unit sphere in this subspace, which is isomorphic to . Another thing to note is that
The previous two calculations established that and are orthogonal, and that the sum of their squared norms is . Now we have Equality holds when either or , or equivalently when and . Lastly, multiplying each side by , we have
Problems
Introductory
Consider the function , where is a positive integer. Show that . (Source)
(APMO 1991 #3) Let , , , , , , , be positive real numbers such that . Show that
Intermediate
Let be a triangle such that
where and denote its semiperimeter and inradius, respectively. Prove that triangle is similar to a triangle whose side lengths are all positive integers with no common divisor and determine those integers. (Source)
Olympiad
is a point inside a given triangle . are the feet of the perpendiculars from to the lines , respectively. Find all for which
is least.
(Source)
Other Resources
Wikipedia entry
Books
The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele.
Problem Solving Strategies by Arthur Engel contains significant material on inequalities.
Retrieved from "
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When a number of objects are created from the same class blueprint, they each have their own distinct copies of instance variables. In the case of the Bicycle class, the instance variables are cadence, gear, and speed. Each Bicycle object has its own values for these variables, stored in different memory locations.
Sometimes, you want to have variables that are common to all objects. This is accomplished with the static modifier. Fields that have the static modifier in their declaration are called static fields or class variables. They are associated with the class, rather than with any object. Every instance of the class shares a class variable, which is in one fixed location in memory. Any object can change the value of a class variable, but class variables can also be manipulated without creating an instance of the class.
For example, suppose you want to create a number of Bicycle objects and assign each a serial number, beginning with 1 for the first object. This ID number is unique to each object and is therefore an instance variable. At the same time, you need a field to keep track of how many Bicycle objects have been created so that you know what ID to assign to the next one. Such a field is not related to any individual object, but to the class as a whole. For this you need a class variable, numberOfBicycles, as follows:
```
public class Bicycle {
private int cadence;
private int gear;
private int speed;
// add an instance variable for the object ID
private int id;
// add a class variable for the
// number of Bicycle objects instantiated
private static int numberOfBicycles = 0;
...
}
```
Class variables are referenced by the class name itself, as in
```
Bicycle.numberOfBicycles
```
This makes it clear that they are class variables.
Note: You can also refer to static fields with an object reference like
```
myBike.numberOfBicycles
```
but this is discouraged because it does not make it clear that they are class variables.
You can use the Bicycle constructor to set the id instance variable and increment the numberOfBicycles class variable:
```
public class Bicycle {
private int cadence;
private int gear;
private int speed;
private int id;
private static int numberOfBicycles = 0;
public Bicycle(int startCadence, int startSpeed, int startGear){
gear = startGear;
cadence = startCadence;
speed = startSpeed;
// increment number of Bicycles
// and assign ID number
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// new method to return the ID instance variable
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return id;
}
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}
```
Class Methods
The Java programming language supports static methods as well as static variables. Static methods, which have the static modifier in their declarations, should be invoked with the class name, without the need for creating an instance of the class, as in
```
ClassName.methodName(args)
```
Note: You can also refer to static methods with an object reference like
```
instanceName.methodName(args)
```
but this is discouraged because it does not make it clear that they are class methods.
A common use for static methods is to access static fields. For example, we could add a static method to the Bicycle class to access the numberOfBicycles static field:
```
public static int getNumberOfBicycles() {
return numberOfBicycles;
}
```
Not all combinations of instance and class variables and methods are allowed:
Instance methods can access instance variables and instance methods directly.
Instance methods can access class variables and class methods directly.
Class methods can access class variables and class methods directly.
Class methods cannot access instance variables or instance methods directly—they must use an object reference. Also, class methods cannot use the this keyword as there is no instance for this to refer to.
Constants
The static modifier, in combination with the final modifier, is also used to define constants. The final modifier indicates that the value of this field cannot change.
For example, the following variable declaration defines a constant named PI, whose value is an approximation of pi (the ratio of the circumference of a circle to its diameter):
```
static final double PI = 3.141592653589793;
```
Constants defined in this way cannot be reassigned, and it is a compile-time error if your program tries to do so. By convention, the names of constant values are spelled in uppercase letters. If the name is composed of more than one word, the words are separated by an underscore (_).
Note: If a primitive type or a string is defined as a constant and the value is known at compile time, the compiler replaces the constant name everywhere in the code with its value. This is called a compile-time constant. If the value of the constant in the outside world changes (for example, if it is legislated that pi actually should be 3.975), you will need to recompile any classes that use this constant to get the current value.
The Bicycle Class
After all the modifications made in this section, the Bicycle class is now:
```
public class Bicycle {
private int cadence;
private int gear;
private int speed;
private int id;
private static int numberOfBicycles = 0;
public Bicycle(int startCadence,
int startSpeed,
int startGear) {
gear = startGear;
cadence = startCadence;
speed = startSpeed;
id = ++numberOfBicycles;
}
public int getID() {
return id;
}
public static int getNumberOfBicycles() {
return numberOfBicycles;
}
public int getCadence() {
return cadence;
}
public void setCadence(int newValue) {
cadence = newValue;
}
public int getGear(){
return gear;
}
public void setGear(int newValue) {
gear = newValue;
}
public int getSpeed() {
return speed;
}
public void applyBrake(int decrement) {
speed -= decrement;
}
public void speedUp(int increment) {
speed += increment;
}
}
```
« Previous
•
Trail
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188137 | https://opentext.uleth.ca/Combinatorics/sect_designs-FisherIneq.html | Fisher's Inequality
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Combinatorics
by Joy Morris
Contents
IndexPrevUpNext
ContentsPrevUpNext
Front Matter
1 What is Combinatorics?
Enumeration
Graph Theory
Ramsey Theory
Design Theory
Coding Theory
I Enumeration
2 Basic Counting Techniques
The product rule
The sum rule
Putting them together
Summing up
3 Permutations, Combinations, and the Binomial Theorem
Permutations
Combinations
The Binomial Theorem
4 Bijections and Combinatorial Proofs
Counting via bijections
Combinatorial proofs
The Arithmetic Triangle (Pascal's Triangle)
5 Counting with Repetitions
Unlimited repetition
Sorting a set that contains repetition
6 Induction and Recursion
Recursively-defined sequences
Basic induction
More advanced induction
7 Generating Functions
What is a generating function?
The Generalised Binomial Theorem
Using generating functions to count things
8 Generating Functions and Recursion
Partial fractions
Factoring polynomials
Using generating functions to solve recursively-defined sequences
9 Some Important Recursively-Defined Sequences
Derangements
Catalan numbers
Bell numbers and exponential generating functions
10 Other Basic Counting Techniques
The Pigeonhole Principle
Inclusion-Exclusion
II Graph Theory
11 Basics of Graph Theory
Background
Basic definitions, terminology, and notation
Subgraphs, complete graphs, and the Handshaking Lemma
Isomorphism of graphs
Random graphs
12 Moving through graphs
Directed graphs
Walks and connectedness
Paths and cycles
Trees
Automorphisms of graphs
13 Euler and Hamilton
Euler tours and trails
Hamilton paths and cycles
14 Graph Colouring
Edge colouring
Ramsey Theory
Vertex colouring
15 Planar graphs
Planar graphs
Euler's Formula
Map colouring
III Design Theory
16 Latin squares
Latin squares and Sudokus
Mutually orthogonal Latin squares (MOLS)
Systems of distinct representatives
17 Designs
Balanced Incomplete Block Designs (BIBD)
Constructing designs, and existence of designs
Fisher's Inequality
18 More designs
Steiner and Kirkman triple systems
t t-designs
Affine planes
Projective planes
19 Designs and Codes
Introduction
Error-correcting codes
Using the generator matrix for encoding
Using the parity-check matrix for decoding
Codes from designs
Back Matter
A Complex Numbers
B Biographical Briefs
List of Entries
Biographies
C Solutions to selected exercises
Solutions for Chapter 2
Solutions for Chapter 3
Solutions for Chapter 4
Solutions for Chapter 5
Solutions for Chapter 6
Solutions for Chapter 7
Solutions for Chapter 8
Solutions for Chapter 9
Solutions for Chapter 10
Solutions for Chapter 11
Solutions for Chapter 12
Solutions for Chapter 13
Solutions for Chapter 14
Solutions for Chapter 15
Solutions for Chapter 16
Solutions for Chapter 17
Solutions for Chapter 18
Solutions for Chapter 19
D List of Notation
Index
HTML produced by PreTeXt
🔗
Section 17.3 Fisher's Inequality
🔗
There is one more important inequality that is not at all obvious, but is necessary for the existence of a BIBD(v,k,λ).(v,k,λ). This is known as Fisher's Inequality, since it was proven by Sir Ronald Aylmer Fisher (1890—1962). The proof we will give is somewhat longer than the standard proof. This is because the standard proof uses linear algebra, which we do not expect to be required background for this course.
🔗
Theorem 17.3.1(Fisher's Inequality).
For any BIBD(v,k,λ),(v,k,λ), we must have b≥v.b≥v.
🔗
Before proving this fact, let's observe the consequences in terms of the usual parameters: v,v,k,k, and λ.λ. We know from Equation(17.1.2) that
b=λ v(v−1)k(k−1),b=λ v(v−1)k(k−1),
🔗
so b≥v b≥v implies
λ v(v−1)k(k−1)≥v.λ v(v−1)k(k−1)≥v.
🔗
Since v v is the number of points of a design, it must be positive, so dividing through by v v does not reverse the inequality. Thus,
λ(v−1)k(k−1)≥1.λ(v−1)k(k−1)≥1.
🔗
Since k k is the number of points in each block, both k k and k−1 k−1 must be positive (we are ignoring the trivial case k=1 k=1), so multiplying through by k(k−1)k(k−1) does not reverse the inequality. Thus,
λ(v−1)≥k(k−1).λ(v−1)≥k(k−1).
🔗
###### Proof of Fisher's Inequality
Suppose we have an arbitrary BIBD((v,k,\lambda)\text{.}) Let (B) be an arbitrary block of this design. For each value of (i) between (0) and (k) (inclusive), let (n_i) denote the number of blocks (B'\neq B) such that (|B'\cap B|=i\text{.}) (When we say (B' \neq B) we allow the blocks to be equal as sets if the block (B) is a repeated block of the design; we are only insisting that (B') not be the exact same block of the design as (B\text{.}))
The following equations involving (n_i) are consequences of easy combinatorial proofs, together with the definition of (n_i\text{:})
\begin{equation} \sum_{i=0}^k n_i = b-1\text{,}\label{eqn_Fisher1}\tag{17.3.1} \end{equation}
because both sides of this equation count every block except (B\text{.})
\begin{equation} \sum_{i=0}^k in_i = k(r-1)\text{,}\label{eqn_Fisher2}\tag{17.3.2} \end{equation}
because both sides of this equation count the number of times elements of (B) appear in some other block of the design.
\begin{equation} \sum_{i=2}^k i(i-1)n_i =k(k-1)(\lambda-1)\text{,} \end{equation}
because both sides of this equation count the number of times all of the ordered pairs of elements from (B) appear together in some other block of the design. Note that when (i=0) or (i=1\text{,}) we have (i(i-1)n_i=0\text{,}) so in fact
\begin{equation} \sum_{i=0}^k i(i-1)n_i =\sum_{i=2}^k i(i-1)n_i =k(k-1)(\lambda-1)\text{.}\label{eqn_Fisher3}\tag{17.3.3} \end{equation}
Adding Equations(17.3.2) and (17.3.3) gives
\begin{equation} \sum_{i=0}^k i^2n_i = k(k-1)(\lambda-1)+k(r-1)\text{.}\label{eqn_Fisher4}\tag{17.3.4} \end{equation}
Now comes the part of the proof where something mysterious happens, and for reasons that are not at all apparent, the result we want will emerge. To fully understand a proof like this one requires deeper mathematics, but even seeing a proof is useful to convince ourselves that the result is true.
Take the polynomial in (x) given by
\begin{equation} \sum_{i=0}^k (x-i)^2 n_i = \sum_{i=0}^k (x^2-2xi+i^2)n_i= x^2\sum_{i=0}^k n_i-2x\sum_{i=0}^k in_i+\sum_{i=0}^k i^2n_i\text{.} \end{equation}
Using Equations(17.3.1),(17.3.2), and (17.3.4), we see that this is equal to
\begin{equation} x^2(b-1)-2xk(r-1)+k(k-1)(\lambda-1)+k(r-1)\text{.} \end{equation}
Notice that the format in which this polynomial started was a sum of squares times non-negative integers, so its value must be non-negative for any (x \in \mathbb R\text{.})
Using the quadratic formula, (ax^2+b'x+c=0) has roots at
\begin{equation} \frac{-b'\pm\sqrt{(b')^2-4ac}}{2a}\text{.} \end{equation}
If a quadratic polynomial has two real roots, then there is a region in which its values are negative. Since this polynomial is non-negative for every (x \in \mathbb R\text{,}) it can have at most one real root, so ((b')^2-4ac \le 0\text{.}) Substituting the actual values from our polynomial, this means that
\begin{equation} (-2k(r-1))^2-4(b-1)(k(k-1)(\lambda-1)+k(r-1)) \le 0\text{.} \end{equation}
Hence,
\begin{equation} k^2(r-1)^2-k(b-1)((k-1)(\lambda-1)+r-1) \le 0\text{.} \end{equation}
Let's rewrite the (b) in terms of (v, r\text{,}) and (k\text{.}) By Theorem 17.1.7, we have (bk=vr\text{,}) so
\begin{equation} k(b-1)=bk-k=vr-k\text{.} \end{equation}
Hence
\begin{equation} k^2(r-1)^2-(vr-k)((k-1)(\lambda-1)+r-1) \le 0\text{.} \end{equation}
Expand the second term slightly, and multiply both sides of the inequality by (v-1\text{:})
\begin{equation} k^2(r-1)^2(v-1)-(vr-k)(k-1)(\lambda-1)(v-1)-(vr-k)(r-1)(v-1) \le 0\text{.} \end{equation}
In the middle expression, we have ((\lambda-1)(v-1)\text{.}) By Theorem 17.1.7, we know that (\lambda=r(k-1)/(v-1)\text{,}) so
\begin{equation} \lambda-1=\frac{r(k-1)-(v-1)}{v-1}\text{.} \end{equation}
Therefore,
\begin{equation} (\lambda-1)(v-1)=r(k-1)-v+1\text{.} \end{equation}
Thus, we have
\begin{equation} k^2(r-1)^2(v-1)-(vr-k)(k-1)(rk-r-v+1)-(vr-k)(r-1)(v-1) \le 0\text{.} \end{equation}
The next step is a lot of work to do by hand. Fortunately there is good math software that can perform routine tasks like this quickly. If we expand this inequality fully, remarkably it has a nice factorisation:
\begin{equation} r(k-r)(v-k)^2 \le 0\text{.} \end{equation}
Now, (r >0) for any design, and ((v-k)^2) is a square, so must be nonnegative. Therefore, this inequality forces (k-r \le 0\text{,}) so (k \le r\text{.}) Hence (r/k\ge 1\text{.}) Using Theorem 17.1.7, we have
\begin{equation} b=vr/k\ge v\text{,} \end{equation}
as desired.
⬛
🔗
Notice that if k k is fixed, then only finitely many values of v v do not meet Fisher's Inequality, so satisfying this inequality did not need to be added as a condition to Wilson's Theorem.
🔗
###### Exercise s 17.3.2.
Find values for (v\text{,}) (k) and (\lambda) that satisfy Theorem 17.1.9 but do not satisfy Fisher's Inequality. What can you say about the existence of a design with these parameters?
Suppose that (\lambda=1) and (k=20\text{.}) How big must (v) be to satisfy Fisher's Inequality? What is the smallest value for (v) that satisfies all of the necessary conditions?
Suppose that (\lambda=2) and (k=20\text{.}) How big must (v) be to satisfy Fisher's Inequality? What is the smallest value for (v) that satisfies all of the necessary conditions?
Explain how you know there does not exist a BIBD with (v = 46\text{,}) (b = 23\text{,}) and (k = 10\text{.})
Explain how you know there does not exist a BIBD with (v = 8\text{,}) (b = 10\text{,}) (k = 4\text{,}) and (r = 5\text{.})
If (\bibd) is a BIBD with (v = 22\text{,}) then what can you say about the value of(b\text{?}) |
188138 | https://brightchamps.com/en-gb/math/math-worksheets/consecutive-integers-worksheets | Table Of Contents
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Consecutive Integers Worksheets
Consecutive integers are numbers that follow each other in order without any gaps. Practicing with consecutive integers worksheets helps students understand patterns and relationships between numbers. These worksheets are designed to help students work with problems involving consecutive numbers.
What are Consecutive Integer Worksheets?
Consecutive integer worksheets are practice materials that help students identify and work with sequences of numbers that follow one after the other without any gaps. These worksheets often include exercises like finding sums of consecutive integers, solving equations involving consecutive integers, and understanding their properties.
Benefits of Consecutive Integer Worksheets
Consecutive integer worksheets offer numerous benefits. Some of them are mentioned below:
Enables practice: These worksheets include a variety of problems, such as sequences and word problems, giving students the opportunity to practice a range of scenarios.
Pattern recognition: They help students develop skills in identifying number patterns, which is crucial for understanding broader mathematical concepts.
Problem-solving skills: Working with consecutive integers enhances students' problem-solving abilities as they learn to approach and solve different types of problems.
Analytical thinking: Worksheets promote analytical thinking by challenging students to use logic and reasoning to find solutions.
Goal-setting: Consecutive integer worksheets encourage students to set small goals and achieve them, helping track progress and boosting motivation. Consecutive integers are fundamental in understanding sequences and patterns in math.
By using these worksheets, students can improve their analytical and problem-solving skills. Understanding consecutive integers is beneficial for both academic and real-life applications.
Download Free Printable Consecutive Integer Worksheets for Students
Consecutive integer worksheets provide an interactive way for students to practice essential math skills. Students can easily download PDF versions of these worksheets to practice at their own pace and strengthen their understanding of consecutive numbers.
Explore Consecutive Integer Worksheets by Grade
We’ve categorized the worksheets by grade level to help students find the appropriate level of challenge for their learning. Click on the respective grade for specialized worksheets:
FAQs for Consecutive Integer Worksheets
1.What is a consecutive integer worksheet?
A consecutive integer worksheet is a learning sheet with exercises that help students practice identifying and solving problems involving consecutive numbers. It often includes sequences, word problems, and equations to build understanding and skills.
A consecutive integer worksheet is a learning sheet with exercises that help students practice identifying and solving problems involving consecutive numbers. It often includes sequences, word problems, and equations to build understanding and skills.
2.How do consecutive integer worksheets help students?
They improve understanding of number sequences, enhance problem-solving skills, and help develop pattern recognition through regular practice.
They improve understanding of number sequences, enhance problem-solving skills, and help develop pattern recognition through regular practice.
3.What age group are consecutive integer worksheets for?
Typically for students in Grades 5 and above, depending on the difficulty level.
Typically for students in Grades 5 and above, depending on the difficulty level.
4.Are there different types of consecutive integer worksheets?
Yes, they can include problems involving sums of sequences, solving equations, identifying patterns, and related word problems.
Yes, they can include problems involving sums of sequences, solving equations, identifying patterns, and related word problems.
5.Can I download consecutive integer worksheets by grade?
Yes, many websites offer grade-specific consecutive integer worksheets aligned to local math curricula like Common Core or National Curriculum.
Yes, many websites offer grade-specific consecutive integer worksheets aligned to local math curricula like Common Core or National Curriculum.
Important Glossaries for Consecutive Integer Worksheets
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188139 | https://openmedscience.com/unlocking-the-nuclear-shell-model-secrets-of-stability/ | Nuclear Shell Model: Exploring Atomic Nuclei | Open Medscience
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Unlocking the Nuclear Shell Model: Secrets of Stability
By Open MedScience / 2024-10-10
The nuclear shell model is a theoretical framework that explains the structure and behaviour of atomic nuclei in terms of energy levels, much like the electron shell model in atomic physics. This model suggests that nucleons (protons and neutrons) within a nucleus occupy discrete energy levels or “shells.” By understanding these energy levels and their organisation, the shell model has been instrumental in explaining nuclear properties such as stability, magic numbers, and nuclear spins. This article explores the historical development of the nuclear shell model, the concepts of energy levels, the significance of magic numbers, and the model’s applications in both nuclear physics and technology.
Introduction
The study of atomic nuclei has been one of the central endeavours of nuclear physics for over a century. While early nuclear models were primarily concerned with the overall properties of nuclei, the nuclear shell model, introduced in the mid-20th century, provided a deeper understanding of the intricate structure within the nucleus. This model, akin to the shell model for electrons in atoms, describes how protons and neutrons, collectively known as nucleons, are arranged within the nucleus in discrete energy levels. The nuclear shell model has not only helped to explain the stability and properties of specific nuclei but has also led to practical applications in nuclear energy and medical imaging.
Historical Development of the Nuclear Shell Model
The nuclear shell model was developed in the late 1940s by physicists Maria Goeppert Mayer and J. Hans D. Jensen, for which they were awarded the Nobel Prize in Physics in 1963. Before this model, the nucleus was understood mainly through the liquid drop model, which treated the nucleus as a collection of nucleons held together by a force analogous to surface tension in liquids. While the liquid drop model successfully explained some nuclear properties, it failed to account for specific patterns in nuclear stability.
One of the most striking features that the liquid drop model could not explain was the occurrence of “magic numbers,” where nuclei with specific numbers of protons or neutrons were found to be unusually stable. It was through the nuclear shell model that these observations were explained. Goeppert Mayer and Jensen proposed that nucleons in a nucleus, like electrons in an atom, occupy energy levels, and when a shell is filled with nucleons, the nucleus becomes particularly stable. These filled shells correspond to the magic numbers: 2, 8, 20, 28, 50, 82, and 126.
See also The Liquid Drop Model: A Cornerstone Of Nuclear Physics
Structure of the Nuclear Shell Model
In the nuclear shell model, nucleons occupy discrete energy levels or “shells,” with each level capable of holding a certain number of protons or neutrons, much like the electron shell model in atomic theory. The arrangement of these energy levels depends on the interplay between the strong nuclear force, which binds nucleons together, and the Pauli exclusion principle, which prevents identical fermions (such as protons or neutrons) from occupying the same quantum state.
The energy levels are derived by solving the Schrödinger equation for a nucleon moving in a potential well, which is typically approximated by a combination of a central potential (representing the overall force holding the nucleons in the nucleus) and a spin-orbit coupling term (which arises from the interaction between a nucleon’s spin and its orbital motion).
Spin-Orbit Coupling
A crucial feature of the nuclear shell model is spin-orbit coupling, which refers to the interaction between a nucleon’s spin and its motion around the nucleus. This interaction causes each energy level to split into two sublevels, one with higher energy and one with lower energy. The strength of the spin-orbit interaction increases with the number of nucleons in the nucleus, leading to more pronounced splitting in heavier nuclei.
Spin-orbit coupling explains the ordering of nuclear energy levels and contributes to the occurrence of magic numbers. When the spin-orbit interaction is taken into account, it becomes clear that certain numbers of protons or neutrons lead to completely filled energy shells, corresponding to particularly stable nuclei.
Magic Numbers and Nuclear Stability
The concept of magic numbers is one of the key successes of the nuclear shell model. These numbers—2, 8, 20, 28, 50, 82, and 126—represent the number of protons or neutrons required to fill a nuclear shell. Nuclei with magic numbers of nucleons exhibit greater stability, as all the available quantum states within a shell are filled, leading to a lower overall energy for the nucleus.
For instance, the nucleus of helium-4, with 2 protons and 2 neutrons, is particularly stable because both the proton and neutron shells are filled. Similarly, lead-208, with 82 protons and 126 neutrons, is a doubly magic nucleus, where both the proton and neutron shells are filled, making it exceptionally stable.
Applications of the Nuclear Shell Model
The nuclear shell model has a wide range of applications in both theoretical and applied physics, including the following areas:
Nuclear Reactions and Decay
The nuclear shell model provides a framework for understanding nuclear reactions, including fission, fusion, and radioactive decay. In particular, the model helps to explain why certain nuclei are more prone to undergoing decay than others. Nuclei with magic numbers of nucleons are generally more resistant to decay, as they are more stable than nuclei with incomplete shells.
See also Radiation Safety: Balancing Benefits and Risks
Nuclear Spin and Magnetic Moments
The nuclear shell model also explains the magnetic moments and spins of nuclei. The total spin of a nucleus is determined by the spins of the individual nucleons, which in turn depend on their arrangement within the energy levels. In some cases, nucleons with paired spins cancel each other out, leading to a nucleus with zero total spin. In other cases, the spins of unpaired nucleons combine to give the nucleus a nonzero spin, which can be detected through its magnetic moment.
Isotopes and Nuclear Mass
The isotopes of an element differ in the number of neutrons they contain, and the nuclear shell model helps explain the variations in stability and mass among different isotopes. For example, certain isotopes are more stable than others because their neutron or proton shells are filled, leading to lower overall energy. The mass of a nucleus can also be understood in terms of the binding energy between nucleons, which is affected by the arrangement of protons and neutrons within the shells.
Limitations of the Nuclear Shell Model
While the nuclear shell model has been highly successful in explaining many nuclear properties, it does have its limitations. The model is most effective for describing nuclei near the magic numbers, where the shells are either completely filled or nearly filled. For nuclei with large numbers of unpaired nucleons, or for those far from the magic numbers, the shell model becomes less accurate, and more sophisticated models are required to explain the observed behaviour.
One such model is the collective model, which combines elements of the shell model with features of the liquid drop model to explain the collective motion of nucleons, such as rotations and vibrations of the nucleus as a whole. This model is particularly useful for describing the properties of deformed nuclei, where the simple shell model breaks down.
The Nuclear Shell Model and Modern Physics
Despite its limitations, the nuclear shell model remains a fundamental part of nuclear physics research. Advances in technology, particularly in particle accelerators and detectors, have allowed physicists to test the predictions of the shell model with greater precision. For example, experiments on unstable, short-lived nuclei have revealed new magic numbers that deviate from those predicted by the original model, leading to a deeper understanding of nuclear forces.
See also Types of Nuclear Reactors: Designs, Benefits, and Future Technologies
The nuclear shell model also plays a crucial role in astrophysics, particularly in the study of stellar nucleosynthesis, the process by which elements are formed in stars. The model helps to explain why certain elements are more abundant in the universe than others and why the production of heavy elements in supernovae follows specific patterns.
Conclusion
The nuclear shell model has provided a powerful framework for understanding the structure and behaviour of atomic nuclei. By treating nucleons as occupying discrete energy levels, the model explains a wide range of nuclear properties, from magic numbers and nuclear stability to the magnetic moments and spins of nuclei. Although the model has its limitations, it has been instrumental in advancing our understanding of nuclear physics and continues to be relevant in modern research.
Applications of the nuclear shell model extend beyond basic science, impacting fields such as nuclear energy, medical imaging, and astrophysics. Its ongoing relevance in both theoretical and applied contexts demonstrates the enduring significance of this foundational model in nuclear physics.
Disclaimer
The content provided in this article, “Unlocking the Nuclear Shell Model: Secrets of Stability”, is intended for informational and educational purposes only. While every effort has been made to ensure the accuracy and reliability of the information presented, Open MedScience makes no representations or warranties regarding the completeness, accuracy, or applicability of the material for any specific purpose.
The scientific concepts and models discussed, including the nuclear shell model, reflect the current understanding within the field of nuclear physics and may be subject to revision as new research emerges. Readers should not interpret the contents of this article as professional scientific advice or guidance.
This article does not replace formal academic instruction or professional consultation in nuclear physics, nuclear engineering, or related disciplines. Open MedScience accepts no liability for any loss, damage, or inconvenience caused as a result of reliance on the information contained herein.
For in-depth research, experimental application, or decision-making based on nuclear models, readers are encouraged to consult peer-reviewed literature and qualified experts in the field.
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This collection is one of our Primary Curriculum collections - tasks that are grouped by topic.
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Torn shapes
Age
7 to 11
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These rectangles have been torn. How many squares did each one have inside it before it was ripped?
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Through the window
Age
7 to 11
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My local DIY shop calculates the price of its windows according to the area of glass and the length of frame used. Can you work out how they arrived at these prices?
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Numerically equal
Age
7 to 11
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Can you draw a square in which the perimeter is numerically equal to the area?
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Fitted
Age
7 to 11
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Nine squares with side lengths 1, 4, 7, 8, 9, 10, 14, 15, and 18 cm can be fitted together to form a rectangle. What are the dimensions of the rectangle?
problem
Making boxes
Age
7 to 11
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Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
problem
Brush loads
Age
7 to 11
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How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well.
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Ribbon squares
Age
7 to 11
Challenge level
What is the largest 'ribbon square' you can make? And the smallest? How many different squares can you make altogether?
problem
Area and perimeter
Age
7 to 11
Challenge level
What can you say about these shapes? This problem challenges you to create shapes with different areas and perimeters.
problem
Twice as big?
Age
7 to 11
Challenge level
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
You may also be interested in this collection of activities from the STEM Learning website, that complement the NRICH activities above. |
188141 | https://pmc.ncbi.nlm.nih.gov/articles/PMC7069578/ | Melanogenic Difference Consideration in Ethnic Skin Type: A Balance Approach Between Skin Brightening Applications and Beneficial Sun Exposure - PMC
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Clin Cosmet Investig Dermatol
. 2020 Mar 9;13:215–232. doi: 10.2147/CCID.S245043
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Melanogenic Difference Consideration in Ethnic Skin Type: A Balance Approach Between Skin Brightening Applications and Beneficial Sun Exposure
Ewa Markiewicz
Ewa Markiewicz
1 Hexis Lab Limited, The Core, Newcastle Helix, Newcastle Upon Tyne NE4 5TF, UK
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1, Olusola Clement Idowu
Olusola Clement Idowu
1 Hexis Lab Limited, The Core, Newcastle Helix, Newcastle Upon Tyne NE4 5TF, UK
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1,✉
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1 Hexis Lab Limited, The Core, Newcastle Helix, Newcastle Upon Tyne NE4 5TF, UK
✉
Correspondence: Olusola Clement Idowu Hexis Lab, Newcastle Helix, The Core, Bath Lane, Newcastle Upon Tyne, NE4 5TF, UK, Phone: Tel +44 191 495 7311 Email sola@hexislab.com
Received 2020 Jan 7; Accepted 2020 Feb 13; Collection date 2020.
© 2020 Markiewicz and Idowu.
This work is published and licensed by Dove Medical Press Limited. The full terms of this license are available at and incorporate the Creative Commons Attribution – Non Commercial (unported, v3.0) License ( By accessing the work you hereby accept the Terms. Non-commercial uses of the work are permitted without any further permission from Dove Medical Press Limited, provided the work is properly attributed. For permission for commercial use of this work, please see paragraphs 4.2 and 5 of our Terms (
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PMCID: PMC7069578 PMID: 32210602
Abstract
Human skin demonstrates a striking variation in tone and color that is evident among multiple demographic populations. Such characteristics are determined predominantly by the expression of the genes controlling the quantity and quality of melanin, which can alter significantly due to the presence of small nucleotide polymorphism affecting various steps of the melanogenesis process and generally linked to the lighter skin phenotypes. Genetically determined, constitutive skin color is additionally complemented by the facultative melanogenesis and tanning responses; with high levels of melanin and melanogenic factors broadly recognized to have a protective effect against the UVR-induced molecular damage in darker skin. Long-term sun exposure, together with a genetic makeup responsible for the ability to tan or the activity of constitutive melanogenic factors, triggers defects in pigmentation across all ethnic skin types. However, sun exposure also has well documented beneficial effects that manifest at both skin homeostasis and the systemic level, such as synthesis of vitamin D, which is thought to be less efficient in the presence of high levels of melanin or potentially linked to the polymorphism in the genes responsible for skin darkening triggered by UVR. In this review, we discuss melanogenesis in a context of constitutive pigmentation, defined by gene polymorphism in ethnic skin types, and facultative pigmentation that is not only associated with the capacity to protect the skin against photo-damage but could also have an impact on vitamin D synthesis through gene polymorphism. Modulating the activities of melanogenic genes, with the focus on the markers specifically altered by polymorphism combined with differential requirements of sun exposure in ethnic skin types, could enhance the applications of already existing skin brightening factors and provide a novel approach toward improved skin tone and health in personalized skincare.
Keywords: ethnic skin types, melanogenesis, hyper-pigmentation, vitamin D
Introduction
Fine skin complexion is defined as clear and unblemished; ideal skin tone is associated with even distribution of the skin pigment –melanin; hyper-pigmentation is characterized by spots or patches that are darker than the skin surrounding them. These specific changes in pigmentation are predominantly linked to the excessive exposure to ultraviolet radiation (UVR) from sunlight, which is, in a broader sense, a well described factor also associated with photo-damage and premature ageing of the skin.1
Skin problems linked to photo-damage are generally considered to affect lighter skin; this is because melanin provides protection against UVR and lighter skin types are more susceptible to its damaging effects. However, the skin types predominantly affected by hyper-pigmentation problems are the darker phenotypes, including Oriental, Indian and occasionally African origin, which are also characterized by higher melanin content.2–5 It is therefore plausible to predict a fine balance in the melanin content required for sufficient protection against photo-damage, which is usually associated with darker skin tone, whilst keeping undesirable hyper-pigmentation responses low. Such balance could be affected if the melanogenesis process was insufficient and its protective effect against UVR exceeded, resulting in the enhanced sensitivity of the skin to sun-induced photo-damage.
Presently, sun protection products are broadly recommended as essential application to counteract the harmful effects of excessive UVR exposure. Such approach has resulted in significant reduction of skin problems caused by photo-damage; leading to further preference toward the sunscreens with high SPF (sun protection factor)which therefore have stronger ability to absorb or reflect the UVR reaching the skin.
However, the parallel evidence that is now emerging also demonstrates that sun exposure has direct and significant benefits in promoting health of the skin, predominantly through stimulation of vitamin D synthesis. Exposure to UVR causes photo-activation of 7-dehydrocholesterol in the skin and synthesis of vitamin D3, which is subsequently circulated and metabolized further to the active form of vitamin D. This mechanism is responsible for more than 90% of the vitamin D production in the body.6,7 Vitamin D is a pro-hormone essential for calcium metabolism, immune function, and skin health.8 In skin, deficient levels of vitamin D are associated with decrease in the optimal differentiation of keratinocytes.9–11
Lighter skin types have the capacity for maximum photo-activation and conversion of 7-dehydrocholesterol into vitamin D that can be stimulated at low intensities of UVR. However, increasing evidence suggests that excessive usage of sunscreens also prevents the synthesis of vitamin D, which can be reduced by 95%.12,13 The prevalence of vitamin D deficiency is also increased in people with darker skin types, such as Indian and African, which need longer time in the sun to produce a similar quantity of vitamin D compared to the Caucasian type. Vitamin D levels are additionally linked to the distribution of the populations across northern and southern latitudes.14
Based on this, the question emerges regarding the available approaches to skincare that would enable sufficient protection against photo-damage and amelioration of hyper-pigmentation whilst simultaneously preserving the beneficial effects of sun exposure. In this review, we discuss this topic in a context of personalized skincare and the possible different approaches required for each skin type. Melanogenesis is a complex process based on a cascade of biochemical reactions regulated by a range of genes.15 Many of these genes are subject to genetic modifications that alter their biological activities and determine the specific melanogenic traits in corresponding skin types.16 The biomarkers fall into several different categories, which are associated with different stages of melanogenesis, including the facultative, UVR-induced tanning responses in addition to the genetically determined traits contributing to specific skin type. Several of these genes are also linked to vitamin D metabolism, particularly in the light Caucasian skin where they could play important roles in the melanogenic responses of the skin to UVR.17 Differences in the constitutive pigmentation traits are also associated with specific hyper-pigmentation problems that are differently manifested in different skin types and could be linked to altered sensitivities to UVR and photo-damage. Finally, we summarize the main applications for skin lightening and improvement of skin tone and propose additional avenues for future considerations regarding the skin-type based approaches toward skin sensitivity to photo-damage and harnessing the benefits of sun exposure.
Melanogenesis: Complex Reactions Behind Skin Color and Tone
Skin color and tone are determined by the presence of melanin, which is a pigment synthesized in the epidermis by neural crest-derived cells, melanocytes, forming an epidermal melanin unit with approximately 40 keratinocytes at the dermal-epidermal junction.18,19 After maturation the melanin is transferred in specialized membrane organelles, melanosomes, into the surrounding keratinocytes and distributed in the supra-basal layers of the epidermis, where they determine the color of the skin and protect against the effects of UVR.20–22 Keratinocytes in the basal layer contain the majority (60–80%) of the total melanosomes, which are localized predominantly over cell nuclei providing photo-protection against UVR-induced DNA damage.23
Melanin is a macromolecular biopolymer derived from tyrosine through series of biochemical reactions.15,24 Melanogenesis is a complex process controlled at different physiological stages, involving a range of enzymes, structural proteins, and intermediate molecules that regulate development of melanocytes, biogenesis and survival of melanosomes as well as synthesis and maturation of melanin and transfer of melanosomes to keratinocytes. Constitutive pigmentation of the skin is also influenced by paracrine regulation of melanogenesis that originates as a result of cross-talk between melanocytes and keratinocytes as well as dermal fibroblasts25–67 (Table 1).
Table 1.
Genes and Biomarkers of Melanogenesis Relevant to Personalized Skincare
| Key Steps of Melanogenesis |
| UV-induced |
| 1 | UV-induced DNA synthesis of POMC (proopiomelanocortin). Processing of POMC to α-MSH (α-melanocyte-stimulating-hormone) and ACTH (adrenocorticotropic hormone)25,26 |
| 2 | Binding of α-MSH or ACTH to MC1R (melanocortin 1 receptor) and its activation25–27 |
| 3 | Activation of ADCY (adenylate cyclase) and increased formation of cAMP25,26,28 |
| 4 | Activation of PKA (protein kinase A) and phosphorylation of CREB (cAMP responsive-element binding) family of transcription factors25,26,28 |
| 5 | CREB-mediated expression of MITF (microphthalmia transcription factor), master regulator of melanocyte development and survival25,26,28 |
| 6 | MITF-induced expression of TYR (tyrosinase), TYRP1 (tyrosinase-related protein 1) and TYRP2/DCT (tyrosinase related protein-2/dopachrome tautomerase) through interactions with M- and E-boxes present in the promoter regions. TYR and TYRP1 are delivered to stage II melanosomes25,26,29 |
| Genetic |
| 7 | Enzymatic oxidation of tyrosine by TYR to DOPA (l-3,4-dihydroxyphenylalanine) and DOPAquinone30,31 |
| Eumelanin pathway | Pheomelanin pathway |
| 8a | Spontaneous conversion of DOPAquinone, via DOPAchrome, to DHI (5,6-dihydroxyindole) and DHICA (5,6-dihydroxyindole-2-carboxylic acid) accelerated by TYRP2/DCT32–34 | 8b | Reaction of DOPAquinone with cysteine to produce 5SCD (5-S-cysteinyldopa) and 2SCD (2-S-cysteinyldopa)32,35 |
| 9a | Oxidization of DHI and DHICA by TYR and TYRP1 to eumelanin polymer32–34 | 9b | Oxidization to intermediates which polymerize to pheomelanin32,35 |
| Additional key players in melanogenesis |
| Melanin synthesis |
| 10 | ASIP (agouti signaling protein); an antagonist of MC1R. Binding to MC1R leads to decreased TYR activity resulting in pheomelanin production27,36 |
| 11 | IRF4 (interferon regulatory factor 4); involved in transcription of TYR, TYRP1 and TYRP2. MITF directly or indirectly regulates IRF4 expression37 |
| 12 | ATRN (attractin) Group XI C-type lectin, trans-membrane protein, functions as accessory receptor for Agouti protein38 |
| 13 | Wnt/β-catenin; activation of nuclear β-catenin by Wnt leads to increased expression of MITF and melanogenesis39 |
| 14 | GSS (glutathione synthetase); involved in GSH (glutathione) biosynthesis. Role in the switch between eumelanogenesis and pheomelanogenesis through interactions with TYR and DOPAquinone40 |
| 15 | GGT7(gamma-glutamyltransferase 7); membrane-associated protein involved in metabolism of glutathione and the trans-peptidation of amino acids40 |
| 16 | RALY (RALY heterogeneous nuclear ribonucleoprotein); RNA binding protein40 |
| 17 | EIF2S2 (eukaryotic translation initiation factor 2); functions in the early steps of protein synthesis40,41 |
| 18 | EIF6 (eukaryotic translation initiation factor 6); role in protein synthesis42 |
| 19 | DRD2 (dopamine receptor D2); signaling shows increase with increasing UV exposure43 |
| Melanocyte biogenesis and survival |
| 20 | KITLG (hyper-pigmentation c-KIT receptor). Development and migration of melanocyte lineages, activates MAPK (mitogen activated protein kinase) leading to up-regulated expression of MITF, which activates keratinocytes to produce factors promoting melanosome phagocytosis44–46 |
| 21 | EDA (ectodysplasin A). Trans-membrane protein of the TNF (tumor necrosis factor) family, cytokine involved in the epithelial-mesenchymal signaling47 |
| 22 | ITCH (itchy homolog); E3 ubiquitin-protein ligase, induces proteasomal degradation40 |
| 23 | HERC2 (HECT and RLD domain containing E3 ubiquitin protein ligase 2); regulates ubiquitin-dependent retention of repair proteins on damaged chromosomes48 |
| 24 | BNC2 (basonuclin 2); zinc finger protein, cell survival after oxidative stress49 |
| 25 | SMARCA2 (SWI/SNF related, matrix associated, actin dependent regulator of chromatin, subfamily A, member 2). Member of the large ATP-dependent chromatin remodeling complex SNF/SWI required for transcriptional activation of repressed genes50 |
| 26 | DDB1 (DNA damage- binding protein 1); involved in DNA nucleotide excision repair, functions as a core component of the E3 ubiquitin ligase complexes51 |
| 27 | EGFR (epidermal growth factor receptor); induces cell proliferation and differentiation52 |
| 28 | FGF7(fibroblasts growth factor); enhances melanin synthesis and melanocyte proliferation, stimulates melanosome transfer after UVB irradiation53 |
| 29 | SCF (stem cell factor); fibroblast factor, enhances melanin synthesis and melanocyte proliferation54 |
| 30 | UQCC (ubiquinol-cytochrome c reductase complex); trans-membrane protein involved in FGF regulated growth control40 |
| 31 | VLDLR (very-low-density-lipoprotein receptor); trans-membrane receptor involved in endocytosis55 |
| 32 | PROCR (protein C receptor); cell survival and proliferation56 |
| 33 | ADAM17 (ADAM metallopeptidase domain 17); involved in cell adhesion and migration57 |
| 34 | ADAMTS 20 (ADAM metallopeptidase with thrombospondin type 1 motif 20); involved in cell adhesion and migration58 |
| Melanosome biogenesis, maturation and trafficking |
| 35 | Pme17(Premelanosome protein); early melanosome development and maturation, fibrils optimize condensation of melanin59 |
| 36 | OCA2 (oculocutaneous albinism type 2); chloride anion channel protein, effector of melanosomal pH, glutathione metabolism, processing and trafficking of tyrosinase to melanosomes60–62 |
| 37 | SLC45A2/MATP (solute carrier family 45 member 2). Membrane transporter; ion transport; increases pH and TYR activity62,63 |
| 38 | SLC24A4 (solute carrier family 24 member 4). Membrane transporter; impact on TYR activity62 |
| 39 | SLC24A5/NCKX5 (solute carrier family 24 member 5). Membrane transporter; a putative NA+/Ca2+ ion exchanger pump, impact on TYR activity62 |
| 40 | MFSD12 (major facilitator superfamily domain containing 12); trans-membrane solute transporter in endosomes and lysosomes in melanocytes. Depletion of MFSD12 increases eumelanin content51 |
| 41 | TMEM38 (trans-membrane protein 38); lysosomal protein, monovalent cation channel, functions in maintenance of intracellular calcium47 |
| 42 | SNX13 (sorting nexin 13); involved in the intracellular trafficking and lysosomal degradation47,55 |
| 43 | EDEM2 (ER degradation enhancing alpha-mannosidase like protein 2), protein folding and trafficking40 |
| 44 | DTNBP1 (dystrobrevin binding protein 1); melanosome biogenesis64 |
| 45 | MAP1LC3A (microtubule associated protein 1 light chain 3 alpha); mediates the physical interactions between microtubules and elements of the cytoskeleton40 |
| 46 | MYO5A (myosin VA); transport of melanosomes in melanocytes, target of MITF65 |
| 47 | LYST (lysosomal trafficking regulator,CHS1); vesicular transport protein, regulated by MITF66 |
| 48 | EXOC2 (exocyst complex component 2); exocytosis, melanosome trafficking, actin remodeling67 |
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Ethnic Skin Phenotypes are Defined by the Genes Controlling Melanogenesis
The color and tone of the skin are determined by the quantities and qualities of the synthesized melanin, which is one of the most variable phenotypes in humans. The geographic patterns of skin pigmentation demonstrate a strong correlation with latitude and UVR intensity; skin tends to be darker in tropical and equatorial regions with higher levels of UVR compared to the regions more distant from the equator.68,69
Constitutive pigmentation depends on the amount of melanin and relative ratio of eumelanin (brown/black pigment) and pheomelanin (yellow-red pigment) as well as the size, quantity, and distribution of melanosomes within the epidermis.16,70 Skin color is however not affected by the differences in melanocyte densities, which remain constant in every skin type.23,71
Skin types show variations in melanosome size and distribution, for example in African skin melanosomes are larger and more dispersed in basal keratinocytes whilst in European skin melanosomes are smaller and clustered together.72–74 In addition, melanosomes derived from dark skin have a neutral pH and higher activity of melanogenic enzymes whilst melanosomes derived from light skin are more acidic and have lower melanogenic activity.75,76
Constitutive skin pigmentation is above all a polygenic trait, with the quantities and type of melanin controlled by the genes with allelic variants through single nucleotide polymorphism (SNP), which is associated with changes in gene activity usually leading to lighter skin phenotype.16 In addition, a number of other genes involved in melanogenesis also demonstrate changes in the levels of expression linked to skin lightening and sun sensitivity, whilst constitutive levels of gene expression are typically higher in dark skin.
Comparison of the major melanogenic genes that are subject to allelic variation or changes in the transcriptional activity reveals the possible functional patterns in four major ethnic skin types. The proportion of the affected genes is predominant in Caucasian skin. This number is significantly decreased in Oriental skin, with evident further reduction in darker skin types of Indian and African origin (Table 2 and Figure 1).
Table 2.
Single Nucleotide Polymorphism (SNP) and Changes in Gene Expression Affecting Melanogenic Traits in Different Skin Types
| Gene | Skin Type | Gene Alterations and Phenotypes |
---
| Melanin synthesis |
| MC1R | Caucasian | Allelic diversity; sun sensitivity and freckles26,45,77,78 |
| Oriental | Allelic diversity; lighter skin reflectance and freckles81,82 |
| MITF | Caucasian | Allelic diversity, polymorphism correlates with the levels of serum 25[OH]D, impact on vitamin D status and deficiency17,64,67 |
| IRF4 | Caucasian | Allelic diversity, reduced skin tanning response, freckling and sun sensitivity79,80 |
| GSS | Caucasian | Allelic diversity40 |
| GGT7 | Caucasian | Lower expression levels in the lightly-pigmented melanocytes40 |
| RALY | Caucasian | Lower expression levels in the lightly-pigmented melanocytes40 |
| EIF2S2 | Caucasian | Lower expression levels in the lightly-pigmented melanocytes40 |
| EIF6 | Caucasian | Lower expression levels in the lightly-pigmented melanocytes40 |
| ASIP | Caucasian | Polymorphism associated with sensitivity to sun and freckling45,66,84,86 |
| Oriental | Allelic diversity66 |
| ATRN | Oriental | Allelic diversity58 |
| DRD2 | Oriental | Allelic diversity83 |
| TYR | Caucasian | Involved in normal variation of pigmentation. Polymorphism correlated with the levels of serum 25[OH]D, impact on vitamin D status and deficiency17,84,88,89 |
| Indian | Accounts for the differences between darkest and lightest skin reflectance90 |
| TYRP1 | Caucasian | Polymorphism correlated with the levels of serum 25[OH]D, impact on vitamin D status and deficiency17,83,84 |
| Oriental | Allelic diversity68,158 |
| Indian | Allelic diversity. Frequently associated with red-bronze skin159 |
| African | Allelic diversity. Frequently associated with red-bronze skin159 |
| TYRP2/DCT | Caucasian | Allelic diversity68,158 |
| Oriental | Allelic diversity68,83,158 |
| Indian | Allelic diversity58,158 |
| African | Allelic diversity158 |
| Melanocyte biogenesis and survival |
| ITCH | Caucasian | Lower expression levels in the lightly-pigmented melanocytes40 |
| UQCC | Caucasian | Lower expression levels in the lightly-pigmented melanocytes40 |
| PROCR | Caucasian | Lower expression levels in the lightly-pigmented melanocytes40 |
| KITLG | Caucasian | Allelic diversity45,83,84 |
| Oriental | Allelic diversity83,84 |
| EDA | Caucasian | Allelic diversity47 |
| EDA | Oriental | Allelic diversity47 |
| BNC2 | Caucasian | Allelic diversity40,66,84,85 |
| Oriental | Allelic diversity66,84 |
| EGFR | Caucasian | Allelic diversity83 |
| Oriental | Allelic diversity83 |
| ADAM17 | Oriental | Allelic diversity58 |
| ADAMTS20 | Oriental | Allelic diversity58 |
| DDB1 | Caucasian | Allelic diversity51 |
| Oriental | Allelic diversity51 |
| Indian | Allelic diversity51 |
| African | Allelic diversity51 |
| HERC2 | Caucasian | Allelic diversity40,51 |
| Indian | Allelic diversity51 |
| African | Allelic diversity51 |
| SMARCA2 | African | Allelic diversity55 |
| VLDLR | African | Allelic diversity55 |
| Melanosome biogenesis and trafficking |
| EDEM2 | Caucasian | Lower expression levels in the lightly-pigmented melanocytes40 |
| DTNBP1 | Caucasian | Allelic diversity. Polymorphism correlated with the levels of serum 25[OH]D64 |
| MAP1LC3 | Caucasian | Lower expression levels in the lightly-pigmented melanocytes40 |
| EXOC2 | Caucasian | Polymorphism correlated with the levels of serum 25[OH]D, impact on vitamin D status and deficiency17,67 |
| MYO5A | Caucasian | Allelic diversity. Polymorphism correlated with the levels of serum 25[OH]D64 |
| LYST | Oriental | Allelic diversity66 |
| SLC24A4 | Caucasian | Allelic diversity45,80,84 |
| Indian | Allelic diversity84 |
| SLC45A2/MATP | Caucasian | Polymorphism associated with olive skin and immature melanosomes63,78,160 |
| Oriental | Allelic diversity78 |
| Indian | Allelic diversity78 |
| African | Allelic diversity78 |
| OCA2 | Caucasian | Allelic diversity45,51,84 |
| Oriental | Allelic diversity; major gene contributing to skin lightening58,84,161,162 |
| Indian | Allelic diversity51,84 |
| African | Allelic diversity51,84 |
| SLC24A5 | Caucasian | Allelic diversity. Mutations disrupt melanosomal maturation and melanin biosynthesis78,84,87 |
| Oriental | Allelic diversity78,87 |
| Indian | Allelic diversity at very high frequencies78,84,87 |
| African | Allelic diversity at very high frequencies51,78,87 |
| MFSD12 | African | Allelic diversity51 |
| SNX13 | African | Allelic diversity55 |
| TMEM38 | African | Allelic diversity51 |
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Figure 1.
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Interactive networks of the major melanogenic genes and gene polymorphism or altered gene expression affecting pigmentation in four ethnic skin types. The biomarkers have been assembled based on the literature;25–90,158–162 the genes affected in each skin type are marked yellow.
The individual genes altered by polymorphism can be further identified as predominant within each skin type and organized in the functional categories based on the role within the melanogenic pathway. In Caucasian skin, such specific group of the polymorphic genes is comprised of the enzymes and transcription factors responsible for the early steps of the melanin synthesis pathway. The melanogenic genes that are specifically affected by polymorphism in Caucasian skin, MC1R (melanocortin 1 receptor),26,45,77,78 MITF (microphthalmia transcription factor),17,64,67 IRF4 (interferon regulatory factor 4)79,80 and GSS (glutathione synthetase)40 play a role in melanin synthesis and in a switch between eumelanin and pheomelanin production. The genes are also involved in the pathways of melanogenic responses to UVR; therefore decreased tanning and increased sun sensitivity would prevail as a result of decreased activity of these factors. Additional genes demonstrate lower expression levels in the lightly-pigmented melanocytes in Caucasian skin, including GGT7 (gamma-glutamyl transferase 7), RALY (heterogenous nuclear ribonucleoprotein), EIF2S2 (eukaryotic translation initiation factor 2) and EIF6 (eukaryotic translation initiation factor 6).40 Caucasian skin also bears a polymorphism in the genes involved in melanosome biogenesis and trafficking such as DTNBL1 (dystrobrevin binding protein 1),64 EXOC2 (exocyst complex component 2),17,67 and MYO5A (myosin VA).64 A number of genes with likely roles in melanocyte biogenesis and survival and melanosome trafficking also demonstrate lower expression in the lightly-pigmented melanocytes, including ITCH (itchy homolog), UQCC(ubiquinol-cytochrome c reductase complex), PROCR (protein C receptor), EDEM2 (ER degradation enhancing alpha-mannosidase like protein 2) and MAP1LC3 (microtubule associated protein 1 light chain 3 alpha).40 In addition to direct association with lighter skin, the polymorphism affecting the genes such as MITF, TYR (tyrosinase), TYRP1 (tyrosinase-related protein 1), EXOC2, MYO5A, and DTNBP1 is also correlated with the levels of serum 25[OH]D, indicating a direct impact on vitamin D status and deficiency in Caucasian individuals.17,64,67
Polymorphic genes specific to Oriental skin fall within the markers of melanin synthesis and responses to UVR exposure such as ATRN (attractin),58 MC1R,81,82 and DRD2 (dopamine receptor 2)83 as well as melanocyte biogenesis and survival, including ADAM17 (ADAM metallopeptidase domain 17), ADAMTS20 (ADAM metallopeptidase with thrombospondin type 1 motif 20),58 and melanosome trafficking, LYST (lysosomal trafficking regulator,CHS1).66 The majority of the polymorphic genes in Oriental skin fall within the group of biomarkers involved in melanocyte biogenesis and survival, with additional genes including KITGL (hyper-pigmentation c-KIT receptor),45,83,84 EDA (ectodysplasin A),47 BNC2 (basonuclin 2),40,66,84,85 and EGFR (epidermal growth factor receptor)83 also harboring SNP modifications in Caucasian skin. Both Oriental and Caucasian skin demonstrate polymorphism in ASIP (agouti signaling protein),45,66,84,86 indicative of potentially enhanced sun sensitivity in these genetic backgrounds.
Majority of the gene polymorphism in Indian skin types falls within a group of melanosome biogenesis and maturation, ion channels and transport membrane proteins SLC45A2/MATP (solute carrier family 45 member 2),78 SLC24A4 (solute carrier family 24 member 4),84 OCA2 (oculocutaneous albinism type 2),51,84 SLC24A5 (solute carrier family 24 member 5)78,84,87 but the SNPs in SLC45A2/MATP, OCA2 and SLC24A5 are also present across all other skin types (Table 2). Both Indian and Caucasian skin share polymorphism in TYR83,84,88-90 and SLC24A4,45,80,84 suggesting that the likely main skin-lightening traits are determined by decrease in the synthesis of melanin and/or the presence of immature melanosomes with altered activity of tyrosinase.
Polymorphic genes in African skin are the biomarkers of melanocyte biogenesis and survival SMARCA2 (SWI/SNF related, matrix associated, actin dependent regulator of chromatin, subfamily A, member 2)55 and VLDLR (very-low-density-lipoprotein receptor)55 as well as melanosome biogenesis and trafficking SNX13 (sorting nexin 13),55 TMEM38 (trans-membrane protein 38)51 and MFSD12 (major facilitator superfamily domain containing 12).51 Genes altered by polymorphism in African skin also represent biomarkers of melanosome biogenesis and maturation including SLC45A2/MATP,78 OCA251,84 and SLC24A551,78,87 but the SNPs are also present across all other skin types (Table 2).
Finally, the strongly positive melanogenic traits are present in both Indian and African skin for ASIP and KITGL, resulting in alleles that would favor higher tyrosinase activity and higher melanin index as a result of significantly enhanced melanogenic responses to UVR characterizing the darker skin. The melanogenic markers can be assembled in the interactive networks representing major genes and polymorphic clusters present in each skin type, together with the emerging links to vitamin D synthesis (Figure 1).
Effect of UVR on Melanogenic Responses and Their Variation in Skin Types
In addition to constitutive pigmentation, which is defined as genetically determined basal melanin production, facultative pigmentation can be described as enhanced production of melanin due to exposure to UVR, the most important environmental factor regulating the melanogenesis process. UVR is also the major environmental stress leading to the development of hyper-pigmentation disorders.
Based on the wavelength, UVR reaching the skin is classified into UVA and UVB. UVA (320–400 nm) is less energetic but can penetrate deep into the skin, reaching the basal layer of the epidermis and dermis. UVA exposure is a major factor in the photo-aging process, leading to the production of reactive oxygen species (ROS), an increase in inflammatory mediators such as IL-1 or IL-6, expression of matrix metalloproteinases (MMPs) and lipid peroxidation.1 UVB (290–320 nm) is more energetic and capable of inducing direct DNA damage through the induction of cyclobutane pyrimidine dimers and 6–4 photoproducts in the epidermis.91 UVB exposure is associated with sunburn and expression of melanogenic enzymes resulting in increased pigmentation.92 Whole UV spectrum is responsible for the cellular responses that are involved in the stimulation of pigmentation and development of pigmentary lesions.
Melanin has protective effect against DNA damage induced by UVA/UVB and there is a clear relationship between these factors in the skin types of different ethnic origins and constitutive pigmentation.23,93,94 For example, predominant type of UVR-induced DNA damage, cyclobutane pyrimidine dimers (CPDs), demonstrate a uniform distribution throughout the epidermis, including melanocytes and basal layer, and the upper dermis in Caucasian skin. In contrast, CPDs are significantly reduced in the epidermis of Indian and African skin and can be mostly detected in the suprabasal layers, with a gradual reduction in the basal layer, indicative of better photo-protection against mutagenesis and faster repair of DNA damage in darkly pigmented skin.95–97 In addition, activation of DNA repair can also be regulated by MC1R, which is frequently affected by SNP in lightly pigmented skin of Caucasian origin.98,99
Exposure to UVR triggers several reactions that ultimately lead to darkening of the skin, including oxidation and polymerization of melanin, redistribution of melanosomes, increase in expression of α-MSH (α-melanocyte-stimulating-hormone) and MITF and transfer of melanin from the lower to upper epidermis.92,100,101 Melanocytes synthesize two types of melanin that are chemically and functionally different.70 Eumelanin is photo-protective through the ability to dissipate >99.9% of UVR and visible light, limiting the extent of UVR penetration within epidermis and scavenging ROS. In contrast, pheomelanin is highly photo-reactive, enhancing the UVR-induced production of ROS and further damage of the cells.102–104
The ratio of eumelanin to pheomelanin is dependent on the catalytic activity of rate-limiting enzyme tyrosinase (TYR) and the availability of cysteine. High TYR activity/low concentrations of cysteine lead to the synthesis of eumelanin, whilst low TYR activity/high concentrations of cysteine lead to the synthesis of pheomelanin. The differences in skin color are not determined by the quantity of melanocytes, which remains constant, but by the activity of melanocytes including the relative levels of eumelanin and pheomelanin.105–107
Decreased ratio of eumelanin to pheomelanin is associated with increased photo-sensitivity, predisposition to freckles, and decreased tanning responses. Direct genetic link to the MC1R gene expression points at its likely role as a major contributing factor affecting the eumelanin/pheomelanin ratio. Consistently, a number of gene polymorphisms and loss-of-function mutations in the MC1R gene result in a decrease in eumelanin production, frequently linked with red hair phenotypes and fair skin that is susceptible to damage.108
The melanogenic response of human skin to UVR occurs in three phases: immediate pigment darkening (IPD), persistent pigmentation (PPD), and delayed tanning (DT). IPD occurs immediately during or after UVR exposure and is transient, whilst PPD lasts longer upon more intense UVR.109,110 Both IPD and PPD rely on oxidation and polymerization of existing melanin or its precursors rather then de novo synthesis, together with redistribution of melanosomes within melanocytes and keratinocytes.111–113 Such responses are not protective against erythema or DNA damage.114–116 In contrast, DT response is detected several days later after UVR exposure and is associated with activation of melanin synthesis pathway.117
Lighter skin phenotypes might have reduced IPD tanning response to UVR, with the threshold of the irradiation dose required to produce IPD/PPD higher than the dose inducing sunburn.118 IPD is associated with reversible oxidation of DHI (5,6-dihydroxyindole) and DHICA (5,6-dihydroxyindole-2-carboxylic acid), whilst PPD develops by irreversible oxidative cleavage of indolequinone to PTCA (pyrrole-2,3,5-tricarboxylic acid) and cross-linking of dihydroxyindole to PTeCA (pyrrole-2,3,4,5-tetracarboxylic acid).119–121 PPD is also associated with photo-degradation of pheomelanin, however its contribution to PPD could be masked by higher content of eumelanin.122
DT involves de novo melanogenesis driven by increased activity of TYR in melanocytes as well as enhanced multiplication and transfer of melanosomes to keratinocytes. Melanin synthesis in DT is initiated by DNA damage in keratinocytes, which leads to up-regulation of POMC (proopiomelanocortin) and its processing into α-MSH. Subsequent binding of α-MSH to MC1R in melanocyte results in activation of melanogenesis, which has a photo-protective effect against further DNA damage. Factors synthesized and secreted by keratinocytes, such as ET-1 (endothelin 1) and IL-1 (interleukin 1) also play a role in a cross-talk between melanogenesis and inflammation.100,123 DT response and tanning abilities are directly related to skin types, showing proportional association with Fitzpatrick classification and constitutive skin color.124 Decreased ability to tan and solar elastosis in Caucasian skin type have been associated with altered activity of IRF4. However, visible changes in pigmentation in tanning typically do not result from significant increase in the melanin content, but rather its re-distribution in the epidermal layers in all skin types.125–127
The entire UVR spectrum is also involved in photo-aging; directly related to the penetration properties in UVA and UVB depending on skin color and melanin content. Enhanced and impaired melanogenic response, particularly linked to TYR activity and inflammation can also lead to the defects in skin pigmentation, which are the primary sign of photo-aging in Indian and Oriental skin.2,4
Defects of Skin Pigmentation and Their Associations with Skin Types
Defects in the pigmentation can be triggered or exacerbated by long-term sun exposure and the type, onset, and frequency of the hyper-pigmented lesions are dictated by the skin complexion and genetic background. For example, hyper-pigmented spots, frequently considered as a sign of photo-aging, develop earlier and are more pronounced in Oriental and Indian skin types compared to the Caucasian skin type.2–4
Hyper-pigmentation can be classified into three main types:
actinic lentigines (AL, lentigo senilis) are light to dark brown spots ranging in size from millimeters to centimeters located mainly on photo-exposed areas such as face, hands, forearms and upper back.128 Actinic lentigines are the clinical signs of photo-aging and considered an indicator of the amount of sun exposure over the course of a life-time.129,130 Changes are characterized by elastosis, hyper-pigmented basal layer due to an increased total content of melanin in the keratinocytes (hypermelaninosis), increased expression of TYR and mitochondria quantities in melanocytes, with unaltered size of melanosomes and melanocytes densities along dermal-epidermal junctions.128,131 Actinic lentigines demonstrate broadened and elongated rete ridges of the dermal-epidermal junction resulting in protrusions of the epidermis into the dermis, together with altered expression of KGF (keratinocyte growth factor), FGF7 (fibroblast growth factor 7), SCF (stem cell factor) and the components of the dermal extra-cellular matrix.132,133 This type of photo-damage affects mainly Caucasian and Indian skin, where it has been associated with variations in MC1R and SLC45A2 genes. Genetic variations in four other genes, namely IRF4, MC1R, ASIP and BNC2 have been associated with lentigines acquired during aging in Caucasian skin.49,134
Additional changes to pigmentation, freckles or ephelides are small, 1–2 mm in diameter, red to light brown spots that are induced by sunlight and most frequently found on the face, arms, neck and chest. Melanocytes in ephelides contain multiple large melanosomes and the genes involved in formation of hyper-pigmented spots include MC1R, IRF4, ASIP, TYR, BNC2 and OCA2. Ephelides are characteristic predominantly in individuals with fair skin and often partially disappear with age.131,135
Post-inflammatory hyper-pigmentation (PIH) occurs as a result of an inflammatory reaction induced by allergic contact such as drug sensitization and endogenous causes such as atopic dermatitis. This type of damage appears as brown patches on photo-exposed areas including face, shoulders and trunk.136 Inflammation in the epidermis results in the release of reactive oxygen species, cytokines, eicosanoids, prostaglandins and leukotrienes that stimulate the melanocytes leading to increased melanin production. The inflammation also causes melanocyte hyperplasia, damage to the basement membrane through collagen IV degradation, leakage of melanins from basal keratinocytes, accumulation of melanophages at the proximity of blood vessels and in the dermis and dermal hyper-pigmentation.137 This type of damage affects all skin types but is more prevalent in Oriental and Indian skin types. Moreover,in both skin types it is more common in darker constitutive pigmentation background due to increased reactivity of melanocytes.136,138,139 Post-inflammatory hyper-pigmentation has a common association with acne in African, Indian and Oriental skin types and can persist after the original acne lesions have been resolved. Similar problems are frequently linked to cosmetic therapies such as laser treatment or chemical peels performed on skin with higher pigmentation levels.140–142
Melasma (M) is hypermelanosis of hormonal origin, which may be stimulated by high levels of estrogen and progesterone and is characterized by large dark brown patches with irregular borders in sun-exposed areas, especially the face.143 On the histological level there is increased elastosis, disruption of basement membrane, flattening of rete ridges, increased micro-vasculature and infiltration of mast cells.144
Enhanced synthesis of melanin is associated with up-regulated expression of TRP1 (tyrosinase-related protein 1), TRP2 (tyrosinase-related protein 2), MITF, melanocyte hypertrophy and activation of a-MSH, corticotrophin and IL-1 in response to UVR.145,146 In addition to hormonal link and inflammation, the most important environmental factor triggering melasma is acute sun exposure. Higher prevalence of melasma onset is observed in darker skin types and the populations living in areas with greater exposure to UVR are more likely to develop this type of pigmentation defect.147 It is also correlated with specific features that depend on skin type, for example accumulation of melanophages and melanin in dermis is more prevalent in darker skin.146 Facial melasma is particularly common in Indian skin types and in the middle- to older age groups.148
Changes in skin pigmentation, particularly those related to UVR exposure, chronic inflammation and immunosuppression, together with constitutive pigmentation background are linked to several types of skin cancer. Compared to Caucasian populations, skin cancers such as basal cell carcinoma (BCC) and melanoma are in general significantly less frequent in darker skin types, as a result of increased photo-protective UVR-filtering effects of epidermal melanin. Major risk factors for melanoma are increased exposure to UVR, fair complexion and freckling. In contrast, squamous cell carcinoma (SCC), is most frequently diagnosed in African skin. Predisposing factors for SCC are burns, chronic inflammation, and scarring.149
Current Approaches and Ingredients for Skin Brightening and Improvement of Skin Tone
Current applications toward management of skin pigmentation defects and improvement of skin tone are important aspects in the field of cosmetics, beauty therapy and dermatology. They can include inhibition of the enzymes at different stages of melanogenesis such as TYR, MITF and MC1R, number and size of melanosomes, interference with melanosomal maturation and transfer of melanin, destruction of melanocytes, exfoliation, dermabrasion, ultrasound and laser therapies. Successful treatments usually combine two or more methods with synergistic effect. Active ingredients are selected from both synthetic and natural sources for the capacity to control pigmentation whilst remaining minimally toxic.150 Identification of new or improved combination of ingredients with a defined mechanism and therapeutic profiles would also be based on the detailed structure–activity relationship studies. Although many of these applications have inhibitory activity against melanogenesis, very few became a commercial product based on cytotoxicity, cutaneous absorption and clinical trials151–157 (Table 3).
Table 3.
Current Applications and Active Ingredients for Improvement of Skin Tone
Synthetic Inhibitors of TYR
Molecules with broad chemical nature and mechanisms of action. The most popular include hydroquinone (HQ) and its derivatives monobenzyl ether of hydroquinone (MBEH), monomethyl ether of hydroquinone (MMEH), benzaldehyde analogs; chalcone analogs; phenolic amines and derivatives of 4-phenylimidazole-2-thione, mequinol, N-acetyl glucosamine, benzimidazole-2-thiol, phenylthiol, phenylthiourea (PTU), p-aminobenzoic acid (PABA), quinazoline, biphenyl derivatives, indole derivatives and thiosemicarbazone.32,151,152
Botanical extracts
Usually contain a combination of active natural ingredients that work in synergy. Some skin-lightening compounds in such extracts include aloesin, anisic acid, arbutin, trans-cinnamaldehyde, p-coumaric acid, cumic acid, epicatechin gallate, ellagic acid, glabridin, hesperidin, kaempferol, 2-oxyresveratrol, resveratrol, azelaic acid, aurone, hydroxystilbenes, hydroxycinnamic acid derivatives, chalcones and trihydroxyflavones, caffeic acid and ginsenoside Rb1.151,152
Derivatives of resorcinol (un-substituted 4-alkylresorcinols)
De-pigmenting derivatives of resorcinol (1,3-benzendiol) include 4-cycloalkylresorcinol, 4-cycloalkylmethylresorcinols, 4-haloresorcinol, 4-(1,3) dithian-2-ylresorcinols and 4-n-butyl resorcinol (rucinol).32
Antioxidants
Reduce the synthesis of melanin by quenching ROS generated through exposure to UVR and oxidization of TYR and DOPA.153,154 Antioxidants with the capacities to interfere with melanogenesis are phytic acid, glutathione, ubiquinone, resveratrol, kojic acid and ferulic acid.151
Vitamins
The capacities to increase turnover rate of melanin (vitamin A), inhibit the transfer of melanosomes from melanocytes to keratinocytes (vitamin B3), interfere with the glycosylation of TYR (vitamin B5), de-activation of UVR-induced ROS and inhibition of TYR (vitamin C) and protection against UVR-induced inflammation (vitamin E).151
Inhibitors of melanosome transfer
Ingredients interfering with melanosome transfer include niacinamide and lectins.47
Molecules cyto-toxic to melanocytes
Molecules with inhibitory effect on melanocyte activity or survival include fomiferin and its derivative fomiferin-3,4-dimethyl ether, fraxidin methyl ether, hernlarin, imperatorin, kuhlmannin, obliquin, osajin and its derivative osajin-4-methyl ether, pachyrrhizin, prenyletin, robustic acid, sphondin, warangalone and xanthyletin.32
Peptides and oligopeptides
Reduce pigmentation through interaction with the protease-activated receptor 2 (PAR-2) in keratinocytes affecting melanin and melanosome uptake by keratinocytes. Custom designed oligopeptides between 6 and 12 amino acids, dipeptides or cyclic peptides have the capacity to translocate to melanosomes and inhibit TYR.32,152
Inhibitors of MC1R/αMSH
The molecules that indirectly inhibit TYR expression through down-regulation of cAMP include glyceollin, methyl and ethyl linoenates, platycodin, bisabolangelone, chrysin, paeonol.152
Alpha hydroxy acids (AHA) and their derivatives
Applications as superficial chemical peels, target stratum corneum and accelerate desquamation of the outer epidermal layers, increasing melanin turnover. Increase the enzymatic activities leading to epidermolysis and promoting the synthesis of elastin fibers. The most commonly used are glycolic, lactic, citric, malic, pyruvic and salicylic acids and their derivatives.151
Inhibitors of wnt/β-catenin signaling pathway
Wnt/β-catenin signaling pathway enhances MITF gene expression and melanogenesis; ingredients inhibiting the pathway through enhanced degradation of β-catenin include cardamonin, fingolimod, pyridinyl imidazoles and andrographolide.155,156
Small interfering RNA (siRNA)
Based on double-stranded, ~21 base pairs RNA with the sequence complementary to the target mRNA. Mediate gene silencing and inhibition of TYR by binding the mRNA and degrading it at the site of application.32
Inhibitors of adaptor protein (AP)-complexes
AP-1 and its interacting partner kinesin family member 13A (KIF13A) are required for the transfer of melanogenic enzymes to melanosomes and their maturation. Inhibition of KIF13A or AP-1 decreases the expression of melanogenic enzymes and synthesis of melanin.32
Anti-inflammatory factors
Dexamethasone, fluocinolone acetonide and tranexamic acid have applications in treatment of hyper-pigmentation defects.47
Physical therapies in association with UV-blocking agents
Based on laser treatment, cryotherapy and dermoabrasion.157
Active ingredients stimulating melanogenesis
The potential capacities to protect the skin from photodamage by increasing the melanin content. The ingredients include pyrazoles, indole alkaloids, cannabinoid derivatives and 2-bromopolymitate derivatives.32
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Proposed Treatments and Solutions toward Personalized Skin Tone Applications
Based on the genetic biomarkers that determine skin types discovered so far, it can be proposed that hyper-pigmentation is linked to the variation in specific traits regulating melanogenic pathways, which are also reflected in skin sensitivity to UVR. Such traits could in turn form a basis for design of skin tone applications compatible with each skin type. Given the variety of the biomarkers, the melanogenic traits would include the quantity and quality of melanin, melanosome biogenesis and maturation, trafficking of melanosomes in melanocyte, transfer of melanosomes to keratinocytes and distribution of melanin in the suprabasal and basal layers. These factors would also be expressed differentially in evenly pigmented and hyper-pigmented skin.
The novel approaches toward hyper-pigmentation in a context of personalized skincare can be based on the following points:
eumelanin is a factor protecting against UVR, the most damaging factor in lighter skin with relatively low levels of melanin.23,93,94
Presence of SNP in the melanogenic gene usually results in down-regulation of the activity, therefore generally associated with skin lightening leading to skin tone variations.
Hyper-pigmentation is linked to increased skin sensitivity and UVR damage, therefore could manifest as deficiency in the protective mechanisms of melanin irrespective of skin type.
Exposure to the sun has beneficial effects on the synthesis of vitamin D; with ~90% synthesized through short 15–20 min daily exposure for lighter skin, which is below the threshold level that induces delayed tanning. Vitamin D3 is produced in the skin through irradiation of 7-dehydrocholesterol, which is then metabolized to blood serum 25-hydroxyvitamin D3 [25(OH)D] and 1α 25-dihydroxyvitamin [D3 (1,25(OH 2D ))].6,7 This mechanism is less efficient in darker skin due to the effects of higher melanin content, which is proposed to have an inhibitory effect on vitamin D synthesis. However, lower activity of the pigmentation genes due to the polymorphism is linked to vitamin D deficiency (with serum level of 25[OH]D as a biomarker) in Caucasian type, indicative of the likely link between vitamin D synthesis and melanogenic capacities related to sun exposure.17,64,67
Based on this, the applications toward hyper-pigmentation can be developed based on the genetic makeup linked to the observed and predicted altered activities of the melanogenic factors specific for each skin type. Such applications, modulating the gene activity would target two mainstreams of melanogenesis, namely facultative and constitutive pigmentation with the specific genetic components considered as unique for each skin type or overlapping between them (Figure 2).
Figure 2.
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Proposed functional cluster targets for enhancement of skin tone and prevention of pigmentation problems in four skin types whilst retaining the capacity for vitamin D synthesis and defense against molecular damage caused by UVR.
Caucasian Skin: Enhancement of DNA Repair, Delayed Tanning and Melanosome Trafficking in Facultative Pigmentation
This skin type demonstrates a significant accumulation of the traits associated with decreased melanogenic responses to UVR and facultative pigmentation. Enhancing UVR-induced melanogenesis as a protective mechanism against molecular damage could be an efficient way of achieving even skin tone in Caucasian skin. Applications would focus on modulating the activity of early melanogenic genes that are activated by UVR and linked to sun sensitivity and DNA repair such as MC1R and its antagonist ASIP. Such applications could affect the activity of MITF, TYR and TYRP1 genes associated with tanning and increased production of melanin through synthesis of melanogenic precursors for oxidation and polymerization. Additional biomarkers of interest could be IRF4 and GSS; involved in regulation of a switch between eumelanin and pheomelanin ratio and sun sensitivity through TYR expression. The potential skin brightening targets could also include the genes involved in melanosome biogenesis and trafficking such as DTNBL1, EXOC2 and MYO5A. The activity of MITF, TYR, TYRP1, DTNBL1, MYO5A and EXOC2 could also correlate with serum 25[OH]D and vitamin D status, therefore enhancement of tanning responses in formulations applied during sun exposure could be beneficial for even skin tone whilst retaining the ability to synthesize vitamin D.
Oriental Skin: Early Melanogenic Traits Associated with Sun Sensitivity and Oxidative Stress, Melanocyte Biogenesis and Survival
Oriental skin demonstrates polymorphism in the traits linked to early melanogenic responses to UVR exposure, sun sensitivity and oxidative stress whilst retaining high activity of TYR. Modulating the activity of genes such as ASIP and ATRN (antagonists of MC1R) in combination with activity of the genes involved in oxidative stress such as BNC2 or DRD2 could be beneficial for sun protection through control of early and facultative melanogenesis. Such applications could be combined with the inhibitors of constitutive TYR to control hyper-pigmentation whilst promoting synthesis of vitamin D in Oriental skin. Other potential targets of skin-brightening applications could also involve the genes associated with melanocyte biogenesis and survival such as KITGL, EDA and EGFR, which are also relevant to Caucasian skin.
Indian Skin: Melanosome Biogenesis and Maturation in Constitutive Pigmentation
Indian skin type does not demonstrate a significant polymorphism in the genes associated with tanning, indicative of the robust responses to UVR. The potential skin tone targets in Indian skin could instead involve the genes responsible for melanosome maturation, ion channels and effectors of melanosomal pH such as SLC45A2/MATP, SLC24A4, OCA2 and SLC24A5. The skin brightening applications could be based on the active ingredients modulating pH, TYR activity as well as processing and trafficking of tyrosinase to melanosomes. The ingredients targeting these genes could be combined with the inhibitors of constitutive TYR expression, to control hyper-pigmentation whilst promoting synthesis of vitamin D. Some aspects of this control could be additionally applicable to Caucasian skin, which also demonstrates the polymorphism in SLC45A2/MATP, SLC24A4, OCA2, SLC24A5 and TYR genes.
African Skin: Melanosomal Trafficking and Dispersion, Lysosomal Targets in Melanocytes
Similar to Indian skin, African skin type would demonstrate significant tanning response to UVR based on the gene polymorphism. The potential skin tone targets could involve the genes involved in melanosome trafficking and dispersion such as MFSD12 as well as lysosomal degradation and maintenance of intracellular calcium such as SNX13 and TMEM38. The ingredients targeting these genes could be combined with the inhibitors of constitutive TYR, to control hyper-pigmentation whilst promoting synthesis of vitamin D (Figure 2).
Summary
This review summarized the main components of the melanogenesis pathways, with a focus on the genes controlling UVR-induced, facultative melanogenesis and genetically determined, constitutive melanogenesis. Skin color is determined by the quantity and quality of melanin, which is regulated differently in ethnic skin types due to small nucleotide polymorphism (SNP) in the specific genes controlling different steps of the melanogenesis process. Gene polymorphism usually results in decreased activity of the gene, generally leading to lighter skin variants among each population. High melanin levels are well recognized as a protective factor against UVR-induced cellular and molecular damage, therefore a significant level of such damage is normally detected in lightly-pigmented skin. Skin responses to UVR are also associated with tanning, which occurs in three phases: immediate pigment darkening (IPD), persistent pigmentation (PPD), and delayed tanning (DT). Decreased ability to tan has been linked to hyper-pigmentation problems such as solar elastosis in lighter skin types. Defects in pigmentation can be triggered or exacerbated by long-term sun exposure and are evident across all skin types. Sun exposure has however a beneficial effect, such as the synthesis of vitamin D, which is thought to be less efficient in darkly pigmented skin due to the inhibitory effect of melanin on vitamin D production. However, gene polymorphism in several major melanogenic enzymes has also been linked to vitamin D deficiency in Caucasian type, indicative of the likely association between vitamin D synthesis and melanogenic capacities related to sun exposure. Given this, novel applications for a treatment or prevention of hyper-pigmentation could be developed based on the differences in the activities of melanogenic genes specific for each skin type. Such treatments could rely on targeting two aspects of melanogenesis separately, with more focus on facultative pigmentation in lighter skin types that are generally more prone to photo-damage whilst retaining high capacity for vitamin D synthesis. Darker skin types would benefit more from targeting expanded network of the genes at the base of constitutive pigmentation, perhaps combined with the already existing inhibitors of main melanogenic enzymes for enhancement of skin capacity toward synthesis of vitamin D. In such context, the skin lightening applications could be extended into personalized skincare and beneficial sun exposure.
Disclosure
The authors report no conflicts of interest in this work.
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Abstract
Introduction
Melanogenesis: Complex Reactions Behind Skin Color and Tone
Ethnic Skin Phenotypes are Defined by the Genes Controlling Melanogenesis
Effect of UVR on Melanogenic Responses and Their Variation in Skin Types
Defects of Skin Pigmentation and Their Associations with Skin Types
Current Approaches and Ingredients for Skin Brightening and Improvement of Skin Tone
Proposed Treatments and Solutions toward Personalized Skin Tone Applications
Summary
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A cone is a solid shape in geometry that tapers smoothly from a flat base to a point called the apex or vertex. A cone can be of different types. A cone is a three-dimensional figure that has a circle as a base and a curved surface that closes off at a point on the top. Such a cone is obtained when we rotate a right-angled triangle by the perpendicular axis. The word “cone” comes from the Greek word “konos”, which means a peak. A cone with its axis perpendicular to the plane of the base and meeting the base at its midpoint is a right circular cone. It can be produced by revolving a right-angled triangle on one of its legs.
What is Right Circular Cone?
Right circular cone is a cone with a circular base in which the axis of the cone is perpendicular to its base and meets the base at its midpoint. The given figure demonstrates a right circular cone, its radius, slant height, vertex and the axis perpendicular to the base of the cone. This axial line that connects the vertex of the cone to the center of the base, is also called the height of the cone, denoted by ‘h’.
Another line that connects the vertex of the cone to the edge of the base is termed as the slant height of the cone and is represented as ‘s’, or ‘l’.
Right Circular Cone Properties
Some of the properties of right circular cone that make it different from any other geometrical figure are:
The axis of a right circular cone is a line that joins its vertex to the center of its circular base.
The distance from its vertex to the edge of its circular base is called the slant height of the cone and is represented by ‘l’ or ‘s’.
It is the same as its height and is represented by ‘h’.
We can construct it by rotating a right-angled triangle with the perpendicular side as an axis.
The surface area that is generated by the hypotenuse of the right-angled triangle while constructing the cone is called the curved or lateral surface area of the right circular cone.
The cross-section of the right circular cone that is parallel to the base of the cone gives a circle.
A section that contains the vertex of the cone and any two points on the base gives an isosceles triangle.
Right Circular Cone Formulas
Some of the general formulae linked with the right circular cone are those associated with its area and volume. Let us note down:
Curved Surface Area of a right circular cone: (\pi rs) or (\pi rl)
Total Surface Area of a right circular cone: (\pi r\left(r+l\right))
Volume of a right circular cone: (\frac{1}{3}\pi r^2h).
Surface Area of Right Circular Cone
Surface Area of any geometrical figure is the area covered by its surface, in the three-dimensional space. The units for surface area of any figure are sq. m, sq. cm, sq. inches, and sq. feet, etc. Let us understand the curved surface area using a figure:
As we can see that when a right circular cone is cut open along its slant height, it forms a sector of a circle. So, its curved surface area is equal to the area of the sector with radius ‘s’. Also, the surface area of a cone can be of two types, namely, curved surface area, and total surface area. As the name suggests, the curved surface area of a right circular cone is the area occupied by the curved surface of the cone. In this case, the area of the base is not included. This can also be termed as the lateral surface area of the cone.
Total surface area of cone is the area occupied by the curved surface along with the area of the base of the cone.
Curved Surface Area of Right Circular Cone
Curved surface area of the cone is equal to the area of the sector with a radius equal to the slant height ‘s’ of the cone. So, the CSA of the right circular cone becomes:
CSA = (\pi rs\ =\ \pi r\sqrt{r^2+h^2})
‘r’ is the radius of the base.
‘h’ is the height of the right circular cone.
and, ‘s’ is the slant height of the cone.
Total Surface Area of Right Circular Cone
In the total surface area of the cone, we include the area of the base of the cone along with the curved surface area of the right circular cone.
So, Total surface area of the cone = Area of the base of the cone + CSA
TSA = (\ \pi r^2+\pi rs\ )
TSA = (\pi r^2+\ \pi r\sqrt{r^2+h^2})
Here,
‘r’ is the radius of the base.
‘h’ is the height of the right circular cone.
And, ‘s’ is the slant height of the cone.
Learn about Frustum of a Cone
Surface Area of Right Circular Cone Formulas
As already discussed, the surfaces of the right circular cone can be of two types, lateral surface area and total surface area. The formula for finding the area of the two surfaces is different. Let us check:
Volume of Right Circular Cone
For any geometrical figure, volume is the space occupied by objects in the three-dimensional space. We can express the volume of a cone with cubic units, like cu. m, cu. cm, cu. feet, and cu. inches.
For a right circular cone with radius ‘r’ and ‘h’, the volume of a cone is represented as one-third of the product of area of base and height of the cone. When the radius of the base and the height of the cone are given, the volume of the cone can be calculated by using the general formula.
From the given image we can conclude that the volume of a right circular cone is one-third the volume of a right circular cylinder.
The formula for finding the volume of a right circular cone is:
Volume of Cone = (\frac{1}{3}\times Area\ ofCircular\ Base\times Height\ of\ the\ Cone)
Volume of Cone = (\frac{1}{3}\times \pi r^2\times h)
Volume of Cone = (\frac{1}{12}\times \pi d^2\times h)
where r = radius of the base
d = diameter of the base
h = height of the right circular cone
Right Circular Cone Equations
A right circular cone with center as origin is represented by given equation:
(\left(x^2+y^2+z^2\right)\cos ^2\theta =\left(lx+my+nz\right)^2)
Here, (\theta) is the semi-vertical angle, and (l, m, n) are the direction cosines of the axis.
Example: Find the equation of a right circular cone with vertex as origin and the line (x=\frac{y}{3}=\frac{z}{2}) as axis that makes a sem-vertical angle of 60-degree.
Solution: The direction cosines of the axis are:
(\left[\left(\frac{1}{\sqrt{1^2+3^2+2^2}}\right),\left(\frac{3}{\sqrt{1^2+3^2+2^2}}\right),\left(\frac{2}{\sqrt{1^2+3^2+2^2}}\right)\right]=\left(\frac{1}{\sqrt{14}},\frac{3}{\sqrt{14}},\frac{2}{\sqrt{14}}\right))
Given that the semi-vertical angle is = 60-degree
So, the equation of a right circular cone with a vertex at the origin is:
(\left(x^2+y^2+z^2\right)\ \frac{1}{4}=\frac{1}{14}\left(x+3y+2z\right)^2)
(7\left(x^2+y^2+z^2\right)=2\left(x^2+9y^2+4z^2+6xy\ +12yz+4xz\right))
(5x^2-11y^2-z^2-12xy-24yz-8xz=0), is the required equation.
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Solved Examples of Right Circular Cone
Example 1: Mary takes a sheet of paper to make a birthday cap in the shape of a right circular cone. The radius of the base of the cap is 3 inches and the height is 4 inches. What will be the slant height of the birthday cap?
Ans: Given that:
Radius of the cap = r = 3 inches
Height of the cap = h = 4 inches
We have to find:
Slant Height of the cap = s
We know that:
(s^2=\ \left(3\right)^2+\left(4\right)^2)
(s^2=\ 9+16)
(s^2=\ 25)
(s=\ \sqrt{25})
s = 5 inches
So, the slant height of the cap is 5 inches.
Example 2: If the radius of the right circular cone is 6 cm and the slant height is 10 cm, find the total surface area of the cone.
Ans: Given that:
Radius of the cone= r = 6 cm
Slant Height of the cone = l = 10 cm
We have to find:
Total Surface Area of the cone = TSA
We know that:
(TSA\ =\ \pi r\left(r+l\right))
(TSA\ =\ \frac{22}{7}\times 6\left(6+10\right))
(TSA\ =\ \frac{22}{7}\times 6\left(16\right))
(TSA\ =\ \frac{2112}{7})
TSA = 301.71 sq. cm.
So, the total surface area of the cone is 301.71 sq. cm.
We hope that the above article is helpful for your understanding and exam preparations. Stay tuned to the Testbook App for more updates on related topics from Mathematics, and various such subjects. Also, reach out to the test series available to examine your knowledge regarding several exams.
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| If you are checking Right Circular Cone article, check related maths articles: |
| Volume of Sphere | Volume of Cylinder |
| Volume of a Frustum | Volume of a Pyramid |
| Volume of a Cuboid | Volume of a Cube |
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Right Circular Cone FAQs
What is right circular cone?
Right circular cone is a cone with a circular base in which the axis of the cone is perpendicular to its base and meets the base at its midpoint.
What is the frustum of a right circular cone?
When a right circular cone is cut by a plane parallel to the base of the cone, we get a frustum of the right circular cone. It is the portion of the cone between the base and the parallel plane for the cone.
How many vertices are there in the right circular cone?
A right circular cone has only one vertex.
How do you find the radius of a right circular cone?
The radius of a cone is basically the radius of the circular base of the cone. If we know the slant height and the height of the cone, the radius of the cone can be calculated using the Pythagorean theorem.(r^2=\sqrt{s^2-h^2})
How is the right circular cone formed?
A cone with its axis perpendicular to the plane of the base is called as the right circular cone. When we revolve a right-angled triangle on one of its legs, we get a right circular cone.
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188143 | https://link.springer.com/article/10.1007/BF00691830 | The inhibition and disposition of intestinal alkaline phosphatase
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Summary
A primary mechanism of amino acid inhibition of intestinal alkaline phosphatase is postulated to be the formation of a dissociable enzyme-amino acid complex at an allosteric zinc site. The degree of inhibition was highly correlated with the Zn2+ stability constant of each amino acid and the inhibition was reversible by the addition of exogenous Zn2+ or by dialysis. This allosteric amino acid inhibition proved to be a useful probe of the membrane arrangement of the enzyme in the intact tissue. The catalytic site appears to face the lumen based on the poor permeability of the substrate, the accumulation of the coproducts in the luminal bath, and the response of the enzyme to luminal pH. Amino acid inhibition of alkaline phosphatase in the intact tissue was only effective in the presence of sodium; whereas sodium was not required in butanol extracted preparations which lacked the sidedness of the intact tissue. Since amino acid uptake from the intestine is sodium dependent, the allosteric inhibitory site is probably intracellular. The results suggest that the intestinal alkaline phosphatase spans the apical membrane with the catalytic site accessible from the lumen and the allosteric inhibitory site from the cytoplasm.
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Kenneth W. Gasser & Leonard B. Kirschner
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Gasser, K.W., Kirschner, L.B. The inhibition and disposition of intestinal alkaline phosphatase.
J Comp Physiol B 157, 461–467 (1987).
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How would the intersection of two uncountable sets be finite?
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$\begingroup$
This is a problem from Discrete Mathematics and its Applications
Here is my book's definition on countable
and definition of having the same cardinality
The only example that my book gave of uncountable set was the set of real numbers. I understand that because if you try listing out all of the members of the set, you would keep going on and on - 1, 1.01, 1.001, etc...... But the intersection of the set of real numbers and itself is the set of real numbers is uncountable as well... Is there another uncountable set that you could use to prove this?
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edited Jan 20, 2015 at 19:35
Lehs
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committedandroidercommittedandroider
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1
1
$\begingroup$ Hint: Any proper interval $[a,b]$ is uncountable. $\endgroup$
Mankind
– Mankind
2015-01-20 19:37:27 +00:00
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$\begingroup$
Here's another idea,
$$(-\infty,0]\cap[0,\infty)={0}$$
What can you say about the cardinality of $(-\infty,0]$ and $[0,\infty)$?
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answered Jan 20, 2015 at 19:44
Tim RaczkowskiTim Raczkowski
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$\endgroup$
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$\begingroup$ The cardinality of (,0] and [0,) would be 1 because 0 is the only element for which they intersect. so the intersection set would just be {0} and the size or cardinality of that set is just 1. $\endgroup$
committedandroider
– committedandroider
2015-01-27 20:46:20 +00:00
Commented Jan 27, 2015 at 20:46
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$\begingroup$
Here's one idea: suppose $C$ and $D$ are disjoint uncountable sets, and $E$ is finite. Consider $A=C \cup E$ and $B=D \cup E$. Then $A \cap B=E$ is finite. Can you come up with two disjoint uncountable sets to set this up?
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answered Jan 20, 2015 at 19:35
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$\begingroup$ Why bother with $E$? $\varnothing$ is finite. $\endgroup$
Brian M. Scott
– Brian M. Scott
2015-01-20 19:36:49 +00:00
Commented Jan 20, 2015 at 19:36
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$\begingroup$ @BrianM.Scott Fair point. This is a little bit nice because it shows the cardinality could be any number of things that are smaller than the cardinality of $C$ and $D$ themselves. $\endgroup$
Ian
– Ian
2015-01-20 19:37:56 +00:00
Commented Jan 20, 2015 at 19:37
$\begingroup$ And disjoint means their intersection is the empty set right? $\endgroup$
committedandroider
– committedandroider
2015-01-20 21:01:57 +00:00
Commented Jan 20, 2015 at 21:01
$\begingroup$ @committedandroider Correct. $\endgroup$
Ian
– Ian
2015-01-20 21:02:25 +00:00
Commented Jan 20, 2015 at 21:02
$\begingroup$ Is the shortcut I used to determine if its countable or not a good one? Try to start to listing them out - 1,2,3,4 vs 1.001, 1.0001. I wasn't sure if this shortcut is a mathematically correct one to use $\endgroup$
committedandroider
– committedandroider
2015-01-20 21:03:30 +00:00
Commented Jan 20, 2015 at 21:03
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$\begingroup$
Can you give an example of an uncountable set $B$ that is disjoint from the uncountable set $A= [0,1)$ ??
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answered Jan 20, 2015 at 19:45
HexedAgainHexedAgain
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$\begingroup$ Two sets are disjoint if their intersection is the empty set. So in this case, I come up with B = (100,101). The intersection of [0,1) and (100,101) would be the empty set {} which has a cardinality or size of 0. $\endgroup$
committedandroider
– committedandroider
2015-01-27 20:50:30 +00:00
Commented Jan 27, 2015 at 20:50
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Find all matrices which satisfy M 2−3 M+3 I=0 M 2−3 M+3 I=0
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I am trying to find all matrices which solve the matrix equation
M 2−3 M+3 I=0 M 2−3 M+3 I=0
Since this doesn't factor I tried expanding this in terms of the coordinates of the matrix. It also occurs to me to put it into "vertex" form:
M 2−3 M+9 4 I+3 4 I=0 M 2−3 M+9 4 I+3 4 I=0
(M−3 2 I)2=−3 4 I(M−3 2 I)2=−3 4 I
but this doesn't look much better.
What I found from expanding by coordinates was, if M=(a c b d)M=(a b c d) then
(a 2+b c−3 a+3 a c+c d−3 c a b+b d−3 b b c+d 2−3 d+3)=(0 0 0 0)(a 2+b c−3 a+3 a b+b d−3 b a c+c d−3 c b c+d 2−3 d+3)=(0 0 0 0)
From the off-diagonal entries I get that either
a+d−3=0 a+d−3=0
or
b=c=0 b=c=0
If a+d−3≠0 a+d−3≠0 then a 2−3 a+3=0 a 2−3 a+3=0 and likewise for d d. Then we get more cases for a a and d d.
If a+d−3=0 a+d−3=0 the upper-left is unchanged and the lower-right is
b c+(3−a)2−3(3−a)+3=0 b c+(3−a)2−3(3−a)+3=0
which simplifies to the same thing from the upper-left and so is redundant. In the off-diagonals
a c+c(a−3)−3 c=0⇒a c+c(a−3)−3 c=0⇒
2 a c−6 c=0 2 a c−6 c=0
We again get cases, and I suppose after chasing cases enough you get the solution set.
However, it just feels like this can't be the intended solution given how tedious and uninformative all of this case-chasing is. Is there some bigger idea I'm missing?
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edited Dec 14, 2018 at 21:53
Mostafa Ayaz
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Have you thought about what the Cayley-Hamilton theorem tells you here?user296602 –user296602 2018-12-14 21:45:39 +00:00 Commented Dec 14, 2018 at 21:45
4 All matrices means all 2×2 2×2 matrices, I guess.egreg –egreg 2018-12-14 22:34:47 +00:00 Commented Dec 14, 2018 at 22:34
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You already found the "completion of the square"
(M−3 2 I)2=−3 4 I(M−3 2 I)2=−3 4 I
Then you can write
(i 2 3–√(M−3 2 I))2=X 2=I(i 2 3(M−3 2 I))2=X 2=I
So we are essentially looking for the square roots of the unit matrix, also complex, or for the square roots of −I−I.
You can find various papers dealing with this subject, for example this related post or this thesis.
-- p.s. --
I thought you were interested in the general case of n×n n×n matrices.
If your question is limited to 2×2 2×2 then the ±I−−−√±I is easily found on the net (e.g.,see the hint on Pauli matrices).
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edited Dec 17, 2018 at 23:35
answered Dec 14, 2018 at 23:36
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The links don't work.YiFan Tey –YiFan Tey 2018-12-17 22:26:23 +00:00 Commented Dec 17, 2018 at 22:26
1 @YiFan I think the Pauli matrices satisfy this condition.Mustafa Said –Mustafa Said 2018-12-17 22:41:33 +00:00 Commented Dec 17, 2018 at 22:41
@MustafaSaid sorry if I wasn't clear. The two links at the bottom of the post don't refer to anything (as viewed on my android phone). I wasn't referring to the math content.YiFan Tey –YiFan Tey 2018-12-17 22:45:23 +00:00 Commented Dec 17, 2018 at 22:45
1 @YiFan you can google "Pauli matrices" and I am pretty sure they satisfy the condition X 2=−I X 2=−I.Mustafa Said –Mustafa Said 2018-12-17 22:48:35 +00:00 Commented Dec 17, 2018 at 22:48
@YiFan: sorry ! I forgot to add the links: added now G Cab –G Cab 2018-12-17 23:14:33 +00:00 Commented Dec 17, 2018 at 23:14
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m 2−3 m+3=0 λ=3 2±i 3√2 m 2−3 m+3=0 λ=3 2±i 3 2
You could say that it is all matrices with eigenvalues equal to 3 2+i 3√2,3 2−i 3√2 3 2+i 3 2,3 2−i 3 2
If we restrict our universe to real 2×2 2×2 matrices.
Then it would be all matrices with characteristic equations equal to:
λ 2−3 λ+3=0 λ 2−3 λ+3=0
We are looking for matrices with trace equal to 3, and determinant 3.
[a−a 2−3 a+3 b b 3−a][a b−a 2−3 a+3 b 3−a]
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edited Dec 14, 2018 at 23:02
answered Dec 14, 2018 at 21:55
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1 How is 0 0 an eigenvalue?Shubham Johri –Shubham Johri 2018-12-14 22:00:43 +00:00 Commented Dec 14, 2018 at 22:00
0 0 is an eigenvalue for n×n n×n matrices n>2 n>2. Rather if m 2−3 m+3=0 m 2−3 m+3=0 then so does m 3−3 m 2+3 m=0 m 3−3 m 2+3 m=0.Yadati Kiran –Yadati Kiran 2018-12-14 22:19:53 +00:00 Commented Dec 14, 2018 at 22:19
@YadatiKiran ? this is obvisouly false.Thinking –Thinking 2018-12-14 22:21:59 +00:00 Commented Dec 14, 2018 at 22:21
@Thinking: And why would that be?Yadati Kiran –Yadati Kiran 2018-12-14 22:31:43 +00:00 Commented Dec 14, 2018 at 22:31
@YadatiKiran The question is rather. Why 0 0 is an eigenvalue of every matrix ? Just take an invertible matrix and 0 0 is never an eigenvalue...Thinking –Thinking 2018-12-14 22:32:55 +00:00 Commented Dec 14, 2018 at 22:32
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Minimal polynomial of M,m M(x),M,m M(x), is a factor of x 2−3 x+3=[x−(3+i 3√2)][x−(3−i 3√2)]x 2−3 x+3=[x−(3+i 3 2)][x−(3−i 3 2)]
Either m M(x)=x−(3+i 3√2)⟹M=[3+i 3√2]m M(x)=x−(3+i 3 2)⟹M=[3+i 3 2]
or m M(x)=x−(3−i 3√2)⟹M=[3−i 3√2]m M(x)=x−(3−i 3 2)⟹M=[3−i 3 2]
or m M(x)=x 2−3 x+3⟹m M(x)=x 2−3 x+3⟹ the eigenvalues of M M are 3±i 3√2 3±i 3 2
In case of 2×2 2×2 matrices, product of eigenvalues =det(M)=3=det(M)=3, sum of eigenvalues =Tr(M)=3=Tr(M)=3
We have M=[a c b 3−a]M=[a b c 3−a] and 3 a−a 2−b c=3;a,b,c∈C 3 a−a 2−b c=3;a,b,c∈C.
You could go for 3×3,4×4,...3×3,4×4,... matrices by defining the same eigenvalues and conditions. In case you are looking for real matrices, you just have to take the real subset of these matrices.
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edited Dec 17, 2018 at 22:06
answered Dec 14, 2018 at 22:03
Shubham JohriShubham Johri
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hint
By Cayley-Hamilton,
if the caracteristic polynom is
x 2−3 x+3 x 2−3 x+3
then
M 2−3 M+3 I=0 M 2−3 M+3 I=0
then
(a−x)(d−x)−b c=x 2−3 x+3(a−x)(d−x)−b c=x 2−3 x+3
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answered Dec 14, 2018 at 21:49
hamam_Abdallahhamam_Abdallah
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Cayley-Hamilton only works in one direction doesn't it? So if the characteristic polynomial is x−3+i 3√2 x−3+i 3 2, the result will also hold. And it can for instance be (x−3+i 3√2)2(x−3+i 3 2)2 as well with certain restrictions.Klaas van Aarsen –Klaas van Aarsen 2018-12-14 22:30:33 +00:00 Commented Dec 14, 2018 at 22:30
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188146 | https://www.cement-co2-protocol.org/en/Content/Internet_Manual/tasks/lower_and_higher_heating_values.htm | You are here:
Lower and Higher Heating Values (LHV and HHV)
There are two different types of heating value, which are the lower heating value (LHVLower heat value) and the higher heating value (HHVHigher heat value). By definition the higher heating value is equal to the lower heating value with the addition of the heat of vaporization of the water content in the fuel. These values can be measured in the laboratory for each type of fuel used in the kiln system. However, higher heating values must be converted, in order to obtain the correct values that should be used in the Plant sheet.
Mathematically the relation between both values can be expressed by the following formula [2006 IPCC Guidelines, Vol. II, Section 1.4.1.2, Box 1.1]
Equation 16: Conversion of higher to lower heating values in GJ/t (= MJ/kg)
where
HHV = Higher heating value
H = Percent hydrogen
M = Percent moisture
Y = Percent oxygen (from an ultimate analysis which determines the amount of carbon, hydrogen, oxygen, nitrogen and sulphur as received (i.e. includes Total Moisture (TM))
In the white cells in line132 to line137 the lower heating values (LHV) of conventional fossil kiln fuels can be entered:
Because in many plants some of the fossil kiln fuels are also used for other purposes, all following sections on LHVs of conventional fossil fuels use the values of the kiln fuels as default values. Those lines are of grey colour an can be overwritten for entering more specific values (as in the previous version of the CSI Protocol):
Drying of raw materials and fuels (line154a to line154f)
Non-Kiln Fuel Lower Heating Values (line311a, line312a and line312b)
Drying of mineral components (line313a to line313f)
On-site power generation (line314a to line314i)
The lines turn into white colour when overwritten.
© 2020 by Global Cement and Concrete Association (GCCA) / ECRA GmbH Cement CO2 and Energy Protocol, Internet Manual, created 27/02/2020 |
188147 | https://math.stackexchange.com/questions/1315109/distributing-groups-of-objects-into-boxes | combinatorics - Distributing groups of objects into boxes - Mathematics Stack Exchange
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Distributing groups of objects into boxes
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How can I enumerate the number of ways of distributing distinct groups of identical objects (but various cardinality) into k k boxes such that at most one box is empty (1)(1) and no combination of objects is repeated between boxes ?
The order of objects inside their box or the order of boxes do not matter : ∘∙∙=∙∘∙∘∙∙=∙∘∙
The only way I found to solve this was to generate all partitions, filter them and finally count the elements. This is impracticable when n n or p increase. Is there a way to count the possibilities without enumerating them ?
Example: There are 7 ways in which {∘,∘,∘,∘,∙,∙} can be distributed in 4 boxes.
∅∘∘∘∘∙∙∅∘∘∘∘∙∙∅∘∙∘∘∘∙∅∘∘∙∘∘∙∅∙∘∘∘∘∙∅∙∘∙∘∘∘∘∙∘∘∘∙
Edit: Ignoring the "one box can be empty"-condition and if all objects are distinct, this is solved with the Stirling number of 2 nd kind {n k}. But since some objects are distinguishable (∘,∙) and some are not (∘,∘) I do not know if this is a real lead.
Edit: Users on the chat indicated me that this problem is similar to counting orbits of a group of actions, counting partitions of a multiset and prime factorization.
Edit: This answer provides a working result using the Polya Enumeration Theorem although it is far from computable for high values (2). Let n=∏r i=1 p a i i, r being the number of type of balls (colors), p i some unique prime (different for each type) and a i the number of balls of this type. Then the given function G(n,k) is my answer.
Edit: Following the idea that this is similar to a factorization problem, this paper may be of use. This function gives the number of unordered factorizations with distinct parts and largest part at most m : g(m,n)=∑d|n d≤m g(d,n/d)
with g(m,1)=1 and g(1,n)=0 for n≠1. And then f(n)=g(n,n) gives the total number of unordered factorizations with distinct parts. For {∘,∘,∘,∘,∙,∙}, it gives us f(2 4×3 2)=14 factorizations. It might be possible to adapt the previous formula to keep only the factorizations of k or k−1 integers.
Edit: The desired function is called P d(k,n) in this paper. Although they don't have a formula for it, they state that H d(n)=∑k k!P d(k,n) where H d(n)=∑⌊log 2 n⌋k=1 H′d(k,n) is the total number of distinct ordered factorizations and H′d(k+1,n)=k!k∑j=0(−1)j(k−1−j)!∑H d(k−j,n d)
where the inside sum is taken over all d such that d|n and for d≥2, d is a (j+1)-st power is the number of ordered factorizations of n into k distinct parts.
(1) This condition could be ignored since we can sum the number of combinations for k and k−1.
(2) I am working with k≈50, r≈1000 and a∈[1;10000].
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edited Oct 24, 2018 at 20:25
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The problem can viewed as enumerating r×k matrices with non-negative integer entries that have (i) row sums a 1,…,a r, and (ii) pairwise distinct columns. The number of such matrices then needs to be divided by k! since the order of columns (boxes) does not matter.
For given ranges of parameters the best shot seems to be using the inclusion-exclusion principle to account for condition (ii). It amounts to summing over the partitions of k in the following formula: (−1)k k!∑p⊢k(−1)|p|π(p)r∏j=1 f(a j,p), where π(p) is the number of permutations of cycle typep, and f(a j,p) for p=(p 1,…,p m) is the number of non-negative integer solutions (n 1,…,n m) to the equation: p 1 n 1+p 2 n 2+⋯+p m n m=a j.
For each fixed partition p=(p 1,…,p m), the values f(a j,p) can be obtained at once for all j∈{1,…,r} by the dynamic-programming algorithm based on the identity: f(a,p)=[x a]1(1−x)e 1⋯(1−x k)e k=⌊a/k⌋∑t=0(−1)t(−e k t)[x a−t k]1(1−x)e 1⋯(1−x k−1)e k−1, where e i is the multiplicity of part i in p, with running time O(k⋅max j a j).
In the given example we have k=4, r=2 and (a 1,a 2)=(4,2), and we need to sum up over the partitions of k, i.e., {(1,1,1,1),(2,1,1),(3,1),(2,2),(4)}. Correspondingly, the summands are
4!1 4 4!f(4,(1,1,1,1))f(2,(1,1,1,1))=1⋅35⋅10=350,
−4!1 2 2!2 1 1!f(4,(2,1,1))f(2,(2,1,1))=−6⋅9⋅4=−216,
4!1 1 1!3 1 1!f(4,(3,1))f(2,(3,1))=8⋅2⋅1=16,
4!2 2 2!f(4,(2,2))f(2,(2,2))=3⋅3⋅2=18,
−4!4 1 1!f(4,(4))f(2,(4))=−6⋅1⋅0=0.
So, we get 1 24(350−216+16+18−0)=7 as expected.
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edited Aug 2, 2021 at 19:54
answered Aug 2, 2021 at 17:46
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188148 | https://www.quora.com/How-do-I-calculate-the-interplanar-spacing-of-a-crystal | Something went wrong. Wait a moment and try again.
Materials Science and Eng...
Interplanar Spacing
Bragg's Law
X-ray Powder Diffraction ...
Crystal Chemistry
Condensed Matter Physics
Materials Science Technol...
Materials Chemistry
5
How do I calculate the interplanar spacing of a crystal?
Rawda Hafez Abolfotoh
Studied Physics at Faculty of Science Ain Shams University
·
7y
Francisco M Neto
PhD in Physics, University of São Paulo (USP) (Graduated 2013)
·
Author has 210 answers and 531K answer views
·
Updated 6y
Interplanar distances depend on which plane you are considering, as well as what kind of unit cell you have. You also need to know the lattice parameters of the unit cell, a, b and c.
Planes are identified in terms of their Miller indices, h, k and l. For example, (210) means the plane with h=2, k=1 and l=0.
Once you have that information, you can use one of the following formulas to calculate the interplanar distances:
Cubic (a=b=c):
1d2hkl=h2+k2+l2a2
Tetragonal (a=b):
1d2hkl=h2+k2a2+l2c2
Orthorhombic:
1d2hkl=
Interplanar distances depend on which plane you are considering, as well as what kind of unit cell you have. You also need to know the lattice parameters of the unit cell, a, b and c.
Planes are identified in terms of their Miller indices, h, k and l. For example, (210) means the plane with h=2, k=1 and l=0.
Once you have that information, you can use one of the following formulas to calculate the interplanar distances:
Cubic (a=b=c):
1d2hkl=h2+k2+l2a2
Tetragonal (a=b):
1d2hkl=h2+k2a2+l2c2
Orthorhombic:
1d2hkl=h2a2+k2b2+l2c2
Hexagonal:
1d2hkl=43h2+hk+k2a2+l2c2
If the angles between the axes are not 90°, you can also have:
Rhombohedral:
1d2hkl=(h2+k2+l2)sin2α+2(hk+kl+hl)(cos2α−cosα)a2(1−3cos2α+2cos3α)
Monoclinic:
1d2hkl=(h2a2+k2sin2βb2+l2c2−2hlcosβac)csc2β
Triclinic:
1d2hkl=h2a2sin2α+k2b2sin2β+l2c2sin2γ+2klbccosα+2hlaccosβ+2hkabcosγ1−cos2α−cos2β−cos2γ+2cosαcosβcosγ
Rajesh Prasad
Professor at Indian Institute of Technology, Delhi (1995–present)
·
Upvoted by
Sumanta Sahoo
, M.sc Physics, Sambalpur University (2018) · Author has 90 answers and 822.3K answer views
·
7y
Related
What do you mean by interplanar distance in crystals?
Actually this a subtle concept which is usually not defined properly.
When we define Miller indices (hkl) we can consider a single plane with well defined intercepts. We do not need a family of parallel planes to define Miller indices of a plane.
But when we define interplanar spacing d_hkl we define it as spacing between ‘adjacent’ planes of (hkl) family consisting of equidistant parallel planes. But this needs further qualification as a single plane with Miller indices (hkl) cannot constitute a family. And any plane parallel to (hkl), however close to it, will also by definition have the same
Actually this a subtle concept which is usually not defined properly.
When we define Miller indices (hkl) we can consider a single plane with well defined intercepts. We do not need a family of parallel planes to define Miller indices of a plane.
But when we define interplanar spacing d_hkl we define it as spacing between ‘adjacent’ planes of (hkl) family consisting of equidistant parallel planes. But this needs further qualification as a single plane with Miller indices (hkl) cannot constitute a family. And any plane parallel to (hkl), however close to it, will also by definition have the same Miller indices. This makes the concept of ‘distance between adjacent planes’ questionable.
A way out of this difficulty is to define the (hkl) family and the interplanar spacing d_hkl as follows.
Consider a single plane (hkl) with intercepts a/h, b/k and c/l on the three crystallographic axes with corresponding lattice parameters a, b and c. This plane, by definition, cannot pass through the origin. Now allow another plane parallel to this plane to pass through the origin. The spacing between these two planes (alternatively, the length of perpendicular drawn from the origin to the (hkl) plane) is defined as the interplanar spacing d_hkl. The family of parallel (hkl) planes are defined by all planes parallel to the first (hkl) plane away from the origin and with spacing d_hkl between adjacent planes.
Philip Howie
materials scientist, crystallographer, academic, researcher
·
Author has 3K answers and 13M answer views
·
7y
Related
What do you mean by interplanar distance in crystals?
Crystals are made up of atoms ordered into neat, periodically-repeating arrangements. We characterise each arrangement into a set of lattice points, which are defined by the symmetry operations possible within the crystal structure, and a motif, which is the arrangement of atoms around a lattice point.
If I pick any three non-collinear lattice points, their positions define a plane. (Note that this plane doesn’t ‘exist’ in any real sense; it’s just a mathematical construct.)
Image source.
Since the arrangement of the atoms/lattice points repeats, so does the plane. I end up with a family of plane
Crystals are made up of atoms ordered into neat, periodically-repeating arrangements. We characterise each arrangement into a set of lattice points, which are defined by the symmetry operations possible within the crystal structure, and a motif, which is the arrangement of atoms around a lattice point.
If I pick any three non-collinear lattice points, their positions define a plane. (Note that this plane doesn’t ‘exist’ in any real sense; it’s just a mathematical construct.)
Image source.
Since the arrangement of the atoms/lattice points repeats, so does the plane. I end up with a family of planes, all parallel to each other, repeating with a period that is defined by the symmetry and periodicity of the crystal structure. There are conventions for unambiguously labelling this family of planes, but that’s not necessary for this discussion.
The interplanar distance (or interplanar spacing) is the perpendicular distance between adjacent planes in this family. It’s the blue distance on this image:
Image source.
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Calculating the interplanar spacing of a crystal involves using Bragg's Law and the Miller indices of the crystal planes. Here’s a step-by-step guide:
Identify Miller Indices (hkl)
The Miller indices (h, k, l) represent the orientation of the crystal planes. You need to determine these indices for the specific planes you are interested in.
Determine the Lattice Parameters
For a crystal, you need to know the lattice parameters (a, b, c) and the angles (α, β, γ) between the axes. For cubic crystals, all sides are equal (a = b = c) and all angles are 90°.
Use the Formula for Interplanar Spaci
Calculating the interplanar spacing of a crystal involves using Bragg's Law and the Miller indices of the crystal planes. Here’s a step-by-step guide:
Identify Miller Indices (hkl)
The Miller indices (h, k, l) represent the orientation of the crystal planes. You need to determine these indices for the specific planes you are interested in.
Determine the Lattice Parameters
For a crystal, you need to know the lattice parameters (a, b, c) and the angles (α, β, γ) between the axes. For cubic crystals, all sides are equal (a = b = c) and all angles are 90°.
Use the Formula for Interplanar Spacing
The general formula for the interplanar spacing d in a crystal is given by:
d=1√(h2a2)+(k2b2)+(l2c2)
For cubic crystals (where a=b=c), this simplifies to:
d=a√h2+k2+l2
Calculate the Interplanar Spacing
Substitute the values of a and the Miller indices h,k,l into the formula to find the interplanar spacing d.
Example Calculation
For a cubic crystal with a lattice parameter and Miller indices :
Identify Miller Indices:
Lattice Parameter:
Calculate:
Conclusion
By following these steps, you can calculate the interplanar spacing for various crystal structures. Adjust the formula according to whether the crystal is cubic, tetragonal, hexagonal, etc., as needed.
Dick Lockwood
Studied Geology (Graduated 1972)
·
Author has 906 answers and 219K answer views
·
4y
Related
What is the interplanar distance for a single crystal?
Depends on the crystallography of the particular substance which composes the crystal and the particular crystal planes which you are trying to find the interplanar distance of. In crystals there may be multiple set of planes, all of which are not parallel or always spaced the same.
Related questions
How can you determine the formula for interplanar spacing?
What is the interplanner distance for a simple cubic crystal?
For cubic crystals, as values of the planar indices increase, does the interplanar spacing increase or decrease?
The lattice constant of a cubic lattice is 4 angston. What is the interplanar spacing for , , planes?
What is the unit of interplanar spacing in a crystal?
Jeanne Paquette
20+ Years Teaching Invertebrate Paleontology at McGill University
·
Author has 2.5K answers and 2.5M answer views
·
6y
Related
What is the interplanar distance of NaCl crystals?
Your question is incomplete. I presume that you are referring to the interplanar distance d(001).
You can get this information by considering the content of the unit cell of NaCl and looking up the length of each edge (a = b = c) of the unit cell. One of many sites where you can look this up is Structure World: NaCl. The edge length would be the interplanar distance d(001) in a primitive unit cell.
The face-centering of the unit cell of NaCl adds a twist.
The atomic pattern is repeated halfway between the base and top of the unit cell (and offset by half of the unit cell edge length, either along
Your question is incomplete. I presume that you are referring to the interplanar distance d(001).
You can get this information by considering the content of the unit cell of NaCl and looking up the length of each edge (a = b = c) of the unit cell. One of many sites where you can look this up is Structure World: NaCl. The edge length would be the interplanar distance d(001) in a primitive unit cell.
The face-centering of the unit cell of NaCl adds a twist.
The atomic pattern is repeated halfway between the base and top of the unit cell (and offset by half of the unit cell edge length, either along the a a or the b axis — thus giving rise to the face-centering arrangement).
As a result, the interplanar distance, labelled (002) or (200), is actually half the value of the interplanar distance d(001).
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Rajesh Prasad
Professor at Indian Institute of Technology, Delhi (1995–present)
·
Author has 90 answers and 822.3K answer views
·
8y
Related
For a given crystal, how do we determine if a particular plane is possible or not and how is it related to systematic absenses?
The question should be rephrased a bit, otherwise it does not make proper sense, because, in a given crystal all planes are possible. Probably what is being asked is
“For a given crystal, how do we determine if, during a diffraction experiment, a REFLECTION from a particular plane is possible or not and how is it related to systematic absenses?
Below, I try to answer this rephrased question.
Now, every plane designated by Miller indices (hkl) has an interplanar spacing d_hkl. And it is possible to have a reflection from this plane if the incoming beam of wavelength lambda makes an angle theta wit
The question should be rephrased a bit, otherwise it does not make proper sense, because, in a given crystal all planes are possible. Probably what is being asked is
“For a given crystal, how do we determine if, during a diffraction experiment, a REFLECTION from a particular plane is possible or not and how is it related to systematic absenses?
Below, I try to answer this rephrased question.
Now, every plane designated by Miller indices (hkl) has an interplanar spacing d_hkl. And it is possible to have a reflection from this plane if the incoming beam of wavelength lambda makes an angle theta with the plane satisfying the famous Bragg’s law
lambda=2 d_hkl sin (theta).
Actually Bragg’s law ensures that scattered radiation from all atoms sitting on successive planes of spacing d_hkl will have a path difference of lambda. Thus there is a constructive interference and one gets a diffracted peak.
Thus for every plane of a given spacing (d_hkl) and a radiation of given wavelength (lambda) you can expect a reflection at an angle theta given by Bragg’s law.
However, in some cases even if the beam is directed at a correct angle theta you do not get any reflection. This is what is called SYSTEMATIC ABSENCES. This happens because sometimes there are intermediate plane of atoms in between the planes spaced at d_hkl. If these atoms are present, they will scatter radiation with a path difference of lambda/2 wrt to those sitting on the original planes of spacing d_hkl. Thus there will be a destructive interference and there will be to diffracted beam from such planes at those angles.
For example, consider a BCC crystal. And consider (001) plane, which is a plane parallel to the cube face. Here, d_001 = a where a is the lattice parameter, or the edge length of the cubic unit cell. So atoms sitting on planes parallel to cube faces and spaced at a will scatter in phase. However, since it is a BCC crystal, there are atoms at the cube centres as well. These atoms will be on planes of spacing d_001/2=a/2. These will produce scattered radiation of path difference lambda/2. They will simply destruct the original radiation coming from planes of spacing a/2. We thus say that (001) peaks are systematically absent from BCC crystals (in fact from any centred lattice, for example, also from body-centred tetragonal).
Let us now think of (002) plane. These planes are also parallel to the faces of the cube but the spacing now is a/2. The central atoms also are included in the successive planes. So when Bragg;s law make path difference between these planes as one lambda, all atoms scatter in phase, nothing is left out. So 002 reflection is present in the BCC crystal. But note that this will be at a different angle from the one calculated for 001.
Now, for BCC crystal it can be shown that whenever h+k+l=odd, there will be an intermediate plane and thus the reflection will be absent. Using this rule we see that 001 is absent as 0+0+1=1 is odd and 002 reflection is present as 0+0+2=2 is even and so is present. These conditions of absences are called EXTINCTION CONDITIONS.
Rajesh Prasad
Professor at Indian Institute of Technology, Delhi (1995–present)
·
Author has 90 answers and 822.3K answer views
·
Updated 5y
Related
How do I get a general distance formula between lattice points in a non cubic crystal structure?
This is really a problem of vector algebra in non Cartesian coordinates. Non cubic crystal system is a good example for non Cartesian coordinates.
The distance between two lattice points is the length of lattice translation vector connecting the two points.
The length of any vector is the square root of its dot product with itself.
The dot product has a simple form in Cartesian system. In non cartesian system the formula is more involved can be expressed in terms of 3x3 metric matrix whose i,j term is given by
[math]\textbf{G}_{ij}=\textbf{a}_{i}.\textbf{a}_{j}[/math]
where ai and aj are the ith and jth basis v
This is really a problem of vector algebra in non Cartesian coordinates. Non cubic crystal system is a good example for non Cartesian coordinates.
The distance between two lattice points is the length of lattice translation vector connecting the two points.
The length of any vector is the square root of its dot product with itself.
The dot product has a simple form in Cartesian system. In non cartesian system the formula is more involved can be expressed in terms of 3x3 metric matrix whose i,j term is given by
[math]\textbf{G}_{ij}=\textbf{a}_{i}.\textbf{a}_{j}[/math]
where ai and aj are the ith and jth basis vectors of the crystal. Thus in terms of the lattice parameters the term [math]G_{12}[/math] for example will be
[math]G_{12}=\textbf{a}_1.\textbf{a}_2=\textbf{a}.\textbf{b}=ab \cos (\gamma)[/math]
In terms of the metric matrix the dot product of two vectors is given by
[math]\textbf{x}.\textbf{y}=(x^T) (G) (y)[/math]
Here (y) is the column vector of components of y and (xT) is the row vector of components of x.
Thus the length of vector is
[math]|\textbf{x}|=[\textbf{x}.\textbf{x}]^{1/2}=[(x^T) (G) (x)]^{1/2}[/math]
For the Cartesian system G is an identity matrix, so you get the simpler formula for the dot product and the length.
I intend to improve the answer by proper formatting and some examples at a later time.
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Gervais Chapuis
Author of more than 300 peer reviewed publications in crystallography
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Author has 267 answers and 395.1K answer views
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Updated 5y
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How can you relate interplanar spacing and lattice constant?
The relation is the following
[math]d(\textbf{h})=\frac{1}{|\textbf{h}|}[/math]
where [math]\textbf{h} = h\textbf{a}^+k\textbf{b}^+l\textbf{c}^[/math] is the reciprocal lattice vector with integer indices (hkl) and basis vectors [math]\textbf{a}^, \textbf{b}^ [/math]and [math] \textbf{c}^[/math]. The reciprocal unit cell vectors are related to the unit cell vectors by the expression
[math]\textbf{a}\cdot \textbf{a}^=1; \textbf{a}\cdot \textbf{b}^=0; \textbf{a}\cdot \textbf{c}^=0, [/math]and circular permutations
or in matrix form
The relation is the following
[math]d(\textbf{h})=\frac{1}{|\textbf{h}|}[/math]
where [math]\textbf{h} = h\textbf{a}^+k\textbf{b}^+l\textbf{c}^[/math] is the reciprocal lattice vector with integer indices (hkl) and basis vectors [math]\textbf{a}^, \textbf{b}^ [/math]and [math] \textbf{c}^[/math]. The reciprocal unit cell vectors are related to the unit cell vectors by the expression
[math]\textbf{a}\cdot \textbf{a}^=1; \textbf{a}\cdot \textbf{b}^=0; \textbf{a}\cdot \textbf{c}^=0, [/math]and circular permutations
or in matrix form
Related questions
What do you mean by interplanar distance in crystals?
What are some examples of crystals with different interplanar spacings?
If an orthorhombic crystal system has “a=2b=3c”, then what will be its interplanar distance?
How do you calculate lattice spacing?
What is interplanar spacing in crystals?
Rajesh Prasad
Professor at Indian Institute of Technology, Delhi (1995–present)
·
Author has 90 answers and 822.3K answer views
·
2y
Related
Can you explain the difference between lattice constant, interplanar spacing, and unit cell dimensions in crystals?
You select a unit cell for a crystal, which is conventionally chosen to be a parallelopiped. Size and shape of the parallelopiped unit cell is completely defined by the lengths (a, b and c) of three edges meeting at a vertex and the three interaxial angles (alpha, beta and gamma) between the three pairs of edges. These six numbers, a, b, c, alpha, beta, gamma, completely define the unit cell dimensions. These are called lattice constants or lattice parameters.
Interplanar spacing is the spacing between two parallel planes defined by some Miller indices (hkl).
Philip Howie
materials scientist, crystallographer, academic, researcher
·
Author has 3K answers and 13M answer views
·
8y
Related
For cubic crystals, as values of the planar indices increase, does the interplanar spacing increase or decrease?
For a cubic crystal, the interplanar spacing is given by
[math]d_{hkl} = \dfrac{a}{\sqrt{h^2 + k^2 + l^2}}[/math]
Therefore, as the indices hkl increase, the interplanar spacing decreases.
This is a reduction of the more general case for any crystal system, which is usually expressed in terms of the reciprocal lattice vectors:
[math]d_{hkl} = \dfrac{1}{ha^ + kb^ + lc^}[/math]
Karan Mehta
Semiconductor device theorist, PhD in EE
·
Author has 1.7K answers and 31M answer views
·
3y
Related
Why does the crystal band structure measured in k-space? Can we convert it to real space as I think it is more convenient to observe its physical meaning that way?
Electrons propagate as waves in a crystal and are described by wave functions in terms of the electron’s crystal momentum
[math]k[/math]
, which is also called the wave vector. According to the Uncertainty Principle, any particle having a well-defined momentum has a poorly defined position in real space, which is consistent with the wave-like nature of electrons. Since electrons are not localized in real space, it does not make sense to describe electrons in terms of real space. But these electron waves do have a well-defined crystal momentum, which is why we choose to work in the
[math]k[/math]
-space or the reciprocal
Electrons propagate as waves in a crystal and are described by wave functions in terms of the electron’s crystal momentum [math]k[/math], which is also called the wave vector. According to the Uncertainty Principle, any particle having a well-defined momentum has a poorly defined position in real space, which is consistent with the wave-like nature of electrons. Since electrons are not localized in real space, it does not make sense to describe electrons in terms of real space. But these electron waves do have a well-defined crystal momentum, which is why we choose to work in the [math]k[/math]-space or the reciprocal lattice space. The [math]k[/math]-space is also referred to as the reciprocal lattice space because it has units of inverse distance.
In a crystal, an electron sees a periodic ionic potential whose period corresponds to the atomic spacing of the crystalline material. Electrons have different properties in different crystalline materials because different materials have different arrangements of atoms and atomic spacings.
If you are familiar with signal analysis, you’d know that working in the frequency domain is often far more useful in formulating and understanding a signal than working in the time domain. Position and momentum have a similar relationship as time and frequency, in that they are conjugate variables which are connected by simple Fourier transformations. Similarly, it is far more convenient and insightful to describe the electronic states in a crystal in terms of [math]k[/math] rather than position [math]x[/math].
Related questions
How can you determine the formula for interplanar spacing?
What is the interplanner distance for a simple cubic crystal?
For cubic crystals, as values of the planar indices increase, does the interplanar spacing increase or decrease?
The lattice constant of a cubic lattice is 4 angston. What is the interplanar spacing for , , planes?
What is the unit of interplanar spacing in a crystal?
What do you mean by interplanar distance in crystals?
What are some examples of crystals with different interplanar spacings?
If an orthorhombic crystal system has “a=2b=3c”, then what will be its interplanar distance?
How do you calculate lattice spacing?
What is interplanar spacing in crystals?
What is the interplanar distance of NaCl crystals?
Is it possible to have two different values for the interplanar distance between planes of atoms in a crystal?
How do you calculate Bragg's law for interplanar spacing?
How do I calculate the Miller indices for a plane or direction in a crystal?
Dear Sir, would you please guide how from the generalized expression of interplanar spacing, we can derive the interplanar spacing for monoclinic, triclinic and other crystal system. Where can I find the derivation?
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Ch. 19 - First Law of Thermodynamics
Ch. 19 - First Law of Thermodynamics
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Chapter 19 The First Law of Thermodynamics PowerPoint® Lectures for University Physics, Thirteenth Edition – Hugh D. Yo
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Citation preview
Chapter 19
The First Law of Thermodynamics PowerPoint® Lectures for University Physics, Thirteenth Edition – Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Copyright © 2012 Pearson Education Inc.
Goals for Chapter 19 • To represent heat transfer and work done in a thermodynamic process and to calculate work
• To relate heat transfer, work done, and internal energy change using the first law of thermodynamics • To distinguish between adiabatic, isochoric, isobaric, and isothermal processes • To understand and use the molar heat capacities at constant volume and constant pressure • To analyze adiabatic processes Copyright © 2012 Pearson Education Inc.
Introduction • A steam locomotive operates using the laws of thermodynamics, but so do air conditioners and car engines.
• We shall revisit the conservation of energy in the form of the first law of thermodynamics.
Copyright © 2012 Pearson Education Inc.
Thermodynamics systems •
A thermodynamic system is any collection of objects that may exchange energy with its surroundings.
•
In a thermodynamic process, changes occur in the state of the system.
•
Careful of signs! Q is positive when heat flows into a system. W is the work done by the system, so it is positive for expansion. (See Figure 19.3 at the right.)
Copyright © 2012 Pearson Education Inc.
Work done during volume changes • Figures 19.4 and 19.5 below show how gas molecules do work when the gas volume changes.
Copyright © 2012 Pearson Education Inc.
Work on a pV-diagram • The work done equals the area under the curve on a pV-diagram. (See Figure 19.6 below.) • Work is positive for expansion and negative for compression. • Follow Example 19.1 for an isothermal (constant-temperature) expansion.
Copyright © 2012 Pearson Education Inc.
Example 19.1
Copyright © 2012 Pearson Education Inc.
Work depends on the path chosen • Figure 19.7 below shows why the work done depends on the path chosen.
Copyright © 2012 Pearson Education Inc.
First law of thermodynamics • First law of thermodynamics: The change in the internal energy U of a system is equal to the heat added minus the work done by the system: U = Q – W. (See Figure 19.9 at the right.) • The first law of thermodynamics is just a generalization of the conservation of energy. • Both Q and W depend on the path chosen between states, but U is independent of the path. • If the changes are infinitesimal, we write the first law as dU = dQ – dW.
Copyright © 2012 Pearson Education Inc.
Work Out!!
Copyright © 2012 Pearson Education Inc.
Cyclic processes and isolated systems • In a cyclic process, the system returns to its initial state. Figure 19.11 below illustrates your body’s cyclic process for one day. • A isolated system does no work and has no heat flow in or out.
Copyright © 2012 Pearson Education Inc.
Example 19.2
Copyright © 2012 Pearson Education Inc.
Example 19.3
Copyright © 2012 Pearson Education Inc.
Example 19.4
Copyright © 2012 Pearson Education Inc.
Example 19.5
Copyright © 2012 Pearson Education Inc.
Four kinds of thermodynamic processes • Adiabatic: No heat is transferred into or out of the system, so Q = 0. • Isochoric: The volume remains constant, so W = 0. • Isobaric: The pressure remains constant, so W = p(V2 – V1). • Isothermal: The temperature remains constant.
Copyright © 2012 Pearson Education Inc.
The four processes on a pV-diagram • Figure 19.16 shows a pV-diagram of the four different processes.
Copyright © 2012 Pearson Education Inc.
Internal energy of an ideal gas • The internal energy of an ideal gas depends only on its temperature, not on its pressure or volume. • The temperature of an ideal gas does not change during a free expansion. (See Figure 19.17 at the right.)
Copyright © 2012 Pearson Education Inc.
Heat capacities of an ideal gas • CV is the molar heat capacity at constant volume. • Cp is the molar heat capacity at constant pressure. • Figure 19.18 at the right shows how we could measure the two molar heat capacities.
Copyright © 2012 Pearson Education Inc.
Relating Cp an CV for an ideal gas
• Figure 19.19 at the right shows that to produce the same temperature change, more heat is required at constant pressure than at constant volume since U is the same in both cases. • This means that Cp > CV. • Cp = CV + R.
Copyright © 2012 Pearson Education Inc.
The ratio of heat capacities • The ratio of heat capacities is = Cp/CV. For ideal gases, = 1.67 (monatomic) and = 1.40 (diatomic). • Table 19.1 shows that theory and experiment are in good agreement for monatomic and diatomic gases. • Follow Example 19.6.
Copyright © 2012 Pearson Education Inc.
Example 19.6
Copyright © 2012 Pearson Education Inc.
Adiabatic processes for an ideal gas • In an adiabatic process, no heat is transferred in or out of the gas, so Q = 0. • Figure 19.20 at the right shows a pV-diagram for an adiabatic expansion. Note that an adiabatic curve at any point is always steeper than an isotherm at that point. • Follow the derivations showing how to calculate the work done during an adiabatic process. Copyright © 2012 Pearson Education Inc.
Adiabatic compression in a diesel engine • Follow Example 19.7 dealing with a diesel engine. Use Figure 19.21 below.
Copyright © 2012 Pearson Education Inc.
Example 19.7
Copyright © 2012 Pearson Education Inc.
Summary
Copyright © 2012 Pearson Education Inc.
Summary
Copyright © 2012 Pearson Education Inc.
Key Equations work done in a volume change
(19.2)
work done in a volume change at constant pressure
(19.3)
W p dV V2
V1
W p V2 V1
U 2 U1 U Q W dU dQ dW
first law of thermodynamics, infinitesimal process
C p CV R
molar heat capacities of an ideal gas Cp CV
ratio of heat capacities
W nCV T1 T2 W
first law of thermodynamics
adiabatic process, ideal gas
CV 1 pV pV p V 1 1 p2V2 R 1 1 1 2 2
Copyright © 2012 Pearson Education Inc.
adiabatic process, ideal gas
(19.4) (19.6) (19.17)
(19.18)
(19.25)
(19.26)
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Procarbazine is defined as a unique anticancer agent used in combination therapy for advanced Hodgkin’s disease, which functions through multiple mechanisms including the inhibition of DNA precursor incorporation and direct DNA damage via methylation reactions.
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2010, Comprehensive ToxicologyK.Y. Yoneda, C.E. Cross
8.24.10.2 Procarbazine
Procarbazine is a methylating agent used in the treatment of Hodgkin’s lymphoma, non-Hodgkin’s lymphoma, brain tumors, melanoma, lung cancer, and multiple myeloma. Pulmonary toxicity has been reported in only nine patients who received combination therapy with mechlorethamine, vincristine, procarbazine, and prednisone (MOPP) for Hodgkin’s lymphoma (Mahmood and Mudad 2002), but appears to be more common when procarbazine is used in combination with vincristine and lomustine with up to 31% experiencing pulmonary involvement in a small series (Coyle et al. 1992). Procarbazine is administered in combination with other chemotherapeutic agents, confounding the assignment of causality to this agent alone. Nevertheless, procarbazine is believed to cause an immune-mediated pulmonary toxicity that occurs after multiple doses and recurs with rechallenge. The common pulmonary symptoms of cough and dyspnea may be accompanied by fever, eosinophilia, and rash. Lung biopsy is nonspecific revealing alveolitis and interstitial fibrosis. Withdrawal of the drug alone may result in improvement, but systemic glucocorticoids may be required with slow improvement over weeks to months (Brooks et al. 1990; Coyle et al. 1992; Mahmood and Mudad 2002).
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Chapter
Clinical Pharmacology of Brain Tumor Chemotherapy
2006, Handbook of Brain Tumor ChemotherapyHerbert B. Newton
Procarbazine.
Procarbazine is a drug that requires hepatic activation to intermediate forms before developing potent activity as an alkylating agent (see Fig. 2.3) [90,91]. PCB is taken orally and is rapidly absorbed by the gastrointestinal tract. Following absorption, it is first metabolized into an azo-PCB derivative, which has similar potency to PCB. Further metabolism by the cytochrome P-450 system converts azo-PCB into two separate azoxy-PCB derivatives, which have significantly greater antitumor activity than PCB or azo-PCB [90,91]. Once activated, PCB alkylates DNA at the O 6 position of guanine . In addition, PCB can induce DNA strand breakage and inhibit DNA, RNA, and protein synthesis. PCB has further pharmacological properties, including activity as a monoamine oxidase-inhibitor and a disulfiram-like effect [8,21]. Potential interactions (i.e., acute hypertension) can occur if PCB is taken concomitantly with sympathomimetic drugs, antihistamines, tricyclic antidepressants, and food, high in tyramine content (e.g., wine, beer, cheese, chocolate, bananas, and yogurt) . Due to the disulfiram-like effect, alcohol should be avoided while taking PCB, or severe gastrointestinal distress will develop. Although water soluble, PCB and its metabolites readily cross the blood–brain barrier, with rapid equilibration between plasma and cerebrospinal fluid.
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FIGURE 2.3.
Procarbazine has been used to treat malignant PBT as a single-agent or in multi-agent regimens [8,21,71–73]. As mentioned above, single-agent PCB was compared to intravenous BCNU, methylprednisolone, and intravenous BCNU plus methylprednisolone in a randomized trial of malignant glioma patients by Green and colleagues . Overall median survival was similar for the PCB (47 weeks) and BCNU (50 weeks) groups. Although the survival percentage was similar for PCB and BCNU at 12 months (44.0 versus 48.5 per cent), long-term survival at 18 and 24 months was superior in the PCB group (28.8 and 22.8 per cent versus 23.8 and 15.6 per cent, respectively). Newton and colleagues found PCB beneficial for glioma patients after failure of irradiation and nitrosourea chemotherapy . In a series of 35 patients they noted 2 complete responses, 7 partial responses, and 11 patients with stable disease. Furthermore, when responses were compared in a cohort of malignant glioma patients who were initially treated with BCNU and then received PCB (after BCNU failure), there was a significant difference in time to progression between the groups . The PCB group had a greater percentage of patients without disease progression at 6 and 12 months as compared to BCNU (48 and 35 per cent versus 26 and 3 per cent, respectively). Other investigators have found single-agent PCB to be less efficacious, with fewer complete and partial responses .
The most commonly used multi-agent regimen that incorporates PCB is PCV (PCB 60 mg/m 2 per days, 8–21, CCNU 110 mg/m 2 day 1, vincristine 1.4 mg/m 2 days 8 and 28; every 8 weeks) [8,21,71–73]. Numerous reports document the efficacy of PCV against anaplastic oligodendrogliomas, anaplastic astrocytomas, recurrent oligodendrogliomas, and mixed gliomas. The most dramatic results have been in patients with anaplastic or recurrent oligodendrogliomas [97–99]. Cairncross and colleagues have determined that anaplastic oligodendrogliomas are relatively chemosensitive, especially to PCV. Several series of patients have reported durable response rates in excess of 50 per cent [73,97–99]. A recent report suggests that anaplastic oligodendrogliomas with a specific molecular genetic profile are more sensitive to PCV . Those patients with tumors that had allelic deletion or loss of heterozygosity of chromosomes 1p and 19q had significantly longer recurrence-free and overall survival than patients that retained 1p and 19q. The use of PCV appears to be beneficial for recurrent typical (i.e., WHO grade II) oligodendroglioma as well, especially after failure of non-PCV regimens . Mixed gliomas with an oligodendroglial component also appear sensitive to PCV [101,102]. PCV has also been used for the treatment of anaplastic and mixed astrocytoma by Levin and colleagues . They found PCV superior to BCNU when comparing time to progression and overall survival. Other multi-agent regimens for malignant gliomas combine PCB with vincristine and either mechlorethamine or etoposide (VP-16) [74,104]. Results with these regimens are similar to that reported for intravenous BCNU.
The most frequent acute toxicity of PCB is nausea and emesis (80–85 per cent) [94,95]. Other less common side effects include fatigue (17 per cent), rash, and neurotoxicity (usually when administered intravenously). For most patients, the dose-limiting toxicity is a combination of neutropenia and thrombocytopenia.
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Book 2006, Handbook of Brain Tumor ChemotherapyHerbert B. Newton
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Neurologic Aspects of Systemic Disease Part III
2014, Handbook of Clinical NeurologyRiccardo Soffietti, ... Roberta Rudà
Procarbazine
Procarbazine is thought to function as an alkylating agent after activation in the liver, and is used for treatment of lymphoma and gliomas. High-dose oral procarbazine can cause CNS toxicity, including somnolence, depression, obtundation, and psychosis (Postma et al., 1998). In addition, due to its weak activity as a monoamine oxidase (MAO) inhibitor, procarbazine can cause hypertensive encephalopathy, headache, and delirium when administered in combination with sympathomimetic agents or after consumption of tyramine-containing foods.
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Handbook2014, Handbook of Clinical NeurologyRiccardo Soffietti, ... Roberta Rudà
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Neuro-Oncology
2010, Blue Books of NeurologyKate Scatchard, Siow Ming Lee
PROCARBAZINE
Procarbazine is a DNA alkylator commonly used in the treatment of Hodgkin lymphoma; it readily crosses the blood brain barrier.69 It causes an acute encephalopathy with confusion and somnolence in 14% to 33% of patients, which is probably secondary to its ability to inhibit monoamine oxidase.109,136,137 Symptoms are therefore worsened by concurrent administration of phenothiazines such as chlorpromazine, which is used frequently as antiemetic prophylaxis. Peripheral neuropathy has also been reported in 2% to 20% of patients receiving procarbazine. Patients typically present with a subjective paresthesia, which usually resolves completely on cessation of treatment.136,137
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Book series2010, Blue Books of NeurologyKate Scatchard, Siow Ming Lee
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Procarbazine
2007, xPharm: The Comprehensive Pharmacology ReferenceKrishna Agrawal
Procarbazine, a methylhydrazine derivative, is an antineoplastic drug. Although not classified as an alkylating agent, procarbazine is bioactivated by cytochrome P-450 to reactive metabolites that methylate DNA …
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Reference work 2007, xPharm: The Comprehensive Pharmacology ReferenceKrishna Agrawal
Chapter
Procarbazine
2016, Meyler's Side Effects of Drugs (Sixteenth Edition)
General information
Procarbazine is an alkylating agent that has been used in the treatment of Hodgkin’s disease in regimens such as MOPP (chlormethine (mechlorethamine), vincristine (Oncovin), procarbazine, and prednisolone) and BEACOPP (bleomycin, etoposide, doxorubicin (Adriamycin), cyclophosphamide, vincristine, procarbazine, and prednisone) . It is also used to treat glioblastoma multiforme. As with many cytotoxic and cytostatic drugs, it can be difficult to attribute adverse events causally to procarbazine, since it is so often used in combinations of this sort.
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Reference work 2016, Meyler's Side Effects of Drugs (Sixteenth Edition)
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A Review of Therapeutic Trials in Multiple Myeloma and Perspectives for Future Trials
1980, Progress in MyelomaMark Vrana, Paul A. Bunn Jr.
Procarbazine
Procarbazine has not been used as a single agent in previously untreated patients. In patients previously failing alkylating agents, responses have been reported in 5 of 28 patients.83
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Book 1980, Progress in MyelomaMark Vrana, Paul A. Bunn Jr.
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188151 | https://gamedev.stackexchange.com/questions/72170/how-simulate-the-return-effect-of-the-wheel-of-fortune-needle | mathematics - How simulate the return effect of the wheel of fortune needle - Game Development Stack Exchange
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How simulate the return effect of the wheel of fortune needle
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Asked 11 years, 6 months ago
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I am trying to create a spinning wheel which slows down after a specific time. I managed to simulate the slowdown of the wheel but I need also to put a needle which go up and down with an angle of 45°.
Here my wheel with the needle :
My wheel turn in clockwise and the needle (object in red in the above scheme) rotate in counter clockwise.
Once done the needle back to his original position slower than the previous movement.
The code i wrote :
```java
// Once 3 seconds are passed, the speed rotation start to decrease
If RealMillisecs() -_fRotateTime > 3000 Then
If _fSpeedRotation > 0 Then
_fSpeedRotation = _fSpeedRotation - 0.001
Else
_fSpeedRotation = 0
EndIf
EndIf
_fWheelRotation -= _fSpeedRotation dt.frametime
// If the needle is not going back to its orginal position then is going up
If not _bAnimBackNeedle Then
_fNeedleRotation = _fWheelRotation Mod 45 -1
/ If its previous rotation greater than actual position
it's mean that the needle has reach its max angle (45)/
If _fNeedleRotation < _fPreviousNeedleRotation Then
_bAnimBackNeedle = True // The going back animation to true
EndIf
_fPreviousNeedleRotation = _fNeedleRotation
EndIf
// If the going back animation is set to true or the wheel has stopped to turn
// get the needle back to its position at constant speed
If _bAnimBackNeedle Or _fSpeedRotation = 0 Then
_fNeedleRotation -= 0.5 dt.frametime
If _fNeedleRotation <= 0 Then
_bAnimBackNeedle = False
EndIf
EndIf
```
My code doesn't make what i hope.
The needle doesn't come back to its original position when the wheel is stopped.
How can I managed the code to have the wheel and the needle with a good behaviour?
mathematics
algorithm
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edited Mar 19, 2014 at 18:11
Dan
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asked Mar 19, 2014 at 17:55
MoerinMoerin
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wow nice, I wish this was available in some form javascript bilogic –bilogic 2025-03-11 14:31:47 +00:00 Commented Mar 11 at 14:31
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How about letting box2d do the work for you? It wasn't too difficult for me to setup everything in RUBE:
There are two major parts: the wheel and the needle.
Wheel
First there is a base. In this case the base is a static body with a square fixture.
The wheel is attached to the base via a revolute joint. This joint allows the wheel to spin freely about the center.
The wheel consists of 33 circle fixtures: 1 big wheel and 32 pegs. I created the pegs using a simple Rubescript. The pegs and wheel are on different bitplanes, allowing the needle to collide only with the pegs and not the wheel. One other ajustment is that the boolean isBullet needs to be true; the reason for this is that the pegs can move very quickly when the wheel is spinning fast and sometimes the peg-needle collision will be missed by box2d. If the isBullet flag is set, box2d resolves those cases correctly.
Needle
The needle is a body with a single fixture in the shape of a kite. I have increased the friction a little on the needle fixture and peg fixtures. This way when the needle rubs against the pegs it will eventually slow the wheel down. We also need to set the bitplane and bitmask of the needle so it only collides with the pegs.
The needle can rotate about the revolute joint. The second body for the revolute joint is the static body from before.
Two opposing distance joints act as springs to try and keep the needle in place. The damping parameter on the distance joints also helps dissipate the wheel's energy.
Result
You get a lot of nice realisms when you go with a physics-based solution. Most notably, you can let the player actually spin the wheel with a swipe of their finger or a click+drag of their mouse. Also, it's easy to do things like add more needles for more players if you want to.
I know there is a lot of stuff here specific to box2d which you might not be familiar with but it is well worth learning IMO. The spinning wheel is an excellent way for you to get started.
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answered Mar 25, 2014 at 2:54
NauticalMileNauticalMile
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Mile. Thanks for this detailed answer. It' works really great with box 2D, i thought that use box 2D would be too difficult to understand and implement with my code. But with RUBE it's seems more easly.Moerin –Moerin 2014-03-27 19:57:05 +00:00 Commented Mar 27, 2014 at 19:57
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As a side-note, your use of EndIf makes me think you're trying to do this in PHP - which makes no sense whatsoever, but whatever. It would help if you could clarify what language you're working with.
But concerning the mathematics alone...
The two events you need to catch are when the edge of the space contacts the needle and when the edge breaks contact with the needle.
cpp
if( (fWheelRotation + 7) % 45 < 6 )
{
fNeedleRotation = ((((367 - fWheelRotation) % 45)^2) 1.8);
} else {
if( fNeedleRotation < 0 )
{
fNeedleRotation = 0;
}
else if( fNeedleRotation < 46 )
{
fNeedleRotation -= 10; // Adjust this value to control return speed
// Be sure to allow a small overshoot
}
}
Due to how close to the threshold you're operating, the offset offers a buffer. Without the buffer, a fast-spinning wheel could bypass the (much smaller) trigger zone(s) without the needle responding.
I could offer a slightly more elegant solution if you included what language you're using.
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answered Mar 22, 2014 at 16:44
Twyla Naythias FoxTwyla Naythias Fox
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188152 | https://math.stackexchange.com/questions/3678124/find-equation-for-all-points-equidistant-to-two-points | Stack Exchange Network
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Find equation for all points equidistant to two points
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0
$\begingroup$
I have two points: A(-2,5,4) and B(6,2,-3). Per the question, I want to find an equation of all points equidistant from those two points. The intuition and visualization is easy, this will be a plane of points such that point C, whatever it may be, creates an isosceles triangle with points A and B. Vectors are still alien to me and I need help writing the problems in mathematical terms.
$A=(-2, 5, 4)$ $B=(6, 2, -3)$ $\vec{AB}=<8,-3,-7>=8\hat{i}-3\hat{j}-7\hat{k}$ $\hat{AB}=\frac{\vec{AB}}{\sqrt{122}}$ $Midpoint=(2, 3.5, 0.5)$
I don't know where to go from here, assuming I did this part correctly. All I know is I need a plane that passes through the midpoint and is perpendicular to $\vec{AB}$. I thought about using $\vec{AB} \cdot \vec{AB^{¥}}=0$ and setting $\vec{AB^{¥}}=$, but WebAssign rejected my end result.
Can somebody walk me through this problem and explain why you took each step?
linear-algebra
multivariable-calculus
vectors
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asked May 16, 2020 at 18:41
DominicTarroDominicTarro
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$\endgroup$
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3 Answers 3
Reset to default
1
$\begingroup$
Let $\vec r$ be the position vector of any general point on this plane. Then, $\vec r - (2, \ 3.5, \ 0.5)$ will represent any general vector lying on the plane. Now, we need all such vectors to be perpendicular to the normal vector of the plane, i.e. the one you obtained $(\vec{AB})$. In other words, equate the dot product to zero: $$\big((\vec r -(2, \ 3.5, \ 0.5)\big) \cdot (8, \ -3, \ -7) =0$$ One can write this equation in cartesian form as $$8(x-2) -3(y-3.5) -7(z-0.5)=0$$
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answered May 16, 2020 at 18:52
VishuVishu
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$\endgroup$
Add a comment |
1
$\begingroup$
A plane perpendicular to $\vec{AB}=\hat{i}-3\hat{j}-7\hat{k}$ and passing through the origin has equation $$8x-3y-7z=0.\tag1$$ If you've learned about dot products, note that the dot product with $\vec{AB}$ of any vector in this plane is $0$. A plane parallel to $(1)$ has equation $$8x-3y-7z=k$$ for some constant $k$, so all that remains to do is to determine the value of $k$ such that midpoint lies on the plane.
Over to you.
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answered May 16, 2020 at 18:55
saulspatzsaulspatz
53.9k77 gold badges3737 silver badges7979 bronze badges
$\endgroup$
Add a comment |
1
$\begingroup$
Why not just work from the original definition of the set? The distance from an arbitrary point to $A$ is $\sqrt{(x+2)^2+(y-5)^2+(z-4)^2}$ and from $B$ is $\sqrt{(x-6)^2+(y-2)^2+(z+3)^2}$. Setting these equal to each other and squaring, we have $$(x+2)^2+(y-5)^2+(z-4)^2 = (x-6)^2+(y-2)^2+(z+3)^2.$$ The squared terms all cancel, leaving $$8x-3y-7z-2=0.$$
Youve got a good start on a solution using vectors. The midpoint is indeed $(2,7/2,1/2)$ and a normal vector to the plane is $(8,-3,-7)$. Theres no particularly good reason to normalize this vector in order to solve this problem, though. Youre just making extra work for yourself by doing that. Now, the normal vector to the plane is orthogonal to the vector between any two point on the plane. So, if you have a normal $\mathbf n$ to the plane and a known point $\mathbf x_0$ on it, an equation of the plane is $\mathbf n\cdot(\mathbf x-\mathbf x_0)=0$. Plug in your values and rearrange as necessary. You should end up with exactly the same equation as above.
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188153 | https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/pages/unit-2-applications-of-differentiation/ | Course Info
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18.01SC | Fall 2010 | Undergraduate
Single Variable Calculus
2. Applications of Differentiation
« Previous | Next »
Introduction
This unit describes techniques for using differentiation to solve many important problems. Part C of this unit presents the Mean Value Theorem and introduces notation and concepts used in the study of integration, the subject of the next two units.
Part A: Approximation and Curve Sketching
Session 23: Linear Approximation
Session 24: Examples of Linear Approximation
Session 25: Introduction to Quadratic Appoximation
Session 26: Using Quadratic Approximations
Session 27: Sketching Graphs I - Polynomials and Rational Functions
Session 28: Sketching Graphs II - General Strategies
Problem Set 3
Part B: Optimization, Related Rates and Newton’s Method
Session 29: Optimization Problems
Session 30: Optimization Problems II
Session 31: Related Rates
Session 32: Ring on a String
Session 33: Newton’s Method
Problem Set 4
Part C: Mean Value Theorem, Antiderivatives and Differential Equations
Session 34: Introduction to the Mean Value Theorem
Session 35: Using the Mean Value Theorem
Session 36: Differentials
Session 37: Antiderivatives
Session 38: Integration by Substitution
Session 39: Introduction to Differential Equations
Session 40: Separation of Variables
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Exam 2
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Session 42: Materials for Exam 2
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188154 | https://brainly.com/question/32320782 | [FREE] Calculate the pKa values for the following acids: a) Phenol (K_a = 1.0 \times 10^{-10}) b) Acrylic acid - brainly.com
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Calculate the pKa values for the following acids:
a) Phenol (K a=1.0×1 0−10)
b) Acrylic acid (K a=5.6×1 0−6)
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The pKa value for phenol is approximately 10 and for acrylic acid is approximately 4.25.
To calculate the pKa values for the following acids, we will use the formula: pKa = -log10(Ka).
a) For phenol (Ka = 1.0 x 10^-10), follow these steps:
Identify the Ka value: 1.0 x 10^-10
Apply the formula: pKa = -log10(1.0 x 10^-10)
Calculate the pKa value: pKa ≈ 10
b) For acrylic acid (Ka = 5.6 x 10^-6), follow these steps:
Identify the Ka value: 5.6 x 10^-6
Apply the formula: pKa = -log10(5.6 x 10^-6)
Calculate the pKa value: pKa ≈ 4.25
Therefore the pKa value for phenol is approximately 10 and for acrylic acid is approximately 4.25.
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The pKa value for phenol is approximately 10, and for acrylic acid, it is approximately 5.25. These values were calculated using the formula pKa = -log(K_a).
Explanation
To calculate the pKa values for phenol and acrylic acid, we use the formula:
pKa=−lo g 10(K a)
A. Phenol
Identify the K a value: K a=1.0×1 0−10
Apply the formula:
pKa=−lo g 10(1.0×1 0−10)
3. Calculate the pKa value:
Using a calculator, −lo g 10(1.0×1 0−10)=10.
Hence, the pKa of phenol is approximately 10.
B. Acrylic Acid
Identify the K a value: K a=5.6×1 0−6
Apply the formula:
pKa=−lo g 10(5.6×1 0−6)
3. Calculate the pKa value:
Using a calculator, −lo g 10(5.6×1 0−6)≈5.25.
Hence, the pKa of acrylic acid is approximately 5.25.
Therefore, the pKa values are:
Phenol: approximately 10
Acrylic Acid: approximately 5.25
Examples & Evidence
For example, acetic acid has a pKa of about 4.76, which is lower than that of phenol, indicating that acetic acid is a stronger acid than phenol. Additionally, comparing acrylic acid's pKa of 5.25 to that of acetic acid shows that acrylic acid is also a relatively stronger acid, although it is weaker than strong acids like hydrochloric acid which has a pKa of around -7.
The calculation of pKa from K_a is commonly used in chemistry to assess acid strength. The relationship between pKa and K_a is well established in acid-base chemistry, providing a reliable method to express acidity in different compounds.
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Community Answer calculate the pka values for the following acids. a) methanol (ka = 2.9 x 10-16) b) citric acid (ka = 7.2 x 10-4)
Community Answer calculate the pka values for the following acids. a) methanol (ka = 2.9 x 10-16) b) lactic acid (ka = 8.3 x 10-4)
Community Answer A 0.50 M solution of an unknown acid has a pH = 4.0. Of the following, which is the acid in the solution? HOCl (Ka = 2.0 x 10-8) HBr (strong acid) HF (Ka = 6.8 x 10-4) C6H5OH (Ka = 1.0 x 10-10)
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Job Description Template
Liaison Job Description Template
April 1, 2025
Are you looking for a perfect candidate to join your business as a Liaison? Managing a hiring process has never been easier! Take this FREE Liaison Job Description Template to start your hiring journey, discover qualified candidates, and let them know what it takes to be a part of your team!
In this article, you will find the most essential skills, responsibilities, and requirements of a Liaison, but you can make it more attractive and relatable by customizing it according to the specific standards of your business.
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What is a Liaison?
A Liaison is a professional who serves as an act of contact between institutions, companies, agencies, individuals, and other parties and facilitates communication and collaboration.
They identify the company’s collaborative and market opportunities and implement actions and strategies to expand the company’s collaborative goals.
Liaisons also resolve communication issues between two sides and maintain positive relationships with people and companies they collaborate with.
Related Job Titles for Liaison
Clinical Liaison
Physician Liaison
Community Liaison
The Top Liaison Skills
| | |
--- |
| Skill | Why it's important |
| Communication | Communication is a primary and foundational skill for Liaisons. This skill is essential because it allows them to interact and collaborate with people of different backgrounds and build and maintain positive working relationships. Communication is also crucial for conflict management, as well as for written correspondence. |
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Liaison Job Description Template (Free)
We are looking for an experienced Liaison to help us improve our business connections and relationships. As a Liaison, you will be responsible for evaluating and identifying our collaborative needs. You will act as a point of contact between various individuals, companies, institutions, and other entities, facilitate collaborative efforts, and implement actions and strategies that will improve and expand our professional relationships. You should do so by reviewing the company’s norms and processes to know the appropriate courses of action.
You will also be responsible for identifying and resolving communication and other issues within the company to maintain healthy relationships and operational efficiency. Furthermore, you will write reports, emails, media releases, and other correspondence to encourage communication and document business updates, progress, and areas of improvement.
As a Liaison, you should have extraordinary communication, negotiation, problem-solving, collaboration, and organizational skills.
Liaison Duties & Responsibilities:
Represent the point of contact between individuals, companies, institutions, and other parties
Research collaborative and market opportunities
Identify the company’s collaborative needs and objectives
Review the company’s norms and processes to stay up-to-date
Implement actions and strategies to expand the company’s collaborative goals
Write reports, emails, media releases, and other correspondence to encourage and facilitate communication
Identify and resolve communication issues between two collaborative sides
Maintain positive relationships with relevant people from companies, organizations, and institutions
Represent the company in media and at events, conferences, and other activities
Write and present reports on business updates, progress, areas of improvement, and other factors
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Bachelor’s degree in business, communications, public relations, business management, or another relevant field
Former working experience as a Liaison or a similar position
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Outstanding conflict resolution and problem-solving skills
Exceptional communication and negotiation
Great organizational skills
Strategic thinking
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Once you gather all the received applications, you can use these sample interview questions for Liaison.
Personal
When and why did you decide to become a Liaison?
What are your most prominent skills?
What activities do you enjoy when you are not working?
Human Resources
How old were you when you first got a job as a Liaison?
What do you think about our current collaborative efforts?
What type of industries and fields do you have the most experience with?
Management
What would you do if you noticed unethical behavior in one of your colleagues?
How would you rate your confidence when it comes to public speaking?
How would you imagine a typical working day in this company?
Technical Skills and Knowledge
How would you establish a collaboration between two companies from completely different industries and backgrounds?
What do you think is more important in this role – communication or negotiation? Explain why.
How do you evaluate the company’s collaborative needs and requirements?
What are the Educational Requirements for a Liaison?
Basic educational requirements for Liaisons include a Bachelor’s degree in business, business management, communications, public relations, or a similar field.
These degrees will provide prospective candidates with the necessary skills and knowledge for this position.
Additionally, candidates should have previous working experience that involves communication and negotiation with various individuals and parties.
Once they get hired, Liaisons would typically receive on-the-job training from their new employer to learn about the job’s specifics and the company’s policies and activities.
How Much to Pay a Liaison when Hiring
Liaisons earn between $28,000 and $72,500 per year.
Their median annual salary is $49,343.
Hourly wages go from $14 to $35.
The median hourly wage is 24.
| | | | | | |
--- --- --- |
| Percentile | 10% | 25% | 50%(Median) | 75% | 90% |
| Hourly Wage | $14 | $18 | $24 | $30 | $35 |
| Annual Wage | $28,000 | $38,000 | $49,343 | $63,000 | $72,500 |
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Frequently Asked Questions about Liaison
What do Liaisons do?
Liaisons are professionals who establish and facilitate communication and collaboration between individuals, companies, institutions, and other entities.
They research and evaluate the company’s collaborative requirements and goals, implement appropriate actions and strategies to expand the company’s collaborative efforts, and resolve communication issues.
What are the different types of Liaisons?
Depending on their working environment, Liaisons can be:
Program Liaisons
Family Liaisons
Insurance Liaisons
Community Health Liaisons
Community Relations Liaisons, etc
How much do Liaisons earn annually?
Liaisons earn between $28,000 and $72,500 per year.
What skills make a successful Liaison?
Successful Liaisons must have extraordinary communication, negotiation, collaboration, problem-solving, organizational, and flexibility skills.
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188156 | https://math.colgate.edu/~integers/u41/u41.pdf | A41 INTEGERS 20 (2020) FINITE SUMS OF THE FLOOR FUNCTIONS SERIES Ryszard Palka West Pomeranian University of Technology, Szczecin, Poland ryszard.palka@zut.edu.pl Zdzislaw Trukszyn West Pomeranian University of Technology, Szczecin, Poland Received: 12/29/18, Revised: 11/11/19, Accepted: 5/3/20, Published: 5/26/20 Abstract This paper deals with new formulas for finite sums of the floor function series. The analogy between the sum of the successive natural numbers and the floor functions is applied to define the sums of a finite series of floor functions of di↵erent powers.
Formulas for di↵erent powers of the floor functions are developed and proved by mathematical induction. The main goal of this paper is the development and proof of the formula that enables the calculation of the sums of finite series of di↵erent powers of the floor functions as a closed-form expression.
1. Introduction The floor function, denoted floor(x) = [x], has been a subject of interest of math-ematicians since the late eighteenth century and continues to attract interest in modern number theory.
From the definition of the floor function, the following can be easily obtained: x y implies [x] [y] , (1) [x + k] = [x] + k, for k 2 Z, (2) [x + y] ≥[x] + [y] , (3) x + 1 2 " + hx 2 i = [x] .
(4) More advanced formulas can be found in the literature [1, 3, 5]: n(n + 1) 4n −2 " = n + 1 4 " , (5) INTEGERS: 20 (2020) 2 an+1 + 1 an−1 + 1 " = a2 −1 for a ≥2, (6) n X i=1 i 4 "2 = n (n + 2) (2n −1) 24 " , (7) n+m−1 X i=n i m " = n.
(8) 2. Sums of Finite Series of the Floor Function In this paper, finite sums of the floor function series will be denoted as: S[ n k ] (r) = n X i=k i k "r , (9) where r, n ≥k ≥1 2 N.
Theorem 2.1. The sums of finite series of di↵erent powers of floor functions can be expressed as: S[ n k ] (r) = hn k ir ⇣ n + 1 −k hn k i⌘ + k r + 1 r X j=0 (−1)j ✓r + 1 j ◆ Bj hn k ir+1−j , (10) where Bj are the Bernoulli numbers of the first kind.
Proof. Each integer n ≥k can be written as n = kp + m, where p = ⇥n k ⇤ , 0 m k −1. And thus it is possible to find an integer 1 l p −1, for k i kp −1, which means that ⇥i k ⇤r = lr. Additionally, for kp i n we have ⇥i k ⇤r = pr. This allows us to express S[ n k ] (r) as a finite sum as follows: S[ n k ] (r) = k (1r + 2r + 3r + ... + (p −1)r)+(m + 1) pr = k p−1 X l=1 lr+(m + 1) pr. (11) The first term is a product of the number k and the Bernoulli sum of next natural numbers in r power lr 1 + lr 2 + ... + lr p−1, thus it calculates the following dependence: Sl (r) = k (1r + 2r + 3r + ... + (p −1)r) (12) in the closed-form expression. Finally, because m = n −kp, we have: S n k = hn k ir ⇣ n + 1 −k hn k i⌘ + k r + 1 r X j=0 (−1)j ✓r + 1 j ◆ Bj hn k ir+1−j .
(13) Expressions (10) and (13) are identical, and thus the statement of the theorem is true.
INTEGERS: 20 (2020) 3 Using formula (13), it is possible to get the expression for the sum of extended floor functions for a 2 Z: S[ n+a k ] (r) = n+a X i=1 i + a k "r − a X i=1 i k "r .
(14) Using the above dependencies, it is possible to define formulas for finite series of the floor function of any power. As an example, we show these formulas for r = 1, 2, 3 and 4: S n k = hn k i⇣ n + 1 −k hn k i⌘ + k ⇣1 2 hn k i2 −1 2 hn k i⌘ = hn k i⇣ n + 1 −k 2 ⇣hn k i + 1 ⌘⌘ , (15) S n k = hn k i2⇣ n + 1 −k hn k i⌘ + k ⇣1 3 hn k i3 −1 2 hn k i2 + 1 6 hn k i⌘ = hn k i2⇣ n + 1 −k 2 ⇣hn k i + 1 ⌘⌘ −1 6k hn k i⇣hn k i2 −1 ⌘ , (16) S n k = hn k i3⇣ n + 1 −k hn k i⌘ + k ⇣1 4 hn k i4 −1 2 hn k i3 + 1 4 hn k i2⌘ = hn k i3⇣ n + 1 −k 2 ⇣hn k i + 1 ⌘⌘ −1 4k hn k i2⇣hn k i2 −1 ⌘ , (17) S n k = hn k i4⇣ n + 1 −k hn k i⌘ + k ⇣1 3 hn k i5 −1 2 hn k i4 + 1 3 hn k i3 −1 30 hn k i⌘ = hn k i4⇣ n + 1 −k 2 ⌘ −1 30k hn k i⇣ 24 hn k i4 −10 hn k i2 + 1 ⌘ .
(18) The above expressions can be easily obtained by using formula (10). Additionally, they can be proved by mathematical induction.
Theorem 2.2. Formula (10) for r = 1 can be written as follows: Sn = n X k hn k i = hn k i n n + 1 −k 2 ⇣hn k i + 1) o .
(19) Proof. We use induction. Each integer n ≥k can be written as n = kp + m, where p = ⇥n k ⇤ , 0 m k −1. Substituting the above expression into (9) we obtain Sn = kp + m k " n kp + m + 1 −k 2 ⇣kp + m k " + 1 ⌘o .
(20) From the definition of the floor function we have h kp+m k i = p, and thus Sn = p ⇣kp 2 + m + 1 −k 2 ⌘ .
(21) INTEGERS: 20 (2020) 4 We can check this formula for n = k (p = 1, m = 0), Sk = 1 · ⇣k 2 + 0 + 1 −k 2 ⌘ = 1.
(22) For the case m k −2, we can write Sn+1 = Skp+m+1 = hkp + m + 1 k in kp + m + 1 + 1 −k 2 ⇣hkp + m + 1 k i + 1 ⌘o , (23) and, after some manipulations, Sn+1 = p(kp 2 + m + 2 −k 2).
(24) From the other side, according to the series definition Sn+1 = Sn+ ⇥n+1 k ⇤ , we obtain Skp+m+1 = hkp + m k in kp + m + 1 −k 2 ⇣hkp + m k i + 1 ⌘o + hkp + m + 1 k i = p ⇣kp 2 + m + 2 −k 2 ⌘ .
(25) Both expressions are identical, which means that formula (19) is true.
For the case m = k −1 we should check the expression Sn+1 = Skp+k. Substi-tuting into (19) we obtain Skp+k = hkp + k k in kp + k + 1 −k 2( hkp + k 2 i + 1) o , (26) and, after some manipulations, this formula can be written as Skp+k = ⇣ p + 1 ⌘⇣kp 2 + 1 ⌘ .
(27) Now we get Sn+1 = Sn + n + 1 k " = Skp+k−1 + kp + k 2 " , (28) and Skp+k = hkp + k −1 k in kp + k −1 + 1 −k 2 ⇣hkp + k −1 k i + 1 ⌘o + hkp + k k i = ⇣ p + 1 ⌘⇣kp 2 + 1 ⌘ .
(29) This proves that formula (19) is true.
Using the same method of the above proof, it is possible to prove formulas for the next sums of the floor functions of higher powers.
INTEGERS: 20 (2020) 5 3. Summary The formulas shown in this paper create the first step in the theory of the sums of the floor functions. The next step will include the expression for the sums of di↵erent forms of the floor functions.
Floor functions are widely applied in informatics (see, e.g. , ), but their application in number theory may lead to solving many significant problems. For example, the problem of “Partitio Numerorum”, which has been analyzed since Euler without satisfactory results, might be solved using new results in the theory of floor functions.
References R. L. Graham, D. E. Knuth, O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 1994.
K. E. Iverson, A Programming Language, Wiley, New York, 1962.
J.-M. De Koninck, A. Mercier, 1001 Problems in Classical Number Theory, AMS, Providence, RI, 2007.
D. Knuth, The Art of Computer Programming, Addison-Wesley, Reading, MA, 1968.
M. A. Nyblom, Some curious sequences involving floor and ceiling functions, Amer. Math.
Monthly 109 (6)(2002), 559-564. |
188157 | https://usaco.guide/CPH.pdf | Competitive Programmer’s Handbook Antti Laaksonen Draft August 19, 2019 ii Contents Preface ix I Basic techniques 1 1 Introduction 3 1.1 Programming languages . . . . . . . . . . . . . . . . . . . . . . . . .
3 1.2 Input and output . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4 1.3 Working with numbers . . . . . . . . . . . . . . . . . . . . . . . . . .
6 1.4 Shortening code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8 1.5 Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10 1.6 Contests and resources . . . . . . . . . . . . . . . . . . . . . . . . . .
15 2 Time complexity 17 2.1 Calculation rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17 2.2 Complexity classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20 2.3 Estimating efficiency . . . . . . . . . . . . . . . . . . . . . . . . . . .
21 2.4 Maximum subarray sum . . . . . . . . . . . . . . . . . . . . . . . . .
21 3 Sorting 25 3.1 Sorting theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25 3.2 Sorting in C++ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29 3.3 Binary search . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
31 4 Data structures 35 4.1 Dynamic arrays . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35 4.2 Set structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37 4.3 Map structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
38 4.4 Iterators and ranges . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39 4.5 Other structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41 4.6 Comparison to sorting . . . . . . . . . . . . . . . . . . . . . . . . . . .
44 5 Complete search 47 5.1 Generating subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47 5.2 Generating permutations . . . . . . . . . . . . . . . . . . . . . . . . .
49 5.3 Backtracking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50 5.4 Pruning the search . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51 5.5 Meet in the middle . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54 iii 6 Greedy algorithms 57 6.1 Coin problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57 6.2 Scheduling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58 6.3 Tasks and deadlines . . . . . . . . . . . . . . . . . . . . . . . . . . . .
60 6.4 Minimizing sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
61 6.5 Data compression . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62 7 Dynamic programming 65 7.1 Coin problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65 7.2 Longest increasing subsequence . . . . . . . . . . . . . . . . . . . . .
70 7.3 Paths in a grid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71 7.4 Knapsack problems . . . . . . . . . . . . . . . . . . . . . . . . . . . .
72 7.5 Edit distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
74 7.6 Counting tilings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
75 8 Amortized analysis 77 8.1 Two pointers method . . . . . . . . . . . . . . . . . . . . . . . . . . .
77 8.2 Nearest smaller elements . . . . . . . . . . . . . . . . . . . . . . . . .
79 8.3 Sliding window minimum . . . . . . . . . . . . . . . . . . . . . . . . .
81 9 Range queries 83 9.1 Static array queries . . . . . . . . . . . . . . . . . . . . . . . . . . . .
84 9.2 Binary indexed tree . . . . . . . . . . . . . . . . . . . . . . . . . . . .
86 9.3 Segment tree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
89 9.4 Additional techniques . . . . . . . . . . . . . . . . . . . . . . . . . . .
93 10 Bit manipulation 95 10.1 Bit representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
95 10.2 Bit operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
96 10.3 Representing sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98 10.4 Bit optimizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
100 10.5 Dynamic programming . . . . . . . . . . . . . . . . . . . . . . . . . .
102 II Graph algorithms 107 11 Basics of graphs 109 11.1 Graph terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
109 11.2 Graph representation . . . . . . . . . . . . . . . . . . . . . . . . . . .
113 12 Graph traversal 117 12.1 Depth-first search . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
117 12.2 Breadth-first search . . . . . . . . . . . . . . . . . . . . . . . . . . . .
119 12.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
121 iv 13 Shortest paths 123 13.1 Bellman–Ford algorithm . . . . . . . . . . . . . . . . . . . . . . . . .
123 13.2 Dijkstra’s algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . .
126 13.3 Floyd–Warshall algorithm . . . . . . . . . . . . . . . . . . . . . . . .
129 14 Tree algorithms 133 14.1 Tree traversal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
134 14.2 Diameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
135 14.3 All longest paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
137 14.4 Binary trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
139 15 Spanning trees 141 15.1 Kruskal’s algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . .
142 15.2 Union-find structure . . . . . . . . . . . . . . . . . . . . . . . . . . . .
145 15.3 Prim’s algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
147 16 Directed graphs 149 16.1 Topological sorting . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
149 16.2 Dynamic programming . . . . . . . . . . . . . . . . . . . . . . . . . .
151 16.3 Successor paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
154 16.4 Cycle detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
155 17 Strong connectivity 157 17.1 Kosaraju’s algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . .
158 17.2 2SAT problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
160 18 Tree queries 163 18.1 Finding ancestors . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
163 18.2 Subtrees and paths . . . . . . . . . . . . . . . . . . . . . . . . . . . .
164 18.3 Lowest common ancestor . . . . . . . . . . . . . . . . . . . . . . . . .
167 18.4 Offline algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
170 19 Paths and circuits 173 19.1 Eulerian paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
173 19.2 Hamiltonian paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
177 19.3 De Bruijn sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . .
178 19.4 Knight’s tours . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
179 20 Flows and cuts 181 20.1 Ford–Fulkerson algorithm . . . . . . . . . . . . . . . . . . . . . . . .
182 20.2 Disjoint paths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
186 20.3 Maximum matchings . . . . . . . . . . . . . . . . . . . . . . . . . . .
187 20.4 Path covers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
190 v III Advanced topics 195 21 Number theory 197 21.1 Primes and factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
197 21.2 Modular arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . .
201 21.3 Solving equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
204 21.4 Other results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
205 22 Combinatorics 207 22.1 Binomial coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . .
208 22.2 Catalan numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
210 22.3 Inclusion-exclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
212 22.4 Burnside’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
214 22.5 Cayley’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
215 23 Matrices 217 23.1 Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
217 23.2 Linear recurrences . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
220 23.3 Graphs and matrices . . . . . . . . . . . . . . . . . . . . . . . . . . .
222 24 Probability 225 24.1 Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
225 24.2 Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
226 24.3 Random variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
228 24.4 Markov chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
230 24.5 Randomized algorithms . . . . . . . . . . . . . . . . . . . . . . . . . .
231 25 Game theory 235 25.1 Game states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
235 25.2 Nim game . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
237 25.3 Sprague–Grundy theorem . . . . . . . . . . . . . . . . . . . . . . . .
238 26 String algorithms 243 26.1 String terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
243 26.2 Trie structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
244 26.3 String hashing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
245 26.4 Z-algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
247 27 Square root algorithms 251 27.1 Combining algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . .
252 27.2 Integer partitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
254 27.3 Mo’s algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
255 28 Segment trees revisited 257 28.1 Lazy propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
258 28.2 Dynamic trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
261 28.3 Data structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
263 28.4 Two-dimensionality . . . . . . . . . . . . . . . . . . . . . . . . . . . .
264 vi 29 Geometry 265 29.1 Complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
266 29.2 Points and lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
268 29.3 Polygon area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
271 29.4 Distance functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
272 30 Sweep line algorithms 275 30.1 Intersection points . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
276 30.2 Closest pair problem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
277 30.3 Convex hull problem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
278 Bibliography 281 vii viii Preface The purpose of this book is to give you a thorough introduction to competitive programming. It is assumed that you already know the basics of programming, but no previous background in competitive programming is needed.
The book is especially intended for students who want to learn algorithms and possibly participate in the International Olympiad in Informatics (IOI) or in the International Collegiate Programming Contest (ICPC). Of course, the book is also suitable for anybody else interested in competitive programming.
It takes a long time to become a good competitive programmer, but it is also an opportunity to learn a lot. You can be sure that you will get a good general understanding of algorithms if you spend time reading the book, solving problems and taking part in contests.
The book is under continuous development. You can always send feedback on the book to ahslaaks@cs.helsinki.fi.
Helsinki, August 2019 Antti Laaksonen ix x Part I Basic techniques 1 Chapter 1 Introduction Competitive programming combines two topics: (1) the design of algorithms and (2) the implementation of algorithms.
The design of algorithms consists of problem solving and mathematical thinking. Skills for analyzing problems and solving them creatively are needed.
An algorithm for solving a problem has to be both correct and efficient, and the core of the problem is often about inventing an efficient algorithm.
Theoretical knowledge of algorithms is important to competitive programmers.
Typically, a solution to a problem is a combination of well-known techniques and new insights. The techniques that appear in competitive programming also form the basis for the scientific research of algorithms.
The implementation of algorithms requires good programming skills. In competitive programming, the solutions are graded by testing an implemented algorithm using a set of test cases. Thus, it is not enough that the idea of the algorithm is correct, but the implementation also has to be correct.
A good coding style in contests is straightforward and concise. Programs should be written quickly, because there is not much time available. Unlike in traditional software engineering, the programs are short (usually at most a few hundred lines of code), and they do not need to be maintained after the contest.
1.1 Programming languages At the moment, the most popular programming languages used in contests are C++, Python and Java. For example, in Google Code Jam 2017, among the best 3,000 participants, 79 % used C++, 16 % used Python and 8 % used Java .
Some participants also used several languages.
Many people think that C++ is the best choice for a competitive programmer, and C++ is nearly always available in contest systems. The benefits of using C++ are that it is a very efficient language and its standard library contains a large collection of data structures and algorithms.
On the other hand, it is good to master several languages and understand their strengths. For example, if large integers are needed in the problem, Python can be a good choice, because it contains built-in operations for calculating with 3 large integers. Still, most problems in programming contests are set so that using a specific programming language is not an unfair advantage.
All example programs in this book are written in C++, and the standard library’s data structures and algorithms are often used. The programs follow the C++11 standard, which can be used in most contests nowadays. If you cannot program in C++ yet, now is a good time to start learning.
C++ code template A typical C++ code template for competitive programming looks like this: #include using namespace std; int main() { // solution comes here } The #include line at the beginning of the code is a feature of the g++ compiler that allows us to include the entire standard library. Thus, it is not needed to separately include libraries such as iostream, vector and algorithm, but rather they are available automatically.
The using line declares that the classes and functions of the standard library can be used directly in the code. Without the using line we would have to write, for example, std::cout, but now it suffices to write cout.
The code can be compiled using the following command: g++ -std=c++11 -O2 -Wall test.cpp -o test This command produces a binary file test from the source code test.cpp. The compiler follows the C++11 standard (-std=c++11), optimizes the code (-O2) and shows warnings about possible errors (-Wall).
1.2 Input and output In most contests, standard streams are used for reading input and writing output.
In C++, the standard streams are cin for input and cout for output. In addition, the C functions scanf and printf can be used.
The input for the program usually consists of numbers and strings that are separated with spaces and newlines. They can be read from the cin stream as follows: int a, b; string x; cin >> a >> b >> x; 4 This kind of code always works, assuming that there is at least one space or newline between each element in the input. For example, the above code can read both of the following inputs: 123 456 monkey 123 456 monkey The cout stream is used for output as follows: int a = 123, b = 456; string x = "monkey"; cout << a << " " << b << " " << x << "\n"; Input and output is sometimes a bottleneck in the program. The following lines at the beginning of the code make input and output more efficient: ios::sync_with_stdio(0); cin.tie(0); Note that the newline "\n" works faster than endl, because endl always causes a flush operation.
The C functions scanf and printf are an alternative to the C++ standard streams. They are usually a bit faster, but they are also more difficult to use. The following code reads two integers from the input: int a, b; scanf("%d %d", &a, &b); The following code prints two integers: int a = 123, b = 456; printf("%d %d\n", a, b); Sometimes the program should read a whole line from the input, possibly containing spaces. This can be accomplished by using the getline function: string s; getline(cin, s); If the amount of data is unknown, the following loop is useful: while (cin >> x) { // code } This loop reads elements from the input one after another, until there is no more data available in the input.
5 In some contest systems, files are used for input and output. An easy solution for this is to write the code as usual using standard streams, but add the following lines to the beginning of the code: freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); After this, the program reads the input from the file ”input.txt” and writes the output to the file ”output.txt”.
1.3 Working with numbers Integers The most used integer type in competitive programming is int, which is a 32-bit type with a value range of −231...231 −1 or about −2·109...2·109. If the type int is not enough, the 64-bit type long long can be used. It has a value range of −263...263 −1 or about −9·1018...9·1018.
The following code defines a long long variable: long long x = 123456789123456789LL; The suffix LL means that the type of the number is long long.
A common mistake when using the type long long is that the type int is still used somewhere in the code. For example, the following code contains a subtle error: int a = 123456789; long long b = aa; cout << b << "\n"; // -1757895751 Even though the variable b is of type long long, both numbers in the expres-sion aa are of type int and the result is also of type int. Because of this, the variable b will contain a wrong result. The problem can be solved by changing the type of a to long long or by changing the expression to (long long)aa.
Usually contest problems are set so that the type long long is enough. Still, it is good to know that the g++ compiler also provides a 128-bit type __int128_t with a value range of −2127...2127 −1 or about −1038...1038. However, this type is not available in all contest systems.
Modular arithmetic We denote by x mod m the remainder when x is divided by m. For example, 17 mod 5 = 2, because 17 = 3·5+2.
Sometimes, the answer to a problem is a very large number but it is enough to output it ”modulo m”, i.e., the remainder when the answer is divided by m (for 6 example, ”modulo 109 +7”). The idea is that even if the actual answer is very large, it suffices to use the types int and long long.
An important property of the remainder is that in addition, subtraction and multiplication, the remainder can be taken before the operation: (a+ b) mod m = (a mod m+ b mod m) mod m (a−b) mod m = (a mod m−b mod m) mod m (a· b) mod m = (a mod m· b mod m) mod m Thus, we can take the remainder after every operation and the numbers will never become too large.
For example, the following code calculates n!, the factorial of n, modulo m: long long x = 1; for (int i = 2; i <= n; i++) { x = (xi)%m; } cout << x%m << "\n"; Usually we want the remainder to always be between 0...m−1. However, in C++ and other languages, the remainder of a negative number is either zero or negative. An easy way to make sure there are no negative remainders is to first calculate the remainder as usual and then add m if the result is negative: x = x%m; if (x < 0) x += m; However, this is only needed when there are subtractions in the code and the remainder may become negative.
Floating point numbers The usual floating point types in competitive programming are the 64-bit double and, as an extension in the g++ compiler, the 80-bit long double. In most cases, double is enough, but long double is more accurate.
The required precision of the answer is usually given in the problem statement.
An easy way to output the answer is to use the printf function and give the number of decimal places in the formatting string. For example, the following code prints the value of x with 9 decimal places: printf("%.9f\n", x); A difficulty when using floating point numbers is that some numbers cannot be represented accurately as floating point numbers, and there will be rounding errors. For example, the result of the following code is surprising: double x = 0.33+0.1; printf("%.20f\n", x); // 0.99999999999999988898 7 Due to a rounding error, the value of x is a bit smaller than 1, while the correct value would be 1.
It is risky to compare floating point numbers with the == operator, because it is possible that the values should be equal but they are not because of precision errors. A better way to compare floating point numbers is to assume that two numbers are equal if the difference between them is less than ε, where ε is a small number.
In practice, the numbers can be compared as follows (ε = 10−9): if (abs(a-b) < 1e-9) { // a and b are equal } Note that while floating point numbers are inaccurate, integers up to a certain limit can still be represented accurately. For example, using double, it is possible to accurately represent all integers whose absolute value is at most 253.
1.4 Shortening code Short code is ideal in competitive programming, because programs should be written as fast as possible. Because of this, competitive programmers often define shorter names for datatypes and other parts of code.
Type names Using the command typedef it is possible to give a shorter name to a datatype.
For example, the name long long is long, so we can define a shorter name ll: typedef long long ll; After this, the code long long a = 123456789; long long b = 987654321; cout << ab << "\n"; can be shortened as follows: ll a = 123456789; ll b = 987654321; cout << ab << "\n"; The command typedef can also be used with more complex types. For example, the following code gives the name vi for a vector of integers and the name pi for a pair that contains two integers.
typedef vector vi; typedef pair pi; 8 Macros Another way to shorten code is to define macros. A macro means that certain strings in the code will be changed before the compilation. In C++, macros are defined using the #define keyword.
For example, we can define the following macros: #define F first #define S second #define PB push_back #define MP make_pair After this, the code v.push_back(make_pair(y1,x1)); v.push_back(make_pair(y2,x2)); int d = v[i].first+v[i].second; can be shortened as follows: v.PB(MP(y1,x1)); v.PB(MP(y2,x2)); int d = v[i].F+v[i].S; A macro can also have parameters which makes it possible to shorten loops and other structures. For example, we can define the following macro: #define REP(i,a,b) for (int i = a; i <= b; i++) After this, the code for (int i = 1; i <= n; i++) { search(i); } can be shortened as follows: REP(i,1,n) { search(i); } Sometimes macros cause bugs that may be difficult to detect. For example, consider the following macro that calculates the square of a number: #define SQ(a) aa This macro does not always work as expected. For example, the code cout << SQ(3+3) << "\n"; 9 corresponds to the code cout << 3+33+3 << "\n"; // 15 A better version of the macro is as follows: #define SQ(a) (a)(a) Now the code cout << SQ(3+3) << "\n"; corresponds to the code cout << (3+3)(3+3) << "\n"; // 36 1.5 Mathematics Mathematics plays an important role in competitive programming, and it is not possible to become a successful competitive programmer without having good mathematical skills. This section discusses some important mathematical concepts and formulas that are needed later in the book.
Sum formulas Each sum of the form n X x=1 xk = 1k +2k +3k +...+ nk, where k is a positive integer, has a closed-form formula that is a polynomial of degree k +1. For example1, n X x=1 x = 1+2+3+...+ n = n(n+1) 2 and n X x=1 x2 = 12 +22 +32 +...+ n2 = n(n+1)(2n+1) 6 .
An arithmetic progression is a sequence of numbers where the difference between any two consecutive numbers is constant. For example, 3,7,11,15 1 There is even a general formula for such sums, called Faulhaber’s formula, but it is too complex to be presented here.
10 is an arithmetic progression with constant 4. The sum of an arithmetic progres-sion can be calculated using the formula a+···+ b | {z } n numbers = n(a+ b) 2 where a is the first number, b is the last number and n is the amount of numbers.
For example, 3+7+11+15 = 4·(3+15) 2 = 36.
The formula is based on the fact that the sum consists of n numbers and the value of each number is (a+ b)/2 on average.
A geometric progression is a sequence of numbers where the ratio between any two consecutive numbers is constant. For example, 3,6,12,24 is a geometric progression with constant 2. The sum of a geometric progression can be calculated using the formula a+ ak + ak2 +···+ b = bk −a k −1 where a is the first number, b is the last number and the ratio between consecu-tive numbers is k. For example, 3+6+12+24 = 24·2−3 2−1 = 45.
This formula can be derived as follows. Let S = a+ ak + ak2 +···+ b.
By multiplying both sides by k, we get kS = ak + ak2 + ak3 +···+ bk, and solving the equation kS −S = bk −a yields the formula.
A special case of a sum of a geometric progression is the formula 1+2+4+8+...+2n−1 = 2n −1.
A harmonic sum is a sum of the form n X x=1 1 x = 1+ 1 2 + 1 3 +...+ 1 n.
An upper bound for a harmonic sum is log2(n)+1. Namely, we can modify each term 1/k so that k becomes the nearest power of two that does not exceed k.
For example, when n = 6, we can estimate the sum as follows: 1+ 1 2 + 1 3 + 1 4 + 1 5 + 1 6 ≤1+ 1 2 + 1 2 + 1 4 + 1 4 + 1 4.
This upper bound consists of log2(n)+1 parts (1, 2·1/2, 4·1/4, etc.), and the value of each part is at most 1.
11 Set theory A set is a collection of elements. For example, the set X = {2,4,7} contains elements 2, 4 and 7. The symbol ; denotes an empty set, and |S| denotes the size of a set S, i.e., the number of elements in the set. For example, in the above set, |X| = 3.
If a set S contains an element x, we write x ∈S, and otherwise we write x ∉S.
For example, in the above set 4 ∈X and 5 ∉X.
New sets can be constructed using set operations: • The intersection A ∩B consists of elements that are in both A and B. For example, if A = {1,2,5} and B = {2,4}, then A ∩B = {2}.
• The union A ∪B consists of elements that are in A or B or both. For example, if A = {3,7} and B = {2,3,8}, then A ∪B = {2,3,7,8}.
• The complement ¯ A consists of elements that are not in A. The interpre-tation of a complement depends on the universal set, which contains all possible elements. For example, if A = {1,2,5,7} and the universal set is {1,2,...,10}, then ¯ A = {3,4,6,8,9,10}.
• The difference A \ B = A ∩¯ B consists of elements that are in A but not in B. Note that B can contain elements that are not in A. For example, if A = {2,3,7,8} and B = {3,5,8}, then A \B = {2,7}.
If each element of A also belongs to S, we say that A is a subset of S, denoted by A ⊂S. A set S always has 2|S| subsets, including the empty set. For example, the subsets of the set {2,4,7} are ;, {2}, {4}, {7}, {2,4}, {2,7}, {4,7} and {2,4,7}.
Some often used sets are N (natural numbers), Z (integers), Q (rational numbers) and R (real numbers). The set N can be defined in two ways, depending on the situation: either N = {0,1,2,...} or N = {1,2,3,...}.
We can also construct a set using a rule of the form {f (n) : n ∈S}, where f (n) is some function. This set contains all elements of the form f (n), where n is an element in S. For example, the set X = {2n : n ∈Z} contains all even integers.
12 Logic The value of a logical expression is either true (1) or false (0). The most impor-tant logical operators are ¬ (negation), ∧(conjunction), ∨(disjunction), ⇒ (implication) and ⇔(equivalence). The following table shows the meanings of these operators: A B ¬A ¬B A ∧B A ∨B A ⇒B A ⇔B 0 0 1 1 0 0 1 1 0 1 1 0 0 1 1 0 1 0 0 1 0 1 0 0 1 1 0 0 1 1 1 1 The expression ¬A has the opposite value of A. The expression A ∧B is true if both A and B are true, and the expression A ∨B is true if A or B or both are true. The expression A ⇒B is true if whenever A is true, also B is true. The expression A ⇔B is true if A and B are both true or both false.
A predicate is an expression that is true or false depending on its parameters.
Predicates are usually denoted by capital letters. For example, we can define a predicate P(x) that is true exactly when x is a prime number. Using this definition, P(7) is true but P(8) is false.
A quantifier connects a logical expression to the elements of a set. The most important quantifiers are ∀(for all) and ∃(there is). For example, ∀x(∃y(y < x)) means that for each element x in the set, there is an element y in the set such that y is smaller than x. This is true in the set of integers, but false in the set of natural numbers.
Using the notation described above, we can express many kinds of logical propositions. For example, ∀x((x > 1∧¬P(x)) ⇒(∃a(∃b(a > 1∧b > 1∧x = ab)))) means that if a number x is larger than 1 and not a prime number, then there are numbers a and b that are larger than 1 and whose product is x. This proposition is true in the set of integers.
Functions The function ⌊x⌋rounds the number x down to an integer, and the function ⌈x⌉ rounds the number x up to an integer. For example, ⌊3/2⌋= 1 and ⌈3/2⌉= 2.
The functions min(x1,x2,...,xn) and max(x1,x2,...,xn) give the smallest and largest of values x1,x2,...,xn. For example, min(1,2,3) = 1 and max(1,2,3) = 3.
13 The factorial n! can be defined n Y x=1 x = 1·2·3·...· n or recursively 0!
= 1 n!
= n·(n−1)!
The Fibonacci numbers arise in many situations. They can be defined recursively as follows: f (0) = 0 f (1) = 1 f (n) = f (n−1)+ f (n−2) The first Fibonacci numbers are 0,1,1,2,3,5,8,13,21,34,55,...
There is also a closed-form formula for calculating Fibonacci numbers, which is sometimes called Binet’s formula: f (n) = (1+ p 5)n −(1− p 5)n 2np 5 .
Logarithms The logarithm of a number x is denoted logk(x), where k is the base of the logarithm. According to the definition, logk(x) = a exactly when ka = x.
A useful property of logarithms is that logk(x) equals the number of times we have to divide x by k before we reach the number 1. For example, log2(32) = 5 because 5 divisions by 2 are needed: 32 →16 →8 →4 →2 →1 Logarithms are often used in the analysis of algorithms, because many ef-ficient algorithms halve something at each step. Hence, we can estimate the efficiency of such algorithms using logarithms.
The logarithm of a product is logk(ab) = logk(a)+logk(b), and consequently, logk(xn) = n·logk(x).
In addition, the logarithm of a quotient is logk ³a b ´ = logk(a)−logk(b).
Another useful formula is logu(x) = logk(x) logk(u), 14 and using this, it is possible to calculate logarithms to any base if there is a way to calculate logarithms to some fixed base.
The natural logarithm ln(x) of a number x is a logarithm whose base is e ≈2.71828. Another property of logarithms is that the number of digits of an integer x in base b is ⌊logb(x)+1⌋. For example, the representation of 123 in base 2 is 1111011 and ⌊log2(123)+1⌋= 7.
1.6 Contests and resources IOI The International Olympiad in Informatics (IOI) is an annual programming contest for secondary school students. Each country is allowed to send a team of four students to the contest. There are usually about 300 participants from 80 countries.
The IOI consists of two five-hour long contests. In both contests, the partic-ipants are asked to solve three algorithm tasks of various difficulty. The tasks are divided into subtasks, each of which has an assigned score. Even if the contestants are divided into teams, they compete as individuals.
The IOI syllabus regulates the topics that may appear in IOI tasks.
Almost all the topics in the IOI syllabus are covered by this book.
Participants for the IOI are selected through national contests. Before the IOI, many regional contests are organized, such as the Baltic Olympiad in Informatics (BOI), the Central European Olympiad in Informatics (CEOI) and the Asia-Pacific Informatics Olympiad (APIO).
Some countries organize online practice contests for future IOI participants, such as the Croatian Open Competition in Informatics and the USA Comput-ing Olympiad . In addition, a large collection of problems from Polish contests is available online .
ICPC The International Collegiate Programming Contest (ICPC) is an annual program-ming contest for university students. Each team in the contest consists of three students, and unlike in the IOI, the students work together; there is only one computer available for each team.
The ICPC consists of several stages, and finally the best teams are invited to the World Finals. While there are tens of thousands of participants in the contest, there are only a small number2 of final slots available, so even advancing to the finals is a great achievement in some regions.
In each ICPC contest, the teams have five hours of time to solve about ten algorithm problems. A solution to a problem is accepted only if it solves all test cases efficiently. During the contest, competitors may view the results of other 2The exact number of final slots varies from year to year; in 2017, there were 133 final slots.
15 teams, but for the last hour the scoreboard is frozen and it is not possible to see the results of the last submissions.
The topics that may appear at the ICPC are not so well specified as those at the IOI. In any case, it is clear that more knowledge is needed at the ICPC, especially more mathematical skills.
Online contests There are also many online contests that are open for everybody. At the moment, the most active contest site is Codeforces, which organizes contests about weekly.
In Codeforces, participants are divided into two divisions: beginners compete in Div2 and more experienced programmers in Div1. Other contest sites include AtCoder, CS Academy, HackerRank and Topcoder.
Some companies organize online contests with onsite finals. Examples of such contests are Facebook Hacker Cup, Google Code Jam and Yandex.Algorithm. Of course, companies also use those contests for recruiting: performing well in a contest is a good way to prove one’s skills.
Books There are already some books (besides this book) that focus on competitive programming and algorithmic problem solving: • S. S. Skiena and M. A. Revilla: Programming Challenges: The Programming Contest Training Manual • S. Halim and F. Halim: Competitive Programming 3: The New Lower Bound of Programming Contests • K. Diks et al.: Looking for a Challenge? The Ultimate Problem Set from the University of Warsaw Programming Competitions The first two books are intended for beginners, whereas the last book contains advanced material.
Of course, general algorithm books are also suitable for competitive program-mers. Some popular books are: • T. H. Cormen, C. E. Leiserson, R. L. Rivest and C. Stein: Introduction to Algorithms • J. Kleinberg and É. Tardos: Algorithm Design • S. S. Skiena: The Algorithm Design Manual 16 Chapter 2 Time complexity The efficiency of algorithms is important in competitive programming. Usually, it is easy to design an algorithm that solves the problem slowly, but the real challenge is to invent a fast algorithm. If the algorithm is too slow, it will get only partial points or no points at all.
The time complexity of an algorithm estimates how much time the algo-rithm will use for some input. The idea is to represent the efficiency as a function whose parameter is the size of the input. By calculating the time complexity, we can find out whether the algorithm is fast enough without implementing it.
2.1 Calculation rules The time complexity of an algorithm is denoted O(···) where the three dots represent some function. Usually, the variable n denotes the input size. For example, if the input is an array of numbers, n will be the size of the array, and if the input is a string, n will be the length of the string.
Loops A common reason why an algorithm is slow is that it contains many loops that go through the input. The more nested loops the algorithm contains, the slower it is.
If there are k nested loops, the time complexity is O(nk).
For example, the time complexity of the following code is O(n): for (int i = 1; i <= n; i++) { // code } And the time complexity of the following code is O(n2): for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { // code } } 17 Order of magnitude A time complexity does not tell us the exact number of times the code inside a loop is executed, but it only shows the order of magnitude. In the following examples, the code inside the loop is executed 3n, n+5 and ⌈n/2⌉times, but the time complexity of each code is O(n).
for (int i = 1; i <= 3n; i++) { // code } for (int i = 1; i <= n+5; i++) { // code } for (int i = 1; i <= n; i += 2) { // code } As another example, the time complexity of the following code is O(n2): for (int i = 1; i <= n; i++) { for (int j = i+1; j <= n; j++) { // code } } Phases If the algorithm consists of consecutive phases, the total time complexity is the largest time complexity of a single phase. The reason for this is that the slowest phase is usually the bottleneck of the code.
For example, the following code consists of three phases with time complexities O(n), O(n2) and O(n). Thus, the total time complexity is O(n2).
for (int i = 1; i <= n; i++) { // code } for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { // code } } for (int i = 1; i <= n; i++) { // code } 18 Several variables Sometimes the time complexity depends on several factors. In this case, the time complexity formula contains several variables.
For example, the time complexity of the following code is O(nm): for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { // code } } Recursion The time complexity of a recursive function depends on the number of times the function is called and the time complexity of a single call. The total time complexity is the product of these values.
For example, consider the following function: void f(int n) { if (n == 1) return; f(n-1); } The call f(n) causes n function calls, and the time complexity of each call is O(1).
Thus, the total time complexity is O(n).
As another example, consider the following function: void g(int n) { if (n == 1) return; g(n-1); g(n-1); } In this case each function call generates two other calls, except for n = 1. Let us see what happens when g is called with parameter n. The following table shows the function calls produced by this single call: function call number of calls g(n) 1 g(n−1) 2 g(n−2) 4 ··· ··· g(1) 2n−1 Based on this, the time complexity is 1+2+4+···+2n−1 = 2n −1 = O(2n).
19 2.2 Complexity classes The following list contains common time complexities of algorithms: O(1) The running time of a constant-time algorithm does not depend on the input size. A typical constant-time algorithm is a direct formula that calculates the answer.
O(logn) A logarithmic algorithm often halves the input size at each step. The running time of such an algorithm is logarithmic, because log2 n equals the number of times n must be divided by 2 to get 1.
O(pn) A square root algorithm is slower than O(logn) but faster than O(n).
A special property of square roots is that pn = n/pn, so the square root pn lies, in some sense, in the middle of the input.
O(n) A linear algorithm goes through the input a constant number of times. This is often the best possible time complexity, because it is usually necessary to access each input element at least once before reporting the answer.
O(nlogn) This time complexity often indicates that the algorithm sorts the input, because the time complexity of efficient sorting algorithms is O(nlogn).
Another possibility is that the algorithm uses a data structure where each operation takes O(logn) time.
O(n2) A quadratic algorithm often contains two nested loops. It is possible to go through all pairs of the input elements in O(n2) time.
O(n3) A cubic algorithm often contains three nested loops. It is possible to go through all triplets of the input elements in O(n3) time.
O(2n) This time complexity often indicates that the algorithm iterates through all subsets of the input elements. For example, the subsets of {1,2,3} are ;, {1}, {2}, {3}, {1,2}, {1,3}, {2,3} and {1,2,3}.
O(n!) This time complexity often indicates that the algorithm iterates through all permutations of the input elements. For example, the permutations of {1,2,3} are (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2) and (3,2,1).
An algorithm is polynomial if its time complexity is at most O(nk) where k is a constant. All the above time complexities except O(2n) and O(n!) are polynomial.
In practice, the constant k is usually small, and therefore a polynomial time complexity roughly means that the algorithm is efficient.
Most algorithms in this book are polynomial. Still, there are many important problems for which no polynomial algorithm is known, i.e., nobody knows how to solve them efficiently. NP-hard problems are an important set of problems, for which no polynomial algorithm is known1.
1A classic book on the topic is M. R. Garey’s and D. S. Johnson’s Computers and Intractability: A Guide to the Theory of NP-Completeness .
20 2.3 Estimating efficiency By calculating the time complexity of an algorithm, it is possible to check, before implementing the algorithm, that it is efficient enough for the problem. The starting point for estimations is the fact that a modern computer can perform some hundreds of millions of operations in a second.
For example, assume that the time limit for a problem is one second and the input size is n = 105. If the time complexity is O(n2), the algorithm will perform about (105)2 = 1010 operations. This should take at least some tens of seconds, so the algorithm seems to be too slow for solving the problem.
On the other hand, given the input size, we can try to guess the required time complexity of the algorithm that solves the problem. The following table contains some useful estimates assuming a time limit of one second.
input size required time complexity n ≤10 O(n!) n ≤20 O(2n) n ≤500 O(n3) n ≤5000 O(n2) n ≤106 O(nlogn) or O(n) n is large O(1) or O(logn) For example, if the input size is n = 105, it is probably expected that the time complexity of the algorithm is O(n) or O(nlogn). This information makes it easier to design the algorithm, because it rules out approaches that would yield an algorithm with a worse time complexity.
Still, it is important to remember that a time complexity is only an estimate of efficiency, because it hides the constant factors. For example, an algorithm that runs in O(n) time may perform n/2 or 5n operations. This has an important effect on the actual running time of the algorithm.
2.4 Maximum subarray sum There are often several possible algorithms for solving a problem such that their time complexities are different. This section discusses a classic problem that has a straightforward O(n3) solution. However, by designing a better algorithm, it is possible to solve the problem in O(n2) time and even in O(n) time.
Given an array of n numbers, our task is to calculate the maximum subar-ray sum, i.e., the largest possible sum of a sequence of consecutive values in the array2. The problem is interesting when there may be negative values in the array. For example, in the array −1 2 4 −3 5 2 −5 2 2J. Bentley’s book Programming Pearls made the problem popular.
21 the following subarray produces the maximum sum 10: −1 2 4 −3 5 2 −5 2 We assume that an empty subarray is allowed, so the maximum subarray sum is always at least 0.
Algorithm 1 A straightforward way to solve the problem is to go through all possible subarrays, calculate the sum of values in each subarray and maintain the maximum sum.
The following code implements this algorithm: int best = 0; for (int a = 0; a < n; a++) { for (int b = a; b < n; b++) { int sum = 0; for (int k = a; k <= b; k++) { sum += array[k]; } best = max(best,sum); } } cout << best << "\n"; The variables a and b fix the first and last index of the subarray, and the sum of values is calculated to the variable sum. The variable best contains the maximum sum found during the search.
The time complexity of the algorithm is O(n3), because it consists of three nested loops that go through the input.
Algorithm 2 It is easy to make Algorithm 1 more efficient by removing one loop from it. This is possible by calculating the sum at the same time when the right end of the subarray moves. The result is the following code: int best = 0; for (int a = 0; a < n; a++) { int sum = 0; for (int b = a; b < n; b++) { sum += array[b]; best = max(best,sum); } } cout << best << "\n"; After this change, the time complexity is O(n2).
22 Algorithm 3 Surprisingly, it is possible to solve the problem in O(n) time3, which means that just one loop is enough. The idea is to calculate, for each array position, the maximum sum of a subarray that ends at that position. After this, the answer for the problem is the maximum of those sums.
Consider the subproblem of finding the maximum-sum subarray that ends at position k. There are two possibilities: 1. The subarray only contains the element at position k.
2. The subarray consists of a subarray that ends at position k −1, followed by the element at position k.
In the latter case, since we want to find a subarray with maximum sum, the subarray that ends at position k −1 should also have the maximum sum. Thus, we can solve the problem efficiently by calculating the maximum subarray sum for each ending position from left to right.
The following code implements the algorithm: int best = 0, sum = 0; for (int k = 0; k < n; k++) { sum = max(array[k],sum+array[k]); best = max(best,sum); } cout << best << "\n"; The algorithm only contains one loop that goes through the input, so the time complexity is O(n). This is also the best possible time complexity, because any algorithm for the problem has to examine all array elements at least once.
Efficiency comparison It is interesting to study how efficient algorithms are in practice. The following table shows the running times of the above algorithms for different values of n on a modern computer.
In each test, the input was generated randomly. The time needed for reading the input was not measured.
array size n Algorithm 1 Algorithm 2 Algorithm 3 102 0.0 s 0.0 s 0.0 s 103 0.1 s 0.0 s 0.0 s 104 > 10.0 s 0.1 s 0.0 s 105 > 10.0 s 5.3 s 0.0 s 106 > 10.0 s > 10.0 s 0.0 s 107 > 10.0 s > 10.0 s 0.0 s 3In , this linear-time algorithm is attributed to J. B. Kadane, and the algorithm is sometimes called Kadane’s algorithm.
23 The comparison shows that all algorithms are efficient when the input size is small, but larger inputs bring out remarkable differences in the running times of the algorithms. Algorithm 1 becomes slow when n = 104, and Algorithm 2 becomes slow when n = 105. Only Algorithm 3 is able to process even the largest inputs instantly.
24 Chapter 3 Sorting Sorting is a fundamental algorithm design problem. Many efficient algorithms use sorting as a subroutine, because it is often easier to process data if the elements are in a sorted order.
For example, the problem ”does an array contain two equal elements?” is easy to solve using sorting. If the array contains two equal elements, they will be next to each other after sorting, so it is easy to find them. Also, the problem ”what is the most frequent element in an array?” can be solved similarly.
There are many algorithms for sorting, and they are also good examples of how to apply different algorithm design techniques. The efficient general sorting algorithms work in O(nlogn) time, and many algorithms that use sorting as a subroutine also have this time complexity.
3.1 Sorting theory The basic problem in sorting is as follows: Given an array that contains n elements, your task is to sort the elements in increasing order.
For example, the array 1 3 8 2 9 2 5 6 will be as follows after sorting: 1 2 2 3 5 6 8 9 O(n2) algorithms Simple algorithms for sorting an array work in O(n2) time. Such algorithms are short and usually consist of two nested loops. A famous O(n2) time sorting 25 algorithm is bubble sort where the elements ”bubble” in the array according to their values.
Bubble sort consists of n rounds. On each round, the algorithm iterates through the elements of the array. Whenever two consecutive elements are found that are not in correct order, the algorithm swaps them. The algorithm can be implemented as follows: for (int i = 0; i < n; i++) { for (int j = 0; j < n-1; j++) { if (array[j] > array[j+1]) { swap(array[j],array[j+1]); } } } After the first round of the algorithm, the largest element will be in the correct position, and in general, after k rounds, the k largest elements will be in the correct positions. Thus, after n rounds, the whole array will be sorted.
For example, in the array 1 3 8 2 9 2 5 6 the first round of bubble sort swaps elements as follows: 1 3 2 8 9 2 5 6 1 3 2 8 2 9 5 6 1 3 2 8 2 5 9 6 1 3 2 8 2 5 6 9 Inversions Bubble sort is an example of a sorting algorithm that always swaps consecutive elements in the array. It turns out that the time complexity of such an algorithm is always at least O(n2), because in the worst case, O(n2) swaps are required for sorting the array.
A useful concept when analyzing sorting algorithms is an inversion: a pair of array elements (array[a],array[b]) such that a < b and array[a] > array[b], i.e., the elements are in the wrong order. For example, the array 26 1 2 2 6 3 5 9 8 has three inversions: (6,3), (6,5) and (9,8). The number of inversions indicates how much work is needed to sort the array. An array is completely sorted when there are no inversions. On the other hand, if the array elements are in the reverse order, the number of inversions is the largest possible: 1+2+···+(n−1) = n(n−1) 2 = O(n2) Swapping a pair of consecutive elements that are in the wrong order removes exactly one inversion from the array. Hence, if a sorting algorithm can only swap consecutive elements, each swap removes at most one inversion, and the time complexity of the algorithm is at least O(n2).
O(nlogn) algorithms It is possible to sort an array efficiently in O(nlogn) time using algorithms that are not limited to swapping consecutive elements. One such algorithm is merge sort1, which is based on recursion.
Merge sort sorts a subarray array[a...b] as follows: 1. If a = b, do not do anything, because the subarray is already sorted.
2. Calculate the position of the middle element: k = ⌊(a+ b)/2⌋.
3. Recursively sort the subarray array[a...k].
4. Recursively sort the subarray array[k +1...b].
5. Merge the sorted subarrays array[a...k] and array[k +1...b] into a sorted subarray array[a...b].
Merge sort is an efficient algorithm, because it halves the size of the subarray at each step. The recursion consists of O(logn) levels, and processing each level takes O(n) time. Merging the subarrays array[a...k] and array[k + 1...b] is possible in linear time, because they are already sorted.
For example, consider sorting the following array: 1 3 6 2 8 2 5 9 The array will be divided into two subarrays as follows: 1 3 6 2 8 2 5 9 Then, the subarrays will be sorted recursively as follows: 1 2 3 6 2 5 8 9 1According to , merge sort was invented by J. von Neumann in 1945.
27 Finally, the algorithm merges the sorted subarrays and creates the final sorted array: 1 2 2 3 5 6 8 9 Sorting lower bound Is it possible to sort an array faster than in O(nlogn) time? It turns out that this is not possible when we restrict ourselves to sorting algorithms that are based on comparing array elements.
The lower bound for the time complexity can be proved by considering sorting as a process where each comparison of two elements gives more information about the contents of the array. The process creates the following tree: x < y?
x < y?
x < y?
x < y?
x < y?
x < y?
x < y?
Here ”x < y?” means that some elements x and y are compared. If x < y, the process continues to the left, and otherwise to the right. The results of the process are the possible ways to sort the array, a total of n! ways. For this reason, the height of the tree must be at least log2(n!) = log2(1)+log2(2)+···+log2(n).
We get a lower bound for this sum by choosing the last n/2 elements and changing the value of each element to log2(n/2). This yields an estimate log2(n!) ≥(n/2)·log2(n/2), so the height of the tree and the minimum possible number of steps in a sorting algorithm in the worst case is at least nlogn.
Counting sort The lower bound nlogn does not apply to algorithms that do not compare array elements but use some other information. An example of such an algorithm is counting sort that sorts an array in O(n) time assuming that every element in the array is an integer between 0... c and c = O(n).
The algorithm creates a bookkeeping array, whose indices are elements of the original array. The algorithm iterates through the original array and calculates how many times each element appears in the array.
28 For example, the array 1 3 6 9 9 3 5 9 corresponds to the following bookkeeping array: 1 0 2 0 1 1 0 0 3 1 2 3 4 5 6 7 8 9 For example, the value at position 3 in the bookkeeping array is 2, because the element 3 appears 2 times in the original array.
Construction of the bookkeeping array takes O(n) time. After this, the sorted array can be created in O(n) time because the number of occurrences of each element can be retrieved from the bookkeeping array. Thus, the total time complexity of counting sort is O(n).
Counting sort is a very efficient algorithm but it can only be used when the constant c is small enough, so that the array elements can be used as indices in the bookkeeping array.
3.2 Sorting in C++ It is almost never a good idea to use a home-made sorting algorithm in a contest, because there are good implementations available in programming languages.
For example, the C++ standard library contains the function sort that can be easily used for sorting arrays and other data structures.
There are many benefits in using a library function. First, it saves time because there is no need to implement the function. Second, the library imple-mentation is certainly correct and efficient: it is not probable that a home-made sorting function would be better.
In this section we will see how to use the C++ sort function. The following code sorts a vector in increasing order: vector v = {4,2,5,3,5,8,3}; sort(v.begin(),v.end()); After the sorting, the contents of the vector will be [2,3,3,4,5,5,8]. The default sorting order is increasing, but a reverse order is possible as follows: sort(v.rbegin(),v.rend()); An ordinary array can be sorted as follows: int n = 7; // array size int a[] = {4,2,5,3,5,8,3}; sort(a,a+n); 29 The following code sorts the string s: string s = "monkey"; sort(s.begin(), s.end()); Sorting a string means that the characters of the string are sorted. For example, the string ”monkey” becomes ”ekmnoy”.
Comparison operators The function sort requires that a comparison operator is defined for the data type of the elements to be sorted. When sorting, this operator will be used whenever it is necessary to find out the order of two elements.
Most C++ data types have a built-in comparison operator, and elements of those types can be sorted automatically. For example, numbers are sorted according to their values and strings are sorted in alphabetical order.
Pairs (pair) are sorted primarily according to their first elements (first).
However, if the first elements of two pairs are equal, they are sorted according to their second elements (second): vector> v; v.push_back({1,5}); v.push_back({2,3}); v.push_back({1,2}); sort(v.begin(), v.end()); After this, the order of the pairs is (1,2), (1,5) and (2,3).
In a similar way, tuples (tuple) are sorted primarily by the first element, secondarily by the second element, etc.2: vector> v; v.push_back({2,1,4}); v.push_back({1,5,3}); v.push_back({2,1,3}); sort(v.begin(), v.end()); After this, the order of the tuples is (1,5,3), (2,1,3) and (2,1,4).
User-defined structs User-defined structs do not have a comparison operator automatically. The operator should be defined inside the struct as a function operator<, whose parameter is another element of the same type. The operator should return true if the element is smaller than the parameter, and false otherwise.
For example, the following struct P contains the x and y coordinates of a point.
The comparison operator is defined so that the points are sorted primarily by the 2Note that in some older compilers, the function make_tuple has to be used to create a tuple instead of braces (for example, make_tuple(2,1,4) instead of {2,1,4}).
30 x coordinate and secondarily by the y coordinate.
struct P { int x, y; bool operator<(const P &p) { if (x != p.x) return x < p.x; else return y < p.y; } }; Comparison functions It is also possible to give an external comparison function to the sort function as a callback function. For example, the following comparison function comp sorts strings primarily by length and secondarily by alphabetical order: bool comp(string a, string b) { if (a.size() != b.size()) return a.size() < b.size(); return a < b; } Now a vector of strings can be sorted as follows: sort(v.begin(), v.end(), comp); 3.3 Binary search A general method for searching for an element in an array is to use a for loop that iterates through the elements of the array. For example, the following code searches for an element x in an array: for (int i = 0; i < n; i++) { if (array[i] == x) { // x found at index i } } The time complexity of this approach is O(n), because in the worst case, it is necessary to check all elements of the array. If the order of the elements is arbitrary, this is also the best possible approach, because there is no additional information available where in the array we should search for the element x.
However, if the array is sorted, the situation is different. In this case it is possible to perform the search much faster, because the order of the elements in the array guides the search. The following binary search algorithm efficiently searches for an element in a sorted array in O(logn) time.
31 Method 1 The usual way to implement binary search resembles looking for a word in a dictionary. The search maintains an active region in the array, which initially contains all array elements. Then, a number of steps is performed, each of which halves the size of the region.
At each step, the search checks the middle element of the active region. If the middle element is the target element, the search terminates. Otherwise, the search recursively continues to the left or right half of the region, depending on the value of the middle element.
The above idea can be implemented as follows: int a = 0, b = n-1; while (a <= b) { int k = (a+b)/2; if (array[k] == x) { // x found at index k } if (array[k] > x) b = k-1; else a = k+1; } In this implementation, the active region is a...b, and initially the region is 0...n−1. The algorithm halves the size of the region at each step, so the time complexity is O(logn).
Method 2 An alternative method to implement binary search is based on an efficient way to iterate through the elements of the array. The idea is to make jumps and slow the speed when we get closer to the target element.
The search goes through the array from left to right, and the initial jump length is n/2. At each step, the jump length will be halved: first n/4, then n/8, n/16, etc., until finally the length is 1. After the jumps, either the target element has been found or we know that it does not appear in the array.
The following code implements the above idea: int k = 0; for (int b = n/2; b >= 1; b /= 2) { while (k+b < n && array[k+b] <= x) k += b; } if (array[k] == x) { // x found at index k } During the search, the variable b contains the current jump length. The time complexity of the algorithm is O(logn), because the code in the while loop is performed at most twice for each jump length.
32 C++ functions The C++ standard library contains the following functions that are based on binary search and work in logarithmic time: • lower_bound returns a pointer to the first array element whose value is at least x.
• upper_bound returns a pointer to the first array element whose value is larger than x.
• equal_range returns both above pointers.
The functions assume that the array is sorted. If there is no such element, the pointer points to the element after the last array element. For example, the following code finds out whether an array contains an element with value x: auto k = lower_bound(array,array+n,x)-array; if (k < n && array[k] == x) { // x found at index k } Then, the following code counts the number of elements whose value is x: auto a = lower_bound(array, array+n, x); auto b = upper_bound(array, array+n, x); cout << b-a << "\n"; Using equal_range, the code becomes shorter: auto r = equal_range(array, array+n, x); cout << r.second-r.first << "\n"; Finding the smallest solution An important use for binary search is to find the position where the value of a function changes. Suppose that we wish to find the smallest value k that is a valid solution for a problem. We are given a function ok(x) that returns true if x is a valid solution and false otherwise. In addition, we know that ok(x) is false when x < k and true when x ≥k. The situation looks as follows: x 0 1 ··· k −1 k k +1 ··· ok(x) false false ··· false true true ··· Now, the value of k can be found using binary search: int x = -1; for (int b = z; b >= 1; b /= 2) { while (!ok(x+b)) x += b; } int k = x+1; 33 The search finds the largest value of x for which ok(x) is false. Thus, the next value k = x+1 is the smallest possible value for which ok(k) is true. The initial jump length z has to be large enough, for example some value for which we know beforehand that ok(z) is true.
The algorithm calls the function ok O(log z) times, so the total time complexity depends on the function ok. For example, if the function works in O(n) time, the total time complexity is O(nlog z).
Finding the maximum value Binary search can also be used to find the maximum value for a function that is first increasing and then decreasing. Our task is to find a position k such that • f (x) < f (x+1) when x < k, and • f (x) > f (x+1) when x ≥k.
The idea is to use binary search for finding the largest value of x for which f (x) < f (x+1). This implies that k = x+1 because f (x+1) > f (x+2). The following code implements the search: int x = -1; for (int b = z; b >= 1; b /= 2) { while (f(x+b) < f(x+b+1)) x += b; } int k = x+1; Note that unlike in the ordinary binary search, here it is not allowed that consecutive values of the function are equal. In this case it would not be possible to know how to continue the search.
34 Chapter 4 Data structures A data structure is a way to store data in the memory of a computer. It is important to choose an appropriate data structure for a problem, because each data structure has its own advantages and disadvantages. The crucial question is: which operations are efficient in the chosen data structure?
This chapter introduces the most important data structures in the C++ stan-dard library. It is a good idea to use the standard library whenever possible, because it will save a lot of time. Later in the book we will learn about more sophisticated data structures that are not available in the standard library.
4.1 Dynamic arrays A dynamic array is an array whose size can be changed during the execution of the program. The most popular dynamic array in C++ is the vector structure, which can be used almost like an ordinary array.
The following code creates an empty vector and adds three elements to it: vector v; v.push_back(3); // v.push_back(2); // [3,2] v.push_back(5); // [3,2,5] After this, the elements can be accessed like in an ordinary array: cout << v << "\n"; // 3 cout << v << "\n"; // 2 cout << v << "\n"; // 5 The function size returns the number of elements in the vector. The following code iterates through the vector and prints all elements in it: for (int i = 0; i < v.size(); i++) { cout << v[i] << "\n"; } 35 A shorter way to iterate through a vector is as follows: for (auto x : v) { cout << x << "\n"; } The function back returns the last element in the vector, and the function pop_back removes the last element: vector v; v.push_back(5); v.push_back(2); cout << v.back() << "\n"; // 2 v.pop_back(); cout << v.back() << "\n"; // 5 The following code creates a vector with five elements: vector v = {2,4,2,5,1}; Another way to create a vector is to give the number of elements and the initial value for each element: // size 10, initial value 0 vector v(10); // size 10, initial value 5 vector v(10, 5); The internal implementation of a vector uses an ordinary array. If the size of the vector increases and the array becomes too small, a new array is allocated and all the elements are moved to the new array. However, this does not happen often and the average time complexity of push_back is O(1).
The string structure is also a dynamic array that can be used almost like a vector. In addition, there is special syntax for strings that is not available in other data structures. Strings can be combined using the + symbol. The function substr(k,x) returns the substring that begins at position k and has length x, and the function find(t) finds the position of the first occurrence of a substring t.
The following code presents some string operations: string a = "hatti"; string b = a+a; cout << b << "\n"; // hattihatti b = ’v’; cout << b << "\n"; // hattivatti string c = b.substr(3,4); cout << c << "\n"; // tiva 36 4.2 Set structures A set is a data structure that maintains a collection of elements. The basic operations of sets are element insertion, search and removal.
The C++ standard library contains two set implementations: The structure set is based on a balanced binary tree and its operations work in O(logn) time.
The structure unordered_set uses hashing, and its operations work in O(1) time on average.
The choice of which set implementation to use is often a matter of taste. The benefit of the set structure is that it maintains the order of the elements and provides functions that are not available in unordered_set. On the other hand, unordered_set can be more efficient.
The following code creates a set that contains integers, and shows some of the operations. The function insert adds an element to the set, the function count returns the number of occurrences of an element in the set, and the function erase removes an element from the set.
set s; s.insert(3); s.insert(2); s.insert(5); cout << s.count(3) << "\n"; // 1 cout << s.count(4) << "\n"; // 0 s.erase(3); s.insert(4); cout << s.count(3) << "\n"; // 0 cout << s.count(4) << "\n"; // 1 A set can be used mostly like a vector, but it is not possible to access the elements using the [] notation. The following code creates a set, prints the number of elements in it, and then iterates through all the elements: set s = {2,5,6,8}; cout << s.size() << "\n"; // 4 for (auto x : s) { cout << x << "\n"; } An important property of sets is that all their elements are distinct. Thus, the function count always returns either 0 (the element is not in the set) or 1 (the element is in the set), and the function insert never adds an element to the set if it is already there. The following code illustrates this: set s; s.insert(5); s.insert(5); s.insert(5); cout << s.count(5) << "\n"; // 1 37 C++ also contains the structures multiset and unordered_multiset that other-wise work like set and unordered_set but they can contain multiple instances of an element. For example, in the following code all three instances of the number 5 are added to a multiset: multiset s; s.insert(5); s.insert(5); s.insert(5); cout << s.count(5) << "\n"; // 3 The function erase removes all instances of an element from a multiset: s.erase(5); cout << s.count(5) << "\n"; // 0 Often, only one instance should be removed, which can be done as follows: s.erase(s.find(5)); cout << s.count(5) << "\n"; // 2 4.3 Map structures A map is a generalized array that consists of key-value-pairs. While the keys in an ordinary array are always the consecutive integers 0,1,...,n−1, where n is the size of the array, the keys in a map can be of any data type and they do not have to be consecutive values.
The C++ standard library contains two map implementations that correspond to the set implementations: the structure map is based on a balanced binary tree and accessing elements takes O(logn) time, while the structure unordered_map uses hashing and accessing elements takes O(1) time on average.
The following code creates a map where the keys are strings and the values are integers: map m; m["monkey"] = 4; m["banana"] = 3; m["harpsichord"] = 9; cout << m["banana"] << "\n"; // 3 If the value of a key is requested but the map does not contain it, the key is automatically added to the map with a default value. For example, in the following code, the key ”aybabtu” with value 0 is added to the map.
map m; cout << m["aybabtu"] << "\n"; // 0 38 The function count checks if a key exists in a map: if (m.count("aybabtu")) { // key exists } The following code prints all the keys and values in a map: for (auto x : m) { cout << x.first << " " << x.second << "\n"; } 4.4 Iterators and ranges Many functions in the C++ standard library operate with iterators. An iterator is a variable that points to an element in a data structure.
The often used iterators begin and end define a range that contains all ele-ments in a data structure. The iterator begin points to the first element in the data structure, and the iterator end points to the position after the last element.
The situation looks as follows: { 3, 4, 6, 8, 12, 13, 14, 17 } ↑ ↑ s.begin() s.end() Note the asymmetry in the iterators: s.begin() points to an element in the data structure, while s.end() points outside the data structure. Thus, the range defined by the iterators is half-open.
Working with ranges Iterators are used in C++ standard library functions that are given a range of elements in a data structure. Usually, we want to process all elements in a data structure, so the iterators begin and end are given for the function.
For example, the following code sorts a vector using the function sort, then reverses the order of the elements using the function reverse, and finally shuffles the order of the elements using the function random_shuffle.
sort(v.begin(), v.end()); reverse(v.begin(), v.end()); random_shuffle(v.begin(), v.end()); These functions can also be used with an ordinary array. In this case, the functions are given pointers to the array instead of iterators: 39 sort(a, a+n); reverse(a, a+n); random_shuffle(a, a+n); Set iterators Iterators are often used to access elements of a set. The following code creates an iterator it that points to the smallest element in a set: set::iterator it = s.begin(); A shorter way to write the code is as follows: auto it = s.begin(); The element to which an iterator points can be accessed using the symbol. For example, the following code prints the first element in the set: auto it = s.begin(); cout << it << "\n"; Iterators can be moved using the operators ++ (forward) and -- (backward), meaning that the iterator moves to the next or previous element in the set.
The following code prints all the elements in increasing order: for (auto it = s.begin(); it != s.end(); it++) { cout << it << "\n"; } The following code prints the largest element in the set: auto it = s.end(); it--; cout << it << "\n"; The function find(x) returns an iterator that points to an element whose value is x. However, if the set does not contain x, the iterator will be end.
auto it = s.find(x); if (it == s.end()) { // x is not found } The function lower_bound(x) returns an iterator to the smallest element in the set whose value is at least x, and the function upper_bound(x) returns an iterator to the smallest element in the set whose value is larger than x. In both functions, if such an element does not exist, the return value is end. These functions are not supported by the unordered_set structure which does not maintain the order of the elements.
40 For example, the following code finds the element nearest to x: auto it = s.lower_bound(x); if (it == s.begin()) { cout << it << "\n"; } else if (it == s.end()) { it--; cout << it << "\n"; } else { int a = it; it--; int b = it; if (x-b < a-x) cout << b << "\n"; else cout << a << "\n"; } The code assumes that the set is not empty, and goes through all possible cases using an iterator it. First, the iterator points to the smallest element whose value is at least x. If it equals begin, the corresponding element is nearest to x. If it equals end, the largest element in the set is nearest to x. If none of the previous cases hold, the element nearest to x is either the element that corresponds to it or the previous element.
4.5 Other structures Bitset A bitset is an array whose each value is either 0 or 1. For example, the following code creates a bitset that contains 10 elements: bitset<10> s; s = 1; s = 1; s = 1; s = 1; cout << s << "\n"; // 1 cout << s << "\n"; // 0 The benefit of using bitsets is that they require less memory than ordinary arrays, because each element in a bitset only uses one bit of memory.
For example, if n bits are stored in an int array, 32n bits of memory will be used, but a corresponding bitset only requires n bits of memory. In addition, the values of a bitset can be efficiently manipulated using bit operators, which makes it possible to optimize algorithms using bit sets.
The following code shows another way to create the above bitset: bitset<10> s(string("0010011010")); // from right to left cout << s << "\n"; // 1 cout << s << "\n"; // 0 41 The function count returns the number of ones in the bitset: bitset<10> s(string("0010011010")); cout << s.count() << "\n"; // 4 The following code shows examples of using bit operations: bitset<10> a(string("0010110110")); bitset<10> b(string("1011011000")); cout << (a&b) << "\n"; // 0010010000 cout << (a|b) << "\n"; // 1011111110 cout << (a^b) << "\n"; // 1001101110 Deque A deque is a dynamic array whose size can be efficiently changed at both ends of the array. Like a vector, a deque provides the functions push_back and pop_back, but it also includes the functions push_front and pop_front which are not avail-able in a vector.
A deque can be used as follows: deque d; d.push_back(5); // d.push_back(2); // [5,2] d.push_front(3); // [3,5,2] d.pop_back(); // [3,5] d.pop_front(); // The internal implementation of a deque is more complex than that of a vector, and for this reason, a deque is slower than a vector. Still, both adding and removing elements take O(1) time on average at both ends.
Stack A stack is a data structure that provides two O(1) time operations: adding an element to the top, and removing an element from the top. It is only possible to access the top element of a stack.
The following code shows how a stack can be used: stack s; s.push(3); s.push(2); s.push(5); cout << s.top(); // 5 s.pop(); cout << s.top(); // 2 42 Queue A queue also provides two O(1) time operations: adding an element to the end of the queue, and removing the first element in the queue. It is only possible to access the first and last element of a queue.
The following code shows how a queue can be used: queue q; q.push(3); q.push(2); q.push(5); cout << q.front(); // 3 q.pop(); cout << q.front(); // 2 Priority queue A priority queue maintains a set of elements. The supported operations are insertion and, depending on the type of the queue, retrieval and removal of either the minimum or maximum element. Insertion and removal take O(logn) time, and retrieval takes O(1) time.
While an ordered set efficiently supports all the operations of a priority queue, the benefit of using a priority queue is that it has smaller constant factors. A priority queue is usually implemented using a heap structure that is much simpler than a balanced binary tree used in an ordered set.
By default, the elements in a C++ priority queue are sorted in decreasing order, and it is possible to find and remove the largest element in the queue. The following code illustrates this: priority_queue q; q.push(3); q.push(5); q.push(7); q.push(2); cout << q.top() << "\n"; // 7 q.pop(); cout << q.top() << "\n"; // 5 q.pop(); q.push(6); cout << q.top() << "\n"; // 6 q.pop(); If we want to create a priority queue that supports finding and removing the smallest element, we can do it as follows: priority_queue,greater\> q; 43 Policy-based data structures The g++ compiler also supports some data structures that are not part of the C++ standard library. Such structures are called policy-based data structures. To use these structures, the following lines must be added to the code: #include using namespace __gnu_pbds; After this, we can define a data structure indexed_set that is like set but can be indexed like an array. The definition for int values is as follows: typedef tree,rb_tree_tag, tree_order_statistics_node_update> indexed_set; Now we can create a set as follows: indexed_set s; s.insert(2); s.insert(3); s.insert(7); s.insert(9); The speciality of this set is that we have access to the indices that the elements would have in a sorted array. The function find_by_order returns an iterator to the element at a given position: auto x = s.find_by_order(2); cout << x << "\n"; // 7 And the function order_of_key returns the position of a given element: cout << s.order_of_key(7) << "\n"; // 2 If the element does not appear in the set, we get the position that the element would have in the set: cout << s.order_of_key(6) << "\n"; // 2 cout << s.order_of_key(8) << "\n"; // 3 Both the functions work in logarithmic time.
4.6 Comparison to sorting It is often possible to solve a problem using either data structures or sorting.
Sometimes there are remarkable differences in the actual efficiency of these approaches, which may be hidden in their time complexities.
Let us consider a problem where we are given two lists A and B that both contain n elements. Our task is to calculate the number of elements that belong 44 to both of the lists. For example, for the lists A = [5,2,8,9] and B = [3,2,9,5], the answer is 3 because the numbers 2, 5 and 9 belong to both of the lists.
A straightforward solution to the problem is to go through all pairs of elements in O(n2) time, but next we will focus on more efficient algorithms.
Algorithm 1 We construct a set of the elements that appear in A, and after this, we iterate through the elements of B and check for each elements if it also belongs to A.
This is efficient because the elements of A are in a set. Using the set structure, the time complexity of the algorithm is O(nlogn).
Algorithm 2 It is not necessary to maintain an ordered set, so instead of the set structure we can also use the unordered_set structure. This is an easy way to make the algorithm more efficient, because we only have to change the underlying data structure. The time complexity of the new algorithm is O(n).
Algorithm 3 Instead of data structures, we can use sorting. First, we sort both lists A and B. After this, we iterate through both the lists at the same time and find the common elements. The time complexity of sorting is O(nlogn), and the rest of the algorithm works in O(n) time, so the total time complexity is O(nlogn).
Efficiency comparison The following table shows how efficient the above algorithms are when n varies and the elements of the lists are random integers between 1...109: n Algorithm 1 Algorithm 2 Algorithm 3 106 1.5 s 0.3 s 0.2 s 2·106 3.7 s 0.8 s 0.3 s 3·106 5.7 s 1.3 s 0.5 s 4·106 7.7 s 1.7 s 0.7 s 5·106 10.0 s 2.3 s 0.9 s Algorithms 1 and 2 are equal except that they use different set structures. In this problem, this choice has an important effect on the running time, because Algorithm 2 is 4–5 times faster than Algorithm 1.
However, the most efficient algorithm is Algorithm 3 which uses sorting.
It only uses half the time compared to Algorithm 2. Interestingly, the time complexity of both Algorithm 1 and Algorithm 3 is O(nlogn), but despite this, Algorithm 3 is ten times faster. This can be explained by the fact that sorting is a 45 simple procedure and it is done only once at the beginning of Algorithm 3, and the rest of the algorithm works in linear time. On the other hand, Algorithm 1 maintains a complex balanced binary tree during the whole algorithm.
46 Chapter 5 Complete search Complete search is a general method that can be used to solve almost any algorithm problem. The idea is to generate all possible solutions to the problem using brute force, and then select the best solution or count the number of solutions, depending on the problem.
Complete search is a good technique if there is enough time to go through all the solutions, because the search is usually easy to implement and it always gives the correct answer. If complete search is too slow, other techniques, such as greedy algorithms or dynamic programming, may be needed.
5.1 Generating subsets We first consider the problem of generating all subsets of a set of n elements. For example, the subsets of {0,1,2} are ;, {0}, {1}, {2}, {0,1}, {0,2}, {1,2} and {0,1,2}.
There are two common methods to generate subsets: we can either perform a recursive search or exploit the bit representation of integers.
Method 1 An elegant way to go through all subsets of a set is to use recursion.
The following function search generates the subsets of the set {0,1,...,n −1}. The function maintains a vector subset that will contain the elements of each subset.
The search begins when the function is called with parameter 0.
void search(int k) { if (k == n) { // process subset } else { search(k+1); subset.push_back(k); search(k+1); subset.pop_back(); } } 47 When the function search is called with parameter k, it decides whether to include the element k in the subset or not, and in both cases, then calls itself with parameter k +1 However, if k = n, the function notices that all elements have been processed and a subset has been generated.
The following tree illustrates the function calls when n = 3. We can always choose either the left branch (k is not included in the subset) or the right branch (k is included in the subset).
search(0) search(1) search(1) search(2) search(2) search(2) search(2) search(3) search(3) search(3) search(3) search(3) search(3) search(3) search(3) ; {2} {1} {1,2} {0} {0,2} {0,1} {0,1,2} Method 2 Another way to generate subsets is based on the bit representation of integers.
Each subset of a set of n elements can be represented as a sequence of n bits, which corresponds to an integer between 0...2n −1. The ones in the bit sequence indicate which elements are included in the subset.
The usual convention is that the last bit corresponds to element 0, the second last bit corresponds to element 1, and so on. For example, the bit representation of 25 is 11001, which corresponds to the subset {0,3,4}.
The following code goes through the subsets of a set of n elements for (int b = 0; b < (1<<n); b++) { // process subset } The following code shows how we can find the elements of a subset that corresponds to a bit sequence. When processing each subset, the code builds a vector that contains the elements in the subset.
for (int b = 0; b < (1<<n); b++) { vector subset; for (int i = 0; i < n; i++) { if (b&(1<<i)) subset.push_back(i); } } 48 5.2 Generating permutations Next we consider the problem of generating all permutations of a set of n elements.
For example, the permutations of {0,1,2} are (0,1,2), (0,2,1), (1,0,2), (1,2,0), (2,0,1) and (2,1,0). Again, there are two approaches: we can either use recursion or go through the permutations iteratively.
Method 1 Like subsets, permutations can be generated using recursion. The following function search goes through the permutations of the set {0,1,...,n −1}. The function builds a vector permutation that contains the permutation, and the search begins when the function is called without parameters.
void search() { if (permutation.size() == n) { // process permutation } else { for (int i = 0; i < n; i++) { if (chosen[i]) continue; chosen[i] = true; permutation.push_back(i); search(); chosen[i] = false; permutation.pop_back(); } } } Each function call adds a new element to permutation. The array chosen indicates which elements are already included in the permutation. If the size of permutation equals the size of the set, a permutation has been generated.
Method 2 Another method for generating permutations is to begin with the permutation {0,1,...,n −1} and repeatedly use a function that constructs the next permu-tation in increasing order.
The C++ standard library contains the function next_permutation that can be used for this: vector permutation; for (int i = 0; i < n; i++) { permutation.push_back(i); } do { // process permutation } while (next_permutation(permutation.begin(),permutation.end())); 49 5.3 Backtracking A backtracking algorithm begins with an empty solution and extends the solution step by step. The search recursively goes through all different ways how a solution can be constructed.
As an example, consider the problem of calculating the number of ways n queens can be placed on an n× n chessboard so that no two queens attack each other. For example, when n = 4, there are two possible solutions: Q Q Q Q Q Q Q Q The problem can be solved using backtracking by placing queens to the board row by row. More precisely, exactly one queen will be placed on each row so that no queen attacks any of the queens placed before. A solution has been found when all n queens have been placed on the board.
For example, when n = 4, some partial solutions generated by the backtrack-ing algorithm are as follows: Q Q Q Q Q Q Q Q Q Q Q Q illegal illegal illegal valid At the bottom level, the three first configurations are illegal, because the queens attack each other. However, the fourth configuration is valid and it can be extended to a complete solution by placing two more queens to the board. There is only one way to place the two remaining queens.
The algorithm can be implemented as follows: 50 void search(int y) { if (y == n) { count++; return; } for (int x = 0; x < n; x++) { if (column[x] || diag1[x+y] || diag2[x-y+n-1]) continue; column[x] = diag1[x+y] = diag2[x-y+n-1] = 1; search(y+1); column[x] = diag1[x+y] = diag2[x-y+n-1] = 0; } } The search begins by calling search(0). The size of the board is n × n, and the code calculates the number of solutions to count.
The code assumes that the rows and columns of the board are numbered from 0 to n−1. When the function search is called with parameter y, it places a queen on row y and then calls itself with parameter y+1. Then, if y = n, a solution has been found and the variable count is increased by one.
The array column keeps track of columns that contain a queen, and the arrays diag1 and diag2 keep track of diagonals. It is not allowed to add another queen to a column or diagonal that already contains a queen. For example, the columns and diagonals of the 4×4 board are numbered as follows: 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 1 2 3 4 2 3 4 5 3 4 5 6 3 4 5 6 2 3 4 5 1 2 3 4 0 1 2 3 column diag1 diag2 Let q(n) denote the number of ways to place n queens on an n× n chessboard.
The above backtracking algorithm tells us that, for example, q(8) = 92. When n increases, the search quickly becomes slow, because the number of solutions increases exponentially. For example, calculating q(16) = 14772512 using the above algorithm already takes about a minute on a modern computer1.
5.4 Pruning the search We can often optimize backtracking by pruning the search tree. The idea is to add ”intelligence” to the algorithm so that it will notice as soon as possible if a partial solution cannot be extended to a complete solution. Such optimizations can have a tremendous effect on the efficiency of the search.
1There is no known way to efficiently calculate larger values of q(n). The current record is q(27) = 234907967154122528, calculated in 2016 .
51 Let us consider the problem of calculating the number of paths in an n × n grid from the upper-left corner to the lower-right corner such that the path visits each square exactly once. For example, in a 7×7 grid, there are 111712 such paths. One of the paths is as follows: We focus on the 7 × 7 case, because its level of difficulty is appropriate to our needs. We begin with a straightforward backtracking algorithm, and then optimize it step by step using observations of how the search can be pruned.
After each optimization, we measure the running time of the algorithm and the number of recursive calls, so that we clearly see the effect of each optimization on the efficiency of the search.
Basic algorithm The first version of the algorithm does not contain any optimizations. We simply use backtracking to generate all possible paths from the upper-left corner to the lower-right corner and count the number of such paths.
• running time: 483 seconds • number of recursive calls: 76 billion Optimization 1 In any solution, we first move one step down or right. There are always two paths that are symmetric about the diagonal of the grid after the first step. For example, the following paths are symmetric: Hence, we can decide that we always first move one step down (or right), and finally multiply the number of solutions by two.
• running time: 244 seconds • number of recursive calls: 38 billion 52 Optimization 2 If the path reaches the lower-right square before it has visited all other squares of the grid, it is clear that it will not be possible to complete the solution. An example of this is the following path: Using this observation, we can terminate the search immediately if we reach the lower-right square too early.
• running time: 119 seconds • number of recursive calls: 20 billion Optimization 3 If the path touches a wall and can turn either left or right, the grid splits into two parts that contain unvisited squares. For example, in the following situation, the path can turn either left or right: In this case, we cannot visit all squares anymore, so we can terminate the search.
This optimization is very useful: • running time: 1.8 seconds • number of recursive calls: 221 million Optimization 4 The idea of Optimization 3 can be generalized: if the path cannot continue forward but can turn either left or right, the grid splits into two parts that both contain unvisited squares. For example, consider the following path: 53 It is clear that we cannot visit all squares anymore, so we can terminate the search. After this optimization, the search is very efficient: • running time: 0.6 seconds • number of recursive calls: 69 million Now is a good moment to stop optimizing the algorithm and see what we have achieved. The running time of the original algorithm was 483 seconds, and now after the optimizations, the running time is only 0.6 seconds. Thus, the algorithm became nearly 1000 times faster after the optimizations.
This is a usual phenomenon in backtracking, because the search tree is usually large and even simple observations can effectively prune the search. Especially useful are optimizations that occur during the first steps of the algorithm, i.e., at the top of the search tree.
5.5 Meet in the middle Meet in the middle is a technique where the search space is divided into two parts of about equal size. A separate search is performed for both of the parts, and finally the results of the searches are combined.
The technique can be used if there is an efficient way to combine the results of the searches. In such a situation, the two searches may require less time than one large search. Typically, we can turn a factor of 2n into a factor of 2n/2 using the meet in the middle technique.
As an example, consider a problem where we are given a list of n numbers and a number x, and we want to find out if it is possible to choose some numbers from the list so that their sum is x. For example, given the list [2,4,5,9] and x = 15, we can choose the numbers [2,4,9] to get 2+4+9 = 15. However, if x = 10 for the same list, it is not possible to form the sum.
A simple algorithm to the problem is to go through all subsets of the elements and check if the sum of any of the subsets is x. The running time of such an algorithm is O(2n), because there are 2n subsets. However, using the meet in the middle technique, we can achieve a more efficient O(2n/2) time algorithm2. Note that O(2n) and O(2n/2) are different complexities because 2n/2 equals p 2n.
2This idea was introduced in 1974 by E. Horowitz and S. Sahni .
54 The idea is to divide the list into two lists A and B such that both lists contain about half of the numbers. The first search generates all subsets of A and stores their sums to a list SA. Correspondingly, the second search creates a list SB from B. After this, it suffices to check if it is possible to choose one element from SA and another element from SB such that their sum is x. This is possible exactly when there is a way to form the sum x using the numbers of the original list.
For example, suppose that the list is [2,4,5,9] and x = 15. First, we divide the list into A = [2,4] and B = [5,9]. After this, we create lists SA = [0,2,4,6] and SB = [0,5,9,14]. In this case, the sum x = 15 is possible to form, because SA contains the sum 6, SB contains the sum 9, and 6+9 = 15. This corresponds to the solution [2,4,9].
We can implement the algorithm so that its time complexity is O(2n/2). First, we generate sorted lists SA and SB, which can be done in O(2n/2) time using a merge-like technique. After this, since the lists are sorted, we can check in O(2n/2) time if the sum x can be created from SA and SB.
55 56 Chapter 6 Greedy algorithms A greedy algorithm constructs a solution to the problem by always making a choice that looks the best at the moment. A greedy algorithm never takes back its choices, but directly constructs the final solution. For this reason, greedy algorithms are usually very efficient.
The difficulty in designing greedy algorithms is to find a greedy strategy that always produces an optimal solution to the problem. The locally optimal choices in a greedy algorithm should also be globally optimal. It is often difficult to argue that a greedy algorithm works.
6.1 Coin problem As a first example, we consider a problem where we are given a set of coins and our task is to form a sum of money n using the coins. The values of the coins are coins = {c1, c2,..., ck}, and each coin can be used as many times we want. What is the minimum number of coins needed?
For example, if the coins are the euro coins (in cents) {1,2,5,10,20,50,100,200} and n = 520, we need at least four coins. The optimal solution is to select coins 200+200+100+20 whose sum is 520.
Greedy algorithm A simple greedy algorithm to the problem always selects the largest possible coin, until the required sum of money has been constructed. This algorithm works in the example case, because we first select two 200 cent coins, then one 100 cent coin and finally one 20 cent coin. But does this algorithm always work?
It turns out that if the coins are the euro coins, the greedy algorithm always works, i.e., it always produces a solution with the fewest possible number of coins.
The correctness of the algorithm can be shown as follows: First, each coin 1, 5, 10, 50 and 100 appears at most once in an optimal solution, because if the solution would contain two such coins, we could replace 57 them by one coin and obtain a better solution. For example, if the solution would contain coins 5+5, we could replace them by coin 10.
In the same way, coins 2 and 20 appear at most twice in an optimal solution, because we could replace coins 2+2+2 by coins 5+1 and coins 20+20+20 by coins 50 + 10. Moreover, an optimal solution cannot contain coins 2 + 2 + 1 or 20+20+10, because we could replace them by coins 5 and 50.
Using these observations, we can show for each coin x that it is not possible to optimally construct a sum x or any larger sum by only using coins that are smaller than x. For example, if x = 100, the largest optimal sum using the smaller coins is 50+20+20+5+2+2 = 99. Thus, the greedy algorithm that always selects the largest coin produces the optimal solution.
This example shows that it can be difficult to argue that a greedy algorithm works, even if the algorithm itself is simple.
General case In the general case, the coin set can contain any coins and the greedy algorithm does not necessarily produce an optimal solution.
We can prove that a greedy algorithm does not work by showing a counterex-ample where the algorithm gives a wrong answer. In this problem we can easily find a counterexample: if the coins are {1,3,4} and the target sum is 6, the greedy algorithm produces the solution 4+1+1 while the optimal solution is 3+3.
It is not known if the general coin problem can be solved using any greedy algorithm1. However, as we will see in Chapter 7, in some cases, the general problem can be efficiently solved using a dynamic programming algorithm that always gives the correct answer.
6.2 Scheduling Many scheduling problems can be solved using greedy algorithms. A classic problem is as follows: Given n events with their starting and ending times, find a schedule that includes as many events as possible. It is not possible to select an event partially. For example, consider the following events: event starting time ending time A 1 3 B 2 5 C 3 9 D 6 8 In this case the maximum number of events is two. For example, we can select events B and D as follows: 1However, it is possible to check in polynomial time if the greedy algorithm presented in this chapter works for a given set of coins .
58 A B C D It is possible to invent several greedy algorithms for the problem, but which of them works in every case?
Algorithm 1 The first idea is to select as short events as possible. In the example case this algorithm selects the following events: A B C D However, selecting short events is not always a correct strategy. For example, the algorithm fails in the following case: If we select the short event, we can only select one event. However, it would be possible to select both long events.
Algorithm 2 Another idea is to always select the next possible event that begins as early as possible. This algorithm selects the following events: A B C D However, we can find a counterexample also for this algorithm. For example, in the following case, the algorithm only selects one event: If we select the first event, it is not possible to select any other events. However, it would be possible to select the other two events.
59 Algorithm 3 The third idea is to always select the next possible event that ends as early as possible. This algorithm selects the following events: A B C D It turns out that this algorithm always produces an optimal solution. The reason for this is that it is always an optimal choice to first select an event that ends as early as possible. After this, it is an optimal choice to select the next event using the same strategy, etc., until we cannot select any more events.
One way to argue that the algorithm works is to consider what happens if we first select an event that ends later than the event that ends as early as possible.
Now, we will have at most an equal number of choices how we can select the next event. Hence, selecting an event that ends later can never yield a better solution, and the greedy algorithm is correct.
6.3 Tasks and deadlines Let us now consider a problem where we are given n tasks with durations and deadlines and our task is to choose an order to perform the tasks. For each task, we earn d −x points where d is the task’s deadline and x is the moment when we finish the task. What is the largest possible total score we can obtain?
For example, suppose that the tasks are as follows: task duration deadline A 4 2 B 3 5 C 2 7 D 4 5 In this case, an optimal schedule for the tasks is as follows: C B A D 0 5 10 In this solution, C yields 5 points, B yields 0 points, A yields −7 points and D yields −8 points, so the total score is −10.
Surprisingly, the optimal solution to the problem does not depend on the deadlines at all, but a correct greedy strategy is to simply perform the tasks sorted by their durations in increasing order. The reason for this is that if we ever perform two tasks one after another such that the first task takes longer than the second task, we can obtain a better solution if we swap the tasks. For example, consider the following schedule: 60 X Y a b Here a > b, so we should swap the tasks: Y X b a Now X gives b points less and Y gives a points more, so the total score increases by a−b > 0. In an optimal solution, for any two consecutive tasks, it must hold that the shorter task comes before the longer task. Thus, the tasks must be performed sorted by their durations.
6.4 Minimizing sums We next consider a problem where we are given n numbers a1,a2,...,an and our task is to find a value x that minimizes the sum |a1 −x|c +|a2 −x|c +···+|an −x|c.
We focus on the cases c = 1 and c = 2.
Case c = 1 In this case, we should minimize the sum |a1 −x|+|a2 −x|+···+|an −x|.
For example, if the numbers are [1,2,9,2,6], the best solution is to select x = 2 which produces the sum |1−2|+|2−2|+|9−2|+|2−2|+|6−2| = 12.
In the general case, the best choice for x is the median of the numbers, i.e., the middle number after sorting. For example, the list [1,2,9,2,6] becomes [1,2,2,6,9] after sorting, so the median is 2.
The median is an optimal choice, because if x is smaller than the median, the sum becomes smaller by increasing x, and if x is larger then the median, the sum becomes smaller by decreasing x. Hence, the optimal solution is that x is the median. If n is even and there are two medians, both medians and all values between them are optimal choices.
Case c = 2 In this case, we should minimize the sum (a1 −x)2 +(a2 −x)2 +···+(an −x)2.
61 For example, if the numbers are [1,2,9,2,6], the best solution is to select x = 4 which produces the sum (1−4)2 +(2−4)2 +(9−4)2 +(2−4)2 +(6−4)2 = 46.
In the general case, the best choice for x is the average of the numbers. In the example the average is (1 + 2 + 9 + 2 + 6)/5 = 4. This result can be derived by presenting the sum as follows: nx2 −2x(a1 + a2 +···+ an)+(a2 1 + a2 2 +···+ a2 n) The last part does not depend on x, so we can ignore it. The remaining parts form a function nx2 −2xs where s = a1 + a2 +···+ an. This is a parabola opening upwards with roots x = 0 and x = 2s/n, and the minimum value is the average of the roots x = s/n, i.e., the average of the numbers a1,a2,...,an.
6.5 Data compression A binary code assigns for each character of a string a codeword that consists of bits. We can compress the string using the binary code by replacing each character by the corresponding codeword. For example, the following binary code assigns codewords for characters A–D: character codeword A 00 B 01 C 10 D 11 This is a constant-length code which means that the length of each codeword is the same. For example, we can compress the string AABACDACA as follows: 000001001011001000 Using this code, the length of the compressed string is 18 bits. However, we can compress the string better if we use a variable-length code where codewords may have different lengths. Then we can give short codewords for characters that appear often and long codewords for characters that appear rarely. It turns out that an optimal code for the above string is as follows: character codeword A 0 B 110 C 10 D 111 An optimal code produces a compressed string that is as short as possible. In this case, the compressed string using the optimal code is 001100101110100, 62 so only 15 bits are needed instead of 18 bits. Thus, thanks to a better code it was possible to save 3 bits in the compressed string.
We require that no codeword is a prefix of another codeword. For example, it is not allowed that a code would contain both codewords 10 and 1011. The reason for this is that we want to be able to generate the original string from the compressed string. If a codeword could be a prefix of another codeword, this would not always be possible. For example, the following code is not valid: character codeword A 10 B 11 C 1011 D 111 Using this code, it would not be possible to know if the compressed string 1011 corresponds to the string AB or the string C.
Huffman coding Huffman coding2 is a greedy algorithm that constructs an optimal code for compressing a given string. The algorithm builds a binary tree based on the frequencies of the characters in the string, and each character’s codeword can be read by following a path from the root to the corresponding node. A move to the left corresponds to bit 0, and a move to the right corresponds to bit 1.
Initially, each character of the string is represented by a node whose weight is the number of times the character occurs in the string. Then at each step two nodes with minimum weights are combined by creating a new node whose weight is the sum of the weights of the original nodes. The process continues until all nodes have been combined.
Next we will see how Huffman coding creates the optimal code for the string AABACDACA. Initially, there are four nodes that correspond to the characters of the string: 5 1 2 1 A B C D The node that represents character A has weight 5 because character A appears 5 times in the string. The other weights have been calculated in the same way.
The first step is to combine the nodes that correspond to characters B and D, both with weight 1. The result is: 5 2 1 1 2 A C B D 0 1 2D. A. Huffman discovered this method when solving a university course assignment and published the algorithm in 1952 .
63 After this, the nodes with weight 2 are combined: 5 2 1 1 2 4 A C B D 0 1 0 1 Finally, the two remaining nodes are combined: 5 2 1 1 2 4 9 A C B D 0 1 0 1 0 1 Now all nodes are in the tree, so the code is ready. The following codewords can be read from the tree: character codeword A 0 B 110 C 10 D 111 64 Chapter 7 Dynamic programming Dynamic programming is a technique that combines the correctness of com-plete search and the efficiency of greedy algorithms. Dynamic programming can be applied if the problem can be divided into overlapping subproblems that can be solved independently.
There are two uses for dynamic programming: • Finding an optimal solution: We want to find a solution that is as large as possible or as small as possible.
• Counting the number of solutions: We want to calculate the total num-ber of possible solutions.
We will first see how dynamic programming can be used to find an optimal solution, and then we will use the same idea for counting the solutions.
Understanding dynamic programming is a milestone in every competitive programmer’s career. While the basic idea is simple, the challenge is how to apply dynamic programming to different problems. This chapter introduces a set of classic problems that are a good starting point.
7.1 Coin problem We first focus on a problem that we have already seen in Chapter 6: Given a set of coin values coins = {c1, c2,..., ck} and a target sum of money n, our task is to form the sum n using as few coins as possible.
In Chapter 6, we solved the problem using a greedy algorithm that always chooses the largest possible coin. The greedy algorithm works, for example, when the coins are the euro coins, but in the general case the greedy algorithm does not necessarily produce an optimal solution.
Now is time to solve the problem efficiently using dynamic programming, so that the algorithm works for any coin set. The dynamic programming algorithm is based on a recursive function that goes through all possibilities how to form the sum, like a brute force algorithm. However, the dynamic programming algorithm is efficient because it uses memoization and calculates the answer to each subproblem only once.
65 Recursive formulation The idea in dynamic programming is to formulate the problem recursively so that the solution to the problem can be calculated from solutions to smaller subproblems. In the coin problem, a natural recursive problem is as follows: what is the smallest number of coins required to form a sum x?
Let solve(x) denote the minimum number of coins required for a sum x.
The values of the function depend on the values of the coins. For example, if coins = {1,3,4}, the first values of the function are as follows: solve(0) = 0 solve(1) = 1 solve(2) = 2 solve(3) = 1 solve(4) = 1 solve(5) = 2 solve(6) = 2 solve(7) = 2 solve(8) = 2 solve(9) = 3 solve(10) = 3 For example, solve(10) = 3, because at least 3 coins are needed to form the sum 10. The optimal solution is 3+3+4 = 10.
The essential property of solve is that its values can be recursively calculated from its smaller values. The idea is to focus on the first coin that we choose for the sum. For example, in the above scenario, the first coin can be either 1, 3 or 4. If we first choose coin 1, the remaining task is to form the sum 9 using the minimum number of coins, which is a subproblem of the original problem.
Of course, the same applies to coins 3 and 4. Thus, we can use the following recursive formula to calculate the minimum number of coins: solve(x) = min(solve(x−1)+1, solve(x−3)+1, solve(x−4)+1).
The base case of the recursion is solve(0) = 0, because no coins are needed to form an empty sum. For example, solve(10) = solve(7)+1 = solve(4)+2 = solve(0)+3 = 3.
Now we are ready to give a general recursive function that calculates the minimum number of coins needed to form a sum x: solve(x) = ∞ x < 0 0 x = 0 minc∈coinssolve(x−c)+1 x > 0 First, if x < 0, the value is ∞, because it is impossible to form a negative sum of money. Then, if x = 0, the value is 0, because no coins are needed to form an 66 empty sum. Finally, if x > 0, the variable c goes through all possibilities how to choose the first coin of the sum.
Once a recursive function that solves the problem has been found, we can directly implement a solution in C++ (the constant INF denotes infinity): int solve(int x) { if (x < 0) return INF; if (x == 0) return 0; int best = INF; for (auto c : coins) { best = min(best, solve(x-c)+1); } return best; } Still, this function is not efficient, because there may be an exponential number of ways to construct the sum. However, next we will see how to make the function efficient using a technique called memoization.
Using memoization The idea of dynamic programming is to use memoization to efficiently calculate values of a recursive function. This means that the values of the function are stored in an array after calculating them. For each parameter, the value of the function is calculated recursively only once, and after this, the value can be directly retrieved from the array.
In this problem, we use arrays bool ready[N]; int value[N]; where ready[x] indicates whether the value of solve(x) has been calculated, and if it is, value[x] contains this value. The constant N has been chosen so that all required values fit in the arrays.
Now the function can be efficiently implemented as follows: int solve(int x) { if (x < 0) return INF; if (x == 0) return 0; if (ready[x]) return value[x]; int best = INF; for (auto c : coins) { best = min(best, solve(x-c)+1); } value[x] = best; ready[x] = true; return best; } 67 The function handles the base cases x < 0 and x = 0 as previously. Then the function checks from ready[x] if solve(x) has already been stored in value[x], and if it is, the function directly returns it. Otherwise the function calculates the value of solve(x) recursively and stores it in value[x].
This function works efficiently, because the answer for each parameter x is calculated recursively only once. After a value of solve(x) has been stored in value[x], it can be efficiently retrieved whenever the function will be called again with the parameter x. The time complexity of the algorithm is O(nk), where n is the target sum and k is the number of coins.
Note that we can also iteratively construct the array value using a loop that simply calculates all the values of solve for parameters 0...n: value = 0; for (int x = 1; x <= n; x++) { value[x] = INF; for (auto c : coins) { if (x-c >= 0) { value[x] = min(value[x], value[x-c]+1); } } } In fact, most competitive programmers prefer this implementation, because it is shorter and has lower constant factors. From now on, we also use iterative implementations in our examples. Still, it is often easier to think about dynamic programming solutions in terms of recursive functions.
Constructing a solution Sometimes we are asked both to find the value of an optimal solution and to give an example how such a solution can be constructed. In the coin problem, for example, we can declare another array that indicates for each sum of money the first coin in an optimal solution: int first[N]; Then, we can modify the algorithm as follows: value = 0; for (int x = 1; x <= n; x++) { value[x] = INF; for (auto c : coins) { if (x-c >= 0 && value[x-c]+1 < value[x]) { value[x] = value[x-c]+1; first[x] = c; } } } 68 After this, the following code can be used to print the coins that appear in an optimal solution for the sum n: while (n > 0) { cout << first[n] << "\n"; n -= first[n]; } Counting the number of solutions Let us now consider another version of the coin problem where our task is to calculate the total number of ways to produce a sum x using the coins. For example, if coins = {1,3,4} and x = 5, there are a total of 6 ways: • 1+1+1+1+1 • 1+1+3 • 1+3+1 • 3+1+1 • 1+4 • 4+1 Again, we can solve the problem recursively. Let solve(x) denote the number of ways we can form the sum x. For example, if coins = {1,3,4}, then solve(5) = 6 and the recursive formula is solve(x) =solve(x−1)+ solve(x−3)+ solve(x−4).
Then, the general recursive function is as follows: solve(x) = 0 x < 0 1 x = 0 P c∈coinssolve(x−c) x > 0 If x < 0, the value is 0, because there are no solutions. If x = 0, the value is 1, because there is only one way to form an empty sum. Otherwise we calculate the sum of all values of the form solve(x−c) where c is in coins.
The following code constructs an array count such that count[x] equals the value of solve(x) for 0 ≤x ≤n: count = 1; for (int x = 1; x <= n; x++) { for (auto c : coins) { if (x-c >= 0) { count[x] += count[x-c]; } } } 69 Often the number of solutions is so large that it is not required to calculate the exact number but it is enough to give the answer modulo m where, for example, m = 109 +7. This can be done by changing the code so that all calculations are done modulo m. In the above code, it suffices to add the line count[x] %= m; after the line count[x] += count[x-c]; Now we have discussed all basic ideas of dynamic programming.
Since dynamic programming can be used in many different situations, we will now go through a set of problems that show further examples about the possibilities of dynamic programming.
7.2 Longest increasing subsequence Our first problem is to find the longest increasing subsequence in an array of n elements. This is a maximum-length sequence of array elements that goes from left to right, and each element in the sequence is larger than the previous element. For example, in the array 6 2 5 1 7 4 8 3 0 1 2 3 4 5 6 7 the longest increasing subsequence contains 4 elements: 6 2 5 1 7 4 8 3 0 1 2 3 4 5 6 7 Let length(k) denote the length of the longest increasing subsequence that ends at position k. Thus, if we calculate all values of length(k) where 0 ≤k ≤n−1, we will find out the length of the longest increasing subsequence. For example, the values of the function for the above array are as follows: length(0) = 1 length(1) = 1 length(2) = 2 length(3) = 1 length(4) = 3 length(5) = 2 length(6) = 4 length(7) = 2 For example, length(6) = 4, because the longest increasing subsequence that ends at position 6 consists of 4 elements.
70 To calculate a value of length(k), we should find a position i < k for which array[i] < array[k] and length(i) is as large as possible. Then we know that length(k) = length(i) + 1, because this is an optimal way to add array[k] to a subsequence. However, if there is no such position i, then length(k) = 1, which means that the subsequence only contains array[k].
Since all values of the function can be calculated from its smaller values, we can use dynamic programming. In the following code, the values of the function will be stored in an array length.
for (int k = 0; k < n; k++) { length[k] = 1; for (int i = 0; i < k; i++) { if (array[i] < array[k]) { length[k] = max(length[k],length[i]+1); } } } This code works in O(n2) time, because it consists of two nested loops. How-ever, it is also possible to implement the dynamic programming calculation more efficiently in O(nlogn) time. Can you find a way to do this?
7.3 Paths in a grid Our next problem is to find a path from the upper-left corner to the lower-right corner of an n × n grid, such that we only move down and right. Each square contains a positive integer, and the path should be constructed so that the sum of the values along the path is as large as possible.
The following picture shows an optimal path in a grid: 3 7 9 2 7 9 8 3 5 5 1 7 9 8 5 3 8 6 4 10 6 3 9 7 8 The sum of the values on the path is 67, and this is the largest possible sum on a path from the upper-left corner to the lower-right corner.
Assume that the rows and columns of the grid are numbered from 1 to n, and value[y][x] equals the value of square (y,x). Let sum(y,x) denote the maximum sum on a path from the upper-left corner to square (y,x). Now sum(n,n) tells us the maximum sum from the upper-left corner to the lower-right corner. For example, in the above grid, sum(5,5) = 67.
We can recursively calculate the sums as follows: sum(y,x) = max(sum(y,x−1),sum(y−1,x))+value[y][x] 71 The recursive formula is based on the observation that a path that ends at square (y,x) can come either from square (y,x−1) or square (y−1,x): → ↓ Thus, we select the direction that maximizes the sum. We assume that sum(y,x) = 0 if y = 0 or x = 0 (because no such paths exist), so the recursive formula also works when y = 1 or x = 1.
Since the function sum has two parameters, the dynamic programming array also has two dimensions. For example, we can use an array int sum[N][N]; and calculate the sums as follows: for (int y = 1; y <= n; y++) { for (int x = 1; x <= n; x++) { sum[y][x] = max(sum[y][x-1],sum[y-1][x])+value[y][x]; } } The time complexity of the algorithm is O(n2).
7.4 Knapsack problems The term knapsack refers to problems where a set of objects is given, and subsets with some properties have to be found. Knapsack problems can often be solved using dynamic programming.
In this section, we focus on the following problem: Given a list of weights [w1,w2,...,wn], determine all sums that can be constructed using the weights.
For example, if the weights are [1,3,3,5], the following sums are possible: 0 1 2 3 4 5 6 7 8 9 10 11 12 X X X X X X X X X X X In this case, all sums between 0...12 are possible, except 2 and 10. For example, the sum 7 is possible because we can select the weights [1,3,3].
To solve the problem, we focus on subproblems where we only use the first k weights to construct sums. Let possible(x,k) = true if we can construct a sum x using the first k weights, and otherwise possible(x,k) = false. The values of the function can be recursively calculated as follows: possible(x,k) = possible(x−wk,k −1)∨possible(x,k −1) 72 The formula is based on the fact that we can either use or not use the weight wk in the sum. If we use wk, the remaining task is to form the sum x−wk using the first k−1 weights, and if we do not use wk, the remaining task is to form the sum x using the first k −1 weights. As the base cases, possible(x,0) = ( true x = 0 false x ̸= 0 because if no weights are used, we can only form the sum 0.
The following table shows all values of the function for the weights [1,3,3,5] (the symbol ”X” indicates the true values): k\x 0 1 2 3 4 5 6 7 8 9 10 11 12 0 X 1 X X 2 X X X X 3 X X X X X X 4 X X X X X X X X X X X After calculating those values, possible(x,n) tells us whether we can con-struct a sum x using all weights.
Let W denote the total sum of the weights. The following O(nW) time dynamic programming solution corresponds to the recursive function: possible = true; for (int k = 1; k <= n; k++) { for (int x = 0; x <= W; x++) { if (x-w[k] >= 0) possible[x][k] |= possible[x-w[k]][k-1]; possible[x][k] |= possible[x][k-1]; } } However, here is a better implementation that only uses a one-dimensional array possible[x] that indicates whether we can construct a subset with sum x.
The trick is to update the array from right to left for each new weight: possible = true; for (int k = 1; k <= n; k++) { for (int x = W; x >= 0; x--) { if (possible[x]) possible[x+w[k]] = true; } } Note that the general idea presented here can be used in many knapsack problems. For example, if we are given objects with weights and values, we can determine for each weight sum the maximum value sum of a subset.
73 7.5 Edit distance The edit distance or Levenshtein distance1 is the minimum number of edit-ing operations needed to transform a string into another string. The allowed editing operations are as follows: • insert a character (e.g. ABC →ABCA) • remove a character (e.g. ABC →AC) • modify a character (e.g. ABC →ADC) For example, the edit distance between LOVE and MOVIE is 2, because we can first perform the operation LOVE →MOVE (modify) and then the operation MOVE → MOVIE (insert). This is the smallest possible number of operations, because it is clear that only one operation is not enough.
Suppose that we are given a string x of length n and a string y of length m, and we want to calculate the edit distance between x and y. To solve the problem, we define a function distance(a,b) that gives the edit distance between prefixes x[0...a] and y[0...b]. Thus, using this function, the edit distance between x and y equals distance(n−1,m−1).
We can calculate values of distance as follows: distance(a,b) = min(distance(a,b −1)+1, distance(a−1,b)+1, distance(a−1,b −1)+cost(a,b)).
Here cost(a,b) = 0 if x[a] = y[b], and otherwise cost(a,b) = 1.
The formula considers the following ways to edit the string x: • distance(a,b −1): insert a character at the end of x • distance(a−1,b): remove the last character from x • distance(a−1,b −1): match or modify the last character of x In the two first cases, one editing operation is needed (insert or remove). In the last case, if x[a] = y[b], we can match the last characters without editing, and otherwise one editing operation is needed (modify).
The following table shows the values of distance in the example case: L O V E M O V I E 0 1 2 3 4 1 1 2 3 4 2 2 1 2 3 3 3 2 1 2 4 4 3 2 2 5 5 4 3 2 1The distance is named after V. I. Levenshtein who studied it in connection with binary codes .
74 The lower-right corner of the table tells us that the edit distance between LOVE and MOVIE is 2. The table also shows how to construct the shortest sequence of editing operations. In this case the path is as follows: L O V E M O V I E 0 1 2 3 4 1 1 2 3 4 2 2 1 2 3 3 3 2 1 2 4 4 3 2 2 5 5 4 3 2 The last characters of LOVE and MOVIE are equal, so the edit distance between them equals the edit distance between LOV and MOVI. We can use one editing operation to remove the character I from MOVI. Thus, the edit distance is one larger than the edit distance between LOV and MOV, etc.
7.6 Counting tilings Sometimes the states of a dynamic programming solution are more complex than fixed combinations of numbers. As an example, consider the problem of calculating the number of distinct ways to fill an n× m grid using 1×2 and 2×1 size tiles. For example, one valid solution for the 4×7 grid is and the total number of solutions is 781.
The problem can be solved using dynamic programming by going through the grid row by row. Each row in a solution can be represented as a string that contains m characters from the set {⊓,⊔,⊏,⊐}. For example, the above solution consists of four rows that correspond to the following strings: • ⊓⊏⊐⊓⊏⊐⊓ • ⊔⊏⊐⊔⊓⊓⊔ • ⊏⊐⊏⊐⊔⊔⊓ • ⊏⊐⊏⊐⊏⊐⊔ Let count(k,x) denote the number of ways to construct a solution for rows 1...k of the grid such that string x corresponds to row k. It is possible to use dynamic programming here, because the state of a row is constrained only by the state of the previous row.
75 A solution is valid if row 1 does not contain the character ⊔, row n does not contain the character ⊓, and all consecutive rows are compatible. For example, the rows ⊔⊏⊐⊔⊓⊓⊔and ⊏⊐⊏⊐⊔⊔⊓are compatible, while the rows ⊓⊏⊐⊓⊏⊐⊓ and ⊏⊐⊏⊐⊏⊐⊔are not compatible.
Since a row consists of m characters and there are four choices for each character, the number of distinct rows is at most 4m. Thus, the time complexity of the solution is O(n42m) because we can go through the O(4m) possible states for each row, and for each state, there are O(4m) possible states for the previous row. In practice, it is a good idea to rotate the grid so that the shorter side has length m, because the factor 42m dominates the time complexity.
It is possible to make the solution more efficient by using a more compact representation for the rows. It turns out that it is sufficient to know which columns of the previous row contain the upper square of a vertical tile. Thus, we can represent a row using only characters ⊓and □, where □is a combination of characters ⊔, ⊏and ⊐. Using this representation, there are only 2m distinct rows and the time complexity is O(n22m).
As a final note, there is also a surprising direct formula for calculating the number of tilings2: ⌈n/2⌉ Y a=1 ⌈m/2⌉ Y b=1 4·(cos2 πa n+1 +cos2 πb m+1) This formula is very efficient, because it calculates the number of tilings in O(nm) time, but since the answer is a product of real numbers, a problem when using the formula is how to store the intermediate results accurately.
2Surprisingly, this formula was discovered in 1961 by two research teams [43, 67] that worked independently.
76 Chapter 8 Amortized analysis The time complexity of an algorithm is often easy to analyze just by examining the structure of the algorithm: what loops does the algorithm contain and how many times the loops are performed. However, sometimes a straightforward analysis does not give a true picture of the efficiency of the algorithm.
Amortized analysis can be used to analyze algorithms that contain opera-tions whose time complexity varies. The idea is to estimate the total time used to all such operations during the execution of the algorithm, instead of focusing on individual operations.
8.1 Two pointers method In the two pointers method, two pointers are used to iterate through the array values. Both pointers can move to one direction only, which ensures that the algorithm works efficiently. Next we discuss two problems that can be solved using the two pointers method.
Subarray sum As the first example, consider a problem where we are given an array of n positive integers and a target sum x, and we want to find a subarray whose sum is x or report that there is no such subarray.
For example, the array 1 3 2 5 1 1 2 3 contains a subarray whose sum is 8: 1 3 2 5 1 1 2 3 This problem can be solved in O(n) time by using the two pointers method.
The idea is to maintain pointers that point to the first and last value of a subarray.
On each turn, the left pointer moves one step to the right, and the right pointer moves to the right as long as the resulting subarray sum is at most x. If the sum becomes exactly x, a solution has been found.
77 As an example, consider the following array and a target sum x = 8: 1 3 2 5 1 1 2 3 The initial subarray contains the values 1, 3 and 2 whose sum is 6: 1 3 2 5 1 1 2 3 Then, the left pointer moves one step to the right. The right pointer does not move, because otherwise the subarray sum would exceed x.
1 3 2 5 1 1 2 3 Again, the left pointer moves one step to the right, and this time the right pointer moves three steps to the right. The subarray sum is 2+5+1 = 8, so a subarray whose sum is x has been found.
1 3 2 5 1 1 2 3 The running time of the algorithm depends on the number of steps the right pointer moves. While there is no useful upper bound on how many steps the pointer can move on a single turn. we know that the pointer moves a total of O(n) steps during the algorithm, because it only moves to the right.
Since both the left and right pointer move O(n) steps during the algorithm, the algorithm works in O(n) time.
2SUM problem Another problem that can be solved using the two pointers method is the following problem, also known as the 2SUM problem: given an array of n numbers and a target sum x, find two array values such that their sum is x, or report that no such values exist.
To solve the problem, we first sort the array values in increasing order. After that, we iterate through the array using two pointers. The left pointer starts at the first value and moves one step to the right on each turn. The right pointer begins at the last value and always moves to the left until the sum of the left and right value is at most x. If the sum is exactly x, a solution has been found.
For example, consider the following array and a target sum x = 12: 1 4 5 6 7 9 9 10 The initial positions of the pointers are as follows. The sum of the values is 1+10 = 11 that is smaller than x.
78 1 4 5 6 7 9 9 10 Then the left pointer moves one step to the right. The right pointer moves three steps to the left, and the sum becomes 4+7 = 11.
1 4 5 6 7 9 9 10 After this, the left pointer moves one step to the right again. The right pointer does not move, and a solution 5+7 = 12 has been found.
1 4 5 6 7 9 9 10 The running time of the algorithm is O(nlogn), because it first sorts the array in O(nlogn) time, and then both pointers move O(n) steps.
Note that it is possible to solve the problem in another way in O(nlogn) time using binary search. In such a solution, we iterate through the array and for each array value, we try to find another value that yields the sum x. This can be done by performing n binary searches, each of which takes O(logn) time.
A more difficult problem is the 3SUM problem that asks to find three array values whose sum is x. Using the idea of the above algorithm, this problem can be solved in O(n2) time1. Can you see how?
8.2 Nearest smaller elements Amortized analysis is often used to estimate the number of operations performed on a data structure. The operations may be distributed unevenly so that most operations occur during a certain phase of the algorithm, but the total number of the operations is limited.
As an example, consider the problem of finding for each array element the nearest smaller element, i.e., the first smaller element that precedes the element in the array. It is possible that no such element exists, in which case the algorithm should report this. Next we will see how the problem can be efficiently solved using a stack structure.
We go through the array from left to right and maintain a stack of array elements. At each array position, we remove elements from the stack until the top element is smaller than the current element, or the stack is empty. Then, we report that the top element is the nearest smaller element of the current element, or if the stack is empty, there is no such element. Finally, we add the current element to the stack.
As an example, consider the following array: 1For a long time, it was thought that solving the 3SUM problem more efficiently than in O(n2) time would not be possible. However, in 2014, it turned out that this is not the case.
79 1 3 4 2 5 3 4 2 First, the elements 1, 3 and 4 are added to the stack, because each element is larger than the previous element. Thus, the nearest smaller element of 4 is 3, and the nearest smaller element of 3 is 1.
1 3 4 2 5 3 4 2 1 3 4 The next element 2 is smaller than the two top elements in the stack. Thus, the elements 3 and 4 are removed from the stack, and then the element 2 is added to the stack. Its nearest smaller element is 1: 1 3 4 2 5 3 4 2 1 2 Then, the element 5 is larger than the element 2, so it will be added to the stack, and its nearest smaller element is 2: 1 3 4 2 5 3 4 2 1 2 5 After this, the element 5 is removed from the stack and the elements 3 and 4 are added to the stack: 1 3 4 2 5 3 4 2 1 2 3 4 Finally, all elements except 1 are removed from the stack and the last element 2 is added to the stack: 1 3 4 2 5 3 4 2 1 2 The efficiency of the algorithm depends on the total number of stack opera-tions. If the current element is larger than the top element in the stack, it is directly added to the stack, which is efficient. However, sometimes the stack can contain several larger elements and it takes time to remove them. Still, each element is added exactly once to the stack and removed at most once from the stack. Thus, each element causes O(1) stack operations, and the algorithm works in O(n) time.
80 8.3 Sliding window minimum A sliding window is a constant-size subarray that moves from left to right through the array. At each window position, we want to calculate some infor-mation about the elements inside the window. In this section, we focus on the problem of maintaining the sliding window minimum, which means that we should report the smallest value inside each window.
The sliding window minimum can be calculated using a similar idea that we used to calculate the nearest smaller elements. We maintain a queue where each element is larger than the previous element, and the first element always corresponds to the minimum element inside the window. After each window move, we remove elements from the end of the queue until the last queue element is smaller than the new window element, or the queue becomes empty. We also remove the first queue element if it is not inside the window anymore. Finally, we add the new window element to the end of the queue.
As an example, consider the following array: 2 1 4 5 3 4 1 2 Suppose that the size of the sliding window is 4. At the first window position, the smallest value is 1: 2 1 4 5 3 4 1 2 1 4 5 Then the window moves one step right. The new element 3 is smaller than the elements 4 and 5 in the queue, so the elements 4 and 5 are removed from the queue and the element 3 is added to the queue. The smallest value is still 1.
2 1 4 5 3 4 1 2 1 3 After this, the window moves again, and the smallest element 1 does not belong to the window anymore. Thus, it is removed from the queue and the smallest value is now 3. Also the new element 4 is added to the queue.
2 1 4 5 3 4 1 2 3 4 The next new element 1 is smaller than all elements in the queue. Thus, all elements are removed from the queue and it will only contain the element 1: 2 1 4 5 3 4 1 2 1 81 Finally the window reaches its last position. The element 2 is added to the queue, but the smallest value inside the window is still 1.
2 1 4 5 3 4 1 2 1 2 Since each array element is added to the queue exactly once and removed from the queue at most once, the algorithm works in O(n) time.
82 Chapter 9 Range queries In this chapter, we discuss data structures that allow us to efficiently process range queries. In a range query, our task is to calculate a value based on a subarray of an array. Typical range queries are: • sumq(a,b): calculate the sum of values in range [a,b] • minq(a,b): find the minimum value in range [a,b] • maxq(a,b): find the maximum value in range [a,b] For example, consider the range [3,6] in the following array: 1 3 8 4 6 1 3 4 0 1 2 3 4 5 6 7 In this case, sumq(3,6) = 14, minq(3,6) = 1 and maxq(3,6) = 6.
A simple way to process range queries is to use a loop that goes through all array values in the range. For example, the following function can be used to process sum queries on an array: int sum(int a, int b) { int s = 0; for (int i = a; i <= b; i++) { s += array[i]; } return s; } This function works in O(n) time, where n is the size of the array. Thus, we can process q queries in O(nq) time using the function. However, if both n and q are large, this approach is slow. Fortunately, it turns out that there are ways to process range queries much more efficiently.
83 9.1 Static array queries We first focus on a situation where the array is static, i.e., the array values are never updated between the queries. In this case, it suffices to construct a static data structure that tells us the answer for any possible query.
Sum queries We can easily process sum queries on a static array by constructing a prefix sum array. Each value in the prefix sum array equals the sum of values in the original array up to that position, i.e., the value at position k is sumq(0,k). The prefix sum array can be constructed in O(n) time.
For example, consider the following array: 1 3 4 8 6 1 4 2 0 1 2 3 4 5 6 7 The corresponding prefix sum array is as follows: 1 4 8 16 22 23 27 29 0 1 2 3 4 5 6 7 Since the prefix sum array contains all values of sumq(0,k), we can calculate any value of sumq(a,b) in O(1) time as follows: sumq(a,b) = sumq(0,b)−sumq(0,a−1) By defining sumq(0,−1) = 0, the above formula also holds when a = 0.
For example, consider the range [3,6]: 1 3 4 8 6 1 4 2 0 1 2 3 4 5 6 7 In this case sumq(3,6) = 8+6+1+4 = 19. This sum can be calculated from two values of the prefix sum array: 1 4 8 16 22 23 27 29 0 1 2 3 4 5 6 7 Thus, sumq(3,6) = sumq(0,6)−sumq(0,2) = 27−8 = 19.
It is also possible to generalize this idea to higher dimensions. For example, we can construct a two-dimensional prefix sum array that can be used to calculate the sum of any rectangular subarray in O(1) time. Each sum in such an array corresponds to a subarray that begins at the upper-left corner of the array.
84 The following picture illustrates the idea: A B C D The sum of the gray subarray can be calculated using the formula S(A)−S(B)−S(C)+ S(D), where S(X) denotes the sum of values in a rectangular subarray from the upper-left corner to the position of X.
Minimum queries Minimum queries are more difficult to process than sum queries. Still, there is a quite simple O(nlogn) time preprocessing method after which we can answer any minimum query in O(1) time1. Note that since minimum and maximum queries can be processed similarly, we can focus on minimum queries.
The idea is to precalculate all values of minq(a,b) where b −a+1 (the length of the range) is a power of two. For example, for the array 1 3 4 8 6 1 4 2 0 1 2 3 4 5 6 7 the following values are calculated: a b minq(a,b) 0 0 1 1 1 3 2 2 4 3 3 8 4 4 6 5 5 1 6 6 4 7 7 2 a b minq(a,b) 0 1 1 1 2 3 2 3 4 3 4 6 4 5 1 5 6 1 6 7 2 a b minq(a,b) 0 3 1 1 4 3 2 5 1 3 6 1 4 7 1 0 7 1 The number of precalculated values is O(nlogn), because there are O(logn) range lengths that are powers of two. The values can be calculated efficiently using the recursive formula minq(a,b) = min(minq(a,a+ w−1),minq(a+ w,b)), 1This technique was introduced in and sometimes called the sparse table method. There are also more sophisticated techniques where the preprocessing time is only O(n), but such algorithms are not needed in competitive programming.
85 where b−a+1 is a power of two and w = (b−a+1)/2. Calculating all those values takes O(nlogn) time.
After this, any value of minq(a,b) can be calculated in O(1) time as a minimum of two precalculated values. Let k be the largest power of two that does not exceed b −a+1. We can calculate the value of minq(a,b) using the formula minq(a,b) = min(minq(a,a+ k −1),minq(b −k +1,b)).
In the above formula, the range [a,b] is represented as the union of the ranges [a,a+ k −1] and [b −k +1,b], both of length k.
As an example, consider the range [1,6]: 1 3 4 8 6 1 4 2 0 1 2 3 4 5 6 7 The length of the range is 6, and the largest power of two that does not exceed 6 is 4. Thus the range [1,6] is the union of the ranges [1,4] and [3,6]: 1 3 4 8 6 1 4 2 0 1 2 3 4 5 6 7 1 3 4 8 6 1 4 2 0 1 2 3 4 5 6 7 Since minq(1,4) = 3 and minq(3,6) = 1, we conclude that minq(1,6) = 1.
9.2 Binary indexed tree A binary indexed tree or a Fenwick tree2 can be seen as a dynamic variant of a prefix sum array. It supports two O(logn) time operations on an array: processing a range sum query and updating a value.
The advantage of a binary indexed tree is that it allows us to efficiently update array values between sum queries. This would not be possible using a prefix sum array, because after each update, it would be necessary to build the whole prefix sum array again in O(n) time.
Structure Even if the name of the structure is a binary indexed tree, it is usually represented as an array. In this section we assume that all arrays are one-indexed, because it makes the implementation easier.
Let p(k) denote the largest power of two that divides k. We store a binary indexed tree as an array tree such that tree[k] = sumq(k −p(k)+1,k), 2The binary indexed tree structure was presented by P. M. Fenwick in 1994 .
86 i.e., each position k contains the sum of values in a range of the original array whose length is p(k) and that ends at position k. For example, since p(6) = 2, tree contains the value of sumq(5,6).
For example, consider the following array: 1 3 4 8 6 1 4 2 1 2 3 4 5 6 7 8 The corresponding binary indexed tree is as follows: 1 4 4 16 6 7 4 29 1 2 3 4 5 6 7 8 The following picture shows more clearly how each value in the binary indexed tree corresponds to a range in the original array: 1 4 4 16 6 7 4 29 1 2 3 4 5 6 7 8 Using a binary indexed tree, any value of sumq(1,k) can be calculated in O(logn) time, because a range [1,k] can always be divided into O(logn) ranges whose sums are stored in the tree.
For example, the range [1,7] consists of the following ranges: 1 4 4 16 6 7 4 29 1 2 3 4 5 6 7 8 Thus, we can calculate the corresponding sum as follows: sumq(1,7) = sumq(1,4)+sumq(5,6)+sumq(7,7) = 16+7+4 = 27 To calculate the value of sumq(a,b) where a > 1, we can use the same trick that we used with prefix sum arrays: sumq(a,b) = sumq(1,b)−sumq(1,a−1).
87 Since we can calculate both sumq(1,b) and sumq(1,a−1) in O(logn) time, the total time complexity is O(logn).
Then, after updating a value in the original array, several values in the binary indexed tree should be updated. For example, if the value at position 3 changes, the sums of the following ranges change: 1 4 4 16 6 7 4 29 1 2 3 4 5 6 7 8 Since each array element belongs to O(logn) ranges in the binary indexed tree, it suffices to update O(logn) values in the tree.
Implementation The operations of a binary indexed tree can be efficiently implemented using bit operations. The key fact needed is that we can calculate any value of p(k) using the formula p(k) = k&−k.
The following function calculates the value of sumq(1,k): int sum(int k) { int s = 0; while (k >= 1) { s += tree[k]; k -= k&-k; } return s; } The following function increases the array value at position k by x (x can be positive or negative): void add(int k, int x) { while (k <= n) { tree[k] += x; k += k&-k; } } The time complexity of both the functions is O(logn), because the functions access O(logn) values in the binary indexed tree, and each move to the next position takes O(1) time.
88 9.3 Segment tree A segment tree3 is a data structure that supports two operations: processing a range query and updating an array value. Segment trees can support sum queries, minimum and maximum queries and many other queries so that both operations work in O(logn) time.
Compared to a binary indexed tree, the advantage of a segment tree is that it is a more general data structure. While binary indexed trees only support sum queries4, segment trees also support other queries. On the other hand, a segment tree requires more memory and is a bit more difficult to implement.
Structure A segment tree is a binary tree such that the nodes on the bottom level of the tree correspond to the array elements, and the other nodes contain information needed for processing range queries.
In this section, we assume that the size of the array is a power of two and zero-based indexing is used, because it is convenient to build a segment tree for such an array. If the size of the array is not a power of two, we can always append extra elements to it.
We will first discuss segment trees that support sum queries. As an example, consider the following array: 5 8 6 3 2 7 2 6 0 1 2 3 4 5 6 7 The corresponding segment tree is as follows: 5 8 6 3 2 7 2 6 13 9 9 8 22 17 39 Each internal tree node corresponds to an array range whose size is a power of two. In the above tree, the value of each internal node is the sum of the corresponding array values, and it can be calculated as the sum of the values of its left and right child node.
3The bottom-up-implementation in this chapter corresponds to that in . Similar structures were used in late 1970’s to solve geometric problems .
4In fact, using two binary indexed trees it is possible to support minimum queries , but this is more complicated than to use a segment tree.
89 It turns out that any range [a,b] can be divided into O(logn) ranges whose values are stored in tree nodes. For example, consider the range [2,7]: 5 8 6 3 2 7 2 6 0 1 2 3 4 5 6 7 Here sumq(2,7) = 6+3+2+7+2+6 = 26. In this case, the following two tree nodes correspond to the range: 5 8 6 3 2 7 2 6 13 9 9 8 22 17 39 Thus, another way to calculate the sum is 9+17 = 26.
When the sum is calculated using nodes located as high as possible in the tree, at most two nodes on each level of the tree are needed. Hence, the total number of nodes is O(logn).
After an array update, we should update all nodes whose value depends on the updated value. This can be done by traversing the path from the updated array element to the top node and updating the nodes along the path.
The following picture shows which tree nodes change if the array value 7 changes: 5 8 6 3 2 7 2 6 13 9 9 8 22 17 39 The path from bottom to top always consists of O(logn) nodes, so each update changes O(logn) nodes in the tree.
Implementation We store a segment tree as an array of 2n elements where n is the size of the original array and a power of two. The tree nodes are stored from top to bottom: 90 tree is the top node, tree and tree are its children, and so on. Finally, the values from tree[n] to tree[2n−1] correspond to the values of the original array on the bottom level of the tree.
For example, the segment tree 5 8 6 3 2 7 2 6 13 9 9 8 22 17 39 is stored as follows: 39 22 17 13 9 9 8 5 8 6 3 2 7 2 6 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Using this representation, the parent of tree[k] is tree[⌊k/2⌋], and its children are tree[2k] and tree[2k +1]. Note that this implies that the position of a node is even if it is a left child and odd if it is a right child.
The following function calculates the value of sumq(a,b): int sum(int a, int b) { a += n; b += n; int s = 0; while (a <= b) { if (a%2 == 1) s += tree[a++]; if (b%2 == 0) s += tree[b--]; a /= 2; b /= 2; } return s; } The function maintains a range that is initially [a+ n,b + n]. Then, at each step, the range is moved one level higher in the tree, and before that, the values of the nodes that do not belong to the higher range are added to the sum.
The following function increases the array value at position k by x: void add(int k, int x) { k += n; tree[k] += x; for (k /= 2; k >= 1; k /= 2) { tree[k] = tree[2k]+tree[2k+1]; } } 91 First the function updates the value at the bottom level of the tree. After this, the function updates the values of all internal tree nodes, until it reaches the top node of the tree.
Both the above functions work in O(logn) time, because a segment tree of n elements consists of O(logn) levels, and the functions move one level higher in the tree at each step.
Other queries Segment trees can support all range queries where it is possible to divide a range into two parts, calculate the answer separately for both parts and then efficiently combine the answers. Examples of such queries are minimum and maximum, greatest common divisor, and bit operations and, or and xor.
For example, the following segment tree supports minimum queries: 5 8 6 3 1 7 2 6 5 3 1 2 3 1 1 In this case, every tree node contains the smallest value in the corresponding array range. The top node of the tree contains the smallest value in the whole array. The operations can be implemented like previously, but instead of sums, minima are calculated.
The structure of a segment tree also allows us to use binary search for locating array elements. For example, if the tree supports minimum queries, we can find the position of an element with the smallest value in O(logn) time.
For example, in the above tree, an element with the smallest value 1 can be found by traversing a path downwards from the top node: 5 8 6 3 1 7 2 6 5 3 1 2 3 1 1 92 9.4 Additional techniques Index compression A limitation in data structures that are built upon an array is that the elements are indexed using consecutive integers. Difficulties arise when large indices are needed. For example, if we wish to use the index 109, the array should contain 109 elements which would require too much memory.
However, we can often bypass this limitation by using index compression, where the original indices are replaced with indices 1,2,3, etc. This can be done if we know all the indices needed during the algorithm beforehand.
The idea is to replace each original index x with c(x) where c is a function that compresses the indices. We require that the order of the indices does not change, so if a < b, then c(a) < c(b). This allows us to conveniently perform queries even if the indices are compressed.
For example, if the original indices are 555, 109 and 8, the new indices are: c(8) = 1 c(555) = 2 c(109) = 3 Range updates So far, we have implemented data structures that support range queries and updates of single values. Let us now consider an opposite situation, where we should update ranges and retrieve single values. We focus on an operation that increases all elements in a range [a,b] by x.
Surprisingly, we can use the data structures presented in this chapter also in this situation. To do this, we build a difference array whose values indicate the differences between consecutive values in the original array. Thus, the original array is the prefix sum array of the difference array. For example, consider the following array: 3 3 1 1 1 5 2 2 0 1 2 3 4 5 6 7 The difference array for the above array is as follows: 3 0 −2 0 0 4 −3 0 0 1 2 3 4 5 6 7 For example, the value 2 at position 6 in the original array corresponds to the sum 3−2+4−3 = 2 in the difference array.
The advantage of the difference array is that we can update a range in the original array by changing just two elements in the difference array. For example, if we want to increase the original array values between positions 1 and 4 by 5, it suffices to increase the difference array value at position 1 by 5 and decrease the value at position 5 by 5. The result is as follows: 93 3 5 −2 0 0 −1 −3 0 0 1 2 3 4 5 6 7 More generally, to increase the values in range [a,b] by x, we increase the value at position a by x and decrease the value at position b +1 by x. Thus, it is only needed to update single values and process sum queries, so we can use a binary indexed tree or a segment tree.
A more difficult problem is to support both range queries and range updates.
In Chapter 28 we will see that even this is possible.
94 Chapter 10 Bit manipulation All data in computer programs is internally stored as bits, i.e., as numbers 0 and 1. This chapter discusses the bit representation of integers, and shows examples of how to use bit operations. It turns out that there are many uses for bit manipulation in algorithm programming.
10.1 Bit representation In programming, an n bit integer is internally stored as a binary number that consists of n bits. For example, the C++ type int is a 32-bit type, which means that every int number consists of 32 bits.
Here is the bit representation of the int number 43: 00000000000000000000000000101011 The bits in the representation are indexed from right to left. To convert a bit representation bk ···b2b1b0 into a number, we can use the formula bk2k +...+ b222 + b121 + b020.
For example, 1·25 +1·23 +1·21 +1·20 = 43.
The bit representation of a number is either signed or unsigned. Usually a signed representation is used, which means that both negative and positive numbers can be represented. A signed variable of n bits can contain any integer between −2n−1 and 2n−1 −1. For example, the int type in C++ is a signed type, so an int variable can contain any integer between −231 and 231 −1.
The first bit in a signed representation is the sign of the number (0 for nonnegative numbers and 1 for negative numbers), and the remaining n−1 bits contain the magnitude of the number. Two’s complement is used, which means that the opposite number of a number is calculated by first inverting all the bits in the number, and then increasing the number by one.
For example, the bit representation of the int number −43 is 11111111111111111111111111010101.
95 In an unsigned representation, only nonnegative numbers can be used, but the upper bound for the values is larger. An unsigned variable of n bits can contain any integer between 0 and 2n −1. For example, in C++, an unsigned int variable can contain any integer between 0 and 232 −1.
There is a connection between the representations: a signed number −x equals an unsigned number 2n −x. For example, the following code shows that the signed number x = −43 equals the unsigned number y = 232 −43: int x = -43; unsigned int y = x; cout << x << "\n"; // -43 cout << y << "\n"; // 4294967253 If a number is larger than the upper bound of the bit representation, the number will overflow. In a signed representation, the next number after 2n−1 −1 is −2n−1, and in an unsigned representation, the next number after 2n −1 is 0.
For example, consider the following code: int x = 2147483647 cout << x << "\n"; // 2147483647 x++; cout << x << "\n"; // -2147483648 Initially, the value of x is 231 −1. This is the largest value that can be stored in an int variable, so the next number after 231 −1 is −231.
10.2 Bit operations And operation The and operation x & y produces a number that has one bits in positions where both x and y have one bits. For example, 22 & 26 = 18, because 10110 (22) & 11010 (26) = 10010 (18) Using the and operation, we can check if a number x is even because x & 1 = 0 if x is even, and x & 1 = 1 if x is odd. More generally, x is divisible by 2k exactly when x & (2k −1) = 0.
Or operation The or operation x | y produces a number that has one bits in positions where at least one of x and y have one bits. For example, 22 | 26 = 30, because 10110 (22) | 11010 (26) = 11110 (30) 96 Xor operation The xor operation x ^ y produces a number that has one bits in positions where exactly one of x and y have one bits. For example, 22 ^ 26 = 12, because 10110 (22) ^ 11010 (26) = 01100 (12) Not operation The not operation ~x produces a number where all the bits of x have been inverted. The formula ~x = −x−1 holds, for example, ~29 = −30.
The result of the not operation at the bit level depends on the length of the bit representation, because the operation inverts all bits. For example, if the numbers are 32-bit int numbers, the result is as follows: x = 29 00000000000000000000000000011101 ~x = −30 11111111111111111111111111100010 Bit shifts The left bit shift x << k appends k zero bits to the number, and the right bit shift x >> k removes the k last bits from the number. For example, 14 << 2 = 56, because 14 and 56 correspond to 1110 and 111000. Similarly, 49 >> 3 = 6, because 49 and 6 correspond to 110001 and 110.
Note that x << k corresponds to multiplying x by 2k, and x >> k corresponds to dividing x by 2k rounded down to an integer.
Applications A number of the form 1 << k has a one bit in position k and all other bits are zero, so we can use such numbers to access single bits of numbers. In particular, the kth bit of a number is one exactly when x & (1 << k) is not zero. The following code prints the bit representation of an int number x: for (int i = 31; i >= 0; i--) { if (x&(1<<i)) cout << "1"; else cout << "0"; } It is also possible to modify single bits of numbers using similar ideas. For example, the formula x | (1 << k) sets the kth bit of x to one, the formula x & ~(1 << k) sets the kth bit of x to zero, and the formula x ^ (1 << k) inverts the kth bit of x.
The formula x & (x−1) sets the last one bit of x to zero, and the formula x & −x sets all the one bits to zero, except for the last one bit. The formula x | (x−1) inverts all the bits after the last one bit. Also note that a positive number x is a power of two exactly when x & (x−1) = 0.
97 Additional functions The g++ compiler provides the following functions for counting bits: • __builtin_clz(x): the number of zeros at the beginning of the number • __builtin_ctz(x): the number of zeros at the end of the number • __builtin_popcount(x): the number of ones in the number • __builtin_parity(x): the parity (even or odd) of the number of ones The functions can be used as follows: int x = 5328; // 00000000000000000001010011010000 cout << __builtin_clz(x) << "\n"; // 19 cout << __builtin_ctz(x) << "\n"; // 4 cout << __builtin_popcount(x) << "\n"; // 5 cout << __builtin_parity(x) << "\n"; // 1 While the above functions only support int numbers, there are also long long versions of the functions available with the suffix ll.
10.3 Representing sets Every subset of a set {0,1,2,...,n−1} can be represented as an n bit integer whose one bits indicate which elements belong to the subset. This is an efficient way to represent sets, because every element requires only one bit of memory, and set operations can be implemented as bit operations.
For example, since int is a 32-bit type, an int number can represent any subset of the set {0,1,2,...,31}. The bit representation of the set {1,3,4,8} is 00000000000000000000000100011010, which corresponds to the number 28 +24 +23 +21 = 282.
Set implementation The following code declares an int variable x that can contain a subset of {0,1,2,...,31}. After this, the code adds the elements 1, 3, 4 and 8 to the set and prints the size of the set.
int x = 0; x |= (1<<1); x |= (1<<3); x |= (1<<4); x |= (1<<8); cout << __builtin_popcount(x) << "\n"; // 4 98 Then, the following code prints all elements that belong to the set: for (int i = 0; i < 32; i++) { if (x&(1<<i)) cout << i << " "; } // output: 1 3 4 8 Set operations Set operations can be implemented as follows as bit operations: set syntax bit syntax intersection a∩b a & b union a∪b a | b complement ¯ a ~a difference a\ b a & (~b) For example, the following code first constructs the sets x = {1,3,4,8} and y = {3,6,8,9}, and then constructs the set z = x∪y = {1,3,4,6,8,9}: int x = (1<<1)|(1<<3)|(1<<4)|(1<<8); int y = (1<<3)|(1<<6)|(1<<8)|(1<<9); int z = x|y; cout << __builtin_popcount(z) << "\n"; // 6 Iterating through subsets The following code goes through the subsets of {0,1,...,n−1}: for (int b = 0; b < (1<<n); b++) { // process subset b } The following code goes through the subsets with exactly k elements: for (int b = 0; b < (1<<n); b++) { if (__builtin_popcount(b) == k) { // process subset b } } The following code goes through the subsets of a set x: int b = 0; do { // process subset b } while (b=(b-x)&x); 99 10.4 Bit optimizations Many algorithms can be optimized using bit operations. Such optimizations do not change the time complexity of the algorithm, but they may have a large impact on the actual running time of the code. In this section we discuss examples of such situations.
Hamming distances The Hamming distance hamming(a,b) between two strings a and b of equal length is the number of positions where the strings differ. For example, hamming(01101,11001) = 2.
Consider the following problem: Given a list of n bit strings, each of length k, calculate the minimum Hamming distance between two strings in the list. For example, the answer for [00111,01101,11110] is 2, because • hamming(00111,01101) = 2, • hamming(00111,11110) = 3, and • hamming(01101,11110) = 3.
A straightforward way to solve the problem is to go through all pairs of strings and calculate their Hamming distances, which yields an O(n2k) time algorithm.
The following function can be used to calculate distances: int hamming(string a, string b) { int d = 0; for (int i = 0; i < k; i++) { if (a[i] != b[i]) d++; } return d; } However, if k is small, we can optimize the code by storing the bit strings as integers and calculating the Hamming distances using bit operations. In particular, if k ≤32, we can just store the strings as int values and use the following function to calculate distances: int hamming(int a, int b) { return __builtin_popcount(a^b); } In the above function, the xor operation constructs a bit string that has one bits in positions where a and b differ. Then, the number of bits is calculated using the __builtin_popcount function.
To compare the implementations, we generated a list of 10000 random bit strings of length 30. Using the first approach, the search took 13.5 seconds, and after the bit optimization, it only took 0.5 seconds. Thus, the bit optimized code was almost 30 times faster than the original code.
100 Counting subgrids As another example, consider the following problem: Given an n× n grid whose each square is either black (1) or white (0), calculate the number of subgrids whose all corners are black. For example, the grid contains two such subgrids: There is an O(n3) time algorithm for solving the problem: go through all O(n2) pairs of rows and for each pair (a,b) calculate the number of columns that contain a black square in both rows in O(n) time. The following code assumes that color[y][x] denotes the color in row y and column x: int count = 0; for (int i = 0; i < n; i++) { if (color[a][i] == 1 && color[b][i] == 1) count++; } Then, those columns account for count(count−1)/2 subgrids with black corners, because we can choose any two of them to form a subgrid.
To optimize this algorithm, we divide the grid into blocks of columns such that each block consists of N consecutive columns. Then, each row is stored as a list of N-bit numbers that describe the colors of the squares. Now we can process N columns at the same time using bit operations. In the following code, color[y][k] represents a block of N colors as bits.
int count = 0; for (int i = 0; i <= n/N; i++) { count += __builtin_popcount(color[a][i]&color[b][i]); } The resulting algorithm works in O(n3/N) time.
We generated a random grid of size 2500×2500 and compared the original and bit optimized implementation. While the original code took 29.6 seconds, the bit optimized version only took 3.1 seconds with N = 32 (int numbers) and 1.7 seconds with N = 64 (long long numbers).
101 10.5 Dynamic programming Bit operations provide an efficient and convenient way to implement dynamic programming algorithms whose states contain subsets of elements, because such states can be stored as integers. Next we discuss examples of combining bit operations and dynamic programming.
Optimal selection As a first example, consider the following problem: We are given the prices of k products over n days, and we want to buy each product exactly once. However, we are allowed to buy at most one product in a day. What is the minimum total price? For example, consider the following scenario (k = 3 and n = 8): product 0 product 1 product 2 0 1 2 3 4 5 6 7 6 9 5 2 8 9 1 6 8 2 6 2 7 5 7 2 5 3 9 7 3 5 1 4 In this scenario, the minimum total price is 5: product 0 product 1 product 2 0 1 2 3 4 5 6 7 6 9 5 2 8 9 1 6 8 2 6 2 7 5 7 2 5 3 9 7 3 5 1 4 Let price[x][d] denote the price of product x on day d. For example, in the above scenario price = 7. Then, let total(S,d) denote the minimum total price for buying a subset S of products by day d. Using this function, the solution to the problem is total({0...k −1},n−1).
First, total(;,d) = 0, because it does not cost anything to buy an empty set, and total({x},0) = price[x], because there is one way to buy one product on the first day. Then, the following recurrence can be used: total(S,d) = min(total(S,d −1), min x∈S (total(S \ x,d −1)+price[x][d])) This means that we either do not buy any product on day d or buy a product x that belongs to S. In the latter case, we remove x from S and add the price of x to the total price.
The next step is to calculate the values of the function using dynamic pro-gramming. To store the function values, we declare an array int total[1<<K][N]; 102 where K and N are suitably large constants. The first dimension of the array corresponds to a bit representation of a subset.
First, the cases where d = 0 can be processed as follows: for (int x = 0; x < k; x++) { total[1<<x] = price[x]; } Then, the recurrence translates into the following code: for (int d = 1; d < n; d++) { for (int s = 0; s < (1<<k); s++) { total[s][d] = total[s][d-1]; for (int x = 0; x < k; x++) { if (s&(1<<x)) { total[s][d] = min(total[s][d], total[s^(1<<x)][d-1]+price[x][d]); } } } } The time complexity of the algorithm is O(n2kk).
From permutations to subsets Using dynamic programming, it is often possible to change an iteration over permutations into an iteration over subsets1. The benefit of this is that n!, the number of permutations, is much larger than 2n, the number of subsets. For example, if n = 20, then n! ≈2.4·1018 and 2n ≈106. Thus, for certain values of n, we can efficiently go through the subsets but not through the permutations.
As an example, consider the following problem: There is an elevator with maximum weight x, and n people with known weights who want to get from the ground floor to the top floor. What is the minimum number of rides needed if the people enter the elevator in an optimal order?
For example, suppose that x = 10, n = 5 and the weights are as follows: person weight 0 2 1 3 2 3 3 5 4 6 In this case, the minimum number of rides is 2. One optimal order is {0,2,3,1,4}, which partitions the people into two rides: first {0,2,3} (total weight 10), and then {1,4} (total weight 9).
1This technique was introduced in 1962 by M. Held and R. M. Karp .
103 The problem can be easily solved in O(n!n) time by testing all possible permu-tations of n people. However, we can use dynamic programming to get a more efficient O(2nn) time algorithm. The idea is to calculate for each subset of people two values: the minimum number of rides needed and the minimum weight of people who ride in the last group.
Let weight[p] denote the weight of person p. We define two functions: rides(S) is the minimum number of rides for a subset S, and last(S) is the minimum weight of the last ride. For example, in the above scenario rides({1,3,4}) = 2 and last({1,3,4}) = 5, because the optimal rides are {1,4} and {3}, and the second ride has weight 5. Of course, our final goal is to calculate the value of rides({0...n−1}).
We can calculate the values of the functions recursively and then apply dynamic programming. The idea is to go through all people who belong to S and optimally choose the last person p who enters the elevator. Each such choice yields a subproblem for a smaller subset of people. If last(S \ p)+weight[p] ≤x, we can add p to the last ride. Otherwise, we have to reserve a new ride that initially only contains p.
To implement dynamic programming, we declare an array pair best[1<<N]; that contains for each subset S a pair (rides(S),last(S)). We set the value for an empty group as follows: best = {1,0}; Then, we can fill the array as follows: for (int s = 1; s < (1<<n); s++) { // initial value: n+1 rides are needed best[s] = {n+1,0}; for (int p = 0; p < n; p++) { if (s&(1<<p)) { auto option = best[s^(1<<p)]; if (option.second+weight[p] <= x) { // add p to an existing ride option.second += weight[p]; } else { // reserve a new ride for p option.first++; option.second = weight[p]; } best[s] = min(best[s], option); } } } 104 Note that the above loop guarantees that for any two subsets S1 and S2 such that S1 ⊂S2, we process S1 before S2. Thus, the dynamic programming values are calculated in the correct order.
Counting subsets Our last problem in this chapter is as follows: Let X = {0...n−1}, and each subset S ⊂X is assigned an integer value[S]. Our task is to calculate for each S sum(S) = X A⊂S value[A], i.e., the sum of values of subsets of S.
For example, suppose that n = 3 and the values are as follows: • value[;] = 3 • value[{0}] = 1 • value[{1}] = 4 • value[{0,1}] = 5 • value[{2}] = 5 • value[{0,2}] = 1 • value[{1,2}] = 3 • value[{0,1,2}] = 3 In this case, for example, sum({0,2}) = value[;]+value[{0}]+value[{2}]+value[{0,2}] = 3+1+5+1 = 10.
Because there are a total of 2n subsets, one possible solution is to go through all pairs of subsets in O(22n) time. However, using dynamic programming, we can solve the problem in O(2nn) time. The idea is to focus on sums where the elements that may be removed from S are restricted.
Let partial(S,k) denote the sum of values of subsets of S with the restriction that only elements 0...k may be removed from S. For example, partial({0,2},1) = value[{2}]+value[{0,2}], because we may only remove elements 0...1. We can calculate values of sum using values of partial, because sum(S) = partial(S,n−1).
The base cases for the function are partial(S,−1) = value[S], because in this case no elements can be removed from S. Then, in the general case we can use the following recurrence: partial(S,k) = ( partial(S,k −1) k ∉S partial(S,k −1)+partial(S {k},k −1) k ∈S 105 Here we focus on the element k. If k ∈S, we have two options: we may either keep k in S or remove it from S.
There is a particularly clever way to implement the calculation of sums. We can declare an array int sum[1<<N]; that will contain the sum of each subset. The array is initialized as follows: for (int s = 0; s < (1<<n); s++) { sum[s] = value[s]; } Then, we can fill the array as follows: for (int k = 0; k < n; k++) { for (int s = 0; s < (1<<n); s++) { if (s&(1<<k)) sum[s] += sum[s^(1<<k)]; } } This code calculates the values of partial(S,k) for k = 0...n−1 to the array sum.
Since partial(S,k) is always based on partial(S,k −1), we can reuse the array sum, which yields a very efficient implementation.
106 Part II Graph algorithms 107 Chapter 11 Basics of graphs Many programming problems can be solved by modeling the problem as a graph problem and using an appropriate graph algorithm. A typical example of a graph is a network of roads and cities in a country. Sometimes, though, the graph is hidden in the problem and it may be difficult to detect it.
This part of the book discusses graph algorithms, especially focusing on topics that are important in competitive programming. In this chapter, we go through concepts related to graphs, and study different ways to represent graphs in algorithms.
11.1 Graph terminology A graph consists of nodes and edges. In this book, the variable n denotes the number of nodes in a graph, and the variable m denotes the number of edges.
The nodes are numbered using integers 1,2,...,n.
For example, the following graph consists of 5 nodes and 7 edges: 1 2 3 4 5 A path leads from node a to node b through edges of the graph. The length of a path is the number of edges in it. For example, the above graph contains a path 1 →3 →4 →5 of length 3 from node 1 to node 5: 1 2 3 4 5 A path is a cycle if the first and last node is the same. For example, the above graph contains a cycle 1 →3 →4 →1. A path is simple if each node appears at most once in the path.
109 Connectivity A graph is connected if there is a path between any two nodes. For example, the following graph is connected: 1 2 3 4 The following graph is not connected, because it is not possible to get from node 4 to any other node: 1 2 3 4 The connected parts of a graph are called its components. For example, the following graph contains three components: {1, 2, 3}, {4, 5, 6, 7} and {8}.
1 2 3 6 7 4 5 8 A tree is a connected graph that consists of n nodes and n−1 edges. There is a unique path between any two nodes of a tree. For example, the following graph is a tree: 1 2 3 4 5 Edge directions A graph is directed if the edges can be traversed in one direction only. For example, the following graph is directed: 1 2 3 4 5 The above graph contains a path 3 →1 →2 →5 from node 3 to node 5, but there is no path from node 5 to node 3.
110 Edge weights In a weighted graph, each edge is assigned a weight. The weights are often interpreted as edge lengths. For example, the following graph is weighted: 1 2 3 4 5 5 1 7 6 7 3 The length of a path in a weighted graph is the sum of the edge weights on the path. For example, in the above graph, the length of the path 1 →2 →5 is 12, and the length of the path 1 →3 →4 →5 is 11. The latter path is the shortest path from node 1 to node 5.
Neighbors and degrees Two nodes are neighbors or adjacent if there is an edge between them. The degree of a node is the number of its neighbors. For example, in the following graph, the neighbors of node 2 are 1, 4 and 5, so its degree is 3.
1 2 3 4 5 The sum of degrees in a graph is always 2m, where m is the number of edges, because each edge increases the degree of exactly two nodes by one. For this reason, the sum of degrees is always even.
A graph is regular if the degree of every node is a constant d. A graph is complete if the degree of every node is n−1, i.e., the graph contains all possible edges between the nodes.
In a directed graph, the indegree of a node is the number of edges that end at the node, and the outdegree of a node is the number of edges that start at the node. For example, in the following graph, the indegree of node 2 is 2, and the outdegree of node 2 is 1.
1 2 3 4 5 111 Colorings In a coloring of a graph, each node is assigned a color so that no adjacent nodes have the same color.
A graph is bipartite if it is possible to color it using two colors. It turns out that a graph is bipartite exactly when it does not contain a cycle with an odd number of edges. For example, the graph 2 3 5 6 4 1 is bipartite, because it can be colored as follows: 2 3 5 6 4 1 However, the graph 2 3 5 6 4 1 is not bipartite, because it is not possible to color the following cycle of three nodes using two colors: 2 3 5 6 4 1 Simplicity A graph is simple if no edge starts and ends at the same node, and there are no multiple edges between two nodes. Often we assume that graphs are simple. For example, the following graph is not simple: 2 3 5 6 4 1 112 11.2 Graph representation There are several ways to represent graphs in algorithms. The choice of a data structure depends on the size of the graph and the way the algorithm processes it. Next we will go through three common representations.
Adjacency list representation In the adjacency list representation, each node x in the graph is assigned an adjacency list that consists of nodes to which there is an edge from x. Adjacency lists are the most popular way to represent graphs, and most algorithms can be efficiently implemented using them.
A convenient way to store the adjacency lists is to declare an array of vectors as follows: vector adj[N]; The constant N is chosen so that all adjacency lists can be stored. For example, the graph 1 2 3 4 can be stored as follows: adj.push_back(2); adj.push_back(3); adj.push_back(4); adj.push_back(4); adj.push_back(1); If the graph is undirected, it can be stored in a similar way, but each edge is added in both directions.
For a weighted graph, the structure can be extended as follows: vector> adj[N]; In this case, the adjacency list of node a contains the pair (b,w) always when there is an edge from node a to node b with weight w. For example, the graph 1 2 3 4 5 7 6 5 2 113 can be stored as follows: adj.push_back({2,5}); adj.push_back({3,7}); adj.push_back({4,6}); adj.push_back({4,5}); adj.push_back({1,2}); The benefit of using adjacency lists is that we can efficiently find the nodes to which we can move from a given node through an edge. For example, the following loop goes through all nodes to which we can move from node s: for (auto u : adj[s]) { // process node u } Adjacency matrix representation An adjacency matrix is a two-dimensional array that indicates which edges the graph contains. We can efficiently check from an adjacency matrix if there is an edge between two nodes. The matrix can be stored as an array int adj[N][N]; where each value adj[a][b] indicates whether the graph contains an edge from node a to node b. If the edge is included in the graph, then adj[a][b] = 1, and otherwise adj[a][b] = 0. For example, the graph 1 2 3 4 can be represented as follows: 1 0 0 0 0 0 0 1 0 0 1 1 0 1 0 0 4 3 2 1 1 2 3 4 If the graph is weighted, the adjacency matrix representation can be extended so that the matrix contains the weight of the edge if the edge exists. Using this representation, the graph 114 1 2 3 4 5 7 6 5 2 corresponds to the following matrix: 2 0 0 0 0 0 0 5 0 0 7 6 0 5 0 0 4 3 2 1 1 2 3 4 The drawback of the adjacency matrix representation is that the matrix contains n2 elements, and usually most of them are zero. For this reason, the representation cannot be used if the graph is large.
Edge list representation An edge list contains all edges of a graph in some order. This is a convenient way to represent a graph if the algorithm processes all edges of the graph and it is not needed to find edges that start at a given node.
The edge list can be stored in a vector vector> edges; where each pair (a,b) denotes that there is an edge from node a to node b. Thus, the graph 1 2 3 4 can be represented as follows: edges.push_back({1,2}); edges.push_back({2,3}); edges.push_back({2,4}); edges.push_back({3,4}); edges.push_back({4,1}); If the graph is weighted, the structure can be extended as follows: 115 vector> edges; Each element in this list is of the form (a,b,w), which means that there is an edge from node a to node b with weight w. For example, the graph 1 2 3 4 5 7 6 5 2 can be represented as follows1: edges.push_back({1,2,5}); edges.push_back({2,3,7}); edges.push_back({2,4,6}); edges.push_back({3,4,5}); edges.push_back({4,1,2}); 1In some older compilers, the function make_tuple must be used instead of the braces (for example, make_tuple(1,2,5) instead of {1,2,5}).
116 Chapter 12 Graph traversal This chapter discusses two fundamental graph algorithms: depth-first search and breadth-first search. Both algorithms are given a starting node in the graph, and they visit all nodes that can be reached from the starting node. The difference in the algorithms is the order in which they visit the nodes.
12.1 Depth-first search Depth-first search (DFS) is a straightforward graph traversal technique. The algorithm begins at a starting node, and proceeds to all other nodes that are reachable from the starting node using the edges of the graph.
Depth-first search always follows a single path in the graph as long as it finds new nodes. After this, it returns to previous nodes and begins to explore other parts of the graph. The algorithm keeps track of visited nodes, so that it processes each node only once.
Example Let us consider how depth-first search processes the following graph: 1 2 3 4 5 We may begin the search at any node of the graph; now we will begin the search at node 1.
The search first proceeds to node 2: 1 2 3 4 5 117 After this, nodes 3 and 5 will be visited: 1 2 3 4 5 The neighbors of node 5 are 2 and 3, but the search has already visited both of them, so it is time to return to the previous nodes. Also the neighbors of nodes 3 and 2 have been visited, so we next move from node 1 to node 4: 1 2 3 4 5 After this, the search terminates because it has visited all nodes.
The time complexity of depth-first search is O(n+ m) where n is the number of nodes and m is the number of edges, because the algorithm processes each node and edge once.
Implementation Depth-first search can be conveniently implemented using recursion. The fol-lowing function dfs begins a depth-first search at a given node. The function assumes that the graph is stored as adjacency lists in an array vector adj[N]; and also maintains an array bool visited[N]; that keeps track of the visited nodes. Initially, each array value is false, and when the search arrives at node s, the value of visited[s] becomes true. The function can be implemented as follows: void dfs(int s) { if (visited[s]) return; visited[s] = true; // process node s for (auto u: adj[s]) { dfs(u); } } 118 12.2 Breadth-first search Breadth-first search (BFS) visits the nodes in increasing order of their distance from the starting node. Thus, we can calculate the distance from the starting node to all other nodes using breadth-first search. However, breadth-first search is more difficult to implement than depth-first search.
Breadth-first search goes through the nodes one level after another. First the search explores the nodes whose distance from the starting node is 1, then the nodes whose distance is 2, and so on. This process continues until all nodes have been visited.
Example Let us consider how breadth-first search processes the following graph: 1 2 3 4 5 6 Suppose that the search begins at node 1. First, we process all nodes that can be reached from node 1 using a single edge: 1 2 3 4 5 6 After this, we proceed to nodes 3 and 5: 1 2 3 4 5 6 Finally, we visit node 6: 1 2 3 4 5 6 119 Now we have calculated the distances from the starting node to all nodes of the graph. The distances are as follows: node distance 1 0 2 1 3 2 4 1 5 2 6 3 Like in depth-first search, the time complexity of breadth-first search is O(n+ m), where n is the number of nodes and m is the number of edges.
Implementation Breadth-first search is more difficult to implement than depth-first search, be-cause the algorithm visits nodes in different parts of the graph. A typical imple-mentation is based on a queue that contains nodes. At each step, the next node in the queue will be processed.
The following code assumes that the graph is stored as adjacency lists and maintains the following data structures: queue q; bool visited[N]; int distance[N]; The queue q contains nodes to be processed in increasing order of their distance. New nodes are always added to the end of the queue, and the node at the beginning of the queue is the next node to be processed. The array visited indicates which nodes the search has already visited, and the array distance will contain the distances from the starting node to all nodes of the graph.
The search can be implemented as follows, starting at node x: visited[x] = true; distance[x] = 0; q.push(x); while (!q.empty()) { int s = q.front(); q.pop(); // process node s for (auto u : adj[s]) { if (visited[u]) continue; visited[u] = true; distance[u] = distance[s]+1; q.push(u); } } 120 12.3 Applications Using the graph traversal algorithms, we can check many properties of graphs.
Usually, both depth-first search and breadth-first search may be used, but in practice, depth-first search is a better choice, because it is easier to implement.
In the following applications we will assume that the graph is undirected.
Connectivity check A graph is connected if there is a path between any two nodes of the graph. Thus, we can check if a graph is connected by starting at an arbitrary node and finding out if we can reach all other nodes.
For example, in the graph 2 1 3 5 4 a depth-first search from node 1 visits the following nodes: 2 1 3 5 4 Since the search did not visit all the nodes, we can conclude that the graph is not connected. In a similar way, we can also find all connected components of a graph by iterating through the nodes and always starting a new depth-first search if the current node does not belong to any component yet.
Finding cycles A graph contains a cycle if during a graph traversal, we find a node whose neighbor (other than the previous node in the current path) has already been visited. For example, the graph 2 1 3 5 4 contains two cycles and we can find one of them as follows: 121 2 1 3 5 4 After moving from node 2 to node 5 we notice that the neighbor 3 of node 5 has already been visited. Thus, the graph contains a cycle that goes through node 3, for example, 3 →2 →5 →3.
Another way to find out whether a graph contains a cycle is to simply calculate the number of nodes and edges in every component. If a component contains c nodes and no cycle, it must contain exactly c −1 edges (so it has to be a tree). If there are c or more edges, the component surely contains a cycle.
Bipartiteness check A graph is bipartite if its nodes can be colored using two colors so that there are no adjacent nodes with the same color. It is surprisingly easy to check if a graph is bipartite using graph traversal algorithms.
The idea is to color the starting node blue, all its neighbors red, all their neighbors blue, and so on. If at some point of the search we notice that two adjacent nodes have the same color, this means that the graph is not bipartite.
Otherwise the graph is bipartite and one coloring has been found.
For example, the graph 2 1 3 5 4 is not bipartite, because a search from node 1 proceeds as follows: 2 1 3 5 4 We notice that the color or both nodes 2 and 5 is red, while they are adjacent nodes in the graph. Thus, the graph is not bipartite.
This algorithm always works, because when there are only two colors avail-able, the color of the starting node in a component determines the colors of all other nodes in the component. It does not make any difference whether the starting node is red or blue.
Note that in the general case, it is difficult to find out if the nodes in a graph can be colored using k colors so that no adjacent nodes have the same color. Even when k = 3, no efficient algorithm is known but the problem is NP-hard.
122 Chapter 13 Shortest paths Finding a shortest path between two nodes of a graph is an important problem that has many practical applications. For example, a natural problem related to a road network is to calculate the shortest possible length of a route between two cities, given the lengths of the roads.
In an unweighted graph, the length of a path equals the number of its edges, and we can simply use breadth-first search to find a shortest path. However, in this chapter we focus on weighted graphs where more sophisticated algorithms are needed for finding shortest paths.
13.1 Bellman–Ford algorithm The Bellman–Ford algorithm1 finds shortest paths from a starting node to all nodes of the graph. The algorithm can process all kinds of graphs, provided that the graph does not contain a cycle with negative length. If the graph contains a negative cycle, the algorithm can detect this.
The algorithm keeps track of distances from the starting node to all nodes of the graph. Initially, the distance to the starting node is 0 and the distance to all other nodes in infinite. The algorithm reduces the distances by finding edges that shorten the paths until it is not possible to reduce any distance.
Example Let us consider how the Bellman–Ford algorithm works in the following graph: 1 2 3 4 6 0 ∞ ∞ ∞ ∞ 5 3 1 3 2 2 7 1The algorithm is named after R. E. Bellman and L. R. Ford who published it independently in 1958 and 1956, respectively [5, 24].
123 Each node of the graph is assigned a distance. Initially, the distance to the starting node is 0, and the distance to all other nodes is infinite.
The algorithm searches for edges that reduce distances. First, all edges from node 1 reduce distances: 1 2 3 4 5 0 5 3 7 ∞ 5 3 1 3 2 2 7 After this, edges 2 →5 and 3 →4 reduce distances: 1 2 3 4 5 0 5 3 4 7 5 3 1 3 2 2 7 Finally, there is one more change: 1 2 3 4 5 0 5 3 4 6 5 3 1 3 2 2 7 After this, no edge can reduce any distance. This means that the distances are final, and we have successfully calculated the shortest distances from the starting node to all nodes of the graph.
For example, the shortest distance 3 from node 1 to node 5 corresponds to the following path: 1 2 3 4 5 0 5 3 4 6 5 3 1 3 2 2 7 124 Implementation The following implementation of the Bellman–Ford algorithm determines the shortest distances from a node x to all nodes of the graph. The code assumes that the graph is stored as an edge list edges that consists of tuples of the form (a,b,w), meaning that there is an edge from node a to node b with weight w.
The algorithm consists of n−1 rounds, and on each round the algorithm goes through all edges of the graph and tries to reduce the distances. The algorithm constructs an array distance that will contain the distances from x to all nodes of the graph. The constant INF denotes an infinite distance.
for (int i = 1; i <= n; i++) distance[i] = INF; distance[x] = 0; for (int i = 1; i <= n-1; i++) { for (auto e : edges) { int a, b, w; tie(a, b, w) = e; distance[b] = min(distance[b], distance[a]+w); } } The time complexity of the algorithm is O(nm), because the algorithm consists of n−1 rounds and iterates through all m edges during a round. If there are no negative cycles in the graph, all distances are final after n−1 rounds, because each shortest path can contain at most n−1 edges.
In practice, the final distances can usually be found faster than in n−1 rounds.
Thus, a possible way to make the algorithm more efficient is to stop the algorithm if no distance can be reduced during a round.
Negative cycles The Bellman–Ford algorithm can also be used to check if the graph contains a cycle with negative length. For example, the graph 1 2 3 4 3 1 5 −7 2 contains a negative cycle 2 →3 →4 →2 with length −4.
If the graph contains a negative cycle, we can shorten infinitely many times any path that contains the cycle by repeating the cycle again and again. Thus, the concept of a shortest path is not meaningful in this situation.
A negative cycle can be detected using the Bellman–Ford algorithm by running the algorithm for n rounds. If the last round reduces any distance, the graph contains a negative cycle. Note that this algorithm can be used to search for a negative cycle in the whole graph regardless of the starting node.
125 SPFA algorithm The SPFA algorithm (”Shortest Path Faster Algorithm”) is a variant of the Bellman–Ford algorithm, that is often more efficient than the original algorithm.
The SPFA algorithm does not go through all the edges on each round, but instead, it chooses the edges to be examined in a more intelligent way.
The algorithm maintains a queue of nodes that might be used for reducing the distances. First, the algorithm adds the starting node x to the queue. Then, the algorithm always processes the first node in the queue, and when an edge a →b reduces a distance, node b is added to the queue.
The efficiency of the SPFA algorithm depends on the structure of the graph: the algorithm is often efficient, but its worst case time complexity is still O(nm) and it is possible to create inputs that make the algorithm as slow as the original Bellman–Ford algorithm.
13.2 Dijkstra’s algorithm Dijkstra’s algorithm2 finds shortest paths from the starting node to all nodes of the graph, like the Bellman–Ford algorithm. The benefit of Dijsktra’s algorithm is that it is more efficient and can be used for processing large graphs. However, the algorithm requires that there are no negative weight edges in the graph.
Like the Bellman–Ford algorithm, Dijkstra’s algorithm maintains distances to the nodes and reduces them during the search. Dijkstra’s algorithm is efficient, because it only processes each edge in the graph once, using the fact that there are no negative edges.
Example Let us consider how Dijkstra’s algorithm works in the following graph when the starting node is node 1: 3 4 2 1 5 ∞ ∞ ∞ 0 ∞ 6 2 5 9 2 1 Like in the Bellman–Ford algorithm, initially the distance to the starting node is 0 and the distance to all other nodes is infinite.
At each step, Dijkstra’s algorithm selects a node that has not been processed yet and whose distance is as small as possible. The first such node is node 1 with distance 0.
2E. W. Dijkstra published the algorithm in 1959 ; however, his original paper does not mention how to implement the algorithm efficiently.
126 When a node is selected, the algorithm goes through all edges that start at the node and reduces the distances using them: 3 4 2 1 5 ∞ 9 5 0 1 6 2 5 9 2 1 In this case, the edges from node 1 reduced the distances of nodes 2, 4 and 5, whose distances are now 5, 9 and 1.
The next node to be processed is node 5 with distance 1. This reduces the distance to node 4 from 9 to 3: 3 4 2 1 5 ∞ 3 5 0 1 6 2 5 9 2 1 After this, the next node is node 4, which reduces the distance to node 3 to 9: 3 4 2 1 5 9 3 5 0 1 6 2 5 9 2 1 A remarkable property in Dijkstra’s algorithm is that whenever a node is selected, its distance is final. For example, at this point of the algorithm, the distances 0, 1 and 3 are the final distances to nodes 1, 5 and 4.
After this, the algorithm processes the two remaining nodes, and the final distances are as follows: 3 4 2 1 5 7 3 5 0 1 6 2 5 9 2 1 127 Negative edges The efficiency of Dijkstra’s algorithm is based on the fact that the graph does not contain negative edges. If there is a negative edge, the algorithm may give incorrect results. As an example, consider the following graph: 1 2 3 4 2 3 6 −5 The shortest path from node 1 to node 4 is 1 →3 →4 and its length is 1. However, Dijkstra’s algorithm finds the path 1 →2 →4 by following the minimum weight edges. The algorithm does not take into account that on the other path, the weight −5 compensates the previous large weight 6.
Implementation The following implementation of Dijkstra’s algorithm calculates the minimum distances from a node x to other nodes of the graph. The graph is stored as adjacency lists so that adj[a] contains a pair (b,w) always when there is an edge from node a to node b with weight w.
An efficient implementation of Dijkstra’s algorithm requires that it is possible to efficiently find the minimum distance node that has not been processed. An appropriate data structure for this is a priority queue that contains the nodes ordered by their distances. Using a priority queue, the next node to be processed can be retrieved in logarithmic time.
In the following code, the priority queue q contains pairs of the form (−d,x), meaning that the current distance to node x is d. The array distance contains the distance to each node, and the array processed indicates whether a node has been processed. Initially the distance is 0 to x and ∞to all other nodes.
for (int i = 1; i <= n; i++) distance[i] = INF; distance[x] = 0; q.push({0,x}); while (!q.empty()) { int a = q.top().second; q.pop(); if (processed[a]) continue; processed[a] = true; for (auto u : adj[a]) { int b = u.first, w = u.second; if (distance[a]+w < distance[b]) { distance[b] = distance[a]+w; q.push({-distance[b],b}); } } } 128 Note that the priority queue contains negative distances to nodes. The reason for this is that the default version of the C++ priority queue finds maximum elements, while we want to find minimum elements. By using negative distances, we can directly use the default priority queue3. Also note that there may be several instances of the same node in the priority queue; however, only the instance with the minimum distance will be processed.
The time complexity of the above implementation is O(n+ mlogm), because the algorithm goes through all nodes of the graph and adds for each edge at most one distance to the priority queue.
13.3 Floyd–Warshall algorithm The Floyd–Warshall algorithm4 provides an alternative way to approach the problem of finding shortest paths. Unlike the other algorithms of this chapter, it finds all shortest paths between the nodes in a single run.
The algorithm maintains a two-dimensional array that contains distances between the nodes. First, distances are calculated only using direct edges between the nodes, and after this, the algorithm reduces distances by using intermediate nodes in paths.
Example Let us consider how the Floyd–Warshall algorithm works in the following graph: 3 4 2 1 5 7 2 5 9 2 1 Initially, the distance from each node to itself is 0, and the distance between nodes a and b is x if there is an edge between nodes a and b with weight x. All other distances are infinite.
In this graph, the initial array is as follows: 1 2 3 4 5 1 0 5 ∞ 9 1 2 5 0 2 ∞ ∞ 3 ∞ 2 0 7 ∞ 4 9 ∞ 7 0 2 5 1 ∞ ∞ 2 0 3Of course, we could also declare the priority queue as in Chapter 4.5 and use positive distances, but the implementation would be a bit longer.
4The algorithm is named after R. W. Floyd and S. Warshall who published it independently in 1962 [23, 70].
129 The algorithm consists of consecutive rounds. On each round, the algorithm selects a new node that can act as an intermediate node in paths from now on, and distances are reduced using this node.
On the first round, node 1 is the new intermediate node. There is a new path between nodes 2 and 4 with length 14, because node 1 connects them. There is also a new path between nodes 2 and 5 with length 6.
1 2 3 4 5 1 0 5 ∞ 9 1 2 5 0 2 14 6 3 ∞ 2 0 7 ∞ 4 9 14 7 0 2 5 1 6 ∞ 2 0 On the second round, node 2 is the new intermediate node. This creates new paths between nodes 1 and 3 and between nodes 3 and 5: 1 2 3 4 5 1 0 5 7 9 1 2 5 0 2 14 6 3 7 2 0 7 8 4 9 14 7 0 2 5 1 6 8 2 0 On the third round, node 3 is the new intermediate round. There is a new path between nodes 2 and 4: 1 2 3 4 5 1 0 5 7 9 1 2 5 0 2 9 6 3 7 2 0 7 8 4 9 9 7 0 2 5 1 6 8 2 0 The algorithm continues like this, until all nodes have been appointed inter-mediate nodes. After the algorithm has finished, the array contains the minimum distances between any two nodes: 1 2 3 4 5 1 0 5 7 3 1 2 5 0 2 8 6 3 7 2 0 7 8 4 3 8 7 0 2 5 1 6 8 2 0 For example, the array tells us that the shortest distance between nodes 2 and 4 is 8. This corresponds to the following path: 130 3 4 2 1 5 7 2 5 9 2 1 Implementation The advantage of the Floyd–Warshall algorithm that it is easy to implement. The following code constructs a distance matrix where distance[a][b] is the shortest distance between nodes a and b. First, the algorithm initializes distance using the adjacency matrix adj of the graph: for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (i == j) distance[i][j] = 0; else if (adj[i][j]) distance[i][j] = adj[i][j]; else distance[i][j] = INF; } } After this, the shortest distances can be found as follows: for (int k = 1; k <= n; k++) { for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { distance[i][j] = min(distance[i][j], distance[i][k]+distance[k][j]); } } } The time complexity of the algorithm is O(n3), because it contains three nested loops that go through the nodes of the graph.
Since the implementation of the Floyd–Warshall algorithm is simple, the algorithm can be a good choice even if it is only needed to find a single shortest path in the graph. However, the algorithm can only be used when the graph is so small that a cubic time complexity is fast enough.
131 132 Chapter 14 Tree algorithms A tree is a connected, acyclic graph that consists of n nodes and n −1 edges.
Removing any edge from a tree divides it into two components, and adding any edge to a tree creates a cycle. Moreover, there is always a unique path between any two nodes of a tree.
For example, the following tree consists of 8 nodes and 7 edges: 1 4 2 3 7 5 6 8 The leaves of a tree are the nodes with degree 1, i.e., with only one neighbor.
For example, the leaves of the above tree are nodes 3, 5, 7 and 8.
In a rooted tree, one of the nodes is appointed the root of the tree, and all other nodes are placed underneath the root. For example, in the following tree, node 1 is the root node.
1 4 2 3 7 5 6 8 In a rooted tree, the children of a node are its lower neighbors, and the parent of a node is its upper neighbor. Each node has exactly one parent, except for the root that does not have a parent. For example, in the above tree, the children of node 2 are nodes 5 and 6, and its parent is node 1.
133 The structure of a rooted tree is recursive: each node of the tree acts as the root of a subtree that contains the node itself and all nodes that are in the subtrees of its children. For example, in the above tree, the subtree of node 2 consists of nodes 2, 5, 6 and 8: 2 5 6 8 14.1 Tree traversal General graph traversal algorithms can be used to traverse the nodes of a tree.
However, the traversal of a tree is easier to implement than that of a general graph, because there are no cycles in the tree and it is not possible to reach a node from multiple directions.
The typical way to traverse a tree is to start a depth-first search at an arbitrary node. The following recursive function can be used: void dfs(int s, int e) { // process node s for (auto u : adj[s]) { if (u != e) dfs(u, s); } } The function is given two parameters: the current node s and the previous node e. The purpose of the parameter e is to make sure that the search only moves to nodes that have not been visited yet.
The following function call starts the search at node x: dfs(x, 0); In the first call e = 0, because there is no previous node, and it is allowed to proceed to any direction in the tree.
Dynamic programming Dynamic programming can be used to calculate some information during a tree traversal. Using dynamic programming, we can, for example, calculate in O(n) time for each node of a rooted tree the number of nodes in its subtree or the length of the longest path from the node to a leaf.
134 As an example, let us calculate for each node s a value count[s]: the number of nodes in its subtree. The subtree contains the node itself and all nodes in the subtrees of its children, so we can calculate the number of nodes recursively using the following code: void dfs(int s, int e) { count[s] = 1; for (auto u : adj[s]) { if (u == e) continue; dfs(u, s); count[s] += count[u]; } } 14.2 Diameter The diameter of a tree is the maximum length of a path between two nodes. For example, consider the following tree: 1 4 2 3 7 5 6 The diameter of this tree is 4, which corresponds to the following path: 1 4 2 3 7 5 6 Note that there may be several maximum-length paths. In the above path, we could replace node 6 with node 5 to obtain another path with length 4.
Next we will discuss two O(n) time algorithms for calculating the diameter of a tree. The first algorithm is based on dynamic programming, and the second algorithm uses two depth-first searches.
Algorithm 1 A general way to approach many tree problems is to first root the tree arbitrarily.
After this, we can try to solve the problem separately for each subtree. Our first algorithm for calculating the diameter is based on this idea.
An important observation is that every path in a rooted tree has a highest point: the highest node that belongs to the path. Thus, we can calculate for each 135 node the length of the longest path whose highest point is the node. One of those paths corresponds to the diameter of the tree.
For example, in the following tree, node 1 is the highest point on the path that corresponds to the diameter: 1 4 2 3 7 5 6 We calculate for each node x two values: • toLeaf(x): the maximum length of a path from x to any leaf • maxLength(x): the maximum length of a path whose highest point is x For example, in the above tree, toLeaf(1) = 2, because there is a path 1 →2 →6, and maxLength(1) = 4, because there is a path 6 →2 →1 →4 →7. In this case, maxLength(1) equals the diameter.
Dynamic programming can be used to calculate the above values for all nodes in O(n) time. First, to calculate toLeaf(x), we go through the children of x, choose a child c with maximum toLeaf(c) and add one to this value. Then, to calculate maxLength(x), we choose two distinct children a and b such that the sum toLeaf(a)+toLeaf(b) is maximum and add two to this sum.
Algorithm 2 Another efficient way to calculate the diameter of a tree is based on two depth-first searches. First, we choose an arbitrary node a in the tree and find the farthest node b from a. Then, we find the farthest node c from b. The diameter of the tree is the distance between b and c.
In the following graph, a, b and c could be: 1 4 2 3 7 5 6 a b c This is an elegant method, but why does it work?
It helps to draw the tree differently so that the path that corresponds to the diameter is horizontal, and all other nodes hang from it: 136 1 4 2 3 7 5 6 a b c x Node x indicates the place where the path from node a joins the path that corresponds to the diameter. The farthest node from a is node b, node c or some other node that is at least as far from node x. Thus, this node is always a valid choice for an endpoint of a path that corresponds to the diameter.
14.3 All longest paths Our next problem is to calculate for every node in the tree the maximum length of a path that begins at the node. This can be seen as a generalization of the tree diameter problem, because the largest of those lengths equals the diameter of the tree. Also this problem can be solved in O(n) time.
As an example, consider the following tree: 1 4 2 3 6 5 Let maxLength(x) denote the maximum length of a path that begins at node x. For example, in the above tree, maxLength(4) = 3, because there is a path 4 →1 →2 →6. Here is a complete table of the values: node x 1 2 3 4 5 6 maxLength(x) 2 2 3 3 3 3 Also in this problem, a good starting point for solving the problem is to root the tree arbitrarily: 1 4 2 3 5 6 The first part of the problem is to calculate for every node x the maximum length of a path that goes through a child of x. For example, the longest path from node 1 goes through its child 2: 137 1 4 2 3 5 6 This part is easy to solve in O(n) time, because we can use dynamic programming as we have done previously.
Then, the second part of the problem is to calculate for every node x the maximum length of a path through its parent p. For example, the longest path from node 3 goes through its parent 1: 1 4 2 3 5 6 At first glance, it seems that we should choose the longest path from p.
However, this does not always work, because the longest path from p may go through x. Here is an example of this situation: 1 4 2 3 5 6 Still, we can solve the second part in O(n) time by storing two maximum lengths for each node x: • maxLength1(x): the maximum length of a path from x • maxLength2(x) the maximum length of a path from x in another direction than the first path For example, in the above graph, maxLength1(1) = 2 using the path 1 →2 →5, and maxLength2(1) = 1 using the path 1 →3.
Finally, if the path that corresponds to maxLength1(p) goes through x, we con-clude that the maximum length is maxLength2(p)+1, and otherwise the maximum length is maxLength1(p)+1.
138 14.4 Binary trees A binary tree is a rooted tree where each node has a left and right subtree. It is possible that a subtree of a node is empty. Thus, every node in a binary tree has zero, one or two children.
For example, the following tree is a binary tree: 1 2 3 4 5 6 7 The nodes of a binary tree have three natural orderings that correspond to different ways to recursively traverse the tree: • pre-order: first process the root, then traverse the left subtree, then traverse the right subtree • in-order: first traverse the left subtree, then process the root, then traverse the right subtree • post-order: first traverse the left subtree, then traverse the right subtree, then process the root For the above tree, the nodes in pre-order are [1,2,4,5,6,3,7], in in-order [4,2,6,5,1,3,7] and in post-order [4,6,5,2,7,3,1].
If we know the pre-order and in-order of a tree, we can reconstruct the exact structure of the tree. For example, the above tree is the only possible tree with pre-order [1,2,4,5,6,3,7] and in-order [4,2,6,5,1,3,7]. In a similar way, the post-order and in-order also determine the structure of a tree.
However, the situation is different if we only know the pre-order and post-order of a tree. In this case, there may be more than one tree that match the orderings. For example, in both of the trees 1 2 1 2 the pre-order is [1,2] and the post-order is [2,1], but the structures of the trees are different.
139 140 Chapter 15 Spanning trees A spanning tree of a graph consists of all nodes of the graph and some of the edges of the graph so that there is a path between any two nodes. Like trees in general, spanning trees are connected and acyclic. Usually there are several ways to construct a spanning tree.
For example, consider the following graph: 1 2 3 4 5 6 3 5 9 5 2 7 6 3 One spanning tree for the graph is as follows: 1 2 3 4 5 6 3 5 9 2 3 The weight of a spanning tree is the sum of its edge weights. For example, the weight of the above spanning tree is 3+5+9+3+2 = 22.
A minimum spanning tree is a spanning tree whose weight is as small as possible. The weight of a minimum spanning tree for the example graph is 20, and such a tree can be constructed as follows: 1 2 3 4 5 6 3 5 2 7 3 141 In a similar way, a maximum spanning tree is a spanning tree whose weight is as large as possible. The weight of a maximum spanning tree for the example graph is 32: 1 2 3 4 5 6 5 9 5 7 6 Note that a graph may have several minimum and maximum spanning trees, so the trees are not unique.
It turns out that several greedy methods can be used to construct minimum and maximum spanning trees. In this chapter, we discuss two algorithms that process the edges of the graph ordered by their weights. We focus on finding minimum spanning trees, but the same algorithms can find maximum spanning trees by processing the edges in reverse order.
15.1 Kruskal’s algorithm In Kruskal’s algorithm1, the initial spanning tree only contains the nodes of the graph and does not contain any edges. Then the algorithm goes through the edges ordered by their weights, and always adds an edge to the tree if it does not create a cycle.
The algorithm maintains the components of the tree. Initially, each node of the graph belongs to a separate component. Always when an edge is added to the tree, two components are joined. Finally, all nodes belong to the same component, and a minimum spanning tree has been found.
Example Let us consider how Kruskal’s algorithm processes the following graph: 1 2 3 4 5 6 3 5 9 5 2 7 6 3 The first step of the algorithm is to sort the edges in increasing order of their weights. The result is the following list: 1The algorithm was published in 1956 by J. B. Kruskal .
142 edge weight 5–6 2 1–2 3 3–6 3 1–5 5 2–3 5 2–5 6 4–6 7 3–4 9 After this, the algorithm goes through the list and adds each edge to the tree if it joins two separate components.
Initially, each node is in its own component: 1 2 3 4 5 6 The first edge to be added to the tree is the edge 5–6 that creates a component {5,6} by joining the components {5} and {6}: 1 2 3 4 5 6 2 After this, the edges 1–2, 3–6 and 1–5 are added in a similar way: 1 2 3 4 5 6 3 5 2 3 After those steps, most components have been joined and there are two components in the tree: {1,2,3,5,6} and {4}.
The next edge in the list is the edge 2–3, but it will not be included in the tree, because nodes 2 and 3 are already in the same component. For the same reason, the edge 2–5 will not be included in the tree.
143 Finally, the edge 4–6 will be included in the tree: 1 2 3 4 5 6 3 5 2 7 3 After this, the algorithm will not add any new edges, because the graph is connected and there is a path between any two nodes. The resulting graph is a minimum spanning tree with weight 2+3+3+5+7 = 20.
Why does this work?
It is a good question why Kruskal’s algorithm works. Why does the greedy strategy guarantee that we will find a minimum spanning tree?
Let us see what happens if the minimum weight edge of the graph is not included in the spanning tree. For example, suppose that a spanning tree for the previous graph would not contain the minimum weight edge 5–6. We do not know the exact structure of such a spanning tree, but in any case it has to contain some edges. Assume that the tree would be as follows: 1 2 3 4 5 6 However, it is not possible that the above tree would be a minimum spanning tree for the graph. The reason for this is that we can remove an edge from the tree and replace it with the minimum weight edge 5–6. This produces a spanning tree whose weight is smaller: 1 2 3 4 5 6 2 For this reason, it is always optimal to include the minimum weight edge in the tree to produce a minimum spanning tree. Using a similar argument, we can show that it is also optimal to add the next edge in weight order to the tree, and so on. Hence, Kruskal’s algorithm works correctly and always produces a minimum spanning tree.
144 Implementation When implementing Kruskal’s algorithm, it is convenient to use the edge list representation of the graph. The first phase of the algorithm sorts the edges in the list in O(mlogm) time. After this, the second phase of the algorithm builds the minimum spanning tree as follows: for (...) { if (!same(a,b)) unite(a,b); } The loop goes through the edges in the list and always processes an edge a–b where a and b are two nodes. Two functions are needed: the function same determines if a and b are in the same component, and the function unite joins the components that contain a and b.
The problem is how to efficiently implement the functions same and unite.
One possibility is to implement the function same as a graph traversal and check if we can get from node a to node b. However, the time complexity of such a function would be O(n+ m) and the resulting algorithm would be slow, because the function same will be called for each edge in the graph.
We will solve the problem using a union-find structure that implements both functions in O(logn) time. Thus, the time complexity of Kruskal’s algorithm will be O(mlogn) after sorting the edge list.
15.2 Union-find structure A union-find structure maintains a collection of sets. The sets are disjoint, so no element belongs to more than one set. Two O(logn) time operations are supported: the unite operation joins two sets, and the find operation finds the representative of the set that contains a given element2.
Structure In a union-find structure, one element in each set is the representative of the set, and there is a chain from any other element of the set to the representative. For example, assume that the sets are {1,4,7}, {5} and {2,3,6,8}: 1 2 3 4 5 6 7 8 2The structure presented here was introduced in 1971 by J. D. Hopcroft and J. D. Ullman .
Later, in 1975, R. E. Tarjan studied a more sophisticated variant of the structure that is discussed in many algorithm textbooks nowadays.
145 In this case the representatives of the sets are 4, 5 and 2. We can find the representative of any element by following the chain that begins at the element.
For example, the element 2 is the representative for the element 6, because we follow the chain 6 →3 →2. Two elements belong to the same set exactly when their representatives are the same.
Two sets can be joined by connecting the representative of one set to the representative of the other set. For example, the sets {1,4,7} and {2,3,6,8} can be joined as follows: 1 2 3 4 6 7 8 The resulting set contains the elements {1,2,3,4,6,7,8}. From this on, the element 2 is the representative for the entire set and the old representative 4 points to the element 2.
The efficiency of the union-find structure depends on how the sets are joined.
It turns out that we can follow a simple strategy: always connect the representa-tive of the smaller set to the representative of the larger set (or if the sets are of equal size, we can make an arbitrary choice). Using this strategy, the length of any chain will be O(logn), so we can find the representative of any element efficiently by following the corresponding chain.
Implementation The union-find structure can be implemented using arrays. In the following implementation, the array link contains for each element the next element in the chain or the element itself if it is a representative, and the array size indicates for each representative the size of the corresponding set.
Initially, each element belongs to a separate set: for (int i = 1; i <= n; i++) link[i] = i; for (int i = 1; i <= n; i++) size[i] = 1; The function find returns the representative for an element x. The represen-tative can be found by following the chain that begins at x.
int find(int x) { while (x != link[x]) x = link[x]; return x; } The function same checks whether elements a and b belong to the same set.
This can easily be done by using the function find: 146 bool same(int a, int b) { return find(a) == find(b); } The function unite joins the sets that contain elements a and b (the elements have to be in different sets). The function first finds the representatives of the sets and then connects the smaller set to the larger set.
void unite(int a, int b) { a = find(a); b = find(b); if (size[a] < size[b]) swap(a,b); size[a] += size[b]; link[b] = a; } The time complexity of the function find is O(logn) assuming that the length of each chain is O(logn). In this case, the functions same and unite also work in O(logn) time. The function unite makes sure that the length of each chain is O(logn) by connecting the smaller set to the larger set.
15.3 Prim’s algorithm Prim’s algorithm3 is an alternative method for finding a minimum spanning tree. The algorithm first adds an arbitrary node to the tree. After this, the algorithm always chooses a minimum-weight edge that adds a new node to the tree. Finally, all nodes have been added to the tree and a minimum spanning tree has been found.
Prim’s algorithm resembles Dijkstra’s algorithm. The difference is that Dijk-stra’s algorithm always selects an edge whose distance from the starting node is minimum, but Prim’s algorithm simply selects the minimum weight edge that adds a new node to the tree.
Example Let us consider how Prim’s algorithm works in the following graph: 1 2 3 4 5 6 3 5 9 5 2 7 6 3 3The algorithm is named after R. C. Prim who published it in 1957 . However, the same algorithm was discovered already in 1930 by V. Jarník.
147 Initially, there are no edges between the nodes: 1 2 3 4 5 6 An arbitrary node can be the starting node, so let us choose node 1. First, we add node 2 that is connected by an edge of weight 3: 1 2 3 4 5 6 3 After this, there are two edges with weight 5, so we can add either node 3 or node 5 to the tree. Let us add node 3 first: 1 2 3 4 5 6 3 5 The process continues until all nodes have been included in the tree: 1 2 3 4 5 6 3 5 2 7 3 Implementation Like Dijkstra’s algorithm, Prim’s algorithm can be efficiently implemented using a priority queue. The priority queue should contain all nodes that can be connected to the current component using a single edge, in increasing order of the weights of the corresponding edges.
The time complexity of Prim’s algorithm is O(n+mlogm) that equals the time complexity of Dijkstra’s algorithm. In practice, Prim’s and Kruskal’s algorithms are both efficient, and the choice of the algorithm is a matter of taste. Still, most competitive programmers use Kruskal’s algorithm.
148 Chapter 16 Directed graphs In this chapter, we focus on two classes of directed graphs: • Acyclic graphs: There are no cycles in the graph, so there is no path from any node to itself1.
• Successor graphs: The outdegree of each node is 1, so each node has a unique successor.
It turns out that in both cases, we can design efficient algorithms that are based on the special properties of the graphs.
16.1 Topological sorting A topological sort is an ordering of the nodes of a directed graph such that if there is a path from node a to node b, then node a appears before node b in the ordering. For example, for the graph 1 2 3 4 5 6 one topological sort is [4,1,5,2,3,6]: 1 2 3 4 5 6 An acyclic graph always has a topological sort. However, if the graph contains a cycle, it is not possible to form a topological sort, because no node of the cycle can appear before the other nodes of the cycle in the ordering. It turns out that depth-first search can be used to both check if a directed graph contains a cycle and, if it does not contain a cycle, to construct a topological sort.
1Directed acyclic graphs are sometimes called DAGs.
149 Algorithm The idea is to go through the nodes of the graph and always begin a depth-first search at the current node if it has not been processed yet. During the searches, the nodes have three possible states: • state 0: the node has not been processed (white) • state 1: the node is under processing (light gray) • state 2: the node has been processed (dark gray) Initially, the state of each node is 0. When a search reaches a node for the first time, its state becomes 1. Finally, after all successors of the node have been processed, its state becomes 2.
If the graph contains a cycle, we will find this out during the search, because sooner or later we will arrive at a node whose state is 1. In this case, it is not possible to construct a topological sort.
If the graph does not contain a cycle, we can construct a topological sort by adding each node to a list when the state of the node becomes 2. This list in reverse order is a topological sort.
Example 1 In the example graph, the search first proceeds from node 1 to node 6: 1 2 3 4 5 6 Now node 6 has been processed, so it is added to the list. After this, also nodes 3, 2 and 1 are added to the list: 1 2 3 4 5 6 At this point, the list is [6,3,2,1]. The next search begins at node 4: 1 2 3 4 5 6 150 Thus, the final list is [6,3,2,1,5,4]. We have processed all nodes, so a topologi-cal sort has been found. The topological sort is the reverse list [4,5,1,2,3,6]: 1 2 3 4 5 6 Note that a topological sort is not unique, and there can be several topological sorts for a graph.
Example 2 Let us now consider a graph for which we cannot construct a topological sort, because the graph contains a cycle: 1 2 3 4 5 6 The search proceeds as follows: 1 2 3 4 5 6 The search reaches node 2 whose state is 1, which means that the graph contains a cycle. In this example, there is a cycle 2 →3 →5 →2.
16.2 Dynamic programming If a directed graph is acyclic, dynamic programming can be applied to it. For example, we can efficiently solve the following problems concerning paths from a starting node to an ending node: • how many different paths are there?
• what is the shortest/longest path?
• what is the minimum/maximum number of edges in a path?
• which nodes certainly appear in any path?
151 Counting the number of paths As an example, let us calculate the number of paths from node 1 to node 6 in the following graph: 1 2 3 4 5 6 There are a total of three such paths: • 1 →2 →3 →6 • 1 →4 →5 →2 →3 →6 • 1 →4 →5 →3 →6 Let paths(x) denote the number of paths from node 1 to node x. As a base case, paths(1) = 1. Then, to calculate other values of paths(x), we may use the recursion paths(x) = paths(a1)+paths(a2)+···+paths(ak) where a1,a2,...,ak are the nodes from which there is an edge to x. Since the graph is acyclic, the values of paths(x) can be calculated in the order of a topological sort. A topological sort for the above graph is as follows: 1 2 3 4 5 6 Hence, the numbers of paths are as follows: 1 2 3 4 5 6 1 1 3 1 2 3 For example, to calculate the value of paths(3), we can use the formula paths(2)+paths(5), because there are edges from nodes 2 and 5 to node 3. Since paths(2) = 2 and paths(5) = 1, we conclude that paths(3) = 3.
152 Extending Dijkstra’s algorithm A by-product of Dijkstra’s algorithm is a directed, acyclic graph that indicates for each node of the original graph the possible ways to reach the node using a shortest path from the starting node. Dynamic programming can be applied to that graph. For example, in the graph 1 2 3 4 5 3 5 4 8 2 1 2 the shortest paths from node 1 may use the following edges: 1 2 3 4 5 3 5 4 2 1 2 Now we can, for example, calculate the number of shortest paths from node 1 to node 5 using dynamic programming: 1 2 3 4 5 3 5 4 2 1 2 1 1 2 3 3 Representing problems as graphs Actually, any dynamic programming problem can be represented as a directed, acyclic graph. In such a graph, each node corresponds to a dynamic programming state and the edges indicate how the states depend on each other.
As an example, consider the problem of forming a sum of money n using coins {c1, c2,..., ck}. In this problem, we can construct a graph where each node corresponds to a sum of money, and the edges show how the coins can be chosen.
For example, for coins {1,3,4} and n = 6, the graph is as follows: 153 0 1 2 3 4 5 6 Using this representation, the shortest path from node 0 to node n corresponds to a solution with the minimum number of coins, and the total number of paths from node 0 to node n equals the total number of solutions.
16.3 Successor paths For the rest of the chapter, we will focus on successor graphs. In those graphs, the outdegree of each node is 1, i.e., exactly one edge starts at each node. A successor graph consists of one or more components, each of which contains one cycle and some paths that lead to it.
Successor graphs are sometimes called functional graphs. The reason for this is that any successor graph corresponds to a function that defines the edges of the graph. The parameter for the function is a node of the graph, and the function gives the successor of that node.
For example, the function x 1 2 3 4 5 6 7 8 9 succ(x) 3 5 7 6 2 2 1 6 3 defines the following graph: 1 2 3 4 5 6 7 8 9 Since each node of a successor graph has a unique successor, we can also define a function succ(x,k) that gives the node that we will reach if we begin at node x and walk k steps forward. For example, in the above graph succ(4,6) = 2, because we will reach node 2 by walking 6 steps from node 4: 4 6 2 5 2 5 2 A straightforward way to calculate a value of succ(x,k) is to start at node x and walk k steps forward, which takes O(k) time. However, using preprocessing, any value of succ(x,k) can be calculated in only O(logk) time.
The idea is to precalculate all values of succ(x,k) where k is a power of two and at most u, where u is the maximum number of steps we will ever walk. This can be efficiently done, because we can use the following recursion: 154 succ(x,k) = ( succ(x) k = 1 succ(succ(x,k/2),k/2) k > 1 Precalculating the values takes O(nlogu) time, because O(logu) values are calculated for each node. In the above graph, the first values are as follows: x 1 2 3 4 5 6 7 8 9 succ(x,1) 3 5 7 6 2 2 1 6 3 succ(x,2) 7 2 1 2 5 5 3 2 7 succ(x,4) 3 2 7 2 5 5 1 2 3 succ(x,8) 7 2 1 2 5 5 3 2 7 ··· After this, any value of succ(x,k) can be calculated by presenting the number of steps k as a sum of powers of two. For example, if we want to calculate the value of succ(x,11), we first form the representation 11 = 8+2+1. Using that, succ(x,11) = succ(succ(succ(x,8),2),1).
For example, in the previous graph succ(4,11) = succ(succ(succ(4,8),2),1) = 5.
Such a representation always consists of O(logk) parts, so calculating a value of succ(x,k) takes O(logk) time.
16.4 Cycle detection Consider a successor graph that only contains a path that ends in a cycle. We may ask the following questions: if we begin our walk at the starting node, what is the first node in the cycle and how many nodes does the cycle contain?
For example, in the graph 5 4 6 3 2 1 we begin our walk at node 1, the first node that belongs to the cycle is node 4, and the cycle consists of three nodes (4, 5 and 6).
A simple way to detect the cycle is to walk in the graph and keep track of all nodes that have been visited. Once a node is visited for the second time, we can conclude that the node is the first node in the cycle. This method works in O(n) time and also uses O(n) memory.
However, there are better algorithms for cycle detection. The time complexity of such algorithms is still O(n), but they only use O(1) memory. This is an important improvement if n is large. Next we will discuss Floyd’s algorithm that achieves these properties.
155 Floyd’s algorithm Floyd’s algorithm2 walks forward in the graph using two pointers a and b.
Both pointers begin at a node x that is the starting node of the graph. Then, on each turn, the pointer a walks one step forward and the pointer b walks two steps forward. The process continues until the pointers meet each other: a = succ(x); b = succ(succ(x)); while (a != b) { a = succ(a); b = succ(succ(b)); } At this point, the pointer a has walked k steps and the pointer b has walked 2k steps, so the length of the cycle divides k. Thus, the first node that belongs to the cycle can be found by moving the pointer a to node x and advancing the pointers step by step until they meet again.
a = x; while (a != b) { a = succ(a); b = succ(b); } first = a; After this, the length of the cycle can be calculated as follows: b = succ(a); length = 1; while (a != b) { b = succ(b); length++; } 2The idea of the algorithm is mentioned in and attributed to R. W. Floyd; however, it is not known if Floyd actually discovered the algorithm.
156 Chapter 17 Strong connectivity In a directed graph, the edges can be traversed in one direction only, so even if the graph is connected, this does not guarantee that there would be a path from a node to another node. For this reason, it is meaningful to define a new concept that requires more than connectivity.
A graph is strongly connected if there is a path from any node to all other nodes in the graph. For example, in the following picture, the left graph is strongly connected while the right graph is not.
1 2 3 4 1 2 3 4 The right graph is not strongly connected because, for example, there is no path from node 2 to node 1.
The strongly connected components of a graph divide the graph into strongly connected parts that are as large as possible. The strongly connected components form an acyclic component graph that represents the deep struc-ture of the original graph.
For example, for the graph 7 3 2 1 6 5 4 the strongly connected components are as follows: 7 3 2 1 6 5 4 157 The corresponding component graph is as follows: B A D C The components are A = {1,2}, B = {3,6,7}, C = {4} and D = {5}.
A component graph is an acyclic, directed graph, so it is easier to process than the original graph. Since the graph does not contain cycles, we can always construct a topological sort and use dynamic programming techniques like those presented in Chapter 16.
17.1 Kosaraju’s algorithm Kosaraju’s algorithm1 is an efficient method for finding the strongly connected components of a directed graph. The algorithm performs two depth-first searches: the first search constructs a list of nodes according to the structure of the graph, and the second search forms the strongly connected components.
Search 1 The first phase of Kosaraju’s algorithm constructs a list of nodes in the order in which a depth-first search processes them. The algorithm goes through the nodes, and begins a depth-first search at each unprocessed node. Each node will be added to the list after it has been processed.
In the example graph, the nodes are processed in the following order: 7 3 2 1 6 5 4 1/8 2/7 9/14 4/5 3/6 11/12 10/13 The notation x/y means that processing the node started at time x and finished at time y. Thus, the corresponding list is as follows: 1According to , S. R. Kosaraju invented this algorithm in 1978 but did not publish it. In 1981, the same algorithm was rediscovered and published by M. Sharir .
158 node processing time 4 5 5 6 2 7 1 8 6 12 7 13 3 14 Search 2 The second phase of the algorithm forms the strongly connected components of the graph. First, the algorithm reverses every edge in the graph. This guarantees that during the second search, we will always find strongly connected components that do not have extra nodes.
After reversing the edges, the example graph is as follows: 7 3 2 1 6 5 4 After this, the algorithm goes through the list of nodes created by the first search, in reverse order. If a node does not belong to a component, the algorithm creates a new component and starts a depth-first search that adds all new nodes found during the search to the new component.
In the example graph, the first component begins at node 3: 7 3 2 1 6 5 4 Note that since all edges are reversed, the component does not ”leak” to other parts in the graph.
159 The next nodes in the list are nodes 7 and 6, but they already belong to a component, so the next new component begins at node 1: 7 3 2 1 6 5 4 Finally, the algorithm processes nodes 5 and 4 that create the remaining strongly connected components: 7 3 2 1 6 5 4 The time complexity of the algorithm is O(n + m), because the algorithm performs two depth-first searches.
17.2 2SAT problem Strong connectivity is also linked with the 2SAT problem2. In this problem, we are given a logical formula (a1 ∨b1)∧(a2 ∨b2)∧···∧(am ∨bm), where each ai and bi is either a logical variable (x1,x2,...,xn) or a negation of a logical variable (¬x1,¬x2,...,¬xn). The symbols ”∧” and ”∨” denote logical operators ”and” and ”or”. Our task is to assign each variable a value so that the formula is true, or state that this is not possible.
For example, the formula L1 = (x2 ∨¬x1)∧(¬x1 ∨¬x2)∧(x1 ∨x3)∧(¬x2 ∨¬x3)∧(x1 ∨x4) is true when the variables are assigned as follows: x1 = false x2 = false x3 = true x4 = true 2The algorithm presented here was introduced in . There is also another well-known linear-time algorithm that is based on backtracking.
160 However, the formula L2 = (x1 ∨x2)∧(x1 ∨¬x2)∧(¬x1 ∨x3)∧(¬x1 ∨¬x3) is always false, regardless of how we assign the values. The reason for this is that we cannot choose a value for x1 without creating a contradiction. If x1 is false, both x2 and ¬x2 should be true which is impossible, and if x1 is true, both x3 and ¬x3 should be true which is also impossible.
The 2SAT problem can be represented as a graph whose nodes correspond to variables xi and negations ¬xi, and edges determine the connections between the variables. Each pair (ai ∨bi) generates two edges: ¬ai →bi and ¬bi →ai.
This means that if ai does not hold, bi must hold, and vice versa.
The graph for the formula L1 is: ¬x3 x2 ¬x4 x1 ¬x1 x4 ¬x2 x3 And the graph for the formula L2 is: x3 x2 ¬x2 ¬x3 ¬x1 x1 The structure of the graph tells us whether it is possible to assign the values of the variables so that the formula is true. It turns out that this can be done exactly when there are no nodes xi and ¬xi such that both nodes belong to the same strongly connected component. If there are such nodes, the graph contains a path from xi to ¬xi and also a path from ¬xi to xi, so both xi and ¬xi should be true which is not possible.
In the graph of the formula L1 there are no nodes xi and ¬xi such that both nodes belong to the same strongly connected component, so a solution exists. In the graph of the formula L2 all nodes belong to the same strongly connected component, so a solution does not exist.
If a solution exists, the values for the variables can be found by going through the nodes of the component graph in a reverse topological sort order. At each step, we process a component that does not contain edges that lead to an unprocessed component. If the variables in the component have not been assigned values, their values will be determined according to the values in the component, and if 161 they already have values, they remain unchanged. The process continues until each variable has been assigned a value.
The component graph for the formula L1 is as follows: A B C D The components are A = {¬x4}, B = {x1,x2,¬x3}, C = {¬x1,¬x2,x3} and D = {x4}.
When constructing the solution, we first process the component D where x4 becomes true. After this, we process the component C where x1 and x2 become false and x3 becomes true. All variables have been assigned values, so the remaining components A and B do not change the variables.
Note that this method works, because the graph has a special structure: if there are paths from node xi to node xj and from node xj to node ¬xj, then node xi never becomes true. The reason for this is that there is also a path from node ¬xj to node ¬xi, and both xi and xj become false.
A more difficult problem is the 3SAT problem, where each part of the formula is of the form (ai ∨bi ∨ci). This problem is NP-hard, so no efficient algorithm for solving the problem is known.
162 Chapter 18 Tree queries This chapter discusses techniques for processing queries on subtrees and paths of a rooted tree. For example, such queries are: • what is the kth ancestor of a node?
• what is the sum of values in the subtree of a node?
• what is the sum of values on a path between two nodes?
• what is the lowest common ancestor of two nodes?
18.1 Finding ancestors The kth ancestor of a node x in a rooted tree is the node that we will reach if we move k levels up from x. Let ancestor(x,k) denote the kth ancestor of a node x (or 0 if there is no such an ancestor). For example, in the following tree, ancestor(2,1) = 1 and ancestor(8,2) = 4.
1 2 4 5 6 3 7 8 An easy way to calculate any value of ancestor(x,k) is to perform a sequence of k moves in the tree. However, the time complexity of this method is O(k), which may be slow, because a tree of n nodes may have a chain of n nodes.
163 Fortunately, using a technique similar to that used in Chapter 16.3, any value of ancestor(x,k) can be efficiently calculated in O(logk) time after preprocessing.
The idea is to precalculate all values ancestor(x,k) where k ≤n is a power of two.
For example, the values for the above tree are as follows: x 1 2 3 4 5 6 7 8 ancestor(x,1) 0 1 4 1 1 2 4 7 ancestor(x,2) 0 0 1 0 0 1 1 4 ancestor(x,4) 0 0 0 0 0 0 0 0 ··· The preprocessing takes O(nlogn) time, because O(logn) values are calculated for each node. After this, any value of ancestor(x,k) can be calculated in O(logk) time by representing k as a sum where each term is a power of two.
18.2 Subtrees and paths A tree traversal array contains the nodes of a rooted tree in the order in which a depth-first search from the root node visits them. For example, in the tree 1 2 3 4 5 6 7 8 9 a depth-first search proceeds as follows: 1 2 3 4 5 6 7 8 9 Hence, the corresponding tree traversal array is as follows: 1 2 6 3 4 7 8 9 5 164 Subtree queries Each subtree of a tree corresponds to a subarray of the tree traversal array such that the first element of the subarray is the root node. For example, the following subarray contains the nodes of the subtree of node 4: 1 2 6 3 4 7 8 9 5 Using this fact, we can efficiently process queries that are related to subtrees of a tree. As an example, consider a problem where each node is assigned a value, and our task is to support the following queries: • update the value of a node • calculate the sum of values in the subtree of a node Consider the following tree where the blue numbers are the values of the nodes. For example, the sum of the subtree of node 4 is 3+4+3+1 = 11.
1 2 3 4 5 6 7 8 9 2 3 5 3 1 4 4 3 1 The idea is to construct a tree traversal array that contains three values for each node: the identifier of the node, the size of the subtree, and the value of the node. For example, the array for the above tree is as follows: node id subtree size node value 1 2 6 3 4 7 8 9 5 9 2 1 1 4 1 1 1 1 2 3 4 5 3 4 3 1 1 Using this array, we can calculate the sum of values in any subtree by first finding out the size of the subtree and then the values of the corresponding nodes.
For example, the values in the subtree of node 4 can be found as follows: node id subtree size node value 1 2 6 3 4 7 8 9 5 9 2 1 1 4 1 1 1 1 2 3 4 5 3 4 3 1 1 To answer the queries efficiently, it suffices to store the values of the nodes in a binary indexed or segment tree. After this, we can both update a value and calculate the sum of values in O(logn) time.
165 Path queries Using a tree traversal array, we can also efficiently calculate sums of values on paths from the root node to any node of the tree. Consider a problem where our task is to support the following queries: • change the value of a node • calculate the sum of values on a path from the root to a node For example, in the following tree, the sum of values from the root node to node 7 is 4+5+5 = 14: 1 2 3 4 5 6 7 8 9 4 5 3 5 2 3 5 3 1 We can solve this problem like before, but now each value in the last row of the array is the sum of values on a path from the root to the node. For example, the following array corresponds to the above tree: node id subtree size path sum 1 2 6 3 4 7 8 9 5 9 2 1 1 4 1 1 1 1 4 9 12 7 9 14 12 10 6 When the value of a node increases by x, the sums of all nodes in its subtree increase by x. For example, if the value of node 4 increases by 1, the array changes as follows: node id subtree size path sum 1 2 6 3 4 7 8 9 5 9 2 1 1 4 1 1 1 1 4 9 12 7 10 15 13 11 6 Thus, to support both the operations, we should be able to increase all values in a range and retrieve a single value. This can be done in O(logn) time using a binary indexed or segment tree (see Chapter 9.4).
166 18.3 Lowest common ancestor The lowest common ancestor of two nodes of a rooted tree is the lowest node whose subtree contains both the nodes. A typical problem is to efficiently process queries that ask to find the lowest common ancestor of two nodes.
For example, in the following tree, the lowest common ancestor of nodes 5 and 8 is node 2: 1 4 2 3 7 5 6 8 Next we will discuss two efficient techniques for finding the lowest common ancestor of two nodes.
Method 1 One way to solve the problem is to use the fact that we can efficiently find the kth ancestor of any node in the tree. Using this, we can divide the problem of finding the lowest common ancestor into two parts.
We use two pointers that initially point to the two nodes whose lowest common ancestor we should find. First, we move one of the pointers upwards so that both pointers point to nodes at the same level.
In the example scenario, we move the second pointer one level up so that it points to node 6 which is at the same level with node 5: 1 4 2 3 7 5 6 8 167 After this, we determine the minimum number of steps needed to move both pointers upwards so that they will point to the same node. The node to which the pointers point after this is the lowest common ancestor.
In the example scenario, it suffices to move both pointers one step upwards to node 2, which is the lowest common ancestor: 1 4 2 3 7 5 6 8 Since both parts of the algorithm can be performed in O(logn) time using precomputed information, we can find the lowest common ancestor of any two nodes in O(logn) time.
Method 2 Another way to solve the problem is based on a tree traversal array1. Once again, the idea is to traverse the nodes using a depth-first search: 1 4 2 3 7 5 6 8 However, we use a different tree traversal array than before: we add each node to the array always when the depth-first search walks through the node, and not only at the first visit. Hence, a node that has k children appears k +1 times in the array and there are a total of 2n−1 nodes in the array.
1This lowest common ancestor algorithm was presented in . This technique is sometimes called the Euler tour technique .
168 We store two values in the array: the identifier of the node and the depth of the node in the tree. The following array corresponds to the above tree: node id depth 1 2 5 2 6 8 6 2 1 3 1 4 7 4 1 1 2 3 2 3 4 3 2 1 2 1 2 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Now we can find the lowest common ancestor of nodes a and b by finding the node with the minimum depth between nodes a and b in the array. For example, the lowest common ancestor of nodes 5 and 8 can be found as follows: node id depth ↑ 1 2 5 2 6 8 6 2 1 3 1 4 7 4 1 1 2 3 2 3 4 3 2 1 2 1 2 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Node 5 is at position 2, node 8 is at position 5, and the node with minimum depth between positions 2...5 is node 2 at position 3 whose depth is 2. Thus, the lowest common ancestor of nodes 5 and 8 is node 2.
Thus, to find the lowest common ancestor of two nodes it suffices to process a range minimum query. Since the array is static, we can process such queries in O(1) time after an O(nlogn) time preprocessing.
Distances of nodes The distance between nodes a and b equals the length of the path from a to b. It turns out that the problem of calculating the distance between nodes reduces to finding their lowest common ancestor.
First, we root the tree arbitrarily. After this, the distance of nodes a and b can be calculated using the formula depth(a)+depth(b)−2·depth(c), where c is the lowest common ancestor of a and b and depth(s) denotes the depth of node s. For example, consider the distance of nodes 5 and 8: 1 4 2 3 7 5 6 8 169 The lowest common ancestor of nodes 5 and 8 is node 2. The depths of the nodes are depth(5) = 3, depth(8) = 4 and depth(2) = 2, so the distance between nodes 5 and 8 is 3+4−2·2 = 3.
18.4 Offline algorithms So far, we have discussed online algorithms for tree queries. Those algorithms are able to process queries one after another so that each query is answered before receiving the next query.
However, in many problems, the online property is not necessary. In this section, we focus on offline algorithms. Those algorithms are given a set of queries which can be answered in any order. It is often easier to design an offline algorithm compared to an online algorithm.
Merging data structures One method to construct an offline algorithm is to perform a depth-first tree traversal and maintain data structures in nodes. At each node s, we create a data structure d[s] that is based on the data structures of the children of s. Then, using this data structure, all queries related to s are processed.
As an example, consider the following problem: We are given a tree where each node has some value. Our task is to process queries of the form ”calculate the number of nodes with value x in the subtree of node s”. For example, in the following tree, the subtree of node 4 contains two nodes whose value is 3.
1 2 3 4 5 6 7 8 9 2 3 5 3 1 4 4 3 1 In this problem, we can use map structures to answer the queries. For example, the maps for node 4 and its children are as follows: 4 1 3 1 1 1 1 3 4 1 2 1 170 If we create such a data structure for each node, we can easily process all given queries, because we can handle all queries related to a node immediately after creating its data structure. For example, the above map structure for node 4 tells us that its subtree contains two nodes whose value is 3.
However, it would be too slow to create all data structures from scratch.
Instead, at each node s, we create an initial data structure d[s] that only contains the value of s. After this, we go through the children of s and merge d[s] and all data structures d[u] where u is a child of s.
For example, in the above tree, the map for node 4 is created by merging the following maps: 4 1 3 1 1 1 3 1 Here the first map is the initial data structure for node 4, and the other three maps correspond to nodes 7, 8 and 9.
The merging at node s can be done as follows: We go through the children of s and at each child u merge d[s] and d[u]. We always copy the contents from d[u] to d[s]. However, before this, we swap the contents of d[s] and d[u] if d[s] is smaller than d[u]. By doing this, each value is copied only O(logn) times during the tree traversal, which ensures that the algorithm is efficient.
To swap the contents of two data structures a and b efficiently, we can just use the following code: swap(a,b); It is guaranteed that the above code works in constant time when a and b are C++ standard library data structures.
Lowest common ancestors There is also an offline algorithm for processing a set of lowest common ancestor queries2. The algorithm is based on the union-find data structure (see Chapter 15.2), and the benefit of the algorithm is that it is easier to implement than the algorithms discussed earlier in this chapter.
The algorithm is given as input a set of pairs of nodes, and it determines for each such pair the lowest common ancestor of the nodes. The algorithm performs a depth-first tree traversal and maintains disjoint sets of nodes. Initially, each node belongs to a separate set. For each set, we also store the highest node in the tree that belongs to the set.
When the algorithm visits a node x, it goes through all nodes y such that the lowest common ancestor of x and y has to be found. If y has already been visited, the algorithm reports that the lowest common ancestor of x and y is the highest node in the set of y. Then, after processing node x, the algorithm joins the sets of x and its parent.
2This algorithm was published by R. E. Tarjan in 1979 .
171 For example, suppose that we want to find the lowest common ancestors of node pairs (5,8) and (2,7) in the following tree: 1 4 2 3 7 5 6 8 In the following trees, gray nodes denote visited nodes and dashed groups of nodes belong to the same set. When the algorithm visits node 8, it notices that node 5 has been visited and the highest node in its set is 2. Thus, the lowest common ancestor of nodes 5 and 8 is 2: 1 4 2 3 7 5 6 8 Later, when visiting node 7, the algorithm determines that the lowest common ancestor of nodes 2 and 7 is 1: 1 4 2 3 7 5 6 8 172 Chapter 19 Paths and circuits This chapter focuses on two types of paths in graphs: • An Eulerian path is a path that goes through each edge exactly once.
• A Hamiltonian path is a path that visits each node exactly once.
While Eulerian and Hamiltonian paths look like similar concepts at first glance, the computational problems related to them are very different. It turns out that there is a simple rule that determines whether a graph contains an Eulerian path, and there is also an efficient algorithm to find such a path if it exists. On the contrary, checking the existence of a Hamiltonian path is a NP-hard problem, and no efficient algorithm is known for solving the problem.
19.1 Eulerian paths An Eulerian path1 is a path that goes exactly once through each edge of the graph. For example, the graph 1 2 3 4 5 has an Eulerian path from node 2 to node 5: 1 2 3 4 5 1.
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1L. Euler studied such paths in 1736 when he solved the famous Königsberg bridge problem.
This was the birth of graph theory.
173 An Eulerian circuit is an Eulerian path that starts and ends at the same node.
For example, the graph 1 2 3 4 5 has an Eulerian circuit that starts and ends at node 1: 1 2 3 4 5 1.
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Existence The existence of Eulerian paths and circuits depends on the degrees of the nodes.
First, an undirected graph has an Eulerian path exactly when all the edges belong to the same connected component and • the degree of each node is even or • the degree of exactly two nodes is odd, and the degree of all other nodes is even.
In the first case, each Eulerian path is also an Eulerian circuit. In the second case, the odd-degree nodes are the starting and ending nodes of an Eulerian path which is not an Eulerian circuit.
For example, in the graph 1 2 3 4 5 nodes 1, 3 and 4 have a degree of 2, and nodes 2 and 5 have a degree of 3. Exactly two nodes have an odd degree, so there is an Eulerian path between nodes 2 and 5, but the graph does not contain an Eulerian circuit.
In a directed graph, we focus on indegrees and outdegrees of the nodes. A directed graph contains an Eulerian path exactly when all the edges belong to the same connected component and • in each node, the indegree equals the outdegree, or 174 • in one node, the indegree is one larger than the outdegree, in another node, the outdegree is one larger than the indegree, and in all other nodes, the indegree equals the outdegree.
In the first case, each Eulerian path is also an Eulerian circuit, and in the second case, the graph contains an Eulerian path that begins at the node whose outdegree is larger and ends at the node whose indegree is larger.
For example, in the graph 1 2 3 4 5 nodes 1, 3 and 4 have both indegree 1 and outdegree 1, node 2 has indegree 1 and outdegree 2, and node 5 has indegree 2 and outdegree 1. Hence, the graph contains an Eulerian path from node 2 to node 5: 1 2 3 4 5 1.
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Hierholzer’s algorithm Hierholzer’s algorithm2 is an efficient method for constructing an Eulerian circuit. The algorithm consists of several rounds, each of which adds new edges to the circuit. Of course, we assume that the graph contains an Eulerian circuit; otherwise Hierholzer’s algorithm cannot find it.
First, the algorithm constructs a circuit that contains some (not necessarily all) of the edges of the graph. After this, the algorithm extends the circuit step by step by adding subcircuits to it. The process continues until all edges have been added to the circuit.
The algorithm extends the circuit by always finding a node x that belongs to the circuit but has an outgoing edge that is not included in the circuit. The algorithm constructs a new path from node x that only contains edges that are not yet in the circuit. Sooner or later, the path will return to node x, which creates a subcircuit.
If the graph only contains an Eulerian path, we can still use Hierholzer’s algorithm to find it by adding an extra edge to the graph and removing the edge after the circuit has been constructed. For example, in an undirected graph, we add the extra edge between the two odd-degree nodes.
Next we will see how Hierholzer’s algorithm constructs an Eulerian circuit for an undirected graph.
2The algorithm was published in 1873 after Hierholzer’s death .
175 Example Let us consider the following graph: 1 2 3 4 5 6 7 Suppose that the algorithm first creates a circuit that begins at node 1. A possible circuit is 1 →2 →3 →1: 1 2 3 4 5 6 7 1.
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After this, the algorithm adds the subcircuit 2 →5 →6 →2 to the circuit: 1 2 3 4 5 6 7 1.
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Finally, the algorithm adds the subcircuit 6 →3 →4 →7 →6 to the circuit: 1 2 3 4 5 6 7 1.
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176 Now all edges are included in the circuit, so we have successfully constructed an Eulerian circuit.
19.2 Hamiltonian paths A Hamiltonian path is a path that visits each node of the graph exactly once.
For example, the graph 1 2 3 4 5 contains a Hamiltonian path from node 1 to node 3: 1 2 3 4 5 1.
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If a Hamiltonian path begins and ends at the same node, it is called a Hamil-tonian circuit. The graph above also has an Hamiltonian circuit that begins and ends at node 1: 1 2 3 4 5 1.
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Existence No efficient method is known for testing if a graph contains a Hamiltonian path, and the problem is NP-hard. Still, in some special cases, we can be certain that a graph contains a Hamiltonian path.
A simple observation is that if the graph is complete, i.e., there is an edge between all pairs of nodes, it also contains a Hamiltonian path. Also stronger results have been achieved: • Dirac’s theorem: If the degree of each node is at least n/2, the graph contains a Hamiltonian path.
• Ore’s theorem: If the sum of degrees of each non-adjacent pair of nodes is at least n, the graph contains a Hamiltonian path.
177 A common property in these theorems and other results is that they guarantee the existence of a Hamiltonian path if the graph has a large number of edges. This makes sense, because the more edges the graph contains, the more possibilities there is to construct a Hamiltonian path.
Construction Since there is no efficient way to check if a Hamiltonian path exists, it is clear that there is also no method to efficiently construct the path, because otherwise we could just try to construct the path and see whether it exists.
A simple way to search for a Hamiltonian path is to use a backtracking algorithm that goes through all possible ways to construct the path. The time complexity of such an algorithm is at least O(n!), because there are n! different ways to choose the order of n nodes.
A more efficient solution is based on dynamic programming (see Chapter 10.5). The idea is to calculate values of a function possible(S,x), where S is a subset of nodes and x is one of the nodes. The function indicates whether there is a Hamiltonian path that visits the nodes of S and ends at node x. It is possible to implement this solution in O(2nn2) time.
19.3 De Bruijn sequences A De Bruijn sequence is a string that contains every string of length n exactly once as a substring, for a fixed alphabet of k characters. The length of such a string is kn + n−1 characters. For example, when n = 3 and k = 2, an example of a De Bruijn sequence is 0001011100.
The substrings of this string are all combinations of three bits: 000, 001, 010, 011, 100, 101, 110 and 111.
It turns out that each De Bruijn sequence corresponds to an Eulerian path in a graph. The idea is to construct a graph where each node contains a string of n−1 characters and each edge adds one character to the string. The following graph corresponds to the above scenario: 00 11 01 10 1 1 0 0 0 1 0 1 An Eulerian path in this graph corresponds to a string that contains all strings of length n. The string contains the characters of the starting node and all characters of the edges. The starting node has n−1 characters and there are kn characters in the edges, so the length of the string is kn + n−1.
178 19.4 Knight’s tours A knight’s tour is a sequence of moves of a knight on an n × n chessboard following the rules of chess such that the knight visits each square exactly once.
A knight’s tour is called a closed tour if the knight finally returns to the starting square and otherwise it is called an open tour.
For example, here is an open knight’s tour on a 5×5 board: 1 4 11 16 25 12 17 2 5 10 3 20 7 24 15 18 13 22 9 6 21 8 19 14 23 A knight’s tour corresponds to a Hamiltonian path in a graph whose nodes represent the squares of the board, and two nodes are connected with an edge if a knight can move between the squares according to the rules of chess.
A natural way to construct a knight’s tour is to use backtracking. The search can be made more efficient by using heuristics that attempt to guide the knight so that a complete tour will be found quickly.
Warnsdorf’s rule Warnsdorf’s rule is a simple and effective heuristic for finding a knight’s tour3.
Using the rule, it is possible to efficiently construct a tour even on a large board.
The idea is to always move the knight so that it ends up in a square where the number of possible moves is as small as possible.
For example, in the following situation, there are five possible squares to which the knight can move (squares a... e): 1 2 a b e c d In this situation, Warnsdorf’s rule moves the knight to square a, because after this choice, there is only a single possible move. The other choices would move the knight to squares where there would be three moves available.
3This heuristic was proposed in Warnsdorf’s book in 1823. There are also polynomial algorithms for finding knight’s tours , but they are more complicated.
179 180 Chapter 20 Flows and cuts In this chapter, we focus on the following two problems: • Finding a maximum flow: What is the maximum amount of flow we can send from a node to another node?
• Finding a minimum cut: What is a minimum-weight set of edges that separates two nodes of the graph?
The input for both these problems is a directed, weighted graph that contains two special nodes: the source is a node with no incoming edges, and the sink is a node with no outgoing edges.
As an example, we will use the following graph where node 1 is the source and node 6 is the sink: 1 2 3 6 4 5 5 6 5 4 1 2 3 8 Maximum flow In the maximum flow problem, our task is to send as much flow as possible from the source to the sink. The weight of each edge is a capacity that restricts the flow that can go through the edge. In each intermediate node, the incoming and outgoing flow has to be equal.
For example, the maximum size of a flow in the example graph is 7. The following picture shows how we can route the flow: 1 2 3 6 4 5 3/5 6/6 5/5 4/4 1/1 2/2 3/3 1/8 181 The notation v/k means that a flow of v units is routed through an edge whose capacity is k units. The size of the flow is 7, because the source sends 3+4 units of flow and the sink receives 5+2 units of flow. It is easy see that this flow is maximum, because the total capacity of the edges leading to the sink is 7.
Minimum cut In the minimum cut problem, our task is to remove a set of edges from the graph such that there will be no path from the source to the sink after the removal and the total weight of the removed edges is minimum.
The minimum size of a cut in the example graph is 7. It suffices to remove the edges 2 →3 and 4 →5: 1 2 3 6 4 5 5 6 5 4 1 2 3 8 After removing the edges, there will be no path from the source to the sink.
The size of the cut is 7, because the weights of the removed edges are 6 and 1.
The cut is minimum, because there is no valid way to remove edges from the graph such that their total weight would be less than 7.
It is not a coincidence that the maximum size of a flow and the minimum size of a cut are the same in the above example. It turns out that a maximum flow and a minimum cut are always equally large, so the concepts are two sides of the same coin.
Next we will discuss the Ford–Fulkerson algorithm that can be used to find the maximum flow and minimum cut of a graph. The algorithm also helps us to understand why they are equally large.
20.1 Ford–Fulkerson algorithm The Ford–Fulkerson algorithm finds the maximum flow in a graph. The algorithm begins with an empty flow, and at each step finds a path from the source to the sink that generates more flow. Finally, when the algorithm cannot increase the flow anymore, the maximum flow has been found.
The algorithm uses a special representation of the graph where each original edge has a reverse edge in another direction. The weight of each edge indicates how much more flow we could route through it. At the beginning of the algorithm, the weight of each original edge equals the capacity of the edge and the weight of each reverse edge is zero.
182 The new representation for the example graph is as follows: 1 2 3 6 4 5 5 0 6 0 5 0 4 0 1 0 2 0 3 0 8 0 Algorithm description The Ford–Fulkerson algorithm consists of several rounds. On each round, the algorithm finds a path from the source to the sink such that each edge on the path has a positive weight. If there is more than one possible path available, we can choose any of them.
For example, suppose we choose the following path: 1 2 3 6 4 5 5 0 6 0 5 0 4 0 1 0 2 0 3 0 8 0 After choosing the path, the flow increases by x units, where x is the smallest edge weight on the path. In addition, the weight of each edge on the path decreases by x and the weight of each reverse edge increases by x.
In the above path, the weights of the edges are 5, 6, 8 and 2. The smallest weight is 2, so the flow increases by 2 and the new graph is as follows: 1 2 3 6 4 5 3 2 4 2 5 0 4 0 1 0 0 2 3 0 6 2 The idea is that increasing the flow decreases the amount of flow that can go through the edges in the future. On the other hand, it is possible to cancel flow later using the reverse edges of the graph if it turns out that it would be beneficial to route the flow in another way.
The algorithm increases the flow as long as there is a path from the source to the sink through positive-weight edges. In the present example, our next path can be as follows: 183 1 2 3 6 4 5 3 2 4 2 5 0 4 0 1 0 0 2 3 0 6 2 The minimum edge weight on this path is 3, so the path increases the flow by 3, and the total flow after processing the path is 5.
The new graph will be as follows: 1 2 3 6 4 5 3 2 1 5 2 3 1 3 1 0 0 2 0 3 6 2 We still need two more rounds before reaching the maximum flow. For exam-ple, we can choose the paths 1 →2 →3 →6 and 1 →4 →5 →3 →6. Both paths increase the flow by 1, and the final graph is as follows: 1 2 3 6 4 5 2 3 0 6 0 5 0 4 0 1 0 2 0 3 7 1 It is not possible to increase the flow anymore, because there is no path from the source to the sink with positive edge weights. Hence, the algorithm terminates and the maximum flow is 7.
Finding paths The Ford–Fulkerson algorithm does not specify how we should choose the paths that increase the flow. In any case, the algorithm will terminate sooner or later and correctly find the maximum flow. However, the efficiency of the algorithm depends on the way the paths are chosen.
A simple way to find paths is to use depth-first search. Usually, this works well, but in the worst case, each path only increases the flow by 1 and the algorithm is slow. Fortunately, we can avoid this situation by using one of the following techniques: 184 The Edmonds–Karp algorithm chooses each path so that the number of edges on the path is as small as possible. This can be done by using breadth-first search instead of depth-first search for finding paths. It can be proven that this guarantees that the flow increases quickly, and the time complexity of the algorithm is O(m2n).
The scaling algorithm uses depth-first search to find paths where each edge weight is at least a threshold value. Initially, the threshold value is some large number, for example the sum of all edge weights of the graph. Always when a path cannot be found, the threshold value is divided by 2. The time complexity of the algorithm is O(m2log c), where c is the initial threshold value.
In practice, the scaling algorithm is easier to implement, because depth-first search can be used for finding paths. Both algorithms are efficient enough for problems that typically appear in programming contests.
Minimum cuts It turns out that once the Ford–Fulkerson algorithm has found a maximum flow, it has also determined a minimum cut. Let A be the set of nodes that can be reached from the source using positive-weight edges. In the example graph, A contains nodes 1, 2 and 4: 1 2 3 6 4 5 2 3 0 6 0 5 0 4 0 1 0 2 0 3 7 1 Now the minimum cut consists of the edges of the original graph that start at some node in A, end at some node outside A, and whose capacity is fully used in the maximum flow. In the above graph, such edges are 2 →3 and 4 →5, that correspond to the minimum cut 6+1 = 7.
Why is the flow produced by the algorithm maximum and why is the cut minimum? The reason is that a graph cannot contain a flow whose size is larger than the weight of any cut of the graph. Hence, always when a flow and a cut are equally large, they are a maximum flow and a minimum cut.
Let us consider any cut of the graph such that the source belongs to A, the sink belongs to B and there are some edges between the sets: A B 185 The size of the cut is the sum of the edges that go from A to B. This is an upper bound for the flow in the graph, because the flow has to proceed from A to B. Thus, the size of a maximum flow is smaller than or equal to the size of any cut in the graph.
On the other hand, the Ford–Fulkerson algorithm produces a flow whose size is exactly as large as the size of a cut in the graph. Thus, the flow has to be a maximum flow and the cut has to be a minimum cut.
20.2 Disjoint paths Many graph problems can be solved by reducing them to the maximum flow problem. Our first example of such a problem is as follows: we are given a directed graph with a source and a sink, and our task is to find the maximum number of disjoint paths from the source to the sink.
Edge-disjoint paths We will first focus on the problem of finding the maximum number of edge-disjoint paths from the source to the sink. This means that we should construct a set of paths such that each edge appears in at most one path.
For example, consider the following graph: 1 2 3 4 5 6 In this graph, the maximum number of edge-disjoint paths is 2. We can choose the paths 1 →2 →4 →3 →6 and 1 →4 →5 →6 as follows: 1 2 3 4 5 6 It turns out that the maximum number of edge-disjoint paths equals the maximum flow of the graph, assuming that the capacity of each edge is one. After the maximum flow has been constructed, the edge-disjoint paths can be found greedily by following paths from the source to the sink.
Node-disjoint paths Let us now consider another problem: finding the maximum number of node-disjoint paths from the source to the sink. In this problem, every node, except 186 for the source and sink, may appear in at most one path. The number of node-disjoint paths may be smaller than the number of edge-disjoint paths.
For example, in the previous graph, the maximum number of node-disjoint paths is 1: 1 2 3 4 5 6 We can reduce also this problem to the maximum flow problem. Since each node can appear in at most one path, we have to limit the flow that goes through the nodes. A standard method for this is to divide each node into two nodes such that the first node has the incoming edges of the original node, the second node has the outgoing edges of the original node, and there is a new edge from the first node to the second node.
In our example, the graph becomes as follows: 1 2 3 4 5 2 3 4 5 6 The maximum flow for the graph is as follows: 1 2 3 4 5 2 3 4 5 6 Thus, the maximum number of node-disjoint paths from the source to the sink is 1.
20.3 Maximum matchings The maximum matching problem asks to find a maximum-size set of node pairs in an undirected graph such that each pair is connected with an edge and each node belongs to at most one pair.
There are polynomial algorithms for finding maximum matchings in general graphs , but such algorithms are complex and rarely seen in programming contests. However, in bipartite graphs, the maximum matching problem is much easier to solve, because we can reduce it to the maximum flow problem.
187 Finding maximum matchings The nodes of a bipartite graph can be always divided into two groups such that all edges of the graph go from the left group to the right group. For example, in the following bipartite graph, the groups are {1,2,3,4} and {5,6,7,8}.
1 2 3 4 5 6 7 8 The size of a maximum matching of this graph is 3: 1 2 3 4 5 6 7 8 We can reduce the bipartite maximum matching problem to the maximum flow problem by adding two new nodes to the graph: a source and a sink. We also add edges from the source to each left node and from each right node to the sink.
After this, the size of a maximum flow in the graph equals the size of a maximum matching in the original graph.
For example, the reduction for the above graph is as follows: 1 2 3 4 5 6 7 8 The maximum flow of this graph is as follows: 1 2 3 4 5 6 7 8 188 Hall’s theorem Hall’s theorem can be used to find out whether a bipartite graph has a matching that contains all left or right nodes. If the number of left and right nodes is the same, Hall’s theorem tells us if it is possible to construct a perfect matching that contains all nodes of the graph.
Assume that we want to find a matching that contains all left nodes. Let X be any set of left nodes and let f (X) be the set of their neighbors. According to Hall’s theorem, a matching that contains all left nodes exists exactly when for each X, the condition |X| ≤|f (X)| holds.
Let us study Hall’s theorem in the example graph. First, let X = {1,3} which yields f (X) = {5,6,8}: 1 2 3 4 5 6 7 8 The condition of Hall’s theorem holds, because |X| = 2 and |f (X)| = 3. Next, let X = {2,4} which yields f (X) = {7}: 1 2 3 4 5 6 7 8 In this case, |X| = 2 and |f (X)| = 1, so the condition of Hall’s theorem does not hold. This means that it is not possible to form a perfect matching for the graph. This result is not surprising, because we already know that the maximum matching of the graph is 3 and not 4.
If the condition of Hall’s theorem does not hold, the set X provides an expla-nation why we cannot form such a matching. Since X contains more nodes than f (X), there are no pairs for all nodes in X. For example, in the above graph, both nodes 2 and 4 should be connected with node 7 which is not possible.
Kőnig’s theorem A minimum node cover of a graph is a minimum set of nodes such that each edge of the graph has at least one endpoint in the set. In a general graph, finding a minimum node cover is a NP-hard problem. However, if the graph is bipartite, K˝ onig’s theorem tells us that the size of a minimum node cover and the size 189 of a maximum matching are always equal. Thus, we can calculate the size of a minimum node cover using a maximum flow algorithm.
Let us consider the following graph with a maximum matching of size 3: 1 2 3 4 5 6 7 8 Now K˝ onig’s theorem tells us that the size of a minimum node cover is also 3.
Such a cover can be constructed as follows: 1 2 3 4 5 6 7 8 The nodes that do not belong to a minimum node cover form a maximum independent set. This is the largest possible set of nodes such that no two nodes in the set are connected with an edge. Once again, finding a maximum independent set in a general graph is a NP-hard problem, but in a bipartite graph we can use K˝ onig’s theorem to solve the problem efficiently. In the example graph, the maximum independent set is as follows: 1 2 3 4 5 6 7 8 20.4 Path covers A path cover is a set of paths in a graph such that each node of the graph belongs to at least one path. It turns out that in directed, acyclic graphs, we can reduce the problem of finding a minimum path cover to the problem of finding a maximum flow in another graph.
190 Node-disjoint path cover In a node-disjoint path cover, each node belongs to exactly one path. As an example, consider the following graph: 1 2 3 4 5 6 7 A minimum node-disjoint path cover of this graph consists of three paths. For example, we can choose the following paths: 1 2 3 4 5 6 7 Note that one of the paths only contains node 2, so it is possible that a path does not contain any edges.
We can find a minimum node-disjoint path cover by constructing a matching graph where each node of the original graph is represented by two nodes: a left node and a right node. There is an edge from a left node to a right node if there is such an edge in the original graph. In addition, the matching graph contains a source and a sink, and there are edges from the source to all left nodes and from all right nodes to the sink.
A maximum matching in the resulting graph corresponds to a minimum node-disjoint path cover in the original graph. For example, the following matching graph for the above graph contains a maximum matching of size 4: 1 2 3 4 5 6 7 1 2 3 4 5 6 7 Each edge in the maximum matching of the matching graph corresponds to an edge in the minimum node-disjoint path cover of the original graph. Thus, the size of the minimum node-disjoint path cover is n−c, where n is the number of nodes in the original graph and c is the size of the maximum matching.
191 General path cover A general path cover is a path cover where a node can belong to more than one path. A minimum general path cover may be smaller than a minimum node-disjoint path cover, because a node can be used multiple times in paths.
Consider again the following graph: 1 2 3 4 5 6 7 The minimum general path cover of this graph consists of two paths. For example, the first path may be as follows: 1 2 3 4 5 6 7 And the second path may be as follows: 1 2 3 4 5 6 7 A minimum general path cover can be found almost like a minimum node-disjoint path cover. It suffices to add some new edges to the matching graph so that there is an edge a →b always when there is a path from a to b in the original graph (possibly through several edges).
The matching graph for the above graph is as follows: 1 2 3 4 5 6 7 1 2 3 4 5 6 7 192 Dilworth’s theorem An antichain is a set of nodes of a graph such that there is no path from any node to another node using the edges of the graph. Dilworth’s theorem states that in a directed acyclic graph, the size of a minimum general path cover equals the size of a maximum antichain.
For example, nodes 3 and 7 form an antichain in the following graph: 1 2 3 4 5 6 7 This is a maximum antichain, because it is not possible to construct any antichain that would contain three nodes. We have seen before that the size of a minimum general path cover of this graph consists of two paths.
193 194 Part III Advanced topics 195 Chapter 21 Number theory Number theory is a branch of mathematics that studies integers. Number theory is a fascinating field, because many questions involving integers are very difficult to solve even if they seem simple at first glance.
As an example, consider the following equation: x3 + y3 + z3 = 33 It is easy to find three real numbers x, y and z that satisfy the equation. For example, we can choose x = 3, y = 3 p 3, z = 3 p 3.
However, it is an open problem in number theory if there are any three integers x, y and z that would satisfy the equation .
In this chapter, we will focus on basic concepts and algorithms in number theory. Throughout the chapter, we will assume that all numbers are integers, if not otherwise stated.
21.1 Primes and factors A number a is called a factor or a divisor of a number b if a divides b. If a is a factor of b, we write a | b, and otherwise we write a ∤b. For example, the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
A number n > 1 is a prime if its only positive factors are 1 and n. For example, 7, 19 and 41 are primes, but 35 is not a prime, because 5·7 = 35. For every number n > 1, there is a unique prime factorization n = pα1 1 pα2 2 ··· pαk k , where p1, p2,..., pk are distinct primes and α1,α2,...,αk are positive numbers.
For example, the prime factorization for 84 is 84 = 22 ·31 ·71.
197 The number of factors of a number n is τ(n) = k Y i=1 (αi +1), because for each prime pi, there are αi +1 ways to choose how many times it appears in the factor. For example, the number of factors of 84 is τ(84) = 3·2·2 = 12.
The factors are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.
The sum of factors of n is σ(n) = k Y i=1 (1+ pi +...+ pαi i ) = k Y i=1 pai+1 i −1 pi −1 , where the latter formula is based on the geometric progression formula. For example, the sum of factors of 84 is σ(84) = 23 −1 2−1 · 32 −1 3−1 · 72 −1 7−1 = 7·4·8 = 224.
The product of factors of n is µ(n) = nτ(n)/2, because we can form τ(n)/2 pairs from the factors, each with product n. For example, the factors of 84 produce the pairs 1 · 84, 2 · 42, 3 · 28, etc., and the product of the factors is µ(84) = 846 = 351298031616.
A number n is called a perfect number if n = σ(n)−n, i.e., n equals the sum of its factors between 1 and n−1. For example, 28 is a perfect number, because 28 = 1+2+4+7+14.
Number of primes It is easy to show that there is an infinite number of primes. If the number of primes would be finite, we could construct a set P = {p1, p2,..., pn} that would contain all the primes. For example, p1 = 2, p2 = 3, p3 = 5, and so on. However, using P, we could form a new prime p1p2 ··· pn +1 that is larger than all elements in P. This is a contradiction, and the number of primes has to be infinite.
Density of primes The density of primes means how often there are primes among the numbers.
Let π(n) denote the number of primes between 1 and n. For example, π(10) = 4, because there are 4 primes between 1 and 10: 2, 3, 5 and 7.
It is possible to show that π(n) ≈ n lnn, which means that primes are quite frequent. For example, the number of primes between 1 and 106 is π(106) = 78498, and 106/ln106 ≈72382.
198 Conjectures There are many conjectures involving primes. Most people think that the con-jectures are true, but nobody has been able to prove them. For example, the following conjectures are famous: • Goldbach’s conjecture: Each even integer n > 2 can be represented as a sum n = a+ b so that both a and b are primes.
• Twin prime conjecture: There is an infinite number of pairs of the form {p, p +2}, where both p and p +2 are primes.
• Legendre’s conjecture: There is always a prime between numbers n2 and (n+1)2, where n is any positive integer.
Basic algorithms If a number n is not prime, it can be represented as a product a· b, where a ≤pn or b ≤pn, so it certainly has a factor between 2 and ⌊pn⌋. Using this observation, we can both test if a number is prime and find the prime factorization of a number in O(pn) time.
The following function prime checks if the given number n is prime. The function attempts to divide n by all numbers between 2 and ⌊pn⌋, and if none of them divides n, then n is prime.
bool prime(int n) { if (n < 2) return false; for (int x = 2; xx <= n; x++) { if (n%x == 0) return false; } return true; } The following function factors constructs a vector that contains the prime factor-ization of n. The function divides n by its prime factors, and adds them to the vector. The process ends when the remaining number n has no factors between 2 and ⌊pn⌋. If n > 1, it is prime and the last factor.
vector factors(int n) { vector f; for (int x = 2; xx <= n; x++) { while (n%x == 0) { f.push_back(x); n /= x; } } if (n > 1) f.push_back(n); return f; } 199 Note that each prime factor appears in the vector as many times as it divides the number. For example, 24 = 23 ·3, so the result of the function is [2,2,2,3].
Sieve of Eratosthenes The sieve of Eratosthenes is a preprocessing algorithm that builds an array using which we can efficiently check if a given number between 2...n is prime and, if it is not, find one prime factor of the number.
The algorithm builds an array sieve whose positions 2,3,...,n are used. The value sieve[k] = 0 means that k is prime, and the value sieve[k] ̸= 0 means that k is not a prime and one of its prime factors is sieve[k].
The algorithm iterates through the numbers 2...n one by one. Always when a new prime x is found, the algorithm records that the multiples of x (2x,3x,4x,...) are not primes, because the number x divides them.
For example, if n = 20, the array is as follows: 0 0 2 0 3 0 2 3 5 0 3 0 7 5 2 0 3 0 5 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 The following code implements the sieve of Eratosthenes. The code assumes that each element of sieve is initially zero.
for (int x = 2; x <= n; x++) { if (sieve[x]) continue; for (int u = 2x; u <= n; u += x) { sieve[u] = x; } } The inner loop of the algorithm is executed n/x times for each value of x. Thus, an upper bound for the running time of the algorithm is the harmonic sum n X x=2 n/x = n/2+ n/3+ n/4+···+ n/n = O(nlogn).
In fact, the algorithm is more efficient, because the inner loop will be executed only if the number x is prime. It can be shown that the running time of the algorithm is only O(nloglogn), a complexity very near to O(n).
Euclid’s algorithm The greatest common divisor of numbers a and b, gcd(a,b), is the greatest number that divides both a and b, and the least common multiple of a and b, lcm(a,b), is the smallest number that is divisible by both a and b. For example, gcd(24,36) = 12 and lcm(24,36) = 72.
The greatest common divisor and the least common multiple are connected as follows: lcm(a,b) = ab gcd(a,b) 200 Euclid’s algorithm1 provides an efficient way to find the greatest common divisor of two numbers. The algorithm is based on the following formula: gcd(a,b) = ( a b = 0 gcd(b,a mod b) b ̸= 0 For example, gcd(24,36) = gcd(36,24) = gcd(24,12) = gcd(12,0) = 12.
The algorithm can be implemented as follows: int gcd(int a, int b) { if (b == 0) return a; return gcd(b, a%b); } It can be shown that Euclid’s algorithm works in O(logn) time, where n = min(a,b). The worst case for the algorithm is the case when a and b are consecu-tive Fibonacci numbers. For example, gcd(13,8) = gcd(8,5) = gcd(5,3) = gcd(3,2) = gcd(2,1) = gcd(1,0) = 1.
Euler’s totient function Numbers a and b are coprime if gcd(a,b) = 1. Euler’s totient function ϕ(n) gives the number of coprime numbers to n between 1 and n. For example, ϕ(12) = 4, because 1, 5, 7 and 11 are coprime to 12.
The value of ϕ(n) can be calculated from the prime factorization of n using the formula ϕ(n) = k Y i=1 pαi−1 i (pi −1).
For example, ϕ(12) = 21 ·(2−1)·30 ·(3−1) = 4. Note that ϕ(n) = n−1 if n is prime.
21.2 Modular arithmetic In modular arithmetic, the set of numbers is limited so that only numbers 0,1,2,...,m−1 are used, where m is a constant. Each number x is represented by the number x mod m: the remainder after dividing x by m. For example, if m = 17, then 75 is represented by 75 mod 17 = 7.
Often we can take remainders before doing calculations. In particular, the following formulas hold: (x+ y) mod m = (x mod m+ y mod m) mod m (x−y) mod m = (x mod m−y mod m) mod m (x· y) mod m = (x mod m· y mod m) mod m xn mod m = (x mod m)n mod m 1Euclid was a Greek mathematician who lived in about 300 BC. This is perhaps the first known algorithm in history.
201 Modular exponentiation There is often need to efficiently calculate the value of xn mod m. This can be done in O(logn) time using the following recursion: xn = 1 n = 0 xn/2 · xn/2 n is even xn−1 · x n is odd It is important that in the case of an even n, the value of xn/2 is calculated only once. This guarantees that the time complexity of the algorithm is O(logn), because n is always halved when it is even.
The following function calculates the value of xn mod m: int modpow(int x, int n, int m) { if (n == 0) return 1%m; long long u = modpow(x,n/2,m); u = (uu)%m; if (n%2 == 1) u = (ux)%m; return u; } Fermat’s theorem and Euler’s theorem Fermat’s theorem states that xm−1 mod m = 1 when m is prime and x and m are coprime. This also yields xk mod m = xk mod (m−1) mod m.
More generally, Euler’s theorem states that xϕ(m) mod m = 1 when x and m are coprime. Fermat’s theorem follows from Euler’s theorem, because if m is a prime, then ϕ(m) = m−1.
Modular inverse The inverse of x modulo m is a number x−1 such that xx−1 mod m = 1.
For example, if x = 6 and m = 17, then x−1 = 3, because 6·3 mod 17 = 1.
Using modular inverses, we can divide numbers modulo m, because division by x corresponds to multiplication by x−1. For example, to evaluate the value 202 of 36/6 mod 17, we can use the formula 2·3 mod 17, because 36 mod 17 = 2 and 6−1 mod 17 = 3.
However, a modular inverse does not always exist. For example, if x = 2 and m = 4, the equation xx−1 mod m = 1 cannot be solved, because all multiples of 2 are even and the remainder can never be 1 when m = 4. It turns out that the value of x−1 mod m can be calculated exactly when x and m are coprime.
If a modular inverse exists, it can be calculated using the formula x−1 = xϕ(m)−1.
If m is prime, the formula becomes x−1 = xm−2.
For example, 6−1 mod 17 = 617−2 mod 17 = 3.
This formula allows us to efficiently calculate modular inverses using the modular exponentation algorithm. The formula can be derived using Euler’s theorem. First, the modular inverse should satisfy the following equation: xx−1 mod m = 1.
On the other hand, according to Euler’s theorem, xϕ(m) mod m = xxϕ(m)−1 mod m = 1, so the numbers x−1 and xϕ(m)−1 are equal.
Computer arithmetic In programming, unsigned integers are represented modulo 2k, where k is the number of bits of the data type. A usual consequence of this is that a number wraps around if it becomes too large.
For example, in C++, numbers of type unsigned int are represented mod-ulo 232. The following code declares an unsigned int variable whose value is 123456789. After this, the value will be multiplied by itself, and the result is 1234567892 mod 232 = 2537071545.
unsigned int x = 123456789; cout << xx << "\n"; // 2537071545 203 21.3 Solving equations Diophantine equations A Diophantine equation is an equation of the form ax+ by = c, where a, b and c are constants and the values of x and y should be found. Each number in the equation has to be an integer. For example, one solution for the equation 5x+2y = 11 is x = 3 and y = −2.
We can efficiently solve a Diophantine equation by using Euclid’s algorithm.
It turns out that we can extend Euclid’s algorithm so that it will find numbers x and y that satisfy the following equation: ax+ by = gcd(a,b) A Diophantine equation can be solved if c is divisible by gcd(a,b), and other-wise it cannot be solved.
As an example, let us find numbers x and y that satisfy the following equation: 39x+15y = 12 The equation can be solved, because gcd(39,15) = 3 and 3 | 12. When Euclid’s algorithm calculates the greatest common divisor of 39 and 15, it produces the following sequence of function calls: gcd(39,15) = gcd(15,9) = gcd(9,6) = gcd(6,3) = gcd(3,0) = 3 This corresponds to the following equations: 39−2·15 = 9 15−1·9 = 6 9−1·6 = 3 Using these equations, we can derive 39·2+15·(−5) = 3 and by multiplying this by 4, the result is 39·8+15·(−20) = 12, so a solution to the equation is x = 8 and y = −20.
A solution to a Diophantine equation is not unique, because we can form an infinite number of solutions if we know one solution. If a pair (x, y) is a solution, then also all pairs (x+ kb gcd(a,b), y− ka gcd(a,b)) are solutions, where k is any integer.
204 Chinese remainder theorem The Chinese remainder theorem solves a group of equations of the form x = a1 mod m1 x = a2 mod m2 ··· x = an mod mn where all pairs of m1,m2,...,mn are coprime.
Let x−1 m be the inverse of x modulo m, and Xk = m1m2 ···mn mk .
Using this notation, a solution to the equations is x = a1X1X1 −1 m1 + a2X2X2 −1 m2 +···+ anXnXn −1 mn.
In this solution, for each k = 1,2,...,n, akXkXk −1 mk mod mk = ak, because XkXk −1 mk mod mk = 1.
Since all other terms in the sum are divisible by mk, they have no effect on the remainder, and x mod mk = ak.
For example, a solution for x = 3 mod 5 x = 4 mod 7 x = 2 mod 3 is 3·21·1+4·15·1+2·35·2 = 263.
Once we have found a solution x, we can create an infinite number of other solutions, because all numbers of the form x+ m1m2 ···mn are solutions.
21.4 Other results Lagrange’s theorem Lagrange’s theorem states that every positive integer can be represented as a sum of four squares, i.e., a2 + b2 + c2 + d2. For example, the number 123 can be represented as the sum 82 +52 +52 +32.
205 Zeckendorf’s theorem Zeckendorf’s theorem states that every positive integer has a unique repre-sentation as a sum of Fibonacci numbers such that no two numbers are equal or consecutive Fibonacci numbers. For example, the number 74 can be represented as the sum 55+13+5+1.
Pythagorean triples A Pythagorean triple is a triple (a,b, c) that satisfies the Pythagorean theorem a2 +b2 = c2, which means that there is a right triangle with side lengths a, b and c. For example, (3,4,5) is a Pythagorean triple.
If (a,b, c) is a Pythagorean triple, all triples of the form (ka,kb,kc) are also Pythagorean triples where k > 1. A Pythagorean triple is primitive if a, b and c are coprime, and all Pythagorean triples can be constructed from primitive triples using a multiplier k.
Euclid’s formula can be used to produce all primitive Pythagorean triples.
Each such triple is of the form (n2 −m2,2nm,n2 + m2), where 0 < m < n, n and m are coprime and at least one of n and m is even. For example, when m = 1 and n = 2, the formula produces the smallest Pythagorean triple (22 −12,2·2·1,22 +12) = (3,4,5).
Wilson’s theorem Wilson’s theorem states that a number n is prime exactly when (n−1)! mod n = n−1.
For example, the number 11 is prime, because 10! mod 11 = 10, and the number 12 is not prime, because 11! mod 12 = 0 ̸= 11.
Hence, Wilson’s theorem can be used to find out whether a number is prime.
However, in practice, the theorem cannot be applied to large values of n, because it is difficult to calculate values of (n−1)! when n is large.
206 Chapter 22 Combinatorics Combinatorics studies methods for counting combinations of objects. Usually, the goal is to find a way to count the combinations efficiently without generating each combination separately.
As an example, consider the problem of counting the number of ways to represent an integer n as a sum of positive integers. For example, there are 8 representations for 4: • 1+1+1+1 • 1+1+2 • 1+2+1 • 2+1+1 • 2+2 • 3+1 • 1+3 • 4 A combinatorial problem can often be solved using a recursive function. In this problem, we can define a function f (n) that gives the number of representations for n. For example, f (4) = 8 according to the above example. The values of the function can be recursively calculated as follows: f (n) = ( 1 n = 0 f (0)+ f (1)+···+ f (n−1) n > 0 The base case is f (0) = 1, because the empty sum represents the number 0. Then, if n > 0, we consider all ways to choose the first number of the sum. If the first number is k, there are f (n−k) representations for the remaining part of the sum.
Thus, we calculate the sum of all values of the form f (n−k) where k < n.
The first values for the function are: f (0) = 1 f (1) = 1 f (2) = 2 f (3) = 4 f (4) = 8 Sometimes, a recursive formula can be replaced with a closed-form formula.
In this problem, f (n) = 2n−1, 207 which is based on the fact that there are n−1 possible positions for +-signs in the sum and we can choose any subset of them.
22.1 Binomial coefficients The binomial coefficient ¡n k ¢ equals the number of ways we can choose a subset of k elements from a set of n elements. For example, ¡5 3 ¢ = 10, because the set {1,2,3,4,5} has 10 subsets of 3 elements: {1,2,3},{1,2,4},{1,2,5},{1,3,4},{1,3,5},{1,4,5},{2,3,4},{2,3,5},{2,4,5},{3,4,5} Formula 1 Binomial coefficients can be recursively calculated as follows: Ã n k !
= Ã n−1 k −1 !
+ Ã n−1 k !
The idea is to fix an element x in the set. If x is included in the subset, we have to choose k −1 elements from n−1 elements, and if x is not included in the subset, we have to choose k elements from n−1 elements.
The base cases for the recursion are à n 0 !
= Ã n n !
= 1, because there is always exactly one way to construct an empty subset and a subset that contains all the elements.
Formula 2 Another way to calculate binomial coefficients is as follows: Ã n k !
= n!
k!(n−k)!.
There are n! permutations of n elements. We go through all permutations and always include the first k elements of the permutation in the subset. Since the order of the elements in the subset and outside the subset does not matter, the result is divided by k! and (n−k)!
Properties For binomial coefficients, Ã n k !
= Ã n n−k !
, 208 because we actually divide a set of n elements into two subsets: the first contains k elements and the second contains n−k elements.
The sum of binomial coefficients is à n 0 !
+ Ã n 1 !
+ Ã n 2 !
+...+ Ã n n !
= 2n.
The reason for the name ”binomial coefficient” can be seen when the binomial (a+ b) is raised to the nth power: (a+ b)n = Ã n 0 !
anb0 + Ã n 1 !
an−1b1 +...+ Ã n n−1 !
a1bn−1 + Ã n n !
a0bn.
Binomial coefficients also appear in Pascal’s triangle where each value equals the sum of two above values: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 ...
...
...
...
...
Boxes and balls ”Boxes and balls” is a useful model, where we count the ways to place k balls in n boxes. Let us consider three scenarios: Scenario 1: Each box can contain at most one ball. For example, when n = 5 and k = 2, there are 10 solutions: In this scenario, the answer is directly the binomial coefficient ¡n k ¢ .
Scenario 2: A box can contain multiple balls. For example, when n = 5 and k = 2, there are 15 solutions: 209 The process of placing the balls in the boxes can be represented as a string that consists of symbols ”o” and ”→”. Initially, assume that we are standing at the leftmost box. The symbol ”o” means that we place a ball in the current box, and the symbol ”→” means that we move to the next box to the right.
Using this notation, each solution is a string that contains k times the symbol ”o” and n−1 times the symbol ”→”. For example, the upper-right solution in the above picture corresponds to the string ”→→o →o →”. Thus, the number of solutions is ¡k+n−1 k ¢ .
Scenario 3: Each box may contain at most one ball, and in addition, no two adjacent boxes may both contain a ball. For example, when n = 5 and k = 2, there are 6 solutions: In this scenario, we can assume that k balls are initially placed in boxes and there is an empty box between each two adjacent boxes. The remaining task is to choose the positions for the remaining empty boxes. There are n−2k +1 such boxes and k +1 positions for them. Thus, using the formula of scenario 2, the number of solutions is ¡ n−k+1 n−2k+1 ¢ .
Multinomial coefficients The multinomial coefficient à n k1,k2,...,km !
= n!
k1!k2!···km!, equals the number of ways we can divide n elements into subsets of sizes k1,k2,...,km, where k1 +k2 +···+km = n. Multinomial coefficients can be seen as a generalization of binomial cofficients; if m = 2, the above formula corresponds to the binomial coefficient formula.
22.2 Catalan numbers The Catalan number Cn equals the number of valid parenthesis expressions that consist of n left parentheses and n right parentheses.
For example, C3 = 5, because we can construct the following parenthesis expressions using three left and right parentheses: • ()()() • (())() • ()(()) • ((())) • (()()) 210 Parenthesis expressions What is exactly a valid parenthesis expression? The following rules precisely define all valid parenthesis expressions: • An empty parenthesis expression is valid.
• If an expression A is valid, then also the expression (A) is valid.
• If expressions A and B are valid, then also the expression AB is valid.
Another way to characterize valid parenthesis expressions is that if we choose any prefix of such an expression, it has to contain at least as many left parenthe-ses as right parentheses. In addition, the complete expression has to contain an equal number of left and right parentheses.
Formula 1 Catalan numbers can be calculated using the formula Cn = n−1 X i=0 CiCn−i−1.
The sum goes through the ways to divide the expression into two parts such that both parts are valid expressions and the first part is as short as possible but not empty. For any i, the first part contains i +1 pairs of parentheses and the number of expressions is the product of the following values: • Ci: the number of ways to construct an expression using the parentheses of the first part, not counting the outermost parentheses • Cn−i−1: the number of ways to construct an expression using the parenthe-ses of the second part The base case is C0 = 1, because we can construct an empty parenthesis expression using zero pairs of parentheses.
Formula 2 Catalan numbers can also be calculated using binomial coefficients: Cn = 1 n+1 Ã 2n n !
The formula can be explained as follows: There are a total of ¡2n n ¢ ways to construct a (not necessarily valid) parenthesis expression that contains n left parentheses and n right parentheses. Let us calculate the number of such expressions that are not valid.
If a parenthesis expression is not valid, it has to contain a prefix where the number of right parentheses exceeds the number of left parentheses. The 211 idea is to reverse each parenthesis that belongs to such a prefix. For example, the expression ())()( contains a prefix ()), and after reversing the prefix, the expression becomes )((()(.
The resulting expression consists of n + 1 left parentheses and n −1 right parentheses. The number of such expressions is ¡ 2n n+1 ¢ , which equals the number of non-valid parenthesis expressions. Thus, the number of valid parenthesis expressions can be calculated using the formula à 2n n !
− Ã 2n n+1 !
= Ã 2n n !
− n n+1 Ã 2n n !
= 1 n+1 Ã 2n n !
.
Counting trees Catalan numbers are also related to trees: • there are Cn binary trees of n nodes • there are Cn−1 rooted trees of n nodes For example, for C3 = 5, the binary trees are and the rooted trees are 22.3 Inclusion-exclusion Inclusion-exclusion is a technique that can be used for counting the size of a union of sets when the sizes of the intersections are known, and vice versa. A simple example of the technique is the formula |A ∪B| = |A|+|B|−|A ∩B|, where A and B are sets and |X| denotes the size of X. The formula can be illustrated as follows: A B A ∩B 212 Our goal is to calculate the size of the union A ∪B that corresponds to the area of the region that belongs to at least one circle. The picture shows that we can calculate the area of A ∪B by first summing the areas of A and B and then subtracting the area of A ∩B.
The same idea can be applied when the number of sets is larger. When there are three sets, the inclusion-exclusion formula is |A ∪B ∪C| = |A|+|B|+|C|−|A ∩B|−|A ∩C|−|B ∩C|+|A ∩B ∩C| and the corresponding picture is A B C A ∩B A ∩C B ∩C A ∩B ∩C In the general case, the size of the union X1 ∪X2 ∪··· ∪Xn can be calcu-lated by going through all possible intersections that contain some of the sets X1, X2,..., Xn. If the intersection contains an odd number of sets, its size is added to the answer, and otherwise its size is subtracted from the answer.
Note that there are similar formulas for calculating the size of an intersection from the sizes of unions. For example, |A ∩B| = |A|+|B|−|A ∪B| and |A ∩B ∩C| = |A|+|B|+|C|−|A ∪B|−|A ∪C|−|B ∪C|+|A ∪B ∪C|.
Derangements As an example, let us count the number of derangements of elements {1,2,...,n}, i.e., permutations where no element remains in its original place. For example, when n = 3, there are two derangements: (2,3,1) and (3,1,2).
One approach for solving the problem is to use inclusion-exclusion. Let Xk be the set of permutations that contain the element k at position k. For example, when n = 3, the sets are as follows: X1 = {(1,2,3),(1,3,2)} X2 = {(1,2,3),(3,2,1)} X3 = {(1,2,3),(2,1,3)} Using these sets, the number of derangements equals n!−|X1 ∪X2 ∪···∪Xn|, 213 so it suffices to calculate the size of the union. Using inclusion-exclusion, this reduces to calculating sizes of intersections which can be done efficiently. For example, when n = 3, the size of |X1 ∪X2 ∪X3| is |X1|+|X2|+|X3|−|X1 ∩X2|−|X1 ∩X3|−|X2 ∩X3|+|X1 ∩X2 ∩X3| = 2+2+2−1−1−1+1 = 4, so the number of solutions is 3!−4 = 2.
It turns out that the problem can also be solved without using inclusion-exclusion. Let f (n) denote the number of derangements for {1,2,...,n}. We can use the following recursive formula: f (n) = 0 n = 1 1 n = 2 (n−1)(f (n−2)+ f (n−1)) n > 2 The formula can be derived by considering the possibilities how the element 1 changes in the derangement. There are n−1 ways to choose an element x that replaces the element 1. In each such choice, there are two options: Option 1: We also replace the element x with the element 1. After this, the remaining task is to construct a derangement of n−2 elements.
Option 2: We replace the element x with some other element than 1. Now we have to construct a derangement of n−1 element, because we cannot replace the element x with the element 1, and all other elements must be changed.
22.4 Burnside’s lemma Burnside’s lemma can be used to count the number of combinations so that only one representative is counted for each group of symmetric combinations.
Burnside’s lemma states that the number of combinations is n X k=1 c(k) n , where there are n ways to change the position of a combination, and there are c(k) combinations that remain unchanged when the kth way is applied.
As an example, let us calculate the number of necklaces of n pearls, where each pearl has m possible colors. Two necklaces are symmetric if they are similar after rotating them. For example, the necklace has the following symmetric necklaces: 214 There are n ways to change the position of a necklace, because we can rotate it 0,1,...,n−1 steps clockwise. If the number of steps is 0, all mn necklaces remain the same, and if the number of steps is 1, only the m necklaces where each pearl has the same color remain the same.
More generally, when the number of steps is k, a total of mgcd(k,n) necklaces remain the same, where gcd(k,n) is the greatest common divisor of k and n. The reason for this is that blocks of pearls of size gcd(k,n) will replace each other. Thus, according to Burnside’s lemma, the number of necklaces is n−1 X i=0 mgcd(i,n) n .
For example, the number of necklaces of length 4 with 3 colors is 34 +3+32 +3 4 = 24.
22.5 Cayley’s formula Cayley’s formula states that there are nn−2 labeled trees that contain n nodes.
The nodes are labeled 1,2,...,n, and two trees are different if either their struc-ture or labeling is different.
For example, when n = 4, the number of labeled trees is 44−2 = 16: 1 2 3 4 2 1 3 4 3 1 2 4 4 1 2 3 1 2 3 4 1 2 4 3 1 3 2 4 1 3 4 2 1 4 2 3 1 4 3 2 2 1 3 4 2 1 4 3 2 3 1 4 2 4 1 3 3 1 2 4 3 2 1 4 Next we will see how Cayley’s formula can be derived using Prüfer codes.
215 Prüfer code A Prüfer code is a sequence of n−2 numbers that describes a labeled tree. The code is constructed by following a process that removes n−2 leaves from the tree.
At each step, the leaf with the smallest label is removed, and the label of its only neighbor is added to the code.
For example, let us calculate the Prüfer code of the following graph: 1 2 3 4 5 First we remove node 1 and add node 4 to the code: 2 3 4 5 Then we remove node 3 and add node 4 to the code: 2 4 5 Finally we remove node 4 and add node 2 to the code: 2 5 Thus, the Prüfer code of the graph is [4,4,2].
We can construct a Prüfer code for any tree, and more importantly, the original tree can be reconstructed from a Prüfer code. Hence, the number of labeled trees of n nodes equals nn−2, the number of Prüfer codes of size n.
216 Chapter 23 Matrices A matrix is a mathematical concept that corresponds to a two-dimensional array in programming. For example, A = 6 13 7 4 7 0 8 2 9 5 4 18 is a matrix of size 3×4, i.e., it has 3 rows and 4 columns. The notation [i, j] refers to the element in row i and column j in a matrix. For example, in the above matrix, A[2,3] = 8 and A[3,1] = 9.
A special case of a matrix is a vector that is a one-dimensional matrix of size n×1. For example, V = 4 7 5 is a vector that contains three elements.
The transpose AT of a matrix A is obtained when the rows and columns of A are swapped, i.e., AT[i, j] = A[j, i]: AT = 6 7 9 13 0 5 7 8 4 4 2 18 A matrix is a square matrix if it has the same number of rows and columns.
For example, the following matrix is a square matrix: S = 3 12 4 5 9 15 0 2 4 23.1 Operations The sum A + B of matrices A and B is defined if the matrices are of the same size. The result is a matrix where each element is the sum of the corresponding elements in A and B.
217 For example, ·6 1 4 3 9 2 ¸ + ·4 9 3 8 1 3 ¸ = ·6+4 1+9 4+3 3+8 9+1 2+3 ¸ = ·10 10 7 11 10 5 ¸ .
Multiplying a matrix A by a value x means that each element of A is multi-plied by x. For example, 2· ·6 1 4 3 9 2 ¸ = ·2·6 2·1 2·4 2·3 2·9 2·2 ¸ = ·12 2 8 6 18 4 ¸ .
Matrix multiplication The product AB of matrices A and B is defined if A is of size a × n and B is of size n × b, i.e., the width of A equals the height of B. The result is a matrix of size a× b whose elements are calculated using the formula AB[i, j] = n X k=1 A[i,k]·B[k, j].
The idea is that each element of AB is a sum of products of elements of A and B according to the following picture: A AB B For example, 1 4 3 9 8 6 · ·1 6 2 9 ¸ = 1·1+4·2 1·6+4·9 3·1+9·2 3·6+9·9 8·1+6·2 8·6+6·9 = 9 42 21 99 20 102 .
Matrix multiplication is associative, so A(BC) = (AB)C holds, but it is not commutative, so AB = BA does not usually hold.
An identity matrix is a square matrix where each element on the diagonal is 1 and all other elements are 0. For example, the following matrix is the 3×3 identity matrix: I = 1 0 0 0 1 0 0 0 1 218 Multiplying a matrix by an identity matrix does not change it. For example, 1 0 0 0 1 0 0 0 1 · 1 4 3 9 8 6 = 1 4 3 9 8 6 and 1 4 3 9 8 6 · ·1 0 0 1 ¸ = 1 4 3 9 8 6 .
Using a straightforward algorithm, we can calculate the product of two n× n matrices in O(n3) time. There are also more efficient algorithms for matrix multiplication1, but they are mostly of theoretical interest and such algorithms are not necessary in competitive programming.
Matrix power The power Ak of a matrix A is defined if A is a square matrix. The definition is based on matrix multiplication: Ak = A · A · A··· A | {z } k times For example, ·2 5 1 4 ¸3 = ·2 5 1 4 ¸ · ·2 5 1 4 ¸ · ·2 5 1 4 ¸ = ·48 165 33 114 ¸ .
In addition, A0 is an identity matrix. For example, ·2 5 1 4 ¸0 = ·1 0 0 1 ¸ .
The matrix Ak can be efficiently calculated in O(n3logk) time using the algorithm in Chapter 21.2. For example, ·2 5 1 4 ¸8 = ·2 5 1 4 ¸4 · ·2 5 1 4 ¸4 .
Determinant The determinant det(A) of a matrix A is defined if A is a square matrix. If A is of size 1×1, then det(A) = A[1,1]. The determinant of a larger matrix is calculated recursively using the formula det(A) = n X j=1 A[1, j]C[1, j], where C[i, j] is the cofactor of A at [i, j]. The cofactor is calculated using the formula C[i, j] = (−1)i+j det(M[i, j]), 1The first such algorithm was Strassen’s algorithm, published in 1969 , whose time complexity is O(n2.80735); the best current algorithm works in O(n2.37286) time.
219 where M[i, j] is obtained by removing row i and column j from A. Due to the coefficient (−1)i+j in the cofactor, every other determinant is positive and negative.
For example, det( ·3 4 1 6 ¸ ) = 3·6−4·1 = 14 and det( 2 4 3 5 1 6 7 2 4 ) = 2·det( ·1 6 2 4 ¸ )−4·det( ·5 6 7 4 ¸ )+3·det( ·5 1 7 2 ¸ ) = 81.
The determinant of A tells us whether there is an inverse matrix A−1 such that A·A−1 = I, where I is an identity matrix. It turns out that A−1 exists exactly when det(A) ̸= 0, and it can be calculated using the formula A−1[i, j] = C[j, i] det(A).
For example, 2 4 3 5 1 6 7 2 4 | {z } A · 1 81 −8 −10 21 22 −13 3 3 24 −18 | {z } A−1 = 1 0 0 0 1 0 0 0 1 | {z } I .
23.2 Linear recurrences A linear recurrence is a function f (n) whose initial values are f (0), f (1),..., f (k− 1) and larger values are calculated recursively using the formula f (n) = c1 f (n−1)+ c2 f (n−2)+...+ ck f (n−k), where c1, c2,..., ck are constant coefficients.
Dynamic programming can be used to calculate any value of f (n) in O(kn) time by calculating all values of f (0), f (1),..., f (n) one after another. However, if k is small, it is possible to calculate f (n) much more efficiently in O(k3logn) time using matrix operations.
Fibonacci numbers A simple example of a linear recurrence is the following function that defines the Fibonacci numbers: f (0) = 0 f (1) = 1 f (n) = f (n−1)+ f (n−2) In this case, k = 2 and c1 = c2 = 1.
220 To efficiently calculate Fibonacci numbers, we represent the Fibonacci formula as a square matrix X of size 2×2, for which the following holds: X · · f (i) f (i +1) ¸ = ·f (i +1) f (i +2) ¸ Thus, values f (i) and f (i +1) are given as ”input” for X, and X calculates values f (i +1) and f (i +2) from them. It turns out that such a matrix is X = ·0 1 1 1 ¸ .
For example, ·0 1 1 1 ¸ · ·f (5) f (6) ¸ = ·0 1 1 1 ¸ · ·5 8 ¸ = · 8 13 ¸ = ·f (6) f (7) ¸ .
Thus, we can calculate f (n) using the formula · f (n) f (n+1) ¸ = X n · ·f (0) f (1) ¸ = ·0 1 1 1 ¸n · ·0 1 ¸ .
The value of X n can be calculated in O(logn) time, so the value of f (n) can also be calculated in O(logn) time.
General case Let us now consider the general case where f (n) is any linear recurrence. Again, our goal is to construct a matrix X for which X · f (i) f (i +1) .
.
.
f (i + k −1) = f (i +1) f (i +2) .
.
.
f (i + k) .
Such a matrix is X = 0 1 0 0 ··· 0 0 0 1 0 ··· 0 0 0 0 1 ··· 0 .
.
.
.
.
.
.
.
.
.
.
.
...
.
.
.
0 0 0 0 ··· 1 ck ck−1 ck−2 ck−3 ··· c1 .
In the first k−1 rows, each element is 0 except that one element is 1. These rows replace f (i) with f (i +1), f (i +1) with f (i +2), and so on. The last row contains the coefficients of the recurrence to calculate the new value f (i + k).
Now, f (n) can be calculated in O(k3logn) time using the formula f (n) f (n+1) .
.
.
f (n+ k −1) = X n · f (0) f (1) .
.
.
f (k −1) .
221 23.3 Graphs and matrices Counting paths The powers of an adjacency matrix of a graph have an interesting property. When V is an adjacency matrix of an unweighted graph, the matrix V n contains the numbers of paths of n edges between the nodes in the graph.
For example, for the graph 1 4 2 3 5 6 the adjacency matrix is V = 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 .
Now, for example, the matrix V 4 = 0 0 1 1 1 0 2 0 0 0 2 2 0 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 contains the numbers of paths of 4 edges between the nodes. For example, V 4[2,5] = 2, because there are two paths of 4 edges from node 2 to node 5: 2 →1 →4 →2 →5 and 2 →6 →3 →2 →5.
Shortest paths Using a similar idea in a weighted graph, we can calculate for each pair of nodes the minimum length of a path between them that contains exactly n edges. To calculate this, we have to define matrix multiplication in a new way, so that we do not calculate the numbers of paths but minimize the lengths of paths.
222 As an example, consider the following graph: 1 4 2 3 5 6 4 1 2 4 1 2 3 2 Let us construct an adjacency matrix where ∞means that an edge does not exist, and other values correspond to edge weights. The matrix is V = ∞ ∞ ∞ 4 ∞ ∞ 2 ∞ ∞ ∞ 1 2 ∞ 4 ∞ ∞ ∞ ∞ ∞ 1 ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ 3 ∞ 2 ∞ .
Instead of the formula AB[i, j] = n X k=1 A[i,k]·B[k, j] we now use the formula AB[i, j] = n min k=1 A[i,k]+B[k, j] for matrix multiplication, so we calculate a minimum instead of a sum, and a sum of elements instead of a product. After this modification, matrix powers correspond to shortest paths in the graph.
For example, as V 4 = ∞ ∞ 10 11 9 ∞ 9 ∞ ∞ ∞ 8 9 ∞ 11 ∞ ∞ ∞ ∞ ∞ 8 ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞ 12 13 11 ∞ , we can conclude that the minimum length of a path of 4 edges from node 2 to node 5 is 8. Such a path is 2 →1 →4 →2 →5.
Kirchhoff’s theorem Kirchhoff’s theorem provides a way to calculate the number of spanning trees of a graph as a determinant of a special matrix. For example, the graph 1 2 3 4 223 has three spanning trees: 1 2 3 4 1 2 3 4 1 2 3 4 To calculate the number of spanning trees, we construct a Laplacean matrix L, where L[i, i] is the degree of node i and L[i, j] = −1 if there is an edge between nodes i and j, and otherwise L[i, j] = 0. The Laplacean matrix for the above graph is as follows: L = 3 −1 −1 −1 −1 1 0 0 −1 0 2 −1 −1 0 −1 2 It can be shown that the number of spanning trees equals the determinant of a matrix that is obtained when we remove any row and any column from L. For example, if we remove the first row and column, the result is det( 1 0 0 0 2 −1 0 −1 2 ) = 3.
The determinant is always the same, regardless of which row and column we remove from L.
Note that Cayley’s formula in Chapter 22.5 is a special case of Kirchhoff’s theorem, because in a complete graph of n nodes det( n−1 −1 ··· −1 −1 n−1 ··· −1 .
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.
.
.
.
...
.
.
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−1 −1 ··· n−1 ) = nn−2.
224 Chapter 24 Probability A probability is a real number between 0 and 1 that indicates how probable an event is. If an event is certain to happen, its probability is 1, and if an event is impossible, its probability is 0. The probability of an event is denoted P(···) where the three dots describe the event.
For example, when throwing a dice, the outcome is an integer between 1 and 6, and the probability of each outcome is 1/6. For example, we can calculate the following probabilities: • P(”the outcome is 4”) = 1/6 • P(”the outcome is not 6”) = 5/6 • P(”the outcome is even”) = 1/2 24.1 Calculation To calculate the probability of an event, we can either use combinatorics or simulate the process that generates the event. As an example, let us calculate the probability of drawing three cards with the same value from a shuffled deck of cards (for example, ♠8, ♣8 and ♦8).
Method 1 We can calculate the probability using the formula number of desired outcomes total number of outcomes .
In this problem, the desired outcomes are those in which the value of each card is the same. There are 13 ¡4 3 ¢ such outcomes, because there are 13 possibilities for the value of the cards and ¡4 3 ¢ ways to choose 3 suits from 4 possible suits.
There are a total of ¡52 3 ¢ outcomes, because we choose 3 cards from 52 cards.
Thus, the probability of the event is 13 ¡4 3 ¢ ¡52 3 ¢ = 1 425.
225 Method 2 Another way to calculate the probability is to simulate the process that generates the event. In this example, we draw three cards, so the process consists of three steps. We require that each step of the process is successful.
Drawing the first card certainly succeeds, because there are no restrictions.
The second step succeeds with probability 3/51, because there are 51 cards left and 3 of them have the same value as the first card. In a similar way, the third step succeeds with probability 2/50.
The probability that the entire process succeeds is 1· 3 51 · 2 50 = 1 425.
24.2 Events An event in probability theory can be represented as a set A ⊂X, where X contains all possible outcomes and A is a subset of outcomes. For example, when drawing a dice, the outcomes are X = {1,2,3,4,5,6}.
Now, for example, the event ”the outcome is even” corresponds to the set A = {2,4,6}.
Each outcome x is assigned a probability p(x). Then, the probability P(A) of an event A can be calculated as a sum of probabilities of outcomes using the formula P(A) = X x∈A p(x).
For example, when throwing a dice, p(x) = 1/6 for each outcome x, so the proba-bility of the event ”the outcome is even” is p(2)+ p(4)+ p(6) = 1/2.
The total probability of the outcomes in X must be 1, i.e., P(X) = 1.
Since the events in probability theory are sets, we can manipulate them using standard set operations: • The complement ¯ A means ”A does not happen”. For example, when throwing a dice, the complement of A = {2,4,6} is ¯ A = {1,3,5}.
• The union A ∪B means ”A or B happen”. For example, the union of A = {2,5} and B = {4,5,6} is A ∪B = {2,4,5,6}.
• The intersection A ∩B means ”A and B happen”. For example, the inter-section of A = {2,5} and B = {4,5,6} is A ∩B = {5}.
226 Complement The probability of the complement ¯ A is calculated using the formula P( ¯ A) = 1−P(A).
Sometimes, we can solve a problem easily using complements by solving the opposite problem. For example, the probability of getting at least one six when throwing a dice ten times is 1−(5/6)10.
Here 5/6 is the probability that the outcome of a single throw is not six, and (5/6)10 is the probability that none of the ten throws is a six. The complement of this is the answer to the problem.
Union The probability of the union A ∪B is calculated using the formula P(A ∪B) = P(A)+ P(B)−P(A ∩B).
For example, when throwing a dice, the union of the events A = ”the outcome is even” and B = ”the outcome is less than 4” is A ∪B = ”the outcome is even or less than 4”, and its probability is P(A ∪B) = P(A)+ P(B)−P(A ∩B) = 1/2+1/2−1/6 = 5/6.
If the events A and B are disjoint, i.e., A ∩B is empty, the probability of the event A ∪B is simply P(A ∪B) = P(A)+ P(B).
Conditional probability The conditional probability P(A|B) = P(A ∩B) P(B) is the probability of A assuming that B happens. Hence, when calculating the probability of A, we only consider the outcomes that also belong to B.
Using the previous sets, P(A|B) = 1/3, because the outcomes of B are {1,2,3}, and one of them is even. This is the probability of an even outcome if we know that the outcome is between 1...3.
227 Intersection Using conditional probability, the probability of the intersection A ∩B can be calculated using the formula P(A ∩B) = P(A)P(B|A).
Events A and B are independent if P(A|B) = P(A) and P(B|A) = P(B), which means that the fact that B happens does not change the probability of A, and vice versa. In this case, the probability of the intersection is P(A ∩B) = P(A)P(B).
For example, when drawing a card from a deck, the events A = ”the suit is clubs” and B = ”the value is four” are independent. Hence the event A ∩B = ”the card is the four of clubs” happens with probability P(A ∩B) = P(A)P(B) = 1/4·1/13 = 1/52.
24.3 Random variables A random variable is a value that is generated by a random process. For example, when throwing two dice, a possible random variable is X = ”the sum of the outcomes”.
For example, if the outcomes are [4,6] (meaning that we first throw a four and then a six), then the value of X is 10.
We denote P(X = x) the probability that the value of a random variable X is x. For example, when throwing two dice, P(X = 10) = 3/36, because the total number of outcomes is 36 and there are three possible ways to obtain the sum 10: [4,6], [5,5] and [6,4].
228 Expected value The expected value E[X] indicates the average value of a random variable X.
The expected value can be calculated as the sum X x P(X = x)x, where x goes through all possible values of X.
For example, when throwing a dice, the expected outcome is 1/6·1+1/6·2+1/6·3+1/6·4+1/6·5+1/6·6 = 7/2.
A useful property of expected values is linearity. It means that the sum E[X1 + X2 + ··· + Xn] always equals the sum E[X1] + E[X2] + ··· + E[Xn]. This formula holds even if random variables depend on each other.
For example, when throwing two dice, the expected sum is E[X1 + X2] = E[X1]+ E[X2] = 7/2+7/2 = 7.
Let us now consider a problem where n balls are randomly placed in n boxes, and our task is to calculate the expected number of empty boxes. Each ball has an equal probability to be placed in any of the boxes. For example, if n = 2, the possibilities are as follows: In this case, the expected number of empty boxes is 0+0+1+1 4 = 1 2.
In the general case, the probability that a single box is empty is ³n−1 n ´n , because no ball should be placed in it. Hence, using linearity, the expected number of empty boxes is n· ³n−1 n ´n .
Distributions The distribution of a random variable X shows the probability of each value that X may have. The distribution consists of values P(X = x). For example, when throwing two dice, the distribution for their sum is: x 2 3 4 5 6 7 8 9 10 11 12 P(X = x) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 229 In a uniform distribution, the random variable X has n possible values a,a+1,...,b and the probability of each value is 1/n. For example, when throwing a dice, a = 1, b = 6 and P(X = x) = 1/6 for each value x.
The expected value of X in a uniform distribution is E[X] = a+ b 2 .
In a binomial distribution, n attempts are made and the probability that a single attempt succeeds is p. The random variable X counts the number of successful attempts, and the probability of a value x is P(X = x) = px(1−p)n−x à n x !
, where px and (1−p)n−x correspond to successful and unsuccessful attemps, and ¡n x ¢ is the number of ways we can choose the order of the attempts.
For example, when throwing a dice ten times, the probability of throwing a six exactly three times is (1/6)3(5/6)7¡10 3 ¢ .
The expected value of X in a binomial distribution is E[X] = pn.
In a geometric distribution, the probability that an attempt succeeds is p, and we continue until the first success happens. The random variable X counts the number of attempts needed, and the probability of a value x is P(X = x) = (1−p)x−1p, where (1−p)x−1 corresponds to the unsuccessful attemps and p corresponds to the first successful attempt.
For example, if we throw a dice until we throw a six, the probability that the number of throws is exactly 4 is (5/6)31/6.
The expected value of X in a geometric distribution is E[X] = 1 p.
24.4 Markov chains A Markov chain is a random process that consists of states and transitions between them. For each state, we know the probabilities for moving to other states. A Markov chain can be represented as a graph whose nodes are states and edges are transitions.
As an example, consider a problem where we are in floor 1 in an n floor building. At each step, we randomly walk either one floor up or one floor down, except that we always walk one floor up from floor 1 and one floor down from floor n. What is the probability of being in floor m after k steps?
In this problem, each floor of the building corresponds to a state in a Markov chain. For example, if n = 5, the graph is as follows: 230 1 2 3 4 5 1 1/2 1/2 1/2 1 1/2 1/2 1/2 The probability distribution of a Markov chain is a vector [p1, p2,..., pn], where pk is the probability that the current state is k. The formula p1 + p2 +···+ pn = 1 always holds.
In the above scenario, the initial distribution is [1,0,0,0,0], because we always begin in floor 1. The next distribution is [0,1,0,0,0], because we can only move from floor 1 to floor 2. After this, we can either move one floor up or one floor down, so the next distribution is [1/2,0,1/2,0,0], and so on.
An efficient way to simulate the walk in a Markov chain is to use dynamic programming. The idea is to maintain the probability distribution, and at each step go through all possibilities how we can move. Using this method, we can simulate a walk of m steps in O(n2m) time.
The transitions of a Markov chain can also be represented as a matrix that updates the probability distribution. In the above scenario, the matrix is 0 1/2 0 0 0 1 0 1/2 0 0 0 1/2 0 1/2 0 0 0 1/2 0 1 0 0 0 1/2 0 .
When we multiply a probability distribution by this matrix, we get the new distribution after moving one step. For example, we can move from the distribu-tion [1,0,0,0,0] to the distribution [0,1,0,0,0] as follows: 0 1/2 0 0 0 1 0 1/2 0 0 0 1/2 0 1/2 0 0 0 1/2 0 1 0 0 0 1/2 0 1 0 0 0 0 = 0 1 0 0 0 .
By calculating matrix powers efficiently, we can calculate the distribution after m steps in O(n3logm) time.
24.5 Randomized algorithms Sometimes we can use randomness for solving a problem, even if the problem is not related to probabilities. A randomized algorithm is an algorithm that is based on randomness.
A Monte Carlo algorithm is a randomized algorithm that may sometimes give a wrong answer. For such an algorithm to be useful, the probability of a wrong answer should be small.
231 A Las Vegas algorithm is a randomized algorithm that always gives the correct answer, but its running time varies randomly. The goal is to design an algorithm that is efficient with high probability.
Next we will go through three example problems that can be solved using randomness.
Order statistics The kth order statistic of an array is the element at position k after sorting the array in increasing order. It is easy to calculate any order statistic in O(nlogn) time by first sorting the array, but is it really needed to sort the entire array just to find one element?
It turns out that we can find order statistics using a randomized algorithm without sorting the array. The algorithm, called quickselect1, is a Las Vegas algorithm: its running time is usually O(n) but O(n2) in the worst case.
The algorithm chooses a random element x of the array, and moves elements smaller than x to the left part of the array, and all other elements to the right part of the array. This takes O(n) time when there are n elements. Assume that the left part contains a elements and the right part contains b elements. If a = k, element x is the kth order statistic. Otherwise, if a > k, we recursively find the kth order statistic for the left part, and if a < k, we recursively find the rth order statistic for the right part where r = k −a. The search continues in a similar way, until the element has been found.
When each element x is randomly chosen, the size of the array about halves at each step, so the time complexity for finding the kth order statistic is about n+ n/2+ n/4+ n/8+··· < 2n = O(n).
The worst case of the algorithm requires still O(n2) time, because it is possible that x is always chosen in such a way that it is one of the smallest or largest elements in the array and O(n) steps are needed. However, the probability for this is so small that this never happens in practice.
Verifying matrix multiplication Our next problem is to verify if AB = C holds when A, B and C are matrices of size n × n. Of course, we can solve the problem by calculating the product AB again (in O(n3) time using the basic algorithm), but one could hope that verifying the answer would by easier than to calculate it from scratch.
It turns out that we can solve the problem using a Monte Carlo algorithm2 whose time complexity is only O(n2). The idea is simple: we choose a random vector X of n elements, and calculate the matrices ABX and CX. If ABX = CX, we report that AB = C, and otherwise we report that AB ̸= C.
1In 1961, C. A. R. Hoare published two algorithms that are efficient on average: quicksort for sorting arrays and quickselect for finding order statistics.
2R. M. Freivalds published this algorithm in 1977 , and it is sometimes called Freivalds’ algorithm.
232 The time complexity of the algorithm is O(n2), because we can calculate the matrices ABX and CX in O(n2) time. We can calculate the matrix ABX efficiently by using the representation A(BX), so only two multiplications of n×n and n×1 size matrices are needed.
The drawback of the algorithm is that there is a small chance that the algorithm makes a mistake when it reports that AB = C. For example, ·6 8 1 3 ¸ ̸= ·8 7 3 2 ¸ , but ·6 8 1 3 ¸·3 6 ¸ = ·8 7 3 2 ¸·3 6 ¸ .
However, in practice, the probability that the algorithm makes a mistake is small, and we can decrease the probability by verifying the result using multiple random vectors X before reporting that AB = C.
Graph coloring Given a graph that contains n nodes and m edges, our task is to find a way to color the nodes of the graph using two colors so that for at least m/2 edges, the endpoints have different colors. For example, in the graph 1 2 3 4 5 a valid coloring is as follows: 1 2 3 4 5 The above graph contains 7 edges, and for 5 of them, the endpoints have different colors, so the coloring is valid.
The problem can be solved using a Las Vegas algorithm that generates random colorings until a valid coloring has been found. In a random coloring, the color of each node is independently chosen so that the probability of both colors is 1/2.
In a random coloring, the probability that the endpoints of a single edge have different colors is 1/2. Hence, the expected number of edges whose endpoints have different colors is m/2. Since it is expected that a random coloring is valid, we will quickly find a valid coloring in practice.
233 234 Chapter 25 Game theory In this chapter, we will focus on two-player games that do not contain random elements. Our goal is to find a strategy that we can follow to win the game no matter what the opponent does, if such a strategy exists.
It turns out that there is a general strategy for such games, and we can analyze the games using the nim theory. First, we will analyze simple games where players remove sticks from heaps, and after this, we will generalize the strategy used in those games to other games.
25.1 Game states Let us consider a game where there is initially a heap of n sticks. Players A and B move alternately, and player A begins. On each move, the player has to remove 1, 2 or 3 sticks from the heap, and the player who removes the last stick wins the game.
For example, if n = 10, the game may proceed as follows: • Player A removes 2 sticks (8 sticks left).
• Player B removes 3 sticks (5 sticks left).
• Player A removes 1 stick (4 sticks left).
• Player B removes 2 sticks (2 sticks left).
• Player A removes 2 sticks and wins.
This game consists of states 0,1,2,...,n, where the number of the state corre-sponds to the number of sticks left.
Winning and losing states A winning state is a state where the player will win the game if they play optimally, and a losing state is a state where the player will lose the game if the opponent plays optimally. It turns out that we can classify all states of a game so that each state is either a winning state or a losing state.
In the above game, state 0 is clearly a losing state, because the player cannot make any moves. States 1, 2 and 3 are winning states, because we can remove 1, 235 2 or 3 sticks and win the game. State 4, in turn, is a losing state, because any move leads to a state that is a winning state for the opponent.
More generally, if there is a move that leads from the current state to a losing state, the current state is a winning state, and otherwise the current state is a losing state. Using this observation, we can classify all states of a game starting with losing states where there are no possible moves.
The states 0...15 of the above game can be classified as follows (W denotes a winning state and L denotes a losing state): L W W W L W W W L W W W L W W W 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 It is easy to analyze this game: a state k is a losing state if k is divisible by 4, and otherwise it is a winning state. An optimal way to play the game is to always choose a move after which the number of sticks in the heap is divisible by 4. Finally, there are no sticks left and the opponent has lost.
Of course, this strategy requires that the number of sticks is not divisible by 4 when it is our move. If it is, there is nothing we can do, and the opponent will win the game if they play optimally.
State graph Let us now consider another stick game, where in each state k, it is allowed to remove any number x of sticks such that x is smaller than k and divides k. For example, in state 8 we may remove 1, 2 or 4 sticks, but in state 7 the only allowed move is to remove 1 stick.
The following picture shows the states 1...9 of the game as a state graph, whose nodes are the states and edges are the moves between them: 1 2 3 4 5 6 7 8 9 The final state in this game is always state 1, which is a losing state, because there are no valid moves. The classification of states 1...9 is as follows: L W L W L W L W L 1 2 3 4 5 6 7 8 9 Surprisingly, in this game, all even-numbered states are winning states, and all odd-numbered states are losing states.
236 25.2 Nim game The nim game is a simple game that has an important role in game theory, because many other games can be played using the same strategy. First, we focus on nim, and then we generalize the strategy to other games.
There are n heaps in nim, and each heap contains some number of sticks.
The players move alternately, and on each turn, the player chooses a heap that still contains sticks and removes any number of sticks from it. The winner is the player who removes the last stick.
The states in nim are of the form [x1,x2,...,xn], where xk denotes the number of sticks in heap k. For example, [10,12,5] is a game where there are three heaps with 10, 12 and 5 sticks. The state [0,0,...,0] is a losing state, because it is not possible to remove any sticks, and this is always the final state.
Analysis It turns out that we can easily classify any nim state by calculating the nim sum s = x1 ⊕x2 ⊕···⊕xn, where ⊕is the xor operation1. The states whose nim sum is 0 are losing states, and all other states are winning states. For example, the nim sum of [10,12,5] is 10⊕12⊕5 = 3, so the state is a winning state.
But how is the nim sum related to the nim game? We can explain this by looking at how the nim sum changes when the nim state changes.
Losing states: The final state [0,0,...,0] is a losing state, and its nim sum is 0, as expected. In other losing states, any move leads to a winning state, because when a single value xk changes, the nim sum also changes, so the nim sum is different from 0 after the move.
Winning states: We can move to a losing state if there is any heap k for which xk ⊕s < xk. In this case, we can remove sticks from heap k so that it will contain xk ⊕s sticks, which will lead to a losing state. There is always such a heap, where xk has a one bit at the position of the leftmost one bit of s.
As an example, consider the state [10,12,5]. This state is a winning state, because its nim sum is 3. Thus, there has to be a move which leads to a losing state. Next we will find out such a move.
The nim sum of the state is as follows: 10 1010 12 1100 5 0101 3 0011 In this case, the heap with 10 sticks is the only heap that has a one bit at the position of the leftmost one bit of the nim sum: 10 1010 12 1100 5 0101 3 0011 1The optimal strategy for nim was published in 1901 by C. L. Bouton .
237 The new size of the heap has to be 10⊕3 = 9, so we will remove just one stick.
After this, the state will be [9,12,5], which is a losing state: 9 1001 12 1100 5 0101 0 0000 Misère game In a misère game, the goal of the game is opposite, so the player who removes the last stick loses the game. It turns out that the misère nim game can be optimally played almost like the standard nim game.
The idea is to first play the misère game like the standard game, but change the strategy at the end of the game. The new strategy will be introduced in a situation where each heap would contain at most one stick after the next move.
In the standard game, we should choose a move after which there is an even number of heaps with one stick. However, in the misère game, we choose a move so that there is an odd number of heaps with one stick.
This strategy works because a state where the strategy changes always appears in the game, and this state is a winning state, because it contains exactly one heap that has more than one stick so the nim sum is not 0.
25.3 Sprague–Grundy theorem The Sprague–Grundy theorem2 generalizes the strategy used in nim to all games that fulfil the following requirements: • There are two players who move alternately.
• The game consists of states, and the possible moves in a state do not depend on whose turn it is.
• The game ends when a player cannot make a move.
• The game surely ends sooner or later.
• The players have complete information about the states and allowed moves, and there is no randomness in the game.
The idea is to calculate for each game state a Grundy number that corresponds to the number of sticks in a nim heap. When we know the Grundy numbers of all states, we can play the game like the nim game.
Grundy numbers The Grundy number of a game state is mex({g1, g2,..., gn}), 2The theorem was independently discovered by R. Sprague and P. M. Grundy .
238 where g1, g2,..., gn are the Grundy numbers of the states to which we can move, and the mex function gives the smallest nonnegative number that is not in the set. For example, mex({0,1,3}) = 2. If there are no possible moves in a state, its Grundy number is 0, because mex(;) = 0.
For example, in the state graph the Grundy numbers are as follows: 0 1 0 2 0 2 The Grundy number of a losing state is 0, and the Grundy number of a winning state is a positive number.
The Grundy number of a state corresponds to the number of sticks in a nim heap. If the Grundy number is 0, we can only move to states whose Grundy numbers are positive, and if the Grundy number is x > 0, we can move to states whose Grundy numbers include all numbers 0,1,...,x−1.
As an example, consider a game where the players move a figure in a maze.
Each square in the maze is either floor or wall. On each turn, the player has to move the figure some number of steps left or up. The winner of the game is the player who makes the last move.
The following picture shows a possible initial state of the game, where @ denotes the figure and denotes a square where it can move.
@ The states of the game are all floor squares of the maze. In the above maze, the Grundy numbers are as follows: 0 1 0 1 0 1 2 0 2 1 0 3 0 4 1 0 4 1 3 2 239 Thus, each state of the maze game corresponds to a heap in the nim game. For example, the Grundy number for the lower-right square is 2, so it is a winning state. We can reach a losing state and win the game by moving either four steps left or two steps up.
Note that unlike in the original nim game, it may be possible to move to a state whose Grundy number is larger than the Grundy number of the current state. However, the opponent can always choose a move that cancels such a move, so it is not possible to escape from a losing state.
Subgames Next we will assume that our game consists of subgames, and on each turn, the player first chooses a subgame and then a move in the subgame. The game ends when it is not possible to make any move in any subgame.
In this case, the Grundy number of a game is the nim sum of the Grundy numbers of the subgames. The game can be played like a nim game by calculating all Grundy numbers for subgames and then their nim sum.
As an example, consider a game that consists of three mazes. In this game, on each turn, the player chooses one of the mazes and then moves the figure in the maze. Assume that the initial state of the game is as follows: @ @ @ The Grundy numbers for the mazes are as follows: 0 1 0 1 0 1 2 0 2 1 0 3 0 4 1 0 4 1 3 2 0 1 2 3 1 0 0 1 2 0 1 2 3 1 2 0 4 0 2 5 3 0 1 2 3 4 1 0 2 1 3 2 4 0 1 2 3 In the initial state, the nim sum of the Grundy numbers is 2⊕3⊕3 = 2, so the first player can win the game. One optimal move is to move two steps up in the first maze, which produces the nim sum 0⊕3⊕3 = 0.
Grundy’s game Sometimes a move in a game divides the game into subgames that are indepen-dent of each other. In this case, the Grundy number of the game is mex({g1, g2,..., gn}), 240 where n is the number of possible moves and gk = ak,1 ⊕ak,2 ⊕...⊕ak,m, where move k generates subgames with Grundy numbers ak,1,ak,2,...,ak,m.
An example of such a game is Grundy’s game. Initially, there is a single heap that contains n sticks. On each turn, the player chooses a heap and divides it into two nonempty heaps such that the heaps are of different size. The player who makes the last move wins the game.
Let f (n) be the Grundy number of a heap that contains n sticks. The Grundy number can be calculated by going through all ways to divide the heap into two heaps. For example, when n = 8, the possibilities are 1+7, 2+6 and 3+5, so f (8) = mex({f (1)⊕f (7), f (2)⊕f (6), f (3)⊕f (5)}).
In this game, the value of f (n) is based on the values of f (1),..., f (n−1). The base cases are f (1) = f (2) = 0, because it is not possible to divide the heaps of 1 and 2 sticks. The first Grundy numbers are: f (1) = 0 f (2) = 0 f (3) = 1 f (4) = 0 f (5) = 2 f (6) = 1 f (7) = 0 f (8) = 2 The Grundy number for n = 8 is 2, so it is possible to win the game. The winning move is to create heaps 1+7, because f (1)⊕f (7) = 0.
241 242 Chapter 26 String algorithms This chapter deals with efficient algorithms for string processing. Many string problems can be easily solved in O(n2) time, but the challenge is to find algorithms that work in O(n) or O(nlogn) time.
For example, a fundamental string processing problem is the pattern match-ing problem: given a string of length n and a pattern of length m, our task is to find the occurrences of the pattern in the string. For example, the pattern ABC occurs two times in the string ABABCBABC.
The pattern matching problem can be easily solved in O(nm) time by a brute force algorithm that tests all positions where the pattern may occur in the string.
However, in this chapter, we will see that there are more efficient algorithms that require only O(n+ m) time.
26.1 String terminology Throughout the chapter, we assume that zero-based indexing is used in strings.
Thus, a string s of length n consists of characters s,s,...,s[n−1]. The set of characters that may appear in strings is called an alphabet. For example, the alphabet {A,B,...,Z} consists of the capital letters of English.
A substring is a sequence of consecutive characters in a string. We use the notation s[a...b] to refer to a substring of s that begins at position a and ends at position b. A string of length n has n(n +1)/2 substrings. For example, the substrings of ABCD are A, B, C, D, AB, BC, CD, ABC, BCD and ABCD.
A subsequence is a sequence of (not necessarily consecutive) characters in a string in their original order. A string of length n has 2n −1 subsequences. For example, the subsequences of ABCD are A, B, C, D, AB, AC, AD, BC, BD, CD, ABC, ABD, ACD, BCD and ABCD.
A prefix is a substring that starts at the beginning of a string, and a suffix is a substring that ends at the end of a string. For example, the prefixes of ABCD are A, AB, ABC and ABCD, and the suffixes of ABCD are D, CD, BCD and ABCD.
A rotation can be generated by moving the characters of a string one by one from the beginning to the end (or vice versa). For example, the rotations of ABCD are ABCD, BCDA, CDAB and DABC.
243 A period is a prefix of a string such that the string can be constructed by repeating the period. The last repetition may be partial and contain only a prefix of the period. For example, the shortest period of ABCABCA is ABC.
A border is a string that is both a prefix and a suffix of a string. For example, the borders of ABACABA are A, ABA and ABACABA.
Strings are compared using the lexicographical order (which corresponds to the alphabetical order). It means that x < y if either x ̸= y and x is a prefix of y, or there is a position k such that x[i] = y[i] when i < k and x[k] < y[k].
26.2 Trie structure A trie is a rooted tree that maintains a set of strings. Each string in the set is stored as a chain of characters that starts at the root. If two strings have a common prefix, they also have a common chain in the tree.
For example, consider the following trie: C T A N A D L Y H E R E This trie corresponds to the set {CANAL,CANDY,THE,THERE}. The character in a node means that a string in the set ends at the node. Such a character is needed, because a string may be a prefix of another string. For example, in the above trie, THE is a prefix of THERE.
We can check in O(n) time whether a trie contains a string of length n, because we can follow the chain that starts at the root node. We can also add a string of length n to the trie in O(n) time by first following the chain and then adding new nodes to the trie if necessary.
Using a trie, we can find the longest prefix of a given string such that the prefix belongs to the set. Moreover, by storing additional information in each node, we can calculate the number of strings that belong to the set and have a given string as a prefix.
A trie can be stored in an array int trie[N][A]; 244 where N is the maximum number of nodes (the maximum total length of the strings in the set) and A is the size of the alphabet. The nodes of a trie are numbered 0,1,2,... so that the number of the root is 0, and trie[s][c] is the next node in the chain when we move from node s using character c.
26.3 String hashing String hashing is a technique that allows us to efficiently check whether two strings are equal1. The idea in string hashing is to compare hash values of strings instead of their individual characters.
Calculating hash values A hash value of a string is a number that is calculated from the characters of the string. If two strings are the same, their hash values are also the same, which makes it possible to compare strings based on their hash values.
A usual way to implement string hashing is polynomial hashing, which means that the hash value of a string s of length n is (sAn−1 +sAn−2 +···+s[n−1]A0) mod B, where s,s,...,s[n−1] are interpreted as the codes of the characters of s, and A and B are pre-chosen constants.
For example, the codes of the characters of ALLEY are: A L L E Y 65 76 76 69 89 Thus, if A = 3 and B = 97, the hash value of ALLEY is (65·34 +76·33 +76·32 +69·31 +89·30) mod 97 = 52.
Preprocessing Using polynomial hashing, we can calculate the hash value of any substring of a string s in O(1) time after an O(n) time preprocessing. The idea is to construct an array h such that h[k] contains the hash value of the prefix s[0...k]. The array values can be recursively calculated as follows: h = s h[k] = (h[k −1]A +s[k]) mod B In addition, we construct an array p where p[k] = Ak mod B: p = 1 p[k] = (p[k −1]A) mod B.
1The technique was popularized by the Karp–Rabin pattern matching algorithm .
245 Constructing these arrays takes O(n) time. After this, the hash value of any substring s[a...b] can be calculated in O(1) time using the formula (h[b]−h[a−1]p[b −a+1]) mod B assuming that a > 0. If a = 0, the hash value is simply h[b].
Using hash values We can efficiently compare strings using hash values. Instead of comparing the individual characters of the strings, the idea is to compare their hash values. If the hash values are equal, the strings are probably equal, and if the hash values are different, the strings are certainly different.
Using hashing, we can often make a brute force algorithm efficient. As an example, consider the pattern matching problem: given a string s and a pattern p, find the positions where p occurs in s. A brute force algorithm goes through all positions where p may occur and compares the strings character by character.
The time complexity of such an algorithm is O(n2).
We can make the brute force algorithm more efficient by using hashing, because the algorithm compares substrings of strings. Using hashing, each comparison only takes O(1) time, because only hash values of substrings are compared. This results in an algorithm with time complexity O(n), which is the best possible time complexity for this problem.
By combining hashing and binary search, it is also possible to find out the lexicographic order of two strings in logarithmic time. This can be done by calculating the length of the common prefix of the strings using binary search.
Once we know the length of the common prefix, we can just check the next character after the prefix, because this determines the order of the strings.
Collisions and parameters An evident risk when comparing hash values is a collision, which means that two strings have different contents but equal hash values. In this case, an algorithm that relies on the hash values concludes that the strings are equal, but in reality they are not, and the algorithm may give incorrect results.
Collisions are always possible, because the number of different strings is larger than the number of different hash values. However, the probability of a collision is small if the constants A and B are carefully chosen. A usual way is to choose random constants near 109, for example as follows: A = 911382323 B = 972663749 Using such constants, the long long type can be used when calculating hash values, because the products AB and BB will fit in long long. But is it enough to have about 109 different hash values?
Let us consider three scenarios where hashing can be used: 246 Scenario 1: Strings x and y are compared with each other. The probability of a collision is 1/B assuming that all hash values are equally probable.
Scenario 2: A string x is compared with strings y1, y2,..., yn. The probability of one or more collisions is 1−(1−1 B)n.
Scenario 3: All pairs of strings x1,x2,...,xn are compared with each other.
The probability of one or more collisions is 1−B ·(B −1)·(B −2)···(B −n+1) Bn .
The following table shows the collision probabilities when n = 106 and the value of B varies: constant B scenario 1 scenario 2 scenario 3 103 0.001000 1.000000 1.000000 106 0.000001 0.632121 1.000000 109 0.000000 0.001000 1.000000 1012 0.000000 0.000000 0.393469 1015 0.000000 0.000000 0.000500 1018 0.000000 0.000000 0.000001 The table shows that in scenario 1, the probability of a collision is negligible when B ≈109. In scenario 2, a collision is possible but the probability is still quite small. However, in scenario 3 the situation is very different: a collision will almost always happen when B ≈109.
The phenomenon in scenario 3 is known as the birthday paradox: if there are n people in a room, the probability that some two people have the same birthday is large even if n is quite small. In hashing, correspondingly, when all hash values are compared with each other, the probability that some two hash values are equal is large.
We can make the probability of a collision smaller by calculating multiple hash values using different parameters. It is unlikely that a collision would occur in all hash values at the same time. For example, two hash values with parameter B ≈109 correspond to one hash value with parameter B ≈1018, which makes the probability of a collision very small.
Some people use constants B = 232 and B = 264, which is convenient, because operations with 32 and 64 bit integers are calculated modulo 232 and 264. How-ever, this is not a good choice, because it is possible to construct inputs that always generate collisions when constants of the form 2x are used .
26.4 Z-algorithm The Z-array z of a string s of length n contains for each k = 0,1,...,n −1 the length of the longest substring of s that begins at position k and is a prefix of 247 s. Thus, z[k] = p tells us that s[0... p −1] equals s[k...k + p −1]. Many string processing problems can be efficiently solved using the Z-array.
For example, the Z-array of ACBACDACBACBACDA is as follows: A C B A C D A C B A C B A C D A – 0 0 2 0 0 5 0 0 7 0 0 2 0 0 1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 In this case, for example, z = 5, because the substring ACBAC of length 5 is a prefix of s, but the substring ACBACB of length 6 is not a prefix of s.
Algorithm description Next we describe an algorithm, called the Z-algorithm2, that efficiently con-structs the Z-array in O(n) time. The algorithm calculates the Z-array values from left to right by both using information already stored in the Z-array and comparing substrings character by character.
To efficiently calculate the Z-array values, the algorithm maintains a range [x, y] such that s[x... y] is a prefix of s and y is as large as possible. Since we know that s[0... y−x] and s[x... y] are equal, we can use this information when calculating Z-values for positions x+1,x+2,..., y.
At each position k, we first check the value of z[k −x]. If k +z[k −x] < y, we know that z[k] = z[k −x]. However, if k +z[k −x] ≥y, s[0... y−k] equals s[k... y], and to determine the value of z[k] we need to compare the substrings character by character. Still, the algorithm works in O(n) time, because we start comparing at positions y−k +1 and y+1.
For example, let us construct the following Z-array: A C B A C D A C B A C B A C D A – ?
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0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 After calculating the value z = 5, the current [x, y] range is [6,10]: A C B A C D A C B A C B A C D A – 0 0 2 0 0 5 ?
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x y 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Now we can calculate subsequent Z-array values efficiently, because we know that s[0...4] and s[6...10] are equal. First, since z = z = 0, we immediately know that also z = z = 0: 2The Z-algorithm was presented in as the simplest known method for linear-time pattern matching, and the original idea was attributed to .
248 A C B A C D A C B A C B A C D A – 0 0 2 0 0 5 0 0 ?
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x y 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Then, since z = 2, we know that z ≥2: A C B A C D A C B A C B A C D A – 0 0 2 0 0 5 0 0 ?
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x y 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 However, we have no information about the string after position 10, so we need to compare the substrings character by character: A C B A C D A C B A C B A C D A – 0 0 2 0 0 5 0 0 ?
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x y 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 It turns out that z = 7, so the new [x, y] range is [9,15]: A C B A C D A C B A C B A C D A – 0 0 2 0 0 5 0 0 7 ?
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x y 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 After this, all the remaining Z-array values can be determined by using the information already stored in the Z-array: A C B A C D A C B A C B A C D A – 0 0 2 0 0 5 0 0 7 0 0 2 0 0 1 x y 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 249 Using the Z-array It is often a matter of taste whether to use string hashing or the Z-algorithm.
Unlike hashing, the Z-algorithm always works and there is no risk for collisions.
On the other hand, the Z-algorithm is more difficult to implement and some problems can only be solved using hashing.
As an example, consider again the pattern matching problem, where our task is to find the occurrences of a pattern p in a string s. We already solved this problem efficiently using string hashing, but the Z-algorithm provides another way to solve the problem.
A usual idea in string processing is to construct a string that consists of mul-tiple strings separated by special characters. In this problem, we can construct a string p#s, where p and s are separated by a special character # that does not occur in the strings. The Z-array of p#s tells us the positions where p occurs in s, because such positions contain the length of p.
For example, if s =HATTIVATTI and p =ATT, the Z-array is as follows: A T T # H A T T I V A T T I – 0 0 0 0 3 0 0 0 0 3 0 0 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 The positions 5 and 10 contain the value 3, which means that the pattern ATT occurs in the corresponding positions of HATTIVATTI.
The time complexity of the resulting algorithm is linear, because it suffices to construct the Z-array and go through its values.
Implementation Here is a short implementation of the Z-algorithm that returns a vector that corresponds to the Z-array.
vector z(string s) { int n = s.size(); vector z(n); int x = 0, y = 0; for (int i = 1; i < n; i++) { z[i] = max(0,min(z[i-x],y-i+1)); while (i+z[i] < n && s[z[i]] == s[i+z[i]]) { x = i; y = i+z[i]; z[i]++; } } return z; } 250 Chapter 27 Square root algorithms A square root algorithm is an algorithm that has a square root in its time complexity. A square root can be seen as a ”poor man’s logarithm”: the complexity O(pn) is better than O(n) but worse than O(logn). In any case, many square root algorithms are fast and usable in practice.
As an example, consider the problem of creating a data structure that sup-ports two operations on an array: modifying an element at a given position and calculating the sum of elements in the given range. We have previously solved the problem using binary indexed and segment trees, that support both operations in O(logn) time. However, now we will solve the problem in another way using a square root structure that allows us to modify elements in O(1) time and calculate sums in O(pn) time.
The idea is to divide the array into blocks of size pn so that each block contains the sum of elements inside the block. For example, an array of 16 elements will be divided into blocks of 4 elements as follows: 5 8 6 3 2 7 2 6 7 1 7 5 6 2 3 2 21 17 20 13 In this structure, it is easy to modify array elements, because it is only needed to update the sum of a single block after each modification, which can be done in O(1) time. For example, the following picture shows how the value of an element and the sum of the corresponding block change: 5 8 6 3 2 5 2 6 7 1 7 5 6 2 3 2 21 15 20 13 Then, to calculate the sum of elements in a range, we divide the range into three parts such that the sum consists of values of single elements and sums of blocks between them: 5 8 6 3 2 5 2 6 7 1 7 5 6 2 3 2 21 15 20 13 251 Since the number of single elements is O(pn) and the number of blocks is also O(pn), the sum query takes O(pn) time. The purpose of the block size pn is that it balances two things: the array is divided into pn blocks, each of which contains pn elements.
In practice, it is not necessary to use the exact value of pn as a parameter, and instead we may use parameters k and n/k where k is different from pn.
The optimal parameter depends on the problem and input. For example, if an algorithm often goes through the blocks but rarely inspects single elements inside the blocks, it may be a good idea to divide the array into k < pn blocks, each of which contains n/k > pn elements.
27.1 Combining algorithms In this section we discuss two square root algorithms that are based on combining two algorithms into one algorithm. In both cases, we could use either of the algorithms without the other and solve the problem in O(n2) time. However, by combining the algorithms, the running time is only O(npn).
Case processing Suppose that we are given a two-dimensional grid that contains n cells. Each cell is assigned a letter, and our task is to find two cells with the same letter whose distance is minimum, where the distance between cells (x1, y1) and (x2, y2) is |x1 −x2|+|y1 −y2|. For example, consider the following grid: A B C A C D E F B A G B D F E A In this case, the minimum distance is 2 between the two ’E’ letters.
We can solve the problem by considering each letter separately. Using this approach, the new problem is to calculate the minimum distance between two cells with a fixed letter c. We focus on two algorithms for this: Algorithm 1: Go through all pairs of cells with letter c, and calculate the minimum distance between such cells. This will take O(k2) time where k is the number of cells with letter c.
Algorithm 2: Perform a breadth-first search that simultaneously starts at each cell with letter c. The minimum distance between two cells with letter c will be calculated in O(n) time.
One way to solve the problem is to choose either of the algorithms and use it for all letters. If we use Algorithm 1, the running time is O(n2), because all cells may contain the same letter, and in this case k = n. Also if we use Algorithm 2, the running time is O(n2), because all cells may have different letters, and in this case n searches are needed.
252 However, we can combine the two algorithms and use different algorithms for different letters depending on how many times each letter appears in the grid.
Assume that a letter c appears k times. If k ≤pn, we use Algorithm 1, and if k > pn, we use Algorithm 2. It turns out that by doing this, the total running time of the algorithm is only O(npn).
First, suppose that we use Algorithm 1 for a letter c. Since c appears at most pn times in the grid, we compare each cell with letter c O(pn) times with other cells. Thus, the time used for processing all such cells is O(npn). Then, suppose that we use Algorithm 2 for a letter c. There are at most pn such letters, so processing those letters also takes O(npn) time.
Batch processing Our next problem also deals with a two-dimensional grid that contains n cells.
Initially, each cell except one is white. We perform n−1 operations, each of which first calculates the minimum distance from a given white cell to a black cell, and then paints the white cell black.
For example, consider the following operation: First, we calculate the minimum distance from the white cell marked with to a black cell. The minimum distance is 2, because we can move two steps left to a black cell. Then, we paint the white cell black: Consider the following two algorithms: Algorithm 1: Use breadth-first search to calculate for each white cell the distance to the nearest black cell. This takes O(n) time, and after the search, we can find the minimum distance from any white cell to a black cell in O(1) time.
Algorithm 2: Maintain a list of cells that have been painted black, go through this list at each operation and then add a new cell to the list. An operation takes O(k) time where k is the length of the list.
We combine the above algorithms by dividing the operations into O(pn) batches, each of which consists of O(pn) operations. At the beginning of each batch, we perform Algorithm 1. Then, we use Algorithm 2 to process the opera-tions in the batch. We clear the list of Algorithm 2 between the batches. At each 253 operation, the minimum distance to a black cell is either the distance calculated by Algorithm 1 or the distance calculated by Algorithm 2.
The resulting algorithm works in O(npn) time. First, Algorithm 1 is per-formed O(pn) times, and each search works in O(n) time. Second, when using Algorithm 2 in a batch, the list contains O(pn) cells (because we clear the list between the batches) and each operation takes O(pn) time.
27.2 Integer partitions Some square root algorithms are based on the following observation: if a positive integer n is represented as a sum of positive integers, such a sum always contains at most O(pn) distinct numbers. The reason for this is that to construct a sum that contains a maximum number of distinct numbers, we should choose small numbers. If we choose the numbers 1,2,...,k, the resulting sum is k(k +1) 2 .
Thus, the maximum amount of distinct numbers is k = O(pn). Next we will discuss two problems that can be solved efficiently using this observation.
Knapsack Suppose that we are given a list of integer weights whose sum is n. Our task is to find out all sums that can be formed using a subset of the weights. For example, if the weights are {1,3,3}, the possible sums are as follows: • 0 (empty set) • 1 • 3 • 1+3 = 4 • 3+3 = 6 • 1+3+3 = 7 Using the standard knapsack approach (see Chapter 7.4), the problem can be solved as follows: we define a function possible(x,k) whose value is 1 if the sum x can be formed using the first k weights, and 0 otherwise. Since the sum of the weights is n, there are at most n weights and all values of the function can be calculated in O(n2) time using dynamic programming.
However, we can make the algorithm more efficient by using the fact that there are at most O(pn) distinct weights. Thus, we can process the weights in groups that consists of similar weights. We can process each group in O(n) time, which yields an O(npn) time algorithm.
The idea is to use an array that records the sums of weights that can be formed using the groups processed so far. The array contains n elements: element k is 1 if the sum k can be formed and 0 otherwise. To process a group of weights, we scan the array from left to right and record the new sums of weights that can be formed using this group and the previous groups.
254 String construction Given a string s of length n and a set of strings D whose total length is m, consider the problem of counting the number of ways s can be formed as a concatenation of strings in D. For example, if s = ABAB and D = {A,B,AB}, there are 4 ways: • A+B+A+B • AB+A+B • A+B+AB • AB+AB We can solve the problem using dynamic programming: Let count(k) denote the number of ways to construct the prefix s[0...k] using the strings in D. Now count(n −1) gives the answer to the problem, and we can solve the problem in O(n2) time using a trie structure.
However, we can solve the problem more efficiently by using string hashing and the fact that there are at most O(pm) distinct string lengths in D. First, we construct a set H that contains all hash values of the strings in D. Then, when calculating a value of count(k), we go through all values of p such that there is a string of length p in D, calculate the hash value of s[k −p +1...k] and check if it belongs to H. Since there are at most O(pm) distinct string lengths, this results in an algorithm whose running time is O(npm).
27.3 Mo’s algorithm Mo’s algorithm1 can be used in many problems that require processing range queries in a static array, i.e., the array values do not change between the queries.
In each query, we are given a range [a,b], and we should calculate a value based on the array elements between positions a and b. Since the array is static, the queries can be processed in any order, and Mo’s algorithm processes the queries in a special order which guarantees that the algorithm works efficiently.
Mo’s algorithm maintains an active range of the array, and the answer to a query concerning the active range is known at each moment. The algorithm processes the queries one by one, and always moves the endpoints of the active range by inserting and removing elements. The time complexity of the algorithm is O(npnf (n)) where the array contains n elements, there are n queries and each insertion and removal of an element takes O(f (n)) time.
The trick in Mo’s algorithm is the order in which the queries are processed: The array is divided into blocks of k = O(pn) elements, and a query [a1,b1] is processed before a query [a2,b2] if either • ⌊a1/k⌋< ⌊a2/k⌋or • ⌊a1/k⌋= ⌊a2/k⌋and b1 < b2.
1According to , this algorithm is named after Mo Tao, a Chinese competitive programmer, but the technique has appeared earlier in the literature .
255 Thus, all queries whose left endpoints are in a certain block are processed one after another sorted according to their right endpoints. Using this order, the algorithm only performs O(npn) operations, because the left endpoint moves O(n) times O(pn) steps, and the right endpoint moves O(pn) times O(n) steps.
Thus, both endpoints move a total of O(npn) steps during the algorithm.
Example As an example, consider a problem where we are given a set of queries, each of them corresponding to a range in an array, and our task is to calculate for each query the number of distinct elements in the range.
In Mo’s algorithm, the queries are always sorted in the same way, but it depends on the problem how the answer to the query is maintained. In this problem, we can maintain an array count where count[x] indicates the number of times an element x occurs in the active range.
When we move from one query to another query, the active range changes.
For example, if the current range is 4 2 5 4 2 4 3 3 4 and the next range is 4 2 5 4 2 4 3 3 4 there will be three steps: the left endpoint moves one step to the right, and the right endpoint moves two steps to the right.
After each step, the array count needs to be updated. After adding an element x, we increase the value of count[x] by 1, and if count[x] = 1 after this, we also increase the answer to the query by 1. Similarly, after removing an element x, we decrease the value of count[x] by 1, and if count[x] = 0 after this, we also decrease the answer to the query by 1.
In this problem, the time needed to perform each step is O(1), so the total time complexity of the algorithm is O(npn).
256 Chapter 28 Segment trees revisited A segment tree is a versatile data structure that can be used to solve a large num-ber of algorithm problems. However, there are many topics related to segment trees that we have not touched yet. Now is time to discuss some more advanced variants of segment trees.
So far, we have implemented the operations of a segment tree by walking from bottom to top in the tree. For example, we have calculated range sums as follows (Chapter 9.3): int sum(int a, int b) { a += n; b += n; int s = 0; while (a <= b) { if (a%2 == 1) s += tree[a++]; if (b%2 == 0) s += tree[b--]; a /= 2; b /= 2; } return s; } However, in more advanced segment trees, it is often necessary to implement the operations in another way, from top to bottom. Using this approach, the function becomes as follows: int sum(int a, int b, int k, int x, int y) { if (b < x || a > y) return 0; if (a <= x && y <= b) return tree[k]; int d = (x+y)/2; return sum(a,b,2k,x,d) + sum(a,b,2k+1,d+1,y); } Now we can calculate any value of sumq(a,b) (the sum of array values in range [a,b]) as follows: int s = sum(a, b, 1, 0, n-1); 257 The parameter k indicates the current position in tree. Initially k equals 1, because we begin at the root of the tree. The range [x, y] corresponds to k and is initially [0,n−1]. When calculating the sum, if [x, y] is outside [a,b], the sum is 0, and if [x, y] is completely inside [a,b], the sum can be found in tree. If [x, y] is partially inside [a,b], the search continues recursively to the left and right half of [x, y]. The left half is [x,d] and the right half is [d +1, y] where d = ⌊x+y 2 ⌋.
The following picture shows how the search proceeds when calculating the value of sumq(a,b). The gray nodes indicate nodes where the recursion stops and the sum can be found in tree.
5 8 6 3 2 7 2 6 7 1 7 5 6 2 3 2 13 9 9 8 8 12 8 5 22 17 20 13 39 33 72 a b Also in this implementation, operations take O(logn) time, because the total number of visited nodes is O(logn).
28.1 Lazy propagation Using lazy propagation, we can build a segment tree that supports both range updates and range queries in O(logn) time. The idea is to perform updates and queries from top to bottom and perform updates lazily so that they are propagated down the tree only when it is necessary.
In a lazy segment tree, nodes contain two types of information. Like in an ordinary segment tree, each node contains the sum or some other value related to the corresponding subarray. In addition, the node may contain information related to lazy updates, which has not been propagated to its children.
There are two types of range updates: each array value in the range is either increased by some value or assigned some value. Both operations can be implemented using similar ideas, and it is even possible to construct a tree that supports both operations at the same time.
Lazy segment trees Let us consider an example where our goal is to construct a segment tree that sup-ports two operations: increasing each value in [a,b] by a constant and calculating 258 the sum of values in [a,b].
We will construct a tree where each node has two values s/z: s denotes the sum of values in the range, and z denotes the value of a lazy update, which means that all values in the range should be increased by z. In the following tree, z = 0 in all nodes, so there are no ongoing lazy updates.
5 8 6 3 2 7 2 6 7 1 7 5 6 2 3 2 13/0 9/0 9/0 8/0 8/0 12/0 8/0 5/0 22/0 17/0 20/0 13/0 39/0 33/0 72/0 When the elements in [a,b] are increased by u, we walk from the root towards the leaves and modify the nodes of the tree as follows: If the range [x, y] of a node is completely inside [a,b], we increase the z value of the node by u and stop. If [x, y] only partially belongs to [a,b], we increase the s value of the node by hu, where h is the size of the intersection of [a,b] and [x, y], and continue our walk recursively in the tree.
For example, the following picture shows the tree after increasing the ele-ments in [a,b] by 2: 5 8 6 3 2 9 2 6 7 1 7 5 6 2 3 2 13/0 9/0 11/0 8/2 8/0 12/0 8/2 5/0 22/0 23/0 20/2 17/0 45/0 45/0 90/0 a b We also calculate the sum of elements in a range [a,b] by walking in the tree from top to bottom. If the range [x, y] of a node completely belongs to [a,b], we add the s value of the node to the sum. Otherwise, we continue the search recursively downwards in the tree.
259 Both in updates and queries, the value of a lazy update is always propagated to the children of the node before processing the node. The idea is that updates will be propagated downwards only when it is necessary, which guarantees that the operations are always efficient.
The following picture shows how the tree changes when we calculate the value of suma(a,b). The rectangle shows the nodes whose values change, because a lazy update is propagated downwards.
5 8 6 3 2 9 2 6 7 1 7 5 6 2 3 2 13/0 9/0 11/0 8/2 8/2 12/2 8/2 5/0 22/0 23/0 28/0 17/0 45/0 45/0 90/0 a b Note that sometimes it is needed to combine lazy updates. This happens when a node that already has a lazy update is assigned another lazy update. When calculating sums, it is easy to combine lazy updates, because the combination of updates z1 and z2 corresponds to an update z1 + z2.
Polynomial updates Lazy updates can be generalized so that it is possible to update ranges using polynomials of the form p(u) = tkuk + tk−1uk−1 +···+ t0.
In this case, the update for a value at position i in [a,b] is p(i −a). For example, adding the polynomial p(u) = u +1 to [a,b] means that the value at position a increases by 1, the value at position a+1 increases by 2, and so on.
To support polynomial updates, each node is assigned k +2 values, where k equals the degree of the polynomial. The value s is the sum of the elements in the range, and the values z0, z1,..., zk are the coefficients of a polynomial that corresponds to a lazy update.
Now, the sum of values in a range [x, y] equals s+ y−x X u=0 zkuk + zk−1uk−1 +···+ z0.
260 The value of such a sum can be efficiently calculated using sum formulas.
For example, the term z0 corresponds to the sum (y−x+1)z0, and the term z1u corresponds to the sum z1(0+1+···+ y−x) = z1 (y−x)(y−x+1) 2 .
When propagating an update in the tree, the indices of p(u) change, because in each range [x, y], the values are calculated for u = 0,1,..., y−x. However, this is not a problem, because p′(u) = p(u + h) is a polynomial of equal degree as p(u).
For example, if p(u) = t2u2 + t1u −t0, then p′(u) = t2(u + h)2 + t1(u + h)−t0 = t2u2 +(2ht2 + t1)u + t2h2 + t1h−t0.
28.2 Dynamic trees An ordinary segment tree is static, which means that each node has a fixed position in the array and the tree requires a fixed amount of memory. In a dynamic segment tree, memory is allocated only for nodes that are actually accessed during the algorithm, which can save a large amount of memory.
The nodes of a dynamic tree can be represented as structs: struct node { int value; int x, y; node left, right; node(int v, int x, int y) : value(v), x(x), y(y) {} }; Here value is the value of the node, [x,y] is the corresponding range, and left and right point to the left and right subtree.
After this, nodes can be created as follows: // create new node node x = new node(0, 0, 15); // change value x->value = 5; Sparse segment trees A dynamic segment tree is useful when the underlying array is sparse, i.e., the range [0,n−1] of allowed indices is large, but most array values are zeros. While an ordinary segment tree uses O(n) memory, a dynamic segment tree only uses O(klogn) memory, where k is the number of operations performed.
A sparse segment tree initially has only one node [0,n −1] whose value is zero, which means that every array value is zero. After updates, new nodes are dynamically added to the tree. For example, if n = 16 and the elements in positions 3 and 10 have been modified, the tree contains the following nodes: 261 [0,15] [0,7] [0,3] [2,3] [8,15] [8,11] [10,11] Any path from the root node to a leaf contains O(logn) nodes, so each operation adds at most O(logn) new nodes to the tree. Thus, after k operations, the tree contains at most O(klogn) nodes.
Note that if we know all elements to be updated at the beginning of the algorithm, a dynamic segment tree is not necessary, because we can use an ordinary segment tree with index compression (Chapter 9.4). However, this is not possible when the indices are generated during the algorithm.
Persistent segment trees Using a dynamic implementation, it is also possible to create a persistent segment tree that stores the modification history of the tree. In such an im-plementation, we can efficiently access all versions of the tree that have existed during the algorithm.
When the modification history is available, we can perform queries in any previous tree like in an ordinary segment tree, because the full structure of each tree is stored. We can also create new trees based on previous trees and modify them independently.
Consider the following sequence of updates, where red nodes change and other nodes remain the same: step 1 step 2 step 3 After each update, most nodes of the tree remain the same, so an efficient way to store the modification history is to represent each tree in the history as a 262 combination of new nodes and subtrees of previous trees. In this example, the modification history can be stored as follows: step 1 step 2 step 3 The structure of each previous tree can be reconstructed by following the pointers starting at the corresponding root node. Since each operation adds only O(logn) new nodes to the tree, it is possible to store the full modification history of the tree.
28.3 Data structures Instead of single values, nodes in a segment tree can also contain data structures that maintain information about the corresponding ranges. In such a tree, the operations take O(f (n)logn) time, where f (n) is the time needed for processing a single node during an operation.
As an example, consider a segment tree that supports queries of the form ”how many times does an element x appear in the range [a,b]?” For example, the element 1 appears three times in the following range: 3 1 2 3 1 1 1 2 To support such queries, we build a segment tree where each node is assigned a data structure that can be asked how many times any element x appears in the corresponding range. Using this tree, the answer to a query can be calculated by combining the results from the nodes that belong to the range.
For example, the following segment tree corresponds to the above array: 3 1 1 1 2 1 3 1 1 1 1 1 1 1 2 1 1 3 1 1 2 3 1 1 1 2 1 2 1 1 1 2 3 1 1 2 1 2 3 1 1 2 3 4 2 2 263 We can build the tree so that each node contains a map structure. In this case, the time needed for processing each node is O(logn), so the total time complexity of a query is O(log2 n). The tree uses O(nlogn) memory, because there are O(logn) levels and each level contains O(n) elements.
28.4 Two-dimensionality A two-dimensional segment tree supports queries related to rectangular sub-arrays of a two-dimensional array. Such a tree can be implemented as nested segment trees: a big tree corresponds to the rows of the array, and each node contains a small tree that corresponds to a column.
For example, in the array 8 5 3 8 3 9 7 1 8 7 5 2 7 6 1 6 the sum of any subarray can be calculated from the following segment tree: 7 6 1 6 13 7 20 8 7 5 2 15 7 22 3 9 7 1 12 8 20 8 5 3 8 13 11 24 15 13 6 8 28 14 42 11 14 10 9 25 19 44 26 27 16 17 53 33 86 The operations of a two-dimensional segment tree take O(log2 n) time, because the big tree and each small tree consist of O(logn) levels. The tree requires O(n2) memory, because each small tree contains O(n) values.
264 Chapter 29 Geometry In geometric problems, it is often challenging to find a way to approach the problem so that the solution to the problem can be conveniently implemented and the number of special cases is small.
As an example, consider a problem where we are given the vertices of a quadrilateral (a polygon that has four vertices), and our task is to calculate its area. For example, a possible input for the problem is as follows: One way to approach the problem is to divide the quadrilateral into two triangles by a straight line between two opposite vertices: After this, it suffices to sum the areas of the triangles. The area of a triangle can be calculated, for example, using Heron’s formula p s(s−a)(s−b)(s−c), where a, b and c are the lengths of the triangle’s sides and s = (a+ b + c)/2.
This is a possible way to solve the problem, but there is one pitfall: how to divide the quadrilateral into triangles? It turns out that sometimes we cannot just pick two arbitrary opposite vertices. For example, in the following situation, the division line is outside the quadrilateral: 265 However, another way to draw the line works: It is clear for a human which of the lines is the correct choice, but the situation is difficult for a computer.
However, it turns out that we can solve the problem using another method that is more convenient to a programmer. Namely, there is a general formula x1y2 −x2y1 + x2y3 −x3y2 + x3y4 −x4y3 + x4y1 −x1y4, that calculates the area of a quadrilateral whose vertices are (x1, y1), (x2, y2), (x3, y3) and (x4, y4). This formula is easy to implement, there are no special cases, and we can even generalize the formula to all polygons.
29.1 Complex numbers A complex number is a number of the form x+ yi, where i = p −1 is the imagi-nary unit. A geometric interpretation of a complex number is that it represents a two-dimensional point (x, y) or a vector from the origin to a point (x, y).
For example, 4+2i corresponds to the following point and vector: (4,2) The C++ complex number class complex is useful when solving geometric problems.
Using the class we can represent points and vectors as complex numbers, and the class contains tools that are useful in geometry.
In the following code, C is the type of a coordinate and P is the type of a point or a vector. In addition, the code defines macros X and Y that can be used to refer to x and y coordinates.
typedef long long C; typedef complex P; #define X real() #define Y imag() 266 For example, the following code defines a point p = (4,2) and prints its x and y coordinates: P p = {4,2}; cout << p.X << " " << p.Y << "\n"; // 4 2 The following code defines vectors v = (3,1) and u = (2,2), and after that calculates the sum s = v+ u.
P v = {3,1}; P u = {2,2}; P s = v+u; cout << s.X << " " << s.Y << "\n"; // 5 3 In practice, an appropriate coordinate type is usually long long (integer) or long double (real number). It is a good idea to use integer whenever possible, because calculations with integers are exact. If real numbers are needed, preci-sion errors should be taken into account when comparing numbers. A safe way to check if real numbers a and b are equal is to compare them using |a−b| < ϵ, where ϵ is a small number (for example, ϵ = 10−9).
Functions In the following examples, the coordinate type is long double.
The function abs(v) calculates the length |v| of a vector v = (x, y) using the formula p x2 + y2. The function can also be used for calculating the distance between points (x1, y1) and (x2, y2), because that distance equals the length of the vector (x2 −x1, y2 −y1).
The following code calculates the distance between points (4,2) and (3,−1): P a = {4,2}; P b = {3,-1}; cout << abs(b-a) << "\n"; // 3.16228 The function arg(v) calculates the angle of a vector v = (x, y) with respect to the x axis. The function gives the angle in radians, where r radians equals 180r/π degrees. The angle of a vector that points to the right is 0, and angles decrease clockwise and increase counterclockwise.
The function polar(s,a) constructs a vector whose length is s and that points to an angle a. A vector can be rotated by an angle a by multiplying it by a vector with length 1 and angle a.
The following code calculates the angle of the vector (4,2), rotates it 1/2 radians counterclockwise, and then calculates the angle again: P v = {4,2}; cout << arg(v) << "\n"; // 0.463648 v = polar(1.0,0.5); cout << arg(v) << "\n"; // 0.963648 267 29.2 Points and lines The cross product a×b of vectors a = (x1, y1) and b = (x2, y2) is calculated using the formula x1y2 −x2y1. The cross product tells us whether b turns left (positive value), does not turn (zero) or turns right (negative value) when it is placed directly after a.
The following picture illustrates the above cases: a b a× b = 6 a b a× b = 0 a b a× b = −8 For example, in the first case a = (4,2) and b = (1,2). The following code calculates the cross product using the class complex: P a = {4,2}; P b = {1,2}; C p = (conj(a)b).Y; // 6 The above code works, because the function conj negates the y coordinate of a vector, and when the vectors (x1,−y1) and (x2, y2) are multiplied together, the y coordinate of the result is x1y2 −x2y1.
Point location Cross products can be used to test whether a point is located on the left or right side of a line. Assume that the line goes through points s1 and s2, we are looking from s1 to s2 and the point is p.
For example, in the following picture, p is on the left side of the line: s1 s2 p The cross product (p −s1)×(p −s2) tells us the location of the point p. If the cross product is positive, p is located on the left side, and if the cross product is negative, p is located on the right side. Finally, if the cross product is zero, points s1, s2 and p are on the same line.
268 Line segment intersection Next we consider the problem of testing whether two line segments ab and cd intersect. The possible cases are: Case 1: The line segments are on the same line and they overlap each other.
In this case, there is an infinite number of intersection points. For example, in the following picture, all points between c and b are intersection points: a d c b In this case, we can use cross products to check if all points are on the same line. After this, we can sort the points and check whether the line segments overlap each other.
Case 2: The line segments have a common vertex that is the only intersection point. For example, in the following picture the intersection point is b = c: a b = c d This case is easy to check, because there are only four possibilities for the intersection point: a = c, a = d, b = c and b = d.
Case 3: There is exactly one intersection point that is not a vertex of any line segment. In the following picture, the point p is the intersection point: c d a b p In this case, the line segments intersect exactly when both points c and d are on different sides of a line through a and b, and points a and b are on different sides of a line through c and d. We can use cross products to check this.
Point distance from a line Another feature of cross products is that the area of a triangle can be calculated using the formula |(a−c)×(b −c)| 2 , 269 where a, b and c are the vertices of the triangle. Using this fact, we can derive a formula for calculating the shortest distance between a point and a line. For example, in the following picture d is the shortest distance between the point p and the line that is defined by the points s1 and s2: s1 s2 p d The area of the triangle whose vertices are s1, s2 and p can be calculated in two ways: it is both 1 2|s2 −s1|d and 1 2((s1 −p)×(s2 −p)). Thus, the shortest distance is d = (s1 −p)×(s2 −p) |s2 −s1| .
Point inside a polygon Let us now consider the problem of testing whether a point is located inside or outside a polygon. For example, in the following picture point a is inside the polygon and point b is outside the polygon.
a b A convenient way to solve the problem is to send a ray from the point to an arbitrary direction and calculate the number of times it touches the boundary of the polygon. If the number is odd, the point is inside the polygon, and if the number is even, the point is outside the polygon.
For example, we could send the following rays: a b The rays from a touch 1 and 3 times the boundary of the polygon, so a is inside the polygon. Correspondingly, the rays from b touch 0 and 2 times the boundary of the polygon, so b is outside the polygon.
270 29.3 Polygon area A general formula for calculating the area of a polygon, sometimes called the shoelace formula, is as follows: 1 2| n−1 X i=1 (pi × pi+1)| = 1 2| n−1 X i=1 (xi yi+1 −xi+1yi)|, Here the vertices are p1 = (x1, y1), p2 = (x2, y2), ..., pn = (xn, yn) in such an order that pi and pi+1 are adjacent vertices on the boundary of the polygon, and the first and last vertex is the same, i.e., p1 = pn.
For example, the area of the polygon (4,1) (7,3) (5,5) (2,4) (4,3) is |(2·5−5·4)+(5·3−7·5)+(7·1−4·3)+(4·3−4·1)+(4·4−2·3)| 2 = 17/2.
The idea of the formula is to go through trapezoids whose one side is a side of the polygon, and another side lies on the horizontal line y = 0. For example: (4,1) (7,3) (5,5) (2,4) (4,3) The area of such a trapezoid is (xi+1 −xi) yi + yi+1 2 , where the vertices of the polygon are pi and pi+1. If xi+1 > xi, the area is positive, and if xi+1 < xi, the area is negative.
The area of the polygon is the sum of areas of all such trapezoids, which yields the formula | n−1 X i=1 (xi+1 −xi) yi + yi+1 2 | = 1 2| n−1 X i=1 (xi yi+1 −xi+1yi)|.
Note that the absolute value of the sum is taken, because the value of the sum may be positive or negative, depending on whether we walk clockwise or counterclockwise along the boundary of the polygon.
271 Pick’s theorem Pick’s theorem provides another way to calculate the area of a polygon provided that all vertices of the polygon have integer coordinates. According to Pick’s theorem, the area of the polygon is a+ b/2−1, where a is the number of integer points inside the polygon and b is the number of integer points on the boundary of the polygon.
For example, the area of the polygon (4,1) (7,3) (5,5) (2,4) (4,3) is 6+7/2−1 = 17/2.
29.4 Distance functions A distance function defines the distance between two points. The usual dis-tance function is the Euclidean distance where the distance between points (x1, y1) and (x2, y2) is q (x2 −x1)2 +(y2 −y1)2.
An alternative distance function is the Manhattan distance where the distance between points (x1, y1) and (x2, y2) is |x1 −x2|+|y1 −y2|.
For example, consider the following picture: (2,1) (5,2) (2,1) (5,2) Euclidean distance Manhattan distance The Euclidean distance between the points is p (5−2)2 +(2−1)2 = p 10 and the Manhattan distance is |5−2|+|2−1| = 4.
The following picture shows regions that are within a distance of 1 from the center point, using the Euclidean and Manhattan distances: 272 Euclidean distance Manhattan distance Rotating coordinates Some problems are easier to solve if Manhattan distances are used instead of Euclidean distances. As an example, consider a problem where we are given n points in the two-dimensional plane and our task is to calculate the maximum Manhattan distance between any two points.
For example, consider the following set of points: A C B D The maximum Manhattan distance is 5 between points B and C: A C B D A useful technique related to Manhattan distances is to rotate all coordinates 45 degrees so that a point (x, y) becomes (x+ y, y−x). For example, after rotating the above points, the result is: A C B D And the maximum distance is as follows: 273 A C B D Consider two points p1 = (x1, y1) and p2 = (x2, y2) whose rotated coordinates are p′ 1 = (x′ 1, y′ 1) and p′ 2 = (x′ 2, y′ 2). Now there are two ways to express the Manhat-tan distance between p1 and p2: |x1 −x2|+|y1 −y2| = max(|x′ 1 −x′ 2|,|y′ 1 −y′ 2|) For example, if p1 = (1,0) and p2 = (3,3), the rotated coordinates are p′ 1 = (1,−1) and p′ 2 = (6,0) and the Manhattan distance is |1−3|+|0−3| = max(|1−6|,|−1−0|) = 5.
The rotated coordinates provide a simple way to operate with Manhattan distances, because we can consider x and y coordinates separately. To maximize the Manhattan distance between two points, we should find two points whose rotated coordinates maximize the value of max(|x′ 1 −x′ 2|,|y′ 1 −y′ 2|).
This is easy, because either the horizontal or vertical difference of the rotated coordinates has to be maximum.
274 Chapter 30 Sweep line algorithms Many geometric problems can be solved using sweep line algorithms. The idea in such algorithms is to represent an instance of the problem as a set of events that correspond to points in the plane. The events are processed in increasing order according to their x or y coordinates.
As an example, consider the following problem: There is a company that has n employees, and we know for each employee their arrival and leaving times on a certain day. Our task is to calculate the maximum number of employees that were in the office at the same time.
The problem can be solved by modeling the situation so that each employee is assigned two events that correspond to their arrival and leaving times. After sorting the events, we go through them and keep track of the number of people in the office. For example, the table person arrival time leaving time John 10 15 Maria 6 12 Peter 14 16 Lisa 5 13 corresponds to the following events: John Maria Peter Lisa We go through the events from left to right and maintain a counter. Always when a person arrives, we increase the value of the counter by one, and when a person leaves, we decrease the value of the counter by one. The answer to the problem is the maximum value of the counter during the algorithm.
In the example, the events are processed as follows: 275 John Maria Peter Lisa + − + − + − + − 3 1 2 2 2 0 1 1 The symbols + and −indicate whether the value of the counter increases or decreases, and the value of the counter is shown below. The maximum value of the counter is 3 between John’s arrival and Maria’s leaving.
The running time of the algorithm is O(nlogn), because sorting the events takes O(nlogn) time and the rest of the algorithm takes O(n) time.
30.1 Intersection points Given a set of n line segments, each of them being either horizontal or vertical, consider the problem of counting the total number of intersection points. For example, when the line segments are there are three intersection points: It is easy to solve the problem in O(n2) time, because we can go through all possible pairs of line segments and check if they intersect. However, we can solve the problem more efficiently in O(nlogn) time using a sweep line algorithm and a range query data structure.
The idea is to process the endpoints of the line segments from left to right and focus on three types of events: (1) horizontal segment begins (2) horizontal segment ends (3) vertical segment 276 The following events correspond to the example: 1 2 1 2 1 2 3 3 We go through the events from left to right and use a data structure that maintains a set of y coordinates where there is an active horizontal segment. At event 1, we add the y coordinate of the segment to the set, and at event 2, we remove the y coordinate from the set.
Intersection points are calculated at event 3. When there is a vertical segment between points y1 and y2, we count the number of active horizontal segments whose y coordinate is between y1 and y2, and add this number to the total number of intersection points.
To store y coordinates of horizontal segments, we can use a binary indexed or segment tree, possibly with index compression. When such structures are used, processing each event takes O(logn) time, so the total running time of the algorithm is O(nlogn).
30.2 Closest pair problem Given a set of n points, our next problem is to find two points whose Euclidean distance is minimum. For example, if the points are we should find the following points: This is another example of a problem that can be solved in O(nlogn) time using a sweep line algorithm1. We go through the points from left to right and maintain a value d: the minimum distance between two points seen so far. At 1Besides this approach, there is also an O(nlogn) time divide-and-conquer algorithm that divides the points into two sets and recursively solves the problem for both sets.
277 each point, we find the nearest point to the left. If the distance is less than d, it is the new minimum distance and we update the value of d.
If the current point is (x, y) and there is a point to the left within a distance of less than d, the x coordinate of such a point must be between [x−d,x] and the y coordinate must be between [y−d, y+ d]. Thus, it suffices to only consider points that are located in those ranges, which makes the algorithm efficient.
For example, in the following picture, the region marked with dashed lines contains the points that can be within a distance of d from the active point: d d The efficiency of the algorithm is based on the fact that the region always contains only O(1) points. We can go through those points in O(logn) time by maintaining a set of points whose x coordinate is between [x−d,x], in increasing order according to their y coordinates.
The time complexity of the algorithm is O(nlogn), because we go through n points and find for each point the nearest point to the left in O(logn) time.
30.3 Convex hull problem A convex hull is the smallest convex polygon that contains all points of a given set. Convexity means that a line segment between any two vertices of the polygon is completely inside the polygon.
For example, for the points the convex hull is as follows: 278 Andrew’s algorithm provides an easy way to construct the convex hull for a set of points in O(nlogn) time. The algorithm first locates the leftmost and rightmost points, and then constructs the convex hull in two parts: first the upper hull and then the lower hull. Both parts are similar, so we can focus on constructing the upper hull.
First, we sort the points primarily according to x coordinates and secondarily according to y coordinates. After this, we go through the points and add each point to the hull. Always after adding a point to the hull, we make sure that the last line segment in the hull does not turn left. As long as it turns left, we repeatedly remove the second last point from the hull.
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285 286 Index 2SAT problem, 160 2SUM problem, 78 3SAT problem, 162 3SUM problem, 79 adjacency list, 113 adjacency matrix, 114 alphabet, 243 amortized analysis, 77 ancestor, 163 and operation, 96 Andrew’s algorithm, 279 antichain, 193 arithmetic progression, 10 backtracking, 50 Bellman–Ford algorithm, 123 binary code, 62 binary indexed tree, 86 binary search, 31 binary tree, 139 Binet’s formula, 14 binomial coefficient, 208 binomial distribution, 230 bipartite graph, 112, 122 birthday paradox, 247 bit representation, 95 bit shift, 97 bitset, 41 border, 244 breadth-first search, 119 bubble sort, 25 Burnside’s lemma, 214 Catalan number, 210 Cayley’s formula, 215 child, 133 Chinese remainder theorem, 205 closest pair, 277 codeword, 62 cofactor, 219 collision, 246 coloring, 112, 233 combinatorics, 207 comparison function, 31 comparison operator, 30 complement, 12 complete graph, 111 complex, 266 complex number, 266 complexity classes, 20 component, 110 component graph, 157 conditional probability, 227 conjuction, 13 connected graph, 110, 121 constant factor, 21 constant-time algorithm, 20 coprime, 201 counting sort, 28 cross product, 268 cubic algorithm, 20 cut, 182 cycle, 109, 121, 149, 155 cycle detection, 155 data compression, 62 data structure, 35 De Bruijn sequence, 178 degree, 111 depth-first search, 117 deque, 42 derangement, 213 determinant, 219 diameter, 135 287 difference, 12 difference array, 93 Dijkstra’s algorithm, 126, 153 Dilworth’s theorem, 193 Diophantine equation, 204 Dirac’s theorem, 177 directed graph, 110 disjunction, 13 distance function, 272 distribution, 229 divisibility, 197 divisor, 197 dynamic array, 35 dynamic programming, 65 dynamic segment tree, 261 edge, 109 edge list, 115 edit distance, 74 Edmonds–Karp algorithm, 184 equivalence, 13 Euclid’s algorithm, 200 Euclid’s formula, 206 Euclidean distance, 272 Euler tour technique, 168 Euler’s theorem, 202 Euler’s totient function, 201 Eulerian circuit, 174 Eulerian path, 173 expected value, 229 extended Euclid’s algorithm, 204 factor, 197 factorial, 14 Faulhaber’s formula, 10 Fenwick tree, 86 Fermat’s theorem, 202 Fibonacci number, 14, 206, 220 floating point number, 7 flow, 181 Floyd’s algorithm, 156 Floyd–Warshall algorithm, 129 Ford–Fulkerson algorithm, 182 Freivalds’ algoritm, 232 functional graph, 154 geometric distribution, 230 geometric progression, 11 geometry, 265 Goldbach’s conjecture, 199 graph, 109 greatest common divisor, 200 greedy algorithm, 57 Grundy number, 238 Grundy’s game, 241 Hall’s theorem, 189 Hamiltonian circuit, 177 Hamiltonian path, 177 Hamming distance, 100 harmonic sum, 11, 200 hash value, 245 hashing, 245 heap, 43 Heron’s formula, 265 heuristic, 179 Hierholzer’s algorithm, 175 Huffman coding, 63 identity matrix, 218 implication, 13 in-order, 139 inclusion-exclusion, 212 indegree, 111 independence, 228 independent set, 190 index compression, 93 input and output, 4 integer, 6 intersection, 12 intersection point, 276 inverse matrix, 220 inversion, 26 iterator, 39 K˝ onig’s theorem, 189 Kadane’s algorithm, 23 Kirchhoff’s theorem, 223 knapsack, 72 knight’s tour, 179 Kosaraju’s algorithm, 158 Kruskal’s algorithm, 142 Lagrange’s theorem, 205 Laplacean matrix, 224 288 Las Vegas algorithm, 231 lazy propagation, 258 lazy segment tree, 258 leaf, 133 least common multiple, 200 Legendre’s conjecture, 199 Levenshtein distance, 74 lexicographical order, 244 line segment intersection, 269 linear algorithm, 20 linear recurrence, 220 logarithm, 14 logarithmic algorithm, 20 logic, 13 longest increasing subsequence, 70 losing state, 235 lowest common ancestor, 167 macro, 9 Manhattan distance, 272 map, 38 Markov chain, 230 matching, 187 matrix, 217 matrix multiplication, 218, 232 matrix power, 219 maximum flow, 181 maximum independent set, 190 maximum matching, 187 maximum query, 83 maximum spanning tree, 142 maximum subarray sum, 21 meet in the middle, 54 memoization, 67 merge sort, 27 mex function, 238 minimum cut, 182, 185 minimum node cover, 189 minimum query, 83 minimum spanning tree, 141 misère game, 238 Mo’s algorithm, 255 modular arithmetic, 6, 201 modular inverse, 202 Monte Carlo algorithm, 231 multinomial coefficient, 210 natural logarithm, 15 nearest smaller elements, 79 negation, 13 negative cycle, 125 neighbor, 111 next_permutation, 49 nim game, 237 nim sum, 237 node, 109 node cover, 189 not operation, 97 NP-hard problem, 20 number theory, 197 or operation, 96 order statistic, 232 Ore’s theorem, 177 outdegree, 111 pair, 30 parent, 133 parenthesis expression, 211 Pascal’s triangle, 209 path, 109 path cover, 190 pattern matching, 243 perfect matching, 189 perfect number, 198 period, 243 permutation, 49 persistent segment tree, 262 Pick’s theorem, 272 point, 266 polynomial algorithm, 20 polynomial hashing, 245 post-order, 139 Prüfer code, 216 pre-order, 139 predicate, 13 prefix, 243 prefix sum array, 84 Prim’s algorithm, 147 prime, 197 prime decomposition, 197 priority queue, 43 probability, 225 programming language, 3 289 Pythagorean triple, 206 quadratic algorithm, 20 quantifier, 13 queen problem, 50 queue, 43 quickselect, 232 quicksort, 232 random variable, 228 random_shuffle, 39 randomized algorithm, 231 range query, 83 regular graph, 111 remainder, 6 reverse, 39 root, 133 rooted tree, 133 rotation, 243 scaling algorithm, 185 segment tree, 89, 257 set, 12, 37 set theory, 12 shoelace formula, 271 shortest path, 123 sieve of Eratosthenes, 200 simple graph, 112 sliding window, 81 sliding window minimum, 81 sort, 29, 39 sorting, 25 spanning tree, 141, 223 sparse segment tree, 261 sparse table, 85 SPFA algorithm, 126 Sprague–Grundy theorem, 238 square matrix, 217 square root algorithm, 251 stack, 42 string, 36, 243 string hashing, 245 strongly connected component, 157 strongly connected graph, 157 subsequence, 243 subset, 12, 47 substring, 243 subtree, 133 successor graph, 154 suffix, 243 sum query, 83 sweep line, 275 time complexity, 17 topological sorting, 149 transpose, 217 tree, 110, 133 tree query, 163 tree traversal array, 164 trie, 244 tuple, 30 typedef, 8 twin prime, 199 two pointers method, 77 two-dimensional segment tree, 264 uniform distribution, 230 union, 12 union-find structure, 145 universal set, 12 vector, 35, 217, 266 Warnsdorf’s rule, 179 weighted graph, 111 Wilson’s theorem, 206 winning state, 235 xor operation, 97 Z-algorithm, 247 Z-array, 247 Zeckendorf’s theorem, 206 290 |
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2015 AIME I Problems/Problem 10
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6 (Finite differences)
8 Solution 7 (Like solution 1 without annoying systems)
9 Solution 8 (First few steps of solution 1)
10 Solution 9 (Cheese Solution)
11 See Also
Problem
Let be a third-degree polynomial with real coefficients satisfying Find .
Solution 1
Let = . Since is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing and , it is easy to see that , and ; otherwise more bends would be required in the graph. Since only the absolute value of is required, there is no loss of generalization by stating that , and . This provides the following system of equations. Using any four of these functions as a system of equations yields
Note: You can use Gaussian elimination to solve these equations.
Solution 2
By drawing the function, and similar to Solution 1, WLOG let . Then, . Set . Then the roots of are . So, . Plug in to find a. We know . So, . Thus, , and then .
Solution 3
Without loss of generality, let . (If , then take as the polynomial, which leaves unchanged.) Because is third-degree, write where clearly must be a permutation of from the given condition. Thus However, subtracting the two equations gives , so comparing coefficients gives and thus both values equal to . As a result, . As a result, and so . Now, we easily deduce that and so removing the without loss of generality gives , which is our answer.
Solution 4
The following solution is similar to solution 3, but assumes nothing. Let . Since has degree 3, has degree 6 and has roots 1,2,3,5,6, and 7. Therefore, for some . Hence . Note that . Since has degree 3, so do and ; and both have the same leading coefficient. Hence and for some (else is not cubic) where is the same as the set . Subtracting the second equation from the first, expanding, and collecting like terms, we have that which must hold for all . Since we have that (1) , (2) and (3) . Since is the sum of 1,2,3,5,6, and 7, we have so that by (1) we have and . We must partition 1,2,3,5,6,7 into 2 sets each with a sum of 12. Consider the set that contains 7. It can't contain 6 or 5 because the sum of that set would already be with only 2 elements. If 1 is in that set, the other element must be 4 which is impossible. Hence the two sets must be and . Note that each of these sets happily satisfy (2). By (3), since the sets have products 42 and 30 we have that . Since is the leading coefficient of , the leading coefficient of is . Thus the leading coefficient of is 4, i.e. . Then from earlier, so that the answer is .
Solution 5
Express in terms of powers of : By the same argument as in the first Solution, we see that is an odd function about the line , so its coefficients and are 0. From there it is relatively simple to solve (as in the above solution, but with a smaller system of equations): and
Solution 6 (Finite differences)
Because a cubic must come in a "wave form" with two points of inflection, we can see that , and . By symmetry, . Now, WLOG let , and . Then, we can use finite differences to get that the third (constant) difference is , and therefore .
Solution 7 (Like solution 1 without annoying systems)
We can rewrite our function as two different cubics, . Note that is the same in both because they must be equal, so their leading terms must be. We must then, following Vieta's, choose our roots such that and verify that the other Vieta's formulas hold. Additionally, a cubic must only cross the x-axis thrice, restricting our choices for roots further. Choosing , , , , , yields: For the constant terms to have a difference of 24 (), we must have , so the constant term of our polynomial is , the absolute value of which is . -- Solution by eiis1000
Solution 8 (First few steps of solution 1)
We can rewrite the function as and . Since we need to find , substitute 0 for x in these two equations.
Isolating in both of the equations, Equating the two and solving for , we see .
~YBSuburbanTea
Solution 9 (Cheese Solution)
Let the leading coefficient of be . Then, it is obvious that . Let us now, let . We then have, after cleaning it up nicely, . We now take the square root of both sides, to obtain . So, . Now, this is the cheese part. Since this is aime, we know that the answer must be an integer, so we assume . Thus, we get .
~~triggod
See Also
2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9Followed by
Problem 11
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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188159 | https://www.amazon.com/Primate-Adaptation-Evolution-John-Fleagle/dp/0123786320 | Skip to
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Primate Adaptation and Evolution 3rd Edition
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Primate Adaptation and Evolution, Third Edition, is a thorough revision of the text of choice for courses in primate evolution. The book retains its grounding in the extant primate groups as the best way to understand the fossil trail and the evolution of these modern forms. However, this coverage is now streamlined, making reference to the many new and excellent books on living primate ecology and adaptation – a field that has burgeoned since the first edition of Primate Adaptation and Evolution.
By drawing out the key features of the extant families and referring to more detailed texts, the author sets the scene and also creates space for a thorough updating of the exciting developments in primate palaeontology – and the reconstruction through early hominid species – of our own human origins. This updated version covers recent developments in primate paleontology and the latest taxonomy, and includes over 200 new illustrations and revised evolutionary trees.
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"...an indispensible reference and must-have textbook for students and scholars with an interest in the evolutionary history of the primates...Hats off to Fleagle for producing a superb book that every primatologist needs and only he could write."--The Quarterly Review of Biology, Primate Adaptation and Evolution, Third Edition
"Fleagle has revised and rewritten each chapter, redone all the tables, added many new figures, and replaced almost all the references. One of the changes is that he now uses a suite of criteria to estimate body mass from fossil skeletons, rather than the single formula he formerly used."--Reference and Research Book News, August 2013
Reviews of the First Edition "Fleagle's book...is a tour de force as well as a tour d'horizon of the primate order....Fleagle's book fills a long-standing need for a comprehensive and up-to-date introductory text in its field." --NATURE "Just occasionally in academic life there comes together in one person research ability, industry, and the willingness to communicate. This is indeed the case for John Fleagle's textbook. An admirable book that deserves the success that it will undoubtedly achieve." --BIOLOGY AND SOCIETY "Professional primatologists and students alike will be delighted with John Fleagle's Primate Adaptation and Evolution, which will brilliantly refresh any university reading list in the subject. For many university courses on primate biology, it will undoubtedly become a primary text." --THE TIMES HIGHER EDUCATION SUPPLEMENT "This is an excellent book on primate comparative anatomy, behavioral ecology, and paleontology." --AMERICAN ANTHROPOLOGIST "John Fleagle has done it! Primate Adaptation and Evolution is the readable and "seeable," but still traditional, textbook that my students have been waiting for someone to write." --AMERICAN JOURNAL OF PHYSICAL ANTHROPOLOGY "This book will be welcomed with a sigh of relief in departments of anatomy and biological anthropology world-wide. Here at last is an up-to-date, comprehensive and reliable textbook for courses in primate and human evolution.... It will, and should, stand as a major teaching resource for primate evolution for some years to come. It is an excellent text and a reflection of Fleagle's major contributions to our field." --PRIMATE EYE "This volume is basically designed as an advanced undergraduate or graduate textbook but is so well illustrated and organized that it can also be used as a general source or reference for less specialized readers. No other single work can match the range of its topics, which matches the interests of its author, one of the leading students of primate paleontology, behavior, morphology, and evolution." --CHOICE
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The market leader in Primate Evolution textbooks―now with a new discussion of molecular systematics and adoption of the latest taxonomy!
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188160 | https://www.scribd.com/presentation/892217144/Blood-Components-and-Clotting | Blood Components and Clotting | PDF | Blood | Blood Cell
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Blood Components and Clotting
The document outlines the components and functions of blood within the circulatory system, highlighting the roles of plasma, red blood cells, white blood cells, and platelets. It explains th…
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Syllabus Statement
Science Understanding - Circulatory system
•
the
transport of materials
within the internal environmen t for exchange with cells is facilitated by the
structure and funcon of the circulatory system
at the cell, ssue and organ levels
•
the
components
of blood
facilitate the transport
of dierent materials around the body (plasma and erythrocytes), play a role in the
clong of blood
(platele ts) and the
protecon
of the body (leucocytes)
adDownload to read ad-free
•
The blood transports oxygen, food and wastes
•
Blood is made up of:
–
Plasma
(liquid part)
•
which is mainly
water
•
55%
of blood volume
–
Formed Elements
(non-liquid)
•
which include
red blood cells
,
white blood cells
,
platelets
and
cell fragments
•
45%
of blood volume
Circulatory System
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F O R M E D E L E M E N T S L I Q U I D P A R T Red Blood Cell White Blood Cell P l a t e l e t P l a s m a SCIENTIFIC NAME Erythrocyte L e u k o c y t e T h r o m b o c y t e
ROLE IN THE BODY Carry O
2
/CO
2
around the body Protecon of the body Blood clong Medium that carries nutrients/wastes around body LIFESPAN 120 days 1 d a y 7 d a y s
Blood Components Summary
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188161 | https://math.stackexchange.com/questions/2407736/linear-algebra-changing-general-form-into-vector-form | Linear Algebra: Changing General form into vector form - Mathematics Stack Exchange
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Linear Algebra: Changing General form into vector form
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I encountered this little rule in one of my answer books: The direction vector of the general equation ax + by = c is d = [b, -a].
I have never seen this rule formally described so I'm wondering if someone could help me find a theorem that describes it fully.
What I also wonder if what happens if I want to convert say a three dimensional plane ax + by + cz = d into vector form? Does the same rule follow where d = [c, -b, -a]?
The context for this was in this chegg textbook answer
linear-algebra
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asked Aug 27, 2017 at 15:15
Frank John LiFrank John Li
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A plane is not a directed line (although it contains many such lines). There is no direction vector in this case.user1551 –user1551 2017-08-27 15:19:26 +00:00 Commented Aug 27, 2017 at 15:19
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If the direction vector of a line is d d, then all points on the line are of the form p 0+t d p 0+t d, where p 0=(x 0,y 0)p 0=(x 0,y 0) is some known point on the line and t∈R t∈R. Set d=(b,−a)d=(b,−a) and plug this into the equation of the line:
a(b t+x 0)+b(−a t+y 0)=a x 0+b y 0=d.a(b t+x 0)+b(−a t+y 0)=a x 0+b y 0=d.
The dot product (a,b)⋅(b,−a)=a b−b a=0(a,b)⋅(b,−a)=a b−b a=0, so the vector (a,b)(a,b) is perpendicular (a.k.a. normal) to the line.
In a similar fashion, the vector (a,b,c)(a,b,c) is perpendicular to the plane a x+b y+c z=d a x+b y+c z=d. A plane is two-dimensional, though, so to convert this into parametric form you need two direction vectors instead of one. There’s an analogous rule to the one you have for a line: Since a a, b b and c c are not all zero, at least two of (a,b,c)×(1,0,0)=(0,c,−b)(a,b,c)×(1,0,0)=(0,c,−b), (a,b,c)×(0,1,0)=(−c,0,a)(a,b,c)×(0,1,0)=(−c,0,a) and (a,b,c)×(0,0,1)=(b,−a,0)(a,b,c)×(0,0,1)=(b,−a,0) are non-zero. These vectors are all orthogonal to (a,b,c)(a,b,c). Choose any two of these vectors and call them u u and v v. Then all of the points on the plane are of the form p 0+s u+t v p 0+s u+t v. I’ll leave verifying this to you.
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answered Aug 28, 2017 at 21:01
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The plane a x+b y+c z=d a x+b y+c z=d is perpendicular to the vector (a,b,c)(a,b,c) ...
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answered Aug 27, 2017 at 16:23
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188162 | https://byjus.com/chemistry/nucleophilic-addition-reactions/ | What is a Nucleophilic Addition Reaction?
A nucleophilic addition reaction is a chemical addition reaction in which a nucleophile forms a sigma bond with an electron-deficient species. These reactions are considered very important in organic chemistry since they enable the conversion of carbonyl groups into a variety of functional groups. Generally, nucleophilic addition reactions of carbonyl compounds can be broken down into the following three steps.
The electrophilic carbonyl carbon forms a sigma bond with the nucleophile.
The carbon-oxygen pi bond is now broken, forming an alkoxide intermediate (the bond pair of electrons are transferred to the oxygen atom).
The subsequent protonation of the alkoxide yields the alcohol derivative.
The carbon-oxygen double bond is directly attacked by strong nucleophiles to give rise to the alkoxide. However, when weak nucleophiles are used, the carbonyl group must be activated with the help of an acid catalyst for the nucleophilic addition reaction to proceed.
The carbonyl group has a coplanar structure and its carbon is sp2 hybridized. However, the attack of the nucleophile on the C=O group results in the breakage of the pi bond. The carbonyl carbon is now sp3 hybridized and forms a sigma bond with the nucleophile. The resulting alkoxide intermediate has a tetrahedral geometry, as illustrated above.
Why do Carbonyl Compounds Undergo Nucleophilic Addition?
In carbonyl compounds, the carbon-oxygen bond is polar. Owing to the relatively higher electronegativity of the oxygen atom, the electron density is higher near the oxygen atom. This leads to the generation of a partial negative charge on the oxygen atom and a partial positive charge on the carbon atom.
Since the carbonyl carbon holds a partial positive charge, it behaves as an electrophile. The partial negative charge on the oxygen atom can be stabilized via the introduction of an acidic group. The proton donated by the acid bonds with the carbonyl oxygen atom and neutralizes the negative charge.
Aldehydes are relatively more reactive towards nucleophilic addition reactions when compared to ketones. This is because the secondary carbocations formed by ketones are stabilized by the adjacent R groups. The primary carbocations formed by aldehydes are less stable than the secondary carbocations formed by ketones and are, therefore, more susceptible to nucleophilic attacks.
Reactions with Hydrogen Cyanide
The nucleophilic addition reaction between hydrogen cyanide (HCN) and carbonyl compounds (generally aldehydes and ketones) results in the formation of cyanohydrins. Base catalysts are often used to increase the rate of the reaction. The cyanide anion (CN–) acts as a powerful nucleophile and attacks the carbonyl carbon to form a new sigma bond, as illustrated below.
The polar nature of the C=O bond makes the carbonyl carbon electrophilic in nature. The cyanide anion executes a nucleophilic attack on the carbonyl carbon, resulting in the formation of an intermediate. This intermediate is now protonated to afford the cyanohydrin product.
Nucleophilic Additions with Monohydric Alcohols
Aldehydes and ketones undergo nucleophilic addition reactions with monohydric alcohols to yield hemiacetals. Upon further reaction with another molecule of the alcohol, an acetal is obtained. Since alcohols are weak nucleophiles, the reaction requires an acid catalyst for the activation of the carbonyl group towards nucleophilic attack.
Since the hemiacetals can undergo hydrolysis to yield the reactants (the alcohol and carbonyl compound), the water formed during the reaction must be removed. In this reaction, the carbonyl oxygen is protonated before the nucleophilic attack is carried out by the alcohol. The nucleophilic alcohol is now deprotonated to form the hemiacetal. This reaction can be repeated to obtain the acetal.
Other Examples
Nucleophilic Addition with Grignard Reagents
The polar nature of Grignard reagents (general formula: R-Mg-X) attributes a partial negative charge to the carbon atom. The following types of alcohols are formed from nucleophilic addition reactions with Grignard reagents.
Primary alcohols are formed when formaldehyde is used.
Other aldehydes yield secondary alcohols upon reacting with Grignard reagents.
The nucleophilic addition reactions between ketones and Grignard reagents yield tertiary alcohols.
The general mechanism of these reactions involves the attack of the nucleophilic carbon (belonging to R-Mg-X) on the carbonyl carbon. A simple acid workup of the resulting alkoxide yields the corresponding alcohol.
Reaction with Primary Amines
The reaction between primary amines and aldehydes/ketones yields imine derivatives along with water. The reaction can be illustrated as follows.
Initially, the nucleophilic nitrogen belonging to the amine attacks the carbonyl carbon. The carbon-oxygen double bond is broken and a new carbon-nitrogen sigma bond is formed. Now a proton is transferred from the amine to the oxygen atom. In the next step of this nucleophilic addition reaction, The OH group is further protonated and water is removed. The carbon atom now forms a double bond with the nitrogen belonging to the amine. This nitrogen is now deprotonated to afford the required imine product.
Frequently Asked Questions
Q1
What is the main feature of the addition reaction?
When an atom is added to a combination with a double or triple bond, an addition reaction occurs. Addition reactions are linked to unsaturated molecules. These are hydrocarbons with two or three double or triple bonds. After an addition reaction is completed, there are no reactant residues left.
Q2
What causes addition reaction?
In organic chemistry, an addition reaction is an organic reaction in which two or more molecules combine to generate a bigger one (the adduct). Molecules with carbon—hetero double bonds, such as carbonyl (C=O) or imine (C=N) groups, can be added because they have double-bond character as well.
Q3
What is a Nucleophilic Addition Reaction?
A nucleophilic addition reaction is a chemical addition reaction in which a nucleophile forms a sigma bond with an electron-deficient species.
Q4
Is hydration an addition reaction?
In the presence of a catalyst, alkenes undergo an addition reaction with water to create alcohol. Hydration is the name for this sort of addition reaction. Water is directly injected into the carbon-carbon double bond.
Q5
Define nucleophiles.
A nucleophile is a word used to refer to substances that tend to donate electron pairs to electrophiles in order to form chemical bonds with them. Any ion or molecule having an electron pair which is free or a pi bond containing two electrons has the ability to behave like nucleophiles.
To learn more about nucleophilic addition reactions and other related concepts, such as SN1 reactions, register with BYJU’S and download the mobile application on your smartphone.
Test Your Knowledge On Nucleophilic Addition Reactions!
Q5
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Select the correct answer and click on the “Finish” button
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Karthika
July 26, 2020 at 9:28 am
It’s really easy to read like a short and sweet answer tq sir…..😊 Like this way you upload so many topics tq sir…
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188163 | https://www.youtube.com/watch?v=hDn7HT5tLog | Sensitivity and Specificity: Avoiding 3 Common Pitfalls
Society of Thoracic Radiology
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Posted: 7 Dec 2023
Dr. David Naeger addresses the often-overlooked but crucial concepts of sensitivity and specificity. Using two cases as examples, Dr. Naeger highlights three common pitfalls. In the first case, involving a fracture diagnosis, the discussion revolves around sensitivity and specificity, emphasizing that these metrics apply only when the disease status is already known. The second case involves a new disease with an expensive diagnostic test, where Dr. Naeger cautions against relying solely on sensitivity and specificity and emphasizes the importance of considering the prevalence of the disease in the population. The third pitfall discussed is the tendency to overlook the significance of prevalence in research populations. Dr. Naeger concludes by stressing the importance of understanding predictive values in real-world scenarios and avoiding these pitfalls for accurate and meaningful diagnostic assessments.
Transcript:
[Music] all right welcome everyone back for our resident boot camp uh session three and this is a pretty cool session uh the topic is skills for the Radiology resident so these are sort of non-interpretive topics that we think are really useful just for like daily practice as a radiologist so I'm going to start us off and we're going to talk about sensitivity and specificity and I actually love talking about this topic because everyone doesn't want to hear about it right they either say oh I have a doctorate I obviously already understand this or I've never gotten it and you know I'm doing okay so what's the big deal or it's boring regardless so I actually think it's pretty interesting so I'm going to show you two cases we're actually going to go through three pitfalls and just to sort of engage gaug you to see if you can figure it out if you happen to avoid all three pitfalls in your hand head you'll be rewarded wondrously all right so here we go let's see if you got this down and make sense to you right we have a non-chest case here but we'll just roll with it let's say that you're doing an ed shift after an amazing chest rotation and you see that you have a fracture so here an obvious scaer fracture so you dutifully call the Ed and say there's a fracture and the other person said hm okay so just curious what is the sensitivity of X-ray for scoid fractures by the way you're like oh boy oh my gosh I don't know 75% of course I knew that yes 75% and he says H well I'm terrified of the orthopod don't really want to call 75 doesn't sound so great maybe we need an MRI you're like look man it's clearly broken why do we need an MRI he said you said you were 75% sure I think we needed the answer so there's your setup what are we going to do do you agree going to protocol that MRI do you say I don't think that's right but I don't know why I know why it's wrong and I'm going to tell you or do you have no idea what's going on ever have an answer do we have a thought anybody want to yell an answer out woo we got some C's and we know why that's bold I like it all right so let if we can get an answer here two ways one we talked about sensitivity what is that and then two what is he really trying to ask us okay so here's a secret if this has never been sorted in your brain clearly sensitivity and specificity are terms that only apply when you already know if they have the the disease or not okay so if you're going to say a sentence this is how I keep it clear in my brain the sentence is a say in someone who's known already to have disease or not Etc you might say that sounds silly why do I need a test characteristic that applies when I already know the answer and I totally agree this is what's used in controlled settings where we have gold standards and we know already like research okay so to get such a number you get a 100 people let's say who you know have disease which means you need to have some way to know that a gold standard okay then we're measuring this on an imperfect test in this case an x-ray does it still work okay so if it finds it in most people then it's 75% sensitive so again this is only in people we already know the answer answer we cannot apply sensitivity directly to an unknown patient presenting so sensitivity is the ability to pick up disease that is there and if it is not sensitive it turns out to be normal too much that's the problem all right so specificity is its mirror we take a 100 people proven to not have disease already and then we say how often is it negative so for example most people then it's 97% specific okay so again this is only what we already know for sure they don't have the disease so it's really just a statistic of a normal of a test being normal and normal people it's not very interesting and if it lacks this then the problem is it's positive too much inaccurate so because it was sered talking about false positives you might might think aha this whole scenario that orthopod thing was about specificity but not quite okay what we're really asking is we have a test an x-ray and we want to know if it is real or fake right in an unknown patient unknown disease status so again the sentences in a person where we don't know if they have disease or not we have a positive test so does that mean they actually have the disease question mark so how would we figure that out what what are the two ways we have a positive test it's real because we can pick it up and also what is the chance of actually having the disease right if it never happens we're not going to have very many real positives and the fake positive is the chance that it incorrectly shows up on a test it's related to specificity and again how much normal is out there in the world you know usually a lot so this term is predictive values that's what we are talking about these say how a test does in patients where we don't know the answer yet so positive predictor value is in someone where we don't know if they actually for real have the disease a positive test confers what chance of having that disease and then the negative is the opposite when we don't know if they have disease or not a negative test confers the chance of having no disease question mark That's a predictive value okay now you might sound aha these are what we should be using in real life and that is totally correct okay this is how tests perform in the real world so we're going to wrap up that orthopod case when it said use seems 75% sure maybe we should get an MRI you say oh I get it you're asking me the chances radiograph confers an actual real fracture positive predictive value it's super high here 98% don't you worry and then off he goes to call the orthopod no MRI all right great so sensor specificity when we know already the disease predictive values when we don't okay and our pitfalls here was thinking about sensitivity and specificity when we care about predictive values and then two we found in this trap of using words like certainty or a sister term rates which don't really mean it's it's just ambiguous what those mean so that's a pitfall in this we use specific language all right case two let's see how you're doing your head with this one we have 100 cases of a brand new disease across the entire planet of a very expensive cure no not the cure for heepsy which is $1,000 a day not the most expensive gene therapy that exists to be cured from this disease we have to buy you an incredibly expensive car so this is very expensive we don't want to do this for everybody so we need to get the diagnosis here and let's say in this magical disease a young Radiology researcher came to the the rescue they found out that in this weird disease it causes high density in the CSF that's not blood that's this disease okay so we found a test and then this young research got some colleagues and published the first paper on these 100 cases he pulled everybody in the world and most had high density all right okay he then found a 100 people so we could have some controls some normal folks and most did not have high density one person who just by a random chance had a car accident or something and had a bleed there a while back okay so the sensitivity and specificity when we know the answer we had long-term followup we know the answer for sure was great so in this paper he said this test has high predictive values too um not only Sensi specificity but High predictive values that means it works in the real world so we should use it let's test everyone all right and how do he get those numbers he or she well we take our study population and they said who has the positive test well looks like most of them have the disease okay good predictive value who is normal looks like most of them are normal folks so that means negative predictive value is great okay so great tests so what should we do now I just set you up do we plan to buy a bunch of cars for a bunch of people like this one look green stripes fancy do we say h maybe the St is not totally right let's do a repeat but perhaps we should start thinking about it you can actually buy Porsche stock now it just ipoed you used to have to buy Volkswagen how interesting or three decide this test is totally useless and throw it away what's the right answer and the answer completely useless test okay but why so so predictive values take into account yes how well it performs but also the likelihood of disease okay so what is the actual likelihood of having this tiny rare disease we have 100 people out of a very big planet right so that means one out of 80 million actually have it so if we could actually scan the whole planet that would be expensive let's say we do a HCT on the whole whole planet 8 billion people we would get almost everyone with our disease and we would get what 80 million normal people who have an abnormal head C because even a tiny percent if you applied to everyone that's a lot of false positives so if you say what happens when you have a positive test fast fast chance is your normal or at least you don't have this disease you have something else right so our prediced value for this rare disease is basically zero total garbage does not help you cuz it's like a lightning strike to have this disease so rare okay so what was the error not taking into account likelihood and the likelihood that they were using was what the likelihood in the research study which was a case control a 100 of each this study has tons of people with disease that was the problem okay so you might say oh researchers they they're all better trained they would never do that here's a paper on an infection it talks about the positive predictive value and it totally calculated it wrong on a case control study what terrible Journal was this the famous new Eng the Journal of Medicine now it's an old study because we've gotten a little better but I have seen this incorrectly put into Radiology conferences right up to a couple years ago so if you are looking at research or doing research you need to know the prevalence in your cohort versus the population out there you're testing on and it does highlight how important our field is with the prevalence underlying our disease right this is why we want someone more likely to have a PE to get the test smokes enough or old enough to get the lung cancer screening or a certain age to get mammographic screening training okay so the pitfall here and our last one was forgetting the important of uh prevalence and how it varies in research populations or other so it's time to drive off into the sunset on this topic we went over three Cas two cases three pitfalls and if you want some more information we actually summarized this in a paper from a while back but it has Radiology examples and goes over this info so it's pretty uh concise to review re-review it and that's it thank you very [Applause] [Music] much |
188164 | https://en.wikipedia.org/wiki/Centriole | Jump to content
Centriole
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From Wikipedia, the free encyclopedia
Organelle in eukaryotic cells
See also: Centrosome
In cell biology a centriole is a cylindrical organelle composed mainly of a protein called tubulin. Centrioles are found in most eukaryotic cells, but are not present in conifers (Pinophyta), flowering plants (angiosperms) and most fungi, and are only present in the male gametes of charophytes, bryophytes, seedless vascular plants, cycads, and Ginkgo. A bound pair of centrioles, surrounded by a highly ordered mass of dense material, called the pericentriolar material (PCM), makes up a structure called a centrosome.
Centrioles are typically made up of nine sets of short microtubule triplets, arranged in a cylinder. Deviations from this structure include crabs and Drosophila melanogaster embryos, with nine doublets, and Caenorhabditis elegans sperm cells and early embryos, with nine singlets. Additional proteins include centrin, cenexin and tektin.
The main function of centrioles is to produce cilia during interphase and the aster and the spindle during cell division.
History
[edit]
The centrosome was discovered jointly by Walther Flemming in 1875 and Edouard Van Beneden in 1876. Edouard Van Beneden made the first observation of centrosomes as composed of two orthogonal centrioles in 1883. Theodor Boveri introduced the term "centrosome" in 1888 and the term "centriole" in 1895. The basal body was named by Theodor Wilhelm Engelmann in 1880. The pattern of centriole duplication was first worked out independently by Étienne de Harven and Joseph G. Gall c. 1950.
Role in cell division
[edit]
Centrioles are involved in the organization of the mitotic spindle and in the completion of cytokinesis. Centrioles were previously thought to be required for the formation of a mitotic spindle in animal cells. However, more recent experiments have demonstrated that cells whose centrioles have been removed via laser ablation can still progress through the G1 stage of interphase before centrioles can be synthesized later in a de novo fashion. Additionally, mutant flies lacking centrioles develop normally, although the adult flies' cells lack flagella and cilia and as a result, they die shortly after birth.
The centrioles can self replicate during cell division.
Cellular organization
[edit]
Centrioles are a very important part of centrosomes, which are involved in organizing microtubules in the cytoplasm. The position of the centriole determines the position of the nucleus and plays a crucial role in the spatial arrangement of the cell.
Fertility
[edit]
Sperm centrioles are important for 2 functions: (1) to form the sperm flagellum and sperm movement and (2) for the development of the embryo after fertilization. The sperm supplies the centriole that creates the centrosome and microtubule system of the zygote.
Ciliogenesis
[edit]
In flagellates and ciliates, the position of the flagellum or cilium is determined by the mother centriole, which becomes the basal body. An inability of cells to use centrioles to make functional flagella and cilia has been linked to a number of genetic and developmental diseases. In particular, the inability of centrioles to properly migrate prior to ciliary assembly has recently been linked to Meckel–Gruber syndrome.
Animal development
[edit]
Proper orientation of cilia via centriole positioning toward the posterior of embryonic node cells is critical for establishing left-right asymmetry, during mammalian development.
Centriole duplication
[edit]
Before DNA replication, cells contain two centrioles, an older mother centriole, and a younger daughter centriole. During cell division, a new centriole grows at the proximal end of both mother and daughter centrioles. After duplication, the two centriole pairs (the freshly assembled centriole is now a daughter centriole in each pair) will remain attached to each other orthogonally until mitosis. At that point the mother and daughter centrioles separate dependently on an enzyme called separase.
The two centrioles in the centrosome are tied to one another. The mother centriole has radiating appendages at the distal end of its long axis and is attached to its daughter at the proximal end. Each daughter cell formed after cell division will inherit one of these pairs. Centrioles start duplicating when DNA replicates.
Origin
[edit]
LECA, the last common ancestor of all eukaryotes was a ciliated cell with centrioles.[citation needed] Some lineages of eukaryotes, such as land plants, do not have centrioles except in their motile male gametes. Centrioles are completely absent from all cells of conifers and flowering plants, which do not have ciliate or flagellate gametes.
It is unclear if the last common ancestor had one or two cilia. Important genes such as those coding for centrins, required for centriole growth, are only found in eukaryotes, and not in bacteria or archaea.
Etymology and pronunciation
[edit]
The word centriole (/ˈsɛntrioʊl/) uses combining forms of centri- and -ole, yielding "little central part", which describes a centriole's typical location near the center of the cell.
Atypical centrioles
[edit]
Typical centrioles are made of 9 triplets of microtubules organized with radial symmetry. Centrioles can vary the number of microtubules and can be made of 9 doublets of microtubules (as in Drosophila melanogaster) or 9 singlets of microtubules as in C. elegans. Atypical centrioles are centrioles that do not have microtubules, such as the Proximal Centriole-Like found in D. melanogaster sperm, or that have microtubules with no radial symmetry, such as in the distal centriole of human spermatozoon. Atypical centrioles may have evolved at least eight times independently during vertebrate evolution and may evolve in the sperm after internal fertilization evolves.
It wasn't clear why the centriole became atypical until recently. The atypical distal centriole forms a dynamic basal complex (DBC) that, together with other structures in the sperm neck, facilitates a cascade of internal sliding that couples tail beating with head kinking. The atypical distal centriole's properties suggest that it evolved into a transmission system that couples the sperm tail motors to the whole sperm, thereby enhancing sperm function.
References
[edit]
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^ Boveri, T. Ueber das Verhalten der Centrosomen bei der Befruchtung des Seeigel-Eies nebst allgemeinen Bemerkungen über Centrosomen und Verwandtes. Verh. d. Phys.-Med. Ges. zu Würzburg, N. F., Bd. XXIX, 1895. link.
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^ La Terra, S; English, CN; Hergert, P; McEwen, BF; Sluder, G; Khodjakov, A (2005). "The de novo centriole assembly pathway in HeLa cells: cell cycle progression and centriole assembly/maturation". The Journal of Cell Biology. 168 (5): 713–22. doi:10.1083/jcb.200411126. PMC 2171814. PMID 15738265.
^ Basto, R; Lau, J; Vinogradova, T; Gardiol, A; Woods, CG; Khodjakov, A; Raff, JW (2006). "Flies without centrioles". Cell. 125 (7): 1375–86. doi:10.1016/j.cell.2006.05.025. PMID 16814722. S2CID 2080684.
^ Feldman, JL; Geimer, S; Marshall, WF (2007). "The mother centriole plays an instructive role in defining cell geometry". PLOS Biology. 5 (6): e149. doi:10.1371/journal.pbio.0050149. PMC 1872036. PMID 17518519.
^ Beisson, J; Wright, M (2003). "Basal body/centriole assembly and continuity". Current Opinion in Cell Biology. 15 (1): 96–104. doi:10.1016/S0955-0674(02)00017-0. PMID 12517710.
^ Avidor-Reiss, T., Khire, A., Fishman, E. L., & Jo, K. H. (2015). Atypical centrioles during sexual reproduction. Frontiers in cell and developmental biology, 3, 21.
Chicago
^ Hewitson, Laura & Schatten, Gerald P. (2003). "The biology of fertilization in humans". In Patrizio, Pasquale; et al. (eds.). A color atlas for human assisted reproduction: laboratory and clinical insights. Lippincott Williams & Wilkins. p. 3. ISBN 978-0-7817-3769-2. Retrieved 9 November 2013.
^ Cui, Cheng; Chatterjee, Bishwanath; Francis, Deanne; Yu, Qing; SanAgustin, Jovenal T.; Francis, Richard; Tansey, Terry; Henry, Charisse; Wang, Baolin; Lemley, Bethan; Pazour, Gregory J.; Lo, Cecilia W. (2011). "Disruption of Mks1 localization to the mother centriole causes cilia defects and developmental malformations in Meckel-Gruber syndrome". Dis. Models Mech. 4 (1): 43–56. doi:10.1242/dmm.006262. PMC 3008963. PMID 21045211.
^ Babu, Deepak; Roy, Sudipto (1 May 2013). "Left–right asymmetry: cilia stir up new surprises in the node". Open Biology. 3 (5): 130052. doi:10.1098/rsob.130052. ISSN 2046-2441. PMC 3866868. PMID 23720541.
^ Tsou, MF; Stearns, T (2006). "Mechanism limiting centrosome duplication to once per cell cycle". Nature. 442 (7105): 947–51. Bibcode:2006Natur.442..947T. doi:10.1038/nature04985. PMID 16862117. S2CID 4413248.
^ Marshall, W.F. (2009). "Centriole Evolution". Current Opinion in Cell Biology. 21 (1): 14–19. doi:10.1016/j.ceb.2009.01.008. PMC 2835302. PMID 19196504.
^ Jump up to: a b Bornens, M.; Azimzadeh, J. (2007). "Origin and Evolution of the Centrosome". Eukaryotic Membranes and Cytoskeleton. Advances in Experimental Medicine and Biology. Vol. 607. pp. 119–129. doi:10.1007/978-0-387-74021-8_10. ISBN 978-0-387-74020-1. PMID 17977464.
^ Rogozin, I. B.; Basu, M. K.; Csürös, M.; Koonin, E. V. (2009). "Analysis of Rare Genomic Changes Does Not Support the Unikont-Bikont Phylogeny and Suggests Cyanobacterial Symbiosis as the Point of Primary Radiation of Eukaryotes". Genome Biology and Evolution. 1: 99–113. doi:10.1093/gbe/evp011. PMC 2817406. PMID 20333181.
^ Avidor-Reiss, Tomer; Gopalakrishnan, Jayachandran (2013). "Building a centriole". Current Opinion in Cell Biology. 25 (1): 72–7. doi:10.1016/j.ceb.2012.10.016. PMC 3578074. PMID 23199753.
^ Blachon, S; Cai, X; Roberts, K. A; Yang, K; Polyanovsky, A; Church, A; Avidor-Reiss, T (2009). "A Proximal Centriole-Like Structure is Present in Drosophila Spermatids and Can Serve as a Model to Study Centriole Duplication". Genetics. 182 (1): 133–44. doi:10.1534/genetics.109.101709. PMC 2674812. PMID 19293139.
^ Fishman, Emily L; Jo, Kyoung; Nguyen, Quynh P. H; Kong, Dong; Royfman, Rachel; Cekic, Anthony R; Khanal, Sushil; Miller, Ann L; Simerly, Calvin; Schatten, Gerald; Loncarek, Jadranka; Mennella, Vito; Avidor-Reiss, Tomer (2018). "A novel atypical sperm centriole is functional during human fertilization". Nature Communications. 9 (1): 2210. Bibcode:2018NatCo...9.2210F. doi:10.1038/s41467-018-04678-8. PMC 5992222. PMID 29880810.
^ Turner, K., N. Solanki, H.O. Salouha, and T. Avidor-Reiss. 2022. Atypical Centriolar Composition Correlates with Internal Fertilization in Fish. Cells. 11:758,
^ Khanal, S., M.R. Leung, A. Royfman, E.L. Fishman, B. Saltzman, H. Bloomfield-Gadelha, T. Zeev-Ben-Mordehai, and T. Avidor-Reiss. 2021. A dynamic basal complex modulates mammalian sperm movement. Nat Commun. 12:3808..
| v t e The centrosome and its components | |
--- |
| Centrioles | SASS6 CENPJ CNTROB CEP104 Centrins (CETN1, CETN2 and CETN3) SFI1 |
| Pericentriolar material | BBS4 Lck PCM1 PCNT TNKS TNKS2 TUBE1 |
| other proteins | CCP110 CEP55 CEP57 CEP63 CEP68 CEP70 CEP72 CEP76 CEP78 CEP90 CEP97 CEP120 CEP135 CEP152 CEP164 CEP170 CEP192 CEP215 CEP250 CEP350 CNTRL NIN NINL |
| v t e Structures of the cell and organelles | |
--- |
| Endomembrane system | Cell membrane Nucleus Endoplasmic reticulum Golgi apparatus Parenthesome Autophagosome Vesicle + Exosome + Lysosome + Endosome + Phagosome + Vacuole + Acrosome Cytoplasmic granule + Melanosome + Microbody + Glyoxysome + Peroxisome + Weibel–Palade body |
| Cytoskeleton | Microfilament Intermediate filament Microtubule Prokaryotic cytoskeleton Microtubule organizing center + Centrosome + Centriole + Basal body + Spindle pole body Myofibril Undulipodium + Cilium + Flagellum + Axoneme + Radial spoke Pseudopodium + Lamellipodium + Filopodium |
| Endosymbionts | Mitochondrion Plastid + Chloroplast + Chromoplast + Gerontoplast + Leucoplast + Amyloplast + Elaioplast + Proteinoplast + Tannosome + Apicoplast Nitroplast |
| Other internal | Nucleolus RNA + Ribosome + Spliceosome + Vault Cytoplasm + Cytosol + Inclusions Proteasome Magnetosome |
| External | Cell wall Extracellular matrix |
| Authority control databases | |
--- |
| National | United States Japan Israel |
| Other | Yale LUX |
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188165 | https://courses.lumenlearning.com/mathforliberalartscorequisite/chapter/writing-proportions/ | Appendix A: Applications
Writing Proportions
Learning Outcomes
Given a statement, write a proportion
Given an equation, determine whether it is a proportion
In the section on Ratios and Rates we saw some ways they are used in our daily lives. When two ratios or rates are equal, the equation relating them is called a proportion.
Proportion
A proportion is an equation of the form [latex]{\Large\frac{a}{b}}={\Large\frac{c}{d}}[/latex], where [latex]b\ne 0,d\ne 0[/latex].The proportion states two ratios or rates are equal. The proportion is read [latex]\text{"}a[/latex] is to [latex]b[/latex], as [latex]c[/latex] is to [latex]d\text{".}[/latex]
The equation [latex]{\Large\frac{1}{2}}={\Large\frac{4}{8}}[/latex] is a proportion because the two fractions are equal. The proportion [latex]{\Large\frac{1}{2}}={\Large\frac{4}{8}}[/latex] is read “[latex]1[/latex] is to [latex]2[/latex] as [latex]4[/latex] is to [latex]8[/latex]“.
If we compare quantities with units, we have to be sure we are comparing them in the right order. For example, in the proportion [latex]{\Large\frac{\text{20 students}}{\text{1 teacher}}}={\Large\frac{\text{60 students}}{\text{3 teachers}}}[/latex] we compare the number of students to the number of teachers. We put students in the numerators and teachers in the denominators.
example
Write each sentence as a proportion:
[latex]3[/latex] is to [latex]7[/latex] as [latex]15[/latex] is to [latex]35[/latex].2. [latex]5[/latex] hits in [latex]8[/latex] at bats is the same as [latex]30[/latex] hits in [latex]48[/latex] at-bats.3. [latex]\text{\$1.50}[/latex] for [latex]6[/latex] ounces is equivalent to [latex]\text{\$2.25}[/latex] for [latex]9[/latex] ounces.
Solution
| |
| 1. |
| [latex]3[/latex] is to [latex]7[/latex] as [latex]15[/latex] is to [latex]35[/latex]. |
| Write as a proportion. | [latex]{\Large\frac{3}{7}}={\Large\frac{15}{35}}[/latex] |
| |
| 2. |
| [latex]5[/latex] hits in [latex]8[/latex] at-bats is the same as [latex]30[/latex] hits in [latex]48[/latex] at-bats. |
| Write each fraction to compare hits to at-bats. | [latex]{\Large\frac{\text{hits}}{\text{at-bats}}}={\Large\frac{\text{hits}}{\text{at-bats}}}[/latex] |
| Write as a proportion. | [latex]{\Large\frac{5}{8}}={\Large\frac{30}{48}}[/latex] |
| |
| 3. |
| [latex]\text{\$1.50}[/latex] for [latex]6[/latex] ounces is equivalent to [latex]\text{\$2.25}[/latex] for [latex]9[/latex] ounces. |
| Write each fraction to compare dollars to ounces. | [latex]{\Large\frac{$}{\text{ounces}}}={\Large\frac{$}{\text{ounces}}}[/latex] |
| Write as a proportion. | [latex]{\Large\frac{1.50}{6}}={\Large\frac{2.25}{9}}[/latex] |
try it
Look at the proportions [latex]{\Large\frac{1}{2}}={\Large\frac{4}{8}}[/latex] and [latex]{\Large\frac{2}{3}}={\Large\frac{6}{9}}[/latex]. From our work with equivalent fractions we know these equations are true. But how do we know if an equation is a proportion with equivalent fractions if it contains fractions with larger numbers?
To determine if a proportion is true, we find the cross products of each proportion. To find the cross products, we multiply each denominator with the opposite numerator (diagonally across the equal sign). The results are called a cross products because of the cross formed. The cross products of a proportion are equal.
Cross Products of a Proportion
For any proportion of the form [latex]{\Large\frac{a}{b}}={\Large\frac{c}{d}}[/latex], where [latex]b\ne 0,d\ne 0[/latex], its cross products are equal.
Cross products can be used to test whether a proportion is true. To test whether an equation makes a proportion, we find the cross products. If they are the equal, we have a proportion.
example
Determine whether each equation is a proportion:
1. [latex]{\Large\frac{4}{9}}={\Large\frac{12}{28}}[/latex]2. [latex]{\Large\frac{17.5}{37.5}}={\Large\frac{7}{15}}[/latex]
Show Solution
SolutionTo determine if the equation is a proportion, we find the cross products. If they are equal, the equation is a proportion.
| |
| 1. |
| [latex]{\Large\frac{4}{9}}={\Large\frac{12}{28}}[/latex] |
| Find the cross products. | [latex]28\cdot 4=1129\cdot 12=108[/latex] |
Since the cross products are not equal, [latex]28\cdot 4\ne 9\cdot 12[/latex], the equation is not a proportion.
| |
| 2. |
| [latex]{\Large\frac{17.5}{37.5}}={\Large\frac{7}{15}}[/latex] |
| Find the cross products. | [latex]15\cdot 17.5=262.5[/latex] [latex]37.5\cdot 7=262.5[/latex] |
Since the cross products are equal, [latex]15\cdot 17.5=37.5\cdot 7[/latex], the equation is a proportion.
try it
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188166 | https://www.teacherspayteachers.com/browse/free?search=classroom%20store%20prices | Classroom Store Prices | TPT
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Classroom Store Price List - Great for Classroom Economy
Created by
Math With Meaning
Starting a classroom economy? This freebie contains an editable price list that you can use for your classroom store! Everything is fully editable so that you can customize the price list to your own classroom! To edit, you will need to use PowerPoint. A ready-to-print PDF version is also included if you want to use it as-is! If you want to see more details about how I use a classroom store in my own middle school math classroom, please read my blog post How to Set Up Your Classroom Store. Be s
Not Grade Specific
For All Subjects
Also included in:Classroom Economy for Middle School Resource Bundle - Classroom Jobs and More
FREE
Rated 4.75 out of 5, based on 51 reviews
4.8(51)
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Classroom Store Prices& Menu - PBIS
Created by
The Helpful Hammer
I use this Menu and wall display to remind my students how they can earn classroom dollars and how they can lose classroom dollars. They reference our classroom store menu/price list every Friday to decide how they want to spend their money! Since second grade teaches currency and economics, a classroom currency and market is a perfect way to reinforce those standards all year long. The incentives are also chosen because they are of little to no cost to the teacher.
Not Grade Specific
Economics , Other (Math)
FREE
Rated 5 out of 5, based on 3 reviews
5.0(3)
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Class Store Price Tags | Printable Classroom Economy Labels
Fourth Grade Frenzy
$1.50 Original Price $1.50
Printable Price Tags for Dramatic Play | Classroom Store, Pretend Play, Math Cen
Print A Toy
$1.50 Original Price $1.50 Rated 5 out of 5, based on 2 reviews 5.0(2)
Classroom Store | Classroom Economy | Editable Class Store Money
Sweet Firstie Fun
$3.50 Original Price $3.50 Rated 4.94 out of 5, based on 54 reviews 4.9(54)
Target inspired classroom store, prize box, treasure box
MrsCreative Teach
$4.00 Original Price $4.00 Rated 4.89 out of 5, based on 57 reviews 4.9(57)
Free Headers for Google Classroom
Created by
Kelly Benefield Teaching Fourth
Make your Google Classroom fun and adorable with these year around Google Classroom headers! It is so simple to upload these images to customize your headers and create a fun virtual classroom space for your students! The 5 headers included in this freebie are formatted and sized for a perfect fit for Google Classroom headers. They will also work great in Google Forms, and they will work in Google Sites as the main header, although a small part of the images may be cut off. Kindly note: These h
Not Grade Specific
For All Subjects
Also included in:Google Classroom™ Headers Bundle
FREE
Rated 4.77 out of 5, based on 26 reviews
4.8(26)
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Superhero Classroom Décor | Editable Binder Covers
Created by
Oh So Random
Superhero Classroom Décor - Printable editable binder cover with a superhero theme. If you like this file, please take a few minutes and leave feedback. Part I BUNDLE can be found HERE! Birthday Hats Calendar Headers Editable Binder Cover Editable Classroom Passes Editable Labels Editable Name Plates Part II BUNDLE can be found HERE! Editable Super News Editable Word Wall Labels Note Bracelets Super Reader Charts Table Signs Welcome Banner PLEASE REMEMBER YOU ARE PURCHASING THIS ITEM FO
PreK - 5 th
For All Subjects
FREE
Rated 4.9 out of 5, based on 10 reviews
4.9(10)
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Superhero Classroom Décor | Editable Binder Cover
Created by
Oh So Random
Superhero Classroom Décor - Printable editable binder cover with a superhero theme. To edit text fields, Ctrl E (PC) and Command E (Mac) will open the text properties toolbar. The COMPLETE Superhero Classroom Décor set can be purchased in two BUNDLES. Part I BUNDLE can be found HERE! Birthday Hats Calendar Headers Editable Binder Cover Editable Classroom Passes Editable Labels Editable Name Plates Part II BUNDLE can be found HERE! Editable Super News Editable Word Wall Labels Note Brac
PreK - 3 rd
For All Subjects
FREE
Rated 4.82 out of 5, based on 10 reviews
4.8(10)
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Free Money Clip Art Set: Coin, Credit Card Reader, Savings Jar, Price Tag, Store
Created by
Dancing Doodles
Grab this free Money Clip Art Pack perfect for creating engaging math, financial literacy, and classroom projects! This collection includes 5 high-quality PNG images: a shiny coin, a credit card reader, a savings jar, a price tag, and a storefront — all designed in a simple, colorful, cartoon style that works well for worksheets, flashcards, presentations, and more. Whether you’re teaching money counting, saving, shopping, or economics, these versatile graphics make your resources visually a
Not Grade Specific
Money Math
FREE
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Sticker Store Template for Classroom Management
Created by
Leahyslittlelearners
Product Description: Transform your classroom management with this easy-to-use Sticker Store Template! This template is perfect for teachers who use stickers as student incentives or classroom currency. Whether you’re building a full classroom economy or just rewarding positive behavior, this store setup makes it fun, organized, and engaging for students. ✨ What’s Included: Sticker Store display templatePre-made price tags for different sticker typesCustomizable blank price cardsOptional stude
Not Grade Specific
Not Subject Specific
FREE
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Discounted French or Classroom Management Customized Bundle
Created by
Allo Mes Amis
YOU SAVE MONEY! - an additional 15%! (This additional 15% is taken from the ORIGINAL listing price of each product. It cannot be combined with any store sales and/or other bundle pricing.) The bundle MUST include at least 5 resources. EMAIL: amanda@allomesamis.com with the products you would like. Subject Line: TPT CUSTOMIZED BUNDLE ✍DON'T FORGET to leave a review on your purchase for TPT credits to be used towards your future purchases! :)©allomesamis
6 th - 12 th
French
FREE
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Classroom Economy FREEBIE
Created by
Chick on the Run
Are you looking to create real world experiences with your kids throughout the year? This product is great to jumpstart a classroom economy! This pack includes: ☆ program overview ☆ standards ☆ vocabulary words ☆ classroom job description ☆ classroom job cards (color and black & white) ☆ reward coupons (color and black & white) ☆ store day price tents (color and black & white) SAVE MONEY AND BUY the Classroom Economy BUNDLE!The bundle includes over 200 pages of tools to implement and ma
2 nd - 5 th
Economics , Math
FREE
Rated 4.8 out of 5, based on 5 reviews
4.8(5)
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Grocery Store Free Budgeting Worksheets Special Education Life Skills
Created by
Miss Lulu
Practice essential money and budgeting skills with this free sample segment from our comprehensive Grocery Store Budgeting Unit. It offers three differentiated levels with worksheets, task cards, and classroom simulation materials. Students learn to locate items, calculate prices, assess affordability, and make purchases, all focused on amounts up to $10. Kristi says, "I absolutely love these! I am in an 18 plus program at our high school and the students are here just for vocational and life
Not Grade Specific
Math
Also included in:Budgeting Units & Worksheets Bundle for Special Education
FREE
Rated 4.65 out of 5, based on 54 reviews
4.7(54)
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Price / Gift TAGS Clip art {Sketch}
Created by
Teacher's Clipart
The Price / Gift TAGS Clip art sketch} Clipart Set has 10 images to represent gift or price tags. PLEASE READ MY TERMS OF USE: This set and/ or graphics are NOT to be sold AS IS! That means, you are not able to sell or give them away as graphics files as I have done. Purchasing the item does not transfer the copyright to you. It means you cannot claim the design as your own design. This license is non-transferable. This means you cannot transfer/ give the license to anyone else, nor can you ch
Not Grade Specific
For All Subjects
FREE
Rated 5 out of 5, based on 13 reviews
5.0(13)
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Pineapple Classroom Decor: Classroom Posters
Created by
The Seeds We Sow
Ready for a quick and easy way to brighten your pineapple-lovin' classroom? Well, here you go! This fresh, eye-catching, bright, and fun poster set will do just that, and best of all, it's free! Included in this set are 8 coordinated pineapple-themed posters ready to print on 8.5 x 11 paper. This poster set is no-prep and was created to bring a cheerful and bright feel to your classroom. It's also my way of saying thank you to you for either following my store, for being a loyal customer, or ju
Not Grade Specific
Also included in:Pineapple Classroom Decor: The Bundle
FREE
Rated 4.95 out of 5, based on 77 reviews
5.0(77)
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Sports Store | Dramatic Play
Created by
Learn and Play Jess
This Sports Store will add to your dramatic play whether you are looking to transform or add to it. It is a great addition to use for the Balls Study. Your students will be eager to engage in the fun. What's Included:Sports Store BannerReal Picture Labels for ShelvesSports Balls Prices List3 Distinct Create Your Own Jersey TemplateCreate Your Own Sports Cap TemplateInspiration PicturesTips for Use:Laminate before useLaminate/Place template inside a pocket sleeveUse templates with washable marker
PreK - 2 nd
Classroom Community, Drama, Social Emotional Learning
FREE
Rated 4.4 out of 5, based on 5 reviews
4.4(5)
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PowerPoint: Turkey Rhythms for the Kodaly or Orff Music Classroom
Created by
Malinda Phillips
Thanks for your download! All items in this store are moving to free resources as of 11/30/2023. I am no longer monitoring this store. If you have any issues, please direct them to TpT's Help Section. Newly Updated 11/09/14 Includes original file and updated version for the same price! This PowerPoint presentation will allow the students to actively review the rhythms of your choice using quarter note and rest, half note and rest, and eighth notes using traditional or stem notation. Student
K - 3 rd
Music Composition, Vocal Music
FREE
Rated 5 out of 5, based on 20 reviews
5.0(20)
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Winter Wonderland Classroom Decor Bunting: Printable PDF ❄️FREEBIE❄️
Created by
Early Bird English Academy
Nice! You've just found some lovely Winter Wonderland Decor to bring the cozy, snowy vibes right into your classroom. If you use this cute bunting, please feel free to take pics and tag @earlybirdenglish.academy on Instagram ❄️☃️ This resource includes 1x PRINTABLE PDF FREEBIE: "Winter Wonderland" BuntingCheck out my other Class Decorations and Seasonal Activities!✔ Halloween Bundle (Bunting, Worksheet, Name Tags) ✔ Christmas Bunting ▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼ Let's Connect!► F
Not Grade Specific
FREE
Rated 4.5 out of 5, based on 4 reviews
4.5(4)
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Classroom Jobs - A classroom economy starter kit
Created by
The Real Teachers of OC
Put your students to work, teach responsibility, promote a sense of community within your classroom. This classroom economy freebie is a great way to start a system in your classroom where your students get paid pretend money for their jobs. Visit our blog at www.therealteachersoforangecounty.blogspot.com to see our economy system in action! Here's what you will find inside: - Description of jobs and responsibilities - checkbook (you will have to copy, cut, and assemble) - price tags to use
2 nd - 6 th
Economics
FREE
Rated 4.81 out of 5, based on 34 reviews
4.8(34)
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El Precio es Correcto (The Price is Right)
Created by
joy quenga
This is a Power Point activity that is loosely based on the gameshow The Price is Right. There are 26 slides and each slide will show an actual clothing item and the store where the clothing item is sold. A click of the mouse will reveal the actual price. Here’s how I play in my classroom, but of course, you can modify to use the activity how it would best work in your classroom. 1) Divide class into groups 2) Choose a group leader 3) Show the first clothing item and ask “¿Cuánto cuesta? 4) Gro
5 th - 12 th
Spanish
FREE
Rated 4.95 out of 5, based on 21 reviews
5.0(21)
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In This Classroom Poster- FREE!
Created by
Amanda Price
Build community in your classroom with this FREE chalkboard poster. This file contains a 300 dpi high resolution JPEG picture for you to print the size you want and display in your classroom. Print it at home, or take it to your local printing store for maximum results. Have a wonderful school year! Some fonts used with Commercial License from Hello Literacy
PreK - 12 th, Adult Education
Character Education
FREE
Rated 4.92 out of 5, based on 63 reviews
4.9(63)
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Price Research Pro: Everyday Life Skills Task Cards for High School SpED
Created by
Growing Minds Inclusively
Introducing "Price Research Pro: Everyday Life Skills Task Cards" — a dynamic resource specially crafted for high school special education life skills classrooms. Dive into the world of practical learning with these engaging task cards designed to empower students in researching prices and mastering fundamental life skills. Key Features:Price Research Prowess: Develop essential skills by actively researching prices for everyday items. Students will learn to navigate stores online websites, compa
9 th - 12 th
FREE
Rated 5 out of 5, based on 1 reviews
5.0(1)
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Grocery Prices Worksheets
Created by
super cool nerd mama
Do your students go to the grocery store (the real one or a play one in your classroom)? If so you need this printable! Have students write down the price of common items at the grocery, circle if the price is per unit or per weight, and determine if the item is on sale. This activity will help you to teach your students the following:Price awarenessDifferentiating between prices “per unit” and prices “by weight”Taking note of which items are on saleIdentifying specific grocery itemsWriting nu
Not Grade Specific
Math
FREE
Rated 4.75 out of 5, based on 4 reviews
4.8(4)
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Thrift Store | Dramatic Play
Created by
Learn and Play Jess
This Thrift Store will add to your dramatic play whether you are looking to transform or add to it. It is a great addition to use for the Reuse, Reduce, Recycle Study. Your students will be eager to engage in the fun. What's Included:Thrift Store BannerOpen/Closed SignHours of Operation SignReal Picture Labels for Shelves/BinsPrices ListTips for Use:Laminate before useLaminate/Place template inside a pocket sleeveUse templates with washable markers
PreK - 2 nd
Classroom Community, Drama, Social Emotional Learning
FREE
Rated 5 out of 5, based on 1 reviews
5.0(1)
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Field Trip to a Grocery Store and Bank
Created by
Suzanne's Classroom
If you study family finances and healthy foods with your class, have you ever considered taking a field trip to a local grocery store and bank? Both need to be prearranged with the bank and grocery store. The bank is usually happy to plan an hour or two with your students. My bank breaks the kids up into groups and different department personnel discuss that aspect of the bank’s service. Very interesting! For the grocery store visit, you usually only need to tell the manager that you will be
4 th - 6 th
Basic Operations, Business, Family Consumer Sciences
FREE
Rated 4.67 out of 5, based on 3 reviews
4.7(3)
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WEEKLY FREEBIE #50: Fruit Stand Shopping
Created by
Dwayne Kohn
Set up a classroom Fruit Stand and the kids won’t even realize that they are learning! Put prices on the fruits and vegetables to practice math abilities and make shopping lists to improve reading skills! Included are the OPEN/CLOSED sign, money (one and five dollar bills), fruit and vegetable cards, and labels. Print out the pages in black and white and have each student make their own store, or print out the pages in full color for an instant classroom learning center! Interested in mor
PreK - 2 nd
Math, Reading
FREE
Rated 5 out of 5, based on 8 reviews
5.0(8)
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FREE Math Vocabulary Words - Primary (grade 1-3) Financial Literacy
Created by
2 Helpful Chicks
Welcome to our TPT store 2HelpfulChicks. In this product we hope that you enjoy how easy it is for you to display the Mathematics vocabulary words that you are using while teaching your Math Strands. This Set of Primary (grade 1-3) Financial Literacy contains 42 words. 28 different words + 14 with American money and spelling. Each of these different math words can be printed and laminated, then displayed on the math wall of your classroom. In this package the Vocabulary Words included are
PreK - 3 rd
Math, Other (Math)
Also included in:Math Vocabulary Words - Primary Grade 1-3 Math Word Wall ALL Strands
FREE
Rated 4.5 out of 5, based on 4 reviews
4.5(4)
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188167 | https://www.youtube.com/watch?v=MGR-_E50Aa0 | Tenemos que hablar sobre los heterogénericos
Español claro y fácil
22400 subscribers
13 likes
Description
402 views
Posted: 5 Apr 2017
Los heterogénericos se refieren a los sustantivos que cambian de género de un idioma a otro, es decir, poseen un género en portugués y otro en español.
Transcript:
bueno chicos otra semana estamos más aquí para intentar los o intentar explicarlos alguno de los temas que siempre resultan de más dificultad muchas gracias también a todos los que nos mandan sus comentarios los que nos hacéis sugerencias de vídeos porque intentamos siempre atender a todos esos pedidos poco a poco vamos a colgar todos los medios que nos pedís vale bueno pues hoy queremos hablar sobre un concepto que tiene una palabra un nombre un poquito raro que son los hetero genérico pero que no es ni más ni menos que aquellos sustantivos que cambian de género cuando están en portugués o cuando están en español o sea sustantivos que van a ser femeninos en el español pero que van a ser este mismo sustantivo será masculino en el portugués o al contrario un sustantivo que es masculino el portugués va a ser femenino en el español y para ello hemos elaborado dos listas con las principales palabras con las que más hay son las que más confusión presentan en la primera lista nos vamos a centrar en aquellas sustantivos aquellas palabras que son masculinas en español y terminan en portugués y en la segunda al contrario empezamos con la primera venga bueno vamos decía vamos a empezar con aquellos sustantivos que en español son siempre masculinos y que en portugués se van a convertir en femenino y ahí tenéis varios ejemplos el árbol que en portugués sería árbol por lo tanto el árbol es masculino árbol es femenina más ejemplos que son todo masculino en español y que se vuelven femeninos en el portugués el color el contestador automático el cuchillo el cutis el desorden el dolor el equipo de fútbol o el equipo de trabajo el estreno de la película el lavaplatos sobre lavavajillas el lavarropas la máquina de lavar la ropa los días de la semana lunes martes miércoles jueves y viernes masculino en español femeninos en portugués el manzano y otros árboles frutales normalmente los árboles frutales son siempre masculino en español y van a ser femeninos en el portugués el mensaje y otras palabras acabadas en aje todas las palabras que acaban en a g son masculinas sin excepción en el portugués femeninas a mensagem el puente el vals el vértigo y de nuevo otra cadena g el viaje vale esto es muy muy importante que lo tengáis en cuenta llevamos ahora justo con lo contrario aquellas palabras que en español van a ser femenino sustantivos en femenino que en el portugués van a ser masculinas o entonces en el masculino las letras a la vela ce son femeninas en español masculinas en portugués la aspiradora del polvo femenino en español masculino portugués o aspirador dup o la computadora que sería un computador la costumbre y todas las demás acabadas en hombre toda la clava que acaba ningún hombre en español son femeninas mucho cuidado con eso la estufa la leche que está siempre os equivocáis la leche femenina en portugués es un late la licuadora la miel también es femenina en español la multiprocesadora la nariz la protesta la radio cuando es el aparato se dice en femenino radio cuando es masculino en español es el hueso la rodilla que suyo ello la sal la sangre la sonrisa y la tiza que es gis masculino en portugués vale chicos bueno pues éstas son las principales palabras no son todas pero son algunas palabras que ya os van a ayudar a empezar a investigar más palabras por vuestra cuenta y por supuesto a que nos preguntáis todas las dudas que tengáis sobre este tema el principio de nuestro blog he venido diciendo todas las semanas que si nos queréis compartir compartáis como diría la gran lola flores y aquí ya tenéis a la genia bueno pues nada con la gran lola flores y con estas palabras que cambian de género en un idioma o en otro os dejamos hasta la próxima semana muchísimas gracias por todo por vuestra atención compartir compartir comentar chao |
188168 | https://www.spectral-ai.com/blog/key-elements-of-burns-assessment-and-evaluation/ | Key Elements of Burns Assessment and Evaluation | Spectral AI
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Home / Blog / Key Elements of Burns Assessment and Evaluation
Burn Wound Healing
Key Elements of Burns Assessment and Evaluation
September 6, 2024
Burns Assessment
Key Elements of Burns Assessment
Evaluation of Burn Depth
Estimating Total Body Surface Area (TBSA)
Assessing Burn Severity
Deepview
Initial Management of Burn Injuries
Fluid Resuscitation of the Injury
Pain Management
Wound Care and Infection Prevention
Long-Term Management and Rehabilitation
Physical Rehabilitation
Psychological Support
Burns Assessment
Assessing and evaluating burns effectively requires a systematic approach to ensure optimal patient outcomes. Clinicians should start by identifying the burn type: thermal, chemical, electrical, or radiation. Next, determine the depth of the injury, classifying it as superficial, partial-thickness, or full-thickness. An accurate survey of the burn extent is critical, this is typically assessed using the “Rule of Nines” or the Lund and Browder chart.
Consider the following key elements:
Burn Depth: Superficial burns affect only the epidermis. Partial-thickness injuries extend into the dermis, while full-thickness burns damage all skin layers and potentially underlying tissues.
Total Body Surface Area (TBSA): Calculate TBSA to estimate fluid resuscitation needs and potential systemic complications.
Location of Burns: Injuries on the face, hands, feet, or perineum require special attention due to functional and aesthetic considerations.
Patient History: Gather comprehensive information on pre-existing conditions, medications, and allergies that may affect healing or treatment options.
Vital Signs and Symptoms: Monitor for signs of shock, respiratory distress, and other systemic responses to severe burns.
Inhalation Injury:Assessing for inhalation injuries in a burn patient involves a thorough clinical evaluation of the airways, signs of respiratory distress, a breathing assessment to indicate upper airway injury and a detailed examination of the face and neck for burns.
Use advanced diagnostic tools such as laser Doppler imaging or ultrasound to assess burn depth more accurately.
Clinicians should re-evaluate burns regularly to monitor healing progress and adjust treatment plans as needed. Early identification of complications, such as infection or compartment syndrome, is vital. Efficient documentation and communication within the care team ensure continuity and quality of care. Employing these methods enhances treatment precision and patient recovery.
Key Elements of Burns Assessment
Accurately assessing burns is vital for effective treatment and optimal recovery. Clinicians must evaluate key elements, such as burn depth, total body surface area (TBSA), and injury location. Burn depth determines the appropriate intervention, with superficial, partial-thickness, and full-thickness injuries requiring different care. Calculating TBSA is essential for fluid resuscitation and predicting complications. Additionally, special attention to burns on the face, hands, feet, and perineum is crucial due to their functional and aesthetic significance.
Evaluation of Burn Depth
Accurate evaluation of burn depth is crucial for effective treatment and optimal recovery. Clinicians utilize several methods to determine the depth of burns, including clinical examination, imaging techniques, and advanced diagnostic tools.
Clinical Examination: Assess characteristics such as color, sensation, and capillary refill. Superficial burns appear red and painful, partial-thickness burns are blistered and moist, while full-thickness burns are leathery and insensate.
Imaging Techniques: Use thermography, ultrasound, or MRI to visualize tissue damage extent.
Advanced Diagnostic Tools: Implement laser Doppler imaging to measure skin perfusion and predict burn depth accurately.
Accurate evaluation of burn depth guides appropriate treatment strategies. Superficial burns often require topical treatments and dressings, whereas partial and full-thickness burns may necessitate surgical intervention, such as debridement or grafting.
Regular reassessment is essential, as burns can evolve, revealing deeper injuries not initially apparent. Early and precise evaluation reduces complications, such as infections and scarring, and enhances healing outcomes.
Furthermore, accurate assessment supports effective communication within the healthcare team, ensuring all members are aligned in their treatment approach. By leveraging these methods, clinicians can provide targeted care, improving recovery rates and overall patient outcomes in burn injuries.
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Estimating Total Body Surface Area (TBSA)
Estimating Total Body Surface Area (TBSA) accurately is crucial for managing burns effectively. Clinicians use various techniques to calculate TBSA, such as the Rule of Nines, the Lund and Browder chart, and the palm method.
Rule of Nines: This method divides the body into sections, each representing 9% (or multiples thereof) of the TBSA, facilitating quick estimates for adults.
Lund and Browder Chart: More precise, this chart adjusts for age, providing detailed assessments by considering different body proportions in children and adults.
Palm Method: Uses the patient’s palm size (approximately 1% of TBSA) to estimate small or scattered burn areas.
Accurate TBSA calculation plays a pivotal role in fluid management. Initiate fluid resuscitation using formulas like the Parkland formula, which bases fluid requirements on TBSA and patient weight. Proper fluid management is vital in preventing shock, maintaining organ perfusion, and stabilizing burn injuries.
Moreover, estimating TBSA guides decisions on burn center referrals, treatment strategies, and potential surgical interventions. Continuous reassessment is necessary, as initial estimates may change with evolving injuries.
In summary, accurate TBSA estimation ensures effective fluid resuscitation and comprehensive patient care. By employing these techniques, clinicians can optimize treatment protocols, improve outcomes, and reduce complications associated with severe burns.
Assessing Burn Severity
Assessing burn severity is essential for guiding treatment and optimizing patient outcomes. Clinicians employ several techniques to evaluate burns, focusing on calculating the Total Body Surface Area (TBSA) and determining burn depth. These assessments directly influence fluid management and overall patient care.
Role in Fluid Management: Accurate TBSA calculation is critical for effective fluid resuscitation. Utilize formulas like the Parkland formula to determine fluid requirements based on TBSA and patient weight. Proper fluid management prevents hypovolemic shock, maintains organ perfusion, and stabilizes burn injuries.
Patient Care Considerations:
Burn Depth: Assess the extent of tissue damage to determine appropriate interventions, such as debridement or grafting.
Location and Type of Burn: Pay special attention to burns on sensitive areas like the face, hands, and joints, which may require specialized treatment.
Continuous Reassessment: Regularly re-evaluate burns to monitor healing and adjust treatment plans as needed. Early and precise assessment reduces complications, such as infections and scarring, enhancing recovery outcomes.
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Initial Management of Burn Injuries
Fluid Resuscitation of the Injury
Fluid resuscitation is critical in managing severe burn injuries and preventing complications like hypovolemic shock and organ failure. Effective fluid management ensures adequate perfusion of vital organs, stabilizes hemodynamics, and supports the healing process. Clinicians must follow established protocols and guidelines to optimize outcomes.
Importance of Fluid Management:
Prevents Hypovolemic Shock: Large burns cause significant fluid loss, leading to decreased blood volume. Timely fluid replacement is essential to prevent shock.
Maintains Organ Perfusion: Adequate fluid resuscitation ensures that organs receive sufficient blood flow, preventing renal and multi-organ failure.
Supports Wound Healing: Proper hydration facilitates tissue repair and reduces complications such as infections.
Protocols and Guidelines:
Parkland Formula: Commonly used to estimate fluid requirements in the first 24 hours post-injury. Calculate 4 mL of lactated Ringer’s solution per kilogram of body weight per percentage of TBSA burned. Administer half of the total volume in the first 8 hours and the remainder over the next 16 hours.
Modified Brooke Formula: Suggests 2 mL of fluid per kilogram per percentage of TBSA burned, offering an alternative approach.
Monitoring and Adjustment: Regularly monitor urine output, vital signs, and hemodynamic parameters to adjust fluid administration as needed.
Continuous Assessment: Reassess fluid needs frequently to account for changes in patient condition and response to treatment. Adjust protocols based on individual patient factors and clinical judgment.
Pain Management
Pain management is a critical component of burn assessment and evaluation, significantly impacting patient recovery and quality of life. Clinicians must adopt a comprehensive approach that includes both pharmacological and non-pharmacological methods to effectively address pain associated with burn injuries.
Pharmacological Approaches:
Analgesics: Administer acetaminophen and NSAIDs for mild to moderate pain. Use opioids for severe pain, carefully monitoring for side effects and dependency.
Anxiolytics and Sedatives: Consider benzodiazepines to manage anxiety and distress, often exacerbating the perception of pain in burn patients.
Local Anesthetics: Apply topical anesthetics such as lidocaine for localized pain relief during wound care procedures.
Non-Pharmacological Methods:
Cognitive Behavioral Therapy (CBT): Implement CBT techniques to help patients cope with pain and anxiety, promoting psychological resilience.
Physical Therapy: Engage patients in gentle exercises and range-of-motion activities to prevent stiffness and promote healing.
Relaxation Techniques: Encourage practices like deep breathing, meditation, and guided imagery to reduce stress and enhance pain tolerance.
Distraction Techniques: Use music therapy, virtual reality, or other distraction methods to shift patient focus away from pain during treatments.
Integrated Pain Management: Combining pharmacological and non-pharmacological strategies ensures a holistic approach to pain management. Regularly assess pain levels using standardized pain scales and adjust treatment plans accordingly. Effective pain management not only enhances patient comfort but also facilitates participation in rehabilitation and improves overall recovery outcomes.
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Wound Care and Infection Prevention
Effective wound care and infection prevention are crucial aspects of burn assessment and evaluation. Clinicians must employ comprehensive techniques to promote healing and minimize complications.
Techniques for Effective Wound Care:
Debridement: Remove necrotic tissue to reduce infection risk and facilitate healing. Use surgical, enzymatic, or autolytic methods based on wound characteristics.
Dressing Selection: Choose appropriate dressings, such as hydrocolloids, foams, or antimicrobial dressings, to maintain a moist environment and protect the wound.
Moisture Management: Ensure wounds remain moist but not overly wet. Use dressings that balance moisture to prevent maceration.
Skin Grafting: Consider autografts or allografts for extensive burns to promote closure and regeneration.
Strategies to Prevent Infection in Burn Injuries:
Aseptic Technique: Strictly adhere to aseptic techniques during wound care procedures to prevent contamination.
Antimicrobial Agents: Apply topical antimicrobial agents like silver sulfadiazine or mafenide acetate to reduce bacterial load.
Systemic Antibiotics: Administer systemic antibiotics based on culture results if signs of systemic infection are present.
Regular Monitoring: Frequently assess wounds for signs of infection, such as increased redness, swelling, or purulent discharge.
Barrier Precautions: Use barrier precautions, including gloves, gowns, and masks, to prevent cross-contamination.
Integrated Care Approach: Combining meticulous wound care with rigorous infection prevention strategies enhances patient outcomes. Monitor and document wound progress regularly, adjusting treatments as needed to address evolving conditions. Educate patients and caregivers on proper wound care techniques and signs of infection to ensure continuity of care at home.
Long-Term Management and Rehabilitation
Physical Rehabilitation
Physical rehabilitation is vital in the long-term management of burn injuries, focusing on restoring function and mobility. Early mobilization is crucial to prevent contractures and improve circulation. Implement range-of-motion exercises and stretching to maintain flexibility and joint function. Strengthening exercises enhance muscle recovery and support overall mobility. Incorporate occupational therapy to help patients regain daily living skills. Additionally, scar management techniques, such as pressure garments and massage, reduce scarring and improve movement. Regularly assess progress and adjust therapy plans to address evolving needs, ensuring comprehensive recovery and optimal outcomes for burn injury patients.
Psychological Support
Psychological support is essential in the long-term management and rehabilitation of burn patients. Early intervention addresses the trauma associated with severe injuries, helping patients cope and recover. Techniques include cognitive-behavioral therapy (CBT) to manage anxiety and depression, and support groups to foster community and shared experiences. Integrate mental health evaluations into routine burn evaluationsto identify psychological needs promptly. Additionally, providing education on the healing process empowers patients, reducing stress and promoting health. Regular follow-ups ensure continuous support, aiding in the holistic recovery of burns patients and enhancing overall well-being in the affected area.
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Key Elements of Burns Assessment
Evaluation of Burn Depth
Estimating Total Body Surface Area (TBSA)
Assessing Burn Severity
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Initial Management of Burn Injuries
Fluid Resuscitation of the Injury
Pain Management
Wound Care and Infection Prevention
Long-Term Management and Rehabilitation
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188169 | https://www.etymonline.com/word/alleviate | Advertisement
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Origin and history of alleviate
alleviate(v.)
early 15c., alleviaten, "to mitigate, relieve (sorrows, suffering, etc.)," from Late Latin alleviatus, past participle of alleviare "lift up, raise," figuratively "to lighten (a burden), comfort, console," from assimilated form of Latin ad "to" (see ad-) + levis "light" in weight (from PIE root legwh- "not heavy, having little weight"). Related: Alleviated; alleviating.
also from early 15c.
Entries linking to alleviate
ad-
word-forming element expressing direction toward or in addition to, from Latin ad "to, toward" in space or time; "with regard to, in relation to," as a prefix, sometimes merely emphatic, from PIE root ad- "to, near, at."
Simplified to a- before sc-, sp- and st-; modified to ac- before many consonants and then re-spelled af-, ag-, al-, etc., in conformity with the following consonant (as in affection, aggression). Also compare ap- (1).
In Old French, reduced to a- in all cases (an evolution already underway in Merovingian Latin), but French refashioned its written forms on the Latin model in 14c., and English did likewise 15c. in words it had picked up from Old French. In many cases pronunciation followed the shift.
Over-correction at the end of the Middle Ages in French and then English "restored" the -d- or a doubled consonant to some words that never had it (accursed, afford). The process went further in England than in France (where the vernacular sometimes resisted the pedantic), resulting in English adjourn, advance, address, advertisement (Modern French ajourner, avancer, adresser, avertissement). In modern word-formation sometimes ad- and ab- are regarded as opposites, but this was not in classical Latin.
legwh-
Proto-Indo-European root meaning "not heavy, having little weight."
It might form all or part of: alleviate; alleviation; alto-rilievo; carnival; elevate; elevation; elevator; leaven; legerdemain; leprechaun; Levant; levator; levee; lever; levity; levy (v.) "to raise or collect;" light (adj.1) "not heavy, having little weight;" lighter (n.1) "type of barge used in unloading;" lung; relevance; relevant; releve; relief; relieve.
It might also be the source of: Sanskrit laghuh "quick, small;" Greek elakhys "small," elaphros "light;" Latin levare "to raise," levis "light in weight, not heavy;" Old Church Slavonic liguku, Russian lëgkij, Polish lekki, Lithuanian lengvas "light in weight;" Old Irish lu "small," laigiu "smaller, worse;" Gothic leihts, Old English leoht "not heavy, light in weight."
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Trends of alleviate
adapted from books.google.com/ngrams/ with a 7-year moving average; ngrams are probably unreliable.
More to explore
ease
c. 1200, "physical comfort, undisturbed state of the body; tranquility, peace of mind," from Old French aise "comfort, pleasure, well-being; opportunity," which is of uncertain origin. According to Watkins is ultimately from Latin adiacens "lying at," present participle of adiace
relieve
late 14c., releven, "alleviate (pain, etc.) wholly or partly, mitigate; afford comfort; allow respite; diminish the pressure of," also "give alms to, provide for;" also figuratively, "take heart, cheer up;" from Old French relever "to raise, relieve" (11c.) and directly from Lati
palliate
early 15c., "alleviate (a disease or its symptoms) without curing," from Medieval Latin palliatus, literally "cloaked," from past participle of Late Latin palliare "cover with a cloak, conceal," from Latin pallium "cloak" (see pall (n.)). Meaning "excuse or extenuate (an offense)
console
"alleviate the grief or mental distress of," 1690s, from French consoler "to comfort, console," from Latin consolari "offer...
assuage
figuratively, of pain, anger, passion, grief, etc., c. 1300, from Anglo-French assuager, Old French assoagier "soften, moderate, alleviate...
allay
confused in Middle English with various senses of Romanic-derived alloy (v.) and especially a now-obsolete verb allege "to alleviate...[OED] Hence the senses "lighten, alleviate; mix, temper, weaken."...
elevation
elevatio) "a lifting up," noun of action from past-participle stem of elevare "lift up, raise," figuratively, "to lighten, alleviate...
soothe
Middle English sothen, from Old English soðian "show to be true, bear witness, offer confirmation" (senses now obsolete), from soð "true" (see sooth). The sense of "quiet, comfort, restore to tranquility," in reference to a person or animal, is by 1690s, via the notion of "to ass
aid
early 15c., aide, "war-time tax," also "help, support, assistance," from Old French aide, earlier aiudha "aid, help, assistance," from Late Latin adiuta, noun use of fem. of adiutus, past participle of Latin adiuvare "to give help to," from ad "to" (see ad-) + iuvare "to help, gi
meliorate
1550s, "to make better, improve" (transitive), a back-formation from melioration or else from Late Latin melioratus, past participle of meliorare "improve," from Latin melior "better," used as comparative of bonus "good," but probably originally meaning "stronger," from PIE root
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188170 | https://www.betterhealth.vic.gov.au/health/healthyliving/placental-abruption | Conditions
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Placental abruption
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Summary
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Placental abruption means the placenta has detached (come away) from the wall of the uterus, either partly or totally. This can cause bleeding in the mother. It may also interfere with the unborn baby’s supply of oxygen and nutrients, which the placenta provides from the mother’s bloodstream through the lining of the uterus.
Doctors cannot reattach the placenta. Without prompt medical treatment, a severe case of placental abruption can have dire consequences for the mother and her unborn child, including death.
Worldwide, placental abruption occurs in about one pregnancy in every 100. In about half of cases, placental abruption is mild and can be managed by ongoing close monitoring of the mother and baby. About 25 per cent of cases are moderate, while the remaining 25 per cent threaten the life of both baby and mother.
Symptoms
Some of the symptoms and signs of moderate to severe placental abruption include:
In some cases, bleeding may occur but the blood may clot between the placenta and the wall of the uterus, so vaginal bleeding may be scanty or even non-existent. This is known as a ‘retroplacental clot’.
The cause is unknown in most cases
In most cases, doctors don’t know the exact cause or causes of placental abruption. It is thought that an abnormal blood supply in the uterus or placenta may play a role, but the cause of the suspected abnormality isn’t clear.
Some of the known causes of placental abruption include:
Risk factors
While the exact cause in most cases is unknown, certain factors make a pregnancy more susceptible to placental abruption. Risk factors may include:
Complications
Complications in severe cases can include:
Diagnosis
The symptoms and signs of placental abruption can mimic those of other pregnancy conditions, such as placenta praevia and preeclampsia. Information that may be used to diagnose placental abruption includes:
Sometimes, the diagnosis of placental abruption can’t be confirmed until childbirth, when the placenta is delivered with an attached blood clot that appears old rather than fresh. The placenta is usually sent to a laboratory for further diagnostic testing.
Treatment
All cases of suspected placental abruption, regardless of severity, should be closely monitored to protect the health and safety of the mother and child. This monitoring is usually done in hospital and should include regular checks of the vital signs of both mother and baby. Treatment depends on the severity of the condition but may include:
Prevention
While it is impossible to prevent placental abruption, the risk can be reduced. Suggestions include:
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Content on this website is provided for information purposes only. Information about a therapy, service, product or treatment does not in any way endorse or support such therapy, service, product or treatment and is not intended to replace advice from your doctor or other registered health professional. The information and materials contained on this website are not intended to constitute a comprehensive guide concerning all aspects of the therapy, product or treatment described on the website. All users are urged to always seek advice from a registered health care professional for diagnosis and answers to their medical questions and to ascertain whether the particular therapy, service, product or treatment described on the website is suitable in their circumstances. The State of Victoria and the Department of Health shall not bear any liability for reliance by any user on the materials contained on this website.
Content on this website is provided for information purposes only. Information about a therapy, service, product or treatment does not in any way endorse or support such therapy, service, product or treatment and is not intended to replace advice from your doctor or other registered health professional. The information and materials contained on this website are not intended to constitute a comprehensive guide concerning all aspects of the therapy, product or treatment described on the website. All users are urged to always seek advice from a registered health care professional for diagnosis and answers to their medical questions and to ascertain whether the particular therapy, service, product or treatment described on the website is suitable in their circumstances. The State of Victoria and the Department of Health shall not bear any liability for reliance by any user on the materials contained on this website.
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188171 | https://www.mathdoubts.com/sin-double-angle-identity-proof/ | Proof of sin2x identity | sin2A formula | sin2θ Rule
[x]
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Proof of Sine double angle identity
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Sine
The sine of double angle identity is a trigonometric identity and used as a formula. It is usually written in the following three popular forms for expanding sine double angle functions in terms of sine and cosine of angles.
(1).sin(2 θ)=2 sinθ cosθ
(2).sin(2 A)=2 sinA cosA
(3).sin(2 x)=2 sinx cosx
The sine double angle rule can be proved in mathematical form geometrically from a right triangle (or right angled triangle). It is your turn to learn the geometric proof of sin double angle formula.
Construction
Let’s consider a right angled triangle. Here, it is Δ E C D. We have to do some geometric settings inside this right triangle by following below steps.
Bisect the angle D C E by drawing a straight line from point C to side D E― and it intersects the side D E― at point F. Each angle is represented by a variable x.
Draw a line to side C E― from point F but it should be perpendicular to side C F―. Assume that it intersects the side C E― at point G.
Draw a perpendicular line to side C D― from point G. It intersects the side C F― at point H and perpendicularly intersects side C D― at point I.
Draw a perpendicular line to side G I― from point F and it intersects the side G I― at point J.
Now, we can start deriving the expansion of the sine of double angle trigonometric identity in mathematical form.
Procedure
When x is used to represent an angle of a right triangle, the sine, cosine and sine of double angle functions are written as sinx, cosx and sin2 x respectively.
Express the Sine of double angle in Ratio form
The side C F― bisects the ∠D C E as two equal angles ∠D C F and ∠F C E. Here, the ∠D C F and ∠F C E are equal and each angle is denoted by x.
∠D C E=∠D C F+∠F C E
⟹∠D C E=x+x
∴∠D C E=2 x
It is cleared that the angle D C E is equal to 2 x. Hence, the sine of angle 2 x is written as sin(2 x) in trigonometric mathematics.
According to the Δ I C G, the sine of double angle can be written in ratio form of the lengths of the sides.
sin2 x=G I C G
The side J F― splits the side G I― as two sides G J― and J I―. So, the length of side G I― can be written as the sum of them mathematically.
G I=G J+J I
Now, replace the length of side G I― by its equivalent value in the expansion of sin(2 x) function.
⟹sin2 x=G J+J I C G
⟹sin2 x=G J C G+J I C G
The sides G I― and D E― are parallel lines. So, the length of the side J I― is exactly equal to the length of the side D F―. Therefore, J I=D F.
⟹sin2 x=G J C G+D F C G
Express trigonometric ratios in functions
The sin(2 x) is expanded in terms of the ratios between the sides. Now, let’s try to express each ratio in the form a trigonometric function.
The side E F― is opposite side of the Δ E C F and its angle is x. Try to write the length of side E F― in terms of a trigonometric function.
sinx=E F C F
⟹E F=C F sinx
Now, substitute the length of perpendicular E F― by its equivalent value in the expansion of sin2 x.
⟹sin2 x=G J C G+C F sinx C G
⟹sin2 x=G J C G+C F C G×sinx
The sides C F― and C G― are sides of the right triangle G C F and the angle of this triangle is also x. According to trigonometry, the ratio of them can be represented by cosx.
cosx=C F C G
Now, replace the ratio of lengths of sides C F― to C G― by its equivalent trigonometric function in the expansion of sin of double angle function.
⟹sin2 x=G J C G+cosx×sinx
⟹sin2 x=G J C G+cosx sinx
⟹sin2 x=G J C G+sinx cosx
Geometric Analysis for unknown angle
The side G J― is one of the sides of the Δ J G F but its angle is unknown. So, it is not possible to express the length of the side G J― in terms of a trigonometric function. Hence, let’s find the angle of this triangle geometrically.
The sides C E― and J F― are parallel lines and C F― is their transversal. ∠E C F and ∠J F C are interior alternate angles and they are equal geometrically.
∠J F C=∠E C F=x
The side F G― is a perpendicular line to side C F―. So, the ∠C F G=90°. but the side J F― splits the ∠C F G as ∠J F C and ∠J F G.
∠J F C+∠J F G=90°
It is proved about that ∠J F C=x.
⟹x+∠J F G=90°
⟹∠J F G=90°−x
Δ J G F is a right triangle in which ∠G J F=90° and ∠J F G=90°−x. The ∠J G F can be calculated by the sum of angles rule of a triangle.
∠J G F+∠G J F+∠J F G=180°
⟹∠J G F+90°+90°−x=180°
⟹∠J G F+180°−x=180°
⟹∠J G F=180°−180°+x
∴∠J G F=x
It is proved that the ∠J G F is also equal to x because it is a similar triangle.
Write trigonometric ratio in functions
It is already derived that
sin2 x=G J C G+sinx cosx
Now, let’s try to write the ratio between sides G J― and C G― in terms of trigonometric functions to get the expansion of sin double angle identity.
G J― is adjacent side of Δ J G F and its angle is derived as x.
cosx=G J G F
⟹G J=G F cosx
Substitute the length of G J― by its equivalent value in the expansion of sin(2 x) function.
sin2 x=G F cosx C G+sinx cosx
⟹sin2 x=G F C G×cosx+sinx cosx
The sides G F― and C G― are sides of Δ G F C. So, consider right triangle G F C once more time.
sinx=G F C G
Therefore, replace the ratio of lengths of sides G F― to C G― by trigonometric function sinx in the sin2 x expansion.
⟹sin2 x=sinx×cosx+sinx cosx
⟹sin2 x=sinx cosx+sinx cosx
∴sin2 x=2 sinx cosx
Therefore, it is proved geometrically that the sin of double angle function can be expanded as two times the product of sin of angle and cos of angle.
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188172 | https://ijmsi.ir/article-1-1522-en.pdf | Iranian Journal of Mathematical Sciences and Informatics Vol. 19, No. 2 (2024), pp 1-12 DOI: 10.61186/ijmsi.19.2.1 On Lah-Ribariˇ c Inequality Involving Averages of Convex Functions Ana Vukeli´ c Faculty of Food Technology and Biotechnology, Mathematics Department, University of Zagreb, Croatia E-mail: ana.vukelic@pbf.unizg.hr Abstract. By using the integral arithmetic mean and the Lah-Ribariˇ c inequality we give the extension of Wulbert’s result from . Also, we obtain inequalities with divided differences using the Lah-Ribariˇ c inequal-ity. As a consequence, the convexity of higher order for function defined by divided difference is proved. Further, we construct a new family of exponentially convex functions and Cauchy-type means by exploring at linear functionals with the obtained inequalities.
Keywords: Lah-Ribariˇ c inequality, Divided differences, n-convex function, (n, m)-convex function, Exponential convexity.
2020 Mathematics subject classification: 26D15, 26D07, 26A51.
1. Introduction Let f be a continous function on an interval I with a nonempty interior.
Then, define: F(x, y) = ( 1 y−x R y x f(t)dt, x, y ∈I, x ̸= y, f(x), x = y ∈I.
(1.1) Wulbert in , proved that the integral arithmetic mean F defined in (1.1) is convex on I2 if f is convex on I. Zhang and Chu, in , rediscovered (with-out referring to and citing Wulbert’s result) that the necessary and sufficient Received 29 January 2019; Accepted 8 August 2019 ©2024 Academic Center for Education, Culture and Research TMU 1 [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] 2 A. Vukeli´ c condition for the convexity of the integral arithmetic mean F is for f to be convex on I.
Let f be a real-valued function defined on the segment [a, b]. The divided difference of order n of the function f at distinct points x0, ..., xn ∈[a, b], is defined recursively (see , ) by f[xi] = f(xi), (i = 0, . . . , n) and f[x0, . . . , xn] = f[x1, . . . , xn] −f[x0, . . . , xn−1] xn −x0 .
The value f[x0, . . . , xn] is independent of the order of the points x0, . . . , xn.
The definition may be extended to include the case that some (or all) of the points coincide. Assuming that f (j−1)(x) exists, we define f[x, . . . , x | {z } j−times ] = f (j−1)(x) (j −1)! .
(1.2) For divided difference the following holds: f[x0, . . . , xn] = n X i=0 f(xi) ω′(xi), where ω(x) = n Y j=0 (x −xj), so we have that f[x0, . . . , xn] = n X i=0 f(xi) Qn j=0,j̸=i(xi −xj).
If the function f has continuous n-th derivative on [a, b], the divided differ-ence f[x0, . . . , xn] can be represented in integral form by f[x0, . . . , xn] = Z ∆n f (n) n X i=0 uixi !
du0 . . . dun−1, where ∆n = ( (u0, . . . , un−1) : ui ≥0, n−1 X i=0 ui ≤1 ) and un = 1 −Pn−1 i=0 ui.
The notion of n-convexity goes back to Popoviciu ().
We follow the definition given by Karlin (): Definition 1.1. A function f : [a, b] →R is said to be n-convex on [a, b], n ≥0, if for all choices of (n + 1) distinct points in [a, b], n-th order divided difference of f satisfies f[x0, ..., xn] ≥0. [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] On Lah-Ribariˇ c Inequality Involving Averages of Convex Functions 3 In fact, Popoviciu proved that each continuous n-convex function on [a, b] is the uniform limit of the sequence of n-convex polynomials. Many related results, as well as some important inequalities due to Favard, Berwald and Steffensen can be found in .
In is proved the following Jensen inequality for divided differences: Theorem 1.2. Let f be an (n + 2)-convex function on (a, b) and x ∈(a, b)n+1.
Then G(x) = f[x0, . . . , xn] is a convex function of the vector x = (x0, . . . , xn). Consequently, f " m X i=0 aixi 0, . . . , m X i=0 aixi n # ≤ m X i=0 aif[xi 0, . . . , xi n] (1.3) holds for all ai ≥0 such that Pm i=0 ai = 1.
Schur polynomial in n + 1 variables x0, . . . , xn of degree d = d0 + . . . + dn (dj’s form nonincreasing sequence non-negative integers, i.e. d0 ≥. . . ≥dn) is defined as S(d0,...,dn)(x0, . . . , xn) = det h xdn−j+j i in i,j=0 det h xj i in i,j=0 .
The numerator consists of alternating polynomials (they change the sign under any transposition of the variables) and so they are all divisible by the denomi-nator which is Vandermonde determinant. Schur polynomial is also symmetric because the numerator and denominator are both alternating.
Using Schur polynomial and Vandermonde determinant (extended with log-arithmic function) V (x; p, q) = det 1 x0 x02 . . .
x0n−1 x0p lnq x0 1 x1 x12 . . .
x1n−1 x1p lnq x1 1 x2 x22 . . .
x2n−1 x2p lnq x2 .
.
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...
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1 xn xn2 . . .
xnn−1 xnp lnq xn we obtain: Proposition 1.3. For monomial function h(x) = xn+k, where k ≥1 is an integer, holds h[x0, . . . , xn] = S(k,0, . . . , 0 | {z } n−times )(x0, . . . , xn) = V (x; n + k, 0) V (x; n, 0) = n X i1=0 i1 X i2=0 · · · ik−1 X ik=0 xi1xi2 · · · xik. [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] 4 A. Vukeli´ c For potential function f(x) = xp = xn+p−n, where p is a real number, holds f[x0, . . . , xn] = V (x; p, 0) V (x; n, 0).
Let f(x, y) be a real-valued function defined on I × J (I = [a, b], J = [c, d]). Then the (l1, l2) divided difference of the function f at distinct points x0, . . . , xl1 ∈I, y0, . . . , yl2 ∈J, is defined by (see ) f x0, . . . , xl1 y0, . . . , yl2 = f([y0, . . . , yl2])[x0, . . . , xl1] = f([x0, . . . , xl1])[y0, . . . , yl2] = l1 X i=0 l2 X j=0 f(xi, yj) ω′(xi)ω′(yj), (1.4) where ω(x) = Ql1 i=0(x −xi), ω(y) = Ql2 j=0(y −yj).
Definition 1.4. A function f : I ×J →R is said to be (l1, l2)-convex or convex of order (l1, l2) if for all distinct points x0, . . . , xl1 ∈I, y0, . . . , yl2 ∈J, f x0, . . . , xl1 y0, . . . , yl2 ≥0.
(1.5) If this inequality is strict, then f is said to be strictly (l1, l2)-convex.
Popoviciu in proved the following theorem: Theorem 1.5. If the partial derivative f (l1+l2) xl1yl2 of f exists, then f is (l1, l2)-convex iff f (l1+l2) xl1yl2 ≥0.
(1.6) If the inequality in (1.6) is strict, then f is strictly (l1, l2)-convex.
The well known Lah-Ribariˇ c inequality is given in the following theorem (see ): Theorem 1.6. Let f be a real valued convex function on [m, M]. Then for m ≤xk ≤M, pk > 0 (1 ≤k ≤n) and Pn k=1 pk = 1 we have n X k=1 pkf(xk) ≤M −¯ x M −mf(m) + ¯ x −m M −mf(M), (1.7) where ¯ x = Pn k=1 pkxk.
The goal of this paper is to give the extension of Wulbert’s result from and also to obtain inequalities with divided differences using the Lah-Ribariˇ c inequality. As a consequence, we will proof the convexity of higher order for function defined by divided difference. In the last section, a new family of exponentially convex functions and Cauchy-type means are constructed by looking to the linear functionals associated with the obtained inequalities. [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] On Lah-Ribariˇ c Inequality Involving Averages of Convex Functions 5 2. Inequalities involving averages The following result is an extension of Wulbert’s results: Theorem 2.1. Let f be a real valued convex function on [m, M] and F is defined in (1.1). Then for m ≤xk, yk ≤M, pk > 0 (1 ≤k ≤n) and Pn k=1 pk = 1 we have n X k=1 pkF(xk, yk) ≤M −1 2(¯ x + ¯ y) M −m F(m, m) + 1 2(¯ x + ¯ y) −m M −m F(M, M), (2.1) where ¯ x = Pn k=1 pkxk and ¯ y = Pn k=1 pkyk.
Consequently, for l1 + l2 = 2 the integral arithmetic mean (1.1) is (l1, l2)-convex on [m, M]2.
Proof. By using the Lah Ribariˇ c inequality (1.7) we get: n X k=1 pkF(xk, yk) = n X k=1 pk Z 1 0 f(syk + (1 −s)xk)ds = Z 1 0 n X k=1 pkf(syk + (1 −s)xk))ds ≤ f(m) M −m Z 1 0 " M − n X k=1 pk(syk + (1 −s)xk) # ds + f(M) M −m Z 1 0 " n X k=1 pk(syk + (1 −s)xk) −m # ds = f(m) M −mM − f(m) M −m Z 1 0 (s¯ y + (1 −s)¯ x)ds + f(M) M −m Z 1 0 (s¯ y + (1 −s)¯ x)ds −m f(M) M −m = M −1 2(¯ x + ¯ y) M −m F(m, m) + 1 2(¯ x + ¯ y) −m M −m F(M, M).
Now, if we put n = 2, x1 = m, x2 = M, y1 = M, y2 = m, p1 = p2 = 1 2, then the inequality (2.1) reduces to F(m, M) + F(M, m) ≤F(m, m) + F(M, M).
Use the definition in (1.4) we get (M −m)2(F[m, M])[M, m] ≥0.
It is known that if this holds for all possible m, M > 0 then F is (1, 1)-convex function (see ).
Wulbert in , proved that the integral arithmetic mean F defined in (1.1) is convex on [m, M]2, so we have F (2+0) x2y0 ≥0 and F (0+2) x0y2 ≥0. So, by using Theorem 1.5 function F is convex of order (2, 0) and (0, 2).
□ [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] 6 A. Vukeli´ c Remark 2.2. Theorem 2.1 is a generalization of the Lah-Ribariˇ c inequality. For xk = yk, k = 1, . . . , n the inequality (2.1) recaptures the Lah-Ribariˇ c inequality (1.7).
The following theorem is the integral version of Theorem 2.1: Theorem 2.3. Let (Ω, A, µ) be a probability space, α, β : Ω→[m, M] be functions from L1(µ) and let f be an convex function on [m, M] and F is defined in (1.1). Then Z Ω F(α(u), β(u))dµ(u) ≤M −1 2(¯ α + ¯ β) M −m F(m, m) + 1 2(¯ α + ¯ β) −m M −m F(M, M), (2.2) where ¯ α = R Ωα(u)dµ(u) and ¯ β = R Ωβ(u)dµ(u).
Proof. By using the integral version of Lah-Ribariˇ c inequality we get: Z Ω F(α(u), β(u))dµ(u) = Z 1 0 Z Ω f(sβ(u) + (1 −s)α(u))dµ(u)ds ≤ f(m) M −m Z 1 0 M − Z Ω (sβ(u) −(1 −s)α(u))dµ(u) ds + f(M) M −m Z 1 0 Z Ω (sβ(u) −(1 −s)α(u))dµ(u) −m ds = M −1 2(¯ α + ¯ β) M −m F(m, m) + 1 2(¯ α + ¯ β) −m M −m F(M, M).
□ 3. Inequalities for divided differences In the following theorem we proof the Lah-Ribariˇ c inequality for divided differences: Theorem 3.1. Let f be an (n+2)-convex function on [m, M] and x ∈[m, M]n+1.
Then l X i=0 aif[xi 0, . . . , xi n] (3.1) ≤ f (n)(m) n!(M −m) M − 1 n + 1 n X j=0 ¯ xj + f (n)(M) n!(M −m) 1 n + 1 n X j=0 ¯ xj −m holds for all ai ≥0 such that Pl i=0 ai = 1 and ¯ xj = Pl i=0 aixi j. Consequently, for n = 1 G(x) = f[x0, x1] is a (l1, l2)-convex function of the vector x = (x0, x1), when l1 + l2 = 2. [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] On Lah-Ribariˇ c Inequality Involving Averages of Convex Functions 7 Proof. Using the Lah-Ribariˇ c inequality for convex function f (n), we have l X i=0 aif[xi 0, . . . , xi n] = l X i=0 ai Z ∆n f (n) n X j=0 ujxi j du0 . . . dun−1 = Z ∆n l X i=0 aif (n) n X j=0 ujxi j du0 . . . dun−1 ≤ f (n)(m) M −m Z ∆n M − l X i=0 ai n X j=0 ujxi j du0 . . . dun−1 +f (n)(M) M −m Z ∆n l X i=0 ai n X j=0 ujxi j −m du0 . . . dun−1 for g(x) = xn n! and h(x) = xn+1 (n + 1)!
= f (n)(m) M −m M Z ∆n g(n) n X j=0 ujxi j du0 . . . dun−1 − l X i=0 ai Z ∆n h(n) n X j=0 ujxi j du0 . . . dun−1 + f (n)(M) M −m l X i=0 ai Z ∆n h(n) n X j=0 ujxi j du0 . . . dun−1 −m Z ∆n g(n) n X j=0 ujxi j du0 . . . dun−1 = f (n)(m) M −m " M · g[xi 0, xi 1, . . . , xi n] − l X i=0 ai · h[xi 0, xi 1, . . . , xi n] # +f (n)(m) M −m " l X i=0 ai · h[xi 0, xi 1, . . . , xi n] −m · g[xi 0, xi 1, . . . , xi n] # = f (n)(m) n!(M −m) M − 1 n + 1 n X j=0 ¯ xj + f (n)(M) n!(M −m) 1 n + 1 n X j=0 ¯ xj −m .
Now, If we put n = 1, x0 0 = m, x0 1 = M, x1 0 = M, x1 1 = m, a1 = a2 = 1 2 and fact that f ′(x) = f[x, x] = G(x, x), similarly as in Theorem 2.1, we can proof that function G is convex function of order (1, 1).
By using Theorem 1.2 and similarly as in Theorem 2.1, we also can proof that function G is convex function of order (2, 0) and (0, 2).
□ [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] 8 A. Vukeli´ c The integral version of Lah-Ribariˇ c inequality for divided differences is given with following theorem: Theorem 3.2. Let p, gi : Ω→[m, M], (i = 0, . . . , n) be functions from L1(µ) and let f be an (n + 2)-convex function on [m, M]. Then Z Ω p(x)f[g0(x), . . . , gn(x)]dµ(x) ≤ f (n)(m) n!(M −m) " M − 1 n + 1 n X i=0 ¯ gi # + f (n)(M) n!(M −m) " 1 n + 1 n X i=0 ¯ gi −m # (3.2) holds for all p(x) ≥0 such that R Ωp(x)dµ(x) = 1 and ¯ gi = R Ωp(x)gi(u)dµ(u).
Proof. Using the integral Lah-Ribariˇ c inequality for convex function f (n), we have the following conclusion Z Ω p(x)f[g0(x), . . . , gn(x)]dµ(x) = Z ∆n Z Ω p(x)f (n) n X i=0 uigi(x) !
dµ(x) !
du0 . . . dun−1 ≤ f (n)(m) M −m Z ∆n M − Z Ω p(x) n X i=0 uigi(x) !
dµ(x) !
du0 . . . dun−1 +f (n)(M) M −m Z ∆n Z Ω p(x) n X i=0 uigi(x) !
dµ(x) −m !
du0 . . . dun−1 = f (n)(m) n!(M −m) " M − 1 n + 1 n X i=0 ¯ gi # + f (n)(M) n!(M −m) " 1 n + 1 n X i=0 ¯ gi −m # .
□ 4. Applications to exponential convexity Motivated by inequalities (2.1), (2.2), (3.1) and (3.2), under the same as-sumptions, we define following functionals: Φ1(f) = M −1 2(¯ x + ¯ y) M −m f(m) + 1 2(¯ x + ¯ y) −m M −m f(M) − n X k=1 pk yk −xk Z yk xk f(t)dt, (4.1) Φ2(f) = M −1 2(¯ α + ¯ β) M −m f(m) + 1 2(¯ α + ¯ β) −m M −m f(M) − Z Ω 1 β(u) −α(u) Z β(u) α(u) f(t)dt !
dµ(u), (4.2) [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] On Lah-Ribariˇ c Inequality Involving Averages of Convex Functions 9 Φ3(f) = f (n)(m) n!(M −m) M − 1 n + 1 n X j=0 ¯ xj + f (n)(M) n!(M −m) 1 n + 1 n X j=0 ¯ xj −m − l X i=0 aif[xi 0, . . . , xi n] (4.3) and Φ4(f) = f (n)(m) n!(M −m) " M − 1 n + 1 n X i=0 ¯ gi # + f (n)(M) n!(M −m) " 1 n + 1 n X i=0 ¯ gi −m # − Z Ω p(x)f[g0(x), . . . , gn(x)]dµ(x).
(4.4) Similarly as in we can construct new families of exponentially convex function and Cauchy type means by looking at these linear functionals. Also, we can proof the monotonicity property of the generalized Cauchy means obtained via these functionals.
Here we present an example for such a family of functions: Example 4.1. Consider a family of functions Ω= {fs : (0, ∞) →R : s ∈R} defined by fs(x) = ( xs s(s−1), s / ∈{0, 1}, xj ln x (−1)1−jj!(1−j)!, s = j ∈{0, 1}.
Here, d2fs dx2 (x) = xs−2 = e(s−2) ln x > 0 which shows that fs is convex for x > 0 and s 7→d2fs dx2 (x) is exponentially convex by definition. Arguing as in we get that the mappings s 7→Φi(fs), i = 1, 2 are exponentially convex. Now we get: µs,q(Φi, Ω) = Φi(fs) Φi(fq) 1 s−q , s ̸= q, exp −Φi(f0fs) Φi(fs) + 1−2s s2−s , s = q / ∈{0, 1}, exp −Φi(f 2 0 ) 2Φi(f0) + 1 , s = q = 0, exp −Φi(f0f1) 2Φi(f1) −1 , s = q = 1, where for s ̸= −1, 0, 1 Φ1(fs) = M −1 2(¯ x + ¯ y) M −m · ms s(s −1) + 1 2(¯ x + ¯ y) −m M −m · M s s(s −1) − 1 s3 −s n X k=1 pk ys+1 k −xs+1 k yk −xk , [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] 10 A. Vukeli´ c Φ1(f−1) = M −1 2(¯ x + ¯ y) M −m · 1 2m + 1 2(¯ x + ¯ y) −m M −m · 1 2M − 1 2 n X k=1 pk ln yk −ln xk yk −xk , Φ1(f0) = 1 2(¯ x + ¯ y) −M M −m · ln m + m −1 2(¯ x + ¯ y) M −m · ln M + n X k=1 pk yk ln yk −xk ln xk yk −xk −1, Φ1(f1) = M −1 2(¯ x + ¯ y) M −m · m ln m + 1 2(¯ x + ¯ y) −m M −m · M ln M − 1 2 n X k=1 pk y2 k ln yk −x2 k ln xk yk −xk + 1 4(¯ x + ¯ y).
For similarly results for Jensen’s inequality involving averages of convex functions see and .
For a family of functions ˜ Ω= n ˜ fs : (0, ∞) →R : s ∈R o defined by ˜ fs (x) = ( xs s(s−1)...(s−(n+1)), s / ∈{0, 1, . . . , n + 1}, xj ln x (−1)n+1−jj!(n+1−j)!, s = j ∈{0, 1, . . . , n + 1}, analogous as above it is easy to prove that s 7→Φi ˜ fs (i = 3, 4) are exponen-tially convex. In this case, we get ˜ µs,q Φi, ˜ Ω (i = 3, 4) as follows ˜ µs,q Φi, ˜ Ω = Φi(fs) Φi(fq) 1 s−q , s ̸= q, exp (−1)n+1(n+1)!Φi(f0fs) Φi(fs) +Pn+1 k=0 1 k−s , s = q / ∈{0, 1, . . . , n + 1}, exp (−1)n+1(n+1)!Φi(f0fs) 2Φi(fs) +Pn+1 k=0 k̸=s 1 k−s , s = q ∈{0, 1, . . . , n + 1}.
For s / ∈{0, 1, . . . , n + 1} Φ3( ˜ fs) = M −1 2(¯ x + ¯ y) M −m ms−n n!(s −n)(s −(n + 1)) + 1 2(¯ x + ¯ y) −m M −m M s−n n!(s −n)(s −(n + 1)) − n+1 Y k=0 1 s −k m X i=0 ai V (xi, s, 0) V (xi, n, 0). [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] On Lah-Ribariˇ c Inequality Involving Averages of Convex Functions 11 For s = j ∈{0, 1, . . . , n + 1} we have ˜ f (n) s (x) = 1 (−1)n+1−j j! (n + 1 −j)!
n−1 X i=0 n−1 Y k=0 k̸=i (j −k)xj−n + n−1 Y i=0 (j −i)xj−n ln x .
So, for s = j ∈{0, 1, . . . , n −1} Φ3( ˜ fj) = M −1 2(¯ x + ¯ y) M −m · mj−n n!
+ 1 2(¯ x + ¯ y) −m M −m · M j−n n!
− n+1 Y k=0 1 j −k m X i=0 ai V (xi, j, 1) V (xi, n, 0) and for l ∈{0, 1} Φ3( ˜ fn+l) = (−1)l+1 M −1 2(¯ x + ¯ y) M −m · ml hPn−1 i=0 1 n+l−1 + ln m i n!
+ 1 2(¯ x + ¯ y) −m M −m · M l hPn−1 i=0 1 n+l−1 + ln M i n!
− n+1 Y k=0 1 n + l −k m X i=0 ai V (xi, n + l, 1) V (xi, n, 0) .
For similarly results for Jensen’s inequality for divided differences see and . See also .
Φi (i = 1, 2, 3, 4) are positive, so then there exists ξi, ˜ ξi ∈[m, M] such that ξs−q i = Φi (fs) Φi (fq), i = 1, 2, ˜ ξs−q i = Φi ˜ fs Φi ˜ fq , i = 3, 4.
Since the function ξi 7→ξs−q i and ˜ ξi 7→˜ ξs−q i are invertible for s ̸= q, we have m ≤ Φi (fs) Φi (fq) 1 s−q ≤M, i = 1, 2, m ≤ Φi ˜ fs Φi ˜ fq 1 s−q ≤M, i = 3, 4, which together with the fact that µs,q (Φi, Ω) and ˜ µs,q Φi, ˜ Ω are continuous, symmetric and monotonous, shows that µs,q (Φi, Ω) and ˜ µs,q Φi, ˜ Ω are means.
Acknowledgments The author wish to thank the referee for the helpful suggestions and com-ments. [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] 12 A. Vukeli´ c References 1. K. E. Atkinson, An Introduction to Numerical Analysis, 2nd ed., Wiley, New York, 1989.
2. V. ˇ Culjak, I. Franji´ c, R. Ghulam, J. Peˇ cari´ c, Schur-convexity of Averages of Convex Function, J. Ineq. Appl., 2011(2011), Article ID 581918.
3. R. Farwig, D. Zwick, Some Divided Difference Inequalities for n-convex Functions, J.
Math. Anal. Appl., 108, (1985), 430–437.
4. J. Jakˇ seti´ c, J. Peˇ cari´ c, G. Roqia, On Jensen’s Inequality Involving Averages of Convex Functions, Sarajevo J. Math., 8(20)(1), (2012), 53–68.
5. S. Karlin, Total Positivity, Stanford Univ. Press, Stanford, 1968.
6. S. Karlin, W. J. Studden, Tchebycheff Systems: with Applications in Analysis and Sta-tistics, Interscience, New York, 1966.
7. P. Lah, M. Ribariˇ c, Converse of Jensen’s Inequality for Convex Functions, Univ. Beograd.
Publ. Elektrotehn. Fak. Ser. Mat. Fiz., 412-460, (1973), 201–205.
8. Z. Pavi´ c, J. Peˇ cari´ c, A. Vukeli´ c, Exponential Convexity and Jensen Inequality for Divided Differences, J. Math. Inequal., 5(2), (2011), 157–168.
9. Z. Pavi´ c, J. Peˇ cari´ c, A. Vukeli´ c, Means for Divided Differences and Exponential Convex-ity, Mediterr. J. Math., 9(1), (2012), 187–198.
10. J. E. Peˇ cari´ c, F. Proschan, Y. L. Tong, Convex Functions, Partial Orderings, and Sta-tistical Applications, Mathematics in science and engineering, vol. 187, Academic Press, 1992.
11. J. Peˇ cari´ c, J. Peri´ c, Improvements of the Giaccardi and the Petrovi´ c Inequality and Related Stolarsky Type Means, An. Univ. Craiova Ser. Mat. Inform., 39, (2012), 65– 75.
12. T. Popoviciu, Sur l’approximation Des Fonctions Convexes D’ordre Superieur, Mathe-matica, 10, (1934), 49–54.
13. T. Popoviciu, Les Functions Convexes, Herman and Cie, ´ Editeurs, Paris (1944).
14. G. Roqia, J. Peˇ cari´ c, A. Vukeli´ c, n—exponential Convexity of Divided Differences and Related Stolarsky Type Menas, Math. Ineq. Appl., 16(4), (2013), 1043–1063.
15. D. E. Wulbert, Favard’s Inequality on Average Values of Convex Functions, Math. Com-put. Model., 37(2003), 1383–1391.
16. X. M. Zhang, Y. M. Chu, Convexity of the Integral Arithmetic Mean of a Convex Func-tion, Rocky Mt. J. Math., 40(3), (2010), 1061–1068. [ DOI: 10.61186/ijmsi.19.2.1 ] [ Downloaded from ijmsi.ir on 2025-09-29 ] Powered by TCPDF (www.tcpdf.org) |
188173 | https://www.cargille.com/about-immersion-oils/ | Services to the sciences since 1924
About Immersion Oils
Immersion Oil & The Microscope – A Paper by John J. Cargille
Cargille’s Expertise with Immersion Oils
Immersion Oil Selection Guide
Immersion Oil Specifications
Immersion Oil & The Microscope – A Paper by John J. Cargille
Most scientists who are using microscopes are not microscopists—they have another field of specialization. In many cases, their understanding of microscopy is limited to on the job training that allows them to get by proficiently. This classic paper by John J. Cargille attempts to broaden scientists’ understanding of the “business area” of the microscope, between the condenser and the objective, as it is affected by the use of oil immersion objectives. The paper also expands on properties of immersion oils and how they can be more fully utilized.
PDF Download – Immersion Oil & the Microscope
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Cargille’s Expertise with Immersion Oils
Cargille Laboratories has consistently led in the development and standardization of immersion oil for over 42 years. Here is the record:
Developed non-drying immersion oils (1940).
First to standardize index and temperature values.
Developed oils of different viscosities.
Developed low fluorescence immersion oil (1940).
First to print key optical properties on every bottle (1942).
First to print full tables of optical and physical properties in Data Sheets (1946).
First nondrying immersion oil adopted by major microscope manufacturers (AO, B&L)
Cargille specifications adopted by U.S. Department of Defense.
Developed the first very high viscosity oils for horizontal, inverted and projecting microscopes in cooperation with US Department of H.E.W. (1950).
Developed the first PCB-free immersion oils (1971).
Supplies oil to meet British immersion oil specifications.
Collaborated with DIN and ISO committees in developing international standards and served as liaison with American microscope manufacturers.
Provides modified formulas for special applications supported by Extended Properties Technical Bulletins.
First to provide Material Safety Data Sheets to insure compliance with OSHA.
First to establish a Refractive Index Laboratory with primary standard, computerized formulation and modeling capabilities for R&D, QC and Customer Service.
First to modify immersion oil formulations for use with various instruments such as calibration, mounting, contact and interface fluids in the fields of fiber optics, lasers, optometry, and optical engineering where special index, temperature, wavelength, viscosity, density and adsorption characteristics are required.
First in worldwide distribution of immersion oils.
Foremost in supplying US and foreign microscope manufacturers with “accessory” oils for new instruments.
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Immersion Oil Selection Guide
For Normal Light Microscopy: Types A and B are virtually interchangeable and are miscible with each other for intermediate viscosities. Produced in larger quantities than other types, Types A and B are the most economical. The deciding factor in choosing between them is the optimum viscosity for your particular application.
Type A, at 150 centistokes, reduces any tendancy to trap air, especially helpful to beginning students. Air bubbles cause image degradation.
Type B, at 1250 cSt, is thick enough for viewing multiple slides with one application. This saves time during batch processing.
Automated Hematology Systems: Use Type 300; Automated Hematology Systems depend on accurate, precisely controlled physical and optical properties of immersion oil for successful imaging and mechanical processing. Type 300 is designed and manufactured to meet the stringent requirements of this equipment, which include specialized viscosity and exacting controls for its consistency.
Inverted, Inclined, Projection, and Long Focus Instruments: Use type NVH or OVH; The greater the gap between the cover glass and objective, or between the slide and condenser, the more desirable high viscosity becomes. The very high viscosities of Type NVH at 21,000 cSt and Type OVH at 46,000 cSt give excellent results for these applications.
Blending Oils from the Miscible Group: The Miscible Group of immersion oils is A, B, 300, NVH and OVH. Users can easily blend any two immersion oils from the Miscible Group to obtain an immersion oil with an intermediate viscosity while maintaining the optical properties common to both.
Fluorescence Microscopy: Extremely low fluorescence is achieved by Type LDF and Type HF. Type FF is virtually fluorescence-free, though not ISO compliant. Type HF is slightly more fluorescent than Type LDF, but is halogen-free. For most non-critical fluorescence microscopy applications, Types A and B are sufficiently low fluorescing. Viscosities for Type LDF, HF, and FF are 500 cSt, 700 cSt and 170 cSt, respectively. Types A and B are 150 cSt and 1250 cSt.
Elevated Temperatures (>23°C to 37°C): Use Type 37. Elevated temperatures can be due to substage illuminators, “hot stage”, or other causes – ideal situations for Cargille Immersion Oil Type 37. Developed specifically for working at human body temperatures, Type 37 has a refractive index of 1.515 and a viscosity of 1250 cSt at 37°C. solving the problem of image degradation above the standard calibration temperature of 23°C. Users can blend for their own working temperature; blending Type B, with a viscosity of 1250 cSt at 23°C with Type 37, 1250 cSt at 37°C maintains a constant 1250 cSt viscosity and optical values and places the temperature of calibration proportionally between 23°C to 37°C.
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Immersion Oil Specifications
PDF Download – Immersion Oil Specifications |
188174 | https://tutorial.math.lamar.edu/classes/de/SecondOrderConcepts.aspx | Paul's Online Notes
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Section 3.1 : Basic Concepts
In this chapter we will be looking exclusively at linear second order differential equations. The most general linear second order differential equation is in the form.
p(t)y′′+q(t)y′+r(t)y=g(t)(1)
In fact, we will rarely look at non-constant coefficient linear second order differential equations. In the case where we assume constant coefficients we will use the following differential equation.
ay′′+by′+cy=g(t)(2)
Where possible we will use (1) just to make the point that certain facts, theorems, properties, and/or techniques can be used with the non-constant form. However, most of the time we will be using (2) as it can be fairly difficult to solve second order non-constant coefficient differential equations.
Initially we will make our life easier by looking at differential equations with g(t)=0. When g(t)=0 we call the differential equation homogeneous and when g(t)≠0 we call the differential equation nonhomogeneous.
So, let’s start thinking about how to go about solving a constant coefficient, homogeneous, linear, second order differential equation. Here is the general constant coefficient, homogeneous, linear, second order differential equation.
ay′′+by′+cy=0
It’s probably best to start off with an example. This example will lead us to a very important fact that we will use in every problem from this point on. The example will also give us clues into how to go about solving these in general.
Example 1 Determine some solutions to
y′′−9y=0
Show Solution
We can get some solutions here simply by inspection. We need functions whose second derivative is 9 times the original function. One of the first functions that I can think of that comes back to itself after two derivatives is an exponential function and with proper exponents the 9 will get taken care of as well.
So, it looks like the following two functions are solutions.
y(t)=e3tandy(t)=e−3t
We’ll leave it to you to verify that these are in fact solutions.
These two functions are not the only solutions to the differential equation however. Any of the following are also solutions to the differential equation.
y(t)=−9e3ty(t)=123e3ty(t)=56e−3ty(t)=149e−3ty(t)=7e3t−6e−3ty(t)=−92e3t−16e−3t
In fact if you think about it any function that is in the form
y(t)=c1e3t+c2e−3t
will be a solution to the differential equation.
This example leads us to a very important fact that we will use in practically every problem in this chapter.
Principle of Superposition
If y1(t) and y2(t) are two solutions to a linear, homogeneous differential equation then so is
y(t)=c1y1(t)+c2y2(t)(3)
Note that we didn’t include the restriction of constant coefficient or second order in this. This will work for any linear homogeneous differential equation.
If we further assume second order and one other condition (which we’ll give in a second) we can go a step further.
If y1(t) and y2(t) are two solutions to a linear, second order homogeneous differential equation and they are “nice enough” then the general solution to the linear, second order homogeneous differential equation is given by (3).
So, just what do we mean by “nice enough”? We’ll hold off on that until a later section. At this point you’ll hopefully believe it when we say that specific functions are “nice enough”.
So, if we now make the assumption that we are dealing with a linear, second order homogeneous differential equation, we now know that (3) will be its general solution. The next question that we can ask is how to find the constants c1 and c2. Since we have two constants it makes sense, hopefully, that we will need two equations, or conditions, to find them.
One way to do this is to specify the value of the solution at two distinct points, or,
y(t0)=y0y(t1)=y1
These are typically called boundary values and are not really the focus of this course so we won’t be working with them here. We do give a brief introduction to boundary values in a later chapter if you are interested in seeing how they work and some of the issues that arise when working with boundary values.
Another way to find the constants would be to specify the value of the solution and its derivative at a particular point. Or,
y(t0)=y0y′(t0)=y′0
These are the two conditions that we’ll be using here. As with the first order differential equations these will be called initial conditions.
Example 2 Solve the following IVP.
y′′−9y=0y(0)=2y′(0)=−1
Show Solution
First, the two functions
y(t)=e3tandy(t)=e−3t
are “nice enough” for us to form the general solution to the differential equation. At this point, please just believe this. You will be able to verify this for yourself in a couple of sections.
The general solution to our differential equation is then
y(t)=c1e−3t+c2e3t
Now all we need to do is apply the initial conditions. This means that we need the derivative of the solution.
y′(t)=−3c1e−3t+3c2e3t
Plug in the initial conditions
2=y(0)=c1+c2−1=y′(0)=−3c1+3c2
This gives us a system of two equations and two unknowns that can be solved. Doing this yields
c1=76c2=56
The solution to the IVP is then,
y(t)=76e−3t+56e3t
Up to this point we’ve only looked at a single differential equation and we got its solution by inspection. For a rare few differential equations we can do this. However, for the vast majority of the second order differential equations out there we will be unable to do this.
So, we would like a method for arriving at the two solutions we will need in order to form a general solution that will work for any linear, constant coefficient, second order homogeneous differential equation. This is easier than it might initially look.
We will use the solutions we found in the first example as a guide. All of the solutions in this example were in the form
y(t)=ert
Note, that we didn’t include a constant in front of it since we can literally include any constant that we want and still get a solution. The important idea here is to get the exponential function. Once we have that we can add on constants to our hearts content.
So, let’s assume that all solutions to
ay′′+by′+cy=0(4)
will be of the form
y(t)=ert(5)
To see if we are correct all we need to do is plug this into the differential equation and see what happens. So, let’s get some derivatives and then plug in.
y′(t)=rerty′′(t)=r2ert
a(r2ert)+b(rert)+c(ert)=0ert(ar2+br+c)=0
So, if (5) is to be a solution to (4) then the following must be true
ert(ar2+br+c)=0
This can be reduced further by noting that exponentials are never zero. Therefore, (5) will be a solution to (4)provided r is a solution to
ar2+br+c=0(6)
This equation is typically called the characteristic equation for (4).
Okay, so how do we use this to find solutions to a linear, constant coefficient, second order homogeneous differential equation? First write down the characteristic equation, (6), for the differential equation, (4). This will be a quadratic equation and so we should expect two roots, r1 and r2. Once we have these two roots we have two solutions to the differential equation.
y1(t)=er1tandy2(t)=er2t(7)
Let’s take a look at a quick example.
Example 3 Find two solutions to
y′′−9y=0
Show Solution
This is the same differential equation that we looked at in the first example. This time however, let’s not just guess. Let’s go through the process as outlined above to see the functions that we guess above are the same as the functions the process gives us.
First write down the characteristic equation for this differential equation and solve it.
r2−9=0⇒r=±3
The two roots are 3 and -3. Therefore, two solutions are
y1(t)=e3tandy2(t)=e−3t
These match up with the first guesses that we made in the first example.
You’ll notice that we neglected to mention whether or not the two solutions listed in (7) are in fact “nice enough” to form the general solution to (4). This was intentional. We have three cases that we need to look at and this will be addressed differently in each of these cases.
So, what are the cases? As we previously noted the characteristic equation is quadratic and so will have two roots, r1 and r2. The roots will have three possible forms. These are
Real, distinct roots,r1≠r2.
Complex root,r1,2=λ±μi.
Double roots,r1=r2=r.
The next three sections will look at each of these in some more depth, including giving forms for the solution that will be “nice enough” to get a general solution.
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188175 | https://search.r-project.org/CRAN/refmans/betafunctions/html/qBeta.4P.html | R: Quantile Given Probability Under the Four-Parameter Beta...
qBeta.4P {betafunctions}R Documentation
Quantile Given Probability Under the Four-Parameter Beta Distribution.
Description
Function for calculating the quantile (i.e., value of x) for a given proportion (i.e., the value of y) under the Four-Parameter Beta Distribution.
Usage
R
qBeta.4P(p, l, u, alpha, beta, lower.tail = TRUE)
Arguments
pA vector (or single value) of proportions or probabilities for which the corresponding value of x (i.e., the quantiles) are to be calculated.
lThe first (lower) location parameter.
uThe second (upper) location parameter.
alphaThe first shape parameter.
betaThe second shape parameter.
lower.tailLogical. Whether the quantile(s) to be calculated is to be under the lower or upper tail. Default is TRUE (lower tail).
Value
A vector of quantiles for specified probabilities or proportions of observations under the four-parameter Beta distribution.
Examples
```R
Assume some variable follows a four-parameter Beta distribution with
location parameters l = 0.25 and u = 0.75, and shape parameters alpha = 5
and beta = 3. To compute the quantile at a specific point of the
distribution (e.g., 0.5) using qBeta.4P():
qBeta.4P(p = 0.5, l = 0.25, u = 0.75, alpha = 5, beta = 3)
```
[Package betafunctions version 1.9.0 Index] |
188176 | https://dev.to/vidyasagarmsc/an-introduction-to-sympy-a-python-library-for-symbolic-mathematics-4gig | Skip to content
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An Introduction to SymPy: A Python Library for Symbolic Mathematics
#python
#libraries
#mathematics
#opensource
SymPy is an open-source Python library for symbolic mathematics.
SymPy — Mathematics Python Library
SymPy aims to become a full-featured computer algebra system (CAS) while keeping the code simple, extensible, and free of external dependencies. With SymPy, you can perform algebraic manipulations, calculus operations, linear algebra, equation solving, discrete mathematics, and much more, all symbolically, rather than numerically.
In this article, we will explore the SymPy library, its capabilities, and some practical examples that demonstrate how it can be used effectively for symbolic computation.
What is SymPy?
At its core, SymPy provides functions for performing symbolic calculations, which means it works with mathematical expressions in their exact form, rather than approximating values as floating-point numbers. SymPy leverages Python’s object-oriented nature to represent mathematical objects, such as variables, functions, and matrices, as Python objects.
SymPy is built in pure Python and does not require any external libraries or software packages. This makes it easy to install and use, and it’s suitable for applications in scientific computing, education, engineering, and mathematics.
Key Features of SymPy
Symbolic computation : Perform algebraic operations like simplification, expansion, and factorization on symbolic expressions.
Calculus : Compute derivatives, integrals, limits, series expansions, and solve differential equations symbolically.
Linear algebra : Manipulate matrices, solve linear systems, and compute determinants, eigenvalues, and eigenvectors.
Equation solving : Solve algebraic equations (linear, polynomial, non-linear) symbolically.
Plotting : Generate plots of functions and equations, both symbolic and numerical.
Installing SymPy
To install SymPy, you can use pip:
```
pip install sympy
```
After installation, you can import the library into your Python script:
```
import sympy as sp
```
Basic Usage of SymPy
Let’s start by using SymPy to define symbolic variables and perform some basic algebraic manipulations.
Defining Symbolic Variables
You can define symbolic variables using the symbols function. For example:
```
import sympy as sp
Define symbols
x, y, z = sp.symbols('x y z')
Display the symbols
print(x, y, z)
```
Simple Algebraic Operations
Once you’ve defined symbolic variables, you can perform various algebraic operations such as addition, multiplication, and exponentiation.
```
Define an expression
expr = x2 + 2x + 1
Simplify the expression
simplified_expr = sp.simplify(expr)
print("Simplified expression:", simplified_expr)
Expand a binomial
expanded_expr = sp.expand((x + 1)2)
print("Expanded expression:", expanded_expr)
Factor an expression
factored_expr = sp.factor(x2 + 2x + 1)
print("Factored expression:", factored_expr)
```
Simplification and Expansion
SymPy has built-in functions for simplifying and expanding expressions:
```
Simplify the expression x2 - 2x + 1 - (x - 1)2
expr_to_simplify = x2 - 2x + 1 - (x - 1)2
simplified = sp.simplify(expr_to_simplify)
print("Simplified expression:", simplified)
```
Solving Equations
SymPy makes it easy to solve equations symbolically. Here’s how to solve a linear equation:
```
Solve x + 3 = 7
solution = sp.solve(x + 3 - 7, x)
print("Solution:", solution)
```
For solving quadratic equations, SymPy will return all possible roots:
```
Solve a quadratic equation
quadratic_solution = sp.solve(x2 - 5x + 6, x)
print("Quadratic solution:", quadratic_solution)
```
Calculus Operations
SymPy excels in symbolic calculus. Let’s look at some common operations, such as differentiation and integration.
Differentiation
To compute the derivative of an expression:
```
Derivative of x 3 + 2x 2 + x with respect to x
derivative = sp.diff(x 3 + 2x 2 + x, x)
print("Derivative:", derivative)
```
Integration
To compute the integral of an expression:
```
Integral of x2 with respect to x
integral = sp.integrate(x2, x)
print("Integral:", integral)
```
Limits
You can also compute limits of expressions:
```
Limit of (sin(x)/x) as x approaches 0
limit = sp.limit(sp.sin(x)/x, x, 0)
print("Limit:", limit)
```
Series Expansion
SymPy allows you to expand an expression as a series around a given point:
```
Series expansion of exp(x) around x = 0
series_expansion = sp.series(sp.exp(x), x, 0, 5)
print("Series expansion:", series_expansion)
```
Solving Differential Equations
SymPy also provides tools for solving ordinary differential equations (ODEs) symbolically.
```
Solve the differential equation y'' - 2y' + y = 0
y = sp.Function('y')
ode_solution = sp.dsolve(y(x).diff(x, x) - 2y(x).diff(x) + y(x), y(x))
print("Differential equation solution:", ode_solution)
```
Linear Algebra with SymPy
SymPy has robust support for linear algebra operations. Let’s see how to perform matrix operations and solve linear systems.
Matrix Operations
```
Define matrices
A = sp.Matrix()
B = sp.Matrix()
Matrix addition
C = A + B
print("Matrix addition:\n", C)
Matrix multiplication
D = A B
print("Matrix multiplication:\n", D)
Determinant of a matrix
det_A = A.det()
print("Determinant of A:", det_A)
Eigenvalues and eigenvectors
eigen = A.eigenvals(), A.eigenvects()
print("Eigenvalues and eigenvectors:", eigen)
```
Solving a Linear System
You can solve systems of linear equations symbolically using the linsolve function.
```
Solve the system of equations: x + y = 2, x - y = 0
system = [x + y - 2, x - y]
solution = sp.linsolve(system, x, y)
print("Solution to the system:", solution)
```
Plotting with SymPy
SymPy includes basic plotting capabilities for visualizing symbolic expressions. You can use the plot function for this purpose:
```
from sympy.plotting import plot
Plot the function sin(x)
plot(sp.sin(x))
```
Advanced Features of SymPy
SymPy has many more advanced features such as:
Mathematical functions : Special functions like gamma, Bessel, and error functions.
Solving Diophantine equations : SymPy can handle integer solutions for certain equations.
SymPy physics module : For symbolic calculations in physics, including mechanics, optics, and quantum mechanics.
Symbolic Fourier transforms : Useful for signal processing and other applications.
Code generation : Generate Python, C, Fortran, or other language code for symbolic expressions.
Applications in Various Fields
SymPy finds applications in numerous domains:
Physics : Quantum mechanics calculations and equation derivations
Engineering : Solving complex mathematical models
Computer Science : Algorithm analysis and optimization
Finance : Option pricing and risk analysis
Code Generation
One of SymPy’s powerful features is its ability to generate code in various programming languages:
```
from sympy.utilities.codegen import codegen
expr = x2 + sin(x) [(c_name, c_code), (h_name, c_header)] = codegen(('f', expr), 'C', 'file', header=False, empty=False)
print(c_code)
```
This feature allows users to convert symbolic expressions into efficient, executable code for numerical computation.
The Jupyter Notebook with the outputs is available on the GitHub repository. You can run the Jupyter Notebook on Colab following the instructions on the GitHub repository.
Conclusion
SymPy is a powerful tool for symbolic mathematics in Python. It can simplify complex expressions, solve equations, perform calculus operations, and handle linear algebra tasks symbolically. Whether you are a researcher, educator, engineer, or student, SymPy provides an easy-to-use yet powerful framework for performing a wide variety of symbolic computations.
With its simple installation process, ease of use, and extensive capabilities, SymPy has become a popular choice in the Python ecosystem for symbolic mathematics. By using the examples in this article as a starting point, you can begin to explore the full range of features SymPy has to offer.
Originally published at
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188177 | https://static.hlt.bme.hu/semantics/external/pages/szingul%C3%A1ris_%C3%A9rt%C3%A9k-felbont%C3%A1s/en.wikipedia.org/wiki/Multiplicative_inverse.html | Multiplicative inverse - Wikipedia
Multiplicative inverse
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The reciprocal function: y = 1/x. For every x except 0, y represents its multiplicative inverse. The graph forms a rectangular hyperbola.
In mathematics, a multiplicative inverse or reciprocal for a number x, denoted by 1/x or x−1, is a number which when multiplied by x yields the multiplicative identity, 1. The multiplicative inverse of a fractiona/b is b/a. For the multiplicative inverse of a real number, divide 1 by the number. For example, the reciprocal of 5 is one fifth (1/5 or 0.2), and the reciprocal of 0.25 is 1 divided by 0.25, or 4. The reciprocal function, the function f(x) that maps x to 1/x, is one of the simplest examples of a function which is its own inverse (an involution).
The term reciprocal was in common use at least as far back as the third edition of Encyclopædia Britannica (1797) to describe two numbers whose product is 1; geometrical quantities in inverse proportion are described as reciprocall in a 1570 translation of Euclid's Elements.
In the phrase multiplicative inverse, the qualifier multiplicative is often omitted and then tacitly understood (in contrast to the additive inverse). Multiplicative inverses can be defined over many mathematical domains as well as numbers. In these cases it can happen that ab ≠ ba; then "inverse" typically implies that an element is both a left and right inverse.
The notation f−1 is sometimes also used for the inverse function of the function f, which is not in general equal to the multiplicative inverse. For example, the multiplicative inverse 1/(sin x) = (sin x)−1 is the cosecant of x, and not the inverse sine of x denoted by sin−1 x or arcsin x. Only for linear maps are they strongly related (see below). The terminology difference reciprocal versus inverse is not sufficient to make this distinction, since many authors prefer the opposite naming convention, probably for historical reasons (for example in French, the inverse function is preferably called bijection réciproque).
[x]
Contents
1 Examples and counterexamples
2 Complex numbers
3 Calculus
4 Algorithms
5 Reciprocals of irrational numbers
6 Further remarks
7 Applications
8 See also
9 Notes
10 References
Examples and counterexamples[edit]
In the real numbers, zerodoes not have a reciprocal because no real number multiplied by 0 produces 1 (the product of any number with zero is zero). With the exception of zero, reciprocals of every real number are real, reciprocals of every rational number are rational, and reciprocals of every complex number are complex. The property that every element other than zero has a multiplicative inverse is part of the definition of a field, of which these are all examples. On the other hand, no integer other than 1 and −1 has an integer reciprocal, and so the integers are not a field.
In modular arithmetic, the modular multiplicative inverse of a is also defined: it is the number x such that ax≡1(mod n). This multiplicative inverse exists if and only ifa and n are coprime. For example, the inverse of 3 modulo 11 is 4 because 4·3≡1(mod 11). The extended Euclidean algorithm may be used to compute it.
The sedenions are an algebra in which every nonzero element has a multiplicative inverse, but which nonetheless has divisors of zero, i.e. nonzero elements x, y such that xy= 0.
A square matrix has an inverse if and only if its determinant has an inverse in the coefficient ring. The linear map that has the matrix A−1 with respect to some base is then the reciprocal function of the map having A as matrix in the same base. Thus, the two distinct notions of the inverse of a function are strongly related in this case, while they must be carefully distinguished in the general case (as noted above).
The trigonometric functions are related by the reciprocal identity: the cotangent is the reciprocal of the tangent; the secant is the reciprocal of the cosine; the cosecant is the reciprocal of the sine.
A ring in which every nonzero element has a multiplicative inverse is a division ring; likewise an algebra in which this holds is a division algebra.
Complex numbers[edit]
As mentioned above, the reciprocal of every nonzero complex number z = a + bi is complex. It can be found by multiplying both top and bottom of 1/z by its complex conjugatez¯=a−b i{\displaystyle {\bar {z}}=a-bi} and using the property that z z¯=‖z‖2{\displaystyle z{\bar {z}}=\|z\|^{2}}, the absolute value of z squared, which is the real number a 2 + b 2:
1 z=z¯z z¯=z¯‖z‖2=a−b i a 2+b 2=a a 2+b 2−b a 2+b 2 i.{\displaystyle {\frac {1}{z}}={\frac {\bar {z}}{z{\bar {z}}}}={\frac {\bar {z}}{\|z\|^{2}}}={\frac {a-bi}{a^{2}+b^{2}}}={\frac {a}{a^{2}+b^{2}}}-{\frac {b}{a^{2}+b^{2}}}i.}
In particular, if ||z||=1 (z has unit magnitude), then 1/z=z¯{\displaystyle 1/z={\bar {z}}}. Consequently, the imaginary units, ±i, have additive inverse equal to multiplicative inverse, and are the only complex numbers with this property. For example, additive and multiplicative inverses of i are −(i) = −i and 1/i = −i, respectively.
For a complex number in polar form z = r(cos φ + i sin φ), the reciprocal simply takes the reciprocal of the magnitude and the negative of the angle:
1 z=1 r(cos(−φ)+i sin(−φ)).{\displaystyle {\frac {1}{z}}={\frac {1}{r}}\left(\cos(-\varphi )+i\sin(-\varphi )\right).}
Geometric intuition for the integral of 1/x. The three integrals from 1 to 2, from 2 to 4, and from 4 to 8 are all equal. Each region is the previous region scaled vertically down by 50%, then horizontally by 200%. Extending this, the integral from 1 to 2 k is k times the integral from 1 to 2, just as ln 2 k = k ln 2.
Calculus[edit]
In real calculus, the derivative of 1/x = x−1 is given by the power rule with the power −1:
d d x x−1=(−1)x(−1)−1=−x−2=−1 x 2.{\displaystyle {\frac {d}{dx}}x^{-1}=(-1)x^{(-1)-1}=-x^{-2}=-{\frac {1}{x^{2}}}.}
The power rule for integrals (Cavalieri's quadrature formula) cannot be used to compute the integral of 1/x, because doing so would result in division by 0:
∫1 x d x=x 0 0+C{\displaystyle \int {\frac {1}{x}}\,dx={\frac {x^{0}}{0}}\ +C}
Instead the integral is given by:
∫1 a 1 x d x=lna,{\displaystyle \int _{1}^{a}{\frac {1}{x}}\,dx=\ln a,}∫1 x d x=lnx+C.{\displaystyle \int {\frac {1}{x}}\,dx=\ln x+C.}
where ln is the natural logarithm. To show this, note that d d x e x=e x{\displaystyle {\frac {d}{dx}}e^{x}=e^{x}}, so if y=e x{\displaystyle y=e^{x}} and x=lny{\displaystyle x=\ln y}, we have:
d y d x=y⇒d y y=d x⇒∫1 y d y=∫1 d x⇒∫1 y d y=x+C=lny+C.{\displaystyle {\frac {dy}{dx}}=y\quad \Rightarrow \quad {\frac {dy}{y}}=dx\quad \Rightarrow \quad \int {\frac {1}{y}}\,dy=\int 1\,dx\quad \Rightarrow \quad \int {\frac {1}{y}}\,dy=x+C=\ln y+C.}
Algorithms[edit]
The reciprocal may be computed by hand with the use of long division.
Computing the reciprocal is important in many division algorithms, since the quotient a/b can be computed by first computing 1/b and then multiplying it by a. Noting that f(x)=1/x−b{\displaystyle f(x)=1/x-b} has a zero at x = 1/b, Newton's method can find that zero, starting with a guess x 0{\displaystyle x_{0}} and iterating using the rule:
x n+1=x n−f(x n)f′(x n)=x n−1/x n−b−1/x n 2=2 x n−b x n 2=x n(2−b x n).{\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}=x_{n}-{\frac {1/x_{n}-b}{-1/x_{n}^{2}}}=2x_{n}-bx_{n}^{2}=x_{n}(2-bx_{n}).}
This continues until the desired precision is reached. For example, suppose we wish to compute 1/17 ≈ 0.0588 with 3 digits of precision. Taking x 0 = 0.1, the following sequence is produced:
x 1 = 0.1(2 − 17 × 0.1) = 0.03 x 2 = 0.03(2 − 17 × 0.03) = 0.0447 x 3 = 0.0447(2 − 17 × 0.0447) ≈ 0.0554 x 4 = 0.0554(2 − 17 × 0.0554) ≈ 0.0586 x 5 = 0.0586(2 − 17 × 0.0586) ≈ 0.0588
A typical initial guess can be found by rounding b to a nearby power of 2, then using bit shifts to compute its reciprocal.
In constructive mathematics, for a real number x to have a reciprocal, it is not sufficient that x ≠ 0. There must instead be given a rational number r such that 0<r<|x|. In terms of the approximation algorithm described above, this is needed to prove that the change in y will eventually become arbitrarily small.
Graph of f(x) = x x showing the minimum at (1/e, e−1/e).
This iteration can also be generalised to a wider sort of inverses, e.g. matrix inverses.
Reciprocals of irrational numbers[edit]
Every number excluding zero has a reciprocal, and reciprocals of certain irrational numbers can have important special properties. Examples include the reciprocal of e (≈0.367879) and the golden ratio's reciprocal (≈0.618034). The first reciprocal is special because no other positive number can produce a lower number when put to the power of itself; f(1/e){\displaystyle f(1/e)} is the global minimum of f(x)=x x{\displaystyle f(x)=x^{x}}. The second number is the only positive number that is equal to its reciprocal plus one:φ=1/φ+1{\displaystyle \varphi =1/\varphi +1}. Its additive inverse is the only negative number that is equal to its reciprocal minus one:−φ=−1/φ−1{\displaystyle -\varphi =-1/\varphi -1}.
The function f(n)=n+(n 2+1),n∈N,n>0{\displaystyle f(n)=n+{\sqrt {(n^{2}+1)}},n\in \mathbb {N} ,n>0} gives an infinite number of irrational numbers that differ with their reciprocal by an integer. For example, f(2){\displaystyle f(2)} is the irrational 2+5{\displaystyle 2+{\sqrt {5}}}. Its reciprocal 1/(2+5){\displaystyle 1/(2+{\sqrt {5}})} is −2+5{\displaystyle -2+{\sqrt {5}}}, exactly 4{\displaystyle 4} less. Such irrational numbers share a curious property: they have the same fractional part as their reciprocal.
Further remarks[edit]
If the multiplication is associative, an element x with a multiplicative inverse cannot be a zero divisor (x is a zero divisor if some nonzero y, xy = 0). To see this, it is sufficient to multiply the equation xy = 0 by the inverse of x (on the left), and then simplify using associativity. In the absence of associativity, the sedenions provide a counterexample.
The converse does not hold: an element which is not a zero divisor is not guaranteed to have a multiplicative inverse. Within Z, all integers except −1, 0, 1 provide examples; they are not zero divisors nor do they have inverses in Z. If the ring or algebra is finite, however, then all elements a which are not zero divisors do have a (left and right) inverse. For, first observe that the map f(x) = ax must be injective: f(x) = f(y) implies x = y:
a x=a y⇒a x−a y=0⇒a(x−y)=0⇒x−y=0⇒x=y.{\displaystyle {\begin{aligned}ax&=ay&\quad \Rightarrow &\quad ax-ay=0\&&\quad \Rightarrow &\quad a(x-y)=0\&&\quad \Rightarrow &\quad x-y=0\&&\quad \Rightarrow &\quad x=y.\end{aligned}}}
Distinct elements map to distinct elements, so the image consists of the same finite number of elements, and the map is necessarily surjective. Specifically, ƒ (namely multiplication by a) must map some element x to 1, ax = 1, so that x is an inverse for a.
Applications[edit]
The expansion of the reciprocal 1/q in any base can also act as a source of pseudo-random numbers, if q is a "suitable" safe prime, a prime of the form 2 p+1 where p is also a prime. A sequence of pseudo-random numbers of length q−1 will be produced by the expansion.
See also[edit]
Division (mathematics)
Fraction (mathematics)
Group (mathematics)
Ring (mathematics)
Division algebra
Exponential decay
Unit fractions – reciprocals of integers
Hyperbola
Repeating decimal
List of sums of reciprocals
Six-sphere coordinates
Notes[edit]
^" In equall Parallelipipedons the bases are reciprokall to their altitudes". OED "Reciprocal" §3a. Sir Henry Billingsley translation of Elements XI, 34.
^Anthony, Dr. "Proof that INT(1/x)dx = lnx". Ask Dr. Math. Drexel University. Retrieved 22 March 2013.
^Mitchell, Douglas W., "A nonlinear random number generator with known, long cycle length," Cryptologia 17, January 1993, 55–62.
References[edit]
Maximally Periodic Reciprocals, Matthews R.A.J. Bulletin of the Institute of Mathematics and its Applications vol 28 pp 147–148 1992
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188178 | http://brennen.caltech.edu/fluidbook/basicfluiddynamics/potentialflow/singularities/cylinder.pdf | An Internet Book on Fluid Dynamics Potential Flow around a Cylinder Superimposing a uniform stream of velocity, U, on the potential flow due a doublet oriented in the x Figure 1: Streamlines in the potential flow of a doublet in a uniform stream.
direction produces the flow and streamlines shown in Figure 1 which has the velocity potential, φ = Ux + Qa π x (x2 + y2) = Ur + Qa πr cos θ (Bgdh1) and the radial velocity ur = ∂φ ∂r = U −Qa πr2 cos θ (Bgdh2) where the polar coordinates, r and θ, are x = r cos θ and y = r sin θ as before. It follows that at the specific radius, r = R = (Qa/πU) 1 2 the radial velocity is zero for all θ. Therefore the radius, r = R, is a streamline and could, if so desired, be replaced by a cylinder of that radius in order to generate the potential flow around a cylinder (shown by the red circle in Figure 1). It follows that the potential flow around a cylinder of radius R is characterized by the velocity potential, velocity components and stream function given by φ = r + R2 r U cos θ (Bgdh3) ur = ∂φ ∂r = 1 −R2 r2 U cos θ (Bgdh4) uθ = 1 r ∂φ ∂θ = − 1 + R2 r2 U sin θ (Bgdh5) ψ = r −R2 r U sin θ (Bgdh6) It will be useful for future purposes to investigate some of the properties of this flow. First note that the tangential velocity on the surface of the cylinder is given by (uθ)r=R = −2U sin θ (Bgdh7) As expected the velocity on the surface increases from zero at the front stagnation point (θ = π) to a maximum of 2U at the ”equator” (θ = π/2) and then decreases again to zero at the rear stagnation point (θ = 0). By Bernoulli’s theorem it follows that, neglecting gravity, the pressure, p, on the surface of the cylinder is given by (p)r=R = p∞+ 1 2ρU2 −1 2 {2U sin θ}2 = p∞+ 1 2ρU2 1 −4 sin2 θ (Bgdh8) where p∞is the pressure far upstream.
It is conventional to define a non-dimensional coefficient of pressure denoted by Cp as Cp = p −p∞ 1 2ρU2 (Bgdh9) and it follows from equation (Bgdh8) that the coefficient of pressure on the surface of the cylinder is given by (Cp)r=R = 1 −4 sin2 θ (Bgdh10) Since the pressure on the surface of the cylinder, (p)r=R, is symmetric fore and aft it must follow that the drag on the cylinder in potential flow is identically zero. This is, again, an example of D’Alembert’s Paradox which states that the drag on any finite body due to potential flow must be zero. We shall revisit this issue in future pages and resolve the apparent conflict with our practical experience. For the present we take note of the sinusoidal pressure distribution on the surface cylinder depicted in Figure 2. Note Figure 2: Pressure distribution on the surface of a cylinder in the potential flow.
that Cp at the front and rear stagnation points is unity as it always is where the velocity is zero. Note the minimum pressure coefficient of −3 at the equator, θ = π/2 and the symmetry of the potential flow pressure distribution that leads to zero drag. Though we jump ahead, we should mention that in the actual flow around a cylinder the pressure over the front between θ = 0 and θ = π/2 is quite close to that of the potential flow. However, the pressure over the rear departs substantially from the potential flow. In practice the main flow leaves the surface at points like SL or ST and, from that point on, the pressure is much lower than the potential flow pressure. This means that the actual drag is far from zero. We will describe and explain these features in more detail on later pages. |
188179 | https://ocw.mit.edu/courses/2-25-advanced-fluid-mechanics-fall-2013/85cfc32996ed711783a020e4569bbbe6_MIT2_25F13_Solution6.4.pdf | MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Kundu & Cohen 6.4 This problem is from “Fluid Mechanics” by P. K. Kundu and I. M. Cohen (a) Take a plane source of strength m at point (−a, 0), a plane sink of equal strength at (a, 0), and superpose a uniform stream U directed along the x-axis. (b) Show that there are two stagnation points located on the x-axis at points 1/2 ±a m + 1 . πaU (c) Show that the streamline passing through the stagnation points is given by ψ = 0. Verify that the line ψ = 0 represents a closed oval-shaped body, whose maximum width h is given by the solution of the equation πUh h = a cot . m The body generated by the superposition of a uniform stream and a source-sink pair is called a Rankine body. It becomes a circular cylinder as the source–sink pair approach each other. c 2.25 Advanced Fluid Mechanics 1 Copyright @ 2010, MIT Potential Flow Kundu & Cohen 6.4 Solution: (a) W (z) = Wuniform flow + Wsource + Wsink where W = φ + iψ, φ is the potential function, and ψ the stream function. Recap from Lecture: W satisfies the Laplace equation which is linear. Therefore, one can superimpose its solutions as above. Wuniform flow = U∞(x + iy) m m mθ iθ Wsource = ln re = ln r + i 2π 2π 2π m m mθ iθ Wsink = − ln re = − ln r + i 2π 2π 2π Substitute expressions for r and θ in terms of x and y (see figure): _ m m mθ r = (x − x0)2 + (y − y0)2 Wsource = ln re iθ = ln r + i 2π 2π 2π y − y0 m m mθ θ = arctan iθ Wsink = − ln re = − ln r + i x − x0 2π 2π 2π 2 m (x + a)2 + y m y m y ⇒ Wtotal = U∞x + ln +i U∞y + arctan − arctan 4π (x − a)2 + y2 2π x + a 2π x − a φ ψ (b) Obtain the velocity field (vx, vy) by invoking v = ∇φ 2 2 ∂φ m (x − a)2 + y 2(x + a) (x + a)2 + y vx = = U∞ + · − · 2(x − a) 2 ∂x 4π (x + a)2 + y2 (x − a) 2 + y 2 [(x − a)2 + y2] m x + a x − a ⇒ vx = U∞ + − 2π (x + a)2 + y2 (x − a)2 + y2 (x − a)2 + y ∂φ m 2 2y (x + a)2 + y2 vy = = · − · 2y 2 ∂y 4π (x + a)2 + y2 (x − a)2 + y 2 [(x − a)2 + y2] my 1 1 ⇒ vy = − 2π (x + a)2 + y2 (x − a)2 + y2 r x0 x y0 y θ location of source/sink arbitrary point (x, y) Alternatively, one can find v by using: vx = ∂ψ ∂y , vy = − ∂ψ ∂x . Find the stagnation point(s) by finding (x, y) such that vx = vy vy = 0 at y = 0, ∀x = 0. c 2.25 Advanced Fluid Mechanics 2 Copyright @ 2010, MIT Potential Flow Kundu & Cohen 6.4 Plug in y = 0 into vx and find x that lets vx = 0: m x + a x − a vx(x, y = 0) = U∞ + − = 0 2π (x + a)2 (x − a)2 m 1 1 U∞ + − = 0 2π x + a x − a 2 − ma OR (after some algebra. . . ) : x 2 − a = 0 πU∞ Using the quadratic formula1 , x = ±a 1 + m aπU∞ (c) Going back to ψ: θ1 θ2 m y m y ψ = U∞y + arctan − arctan 2π x + a 2π x − a m 2ay = U∞y − arctan 2π x2 + y2 − a2 −(θ1−θ2) A “Rankine oval” is defined by a curve ψ = 0, or m 2ay U∞y − arctan = 0 (1) 2π x2 + y2 − a2 Maximum half-width, h, is obtained when x = 0: θ1 θ2 a a (x, y) m 2ah U∞h = 2π arctan h2 − a2 2πU∞h h2 − a2 m = arccot 2ah −a a h ⇒ h 1 − a 2 = 2a cot 2πU∞h (2) h m D Problem Solution by Sungyon Lee, Fall 2005 √ −b± b2−4ac ax2 + bx + c = 0, x = 2a c 2.25 Advanced Fluid Mechanics 3 Copyright @ 2010, MIT 1 Potential Flow Kundu & Cohen 6.4 Figure 1: MATLAB® plot of streamlines for a Rankine oval. c 2.25 Advanced Fluid Mechanics 4 Copyright @ 2010, MIT MIT OpenCourseWare 2.25 Advanced Fluid Mechanics Fall 2013 For information about citing these materials or our Terms of Use, visit: |
188180 | https://www.bcbst.com/docs/providers/B-Hemolytic-Streptococcus-Testing-G2159.pdf | Policy Reimbursement Policy This document has been classified as public information Page 1 of 23 β-Hemolytic Streptococcus Testing POLICY DESCRIPTION | INDICATIONS AND/OR LIMITATIONS OF COVERAGE | TABLE OF TERMINOLOGY | SCIENTIFIC BACKGROUND | GUIDELINES AND RECOMMENDATIONS | APPLICABLE STATE AND FEDERAL REGULATIONS | APPLICABLE CPT/HCPCS PROCEDURE CODES | EVIDENCE-BASED SCIENTIFIC REFERENCES I. Policy Description Streptococcus are Gram-positive, catalase-negative bacteria that are further divided into α-hemolytic, such as S. pneumoniae and S. mutans; β-hemolytic, such as S. pyogenes (Group A), S. agalactiae (Group B), and S. dysgalactiae subsp equisimilis (Groups C and G); and γ-hemolytic, such as Enterococcus faecalis and E. faecium (Wessels, 2022). Streptococcal infections can be manifested in a variety of pathologies, including cutaneous infections, pharyngitis, acute rheumatic fever, pneumonia, postpartum endometritis, and toxic shock syndrome to name a few. Streptococcal infections can be identified using bacterial cultures obtained from blood, saliva, pus, mucosal, and skin samples as well as rapid antigen diagnostic testing (RADT) and nucleic acid-based methodologies (Chow, 2022; Wessels, 2022). For prenatal screening of Group B Streptococcus, please review policy AHS-G2035. II. Indications and/or Limitations of Coverage Application of coverage criteria is dependent upon an individual’s benefit coverage at the time of the request. Specifications pertaining to Medicare and Medicaid can be found in the “Applicable State and Federal Regulations” section of this policy document. 1) For the detection of a streptococcal infection causing respiratory illness, bacterial culture testing from a throat swab MEETS COVERAGE CRITERIA when one of the following conditions is met: a) When the individual has a modified Centor criteria score of 3 or greater (see Note 1 below). b) When the individual is suspected of having bacterial pharyngitis in the absence of viral features, (e.g., cough, oral ulcers, rhinorrhea). c) Following a negative rapid antigen diagnostic test (RADT) in a symptomatic child or adolescent. 2) Blood culture testing for a streptococcal infection MEETS COVERAGE CRITERIA when one of the following conditions is met: a) For individuals who fail to demonstrate clinical improvement and in those who have progressive symptoms or clinical deterioration after initiation of antibiotic therapy. b) For individuals who have progressive symptoms or clinical deterioration after the initiation of antibiotic therapy. c) In cases of suspected prosthetic joint infection. Policy Reimbursement Policy This document has been classified as public information Page 2 of 23 3) In cases of skin and/or soft tissue infections, bacterial culture testing for a streptococcal infection from a skin swab or from pus MEETS COVERAGE CRITERIA. 4) For individuals with suspected acute rheumatic fever (ARF) or post-streptococcal glomerulonephritis (PSGN), the following testing MEETS COVERAGE CRITERIA: a) Serological titer testing. b) Anti-streptolysin O immunoassay. c) Hyaluronidase activity or anti-hyaluronidase immunoassay. d) Streptokinase activity or anti-streptokinase immunoassay. 5) In cases of suspected viral pharyngitis, bacterial culture testing for streptococci from a throat swab DOES NOT MEET COVERAGE CRITERIA. 6) Except in cases of asymptomatic children under the age of three years who have a mitigating circumstance (including a symptomatic family member), RADT for a streptococcal infection DOES NOT MEET COVERAGE CRITERIA in any of the following situations: a) As a follow-up test for individuals who have had bacterial culture test for a streptococcal infection. b) As a screening method in an asymptomatic patient. c) For individuals with suspected viral pharyngitis. 7) For all situations not descried above, serological titer testing DOES NOT MEET COVERAGE CRITERIA. 8) Simultaneous ordering of both direct probe and amplification probe for the same organism in a single encounter DOES NOT MEET COVERAGE CRITERIA. The following does not meet coverage criteria due to a lack of available published scientific literature confirming that the test(s) is/are required and beneficial for the diagnosis and treatment of an individual’s illness. 9) For all situations not described above, testing with an anti-streptolysin O immunoassay, a hyaluronidase activity or anti-hyaluronidase immunoassay, or a streptokinase activity or anti-streptokinase immunoassay DOES NOT MEET COVERAGE CRITERIA. 10) For all situations the following tests DO NOT MEET COVERAGE CRITERIA: a) Panel tests that screen and identify multiple streptococcal strains (S. pyogenes [group A], S. agalactiae [group B], S. dysgalactiae [groups C/G], a-hemolytic streptococcus, and/or g-hemolytic streptococcus), using either immunoassay or nucleic acid-based assays (e.g., Solana Strep Complete Assay, Lyra Direct Strep Assay). Policy Reimbursement Policy This document has been classified as public information Page 3 of 23 b) MALDI-TOF identification of streptococcus. c) The quantification of any strain of streptococcus using nucleic acid amplification, including PCR. d) Nicotinamide-adenine dinucleotidase activity or anti-nicotinamide-adenine immunoassay. Note: Centor criteria includes tonsillar exudates, tender anterior cervical lymphadenopathy, fever, and absence of cough with each criterion being worth one point (Chow, 2023). III. Table of Terminology Term Definition AAOS American Academy of Orthopaedic Surgeons AAP American Association of Pediatrics ACOG American College of Obstetricians and Gynecologists ADB Anti-DNase B AHA American Heart Association ARF Acute rheumatic fever ASK Anti-streptokinase ASM American Society for Microbiology ASO Anti-streptolysin O ATS American Thoracic Society C3 Complement component 3 CAP Community-acquired pneumonia CDC Centers for Disease Control and Prevention CMS Centers for Medicare and Medicaid Services CNS Central nervous system CSF Cerebrospinal fluid DNA Deoxyribose nucleic acid DNases Deoxyribonucleases EIA Enzyme immunoassays EOS Early-onset bacterial sepsis FDA Food and Drug Administration GAS Group A Streptococcus GBS Group B Streptococcus GCS Group C Streptococcus GGS Group G Streptococcus HDA Helicase-dependent amplification ICSI Institute for Clinical Systems Improvement IDSA Infectious Diseases Society of America Policy Reimbursement Policy This document has been classified as public information Page 4 of 23 LDTs Laboratory developed Tests LR Likelihood ratio MALDI-TOF Matrix-assisted laser desorption/ionization-Time of flight NAAT Nucleic acid amplification test NADase Nicotinamide adenine dinucleotidase NADTs Rapid antigen detection tests NICE National Institute for Health and Care Excellence OIA Optical immunoassays PCR Polymerase chain reaction PIDS Pediatric Infectious Diseases Society PJI Prosthetic joint infection POC Point of care PSGN Post-streptococcal glomerulonephritis PYR Pyrrolidonyl aminopeptidase qPCR Quantitative PCR RADT Rapid antigen diagnostic testing RIDT Rapid in vitro diagnostic tests RNA Ribonucleic acid RNATs Rapid nucleic acid tests rt-PCR Real-time polymerase chain reaction SDSE Streptococcus dysgalactiae subspecies equisimilis TSA Trypticase soy agar IV. Scientific Background Bacterial acute pharyngitis is caused most often by a Group A Streptococcus (S. pyogenes or GAS), accounting for 5-15% of all acute pharyngitis cases in adults. Group C or Group G Streptococcus (S. dysgalactiae subsp equisimilis or GCS/GGS) is believed to be a causative agent in 5-10% of the cases of pharyngitis; however, “pharyngitis cause group C or G Streptococcus is clinically indistinguishable from GAS pharyngitis” but is more common in young adults and college students (Chow, 2023). “Diagnosis of infection due to group C streptococci (GCS) and group G streptococci (GGS) depends on identification of the organism in a culture from a clinical specimen. In general, a positive culture from a normally sterile site, such as blood, synovial fluid, or cerebrospinal fluid (CSF), can be considered definitive evidence of infection in the setting of a compatible clinical syndrome. The interpretation of positive cultures for GCS or GGS from the pharynx or from cutaneous sites such as open ulcers or wounds is less straightforward since asymptomatic colonization of the upper airway and skin also occurs (Wessels, 2022). GAS occurs most frequently in the very young and the elderly; although, GAS infections can occur in any age-group. The rates of severe GAS infections have been increasing in the United States as well as in other developed nations (Schwartz et al., 1990). Policy Reimbursement Policy This document has been classified as public information Page 5 of 23 The Centor criteria can be used to gauge the likelihood of pharyngitis due to a GAS infection. The four components of the Centor criteria are tonsillar exudates, tender anterior cervical lymphadenopathy, fever, and absence of cough with each criterion being worth one point. Patients who score less than three according to the Centor criteria are unlikely to have pharyngitis due to GAS and do not require strep testing or antibiotics; patients scoring ≥3 can be tested for GAS pharyngitis (Chow, 2023) Group A Streptococcus is associated with bacterial pharyngitis, scarlet fever, acute rheumatic fever, and post-streptococcal glomerulonephritis. Group A strep pharyngitis presents as a sudden onset of sore throat with odynophagia and fever; it is commonly referred to as “strep throat.” In children, additional symptoms can include abdominal pain, nausea, and vomiting. Viral pharyngitis, which accounts for more than 80% of pharyngitis, typically presents with cough, rhinorrhea, hoarseness, oral ulcers, and conjunctivitis unlike GAS pharyngitis. Rare cases of mucopurulent rhinitis caused by GAS has been reported in children under the age of three (CDC, 2022b). Scarlet fever can accompany strep throat. Besides the typical erythematous rash that typically begins on the trunk before spreading outward, scarlet fever can also present as a flushed face, “and the area around the mouth may appear pale (i.e., circumoral pallor).” “Strawberry tongue” can occur due to “yellowish white coating with red papillae” (CDC, 2022d). Scarlet fever is more easily transmitted than asymptomatic carriers through saliva and nasal secretions. Acute Rheumatic Fever (AFR), besides the characteristic fever, can affect the cardiovascular system (carditis and valvulitis), the musculoskeletal system (arthritis), the integumentary system (subcutaneous nodules and erythema marginatum), and the central nervous system (chorea). “Inadequate or lack of antibiotic treatment of streptococcal pharyngitis increases the risk of someone developing acute rheumatic fever. In approximately one-third of patients, acute rheumatic fever follows subclinical streptococcal infections or infections for which medical attention was not sought” (CDC, 2022a). Post-streptococcal glomerulonephritis (PSGN) presents with edema, hypertension, proteinuria, macroscopic hematuria, lethargy, and, at times, anorexia. “Laboratory examination usually reveals mild normocytic normochromic anemia, slight hypoproteinemia, elevated blood urea nitrogen and creatinine, elevated erythrocyte sedimentation rate, and low total hemolytic complement and C3 complement.” Urine output is usually decreased, and urine examination “often reveals protein (usually <3 grams per day) and hemoglobin with red blood cell casts” (CDC, 2022c). The virulence factors of GAS include M proteins, a group of more than 80 known proteins that protein the bacteria against phagocytosis; streptolysin O, a thiol-activated cytolysin; hyaluronidase, which hydrolyzes hyaluronic acid within the host tissue; streptokinase, an enzyme that activates plasmin; nicotinamide-adenine dinucleotidase (NADase), a glycohydrolase of uncertain function; and deoxyribonucleases (DNases) A, B, C, and D. Streptolysin O bind to the eukaryotic membrane’s cholesterol to facilitate the characteristic cellular lysis of a GAS infection. Cholesterol and anti-streptolysis O (ASO) antibodies can mitigate streptolysin O damage, and ASO titers often increase following an infection with the peak occurring around four to five weeks post-infection. “Nonsuppurative complications such as rheumatic fever and poststreptococcal glomerulonephritis generally develop during the second or third week of illness… About 80 percent of patients with acute rheumatic fever or poststreptococcal glomerulonephritis demonstrate a rise in ASO titer; however, the degree of ASO titer elevation does not correlate with severity of disease. In patients with suspected rheumatic fever or glomerulonephritis but with an undetectable ASO titer, prompt testing for other antistreptococcal antibodies such as anti-DNase B (detectable for six to nine months following infection), streptokinase, and antihyaluronidase should be performed” (Stevens & Bryant, 2022). Policy Reimbursement Policy This document has been classified as public information Page 6 of 23 Acute rheumatic fever (ARF) can occur two to four weeks following GAS pharyngitis. The five major manifestations of ARF are carditis and valvulitis (up to 70% of patients exhibit this condition with ARF), arthritis (up to 66%), CNS system involvement (10-30%), subcutaneous nodules (0-10%), and erythema marginatum (<6%) (Steer & Gibofsky, 2022). A diagnosis of ARF is not predicated by confirmation of a preceding GAS infection; however, it is helpful, especially in diagnosing children and young adults with arthritis and/or carditis. Evidence of GAS should include either a positive throat culture, a positive RADT, or an elevated or rising titer of either ASO or anti-DNase B. These two antibodies are used frequently in clinical practice due to their high sensitivity in diagnosing streptococcal infections (Steer & Gibofsky, 2022; Steer et al., 2015). A study by Blyth and Robertson demonstrated that the sensitivity of using only a single antibody in the diagnosis of streptococcus ranged from 70.5-72.7%; however, the combination of ASO and anti-DNase B increased the specificity to 88.6% with a sensitivity of 95.5%. The addition of anti-streptokinase (ASK) did not increase either the sensitivity or specificity of testing (Blyth & Robertson, 2006). A study in Norway in 2013 show that necrotizing soft tissue infections can be caused by GAS or GGS/GCS. The mean annual incidence rate is 1.4 per 100,000. During the time period studied (2000-2009), 61 cases of necrotizing soft tissue infections in Norway were due to GAS while nine cases were due to GCS/GGS. “Our findings indicate a high frequency of streptococcal necrotizing fasciitis in our community. GCS/GGS infections contribute to the disease burden but differ from GAS cases in frequency and predisposing factors.” They note that “the GCS/GGS patients were older, had comorbidities more often and had anatomically more superficial disease than the GAS patients (Bruun et al., 2013).” A review in 2014 also noted the population most affected by GCS/GGS, but they note that “the case fatality in bacteremia has been reported to be 15-18%” (Rantala, 2014). Group B Streptococcus (GBS) is frequently found in human gastrointestinal tracts and genitalia and can be spread to the upper respiratory tract of newborns. In neonates, a GBS infections can cause bacteremia, pneumonia, meningitis, and sepsis. GBS can also cause complications in pregnancy, such as urinary tract infections and chorioamnionitis. GBS, in pregnant and postpartum women, is of special concern since it is implicated in up to 31% of cases of bacteremia without a focus, 8% of postpartum endometritis, and 2% of pneumonia; moreover, if left unchecked, GBS can also result in preterm labor and miscarriage. In the adult population at large, GBS infections can be manifest as soft tissue infections, sepsis, and bacteremia (Barshak, 2022; Puopolo et al., 2022). “Invasive disease in infants is categorized on the basis of chronologic age at onset. Early-onset disease usually occurs within the first 24 hours of life (range, 0 through 6 days) and is characterized by signs of systemic infection, respiratory distress, apnea, shock, pneumonia, and less often, meningitis (5%–10% of cases). Late-onset disease, which typically occurs at 3 to 4 weeks of age (range, 7 through 89 days), commonly manifests as occult bacteremia or meningitis (approximately 30% of cases); other focal infections, such as osteomyelitis, septic arthritis, necrotizing fasciitis, pneumonia, adenitis, and cellulitis, occur less commonly. Nearly 50% of survivors of early- or late-onset meningitis have long-term neurologic sequelae (encephalomalacia, cortical blindness, cerebral palsy, visual impairment, hearing deficits, or learning disabilities). Late, late-onset disease occurs at 90 days of age and beyond, usually in very preterm infants requiring prolonged hospitalization” (Pediatrics, 2018). Type of Testing Policy Reimbursement Policy This document has been classified as public information Page 7 of 23 Test Description Rationale Culture Cultures can be taken from a swab of the affected tissue when possible, such as the back of the throat and tonsils (1). The cultures are typically grown on a solid, complex rich medium such as Trypticase Soy Agar (TSA) supplemented with 5% sheep blood so that the zone of b-hemolysis can easily be visualized (2). Culture testing can be supplemented with additional conventional identification tests, such as the Lancefield antigen determination test and the PYR test (3). The CDC considers the throat culture the ‘gold standard’ (4). This testing method can be time-intensive. “Throat culture also can identify other bacteria that cause pharyngitis less commonly than GAS (eg, group C and group G streptococci, Arcanobacterium haemolyticum). However, most laboratories do not routinely identify these pathogens in throat cultures unless specifically requested to do so (5).” Serology Many possible serological tests can be performed, including a measurement of the antibody titers associated with a streptococcal infection. Virulence factors that can be monitored include hyaluronidase, streptokinase, nicotinamide-adenine dinucleotidase, DNase B, and streptolysin O. DNase B and streptolysin O are more frequently used in clinical practice (6). Anti-streptococcal antibody titers represent past infections and should not be used to routinely diagnose an acute infection (7). Antistreptolysin O (ASO) and/or anti-DNase B (ADB) testing can be used to determine prior streptococcal infection associated with disorders such as rheumatic fever and glomerulonephritis. “An increase in titer from acute to convalescent (at least two weeks apart) is considered the best evidence of antecedent GAS infection. The antibody response of ASO peaks at approximately three to five weeks following GAS pharyngitis, which usually is during the first to third week of ARF, while ADB titers peak at six to eight weeks (8).” Antibody titers are dependent on the age of the patients with children having considerably higher ‘normal’ levels than adults due to frequent exposure to S. pyrogenes (3). Rapid Antigen Diagnostic Testing (RADT) RADTs can be performed on a swab at the point of care or can be transported to a lab for testing (9). Numerous RADTs directly detect antigens through an agglutination method or the use of immunoassays, including enzyme-based assays, optical assays, and liposome-Many RADTs are commercially available but can vary considerably in specificity, sensitivity, and ease of use. “In pediatric patients, if the direct antigen test is negative, and if the direct antigen test is known to have a sensitivity of <80%, a second throat swab should be examined by a more Policy Reimbursement Policy This document has been classified as public information Page 8 of 23 based assays that are commercially available (3). sensitive direct NAAT or by culture as a means of arbitrating possible false-negative direct antigen test results. This secondary testing is not necessarily required in adults. A convenient means of facilitating this 2-step algorithm of testing for Streptococcus pyogenes in pediatric patients is to collect a dual swab initially, recognizing that the second swab will be discarded if the direct antigen test is positive (9).” Nucleic Acid Amplification Tests (NAATs) NAATs amplify DNA or RNA to detect the presence of microorganisms. Some are offered as point-of-care (POC) rapid diagnostic tests while others require special laboratory equipment (9). Some NAATs utilize real-time polymerase chain reaction (rt-PCR), such as the Lyra Direct Strep Assay, while others use a helicase-dependent amplification (HDA)-based methodology like the Solana Strep Complete assay. NAATs are often qualitative but specific NAATs can be quantitative. NAATs can vary in their selectivity, sensitivity, and ability to differentiate between strains of streptococci. More sensitive than antibody-based testing for streptococcus. Direct NAATs usually require the use of enriched broth cultures. “Negative direct NAAT results do not have to be arbitrated by a secondary test (9).” Matrix-Assisted Laser Desorption Ionization-Time of Flight (MALDI-TOF) MALDI-TOF mass spectrometry can be used to quickly identify both gram-negative and gram-positive bacteria once the organism is available in a pure culture on solid medium. The results of the MALDI-TOF test is compared to a known database of spectra of microorganisms for identification (10). “For less common organisms, the MALDI-TOF result may not be conclusive, and additional bench tests or molecular tests may be required (10).” (1) (AACC, 2021);(2) (Gera & McIver, 2013); (3) (Spellerberg & Brandt, 2016); (4) (CDC, 2022b); (5) (Wald, 2022); (6) (Stevens & Bryant, 2022); (7) (Shulman et al., 2012); (8) (Steer & Gibofsky, 2022); (9) (Miller et al., 2018); (10) (Freeman & Roberts, 2023). Clinical Utility and Validity Policy Reimbursement Policy This document has been classified as public information Page 9 of 23 Rapid in vitro diagnostic tests (RIDT), such as the Alere I Strep A, have been CLIA-waived by the FDA. These tests provide results more quickly than the traditional “gold standard” bacterial culture testing. A 2018 study comparing rapid antigen GAS testing, the Alere I Strep A test—an RIDT using isothermal nucleic acid amplification, and throat cultures. “The sensitivity and specificity of the molecular test were 98% and 100%, respectively, compared with culture. There was a 9% false-positive rate with the rapid antigen-based testing…. The Alere test is sufficiently sensitive and specific for definitive GAS testing in a pediatric urgent care setting” (Weinzierl et al., 2018). In 2016, Cohen et al., extensively reviewed the use of rapid antigen detection tests (RADT) for GAS in children. They reviewed 98 unique studies consisting of a total of 101,121 participants and compared both major types of RADTs—enzyme immunoassays (EIA) and optical immunoassays (OIA). “RADT had a summary sensitivity of 85.6%...There was substantial heterogeneity in sensitivity across studies; specificity was more stable. There was no trade-off between sensitivity and specificity….The sensitivity of EIA and OIA tests was comparable (summary sensitivity 85.4% versus 86.2%)… Based on these results, we would expect that amongst 100 children with strep throat, 86 would be correctly detected with the rapid test while 14 would be missed and not receive antibiotic treatment” (Cohen et al., 2016). Another multicenter study using the Alere I Strep A test on cultures obtained from 481 patients of all ages show that the RIDT had 96.0% sensitivity and 94.6% specificity. The authors conclude that this “could provide a one-step, rapid, point-of-care testing method for GAS pharyngitis and obviate backup testing on negative results” (Cohen et al., 2015). This study did note that there are newer tests available that have higher sensitivity, but these tests require more time than the Alere I Strep A method. Due to the time constraints of clinical laboratories and the variability of RADTs, nucleic acid amplification test (NAAT) use has been increasing in clinical settings. The FDA has approved multiple NAATs for the detection of Streptococcus. The Lyra Direct strep assay is an FDA-approved, NAAT that uses real-time PCR to qualitatively detect the presence of GAS and GGS/GCS in throat swab samples. It should be noted, though, that this assay does not distinguish between GGS and GCS. A study by Boyanton et al., (2016) evaluated the efficacy of the Lyra Direct method as compared to the traditional, time-consuming culture test for GAS and GGS/GCS. The sample sizes were not large (n = 19 for GAS and n = 5 for GGS/GCS out of a total of 161 samples submitted); however, the Lyra Direct strep assay did correctly detect “all b-hemolytic streptococci...” and “in batch mode, the Lyra assay reduced intra-laboratory turnaround time by 60% (18.1 h versus 45.0 h) but increased hands-on time by 96% (3 min 16 s versus 1 min 40 s per specimen)” (Boyanton et al., 2016). The authors note that the RADTs “have largely augmented bacterial culture (the gold standard). However, the performance of commercially available [RADTs] varies greatly depending upon the manufacturer, methodology used (i.e., optical immunoassay, immunochromatographic, or enzyme immunoassay), and the patient population (i.e., pediatric versus adult) being tested. Due to these limitations, nucleic acid amplification tests (NAATs) are being implemented in clinical laboratories (Boyanton et al., 2016).” The Solana method is also an FDA-approved NAAT, but it uses a rapid helicase-dependent amplification (HDA) methodology. Solana is available for either GAS testing or as a panel testing for GAS, GCS, and GGS. A study by Uphoff et al., (2016) colleagues compared the Solana GAS testing to that of conventional culture testing. Their research used 1082 throat swab specimens. The traditional culture tested positive in 20.7% of the samples as compared to 22.6% positive values in the HDA-based methodology. The Solana assay in their results had 98.2% sensitivity and 97.2% specificity. “In 35 min, the HDA method provided rapid, sensitive GAS detection, making culture confirmation unnecessary” (Uphoff et al., 2016). Recently, another study compared an HDA-based method to the Simplex GAS Direct PCR-based method, which is another FDA- Policy Reimbursement Policy This document has been classified as public information Page 10 of 23 approved diagnostic test. The Simplex GAS Direct method does not require initial DNA extraction from the sample, a potential time-saving benefit. The study used 289 throat swabs. The HDA- based method “compared to Simplexa qPCR had sensitivity, specificity, positive predictive value and negative predictive value of 93.1% vs 100%, 100% vs. 100%, 100% vs. 100% and 98.31% vs. 100% respectively… Simplexa qPCR has improved performance and diagnostic efficiency in a high-volume laboratory compared to [HDA-based method] for GAS detection in throat swabs” (Church et al., 2018). The Solana® Strep Complete Assay by Quidel received FDA clearance in 2016. According to Quidel’s FDA application, it is defined as “a rapid in vitro diagnostic test, using isothermal amplification technology (helicase-dependent amplification, HDA) for the qualitative detection and differentiation of Streptococcus pyogenes (Group A β-hemolytic Streptococcus) and Streptococcus dysgalactiae (pyogenic Group C and G β-hemolytic Streptococcus) nucleic acids isolated from throat swab specimens obtained from patients with signs and symptoms of pharyngitis, such as sore throat” (Lollar, 2016). This test must be performed using Quidel’s Solana proprietary equipment. According to the 510(k) application, the Solana Strep Complete Assay panel has a clinical sensitivity and specificity for GAS of 98.8% and 98.9%, respectively, as compared to the Lyra Direct Strep Assay’s reported 96.5% sensitivity and 98.0% specificity for GAS. The Lyra Direct Strep Assay is a real-time PCR-based assay that cannot differentiate between the pyogenic strains of streptococci. Concerning the pyrogenic GCS/GGS, the Solana Strep Complete Assay panel has a clinical sensitivity of 100% with a specificity of 99.5% as compared to Lyra Direct Strep Assay’s reported 95.7% sensitivity and 98.3% specificity for GCS/GGS strains. The reported testing time also varies between the two assays with Solana requiring 25 minutes versus 60-70 minutes for the Lyra Direct Strep Assay (Lollar, 2016). A recent study by Helmig and Gertsen (2017) evaluated the accuracy of PCR-based testing for GBS in pregnant women. Their study used rectovaginal swabs from 106 women in gestational weeks 35-37. For each, both a GBC culture and a PCR-based molecular GBS test (Xpert GBS of Cepheid Ltd) were performed. Only one PCR test yielded no result, so the invalid PCR-based test rate is <1%. 25/106 of the GBS cultures tested positive as compared to 27/105 of the PCR-based test. The specificity of the PCR-based test was 97.5% with a 100% sensitivity and a 92.6% positive predictive value. The authors conclude that “the PCR test has sufficient accuracy to direct intrapartum antibiotic prophylaxis for GBS transmission during delivery (Helmig & Gertsen, 2017).” A preliminary study in France of 1416 mothers with newborns compared swab cultures and GBS PCR assay for their predictive value of early-onset bacterial sepsis (EOS) in newborns since GBS is the most common cause of EOS. The results show that “the diagnostic values of the two tests highlighted a nonsignificant superiority of intrapartum GBS PCR assay” but that “the negative predictive value was improved with intrapartum PCR assay (negative likelihood ratio [LR]: 0.3 [0.1-0.9] vs. 0.6 [0.4-1.1])…. These results suggest that the intrapartum GBS PCR assay offers a better predictive value of GBS EOS that the usual vaginal culture swab at the 9th month but requires confirmation by large studies” (Raignoux et al., 2016). Luo et al., (2019) “evaluated the overall diagnosis and treatment of acute pharyngitis in the United States, including predictors of test type and antibiotic prescription.” Five categories of tests were identified, which were RADT [rapid antigen detection test], RADT plus culture, other tests, nucleic acid amplification testing (NAAT), and no test. Pharyngitis events from 2011-2015 were examined and a total of 18.8 million pharyngitis events across 11.6 million patients were included. Overall, 68.2% of events were found to occur once, with 29.1% requiring further follow-up. Furthermore, 43% of events were Policy Reimbursement Policy This document has been classified as public information Page 11 of 23 diagnosed by RADT and 20% were diagnosed by RADT plus culture. NAAT testing also increased 3.5-fold from 2011-2015 (going from 0.06% to 0.27%). Antibiotics were used in 49.3% of events as a whole. For RADT plus culture, antibiotics were used 31.2% of the time, for NAAT alone, 34.5%, for RADT alone, 54.2%, for no test, 57.1%. The authors concluded that “Diagnostic testing can help lower the incidence of inappropriate antibiotic use, and inclusion of NAAT in the clinical guidelines for GAS pharyngitis warrants consideration” (Luo et al., 2019). O. Luiz et al., (2019) evaluated the “prevalence and persistence of beta-haemolytic streptococci throat carriage and type the bacterial population.” A total of 121 children and 127 young adult volunteers contributed throat swabs (for culture), and these volunteers were screened quarterly for beta-haemolytic bacterial species. Carriage was detected in 34 volunteers (13.7%). Seventeen children were found to carry Group A Streptococcus, while seventeen young adults were found to carry four separate subspecies (Streptococcus dysgalactiae subsp. equisimilis (SDSE), Streptococcus pyogenes, Streptococcus agalactiae and the Streptococcus anginosus group). The authors also identified persistent carriage for as long as six months in two children and for as long as one year in three young adults. The authors concluded that “prevalence was slightly greater among children, but persistent carriage was greater among young adults, with SDSE being the species most associated with persistence” (O. Luiz et al., 2019). Fraser et al., (2020) performed a meta-analysis to assess the cost-effectiveness of point-of-care testing for detection of Group A Streptococcus. The authors remarked that this type of testing has seen increased use as an adjunct for managing care, such as for prescribing antibiotics. Thirty-eight studies of clinical effectiveness were included, along with three studies of cost-effectiveness. Twenty-six articles “reported on the test accuracy of point-of-care tests and/or clinical scores with biological culture as a reference standard”. Overall, 21 point-of-care tests were evaluated. The authors identified two populations of interest; “patients with Centor/McIsaac scores of ≥ 3 points or FeverPAIN scores of ≥ 4 points.” Test sensitivity for these populations ranged from 0.829-0.946 while test specificity ranged from 0.849-0.991. However, the authors did note there was significant heterogeneity and expressed doubts that any single study “accurately captured a test's true performance.” The authors developed an economic model to explore the cost-effectiveness of this type of testing, and 14 of the 21 tests were included in this model. Per the current National Institute for Health and Care Excellence's cost-effectiveness thresholds, these tests were not found to be cost-effective. The authors acknowledged significant uncertainties in the estimates, such as penalties for antibiotic over-prescriptions. The authors concluded that “the systematic review and the cost-effectiveness models identified uncertainties around the adoption of point-of-care tests in primary and secondary care settings. Although sensitivity and specificity estimates are promising, we have little information to establish the most accurate point-of-care test.” (Fraser et al., 2020; Kim et al., 2019) Bilir et al., (2021) studied the cost-effectiveness of point of care (POC) nucleic acid amplification tests (NAAT) for streptococcus in the US. Point of care NAAT was compared to rapid antigen detection tests (RADT) and culture. Costs, clinical effects, antibiotic complications, number of patients treated, and antibiotic utilization were studied. Analysis showed that the POC NAAT method would cost $44 per patient while RADT and culture would cost $78 per patient. "Compared with RADT + culture, POC NAAT would increase the number of appropriately treated patients and avert unnecessary use of antibiotics.” According to the results, “POC NAAT would be less costly and more effective than RADT Policy Reimbursement Policy This document has been classified as public information Page 12 of 23 + culture; POC NAAT adoption may yield cost savings to US third-party payers. Access to POC NAAT is important to optimize GAS diagnosis and treatment decisions in the United States" (Bilir et al., 2021). In a metanalysis, Dubois et al., (2021) studied the diagnostic accuracy of rapid antigen detection tests (NADTs) vs rapid nucleic acid tests (RNATs) for diagnosis of group A streptococcal pharyngitis. A total of 38 studies using RNAT were included, with a sensitivity of 97.5% and specificity of 95.1%. RADTs had a sensitivity of 82.3%, but specificity was similar to the sensitivity of RNATs. Overall, RNATs were more sensitive than RADTs. The authors conclude that "the high diagnostic accuracy of RNATs may allow their use as stand-alone tests to diagnose group A streptococcus pharyngitis" (Dubois et al., 2021). McCarty et al., (2022) studied the clinical utility on the GenMark Dx ePlex® blood culture identification gram-positive panel. The panel results were evaluated and compared to MALDI-TOF mass spectrometry and traditional antimicrobial susceptibility testing. One hundred Gram-Positive bacteria were represented. “The positive percent agreement (PPA) was 97/97 with two false positives.” The study included chart reviews of 80 patients. The average time for organism identification was 24.4 hours faster and the average time for optimization was 29.2 hours faster for the eight patients identified with organisms such as streptococci. The authors “confirm high sensitivity and specificity of the FDA-cleared GenMark Dx ePlex BCID-GP Panel compared to MALDI-TOF MS on bacterial isolates and identify opportunities for earlier optimization of antimicrobial therapy that may also be accompanied by potential cost savings (McCarty et al., 2022). V. Guidelines and Recommendations Centers for Disease Control and Prevention Acute Pharyngitis (CDC, 2022b): Most cases of acute pharyngitis are viral. Only 20-30% of sore throats in children and 5-15% in adults are due to group A Streptococcus (GAS). History and clinical examination can be used to diagnosis viral pharyngitis when clear viral symptoms (e.g., cough, rhinorrhea, hoarseness, oral ulcers, conjunctivitis) are present; these patients do not need testing for group A strep. However, clinical examination cannot be used to differentiate viral and group A strep pharyngitis in the absence of viral symptoms, even for experienced clinicians. The diagnosis of group A strep pharyngitis is confirmed by either a rapid antigen detection test (RADT) or a throat culture. RADTs have high specificity for group A strep but varying sensitivities when compared to throat culture, which is considered the gold standard diagnostic test. Testing for group A strep pharyngitis is not routinely indicated for children younger than three years of age or for adults. Clinicians can use a positive RADT as confirmation of group A strep pharyngitis in children, though it also notes that a negative RADT should be followed with a throat culture in children with symptoms of pharyngitis (CDC, 2022b). The CDC also comments on asymptomatic Group A carriers, stating that these carriers usually do not require treatment. The CDC defines carriers as having “positive throat cultures or are RADT positive, but do not have clinical symptoms or an immunologic response to group A strep antigens on laboratory testing” (CDC, 2022b). Scarlet Fever (CDC, 2022d): Scarlet fever (scarlatina) consists of an erythematous rash caused by GAS and can occur along with acute pharyngitis. “The differential diagnosis of scarlet fever with pharyngitis Policy Reimbursement Policy This document has been classified as public information Page 13 of 23 includes multiple viral pathogens that can cause acute pharyngitis with a viral exanthema. Clinicians need to use either a rapid antigen detection test (RADT) or throat culture to confirm scarlet fever with pharyngitis. RADTs have high specificity for group A strep but varying sensitivities when compared to throat culture. Throat culture is the gold standard diagnostic test. Clinicians should follow up a negative RADT in a child with symptoms of scarlet fever with a throat culture. Clinicians should have a mechanism in place to contact the family and initiate antibiotics if the back-up throat culture is positive” (CDC, 2022d). Post-Streptococcal Glomerulonephritis (PSGN) (CDC, 2022c): PSGN is primarily due to a GAS infection, but rare cases of GCS-induced PSGN have been reported. Clinical features include edema, hypertension, proteinuria, macroscopic hematuria, and lethargy. As such, “The differential diagnosis of PSGN includes other infectious and non-infectious causes of acute glomerulonephritis. Clinical history and findings with evidence of a preceding group A strep infection should inform a PSGN diagnosis. Evidence of preceding group A strep infection can include: • Isolation of group A strep from throat • Skin lesions • Elevated streptococcal antibodies” (CDC, 2022c). Acute Rheumatic Fever (CDC, 2022a): “The differential diagnosis of acute rheumatic fever is broad due to the various symptoms of the disease. The differential diagnosis may include but is not limited to: rheumatoid arthritis, juvenile idiopathic arthritis, septic arthritis, systemic lupus erythematosus, serum sickness, Lyme disease, infective endocarditis, viral myocarditis, Henoch-Schonlein purpura, gout, sarcoidosis, leukemia, and Hodgkin’s disease.” The CDC notes that no definitive diagnostic test exists for acute rheumatic fever and recommends using the Jones criteria (endorsed by the American Heart Association) to make a clinical diagnosis, which “now includes the addition of subclinical carditis as a major criteria and stratification of the major and minor criteria based upon epidemiologic risk (e.g., low, moderate, or high risk populations)” (CDC, 2022a). American Association of Pediatrics (AAP) The AAP has published the Red Book (Kimberlin et al., 2021) as guidance for infectious diseases in the pediatric population. Their relevant comments and recommendations include: • “Children with pharyngitis and obvious viral symptoms (e.g., rhinorrhea, cough, hoarseness, oral ulcers) should not be tested or treated for GAS [Group A Streptococcus] infection; testing also generally is not recommended for children younger than 3 years.” • “Several rapid diagnostic tests for GAS pharyngitis are available…Specificities of these tests generally are high (very few false-positive results), but the reported sensitivities vary considerably (i.e., false-negative results occur).” • “The US Food and Drug Administration (FDA) has cleared a variety of rapid tests for use in home settings. Parents should be informed that home use is discouraged because of the risk of false-positive testing that represents colonization.” • “Because of the very high specificity of rapid tests, a positive test result does not require throat culture confirmation. Rapid diagnostic tests using techniques such as polymerase chain reaction Policy Reimbursement Policy This document has been classified as public information Page 14 of 23 (PCR), chemiluminescent DNA probes, and isothermal nucleic acid amplification tests have been developed…Some studies suggest that these tests may be as sensitive as standard throat cultures on sheep blood agar.” • “Children with manifestations highly suggestive of viral infection, such as coryza, conjunctivitis, hoarseness, cough, anterior stomatitis, discrete ulcerative oral lesions, or diarrhea, are very unlikely to have true GAS pharyngitis and should not be tested.” • “Testing children younger than 3 years generally is not indicated. Although small outbreaks of GAS pharyngitis have been reported in young children in child care settings, the risk of ARF is so remote in young children in industrialized countries that diagnostic studies for GAS pharyngitis generally are not indicated for children younger than 3 years.” • “In contrast, children with acute onset of sore throat and clinical signs and symptoms such as pharyngeal exudate, pain on swallowing, fever, and enlarged tender anterior cervical lymph nodes, without concurrent viral symptoms and/or exposure to a person with GAS pharyngitis, are more likely to have GAS infection and should have a rapid antigen test and a throat culture if the rapid test result is negative, with treatment initiated if a test result is positive.” • “Testing asymptomatic household contacts for GAS infection is not recommended except when the contacts are at increased risk of developing sequelae of GAS infection, such as ARF or acute glomerulonephritis; if test results are positive, such contacts should be treated.” • “Testing asymptomatic household contacts usually is not helpful. However, if multiple household members have pharyngitis or other GAS infections, simultaneous cultures of all household members and treatment of all with positive cultures or rapid antigen test results may be of value.” • “In suspected invasive GAS infections, cultures of blood and of focal sites of possible infection are indicated.” • “Laboratory evidence of antecedent GAS infection should be confirmed in all cases of suspected ARF [acute rheumatic fever], and evidence includes an increased or rising ASO or anti-DNAase B titer, or a positive rapid antigen or streptococcal throat culture. Because of the long latency between GAS infection and presentation with chorea, such laboratory evidence may be lacking in cases where chorea is the major criteria.” • “Post-treatment throat swab cultures are indicated only for patients who are at particularly high risk of ARF [acute rheumatic fever] (e.g., those living in an area with endemic infection).” Regarding the management of infants at risk of group B streptococcal disease, a list of recommendations was provided. The relevant points are included below: • “Early-onset GBS infection is diagnosed by blood or CSF culture. Common laboratory tests such as the complete blood cell count and C-reactive protein do not perform well in predicting early-onset infection, particularly among well-appearing infants at lowest baseline risk of infection.” • “Evaluation for late-onset GBS disease should be based on clinical signs of illness in the infant. Diagnosis is based on the isolation of group B streptococci from blood, CSF, or other normally sterile sites. Late-onset GBS disease occurs among infants born to mothers who had positive GBS screen results as well as those who had negative screen results during pregnancy. Adequate IAP does not protect infants from late-onset GBS disease” (Puopolo et al., 2019). American Heart Association (AHA) Policy Reimbursement Policy This document has been classified as public information Page 15 of 23 The AHA published a revision to the Jones criteria for diagnosis of acute rheumatic fever in 2015. In it, they note the importance of identifying laboratory evidence of a group A streptococcal infection. The AHA lists three clinical features that can serve as evidence for a preceding Group A Streptococcus infection, which are as follows: • “Increased or rising anti-streptolysin O titer or other streptococcal antibodies (anti-DNASE B). A rise in titer is better evidence than a single titer result.” • “A positive throat culture for group A β-hemolytic streptococci.” • “A positive rapid group A streptococcal carbohydrate antigen test in a child whose clinical presentation suggests a high pretest probability of streptococcal pharyngitis” (Gewitz et al., 2015). Institute for Clinical Systems Improvement (ICSI) In 2017, the ICSI updated their guidelines titled Diagnosis and treatment of respiratory illness in children and adults. They give the following consensus recommendation: “It is the consensus of the ICSI work group to NOT test for Group A Streptococcal (GAS) pharyngitis in patients with modified Centor criteria scores less than three or when viral features like rhinorrhea, cough, oral ulcers and/or hoarseness are present. Testing should generally be reserved for patients when there is a high suspicion for GAS and for whom there is intention to treat with antibiotics” (Short et al., 2017). The Centor criteria include age of patient, physical state of the tonsils and lymph nodes, temperature, and presence or absence of cough (Centor & McIsaac, 2022). American Thoracic Society (ATS) and Infectious Diseases Society of America (IDSA) The ATS and IDSA published a joint guideline on the diagnosis and treatment of community-acquired pneumonia in adults. The guideline notes that group A Streptococcus may be associated with influenza pneumonia. Their relevant recommendations are listed below: • “We recommend not obtaining sputum Gram stain and culture routinely in adults with CAP managed in the outpatient setting (strong recommendation, very low quality of evidence).” • “We recommend not obtaining blood cultures in adults with CAP managed in the outpatient setting (strong recommendation, very low quality of evidence)” (Metlay et al., 2019). Infectious Diseases Society of America (IDSA) The 2014 update of the IDSA’s guidelines concerning skin and soft tissue infections included a recommendation (strong; moderate-quality evidence) of “Gram stain and culture of the pus or exudates from skin lesions of impetigo and ecthyma are recommended to help identify whether Staphylococcus aureus and/or -hemolytic Streptococcus is the cause, but treatment without these studies is reasonable in typical cases.” They make a similar recommendation in the cases of pus from carbuncles and abscesses as well as pyomyositis; however, they do not recommend (strong, moderate) a “Gram stain and culture of pus from inflamed epidermoid cysts”. As for erysipelas and cellulitis, “cultures of blood or cutaneious aspirates, biopsies, or swabs are not routinely recommended (strong, moderate) …cultures of blood are recommended (strong, moderate), and cultures and microscopic examination of cutaneious aspirates, biopsies, or swabs should be considered in patients with malignancy on chemotherapy, neutropenia, Policy Reimbursement Policy This document has been classified as public information Page 16 of 23 severe cell-mediated immunodeficiency, immersion injuries, and animal bites (weak, moderate)” (Stevens et al., 2014.) The IDSA and the American Society for Microbiology (ASM) published a guideline in 2018 titled “A Guide to Utilization of the Microbiology Laboratory for Diagnosis of Infectious Diseases.” This guideline includes items on the laboratory diagnosis of pharyngitis, which are as follows: • For Streptococcus pyogenes, direct NAAT, nucleic acid probe tests, or a rapid direct antigen test (followed by a culture or NAAT test if negative) may all be performed. • For Groups C and G β-hemolytic streptococci, a NAAT may be performed, or a combination of throat culture and antigen tests on isolates for groups C and G streptococci may be performed. Other relevant comments include: • “A rapid antigen test for Streptococcus pyogenes may be performed at the point of care by healthcare personnel or transported to the laboratory for performance of the test…in pediatric patients, if the direct antigen test is negative, and if the direct antigen test is known to have a sensitivity of <80%, a second throat swab should be examined by a more sensitive direct NAAT or by culture as a means of arbitrating possible false-negative direct antigen test results…this secondary testing is not necessarily required in adults” • “Direct and amplified NAATs for Streptococcus pyogenes are more sensitive than direct antigen tests and, as a result, negative direct NAAT results do not have to be arbitrated by a secondary test.” • “Detection of group C and G β-hemolytic streptococci is accomplished by throat culture in those patients in whom there exists a concern for an etiologic role for these organisms. Only large colony types are identified, as tiny colonies demonstrating groups C and G antigens are in the Streptococcus anginosus (S. milleri) group” (Miller et al., 2018). American Academy of Otolaryngology-Head and Neck Surgery Foundation Although the focus of this guideline is the tonsillectomy procedure in children, there are some relevant comments. The Academy notes that “In practice, streptococcal carriage is strongly suggested by positive strep cultures or other strep tests when the child lacks signs or symptoms of acute pharyngitis.” (Mitchell et al., 2019) IDSA endorsed this guideline in February 2019 (IDSA, 2019a). American Academy of Orthopaedic Surgeons Although this guideline focuses on management of periprosthetic joint infections, there is a relevant recommendation, which states that “synovial fluid aerobic and anaerobic bacterial cultures” have moderate evidence to support their use to “aid in the diagnosis of prosthetic joint infection (PJI)” (AAOS, 2019). IDSA endorsed this guideline in March 2019 (IDSA, 2019b). 2011 Pediatric Infectious Diseases Society (PIDS) and Infectious Diseases Society of America (IDSA) The 2011 joint PIDS-IDSA guidelines concerning pediatric community-acquired pneumonia (CAP) recommended (strong recommendation; moderate-quality evidence) that “blood cultures should not be routinely performed in nontoxic, fully immunized children with CAP managed in the outpatient setting” Policy Reimbursement Policy This document has been classified as public information Page 17 of 23 and that “blood cultures should be obtained in children who fail to demonstrate clinical improvement and in those who have progressive symptoms or clinical deterioration after initiation of antibiotic therapy”. Concerning inpatient services, they recommend (strong recommendation; low-quality evidence) that “blood cultures should be obtained in children requiring hospitalization for presumed bacterial CAP that is moderate to severe, particularly those with complicated pneumonia”; however, “in improving patients who otherwise meet criteria for discharge, a positive blood culture with identification or susceptibility results pending should not be routinely preclude discharge of that patient with appropriate oral or intravenous antimicrobial therapy. The patient can be discharged if close follow-up is assured (weak recommendation; low-quality evidence).” For pneumococcal bacteremia, they do not recommend repeated blood cultures to document resolution (weak recommendation; low-quality evidence), but they do recommend “repeated blood cultures to document resolution of bacteremia…caused by S. aureus, regardless of clinical status (strong recommendation; low-quality evidence).” With respect to sputum gram stain and culture, “sputum samples for culture and Gram stain should be obtained in hospitalized children who can produce sputum” (weak recommendation; low-quality evidence). They do not recommend using urinary antigen detection testing “for the diagnosis of pneumococcal pneumonia in children; false-positive tests are common (strong recommendation; high-quality evidence)” (Bradley et al., 2011). American College of Obstetricians and Gynecologists (ACOG) The ACOG issued Committee Opinion #797 in 2020. ACOG recommends that “Regardless of planned mode of birth, all pregnant women should undergo antepartum screening for GBS at 36 0/7–37 6/7 weeks of gestation, unless intrapartum antibiotic prophylaxis for GBS is indicated because of GBS bacteriuria during the pregnancy or because of a history of a previous GBS-infected newborn” (ACOG, 2020). This committee opinion was reaffirmed in 2022. American Society for Microbiology The ASM endorsed the above ACOG recommendation, stating that “The recommended screening interval has changed from 35-37 weeks (per CDC 2010 guidelines) to 36 0/7 to 37 6/7 weeks (ACOG 2019 recommendations).” Concerning identification of group B streptococcus, the ASM propounds the following: “Recommendation: Acceptable phenotypic and proteomic methods of identification of candidate isolates include CAMP test, latex agglutination, and mass spectrometry.” “Recommendation: Nucleic acid amplification-based identification of GBS from enrichment broth is acceptable, but not sufficient for all patients.” “Recommendation: Latex agglutination directly from enrichment broth and direct-from-specimen immunoassays are unacceptable methods for GBS detection.” The guideline also recommends performing “antimicrobial susceptibility testing on all GBS [Group B Streptococcus] isolates from pregnant women with penicillin allergy”, and most recently the ASM included options for vancomycin reporting (Filkins et al., 2021). Policy Reimbursement Policy This document has been classified as public information Page 18 of 23 National Institute for Health and Care Excellence The NICE published an update on “rapid tests for group A streptococcal infections in people with a sore throat.” They stated that “Rapid tests for strep A infections are not recommended for routine adoption for people with a sore throat. This is because their effect on improving antimicrobial prescribing and stewardship, and on patient outcomes, as compared with clinical scoring tools alone, is likely to be limited” (NICE, 2019). VI. Applicable State and Federal Regulations DISCLAIMER: If there is a conflict between this Policy and any relevant, applicable government policy for a particular member [e.g., Local Coverage Determinations (LCDs) or National Coverage Determinations (NCDs) for Medicare and/or state coverage for Medicaid], then the government policy will be used to make the determination. For the most up-to-date Medicare policies and coverage, please visit the Medicare search website: For the most up-to-date Medicaid policies and coverage, visit the applicable state Medicaid website. Food and Drug Administration (FDA) The FDA approved the Lyra Direct Strep Assay (k133833) on 04/16/2014 and reclassified it on 07/11/2014. It is a “Real-Time PCR in vitro diagnostic test for the qualitative detection and differentiation of Group A β -hemolytic Streptococcus (Streptococcus pyogenes) and pyogenic Group C and G β -hemolytic Streptococcus nucleic acids isolated from throat swab specimens obtained from patients with signs and symptoms of pharyngitis, such as sore throat. The assay does not differentiate between pyogenic Groups C and G β-hemolytic Streptococcus” (Hojvat, 2014). The FDA has also approved the Solana Strep Complete Assay by Quidel that is “an in vitro diagnostic test for the detection of Group A, C and G beta- hemolytic Streptococcus in throat swab specimens from symptomatic patients” on 10/25/2016 (K162274) (FDA, 2016). On 03/06/2019, the FDA approved GenePOC’s Strep A assay to be performed using GenePOC’s Revogene instrument as a “single-use test for qualitative detection of Streptococcus pyogenes (group A Streptococcus-GAS) nucleic acids from throat swab specimens obtained from patients with signs and symptoms of pharyngitis”. (FDA, 2019) On November 9, 2020, the FDA approved Mesa Biotech, Inc.’s Accula™ Strep A Test, which is a semi-automated, colorimetric polymerase chain reaction (PCR) nucleic acid amplification test “to qualitatively detect Streptococcus pyogenes (Group A βhemolytic Streptococcus, Strep A) bacterial nucleic acid from unprocessed throat swabs that have not undergone prior nucleic acid extraction”. (FDA, 2020) Many labs have developed specific tests that they must validate and perform in house. These laboratory-developed tests (LDTs) are regulated by the Centers for Medicare and Medicaid (CMS) as high-complexity tests under the Clinical Laboratory Improvement Amendments of 1988 (CLIA ’88). LDTs are not approved or cleared by the U. S. Food and Drug Administration; however, FDA clearance or approval is not currently required for clinical use. Policy Reimbursement Policy This document has been classified as public information Page 19 of 23 VII. Applicable CPT/HCPCS Procedure Codes CPT Code Description 86060 Antistreptolysin 0; titer 86063 Antistreptolysin 0; screen 86215 Deoxyribonuclease, antibody 86317 Immunoassay for infectious agent antibody, quantitative, not otherwise specified 86318 Immunoassay for infectious agent antibody(ies), qualitative or semiquantitative, single step-method (e.g., reagent strip); 87040 Culture, bacterial; blood, aerobic, with isolation and presumptive identification of isolates (includes anaerobic culture, if appropriate) 87070 Culture, bacterial; any other source except urine, blood or stool, aerobic, with isolation and presumptive identification of isolates 87071 Culture, bacterial; quantitative, aerobic with isolation and presumptive identification of isolates, any source except urine, blood or stool 87077 Culture, bacterial; aerobic isolate, additional methods required for definitive identification, each isolate 87081 Culture, presumptive, pathogenic organisms, screening only; 87430 Infectious agent antigen detection by immunoassay technique, (e.g., enzyme immunoassay [EIA], enzyme-linked immunosorbent assay [ELISA], fluorescence immunoassay [FIA], immunochemiluminometric assay [IMCA]) qualitative or semiquantitative; Streptococcus, group A 87650 Infectious agent detection by nucleic acid (DNA or RNA); Streptococcus, group A, direct probe technique 87651 Infectious agent detection by nucleic acid (DNA or RNA); Streptococcus, group A, amplified probe technique 87652 Infectious agent detection by nucleic acid (DNA or RNA); Streptococcus, group A, quantification 87797 Infectious agent detection by nucleic acid (DNA or RNA), not otherwise specified; direct probe technique, each organism 87798 Infectious agent detection by nucleic acid (DNA or RNA), not otherwise specified; amplified probe technique, each organism 87799 Infectious agent detection by nucleic acid (DNA or RNA), not otherwise specified; quantification, each organism 87880 Infectious agent antigen detection by immunoassay with direct optical (ie, visual) observation; Streptococcus, group A Current Procedural Terminology© American Medical Association. All Rights reserved. Procedure codes appearing in Medical Policy documents are included only as a general reference tool for each policy. They may not be all-inclusive. Policy Reimbursement Policy This document has been classified as public information Page 20 of 23 VIII. Evidence-based Scientific References AACC. (2015, 12/30/2017). Strep Throat Test. American Association for Clinical Chemistry. Retrieved 08/03/2018 from AAOS. (2019). DIAGNOSIS AND PREVENTION OF PERIPROSTHETIC JOINT INFECTIONS CLINICAL PRACTICE GUIDELINE. ACOG. (2020). Prevention of Group B Streptococcal Early-Onset Disease in Newborns. Barshak, M. B. (2022, 20/16/2021). Group B streptococcal infections in nonpregnant adults. Wolters Kluwer Bilir, S. P., Kruger, E., Faller, M., Munakata, J., Karichu, J. K., Sickler, J., & Cheng, M. M. (2021). US cost-effectiveness and budget impact of point-of-care NAAT for streptococcus. Am J Manag Care, 27(5), e157-e163. Blyth, C. C., & Robertson, P. W. (2006). Anti-streptococcal antibodies in the diagnosis of acute and post-streptococcal disease: streptokinase versus streptolysin O and deoxyribonuclease B. Pathology, 38(2), 152-156. Boyanton, B. L., Jr., Darnell, E. M., Prada, A. E., Hansz, D. M., & Robinson-Dunn, B. (2016). Evaluation of the Lyra Direct Strep Assay To Detect Group A Streptococcus and Group C and G Beta-Hemolytic Streptococcus from Pharyngeal Specimens. J Clin Microbiol, 54(1), 175-177. Bradley, J. S., Byington, C. L., Shah, S. S., Alverson, B., Carter, E. R., Harrison, C., Kaplan, S. L., Mace, S. E., McCracken, J. G. H., Moore, M. R., St Peter, S. D., Stockwell, J. A., & Swanson, J. T. (2011). The Management of Community-Acquired Pneumonia in Infants and Children Older Than 3 Months of Age: Clinical Practice Guidelines by the Pediatric Infectious Diseases Society and the Infectious Diseases Society of America. Clinical Infectious Diseases, 53(7), e25-e76. Bruun, T., Kittang, B. R., de Hoog, B. J., Aardal, S., Flaatten, H. K., Langeland, N., Mylvaganam, H., Vindenes, H. A., & Skrede, S. (2013). Necrotizing soft tissue infections caused by Streptococcus pyogenes and Streptococcus dysgalactiae subsp. equisimilis of groups C and G in western Norway. Clin Microbiol Infect, 19(12), E545-550. CDC. (2022a, 06/27/2022). Acute Rheumatic Fever. Centers for Disease Control and Prevention. Retrieved 8/10/2022 from CDC. (2022b, 06/27/2022). Pharyngitis (Strep Throat). Centers for Disease Control and Prevention. Retrieved 8/10/2022 from CDC. (2022c, 06/27/2022). Post-Streptococcal Glomerulonephritis. Centers for Disease Control and Prevention. Retrieved 8/15/2022 from CDC. (2022d, 06/27/2022). Scarlet Fever. Centers for Disease Control and Prevention. Retrieved 8/10/2022 from Centor, R. M., & McIsaac, W. (2022). Centor Score (Modified/McIsaac) for Strep Pharyngitis. MDCalc. Chow, A. W. (2023, 08/24/2023). Evaluation of acute pharyngitis in adults. Policy Reimbursement Policy This document has been classified as public information Page 21 of 23 Church, D. L., Lloyd, T., Larios, O., & Gregson, D. B. (2018). Evaluation of Simplexa Group A Strep Direct Kit Compared to Hologic Group A Streptococcal Direct Assay for Detection of Group A Streptococcus in Throat Swabs. J Clin Microbiol, 56(3). Cohen, D. M., Russo, M. E., Jaggi, P., Kline, J., Gluckman, W., & Parekh, A. (2015). Multicenter Clinical Evaluation of the Novel Alere i Strep A Isothermal Nucleic Acid Amplification Test. J Clin Microbiol, 53(7), 2258-2261. Cohen, J. F., Bertille, N., Cohen, R., & Chalumeau, M. (2016). Rapid antigen detection test for group A streptococcus in children with pharyngitis. Cochrane Database Syst Rev, 7, Cd010502. Dubois, C., Smeesters, P. R., Refes, Y., Levy, C., Bidet, P., Cohen, R., Chalumeau, M., Toubiana, J., & Cohen, J. F. (2021). Diagnostic accuracy of rapid nucleic acid tests for group A streptococcal pharyngitis: systematic review and meta-analysis. Clinical Microbiology and Infection. FDA. (2016, 06/18/2018). Product Classification. U.S. Department of Health & Human Services. Retrieved 06/22/2018 from FDA. (2019). 510(k) Substantial Equivalence Determination Desion Summary (K183366). FDA. (2020). Groups A, C And G Beta-Hemolytic Streptococcus Nucleic Acid Amplification System. Filkins, L., Hauser, J., Robinson-Dunn, B., Tibbetts, R., Boyanton, B., & Revell, P. (2021, 7/23/2021). Guidelines for the Detection and Identification of Group B Streptococcus. Fraser, H., Gallacher, D., Achana, F., Court, R., Taylor-Phillips, S., Nduka, C., Stinton, C., Willans, R., Gill, P., & Mistry, H. (2020). Rapid antigen detection and molecular tests for group A streptococcal infections for acute sore throat: systematic reviews and economic evaluation. Health Technol Assess, 24(31), 1-232. Freeman, J., & Roberts, S. (2023, 9/21/2021). Approach to Gram stain and culture results in the microbiology laboratory. Wolters Kluwer. Gera, K., & McIver, K. S. (2013). Laboratory Growth and Maintenance of Streptococcus pyogenes (The Group A Streptococcus, GAS). Curr Protoc Microbiol, 30, 9d.2.1-9d.2.13. Gewitz, M., H., Baltimore, R., S., Tani, L., Y., Sable, C., A., Shulman, S., T., Carapetis, J., Remenyi, B., Taubert, K., A., Bolger, A., F., Beerman, L., Mayosi, B., M., Beaton, A., Pandian, N., G., & Kaplan, E., L. (2015). Revision of the Jones Criteria for the Diagnosis of Acute Rheumatic Fever in the Era of Doppler Echocardiography. Circulation, 131(20), 1806-1818. Helmig, R. B., & Gertsen, J. B. (2017). Diagnostic accuracy of polymerase chain reaction for intrapartum detection of group B streptococcus colonization. Acta Obstet Gynecol Scand, 96(9), 1070-1074. Hojvat, S. A. (2014). Evaluation of Class III Designation--De Novo Request. Silver Spring, MD: Food and Drug Administration Retrieved from IDSA. (2019a). Clinical Practice Guideline: Tonsillectomy in Children (Update) (Endorsed). Policy Reimbursement Policy This document has been classified as public information Page 22 of 23 IDSA. (2019b). Diagnosis and Prevention of Periprosthetic Joint Infections (Endorsed). Kim, H. N., Kim, J., Jang, W. S., Nam, J., & Lim, C. S. (2019). Performance evaluation of three rapid antigen tests for the diagnosis of group A Streptococci. BMJ Open, 9(8), e025438. Kimberlin, D. W., Barnett, E. D., Lynfield, R., & Sawyer, M. H. (2021). Group A Streptococcal Infections. Lollar, R. (2016). K162274 510(k) premarket notification of intent to market Solana Strep Complete Assay. FDA Retrieved from Luo, R., Sickler, J., Vahidnia, F., Lee, Y.-C., Frogner, B., & Thompson, M. (2019). Diagnosis and Management of Group a Streptococcal Pharyngitis in the United States, 2011–2015. BMC Infectious Diseases, 19(1), 193. McCarty, T., White, C., Meeder, J., Moates, D., Piece, H., Edwards, W., Hutchinson, J., Lee, ., & Leal J, S. (2022). Analytical perfomance and potential clinical utility of the GenMark Dx eplex blood culture identification gram-positive panel. Diagnostic Microbiology and Infectious Disease, 104(3), 115762. Metlay, J. P., Waterer, G. W., Long, A. C., Anzueto, A., Brozek, J., Crothers, K., Cooley, L. A., Dean, N. C., Fine, M. J., Flanders, S. A., Griffin, M. R., Metersky, M. L., Musher, D. M., Restrepo, M. I., & Whitney, C. G. (2019). Diagnosis and Treatment of Adults with Community-acquired Pneumonia. An Official Clinical Practice Guideline of the American Thoracic Society and Infectious Diseases Society of America. Am J Respir Crit Care Med, 200(7), e45-e67. Miller, J. M., Binnicker, M. J., Campbell, S., Carroll, K. C., Chapin, K. C., Gilligan, P. H., Gonzalez, M. D., Jerris, R. C., Kehl, S. C., Patel, R., Pritt, B. S., Richter, S. S., Robinson-Dunn, B., Schwartzman, J. D., Snyder, J. W., Telford, S., III, Theel, E. S., Thomson, R. B., Jr., Weinstein, M. P., & Yao, J. D. (2018). A Guide to Utilization of the Microbiology Laboratory for Diagnosis of Infectious Diseases: 2018 Update by the Infectious Diseases Society of America and the American Society for Microbiology. Clinical Infectious Diseases, 67(6), e1-e94. Mitchell, R. B., Archer, S. M., Ishman, S. L., Rosenfeld, R. M., Coles, S., Finestone, S. A., Friedman, N. R., Giordano, T., Hildrew, D. M., Kim, T. W., Lloyd, R. M., Parikh, S. R., Shulman, S. T., Walner, D. L., Walsh, S. A., & Nnacheta, L. C. (2019). Clinical Practice Guideline: Tonsillectomy in Children (Update). Otolaryngol Head Neck Surg, 160(1_suppl), S1-s42. NICE. (2019). Rapid tests for group A streptococcal infections in people with a sore throat. O. Luiz, F., Alves, K. B., & Barros, . R. (2019). Prevalence and long-term persistence of beta-haemolytic streptococci throat carriage among children and young adults. J Med Microbiol, 68(10), 1526-1533. Pediatrics, A. A. o. (2018). Group B Streptococcal Infections. In D. Kimberlin, M. Brady, M. Jackson, & S. Long (Eds.), Red Book: 2018 Report of the Committee on Infectious Diseases (pp. 762-768). American Academy of Pediatrics. Puopolo, K. M., Lynfield, R., & Cummings, J. J. (2019). Management of Infants at Risk for Group B Streptococcal Disease. Pediatrics, 144(2), e20191881. Puopolo, K. M., Madoff, L. C., & Baker, C. J. (2021, 12/1/2021). Group B streptococcal infection in pregnant women. Wolters Kluwer. Policy Reimbursement Policy This document has been classified as public information Page 23 of 23 Raignoux, J., Benard, M., Huo Yung Kai, S., Dicky, O., Berrebi, A., Bibet, L., Chetouani, A. S., Marty, N., Cavalie, L., Casper, C., & Assouline-Azogui, C. (2016). [Is rapid intrapartum vaginal screening test of group B streptococci (GBS) during partum useful in identifying infants developing early-onset GBS sepsis in postpartum period?]. Arch Pediatr, 23(9), 899-907. (Test de depistage rapide intra partum du portage vaginal de streptocoque du groupe B (SGB) pour le reperage des nouveau-nes a risque d'infection neonatale precoce a SGB. Etude observationnelle analytique dans une maternite de type III.) Rantala, S. (2014). Streptococcus dysgalactiae subsp. equisimilis bacteremia: an emerging infection. Eur J Clin Microbiol Infect Dis, 33(8), 1303-1310. Schwartz, B., Facklam, R. R., & Breiman, R. F. (1990). Changing epidemiology of group A streptococcal infection in the USA. Lancet, 336(8724), 1167-1171. Short, S., Bashir, H., Marshall, P., Miller, N., Olmschenk, D., Prigge, K., & Solyntjes, L. (2017). Diagnosis and Treatment of Respiratory Illness in Children and Adults (5th ed.). Institute for Clinical Systems Improvement. Shulman, S. T., Bisno, A. L., Clegg, H. W., Gerber, M. A., Kaplan, E. L., Lee, G., Martin, J. M., & Van Beneden, C. (2012). Clinical practice guideline for the diagnosis and management of group A streptococcal pharyngitis: 2012 update by the Infectious Diseases Society of America. Clin Infect Dis, 55(10), e86-102. Spellerberg, B., & Brandt, C. (2016). Laboratory Diagnosis of Streptococcus pyogenes (group A streptococci). In J. J. Ferretti, D. L. Stevens, & V. A. Fischetti (Eds.), Streptococcus pyogenes : Basic Biology to Clinical Manifestations. University of Oklahoma Health Sciences Center. Steer, A., & Gibofsky, A. (2022, 3/10/2022). Acute rheumatic fever: Clinical manifestations and diagnosis. Steer, A. C., Smeesters, P. R., & Curtis, N. (2015). Streptococcal Serology: Secrets for the Specialist. Pediatr Infect Dis J, 34(11), 1250-1252. Stevens, D. L., Bisno, A. L., Chambers, H. F., Dellinger, E. P., Goldstein, E. J. C., Gorbach, S. L., Hirschmann, J. V., Kaplan, S. L., Montoya, J. G., & Wade, J. C. (2014). Practice Guidelines for the Diagnosis and Management of Skin and Soft Tissue Infections: 2014 Update by the Infectious Diseases Society of America. Clinical Infectious Diseases, 59(2), e10-e52. Stevens, D. L., & Bryant, A. (2022, 4/6/2022). Group A streptococcus: Virulence factors and pathogenic mechanisms. Uphoff, T. S., Buchan, B. W., Ledeboer, N. A., Granato, P. A., Daly, J. A., & Marti, T. N. (2016). Multicenter Evaluation of the Solana Group A Streptococcus Assay: Comparison with Culture. J Clin Microbiol, 54(9), 2388-2390. Wald, E. R. (2022, 2/1/2022). Group A streptococcal tonsillopharyngitis in children and adolescents: Clinical features and diagnosis. Wolters Kluwer. Weinzierl, E. P., Jerris, R. C., Gonzalez, M. D., Piccini, J. A., & Rogers, B. B. (2018). Comparison of Alere i Strep A Rapid Molecular Assay With Rapid Antigen Testing and Culture in a Pediatric Outpatient Setting. American Journal of Clinical Pathology, aqy038-aqy038. Wessels, M. R. (2022, 08/23/2022). Group C and group G streptococcal infection. |
188181 | https://calculator-online.org/inequation/e/logarithm_one_divide_three_logarithm_three_x_less_than_or_equal_to_minus_one | Solve the inequality log(1/3,(log(3,x)))<=-1 (logarithm of (1 divide by 3,(logarithm of (3,x))) less than or equal to minus 1) - Specify the set of solutions of the inequality in detail step by step. [THERE'S THE ANSWER!] online
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Inequation:
½x<6
1-3x<-1+4
(log(x^2+9)/log(3))log((4/5))(3x)/(x+2)-log((4/5))(6-x)<=0
17x+7<7x−12
Identical expressions
log(one / three ,(log(three ,x)))<=- one
logarithm of (1 divide by 3,( logarithm of (3,x))) less than or equal to minus 1
logarithm of (one divide by three ,( logarithm of (three ,x))) less than or equal to minus one
log1/3,log3,x<=-1
log(1 divide by 3,(log(3,x)))<=-1
Similar expressions
log(1/3,(log(3,x)))<=+1
Solving inequalities/ log(1/3,(log(3,x)))<=-1
log(1/3,(log(3,x)))<=-1 inequation
inequation
Solve the inequality!
A inequation with variable
The solution
You have entered [src]
/ log(3)\
log|1/3, ------| <= -1
\ log(x)/
log(1 3)≤−1\log{\left(\frac{1}{3} \right)} \leq -1 lo g(3 1)≤−1
log(1/3, log(3)/log(x)) <= -1
Detail solution
Given the inequality:
log(1 3)≤−1\log{\left(\frac{1}{3} \right)} \leq -1 lo g(3 1)≤−1
To solve this inequality, we must first solve the corresponding equation:
log(1 3)=−1\log{\left(\frac{1}{3} \right)} = -1 lo g(3 1)=−1
Solve:
x 1=3 3 x_{1} = \sqrt{3}x 1=3 3
x 1=3 3 x_{1} = \sqrt{3}x 1=3 3
This roots
x 1=3 3 x_{1} = \sqrt{3}x 1=3 3
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
x 0≤x 1 x_{0} \leq x_{1}x 0≤x 1
For example, let's take the point
x 0=x 1−1 10 x_{0} = x_{1} - \frac{1}{10}x 0=x 1−10 1
=
−1 10+3 3- \frac{1}{10} + \sqrt{3}−10 1+3 3
=
−1 10+3 3- \frac{1}{10} + \sqrt{3}−10 1+3 3
substitute to the expression
log(1 3)≤−1\log{\left(\frac{1}{3} \right)} \leq -1 lo g(3 1)≤−1
log(1 3)≤−1\log{\left(\frac{1}{3} \right)} \leq -1 lo g(3 1)≤−1
-log(3)
/ log(3) \
log <= -1
| / 1 3 ___\|
|log|- -- + \/ 3 ||
\ \ 10 //
but
-log(3)
/ log(3) \
log >= -1
| / 1 3 ___\|
|log|- -- + \/ 3 ||
\ \ 10 //
Then
x≤3 3 x \leq \sqrt{3}x≤3 3
no execute
the solution of our inequality is:
x≥3 3 x \geq \sqrt{3}x≥3 3
_____
/
-------•-------
x1
Solving inequality on a graph
Rapid solution [src]
/3 ___ \
And\/ 3 <= x, x < 3/
3 3≤x∧x<3\sqrt{3} \leq x \wedge x < 3 3 3≤x∧x<3
(x < 3)∧(3^(1/3) <= x)
Rapid solution 2 [src]
3 ___
[\/ 3 , 3)
x i n[3 3,3)x\ in\ \left[\sqrt{3}, 3\right)x in[3 3,3)
x in Interval.Ropen(3^(1/3), 3)
© Mister Exam – Calculator |
188182 | https://www.soa.org/globalassets/assets/Files/Edu/2018/2018-10-exam-fm-sample-solutions.pdf | 1 SOCIETY OF ACTUARIES EXAM FM FINANCIAL MATHEMATICS EXAM FM SAMPLE SOLUTIONS This set of sample questions includes those published on the interest theory topic for use with previous versions of this examination. Questions from previous versions of this document that are not relevant for the syllabus effective with the October 2022 administration have been deleted. The questions have been renumbered. Some of the questions in this study note are taken from past SOA examinations. These questions are representative of the types of questions that might be asked of candidates sitting for the Financial Mathematics (FM) Exam. These questions are intended to represent the depth of understanding required of candidates. The distribution of questions by topic is not intended to represent the distribution of questions on future exams. The following model solutions are presented for educational purposes. Alternative methods of solution are acceptable. In these solutions, m s is the m-year spot rate and m t f is the m-year forward rate, deferred t years. Update history: October 2022: Questions 208-275 were added January 2023: Question 204 was deleted June 2023 Questions 276-385 were added August 2024: Questions 386-462 were added Copyright 2024 by the Society of Actuaries. 2 1. Solution: C Given the same principal invested for the same period of time yields the same accumulated value, the two measures of interest (2) 0.04 i = and δ must be equivalent, which means: 2 (2) 1 2 i eδ + = over a one-year period. Thus, 2 (2) 2 1 1.02 1.0404 2 ln(1.0404) 0.0396.
i eδ δ = + = = = = 2. Solution: E From basic principles, the accumulated values after 20 and 40 years are 4 24 20 16 4 4 4 44 40 36 4 4 (1 ) (1 ) 100[(1 ) (1 ) (1 ) ] 100 1 (1 ) (1 ) (1 ) 100[(1 ) (1 ) (1 ) ] 100 .
1 (1 ) i i i i i i i i i i i i + − + + + + + + + = − + + − + + + + + + + = − + The ratio is 5, and thus (setting 4 (1 ) x i = + ) 4 44 11 4 24 6 6 11 5 10 10 5 5 5 (1 ) (1 ) 5 (1 ) (1 ) 5 5 5 5 1 5 4 0 ( 1)( 4) 0.
i i x x i i x x x x x x x x x x x x + − + − = = + − + − − = − − = − − + = − − = Only the second root gives a positive solution. Thus 5 11 4 1.31951 1.31951 1.31951 100 6195.
1 1.31951 x x X = = − = = − 3 Annuity symbols can also be used. Using the annual interest rate, the equation is 40 20 4 4 40 20 40 20 20 100 5(100) (1 ) 1 (1 ) 1 5 (1 ) 5(1 ) 4 0 (1 ) 4 s s a a i i i i i i i = + − + − = + − + + = + = and the solution proceeds as above. 3. Solution: C Eric’s (compound) interest in the last 6 months of the 8th year is 15 100 1 2 2 i i + . Mike’s (simple) interest for the same period is 200 2 i . Thus, 15 15 100 1 200 2 2 2 1 2 2 1 1.047294 2 0.09459 9.46%.
i i i i i i + = + = + = = = 4 4. Solution: C ( ) 1 1 1 1 2 77.1 1 1 0.011025 0.85003 1 1.105 0.14997 ln(0.14997) 19.
ln(1.105) n n n n n n n n n n n n n nv v Ia i a nv nv v i i a nv nv i i i a v v i i v n + + + + − = + − = + = − + − − = = = = − = = − = To obtain the present value without remembering the formula for an increasing annuity, consider the payments as a perpetuity of 1 starting at time 2, a perpetuity of 1 starting at time 3, up to a perpetuity of 1 starting at time n + 1. The present value one period before the start of each perpetuity is 1/i. The total present value is 2 (1/ )( ) (1/ ) .
n n i v v v i a + + + = 5. Solution: C The interest earned is a decreasing annuity of 6, 5.4, etc. Combined with the annual deposits of 100, the accumulated value in fund Y is ( ) ( ) 10 0.09 10 0.09 10 10 0.09 6( ) 100 10 1.09 6 100 15.19293 0.09 565.38 1519.29 2084.67.
Ds s s + − = + = + = 6. Solution: D For the first 10 years, each payment equals 150% of interest due. The lender charges 10%, therefore 5% of the principal outstanding will be used to reduce the principal. At the end of 10 years, the amount outstanding is ( ) 10 1000 1 0.05 598.74 − = . Thus, the equation of value for the last 10 years using a comparison date of the end of year 10 is 10 10% 598.74 6.1446 97.44.
Xa X X = = = 5 7. Solution: B The book value at time 6 is the present value of future payments: 4 6 4 0.06 10,000 800 7920.94 2772.08 10,693.
BV v a = + = + = The interest portion is 10,693(0.06) = 641.58. 8. Solution: A The value of the perpetuity after the fifth payment is 100/0.08 = 1250. The equation to solve is: 2 24 25 1250 ( 1.08 1.08 ) ( ) (25) /1.08 50(1.08) 54.
X v v v X v v v X X = + + + = + + + = = = 9. Solution: C Equation of value at end of 30 years: 40 40 30 40 30 40 1/40 10(1 / 4) (1.03) 20(1.03) 100 10(1 / 4) [100 20(1.03) ]/1.03 15.7738 1 / 4 1.57738 0.98867 4(1 0.98867) 0.0453 4.53%.
d d d d − − − − + = − = − = − = = = − = = 10. Solution: E The accumulation function is 2 3 0 ( ) exp ( /100) exp( / 300).
t a t s ds t = = ∫ The accumulated value of 100 at time 3 is 3 100exp(3 / 300) 109.41743.
= The amount of interest earned from time 3 to time 6 equals the accumulated value at time 6 minus the accumulated value at time 3. Thus ( ) 109.41743 [ (6) / (3) 1] (109.41743 )(2.0544332 /1.0941743 1) (109.41743 )0.877613 96.026159 0.122387 784.61.
X a a X X X X X X X + − = + − = + = = = 6 11. Solution: A 5 5 9.2% 1 (1 ) 167.50 10 10(1.092) 1.092 (1 ) /1.092 167.50 38.6955 6.440011 (1 ) /1.092 (167.50 38.6955)[1 (1 ) /1.092] 6.44001(1 ) /1.092 128.8045 135.24451(1 ) /1.092 1 1.0400 0.0400 4.0%.
t t k a k k k k k k k K ∞ − = + = + + = + − + − − + = + = + + = = ⇒ = ∑ 12. Solution: B 10 0.0807 Option 1: 2000 299 Total payments 2990 Option 2: Interest needs to be 2990 2000 990 990 [2000 1800 1600 200] 11,000 0.09 9.00% Pa P i i i = = ⇒ = − = = + + + + = = = 13. Solution: B Monthly payment at time t is 1 1000(0.98)t−. Because the loan amount is unknown, the outstanding balance must be calculated prospectively. The value at time 40 months is the present value of payments from time 41 to time 60: 40 1 59 20 40 40 1 60 21 1000[0.98 0.98 ] 0.98 0.98 1000 , 1/ (1.0075) 1 0.98 0.44238 0.25434 1000 6888.
1 0.97270 OB v v v v v v = + + − = = − − = = − 7 14. Solution: C The equation of value is ( ) 3 2 3 2 98 98 8000 (1 ) 1 (1 ) 1 81.63 1 2 8 1 4 1 81.63 10 81.63 12.25% n n n n n S S i i i i i i i i i + = + − + − + = + = − − + = = = 15. Solution: B Convert 9% convertible quarterly to an effective rate of j per month: 3 0.09 (1 ) 1 4 j + = + or j = 0.00744. Then 60 60 0.00744 60 0.00744 60 48.6136 38.4592 2( ) 2 2 2729.7.
0.00744 0.00744 a v Ia − − = = = 16. Solution: A Equating present values: 2 10 2 10 10 10 100 200 300 600 100 200(0.76) 300(0.76) 600 425.28 600 0.7088 0.96617 1.03501 1 0.035 3.5%.
n n v v v v v v v i i + + = + + = = = = = + = = 8 17. Solution: A The accumulation function is: ( ) 0 0 1 ln 8 8 8 ( ) .
8 t t dr r r t a t e e + + + ∫ = = = Using the equation of value at end of 10 years: ( ) 10 10 10 0 0 0 (10) 18 / 8 20,000 8 (8 ) 18 ( ) (8 ) / 8 20,000 180 111.
180 a k tk dt k t dt k dt a t t k k = + = + = + = ⇒ = = ∫ ∫ ∫ 18. Solution: D Let C be the redemption value and 1/ (1 ) v i = + . Then 2 2 2 2 1000 1 1000 381.50 1000(1.03125)(1 0.5889 ) 381.50 1055.11.
n ni n X ra Cv v r i = + − = + = − + = 19. Solution: D Equate net present values: 2 2 4000 2000 4000 2000 4000 4000 2000 6000 1.21 1.1 5460.
v v v Xv X X − + + = + − + = + = 20. Solution: D The present value of the perpetuity = X/i. Let B be the present value of Brian’s payments. 2 0.4 0.4 0.4 1 0.6 0.36 , n n n n n X B Xa i a v v i X K v i X K i = = = ⇒ = − ⇒ = = = Thus the charity’s share is 36% of the perpetuity’s present value. 9 21. Solution: D The given information yields the following amounts of interest paid: 10 10 6% 0.12 Seth 5000 1 1 8954.24 5000 3954.24 2 Janice 5000(0.06)(10) 3000.00 5000 Lori (10) 5000 1793.40 where = 679.35 The sum is 8747.64.
P P a = + − = − = = = = − = = 22. Solution: E For Bruce, 11 10 10 100[(1 ) (1 ) ] 100(1 i) X i i i = + − + = + . Similarly, for Robbie, 16 50(1 ) X i i = + .Dividing the second equation by the first gives 6 1 0.5(1 ) i = + which implies 1/6 2 1 0.122462 i = −= . Thus 10 100(1.122462) (0.122462) 38.879.
X = = 23. Solution: D Year t interest is 1 1 1 n t n t i ia v −+ −+ = − . Year t+1 principal repaid is 1 (1 ) n t n t v v − − − − = . 1 1 1 (1 ) 1 .
n t n t n t n t X v v v v v d −+ − − − = − + = + − = + 24. Solution: B For the first perpetuity, 3 6 3 3 3 3 3 32 10( ) 10 / (1 ) 32 32 10 32 / 42.
v v v v v v v = + + = − − = = For the second perpetuity, 1/3 2/3 1/3 1/3 1/9 1/9 / (1 ) (32 / 42) /[1 (32 / 42) ] 32.599.
X v v v v = + + = − = − = 25 Solution: D Under either scenario, the company will have 822,703(0.05) = 41,135 to invest at the end of each of the four years. Under Scenario A these payments will be invested at 4.5% and accumulate to 4 0.045 41,135 41,135(4.2782) 175,984.
s = = Adding the maturity value produces 998,687 for a loss of 1,313. Note that only answer D has this value. The Scenario B calculation is 4 0.055 41,135 41,135(4.3423) 178,621 822,703 1,000,000 1,324.
s = = + − = 10 26. Solution: D. The present value is 2 2 20 20 21 21 5000[1.07 1.07 1.07 ] 1.07 1.07 1.01905 1.48622 5000 5000 122,617.
1 1.07 1 1.01905 v v v v v v + + + − − = = = − − 27. Solution: C. The first cash flow of 60,000 at time 3 earns 2400 in interest for a time 4 receipt of 62,400. Combined with the final payment, the investment returns 122,400 at time 4. The present value is 4 122,400(1.05) 100,699.
−= The net present value is 699. 28. Solution: B. Using spot rates, the value of the bond is: 2 3 60 /1.07 60 /1.08 1060 /1.09 926.03.
+ + = 29. Solution: E. Using spot rates, the value of the bond is: 2 3 60 /1.07 60 /1.08 1060 /1.09 926.03.
+ + = The annual effective rate is the solution to 3 3 926.03 60 1000(1 ) i a i − = + + . Using a calculator, the solution is 8.9%. 30. Solution: C. Duration is the negative derivative of the price multiplied by one plus the interest rate and divided by the price. Hence, the duration is –(–700)(1.08)/100 = 7.56. 31. Solution: C The size of the dividend does not matter, so assume it is 1. Then the duration is 1 1 ( ) / 1/ ( ) 1 1.1 11.
1/ 1/ 0.1 t t t t tv Ia a i di a i i d v ∞ = ∞ ∞ ∞ ∞ = = = = = = = ∑ ∑ 11 32. Solution: B 1 1 1 1 1.02 ( ) / 1 Duration = .
1/ 1.02 t t t t j j t t t t t j t t t tv R tv Ia a j a j d v R v ∞ ∞ ∞ ∞ = = ∞ ∞ ∞ = = = = = = ∑ ∑ ∑ ∑ The interest rate j is such that 1 (1 ) 1.02 1.02 /1.05 0.03/1.02.
j v j − + = = ⇒ = Then the duration is 1/ (1 ) / (1.05 /1.02) / (0.03/1.02) 1.05 / 0.03 35.
d j j = + = = = 33. Solution: A The outstanding balance is the present value of future payments. With only one future payment, that payment must be 559.12(1.08) = 603.85. The amount borrowed is 4 0.08 603.85 2000.
a = The first payment has 2000(0.08) = 160 in interest, thus the principal repaid is 603.85 – 160 = 443.85. Alternatively, observe that the principal repaid in the final payment is the outstanding loan balance at the previous payment, or 559.12. Principal repayments form a geometrically decreasing sequence, so the principal repaid in the first payment is 3 559.12 /1.08 443.85.
= 34. Solution: B Because the yield rate equals the coupon rate, Bill paid 1000 for the bond. In return he receives 30 every six months, which accumulates to 20 30 j s where j is the semi-annual interest rate. The equation of value is 10 20 20 1000(1.07) 30 1000 32.238.
j j s s = + ⇒ = Using a calculator to solve for the interest rate produces j = 0.0476 and so 2 1.0476 1 0.0975 9.75%.
i = −= = 35. Solution: A To receive 3000 per month at age 65 the fund must accumulate to 3,000(1,000/9.65) = 310,880.83. The equation of value is 300 0.08/12 310,880.83 957.36657 324.72.
Xs X = = ⇒ 36. Solution: D (A) The left-hand side evaluates the deposits at age 0, while the right-hand side evaluates the withdrawals at age 17. (B) The left-hand side has 16 deposits, not 17. (C) The left-hand side has 18 deposits, not 17. (D) The left-hand side evaluates the deposits at age 18 and the right-hand side evaluates the withdrawals at age 18. (E) The left-hand side has 18 deposits, not 17 and 5 withdrawals, not 4. 12 37. Solution: D Because only Bond II provides a cash flow at time 1, it must be considered first. The bond provides 1025 at time 1 and thus 1000/1025 = 0.97561 units of this bond provides the required cash. This bond then also provides 0.97561(25) = 24.39025 at time 0.5. Thus Bond I must provide 1000 – 24.39025 = 975.60975 at time 0.5. The bond provides 1040 and thus 975.60975/1040 = 0.93809 units must be purchased. 38. Solution: C Because only Mortgage II provides a cash flow at time two, it must be considered first. The mortgage provides 2 0.07 / 0.553092 Y a Y = at times one and two. Therefore, 0.553092Y = 1000 for Y = 1808.02. Mortgage I must provide 2000 – 1000 = 1000 at time one and thus X = 1000/1.06 = 943.40. The sum is 2751.42. 39. Solution: A Bond I provides the cash flow at time one. Because 1000 is needed, one unit of the bond should be purchased, at a cost of 1000/1.06 = 943.40. Bond II must provide 2000 at time three. Therefore, the amount to be reinvested at time two is 2000/1.065 = 1877.93. The purchase price of the two-year bond is 2 1877.93/1.07 1640.26 = . The total price is 2583.66. 40. Solution: C Given the coupon rate is greater than the yield rate, the bond sells at a premium. Thus, the minimum yield rate for this callable bond is calculated based on a call at the earliest possible date because that is most disadvantageous to the bond holder (earliest time at which a loss occurs). Thus, X, the par value, which equals the redemption value because the bond is a par value bond, must satisfy Price = 30 0.03 30 0.03 1722.25 0.04 1.196 1440.
Xa Xv X X = + = ⇒ = 41. Solution: B Because 40/1200 is greater than 0.03, for early redemption the earliest redemption should be evaluated. If redeemed after 15 years, the price is 30 30 0.03 40 1200 /1.03 1278.40 a + = . If the bond is redeemed at maturity, the price is 40 40 0.03 40 1100 /1.03 1261.80 a + = . The smallest value should be selected, which is 1261.80. (When working with callable bonds, the maximum a buyer will pay is the smallest price over the various call dates. Paying more may not earn the desired yield.) 13 42. Solution: E Given the coupon rate is less than the yield rate, the bond sells at a discount. Thus, the minimum yield rate for this callable bond is calculated based on a call at the latest possible date because that is most disadvantageous to the bond holder (latest time at which a gain occurs). Thus, X, the par value, which equals the redemption value because the bond is a par value bond, must satisfy Price = 20 0.03 20 0.03 1021.50 0.02 0.851225 1200.
Xa Xv X X = + = ⇒ = 43. Solution: B Given the price is less than the amount paid for an early call, the minimum yield rate for this callable bond is calculated based on a call at the latest possible date. Thus, for an early call, the effective yield rate per coupon period, j, must satisfy Price = 19 19 1021.50 22 1200 j j a v = + . Using the calculator, j = 2.86%. We also must check the yield if the bond is redeemed at maturity. The equation is 20 20 1021.50 22 1100 j j a v = + . The solution is j = 2.46% Thus, the yield, expressed as a nominal annual rate of interest convertible semiannually, is twice the smaller of the two values, or 4.92%. 44. Solution: C First, the present value of the liability is 15 6.2% 35,000 335,530.30.
PV a = = The duration of the liability is: 2 15 35,000 2(35,000) 15(35,000) 2,312,521.95 6.89214.
335,530.30 335,530.30 t t t t tv R v v v d v R + + + = = = = ∑ ∑ Let X denote the amount invested in the 5 year bond. Then, (5) 1 (10) 6.89214 208,556.
335,530.30 335,530.30 X X X + − = => = 45 . Solution: A The present value of the first eight payments is: 8 9 2 7 8 2000 2000(1.03) 2000 2000(1.03) ...
2000(1.03) 13,136.41.
1 1.03 v v PV v v v v − = + + + = = − The present value of the last eight payments is: 7 9 7 2 10 7 8 16 7 9 7 9 17 2000(1.03) 0.97 2000(1.03) (0.97) 2000(1.03) (0.97 ) 2000(1.03) 0.97 2000(1.03) (0.97) 7,552.22.
1 0.97 PV v v v v v v = + + + − = = − Therefore, the total loan amount is L = 20,688.63. 14 46. Solution: E 1 2 2 3 0 2 3 3 0 0 3 3 3 100 2000 500exp 3 150 50 4 exp 0.5 exp 0.5ln 3 150 3 150 4 exp 0.5ln 1 1 450 450 16 1 450 18.8988 t t t r dr r r r dr r t t t t = + = = + + = + = + = + = ∫ ∫ 47. Solution: E Let F, C, r, and i have their usual interpretations. The discount is ( ) n Ci Fr a − and the discount in the coupon at time t is 1 ( ) n t Ci Fr v −+ − . Then, 26 21 5 26 40 0.095 194.82 ( ) 306.69 ( ) 0.63523 0.91324 0.095 ( ) 194.82(1.095) 2062.53 Discount 2062.53 21,135 Ci Fr v Ci Fr v v v i Ci Fr a = − = − = ⇒ = ⇒= − = = = = 48. Solution: A 8 5 1 1 4 1 699.68 842.39 (annualpayment) 699.68 581.14 1.0475 842.39 581.14 261.25 261.25 5500 (loan amount) 0.0475 Total interest = 842.39(8) 5500 1239.12 Pv P P I L −+ = = = = = − = = = − = 15 49. Solution: D 18 18 18 0.007 24 0.004 22,000(1.007) 450.30 16,337.10 16,337.10 715.27 OB s Pa P = − = = = 50. Solution: C If the bond has no premium or discount, it was bought at par so the yield rate equals the coupon rate, 0.038. ( ) ( ) 2 14 14 2 14 14 14 14 14 14 1 1(190) 2(190) 14(190) 14(5000) 2 190 190 190 5000 95 7(5000) 190 5000 5.5554 v v v v d v v v v Ia v d a v d + + + + = + + + + + = + = Or, taking advantage of a shortcut: 14 0.038 11.1107.
d a = = This is in half years, so dividing by two, 11.1107 5.5554 2 d = = . 51. Solution: A [ ] [ ] 7.959 7.425 1.072 (0.08) (0.072) 1 ( ) (0.08) 1000 1 (0.008)(7.425) 940.60 v P P i v P = = = −∆ = − = 52. Solution: E 3 2 3 2 1 2 3 3 3 2 2 2 3 2 1 2 1 2 (1 ) (1 ) (1 ) 1 0.85892 , 0.052 (1 s ) 1 0.90703 , 0.050 (1 ) 1.052 1.050 (1 ) 0.056 s s f s s s f f + = + + = = + = = + = + = 16 53. Solution: C Let 0 d be the Macaulay duration at time 0. 0 8 0.05 1 0 2 7 0.05 1 2 6.7864 1 5.7864 6.0757 5.7864 0.9524 6.0757 d a d d d a d d = = = −= = = = = This solution employs the fact that when a coupon bond sells at par the duration equals the present value of an annuity-due. For the duration just before the first coupon the cash flows are the same as for the original bond, but all occur one year sooner. Hence the duration is one year less. Alternatively, note that the numerators for 1 d and 2 d are identical. That is because they differ only with respect to the coupon at time 1 (which is time 0 for this calculation) and so the payment does not add anything. The denominator for 2 d is the present value of the same bond, but with 7 years, which is 5000. The denominator for 1 d has the extra coupon of 250 and so is 5250. The desired ratio is then 5000/5250 = 0.9524. 54. Solution: A Let N be the number of shares bought of the bond as indicated by the subscript. (105) 100, 0.9524 (100) 102 0.9524(5), 0.9724 (107) 99 0.9524(5), 0.8807 C C B B A A N N N N N N = = = − = = − = 55. Solution: B All are true except B. Immunization requires frequent rebalancing. 56. Solution: D Set up the following two equations in the two unknowns: 2 2 1 3 (1.05) (1.05) 6000 2 (1.05) 2 (1.05) 0.
A B A B − − + = − = Solving simultaneously gives: 2721.09 3307.50 586.41.
A B A B = = − = 17 57. Solution: A Set up the following two equations in the two unknowns. 3 3 (1) 5000(1.03) (1.03) 12,000 5463.635 (1.03) 12,000 (1.03) 6536.365 (2) 3(5000)(1.03) (1.03) 0 16,390.905 6536.365 0 2.5076 7039.27 2807.12 b b b b B B B bB b b B B b − − − − + = ⇒ + = ⇒ = − = ⇒ − = = = = 58. Solution: D 2 9 5 3 10 6 (1 ) (1 ) 95,000(1 ) 2 (1 ) 9 (1 ) 5(95,000)(1 ) A L A L P A i B i P i P A i B i P i − − − − − − = + + + = + ′ = − + − + ′ = − + Set the present values and derivatives equal and solve simultaneously. 0.92456 0.70259 78,083 1.7780 6.0801 375,400 78,083(1.7780 / 0.92456) 375,400 47,630 0.70259(1.7780 / 0.92456) 6.0801 [78,083 0.70259(47,630)]/ 0.92456 48,259 1.0132 A B A B B A A B + = − − = − − = = − = − = = 59. Solution: D Throughout the solution, let j = i/2. For bond A, the coupon rate is (i + 0.04)/2 = j + 0.02. For bond B, the coupon rate is (i – 0.04)/2 = j – 0.02. The price of bond A is 20 20 10,000( 0.02) 10,000(1 ) A j P j a j − = + + + . The price of bond B is 20 20 10,000( 0.02) 10,000(1 ) B j P j a j − = − + + . Thus, 20 20 20 5,341.12 [200 ( 200)] 400 5,341.12 / 400 13.3528.
A B j j j P P a a a − = = −− = = = Using the financial calculator, j = 0.042 and i =2(0.042)=0.084. 18 60. Solution: D The initial level monthly payment is 15 12 0.09/12 180 0.0075 400,000 400,000 4,057.07.
R a a × = = = The outstanding loan balance after the 36th payment is 36 180 36 0.0075 144 0.0075 4,057.07 4,057.07(87.8711) 356,499.17.
B Ra a − = = = = The revised payment is 4,057.07 – 409.88 = 3,647.19. Thus, 144 /12 144 /12 356,499.17 3,647.19 356,499.17 / 3,647.19 97.7463.
j j a a = = = Using the financial calculator, j/12 = 0.575%, for j = 6.9%. 61. Solution: D The price of the first bond is 30 2 60 30 2 0.05/2 60 0.025 1000(0.05 / 2) 1200(1 0.05 / 2) 25 1200(1.025) 772.72 272.74 1,045.46.
a a − × − × + + = + = + = The price of the second bond is also 1,045.46. The equation to solve is 60 60 /2 1,045.46 25 800(1 / 2) .
j a j − = + + The financial calculator can be used to solve for j/2 = 2.2% for j = 4.4%. 62. Solution: E Let n = years. The equation to solve is 2 12 1000(1.03) 2(1000)(1.0025) 2 ln1.03 ln1000 12 ln1.0025 ln 2000 0.029155 0.69315 23.775.
n n n n n n = + = + = = This is 285.3 months. The next interest payment to Lucas is at a multiple of 6, which is 288 months. 19 63. Solution: A Equating the accumulated values after 4 years provides an equation in K. 4 4 0 4 4 0 0 2 1 10 1 10exp 25 0.25 1 1 4ln(1 0.04 ) 4ln(K 0.25t) 4ln( 1) 4ln( ) 4ln 0.25 1 1 0.04 0.04 1 5.
K dt K t K K dt K K K t K K K K K K + = + + + = = + = + − = + + + = = = ∫ ∫ Therefore, 4 10(1 5 / 25) 20.74.
X = + = 64. Solution: D The outstanding balance at time 25 is 25 25 25 100( ) 100 .
a Da i − = The principle repaid in the 26th payment is 25 25 25 25 2500 (100) 2500 2500 100 100 .
a X i a a i − = − = − + = The amount borrowed is the present value of all 50 payments, 25 25 25 2500 100( ) .
a v Da + Interest paid in the first payment is then 25 25 25 25 25 25 25 25 25 25 25 2500 100( ) 2500(1 ) 100 (25 ) 2500 2500 2500 100 2500 .
i a v Da v v a v v v a Xv + = − + − = − + − = − 65. Solution: C The accumulated value is 20 0.0816 1000 50,382.16.
s = This must provide a semi-annual annuity-due of 3000. Let n be the number of payments. Then solve 0.04 3000 50,382.16 n a = for n = 26.47. Therefore, there will be 26 full payments plus one final, smaller, payment. The equation is 26 26 0.04 50,382.16 3000 (1.04) a X − = + with solution X = 1430. Note that the while the final payment is the 27th payment, because this is an annuity-due, it takes place 26 periods after the annuity begins. 20 66. Solution: D For the first perpetuity, ( ) ( ) 2 2 1 1 7.21 1 1 1 1 1 6.21 0.0775.
i i i + = + − = + − = For the second perpetuity, ( ) ( ) 1 3 1 1 (1.0875) 7.21 1.0775 0.01 1 1.286139 7.21(1.0875) 0.286139 1.74.
R R R − + = + − = = 67. Solution: E 5 5 5 5 5 15 15 5 10,000 100( ) 100 0.05 10,000 1256.64 8.13273 1075 a v Ia Xv a Xv a X X − = + = + = + = 68. Solution: C 5 10 0.06 5000 (1.05) 5000 297.22 13.1808(1.2763) Xs X = = = 69. Solution: E The monthly payment on the original loan is 180 8/12% 65,000 621.17 a = . After 12 payments the outstanding balance is 168 8/12% 621.17 62,661.40 a = . The revised payment is 168 6/12% 62,661.40 552.19.
a = 21 70. Solution: E At the time of the final deposit the fund has 18 0.07 750 25,499.27.
s = This is an immediate annuity because the evaluation is done at the time the last payments is made (which is the end of the final year). A tuition payment of 17 6000(1.05) 13,752.11 = is made, leaving 11,747.16. It earns 7%, so a year later the fund has 11,747.16(1.07) = 12,569.46. Tuition has grown to 13,752.11(1.05) = 14,439.72. The amount needed is 14,439.72 – 12,569.46 = 1,870.26 71. Solution: B The coupons are 1000(0.09)/2 = 45. The present value of the coupons and redemption value at 5% per semiannual period is 40 40 0.05 45 1200(1.05) 942.61.
P a − = + = 72. Solution: A For a bond bought at discount, the minimum price will occur at the latest possible redemption date. 20 20 0.06 50 1000(1.06) 885.30.
P a − = + = (When working with callable bonds, the maximum a buyer will pay is the smallest price over the various call dates. Paying more may not earn the desired yield.) 73. Solution: C 5 4 1.095 1 11.5% 1.090 −= 74. Solution: D The accumulated value of the first year of payments is 12 0.005 2000 24,671.12.
s = This amount increases at 2% per year. The effective annual interest rate is 12 1.005 1 0.061678.
−= The present value is then 25 25 1 1 1 26 1 1.02 24,671.12 1.02 (1.061678) 24,671.121.02 1.061678 0.960743 0.960743 24,187.37 374,444.
1 0.960743 k k k k k P − − = = = = − = = − ∑ ∑ This is 56 less than the lump sum amount. 22 75. Solution: A The monthly interest rate is 0.072/12 = 0.006. 6500 five years from today has value 60 6500(1.006) 4539.77 − = . The equation of value is 2 4539.77 1700(1.006) 3400(1.006) .
n n − − = + Let 1.006 n x − = . Then, solve the quadratic equation 2 2 3400 1700 4539.77 0 1700 1700 4(3400)( 4539.77) 0.93225.
2(3400) x x x + − = − + − − = = Then, 1.006 0.9325 ln(1.006) ln(0.93225) 11.73.
n n n −= ⇒− = ⇒ = To ensure there is 6500 in five years, the deposits must be made earlier and thus the maximum integral value is 11. 76. Solution: C ( ) ( ) ( ) ( ) 4 4 4 2 2 1 2 39 1 2 38 39 39( 2) 38 38( 4) 38 1 4 39 1 / 4 39 2 38 4 39 38 1/ (19.5 9.5) 0.1 1 1 / 2 .95 1.108 10.8%.
d d d d d d d d i d i − − − − − − = ⇒ = ⇒ − = − − − − = − = − = + = − = = ⇒= 77. Solution: C The monthly interest rate is 0.042/12 = 0.0035. The quarterly interest rate is 3 1.0035 1 0.0105 −= . The investor makes 41 quarterly deposits and the ending date is 124 months from the start. Using January 1 of year y as the comparison date produces the following equation: 41 124 1 100 1.9 1.0105 1.0035 k k X X = + = ∑ Substituting 3 1.0105 1.0035 = gives answer (C). 23 78. Solution: D Convert the two annual rates, 4% and 5%, to two-year rates as 2 1.04 1 0.0816 −= and 2 1,05 1 0.1025 −= . The accumulated value is 4 3 0.0816 2 0.1025 100 (1.05) 100 100(3.51678)(1.21551) 100(2.31801) 659.269 s s + = + = . With only five payments, an alternative approach is to accumulate each one to time ten and add them up. The two-year yield rate is the solution to 5 100 659.269 i s = . Using the calculator, the two-year rate is 0.093637. The annual rate is 0.5 1.093637 1 0.04577 −= which is 4.58%. 79. Solution: C ( ) 1 12 15 4 8% 216 0.6434% 1.08 1 0.006434 1 25,000 1.08 25,000(3.57710) 240.38 3.17217(117.2790) ä Xä X −= = = = 80. Solution: B perp.
10 15 0.08 10 10 0.10 10 1 1 1 0.08 0.1 PV (15,000) 15,000 0.1 1.1 164,457.87 15,000 179,457.87 179,458 1.10 9.244 6.759 179,458 1.10 17,384 a X a X X − = + + = + = + = + = = 81. Solution: A 14 14 0.03 14 0.03 14 14 14 0.03 14 0.03 1050.50 22.50 ( ) 300(1.03) 14(1.03) 22.50 300(1.03) 0.03 22.50(11.2961) (79.3102) 198.3353 7.54 a X Ia a a X X X − − − = + + − = + + = + + = 24 25 82. Solution: D The amount of the loan is the present value of the deferred increasing annuity: 30 10 10 30 0.05 30 0.05 30 0.05 30 0.05 30(1.05) (1.05) 500 500( ) (1.05 )(500) 64,257.
0.05 /1.05 a a Ia a − − − − + = + = 83. Solution: C 30 30 30 30 30 29 29 1/29 (1 ) (1.03) (1 ) (1.03) 50,000 (1 ) 5,000 (1 ) ( 0.03) 0.03 50,000 / (1 ) 5,000 (1 ) 10 10 1 0.082637 i i i i i i i i i + − + − + = + − − + = + = = −= The accumulated amount is 30 30 30 (1.082637) (1.03) 50,000 (1.082637) 797,836.82 (1.082637) (0.082637 0.03) − = − 84. Solution: D The first payment is 2,000, and the second payment of 2,010 is 1.005 times the first payment. Since we are given that the series of quarterly payments is geometric, the payments multiply by 1.005 every quarter. Based on the quarterly interest rate, the equation of value is 2 2 3 3 2,000 100,000 2,000 2,000(1.005) 2,000(1.005) 2,000(1.005) 1 1.005 1 1.005 2,000 /100,000 0.98 /1.005.
v v v v v v = + + + + = − − = ⇒ = . The annual effective rate is ( ) 4 4 1 0.98 /1.005 1 0.10601 10.6% v − −−= −= = . 85. Solution: A Present value for the first 10 years is ( ) ( ) 10 1 1.06 7.58 ln 1.06 − − = Present value of the payments after 10 years is ( ) ( ) ( ) ( ) ( ) 10 0 0.5584 1.06 1.03 1.06 19.45 ln 1.06 ln 1.03 s s ds ∞ − − = = − ∫ Total present value = 27.03 26 86. Solution: C ( ) ( ) ( ) 10 5 1 5 2 1 10,000 1.06 1.06 75,000 11 13,382.26 1.1236 75,000 6 1.1236 27,526.83 24,498.78 dt t X e X X X + ∫ + = + = = = 87. Solution: D The effective annual interest rate is 1 1 (1 ) 1 (1 0.055) 1 5.82% i d − − = − −= − −= The balance on the loan at time 2 is 2 15,000,000(1.0582) 16,796,809.
= The number of payments is given by | 1,200,000 16,796,809 n a = which gives n = 29.795 => 29 payments of 1,200,000. The final equation of value is 30 29 1,200,000 (1.0582) 16,796,809 (16,796,809 16,621,012)(5.45799) 959,490.
a X X − + = = − = 88. Solution: C 2 4 2 2 2 1 0.525(1 ) 1 0.525(1 ) 0.90476 0.95119 1 0.1427(1 ) 1 v (1 0.90476) / 0.1427 0.667414 0.332596 ln(0.332596) / ln(0.95119) 22 n n n v v v v v v v v n − = − ⇒= + ⇒ = ⇒ = − = − ⇒− = − = ⇒ = = = 89. Solution: C The monthly payment is 360 0.005 200,000 / 1199.10 a = . Using the equivalent annual effective rate of 6.17%, the present value (at time 0) of the five extra payments is 41,929.54 which reduces the original loan amount to 200,000 – 41,929.54 = 158,070.46. The number of months required is the solution to 0.005 158,070.46 1199.10 n a = . Using calculator, n = 215.78 months are needed to pay off this amount. So there are 215 full payments plus one fractional payment at the end of the 216th month, which is December 31, 2020. 90. Solution: D The annual effective interest rate is 0.08/(1 – 0.08) = 0.08696. The level payments are 5 0.08696 500,000 / 500,000 / 3.9205 127,535.
a = = This rounds up to 128,000. The equation of value for X is 5 4 0.08696 128,000 (1.08696) 500,000 (500,000 417,466.36)(1.51729) 125,227.
a X X − + = = − = 27 91. Solution: B The accumulated value is the reciprocal of the price. The equation is X[(1/0.94)+(1/0.95)+(1/0.96)+(1/0.97)+(1/0.98)+(1/0.99)] = 100,000. X= 16,078 92. Solution: D Let P be the annual payment. The fifth line is obtained by solving a quadratic equation. 10 10 6 1 10 5 10 5 5 10 1/10 10 (1 ) 3600 4871 1 3600 4871 1 0.739068 0.69656 0.485195 0.485195 1 0.075 1 3600 48,000 0.075 P v Pv v v v v v v i v X P i −+ − − = = − = − = = = = −= − = = = 93. Solution: A Let j = periodic yield rate, r = periodic coupon rate, F = redemption (face) value, P = price, n = number of time periods, and 1 1 j v j = + . In this problem, ( ) 1 2 1.0705 1 0.03465 j = −= , r = 0.035, P = 10,000, and n = 50. The present value equation for a bond is n j n j P Fv Fra = + ; solving for the redemption value F yields 50 50 0.03465 10,000 10,000 9,918.
(1.03465) 0.035 0.18211 0.035(23.6044) n j n j P F v ra a − = = = = + + + . 94. Solution: B Jeff’s monthly cash flows are coupons of 10,000(0.09)/12 = 75 less loan payments of 2000(0.08)/12 = 13.33 for a net income of 61.67. At the end of the ten years (in addition to the 61.67) he receives 10,000 for the bond less a 2,000 loan repayment. The equation is (12) (12) 120 120 /12 (12) 12 8000 61.67 8000(1 /12) /12 0.00770875 1.00770875 1 0.0965 9.65%.
i a i i i − = + + = = −= = 28 95. Solution: B The present value equation for a par-valued annual coupon bond is n i n i P Fv Fra = + ; solving for the coupon rate r yields 1 n n i i n i n i n i P Fv v P r Fa a F a − = = − . All three bonds have the same values except for F. We can write r = x(1/F) + y. From the first two bonds: 0.0528 = x/1000 + y and 0.0440 = x/1100 + y. Then, 0.0528 – 0.044 = x(1/1000 – 1/1100) for x = 96.8 and y = 0.0528 – 96.8/1000 = –0.044. For the third bond, r = 96.8/1320 – 0.044 = 0.2933 = 2.93%. 96. Solution: A The effective semi-annual yield rate is 2 (2) (2) 1.04 1 1.9804% 2 2 i i = + => = . Then, 2 12 12 13 12 582.53 (1.02) (1.02 ) (1.02 ) 250 1.02 (1.02 ) 250 12.015 197.579 32.04.
1 1.02 c v c v c v v v v c v c c v = + + + + − = + = + => = − 13 12 1.02 (1.02 ) 582.53 250 12.015 197.579 32.04 1 1.02 v v c v c c v − = + = + => = − 97. Solution: E Book values are linked by BV3(1 + i) – Fr = BV4. Thus 1254.87(1.06) – Fr = 1277.38. Therefore, the coupon is Fr = 52.7822. The prospective formula for the book value at time 3 is ( 3) ( 3) ( 3) 1 1.06 1254.87 52.7822 1890(1.06) 0.06 375.1667 1010.297(1.06) ln(375.1667 /1010.297) 3 17.
ln(1.06) n n n n − − − − − − − = + = − = = − Thus, n = 20. Note that the financial calculator can be used to solve for n – 3. 29 98. Solution: A Book values are linked by BV3(1 + i) – Fr = BV4. Thus BV3(1.04) – 2500(0.035) = BV3 + 8.44. Therefore, BV3 = [2500(0.035) + 8.44]/0.04 = 2398.5. The prospective formula for the book value at time 3 is, where m is the number of six-month periods. ( 3) ( 3) ( 3) 1 1.04 2398.5 2500(0.035) 2500(1.04) 0.04 211 312.5(1.04) ln(211/ 312.5) 3 10.
ln(1.04) m m m m − − − − − − − = + = − = = − Thus, m = 13 and n = m/2 = 6.5. Note that the financial calculator can be used to solve for m – 3. 99. Solution: C ( ) ( ) ( ) ( ) 1 1 0 2 2 1 1 2 1 3 3 2 1/3 1 2 3 2 2 0.04 1 0.06 1 (1.06)(1.04) 1 0.04995 1 1 0.08 1 [(1.08)(1.04995) ] 1 0.05987 6%. 1 s f s f s s s f s s = = + = = − ⇒ = −= + + = = − ⇒ = −= = + 100. Solution: B The Macaulay duration of Annuity A is 2 2 2 2 0(1) 1( ) 2( ) 2 0.93 1 1 v v v v v v v v + + + = = + + + + , which leads to the quadratic equation 2 1.07 0.07 0.93 0 v v + − = . The unique positive solution is v = 0.9. The Macaulay duration of Annuity B is 2 3 2 3 0(1) 1( ) 2( ) 3( ) 1.369.
1 v v v v v v + + + = + + + 101. Solution: D With v =1/1.07, 2 3 4 2 3 4 2(40,000) 3(25,000) 4(100,000) 3.314.
40,000 25,000 100,000 v v v D v v v + + = = + + 30 102. Solution: C 2 0 0 1/ ( ) (1 ) / 1 30 1/ (1 ) / n n n n nv Ia di i i MacD a d i i i v ∞ = ∞ ∞ ∞ = + = = = = = = + ∑ ∑ and so i = 1/30. 30 29.032.
1 1 1 30 MacD ModD i = = = + + 103. Solution: B I) False. The yield curve structure is not relevant. II) True. III) False. Matching the present values is not sufficient when interest rates change. 104. Solution: A The present value function and its derivatives are 3 1 4 4 2 5 5 3 6 ( ) (1 ) 500(1 ) 1000(1 ) ( ) 3 (1 ) 500(1 ) 4000(1 ) ( ) 12 (1 ) 1000(1 ) 20,000(1 ) .
P i X Y i i i P i Y i i i P i Y i i i − − − − − − − − − = + + − + − + ′ = − + + + + + ′′ = + − + − + The equations to solve for matching present values and duration (at i = 0.10) and their solution are (0.1) 0.7513 1137.56 0 (0.1) 2.0490 2896.91 0 2896.91/ 2.0490 1413.82 1137.56 0.7513(1413.82) 75.36.
P X Y P Y Y X = + − = ′ = − + = = = = − = The second derivative is 5 3 6 (0.1) 12(1413.82)(1.1) 1000(1.1) 20,000(1.1) 1506.34.
P − − − ′′ = − − = − Redington immunization requires a positive value for the second derivative, so the condition is not satisfied. 31 105. Solution: D This solution uses time 8 as the valuation time. The two equations to solve are 2 8 7 ( ) 300,000(1 ) (1 ) 1,000,000 0 ( ) 600,000(1 ) (8 ) (1 ) 0.
y y P i i X i P i i y X i − − = + + + − = ′ = + + − + = Inserting the interest rate of 4% and solving: 2 8 7 2 8 7 7 300,000(1.04) (1.04) 1,000,000 0 600,000(1.04) (8 ) (1.04) 0 (1.04) [1,000,000 300,000(1.04) ]/1.04 493,595.85 624,000 (8 )(1.04) (493,595.85) 0 8 624,000 /[493,595.85(1.04) ] 8.9607 X 493,5 y y y X y X X y y − − − + − = + − = = − = + − = = + = = 8.9607 95.85(1.04) 701,459.
= 106. Solution: A This solution uses Macaulay duration and convexity. The same conclusion would result had modified duration and convexity been used. The liabilities have present value 2 5 573/1.07 701/1.07 1000.
+ = Only portfolios A, B, and E have a present value of 1000. The duration of the liabilities is 2 5 [2(573) /1.07 5(701) /1.07 ]/1000 3.5.
+ = The duration of a zero coupon bond is its term. The portfolio duration is the weighted average of the terms. For portfolio A the duration is [500(1) + 500(6)]/1000 = 3.5. For portfolio B it is [572(1) + 428(6)]/1000 = 3.14. For portfolio E it is 3.5. This eliminates portfolio B. The convexity of the liabilities is 2 5 [4(573) /1.07 25(701) /1.07 ]/1000 14.5.
+ = The convexity of a zero-coupon bond is the square of its term. For portfolio A the convexity is [500(1) + 500(36)]/1000 = 18.5 which is greater than the convexity of the liabilities. Hence portfolio A provides Redington immunization. As a check, the convexity of portfolio E is 12.25, which is less than the liability convexity. 107. Solution: D The present value of the liabilities is 1000, so that requirement is met. The duration of the liabilities is 1 2 3 402.11[1.1 2(1.1) 3(1.1) ]/1000 1.9365.
− − − + + = Let X be the investment in the one-year bond. The duration of a zero-coupon is its term. The duration of the two bonds is then [X + (1000 – X)(3)]/1000 = 3 – 0.002X. Setting this equal to 1.9365 and solving yields X = 531.75. 32 108. Solution: A Let x, y, and z represent the amounts invested in the 5-year, 15-year, and 20-year zero-coupon bonds, respectively. Note that in this problem, one of these three variables is 0. The present value, Macaulay duration, and Macaulay convexity of the assets are, respectively, 2 2 2 5 15 20 5 15 20 , , x y z x y z x y z x y z x y z + + + + + + + + + + . We are given that the present value, Macaulay duration, and Macaulay convexity of the liabilities are, respectively, 9697, 15.24, and 242.47. Since present values and Macaulay durations need to match for the assets and liabilities, we have the two equations 5 15 20 9697, 15.24 x y z x y z x y z + + + + = = + + . Note that 5 and 15 are both less than the desired Macaulay duration 15.24, so z cannot be zero. So try either the 5-year and 20-year bonds (i.e. y = 0), or the 15-year and 20-year bonds (i.e. x = 0). In the former case, substituting y = 0 and solving for x and z yields (20 15.24)9697 3077.18 20 5 x − = = − and (15.24 5)9697 6619.82 20 5 z − = = − . We need to check if the Macaulay convexity of the assets exceeds that of the liabilities. The Macaulay convexity of the assets is 2 2 5 (3077.18) 20 (6619.82) 281.00 9697 + = , which exceeds the Macaulay convexity of the liabilities, 242.47. The company should invest 3077 for the 5-year bond and 6620 for the 20-year bond. Note that setting x = 0 produces y = 9231.54 and z = 465.46 and the convexity is 233.40, which is less than that of the liabilities. 109. Solution: E The correct answer is the lowest cost portfolio that provides for $11,000 at the end of year one and provides for $12,100 at the end of year two. Let H, I, and J represent the face amount of each purchased bond. The time one payment can be exactly matched with H + 0.12J = 11,000. The time two payment can be matched with I + 1.12J = 12,100. The cost of the three bonds is H/1.1 + I/1.2321 + J. This function is to be minimized under the two constraints. Substituting for H and I gives (11,000 – 0.12J)/1.1 + (12,100 – 1.12J)/1.2321 + J = 19,820 – 0.0181J. This is minimized by purchasing the largest possible amount of J. This is 12,100/1.12 = 10,803.57. Then, H = 11,000 – 0.12(10,803.57) = 9703.57. The cost of Bond H is 9703.57/1.1 = 8,821.43. 33 110. Solution: C The strategy is to use the two highest yielding assets: the one-year bond and the two-year zero-coupon bond. The cost of these bonds is 2 25,000 /1.0675 20,000 /1.05 41,560.
+ = 111. Solution: E Let P be the annual interest paid. The present value of John’s payments is 0.05 X Pa . The present value of Karen’s payments is 0.05 (1.05) (1.05) / 0.05 X X P a P − − ∞ = . Then, 0.05 (1.05) / 0.05 1.59 1.05 1 1.05 1.59 0.05 0.05 1.59 2.59(1.05) ln1.59 ln 2.59 ln1.05 10.
X X X X X P Pa X X − − − − = − = = = − = 112. Solution: A Cheryl’s force of interest at all times is ln(1.07) = 0.06766. Gomer’s accumulation function is from time 3 is 1 + yt and the force of interest is y/(1 + yt). To be equal at time 2, the equation is 0.06766 = y/(1 + 2y), which implies 0.06766 + 0.13532y = y for y = 0.07825. Gomer’s account value is 1000(1 + 2x0.07825) = 1156.5. 113. Solution: D One way to view these payments is as a sequence of level immediate perpetuities of 1 that are deferred n-1, n, n+1,… years. The present value is then 1 1 2 2 3 2 2 / / / ( / )( ) / .
n n n n n v i v i v i v i v v v v i − + − − + + + = + + + = Noting that only answers C, D, and E have this form and all have the same numerator, 2 2 2 2 / / ( ) / .
n n n v i v vi v d − = = 114. Solution: B The monthly interest rate is 1/12 (1.08) 1 0.643%.
j = −= Then, 4 0.08 252 0.00643 20,000 , 90,122.24 630.99 , 142.83.
s Xs X X = = = 34 115. Solution: D 20 10 20 10 20 10 1 1 1.5 , 1.5 , 1.5 0.5 0.
e e a a e e δ δ δ δ δ δ − − − − − − = = − + = Let 10 .
X e δ − = We then have the quadratic equation 2 1.5 0.5 0 X X − + = with solution X = 0.5 for ln 0.5 / ( 10) 0.069315.
δ = − = Then, the accumulated value of a 7-year continuous annuity of 1 is 7(0.069315) 7 1 9.01.
0.069315 e s − = = 116. Solution: B The present value is 3 10 17 4 7 3 3 7 3 7 3 3 7 3 7 7 7 (1 ) (1 ) .
1 1 n n n n v v v v a a v v v v a v v −+ + + + + + + + − − − − − = = = − − 117. Solution: C From the first annuity, 0.109 1.109 1 21.8 21.8 200[1.109 1].
0.109 n n n X s − = = ⋅ = − From the second annuity, 2 1 19,208( ) 19,208 19,208 .
1 1.109 1 n n n n n v X v v v = + + = = − − Hence, 2 1 200[1.109 1] 19,2081.109 1 [1.109 1] 19,208 / 200 96.04 1.109 1 9.8 200(9.8) 1960.
n n n n X − = − − = = −= = = 118. Solution: C 60 60 60 1% 60 45.4 33.03 2( ) 2 2 2,474.60.
0.01 0.01 a v Ia − − = = = 35 119. Solution: E Let j be the semi-annual interest rate. Then, 1 2 2 2 2 475,000 300 300 (1 ) 200( ) 300 300 / 200 / 474,700 300 200 0 300 300 4(474,700)( 200) 0.02084 2(474,700) (1 ) 1 0.04212 4.21%.
j j a j Ia j j j j j i j − ∞ ∞ = + + + = + + − − = + − − = = = + −= = 120. Solution: B The present value is 2 0.06 0.06 4 2( ) 4 / 0.06 2(1.06) / 0.06 655.56.
a Ia ∞ ∞ + = + = 121. Solution: A The present value of the income is 0.1025 100 100 / 0.1025 975.61.
a∞ = = The present value of the investment is 2 3 4 5 6 1 2 3 4 5 1 1 1.05 /1.1025 (1.05 /1.1025) (1.05 /1.1025) (1.05 /1.1025) (1.05 /1.1025) 1 1.05 [1 1.05 1.05 1.05 1.05 1.05 ] 5.3295 .
1 1.05 X X X X − − − − − − − + + + + + − = + + + + + = = − Then 975.61=5.3295X for X = 183.06. 122. Solution: A The present value of the ten level payments is 10 0.05 8.10782 .
Xa X = The present value of the remaining payments is 10 10 10 11 2 1.015 1.015 /1.05 ( 1.015 1.015 ) 18.69366 .
1 1.015 1 1.015 /1.05 v X v v X X X v + + = = = − − Then, 45,000 = 8.10782X + 18.69366X = 26.80148X for X = 1679. 123. Solution: D The equation of value is 0.06 2 3 2 0.06 10,000 ( 0.996 0.996 ) 15.189 .
1 0.996 1 0.996 v e X v v v X X X v e − − = + + + = = = − − The solution is X = 10,000/15.189 = 658.37. 36 124. Solution: D Discounting at 10%, the net present values are 4.59, –2.36, and –9.54 for Projects A, B, and C respectively. Hence, only Project A should be funded. Note that Project C’s net present value need not be calculated. Its cash flows are the same as Project B except being 50 less at time 2 and 50 more at time 4. This indicates Project C must have a lower net present value and therefore be negative. 125. Solution: D The loan balance after 10 years is still 100,000. For the next 10 payments, the interest paid is 10% of the outstanding balance and therefore the principal repaid is 5% of the outstanding balance. After 10 years the oustanding balance is 10 100,000(0.95) 59,874.
= Then, 10 0.1 59,874 / 59,874 / 6.14457 9,744.
X a = = = 126. Solution: B First determine number of regular payments: 4 4 0.06 0.06 4000 600 , (4000 / 600)1.06 8.4165.
n n v a a = = = Using the calculator, n = 12.07 and thus there are 11 regular payments. The equation for the balloon payment, X, is: 4 16 11 0.06 4000 600 3748.29 0.39365 , 639.43.
v a Xv X X = + = + = 127. Solution: C ( ) 5 5 0.11 5 0.12 20,000 1.11 (3.69590 3.60478 /1.68506) 5.83516X X 20,000 / 5.83516 3427.50.
X a a X − = + = + = = = 128. Solution: A The principal repaid in the first payment is 100 – iL. The outstanding principal is L – 100 + iL = L + 25. Hence, iL = 125. Also, 16 8 16 8 8 16 16 8 2 8 300(1 ) 200(1 ) 300 200 125 100 200 300 300 200 25 0 200 200 4(300)(25) 200 100 0.5.
600 600 v v L a a i iL v v v v v − − − = − = = = + − − + = ± − ± = = = The larger of the two values is used due to the value being known to exceed 0.3. The outstanding valance at time eight is the present value of the remaining payments: 8 1/8 1 0.5 300 300 1657.
2 1 a − = = − 37 129. Solution: E Let j be the monthly rate and X be the level monthly payment. The principal repaid in the first payment is 1400 = X – 60,000j. The principal repaid in the second payment is 1414 = X – (60,000 – 1400)j. Substituting X = 1400 + 60,000j from the first equation gives 1414 = 1400 + 60,000j – 58,600j or 14 = 1400j and thus j = 0.01 and X = 2000. Let n be the number of payments. Then 0.01 60,000 2000 n a = and the calculator (or algebra) gives n = 35.8455. The equation for the drop payment, P, is 36 35 0.01 60,000 2000 58,817.16 0.698925 a Pv P = + = + for P = 1692. 130. Solution: C The accumulated value is ( ) 24 24 0.06/12 24 0.08/12 1000 (1 0.08 /12) 1000(25.4320(1.1729) 25.9332) 55,762.
s s + + = + = 131. Solution: C Each month the principal paid increases by 1/12 1.1 . Thus, the amount of principal paid increases to 1/12 30 6 2 500(1.1 ) 500(1.1) 605.
−= = 132. Solution: C 20 10 10 20 11 20 10 10 21 10 10 20 10 11 21 20 10 10 10 21 Int 900 300 900(1 ) 300(1 ) 1200 300 900 Int 900 900(1 ) Int 2Int 1200 300 900 1800 1800 9 15 6 0 2 / 3 Int 900(1 ) 300 i i i i a a v v v v i a v v v v v v v v = ⋅ ⋅ + = − + − = − − = ⋅ = − = ⇒ − − = − ⇒ − + = ⇒ = = − = 133. Solution: C The original monthly payment is 240 0.005 85,000 / 85,000 /139.5808 608.97 a = = . On July 1, 2009 there has been 4 years of payments, hence 16x12 = 192 remaining payments. The outstanding balance is 192 0.005 608.97 608.97(123.2380) 75,048.24 a = = . The number of remaining payments after refinancing is determined as 0.0045 1 1.0045 75,048.24 500 500 0.0045 0.67543 1 1.0045 ln(0.32457) / ln(1.0045) 250.62.
n n n a n − − − = = = − = − = Thus the final payment will be 251 months from June 30, 2009. This is 20 years and 11 months and so the final payment is May 31, 2030. 38 134. Solution: B Just prior to the extra payment at time 5, the outstand balance is 20 0.07 1300 1300(10.5940) 13,772.20 a = = . After the extra payment it is 11,172.20. Paying this off in 15 years requires annual payments of 15 0.07 11,172.20 / 11,172.20 / 9.1079 1226.65 a = = . 135. Solution: C During the first redemption period the modified coupon rate is 1000(0.035)/1250 = 2.80% which is larger than the desired yield rate. If redeemed during this period, bond sells at a premium and so the worst case for the buyer is the earliest redemption. The price if called at that time is 20 20 0.025 35 1250(1.025) 35(15.5892) 762.84 1308.46 a − + = + = . During the second redemption period the modified coupon rate is 1000(0.035)/1125 = 3.11% which is also larger than the desired yield rate and the worst case for the buyer is again the earliest redemption. The price if called at that time is 40 40 0.025 35 1125(1.025) 35(25.1028) 418.98 1297.58 a − + = + = . Finally, if the bond is not called, its value is 60 60 0.025 35 1000(1.025) 35(30.9087) 227.28 1309.08 a − + = + = . The appropriate price is the lowest of these three, which relates to the bond being called after the 40th coupon is paid. 136. Solution: B Because the yield is less than the coupon rate, the bond sells at a premium and the worst case for the buyer is an early call. Hence the price should be calculated based on the bond being called at time 16. The price is 16 16 0.05 100 1000(1.05) 100(10.0378) 458.11 1542 a − + = + = . (When working with callable bonds, the maximum a buyer will pay is the smallest price over the various call dates. Paying more may not earn the desired yield.) 137. Solution: A All calculations are in millions. For the ten-year bond, at time ten it is redeemed for 10 2(1.08) 4.31785 = . After being reinvested at 12% it matures at time twenty for 10 4.31785(1.12) 13.4106 = . The thirty-year bond has a redemption value of 30 4(1.08) 40.2506 = . For the buyer to earn 10%, it is sold for 10 40.2506(1.1) 15.5184 − = . The gain is 13.4106 + 15.5184 – 6 = 22.9290. 138. Solution: A The book value after the third coupon is 37 37 0.0265 7500(0.037) (1.0265) 6493.05 0.379943 a C C − + = + and after the fourth coupon it is 36 36 0.0265 7500(0.037) (1.0265) 6387.61 0.390012 a C C − + = + . Then, 39 6493.05 0.379943 (6387.61 0.390012 ) 28.31 105.44 0.010069 28.31 7660.15.
C C C C + − + = − = = 40 139. Solution: C The semiannual yield rate is 1/2 1.1 1 0.0488 −= . Assuming the bond is called for 2900 after four years, the purchase price is 8 8 0.0488 150 2900(1.0488) 150(6.4947) 1980.87 2955.08 a − + = + = . With a call after the first coupon, the equation to solve for the semi-annual yield rate (j) and then the annual effective rate (i) is 2 2955.08 (150 2960) / (1 ) 1 1.05242 1.05242 1 0.10759.
j j i = + + + = = −= 140. Solution: C The book value after the sixth coupon is 34 34 0.036 1000( / 2) 1000(1.036) 9716.01 300.45 r a r − + = + . After the seventh coupon it is 33 33 0.036 1000( / 2) 1000(1.036) 9565.79 311.26 r a r − + = + . Then, 4.36 9565.79 311.26 (9716.01 300.45) 10.81 150.22 (10.81 4.36) /150.22 0.0429.
r r r r = + − + = − = − = 141. Solution: B The two equations are: 5 5 0.04 5 5 0.04 (10,000 ) 9,000(1.04) 44,518.22 7,397.34 1.2 [10,000( 0.01)] 11,000(1.04) 44,518.22 9,486.38.
P r a r P r a r − − = + = + = + + = + Subtracting the first equation from the second gives 0.2P = 2089.04 for P = 10,445.20. Inserting this in the first equation gives r = (10,445.20 – 7,397.34)/44,518.22 = 0.0685. 142. Solution: C When the yield is 6.8% < 8%, the bond is sold at a premium and hence an early call is most disadvantageous. Therefore, 10 10 0.034 40 1000(1.034) 1050.15 P a − = + = . When the yield is 8.8% > 8%, the bond is sold at discount. Hence, Q < 1000 < P. and thus Q = 1050.15 – 123.36 = 926.79. Also, because the bond is sold at a discount, the latest call is the most disadvantageous. Thus, 2 2 2 2 0.044 2 40 40 926.79 40 1000(1.044) (1.044) 1000 909.09 90.90(1.044) 0.044 .044 17.70 90.90(1.044) 2 ln(17.70 / 90.90) / ln(1.044) 38 19.
n n n n n a n n − − − − = + = + − = + = = − = = 41 143. Solution: B The fund will have 4 4 0.05 500(1.05) 100 176.74 s − = after four years. After returning 75% to the insured, the insurer receives 0.25(176.74) = 44.19. So the insurer’s cash flows are to pay 100 at time 0, receive 125 at time 2, and receive 44.19 at time four. The equation of value and the solution are: 4 2 2 2 100(1 ) 125(1 ) 44.19 0 125 ( 125) 4(100)( 44.19) (1 ) 1.5374 200 1 1.2399 24%.
i i i i i + − + − = ± − − − + = = + = = 144. Solution: B The Macaulay duration of the perpetuity is 2 1 1 ( ) (1 ) / 1 1 1/ 17.6.
1/ n n n n nv Ia i i i i a i i v ∞ = ∞ ∞ ∞ = + + = = = = + = ∑ ∑ This implies that i = 1/16.6. With i = 2i = 2/16.6, the duration is 1 + 16.6/2 = 9.3. 145. Solution: A Because the interest rate is greater than zero, the Macaulay duration of each bond is greater than its modified duration. Therefore, the bond with a Macaulay duration of c must be the bond with a modified duration of a and a = c/(1 + i) which implies 1 + i = c/a. The Macaulay duration of the other bond is b(1 + i) =bc/a. 146. Solution: B 11 1.10 (0.1025) (0.10) 0.97534 (0.10).
1.1025 P P P ≈ = Therefore, the approximate percentage price change is 100(0.97534 – 1) = –2.47%. 147. Solution: B Cash-flow matching limits the number of investment choices available to the portfolio manager to a subset of the choices available for immunization. 42 148. Solution: C Options for full immunization are: 2J (cost is 3000), K+2L (cost is 2500), and M (cost is 4000). The lowest possible cost is 2500. Another way to view this is that the prices divided by total cash flows are 0.6, 0.5, 0.5, and 0.8. The cheapest option will be to use K and L, if possible. 149. Solution: B The present value of the assets is 15,000 + 45,000 = 60,000 which is also the present value of the liability. The modified duration of the assets is the weighted average, or 0.25(1.80) + 0.75Dmod. The modified duration of the liability is 3/1.1 and so Dmod = (3/1.1 – 0.45)/0.75 = 3.04. 150. Solution: C Let A be the redemption value of the zero-coupon bonds purchased and B the number of two-year bonds purchased. The total present value is: 2 1783.76 /1.05 (100 /1.06 1100 /1.06 ) 0.95238 1073.3357 .
A B A B = + + = + To exactly match the cash flow at time one, A + 100B = 1000. Substituting B = 10 – 0.01A in the first equation gives 1783.76 = 0.95238A + 10733.357 – 10.733357A for A = 8949.597/9.780977 = 915. The amount invested is then 915/1.05 = 871. 151. Solution: B The company must purchase 4000 in one-year bonds and 6000 in two-year bonds. The total purchase price is 2 4000 /1.08 6000 /1.11 8573.
+ = 152. Solution: C The modified duration is 11/1.10 = 10. Then, (0.1025) (0.10)[1 (0.1025 0.10)10] 0.975 (0.10).
P P P ≈ − − = Therefore, the approximate percentage price change is 100(0.975 – 1) = –2.50%. 153. Solution: B 7.959 1.072 (0.08) 1000 942.54.
1.08 P ≈ = 154. Solution: E Modified duration = (Macaulay duration)/(1 + i) and so Macaulay duration = 8(1.064) = 8.512. 8.512 1.064 112,955 107,676 1.07 MAC E = = and 112,955[1 (0.07 0.064)(8)] 107,533.
MOD E = − − = Then, 107,676 107,533 143.
MAC MOD E E − = − = 43 155. Solution: C The Macaulay duration of the portfolio is 35,000(7.28) 65,000(12.74) 10.829.
35,000 65,000 + = + Then, 1/10.829 10.829 1.0432 1.0432 105,000 105,000 100,000 1.004516 0.0385.
1 1 100,000 i i i = ⇒ = = ⇒= + + 156. Solution: A 1.05 ln(121,212 /123,000) 121,212 123,000 3.8512.
1.054 ln(1.05 /1.054) MAC D MAC D = ⇒ = = Then, 3.8512 /1.05 3.67.
MOD D = = 157. Solution: A I provides cash flows that exactly matches the liabilities. II only has PV(A) = PV(B), which is not sufficient for exact matching. III describes Redington immunization, not exact matching. 158. Solution: D Let F be the face amount of Bond X. Then, 15 15 15 15 2695.39 200 and 3490.78 200 2 a Fv a Fv = + = + . Subtract the first equation from the second to obtain 15.
795.39 Fv = Then for bond X, 15 15 2695.39 200 795.39 (2695.39 795.39) / 200 9.5 a a = + ⇒ = − = . This implies i =0.0634. Then 15 15 9.5 (1 ) / 0.0634 1 0.0634(9.5) 0.3977 v v = − ⇒ = − = and 795.39 / 0.3977 2000.
F = = The coupon rate is 200/2000 = 10.0%. 159. Solution: D The value of 1 invested with bank P after three years is 3 1.04 0.02 1.144864.
+ = The yield from Bank Q satisfies 3 1/3 1.144864 (1 ) 1.144864 1 0.04613 4.6%.
i i = + ⇒= −= = 44 160. Solution: D With a continuously compounded annual interest rate of 6%, 0.06.
v e− = The value of the first annuity is 1.2 1.2 20 0.06 20 0.06 20 1 20 20 1 600,000 ( ) 102.614 .
1 e e a v e X Ia X X X d e − − − − − − − − = = = = − Hence, X = 600,000/102.614 = 5847.155. Then the value of the second annuity is 1.5 1.5 25 0.06 25 0.06 1 25 25 1 5847.155 5847.155 779,366.
1 e e a v e d e − − − − − − − − = = − 161. Solution: B The amount of principal repaid at payment 15 is (where R is the quarterly payment) 40 15 1 26 26 10,030.27 (1.03) 10,030.27(1.03) 21,631.19.
Rv R R − + − = = ⇒ = = The amount of interest in payment 25 is 40 25 1 16 21,631.19(1 ) 21,631.19(1 1.03 ) 8,151.35.
v − + − − = − = 162. Solution: E The present value of the payments (4000 at month 36 plus the payments of X) must match the present value of the present value of the amounts borrowed (4000 at month 0 plus the payments of 800). The quarterly interest rate is 0.264/4 and all payment times should be in quarters of a year. On that time scale, the 4000 at month 36 is at time 12. The payments of 4000 are at times 1/6, 3/6, 5/6, …, 71/6 and there are 36 such payments. One way to write the present value of these payments is 36 12 0.5 1 3 4000 .
0.264 0.264 1 1 4 4 n n X − = + + + ∑ The payments of 800 are at times 1, 3, 5, 7, 9, and 11,, in quarters. One way to write the present value of these payments plus the initial debt of 4000 is 6 2 1 1 800 4000 .
0.264 1 12 n n − = + + ∑ These are the two sides of equation in answer choice E. 45 163. Solution: B Let r be the coupon rate for Bond A. The coupon rate for Bond B is then r + 0.01. Then, 20 20 20 0.1 20 0.1 20 20 0.1 20 0.1 1 1 1600 1000 ( 0.01) (1.1) (1.1) 2 1.6 2 0.01 0.29729 17.02713 0.08514 (1.1) 1.6 0.29729 0.08514 0.0715 7.15%.
17.02713 ra r a ra a r r = + + + + = + + = + + − − = = = 164. Solution: E Let n be the number of payments and let j be the interest rate per half-year. Because the given values are n – 1 half-years apart, 1 7,968.89(1 ) 19,549.25.
n j − + = Also, 1 1 1 1 7,968.89 /19,549.25 7,968.89 1,000 1,000( 1) 1,000 1 1,000 1 n n n v a a j j − − − − = = + = + = + . Then, 1 7,968.89 /19,549.25 0.085 7,968.89 1 1,000 j − = = − for 2 (1.085) 1 0.1772 17.7%.
i = −= = 165. Solution: B The denominator of the duration is the present value of the annuity: 20 20 0.02 30 0.02 4 78.1729 .
Xa Xv a X + = The numerator is the time-weighted present value of the annuity. In units of X we need the present value of 0, 1, …, 19, 80, 84, …, 196. One way to view this is as four times a 49-year increasing immediate annuity (so payments of 4, 8, …, 76, 80, 84, …, 196) less three times a 19-year increasing immediate annuity (so payments of 3, 6, …, 57). The present value is: 49 0.02 19 0.02 4 ( ) 3 ( ) [4(655.2078) 3(147.4923)] 2,178.3542 .
X Ia X Ia X X − = − = The duration is the ratio, 2,178.3542/78.1729 = 27.87. 166. Solution: A ( ) 22 2 8 2 14 1/14 100 1 1 1 2 100 1 1 1 2 2 2 2 2 1 2 (2 1)2 0.1015 10.15%.
i i i i i i + + − = + + − = + = − = = 46 167. Solution: D Let C be the amount of the semiannual coupon for bond B. 10 10 0.03 10 10 0.035 40 1000(1.03) 1085.30 1085.30 1000(1.035) 8.3166 708.9188 (1085.30 708.9188) / 8.3166 45.2566 45.2566 2 0.0905 9.05%.
1000 X a X Ca C C y − − = + = = = + = + = − = × = = = 168. Solution: C Let X be the original loan value. From the original loan terms, 15 50 X a = . Under the revised repayment plan, 5 10 5 50 30 X a v a = + . Equating the two gives 5 15 10 5 50 50 30 a a v a = + which does not match answer A. All the other choices use s. Multiplying both sides by 10 (1 ) i + gives 5 15 10 5 50 50 30 v s s s = + , which is answer C. This can also be obtained by equating the values of the two payment streams at time 10 rather than time 0. 169. Solution: A The effective monthly rate is 1/12 1.065 1 0.0052617 −= . The accumulated value is 180 0.0052617 180 0.0052617 180 0.0052617 1097 5( ) 180 1097(298.733) 5 0.0052617 300.3049 180 327,710 5 442,031.
0.0052617 s Is s + − = + − = + = 170. Solution D Fund K receives 1000 at the end of each year and also receives interest payments of 1300, 1235, 1170, …, 65. The accumulated value is 20 0.0825 20 0.0825 20 20 0.0825 1000 65( ) 20(1.0825) 1000(47.0491) 65 0.0825 97.6311 47.0491 47,049.1 65 86,902.
0.0825 s Ds s + − = + − = + = 47 171. Solution: D For Q the accumulated value is 25 0.09 84.7009 Xs X = . For R the accumulated value is 24 0.08 24 0.08 24 72.1059 24 100(25) 9( ) 2500 9 2500 9 7911.91.
0.08 0.08 s Is − − + = + = + = Then X = 7911.91/84.7009 = 93.41. 172. Solution: E The accumulated values for Funds X and Y are 10 1000 1 2 k + and 10 921.90 1 2 k − − respectively. Equating them and solving for k: 10 10 10 10 2 2 2 1000 1 921.90 1 2 2 0.9219 1 1 1 2 2 4 1 0.9919 4 0.0324 0.18.
k k k k k k k k − + = − = + − = − − = = = 10 0.18 1000 1 2367.36.
2 P = + = 173. Solution: A The 3-year interest rate is 3 1.07 1 0.225043 −= . Then, 0.225043 1.225043 735 5.443595 0.225043 Xa X X ∞ = = = and X = 735/5.443595 = 135.02. 174. Solution: B 0.06 0.06 2 2600 9( ) 1 1 1.06 9 0.06 1.06 0.06 2500 0.06 (2600 2500)(0.06) 6.
Pa v Ia P P P ∞ ∞ = + = + = + = − = 48 175. Solution: B ( ) ( ) ( ) 10 10 5 10 5 10 10 5 10 10 5 2 5 1/5 475 400 1 1 475 400 475 1 400 1 875 400 75 0 400 400 4(875)(75) 0.6 2(875) (1/ 0.6) 1 0.1076 10.76% i i a a v a v v v i i v v v v v v i ∞ = + − − + = − = − + − − = ± + = = = −= = 176. Solution: E Based on the effective yield rate, 2 100 /1.1 2 /1.1 39.03 X X X = + ⇒ = . After one year, the outstanding loan balance is 100 + 8 – 39.03 = 68.97. For the balance to be zero after two years, 68.97(1 ) 2(39.03) 0 78.06 / 68.97 1 0.1318 13.2%.
i i + − = ⇒= −= = 177. Solution: D The amount borrowed is 5 5 0.1 5 0.1 1000 2000 8498.35.
a v a + = The outstanding balance after five years is 5 0.1 2000 7581.57 a = . The principal repaid is 8498.35 – 7581.57 = 916.78. The interest paid is 5000 – 916.78 = 4083.22. 178. Solution: B The accumulation is 10 0.08 10 0.08 10 15.6455 10 10 0.12 ( ) 10 0.12 10 0.12 18.4683 .
0.08 0.08 s X X Is X X X X X − − + = + = + = Then, X = 10,000/18.4683 = 541.47 179. Solution: C The equation of value is 30 30 0.05 30 0.05 30 0.05 30 0.05 30 10,000 ( 5) 5( ) ( 5) 5 0.05 16.14107 6.94132 ( 5)(15.37245) 5 15.37245 843.11275 0.05 (10,000 843.11275) /15.37245 705.36.
a v X a Ia X a X X X − = + − = + − − = + − = − = + = 49 50 180. Solution: A 2 3 4 5 150 100 150 50 150 150 150 600 221.94 1.15 1.15 1.15 1.15 1.15 NPV − − = − + + + + + = − 181. Solution: A I is true. II is false, the price sensitivity of assets and liabilities must be equal. III is false, the convexity of assets should be greater than the convexity of liabilities. 182. Solution: D The effective annual rate of interest is 12 (1.005) 1 0.06168 −= . The present value of the tuition payments six months before the first payment is 6 4 0.06168 25,000(1.005) 24,262.95(3.66473) 88,917.16 a − = = . The accumulated value of the deposits at that time is 0.005 1000 n s . Equating the two amounts: 1.005 1 88,917.16 1000 0.005 1.44459 1.005 ln(1.44459) / ln(1.005) 73.75.
n n n − = = = = Therefore, at least 74 payments will be required. 183. Solution: E Let x be the annual payment amount. Macaulay duration is 2 3 7 2 3 7 1 2 3 7 17.6315 1.1 1.1 1.1 1.1 3.62 4.8684 1.1 1.1 1.1 1.1 x x x x x x x x + + + + = = + + + + . Alternatively, the duration can be calculated as 7 0.1 7 0.1 ( ) / .
Ia a 184. Solution: E PV of liabilities is 2 3 402.11(1/1.1 1/1.1 1/1.1 ) 1000.
+ + = Duration of liabilities is 2 3 402.11(1/1.1 2 /1.1 3/1.1 ) /1000 1.93653.
+ + = Let X be the investment in one-year bonds. To match duration, since zero-coupon bonds have duration = maturity, 1.93653 = [X + 3(1000 – X)]/1000. Then, 2X = 3000 – 1936.53 = 1063.47 and X = 532. 51 185. Solution: E A change in face value multiplies all cash flows by the same amount. Therefore, there is no change in the duration. If the coupon rate increases, the coupons become larger, but the redemption value stays the same. This causes payments prior to redemption to receive more weight relative to the payment at redemption and thus the duration will decrease. 186. Solution: A We are given (4) 8% i = and want to determine (12) /12 i . The equation that links the two and its solution is: 12 4 4 (12) (4) 4/12 (12) 1/3 (12) 8% 1 1 1 12 4 4 8% 1 1 12 4 8% 1 1.
12 4 i i i i + = + = + + = + = + − 187. Solution: E Let m be the monthly payment and i be the monthly interest rate. The interest in the first payment is 125,000i and the principal repaid is 125,000 – 124,750 = 250. Thus m = 125,000i + 250. Similarly, for the second payment, m = 124,750i +252. Thus, 250i = 2 for i = 2/250 = 0.008 and then m = 1250. To obtain the number of payments, the equation to solve is 0.008 125,000 1250 1 1.008 100 0.008 0.2 1.008 ln(0.2) / ln(1.008) 202.
n n n a n − − = − = = = − = 188. Solution: C Let x and y be the amount invested in the five and twenty year bonds respectively. To match the present values: 0.07(10) 0.07(15) 500,000 500,000 423,262.
x y e e − − + = + = To match the durations, noting that the denominators of the durations for assets and liabilities are the same, 0.07(10) 0.07(15) 5 20 500,000(10) 500,000(15) 5,107,460.
x y e e − − + = + = Subtracting five times the first equation from the second one gives 15y = 2,991,150 for y = 199,410 and x = 423,262 – 199,410 = 223,852. 52 189. Solution: C Let n be the term of the bond in half-years. We know that 601 1080 n v = and thus 601/1080 n v = . Then 0.05 1 1 601/1080 8.87037.
0.05 0.05 n n v a − − = = = The purchase price of the bond is 0.05 40 1080 40(8.87037) 610 956.
n n a v + = + = 190. Solution: B Principal repaid in the first payment is 1000 – 10,000(0.04) = 600. Therefore, the principle repaid in the tenth payment is 9 600(1.04) 854 = and the interest paid is 1000 – 854 = 146. 191. Solution: C After one year the outstanding balance is 48 0.025 500 13,886.58 a = . This must match the present value of the revised payments: 6 12 6 0.025 36 0.025 13,886.58 500 4.74964 8,757.69 (13,886.58 8,757.69) / 4.74964 1,079.85.
Xv a v a X X = + = + = − = Alternatively, each missing payment is being replaced with a larger payment six months later. The larger payment should be the payment due plus the missed payment with interest, or 6 500 500(1.025) 1,079.85.
+ = 192. Solution: A Only Bond III can match the liability at time 3. The bond must mature for 1000. Only Bond II can match the liability at time 2. The face value and coupon must total 1000. If X is the face value, then X + 0.02X = 1000 and thus X = 980.39. Only Answer A has these to values. To check, Bond II also provides a coupon of 0.02(980.39) = 19.61 at time 1. Therefore, Bond I must provide the remaining 980.39 from its coupon and redemption value. If Y is the face value, then Y + 0.01Y = 980.39 for Y = 970.68. 193. Solution: A Cash flows (in thousands) are 12, 12, 12, 12, and 162. The first bond provides payments of 10, 10, 10, 10, and 110. Therefore, the second bond must provide 2, 2, 2, 2, and 52. This implies a coupon rate of 2/50 = 4% and a face amount of 50. Only Answers A and B provide these. At an 8% yield, the price of this bond is 42.015 (or 42,015). 53 194. Solution: C Let i be the yield rate. Then, 30 30 30 30 30 1/30 3609.29 2000(2 ) 2250(1 ) 4000[1 (1 ) ] 2250(1 ) (1 ) (4000 3609.29) / (4000 2250) 0.22326 0.22326 1 0.051251.
i i a i i i i i − − − − − = + + = − + + + + = − − = = −= Modified duration is Macaulay duration divided by one plus the yield rate: 14.14/1.051251 = 13.71. 195. Solution: A The amount of the dividends does not matter, so they will be assumed to be 1. First, calculate the Macaulay duration. The present value of the dividends is 4 4 1.1 (1/ 0.1) 6.83013.
v a − ∞= = The numerator is the present value of “payments” of 5, 6, 7, … starting five years from now. This can be decomposed as a level of annuity of 4 and an increasing annuity of 1, 2, 3, … . The present value is 4 4 2 4 2 4 1 1 4 1.1 [4 ( ) ] 102.452.
0.1 1.1 0.1 i v a Ia v i i ∞ ∞ + + = + = + = The Macaulay duration is 102.452/6.83013 = 15. The modified duration is 15/1.1 = 13.64. 196. Solution: A Let r be the semiannual coupon rate. For the original bond, 6 6 0.05 1000 1000 5075.692 746.215.
P ra v r = + = + For the modified bond, 12 12 0.05 49 1000 1000 8863.252 556.837.
P ra v r − = + = + Subtracting the second equation from the first gives 49 = –3787.56r + 189.378. The solution is r = 0.037 and the coupon is 37. 197. Solution: A Let h(i) be the present value of the cash flows. For Redington immunization, the value of the function and its first derivative at 25% must be zero and the second derivative must be positive. X is immunized because: 3 2 4 3 5 (0.25) 102,400 192,000 /1.25 100,000 /1.25 0 (0.25) 192,000 /1.25 100,000(3) /1.25 0 (0.25) 192,000(2) /1.25 100,000(3)(4) /1.25 196,608 0 h h h = − + = ′ = − = ′′ = − + = > Y is not immunized because: 2 3 2 3 4 (0.25) 158,400 342,000 /1.25 100,000 /1.25 100,000 /1.25 0 (0.25) 342,000 /1.25 100,000(2) /1.25 100,000(3) /1.25 6,400 0 h h = − + + = ′ = − − = − ≠ Z is not immunized because 2 3 (0.25) 89,600 288,000 /1.25 100,000 /1.25 300,000 /1.25 51,200 0 h = − + + − = ≠ 54 198. Solution: E The bond sells at a premium, so the worst-case scenario is redemption at time six. Then, 6 6 6 6 6 1/6 1023 1000 1000(1 ) 0.96 1000[1 (1 ) ] 1000(1 ) 0.96 (1 ) 0.448 0.448 1 0.1432 14.32%.
i i a i i i i i − − − − − = + + = − + + + + = = −= 199. Solution: E Answer E is false because the convexity of the assets must be greater than the convexity of the liabilities. 200. Solution: E The accumulated value to time 4 is 3 0.42 0.32 1.26 0.42 3 3 0.5 0.08(4 ) 0.32 0.42 0.32 1 1 1 100 ( ) 100 100 100 657.
0.42 0.42 t t t t e e e e e e dt e e dt e − − = = = = ∫ ∫ 201. Solution: E The value at time 17 of the payments beginning at time 18 is 2 2 1 1 (1 ) 1 1.035 2500 2500 2500 .
1 1.035 0.035 0.035 1 1.035 k k k k k k + + + + + + = = + − − The total present value is 2 17 15 0.035 1 115,000 2500(1.035 ) 2500 0.035 46(0.035 ) 10.7516(0.035 ) 0.55720(1 ) 1.61 0.37631 0.55720 0.01889 1.89%.
46 10.7516 0.55720 k a v k k k k k − + = + − − = − + + − − = = = − + 55 202. Solution: B The initial payment, X is 19 20 21 2 20 1 1.02 1.02 1/1.03 1.02 /1.03 200,000 17.7267 1.03 1 1.02 /1.03 1.03 1.03 11,282.42.
X X X − = + + + = = − = The final payment is 19 11,282.42(1.02) 16,436.36.
= 203. Solution: E The annual payment is 10 0.1 10,000 / 10,000 / 6.14457 1627.45.
a = = The balance at time 3 is 7 0.1 1627.45 1627.45(4.8684) 7923.08.
a = = With one-half year at simple interest, the balance at time 3.5 is 7923.08(1.05) = 8319.23. 204. DELETED 205. Solution: E Let x be the amount invested in Bond A and y the amount invested in Bond B. Then 2y is invested in Bond C. To match the present value of the assets and liabilities: 20.5 2 190,000(1.07) 3 47,466.39.
x y y x y − + + = + = To match the Macauley durations, 10 15 30(2 ) 20.5 47,466.39 x y y + + = . Then, 20.5(47,466.39) 10(47,466.39 3 ) 75 20.5(47,366.39) 10(47,466.39) 11,075.49 75 30 y y y = − + − = = − and X = 47,466.39 – 3(11,075.49) = 14,239.92. 56 206. Solution: B Let X be the price of Bond X. Then, for the two bonds: 2 2 0.035 2 2 0.035 10,000(0.03) (1.035) 969.52 10,000(0.025) ( 50)(1.035) .
n n n n X a c X a c − − = + − = + + Subtracting the second equation from the first gives 2 2 0.035 2 2 2 969.52 50 50(1.035) 50 969.52 [1 (1.035) ] 50(1.035) 0.035 1.035 459.05 /1478.57 0.310469 (0.5)ln(0.310469) / ln(1.035) 17.
n n n n n a n − − − − = − = − − = = = − = 207. Solution: B With simple interest, the deposit in Bank X earns 1000(0.07) = 70 in year 8. With compound interest, the earning in Bank Y in year 8 is 32 28 1000(1.0125) 1000(1.0125) 72.14.
− = The absolute difference is 2.14. 208. Solution: E Split this into two perpetuities. One starts at time 0.5 at 500 increasing by 10 every year. The other starts at time 1 at 500 with payments increasing by 10 every year. The semiannual interest rate is 1.0750.5-1=0.0368221. The present value of an increasing perpetuity immediate is found using the formula: 2 P Q i i + , where P is the initial amount and Q is the increase amount. The first perpetuity, valued at time 0: 2 500 10 (1.0368221) 8755.39 0.075 0.075 + = The second perpetuity, valued at time 0: 2 500 10 8444.44 0.075 0.075 + = The total is 8755.39 + 8444.44 = 17,199.83. 209. Solution: B 10 0.05 10 0.05 5000(10) 5000 ( ) 100,000 10 5000 50,000 0.05 13.206787 10 10 0.05 0.15592 i Is s i i i + = − = − = = 57 210. Solution: C ( ) 10 0.08 10 0.10 Payment equals: 10,000 / 1,490.29 Accumulated total equals: 1,490.29 23,751.46 23,751.46 10,000 13,751.46 a s = = − = 211. Solution: D 120 120 0.01 Value of fund after 20 years: 500 (1.01) 383,404.42 383,404.42 (1 ) (1.01) 0.01 3796.08 s X X X i d i X = = = + = = 212. Solution: D 12 12 12 0.10 Using the retrospective method, OB 12,000(1.10) 1000 37,661.14 1000(21.38428) 16,276.86 s = − = − = 213. Solution: E Using the BAII Plus calculator: n = 12 PV = 911.37 PMT = -40 FV = -1000 CPT I/Y and you get 5.0% is the half-year rate: (2) 2 . (1 ) (1 0.05) ; 0.1025.
2 i i i + = + = 214. Solution: B ( ) ( ) 4 0.08 0.12 100 1 200 4 4 ln 0.97 ln(200 /100) 0.08 0.0418 ln 2 16.57 n n e n n n n − − = − = + = = 58 215. Solution: D Equating present values gives 10 10 1.092 1 5000 1.04 .04 .092 1.04 3377.82 (12.094127) 279.29 X X X − = − = = 216. Solution: E 20 0.05 ( ) 962.92 (0.04 0.05 ) 962.92 ( 0.01 )12.46221 962.92 0.875378 1100 n P C Fr Ci a C C C a C C C C = + − = + − = + − = = The discount is 1100 – 962.92 = 137.08. 217. Solution: D Equate the accumulated value of the deposits to the present value of the perpetuity: ( ) 10 10 10 10 10 10 10 10 20 (using the BAII Plus) 12.3% s Is i i s s i − = = − = = ⇒= The PV of the perpetuity is 10/0.123 = 81.30. 218. Solution: C 20 0.08 10 0.08 1300 4000 12,763.59 4000 (6.71008) 1306.03 a Xa X − = − = = 219. Solution: C 3 2 3 2 5 5 2 3 1 2 6.16078 2.6452 2.32908 2 1 5 4.0137 2.6233 1.5300 2.6452 2.6233 0.0219 L A v v d v v Xv Xv d Xv Xv + = = = + + = = = + − = 59 220. Solution: D The bank’s accumulated value at the end of 30 years is: ( ) 30 0.04 30 0.05 30 100,000 364,841 100,000 1 364,841 0.044 s a i i = + = = 221. Solution: E 999.35 x 1.06 = 1059.31 will be available to make the first payment of 1000, leaving 59.31 to be reinvested at X%. 817.65 x 1.072 = 936.13 will be available from the second bond to make the second payment of 1000, leaving 63.87 to come from the reinvestment of 59.31. X = 100(63.87 / 59.31 – 1) = 7.69. 222. Solution: E 4 5 4 1 4 5 4 5 1 4 1 4 (1 ) (1 ) (1 ) (1.09) (1 ) (1.095) 0.1152 s f s f f + + = + + = = 223. Solution: C 1 1 mod (1 ) (1 ) (1 ) (1 ) (1 ) n n n n d i n i di D n i i i − −− − − − + + = − = = + + + 224. Solution: E The amount invested in three-year bond is equal to the PV of the third year’s payout, ( ) 3 1000 1.1 751.31 = The amount invested in one-year bond is equal to the PV of the first year’s payout, 1000 1.08 925.93 = 925.93 751.31 174.61 − = 225. Solution: B PV = 04 .
1 10 + ( ) 2 045 .
1 12 + ( ) 3 055 .
1 15 + ( ) 4 07 .
1 20 = 9.615 + 10.989 + 12.774 + 15.258 = 48.64 60 226. Solution: D Let i = yield rate, r = coupon rate (if any), F = face value, P = price, n = # of years. For the first bond: 36 36 0.8 0.8 0.006218 P F Fv v i = = = = For the second bond: 0.006218 0.006218 4 0.8 (0.006218) 9 0.8 (0.0027634) n n n n P F Fv Fa v a = = + = + Using the BAII Plus, where PV=0.8, I/Y=.6218, PMT=0.0027634, FV=1 CPT N results in n=72. 227. Solution: B Since the bond has no coupons, the Macaulay duration is the same as the amount of time until maturity, namely 4 years. Thus, the effective annual yield rate, y, is 1/4 1200 1 0.046635 1000 −= . The modified duration equals the Macaulay duration divided by (1 + y). Thus the modified duration is 4 3.82177 1.046635 = years. 228. Solution: C Using the general Macaulay duration formula: t t t t R v t R v ∑ ∑ where R is the cashflow: Period Cashflow PV at 8% Period ×PV 1 10 9.26 9.26 2 12 10.29 20.58 3 15 11.91 35.73 4 20 14.70 58.80 5 30 20.42 102.10 Total 66.58 226.47 Macaulay duration = 226.47/66.58 = 3.401472 years 229. Solution: C When a company’s position is Redington immunized, its position is definitely protected from sufficiently small changes in yield rate, in either direction. However, its position may or may not be protected from large changes in yield rate. 61 62 230. Solution: D Amount of loan = L 10 10% 6 10 10% 4 10% 6 7% Initial expected yield rate 10.00% Annual payment / Accumulated value at time 10 ( / )( 1.07 ) L a L a s s = = = + 1/10 1/10 6 4 10% 6 7% 10 10% 1/10 Accum Value Yield rate = 1 1.07 1 4.6410(1.5007) 7.1533 1 6.1446 8.67% L s s a − + = − + = − = 231. Solution: E Let L = the loan amount. Note that 5 1 (1 ) i j + = + . The equation of value is 5 120 k i k j P a L a ⋅ = = ⋅ so that 5 5 5 120 1 (1 ) 120 1 (1 ) 1 (1 ) 120 1 (1 ) 120 (1 ) 1 120 k j k i k k k k a P a j i j i i i j i i j j j − − − − = − + = − + − + = − + = + − = Next, using the fact that 0 < j < 0.04, we get 5 (1 ) 1 5 5.41633 j j + − < < by plugging in a small value like 0.000000001 and 0.04 resulting in P equaling more than 600 but less than 650. 63 232. Solution: B Using the retrospective method: 6 6 0.05 4000(1.05) 250 5360.38 1700.48 3659.90 s − − = 233. Solution: D Let t represent the number of years since the beginning of year 1. Since the annual effective interest rate is 3% in each of years 1 through 10, and 2% each year thereafter, the present value of an amount is calculated by multiplying it by a discounting factor of 1 (1.03)t if 0 10 t ≤≤ , and 10 10 1 (1.03) (1.02)t− if 10 t > . The balance is initially 0 (the account is new before the first deposit). Deposits of X are made at times t = 0, 1, 2, 3,…, 25, or equivalently at time 1 t k = − for each whole number k from 1 to 26 inclusive. For the final balance to become 0, a withdrawal of 100,000 at time t = 25 would be needed. Since the net present value of the cash flows (withdrawals minus deposits) must be zero, in a time period from a zero balance to another zero balance, we have 11 26 10 25 10 1 10 1 10 1 12 100,000 1 1 0 (1.03) (1.02) (1.03) (1.03) (1.02) k k k k X X − − −− = = − − = ∑ ∑ 11 26 10 15 1 10 11 1 12 100,000 1 1 (1.03) (1.02) (1.03) (1.03) (1.02) k k k k X X − − = = = + ∑ ∑ . 234. Solution: A For the first bond: 20 20 0.065 (0.076)(6000) 6000 P a v = + P = 6727.22. For the second bond: 20 20 0.065 6727.22 ( )(7500) 7500 r a v = + 7500r = 417.37, so r = 0.0556. 64 235. Solution: E 40 40 40 40 40 40 | 40 40 40 40| 75 257.18 60 ( ) 257.18 15 15 257.18 (0.252572) 320.33 257.18 250.01 X Y X X Y P Fra Cv a Cv P P a C K v P P a Kv Kv a K K = + = + = − = + + − = = − = − = − = 236. Solution: C The PV of the first twenty payments is: 20 1.04 1 1.10 14,000 157,337.48 0.10 0.04 − = − The PV of the remaining payments starting at time 21 is: 19 20 1 14,000(1.04) (1.01) (1.10) 49,202.44 0.10 0.01 − = − Total equals 206,539.92. 237. Solution: B 1 16,000 1,000 0.057 ( /100) 1 16 0.057 /100 1 0.057 /100 16 0.55 r r r r = −− = + + = = 238. Solution: E Let j equal the five-year interest rate. 5 2 (1 ) (1.09) 0.538624 2 10 38.18 0.538624 (0.538624) j j PV + = = = + = 65 239. Solution: A From the first bond: 8 8 0.03 25 P a Cv = + From the second bond: 4 4 0.03 0.93 25 P a Cv = + Multiply the first equation by 0.93 and plug into the second equation: 4 8 0.03 4 0.03 0.93(25) 0.93 25 163.2078 0.73415 92.9275 0.88849 70.2804 0.15434 455.37 a C a Cv C C C C + = + + = + = = 240. Solution: D The PV of the liability is 2 600,000 548,387.92 1.046 = and its Macaulay duration is 2. Then, equating present values: 92 .
387 , 548 046 .
1 046 .
1 4 = + y x And equating durations: 4 ( /1.046) ( /1.046 ) (1) (4) 2 548,387.92 548,387.92 x y + = Solving the system of equations results in x = 382,409 241. Solution: E Solution: 3 2 4000(1.04 ) 1400[1 (1 ) (1 ) ], i i = + + + + or i solves the quadratic equation: i2 + 3i −0.2139 = 0. Thus, i = 6.97% because the other root is negative. 242. Solution: B 4 10 0.01 0.01 6 0.01 100,000 0.05 100,000 (5.79548)(0.96098) (20)(0.90529) 100,000 (23.6751) 4223.85 X Xa v v X X X X = + = + = = 66 243. Solution: D Using BA II Plus: 180 0.075/12 0.075/12 60,000 556.21 49,893 556.21 132 m Xa X a m = = = = So, (180 – 132) = 48 payments have been made so far: 0.6/12 49,893 556.21 119.3 n a n = = Use 120 future payments including the smaller one. 48+120=168. 244. Solution: C 20 20 0.10 20 0.10 20 2000 110 0.10 2000 110(8.51356) (55.40691) 1063.51 (55.40691) 19.1945 a v a X X X X − = + = + = = 245. Solution: E 10 1.15 1 1.05 100 1483.62 0.05 0.15 PV − = = − 246. Solution: D ( ) 1 1 4 1 5 4 5 4 4 1 1144.5 865 865 0.75579 1144.5 0.75579, 0.93240, 0.07251 (0.07251) 797.50 10,999.02 n n n n OB Ra Rv P Rv Rv Rv v Rv v v i I X X − − − + − = = = = = = = = = = = = = = = 67 247. Solution: C Given that the problem states that the inequality is true for all interest rates from 0% to 10% and all values of Y, it is sufficient to determine it for one set of values. Select i = 7% and Y = 121. Then, 3 3 121/ (1 3(0.07)) 100 121/ (1.07) 98 / 77 121(1 0.07(3)) 95.59 121(0.93) 97.33 Q R S T = + = = = = − = = = Hence, S < T < R < Q 248. Solution: C The yield rate on Kate’s bond is (2) 10 10 2 (2) (1000 100) 25 1000 0.0371551 2 i a v i − = + = The discount on Wallace’s bond is 8 8 0.05 (1000 ) 25 1000 1000 838.42, 161.58 D a v D D − = + − = = The book value of Kate’s bond at time 1 is 8 8 0.0371551 25 1000 917.19 B a v B = + = The difference is B – D =917.19 – 161.58 = 755.61 68 249. Solution: B 2 4 3 5 1 3 2 4 1000(1 ) 300(1 ) 1153.84 2000(1 ) 1200(1 ) 2667.91 (1 ) (1 ) 1153.84 0.95238 0.86384 (1 ) 3 (1 ) 2667.91 0.90703 2.46811 L L L L A L P i i P P i i P P X i Y i X Y P X i Y i X Y − − − − − − − − = + + + = ′ = − + − + ′ = − = + + + = + ′ = − + − + − = − − So, we have two equations and two unknowns. Solving simultaneously, we get: 953.57, 346.61, 2.75.
Y Y X X = = = 250. Solution: C 20 0.03 1000 67.22 67.22(20) 1344.31 1344.31 1000 (1000 950 900 50) 344.31 (10,500) 0.03279 Pa P i i i = = = = + + + + + = = 251. Solution: B (12) 1/12 120 6 1 120 24 1 24 1.08 1 0.006434 12 600 (1/1.006434) 1254.47 1254.47 673.42 i P P P v −+ − + = −= = = = = 252. Solution: B 20 20 0 0 2 ln(1 2 ) 1 2 20 5 41 41 (1 ) 0.204035 (1 0.204035) 2.53 dt t t e e i i + + ∫ = = = + = + = 69 253. Solution: B ( ) ( ) 10 20 20 10 20 10 20 10 10 600 600 1000 1 3 5 1 5 5 3 3 0 5 3 2 3 9 4(5)(2) 3 7 Let 0.4 9.59582% 2(5) 10 600 1 0.4 8753.8 0.0959582 i v a i i v v i i v v v v x v i X ⋅ = + − = + − = + = + − −± + −± = = = = ⇒= = + = 254. Solution: B 2 2 2 150 10 5000 5000 150 10 0 150 ( 150) 4(5000)( 10) 10,000 0.06217 i i i i i i = + − − = ± − − − = = 255. Solution: D If paid in one lump sum, the total interest paid is 20 (1.05 1) 1.65330 .
X X − = With level payments for 10 years, the total interest paid is 10 0.05 10 0.29505 .
X X X a − = Then, 1.65330 1000 0.29505 736.24.
X X X = + = 70 256. Solution: B 20 1 20 1 20 1 21 21 (1 ) 2 1 0.5 1.05 2 (21 )ln(1.05) ln 2 7 t t t t t t t P I v v v v t t −+ −+ −+ − − = = − = = = − = = 257. Solution: B 2 2 10,000 10,815 20,800 20,800 20,800 4(10,815)(10,000) 21,630 0.970873 or 0.952381 0.03 or 0.05 0.03 0.05 0.02 v v v v i + = ± − = = = − = 258. Solution: A 30 30 0.042 4 4 0.021 45 1200 1108.85 1108.85 20 1376.69 A A n n P a v P a v = + = = + Using the BA II Plus: PV = 1108.85 PMT = 20 FV = 1376.69 I/Y = 2.1 Solve for 4n and get 4n = 48, n = 12. Or solve the equation to get 4 0.36876 n v = and then solve for n. 259. Solution: C Statement I should have an 60 s on the left. Statement II has an annuity-due rather than an annuity-immediate on the right. Statement III is correct. 71 260. Solution: D 15(961.54)(1) 20(966.14)(2) 30(878.41)(3) 2.198495 15(961.54) 20(966.14) 30(878.41) + + = + + 261. Solution: B 17 18 0.10 1,000,000(1.10) 1,000,000 5,054,470.285 (45.59917) 1,000,000 4,054,470.285 (45.59917) 88,915.43 Ps P P P − = − = = = 262. Solution: D Using time 5 as the first reference point, then bringing that value back to time 0: ( ) 5 5 5 500 500 i i v a Ia + This combines a five-year level annuity-due of 500 plus an increasing annuity-due starting with 500 and increasing by 500. 263. Solution: B Let face amount equal 1. ( ) 18 18 18 18 18 1 1.61 2.25 1.61 2.25(1 ) 1.25 0.64 0.96342 1 1.45 2.25 1.45 2.25 1 1.25 0.8 0.64 ln 0.963492 ln 0.64 12 n n n n n n v i v i v v v v v i v i v v v v n n − = + = − + = = − = + = − + = = = = 72 264. Solution: C 10 10 0.10 2 20 20 0.048881 450 1000 Using BA II Plus calculator: 10.50 1.10 (1 ) , 0.048881 5.25 1000 451.64 Xa v X j j P a v P = + = = + = = + = 265. Solution: E Let I be the amount of interest in the first month. 0 1 1 0 , ( ) P m I P I P P m − + = = − − In the first month, the interest 1 0 ( ) P P m − − was charged on a principal of 0 P , so the effective monthly interest rate (expressed as a decimal) of the first loan is 1 0 1 0 0 0 ( ) P P m P P m P P − − − + = . The nominal annual interest rate (expressed as a decimal) for both loans is therefore 1 0 0 12 P P m P − + , so the effective daily rate (expressed as a decimal) for the second loan is 1 0 0 12 365 P P m P − + . Finally, the effective monthly rate (expressed as a decimal) for the second loan is 365/12 1 0 0 12 1 1 365 P P m P − + + − . 266. Solution: D. (0, ) (1 ) (0, ) (1 ) (1 ) (1 ) (1 ) (0, ) (0, ) m n m m n n P m i P n i i X i i P m X P n − − − − + − = + = + + = = + + = 73 267. Solution: A 180 120 0.005 180 0.005 360 0.005 2000 180,146.91 180,146.91 1520.18 1520.18 253,553.61 OB a Pa P L a = = = = = = 268. Solution: E The conditions 1) assets and liabilities have equal present values and equal modified durations, and 2) the convexity of its assets exceeds the convexity of its liabilities are precisely what is required for Redington immunization. 269. Solution: E 6 6 6 6 90.17 4 132.47 4 1.6 j j a Xv a Xv = + = + Multiply the first equation by 1.6: 6 6 144.272 6.4 1.6 j a Xv = + Subtract the second equation: 6 11.802=2.4 j a Use annuity calculation on BA II Plus: (2) (2) 6% , 12% 2 i j i = = = 270. Solution: B 2 0.05 2 9297 5000 5000 4535.12 1.05 Pa P = = = 74 271. Solution: B Using the BA II Plus calculator: 2 (2) (2) 10 10 0.019804 5 5 1.04= 1 2 0.019804 2 25 1000 1046.72 1046.72 100 30 1000 4.2036% A A j j i i P a v P a v j + = = + = − = + = 272. Solution: E Value at time 5 years: 20 20 0.02 0.02 100,000(1.02) 2500 87,851.32 5000 m s a − = m = 21.86, using the BA II Plus. Since we want a balloon payment, use m = 21. 21 21 0.02 87,851.32 5000a Bv = + B = 4236.70, so balloon equals 5000 + 4236.70 = 9236.70 273. Solution: B 40 40 0.05 20 20 0.04 20 20 0.04 900 1000 900 Using BA II Plus: 1000 45 45 900 1022.31 1022.31 1100 38.28 ra v r P a v P Fra v Fr = + = = + = = + = 75 274. Solution: C Let Y indicate the nominal value of the two-year bond, then: 2 0.05 1.05 9465= + , so 10,000.
1.08 1.08 Y Y Y = Thus, the amount of liability at the end of the second year is 10,500. Hence, the liability at the end of the first year is: 10,500 = 5250.
2 So, the amount invested in the one-year bond is: 5250 10,000(0.05) 4481.
1.06 − = 275. Solution: C Macaulay duration of the liability is 3. Asset duration must equal 3. Let P1 and P4 be the present values of the two assets. 1 4 4 1 1 4 3 1 1 1 4 4 4 1 4 3 1 4 3, then 2 20,000 20,000 50,000 1 1 , , : 1 0.8(1 ) : 50,000 1 2 2 (1 ) (1 ) (1 ) 1.25; (1 ) 1.077217 P P P P P P P P i P P i i P P i i i i + = = + + = = = = = = + + + + + = + = PV of assets must equal PV of liabilities. So, PV of assets equals: 1 1 1 20,000 2 3 3 55,699.07.
1.077217 P P P + = = = Amount of liability equals: 55,699.07(1.077217)3 = 69,623.83. 276. Solution: D ( ) 10 29 0.06 10 19 0.06 10 29 1 20 0.06 147.20 130.70 16.67 147.20 72.98 3.08 223.26 a v Da v v v + + = + + = + + = 76 277. Solution: C 14 13 2 12 14 14 15 2000 (1.08) (1.03)(1.08) (1.03) (1.08) (1.03) 1.08 1.03 /1.08 1 1.03/1.08 32.284 61.95 X X X X X X X = + + + + − = − = = 278. Solution: E The present value of annuity X is 10 10 1 1.0331 1.0331 .
v a i − = The present value of annuity Y is 2 12 10 2 10 2 2 1 ( ) .
1 (1 ) 1 v v v P v v P P v i − − + + = = − + − Equating the present values and solving, 2 2 (1 ) 1 1.0331 1.0331 1.0331(2.075) 2.14.
i P s i + − = = = = 279. Solution: D 12 (12) (12) 60 3 1 60 33 1 33 1.12 1 12 0.00948879 12 1 900 1.00948879 1556.43 1 1556.43 1194.78 1.00948879 i i P P P −+ − + = + = = = = = 280. Solution: C ( )10 0.05 10 0.05 5000 10 0.08 10 5000 10 0.08 0.05 75,654.30 X Is s X X = + − = + = 77 281. Solution: B 8 12 8 0.03 38 1000 1056.16 BV a v = + = 282. Solution: D 2 3 4 1 2 2 2 4 3 2 2 4 3 2 2 4 3 2 2 2 2 2 2 2 2 2 2 (1 ) ( ) (1 ) (1 ) (1 ) ( ) (1 1 ) (1 ) ( ) (1 1 ) ( ) 2 ( ) ( ) 2 2 ( ) 1 1 0 0.61803 1 1.61803 0.61803 I I i P P v v i v v v v i v v v v v v v v v v v v v v v v v v v v v v v i i + = + + − + − = + + − + − = + + − + − = + − + = + = + = + + −= = + = = 283. Solution: C 2 2 2 2 2 500(1 ) 600 0.095445 100(1 ) 100(1 ) 600(1.10) 100(1.10) 600 236 100(1 ) 100(1 ) 236 (1 ) (1 ) 2.36 0 1 1 9.44 (1 ) 2 0.115549 0.0201 X X Y Y Z Z Z Y Y Y Y Y Y Y X + = = + + + = + = + = + + + = + + + − = −± + + = = − = 284. Solution: B Assuming the loss of interest is the loss of the last 9 months interest: ( ) ( ) ( ) 3 0.75 2.25 3 1.08 1.08 1.189 1 5.94% j j − = = = + = 78 285. Solution: C 20 0.0925 10 360 % 12 1000 57,485.26 500 57,450.21 57,485.26 57,450.21 35.05 s a = = − = 286. Solution: B Let j equal the quarterly rate and let k equal the four-year rate. ( ) 4 1 4 1 300 (1 ) 1 299 1 1 1 299 0.054875 (1 ) 300 15.61 j j j k k X k k X = + = + = + = + = = 287. Solution: A The outstanding loan balance at any point in time is equal to the present value (at that same point in time) of the remaining installment payments. 20 40 40 0.98 1 1.0075 1000(0.98) 6889.11 0.0075 0.02 B − = = + 288. Solution: C 5 10 0.06 5 400 1000 1944.03 1944.03 0.06 461.51 OB a Xa X = − = = = 79 289. Solution: C Option 1 20 0.05 1500 120.3639 Total Payment 120.3639(20) 2407.28 Pa P = = = = Option 2 ( ) ( ) ( ) 19 0 Total Payment 7520 1500 1500 75 1500 275 (1500 1975) 1500 1500 (20) 75 19 20 1500 30,000 75 2 1500 15,750 k i i i i i i k i i i = = + + − + − ⋅+ + − = + − ⋅ = + − ⋅ = + ∑ 2407.28 1500 15,750 0.0576 i i = + = 290. Solution: C (2) 15 5 15 /2 (2) (2) 1,100,000 1,000,000 40,000 .03153 2 0.06306 i MV v a i i = = + = = 291. Solution: B 2 2 2 4 (4) (4) 150 100 80 80 100 150 0 100 100 4(80)( 150) 160 0.880199 0.136106 1.136106 1 4 0.129664 v v v v v v i i i = + + − = − ± − − = = = = + = 80 292. Solution: B Since the bond is bought at a premium and redemption will occur to the investor’s greatest disadvantage, assume the bond is called at the earliest possible redemption date, or simultaneous with the coupon payment occurring at the end of the 15th year. The cash flows then become a bond paying semiannual coupons of 40 for 15 years and returning 1000 at the end of the 15th year. At a yield of 7% effective annually: 2 (2) (2) 30 30 1.07 1 2 0.034408 2 40 1000 1103.61 i i P a v P = + = = + = 293. Solution: A ( ) ( ) ( ) ( ) ( ) 4 5 3 4 1500 1 4000 1 ( ) 100 500 1 1000 1 P i i i P i i i − − − − ′ + − + − = + + − + 294. Solution: E 4 2 2 1.05 1.04 (1 ) 0.0601 f f = + = 295. Solution: B Besides present values of assets and liabilities matching, 1) their modified durations must also match, and 2) the convexity of the assets must exceed the convexity of the liabilities, in order for the company’s position to be immunized against small changes in interest rate. Only company V satisfies all these conditions. 296. Solution: B Let j be the monthly interest rate. 0.012 50,000(1 ) 800 49,800 0.012 50,000 800 116.22 n j j a n + − = = = = Drop payment at payment number 117. 81 297. Solution: C 27 27 27 27 27 21 0.75 37 21 0.75 (1 ) 37 16 0.18243 37 0.03248 0.75 4 (1 ) 3 9 i n n F iFa Fv v v v i F Fv i n = + = − + = = = + = = 298. Solution: A 5 5 1.04 1 1.04 1 1.12 3.87 17.26 21.13 0.12 0.04 1.12 0.12 0.08 − + = + = − − 299. Solution: B Let x be the amount invested in the one-year bond and y the amount invested in the four-year bond. First match the present value of assets and liabilities: 2000 A L PV PV x y = + = Second, the durations of assets and liabilities should also match: 1 4 1 4(2000 ) 3 2000 666.67 A A L x y D x y x x D D x + = + + − = = = = Convexity of the assets is: 2 2 666.67(1 ) 1333.33(4 ) 11 2000 + = Convexity of the liability is: 2 3 9 = . Convexity of assets is greater than convexity of liabilities so Reddington immunization is met. 82 300. Solution: C 40 0.02 20 20 20 0.02 20 0.02 5,000,000 182,778.74 5,000,000(1.02) 183,000 2,983,318.31 2,983,318.31 200,000 17.89 n Xa X OB s OB a n = = = − = = = 20 original payments plus 18 with the drop payment equals 38 total payments. 301. Solution: A Since the coupon rate per coupon payment period 4% is greater than the effective rate of interest per coupon payment period 2.9563%, it is to the disadvantage of the bond holder to have the bond redeemed at an early date. Hence, we only need to calculate the present value of such a bond at the worst-case scenario, which is that the bond is called at the end of the 5th year. 10 10 0.029563 4 100 108.92 P a v P = + = 302. Solution: D 20 20 0.9 1 1.03 100 (1.03) 0.03 0.10 1295.80 FV FV − = + = Alternatively, 19 20 1 19 18 19 0 1 (1.03) (0.9) (1.03) 100[(1.03) 0.9(1.03) (0.9) (1.03) ] 100 1295.80.
1 0.9(1.03) FV − − − = + + + = = − 303. Solution: D First, find ( ) a t 2 0 0 ( ) exp exp exp /100 50 t t r r a t dr dr t δ = = = ∫ ∫ The balance in the account at time 10 is: X a a X a 31637 .
2 48 .
815 ) 4 ( ) 10 ( ) 10 ( 300 + = + The total interest earned from 0 10 t to t = = is: 815.48 2.31637 (300 ) 4 192.08.
X X X X + − + = => = 83 84 304. Solution: A Since Asset Y provides a cash flow at the same time that the liability is due (t = 4), we can apply its 250,000 value to reducing the liability amount from 750,000 to 500,000. Then, we can establish the following two equations, both using t = 4 as the reference point for all cash flows. 2 2 500,000 (0.95) (0.95) X Z X Z A v A v A A − − = + = + Second, taking the derivative (with respect to v) of both sides of the first equation, we have: 3 3 2 0 2 2 (0.95) 0 2 (0.95) (0.95) X Z X Z X Z A v A A A A A − − − = − + = − + = − + Then, subtracting the second equation from the first equation yields: 2 500,000 3 (0.95) 150,416.67 X X A A − = = 305. Solution: D 4 (4) (4) 5 1.06 1 4 0.014674 4 50,000(1.014674) 2,125 31.86 n i i a n = + = = = There will be 31 payments of 2125. 306. Solution: A The PV and duration of the liability payments using 7% rate are 12 1,750,000 777,021 PV v = = and duration 12. The amount invested in the 5-year bond is 5 242,180 172,671 1.07 = , Thus, the amount invested in the 14 year bond is 777,021-172,671=604,350. The maturity value of the 14-year bond is 14 604,350(1.07) 1,558,337.
= The surplus if the interest rate moves to 4% is: 5 14 12 242,180 1,558,337 1,750,000 5,910 1.04 1.04 1.04 A L PV PV − = + − = . 85 307. Solution: A Let ( ) ( ) ( ) A L h i PV i PV i = − . Full immunization of a single liability requires both equations: ( ) 0, '( ) 0 h i h i = = 3 2 1 3 2 4 3 1 3 20,000 0 3 40,000 0 Av A v v Av A v v + − = + − = 1 .
1.055 v = Solve these two equations in two unknowns to get 1 A = 9478.67 308. Solution: D Let 1 2 3 , , F F F be the redemption amounts of each bond for purchase. To exactly match the liabilities with cash income: 3 3 2 2 1 1 1000 1000 1000 (1.02) 980.39 1000 980.39(0.02) (1.01) 970.69 F F F F F F = = = = − = = The total purchase price is 2 3 970.69(1.01) 980.39(0.02) 980.39(1.02) 1000 2241.82.
1.14 1.15 1.15 1.18 + + + = 309. Solution: B The PV of the annuity following the 11th payment is: 9 0.06 10 68.0169.
a = The effective semi-annual rate is (2) 1/2 1.06 1 0.02956301.
2 i j = = −= Next, 1 68.0169 0.02956301 0.005 1.67 PV K K = = − = 86 310. Solution: B Split this into three perpetuities with payments 3 years apart. Find the three-year interest rate: ( ) 1 3 1.125 1 , 0.423828 j j = + = . The present value of the three perpetuities, starting at times 0, 1, and 2 is: 2 100(1.423828) (1.423828) 100(1.423828) 9450 0.423828 0.423828(1.125) 0.423828(1.125) 2963.19.
X X + + = = 311. Solution: E Note that had the borrower 1) charged X at the end of month 0, and 2) paid off the remaining 3000 at the beginning of month 16, then the initial and final balances would have become 0. In that situation, the present value of the amounts charged to the credit card, minus the present values of the payments, would have been 0 in the 15-month period. The amounts charged to the card were 79.99 at each of times 0.5 12 n t − = , for each whole number of n from 1 to 15. The monthly payments were 250 at each of times 12 n t = , for each whole number n from 1 to 15 inclusive. Then to make the final balance 0, the final (additional) payment would have been 3000 at time 15 5 12 4 t = = . Therefore, we have ( ) ( ) ( ) ( ) ( ) ( ) 15 15 0.5 5 1 1 12 12 4 15 15 0.5 5 1 1 12 4 12 79.99 250 3000 0 1.168 1.168 1.168 79.99 3000 250 1.168 1.168 1.168 n n n n n n n n X X − = = − = = + − − = + = + ∑ ∑ ∑ ∑ . 312. Solution: A The accumulated value is 2 0 ( ) exp 0.03 0.005 exp[0.03 0.0025 ] t a t r dr t t = + = + ∫ , The account balance is (2) 1.0725 100 (2) 100 100(1.0725) 100 211.07.
(1) 1.0330 a a a + = + = 87 88 313. Solution: E Break this into two parts – the first 30 increasing payments and the remaining level perpetuity. ( ) ( ) 30 29 29 1.03 1 1 1.07 350 1.07 350 1.03 8033.38 0.07 0.03 0.07 v − + = − 314. Solution: E Let n index the payment times in months. Then for a payment at time n, the discount factor is ( ) /6 1 1.045 n . The end-of-month payment at month n is be 500 ( 1) + − n X . Therefore, we have ( ) 60 /6 1 500 ( 1) 30,000 1.045 n n n X = + − =∑ . 315. Solution: A Let i be the yield rate, v = 1/(1 + i), and let n be the term. For Bond A, 20,000 10,000 n v = and so 0.5 n v = . For Bond B,10,835.58( 0.04 ) 10,000 n n i v a + = and so 10,000 0.5 / 0.04 10.5721 10,835.58 ni a = − = . For Bond C, 10,000 ( 0.03 ) 0.81716 12,237.51.
n ni X v a X X = + = ⇒ = 316. Solution: C [ ] 10 2 10 0.052 1 1 50,000 ( 200) 200 0.052 0.052 50,000 (7.647284) (0.602341) 19.230769 3846.153846 73964.50 50,000 19.230765 46,868.54 162.83 ka v k k k k k = + + + = + + + = + = 89 317. Solution: B Let A r represent the coupon rate of bond A. The coupon rate of bond B is then 0.005 + A r . From the given information, ( ) 30 30 30 0.07 30 0.07 30 30 0.07 30 0.07 1 1 3000 1000 0.005 (1.07) (1.07) 2 3 2 0.005 (1.07) 3 0.26273 24.81808 0.06205 3 0.26273 0.06205 0.1078 10.78% 24.81808 A A A A A r a r a r a a r r = + + + + = + + = + + − − = = = . 318. Solution: C [ ] 20 12 1 12 9 7 ( ) 1 7 0.0375 9.7497 9.7497 (0.0375) 35 1193.33 P C i g v Ci Cg Ci Fr C C − + = = − = − = − = − = 319. Solution: C 40 40 0.025 0.025 350 1.1 8785.97 (0.409674) 14,883.24 1.1 16,371.57 14,724.91 350 16,371.57 48 24 A A A A A A m m P a P v P P P R P a v m n = + = + = = = = + = = 90 320. Solution: A Let i represent the common yield rate of the two bonds. Since the modified duration is the Macaulay duration divided by (1 + i) and i > 0, the Macaulay duration of each bond is greater than its modified duration. Since a < d < b, the Macaulay duration of d years must be associated with the bond with modified duration a years. Since the bonds have the same yield rate, the ratio of the two types of duration is the same for each bond. So if x represents the Macaulay duration of the other bond in years, we have d / a = x / b implies ax = bd implies x = bd / a. The Macaulay duration of the other bond is bd / a years. 321. Solution: C ( ) ( ) 3x4 2 4 0.05 3 0.04 0.06 1000 1 1 4 2 1670.42 x e − + − = 322. Solution: C ( ) ( ) ( ) 45 45 45 45 45 0.0525 75.6 7 1.2 7 1 1 75.6 1.2 7 1 7 63 1 10 0.0525 7 1200 i s i i i i i i i X s = + − = + − = + = = = = 91 323. Solution: C 10 8 9 8 2 1.06 1 1.10 20,000 0.10 0.06 2584.39 1.06 1.06 7353 1.10 1.10 X X LB X − = − = = + = 324. Solution: D [ ] 0 0 5 1 ( ) exp 8 exp ln( 8) exp ln( 8) ln(8) 8 8 exp ln 8 8 (5) 13/ 8 1.08333 (4) 12 / 8 0.08333 t t a t dr r t t t t a a i = + = + = + − + + = = = = = ∫ 325. Solution: D 2 2 /2 98.5 3 100 0.037929 2 0.07586 j a v j j = + = = 326. Solution: D ( ) ( ) ( ) 1 13 0.04 20 4 20 0.04 20 4 20 0.04 Adjustment in book value 0.08 0.04 43.24 1800 1800 0.08 1800 144 1800 2778.50 n t i g i v F v F F P a v a v −+ = − = − = = = + = + = 92 327. Solution: A 1 1 1.0614 5.5554 0.0614 7 n n − = = 328. Solution: C 36 36 0.01 36 0.01 8000 15,000 1000 1.01 8408.60 8408.60 (30.10751) 279.29 Xa Xa X X − = + = = = 329. Solution: B 1 10 20000 20000 9.562 1.1 1 1.09 0.09 0.1 2091.61 R = = − − = The fourth payment is ( ) 2091.61 1.1 2300.71 = The principal outstanding before the fourth payment is 9 1.1 1 1.09 2300.71 19708.94 0.09 0.1 − = − Interest in the fourth payment is ( ) 19708.94 0.09 1773.80 = = Principal repaid 2300.71 1773.80 526.91 X = = − = 330. Solution: E Interest earned on the first account is ( ) ( ) 3 2 0 0 1 1 3 3 1 3 1 2 3.
dt dt t t e e + + ∫ ∫ − = + − + = Interest on the second account is ( ) 3 2 1.05 1.05 0.055125 X X − = . 3 0.055125 54.42 X X = ⇒ = . 93 331. Solution: D 5 3 2 1.0725 1.0575 (1 ) 0.0954 f f = + = 332. Solution: C ( ) 5 5 0.08 2 0.08 150,000 14,000 1.08 150,000 (88,703.01 2.2464 )1.46933 5,958 s Xs X X = + = + = 333. Solution: A I is false as it is true for small parallel changes in interest rates only. II is false, as the durations will be equal. III is false as the convexity of assets should be greater than the convexity of liabilities. 334. Solution: B 5 20 0.04 5 5 9550(1 ) 756.97 9550(1 ) 10,698.97 (1 ) 1.12031 0.02298 i a i i i + = + = + = = 335. Solution: B ( ) 600(1 ) 1000 10 (1 ) 6 0.6 600 2 1 600 2 1 600 1 0.6 0.6 2 750 n n n n ni n n i i v i F a Fv i v F v i F F + = + = = = + − = + = − + = 94 336. Solution: B 29 0.03 29 0.03 29 40( ) 30(1000) 40 30(1000) 53,433.89 0.03 s Is − + = + = 337. Solution: C 0.0625 33 33 0.0625 33 33 5020 360 33.85 Use 33 5020 360 5020 4980.95 39.05 288.75 360 288.75 648.75 n a n n a Xv Xv Xv X B = = = = + = + = = = + = 338. Solution: A For a 7.5% yield rate, the present value and Macaulay duration of the assets are, respectively, 30,000 + 20,000 = 50,000 and 30,000(28) 20,000(35) 30.8 30,000 20,000 + = + . The present value and Macaulay duration, of the liabilities are, respectively, 50,000(1.075) 50,000 (1.075) y y = and y. Note that the present values of assets and liabilities already match. Since Macaulay durations must match, y = 30.8. 339. Solution: B Use the full immunization equations and let N be the maturity value of the asset maturing in n years. 7 ( 12) 7 ( 12) 242,180(1.07) (1.07) 1,750,000 0 242,180(7)(1.07) ( 12)(1.07) 0 n n N N n − − − − + − = − − = From the first equation: ( 12) 7 (1.07) 1,750,000 242,180(1.07) 1,361,112.
n N − − = − = Substituting this in the second equation: 7 12 242,180(7)(1.07) /1,361,112 2 n − = = and so n = 14. 95 340. Solution: B 2 2 3 29 30 30 30 30 2895.28 (1.01) (1.01) ...
(1.01) 1000 1.01 1 1.031 1000 2895.28 0.031 0.01 113.75 Rv R v R v R v v R v R = + + + + + − => + = − => = 341. Solution: C ( ) 0.12 1 0.12749685 1 3000 1.12749685 0.12749685 0.07 58,829.14 i e i = − = − = 342. Solution: E If accumulating the payments, they accumulate for 0, 1, …, 39 periods. This accumulation is reflected in the left-hand sides of (A) and (D). Since the accumulated value is X, neither right-hand side is correct. If discounting the payments to time zero, the payments are discounted for 1, 2, …, 40 periods as reflected in the left-hand side of (E). The right-hand side of (E) discounts the accumulated value of X for 40 periods and hence is the correct value. Answers (B) and (C) do not reflect the time-zero present value. 343. Solution: D The cost to purchase the bond at the end of year 3 is 19 .
1885 ) 03 .
1 ( 2000 2 = × − . Subtracting this cost from 2260.19 we get 375, the amount of interest paid which is 3.75% of 10000. Thus the interest rate is 3.75%. 96 344. Solution: A Let a, b, and c represent the face values of the three bonds. One, two, three, and four years from now, respectively: the 1-year bond provides payments of 1.01a, 0, 0, 0; the 3-year bond provides payments of 0.05b, 0.05b, 1.05b, 0; and the 4-year bond provides payments of 0.07c, 0.07c, 0.07c, 1.07c. The total payments one, two, three, and four years from now must match the liabilities. Therefore, we have 1.01 0.05 0.07 5766 0.05 0.07 1.05 0.07 15421 1.07 7811 + + = + = + = = a b c b c X b c c Note that to find X, we do not need the first equation. Solving the fourth equation for c yields 7811 7300 1.07 = = c . Substituting this value of c into the third equation and solving for b yields 15421 0.07(7300) 14200 1.05 − = = b . Finally, substituting these values of b and c into the second equation yields 0.05(14200) 0.07(7300) 1221 = + = X . 345. Solution: E (2) (2) 2 2 0.059 49 (2) 0.10 980 0.10 0.018 0.118 0.059 2 915.70 49 1000 2 12 6 n n r i i a v n n = = = + = = = + = = 97 346. Solution: B The price of a one-year bond for 1000 is 2 30 1030 1017.45 1.021 1.021 + = . Therefore, to match the payment at time 2 we need to invest 15001017.45 1481.72 1030 = in the one-year bond. The one-year bond gives a payment of 1500 30 43.69 1030 = at time 0.5. Therefore, the amount that needs to be invested in the six month zero coupon bond is = 66 .
1922 0175 .
1 69 .
43 2000 = − . The total cost of the dedicated portfolio is: 1481.72 + 1922.66 = 3404.38. 347. Solution: A The current value for couch #2 is 0.1(4/12) 260 1500 1450.82 1190.82 e X X X − = − = − ⇒ = . The current value for couch #1 is 0.1(6/12) 0.1(2/12) 1500 1190.82 1426.84 1210.83 216.
e e − − = − = 348. Solution: A 20 1.05 1 1.04 2500 52,732.61 0.04 0.05 X PV − = = − Annuity Y has the same increasing percentage as interest rate, so: 30 Y PV k = 1757.75 X Y PV PV k = => = 349. Solution: A Solution: 0.5(0.4) (1/2) 2.4 2.0 2 1 exp 0.5 10 d dt t − − = − ∫ ( ) ( ) 0.2 (1/2) 2.4 2.0 2 0.2 (1/2) (1/2) 1 2 exp 2ln(10 ) | 8 1 2 7.6 0.20063 d t d d − − − = − − − = = 98 350. Solution: D The price of Bond A is 1 2 3 3 60(1.04 1.04 1.04 ) 1000(1.04 ) 1055.50, − − − − + + + = while the Macaulay duration of Bond A is 1 2 3 3 60[1.04 2(1.04 ) 3(1.04 )] 3(1000)(1.04 ) 2.838.
1055.50 − − − − + + + = Note that the one-year zero-coupon bond has duration 1. Let w denote the proportion of wealth to invest in Bond A; then, 1 w − is the proportion of wealth invested in Bond B. Then 2 2.838 1(1 ), w w = + − or w = 0.5440. 351. Solution: A Let K be the amount that Bank B paid. Equating the amount borrowed (the 16 payments discounted at 7%) to the actual payments received (using the 6% yield rate) gives the equation 8 16 7% 8 6% 1000 1000 1.06 K a a = + . Then, 8 16 7% 8 6% 8 1000( )1.06 1000(9.44665 6.20979)1.06 3236.86(1.59385) 5159.06.
K a a = − = − = = 352. Solution: A ( ) 640(1 ) 1000 (1 ) 1.5625 0.64 640 2 1 640 2 1 640 1 0.64 0.64 2 780.49 n n n n n i n n i i v i F a Fv i v F v i F F + = + = = = + − = + = − + = 353. Solution: E The price is P = 950. The modified duration is ' 4750 5 950 P P − − = − = . Macaulay duration is (1.09)(5) 5.45 = 99 354. Solution: B Because Bond A sold for its fact amount, the yield rate is the coupon rate of 6% per year. The present values of the three bonds are: Bond A: 1000 (given) Bond B: 5 1000(1.06) 747.26 −= Bond C: 10 1000(1.06) 558.39 − = The durations are: Bond A: 7.8017 Bond B: 5 Bond C: 10 The portfolio duration is the average of these three durations, weighted by the bond prices. Duration = 426 .
7 39 .
558 26 .
747 1000 ) 10 ( 39 .
558 ) 5 ( 26 .
747 ) 8017 .
7 ( 1000 = + + + + 355. Solution: A The present value of the two payments is: 5 1,000,000 1,000,000 1,821,927.07 1.04 + = The present value of the perpetuity is: 2 2 1000 (1.04) 1,821,927.07 0.04 0.04 1000 1,751,852.99 0.04 0.04 1,126,852.99 0.04 45,074.12 X X X X + = + = = = 356. Solution: C ( ) 6 2 6 2 4(2) 0.5 1000exp 5 0.5 1000exp ln(5 0.5 ) | 8 1000exp ln8 ln 6 1000 1333.33 6 0.08 1333.33 1 4 1134.35 X dt t t Y Y − = + = + = − = = = − = ∫ 100 357. Solution: D The quarterly interest rate is 1/4 1.08 1 0.01943 −= ( ) 1 3 5 24 0.01943 1000 1.08 1.08 1.08 19,406.51 2.40034 8085 a X X X − − − = + + = = 358. Solution: E 365 20 1 1 44.6% 0.98 −= 359. Solution: C First calculate the three-year interest rate. 3 1.06 1 0.191016 −= This is an arithmetic increasing perpetuity. 2 1 1 655.56 0.191016 0.191016 20.08 X X + = = 360. Solution: D The present value of the first year’s payments is: 1 12 12 0.006434 1.08 1 0.00643403 500 5756.43 a −= = This perpetuity can be thought of as a geometrically increasing perpetuity-due with first payment 5756.43. The present value is: 1 5756.43 1.08 0.08 0.05 207,231.44 − = 361. Solution: D ( ) ( ) ( ) 12 12 60 5000 1.01 5000 5000 1 1.01 1 44,423.95 81.67 543.94 i Xs i d i X X = = + = − = = 101 362. Solution: B 10 0.04 10 0.04 10 10 10,000(10) 600 600 137,295.27 0.04 55,400(1 ) 137,295.27 0.095.
s s i i − + + = + = = 363. Solution: E 5 0.04 5 0.04 5 1000(5) 50 50 5791.22 0.04 s s − + + = 364. Solution: A The total amount paid is n. The initial loan amount is n i a The total interest paid equals n i n a − . 365. Solution: C ( ) ( ) ( ) ( ) 5 5 2 2 2 5 5 1 100 1 0.05 exp 100 1.1025 exp ln(1 ) 1 6 110.25 220.50 3 100 220.50 1 100 1 0.453515 220.50 1 0.853726 0.1463 t t d d d d + = + + = = = − − = = − = = ∫ 366. Solution: A ( ) 5 5 0.13 3 0.13 1 2 100 50 1.13 1703.81 1703.81 1.12 1 99.33 s s x X + = = − = 102 367. Solution: B 12 240 0.006434 30 0.07 (1 ) 1.08 0.006434 225 128,848.51 128,848.51 128,848.51 (13.27767) 9704.15 j j s Xa X X + = = = = = = 368. Solution: C 20 0.015 4 16 0.015 4 10,000 582.46 582.46 8230.86 Pa P OB a OB = = = = Remaining payments total: 16(582.46) 9319.32 9319.32 8230.86 1088.46.
= − = 369. Solution: A 12 180 0.0073 59 121 0.0073 59 1.0912 (1 ) 0.0073 100,000 1000.02 1000.02 80,174.59 j j Pa P OB a OB = + = = = = = 12 1.0576 (1 ) 0.004678 j j = + = Interest portion of first revised payment: 80,174.59(0.004678) 375.04.
= 103 370. Solution: A Define i′ as the quarterly effective interest rate for the loan Solve for i′ 20 291.23 5000 1.5% i a i ′ = ′ = Then, solve for j using the formula ( ) 12 4 1 1 0.015 12 0.0597.
j j + = + = 371. Solution: B All prices imply an interest rate of 6% indicating that the yield rate does not change with duration, that is. the yield curve is flat. 372. Solution: E Macaulay duration at an interest rate of 5% is ( ) 5 5 5 5 100 5000 100 1000 100(12.56639) 3917.63 4.2535 100(4.32947) 783.53 Ia v a v + + + = = + 373. Solution: C 12 24 0.0075 0.0075 1000 21,889.15 21,889.15 2000 11.46 n OB a a n = = = = Drop payment will be made at the 12th month. 374. Solution: E 120 0.0075 0.0075 30,000 380.03 33,000 380.03 141 n Pa P a n = = = = 104 375. Solution: B 1 6 5.76 1 0.041667 5 4.8 1.041666 d v r r r v = + = + = = = 376. Solution: D ( )10 10 10 9.26 1 0.079914 32.70395 4.8720.
6.71270 i i Ia d a = + = = = = 377. Solution: E 0.01 200 0.0201 0.01 200 0.0201 0.01 2.01 2.01 98.13168 400 n n n Xa Xa a a a n = = = = 378. Solution: C Let r be the coupon rate. 20 20 0.07 2300 2000 2000 2000 168.32 ra v r = + = Bond is bought at a premium, so assume called as early as possible at year 18. 18 18 0.07 168.32 2000 2284.85 P a v P = + = 105 379. Solution: C Since the annual effective discount rate is 3.2%, the present value of an amount is calculated by multiplying it by a discounting factor of (1 0.032) (0.968) t t − = , where t is the number of years since the deposit. At time t = 0, an initial deposit of 50,000 is made just after the balance is 0 (the account is new just before the deposit). The withdrawals are then X at each of times t = 2, 4, 6, 8, 10, 12 or equivalently at time 2 t k = for each whole number k from 1 to 6 inclusive. Then to make the final balance 0, an additional withdrawal of 45,000 at time t = 12 would be needed. Since the net present value of the cash flows (withdrawals minus deposits) must be zero, in a time period from a zero balance to another zero balance, we have 6 12 2 1 45,000(0.968) (0.968) 50,000 0 k k X = + − = ∑ 6 12 2 1 45,000(0.968) (0.968) 50,000 k k X = + = ∑ . 380. Solution: E 0.045 3400 240 23.05 n a n = = Use 23 for the number of full payments of 240. X will be paid at time 24. 24 23 0.045 24 24 3400 240 3400 3395.47 4.53 13.04 a Xv Xv Xv X = + = + = = 381. Solution: C The minimum yield will occur at the earliest redemption date since the bond is brought at a premium. Therefore n = 18. 18 18 1321 111.10 1111 0.07985 i a v i = + = 106 382. Solution: A Conditions for full immunization of a single liability cash flow are: 1. PV (Assets) = PV (Liability) 2. Duration (Assets) = Duration (Liability) 3. The asset cash flows occur before and after the liability cash flow. Checking Condition 1, only I and II are present value-matched, so can rule out III at this point. Checking Condition 2, only I is duration-matched, so can rule out II at this point. Checking Condition 3, the asset cash flows do occur before and after the liability for I. So only I fully immunizes the liability. 383. Solution: B Let i represent the effective market annual yield rate and 1 1 v i = + . The Macaulay duration is 3.70 years, which is equal to the present-value-weighted times of the liabilities. Therefore, we have 5 5 5 5 5 5 5 20,000(0) 100,000 (5) 25 3.70 20,000 100,000 1 5 3.70 18.5 25 3.70 6.5 0.89342 1 1.11929 v v v v v v v v i + = = + + + = = = + = . Modified duration equals Macaulay duration divided by (1 + i), so the modified duration is 3.70 3.30567 1.11929 = years. 384. Solution: A Using the basic formula: ( ) ( ) 2 3 58 57 57| 58| 6.88 87.5 87.5 6.88 39.0667 0.00625 6.88 5150 BV BV a Cv a Cv C C − = + − + = − = = 107 385. Solution: E 12 12 0.02 12 12 0.02 1000 500 2 1000 P Xa v P Xa v = + + = + Subtract the first equation from the second. 12 0.02 500 47.28 Xa X = = 386. Solution: C First, note that the monthly rate of interest is 0.072/12 = 0.006. Also, the first payment occurs 60 months out, and the last payment occurs 240 months out. Thus, there are 181 payments in total. Then, the present value of the annuity is 180 60 181 180 180 60 60 0 750 1.01 1.01 1 ...
(1.006) 1.006 1.006 1.01 1 750 1.01 750 1.006 1.01 (1.006) 1.006 (1.006) 1 1.006 523.82(1.05085 / 0.0039761) 138,440.
k= + + + − = = − = = ∑ 387. Solution: C The price of the geometric perpetuity-due, as a function of the annual effective interest rate i, is given by 2 3 0.99 0.99 0.99 1 1 0.99 ( ) 1 ...
1 0.99 1 1 1 0.01 0.01 1 1 i P i i i i i i i + = + + + + = = = + + + + + + − + . Therefore, the modified duration is given by 2 0.99 ( ) 0.99 (0.01 ) 1 ( ) (0.01 )(1 ) 0.01 P i i i P i i i i − ′ + − = − = + + + + . Since Macaulay duration is modified duration times (1 ) i + , the Macaulay duration is 0.99 12 0.01 i = + . Solving for i gives 0.0725 i = . Therefore, the modified duration is 12 12 11.18881 1 1 0.0725 i = = + + years. 108 388. Solution: E Since the annual nominal interest rate is 6%, compounded monthly, the monthly effective interest rate is 6% 0.5% 12 = and the monthly discounting factor is 1 1.005 . The 10,000 borrowed today is the present value of a 30-month annuity immediate with monthly payments of P each, followed by a 40-month annuity immediate with monthly payments of 1.1P but with payments deferred by 30 months, followed by a 50-month annuity immediate with monthly payments of 1.2P but with payments deferred by 30 40 70 + = months. Therefore, ( ) ( ) ( ) ( ) 30 70 30 0.005 40 0.005 50 0.005 10,000 1.005 1.1 1.005 1.2 Pa Pa Pa − − = + + . 389. Solution: C Since the Macaulay duration of a set of cash flows is a weighted sum of the payment times, deferring by six years must add exactly 6 to this duration. The Macaulay duration is (1 + j) times the modified duration, so that (1 )(9 4) 6 j + − = , which implies j = 0.2. If P(i) denotes the present value of the original set of cashflows at rate i, we know that the modified duration '( ) '( ) 4 '( ) 40 ( ) 10 P j P j P j P j − − = = ⇒ = − . The modified duration of the altered payments is 0 '( ) 40 3.077.
3 ( ) 13 P j P j + − = = + The Macaulay duration is 1.2(3.077) = 3.69. 390. Solution: B For all 3 investments, convexity is ( ) / ( ) P i P i ′′ . Let C be the redemption value of the bond and the initial payments of the perpetuities (it turns out the value is irrelevant). For i) 50 51 52 52 50 2 ( ) (1 ) , ( ) 50 (1 ) , ( ) 2550 (1 ) Convexity 2550(1.05) /1.05 2550 /1.05 2313.
P i C i P i C i P i C i − − − − − ′ ′′ = + = − + = + = = = For ii) 1 2 3 3 1 2 ( ) , ( ) , ( ) 2 Convexity 2(0.05) / 0.05 2 / 0.05 800.
P i Ci P i Ci P i Ci − − − − − ′ ′′ = = − = = = = For iii) 1 2 3 3 1 2 ( ) ( 0.03) , ( ) ( 0.03) , ( ) 2 ( 0.03) Convexity 2(0.05 0.03) / (0.05 0.03) 2 / 0.02 5000.
P i C i P i C i P i C i − − − − − ′ ′′ = − = − − = − = − − = = Thus, convexity when ranked from lowest to highest is: 800 < 2,313 < 5,000, or y < x < z. 109 391. Solution: E ( ) ( ) 2 4 6 4 2 2 2 2 2 2 2 2 2 2 2 0 0 4 2 2 1 4 2 1 4 2 1 4 Xv Yv Xv Xv Yv X Y Y X v X X i Y Y X X i Y Y X X i Y Y X − + = − + = ± − = + = ± − + = ± − = − ± − 392. Solution: A ( ) ( ) 5 0.06 5 0.06 5 1260 200(5) 200 1000 200 0.06 1000 200 16.2553 0.08 s i Is i i i − = + = + = + = 393. Solution: C 5 5 0.03 5 1,000,000 20,000 (1.03) 50,000 1 1,000,000 94,341.97 (1.03) 50,000 50,000(1 ) 1,049,905.88 0.05 i a a i i i i i − ∞ − = + + = + + = = 394. Solution: D 5 4 5 4 1 8 5 ln( 8) 5 1 13 12 1.08333 8.3% dt t t i e e i + + ∫ + = = = = = 110 395. Solution: D ( ) 2 1 1 1.05 441.
0.05 0.05 + = 396. Solution: E If j is the semi-annual effective interest rate, then 1 + (1/j) = 23 j = 1/22 Then the annual effective interest rate is 2 (1 1/ 22) 1 0.093.
+ −= 397. Solution: E First determine the present value of the four payments within a year. Then determine the present value of ten of these. 4 4 0.03 10 0.12551 0.012 1 1 0.12551 4 100( ) P Ia a + −= = 398. Solution: D ( ) ( ) ( ) ( ) ( ) 2 2 10 15 1 1 1 1 134.5679 0.09 0.09 X 2 134.5679 1 2 0.42241 0.274538 57.83 Ia i i Ia v Ia v Ia ∞ ∞ ∞ ∞ = + = + = = − + = −× + = 399. Solution: A Accumulate values to time ten. ( ) 10 10 0.08 10 0.08 10 10 1.085 500 10,000 500 10,000 1.085 7,243.28 10,000 1.085 7626.45 X s s X = + + = + = = 111 400. Solution: D Repayment = 120 .08/12 1000 12.13276 a Accumulated loan repayments at 6% = 120 .005 12.13276 1988.31 s 10 1000(1 ) 1988.31 0.0711 j j + = ⇒ = 401. Solution: A 1 1 1 65.42 (1 ) (1 1/1.07) 1000 13,650(0.07) 955.50 1000 955.50 44.50 n n P v P P I P −+ = − = − = = = = − = 402. Solution: B ( )( ) ( ) 1 ' 1 1 1 ' ' 1 ' ' i r i i r i r i i i r i r + + = + + + + = + = + + 403. Solution: B 0.09 4 0.0225 1.0225^ 4 1.09308 0.09308310,000 930.83 = = = 5 5461 930.83 8.00% i s i = = 404. Solution: D 2 3 3 3 100 100 1100 1168.33 1.03 1.035 1.04 1168.33 100 1000 3.94% i MV a v i = + + = = + = 112 405. Solution: A Let P be the initial payment. Then, 60 1 1 12,000 (1.01) / (1 ) 60 /1.01 202.
t t t P i P P − = = + = = ∑ The interest in the first month is 0.01(12,000) = 120. The principal being repaid in the first month is 202 – 120 = 82. 406. Solution: B 1.09 1 0.038 1.05 −= 407. Solution: E At the end of 5 years and 60 payments, the borrower still owes 150,000. It will take n monthly payments of 1000 to retire the loan where 0.005 1000 150,000 278 n a n = = Thus, it will take 338 total payments. 408. Solution: C Suppose level payment is 500. 2 3 500 500 500 500 1.0425 1.045 1.0475 1.054 479.62 457.86 435.02 411.35 1783.85 PV = + + + = + + + = 4 500 1783.85 4.74% i a i = = 409. Solution: B ( )10 0.06 10 0.06 36.96241 5.02201 7.36009 Ia d a = = = 113 410. Solution: A The percentage by which the balance accumulates each year is the effective interest rate. 411. Solution: E We have 2 mod mod ( ) 1/ , ( ) 1/ , 1/ 15 1/15 (1 ) 15(1 1/15) 16 mac P i i P i i d i i d d i ′ = = − = = ⇒= = + = + = 412. Solution: A The Macaulay duration is 2 2 2 1 2 1 .
e e e e e e −δ −δ −δ −δ −δ −δ + = + + + 413. Solution: E The effective rate per month is 0025 .
0 12 03 .
0 = = i . Hence the original loan amount is 60 60 359.37 20,000 i i L Pa a . Using the retrospective approach, the outstanding loan balance is given by 36 11 10 36 36 36 11 10 36 (1 ) { (1 ) (1 ) } 20,000 (1.0025) {359.37 359.37 (1.0025) 359.37 (1.0025) } 9099.17 i i OLB L i Ps P i P i s 414. Solution: C The total interest payment of 400 in the 1-year period is 5% of the new balance of 8000, so by definition, 5% is the effective discount rate. 415. Solution: A The effective monthly interest rate is 0.072 0.006 12 i = = . Let A L = the amount of Loan A and B L = the amount of Loan B. We then have 48 A 48 48 1.5 / (1 ) i i L m a m a i = ⋅ + ⋅ ⋅ + 48 B 48 48 1.2 0.9 / (1 ) i i L m a m a i = ⋅ ⋅ + ⋅ ⋅ + Hence 48 48 48 48 A 48 48 B 48 48 1.5 / (1 ) 1 1.5 /1.006 1.12668 1.2 0.9 / (1 ) (1.2 0.9 /1.006 )1.006 i i i i ma ma i L L ma ma i + + + = = = + + + 114 416. Solution: C Let i = yield rate. Then, for bond A: 800 1000( / 2) 1000 500(1 ) 1000 0.6.
m m m m mi i a v v v v = + = − + ⇒ = The price of bond B is: 3 3 3 3 3 3 1000( / 2) 1000 500(1 ) 1000 500(1 ) 500(1 0.6 ) 608.
m m m m mi i a v v v v + = − + = + = + = 417. Solution: D No computations are needed. Since the desired yield rate of 6% is higher than the coupon rate of 5%, the investor is at the greatest disadvantage when the bond is called at maturity. Note that the two bonds are alike in all other ways, with the same maturity date, face value, desired yield rate, and coupon rate. So, the maximum price the investor should be willing to pay for the callable bond matches the price of the non-callable bond, namely 4361. 418. Solution: A The amount charged to the card was 1500 immediately. The annual fees were 20 after 1 year and 20 after 2 years. The monthly payments were X after 12 k years, for each whole number k from 1 to 24 inclusive. Then to make the final balance 0, the final payment would have been 200 after 2 years. Therefore, 24 0.18 0.18(1) 0.18(2) 0.18(2) 12 1 24 0.18 0.36 0.015 0.36 1 1500 20 20 200 0 1500 20 20 200 k k k k e e X e e e e X e e − − − − = − − − − = + + − − = + + − = ∑ ∑ . 419. Solution: D Let j be the quarterly interest rate. Then, 40 60 60 0.01285 3,000 5,000 .01285 1,000 950 950 39,565.26 441.59 (1,000 441.59) / 39,565.26 0.014114.
v j ra v r r = = = + = + = − = This is the quarterly rate. The annual rate is 0.014114(4) = 0.056456 = 5.65%. 115 420. Solution: D 5 1.03 1 1.07 4 4 0.07 0.03 1 350(1.07) 350(1.03) 0.07 1623.96 4293.23 5917.19.
X v X − − = + = + = 421. Solution: E The definition of “purchased at premium” is P > C; purchase price greater than redemption value. 422. Solution: A 180 360 0.0054167 225,000 1960 0.0054167 225,000 1414.50 i a i Xa X = = = = 423. Solution: A The Macaulay duration of bond B is 10 = B MacD . The Macaulay duration of bond A is 20 2 = = B A MacD MacD The price of bond A is i P 1 = , where i is the annual yield. So, the modified duration of bond A is ( ) 1 ( ) A P i ModD P i i ′ = − = . 1 (1 ) (1 ) 20 1/19 A A MacD ModD i i i i = × + = + = ⇒= , therefore 1 19 A ModD i = = . 424. Solution: D 1 1500 0.087 ( 0.02) 14,018.69 1 14,018.69 0.078 ( 0.02) 1373.83 PV PV P P = −− = = −− = . 116 425. Solution: C 2 4 3 3 5 4 3 5 4 0 1000(1 ) 750(1 ) (1 ) ' 0 2000(1 ) 3000(1 ) 3 (1 ) 2000(1.06) 3000(1.06) 3 (1.06) 1679.24 2241.77 2.37628 1650.06 PV i i X Y i PV i i Y i Y Y Y − − − − − − − − − = = + + + − − + = = − + − + + + + = + = = 426. Solution: B Matching the asset and liability present values and durations, we have the system of equations 0.5 0.75 0.5 0.75 0.5 0.75 0.25 0.25 10,000 0.5 7,500 0.5 2,500 5,000 5,000(1.02) 4,975 Xv Yv v Xv Yv v Xv v X v − + = + = = = = = 427. Solution: D Each coupon payment is 7000(0.06) 420 = . The accumulated value of the reinvested coupons after 18 years is: 18 0.057 420 12,617.50.
s = Therefore, 18 (1.0518) 12,617.50 7,500 8105.45 P P = + => = Finally, 18 18 8105.45 420 7500 4.91% i P a v i = = + => = 428. Solution: A The PV of perpetuity X is 2 1 1 3000 3000 125,939.41, where (1.0494) (1 ) j j + = = + The PV of perpetuity Y is 2 2 2 1.0494 10.36543134 , 1.0494 1 k k where j j = = + Setting PV’s equal gives k = 12,149.94. 117 429. Solution: A 10 10 0.10 10 10 0.10 0.09( ) 10 0.09 2.61323 3.85543 6.8922 0.55301 0.38554 6.8922 6.2656 1.10 Ia v d a v d v + = + + = = + = = 430. Solution: B Let i = yield rate. Let n be the term in years of the two bonds. Let F = the face value of the second bond. Then the coupon rate for the second bond = i/2. The equation of value for the first bond is 890 1000 0.89.
n n v v For the second bond we have 1 890 (1 2 ) (1 ) 2 2 2 2 1780 1780 941.80.
1 1.89 n n n n n n n i n i i v F F Fa Fv F Fv v v v i F v 431. Solution. E 1000(0.03) 30 Fr = = 1/2 (1.05) 1 0.024695 j = −= 9 11 9 1000 30 1042.35 j j B v a = + = 8 12 8 1000 30 1038.09 j j B v a = + = 1042.35 1038.09 4.26 − = 432. Solution: E (A) It does need to be satisfied (B) Requires more rebalancing (C) Convexity of assets must be greater than for liabilities (D) Convexity is greater than 0 (E) True 118 433. Solution: A OB, after year 5 10 5 100 378.17 v a = = OB, after year 5 15 100 378.11 a = = |378.17 – 378.11| = 0.06, round to 0 434. Solution: E First, use bond A to determine the interest rate. Because the equation for the interest rate cannot be solved algebraically, use a financial calculator to get i/2, which yields i/2 = 0.035, or (2) 0.07 i = . For bond B, 30 60 30 0.035 30 0.035 1000(0.08 / 2) 1000(0.10 / 2)1.035 1000(1.035) 735.68 327.63 126.93 1190.24, P a a − − = + + = + + = 435. Solution: A 40 40 0.025 30 30 0.025 20 20 0.025 Call after 20 years:30 1000 1125.51 Call after 15 years :30 1030 1118.95 Call after 10 years :30 1060 1114.56 a v a v a v + = + = + = The correct price is then the smallest or 1115. 436. Solution: B First, calculate the initial payment P as follows: 2 2 (1.10) 1,000,000 1000 909,090.91 0.1 0.1 80,909.09.
P Q i i P P + = + = = Thus, the payment on January 1, 2020 will be P + 20Q = 80,909 + 20(1000) = 100,909. The balance on 1 January 2020, immediately after the payment at that time, will be 2 100,909.09 1000 1000 1,119,090.90.
0.10 0.10 + + = 119 437. Solution: D The accumulation of discount at time 8 is 491.51, which is equal to 50 8 i s . Using a BA-II calculator, i can be computed as 5.8099441%. The accumulation of discount at time 25 is 50 25 i s , or 2670.91, the bond’s redemption value is therefore 20,000+2671=22,671. 438. Solution: E (60 15)/365 45/365 0.975 250 ( 250) 0.975 250 ( 250)(1.242) 4827.18 X X v X X X − − = + − = + − = 439. Solution: D ( ) 40 0.08 20 0.08 20 0.06 20,000 1677.20 1677.20 16,467.03 16,407.03 1435.67 20 1677.20 1435.67 62,257 a a a = = = + = 440. Solution: C The present value of the perpetuity-due, at annual effective interest rate i, is given by 2 2 1 1.0625 ( ) 17 0.0625 1 '( ) 1 '( ) 1 1 15.058824.
1 ( ) (1 ) 0.0625(1 0.0625) i P i i P i i P i i v i P i i i i + = = = − = − = = = = = + + + . The first-order modified approximation of the present value, at an interest rate i near an initial interest rate 0 i , is therefore given by ( ) ( ) ( ) ( ) approx 0 0 mod 0 ( ) 1 17 1 0.005 (15.058824) 18.28.
P i P i i i D i = − − = −− = Actual new price is 1.0575 18.39130.
.0575 = 18.28 18.39130 0.00605 0.61%.
18.39130 X Y Y − − = = − = − 120 441. Solution: E Let denote the discount factor applicable to a period of one month and let X be the constant periodic amount received by both charities. For Charity A, 198,000 .
1 X v = − Since is the discount factor applicable to a two-month period, for Charity B, 2 100,000 .
1 X v = − Dividing the first equation by the second gives so that The yearly discount factor is so that the annual effective interest rate is 1 1 0.274 27.4%.
0.78472 −= = 442. Solution: C 5 5 4 4 1 25 1 exp exp ln(20 ) exp(ln 25 ln 24) 20 24 1 0.041667 4.17% 24 i dt t t i + = = + = − = + = = = ∫ 443. Solution: C Let X be the monthly deposit of year 1. Then determine the present value of the first year’s monthly payments and then use geometric formula to accumulate to year 10. 12 0.024 1 1 0.024266.
12 i = + −= 10 10 12 0.002 1.10 1 1.024266 (1.024266)(1.024266) 21,234.05 0.024266 0.10 (11.869136)[13.74278] 16,311.46 100 Xa X X − = − = = The payments in year 5 will be 4 100(1.1) 146.41.
= 121 444. Solution: D The yield rate is equal to the coupon rate for Bond Z, so the price of Bond Z is 1000. This implies the price of Bond Y is 1,169.30. 4 2 4 19 2 19 1169.30 20 1000 , where 1.0325 (1 ) 4 76 19 1169.30 401.50 1000 , where 1.0325 1 44.25 n j n j i i a v j n n Ra v i R = + = + => = => = − = + = + => = 445. Solution: E Use Bond 1 to find information about 0 1 1 i i + + : 11.735(1 ) 0 0 0 11.735(1 ) 0 1 1 ( ) ( ) 21,635.83 20,400 1 1 1 1.060579902 1 mac D i i i i P i P i i i i i + + + + ≈ => = => + + + = + Next, 13.101(1 ) 13.101/11.735 0 1 1.060579902 1.067865963 1 i i i + + = = + The first-order Macaulay approximation for Bond 2 is: 13.101(1 ) 0 1 ( ) 20,400 21,784.47 1 i i P i i + + = = + 446. Solution: D There are two blocks of 42 payments (paid every two months) increasing in an arithmetic progression. The value of each of these blocks two months before the first payment is: 42 42| 42| 2 42 35.54114068 30.03429308 100 5 3554.1141 5 6989.024, 0.008016 where (1 0.048 /12) 1 0.008016 j j a v a j j − − + = + = = + −= The present value is then: 6989.024(1.008016) 6989.024(1 .048 /12) 14,062.03.
+ + = 122 447. Solution: A First, the size of the level payment does not have an impact on the duration. So, for the perpetuity-immediate: 2 1 1 1 1 ( ) 32.25 3.2% 1 32.25 mac i i i D i d i i i + + = = = => = => = Next, the Macaulay duration of a perpetuity-due is: 2 2 mod 1 1 1 1 1 31.25 ( ) 31.25 (0.032) 30.281 1 1 0.032 1.032 mac i i i i D i D i i d i + + = = = = = => = = + 448. Solution: D Using the financial calculator to solve for the yield rate with 15, 2092.88, 30.25 [because 2500(.0484) / 4 30.25], 2316.20 N PV PMT FV = = = − = = − , i = 2.06% is the nominal quarterly yield rate. 2316.3(0.0206) 30.25 17.46578 − = is the amount for accumulation of discount in the 20th coupon and 28 17.46578(1.0206) 30.91 = is the amount for accumulation of discount in the last coupon. 449. Solution: C Just after the 20th coupon is paid, there are still 40 coupon periods left, and the book value of the bond is B20 = PV(remaining cash flows) = PV(10 coupons of 35, 30 coupons of 45, and 1 redemption value of 1,000) = 10 40 10 0.04 30 0.04 35 45 (1.04) 1000(1.04) 283.88 515.68 202.29 1017.85.
a a − − + + = + + = 123 450. Solution: B The PV and duration of the assets must match the liabilities, 24,000 + 1,000 = 25,000 and [24,000(7.5) + 1,000(20)]/25,000 = 8. The convexity of the assets must be greater than [24,000(7.52) + 1,000(202)]/25,000 = 70. Let f represent the fraction of 25,000 that is invested in the n-year bond. The duration of the assets is fn + (1 – f)2n and the convexity is fn2 + (1 – f)4n2. For duration, 8 = fn + (1 – f)2n which implies f = 2 – 8/n. Since f must be between 0 and 1, n must be between 4 and 8 (inclusive). Substituting for f, for convexity: 2 2 2 2 (2 8 / ) (1 2 8 / )4 70 2 24 70 12 35 0 ( 7)( 5) 0.
n n n n n n n n n n − + − + > − + > − + < − − < This forces n to be between 5 and 7 which is the answer. 451. Solution: E The quarterly rate of interest is 0.05/4 = 0.0125. There are 20 deposits of 1,000, and 60deposits that are geometrically increasing above 1,000. Then, AV20 = 59 60 60 20 0.0125 1.02 1.02 1,000 (1.0125) 1,000(1.02)(1.0125) 1 1.0125 1.0125 s + + + + 60 60 60 20 .0125 1.02 1 1.0125 1,000 (1.0125) 1,000(1.02)(1.0125) 1.02 1 1.0125 48,138.59 161,639.07 209,778.
s − = + − = + = 124 452. Solution B Because the outlays all occur before the inflows, there is a unique yield rate. At a 50% interest rate, the equation is 2 30 40 /1.5 60 /1.5 90 /1.5 0 1.5 3 n n − − + + = ⇒ = for ln(3) / ln(1.5) 2.71 3 n = = < .. For the final payment to be later, the yield rate must less than 50%. 453. Solution: D The accumulation of discount of at time 8 is 491.51, which is equal to 8 50 i s . Using a BA-II Plus calculator, i can be computed as 5.8099441%. The accumulation of discount at time 25 is 25 50 i s , or 2670.91, the bond’s redemption value is therefore 20,000+2671=22,671. There is also an algebraic solution for the interest rate: 8 8 16 16 16 8 8 8 8 8 8 (1 ) 1 491.51 (1 ) 1 1263.74 (1 ) 1 [(1 ) 1][(1 ) 1] 2.57114 (1 ) 1 (1 ) 1 (1 ) 1 (1 ) 1.57114 0.0581.
i s i i s i i i i i i i i i + − = = + − = = + − + + + − = = = + + + − + − + = = 454. Solution: E The amounts for accumulation of the discount in the annual coupon payments (amounts of increase in book value) form a geometric progression with common ratio (1 ) i + , where i is the annual effective yield rate. The amount for accumulation of discount in the 9th coupon payment is 4730 4478 252 − = . Since the redemption value is the book value just after the last (in this case 10th) coupon payment but before the redemption payment, the amount for accumulation of the discount in the 10th coupon payment is 5000 4730 270 − = . Therefore, 270 1 252 i + = , which leads to the conclusion that the annual effective yield rate is 270 1 0.071429.
252 i = −= . 125 455. Solution: E First note that the yield rate is equal to the coupon rate for Bond Z, so the price of Bond Z is 1000. This implies the price of Bond Y is 1,073.78. The semiannual yield rate is 2 1.015 1 0.030225.
−= Then, 2 2 0.030225 1073.78 34 1000(1.030225) 2 30 15.
n n a n n − = + ⇒ = ⇒ = The price of bond X is 1073.78 – 138.88 = 934.90. The annual yield rate is 4 1.015 1 0.06136.
−= Then, 15 15 0.06136 934.90 1000(1.06136) 9.62652 409.3165 54.60.
Ra R R − = + = + ⇒ = 456. Solution: E Assume the face amount is 1. The price of the bond is 40 40 0.01 0.015 1.01 1.1642.
a − + = At the given coupon and return rates, the premium is 1.1642 – 1 = 0.1642. We are looking for the time at which the book value is 1 + 0.1642/2 = 1.0821. Then, (40 ) 40 0.01 (40 ) (40 ) (40 ) (40 ) 1.0821 0.015 1.01 1 1.01 0.015 1.01 0.01 1.5 0.5(1.01) 0.8328 1.01 ln 0.8328 40 18.03 ln1.01 21.97.
k k k k k k a k k − − − − − − − − − − − = + − = + = − = − = − = = For the book value to exceed 1.0821, round up to k = 22. 457. Solution: B Let x be the amount of a coupon payment. The book value right after the nth coupon is 0.02 95.5087 100 n n xa v = + The price of the bond is the present value of the time n book value plus the present value of the first n coupons. 0.02 91.8243 95.5087 95.5087 95.5087 100 3.6844 4.4913 ln(3.6844 / 4.4913) / ln(1/1.02) 10.
n n n n n v xa v v v n = + = + − = = − = 126 458. Solution: B 2 3 5 3 4 6 3 ( ) 1000(1 ) 3000(1 ) (1 ) (0.1) 3089.39 0.62092 ' ( ) 2000(1 ) 9000(1 ) 5 (1 ) ' (0.1) 7649.75 2.82237 7649.75 2.82237 4 3089.39 0.62092 13,793.98 ( ) 1000(1 ) 3000(1 A A A A B P i i i X i P X P i i i X i P X X X X P i i i − − − − − − − = + + + + + = + − = + + + + + − = + + = + = = + + + 4 7 4 5 8 ) 13,793.98(1 ) (0.1) 9878.85 ' ( ) 3000(1 ) 12,000(1 ) 7(13,793.98) (1 ) ' (0.1) 54,545.05 54,545.05 5.5214 9878.85 B B B i P P i i i X i P Y − − − − − + + = − = + + + + + − = = = 459. Solution: A 10,000 (1.07) 152,857.14 0.07 L = = 2 1 1 ( ) 0.07 (0.07) 14.2857 1 1.07 0.07 14.285714 13.351135 1.07 n n Ia d a v + = = = = = [ ] 152,857.14 1 ( 0.02)(13.351135 193,673.47 M = −− = 193,673.47 152,857.14 40,816.33 M L − = − = 127 460. Solution: B The present value of the geometric payment perpetuity-immediate, at annual effective interest rate i and with payment growth rate 0.005 r = − , is given by 1 1 ( ) 0.005 P i i r i = = − + . The modified duration is therefore given by 2 1 ( ) 1 ( 0.005) 1 ( ) 0.005 0.005 P i i v P i i i − ′ + = − = − = + + . The first-order modified approximation of the present value, at an interest rate i near an initial interest rate 0 i , is therefore given by ( ) ( ) 0 0 0 0 0 1 1 1 0.005 0.005 i i E P i i i v i i − = − − = − + + . Therefore, using 0.065 for 0 i and 0.055 for i, we find that the percentage error in this approximation is 1 0.055 0.065 1 1 0.065 0.005 0.065 0.005 0.055 0.005 0.0204 2.04%.
1 0.055 0.005 E P P − − − − + + + = = − = − + 128 461. Solution: D The present value of the geometric payment perpetuity-due, at annual effective interest rate i and with payment growth rate 0.02 r = − , is given by 1 1 ( ) 0.02 i i P i i r i + + = = − + . The Macaulay duration is therefore given by ( ) 2 2 0.02 1 1 0.02 (1 ) (1 ) '( ) ( 0.02) 0.98 ( 0.02) (1 ) 1 1 ( ) 0.02 0.02 0.02 i i P i i i d i v i P i i i i − − + + − + + = + = = − = = + + + + . The first-order Macaulay approximation of the present value, at an interest rate i near an initial interest rate 0 i , is therefore given by ( ) ( ) mac 0 0 1 0.02 0.02 0 0 0 0 0 1 1 1 1 0.02 1 D i i i i i E P i i i i − + + + + = = + + + . Using 0.08 for 0 i and 0.07 for i, we find that the percentage error in this approximation is 0.98 0.08 0.02 1.08 1.08 1.07 0.08 0.02 1.07 0.07 0.02 0.004883 0.49%.
1.07 0.07 0.02 E P P + − + + − = = − = − + 462. Solution: D The outstanding loan balance here can be calculated using either the prospective approach or the retrospective approach, although the retrospective approach is easier, since it does not require the value of n. 10 10 0.0325 5,000,000(1.0325) 303,244.53 3,367,821.18.
s − = |
188183 | https://pubs.acs.org/doi/10.1021/mp400170b | Drug Delivery across the Blood–Brain Barrier | Molecular Pharmaceutics
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Editorial May 6, 2013
Drug Delivery across the Blood–Brain Barrier
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Leslie A. Khawli
Saileta Prabhu
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Genentech Research & Early Development, One DNA Way, South San Francisco, California 94080, United States
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Molecular Pharmaceutics
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Central nervous system
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SPECIAL ISSUE
This article is part of the Drug Delivery across the Blood-Brain Barrier special issue.
The delivery of drugs to the central nervous system (CNS) remains a challenge in the treatment of neurological diseases such as Alzheimer’s disease, Parkinson’s disease, and stroke. The major challenge to CNS drug delivery is the presence of the blood–brain barrier (BBB), which limits the access of drugs to the brain parenchyma. The BBB limits drug delivery by generally allowing only those molecules that are lipophilic and have low molecular weight (less than 400–500 Da) to enter the brain from the bloodstream through the transcellular route. (3) In this context, it has been reported that approximately 98% of small molecules and nearly all large therapeutic molecules, such as monoclonal antibodies, antisense oligonucleotides, or viral vectors, cannot pass through this barrier. (4) For these reasons, delivery of drugs to the brain is still a major challenge, and recent reports indicate that less than 10% of therapeutic agents for neurological diseases enter into clinical trials because of poor brain penetration. (5) Attempts to overcome this barrier involve increasing drug delivery of intravascularly administered drugs by manipulating either the drug or capillary permeability, (1) and/or by local administration into brain fluids, such as the cerebrospinal fluid of brain ventricles or the interstitial fluid of brain tissue. (2)
In recognition to the difficulties associated with CNS delivery across the BBB, research efforts have focused on the development of new strategies to more effectively deliver drugs to the brain. These strategies include manipulating the characteristics of the drugs or utilizing endogenous transporters or receptors at the BBB. Successful delivery into the brain tissue has been achieved in some cases through the use of nanoparticulate drug carriers (6) or by antibodies that target receptors at the BBB, such as transferrin, insulin, and low density lipoprotein receptors, that facilitate their transport across the BBB via receptor mediated transcytosis (RMT). (7-11) As such, the opportunities for delivery of drugs to the CNS through endogenous RMT systems are exciting, (12) but greater understanding of CNS physiology and pathophysiology of the CNS as well as further validation in relevant preclinical disease models of the quantitative brain distribution kinetics and efficacy of targeted drugs is needed for translation to the clinic. (13)
Advances in both drug delivery strategies across the BBB and in vivo brain uptake measurement methods will be important for achieving successful end results. With this in mind, this special issue provides a snapshot of some of the ongoing research efforts in the CNS field. Although the topics adopted by the ten different papers are diverse, the common theme that unifies these papers is the need to solve the fundamental problems of drug delivery to the brain by using various strategies. Some of the problems addressed by the papers in this thematic issue have a long history, such as the need to better understand the physiology and anatomy of the BBB or the need to improve our knowledge of changes in brain proteomic profile and in BBB transport systems under pathological conditions. Other problems have also emerged as significant, such as the need to develop more predictable preclinical disease models to guide compound selection and clinical trial design. In recognition of this continuum of approaches and challenges, this special issue highlights the recent work of investigators as they address the barrier issue in order to provide insight into the restrictive properties of the BBB. Novel methods, such as the use of image-based quantitative kinetics, are also presented and provide thoughtful and innovative strategies for improving the delivery of macromolecules to the brain. One such work tends to focus on how the structure and function of the brain allows small and large molecules to cross the BBB. Moreover, other papers tend to address some of the newer topics, such as the underlying mechanism of antibody-based drug transport and distribution to the brain through the use of preclinical studies that provide direct insight into the relationship between pathological conditions, BBB transport, and drug effects in relevant in vitro and in vivo models. The collection of four different reviews and six original research articles in this issue provide key information on current experimental methods of crossing the BBB to successfully deliver drugs to the CNS, making it a particularly exciting time to translate research to the clinic.
The first four articles in this issue are focused on understanding the mechanism of drug delivery across the BBB and the physiology of the BBB. The first article, authored by Strazielle and Ghersi-Egea, is very effective at comparing the biological and physiological properties of the two main CNS barriers, the BBB and the blood–CSF barrier (BCSFB). This article tackles the structural and biochemical features of brain barriers relevant to the cerebral delivery of small and macromolecular drugs, and gives quantitative information about the possible transport mechanisms at the developing blood–CSF interface. A second review article, authored by Wolak and Thorne, introduces the basic nature of macromolecular diffusion inside the extracellular space (ECS) of the brain together with commonly utilized methodologies for analysis of diffusion in the brain ECS. This article discusses these issues in the context of drug delivery into CNS disease/injury models, and provides mathematical models to estimate macromolecule diffusion. The third article, by Abuqayyas and Balthasar, evaluates the role of neonatal Fc receptor and Fc gamma receptor in the transport of antibodies into the brain and the implications of these findings. This is an important study for the antibody biologics field and contributes specifically to an area that is puzzled with conflicting data. This is followed by an article (Boswell et al.) describing the use of a variety of molecular probes to characterize the vascular biology of a preclinical model of neurodegenerative disease. It is believed that this experimental multiprobe approach can play a fundamental role in advancing our understanding of the BBB physiology and delivery of drugs to brain in mouse models of BBB disruption.
The next three articles in this issue focus on CNS drug delivery systems involving different approaches to enhance drug delivery to the brain. A timely review by Papisov et al. concentrates on evaluating the impact of administration route on the delivery of macromolecular drugs through the use of image-based quantitative kinetics. It includes a general discussion of the anatomy of the subarachnoid space, the pathways and mechanisms of CSF secretion and reabsorption, and describes the exchange between the subarachnoid CSF compartments and the brain interstitium. The review concludes with a brief discussion on potential changes to the function of these pathways in pathophysiological conditions. Another article, by Lindqvist et al., presents nanocarriers as a promising delivery system to enhance transport at the BBB. One of the key innovations of this approach is to change the systemic pharmacokinetics (PK) of the released drug using liposomes and polymeric nanoparticles that can potentially facilitate the transport of large molecules. Indeed, the results showing the enhanced transport to the brain are impressive, for which both the PK and pharmacological response between plasma and brain were measured with microdialysis. This is followed by an article (Haqqani et al.) suggesting that the use of selected peptides as a carrier system for protein delivery may enhance the delivery of protein drugs to the brain. This research article describes the development of a sensitive and selective mass spectrometry-based method to monitor and quantify panels of unlabeled single domain antibodies in body fluids, including CSF. Therefore, these techniques represent an elegant approach for dealing with the difficult problem of quantitatively assessing biologic penetration into the CNS.
Two additional research articles are of interest to the readers in this special issue and provide unique insights into the CNS field which create even more valuable knowledge critical for developing therapies to treat CNS disorders. One such work, by Agyare et al., presents an experimentally verifiable mechanism describing how Alzheimer’s disease amyloid β (Aβ) proteins accumulate in the cerebral vasculature to cause cerebral amyloid angiopathy (CAA). It is widely believed that the accumulation of Aβ40 and DutchAβ40 in the cerebral vasculature is due to their inefficient clearance from the brain. These investigators provide experimental evidence in support of this theory and isolate the key kinetic events that promote cerebrovascular versus parenchymal Aβ deposition. Moreover, this study establishes that the accumulation of Aβ proteins in the cerebrovascular endothelium is the earliest pathological event that triggers CAA. In another article, by Harati et al., the interconnected features related to the regulation of ABC and SLC transporters occurring during BBB maturation was addressed. These investigators have characterized the brain capillary expression and activity profiles of 4 proteins implicated in transport of drugs across the BBB, and also decipher a role for β-catenin and endothelin-1 (ET-1). These findings provide novel insight into the functional expression of ABC and SLC transporters under β-catenin regulation and illustrate a major role of the inflammatory component ET-1 in the functional regulation of both ABC and SLC transporters at the BBB. Such work uncovers new knowledge on regulation of drug transporters at the BBB.
The final article, by Deo et al., presents the challenges in the translation of preclinical findings in CNS drug development to the clinic, particularly focusing on the predictive value of preclinical rodent models. This review describes different approaches to measuring drug disposition in the brain with special emphasis given on the proposed impact of disease states and species on the BBB function.
As a whole, this thematic issue demonstrates how the field of CNS delivery has evolved into a real mainstream scientific domain. Exploration of a variety of strategies for enhancing drug transport across the BBB along with more rigorous determination of kinetic measures at the site of action can offer not only solutions to pressing existing challenges but also revolutionary approaches to growing and emerging needs. The diversity of published papers reinforced our belief that the domain of CNS drug delivery is better positioned for designing and developing novel therapeutic agents in order to increase drug exposure to the brain.
We would like to extend our thanks to all the authors, who had high quality submissions and provided great insight into the current state of the field, which has put us ahead in the quest for understanding CNS anatomy and physiology. We would particularly like to thank the special issue reviewers, whose efforts substantially contributed to the improvement of the overall quality of this thematic issue. Special thanks to our colleagues at Genentech, Ryan Watts, Daniela Bumbaca, C. Andrew Boswell, and Joseph Ware for scientific exchange and discussions. And of course we would like to thank the journal staff, for providing support and input, and for giving us the opportunity to showcase the recent advances in the field of CNS drug delivery systems.
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Leslie A. Khawli,Guest Editor,Genentech Research & Early Development, One DNA Way, South San Francisco, California 94080, United States
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Saileta Prabhu,Genentech Research & Early Development, One DNA Way, South San Francisco, California 94080, United States
Notes Views expressed in this editorial are those of the author and not necessarily the views of the ACS.
References
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This article references 13 other publications.
1Siegal, T.; Zylber-Katz, E.Clin. Pharmacokinet.2002, 41, 171–86 Google Scholar1 Strategies for increasing drug delivery to the brain. Focus on brain lymphoma Siegal, Tali; Zylber-Katz, Ester Clinical Pharmacokinetics (2002), 41 (3), 171-186 CODEN: CPKNDH; ISSN:0312-5963. (Adis International Ltd.) A review. The blood-brain barrier (BBB) is a gate that controls the influx and efflux of a wide variety of substances and consequently restricts the delivery of drugs into the central nervous system (CNS). Brain tumors may disrupt the function of this barrier locally and nonhomogeneously. Therefore, the delivery of drugs to brain tumors has long been a controversial subject. The current concept is that inadequate drug delivery is a major factor that explains the unsatisfactory response of chemosensitive brain tumors. Various strategies have been devised to circumvent the BBB in order to increase drug delivery to the CNS. The various approaches can be categorized as those that attempt to increase delivery of intravascularly administered drugs and those that attempt to increase delivery by local drug administration. Strategies that increase delivery of intravascularly injected drugs can manipulate either the drugs or the capillary permeability of the various barriers (BBB or blood-tumor barrier), or they may attempt to increase plasma concns. or the fraction of the drug reaching the tumor (high-dose chemotherapy, intra-arterial injection). Neurotoxicity is a major concern with increased penetration of drugs into the CNS or when local delivery is practised. Systemic toxicity remains the limiting factor for most methods that use intravascular delivery. This review evaluates the strategies used to increase drug delivery in view of current knowledge of drug pharmacokinetics and its relevance to clin. studies of chemosensitive brain tumors. The main focus is on primary CNS lymphoma, as this is a chemosensitive brain tumor and its management routinely utilizes specialized strategies to enhance drug delivery to the affected CNS compartments.
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3Pardridge, W. M.Drug Discovery Today 2007, 12, 54–61 Google Scholar3 Blood-brain barrier delivery Pardridge, William M. Drug Discovery Today (2006), 12 (1&2), 54-61 CODEN: DDTOFS; ISSN:1359-6446. (Elsevier B.V.) A review. Neuropharmaceutics is the largest potential growth sector of the pharmaceutical industry. However, this growth is blocked by the problem of the blood-brain barrier (BBB). Essentially 100% of large-mol. drugs and >98% of small-mol. drugs do not cross the BBB. The BBB can be traversed because there are multiple endogenous transporters within this barrier. Therefore, brain drug development programs of the future need to be re-configured so that drugs are formulated to enable transport into the brain via endogenous BBB transporters.
4Gabathuler, R.Neurobiol. Dis.2010, 37, 48–57 Google Scholar4 Approaches to transport therapeutic drugs across the blood-brain barrier to treat brain diseases Gabathuler, Reinhard Neurobiology of Disease (2010), 37 (1), 48-57 CODEN: NUDIEM; ISSN:0969-9961. (Elsevier B.V.) A review. The central nervous system is protected by barriers which control the entry of compds. into the brain, thereby regulating brain homeostasis. The blood-brain barrier, formed by the endothelial cells of the brain capillaries, restricts access to brain cells of blood-borne compds. and facilitates nutrients essential for normal metab. to reach brain cells. This very tight regulation of the brain homeostasis results in the inability of some small and large therapeutic compds. to cross the blood-brain barrier (BBB). Therefore, various strategies are being developed to enhance the amt. and concn. of therapeutic compds. in the brain. In this review, we will address the different approaches used to increase the transport of therapeutics from blood into the brain parenchyma. We will mainly conc. on the physiol. approach which takes advantage of specific receptors already expressed on the capillary endothelial cells forming the BBB and necessary for the survival of brain cells. Among all the approaches used for increasing brain delivery of therapeutics, the most accepted method is the use of the physiol. approach which takes advantage of the transcytosis capacity of specific receptors expressed at the BBB. The low d. lipoprotein receptor related protein (LRP) is the most adapted for such use with the engineered peptide compd. (EPiC) platform incorporating the Angiopep peptide in new therapeutics the most advanced with promising data in the clinic.
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7Lee, H. J.; Engelhardt, B.; Lesley, J.; Bickel, U.; Pardridge, W. M.J. Pharmacol. Exp. Ther.2000, 292, 1048–52Google Scholar7 Targeting rat anti-mouse transferrin receptor monoclonal antibodies through blood-brain barrier in mouse Lee, Hwa Jeong; Engelhardt, Britta; Lesley, Jayne; Bickel, Ulrich; Pardridge, William M. Journal of Pharmacology and Experimental Therapeutics (2000), 292 (3), 1048-1052 CODEN: JPETAB; ISSN:0022-3565. (American Society for Pharmacology and Experimental Therapeutics) Drug targeting through the brain capillary endothelium, which forms the blood-brain barrier (BBB) in vivo, may be achieved with peptidomimetic monoclonal antibodies that target peptide transcytosis systems on the BBB in vivo. Murine monoclonal antibodies to the rat transferrin receptor, such as the OX26 monoclonal antibody, are targeted through the BBB on the transferrin receptor in the rat. However, the present studies show the OX26 monoclonal antibody is not an effective brain delivery vector in mice. The emergence of transgenic mouse models creates a need for brain drug-targeting vectors for this species. Two rat monoclonal antibodies, 8D3 and R17-217, to the mouse transferrin receptor were evaluated in the present studies. Both the R17-217 and the 8D3 antibody had comparable permeability-surface area products at the mouse BBB in vivo. However, owing to a higher plasma area under the concn. curve, the mouse brain uptake of the 8D3 antibody was higher, 3.1 ± 0.4% of injected dose [(ID)/g] compared with the brain uptake of the R17 antibody 1.6 ± 0.2% ID/g, at 60 min after i.v. injection. Conversely, the mouse brain uptake of the OX26 antibody, which does not recognize the mouse transferrin receptor, was negligible, 0.06 ± 0.01% ID/g. The R17-217 antibody was more selective for brain because this antibody was not measureably taken up by liver. The capillary depletion technique demonstrated transcytosis of the R17-217 antibody through the mouse BBB in vivo. The brain uptake of the 8D3 antibody was saturable, consistent with a receptor-mediated transport process. In conclusion, these studies indicate rat monoclonal antibodies to the mouse transferrin receptor may be used for brain drug-targeting studies in mice such as transgenic mouse models.
8Yu, Y. J.; Zhang, Y.; Kenrick, M.; Hoyte, K.; Luk, W.; Lu, Y.; Atwal, J.; Elliott, J. M.; Prabhu, S.; Watts, R. J.; Dennis, M. S.Sci. Transl. Med.2011, 3 (84) 84ra44 Google ScholarThere is no corresponding record for this reference.
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11. 11Bertrand, Y.; Currie, J. C.; Demeule, M.; Régina, A.; Ché, C.; Abulrob, A.; Fatehi, D.; Sartelet, H.; Gabathuler, R.; Castaigne, J. P.; Stanimirovic, D.; Béliveau, R.J. Cell. Mol. Med.2009, 14 (12) 2827–39 Google ScholarThere is no corresponding record for this reference.
12. 12Pardridge, W. M.J. Cerebral Blood Metab.2012, 32 (8) 1959–72 Google Scholar12 Drug transport across the blood-brain barrier Pardridge, William M. Journal of Cerebral Blood Flow & Metabolism (2012), 32 (11), 1959-1972 CODEN: JCBMDN; ISSN:0271-678X. (Nature Publishing Group) A review. The blood-brain barrier (BBB) prevents the brain uptake of most pharmaceuticals. This property arises from the epithelial-like tight junctions within the brain capillary endothelium. The BBB is anatomically and functionally distinct from the blood-cerebrospinal fluid barrier at the choroid plexus. Certain small mol. drugs may cross the BBB via lipid-mediated free diffusion, providing the drug has a mol. wt. <400 Da and forms <8 hydrogen bonds. These chem. properties are lacking in the majority of small mol. drugs, and all large mol. drugs. Nevertheless, drugs can be reengineered for BBB transport, based on the knowledge of the endogenous transport systems within the BBB. Small mol. drugs can be synthesized that access carrier-mediated transport (CMT) systems within the BBB. Large mol. drugs can be reengineered with mol. Trojan horse delivery systems to access receptor-mediated transport (RMT) systems within the BBB. Peptide and antisense radiopharmaceuticals are made brain-penetrating with the combined use of RMT-based delivery systems and avidin-biotin technol. Knowledge on the endogenous CMT and RMT systems expressed at the BBB enable new solns. to the problem of BBB drug transport. Journal of Cerebral Blood Flow & Metab. (2012) 32, 1959-1972; doi:10.1038/jcbfm.2012.126; published online 29 August 2012.
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1. 1Siegal, T.; Zylber-Katz, E.Clin. Pharmacokinet.2002, 41, 171–86 1 Strategies for increasing drug delivery to the brain. Focus on brain lymphoma Siegal, Tali; Zylber-Katz, Ester Clinical Pharmacokinetics (2002), 41 (3), 171-186 CODEN: CPKNDH; ISSN:0312-5963. (Adis International Ltd.) A review. The blood-brain barrier (BBB) is a gate that controls the influx and efflux of a wide variety of substances and consequently restricts the delivery of drugs into the central nervous system (CNS). Brain tumors may disrupt the function of this barrier locally and nonhomogeneously. Therefore, the delivery of drugs to brain tumors has long been a controversial subject. The current concept is that inadequate drug delivery is a major factor that explains the unsatisfactory response of chemosensitive brain tumors. Various strategies have been devised to circumvent the BBB in order to increase drug delivery to the CNS. The various approaches can be categorized as those that attempt to increase delivery of intravascularly administered drugs and those that attempt to increase delivery by local drug administration. Strategies that increase delivery of intravascularly injected drugs can manipulate either the drugs or the capillary permeability of the various barriers (BBB or blood-tumor barrier), or they may attempt to increase plasma concns. or the fraction of the drug reaching the tumor (high-dose chemotherapy, intra-arterial injection). Neurotoxicity is a major concern with increased penetration of drugs into the CNS or when local delivery is practised. Systemic toxicity remains the limiting factor for most methods that use intravascular delivery. This review evaluates the strategies used to increase drug delivery in view of current knowledge of drug pharmacokinetics and its relevance to clin. studies of chemosensitive brain tumors. The main focus is on primary CNS lymphoma, as this is a chemosensitive brain tumor and its management routinely utilizes specialized strategies to enhance drug delivery to the affected CNS compartments.
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3. 3Pardridge, W. M.Drug Discovery Today 2007, 12, 54–61 3 Blood-brain barrier delivery Pardridge, William M. Drug Discovery Today (2006), 12 (1&2), 54-61 CODEN: DDTOFS; ISSN:1359-6446. (Elsevier B.V.) A review. Neuropharmaceutics is the largest potential growth sector of the pharmaceutical industry. However, this growth is blocked by the problem of the blood-brain barrier (BBB). Essentially 100% of large-mol. drugs and >98% of small-mol. drugs do not cross the BBB. The BBB can be traversed because there are multiple endogenous transporters within this barrier. Therefore, brain drug development programs of the future need to be re-configured so that drugs are formulated to enable transport into the brain via endogenous BBB transporters.
4. 4Gabathuler, R.Neurobiol. Dis.2010, 37, 48–57 4 Approaches to transport therapeutic drugs across the blood-brain barrier to treat brain diseases Gabathuler, Reinhard Neurobiology of Disease (2010), 37 (1), 48-57 CODEN: NUDIEM; ISSN:0969-9961. (Elsevier B.V.) A review. The central nervous system is protected by barriers which control the entry of compds. into the brain, thereby regulating brain homeostasis. The blood-brain barrier, formed by the endothelial cells of the brain capillaries, restricts access to brain cells of blood-borne compds. and facilitates nutrients essential for normal metab. to reach brain cells. This very tight regulation of the brain homeostasis results in the inability of some small and large therapeutic compds. to cross the blood-brain barrier (BBB). Therefore, various strategies are being developed to enhance the amt. and concn. of therapeutic compds. in the brain. In this review, we will address the different approaches used to increase the transport of therapeutics from blood into the brain parenchyma. We will mainly conc. on the physiol. approach which takes advantage of specific receptors already expressed on the capillary endothelial cells forming the BBB and necessary for the survival of brain cells. Among all the approaches used for increasing brain delivery of therapeutics, the most accepted method is the use of the physiol. approach which takes advantage of the transcytosis capacity of specific receptors expressed at the BBB. The low d. lipoprotein receptor related protein (LRP) is the most adapted for such use with the engineered peptide compd. (EPiC) platform incorporating the Angiopep peptide in new therapeutics the most advanced with promising data in the clinic.
5. 5Pangalos, M. N.; Schechter, L. E.; Hurko, O.Nat. Rev. Drug Discovery 2007, 6, 521–32 There is no corresponding record for this reference.
6. 6Muller, R. H.; Keck, C. M.J. Nanosci. Nanotechnol.2004, 4 (5) 471–83 There is no corresponding record for this reference.
7. 7Lee, H. J.; Engelhardt, B.; Lesley, J.; Bickel, U.; Pardridge, W. M.J. Pharmacol. Exp. Ther.2000, 292, 1048–52 7 Targeting rat anti-mouse transferrin receptor monoclonal antibodies through blood-brain barrier in mouse Lee, Hwa Jeong; Engelhardt, Britta; Lesley, Jayne; Bickel, Ulrich; Pardridge, William M. Journal of Pharmacology and Experimental Therapeutics (2000), 292 (3), 1048-1052 CODEN: JPETAB; ISSN:0022-3565. (American Society for Pharmacology and Experimental Therapeutics) Drug targeting through the brain capillary endothelium, which forms the blood-brain barrier (BBB) in vivo, may be achieved with peptidomimetic monoclonal antibodies that target peptide transcytosis systems on the BBB in vivo. Murine monoclonal antibodies to the rat transferrin receptor, such as the OX26 monoclonal antibody, are targeted through the BBB on the transferrin receptor in the rat. However, the present studies show the OX26 monoclonal antibody is not an effective brain delivery vector in mice. The emergence of transgenic mouse models creates a need for brain drug-targeting vectors for this species. Two rat monoclonal antibodies, 8D3 and R17-217, to the mouse transferrin receptor were evaluated in the present studies. Both the R17-217 and the 8D3 antibody had comparable permeability-surface area products at the mouse BBB in vivo. However, owing to a higher plasma area under the concn. curve, the mouse brain uptake of the 8D3 antibody was higher, 3.1 ± 0.4% of injected dose [(ID)/g] compared with the brain uptake of the R17 antibody 1.6 ± 0.2% ID/g, at 60 min after i.v. injection. Conversely, the mouse brain uptake of the OX26 antibody, which does not recognize the mouse transferrin receptor, was negligible, 0.06 ± 0.01% ID/g. The R17-217 antibody was more selective for brain because this antibody was not measureably taken up by liver. The capillary depletion technique demonstrated transcytosis of the R17-217 antibody through the mouse BBB in vivo. The brain uptake of the 8D3 antibody was saturable, consistent with a receptor-mediated transport process. In conclusion, these studies indicate rat monoclonal antibodies to the mouse transferrin receptor may be used for brain drug-targeting studies in mice such as transgenic mouse models.
8. 8Yu, Y. J.; Zhang, Y.; Kenrick, M.; Hoyte, K.; Luk, W.; Lu, Y.; Atwal, J.; Elliott, J. M.; Prabhu, S.; Watts, R. J.; Dennis, M. S.Sci. Transl. Med.2011, 3 (84) 84ra44 There is no corresponding record for this reference.
9. 9Karkan, D.; Pfeifer, C.; Vitalis, T. Z.; Arthur, G.; Ujiie, M.; Chen, Q.; Tsai, S.; Koliatis, G.; Gabathuler, R.; Jefferies, W. A.PLoS One 2008, 3 (6) e2469 updatesThis document has been updated
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10. 10Coloma, M. J.; Lee, H. J.; Kurihara, A.; Landaw, E. M.; Boado, R,J.; Morrison, S. L.; Pardridge, W. M.Pharm. Res.2000, 17, 266–74 There is no corresponding record for this reference.
11. 11Bertrand, Y.; Currie, J. C.; Demeule, M.; Régina, A.; Ché, C.; Abulrob, A.; Fatehi, D.; Sartelet, H.; Gabathuler, R.; Castaigne, J. P.; Stanimirovic, D.; Béliveau, R.J. Cell. Mol. Med.2009, 14 (12) 2827–39 There is no corresponding record for this reference.
12. 12Pardridge, W. M.J. Cerebral Blood Metab.2012, 32 (8) 1959–72 12 Drug transport across the blood-brain barrier Pardridge, William M. Journal of Cerebral Blood Flow & Metabolism (2012), 32 (11), 1959-1972 CODEN: JCBMDN; ISSN:0271-678X. (Nature Publishing Group) A review. The blood-brain barrier (BBB) prevents the brain uptake of most pharmaceuticals. This property arises from the epithelial-like tight junctions within the brain capillary endothelium. The BBB is anatomically and functionally distinct from the blood-cerebrospinal fluid barrier at the choroid plexus. Certain small mol. drugs may cross the BBB via lipid-mediated free diffusion, providing the drug has a mol. wt. <400 Da and forms <8 hydrogen bonds. These chem. properties are lacking in the majority of small mol. drugs, and all large mol. drugs. Nevertheless, drugs can be reengineered for BBB transport, based on the knowledge of the endogenous transport systems within the BBB. Small mol. drugs can be synthesized that access carrier-mediated transport (CMT) systems within the BBB. Large mol. drugs can be reengineered with mol. Trojan horse delivery systems to access receptor-mediated transport (RMT) systems within the BBB. Peptide and antisense radiopharmaceuticals are made brain-penetrating with the combined use of RMT-based delivery systems and avidin-biotin technol. Knowledge on the endogenous CMT and RMT systems expressed at the BBB enable new solns. to the problem of BBB drug transport. Journal of Cerebral Blood Flow & Metab. (2012) 32, 1959-1972; doi:10.1038/jcbfm.2012.126; published online 29 August 2012.
13. 13de Boer, A. G.; Gaillard, P. J.Annu. Rev. Pharmacol. Toxicol.2007, 47, 323–55 There is no corresponding record for this reference.
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188184 | https://www.ibm.com/docs/SSLVMB_29.0.0/nl/zh/CN/pdf/IBM_SPSS_Statistics_Core_System_User_Guide.pdf | IBM SPSS Statistics 29 Core System 用户 指南 IBM 注 使用本信息及其支持的产品前，请先阅读 第239 页的『声明』 中的信息。 产品信息 本版本适用于 IBM® SPSS Statistics V 29，发行版 0 ，修订版 1 以及所有后续发行版和修订版，直到在新版本中另有声明为 止。 © Copyright International Business Machines Corporation .
内容 第 1 章 概况............................................................................................................1 Windows.......................................................................................................................................................1 指定的窗口和活动窗口............................................................................................................................1 对话框列表中的变量名称和变量标签........................................................................................................... 1 数据类型、测量级别和变量列表图标........................................................................................................... 2 获得关于对话框中的变量的信息...................................................................................................................2 自动恢复....................................................................................................................................................... 2 复原点..................................................................................................................................................... 2 了解更多信息................................................................................................................................................3 第 2 章 获得帮助..................................................................................................... 5 第 3 章 数据文件..................................................................................................... 7 打开数据文件................................................................................................................................................7 打开数据文件.......................................................................................................................................... 7 数据文件类型.......................................................................................................................................... 7 读取 Excel 文件....................................................................................................................................... 8 读取旧 Excel 文件和其他电子表格..........................................................................................................8 读取 dBASE 文件.....................................................................................................................................9 读取 Stata 文件....................................................................................................................................... 9 读取 CSV 文件......................................................................................................................................... 9 文本向导................................................................................................................................................10 读取数据库文件.....................................................................................................................................13 读取 Cognos BI 数据.............................................................................................................................17 读取 Cognos TM1 数据......................................................................................................................... 19 文件信息..................................................................................................................................................... 20 保存数据文件..............................................................................................................................................20 保存已修改的数据文件..........................................................................................................................20 以代码页字符编码保存数据文件...........................................................................................................20 以外部格式保存数据文件......................................................................................................................21 以 Excel 格式保存数据文件...................................................................................................................23 以 SAS 格式保存数据文件.....................................................................................................................23 以 Stata 格式保存数据文件...................................................................................................................24 保存变量子集........................................................................................................................................ 25 加密数据文件........................................................................................................................................ 25 导出到数据库........................................................................................................................................ 26 导出到 Cognos TM1..............................................................................................................................30 比较数据集................................................................................................................................................. 32 比较数据集：“比较”选项卡.................................................................................................................. 32 比较数据集：“属性”选项卡.................................................................................................................. 32 比较数据集：“输出”选项卡.................................................................................................................. 33 保护原始数据..............................................................................................................................................33 虚拟活动文件..............................................................................................................................................33 创建数据高速缓存.................................................................................................................................34 第 4 章 分布式分析模式..........................................................................................37 服务器登录................................................................................................................................................. 37 添加或编辑服务器登录设置.................................................................................................................. 37 选择、切换或添加服务器......................................................................................................................37 搜索可用服务器.....................................................................................................................................38 iii 从远程服务器打开数据文件........................................................................................................................38 自动重新编码..............................................................................................................................................39 分布式分析模式下过程的可用性................................................................................................................ 40 绝对和相对路径指定...................................................................................................................................40 第 5 章 数据编辑器................................................................................................ 41 概述............................................................................................................................................................ 41 概述-从对话框中选择变量.................................................................................................................... 41 数据视图..................................................................................................................................................... 43 变量视图..................................................................................................................................................... 44 显示或定义变量属性............................................................................................................................. 44 变量名称................................................................................................................................................44 变量测量级别........................................................................................................................................ 45 变量类型................................................................................................................................................45 变量标签................................................................................................................................................47 值标签................................................................................................................................................... 47 在标签中插入换行符............................................................................................................................. 47 缺失值................................................................................................................................................... 47 角色.......................................................................................................................................................47 列宽.......................................................................................................................................................48 变量对齐................................................................................................................................................48 将变量定义属性应用于多个变量...........................................................................................................48 定制变量属性........................................................................................................................................ 49 定制变量视图........................................................................................................................................ 50 拼写检查................................................................................................................................................51 输入数据..................................................................................................................................................... 51 输入数值数据........................................................................................................................................ 52 输入非数值数据.....................................................................................................................................52 使用值标签进行数据输入......................................................................................................................52 数据编辑器中的数据值限制.................................................................................................................. 52 编辑数据..................................................................................................................................................... 52 替换或修改数据值.................................................................................................................................52 剪切、复制并粘贴数据值......................................................................................................................53 插入新个案............................................................................................................................................53 插入新变量............................................................................................................................................53 更改数据类型........................................................................................................................................ 54 查找个案、变量或插补............................................................................................................................... 54 查找并替换数据和属性值........................................................................................................................... 54 为选定变量获取描述统计........................................................................................................................... 55 数据编辑器中的个案选择状态....................................................................................................................55 数据编辑器显示选项...................................................................................................................................56 数据编辑器打印.......................................................................................................................................... 56 打印数据编辑器内容............................................................................................................................. 56 第 6 章 使用多数据源............................................................................................. 57 多数据源的基本处理...................................................................................................................................57 使用命令语法中的多个数据集....................................................................................................................57 在数据集之间复制和粘贴信息....................................................................................................................57 重命名数据集..............................................................................................................................................57 不显示多个数据集...................................................................................................................................... 58 第 7 章 数据准备................................................................................................... 59 变量属性..................................................................................................................................................... 59 定义变量属性..............................................................................................................................................59 定义变量属性........................................................................................................................................ 59 定义值标签和其他变量属性.................................................................................................................. 60 指定测量级别........................................................................................................................................ 61 iv 定制变量属性........................................................................................................................................ 61 复制变量属性........................................................................................................................................ 61 为测量级别未知的变量设置测量级别.........................................................................................................61 多重响应集(R).............................................................................................................................................62 定义多响应集........................................................................................................................................ 62 复制数据属性..............................................................................................................................................63 复制数据属性........................................................................................................................................ 63 标识重复个案..............................................................................................................................................65 可视分箱化................................................................................................................................................. 66 分箱化变量............................................................................................................................................67 分箱化变量............................................................................................................................................67 自动生成分箱化类别............................................................................................................................. 68 复制分箱化类别.....................................................................................................................................69 可视分箱化中的用户缺失值.................................................................................................................. 70 第 8 章 数据转换................................................................................................... 71 数据转换..................................................................................................................................................... 71 计算变量..................................................................................................................................................... 71 计算变量：If 个案................................................................................................................................. 71 计算变量：类型和标签..........................................................................................................................71 功能............................................................................................................................................................ 72 函数中的缺失值.......................................................................................................................................... 72 随机数字生成器.......................................................................................................................................... 72 计算个案中值的出现次数........................................................................................................................... 73 统计个案内的值: 要统计的值................................................................................................................ 73 统计出现次数: If 个案........................................................................................................................... 73 转换值.........................................................................................................................................................73 对值重新编码..............................................................................................................................................74 重新编码到相同的变量中........................................................................................................................... 74 重新编码为相同变量：旧值和新值....................................................................................................... 74 重新编码为不同变量...................................................................................................................................74 重新编码为不同变量：旧值和新值....................................................................................................... 75 自动重新编码..............................................................................................................................................75 个案排秩..................................................................................................................................................... 76 个案排秩：类型.....................................................................................................................................77 个案排秩: 结..........................................................................................................................................77 日期和时间向导.......................................................................................................................................... 78 IBM SPSS Statistics 中的日期和时间 .................................................................................................. 78 从字符串中创建一个日期/时间变量......................................................................................................79 从变量组中创建一个日期/时间变量......................................................................................................79 从日期/时间变量中加减值.................................................................................................................... 79 提取部分日期/时间变量........................................................................................................................ 81 时间序列数据转换...................................................................................................................................... 82 定义日期................................................................................................................................................82 创建时间序列........................................................................................................................................ 83 替换缺失值............................................................................................................................................84 第 9 章 文件处理和文件转换................................................................................... 85 文件处理和文件转换...................................................................................................................................85 排序个案..................................................................................................................................................... 85 变量排序..................................................................................................................................................... 86 交换............................................................................................................................................................ 86 合并数据文件..............................................................................................................................................86 添加个案................................................................................................................................................87 添加变量................................................................................................................................................88 聚集数据..................................................................................................................................................... 90 汇总数据: 汇总函数...............................................................................................................................91 v 汇总数据：变量名和标签......................................................................................................................91 拆分文件..................................................................................................................................................... 91 选择个案(L).................................................................................................................................................92 选择个案：If......................................................................................................................................... 93 选择个案：随机样本............................................................................................................................. 93 选择个案：范围.....................................................................................................................................93 对个案进行加权.......................................................................................................................................... 93 重组数据..................................................................................................................................................... 94 重组数据................................................................................................................................................94 重组数据向导：选择类型......................................................................................................................94 重组数据向导（变量到个案）：变量组数............................................................................................ 96 重组数据向导（变量到个案）：选择变量............................................................................................ 96 重组数据向导（变量到个案）：创建索引变量.....................................................................................97 重组数据向导（变量到个案）：创建一个索引变量............................................................................. 98 重组数据向导（变量到个案）：创建多个索引变量............................................................................. 98 重组数据向导（变量到个案）：选项................................................................................................... 99 重组数据向导（个案到变量）：选择变量............................................................................................ 99 重组数据向导（个案到变量）：对数据进行排序................................................................................. 99 重组数据向导（个案到变量）：选项................................................................................................. 100 重构数据向导：完成...........................................................................................................................100 第 10 章 处理输出................................................................................................103 使用输出...................................................................................................................................................103 查看器 - 经典............................................................................................................................................103 更改概要级别...................................................................................................................................... 103 将项添加到查看器...............................................................................................................................103 查找和替换查看器中的信息................................................................................................................104 保存输出............................................................................................................................................. 105 交互式输出..........................................................................................................................................106 导出输出............................................................................................................................................. 107 查看器打印..........................................................................................................................................111 查看器 - 工作簿........................................................................................................................................ 113 创建、打开和保存工作簿输出............................................................................................................ 114 将项添加到工作簿...............................................................................................................................115 查看器概要............................................................................................................................................... 115 折叠和展开概要视图...........................................................................................................................116 更改概要项的大小...............................................................................................................................116 更改概要中的字体...............................................................................................................................116 显示和隐藏结果........................................................................................................................................116 移动、删除和复制输出.............................................................................................................................117 更改初始对齐方式.................................................................................................................................... 117 更改各输出项的对齐方式......................................................................................................................... 117 复制输出到其他应用程序中......................................................................................................................117 第 11 章 透视表...................................................................................................119 透视表.......................................................................................................................................................119 操作透视表............................................................................................................................................... 119 激活透视表..........................................................................................................................................119 透视表................................................................................................................................................. 120 更改元素在维度内的显示顺序............................................................................................................ 120 在维度元素中移动行和列....................................................................................................................120 交换行和列..........................................................................................................................................120 对行或列分组...................................................................................................................................... 120 对行或列取消分组...............................................................................................................................120 旋转行标签或列标签...........................................................................................................................120 对行排序............................................................................................................................................. 121 插入行和列..........................................................................................................................................121 vi 控制变量和值标签的显示。................................................................................................................121 更改输出语言...................................................................................................................................... 122 导航大型表..........................................................................................................................................122 撤销更改............................................................................................................................................. 122 使用层.......................................................................................................................................................123 创建并显示层...................................................................................................................................... 123 转至层类别..........................................................................................................................................123 显示和隐藏项目........................................................................................................................................123 隐藏表中的行和列...............................................................................................................................123 显示表中的隐藏行和列....................................................................................................................... 123 隐藏和显示维度标签...........................................................................................................................123 隐藏和显示表标题...............................................................................................................................123 表格外观...................................................................................................................................................124 应用表格外观...................................................................................................................................... 124 编辑或创建表格外观...........................................................................................................................124 表属性.......................................................................................................................................................124 更改透视表属性.................................................................................................................................. 124 表属性：常规...................................................................................................................................... 125 表属性：注释...................................................................................................................................... 125 表属性：单元格格式...........................................................................................................................125 表属性：边框...................................................................................................................................... 126 表属性：打印...................................................................................................................................... 126 单元格属性............................................................................................................................................... 127 字体和背景..........................................................................................................................................127 格式值................................................................................................................................................. 127 对齐和页边距...................................................................................................................................... 127 脚注和文字说明........................................................................................................................................127 添加脚注和文字说明...........................................................................................................................127 隐藏或显示文字说明...........................................................................................................................127 隐藏或显示表中的脚注....................................................................................................................... 127 脚注标记............................................................................................................................................. 128 对脚注重新编号.................................................................................................................................. 128 编辑遗存表中的脚注...........................................................................................................................128 数据单元格宽度........................................................................................................................................129 更改列宽...................................................................................................................................................129 显示透视表中的隐藏边框......................................................................................................................... 129 在透视表中选择行、列和单元格.............................................................................................................. 129 打印透视表............................................................................................................................................... 129 控制宽表和长表的表分隔符................................................................................................................129 从透视表创建图表.................................................................................................................................... 130 色阶.......................................................................................................................................................... 131 遗存表.......................................................................................................................................................131 第 12 章 模型...................................................................................................... 133 与模型进行交互作用.................................................................................................................................133 使用模型查看器.................................................................................................................................. 133 打印模型...................................................................................................................................................134 探索模型...................................................................................................................................................134 将模型中使用的字段保存到新的数据集................................................................................................... 134 根据重要性将预测变量保存到新的数据集............................................................................................... 134 整体查看器............................................................................................................................................... 135 整体模型............................................................................................................................................. 135 拆分模型查看器........................................................................................................................................136 第 13 章 自动输出修改......................................................................................... 137 样式输出：选择........................................................................................................................................137 样式输出...................................................................................................................................................138 vii 样式输出：标签和文本....................................................................................................................... 139 样式输出：索引.................................................................................................................................. 140 样式输出：表格外观...........................................................................................................................140 样式输出：大小.................................................................................................................................. 140 表格样式...................................................................................................................................................140 表格样式：条件.................................................................................................................................. 141 表格样式：格式.................................................................................................................................. 141 第 14 章 使用命令语法......................................................................................... 143 语法规则...................................................................................................................................................143 从对话框粘贴语法.................................................................................................................................... 144 从对话框粘贴语法...............................................................................................................................144 从输出日志复制语法.................................................................................................................................144 从输出日志复制语法...........................................................................................................................144 使用语法编辑器........................................................................................................................................145 语法编辑器窗口.................................................................................................................................. 145 术语.....................................................................................................................................................146 自动完成............................................................................................................................................. 146 颜色编码............................................................................................................................................. 146 断点.....................................................................................................................................................147 书签.....................................................................................................................................................148 注释或取消注释文本...........................................................................................................................149 设置语法格式...................................................................................................................................... 149 运行命令语法...................................................................................................................................... 150 语法文件中的字符集编码....................................................................................................................150 多条执行命令...................................................................................................................................... 151 语法文件中的字符集编码......................................................................................................................... 151 多条执行命令............................................................................................................................................152 加密语法文件............................................................................................................................................152 第 15 章 图表工具的概述......................................................................................153 生成和编辑图表........................................................................................................................................153 生成图表............................................................................................................................................. 153 编辑图表............................................................................................................................................. 153 图表定义选项............................................................................................................................................154 添加和编辑标题和脚注....................................................................................................................... 154 设置一般选项...................................................................................................................................... 155 第 16 章 使用预测模型对数据评分......................................................................... 157 评分向导...................................................................................................................................................157 将模型字段匹配到数据集字段............................................................................................................ 157 选择评分函数...................................................................................................................................... 159 对活动数据集进行评分....................................................................................................................... 159 合并模型和转换 XML 文件........................................................................................................................159 第 17 章 实用程序(U)........................................................................................... 161 实用程序...................................................................................................................................................161 变量信息...................................................................................................................................................161 数据文件注释............................................................................................................................................161 变量集.......................................................................................................................................................161 定义变量集............................................................................................................................................... 162 使用变量集合显示和隐藏变量..................................................................................................................162 重新排序目标变量列表.............................................................................................................................162 第 18 章 选项...................................................................................................... 163 选项.......................................................................................................................................................... 163 一般选项...................................................................................................................................................163 viii 语言选项...................................................................................................................................................165 查看器选项............................................................................................................................................... 166 数据选项...................................................................................................................................................166 更改缺省变量视图...............................................................................................................................167 货币选项...................................................................................................................................................167 创建定制货币格式...............................................................................................................................168 输出选项...................................................................................................................................................168 图表选项...................................................................................................................................................168 数据元素颜色...................................................................................................................................... 169 数据元素线..........................................................................................................................................169 数据元素标记...................................................................................................................................... 169 数据元素填充...................................................................................................................................... 170 透视表选项............................................................................................................................................... 170 文件位置选项............................................................................................................................................171 脚本选项...................................................................................................................................................172 多重插补选项............................................................................................................................................173 语法编辑器选项........................................................................................................................................173 隐私选项...................................................................................................................................................174 第 19 章 定制菜单和工具栏...................................................................................175 定制菜单和工具栏.................................................................................................................................... 175 菜单编辑器............................................................................................................................................... 175 定制工具栏............................................................................................................................................... 175 显示工具栏............................................................................................................................................... 175 定制工具栏............................................................................................................................................... 175 工具栏属性..........................................................................................................................................176 编辑工具栏..........................................................................................................................................176 创建新工具..........................................................................................................................................176 第 20 章 扩展...................................................................................................... 177 扩展中心...................................................................................................................................................177 “浏览”选项卡...................................................................................................................................... 177 “已安装”选项卡...................................................................................................................................178 设置.....................................................................................................................................................178 扩展详细信息...................................................................................................................................... 178 安装本地扩展束........................................................................................................................................179 扩展的安装位置.................................................................................................................................. 179 必需的 R 软件包..................................................................................................................................180 扩展束的批处理安装...........................................................................................................................181 创建并管理定制对话框.............................................................................................................................181 定制对话框构建器布局....................................................................................................................... 182 构建定制对话框.................................................................................................................................. 182 对话框属性..........................................................................................................................................182 为定制对话框指定菜单位置................................................................................................................183 在画布上布置控件...............................................................................................................................184 构建语法模板...................................................................................................................................... 184 预览定制对话框.................................................................................................................................. 186 控件类型............................................................................................................................................. 186 扩展属性............................................................................................................................................. 201 管理定制对话框.................................................................................................................................. 204 扩展命令的自定义对话框....................................................................................................................207 创建定制对话框的本地化版本............................................................................................................ 207 创建和编辑扩展束.................................................................................................................................... 208 第 21 章 生产作业................................................................................................211 语法文件...................................................................................................................................................211 输出：.......................................................................................................................................................212 ix HTML 选项 ......................................................................................................................................... 212 PowerPoint 选项 ................................................................................................................................ 212 PDF 选项 ............................................................................................................................................ 213 文本选项............................................................................................................................................. 213 使用 OUTPUT 命令的生产作业........................................................................................................... 213 运行时值...................................................................................................................................................213 运行选项...................................................................................................................................................214 服务器登录............................................................................................................................................... 214 添加或编辑服务器登录设置................................................................................................................214 用户提示...................................................................................................................................................214 后台作业状态............................................................................................................................................214 从命令行运行生产作业.............................................................................................................................215 转换生产工具文件.................................................................................................................................... 216 第 22 章 输出管理系统......................................................................................... 217 输出对象类型............................................................................................................................................218 命令标识和表子类型.................................................................................................................................218 标签.......................................................................................................................................................... 219 OMS：选项...............................................................................................................................................219 日志记录...................................................................................................................................................221 从查看器排除输出显示.............................................................................................................................222 将输出转到 IBM SPSS Statistics 数据文件.............................................................................................. 222 从多个表创建的数据文件....................................................................................................................222 控制列元素转换为数据文件中的控制变量..........................................................................................222 OMS 生成的数据文件中的变量名称....................................................................................................223 OXML 表结构............................................................................................................................................ 223 OMS 标识..................................................................................................................................................225 从查看器概要复制 OMS 标识..............................................................................................................225 第 23 章 脚本编写工具......................................................................................... 227 自动脚本...................................................................................................................................................227 创建自动脚本...................................................................................................................................... 228 关联已有脚本与查看器对象................................................................................................................228 以 Python 编程语言编写脚本................................................................................................................... 228 运行 Python 脚本与 Python 程序........................................................................................................229 “Python 编程语言”的“脚本编辑器”....................................................................................................230 Basic 语言中的脚本编写.......................................................................................................................... 230 与 16.0 之前版本的兼容性..................................................................................................................230 脚本上下文对象.................................................................................................................................. 232 启动脚本...................................................................................................................................................232 第 24 章 TABLES 和 IGRAPH 命令语法转换器........................................................ 235 第 25 章 对数据文件、输出文档和语法文件进行加密............................................... 237 声明...................................................................................................................239 商标.......................................................................................................................................................... 240 索引...................................................................................................................241 x 第 1 章 概况 Windows IBM SPSS Statistics 中有一些不同类型的窗口： 数据编辑器。 数据编辑器显示数据文件的内容。 您可以用数据编辑器创建新的数据文件或修改现有的数据 文件。 如果打开了多个数据文件，则每个数据文件都有一个单独的数据编辑器窗口。 查看器。 所有的统计结果、表格和图表都显示在“查看器”中。 您可以编辑输出并进行保存，便于以后使 用。 “查看器”窗口在您第一次运行生成输出的过程时自动打开。 透视表编辑器。 使用透视表编辑器可以通过多种方法修改显示在透视表中的输出。 您可以编辑文本，交换 行中和列中的数据，添加颜色，创建多维表，以及选择隐藏和显示结果。 图表编辑器。 您可以修改图表窗口中的高分辨率图表以及绘图。 您可以更改颜色，选择不同类型的字体或 大小，切换水平轴和垂直轴，旋转三维散点图，甚至更改图表类型。 文本输出编辑器。 没有显示在透视表中的文本输出可以用文本输出编辑器进行修改。 可以编辑输出并更改 字体特征（类型、样式、颜色、大小）。 语法编辑器。 可以将对话框中的选择内容粘贴到语法窗口，在语法窗口中您的选择显示为命令语法格式。 然后可以编辑命令语法，以使用不能通过对话框使用的特殊功能。 您可以将这些命令保存在文件中，便于在 以后的会话中使用。 指定的窗口和活动窗口 如果您打开了多个“查看器”窗口，输出会转到指定的“查看器”窗口。 如果打开了多个“语法编辑器”窗口，那 么命令语法将粘贴到指定的“语法编辑器”窗口中。 指定的窗口在标题栏中用加号图标表示。 您可以随时更 改指定的窗口。 指定的窗口不应与活动窗口相混淆，活动窗口是当前选中的窗口。 如果您有重叠的窗口，则活动窗口显示在 最前面。 如果您打开一个窗口，该窗口就自动成为活动窗口和指定的窗口。 更改指定的窗口 1. 使您要指定的窗口成为活动窗口（单击窗口中的任意位置）。 2. 从菜单中选择： 实用程序 > 指定窗口 注：对于数据编辑器窗口，活动的数据编辑器窗口确定在后续计算或分析中使用的数据集。 没有“指定的”数 据编辑器窗口。 请参阅主题第57 页的『多数据源的基本处理』，了解更多信息。 对话框列表中的变量名称和变量标签 可以在对话框列表中显示变量名称或变量标签，并且可以控制变量列表中的变量排序顺序。 要控制源列表中 变量的缺省显示属性，请从“编辑”菜单中选择选项。 有关更多信息，请参阅第163 页的『一般选项』主 题。 您也可以更改对话框中的变量列表显示属性。 更改显示属性的方法因对话框而异： • 如果对话框在源变量列表上方提供有排序和显示控件，则使用这些控件更改显示属性。 • 如果对话框在源变量列表上方未包含排序控件，请右键单击源列表中的任何变量，然后从弹出菜单中选择 显示属性。 可以显示变量名称或变量标签（对于未定义标签的变量，显示变量名称），并且可以按文件顺序、字母顺序 或测量级别对源列表进行排序。 （在源变量列表上方具有排序控件的对话框中，缺省选择无将按文件顺序对 列表排序。） 数据类型、测量级别和变量列表图标 显示在对话框列表中的变量旁边的图标提供有关变量类型和测量级别的信息。 表 1: 测量级别图标 数字 String 日期 时间 刻度（连续） 不适用 序数 名义 • 有关测量级别的更多信息，请参阅第45 页的『变量测量级别』。 • 有关数字、字符串、日期和时间数据类型的更多信息，请参阅第45 页的『变量类型』。 获得关于对话框中的变量的信息 许多对话框提供有查看有关在变量列表中所显示变量的更多信息的功能。 1. 右键单击源变量列表或目标变量列表中的变量。 2. 选择变量信息。 自动恢复 如果活动会话中有未保存的数据并且 IBM SPSS Statistics 意外退出，那么将在实例中显示“自动恢复”对话框 （在重新启动 SPSS Statistics 后显示对话框）。 该对话框提供选项以供从意外退出的先前会话复原数据以 及删除已保存的会话数据。 注: 保存的会话数据将保持备份状态，直至恢复或删除数据。 每次启动 SPSS Statistics 时，“文档恢复”对话 框将继续显示，直到所有保存的会话都已恢复或删除为止。 请参阅第163 页的『一般选项』对话框中的自动恢复部分以获取有关可用自动会话设置的信息。 “自动恢复”所用时间过长 如果“自动恢复”功能需要长时间才能完成，将显示“自动恢复”对话框的变体，其中提供选项以完全或者针对 特定文件禁用“自动恢复”。 您还可以选择不再针对活动会话显示此对话框。 如果选择禁用自动恢复，则可以通过编辑 > 选项重新激活该功能... > 常规 > 自动恢复。 有关更多信息，请参 阅第163 页的『一般选项』。 复原点 复原点保存意外退出（自动恢复）或显式保存的活动会话的数据。 每个复原点是一个 SPSS Statistics 会话 快照。 每个复原点都包含会话意外退出时处于活动状态或显式保存的数据编辑器、语法和输出文件信息。 保存的复原点将保持备份状态，直至恢复或删除。 注: 请参阅第163 页的『一般选项』对话框中的自动恢复部分以获取有关可用自动会话设置的信息。 打开复原点 每次启动 SPSS Statistics 时，可用复原点自动在 SPSS Statistics 欢迎对话框中显示。 还可以通过 SPSS Statistics 文件菜单访问复原点： 2 IBM SPSS Statistics 29 Core System 用户指南 1. 单击文件 > 打开复原点...。 “自动恢复”对话框显示所有可用复原点。 2. 选择要打开的复原点，然后单击确定。 保存复原点 当活动会话意外退出时，将自动创建 SPSS Statistics 会话复原点。 您可以通过文件 > 保存复原点... 手动保 存复原点 删除复原点 可以从 SPSS Statistics 欢迎页面或者从文件 > 打开复原点... 对话框删除现有会话复原点。 选择复原点，然 后单击“删除”图标 。 了解更多信息 要获得基础知识的综合概述，请参阅联机教程。 从任意 IBM SPSS Statistics 菜单中选择： 帮助 > 教程 第 1 章 概况 3 4 IBM SPSS Statistics 29 Core System 用户指南 第 2 章 获得帮助 IBM Documentation 包含若干不同部分。 帮助 您当前位于的部分。 提供有关用户界面的信息。 每个可选模块都有独立的一部分。 参考 命令语言和 GPL 图形语言的参考信息。 命令语言的参考材料还以 PDF 形式提供：帮助 > 命令语法参 考。 教程 有关使用众多基本功能的分步说明。 案例研究 用于创建各种类型的统计分析以及如何解释结果的实践示例。 统计指导 指导您完成查找要使用的过程。 集成插件 Python 和 R 插件的独立部分。 上下文相关帮助 在用户界面中的许多地方，都可以获得上下文相关的帮助。 • 对话框中的“帮助”按钮可将您直接转到该对话框的帮助主题。 • 右键单击查看器中已激活的透视表内的项，并在弹出菜单中选择这是什么？以显示这些项的定义。 • 在命令语法窗口中，将光标放在命令的语法块中的任意位置，然后按键盘上的 F1。 显示该命令的帮助。 其他资源 在以下网站可以找到许多常见问题的解答： 如果您是使用任何学生版 IBM SPSS 软件产品的学生，请访问我们专为学生提供的在线教育解决方案页面。 如果您是使用大学提供的 IBM SPSS 软件副本的学生，请联系所在大学的 IBM SPSS 产品协调员。 IBM SPSS Statistics 社区包含适合各级别的用户和应用程序开发者的资源。 下载实用程序、图形示例、新统 计模块和文章。 请在下列地址访问 IBM SPSS Statistics 社区： 提供了统计算法和命令语法 的 PDF 格式文档。 6 IBM SPSS Statistics 29 Core System 用户指南 第 3 章 数据文件 数据文件有多种格式，而本软件被设计为可以处理其中的许多格式，包括： • Excel 电子表格 • 来自许多数据库源（包括 Oracle、SQLServer、DB2 等等）的数据库表 • 制表符分隔类型、CSV 类型和其他类型的简单文本文件 • SAS 数据文件 • Stata 数据文件 打开数据文件 除了以 IBM SPSS Statistics 格式保存的文件以外，还可以打开 Excel、SAS、Stata、制表符分隔文件和其他 文件，而无需将文件转换为中间格式或输入数据定义信息。 • 打开数据文件会使其成为活动数据集。 如果已经打开了一个或多个数据文件，那么它们将保持打开状态， 并可在以后的会话中使用。 单击“数据编辑器”窗口中的任意位置会使打开的数据文件成为活动数据集。 请 参阅主题第57 页的『第 6 章 使用多数据源』，了解更多信息。 • 在分布式分析中，使用远程服务器处理命令和运行过程的模式、可用的数据文件、文件夹和驱动器取决于 远程服务器上可用的内容。 当前服务器名称在对话框的顶部指明。 除非将驱动器指定为共享设备，或者将 包含数据文件的文件夹指定为共享文件夹，否则将不能访问本地计算机上的数据文件。 请参阅主题第37 页的『第 4 章 分布式分析模式』，了解更多信息。 打开数据文件 1. 从菜单中选择： 文件 > 打开 > 数据...
2. 在“打开数据”对话框中，选择要打开的文件。 3. 单击打开。 （可选）您可以执行以下操作： • 根据观测值，最小化字符串宽度将每个字符串变量宽度自动设置为该变量的最长观测值。 在 Unicode 方式 下读取代码页数据文件时特别有用。 有关更多信息，请参阅第163 页的『一般选项』主题。 • 从电子表格文件的第一行读取变量名称。 • 指定电子表格文件中要读取的单元格范围。 • 指定 Excel 文件中要读取的工作表（Excel 95 或更高版本）。 有关从数据库中读取数据的信息，请参阅第13 页的『读取数据库文件』。 有关从文本数据文件中读取数 据的信息，请参阅第10 页的『文本向导』。 有关读取 IBM Cognos 数据的信息，请参阅第17 页的『读 取 Cognos BI 数据』。 数据文件类型 SPSS Statistics. 打开以 IBM SPSS Statistics 格式保存的数据文件以及 DOS 产品 SPSS/PC+。 SPSS Statistics 压缩。 打开以 IBM SPSS Statistics 压缩格式保存的数据文件。 SPSS/PC+. 打开 SPSS/PC+ 数据文件。 此选项只在 Windows 操作系统中可用。 便携。 打开以可移植格式保存的数据文件。 以可移植格式保存文件比以 IBM SPSS Statistics 格式保存文件 所耗费的时间要长得多。 Excel。 打开 Excel 文件。 Lotus 1-2-3. 打开以 1-2-3 格式（对于 Lotus R3.0、2.0 或 1A）保存的数据文件。 SYLK。 打开以 SYLK（符号链接）格式保存的数据文件，SYLK 格式是部分电子表格应用程序所使用的格 式。 dBASE。 打开 dBASE 格式文件（dBASE IV、dBASE III 或 III PLUS 或者 dBASE II）。 每个个案均是一条 记录。 当您以这种格式保存文件时，变量和值标签以及缺失值的指定会丢失。 SAS。 SAS V6 - V9 和 SAS 传输文件。 Stata。 Stata V4-V13。 读取 Excel 文件 本主题适用于 Excel 95 和更高版本文件。 要读取 Excel 4 或更低版本，请参阅第8 页的『读取旧 Excel 文 件和其他电子表格』主题。 要导入 Excel 文件 1. 从菜单中选择： 文件 > 导入数据 > Excel...，或者将现有 Excel 文件直接拖放到打开的 IBM SPSS Statistics 实例。 2. 选择相应的导入设置。 工作表 Excel 文件可以包含多个工作表。 缺省情况下，数据编辑器读取第一张工作表。 要读取其它工作表， 请从列表中选择工作表。 范围 也可以读取某个范围的单元格。 请使用与在 Excel 中相同的方法指定单元格范围。 例如：A1:D10。 从第一行数据读取变量名称 可以从文件的第一行或所定义的范围的第一行读取变量名称。 不符合变量命名规则的值会转换为有效 的变量名称，原始名称将用作变量标签。 用于确定数据类型的值的百分比 每个变量的数据类型由符合相同格式的值的百分比确定。 • 值必须大于 50。 • 用于确定百分比的分母为每个变量的非空值的数目。 • 如果指定百分比的值使用了不一致格式，那么会为变量指定字符串数据类型。 • 对于要买百分比值分配数字格式（包括日期和时间格式）的变量，不符合该格式的值将分配系统缺 失值。 忽略隐藏的行和列 不会包含 Excel 文件中隐藏的行和列。 此选项仅适用于 Excel 2007 和更高版本的文件（XLSX 和 XLSM）。 从字符串值中移除前导空格 将移除字符串值开头的任何空格。 从字符串值中移除尾部空格 将移除字符串值尾部的空格。 此设置影响字符串变量的定义宽度的计算。 3. 单击确定。 读取旧 Excel 文件和其他电子表格 本主题适用于读取 Excel 4 或更低版本的文件、Lotus 1-2-3 文件和 SYLK 格式电子表格文件。 有关读取 Excel 95 或更高版本文件的信息，请参阅主题第8 页的『读取 Excel 文件』。 读取变量名称。 对于电子表格，您可以从文件的第一行或定义范围的第一行读取变量名称。 根据需要转换 值以创建有效的变量名，包括将空格转换为下划线。 范围。 对于电子表格数据文件，您还可以读取某个单元格范围。 请使用与在电子表格应用程序中相同的方 法指定单元格范围。 8 IBM SPSS Statistics 29 Core System 用户指南 如何读取电子表格 • 每个变量的数据类型和宽度由列中第一个数据单元格的列宽和数据类型确定。 其他类型的值会转换为系统 缺失值。 如果列中的第一个数据单元格是空白的，则使用该电子表格的全局缺省数据类型（通常为数 值）。 • 对于数值变量，空白单元格会转换为系统缺失的值，用句点表示。 对于字符串变量，空格是有效的字符串 值，空白单元格被视为有效的字符串值。 • 如果不从电子表格读取变量名称，则列字母（A、B、C...）用于 Excel 和 Lotus 文件的变量名称。 对于以 R1C1 显示格式保存的 SYLK 文件和 Excel 文件，本软件使用以字母 C 开头的列号作为变量名称（C1、 C2、C3 等）。 读取 dBASE 文件 数据库文件在逻辑上与 IBM SPSS Statistics 数据文件非常相似。 以下一般规则适用于 dBASE 文件： • 字段名将转换为有效的变量名。 • 用于 dBASE 字段名称的冒号会转换为下划线。 • 包含标记为要删除但未实际清除的记录。 本软件创建一个新的字符串变量 D_R，该变量对标记为要删除的 个案包含一个星号。 读取 Stata 文件 以下一般规则适用于 Stata 数据文件： • 变量名称。 Stata 变量名称以区分大小写的形式转换为 IBM SPSS Statistics 变量名称。 通过附加下划线和 顺序字母（A、_B、_C、...、_Z、_AA、_AB、...等等），将只有大小写不同的 Stata 变量名称转换为有 效的变量名称。 • 变量标签。 Stata 变量标签转换为 IBM SPSS Statistics 变量标签。 • 值标签。 Stata 值标签转换为 IBM SPSS Statistics 值标签，但归为“扩展”缺失值的 Stata 值标签除外。 长 度超过 120 字节的值标签将被截断。 • 字符串变量。 Stata strl 变量将转换为字符串变量。 长度超过 32K 字节的值将被截断。 包含 BLOB（二进 制大对象）的 Stata strl 值将转换为空字符串。 • 缺失值。 Stata“扩展”缺失值转换为系统缺失值。 • 日期转换。 Stata 日期格式值转换为 IBM SPSS Statistics DATE 格式 (d-m-y) 值。 Stata“时间序列”日期格 式值（周数、月数、季度数等）转换为简单数值 (F) 格式，同时保留原始内部整数值，即从 1960 年开始 算起的周数、月数、季度数等等。 读取 CSV 文件 要读取 CSV 文件，请从菜单选择：文件 > 导入数据 > CSV 注: “导入数据”功能不支持在带引号文本中包含嵌入换行符的 CSV 数据。 可能的变通方法是将 CSV 文件（包 含带引号的嵌入式换行符）另存为 .xls/.xlsx 文件，然后使用导入 Excel 功能。 “读取 CSV 文件”对话框读取 CSV 格式文本数据文件，这些文件使用逗号、分号或制表符作为值之间的定界 符。 如果文本文件使用其他定界符，在文件开头部分包含不是变量名称或数据值的文本，或具有其他特殊注意事 项，请使用文本向导读取文件。 第一行包含变量名 文件中的第一个非空白行包含用作变量名称的标签文本。 作为无效变量名称的值会自动转换为有效变量 名称。 从字符串值中移除前导空格 将移除字符串值开头的任何空格。 从字符串值中移除尾部空格 将移除字符串值尾部的空格。 此设置影响字符串变量的定义宽度的计算。 第 3 章 数据文件 9 值之间的定界符 定界符可以为逗号、分号或制表符。 如果定界符为任何其他字符或空格，请使用文本向导读取文件。 小数符号 用于指示文本数据文件中小数的符号。 符号可以为句点或逗号。 文本限定符 用于包括包含定界字符的值的字符。 限定符出现在值的开头和结尾。 限定符可以为双引号、单引号或 无。 用于确定数据类型的值的百分比 每个变量的数据类型由符合相同格式的值的百分比确定。 • 值必须大于 50。 • 如果指定百分比的值使用了不一致格式，那么会为变量指定字符串数据类型。 • 对于要买百分比值分配数字格式（包括日期和时间格式）的变量，不符合该格式的值将分配系统缺失 值。 在本地缓存数据 数据高速缓存是数据文件的完整副本，它存储在临时磁盘空间中。 高速缓存数据文件可以改进性能。 文本向导 文本向导可以读取多种格式的文本数据文件： • 制表符分隔文件 • 空格分隔文件 • 逗号分隔文件 • 固定字段格式的文件 对于分隔的文件，也可以将其他字符指定为值之间的分隔符，并且可以指定多个分隔符。 读取文本数据文件 1. 从菜单中选择： 文件 > 导入数据 > 文本数据...
2. 在“打开数据”对话框中选择文本文件。 3. 如有必要，选择文件的编码。 4. 按照文本向导中的步骤来定义如何读取数据文件。 编码 文件的编码影响读取字符数据的方式。 Unicode 数据文件通常包含一个标识字符编码的字节顺序标记。 一些 应用程序创建不包含字节顺序标记的 Unicode 文件，代码页数据文件未包含任何编码标识。 • Unicode (UTF-8)。 以 Unicode UTF-8 编码读取文件。 • Unicode (UTF-16)。 按操作系统的字节顺序以 Unicode UTF-16 编码读取文件。 • Unicode (UTF-16BE)。 以 Unicode UTF-16 Big Endian 格式读取文件。 • Unicode (UTF-16LE)。 以 Unicode UTF-16 Little Endian 格式读取文件。 • 本地编码。 以当前语言环境代码页字符编码读取文件。 如果文件包含 Unicode 字节顺序标记，那么将以该 Unicode 编码读取该文件，而无论您选择了哪种编码。 如果文件未包含 Unicode 字节顺序标记，那么除非您选择了某种 Unicode 编码，否则缺省情况下假定编码为 当前语言环境代码页字符编码。 要更改采样另一种代码页字符编码的数据文件的当前语言环境，请从菜单中选择“编辑”>“选项”，然后在“语 言”选项卡上更改语言环境。 10 IBM SPSS Statistics 29 Core System 用户指南 文本向导：步骤 1 文本文件显示在一个预览窗口中。 可以应用预定义的格式（以前在文本向导中保存的），或者按照文本向导 中的步骤来指定如何读取数据。 文本向导：步骤 2 此步骤提供变量的信息。 变量类似于数据库中的字段。 例如，问卷中的每一项都是一个变量。 变量是如何排列的？ 变量的安排定义用于将一个变量与另一个变量区分开来的方法。 定界 使用空格、逗号、制表符和其他字符分隔变量。 变量为每个个案按照同样的顺序进行记录，但不一 定在相同的列位置。 固定宽度 对于数据文件中的每个个案，每个变量都记录在同一个记录（行）上的相同列位置。 变量之间不需 要分隔符。 列位置确定要读取的是哪个变量。 注：文本向导无法读取固定宽度的 Unicode 文本文件。 您可以使用 DATA LIST 命令来读取固定宽 度的 Unicode 文本文件。 文件开头是否包括变量名？ 指定行号上的值用于创建变量名称。 不符合变量命名规则的值会转换为有效的变量名。 小数符号是什么？ 用于指示小数符号可为句点或逗号的字符。 文本向导：步骤 3（定界的文件） 这一步提供有关个案的信息。 个案类似于数据库中的记录。 例如，问卷的每个响应者都是一个个案。 第一个数据个案从哪个行号开始？(F) 表示包含数据值的数据文件的第一行。 如果数据文件的顶行包含描述 性标签或者包含不代表数据值的其他文本，这就不是第 1 行。 如何表示个案？ 控制文本向导如何确定每个个案结束、下一个个案开始的位置。 • 每一行表示一个个案。 每一行仅包含一个个案。 每个个案通常包含在一个单行中，即使这一行对于有大量 变量的数据文件会很长。 如果不是所有的行都包含相同数量的数据值，则每个个案的变量数由数据值的个 数最多的行决定。 对于数据值较少的个案，多出来的变量指定为缺失值。 • 变量的特定编号表示一个个案。 每个实例的指定变量数告诉文本向导在哪里停止读取某个个案，并开始读 取下一个个案。 同一行可以包含多个个案，个案可以在一行的中间开始，并在下一行继续。 文本向导按照 读取的值的数量确定每个个案的结束，不管有多少行。 每个个案必须包含所有变量的数据值（或者由分隔 符表示的缺失值），否则数据文件将无法正确读取。 您要导入多少个个案？ 您可以导入数据文件中的所有个案，可以导入前 n 个个案（n 是您指定的数字），也 可以随机导入指定百分比的样本。 因为随机抽样程序对每个个案都作出独立的假随机决策，所以选定的个案 的百分比可能只与指定的百分比相近。 数据文件中的个案越多，选定个案的百分比与指定百分比就越接近。 文本向导：步骤 3（固定宽度的文件） 这一步提供有关个案的信息。 个案类似于数据库中的记录。 例如，问卷中的每个响应者都是一个个案。 第一个数据个案从哪个行号开始？(F) 表示包含数据值的数据文件的第一行。 如果数据文件的顶行包含描述 性标签或者包含不代表数据值的其他文本，这就不是第 1 行。 多少行表示一个个案？(L) 控制文本向导如何确定每个个案结束、下一个个案开始的位置。 每个变量由其个 案内的行数及其列位置定义。 需要指定每个个案的行数，才能正确读取数据。 您要导入多少个个案？ 您可以导入数据文件中的所有个案，可以导入前 n 个个案（n 是您指定的数字），也 可以随机导入指定百分比的样本。 因为随机抽样程序对每个个案都作出独立的假随机决策，所以选定的个案 的百分比可能只与指定的百分比相近。 数据文件中的个案越多，选定个案的百分比与指定百分比就越接近。 第 3 章 数据文件 11 文本向导：步骤 4（定界的文件） 此步骤指定在文本数据文件中使用的定界符和文本限定符。 您还可在字符串值中指定前导空格和尾部空格的 处理。 变量之间存在哪些定界符？ 用于分隔数据值的字符或符号。 可以选择空格、逗号、分号、制表符或其他字符的任意组合。 中间没有 插入数据值的多个连续的分隔符被视为缺失值。 文本限定符是什么？ 用于包括包含分隔符字符的值的字符。 文本限定符出现在值的开头和结尾，封装了整个值。 前导空格和尾部空格 控制字符串值中前导空格和尾部空格的处理。 从字符串值中移除前导空格 将移除字符串值开头的任何空格。 从字符串值中移除尾部空格 计算字符串变量的定义宽度时将忽略值尾部的空格。 如果将空格选作定界符，那么将多个连续空格 作为多个定界符处理。 文本向导：步骤 4（固定宽度的文件） 此步骤显示文本向导对于如何读取数据文件的最佳猜测，并使您可以修改文本向导从数据文件读取变量的方 式。 预览窗口中的垂直线表示文本向导当前认为每个变量在文件中开始的位置。 必要时插入、移动和删除变量换行符以分隔变量。 如果每个个案使用多行，则数据将按每个个案一行的方式 显示，后续行附加在行的末尾。 注意： 对于计算机生成的数据文件，其所生成的一连串连续的数据值没有插入空格或其他明显特征，这样就很难确 定每个变量开始的位置。 这样的数据文件通常依赖于数据定义文件或其他一些指定每个变量的行和列位置的 书面说明。 文本向导：步骤 5 此步骤控制变量名称和用于读取每个变量的数据格式。 还可指定要排除的变量。 变量名称 可以用自己的变量名称覆盖缺省的变量名称。 如果从数据文件读取变量名称，那么将自动修改不符合变 量命名规则的名称。 在预览窗口中选择一个变量，然后输入变量名。 数据格式 在预览窗口选择一个变量，然后从列表选择一种格式。 • 自动将跟踪所有数据值的求值确定数据格式。 • 要排除某个变量，请选择请勿导入。 用于确定自动数据格式的值所占的百分比(P) 针对自动格式，每个变量的数据格式由符合相同格式的值的百分比确定。 • 值必须大于 50。 • 用于确定百分比的分母为每个变量的非空值的数目。 • 如果指定百分比的值使用了不一致格式，那么会为变量指定字符串数据类型。 • 对于要买百分比值分配数字格式（包括日期和时间格式）的变量，不符合该格式的值将分配系统缺失 值。 格式化选项 用于读取变量的格式化选项包括： 自动 格式根据所有数据值的求值进行确定。 12 IBM SPSS Statistics 29 Core System 用户指南 数值 有效值包括数字、前导加号或减号以及小数指示符。 String 有效值包括几乎任何键盘字符和嵌入的空格。 对于分隔的文件，可以指定值的字符数量，最多可以指定 为 32,767 个。 缺省情况下，该值设置为选定的变量在文件前 250 行中遇到过的最长的字符串值。 对于 固定宽度的文件，字符串值的字符的数量由变量换行符的位置定义。 日期/时间 有效值包括常用的日期格式 dd-mm-yyyy、mm/dd/yyyy、dd.mm.yyyy、yyyy/mm/dd、hh:mm:ss 以及 其他各种日期和时间格式。 月份可以用数字、罗马数字或三个字母的缩写形式表示，也可以使用全拼的 格式。 从列表中选择一个日期格式。 美元符号 有效值为数字，前导美元符是可选的，作为千位分隔符的逗号也是可选的。 逗号 有效值包括将句点用作小数指示符和将逗号用作千位分隔符的数字。 点 有效值包括将逗号用作小数指示符和将句点用作千位分隔符的数字。 请勿导入 省略在导入的数据文件中选择的变量。 注: 包含对选定的格式无效的字符的值将视为缺失值。 包含任何指定分隔符的值将视为多个值。 文本向导：步骤 6 这是文本向导的最后一步。 您可以将规范保存在文件中，以在导入相似文本数据文件时使用。 您还可以将 文本向导生成的语法粘贴到语法窗口中。 然后就可以自定义和／或保存语法，以便用于其他对话或生产作业 中。 在本地缓存数据. 数据高速缓存是数据文件的完整副本，存储在临时磁盘空间中。 高速缓存数据文件可以改 进性能。 读取数据库文件 只要有某种数据库格式的数据库驱动程序，就可以读取该数据库格式的数据。 在本地分析方式下，必需的驱 动程序必须安装在本地计算机上。 用分布式分析模式时（IBM SPSS Statistics Server 提供），远程服务器 上必须安装这些驱动程序。 有关更多信息，请参阅第37 页的『第 4 章 分布式分析模式』主题。 注：如果您运行的是 IBM SPSS Statistics 的 Windows 64 位版本，那么将无法读取 Excel、Access 或 dBASE 数据库源，即使它们可能出现在可用数据库源的列表上也是如此。 这些产品的 32 位 ODBC 驱动程序 不兼容。 读取数据库文件 1. 从菜单中选择： 文件 > 导入数据 > 数据库 > 新建查询...
2. 选择数据源。 3. 如果需要（取决于数据源），可选择数据库文件和/或输入登录名、密码和其他信息。 4. 选择表和字段。 对于 OLE DB 数据源（仅在 Windows 操作系统上可用）只能选择一个表。 5. 指定表之间的关系。 6. 选择性地执行下列操作： • 为数据指定任何选择条件。 • 添加一个提示，供用户输入信息以创建参数查询。 • 运行构建的查询之前请先保存。 第 3 章 数据文件 13 连接汇聚 如果在同一会话或作业中多次访问同一数据库源，可以通过连接汇聚提高性能。 1. 在向导的最后一步，将命令语法粘贴到语法窗口中。 2. 在带引号的 CONNECT 字符串末尾，添加 Pooling=true。 编辑已保存的数据库查询 1. 从菜单中选择： 文件 > 导入数据 > 数据库 > 编辑查询...
2. 选择要编辑的查询文件 (.spq)。 3. 请按照创建新查询的说明操作。 使用已保存的查询读取数据库文件 1. 从菜单中选择： 文件 > 导入数据 > 数据库 > 运行查询...
2. 选择要运行的查询文件 (.spq)。 3. 如果需要（取决于数据库文件），输入登录名和密码。 4. 如果查询包含嵌入的提示，则根据需要输入其他信息（例如，要检索销售数据的季度）。 选择数据源 使用“数据库向导”的第一个屏幕选择要读取的数据源类型。 ODBC 数据源 如果没有配置任何 ODBC 数据源，或者要添加新的数据源，请单击添加 ODBC 数据源。 • 在 Linux 操作系统中，该按钮不可用。 在 odbc.ini 中指定 ODBC 数据源，并且 ODBCINI 环境变量必须设 定为该文件的位置。 有关更多信息，请参阅数据库驱动程序文档。 • 用分布式分析模式时（IBM SPSS Statistics Server 提供），该按钮不可用。 要用分布式分析模式添加数据 源，请咨询系统管理员。 ODBC 数据源包含两部分重要的信息：将用于访问数据的驱动程序以及要访问的数据库的位置。 要指定数据 源，必须装有适当的驱动程序。 针对不同数据库格式的驱动程序包括在安装介质上。 选择数据字段 “选择数据”步骤控制将读取哪些表和字段。 数据库字段（列）读取为变量。 如果在表中选择了任何字段，则其所有字段将显示在以下“数据库向导”窗口中，但只有在此步骤中选中的字 段才作为变量导入。 这允许您创建表连接，并使用未导入的字段指定条件。 显示字段名称。 要列示表中的字段，请单击表名左边的加号 (+)。 要隐藏字段，请单击表名左边的减号 (-)。 要添加字段。 双击“可用的表”列表中的任何字段，或者将其拖到“按此顺序检索字段”列表中。 在字段列表中 拖放字段可以对其重新排序。 删除字段。 双击“按此顺序检索字段”列表中的任何字段，或者将其拖到“可用的表”列表。 将字段名称排序。 如果选中此复选框，“数据库向导”会按照字母顺序显示可用的字段。 缺省情况下，可用表的列表只显示标准数据库表。 您可以控制列表中显示的项的类型： • 表。 标准数据库表。 • 视图。 视图是由查询定义的虚拟的或动态的“表”。 视图中可以包含根据其他字段值计算得出的多个表和/ 或字段的连接。 • 同义词。 同义词是表或视图的别名，通常在查询中定义。 14 IBM SPSS Statistics 29 Core System 用户指南 • 系统表。 系统表定义数据库的属性。 在某些情况下，标准数据库表可能会被分类成系统表，并且仅在选中 此选项后才会显示。 通常只有数据库管理员才具有访问真正的系统表的权限。 注：对于 OLE DB 数据源（仅在 Windows 操作系统上可用），只能从单个表中选择字段。 OLE DB 数据源不 支持多个表连接。 创建表之间的关系 “指定关系”步骤使您可以定义 ODBC 数据源的表之间的关系。 如果选择的字段来自一个以上的表，那么必须 定义至少一个连接。 建立关系。 要创建关系，请将任意表上的字段拖到要连接的字段上。 “数据库向导”将在两个字段之间画一 条连接线，表明它们的关系。 这些字段的数据类型必须相同。 自动连接表。 可尝试按照主/外键或匹配的字段名和数据类型来自动连接表。 连接类型。 如果驱动程序支持外部连接，那么可以指定内部连接、左边外连接或右边外连接。 • Inner joins. 内部连接仅包括相关字段相等的行。 在此示例中，将包括两个表中具有匹配的 ID 值的所有 行。 • 外部连接。 除了内部连接的一对一匹配外，还可以使用外部连接通过一对多匹配方案来合并表。 例如，您 可以将其中只包含少量代表数据值和关联描述性标签的记录的表，和包含上百个或上千个代表调查响应者 的记录的表中的值相匹配。 左边外连接包括左边的表中的所有记录，而仅包括右边的表中相关字段相等的 记录。 在右边外连接中，连接从右边的表导入所有记录，而仅从左边的表导入相关字段相等的记录。 计算新字段 如果处于分布式模式中，并已连接到远程服务器（IBM SPSS Statistics Server 提供），那么可以先计算新字 段，然后再将数据读入 IBM SPSS Statistics 中。 您也可以在将数据读入 IBM SPSS Statistics 后计算新字段，但对于大型数据源，在数据库中计算新字段可以 节省时间。 新字段名。 名称必须遵守 IBM SPSS Statistics 变量名称规则。 表达式。 输入用于计算新字段的表达式。 可以从字段列表拖动现有字段名称，从函数列表拖动函数。 限制检索的个案 “限制检索的案例”步骤允许您指定选择个案（行）的子集的条件。 限制个案通常包括用条件填充条件网格。 条件由两个表达式以及它们之间的某种关系组成。 该表达式返回每个个案的 true、false 或 missing 值。 • 如果结果是 true，那么选中该个案。 • 如果结果是 false 或 missing，那么不选中该个案。 • 大多数条件使用六个关系运算符（<、>、<=、>=、= 和 <>）中的一个或多个。 • 表达式可以包括字段名、常数、算术运算符、数字和其他函数以及逻辑变量。 您可将未计划导入的字段用 作变量。 要建立条件，至少需要两个表达式和一种连接表达式的关系。 1. 要建立表达式，请选择下列一种方法： • 在“表达式”单元格中，输入字段名、常数、算术运算符、数字和其他函数或逻辑变量。 • 双击“字段”列表中的一个字段。 • 将字段从“字段”列表中拖到“表达式”单元格。 • 从任何活动的“表达式”单元格的下拉菜单中选择一个字段。 2. 要选择关系运算符（如 = 或 >），请将光标放在“关系”单元格上，然后输入运算符或从下拉菜单中进行选 择。 如果 SQL 包含具有个案选择表达式的 WHERE 子句，则表达式中的日期和时间需要以特殊方式指定（包括 示例中显示的花括号）： 第 3 章 数据文件 15 • 应使用一般形式 {d 'yyyy-mm-dd'} 来指定日期文本。 • 应使用一般格式 {t 'hh:mm:ss'} 来指定时间文本。 • 应使用一般格式 {ts 'yyyy-mm-dd hh:mm:ss'} 来指定日期/时间文本（时间戳）。 • 整个日期和/或时间值都必须用单引号括起。 年份必须以四位数的形式表示；日期和时间的值的每个部 分都必须包含两位数。 例如，2005 年 1 月 1 日上午 1:05 应表示为以下形式： {ts '2005-01-01 01:05:00'} 函数。 可选择内置算术、逻辑、字符串、日期和时间 SQL 函数。 可将函数从列表中拖到表达式中，或者 输入任何有效的 SQL 函数。 有关有效 SQL 函数，请参阅您的数据库文档。 使用随机抽样。 该选项从数据源选择个案的随机样本。 对于大数据源，您可能需要将个案数限制为小 的、具有代表性的样本，这可以显著减少其运行程序所需的时间。 本机随机抽样（如果对该数据源可 用）速度比 IBM SPSS Statistics 随机抽样要快，因为 IBM SPSS Statistics 随机抽样必须读取整个数据源 才能抽取随机样本。 • 近似于 (Approximately). 生成近似于指定个案百分比的随机样本。 由于此例程为每个个案作出独立的伪 随机决策，因此选定个案的百分比只能近似于指定的百分比。 数据文件中的个案越多，选定个案的百 分比与指定百分比就越接近。 • 确切 (Exactly). 从指定的个案总数中选择指定个案数的随机样本。 如果指定的个案总数超过数据文件中 的个案总数，那么样本将按比例包含比请求数目少的个案。 注：如果使用随机抽样，那么汇总（IBM SPSS Statistics Server 中的分布式模式提供）不可用。 输入值提示。 可以在查询中嵌入一个提示来创建参数查询。 用户运行查询时，系统将要求用户输入信息 （根据此处指定的信息）。 如果要查看同一数据的不同视图，那么您可能需要这样做。 例如，您可能想 要运行相同的查询来查看不同财政季度的销售数据。 3. 将光标放在任何“表达式”单元格中，然后单击输入值提示来创建提示。 创建参数查询 使用“输入值提示”步骤来创建一个对话框，在每次有人运行查询时请求用户提供信息。 如果要用不同的条件 查询相同的数据源，则该功能将很有用。 要建立提示，请输入提示字符串和缺省值。 每次用户运行查询时，该提示字符串都会显示。 该字符串应指 定要输入的信息类型。 如果用户不从列表中进行选择，那么该字符串应给出有关如何设置输入格式的提示。 示例如下：输入季度（Q1、Q2、Q3...）。 允许用户从列表中选择值。 如果选中该复选框，您可以限制用户选择您放在此处的值。 确保使用回车分隔 值。 数据类型。 在此处选择数据类型（数字、字符串或日期）。 日期和时间值必须以特定格式输入。 • 日期值必须使用一般格式 yyyy-mm-dd。 • 时间值必须使用一般格式 hh:mm:ss。 • 日期/时间值（时间戳记）必须使用一般格式 yyyy-mm-dd hh:mm:ss。 汇总数据 如果处于分布式模式中，并已连接到远程服务器（IBM SPSS Statistics Server 提供），那么可以先对数据进 行汇总，然后再将其读入 IBM SPSS Statistics 中。 还可以在将数据读到 IBM SPSS Statistics 中之后再对其进行汇总，但对于大数据源来说，预先进行汇总可以 节省时间。 1. 要创建汇总数据，请选择一个或多个定义如何分组个案的分组变量。 2. 选择一个或多个汇总变量。 3. 为每个汇总变量选择一个汇总函数。 4. 或者，创建包含每个分类组中的个案数的变量。 16 IBM SPSS Statistics 29 Core System 用户指南 注：如果使用 IBM SPSS Statistics 随机抽样，汇总将不可用。 定义变量 变量名称和标签。 完整的数据库字段（列）名用作变量标签。 除非您修改变量名称，否则“数据库向导”将 按照以下两种方法中的一种将变量名称指定给数据库的每一列： • 如果数据库字段的名称是有效的、唯一的变量名，那么该名称将用作变量名。 • 如果数据库字段的名称未形成有效的唯一变量名，那么将自动生成一个新的唯一名称。 单击任何单元格来编辑变量名称。 将字符串转换为数字值。 如果要自动将字符串变量转换为数值型变量，请针对该字符串变量选择重新编码为 数值型框。 字符串值按照原始值的字母顺序转换为连续的整数值。 原始值保留为新的变量的值标签。 变量宽度字符串字段的宽度。 该选项控制变量宽度字符串值的宽度。 缺省情况下，宽度为 255 个字节，并 且只读取前 255 个字节（通常指单字节语言中的 255 个字符）。 宽度至多可以为 32,767 个字节。 尽管您 可能不需要截断字符串值，但也不要指定不必要的长值，这会导致处理效率很低。 根据观察值，最小化字符串宽度。 自动将每个字符串变量的宽度设置为最长观测值。 排序个案 如果处于分布式模式中，并已连接到远程服务器（IBM SPSS Statistics Server 提供），那么可以先对数据进 行分类，然后再将其读入 IBM SPSS Statistics 中。 还可以在将数据读到 IBM SPSS Statistics 中之后再对其进行排序，但对于大数据源来说，预先进行排序可以 节省时间。 结果 “结果”步骤显示查询的 SQL Select 语句。 • 可以在运行查询前编辑该 SQL Select 语句，但是如果单击上一步按钮在前面的步骤中进行更改，那么对 Select 语句所做的更改将丢失。 • 要保存查询以供将来使用，可使用将查询保存到文件区段。 • 要将完整的 GET DATA 语法粘贴到语法窗口，请选择将其粘贴到语法编辑器以供将来修改。 从“结果”窗口 复制和粘贴 Select 语句将不会粘贴所需的命令语法。 注：粘贴的语法在由向导生成的每一行 SQL 中的结束引号之前包含一个空格。 这些空格不是多余的。 在处 理命令时，SQL 语句的所有行将以一种文本形式合并在一起。 如果没有空格，在一行的最后一个字符和下一 行的第一个字符之间就不会存在空格。 读取 Cognos BI 数据 如果您拥有对 IBM Cognos Business Intelligence 服务器的访问权，您可将 IBM Cognos Business Intelligence 数据包和列表报告读入 IBM SPSS Statistics。 要读取 IBM Cognos Business Intelligence 数据： 要点: 如果不存在 Cognos Analytics Administrator 许可证，那么将不会完整导入 Cognos BI 数据。 您必须 在导入 Cognos BI 数据前已拥有或购买 Cognos Analytics Administrator 许可证。 导入 Cognos BI 数据的用 户必须将其角色设置为“系统管理员”。 有关 Cognos Analytics 中许可证角色的更多信息，请参阅 How do you restrict users based on their License Roles in Cognos Analytics (versions 11.0.0 to 11.0.6)。 1. 从菜单中选择： 文件 > 导入数据 > Cognos Business Intelligence 2. 指定 IBM Cognos Business Intelligence 服务器连接的 URL。 3. 指定数据包或报告的位置。 4. 选择您想读取的数据字段或报告。 （可选）您可以执行以下操作： 第 3 章 数据文件 17 • 选择数据包的过滤器。 • 导入汇总数据而非原始数据。 • 指定参数值。 众数。 指定要读取的信息类型：数据或报告。 可以读取的唯一报告类型是列表报告。 连接。 Cognos Business Intelligence 服务器的 URL。 单击编辑按钮定义用于导入数据或报告的新 Cognos 连接详细信息。 有关更多信息，请参阅第18 页的『Cognos 连接』主题。 位置。 您想读取的软件包或报告的位置。 单击编辑按钮以显示用于导入内容的可用源列表。 有关更多信 息，请参阅第18 页的『Cognos 连接』主题。 内容。 对于数据，显示可用的数据包和过滤器。 对于报告，显示可用的报告。 要导入的字段。 对于数据包，选择要包含的字段并将其移至此列表。 要导入的报告。 对于报告，选择要导入的列表报告。 报告必须为列表报告。 要应用的过滤器。 对于数据包，选择您要应用的过滤器并将其移至此列表。 参数。 如果启用此按钮，则会为选定的对象定义参数。 在导入数据之前，您可以使用参数做出调整（例 如，执行参数化计算)。 如果定义了参数但未提供缺省值，那么此按钮将显示一个警告三角形。 执行导入前汇总数据。 对于数据包，如果已在数据包中定义汇总，您可以导入汇总的数据而非原始数据。 Cognos 连接 “Cognos 连接”对话框指定IBM Cognos Business Intelligence 服务器 URL 及任何所需的额外凭证。 Cognos 服务器 URL。 IBM Cognos Business Intelligence 服务器的 URL。 这是在服务器上 IBM Cognos 配 置 的“外部调度程序 URI”环境属性的值。 请联系系统管理员以获取更多信息。 众数。 如果需要使用特定命名空间、用户名和密码（例如，以管理员身份）登录，选择设置凭证。 选择用 户匿名连接在没有用户凭证的情况下登录，在该情况下不用填写其他字段。 选择存储的凭证以使用存储的凭 证中的登录信息。 要使用存储的凭证，您必须连接到包含此凭证的 IBM SPSS 协作和部署服务存储库。 连接 到该存储库之后，请单击浏览以查看可用凭证的列表。 名称空间 ID。 用于登录到服务器的安全验证提供程序。 验证提供程序可用于定义和维持用户、组及角色， 并控制验证过程。 用户名。 键入用于登录到服务器的用户名。 密码。 输入与指定用户名关联的密码。 另存为缺省值。 将这些设置保存为您的缺省值，以避免每次重新输入。 Cognos 连接 “指定位置”对话框使您可以选择导入数据的程序包，或导入报告的程序包或文件夹。 其中显示您可使用的公 共文件夹。 如果在主对话框选择数据，列表将显示包含数据包的文件夹。 如果在主对话框选择报告，列表 将显示包含列表报告的文件夹。 通过导航文件夹结构选择您想要的位置。 指定数据或报告的参数。 如果参数已定义，在导入数据或报告之前，您可以指定数据对象或报告的参数值。 报告的参数示例是报告起 始内容的起始与结束日期。 名称。 在 IBM Cognos Business Intelligence 数据库指定的参数名称。 类型。 参数说明。 值。 指定给参数的值。 要输入或编辑值，在表中双击该值的单元格。 此处的值未验证；任何无效值在运行 时将受检测。 自动删除表中的无效参数。 此选项在缺省情况下选定，并将删除在数据对象或报告中找到的任何无效参数。 18 IBM SPSS Statistics 29 Core System 用户指南 更改变量名称 对于 IBM Cognos Business Intelligence 数据包，其包字段名称会自动换为有效的变量名称。 您可以使用 “读取 Cognos 数据”对话框的“字段”选项卡以覆盖缺省名称。 名称必须是唯一的并必须符合变量命名规则。 请参阅主题第44 页的『变量名称』，了解更多信息。 读取 Cognos TM1 数据 如果您有权访问 IBM Cognos TM1 数据库，那么可以将 TM1 数据从指定视图导入到 IBM SPSS Statistics 中。 TM1 中的多维 OLAP 立方体数据在读取到 SPSS Statistics 中时将被序列化。 要点: 要启用 SPSS Statistics 与 TM1 之间的数据交换，必须将以下三个进程从 SPSS Statistics 复制到 TM1 服务器：ExportToSPSS.pro、ImportFromSPSS.pro 和 SPSSCreateNewMeasures.pro。 要将这些 进程添加到 TM1 服务器，必须将它们复制到 TM1 服务器的数据目录并重新启动 TM1 服务器。 这些文件在 SPSS Statistics 安装目录下的 common/scripts/TM1 目录中可用。 限制: • 从中导入的 TM1 视图必须包含测量维度中的一个或多个元素。 • 要从 TM1 导入的数据必须采用 UTF-8 格式。 将导入指定 TM1 视图中的所有数据。 因此，最好将视图限制为分析所需的数据。 最好在 TM1 中完成所有必 需的数据过滤，例如，使用 TM1 子集编辑器进行过滤。 要读取 TM1 数据，请执行以下操作： 1. 从菜单中选择： 文件 > 导入数据 > Cognos TM1 2. 连接到 TM1 Performance Management 系统。 3. 登录到 TM1 服务器。 4. 选择 TM1 立方体并选择要导入的视图。 （可选）您可以覆盖根据 TM1 维度名称和度量名称创建的 SPSS Statistics 变量的缺省名称。 PM 系统 这是 Performance Management 系统的 URL，该系统包含要连接到的 TM1 服务器。 Performance Management 系统定义为所有 TM1 服务器的单个 URL。 通过此 URL，您可以发现和访问环境中已安装 且正在运行的所有 TM1 服务器。 请输入此 URL 并单击连接。 TM1 Server 建立与 Performance Management 系统的连接后，请选择包含要导入的数据的服务器，然后单击登录。 如果您先前未连接到此服务器，那么系统将提示您登录。 用户名和密码 选择此选项可以使用特定的用户名和密码登录。 如果此服务器使用认证方式 5（IBM Cognos 安全 性），请从可用列表中选择标识安全认证提供程序的名称空间。 存储凭证 选择此选项可以使用存储的凭证中的登录信息。 要使用存储的凭证，您必须连接到包含此凭证的 IBM SPSS 协作和部署服务存储库。 连接到该存储库之后，请单击浏览以查看可用凭证的列表。 选择要导入的 TM1 立方体视图 用于列出 TM1 服务器中您可以从中导入数据的立方体的名称。 双击立方体将显示可以导入的视图的列 表。 选择所需视图并单击向右箭头可以将其移入要导入的视图字段中。 列维度 用于列出所选视图中的列维度的名称。 行维度 用于列出所选视图中的行维度的名称。 上下文维度 用于列出所选视图中的上下文维度的名称。 注: 第 3 章 数据文件 19 • 导入数据时，将为每个常规维度以及测量维度中的每个元素创建一个单独的 SPSS Statistics 变量。 • TM1 中的空单元格和值为 0 的单元格将转换为系统缺失值。 • 包含无法转换为数字值的字符串值的单元格将转换为系统缺失值。 更改变量名称 缺省情况下，有效 IBM SPSS Statistics 变量名称将根据维度名称和所选 IBM Cognos TM1 多维数据集视图 的度量维度中元素的名称自动生成。 您可以使用“从 TM1 导入”对话框的“字段”选项卡覆盖缺省名称。 名称 必须唯一，并且必须符合变量命名规则。 文件信息 数据文件包含的内容远不止是原始数据。 它还包含所有变量定义信息，包括： • 变量名称 • 变量格式 • 描述性的变量标签和值标签 这些信息存储在数据文件的字典部分。 数据编辑器提供一种查看变量定义信息的方法。 还可以显示活动数 据集或任何其他数据文件的完整字典信息。 显示数据文件信息 1. 从数据编辑器窗口的菜单中选择： 文件 > 显示数据文件信息 2. 对于当前打开的数据文件，请选择工作文件。 3. 对于其他数据文件，选择外部文件，然后选择数据文件。 查看器中会显示数据文件信息。 保存数据文件 除了以 IBM SPSS Statistics 格式保存数据文件之外，还可以用多种外部格式保存数据，这些格式包括： • Excel 和其他电子表格格式 • Tab 分隔和 CSV 文本文件 • SAS • Stata • 数据库表 保存已修改的数据文件 1. 使数据编辑器成为活动窗口（单击窗口的任何部位即可使其成为活动窗口）。 2. 从菜单中选择： 文件 > 保存 已修改的数据文件被保存，并覆盖文件的上一版本。 以代码页字符编码保存数据文件 v16.0 之前的 IBM SPSS 统计信息 版本无法读取 Unicode 数据文件。 在 Unicode 方式下，要以代码页字符 编码保存数据文件： 1. 使数据编辑器成为活动窗口（单击窗口的任何部位即可使其成为活动窗口）。 2. 从菜单中选择： 文件 > 另存为 20 IBM SPSS Statistics 29 Core System 用户指南 3. 从“保存数据”对话框的保存类型下拉列表中，选择 SPSS Statistics 本地编码。 4. 输入新数据文件的名称。 修改后的数据文件将以当前语言环境代码页字符编码进行保存。 此操作不会影响活动数据集。 活动数据集 的编码未更改。 以代码页字符编码保存文件类似于以外部格式（例如制表符分隔文本或 Excel）保存文件。 以外部格式保存数据文件 1. 使数据编辑器成为活动窗口（单击窗口的任何部位即可使其成为活动窗口）。 2. 从菜单中选择： 文件 > 另存为...
3. 从下拉列表中选择文件类型。 4. 输入新数据文件的文件名。 选项 根据文件类型，提供了附加选项。 编码 可用于 SAS 文件和文本数据格式：选项卡定界文本、逗号定界文本和固定 ASCII 文本。 将变量名称写入文件 可用于 Excel、选项卡定界、逗号定界、1-2-3 和 SYLK 格式。 针对 Excel 97 和更高版本，可以编写变 量名称或标签。 对于不具有定义的变量标签的变量，将使用变量名称。 工作表名称 针对 Excel 2007 和更高版本，可指定工作表名称。 还可向现有文件追加工作表。 将值标签保存到 .sas 文件 SAS 6 和更高版本。 有关将数据导出到数据库表的信息，请参阅第26 页的『导出到数据库』。 保存数据：数据文件类型 您可以用以下格式保存数据： SPSS Statistics (.sav). IBM SPSS Statistics 格式。 • 7.5 版之前的软件无法读取以 IBM SPSS Statistics 格式保存的数据文件。 V16.0 之前的 IBM SPSS Statistics 发行版无法读取以 Unicode 编码保存的数据文件 • 在 V10.x 或 V11.x 中使用变量名称长度超出 8 个字节的数据文件时，将使用变量名称的唯一 8 字节版本， 但是在发行版 12.0 或更高版本中将保留原变量名称。 在 10.0 之前的发行版中，在您保存数据文件时原来 的长变量名称会丢失。 • 在 13.0 版之前使用字符串变量长于 255 字节的数据文件时，会将这些字符串变量分解为多个 255 字节的 字符串变量。 SPSS Statistics 压缩 (.zsav)。 压缩的 IBM SPSS Statistics 格式。 • ZSAV 文件具有与 SAV 文件相同的特征，但它们占用较少的磁盘空间。 • 根据文件的大小和系统配置，ZSAV 文件需要更多或更少的时间来打开和保存。 需要额外的时间解压缩和 压缩 ZSAV 文件。 但是，因为 ZSAV 文件在磁盘上较小，可减少从磁盘读取和写入所需的时间。 由于文件 大小变大，节省的时间超过解压缩和压缩文件所需的额外时间。 • 只有 IBM SPSS Statistics 版本 21 或更高的版本可以打开 ZSAV 文件。 • 保存带有您本地代码页编码的数据文件的选项对 ZSAV 文件不可用。 这些文件总是以 UTF-8 编码保存。 SPSS Statistics 本地编码 (.sav)。 在 Unicode 方式下，此选项以当前语言环境代码页字符编码保存数据文 件。 在代码页方式下，此选项不可用。 SPSS 7.0 (.sav)。 7.0 版格式。 7.0 版和较早版本的 Windows 版可以读取以 7.0 版格式保存的数据文件， 但是不包括已定义的多响应集或 Data Entry for Windows 信息。 第 3 章 数据文件 21 SPSS/PC+ (.sys)。 SPSS/PC+ 格式。 如果数据文件包含的变量多于 500 个，将仅保存前 500 个。 对于具 有多个已定义用户缺失值的变量，将把其他的用户缺失值记录到第一个已定义用户缺失值中。 此格式只在 Windows 操作系统上可用。 可移植格式 (.por)。 可移植格式，其他版本的IBM SPSS Statistics 以及其他操作系统上的版本都可以读取 此格式。 变量名称限制为八字节，并自动转换成唯一的八字节名称（如果必要）。 在多数情况下，不再需 要以便携格式保存数据，因为 IBM SPSS Statistics 数据文件应该独立于平台/操作系统。 您无法在 Unicode 方式下以可移植文件来保存数据文件。 请参阅第163 页的『一般选项』主题以获取更多信息。 制表符分隔 (.dat)。 用制表符分隔值的文本文件。 （注：嵌入在字符串值中的制表符将作为制表符保留在 制表符分隔文件中。 不对嵌入值中的 Tab 字符和分隔值的 Tab 字符进行区分。） 您可以使用 Unicode 编码 或本地代码页编码保存文件。 逗号分隔 (.csv)。 用逗号或分号分隔值的文本文件。 如果当前 IBM SPSS Statistics 小数指示符为句点，那 么用逗号分隔各值。 如果当前小数指示符为逗号，那么用分号分隔各值。 您可以使用 Unicode 编码或本地 代码页编码保存文件。 固定 ASCII (.dat)。 固定格式的文本文件，对所有变量使用缺省的书写格式。 在变量字段之间没有 tab 或 空格。 您可以使用 Unicode 编码或本地代码页编码保存文件。 Excel 2007 (.xlsx)。 Microsoft Excel 2007 XLSX 格式工作表。 最大变量数为 16,000；删除超过 16,000 的任何其他变量。 如果数据集包含一百万个个案，在工作表中创建多页。 Excel 97 到 2003 (.xls)。 Microsoft Excel 97 工作表。 最大变量数为 256；将删除前 256 个变量后面的任 何其他变量。 如果数据集包含 65,356 个个案，在工作表中创建多页。 Excel 2.1 (.xls)。 Microsoft Excel 2.1 电子表格文件。 最大变量数为 256，最大行数为 16,384。 1-2-3 R3.0 (.wk3)。 Lotus 1-2-3 电子表格文件，版本 3.0。 可以保存的最大变量数为 256。 1-2-3 R2.0 (.wk1)。 Lotus 1-2-3 电子表格文件，发行版 2.0。 可以保存的最大变量数为 256。 1-2-3 R1.0 (.wks)。 Lotus 1-2-3 电子表格文件，版本 1A。 可以保存的最大变量数为 256。 SYLK (.slk)。 Microsoft Excel 和 Multiplan 电子表格文件的符号链接格式。 可以保存的最大变量数为 256。 dBASE IV (.dbf). dBASE IV 格式。 dBASE III (.dbf). dBASE III 格式。 dBASE II (.dbf). dBASE II 格式。 SAS v9+ Windows (.sas7bdat)。 SAS v9 Windows 版。 您可以使用 Unicode (UTF-8) 或本地代码页编码 保存。 SAS v9+ UNIX (.sas7bdat)。 SAS v9 UNIX 版。 您可以使用 Unicode (UTF-8) 或本地代码页编码保存。 SAS v7-8 Windows 短扩展名 (.sd7)。 SAS V7 - 8 for Windows 短文件名格式。 SAS v7-8 Windows 长扩展名 (.sas7bdat)。 SAS V7 - 8 for Windows 长文件名格式。 SAS v7-8 UNIX 版 (.sas7bdat)。 SAS v8 UNIX 版。 SAS v6 Windows 版 (.sd2)。 用于 Windows/OS2 的 SAS V6 文件格式。 SAS v6 UNIX 版 (.ssd01)。 用于 UNIX（Sun、HP、IBM）的 SAS V6 文件格式。 SAS v6 Alpha/OSF 版 (.ssd04)。 用于 Alpha/OSF (DEC UNIX) 的 SAS V6 文件格式。 SAS Transport (.xpt)。 SAS 传输格式文件。 Stata V13 Intercooled 版 (.dta)。 Stata V13 SE 版 (.dta)。 Stata V12 Intercooled 版 (.dta)。 Stata V12 SE 版 (.dta)。 Stata V11 Intercooled 版 (.dta)。 22 IBM SPSS Statistics 29 Core System 用户指南 Stata V11 SE 版 (.dta)。 Stata V10 Intercooled 版 (.dta)。 Stata V10 SE 版 (.dta)。 Stata V9 Intercooled 版 (.dta)。 Stata V9 SE 版 (.dta)。 Stata V8 Intercooled 版 (.dta)。 Stata V8 SE 版 (.dta)。 Stata V7 Intercooled 版 (.dta)。 Stata V7 SE 版 (.dta)。 Stata V6 (.dta)。 Stata V4–5 (.dta)。 注：SAS 数据文件名称的最大长度可以为 32 个字符。 不允许使用空格以及除下划线（“”）以外的非字母 数字字符，并且名称必须以字母或下划线开头，后面可以跟有数字。 以 Excel 格式保存数据文件 您可以用三种 Microsoft Excel 文件格式之一来保存数据。 Excel 2.1、Excel 97 和 Excel 2007。 • Excel 2.1 和 Excel 97 限于 256 个列；所以只能包括前 256 个变量。 • Excel 2007 限于 16,000 个列；所以只能包括前 16,000 个变量。 • Excel 2.1 限于 16,384 个行；所以只能包括前 16,384 个个案。 • Excel 97 和 Excel 2007 每页有行数的限制，但是工作表可以有多页，如果已经超过单页最大数，可以创 建多页。 选项 • 对于 Excel 的所有版本，可将变量名称作为 Excel 文件的第一行包含。 • 针对 Excel 97 和更高版本，可以编写变量名称或标签。 对于不具有定义的变量标签的变量，将使用变量名 称。 • 针对 Excel 2007 和更高版本，可指定工作表名称。 还可向现有文件追加工作表。 变量类型 下表显示 IBM SPSS Statistics 中的原始数据与 Excel 中的导出数据之间的变量类型匹配情况。 表 2: Excel 数据格式如何映射为 IBM SPSS Statistics 变量类型和格式 IBM SPSS Statistics 变量类型 Excel 数据格式 数值 0.00; #,##0.00; ...
逗号 0.00; #,##0.00; ...
美元符号 $#,##0_); ...
日期 d-mmm-yyyy 时间 hh:mm:ss String 常规 以 SAS 格式保存数据文件 当将您的数据保存为 SAS 文件时，将对数据的各个方面进行特殊处理。 这些情况包括： 第 3 章 数据文件 23 • IBM SPSS Statistics 变量名中允许的某些字符在 SAS 中无效，例如 @、# 和 $。导出数据时，这些非法字 符将替换为下划线。 • 包含多字节字符（例如，日语或中文字符）的 IBM SPSS Statistics 变量名称将转换为一般形式的变量名 称，即 Vnnn，其中 nnn 是整数值。 • 包含多于 40 个字符的 IBM SPSS Statistics 变量标签在导出至 SAS v6 文件时会被截断。 • 只要存在，IBM SPSS Statistics 变量标签，就将它们映射为 SAS 变量标签。 如果在 IBM SPSS Statistics 数据中不存在任何变量标签，则将变量名称映射为 SAS 变量标签。 • SAS 仅允许一个值为系统缺失值，而 IBM SPSS Statistics 除系统缺失值外还允许有许多用户缺失值。 因 此，IBM SPSS Statistics 中的所有用户缺失值均映射为 SAS 文件中的单个系统缺失值。 • SAS 6-8 数据文件以当前 IBM SPSS Statistics 本地编码保存，不考虑当前模式（Unicode 或代码页）。 在 Unicode 方式下，SAS 9 文件以 UTF-8 格式保存。 在代码页模式下，SAS 9 文件以当前本地编码保存。 • 最多可以保存 32,767 个变量到 SAS 6-8。 • SAS 数据文件名称的最大长度可以为 32 个字符。 不允许使用空格以及除下划线（“”）以外的非字母数字 字符，并且名称必须以字母或下划线开头，后面可以跟有数字。 保存值标签 您可以选择将与数据文件相关联的值和值标签保存到 SAS 语法文件中。 该语法文件包含可在 SAS 中运行以 创建 SAS 格式目录文件的 proc format 和 proc datasets 命令。 SAS 传输格式文件不支持此功能。 变量类型 下表显示 IBM SPSS Statistics 中的原始数据与 SAS 中的导出数据之间的变量类型匹配情况。 表 3: SAS 变量类型和格式如何映射为 IBM SPSS Statistics 类型和格式 IBM SPSS Statistics 变量类型 SAS 变量类型 SAS 数据格式 数值 数值 12 逗号 数值 12 点 数值 12 科学记数法 数值 12 日期 数值 （日期）例如：MMDDYY10 ...
日期（时间） 数值 Time18 美元 数值 12 定制货币 数值 12 String 字符 $8 以 Stata 格式保存数据文件 • 可以采用 Stata 5-13 格式，也可以同时采用 Intercooled 和 SE 格式（V7 或更高版本）写入数据。 • 以 Stata 5 格式保存的数据文件可以由 Stata 4 读取。 • 变量标签的前 80 个字节保存为 Stata 变量标签。 • 对于 Stata R4 - R8，数字变量的值标签的前 80 个字节保存为 Stata 值标签。 对于 Stata R9 或更高版本， 保存数字变量的完整值标签。 对于字符串变量、非整数值和绝对值大于 2,147,483,647 的数值，将去除值 标签。 • 对于 V7 及更高版本，区分大小写形式的变量名称的前 32 个字节保存为 Stata 变量名称。 对于早期版本， 变量名称的前 8 个字节保存为 Stata 变量名称。 字母、数字和下划线以外的任何字符都转换为下划线。 24 IBM SPSS Statistics 29 Core System 用户指南 • 包含多字节字符（例如，日语或中文字符）的 IBM SPSS Statistics 变量名称将转换为一般单字节变量名 称。 • 对于 V5 - V6 和 Intercooled V7 及更高版本，保存字符串值的前 80 个字节。 对于 Stata SE 7-12，保存字 符串值的前 244 个字节。 对于 Stata SE 13 或更高版本，无论长度是多少，都保存完整的字符串值。 • 对于 V5 - V6 和 Intercooled V7 及更高版本，仅保存前 2,047 个变量。 对于 Stata SE 7 或更高版本，仅保 存前 32,767 个变量。 表 4: Stata 变量类型和格式如何映射到 IBM SPSS Statistics 类型和格式 IBM SPSS Statistics 变量类型 Stata 变量类型 Stata 数据格式 数值 数值 g 逗号 数值 g 点 数值 g 科学记数法 数值 g Date、Datetime 数值 D_m_Y Time、DTime 数值 g（秒数） Wkday 数值 g (1-7) 月 数值 g (1-12) 美元 数值 g 定制货币 数值 g String String s Date、Adate、Edate、SDate、Jdate、Qyr、Moyr、Wkyr 保存变量子集 使用“将数据保存为变量”对话框可以选择要在新数据文件中保存的变量。 缺省情况下将保存所有变量。 取 消选择您不想保存的变量，或单击全部丢弃然后选择想要保存的变量。 仅可视。 仅选择当前使用的变量集中的变量。 有关更多信息，请参阅第162 页的『使用变量集合显示和隐 藏变量』主题。 保存变量子集 1. 使数据编辑器成为活动窗口（单击窗口的任何部位即可使其成为活动窗口）。 2. 从菜单中选择： 文件 > 另存为...
3. 单击变量。 4. 选择想要保存的变量。 加密数据文件 针对 IBM SPSS Statistics 数据文件，您可以通过使用密码加密文件以保护存储在数据文件中的机密信息。 一旦加密，文件只能通过密码打开。 1. 使数据编辑器成为活动窗口（单击窗口的任何部位即可使其成为活动窗口）。 2. 从菜单中选择： 文件 > 另存为...
3. 在“将数据保存为”对话框中，选择用密码加密文件。 4. 单击保存。 第 3 章 数据文件 25 5. 在“对文件进行加密”对话框中，请提供密码，并在“确认密码”文本框中重新输入该密码。 密码限制在 10 个字符并区分大小写。 警告：密码丢失后将无法恢复。 如果密码丢失，那么将无法打开文件。 创建强密码 • 至少使用八个字符。 • 在密码中使用数字、符号甚至标点符号。 • 避免使用数字序列或字符序列（例如 "123" 和 "abc"）并避免重复，例如 "111aaa"。 • 不要创建使用个人信息（例如生日或昵称）的密码。 • 定期更改密码。 注：不支持将已加密的文件存储到 IBM SPSS 协作和部署服务存储库 中。 修改已加密的文件 • 如果打开加密文件，对其进行修改并选择 “文件”>“保存”，修改后的文件将以相同的密码保存。 • 您可以通过打开文件、重复加密步骤并在“加密文件”对话框中指定不同的密码，在加密的文件上更改密 码。 • 通过打开文件，选择“文件 > 另存为”并在“将数据另存为”对话框中取消选择使用密码加密文件，可以保存 加密文件的未加密版本。 注意：在 V21 之前的 IBM SPSS Statistics 版本中，无法打开经过加密的数据文件和输出文档。 在 V22 之前 的版本中，无法打开经过加密的语法文件。 导出到数据库 可使用导出到数据库向导完成以下任务： • 替换现有数据库表字段（列）中的值或为表添加新字段。 • 将新记录（行）追加到数据库表。 • 完全替换数据库表或创建新表。 要将数据导出到数据库，请执行下列操作： 1. 在包含要导出数据的数据集的“数据编辑器”窗口中，从菜单中选择： 文件 > 导出 > 数据库 2. 选择数据库源。 3. 按照导出向导中的说明操作以导出数据。 从 IBM SPSS Statistics 变量创建数据库字段 创建新字段（向现有数据库表添加字段、创建新表或替换表）时，可以指定字段名、数据类型和宽度（适用 的情况下）。 字段名称。 缺省字段名称与 IBM SPSS Statistics 变量名称相同。 可以将字段名称更改为数据库格式允许的 任何名称。 例如，很多数据库允许字段名称中包含 变量名称中不允许的字符，包括空格。 因此，类似于 CallWaiting 的 变量名称可以更改为字段名称 Call Waiting。 类型。 导出向导基于标准 ODBC 数据类型或选定的数据库格式允许的与定义的 IBM SPSS Statistics 数据格 式最匹配的数据类型进行首次数据类型指定，但是数据库可对在 IBM SPSS Statistics 中没有直接对应类型的 类型进行区分。 例如，IBM SPSS Statistics 中的很多数值都以双精度浮点值保存，而数据库数值类型包含浮 点（双精度）、整数、实数等等。 此外，很多数据库没有与 IBM SPSS Statistics 对应的时间格式类型。 可 以将数据类型更改为下拉列表中可用的任何类型。 作为一个通常的规则，变量的基本数据类型（字符串或数值）应与数据库字段的基本数据类型相匹配。 如果 出现了数据库无法解决的数据类型不匹配，那么会出现错误结果，并且不会将任何数据导出到数据库中。 例 如，如果将 字符串变量导出到数值数据类型的数据库字段，那么当字符串变量的任何值包含非数字字符时， 结果将出错。 26 IBM SPSS Statistics 29 Core System 用户指南 宽度。 可以对 string (char, varchar) 字段类型的已定义宽度进行更改。 数字字段宽度由数据类型决定。 缺省情况下，IBM SPSS Statistics 变量格式根据下列总体原则映射到数据库字段类型。 实际数据库字段类型 可能取决于数据库。 表 5: 数据库的格式转换 IBM SPSS Statistics 变量格式 数据库字段类型 数值 浮点数或双精度数 逗号 浮点数或双精度数 点 浮点数或双精度数 科学记数法 浮点数或双精度数 数据 日期、日期时间或时间戳 日期时间 日期时间或时间戳 Time、DTime 浮点数或双精度数（秒数） Wkday 整数 (1 - 7) 月 整数 (1 - 12) 美元 浮点数或双精度数 定制货币 浮点数或双精度数 String Char 或 Varchar 用户缺失值 将来自变量的数据导出到数据库字段时，有两个选项可用于处理用户缺失值： • 导出为有效值。 用户缺失值当作常规的、有效的非缺失值处理。 • 将数值型的用户缺失值作为 Null 导出，并将字符串类型的用户缺失值作为空格导出。 数值型的用户缺失 值被视为与系统缺失值相同。 字符串类型的用户缺失值被转换为空格（字符串不能是系统缺失的）。 选择数据源 在导出到数据库向导的第一个面板中，选择要向其导出数据的数据源。 可以将数据导出到具有相应 ODBC 驱动程序的任何数据库源。 （注：不支持将数据导出到 OLE DB 数据 源。） 如果没有配置任何 ODBC 数据源，或者要添加新的数据源，请单击添加 ODBC 数据源。 • 在 Linux 操作系统中，该按钮不可用。 在 odbc.ini 中指定 ODBC 数据源，并且 ODBCINI 环境变量必须设 定为该文件的位置。 有关更多信息，请参阅数据库驱动程序文档。 • 用分布式分析模式时（IBM SPSS Statistics Server 提供），该按钮不可用。 要用分布式分析模式添加数据 源，请咨询系统管理员。 ODBC 数据源包含两部分重要的信息：将用于访问数据的驱动程序以及要访问的数据库的位置。 要指定数据 源，必须装有适当的驱动程序。 针对不同数据库格式的驱动程序包括在安装介质上。 有些数据源可能要求登录 ID 和密码才能进行到下一步。 选择如何导出数据 选择了数据源后，需要指明导出数据所采用的方式。 以下选择可供将数据导出到数据库： • 替换现有字段中的值。 将现有表中选定字段的值用活动数据集中选定变量的值替换。 请参阅主题第28 页的『替换现有字段中的值』，了解更多信息。 第 3 章 数据文件 27 • 向现有表中添加新字段。 在现有表中创建新字段，这些字段将包含活动数据集中选定变量的值。 请参阅主 题第29 页的『添加新字段』，了解更多信息。 对于 Excel 文件，此选项不可用。 • 向现有表中追加新记录。 向现有表添加新记录（行），这些记录将包含活动数据集中的个案的值。 请参阅 主题第29 页的『追加新记录（个案）』，了解更多信息。 • 丢弃现有表并创建同名的新表。 删除指定的表并创建同名的新表，新表中将包含活动数据集中的选定变 量。 原始表中包括字段属性定义（例如主键和数据类型）在内的所有信息都将丢失。 请参阅主题第29 页的『创建新表或替换表』，了解更多信息。 • 创建新表。 在数据库中创建新表，其中包含来自活动数据集中的选定变量的数据。 名称可以是数据源允许 作为表名的任何值。 此名称不能与数据库中现有表或视图名称重复。 请参阅主题第29 页的『创建新表 或替换表』，了解更多信息。 选择表 修改或替换数据库中的表时，需要选择要修改或替换的表。 导出到数据库向导中的此面板显示了一个列表， 其中列出选定数据库中的表和视图。 缺省情况下，此列表仅显示标准数据库表。 您可以控制列表中显示的项的类型： • 表。 标准数据库表。 • 视图。 视图是由查询定义的虚拟的或动态的“表”。 视图中可以包含根据其他字段值计算得出的多个表和/ 或字段的连接。 虽然可以对视图追加记录，或替换其现有字段的值，但是您能够修改的字段是受限制的， 这取决于视图的结构。 例如，不能修改派生的字段、为视图添加字段或替换视图。 • 同义词。 同义词是表或视图的别名，通常在查询中定义。 • 系统表。 系统表定义数据库的属性。 在某些情况下，标准数据库表可能会被归类为系统表，并且仅在选择 了该选项后才会显示。 通常只有数据库管理员才具有访问真正的系统表的权限。 选择要导出的个案 导出到数据库向导中对个案的选择仅限于所有个案以及使用先前定义的过滤条件选择的个案。 如果没有有效 的个案过滤，此面板将不显示，并将导出活动数据集中的所有个案。 有关定义用于个案选择的过滤条件的信息，请参阅第92 页的『选择个案(L)』。 将个案匹配到记录 在向现有表添加字段（列）或替换现有字段中的值时，需确保活动数据集中的每个个案（行）与数据库中相 应的记录能够正确匹配。 • 在数据库中，唯一标识每个记录的字段或字段组通常被指定为主键。 • 需要确定与主键字段或其他能唯一标识每个记录的字段组对应的变量。 • 字段不一定必须是数据库中的主键，但是字段值或字段值的组合对于每个个案必须是唯一的。 要使变量与数据库中唯一标识每个记录的字段匹配，请执行下列操作： 1. 将变量拖放至相应的数据库字段上。 或 2. 从变量列表中选择变量，再选择数据库表中相应的字段，然后单击连接。 要删除连接线，请执行下列操作： 3. 选择连接线并按 Delete 键。 注：变量名称和数据库字段名不一定相同（因为数据库字段名可能包含 IBM SPSS Statistics 变量名称所不允 许的字符），但是如果活动数据集是从所修改的数据库表创建的，那么变量名称或变量标签两者之一通常将 至少与数据库字段名相似。 替换现有字段中的值 要替换数据库中现有字段的值，请执行下列操作： 28 IBM SPSS Statistics 29 Core System 用户指南 1. 在导出到数据库向导的选择如何导出数据面板中，选择替换现有字段中的值。 2. 在选择表或视图面板中，选择数据库表。 主题。 3. 在将个案匹配到记录面板中，将唯一标识每个个案的变量与相应的数据库字段名称相匹配。 主题。 4. 对于要替换其值的每个字段，将包含新值的变量拖放到相应数据库字段名旁的值的源列。 • 作为一个通常的规则，变量的基本数据类型（字符串或数值）应与数据库字段的基本数据类型相匹配。 如 果出现了数据库无法解决的数据类型不匹配，那么会出现错误结果，并且不会将任何数据导出到数据库 中。 例如，如果将字符串变量导出到数值数据类型（例如双精度数、实数或整数）的数据库字段，则当字 符串变量的任何值包含非数字字符时，结果将出错。 变量旁的图标中的字母 a 指示该变量为字符串变量。 • 不能修改字段名、类型或宽度。 原始数据库字段属性将被保留，只有值被替换。 添加新字段 要向现有数据库表添加新字段，请执行下列操作： 1. 在导出到数据库向导的选择如何导出数据面板中，选择向现有表中添加新字段。 2. 在选择表或视图面板中，选择数据库表。 主题。 3. 在将个案匹配到记录面板中，将唯一标识每个个案的变量与相应的数据库字段名称相匹配。 主题。 4. 将要作为新字段添加的变量拖放到值的源列。 有关字段名称和数据类型的信息，请参阅关于通过 IBM SPSS Statistics 变量创建数据库字段的部分（位于 第26 页的『导出到数据库』）。 值标签。 如果变量定义了变量标签，请导出值标签文本而不是值。 针对未定义值标签的值，数据值将导出 为文本字符串。 此选项不适用于日期格式变量或未定义任何值标签的变量。 显示现有字段。 选择此选项以显示现有字段的列表。 不能使用导出到数据库向导中的此面板替换现有字 段，但是了解表中已有的字段会有帮助。 如果要替换现有字段中的值，请参阅第28 页的『替换现有字段中 的值』。 追加新记录（个案） 要将新记录（个案）追加到数据库表，请执行下列操作： 1. 在导出到数据库向导的选择如何导出数据面板中，选择向现有表中追加新记录。 2. 在选择表或视图面板中，选择数据库表。 主题。 3. 将活动数据集中的变量与表字段匹配，方法是将变量拖放到值的源列。 导出到数据库向导将根据存储在活动数据集（如可用）中的原始数据库表和/或与字段名同名的变量名称的信 息，自动选择与现有字段匹配的所有变量。 此初始自动匹配功能的目的仅是提供一个指导，并不妨碍您更改 变量与数据库字段匹配的方式。 向现有表中添加新字段时，将应用以下基本规则/限制： • 活动数据集中的所有个案（或所有选定个案）都将添加到表中。 如果这些个案中任何一个与数据库中的现 有记录重复，那么当遇到重复键值时，会发生错误。 有关仅导出选定个案的信息，请参阅第28 页的『选 择要导出的个案』。 • 可以使用会话中创建的新变量的值作为现有字段的值，但是不能添加新字段或更改现有字段的名称。 要向 数据库表添加新字段，请参阅第29 页的『添加新字段』。 • 对于任何排除的数据库字段或未匹配到变量的字段，数据库表中的新增记录将没有相应的值。 （如果值的 源单元格为空，那么没有任何变量匹配到字段。） 创建新表或替换表 要创建新的数据库表，或替换现有的数据库表，请执行下列步骤： 1. 在导出向导的选择如何导出数据面板中，选择丢弃现有表并创建同名的新表或选择创建新表并为新表输 入一个名称。 如果表名称包含除字母、数字或下划线以外的任何其他字符，名称必须用双引号括起。 2. 如果要替换现有的表，则在选择表或视图面板中选择数据库表。 主题。 第 3 章 数据文件 29 3. 将变量拖放到要保存的变量列。 4. （可选）您可以指定定义主键的变量/字段，更改字段名以及更改数据类型。 主键。 要将变量指定为数据库表中的主键，请选择由键图标标识的列中的框。 • 主键的所有值必须是唯一的，否则将发生错误。 • 如果选择了单个变量作为主键，那么每个记录（个案）对于该变量必须具有唯一值。 • 如果选择了多个变量作为主键，这定义了组合主键，则选定变量的值的组合对每个个案来说必须是唯一 的。 有关字段名称和数据类型的信息，请参阅关于通过 IBM SPSS Statistics 变量创建数据库字段的部分（位于 第26 页的『导出到数据库』）。 值标签。 如果变量定义了变量标签，请导出值标签文本而不是值。 对于没有已定义的值标签的值，数据值 将作为文本字符串导出。 此选项不适用于日期格式变量或未定义任何值标签的变量。 完成数据库导出向导 “导出到数据库”向导的最后一步提供了导出指定项的摘要。 目录 • 数据集。 用于导出数据的数据集的 IBM SPSS Statistics 会话名称。 如果有多个打开的数据源，此信息将 非常有用。 只有在显式指定了数据集名称时，使用命令语法打开的数据源才有数据集名称。 • 表格。 要修改或创建的表的名称。 • 要导出的个案。 导出所有个案，或者导出通过先前定义的过滤条件选择的个案。 • 操作。 指示对数据库进行修改的方式（例如，创建新表或者向现有表添加字段或记录）。 • 用户缺失值。 用户缺失值可作为有效值导出；或者，对于数字变量与系统缺失值同等对待，对于字符串变 量转换为空格。 此设置在选择要导出的变量的面板中控制。 批量装入 批量装入。 将数据批量提交到数据库，而不是一次提交一条记录。 此操作可以使操作速度更快，对于大型 数据文件尤其如此。 • 批量大小。 指定每个批处理操作提交的记录数。 • 批量落实。 将指定批量大小的记录落实到数据库。 • ODBC 绑定。 使用 ODBC 绑定方法落实指定批量大小的记录。 只有在数据库支持 ODBC 绑定时，此选项 才可用。 在 macOS 上，此选项不可用。 – 逐行绑定。 通常，与逐条插入数据的参数化插入相比，逐行绑定的速度更快。 – 逐列绑定。 通过将每个数据库列与 n 个值所构成的数组绑定，逐列绑定可提高性能。 您希望做什么？ • 导出数据。 将数据导出到数据库中。 • 粘贴语法。 将用于导出数据的命令语法粘贴到语法窗口中。 您可以修改并保存所粘贴的命令语法。 导出到 Cognos TM1 如果您有权访问 IBM Cognos TM1 数据库，那么可以将数据从 IBM SPSS Statistics 导出到 TM1。 如果您要 从 TM1 导入数据、在 SPSS Statistics 中对此数据进行转换或评分，然后再将结果导出回 TM1，那么此功能 特别有用。 要点: 要启用 SPSS Statistics 与 TM1 之间的数据交换，必须将以下三个进程从 SPSS Statistics 复制到 TM1 服务器：ExportToSPSS.pro、ImportFromSPSS.pro 和 SPSSCreateNewMeasures.pro。 要将这些 进程添加到 TM1 服务器，必须将它们复制到 TM1 服务器的数据目录并重新启动 TM1 服务器。 这些文件在 SPSS Statistics 安装目录下的 common/scripts/TM1 目录中可用。 30 IBM SPSS Statistics 29 Core System 用户指南 要将数据导出到 TM1，请执行以下操作： 1. 从菜单中选择： 文件 > 导出 > Cognos TM1 2. 连接到 TM1 Performance Management 系统。 3. 登录到 TM1 服务器。 4. 选择要将数据导出到的 TM1 立方体。 5. 指定从活动数据集中的字段到该 TM1 立方体中的维度和度量的映射。 PM 系统 这是 Performance Management 系统的 URL，该系统包含要连接到的 TM1 服务器。 对于所有 TM1 服务 器，Performance Management 系统定义为单一 URL。 通过此 URL，您可以发现和访问环境中已安装且 正在运行的所有 TM1 服务器。 请输入此 URL 并单击连接。 TM1 服务器 建立与 Performance Management 系统的连接后，请选择包含要将数据导出到的立方体的服务器，然后 单击登录。 如果您先前未连接到此服务器，系统将提示您输入用户名和密码。 如果此服务器使用认证方 式 5（IBM Cognos 安全性），请从可用列表中选择标识安全认证提供程序的名称空间。 选择要导出的 TM1 立方体 用于列出 TM1 服务器中您可以将数据导出到的立方体的名称。 选择所需立方体并单击向右箭头可以将其 移入导出到立方体字段中。 注: • 对于映射到 TM1 立方体的度量维度中的元素的字段，导出时将忽略其系统缺失值和用户缺失值。 TM1 立 方体中的关联单元格保持不变。 • 值为 0 且映射到度量维度中的元素的字段将导出为有效值。 将字段映射到 TM1 维度 使用“导出到 TM1”对话框中的“映射”选项卡可以将 SPSS Statistics 字段映射到关联的 IBM Cognos TM1 维 度和度量。 您可以映射到度量维度中的现有元素，也可以在 TM1 立方体的度量维度中创建新元素。 • 对于指定 TM1 立方体中的每个常规维度，必须将活动数据集中的某个字段映射到该维度，或者必须为该维 度指定一个分区。 分区指定维度的单个叶元素，以使所有导出的案例都与指定的叶元素关联。 • 对于映射到常规维度的字段，将不会导出字段值与指定维度中的叶元素不匹配的案例。 在这种情况下，您 只能导出到叶元素。 • 只能将活动数据集中的字符串字段映射到常规维度。 只能将活动数据集中的数字字段映射到立方体的度量 维度中的元素。 • 导出到测量维度中的现有元素的值将覆盖 TM1 多维数据集中的关联单元格。 要将 SPSS Statistics 字段映射到常规 TM1 维度，或者映射到度量维度中的现有元素，请执行以下操作： 1. 从“字段”列表中选择 SPSS Statistics 字段。 2. 从“TM1 维度”列表中选择关联的 TM1 维度或度量。 3. 单击映射。 要将 SPSS Statistics 字段映射到度量维度中的新元素，请执行以下操作： 1. 从“字段”列表中选择 SPSS Statistics 字段。 2. 从“TM1 维度”列表中选择度量维度的项。 3. 单击新建，在“TM1 度量名称”对话框中输入度量元素的名称，然后单击确定。 要为常规维度指定分区，请执行以下操作： 1. 从“TM1 维度”列表中选择维度。 2. 单击分区。 第 3 章 数据文件 31 3. 在“选择叶成员”对话框中，选择指定分区的元素，然后单击确定。 您可以通过在“查找”文本框中输入搜 索字符串并单击查找下一个来搜索特定元素。 如果某个元素的任何部分与搜索字符串匹配，那么表明找 到匹配项。 • 搜索字符串中包含的空格将包括在搜索中。 • 搜索字符串不区分大小写。 • 系统对星号 () 的处理与其他任何字符相同，不作通配符搜索。 您可以通过从“TM1 维度”列表中选择映射的项并单击取消映射来除去映射定义。 可以通过从“TM1 维度”列 表中选择度量并单击删除来删除指定的新度量。 比较数据集 “比较数据集”会将活动数据集与当前会话中另一个数据集比较，或与 IBM SPSS Statistics 格式的外部文件比 较。 要比较数据集 1. 打开一个数据文件并确保它是活动数据集。 （您可以通过单击数据集的“数据编辑器”窗口使该数据集成 为活动数据集）。 2. 从菜单中选择： 数据 > 比较数据集 3. 选择打开数据集，或您想与活动数据集比较的 IBM SPSS Statistics 数据文件。 4. 选择一个或多个您想比较的字段（变量）。 （可选）您可以执行以下操作： • 根据一个或多个个案 ID 值匹配个案（记录）。 • 比较数据字典属性（字段和值标签、用户缺失值、测量级别等）。 • 在识别不匹配个案的活动数据集中，创建标记字段。 • 创建只包含匹配个案或者不匹配个案的新数据集。 比较数据集：“比较”选项卡 “匹配”字段列表显示在两个数据集中具有相同名称和相同基本类型（字符串或数字）的字段列表。 1. 选择一个或多个要比较的字段（变量）。 两个数据集的比较只基于已选的字段。 2. 要查看在两个数据集中没有匹配的名称或者没有相同基本类型的字段列表，请单击不匹配的字段。 不匹 配的字段将从两个数据集的比较中排除。 3. 或者，选择一个或多个能识别每个个案的个案（记录）ID 字段。 • 如果您指定多个个案 ID 字段，每个唯一组合值可识别一个个案。 • 两个数据文件必须以个案 ID 字段的升序方式排序。 如果数据集尚未排序，选择（选中）排序个案以个案 ID 顺序对两个数据集进行排序。 • 如果您不包含任何个案 ID 字段，个案将按文件顺序排序。 也就是说，将活动数据集中的第一个个案（记 录）与其他数据集中的第一个个案比较，依此类推。 比较数据集：不匹配的字段 “不匹配字段”对话框显示在两个数据集中视为未匹配的字段（变量）列表。 不匹配字段是指在其中一个数据 集缺失，或在两个文件中不是相同基本类型（字符串或数字）的字段。 不匹配的字段将从两个数据集的比较 中排除。 比较数据集：“属性”选项卡 缺省情况下，只将数据值进行比较，但字段属性（数据字典属性），例如，值标签、用户缺失值及测量级别 则不进行比较。 要比较字段属性： 32 IBM SPSS Statistics 29 Core System 用户指南 1. 在“比较数据集”对话框中，请单击属性选项卡。 2. 单击“比较数据字典”。 3. 选择您要想比较的属性。 • 宽度。 对于数值字段，为显示的最大字符数量（数字加上格式字符，例如，货币符号、分组符号和小数指 示符）。 对于字符串字段，为允许的最大字节数。 • 标签 。 描述性字段标签。 • 值标签。 描述性值标签。 • 缺失。 已定义的用户缺失值。 • 列。 数据编辑器的数据视图中的列宽。 • 对齐。 在数据编辑器的数据视图中的对齐。 • 度量。 测量级别。 • 角色。 字段角色。 • 属性。 用户定义的自定义字段属性。 比较数据集：“输出”选项卡 缺省情况下，比较数据集会在活动数据集中创建新字段，该字段可识别不匹配项并生成一个提供前 100 个不 匹配项的详细信息表。 您可以使用“输出”选项卡以更改输出选项。 在新字段标记不匹配项。 在活动数据集中创建识别不匹配项的新字段。 • 如果新字段值不同，其值为 1，如果所有值皆相同，则为 0。 如果活动数据集中的个案（记录）不在其他 数据集中，其值为 -1。 • 新字段的缺省名称为 CasesCompare。 您可以指定不同字段名称。 名称必须符合字段（变量）命名规则。 有关更多信息，请参阅第44 页的『变量名称』主题。 将匹配的个案复制到新数据集。 创建新数据集，其中只包含活动数据集中与其他数据集的值匹配的个案（记 录）。 数据集名称必须符合字段（变量）命名规则。 如果数据集已存在，将被覆盖。 将不匹配的个案复制到新数据集。 创建新数据集，其中只包含活动数据集中与其他数据集的值不同的个案。 数据集名称必须符合字段（变量）命名规则。 如果数据集已存在，将被覆盖。 限制逐案表格。 对于存在于活动数据集中，也存在于其他数据集，并在两个数据集中也有相同基本类型（数 字或字符串）的个案（记录），逐案表格提供了每个个案不匹配值的详细信息。 缺省情况下，该表限制为前 100 个不匹配项。 您可以指定不同值或取消选择（取消选中）此项以包括所有不匹配项。 保护原始数据 为了防止原始数据被意外修改或删除，您可以将该文件标记为只读。 1. 从“数据编辑器”菜单中选择： 文件 > 将文件标记为只读 如果随后对数据进行修改，然后尝试保存数据文件，则只能用其他文件名保存数据；这样原始数据就不会受 影响。 通过从“文件”菜单中选择将文件标记为读写您可以将文件权限改回为读写。 虚拟活动文件 虚拟活动文件使您能够处理大型数据文件，而无需足够大（或更大）的临时磁盘空间。 对于大多数分析和绘 图过程，会在每次您运行不同的过程时重新读取初始数据源。 修改数据的过程需要确定的临时磁盘空间量来 记录更改，而且一些操作始终需要足够的磁盘空间容纳数据文件的至少一个完整副本。 不需要任何临时磁盘空间的操作包括： • 读取 IBM SPSS Statistics 数据文件 第 3 章 数据文件 33 • 合并两个或更多 IBM SPSS Statistics 数据文件 • 使用数据库向导读取数据库表 • 将 IBM SPSS Statistics 数据文件与数据库表合并 • 运行读取数据的过程（例如“频率”、“交叉表”和“探索”） 在临时磁盘空间中创建一列或多列数据的操作包括： • 计算新变量 • 对现有变量重新编码 • 运行创建或修改变量的过程（例如在“线性回归”中保存预测值） 在临时磁盘空间中创建数据文件的整个副本的操作包括： • 读取 Excel 文件 • 运行对数据排序的过程（例如“对个案进行排序”和“拆分文件”） • 使用 GET TRANSLATE 或 DATA LIST 命令读取数据 • 使用“高速缓存数据”工具或 CACHE 命令 • 从 IBM SPSS Statistics 启动其他读取数据文件的应用程序（例如 AnswerTree 和 DecisionTime） 注：GET DATA 命令提供类似于 DATA LIST 的功能，但不在临时磁盘空间中创建数据文件的完整副本。 命 令语法中的 SPLIT FILE 命令不对数据文件进行排序，因此不创建数据文件的副本。 但是此命令需要已排 序的数据才能正确运行，此过程的对话框接口将自动对数据文件进行排序，生成数据文件的完整副本。 （命 令语法在学生版中不可用。） 缺省情况下创建数据文件的完整副本的操作： • 使用数据库向导读取数据库 • 使用文本向导读取文本文件 文本向导提供了可选的设置以自动高速缓存数据。 缺省情况下已选择此选项。 您可以通过取消选择在本地 高速缓存数据关闭此选项。 对于数据库向导，您可以粘贴生成的命令语法并删除 CACHE 命令。 创建数据高速缓存 尽管虚拟活动文件可以显著地减少所需的临时磁盘空间量，但是缺少“活动”文件的临时副本意味着必须为每 个过程均重复读取初始数据源。 对于从外部源读取的大型数据文件，创建数据的临时副本可以改善性能。 例如，对于从数据库源读取的数据表，必须为需要读取数据的任何命令或过程重复执行从数据库读取信息的 SQL 查询。 由于实际上所有统计分析过程和图表绘制过程均需要读取数据，因此会为您运行的每个过程重复 执行 SQL 查询，如果您运行大量过程，这会导致处理时间的显著增加。 如果您在执行分析的计算机（本地计算机或远程计算机）上有足够的磁盘空间，那么可以通过创建活动文件 的数据高速缓存来消除多次 SQL 查询，并减少处理时间。 此数据高速缓存是完整数据的临时副本。 注：缺省情况下，数据库向导自动创建数据高速缓存，但是如果您在命令语法中使用 GET DATA 命令读取数 据库，那么不会自动创建数据高速缓存。 （命令语法在学生版中不可用。） 创建数据高速缓存 1. 从菜单中选择： 文件 > 高速缓存数据...
2. 单击确定或立即高速缓存。 单击确定会在程序下次读取数据（例如，下次您运行统计程序）时创建数据高速缓存，这通常是您想要的， 因为它不要求额外的数据传递。 立即高速缓存会立即创建数据高速缓存，这在大多数情况下是不必要的。 立即高速缓存主要用于以下两种情况： • 数据源被“锁定”，在您结束会话、打开不同的数据源或高速缓存数据之前，任何人都无法更新该数据源。 • 对于大型数据源，如果您高速缓存了数据，那么在数据编辑器的“数据视图”选项卡的内容中进行滚动会快 得多。 34 IBM SPSS Statistics 29 Core System 用户指南 自动高速缓存数据 可以使用 SET 命令，在活动数据文件中发生指定数量的更改之后自动创建数据高速缓存。 缺省情况下，在 活动数据文件发生 20 个更改之后，会自动高速缓存该活动数据文件。 1. 从菜单中选择： 文件 > 新建 > 语法 2. 在语法窗口中键入 SET CACHE n（其中 n 表示数据文件高速缓冲前活动数据文件中更改的数量）。 3. 从语法窗口中的菜单选择： 运行 > 全部 注：高速缓存设置并不是在各个会话之间持续存在。 每次启动新会话时，该值均重置为缺省值 20。 第 3 章 数据文件 35 36 IBM SPSS Statistics 29 Core System 用户指南 第 4 章 分布式分析模式 分布式分析模式允许您使用本地（或桌面）计算机以外的计算机以进行内存密集型工作。 由于用于分布式分 析的远程服务器通常比本地计算机性能更强，速度更快，因此使用分布式分析模式可以显著地缩短计算机处 理时间。 如果工作涉及到以下内容，那么通过远程服务器进行分布式分析可能会有用： • 大型数据文件，尤其是从数据库源中读取的数据。 • 内存密集型任务。 任何在本地分析模式下耗时很长的任务均是适合于进行分布式分析的对象。 分布式分析仅影响与数据相关的任务，例如读取数据、转换数据、计算新变量和计算统计。 分布式分析对与 编辑输出相关的任务没有影响，例如操作透视表或修改图表。 注：只有在既有该软件的本地版本，又能访问安装在远程服务器上的该软件的许可服务器版本的情况下，才 能使用分布式分析。 服务器登录 “服务器登录”对话框允许您选择处理命令和运行过程的计算机。 您可以选择本地计算机，也可以选择远程服 务器。 您可以在列表中添加、修改或删除远程服务器。 远程服务器通常需要用户标识和密码，可能还需要域名。 如果许可您使用 Statistics Adapter 并且站点正在运行 IBM SPSS 协作和部署服务，那么您可能可以使用单点 登录连接到远程服务器。 单点登录允许用户连接到远程服务器，而不必明确提供用户标识和密码。 这将对 用户在当前计算机上的现有凭证执行必要验证，例如，通过 Windows Active Directory 执行验证。 请联系系 统管理员以获取有关可用服务器、用户标识和密码、域名的信息以及其他连接信息（包括站点是否支持单点 登录）。 您可以选择缺省服务器并保存与任何服务器相关联的用户标识、域名和密码。 启动新会话时，会自动连接到 缺省服务器。 重要：您可以连接到与客户端处于不同发行版级别的服务器。 与客户端比较，服务器可以是一个或两个较新 或较旧的发行版本。 然而，不建议长时间保持此配置。 如果服务器比客户端更新，服务器可能会创建客户 端无法读取的输出。 如果客户端比服务器更新，服务器可能无法识别通过客户端提交的语法。 因此，您应 该与您的管理员谈论有关连接到处于同版本级别客户端上的服务器。 如果您被许可使用 Statistics Adapter，且站点正在运行 IBM SPSS 协作和部署服务 3.5 或更新版本，那么可 单击搜索...以查看网络上可用的服务器列表。 如果尚未登录到 IBM SPSS 协作和部署服务存储库，那么您将 被提示输入连接信息，然后才能查看服务器列表。 添加或编辑服务器登录设置 使用“服务器登录设置”对话框可以添加或编辑远程服务器的连接信息，以便在分布式分析模式下使用。 请联系系统管理员获取可用服务器列表、服务器端口号以及其他连接信息。 除非有管理员指示，否则请勿使 用“安全套接字层”。 服务器名称。 服务器“名称”可以是分配给计算机的字母数字名称（例如 NetworkServer），也可以是分配给 计算机的 IP 地址（例如 202.123.456.78）。 端口号。 端口号是服务器软件用于通信的端口。 描述。 可以输入可选的描述以显示在服务器列表中。 使用安全套接字层连接。 安全套接字层 (SSL) 在分布式分析请求发送到远程服务器时加密请求。 使用 SSL 之 前，请与管理员协商。 要启用此选项，必须在桌面计算机和服务器上配置 SSL。 选择、切换或添加服务器 1. 从菜单中选择： 文件 > 切换服务器...
选择缺省服务器： 2. 在服务器列表中，选择要使用的服务器旁边的框。 3. 如果针对单点登录对服务器进行了配置，那么只需确保未选中设置凭证。 否则，请选中设置凭证，并输 入管理员提供的用户标识、域名和密码。 注：启动新会话时，会自动连接到缺省服务器。 切换到其他服务器： 4. 从列表中选择服务器。 5. 如果针对单点登录对服务器进行了配置，那么只需确保未选中设置凭证。 否则，请选中设置凭证，并输 入用户标识、域名和密码（如有必要）。 注：在会话期间切换服务器时，所有打开的窗口均会关闭。 在窗口关闭之前会提示您保存更改。 添加服务器： 6. 从管理员处获取服务器连接信息。 7. 单击添加打开“服务器登录设置”对话框。 8. 输入连接信息和可选设置，然后单击确定。 编辑服务器： 9. 从管理员处获取修订的连接信息。 10. 单击编辑打开“服务器登录设置”对话框。 11. 输入更改并单击确定。 搜索可用的服务器： 注：仅当您被许可使用 Statistics Adapter，且站点正在运行 IBM SPSS 协作和部署服务 3.5 或更新版本 时，才可使用搜索可用服务器功能。 12. 单击搜索...以打开“搜索服务器”对话框。 如果您尚未登录到 IBM SPSS 协作和部署服务存储库，将提示 您连接信息。 13. 选择一个或多个服务器，并单击确定。 服务器现在将出现在“服务器登录”对话框中。 14. 要连接到其中一个服务器，请按照“切换到其他服务器”的说明进行操作。 搜索可用服务器 使用“搜索服务器”对话框，选择网络上可用的一个或多个服务器。 在您从“服务器登录”对话框中单击搜索...
时，该对话框出现。 选择一个或多个服务器，单击确定以将其添加到“服务器登录”对话框中。 尽管您可在“服务器登录”对话框中 手动添加服务器，但搜索可用服务器功能允许您无需知道正确的服务器名称和端口号，就能连接到服务器。 这信息会自动提供。 不过，您仍然需要提供正确的登录信息，例如用户名、域名和密码等。 从远程服务器打开数据文件 在分布式分析模式下，“打开远程文件”对话框替换了标准的“打开文件”对话框。 • 可用文件、文件夹和驱动器的列表内容取决于远程服务器上可用的，或可从该服务器访问的文件、文件夹 和驱动器。 当前服务器名称在对话框的顶部指明。 • 在分布式分析模式下，除非将驱动器指定为共享设备，或者将包含数据文件的文件夹指定为共享文件夹， 否则将不能访问本地计算机上的数据文件。 有关如何与服务器网络“共享”本地计算机上的文件夹的信息， 请参阅操作系统文档。 • 如果服务器在运行不同的操作系统（例如您在运行 Windows 而服务器在运行 UNIX），那么即使本地数据 文件位于共享文件夹中，您在分布式分析模式下也可能无法访问它们。 38 IBM SPSS Statistics 29 Core System 用户指南 自动重新编码 使用“自动重新编码”对话框可以将字符串值和数值转换为连续整数。 当类别代码不连续时，对许多过程来 说，生成的空单元格将降低性能并增加内存要求。 此外，某些过程不能使用字符串变量，某些过程要求因子 级别为连续的整数值。 • “自动重新编码”创建的新变量保留了旧变量中任何已定义的变量标签和值标签。 对没有已定义值标签的任 何值，将使用原值作为重新编码后的值的标签。 一个表显示了旧值、新值以及值标签。 • 字符串值将按字母顺序重新编码，其中大写字母将排在相应的小写字母之前。 • 缺失值被重新编码为高于任何非缺失值的缺失值，并保留它们的原有顺序。 例如，如果原变量有 10 个非 缺失值，最低的缺失值将被重新编码为 11，值 11 将作为新变量的缺失值。 对所有变量使用相同的重新编码方案。 使用此选项可以将一个自动重新编码方案应用于所有选定变量，从而 为所有新变量生成一致的编码方案。 如果选择此选项，那么下列规则和限制适用： • 所有变量必须为相同类型（数值或字符串）。 • 所有选定变量的观测值用于创建要重新编码为连续整数的值的排序顺序。 • 新变量的用户缺失值基于具有已定义用户缺失值的列表中的第一个变量。 将来自其他原变量的所有其他值 （系统缺失值除外）都视为有效。 将空字符串值视为用户缺失值。 对于字符串变量，不将空白值或空值视为系统缺失值。 此选项会将空字符 串自动重新编码为高于最高非缺失值的用户缺失值。 模板 可以将自动重新编码方案保存在模板文件中，然后将其应用于其他变量和其他数据文件。 例如，每月可能有大量的字母数值产品代码要自动重新编码为整数，但在某些月份中添加了新产品代码，这 会更改原始的自动重新编码方案。 如果将原始设计保存在模板中，然后将其应用于包含一组新代码的新数 据，那么数据中遇到的任何新代码都会被自动重新编码为高于模板中最后一个值的值，从而保留了原始产品 代码的原始自动重新编码方案。 将模板另存为。 将所选变量的自动重新编码方案保存在外部模板文件中。 • 该模板包含将原始非缺失值映射到重新编码的值的信息。 • 模板中仅保存非缺失值的信息。 不保留用户缺失值的信息。 • 如果选择了多个要进行重新编码的变量，但尚未选择对所有变量使用相同的自动重新编码方案，或者不应 用现有模板作为自动重新编码的一部分，那么模板将基于列表中的第一个变量。 • 如果选择了多个要进行重新编码的变量，还选择了对所有变量使用相同的重新编码方案和/或选择了应用模 板，那么模板将包含用于所有变量的组合自动重新编码方案。 从文件应用模板。 将以前保存的自动重新编码模板应用于所选要进行重新编码的变量，从而将变量中的任何 其他值附加到设计的末尾，并保留所保存设计中存储的原始值和自动重新编码的值的关系。 • 所选要进行重新编码的所有变量必须是相同类型（数值或字符串），并且该类型必须与模板中定义的类型 匹配。 • 模板不包含有关用户缺失值的任何信息。 目标变量的用户缺失值基于具有已定义用户缺失值的原变量列表 中的第一个变量。 将来自其他原变量的所有其他值（系统缺失值除外）都视为有效。 • 首先应用来自模板的值映射。 将所有剩余的值重新编码为高于模板中最后一个值的值，同时将用户缺失值 （基于具有已定义用户缺失值的列表中的第一个变量）重新编码为高于最后一个有效值的值。 • 如果选择了多个要进行自动重新编码的变量，那么首先应用模板，然后对所选变量中的所有其他值进行通 用的组合式自动重新编码，从而生成一个可用于所有所选变量的通用自动重新编码方案。 将字符串值或数字值重新编码为连续整数 1. 从菜单中选择： 转换 > 自动重新编码...
2. 选择一个或多个要重新编码的变量。 第 4 章 分布式分析模式 39 3. 对每个选定的变量，为新变量输入一个名称并单击新名称。 分布式分析模式下过程的可用性 在分布式分析模式下，只有同时安装在本地版本和远程服务器上的版本中的过程才可用。 如果在本地安装远程服务器上未安装的可选组件，并从本地计算机切换到远程服务器，那么受影响的程序将 从菜单中移去，相应的命令语法将导致出现错误。 切换回本地方式可恢复所有受影响的过程。 绝对和相对路径指定 在分布式分析模式下，数据文件和命令语法文件的相对路径指定是相对于当前服务器的，而不是相对于本地 计算机。 相对路径指定（如 /mydocs/mydata.sav）并不指向您的本地驱动器上的目录和文件；它指向远程 服务器的硬盘上的目录和文件。 Windows UNC 路径指定 如果在使用 Windows 服务器版本的程序，则在使用命令语法访问数据和语法文件时，可以使用通用命名约 定 (UNC) 指定。 UNC 指定的一般形式为： \servername\sharename\path\filename • Servername 是包含数据文件的计算机的名称。 • Sharename 是该计算机上指定为共享文件夹的文件夹（目录）。 • Path 是共享文件夹下的任何其他文件夹（子目录）路径。 • Filename 是数据文件的名称。 示例如下： GET FILE='\hqdev001\public\july\sales.sav'。 如果计算机没有为其指定的名称，则可以使用其 IP 地址，例如： GET FILE='\204.125.125.53\public\july\sales.sav'。 即使是通过 UNC 路径指定，您也只能访问指定为共享的驱动器和文件夹中的数据与语法文件。 使用分布式 分析模式时，此情况包括本地计算机上的数据与语法文件。 UNIX 绝对路径指定 在 UNIX 服务器版本中，不存在与 UNC 路径等同的路径指定法，所有目录路径必须是以服务器根目录开始的 绝对路径，不允许使用相对路径。 例如，如果数据文件位于 /bin/data 且当前目录也是 /bin/data，则GET FILE='sales.sav' 无效；必须指定整个路径，例如： GET FILE='/bin/sales.sav'.
INSERT FILE='/bin/salesjob.sps'.
40 IBM SPSS Statistics 29 Core System 用户指南 第 5 章 数据编辑器 数据编辑器提供一种类似电子表格的便利方法来创建和编辑数据文件。 当您启动会话时，“数据编辑器”窗口 自动打开。 数据编辑器提供了数据的三个视图: • 过视图- 此视图提供数据集的整体构想。 • 数据视图- 此视图显示实际数据值或定义的值标签。 • 变量视图- 此视图显示变量定义信息，包括定义的变量和值标签，数据类型 (例如，字符串，日期或数 字) ，测量级别 (名义，有序或刻度) 以及用户定义的缺失值。 在 "数据" 和 "变量" 视图中，可以添加，更改和删除数据文件中包含的信息。 在数据文件中进行更改后，单击左上角的 "更新" 以获取数据的最新表示。 概述 "概述" 包含两个部分。 顶部部分提供有关整个数据集或文件的信息: • 在左上角，只要数据集或文件中的信息已更改，就会启用 "更新" 按钮，并且您需要更新视图，以便与更改 保持同步。 • 在右上角的两个 "图表格式" 按钮中，您可以在选项卡上的所有图表的查看计数和百分比之间进行切换。 • "更新" 按钮正下方是一个磁贴，用于提供数据集名称和文件名 (如果适用) 以及变量和个案数。 • 下一个区域显示有关变量的信息。 您可以查看数据集或文件中变量的变量类型或已分配测量级别的频率分 布。 图表下方左下方的两个按钮允许您在查看变量类型和测量级别之间进行切换。 可以选择一个箭头按 钮，将您转至 "变量视图"。 • 右侧区域提供缺失值的摘要: 变量，个案和个别值的饼图中的完整数据和缺失数据。 如果 "缺失值" 选项已 获得许可，那么将显示一个箭头按钮，当单击此按钮时，将调用 "多重插补分析模式" 对话框。 底部部分显示有关所选变量的基本信息: • 在左侧， "变量报告" 下拉列表允许您选择要显示的特定变量。 • 对于具有 100 个或更少相异值的分类变量 (分配的名义或有序测量级别) ，将显示一个条形图或一个计数或 唯一值百分比的饼图。 • 在右侧，将显示一个频率表，其中显示唯一值，计数和百分比。 箭头按钮调用 "频率" 对话框。 • 对于刻度变量，直方图显示在左侧，频率计数或百分比显示在垂直轴上。 • 右侧显示一组描述性统计，包括平均值，标准差，最小值和最大值以及四分位数 (第 25 个，第 50 个和第 75 个 百分位数)。 箭头按钮将调用 "探索" 对话框。 概述-从对话框中选择变量 "选择变量" 设置提供了一个选项，用于选择要在变量报告中显示的变量。 选择变量 单击下图中的箭头所指示的链接以打开 "选择变量" 对话框。 图 1: "概述" 选项卡-"链接到变量" 对话框 42 IBM SPSS Statistics 29 Core System 用户指南 从下拉列表中选择要查看其报告的其中一个变量，然后按 继续。 图 2: "概述" 选项卡-变量 对话框 您可以按分配的测量级别和/或通过在 "搜索" 框中输入字符串来过滤变量。 如果显示了标签，那么搜索框的 内容将应用于变量名称或名称和标签。 数据视图 数据视图的许多功能类似于电子表格应用程序中提供的功能。 不过，有一些重要的不同之处： • 行为个案。 每一行代表一个个案或一个观察值。 例如，问卷的每个响应者都是一个个案。 • 列为变量。 每一列代表一个要度量的变量或特征。 例如，问卷中的每一项都是一个变量。 • 单元格包含值。 每个单元格均包含某个个案的某个变量的单个值。 单元格是个案和变量相交的位置。 单 元格仅包含数据值。 与电子表格程序不同，数据编辑器中的单元格不能包含公式。 • 数据文件是矩形的。 数据文件的尺寸由个案和变量的个数确定。 您可以在任何单元格中输入数据。 如果 在定义的数据文件的界限以外的单元格中输入数据，则数据矩形将扩展以包含该单元格与文件界限之间的 任何行和/或列。 数据文件的界限内没有“空”单元格。 对于数值变量，空白单元格会转换成系统缺失值。 对于字符串变量，则将空白单元格视为有效值。 第 5 章 数据编辑器 43 变量视图 “变量视图”包含对数据文件中每个变量的属性的描述。 在“变量视图”中： • 行为变量。 • 列为变量属性。 您可以添加或删除变量，以及修改变量属性，包括： • 变量名 • 数据类型 • 数字位数或字符个数 • 小数位数 • 描述性的变量标签和值标签 • 用户定义的缺失值 • 列宽 • 测量级别 当您保存数据文件时，将保存所有这些属性。 除了在“变量视图”中定义变量属性以外，另外还有两种方法可用于定义变量属性： • “复制数据属性向导”使您能够使用外部 IBM SPSS Statistics 数据文件或当前会话中的另一数据集作为模 板，用于定义活动数据集中的文件和变量的属性。 您也可以将活动数据集中的变量用作活动数据集中其他 变量的模板。 “复制数据属性”可以在数据编辑器窗口的“数据”菜单上找到。 • “定义变量属性”（也在数据编辑器窗口的“数据”菜单上提供）扫描您的数据并列出任何选定变量的所有唯 一数据值、标识未标注的值以及提供自动标注功能。 这对于使用数值代码表示类别（例如，0 = Male，1 = Female）的分类变量特别有用。 显示或定义变量属性 1. 使数据编辑器成为活动窗口。 2. 双击位于“数据视图”中列顶端的变量名称，或单击变量视图选项卡。 3. 要定义新变量，请在任意空白行中输入一个变量名称。 4. 选择您想要定义或修改的属性。 变量名称 以下规则适用于变量名称： • 每个变量名称必须是唯一的；不允许重复。 • 变量名可长达 64 字节，第一个字符必须是字母或字符 @、#或 $。后续字符可以是字母、数字、非标点字 符和句点 (.) 的任意组合。在代码页方式中，64 字节通常是单字节语言的 64 个字符（例如，英语、法语、 德语、西班牙语、意大利语、希伯来语、俄语、希腊语、阿拉伯语和泰语）和 32 个字符的双字节语言 （例如日语、中文和韩语）。 许多在代码页模式中只占一个字节的字符串在 Unicode 方式下则会占到两个 或更多字节。 例如，在代码页格式中，é 占一个字节，而在 Unicode 格式中，它占两个字节；因此， résumé 在代码页文件中占 6 个字节，而在 Unicode 方式下占 8 个字节。 注: 字母包括平台字符集所支持的语言中，书写日常文字所用的任何非标点字符。 • 变量名称不能包含空格。 • 变量名称第一个位置中的 # 字符将变量定义为临时变量。 只能使用命令语法创建临时变量。 在创建新变 量的对话框中，无法指定 # 作为变量的第一个字符。 • 第一个位置中的 $ 符号表示变量为系统变量。 $ 符号不能作为用户定义的变量的第一个字符。 • 可在变量名称中使用句点、下划线和字符 $、# 以及 @。 例如，A._$@#1 是一个有效的变量名称。 • 变量名称不能以句点开头或结尾。 以句点开头的名称无效；以句点结尾的名称可能被解释为命令终止符。 在创建新变量的对话框中，无法创建以句点开始或结束的变量。 44 IBM SPSS Statistics 29 Core System 用户指南 • 应避免使用下划线结束变量名称，因为这样的名称可能与命令和过程自动创建的变量名称冲突。 • 不能将保留关键字用作变量名称。 保留关键字有：ALL、AND、BY、EQ、GE、GT、LE、LT、NE、NOT、 OR、TO 和 WITH。 • 可以用任意混合的大小写字符来定义变量名称，大小写将为显示目的而保留。 • 当长变量名称需要在输出中换行为多行时，会在下划线、句点和内容从小写变为大写的位置进行换行。 变量测量级别 您可以将测量级别指定为刻度（定距或者定比刻度上的数值数据）、有序或名义。 名义数据和有序数据可以 是字符串（字母数字）或数值。 • 名义 (Nominal). 当变量的值表示没有内部排名的类别 (例如，员工工作所在公司的部门) 时，可将该变量视 为名义变量。 名义变量的示例包括地区、邮政编码和宗教信仰。 • 有序 (Ordinal). 当变量的值表示具有某种内在等级的类别 (例如，从高度不满意到高度满意的服务满意度级 别) 时，可将该变量视为有序变量。 有序变量的示例包括表示满意度或可信度的态度分数和优先选择评 分。 • 标度 (scale). 当变量的值表示具有有意义度量的有序类别时，可以将该变量视为刻度 (连续) ，以便在值之 间进行适当的距离比较。 刻度变量的示例包括以年为单位的年龄和以千美元为单位的收入。 注：对于有序字符串变量，将假定字符串值的字母顺序反映了类别的真实顺序。 例如，对于具有 low、 medium、high 值的字符串变量，类别的顺序将解释为 high、low、medium，这个顺序是错误的。 通常，使 用数值代码代表有序数据更为可靠。 对于通过转换创建的新数字变量、来自外部源的数据以及在版本 8 之前创建的 IBM SPSS Statistics 数据文 件，缺省测量级别由以下表格中的条件决定。 将以条件在表格中的排序顺序对其进行评估。 将应用与数据 匹配的第一个条件的测量级别。 表 6: 用于确定测量级别的规则 条件 测量级别 变量的所有值均缺失 名义 格式为美元或定制货币 连续 格式为日期或时间（不包括月份和星期） 连续 变量包含至少一个非整数值 连续 变量包含至少一个负值 连续 变量不包含少于 10,000 的有效值 连续 变量具有 N 个或更多唯一有效值 连续 变量不包含少于 10 的有效值 连续 变量具有少于 N 个唯一有效值 名义 N 是用户指定的分界值。 缺省值为 24。 • 您可以在“选项”对话框中更改分界值。 有关更多信息，请参阅第166 页的『数据选项』主题。 • “数据”菜单中的“定义变量属性”对话框可帮助您指派正确的测量级别。 有关更多信息，请参阅第61 页的 『指定测量级别』主题。 变量类型 “变量类型”指定每个变量的数据类型。 缺省情况下，假定所有新变量都为数值变量。 您可以使用“变量类型” 来更改数据类型。 “变量类型”对话框的内容取决于选定的数据类型。 对于某些数据类型，存在有关宽度和 小数位数的文本框；对于其他数据类型，只需从可滚动的示例列表中选择一种格式即可。 以下数据类型可用。 第 5 章 数据编辑器 45 数值 值为数字的变量。 值以标准数值格式显示。 数据编辑器接受以标准格式或科学记数法表示的数值。 逗号 变量值显示为每三位用逗号分隔，并用句点作为小数分隔符的数字变量。 数据编辑器为逗号变量接受带 或不带逗号的数值，或以科学记数法表示的数值。 值的小数指示符右侧不能包含逗号。 点 变量值显示为每三位用句点分隔，并带有逗号作为小数分隔符的数字变量。 数据编辑器为点变量接受带 或不带点的数值，或以科学记数法表示的数值。 值的小数指示符右侧不能包含句点。 科学记数法 一个数字变量，它的值以嵌入的 E 以及带符号的 10 次幂指数形式显示。 无论此类变量是否带有指数， 数据编辑器都接受这些变量的数字值。 指数可以先于 E 或 D，带有可选符号或符号（例如，123、 1.23E2、1.23D2、1.23E+2 和 1.23+2）。 日期 一种数字变量，其值以若干种日历-日期或时钟-时间格式中的一种显示。 从列表中选择一种格式。 输入 日期时可以用斜杠、连字符、句号、逗号或空格作为分隔符。 两位数年份值的世纪范围由您的“选项”设 置确定（从“编辑”菜单中，选择选项然后单击数据选项卡）。 美元符号 数字变量，显示时前面带美元符 ($)，每三位用逗号分隔，并用句点作为小数分隔符。 可以输入带有或 不带有前导美元符号的数据值。 自定义货币 一种数字变量，其值以定制货币格式中的一种显示，定制货币格式是在“选项”对话框的“货币”选项卡中 定义的。 定义的自定义货币字符不能用于数据输入，但显示在数据编辑器中。 String 字符串变量的值不是数值，因此不用在计算中。 这些值可以包含受支持的字符（字母、数字、非标点字 符和句点的组合）直到定义的字节长度。 字符串变量区分大小写字母。 此类型又称为字母数字变量。 受限数值 值限于非负整数的变量。 在显示值时，填充先导 0 以达到最大变量宽度。 可以以科学记数法输入值。 定义变量类型 1. 单击您想要定义的变量的类型单元格中的按钮。 2. 在“变量类型”对话框中选择数据类型。 3. 单击确定。 输入格式与显示格式 根据格式，值在“数据视图”中的显示可能与输入的及内部存储的实际值不同。 以下是一些通用准则： • 对于数值、逗号和点格式，您可以输入具有任意小数位数（最多 16 位）的值，整个值会存储在内部。 “数 据视图”仅显示定义的小数位数，并且对具有更多小数位的值进行舍入。 不过，在所有的计算中都使用完 整的值。 • 对于字符串变量，所有值都向右填充到最大宽度。 对于最大宽度为三的字符串变量，No 值在内部存储为 “No”，并不等于“No”。 • 对于日期格式，您可以使用斜杠、短划线、空格、逗号或句点作为日、月、年值之间的分隔符，并且可以 为月值输入数字、三字母缩写或完整名称。 以通用格式 dd-mmm-yy 显示日期，其中以短划线作为分隔符 并对月份使用三字母缩写。 以通用格式 dd/mm/yy 和 mm/dd/yy 显示日期，其中以斜杠作为分隔符并对月 份使用数字。 日期在内部存储为自 1582 年 10 月 14 日以来的累计秒数。 两位数年份值的日期的世纪范 围由您的“选项”设置确定（从“编辑”菜单中，选择选项然后单击数据选项卡）。 • 在时间格式中，您可以使用冒号、句号或空格作为小时、分钟和秒数间的分隔符。 时间以冒号分隔显示。 时间被内部储存为表示时间间隔的秒数。 例如，10:00:00 被内部储存为 36000，即 60（秒每分钟）x 60 （分钟每小时）x 10（小时）。 46 IBM SPSS Statistics 29 Core System 用户指南 变量标签 您可以为描述性变量标签分配最多可达 256 个的字符（在双字节语言中则为 128 个字符)。 变量标签可以包 含空格和变量名称中所不允许的保留字符。 指定变量标签 1. 使数据编辑器成为活动窗口。 2. 双击位于“数据视图”中列顶端的变量名称，或单击变量视图选项卡。 3. 在变量的标签单元格中，输入描述性变量标签。 值标签 您可以为每个变量值分配描述值标签。 当您的数据文件使用数值代码表示非数值类别时（例如：代码 1 和 2 代表 male 和 female），此过程特别有用。 指定值标签 1. 单击您想要定义的变量的值单元格中的按钮。 2. 在每个值中，输入值和标签。 3. 单击添加输入值标签。 4. 单击确定。 在标签中插入换行符 若单元格或区域没有足够宽度在一行内显示整个标签，则透视表和图表中的变量标签和值标签会自动换行至 多行，而如果您想将标签换行至不同点，可以编辑结果，插入手动换行符。 您还可以创建变量标签和值标 签，这将一直换行至指定点并以多行显示。 1. 对于变量标签，在数据编辑器的变量视图中选择标签变量单元格。 2. 对于值标签，在数据编辑器的变量视图中选择值变量单元格，单击单元格中的按钮，选择您想在“值标签” 对话框中修改的标签。 3. 在标签中您想让标签换行的地方，输入\n。 \n 不在透视表或图表中显示；它被翻译为换行字符。 缺失值 缺失值将指定数据值定义为用户缺失值。 例如，您想要区分因响应者拒绝回答问题造成的数据缺失与由于问 题不适于该响应者而未回答所引起的数据缺失。 将指定为用户缺失值的数据值标记为进行特殊处理，并将其 从大多数计算中排除。 定义缺失值 1. 单击您想要定义的变量的缺失单元格中的按钮。 2. 输入表示缺失数据的值或值范围。 角色 某些对话框支持可用于预先选择分析变量的预定义角色。 当打开其中一个对话框时，满足角色要求的变量将 自动显示在目标列表中。 可用角色包括： 输入。 变量将用作输入（例如，预测变量、自变量）。 目标。 变量将用作输出或目标（例如，因变量）。 两者。 变量将同时用作输入和输出。 无。 变量没有角色分配。 第 5 章 数据编辑器 47 分区。 变量将用于将数据划分为单独的训练、检验和验证样本。 拆分。 包括以便与 IBM SPSS Modeler 相互兼容。 具有此角色的变量不会在 IBM SPSS Statistics 中用作拆 分文件变量。 • 缺省情况下，为所有变量分配输入角色。 这包括外部文件格式的数据和 IBM SPSS Statistics 18 之前版本 的数据文件。 • 角色分配只影响支持角色分配的对话框。 它对命令语法没有影响。 指定角色 1. 从变量的角色单元格的列表中选择角色。 列宽 您可以为列宽指定一些字符。 也可以通过单击并拖拽列边框来更改数据视图中的列宽。 • 比例字体的列宽是基于平均字符宽度。 根据值中使用的字符，在指定宽度内可以显示较多或较少的字符。 • 列宽度只影响数据编辑器中的值显示。 更改列宽不会改变变量的已定义宽度。 变量对齐 对齐控制着数据视图中数据值和/或值标签的显示。 缺省对齐方式为数字变量在右边，字符串变量在左边。 此设置只影响数据编辑器中的显示。 将变量定义属性应用于多个变量 在定义了变量的变量定义属性之后,可以复制一个或多个属性并将其应用到一个或多个变量。 基础复制粘贴操作被用于变量定义属性的应用。 您可以执行以下操作： • 复制一个单一属性（例如，值标签）并将其粘贴至一个或多个变量的相同属性单元格中。 • 从一个变量中复制所有属性并将其粘贴至一个或多个其他变量。 • 通过一个已复制变量的所有属性创建多个新变量。 将变量定义属性应用于其他变量 从已定义变量中应用单个属性 1. 在变量视图中，选择您想应用于其他变量的属性单元格。 2. 从菜单中选择： 编辑 > 复制 3. 选择您想应用其属性的属性单元格。 （您可以选择多个目标变量。） 4. 从菜单中选择： 编辑 > 粘贴 如果您将属性粘贴至空白行，那么新变量就与所有属性中除选定属性以外的缺省属性一同被创建。 应用已定义变量中的全部属性 1. 在变量视图中，与您想使用的属性一同选择变量行号。 （突出显示整行。） 2. 从菜单中选择： 编辑 > 复制 3. 与您想使用的属性一同选择变量行号。 （您可以选择多个目标变量。） 4. 从菜单中选择： 编辑 > 粘贴 48 IBM SPSS Statistics 29 Core System 用户指南 通过相同属性生成多个新变量 1. 在变量视图中，单击具有您想用于新变量属性的变量。 （突出显示整行。） 2. 从菜单中选择： 编辑 > 复制 3. 在数据文件中单击最后一个已定义变量下面的空行号码。 4. 从菜单中选择： 编辑 > 粘贴变量...
5. 在“粘贴变量”对话框中，输入您想创建的变量数。 6. 输入新变量的前缀和起始数字。 7. 单击确定。 新变量名称将包括指定前缀和由指定数字开始的序列数。 定制变量属性 除了标准变量属性（如值标签、缺失值、测量级别）之外，还可以自己创建定制的变量属性。 与标准变量属 性一样，这些定制变量属性也随 IBM SPSS Statistics 数据文件一同保存。 因此，您可以创建识别调查问题 响应类型的变量属性（例如，单选、多选、填空）或计算变量使用的公式。 创建定制变量属性 创建新的定制属性： 1. 在“变量视图”中，从菜单中选择： 数据 > 新建定制属性...
2. 将要向其分配新属性的变量拖放到“选定的变量”列表中。 3. 为新属性输入名称。 属性名必须遵循与变量名称相同的规则。 请参阅主题第44 页的『变量名称』，了 解更多信息。 4. 为新属性输入一个可选值。 如果选中多个变量，那么该值将指定给所有选定的变量。 可以将此字段保留 为空，然后在“变量视图”中输入各变量的值。 在数据编辑器中显示属性。 在数据编辑器的“变量视图”中显示属性。 有关控制定制属性的显示的信息，请 参阅下面的第49 页的『显示和编辑定制变量属性』。 显示已定义的属性列表。 显示已经为数据集定义的定制属性列表。 以美元符号 ($) 开头的属性名称是不能修 改的保留属性。 显示和编辑定制变量属性 可以在数据编辑器的“变量视图”中显示和编辑定制变量属性。 • 定制变量属性的名称用方括号括起。 • 以美元符号开头的属性名是保留名称，不能修改这些名称。 • 空单元格表示该变量没有属性；单元格中显示为 Empty 的文本表示该变量具有属性，但还没有为该变量的 属性赋值。 在单元格中输入文本后，该变量即拥有了具有您所输入的值的属性。 • 在单元格中显示的数组... 表示此属性是属性数组，即包含多个值的属性。 单击单元格中的按钮以显示值列 表。 显示和编辑定制变量属性 1. 在“变量视图”中，从菜单中选择： 查看 > 定制变量视图...
2. 选择（选中）要显示的定制变量属性。 （定制变量属性用方括号括起。） 一旦在“变量视图”中显示属性，您即可直接在数据编辑器中对其进行编辑。 第 5 章 数据编辑器 49 变量属性数组 文本数组...（显示在“变量视图”中定制变量属性的单元格内，或者显示在“定义变量属性”中的“定制变量属 性”对话框内）表示此为属性数组，即包含多个值的属性。 例如，可以有一个属性数组，它标识用于计算派 生变量的所有源变量。 单击单元格中的按钮可以显示和编辑值列表。 定制变量视图 您可以使用定制变量视图控制变量视图中显示的属性（例如，名称、类型、标签）及其显示顺序。 • 任何与数据库有关的定制变量属性都以方括号括起。 有关更多信息，请参阅第49 页的『创建定制变量属 性』主题。 • 自定义显示设置保存在 IBM SPSS Statistics 数据文件中。 • 您还可以控制变量视图中的缺省显示和属性顺序。 有关更多信息，请参阅第167 页的『更改缺省变量视 图』主题。 定制变量视图 1. 在“变量视图”中，从菜单中选择： 查看 > 定制变量视图...
2. 选择（选中）要显示的变量属性。 3. 使用向上和向下箭头按钮更改属性的显示顺序。 恢复缺省。 应用缺省显示与顺序设置。 拼写检查 变量和值标签 检查变量标签与值标签的拼写： 1. 在“数据编辑器”窗口中，选择“变量视图”选项卡。 2. 右键单击标签或值列，并从弹出菜单中选择： 拼写 或 3. 在“变量视图”中，从菜单中选择： 实用程序 > 拼写 或 4. 在对话框中单击拼写。 （这将值标签的拼写检查限制为特定变量。） 拼写检查仅限于数据编辑器的变量视图中的变量标签和值标签。 字符串数据值 检查字符串数据值的拼写： 1. 在“数据编辑器”窗口中，选择“数据视图”选项卡。 2. （可选）选择一个或多个待检查的变量（列）。 要选择变量，请单击列顶部的变量名称。 3. 从菜单中选择： 实用程序 > 拼写 • 如果在“数据视图”中没有选定的变量，则会选中所有字符串变量。 • 如果数据集不包含字符串变量，或选定变量均非字符串变量，那么“实用程序”上的“拼写”选项被禁用。 定制变量视图 您可以使用定制变量视图控制变量视图中显示的属性（例如，名称、类型、标签）及其显示顺序。 50 IBM SPSS Statistics 29 Core System 用户指南 • 任何与数据库有关的定制变量属性都以方括号括起。 有关更多信息，请参阅第49 页的『创建定制变量属 性』主题。 • 自定义显示设置保存在 IBM SPSS Statistics 数据文件中。 • 您还可以控制变量视图中的缺省显示和属性顺序。 有关更多信息，请参阅第167 页的『更改缺省变量视 图』主题。 定制变量视图 1. 在“变量视图”中，从菜单中选择： 查看 > 定制变量视图...
2. 选择（选中）要显示的变量属性。 3. 使用向上和向下箭头按钮更改属性的显示顺序。 恢复缺省。 应用缺省显示与顺序设置。 拼写检查 变量和值标签 检查变量标签与值标签的拼写： 1. 在“数据编辑器”窗口中，选择“变量视图”选项卡。 2. 右键单击标签或值列，并从弹出菜单中选择： 拼写 或 3. 在“变量视图”中，从菜单中选择： 实用程序 > 拼写 或 4. 在对话框中单击拼写。 （这将值标签的拼写检查限制为特定变量。） 拼写检查仅限于数据编辑器的变量视图中的变量标签和值标签。 字符串数据值 检查字符串数据值的拼写： 1. 在“数据编辑器”窗口中，选择“数据视图”选项卡。 2. （可选）选择一个或多个待检查的变量（列）。 要选择变量，请单击列顶部的变量名称。 3. 从菜单中选择： 实用程序 > 拼写 • 如果在“数据视图”中没有选定的变量，则会选中所有字符串变量。 • 如果数据集不包含字符串变量，或选定变量均非字符串变量，那么“实用程序”上的“拼写”选项被禁用。 输入数据 在数据视图中，您可以在数据编辑器里直接输入数据。 您可以以任何顺序输入数据。 您可以在选定区域或 个别单元格中，按个案或按变量输入数据。 • 突出显示活动单元格。 • 在数据编辑器左上角显示变量名称和活动单元格的行号。 • 当您选择一个单元格输入数据值时，该值显示在数据编辑器顶端的单元格编辑器中。 • 当按下回车键或选择另一个单元格时，数据值方被保存。 • 欲输入除简单数值数据外的任何其他数据时，您必须先定义变量类型。 如果在空列中输入一个值，数据编辑器将自动创建一个新变量并赋予一个变量名称。 第 5 章 数据编辑器 51 输入数值数据 1. 在数据视图中选择一个单元格。 2. 输入数据值。 （值被显示在数据编辑器顶端的单元格编辑器中。） 3. 按回车键或选择另一个单元格，即可保存值。 输入非数值数据 1. 双击位于“数据视图”中列顶端的变量名称，或单击变量视图选项卡。 2. 单击变量的类型单元格。 3. 在“变量类型”对话框中选择数据类型。 4. 单击确定。 5. 双击行号或单击数据视图选项卡。 6. 在新近定义的变量列中输入数据。 使用值标签进行数据输入 1. 若值标签没有显示在当前数据视图中，就从菜单中选择： 视图 > 值标签 2. 单击您想输入值的单元格。 3. 从下拉列表中选择值标签。 值被输入后，值标签就显示在单元格里。 注：此过程仅当您已对变量的值标签定义时才会运行。 数据编辑器中的数据值限制 既定变量类型和宽度决定了在数据视图单元格中可以输入的值类型。 • 如果输入了一个既定变量类型不允许的字符，那么该字符无法输入。 • 对于字符串变量，不允许字符超出既定宽度。 • 对于数字变量，可以输入超出既定宽度的整数值，但数据编辑器既显示科学记数法又显示一部分值，其后 用省略号（...）指明该值已超出既定宽度。 更改变量的既定宽度即可在单元格中显示该值。 注：更改列宽不会影响变量宽度。 编辑数据 通过数据编辑器，您可以在数据视图中以多种方法修改数据值。 您可以执行以下操作： • 更改数据值 • 剪切、复制并粘贴数据值 • 添加删除个案 • 添加删除变量 • 更改变量顺序 替换或修改数据值 删除原有值，输入新值 1. 在数据视图中双击单元格。 （单元格值显示在单元格编辑器中。） 2. 在单元格或单元格编辑器中直接编辑值。 3. 按回车键或选择另一个单元格保存新值。 52 IBM SPSS Statistics 29 Core System 用户指南 剪切、复制并粘贴数据值 在数据编辑器中您可以对单个单元格值或组值进行剪切、复制和粘贴。 您可以执行以下操作： • 将一个单一单元格值移动或复制到另一个单元格 • 将一个单一单元格值移动或复制到一组单元格 • 将单个个案（行）的值移动或复制到多个个案 • 将单个变量（列）的值移动或复制到多个变量 • 将一组单元格值移动或复制到另一组单元格 数据编辑器中已粘贴值的数据转换 若源单元格与目标单元格的既定变量类型不同，数据编辑器会试图将该值转换。 如果无法转换，将在目标单 元格中插入系统缺失值。 将数值或日期转换为字符串。 数值（例如，数值、美元符、点或逗号）和日期格式若被粘贴至字符串变量单 元格，那么将被转换为字符串。 在单元格中显示时，字符串值为数值型值。 例如，对于一个美元符号格式 的变量，显示的美元符号图标则变成字符串值的一部分。 超出既定字符串值宽度的值将被截去。 将字符串转换为数值或日期。 包含被目标单元格的数值或日期格式所接受字符的字符串值将被转换为同等的 数值或日期值。 例如，如果目标单元格的格式类型为日-月-年格式中的一种，那么字符串 25/12/91 将被转 换为有效日期，但如果目标单元格的格式类型为月-日-年格式中的一种，那么该值将被转换为系统缺失值。 将日期转换为数值。 如果目标单元格为数值格式中的一种，那么日期和时间值将被转换为秒数（例如，数 值、美元符号、点或逗号）。 因为日期从 1582 年 10 月 14 日开始被内部储存为秒数，且将日期转换为数 值可以产生一些非常大的数字。 例如，日期 10/29/91 被转换为数值 12,908,073,600。 将数值转换为日期或时间。 如果该值表示能产生有效日期或时间的秒数，那么该数值将被转换为日期或时 间。 对于日期，小于 86,400 的数值将被转换为系统缺失值。 插入新个案 在空行的单元格值输入数据会自动创建一个新个案。 数据编辑器将为所有个案中的所有其他变量插入系统缺 失值。 若在新个案和已有个案间存在任何空行，则该空行将成为具有用于所有变量的系统缺失值的新个案。 您也可以在已有个案之间插入新个案。 在已有个案之间插入新个案 1. 在数据视图中，在想要插入新个案地方的下方个案（行）中任选一个单元格。 2. 从菜单中选择： 编辑 > 插入个案 在个案中插入新行，而所有变量都被赋予系统缺失值。 插入新变量 在数据视图中的空列中或在变量视图中输入数据将自动创建一个具有缺省变量名称（前缀 var 和序列数）和 缺省数据格式类型（数值）的新变量。 数据编辑器将为新变量插入用于所有个案的系统缺失值。 若在数据 视图中存在任何空列或在变量视图中的新变量和已有变量间存在任何空行，该行或列都将成为具有用于所有 个案的系统缺失值的新变量。 您也可以在已有变量之间插入新变量。 在已有变量之间插入新变量 1. 在（数据视图）变量右侧或想要插入新变量地方的下方（变量视图）任选一个单元格。 2. 从菜单中选择： 编辑 > 插入变量 新变量被插入了用于所有个案的系统缺失值。 第 5 章 数据编辑器 53 移动变量 1. 要选取变量，请单击数据视图中的变量名称或变量视图中的变量行号。 2. 将变量拖放到新位置中。 3. 如果要将变量放在两个现有变量之间：在“数据视图”中，将该变量放置在您希望的位置右侧的变量列中， 或者在“变量视图”中，将该变量放在您希望的位置下方的变量行中。 更改数据类型 您随时可以通过使用变量视图中的“变量类型”对话框来更改变量的数据类型。 数据编辑器将试图将已有值转 换为新类型。 若无法转换，那么赋予其系统缺失值。 该转换规则与将不同格式类型的数据值粘贴至变量的 规则相同。 若数据格式的更改会引起缺失值规格或值标签的丢失，数据编辑器将显示提示对话框并询问您是 否想要继续此更改或将其取消。 查找个案、变量或插补 在数据编辑器中进入“转到”对话框查找指定个案（行）号或变量名称。 -cases 1. 对于个案，从菜单中选择： 编辑 > 转至案例...
2. 在数据视图中输入一个表示当前行号的整数值。 注：个别个案的当前行号可以因排序和其他操作而更改。 变量 1. 对于变量，从菜单中选择： 编辑 > 转至变量...
2. 输入变量名称或从下拉列表中选择变量。 插补 1. 从菜单中选择： 编辑 > 转至插补...
2. 从下拉列表中选择插补（或原始数据）。 或者，也可从“数据编辑器”的数据视图上编辑栏的下拉列表中选择插补。 在选择插补时相对个案位置被保留。 例如，如果在源数据集中有 1000 个个案，个案 1034，即第 1 个插补 中的第 34 个个案，显示在网格顶部。 如果您从下拉列表中选择插补 2，那么个案 2034，即第 2 个插补中 的第 34 个个案，将显示在网格顶部。 如果您从下拉列表中选择原始数据，则个案 34 将显示在网格顶部。 在两个插补之间浏览时，列位置也将被保留，这样可以方便地进行值比较。 查找并替换数据和属性值 在数据视图中查找和/或替换数据值或者在变量视图中查找和/或替换属性值： 1. 在列中单击想要搜索的单元格。 （禁止对单列的值进行查找和替换。） 2. 从菜单中选择： 编辑 > 查找 或 编辑 > 替换 数据视图 • 不得在数据视图中搜索。 搜索向导一般为不可按。 54 IBM SPSS Statistics 29 Core System 用户指南 • 对于日期和时间，将搜索“数据视图”中显示的格式化值。 例如，显示为 10/28/2007 的日期若以 10-28-2007 日期的形式就无法被找到。 • 对于其他数值变量，包含、开始以和结束以将搜索格式化值。 例如，使用以 开头为选项，Dollar 格式变量 的搜索值 $ 123 将找到 $ 123.00 和 $123.40，但找不到 $1234。 通过 整个单元格选项，搜索值可以格式 化或未格式化（简单的 F 数字格式），但只有精确的数字值（在数据编辑器中显示的精度）匹配。 • 数值系统缺失值由单个句点 (.) 代表，要查找系统缺失值，请输入单个句点作为搜索值并选择整个单元格。 • 若值标签显示为选定变量列，则标签文本而不是潜在数据值被找到，且您不能替换该标签文本。 变量视图 • 查找仅用于名称、标签、值、丢失以及定制变量属性列。 • 替换仅用于标签、值以及自定义变量属性列。 • 在值（值标签）列中，搜索字符串可与数据值或值标签匹配。 注：对数据值的替换将删除任何与该值有关的原有值标签。 为选定变量获取描述统计 要为选定变量获取描述统计： 1. 右键单击“数据视图”或“变量视图”中的选定变量。 2. 从弹出菜单中选择描述统计。 缺省情况下，频率表（计数表）显示所有具有 24 个或较少唯一值的变量。 汇总统计由变量测量级别和数据 类型（数值或字符串）决定： • 字符串。 不为字符串变量计算汇总统计。 • 数值、名义或未知测量级别。 范围、最小值、最大值、模式。 • 数值、有序测量级别。 范围、最小值、最大值、模式、平均值、中位数。 • 数值、连续（刻度）测量级别。 范围、最小值、最大值、模式、平均值、中位数、标准差。 还可以获取名义变量和有序变量的条形图、连续（标度）变量的直方图，以及更改确定何时显示频率表的临 界值。 请参阅第168 页的『输出选项』主题以获取更多信息。 数据编辑器中的个案选择状态 缺省情况下，在选定个案子集且未废弃未选中的个案时，未选中的个案将隐藏在数据编辑器中。 从数据编辑 器复制行时，不会提取隐藏的个案。 您可以通过取消选择编辑 > 隐藏排除的个案，或右键单击“数据编辑器”并取消选择隐藏排除的个案选项，来 选择显示隐藏的个案。 未隐藏未选中的案例时，将在数据编辑器中通过行号标记对角线（斜杠）。 图 3: 数据编辑器中的已过滤个案 第 5 章 数据编辑器 55 数据编辑器显示选项 视图菜单为数据编辑器提供了多个显示选项： 字体。 此选项控制数据显示的字体特征。 网格线。 此选项可在网格线的显示之间转换。 值标签。 此选项可在实际数据值的显示与用户定义的描述性值标签之间转换。 此选项只在数据视图中可 用。 使用多个视图 在数据视图中，您可以通过使用位于水平横轴下端、垂直数轴右侧的拆分器创建多个视图（窗格）。 您也可以使用窗口菜单插入并删除窗格拆分。 插入拆分： 1. 在数据视图中，从菜单中选择： 窗口 > 分割 拆分器被插入到上方及选定单元格的左侧。 • 若左上角的单元格被选中，则拆分器的插入将当前视图水平垂直大致平分。 • 若除第一列中的顶端单元格外的任何单元格被选中，则水平窗格拆分器被插入于选定单元格的上方。 • 若除顶端行中的第一个单元格外的任何单元格被选中，那么垂直窗格拆分器被插入于选定单元格的左侧。 数据编辑器打印 数据文件将按照其在屏幕上的显示打印。 • 当前显示视图中的信息被打印。 在数据视图中，数据被打印。 在变量视图中，数据定义信息被打印。 • 若网格线在当前选中视图中有所显示，就会被打印。 • 若数据视图中的值标签在当前有所显示，就会被打印。 此外，实际数据值被打印。 使用数据编辑器中的视图菜单显示或隐藏网格线并在数据值和值标签的显示间转换。 打印数据编辑器内容 1. 使数据编辑器成为活动窗口。 2. 单击要打印视图的选项卡。 3. 从菜单中选择： 文件 > 打印...
56 IBM SPSS Statistics 29 Core System 用户指南 第 6 章 使用多数据源 从版本 14.0 开始，可以同时打开多个数据源，从而使得以下操作更加轻松： • 在数据源之间来回切换。 • 比较不同数据源的内容。 • 在数据源之间复制和粘贴。 • 创建个案和/或变量的多个子集以进行分析。 • 在不事先保存每个数据源的情况下，合并来自各种数据格式（例如：电子表格、数据库、文本数据）的多 个数据源。 多数据源的基本处理 缺省情况下，所打开的每个数据源都显示在新的数据编辑器窗口中。 （请参阅第163 页的『一般选项』以 获取有关通过更改缺省行为来在单一“数据编辑器”窗口中一次仅显示一个数据集的信息。） • 以前打开的任何数据源都保持打开状态，并可供未来使用。 • 第一次打开数据源时，它自动成为活动数据集。 • 只需通过单击要使用的数据源的数据编辑器窗口中任意位置，或从“窗口”菜单中选择该数据源的数据编辑 器窗口，就可以更改活动数据集。 • 只有活动数据集中的变量才能用于分析。 • 访问数据的任何对话框打开（包括显示变量列表的所有对话框）时，不能更改活动数据集。 • 在会话过程中，必须至少打开一个数据编辑器窗口。 关闭最后一个打开的数据编辑器窗口时，IBM SPSS Statistics 将自动关闭，提示首先保存更改。 使用命令语法中的多个数据集 如果使用打开数据源的命令语法（例如 GET FILE、GET DATA），则需要使用 DATASET NAME 命令显式 地命名每个数据集，以便同时打开多个数据源。 主题以获取更多信息。 在使用命令语法时，活动数据集名称显示在语法窗口的工具栏上。 所有以下操作均可更改活动数据集： • 使用 DATASET ACTIVATE 命令。 主题。 • 单击数据集的“数据编辑器”窗口中的任意位置。 • 在语法窗口中工具栏上的活动下拉列表中，选择数据集名称。 在数据集之间复制和粘贴信息 可以使用与在单个数据文件中复制和粘贴信息基本相同的方法，将数据和变量定义属性从一个数据集复制到 另一数据集。 • 在“数据视图”中复制和粘贴所选数据单元格将仅粘贴数据值，而不包括变量定义属性。 • 通过选择在列顶部的变量名称来复制和粘贴“数据视图”中的整个变量，会粘贴该变量的所有数据和所有数 据定义属性。 • 在“变量视图”中复制和粘贴变量定义属性或整个变量将粘贴所选属性（或整个变量定义），但不粘贴任何 数据值。 重命名数据集 通过菜单和对话框打开数据源时，向每个数据源都自动分配 DataSetn 的数据集名，其中 n 是连续整数值； 而使用命令语法打开数据源时，不会分配数据集名，除非用 DATASET NAME 显式地指定名称。 提供更具描 述性的数据集名： 1. 从要更改名称的数据集的“数据编辑器”窗口中的菜单，选择： 文件 > 重命名数据集...
2. 输入符合变量命名规则的新数据集名。 有关更多信息，请参阅第44 页的『变量名称』主题。 不显示多个数据集 如果您希望一次只提供一个数据集，并取消多个数据集功能： 1. 从菜单中选择： 编辑 > 选项...
2. 单击常规选项卡。 选择（选中）一次只能打开一个数据集。 请参阅主题第163 页的『一般选项』，了解更多信息。 58 IBM SPSS Statistics 29 Core System 用户指南 第 7 章 数据准备 一旦您打开了某个数据文件，或在数据编辑器中输入了数据，您就可以开始创建报表、图表和分析，而无需 任何附加的预备工作。 不过，您可能会觉得某些附加的数据准备功能很有用，包括可以： • 指定可以描述数据和确定应如何处理特定值的变量属性。 • 确定可能包含重复信息的案例，并从分析中排除这些案例，或从数据文件中删除这些案例。 • 创建具有少数不同类别的新变量，这些类别表示具有大量可能值的变量的值范围。 变量属性 在数据编辑器的数据视图中输入的数据或从外部文件格式（例如，Excel 电子表格或文本数据文件）读入到 的数据，缺乏某些您可能觉得非常有用的变量属性，包括： • 数值代码的描述值标签的定义（例如：0 = Male 和 1 = Female）。 • 缺失值代码的标识（例如：99 = Not applicable）。 • 指定测量级别（名义、有序或刻度）。 所有这些变量属性（及其他属性）均可以在数据编辑器的“变量视图”中指定。 还有多个实用程序可在此过程 中对您有所帮助： • 定义变量属性可帮助您定义描述值标签和缺失值。 对于带有用作类别值的数值代码的分类数据，这就特别 有用。 请参阅主题第59 页的『定义变量属性』，了解更多信息。 • 设置未知测量级别标识未定义测量级别的变量（字段），并能够为这些变量设置测量级别。 这对于测量级 别能够影响结果或决定可用功能的过程非常重要。 请参阅主题第61 页的『为测量级别未知的变量设置测 量级别』，了解更多信息。 • 复制数据属性使您能够使用现有的 IBM SPSS Statistics 数据文件作为当前数据文件中的文件和变量属性的 模板。 如果您经常使用包含相似内容的外部格式数据文件（如 Excel 格式的月度报表），这就特别有用。 请参阅主题第63 页的『复制数据属性』，了解更多信息。 定义变量属性 “定义变量属性”在为变量指定属性，包括为分类（名义、有序）变量创建描述性值标签的过程中为您提供帮 助。 定义变量属性： • 扫描实际数据值并为每个选定的变量列出所有的唯一数据值。 • 识别未标注的值并提供“自动标注”功能。 • 提供将已定义的值标签和其他属性从另一变量复制到选定的变量，或是从选定的变量复制到多个其他变量 的功能。 注：要在不先扫描个案的情况下使用“定义变量属性”，那么对要扫描的个案数输入 0。 定义变量属性 1. 从菜单中选择： 数据 > 定义变量属性...
2. 选择要为其创建值标签或要定义或更改其他变量属性（例如缺失值或描述性变量标签）的数值或字符串 变量。 3. 指定要进行扫描以生成唯一值列表的个案数。 这对具有大量个案的数据文件特别有用，因为这种情况下 扫描完整数据文件可能需要很长时间。 4. 指定要显示的唯一值数目的上限。 这主要为了防止出现为标度（连续区间、比率）变量列出上百、上千 甚至上百万个值的情况。 5. 单击继续以打开“定义变量属性”主对话框。 6. 选择要为其创建值标签或要定义或更改其他变量属性的变量。 7. 为在“值标签”网格中显示的任何未标记的值输入标签文本。 8. 如果存在要为其创建值标签的值，但是并未显示这些值，可在位于最后一个扫描值之下的值列中输入 值。 9. 对要为其创建值标签的每个列出的变量重复该过程。 10. 单击确定以应用值标签和其他变量属性。 定义值标签和其他变量属性 “定义变量属性”主对话框提供已扫描变量的下列信息： 已扫描的变量列表。 对每一个已扫描的变量，未标记 (U.) 列中的复选标记指示该变量包含未指定值标签的 值。 要对变量列表排序以在列表顶部显示所有含未标记值的变量： 1. 单击“已扫描的变量列表”下的未标记列标题。 您也可单击“已扫描的变量列表”下相应的列标题，按变量名称和测量级别进行排序。 值标签网格 • 标签 。 显示任何已定义的值标签。 您可在该列添加或更改标签。 • 值。 每个选定的变量的唯一值。 该唯一值列表基于已扫描的个案数。 例如，如果您仅扫描了数据文件中 的前 100 个个案，那么列表仅反映这些个案中的唯一值。 如果数据文件已按变量排序，而您想要为该变量 指定值标签，列表中显示的唯一值可能远远少于数据中实际有的唯一值。 • 计数。 每一个值出现在已扫描个案中的次数。 • 缺失。 定义为代表缺失数据的值。 您可以单击复选框来更改类别的缺失值指定。 复选标记指示该类别已 被定义为用户缺失类别。 如果某个变量具有定义为用户缺失值的值范围（例如，90-99），那么不能使用 “定义变量属性”为该变量添加或删除缺失值类别。 您可使用数据编辑器中的“变量视图”为具有缺失值范围 的变量修改缺失值类别。 有关更多信息，请参阅第47 页的『缺失值』主题。 • 已更改。 指示您已添加或更改值标签。 注：如果在初始对话框中指定要扫描的个案数为 0，除预先存在的值标签和/或为选定变量定义的缺失值类别 之外，“值标签”网格初始情况下将为空白。 此外，测量级别的建议按钮将被禁用。 测量级别。 值标签主要对分类（标定和有序）变量很有用，某些过程区别对待分类变量和刻度变量；因此， 指定正确的测量级别有时很重要。 但是，在缺省情况下，所有新的数字变量都将被指定该刻度测量级别。 所以，许多事实上为分类变量的变量可能最初显示为刻度变量。 如果您不确定要将哪一种测量级别指定给变量，单击建议。 角色。 某些对话框支持根据定义的角色预先选择用于分析的变量。 有关更多信息，请参阅第47 页的『角 色』主题。 复制属性。 您可以将值标签和其他变量属性从另一变量复制到当前选定的变量，或是从当前选定的变量复制 到一个或多个其他变量。 未标记的值。 要为未标记的值自动创建标签，请单击自动标记。 变量标签和显示格式 您可更改描述性变量标签和显示格式。 • 您不能更改变量的基本类型（字符串或数值）。 • 对字符串变量，您只可更改变量标签，而不能更改显示格式。 • 对数字变量，您可以更改数值类型（例如数值、日期、美元或定制货币）、宽度（最大位数，包括小数和/ 或分组指示符）和小数位数。 • 对数值日期格式，您可选择一个特定日期格式（例如 dd-mm-yyyy、mm/dd/yy 和 yyyyddd） • 对数值定制格式，您可选择五个定制货币格式中的一个（CCA 到 CCE）。 有关更多信息，请参阅第167 页的『货币选项』主题。 60 IBM SPSS Statistics 29 Core System 用户指南 • 如果指定宽度小于预先存在的已定义值标签或缺失值类别的扫描值或显示值的宽度，在值列中将显示星 号。 • 预先存在的已定义值标签或缺失值类别的扫描值或显示值对选定的显示格式类型无效，则显示一个句点 (.)。 例如，小于 86,400 的内部数值对日期格式变量是无效的。 指定测量级别 当您对“定义变量属性”主对话框中的测量级别单击建议时，将基于已扫描的个案和定义的值标签计算当前变 量，打开的“建议测量级别”对话框将建议一个测量级别。 “说明”区域提供了用于提供测量级别建议的标准的 简要描述。 注：测量级别计算中不包含定义为代表缺失值的值。 例如，测量级别建议的说明可能表示该建议的部分依据 是变量不包含负值这一事实，然而，事实上变量可能包含负值，但是这些负值已被定义为缺失值。 1. 单击继续以接受建议的测量级别，或单击取消，不更改测量级别。 定制变量属性 使用“定义变量属性”中的属性按钮可打开“定制变量属性”对话框。 除标准变量属性（例如，值标签、缺失值 和测量级别）外，还可以创建自己的定制变量属性。 与标准变量属性一样，这些定制变量属性也随 IBM SPSS Statistics 数据文件一同保存。 名称。 属性名必须遵循与变量名称相同的规则。 有关更多信息，请参阅第44 页的『变量名称』主题。 值。 指定给选定变量的属性的值。 • 以美元符号开头的属性名是保留名称，不能修改这些名称。 您可以单击所需单元格中的按钮以查看保留属 性的内容。 • 在“值”单元格中显示的数组... 表示此属性是属性数组，即包含多个值的属性。 单击单元格中的按钮以显示 值列表。 复制变量属性 当您单击“定义变量属性”主对话框中的从另一个变量或到其他变量时，将显示“应用标签和水平”对话框。 将 显示与当前变量类型（数值或字符串）匹配的所有已扫描的变量。 对字符串变量，定义的宽度也必须匹配。 1. 选择要从中复制值标签和其他变量属性（变量标签除外）的单个变量。 或 2. 选择要将值标签和其他变量属性复制到其中的一个或多个变量。 3. 单击复制以复制值标签和测量级别。 • 此操作不替换目标变量的现有值标签和缺失值类别。 • 没有为目标变量定义的值标签和缺失值类别将被添加至目标变量的值标签和缺失值类别集中。 • 此操作总是替换目标变量的测量级别。 • 此操作总是替换目标变量的角色。 • 如果源或目标变量中的任何一个含有定义的缺失值范围，那么将不复制缺失值定义。 为测量级别未知的变量设置测量级别 对于某些过程，测量级别可能影响结果，或决定哪些功能可用，并且您将无法访问这些过程的对话框，除非 所有变量均已定义测量级别。 “设置未知测量级别”对话框允许您为任何测量级别未知的变量定义测量级别， 而无需执行数据传递（对于大型数据文件这可能非常耗时）。 在特定情况下，文件中的某些或全部数字变量（字段）的测量级别可能未知。 这些情况包括： • 在首次数据传递之前，来自 Excel 95 或更高版本文件、文本数据文件或数据库源的数值变量。 • 在这些变量创建之后但在首次数据传递之前，通过转换命令创建的新数值变量。 第 7 章 数据准备 61 这些情况主要适用于通过命令语法读取数据或创建新变量。 读取数据和创建新转换变量的对话框会自动执行 数据传递，以根据缺省测量级别规则来设置测量级别。 为测量级别未知的变量设置测量级别 1. 在为过程显示的警报对话框中，单击手动指定。 或 2. 从菜单中选择： 数据 > 设置未知的测量级别 3. 将变量（字段）从源列表移动到适当的测量级别目标列表。 • 名义 (Nominal). 当变量的值表示没有内部排名的类别 (例如，员工工作所在公司的部门) 时，可将该变量视 为名义变量。 名义变量的示例包括地区、邮政编码和宗教信仰。 • 有序 (Ordinal). 当变量的值表示具有某种内在等级的类别 (例如，从高度不满意到高度满意的服务满意度级 别) 时，可将该变量视为有序变量。 有序变量的示例包括表示满意度或可信度的态度分数和优先选择评 分。 • 连续。当变量的值表示具有有意义度量的有序类别时，可以将该变量视为刻度 (连续) ，以便在值之间进行 适当的距离比较。 刻度变量的示例包括以年为单位的年龄和以千美元为单位的收入。 多重响应集(R) 定制表和图表构建器支持一种称为多响应集的特殊“变量”。 多响应集不是通常意义上真正的“变量”。 多响应 集不显示在数据编辑器中，也不能由其他过程识别。 多响应集使用多个变量记录对问题的答复，其中响应者 可以给出多个答案。 多响应集以与分类变量相同的方式对待，可以对分类变量执行的大多数操作也可以对多 响应集执行。 多响应集根据数据文件中的多个变量进行构造。 多响应集是数据文件中的特殊结构。 可以在 IBM SPSS Statistics 数据文件中定义和保存多响应集，但不能从其他文件格式导入或向其导出多响应集。 可以使用“复 制数据属性”从其他 IBM SPSS Statistics 数据文件复制多响应集，从“数据编辑器”窗口的“数据”菜单可以访 问该功能。 主题以获取更多信息。 定义多响应集 定义多响应集： 1. 从菜单中选择： 数据 > 定义多个响应集...
2. 选择两个或更多个变量。 如果变量以二分法进行编码，那么请指明要计数的值。 3. 为每个多响应集输入唯一名称。 名称最多可以包含 63 个字节。 将自动向集名称的开头添加一个美元符 号。 4. 为集合输入描述性标签。 （此为可选。） 5. 单击添加将多响应集添加到定义的集的列表中。 二分法 多二分集通常包含多个二分变量：仅具有两个可能值（是/否、存在/不存在、选中/未选中性质）的变量。 尽 管变量可能不是严格二分的，但集中的所有变量都以相同方式进行编码，且“已计算的值”表示肯定/存在/选 中条件。 例如，有这样一项问卷调查：“您的生活是否离不开新闻?”，并提供了五个可能的答复。 响应者可以通过选 中每个选择旁的框来进行多项选择。 五个回答成为数据文件中的五个变量，代码 0 表示否（未选中），代码 1 表示是（选中）。 在多二分集中，“已计算的值”为 1。 样本数据文件 survey_sample.sav 已定义了三个多重响应集。$mltnews 是多二分集。 主题。 1. 选择（单击）多响应集列表中的 $mltnews 。 此操作会显示用于定义此多响应集的变量和设置。 62 IBM SPSS Statistics 29 Core System 用户指南 • “集合中的变量”列表显示用于构成多响应集的五个变量。 • “变量编码”组表示变量是二分的。 • “已计算的值”为 1。 2. 在“集合中的变量”列表中选择（单击）其中一个变量。 3. 右键单击该变量并从弹出菜单中选择变量信息。 4. 在“变量信息”窗口中，单击“值标签”下拉列表上的箭头，以显示已定义的值标签的完整列表。 值标签指明变量是二分的，值 0 和 1 分别表示 No 和 Yes。 列表中的所有五个值都以相同方式进行编码，值 1（Yes 的代码）是多二分集的已计算的值。 类别 多类别集由多个变量组成，所有这些变量都以相同方式进行编码，通常具有许多可能的响应类别。 例如，某 个调查项目为“请列举最能描述您的种族血统的民族，最多三个”。 可能有上百种回答，但为了进行编码，列 表限制为 40 个最常见的民族，任何其他回答都归为“其他”类别。 在数据文件中，三种选择成为三个变量， 每个变量有 41 个类别（40 个已编码的民族和一个“其他”类别）。 在样本数据文件中，$ethmult 和 $mltcars 是多类别集。 类别标签源 对于多个二分，可以控制标记集合的方式。 • 变量标签。 使用已定义的变量标签（或不带已定义变量标签的变量的变量名称）作为集类别标签。 例如， 如果集中的所有变量对于已计算的值（例如，Yes）具有相同的值标签（或未定义值标签），那么应使用变 量标签作为集类别标签。 • 已计算的值的标签。 使用已计算的值的已定义值标签作为集类别标签。 只有在所有变量对于已计算的值都 定义了值标签，且已计算的值的值标签对于每个变量都不相同时选择此选项。 • 使用变量标签作为集标签。 如果选择已计算的值的标签，那么也可以使用集中具有已定义变量标签的第一 个变量的变量标签作为集标签。 如果集中的变量都未定义变量标签，那么将集中第一个变量的名称作为集 标签。 复制数据属性 复制数据属性 通过复制数据属性向导可以使用外部 IBM SPSS Statistics 数据文件作为模板来定义活动数据集中的文件和变 量属性。 您也可以将活动数据集中的变量用作活动数据集中其他变量的模板。 您可以执行以下操作： • 将所选文件属性从外部数据文件或打开的数据集复制到活动数据集。 文件属性包括文档、文件标签、多响 应集、变量集和加权。 • 将所选变量属性从外部数据文件或打开的数据集复制到活动数据集中匹配的变量。 变量属性包括值标签、 缺失值、测量级别、变量标签、打印和书写格式、对齐方式和列宽（在数据编辑器中）。 • 将所选变量属性从外部数据文件、打开的数据集或活动数据集中的一个变量复制到活动数据集中的多个变 量。 • 根据外部数据文件或打开的数据集中的所选变量，在活动数据集中创建新变量。 在复制数据属性时，适用以下一般规则： • 如果您使用外部数据文件作为源数据文件，那么该文件必须为 IBM SPSS Statistics 格式的数据文件。 • 如果使用活动数据集作为源数据文件，则其必须包含至少一个变量。 不能使用完全空白的活动数据集作为 源数据文件。 • 源数据集中的未定义（空）属性不会覆盖活动数据集中的已定义属性。 • 源变量中的变量属性将仅复制到类型匹配的目标变量 -- 字符串（字母数值）或数值（包括数值、日期和货 币）。 注：“复制数据属性”替换了以前在“文件”菜单上提供的“应用数据字典”。 第 7 章 数据准备 63 复制数据属性 1. 从数据编辑器窗口的菜单中选择： 数据 > 复制数据属性...
2. 选择具有您想要复制的文件和/或变量属性的数据文件。 这可以是当前打开的数据集、外部 IBM SPSS Statistics 数据文件或活动数据集。 3. 按照复制数据属性向导中的逐步操作说明进行操作。 选择源变量和目标变量 在这一步中，您可以指定包含想要复制的变量属性的源变量，以及将接收这些变量属性的目标变量。 将所选源数据集变量的属性应用于匹配的活动数据集变量。 将变量属性从一个或多个所选源变量复制到活动 数据集中匹配的变量。 如果变量名称和类型（字符串或数值）都相同，那么变量是“匹配的”。 对于字符串 变量，定义的长度也必须相同。 缺省情况下，在两个变量列表中仅显示匹配的变量。 • 如果尚不存在匹配变量，那么在活动数据集中创建匹配变量。 这将更新源列表以显示源数据文件中的所有 变量。 如果活动数据集中没有所选的源变量（基于变量名称），那么将在活动数据集中创建新变量，变量 名称和属性都出自源数据文件。 如果活动数据集不包含任何变量（空白的新数据集），则显示源数据文件中的所有变量，并基于所选源变量 在活动数据集中自动创建新变量。 选择源列表中将复制属性的一个变量以及目标列表中将应用属性的一个或多个变量。 可以将源列表中的单个 所选变量中的变量属性应用于活动数据集列表中的一个或多个所选变量。 活动数据集列表中仅示与源列表中 所选变量类型相同（数值或字符串）的变量。 对于字符串变量，仅显示定义长度与源变量相同的字符串。 如果活动数据集不包含变量，那么此选项不可用。 注：不能使用此选项在活动数据集中创建新变量。 仅应用数据集属性 -- 未选定变量。 仅将文件属性（例如：文档、文件标签、权重）应用于活动数据集。 将 不应用任何变量属性。 如果活动数据集同时也是源数据文件，那么此选项不可用。 选择要复制的变量属性 您可以将选定的变量属性从源变量复制到目标变量。 源变量中未定义的（空）属性不会覆盖目标变量中的已 定义属性。 值标签。 值标签是与数据值相关联的描述性标签。 当数据值用于代表非数值类别时（例如：代码 1 和 2 代 表 Male 和 Female），常常使用值标签。 您可以在目标变量中替换或合并值标签。 • 替换删除目标变量的任何已定义的值标签，并用源变量中定义的值标签替换它们。 • 合并将源变量中的已定义值标签与目标变量的任何现有已定义值标签进行合并。 如果相同的值在源变量和 目标变量中都有一个已定义值标签，那么目标变量中的值标签保持不变。 定制属性。 用户定义的定制变量属性。 有关更多信息，请参阅第49 页的『定制变量属性』主题。 • 替换删除目标变量的任何定制属性，并用源变量中定义的属性替换它们。 • 合并将源变量中定义的属性与目标变量的任何现有已定义属性进行合并。 缺失值。 缺失值是被标识为代表缺失数据的值（例如：98 代表 Do not know，99 代表 Not applicable）。 通常，这些值也具有已定义的值标签，描述缺失值代码代表的含义。 删除目标变量的所有现有的已定义缺失 值，并用源变量中的已定义缺失值替换它们。 变量标签。 描述性变量标签可以包含空格和变量名称中所不允许的保留字符。 如果正在将单个源变量中的 变量属性复制到多个目标变量，那么在选择此选项前可能需要三思。 测量级别。 测量级别可以是名义、有序或刻度。 角色。 某些对话框支持基于定义的角色预先选择分析变量的功能。 有关更多信息，请参阅第47 页的『角 色』主题。 格式。 对于数字变量，此选项控制数值类型（如数值、日期或货币）、宽度（显示字符的总数，包括前导和 拖尾字符以及小数指示符）和显示的小数位数。 对于字符串变量，忽略此选项。 64 IBM SPSS Statistics 29 Core System 用户指南 对齐。 此选项仅影响数据编辑器的“数据视图”中的对齐方式（左对齐、右对齐、居中）。 数据编辑器列宽度。 此选项仅影响数据编辑器的“数据视图”中的列宽。 复制数据集（文件）属性 可以将源数据文件中所选的全局数据集属性应用于活动数据集。 （如果活动数据集是源数据文件，则此选项 不可用。） 多响应集。 将源数据文件中的多响应集定义应用于活动数据集。 • 忽略在活动数据集中没有包含变量的多响应集，除非将根据“复制数据属性向导”的第 2 步（选择源变量和 目标变量）中的规范来创建这些变量。 • 替换删除活动数据集中的所有多响应集，并用源数据文件中的多响应集替换它们。 • 合并将源数据文件中的多响应集添加到活动数据集中多响应集的集合中。 如果在两个文件中都存在相同名 称的集合，那么活动数据集中的现有集合保持不变。 变量集。 变量集用于控制在对话框中显示的变量列表。 变量集是通过从“实用程序”菜单中选择“定义变量集” 来定义的。 • 忽略源数据文件中包含活动数据集中没有的变量的集合，除非将根据复制数据属性向导的第 2 步（选择源 变量和目标变量）中的规范来创建这些变量。 • 替换删除活动数据集中任何现有变量集，并用源数据文件中的变量集替换它们。 • 合并将源数据文件中的变量集添加到活动数据集中变量集的集合中。 如果在两个文件中都存在相同名称的 集合，那么活动数据集中的现有集合保持不变。 文档。 通过 DOCUMENT 命令追加到数据文件中的注释。 • 替换删除活动数据集中的任何现有文档，并用源数据文件中的文档替换它们。 • 合并可用于合并源数据集和活动数据集中的文档。 将在活动数据集中没有的源文件中的唯一文档添加到活 动数据集。 然后对所有文档按日期排序。 定制属性。 定制数据文件属性，通常使用命令语法中的 DATAFILE ATTRIBUTE 命令创建这些属性。 • 替换删除活动数据集中的任何现有定制数据文件属性，并用源数据文件中的数据文件属性替换它们。 • 合并可用于合并源数据集和活动数据集中的数据文件属性。 将活动数据集中没有的源文件中的唯一属性名 添加到活动数据集。 如果在两个数据文件中都存在相同的属性名，则活动数据集中的已命名属性保持不 变。 权重指定。 如果活动数据集中有匹配的变量，那么按源数据文件中的当前权重变量对个案进行加权。 这会 覆盖活动数据集中当前生效的任何加权。 文件标签。 通过 FILE LABEL 命令向数据文件应用的描述性标签。 结果 复制数据属性向导中的最后一步提供如下信息：变量数目（将从源数据文件中复制这些变量的变量属性）、 将创建的新变量的数目和将复制的数据集（文件）属性的数目。 也可以选择将所生成的命令语法粘贴到语法窗口中，并保存该语法以供以后使用。 标识重复个案 在数据中出现“重复”个案有多种原因，包括： • 数据输入错误，意外地多次输入了同一个案。 • 多个个案具有相同的主标识值，但它们有不同的次标识值，就像居住在同一间屋子的多个家庭成员。 • 多个个案代表同一个案，但是对于除标识该个案的变量之外的其他变量有不同值，例如由同一个人或公司 在不同时间购买的不同产品。 “标识重复个案”允许您自由定义重复，并在一定程度上控制对主个案和重复个案的自动确定。 标识和标记重复个案 第 7 章 数据准备 65 1. 从菜单中选择： 数据 > 识别重复案例...
2. 选择标识匹配个案的一个或多个变量。 3. 在“要创建的变量”组中选择一个或多个选项。 （可选）您可以执行以下操作： 4. 选择一个或多个变量对由选定的匹配个案变量定义的组内的个案排序。 由这些变量定义的排序顺序决定 了每一组中的“第一个”和“最后一个”个案。 否则，将使用原文件顺序。 5. 自动过滤重复个案使其不包含在报告、图表或统计计算中。 定义匹配个案的依据。 如果个案的值匹配所有选定的变量，则它们视为是重复个案。 如果您只想标识在所 有方面 100% 匹配的个案，请选中所有变量。 在匹配组内的排序标准。 将按定义匹配个案的变量自动对个案排序。 您可以选择将决定每一匹配组中个案 的顺序的其他排序变量。 • 对每个排序变量，您可按升序或降序排序。 • 如果您选择多个排序变量，将按列表中前一个变量类别中的每个变量对个案进行排序。 例如，如果您选择 日期作为第一排序变量，数量作为第二排序变量，则在每一日期中将按数量对个案排序。 • 使用列表右边的向上和向下箭头按钮更改变量的排序顺序。 • 排序顺序决定了每个匹配组内的“第一个”和“最后一个”个案，这些个案将决定可选的主指示符变量的值。 例如，如果您想要过滤每一匹配组中除最近个案之外的所有个案，您可以按日期变量的升序对组内个案排 序，这将使得组中最近的日期排在最后。 基本个案指示符。 为所有唯一个案和每一匹配个案组中标识为主个案的个案创建值为 1 的变量，为每一组中 非主要的重复个案创建值为 0 的变量。 • 由匹配组内的排序顺序确定，主个案可为每一匹配组中的最后一个或是第一个个案。 如果未指定任何排序 变量，原文件顺序将决定每一组内的个案顺序。 • 您可将指示符变量作为过滤变量，在报告和分析中排除非主要的重复个案，而不从数据文件中删除这些个 案。 每组中匹配个案的连续计数。 在每一匹配组中为个案创建序列值为 1 到 n 的变量。 该序列基于每一组中当 前个案的顺序，可以是原文件顺序，也可以是由任何指定的排序变量决定的顺序。 将匹配个案移至文件顶端。 对数据文件排序使所有匹配个案组位于数据文件顶部，这样方便在数据编辑器中 查看匹配个案。 显示已创建变量的显示频率。 包含已创建变量的每一个值的计数的频率表。 例如，对主指示符变量，表中 显示变量值为 0 的个案数（表示重复个案数）和变量值为 1 的个案数（表示唯一主个案数）。 缺失值 对于数值变量，会将系统缺失值视为与任何其他值一样，即，将具有标识变量的系统缺失值的个案视为具有 该变量的匹配值的个案。 对字符串变量，标识变量没有值的个案将作为该变量具有匹配值的个案。 过滤后的案例 忽略过滤条件。 在重复个案的评估中包含已过滤的个案。 如果要排除个案，请使用数据 > 选择个案并选择 删除未选定个案来定义选择规则。 可视分箱化 可视分箱化设计的目的在于，帮助您在将现有变量的连续值进行分组的基础上，将新变量创建到数目有限的 不同类别中。 可以将可视分箱化用于： • 从连续刻度变量创建分类变量。 例如，您可以使用刻度收入变量创建包含收入范围的新的分类变量。 • 将大量有序类别拼并到一小组类别中。 例如，您可以将具有 9 个级别的等级标度拼并为分别代表低、中、 高的三个类别。 66 IBM SPSS Statistics 29 Core System 用户指南 在第一步中： 1. 选择要为其创建新分类（分箱化）变量的数值标度和/或序数变量。 您可以选择限制要扫描的个案数。 对于含有大量个案的数据文件，限制扫描个案的数目可以节省时间，但您 应尽可能避免限制此数目，因为这会影响可视分箱化中的后续计算使用的值的分布。 注：字符串变量和标定数字变量不显示在源变量列表中。 可视分箱化需要按标度或序数级别度量的数字变 量，因为它假定数据值代表某种逻辑顺序，该顺序可用于以有意义的方式对值分组。 您可以在数据编辑器的 变量视图中更改变量的定义测量级别。 请参阅第45 页的『变量测量级别』主题以获取更多信息。 分箱化变量 1. 从数据编辑器窗口的菜单中选择： 转换 > 可视分箱...
2. 选择要为其创建新分类（分箱化）变量的数值刻度和/或序数变量。 3. 在“已扫描的变量”列表中选择一个变量。 4. 为新的分箱化变量输入一个名称。 变量名称必须是唯一的，并且必须遵循变量命名规则。 请参阅主题第 44 页的『变量名称』，了解更多信息。 5. 为新变量定义分箱化标准。 请参阅主题第67 页的『分箱化变量』，了解更多信息。 6. 单击确定。 分箱化变量 可视分箱化主对话框提供已扫描变量的以下信息： 已扫描变量列表(C) 显示您在初始对话框中选择的变量。 通过单击列标题，您可以按测量级别（标度或有序）、变量标签或 名称对列表排序。 扫描个案数 指示已扫描个案的数目。 将使用选定变量的所有不带用户缺失值或系统缺失值的已扫描个案，来生成在 可视分箱化的计算中使用的值的分布，包括在主对话框中显示的直方图，以及基于百分位或标准差单位 的分割点。 缺失值 指示带有用户缺失值或系统缺失值的已扫描个案的数目。 缺失值不包含在任何分箱化类别中。有关更多 信息，请参阅主题 第70 页的『可视分箱化中的用户缺失值』。 当前变量 当前选定变量的名称和变量标签（如果有），该变量将用作新的分箱化变量的基础。 分箱化变量 新的分箱化变量的名称和可选的变量标签。 名称 必须为新变量输入一个名称。 变量名称必须是唯一的，并且必须遵循变量命名规则。有关更多信 息，请参阅主题 第44 页的『变量名称』。 标签(L) 您可以输入长度不超过 255 个字符的描述性变量标签。 缺省的变量标签是标签末尾附有（分箱化） 的源变量的变量名称或变量标签（如果有）。 最小值和最大值 当前选定变量的最小值和最大值，它基于已扫描的个案，并且不包括定义为用户缺失的值。 非缺失值 直方图根据扫描的个案显示当前所选变量的非缺失值的分布。 • 在为新变量定义了分箱之后，会在直方图上显示垂直线表示定义分箱的分割点。 • 您可以单击分割点线并拖到直方图上的其他位置，以更改分箱范围。 • 您还可以通过将分割点线拖到直方图外而移去分箱。 第 7 章 数据准备 67 注: 直方图（显示非缺失值）、最小值和最大值基于已扫描的值。 如果不在扫描中包含所有个案，就可能不 能精确地反映真实的分布，尤其是对于数据文件已按选定变量排序的情况。 如果不扫描任何个案，那么不提 供任何关于值的分布的信息。 网格 显示定义每个分箱的上端点的值，以及每个分箱的可选值标签。 值 定义每个分箱的上端点的值。 您可以输入值，或者使用生成分割点以基于选定的标准自动创建分 箱。 缺省情况下，会自动包含值为 HIGH 的分割点。 此分箱将包含其他分割点以上的所有非缺失 值。 由最低分割点定义的分箱将包含所有低于或等于该值的非缺失值（或者只包含低于该值的非缺 失值，这取决于您如何定义上端点）。 标签(L) 新的分箱化变量的值的可选描述性标签。 由于新变量的值只是从 1 到 n 的连续整数，所以描述值所 代表的内容的标签非常有用。 您可以输入标签，或者使用生成标签以自动创建值标签。 从网格中删除分箱 1. 右键单击分箱的值或标签单元格。 2. 从弹出菜单中，选择删除行。 注: 如果删除 HIGH 分箱，那么将为任何值高于最后一个指定的分割点值的个案分配新变量的系统缺失值。 删除所有标签或所有已定义的分箱 1. 右键单击网格中的任意位置。 2. 从弹出菜单中，选择删除所有标签或删除所有分割点。 上端点 控制在网格的值列输入的上端点值的处理。 包含 (<=) 具有在值单元格中指定的值的个案包含在分箱化类别中。 例如，如果您指定值 25、50 和 75，那么 值正好为 25 的个案将进入第一个分箱，因为它将包含值小于等于 25 的所有个案。 已排除 (<) 具有在值单元格中指定的值的个案不包含在分箱化类别中。 而是包含在下一个分箱中。 例如，如果 您指定值 25、50 和 75，那么值正好为 25 的个案将进入第二个分箱而非第一个分箱，因为第一个分 箱将仅包含值小于 25 的个案。 生成分割点 自动为等宽度间隔、具有相等个案数的区间或者基于标准差的区间生成分箱化类别。如果扫描的个案为 零，那么此值不可用。 生成标签(A) 基于网格中的值和指定的上端点的处理（包含或排除），为新的分箱化变量的连续整数值生成描述性标 签。 反转刻度(S) 缺省情况下，新分箱化变量的值为从 1 到 n 的升序整数。 反转刻度使值从 n 整数递减到 1。 复制分箱 您可以将分箱化规范从另一个变量复制到当前所选变量，或从所选变量复制到多个其他变量。 自动生成分箱化类别 “生成分割点”对话框允许您基于选定的标准自动生成分箱化类别。 使用“生成分割点”对话框 1. 在“已扫描的变量列表”中选择（单击）一个变量。 2. 单击生成分割点。 3. 选择用于生成将定义分箱化类别的分割点的标准。 68 IBM SPSS Statistics 29 Core System 用户指南 4. 单击应用。 注：如果未扫描到任何个案，那么“生成分割点”对话框不可用。 等宽度间隔。 根据以下三条标准中的任意两条，生成等宽（例如 1 - 10、11 - 20 和 21 - 30）的分箱化类 别： • 第一个分割点的位置。 定义最下面的分箱化类别的上端点的值（例如，值 10 表示包含所有不超过 10 的 值的范围）。 • 分割点数量。 分箱化类别数是分割点数量加一。 例如，9 个分割点会生成 10 个分箱化类别。 • 宽度。 每个区间的宽度。 例如，值 10 会将年龄 分箱化为长度为 10 年的区间。 基于已扫描个案的等百分位。 基于以下标准之一，生成一些分箱化类别，使得每个分箱中的个案数相等（对 于百分位，使用 Aempirical 算法）： • 分割点数量。 分箱化类别数是分割点数量加一。 例如，三个分割点会生成四个百分位分箱（四分位数）， 每个分箱包含 25% 的个案。 • 宽度 (%)。每个时间间隔的宽度，被表示为个案总数的百分比。 例如，值 33.3 将生成三个分箱化类别 （两个分割点），每个类别包含 33.3% 的个案。 如果源变量包含的不同值相对较少，或者有大量个案具有相同的值，那么获取的分箱数可能少于请求的分箱 数。 如果在分割点处有多个相同的值，那么它们都将转到相同的区间；因此实际百分比并不总是完全相同 的。 基于已扫描个案的平均和选定标准差处的分割点。 基于变量分布的平均值和标准差的值生成分箱化类别。 • 如果不选择任何标准差区间，则将使用平均值作为分割点来划分分箱，从而创建两个分箱化类别。 • 您可以基于一倍、两倍和/或三倍标准差选择标准差区间的任意组合。 例如，选择所有三个标准差将生成 八个分箱化类别 -- 每个标准差区间内两个分箱（三个区间共六个分箱），平均值上下超过三倍标准差的个 案两个分箱。 在正态分布中，68% 的个案落在平均值的一倍标准差范围内；95% 的个案落在平均值的两倍标准差范围 内；99% 的个案落在平均值的三倍标准差范围内。 基于标准差创建分箱化类别可能导致一些定义的分箱落 在实际数据范围以外，甚至落在可能的数据值范围以外（例如，负的薪金范围）。 注：百分位和标准差的计算是基于已扫描个案的。 如果您限制已扫描个案的数目，则生成的分箱可能不包含 一部分您本想在分箱中包含的个案，尤其是对于数据文件按源变量排序的情况。 例如，如果您将扫描仅限于 具有 1000 个个案的数据文件的前 100 个个案，且该数据文件按响应者年龄升序排序，那么，您会发现前三 个分箱每个仅包含约 3.3% 的个案，而最后一个分箱包含 90% 的个案，而非四个百分位年龄分箱的每一个 均包含 25% 的个案。 复制分箱化类别 为一个或多个变量创建分箱化类别时，您可以将分箱化规范从另一个变量复制到当前选定变量，或者从选定 变量复制到多个其他变量。 复制分箱规范 1. 为至少一个变量定义分箱化类别，请勿 单击确定或粘贴。 2. 在您已为其定义了分箱化类别的“已扫描的变量列表”中，选择（单击）一个变量。 3. 单击到其他变量。 4. 选择您想要为其以相同的分箱化类别创建新变量的变量。 5. 单击复制。 或 6. 在您想要向其复制定义的分箱化类别的“已扫描的变量列表”中，选择（单击）一个变量。 7. 单击从另一个变量。 8. 选择您想要复制的具有定义的分箱化类别的变量。 9. 单击复制。 如果对于您从中复制分箱化规范的变量，有指定的值标签，那么也复制这些指定的值标签。 第 7 章 数据准备 69 注：在“可视分箱化”主对话框中单击确定新建分箱化变量（或者以任何其他方式关闭该对话框）后，就不能 使用“可视分箱化”将这些分箱化类别复制到其他变量。 可视分箱化中的用户缺失值 源变量中定义为用户缺失的值（标识为缺失数据代码的值）不包含在新变量的分箱化类别中。 源变量的用户 缺失值复制为新变量的用户缺失值，同时还复制缺失值代码的任何已定义值标签。 如果某个缺失值代码与新变量的其中一个分箱化类别值相冲突，则通过为最高分箱化类别值加上 100，将新 变量的缺失值代码重新编码为不冲突的值。 例如，如果值 1 定义为源变量的用户缺失值，且新变量将有六个 分箱化类别，则对于源变量值为 1 的任何个案，其新变量的值将为 106，且 106 将定义为用户缺失值。 如 果源变量的用户缺失值已有定义的值标签，则将保留该标签作为新变量的重新编码值的值标签。 注：如果源变量有定义的用户缺失值范围，其形式为 LO-n，其中 n 是正数，那么新变量相应的用户缺失值将 是负数。 70 IBM SPSS Statistics 29 Core System 用户指南 第 8 章 数据转换 数据转换 在理想情况下，原始数据非常适用于要执行分析的类型，并且，变量间的任何关系都是合适的线性或切合的 正交关系。 不幸的是，这种情况非常少。 初步分析可能会暴露出编码方案不方便或编码错误，或者可能需 要数据转换以揭示变量间的真实关系。 您可以执行从简单任务（比如拼并类别以进行分析）到更高级任务（比如基于复杂方程和条件语句创建新的 变量）的数据转换。 计算变量 使用“计算”对话框可以基于其他变量的数值转换来计算变量的值。 • 您可以计算数值或字符串（字母数值）变量的值。 • 您可以创建新的变量或替换现有变量的值。 对于新的变量，您也可以指定变量类型和标签。 • 您可以基于逻辑条件有选择的计算数据子集的值。 • 您可以使用多种内置函数，包括算术函数、统计函数、分布函数和字符串函数。 计算变量 1. 从菜单中选择： 转换 > 计算变量...
2. 输入一个目标变量的名称。 它可以是一个现有变量或是一个要添加到活动数据集中的新变量。 3. 要构建一个表达式，可以将成分粘贴到“表达式”字段中或是在“表达式”字段中直接输入。 • 通过从“函数组”列表中选择组，然后双击“函数和特殊变量”列表中的函数或变量（或选择函数或变量，然 后单击“函数组”列表相邻的箭头），可以粘贴函数或常用的系统变量。 填充问号指示的任何参数（仅适用 于函数）。 标记为所有的函数组提供所有可用函数和系统变量的列表。 对话框的保留区域中显示对当前所 选函数或变量的简要描述。 • 字符串常量必须包含在引号或撇号中。 • 如果值包含小数，则必须使用句号 (.) 作为小数指示符。 • 对于新字符串变量，还必须选择 类型和标签以指定数据类型。 • 在 过滤依据 字段中，可以输入搜索项以过滤功能。 如果选择 包含描述，那么搜索还会按功能描述进行过 滤。 计算变量：If 个案 “If 个案”对话框允许您使用条件表达式对选定的个案子集应用数据转换。 条件表达式对每个个案都返回一个 值：true、false 或 missing。 • 如果条件表达式的结果为 true，那么所选子集中将包含该个案。 • 如果条件表达式结果为 false 或 missing，则所选子集中不包含该个案。 • 大多数条件表达式在计算器键盘上使用一个或多个关系运算符（<、>、<=、>=、= 和 ~=）。 • 条件表达式可以包含变量名称、常数、算术运算符、数值（和其他）函数、逻辑变量以及关系运算符。 计算变量：类型和标签 缺省情况下，新的计算变量是数值变量。 要计算新的字符串变量，必须指定数据类型和宽度。 标签 。 描述性变量标签最多可以包含 255 个字节。 您可以输入标签或使用计算表达式的前 110 个字符作为 标签。 类型。 计算变量可以是数值或字符串（字母数值）。 字符串变量不能用于计算。 功能 支持很多类型的函数，包括： • 算术函数 • 统计函数 • 字符串函数 • 日期和时间函数 • 分布函数 • 随机变量函数 • 缺失值函数 • 评分函数 若要了解上述各函数的更多信息和详细说明，请在帮助系统的“索引”选项卡中键入函数。 函数中的缺失值 函数和简单算术表达式以不同的方式处理缺失值。 在以下表达式中： (var1+var2+var3)/3 如果一个个案里三个变量中的任何一个有缺失值，结果将缺失。 在以下表达式中： MEAN(var1, var2, var3) 只有当个案中三个变量都有缺失值的时候，结果才将缺失。 对于统计函数，您可以指定必须包含非缺失值的参数的最小数目。 为此，要在函数名称之后输入一个句点和 最小数目，例如： MEAN.2(var1, var2, var3) 随机数字生成器 使用“随机数生成器”对话框可以选择随机数生成器并设置起始序列值，以便重新生成随机数的序列。 活动生成器。 有两种不同的随机数字生成器可供使用： • 兼容的版本 12. 在 V12 和先前发行版中使用的随机数生成器。 如果您要重新生成在早期版本中根据指定的 种子值生成的随机结果，可使用此随机数字生成器。 • Mersenne 旋转 (Mersenne Twister). 较新的随机数生成器，用于模拟目的更可靠。 如果从版本 12 或早期版 本再现随机结果不是问题，那么可使用此随机数生成器。 活动生成器初始化。 每次生成随机数以用于转换（例如随机分布函数）、随机抽样或是个案加权时，随机数 种子都随之更改。 要复制随机数的序列，请先设置初始化起始点值，然后再进行使用这些随机数的每个分 析。 此值必须是正整数。 某些过程（例如线性模型）具有内部随机数字生成器。 选择随机数生成器和/或设置初始化值： 1. 从菜单中选择： 转换 > 随机数字生成器 72 IBM SPSS Statistics 29 Core System 用户指南 计算个案中值的出现次数 该对话框将创建一个变量，该变量统计每个个案的变量列表中相同值的出现次数。 例如，某调查可能包含一 个杂志列表，并使用是/否复选框来表示每个响应者阅读哪些杂志。 您可以计算响应者回答是的数目以创建 包含他所阅读的杂志总数的一个新变量。 计算个案内值的出现次数 1. 从菜单中选择： 转换 > 对个案内的值进行计数...
2. 输入一个目标变量名称。 3. 选择同类型（数值或字符串）的两个或更多变量。 4. 单击定义值并指定应计数的一个或多个值。 或者，可以定义要为其计算值出现次数的个案子集。 统计个案内的值: 要统计的值 每当选定变量之一与此处的“要计数的值”列表中指定的值匹配时，目标变量（主对话框中）的值就增加 1。 如果一个个案匹配任何变量的多个指定值，那么目标变量就为该变量递增多次。 指定的值可包括个别值、缺失值或系统缺失值和范围。 范围包括其端点和范围内的任何用户缺失值。 统计出现次数: If 个案 通过“If 个案”对话框可以使用条件表达式统计所选个案子集的值的出现次数。 条件表达式对每个个案都返回 一个值：true、false 或 missing。 转换值 转换值创建包含来自前面或后面个案的现有变量值的新变量。 名称。 新变量的名称。 这必须是活动数据集中尚未存在的名称。 从早期个案（延迟）获取值。 从活动数据集中前一个个案获取值。 例如，当缺省个案数是 1 时，新变量的 每个个案有来自前一个个案的原始变量的值。 从以下个案（提前）获取值。 从活动数据集中后一个个案获取值。 例如，当缺省个案数是 1 时，新变量的 每个个案有来自下一个个案的原始变量的值。 要转换的个案数。 从前面或后面第 n 个个案获取值，其中 n 是指定的值。 值必须是一个非负整数。 • 如果拆分文件处理正在进行，转换的范围限于每个拆分组。 无法从前面或后面拆分组中的个案获取转换 值。 • 忽略过滤状态。 • 结果变量的值被设置为数据集或拆分组中最开始或最后 n 个个案的系统缺失，其中 n 是为要转换的个案数 指定的值。 例如，使用值为 1 的延迟方法会将结果变量设置为数据集中第一个个案（或每个拆分组中第一 个个案）的系统缺失。 • 保留用户缺失值。 • 来自原始变量的字典信息（包括定义值标签和用户缺失值分配）被应用于新变量。 （注：不包括定制变量 属性。） • 对于描述创建变量的转换操作的新变量，自动生成变量标签。 创建具有转换值的新变量 1. 从菜单中选择： 转换 > 转换值 2. 选择变量，用作新变量的值源。 3. 输入新变量的名称。 第 8 章 数据转换 73 4. 选择转换方法（延迟或提前）和要转换的个案数。 5. 单击更改。 6. 为您想创建的每个新变量重复该操作。 对值重新编码 您可通过对值重新编码来修改数据值。 这对拼并类别或组合类别特别有用。 您可对现有变量中的值重新编 码，或基于现有变量的重新编码后的值创建新的变量。 重新编码到相同的变量中 使用“重新编码到相同的变量中”对话框可以重新指派现有变量的值或将现有值的范围拼并为新值。 例如，可 将薪金拼并到薪金范围类别中。 您可以对数字变量和字符串变量重新编码。 如果选择多个变量，则它们必须为相同类型。 不可将数字变量 和字符串变量放在一起重新编码。 对变量的值重新编码 1. 从菜单中选择： 转换 > 重新编码为相同变量...
2. 选择要重新编码的变量。 如果选择多个变量，则它们必须为相同类型（数值或字符串）。 3. 单击旧值和新值并指定如何对值重新编码。 （可选）可以定义要重新编码的案例子集。 执行此操作的“If 案例”对话框与“盘点出现次数”的描述相同。 重新编码为相同变量：旧值和新值 您可在该对话框中定义要重新编码的值。 所有指定的值必须与主对话框中选择的变量的数据类型（数值或字 符串）相同。 旧值。 要重新编码的值。 您可对单个值、值范围和缺失值重新编码。 不可为字符串变量选择系统缺失值和 范围，因为这两个概念都不适用于字符串变量。 范围包括其端点和范围内的任何用户缺失值。 • Value. 要重新编码为新值的单个旧值。 该值必须具有与所重新编码的变量相同的数据类型（数值或字符 串）。 • 系统缺失值。当未根据您指定的格式类型定义数据中的值时，当数字字段为空时，或者当未定义由转换命 令生成的值时，由程序指定的值。 数值型的系统缺失值显示为句号。 字符串变量不能具有系统缺失值，因 为任何字符在字符串变量中均是合法的。 • 系统或用户缺失值。含有用户缺失值定义、本身未知或由系统缺失值分配的值的观测值，使用句点 (.) 表 示。 • 范围 (Range). 值的包含范围。 不适用于字符串变量。 包含范围内的所有用户缺失值。 • 所有其他值。未包含在 "旧-新" 列表的其中一个规范中的任何剩余值。 这在“旧-新”列表上显示为“ELSE”。 新值。 由每个旧值或值范围重新编码获得的单个值。 您可输入一个值或指定系统缺失值。 • 值。一个或多个旧值将重新编码到其中的值。 该值必须具有与旧值相同的数据类型（数值或字符串）。 • 系统缺失值。将指定的旧值重新编码到系统缺失值中。 系统缺失值不用在计算中，并且许多过程中都排除 带系统缺失值的个案。 不适用于字符串变量。 旧–>新。 将用于对变量重新编码的指定值列表。 可以在列表中添加、更改和删除指定值。 列表将根据指定 的旧值按下列顺序自动排序：单值、缺失值、范围以及所有其他值。 如果更改列表中的重新编码指定，则过 程自动对列表重新排序（如果必要）以保持此顺序。 重新编码为不同变量 使用“重新编码为不同变量”对话框可以重新指派现有变量的值或将现有值的范围拼并为新变量的新值。 例 如，您可将薪金拼并到包含薪金范围类别的新变量中。 74 IBM SPSS Statistics 29 Core System 用户指南 • 您可以对数字变量和字符串变量重新编码。 • 您可以将数值变量重新编码为字符串变量，反之亦然。 • 如果选择多个变量，则它们必须为相同类型。 不可将数值变量和字符串变量放在一起重新编码。 将变量的值重新编码为新变量 1. 从菜单中选择： 转换 > 重新编码为不同变量...
2. 选择要重新编码的变量。 如果选择多个变量，则它们必须为相同类型（数值或字符串）。 3. 为每一新变量键入输出（新）变量名称并单击更改。 4. 单击旧值和新值并指定如何对值重新编码。 （可选）您可以定义要重新编码的案例子集。 执行此操作的“If 个案”对话框与“盘点出现次数”的描述相同。 重新编码为不同变量：旧值和新值 您可在该对话框中定义要重新编码的值。 旧值。 要重新编码的值。 您可对单个值、值范围和缺失值重新编码。 不可为字符串变量选择系统缺失值和 范围，因为这两个概念都不适用于字符串变量。 旧值必须与原变量是相同的数据类型（数值或字符串）。 范围包括其端点和范围内的任何用户缺失值。 • Value. 要重新编码为新值的单个旧值。 该值必须具有与所重新编码的变量相同的数据类型（数值或字符 串）。 • 系统缺失值。当未根据您指定的格式类型定义数据中的值时，当数字字段为空时，或者当未定义由转换命 令生成的值时，由程序指定的值。 数值型的系统缺失值显示为句号。 字符串变量不能具有系统缺失值，因 为任何字符在字符串变量中均是合法的。 • 系统或用户缺失值。含有用户缺失值定义、本身未知或由系统缺失值分配的值的观测值，使用句点 (.) 表 示。 • 范围 (Range). 值的包含范围。 不适用于字符串变量。 包含范围内的所有用户缺失值。 • 所有其他值。未包含在 "旧-新" 列表的其中一个规范中的任何剩余值。 这在“旧-新”列表上显示为“ELSE”。 新值。 由每个旧值或值范围重新编码获得的单个值。 新值可为数值或字符串。 • 值。一个或多个旧值将重新编码到其中的值。 该值必须具有与旧值相同的数据类型（数值或字符串）。 • 系统缺失值。将指定的旧值重新编码到系统缺失值中。 系统缺失值不用在计算中，并且许多过程中都排除 带系统缺失值的个案。 不适用于字符串变量。 • 复制旧值。保留旧值。 如果某些值不需要重新编码，那么使用此项以包含旧值。 任何未指定的旧值不包含 在新变量中，具有那些值的个案将分配新变量的系统缺失值。 输出变量为字符串 (Output variables are strings). 将重新编码的新变量定义为字符串 (字母数字) 变量。 旧变 量可能为数值型或字符串变量。 将数字字符串转换为数字。将包含数字的字符串值转换为数字值。 包含除数字和可选的符号（+ 或 -）以外 的任何字符的字符串会分配系统缺失值。 旧–>新。 将用于对变量重新编码的指定值列表。 可以在列表中添加、更改和删除指定值。 列表将根据指定 的旧值按下列顺序自动排序：单值、缺失值、范围以及所有其他值。 如果更改列表中的重新编码指定，则过 程自动对列表重新排序（如果必要）以保持此顺序。 自动重新编码 使用“自动重新编码”对话框可以将字符串值和数值转换为连续整数。 当类别代码不连续时，对许多过程来 说，生成的空单元格将降低性能并增加内存要求。 此外，某些过程不能使用字符串变量，某些过程要求因子 级别为连续的整数值。 • “自动重新编码”创建的新变量保留了旧变量中任何已定义的变量标签和值标签。 对没有已定义值标签的任 何值，将使用原值作为重新编码后的值的标签。 一个表显示了旧值、新值以及值标签。 第 8 章 数据转换 75 • 字符串值将按字母顺序重新编码，其中大写字母将排在相应的小写字母之前。 • 缺失值被重新编码为高于任何非缺失值的缺失值，并保留它们的原有顺序。 例如，如果原变量有 10 个非 缺失值，最低的缺失值将被重新编码为 11，值 11 将作为新变量的缺失值。 对所有变量使用相同的重新编码方案。 使用此选项可以将一个自动重新编码方案应用于所有选定变量，从而 为所有新变量生成一致的编码方案。 如果选择此选项，那么下列规则和限制适用： • 所有变量必须为相同类型（数值或字符串）。 • 所有选定变量的观测值用于创建要重新编码为连续整数的值的排序顺序。 • 新变量的用户缺失值基于具有已定义用户缺失值的列表中的第一个变量。 将来自其他原变量的所有其他值 （系统缺失值除外）都视为有效。 将空字符串值视为用户缺失值。 对于字符串变量，不将空白值或空值视为系统缺失值。 此选项会将空字符 串自动重新编码为高于最高非缺失值的用户缺失值。 模板 可以将自动重新编码方案保存在模板文件中，然后将其应用于其他变量和其他数据文件。 例如，每月可能有大量的字母数值产品代码要自动重新编码为整数，但在某些月份中添加了新产品代码，这 会更改原始的自动重新编码方案。 如果将原始设计保存在模板中，然后将其应用于包含一组新代码的新数 据，那么数据中遇到的任何新代码都会被自动重新编码为高于模板中最后一个值的值，从而保留了原始产品 代码的原始自动重新编码方案。 将模板另存为。 将所选变量的自动重新编码方案保存在外部模板文件中。 • 该模板包含将原始非缺失值映射到重新编码的值的信息。 • 模板中仅保存非缺失值的信息。 不保留用户缺失值的信息。 • 如果选择了多个要进行重新编码的变量，但尚未选择对所有变量使用相同的自动重新编码方案，或者不应 用现有模板作为自动重新编码的一部分，那么模板将基于列表中的第一个变量。 • 如果选择了多个要进行重新编码的变量，还选择了对所有变量使用相同的重新编码方案和/或选择了应用模 板，那么模板将包含用于所有变量的组合自动重新编码方案。 从文件应用模板。 将以前保存的自动重新编码模板应用于所选要进行重新编码的变量，从而将变量中的任何 其他值附加到设计的末尾，并保留所保存设计中存储的原始值和自动重新编码的值的关系。 • 所选要进行重新编码的所有变量必须是相同类型（数值或字符串），并且该类型必须与模板中定义的类型 匹配。 • 模板不包含有关用户缺失值的任何信息。 目标变量的用户缺失值基于具有已定义用户缺失值的原变量列表 中的第一个变量。 将来自其他原变量的所有其他值（系统缺失值除外）都视为有效。 • 首先应用来自模板的值映射。 将所有剩余的值重新编码为高于模板中最后一个值的值，同时将用户缺失值 （基于具有已定义用户缺失值的列表中的第一个变量）重新编码为高于最后一个有效值的值。 • 如果选择了多个要进行自动重新编码的变量，那么首先应用模板，然后对所选变量中的所有其他值进行通 用的组合式自动重新编码，从而生成一个可用于所有所选变量的通用自动重新编码方案。 将字符串值或数字值重新编码为连续整数 1. 从菜单中选择： 转换 > 自动重新编码...
2. 选择一个或多个要重新编码的变量。 3. 对每个选定的变量，为新变量输入一个名称并单击新名称。 个案排秩 使用“个案排秩”对话框可以为数值变量创建包含秩、常规得分和 Savage 得分以及百分位值的新变量。 将基于原变量名称和选定的测量自动生成新变量名称和描述性变量标签。 一个摘要表将列出原变量、新变量 和变量标签。 （注：自动生成的新变量名称的最大长度限制为 8 个字节。） 76 IBM SPSS Statistics 29 Core System 用户指南 （可选）您可以执行以下操作： • 按升序或降序对个案排秩。 • 通过在“依据”列表中选择一个或多个分组变量而将排秩组织为子组。 将计算每个组中的秩。 组通过分组变 量的值组合定义。 例如，如果您选择 gender 和 minority 作为分组变量，则将为 gender 和 minority 的每一 组合计算秩。 对个案排秩 1. 从菜单中选择： 转换 > 个案排秩...
2. 选择一个或多个要排秩的变量。 您可以只对数值变量排秩。 或者，您可按升序或降序对个案排秩并将秩组织为子组。 个案排秩：类型 您可以选择多种排秩方法。 将为每种方法创建单独的排秩变量。 排秩方法包含单秩、Savage 得分、分数秩 和百分位数。 您也可以基于比例估计和正态得分创建排秩。 排秩。简单秩。 新变量的值等于它的等级。 Savage 得分。包含基于指数分布的 Savage 评分的新变量。 分数排秩。新变量的值等于等级除以非缺失观测值的权重总和。 分数排秩（百分比）。每个排序值除以具有有效值的观测值数再乘以 100。 个案权重总和。新变量的值等于观测值权重的总和。 对于同一组中的所有个案，该新变量是一个常数。 Ntiles。等级基于百分位数组，每个组包含大约相同数量的观测值。 例如，4 个 Ntile 会将等级 1 指定给第 25 个百分位以下的个案，将等级 2 指定给第 25 个与第 50 个百分位之间的个案，将等级 3 指定给第 50 个 与第 75 个百分位之间的个案，将等级 4 指定给第 75 个百分位以上的个案。 比例估计。对应于特定排序的分布的累积比例的估计。 正态得分。对应于估计累积比例的 z 评分。 比例估计公式。 对于比例估计和正态得分，您可以选择比例估计公式：Blom、Tukey、Rankit 或 Van der Waerden。 • Blom。基于使用公式 (r-3/8) / (w+1/4) 估计的比例值新建排秩变量，其中 w 是观测值的权重和，而 r 是排 秩等级。 • Tukey。使用公式 (r-1/3) / (w+1/3)，其中 r 是排序，w 是观测值权重的总和。 • Rankit。使用公式 (r-1/2) / w 计算，其中 w 是观测值数，r 是序列，范围是从 1 到 w。 • Van der Waerden。Van der Waerden 的变换由公式 r/(w+1) 定义，其中 w 为观测值权重的和，r 为从 1 到 w 的秩。 个案排秩: 结 该对话框控制为原变量上具有相同值的个案指定排秩的方法。 下表显示了不同方法是如何将秩指定给同数的值的： 表 7: 方法和结果排秩 值 中间 最低价 高 顺序 10 1 1 1 1 15 3 2 4 2 15 3 2 4 2 15 3 2 4 2 第 8 章 数据转换 77 表 7: 方法和结果排秩 (继续) 值 中间 最低价 高 顺序 16 5 5 5 3 20 6 6 6 4 日期和时间向导 日期和时间向导简化了与日期和时间变量关联的许多常见任务。 使用日期和时间向导 1. 从菜单中选择： 转换 > 日期和时间向导...
2. 选择要完成的任务，并遵循相应的步骤定义该任务。 • 学习日期和时间的表示方式。 此选择将打开一个屏幕，其中提供 IBM SPSS Statistics 中日期/时间变量的 简要概述。 通过单击帮助按钮后，此屏幕还可提供指向更详细信息的链接。 • 从包含日期或时间的字符串中创建日期/时间变量。 使用此选项从可用字符串变量中创建日期/时间变量。 例如，某个字符串变量以 mm/dd/yyyy 的形式表示日期，您想要据此创建日期/时间变量。 • 在包括部分日期或次数的变量中创建一个日期/时间变量。 使用此选项可以从一组现有变量中构建日期/时 间变量。 例如，一个变量表示月（作为整数），第二个变量表示一月中的一天，第三个变量表示年。 可以 将这三个变量组合成单个日期/时间变量。 • 使用日期和时间进行计算。 使用此选项在日期/时间变量中加上或减去值。 例如，可以通过从表示某过程 的结束时间的变量中减去表示该过程的开始时间的另一变量，来计算该过程的期间。 • 提取日期或时间变量的一部分。 使用此选项可以提取日期/时间变量的组成部分，如从一个日期/时间变量 （形式为 mm/dd/yyyy）中提取一月中的一天。 • 为数据集指定周期性。 此选项会打开“定义日期”对话框，用于创建由一组连续日期组成的日期/时间变 量。 此功能通常用于将日期与时间序列数据相关联。 注：如果数据集缺少完成任务所需的变量类型，那么将禁用任务。 例如，如果数据集不包含字符串变量，则 从字符串创建日期/时间变量的任务不适用，因而被禁用。 IBM SPSS Statistics 中的日期和时间 IBM SPSS Statistics 中表示日期和时间的变量为数字变量，显示格式对应于特定的日期/时间格式。 这些变 量统称日期/时间变量。 将实际表示日期的日期/时间变量与表示独立于任何日期的持续时间的变量（如 20 小时、10 分钟和 15 秒钟）区分。 后者称为持续时间变量，前者称为日期或日期/时间变量。 有关显示格式 的完整列表，请参阅“命令语法参考”的“通用”部分中的“日期和时间”。 日期和日期/时间变量。 日期变量具有表示日期的格式，如 mm/dd/yyyy。 日期/时间变量具有表示日期和时 间的格式，如 dd-mmm-yyyy hh:mm:ss。 日期和日期/时间变量在内部存储为自 1582 年 10 月 14 日以来的 累计秒数。 日期和日期/时间变量有时称为日期格式的变量。 • 两位和四位数字年份规格都已经识别。 缺省情况下，两位数字年份表示从当前日期的前 69 年到后 30 年 这一范围。 这一范围由您的选项设置确定且可被设定（从编辑菜单中，选择选项并单击日期选项卡）。 • 破折号、句号、逗号、斜杠或空格都可在日-月-年格式中用作分隔符。 • 月份可以用数字、罗马数字或三个字符的缩写形式表示，也可以使用全拼的格式。 三个字母的缩写和全拼 的月份名称必须使用英文；使用其它语言的月份名称无法被识别。 持续时间变量。 持续时间变量具有表示持续时间的格式，例如 hh:mm。 持续时间变量在内部存储为不引用 特定日期的秒数。 • 在时间规格中，（适用于日期/时间和持续时间变量），冒号可用作小时、分钟和秒数之间的分隔符。 必 须有小时和分钟，但秒有没有都可以。 必须用句号分隔秒数与小数秒数。 小时数可无限度大，但分钟数最 大值为 59 而秒数为 59.999...。 78 IBM SPSS Statistics 29 Core System 用户指南 当前日期和时间。 系统变量 $TIME 包括当前日期和时间。 它表示从 1582 年 10 月 14 日到使用它的转换命 令执行时的日期和时间期间的秒数。 从字符串中创建一个日期/时间变量 从字符串变量中创建一个日期/时间变量： 1. 在日期和时间向导简介屏幕上选择从包含日期或时间的字符串中创建一个日期/时间变量。 选择字符串变量转换至日期/时间变量 1. 选择字符串变量在变量列表转换。 注意，该列表只显示字符串变量。 2. 从与通过字符串变量表示的日期格式匹配的模式列表中选择模式。 样本值列表在数据文件夹中显示所选 变量的实际值。 不适用于所选模式的字符串变量值将导致新变量的系统缺失值。 指定将字符串变量转换至日期/时间变量的结果 1. 为结果变量输入名称。 不得为现有变量的名称。 （可选）您可以执行以下操作： • 从输出格式列表中为新变量选择一个日期/时间格式。 • 为新变量指定一个描述性变量标签。 从变量组中创建一个日期/时间变量 将一组现有变量合并到单个日期/时间变量： 1. 在日期和时间向导的介绍屏幕上选择从包含部分日期或时间的变量中创建一个日期/时间变量。 选择变量合并到单个日期/时间变量 1. 选择表示日期/时间的不同部分的变量。 • 不允许某些选择组合。 例如，从年、月、日中创建一个日期/时间变量无效，因为一旦年份被选中，就需 要一个完整日期。 • 您不能使用现有的日期/时间变量作为要创建的最终日期/时间变量的组成部分。 组成新日期/时间变量的变 量必须为整数。 允许使用现有日期/时间变量作为新变量的秒数部分是个例外。 由于允许使用小数秒数， 用于秒数的变量不必为整数。 • 对于新变量的任何部分，超出允许范围的值将产生新变量的系统缺失值。 例如，如果您不小心对“月”使用 了表示一个月中某天的变量，那么由于 IBM SPSS Statistics 中月份的有效范围是 1 - 13，因此将为新变量 的系统缺失值指定范围 14 - 31 内任何具有一月中某天值的个案。 指定由合并变量创建的日期/时间变量 1. 为结果变量输入名称。 不得为现有变量的名称。 2. 从输出格式列表中选择一个日期/时间格式。 （可选）您可以执行以下操作： • 为新变量指定一个描述性变量标签。 从日期/时间变量中加减值 在日期/时间变量中加减值： 1. 在日期和时间向导的介绍屏幕上选择使用日期和时间进行计算。 第 8 章 数据转换 79 选择计算类型以使用日期/时间变量执行 • 从日期中添加或提取持续时间。 使用此选项在日期格式变量中加上或减去。 您可以增加或减少为固定值的 持续时间，如 10 天，或数值变量值，如表示年份的变量。 • 计算两个日期之间的时间数。 使用此选项获取在所选单元格测量的两个日期间的差别。 例如，您可以获取 分隔两个日期的年数或天数。 • 提取两个持续时间。 使用此选项获取两个具有持续时间格式的变量间的差别。如 hh:mm 或 hh:mm:ss。 注：如果数据集缺少完成任务所需的变量类型，则会禁用任务。 例如，如果数据集缺少两个具有持续时间格 式的变量，那么提取两个持续时间的任务将不适用并被禁用。 从日期中添加或提取持续时间 从日期格式变量中添加或提取持续时间： 1. 在标有在日期上计算的日期和时间向导的介绍屏幕上选择从日期中增加或减少持续时间。 选择日期/时间变量和要加减的持续时间 1. 选择日期（或时间）变量。 2. 选择一个持续时间变量或为持续时间常量输入一个值。 用于持续时间的变量不得为日期或日期/时间变 量。 它们可以为持续时间变量或简单的数值变量。 3. 从下拉列表中选择持续时间表示的单位。 如果使用一个变量且该变量为持续时间格式，如 hh:mm 或 hh:mm:ss 时，选择持续时间。 从日期/时间变量中指定增加或减少持续时间的结果 1. 为结果变量输入名称。 不得为现有变量的名称。 （可选）您可以执行以下操作： • 为新变量指定一个描述性变量标签。 提取日期格式变量 提取两个日期格式变量： 1. 在标有在日期上计算的日期和时间向导的介绍屏幕上选择计算两个日期之间的时间数。 选择要提取的日期格式变量 1. 选择要提取的变量。 2. 从下拉列表中为结果选择单位。 3. 选择计算结果的方法（结果处理）。 结果处理 以下选项可用于计算结果的方法： • 舍零取整。 忽略结果的任何小数比例。 例如，从 10/21/2007 中减去 10/28/2006 将返回年份为 0 月份 为 11 的结果。 • 取整。 结果四舍五入为最接近的整数。 例如，从 10/21/2007 中减去 10/28/2006 将返回年份为 0 月份 为 12 的结果。 • 保留小数部分。 保留完整值；任何舍入或取整都不适用。 例如，从 10/21/2007 中提取 10/28/2006 返回 年份为 .98 和月份为 11.76 的结果。 为保留舍入和小数，年份结果基于一年中平均天数（365.25），而月份基于一月中平均天数（30.4375）。 例如，从 3/1/2007（m/d/y 格式）中提取 2/1/2007 返回月份为 0.92 的小数结果，反之从 2/1/2007 中提 取 3/1/2007 返回月份为 1.02 的小数差。 这同样会影响到包括闰年在内的时间跨度的计算值。 例如，与在 非闰年的相同时间跨度 0.92 相比，从 3/1/2008 中提取 2/1/2008 返回一个月份为 0.95 的小数差。 80 IBM SPSS Statistics 29 Core System 用户指南 表 8: 年份的日期差异 日期 1 日期 2 截断 舍入 小数 10/21/2006 10/28/2007 1 1 1.02 10/28/2006 10/21/2007 0 1 .98 2/1/2007 3/1/2007 0 0 .08 2/1/2008 3/1/2008 0 0 .08 3/1/2007 4/1/2007 0 0 .08 4/1/2007 5/1/2007 0 0 .08 表 9: 月份的日期差异 日期 1 日期 2 截断 舍入 小数 10/21/2006 10/28/2007 12 12 12.22 10/28/2006 10/21/2007 11 12 11.76 2/1/2007 3/1/2007 1 1 .92 2/1/2008 3/1/2008 1 1 .95 3/1/2007 4/1/2007 1 1 1.02 4/1/2007 5/1/2007 1 1 .99 指定提取两个日期格式变量的结果 1. 为结果变量输入名称。 不得为现有变量的名称。 （可选）您可以执行以下操作： • 为新变量指定一个描述性变量标签。 提取持续时间变量 提取两个持续时间变量： 1. 在标有在日期上计算的日期和时间向导屏幕上选择提取两个持续时间。 选择要提取的持续时间变量 1. 选择要提取的变量。 指定提取两个持续时间变量的结果 1. 为结果变量输入名称。 不得为现有变量的名称。 2. 从输出格式列表中选择一个持续时间格式。 （可选）您可以执行以下操作： • 为新变量指定一个描述性变量标签。 提取部分日期/时间变量 从一个日期/时间变量中提取一个成分，如年份： 1. 在日期和时间向导的介绍屏幕上选择提取一个日期或时间变量的一部分。 第 8 章 数据转换 81 从日期/时间变量中选择要提取的成分 1. 选择要提取的包含日期或时间部分的变量。 2. 从下拉列表中选择要提取的部分变量。 您可以从明确不是显示日期部分的日期中提取信息，如一周中的 一天。 指定从日期/时间变量中提取成分的结果 1. 为结果变量输入名称。 不得为现有变量的名称。 2. 如果您正要提取日期/时间变量的日期或时间部分，那么您必须从输出格式列表中选择一种格式。 一旦不 需要输出格式，输出格式列表将被禁用。 （可选）您可以执行以下操作： • 为新变量指定一个描述性变量标签。 时间序列数据转换 提供了一些在时间序列分析中有用的数据转换： • 生成日期变量来建立周期性并区分历史周期、验证周期和预测周期。 • 创建新的时间序列变量作为现有时间序列变量的函数。 • 使用基于一个方法的估计值替换系统缺失值和用户缺失值。 时间序列可通过定期度量某个变量（或变量集）来获得。 时间序列数据转换假定这样一种数据文件结构：其 中每个个案（行）代表不同时间的一组观察值，而个案之间的时间长度是均匀的。 定义日期 使用“定义日期”对话框可以生成日期变量，用于建立时间序列周期性和标注来自时间序列分析的输出。 个案为。 定义用于生成日期的时间区间。 • 未注日期将移去以前定义的所有日期变量。 将删除任何具有下列名称的变量：year_、quarter_、 month_、week_、day_、hour_、minute_、second_ 和 date_。 • 自定义表示存在由命令语法创建的自定义日期变量（例如四天制工作周）。 此项仅反映活动数据集的当前 状态。 在列表中选择它不会有任何影响。 第一个个案为。 定义指定给第一个个案的起始日期值。 基于时间区间的序列值将指定给后面的个案。 较高级别的周期性。 表示重复性循环变动，例如一年中的月数或者一周中的天数。 显示的值表示您可以输 入的最大值。 对于小时、分钟和秒数，最大值为显示值减去 1。 将为用于定义日期的每个成分创建新的数字变量。 新变量名称以下划线结尾。 还将从成分中创建一个描述 性字符串变量 date_。 例如，如果您已选择 Weeks, days, hours，那么将创建下列四个新变量：week_、 day_、hour_ 和 date_。 如果已经定义了日期变量，当您定义的新日期变量与现有日期变量同名时，现有日期变量将被替换。 为时间序列数据定义日期 1. 从菜单中选择： 数据 > 定义日期...
2. 从“个案为”列表中选择一个时间区间。 3. 输入定义“第一个个案为”的起始日期的值，该值将确定指定给第一个个案的日期。 日期变量与日期格式变量 使用“定义日期”创建的日期变量不应与数据编辑器的“变量视图”中定义的日期格式变量混淆。 日期变量用于 为时间序列数据创建周期性。 日期格式变量代表以各种日期/时间格式显示的日期和/或时间。 日期变量是简 82 IBM SPSS Statistics 29 Core System 用户指南 单的整数，代表距离用户指定的起始点的天数、周数、小时数等。 多数日期格式变量在内部存储为自 1582 年 10 月 14 日以来的累计秒数。 创建时间序列 “创建时间序列”对话框允许您基于现有数值型时间序列变量的函数创建新的变量。 这些转换后的值在时间序 列分析中非常有用。 新变量缺省名称为用来创建它的现有变量的前六个字符，然后紧跟一个下划线和序列号。 例如，对于变量 price，新变量名称可以是 price_1。 新变量保留来自原始变量的已定义值标签。 创建时间序列的可用函数包括差分、移动平均值、移动中位数、延迟和提前函数。 要创建新的时间序列变量 1. 从菜单中选择： 转换 > 创建时间序列...
2. 选择要用来转换原始变量的时间序列函数。 3. 选择要从中创建新的时间序列变量的变量。 仅可使用数字变量。 （可选）您可以执行以下操作： • 输入变量名称以覆盖缺省的新变量名称。 • 为选定变量更改函数。 时间序列转换函数 差分。 序列中相邻值之间的非季节性差异。 阶数为用于计算差分的以前值个数。 由于每阶差分丢失一个观 察值，因此系统缺失值会出现在序列开头。 例如，如果差分阶数为 2，则值，则前两个个案会包含新变量的 系统缺失值。 季节性差分。 相隔恒定距离的序列值之间的差分。 该跨度基于当前定义的周期性。 要计算季节性差分，您 必须定义日期变量（“日期”菜单，“定义日期”），其中包括周期性成分（例如一年中的月份）。 阶数为用于 计算差分的季节性周期个数。 在序列开头，带有系统缺失值的个案个数，等于阶数乘以周期性。 例如，如 果当前周期性为 12，且阶数为 2，则前 24 个个案会包含新变量的系统缺失值。 中心移动平均值。 当前序列值与其周围某个跨度内序列值的平均值。 跨度为用于计算平均值的序列值个 数。 如果跨度为偶数，则移动平均值通过对每组非中心平均值求平均值而得出。 在跨度为 n 的序列的开头 和末尾，带有系统缺失值的个案的数量等于 n/2（偶数跨度值）和 (n–1)/2（奇数跨度值）。 例如，如果跨 度为 5，则在序列开头和末尾带有系统缺失值的个案个数为 2。 前移动平均值。 当前序列值之前的序列值的平均值。 跨度为用于计算平均值的前面序列值个数。 在序列开 头，带有系统缺失值的个案个数，等于跨度值。 移动中位数。 当前序列值与其周围某个跨度内序列值的中位数。 跨度为用于计算中位数的序列值个数。 如 果跨度为偶数，则中位数通过对每组非中心中位数求平均值而得出。 在跨度为 n 的序列的开头和末尾，带有 系统缺失值的个案的数量等于 n/2（偶数跨度值）和 (n–1)/2（奇数跨度值）。 例如，如果跨度为 5，则在 序列开头和末尾带有系统缺失值的个案个数为 2。 累积和。 当前序列值与其周围序列值的累积和。 延迟。 根据指定的延迟阶数，上一个个案的值。 阶数为从中获取值的当前个案之前的个案个数。 在序列开 头，带有系统缺失值的个案个数，等于阶数值。 提前。 根据指定的提前阶数，后一个个案的值。 阶数为从中获取值的当前个案之后的个案个数。 在序列末 尾，带有系统缺失值的个案个数，等于阶数值。 平滑。 基于复合数据平滑器的新序列值。 更平滑的开始的中位数为 4，以 2 的运行中位数为中心。 然后， 它通过应用运行中位数 5，运行中位数 3 和汉宁（运行加权平均值）来重新计算这些值。 从原始序列中减去 平滑后的序列，计算得出残差。 然后对计算得出的残差重复这整个过程。 最后，减去该过程首次获得的平 滑值，得到平滑残差。 这有时也称为 T4253H 平滑。 第 8 章 数据转换 83 替换缺失值 缺失观察值可能会在分析中导致问题，如果序列中缺少值，会无法计算某些时间序列度量。 有时，仅仅是特 定的观测值未知。 此外，数据缺失的原因可能包括： • 每个差分度会使序列长度减 1。 • 每个季节性差分度会使序列长度减少 1 个季节。 • 如果您创建的新序列包含在现有序列末尾之外的预测（单击保存按钮，选择合适选项），原始序列和产生 的残差序列会缺少新观察值的数据。 • 某些转换（例如，对数转换）会导致缺失原始序列的特定值数据。 序列开头和末尾的缺失数据不会引发特殊的问题，只会缩短序列的有效长度。 序列中间的缺口（内嵌缺失数 据）是更为严重的问题。 根据使用的分析过程，问题程度也有所不同。 “替换缺失值”对话框允许您从现有变量创建新的时间序列变量，将缺失值替换为根据多种方法之一估计的 值。 新变量缺省名称为用来创建它的现有变量的前六个字符，然后紧跟一个下划线和序列号。 例如，对于 变量 price，新变量名称可以是 price_1。 新变量保留来自原始变量的已定义值标签。 要替换时间序列变量的缺失值 1. 从菜单中选择： 转换 > 替换缺失值...
2. 选择要用于替换缺失值的估算方法。 3. 选择要替换其缺失值的变量。 （可选）您可以执行以下操作： • 输入变量名称以覆盖缺省的新变量名称。 • 为选定变量更改估算方法。 替换缺失值的估算方法 序列平均值。 将缺失值替换为整个序列的平均值。 邻近点的平均值。 使用有效周围值的平均值替换缺失值。 邻近点的跨度为缺失值上下用于计算平均值的有 效值个数。 邻近点的中位值。 使用有效周围值的中值替换缺失值。 邻近点的跨度为缺失值上下用于计算中位值的有效 值个数。 线性插值。 使用线性插值替换缺失值。 缺失值之前的最后一个有效值以及之后的第一个有效值用于插值。 如果序列中的第一个或最后一个个案具有缺失值，则不必替换。 该点的线性趋势。 使用该点的线性趋势替换缺失值。 现有序列在标度为 1 到 n 的索引变量上回归。 缺失值 将被替换为其预测值。 84 IBM SPSS Statistics 29 Core System 用户指南 第 9 章 文件处理和文件转换 文件处理和文件转换 数据文件的组织形式对于特定需要而言并非总是理想的。 您可能想要合并数据文件、以不同的顺序对数据排 序、选择个案的子集、或通过对个案进行分组来更改分析的单位。 有大量可用的文件转换功能，包括： 对数据排序。 您可根据一个或多个变量的值对个案排序。 变换个案和变量。 IBM SPSS Statistics 数据文件格式将行作为个案读取，将列作为变量读取。 对于顺序相 反的数据文件，您可交换行和列并以正确的格式读取数据。 合并文件。 您可合并两个或更多个数据文件。 您可将变量相同但个案不同或个案相同但变量不同的文件进 行合并。 选择个案的子集。 您可以将分析限制为个案的子集或对不同的子集执行同步分析。 对数据汇总。 您可通过基于一个或多个分组变量的值对个案汇总而更改分析的单位。 对数据加权。 您可以基于权重变量的值对个案加权以进行分析。 重组数据。 您可重组数据以从多个个案中创建单个个案（记录），或从单个个案中创建多个个案。 排序个案 此对话框基于一个或多个排序变量的值对活动数据集的个案（行）排序。 您可以按升序或降序对个案排序。 • 如果您选择了多个排序变量，个案按“排序”列表的前一个变量的类别中的每个变量排序。 例如：如果您将 选择性别作为第一个排序变量，选择少数民族作为第二个排序变量，则将按每个性别类别中的少数民族分 类对个案排序。 • 排序顺序基于语言环境定义的顺序（不一定与字符代码的数值顺序相同）。 缺省的语言环境是操作系统的 语言环境。 可以控制语言环境，方法是使用“选项”对话框的“常规”选项卡上的“语言”设置（“编辑”菜 单）。 要对个案排序，请执行下列操作 1. 从菜单中选择： 数据 > 排序个案...
2. 选择一个或多个排序变量。 此外，您还可以进行下列操作： 为保存文件编制索引。 当数据文件与 STAR JOIN 合并时，为表查找文件编制索引可以提高性能。 保存已排序文件。 您可以使用保存为加密的选项保存已排序文件。 加密允许您保护存储在文件的密件信 息。 一旦加密，文件只能通过提供指定给文件的密码打开。 要使用加密保存已排序文件： 3. 选择保存带分类数据的文件，然后单击文件。 4. 在“将已排序数据保存为”对话框中，选择使用密码加密文件。 5. 单击保存。 6. 在“加密文件”对话框中，提供密码并重新在“确认密码”文本框中输入。 密码限制在 10 个字符并区分大小 写。 警告：密码丢失后将无法恢复。 如果密码丢失，那么将无法打开文件。 创建强密码 • 至少使用八个字符。 • 在密码中使用数字、符号甚至标点符号。 • 避免使用数字序列或字符序列（例如 "123" 和 "abc"）并避免重复，例如 "111aaa"。 • 不要创建使用个人信息（例如生日或昵称）的密码。 • 定期更改密码。 注：不支持将已加密的文件存储到 IBM SPSS 协作和部署服务存储库 中。 注意：在 V21 之前的 IBM SPSS Statistics 版本中，无法打开经过加密的数据文件和输出文档。 在 V22 之前 的版本中，无法打开经过加密的语法文件。 变量排序 您可以根据任何变量属性值（例如，变量名称、数据类型、测量级别），包括自定义变量属性，对活动数据 集的变量进行排序。 • 可以按升序或降序对值排序。 • 您可以在自定义变量属性中保存原始（预先分类）变量顺序。 • 根据自定义变量属性的值进行排序，只限于当前显示在变量视图中的自定义变量属性。 有关定制变量属性的更多信息，请参阅第49 页的『定制变量属性』。 要排列变量 在数据编辑器的变量视图中： 1. 右键单击属性列标题，并从弹出菜单中选择按升序排序或按降序排序。 或 2. 在“变量视图”或“数据视图”中，从菜单中选择： 数据 > 排序变量 3. 选择用于排序变量的属性。 4. 选择排序顺序（升序或降序）。 • 变量属性列表与数据编辑器的“变量视图”中显示的属性列名称相匹配。 • 您可以在自定义变量属性中保存原始（预先分类）变量顺序。 对于每个变量，该属性值是一个整数值，表 示它在排序前的位置；因此，在根据该定制属性值对变量进行排序时，您可以恢复原来的变量顺序。 交换 “变换”创建一个新的数据文件，初始数据文件中的行列被变换，使个案（行）成为变量同时变量（列）成为 个案。 “变换”自动创建新的变量名称，并显示新的变量名称列表。 • 自动创建包含初始变量名称 case_lbl 的新字符串变量。 • 如果活动数据集包含具有唯一值的 ID 或名称变量，那么可以使用其作为名称变量，且其值在转置后的数据 文件中将用作变量名称。 如果它是数字变量，变量名称将以字母 V 开头，后跟数字值。 • 用户缺失值会转换为转置数据文件中的系统缺失值。 要保留所有这些值，请在“数据编辑器”的“变量视图” 中更改缺失值的定义。 要变换变量和个案，请执行下列操作 1. 从菜单中选择： 数据 > 变换...
2. 选择要转置为观测值的一个或多个变量。 合并数据文件 可以使用两种不同的方式合并两个文件中的数据。 您可以执行以下操作： 86 IBM SPSS Statistics 29 Core System 用户指南 • 合并活动数据集和另一个打开的数据集，或包含相同变量但不同个案的 IBM SPSS Statistics 数据文件。 • 合并活动数据集和另一个打开的数据集，或包含相同个案但不同变量的 IBM SPSS Statistics 数据文件。 合并文件 1. 从菜单中选择： 数据 > 合并文件 2. 选择添加个案或添加变量。 添加个案 “添加个案”将活动数据集与另外一个数据集或包含相同变量（列）但不同个案（行）的外部 IBM SPSS Statistics 数据文件合并在一起。 例如，您可能在两个不同的销售区域中记录客户的相同信息，并以单独文 件的形式维护每个区域的数据。 另外的数据集可以是外部 IBM SPSS Statistics 数据文件，也可以是当前会 话中可用的数据集。 非成对变量。 要从新的合并的数据文件中排除的变量。 使用星号 () 标识活动数据集中的变量。 其他数据集 中的变量使用加号 (+ ) 进行标识。 缺省情况下，此列表包含： • 两个数据文件中的变量名称互不匹配的变量。 可以在不成对的变量中创建对，并将其包含在新的合并文件 中。 • 在一个文件定义为数值数据而在另一文件中被定义为字符串数据的变量。 数字变量不能与字符串变量合 并。 • 宽度不相等的字符串变量。 两个数据文件中的字符串变量的定义宽度必须相同。 新的活动数据集中的变量。 要包含在新的合并的数据文件中的变量。 缺省情况下，在列表上包含所有名称 和数据类型（数字或字符串）都匹配的变量。 • 如果不想将某些变量包含在合并文件中，可从列表中删除这些变量。 • 包含在合并文件中的任何不成对的变量，都包含一些个案的缺失数据，而这些个案是来自于不包含该变量 的文件。 将个案源表示为变量。指示每个观测值的源数据文件。 对于来自活动数据集的个案，此变量具有 0 值；对于 来自外部数据文件的个案，此变量具有 1 值。 1. 打开至少一个要合并的数据文件。 如果打开了多个数据集，那么使其中一个要合并的数据集成为活动数 据集。 此文件中的个案将首先出现在新的合并数据文件中。 2. 从菜单中选择： 数据 > 合并文件 > 添加个案...
3. 选择要与活动数据集合并的数据集或外部 IBM SPSS Statistics 数据文件。 4. 在“新活动数据集”列表的“变量”中删除不想要的任何变量。 5. 任何从“不成对的变量”列表中添加的变量对都表示相同信息，这些信息以不同变量名称在两个文件中记 录。 例如，出生日期可能在某一文件中的变量名称是 brthdate，而在另一个文件中的变量名称是 datebrth。 选择不成对的变量对 1. 在“不成对的变量”列表中单击其中一个变量。 2. 按住 Ctrl 并单击列表上的其他变量。 （按 Ctrl 键并同时单击鼠标左键。） 3. 单击对将变量对移到“新活动数据集”列表中的“变量”中。 （活动数据集中的变量名称用作合并文件中的 变量名称。） 添加个案：重命名 可以重命名活动数据集或其他数据集中的变量，然后将这些变量从不成对列表移动到要包含在合并数据文件 的变量列表中。 重命名变量使您可以： • 对于变量对，使用外部文件的变量名称，而不是活动数据集中的变量名称。 第 9 章 文件处理和文件转换 87 • 包含两个名称相同但类型不匹配或字符串宽度不同的变量。 例如，要同时包含活动数据集中的数值变量性 别和外部文件中的字符串变量性别，则首先必须重命名其中一个变量。 添加个案：字典信息 活动数据集中的所有现有字典信息（变量和值标签、用户缺失值、显示格式）会应用到合并数据文件中。 • 如果在活动数据集中没有定义变量的任何字典信息，那么会使用另一个数据集中的字典信息。 • 如果活动数据集包含一个变量的任何已定义值标签或用户缺失值，那么会忽略其他数据集中该变量的所有 其他值标签或用户缺失值。 合并两个以上的数据源 使用命令语法最多可以合并 50 个数据集和/或数据文件。 有关更多信息，请参阅命令语法参考（可从“帮助” 菜单获得）中的 ADD FILES 命令。 添加变量 “添加变量”将活动数据文件与另一个打开的数据文件或包含相同个案（行）但不同变量（列）的外部 IBM SPSS Statistics 数据文件合并在一起。 例如，可能要将一个包含检验前结果的数据文件与包含检验后结果的 数据文件合并。 “合并方法”选项卡 使用合并方法选项卡定义合并类型。 基于文件顺序一对一合并 文件中的个案顺序决定了个案的匹配方式。 这是两个文件中不存在具有相同名称和基本类型（字符串或 数字）的变量时的缺省设置。 该设置以下列各式生成 MATCH FILES 命令语法（其中 [name] 是数据集 或外部文件规范（以引号括起））： MATCH FILES FILE= /FILE="[name]" 基于键值一对一合并 基于一个或多个键变量的值匹配个案。 这是两个文件中存在具有相同名称和基本类型（字符串或数字） 的一个或多个变量时的缺省设置。 该设置以下列各式生成 MATCH FILES 命令语法（其中 [name] 是数 据集或外部文件规范（以引号括起））： MATCH FILES FILE= /FILE="[name]" /BY [key varlist] 基于键值一对多合并 一个文件包含个案数据，一个文件是查找表。 来自查找表的个案与个案数据文件中具有匹配键值的个案 合并。 相同键值在个案数据文件中可以多次出现。 查找表中的一个个案可以与个案数据文件中的多个个 案合并。 • 个案数据文件中的所有个案都包含在合并文件中。 • 如果查找表文件中的个案在个案数据文件中没有对应的具有匹配键值的个案，那么不包含此类个案。 • 查找表文件不能包含重复的键值。 如果文件包含多个键变量，那么键值是这些值的组合。 该设置对查找表使用 TABLE 子命令来生成 MATCH FILES 语法。 选择查找表 仅当选择了基于键值一对多合并时，才会启用下列设置。 [活动数据集名称] 活动数据集名称后跟星号。 • 缺省情况下，不会选择该选项。 如果选中该选项，该设置以下列各式生成语法（其中 [name] 是 数据集或外部文件规范（以引号括起））： MATCH FILES TABLE= /FILE=”[name]” /BY [key varlist] 88 IBM SPSS Statistics 29 Core System 用户指南 [第二个数据集名称或文件名] 第二个数据集或文件的名称。 • 这是缺省设置。 • 如果文件是外部文件，那么仅提供文件名（而不是整个路径），完整路径将包含在生成的语法中。 • 如果第二个文件是外部文件（且选择了排序，或者它包含的字符串键要求更改定义的长度），那么 必须先打开该文件，并为其分配唯一名称。 • 如果第二个文件是需要打开的外部文件（基于前一个条件），且未命名活动数据集，那么必须先命 名活动数据集，然后才能打开第二个数据文件。 • 当打开第二个文件进行排序或更改字符串键长度时，在合并操作后仍保持打开状态。 由于文件已 更改，因此在关闭文件时，将提示您保存更改。 • 如果选中该选项，该设置以下列各式生成语法（其中 [name] 是数据集或外部文件规范（以引号括 起））： MATCH FILES FILE= /TABLE=”[name]” /BY [key varlist] 合并之前按键值对文件排序 对于键值合并，两个文件都必须按照键变量值排序。 • 仅当选择了其中一个键值合并选项时，才会启用该设置。 • 如果其中一个文件是外部文件，将打开此文件并进行排序。 除非显式保存文件，否则将不会保存已排 序的文件。 • 如果文件已经排序，可以取消选择该选项来节省时间。 • 当选择该选项时，该设置会生成 SORT CASES 语法。 键变量 对于键值合并，缺省情况下，具有相同名称和基本数据类型（字符串或数字）的变量或作为键变量包含 在内。 使用变量选项卡添加、移除或更改键变量的顺序。 注: • 所选合并始终生成 MATCH FILES 命令语法（从不生成 STAR JOIN 语法）。 • SORT CASES 和 ALTER TYPE 命令语法优先于 MATCH FILES 命令语法。 • 是否包含 DROP 子命令语法和可选的 RENAME 子命令语法取决于在变量选项卡上所做的选择。 • 如果选择了其中一个键值合并选项，则会包括 BY 子命令语法。 • 当字符串键具有不同的定义长度时，会自动生成 ALTER TYPE 语法来确保定义相同的长度。 “变量”选项卡 使用变量选项卡添加、移除和重命名变量以包含在合并后的文件中。 • 活动数据文件中的变量使用星号 () 标识。 • 其他数据文件中的变量使用加号 (+) 标识。 排除的变量 要从新合并的数据文件中排除的变量。 • 当在合并方法选项卡上选择了基于文件顺序一对一合并设置时，第二个数据集中与活动数据集中的变 量同名的变量将被排除。 • 当在合并方法选项卡上选择了基于键值一对一合并或基于键值一对多合并设置时，排除第二个数据集 中所有重复名称变量（与活动数据集中相同变量名称具有不同基本类型（字符串/数字）的变量）。 包含的变量 要包含在新合并数据文件中的变量。 如果想要在合并后数据文件中包含有重复名称的已排除变量，请使 用重命名来更改名称。 • 当在合并方法选项卡上选择了基于文件顺序一对一合并设置时，将包含活动数据集中的所有变量和第 二个数据集中所有唯一命名的变量。 第 9 章 文件处理和文件转换 89 • 当在合并方法选项卡上选择了基于键值一对一合并或基于键值一对多合并设置时，将包含两个数据集 中所有唯一命名的变量。 对于活动数据集中在第二个数据集中具有重复名称但具有不同基本类型（字 符串/数字）的变量，也会包含在内。 键变量 对于键值合并，基于键变量值合并个案。 • 缺省列表分配由所选合并方法确定。 变量不会自动分配到键变量列表。 – 当在合并方法选项卡上选择了基于文件顺序一对一合并设置时，不会包含变量，移动控件处于禁用 状态，且手动拖放变量也无任何效果。 – 当在合并方法选项卡上选择了基于键值一对一合并或基于键值一对多合并设置时，将包含两个数据 集中所有具有相同名称和相同基本类型（字符串/数字）的变量。 • 每一个键变量在两个文件中都必须具有相同名称和相同基本数据类型（字符串/数字）。 • 如果某个键变量在两个文件中具有不同的名称，使用重命名来更改其中一个名称。 • 如果某字符串键变量在两个文件中具有不同的定义长度，具有较短长度的变量会自动调整为较长的长 度。 除非显式保存原始文件的修改后版本，否则不会保存修改后文件。 添加变量：重命名 您可以重命名活动数据集或其他数据文件中的变量，然后将这些变量移动到要包含在合并数据文件中的变量 列表。 如果您要在两个文件中包含两个具有不同信息但名称相同的变量，或在两个文件中的关键字变量具有 不同名称，此选项很有用。 合并两个以上的数据源 使用命令语法可以合并两个以上的数据文件。 • 使用匹配文件以合并多个不含关键字变量的文件，或多个已经以关键字变量值排序的文件。 主题以获取更 多信息。 • 使用 STAR JOIN 以合并多个文件，其中有一个个案数据文件和多个表查找文件。 不需要以关键字变量值 排序文件，并且每个表查找文件可以使用不相同的关键字变量。 主题以获取更多信息。 聚集数据 “汇总数据”将活动数据集中的个案组汇总为单个个案并创建新的汇总文件，或在活动数据集中创建包含汇总 数据的新变量。 基于零个或多个中断（分组）变量的值汇总个案。 如果未指定分组变量，那么整个数据集 将成为单个分类组。 • 如果创建新的汇总数据文件，则新数据文件对由分组变量定义的每个组都包含一个个案。 例如，如果分组 变量有两个值，则新的数据文件将仅包含两个个案。 如果未指定分组变量，则新数据文件将包含一个个 案。 • 如果将汇总变量添加到活动数据集，那么不汇总数据文件本身。 分组变量值相同的每个个案对新汇总变量 都得到相同的值。 例如，如果性别是唯一的分组变量，那么所有男性对于表示平均年龄的新汇总变量将得 到相同的值。 如果未指定分组变量，那么对于代表平均年龄的新汇总变量，所有个案将收到相同值。 分组变量。 个案会根据分组变量的值分组在一起。 分组变量的每个唯一组合定义一个组。 创建新的汇总数 据文件时，所有分组变量保存在新文件中，同时保留其现有名称和字典信息。 分组变量（如果已指定）可以 是数值型或字符串型。 汇总变量。 源变量用于汇总函数，以创建新的汇总变量。 汇总变量名称紧跟一个可选的变量标签、汇总函 数的名称和源变量名称（用括号括起来）。 您可使用新的变量名称覆盖缺省汇总变量名称，提供描述性变量标签，以及更改用于计算汇总数据值的函 数。 您还可以创建包含每个分类组中的个案数的变量。 汇总数据文件 1. 从菜单中选择： 数据 > 汇总...
90 IBM SPSS Statistics 29 Core System 用户指南 2. 根据需要，选择分组变量，以定义如何对个案分组以创建汇总数据。 如果未指定分组变量，那么整个数 据集将成为单个分类组。 3. 选择一个或多个汇总变量。 4. 为每个汇总变量选择一个汇总函数。 保存汇总结果 可以将汇总变量添加到活动数据集，或创建新的汇总数据文件。 • 将汇总变量添加至活动数据集 (Add aggregated variables to active dataset). 基于聚集函数的新变量会添加 到活动数据集中。 数据文件本身并不汇总。 分隔变量值相同的每个观测值对新聚合变量都得到相同的值。 • 创建仅包含聚集变量的新数据集。将聚集的数据保存到当前会话中的新数据集。 该文件包含定义汇总个案 的分组变量以及由汇总函数定义的所有汇总变量。 活动数据集不受影响。 • 写入仅包含聚集变量的新数据文件。将聚集的数据保存到外部数据文件。 该文件包含定义汇总个案的分组 变量以及由汇总函数定义的所有汇总变量。 活动数据集不受影响。 用于大数据文件的排序选项 对于非常大的数据文件，汇总已预先排序的数据可能更有效。 文件已按分类变量排序 (File is already sorted on break variables). 如果已按中断变量的值对数据进行排序， 那么此选项使过程能够更快地运行并使用更少的内存。 应谨慎使用该选项。 • 数据必须按照分组变量的值进行排序，且分组变量的顺序必须与为“汇总数据”过程指定的分组变量的顺序 相同。 • 如果要将变量添加到活动数据集，那么只有在按照分组变量的升序值对数据进行了排序的情况下才选择此 选项。 在聚合前对文件进行排序 (Sort file before aggregating). 在具有大型数据文件的极少数情况下，您可能发现需 要在聚集之前按分类变量的值对数据文件进行排序。 建议不要使用该选项，除非您遇到了内存或性能问题。 汇总数据: 汇总函数 此对话框指定的函数用于计算“汇总数据”对话框中的“汇总变量”列表中选定变量的汇总数据值。 汇总函数包 括： • 数字变量的汇总函数，包括平均值、中位数、标准差和总和 • 个案数，包括未加权、加权、非缺失和缺失 • 在指定值以上或以下的值的百分比、分数或计数 • 在指定范围之内或之外的值的百分比、分数或计数 汇总数据：变量名和标签 “汇总数据”指定新的数据文件中汇总变量的缺省变量名称。 此对话框使您能够更改“汇总变量”列表上选定变 量的名称并提供描述性变量标签。 请参阅主题第44 页的『变量名称』，了解更多信息。 拆分文件 “拆分文件”将数据文件拆分为单独的组，以根据一个或多个分组变量的值进行分析。 如果您选择了多个分组 变量，个案按“分组依据”列表的前一个变量的类别中的每个变量进行分组。 例如：如果您选择性别作为第一 个分组变量，选择少数民族作为第二个分组变量，将按每个性别类别中的少数民族分类对个案进行分组。 • 您最多可指定 8 个分组变量。 • 长字符串变量（长于 8 个字节的字符串变量）的每 8 个字节计为一个变量，直到达到 8 个分组变量的限 制。 • 个案应按分组变量的值并以“分组依据”列表中列出变量的相同顺序排序。 如果数据文件未准备好进行排 序，请选择按分组变量排序文件。 第 9 章 文件处理和文件转换 91 比较各组。 将拆分文件组显示在一起以用于比较目的。 对于透视表，将创建单个数据透视表，且可以将每 个拆分文件变量在表的维度之间移动。 对于图表，为每个拆分文件组分别创建图表，并在“查看器”中将图表 显示在一起。 按组组织输出。 为每个拆分文件组分别显示每个过程中的所有结果。 要拆分数据文件以进行分析，请执行下列操作 1. 从菜单中选择： 数据 > 拆分文件...
2. 选择比较各组或按组来组织输出。 3. 选择一个或多个分组变量。 选择个案(L) “选择个案”提供了几种基于包含变量和复杂表达式的条件来选择个案子组的方法。 您也可以选择个案的随机 样本。 用于定义子组的条件可包括： • 变量值和范围 • 日期和时间范围 • 个案（行）编号 • 算术表达式 • 逻辑表达式 • 函数 全部个案。关闭观测值过滤，并使用全部观测值。 如果满足条件。使用条件表达式来选择个案。 如果条件表达式的结果为 true，那么选择该个案。 如果结果 为 false 或 missing，那么不选择该个案。 随机个案样本。根据近似的百分比或确切的观测值数选择随机样本。 基于时间或个案范围。基于观测值编号范围或日期/时间范围选择观测值。 使用过滤变量。使用数据文件中的所选数字变量作为过滤器变量。 选择过滤变量值不为 0 或 missing 的个 案。 输出 此部分控制对未选定个案的处理。 可以选择以下可选项之一来处理未选定个案： • 过滤掉未选定的个案。 未选定的个案不包括在分析中，但保留在数据集中。 如果关闭过滤，则在会话中稍 后可以使用未选定个案。 如果选择随机样本，或者基于条件表达式选择个案，则此项生成名为 filter$ 的 变量，对选定个案该变量的值为 1，对未选定个案该变量的值为 0。 • 将选定的个案复制到新数据集。 选定的个案复制到新数据集，原始数据集未受影响。 未选中个案不包括在 新数据集中，而在初始数据集中保持其初始状态。 • 删除未选定个案。 从数据集删除未选定个案。 只有退出文件而不保存任何更改，然后重新打开文件，才能 恢复删除的个案。 如果保存对数据文件的更改，则会永久删除个案。 注：如果删除未选定个案并保存文件，则无法恢复个案。 要选择个案的子集，请执行下列操作 1. 从菜单中选择： 数据 > 选择个案...
2. 选择一个选择个案的方法。 3. 指定选择个案的条件。 92 IBM SPSS Statistics 29 Core System 用户指南 选择个案：If 此对话框允许您使用条件表达式选择个案的子集。 条件表达式对每个个案都返回一个值：true、false 或 missing。 • 如果条件表达式的结果为 true，那么所选子集中将包含该个案。 • 如果条件表达式结果为 false 或 missing，则所选子集中不包含该个案。 • 大多数条件表达式在计算器键盘上使用一个或多个关系运算符（<、>、<=、>=、= 和 ~=）。 • 条件表达式可以包含变量名称、常数、算术运算符、数值（和其他）函数、逻辑变量以及关系运算符。 选择个案：随机样本 此对话框允许您基于近似的百分比或准确的个案数来选择随机样本。 抽样是在不进行替换的情况下执行的； 所以同一个案不会被选择多次。 近似。 生成近似于指定个案百分比的随机样本。 由于此例程为每个个案作出独立的伪随机决策，因此选定 个案的百分比只能近似于指定的百分比。 数据文件中的个案越多，选定个案的百分比与指定百分比就越接 近。 精确。 用户指定的个案数。 您还必须指定用于生成样本的个案数。 第 2 个数目应该小于或等于数据文件中 个案的总数。 如果该数字超过数据文件中的个案总数，那么样本所包含的个案数将按请求的数目成比例减 少。 选择个案：范围 此对话框根据个案数目范围或日期/时间范围来选择个案。 • 个案范围基于在数据编辑器中显示的行号。 • 日期和时间范围仅可用于定义有日期变量（“日期”菜单，“定义日期”）的时间序列数据。 注：如果未选定个案被过滤掉（而不是被删除），随后对数据集排序将关闭由此对话框应用的过滤。 对个案进行加权 “个案加权”可以通过模拟复制来为个案分配不同的权重，以便进行统计分析。 • 加权变量的值应指出数据文件中单个个案所表示的观测值数量。 • 加权变量值为 0、负值或缺失值的个案排除在分析之外。 • 小数值有效，在一些过程如“频率”、“交叉表”和“自定义表”中会使用小数权重值。 然而，大多数过程将权 重变量视为重复权重，并将小数权重简单地四舍五入为最接近的整数。 某些过程完全忽略权重变量，这一 限制在特定过程文档中加以说明。 一旦应用了一个权重变量，该权重变量始终保持有效，直到您选择另一个权重变量或关闭加权。 如果您保存 加权后的数据文件，加权信息随数据文件一起保存。 您可随时关闭加权，即使在文件以加权形式保存之后也 可以。 交叉表中的权重。 “交叉表”过程有多个处理个案权重的选项。 主题。 散点图和直方图中的权重。 散点图和直方图具有打开和关闭个案权重的选项，但对权重变量值为 0、负值或 缺失值的个案没有影响。 即使您在图表中关闭加权，这些个案仍将被排除在图表之外。 要对个案进行加权，请执行下列操作 1. 从菜单中选择： 数据 > 对个案进行加权...
2. 选择加权个案依据。 3. 选择一个频率变量。 该频率变量的值用作个案权重。 例如，频率变量值为 3 的个案将在加权数据文件中代表 3 个个案。 第 9 章 文件处理和文件转换 93 重组数据 使用重组数据向导可为要使用的过程重组数据。 该向导会使用新的重组过的文件替代当前文件。 该向导可 以： • 将选定变量重组为个案(C) • 将选定个案重构为变量(V) • 变换所有数据 重组数据 1. 从菜单中选择： 数据 > 重组...
2. 选择要执行的重组的类型。 3. 选择要重组的数据。 （可选）您可以执行以下操作： • 创建标识变量，这些变量允许您将新文件中的值追溯到原始文件中的值。 • 在重构之前对数据排序 • 为新文件定义选项 • 将命令语法粘贴到语法窗口中 重组数据向导：选择类型 使用重组数据向导可重组数据。 在第一个对话框中，选择要执行的重组的类型。 • 将选定变量重组为个案。 如果数据中是相关列构成的组，并且希望其在新数据文件中显示为行组，可选择 此项。 如果选择此项，向导将显示变量到个案的步骤。 • 将选定个案重组为变量。 如果数据中是相关行构成的组，并且希望其在新数据文件中显示为列组，可选择 此项。 如果选择此项，向导将显示个案到变量的步骤。 • 变换所有数据。 要变换数据时选择此项。 在新数据中，所有行将成为列，而所有列将成为行。 此选项会 关闭重组数据向导并打开“变换数据”对话框。 决定如何重组数据 变量包含要分析的信息，例如度量或得分。 个案是观测值，例如个体。 在简单数据结构中，每个变量是数 据中的单独一列，每个个案是单独一行。 因此，举个例子，如果度量班里所有学生的考试得分，所有得分将 仅显示在一列中，而对于每个学生均有一行。 分析数据时，您常常根据某个条件分析变量如何变化。 该条件可以是特定的试验处理、人口统计、时间点或 者其他。 在数据分析中，感兴趣的条件常常称为因子。 分析因子时，数据结构是复杂的。 在数据中，关于 变量的信息可能位于多个列中（例如，对于每个因子级别均有一列），或者，关于个案的信息可能位于多个 行中（例如，对于每个因子级别均有一行）。 重组数据向导可帮助您重组具有复杂数据结构的文件。 当前文件的结构以及要在新文件中使用的结构决定了您在向导中选择的选项。 数据在当前文件中是如何排列的？ 当前数据的排列可能是因子记录在单独的变量中（在个案组中），也可能 是因子与变量记录在一起（在变量组中）。 • 个案组。 当前文件的变量和条件记录在单独的列中吗？ 例如： 表 10: 变量和条件位于不同列中的数据 变量 因子 8 1 9 1 3 2 94 IBM SPSS Statistics 29 Core System 用户指南 表 10: 变量和条件位于不同列中的数据 (继续) 变量 因子 1 2 在此示例中，开头两行是个案组，因为它们相关。 它们包含相同因子级别的数据。 在 IBM SPSS Statistics 数据分析中，当数据以这种方式构造时，因子常常称为分组变量。 • 列组。 当前文件的变量和条件记录在同一列中吗？ 例如： 表 11: 变量和条件位于同一列中的数据 var_1 var_2 8 3 9 1 在此示例中，这两列是变量组，因为它们相关。 它们包含相同变量的数据 --var_1 表示因子级别 1，var_2 表示因子级别 2。 在 IBM SPSS Statistics 数据分析中，当数据以这种方式结构化时，该因子通常称为重复测 量 。 数据在新文件中应如何排列？ 这通常由要用来分析数据的过程决定。 • 要求使用个案组的过程。 要执行需要分组变量的分析，数据必须按个案组构造。 示例包括“一般线性模型” 的单变量、多变量和方差成分；“混合模型”；“OLAP 多维数据集”；以及“T 检验”或“非参数检验”的独立样 本。 如果当前数据结构是变量组，要执行这些分析，请选择将选定变量重组为个案。 • 要求使用变量组的过程。 要分析重复测量，数据必须按变量组构造。 示例包括“一般线性模型”的重复测 量、“Cox 回归分析”的依时协变量 分析、“T 检验”的配对样本 或“非参数检验”的相关样本。 如果当前数据 结构是个案组，要执行这些分析，请选择将选定个案重组为变量。 变量到个案的示例 在此示例中，每个因子的检验得分记录在单独的列（A 和 B）中。 表 12: 每个因子的检验得分记录在单独的列中 score_a score_b 1014 864 684 636 810 638 您想要执行独立样本 T 检验。 您有一个由 score_a 和 score_b 组成的列组，但没有该过程需要的分组变量。 在重构数据向导中选择将选定变量重构为个案，将一个变量组重组为名为 score 的变量，并创建名为 group 的索引。 新的数据文件显示在下图中。 表 13: 变量到个案的新的、重组的数据 组 评分 SCORE_A 1014 SCORE_B 864 SCORE_A 684 SCORE_B 636 SCORE_A 810 SCORE_B 638 第 9 章 文件处理和文件转换 95 如果运行独立样本 T 检验，则现在可以使用 group 作为分组变量。 个案到变量的示例 在此示例中，将为每个主体记录两次检验得分，即处理之前和之后的得分。 表 14: 个案到变量的当前数据 标识 得分 time 1 1014 之前 1 864 之后 2 684 之前 2 636 之后 您想要执行成对样本 T 检验。 数据结构是个案组，但没有过程需要的成对变量的重复测量。 在重组数据向 导中选择将选定个案重组为变量，使用 id 标识当前数据中的组，并使用 time 创建新文件中的变量组。 表 15: 个案到变量的新的、重组的数据 标识 之后 之前 1 864 1014 2 636 684 如果运行成对样本 T 检验，那么现在可将 bef 和 aft 用作变量对。 重组数据向导（变量到个案）：变量组数 注：如果选择将变量组重组为行，则向导会提供此步骤。 在此步骤中，决定要将当前文件中的多少变量组重组到新文件中。 当前文件中有多少个变量组？ 考虑有多少变量组存在于当前数据中。 相关列组（称为变量组）在单独的列 中记录同一个变量的重复测量。 例如，如果当前数据中有三列（w1、w2 和 w3）记录宽度，则有一个变量 组。 如果有另外三列（h1、h2 和 h3）记录高度，则有两个变量组。 新文件中应有多少个变量组？ 考虑您想要让新数据文件中出现多少个变量组。 您无需将所有变量组重构到 新文件中。 • 一个。 向导将从当前文件中的一个变量组在新文件中创建单个重组的变量。 • 多个。 向导将在新文件中创建多个重组的变量。 所指定的个数会影响下一步，在下一步中向导将自动创建 指定个数的新变量。 重组数据向导（变量到个案）：选择变量 注：如果选择将变量组重组为行，则向导会提供此步骤。 在此步骤中，提供关于应如何在新文件中使用当前文件中的变量的信息。 您还可以创建在新文件中标识行的 变量。 应如何标识新行？ 您可以在新数据文件中创建一个变量，该变量标识当前数据文件中用来创建一组新行的 行。 该标识可以是连续的个案号，也可以是变量的值。 使用“个案组标识”中的控件可定义新文件中的标识 变量。 单击单元格可更改缺省变量名称并为标识变量提供描述性变量标签。 新文件中应重构哪些内容？ 在前一步中，您告诉了向导要重组多少个变量组。 向导已为每个组分别创建了 一个新变量。 变量组的值将出现在新文件的该变量中。 使用“要变换的变量”中的控件可定义新文件中的重 组的变量。 指定一个重组的变量 96 IBM SPSS Statistics 29 Core System 用户指南 1. 将组成要转换的变量组的变量放到“要变换的变量”列表中。 组中的所有变量必须是同一类型（数值或字 符串）。 在变量组中可以将同一个变量包含多次（变量是从源变量列表复制而非移动），该变量的值会在新文件中重 复。 指定多个重组的变量 1. 从“目标变量”下拉列表中选择要定义的第一个目标变量。 2. 将组成要转换的变量组的变量放到“要变换的变量”列表中。 组中的所有变量必须是同一类型（数值或字 符串）。 可以在变量组中将同一个变量包含多次。 （变量是从源变量列表复制而非移动，其值在新文件 中重复。） 3. 选择下一个要定义的目标变量，并为提供的所有目标变量重复变量选择过程。 • 尽管可以在同一个目标变量组中将同一个变量包含多次，但不能将同一个变量包含在多个目标变量组中。 • 每个目标变量组列表必须包含相同个数的变量。 （多次列出的变量包含在计数中。） • 目标变量组的个数由您在前一步中指定的变量组的个数决定。 在此处您可以更改缺省变量名称，但必须返 回到前一步才能更改要重组的变量组的个数。 • 必须为提供的所有目标变量定义变量组（方法是在源列表中选择变量），然后才能继续下一步。 应将哪些内容复制到新文件中？ 未重组的变量可以被复制到新文件中。 它们的值将复制到新的行。 将要复 制到新文件中的变量移至“固定变量”列表中。 重组数据向导（变量到个案）：创建索引变量 注：如果选择将变量组重组为行，则向导会提供此步骤。 在此步骤中，决定是否创建索引变量。 索引是一个新变量，它根据从中创建新行的原始变量按顺序标识行 组。 新文件中应有多少个索引变量？ 索引变量可以用作过程中的分组变量。 在大多数情况下，单个索引变量就 已足够，但是，如果当前文件中的变量组反映多个因子级别，则使用多个索引可能比较合适。 • 一个。 向导将创建单个索引变量。 • 多个。 向导将创建多个索引，请输入要创建的索引数。 所指定的个数会影响下一步，在下一步中向导将自 动创建指定个数的索引。 • 无。 如果不想在新文件中创建索引变量，则选择此项。 变量到个案的单索引示例 在当前数据中，有一个变量组宽度和一个因子时间。 度量宽度三次，并记录到 w1、w2 和 w3 中。 表 16: 单索引的当前数据 主体 w1 w2 w3 1 6.7 4.3 5.7 2 7.1 5.9 5.6 我们会将变量组重构为单个变量宽度，并创建单个数字索引。 新的数据显示在下表中。 表 17: 带有一个索引的新的重构的数据 主体 索引 宽度 1 1 6.7 1 2 4.3 1 3 5.7 2 1 7.1 第 9 章 文件处理和文件转换 97 表 17: 带有一个索引的新的重构的数据 (继续) 主体 索引 宽度 2 2 5.9 2 3 5.6 索引从 1 开始，对组中的每个变量递增。 每次遇到原始文件中的新行时，它会重新开始。 现在就能在需要 分组变量的过程中使用索引了。 变量到个案的双索引示例 当变量组记录多个因子时，您可以创建多个索引；但是，当前数据的排列必须使第一个因子的水平为主索 引，后续因子的水平在主索引内循环。 在当前数据中，存在一个变量组，width 和两个因子 A 和 B。 这些数 据的排列方式决定了因子 B 的级别在因子 A 的级别内。 表 18: 双索引的当前数据 主体 w_a1b1 w_a1b2 w_a2b1 w_a2b2 1 5.5 6.4 5.8 5.9 2 7.4 7.1 5.6 6.7 我们会将变量组重构到单个变量宽度，并创建两个索引。 新的数据显示在下表中。 表 19: 带有两个索引的新的重构的数据 主体 index_a index_b 宽度 1 1 1 5.5 1 1 2 6.4 1 2 1 5.8 1 2 2 5.9 2 1 1 7.4 2 1 2 7.1 2 2 1 5.6 2 2 2 6.7 重组数据向导（变量到个案）：创建一个索引变量 注：如果选择将变量组重组为行并创建一个索引变量，则向导会提供此步骤。 在此步骤中，决定要让索引变量取什么值。 值可以是顺序编号或原始变量组中变量的名称。 还可以为新的 索引变量指定名称和标签。 请参阅第97 页的『变量到个案的单索引示例』主题以获取更多信息。 • 有序数。 向导会自动将顺序编号指定为索引值。 • 变量名称。 向导将使用选定变量组的名称作为索引值。 请从列表中选择一个变量组。 • 名称和标签。 单击单元格可更改缺省变量名称并为索引变量提供描述性变量标签。 重组数据向导（变量到个案）：创建多个索引变量 注：如果选择将变量组重组到行中并创建多个索引变量，则向导会提供此步骤。 在此步骤中，为每个索引变量指定水平数。 还可以为新的索引变量指定名称和标签。 98 IBM SPSS Statistics 29 Core System 用户指南 请参阅第98 页的『变量到个案的双索引示例』主题以获取更多信息。 当前文件中记录了多少个水平？ 考虑当前数据中记录了多少个因子级别。 一个水平定义一组具有完全相同 的条件的个案。 如果有多个因子，则当前数据的排列必须使得第一个因子的水平是主索引，后续因子的水平 在其内循环。 新文件中应有多少个水平？ 输入每个索引的水平数。 多个索引变量的值始终是顺序编号。 这些值从 1 开 始，对每个水平递增。 第一个索引递增最慢，最后一个索引递增最快。 总计组合水平。 创建的水平数不能超过当前数据中存在的水平数。 因为重构的数据中每个处理组合将包含 一行，所以向导会检查您创建的水平数。 它会将您创建的水平的乘积与变量组中的变量数进行比较。 这两 者必须匹配。 名称和标签。 单击单元格可更改缺省变量名称并为索引变量提供描述性变量标签。 重组数据向导（变量到个案）：选项 注：如果选择将变量组重组为行，则向导会提供此步骤。 在此步骤中，指定新的重组的文件的选项。 删除未选中的变量？ 在选择变量的步骤中（第 3 步），从当前数据中选择了要重组的变量组、要复制的变量 以及一个标识变量。 选定变量中的值将出现在新文件中。 如果当前数据中有其他变量，您可以选择是丢弃 它们还是保留它们。 保留缺失值？ 向导在每个可能的新行中检查是否有空值。 空值是系统缺失值或空白值。 您可以选择是保留 还是丢弃只包含空值的行。 创建计数变量？ 向导可在新文件中创建一个计数变量。 它包含由当前数据中的行所生成的新行的数目。 如 果选择丢弃新文件中的空值，那么计数变量可能会有用，因为这样可对当前文件中的给定行生成不同数目的 新行。 单击单元格可更改缺省变量名称并为计数变量提供描述性变量标签。 重组数据向导（个案到变量）：选择变量 注：如果选择将个案组重组为列，则向导会提供此步骤。 在此步骤中，提供关于应如何在新文件中使用当前文件中的变量的信息。 什么标识当前数据中的个案组？ 一个个案组是一组行，这些行由于度量同一个观察单元格（例如个体或机 构）而相关。 向导需要知道当前文件中哪些变量标识个案组，以便能将每个组合并到新文件的单独一行中。 将在当前文件中标识个案组的变量移至“标识变量”列表中。 用来拆分当前数据文件的变量被自动用来标识个 案组。 每次遇到标识值的新组合时，向导将创建一个新行，因此当前文件中的个案应按标识变量的值排序， 顺序与在“标识变量”列表中列出变量的顺序相同。 如果当前文件尚未排序，可以在下一步对其进行排序。 应如何在新文件中创建新的变量组？ 在原始数据中，变量出现在单独一列中。 在新的数据文件中，该变量 将出现在多个新列中。 索引变量是向导应用来创建新列的当前数据中的变量。 重组的数据中将包含与这些 列中的唯一值一一对应的新变量。 将应该用来组成新变量组的变量移至“索引变量”列表中。 当向导提供选 项时，还可以选择按索引对新列排序。 其他列会怎样？ 向导将自动决定如何处理保留在“当前文件”列表中的变量。 它检查每个变量，查看数据值 在个案组中是否变化。 如果有变化，向导会将这些值重组到新文件中的一个变量组中。 如果无变化，向导 会将这些值复制到新文件。 当确定某个变量在组中是否变化时，用户缺失值当作有效值处理，但系统缺失值 则不然。 如果该组包含一个有效或用户缺失值和系统缺失值，则会将其当作在组中无变化的变量来处理，向 导会将这些值复制到新文件。 重组数据向导（个案到变量）：对数据进行排序 注：如果选择将个案组重组为列，则向导会提供此步骤。 在此步骤中，决定在重组当前文件之前是否对其排序。 每次向导遇到新的标识值组合时，都会创建新行，因 此按标识个案组的变量对数据排序就很重要。 行在当前文件中是如何排序的？ 考虑当前数据如何排序，以及使用哪些变量标识个案组（在前一步中指 定）。 第 9 章 文件处理和文件转换 99 • 是。 向导将自动按标识变量对当前数据排序，其顺序与前一步中变量在“标识变量”列表中列出的顺序相 同。 当数据未按标识变量排序，或者您不确定时选择此项。 此选项要求单独遍历数据，但可保证对行正确 排序以便重构。 • 否。 向导不会对当前数据进行排序。 如果确信当前数据已按标识个案组的变量进行了排序，则选择此项。 重组数据向导（个案到变量）：选项 注：如果选择将个案组重组为列，则向导会提供此步骤。 在此步骤中，指定新的重组的文件的选项。 应如何在新文件中对新变量组进行排序？ • 按变量。 向导将从一个原始变量创建的新变量分组在一起。 • 按索引。 向导根据索引变量的值对变量分组。 示例。 要重组的变量是 w 和 h，索引是月份： w、h 和 month 按变量分组的结果是： w.jan、w.feb 和 h.jan 按索引分组的结果是： w.jan、h.jan 和 w.feb 创建计数变量？ 向导可在新文件中创建一个计数变量。 它包含用于在新数据文件中创建行的当前数据中的 行数。 创建指示符变量？ 向导可以使用索引变量在新的数据文件中创建指示符变量。 它为索引变量的每个唯一值 创建一个新变量。 指示符变量指示个案存在值还是缺少值。 如果个案有值，那么指示符变量的值为 1；否则 为 0。 示例。 索引变量为产品。 它记录客户购买的产品。 原始数据为： 表 20: 在单一变量（列）中记录的所有产品 客户 产品 (product) 1 小鸡 1 鸡蛋 2 鸡蛋 3 小鸡 创建指示符变量会生成与产品的唯一值一一对应的新变量。 重组的数据为： 表 21: 分隔每个产品类型的指示符变量 客户 indchick indeggs 1 1 1 2 0 1 3 1 0 在此示例中，重组的数据可以用来获取客户所购买产品的频率计数。 重构数据向导：完成 这是重组数据向导的最后一步。 决定如何处理您的指定。 • 立即重组。 向导将创建新的重组的文件。 如果想要立即替换当前文件，则选择此项。 100 IBM SPSS Statistics 29 Core System 用户指南 注：如果原始数据已加权，则除非已重组了用作权重的变量或者已将其从新文件删除，否则新数据也会加 权。 • 粘贴语法。 向导会将其生成的语法粘贴到语法窗口。 当尚未准备好替换当前文件时，想要修改语法时，或 者想要保存语法以供将来使用时，选择此项。 第 9 章 文件处理和文件转换 101 102 IBM SPSS Statistics 29 Core System 用户指南 第 10 章 处理输出 运行过程时，结果显示在称为“查看器”的窗口中。 “查看器”窗口可以通过两种方式运作。 经典（语法和输出） 缺省方式以经典查看器格式显示输出。 在该方式下，可以很容易地浏览到您要查看的输出。 还可以操作 输出并创建刚好包含您需要的输出的文档。 工作簿 工作簿方式可将 SPSS Statistics 语法编辑功能与笔记本方法联系起来，其中，笔记本方法提供交互式方 法来运行语法和查看相应的输出。 工作簿文档 (.spwb) 是由各个段落组成。 这些段落包含各个输出元 素（语法、过程、图表，等等）。 语法段落提供了完整的语法编辑和运行功能。 非语法段落提供了完整 的富文本格式编辑功能。 有关切换输出方式的信息，请参阅第163 页的『一般选项』。 使用输出 运行过程时，结果显示在称为“查看器”的窗口中。 在该窗口中，可以很容易地浏览到您要查看的输出。 您 也可以使用输出并创建包含所需的确切输出内容的文档。 查看器 - 经典 缺省情况下，输出结果显示在查看器中。 您可以使用查看器来完成下列任务： • 浏览结果 • 显示或隐藏选中的表格和图表 • 通过移动选中的项更改结果的显示顺序 • 在查看器和其他应用程序间移动项 查看器分为两个窗格： • 左窗格包含内容的概要视图。 • 右窗格包含统计表、图表和文本输出。 您可以单击概要中的项，以直接转到相应的表或图表。 可以单击并拖动概要窗格的右边框来更改概要窗格的 宽度。 更改概要级别 注: 仅当查看器处于经典（语法和输出）模式（编辑 > 选项 > 常规 > 应用程序模式 > 经典（语法和输出）） 时，以下指导信息才适用。 1. 在概要窗格中单击项。 2. 从菜单中选择： 编辑 > 概要 > 上升 或 编辑 > 概要 > 降级 将项添加到查看器 注: 仅当查看器处于经典（语法和输出）模式（编辑 > 选项 > 常规 > 应用程序模式 > 经典（语法和输出）） 时，以下指导信息才适用。 在查看器中，您可以添加项，如标题、新文本、图表或来自其他应用程序的材料。 添加标题或文本 可以将没有连接到表格或图表的文本项添加到查看器。 1. 单击将放在标题或文本前面的表、图表或其他对象。 2. 从菜单中选择： 插入 > 新标题 或 插入 > 新文本 3. 双击新对象。 4. 输入文本。 添加文本文件 1. 在查看器的概要窗格或内容窗格中，单击要放在文本前面的表格、图表或其他对象。 2. 从菜单中选择： 插入 > 文本文件...
3. 选择文本文件。 要编辑文本，请双击该文本。 将对象粘贴到查看器 来自其他应用程序的对象可以粘贴到查看器中。 您可以使用“粘贴于后”或“选择性粘贴”。 两种类型的粘贴都 把新对象放在查看器中当前选中的对象的后面。 要选择粘贴对象的格式，请使用“选择性粘贴”。 查找和替换查看器中的信息 1. 要在查看器中查找或替换信息，请从菜单中选择： 编辑 > 查找 或 编辑 > 替换 您可使用“查找和替换”来： • 搜索整个文档或仅选中项。 • 从当前位置向下或向上搜索。 • 搜索窗格或限制搜索内容或概要窗格。 • 搜索隐藏项。 这些项包括内容窗格中所有处于隐藏状态的项（例如，缺省情况下处于隐藏状态的“注释” 表）以及透视表中处于隐藏状态的行和列。 • 限制搜索条件，以进行区分大小写的匹配。 • 限制透视表中的搜索标准至整个单元格内容匹配。 • 限制透视表中的搜索条件，以便仅与脚注标记进行匹配。 如果在查看器中选择了除透视表以外的任何其他 内容，那么此选项不可用。 隐藏的项和透视表层 • 多维透视表的当前可见层下方的层并不会被视为处于隐藏状态，并且将包括在搜索区域中，即使搜索未包 括隐藏项也是如此。 • 隐藏项目包括内容窗格中的隐藏项（概要窗格中有合上的书形图标的项，或包含在概要窗格的折叠块中的 项）和透视表中缺省情况下隐藏的行和列（例如，缺省情况下隐藏的空行和空列）或手动隐藏的行和列 （通过编辑表并选择隐藏特定行或列）。 隐藏项目仅在您明确选择包含隐藏项目时包含在搜索中。 104 IBM SPSS Statistics 29 Core System 用户指南 • 在两种情况下，包含搜索文本或值的隐藏或不可见元素会在找到时显示，但项目随后即返回其原始状态。 在透视表中查找值范围 要在透视表中查找属于指定范围值中的值： 1. 在查看器中激活透视表或者选择一个或多个透视表。 确保只选中了透视表。 如果选中了其他任何对象， “范围”选项不可用。 2. 从菜单中选择： 编辑 > 查找 3. 单击范围选项卡。 4. 指定范围类型：之间、大于或等于、小于或等于。 5. 选择定义范围的一或多个值。 • 如果任一值包含非数值的字符，將两个值视为字符串。 • 如果两个值都是数字，只搜索数值。 • 使用“范围”选项卡无法替换值。 此功能对遗存表不可用。 有关更多信息，请参阅第131 页的『遗存表』主题。 保存输出 注: 仅当查看器处于经典（语法和输出）模式（编辑 > 选项 > 常规 > 应用程序模式 > 经典（语法和输出）） 时，以下信息才适用。 可以将查看器的内容保存为以下格式： • 查看器文件 (.spv)。 用于在“查看器”窗口中显示文件的格式。 要以其他格式（例如，文本、Word、Excel）保存结果，请使用文件 > 导出。 保存查看器文档 注: 仅当查看器处于经典（语法和输出）模式（编辑 > 选项 > 常规 > 应用程序模式 > 经典（语法和输出）） 时，以下指导信息才适用。 1. 从查看器窗口菜单中选择： 文件 > 保存 2. 输入文档的名称，然后单击保存。 您可以选择性地执行下列操作： 锁定文件以防止在 IBM SPSS Smartreader 中进行编辑 如果锁定了查看器文档，您可以操作透视表（对调行与列，更改显示层等），但无法在 IBM SPSS Smartreader 中编辑任何输出或保存任何更改到查看器文档（使用查看器文档的独立产品）。 此设置 对在 IBM SPSS Statistics 或 IBM SPSS Modeler 中打开的查看器文档没有影响。 使用密码对文件进行加密 通过使用密码对文档进行加密，可以对查看器文档中存储的机密信息进行保护。 一旦加密，文档只能 通过提供密码打开。 还需要 IBM SPSS Smartreader 用户提供密码才能打开文件。 要对查看器文档进行加密，请完成下列步骤： a. 在“将输出另存为”对话框中，选择使用密码加密文件。 b. 单击保存。 c. 在“对文件进行加密”对话框中，请提供密码，并在“确认密码”文本框中重新输入该密码。 密码限 制在 10 个字符并区分大小写。 警告：密码丢失后将无法恢复。 如果密码丢失，那么将无法打开文件。 创建强密码 第 10 章 处理输出 105 • 至少使用八个字符。 • 在密码中使用数字、符号甚至标点符号。 • 避免使用数字序列或字符序列（例如 "123" 和 "abc"）并避免重复，例如 "111aaa"。 • 不要创建使用个人信息（例如生日或昵称）的密码。 • 定期更改密码。 注：不支持将已加密的文件存储到 IBM SPSS 协作和部署服务存储库 中。 修改已加密的文件 • 如果打开加密文件，对其进行修改并选择“文件” > “保存”，修改后的文件将以相同的密码保存。 • 您可以通过打开文件、重复加密步骤并在“加密文件”对话框中指定不同的密码，在加密的文件上更 改密码。 • 通过打开文件，选择“文件 > 另存为”并在"将数据“另存为”对话框中取消选择使用密码加密文件，可 以保存加密文件的未加密版本。 注意：在 V21 之前的 IBM SPSS Statistics 版本中，无法打开经过加密的数据文件和输出文档。 在 V22 之前的版本中，无法打开经过加密的语法文件。 请使用输出文档存储所需的模型信息(O) 只有在输出文档中存在需要辅助信息来启用一些交互式功能的模型查看器项时，此选项才适用。 单击 更多信息将显示这些模型查看器项和需要辅助信息的交互式功能的列表。 随输出文档一起存储此信息 可能会显著增大文档大小。 即使您选择不存储此信息，也仍然可以打开这些输出项，但指定的交互式 功能将不可用。 交互式输出 交互式输出对象包含多个相关的输出对象。 在一个对象中所作的选择可以更改其他对象中显示或突出显示的 内容。 例如，在表中选择一行可能会在地图中突出显示一个区域或者针对另一个类别显示图表。 交互式输出对象不支持编辑功能，例如更改文本、颜色、字体或表边框。 您可以将各个对象从交互式对象复 制到查看器。 在透视表编辑器中，可以对那些从交互式输出复制的表进行编辑。 从交互式输出复制对象 文件 > 复制到查看器将各个输出对象复制到“查看器”窗口。 • 可用的选项视交互式输出的内容而定。 • 图表和地图将创建图表对象。 • 表将创建透视表，您可以在透视表编辑器中编辑这个表。 • 快照将创建当前视图的图像。 • 模型将创建当前交互式输出对象的副本。 编辑 > 复制对象将各个输出对象复制到剪贴板。 • 将复制的对象粘贴到查看器相当于文件 > 复制到查看器。 • 将对象粘贴到另一个应用程序时，将以图像形式粘贴该对象。 缩放和平移 对于地图，您可以使用查看 > 缩放来缩放地图视图。 在缩放后的地图视图中，可以使用查看 > 平移来移动视 图。 打印设置 文件 > 打印设置控制如何打印交互式对象。 • 仅打印可见视图。 仅打印当前显示的视图。 此选项是缺省设置。 106 IBM SPSS Statistics 29 Core System 用户指南 • 打印所有视图。 打印交互式输出中包含的所有视图。 • 选择的选项还将确定用于导出输出对象的缺省操作。 导出输出 导出输出以 HTML、文本、Word、Excel 和 PDF 格式保存查看器输出。 图表也可以以多种不同图形格式导 出。 要导出输出 1. 使查看器成为活动窗口（单击该窗口中的任意位置）。 2. 从菜单中选择： 文件 > 导出...
3. 输入一个文件名（或图表前缀）并选择一种导出格式。 要导出的对象 您可以导出查看器中的所有对象、所有可视对象或者仅选定的对象。 文档类型 可用选项包括： • Word/RTF (.doc)。 透视表导出为 Word 表，并带有所有完整的格式属性（例如，单元格边框、字体 样式和背景色）。 文本输出导出为格式化的 RTF。 图表、树形图和模型视图包括在 PNG 格式中。 请 注意，Microsoft Word 可能无法正确显示过宽的表格。 注: 粘贴选项不可用于 Word/RTF (doc) 设置。 • Word (docx)。 透视表导出为 Word 表，并带有所有完整的格式属性（例如，单元格边框、字体样式 和背景色）。 .docx 文件是 XML 文件的归档。 文本输出为 XML。 图表、树图和模型视图包含为高 分辨率图像（.eps（对于 macOS）和 .emf（对于 Windows）。 请注意，Microsoft Word 可能无 法正确显示过宽的表格。 • Excel。 透视表行、列和单元格导出为 Excel 行、列和单元格，并带有所有完整的格式属性（例如单元 格边框、字体样式和背景色）。 文本输出连同所有字体属性一起导出。 文本输出中的一行即为 Excel 文件中的一行，文本输出中行的所有内容包含在单个单元格中。 图表、树形图和模型视图包括在 PNG 格式中。 输出可以作为 Excel 97-2004 或 Excel 2007 及更高版本导出。 • HTML (.htm)。 透视表导出为 HTML 表。 文本输出导出为预设置格式的 HTML。 图表、树形图和模 型视图都按所选图形格式嵌入文档中。 需要使用与 HTML 5 兼容的浏览器才能查看以 HTML 格式导出 的输出。 • 便携文档格式 (.pdf)。 所有输出都将按?打印预览?中的显示导出，所有格式设置特性都不变。 • 文本。 文本输出格式包括纯文本、UTF-8 和 UTF-16。 透视表可以制表符分隔格式或空格分隔格式导 出。 所有文本输出以空格分隔格式导出。 对于图表、树形图和模型视图，文本文件中将针对每个图片 插入一个相应的行，指示图像文件名称。 • 无（仅图形）。 可用的导出格式包括：BMP、EPS、SVG、JPEG、PNG、TIF 和 PDF。 在 Windows 操作系统上还可使用 EMF（增强型图元文件）格式。 打开所在文件夹 打开导出操作所创建的文件所在的文件夹。 输出管理系统。 还可以将所有输出或用户指定的输出类型自动导出为 Word、Excel、PDF、HTML、文本或 IBM SPSS Statistics 格式的数据文件。 请参阅第217 页的『第 22 章 输出管理系统』以获取更多信息。 HTML 选项 HTML 导出需要与 HTML 5 兼容的浏览器。 下列选项可用于以 HTML 格式导出输出： 透视表中的层。 缺省情况下，包括或排除透视表层由各个透视表的表属性控制。 您可以覆盖此设置，包含 所有层或排除当前可视层之外的所有层。 有关更多信息，请参阅第126 页的『表属性：打印』主题。 第 10 章 处理输出 107 将分层的表格导出为交互式。 分层表将按其在查看器中的外观显示，而且您可以在浏览器中以交互方式更改 显示的层。 如果不选择此选项，那么每个表层都会显示为一个独立的表。 HTML 表格。 此选项控制导出的透视表所包括的样式信息。 • 使用样式及固定宽度列导出。 保留所有透视表样式信息（字体样式、背景色等）和列宽。 • 不使用样式导出。 透视表转换为缺省 HTML 表格。 不保留样式属性。 自动确定列宽。 包括脚注及文字说明。 控制包含或排除所有透视表脚注及文字说明。 模型视图。 缺省情况下，包含或排除模型视图由每个模型的模型属性控制。 您可以覆盖此设置，包含所有 视图或排除当前可见视图之外的所有视图。 请参阅第133 页的『模型属性』主题以获取更多信息。 （注： 包括表在内的所有模型视图将导出为图形。） 注：对于 HTML，您还可以控制导出图表的图像文件格式。 有关更多信息，请参阅第110 页的『图形格式选 项』主题。 设置 HTML 导出选项 1. 选择 HTML 作为导出格式。 2. 单击更改选项。 导出输出：Word/DOCX 和 Word/RTF 选项 下列选项可用来以 Word 文档 (.docx) 和 Word/RTF (.doc) 格式导出输出： 透视表中的层。 缺省情况下，包括或排除透视表层由各个透视表的表属性控制。 您可以覆盖此设置，包含 所有层或排除当前可视层之外的所有层。 有关更多信息，请参阅第126 页的『表属性：打印』主题。 宽透视表。 控制当表相对于定义文档宽度时过宽的处理方法。 缺省情况下，表将换行以适合宽度。 表将划 分为不同的部分，且行标签对每个表部分重复。 另外，您还可以缩小宽表，或者不改动宽表，而允许其延伸 到定义的文档宽度以外。 保留分界点。 如果您已定义分界点，那么这些设置将保留在 Word 表格中。 包括脚注及文字说明。 控制包含或排除所有透视表脚注及文字说明。 模型视图。 缺省情况下，包含或排除模型视图由每个模型的模型属性控制。 您可以覆盖此设置，包含所有 视图或排除当前可见视图之外的所有视图。 请参阅第133 页的『模型属性』主题以获取更多信息。 （注： 包括表在内的所有模型视图将导出为图形。） 导出页面设置。 这将打开一个对话框，在此可以定义导出文档的页面大小和边距。 用于确定换行和缩小行 为的文档宽度是页面宽度减去左右边距。 设置 Word 导出选项 1. 选择 Word 文档 (.docx) 或 Word/RTF (.doc) 作为导出格式。 2. 单击更改选项...。 Excel 选项 以下选项可用于以 Excel 格式导出输出： 创建一个工作表或工作簿，或修改一个现有工作表。 缺省情况下，将创建一个新工作簿。 如果拥有指定名 称的文件已存在，将被覆盖。 如果您选择创建工作表的选项，且拥有指定名称的工作表已存在于指定文件 中，那么将被覆盖。 如果您选择修改现有工作表的选项，您还必须指定工作表名称。 （这对创建工作表为 可选选项。） 工作表名称不能超过 31 个字符，不能包含正斜杠或反斜杠、方括号、问号或星号。 导出到 Excel 97-2004 时，如果您修改一个现有工作表，图表、模型视图和树形图则不包括在导出输出中。 工作表中的位置。 控制导出输出在工作表中的位置。 缺省情况下，导出输出将添加到拥有任何内容的最后 一列之后，从第一行开始，而不修改任何现有内容。 这对于将新列添加到现有工作表是一个不错的选择。 将导出输出添加到最后一行之后对于将新行添加到现有工作表是一个不错的选择。 将导出输出添加到特定单 元格位置的开始将覆盖添加导出输出的区域中的任何现有内容。 108 IBM SPSS Statistics 29 Core System 用户指南 透视表中的层。 缺省情况下，包括或排除透视表层由各个透视表的表属性控制。 您可以覆盖此设置，包含 所有层或排除当前可视层之外的所有层。 有关更多信息，请参阅第126 页的『表属性：打印』主题。 包括脚注及文字说明。 控制包含或排除所有透视表脚注及文字说明。 模型视图。 缺省情况下，包含或排除模型视图由每个模型的模型属性控制。 您可以覆盖此设置，包含所有 视图或排除当前可见视图之外的所有视图。 请参阅第133 页的『模型属性』主题以获取更多信息。 （注： 包括表在内的所有模型视图将导出为图形。） 设置 Excel 导出选项 1. 选择 Excel 97-2004 (.xls)、Excel 2007 和更高版本 (.xlsx) 或启用宏的 Excel 2007 和更高版本 (.xlsm) 作为导出类型。 2. 单击更改选项。 PowerPoint 选项 对于 PowerPoint，可用的选项如下所示： 透视表中的层。 缺省情况下，包括或排除透视表层由各个透视表的表属性控制。 您可以覆盖此设置，包含 所有层或排除当前可视层之外的所有层。 有关更多信息，请参阅第126 页的『表属性：打印』主题。 宽透视表。 控制当表相对于定义文档宽度时过宽的处理方法。 缺省情况下，表将换行以适合宽度。 表将划 分为不同的部分，且行标签对每个表部分重复。 另外，您还可以缩小宽表，或者不改动宽表，而允许其延伸 到定义的文档宽度以外。 包括脚注及文字说明。 控制包含或排除所有透视表脚注及文字说明。 使用查看内略图项作为幻灯片标题。 在导出操作创建的每个幻灯片上包括标题。 每个幻灯片都包含从查看 器中导出的单个项。 标题是根据查看器概要窗格中的项的概要条目构造而成。 模型视图。 缺省情况下，包含或排除模型视图由每个模型的模型属性控制。 您可以覆盖此设置，包含所有 视图或排除当前可见视图之外的所有视图。 请参阅第133 页的『模型属性』主题以获取更多信息。 （注： 包括表在内的所有模型视图将导出为图形。） 导出页面设置。 这将打开一个对话框，您可以在其中定义导出的文档的页面大小和边距。 用于确定换行和 缩小行为的文档宽度是页面宽度减去左右边距。 设置 PowerPoint 导出选项 1. 选择 PowerPoint 作为导出格式。 2. 单击更改选项。 注：导出到 PowerPoint 仅在 Windows 操作系统上可用。 PDF 选项 对于 PDF，可用选项有： 嵌入书签。 此选项提供 PDF 文档中对应于查看器概要条目的书签。 与查看器概要窗格一样，通过书签可以 更轻松地浏览具有大量输出对象的文档。 内嵌字体。 内嵌字体确保 PDF 文档在所有计算机上外观相同。 否则，如果文档中使用的一些字体在用于查 看（或打印）PDF 文档的计算机上不可用，则替换字体的效果可能欠佳。 透视表中的层。 缺省情况下，包括或排除透视表层由各个透视表的表属性控制。 您可以覆盖此设置，包含 所有层或排除当前可视层之外的所有层。 有关更多信息，请参阅第126 页的『表属性：打印』主题。 模型视图。 缺省情况下，包含或排除模型视图由每个模型的模型属性控制。 您可以覆盖此设置，包含所有 视图或排除当前可见视图之外的所有视图。 请参阅第133 页的『模型属性』主题以获取更多信息。 （注： 包括表在内的所有模型视图将导出为图形。） 设置 PDF 导出选项 1. 选择可移植文档格式作为导出格式。 2. 单击更改选项。 第 10 章 处理输出 109 其他影响 PDF 输出的设置 页面设置/页面属性。 PDF 文档中的页面大小、方向、边距、页眉及页脚的内容和显示以及打印的图表大小 均由页面设置选项和页面属性选项控制。 表格属性/表格外观。 宽表和/或长表的比例调整以及表层打印由每个表的表属性控制。 这些属性也可保存在 表格外观中。 缺省/当前打印机。 PDF 文档的分辨率 (DPI) 是缺省打印机或当前选定的打印机的当前分辨率设置（可以使 用“页面设置”进行更改）。 最高分辨率为 1200 DPI。 如果打印机的设置更高，则 PDF 文档的分辨率将为 1200 DPI。 注意：在低分辨率的打印机上打印高分辨率文档时，打印的效果可能会比较差。 文本选项 将输出导出到文本 (.txt) 时，有三个可用的选项。 文本 - 纯文本 (.txt) 创建一个文本文件，其中包含文本输出以及用于分隔图形文件的路径。 多种图形文件格式可供使用。 文本 - UTF8 (.txt) 创建一个文本文件，其中包含 Unicode 文本输出以及用于分隔图形文件的路径。 使用 UTF-8 编码。 多 种图形文件格式可供使用。 文本 - UTF16 (.txt) 创建一个文本文件，其中包含 Unicode 文本输出以及用于分隔图形文件的路径。 使用 UTF-16 编码。 多 种图形文件格式可供使用。 对于文本输出，可用选项有： 透视表格式。 透视表可以制表符分隔格式或空格分隔格式导出。 对于空格分隔格式，您还可以控制： • 列宽。 自动适应不会对任何列内容换行，每列宽度为该列中最宽标签或值的宽度。 定制为表中的所有列设 置最大列宽，超过该宽度的值将换行，从而在该列的下一行中继续显示。 • 行/列边框字符。 控制用于创建行和列边框的字符。 如果您不希望显示行边框和列边框，请输入空格作为 值。 透视表中的层。 缺省情况下，包括或排除透视表层由各个透视表的表属性控制。 您可以覆盖此设置，包含 所有层或排除当前可视层之外的所有层。 有关更多信息，请参阅第126 页的『表属性：打印』主题。 包括脚注及文字说明。 控制包含或排除所有透视表脚注及文字说明。 模型视图。 缺省情况下，包含或排除模型视图由每个模型的模型属性控制。 您可以覆盖此设置，包含所有 视图或排除当前可见视图之外的所有视图。 请参阅第133 页的『模型属性』主题以获取更多信息。 （注： 包括表在内的所有模型视图将导出为图形。） 设置文本导出选项 1. 选择文本 - 纯文本 (.txt)、文本 - UTF8 (.txt) 或文本 - UTF16 (.txt) 作为导出格式。 2. 单击更改选项...。 适用于只导出图形的选项 对于只导出图形，可用选项有： 模型视图。 缺省情况下，包含或排除模型视图由每个模型的模型属性控制。 您可以覆盖此设置，包含所有 视图或排除当前可见视图之外的所有视图。 请参阅第133 页的『模型属性』主题以获取更多信息。 （注： 包括表在内的所有模型视图将导出为图形。） 图形格式选项 对于 HTML 与文本文档，如果仅导出图表，您可以选择图形格式，并为每种图形格式控制不同可选设置。 要选择导出图表的图形格式和选项： 1. 选择 HTML、文本或无（只有图形）作为文档类型。 110 IBM SPSS Statistics 29 Core System 用户指南 2. 从下拉列表中选择图形文件格式。 3. 单击更改选项更改选定图形文件格式选项。 JPEG 图表导出选项 • 图像大小。 原始图表大小的百分比，可达 200％。 • 转换为灰度。 将颜色转换成灰色阴影。 BMP 图表导出选项 • 图像大小。 原始图表大小的百分比，最高可达 200%。 • 压缩图像，以减小文件大小。 这是一种无损压缩技术，这种技术能够创建较小的文件，而不会影响图像质 量。 PNG 图表导出选项 图像大小。 原始图表大小的百分比，可达 200％。 颜色深度。 决定导出图表中的颜色的数量。 在任何深度下保存的图表最少的颜色数是实际使用的颜色数， 最大则是深度允许的颜色数。 例如，如果图表包含三种颜色：红色、白色和黑色，而您把它保存为 16 色， 那么图表仍将保持三种颜色。 • 如果图表中的颜色数超出了该深度的颜色数，那么颜色将进行仿色处理，以复制图表中的颜色。 • 当前屏幕深度是当前显示在计算机显示器上的颜色数。 EMF 和 TIFF 图表导出选项 图像大小。 原始图表大小的百分比，最高可达 200%。 注意：EMF（增强型图元文件）格式只在 Windows 操作系统上可用。 EPS 图表导出选项 图像大小。 您可以将大小指定为原图像大小的百分比（可达 200％），或指定图像的像素宽度（高度取决于 宽值和宽高比）。 导出的图像总是与原始图像成比例。 包括 TIFF 预览图像。 将包含 EPS 图像的预览保存为 TIFF 格式，便于在不能在屏幕上显示 EPS 图像的应用 程序中进行显示。 字体。 控制 EPS 图像中的字体处理。 • 使用字体参考。 如果图表中使用的字体在输出设备上可用，那么将使用这些字体。 否则，输出设备将使用 备用字体。 • 将字体替换为曲线。 将字体转换为 PostScript 曲线数据。 文本本身不再像可以编辑 EPS 图形的应用程序 中的文本那样可编辑。 如果图表中使用的字体对输出设备不可用，则该选项十分有用。 查看器打印 有两个选项可用于打印“查看器”窗口的内容： 所有可见输出。 仅打印当前显示在内容窗格中的项。 不打印隐藏的项（概要窗格中有合上的书形图标的 项，或者折叠的概要层中隐藏的项）。 选择。 仅打印概要窗格和/或内容窗格中当前选中的项。 打印输出和图表 1. 使查看器成为活动窗口（单击该窗口中的任意位置）。 2. 从菜单中选择： 文件 > 打印...
3. 选择所需的打印设置。 第 10 章 处理输出 111 4. 单击确定进行打印。 打印预览 “打印预览”显示查看器文档的每一页上将会打印的内容。 我们建议您在实际打印查看器文档前查看“打印预 览”，这是因为“打印预览”将显示查看器的内容窗格中可能不显示的项，其中包括： • 分页符 • 透视表隐藏的层 • 宽表中的分界符 • 打印在每一页上的页眉和页脚 如果在查看器中当前选中了任何输出，那么预览将仅显示选中的输出。 要查看所有输出的预览，请确保查看 器中没有选择任何输出。 页面属性：页眉和页脚 页眉和页脚是打印在每一页的顶部和底部的信息。 您可以输入任何文本用作页眉和页脚。 还可以使用对话 框中间的工具栏插入： • 日期和时间 • 页码 • 查看器文件名 • 概要标题标签 • 页面标题和子标题 • 设为缺省使用此处指定的设置作为新查看器文档的缺省设置。 （注：此选项会将“页眉/页脚”选项卡和“选 项”选项卡上的当前设置变为缺省设置。） • 概要标题标签表示每一页上的第一项的第一、第二、第三和/或第四层的概要标题。 • 页面标题和子标题打印当前的页面标题和子标题。 它们是用查看器“插入”菜单上的“新建页面标题”或 TITLE 和 SUBTITLE 命令创建的。 如果没有指定任何页面标题或子标题，则该设置将被忽略。 注意：新页面标题和子标题的字体特征由“选项”对话框（通过选择“编辑”菜单中的“选项”来访问）的“查看 器”选项卡控制。 现有页面标题和子标题的字体特征可以通过在查看器中编辑标题进行更改。 要查看页眉和页脚在打印页面上的效果，可选择“文件”菜单上的打印预览。 插入页面的页眉和页脚 1. 使查看器成为活动窗口（单击该窗口中的任意位置）。 2. 从菜单中选择： 文件 > 页面属性...
3. 单击页眉/页脚选项卡。 4. 输入要显示在每一页上的页眉和/或页脚。 页面属性：选项 该对话框控制打印图表大小、打印输出项之间的空格和页编号。 • 打印图表尺寸。 控制相对于定义的页面大小的打印图表的大小。 图表的宽高比（宽与高的比）不受打印图 表大小的影响。 图表的整体打印大小受其高度和宽度的限制。 当图表的外边框达到页面的左右边框时，图 表大小就无法再扩展以填充更多页面高度。 • 各项之间的间距。 控制打印的各项之间的间距。 每个透视表、图表和文本对象都是一个独立的项。 该设 置不影响各项在查看器中的显示。 • 起始页码。 用指定的数字开始按顺序编排页码。 • 设为缺省。 此选项使用此处指定的设置作为新查看器文档的缺省设置。 （请注意，此选项将当前页眉/页 脚设置和选项设置设定为缺省值。） 112 IBM SPSS Statistics 29 Core System 用户指南 更改打印图表大小、页编号和打印项之间的空格 1. 使查看器成为活动窗口（单击该窗口中的任意位置）。 2. 从菜单中选择： 文件 > 页面属性...
3. 单击选项选项卡。 4. 更改设置并单击确定。 查看器 - 工作簿 以工作簿方式查看输出可将 SPSS Statistics 语法编辑功能与笔记本方法联系起来，其中，笔记本方法提供交 互式方法来运行语法和查看相应的输出。 工作簿文档 (.spwb) 是由各个段落组成。 这些段落包含输出元素 （语法、表、图表，等等）。 语法段落提供了完整的语法编辑和运行功能。 富文本格式段落提供了完整的 富文本格式编辑功能。 每个工作簿提供以下方面的选项： • 浏览结果 • 生成和运行语法 • 创作和编辑富文本格式 • 编辑输出元素 • 剪切、复制、粘贴和删除工作簿元素 • 撤销和重做操作 • 显示或隐藏选中的表格和图表 • 通过移动选中的项更改结果的显示顺序 • 在查看器和其他应用程序间移动各项 工作簿包含五个区域： 菜单栏 提供一般产品功能。 工具栏 提供对常用功能的快捷访问。 导航窗格 包含工作簿内容的概要视图。 内容窗格 包含段落工作空间，其中包含语法编辑器、统计表、图表和文本输出。 状态栏 提供产品状态消息。 您可以单击导航窗格中的项目，以直接转至内容窗格中的相应项。 您可以单击并拖动导航窗格的右边框，以 更改窗格的宽度。 工作簿工具栏 处于工作簿方式时，查看器以经典方式共享大部分工具栏选项。 以下工具栏选项是查看器在处于工作簿方式 时所特有的。 表 22: 工作簿工具栏选项 工具栏选项 描述 创建新工作簿语法段落。 第 10 章 处理输出 113 表 22: 工作簿工具栏选项 (继续) 工具栏选项 描述 创建新工作簿富文本格式段落。 显示活动数据集。 如果打开了其他数据集，可以从 下拉列表中进行选择。 不受支持的功能 以下工作簿功能不受支持： • 工作簿方式不支持 COM 脚本编制。 • 剪切、复制、粘贴或删除单个语法或富文本格式段落 • 在不同段落之间剪切、复制、粘贴或删除节点 • 生产设施作业 • 以下语法命令（您必须针对这些命令切换为经典方式）： – OUTPUT ACTIVATE – OUTPUT CLOSE – OUTPUT DISPLAY – OUTPUT EXPORT – OUTPUT MODIFY – OUTPUT NAME – OUTPUT NEW – OUTPUT OPEN – OUTPUT SAVE – OMS 命令 • 自动恢复 • 脚本编制 创建、打开和保存工作簿输出 注: • 仅当查看器处于工作簿方式（编辑 > 选项 > 通用 > 应用方式 > 工作簿）时，以下信息才适用。 • 处于工作簿方式时，无法创建或打开单独的语法或输出文档。 创建新工作簿 1. 从 SPSS Statistics 菜单中选择： 文件 > 新建 > 工作簿 新的查看器窗口将以工作簿方式打开，提供交互式方法来运行语法和查看相应的输出。 缺省情况下，新 工作簿包含空的语法段落。 2. 有关处理工作簿的信息，请参阅第115 页的『将项添加到工作簿』。 打开工作簿 1. 从 SPSS Statistics 菜单中选择： 114 IBM SPSS Statistics 29 Core System 用户指南 文件 > 打开 > 工作簿...
2. 选择现有工作簿文件 (.spwb) 并单击打开。 3. 有关处理工作簿的信息，请参阅第115 页的『将项添加到工作簿』。 保存工作簿 工作簿的内容可以保存到 IBM SPSS Statistics 工作簿文件 (.spwb)。 1. 从 SPSS Statistics 菜单中选择： 文件 > 保存或文件 > 另存为...
2. 要以其他格式（例如，文本、Word、Excel）保存结果，请选择文件 > 导出。 将项添加到工作簿 注: 仅当查看器处于工作簿方式（编辑 > 选项 > 通用 > 应用方式 > 工作簿）时，以下指示信息才适用。 查看器处于工作簿方式时，您可以添加项语法段落、富文本格式段落或来自其他应用程序的材料。 添加语法段落 语法段落允许您创建新语法或修改现有 IBM SPSS Statistics 语法。 1. 在概要或段落工作空间中选择一个位置。 2. 从菜单中选择： 插入 > 新语法段落 或 单击工具栏上的创建新语法段落控件。 3. 在新语法段落中输入或编辑 SPSS Statistics 语法。 4. 单击运行语法控件以运行对应段落中的语法。 单击全部控件以运行所有可用段落中的语法。 添加富文本格式段落 1. 在概要窗格或段落工作空间中，单击要放在富文本格式段落前面的表格、图表或其他对象。 2. 从菜单中选择： 插入 > 新富文本格式段落 3. 在新富文本格式段落中输入所需文本。 每个富文本格式段落提供了一个编辑器，其中用户可以输入格式化文本（例如，字体大小、字体系列、文本 颜色等）。 您可以利用编辑器为工作簿语法和输出元素编写注释或描述。 将对象粘贴到工作簿中 来自其他应用程序的对象可以粘贴到工作簿中。 可以使用之后粘贴或选择性粘贴。 两种类型的粘贴都把新 对象放在工作簿中当前选中的对象的后面。 如果您要选择所粘贴对象的格式，请使用选择性粘贴。 查看器概要 概要窗格提供查看器文档的目录。 您可以使用概要窗格来浏览结果并控制显示。 概要窗格中的大部分操作 对内容窗格有相应的影响。 • 在概要窗格中选择一项将在内容窗格中显示相应的项。 • 在概要窗格中移动一项将移动内容窗格中相应的项。 • 折叠概要视图将隐藏所折叠级别中所有的项的结果。 第 10 章 处理输出 115 控制概要显示 要控制概要显示，您可以执行下列操作： • 展开和折叠概要视图 • 更改选定各项的概要级别 • 更改概要显示中的各项的大小 • 更改概要显示中使用的字体 折叠和展开概要视图 1. 单击要折叠或展开的概要项左边的框。 或 2. 单击概要中的项。 3. 从菜单中选择： 视图 > 折叠 或 视图 > 展开 更改概要项的大小 1. 从菜单中选择： 查看 > 概要大小 2. 选择概要大小（小、中或大）。 更改概要中的字体 1. 从菜单中选择： 查看 > 概要字体...
2. 选择一种字体。 显示和隐藏结果 在查看器中，可以选择性地显示和隐藏单个表或者整个过程的结果。 当您想减少内容窗格中的可视输出量 时，此过程十分有用。 隐藏表格和图表 1. 在查看器的概要窗格中，双击该项的书形图标。 或 2. 单击某项将其选中。 3. 从菜单中选择： 视图 > 隐藏 或 4. 单击“概要”工具栏上的合上的书形（隐藏）图标（仅限经典方式）。 打开的书形（显示）图标变为活动图标，表示该项现在已隐藏。 隐藏过程结果 1. 在概要窗格中单击过程名称左侧的框。 116 IBM SPSS Statistics 29 Core System 用户指南 这将隐藏此程序的所有结果，并将概要视图折叠。 移动、删除和复制输出 通过复制、移动或删除一个项或一组项，可以重新排列结果。 在查看器中移动输出 1. 选择概要窗格或内容窗格中的某项。 2. 右键单击并从菜单中选择剪切。 3. 选择用于放置这些项的位置，右键单击，然后选择之后粘贴。 或 4. 将选中的项拖放到不同的位置（仅限经典方式）。 在查看器中删除输出 1. 选择概要窗格或内容窗格中的某项。 2. 按 Delete 键。 或 3. 从菜单中选择： 编辑 > 删除 更改初始对齐方式 缺省情况下，所有的结果最初都是左对齐的。 要更改新输出项的初始对齐方式： 1. 从菜单中选择： 编辑 > 选项 2. 单击查看器选项卡。 3. 在“初始输出状态”组中，选择项类型（例如，透视表、图表、文本输出）。 4. 选择所需的对齐选项。 更改各输出项的对齐方式 1. 在概要或内容窗格中，单击要对齐的项。 2. 从菜单中选择： 格式 > 左对齐 或 格式 > 居中 或 格式 > 右对齐 复制输出到其他应用程序中 输出对象可以复制并粘贴到其他应用程序中，如字处理程序或电子表格。 可以粘贴多种格式的输出。 根据 目标应用程序和所选输出对象不同，可以使用下面的部分或所有格式： 图元文件 WMF 和 EMF 图元文件格式。 这些格式只在 Windows 操作系统上可用。 第 10 章 处理输出 117 RTF（富文本格式） 可以 RTF 格式复制和粘贴多个选定对象、文本输出、透视表。 对于透视表，在大多数应用程序中，这意 味着将表作为表格粘贴，然后可在其他应用程序中编辑它。 根据透视表选项设置，将对文档宽度过宽的 透视表进行换行、按比例缩小以适合文档刻度或不进行更改。 请参阅主题第170 页的『透视表选项』， 了解更多信息。 注释: • Microsoft Word 可能无法正确显示过宽的表格。 • 在复制并粘贴箱图和直方图时需要 Microsoft Office V16（或更高版本）。 图像 JPG 和 PNG 图像格式。 BIFF 可以 BIFF 格式粘贴透视表和文本输出到电子表格。 透视表中的数字保留数值精确度。 此格式仅在 Windows 操作系统上可用。 文本 可以文本格式复制和粘贴透视表和文本输出。 此过程对于电子邮件这样的应用程序很有用，因为这种应 用程序只接受或传送文本。 Microsoft Office 图形对象 支持该格式的图表可以复制到 Microsoft Office 应用程序，并向原生 Microsoft Office 图表一样在这些应 用程序中进行编辑。 由于 SPSS Statistics/SPSS Modeler 图表和 Microsoft Office 图表之间存在差异， SPSS Statistics/SPSS Modeler 图表的一些功能不会保留在复制的版本中。 不支持以 Microsoft Office 图 形对象格式复制多个选中的图表。 如果目标应用程序支持多种可用格式，则可能包含允许您选择格式“选择性粘贴”菜单项，或自动显示可用格 式列表。 复制和粘贴多个输出对象 当粘贴多个输出对象到其他应用程序时，以下限制适用： RTF 格式 在大部分应用程序中，透视表将粘贴为可以在该应用程序中进行编辑的表。 以图像格式粘贴图表、树和 模型视图。 图元文件和图像格式 所有选定输出对象都将粘贴为其他应用程序中的单个对象。 BIFF 格式 排除图表、树和模型视图。 您还可使用 第107 页的『导出输出』 导出多个输出对象到其他应用程序或格式中。 复制 您可以在“输出查看器”中选择一个或多个对象并右键单击鼠标，然后选择复制以复制所有可用格式的选定 项。 或者，您可以在“输出查看器”中选择一个或多个对象，然后选择编辑 > 复制。 复制为 您可以在“输出查看器”中选择对象，并选择编辑 > 复制为，以复制最流行的格式（例如，Microsoft Office 图形对象或图像）。 请注意，如果复制为呈灰色或者对于选中的对象不存在，那么此复制格式不适用于此特 定对象。 118 IBM SPSS Statistics 29 Core System 用户指南 第 11 章 透视表 透视表 在可交互旋转的表中显示很多结果。 即，可以排列行、列和层。 注：如果您需要与 20 之前的 IBM SPSS Statistics 版本兼容的表，则建议您将其作为遗存表来呈现。 有关更 多信息，请参阅第131 页的『遗存表』主题。 操作透视表 用于操纵透视表的选项包括： • 交换行和列 • 移动行和列 • 创建多维层 • 对行和列进行分组和取消分组 • 显示和隐藏行、列和其他信息 • 旋转行标签和列标签 • 查找术语的定义 透视表编辑窗格 透视表编辑器提供了滑出窗格，其中提供了用于操作透视表外观的各种选项。 通过单击透视表选项控件可激 活滑出窗格。 可用的透视表选项有： “常规”选项卡 提供多个选项，用于隐藏空行和空列、放置行维度标签以及设置列和行标签的宽度。 “区域格式”选项卡 提供用于各种透视表文本区域的文本编辑选项。 “边框”选项卡 提供用于定义透视表边框设置的选项。 “单元格格式”选项卡 提供用于指定透视表单元格格式的选项。 “注释”选项卡 提供用于指定脚注格式和用于输入透视表注释文本的选项。 激活透视表 在操作或修改透视表之前，您需要激活表。 要激活表： 1. 双击该表。 或者 2. 右键单击该表，然后从弹出菜单中选择编辑。 透视表 1. 激活透视表。 2. 从菜单中选择： 透视 > 透视托盘 表具有三个维度：行、列和层。 维度中可包含多个元素（或不包含元素）。 可以在维度之间或内部移动元 素，以更改表的组织。 要移动元素，只需将其拖放到所需的位置。 更改元素在维度内的显示顺序 要更改表维度（行、列或层）中元素的显示顺序： 1. 如果透视托盘尚未打开，则从“透视表”菜单中选择： 透视 > 透视托盘 2. 将维度内的元素拖放到透视托盘中。 在维度元素中移动行和列 1. 在表（而非透视托盘）上，单击您想要移动的行或列的标签。 2. 将标签拖到新位置。 交换行和列 如果只想翻转行和列，则可以简单地使用透视表托盘： 1. 从菜单中选择： 透视 > 转置行和列 这与将所有行元素拖放到列维并将所有列元素拖放到行维所产生的效果相同。 对行或列分组 1. 选择您想要分组在一起的行或列的标签（单击并拖动，或者按住 Shift 并单击以选择多个标签）。 2. 从菜单中选择： 编辑 > 组 将自动插入一个组标签。 双击组标签可编辑标签文本。 注意：要将行或列添加到现有的组，必须先对当前组中的项取消分组。 然后创建包含附加项的新组。 对行或列取消分组 1. 单击您想要取消分组的行或列的组标签的任意位置。 2. 从菜单中选择： 编辑 > 取消分组 取消分组会自动删除组标签。 旋转行标签或列标签 您可以在表中最内部列标签和最外部行标签的水平和垂直显示之间旋转标签。 1. 从菜单中选择： 格式 > 旋转内部列标签 或 格式 > 旋转外部行标签 120 IBM SPSS Statistics 29 Core System 用户指南 仅可旋转最内部的列标签和最外部的行标签。 对行排序 排序透视表行： 1. 激活该表。 2. 在列中选择任何用作排序依据的单元格。 要仅对一组选定的行进行排序，请从列中选择两个或两个以上 要用作排序依据的连续单元格。 3. 从菜单中选择： 编辑 > 对行排序 4. 从子菜单选择升序或降序。 • 如果行维度包含组，那么排序仅影响包含所选内容的组。 • 无法跨组边界进行排序。 • 无法对行维度中存在多个项的表进行排序。 注：在遗存表中此功能不可用。 相关信息 第131 页的『遗存表』 插入行和列 要在透视表中插入行或列： 1. 激活该表。 2. 选择表中的任何单元格。 3. 从菜单中选择： 在前面插入 或 在后面插入 从子菜单中选择： 行 或 列 • 加号 (+) 在新行或新列的每个单元格中插入，以防止新行或新列因为空白而自动隐藏。 • 在具有嵌套或层维度的表中，会在每个对应的维度插入列和行。 注：在遗存表中此功能不可用。 相关信息 第131 页的『遗存表』 控制变量和值标签的显示。 如果变量包含描述性的变量标签或值标签，那么您可以对透视表中变量名称、变量标签、数据值和值标签的 显示进行控制。 1. 激活透视表。 2. 从菜单中选择： 视图 > 变量标签 或 第 11 章 透视表 121 视图 > 值标签 3. 从子菜单选择下列选项之一： • 名称或值。 仅显示变量名称（或值）。 描述性标签不会显示。 • 标签。 仅显示描述性标签。 不显示变量名称（或值）。 • 两者。 显示名称（或值）及描述性标签。 注：此功能不适用于遗存表。 要控制透视表和其他输出对象的缺省标签显示，请使用“编辑>选项>输出”。 相关信息 第131 页的『遗存表』 第168 页的『输出选项』 更改输出语言 要更改透视表中的输出语言： 1. 激活透视表 2. 从菜单中选择： 查看 > 语言 3. 选择下列可用语言之一： 更改语言只影响应用程序生成的文本，如表标题、行和列标签以及脚注文本。 变量名称、描述性变量及值标 签不受影响。 注：对于遗存表，此功能不可用。 要控制透视表和其他输出对象的缺省语言，请使用“编辑”>“选项”>“语言”。 相关信息 第131 页的『遗存表』 第165 页的『语言选项』 导航大型表 要使用导航窗格导航大型表： 1. 激活该表。 2. 从菜单中选择： 视图 > 导航 撤销更改 您可以撤销最近对激活的透视表进行的更改，也可以撤销对其进行的所有更改。 这两种操作仅适用于自最近 激活该表以来所作的更改。 要撤销最近的更改，请完成下列步骤： 1. 从菜单中选择： 编辑 > 撤销 要撤销所有更改，请完成下列步骤： 2. 从菜单中选择： 编辑 > 恢复 注：“编辑 > 复原”不可用于旧表。 122 IBM SPSS Statistics 29 Core System 用户指南 使用层 您可以为每个类别或类别的组合显示一个单独的二维表。 可将该类表视为层的堆积，并且只显示最上层。 创建并显示层 要创建层： 1. 激活透视表。 2. 如果透视托盘尚未开启，请从透视托盘菜单中选择： 透视 > 透视托盘 3. 将一个元素从行维度或列维度拖到层维度。 将元素移至层维度中，会创建一个多维表，但只会显示单个二维分区。 所显示的表为最上层的表。 例如， 如果层维度中有一个“是/否”分类变量，那么多维表将有两个层：一个用于“是”类别，而另一个用于“否”类 别。 更改显示的层 1. 从层（在透视表本身中，而非透视表托盘中）的下列列表中选择类别。 转至层类别 “转到层类别”使您可以在透视表中更改层。 当存在许多层或选定层包含许多类别时，此对话框尤其有用。 显示和隐藏项目 许多类型的单元格都可以隐藏，其中包括： • 维度标签 • 类别，包括行或列中的标签单元格和数据单元格 • 类别标签（无需隐藏数据单元格） • 脚注、标题和标注 隐藏表中的行和列 显示表中的隐藏行和列 1. 从菜单中选择： 视图 > 显示全部类别 这将显示表中的所有隐藏行和列。 （如果在此表的“表格属性”中选择了隐藏空行列，则完全为空的行或列仍 会隐藏。） 隐藏和显示维度标签 1. 在维数中选择维数标签或任何类别标签。 2. 从“查看”菜单或弹出菜单中，选择隐藏维度标签或显示维度标签。 隐藏和显示表标题 要隐藏标题： 1. 激活透视表。 2. 选择标题。 3. 从“视图”菜单中选择隐藏。 第 11 章 透视表 123 要显示隐藏的标题： 4. 从“查看”菜单中选择全部显示。 表格外观 表格外观是定义表的外观的一组属性。 您可以选择先前定义的表格外观，或创建自己的表格外观。 • 在应用表格外观前后，可以使用单元格属性更改个别单元格或单元格组的单元格格式。 即使您应用新的表 格外观，编辑的单元格格式仍保持不变。 • 您还可以选择将所有单元格重置为由当前表格外观定义的单元格格式。 此选项将重置已编辑的所有单元 格。 如果在“表格外观文件”列表中选择了按显示，则任何已编辑的单元格都将重置为当前的表格属性。 • 仅“表格属性”对话框中定义的表格属性将保存在表格外观中。 TableLook 不包括单个单元格修改。 注：在 IBM SPSS Statistics 的早期版本中创建的表格外观不能用于 V16.0 或更高版本。 应用表格外观 1. 激活数据透视表。 2. 从菜单中选择： 格式 > 表格外观...
3. 从文件列表中选择一个表格外观。 要从另一个目录中选择文件，请单击浏览。 4. 单击确定将表格外观应用到选定的透视表。 编辑或创建表格外观 1. 在“表格外观”对话框中，从文件列表中选择一个表格外观。 2. 单击编辑外观。 3. 为需要的属性调整表格属性，然后单击确定。 4. 单击保存外观保存已编辑的表格外观，或者单击另存为将其保存为新的表格外观。 • 编辑表格外观仅影响选定的透视表。 已编辑的表格外观不会应用到使用该表格外观的任何其他表，除非您 选择这些表并重新应用表格外观。 • 仅“表格属性”对话框中定义的表格属性将保存在表格外观中。 表格外观不包括单个单元格修改。 表属性 “表属性”允许您设置表的常规属性，并且允许设置表的各个部分的单元格样式，以及将其中的一组属性保存 为 TableLook。 您可以执行以下操作： • 控制常规属性，例如隐藏空行或列，以及调整打印属性。 • 控制脚注标记的格式和位置。 • 确定用于下列内容的特定格式：数据区域中的单元格、行和列标签以及表的其他区域。 • 控制形成表的每个区域的边框的线条的宽度和颜色。 更改透视表属性 1. 激活透视表。 2. 从菜单中选择： 格式 > 表属性...
3. 选择一个选项卡（常规、脚注、单元格格式、边框或打印）。 4. 选择您想要的选项。 5. 单击确定或应用。 124 IBM SPSS Statistics 29 Core System 用户指南 新的属性将应用到选定的透视表。 要将新的表格属性应用到表格外观而不仅仅是选定的表，请编辑表格外观 （“格式”菜单中的“表格外观”）。 表属性：常规 多个属性作为一个整体应用到表。 您可以执行以下操作： • 显示或隐藏空的行与列。 （空的行或列在任何数据单元格中均不存在任何内容。） • 控制在长表中显示的缺省行数。 要在表中显示所有行，而不考虑表的长度，请取消选中按行显示表。 注： 此功能仅适用于遗存表。 • 控制行标签的位置，它们可以位于左上角或嵌套。 • 控制最大和最小列宽（以磅表示）。 要更改常规表的属性： 1. 单击常规选项卡。 2. 选择您想要的选项。 3. 单击确定或应用。 设置待显示行 注: 此功能仅适用于遗存表。 缺省情况下，行数较多的表显示在多个部分（每个部分 100 行）中。 控制在表中显示的行数： 1. 选择按行显示表格。 2. 单击设置待显示行。 或 3. 从已激活透视表的“视图”菜单中，选择按行显示表格和设置待显示行。 待显示行。 控制一次显示的最大行数。 导航控件允许您移动到表格的不同部分。 最小值为 10。 缺省值为 100。 短行/孤立行容差。 控制表格最内层行维度的最大行数，以便在表格的显示视图之间拆分。 例如，如果最内 层行维度的每个组中有 6 个类别，则指定值 6 可以防止任何组在显示视图之间拆分。 该设置可能导致某个显 示视图中的总行数超过指定的最大待显示行数。 表属性：注释 “表属性”对话框的“注释”选项卡控制脚注格式编排和表注释文本。 脚注。 脚注标记的属性包括样式和相对于文本的位置。 • 脚注标记的样式为数字（1、2、3...）或字母（a、b、c ...）。 • 脚注标记可以作为上标或下标附加在文本上。 注释文本。 您可以向每个表格添加注释文本。 • 如果您在查看器中将光标悬停在表格上，注释文本就会显示在工具提示中。 • 当焦点位于表上时，屏幕朗读器将朗读注释文本。 • 查看器中的工具提示仅显示注释的前 200 个字符，但屏幕朗读器会朗读全部文本。 • 如果将输出导出到 HTML，会使用注释文本作为替代文本。 可以在所有表格创建时自动添加注释（“编辑”菜单 >“选项”>“数据透视表”）。 表属性：单元格格式 表可以按不同的区域设置格式：标题、层、角标签、行标签、列标签、数据、文字说明和脚注。 对于表的每 个区域，您都可以修改相关联的单元格格式。 单元格格式包括文本特征（如字体、大小、颜色以及样式）、 水平和垂直对齐、背景色以及内部单元格边距。 第 11 章 透视表 125 将单元格格式应用到区域（信息的类别）。 它们不是个别单元格的特征。 当对表进行旋转时，此区别是重 要的考虑因素。 例如： • 如果指定粗体作为列标签的单元格格式，那么不管列维度中当前显示的信息是什么，列标签总是显示为粗 体。 如果您将一项从列维度移至其他维度，它不会保留列标签的粗体特征。 • 如果您仅是通过突出显示已激活的透视表中的单元格，并单击工具栏上的“粗体”按钮而将列标签变为粗 体，则不管您将那些单元格移至哪个维度，其内容总是保留粗体，而对于其他移至列维度的项，列标签不 会保留粗体特征。 要更改单元格格式： 1. 选择单元格格式选项卡。 2. 从下拉列表中选择一个区域，或单击样本的一个区域。 3. 选择该区域的特征。 所选区域会显示在示例中。 4. 单击确定或应用。 交替行颜色 将另一种背景和/或文本颜色应用于表的“数据”区域中的交替行： 1. 从“区域”下拉列表中选择数据。 2. 选择（选中）背景颜色组中的交替行颜色。 3. 选择用于交替行背景和文本的颜色。 交替行颜色仅影响表格的数据区域。 它们不会影响行或列标签区域。 表属性：边框 对于表中的每个边框位置，您都可以选择线样式和颜色。 如果选择无作为线样式，则在选定位置不会出现线 条。 要更改表边框： 1. 单击边框选项卡。 2. 选择边框位置，方法是在列表中单击边框的名称，或者在“样本”区域中单击线。 3. 选择一种线样式或选择无。 4. 选择一种颜色。 5. 单击确定或应用。 表属性：打印 您可以控制所打印透视表的以下属性： • 打印表的所有层或仅打印顶层，并在单独页面上打印每一层。 • 水平或垂直地收缩表，以适合页面打印。 • 如果表对于定义的页面大小太宽和/或太长，则通过控制包含在表的任何打印部分的最小行数和列数来控制 短行/孤立行。 注意：如果由于在表的上方存在其他输出而导致该表在当前页面中放不下，但是定义的页面长度能够容纳 该表，那么将自动在新页面中打印该表，而不考虑短行/孤立行设置。 • 包含无法容纳在单页中的表的接续文本。 您可以在每页的底部和顶部显示接续文本。 如果未选择任一选 项，则不会显示接续文本。 要控制透视表打印属性： 1. 单击打印选项卡。 2. 选择您想要的打印选项。 3. 单击确定或应用。 126 IBM SPSS Statistics 29 Core System 用户指南 单元格属性 单元格属性会应用到选定的单元格。 您可以更改字体、值的格式、对齐方式、边距和颜色。 单元格属性会 覆盖表格属性；所以，更改表格属性不会影响到任何单独应用的单元格属性。 要更改单元格属性： 1. 激活一个表并在表中选择单元格。 2. 从“格式”菜单或弹出菜单中选择单元格属性。 字体和背景 “字体及背景”选项卡控制表中选定单元格的字体样式、颜色和背景颜色。 格式值 “格式值”选项卡控制选定单元格的值格式。 您可以选择数字、日期、时间或货币的格式，并且可以调整显示 的小数位数。 注：货币格式列表包含美元格式（带有前导美元符号的数字）和五个定制货币格式。 缺省情况下，所有自定 义货币格式设置为缺省的数字格式，其中不包含货币或其他自定义符号。 使用编辑>选项>货币来定义定制 货币格式。 对齐和页边距 “对齐与页边距”选项卡控制选定单元格的值的水平和垂直对齐方式，以及上下左右边距。 混合水平对齐按照 单元格的类型对齐其内容。 例如，日期为右对齐，而文本值为左对齐。 脚注和文字说明 您可以向表添加脚注和文字说明。 还可以隐藏脚注或标注，更改脚注标记以及重新对脚注编号。 添加脚注和文字说明 要向表添加文字说明： 1. 从“插入”菜单中选择文字说明。 可以将脚注附加到表中的任意项。 要添加脚注： 1. 在激活的透视表中，单击标题、单元格或标注。 2. 从“插入”菜单中选择脚注。 3. 在提供的区域中插入脚注文本。 隐藏或显示文字说明 要隐藏文字说明： 1. 选择文字说明。 2. 从“视图”菜单中选择隐藏。 要显示隐藏的文字说明： 1. 从“查看”菜单中选择全部显示。 隐藏或显示表中的脚注 要隐藏脚注： 1. 右键单击包含脚注参考的单元格，然后从弹出菜单中选择隐藏脚注。 或 第 11 章 透视表 127 2. 从表的脚注区域中选择脚注，并从弹出菜单中选择隐藏。 注意：对于旧表，选择表的脚注区域，并从弹出菜单中选择编辑脚注，然后取消选择（取消选中）您要 隐藏的任何脚注的“可见”属性。 如果单元格包含多个脚注，则使用后一种方法选择隐藏脚注。 要隐藏表中的所有脚注，请完成下列步骤： 1. 在表的脚注区域中选择所有脚注（通过单击并拖动或者在按住 Shift 键的情况下单击来选择脚注），然后 从“查看”菜单中选择隐藏。 注：对于遗存表，请选择表的脚注区域，然后从“视图”菜单中选择隐藏。 要显示隐藏的脚注： 1. 从“视图”菜单中选择显示全部脚注。 脚注标记 “脚注标记”更改可用于对脚注进行标记的字符。 缺省情况下，标准脚注标记为顺序字母或数字，具体取决于 表属性设置。 您也可以指定特殊标记。 在您对脚注重新编号，或者切换用于标准标记的数字和字母时，不 会影响特殊标记。 标准标记的数字或字母显示以及脚注标记的下标或下标位置由“表属性”对话框的“脚注”选 项卡控制。 注：要更改遗存表中的脚注标记，请参阅第128 页的『编辑遗存表中的脚注』。 要更改脚注标记： 1. 选择脚注。 2. 从格式菜单中选择脚注标记。 特殊标记限制为 2 个字符。 在表的脚注区域中，带有特殊标记的脚注位于具有顺序字母或数字的字母或数字 之前；因此更改特殊标记可能会对脚注列表进行重新排序。 对脚注重新编号 当通过交换行、列和层对表进行旋转之后，可能会打乱脚注的顺序。 要对脚注重新编号： 1. 从“格式”菜单中选择对脚注重新编号。 编辑遗存表中的脚注 对于遗存表，您可以使用“编辑脚注”对话框来输入和修改脚注文本和字体设置，更改脚注标记，以及选择隐 藏或删除脚注等。 当您在遗存表中插入新的脚注时，“编辑脚注”对话框将自动打开。 要使用“编辑脚注”对话框来编辑现有脚注 （而不创建新脚注），请完成下列步骤： 1. 双击表的脚注区域或从菜单中选择：格式 > 编辑脚注。 标记。 缺省情况下，标准脚注标记为顺序字母或数字，具体取决于表属性设置。 要指定特殊标记，只需在 “标记”列中输入新的标记值。 在您对脚注重新编号，或者切换用于标准标记的数字和字母时，不会影响特殊 标记。 标准标记的数字或字母显示以及脚注标记的下标或下标位置由“表属性”对话框的“脚注”选项卡控制。 请参阅第125 页的『表属性：注释』主题以获取更多信息。 要将特殊标记改回标准标记，请右键单击“编辑脚注”对话框中的标记，并从弹出菜单中选择脚注标记，然后 在“脚注标记”对话框中选择“标准”标记。 脚注。 脚注内容。 其显示反映当前字体和背景设置。 您可以使用“格式”子对话框来更改各个脚注的字体设 置。 请参阅第129 页的『脚注字体和颜色设置』主题以获取更多信息。 将对所有脚注应用单一背景色，您 可以在“单元格属性”对话框的“字体和背景”选项卡中更改该背景色。 有关更多信息，请参阅第127 页的 『字体和背景』主题。 可视。 所有脚注缺省情况下可见。 取消选择（取消选中）“可见”复选框将隐藏脚注。 128 IBM SPSS Statistics 29 Core System 用户指南 脚注字体和颜色设置 对于遗存表，您可以使用“格式”对话框来更改一个或多个选定脚注的字体系列、样式、大小和颜色： 1. 在“编辑脚注”对话框的“脚注”网格中，选择（单击）一个或多个脚注。 2. 单击格式按钮。 选定的字体系列、样式、大小和颜色将应用于所有选定的脚注。 背景颜色、对齐和边距可以在“单元格属性”对话框中设置，并应用于所有脚注。 您无法对各单独的脚注更改 这些设置。 有关更多信息，请参阅第127 页的『字体和背景』主题。 数据单元格宽度 “设置数据单元格宽度”用于将所有数据单元格设置为相同的宽度。 要为所有数据单元格设置宽度： 1. 从菜单中选择： 格式 > 设置数据单元格宽度...
2. 输入一个值作为单元格宽度。 更改列宽 1. 单击并拖动列边框。 显示透视表中的隐藏边框 对于不含很多可视边框的表，可以显示隐藏边框。 这可以简化更改列宽之类的任务。 1. 从“视图”菜单中选择网格线。 在透视表中选择行、列和单元格 您可以选择整行或整列，也可以选择一组指定的数据单元格和标签单元格。 要选择多个单元格，请完成下列步骤： 选择 > 数据单元格和标签单元格 打印透视表 有多个因素会影响打印透视表外观的方式，可以通过更改透视表属性控制这些因素。 • 对于多维透视表（带有多层的表），您可以打印所有层，也可以只打印顶层（可视层）。 有关更多信息， 请参阅第126 页的『表属性：打印』主题。 • 对于长或宽的透视表，可以自动调整表的大小以适合页面，也可以控制表分隔符和分页符的位置。 有关更 多信息，请参阅第126 页的『表属性：打印』主题。 • 如果有的表太宽或太长而无法在单个页面中打印，您可以控制页间的表分隔符位置。 可以使用“文件”菜单上的“打印预览”查看打印的透视表的外观。 控制宽表和长表的表分隔符 有的透视表太宽或太长，无法在定义的页面大小中打印，程序会自动分割它们，并将它们打印在多个部分 中。 您可以执行以下操作： • 控制拆分大型表的行或列的位置。 • 指定分割表时应该保留在一起的行和列。 • 重定大型表的比例以适应定义的页面大小。 第 11 章 透视表 129 为打印的透视表指定行和列分隔符 1. 激活透视表。 2. 单击要插入分隔符位置左边的列中的任何单元格，或者单击要插入分隔符的行之前的行中的任何单元 格。 注意：对于遗存表，您必须单击列标签或行标签单元格。 3. 从菜单中选择： 格式 > 分界点 > 垂直分界点 或 格式 > 分界点 > 水平分界点 注：对于旧表，请为垂直和水平断点选择 格式 > 此处断开。 要指定要保留在一起的行或列： 1. 选择要保留在一起的行或列的标签。 单击并拖动或按住 Shift 键的同时单击来选择多个行或列标签。 2. 从菜单中选择： 格式 > 分界点 > 保持在一起 注: 对于旧表，请选择 格式 > 保持在一起。 查看分界点和“保留在一起”分组： 1. 从菜单中选择： 格式 > 分界点 > 显示分界点 分界点显示为垂直线或水平线。 “保留在一起”分组显示为采用深色边框围起来的灰色矩形区域。 注意：遗存表不支持显示分界点和“保留在一起”分组。 清除分界点和“保留在一起”分组 清除分界点： 1. 单击垂直分界点左侧列中的任何单元格，或单击垂直分界点上方行中的任何单元格。 注意：对于遗存表，您必须单击列标签或行标签单元格。 2. 从菜单中选择： 格式 > 分界点 > 清除分界点或组 注: 对于旧表，请选择格式 > 在此处除去分隔符。 清除“保留在一起”分组： 3. 选择用于指定分组的列或行标签。 4. 从菜单中选择： 格式 > 分界点 > 清除分界点或组 注: 对于旧表，请选择格式 > 除去保持在一起。 在您旋转或重新排序任何行或列时，所有分界点和“保留在一起”分组将自动清除。 此行为不适用于遗存表。 从透视表创建图表 1. 双击透视表以将其激活。 2. 选择想要在图中显示的行、列或单元格。 3. 右键单击选定区域中的任意位置。 4. 从弹出菜单中选择创建图形并选择一个图表类型。 130 IBM SPSS Statistics 29 Core System 用户指南 色阶 大多数表包含混合值，将热图应用于整个表通常会生成变化范围很大的表。 透视表编辑器包含色阶菜单选 项，该选项提供热图样式设置，根据单元格值以不同颜色来显示所选择的表单元格。 指定色阶值 1. 双击透视表以将其激活。 2. 选择您想要以热图样式颜色格式表示的单元格（必须至少选择两个单元格）。 3. 右键单击所选单元格区域中的任意位置，然后从菜单中选择色阶。 4. 色阶对话框提供下列设置。 使用绝对值 选择了此项时，负值将被视为正值。 缺省情况下已启用该设置。 低颜色值 设置低值颜色。 单击颜色选取器以选择颜色。 字段旁边的数字值指示所选颜色的 RBG 值。 高颜色值 设置高值颜色。 单击颜色选取器以选择颜色。 字段旁边的数字值指示所选颜色的 RBG 值。 5. 选择色阶设置之后，单击确定。 所选择的透视表单元格以所选择的低颜色值与高颜色值之间的颜色范围 内的颜色显示。 注: 单击重置以将色阶设置设定为其缺省值。 遗存表 您可以选择将表作为遗存表来呈现（称为版本 19 中的全功能表），它们与 20 之前的版本 IBM SPSS Statistics 完全兼容。 遗存表呈现可能较为缓慢，仅在需要与 20 之前的版本兼容时推荐使用。 有关如何创 建遗存表的信息，请参阅第170 页的『透视表选项』。 第 11 章 透视表 131 132 IBM SPSS Statistics 29 Core System 用户指南 第 12 章 模型 一些结果显示为模型，在输出查看器中显示为一种特殊类型的可视化。 在输出查看器中显示的可视化不是唯 一可用的模型视图。 单个模型包含许多不同视图。 您可以在模型查看器中激活模型，并直接与该模型进行 交互以显示可用的模型视图。 也可以选择打印和导出模型中的所有视图。 与模型进行交互作用 与模型进行交互作用，请首先将其激活： 1. 双击该模型。 或 2. 右键单击该模型，然后从弹出菜单中选择编辑内容。 3. 从子菜单中。选择在单独的窗口中。 激活模型将在模型查看器中显示模型。 有关更多信息，请参阅第133 页的『使用模型查看器』主题。 使用模型查看器 模型查看器是一种显示可用模型视图和编辑模型视图外观的交互式工具。 （有关显示“模型查看器”的信息， 请参阅第133 页的『与模型进行交互作用』。） 有两种不同样式的模型查看器： • 拆分成主/辅助视图。 在此样式中，主视图出现在模型查看器的左部。 主视图显示模型的一些一般可视化 （例如网络图片）。 主视图本身可有多个模型视图。 主视图下方的下拉列表允许您从可用主视图中选择。 辅助视图出现在模型查看器的右部。 与主视图中的一般可视化相比，辅助视图通常显示模型的更多详细可 视化（包括表格）。 与主视图一样，辅助视图可有多个模型视图。 辅助视图下方的下拉列表允许您从可用 主视图中选择。 辅助视图还可为主视图中选择的元素显示特定可视化。 例如，根据模型类型，您可以在主 视图中选择一个变量节点，以在辅助视图中显示该变量的表格。 • 通过缩略图，一次打开一个视图。 在此样式中，只有一个视图可见，而其他视图则通过模型查看器左部的 缩略图进行访问。 每个视图显示模型的一些可视化。 显示的特定可视化取决于创建模型的过程。 有关使用特定模型的信息，请参考创建模型的过程文档。 模型视图表格 在模型查看器中显示的表格不是透视表。 您无法像操作透视表一样操作这些表格。 设置模型属性 在模型查看器中，您可以设置特定模型属性。 请参阅第133 页的『模型属性』主题以获取更多信息。 复制模型视图 您还可复制模型查看器中的单个模型视图。 请参阅第134 页的『复制模型视图』主题以获取更多信息。 模型属性 根据“模型查看器”选择： 文件 > 属性 或 文件 > 打印视图 每个模型都有相关属性，让您指定从输出查看器中打印哪些视图。 缺省情况下，只打印在输出查看器中可见 的视图。 一般总是主视图，且只有一个主视图。 您还可以指定打印所有可用的模型视图。 这些包括所有主 视图以及所有辅助试图（基于主视图选择的辅助视图除外；这些是不可打印的）。 请注意，还可以打印模型 查看器本身中的各个模型视图。 请参阅第134 页的『打印模型』主题以获取更多信息。 复制模型视图 从模型查看器的“编辑”菜单中，您可以复制当前显示的主视图或当前显示的辅助视图。 只复制一个模型视 图。 您可以将模型视图粘贴到输出查看器中，在这里单个模型视图随后以可以在图形板编辑器中编辑的可视 化显示。 “粘贴到输出查看器”允许您同时显示多个模型视图。 您还可粘贴到其他应用程序中，视图会根据 目标应用程序显示为一个图像或表格。 打印模型 从模型查看器中打印 您可以打印模型查看器自带的单个模型视图。 1. 在模型查看器中激活模型。 有关更多信息，请参阅第133 页的『与模型进行交互作用』主题。 2. 从菜单中选择“查看”>“编辑方式”（如果可用）。 3. 在主视图或辅助视图（根据您想打印哪一个）中的“一般工具栏”调色板上，单击打印图标。 （如果未显 示此调色板，请从“查看”菜单中选择调色板>常规。） 注：如果模型查看器不支持打印图标，请选择“文件”>“打印”。 从输出查看器中打印 当您从输出查看器中打印时，为特定模型所打印的视图数取决于模型属性。 模型可以设置为只打印显示的视 图或所有可用模型视图。 有关更多信息，请参阅第133 页的『模型属性』主题。 探索模型 缺省情况下，当您从输出查看器中导出模型时，包括或排除模型视图由每个模型的模型属性控制。 有关模型 属性的更多信息，请参阅第133 页的『模型属性』。 在导出时，您可以覆盖此设置，并包括所有模型视图 或只包括当前可见的模型视图。 在“导出输出”对话框中，单击“文档”组中的更改选项...。 有关导出以及此对 话框的更多信息，请参阅第107 页的『导出输出』。 请注意，包括表在内的所有模型视图，均导出为图 形。 还请注意，基于主视图选择的辅助视图不可导出。 将模型中使用的字段保存到新的数据集 您可以将模型中使用的字段保存到新的数据集。 1. 在模型查看器中激活模型。 有关更多信息，请参阅第133 页的『与模型进行交互作用』主题。 2. 从菜单中选择： 生成 > 字段选择（模型输入和目标） 数据集名称。 请指定一个有效的数据集名称。 可以在同一会话中继续使用数据集，但不会将其另存为文 件，除非在会话结束之前明确将其保存为文件。 数据集名称必须符合变量命名规则。 请参阅主题第44 页的 『变量名称』，了解更多信息。 根据重要性将预测变量保存到新的数据集 您可以根据预测变量重要性图表中的信息，将预测变量保存到新的数据集。 1. 在模型查看器中激活模型。 有关更多信息，请参阅第133 页的『与模型进行交互作用』主题。 2. 从菜单中选择： 生成 > 字段选择（预测变量重要性） 变量总数。 包括或排除等于指定数字的最重要预测变量。 重要性大于。 包括或排除所有相对重要性高于指定值的预测变量。 3. 单击确定后，将显示“新数据集”对话框。 134 IBM SPSS Statistics 29 Core System 用户指南 数据集名称。 请指定一个有效的数据集名称。 可以在同一会话中继续使用数据集，但不会将其另存为文 件，除非在会话结束之前明确将其保存为文件。 数据集名称必须符合变量命名规则。 请参阅主题第44 页的 『变量名称』，了解更多信息。 整体查看器 整体模型 整体模型提供了有关整体中的组件模型和整体性能的信息。 主 (独立视图) 工具栏允许您选择使用整体或参考模型来进行评分。 如果使用整体进行评分，您还可以选择组 合规则。 这些更改不需要重新执行模型；但是，这些选择将保存到模型 以供评分和/或下游模型评估。 它们 也会影响从整体查看器导出的 PMML。 组合规则。 对整体进行评分时，此规则用于组合来自基本模型的预测值，以计算整体评分值。 • 可以使用投票、最高概率或最高平均概率来组合分类目标的整体预测值。 投票选择在基本模型中最常具有 最高概率的类别。 最高概率选择在所有基本模型中取得单个最高概率的类别。 最高平均概率选择当类别概 率在基本模型中取平均值时具有最高值的类别。 • 可以通过对来自基本模型的预测值取平均值或中位数，对连续目标的整体预测值进行组合。 缺省值取自在建模过程中生成的指定。 更改组合规则会重新计算模型精确性并更新模型精确性的所有视图。 也会更新预测变量重要性图表。 如果选择参考模型用于评分，那么此控件将被禁用。 显示所有组合规则。 选择该选项时，所有可用组合规则的结果将显示在模型质量图表中。 组件模型精确性 图表也将更新以显示每种投票方式的参考线。 模型摘要 "模型摘要"视图是整体质量和差异性的快照摘要。 质量。 该图表显示与参考模型和 navie 模型相比较的最终模型精确性。 精确性采用“越大越好”的格式呈现； “最佳”模型将具有最高精确性。 对于分类目标，精确性就是预测值与观测值匹配的记录百分比。 对于连续 目标，精确性为 1 减去预测中的平均绝对误差（预测值的绝对平均值减去观测值）与预测值范围 （最大预测 值减去最小预测值）的比值。 对于 bagging 整体，参考模型是基于整个培训分区构建的标准模型。 对于增强型整体，参考模型是第一个组 件模型。 如果未构建模型，那么由 Naive 模型代表精确性，并将所有记录分配给模态类别。 不会为连续目标计算 Naive 模型。 差异性。 此图表显示用于构建整体的组件模型中的“不同意见”，格式越多样化，显示的图表越大。 这是一 种基本模型间预测差异程度的测量。 差异性对增强型整体模型不可用，同时也不会对连续目标显示。 预测变量重要性 通常，您将需要将建模工作专注于最重要的预测变量字段，并考虑删除或忽略那些最不重要的变量。 预测变 量重要性图表可以在模型估计时指示每个预测变量的相对重要性，从而帮助您实现这一点。 由于它们是相对 值，因此显示的所有预测变量的值总和为 1.0。 预测变量重要性与模型精确性无关。 它只与每个预测变量在 预测中的重要性有关，而不涉及预测是否精确。 预测变量重要性对所有整体模型均不可用。 预测变量集在组件模型之间可能会有所不同，但可以为至少在一 个组件模型中使用的预测变量计算重要性。 预测变量频率 由于选择的建模方法或预测变量选择不同，预测变量集在组件模型之间也可能不同。 “预测变量频率”图是一 个点图，显示了预测变量在组件模型中的总体分布。 每个点表示一个或多个包含预测变量的组件模型。 预 测变量绘制在 y 轴上，并按频率的降序排序；因此，最上面的预测变量是在最多组件模型中使用的预测变 量，而最下面的预测变量是在最少组件模型中使用的预测变量。 将显示前 10 个预测变量。 第 12 章 模型 135 出现频率最高的预测变量通常是最重要的预测变量。 此图对于预测变量集在组件模型之间无法改变的方法没 用。 组件模型精确性 该图表是组件模型预测精确性的点图。 每个点代表在 y 轴上绘制了精确性水平的一个或多个组件模型。 悬 停在任意点上可获得对应的单独组件模型的信息。 参考线。 该图显示整体的颜色编码线以及参考模型和 naïve 模型。 对应于要用于评分的模型的线的旁边会显 示一个复选标记。 互动。 该图表会在您更改组合规则时更新。 Boosted 整体。 为 boosted 整体显示一个线图。 组件模型详细信息 该表显示关于组件模型的信息，按行列出。 缺省情况下，组件模型按模型编号的升序排序。 您可以按任意 列的值对这些行以升序或降序排序。 模型。 代表组件模型创建顺序的数字。 精确性。 总体准确性，以百分比表示。 方法。 建模方法。 预测变量。 组件模型中使用的预测变量的数目。 模型大小。 模型大小取决于建模方法：对于树，这是树中的节点数；对于线性模型，这是系数的数目；对于 神经网络，这是突触的数目。 记录。 训练样本中输入记录的加权数。 自动数据准备 此视图显示有关排除了哪些字段及如何在自动数据准备 (ADP) 步骤中转换字段的信息。 对于每个已转换或排 除的字段，此表中列出了字段名、字段在分析中的角色以及 ADP 步骤所采取的操作。 字段是按照字段名称 的字母升序排列的。 操作 Trim outliers（如果显示）表示位于分界值（平均值的 3 个标准差）之外的连续预测变量值被设为分界 值。 拆分模型查看器 拆分模型查看器列出了每个拆分的模型，并提供了有关拆分模型的摘要。 拆分。 列标题显示用于创建拆分的字段，而单元格则是拆分值。 双击任何拆分可以打开为该拆分构建的模 型的“模型查看器”。 精确性。 总体准确性，以百分比表示。 模型大小。 模型大小取决于建模方法：对于树，这是树中的节点数；对于线性模型，这是系数的数目；对于 神经网络，这是突触的数目。 记录。 训练样本中输入记录的加权数。 136 IBM SPSS Statistics 29 Core System 用户指南 第 13 章 自动输出修改 自动输出修改将格式编排和其他更改应用到活动查看器窗口的内容。 可应用的更改包括： • 所有或选定的查看器对象 • 选定的输出对象类型（例如图表、日志、透视表） • 基于条件表达式的透视表内容 • 概要（导航）窗格内容 可以进行的更改类型包括： • 删除对象 • 为对象建立索引（添加顺序编号方案） • 更改对象的可视属性 • 更改概要标签文本 • 变换透视表中的行和列 • 更改透视表的选定层 • 根据条件表达式更改透视表中选定区域或特定单元格的格式编排（例如，将所有小于 0.05 的显著性值设为 粗体） 要指定自动输出修改： 1. 从菜单选择：实用程序 > 样式输出 2. 在查看器中选择一个或多个对象。 3. 在“选择”对话框中选择需要的选项。 （也可在打开对话框前选择对象。） 4. 在“样式输出”对话框中选择需要的输出更改。 样式输出：选择 样式输出：选择对话框可针对“样式输出”对话框上指定的更改，确定基本选择条件。 此外，您也可以在打开样式输出：选择对话框后，在“查看器”或“工作簿”中选择对象。 仅限选定对象(S) 更改仅应用于符合指定条件的选定对象。 选择作为最后的命令 如果选择此项，更改仅应用于最后一个过程的输出。 如果不选择此项，更改会应用于过程的特定实 例。 例如，如果“频率”过程有三个实例，而您选择第二个实例，那么更改仅应用于该实例。 如果根 据选择粘贴语法，此选项会选择该过程的第二个实例。 如果选择了多个过程的输出，那么此选项仅 适用于这样的过程：相应的输出是“查看器”或“工作簿”中的最后一个输出块。 选择作为一个组(G) 如果选择此项，在主“样式输出”对话框上将把选择中的所有对象作为一个组来处理。 如果不选择此 项，将把选定对象作为个别选择来处理，您可以分别设置每个对象的属性。 此类型的所有对象(A) 更改应用于符合指定条件的选定类型的所有对象。 此选项仅适用于这样的情况：在“查看器”或“工作簿” 中选择单一对象类型。 对象类型包括表格、警告、日志、图表、树形图、文本、模型和概要标题。 此子类型的所有对象(S) 更改应用于与符合指定条件的选定表格属于同一子类型的所有表格。 此选项仅适用于这样的情况：在 “查看器”或“工作簿”中选择单一的表子类型。 例如，选择可以包含两个独立的频率表，但不能包含一个 频率表和一个描述表。 名称相似的对象(N) 更改应用于有符合指定条件的类似名称的所有对象。 条件 选项为“包含”、“精确”、“开头”和“结尾”。 值 显示在“查看器”或“工作簿”的概要窗格中的名称。 更新 选择“查看器”或“工作簿”中符合指定值的具体条件的所有对象。 样式输出 “样式输出”对话框指定您在查看器中要对选定的输出对象所作的更改。 创建输出的备份。 自动输出修改过程所作的更改无法撤销。 要保留原始查看器文档，请创建备份副本。 选项和属性 您可以修改的对象或组的列表由您在查看器中选择的对象和在“样式输出：选择”对话框中所作的选择决定。 选择。 选定的过程或对象类型组的名称。 如果选择文本后面的括号中有整数，那么更改仅按查看器中对象 的顺序应用于该过程的实例。 例如，“频率(2)”仅将更改应用于查看器输出中“频率”过程的第二个实例。 类型。 对象的类型。 例如：日志、标题、表格、图表。 对于个别表类型，还会显示表子类型。 删除。 指定是否应该删除选择。 可视。 指定选择应该可视还是隐藏。 缺省选项为“按原样”，意味着保留选择的当前可视性属性。 属性。 要应用到选择的更改摘要。 添加。 向列表添加一行并打开“样式输出：选择”对话框。 可以在指定中选择其他对象并指定选择条件。 复制。 复制选定行。 上移和下移。 在列表中上移或下移选定行。 顺序可能很重要，因为在后续行中指定的更改可能覆盖在先前 行中指定的更改。 创建属性更改的报告。 在查看器中显示汇总更改的表。 对象属性 在“对象属性”部分中可指定要对“选择和属性”部分中每个选择所作的更改。 可用属性由“选择和属性”部分中 的选定行决定。 命令。 如果选择引用单个过程，则为该过程的名称。 选择可以包含同一过程的多个实例。 类型。 对象的类型。 子类型。 如果选择引用单个表类型，则显示表子类型名称。 概要标签。 概要窗格中与选择关联的标签。 可以替换标签文本或向标签添加信息。 有关更多信息，请参阅 主题“样式输出：标签和文本”。 索引格式。 向选择中的对象添加顺序数字、字母或罗马数字。 有关更多信息，请参阅主题“样式输出：索 引”。 表格标题。 表格的标题。 可以替换标题或向标题添加信息。 有关更多信息，请参阅主题“样式输出：标签和 文本”。 表格外观。 用于表格的表格外观。 有关更多信息，请参阅主题“样式输出：表格外观”。 变换。 变换表格中的行和列。 顶层。 对于带分层的表格，是为每层显示的类别。 条件样式。 表格的条件样式更改。 有关更多信息，请参见主题“表格样式”。 排序。 按选定的列标签值对表格内容排序。 可用列标签显示在下拉列表中。 该选项仅在选择包含单个表子 类型时可用。 138 IBM SPSS Statistics 29 Core System 用户指南 排序方向。 指定表格的排序方向。 注释文本。 您可以向每个表格添加注释文本。 • 如果您在查看器中将光标悬停在表格上，注释文本就会显示在工具提示中。 • 屏幕朗读器会在焦点位于表格上时朗读注释文本。 • 查看器中的工具提示仅显示注释的前 200 个字符，但屏幕朗读器会朗读全部文本。 • 如果将输出导出到 HTML，会使用注释文本作为替代文本。 内容。 日志、标题和文本对象的文本。 可以替换文本或向文本添加信息。 有关更多信息，请参阅主题“样式 输出：标签和文本”。 字体。 日志、标题和文本对象的字体。 字体大小。 日志、标题和文本对象的字体大小。 文本颜色。 日志、标题和文本对象的文本颜色。 图表模板。 用于图表的图表模板，但使用图形画板模板选择器创建的图表除外。 图形画板样式表。 用于使用图形画板模板选择器创建的图表的样式表。 大小。 图表和树形图的大小。 注释文本中的特殊变量 可以在注释文本字段中加入特殊变量以插入日期、时间和其他值。 )DATE 格式为 dd-mmm-yyyy 的当前日期。 )ADATE 格式为 mm/dd/yyyy 的当前日期。 )SDATE 格式为 yyyy/mm/dd 的当前日期。 )EDATE 格式为 dd.mm.yyyy 的当前日期。 )TIME 格式为 hh:mm:ss 的 12 小时制当前时间。 )ETIME 格式为 hh:mm:ss 的 24 小时制当前时间。 )INDEX 定义的索引值。 有关更多信息，请参阅主题“样式输出：索引”。 )TITLE 表格的概要标签的文本。 )PROCEDURE 创建表格的过程的名称。 )DATASET 用于创建表格的数据集的名称。 \n 插入换行符。 样式输出：标签和文本 “样式输出：标签和文本”对话框在概要标签、文本对象和表格标题中替换或添加文本。 它还指定概要标签、 文本对象和表格标题的索引值的包含与放置。 将文本添加到标签或文本对象。 可以将文本添加到现有文本的前面或后面，或者替换现有文本。 第 13 章 自动输出修改 139 添加索引。 添加顺序字母、数字或罗马数字。 可以将索引放在文本前面或后面。 还可以指定一个或多个字 符作为文本和索引之间的分隔符。 有关索引格式设置的信息，请参见主题“样式输出：索引”。 样式输出：索引 “样式输出：索引”对话框指定索引格式和起始值。 类型：顺序索引值可以是数字、小写或大写字母、小写或大写罗马数字。 起始值。 起始值可以是任何对所选类型有效的值。 要在输出中显示索引值，必须在所选对象类型的“样式输出：标签和文本”对话框中选择添加索引。 • 对于概要标签，在“样式输出”对话框的“属性”列中选择概要标签。 • 对于表格标题，在“样式输出”对话框的“属性”列中选择表格标题。 • 对于文本对象，在“样式输出”对话框的“属性”列中选择内容。 样式输出：表格外观 表格外观是定义表的外观的一组属性。 您可以选择先前定义的表格外观，或创建自己的表格外观。 • 在应用表格外观前后，可以使用单元格属性更改个别单元格或单元格组的单元格格式。 即使您应用新的表 格外观，编辑的单元格格式仍保持不变。 • 您还可以选择将所有单元格重置为由当前表格外观定义的单元格格式。 此选项将重置已编辑的所有单元 格。 如果在“表格外观文件”列表中选择了按显示，则任何已编辑的单元格都将重置为当前的表格属性。 • 仅“表格属性”对话框中定义的表格属性将保存在表格外观中。 TableLook 不包括单个单元格修改。 样式输出：大小 “样式输出：大小”对话框控制图表和树形图的大小。 可以厘米、英寸或点为单位指定高度和宽度。 表格样式 “表格样式”对话框指定根据具体条件自动更改透视表属性的条件。 例如，可以将所有小于 0.05 的显著性值 设为粗体和红色。 可以从“样式输出”对话框或从特定统计过程的对话框访问“表格样式”对话框。 • 可以从“样式输出”对话框或从特定统计过程的对话框访问“表格样式”对话框。 • 支持“表格样式”对话框的统计过程对话框是“双变量相关性”、“交叉表格”、“定制表格”、“描述”、“频 率”、“Logistic 回归”、“线性回归”和“平均值”。 表。 应用条件的表格。 如果从“样式输出”对话框访问此对话框，唯一的选项是“所有适用表格”。 如果从统 计过程对话框访问此对话框，可以从过程特定表格的列表选择表格类型。 值。 行或列标签值，定义表中用于搜索符合条件的值的区域。 可以从列表中选择值或输入值。 列表中的值 不受输出语言影响，适用于值的众多变体。 列表中提供的值取决于表格类型。 • 计数。 具有以当前输出语言表达的任何下列标签或等价标签的行或列：“Frequency”、“Count”、“N”。 • 平均值。 具有以当前输出语言表达的“Mean”标签或等价标签的行或列。 • 中位数。具有以当前输出语言表达的“Median”标签或等价标签的行或列。 • 百分比。 具有以当前输出语言表达的“Percent”标签或等价标签的行或列。 • 残差。 具有以当前输出语言表达的任何下列标签或等价标签的行或列：“Resid”、“Residual”和“Std.
Residual”。 • 相关性。 具有以当前输出语言表达的任何下列标签或等价标签的行或列：“Adjusted R Square”、 “Correlation Coefficient”、“Correlations”、“Pearson Correlation”、“R Square”。 • 显著性。具有任何这些标签的行或列或当前输出语言中的等价项：“Approx Sig.”、 “Asymp. Sig (2-sided)”、“Exact Sig.”、 “Exact Sig. (1-sided)”、“Exact Sig. (2-sided)”、“Sig.”、 “Sig. (1-tailed)”、“Sig.
(2-tailed)” • 所有数据单元格。 包括所有数据单元格。 140 IBM SPSS Statistics 29 Core System 用户指南 维度。 指定是搜索有指定值标签的行还是列，或者行列都搜索。 条件。 指定要查找的条件。 有关更多信息，请参见主题“表格样式：条件”。 格式编排。 指定要应用到符合条件的表单元格或区域的格式设置。 有关更多信息，请参见主题“表格样式： 样式”。 添加。 向列表添加一行。 复制。 复制选定行。 上移和下移。 在列表中上移或下移选定行。 顺序可能很重要，因为在后续行中指定的更改可能覆盖在先前 行中指定的更改。 创建条件样式的报告。 在查看器中显示汇总更改的表。 如果从统计过程对话框访问“表格样式”对话框，则 此选项可用。 “样式输出”对话框另有一个用于创建报告的选项。 表格样式：条件 “表格样式：条件”对话框指定应用更改的条件。 有两个选项： • 到此类型的所有值。 唯一的条件是“表格样式”对话框中的“值”列中指定的值。 这是缺省选项。 • 基于下列条件。 在“表格样式”对话框中“值”和“维度”列所指定的表格区域中，查找符合指定条件的值。 值。 列表包含比较表达式，例如精确、小于、大于和介于。 • 绝对值可用于只需要一个值的比较表达式。 例如，可以查找绝对值大于 0.5 的相关性。 • 上限和下限是指定表格区域中最高和最低的 n 值。 数目的值必须是整数。 • 系统缺失值。 在指定的表格区域中查找系统缺失值。 表格样式：格式 “表格样式：格式”对话框指定要根据“表格样式：条件”对话框中指定的条件应用的更改。 使用表格外观缺省值。 如果先前没有手动或通过自动输出修改更改格式，则此项等价于不作格式更改。 如 果先前更改过格式，则此项会消除那些更改，使有关的表格区域恢复为缺省格式。 应用新格式编排。 应用指定的格式更改。 格式更改包括字体样式及颜色、背景颜色、数字值的格式（包括 日期和时间），以及显示的小数位数。 应用到。 指定要应用更改的表格区域。 • 仅限单元格。 仅将更改应用到符合条件的表单元格。 • 整列。 将更改应用到包含一个符合条件的单元格的一整列。 此选项包括列标签。 • 整行。 将更改应用到包含一个符合条件的单元格的一整行。 此选项包括行标签。 替换值。 将值替换为指定的新值。 对于仅限单元格，此选项替换各个符合条件的单元格值。 对于整行和整 列，此选项替换行或列中的所有值。 第 13 章 自动输出修改 141 142 IBM SPSS Statistics 29 Core System 用户指南 第 14 章 使用命令语法 这种功能强大的命令语言，使您可以保存和自动执行许多常见任务。 它还提供一些在菜单和对话框中没有的 功能。 大多数命令可以从菜单和对话框访问。 但是，某些命令和选项只能通过命令语言使用。 命令语言还允许您 将作业保存在语法文件中，以便将来可以重复进行分析，或者通过生产作业在自动化作业中运行该语法文 件。 语法文件就是一个包含命令的文本文件。 虽然可以打开语法窗口并键入命令，但如果您使用以下一种方法， 让软件帮助您构建语法文件，那么事情就会简单很多： • 从对话框粘贴命令语法 • 从输出日志复制语法 • 从日志文件复制语法 详细命令语法参考信息以两种形式提供：集成到整体“帮助”系统中；在“帮助”菜单中，作为名为命令语法参 考的单独 PDF 文件提供。 按 F1 键可获得语法窗口中当前命令的上下文相关的帮助。 语法规则 当您在会话过程中从命令语法窗口运行命令时，将以交互模式运行命令。 以下规则适用于交互模式下的命令规范： • 每个命令必须从新行开始。 命令可在命令行的任何列中开始，并持续所需任意数量的行。 END DATA 命令 是个例外，该命令必须在数据结尾之后第一行的第一列中开始。 • 每个命令应该以句点为命令终止符。 但是，最好省略 BEGIN DATA 中的终止符，以便将内联数据视为一 个连续指定。 • 命令终止符必须是命令中的最后一个非空白字符。 • 如果没有句点作为命令终止符，那么将空行解释为命令终止符。 注：要兼容命令执行的其他方式（包括在交互式会话中使用 INSERT 或 INCLUDE 命令运行的命令文件）， 命令语法的每行不应超过 256 个字符。 • 大多数子命令用斜杠 (/) 分隔。 命令中第一个子命令之前的斜杠通常是可选的。 • 变量名称必须使用全名拼写。 • 单引号或双引号内包含的文本必须包含在单独一行上。 • 区域设置或语言环境设置如何，必须使用句点 (.) 指示小数。 • 以句点结束的变量名称可能在由对话框创建的命令中导致错误。 不能在对话框中创建这样的变量名称，并 且通常应避免这样的变量名称。 命令语法不区分大小写，三个字母或四个字母的缩写可用于许多命令规范。 可以使用任意多行指定单独一条 命令。 可以在几乎所有允许单个空格的地方添加空格或换行符，例如斜杠、括号和算术运算符两端或变量名 称之间。 例如： FREQUENCIES VARIABLES=JOBCAT GENDER /PERCENTILES=25 50 75 /BARCHART.
和 freq var=jobcat gender /percent=25 50 75 /bar.
都是可接受的替代形式，它们生成相同的结果。 INCLUDE 文件 对于通过 INCLUDE 命令运行的命令文件，可使用批处理模式语法规则。 以下规则适用于批处理模式下的命令规范： • 命令文件中的所有命令都必须从列 1 开始。 如果要缩进命令规范以使命令文件更可读，那么可以在第一列 中使用加号 (+) 或减号 (-) 。 • 如果对命令使用多行，那么每个连续行的列 1 必须为空。 • 命令终止符是可选的。 • 一行不能超过 256 个字符；将截断任何额外的字符。 除非现有的命令文件已使用了 INCLUDE 命令，否则应改用 INSERT 命令，因为它适用于符合任意规则的命 令文件。 如果通过将对话框选择粘贴到语法窗口中来生成命令语法，则这些命令的格式适用于任何操作模 式。 有关更多信息，请参阅命令语法参考（可从“帮助”菜单获得 PDF 格式的文件）。 从对话框粘贴语法 建立命令语法文件的最简单方法是在对话框中进行选择，然后将选择的语法粘贴到语法窗口中。 通过在冗长 分析的每一步粘贴语法，您可以构建作业文件，该文件使您可以在将来重复进行分析，或者通过生产工具运 行自动化作业。 在语法窗口中，您可以运行粘贴的语法，对其进行编辑，并将其保存到语法文件。 从对话框粘贴语法 1. 打开对话框并选择所需的语法。 2. 单击粘贴。 命令语法会粘贴到指定的语法窗口中。 如果没有打开语法窗口，则会自动打开新的语法窗口，语法会粘贴在 那里。 缺省情况下，语法会粘贴到最后一条命令的后面。 您可以选择将语法粘贴到光标位置，或者覆盖选 中的语法。 该设置在“选项”对话框的“语法编辑器”选项卡上指定。 从输出日志复制语法 您可以通过从显示在查看器中的日志复制命令语法来建立语法文件。 要使用此方法，必须在运行分析之前， 在“查看器”设置（“编辑”菜单，“选项”，“查看器”选项卡）中选择在日志中显示命令。 然后每条命令将随同 分析的输出一起显示在查看器中。 在语法窗口中，您可以运行粘贴的语法，对其进行编辑，并将其保存到语法文件。 从输出日志复制语法 1. 在运行分析之前，从菜单中选择： 编辑 > 选项...
2. 在“查看器”选项卡上，选择在日志中显示命令。 运行分析的同时，在对话框中选择的命令将记录到日志中。 3. 打开先前保存的语法文件或新建一个语法文件。 要新建语法文件，从菜单中选择： 文件 > 新建 > 语法 4. 在查看器中，双击日志项以将其激活。 5. 选择要复制的文本。 6. 从“查看器”菜单中选择： 编辑 > 复制 7. 在语法窗口中，从菜单中选择： 编辑 > 粘贴 144 IBM SPSS Statistics 29 Core System 用户指南 使用语法编辑器 语法编辑器提供一个专为创建、编辑和运行命令语法而设计的环境。 语法编辑器有以下特色： • 自动完成。 随着您的输入，您可以从上下文敏感列表中选择命令、子命令、关键字和关键字值。 您可以选 择自动提示列表或按需要显示列表。 • 颜色编码。 命令语法（命令、子命令、关键字和关键字值）的识别元素是颜色编码，因此浏览一下即可定 位未识别项。 另外，一些常见语法错误--如未匹配的引号--都经过颜色编码以供快速识别。 • 分界点。 您可以在指定点停止执行命令语法，这允许您在查看数据或输出后再继续。 • 书签。 您可以设置书签，以允许您快速导航大型命令语法文件。 • 自动缩进。 您可以使用与从对话框粘贴的语法类似的缩进样式来自动设置您的语法格式。 • 逐步执行。 您可以一次一个命令地逐步执行命令语法，使用单击前进到下一个命令。 注：当使用从右至左语言时，建议在“选项”对话框中的“语法编辑器”选项卡上选中“针对从右至左语言优化” 复选框。 语法编辑器窗口 语法编辑器窗口分为四个区域： • 编辑器窗格是语法编辑器窗口的主要部分，是您输入和编辑命令语法的地方。 • 装订线与编辑器窗格相邻，显示行号和分界点位置等信息。 • 导航窗格位于装订线和编辑器窗格左侧，在语法编辑器窗口中显示所有命令列表，单击即可导航到任何命 令。 • 错误窗格位于编辑器窗格下方，显示运行时间错误。 装订线内容 行号、分界点、书签、命令跨度和进度指示符显示在语法窗口中编辑窗格的左侧。 • 行号不考虑 INSERT 和 INCLUDE 命令中引用的任何外部文件。您可以通过从菜单中选取“视图”>“显示行 号”来显示或隐藏行号。 • 分界点在指定点停止执行，表示为一个与设置分界点的命令相邻的红圈。 • 书签在命令语法文件中标记特定行，表示为包含分配到书签的数字（1-9）的正方形。 悬停在书签图标上 将显示分配到书签的书签编号以及名称（如果有）。 • 命令跨度是以可视的方式标识命令开始和结束位置的图标。 您可以通过从菜单中选取“视图”>“显示命令跨 度”来显示或隐藏命令跨度。 • 给定语法运行的进度在装订线中使用向下箭头表示，从第一个命令运行扩展到最后一个命令运行。 这在运 行包含分界点的命令语法和逐步执行命令语法时最有用。 有关更多信息，请参阅第150 页的『运行命令语 法』主题。 导航窗格 导航窗格在语法窗口中包含所有已识别命令列表，以它们在窗口中发生的顺序显示。 单击导航窗格中的命令 会将光标置于命令开始。 • 您可以使用向上和向下箭头键在命令列表中移动，或者单击命令以导航到该命令。 双击将选择命令。 • 包含某些类型的语法错误--如未匹配的引号--的命令名称缺省情况下显示为红色加粗文本。 有关更多信 息，请参阅第146 页的『颜色编码』主题。 • 未识别文本的每行第一个单词显示为灰色。 • 您可以通过从菜单中选取“视图”>“显示导航窗格”来显示或隐藏导航窗格。 错误窗格 错误窗格显示最近运行中的运行时间错误。 • 每个错误的信息包含存在错误的命令的起始行号。 • 您可以使用向上和向下箭头移动通过错误列表。 第 14 章 使用命令语法 145 • 单击列表中的一个条目会将光标置于生成该错误的命令的第一行上。 • 您可以通过从菜单中选取"视图">"显示错误窗格"来显示或隐藏错误窗格。 使用多个视图 您可以将编辑器窗格拆分为上下两个窗格。 1. 从菜单中选择： 窗口 > 拆分 在导航和错误窗格中的操作，例如单击某个错误，将对光标所在的窗格起作用。 您可以通过双击拆分器或选取“窗口”>“移除拆分”来移除拆分器。 术语 命令。 语法的基本单位是命令。 每个命令以命令名称开头，命令名称包括一个、两个或三个单词--例如， DESCRIPTIVES、SORT CASES 或 ADD VALUE LABELS。 子命令。 大多数命令包含子命令。 子命令提供附加规范，以正斜杠开头，后面跟子命令名称。 关键字。 关键字是通常在子命令中使用的固定术语，用来指定子命令的可用选项。 关键字值。 关键字可以拥有如指定一个选项或数字值的固定术语之类的值。 示例 CODEBOOK gender jobcat salary /VARINFO VALUELABELS MISSING /OPTIONS VARORDER=MEASURE.
• 命令名称是 CODEBOOK。 • VARINFO 和 OPTIONS 是子命令。 • VALUELABELS、MISSING 和 VARORDER 是关键字。 • MEASURE 是与 VARORDER 相关的关键字值。 自动完成 语法编辑器以自动完成命令、子命令、关键字和关键字值的形式提供帮助。 缺省情况下，它将在您输入时提 示一个可用术语的上下文敏感列表。 按 Enter 或 Tab 键将在光标位置插入列表中当前突出显示的项目。 您 可以通过按 Ctrl+空格按需要显示列表，也可以通过按 Esc 键关闭列表。 “工具”菜单上的“自动完成”菜单项可以在自动显示或不自动显示自动完成列表之间进行切换。 您还可从“选 项”对话框中的“语法编辑器”选项卡启用或禁用自动显示列表。 切换自动完成菜单项会覆盖“选项”对话框上 的设置，但不会在会话之间保持不变。 注：如果输入一个空格，自动完成列表将关闭。 对于由多个单词组成的命令（例如 ADD FILES），请先选 择该命令，然后再输入空格。 颜色编码 语法编辑器颜色代码识别命令语法元素，如命令和子命令，以及各种语法错误，如未匹配的引号或圆括号。 未识别的文本未经颜色编码。 命令。 缺省情况下，识别的命令显示为蓝色粗体文本。 但是，如果在命令中有一个识别的语法错误--如缺失 的圆括号--命令名称缺省情况下显示为红色加粗文本。 注：命令名称的简写--如 FREQUENCIES 简写为 FREQ --不标为彩色，但此简写有效。 子命令。 识别的子命令缺省情况下显示为绿色。 但是，如果子命令缺失一个所需等号或随后跟的是一个无 效等号，子命令名称缺省情况下显示为红色。 关键字。 识别的关键字缺省情况下显示为褐紫红色。 但是，如果关键字缺失一个所需等号或随后跟的是一 个无效等号，关键字缺省情况下显示为红色。 146 IBM SPSS Statistics 29 Core System 用户指南 关键字值。 识别的关键字值缺省情况下显示为橙色。 用户指定的关键字值未经颜色编码，如整数、实数和 加引号的字符串。 注释。 注释中的文本缺省情况下显示为灰色。 引号。 引号及其中的文本缺省情况下显示为黑色。 语法错误。 与以下语法错误相关的文本缺省情况下显示为红色。 • 未匹配的圆括号、方括号和引号。 未检测到注释和加引号的字符串中未匹配的圆括号和方括号。 在加引号 的字符串中未匹配的单引号或双引号在语法上无效。 某些命令包含不是命令语法的文本块--如 BEGIN DATA-END DATA、BEGIN GPL-END GPL 和 BEGIN PROGRAM-END PROGRAM。 在这种块中未检测到未匹配的值。 • 长行。 长行是包含的字符数超过 251 的行。 • 结束语句。 一些命令在命令终止符之前需要 END 语句（例如，BEGIN DATA-END DATA）或在命令流中 的稍后某个时刻需要匹配的 END 命令（例如，LOOP-END LOOP）。 在这两种情况下，命令缺省情况下显 示为红色，直到添加了所需 END 语句。 注：通过从“工具”菜单的“验证错误”子菜单中选择“下一个错误”或“上一个错误”，您可以导航到下一个或上 一个语法错误。 从“选项”对话框中的“语法编辑器”选项卡，您可以更改缺省颜色和文本样式，也可以关闭或打开颜色编码。 还可以通过从菜单中选择“工具”>“颜色编码”来开启或关闭针对命令、子命令、关键字和关键字值的颜色编 码。 可以通过选择“工具”>“验证”来开启或关闭针对语法错误的颜色编码。 在“工具”菜单上进行的选择会覆 盖“选项”对话框中的设置，但不会在会话之间保持不变。 注：不支持“宏”中的命令语法颜色编码。 断点 分界点允许您在语法窗口中的指定点停止执行命令语法，并在准备好时继续执行。 • 分界点设置在命令级别，并在运行命令之前停止执行。 • 分界点无法在 LOOP-END LOOP、DO IF-END IF、DO REPEAT-END REPEAT、INPUT PROGRAM-END INPUT PROGRAM 和 MATRIX-END MATRIX 块中发生。 但是，它们可以设置在这种块的开始，并在运行 块之前停止执行。 • Breakpoints cannot be set on lines containing non-IBM SPSS Statistics command syntax, such as occur within BEGIN PROGRAM-END PROGRAM, BEGIN DATA-END DATA, and BEGIN GPL-END GPL blocks.
• 分界点不与命令语法文件一起保存，不包括在复制的文本中。 • 缺省情况下，在执行过程中接受分界点。 可以在“工具”>“荣誉断点”中切换是否使用断点。 插入分界点 1. 单击命令文本左侧装订线的任意位置。 或 2. 将光标置于命令中。 3. 从菜单中选择： 工具 > 切换分界点 分界点在命令文本左侧装订线中以及与命令名称相同的行上表示为红圈。 清除分界点 清除单个分界点： 1. 在命令文本左侧装订线中单击表示分界点的图标。 或 2. 将光标置于命令中。 3. 从菜单中选择： 第 14 章 使用命令语法 147 工具 > 切换分界点 清除所有分界点： 4. 从菜单中选择： 工具 > 清除所有分界点 请参阅第150 页的『运行命令语法』以获取有关在存在分界点的情况下运行时行为的信息。 书签 书签允许您快速导航到命令语法文件中的指定位置。 您在给定文件中可以拥有最多 9 个书签。 书签与文件 一起保存，但在复制文本时不包括。 插入书签 1. 将光标定位在要插入书签的行上。 2. 从菜单中选择： 工具 > 切换书签 给新书签在从 1 到 9 的范围内分配下一个可用的编号。 将用一个把分配的编号包围起来的正方形来表示书 签，并显示在命令文本左侧的装订线中。 清除书签 清除单个书签： 1. 将光标置于包含书签的行上。 2. 从菜单中选择： 工具 > 切换书签 清除所有书签： 1. 从菜单中选择： 工具 > 清除所有书签 重命名书签 您可以将一个名称与一个书签关联起来。 这是除在创建时分配给书签的数字（1-9）以外的另一名称。 1. 从菜单中选择： 工具 > 重命名书签 2. 为书签输入名称并单击确定。 指定的名称将替换书签的任何现有名称。 导航书签 导航到下一个或上一个书签： 1. 从菜单中选择： 工具 > 下一个书签 或者 工具 > 上一个书签 导航到特定书签： 1. 从菜单中选择： 工具 > 转至书签 2. 选择书签。 148 IBM SPSS Statistics 29 Core System 用户指南 注释或取消注释文本 您可以注释整个命令以及未识别为命令语法的文本，还可以取消注释先前被注释的文本。 注释文本 1. 选择文本。 注意，如果选择了命令的任何部分，将对命令进行注释。 2. 从菜单中选择： 工具 > 切换注释选择 可通过以下方法来注释掉单个命令：将光标定位在命令中的任意位置并选取“工具”>“切换注释选择”。 取消注释文本 1. 选择要取消注释的文本。 注意，如果选择了命令的任何部分，将对命令取消注释。 2. 从菜单中选择： 工具 > 切换注释选择 可通过以下方法来取消注释单个命令：将光标定位在命令中的任意位置并选取“工具”>“切换注释选择”。 注 意，此功能不会从命令中删除注释（以 / 和 / 开始的文本）或通过 COMMENT 关键字创建的注释。 设置语法格式 您可以增加或减少选定语法行的缩进，还可以自动调整选定内容的缩进，以使语法格式与从对话框粘贴的语 法格式类似。 • 缺省缩进为四个空格，应用于选定语法行的缩进，以及自动缩进。 您可以在“选项”对话框的“语法编辑器” 选项卡上更改缩进大小。 • 注意，在“语法编辑器”中使用 Tab 键不会插入制表符。 它会插入一个空格。 缩进文本 1. 选择文本或将光标定位到您要缩进的单行上。 2. 从菜单中选择： 工具 > 缩进语法 > 缩进 您还可以按 Tab 键缩进选定内容或行。 减少缩进 1. 选择文本或将光标定位到您要减少缩进的单行上。 2. 从菜单中选择： 工具 > 缩进语法 > 减少缩进 自动缩进文本 1. 选择文本。 2. 从菜单中选择： 工具 > 缩进语法 > 自动缩进 在您自动缩进文本时，任何现有缩进将被删除，并替换为自动产生的缩进。 注意，在 BEGIN PROGRAM 块 中的自动缩进代码可能会破坏那些依赖于特定缩进而工作的代码，例如包含循环和条件块的 Python 代码。 采用自动缩进功能的格式语法不能在批处理模式下运行。 例如，自动缩进 INPUT PROGRAM-END INPUT PROGRAM、LOOP-END LOOP、DO IF-END IF 或 DO REPEAT-END REPEAT 块，将导致在批处理模式中 语法失败，因为块中的命令将被缩进，而不会从列 1 开始，这与批处理模式的要求不符。 不过，您可以在批 处理模式下使用 -i 开关，强制 Batch Facility 使用交互语法规则。 请参阅第143 页的『语法规则』主题以获 取更多信息。 第 14 章 使用命令语法 149 运行命令语法 1. 在语法窗口突出显示想要运行的命令。 2. 单击语法编辑器工具栏上的“运行”按钮（指向右边的三角形）。 运行所选命令，或者如果没有选择，运 行光标所在命令。 或 3. 从“运行”菜单中选择一项。 • 全部。 运行语法窗口中的所有命令，接受任何分界点。 • 选择。 运行当前选择的命令，接受任何分界点。 这包括任何只有部分突出显示的命令。 如果没有选择， 则运行光标所在的命令。 • 至结束。 运行从当前选择中的第一个命令到语法窗口中的最后一个命令的所有命令，接受任何分界点。 如 果没有选择，则从光标所在的命令开始运行。 • 逐步执行。 一次一个命令地从语法窗口中的第一个命令（从开始逐步执行）或从当前选择中的光标所在的 命令（从当前逐步执行）运行命令语法。 如有选择的文本，则从选择中的第一个命令开始运行。 在给定命 令运行之后，光标前进到下一个命令，您可以通过选择“继续”继续逐步执行。 LOOP-END LOOP、DO IF-END IF、DO REPEAT-END REPEAT、INPUT PROGRAM-END INPUT PROGRAM 和 MATRIX-END MATRIX 块在使用“逐步执行”时被视为单个命令。 您无法步进到其中任何一个 块。 • 继续。 继续被分界点或逐步执行停止的运行。 进度指示符 给定语法运行的进度在装订线中使用向下箭头表示，跨到最后一组命令运行。 例如，您选择运行包含分界点 的语法窗口中的所有命令。 在第一个分界点，箭头将从窗口中的第一个命令跨到包含该分界点之前的命令。 在第二个分界点，箭头将从包含第一个分界点的命令扩展到包含第二个分界点之前的命令。 带有分界点的运行时间行为 • 当运行包含分界点的命令语法时，在每个分界点执行停止。 具体而言，将命令语法的块从给定的断点（或 开始运行）到下一个断点（或结束运行）提交给执行，就像选择了该语法并选择了“运行” > “选择”一样。 • 您可以使用多个语法窗口，每个都有自己的一组分界点，但是执行命令语法的队列只有一个。 一旦提交了 一个命令语法块（如到第一个分界点的命令语法块），则不会执行其他命令语法块，直到前一个块已完 成，无论块是否在相同或不同的语法窗口中。 • 执行在一个分界点处停止，您可以在其他语法窗口中运行命令语法，并检查数据编辑器或查看器窗口。 但 是，修改包含分界点的语法窗口的内容或更改光标在该窗口中的位置将取消运行。 语法文件中的字符集编码 语法文件的字符集编码可以是 Unicode 或代码页编码。 Unicode 文件可以包含许多不同字符集中的字符。 代码页文件限制为在特定语言或语言环境中受支持的字符。 例如，采样西欧编码的代码页文件不能包含日语 或中文字符。 读取语法文件 要正确地读取语法文件，语法编辑器需要知道文件的字符编码。 • 包含 Unicode UTF-8 字节顺序标记的文件将作为 Unicode UTF-8 编码进行读取，而无论您选择的是哪种编 码。 此字节顺序标记位于文件开头，但未显示出来。 • 缺省情况下，未包含任何编码信息的文件将作为 Unicode UTF-8（在 Unicode 方式下）或当前语言环境字 符编码（在代码页方式下）进行读取。 要覆盖缺省行为，请选择 Unicode (UTF-8) 或本地编码。 • 如果语法文件顶部包含代码页编码标识，那么按照声明将处于启用状态。 从 R23 开始，将自动在以代码页 编码进行保存的语法文件中插入注释。 例如，文件的第一行可能如下所示： Encoding: en_US.windows-1252.
如果您选择了按照声明，那么该编码将用于读取文件。 150 IBM SPSS Statistics 29 Core System 用户指南 保存语法文件 缺省情况下，语法文件将以 Unicode UTF-8（在 Unicode 方式下）或当前语言环境字符编码（在代码页方式 下）进行保存。 要覆盖缺省行为，请在“将语法另存为”对话框中选择 Unicode (UTF-8) 或本地编码。 • 如果您保存新的语法文件或者以其他编码保存文件，那么文件顶部将插入一个标识编码的注释。 如果编码 注释已存在，它将被替换掉。 • 如果您在保存语法文件后未将其关闭就再次保存，那么该文件将以同一编码进行保存。 多条执行命令 从对话框粘贴或者从日志或日志文件中复制的语法可能包含 EXECUTE 命令。 从语法窗口运行命令时，通常 不需要 EXECUTE 命令，并且该命令可能会降低性能，尤其是对于大型数据文件，因为每个 EXECUTE 命令都 读取整个数据文件。 和 。 有关更多信息，请参阅命令语法引用中的 EXECUTE 命令（可从任何IBM SPSS Statistics 窗口的“帮助”菜单获取）。 延迟函数 一个值得注意的例外情况是包含延迟函数的转换命令。 在没有任何干扰性 EXECUTE 命令或者其他读取数据 的命令的一系列转换命令中，延迟函数是在所有其他转换之后计算的，与命令顺序无关。 例如， COMPUTE lagvar=LAG(var1).
COMPUTE var1=var12.
和 COMPUTE lagvar=LAG(var1).
EXECUTE.
COMPUTE var1=var12.
对于 lagvar 的值生成不同的结果，因为前者使用转换后的 var1 值，而后者使用原始值。 语法文件中的字符集编码 语法文件的字符集编码可以是 Unicode 或代码页编码。 Unicode 文件可以包含许多不同字符集中的字符。 代码页文件限制为在特定语言或语言环境中受支持的字符。 例如，采样西欧编码的代码页文件不能包含日语 或中文字符。 读取语法文件 要正确地读取语法文件，语法编辑器需要知道文件的字符编码。 • 包含 Unicode UTF-8 字节顺序标记的文件将作为 Unicode UTF-8 编码进行读取，而无论您选择的是哪种编 码。 此字节顺序标记位于文件开头，但未显示出来。 • 缺省情况下，未包含任何编码信息的文件将作为 Unicode UTF-8（在 Unicode 方式下）或当前语言环境字 符编码（在代码页方式下）进行读取。 要覆盖缺省行为，请选择 Unicode (UTF-8) 或本地编码。 • 如果语法文件顶部包含代码页编码标识，那么按照声明将处于启用状态。 从 R23 开始，将自动在以代码页 编码进行保存的语法文件中插入注释。 例如，文件的第一行可能如下所示： Encoding: en_US.windows-1252.
如果您选择了按照声明，那么该编码将用于读取文件。 保存语法文件 缺省情况下，语法文件将以 Unicode UTF-8（在 Unicode 方式下）或当前语言环境字符编码（在代码页方式 下）进行保存。 要覆盖缺省行为，请在“将语法另存为”对话框中选择 Unicode (UTF-8) 或本地编码。 • 如果您保存新的语法文件或者以其他编码保存文件，那么文件顶部将插入一个标识编码的注释。 如果编码 注释已存在，它将被替换掉。 • 如果您在保存语法文件后未将其关闭就再次保存，那么该文件将以同一编码进行保存。 第 14 章 使用命令语法 151 多条执行命令 从对话框粘贴或者从日志或日志文件中复制的语法可能包含 EXECUTE 命令。 从语法窗口运行命令时，通常 不需要 EXECUTE 命令，并且该命令可能会降低性能，尤其是对于大型数据文件，因为每个 EXECUTE 命令都 读取整个数据文件。 和 。 有关更多信息，请参阅命令语法引用中的 EXECUTE 命令（可从任何IBM SPSS Statistics 窗口的“帮助”菜单获取）。 延迟函数 一个值得注意的例外情况是包含延迟函数的转换命令。 在没有任何干扰性 EXECUTE 命令或者其他读取数据 的命令的一系列转换命令中，延迟函数是在所有其他转换之后计算的，与命令顺序无关。 例如， COMPUTE lagvar=LAG(var1).
COMPUTE var1=var12.
和 COMPUTE lagvar=LAG(var1).
EXECUTE.
COMPUTE var1=var12.
对于 lagvar 的值生成不同的结果，因为前者使用转换后的 var1 值，而后者使用原始值。 加密语法文件 可以通过用密码将语法文件加密的方法保护语法文件。 只有提供密码才能打开已加密的文件。 注: 已加密的语法文件无法用于生产作业或与 IBM SPSS Statistics 批处理工具（随 IBM SPSS Statistics Server 提供）配合使用。 要将语法编辑器的内容保存为加密的语法文件： 1. 使语法编辑器成为活动窗口（单击窗口的任何部位即可使其成为活动窗口）。 2. 从菜单中选择： 文件 > 另存为...
3. 从“另存为类型”下拉列表选择加密语法。 4. 单击保存。 5. 在“加密文件”对话框中提供密码并重新在“确认密码”文本框中输入。 密码限制在 10 个字符并区分大小 写。 警告：密码丢失后将无法恢复。 如果密码丢失，那么将无法打开文件。 创建强密码 • 至少使用八个字符。 • 在密码中使用数字、符号甚至标点符号。 • 避免使用数字序列或字符序列（例如 "123" 和 "abc"）并避免重复，例如 "111aaa"。 • 不要创建使用个人信息（例如生日或昵称）的密码。 • 定期更改密码。 注：不支持将已加密的文件存储到 IBM SPSS 协作和部署服务存储库 中。 修改加密文件 • 如果打开加密文件，修改文件并选择“文件” > “保存”，那么修改的文件以相同的密码保存。 • 您可以通过打开文件、重复加密步骤并在“加密文件”对话框中指定不同的密码，在加密的文件上更改密 码。 • 通过打开文件，选择“文件 > 另存为”并从“另存为类型”下拉列表中选择语法 ，可以保存已加密文件的未加 密版本。 注: 在 V22 之前的 IBM SPSS Statistics 版本中，无法打开已加密的语法文件。 152 IBM SPSS Statistics 29 Core System 用户指南 第 15 章 图表工具的概述 通过“图形”菜单和“分析”菜单中的多个过程创建高分辨率图表和图。 本章提供图表工具的概述。 生成和编辑图表 创建图表之前，您需要从数据编辑器中获取数据。 您可以将数据直接输入数据编辑器，打开先前保存的数据 文件，或读取一个电子表格、制表符分隔的数据文件或数据库文件。 “帮助”菜单中的“教程”选择有创建和修 改图表的联机示例，联机帮助系统提供有关创建和修改所有图表类型的信息。 生成图表 使用图表构建器可以根据预定义的图库图表或图表的单独部分（例如，轴和条形）生成图表。 可以通过将图 库图表或基本元素拖放到画布（“图表构建器”对话框上“变量”列表右侧的较大区域）上来生成图表。 以了解 对“图表构建器”对话框的说明。 生成图表时，画布会显示图表的预览。 虽然预览使用已定义的变量标签和测量级别，但预览并不会显示实际 的数据。 它使用随机生成的数据简略勾勒出图表的外观。 。 对于新用户，使用图库是首选方法。 有关使用库的信息，请参阅第153 页的『根据图库生成图表』。 如何启动图表构建器 1. 从菜单中选择： 图形 > 图表构建器 根据图库生成图表 构建图表的最简单方法是使用图库。 下列是从库构建图表的一般步骤。 ，以获取不同图表类型的特定信 息。 1. 如果尚未显示图库选项卡，请单击它。 2. 在“选择范围”列表中，选择一个图表类别。 每个类别都提供了多种类型。 3. 将所需图表的图片拖到画布上。 也可以双击该图片。 如果画布已显示了一个图表，那么图库图表会替换 该图表上的轴系和图形元素。 a. 将变量从“变量”列表拖放至轴放置区以及分组放置区（如果有）。 如果轴放置区已显示统计，而您要 使用该统计，那么不必将变量拖到放置区中。 只有区域中的文本为蓝色时才需要向该区域添加变量。 如果文本为黑色，那么表明该区域已包含变量或统计。 以获取有关可用统计的信息。 注：变量的测量级别很重要。 生成图表时，图表构建器根据测量级别设置缺省值。 此外，测量级别不 同，产生的图表看上去也会不同。 可以临时更改变量的测量级别，方法是右键单击该变量，然后选择 选项。 主题以获取更多信息。 b. 如果需要更改统计或修改轴或图注的属性（如刻度范围），请单击图表构建器侧边栏中的元素属性。 （如果未显示侧边栏，请单击图表构建器右上角中的按钮以显示侧边栏。） c. 在“编辑属性”列表中，选择要更改的项。 （有关特定属性的信息，请单击帮助。） d. 如果需要将更多变量添加到图表中（例如要进行汇总或将变量放入面板），可在“图表构建器”对话框 中单击组/点 ID 选项卡并选择一个或多个选项。 然后，将分类变量拖放至画布上显示的新放置区。 主 题以获取更多信息。 4. 如果要变换图表（例如，使条变为水平的），请单击基本元素选项卡，然后单击变换。 5. 单击确定以创建图表。 图表将显示在查看器中。 编辑图表 图表编辑器提供功能强大、易于使用的环境，供您定制图表和探索数据。 图表编辑器有以下特色： • 简单、直观的用户界面。 使用菜单和工具栏可以快速选择并编辑图表的各个部分。 也还可以直接在图表上 输入文本。 • 众多格式和统计选项。 您可以选择各种风格和统计选项。 • 强大的探索工具。 您可以用各种方式探索数据，如标记数据，对数据进行重新排序，或者旋转数据。 可以 更改图表类型和图表中的变量角色。 还可以添加分布曲线、拟合线、内插线和参考线。 • 灵活的模板，使外观和操作保持一致。 可以创建定制的模板，并用这些模板轻松创建具有所希望的外观和 选项的图表。 例如，如果希望坐标轴标签始终朝向特定方向，就可以在模板中指定这个方向，然后将模板 应用于其他图表。 如何查看图表编辑器 1. 双击查看器中的一个图表。 图表编辑器基础 图表编辑器提供多种操作图表的方法。 菜单 图表编辑器中可以执行的很多操作都是用菜单完成的，特别是将某项添加到图表的时候。 例如，使用菜单将 拟合线添加到散点图。 将某项添加到图表后，经常需要使用“属性”对话框来指定对应所添加的项的选项。 “属性”对话框 在“属性”对话框可以找到对应图表及其图表元素的选项。 要查看“属性”对话框，您可以： 1. 双击图表元素。 或 2. 选择图表元素，然后在菜单中选择： 编辑 > 属性 此外，“属性”对话框在您将某项添加到图表时自动出现。 使用“属性”对话框的选项卡可以设置选项，并对图表作出其他更改。 在“属性”对话框显示的选项卡取决于您 当前的选择。 有些选项卡包含预览，可使您了解应用所做的更改后对当前选择的影响。 但是图表本身并不反映您所作的更 改，只有单击应用才会反映出来。 可以在多个选项卡上作出更改，然后单击应用。 如果您必须更改选择以 修改图表上不同的元素，那么在更改选择前单击应用。 如果没有在更改选择前单击应用，那么以后单击应用 只能将更改应用于当前选定的元素。 只有特定设置可用，这取决于您的选择。 各个选项卡的帮助指定查看选项卡需要选择的内容。 如果选择了 多个元素，那么只能更改所有元素共有的设置。 工具栏 工具栏提供“属性”对话框中某些功能的快捷方式。 例如，可以不用“属性”对话框的“文本”选项卡，而使用 “编辑”工具栏来更改文本的字体和样式。 保存更改 关闭图表编辑器时，将保存图表修改。 修改过的图表将随后显示在查看器中。 图表定义选项 在图表构建器中定义图表时，可以为图表创建添加标题和更改选项。 添加和编辑标题和脚注 可以向图表添加标题和脚注，以帮助查看器解释图表。 图表构建器还会自动在脚注中显示误差条形图信息。 154 IBM SPSS Statistics 29 Core System 用户指南 添加或修改标题和脚注 1. 单击标题/脚注选项卡，或者单击预览中显示的标题或脚注。 2. 使用元素属性选项卡以编辑标题/脚注文本。 对于标题 1，将基于图表中使用的图表类型和变量生成自动标题。 除非选择自定义或无，否则将使用此自动 标题。 如何移去标题或脚注 1. 单击标题/脚注选项卡。 2. 取消选择要移去的标题或脚注。 如何编辑标题或脚注文本 在添加标题和脚注时，不能直接在图表上编辑与其关联的文本。 图表构建器中的其他项相同，可以使用元素 属性选项卡进行编辑。 1. 在“编辑属性”列表中，选择标题、子标题或脚注（如标题 1）。 2. 在“内容”框中，键入与标题、子标题或脚注关联的文本。 设置一般选项 图表构建器为图表提供一般选项。 这些选项适用于整个图表，而不是图表上的特定项。 一般选项包含缺失 值处理、图表大小和面板换行。 1. 单击图表构建器侧边栏中的选项选项卡。 （如果未显示侧边栏，请单击图表构建器右上角中的按钮以显 示侧边栏。） 2. 修改一般选项。 随后是有关这些选项的详细信息。 用户缺失值 分组变量。 如果用于定义类别或子组的变量有缺失值，那么选择包含，以便在图表中包含用户缺失值（由用 户标识为缺失的值）的类别。 这些“缺失”类别还在计算统计时作为分组变量。 “缺失”类别显示在分类轴上或 图例中，例如，添加额外的条，或向饼图添加分区。 如果没有缺失值，那么不显示“缺失”类别。 如果选择此选项，并要在绘制完图表后取消显示，请在图表编辑器中打开图表，从“编辑”菜单中选择属性。 使用“类别”选项卡将要取消显示的类别移到“排除”列表中。 但是，请注意，如果隐藏“缺失”类别，则不会重 新计算统计。 因此，像百分比统计一类的内容仍将考虑“缺失”类别。 注：此控制不影响系统缺失值。 始终从图表中排除这些值。 汇总统计和个案值。 可以选择以下选项之一，以排除具有缺失值的个案： • 按列表排除以获取图表的一致基数。 如果图表中的任何变量对给定个案有缺失值，则从图表中排除整个个 案。 • 逐个变量排除以最大限度地使用数据。 如果所选变量有缺失值，那么在分析该变量时排除含有这些缺失值 的个案。 图表大小和面板 图表大小。 指定大于 100 的百分比可将图表放大，指定小于 100 的百分比可将图表缩小。 此百分比相对于 缺省的图表大小。 面板。 有很多面板列时，选择使面板换行可允许面板跨行换行，而不是强制放入一个特定的行中。 除非选 中此选项，否则面板将缩小以强制放入一行中。 第 15 章 图表工具的概述 155 156 IBM SPSS Statistics 29 Core System 用户指南 第 16 章 使用预测模型对数据评分 将预测模型应用于一组数据的过程称为计分数据。IBM SPSS Statistics 具有用于构建预测模型（如回归、聚 类、树和神经网络模型）的过程。 构建模型后，可以将模型指定项保存在文件中，该文件包含重建模型所需 的所有信息。 然后，您可以使用该模型文件在其他数据集中生成预测得分。 注：某些过程生成模型 XML 文 件，而某些过程生成压缩文件归档（.zip 文件）。 示例。 公司直销部门使用试验邮寄的结果，为其联系人数据库的其余部分指定倾向得分，他们使用各种人口 统计学特征来标识最有可能做出响应和购买产品的联系人。 可以将评分视为数据转换。 模型在内部表现为一组数值转换，这组转换将应用于一组给定字段/变量（模型 中指定的预测变量）以获得预测结果。 在此意义下，用给定模型对数据评分的过程与对一组数据应用任何函 数（例如平方根函数）在实质上相同。 评分过程包含两个基本步骤： 1. 构建模型并保存模型文件。 使用数据集构建兴趣结果（通常被称为目标）已知的模型。 例如，如果您希 望构建可预测谁可能会响应直接邮寄活动的模型，那么需要从已包含响应人和未响应人信息的数据集开 始。 例如，这可能是对一小组客户发送的试验邮寄的结果或来自过去类似活动的响应信息。 注：对于一些模型类型，没有兴趣的目标结果。 例如，聚类模型没有目标，而一些最近邻模型也没有目 标。 2. 应用该模型到其他数据集（其中兴趣结果未知）以获取预测结果。 评分向导 您可以使用评分向导将通过一个数据集创建的模型应用到另一个数据集，并生成得分，如兴趣结果的预测值 和/或预测概率。 要使用预测模型对数据集进行评分 1. 打开您要评分的数据集。 2. 打开评分向导。 从菜单中选择： 实用程序 > 评分向导。 3. 选择模型 XML 文件或压缩文件归档（.zip 文件）。 使用浏览按钮导航到其他位置以选择模型文件。 4. 匹配存档数据集中的字段到模型中使用的字段。 请参阅第157 页的『将模型字段匹配到数据集字段』主 题以获取更多信息。 5. 选择您要使用的评分函数。 请参阅第159 页的『选择评分函数』主题以获取更多信息。 选择评分模型。 模型文件可以是包含模型 PMML 的 XML 文件或压缩文件归档（.zip 文件）。 该列表只显示 扩展名为 .zip 或 .xml 的文件；列表中不显示文件扩展名。 您可以使用 IBM SPSS Statistics 创建的任何模型 文件。 您还可以使用由其他应用程序（如 IBM SPSS Modeler）创建的模型文件，但其中一些模型文件无法 被 IBM SPSS Statistics 读取，包括任何具有多个目标字段（变量）的模型。 模型详细信息。 这里显示所选模型的基本信息，如模型类型、目标（如果有）和用于构建模型的预测变量。 由于必须读取模型文件才能获取该信息，因此在显示所选模型的此类信息之前可能会有延迟。 如果 XML 文 件或 .zip 文件无法识别为 IBM SPSS Statistics 可以读取的模型，那么将显示一条指示该文件无法读取的消 息。 将模型字段匹配到数据集字段 为了对活动数据集进行评分，该数据集必须包含对应于模型中的所有预测变量的字段（变量）。 如果模型还 包含拆分字段，那么该数据集还必须包含对应于模型中所有拆分字段的字段。 • 缺省情况下，自动匹配活动数据集中任何与模型中的字段具有相同名称和类型的字段。 • 使用下拉列表匹配数据集字段到模型字段。 模型和数据集中每个字段的数据类型必须相同才能匹配字段。 • 在模型中的所有预测变量（以及拆分字段，如果有的话）与活动数据集中的字段匹配之前，您无法继续向 导或对活动数据集进行评分。 数据集字段。 下拉列表包含活动数据集中所有字段的名称。 无法选择不匹配相应模型字段数据类型的字 段。 模型字段。 模型中使用的字段。 角色。 显示的角色可以是以下角色之一： • 预测变量。 该字段在模型中用作预测变量。 即，预测变量的值用于“预测”兴趣目标结果的值。 • 拆分。 拆分字段的值用于定义分组，其中每个分组单独评分。 拆分字段值的每个唯一组合有一个单独分 组。 （注：拆分仅可用于某些模型。） • 记录标识。 记录（个案）标识。 度量。 模型中定义的字段的测量级别。 对于测量级别会影响得分的模型，将使用模型中定义的测量级别， 而不是活动数据集中定义的测量级别。 有关测量级别的更多信息，请参阅第45 页的『变量测量级别』。 类型。 模型中定义的数据类型。 活动数据集中的数据类型必须匹配模型中的数据类型。 数据类型可以是以 下类型之一： • 字符串。 活动数据集中，数据类型为字符串的字段匹配模型中的字符串数据类型。 • 数值。 活动数据集中，显示格式不是日期或时间格式的数值字段匹配模型中的数值数据类型。 其中包括 F （数值）、Dollar、Dot、Comma、E（科学记数法）和自定义货币格式。 具有 Wkday（一周中的某天） 和 Month（一年中的某月）格式的字段也被视为数值，而不是日期。 对于一些模型类型，活动数据集中的 日期和时间字段也被视为与模型中的数值数据类型匹配。 • 日期。 活动数据集中，显示格式包含日期但不包含时间的数值字段匹配模型中的日期类型。 其中包括 Date (dd-mm-yyyy)、Adate (mm/dd/yyyy)、Edate (dd.mm.yyyy)、Sdate (yyyy/mm/dd) 和 Jdate (dddyyyy)。 • 时间。 活动数据集中，显示格式包含时间但不包含日期的数值字段匹配模型中的时间数据类型。 其中包括 Time (hh:mm:ss) 和 Dtime (dd hh:mm:ss) • 时间戳。 活动数据集中，显示格式同时包含日期和时间的数值字段匹配模型中的时间戳数据类型。 这对应 于活动数据集中的 Datetime 格式 (dd-mm-yyyy hh:mm:ss)。 注：除了字段名称和类型外，您还应该确保要评分的数据集中的实际数据值的记录方式与用于构建模型的数 据集中的数据值相同。 例如，如果模型使用 Income 字段构建，后者将收入划分为四种类别，而活动数据集 中的 IncomeCategory 则将收入划分为六种类别或四种不同的类别，因此这些字段实际上彼此并不匹配，结 果得分将不可靠。 缺失值 此组选项控制评分过程中遇到的模型中定义的预测变量缺失值的处理。 评分过程中的缺失值指下列值之一： • 预测变量不包含值。 对于数值字段（变量），预测变量表示系统缺失值。 对于字符串字段，预测变量表示 空字符串。 • 在模型中，已将给定预测变量的值定义为用户缺失值。 在活动数据集中（而未在模型中）定义为用户缺失 的值在得分过程中并不被视为缺失值。 • 预测变量是分类变量，且其值不是模型中所定义的某个类别。 使用值替换。 对具有缺失值的个案评分时尝试使用值替换。 确定替换缺失值的值的方法取决于预测模型的 类型。 • 线性回归和判别模型。 对于线性回归和判别模型中的自变量，如果构建和保存模型时指定了缺失值的平均 值替换，则在评分计算中将使用此平均值代替缺失值，评分过程继续。 如果平均值不可用，那么返回系统 缺失值。 • 决策树模型。 对于 CHAID 和穷举 CHAID 模型，将为缺失的拆分变量选择最大的子节点。 最大的子节点是 使用学习样本个案的子节点中具有最大总体的子节点。 对于 C&RT 和 QUEST 模型，首先使用替代拆分变 量（如果有）。 （替代拆分是使用替代预测变量尽可能匹配原始拆分的拆分。） 如果没有指定替代拆分或 者所有替代拆分变量都缺失，那么使用最大的子节点。 158 IBM SPSS Statistics 29 Core System 用户指南 • Logistic 回归模型。 对于 Logistic 回归模型中的协变量，如果预测变量的平均值作为保存的模型的一部分 包含在其中，那么在评分计算中将使用此平均值代替缺失值，评分过程继续。 如果预测变量是分类变量 （例如，Logistic 回归模型中的因子），或者如果平均值不可用，则返回系统缺失值。 使用系统缺失值。 对具有缺失值的个案进行评分时返回系统缺失值。 选择评分函数 评分函数是所选模型可用的“得分”类型。 例如，目标的预测值、预测值的概率或所选目标值的概率。 评分函数。 可用的评分函数取决于模型。 下列值中的一个或多个将在列表中可用： • 预测值。 兴趣目标结果的预测值。 这对所有模型可用，没有目标的模型除外。 • 预测值的概率。 预测值的概率是以比例表示的正确值。 这对具有分类目标的大部分模型可用。 • 所选值的概率。 所选值的概率是以比例表示的正确值。 从“值”列中的下拉列表选择一个值。 可用值由模 型定义。 这对具有分类目标的大部分模型可用。 • 置信。 与分类目标的预测值关联的概率测量。 对于二元 Logistic 回归、多项 Logistic 回归和朴素贝叶斯模 型，其结果与预测值的概率相同。 对于树和 Ruleset 模型，置信可以被解释为预测类别的调整概率，而且 始终比预测值的概率小。 对于这些模型，置信值比预测值的概率更加可靠。 • 节点编号。 树模型的预测终端节点编号。 • 标准误差。预测值的标准误差。 对带刻度目标的线性回归模型、一般线性模型和广义线性模型可用。 这只 在模型文件中保存了协方差矩阵时可用。 • 累积风险 (Cumulative Hazard). 估计的累积风险函数。 该值指示了在给定预测变量值的前提下，在指定时 间或该时间之前观察到事件的概率。 • 最近邻元素。 最近邻元素的 ID。 如果提供了值，则标识是个案标签变量的值，否则，标识是个案编号。 仅适用于最近相邻元素模型。 • 第 K 个最近邻元素。 第 k 个最近邻元素的 ID。 在“值”列中输入一个整数作为 k 值。 如果提供了值，则标 识是个案标签变量的值，否则，标识是个案编号。 仅适用于最近相邻元素模型。 • 到最近邻元素的距离。 到最近邻元素的距离。 根据模型，将使用 Euclidean 或城市街区距离。 仅适用于 最近相邻元素模型。 • 到第 k 个最近邻元素的距离。 到第 k 个最近邻元素的距离。 在“值”列中输入一个整数作为 k 值。 根据模 型，将使用 Euclidean 或城市街区距离。 仅适用于最近相邻元素模型。 字段名称。 每个选定的评分函数在活动数据集中保存一个新的字段（变量）。 您可以使用缺省名称或输入 新名称。 如果活动数据集中已存在具有相同名称的字段，它们将被替换。 有关字段命名规则的信息，请参 阅第44 页的『变量名称』。 值。 有关使用“值”设置的函数的说明，请参阅评分函数的说明。 对活动数据集进行评分 在向导的最后步骤，您可以对活动数据集进行评分或粘贴所生成的命令语法到语法窗口。 然后，您可以修改 和/或保存所生成的命令语法。 合并模型和转换 XML 文件 一些预测模型使用以不同方式修改或转换的数据构建。 为了以有意义的方式将这些模型应用到其他数据集， 还必须在要评分的数据集上执行相同的转换，或将该转换同时反映在模型文件中。 在模型文件中包括转换的 过程分为两个步骤： 1. 保存转换到转换 XML 文件中。 这只能通过在命令语法中使用 TMS BEGIN 和 TMS END 来完成。 主题。 2. 将模型文件（XML 文件或 .zip 文件）和转换 XML 文件合并到新的合并模型 XML 文件中。 要将模型文件和转换 XML 文件合并到新的合并模型文件中，请执行下列操作： 3. 从菜单中选择： 实用程序 > 合并模型 XML 第 16 章 使用预测模型对数据评分 159 4. 选择模型 XML 文件。 5. 选择转换 XML 文件。 6. 输入新的合并模型 XML 文件的路径和名称，或使用浏览选择位置和名称。 注：您不能合并包含拆分（每个拆分组的单独模型信息）的模型或具有转换 XML 文件的整体模型的模型 .zip 文件。 160 IBM SPSS Statistics 29 Core System 用户指南 第 17 章 实用程序(U) 实用程序 本章描述可在“实用程序”菜单上找到的功能，以及重新排列目标变量列表的能力。 • 有关评分向导的信息，请参阅第157 页的『第 16 章 使用预测模型对数据评分』。 • 有关合并模型和转换 XML 文件的信息，请参阅第159 页的『合并模型和转换 XML 文件』。 变量信息 “变量”对话框为当前选定的变量显示变量定义信息，包括： • 变量标签(L) • 数据格式 • 用户缺失值 • 值标签 • 测量级别 可视。 变量列表中的“可视”列指示变量当前在数据编辑器或对话框变量列表中是否可见。 转到。 转至数据编辑器窗口中选定的变量。 粘贴。 将选定的变量粘贴到指定语法窗口中光标所在的位置。 要修改变量定义，可使用数据编辑器的变量视图。 可通过两种方法来打开“变量”对话框。 • 从菜单选择实用程序 > 变量...。 • 在数据编辑器中右键单击变量名称，并从上下文菜单选择变量信息...。 数据文件注释 您可以随数据文件包含描述性注释。 对于 IBM SPSS Statistics 数据文件，这些注释随数据文件一起保存。 添加、修改、删除或显示数据文件注释 1. 从菜单中选择： 实用程序 > 数据文件注释...
2. 要在“查看器”中显示注释，请选择在输出中显示注释。 注释可以是任何长度，但每行最多 80 个字节（在单字节语言中通常为 80 个字符），满 80 个字符将自动换 行。 注释显示在与文本输出相同的字体中，以准确反映它们在查看器中显示时的显示方式。 只要添加或修改注释，日期戳记（括号中的当前日期）就会自动附加到注释列表的末尾。 如果您修改现有注 释，或在现有注释之间插入新注释，这可能混淆与注释相关联的日期。 变量集 您可以通过定义和使用变量集，限制数据编辑器和对话框变量列表中显示的变量。 对于含有大量变量的数据 文件，这一点尤其有用。 小型变量集更容易找到和选择变量供您分析。 定义变量集 “定义变量集”可创建要显示在数据编辑器和对话框变量列表中的变量子集。 定义的变量集保存在 IBM SPSS Statistics 数据文件中。 集合名称。 集合名称最多可以包含 64 个字节。 可以使用任意字符，包括空白。 集合中的变量。 集合中可包含数字变量和字符串变量的任何组合。 变量在集合中的顺序对变量在数据编辑 器或对话框变量列表中显示的顺序没有任何影响。 一个变量可以属于多个集合。 定义变量集 1. 从菜单中选择： 实用程序 > 定义变量集...
2. 选择想要在集合中包含的变量。 3. 为集合输入一个名称（不超过 64 个字节）。 4. 单击添加集合。 使用变量集合显示和隐藏变量 “使用变量集”将数据编辑器和对话框变量列表中显示的变量限制为所选（选中）集合中的变量。 • 数据编辑器和对话框变量列表中显示的变量集合是所有选择的集合的并集。 • 一个变量可以包含在多个选定集合中。 • 变量在选定集合中的顺序以及选定集合的顺序对变量在数据编辑器或对话框变量列表中显示的顺序没有任 何影响。 • 尽管定义的变量集保存在 IBM SPSS Statistics 数据文件中，但当前选择的集的列表在您每次打开数据文件 时都重置为缺省内置集。 可用变量集列表包括为活动数据集定义的所有变量集，加上两个内置集： • ALLVARIABLES。 此集合包含数据文件中的所有变量，包括在会话期间新建的变量。 • NEWVARIABLES。 此集合仅包含在会话期间新建的变量。 注：即使在创建新变量后保存数据文件，这些新变量仍然包括在 NEWVARIABLES 集合中，直到您关闭并 重新打开数据文件。 必须至少选择一个变量集。 如果选择 ALLVARIABLES，则任何其他选择的集将没有任何视觉效果，因为该集 包含所有变量。 1. 从菜单中选择： 实用程序 > 使用变量集...
2. 选择定义的变量集，其中包含您想要显示在数据编辑器和对话框变量列表中的变量。 显示所有变量 1. 从菜单中选择： 实用程序 > 显示所有变量 重新排序目标变量列表 变量在对话框目标列表中出现的顺序就是从源列表中选择它们的顺序。 如果要更改变量在目标列表中的顺 序，但不希望取消选择所有变量并以新的顺序重新进行选择，那么您可以使用 Ctrl 键（Macintosh：命令 键）以及向上/向下箭头键在目标列表中向上/向下移动变量。 如果多个变量连续（组在一起），您可以同时 移动它们。 不能移动不连续的变量组。 162 IBM SPSS Statistics 29 Core System 用户指南 第 18 章 选项 选项 选项控制各种设置： • 会话日志，记录每次会话中运行的所有命令 • 显示对话框源列表中的变量的顺序 • 在新输出结果中显示和隐藏的项 • 新透视表的 TableLook • 自定义货币格式 选项控制各种设置。 更改选项设置 1. 从菜单中选择： 编辑 > 选项...
2. 单击要更改的设置所在的选项卡。 3. 保存设置。 4. 单击确定或应用。 一般选项 应用程序方式 提供确定如何显示输出的设置。 经典（语法和输出） 缺省设置以经典查看器格式显示输出。 有关更多信息，请参阅第103 页的『查看器 - 经典』。 工作簿 工作簿选项将 SPSS Statistics 后端功能与笔记本方法联系起来，其中，笔记本方法提供交互式方法来运 行语法和查看相应的输出。 工作簿文档 (.spwb) 是由各个段落组成。 这些段落包含各个输出元素（语 法、过程、图表，等等）。 语法段落提供了完整的语法编辑和运行功能。 非语法段落提供了完整的富文 本格式编辑功能。 有关更多信息，请参阅第113 页的『查看器 - 工作簿』。 变量列表 这些设置控制对话框列表中变量的显示。 可以显示变量名称或变量标签。 名称或标签可按字母顺序或文件 顺序显示，或按测量级别进行分组。 显示顺序仅影响源变量列表。 目标变量列表总是反映变量选择的顺 序。 角色 某些对话框支持基于定义的角色预先选择分析变量的功能。 请参阅主题第47 页的『角色』，了解更多信 息。 使用预定义角色 缺省情况下，基于定义的角色预先选择变量。 使用定制分配 缺省情况下，不要使用角色预先选择变量。 在支持此功能的对话框中，您还可以在预定义角色和自定义分配之间切换。 此处的设置仅控制每个数据集的 初始缺省有效行为。 最大线程数 多线程过程在计算结果时使用的线程数。 自动设置基于可用处理核心数。 如果在多线程过程运行时希望更 多处理资源可用于其他应用程序，请指定较小的值。 在分发分析方式中将禁用此选项。 输出 表格中较小的数值没有科学记数法(O) 对于输出中较小的小数值不显示科学记数法。 非常小的小数值将显示为 0（或者 0.000）。 此设置仅影 响具有“常规”格式的输出，该格式由应用程序确定。 此设置不会影响显著性水平或其他具有标准值范围 的统计的显示。 透视表中的许多数字值的格式基于与数字值关联的变量的格式。 当显著性值小于 .001 时，显示“<.001”（观察到的观察值） 选择此设置时，较小的显著性值在表输出中显示为“<.001”。 将语言环境数字分组格式应用到数值(L) 将当前语言环境的数字分组格式应用于透视表和图表中的数值以及“数据编辑器”中的数值。 例如，在启 用了该设置的法语语言环境中，值 34419,57 将显示为 34 419,57。 分组格式不适用于树、模型查看器项目、DOT 或 COMMA 格式的数值以及 DOLLAR 或自定义货币格式的数 值。 但它适用于以 DTIME 格式数值来显示日期值--例如，以 ddd hh:mm 的格式显示 ddd 的值。 显示小数值的前导零。 显示仅包含小数部分的数字值的前导零。 例如，显示前导零时，值 .123 显示为 0.123。 此设置不适用 于具有货币或百分比格式的数字值。 除了固定 ASCII 文件 (.dat)，将数据保存到外部文件时不会包含前 导零。 测量系统 用于指定属性（如透视表单元格边距、单元格宽度和打印时表之间的空白）的测量系统（点、英寸或厘 米）。 通知 控制程序通知您已经完成运行过程或者可以在查看器中获得结果的方式。 Windows 观感 控制窗口和对话框的基本外观。 如果您在更改外观后发现任何显示问题，请尝试关闭并重新启动应用程 序。 在启动时打开语法窗口 语法窗口是用于输入、编辑和运行命令的文本文件窗口。 如果您经常使用命令语法，请选择该选项，在 每次会话开始时自动打开语法窗口。 对于喜欢用命令语法而不喜欢用对话框的熟练用户来说，此选项很 有用。 一次只能打开一个数据集 每次使用菜单和对话框打开其他数据源时，请关闭当前打开的数据源。 缺省情况下，每次使用菜单和对 话框打开新数据源时，该数据源会在新的“数据编辑器”窗口中打开，并且在其他“数据编辑器”窗口中打 开的任何其他数据源都会保持打开状态，并在会话期间可用，直到明确关闭。 选择此选项会立即生效，但不会关闭在更改设置时所打开的任何数据集。 此设置对使用命令语法打开的 数据源没有影响，该语法依赖于 DATASET 命令来控制多个数据集。 请参阅主题第57 页的『第 6 章 使 用多数据源』，了解更多信息。 显示本机 macOS 文件对话框 选择时，将以本机 macOS 格式显示 macOS 版本的 SPSS Statistics 中的文件选择对话框。 以前，文件 选择对话框以进行大量定制以容纳特定 SPSS Statistics 文件功能。 此设置允许您将文件选择对话框设置 为缺省本机 macOS 格式。 注: 此功能仅在 macOS 环境中可用。 164 IBM SPSS Statistics 29 Core System 用户指南 自动恢复 这些设置控制 IBM SPSS Statistics 自动恢复功能。 自动恢复设计为在应用程序意外退出的实例中恢复未保 存的文件和内容。 您可以选择以启动/禁用自动恢复功能（缺省情况下，启用此功能），选择保存文件之间 的时间间隔（分钟），设置要保存的复原点的最大数量，以及查看或更改自动恢复文件位置。 注: 选择以过高频率保存文件（例如，每 2 分钟一次）可能导致降低系统性能。 在意外退出后重新启动 SPSS Statistics，将向您显示 IBM SPSS Statistics 错误报告，允许您输入意外退出前 会话的相关信息。 离开退出报告后，将向您显示“文档恢复”对话框，该对话框提供用于恢复先前会话数据或 删除已保存的会话数据的选项。 有关更多信息，请参阅第2 页的『自动恢复』。 语言选项 语言 输出语言。 控制输出中使用的语言。 不适用于简单文本输出。 可用语言的列表取决于当前安装的语言文 件。 （注：此设置不会影响用户界面语言。） 根据所选的语言，您可能还需要使用 Unicode 字符编码以正 确显示字符。 注：依赖于输出中语言特定的文本字符串的定制脚本在更改输出语言后可能无法正常运行。 有关更多信息， 请参见主题第172 页的『脚本选项』。 用户界面。 此设置控制菜单、对话框和其他用户界面功能部件中使用的语言。 （注：此设置不会影响输出 语言。） 字符编码和语言环境 该选项控制确定读写数据文件和语法文件的编码方式的缺省行为。 您只能在未打开数据源时更改这些设置， 并且这些设置对后续会话继续有效，直到明确更改。 Unicode（通用字符集）。 使用 Unicode 编码 (UTF-8) 来读写文件。 此模式也称为 Unicode 模式。 语言环境的写入系统。 使用当前语言环境选择确定读写文件的编码方式。 此模式也称为代码页模式。 语言环境。 可以从列表中选择语言环境或输入任何有效的语言环境值。 操作系统写入系统会将语言环境设 置为操作系统的写入系统。 在您每次启动应用程序时，应用程序从操作系统获取此信息。 语言环境设置主 要在代码页模式下有意义，但它也会影响一些字符在 Unicode 模式下的呈现方式。 Unicode 模式和 Unicode 文件的一些重要隐含信息： • IBM SPSS Statistics data files and syntax files that are saved in Unicode encoding should not be used in releases of IBM SPSS Statistics before 16.0. 对于语法文件，您可在保存文件时指定编码方式。 对于数据 文件，如果要使用较早的版本读取文件，则应该以代码页模式打开数据文件，然后重新保存。 • 以 Unicode 方式读取代码页数据文件时，所有字符串变量的限定宽度扩大为原来的三倍。 要将每个字符串 变量的宽度自动设置为该变量的最长观测值，请选择“打开数据”对话框中的根据观测值来最小化字符串宽 度。 双向文本 如果混合使用从右到左的语言（如阿拉伯语或希伯来语）与从左到右的语言（如英语），请选择文本排列的 方向。 单个单词将仍按其语言所决定的正确方向排列。 此选项仅控制整段文本（例如在一个编辑字段中输 入的所有文本）的文本排列。 • 自动。 文本排列由每个单词中使用的字符决定。 这是缺省选项。 • 从右到左。 文本从右到左排列。 • 从左到右。 文本从左到右排列。 第 18 章 选项 165 查看器选项 查看器输出显示选项仅影响更改设置后生成的新输出。 已经显示在查看器中的输出不受这些设置更改的影 响。 初始输出状态 控制每次运行过程时自动显示或隐藏的项，以及各项的初始对齐方式。 您可以控制下列项目的显示：日 志、警告、注释、标题、透视表、图表、树图和文本输出。 您还可以在登录或注销时显示命令。 可以从 日志复制命令语法，并将其保存在语法文件中。 注: 在查看器中所有输出项都按照左对齐方式显示。 只有打印输出的对齐方式受对齐方式设置的影响。 居中和右对齐项用一个小符号表示。 标题 控制新输出标题的字体样式、大小和颜色。 字体大小列表提供了一组预定义的大小，但您可以手动输入 其他支持的大小值。 页面标题 控制新页面标题以及由 TITLE 和 SUBTITLE 命令语法生成的或由插入菜单上的新页面标题创建的页面 标题的字体样式、大小和颜色。 字体大小列表提供了一组预定义的大小，但您可以手动输入其他支持的 大小值。 文本输出(O) 用于文本输出的字体。 文本输出设计成与等宽（固定间距）字体一起使用。 如果选择比例字体，表格输 出将不会正确对齐。 字体大小列表提供了一组预定义的大小，但您可以手动输入其他支持的大小值。 缺省页面设置 控制用于打印的方向和页边距缺省选项。 数据选项 转换与合并选项。 程序每次执行命令时，都要读取数据文件。 某些数据转换（例如，“计算”和“重新编码”） 和文件转换（例如，“添加变量”和“添加个案”）不要求单独遍历数据，这些命令的执行可以延迟到程序读取 数据以执行其他命令（例如，统计或制图过程）时进行。 • 对于读取数据要花费一定时间的大型数据文件，您可以选择使用前计算值，以延迟执行并节省处理时间。 选择此选项后，使用“计算变量”等对话框进行的转换的结果不会立即显示在数据编辑器中；这些转换所创 建的新变量显示时不含任何数据值；存在暂挂的转换时，数据编辑器中的数据值不可更改。 任何读取数据 的命令（例如，统计或制图过程）都将执行暂挂的转换，并更新数据编辑器中显示的数据。 或者，您可使 用“转换”菜单上的运行暂挂的转换。 • 如果使用立即计算值这一缺省设置，那么从对话框粘贴命令语法时，在每个转换命令后都会粘贴 EXECUTE 命令。 请参阅主题第151 页的『多条执行命令』，了解更多信息。 显示新数值变量的格式。 控制新数字变量的缺省显示宽度和小数位数。 新字符串变量没有缺省的显示格 式。 如果某个值对于指定的显示格式而言过大，那么前几个小数位会进行四舍五入，值转换为科学记数法。 显示格式不影响内部数据值。 例如，值 123456.78 在显示时可能四舍五入为 123457，但是在任何计算中 都会使用未经四舍五入的原始值。 设置两位数年份的世纪范围。 定义以两位数年份（例如，10/28/86 和 29-OCT-87）输入和/或显示的日期 格式变量的年份范围。 自动范围设置基于当前年份，从 69 年前开始，直至 30 年后结束（加上当年，整个 范围为 100 年）。 对于定制范围，结束年份根据您输入的开始年份值自动确定。 随机数字生成器。 有两种不同的随机数字生成器可供使用： • 兼容的版本 12. 在 V12 和先前发行版中使用的随机数生成器。 如果您要重新生成在早期版本中根据指定的 种子值生成的随机结果，可使用此随机数字生成器。 • Mersenne 旋转 (Mersenne Twister). 较新的随机数生成器，用于模拟目的更可靠。 如果从版本 12 或早期版 本再现随机结果不是问题，那么可使用此随机数生成器。 指定测量级别。 对于从外部文件格式读取的数据，较旧版本的 IBM SPSS Statistics 数据文件（8.0 以前的版 本）以及在会话中创建的新字段，数值字段的测量级别取决于一组规则，包括唯一值的数量。 您可以为数字 变量指定数据值的最小数量，用于将该变量分类为连续（标度）变量或名义变量。 唯一值的数量少于指定数 量的变量将归类为名义变量。 166 IBM SPSS Statistics 29 Core System 用户指南 当决定应用连续（刻度）或名义测量级别时，会在应用最小数据值规则数之前评估许多其他条件。 将以条件 在下表格中的排列顺序对其进行评估。 将应用与数据匹配的第一个条件的测量级别。 表 23: 用于确定缺省测量级别的规则 条件 测量级别 格式为美元或定制货币 连续 格式为日期或时间（不包括月份和星期） 连续 变量的所有值均缺失 名义 变量包含至少一个非整数值 连续 变量包含至少一个负值 连续 变量不包含少于 10,000 的有效值 连续 变量具有 N 个或更多唯一有效值 连续 变量不包含少于 10 的有效值 连续 变量具有少于 N 个唯一有效值 名义 N 是用户指定的分界值。 缺省值为 24。 数值的四舍五入和取整。 对于 RND 和 TRUNC 函数，此设置控制用来将接近四舍五入边界的值四舍五入的缺 省阈值。 此设置指定为位数，在安装时设置为 6，这对于大部分应用程序应已足够。 位数设为 0 会产生与版 本 10 相同的结果。 将位数设为 10 会产生与版本 11 和 12 相同的结果。 • 对于 RND 函数，此设置指定最不显著的位数，要四舍五入的值可能未达到四舍五入的阈值，但是仍然被四 舍五入。 例如，将介于 1.0 与 2.0 之间的值四舍五入为最接近的整数时，此设置指定该值距离 1.5（四舍 五入为 2.0 的阈值）还差多少时仍四舍五入为 2.0。 • 对于 TRUNC 函数，此设置指定最不显著的位数，要取整的值可能未达到最接近的四舍五入边界，在取整之 前要被四舍五入。 例如，将介于 1.0 与 2.0 之间的值截断为最接近的整数时，此设置指定该值距离 2.0 还 差多少时四舍五入为 2.0。 自定义变量视图。 控制变量视图中各个属性的缺省显示和顺序。 请参阅主题第167 页的『更改缺省变量视 图』，了解更多信息。 更改字典。 控制用于检查变量视图中项目拼写的字典语言版本。 请参阅主题第50 页的『拼写检查』，了解 更多信息。 更改缺省变量视图 缺省情况下，您可以使用“定制变量视图”来控制“变量视图”中显示的属性（例如，名称、类型、标签）及其 显示顺序。 单击定制变量视图。 1. 选择（选中）要显示的变量属性。 2. 使用向上和向下箭头按钮更改属性的显示顺序。 货币选项 最多可以创建五种自定义货币显示格式，格式可以包括特殊的前缀和后缀字符以及对负值的特殊处理方式。 五种自定义货币格式名称为 CCA、CCB、CCC、CCD 和 CCE。 您不能更改格式名或添加新的格式名。 要修 改自定义货币格式，请从源列表中选择格式名，然后进行所需的更改。 为自定义货币格式定义的前缀、后缀和小数指示符仅用于显示目的。 您不能用自定义货币字符在数据编辑器 中输入值。 第 18 章 选项 167 创建定制货币格式 1. 单击货币选项卡。 2. 从列表（CCA、CCB、CCC、CCD 和 CCE）中选择一种货币格式。 3. 输入前缀、后缀和小数指示符的值。 4. 单击确定或应用。 输出选项 输出选项控制某些输出选项的缺省设置。 大纲标注。 控制查看器概要窗格中变量名称、变量标签、数据值和值标签的显示。 透视表标注。 控制透视表中变量名称、变量标签、数据值和值标签的显示。 描述性变量标签和值标签（数据编辑器、标签和值列中的变量视图）通常可使结果更容易看懂。 但是，长标 签在某些表中可能很难处理。 在此更改标签选项只影响新输出的缺省显示。 单击描述。 控制数据编辑器中为选定变量生成的描述统计选项。 请参阅主题第55 页的『为选定变量获取描 述统计』，了解更多信息。 • 排除具有多个类别的表。 对于唯一值的个数多于指定数量的变量，不显示频率表。 • 在输出中包括图表。 对于名义变量和有序变量及具有未知测量级别的变量，显示条形图。 对于连续字段 （标度），显示直方图。 输出显示。 对于可以生成模型查看器输出或者透视表和图表输出的过程，控制生成的输出类型。 对于 V29，此设置仅应用于广义线性混合模型过程。 屏幕朗读器辅助功能。 控制屏幕朗读器如何朗读透视表行标签和列标签。 可以朗读每个数据单元格的行标 签和列标签，也可以只朗读在表中数据单元格之间移动时更改的标签。 图表选项 图表模板 新的图表可以使用此处选择的设置或图表模板文件中的设置。 下拉列表中提供了可用的图表模板。 样本 部分进行更新以反映所选图表模板的样式。 在从下拉列表中选择用户指定时，单击浏览以选择一个图表模板文件。 要创建图表模板文件，请用所需 的属性创建一个图表并将其保存为模板（从“文件”菜单选择保存图表模板）。 当前设置 可用设置包括： 字体 用于新图表中所有文本的字体。 样式循环首选项 新图表的颜色和图案的初始分配。 仅在颜色之间循环仅使用颜色区分图表元素，而不使用图案。 仅 在图案之间循环仅使用线条样式、标记符号或填充图案来区分图表元素，而不使用颜色。 机架 控制新图表上的内框和外框的显示。 网格线 控制新图表上的刻度轴网格线和类别轴网格线的显示。 样式循环 自定义新图表的颜色、线条样式、标记符号和填充图案。 可以更改在创建新图表时使用的颜色和图案的 顺序。 图表宽高比 新图表的外框的宽和高的比率。 可以将宽高比指定为 0.1 到 10.0。 小于 1 的值使图表的高度比宽度 大。 大于 1 的值使图表的宽度比高度大。 等于 1 的值生成的是正方形图表。 图表创建后，其宽高比就 无法更改。 168 IBM SPSS Statistics 29 Core System 用户指南 样本设置 可用设置提供选项以供将样本图表显示为条形图或散点图。 数据元素颜色 指定新图表中的数据元素（如条和标记）应该使用的颜色的顺序。 在“图表选项”主对话框的“样式循环首选 项”组中选择了包括颜色的选项后，就会使用颜色。 例如，如果您创建了有两个组的簇状条形图，并且选择了“图表选项”主对话框中的在颜色之间循环，然后在 图案之间循环，那么“分组图表”列表中的最前面两种颜色将用作新图表上的条的颜色。 更改颜色使用的顺序 1. 选择简单图表，然后选择一种颜色，用于没有类别的图表。 2. 选择分组图表，更改有类别的图表的颜色循环。 要更改类别的颜色，请选择一个类别，然后从调色板中 为该类别选择一种颜色。 （可选）您可以执行以下操作： • 在选定的类别上面插入一个新的类别。 • 移动选定的类别。 • 移去选定的类别。 • 将顺序重置为缺省顺序。 • 选择一种颜色井并单击编辑来编辑该颜色。 数据元素线 指定新图表中的线数据元素应使用的样式的顺序。 如果图表包括线数据元素，并且在“图表选项”主对话框的 “样式循环首选项”组中选择了包括图案的选项，就会使用线样式。 例如，如果您创建了有两个组的线图，并且选择了“图表选项”主对话框中的仅在图案之间循环，那么“分组 图表”列表中的最前面两种样式将用作新图表上的线的图案。 更改线样式使用的顺序 1. 选择简单图表，然后选择一种线样式，用于没有类别的线图。 2. 选择分组图表，更改有类别的线图的图案循环。 要更改类别的线样式，请选择一个类别，然后从调色板 中为该类别选择一种线样式。 （可选）您可以执行以下操作： • 在选定的类别上面插入一个新的类别。 • 移动选定的类别。 • 移去选定的类别。 • 将顺序重置为缺省顺序。 数据元素标记 指定新图表中的标记数据元素应使用的符号的顺序。 如果图表包括标记数据元素，并且在“图表选项”主对话 框的“样式循环首选项”组中选择了包括图案的选项，就会使用标记样式。 例如，如果您创建了有两个组的散点图，并且选择了“图表选项”主对话框中的仅在图案之间循环，那么“分 组图表”列表中的最前面两种符号将用作新图表上的标记。 更改标记样式使用的顺序 1. 选择简单图表，然后选择一种标记符号，用于没有类别的图表。 2. 选择分组图表，更改有类别的图表的图案循环。 要更改类别的标记符号，请选择一个类别，然后从调色 板中为该类别选择一种符号。 （可选）您可以执行以下操作： 第 18 章 选项 169 • 在选定的类别上面插入一个新的类别。 • 移动选定的类别。 • 移去选定的类别。 • 将顺序重置为缺省顺序。 数据元素填充 指定新图表中的条和区域数据元素应使用的填充样式的顺序。 如果图表包括条或区域数据元素，并且在“图 表选项”主对话框的“样式循环首选项”组中选择了包括图案的选项，就会使用填充样式。 例如，如果您创建了有两个组的簇状条形图，并且选择了“图表选项”主对话框中的仅在图案之间循环，那么 “分组图表”列表中的最前面两种样式将用作新图表上的条填充图案。 更改填充样式使用的顺序 1. 选择简单图表，然后选择一种填充图案，用于没有类别的图表。 2. 选择分组图表，更改有类别的图表的图案循环。 要更改类别的填充图案，请选择一个类别，然后从调色 板中为该类别选择一种填充图案。 （可选）您可以执行以下操作： • 在选定的类别上面插入一个新的类别。 • 移动选定的类别。 • 移去选定的类别。 • 将顺序重置为缺省顺序。 透视表选项 “透视表”选项可设置与透视表显示有关的各种选项。 表格外观(T) 从文件列表中选择一个表格外观，并单击确定或应用。 可以使用随 IBM SPSS Statistics 一起提供的表格外 观，或者在透视表编辑器（从“格式”菜单选择表格外观）中创建自己的表格外观。 浏览 使您可以从另一个目录中选择表格外观。 设置表格外观目录(S) 使您可以更改缺省的表格外观目录。 使用浏览导航到您要使用的目录，选择该目录中的“表格外观”，然 后选择设置表格外观目录。 注: 以 IBM SPSS Statistics 的早期版本创建的表格外观不能用于 16.0 或更高版本。 列宽度(W) 这些选项控制透视表中列宽的自动调整。 仅标签(A) 将列宽调整为列标签的宽度。 这会生成结构更紧凑的表，但宽度超过标签的数据值可能会被截去。 调整所有表格的标签和数据(U) 将列宽调整为列标签的宽度或最大数据值的宽度中的较大者。 这会生成较宽的表，但可以保证显示所有 的值。 表注释 您可以为每个表格自动添加注释。 • 如果您在查看器中将光标悬停在表格上，注释文本就会显示在工具提示中。 • 屏幕朗读器会在焦点位于表格上时朗读注释文本。 170 IBM SPSS Statistics 29 Core System 用户指南 • 查看器中的工具提示仅显示注释的前 200 个字符，但屏幕朗读器会朗读全部文本。 • 如果将输出导出到 HTML，会使用注释文本作为替代文本。 title 在注释中包含表格标题。 过程 包含创建表格的过程的名称。 日期 包含创建表格的日期。 数据集 包含用于创建表格的数据集的名称。 样本 “样本”部分提供所选表格外观的可视化表示。 以丰富文本格式复制宽表格至剪贴板 以 Word 格式粘贴透视表时，对文档宽度太宽的表将被包装，缩小以适合文档宽度，或者保持不变。 在表中显示脚注和文字说明 使用此设置可在表中显示或隐藏脚注和文字说明。 文件位置选项 “文件位置”选项卡上的选项控制应用程序在每个会话开始时打开和保存文件的缺省位置、日志文件位置、临 时文件夹位置，以及出现在最近使用的文件列表中的文件数量和 IBM SPSS Statistics- Python 的集成插件 使 用的 Python 3.10 的安装。 打开和保存对话框的启动文件夹 指定文件夹 指定的文件夹用作每个会话开头的缺省位置。 您可为数据和其他（非数据）文件指定不同的缺省位置。 最后使用的文件夹 在上一次会话中打开或保存文件的最后一个文件夹，用作下一次会话的缺省文件夹。 这适用于数据和其 他（非数据）文件。 仅对用于打开和保存文件的对话框应用这些设置，并且“最后使用的文件夹”由用于打开或保存文件的最后一 个对话框确定。 通过命令语法打开或保存的文件不会影响这些设置，并且不受这些设置的影响。 这些设置 仅在本地分析模式下可用。 在已连接到远程服务器的分布式分析模式中（需要 IBM SPSS Statistics Server），您不能控制启动文件夹位置。 会话日志 您可以使用会话日志自动记录会话中运行的命令。 这包括在语法窗口中输入和运行的命令以及由对话框选项 生成的命令。 可以编辑日志文件，并在其他会话中再次使用命令。 可以打开或关闭日志记录，追加或覆盖 日志文件，以及选择日志文件的名称和位置。 可以从日志文件复制命令语法，并将其保存在语法文件中。 临时文件夹 这将指定在会话过程中创建的临时文件的位置。 在分布式模式（随服务器版本一起提供）中，这不会影响临 时数据文件的位置。 在分布式模式中，临时数据文件的位置由环境变量 SPSSTMPDIR 控制，该变量只能在 运行软件的服务器版本的计算机上设置。 如果需要更改临时目录的位置，请联系您的系统管理员。 第 18 章 选项 171 列出最近使用的文件数量 该选项控制出现在“文件”菜单上的最近使用文件的数量。 Python 3 位置 这些设置指定在您从 SPSS Statistics 运行 Python 时，IBM SPSS Statistics- Python 的集成插件 使用的 Python 3.10.4 的安装。 缺省情况下，将使用随 SPSS Statistics 一起安装的 Python 分发 3.10.4 。 分发位于 安装了 SPSS Statistics 的目录下的 Python3 目录中。 要使用计算机上不同的 Python 安装，请指定该 Python 安装的根目录的路径。 该设置在当前会话中生效，除非已通过运行 Python 程序、Python 脚本或在 Python 中实现的扩展命令启动了 Python。 在这种情况下，设置将在下一个会话中生效。 此设置仅在本地分 析模式下可用。 在分布式分析模式下（需要 IBM SPSS Statistics Server）远程服务器上的 Python 位置是从 IBM SPSS Statistics Administration Console 设置的。 请联系系统管理员以获取帮助。 R 位置 这些设置指定从 SPSS Statistics 运行 R 代码时使用的 R4.2 的安装。 缺省情况下，将使用随 SPSS Statistics 一起安装的 R 4.2 分发。 要在计算机上使用不同的 R 安装，请在提供的字段中指定 R 主目录。 此设置在当 前会话中生效，除非已通过运行 R 扩展命令启动了 R。 在这种情况下，设置将在下一个会话中生效。 要点: • 在 MacOS 上，如果存在外部 R 安装，那么不能使用内部 R 安装。 • 在 macOS 上，必须先删除现有 /Library/Frameworks/R.framework/Versions/4.0 文件夹，然后 再安装其他 R 版本。 • 如果选择使用与缺省安装版本不同的 R 版本，那么必须安装并配置 64 位 R 版本。 SPSS Statistics 不支持 32 位版本的 R。 脚本选项 使用“脚本”选项卡可以指定缺省脚本语言，以及要使用的任何自动脚本。 可以使用脚本自动处理许多功能， 包括自定义透视表。 注: Legacy Sax Basic 用户必须手动转换任何自定义自动脚本。 随 16.0 以前的版本安装的自动脚本作为一组 单独的脚本文件提供，位于 IBM SPSS Statistics 安装目录的 Samples 子目录中。 缺省情况下，输出项与自 动脚本无关联。 您必须手动将所有自动脚本与输出项相关联，如下所述。 有关转换遗存自动脚本的信息， 请参阅第230 页的『与 16.0 之前版本的兼容性』。 缺省脚本语言 缺省脚本语言确定创建新脚本时启动的脚本编辑器。 它还指定了其可执行程序将用于运行自动脚本的缺 省语言。 可用脚本语言取决于使用的平台。 对于 Windows 平台，可用脚本语言为 Basic（随 Core System 一起安装）和 Python 编程语言。 对于所有其他平台，脚本以 Python 编程语言提供。 您可以选 择 Python 3 作为 Python 编程的缺省脚本语言。 要将脚本编制与 Python 编程语言配合使用，您需要 Python 功能，这是 IBM SPSS Statistics 产品的一部 分。 启用自动脚本 该复选框允许您启用或禁用自动脚本。 缺省情况下，已启用自动脚本。 基础自动脚本 在应用任何其他自动脚本之前，应用于所有新查看器对象的可选脚本。 指定用作基础自动脚本的脚本文 件，以及其可执行程序将用于运行脚本的语言。 将自动脚本应用于输出项 1. 在“命令标识”网格中，选择一条命令以生成将对其应用自动脚本的输出项。 在“对象和脚本”网格的对象列中，显示与选定命令关联的对象列表。 脚本列显示选定命令的任何现有脚 本。 172 IBM SPSS Statistics 29 Core System 用户指南 2. 为对象列中显示的任何项指定脚本。 单击相应的脚本单元格。 输入脚本路径，或单击省略号 (...) 按钮以 浏览脚本。 3. 指定其可执行程序将用于运行脚本的语言。 注: 选定语言不受更改缺省脚本语言的影响。 4. 单击应用或确定。 除去自动脚本关联 1. 在“对象和脚本”网格中，单击“脚本”列中与要取消关联的脚本对应的单元格。 2. 删除脚本路径，然后单击“对象和脚本”网格中的任何其他单元格。 3. 单击应用或确定。 多重插补选项 “多重插补”选项卡控制与多重插补相关的两类首选项： 插补数据标记。 缺省情况下，包含插补数据的单元格与包含非插补数据的单元格具有不同的背景颜色。 插 补数据的直观显示有助于您滚动数据集并找到这些单元格。 还可以更改缺省单元格背景颜色、字体，以及使 插补数据粗体显示等。 分析输出。 这组首选项控制在分析多重插补数据集时产生的查看器输出类型。 缺省情况下，将为每个原始 （插补前）数据集和插补数据集产生输出。 此外，还为那些支持插补数据汇聚的过程生成最终的汇聚结果。 当执行单变量汇聚时，还会显示汇聚诊断结果。 不过，您可以隐藏那些不愿看到的输出。 设置多重插补选项 从菜单中选择： 编辑 > 选项 单击“多重插补”选项卡。 语法编辑器选项 语法颜色编码 您可以关闭或打开命令、子命令、关键字、关键字值和注释的颜色编码，您可以指定每个的字体样式和颜 色。 错误颜色编码 您可以关闭或打开某个语法错误的颜色编码，您可以指定使用的字体样式和颜色。 包含错误的命令名称和文 本（在命令内）都经过颜色编码，您可以选择每个的不同样式。 请参阅主题第146 页的『颜色编码』，了 解更多信息。 自动补全设置 您可以关闭或打开自动完成控制的自动显示。 始终可以通过按 Ctrl+空格键显示自动完成控制。 请参阅主题 第146 页的『自动完成』，了解更多信息。 缩进大小 指定缩进中的空格数量。 该设置适用于缩进选定的语法行以及自动缩进。 装订线 这些选项指定在语法编辑器的装订线（为行号、书签、分界点和命令跨度保留的文本窗格左侧区域）内显示 或隐藏行号和命令跨度的缺省行为。 命令跨度是提供命令开始和结束的可视指示符的图标。 窗格 显示导航窗格。 这指定显示或隐藏导航窗格的缺省值。 导航窗格在语法窗口中包含所有已识别命令的列 表，以它们发生的顺序显示。 单击导航窗格中的命令会将光标置于命令开始。 第 18 章 选项 173 当发现错误时，自动打开“错误跟踪”窗格。 这指定当发现运行时间错误时显示或隐藏错误跟踪窗格的缺省 值。 针对从右至左语言优化。 使用从右至左语言时，选中此复选框可获得最佳用户体验。 从对话框粘贴语法。 指定在从对话框粘贴语法时，语法在指定语法窗口中的插入位置。 在最后一个命令之 后会在最后一个命令之后插入粘贴的语法。 在光标位置或所选内容处会在光标所在位置插入粘贴的语法；或 者，如果已选择语法块，那么将用粘贴的语法替换所选语法块。 隐私选项 “隐私”选项卡提供以下方面的选项： 允许 SPSS Statistics 与 IBM 共享您的信息 启用时，SPSS Statistics 与 IBM 共享您的使用情况和性能数据。 此设置可帮助改进 SPSS Statistics。 将以安全方式传输所有收集的数据。 禁用 SPSS Statistics 向 IBM 发送错误报告 启用时，SPSS Statistics 不会自动向 IBM 发送错误报告。 设置“隐私”选项 1. 从菜单中，选择：编辑 > 选项...
2. 单击“隐私”选项卡。 174 IBM SPSS Statistics 29 Core System 用户指南 第 19 章 定制菜单和工具栏 定制菜单和工具栏 菜单编辑器 可以使用菜单编辑器定制您的菜单。 通过菜单编辑器，您可以： • 添加运行定制脚本的菜单项。 • 添加运行命令语法文件的菜单项。 • 添加启动其他应用程序并自动将数据发送到其他应用程序的菜单项。 您可以采用下列格式向其他应用程序发送数据：IBM SPSS Statistics、Excel、Lotus 1-2-3、制表符分隔格 式以及 dBASE IV。 向菜单添加项 1. 从菜单中选择： 查看 > 菜单编辑器...
2. 在菜单编辑器对话框中，双击想要对其添加新项的菜单（或单击加号图标）。 3. 选择要在其上显示新项的菜单项。 4. 单击插入项以插入新的菜单项。 5. 为新项输入文本。 在 Windows 操作系统上，字母前面的 & 符号指定该字母应用作带下划线的助记键。 6. 为新项选择文件类型（脚本文件、命令语法文件或外部应用程序）。 7. 单击浏览以选择要附加到菜单项的文件。 您还可以在菜单项之间添加全新的菜单和分隔符。 另外，当您在菜单上选择另一个应用程序时，还可以自动 将数据编辑器的内容发送到该应用程序。 定制工具栏 您可以定制工具栏并创建新的工具栏。 工具栏可以包含任何可用的工具，包括所有菜单操作的工具。 它们 还可以包含启动其他应用程序、运行命令语法文件或运行脚本文件的定制工具。 显示工具栏 使用“显示工具栏”可显示或隐藏工具栏、定制工具栏和创建新工具栏。 工具栏可以包含任何可用的工具，包 括所有菜单操作的工具。 它们还可以包含启动其他应用程序、运行命令语法文件或运行脚本文件的定制工 具。 定制工具栏 1. 从菜单中选择： 视图 > 工具栏 > 定制 2. 选择想要定制的工具栏并单击编辑，或单击新建创建新的工具栏。 3. 对于新的工具栏，请为工具栏输入名称，选择您想让工具栏出现的窗口，并单击编辑。 4. 在“类别”列表中选择一项以显示该类别中的可用工具。 5. 将需要的工具拖放到对话框中显示的工具栏上。 6. 要从工具栏移去工具，那么将其从对话框中显示的工具栏中拖走。 创建打开文件、运行命令语法文件或运行脚本的定制工具： 7. 在“编辑工具栏”对话框中单击新工具。 8. 为工具输入描述性标签。 9. 选择需要该工具执行的操作（打开文件、运行命令语法文件或运行脚本）。 10. 单击浏览选择要与此工具相关联的文件或应用程序。 新工具显示在用户定义的类别中，该类别还包含用户定义的菜单项。 工具栏属性 使用“工具栏属性”可选择想让选定的工具栏出现在其中的窗口类型。 此对话框还用于为新工具栏创建名称。 设置工具栏属性 1. 从菜单中选择： 视图 > 工具栏 > 定制 2. 对于现有工具栏，单击编辑，然后在“编辑工具栏”对话框中单击属性。 3. 对于新的工具栏，单击新工具。 4. 选择想要让工具栏出现在其中的窗口类型。 对于新的工具栏，还请输入工具栏名称。 编辑工具栏 使用“编辑工具栏”对话框可定制现有工具栏和创建新工具栏。 工具栏可以包含任何可用的工具，包括所有菜 单操作的工具。 它们还可以包含启动其他应用程序、运行命令语法文件或运行脚本文件的定制工具。 更改工具栏图像 1. 选择想要更改其在工具栏上显示图像的工具。 2. 单击更改图像。 3. 选择要用于工具的图像文件。 支持下列图像格式：BMP、PNG、GIF 和 JPG。 • 图像必须为正方形。 否则，图像会被切割为正方形形状。 • 图像会被自动缩放以适应要求。 为获得最佳显示效果，请使用 16x16 像素图像作为小工具栏图像，或使 用 32x32 像素图像作为大工具栏图像。 创建新工具 使用“创建新工具”对话框可创建启动其他应用程序、运行命令语法文件和运行脚本文件的定制工具。 176 IBM SPSS Statistics 29 Core System 用户指南 第 20 章 扩展 扩展是扩展 IBM SPSS Statistics 的功能的定制组件。 扩展打包在扩展束（.spe 或 .spxt 文件）中，并 安装到 IBM SPSS Statistics。 任何用户都可创建扩展，并可通过共享关联扩展束与其他用户共享。 提供了以下实用程序来处理扩展： • 第177 页的『扩展中心』（可从扩展 > 扩展中心访问）是用于从 GitHub 上的 IBM SPSS 预测性分析集合 搜索、下载和安装扩展的接口。 从“扩展中心”对话框，还可查看计算机上安装的扩展的详细信息，获取安 装的扩展的更新以及移除扩展。 • 您可以从扩展 > 安装本地扩展束安装本地计算机上存储的扩展束。 • 您可以使用针对扩展的定制对话框构建器创建包含用户界面的扩展，此用户界面称为定制对话框。 定制对 话框会生成SPSS Statistics 命令语法，用于执行与此扩展关联的任务。 在设计定制对话框时设计生成的语 法 • 如果无法使用“针对扩展的定制对话框构建器”，可以使用“创建扩展束”和“编辑扩展束”对话框。 有关更多 信息，请参见主题第208 页的『创建和编辑扩展束』。 扩展中心 从“扩展中心”对话框，可执行以下任务： • 浏览 GitHub 上的 IBM SPSS 预测性分析 集合提供的扩展。 您可以选择扩展以立即安装，也可以下载选定 的扩展并在以后安装。 • 获取计算机上已经安装的扩展的更新版本。 • 查看有关计算机上已经安装的扩展的详细信息。 • 除去计算机上安装的扩展。 要下载或除去扩展： 1. 从菜单选择：扩展 > 扩展中心 2. 选择要下载或移除的扩展并单击确定。 在单击确定时，将处理在“探索”和“已安装”选项卡上生成的所有 选择。 缺省情况下，将在计算机上下载和安装针对下载选中的扩展。 从“设置”选项卡，可选择将选定扩展下载到指 定位置而不进行安装。 您可在以后选择扩展 > 安装本地扩展束安装它们。 对于 Windows 和 Mac，可以通过 双击扩展束文件来安装扩展。 要点: • 如果在应用程序中连接到 Extension Hub 时遇到问题，那么可以手动访问 GitHub 上的 Extension Hub ( • 扩展总是会安装或下载到您的本地计算机。 如果以分布式分析方式工作，请参阅主题 第179 页的『安装 本地扩展束』 以获取更多详细信息。 • 对于 Windows 7 和更新版本，安装更新版本的现有扩展束可能需要使用管理员特权运行 IBM SPSS Statistics。 您可以通过右键单击 IBM SPSS Statistics 的图标并选择以管理员身份运行来使用管理员权限 启动 IBM SPSS Statistics。 尤其是，如果收到指示无法安装一个或多个扩展束的错误消息时，那么尝试使 用管理员特权运行。 注: 以后可随时通过单击“已安装”选项卡上扩展的更多信息... 查看安装扩展时同意的许可。 “浏览”选项卡 “浏览”选项卡显示在 GitHub( 上的 IBM SPSS 预测性分析 集合中提 供的所有扩展。在“探索”选项卡中，可以选择要下载和安装的新扩展，并且可以为计算机上已安装的扩展选 择更新。 “浏览”选项卡需要因特网连接。 • 对于每个扩展，会显示最新版本号和该版本的关联日期。 还提供扩展的简述。 针对计算机上已安装的扩 展，还会显示安装的版本号。 • 您可以通过单击更多信息查看有关扩展的详细信息。 更新可用时，更多信息会显示有关更新的信息。 • 您可以通过单击先决条件，查看运行扩展的先决条件，例如，是否需要 IBM SPSS Statistics-Integration Plug-in for R 。 更新可用时，先决条件会显示有关更新的信息。 优化依据 您可以优化显示的扩展集。 您可以按以下项进行优化：扩展的常规类别、实现扩展所采用的语言、提供扩展 的组织的类型以及扩展的状态。 针对每个组（例如，类别），可以选择按其优化显示的扩展列表的多个项。 您还可以按搜索项进行优化。 搜索不区分大小写，星号 () 将作为任何其他字符，并且不指示通配符搜索。 • 要优化显示的扩展列表，请单击应用。 在光标置于搜索框中时按 Enter 键与单击应用的效果相同。 • 要重置列表以显示所有可用扩展，请删除搜索框中的任何文本，取消选择所有项，并单击应用。 “已安装”选项卡 “已安装”选项卡会显示计算机上已安装的所有扩展。 从“已安装”选项卡中可以选择 GitHub 上的 IBM SPSS 预测性分析集合提供的已安装扩展的更新和移除扩展。 要获取安装的扩展的更新，必须具有因特网连接。 • 对于每个扩展，将显示已安装的版本号。 提供了因特网连接时，将显示最新版本数以及此版本的关联日 期。 还提供扩展的简述。 • 您可以通过单击更多信息查看有关扩展的详细信息。 更新可用时，更多信息会显示有关更新的信息。 • 您可以通过单击先决条件，查看运行扩展的先决条件，例如，是否需要 IBM SPSS Statistics-Integration Plug-in for R 。 更新可用时，先决条件会显示有关更新的信息。 优化依据 您可以优化显示的扩展集。 您可以按以下项进行优化：扩展的常规类别、实现扩展所采用的语言、提供扩展 的组织的类型以及扩展的状态。 针对每个组（例如，类别），可以选择按其优化显示的扩展列表的多个项。 您还可以按搜索项进行优化。 搜索不区分大小写，星号 () 将作为任何其他字符，并且不指示通配符搜索。 • 要优化显示的扩展列表，请单击应用。 在光标置于搜索框中时按 Enter 键与单击应用的效果相同。 • 要重置列表以显示所有可用扩展，请删除搜索框中的任何文本，取消选择所有项，并单击应用。 专用扩展 专用扩展是计算机上已经安装但是 GitHub 上的 IBM SPSS 预测性分析集合尚未提供的扩展。 用于优化显示 的扩展集和查看运行扩展的先决条件的功能不适用于专用扩展。 注: 在没有因特网连接的情况下使用“扩展中心”时，“已安装”选项卡的部分功能可能不可用。 设置 “设置”选项卡指定是下载并安装针对下载选中的安装，还是下载担不安装。 此设置适用于新扩展和现有扩展 的更新。 如果您正下载扩展以分发给组织内的其他用户，那么可以选择下载扩展而不进行安装。 如果不具 有用于运行扩展的必备软件但计划获取必备软件，那么还可以选择下载但不安装扩展。 如果选择下载扩展而不进行安装，那么稍后可通过选择扩展 > 安装本地扩展束进行安装。 对于 Windows 和 Mac，可以通过双击扩展束文件来安装扩展。 扩展详细信息 “扩展详细信息”对话框中显示扩展作者提供的信息。 除必需信息（例如“摘要”和“版本”）外，作者还可提供 指向相关位置（例如作者的主页）的 URL。 如果扩展是从“扩展中心”下载的，那么它包含许可证，可以通过 单击查看许可证来查看。 178 IBM SPSS Statistics 29 Core System 用户指南 组件。 “组件”组列出扩展包含的定制对话框（如果有）和任何扩展命令的名称。 扩展命令可以从语法编辑 器中运行，其运行方式与内置 IBM SPSS Statistics 命令相同。 可通过将光标放在命令中（在语法窗口中） 并按 F1 键获取扩展命令的帮助。 在语法编辑器中，还可以通过运行 CommandName /HELP 获取帮助。 注: 安装包含定制对话框的扩展可能需要重新启动 IBM SPSS Statistics ，以便在组件表中查看对话框的输 入。 相关性。 “相关性”组列出运行扩展中包含的组件所需的附加组件。 • Python 和 R 的集成插件 扩展的组件可能需要 Integration Plug-in for Python、R 的集成插件 或 。 Integration Plug-in for Java™ 随 Core System 一起安装并且不需要单独安装。 • R 程序包。 列出扩展所需的任何 R 程序包。 在安装扩展期间，安装程序尝试在机器上下载和安装必需的软 件包。 如果此过程失败，那么将发出提醒，然后您将需要手动安装软件包。 请参阅主题第180 页的『必 需的 R 软件包』，了解更多信息。 • Python 模块。 列出扩展所需的任何 Python 模块。 所有此类模块均可从 IBM SPSS 预测性分析社区 ( ) 获取。 将模块复制到为 SHOW EXTPATHS 命令的输 出中显示的扩展命令指定的位置。 或者，您可以将模块复制到 Python 搜索路径中的某个位置，例如 Python site-packages 目录。 • 扩展。 列出当前扩展所需的所有扩展的名称。 可从“扩展中心”下载所有此类扩展。 安装本地扩展束 要安装存储在本地计算机上的扩展束： 1. 从菜单中选择： 扩展 > 安装本地扩展束...
2. 选择扩展束。 扩展束的文件类型为 .spe 或 .spxt 您还可以使用命令行实用程序安装扩展束，这样可以一次安装多个扩展束。 请参阅 第181 页的『扩展束的 批处理安装』 主题以获取更多信息。 以分布方式安装扩展束 如果您在分布模式下工作，那么必须使用命令行实用程序，将扩展束安装在相关的 IBM SPSS Statistics 服务 器上。 如果扩展束包含自定义对话框，则还必须在本地计算机上安装扩展束（如前所述，从菜单中安装）。 如果您不知道扩展束是否包含定制对话框，那么最好在如果您不知道扩展束是否包含定制对话框，那么最好 在 IBM SPSS Statistics 服务器计算机和本地计算机上同时安装扩展束。 有关使用命令行实用程序的信息， 请参阅第181 页的『扩展束的批处理安装』。 要点: 对于 Windows 7 以及 Windows 更高版本的用户，安装现有扩展束的更新后版本可能需要使用管理员 特权运行 IBM SPSS Statistics。 您可以通过右键单击 IBM SPSS Statistics 的图标并选择以管理员身份运行 来使用管理员权限启动 IBM SPSS Statistics。 尤其是，如果收到指示无法安装一个或多个扩展束的错误消息 时，那么尝试使用管理员特权运行。 扩展的安装位置 缺省情况下，扩展安装到操作系统的一般用户可写入的位置。 要查看位置，请运行 SHOW EXTPATHS 语法 命令。 输出将在“Locations for extension commands”标题下显示扩展安装位置的列表。 扩展将安装到列表 中第一个可写入的位置。 通过使用 SPSS_EXTENSIONS_PATH 环境变量来定义路径，您可以覆盖缺省位置。 如果要指定多个备用位 置，对于 Windows 请使用分号分隔，对于 Linux 和 Mac，那么使用冒号分隔。 指定的位置在目标计算机上 必须存在。 设置 SPSS_EXTENSIONS_PATH 后，必须重新启动 IBM SPSS Statistics 以使更改生效。 扩展束中包含的定制对话框将安装到 SHOW EXTPATHS 语法命令的输出中“定制对话框的位置”标题下的列表 中的第一个可写入位置。 您可以通过使用 SPSS_CDIALOGS_PATH 环境变量定义路径来覆盖缺省位置（与 SPSS_EXTENSIONS_PATH 环境变量的内容相同）。 注: 对于以 Python 3 实现的扩展，可以使用 SPSSEX_EXTENSIONS_PATH 和 SPSSEX_CDIALOGS_PATH 环 境变量来指定替代安装位置。 SPSSEX_EXTENSIONS_PATH 和 SPSSEX_CDIALOGS_PATH 变量的规范与 第 20 章 扩展 179 SPSS_EXTENSIONS_PATH 和 SPSS_CDIALOGS_PATH 的规范相同，并且优先于后者和扩展的缺省安装位 置。 仅当指定 Python 3 作为扩展的 Python 版本时，才适用 SPSSEX_EXTENSIONS_PATH 和 SPSSEX_CDIALOGS_PATH 环境变量。 您可以从“扩展详细信息”对话框（可通过单击“扩展中心”对话框中关 联扩展的更多信息视图来访问）查看 Python 版本。 要在 Windows 上创建环境变量，请从控制面板： Windows 7 1. 选择“用户帐户”。 2. 选择更改我的环境变量。 3. 单击新建，在变量名称字段中输入环境变量的名称（例如，SPSS_EXTENSIONS_PATH），在“变量值”字 段中输入一条或多条路径。 Windows 8 1. 选择“系统”。 2. 选择“高级”选项卡并单击环境变量。 从“高级系统设置”访问“高级”选项卡。 3. 在“用户变量”部分中，单击新建，在变量名称字段中输入环境变量的名称（例如， SPSS_EXTENSIONS_PATH），并在变量值字段中输入一条或多条路径。 Windows 10 1. 在 Windows 搜索字段中输入环境变量，然后单击提供的编辑系统环境变量结果。 “系统属性”对话框将打 开至“高级”选项卡，其中提供环境变量... 控件。 2. 单击环境变量... 以打开“环境变量”对话框。 3. 在 部分的用户变量中，单击新建，在变量名字段中输入环境变量的名称（例如， SPSS_EXTENSIONS_PATH），并在变量值 字段中输入路径或路径。 要点: 对于 Windows 7 以及 Windows 更高版本的用户，安装现有扩展束的更新后版本可能需要使用管理员 特权运行 IBM SPSS Statistics。 您可以通过右键单击 IBM SPSS Statistics 的图标并选择以管理员身份运行 来使用管理员权限启动 IBM SPSS Statistics。 尤其是，如果收到指示无法安装一个或多个扩展束的错误消息 时，那么尝试使用管理员特权运行。 必需的 R 软件包 扩展安装程序尝试下载并安装扩展需要但在您的计算机上未找到的 R 程序包。 如果您的计算机无法访问 Internet，则需要向其他拥有所需程序包的人员索取。 如果软件包安装失败，那么将向您提醒所需软件包列 表。 安装扩展后，还可以从“扩展详细信息”对话框查看列表。 请参阅主题第178 页的『扩展详细信息』以 获取更多信息。 软件包可以从 R 内安装。 有关详细信息，请参阅随 R 分发的 R 安装和管理指南。 注: 对于 UNIX（包括 Linux）用户，以源格式下载软件包，然后进行编译。 这需要您在机器上安装相应的工 具。 请参阅《R 安装和管理指南》以获取详细信息。 尤其是，Debian 用户应从 apt-get install r-base-dev 安装 r-base-dev 软件包。 许可权 缺省情况下，所需的 R 程序包将被安装到 R 安装位置下的 library 文件夹中（例如，Windows 上的 C:\Program Files\R\R-4.2.0\library）。 如果对此位置没有写入权限，或打算在其他位置保存为扩展安装的 R 程序包，则可通过定义 SPSS_RPACKAGES_PATH 环境变量以指定一个或多个备用位置。 如果在 SPSS_RPACKAGES_PATH 中指定了路径，则它将被添加到 R 库搜索路径，并优先于缺省位置。 R 软件包将 安装到第一个可写位置。 对于多个位置，请使用分号（在 Windows 上）和冒号（在 Linux 和 Mac 上）分隔 每个位置。 指定的位置在目标机器上必须存在。 在设置 SPSS_RPACKAGES_PATH 后，您将需要重新启动 IBM SPSS Statistics 以使更改生效。 有关如何在 Windows 上设置环境变量的信息，请参阅第179 页的『扩 展的安装位置』。 180 IBM SPSS Statistics 29 Core System 用户指南 扩展束的批处理安装 您可以使用位于 IBM SPSS Statistics 安装目录下的批处理实用程序 installextbundles.bat（在 Mac 和 UNIX 系统中为 installextbundles.sh），一次安装多个扩展束。 对于 Windows 和 Mac，实用程序 位于此安装目录的根目录中。 对于 Linux 和用于 UNIX 版本的 IBM SPSS Statistics 服务器，该实用程序位于 安装目录的 bin 子目录。 该实用程序设计为从命令提示符运行，且必须从其安装位置运行。 命令行格式 为： installextbundles [–statssrv] [–download no|yes ] –source | ...
-statssrv。 指定您在 IBM SPSS Statistics 服务器上运行此实用程序。 您还应该在连接到服务器的客户端机 器上安装相同扩展束。 -download no|yes。 指定此实用程序是否有权访问 Internet，以便下载指定扩展束所需的任何 R 程序包。 缺省值为 no。 如果保留缺省值或不具有因特网访问权，那么需要手动安装任何必需的 R 软件包。 请参阅主 题第180 页的『必需的 R 软件包』，了解更多信息。 –source | ... 指定要安装的扩展束。 您可以指定包含扩展束的文件夹路径，或指定扩 展束文件名的列表。 如果指定文件夹，那么将安装文件夹中找到的所有扩展束（类型为 .spe 或 .spxt 的文件）。 多个文件名之间以一个或多个空格分开。 如果路径包含空格，那么需要使用双引号将此路径引 起。 注: 当在用于 UNIX 版本的 IBM SPSS Statistics 服务器上运行 installextbundles.sh 时，Bash shell 必 须存在于服务器计算机上。 创建并管理定制对话框 用于扩展的定制对话框构建器定制对话框，以生成要在内部 使用定制对话框构建器，您可以： • 针对内置 IBM SPSS Statistics 过程创建您自己的对话框版本。 例如，可以为“频率”过程创建对话框，以允 许用户选择变量集，然后通过预设的输出标准化选项生成命令语法。 • 创建用于为扩展命令生成命令语法的用户界面。 扩展命令是用户定义的 IBM SPSS Statistics 命令，采用 Python 编程语言、R 或 Java 进行实现。 有关更多信息，请参阅第207 页的『扩展命令的自定义对话框』 主题。 • 打开包含定制对话框（此对话框可能由其他用户创建）规范的文件，并将此对话框添加到您的 IBM SPSS Statistics 安装中，从而选择性地进行您自己的修改。 • 保存定制对话框的规范，以便其他用户可以将其添加到他们的 IBM SPSS Statistics 安装中。 “定制对话框构建程序”支持两种方式的操作： 扩展的定制对话框构建器(U) 此方式在新扩展或现有扩展中创建和修改定制节点。 以此方式打开“定制对话框构建器”时，会创建包含 空的定制对话框的新扩展。 以此方式保存或安装定制对话框时，会将其作为扩展的一部分进行保存或安 装。 以此方式创建的对话框称为增强的定制对话框。 增强的定制对话框包含 R24 之前的 IBM SPSS Statistics 发行版中未提供的许多新功能。 因此，增强的对话框与 R24 之前的版本不兼容。 要以此方式打开“定制对话框构建器”，请从菜单中选择; 扩展 > 用于扩展的定制对话框构建器 定制对话框构建器 - 兼容性方式 此方式可创建和修改与 IBM SPSS Statistics 的所有发行版兼容的定制对话框，称为兼容的定制对话框。 兼容型定制对话框不具有随增强型对话框提供的附加功能。 • 兼容性方式中支持的控件和控件属性是 R24 之前的发行版中定制对话框构建器中支持的控件和属性。 • 以兼容性方式保存对话框时，会将其保存为兼容型定制对话框包 (.spd) 文件，您可以随后将此文件添 加到扩展中。 第 20 章 扩展 181 • 从 IBM SPSS Statistics 26 开始，使用“用于扩展的定制对话框构建器”来创建 .spe 或 .spxt 文件 （不是 .spd 文件）。 要以兼容性方式打开“定制对话框构建器”，请从菜单中选择： 扩展 > 实用程序 > 定制对话框构建器（兼容性方式）...
定制对话框构建器布局 画布 画布是定制对话框构建器的一个区域，您可以在其中设计对话框的布局。 “属性”窗格 属性窗格是定制对话框构建器的一个区域，您可以在其中指定对话框中包含的控件的属性以及对话框自身的 属性，例如菜单位置。 工具选用板 工具选用板提供了一组可以包括在定制对话框中的控件。 您可以从视图菜单中选择“工具选用板”，以显示或 隐藏工具选用板。 语法模板 语法模板可指定由定制对话框生成的命令语法 。 您可以通过单击移至新窗口，将语法模板窗格移至单独的 窗口。 要将单独的“语法模板”窗口移动回“定制对话框构建器”，请单击复原到主窗口。 构建定制对话框 构建定制对话框时涉及的基本步骤如下所示： 注: 如果在兼容性方式下工作，则下面提到的一些特殊注意事项将适用。 1. 指定对话框本身的属性，如启动对话框时出现的标题以及 IBM SPSS Statistics 菜单中对话框新菜单项的 位置。 请参阅 第182 页的『对话框属性』主题以获取更多信息。 2. 指定构成对话框和子对话框的控件，例如源和目标变量列表。 有关更多信息，请参阅第186 页的『控件 类型』主题。 3. 创建语法模板以指定由对话框产生的命令语法。 有关更多信息，请参阅 "构建语法模板" 主题。 4. 指定包含对话框的扩展的属性。 有关更多信息，请参阅第201 页的『扩展属性』主题。 注: 此步骤在兼容性方式下不适用。 5. 将包含对话框的扩展安装到 IBM SPSS Statistics，并且/或者将扩展保存到扩展束（.spe 或 .spxt） 文件。 请参阅 第204 页的『管理定制对话框』主题以获取更多信息。 注: 在兼容性模式下，将对话框安装到 IBM SPSS Statistics，并且/或将对话框的规范保存到兼容的定制 对话框包 (.spd) 文件中。 有关更多信息，请参阅 "以兼容性方式管理定制对话框" 主题。 您可以在构建对话框时进行预览。 有关更多信息，请参阅第186 页的『预览定制对话框』主题。 对话框属性 查看和设置“对话框属性”： 1. 单击任何控件外部区域中的画布。 如果画布上没有控件，“对话框属性”将始终可见。 对话框名称。 “对话框名称”属性是必填项，它指定要与对话框相关联的唯一名称。 为了尽量降低名称冲突 可能性，您可以使用贵公司的标识（例如 URL）作为对话框名称的前缀。 菜单位置。 单击省略符 (...) 按钮可以打开“菜单位置”对话框，使用该对话框可以指定定制对话框的菜单项的 名称和位置。 182 IBM SPSS Statistics 29 Core System 用户指南 标题。 “标题”属性指定要显示在对话框的标题栏中的文本。 帮助文件。 “帮助文件”属性是可选项，用于指定对话框的帮助文件的路径。 这是用户单击对话框中的帮助 按钮时将要启动的文件。 帮助文件必须为 HTML 格式。 安装或保存对话框后，对话框的规范随附指定帮助 文件的副本。 如果没有关联的帮助文件，运行时间对话框的帮助按钮将被隐藏。 • 当您添加帮助文件时，和帮助文件位于相同目录中的帮助文件本地化版本将会自动添加到对话框。 帮助文 件本地化版本的名称为 .htm。 有关更多信息，请参见主题 第207 页的『创建定制对话框的本地化版本』。 • 可以将图像文件和样式表等支持文件添加到对话框，方法是先保存对话框， 然后，您可以将支持文件手动 添加到对话框文件（.cfe 或 .spd）。 有关访问和手动修改定制对话框文件的信息，请参见主题第207 页的『创建定制对话框的本地化版本』中题为“本地化对话框字符串”的部分。 Web 部署属性。 允许您将属性文件关联到此对话框，以便在构建采用 Web 方式部署的瘦客户端应用程序时 使用。 无模式。 指定对话框为模式还是无模式对话框。 如果为模式对话框，则它必须关闭后用户方可与主应用程 序窗口（“数据”、“输出”和“语法”）或其它打开的对话框进行交互作用。 无模式对话框没有此限制。 缺省为 无模式。 数据源。 指定源列表、目标列表和字段选择器控件中显示的字段的源。 缺省情况下，活动数据集为数据 源。 您可以将数据集选择器控件指定为数据源。 在运行时，在控件中选择的数据集为数据源。 单击省略符 (...) 按钮以打开对话框并指定数据源。 必需附加组件。 指定运行对话框生成的命令语法所需的一个或多个附加组件，例如 Integration Plug-in for Python 或 R 的集成插件。 例如，如果对话框生成以 R 语言实现的扩展命令的命令语法，则选中 R 的集成插 件 对应的复选框。 用户在安装或运行对话框时将得到缺少所需插件的警告消息。 注: “所需附加组件”属性仅在兼容性方式下适用。 在使用用于扩展的自定义对话框构建程序时，需要的任何 附加组件都是作为扩展的一部分指定的。 为对话框指定数据源 “数据源”对话框指定源列表、目标列表和字段选择器控件中显示的字段的源。 缺省情况下，活动数据集为数 据源。 您可以将“数据集选择器”控件指定为数据源。 在运行时，在控件中选择的数据集是数据源。 为定制对话框指定菜单位置 使用“菜单位置”对话框，可为定制对话框指定菜单项的名称与位置。 定制对话框的菜单项不会出现在 IBM SPSS Statistics 的“菜单编辑器”中。 如果正在兼容性方式中工作，请参阅“兼容性方式”部分以获取指示信 息。 1. 双击想要对其添加新对话框项的菜单（或单击加号图标）。 缺省情况下，对话框安装到所有窗口类型 （“数据编辑器”、“语法”和“查看器”），且会显示对所有窗口类型都通用的菜单。 从安装到列表中选定 窗口类型后，您可以选择将这个对话框安装到特定的窗口类型中，并显示适合该窗口类型的菜单。 2. 选择要在其上显示新对话框项的菜单项，并单击添加。 在添加项之后，可以使用上移和下移调整其位 置。 还可通过单击移动到新菜单下，将项移动到新子菜单下。 安装对话框时，如果不存在子菜单，那么 会将其添加到 IBM SPSS Statistics 菜单系统。 3. 输入菜单项的名称。 特定菜单或子菜单中的名称必须唯一。 兼容性方式 1. 双击想要对其添加新对话框项的菜单（或单击加号图标）。 如果要创建定制菜单或子菜单，请使用“菜单编辑器”。 有关更多信息，请参阅第175 页的『菜单编辑 器』，另请参阅“注释”部分中的关联注释。 2. 选择要在其上显示新对话框项的菜单项。 在添加项之后，可以使用上移和下移按钮调整其位置。 3. 为菜单项输入标题。 特定菜单或子菜单中的标题必须唯一。 4. 单击添加。 第 20 章 扩展 183 选项 • 在新菜单项上面或下面添加分隔符。 • 为定制对话框指定其菜单项旁边所出现图像的路径。 支持的图像格式为 gif 和 png。 图像不得大于 16 x 16 像素。 注: • “菜单位置”对话框显示所有附加模块的菜单。 确保将您的定制对话框的菜单项添加到可供您或对话框的其 他用户使用的菜单。 • 如果将对话框添加到从菜单编辑器（视图 > 菜单编辑器）创建的菜单，那么对话框的用户必须从其菜单编 辑器手动创建相同菜单。 否则，会将此对话框添加到其扩展菜单中。 在画布上布置控件 通过将控件从工具选用板拖放到画布上，可以将这些控件添加到定制对话框。画布分为四个功能列，您可以 在其中放置控件。 适用以下注意事项： • 第一列（最左侧的列）主要用于放置源列表控件。源列表控件实际上必须在第一列中。 • 目标列表控件必须位于第二列中。 • “子对话框”按钮必须位于左右侧列中（例如，如果仅使用三个列，那么位于第三列），其他控件可位于与 “子对话框”按钮相同的列中。 在此方面，第四列仅包含子对话框按钮。 注: 在兼容性方式下，仅提供三个列以布置控件。 尽管未在画布上显示，每个定制对话框均包含有确定、粘贴、取消和帮助按钮，它们位于对话框的底部。 这 些按钮的存在及其位置均为自动确定。但如果没有帮助文件关联到对话框（由“对话框属性”中的“帮助文件” 属性指定），那么帮助按钮将被隐藏。 通过向上或向下拖动控件，可以更改这些控件在一列中的垂直顺序，但是各个控件的确切位置将自动确定。 在运行时，当对话框自身调整大小时，各个控件将相应地调整大小。 源列表和目标列表之类的控件将自动展 开，以填充其下方的可用空间。 构建语法模板 语法模板指定将由定制对话框产生的命令语法。 单个定制对话框可为任意数量的内置 IBM SPSS Statistics 命令或扩展命令生成命令语法。 语法模板可以包含需要始终生成的静态文本，以及在运行时间被替换为关联定制对话框控件值的控件标识。 例如，不依赖于用户输入的命令名称和子命令指定可以作为静态文本。 而在目标列表中指定的变量集可使用 目标列表控件的控件标识来表示。 构建语法模板 1. 对于不依赖于用户指定值的静态命令语法，请按照?语法编辑器?中那样输入语法。 “语法模板”支持语法 编辑器的自动完成和颜色编码功能。 请参阅主题第145 页的『使用语法编辑器』，了解更多信息。 2. 在需要插入控件所生成命令语法的位置，以 %%Identifier%% 形式插入控件标识，其中 Identifier 为控件的“标识”属性的值。 • 您可以通过选择标识的表中一行，右键单击并选择添加到语法模板，来插入控件标识。 还可通过右键 单击画布中的控件并选择添加到语法模板，来插入控件标识。 • 您还可以按 Ctrl + 空格键来从可用控件标识列表中进行选择。 该列表包含的控件标识后接可以使用语 法自动完成功能的项。 如果您手动输入标识，请保留所有空格，这是因为标识中的所有空格都有意义。 在运行时，对于除复选框和复选框组以外的所有控件，每个标识均被替换为关联控件中“语法”属性的当前 值。 对于复选框和复选框组，标识将被替换为相关控件中“选中语法”或“未选中语法”属性的当前值，具 体取决于控件的当前状态是选中还是未选中。 请参阅主题第186 页的『控件类型』，了解更多信息。 注: 如果没有终止符，在运行时生成的语法将自动包括命令终止符（句点）作为最后一个字符。 184 IBM SPSS Statistics 29 Core System 用户指南 示例：在语法模板中包含运行时值 考虑“频率”对话框的简化版本，即只包含源列表控件与目标列表控件，并生成以下形式的命令语法： FREQUENCIES VARIABLES=var1 var2... /FORMAT = NOTABLE /BARCHART.
生成此语法的语法模板应形如： FREQUENCIES VARIABLES=%%target_list%% /FORMAT = NOTABLE /BARCHART.
• %%target_list%% 为目标列表控件的“标识”属性值。 在运行时，它将被替换为控件的“语法”属性的当前 值。 • 将目标列表控件的“语法”属性定义为 %%ThisValue%%，指定在运行时属性的当前值将为控件的值，即目 标列表中的变量集。 示例：包含容器控件中的命令语法 在前一个示例基础上，考虑添加一个包含单组复选框的“统计”子对话框，以允许用户指定平均值、标准差、 最小值和最大值。 假设复选框包含在项目组控件中。 生成的命令语法示例为： FREQUENCIES VARIABLES=var1 var2... /FORMAT = NOTABLE /STATISTICS MEAN STDDEV /BARCHART.
生成此语法的语法模板应形如： FREQUENCIES VARIABLES=%%target_list%% /FORMAT = NOTABLE %%stats_group%% /BARCHART.
• %%target_list%% 为目标列表控件的“标识”属性值，而 %%stats_group%% 则为项组控件的“标识”属 性值。 以下定义说明了一种指定项目组及其中包含的复选框的“语法”属性以生成正确结果的方法。 如前一个示例中 所述，目标列表的“语法”属性将被设为 %%ThisValue%%。 项组的“语法”属性：/STATISTICS %%ThisValue%% 平均值复选框的“选中语法”属性：MEAN 标准差复选框的“选中语法”属性：STDDEV 最小值复选框的“选中语法”属性：MINIMUM 最大值复选框的“选中语法”属性：MAXIMUM 在运行时，%%stats_group%% 将被替换为项目组控件的“语法”属性的当前值。 具体地，%%ThisValue%% 将被替换为每个复选框的“选中语法”或“未选中语法”属性（取决于其状态为选中还是未选中）的值所组成的 空格分隔列表。 由于只为?选中语法?属性指定了值，因此只有选中的复选框才在 %%ThisValue%% 中出 现。 例如，如果用户选中了平均值和标准差复选框，那么项目组控件的“语法”属性的运行时值将为 / STATISTICS MEAN STDDEV。 如果未选中任何复选框，那么项组控件的?语法?属性将为空，生成的命令语法也不会包含任何对 % %stats_group%% 的引用。 这可能是需要的也可能不需要。 例如，即使未选中任何复选框，您可能仍然希 望生成 STATISTICS 子命令。 这可以通过在语法模板中直接引用复选框的标识来实现，即： FREQUENCIES VARIABLES=%%target_list%% /FORMAT = NOTABLE /STATISTICS %%stats_mean%% %%stats_stddev%% %%stats_min%% %%stats_max%% /BARCHART.
第 20 章 扩展 185 预览定制对话框 您可以预览当前在定制对话框构建器中打开的对话框。 从IBM SPSS Statistics 中的从菜单运行时，按原样 显示对话框。 • 源变量列表中被填入虚变量，并可将其移动到目标列表。 • 粘贴按钮将命令语法粘贴到指定的语法窗口中。 • 单击确定可关闭预览。 • 如果指定了帮助文件，则会启用帮助按钮，并可用它来打开指定文件。 如果未指定帮助文件，则预览时会 禁用帮助按钮；当实际对话框运行时，此按钮也将被隐藏。 预览定制对话框： 1. 从“自定义对话框构建程序”的菜单中选择： 文件 > 预览对话框 控件类型 工具调色板提供了可添加到定制对话框中的控件。 源列表 活动数据集中的源变量列表。 有关更多信息，请参阅 "源列表"。 目标列表 从源列表传输的变量的目标。 有关更多信息，请参阅 "目标列表"。 字段选择器 活动数据集中的所有字段的列表。 有关更多信息，请参阅第188 页的『字段选择器』。 数据集选择器 当前打开的数据集的列表。 有关更多信息，请参阅 "数据集选择器"。 复选框 单个复选框。 有关更多信息，请参阅第189 页的『复选框』。 组合框 用于创建下拉列表的组合框。 有关更多信息，请参阅第190 页的『组合框』。 列表框 用于创建单个选择或多个选择列表的列表框。 有关更多信息，请参阅第191 页的『列表框』。 文本控件 接受任意文本作为输入的文本框。 有关更多信息，请参阅第192 页的『“文本”控件』。 数字控件 限制为将数字值作为输入的文本框。 有关更多信息，请参阅第193 页的『数字控件』。 日期控件 用于指定日期/时间值的微调控件，包括日期、时间和日期时间。 有关更多信息，请参阅第194 页的 『日期控件』。 受保护的文本 使用星号屏蔽用户输入的文本框。 有关更多信息，请参阅第194 页的『安全文本』。 静态文本控件 用于显示静态文本的控件。 有关更多信息，请参阅第195 页的『静态文本控件』。 颜色选取器 用于指定颜色并生成相关联的 RGB 值的控件。 有关更多信息，请参阅第195 页的『颜色选取器』。 表控件 具有固定数量的列和在运行时添加的可变行数的表。 有关更多信息，请参阅第196 页的『表控件』。 项目组 用于分组一组控件（例如一组复选框）的容器。 有关更多信息，请参阅第197 页的『项目组』。 单选按钮组 一组单选按钮。 有关更多信息，请参阅第198 页的『单选组』。 186 IBM SPSS Statistics 29 Core System 用户指南 复选框组 由单个复选框启用或禁用为组的一组控件的容器。 有关更多信息，请参阅第199 页的『复选框组』。 文件浏览器 用于浏览文件系统以打开或保存文件的控件。 有关更多信息，请参阅第199 页的『文件浏览器』。 制表符 单个选项卡。 有关更多信息，请参阅第200 页的『选项卡』。 子对话框按钮 用于启动子对话框的按钮。 有关更多信息，请参阅第200 页的『子对话框按钮』。 源列表 “源变量列表”控件显示来自活动数据集，且对对话框的最终用户可用的变量列表。 可以显示所有来自活动数 据集的变量（缺省情况下），或根据类型和测量级别来过滤列表，例如，具有刻度测量级别的数值变量。 使 用“源列表”控件暗示使用一个或多个目标列表控件。 源列表控件具有下列属性： 标识。 控件的唯一标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。指定文本仅当悬停在控件标题区域上时显 示。 悬停在列出的变量上将显示变量名称和标签。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上不支持“助记键”属性。 变量转移。 指定从源列表转移到目标列表的变量是保留在源列表中（复制变量），还是从源列表中删除（移 动变量）。 变量过滤。 可以过滤在控件中显示的变量集。 可以按类型与测量级别过滤，并可指定在变量列表中包括多 响应集。 单击省略符 (...) 按钮将打开“过滤器”对话框。 还可通过双击画布上的“源列表”控件以打开“过滤”对 话框。 有关更多信息，请参阅第189 页的『对变量列表进行过滤』主题。 注: 源列表控件不能添加到子对话框中。 目标列表 源列表控件提供从源列表中转移变量的目标。 “目标列表”控件的使用假定存在源列表控件。 可以指定仅单 个变量能够被转移到控件，或能够转移多个变量；还可以限制转移到控件的变量类型，例如，只转移具有名 义或有序测量级别的数值变量。 目标列表控件具有下列属性： 标识符。控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。指定文本仅当悬停在控件标题区域上时显 示。 悬停在列出的变量上将显示变量名称和标签。 目标列表类型。 指定可向控件传输多个变量还是只能传输单个变量。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上不支持“助记键”属性。 分隔符类型。 指定生成的语法中的所选字段之间的定界符。 允许的分隔符是空格、逗号和加号 (+)。 您还可 以输入任意单个字符作为分隔符。 最小字段数。 必须为控件指定的最小字段数（如果具有）。 最大字段数。 可以为控件指定的最大字段数（如果具有）。 执行所必需。指定该控件中是否需要值，以便继续执行操作。 如果指定了 True，那么在为此控件指定值之 前，确定和粘贴按钮将处于禁用状态。 如果指定了 False，那么此控件中缺少值并不会对确定和粘贴按钮的 状态造成任何影响。 缺省值为 True。 第 20 章 扩展 187 变量过滤。 允许约束可向控件传输的变量类型。 可以按类型与测量级别进行限制，并可指定是否能够转移 多响应集到控件。 单击省略符 (...) 按钮将打开“过滤器”对话框。 还可通过双击画布上的“目标列表”控件以打 开“过滤”对话框。 请参阅主题第189 页的『对变量列表进行过滤』，了解更多信息。 语法。 指定该控件在运行时间生成的并可插入到语法模板的命令语法。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定控件的运行时值，即传输到控件的变量的列表。 这是缺省值。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 注: 目标列表控件不能添加到子对话框中。 字段选择器 " 字段选择器 " 控件显示可用于对话框的最终用户的字段列表。您可以根据类型和测量级别过滤列表，例 如，具有刻度测量级别的数字字段。 还可以将任何其他“字段选择器”或“目标列表”控件指定为当前“字段选 择器”的字段源。 “字段选择器”控件具有下列属性： 标识。 控件的唯一标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 标题位置。 指定标题相对于控件的位置。 值为“顶部”和“左侧”，其中“顶部”是缺省值。 仅当选择器类型设 置为选择单个字段时，此属性才适用。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。仅当鼠标指针悬停在此控件的标题区域上 时，才会显示指定的文本。 将鼠标悬停在某个列示字段上将显示字段名称和标签。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 选择器类型。 指定可使用定制对话框中的字段选择器从字段列表选择单个字段还是多个字段。 分隔符类型。 指定生成的语法中的所选字段之间的定界符。 允许的分隔符是空格、逗号和加号 (+)。 您还可 以输入任意单个字符作为分隔符。 最小字段数。必须为控件指定的最小字段数（如果具有）。 最大字段数。可以为控件指定的最大字段数（如果具有）。 执行时必需。指定该控件中是否需要值，以便继续执行操作。 如果指定了 True，那么在为此控件指定值之 前，确定和粘贴按钮将处于禁用状态。 如果指定了 False，那么此控件中缺少值并不会对确定和粘贴按钮的 状态造成任何影响。 变量过滤。 允许过滤在控件中显示的字段集。 可以按字段类型与测量级别过滤，并可指定在字段列表中包 括多个响应集。 单击省略符 (...) 按钮将打开“过滤器”对话框。 您也可以通过双击画布上的“字段选择器”控件 来打开“过滤器”对话框。 请参阅主题第189 页的『对变量列表进行过滤』，了解更多信息。 资源源。 指定另一个“字段选择器”或“目标列表控件”是当前“字段选择器”的字段源。 未设置“字段源”属性 时，字段的源为对话框的“数据源”属性确定的当前数据源。 单击省略符 (...) 按钮以打开对话框并指定字段 源。 语法。 指定由此控件生成并运行命令语法，并且可插入到语法模板中。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定控件的运行时值，它是字段列表。 这是缺省选项。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 注: “字段选择器”控件在兼容性方式中不可用。 188 IBM SPSS Statistics 29 Core System 用户指南 为“字段选择器”指定“字段源” “字段源”对话框可指定“字段选择器”中显示的资源的源，并覆盖此对话框的“数据源”属性的设置。 源可以是 任何其他“字段选择器”或“目标列表”控件。 您可以从当前数据源选择显示包含在选定控件中的字段或不包含 在选定控件中的的字段。 数据源由对话框的“数据源”属性确定。 对变量列表进行过滤 通过使用与源列表、目标列表和字段选择器控件相关联的“过滤器”对话框，可以对活动数据集中能够在列表 中显示的变量类型进行过滤。 还可以指定是否包含与活动数据集关联的多响应集。 数字变量包括除日期格 式和时间格式以外的所有数字格式。 数据集选择器 数据集选择器控件显示当前打开的数据集的列表。 您可以使用数据集选择器控件捕获所选数据集的名称。 还可将其与对话框的“数据源”属性配合使用，以设置源列表和字段选择器控件使用的数据源。 标识符。控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 执行需要。 指定该控件中是否需要值，以便继续执行操作。 如果指定了 True，那么在为此控件指定值之 前，确定和粘贴按钮将处于禁用状态。 如果指定了 False，那么此控件中缺少值并不会对确定和粘贴按钮的 状态造成任何影响。缺省为 False。 语法。 指定该控件在运行时生成并可插入到语法模板中的命令语法。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定控件的运行时值，它是所选数据集的名称。 这是缺省值。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 注: 数据集选择器控件不支持兼容性方式。 复选框 “复选框”控件是一个简单的复选框，它可以针对选中和未选中状态生成不同的命令语法。 复选框控件具有下 列属性： 标识。 控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 缺省值。 复选框的缺省状态为选中或未选中。 选中/未选中语法。 指定在选中控件或取消选中控件时生成的命令语法。 要在语法模板中包含命令语法，请 使用“标识”属性的值。 所生成的语法将在该标识的指定位置进行插入，而与它是根据“选中语法”属性还是 “未选中语法”属性生成无关。 例如，如果标识为 checkbox1，那么在运行时，如果选中该复选框，则语法模 板中的 %%checkbox1%% 实例将被替换为选中的语法 属性的值；如果未选中该复选框，则替换为未选中的 语法 属性的值。 第 20 章 扩展 189 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 组合框 使用“组合框”控件可以创建下拉列表，此下拉列表可以生成特定于所选列表项的命令语法。 它限制为单选。 组合框控件具有下列属性： 标识。 控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 标题位置。 指定标题相对于控件的位置。 值为“顶部”和“左侧”，其中“顶部”是缺省值。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 列表项。 单击省略号 (...) 按钮以打开 列表项属性对话框，该对话框允许您指定控件的列表项。 您还可以通 过双击画布上的“组合框”控件来打开“列表项属性”对话框。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 快捷键 通过按下 Alt+[助记键] 激活。在 Mac 上，不支持“助记键”属性。 可编辑。 指定组合框是否可编辑。 在控件可编辑时，可在运行时输入定制值。 语法。 指定由此控件在运行时生成且可以插入到语法模板中的命令语法。 • 值 %%ThisValue%% 指定控件的运行时值，也是缺省值。 如果手动定义列表项，那么运行时值是所选列表 项的“语法”属性的值。 如果列表项基于目标列表控件，那么运行时值为选定列表项的值。 对于多选列表框 控件，运行时值为以空格分隔的选定项列表。 请参阅主题第190 页的『为组合框和列表框指定列表项』， 了解更多信息。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 报价处理。当语法属性包含 %%ThisValue%% 作为带引号字符串的一部分时，指定处理运行时值 % %ThisValue%% 中的引号。 在此上下文中，加引号的字符串是用单引号或双引号引起来的字符串。 引号处 理仅适用于类型与用于引起 %%ThisValue%% 的引号相同的引号。 提供了以下类型的引号处理。 语法 与括起的引号匹配的 %%ThisValue%% 运行时值中引号将变为双引号。 例如，如果“语法”属性为 '%%ThisValue%%'，且组合框的运行时值为 Combo box's value，那么生成的语法为 'Combo box''s value'。 Python 运行时值 %%ThisValue%% 中与所含引号匹配的引号将使用反斜杠 () 字符进行转义，例如，如果语 法属性为 '%%ThisValue%%'，并且组合框的运行时值为组合框的值，则生成的语法为 'Combo box\'s value'。 请注意，%%ThisValue%% 使用三重引号引起时，不会执行引号处理。 R 运行时值 %%ThisValue%% 中与所含引号匹配的引号将使用反斜杠 () 字符进行转义，例如，如果语 法属性为 '%%ThisValue%%'，并且组合框的运行时值为组合框的值，则生成的语法为 'Combo box\'s value'。 无 与括起的引号匹配的 %%ThisValue%% 运行时值中引号将保留，不进行任何修改。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 为组合框和列表框指定列表项 使用“列表项属性”对话框可以指定组合框或列表框控件的列表项。 190 IBM SPSS Statistics 29 Core System 用户指南 手动定义的值。 允许您显式指定各个列表项。 • 标识。 列表项的唯一标识。 • 名称。 在列表中出现的该项的名称。 名称为必需的字段。 • 缺省值。 对于组合框，指定列表项是否为组合框中显示的缺省项。 对于列表框，指定列表项是否缺省选 中。 • 语法。 指定选择此列表项时生成的命令语法。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 注: 可以在现有列表底部的空白行中添加新的列表项。 输入标识以外的任何属性会生成唯一的标识，您可以 保留或修改此标识。 要删除列表项，请单击其标识单元格，并按“删除”。 基于目标列表或字段选择器控件内容的值。 指定使用选定目标列表或字段选择器控件中变量的关联值动态填 充列表项。 选择现有的目标列表或字段选择器控件作为列表项的源，或者在目标列表或字段选择器组合框的 文本区域中输入目标列表或字段选择器控件的“标识”属性值。 使用后一种方法可以输入您打算稍后添加的目 标列表或字段选择器控件的标识。 • 变量名称。 使用指定目标列表或字段选择器控件中的变量名称填充列表项。 • 值标签。 使用指定目标列表控件中变量的关联值标签集合填充列表项。 您可以选择由关联组合框或列表框 控件生成的命令语法是否包含选定值标签或其相关值。 值标签选项对字段选择器不可用。 • 定制属性。 使用包含指定定制属性的目标列表控件中变量的关联属性值集合填充列表项。 定制属性选项对 字段选择器不可用。 • 语法。 显示相关组合框或列表框控件的“语法”属性，从而使您能够对此属性进行更改。 有关更多信息，请 参阅第190 页的『组合框』主题。 列表框 使用“列表框”控件可以显示支持单选或多选的项列表，并且可以生成特定于所选项的命令语法。 列表框控件 具有下列属性： 标识。 控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 列表项。 单击省略号 (...) 按钮以打开 列表项属性对话框，该对话框允许指定控件的列表项。 还可通过双击 画布上的列表框控件以打开“列表项属性”对话框。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 快捷键 通过按下 Alt+[助记键] 激活。在 Mac 上，不支持“助记键”属性。 列表框类型。 指定列表框是仅支持单选还是支持多选。 还可指定列表项显示为复选框列表。 分隔符类型。 在生成的语法中的所选列表项之间指定定界符。允许的分隔符是空格、逗号和加号 (+)。还可 以输入任意单个字符作为分隔符。 所选最小值。 必须在控件中选择的最小项数（如果具有）。 最大所选项数。 在控件（如果有）中可以选择的最大项数。 语法。 指定由此控件在运行时生成且可以插入到语法模板中的命令语法。 • 值 %%ThisValue%% 指定控件的运行时值，也是缺省值。 如果手动定义列表项，那么运行时值是所选列表 项的“语法”属性的值。 如果列表项基于目标列表控件，那么运行时值为选定列表项的值。 对于多选列表框 控件，运行时值为以指定分隔符类型分隔（缺省为空格分隔）的选定项列表。 请参阅主题第190 页的『为 组合框和列表框指定列表项』，了解更多信息。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 第 20 章 扩展 191 报价处理。当语法属性包含 %%ThisValue%% 作为带引号字符串的一部分时，指定处理运行时值 % %ThisValue%% 中的引号。 在此上下文中，加引号的字符串是用单引号或双引号引起来的字符串。 引号处 理仅适用于类型与用于引起 %%ThisValue%% 的引号相同的引号。 提供了以下类型的引号处理。 语法 与括起的引号匹配的 %%ThisValue%% 运行时值中引号将变为双引号。 例如，如果语法属性为“% %ThisValue%%”，并且所选列表项为 List item's value，那么生成的语法为 'List item''s value'。 Python 运行时值 %%ThisValue%% 中与所含引号匹配的引号将使用反斜杠 () 进行转义。例如， 语法属性 为 '%%ThisValue%%' 且选定的列表项为列表项的值，那么将生成的语法为 'List item\'s value'。 请注意，%%ThisValue%% 使用三重引号引起时，不会执行引号处理。 R 运行时值 %%ThisValue%% 中与所含引号匹配的引号将使用反斜杠 () 进行转义。例如， 语法属性 为 '%%ThisValue%%' 且选定的列表项为列表项的值，那么将生成的语法为 'List item\'s value'。 无 与括起的引号匹配的 %%ThisValue%% 运行时值中引号将保留，不进行任何修改。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 “文本”控件 “文本”控件是一个简单的文本框，它可以接受任意输入并具有下列属性： 标识。 控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 标题位置。 指定标题相对于控件的位置。 值为“顶部”和“左侧”，其中“顶部”是缺省值。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 文本内容。 指定允许的内容。 值 Any 指定内容是任意的。 值 Variable Name 指定文本框必须包含遵守 IBM SPSS Statistics 变量名称的规则的字符串。 请参阅主题第44 页的『变量名称』，了解更多信息。 值 New dataset name 指定文本框必须包含有效的 IBM SPSS Statistics 数据集名称，且不能与当前打开的数据 集名称相同。 缺省值。 文本框的缺省内容。 宽度。指定控件的文本区域的宽度（字符）。 允许的值为正整数。 空值表示将自动确定宽度。 执行需要。 指定该控件中是否需要值，以便继续执行操作。 如果指定了 True，那么在为此控件指定值之 前，确定和粘贴按钮将处于禁用状态。 如果指定了 False，那么此控件中缺少值并不会对确定和粘贴按钮的 状态造成任何影响。缺省为 False。 语法。 指定命令语法，由此控件在运行时生成，可插入到语法模板中。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定控件的运行时值，它是文本框的内容。 这是缺省值。 • 如果“语法”属性包含 %%ThisValue%%，并且文本框的运行时值为空，那么文本框控件不会生成任何命令 语法。 引号处理。 当语法属性包含 %%ThisValue%% 作为带引号字符串的一部分时，指定处理运行时值 % %ThisValue%% 中的引号。 在此上下文中，加引号的字符串是用单引号或双引号引起来的字符串。 引号处 理仅适用于类型与用于引起 %%ThisValue%% 的引号相同的引号。 提供了以下类型的引号处理。 192 IBM SPSS Statistics 29 Core System 用户指南 语法 与括起的引号匹配的 %%ThisValue%% 运行时值中引号将变为双引号。 例如，如果语法属性为 '% %ThisValue%%' 并且文本控件的运行时值为 Text box's value，那么生成的语法为 'Text box''s value'。 python 使用反斜杠字符 () 对运行时值 %%ThisValue%% 中与括起的引号匹配的引号进行转义。例如，如果 语法属性为 '%%ThisValue%%'，文本控件的运行时值为 Text box's value，那么生成的语法 将为 'Text box\'s value'。 %%ThisValue%% 使用三重引号引起时，不会执行引号处理。 R 使用反斜杠字符 () 对运行时值 %%ThisValue%% 中与括起的引号匹配的引号进行转义。例如，如果 语法属性为 '%%ThisValue%%'，文本控件的运行时值为 Text box's value，那么生成的语法 将为 'Text box\'s value'。 无 与括起的引号匹配的 %%ThisValue%% 运行时值中引号将保留，不进行任何修改。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 数字控件 “数字”控件为文本框，它用于输入数值，并具有下列属性： 标识符。控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 标题位置。 指定标题相对于控件的位置。 值为“顶部”和“左侧”，其中“顶部”是缺省值。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 数值类型。 指定对输入内容的任何限制。 值“Real”指定对输入值除了必须为数值外，没有其它限制。 如果 值为“Integer”，表示输入值必须为整数。 微调输入。 指定控件是否显示为微调器 (spinner)。 缺省值为 False。 增量。 控件显示为微调器 (spinner) 时的增量。 缺省值。 缺省值，如果有的话。 最小值。 最小允许值，如果有的话。 最大值。 最大允许值，如果有的话。 宽度。 指定控件的文本区域的宽度（字符）。 允许的值为正整数。 空值表示将自动确定宽度。 执行需要。 指定该控件中是否需要值，以便继续执行操作。 如果指定了 True，那么在为此控件指定值之 前，确定和粘贴按钮将处于禁用状态。 如果指定了 False，那么此控件中缺少值并不会对确定和粘贴按钮的 状态造成任何影响。缺省为 False。 语法。 指定命令语法，由此控件在运行时生成，可插入到语法模板中。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定控件的运行时值，它是数值。 这是缺省值。 • 如果“语法”属性包含 %%ThisValue%%，并且数字控件的运行时值为空，那么数字控件不会生成任何命令 语法。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 第 20 章 扩展 193 日期控件 日期控件是用于指定日期/时间值（包括日期、时间和日期时间）的微调器 (spinner) 控件。 日期控件具有下 列属性： 标识符。控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 标题位置。 指定标题相对于控件的位置。 值为“顶部”和“左侧”，其中“顶部”是缺省值。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 类型。 指定控件是用于日期、时间还是日期时间值。 日期 控件指定格式为 yyyy-mm-dd 的日历日期。 缺省运行时值通过“缺省值”属性指定。 时间 控件指定格式为 hh:mm:ss 的时间。 缺省运行时值为当前的时间。 日期时间 控件指定格式为 yyyy-mm-dd hh:mm:ss 的日期和时间。 缺省运行时值为当前日期和时间。 缺省值。 在类型为“日期”时控件的缺省运行时值。 您可以指定以显示当前日期或特定日期。 语法。 指定命令语法，由此控件在运行时生成，可插入到语法模板中。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定控件的运行时值。 这是缺省值。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 注: 日期控件不支持兼容性方式。 安全文本 安全文本控件是使用星号掩饰用户输入的文本框。 标识符。控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 标题位置。 指定标题相对于控件的位置。 值为“顶部”和“左侧”，其中“顶部”是缺省值。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 加密传递的值。 指定是否加密生成的命令语法中的值。 值为 True，指定值已加密。 缺省值为 False，指定 不会加密值。 仅可通过 IBM SPSS Statistics 命令对加密值进行解密，这些命令可处理加密密码，例如，GET 和 SAVE 命令。 宽度。指定控件的文本区域的宽度（字符）。 允许的值为正整数。 空值表示将自动确定宽度。 执行所必需。 指定该控件中是否需要值，以便继续执行操作。 如果指定了 True，那么在为此控件指定值之 前，确定和粘贴按钮将处于禁用状态。 如果指定了 False，那么此控件中缺少值并不会对确定和粘贴按钮的 状态造成任何影响。缺省值为 False。 语法。 指定命令语法，由此控件在运行时生成，可插入到语法模板中。 194 IBM SPSS Statistics 29 Core System 用户指南 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定控件的运行时值，它是文本框的内容。 这是缺省值。 • 如果语法属性包含 %%ThisValue%% 并且受保护的文本控件的运行时值为空，那么受保护的文本控件不会 生成任何命令语法。 引号处理。 指定在以下情况下处理运行时值 %%ThisValue%% 中的引号：当语法属性在引起来的字符串中 包含 %%ThisValue%% 时。 在此上下文中，加引号的字符串为使用单引号或双引号引起的字符串。 引号处 理仅适用于类型与用于引起 %%ThisValue%% 的引号相同的引号，且仅当 Encrypt passed value=False 时适用。 提供了以下类型的引号处理。 语法 匹配嵌套引号的 %%ThisValue%% 运行时值中的引号将加倍。 例如，如果语法属性为 '% %ThisValue%%' 并且控件的运行时值为 SecuredText's value，那么生成的语法为 'Secured Text''s value'。 Python 使用反斜杠字符 () 对运行时值 %%ThisValue%% 中与括起的引号匹配的引号进行转义。例如，如果 语法属性为 '%%ThisValue%%'，控件的运行时值为 Secured Text's value，那么生成的语法 为 'Secured Text\'s value'。 在三重引号括起 %%ThisValue%% 时，不执行引号处理。 R 使用反斜杠字符 () 对运行时值 %%ThisValue%% 中与括起的引号匹配的引号进行转义。例如，如果 语法属性为 '%%ThisValue%%'，控件的运行时值为 Secured Text's value，那么生成的语法 为 'Secured Text\'s value'。 无 与括起的引号匹配的 %%ThisValue%% 运行时值中引号将保留，不进行任何修改。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 注: 安全文本控件不支持兼容性方式。 静态文本控件 使用“静态文本”控件可以向对话框添加文本块，此控件具有下列属性： 标识符。控件的唯一标识。 标题。 文本块的内容。 针对多行文本或长文本行，请单击省略符 (...) 按钮，并在“标题属性”对话框中输入文 本。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 颜色选取器 颜色选取器控件为用于指定颜色和生成关联 RGB 值的用户界面。 颜色选取器控件具有下列属性： 标识符。控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 标题位置。 指定标题相对于控件的位置。 值为“顶部”和“左侧”，其中“顶部”是缺省值。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 语法。 指定命令语法，由此控件在运行时生成，可插入到语法模板中。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 第 20 章 扩展 195 • 值 %%ThisValue%% 指定控件的运行时值，即选中的颜色的 RGB 值。 RGB 值表示为采用以下顺序的整数 的空格分隔列表：R 值、G 值和 B 值。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 注: 颜色选取器控件不支持兼容性方式。 表控件 表控件用于创建具有固定数目列以及在运行时添加的可变数目行的表。 表控件具有下列属性： 标识符。控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 快捷键 通过按下 Alt+[助记键] 激活。在 Mac 上，不支持“助记键”属性。 重新排序按钮。 指定是否将上移和下移按钮添加到表中。 在运行时会使用这些按钮来对表的行进行重新排 序。 表列。 单击省略号 (...) 按钮以打开 表列对话框，您可以在其中指定表的列。 最小行数。 必须位于表中的最小行数。 最大行数。 可位于表中的最大行数。 执行时必需。指定该控件中是否需要值，以便继续执行操作。 如果指定了 True，那么在为此控件指定值之 前，确定和粘贴按钮将处于禁用状态。 如果指定了 False，那么此控件中缺少值并不会对确定和粘贴按钮的 状态造成任何影响。 语法。 指定由此控件在运行时生成且可以插入到语法模板中的命令语法 • 值 %%ThisValue%% 指定控件的运行时值，也是缺省值。 运行时值是表中每个列（从最左边列开始）生成 的语法的空格分隔列表。 如果语法属性包含 %%ThisValue%%，且列都未生成语法，那么整个表不会生成 任何语法 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 注: 表控件不支持兼容性方式。 指定表控件的列 “表列”对话框指定表控件的列的属性。 标识。 列的唯一标识。 列名。 表中显示的列名。 内容。 指定列的数据类型。 值实数指定输入的值必须是数字，此外没有任何限制。 值 Integer 指定输入值 必须为整数。 值任何指定对输入值没有任何限制。 值为 Variable Name 指定此值必须满足 IBM SPSS Statistics 中有效变量名称的需求。 缺省值。 在运行时将新行添加到表时此列的缺省值（如果具有）。 分隔符类型。 在生成的语法中列的值之间指定定界符。 允许的分隔符是空格、逗号和加号 (+)。还可以输入 任意单个字符作为分隔符。 加上引号。 指定在生成的语法中是否使用双引号将列中每个值引起。 引号处理。 指定“加引号”属性为 true 时处理列的单元格输入的引号。 引号处理仅适用于单元格值中双引 号。 提供了以下类型的引号处理。 196 IBM SPSS Statistics 29 Core System 用户指南 语法 将对单元格值中双引号添加双引号。 例如，如果单元格值为 This "quoted" value，那么生成的 语法为 "This ""quoted"" value"。 python 单元格值中的双引号用反斜杠 () 转义。例如，如果单元格值是“This "quoted" value”，则生 成的语法为 “This \"quoted\" value”。 R 单元格值中的双引号用反斜杠 () 转义。例如，如果单元格值是“This "quoted" value”，则生 成的语法为 “This \"quoted\" value”。 无 单元格值中的双引号将保留不做修改。 宽度（字符）。 指定列的宽度（字符数）。 允许的值为正整数。 语法。 指定此列在运行时生成的命令语法 针对表整体生成的语法是一个表中每个列（从最左侧的列开始） 生成的语法的以空格分隔的列表。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定列的运行时值，它是列中值的列表，以指定分隔符分隔。 • 如果列的“语法”属性包含 %%ThisValue%%，并且列的运行时值为空，那么列不会生成任何语法 注: 您可以在“表列”对话框中现有列表底部的空行中，针对一个新表列添加一行。 输入标识以外的任何属性 会生成唯一的标识，您可以保留或修改此标识。 您可以通过单击“表”列的标识单元并按 Delete 键来删除 “表”列。 链接到控件 您可以将“表”控件链接至“字段选择器”控件。 将表控件链接到字段选择器时，针对字段选择器中每个字段， 表中会具有一行。 通过向“字段选择器”添加字段将行添加到表。 通过从字段选择器移除字段会从表删除 行。 例如，已链接的“表”控件可用于指定“字段选择器”中选中的字段的属性。 要启用链接，表必须具有包含“内容的变量名称”属性的列，并且画布上至少有一个字段选择器控件。 要将表控件链接到字段选择器，请从“表列”对话框的“链接到控件”组中的“可用控件”列表中指定字段选择 器。 然后，选择称为链接的列表列，该列定义链接。 在呈现表时，链接的列显示字段选择器中的当前字 段。 仅可以链接至多字段字段选择器。 项目组 项目组控件是包含其他控件的容器，使用此控件可以对多个控件生成的语法进行分组和控制。 例如，您有一 组用于为子命令指定可选设置的复选框，但希望只有在至少选中了一个复选框的情况下，才生成该子命令的 语法 这通过使用“项目组”控件作为复选框控件的容器来实现。 可以在项目组中包含以下类型的控件：字段 选择器、数据集选择器、复选框、组合框、列表框、文本控件、数字控件、日期控件、安全文本、静态文 本、颜色选取器、表控件、单选组和文件浏览器。。 “项目组”控件具有下列属性： 标识符。控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 这个组的可选标题。 对于多行标题或长标题，单击省略号 (...) 按钮，并在“标题属性”对话框中输入标 题。 语法。 指定命令语法，由此控件在运行时生成，可插入到语法模板中。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 可以将标识包含在“项组”中的任何控件中。 在运行时，这些标识将替换为控件所生成的语法 • 值 %%ThisValue%% 将生成项目组中各个控件所生成的语法的列表（以空格分隔），并按其相应控件在此 组中的显示顺序排列（从上到下）。 这是缺省值。 如果“语法”属性包含 %%ThisValue%%，并且此项目组 中的所有控件均未生成语法，那么项目组整体不生成任何命令语法 第 20 章 扩展 197 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 单选组 “单选按钮组”控件是一组单选按钮的容器，每个单选按钮又可包含一组嵌套控件。 单选按钮组控件具有下列 属性： 标识符。 控件的唯一标识。这是在语法模板中引用控件时要使用的标识。 标题。 适用于组的可选标题。 对于分布在多行的标题或长标题，单击省略号 (...) 按钮，并在“标题属性”对话 框中输入标题。 工具提示。当用户悬停在控件上时，显示的可选工具提示文本。 单选按钮。 单击省略号 (...) 按钮可打开 单选按钮组属性对话框，该对话框允许您指定单选按钮的属性，以 及添加或移除按钮组中的按钮。 在给定单选按钮下嵌套控件的功能是单选按钮的属性，在“单选按钮组属性” 对话框中进行设置。 请注意，还可通过双击画布上的“单选组”控件以打开“单选组属性”对话框。 语法。 指定由此控件在运行时生成且可以插入到语法模板中的命令语法。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定单选按钮组的运行时值，它是所选单选按钮的语法属性的值。 这是缺省值。 如 果“语法”属性包括 %%ThisValue%%，且所选单选按钮未生成语法，那么单选按钮组也不会生成任何命令 语法。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 定义单选按钮 使用“单选按钮组属性”对话框中可以指定一组单选按钮。 标识。 单选按钮的唯一标识。 列名。 在单选按钮旁边显示的名称。 名称为必需的字段。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 在名称中用作记忆键的可选字符。 指定字符必须存在于名称中。 嵌套组。 指定是否可以将其他控件嵌套在此单选按钮之下。 缺省为 False。 如果嵌套组属性设置为 True， 那么会在相关单选按钮下显示矩形放置区，并且可以嵌套和缩进。 可以在单选按钮下嵌套以下控件：字段选 择器、数据集选择器、复选框、组合框、列表框、文件控件、数字控件、日期控件、安全文本、静态文本、 颜色选取器、表控件和文件浏览器。 缺省。 指定此单选按钮是否为缺省选择。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 语法。 指定选择单选按钮时生成的命令语法。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 对于包含嵌套控件的单选按钮，值 %%ThisValue%% 将生成各个嵌套控件所生成的语法的列表（以空格分 隔），并按其相应控件在此单选按钮下的显示顺序排列（从上到下）。 可以在现有列表底部的空白行中添加新的单选按钮。 输入标识以外的任何属性会生成唯一的标识，您可以保 留或修改此标识。 通过单击单选按钮的标识单元格并按 Delete 键，可以删除该按钮。 198 IBM SPSS Statistics 29 Core System 用户指南 复选框组 “复选框组”控件是包含一组控件的容器，这些控件由单个复选框成组启用或禁用。 “复选框组”中可包含以下 类型的控件：字段选择器、数据集选择器、复选框、组合框、列表框、文本控件、数字控件、日期控件、安 全文本、静态文本、颜色选取器、表控件、单选组和文件浏览器。。。 复选框组控件具有下列属性： 标识符。控件的唯一标识。这是在语法模板中引用控件时要使用的标识。 标题。 这个组的可选标题。 对于多行标题或长标题，单击省略号 (...) 按钮，并在“标题属性”对话框中输入标 题。 复选框标题。 随控制复选框一起显示的可选标签。 支持使用 \n 来指定换行符。 工具提示。当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 缺省值。 这是控制复选框的缺省状态 - 选中或未选中。 选中/未选中语法。 指定在选中以及未选中此控件的情况下，生成的命令语法。 要将命令语法包括在语法模 板中，请使用“标识”属性的值。 生成的语法将插入到该标识的指定位置，而与它是根据“选中语法”属性还是 “未选中语法”属性生成无关。 例如，如果标识符为 checkboxgroup1，那么在运行时，如果选中该复选框， 则语法模板中的 %%checkboxgroup1%% 实例将被替换为选中的语法 属性的值；如果未选中该复选框，则 替换为未选中的语法 属性。 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 可以将标识包含在“复选框组”中的任何控件中。 在运行时，这些标识将替换为控件所生成的语法。 • 值 %%ThisValue%% 可以用于“选中语法”属性，也可以用于“未选中语法”属性。 它将生成复选框组中各个 控件所生成的语法的列表（以空格分隔），并按其相应控件在此组中的显示顺序排列（从上到下）。 • 缺省情况下，“选中语法”属性的值为 %%ThisValue%%，而“未选中语法”属性的值为空。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 文件浏览器 “文件浏览器”控件包含一个文件路径文本框和一个浏览器按钮，它可打开一个标准 IBM SPSS Statistics 对话 框以打开或保存文件。 “文件浏览器”控件具有下列属性： 标识符。控件的唯一标识。 这是在语法模板中引用控件时要使用的标识。 标题。 与控件一起显示的可选标题。 对于多行标题或长标题，单击省略符 (...) 按钮并在“标题属性”对话框中 输入标题。 标题位置。 指定标题相对于控件的位置。 值为“顶部”和“左侧”，其中“顶部”是缺省值。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 文件系统操作。 指定由浏览器按钮打开的对话框是用来打开文件还是保存文件。 值“打开”表示浏览对话框 将验证指定的文件是否存在。 如果值为“Save”，表示浏览器对话框不会验证指定文件是否存在。 浏览器类型。 指定浏览对话框是用于选择文件（查找文件）还是选择文件夹（查找文件夹）。 文件过滤器。 单击省略号 (...) 按钮以打开文件过滤对话框，您可以从中指定打开或保存对话框的可用文件类 型。 缺省情况下允许所有文件类型。 请注意，您也可以通过双击画布上的“文件浏览器”控件来打开“文件过 滤器”对话框。 文件系统类型。 在分布式分析方式下，此属性指定打开或保存对话框是浏览运行 IBM SPSS Statistics Server 的文件系统还是浏览本地计算机的文件系统。 选择服务器浏览服务器的文件系统，或选择客户端浏览 本地计算机的文件系统。 该属性对本地分析模式没有影响。 第 20 章 扩展 199 执行需要。 指定该控件中是否需要值，以便继续执行操作。 如果指定了 True，那么在为此控件指定值之 前，确定和粘贴按钮将处于禁用状态。 如果指定了 False，那么此控件中缺少值并不会对确定和粘贴按钮的 状态造成任何影响。缺省值为 False。 缺省值。 控件的缺省值。 语法。 指定由此控件在运行时生成且可以插入到语法模板中的命令语法 • 您可以指定过任何有效命令语法。 对于多行语法或长语法，请单击省略号 (...) 按钮，并在语法属性对话框 中输入语法。 • 值 %%ThisValue%% 指定文本框的运行时值，此值是文件路径，由用户手动指定或者由浏览对话框填写。 这是缺省值。 • 如果“语法”属性包含 %%ThisValue%%，并且文本框的运行时值为空，那么文件浏览器控件不会生成任何 命令语法 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 文件类型过滤器 使用“文件过滤器”对话框可以指定一些文件类型，这些文件类型将显示在通过“文件系统浏览器”控件访问的 打开及保存对话框的“文件类型”和“另存为类型”下拉列表中。 缺省情况下允许所有文件类型。 要指定未在对话框中显式列出的文件类型，请完成下列步骤： 1. 选择“其他”。 2. 输入文件类型名称。 3. 以 .后缀名形式输入文件类型，例如 .xls。 您可以指定多种文件类型，各种文件类型之间以分号分 隔。 选项卡 选项卡控件用于向对话框添加选项卡。 可以向新选项卡添加任何其他控件。 选项卡控件具有下列属性： 标识。 控件的唯一标识。 标题。 选项卡的标题。 位置。 指定此选项卡在对话框上的位置，此位置与对话框上的其他选项卡位置相对。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 注: “选项卡”控件在兼容性方式中不可用。 子对话框按钮 “子对话框按钮”指定用于打开子对话框的按钮，以便在对话框生成器中访问子对话框。 子对话框按钮控件具 有下列属性： 标识。 控件的唯一标识。 标题。 在按钮中显示的文本。 工具提示。 当用户悬停在控件上时，显示的可选工具提示文本。 子对话框。 单击省略号 (...) 按钮可打开子对话框的“自定义对话框构建程序”。 还可双击子对话框按钮，以打 开生成器。 助记键。 这是标题中的可选字符，用作控件的键盘快捷键。 这些字符在标题中显示时带有下划线。 按 Alt + [助记键] 可激活快捷方式。 在 Mac 上，不支持“助记键”属性。 启用规则。 指定用于确定何时启用当前控件的规则。 单击省略符 (...) 按钮以打开对话框并指定规则。 仅当 画布上存在可用于指定启用规则的控件时，才会显示“启用规则”属性。 注: “子对话框按钮”控件不能添加到子对话框。 200 IBM SPSS Statistics 29 Core System 用户指南 子对话框的对话框属性 查看和设置子对话框属性： 1. 在主对话框中双击子对话框按钮以打开子对话框，或单击子对话框按钮并单击省略号 (...) 按钮以显示子对 话框属性。 2. 在子对话框中，单击任何控件外部区域中的画布。 如果画布上没有控件，子对话框属性将始终可见。 子对话框名称。 子对话框的唯一标识。 “子对话框名称”为必需字段。 注: 如果您在语法模板（如在 %%My Sub-dialog Name%%）中将子对话框名称指定为标识，那么它将在运 行时被替换为子对话框中每个控件生成的语法的空格分隔列表，按顺序显示（从上到小，从左到右）。 标题。 指定在子对话框的标题栏中显示的文本。 “标题”属性为可选，但建议填入值。 帮助文件。 为子对话框指定可选帮助文件的路径。 这是用户单击子对话框中的帮助按钮时启动的文件，并 可以就是对主对话框指定的帮助文件。 帮助文件必须为 HTML 格式。 请参阅对话框属性中“帮助文件”属性 的描述以获取更多信息。 为控件指定“启用规则” 您可以指定用于确定何时启用控件的规则。 例如，您可以指定填充目标列表时启用单选按钮组。 用于指定 启用规则的可用选项取决于用于定义规则的控件的类型。 目标列表或字段选择器 您可以指定在使用至少一个字段（非空字段）填充目标列表或字段选择器时启用当前控件。 还可以指定 在未填充目标列表或字段选择器时启用当前控件。 复选框或复选框组 您可以指定在选中复选框或复选框组时启用当前控件。 还可以指定在未选中复选框或复选框组时启用当 前控件。 组合框或单选项列表框 您可以指定在组合框或单选列表框中选择特定值时启用当前控件。 还可以指定在组合框或单选列表框中 未选择特定值时启用当前控件。 多选列表框 您可以指定“多选列表框”中的选定值中包含某个特定值时启用当前控件。 还可以指定特定值不属于多选 列表框中选定值时启用当前控件。 单选按钮组 您可以指定在选择特定单选按钮时启用当前控件。 或者，您可以指定未选中某个特定单选按钮时启用当 前控件。 可以为其指定启用规则的控件具有关联“启用规则”属性。 注: • 不管是否启用了用于定义规则的控件，启用规则都适用。 例如，请考虑指定填充目标列表时启用单选按钮 组的规则。 只要填充目标列表，就会启用单选按钮组，无论是否启用目标列表都是如此。 • 禁用选项卡控件时，会禁用选项卡上的所有控件，不管这些控件中任何控件是否具有达到需求的启用规 则。 • 禁用复选框组时，就会禁用该组中的所有控件，不管是否已选中控制复选框。 扩展属性 “扩展属性”对话框指定用于扩展的自定义对话框构建程序中的当前扩展的相关信息，例如扩展的名称和扩展 中的文件。 注: “扩展属性”对话框在兼容性方式下不适用。 • 在用于扩展的自定义对话框构建程序中创建的所有定制对话框都是扩展的一部分。 • 必须先指定“扩展属性”对话框的“必需”选项卡上的字段，才能安装扩展和其中包含的定制对话框。 要指定扩展的属性，请从用于扩展的自定义对话框构建程序中的菜单选择： 第 20 章 扩展 201 扩展 > 属性 扩展的必需属性 名称 与扩展关联的唯一名称。 它可以包含最多三个单词，且不区分大小写。 字符限制为 7 位 ASCII。 为降低 名称冲突可能性，您可能需要使用多个单词的名称，其中首个单词包含您组织的标识（例如 URL）。 目录 关于扩展的简短描述，用于在单行中显示。 版本 x.x.x 格式的版本标识，其中，标识的每个组件必须是一个整数，例如，1.0.0。 如果未提供，那么将隐 含零。 例如，版本标识 3.1 暗示 3.1.0。 版本标识与 IBM SPSS Statistics 版本无关。 最低 SPSS Statistics 版本 运行扩展所需的最低 SPSS Statistics 版本 文件 “文件”列表显示扩展中当前包含的文件。 单击添加以将文件添加到扩展中。 还可以从扩展中除去文件， 并将文件抽取到指定文件夹中。 • 增强的定制对话框的文件类型为 .cfe，兼容的定制对话框的文件类型为 .spd。 每个扩展可包含多 个 .cfe 文件，但是只能包含一个 .spd 文件。 • 一个扩展必须至少包含扩展命令的一个定制对话框规范（.cfe 或 .spd）文件或一个 XML 规范文件。 如果包含 XML 规范文件，那么扩展必须至少包含一个 Python、R 或 Java 代码文件，具体来说 是 .py、.pyc、.pyo、.R、.class 或 .jar 类型的文件。 • 从“可选”选项卡上的“本地化”设置中添加扩展组件的翻译文件。 • 您可以在扩展中添加自述文件。 将文件名指定为 ReadMe.txt。 用户可以从显示扩展的详细信息的对 话框访问自述文件。 可以包含自述文件的本地化版本，指定为 ReadMe.txt，例如法语版本为 ReadMe_fr.txt。 扩展的可选属性 常规属性 描述 比为摘要字段提供的更详细的扩展描述。 例如，您可以列出扩展提供的主要功能。 日期 扩展的当前版本的可选日期。 未提供格式化。 作者 扩展的作者。 您可能希望包括一个电子邮件地址。 链接 要与扩展关联的一组 URL；例如，作者的主页。 此字段的格式不受限制，因此请确保使用空格、逗号或 其他有效的分隔符来分隔多个 URL。 关键词 与扩展关联的一组关键字。 平台 在特定操作系统平台上使用扩展时应用的任何限制的相关信息。 相关性 最高 SPSS Statistics 版本 可以在其之上运行扩展的 IBM SPSS Statistics 的最高版本。 所需的 Integration Plug-in for Python 指定是否需要 Integration Plug-in for Python。 • 如果 Python 实现代码指定为以 Python 3 运行，那么针对 Python 版本请选择 Python 3。 202 IBM SPSS Statistics 29 Core System 用户指南 • 如果扩展需要扩展中未明确包含的 Python 模块，那么在“必需的 Python 模块”控件中输入名称。 任何 此类模块都应添加到IBM SPSS 预测性分析社区 ( )。要添加第一个模块，请单击“必需的 Python 模块”控件中的任意位置以突出显示该输入字 段。 按 Enter 键（光标位于给定行中）将创建新行。 通过选择一行并按删除可删除该行。 扩展的用户负责下载任何所需的 Python 模块，并将其复制到针对扩展命令指定的位置，如 SHOW EXTPATHS 命令的输出中所示。 或者，可以将模块复制到 Python 搜索路径上的位置，例如，Python site-packages 目录。 需要 R 的集成插件 指定是否需要 R 的集成插件。 如果扩展需要来自 CRAN 程序包存储库的任何 R 程序包，那么在“所需 R 程序包”控件中输入这些程序包 的名称。 名称区分大小写。 要添加第一个程序包，在“所需 R 程序包”控件中单击任意位置以突出显示输 入字段。 按 Enter 键（光标位于给定行中）将创建新行。 通过选择一行并按删除可删除该行。 安装扩 展时，IBM SPSS Statistics 将检查是否安装了所需的 R 软件包，并尝试下载并安装缺少的任何软件包。 所需扩展 输入当前扩展所需的任何扩展的名称。 要添加第一个扩展，请单击“所需扩展”控件中的任意位置以突出 显示输入字段。 保持光标在特定行中，按 Enter 键将创建新行。 通过选择一行并按删除可删除该行。 扩展的用户负责安装任何所需的扩展。 注: 没有用于指定 Integration Plug-in for Java 的选项，因为它始终随 IBM SPSS Statistics 进行安装。 本地化 定制对话框 可以为扩展内的定制对话框添加属性文件（指定对话框中显示的所有字符串）的已转换版本。 要为特定 对话框添加翻译，请选中此对话框，单击添加翻译，并选择包含已翻译的版本的文件夹。 特定对话框的 所有已转换文件必须在同一文件夹中。 有关创建转换文件的指示信息，请参阅主题第207 页的『创建定 制对话框的本地化版本』。 转换目录文件夹 您可以包含含有转换目录的文件夹。 这允许您为扩展所包含扩展命令的实现程序（Python 或 R）提供本 地化消息和本地化输出。 还可以为最终用户从“扩展中心”查看扩展详细信息时显示的扩展的摘要和描述 字段提供本地化版本。 扩展的所有本地化文件集都必须在名为 lang 的文件夹中。 请浏览至包含本地化 文件的 lang 文件夹，然后选择该文件夹。 有关本地化 Python 和 R 程序输出的信息，请参阅帮助系统中关于 Integration Plug-in for Python 和 R 的 集成插件 的主题。 要提供摘要和描述字段的本地化版本，请为正在提供翻译的每种语言创建名为 .properties 的文件。 在运行时，如果找不到当前用户界面语言 的 .properties 文件，那么会使用“必需”和“可选”选项卡上指定的摘要与描述字段的值。 • 是扩展的名称字段的值，任何空格都替换为下划线字符。 • 是特定语言的标识。 IBM SPSS Statistics 支持的语言的标识如下所示。 例如，名为 MYORG MYSTAT 的扩展的法语转换存储在文件 MYORG_MYSTAT_fr.properties 中。 .properties 文件必须包含以下两行，它们指定两个字段的本地化文本： Summary= Description= • 关键字 Summary 和 Description 必须为英语，并且本地化文本必须与关键字在同一行上且没有换行 符。 • 文件必须采用 ISO 8859-1 编码形式。 无法在此编码中直接表示的字符必须带 Unicode 转义符 （“\u”）。 包含本地化文件的 lang 文件夹必须具有名为 的子文件夹，其中包含特定 语言的本地化 .properties 文件。 例如，法语 .properties 文件必须在 lang/fr 文件夹中。 语言标识 第 20 章 扩展 203 de。 德语 en。 英语 es。 西班牙语 fr。 法语 it。 意大利语 ja。 日语 ko。 韩国语 pl。 波兰语 pt_BR。 巴西葡萄牙语 ru。 俄语 zh_CN。 简体中文 zh_TW。 繁体中文 管理定制对话框 通过用于扩展的自定义对话框构建程序可以管理由您或其他用户在扩展中创建的定制对话框。 务必将定制对 话框安装在，只有这样才可以使用这些对话框。 注: 如果您在兼容性方式下工作，那么请参见主题第205 页的『在兼容性方式下管理定制对话框』。 打开包含定制对话框的扩展 您可以打开扩展束文件（.spe 或 .spxt），这些文件包含一个或多个定制对话框的规范，或者可以打开 已安装的扩展。 可以修改扩展中的任何增强型对话框，然后保存或安装扩展。 安装扩展也会安装扩展中包 含的对话框。 保存扩展也会保存已对扩展中任何对话框作出的更改。 要打开扩展束文件，请从用于扩展的自定义对话框构建程序中的菜单选择： 文件 > 打开 要打开已安装的扩展，请从“扩展的自定义对话框构建程序”中的菜单选择： 文件 > 打开已安装的 注: 如果打开已安装的扩展以对其进行修改，选择文件 > 安装将重新安装该扩展，并替换现有版本。 保存到扩展束文件 保存在用于扩展的自定义对话框构建程序中打开的扩展也会保存扩展中包含的定制对话框。 扩展将保存到扩 展束文件（.spe 或 .spxt）中。 在“扩展的定制对话框构建器”的菜单中，选择： 文件 > 保存 安装扩展 安装在用于扩展的自定义对话框构建程序中打开的扩展也会安装扩展中包含的定制对话框。 安装现有扩展将 会替换现有版本，包括替换已安装的扩展中的所有定制对话框。 要安装当前打开的扩展，请从“扩展的自定义对话框构建程序”中的菜单选择： 文件 > 安装 缺省情况下，扩展安装到操作系统的一般用户可写入的位置。 有关更多信息，请参见主题第179 页的『扩 展的安装位置』。 204 IBM SPSS Statistics 29 Core System 用户指南 卸载扩展 在“扩展的定制对话框构建器”的菜单中，选择： 文件 > 卸载 卸载扩展也会卸载扩展中包含的所有定制对话框。 您也可以从“扩展中心”卸载扩展。 将兼容定制对话框转换为增强型定制对话框 如果当前打开的扩展包含兼容定制对话框 (.spd) 文件，您可以将该兼容对话框转换为增强型对话框。 在扩 展中会保留定制对话框的原始兼容版本。 在“扩展的定制对话框构建器”的菜单中，选择： 扩展 > 转换兼容对话框 将定制对话框添加到扩展 可以将新的增强型定制对话框添加到扩展。 在“扩展的定制对话框构建器”的菜单中，选择： 扩展 > 新建对话框 在扩展中的多个定制对话框之间切换 如果当前扩展包含多个定制对话框，您可以在这些对话框之间切换。 在“扩展的定制对话框构建器”的菜单中，选择： 扩展 > 编辑对话框并选择要使用的定制对话框。 注: 不能在用于扩展的自定义对话框构建程序中编辑兼容定制对话框 (.spd) 文件。 要修改兼容性定制对话 框，您必须在兼容性方式中使用“自定义对话框构建程序”。 如果您没有兼容定制对话框的单独副本，那么可 使用“扩展属性”对话框进行抽取，该对话框可从用于扩展的定制对话框构建程序中的扩展 > 属性访问。 创建新扩展 在用于扩展的自定义对话框构建程序中创建新扩展时，系统会将新的空定制对话框添加到扩展。 要新建扩展，请从“扩展的自定义对话框构建程序”中的菜单选择： 文件 > 新建 在兼容性方式下管理定制对话框 在兼容性方式下，您可以创建并修改与 IBM SPSS Statistics 的所有发行版兼容的定制对话框，它们被称为兼 容定制对话框。 要以兼容性方式打开“定制对话框构建器”，请从菜单中选择： 扩展 > 实用程序 > 定制对话框构建器（兼容性方式）...
打开兼容定制对话框 可以打开包含兼容定制对话框规范的兼容定制对话框包 (.spd) 文件，也可以打开已安装的兼容定制对话 框。 要打开兼容定制对话框包文件，请从“自定义对话框构建程序”中的菜单选择： 文件 > 打开 要打开已安装的兼容定制对话框，请从“自定义对话框构建程序”中的菜单选择： 文件 > 打开已安装的 第 20 章 扩展 205 注: 如果打开已安装的对话框并对其进行修改，选择文件 > 安装将重新安装该对话框，并替换现有版本。 保存至兼容定制对话框包文件 您可以将兼容定制对话框的规范保存至外部文件。 规范将被保存到兼容定制对话框包 (.spd) 文件。 从定制对话框构建程序的菜单中选择： 文件 > 保存 将兼容定制对话框转换为增强型定制对话框 您可以将以兼容性方式打开的对话框转换为增强对话框，以利用可用于增强对话框的功能。 然后对话框的增 强版本会成为新扩展的一部分。 从定制对话框构建程序的菜单中选择： 文件 > 转换为增强型 注: 无法转换为增强对话框。 安装兼容定制对话框 可以安装在自定义对话框构建程序中打开的对话框，也可以从兼容定制对话框包 (.spd) 文件安装对话框。 重新安装现有对话框将替换已有版本。 要安装当前打开的对话框，请从自定义对话框构建程序中的菜单选择： 文件 > 安装 要从兼容定制对话框包文件安装，请从 IBM SPSS Statistics 菜单选择： 扩展 > 实用程序 > 安装定制对话框（兼容性方式）...
对于 Windows 7 和更高版本及 Mac，对话框将被安装至常规用户可写位置。 要查看定制对话框的当前安装 位置，请运行以下命令语法：SHOW EXTPATHS。 对于 Linux，缺省情况下，安装对话框需要对 IBM SPSS Statistics 安装目录具有写入权限。 如果对所需位置没有写入权限，或打算在其它位置保存安装的对话框，则可通过定义 SPSS_CDIALOGS_PATH 环境变量以指定一个或多个备用位置。 如果在 SPSS_CDIALOGS_PATH 中指定了路 径，则它优先于缺省位置。 定制对话框将被安装到第一个可写入位置。 注意， Mac 用户还可以使用 SPSS_CDIALOGS_PATH 环境变量。 如果要指定多个备用位置，对于 Windows 请使用分号分隔，对于 Linux 和 Mac，则使用冒号分隔。 指定的位置在目标机器上必须存在。 设置 SPSS_CDIALOGS_PATH 后，您必须重 新启动 IBM SPSS Statistics 以使更改生效。 要在 Windows 上创建 SPSS_CDIALOGS_PATH 环境变量，从控制面板中： Windows 7 1. 选择“用户帐户”。 2. 选择“更改我的环境变量”。 3. 单击新建，并在“变量名称”字段中输入 SPSS_CDIALOGS_PATH ，然后在“变量值”字段中输入路径。 Windows 8.1 和更高版本 1. 选择“系统”。 2. 选择可从“高级”系统设置访问的“高级”选项卡，然后单击环境变量。 3. 在“用户变量”部分，单击新建，并在“变量名称”字段中输入 SPSS_CDIALOGS_PATH，然后在“变量值”字 段中输入路径。 将兼容定制对话框作为扩展共享 您可以创建包含兼容定制对话框的扩展，使您能够和其他用户共享该对话框。 有关更多信息，请参阅第208 页的『创建和编辑扩展束』主题。 206 IBM SPSS Statistics 29 Core System 用户指南 卸载兼容定制对话框 从定制对话框构建程序的菜单中选择： 文件 > 卸载 新建兼容定制对话框 要新建兼容定制对话框，请从“自定义对话框构建程序”中的菜单选择： 文件 > 新建 扩展命令的自定义对话框 扩展命令是用户定义的 IBM SPSS Statistics 命令，采用 Python 编程语言、R 或 Java 进行实现。 部署到 IBM SPSS Statistics 实例之后，扩展命令的运行方式与任何内置的 IBM SPSS Statistics 命令相同。 可以使 用自定义对话框构建程序来创建用于扩展命令的对话框，方法是指定对话框的语法模板，使其生成用于扩展 命令的命令语法。 在用于扩展的自定义对话框构建程序中构建定制对话框时，您要将用于扩展命令的文件（指定扩展命令的语 法的 XML 文件和用 Python、R 或 Java 编写的实现文件）添加到包含该对话框的扩展。 在用于扩展的自定 义对话框构建程序中，您要从“扩展属性”对话框将文件添加到扩展，该对话框可从扩展 > 属性访问。 然后， 可以安装扩展或将其保存到扩展束（.spe 或 .spxt）文件，以便您可以将其与其他用户共享。 如果您在兼容性方式下工作，首先要将定制对话框保存到兼容定制对话框包 (.spd) 文件。 然后，您可以新 建扩展束，并将 .spd 文件和扩展命令的文件（指定扩展命令语法的 XML 文件和以 Python、R 或 Java 编写 的实施文件）添加至扩展束。 然后，可以安装扩展束或与其他用户共享。 请参阅主题第208 页的『创建和 编辑扩展束』，了解更多信息。 创建定制对话框的本地化版本 您可以为 IBM SPSS Statistics 支持的任何语言，创建定制对话框的本地化版本。 可以将定制对话框中显示 的任何字符串本地化，并可以将可选的帮助文件本地化。 要本地化对话框字符串 您必须为计划部署的每种语言创建与定制对话框相关联的属性文件的副本。 此属性文件包含所有与对话框相 关联的可本地化字符串。 从扩展中抽取定制对话框文件 (.cfe)，方法是在（用于扩展的自定义对话框构建程序中的）“扩展属性”对话 框中选择该文件，然后单击抽取。 然后，抽取 .cfe 文件的内容。 .cfe 文件就是 .zip 文件。 抽取 的 .cfe 文件内容包括每种受支持语言的属性文件，其中特定语言的文件是按照 .properties 的方式来命名（请参阅下表中遵循的语言标识符）。 注: 如果您在兼容性方式下工作，那么应该将定制对话框保存到外部 .spd 文件，然后抽取 .spd 文件的内 容。 .spd 文件就是 .zip 文件。 为计划部署的每种语言生成 .properties 文件的一个 副本，并将其重命名为 .properties。 请使用下面的表格 中的语言标识。 例如，如果对话框名称为 mydialog，而您打算创建日语版本的对话框，则本地化后的属性 文件必须命名为 mydialog_ja.properties。 1. 使用支持 UTF-8 的文本编辑器（例如 Windows 上的记事本或者 Mac 上的 TextEdit 应用程序打开您计划 翻译的每个属性文件。 修改需要本地化的任何属性的相关值，但不要更改属性名称。 与特定控件关联的 属性是以控件的标识为前缀。 例如，某个标识为 options_button 的控件的“工具提示”属性为 options_button_tooltip_LABEL。 标题属性是按照 _LABEL 的规则来简单地命名，例如 options_button_LABEL。 2. 从“扩展属性”对话框的“可选”选项卡上的“本地化”设置将属性文件的本地化版本重新添加到定制对话框文 件 (.cfe)。有关更多信息，请参阅主题 第202 页的『扩展的可选属性』。如果您正在使用 .spd 文件， 那么必须将本地化的属性文件手动添加回 .spd 文件中。 启动对话框时，IBM SPSS Statistics 将搜索语言标识与当前语言相匹配的属性文件，其中，当前语言由“选 项”对话框中“常规”选项卡上的“语言”下拉菜单指定。 如果找不到此类属性文件，则使用缺省文件 .properties。 第 20 章 扩展 207 对帮助文件进行本地化 1. 创建与定制对话框相关联的帮助文件的副本，然后针对所需语言将文本本地化。 2. 使用下表中的语言标识符，将副本重命名为 .htm。 例如， 如果帮助文件为 myhelp.htm，且您打算创建德语版本的文件，则本地化后的帮助文件应命名为 myhelp_de.htm。 将帮助文件的所有本地化版本存储在与非本地化版本相同的目录中。 当您从对话框属性的“帮助文件”属性添 加非本地化的帮助文件时，本地化版本会自动添加到对话框。 如果存在其它需要本地化的辅助文件，例如图像文件，则必须手动修改主帮助文件中的相应路径，以指向本 地化后的版本。 必须将包括本地化版本在内的辅助文件手动添加到定制对话框（.cfe 或 .spd）文件。 请 参见前面题为“本地化对话框字符串”的部分，了解关于访问和手动修改定制对话框文件的信息。 启动对话框时，IBM SPSS Statistics 将搜索语言标识与当前语言相匹配的帮助文件，其中，当前语言由“选 项”对话框中“常规”选项卡上的“语言”下拉菜单指定。 如果找不到这样的帮助文件，那么将使用对对话框指 定的帮助文件（对话框属性的“帮助文件”属性中指定的文件）。 语言标识 de。 德语 en。 英语 es。 西班牙语 fr。 法语 it。 意大利语 ja。 日语 ko。 韩国语 pl。 波兰语 pt_BR。 巴西葡萄牙语 ru。 俄语 zh_CN。 简体中文 zh_TW。 繁体中文 注意：定制对话框及相关帮助文件中的文本不限于 IBM SPSS Statistics 所支持的语言。 您可以采用任何语 言任意编写对话框和帮助文本，而不必创建特定于语言的属性文件和帮助文件。 随后，所有对话框用户都将 以该语言查看文本。 创建和编辑扩展束 您可以从“创建扩展束”或“编辑扩展束”对话框创建或编辑任何扩展束。 • 如果要创建或编辑包含增强的定制对话框的扩展束，那么使用“针对扩展的定制对话框构建器”可能更适 合。 使用“针对扩展的定制对话框构建器”，可以在创建或编辑包含对话框的扩展的同时创建或修改增强的 定制对话框。 并且可以从“针对扩展的定制对话框构建器”中编辑已安装的扩展。 • 如果扩展束不包含增强的定制对话框，那么必须使用“创建扩展束”或“编辑扩展束”对话框来创建或编辑扩 展束。 例如，如果要创建包含兼容的定制对话框包 (.spd) 文件并且不包含增强的对话框的扩展，或者创 建仅包含扩展命令的扩展，那么必须使用“创建扩展束”对话框。 创建扩展束 1. 从菜单中选择： 扩展 > 实用程序 > 创建扩展束...
2. 在“必需”选项卡上为所有字段输入值。 3. 在“可选”选项卡上，为您的扩展需要的字段输入值。 208 IBM SPSS Statistics 29 Core System 用户指南 4. 为扩展束指定目标文件。 5. 单击保存将扩展束保存到指定的位置。 此操作将关闭“创建扩展束”对话框。 编辑扩展束 1. 从菜单中选择： 扩展 > 实用程序 > 编辑扩展束...
2. 打开扩展束。 3. 修改“必需”选项卡上任何字段的值。 4. 修改“可选”选项卡上任何字段的值。 5. 为扩展束指定目标文件。 6. 单击保存将扩展束保存到指定的位置。 此操作将关闭“编辑扩展束”对话框。 有关“必需”选项卡和“可选”选项卡上的字段的详细信息，请参阅第202 页的『扩展的必需属性』和第202 页的『扩展的可选属性』主题。 第 20 章 扩展 209 210 IBM SPSS Statistics 29 Core System 用户指南 第 21 章 生产作业 生产作业以自动化方式运行 IBM SPSS Statistics。 该程序以无人看管方式运行，并且在最后一个命令运行之 后结束。 您也可以将生产作业安排在预定时间自动运行。 如果常常运行相同的一组耗时的分析（例如周报 告），则生产作业很有用。 您可以通过两种不同的方式来运行生产作业： 交互式。 该程序在本地计算机或远程服务器上的单独会话中以无人看管方式运行。 本地计算机必须保持打 开（并连接到远程服务器，如果适用），直到作业完成。 在服务器后台中。 该程序在远程服务器上的单独会话中运行。 本地计算机不必保持打开，也不必保持连接 到远程服务器。 您可以断开连接，稍后再检索结果。 注：从远程服务器上运行生产作业，需要对运行 IBM SPSS Statistics Server 的服务器拥有访问权限。 创建并运行生产作业 创建并运行生产作业： 1. 从菜单中选择： 实用程序 > 生产作业 2. 单击新建以创建新的生产作业。 或 3. 从列表中选择要运行或修改的生产作业。 单击浏览以更改列表中显示的文件的目录位置。 注：在 R16.0 之前的发行版中创建的生产工具文件 (.spp) 无法在 R16.0 或更高版本中运行。 现在提供了 一个转换实用程序，可将 Windows 和 Macintosh 生产工具作业转换成生产作业 (.spj)。 有关更多信息， 请参见主题第216 页的『转换生产工具文件』。 4. 指定要在作业中包含的一个或多个命令语法文件。 单击加号 (+) 图标以选择命令语法文件。 5. 选择输出文件名称、位置和格式。 6. 单击运行以交互方式运行生产作业或在服务器上后台运行。 缺省编码 缺省情况下，IBM SPSS Statistics 在 Unicode 方式下运行。 您可以在 Unicode 方式下或以当前语言环境编 码运行生产作业。 编码影响数据和语法文件的读取方式。 有关更多信息，请参见主题第163 页的『一般选 项』。 • Unicode (UTF-8)。 生产作业在 Unicode 方式下运行。 缺省情况下，文本数据文件和命令语法文件将作为 Unicode UTF-8 编码进行读取。 您可以在 GET DATA 命令中使用 ENCODING 子命令来指定文本数据文件 的代码页编码。 可以在 INCLUDE 或 INSERT 命令中使用 ENCODING 子命令来指定语法文件的代码页编 码。 – 语法文件的本地编码。 如果语法文件未包含 UTF-8 字节顺序标记，将以当前语言环境编码读取语法文 件。 此设置将覆盖对 INCLUDE 或 INSERT 指定的任何 ENCODING。 另外，它还忽略文件中的所有代码 页标识。 • 本地编码。 生产作业以当前语言环境编码运行。 除非在读取文本数据的命令（例如 GET DATA）中显式指 定了其他编码，否则将以当前语言环境编码读取文本数据文件。 包含 Unicode UTF-8 字节顺序标记的语法 文件将作为 Unicode UTF-8 编码进行读取。 所有其他语法文件将以当前语言环境编码进行读取。 语法文件 生产作业使用命令语法文件来告知 IBM SPSS Statistics 要执行的任务。 命令语法文件是包含命令语法的简 单文本文件。 可以使用语法编辑器或任何文本编辑器来创建文件。 您也可以通过将对话框选择粘贴到语法 窗口中来生成命令语法。 有关更多信息，请参阅第143 页的『第 14 章 使用命令语法』主题。 如果包括多个命令语法文件，那么它们将按其在列表中的显示顺序连接在一起，并作为单个作业来运行。 语法格式。 控制用于作业的语法规则的形式。 • 交互式。 每条命令必须以句点结尾。 句点可出现在命令中的任何位置，命令可延续多行，但作为某行最后 一个非空格字符的句点被解释为命令结尾。 接续行和新命令可在新行的任何位置开始。 当您在语法窗口中 选择和运行命令时，这些是生效的“交互式”规则。 • 批处理。 每个命令必须在新的一行开始（在命令开头没有空格），接续行必须缩进至少一个空格。 如果您 要缩进新命令，可以使用加号、短划线或句点作为行开头处的第一个字符，然后再缩进实际命令。 命令结 尾处的句点是可选的。 此设置与 INCLUDE 命令包括的命令文件的语法规则相兼容。 注：如果语法文件包含 GGRAPH 命令语法（包括 GPL 语句），那么请勿使用“批处理”选项。 GPL 语句仅 可在交互式规则下运行。 错误处理： 控制作业中错误条件的处理。 • 错误后继续处理。 作业中的错误不会自动停止命令处理。 生产作业文件中的命令被当作正常命令流的一部 分，命令处理以正常方式继续进行。 • 立即停止处理。 当在生产作业文件中遇到第一个错误时，命令处理即停止。 此设置与 INCLUDE 命令包括 的命令文件的行为相兼容。 输出： 这些选项控制生产作业结果的名称、位置和格式。 可用格式选项有： • 查看器文件 (.spv)。 结果以 IBM SPSS Statistics 查看器格式保存在指定的文件位置。 您可以存储到磁盘 或 IBM SPSS 协作和部署服务存储库。 存储到 IBM SPSS 协作和部署服务存储库 需要 Statistics Adapter。 • Word (docx)。 透视表导出为 Word 表，并带有所有完整的格式属性（例如，单元格边框、字体样式和背 景色）。 .docx 文件是 XML 文件的归档。 文本输出为 XML。 图表、树图和模型视图包含为高分辨率图 像（.eps（对于 macOS）和 .emf（对于 Windows）。 请注意，Microsoft Word 可能无法正确显示过 宽的表格。 • Excel。 透视表行、列和单元格导出为 Excel 行、列和单元格，并带有所有完整的格式属性（例如单元格 边框、字体样式和背景色）。 文本输出连同所有字体属性一起导出。 文本输出中的一行即为 Excel 文件中 的一行，文本输出中行的所有内容包含在单个单元格中。 图表、树形图和模型视图包括在 PNG 格式中。 输出可以作为 Excel 97-2004 或 Excel 2007 及更高版本导出。 • HTML (.htm)。 透视表导出为 HTML 表。 文本输出导出为预设置格式的 HTML。 图表、树形图和模型视 图都按所选图形格式嵌入文档中。 需要使用与 HTML 5 兼容的浏览器才能查看以 HTML 格式导出的输出。 • 便携文档格式 (.pdf)。 所有输出都将按?打印预览?中的显示导出，所有格式设置特性都不变。 • 文本。 文本输出格式包括纯文本、UTF-8 和 UTF-16。 透视表可以制表符分隔格式或空格分隔格式导出。 所有文本输出以空格分隔格式导出。 对于图表、树形图和模型视图，文本文件中将针对每个图片插入一个 相应的行，指示图像文件名称。 完成时打印查看器文件。 在生产作业完成时，将最终查看器输出文件发送到打印机。 当在远程服务器上后 台运行生产作业时，此选项不可用。 HTML 选项 表选项 HTML 格式无表格选项可用。 所有透视表均转换为 HTML 表。 图像选项 可用图像类型为：JPEG、PNG 和 BMP。 图像大小的缩放范围是从 1% 到 200%。 PowerPoint 选项 表格选项。 可以使用查看器略图项作为幻灯片标题。 每个幻灯片包含单个输出项。 标题是根据查看器概要 窗格中的项的概要条目构造而成。 图像选项。 图像大小的缩放范围是从 1% 到 200%。 （所有图像以 TIFF 格式导出到 PowerPoint。） 212 IBM SPSS Statistics 29 Core System 用户指南 注：PowerPoint 只在 Windows 操作系统上可用，且需要 PowerPoint 97 或更新版本。 PDF 选项 嵌入书签。 此选项提供 PDF 文档中对应于查看器概要条目的书签。 与查看器概要窗格一样，通过书签可以 更轻松地浏览具有大量输出对象的文档。 内嵌字体。 内嵌字体确保 PDF 文档在所有计算机上外观相同。 否则，如果文档中使用的一些字体在用于查 看（或打印）PDF 文档的计算机上不可用，则替换字体的效果可能欠佳。 文本选项 表格选项。 透视表可以制表符分隔格式或空格分隔格式导出。 对于空格分隔格式，您还可以控制： • 列宽。 自动适应不会对任何列内容换行，每列宽度为该列中最宽标签或值的宽度。 定制为表中的所有列设 置最大列宽，超过该宽度的值将换行，从而在该列的下一行中继续显示。 • 行/列边框字符。 控制用于创建行和列边框的字符。 如果您不希望显示行边框和列边框，请输入空格作为 值。 图像选项。 可用图像类型为：EPS、JPEG、TIFF、PNG 和 BMP。 在 Windows 操作系统上还可使用 EMF （增强型图元文件）格式。 图像大小的缩放范围是从 1% 到 200%。 使用 OUTPUT 命令的生产作业 生产作业使用 OUTPUT 命令，例如 OUTPUT SAVE、OUTPUT ACTIVATE 和 OUTPUT NEW。 生产作业期间 执行的 OUTPUT SAVE 命令将把指定输出文档的内容写入到指定位置。 这可作为生产作业创建的输出文件的 补充。 使用 OUTPUT NEW 创建新输出文档时，建议使用 OUTPUT SAVE 命令显式保存文档。 生产作业输出文件由作业结束时的活动输出文档的内容组成。 对于包含 OUTPUT 命令的作业，输出文件可能 不包含会话中创建的所有输出。 例如，假设生产作业包括一些过程，后接 OUTPUT NEW 命令，然后是更多 过程但没有 OUTPUT 命令。 OUTPUT NEW 命令会定义一个新的活动输出文档。 生产作业结束时，将仅包含 OUTPUT NEW 命令后执行的过程的输出。 运行时值 在生产作业文件中定义、在命令语法文件中使用的运行时值可简化一些任务，如对不同数据文件执行相同分 析或对不同变量集运行相同的命令集等任务。 例如，您可以定义运行时值 @datafile，以便在每次运行使用 字符串 @datafile 的生产作业时，提示您输入数据文件名来代替命令语法文件中的文件名。 • 运行时值替换使用宏设施 (DEFINE-!ENDDEFINE) 用于创建字符串替换值。 • 命令语法文件中括在引号内的运行时值将被忽略。 如果需要将运行时值括在引号内，请选择将值括在引号 内。 如果运行时值只是引用字符串的一部分，那么可以将运行时值包含在具有 的宏中！取消引用 和 ！ EVAL 参数。 符号。 在命令语法文件中使用的字符串，可触发生产作业提示用户输入值。 所有符号名称必须以 @ 开头且 必须符合变量命名规则。 有关更多信息，请参阅第44 页的『变量名称』主题。 缺省值。 在您不输入其他值的情况下，生产作业缺省提供的值。 当生产作业提示您输入信息时会显示该 值。 您可以在运行时替换或修改该值。 如果未提供缺省值，则在使用命令行开关运行生产作业时，不要使 用 silent 关键字，除非另外使用了 -symbol 开关来指定运行时值。 有关更多信息，请参阅第215 页的 『从命令行运行生产作业』主题。 用户提示。 在生产作业提示您输入信息时显示的说明性标签。 例如，您可以使用短语“使用什么数据文 件？”来标识一个需要填写数据文件名的字段。 将值括在引号内。 使用引号将缺省值或用户输入的值括起来。 例如，文件规范必须以引号括起。 包含用户提示符的命令语法文件 GET FILE @datafile. /check the Quote value option to quote file specifications.
FREQUENCIES VARIABLES=@varlist. /do not check the Quote value option 第 21 章 生产作业 213 使用宏替换字符串值的一部分 如果整个替换字符串都括在引号内，您可以使用将值括在引号内选项。 如果替换字符串只是引用字符串的一 部分，那么可以在宏中包含运行时值，使用 ！取消引用 和 ！EVAL 函数。 DEFINE !LabelSub() VARIABLE LABELS Var1 !QUOTE(!concat(!UNQUOTE('First part of label - '), !UNQUOTE(!EVAL(@replace)), !UNQUOTE(' - rest of label'))).
!ENDDEFINE.
!LabelSub.
运行选项 您可以通过两种不同的方式来运行生产作业： 交互式。 该程序在本地计算机或远程服务器上的单独会话中以无人看管方式运行。 本地计算机必须保持打 开（并连接到远程服务器，如果适用），直到作业完成。 在服务器后台中。 该程序在远程服务器上的单独会话中运行。 本地计算机不必保持打开，也不必保持连接 到远程服务器。 您可以断开连接，稍后再检索结果。 注：从远程服务器上运行生产作业，需要对运行 IBM SPSS Statistics Server 的服务器拥有访问权限。 Statistics 服务器。 如果选择在远程服务器上后台运行生产作业，则必须指定其运行所在的服务器。 单击选 择服务器以指定服务器。 这仅适用于在远程服务器上后台运行的作业，而不适用于在远程服务器上交互式运 行的作业。 服务器登录 使用“服务器登录”对话框以添加或修改远程服务器，并选择要用于运行当前生产作业的服务器。 远程服务器 通常需要用户标识和密码，可能还需要域名。 请联系系统管理员以获取关于可用服务器、用户标识和密码、 域名的信息以及其他连接信息。 如果站点正在运行 IBM SPSS 协作和部署服务 3.5 或更新版本，则可单击搜索....以查看网络上可用的服务器 列表。 如果尚未登录到 IBM SPSS 协作和部署服务存储库，那么您将被提示输入连接信息，然后才能查看服 务器列表。 添加或编辑服务器登录设置 使用“服务器登录设置”对话框可以添加或编辑远程服务器的连接信息，以便在分布式分析模式下使用。 请联系系统管理员获取可用服务器列表、服务器端口号以及其他连接信息。 除非有管理员指示，否则请勿使 用“安全套接字层”。 服务器名称。 服务器“名称”可以是分配给计算机的字母数字名称（例如 NetworkServer），也可以是分配给 计算机的 IP 地址（例如 202.123.456.78）。 端口号。 端口号是服务器软件用于通信的端口。 描述。 可以输入可选的描述以显示在服务器列表中。 使用安全套接字层连接。 安全套接字层 (SSL) 在分布式分析请求发送到远程服务器时加密请求。 使用 SSL 之 前，请与管理员协商。 要启用此选项，必须在桌面计算机和服务器上配置 SSL。 用户提示 只要您运行包含已定义运行时间符号的生产作业，生产作业就会提示您提供值。 可以替换或修改显示的缺省 值。 然后这些值将替换与生产作业相关联的所有命令语法文件中的运行时间符号。 后台作业状态 “后台作业状态”选项卡显示已被提交到在远程服务器上后台运行的生产作业状态。 214 IBM SPSS Statistics 29 Core System 用户指南 服务器名称。 显示当前选定的远程服务器的名称。 在列表中仅显示被提交给此服务器的作业。 要显示被提 交给其他服务器的作业，单击选择服务器。 作业状态信息。 包括生产作业名称、当前作业状态以及开始与结束时间。 刷新。 更新作业状态信息。 获取作业输出。 从选定的生产作业检索输出。 每个作业的输出位于作业运行所在的服务器上；因此您必须 切换到该服务器的状态，以便选择作业并检索输出。 如果作业状态为运行，将禁用此按钮。 取消作业。 取消选定的生产作业。 此按钮仅在作业状态为运行时才可用。 删除作业。 删除选定的生产作业。 这将从列表中删除作业，并从远程服务器上删除相关文件。 如果作业状 态为运行，将禁用此按钮。 注：后台作业状态不反映那些在远程服务器上交互式运行的作业的状态。 从命令行运行生产作业 命令行开关使您能够使用操作系统上提供的调度实用程序将生产作业调度为在特定时间自动运行。 命令行参 数的基本格式为: Windows: stats --production filename.spj MacOS： ./stats --production filename.spj 根据调用生产作业的方式，可能需要包含应用程序的可执行文件 (位于安装应用程序的目录中) 和/或生产作 业文件的目录路径。 注: 您无法在 IBM SPSS Statistics Subscription 应用程序运行时运行命令行生产作业。 可以使用以下开关从命令行运行生产作业： -production [prompt|silent]. 在生产模式下启动应用程序。 prompt 和 silent 关键字指定是否显示对话 框，以提示输入在作业中指定的运行时值。 prompt 关键字为缺省值，即显示对话框。 silent 关键字不显示 对话框。 如果您使用 silent 关键字，那么可使用 -symbol 开关定义运行时间符号。 否则，将使用缺省 值。 在使用 -production 开关时，会忽略 -switchserver 和 -singleseat 开关。 -symbol . 生产作业中使用的符号/值对的列表。 每个符号名均以 @ 开头。 包含空格的值必须以引 号括起。 对于不同的操作系统，在字符串字面值中包括引号或撇号的规则有所不同，但将包含单引号或撇号 的字符串括在双引号内的做法通常可行（例如，"'a quoted value'"）。 这些符号必须使用“运行时值” 选项卡在生产作业中定义。 请参阅主题第213 页的『运行时值』，了解更多信息。 -background。 在远程服务器上后台运行生产作业。 本地计算机不必保持打开，也不必保持连接到远程服 务器。 您可以断开连接，稍后再检索结果。 您还必须包括 -production 开关并使用 -server 开关指定服 务器。 要从远程服务器上运行生产作业，还需要指定服务器登录信息： -server 或 -server 。服务器的名称或 IP 地址和端口号。 仅 Windows。 -user 。有效用户名。 如果需要域名，请在用户名前面加上域名和反斜杠 ()。仅限 Windows。 -password 。用户的密码。 示例 Windows: stats --production /Users/Simon/job.spj silent --symbol @sex male 第 21 章 生产作业 215 MacOS： ./stats --production /Users/Simon/job.spj silent --symbol @sex male • 本例假设您从安装目录运行命令行，因此无需为 IBM SPSS Statistics Subscription 可执行文件指定路径。 缺省安装路径如下所示： Windows：$InstalledPath$ MacOS: /Applications/IBM SPSS Statistics/SPSS Statistics.app/Contents/MacOS • 生产作业位置的目录路径使用 Windows 反斜杠约定。 在 Macintosh 和 Linux 上，那么使用正斜杠。 在引 用数据文件规范中的正斜杠可适用于所有操作系统，因为该加引号的字符串会插入命令语法文件中。在所 有操作系统上，对于包含文件规范的命令（例如，GET FILE、GET DATA、SAVE），正斜杠是可以接受 的。 • silent 关键字禁止生产作业中的任何用户提示，并且 --symbol 开关将插入提供的字符串 ("男性") ，只 要运行时符号 @sex 出现在生产作业中包含的命令语法文件中。 转换生产工具文件 用低于 16.0 的版本创建的生产工具作业文件 (.spp) 在 16.0 或更高的版本中不能运行。 对于在之前版本中创 建的 Windows 和 Macintosh 生产工具作业文件，可以使用安装目录下的 prodconvert 将其转换成新的生产 作业文件 (.spj)。 使用以下规格从命令窗口中运行 prodconvert： [installpath]\prodconvert [filepath]\filename.spp 其中 [installpath] 是安装 IBM SPSS Statistics 的文件夹位置，[filepath] 是原始生产作业文件所在的文件夹。 在与原始文件相同的文件夹下，将创建具有相同名称的新文件，但其扩展名为 .spj。 （注：如果路径包含空 格，请使用双引号将每个路径和文件规范括起来。 在 Macintosh 操作系统上，使用正斜杠而非反斜杠。） 限制 • 不支持 WMF 和 EMF 图表格式。 而使用 PNG 格式来替代这些格式。 • 不支持导出选项输出文档（无图表）、仅图表和无。 包括选定格式支持的所有输出对象。 • 忽略远程服务器设置。 要为分布式分析指定远程服务器设置，需要从命令行运行生产作业，并通过命令行 开关指定服务器设置。 有关更多信息，请参阅第215 页的『从命令行运行生产作业』主题。 216 IBM SPSS Statistics 29 Core System 用户指南 第 22 章 输出管理系统 输出管理系统 (OMS) 能够用不同的格式将选中的输出类别自动写入不同的输出文件。 格式包括：Word、 Excel、PDF、IBM SPSS Statistics 数据文件格式 (.sav)、查看器文件格式 (.spv)、XML、HTML 和文本。 请 参阅主题第219 页的『OMS：选项』，了解更多信息。 使用输出管理系统控制面板 1. 从菜单中选择： 实用程序 > OMS 控制面板...
可以使用该控制面板启动和停止输出到不同目标的路由。 • 每个 OMS 请求将保持活动状态，直到明确停止或直到会话结束为止。 • 在 OMS 请求中指定的目标文件对其他过程和其他应用程序不可用，直到该 OMS 请求终止为止。 • 如果一个 OMS 请求处于活动状态，那么指定的目标文件存储在内存 (RAM) 中，因此将大量输出写入外部 文件的活动 OMS 请求可能要消耗大量内存。 • 多个 OMS 请求是彼此独立的。 相同的输出可以不同格式转到不同位置，这取决于不同 OMS 请求的指定。 • 任何特定目标中的输出对象的顺序都是创建这些对象的顺序，这是由生成输出的过程的顺序和操作确定 的。 限制 • 对于输出 XML 格式，有关标题输出类型的指定不起任何作用。 如果包含某个过程的任何输出，那么应包 含过程标题输出。 • 如果 OMS 指定导致对于过程只包含标题对象或注释表，那么不会为此过程包含任何内容。 添加新 OMS 请求 1. 选择要包括的输出类型（表、图形等等）。 请参阅第218 页的『输出对象类型』主题以获取更多信息。 2. 选择要包括的命令。 如果要包括所有输出，请选择列表中的所有项。 请参阅第218 页的『命令标识和表 子类型』主题以获取更多信息。 3. 对于生成透视表输出的命令，选择要包括的特定表类型。 列表仅显示在选定命令中可用的表；在一个或多个选定命令中可用的任何表格类型都显示在列表中。 如 果没有选择任何命令，那么显示所有表格类型。 请参阅第218 页的『命令标识和表子类型』主题以获取 更多信息。 4. 要根据文本标签而不是子类型选择表，请单击标签。 请参阅第219 页的『标签』主题以获取更多信息。 5. 单击选项指定输出格式（例如，IBM SPSS Statistics 数据文件、XML 或 HTML）。 缺省情况下将使用输 出 XML 格式。 请参阅主题第219 页的『OMS：选项』，了解更多信息。 6. 指定输出目标： • 文件。 所有选定的输出将转到单个文件。 • 根据对象名称。 输出将根据对象名称转到多个目标文件。 将为每个输出对象分别创建一个文件，其文 件名基于表子类型名称或表标签。 输入目标文件夹名称。 • 新数据集。 对于 IBM SPSS Statistics 数据文件格式输出，可以将输出转到数据集。 可在相同的会话中 继续使用数据集，但不会保存它，除非您在会话结束之前明确将其保存为文件。 该选项仅适用于 IBM SPSS Statistics 数据文件格式输出。 数据集名称必须符合变量命名规则。 有关更多信息，请参阅第44 页的『变量名称』主题。 7. 选择性地执行下列操作： • 从查看器排除选定的输出。 如果选择排除在查看器之外，那么不会在“查看器”窗口中显示 OMS 请求的输 出类型。 如果多个活动的 OMS 请求包括相同的输出类型，那么这些输出类型在查看器中的显示由包含这 些输出类型的最近的 OMS 请求确定。 请参阅第222 页的『从查看器排除输出显示』主题以获取更多信 息。 • 将 ID 字符串分配到请求。 所有请求都自动分配有一个 ID 值，您可以使用描述性 ID 覆盖系统缺省的 ID 字 符串，如果您有多个活动请求并且想轻松标识它们，那么此方法很有用。 分配的 ID 值不能以美元符号 ($) 开头。 选择列表中的多个项的提示 下列提示适用于选择列表中的多个项： • 按 Ctrl 和 A 可选择列表中的所有项。 • 按住 Shift 并单击可选择多个连续的项。 • 按住 Ctrl 并单击可选择多个不连续的项。 终止和删除 OMS 请求 “请求”列表中显示活动的和新的 OMS 请求，最新的请求显示在顶部。 您可以通过单击并拖动边界来更改信 息列的宽度，可以水平滚动列表来查看有关特定请求的更多信息。 状态列中的词Active 后面的星号 () 指示使用命令语法创建的 OMS 请求，其中包含在“控制面板”中不可用 的功能部件。 终止特定的活动 OMS 请求： 1. 在“请求”列表中，单击请求的行中的任意单元格。 2. 单击终止。 终止所有的活动 OMS 请求： 1. 单击全部终止。 删除新请求（已添加但尚未成为活动状态的请求）： 1. 在“请求”列表中，单击请求的行中的任意单元格。 2. 单击删除。 注：单击确定后，活动 OMS 请求才会终止。 输出对象类型 有不同类型的输出对象： 图表。 这包括使用图表构建器创建的图表、图表过程以及通过统计过程创建的图表（例如，通过频率过程创 建的条形图）。 标题。 在查看器的概要窗格中标记为标题的文本对象。 日志。 日志文本对象。 日志对象包含特定类型的错误和警告消息。 根据“选项”设置（“编辑”菜单，“选 项”，“查看器”选项卡），日志对象还可以包含在会话期间执行的命令语法。 日志对象在查看器的概要窗格 中标记为日志。 模型。 模型查看器中显示的输出对象。 单个模型对象可以包含多个模型视图，包括表格和图表。 表。 在查看器中作为透视表的输出对象（包括“注释”表）。 表是可以转到 IBM SPSS Statistics 数据文件 (.sav) 格式的唯一输出对象。 文本。 不是日志也不是标题的文本对象（包括在查看器的概要窗格中标记为文本输出的对象）。 树。 由“决策树”选项生成的树模型图。 警告。 警告对象包含特定类型的错误和警告消息。 命令标识和表子类型 命令标识 命令标识可用于所有统计和绘图过程，以及在查看器的概要窗格中生成包含自己的可识别标题的输出区的任 何其他命令。 这些标识通常（但不总是）与菜单上的过程名称和对话框标题相同或类似，这些名称和标题通 218 IBM SPSS Statistics 29 Core System 用户指南 常（但不总是）与基础命令名称相同。 例如，“频率”过程的命令标识为“Frequencies”，基础命令名称也相 同。 但是，有时候过程名称和命令标识和/或命令名称并不相同。 例如，“非参数检验”子菜单（位于“分析”菜单 中）上的所有过程使用相同的基础命令，并且命令标识与基础命令名称 Npar Tests 相同。 表子类型 表子类型是可以生成的透视表的不同类型。 某些子类型仅由一个命令生成；其他子类型可由多个命令生成 （尽管表看上去并不相同）。 虽然表子类型名称通常是描述性的，但也可以选择其他名称（尤其是选择了大 量命令时）；并且，两个子类型的名称可以非常类似。 查找命令标识和表子类型 如果不确定，可以在“查看器”窗口中查找命令标识和表子类型名称： 1. 运行过程以在查看器中生成一些输出。 2. 右键单击查看器概要窗格中的项。 3. 选择复制 OMS 命令标识或复制 OMS 表子类型。 4. 将复制的命令标识或表子类型名称粘贴到任意文本编辑器中（如语法编辑器窗口）。 标签 作为表子类型名称的替代项，主题以获取更多信息。）您可以根据“查看器”的概要窗格中显示的文本来选择 表。 还可以根据其他对象类型的标签来选择它们。 标签可以用于区分同一类型的多个表，在这些表中，概 要文本反映特定输出对象（如变量名称或标签）的某个属性。 但是有一些因素会影响标签文本： • 如果正在处理拆分文件，那么拆分文件组标识将会附加到标签上。 • 包含有关变量或值的信息的标签受当前输出标签选项设置（“编辑”菜单，“选项”，“输出标签”选项卡）的 影响。 • 标签受当前输出语言设置（“编辑”菜单，“选项”，“常规”选项卡）的影响。 指定用于标识输出对象的标签 1. 在输出管理系统控制面板中，选择一个或多个输出类型，然后选择一个或多个命令。 2. 单击标签。 3. 输入与显示在“查看器”窗口的概要窗格中的标签完全相同的标签。 （也可以右键单击概要中的项，选择 复制 OMS 标签，然后将复制的标签粘贴到“标签”文本字段中。） 4. 单击添加。 5. 对要包括的每个标签重复该过程。 6. 单击继续。 通配符 可以使用星号 () 作为通配符字符，用作标签字符串的最后一个字符。 将选择以指定字符串（星号除外）开 头的所有标签。 此过程只在星号是最后一个字符时才起作用，因为星号可显示为标签中的有效字符。 OMS：选项 您可以使用“OMS: 选项”对话框来： • 指定输出格式。 • 指定图像格式（对于 HTML 和输出 XML 输出格式）。 • 指定哪些表维度元素应转到行维度中。 • 对于 IBM SPSS Statistics 数据文件格式，将包括标识作为每个个案的源的表序列号的变量。 指定 OMS 选项 1. 单击输出管理系统控制面板中的选项。 第 22 章 输出管理系统 219 格式 Excel Excel 97-2004 和 Excel 2007 及更高版本的格式。透视表行、列和单元格将导出为 Excel 行、列和单元 格，所有格式属性保持不变，例如，单元格边框、字体样式和背景色。 文本输出连同所有字体属性一起 导出。 文本输出中的一行即为 Excel 文件中的一行，文本输出中行的所有内容包含在单个单元格中。 图 表、树形图和模型视图包括在 PNG 格式中。 HTML 将在查看器中作为透视表的输出对象被转换为 HTML 表。 文本输出对象在 HTML 中标记为 。 图 表、树形图和模型视图都按所选格式嵌入文档中。 输出 XML 符合 spss 输出架构的 XML。 PDF 输出将按“打印预览”中的显示导出，所有格式设置特性都不变。 PDF 文件包含对应于查看器概要窗格中 条目的书签。 IBM SPSS Statistics 数据文件 此格式是二进制文件格式。 将排除表以外的所有输出对象类型。 表的每列成为数据文件中的一个变量。 要使用在相同会话中通过 OMS 创建的数据文件，必须先终止活动的 OMS 请求，然后才能打开数据文 件。请参阅第222 页的『将输出转到 IBM SPSS Statistics 数据文件』主题以获取更多信息。 文本 空格分隔的文本。 输出写为文本，具有与固定间距字体的间距对齐的表格输出。排除图表、树形图和模 型视图。 制表符分隔的文本 制表符分隔的文本。 对于在查看器中显示为透视表的输出，制表符分隔表列元素。 文本区线按原样写 出；不尝试在有用的位置使用制表符分隔它们。排除图表、树形图和模型视图。 查看器文件 这与保存“查看器”窗口内容时使用的格式相同。 词 透视表和文本将导出为 Word XML (.docx)，其中包含完整的格式属性（例如，单元格边框、字体样式 和背景色）。 图表、树图和模型视图将以高分辨率图像（.eps for macOS 和 .emf for Windows）的 形式包括在其中。 图形图像 对于 HTML 和输出 XML 格式，可以包括图表、树形图和模型视图作为图像文件。 为每个图表和/或树创建单 独的图像文件。 • 对于 HTML 文档格式，将为每一个图像文件，在 HTML 文档中包括标准的 标 记。 • 对于输出 XML 文档格式，XML 文件将为每个图像文件包含一个 chart 元素，其中包括指示了通用表单的 ImageFile 属性：。 • 图像文件保存在单独的子目录（文件夹）中。 子目录的名称是目标文件的名称，没有任何扩展名，在末尾 附加有 files。 例如，如果目标文件是 julydata.htm，则图像子目录的名称将是 julydata_files。 格式 可用的图像格式有 PNG、JPG 和 BMP。 大小 图像大小的缩放范围是从 10% 到 200%。 包含图像地图 对于 HTML 文档格式，该选项可创建图像地图工具提示，它显示某些图表元素例如线图上选定点或条形 图上选定条的值信息。 220 IBM SPSS Statistics 29 Core System 用户指南 透视表 对于透视表输出，可以指定应显示在列中的维度元素。 所有其他维度元素显示在行中。 对于 IBM SPSS Statistics 数据文件格式，表列成为变量，行成为个案。 • 如果指定列的多个维度元素，它们将以列出的顺序嵌套在列中。 对于 IBM SPSS Statistics 数据文件格式， 变量名称由内嵌列元素构成。 请参阅第223 页的『OMS 生成的数据文件中的变量名称』主题以获取更多 信息。 • 如果表不包含任何列出的维度元素，则该表的所有维度元素将显示在行中。 • 在此处指定的表旋转对显示在查看器中的表没有影响。 表的每个维度 -- 行、列、层 -- 可包含零个元素或多个元素。 例如，简单二维交叉制表包含一个行维度元素 和一个列维度元素，每个元素包含表中使用的一个变量。 您可以使用位置参数或维度元素“名称”来指定要放 在列维度中的维度元素。 行中的所有维度 为每个表创建一行。 对于 IBM SPSS Statistics 格式数据文件，这意味着每个表是一个个案，所有表元素 都是变量。 位置列表 位置参数的一般形式是一个字母，表示元素的缺省位置，即 C 表示列，R 表示行，L 表示层；后面是一 个正整数，表示该维度中的缺省位置。 例如，R1 表示最外面的行维度元素。 • 要指定多个维度中的多个元素，请使用空格分隔每个维度，例如 R1 C2。 • 跟有 ALL 的维度字母表示该维度中按照缺省顺序排列的所有元素。 例如，CALL 与缺省行为相同（使 用按照缺省顺序排列的所有列元素来创建列）。 • CALL RALL LALL（或 RALL CALL LALL 等等）将把所有维度元素放在列中。 对于 IBM SPSS Statistics 数据文件格式，此操作将在数据文件的每个表中创建一行/个案。 维数名称列表 作为位置参数的替代选择，可以使用维度元素“名称”，这些名称是显示在表中的文本标签。 例如，简单 二维交叉制表包含单个行维度元素和单个列维度元素，每个元素都具有基于维度中的变量的标签，加上 标记为 Statistics 的单个层维度元素（如果英语是输出语言）。 • 维度元素名称可能不同，这取决于影响表中变量名称和/或标签的显示的输出语言和/或设置。 • 每个维度元素名称必须用单引号或双引号括起来。 要指定多个维度元素名称，请在每个用引号括起的 名称之间加一个空格。 与维度元素关联的标签不总是明显的。 查看透视表的所有维度元素及其标签 1. 激活（双击）查看器中的表。 2. 从菜单中选择： 视图 > 全部显示 或 3. 如果未显示透视托盘，请从菜单中选择： 透视 > 透视托盘 元素标签显示在透视托盘中。 日志记录 您可以将 OMS 活动以 XML 或文本格式记录在日志中。 • 该日志将跟踪会话的所有新 OMS 请求，但不包括在请求记录前已激活的 OMS 请求。 • 如果指定新的日志文件或者取消选择（取消选中）记录 OMS 活动，那么当前日志文件将结束。 如何指定 OMS 日志记录 第 22 章 输出管理系统 221 指定 OMS 日志记录： 1. 单击输出管理系统控制面板中的日志记录。 从查看器排除输出显示 通过禁止在“查看器”窗口中显示 OMS 请求中选择的所有输出，排除在查看器之外复选框将影响这些输出。 对于生成大量输出的生产作业，以及当您不需要查看器文档形式（.spv 文件）的结果时，此过程通常很有 用。 还可以使用此功能取消显示不想看到的特定输出对象，而不会将任何其他输出转到某些外部文件和格 式。 取消显示某些输出对象而不将其他输出转到外部文件： 1. 创建标识不需要的输出的 OMS 请求。 2. 选择排除在查看器之外。 3. 对于输出目标，选择文件，但将“文件”字段留空。 4. 单击添加。 选定的输出将从查看器中排除，而所有其他输出将以正常形式显示在查看器中。 注：该设置对保存到外部格式或文件（包括查看器 SPV 格式）的 OMS 输出没有影响。 它对在通过批处理工 具（随 IBM SPSS Statistics 服务器提供）执行的批处理作业中保存到 SPV 格式的输出也没有影响。 将输出转到 IBM SPSS Statistics 数据文件 IBM SPSS Statistics 格式的数据文件由列中的变量和行中的个案组成，这是透视表转换为数据文件所用的基 本格式： • 表中的列是数据文件中的变量。 有效变量名称从列标签构成。 • 表中的行标签成为数据文件中具有一般变量名称的变量（Var1、Var2、Var3，等等）。 这些变量的值是 表中的行标签。 • 三个表标识变量将自动包含在数据文件中：Command、Subtype_ 和 Label_。这三个变量都是字符串变 量。 前两个变量与命令和子类型标识对应。 请参阅主题第218 页的『命令标识和表子类型』，了解更多 信息。 Label_ 包含表标题文本。 • 表中的行成为数据文件中的个案。 从多个表创建的数据文件 将多个表转到相同的数据文件时，每个表以类似于合并数据文件的方式添加到数据文件，方法是将一个数据 文件的个案添加到另一个数据文件（“数据”菜单，“合并文件”，“添加个案”）。 • 每个后续的表将始终向数据文件添加个案。 • 如果表中的列标签不同，每个表也可以将变量添加到数据文件，对于没有相同列标签的其他表包含个案缺 失值。 • 如果表的行元素的数量互不相同，则不会创建任何数据文件。 行数不必相同；成为数据文件中的变量的行 元素的数量必须相同。 例如，双变量交叉制表和三变量交叉制表包含不同数量的行元素，因为“层”变量实 际上嵌套在缺省的三变量交叉制表显示中的行变量内。 控制列元素转换为数据文件中的控制变量 在输出管理系统控制面板的“选项”对话框中，可以指定列中应包括的维度元素，以便用于创建生成的数据文 件中的变量。 此过程等同于在查看器中旋转表。 例如，“频率”过程生成一个描述性统计表，其中统计在行中，而“描述统计”过程生成描述性统计表，其中统 计在列中。 要以有意义的方式同时包括相同数据文件中的这两个表类型，您需要更改一种表类型的列维度。 因为这两个表类型使用元素名称“统计”作为统计维度，所以可将“频率”统计表中的统计放在列中，方法是在 “OMS: 选项”对话框的维度名称列表中指定“统计”（在引号中）。 某些变量将具有缺失值，因为表结构仍与列中的统计不完全相同。 222 IBM SPSS Statistics 29 Core System 用户指南 OMS 生成的数据文件中的变量名称 OMS 从列标签构建有效的唯一变量名称： • 为行元素和层元素分配了通用变量名称：前缀 Var 后跟一个序列号。 • 将移去变量名称中不允许的字符（空格、括号等等）。 例如，“This (Column) Label”将成为名为 ThisColumnLabel 的变量。 • 如果标签以变量名称中允许使用但不允许用作第一个字符的字符开头（例如数字），那么将插入“@”作为 前缀。 例如，“2nd”将成为名为 @2nd 的变量。 • 将导致重复变量名称的列标签通过追加下划线和顺序字母进行解析。 例如，“Count”的第二个实例将变为 名为 Count_A 的变量。 • 将从生成的变量名称中移去标签末尾的下划线或句点。 不移去位于自动生成的变量 Command_、 Subtype_ 和 Label_ 末尾的下划线。 • 如果列维度中有多个元素，则可通过在类别标签之间添加下划线来组合类别标签，以此构建变量名称。 不 包括组标签。 例如，如果 VarB 嵌套在列中 VarA 的下面，则可获得的变量如 CatA1_CatB1，而不是 VarA_CatA1_VarB_CatB1。 OXML 表结构 输出 XML (OXML) 是符合 spss 输出架构的 XML。 有关该架构的详细说明，请参阅“帮助”系统的“输出架构” 部分。 • OMS 命令和子类型标识用作 OXML 中的 command 和 subType 属性的值。 示例如下： • OMS command 和 subType 属性值不受输出语言或变量名称/标签或值/值标签的显示设置的影响。 • XML 区分大小写。 “frequencies”的 subType 属性不同于“Frequencies”的 subType 属性。 • 显示在表中的所有信息都包含在 OXML 的属性值中。 在各个单元格级别，OXML 由“空”元素组成，这些元 素包含属性，但不包含属性值中包含的内容以外的“内容”。 • OXML 中的表结构逐行表示；表示列的元素嵌套在行中，单个单元格嵌套在列元素中： ... 上述示例是结构的简化表示形式，显示这些元素的后代/前辈关系。 但是，该示例不必显示父/子关系，因为 存在通常会产生干扰作用的内嵌元素级别。 以下示例显示了简单频率表和该表的完整的输出 XML 表示形式。 表 24: 简单频率表 性别 频率 百分比 有效百分比 累积百分比 有效 女性 216 45.6 45.6 45.6 男性 258 54.4 54.4 100.0 合计 474 100.0 100.0 第 22 章 输出管理系统 223 <?xml version="1.0" encoding="UTF-8" ?> 图 4: 简单频率表的输出 XML 您可能注意到，一个简单的小型表生成了大量的 XML。 部分原因是，XML 包含一些在原始表中不明显的信 息、一些甚至在原始表中不可用的信息以及一些冗余信息。 • 显示（或应显示）在查看器的透视表中的表内容包含在文本属性中。 示例如下： • 文本属性受输出语言和影响变量名称/标签和值/值标签显示的设置的影响。 在此示例中，文本属性值不相 同，这取决于输出语言，而命令属性值保持相同，不管输出语言是什么。 • 只要在行标签或列标签中使用了变量或变量值，XML 都将包含一个文本属性以及一个或多个附加属性值。 示例如下： ... • 对于数值变量，将存在数字属性而不是字符串属性。 只有在变量或值具有已定义的标签时，才有标签属 性。 224 IBM SPSS Statistics 29 Core System 用户指南 • 包含数字单元格值的 元素将涵盖文本属性和一个或多个其他属性值。 示例如下： 数字属性是实际的、未经过四舍五入的数值，小数属性指示表中显示的小数位数。 • 因为列嵌套在行中，所以标识每列的类别元素将对每行重复。 例如，由于统计信息显示在多个列中，因此 元素 会在 XML 中出现三次：一次针对男性行，一次针对女性行，还 有一次用于合计行。 OMS 标识 “OMS 标识”对话框用于帮助您编写 OMS 命令语法。 可以使用此对话框将所选命令和子类型标识粘贴到命令 语法窗口中。 有关更多信息， 使用“OMS 标识”对话框 1. 从菜单中选择： 实用程序 > OMS 标识...
2. 选择一个或多个命令标识或子类型标识。 （按住 Ctrl 再单击可选择每个列表中的多个标识。） 3. 单击粘贴命令和/或粘贴子类型。 • 可用子类型的列表是根据当前选定的命令得出来的。 如果选择了多个命令，可用子类型的列表就是选定的 任何命令的所有可用子类型的集合。 如果没有选择任何命令，则列出所有子类型。 • 标识粘贴到指定命令语法窗口的当前光标位置。 如果没有打开的命令语法窗口，则自动打开新的语法窗 口。 • 每个命令和/或子类型标识在粘贴时都用引号引起来，因为 OMS 命令语法要求使用引号。 • COMMANDS 和 SUBTYPES 关键字的标识列表必须用方括号括起来，如： /IF COMMANDS=['Crosstabs' 'Descriptives'] SUBTYPES=['Crosstabulation' 'Descriptive Statistics'] 从查看器概要复制 OMS 标识 您可以从“查看器”概要窗格复制并粘贴 OMS 命令和子类型标识。 1. 在概要窗格中，右键单击项的概要条目。 2. 选择复制 OMS 命令标识或复制 OMS 表子类型。 这种方法与“OMS 标识”对话框方法有一点区别：所复制的标识不会自动粘贴到命令语法窗口中。 标识只复 制到剪贴板，然后可以将它粘贴到任何需要的地方。 因为命令和子类型标识的值与相应的输出 XML 格式 (OXML) 的命令和子类型的属性值相同，所以在编写 XSLT 转换时可能会觉得这种复制/粘贴方法很有用。 复制 OMS 标签 在使用 LABELS 关键字时，您可以复制标签（而不是标识）。 标签可以用于区分同一类型的多个图形或表 格，在这些图形或表格中，概要文本反映特定输出对象（如变量名称或标签）的某个属性。 但是有一些因素 会影响标签文本： • 如果正在处理拆分文件，那么拆分文件组标识将会附加到标签上。 • 包括关于变量或值的信息的标签受概要窗格（“编辑”菜单，“选项”，“输出标签”选项卡）中的变量名称/标 签和值/值标签的显示设置的影响。 • 标签受当前输出语言设置（“编辑”菜单，“选项”，“常规”选项卡）的影响。 复制 OMS 标签 1. 在概要窗格中，右键单击项的概要条目。 2. 选择复制 OMS 标签。 第 22 章 输出管理系统 225 与命令标识和子类型标识一样，标签必须用引号引起来，整个列表必须用方括号括起来，如： /IF LABELS=['Employment Category' 'Education Level'] 226 IBM SPSS Statistics 29 Core System 用户指南 第 23 章 脚本编写工具 脚本编写工具使您可以将一些任务自动化，包括： • 打开并保存数据文件。 • 以各种格式将图表导出为图形文件。 • 在查看器中自定义输出。 可用脚本语言取决于使用的平台。 对于 Windows 平台，可用脚本语言为 Basic（随 Core System 一起安 装）和 Python 编程语言。 对于所有其他平台，脚本以 Python 编程语言提供。 要将脚本编制与 Python 编程语言配合使用，您需要 Python 功能，这是 IBM SPSS Statistics 产品的一部 分。 缺省脚本语言 缺省脚本语言确定创建新脚本时启动的脚本编辑器。 它还指定其可执行文件用于运行自动脚本的缺省语言。 在 Windows 中，缺省脚本语言为 Basic 语言。 可以从“选项”对话框中的“脚本”选项卡中更改缺省语言。 有 关更多信息，请参阅第172 页的『脚本选项』主题。 样本脚本 在安装 IBM SPSS Statistics 的目录的 Samples 子目录中包含一些脚本。 可以不做任何更改使用这些脚本， 或者可以按照您的需要自定义。 创建新脚本 1. 从菜单中选择： 文件 > 新建 > 脚本 将打开与缺省脚本语言关联的编辑器。 运行脚本 1. 从菜单中选择： 实用程序 > 运行脚本...
2. 选择所需的脚本。 3. 单击运行。 除了通过”实用程序“>”运行脚本“，Python 脚本可按照多种方式运行。 有关更多信息，请参阅第228 页的 『以 Python 编程语言编写脚本』主题。 编辑脚本 1. 从菜单中选择： 文件 > 打开 > 脚本...
2. 选择所需的脚本。 3. 单击打开。 将在与脚本所使用的语言相关联的编辑器中打开该脚本。 自动脚本 自动脚本即当被从已选过程中创建的特定输出触发时自动运行脚本。 例如，可以使用自动脚本在“双变量相 关性”过程生成“相关”表时自动移去上对角线并突出显示低于某个显著性水平的相关系数。 自动脚本可以特定于一个给定过程和输出类型或应用于不同过程中的特定输出类型。 例如，您可能拥有一个 既能为单因素 ANOVA 生成的 ANOVA 表设置格式，又能为由其他统计过程生成的 ANOVA 表设置格式的脚 本。 另一方面，“频率”过程生成频率表和统计表，而您可以选择两个表各自具有不同的自动脚本。 给定过程的每个输出类型只可与一个单一自动脚本相关。但是，您可以在为特定输出类型应用任何自动脚本 之前，创建应用于所有新查看器项的基本自动脚本。 “选项”对话框中的“脚本”选项卡（从“编辑”菜单进入）显示您系统中配置的自动脚本并使您能创建新的自动 脚本或修改已有自动脚本的设置。 或者，还可以从“查看器”中直接创建并配置输出项的自动脚本。 触发自动脚本的事件 以下事件可以触发自动脚本： • 透视表的创建 • “注释”对象的创建 • “警告”对象的创建 也可以使用脚本间接触发一个自动脚本。 例如，可以编写调用 Correlations 过程的脚本，而 Correlations 过 程反过来触发注册到产生的 Correlations 表中的自动脚本。 创建自动脚本 您可以通过从要用作触发器的输出对象（例如，频率表）开始创建自动脚本。 1. 在“查看器”中，选择将要触发自动脚本的对象。 2. 从菜单中选择实用程序 > 创建/编辑自动脚本...
若已选对象并不具有关联自动脚本，那么“打开”对话框将提示您输入新脚本的位置与名称。 3. 浏览至新脚本将被保存的位置，输入文件名并单击打开。 缺省脚本语言编辑器打开。您可以从“选项”对 话框的“脚本”选项卡更改缺省脚本语言。 4. 键入代码。 注: 缺省情况下，与缺省脚本语言相关联的可执行文件将用于运行自动脚本。 可以从“选项”对话框中的“脚 本”选项卡中更改可执行文件。 若已选对象已与自动脚本关联，那么该脚本在脚本编辑器中以与之相关的脚本编写语言打开。 关联已有脚本与查看器对象 可以通过将其与“查看器”中的已选对象关联 -- 例如，频率表，将已有脚本用作自动脚本。 1. 在“查看器”中，选择一个对象与自动脚本关联（多个“查看器”对象可以触发同一自动脚本，但每个对象只 能与单一自动脚本关联）。 2. 从菜单中选择： 实用程序 > 关联自动脚本...
若已选对象并不具有关联自动脚本，则“选择自动脚本”对话框打开。 3. 浏览所需脚本并选中。 4. 单击应用。 若已选对象已与自动脚本关联，那么提示您对所需更改的关联进行验证。 单击确定打开“选择自动脚本”对话 框。 或者，还可以在“选项”对话框中的“脚本”选项卡中将已有脚本配置为自动脚本。 自动脚本可被应用于已选输 出类型集或特定为应用于所有新建“查看器”项的基础自动脚本。 请参阅主题第172 页的『脚本选项』，了 解更多信息。 以 Python 编程语言编写脚本 IBM SPSS Statistics 提供两个单独的接口，用于在 Windows、Linux、macOS 和 IBM SPSS Statistics Server 上使用 Python 语言进行编程。 使用这些接口需要 IBM SPSS Statistics- Python 的集成插件，缺省情 况下，此插件随 IBM SPSS Statistics 产品一起安装。 有关 Python 编程语言的入门帮助，请参阅 Python 教 程 ( )。 228 IBM SPSS Statistics 29 Core System 用户指南 Python 脚本 Python 脚本利用 Python SpssClient 模块提供的界面。 它们在用户界面和输出对象上操作并能运行命令 语法。 例如，可以使用 Python 脚本自定义透视表。 • 从“实用程序>运行脚本”、从通过 IBM SPSS Statistics 启动的 Python 编辑器（从“文件>打开>脚本”访 问）或者从外部 Python 进程（例如 Python IDE 或 Python 解释器）运行 Python 脚本。 • Python 脚本可作为自动脚本运行。 • Python 脚本在运行 IBM SPSS Statistics 客户端的机器上运行。 可以在《IBM SPSS Statistics 脚本编写指南》中找到可用于 Python 脚本的 IBM SPSS Statistics 类以及方法 的完整文档，该指南位于“帮助”系统中的 Integration Plug-in for Python 下。 Python 程序 Python 程序利用 Python spss 模块提供的界面。 它们在 IBM SPSS Statistics 处理器上操作并被用于控制命 令语法工作流、读取及写入活动数据集、创建新数据集及创建能生成其自有透视表输出的客户过程。 • Python 程序可从命令语法中与 BEGIN PROGRAM-END PROGRAM 块共同运行，或从外部 Python 过程，如 Python IDE 或 Python 翻译器中运行。 • Python 程序不可作为自动脚本运行。 • 在分布式分析模式下（可用于 IBM SPSS Statistics 服务器），Python 程序在 IBM SPSS Statistics 服务器 运行的电脑上执行。 有关 Python 程序的更多信息（包括可用于这些程序的 IBM SPSS Statistics 函数和类的完整文档），您可以 在 Python Integration Package for IBM SPSS Statistics 的文档中找到，该文档位于“帮助”系统中的 Integration Plug-in for Python 下。 运行 Python 脚本与 Python 程序 Python 脚本和 Python 程序都可以在 IBM SPSS Statistics 中运行，也可以从外部 Python 进程运行，例如 Python IDE 或 Python 解释器。 Python 脚本 从 IBM SPSS Statistics 中运行“Python 脚本”。 您可以通过“实用程序>运行脚本”或从 Python 脚本编辑器 （从“文件>打开>脚本”打开 Python 文件 (.py) 时启动）运行 Python 脚本。 从 Python 编辑器中运行的脚本 在启动编辑器的 IBM SPSS Statistics 客户端上操作，该编辑器从 IBM SPSS Statistics 中启动。 这使您能从 Python 编辑器中调试您的 Python 代码。 从“外部 Python 过程”中运行 Python 脚本。 可以从任何外部 Python 进程中运行 Python 脚本，例如不从 IBM SPSS Statistics 中启动的 Python IDE 或 Python 翻译器。 脚本将尝试连接到现有 IBM SPSS Statistics 客户端。 如果找到多个客户端，将连接到最近启动的客户端。 如果现有客户端未找到, Python 脚本启动 IBM SPSS Statistics 客户端的新实例。 缺省状态下，“数据编辑器”与“查看器”在新客户端中均不可视。 可 以选择使其可视或在不可视模式下使用数据集与输出文档。 Python 程序 从“命令语法”中运行 Python 程序。 可以通过在命令语法中向 BEGIN PROGRAM-END PROGRAM 块中嵌入 Python 代码运行 Python 程序。 可以从 IBM SPSS Statistics 客户端中或 IBM SPSS Statistics 批处理工具 --即由 IBM SPSS Statistics Server 提供的一个单独可执行文件中运行命令语法。 从“外部 Python 过程”中运行 Python 程序。 可以从任何外部 Python 过程中运行 Python 程序，例如 Python IDE 或 Python 翻译器。 在此模式下，Python 程序无需 IBM SPSS Statistics 客户端的相关实例即可 启动 IBM SPSS Statistics 处理器的新实例。 可以通过所选的 Python IDE，使用此模式调试您的 Python 程 序。 从“Python 程序”中调用“Python 脚本”，反之亦然 从“Python 程序”中运行“Python 脚本”。 您可以通过导入包含脚本的 Python 模块并调用执行脚本的模块中 的函数从 Python 程序中运行 Python 脚本。 您还可直接从 Python 程序中调用 Python 脚本方法。 从外部 Python 进程运行 Python 程序，或从 IBM SPSS Statistics 批处理工具（随 IBM SPSS Statistics 服务器提 供）运行 Python 程序时，这些功能不可用。 第 23 章 脚本编写工具 229 从“Python 程序”中触发的“Python 自动脚本”。 特定为自动脚本的 Python 脚本将在 Python 程序执行过程 时被触发，该过程包含与自动脚本关联的输出项。 例如，将一个自动脚本与由“描述”过程生成的“可描述统 计数据”表关联。 然后运行执行“描述”过程的 Python 程序。 Python 自动脚本将被执行。 从“Python 脚本”中运行“Python 程序”。 Python 脚本可以运行命令语法，这就意味着它们能运行包含 Python 程序的命令语法。 限制与警告 • 从由 IBM SPSS Statistics 启动的 Python 编辑器中运行 Python 程序将启动 IBM SPSS Statistics 处理器的 新实例且不与由编辑器启动的 IBM SPSS Statistics 实例产生交互作用。 • Python 程序不打算从“实用程序” > “运行脚本”运行。 • Python 程序不可作为自动脚本运行。 • 由 spss 模块提供的界面无法在 Python 脚本中使用。 “Python 编程语言”的“脚本编辑器” 对于 Python 编程语言而言，缺省编辑器为 Python 提供的 IDLE。 IDLE 提供有限功能集的集成开发环境 （IDE）。 许多 IDE's 均适用于 Python 编程语言。 例如，在 Windows 中，可以自由选择使用 PythonWin IDE。 更改 Python 编程语言的脚本编辑器： 1. 打开位于其安装 IBM SPSS Statistics 目录中的文件 clientscriptingcfg.ini。 注: 必须使用支持 UTF-16 的文本编辑器来编辑 clientscriptingcfg.ini，例如 Windows 上的 SciTE，或 Mac 上的 TextEdit 应用程序。 2. 在标有 [Python3] 的部分下，更改 EDITOR_PATH 的值以指向用于启动编辑器的文件。 Python 3 的规 范是独立的。 3. 同一部分中，更改 EDITOR_ARGS 值以处理需传递至编辑器的任何参数。 若无需任何参数，移除一切已 有值。 Basic 语言中的脚本编写 Basic 语言中的脚本编写只适用于 Windows 并以 Core System 安装。 关于 Basic 语言中脚本编写的拓展在 线帮助可从 IBM SPSS Statistics “Basic 语言脚本编辑器”获取。 可通过单击“文件”>“新建”>“脚本”，并将缺 省的脚本语言（通过“选项” 对话框中的“脚本”选项卡进行设置）设置为 Basic（Windows 上的系统缺省值） 来访问该编辑器。 另外，也可以通过单击”文件“>“打开”>“脚本”，并在“文件类型”列表中选取“Basic （wwd、sbs）“来访问该编辑器。 注：对于 Windows 7 和更高版本，需要安装您的系统中可能不存在的 Windows Help 程序 (WinHlp32.exe)，才能访问关于 Basic 语言脚本编制的联机帮助。 如果您不能查看在线帮助，请联系 Microsoft 以获取 Windows Help 程序 (WinHlp32.exe) 的指示信息。 与 16.0 之前版本的兼容性 废弃方法与属性 一些自动化方法与属性已被版本 16.0 及以上版本废弃。 就常规功能而言，它包括了所有与互动图像相关的 对象、“草稿文档”对象及与地图相关的方法与属性。 更多详情，参见“IBM SPSS Statistics Basic 语言脚本编 辑器”所提供帮助系统中的“16.0 版本的发布注释”。 在脚本编辑器中，从“帮助 >IBM SPSS Statistics 对象帮 助”访问特定于 IBM SPSS Statistics 的帮助。 全局过程 16.0 之前的版本中，脚本编写工具包含全局过程文件。 即使 16.0 之前的 Global.sbs (更名为 Global.wwd） 版本中安装了向后兼容性，16.0 及以上版本中，脚本编写工具并未使用全局过程文件。 要迁移在全局过程文件中调用函数的脚本的 pre-16.0 版本，请将语句 '#Uses "\Samples\Global.wwd" 添加到脚本的声明部分，其中 是 IBM SPSS Statistics 的 230 IBM SPSS Statistics 29 Core System 用户指南 安装目录。"#Uses 是 Basic 脚本处理器识别的特殊注释。 若不确定脚本是否使用了全局过程文件，则应添 加 '#Uses 语句。 您也可以使用 '$Include: 来代替 '#Uses。 Legacy 自动脚本 16.0 之前的版本中，脚本编写工具包括包含了所有自动脚本的单一自动脚本文件。 16.0 及以上版本则无单 一自动脚本文件。 现在每个自动脚本均被保存在单独文件中并可应用于一个或多个输出项中，相反在 16.0 之前版本中每个自动脚本均特定于一个个别输出项。 随 16.0 以前版本安装的一些自动脚本位于 IBM SPSS Statistics 安装目录的 Samples 子目录中，作为一组单 独的脚本文件提供。 它们由以 Autoscript 结尾的文件名和 wwd 的文件类型标识。 缺省情况下，它们与任何 输出项均无关联。 从“选项”对话框的“脚本”选项卡中进行关联。 有关更多信息，请参阅第172 页的『脚本 选项』主题。 任何用于 16.0 之前版本的自定义自动脚本必须手动转换并从“选项”对话框的“脚本”选项卡中关联至一个或 多个输出项。 转换过程包括以下步骤： 1. 从 legacy Autoscript.sbs 文件中提取特定自动脚本的子例程并以扩展名 wwd 或 sbs 将其存为新建文件。 文件名随意。 2. 将子例集名称改为 Main 并删除参数规范，保持对脚本所需参数，如触发自动脚本的透视表的追踪。 3. 使用 scriptContext 对象（总是可用）获取自动脚本所需值，如触发自动脚本的输出项。 4. 从“选项”对话框的“脚本”选项卡中，将脚本文件与输出对象相关联。 为说明转换代码，请考虑来自 legacy Autoscript.sbs 文件的自动脚本 Descriptives_Table_DescriptiveStatistics_Create。 Sub Descriptives_Table_DescriptiveStatistics_Create _ (objPivotTable As Object,objOutputDoc As Object,lngIndex As Long) 'Autoscript 'Trigger Event: DescriptiveStatistics Table Creation after running ' Descriptives procedure.
'Purpose: Swaps the Rows and Columns in the currently active pivot table. 'Assumptions: Selected Pivot Table is already activated.
'Effects: Swaps the Rows and Columns in the output 'Inputs: Pivot Table, OutputDoc, Item Index Dim objPivotManager As ISpssPivotMgr Set objPivotManager=objPivotTable.PivotManager objPivotManager.TransposeRowsWithColumns End Sub 以下为已转换脚本： Sub Main 'Purpose: Swaps the Rows and Columns in the currently active pivot table. 'Effects: Swaps the Rows and Columns in the output Dim objOutputItem As ISpssItem Dim objPivotTable as PivotTable Set objOutputItem = scriptContext.GetOutputItem() Set objPivotTable = objOutputItem.ActivateTable Dim objPivotManager As ISpssPivotMgr Set objPivotManager = objPivotTable.PivotManager objPivotManager.TransposeRowsWithColumns objOutputItem.Deactivate End Sub • 注意，已转换脚本中并未指明该脚本会应用哪个对象。 输出项与自动脚本间的关联从“选项”对话框的“脚 本”选项卡中设置并在会话之间保持不变。 • scriptContext.GetOutputItem 获得触发自动脚本的输出项（一个 ISpssItem 对象）。 • 以 scriptContext.GetOutputItem 返回的对象未被激活。 若您的脚本需要激活对象，则需要将其激 活，如本示例中所示的 ActivateTable 方法。 当一切表操作完成，调用取消激活方法。 对于版本 16.0，在作为自动脚本运行的脚本与未作为自动脚本运行的脚本间并无差别。 任何进行过相应编 码的脚本均可用于所有上下文中。 有关更多信息，请参阅第232 页的『脚本上下文对象』主题。 注：要从应用程序创建事件触发脚本，请参阅第232 页的『启动脚本』。 脚本编辑器 对于 16.0 及以上版本，用于 Basic 语言的脚本编辑器不再支持以下 16.0 之前功能： 第 23 章 脚本编写工具 231 • 脚本、分析、图形、实用程序、插件菜单。 • 将命令语法粘贴至脚本窗口的能力。 IBM SPSS Statistics Basic Script 编辑器是从 IBM SPSS Statistics 中通过文件 > 新建 > 脚本，文件 > 打开 > 脚本或实用程序 > 创建/编辑自动脚本（从查看器窗口）启动的独立应用程序。 使您能针对从中启动的 IBM SPSS Statistics 实例运行脚本。 已经打开，编辑器将一直开启直至退出 IBM SPSS Statistics，但是使用 IBM SPSS Statistics 对象的脚本将不再运行。 文件类型 对于 16.0 及以上版本，脚本编写工具将继续支持 sbs 文件类型的脚本运行与编辑。 缺省状态下，通过“IBM SPSS Statistics Basic 语言脚本编辑器”创建的新 Basic 语言脚本的文件类型为 wwd。 使用“外部 COM 客户端” 对于 16.0 及以上版本，用于例示“外部 COM 客户端”中 IBM SPSS Statistics 的程序标识为 SPSS.Application16。 应用程序对象应声明为 spsswinLib.Application16。 例如： Dim objSpssApp As spsswinLib.Application16 Set objSpssApp=CreateObject("SPSS.Application16") 要连接到一个来自外部 COM 客户端的 IBM SPSS Statistics 客户端运行实例，请使用： Dim objSpssApp As spsswinLib.Application16 Set objSpssApp=GetObject("","SPSS.Application16") 如果多个客户端正在运行，GetObject 将连接到最近启动的客户端。 注：对于 16.0 之后的版本，标识仍然是 Application16。 脚本上下文对象 脚本作为自动脚本运行时的检测 利用 scriptContext 对象，可以在脚本作为自动脚本运行时进行检测。 这使您可以对脚本进行编码以便 其在所有上下文中可以运行（无论自动脚本与否）。 此普通脚本说明了方法。 Sub Main If scriptContext Is Nothing Then MsgBox "I'm not an autoscript" Else MsgBox "I'm an autoscript" End If End Sub • 当脚本不作为自动脚本运行时，scriptContext 对象将具有无值。 • 本示例中被赋予 If-Else 逻辑，应在 Else 子句中包含自动脚本特定代码。 任何不在自动脚本上下文中 运行的代码均应包含在 If 子句中。 当然也可以包含将要在所有上下文中运行的代码。 获区自动脚本所需值 scriptContext 对象提供自动脚本所需值的使用，如触发当前自动脚本的输出项。 • scriptContext.GetOutputItem 方法返回至触发当前自动脚本的输出项（一个 ISpssItem 对象）。 • scriptContext.GetOutputDoc 方法返回至与当前自动脚本关联的输出文档（一个 ISpssOutputDoc 对象）。 • scriptContext.GetOutputItemIndex 方法返回至相关输出文档中触发当前自动脚本的输出项索引。 注：以 scriptContext.GetOutputItem 返回的对象未被激活。 若您的脚本需要激活对象，则需要将其 激活 -- 例如通过 ActivateTable 方法。 当一切操作完成，调用取消激活方法。 启动脚本 您可以创建在每次会话启动时运行的脚本，以及在每次切换服务器时运行的单独脚本。 在 Windows 上，您 可以使用这些脚本的 Python 和 Basic 版本。 对于所有其他平台，脚本只能在 Python 中。 • 启动脚本必须命名为 StartClient_.py(Python) 或 StartClient_.wwd(Basic)。 232 IBM SPSS Statistics 29 Core System 用户指南 • 在切换服务器时运行的脚本必须命名为 StartServer_.py(Python) 或 StartServer_.wwd(Basic)。 • 这些脚本必须位于安装目录的脚本目录中 -- 位于 Windows 的安装目录的根目录下，并且位于 MacOS 的 / Applications/IBM SPSS Statistics/Resources 目录下。 注意，不论您是否在分布模式下工作，所有脚本 （包括 StartServer_ 脚本）必须位于客户端计算机上。 • 在 Windows 上，如果 scripts 目录同时包含 Python 和 Basic 版本的 StartClient_ 或 StartServer_，则两种 版本的脚本都会执行。 执行顺序为 Python 版本优先于 Basic 版本。 • 如果您的系统被配置成以分布式模式启动，则当每次会话启动时，先运行 StartClient_ 脚本，然后再运行 StartServer_ 脚本。 注：StartServer_ 脚本还会在每次切换服务器时运行，但 StartClient_ 脚本仅在会话启 动时运行。 示例 这是将驱动器盘符映射到由 UNC 标识指定的共享网络资源的 StartServer_ 脚本示例。 它允许在分布式模式 中工作的用户从“打开远程文件”对话框中访问网络资源上的数据文件。 #StartServer_.py import SpssClient SpssClient.StartClient() SpssClient.RunSyntax(r""" HOST COMMAND=['net use y: \myserver\data']. """) SpssClient.StopClient() SpssClient.RunSyntax 方法用于运行 HOST 命令，该命令将调用 Windows 命令 net use 以执行映射。 当 StartServer_ 脚本运行时， IBM SPSS Statistics 处于分布式方式，因此 HOST 命令在 IBM SPSS Statistics Server 机器上运行。 第 23 章 脚本编写工具 233 234 IBM SPSS Statistics 29 Core System 用户指南 第 24 章 TABLES 和 IGRAPH 命令语法转换器 如果您的命令语法文件包含 TABLES 语法，您要将其转换为 CTABLES 语法和/或您要将 IGRAPH 语法转换为 GGRAPH 语法，可以使用一个简单实用程序进行转换。 但是，TABLES 与 CTABLES 以及 IGRAPH 与 GGRAPH 在功能上有很大差别。 您可能会发现该实用程序无法转换某些 TABLES 和 IGRAPH 语法作业，或者 转换得到的 CTABLES 和 GGRAPH 语法所生成的表格和图形可能与 TABLES 及 IGRAPH 命令所生成的原始表 格和图形差别较大。 对于大多数的表，您可以编辑转换后的语法，使之生成与原始表十分相似的表。 该实用程序设计用于： • 根据现有语法文件创建新的语法文件。 不更改原始语法文件。 • 只转换语法文件中的 TABLES 及 IGRAPH 命令。 不更改该文件中的其他命令。 • 以注释形式保留原始 TABLES 及 IGRAPH 语法。 • 用注释标识每个转换块的开头和结尾。 • 标识无法转换的 TABLES 和 IGRAPH 语法命令。 • 转换符合交互式或生产模式语法规则的命令语法文件。 此实用程序无法转换包含错误的命令。 还具有以下其他限制。 TABLES 限制 在某些情况下，该实用程序对 TABLES 命令的转换可能不正确，例如具有以下特点的 TABLES 命令： • 如果变量自身带括号（例如 var1 by (statvar) by (labvar)），那么需要在 TABLES 子命令中使 用具有首字母“sta”或“lab”的带括号变量名称。 这些变量将被解释为 (STATISTICS) 和 (LABELS) 关 键字。 • 包含的 SORT 子命令使用缩写字母 A 或 D 指示升序或降序。 这些缩写字母将被解释为变量名称。 该实用程序无法转换具有以下特点的 TABLES 命令： • 包含语法错误。 • 包含使用 TO 关键字引用变量范围的 OBSERVATION 子命令（例如，var01 TO var05）。 • 包含用加号分隔成多部分的字符串（例如，TITLE "My" + "Title"）。 • 包含在没有宏扩展时会导致 TABLES 语法无效的宏调用。 由于转换器不扩展宏调用，因此会将宏调用简单 地视为标准 TABLES 语法的一部分。 该实用程序不转换包含在宏中的 TABLES 命令。 所有宏都不受转换过程的影响。 IGRAPH 限制 IGRAPH 在发行版 16 中进行了重大更改。 由于这些更改，在该发行版之前创建的 IGRAPH 语法中的一些子 命令和关键字可能不受支持。 有关完整的修订历史记录，请参阅命令语法参考中的 IGRAPH 部分。 转换实用程序可能会在 GGRAPH 语法中生成存储在 INLINETEMPLATE 关键字中的其他语法。 此关键字只能 通过转换程序创建。 用户无法编辑其语法。 使用转换实用程序 转换实用程序 SyntaxConverter.exe 可以在安装目录中找到。 该实用程序设计为从命令提示符处运行。 命令 的一般形式为： syntaxconverter.exe [path]/inputfilename.sps [path]/outputfilename.sps 您必须从安装目录运行此命令。 如果有任何目录名中包含空格，应当用引号将路径和文件名整体括起来，如下所示： syntaxconverter.exe /myfiles/oldfile.sps "/new files/newfile.sps" 交互式与生产模式命令语法规则 转换实用程序可以转换使用交互式或生产模式语法规则的命令文件。 交互式。 交互式语法规则为： • 每条命令都另起一行。 • 每个命令都以句点 (.) 结束。 生产模式。 生产工具和文件中的命令（通过在其他命令文件中使用 INCLUDE 命令访问）使用生产模式语法 规则： • 每条命令必须从新的一行的第一列开始。 • 连续行至少须缩进一个空格。 • 命令结尾处的句点是可选的。 如果命令文件使用生产方式语法规则，并且在每个命令末尾都不包含句点，那么在运行 SyntaxConverter.exe 时需要包含命令行开关 -b（或 /b），如下所示： syntaxconverter.exe -b /myfiles/oldfile.sps /myfiles/newfile.sps SyntaxConverter 脚本（仅 Windows） 在 Windows 上，您也可以通过脚本 SyntaxConverter.wwd 运行语法转换器，该脚本位于安装目录中的 Samples 目录中。 1. 从菜单中选择： 实用程序 > 运行脚本...
2. 浏览至 Samples 目录并选择 SyntaxConverter.wwd。 这将打开一个简单的对话框，在该对话框中可以指定新旧命令语法文件的名称和位置。 236 IBM SPSS Statistics 29 Core System 用户指南 第 25 章 对数据文件、输出文档和语法文件进行加密 您可以通过使用密码对文件进行加密来保护存储在数据文件、输出文档或语法文件中的机密信息。 一旦加 密，文件只能通过密码打开。 数据文件、输出文档和语法文件的“另存为”对话框中提供了用于对文件进行加 密的选项。 另外，您也可以在对数据文件进行排序以及保存排序后的文件时对该文件进行加密。 • 密码丢失后将无法恢复。 如果密码丢失，那么将无法打开文件。 • 密码限制在 10 个字符并区分大小写。 创建强密码 • 至少使用八个字符。 • 在密码中使用数字、符号甚至标点符号。 • 避免使用数字序列或字符序列（例如 "123" 和 "abc"）并避免重复，例如 "111aaa"。 • 不要创建使用个人信息（例如生日或昵称）的密码。 • 定期更改密码。 修改加密文件 • 如果打开加密文件，对其进行修改并选择“文件” > “保存”，修改后的文件将以相同的密码保存。 • 您可以通过打开文件、重复加密步骤并在“加密文件”对话框中指定不同的密码，在加密的文件上更改密 码。 • 通过打开文件，选择“文件 > 另存为”并在关联“另存为”对话框中取消选择 使用密码加密文件 ，可以保存已 加密数据文件或输出文档的未加密版本。 对于已加密的语法文件，从“另存为类型”下拉列表中选择语法可 以保存该文件的未加密版本。 注意：在 V21 之前的 IBM SPSS Statistics 版本中，无法打开经过加密的数据文件和输出文档。 在 V22 之前 的版本中，无法打开经过加密的语法文件。 238 IBM SPSS Statistics 29 Core System 用户指南 声明 本信息是为在美国提供的产品和服务编写的。 IBM 提供了本资料的其他语言版本。 但是，您可能需要拥有 该语言的产品副本或产品版本才能访问这些资料。 IBM 可能在其他国家或地区不提供本文档中所讨论的产品、服务或功能特性。 有关您当前所在区域的产品和 服务的信息，请向您当地的 IBM 代表咨询。 任何对 IBM 产品、程序或服务的引用并非意在明示或暗示只能 使用 IBM 的产品、程序或服务。 只要不侵犯 IBM 的知识产权，任何同等功能的产品、程序或服务都可以代 替 IBM 产品、程序或服务。 但是，评估和验证任何非 IBM 产品、程序或服务，则由用户自行负责。 IBM 可能已拥有或正在申请与本文档内容有关的各项专利。 提供本文档并未授予用户使用这些专利的任何许 可。 您可以通过书面方式将许可查询寄往： IBM Director of Licensing IBM Corporation North Castle Drive, MD-NC119 Armonk, NY 10504-1785 US 有关双字节 (DBCS) 信息的许可查询，请与您所在国家或地区的 IBM 知识产权部门联系，或以书面形式将查 询寄往： Intellectual Property Licensing Legal and Intellectual Property Law IBM Japan Ltd.
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商标 IBM、IBM 徽标和 ibm.com 是 International Business Machines Corp. 在全球许多司法辖区的商标或注册商 标。 其他产品和服务名称可能是 IBM 或其他公司的商标。 最新的 IBM 商标列表可以在 Web 上的 “Copyright and trademark information”中获取，地址为：www.ibm.com/legal/copytrade.shtml。 Adobe、Adobe 徽标、PostScript 和 PostScript 徽标是 Adobe Systems Incorporated 在美国和/或其他国家 或地区的注册商标或商标。 Intel、Intel 徽标、Intel Inside、Intel Inside 徽标、Intel Centrino、 Intel Centrino 徽标、Celeron、Intel Xeon、Intel SpeedStep、Itanium 和 Pentium 是 Intel Corporation 或其子公司在美国和其他国家或地区的 商标或注册商标。 Linux 是 Linus Torvalds 在美国和/或其他国家或地区的注册商标。 Microsoft、Windows、Windows NT 和 Windows 徽标是 Microsoft Corporation 在美国和/或其他国家或地区 的商标。 UNIX 是 The Open Group 在美国和/或其他国家或地区的注册商标。 Java 和所有基于 Java 的商标和徽标是 Oracle 和/或其子公司的商标或注册商标。 240 IBM SPSS Statistics 29 Core System 用户指南 索引 Special Characters - 时间序列数据 数据转换 82 -cases 查找重复 65 在数据编辑器中查找 54 “帮助”窗口 5 保存 复原点 2 保存输出 文本格式 107, 110 Excel 格式 107, 108 HTML 107 HTML 格式 107 PDF 格式 107, 109 PowerPoint 格式 107, 109 Word 格式 107, 108 保存图表 元文件 107 BMP 文件 107, 111 EMF 文件 107 EPS 文件 107, 111 JPEG 文件 107, 111 PICT 文件 107 PNG 文件 111 PostScript 文件 111 TIFF 文件 111 保存文件 控制缺省文件位置 171 数据库文件查询 17 数据文件 21 IBM SPSS Statistics 数据文件 20 背景颜色 127 本地编码 150, 151 比较数据集 32 比例估计 在个案排秩中 77 边框 显示隐藏的边框 129 编辑数据 52, 53 变换变量和个案 86 变换行和列 120 变量 插入新变量 53 定义 44 定义变量集 162 定义信息 161 对话框中的变量信息 2 对话框中的显示顺序 163 排序 86 移动 54 在数据编辑器中查找 54 重命名合并的数据文件 87 重新编码 39, 74, 75 重组为个案 94 变量标签 变量标签 (继续) 插入换行符 47 在对话框中 1, 163 在概要窗格中 168 在合并数据文件中 88 在透视表中 168 变量对 创建 94 变量集 定义 162 使用 162 变量列表 重新排序目标列表 162 变量名称 便携格式文件 21 规则 44 混合个案变量名称 44 截断早期发行版中的长变量名称 21 由 OMS 生成 223 在对话框中 1, 163 在输出中使长变量名称换行 44 变量视图 自定义 50, 167 变量属性 定制 49 复制和粘贴 49 复制粘贴 48 变量信息 161 便携格式文件 变量名称 21 标度变量 分箱化以创建分类变量 66 标签 插入组标签 120 删除 120 与 OMS 中的子类型名称 219 标题 图表 154 表 背景颜色 127 边距 127 单元格属性 127 对齐 127 将 TABLES 命令语法转换为 CTABLES 235 控制表分隔符 129 字体 127 表格外观 应用 124 表图 130 表子类型 与标签 219 采样 随机样本 93 菜单 定制 175 测量级别 定义 45 索引 241 测量级别 (继续) 对话框中的图标 2 缺省测量级别 166 未知测量级别 61 测量系统 163 层 创建 123 打印 111, 125, 126 显示 123 在透视表中 123 插补 在数据编辑器中查找 54 插入组标签 120 查看器 保存文档 105 查找和替换信息 104 概要 115 概要窗格 103 更改概要大小 116 更改概要级别 103 更改概要字体 116 工作簿 113 结果窗格 103 使用 OMS 排除输出类型 222 输出项之间的空格 112 搜索和替换信息 104 显示选项 166 隐藏结果 116 展开概要 116 折叠概要 116 查看者 保存工作簿 114 创建工作簿 114 打开工作簿 114 显示变量标签 168 显示变量名称 168 显示数据值 168 显示值标签 168 查找和替换 查看器文档 104 差分函数 83 拆分模型查看器 136 拆分文件处理 91 触发事件 自动脚本 227 窗格拆分 数据编辑器 56 语法编辑器 145 窗口拆分 数据编辑器 56 语法编辑器 145 垂直标签文本 120 磁盘空间 临时 33, 34 打开 复原点 2 打开文件 电子表格文件 7 控制缺省文件位置 171 数据文件 7 文本数据文件 10 制表符分隔文件 7 dBASE 文件 7, 9 Excel 文件 7 打开文件 (继续) Lotus 1-2-3 文件 7 Stata 文件 9 SYSTAT 文件 7 打印 层 111, 125, 126 打印预览 112 调整表的大小 125, 126 控制表分隔符 129 模型 134 输出项之间的空格 112 数据 56 透视表 111 图表 111 图表大小 112 文本输出 111 页码 112 页眉和页脚 112 大小 在概要中 116 单元格属性 127 导出 模型 134 导出输出 文本格式 219 Excel 格式 107, 108, 219 HTML 107 HTML 格式 107 OMS 217 PDF 格式 107, 109, 219 PowerPoint 格式 107 Word 格式 107, 108, 219 导出数据 添加导出数据的菜单项 175 导出图表 自动化生产 211 导入数据 7, 13 登录到服务器 37 电子表格文件 读取 8 调整 透视表 125 调整比例 透视表 126 定义变量 变量标签 47 复制粘贴属性 48, 49 模板 48, 49 缺失值 47 数据类型 45 应用数据字典 63 值标签 47, 59 定制变量属性 49 定制表 将 TABLES 命令语法转换为 CTABLES 235 定制对话框构建器 安全文本 194 安装包含对话框的扩展 204 安装兼容定制对话框 205 帮助文件 182 保存包含对话框的扩展 204 保存兼容定制对话框 205 本地化对话框和帮助文件 207 表控件 196 242 IBM SPSS Statistics 29 Core System 用户指南 定制对话框构建器 (继续) 表控件列 196 布局规则 184 菜单位置 183 打开包含对话框的扩展 204 打开兼容定制对话框包文件 205 单选按钮组 198 对话框属性 182 复选框 189 复选框组 199 过滤变量列表 189 兼容定制对话框包 (spd) 文件 205 静态文本控件 195 扩展命令的定制对话框 207 扩展束文件 204 列表框 191 列表框列表项 190 目标列表 187 启用规则 201 日期控件 194 数据集选择器 189 数据源 183 数字控件 193 文本控件 192 文件类型过滤器 200 文件浏览器 199 项目组控件 197 修改兼容定制对话框 205 修改已安装的扩展中的对话框 204 选项卡 200 颜色选取器 195 语法模板 184 预览 186 源列表 187 转换为增强对话框 205 字段选择器 188 字段源 189 子对话框按钮 200 子对话框属性 201 组合框 190 组合框列表项 190 定制属性 49 逗号分隔文件 10 端口号 37, 214 断点 语法编辑器 147 对个案进行加权 交叉表中的分数权重 93 对个案排序 85 对话框 变量图标 2 变量显示顺序 163 变量信息 2 定义变量集 162 使用变量集 162 显示变量标签 1, 163 显示变量名称 1, 163 重新排序目标列表 162 对齐 在数据编辑器中 48 output 117, 166 对齐方式 output 117, 166 对行或列分组 120 对值进行重新编码 75 对值重新编码 39, 66, 74, 75 多个打开的数据文件 消除 58 多个视图/窗格 数据编辑器 56 语法编辑器 145 多响应集 定义 62 多二分 62 多类别 62 分表符 129 分布模式 37 分布式模式 可用过程 40 数据文件访问 38 相对路径 40 分段 66 分割表 控制表分隔符 129 分类数据 将定距数据转化为离散的类别 66 分箱化 66 分组变量 创建 94 在“汇总数据”中 90 服务器 编辑 37, 214 登录 37 端口号 37, 214 名称 37, 214 添加 37, 214 复原点 保存 2 打开 2 复制和粘贴输出到其他应用程序 117 概要 更改级别 103 在查看器中 115 展开 116 折叠 116 个案 插入新个案 53 加权 93 排序 85 选择子集 92, 93 重组为变量 94 个案的子集 随机样本 93 选择 92, 93 个案排秩 百分位数 77 分数排秩 77 同数的值 77 Savage 得分 77 工具栏 创建 175, 176 创建新工具 176 定制 175 显示和隐藏 175 显示在不同窗口中 176 自定义 176 工作簿 保存 114 索引 243 工作簿 (继续) 加密 114 应用程序方式 163 固定格式 10 过视图 41 函数 缺失值处理 72 合并数据文件 具有不同变量的文件 88 具有不同个案的文件 87 重命名变量 87 字典信息 88 后处理输出 137–141 环境变量 SPSSTMPDIR 171 缓存 活动文件 34 换行 变量和值标签 47 控制换行文本的列宽 125 换行符 变量和值标签 47 汇总数据 变量名称和标签 91 汇总函数 91 会话日志 171 活动窗口 1 活动文件 创建临时活动文件 34 缓存 34 虚拟活动文件 33 货币格式 167 基于关键字的表 88 计算变量 计算新的字符串变量 71 计算出现次数 73 季节性差分函数 83 加权数据 和重构的数据文件 100 兼容的定制对话框 181 将输出居中 117, 166 交叉表 分数权重 93 交互式输出 106 交替行颜色 透视表 125 角色 数据编辑器 47 脚本 编辑 227 创建 227 基本 230 启动脚本 232 缺省语言 172, 227 使用工具栏按钮运行 176 添加到菜单 175 语言 227 正在运行 227 自动脚本 227 Python 228 脚注 标记 125 图表 154 重新编号 128 接续文本 透视表的 126 经典 应用程序方式 163 聚集数据 90 科学记数法 输出中取消 163 可视分段器 66 刻度 测量级别 45 空格分隔的数据 10 控制显示行数 125 快速透视表 170 宽表 粘贴到 Microsoft Word 中 117 宽高比 168 扩展 安装扩展的更新 177 测量级别 61 查看安装的扩展 178 查找和安装新扩展 177 除去扩展 178 扩展详细信息 178 扩展的定制对话框构建器(U) 181 扩展命令 定制对话框 207 扩展束 安装扩展束 179 安装在 Statistics Server 上 181 创建扩展束 208 批量安装 181 累积和函数 83 联机帮助 5 列 更改透视表中的宽度 129 在透视表中选择 129 列宽 控制换行文本的宽度 125 控制缺省宽度 170 控制最大宽度 125 透视表 129 在数据编辑器中 48 临时磁盘空间 33, 34 临时活动文件 34 临时目录 本地模式的设置位置 171 SPSSTMPDIR 环境变量 171 描述统计 数据编辑器 55 名义 测量级别 45, 61 命令标识 218 命令行开关 生产作业 215 命令语法 访问《命令语法参考》 5 日志文件 151, 152 生产作业规则 211 使用工具栏按钮运行 176 输出日志 144 添加到菜单 175 语法规则 143 粘贴 144 正在运行 150 244 IBM SPSS Statistics 29 Core System 用户指南 命令语法编辑器 断点 147, 150 多个视图/窗格 145 分界点 145 命令跨度 145 设置语法格式 149 书签 145, 148 行号 145 颜色编码 146 语法缩进 149 注释或取消注释文本 149 自动完成 146 options 173 命令语法文件 150, 151 命令语言 143 模板 变量定义 48, 49 使用外部数据文件作为模板 63 图表 155 在图表中 168 模型 打印 134 导出 134 复制 134 合并模型和转换文件 159 激活 133 交互 133 模型查看器 133 评分 157 属性 133 支持导出和评分的模型 157 模型查看器 拆分模型 136 目标列表 162 内部连接 15 内存 163 年份 两位数的值 166 排列变量 86 排序 变量 86 透视表行 121 拼并类别 66 拼写 字典 166 平滑函数 83 评分 合并模型和转换 XML 文件 159 匹配数据集字段到模型字段 157 评分函数 159 缺失值 157 支持导出和评分的模型 157 前移动平均值函数 83 缺省文件位置 171 缺失值 定义 47 函数中的 72 评分模型 157 图表 155 在时间序列数据中替换 84 字符串变量 47 日期变量 为时间序列数据定义的 82 日期格式 日期格式 (继续) 两位数的年份 166 日期格式变量 根据变量集创建日期/时间变量 78 根据字符串创建日期/时间变量 78 提取日期/时间变量的组成部分 78 在日期/时间变量中加上或减去 78 日志文件 171 色阶 131 生产工具 转换文件到生产作业 216 生产设施 使用日志文件的命令语法 163 生产作业 导出图表 211 调度生产作业 215 命令行开关 215 输出文件 211 语法规则 211 运行多个生产作业 215 转换生产工具文件 216 时间序列数据 创建新的时间序列变量 83 定义日期变量 82 替换缺失值 84 转换函数 83 使用 OMS 从查看器排除输出 222 使用 OMS 从查看器隐藏（排除）输出 222 使用命令语言编程 143 受限数值格式 45 书签 语法编辑器 148 输出 保存 105 查看器 103 复制 117 工作簿 113 加密 105 删除 117 显示 116 移动 117 隐藏 116 粘贴到其他应用程序中 117 输出对象类型 在 OMS 中 218 输出管理系统 (OMS) 217, 225 输入格式 46 输入数据 非数值 52 数字 52 用于值标签 52 属性 表 125 定制变量属性 49 透视表 124 数据编辑器 编辑数据 52, 53 变量视图 44 插入新变量 53 插入新个案 53 打印 56 定义变量 44 对齐 48 多个打开的数据文件 57, 163 索引 245 数据编辑器 (继续) 多个视图/窗格 56 更改数据类型 54 过视图 41 将数据发送到其他应用程序 175 角色 47 列宽 48 描述统计 55 描述统计选项 168 输入非数值数据 52 输入数据 51 输入数字数据 52 数据视图 43 数据值限制 52 显示选项 56 移动变量 54 已过滤的个案 55 数据集 比较 32 重命名 57 数据库 保存 26 保存查询 17 表连接 15 参数查询 15, 16 创建关系 15 创建新表 29 定义变量 17 读取 13, 14 更新 26 将字符串转换为数值型变量 17 随机抽样 15 提示值 16 替换表 29 替换现有字段中的值 28 条件表达式 15 向表添加新字段 29 向表追加记录（个案） 29 选择数据源 14 选择数据字段 14 验证结果 17 正在读取 14 指定条件 15 Microsoft Access 14 SQL 语法 17 Where 子句 15 数据类型 定义 45 更改 54 输入格式 46 显示格式 46 自定义货币 45, 167 数据视图 43 数据输入 51 数据文件 保存 20, 21 保存变量子集 25 变换 86 打开 7 多个打开的数据文件 57, 163 翻转 86 防止 33 改善大型文件的性能 34 加密 25 数据文件 (继续) 将输出保存为 IBM SPSS Statistics 数据文件 217 添加注释 161 文本 10 文件信息 20 远程服务器 38 重组 94 字典信息 20 数据转换 对值进行重新编码 75 对值重新编码 39, 74, 75 个案排秩 76 函数 72 计算变量 71 时间序列 82, 83 条件转换 71 延迟执行 166 字符串变量 71 数据字典 从另一个文件应用 63 数字格式 45, 46 双向文本 165 搜索和替换 查看器文档 104 速度 高速缓存数据 34 算法 5 随机数种子 72 随机样本 数据库 15 随机数种子 72 选择 93 提前函数 73, 83 题注 127 替换缺失值 邻近点的平均值 84 邻近点的中位数 84 线性插值 84 线性趋势 84 序列平均值 84 添加组标签 120 条件转换 71 透视表 背景颜色 127 边距 127 边框 126 编辑 119 变换行和列 120 变量标签 121 操作 119 层 123 插入行和列 121 插入组标签 120 常规属性 125 撤销更改 122 从表创建图表 130 打印层 111 打印大型表 129 单元格格式 125 单元格宽度 129 单元格属性 127 导出为 HTML 107 调整比例以适合页面 125, 126 对齐 127 246 IBM SPSS Statistics 29 Core System 用户指南 透视表 (继续) 对行或列分组 120 对行或列取消分组 120 对行排序 121 更改外观 124 更改显示顺序 120 交替行颜色 125 脚注 127–129 脚注属性 125 接续文本 126 控制表分隔符 129 控制显示行数 125 快速呈现表格 170 快速透视表 170 缺省列宽的调整 170 色阶 131 删除组标签 120 使用图标 120 属性 124 题注 127 透视 119, 120 网格线 129 文字说明 127 显示和隐藏单元格 123 显示隐藏的边框 129 新表的缺省外观 170 旋转标签 120 选择行和列 129 移动行和列 120 遗存表 131 语言 122 粘贴到其他应用程序中 117 粘贴为表 117 值标签 121 字体 127 透视表中的单元格 格式 125 宽度 129 显示 123 选择 129 隐藏 123 透视表中的颜色 边框 126 图标 在对话框中 2 图表 从透视表创建 130 大小 155 导出 107 概述 153 宽高比 168 面板换行 155 模板 155, 168 缺失值 155 图表编辑器 属性 154 图表构建器 图库 153 图表选项 168 外部连接 15 网格线 透视表 129 未知测量级别 61 文本 文本 (继续) 将输出导出为文本 107, 110, 219 数据文件 10 文档恢复 2 文件 打开 7 文件变换 拆分文件处理 91 文件位置 控制缺省文件位置 171 文件信息 20 文件转换 变换变量和个案 86 对个案进行加权 93 对个案排序 85 合并数据文件 87, 88 聚集数据 90 重组数据 94 文字说明 127 显示 标题 123 工具栏 175 脚注 127 结果 116 维度标签 123 文字说明 127 行或列 123 显示格式 46 显示顺序 120 行数 在透视表中选择 129 性能 高速缓存数据 34 虚拟活动文件 33 旋转 使用 OMS 控制导出的输出 222 旋转标签 120 选项 查看器 166 常规 163 脚本 172 两位数的年份 166 数据 166 透视表外观 170 语言 165 选择方法 在透视表中选择行和列 129 选择个案 个案范围 93 基于选择条件 93 日期范围 93 时间范围 93 随机样本 93 延迟函数 73, 83 颜色编码 语法编辑器 146 样式输出 138 页编号 112 页脚 112 页眉 112 页面设置 图表大小 112 页眉和页脚 112 移动行和列 120 索引 247 移动中位数函数 83 移去组标签 120 遗存表 131 已过滤的个案 在数据编辑器中 55 隐藏 标题 123 工具栏 175 脚注 127 维度标签 123 文字说明 127 行和列 123 隐藏变量 对话框列表 162 数据编辑器 162 应用程序方式 工作簿 163 经典 163 用户缺失值 47 有序 测量级别 45, 61 语法 访问《命令语法参考》 5 日志文件 151, 152 生产作业规则 211 使用工具栏按钮运行命令语法 176 输出日志 144 语法规则 143 粘贴 144 正在运行 150 Unicode 命令语法文件 150, 151 语法编辑器 断点 147, 150 多个视图/窗格 145 分界点 145 命令跨度 145 设置语法格式 149 书签 145, 148 行号 145 颜色编码 146 语法缩进 149 注释或取消注释文本 149 自动完成 146 options 173 语法文件 加密 152 语法转换器 235 语言 更改输出语言 122 语言设置 165 元文件 导出图表 107 远程服务器 编辑 37, 214 登录 37 可用过程 40 数据文件访问 38 添加 37, 214 相对路径 40 在语法文件中删除多条 EXECUTE 151, 152 增强型定制对话框 181 粘贴输出到其他应用程序中 117 整体查看器 模型摘要 135 整体查看器 (继续) 预测变量频率 135 预测变量重要性 135 自动数据准备 136 组件模型精确性 136 组件模型详细信息 136 正态得分 在个案排秩中 77 值标签 插入换行符 47 复制 61 应用到多个变量 61 用于数据输入 52 在 Excel 文件中保存。 21 在概要窗格中 168 在合并数据文件中 88 在数据编辑器中 56 在透视表中 168 指定的窗口 1 制表符分隔文件 保存 21 打开 7 中心移动平均值函数 83 重复个案（记录） 查找和过滤 65 重构数据 为变量到个案创建多个索引变量 98 重命名数据集 57 重新排序行和列 120 重组数据 变量到个案的变量组 96 变量到个案的单索引示例 97 变量到个案的示例 95 变量到个案的双索引示例 98 变量到个案的选项 99 概述 94 个案到变量的示例 96 个案到变量的选项 100 和加权数据 100 为变量到个案创建单个索引变量 98 为变量到个案创建索引变量 97 为个案到变量排序数据 99 选择变量到个案的数据 96 选择个案到变量的数据 99 重组的类型 94 转换值 73 字典 20 字符编码 165 字符串变量 计算新的字符串变量 71 截断早期发行版中的长字符串 21 缺失值 47 输入数据 52 重新编码为连续整数 39, 75 字符串格式 45 字体 在概要窗格中 116 在数据编辑器中 56 自定义货币格式 45, 167 自动化生产 211 自动恢复 2, 163 自动脚本 触发事件 227 创建 228 248 IBM SPSS Statistics 29 Core System 用户指南 自动脚本 (继续) 关联查看器对象 228 基本版 232 自动输出修改 137–141 自由字段格式 10 子标题 图表 154 子类型 与标签 219 组标签 120 A Access (Microsoft) 14 B Blom 估计 77 BMP 文件 导出图表 107, 111 C Cognos 导出到 Cognos TM1 30 读取 Cognos Business Intelligence 数据 17 读取 Cognos TM1 数据 19 COMMA 格式 45, 46 CSV 格式 保存数据 21 读取数据 9, 10 CTABLES 将 TABLES 命令语法转换为 CTABLES 235 D DATA LIST 与 GET DATA 命令 33 dBASE 文件 保存 21 读取 7, 9 DOLLAR 格式 45, 46 DOT 格式 45, 46 E EPS 文件 导出图表 107, 111 Excel 格式 导出输出 107, 108, 219 Excel 文件 保存 21 保存值标签，而非值 21 打开 7 读取 8 添加将数据发送到 Excel 的菜单项 175 EXECUTE（命令） 从对话框粘贴 151, 152 G GET DATA GET DATA (继续) 与 DATA LIST 命令 33 与 GET CAPTURE 命令 33 GGRAPH 将 IGRAPH 转换为 GGRAPH 235 H HTML 导出输出 107 I IBM SPSS Statistics 数据文件格式 将输出转到数据文件 219, 222 IGRAPH 将 IGRAPH 转换为 GGRAPH 235 J JPEG 文件 导出图表 107, 111 L Lotus 1-2-3 文件 保存 21 打开 7 添加将数据发送到 Lotus 的菜单项 175 M Microsoft Access 14 O OMS 表子类型 218 从查看器排除输出 222 将 XSLT 与 OXML 一起使用 225 控制表旋转 219, 222 命令标识 218 输出对象类型 218 文本格式 219 Excel 格式 219 IBM SPSS Statistics 数据文件格式 219, 222 PDF 格式 219 SAV 文件格式 219, 222 SAV 文件中的变量名称 223 Word 格式 219 XML 219, 223 options 变量视图 167 货币 167 输出标签 168 数据编辑器中的描述统计 168 图表 168 语法编辑器 173 Options 临时目录 171 output 导出 107 索引 249 output (继续) 对齐 117, 166 更改输出语言 122 集中 117, 166 交互式 106 修改 137–141 OXML 225 P PDF 导出输出 107, 109 PDF 格式 导出输出 219 PNG 文件 导出图表 107, 111 PostScript 文件（压缩） 导出图表 107, 111 PowerPoint 将输出导出为 PowerPoint 109 PowerPoint 格式 导出输出 107 Python 脚本 228 R Rankit 估计 77 S SAS 文件 保存 21 打开 7 读取 7 SAV 文件格式 将输出转到 IBM SPSS Statistics 数据文件 219, 222 Savage 得分 77 spp 文件 转换到 spj 文件 216 SPSSTMPDIR 环境变量 171 Stata 文件 保存 21 打开 7, 9 读取 7 SYSTAT 文件 打开 7 T T4253H 平滑 83 TableLooks 创建 124 TIFF 文件 导出图表 107, 111 TM1 导出到 Cognos TM1 30 读取 Cognos TM1 数据 19 Tukey 估计 77 U Unicode 7, 20, 165 Unicode 命令语法文件 150, 151 V Van der Waerden 估计 77 W windows 活动窗口 1 指定的窗口 1 Word 格式 导出输出 107, 108, 219 宽表 107 X XML 从 OMS 输出 OXML 225 将输出保存为 XML 217 将输出转到 XML 219 OXML 中的表结构 223 XSLT 与 OXML 一起使用 225 Z z 得分 在个案排秩中 77 250 IBM SPSS Statistics 29 Core System 用户指南 IBM® |
188185 | https://courses.lumenlearning.com/suny-fmcc-macroeconomics/chapter/price-elasticity-of-demand-and-price-elasticity-of-supply/ | Price Elasticity of Demand and Price Elasticity of Supply | OpenStax Macroeconomics 2e
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OpenStax Macroeconomics 2e
Elasticity
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Price Elasticity of Demand and Price Elasticity of Supply
Learning Objectives
By the end of this section, you will be able to:
Calculate the price elasticity of demand
Calculate the price elasticity of supply
Both the demand and supply curve show the relationship between price and the number of units demanded or supplied. Price elasticity is the ratio between the percentage change in the quantity demanded (Qd) or supplied (Qs) and the corresponding percent change in price. The price elasticity of demand is the percentage change in the quantity demanded of a good or service divided by the percentage change in the price. The price elasticity of supply is the percentage change in quantity supplied divided by the percentage change in price.
We can usefully divide elasticities into three broad categories: elastic, inelastic, and unitary. An elastic demand or elastic supply is one in which the elasticity is greater than one, indicating a high responsiveness to changes in price. Elasticities that are less than one indicate low responsiveness to price changes and correspond toinelastic demand or inelastic supply. Unitary elasticities indicate proportional responsiveness of either demand or supply, as [link] summarizes.
Elastic, Inelastic, and Unitary: Three Cases of Elasticity| If . . . | Then . . . | And It Is Called . . . |
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| % change in quantity>% change in price% change in quantity>% change in price | % change in quantity% change in price>1% change in quantity% change in price>1 | Elastic |
| 2 π r 2 π r | π r 2 π r 2 | Unitary |
| % change in quantity<% change in price% change in quantity<% change in price | % change in quantity% change in price<1% change in quantity% change in price<1 | Inelastic |
Before we delve into the details of elasticity, enjoy this article on elasticity and ticket prices at the Super Bowl.
To calculate elasticity along a demand or supply curve economists use the average percent change in both quantity and price. This is called the Midpoint Method for Elasticity, and is represented in the following equations:
% change in quantity=Q 2−Q 1(Q 2+Q 1)/2×100% change in price=P 2−P 1(P 2+P 1)/2×100% change in quantity=Q 2−Q 1(Q 2+Q 1)/2×100% change in price=P 2−P 1(P 2+P 1)/2×100
The advantage of the Midpoint Method is that one obtains the same elasticity between two price points whether there is a price increase or decrease. This is because the formula uses the same base (average quantity and average price) for both cases.
Calculating Price Elasticity of Demand
Let’s calculate the elasticity between points A and B and between points G and H as [link] shows.
Figure 5.2 Calculating the Price Elasticity of Demand We calculate the price elasticity of demand as the percentage change in quantity divided by the percentage change in price.
First, apply the formula to calculate the elasticity as price decreases from $70 at point B to $60 at point A:
% change in quantity=3,000−2,800(3,000+2,800)/2×100=200 2,900×100=6.9% change in price=60−70(60+70)/2×100=−10 65×100=−15.4 Price Elasticity of Demand=6.9−15.4=0.45% change in quantity=3,000−2,800(3,000+2,800)/2×100=200 2,900×100=6.9% change in price=60−70(60+70)/2×100=−10 65×100=−15.4 Price Elasticity of Demand=6.9−15.4=0.45
Therefore, the elasticity of demand between these two points is 6.9−15.4 6.9−15.4 which is 0.45, an amount smaller than one, showing that the demand is inelastic in this interval. Price elasticities of demand are always negative since price and quantity demanded always move in opposite directions (on the demand curve). By convention, we always talk about elasticities as positive numbers. Mathematically, we take the absolute value of the result. We will ignore this detail from now on, while remembering to interpret elasticities as positive numbers.
This means that, along the demand curve between point B and A, if the price changes by 1%, the quantity demanded will change by 0.45%. A change in the price will result in a smaller percentage change in the quantity demanded. For example, a 10% increase in the price will result in only a 4.5% decrease in quantity demanded. A 10% decrease in the price will result in only a 4.5% increase in the quantity demanded. Price elasticities of demand are negative numbers indicating that the demand curve is downward sloping, but we read them as absolute values. The following Work It Out feature will walk you through calculating the price elasticity of demand.
work it out
Finding the Price Elasticity of Demand
Calculate the price elasticity of demand using the data in [link] for an increase in price from G to H. Has the elasticity increased or decreased?
Step 1. We know that:
Price Elasticity of Demand=% change in quantity% change in price Price Elasticity of Demand=% change in quantity% change in price
Step 2. From the Midpoint Formula we know that:
change in quantity=Q 2−Q 1(Q 2+Q 1)/2×100 change in price=P 2−P 1(P 2+P 1)/2×100 change in quantity=Q 2−Q 1(Q 2+Q 1)/2×100 change in price=P 2−P 1(P 2+P 1)/2×100
Step 3. So we can use the values provided in the figure in each equation:
% change in quantity=1,600−1,800(1,600+1,800)/2×100=−200 1,700×100=−11.76% change in price=130−120(130+120)/2×100=10 125×100=8.0% change in quantity=1,600−1,800(1,600+1,800)/2×100=−200 1,700×100=−11.76% change in price=130−120(130+120)/2×100=10 125×100=8.0
Step 4. Then, we can use those values to determine the price elasticity of demand:
Price Elasticity of Demand=% change in quantity% change in price=−11.76 8=1.47 Price Elasticity of Demand=% change in quantity% change in price=−11.76 8=1.47
Therefore, the elasticity of demand from G to is H 1.47. The magnitude of the elasticity has increased (in absolute value) as we moved up along the demand curve from points A to B. Recall that the elasticity between these two points was 0.45. Demand was inelastic between points A and B and elastic between points G and H. This shows us that price elasticity of demand changes at different points along a straight-line demand curve.
Calculating the Price Elasticity of Supply
Assume that an apartment rents for $650 per month and at that price the landlord rents 10,000 units are rented as [link] shows. When the price increases to $700 per month, the landlord supplies 13,000 units into the market. By what percentage does apartment supply increase? What is the price sensitivity?
Figure 5.3 Price Elasticity of Supply We calculate the price elasticity of supply as the percentage change in quantity divided by the percentage change in price.
Using the Midpoint Method,
% change in quantity=13,000−10,000(13,000+10,000)/2×100=3,000 11,500×100=26.1% change in price=$700−$650($700+$650)/2×100=50 675×100=7.4 Price Elasticity of Supply=26.1 7.4=3.53% change in quantity=13,000−10,000(13,000+10,000)/2×100=3,000 11,500×100=26.1% change in price=$700−$650($700+$650)/2×100=50 675×100=7.4 Price Elasticity of Supply=26.1 7.4=3.53
Again, as with the elasticity of demand, the elasticity of supply is not followed by any units. Elasticity is a ratio of one percentage change to another percentage change—nothing more—and we read it as an absolute value. In this case, a 1% rise in price causes an increase in quantity supplied of 3.5%. The greater than one elasticity of supply means that the percentage change in quantity supplied will be greater than a one percent price change. If you’re starting to wonder if the concept of slope fits into this calculation, read the following Clear It Up box.
clear it up
Is the elasticity the slope?
It is a common mistake to confuse the slope of either the supply or demand curve with its elasticity. The slope is the rate of change in units along the curve, or the rise/run (change in y over the change in x). For example, in [link], at each point shown on the demand curve, price drops by $10 and the number of units demanded increases by 200 compared to the point to its left. The slope is –10/200 along the entire demand curve and does not change. The price elasticity, however, changes along the curve. Elasticity between points A and B was 0.45 and increased to 1.47 between points G and H. Elasticity is the percentage change, which is a different calculation from the slope and has a different meaning.
When we are at the upper end of a demand curve, where price is high and the quantity demanded is low, a small change in the quantity demanded, even in, say, one unit, is pretty big in percentage terms. A change in price of, say, a dollar, is going to be much less important in percentage terms than it would have been at the bottom of the demand curve. Likewise, at the bottom of the demand curve, that one unit change when the quantity demanded is high will be small as a percentage.
Thus, at one end of the demand curve, where we have a large percentage change in quantity demanded over a small percentage change in price, the elasticity value would be high, or demand would be relatively elastic. Even with the same change in the price and the same change in the quantity demanded, at the other end of the demand curve the quantity is much higher, and the price is much lower, so the percentage change in quantity demanded is smaller and the percentage change in price is much higher. That means at the bottom of the curve we’d have a small numerator over a large denominator, so the elasticity measure would be much lower, or inelastic.
As we move along the demand curve, the values for quantity and price go up or down, depending on which way we are moving, so the percentages for, say, a $1 difference in price or a one unit difference in quantity, will change as well, which means the ratios of those percentages and hence the elasticity will change.
Key Concepts and Summary
Price elasticity measures the responsiveness of the quantity demanded or supplied of a good to a change in its price. We compute it as the percentage change in quantity demanded (or supplied) divided by the percentage change in price. We can describe elasticity as elastic (or very responsive), unit elastic, or inelastic (not very responsive). Elastic demand or supply curves indicate that quantity demanded or supplied respond to price changes in a greater than proportional manner. An inelastic demand or supply curve is one where a given percentage change in price will cause a smaller percentage change in quantity demanded or supplied. A unitary elasticity means that a given percentage change in price leads to an equal percentage change in quantity demanded or supplied.
Self-Check Questions
From the data in [link] about demand for smart phones, calculate the price elasticity of demand from: point B to point C, point D to point E, and point G to point H. Classify the elasticity at each point as elastic, inelastic, or unit elastic.
| Points | P | Q |
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| A | 60 | 3,000 |
| B | 70 | 2,800 |
| C | 80 | 2,600 |
| D | 90 | 2,400 |
| E | 100 | 2,200 |
| F | 110 | 2,000 |
| G | 120 | 1,800 |
| H | 130 | 1,600 |
Show Solution
From point B to point C, price rises from $70 to $80, and Qd decreases from 2,800 to 2,600. So:
% change in quantity=2600−2800(2600+2800)÷2×100=−200 2700×100=−7.41% change in price=80−70(80+70)÷2×100=10 75×100=13.33 Elasticity of Demand=−7.41 13.33=0.56% change in quantity=2600−2800(2600+2800)÷2×100=−200 2700×100=−7.41% change in price=80−70(80+70)÷2×100=10 75×100=13.33 Elasticity of Demand=−7.41 13.33=0.56
The demand curve is inelastic in this area; that is, its elasticity value is less than one.
Answer from Point D to point E:
% change in quantity=2200−2400(2200+2400)÷2×100=−200 2300×100=−8.7% change in price=100−90(100+90)÷2×100=10 95×100=10.53 Elasticity of Demand=−8.7%10.53%=0.83% change in quantity=2200−2400(2200+2400)÷2×100=−200 2300×100=−8.7% change in price=100−90(100+90)÷2×100=10 95×100=10.53 Elasticity of Demand=−8.7%10.53%=0.83
The demand curve is inelastic in this area; that is, its elasticity value is less than one.
Answer from Point G to point H:
% change in quantity=1600−1800 1700×100=−200 1700×100=−11.76% change in price=130−120 125×100=10 125×100=8.00 Elasticity of Demand=−11.76%8.00%=−1.47% change in quantity=1600−1800 1700×100=−200 1700×100=−11.76% change in price=130−120 125×100=10 125×100=8.00 Elasticity of Demand=−11.76%8.00%=−1.47
The demand curve is elastic in this interval.
From the data in [link] about supply of alarm clocks, calculate the price elasticity of supply from: point J to point K, point L to point M, and point N to point P. Classify the elasticity at each point as elastic, inelastic, or unit elastic.
| Point | Price | Quantity Supplied |
---
| J | $8 | 50 |
| K | $9 | 70 |
| L | $10 | 80 |
| M | $11 | 88 |
| N | $12 | 95 |
| P | $13 | 100 |
Show Solution
From point J to point K, price rises from $8 to $9, and quantity rises from 50 to 70. So:
% change in quantity=70−50(70+50)÷2×100=20 60×100=33.33% change in price=$9−$8($9+$8)÷2×100=1 8.5×100=11.76 Elasticity of Supply=33.33%11.76%=2.83% change in quantity=70−50(70+50)÷2×100=20 60×100=33.33% change in price=$9−$8($9+$8)÷2×100=1 8.5×100=11.76 Elasticity of Supply=33.33%11.76%=2.83
The supply curve is elastic in this area; that is, its elasticity value is greater than one.
From point L to point M, the price rises from $10 to $11, while the Qs rises from 80 to 88:
% change in quantity=88−80(88+80)÷2×100=8 84×100=9.52%change in price=$11−$10($11+$10)÷2×100=1 10.5×100=9.52 Elasticity of Demand=9.52%9.52%=1.0% change in quantity=88−80(88+80)÷2×100=8 84×100=9.52%change in price=$11−$10($11+$10)÷2×100=1 10.5×100=9.52 Elasticity of Demand=9.52%9.52%=1.0
The supply curve has unitary elasticity in this area.
From point N to point P, the price rises from $12 to $13, and Qs rises from 95 to 100:
% change in quantity=100−95(100+95)÷2×100=5 97.5×100=5.13% change in price=$13−$12($13+$12)÷2×100=1 12.5×100=8.0 Elasticity of Supply=5.13%8.0%=0.64% change in quantity=100−95(100+95)÷2×100=5 97.5×100=5.13% change in price=$13−$12($13+$12)÷2×100=1 12.5×100=8.0 Elasticity of Supply=5.13%8.0%=0.64
The supply curve is inelastic in this region of the supply curve.
Review Questions
What is the formula for calculating elasticity?
What is the price elasticity of demand? Can you explain it in your own words?
What is the price elasticity of supply? Can you explain it in your own words?
Critical Thinking Questions
Transatlantic air travel in business class has an estimated elasticity of demand of 0.62, while transatlantic air travel in economy class has an estimated price elasticity of 0.12. Why do you think this is the case?
What is the relationship between price elasticity and position on the demand curve? For example, as you move up the demand curve to higher prices and lower quantities, what happens to the measured elasticity? How would you explain that?
Problems
The equation for a demand curve is P = 48 – 3Q. What is the elasticity in moving from a quantity of 5 to a quantity of 6?
The equation for a demand curve is P = 2/Q. What is the elasticity of demand as price falls from 5 to 4? What is the elasticity of demand as the price falls from 9 to 8? Would you expect these answers to be the same?
The equation for a supply curve is 4P = Q. What is the elasticity of supply as price rises from 3 to 4? What is the elasticity of supply as the price rises from 7 to 8? Would you expect these answers to be the same?
The equation for a supply curve is P = 3Q – 8. What is the elasticity in moving from a price of 4 to a price of 7?
Glossary
elastic demand when the elasticity of demand is greater than one, indicating a high responsiveness of quantity demanded or supplied to changes in price elastic supply when the elasticity of either supply is greater than one, indicating a high responsiveness of quantity demanded or supplied to changes in price elasticity an economics concept that measures responsiveness of one variable to changes in another variable inelastic demand when the elasticity of demand is less than one, indicating that a 1 percent increase in price paid by the consumer leads to less than a 1 percent change in purchases (and vice versa); this indicates a low responsiveness by consumers to price changes inelastic supply when the elasticity of supply is less than one, indicating that a 1 percent increase in price paid to the firm will result in a less than 1 percent increase in production by the firm; this indicates a low responsiveness of the firm to price increases (and vice versa if prices drop)price elasticity the relationship between the percent change in price resulting in a corresponding percentage change in the quantity demanded or supplied price elasticity of demand percentage change in the quantity demanded of a good or service divided the percentage change in price price elasticity of supply percentage change in the quantity supplied divided by the percentage change in price unitary elasticity when the calculated elasticity is equal to one indicating that a change in the price of the good or service results in a proportional change in the quantity demanded or supplied
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尽管有的教材版本中略去了《命题》部分，在部分高考卷中也不是考察的重点甚至不做考察，但对命题和集合的学习有助于建立初步的逻辑思维，这对数理学科的学习非常有帮助。
命题
命题部分需要掌握的核心内容较少并且较为简单，但由于是初次接触这种略显公式化地将“条件”和“结论”明确分开的语句模式，因而需要进行一定量的训练来形成思维模式。好在这部分的练习方式非常灵活、生动有趣。
（一）
基本概念
关于命题，最需要明确的就是，命题是一种推出（推断）关系，不是陈述、疑问、命令。
既然是推出（推断），它就可能是正确的（真命题），也可能是错误的（假命题）。
命题一定能判断真假。但是能判断真假的语句不一定是命题，比如虚假陈述。
举例
（1）某天你的朋友对你讲：“今天我买了巧克力。”
这句话陈述了一个事实（也可能是骗人的），没有推断，不是命题。
（2）紧接着他问你：“你要吃我买的巧克力吗？”
这是疑问句，抛出了问题，没有推断，也不是命题。
（3）看你没有反应，他继续说：“吃掉这块巧克力吧！”
这是祈使句，未作推断，也不是命题。
（4）你告诉他为什么不吃：“巧克力吃多了的话，会容易发胖的。”
这句话作了推出（推断），它根据“巧克力吃多了”，推断出“容易发胖”，因而是命题。
练习：对于一句话是否是命题，可以在日常生活中经常练习，比如专门抽出一个空闲的时间段，对自己讲出的每句话和朋友或家人对你讲的每句话，都进行判断是否是命题。
（二）
在学习命题时，必须要学会把命题拆分为明确且独立的“条件”和“结论”两部分，用“如果……那么……”的句式来表达。
很多命题乍看上去，条件和结论两部分区分得并不是很明确，因而需要专门地训练，
比如“巧克力吃多了容易发胖。”这句话，拆分为“如果……那么……”就是“如果（我）吃多了巧克力，那么（我）会容易发胖。”
练习：对于命题的拆分，可以用上面讲到的方法中一并练习：在某个时间段内，与朋友或家人约好，对讲出和听到的每个命题，都用“如果……那么……”的句式再复述一遍，并讨论在复述的过程中，意思是否产生了偏差，直到熟练为止。
（三）
练习：对初中、小学曾经学过的数学、物理、化学、生物等各科课本中的定理、公理、推论都进行是否是命题的判断，对其中的全部命题全部进行“如果……那么……”的改写形式。
比如“有两条边和它们的夹角相等的两个三角形全等”
可以写为：
如果有两个三角形它们有两条边和这两条边的夹角相等，
那么这两个三角形全等。
这样可以同时回顾复习之前学过的重要定理，同时更加明晰它们的推出关系。
（四）
对于给定命题，能够写出它的逆命题、否命题、逆否命题，并判断真假。
在能够熟练拆分命题的情况下，写出给定命题的逆命题、否命题、逆否命题都很简单，不做过多阐述。
值得注意的是，原命题和它的逆否命题真假一致，逆命题和否命题真假一致（逆命题和否命题互为逆否命题），原命题和逆命题（以及否命题）的真假无必然联系。
练习：在前文中回顾之前所学的定理公理推论时，再全部陈述出它们的逆命题、否命题和逆否命题，并判断它们的真假。
充分条件和必要条件
充分条件和必要条件的内容较少且相对简单，不做过多阐述。由于是初次接触，也需要一定量的练习来熟悉这套逻辑语言。
集合
《集合》是高中数学中非常基础的内容。尽管在高考中纯粹关于集合的题目很少，但是集合的基本思想和应用是贯穿于高中数学的各个部分的，因此这部分也要扎实掌握。
（一）
要理解集合是指一个群体，这个群体有可能能用通用的公式来表达，也可能只能一一列举。无论如何，这些元素组成了一个群体。这个群体（集合）可能有很多个元素，可能有无穷多个元素，也可能只有一个元素，或者没有元素。
要注意辨析的有：空集是没有任何元素的集合。
只有一个元素的集合和元素是不同的，前者是集合，后者是元素。
比如{1}就是一个集合，它只有一个元素：1。而1本身只是一个数字或者元素。
（二）
集合之间的关系和运算
熟练掌握全集、交集、并集、补集、和集、差集、（真）包含于的概念。
注意辨别“包含于”（⊆）是集合间的关系，“属于”（∈）是元素和集合的关系。
维恩图可以帮助直观理解集合间的关系，特别是集合运算部分，刚开始练习时，每个都要画维恩图。
在计算集合元素个数时，要理解计算式分别减去的各项和加上的各项是为什么？（因为被重复计算或被重复减去）
举例
card(AUBUC)=card(A)+card(B)+card(C)-card(A∩B)-card(B∩C)-ard(C∩A)+card(A∩B∩C)
注：card(X)表示集合X中元素的个数。
第一，全部相加：card(A)+card(B)+card(C)，
第二，由于A∩B、B∩C、C∩A中的元素，分别同时出现在A和B、B和C、C和A中，总共被算进去了2次（多了1次），因此要减去重复的，于是：
card(A)+card(B)+card(C)-card(A∩B)-card(B∩C)-card(C∩A)
第三，由于A∩B∩C中的元素，在单独的A+B+C时被算了3次，在A∩B、B∩C、C∩A中都出现了，在减去A∩B和B∩C和C∩A的元素个数时，A∩B∩C中的元素又都被减去了3次，相当于又根本没有算进来，因此要再加回来，因此得到原公式。
练习：用韦恩图和上述思路，推导出求4、5、6个集合的并集的元素个数的公式，并得出对任意的n个集合求它们并集的元素个数的公式。
（三）
子集与推出关系
这部分涉及到整个高中数学中非常基础的应用，主要有判断定义域和值域、解各类不等式等，因此要熟练掌握。
尤其是用数轴来表示数集的一部分以及数集间的关系。这在整个高中数学都会经常用到。
命题和集合部分的内容就这些，主要是多练习，养成思维模式。
下篇预定讲函数初步，是高中数学的重中之重，敬请期待。
编辑于 2022-02-25 09:07
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112 条评论
默认
最新
逐光
感谢作者哥哥。真心觉得基础知识太重要了，看似简单，但是理解不透彻，遇到难题就不会做了。我高中有4个复读的同学，第一次高考都是4百多分，3本太贵上不了，只能上专科，但是不甘心就去另外一所重点高中复读，复读第一天物理老师兼班主任就说：“数学物理遇到难题你们不会做，只会做简单一点的题目，根本原因是基础不扎实，就算刷题再多，也提高不了的。看似简单的公式，定理，一定要明白背后的原理，只有这样，遇到难题就不怕了”。。。后来，那4个同学全部考上985了，我只知道一个是重庆大学，一个是武汉大学，还有其他的同学也考上一本，二本的。09年湖北高考
2019-10-29
回复61
视察者
Lucky1
比起前途来说学费不值一提
2021-05-05
回复7
心之所向
Lucky1
那就看你钱花在哪里更值了。
2021-06-25
回复1
展开其他 1 条回复
是庸
“对基础概念深入透彻理解”看了你的讲义我也许也能上个211哈哈
2019-08-31
回复39
rq cen
作者
我加油更，你加油学
2019-08-31
回复73
呐喊
rq cen
好呀好呀，您辛苦了
2020-02-04
回复11
我很会赚钱
辛苦了谢谢
2019-09-21 · 热评
回复31
年年
可以将这些内容作为高考复习的基础梳理么~
2019-08-20
回复20
Galen
“第二，由于A∩B、B∩C、C∩A中的元素（包括A∩B∩C的元素在内），分别同时出现在A和B、B和C、C和A中，总共被算进去了3次（多了2次），因此要减去重复的，于是.......”
（我觉得是A∩B 、B∩C、C∩A中的元素算了两次，A∩B∩C算了三次，这样才符合题意啊）
2019-11-17
回复7
rq cen
作者
谢谢，已更正
2019-11-17
回复4
rq cen
作者
启锁
再推一次就只要几分钟了
2021-03-22
回复3
查看全部 6 条回复
onesagkn
太棒了，马上转给孩子看看
2019-08-20
回复8
兄友弟恭
我不赞同您对“命题”下的定义，尽管它在高中阶段通常是适用的。以下是我所了解的定义（不是给您看的，因为您应该比我还懂。）
命题是对事物的陈述并且有真值。理解命题是推理、判断的基础。举例（例子就在你文章里找了）:如果一个人说:今天我买了巧克力。这是一个命题，通过询问等手段可得出真值——如果你买了巧克力，那么命题真值为真；如果你没买巧克力，命题真值为假。
你您举例的，“如果我巧克力吃多了，那么我会发胖”，是蕴涵关系的复合命题，有四种可能，只有当“你”巧克力真的吃多了，并且“你”没有发胖时，命题的真值为假。最终是真还是假，需要您亲自告诉我您吃的是多还是少，您胖了没有。
高中数学确实通常多研究这种蕴涵关系的命题，但以后呢？命题形式很多种，以偏概全是不妥的。
2020-09-17
回复8
石波清
“并且有真值”应该怎么理解？是说其值集中至少有一个为真？
2023-12-02
回复喜欢
蕙蒽
答主可以更一下对练习的答案吗，这样方便自己对照写的对不对
2020-11-21
回复4
sciphoton
能判断真假的陈述句都是命题。比如“0是自然数”。又比如“今天我吃了巧克力”的意思是“在某个具体的时间某个具体的人吃了巧克力”。
2020-05-02
回复4
大橘为重
谢谢大佬
2020-09-06
回复1
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rq cen
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巴黎综合理工学院 分子化学硕士
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发表时间：2021-09-10 11:23
中国国际经济咨询有限公司 博士后工作站 刘斐然
随着政府投资审批权限下放、民营企业自我诊断能力加强以及信息技术膨胀式发展，以往程序型、低价值、模仿性和依靠专有数据库垄断的咨询业务将迅速萎缩。先进方法论是咨询机构在“十四五”时期参与行业内高端竞争、打造重大咨询项目、形成代表性咨询服务的关键武器。公司设立时就高度重视咨询理论研究和成果质量，亦是最早引进项目投资可行性研究方法论的咨询机构。研发能力构建与核心产品打造是公司“十四五”时期内功修炼方面的重点工作。本文围绕项目投资决策中的多目标优化问题，基于实际案例介绍多目标优化方法在此类问题上的应用。
一、常见的多目标决策问题
在进行项目投资决策时可能需要考虑多个目标，并且这些目标之间可能具有冲突且相互影响。以政府与社会资本合作（PPP）项目为例，政府方希望节约成本，社会投资人希望追求收益最大化，用户希望收费更低。通常在此类项目中，还包括公共资源作为项目补充资源的使用最小化等子目标。一个子目标的改善可能会引起另一个子目标上效益的降低（或成本的升高），通常不存在唯一的使所有目标函数同时达到最优值的绝对最优方案（解），而是存在多个相互之间无法比较绝对优劣的帕累托解（Pareto solutions）。
以下示例来自于互联网公开内容。例如有一个PPP项目中政府和社会资本方效益最大化（成本最小化）的多目标优化问题，包括两个目标函数（目标1和目标2），它们各自的决策结果空间如下图中上半部分所示。这两个目标的决策结果空间重叠部分如下图的左下角子图所示。其中，在红色点构成的区域上，要提升一个目标的效益，就要以牺牲另一个目标的效益为代价，因而无法绝对地比较这些方案的优劣。结果空间中的红色点在决策方案空间上的分布如下图的右下角子图所示，即帕累托前沿。
图片来源：知乎，原作者木木松
现实中，最终方案的决策权被交给出资人，参与咨询或项目规划的人员负责生成这些备选方案，其中主要任务是找到这些帕累托解。此种决策方法省去了政府方和社会投资人要事先确定目标优先次序或加权比重的工作，因而为待解决问题提供了充分的决策空间。和传统的数学规划法相比，智能优化算法更适合求解多目标优化问题，因其对Pareto最优前端（Pareto Front，简称PF）的形状和连续性不敏感，能很好地刻画分布不均匀和不规则的帕累托解。
二、投资案例及分析条件设定
以下案例均来自于财政部政府和社会资本合作中心（CPPPC）管理库，有关信息和数据均摘自各案例项目的可行性研究报告、PPP实施方案报告、物有所值评价报告以及财政承受能力论证报告。
1、白云区黑石头片区污水处理厂工程
白云区黑石头片区污水处理厂工程PPP 项目（以下简称“项目”或“此项目”），系指以PPP 模式实施白云区黑石头片区污水处理厂工程的投资、建设、运营、维护、更新和移交。此项目属于新建项目，拟建厂址位于北二环黑石头大桥西北侧。项目所属行业为市政工程（污水处理）。此项目由贵阳市白云区人民政府（以下简称“区政府）发起。区政府授权白云区住房和城乡建设局（以下简称“区住建局”）作为此项目实施机构，由其依法和PPP 项目合同约定履行实施机构职责。区政府授权白云城市建设投资有限公司（以下简称：“白云城投公司”）为政府方出资代表。
此项目提供的公共产品或服务为：本工程服务范围为黑石头片区，俊发城地块，服务面积168.92 公顷。
此项目实施内容为：以“建设-运营-移交（Build-Operate-Transfer，BOT）”方式实施此项目生活污水处理设施的特许经营市场化运作。即由项目实施机构采用公开竞争方式选择确定社会资本和政府授权出资代表在白云区共同设立项目公司，授予项目公司指定期限的特许经营权，在特许经营期内由项目公司自筹资金负责此项目所涉全部污水处理设施、设备的投资、建设、运营、维护和更新，向政府方（实施机构）提供污水处理服务，收取可用性服务费和运维绩效服务费。
此项目前置条件与主要经济参数如下：
1）项目建设期为1年，运营期最长不超过29年，项目全生命周期不超过30年；
2）项目总投资6776.22万元，资本金比例为20%，余下建设投资通过贷款方式筹措；贷款利率为中长期基准率4.9%上浮30%，为6.37%；
3）计算政府方支付成本所用折现率为5.5%；
4）可能的风险成本率为3%，建设期计算基数为资本金，运营期计算基数为年运维成本（万元），估计的运维成本如下表：
5）项目回报机制为政府付费方式，即不向终端用户收取费用，运营期使用者付费部分为零；
6）项目若采用PPP模式运行，其竞争中立调整值为624.26万元。
此项目原方案内容为：
1）政府方的资本金出资比例为10%，社会投资人为90%，项目公司股权比例以此划分；
2）对于社会投资人投入的建设成本和运营成本，按合理利润率6%给予回报；
3）运营期设为29年。
2、清镇市农村环境综合整治项目
清镇市农村环境综合整治PPP项目包括河道综合治理工程、污水处理工程、垃圾收运系统工程三类项目，共20个子项（包括4个河道综合治理工程、7个污水处理工程和9个垃圾转运站工程）。清镇市水务管理局受清镇市人民政府授权担任此项目实施机构，负责：项目识别、项目准备、项目采购，与中标社会资本及后续成立的项目公司签署PPP 项目合同，并牵头市财政局、发改局等相关部门对项目公司进行监管和考核，合作期满项目移交等工作。
河道综合治理工程类项目包含民联河、中山沟、毛家寨沟、右七河综合治理4个子项，治理总长度为6184m。其中民联河治理长度1353m，中山沟治理长度1900m，毛家寨沟治理长度2181m，右七河治理长度750m。
污水处理工程类项目包括乡镇污水处理工程（暗流镇污水处理工程、犁倭镇污水处理工程、流长乡污水处理工程、王庄乡污水处理工程、王庄新城污水处理站、麦格乡污水处理工程）及蔡水村一、二组污水治理工程，共计 7 个污水处理工程子项。
垃圾收运系统主要包括源头垃圾收集点、垃圾收集车、垃圾转运站及配套设施、垃圾运输车、保洁清扫车辆设施。根据经批复的可研报告，垃圾收运系统工程含9个新建转运站子项，设计规模为900吨天。
清镇市农村综合环境整治属于系统性工程，其中河道治理、污水处理及垃圾收运属于较为明显的基础设施短板，这三类基础设施的缺失给居民的生活带来较大的不便。此项目三类项目的服务范围在空间上存在重叠，且存在功能互相促进的内在联系：河道治理和垃圾的有序收集及清运能有效减少进厂污水的含沙量及大颗粒悬浮物，起到预处理的效果，提高污水处理厂的运行效率。同时污水经处理后排放也能进一步改善和保持河道的水质。因而，此项目的项目内容存在内在联系。下图为此项目交易结构图。
此项目前置条件与主要经济参数如下：
1）项目建设期为2年，运营期最长不超过28年，项目全生命周期不超过30年；
2）项目总投资25355.53万元，资本金比例为20%，余下建设投资通过贷款方式筹措；贷款利率为6.86%；
3）计算政府方支付成本所用折现率为7.5%；
4）可能的风险成本率约为3%（为转移风险与自留风险成本之和），为保持计算口径一致，假设此项目风险成本计算基础与前一项目相同，所估计的运维成本（万元）见下表：
5）项目运营期向终端用户收取少量服务费用，具体使用者付费情况见下表:
6）项目若采用PPP模式运行，其竞争中立调整值为2324.27万元。
此项目原方案内容为：
1）政府方的资本金出资比例为5%，社会投资人为95%，项目公司股权比例以此划分；
2）对于社会投资人投入的建设成本和运营成本，按合理利润率7.5%给予回报；
3）运营期设为18年。
3、乌当区教育设施、设备及后勤服务外包项目
贵阳市乌当区教育设施、设备及后勤服务外包PPP项目由贵阳市乌当区人民政府发起，贵阳市乌当区教育局为本项目实施机构。贵阳市乌当区人民政府授权区属国有企业贵阳泉城五韵旅游文化投资开发有限公司为政府方出资代表。
此项目所涉三个工程子项目的建设地点为：贵州省贵阳市乌当区水田镇、下坝镇和振新社区。乌当区下坝镇九年制学校建设项目总建筑面积36824.2㎡，土地属国有土地划拨；乌当区水田中学建设项目总占地面积66667.19㎡，总建筑面积58858.18㎡土地拟以划拨方式取得乌当区振新幼儿园建设项目总占地面积6630.72㎡，规划新建总建筑面积3364.68㎡，土地拟以划拨方式取得。此项目的教育设施建设项目内容包括：乌当区下坝镇九年制学校建设项目、乌当区水田中学建设项目、乌当区振新幼儿园建设项目、教学仪器、设备等器材、后勤服务。
下图为此项目的交易结构。
此项目前置条件与主要经济参数如下：
1）项目建设期为1年，运营期最长不超过29年，项目全生命周期不超过30年；
2）项目总投资32546.56万元，资本金比例为20%，余下建设投资通过贷款方式筹措；贷款利率为6.86%；
3）计算政府方支付成本所用折现率为7.5%；
4）可能的风险成本率约为5%，为保持计算口径一致，假设此项目风险成本计算基础与前一项目相同，即以建设期和运营期资金投入为基础计算风险成本，项目所估计的运维年成本为2139.42万元。
5）项目运营期向终端用户收取少量服务费用，具体使用者付费情况见下表:
6）项目若采用PPP模式运行，其竞争中立调整值为1504.6万元。
此项目原方案内容为：
1）政府方的资本金出资比例为15%，社会投资人为85%，项目公司股权比例以此划分；
2）对于社会投资人投入的建设成本和运营成本，按合理利润率9.5%给予回报；
3）运营期设为12年。
4、多目标优化分析条件设定
决策变量。根据各项目的PPP实施方案报告，将股权比例（资本金投入比例）、建设及运营成本资金回报率（合理回报率）、社会投资人收益率以及运营期年数作为模型的决策变量。这些变量的取值将决定一个投资方案的最终结果。
风险估计。案例项目风险波动情况通过几何布朗运动模拟，成本波动率通过2011-2020年工业生产者采购价格指数确定。单次优化中，决策方案的模拟数量设为200个，迭代次数为50代。
风险偏好假设。为更加符合政策要求以及贴合实际经验，在各案例贷款利率不变的前提下，假设还贷期与运营期保持一致，各项目运营期不低于10年。同时，设定社会投资人的风险偏好为时间风险厌恶型，即更倾向于较快回本而不是一味地追求回报总额。
模型实现及有效性说明。优化结果根据建立的多目标优化模型得出，具体代码于软件Matlab2020b中实现。需要说明的是，因案例项目均已经落地，优化模型通过原方案数据测算出的结果不作为项目实施方案的参考，仅用于对照模型的优化效果。
特别地，关于社会投资人回报率。社会投资人最终的回报率由采购结果确定，因而在前期决策中，这一信息相对于决策者是不可完全掌握的。现实中，通常社会投资人可接受的收益率不会超过10%，因而将社会投资人收益率设为一个取值在0~10%范围内的决策变量，用于计算其收益现值。
三、案例优化结果的统计描述
（1）白云区黑石头片区污水处理厂工程PPP项目
下图为200个生成的决策方案在四个决策变量取值范围上的分布图。其中，政府方股权比重在10%（0.1）以下的方案约140个左右，其余约60个方案在这一变量的取值上介于10%~49%之间。合理回报率在1%以下的方案约有120个，数目第二多的区间为4%~5%。在社会投资人折现率变量上，几乎全部200个方案的取值都极为接近零，说明这些优化方案给出的是社会投资人盈亏平衡状态下的结果。在运营年数上，近150个方案的运营年数都为10，剩余四分之一的方案年数都在11~15年之间，说明优化结果相对于原方案大大地缩短了运营周期，使得政府方总体支付水平明显下降。
（2）清镇市农村环境综合整治PPP项目
下图为清镇市农村环境综合整治PPP项目200个生成的决策方案在四个决策变量取值范围上的分布图。政府方股权比重在0~30%的区间上的方案数约占总数（200个）的五分之二，剩余五分之三集中于30%~49%的区间上。在合理回报率方面，约有190个方案集中于0~0.1%的区间上。在社会投资人收益率（折现率）方面，与上一案例情况类似地，几乎全部200个优化方案的取值都处于0.01%（0.0001）的区间上。运营年数上，绝大多数优化方案分布于10~13年的区间上。综合以上结果来看，项目运营期成本占政府方支付总成本比重较大，因而优化方案通过提升政府方资本金投入比重、压缩回报率以及降低运营年数来使得政府方支付水平降低，同时确保社会投资人能够收回成本。
（3）乌当区教育设施、设备及后勤服务外包PPP项目
下图为乌当区教育设施、设备及后勤服务外包PPP项目200个生成的决策方案在四个决策变量取值范围上的分布图。图中显示，在股权比重方面，全部200个优化方案均集中于40%以下的区间，其中约超过160个方案集中于10%以下的区间。在合理回报率上，约半数优化方案集中0~0.1%的区间上，剩余方案在0.2~0.8%区间上分布较为均匀，最大不超过1.4%。和上述两案例类似地，社会投资人折现率（收益率）均被控制在0.01%水平以下。从运营年数来看，全部优化方案较为均匀地分布于10，11和12年三个水平上。
综合上述结果来看，优化方案给出的结果均为社会投资人“保本”策略（收益率接近于零），能够为政府方决策者提供“拧干”后的项目投资数据。因而，为客观地反映优化效果，在以项目原方案数据进行比较时，将社会投资人财务基准收益率统一设置为一个非常小的值，例如0.0001，用作社会投资人计算其效益现值的折现率。当其现值大于零，意味着社会投资人参与该项目能够收回成本，不至亏损。
四、最优投资边界与原方案比较
（1）白云区黑石头片区污水处理厂工程PPP项目
该项目的帕累托前沿如下图所示。其中，优化方案在政府方成本现值上的范围约为9400万元~1.06亿元，在社会投资人收益现值上的范围约为500万元~1500万元。若采用原方案数据计算，政府方支出的现值总成本在1.3亿元~1.35亿元之间，社会投资人最总效益大于5000万。项目的PPP实施方案报告中，白云区财政29年的支付总成本现值上限22538.22 万元，其原因在于提高了政府方对于社会投资人资本金投入的回报，设定了贷款期为10年，对政府支付成本进行了较为保守的估计。
从图中可看出，优化决策方案的最优边界在其范围内是凸向左上角坐标原点的，表示最理想的情况是政府以尽可能小的成本投入实现了社会投资人尽可能大的收益。经原方案数据测算得出的结果显示，政府方通过付出相对较多的成本，使得社会投资人收益水平更靠近其预期水平，可能产生更强的引资效果。
（2）清镇市农村环境综合整治PPP项目
该项目的帕累托前沿如下图所示。其中，优化方案在政府方成本现值上的范围约为5.5亿元~6.3亿元，在社会投资人收益现值上的范围约为1.5亿元~2.5亿元。若采用原方案数据计算，政府方支出的现值总成本约为7.5亿元，社会投资人最总效益约为4.5亿元。
从图中可看出，优化决策方案的最优边界在其范围内分段斜切于左上角原点坐标系二维平面，表示政府方在不同的支付水平上，较高的支付水平可以实现社会投资人较高的回报，然而超出6.3亿元现值的界限时，项目可能无法满足“物有所值”定量评估的标准（VFM值大于零），这将意味着项目采用PPP模式是不经济的。与上一案例情况类似地，此项目原方案数据的测算结果表明，存在政府放大投入使得社会投资人获取高额回报的可能。
（3）乌当区教育设施、设备及后勤服务外包PPP项目
该项目的帕累托前沿如下图所示。其中，优化方案在政府方成本现值上的范围分段较为明显，集中于5.5亿元~5.6亿元，5.7亿元~5.75亿元，5.85亿元~5.9亿元的区间上。这种现象产生的可能原因是项目原始数据中存在较明显的阶梯状趋势。优化方案在社会投资人收益现值上的范围约为5000万元~7000万元。若采用原方案数据计算，政府方支出的现值总成本约为6.15亿元，社会投资人最总效益约为1亿元。
从图中可看出，优化决策方案的最优边界在其范围内分段斜切于左上角原点坐标系二维平面，且斜率较高，表示在一定区间上社会投资人的收益相对于政府的投入具有较高的弹性，政府可适当增加投入以增强项目对社会投资人吸引力。与上两个案例情况类似地，此项目原方案数据的测算结果表明，政府方放大了投入，可能致使社会投资人获取超额回报。
五、优化方案比选分析
从200个生成的优化方案中，选择政府方最小资本金投入比例（股权）、最小合理回报率（利润率）方案，政府方最小总成本方案以及社会投资人最大效益方案进行比选分析。
（1）白云区黑石头片区污水处理厂工程PPP项目
（2）清镇市农村环境综合整治PPP项目
（3）乌当区教育设施、设备及后勤服务外包PPP项目
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188188 | https://scienceatcchs.weebly.com/science-practices-blog/dimensional-analysis-disasters-homework | Dimensional Analysis & Disasters - HOMEWORK - HS SCIENCE @ CCHS
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Dimensional Analysis & Disasters - HOMEWORK
8/21/2015
11 Comments
Using the Internet as your resource, find at least 2 catastrophe’s that have occurred because of an error in Dimensional Analysis.
Show your resources.
Write a clear, brief summary explaining and/or describing the catastrophe and in what way that dimensional analysis was involved in the error, in the comments section below.
Be prepared to share your homework NEXT CLASS.
11 Comments
Lil Marie Alvarez
8/27/2015 11:15:35 am
Catastrophes:
The Tacoma Narrows Bridge Disaster (1940)
The Space Shuttle Disaster
The Tacoma Narrows Bridge had low torsional stiffness in the bridge road bed construction.
The Space Shuttle Disaster occurred because the space agency was ignoring/misinterpreting seal performance at low temperatures.
Sources:
Reply
Gabriela domene
8/27/2015 11:26:04 am
The tacoma narrows disaster: It was a disaster in 1940 that a bridge road bed construction had torsional stiffness problems
Comet air disaster:The windows started to crack and began to have eventual failure because of hight stress levels in the window surround structure.
Reply
Marta Nicolau
8/27/2015 11:31:56 am
The Titanic Disaster (1912)
The Comet Air Disaster (1954)
The Titanic Disaster had an error in the analysis of its sink ability. The ships compartments were open top, and the water would fill up and spill into the next compartment.
The Comet Air Disaster occurred due to too high stress levels in the window’s structure. This resulted in cracking and eventual failure.
Sources:
Reply
Stephanie Marie
8/27/2015 02:45:48 pm
1.The Vancouver Stock Exchange: The Vancouver Stock Exchange started with a value of 1000.000 and after each bargain they updated it. The value was shorten and not rounded. If it would have been rounded the value would have been 1098.892.
2.Explosion of the Ariane 5: There was a software failure in the inertial reference system, the 64 floating point number of the rocket was changed to a 16 bit signed integer, and because of that change the explosion occurred.
Source:
Reply
Miranda
8/30/2015 03:51:56 pm
The Patriot Missile Failure
An american missile battery failed to track an incoming iraqui missile due to inaccurate calculation of time
Mars probe lost
NASA lost its 125 million mars climate orbiter because "engineers mistook acceleration readings measured in english units of pounds-seconds for newton- seconds." (Lee Holtz)
Reply
anonymous
10/1/2018 05:32:05 pm
Thank you for the financial algebra amswere
Reply
Anomouyis
10/1/2018 11:30:44 pm
You live in fort myers
Nicolas Caimari
8/30/2015 04:26:32 pm
1) Columbus miscalculated the circumference of the earth, using Roman miles instead of nautical miles. He ended up in The Bahamas in October 12, 1492, and assumed it was Asia. (
2)In 1994 an aircraft was 30,000 pounds overweight because of a kilogram-to-pounds conversion error. (kilogram-to-pounds conversion)
Reply
Luis Emilio
8/31/2015 01:06:42 am
1.)
The explosion of the Ariane 5 rocket, on June 4, 1996, was ultimately the consequence of a simple overflow.
The sinking of the Sleipner A offshore platform in Gandsfjorden near Stavanger, Norway, on August 23, 1991, resulted in a loss of nearly one billion dollars.
3.)
Reply
Big Pops
9/22/2020 02:56:10 pm
yoo you gay boy
Reply
mari
11/4/2022 08:57:15 am
i fucking hate school
Reply
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188189 | https://www.sohu.com/a/749758723_120471245 | 高中数学——立体几何！
2024-01-05 15:18
来源:
成都教育帮帮帮
链接复制成功
发布于：四川省
高中数学——立体几何
知识点基本原理：
高中立体几何主要研究三维空间中的点、直线、平面以及多面体等几何对象的性质和相互关系。以下是一些核心概念和原理：
空间直角坐标系：通过设立三个互相垂直的数轴x、y、z，可以确定空间中任意一点的位置，并且能够方便地进行距离、夹角、平行、垂直等问题的计算。
向量在立体几何中的应用：利用向量的概念，可以描述空间中的线段方向和长度，判断线线、线面、面面的平行与垂直关系，计算异面直线所成角和二面角的大小。
空间直线和平面的一般方程：如直线的一般式为Ax + By + Cz = D，平面的一般式为Ax + By + Cz + D = 0，这些方程用于解决直线与平面的位置关系问题。
常见立体图形（棱柱、棱锥、球体等）的性质：包括表面积、体积的计算公式，以及各种特殊位置关系下的求解问题。
截面问题：分析立体图形被平面切割后所得截面的形状和面积。
题目示例及答案：
题目1（填空题）： 已知正方体ABCD-A'B'C'D'的棱长为a，则此正方体的表面积S为______。
答案：6a²
题目2（解答题）： 在空间直角坐标系中，已知平面α过原点，其法向量为n=(1,2,-1)，直线l的方向向量为d=(3,2,2)且过点P(1,1,1)，判断直线l与平面α的位置关系，并求出它们的夹角θ。
答案：首先求得点P到平面α的距离，若距离为0，则直线l在平面α内；若不为0，则直线l与平面α相交。然后利用向量的点积计算cosθ = |n·d| / (|n| |d|)，得到夹角θ。
题目3（证明题）： 在四棱锥P-ABCD中，底面ABCD是矩形，PD⊥平面ABCD，E是PC的中点，F是AD的中点，证明EF∥平面PBD。
证明：连接BF并延长交PD于G，因为底面ABCD为矩形，所以BF//PD，又因PD⊥平面ABCD，故BF⊥平面ABCD。而CF是平面PBC内的直线，所以BF⊥CF，即BF⊥平面PDC。由于E是PC的中点，所以GE是△PCD的中位线，因此GE//PD，所以GE⊥平面ABCD，从而EG⊥CF。根据线面平行判定定理，可知EF∥平面PBD。返回搜狐，查看更多
平台声明：该文观点仅代表作者本人，搜狐号系信息发布平台，搜狐仅提供信息存储空间服务。
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188190 | https://www.quora.com/What-is-the-relationship-between-the-centers-of-two-circles-that-are-tangent-to-each-other-at-a-point | Something went wrong. Wait a moment and try again.
Chords and Tangents
Mathematical Concepts
Geometric Tangency
Arcs and Curves
Tangent Law
Geometric Mathematics
Circles and Distances
5
What is the relationship between the centers of two circles that are tangent to each other at a point?
·
When two circles are tangent to each other at a point, the relationship between their centers can be described as follows:
Distance Between Centers
: The distance between the centers of the two circles is equal to the sum of their radii. If we denote the radii of the two circles as
r
1
and
r
2
, and the distance between their centers as
d
, then the relationship can be expressed mathematically as:
d
=
r
1
+
r
2
2. Point of Tangency
: The point where the circles are tangent lies on the line connecting their centers. This point is the closest point between the two circles.
3. Internal and External Tangent
When two circles are tangent to each other at a point, the relationship between their centers can be described as follows:
Distance Between Centers
: The distance between the centers of the two circles is equal to the sum of their radii. If we denote the radii of the two circles as
r
1
and
r
2
, and the distance between their centers as
d
, then the relationship can be expressed mathematically as:
d
=
r
1
+
r
2
2. Point of Tangency
: The point where the circles are tangent lies on the line connecting their centers. This point is the closest point between the two circles.
3. Internal and External Tangents
: If the circles are tangent externally, they touch at one point, and the distance between their centers is exactly the sum of their radii. If they are tangent internally, one circle lies inside the other, and the distance between their centers is the absolute difference of their radii:
d
=
|
r
1
−
r
2
|
In summary, for two externally tangent circles, their centers are separated by a distance equal to the sum of their radii, while for internally tangent circles, the distance equals the absolute difference of their radii.
Robert Anderson
Former Mediocre Engineer
·
Author has 589 answers and 233.1K answer views
·
2y
The centers and point of tangency are collinear. The line containing the centers and point of tangency are perpendicular to the tangent line.
The centers and point of tangency are collinear. The line containing the centers and point of tangency are perpendicular to the tangent line.
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There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you?
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Jack Vogt
Former Rocket Scientist (1956–1990)
·
Author has 1.8K answers and 255.1K answer views
·
2y
One relationship is
the distance between the centers is equal to the sum of the radii of the circles.
Another is
the point of tangency and the two centers are colinear (all on a line).
Related questions
How do you find the center of two circles that are tangent internally?
Does a pair of common tangents of two circles meet the line joining centres of the two circles at one point?
What is the common tangent of two circles in which one is inside the other and they don't touch each other?
What is the relationship between the areas of two circles that are tangent to each other?
What is the relationship between the radii and centres of three mutually tangent circles?
Chris Hasemore
Studied Mathematics
·
Author has 568 answers and 63.3K answer views
·
1y
Related
What is the relationship between the centers of two circles that have a common tangent and touch each other at exactly one point?
The centres of both circles lie on the perpendicular line to the tangent through the point of contact. And the distance between those centres is R+/-r (depending on whether the circles touch internally (a small circle INSIDE the bigger circle) or externally) where R=radius of larger circle & r=rad of smaller circle.
Jan Bolluyt
Studied Mathematics & Physics (till 12th) (Graduated 1970)
·
Author has 959 answers and 71.6K answer views
·
Jun 17
Related
What is the angle between two circles that are tangent to each other?
When two circles are tangent to each other, the “angle” is 180 degrees formed by the two radii that are the same line segment.
When two circles are tangent to each other, the “angle” is 180 degrees formed by the two radii that are the same line segment.
Related questions
How do I find the point of intersection of direct common tangents for two circles which touch each other?
What is the relationship between the radii of two circles that are tangent to each other?
Do concentric circles have common tangents?
What is the relationship between two circles if they do not have a common tangent?
What is the distance between the top points of two intersecting circles when the tangents from each circle meet at a point at a given angle?
Mariano Zomeño
Studied Physics at Complutense University of Madrid
·
Author has 300 answers and 1.3M answer views
·
8y
Related
Is it necessary that when two circles intersect then the tangents at the point of intersection passes though the centres of the circles?
Nope, and here’s an example: two circles intersecting in I, whose tangent doesn’t go through the centers.
Please excuse those disgusting circles, they’re what happens when you use tape as a compass.
Antway, here’s the trick to draw tangent line: pick the point P from the circumference in which you want the tangent line. Then, join the centre O with P. Finally, draw a line perpendicular to OP that goes through P. That’s your tangent.
As you can see, the tangent only goes through another circumference’s centre if said centre gets in the way, that is, if the centres and the point P form a ninety deg
Nope, and here’s an example: two circles intersecting in I, whose tangent doesn’t go through the centers.
Please excuse those disgusting circles, they’re what happens when you use tape as a compass.
Antway, here’s the trick to draw tangent line: pick the point P from the circumference in which you want the tangent line. Then, join the centre O with P. Finally, draw a line perpendicular to OP that goes through P. That’s your tangent.
As you can see, the tangent only goes through another circumference’s centre if said centre gets in the way, that is, if the centres and the point P form a ninety degree angle.
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Arega Tayu
Author has 2.6K answers and 1.1M answer views
·
1y
Related
What is the relationship between the centers of two circles that have a common tangent and touch each other at exactly one point?
The relationship between the centers of two circles with a common tangent is that their centers are lying on the straight line perpendicular to the tangent at the point of contact.
David Joyce
Dave's Short Course in Trig,
·
Author has 9.9K answers and 68.1M answer views
·
8y
Related
What is the relationship with a tangent line and tangent as a trigonometric ratio?
The tangent is the length of a line DE tangent to the unit circle cut off by the angle EAB in standard position.
Also the secant is the length of AD that cuts the circle.
The tangent is the length of a line DE tangent to the unit circle cut off by the angle EAB in standard position.
Also the secant is the length of AD that cuts the circle.
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Mindy Dyar
Math whiz and tutor, IT/EE superstar, Mom of 4, total badass
·
Author has 184 answers and 234.1K answer views
·
4y
Related
What is the angle made by the radius of a circle with the tangent of the circle at the point of contact?
The tangent is the slope of the line that goes through a given point on a curve that gives a linear approximation of the steepness of the curve at that point.
In a circle, if you take any point on the circumference, the tangent will be the same. It is the same at all points because a circle is uniform, meaning its curve is exactly the same all the way around.
Now the radius of a circle is a line segment drawn from the center of the circle to any point on the circumference.
The angle made between the radius and the tangent at the point of contact on a circle is 90°.
The tangent line is purple. Forg
The tangent is the slope of the line that goes through a given point on a curve that gives a linear approximation of the steepness of the curve at that point.
In a circle, if you take any point on the circumference, the tangent will be the same. It is the same at all points because a circle is uniform, meaning its curve is exactly the same all the way around.
Now the radius of a circle is a line segment drawn from the center of the circle to any point on the circumference.
The angle made between the radius and the tangent at the point of contact on a circle is 90°.
The tangent line is purple. Forgive my lack of art skills!
Okay, so how do we know 90° is the correct angle? It looks correct, right? Sounds good! Final answer! 90°!
Anyway, just because it looks to be 90° doesn't actually mean it is. Figures are oftentimes not drawn to scale. There is more than one way to solve this or prove the angle is 90°. Here is one way:
Situate your circle on a coordinate plane and using the center of your circle and one point on the circumference of your circle, calculate the slope of the line that would connect the two points. This is the slope of the radius. Also, write down the equation of the circle you drew on the coordinate plane.
For any continuous function, taking the 1st derivative of that function at a specific point, will give the slope of the tangent line at that point. Note that a circle is not a function, but if you solve your circle equation for y, you split the circle into two functions: the positive square root and the negative square root. Together, those two functions make a circle.
So take the derivative of the positive square root wrt x and evaluate the derivative at the point on the circumference that the radius touches. This must be the same point as in step 1. You will get a plain old number for the result. This number is the slope of the tangent line at that point on the circumference — or really, any point on the circumference, because they are all the same, as mentioned above.
Now, compare the slope of the radius from step 1 with the slope of the tangent line from step 3. The two slopes should be perpendicular to each other. If the slope of the radius is a/b, then the slope of the tangent line should be -b/a (negative b/a). By verifying with calculations that the two slopes are negative reciprocals of each other, you are proving that the two lines are perpendicular. Lastly, two lines that are perpendicular to each other have an angle of 90° between them. The previous sentence is a Geometry law, so if writing a proof, you can reference the specific law, but you have to look it up!
Phew! That was a bit of writing! So there is one way to show that the angle between the radius and the tangent line is 90°. There are other methods, but this one is reliable :)
Paul Dunkley
Associate
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Author has 984 answers and 1.9M answer views
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Updated 2y
Related
How do you find the center of two circles that are tangent internally?
The two given circles will have their centres on a perpendicular drawn through the common point of tangency T, bisect the right angle to form a 45° angle between chord and tangent, that bisctor will meet the circles at A and B.
Draw through A and B parallel to the given tangent meeting the perpendicular in C and D. Now C and D are the required centres.
Since in both circles CA=CT and DB=DT so C and D are the centres.
The two given circles will have their centres on a perpendicular drawn through the common point of tangency T, bisect the right angle to form a 45° angle between chord and tangent, that bisctor will meet the circles at A and B.
Draw through A and B parallel to the given tangent meeting the perpendicular in C and D. Now C and D are the required centres.
Since in both circles CA=CT and DB=DT so C and D are the centres.
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Dean Rubine
Former Faculty at Carnegie Mellon School Of Computer Science (1991–1994)
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Author has 10.5K answers and 23.4M answer views
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Feb 7
Related
What is the relationship between the centers of two intersecting circles?
Bot question.
Let’s call the circles and their centers A and B, and without loss of generality, let’s say circle A is at least as big as circle B (in area, radius, diameter and circumference).
If the circles intersect in one place T, that’s a tangent, and there are two cases, internal tangent and external tangent.
For the internal tangent we have AB=AT−BT; we’d have to write AB=|AT−BT| if we weren’t assured AT≥BT.
For the external tangent we have AB=AT+BT.
For the case where there are two intersections, T and U, we get radii AT,AU,BT,BU, and congruent triangles ATB and AUB, so by the
Bot question.
Let’s call the circles and their centers A and B, and without loss of generality, let’s say circle A is at least as big as circle B (in area, radius, diameter and circumference).
If the circles intersect in one place T, that’s a tangent, and there are two cases, internal tangent and external tangent.
For the internal tangent we have AB=AT−BT; we’d have to write AB=|AT−BT| if we weren’t assured AT≥BT.
For the external tangent we have AB=AT+BT.
For the case where there are two intersections, T and U, we get radii AT,AU,BT,BU, and congruent triangles ATB and AUB, so by the triangle inequality, AB|AT−BT|. That applies whether center B is in the interior, on, or in the exterior of circle A.
I think if we put all these together we get
|AT−BT|≤AB≤AT+BT
where we could lose the absolute value if we knew AT≥BT.
In english,
Answer: The distance between the two centers is between the absolute difference of the radii and the sum of the radii.
Jan Bolluyt
Studied Mathematics & Physics (till 12th) (Graduated 1970)
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Author has 959 answers and 71.6K answer views
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Mar 25
Related
Suggest a diameter of circle k. If tangents to circle k are constructed through points L and J, what relationship would exist between the two tangents?
The two points (L and J) are on or outside circle k. The possible relationships of the two tangents (or possible 3 or 4 tangents).
two parallel lines.
Four tangents with two intersections outside the circle.
Three intersecting lines.
Three intersecting line and two parallel lines
2 Intersecting lines.
Inscribed circle
And an extra circle k for doodling if you find more!
The two points (L and J) are on or outside circle k. The possible relationships of the two tangents (or possible 3 or 4 tangents).
two parallel lines.
Four tangents with two intersections outside the circle.
Three intersecting lines.
Three intersecting line and two parallel lines
2 Intersecting lines.
Inscribed circle
And an extra circle k for doodling if you find more!
Dave Carlson
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Author has 466 answers and 768.6K answer views
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Feb 23
Related
What is the angle between two circles that are tangent to each other?
If you mean two circles touching at a single point, forming an “8” or infinity sign, the answer is 0 degrees.
Technically if you draw a line from the center of a circle to a point on the edge of the circle, the tangent line at that point will make a 90 degree angle with the line that connects the edge to the center. Meaning the circle’s “arc” at that point is “directed” at 90 degrees, with the radius at that point. So when you connect two circles, both their “arcs” are pointed in the same direction, 90 degrees to their radiuses. (Their radiuses at this point will connect to form one continuous
If you mean two circles touching at a single point, forming an “8” or infinity sign, the answer is 0 degrees.
Technically if you draw a line from the center of a circle to a point on the edge of the circle, the tangent line at that point will make a 90 degree angle with the line that connects the edge to the center. Meaning the circle’s “arc” at that point is “directed” at 90 degrees, with the radius at that point. So when you connect two circles, both their “arcs” are pointed in the same direction, 90 degrees to their radiuses. (Their radiuses at this point will connect to form one continuous line - try drawing on a sheet of paper and you’ll see what I mean). So since their arcs are “pointing” in the same direction, the angle between them is 0 degrees.
Hope this helps.
Related questions
How do you find the center of two circles that are tangent internally?
Does a pair of common tangents of two circles meet the line joining centres of the two circles at one point?
What is the common tangent of two circles in which one is inside the other and they don't touch each other?
What is the relationship between the areas of two circles that are tangent to each other?
What is the relationship between the radii and centres of three mutually tangent circles?
How do I find the point of intersection of direct common tangents for two circles which touch each other?
What is the relationship between the radii of two circles that are tangent to each other?
Do concentric circles have common tangents?
What is the relationship between two circles if they do not have a common tangent?
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What is the relationship of a tangent to a circle?
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188191 | https://chem.libretexts.org/Courses/Palomar_College/PC%3A_CHEM100_-_Fundamentals_of_Chemistry/15%3A_Chemical_Bonding/15.4%3A_Lewis_Structures%3A_Counting_Valence_Electrons | Skip to main content
15.4: Lewis Structures: Counting Valence Electrons
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: Mar 24, 2021
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15.3: Lewis Structures: Electrons Shared
15.5: Predicting the Shapes of Molecules
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: 105350
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Learning Objectives
Draw Lewis structures for covalent compounds.
The following procedure can be used to construct Lewis electron structures for more complex molecules and ions:
How-to: Constructing Lewis electron structures
1. Determine the total number of valence electrons in the molecule or ion.
Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.)
If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion.
For CO32−, for example, we add two electrons to the total because of the −2 charge.
2. Arrange the atoms to show specific connections.
When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl4 and CO32−, which both have C as the central atom), which is another clue to the compound’s structure.
Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central.
3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond.
In H2O, for example, there is a bonding pair of electrons between oxygen and each hydrogen.
Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen).
These electrons will usually be lone pairs.
If any electrons are left over, place them on the central atom.
We will explain later that some atoms are able to accommodate more than eight electrons.
6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet.
This will not change the number of electrons on the terminal atoms.
Final check
Always make sure all valence electrons are accounted for and each atom has an octet of electrons except for hydrogen (with two electrons).
The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal.
Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed.
Example 15.4.1: Water
Write the Lewis Structure for H2O.
Solution
| Steps for Writing Lewis Structures | Example 15.4.1 |
--- |
| 1. Determine the total number of valence electrons in the molecule or ion. | Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons. |
| 2. Arrange the atoms to show specific connections. | H O H Because H atoms are almost always terminal, the arrangement within the molecule must be HOH. |
| 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). | Placing one bonding pair of electrons between the O atom and each H atom gives H -O- H with 4 electrons left over. Each H atom has a full valence shell of 2 electrons. |
| 5. If any electrons are left over, place them on the central atom. | Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure: |
| 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. | Not necessary |
| 7. Final check | The Lewis structure gives oxygen an octet and each hydrogen two electrons, |
Example 15.4.2
Write the Lewis structure for the CH2O molecule
Solution
| Steps for Writing Lewis Structures | Example 15.4.2 |
--- |
| 1. Determine the total number of valence electrons in the molecule or ion. | Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons. |
| 2. Arrange the atoms to show specific connections. | Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. |
| 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. | Placing a bonding pair of electrons between each pair of bonded atoms gives the following: Six electrons are used, and 6 are left over. |
| 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). | Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following: Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons. |
| 5. If any electrons are left over, place them on the central atom. | Not necessary There are no electrons left to place on the central atom. |
| 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. | To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond: |
| 7. Final check | Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid. |
Exercise 15.4.1
Write Lewis electron structures for CO2 and SCl2, a vile-smelling, unstable red liquid that is used in the manufacture of rubber.
Answer CO2
: .
Answer SCl2
: .
The United States Supreme Court has the unenviable task of deciding what the law is. This responsibility can be a major challenge when there is no clear principle involved or where there is a new situation not encountered before. Chemistry faces the same challenge in extending basic concepts to fit a new situation. Drawing of Lewis structures for polyatomic ions uses the same approach, but tweaks the process a little to fit a somewhat different set of circumstances.
Writing Lewis Structures for Polyatomic Ions
Recall that a polyatomic ion is a group of atoms that are covalently bonded together and which carry an overall electrical charge. The ammonium ion, NH+4, is formed when a hydrogen ion (H+) attaches to the lone pair of an ammonia (NH3) molecule in a coordinate covalent bond.
When drawing the Lewis structure of a polyatomic ion, the charge of the ion is reflected in the number of total valence electrons in the structure. In the case of the ammonium ion:
1N atom =5 valence electrons
4H atoms =4×1=4 valence electrons
subtract 1 electron for the 1+charge of the ion
total of 8 valence electrons in the ion
It is customary to put the Lewis structure of a polyatomic ion into a large set of brackets, with the charge of the ion as a superscript outside the brackets.
Exercise 15.4.2
Draw the Lewis electron dot structure for the sulfate ion.
Answer
Exceptions to the Octet Rule
As important and useful as the octet rule is in chemical bonding, there are some well-known violations. This does not mean that the octet rule is useless—quite the contrary. As with many rules, there are exceptions, or violations.
There are three violations to the octet rule. Odd-electron molecules represent the first violation to the octet rule. Although they are few, some stable compounds have an odd number of electrons in their valence shells. With an odd number of electrons, at least one atom in the molecule will have to violate the octet rule. Examples of stable odd-electron molecules are NO, NO2, and ClO2. The Lewis electron dot diagram for NO is as follows:
Although the O atom has an octet of electrons, the N atom has only seven electrons in its valence shell. Although NO is a stable compound, it is very chemically reactive, as are most other odd-electron compounds.
Electron-deficient molecules represent the second violation to the octet rule. These stable compounds have less than eight electrons around an atom in the molecule. The most common examples are the covalent compounds of beryllium and boron. For example, beryllium can form two covalent bonds, resulting in only four electrons in its valence shell:
Boron commonly makes only three covalent bonds, resulting in only six valence electrons around the B atom. A well-known example is BF3:
The third violation to the octet rule is found in those compounds with more than eight electrons assigned to their valence shell. These are called expanded valence shell molecules. Such compounds are formed only by central atoms in the third row of the periodic table or beyond that have empty d orbitals in their valence shells that can participate in covalent bonding. One such compound is PF5. The only reasonable Lewis electron dot diagram for this compound has the P atom making five covalent bonds:
Formally, the P atom has 10 electrons in its valence shell.
Example 15.4.3: Octet Violations
Identify each violation to the octet rule by drawing a Lewis electron dot diagram.
ClO
SF6
Solution
a. With one Cl atom and one O atom, this molecule has 6 + 7 = 13 valence electrons, so it is an odd-electron molecule. A Lewis electron dot diagram for this molecule is as follows:
b. In SF6, the central S atom makes six covalent bonds to the six surrounding F atoms, so it is an expanded valence shell molecule. Its Lewis electron dot diagram is as follows:
Exercise 15.4.3: Xenon Difluoride
Identify the violation to the octet rule in XeF2 by drawing a Lewis electron dot diagram.
Answer:
: The Xe atom has an expanded valence shell with more than eight electrons around it.
Summary
Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. In Lewis electron structures, we encounter bonding pairs, which are shared by two atoms, and lone pairs, which are not shared between atoms. Lewis structures for polyatomic ions follow the same rules as those for other covalent compounds. There are three violations to the octet rule: odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules
Contributors and Attributions
Modified by Joshua Halpern (Howard University)
Marisa Alviar-Agnew (Sacramento City College)
Henry Agnew (UC Davis)
15.3: Lewis Structures: Electrons Shared
15.5: Predicting the Shapes of Molecules |
188192 | https://www.ablebits.com/office-addins-blog/fv-function-excel-calculate-future-value/ | Ablebits blog
Excel
Financial functions
Excel FV function to calculate future value
This tutorial looks at how to use the FV function in Excel to find the future value of a series of periodic payments and a single lump-sum payment.
Building your personal and corporate finances requires thorough planning. One of the most important factors of success is understanding how much an investment made today will grow to in the future. That is called the future value of investment, and this tutorial will teach you how to calculate it in Excel.
Future value in Excel
The future value (FV) is one of the key metrics in financial planning that defines the value of a current asset in the future. In other words, FV measures how much a given amount of money will be worth at a specific time in the future.
Normally, the FV calculation is based on an anticipated growth rate, or rate of return. When the money is deposited in a saving account with a predefined interest rate, determining a future value is quite easy. The FV of investments in stocks, bonds or other securities may be hard to calculate accurately because of a volatile rate of return.
Luckily, Microsoft Excel provides a special function that does all the math behind the scenes based on the arguments that you specify.
Excel FV function
FV is an Excel financial function that returns the future value of an investment based on a fixed interest rate. It works for both a series of periodic payments and a single lump-sum payment.
The function is available in all versions Excel 365, Excel 2019, Excel 2016, Excel 2013, Excel 2010 and Excel 2007.
The FV syntax is as follows:
FV(rate, nper, pmt, [pv], [type])
Rate (required) - the interest rate per period. If you pay once a year, supply an annual interest rate; if you pay each month, then you should specify a monthly interest rate, and so on.
Nper (required) - the total number of payment periods for the lifetime of an annuity.
Pmt (optional) - the constant amount paid each period. Should be expressed as a negative number. If omitted, it is assumed to be 0, and the pv argument must be included.
Pv (optional) - the present value of the investment. Should be represented by a negative number. If omitted, it defaults to 0, and the pmt argument must be included.
Type (optional) - indicates when the payments are made:
0 or omitted (default) - at the end of a period (regular annuity)
1 - at the beginning of a period (annuity due)
4 things to remember about Excel FV function
To correctly build a FV formula in your worksheets and avoid common errors, please keep in mind these usage notes:
For any inflows such as dividends or other earnings, use positive numbers. For any outflows such as deposits to a saving or investing account, use negative numbers.
If the present value (pv) is zero or omitted, the payment amount (pmt) must be included, and vice versa.
The rate argument can be expressed as a percentage or decimal number, e.g. 8% or 0.08.
To get the correct future value, you must be consistent with nper and rate. For instance, if you make 3 yearly payments at an annual interest rate of 5%, use 3 for nper and 5% for rate. If you do a series of monthly investments for a period of 3 years, then use 312 (a total of 36 payments) for nper and 5%/12 for rate.
Basic future value formula in Excel
This example shows how to use the FV function in Excel in its simplest form to calculate future value, given a periodic interest rate, the total number of periods, and a constant payment amount per period.
Periodic interest rate (rate): C2
Number of periods (nper): C3
Payment amount (pmt): C4
Let's say you are going to make a yearly $1,000 payment for 10 years with an annual interest rate of 6%. It is assumed to be a regular annuity where all payments are made at the end of the year.
To find the future value, configure the FV function in this way:
=FV(C2, C3, C4)
Please notice that pmt is a negative number because this money is paid out.
If the payment is represented by a positive number, don't forget to put the minus sign right before the pmt argument:
=FV(C2, C3, -C4)
How to calculate future value in Excel - formula examples
The basic Excel FV formula is very simple, right? Now, let's have a look at how to tweak it to handle a couple of most common scenarios.
FV formula for periodic payments
When investing money through a series of regular savings, it often happens that you are provided with an annual interest rate and the investment term defined in years, whereas the payments are to be made weekly, monthly, quarterly or semiannually. In such situations, it is very important that the rate and nper units be consistent.
To convert an annual interest rate to a periodic rate, divide the annual rate by the number of periods per year:
Monthly payments: rate = annual interest rate / 12
Quarterly payments: rate = annual interest rate / 4
Semiannual payments: rate = annual interest rate / 2
To get the total number of periods, multiply the term in years by the number of periods per year:
Monthly payments: nper = no. of years 12
Quarterly payments: nper = no. of years 4
Semiannual payments: nper = no. of years 2
Now, let's see how it works in practice. Suppose you monthly invest $200 for 3 years with an annual interest rate of 6%. The source data is input in these cells:
Annual interest rate (B2): 6%
No. of years (B3): 3
Monthly payment (B4): -200
Periods per year (B5): 12
To calculate the future value of this investment, the formula in B7 is:
=FV(B2/B5, B3B5, B4)
As shown in the image below, the same formula determines the future value based on quarterly savings equally well:
FV formula for lump-sum investment
If you choose to invest money as a one-time lump sum payment, the future value formula is based on the present value (pv) rather than periodic payment (pmt).
So, we set up our sample data as follows:
Annual interest rate (C2): 7%
No. of years (C3): 5
Present value (C4): -1000
The formula to calculate the future value of the investment is:
=FV(C2, C3, ,C4)
Please notice that:
The investment amount (pv) is a negative number because it's an outflow.
The pmt argument is 0 or omitted.
If the compounding periods for your investment are not annual, then to determine the future value accurately, you need to make the following adjustments to the formula:
For rate, divide an annual interest rate by the number of compounding periods per year.
For nper, multiply the number of years by the number of compounding periods per year.
As an example, let's find the future value of the above investment with an interest rate compounded monthly. For this, we divide an annual interest rate (C2) by 12 and multiply the number of years (C3) by 12:
=FV(C2/12, C312, ,C4)
or
=FV(C2/C5, C3C5, ,C4)
Where C5 is the number of compounding periods per year:
Get future value for different compounding periods
To compare the amount of growth generated by various compounding periods, you need to supply different rate and nper to the FV function.
To have all calculations performed with a single formula, do the following:
Input the number of compounding periods per year in B2.
Arrange your data like shown in the image below.
Enter the following formula in C2 and drag it down through C6:
=FV($F$2/B2, $F$3B2, ,$F$4)
Please pay attention that we lock the annual interest rate ($F$2), the number of years ($F$3) and the investment amount ($F$4) references with the dollar sign ($) so they won't shift when copying down the formula.
Make a future value calculator in Excel
If your goal is to build a universal FV calculator that works for both periodic and lump-sum payments with either annuity type, then you will need to use the Excel FV function in its full form.
For starters, allocate cells for all the arguments, including the optional ones like shown in the screenshot below. And then, define the arguments in this way:
Rate (periodic interest rate): B2/ B7 (annual interest rate / periods per year)
Nper (total number of payment periods): B3B7 (number of years periods per year)
Pmt (periodic payment amount): B4
Pv (initial investment): B5
Type (when payments are due): B6
Compounding periods per year: B7
Putting the arguments together, we get this formula:
=FV(B2/B7, B3B7, B4, B5, B6)
Suppose you wish to save some money for renovating your house in 5 years. You deposit $3,000 to your saving account at an interest rate of 7% compounded monthly. Furthermore, you are going to add $100 at the beginning of each month. How much money will there be in your saving account in 5 years? According to our Excel FV calculator - around $11,500.
When setting up a future value calculator for other users, there are a few things to take notice of:
Both pmt and pv should be negative numbers because they represent an outflow. If positive numbers are entered in the corresponding cells, then put the minus sign before these arguments directly in the formula.
If pmt is zero or omitted, be sure to specify the present value (pv) and vice versa.
For type, consider creating a drop-down list to only allow 0 and 1 values. This will help you prevent accidental mistakes that users could make.
The Compounding periods per year cell (B7) must have a number in it other than zero, otherwise the formula will return a #DIV/0 error. If an interest rate is compounded annually, enter 1 in that cell.
Excel FV function not working
If a FV formula results in an error or yields a wrong result, in all likelihood, that will be one of the following.
#VALUE! error
May occur if one or more arguments are non-numeric. To fix the error, check if any of the numbers referenced in your formula are formatted as text. If some are, then convert text values to numbers.
FV function returns an incorrect future value
If the returned future value is negative or much lower than expected, most likely, either the pmt or pv argument, or both, are represented by positive numbers. Please remember that negative numbers should be used for all outgoing payments.
That's how to how to calculate future value of annuity in Excel. I thank you for reading and hope to see you on our blog next week!
Practice workbook for download
Future value formula in Excel (.xlsx file)
You may also be interested in:
How to use PV function in Excel to calculate present value
Find net present value (NPV) in Excel
Excel NPER function with formula examples
Excel RATE function to calculate interest rate
Calculating compound interest in Excel
Excel PMT function to calculate amount paid each period
2 comments
Sweta Jain says:
How to calculate future value in excel if my periodic payment is different in different years. please help
Alexander Trifuntov (Ablebits Team) says:
Hi! If I understand your task correctly, Excel's NPV function calculates the present value of an investment based on a discount rate and various future payments: How to calculate NPV in Excel - net present value formula examples.
Post a comment
Click here to cancel reply. |
188193 | https://msuweb.montclair.edu/~lebelp/BaumolEmpiricalDemand1965.pdf | The Empirical Determination of Demand Relationships William J. Baumol William Baumol is Professor of Economics at Princeton and New York Universities. This piece is taken from his book Economic Theory and Operations Analysis, 1965.
I. Why Demand Functions?
Demand functions, as they are, defined in economic analysis, are rather queer creatures, somewhat abstract, containing generous elements of the hypothetical, and, in general, marked by an aura of unreality. The peculiarity of the concept is well illustrated by the fact that only one point on a demand curvy can ever be observed directly with any degree of confidence because by the time we can obtain the data with which to plot a second point, the entire curve may well have shifted without our knowing it. A more fundamental but related source of our discomfort with the idea is the fact that the demand relationship is defined as the answer to the set of hypothetical questions which begin, "What would consumers do if price (or advertising outlay, or some other type of marketing effort) were different than it is in fact?" We are, then, dealing with information about potential consumer behavior in situations, which consumers may never have experienced. And, since we have very little confidence in the constancy of consumer tastes and desires, all of these data are taken to refer to possible events at just one moment of time-e.g., consumer reactions to alternative possible prices if any of them were to occur tomorrow at 2:47 P.m.
In view of all this, there should be little wonder that people with an orientation toward applied economics occasionally become somewhat impatient with the economic theorist's demand function. Yet no matter how ingenious the circumlocutions, which may have been employed, they have been unable to find an acceptable substitute for the concept. For the demand function must ultimately play a critical role in any probing marketing decision process, and there is really no way to get away from it.
For example, to decide on the number of salesmen which will best serve the interests of the firm, it is first necessary to know what difference in consumer purchases would result from alternative sales force sizes. But this is precisely the sort of odd and hypothetical information, which goes to make up the demand relationship. It is for exactly the same reason that many large and reputable firms in diverse fields of industry are conducting ambitious research programs whose aim is the determination of their advertising-demand curves, that is, the relationship between their advertising outlays and their sales. So far, these efforts have met with varying degrees of success, and it must be admitted that many of them have not come up with very meaningful results. For the empirical determination of demand relationships is no simple matter and there are many booby traps for the amateur investigator and the unwary. It is no trick at all, on looking over a small sample of the published demand studies, to come up with horrible examples of just about every available type of misstep.
This chapter is designed primarily to point out some of the pitfalls, which threaten the investigator of demand relationships. Its aim is to warn the reader to proceed with extreme caution in any such enterprise. No cut-and-dried solutions are offered to the problems, which are discussed. This is true for two reasons. First, because many of the methods for dealing with these difficulties are highly technical matters of specialized econometric analysis and so are completely outside the scope of this volume. Second, and more important, solutions are not listed mechanically because there simply are no panaceas; the problems must be dealt with case by case as they arise, and the effectiveness with which they can be handled is still highly dependent on the skill, experience, and judgment of the specialist investigator.
If after reading the chapter the reader is left somewhat worried and uncomfortable, it will have accomplished its purpose. However, it should be emphasized that the problems which are raised, serious and difficult though they be, are not totally intractable and beyond the power of our statistical techniques.
- 2 -II. Interview Approaches to Demand Determination Before turning to statistical methods for the finding of demand functions, it is appropriate to say a few words about a more direct method for dealing with the problem-the consumer interview approach. In its most blatant and naive form, consumers are simply collared by the interviewer and asked how much they would be willing to purchase of a given product at a number of alternative product price levels.
It should be obvious enough that this is a dangerous and unreliable procedure. People just have not thought out in advance what they would do in these hypothetical situations, and their snap judgments thrown up at the request of the interviewer cannot inspire a great deal of confidence.
Even if they attempt to offer honest answers, even if they had thought about their decisions in advance, consumers might well find that when confronted with the harsh realities of the concrete situation, they behave in a manner, which belies their own expectations. When we get to the effects of advertising on demand, the problems of such a direct interview approach become even more apparent. What is the consumer to be asked - how much more of the company's product he would buy if it were to institute a 1 per cent increase in its spot announcements to its television budget?
Much more subtle and effective approaches to consumer interviewing are indeed possible.
Indirect, but far more revealing questions can be asked. Consumers may, for example, be asked about the difference in price between two competing products, and if it turns out that they simply do not know the facts of the matter, one may be led to infer that a lower product price may have a relatively limited influence on consumer behavior just because few consumers are likely to be aware of its existence. A clever interview designer may in this way build up a strategy of indirect questions, which gradually isolates the required facts.
Alternatively, consumers may be placed in simulated market situations, so-called consumer clinics, in which changes in their behavior can be observed as the circumstances of the experiment are varied. An obvious approach to this matter is to get groups of housewives together, give them small amounts of money with which they are offered the opportunity to purchase one of, say, several brands of dishwasher soap which are put on display at the clinic, and observe what happens as the posted prices on the displays are varied from group to group.
Here again, much more subtle variants in experimental design are clearly possible.
But even the best of these procedures has its limitations for our purpose, which is the determination of the precise form of a demand relationship. Artificial consumer clinic experiments inevitably introduce some degree of distortion because subjects cannot be kept from realizing that they are in an experimental situation. In any event, such clinics are rather expensive and so the samples involved are usually extremely small-too small for confidence in any inferences, which are drawn about the magnitudes of the parameters of the demand relationships for the body of consumers as a whole. And large sample interviews, which approach the determination of consumer demand patterns by subtle and indirect questions are often highly revealing, but they rarely can supply the quantitative information required for the estimation of a demand equation.
III. Direct Market Experiments A second alternative approach, which is sometimes considered as a means for finding demand relationship information is the direct market experiment. A company engages in a deliberate program of price or advertising level variation. Suppose it increases its newspaper advertising outlay in one city by 5 per cent, in another city it increases this outlay by 10 per cent, and in still a third metropolis a 10 per cent reduction is undertaken.
In some ways such a direct experimental approach must always be the most revealing. It gives real answers to our formerly hypothetical questions and does so without subjecting the consumer to the artificial atmosphere of the interview situation or the consumer clinic.
However, direct experimentation has its serious limitations as well.
- 3 -1. It can be very expensive or extremely risky for the firm. Customers lost by an experimental price increase may never be regained from competitive products, which they might otherwise never have tried, and a 10 per cent increase in advertising outlay for any protracted period may be no trivial matter.
2. Market experiments are almost never controlled experiments, so that the observations, which they yield are likely to be colored by all sorts of fortuitous occurrences-coincidental changes in consumer incomes or in competitive advertising programs, peculiarities of the weather during the period of the experiment, etc.
3. Because of the high cost of the experiments and because it is often simply physically impossible to try out a large number of variations, the number of observations is likely to be unsatisfactorily small. If, for example, it is desired to determine the effects of varied advertising outlay in a national periodical, the company cannot increase the size of its ads, which are seen by Nashville readers and simultaneously reduce those, which are seen in Lexington, Kentucky. This difficulty has been eased to some extent by the fact that a number of national magazines now put out several regional editions, but by and large the problem remains: market experiments usually supply information only about a very limited number of alternatives.
4. For similar reasons, market experiments are often of only relatively brief duration.
Companies cannot afford to permit them to run long enough to display much more than impact effects. And yet the distinction between impact effects and long-run effects of a change are often extremely significant, as was so clearly demonstrated by the sharp but very temporary drop in cigarette sales when the first announcement was made about the association between smoking and the incidence of cancer. How often has a rise in the price of a product caused a major reduction in purchases for a few weeks, with customers then gradually but steadily drifting back?
Market experiments do have a role to play in demand relationship determination. They can be important as a check on the results of a statistical study. Or they can provide some critical information about a few points on the demand curve in which past experience is entirely lacking. In some special circumstances experimentation is particularly convenient and has been used in the past, apparently with a considerable degree of success. For example, some mail-order houses have employed systematic programs in which a few special experimental pages were bound inconspicuously into the catalogues distributed to customers within restricted geographic regions, thus permitting observation of the effects of price, product, or even catalogue display variations. However, it should also be clear that market experiments cannot by themselves be relied upon universally to provide the demand information needed by management. Economics is just not a subject which lends itself readily to experimentation, largely because there are always too many elements beyond the control of the investigator and because economic experimentation is often inherently too expensive, risky, and difficult.
IV. Standard Statistical Approaches The third, and generally most attractive, approach to demand function determination attempts to squeeze its information out of sources such as the accumulated records of the past (a time-series analysis), or a comparative evaluation of the performance of different sectors of the market (a cross-sectional analysis). The available statistics on sales, prices, advertising outlays of the most relevant varieties, and other marketing data are gathered together and then analyzed with the aid of 'the standard statistical techniques.
The basic procedure is simple enough; in fact, as we shall see presently, it is often far too simple. Suppose, for example, that the following data on company sales and advertising outlays have been accumulated: TABLE I.
Year 1950 1951 1952 1953 1954 1955 1956 1957 Sales .....................................67 73 54 62 70 75 79 83 (millions of dollars) Advertising ...................................12 15 13 14 18 17 19 15 (millions of dollars) - 4 -Once the figures have been plotted, the pattern formed by the dots can be used in an obvious manner to fit a straight line (see Figure 1) or a curve to them. This line is then taken as the desired advertising-demand curve. Its slope can be used as a measure of advertising effectiveness, that is, it measures the marginal sales productivity of an advertising dollar, 0 sales/ advertising outlay. This line can be determined impressionistically simply by drawing in a line that appears to fit the dots fairly well, or any one of a variety of more systematic methods can be used.
The most widely employed and best known of these techniques is the method of least squares,' in which the object is to find that line which makes the sum of the (squared) vertical deviations between our dots and the fitted line as small as possible, where the deviations are defined as the vertical distances such as AB or CD in Figure 1. The idea is inherently attractive. We wish to minimize deviations because a line, which involves very substantial deviations from the dots representing our data surely does not represent the information in a very satisfactory way. But if, in our addition process, a large negative deviation such as AB (that is, a case where the line underestimates the vertical coordinate of our dot) happens to be largely cancelled out by a positive deviation, CD, the sum of the deviations can turn out to be small. This is surely not what we want in looking for a line, which does not deviate much from the dots. One can avoid ending up with a line, which fits the facts rather badly but in which the positive and negative deviations add up to a rather small number, by squaring all the deviation figures before adding them together. Since the square of a negative real number as well as that of a positive real number is always positive, large, squared negative deviations cannot offset large, squared positive deviations, and the sum of squares deviations will never add up to a small number unless our line happens to fit the dots closely.2 There exist still more sophisticated techniques for fitting our advertising demand curve from the data. Although it is often too complex and expensive to employ in practice, professional statisticians usually consider the method of maximum likelihood as their ideal. This method requires some information about the probability distribution of the random elements, which influence sales. From this probability distribution the statistician determines a likelihood function L=f (xt, ,yt,, a, b) where xt and yt represent, respectively, advertising expenditures and sales in year t, and a and b are 2 Other devices (such as the absolute value or the fourth power of the deviation) might accomplish the objective discussed. The reason one chooses to minimize the sum of the squares is that under very simple assumptions such estimates have several extremely desirable technical properties, among them, that these estimated parameter values are "best" in the sense that they minimize variance of the estimate and are unbiased.
To find the straight-line equation which satisfies our least squares requirement we employ the symbol xt to represent sales in year t and yt to represent advertising outlay in that year and let the equation of the line to be fitted be written yct = a + bxt where the subscript c in yct is there to remind us that in our equation the y is a figure calculated from the formula rather than observation. Now we proceed as follows: Step 1. Define a deviation from our line as yt - yct = yt - (a + bxt ) = yt - a - bxt.
Step 2. Define a squared deviation as (yt - y ct ) 2 = y 2 t + a 2 + b 2x 2 t - 2ayt - 2bxtyt + 2abxt.
Step 3. Add the squared deviations Σ(yt - yct) 2 = Σy 2 t + na 2 + b 2 Σx 2 t - 2a Σ yt - 2b Σxtyt + 2ab Σxt where, since a is a constant, a 2 = a 2 + a 2 + a 2 . . . ( n equal terms) = na 2.
Step 4. Find the values of a and b (the parameters of our equation) which minimize the sum of the squared deviations.
Vie do this with the aid of the usual calculus procedure by taking partial derivatives with respect to a and b and setting them equal to zero, thus: - 5 - ∂ Σ (yt – yct) 2 = 2an – 2 Σ yt + 2b Σ xt = 0 ∂ a and ∂ Σ (yt – ytc) 2 = 2b Σ x2 t - 2 Σ xtyt + 2a Σ xt = 0 ∂ a These last two equations contain, in addition to a and b, only known statistical figures xt and yt. The equations can therefore be solved simultaneously to obtain the desired parameter values, a and b, i.e., they determine for us the least squares line yct = a +bxt. These two equations are usually referred to as the normal equations of the least squares method in this most elementary (two-variable straight line) case. The procedure employed in fitting many variable equations or curvilinear equations is a simple and obvious extension of that which has just been described. the constants in our advertising demand equation yt = a + bxt.. This likelihood function is defined as an answer to the following type of question: "Given any specific values of the parameters in our equation, say a = 5 and b = 63, how likely is it that the demand situation would have generated the statistics x1950 = 12, y1950 = 67, etc.?" (Note that these values of sales and advertising are in fact our observed statistical figures taken from Table I.) Considering all possible values of a and b, we can then employ the differential calculus to find the a and b combination, which maximizes the value of the likelihood, L. We will then have found the a and b which provide, in this sense, the best possible explanation of the observed facts, i.e., we will have found that equation yt = a + bxt whose parameters a and b are most likely to be the correct values of the true but unknown parameters, given the facts which were actually observed by the data collector.
It is of interest to note that in some special cases the least squares method turns out to be identical with maximum likelihood. That is, in these fortunate circumstances the least squares calculation becomes equivalent to the maximum likelihood procedure. We shall presently discuss one of the things which may go wrong if the least squares method is employed in situations where it does not yield the same results as the maximum likelihood calculation.
Having described now in highly general and impressionistic terms the methods which are most commonly employed by the statistician to determine relationships, let us now see some of the problems to which they give rise.
V. Omission of Important Variables Clearly, sales are affected by other variables in addition to the company's advertising expenditure. Prices, competitive advertising, consumer income variations, and other variables also play an important role in any demand relationship. If, therefore, we try to extract from our statistics a simple equation relating sales to advertising outlay alone, and in the process we ignore all other variables, our results are likely to be very badly distorted. We may ascribe to the company's advertising outlays sales trends, which are really the result of the behavior of other economic changes. The behavior of other variables can thus conceal and even offset the effects of advertising. To show how serious the results can be, consider the illustrative demand equation (1) S = 50 + 4A + 0.02Y - 6 -where S represents sales, A advertising expenditure, and Y consumer income. The values given in Table II can easily be seen to satisfy the equation precisely, and any standard estimation procedure based on such information can be expected to yield the correct equation.
TABLE II.
Date 1956 1957 1958 Y 3,000 4,000 3,500 A 2 3 2.5 S 118 142 130 But the standard calculation shows that a two-variable, straight, least squares line which gives us a (perfect!) correlation between S and A alone (ignoring Y) and which is based on these same values will yield the equation (2) S - 24A + 70.
This equation asserts that each added dollar of advertising expenditure brings in $24 in sales, instead of the true $4 return shown by equation (1). in addition, because of the perfect correlation there is, in this case, no residual unexplained variation in S which is left to be accounted for by a subsequent correlation between S and Y, i.e., this incorrect procedure appears to show that consumer income has absolutely no influence on demand! This advertising coefficient has been inflated by usurping to itself the influence of Y on sales.
Incidentally, if, instead of proceeding as we just did, we had started off by finding a least squares equation relating sales to consumer income alone, we would have obtained from the same statistics the equation S - 0.024Y + 46 which this time overvalues the influence of income on sales and ascribes absolutely no effectiveness to advertising.
It is clear, then, that more than two variables must usually be taken into account in the statistical estimation of a demand relationship. And, in fact, this is ordinarily done, the estimation usually employing what is called a least squares multiple regression technique.
However, it should be remembered that even if we include five variables in our analysis but omit a sixth rather important variable, precisely the same difficulties will be encountered. That is, the omission of any important variable, however defined, from the statistical procedure can lead to serious distortions in its results.
This might appear to constitute an argument for the inclusion in the analysis of every variable, which comes to the statistician's mind as a factor of possible importance, just as a matter of insurance. Unfortunately, however, we are not at liberty to go on adding variables willy-nilly.
The more variables whose influence we want to take into account the more data we require as a basis for the estimation. If we only have statistical information pertaining to three points in time, it is ridiculous to try to disentangle the influence of fifteen variables. In fact, the statistician requires many pieces of information for every variable he includes in his analysis, if he is to estimate his relationship with a clear conscience.
However, large masses of marketing data are not easily come by. Records are often woefully incomplete; additional data can sometimes be acquired only at considerable expense, and in any event, statistics, which go too far back in time, are apt to be obsolete and irrelevant for the company's current circumstances. We must; therefore, very frequently be contented with - 7 -skimpy figures, which force us to be extremely niggardly in the number of variables which we take into account, despite the very great dangers involved.
VI. Inclusion of Mutually Correlated Variables Another difficulty which, to some extent, can help to make life easier as far as the problem of the preceding section is concerned arises when a number of the relevant variables are themselves closely interrelated. For example, one encounters advertising effectiveness studies in which income and years of education per inhabitant are both included as variables.
Now, education is itself very closely related to income level both because higher-income families can afford to provide more education and larger inheritances to their children and because a more educated person is often in a position to earn a higher income.
It may nevertheless be true that education and income do have different consequences for advertising effectiveness. For example, an increase in income without any change in educational level could increase the person's willingness to purchase more in response to an ad, whereas more education not backed up by larger purchasing power might have the reverse effect. But, in general, there, is no statistical method whereby these two consequences can be separated, because, for the bulk of the population, whenever one of these variables increases in value, so does the other. Hence, the statistics which can merely exhibit directions of variation might show that, other things remaining equal, whenever sales increased, income also increased, and so (as a consequence?) did education.
In such circumstances if we include both the income and the educational level variables in the statistical demand-fitting procedure, the chances are that the mechanics of the procedure will provide a perfectly arbitrary ascription of the sales changes to our two causal variables. And sometimes the results may turn out completely nonsensical because the standard computational procedure has no way to apply common sense in imputing the total sales change to the separate influences of education and income changes. Therefore, if in a demand relationship there occur several variables, which are themselves highly correlated, it is usually wise to omit all but one of any such set of variables in a statistical study. If this is not done, another powerful source of nonsense results is introduced.
VII. Simultaneous Relationship Problems The difficulties, which have so far been discussed, while they can be extremely important and are often overlooked in practice (with rather sad consequences) may, by and large, be considered rather routine and in retrospect, fairly obvious matters.
We come now to a far more subtle and perhaps a far more serious problem, which was only brought to our attention in 1927 by E. J. Working and which has only received serious and systematic attention quite recently, largely as a result of the work of the Cowles Foundation.
The problem in question, in a sense, follows from the difficulty, which was discussed in the previous section. If there is a close correlation between two variables, it is likely to mean that they are not independent of one another and that there is at least one other relevant equation in the system, which expresses the relationship between them. For example, in our illustrative case there f might be an equation indicating how income level is ordinarily increased by a person's education. We then end up having to deal with not just a single demand equation, but with a system of several equations in which a number of the variables interact mutually and are determined simultaneously.
Economics is characterized by such simultaneous relationships. The standard example is the price determination process in which a supply equation is involved as well as our demand relationship. Similarly, simultaneous relationships constitute the core of national income analysis. National income depends on the demand for consumers' goods which helps - 8 -determine the level of profitable production. But the consumption demand equation, in turn, involves national income (as a measure of the public's purchasing power) as a variable. To mention another simultaneous relationship example, the coal mining industry is a customer for steel whose volume of demand depends on coal sales, but the demand for coal itself depends heavily on the amount of coal to be used in producing steel. It is possible to expand the list of simultaneous relationships in economics indefinitely.
The empirical data, which are generated by such a set of equations, are the information source on which the statistician must base his estimates of the relationships. But since these data are the result of a number of such relationships, the difficult problem arises of separating out the relationships from the observed statistics.
Unless steps are taken to make sure that the influences of the several simultaneous relationships on the data can be and have been separated, there is not the slightest justification for the use of any estimation procedure, such as that depicted in Figure I, to compute a statistical relationship. Yet it will readily be recognized how frequently this completely falla-cious procedure is employed in practice in the form of simple or multiple correlations computed without any attempt to cope with the simultaneous relationship problem. Let us see now how serious are the distortions, which can be expected to result.
VIII. The Identification Problem In rather general terms our basic problem can conveniently be divided into two parts: 1. In some circumstances the simultaneous relationships (equations) will be so similar in character that it will be impossible to unscramble them (or at least some of them) from the statistics. Such relationships are said to be unidentifiable. Presently it will be shown how such an unhappy situation can arise, and it will be indicated that it is unfortunately not unheard of in marketing problems. Clearly, in such a case, we are wasting our time in a statistical investigation of the equation in question. There do exist some mathematical tests, which show whether or not an equation is identified (i.e., whether or not it is in principle possible to separate it from the other relationships in the system). These tests should always be applied before embarking on the type of statistical investigation under discussion. It must be emphasized that if an equation happens not to be identified, it is impossible even to approximate the true equation from statistical data alone.
Market experiments or other substitute approaches must be employed to obtain this information.
2. Even if an equation turns out to be identified, precautions must be taken to insure that a statistically estimated equation is not distorted by the presence of the simultaneous relationships. We will see in the next section that an ordinary least-squares procedure is likely to lead to precisely this sort of distortion.
In this section we deal with the first of these, the identification problem-the circumstances under which it is, at least in principle, possible to unscramble our simultaneous relationships statistically. To illustrate, let us consider what is involved in finding statistically an advertising-demand curve such as the one, which Figure I attempted to construct in a rather primitive fashion. Now while sales are doubtless affected by advertising, as the advertising-demand function assumes, this function is often accompanied by a second relationship in which what we might call the direction of causation is reversed. It is well known that a firm's advertising budget is frequently affected by its sales volume. In fact, many - 9 -businesses operate on a rule of thumb, which allocates to advertising expenditure a fixed proportion of their total revenues. For such a business, then, we will have two advertising expenditure demand relationships: (1) the demand function which shows how quantity demanded, Q, is affected by a firm's advertising budget, A : Q = f (A) and (2) the budgeting equation which shows how the firm's advertising decisions are affected by the demand for its product: A = g (Q).
Both of these relationships may actually be of interest to the businessman. The first, as already stated, is directly relevant to his own optimal expenditure decision. The second, if obtained from industry records, will give him vital information about the behavior patterns of his competitors.
The firm's actual sales and its actual advertising expenditure will, of course, depend on both its advertising budgeting practices (the budgeting equation) and on the demand-advertising relationships. In Figure 2 the graphs of two such hypothetical relationships are depicted.
In Figure 2A we show the two curves which the statistician is seeking. We make ourselves, as it were, momentarily omniscient and thus have no difficulty envisioning the true relationships.
However, the information available to the statistician is much more restricted, as we shall now see. In our situation the actual advertising expenditure, A, and the volume of sales, Q, are determined, as for any simultaneous equation, by the point of intersection, P, of the two curves.
We now can describe two cases of non-identification: Case 1. Neither curve identified. If the two curves were to retain their shape from year to year, that is, i f neither o f them ever shifted, all the intersection points P would coincide or at least lie very close together (Figure 2B). There would only be a single observed point, as in the figure, or the tightly clustered points would form no discernible pattern, and so the shape of neither curve could even approximately be found from the data. We see then, though it may be a bit surprising, that curves, which never shift are from this point of view the worst of all possibilities.
Case 2. One of the curves not identified (but the other curve identifiable). This is a case frequently encountered in practice when the demand curve of one firm is investigated. The data form a neat and simple pattern, but what they describe is the firm's inflexible advertising - 10 -budgeting practices rather than the nature of the demand for its product. In such circum-stances what happens is that the budget curve never shifts but the demand curve does. There will then be a number of different intersection points, such as P, P, and P", but they will always describe only the shape of the advertising budget line (Figure 3) . The reader can well imagine how often statistical attempts to find the advertising demand curve have produced neat linear relationships (and spectacularly high correlation coefficients), though what the triumphant investigator has located (without his knowing it) is a totally different curve from the one he was seeking. The situation, which we have just examined, is really ideal from the point of view of the statistician, provided the relationship, which is not shifting happens to be the one he is seeking. But the question remains: how is he to know when one relationship is standing still, and even if he somehow knows this, how does he determine which one it is?
We will see that in the answers to these questions lies the key to the solution of the identification problem.
It will be shown presently that only where both curves shift over time or from firm to firm or from geographical territory to territory can they ordinarily both be identified. However, in this case the difficult task of unscrambling the two relationships becomes particularly acute.
Figure 4 illustrates how three points, A, B, and C, in a diagram similar to Figure 1 might have been generated by three different (shifted) pairs of our curves. It is, noteworthy that the negatively sloping (!) "advertising curve" FP estimated statistically from these points bears not the slightest resemblance to any of the true curves. Nor, since it is merely a recording of points - 11 -of intersection, is there any reason why it should. The shape of FF' is not even any sort of "compromise" between those o f the budget and advertising demand curves! We conclude that where simultaneous relationships are present the standard curve-fitting techniques described in Section 4 and Figure 1 may well break down completely. Their results are likely to bear absolutely no resemblance to the equations, which are being sought! Such a naive approach may therefore well be worse than no investigation because misleading information is usually worse than no information at all.
Let us now see how one can, in principle, test whether the relationship we are seeking is identified (potentially discoverable by statistical means). First we note that, as the model has so far been described, there is no way of accounting for any shifts in either relationship, which as we have observed, are crucial for our problem. The reason is that only two variables, A and Q, have been considered in the relationships Q = f (A) (the demand relationship) and A = g (Q ) ( the advertising budget equation).
There must, in fact, be some other influences (other variables), which disturb the relationships between Q and A and produce the shifts in their graphs. These additional variables must be taken explicitly into account. As we know, the demand relationship is likely to involve many variables in addition to A. For example, consumer's disposable income is a variable which affects the volume of sales resulting from a given level of advertising expenditure though, very likely, it does not enter the firm's budget calculation explicitly but only indirectly via the effects of income on the sales of the company's product. Similarly, the firm's budget policy may be affected by its past dividend payments, which determine how much it can currently spare for advertising expenditure, but this dividend policy will have little or no effect on the demand curve for its products. Suppose, for the sake of simplicity, that the four variables Q, A, Y (the disposable income), and D (the total dividend payments in the preceding year) are the only ones that are relevant to the problem. Our two relationships then become: (3) the advertising demand function Q = f ( A, Y ) and (4) the advertising budget equation A = g(Q, D).
Here changes in the value of Y are what produce the shifts in the graph of the demand equation, which have been discussed. Similarly, changes in D produce shifts in the advertising budget curve.
Now that we have examined how shifts in the two curves are produced we can return to the question of identification. Let us see, intuitively, how the presence of the shift variables in - 12 -equations ( 3) and (4) makes it possible, in principle, to separate the relationships from the statistics (i.e., how the shift variables identify the equations). It will be shown now that Y and D permit the statistician, at least conceptually, to divide up the statistical information in such a way that he is left with situations like that depicted in Figure 3. Such a situation gives him the information that permits him to infer which of the relationships is shifting and which is stand-ing still. That is, he can determine when one graph is not moving while the other shifts around, so that the resulting dots trace out the graph of the equation which is not shifting, the equation he is trying to estimate. The reader should first be warned, however, that the procedure, which is about to be described is not usually a practical estimation (curve finding) procedure and that other, more sophisticated measures are normally employed fox the purpose.
In Figure 5 we replot the data of Figure 1. Let us, in addition, determine the level of income for each point, Y, for that particular year. Suppose this information is as shown in Table III.
TABLE III.
Advertising Demand point 1950 1951 1952 1953 1954 1955 1956 1957 Disposable Income Y 360 297 295 307 428 381 420 300 ( $ billions) We note that the income values for the points representing 1951, 1952, 1953, and 1957 are fairly close together. Hence, if we are convinced that Y is the only variable which makes for sizable shifts in the advertising demand curve, it is reasonable to assume that all four points lie on ( or close to) the same curve, that is, among these points there has occurred little or no shift in the curve. We may therefore use these four points (ignoring the others) to locate a demand curve UU' (for income level approximately $300 billion) as shown. Similarly, we can use points for years 1954 and 1956 alone to find the shape of the advertising demand curve W', which pertains to income level approximately $420 billion, etc. In other words, the additional information on the value of Y for each point has permitted us, in principle, to ignore all points which contain information irrelevant to a given advertising demand curve.
We see, then, that if variable Y is present in one equation but not in the other, it permits us, in principle, to discover statistical points over which the budget line has shifted but through which the demand curve remains unchanged. In this same way we were able to trace out a - 13 -budget line in Figure 3. But it will be remembered that in the situation shown in Figure 3, the demand curve is unidentifiable because the budget curve never shifts. There is no variable such as D in the budget relationship, which will move the budget line about and yet permit the demand curve to stay still. This gives us the following result: one of a pair of simultaneous relationships will be identified if it lacks a variable, which is present in the other relationship.
The relevance of the shift variables Y and D for identification can also be seen in another way.
Suppose we use some correlation procedure to find a statistical relationship among variables Q, A, Y, and D. The system is identified if it is possible in principle for this statistical relationship to be an approximation to either equation (3) (the demand function) or to (4) (the budget function) and if it is possible to find out whether the statistical curve represents (3), (4), or neither. There are three possibilities: 1. The statistical relationship turns out to take the form Q = F(A, Y, D) in which all four variables are present (their coefficients are significantly different from zero). In that case we know that the statistics have given us a mongrel function, which resembles neither of the relationships, which we are seeking, for neither of the relationships contains both variables, Y and D.
2. Suppose now that the statistical relationship turns out to have an equation of the form F (A. Y) = Q; i.e., D plays no role in the equation. Then the statistical equation cannot involve any advertising budget function component, for if it did, any change in D would have influenced the relationship via its effect on the budget equation. In this case the sta-tistical equation must be an estimate of the demand relationship (3) alone.
3. Similarly, if the form of the statistical equation is F (A, D) = Q, it must represent the budget relationship (4) alone.
Thus the two variables Y and D, each of which appears in one and only one of the two relationships, have permitted us to identify both equations. For example, the presence of the variable D, which occurs only in the budget equation, acts as a warning signal, which notifies us at once when the budget equation has somehow gotten itself mixed in with our demand information. We conclude, again, that two simultaneous equations are normally identified (they can, at least in principle, be unscrambled from the statistics) if each equation contains at least one variable, which is absent from the other.
Of course, some more powerful identification criteria exist. For example, an obvious extension of the preceding result is the theorem that in a system of n simultaneous equations, a necessary condition for identification of any one of the equations, say the ith, is that every other equation in the system contain at least one variable which is missing from equation i.
This is hardly the place for a systematic discussion of the identification problem, and most of what has been said on the subject has been intended to be intuitive rather than rigorous.
However, the reader should have gathered that it is an extremely serious problem and that inadequate attention on the part of the analyst to this problem can easily invalidate his statistical results in their, entirety. |
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绪论及预备知识
一、数学到底考什么？
综合能力考试中数学部分主要考察考生的运算能力、逻辑推理能力、空间想象能力和 数据处理能力，通过问题求解和条件充分性判断两种形式测试，涉及的数学知识范围有：
（一） 算术
1．整数
(1) 整数及其运算 (2) 整除、公倍数、公约数 (3) 奇数、偶数 (4) 质数、合数
2．分数、小数、百分数 3．比与比例 4．数轴与绝对值
（二）代数
1．整式
(1) 整式及其运算 (2) 整式的因式与因式分解
2．分式及其运算 3．函数
(1) 集合 (2) 一元二次函数及其图像 (3) 指数函数、对数函数
4．代数方程
(1) 一元一次方程 (2) 一元二次方程 (3) 二元一次方程组
5．不等式
(1) 不等式的性质 (2) 均值不等式 (3) 不等式求解
一元一次不等式（组），一元二次不等式，简单绝对值不等式，简单 分式不等式. 24 小时热线电话：4008-219-288
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数列、等差数列、等比数列
（三）几何
1．平面图形
(1)三角形 (2)四边形 矩形、平行四边形、梯形 (3)圆与扇形
2．空间几何体
(1)长方体 (2)柱体 (3)球体
3．平面解析几何
(1)平面直角坐标系 (2)直线方程与圆的方程 (3)两点间距离公式与点到直线的距离公式
(四)数据分析
l.计数原理
(1)加法原理、乘法原理 (2)排列与排列数 (3)组合与组合数
2．数据描述
(1)平均值 (2)方差与标准差 (3)数据的图表表示:直方图，饼图，数表
3．概率
(1)事件及其简单运算 (2)加法公式 (3)乘法公式 (4)古典概型 (5)伯努利概型
二、数学命题特点是什么？ 24 小时热线电话：4008-219-288
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数学涵盖初中和高中六年的知识，面大，量广，范围广，考生复习很难抓住重点，同 时真题解题技巧性极强，因此加大技巧性训练越来越重要.
三、充分性判断解题规则
1.充分性命题定义
对两个命题 A 和 B 而言，若由命题 A 成立，肯定可以推出 B 也成立（即 A B 为真命 题），则称命题 A 是命题 B 成立的充分条件，或称命题 B 是命题 A 成立的必要条件. 【注意】A 是 B 的充分条件可以巧妙地理解为：有 A 必有 B，无 A 时 B 不定.
2.解题说明与各选项含义
本类题要求判断所给出的条件能否充分支持题干中陈述的结论，即只要分析条件是否 充分即可，而不必考虑条件是否必要.阅读条件（1）和（2）后选择： （A） 条件（1）充分，但条件（2）不充分. （B） 条件（2）充分，但条件（1）不充分. （C） 条件（1）和（2）单独都不充分，但条件（1）和条件（2）联合起来充分. （D） 条件（1）充分，条件（2）也充分. （E） 条件（1）和（2）单独都不充分，条件（1）和条件（2）联合起来也不充分. ▲以上规定全书都适用，以后不再重复说明.
3.图示描述
（1）条件一 （2）条件二 √ × （A） × √ （B） × ＋ ×＝√ （C） √ √ （D） × ＋ ×＝× （E） “√”表示充分，“×”表示不充分，“＋”表示两条件需要联合.
4.常用的求解方法
（1）解法一 直接定义分析法（即由 A 推导 B） 若由 A 可推导出 B，则 A 是 B 的充分条件；若由 A 推导出与 B 矛盾的结论，则 A 不是 B 的充分条件.解法一是解“条件充分性判断”型题的最基本的解法，应熟练掌握. （2）解法二 题干等价推导法（寻求题干结论的充分必要条件）即，要判断 A 是否是 B 的充分条件，可找出 B 的充要条件 C，在判断 A 是否是 C 的充分条件. （3）解法三 特殊反例法 由条件中的特殊值或条件的特殊情况入手，推导出与题干矛盾的结论，从而得出条件
不充分的选择. 24 小时热线电话：4008-219-288
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【注意】此种方法不能用在条件具有充分性的肯定性 ．．． 的判断上.因为某一个特值充分，不能 说明其他数值也充分.
解题相应的技巧
（1）当条件给定的参数范围落入题干成立范围时，即判断该条件是充分. （2）对条件做不同标记，这样方便答题. （3）当发现所给的两个条件是矛盾关系时，备选答案范围为 A,B,D,E. （4）当发现所给的条件是包含关系时，比如条件二的范围包含条件一的范围，则备选答案 范围为 A,D,E. （5）当确定条件 1（2）具备充分性，条件 2（1）未定的情况时，备选答案范围为 A(B),D. （6） 当确定条件 1 （2） 不具备充分性， 条件 2 （1） 未定的情况时， 备选答案范围为 B(A),C,E. 【注意】考试中，很多考生不敢选 E 而导致丢掉应该得到的分数，所以在确定无误的情 况下，要能够果断地选 E. 24 小时热线电话：4008-219-288
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第一章 实数 第一节 考试要点剖析
一、数的概念与性质
1.整数与自然数
整数 Z ： , 2, 1, 0,1, 2,3,
自然数 N ： 0,1, 2,3,
+
00
Z NZ
Z
正整数 自然数 最小的自然数为 整数 负整数
2.质数与合数
质数:如果一个大于 1的正整数，只能被 1 和它本身整除（只有 1和其本身两个约数），那 么这个正整数叫做质数（质数也称素数）. 合数：一个正整数除了能被 1和本身整除外，还能被其他的正整数整除（除了 1和其本身之
外，还有其他约数），这样的正整数叫做合数.
【大纲考点】1.整数 （1）整数及其运算，（2）整除、公倍数、公约数，（3）奇数、 偶数，（4）质数、合数；2.分数、小数，百分数；3.比与比例；4.数轴与绝对值. 【命题剖析】对于实数的计算，不仅要掌握整数、分数的运算的技巧、比例运算技巧等， 更要一定高度对各部分数学知识做一个综合归纳，例如“裂项求和公式”、“大量重复 部分设 t 法”“约分技巧”等.本章命题主要体现在以下六个方面： 1.概念型题目：主要围绕奇数、偶数、质数、合数、公约数和公倍数展开； 2.计算型题目：主要围绕很长串的数字的计算和化简计算； 3.比例的相关定理：等比定理、合比定理、分比定理； 4.有理数与无理数的性质几其化简； 5.利用绝对值的几何意义进行化简计算； 6.平均值和方差定义及其应用，特别是均值不等式的运用. 24 小时热线电话：4008-219-288
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▲质数与合数有如下重要性质： ①质数和合数都在正整数范围，且有无数多个. ②2 是唯一的既是质数又是偶数的整数，即是唯一的偶质数.大于 2 的质数必为奇数. 质数中只有一个偶数 2，最小的质数为 2. ③若质数 ,|p a b 则必有 | | .p a p b 或 （注： |p a 表示 p 是 a 的约数）. ④若正整数 a b， 的积是质数 p ，则必有 .a p b p 或
⑤1 既不是质数也不是合数. ⑥如果两个质数的和或差是奇数，那么其中必有一个是 2；如果两个质数的积是偶数，那 么其中也必有一个是 2. ⑦最小的合数为 4.任何合数都可以分解为几个质数的积，能写成几个质数的积的正整数就 是合数. ⑧互质数：公约数只有 1的两个数称为互质数，如 9 和 16.
3.奇数与偶数
奇数：不能被 2 整除的数. 偶数：能被 2 整除的数.【注意】 0 是属于偶数.整数可以分成奇数和偶数 【注意】两个相邻整数必为一奇一偶.除了最小质数 2 是偶数外，其余质数均为奇数.
4.分数与小数
分数：将单位“1”平均分成若干份，表示这样的一份或几份的数 叫做分数. 小数：实是实数的一种特殊的表现形式.所有分数都可以表示成小数，小数中的圆点叫做小 数点，它是一个小数的整数部分和小数部分的分界号.其中整数部分是零的小数叫做纯小 数，整数部分不是零的小数叫做带小数.
5.整除、倍数、约数
数的整除： 当整数 a 除以非零整数 b ，商正好是整数而无余数时， 则称 a 能被 b 整除或 b 能整除 a .倍数和约数：当 a 能被 b 整除时，称 a 是 b 的倍数， b 是 a 的约数. 最小公倍数：几个数公有的倍数叫做这几个数的公倍数，其中最小的一个叫做这几个数的 最小公倍数. 最小公倍数的表示：数学上常用方括号表示.如 [12,18, 20] 即 12、18 和 20 的最小公倍数. 最小公倍数的求法：求几个自然数的最小公倍数，有两种方法： ①分解质因数法.先把这几个数分解质因数， 再把它们一切公有的质因数和其中几个数公有 的质因数以及每个数的独有的质因数全部连乘起来，所得的积就是它们的最小公倍数. ②
②公式法.由于两个数的乘积等于这两个数的最大公约数与最小公倍数 24 小时热线电话：4008-219-288
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的积.即 , [ . ] a b a b a b .所以，求两个数的最小公倍数，就可以先求出它们的最大公 约数 然后用上述公式求出它们的最小公倍数.
二、实数的分类
1.实数包括有理数和无理数
0
正整数 正有理数 正分数 有理数 有限小数，无限循环小数 负整数 实数 负有理数 负分数 正无理数 无理数 无限不循环小数 负无理数
2.按性质符号分类
实数
负无理数 负分数 负整数 负有理数 负实数 正无理数 正分数 正整数 正有理数 正实数
0
3.常见整除的特点
能被 2 整除的数：个位 0,2,4,6,8. 能被 3 整除的数：各数位数字之和必能被 3 整除. 能被 4 整除的数：末两位（个位和十位）数字必能被 4 整除. 能被 5 整除的数：个位为 0 或 5. 能被 6 整除的数：同时满足能被 2 和 3 整除的条件.
能被 8 整除的数：末三位（个位、十位和百位）数字必能被 8 整除. 24 小时热线电话：4008-219-288
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能被 9 整除的数：各数位数字之和必能被 9 整除. 能被 10 整除的数：个位必为 0. 能被 11 整除的数： 从右向左,奇数位数字之和减去偶数位数字之和能被 11 整除 （包括 0） .
三、绝对值
1.定义
正数的绝对值是它本身;负数的绝对值是它相反数；零的绝对值还是零.
2.数学描述
实数 a 的绝对值定义为：
00 00
a aa aa a
,其几何意义是一个实数 a 在数轴上所对应的点到原点的距离值.
3.绝对值的性质
（1）对称性： a a 【注意】平方和根号也具有非负性 （2）等价性：
2222
,a a a a a a R
（3）自比性： a a a ，推而广之， 1, 0, 1, 0.
xx xxx x
（4）非负性：即 0a ，任何实数 a 的绝对值非负. （5）乘法运算性质： , aaab a b b b
.（6）不等式性质： , 0x a a x a a , , ,x a x a or x a , 0a
推而广之，具有非负性的数还有正偶数次方（根式），如 2 4 4
, , ,a a a a… …
4.三角不等式
a b a b a b
【注意】考试要求掌握等号成立条件的判断.
四、比和比例
1.比 24 小时热线电话：4008-219-288
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两个数相除，又称为这两个数的比，即 : aa b b
.其中 a 叫做比的前项， b 叫做比的后 项.相除所得商叫做比值，记作 : aa b kb
.在实际应用中，常将比值表示成百分数，称 为百分比.
2.比 例
相等的比称为比例，记作 : : a ca b c d b d
或 .其中 a 和 d 称为比例外项， b 和 c 称为比 例内项.当 : :a b b c : :a b b c 时， 称 b 为 a 和 c 的比例中项， 显然当 , ,a b c 均为正数时，
b 为 a 和 c 的几何平均值.
3.正 比
若 , 0y kx k ，则称 y 与 x 成正比， k 称为比例系数. 【注意】 并不是 x 和 y 同时增大或减小才称为正比.比如当 0k 时， x 增大时， y 反而减小.
4.反比
若 , 0
ky kx
，则称 y 与 x 成反比， k 称为比例系数.
5.比例的基本性质
（1） : : .a b c d ad bc
（2） : : : : : : : :a b c d b a d c b d a c d b c a
6.重要定理
（1）更比定理： .
a c a bb d c d
（2）反比定理： .
a c b db d a c
（3）合比定理： .
a c a b c db d b d
（4）分比定理： .
a c a b c db d b d
（5）合分比定理：
1m
a c a mc a cb d b md b d
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（6）等比定理： .( 0)
a c e a c e b d fb d f b d f
增减性变化关系 ( , , 0) a b m ＞
（1） .,1 bambmaba 则若 反之也成立. （2） .,10 bambmaba 则若 反之也成立.
四、平均值
1.算术平均值
n 个数 1 2, ,..., nx x x ，则称 1 2 ... nx x xx n
为这 n 个数的算术平均值，简记为 1
nii
xx n
.
2.几何平均值
设 n 个 正 数 1 2, , , nx x x ， 称 1 2
ngn
x x x x 为 这 n 个 正 数 的 几 何 平 均 值 ， 简 记 为
nniig
xx
1
.【注意】几何平均值是对于正数而言.
3.均值不等式
（1）当 1 2 3, , nx x x x 为整数时，它们的算术平均值大于或等于它们的几何平均值，用符 号可表示为： 1 2 31 2 3 ( 0 1, 2, )n n n i
x x x x x x x x x i nn
其中 ＞ ，
当且仅当“ 1 2 3 nx x x x ”时，可以取等号. （2）当 2n 时，正数 1 2,x x 是几何平均数 1 2x x 称作 1 2x x和 的比例中项. （3）常用变式 1： 2a b ab （ , 0a b ＞ ）即对于正数而言，两个数之和大于或等于 2
倍的 a 和 b 的几和平均数，当且仅当 a b 时取得最小值. 24 小时热线电话：4008-219-288
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（4）常用变式 2： 1 2( 0) a aa
＞ （ , 0a b ＞ ），即对于正数而言，互为倒数的两个数之 和大于或等于 2，当且仅当 1a 时取得最小值 2.
第二节 基础过关题型
一、有理数问题
【题型 1】有理数基本性质 【例 1.1】（2008-1）一辆出租车有段时间的营运全在东西走向的一条大道上，若规定向 东为正，向西为负，且该车行驶公里数依次为-10，+6，+5，-8，+9，-15，+12，则将最后 最后一名乘客送到目的地时，该车的位置（ ）. （A）在首次出发地东面 1 公里处 （B）在首次出发地西面 1 公里处 （C）在首次出发地东面 2 公里处 （D）在首次出发地西面 2 公里处 （E）仍在首次出发地 【题型 2】整除问题 【例 1.2】（2007-10） m 是一个整数. （1）若 pm q
，其中 p 与 q 为非零整数，且 2
m 是一个整数 （2）若 pm q
，其中 p 与 q 为非零整数，且 2 41
mm
是一个整数
【思路电拨】 1. 对于代数式问题，常用的方法有：①特值法②设 k 法③因式分解法④拆项法. 2. 对于数的整除，重点掌握被 2,3,5,9 整除的.
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【题型 3】带余除法 【例 1.3-1】 (2015) 几个朋友外出游玩， 购买了一些瓶装水， 则能确定购买的瓶装水数量. （1）若每人分 3 瓶，则剩余 30 瓶 （2）若每人分 10 瓶，则只有一人不够 【例 1.3-2】（2019） n 为正整数，则能确定 5n 的余数. （1）已知 2n 的余数 （2）已知 3n 的余数
【思路电拨】 1.重要等量关系： 7 3 2 1 可以改写成 1327 .同理：若 a 除以 b 余 r ，设商为 k ，则 rbk a .2. 同余问题：用含 n 的代数式表示下列数： （1）一个数除以 4 余 1，除以 5 余 1，除以 6 余 1，那么这个数为 160 n .（2）一个数除以 4 余 3，除以 5 余 2，除以 6 余 1，那么这个数为 760 n .（3）一个数除以 4 余 1，除以 5 余 2，除以 6 余 3，那么这个数为 360 n .3.不同余问题:转化为不定方程求解. 24 小时热线电话：4008-219-288
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【题型 4】倍数约数分析法 【例 1.4-1】（2017）将长、宽、高分别为 12、9 和 6 的长方体切割成正方体，且切割后 无剩余，则能切割成相同正方体的最少个数为（ ）个. （A）3 （B）6 （C）24 （D）96 （E）648 【例 1.4-2】（2009-10） a b c d e 的最大值是 133. （1） edcba ,,,, 是大于 1 的自然数,且 2700 abcde
（2） edcba ,,,, 是大于 1 的自然数,且 2000 abcde
【题型 5】奇偶分析法 【例 1.5】（2014-10） 2 2
m n 是 4 的倍数.
（1） m ， n 都是偶数 （2） m ， n 都是奇数
【思路电拨】奇数偶数问题常用技巧有： 1. 表示方法：奇数 2 1, n n Z ；偶数 2 , n n Z ；（用含 n 的式子表示） 【注】0 也是偶数. 2.四则运算：奇数 奇数= 偶数；奇数 偶数= 奇数 ；偶数 偶数= 偶数 ； 奇数 奇数= 奇数 ；奇数 偶数= 偶数 ；偶数 偶数= 偶数 ；
【思路电拨】倍数与约数问题应注意以下几点： 1.如何用分解质因数法求最大公约数和最小倍数？ 2.设未知数的方法：若两个数的最大公约数是 k,则这两数分别为 ,ak bk .3.常用定理：两个正整数的的乘积等于这两个数的最大公因数和最小公倍数的积. 即 ( , ) [ , ]ab a b a b ，其中（ ）表示最大公约数，[ ]表示最小公倍数. 24 小时热线电话：4008-219-288
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【题型 6】质数合数分析法 【例 1.6-1】（2010-1）三名小孩中有一名学龄前儿童（年龄不足 6 岁），他们的年龄都 是质数（素数），且依次相差 6 岁，他们的年龄之和为（ ）. （A） 21 （B） 27 （C）33 （D）39 （E）51 【例 1.6-2】 （2015）设 nm, 是小于 20 的质数，满足条件 2 nm 的 nm, 共有（ ）.（A）2 组 （B）3 组 （C）4 组 （D）5 组 （E）6 组
【思路电拨】质数合数问题常用以下方法： 1.质数定义：如果一个大于 1 的正整数，只能被 1和本身整除，那么这个正整数叫做质 数.（质数也成素数）. 2.30 以内的数：2,3,5,7,11,13,17,19,23,29. 3.数字“2”突破法 ：在所有的质数中，只有 2 是唯一的偶数. 24 小时热线电话：4008-219-288
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二、无理数问题
【题型 7】无理数的基本概念 【例 1.7-1】在 1 ， 0 ， 31 ， 2 ，
， 101101110 .0 中任取一个数，取到无理数的概率 是（ ）. （A） 61 （B） 31 （C） 21 （D） 32 （E） 65
【例 1.7-2】（2008-10）一个大于 1 的自然数的算术平方根为 a ，则与该自然数左右相邻 的两个自然数的算术平方根分别为（ ）. （A） 1, 1a a （B） 1, 1a a （C） 1, 1a a
（D） 2 2
1, 1a a （E） 2 2
1, 1a a
【例 1.7-3】(2004-10) baba 2
(1) 0a , 0b (2) 0a , 0b
【思路电拨】 1.什么是有理数？整数和分数统称为有理数. 2.什么是无理数？无限不循环小数是无理数. 3.无限循环小数是无理数吗？无限循环小数是分数，属于有理数，而无限不循环小 数是无理数. 4.二次根式基本概念：①一个正数有两个平方根，其中那个正的叫做算术平方根， 用符号“ ”表示.②二次根式化简： 2
a a .24 小时热线电话：4008-219-288
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【题型 8】无理数四则运算 【例 1.8-1】设 a 是一个无理数，且 ,a b 满足 1 0ab a b ，则 b 是一个（ ）. （A）小于 0 的有理数 （B）大于 0 的有理数 （C）小于 0 的无理数 （D）大于 0 的无理数 （E）以上均不对 【例 1.8-2】设 5 15 1
的整数部分为 a ，小数部分为 b ，则 5ab =（ ）. （A）-2 （B）-1 （C）0 （D）1 （E）2
【思路电拨】 1.无理数的四则运算： 有理数+无理数= 无理数； 无理数+无理数=无理数或有理数； 有理数×无理数= 0 或无理数； 无理数×无理数=无理数或有理数； 2.常用性质：若 ,a b 为有理数，
为无理数， 00 0
aa b b
.2. 小数部分与整数部分：关键在于先确定整数部分，然后再求小数部分，数部分是 原数减去整数部分. 3. 分母有理化:指的是将该原为无理数的分母化为有理数的过程 .24 小时热线电话：4008-219-288
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三、实数运算技巧
【题型 9】长串式子计算技巧 【例 1.9-1】（2009-1）设直线 1)1( ynnx ( n 为正整数)与两坐标轴围成的三角形面 积 nS （ n 1,2, ,2009），则 2009 21 SSS （ ）. （A） 1 2009 2 2008
（B） 2009 2008 21 （C） 2010 2009 21
（D） 2009 2010 21 （E）以上结论均不正确 【例 1.9-2】（2015）已知 1 2 1 2 3n nM a a a a a a ，
1 2 2 3 1n nN a a a a a a ，则 M N .（1） 1 0a （2） 1 0na a
【思路电拨】常用运算技巧有 :1. 约分化简技巧 2. 裂项技巧 3 . 妙用平方差公式 4. 分母有理化技巧
大量重复部分设 t 法 6. 数列求和套公式 7. 求相同数字数的和 8. 阶层裂项公式 .24 小时热线电话：4008-219-288
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【例 1.9-3】（2007-10）
238
1 1 1 1( ) ( ) ( )2 2 2 20.1 0.2 0.3 0.4 0.9
（ ）.（A） 255 768 （B） 85 512 （C） 85 384 （D） 255 256 （E）以上均不对
四、比例问题
【题型 10】比例基本问题 【例 1.10-1】（2001-10）若 0 0a b k＞ ＞ ， ＞ ，则下列不等式能够成立的是（ ）. （A） b b ka a k
＜- （B） a a kb b k
＞ （C） b b ka a k
＞-
（D） a b kb a k
＜ （E）以上结果均不正确
1.什么情况下使用“见比设 k 法”? 2.如何合成连比? 若 : 2 : 3 a b ， : 4 : 5 b c ，则 : :a b c =3.什么是正比例和反比例关系？ 4.比例关系的增减性. 若 , , 0a b m ＞ ：①若 1
ab ，则 a a mb b m
（分母增幅较大） ②若 0 1
ab
，则 a a mb b m
（分子增幅较大） 24 小时热线电话：4008-219-288
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【例 1.10-2】（2002-1）设 1 1 1: : 4 : 5 : 6
x y z ，则使 74 x y z 成立的 y 的值是（ ）. （A）24 （B）36 （C） 74 3 (D) 37 2 (E )以上结果均不对 【题型 11】比例定理 【例 1.11-1】若 0a b c ， 2 2 2a b b c c a kc a b
，则 k 的值为（ ）. （A）2 （B）3 （C）-3 （D）1 （E）以上选项均不正确 【例 1.11-2】（2004-10） c a ba b b c c a
.（1） 0 c a b （2） 0 a b c
【思路电拨】 1.什么是等比定理？ 2.什么是分比定理？ 3.什么是合比定理？ 24 小时热线电话：4008-219-288
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五、绝对值问题
【题型 12】去绝对值 【例 1.12-1】（2011-10） 已知 1, 01, 0
xg x x
，
1 1 2 2f x x g x x x x ，则 f x 是与 x 无关的常数. （1） 1 0x
（2） 1 2x
【题型 13】绝对值方程与不等式 【例 1.13-1】（2017）不等式 1 2x x 的解集为（ ）. （A） ( ,1] （B） 3( , ]2
（C） 3[1, ]2 （D） [1, ) （E） 3[ , )2
【思路电拨】常用方法： （1）分类讨论法（2）几何意义法（3）平方法（4）选项代入法.
【思路电拨】绝对值定义：数 a 到原点的距离叫做 a 的绝对值，可表示为 a .
,aaaaa
0，＜ 0
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【例 1.13-2】（2019）设实数 ,a b 满足 6ab ， 6a b a b ，则 2 2
a b （ ）.
（A）10 （B）11 （C）12 （D）13 （E）14
【题型 14】非负数和为零 【例 1.14-1】(1997-1）若 2 10
( 60) 90 ( 130) 0a b c ，则 a b c 的值是（ ）. （A）0 （B）280 （C）100 （D）-100 （E）无法确定 【例 1.14-2】 （2009-1） 已知实数 , , ,a b x y 满足 2
2 1y x a 和 2
2 1x y b ，则 3 3x y a b
( ). （A） 25 （B） 26 （C） 27 （D） 28 （E） 29
【思路电拨】常见的形式有： 1.基本原理（标准型）：若 2 0a b c ，则
000
abc
2.常见的三种变式：（1）配方型.（2）两式相加减型.（3）隐含定义域型. 24 小时热线电话：4008-219-288
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【题型 15】绝对值自比性问题 【例 1.15-1】（2005-1）实数 ba, 满足 baabaa ＞)( .（1） 0＜a （2） ab ＞
【例 1.15-2】已知 1
ba ca b c
，则 2019
( ) ( )
abc bc ac ab ab bc ac abc
（ ）. （A）1 （B）-1 （C）2 (D)-2 (E) 12
【题型 16】三角不等式 【例 1.16-1】 （2015） 已知 1 2 3, ,x x x 为实数， x 为 1 2 3, ,x x x 的平均值. 则 1, 1, 2,3 kx x k .(1) 1, 1, 2,3 kx k （2） 1 0x
【思路电拨】绝对值自比性： 1. a a .2. 1, 0. 1 0.
a a aaa a
＞， ＜
3 一般地，绝对值自比性问题常综合代数式符号判断进行考察.
【思路电拨】 三角不等式： a b a b a b ， a b a b a b .【注意】考试要求掌握等号成立条件的判断. 24 小时热线电话：4008-219-288
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【题型 17】绝对值函数的最值问题（难点） 【思路电拨】三大常见模型 模型一： 两绝对值和 1 3y x x . 模型二： 两个绝对值差 1 3y x x .
模型三：三个绝对值之和 1 2 3y x x x .24 小时热线电话：4008-219-288
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【例 1.17-1】（2008-10） 5216 8|1| 2 xxxx .（1） 2x＞ . （2） 3x .【例 1.17-2】.（2008-1） ( ) f x 有最小值 2. （1） 5 1( ) 12 12
f x x x （2） ( ) 2 4f x x x
【例 1.17-3】 （2009-10）设 |20 ||20 ||| axxaxy ,其中 20 0 a ,则对于 满足 20 xa 的 x 值, y 的最小值是（ ）. （A）10 （B）15 （C）20 （D）25 （E） 30 24 小时热线电话：4008-219-288
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【题型 18】平均值与均值不等式 【例 1.18-1】（2005-10） , , a b c 的算术平均值是 14 3 ，则几何平均值是 4. （1） , , a b c 是满足 1a b c 的三个整数，且 4b
（2） , , a b c 是满足 1a b c 的三个整数，且 2b
【例 1.18-2】（2009-10） cbacba 111 .（1） 1abc （2） cba ，， 为不全相等的正数
【思路电拨】 1. 算术平均值： n 个数 nxxxx ，， 321 ,, 的算术平均值为为 ,321
nxxxx n
记为 ini
xnx 1
1
几何平均值： n 个正数 nxxxx ，， 321 ,, 的几何平均值为为 1 2 3 ,n nx x x x 记为
nini
x
1
G
【注意】只有正数才有几何平均值 3.均值不等式：几个正数的算术平均值大于或等于几何平均值 （1）口诀：“一正二定三相等” “正”是使用均值不等式的前提；“定”是均值不等式的目标;“相等”是取最值得 条件 （2）常用方法：①直接套用②拆项和添项 ③主动乘某式 24 小时热线电话：4008-219-288
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【例 1.18-3】 （2019） 设函数 2
( ) 2 af x x x
（ 0a＞ ）在 (0, ) 内的最小值为 0( )f x =12， 则 0x
（ ）. （A）5 （B）4 （C）3 （D）2 （E）1 24 小时热线电话：4008-219-288
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第二章 整式与分式 第一节 考试要点剖析
一、基本定义
1.单项式
数与字母的积这样的代数式叫做单项式，如 2
3x ；单独一个数或一个字母也是单项式. 其中单项式中的字母因数叫做单项式的系数； 所有字母的指数的和叫做这个单项式的次数； 若单项式表示 m n p
ax y z ，那么 a 称为单项式 m n p
ax y z 的系数， m n p 叫做这个单项式 的次数. 【注意】数与字母之间是乘积关系.
2.多项式
几个单项式的和叫做多项式.在多项式中，每个单项式叫做多项式的项，其中不含字母 的项叫做常数项.一个多项式有几项就叫做几项式.多项式中的符号， 看作各项的性质符号； 多项式中，次数最高项的次数，就是这个多项式的次数.例如， 31 1 2 2 nm n m n
ax y bx y cz ，此为 3 项式，若 31122 nmnmn ，则此多项式为 2 2m n 次式. （1）把一个多项式按某一个字母的指数从大到小的顺序排列起来，叫做把多项式按这 个字母降幂排列. （2）把一个多项式按某一个字母的指数从小到大的顺序排列起来，叫做把多项式按这 个字母升幂排列. 有两个或两个以上字母的多项式，排列时，要注意：要先确认按照哪个字母的指数来 排列；然后再根据次字母的升幂还是降幂进行排列.
【大纲考点】1.整式（1）整式及其运算（2）整式的因式与因式分解；2.分式及其运算. 【命题剖析】整式与分式在考察中不仅仅单独作为题目来考察，更重要的是为方程，函 数，数列起到了重要的铺垫作用. 核心考察的题目类型主要集中于考察乘法公式、整式 化简，整式的因式分解、分式的恒等变形（通分、约分），以及用函数观点看代数式， 例如取值法等等. 24 小时热线电话：4008-219-288
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3.整式
单项式和多项式统称为整式.
4.分式
分式定义：用 ,A B 表示两个整式, BA 就可以表示成 BA 的形式，如果除式Ｂ中含有 字母，式子 BA 就叫做分式．
分式基本性质
性质 分式的分子与字母都乘以（或除以）同一个不为零的整式，分式 的值不变
表示
A AM A MB BM B M
（ M 为不等于零的整式）
应用
符号 法则
分子、分母与分式本身的符号，改变其中任何两个，分 式的值不变 ,
a a a ab b b b
约分 把一个分式的分子与分母的所有公因式约去叫做约分
通分 把几个异分母的分式分别化成与原本的分式相等的同 分母的分式叫做通分
5.最简分式
分式的分子与分母没有公因式时，叫做最简分式.
6.有理式
整式和分式统称为有理式.
二、整式的运算公式
222223322322332222222222
( )( ); 2. 2 ( ) ; 3. ( )( ); 4. 3 3 ( ) ; 5. 2 2 2 ( ) ; 16. ;2
a b a b a ba ab b a ba b a b a ab ba a b ab b a ba b c ab bc ac a b ca b c ab bc ac a b a c b c
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变形公式有：
22 2
2 ,a b a b ab .babababaab 22
222222
2222
2 2a b a b a b ，
22
4 .a b a b ab
2222
2 .a b c a b c ab bc ac
三、整式的除法
1.因式定理
)(xf 含有 ax b 因式 )(xf 能被 ax b 整除 ( ) 0
bf a ；尤其， )(xf 含有 x a 因式 )(xf 能被 x a 整除 0)( af .
2.余式定理
多项式 )(xf 除以 ax b 的余式为 )( abf ；尤其，多项式 )(xf 除以 x a 的余式为
f a .
四、分解因式
1.分解因式的概念：
把一个多项式化成几个整式的积的形式，这种变形叫做分解因式（又叫因式分解）. （1）因式分解的实质是一种恒等变形，是一种化和为积的变形. （2）因式分解与整式乘法是互逆的. （3）在因式分解的结果中，每个因式都必须是整式. （4）因式分解要分解到不能再分解为止.
2.因式分解的基本方法：
(1)提取公因式 （2）运用公式法；（3）分组分解法（4）十字相乘法（5）双十字相乘法.
3.因式分解的一般步骤：一提二套三分组. 24 小时热线电话：4008-219-288
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第二节 基础过关题型
一、整式
【题型 1】乘法公式 【例 2.1-1】（2013-10）已知 2 2
, 1f x y x y x y .则 , 1f x y
（1） x y （2） 1x y
【例 2.1-2】 （2002-1）已知 , , a b c 是不完全相等的任意实数，若 2
x a bc ， 2
y b ac ，
2
z c ab ，则 , ,x y z （ ）.
（A）都不大于 0 （B）至少有一个大于 0 （C） 至少有一个小于 0
【思路电拨】乘法公式是在多项式乘法的基础上，将一般法则应用于特殊形式的多项式 相乘，得出的既有特殊性、又有实用性的具体结论，在代数式的化简求值、恒等变形等 方面有广泛的应用，在学习乘法公式时，做到以下几点： 1.熟悉每个公式的结构特征；2.根据待求式的特点，模仿套用公式；3.逆向运用公式 乘法核心六公式有：
222223322322332222222222
1 ( )( ); 2 2 ( ) ; 3 ( )( ); 4 3 3 ( ) ; 5 2 2 2 ( ) ; 16 ;2
a b a b a ba ab b a ba b a b a ab ba a b ab b a ba b c ab bc ac a b ca b c ab bc ac a b a c b c
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（D）都不小于 0 （E）以上均不对 【题型 2】因式分解 【例 2.2-1】（2010-1）多项式 623 bx ax x 的两个因式是 1x 和 2x ，则其第三个 一次因式为（ ）. （A） 6x （B） 3x （C） 1x （D） 2x （E） 3x
【题型 3】双十字相乘法因式分解 【例 2.3】（2008-10） 2 2
6 10 4 0x mxy y y 的图像时两条直线. （1） 7m （2） 7m
【思路电拨】因式分解是把几个整式的积的形式的运算.它与整式乘法是互逆运算.常用 方法：① 提公因式法 ② 公式法 ③ 十字相乘法 ④分组分解法
【思路电拨】形如 2 2
ax bxy cy dx ey f 式子因式分解： 1.分解步骤：先分解 2
x 项、 2
y 项、常数项，然后去凑 xy 项、 x 项、 y 项的系数. 2.口诀：“左十字、右十字、大十字”. 24 小时热线电话：4008-219-288
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【题型 4】多项式展开系数 【例 2.4-1】（2009-1） 2 21 2(1 ) (1 ) (1 ) ( 1) 2 ( 1) ( 1) n nnx x x a x a x na x ，则 1 2 32 3 na a a na ( ). （A） 3 12
n
（B）
1
3 12
n
（C）
1
3 32
n
（D） 3 32
n
（E） 3 34
n
【例 2.4-2】（2010-10） 已知 3 2 3 41 2 3 41x kx a x a x a x a x 对所有实数 x 都成立， 则 1 2 3 4 8a a a a .（1） 2 9a （2） 3 27 a
【思路电拨】多项式展开应注意一下几点： 1.遇到多项式相等只要保证多项式对应系数相等即可. 2.常用方法有：待定系数法、赋值法. 24 小时热线电话：4008-219-288
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【题型 5】代数式最值问题 【例2.5】设实数 ,x y 满足等式 2 2
4 4 3 3 6 0x xy y x y ，则 x y 的最大值为 （ ）. （A）2 （B） 3 （C） 2 3 （D） 3 2 （E） 3 3
【题型 6】整式的除法和因式定理、余式定理 【例 2.6】（2009-10）二次三项式 2 6x x 是多项式 4 3 2
2 1x x ax bx a b 的一个 因式. （1） 16 a （2） 2b
【思路电拨】遇到代数式最值问题一般有以下三种思路: 1.凑完全平方式.2.均值不等式.3.二次函数求最值.
因式定理 1.因式定理：多项式 ( ) f x 含有因式 ( ) 0x a f a .证明：设 ( ) ( ) ( ) f x x a g x ，则令 x a ，易得： ( ) 0 ( ) 0f a g x
2.余式定理：多项式 ( ) f x 除以 x a 的余式为 ( ) f a
证明：设 ( ) ( ) ( ) ( ) f x x a g x r x ，则令 x a ，易得： ( ) ( ) f a r a
【评注】以上证明的核心是等式的赋值法. 24 小时热线电话：4008-219-288
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二、分式
【题型 7】齐次分式求值问题 【例 2.7-1】（2008-10）若 1 1: :3 4
a b ，则 12 16 12 8
a ba b
（ ）. （A）2 （B）3 （C）4 （D）-3 （E）-2 【例 2.7-2】 （2013-1）设 , ,x y z 为非零实数，则 2 3 4 12
x y zx y z
.（1） 3 2 0x y （2） 2 0y z
【题型 8】特殊形式求值 【例 2.8-1】设 x 是非零实数，则 33
1 18 x x
（1） 1 3x x
（2） 22
1 7x x
【思路电拨】分式求值问题常使用一下策略： 1. 齐次分式求值问题可使用赋值法. 2. 非齐次分式求值可使用“整体代换”策略.
【思路电拨】 1.已知 1
x ax
或 2 1 0x ax ，代数式的值. 2.将 1
x ax
直接套用完成平方公式可得： 2 22
1 2x ax
;【注意】若已知 2 1 0x ax ，则可它两遍同时除 x 变形为 1
x ax
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【例 2.8-2】(2009-1） 22
32 5 2 11
a a a
（1） a 是方程 2 3 1 0x x 的根 （2） 1a
【题型 9】解分式方程 【例 1.9-1】（2007-10）方程 2
1 1 01 1 1
ax x x
有实根. （1）实数 2a （2） 实数 2a
【例 1.9-2】（2009-10）关于 x 的方程 1 1 1 33 22 2
x xx x x a a x
与 有相同的增根. （1） 2a （2） 2a
【思路电拨】解分式方程的一般步骤：
去分母 . 即将分式方程两边同时乘以最简公因式，将其化为整式方程 .2. 解整式方程 .3. 验根 . 即检查解出的根是否会使原方程分母的为 0，若分母为 0，则这个解是原分 式方程的增根；若分母不为 0，则这解是分式方程的解 .
【注意】增根代入分式方程，分式方程无意义；但增根代入整式方程，整式成立 .24 小时热线电话：4008-219-288
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【题型 10】穿针引线法解高次不等式 【例 2.10-1】（2009-1） 2 2
2 8)(2 )(2 2 6) 0x x x x x （ .（1） ( 3, 2) x （2） 2,3 x
【例 2.10-2】（2013-10）不等式
22
2 3 05 6
x xx x
的解集是（ ）（A） (2,3) （B） ( ,2] （C） [3, )
（D） ( , 2] [3, ) （E） ( , 2) (3, )
【思路电拨】操作指南： 1.移项.使得式子右边为 0. 2.因式分解，并且需要保证每个因式的最高次项系数均为正. 3.在数轴上标注出所有零点，如果有恒大于 0 的式子直接“丢弃”. 4.从右上方开始起针，依次去穿每一个零点，并且把握“奇过偶不过”原则. 5.注意零点是实心点还是空心点，并根据图直接写出解集. 24 小时热线电话：4008-219-288
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第三章 方程和不等式 第一节 考试要点剖析
一、方程 （一）基本概念
1、方程与方程的解：
（1）含有未知数的等式叫做方程，方程中的未知数称为元.使方程成立的未知数的值叫做 方程的解，一元方程的解也叫做方程的根. （2）求方程的解或者确定方程无解的过程叫做解方程. （3）组成方程组的所有方程的公共解，叫做方程组的解.
2、方程的元和次：
所谓“几元”是指方程中含有几个未知数，“次”是指方程中未知数的最高次数
（二）常见的方程（组）
1、一元一次方程：
只含有一个未知数，且未知数的最高次数是 1 的方程，称为一元一次方程，其一般 形式为： ( 0) ax b a ，方程的解为 bx a
.
2.二元一次方程组：
形如 1 1 12 2 2
a x b y ca x b y c
（其中 1 1a b与 不能同时为 0， 2a 与 2b 不能同时为 0） ，
【大纲考点】 1．一元二次函数及其图像； 2．代数方程(1)一元一次方程，(2)一元二次 方程，(3)二元一次方程组； 3．不等式(1)不等式的性质，(2)均值不等式，(3)不等式 求解：一元一次不等式 (组)，一元二次不等式，简单绝对值不等式，简单分式不等式 .
【命题剖析】本章内容中二次函数是核心 .其中，二次方程，二次不等式等等内容都是 围绕二次函数展开的 .学习过程中，一是要学会解方程，解不等式，更加要明白他们的一 些性质，比如判别式，韦达定理等等 .对于超越方程和不等式往往不是考察的重点，但是 也要清楚 .24 小时热线电话：4008-219-288
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若如果 1 12 2
a ba b
，则方程组有唯一解.
3、一元二次方程：
（1）定义： 只含有一个未知数，且未知数的最高次数是 2 的方程，称为一元二次方程， 其标准形式为： 2 0, ( 0) ax bx c a .（2）根的判别式： 2 4b ac ，①若 0 ＞ 有两个不相等的实数根
②若 0 = 有两个相等的实数根
③若 0 ＜ 无实数根
（3）一元二次方程： 2 0, ( 0) ax bx c a 的解法： ①因式分解法 ②求根公式法：当 0 时， 1 2, 2
bx x a
③配方法 （4）根与系数的关系----韦达定理 若 1 2,x x 是一元二次方程 2 0, ( 0) ax bx c a 的根， 则必 1 2 1 2,
b cx x x x a a
，反之 亦然. 韦达定理的常用变形公式：
1212122121222212122212121212
1 1121 123 4
x xx x x x x x x x x x x x x x x x x x x x
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222121212224422121212
4 25 2
x x x x x x x x x x x x
二、不等式 （一）不等式基本性质
1、不等式的定义： 把两个解析式用不等号连接起来就构成不等式，使不等式成立的未 知数的值称为不等式的解.（不等号包括 、 、 、 、 五种）
2.不等式的基本性质
（1）不等式两边同时加（或减）一个数，不等号方向不变； （2）若不等式两边同时乘一个正数，不等号方向不变；若不等式两边同时乘一个负数，不 等号方向要改变；若 , 0a b c ，则 a b ， 0d ，则 ac bd .（3）传递性：若 a b＞ ， b c＞ ，则 a c＞ .（4）同向相加性：若 a b＞ ， c d＞ ，则 a c b d ＞ .（5）同向皆正相乘性： 00
a b ac bd c d ＞ ＞ ＞＞ ＞ .（6）皆正乘方性：若 0a b＞ ＞ ，则 0n n
a b＞ ＞ （ n 为正整数）. （7）皆正开方性：若 0a b＞ ＞ ，则 0n n
a b＞ ＞ （ n 为正整数）.
（二）不等式常见形式
1.一元一次不等式
（1）.定义：含有一个未知数且未知数的最高次数为一次的不等式叫做一元一次不等式. 般形式为 0, ( 0) ax b a ，（2）解法：解集可根据不等式基本性质直接求出.
即 ( 0) ax b a ＞
00 .
ba x aba x a
＞ 时， ＞ ，＜ 时， ＜ 24 小时热线电话：4008-219-288
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2.一元一次不等式组
先分别求出每一个不等式的解集，然后求这几个不等式的交集即可. 求交集常用的技巧有： 技巧 1：运用数轴直观的表示出解集. 技巧 2：利用口诀.“大大取大，小小取小，大于小的小于大的取中间，小小大大是无解”
3.一元二次不等式：
定义：含有一个未知数且未知数的最高次数为二次的不等式叫做一元二次不等式.
4.特殊不等式
根据其形式转化为相应的不等式组，并且要特别注意隐含定义域. （1） ( ) ( ) f x g x ＜ 2
( ) 0( ) 0( ) ( )
f xg x f x g x
＞＜
（2） ( ) ( ) f x g x ＞ 2
( ) 0( )( ) 0( ) ( )
f xg x f x g x
可舍去 ＞
或 ( ) 0( ) 0
g x f x
＜
（3） ( ) ( ) f x g x ＜
( ) 0( ) 0( )( ) ( )
f xg x f x g x
可舍去 ＜
三、函数
1.一元二次函数基本性质
（1） 标准形式： 一元二次方程 2 0, ( 0) ax bx c a 的根与二次函数 2
y ax bx c 图像的关系： 二次函数 2
y ax bx c 的图像是一条抛物线，它的对称轴为 2
bx a
，顶点坐标 为
2
4( , )2 4
b ac ba a
，当 0a 时开口向上，当 0a 时开口向下. 24 小时热线电话：4008-219-288
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（2）顶点式：将标准形式配方后，可将其转为：
22
4( )2 4
b ac by a x a a
，由此，可以 观察出该二次函数开口方向，对称轴以及最值. （3）交点式：如果一元二次方程 2 0, ( 0) ax bx c a 的根为 1 2x x和 ，则二次函数
2
y ax bx c 的图像与 x 轴交于点 1 2, 0 , 0 x x和
（4）用函数的观点看一元二次方程（不等式）
2.指数函数和对数函数
（1）指数和对数运算公式
指 数 对 数
定 义 b
a N log a N b （b 叫做以 a 为底 N 的对数）
关系式 log , ( 0, 1, 0) baa N N b a a N
2
4b ac 0 0 0
2
0
f x ax bx ca
0f x 的根
1,2
2
bx a
1,2 2
bx a
无实根
0f x 的解集 1 2x x x x 或 2
bx a
R
0f x 的解集 1 2x x x
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运算性质
（1） r s r s
a a a
（2） sr rs
a a
（3） r r r
ab a b
（4） 0 11, pp
a a a
( 0) a
（1） log log log ( )a a aM N MN
（2） log log log a a a
MM N N
（3） log log na aM n M
( 0, 0, 0, 1) M N a a
（4）（换底公式）
log log log
bab
NN a
（2）图像及性质
名称 指数函数 对数函数
表达式 x
y a ( 0, 1) a a log ay x ( 0, 1) a a
图像 x
y a x
y a
(0 1) a ( 1) a
log ay x ( 1) a
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性质 （1）定义域：R （2）值域： (0, )
（3）恒过点（0，1） （4）当 1a 时，在 R 上递增； 当 0 1a 时， 在 R 上递减.
（1）定义域： (0, )
（2）值域：R （3）恒过点（1，0） （4）当 1a 时，在 (0, ) 上递增 当 0 1a 时，在 (0, ) 递减
关系 x
y a 与 log ay x 互为反函数，两者图像关于 =y x 对称.
第二节 基础过关题型
一 、一次形式
【题型 1】一次方程(组) 【例 3.1-1】（2014-10） 12 3 6
x x x
，则 x （ ）. （A）-2 （B）-1 （C）0 （D）1 （E）2
【思路电拨】对于方程组，最常考的是二元一次方程组和三元一次方程组，尤其在解应 用题时，应用更广泛 .
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【例 3.1-2】（2004-1）方程组
24
xzzyayx
，得 zyx ,, 等差. （1） 1a (2) 0a
【题型 2】不等式基本性质 【例 3.2-1】（2008-1） 2 2
ab cb .（1）实数 , , a b c 满足 0a b c （2）实数 , , a b c 满足 a b c
【思路电拨】不等式性质常在条件充分性判断中进行考察，要想快速解题不仅要掌握 不等式基本性质，而且还要熟练运用充分性判断“特值法” .24 小时热线电话：4008-219-288
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【例 3.2-2】（2015） 已知 ,a b 为实数.则 2a 或 2b
（1） 4a b （2） 4ab
【例 3.2-3】（2016）设是 yx, 实数，则 .4,6 yx
（1） 2x y (2) 2 2y x
【题型 3】一次函数 【例 3.3-1】(2008-1)两直线 1 xy ， 7 ax y 与 x 轴所围成的面积是 427 .（1） 3a （2） 2a
【思路电拨】 一次函数的一般形式为 y kx b （ 0k ），特别地， 当 0b 时， y kx
称为正比例函数 .其性质主要有：
函数图形是一条直线 .因此画一次函数图像只需要找两个点即可 .2. 0k 表示图像从左往右呈上升趋势， 0k 表示图像从左往右呈下降趋势；
0b 表示图像与 y 交于正半轴， 0k 表示图像与 y 交于负半轴； 24 小时热线电话：4008-219-288
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【例3.3-2】（2013-10）设直线 y x b 分别在第一和第三象限与曲线 4
y x
相交于点A、 点B．则能确定 b 的值. （1）已知以AB为对角线的正方形的面积 （2）点A的横坐标小于纵坐标
二、二次形式
【题型 4】一元二次方程 【例 3.4-1】（2006-1）方程 2 2 0x ax 与 2 2 0x x a 有一个公共解. （1） 3a （2） 2a
【思路电拨】一元二次方程的解法有：配方法、公式法、因式分解、十字相乘法 .
【注意】①若二次项系数是字母形式，注意讨论是否为 0. ②若 0＜ ，则方程无实根 .24 小时热线电话：4008-219-288
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【题型 5】一元二次不等式 【例 3.5】（2002-10）不等式 xx 2
54 的解集是（ ）（A）全体实数 （B） )1,5( （C） )2,4( （D）空集 （E）以上均不对 【题型 6】二次函数（重点） 【例 3.6-1】 （2000-10） 一抛物线以 y 轴为对称轴， 且过点 )21,1( 及原点， 一直线 l 过点 )25,1(
和点 )23,0( ，则直线 l 被抛物线截得的线段长度为（ ）. （A） 24 （B） 23 （C） 34 （D） 33 （E） 32
【思路电拨】主要掌握两方面的技能：一方面根据抛物线的图像来分析系数的符号；另 一方面告知系数的符号，能够画出抛物线，并能判断所经过的象限 .
【思路电拨】
一元二次方程的解法有：配方法、公式法、因式分解、十字相乘法 .2. 求一元二次不等式的解集：首先将二次项系数化为正，然后解出对应方程的根，最 后根据“大于两边分，小于取中间”可快速写出解集 .
【注意】一般地，方程解问题可以转化为“恒成立问题”处理 .24 小时热线电话：4008-219-288
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【例 3.6-2】（2011-10）抛物线 2 2 2y x a x a 与 x 轴相切 （1） 0a （2） 2 6 0a a
【例 3.6-3】（2014-1）已知二次函数 2
f x ax bx c ，则能确定 , ,a b c 的值. （1）曲线 y f x 经过点（0,0）和点（1,1） （2）曲线 y f x 与直线 y a b 相切 【题型 7】根的判别式 【例 3.7-1】（2000-10）已知 , , a b c 是 ABC 的三条边长，并且 1a c ,
若 2
4( )( ) 0b x a x c x 有两个相同实根，则 ABC 为（ ）.
【思路电拨】已知方程 2 0ax bx c 的根的情况 .1. 有两个不相等的实数根，则 2
04 0
ab ac
＞
有两个相等的实数根，则 2
04 0
ab ac
=
没有实数根，则 2
04 0
ab ac
＜ 或 00
a bc
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（A）等边三角形 （B）等腰三角形 （C）直角三角形 （D）钝角三角形 （E）锐角三角形 【例 3.7-2】（2013-1）已知二次函数 2
f x ax bx c ，则方程 0f x 有两不同实 根. （1） 0a c （2） 0a b c
【例 3.7-3】已知关于 x 的方程 2 6 ( 2) 3 9 2 0x x a x a 有两个不等的实根，则 系数 a 的取值范围是（ ）. （A） 2a 或 0a （B） 0a （C） 0a 或 2a
（D） 2a （E）以上答案均不正确 【题型 8】韦达定理 【例 3.8-1】 （1997-1） 若 2 1 0x bx 的两根为 1x 和 2x ，且
12
1 1 5
x x
，则 b 的值为 （ ）（A）-10 （B）-5 （C）3 (D)5 (E)10
【思路电拨】一元二次方程韦达定理常见的两种类型 :
类型 1：若 2 0( 0) ax bx c a 的两根为 1 2,x x ，则有 1 21 2
bx x acx x a
类型 2：倒数根问题 .24 小时热线电话：4008-219-288
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【例 3.8-2】 （2009-1） 2
3 0( 0) x bx c c 的两个根为
、 .如果又以 +
、 为根的一元二次方程是 2
3 0x bx c .则 b 和 c 分别为( ). （A） 2 、 6 （B） 3 、 4 （C） 2 、 6- （D） 3 、 6- （E）以上结论均不正确 【例 3.8-3】（2003-1）一元二次方程 2 0x bx c 的两根之差的绝对值为 4. （1） 4, 0b c
（2） 2 4 16 b c
【题型 9】二次函数最值问题 【例 3.9】（2007-10）一元二次函数 (1 )y x x 的最大值为（ ）.
（A） 0.05 （B） 0.10 （C） 0.15 （D） 0.20 （E） 0.25
【思路电拨】若 2 ( 0) y ax bx c a 为一元二次函数，则
当 x R 时，若 0a＞ ，函数图像开口向上， y 有最小值，
2min
44
ac by a
当 x R 时，若 0a＜ ，函数图像开口向下， y 有最大值，
2max
44
ac by a
若 x 的定义域不是全体实数，则需要结合函数图像进行分析 .24 小时热线电话：4008-219-288
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【题型 10】二次函数恒成立问题 【例 3.10-1】（2011-10）不等式 2 6 2 0ax a x 对所有实数 x 都成立. （1） 0 3a （2） 1 5a
【例 3.10-2】（2012-10）若不等式
22
( ) ( ) 4
x a x ax
对 ),0( x 恒成立， 则常数 a 的取值范围是（ ）. （A） ( , 1) （B） (1, ) （C） ( 1,1)
（D） ( 1, ) （E） ( , 1) (1, )
【思路电拨】若 2 ( 0) y ax bx c a 为一元二次函数，则 1.对于一元二次不等式： 2 0ax bx c ＞ 恒成立 00
a
＞＜
对于一元二次不等式： 2 0ax bx c ＜ 恒成立 00
a
＜＜
【注意】若二次项系数含有参数，不要忘记讨论系数 a 为 0 的情况 .24 小时热线电话：4008-219-288
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【例 3.10-3】（2008-10）若 2 12 3 0y x yx
对一切实数 x 恒成立，则 y 的取值 范围是（ ）. (A) 1 3y (B) 2 4y (C) 1 4y
(D) 3 5y (E) 2 5y
【题型 11】根的分布 【例 3.11-1】（2005-10）方程 2 0x ax b 有一正一负两实根.
（1） 34b C （2） 57b C
【思路电拨】若 2 ( 0) y ax bx c a 为一元二次函数，则 :
正负根 （1）方程有两个不相等的正根 1 21 2
000
x xx x
＞＞＞
（2）方程有两个不相等的负根 1 21 2
000
x xx x
＞＜＞
（3）方程有一正根和一个负根 1 2 0 0x x ac ＜ ＜
【注意】若题目进一步要求比较正根和负根的绝对值的大小，则可根据韦达定理两根之 和进一步约束 .
区间根：区间根问题可通过开口方向、端点位置、对称轴位置、判别式进行约束范围 .24 小时热线电话：4008-219-288
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【例 3.11-2】（2008-1）关于 x 的方程 2
2 2 3 5 0ax x a 的一个根大于 1，另一个根小 于 1. （1） 3a （2） 0a＜
【例 3.11-3】 （1998-1）要使方程 2 2
3 ( 5) 2 0x m x m m 的两根分别满足 10 1x
和 21 2x ，实数 m 的取值范围应是（ ）. （A） 2 1m （B） 4 1m （C） 4 2m
（D） 1 65 12 m
（E） 3 1m ＜ ＜24 小时热线电话：4008-219-288
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三、根式、指数、对数方程（不等式）
【题型 12】根式方程与不等式 【例 3.12-1】（2007-1）方程 x p x 有两个不相等正根. （1） 0p （2） 14
p
【例 3.12-2】（2007-10） 2
1 1x x
（1） [ 1,0] x
（2） 1(0, ]2
x
【思路电拨】1.根式方程 （1）根式方程常用的方法有：平方法、配方法、换元法、两边取对数、图像法 （2） ( ) ( ) f x g x
2
( ) ( ) ( ) 0( ) 0
f x g xf xg x
【易错点】根式方程的隐含定义域. 2.根式不等式 （1） ( ) ( ) f x g x 2
( ) 0( ) 0( ) ( )
f xg x f x g x
（2） ( ) ( ) f x g x 2
( ) 0( )( ) 0( ) ( )
f xg x f x g x
可舍去 或 ( ) 0( ) 0
g x f x
（3） ( ) ( ) f x g x ( ) 0( ) 0( )( ) ( )
f xg x f x g x
可舍去 24 小时热线电话：4008-219-288
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【题型 13】指数的计算 【例 3.13-1】（2008-10） a b
（1） ,a b 为实数，且 2 2
a b
（2） ,a b 为实数，且 1 1( ) ( )2 2
ab
【例 3.13-2】（2000-1）解方程
12
4 2 1
xx
，则（ ）（A）方程有两个正实数根 （B）方程只有一个正实根 （C）方程只有一个负实根 （D）方程一正一负两个实根 （E）方程有两个负实根 【题型 14】对数的计算 【例 3.14-1】（2001-1）若 32,12,2 xx 成等比数列，则 x （ ）.
（A） 5log 2 （B） 6log 2 （C） 7log 2 （D） 8log 2
【思路电拨】指数的核心在于两点：
指数的核心在于两点：一个是指数基本公式的应用；另一个是转化形式，比如统一底 数或指数，然后进行比较大小 .2. 常用方法：化同底、换元法、两边取对数、图像法
【思路电拨】对数问题应注意：
对数是指数的逆运算，两者可以结合起来记忆 .2. 解对数方程（不等式）常用方法：化同底、换元法 .3. 要熟练掌握并运用常用的对数公式 .24 小时热线电话：4008-219-288
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（E） 9log 2
【例 3.14-2】（2009-1） log 1a x
（1） 12, 4 , 12
x a （2） 4,6 ,1 2x a
第四章 数列 24 小时热线电话：4008-219-288
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第一节 考试要点剖析
一. 数列基本知识
数列定义：数列是按一定次序排列的一列数.数列中单独的每一个数都叫做数列的项. 一般形式： 1 2 3, , ,na a a a，…， … 简记为 na .
1.通项公式：
如果一个数列 na 的第 n 项 na 与项数 n 满足某个函数关系式， 那么我们就可以把这个关系 式叫做数列的通项公式. 例如：数列：1,3,5,7,9,11，…可以表示为： 2 1( 1) na n n .特别地，常数列 , , ,..., ,... c c c c 是公差 0d 的等差数列 【注意】不是所有的数列都有通项公式，同一个数列也可能有多个通项公式.
2.递推公式
n
a 与其后项之间的关系称为递推公式.
3.数列的前 n 项和公式：
数列 na 的前 n 项和记作 nS ，对于数列 na 显然有 1 2 3n nS a a a a …
4.数列的分类
（1）按项数分类 有穷数列（项数有限）、无穷数列（项数无穷） （2）按单调性分类 数列按照单调性可分为递增数列、递减数列、常数数列、摆动数列.
二、等差数列
【大纲考点】1.数列基本性质；2.等差数列；3.等比数列. 【命题剖析】等差和等比数列的考察主要是在数列通项与和项的基本性质上面，尤其 是性质的巧用以及计算方面.数列一般性质突出考察数列的函数性质和数列的递推. 24 小时热线电话：4008-219-288
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定义：若数列 na 从第 2 项开始，每一项与它的前一项的差等于同一个定值，则称此数列 为等差数列，称此常数为等差数列的公差，公差通常用字母 d 来表示，即
1
( )n na a d n N ，
1.等差数列的通项公式： 1 1na a n d ( )n N
2.前 n 项和公式：
公式 1： 1
2
nn
n a aS 公式 2： 1 12
n
dS na n n 公式 3： 21
2 2
n
d dS n a n
【评注】根据公式 3 可知：① 若 0d 时， nS 和 n 构成二次函数关系，且没有常数项， 换言之， nSn 是一次函数形式.②若等差数列前 n 项和表达式有常数项，那么该表达式是从 第二项开始的等差数列，其通项公式是分段函数形式.
3.下标和性质
若 m n p q ，则 m n p qa a a a .
4． 等差中项： 若 , , a b c 成等差数列，则 b 叫做 a 与 c 的等差中项，且 2b a c
5．等长片段和性质：
若 nS 是等差数列 na 的前 n 项和，则 2 3 2, , ,... n n n n nS S S S S 仍成为等差数列，新公差为
2
n d .
6.前 n 项和最值：
（1）通项公式 na 判断法：最值取在变号前. （2）前 n 项和 nS 判断法：① 21
2 2
n
d dS n a n
的最大最小项的讨论.
三、等比数列 24 小时热线电话：4008-219-288
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定义：若数列 na 中，从第 2 项起，后一项与它的前一项的比是一个定值，则称此数列为 等比数列，公比通常用字母 q 来表示，即 1 ( )nn
a q n Na
特别地，非零常数列 ,.., c c ，是公比 1q 的等比数列
1.通项公式: 11
nn
a a q n N
1 0, 0a q
2.前 n 项和公式：
运用求和公式前，首先需要分清 q 的取值. ①当 1q 时， 1nS na ； ②当 1q 时， 1 11
nn
a qS q
；【易错点】当无法确定 q 的值时，应分分 1q 或 1q 两种情况进行讨论.
3.下标和性质： 若 m n p q ，则 m n p qa a a a .
4.等比中项： 若 , , a b c 三者成等比数列，那么 b 叫做 ,a c 的等比中项，且有 2
b ac
5.等比数列等长片段和： 若 nS 是等比数列 na 的前 n 项和， 则 2 3 2, , ,... n n n n nS S S S S
仍成为等比数列.
6.无穷等比递缩数列： 若 1q ， na 的公比为 q ，则该数列的各项和 1
1
n
aS q
【注意】1. 等比数列奇数项和偶数项同号. 2．等比中项可以是正的或者是负的. 3． nS 的单调性的讨论，如果 10, 0q a ， nS 单调增加，如果 11, 0q a ， nS 单调减 少如果 0q 则呈现振荡.讨论的方法可以通过数列通项的正负性. 4. nnS aq b 和形式的通项为等比数列当且仅当 0a b .
第二节 基础过关题型
扫码查看答案
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一、等差数列
【题型 1】等差数列通项公式： 【例 4.1-1】（2015）设 na 是等差数列.则能确定数列 na .（1） 1 6 0a a （2） 1 6 1a a
【例 4.1-2】（2008-10）下列通项公式表示的数列为等差数列的是（ ）. (A) 1
n
na n
（B） 2 1na n （C） 5 ( 1) n
n
(D) 3 1na n (E) 3
n
a n n
【例 4.1-3】（2011-10） 若等差数列 na 满足 7 35 12 0a a ，则
【思路电拨】 1.等差数列通项公式： （1）基本公式： 1 ( 1) na a n d . （2）变异公式： ( )n ma a n m d .2.等差数列求和公式 前 n 项和公式： 公式 1： 1
2
nn
n a aS . 公式 2： 1 12
n
dS na n n .公式 3： 21( )2 2
n
d dS n a n （常用于最值判断）. 24 小时热线电话：4008-219-288
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15 1
kk
a
（ ）. （A）15 （B）24 （C）30 （D）45 （E）60 【题型 2】等差数列基本性质 【例 4.2-1】（2018）设 na 为等差数列，则能确定 1 2 9a a a 的值. (1)已知 1a 的值 (2)已知 5a 的值 【例 4.2-2】 （2006-1） 若 6， a ，c 成等差数列， 且 22 ,36 ca ， 也成等差数列， 则 c 为（ ）.（A）-6 （B）2 （C）3 或 2 （D） 6 或 2 （E）以上均不对
【思路电拨】 1. 下标和性质：在等差数列 na 中，若 m n p q ，则 m n p qa a a a .2. 等比中项：若 , , a b c 成等差数列，则 b 叫做 a 与 c 的等差中项，且 2b a c .24 小时热线电话：4008-219-288
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【题型 3】连续等长片段和 【例 4.3】（1998-10）若等差数列中前 5 项和 5 15 S ，前前 15 项和 15 120 S ，则前 10 项和 10 S =（ ）. （A）40 （B）45 （C）50 （D）55 （E）60 【题型 4】等差数列最值问题 【例 4.4-1】 （2001-10)等差数列 na 中, 0,0 65 aa ,且 56 aa , nS 是前 n 项之和, 则（ ）. （A） 321 ,, SSS 均小于 0,而 ,...... , 54 SS 均大于 0
（B） 521 ,......., , SSS 均小于 0,而 ,......, , 76 SS 均大于 0
【思路点拨】 若 nS 是等差数列 na 的前 n 项和，则 2 3 2, , ,... n n n n nS S S S S 仍成为等差数列，新公 差为 2
n d .
【注意】① 2 3, , ,... n n nS S S 不是连续等长片段 . ②有时可以令 1n 快速解题.
【思路电拨】等差数列 nS 的最值问题常用以下两种方法： 1.通项公式 na 判断法：最值取在变号前，令 0na ，解出 n 取变号之前即可 .特别地， 若 n 解出恰好为整数，则 nS 和 1nS 均为最值 .
2.前 n 项和 nS 判断法：将 nS 看作二次函数，利用对称轴求解最值 .24 小时热线电话：4008-219-288
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（C） 921 ........, , SSS 均小于 0,而 ,..... , 11 10 SS 均大于 0 （D） 10 21 ,......, , SSS 均小于 0,而 ,..... , 12 11 SS 均大于 0 （E）以上答案均不对 【例 4.4-2】（2015-1）已知数列 na 是公差大于零的等差数列， nS 是 na 的前 n 项和. 则 10 ,nS S 1, 2, n .（1） 10 0a （2） 11 10 0a a
【题型 5】等差数列应用题 【例 4.5】 （2011-1） 一所四年制大学每年的毕业生七月份离校， 新生九月份入学.该校 2001 年招生 2000 名，之后每年比上一年多招 200 名，则该校 2007 年九月底的在校学生有 （A）14000 名 （B）11600 名 （C）9000 名 （D）6200 名 （E）3200 名 24 小时热线电话：4008-219-288
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【题型 6】等差数列比值问题 【例 4.6-1】一个等差数列的前 12 项和为 354， 前 12 项中偶数项之和与奇数项之和的比是 32:57，则公差 d =（ ）. （A）3 （B）4 （C）5 （D）6 （E）7 【例 4.6-2】（2009-1） na 的前 n 项和 nS 与 nb 的前 n 项和 nT 满足 19 19 : 3 : 2 S T .（1） na 和 nb 是等差数列 （2） 10 10 : 3: 2 a b
【思路电拨】常见以下形式： 类型一：同一个等差数列奇数项与偶数项比值 （1）若等差数列一共有 2n 项，则
1
= nn
S aS a
奇偶
（2）若等差数列一共有 2 1n 项，则 1=
S nS n
奇偶
类型二：两个数列相同奇数项之比 若等差数列 na 和 nb 的前 2 1k 项的和分别为 2 1kS 和 2 1kT ，则 2 12 1
kkkk
S aT b
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二、等比数列
【题型 7】通项公式、求和公式 【例 4.7-1】（2007-10） 6 126 S .（1）数列 na 的通项公式是 10(3 4),( )na n n N
（2）数列 na 的通项公式是 2 ( )nna n N
【例 4.7-2】（2012-10）设 na 是非负等比数列.若
8351
1 11, ,4 n n
a a a
（ ）. （A）255 （B） 255 4 （C） 255 8 （D） 255 16 （E） 255 32
【思路电拨 1. 通项公式：基本公式: 11
nn
a a q n N
， 1 0, 0a q .变异公式: ,n mn ma a q m n N
1 0, 0a q .2.求和公式： 运用求和公式前，首先需要分清 q 的取值 .
①当 1q 时， 1nS na ； ②当 1q 时， 1 11
nn
a qS q
；【易错点】当无法确定 q 的值时，应分 1q 或 1q 两种情况进行讨论 .24 小时热线电话：4008-219-288
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【题型 8】等比数列下标和性质、中项性质 【例 4.8】 （2011-10）若等比数列 na 满足： 2 4 3 5 2 82 25 a a a a a a ，且 1 0a ＞ ，则 3 5a a
（ ）. （A）8 （B）5 （C）3 （D）2 （E）1 【例 4.8】（2002-1）设有两个数列
62 1 3, 2 1 2 1 , 2 12
aa ， 和 ， ，则使前 者成为等差数列、后者成为等比数列的实数 a 的值（ ）. （A）0 个 （B）1 个 （C）2 个 （D）3 个 （E）4 个
【思路电拨】 1.在等比数列 na 中，若 m n p q ，则 m n p qa a a a .
等比中项：若 , , a b c 成等比数列，则 b 叫做 ,a c 的等比中项，且有 2
b ac .24 小时热线电话：4008-219-288
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1
A
1
B 1C
1
D
2
A
2
B
2
C
3
A 3B
3
C3D 2D
【题型 9】连续等长片段和 【例 4.9】在等比数列 na 中， 4 1S ， 8 3S ，则 17 18 19 20 a a a a 的值是（ ）. （A）8 （B）10 （C）12 （D）16 （E）以上均不对 【题型 10】无穷等比递缩数列 【例 4.10】如图，四边形 1 1 1 1A B C D 是平行四边形， 2 2 2 2, , ,A B C D 分别是 1 1 1 1A B C D 四边的中点，
3333
, , ,A B C D 分别是四边形 2 2 2 2, , ,A B C D 四边的中点，依次下去，得到四边形序列
( 1,2,3...) m m m mA B C D m ，设 m m m mA B C D 的面积为 mS 且 1 12, S 则 1 2 3 .... S S S （ ）. （A）16 （B）20 （C）24 （D）28 （E）30
【思路电拨】 等比数列 nS 中， 2 3 2, ,n n n n nS S S S S 仍成等比数列，新公比为 n
q .【注意】 ① 2 3, , ,... n n nS S S 不是连续等长片段 ②有时可以令 1n 快速解题.
【思路电拨】若 1q ， na 的公比为 q ， n ，则该数列的各项和 1
1
n
aS q
【评注】无穷表示： n ，递缩表示： 1q ＜24 小时热线电话：4008-219-288
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【题型 11】等比数列文字应用题 【例 4.11】（2010-10）某地震灾区现居民住房的总面积为 a 平方米，当地政府计划每年 以10% 的住房增长率建设新房，并决定每年拆除固定数量的危旧房.如果 10 年后该地的住 房总面积正好比现有住房面积增加一倍， 那么， 每年应该拆除危旧房的面积是 （ ） 平方米. （注： 9
1.1 2.4 ， 10
1.1 2.6 ， 11
1.1 2.9 精确到小数点后一位） （A） 180 a （B） 140 a （C） 380 a （D） 120 a （E）以上结论都不正确
三、数列综合题
【题型 12】数列综合题 【例 4.12-1】（2001-1）在等差数列 na 中， 3 2a ， 11 6a ；数列 nb 是等比数列.若
23
b a ， 32
1
b a
，则满足
26
1
n
b a
＞ 的最大的 n 为（ ）. （A）3 (B)4 (C)5 (D)6 （E）7
【思路电拨】本类问题将等差和等比数列结合出题，属于考试热点内容，常应该注 意一下情况： 1.灵活运用等于等比数列的基本公式. 2.既是等差又是等比数列的数列，是非零常数列. 3.注意等比数列的奇数项同号且偶数项同号. 4.等差等比数列一般不单独命题，常常作为条件充分性判断的两个条件. 5.等差、等比数列一般结合、韦达定理、根的判别式、二次函数、指数与对数考察. 24 小时热线电话：4008-219-288
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【例4.12-3】（2013-10）设 ,a b 为常数.则关于x的二次 2 2 2
( 1) 2( ) 1 0a x a b x b 具有重实根. （1） ,1, a b 成等差数列 （2） ,1, a b 成等比数列
四、其它数列问题
【题型 13】递推公式求通项 【例 4.13-1】（2010-10） 11 ( 1, 2, ). 2
nn
x n
（1） 1
12
x ， 1
1 (1 )2
nn
x x ( 1, 2, )n
（2） 1
12
x ， 1
1 (1 )2
nn
x x ( 1, 2, )n
【例 4.13-2】（2013-10）设数列 na 满足: 1 11, ( 1). 3
nn
na a a n 则 100 a =（ ）. （A）1650 （B）1651 （C） 5050 3 （D）3300 （E）3301 24 小时热线电话：4008-219-288
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变式思维训练
1.（2019）设数列 na 满足 1 0a ， 1 2 1n na a ，则 100 a （ ）. （A） 99
2 1 （B） 99
2 （C） 99
2 1 （D） 100
2 1 （E） 100
2 1
【题型 14】已知 S n 求 a n 问题 【例 4.14】 （2003-10）若数列 na 的前 n 项和 2
4 2nS n n ，则它的通项公式是（ ）.（A） 8 3n （B） 4 1n （C） 8 2n
（D） 8 5n （E）以上答案均不正确
【思路电拨】求通项公式的一般方法
若已知数列 na 和前 n 项和 nS ，则 11
, 1, 2n n n
S na S S n
必要时可以采用特值法令 1, 2,3 n 快速选择答案 .. 24 小时热线电话：4008-219-288
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第五章 应用题 第一节 考试要点剖析
一、比、百分比、比例问题
0 0 00 0 0100 100 1 100
变化量 现值-原值 现值 变化率 变前量 原值 原值
【注意】变化率包括增长率和下降率两个，所以上式用绝对值表示. 2.回复原值：原值现降 0 0p ，再增
0 00 0
1
pp 才能回复原值；或者先增 0 0p 再降
0 00 0
1
pp 才能 恢复原值. 3.甲比乙大 0 0p 0 0p
甲-乙 乙
0 01 p 甲 乙 ；甲是乙的 0 0p 0 0p 甲 乙
【注意】甲比乙大 0 0p ≠乙比甲小 0 0p （因为基准量不同），甲比乙大 0 0p 乙比甲小
0 00 0
1
pp .4.比例性质：如果 a cb d
，则 ad bc .5.等比定理： 0
a c e a c e b d fb d f b d f
对应占的比例 部分量 总量
二、增长率问题
1.增长率 增长率 a0 00 01p a p 原值 现值 ； 0 00 01a
p a p 原值
下降率 现值
【注意】一件商品先提价 0 0p 再降价 0 0p ，或者先降价 0 0p 再提价 24 小时热线电话：4008-219-288
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0 0
p ，回不到原价，应 该比原价小，因为： 0 0 0 00 0 0 01 1 1 1a p p a p p a .2.平均增长率模型
(1 )n
a x b （其中 a 表示变化前的量， b 变化后的量， n 表示增长次数.）
三、利润问题
进价 售价 利润
0000
100% 100 1 100
利润 售价-进价 售价 进价 进价 价利 进润率
利润 进价 利润率 进价 售价 1
四、浓度问题
1． = +溶液 溶质 溶剂 ， = 100%= 100% +
溶质 溶质 浓度 溶液 溶质 溶剂 ；2．重要等量关系． (1)浓度不变准则：将溶液分成若干份，每份的溶液相等，都等于原来溶液的溶度；将 溶度倒掉一部分后，剩余的溶液的浓度与原溶液的浓度相等． (2)物质守恒准则：物质（无论是溶质、溶剂，还是溶液）不会增多也不会减少，前后 都是守恒的． 3．重要命题思路． (1)“清水稀释”问题：特点是加溶剂，溶质不变，以溶质为基准进行求解． (2)“浓缩”问题：也称“蒸发”问题，特点是减少溶剂，溶质不变，以溶质为基准进 行求解 (3)“加浓”问题：特点是增加溶质，溶剂不变，以溶剂为基准进行求解． (4)“混合”问题：用两种或多种溶液混合在一起，采用溶质或溶剂质量守恒分析，也 可利用杠杆原理分析．
(5)“置换”问题：一般是用溶剂等量置换溶液，可以记住结论， 24 小时热线电话：4008-219-288
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原来溶液 V 升，倒出 a 升，再补等量的溶剂（水），则浓度为原来的 VaV ．（6）浓度问题常用“杠杆原理--交叉法” 当一个整体按照某个标准分为两部分时（或有两部分混合成一个整体），可以根据杠 杆原理得到一种巧妙的求解方法;若分为甲乙两部分，则甲的数量：乙的数量=乙到支点的 就距离：甲到支点的距离.
五、平均值问题
（1）算术平均值： 1 2 3 nx x x xx n
…+
（2）加权平均数：若 n 个数 1 2, , nx x x…， 的全分别是 1 2, , nf f f…， ，则
112233nn
x f x f x f x f x n
…+
（3）常用杠杆原理--交叉法 （4）如涉及到最值问题，常用极端假设法
六、不定方程
当方程或方程组中未知数较多，而无法通过解方程分角度来确定数值，这种方程称为 不定方程.不定方程必须结合所给的一些性质，如整除、奇数偶数、质数合数、范围大小等 特征才能确定答案.
七、工程问题
1.工作量 s 、工作效率 v 、工作时间 t 三者的关系：
s vt 工作量 工作效率 工作时间 ； st v
工作量 工作时间 工作效率
；
sv t
工作量 工作效率 工作时间
2.重要说明：工作量：对于一个题，工作量往往是一定的，可以将总的工作量看做“1”； 工作效率：合作时，总的效率等于各效率的代数和. 3.重要结论 若甲单独完成需要 m 天，以单独完成需要 n 天；则： （1）甲的效率为 m
1 ，乙的效率为 n
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（2）甲乙合作的效率为 nm
11 ； （3）甲乙合作完成需要的时间为 nmmn nm
111 .
八、路程问题（与工程问题相似）
1.路程 s 、速度 v 、时间 t 之间的关系：
vt s ， vst ， tsv
2.对于直线型的路程问题： （1）相遇: 1 2 1 2 1 2= ( )S S S v t v t v v t 相遇
（2）追及: tvvtvtvS 212121 SS 追及
3.对于圆圈型的路程问题： （从同一起点同时出发，周长为 s ， 相遇一
次的时间为 t ）
（1）反向运动（类似直线型相遇）：
121212
= ( )S S S v t v t v v t
即：每相遇一次，甲与乙路程之和为一圈，若相遇 n 次， 则有
① SnSS 乙甲 ② 1S
乙乙乙乙甲乙甲
SnSSSnSSVV
（2）同向运动（类似于直线追及）：
121212
( )S S S v t v t v v t
即：甲乙每相遇一次，甲比乙多跑一圈，若相遇 n 次，则有 ① SnSS 乙甲 ②
乙乙乙乙甲乙甲
SnSSnSSSVV S1
【解题技巧】在做圆圈型追及相遇问题时，在求第 k 次相遇情况时，可以将 1k 次相遇看 成点进行分析考虑.
4.顺水、逆水问题： 水船顺水 vvv ； 水船逆水 vvv 24 小时热线电话：4008-219-288
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5.相对速度（两个物体运动时，可将一个作为参照物，看成相对静止的） 同向运动： 21 vvv 同向 ；相向运动： 21 vvv 相向
九、集合问题
1．两个集合 公式： A B A B A B A B 全集
2.三个集合
A B C A B C A B B C A C A B C A B C 公式： 全集-
十、分段计费
分段计费是指不同的范围对应着不同的计费方式，在实际中应用很广泛，比如电费、 水费、邮费、个税、话费、出租车费、销售提成等等．解题思路的关键点有两个，一个是 先计算每个分界点的值，确定所给的数值落在哪个范围；另外，对应选取正确的计费表达 式，按照所给的标准进行求解.
十一、最值问题
最值问题是文字应用题的延伸部分， 是将定制问题转化为动态问题的过程.解数学问题 应用题的重点在过好三关：（1）事理关──阅读理解，知道命题所表达的内容；（2）文 理关──将“问题情境”中的文字语言转化为符号语言，用数学关系式表述事件；（3）数 理关──由题意建立相关的数学模型，将实际问题数学化，并解答这一数学模型，得出符 合实际意义的解答.
十二、线性规划问题
线性规划问题往往要根据题目需求建立可行性区域，然后算得最值.
十三、年龄问题
年龄问题的特点有两个，一个是年龄的差值恒定；另一个是年龄同步增长. 【注意】年龄要选好参照年份,如果年龄计算得到矛盾，看看几年前是否还未出生,因 为出生后才对年龄有影响.
十四、植树问题
A
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这类应用题是以“植树”为背景，凡是研究总路程、株距、段数、棵树四种数量关系 的应用题叫做植树问题.
第二节 基础过关题型
【题型 1】简单方程（组）问题 【例 5.1】（2010-1）售出一件甲商品比售出一件乙商品利润要高. （1）售出 5 件甲商品，4 件乙商品共获利 50 元 （2）售出 4 件甲商品，5 件乙商品共获利 47 元 【题型 2】简单比例问题 【例 5.2-1】（1997-10）某地连续举办三场国际商业足球比赛, 第二场观众比第一场少了 80%, 第三场观众比第二场减少了 50%,若第三场观众仅有 2500 人, 则第一场观众有 （ ）.（A）15000 人 （B）20000 人 （C）22500 人 （D）25000 人 （E）27500 人
【思路电拨】 简单比例问题常见以下考点：
1.比例化简问题：遇到分数比例关系，可扩大相应的倍数将其化为整数比例 2.遇到连比问题：可将单独的两个联合关系合成为一个比例关系 3.统一比例法：针对多比例问题，若比例和比例之间有不变量，则统一其份数，然后根 据比例变化，建立比例份数和数量之间的关系. 4.还原问题：遇到“余下的 /n m 又 k 个”,最好从后往前倒着计算,否则直接计算运算 量较大. 5.熟练使用比例定理.
扫码查看答案
【思路电拨】根据题目含义，列出方程（方程组）进行求解，该题型一般较为简单. 24 小时热线电话：4008-219-288
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【例 5.2-2】（2016）某家庭在一年总支出中，子女教育支出与生活资料支出的比为 3：8， 文化娱乐支出与子女教育支出为 1：2.已知文化娱乐支出占家庭总支出的 10.5%，则生活资 料支出占家庭总支出的（ ）. （A）40% （B）42% （C）48% （D）56% （E）64% 【例 5.2-3】（2009-1）某国参加北京奥运会的男女运动员比例原为 19:12.由于先增加若 干名女运动员,使男女运动员比例变为 20:13,后又增加了若干名男运动员,于是男女运动 员比例最终变为 30:19.如果后增加的男运动员比先增加的女运动员多 3 人,则最后运动 员的总人数为（ ）. （A）686 （B）637 （C）700 （D）661 （E）600 【例 5.2-4】（2013-10）甲、乙、丙三个容器中装有盐水.现将甲容器中盐水的 13 倒入乙 容器， 摇匀后将乙容器中盐水的 14 倒入丙容器， 摇匀后再将丙容器中盐水的 110 倒回甲容器， 此时甲、乙、丙三个容器中盐水的含盐量都是 9 千克.则甲容器中原来的盐水含盐量是（ ） 千克. （A）13 （B）12.5 （C）12 （D）10 （E）9.5 24 小时热线电话：4008-219-288
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【题型 3】增长率问题 【例 5.3-1】 （1997-10）银行的一年期定期存款利率为 10%, 某人于 1991 年 1 月 1 日存入 1000 元, 1994 年 1 月 1 日取出, 若按复利计算, 他取出时所得的本金和利息共计是 （ ）.（A）10300 元 （B）10303 元 （C）13000 元 （D）13310 元 （E）14641 元 【例 5.3-2】某商品在促销期间销售价比上月下跌 10%，销量比上月增加 10%.问：该商品 在促销期间的营业额与上月相比变化幅度是（ ）. (A)增加 10% (B)增加 1% (C)下跌 10% (D)下跌 1% (E)不变 【例 5.3-3】能确定某企业产值的月平均增长率. （1）已知一月份的产值 （2）已知全年的总产值
【思路电拨】 1. = 100%
变化量 变化率 变前量 （关键在于把握比例的基数） 2.平均增长率模型： (1 )n
a x b 24 小时热线电话：4008-219-288
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【题型 4】利润问题 【例 5.4-1】（2001-1）一商店把某商品按标价的九折出售，仍可获利 20%，若该商品的进 价为 21 元，则该商品每件的标价为（ ）. （A）26 元 （B）28 元 （C）30 元 （D）32 元 （E）34 元 【例 5.4-2】（2006-1）某电子产品一月份按原定价的 80%出售，能获利 20%；二月份由于 进价降低，二月份由于进价降低，按照原定价的 75%出售，能获利 25%，那么二月份进价是 一月份进价的百分之（ ）. （A）92 （B）90 （C）85 （D）80 （E）75 【例 5.4-3】（2009-10）甲、乙两商店某种商品的进货价格都是 200 元,甲店以高于进货
价格 20%的价格出售,乙店以高于进货价格 15%的价格出售,结果乙店的
【思路电拨】利润问题要选对基准量，注意折扣的变化与利润的关系，解题关键是要分 清成本价，原销售价、“优惠价”和利润这几个概念，有些题目会给出利润所占的百分 比.此外，还要掌握常用等量关系： 1. = 100%
变化量 变化率 变前量 （关键在于把握比例的基数）. 2. 进价 售价 利润 ； 0 00 0100% 100 1 100
利润 售价-进价 售价 进价 进价 价利 进润率 .3. 利润 进价 利润率 进价 售价 1 .24 小时热线电话：4008-219-288
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售出件数是甲店的 2 倍.扣除营业税后乙店的利润比甲店多 5400 元.若设营业税率是营业额 的 5%,那么甲、乙两店售出该商品各为（ ）件. （A）450,900 （B）500,1000 （C）550,1100 （D）600,1200 （E）650,1300 【题型 5】浓度问题 【例 5.5-1】当含盐 30%的 60 千克盐水蒸发为含盐 40%的盐水时，盐水重量为（ ）千克. （A）45 （B）50 （C）55 （D）60 （E）以上结果均不对 【例 5.5-2】（2011-10）某种新鲜水果的含水量为 98%，一天后的汗水量降为 97.5%，某 商店以每斤 1 元的价格购进了 1000 斤新鲜水果，预计当天能售出 60%，两天内售完.要 使利润维持在 20%，则每斤水果的平均售价应定为（ ）. （A）1.20 （B）1.25 （C）1.30 （D）1.35 （E）1.40 【例 5.5-3】（2008-1）若用浓度为 30% 和 20% 的甲、乙两种食盐溶液配成浓度为 24%的 食盐 500 克，则甲、乙溶液应各取（ ）. （A）180 克和 320 克 （B）185 克和 315 克 （C）190 克和 310 克 （D）195 克和 305 克 （E）200 克和 300 克
【思路电拨】1． = +溶液 溶质 溶剂 ， = 100%= 100% +
溶质 溶质 浓度 溶液 溶质 溶剂 .2.重要解题思路：准确把握题目的等量关系. 24 小时热线电话：4008-219-288
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【例 5.5-4】（2012-10）一满桶纯酒精倒出 10 升后，加满水搅匀，再倒出 4 升后，再加 满水，此时，桶中的纯酒精与水的体积之比是 2:3，则该桶的容积是（ ）升. （A）15 （B）18 （C）20 （D）22 （E）25 【题型 6】平均值问题 【例 5.6-1】（2003-1）车间共有 40 人，某技术操作考核的平均成绩为 80 分，其中男工 平均成绩为 83 分，女工平均成绩为 78 分.该车间有女工（ ）. （A）16 人 （B）18 人 （C）20 人 （D）24 人 （E）28 人 【例 5.6-2】（2011-1）在一次英语考试中，某班的及格率为 80%. （1）男生及格率为 70%，女生及格率为 90% （2）男生的平均分与女生的平均分相等
【思路电拨】常用以下策略 1. 若已知总体分成二类的平均数以及总体的平均数， 可以利用杠杆交叉法求求二者比 例. 2. 若已知二者平均数以及二者之间的比例可以求总体的平均数.（不需要知道二者具 体数量） 3. 若遇到最值问题，常用极端假设法. 4. 若以上方法均不适用，可采取万能方法---方程思想. 24 小时热线电话：4008-219-288
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【例 5.6-3】（2015）在某次考试中，甲、乙、丙三个班的平均成绩分别为 80,81 和 81.5， 三个班的学生得分之和为 6952，三个班共有学生（ ）. （A）85 名 （B）86 名 （C）87 名 （D）88 名 （ E） 90 名 【例 5.6-4】已知三种水果的平均价格为 10 元/千克，则每种水果的价格均不超过 18 元. （1）三种水果中价格最低的为 6 元/千克 （2）购买质量分别是 1 千克、1 千克、2 千克的三中水果共用 46 元 【题型 7】整数不定方程（不等式） 【例 5.7-1】（2011-1） 在年底的献爱心活动中，某单位共有 100 人参加捐款，经统计， 捐款总额是 19000 元，个人捐款数额有 100 元、500 元和 2000 元三种，该单位捐款 500 元 的人数为（ ）. （A）13 （B）18 （C）25 （D）30 （E）28
【思路电拨】 1. 不定方程： 不定方程， 是指未知数的个数多于方程个数， 且未知数受到某些限制 （如 要求整数或正整数等等）的方程或方程组. 2. 不定方程常用方法：①穷举法②利用整除以及奇偶分析法缩小范围③选项代入法 3. 不定不等式问题：列出相应不等式即可锁定整数解. 24 小时热线电话：4008-219-288
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【题型 8】路程问题 【例 5.8-1】 （2015）某人驾车从 A 地赶往 B 地，前一半路程比计划多用时 45 分钟，平 均速度只有计划的 80%.若后一半路程的平均速度为 120 千米/小时，此人还能按原定时间 到达 B 地.A，B 两地的距离为（ ）. （A）450 千米 （B）480 千米 （C）520 千米 （D）540 千米 （D）600 千米 【例 5.8-2】（2009-10）甲、乙两人在环形跑道上跑步,他们同时从起点出发,当方向相反 时每隔 48 秒相遇一次,当方向相同时每隔 10 分钟相遇一次.若甲每分钟比乙快 40 米,则甲、
乙两人的跑步速度分别是（ ）米/分.
【思路电拨】 1.基本公式：路程=速度 ×时间，即 s vt
2.直线型问题： 1 21 2
v t v t Sv t v t S
相遇 追及
相遇： 追及： （追及问题：必须速度快者追慢者） 3.环形问题： 1 21 2
v t v t ns v t v t ns
相遇： 追及： （同时同地出发） 4. 流水行船问题：需要考虑水流速度
==
v v vv v v
顺静水逆静水
5.火车过桥问题：需要考虑车身的长度 24 小时热线电话：4008-219-288
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（A）470,430 （B）380,340 （C）370,330 （D）280,240 （E）270,230 【例 5.8-3】（2009-1）一艘轮船往返航行于甲、乙两码头之间，船在静水中的速度不变， 则当这条河的水流速度增加 50%时，往返一次所需的时间比原来将（ ）. （A）增加 （B）减少半个小时 （C）不变 （D）减少 1 个小时 （E）无法判断 【例 5.8-4】（2010-10）在一条与铁路平行的公路上有一行人与一骑车人同向行进,行人 速度 3.6 千米/小时,骑车人速度为 10.8 千米/小时,如果一列火车从他们的后面同向匀速驶 来,它通过行人的时间是 22 秒,通过骑车人的时间是 26 秒,则这列火车的车身长为 （ ）米. （A）186 （B）268 （C）168 （D）286 （E）188 【题型 9】工程问题 【例 5.9-1】 某单位春季植树 100 颗，前 2 天安排乙组植树，其余任务由甲、乙两组用 3 天完成，已知甲组每天比乙组多植树 4 棵，则甲组每天植树（ ）. （A） 11 棵 （B） 12 棵 （C） 13 棵 （D）15 棵 （E） 17 棵
【思路电拨】遇到此类问题，通常将整个工程量（放水量）看成单位“1”，然后根据 题干条件按比例求解，通常假设总量（工程量，放水量）=1 进行分析. 【重要公式】1.各效率代数和=总效率.
工作量 部分量 2.工作效率 ，总量 工作时间 其对应的比例
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【例 5.9-2】一件工程要在规定时间内完成,若甲单独做则比规定的时间推迟 4 天,若乙单 独做则比规定的时间提前 2 天完成.若甲、 乙合作了 3 天,剩下的部分由甲单独做,恰好在规 定时间内完成,则规定时间为（ ）天. （A）19 （B）20 （C）21 （D）22 （E）24 【题型 10】分段计费问题 【例 5.10】某单位采取分段收费的方式收取网络流量（单位：GB）费用：每月流量 20（含） 以内免费，流量 20 到 30（含）的每 GB 收费 1 元，流量 30 到 40（含）的每 GB 收费 3 元， 流量 40 以上的每 GB 收费 5 元，小王这个月用了 45GB 的流量，则他应该交费（ ）. （A）45 元 （B）65 元 （C）75 元 （D）85 元 （E）135
【思路点拨】分段计费问题一般涉及两种类型：（以阶梯电费为例） 类型一：已知用电量，分段汇总求和即可求出应交电费. 类型二：已知所交电费，反向推断用电量.关键在于定位电费位于哪一档. 24 小时热线电话：4008-219-288
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变式思维训练
1.为了调节个人收入，减少中低收入者的赋税负担，国家调整了个人工资薪金所得税的征 收方案.已知原方案的起征点为 2000 元/月，税费分九级征收，前四级税率见下表：
级数 全月应纳税所得额 q （元） 税率（%）
1 0 500 q 5
2 500 2000 q 10
3 2000 5000 q 15
4 5000 20000 q 20
新方案的起征点为 3500 元/月，税费分七级征收，前三级税率见下表： 若某人在新方案下每月缴纳的个人工资薪金所得税是 345 元， 则此人每月缴纳的个人工 资薪金所得税比原方案减少了（ ）元. （A）825 （B）480 （C）345 （D）280 （E）135
级数 全月应纳税所得额 q （元） 税率（%）
1 0 1500 q 3
2 1500 4500 q 10
3 4500 9000 q 20 24 小时热线电话：4008-219-288
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【题型 11】集合问题 【例 5.11-1】（2008-1）申请驾照必须参加理论考试和路考，且两种考试均通过，若在同 一批学员中有 70%的人通过了理论考试，80%的人通过了路考，则最后领到驾照的人有 60%. （1）10%的人两种考试都没有通过 （2）20%的人仅通过了路考
1.标准型： ①两个集合： A B A B A B A B 全集
②三个集合：
A B C A B C A B B C A C A B C A B C 全集
2.变异型：
1 2A B C A B C 满足两个条件 满足三个条件
3.画图分析型：若不符合以上模型，则可采用画图分析法.
A B A B
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【例 5.11-2】（2010-1）某公司的员工中，拥有本科毕业证、计算机登记证、汽车驾驶证 的人数分别为 130，110，90.又知只有一种证的人数为 140，三证齐全的人数为 30，则恰 有双证的人数为（ ）. （A）45 （B）50 （C）52 （D）65 （E）100 【例 5.11-3】（2017）老师问班上 50 名同学周末复习情况，结果有 20 人复习过数学、30 人复习过语文、6 人复习过英语，且同时复习过数学和语文的有 10 人、同时复习过语文和 英语的有 2 人、同时复习过英语和数学的有 3 人若同时复习过这三门课的人为 0，则没有 复习过这三门课程的学生人数为（ ）. （A）7 （B）8 （C）9 （D）10 （E）11 【例 5.11-4】（2018）有 96 位顾客至少购买了甲、乙、丙三种商品中的一种，经调查： 同时购买了甲、乙两种商品的有 8 位，同时购买甲、丙两种商品的有 12 位，同时购买了 乙、丙两种商品的有 6 位，同时购买了三种商品的有 2 位，则仅购买一种商品的顾客有 ( ). （A）70 位 （B）72 位 （C）74 位 （D）76 位 （E）82 位 24 小时热线电话：4008-219-288
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【题型 12】最值问题 （1）二次函数最值 【例 1.12-1】（2016）某商场将每台进价为 2000 元的冰箱以 2400 元销售时，每天销售 8 台， 调研表明这种冰箱的售价每降低 50 元， 每天就能多销售 4 台， 若要每天销售利润最大， 则该冰箱的定价应为（ ）元. （A）2200 （B）2250 （C）2300 （D）2350 （E）2400 （2）均值最值 【例 1.12-2】已知某厂生产 x 件产品的成本为 2125000 200 40
c x x （元），要使平 均成本最小，则所应生产的产品件数为（ ）. （A）100 件 （B）200 件 （C）1000 件 （D）2000 件 （E）以上均不正确
【思路点拨】最值问题有以下类型： 类型一：一元二次函数最值问题. 类型二：均值不等式最值问题. 类型三：极端假设法. 24 小时热线电话：4008-219-288
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（3）极端假设法 【例 1.12-3】（2013-1）某单位年终共发了 100 万元奖金，奖金金额分别是一等奖 1.5 万 元、二等奖 1 万元、 三等奖 0.5 万元，则该单位至少有 100 人. （1）得二等奖的人数最多 （2）得三等奖的人数最多 【题型 13】线性规划问题 【例 5.13】（2010-1）某居民小区决定投资 15 万元修建停车位，据测算，修建一个室内 车位的费用为 5000 元，修建一个室外车位的费用为 1000 元，考虑到实际因素，计划室外 车位的数量不少于室内车位的 2 倍，也不多于室内车位的 3 倍，这笔投资最多可建车位的 数量为（ ）. （A）78 （B）74 （C）72 （D）70 （E）66 【题型 14】植树问题 【例 5.14-1】某道路一侧原有路灯 106 盏，相邻两盏灯的距离为 36 米，现计划全部换为
新型节能灯，两灯距离变为 70 米，共需新型节能灯（ ）盏.
【思路电拨】 类型一：直线型.若道路长度为 l 米，每隔 n 米植树，两端都种，则共有 1
ln 棵树. 类型二：圆圈型.若道路周长为 l 米，每隔 n 米植树，则共有 ln 棵树.
【思路点拨】线性规划有以下解题思路： 1. 根据已知条件列出约束条件（不等式组）和目标函数，画图图像通过目标函数平移 确定最优解. 2. 将约束条件中的不等号改写成等号，然后从边界附近找最优解. 24 小时热线电话：4008-219-288
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（A）54 （B）55 （C）108 （D）110 （E）112 【例 5.14-2】（2019）将一批树苗种在一个正方形花园边上，四角都种，如果每隔 3 米 种一课，那么剩下 10 棵树苗，如果每隔 2 米种一棵，那么恰好种满正方形的 3 条边，则这 批树苗有（ ）棵. （A） 54 （B）60 （C）70 （D） 82 （E） 94 【题型 15】年龄问题 【例 5.14-1】甲、乙、丙三人在 2008 年的年龄之和为 60 岁，2010 年甲是丙年龄的两倍， 2011 年乙是丙年龄的两倍，问甲是哪年出生（ ）. （A）1988 （B）1986 （C）1984 （D）1982 （E）以上均不对 【例 5.14-2】妈妈对女儿说：我像你这么大的时候你才两岁.女儿对妈妈说：等我到了您 这个年龄的时候您就 74 岁了，则女儿今年（ ）岁. （A）25 （B）26 （C）27 （D）28 （E）30
【思路电拨】年龄问题的关键是选取参照年份.年龄问题的特点有两个：一个是差值恒 定，另一个是同步增长. 24 小时热线电话：4008-219-288
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变式思维训练
1.已知赵先生的年龄是钱先生年龄的 2 倍，钱先生比孙先生小 7 岁，三位先生的年龄之和 是小于 70 的质数，且这个质数各位数字之和为 13，那么钱先生的年龄为（ ）. （A ）15 （B）18 （C）14 （D）7 （E）以上均不对 【题型 16】牛吃草问题、给水排水问题 【例 1.16】（2001-1）一艘轮船发生漏水事故.当漏进水 600 桶时，两部抽水机开始排水， 甲机每分钟能排水 20 桶，乙机每分钟能排水 16 桶，经 50 分钟刚好将全部排完.则每分钟 漏进的水有（ ）. （A）12 桶 （B）18 桶 （C）24 桶 （D）30 桶 （E）35 桶 24 小时热线电话：4008-219-288
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第六章 平面几何 第一节、考试要点剖析
一、平行直线
1.直线和平行线的夹角
如下图所示： 1 与 4 是同位角，同位角相等；
2 与 4 是内错角，内错角相等；
3 与 4 是同旁内角，同旁内角互补.
2.直线被一组平行线截得的线段成比例
【大纲考点】平面图形：三角形、四边形（矩形、平行四边形、梯形）、圆形与扇形. 【命题剖析】几何是建立在空间想象能力上的演绎推理和数学计算.平面几何主要考察 三角形、四边形及圆形等平面图形的角度、周长、面积等的计算和运用，命题主要围绕 几何图形的面积计算.所考查的图形一般不会是简单的三角形、四边形或圆，而是由这 些基本图形所构成的组合图形，只要能快速地把所求面积图形分解为熟悉的图形，问题 就迎刃而解了.因此考试的难点是对图形的拆分， 考试一般考组合图形 （阴影部分面积） ，这就要求考生能够快速地把所求复杂图形或拆分或割补成几个简单熟悉的图形. 24 小时热线电话：4008-219-288
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: :a b c d : :a b c d
二、三角形
1.三角形的角
内角之和为 180 ，外角等于不相邻的两个内角之和.
2.三边的关系
任意两边之和大于第三边，即： a b c ;任意两边之差小于第三边，即 a b c .
3.面积公式
1 1,2 2
S ah p p a p b p c p a b c .其中， h 是 a 边上的高， p 是三 角形的半周长.即 2
cbap
4.特殊三角形（直角、等腰、等边）
（1）直角三角形 ①直角三形的两个锐角互余； ②勾股定理：直角三角形两直角边的平方和等于斜边的平方；
222
a b c
③勾股定理逆定理：如果一个三角形的一条边的平方等于另外两条边的平方和，那么这个 三角形是直角三角形． 推论： 2 2 2
a b c Rt 2 2 2
a b c 锐角三角形
222
a b c 钝角三角形 24 小时热线电话：4008-219-288
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④直角三角形斜边上的中线等于斜边的一半． ⑤在直角三角形中，如果一个锐角等于 30°，那么它所对的直角边等于斜边的一半． ⑥常见勾股数： 3, 4,5 , 5,12,13 , 7, 24, 25 , 8,15,17 ；⑦(i)等腰直角三角形三边之比为： 1:1: 2 ；(ii)内角为 30 ,60 ,90 o o o 的直角三角形三边之比为： 1: 3 : 2 ；（2）等腰三角形 ①等腰三角形的两个底角相等．(简写成“等边对等角”) ②如果一个三角形有两个角相等，那么这两个角所对的边也相等．(简写成“等角对等边”) ③等腰三角形的“三线合一”定理：等腰三角形的顶角平分线、底边上的中线和底边上的 高互相重合，简称“三线合一”． （3）等边三角形 ①等边三角形的各个内角都相等，并且每一个内角都等于 60°． ②三个角都相等的三角形是等边三角形． ③有一个角是 60°的等腰三角形是等边三角形． ④等边三角形四线合一、四心合一. ⑤等边三角形面积= 43 2
a ，其中 a 为边长. ⑥等边三角形高与边长之比为： 33 : 2 :1 2
，
5.三角形的全等和相似
三角形的全等，就是两个三角形完全相同，具有相同的边长、角和面积等. 三角形的相似，重点考查的不是判断两个三角形相似与否，而是相似的性质，并且相 似三角形的性质完全可以延续到其他相似图形. （1）相似三角形对应边成比例，称为相似比. （2）相似三角形的高、中线、角平分线的比也等于相似比. （3）相似三角形的周长之比也等于相似比. （4）相似三角形的面积之比等于相似比的平方. 如图所示：
①金字塔模型 24 小时热线电话：4008-219-288
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若 / / DE BC ，则 ~ADE ABC ，即 （1） ADE ABC
CAD AE DE AG kAB AC BC AH C
，（边长比等于相似比，这里 k 称作相似比） （2） 2
1212
ADE ABC
DE AG S kS BC AH
（面积比等于相似比的平方） 【注意】①相似三角形的对应边的比，三角形的高、中线、角平分线的比， 周长之比都等于相似比，②面积之比等于相似比的平方相似. ②沙漏模型 若 / / AB DE ，则 ~CAB CED ，即 （1） ABC CDE
CAB AC BC CH kDE CE DC CG C
，（边长比等于相似比，这里 k 称作相似比） （2） 2
1212
ABC CDE
AB CH S kS DE CG
（面积比等于相似比的平方） ③等角反向模型
在 ABC Rt 中， 90 o
B ，若 DE AC ，则 ~AED ABC ，即 24 小时热线电话：4008-219-288
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（1） ABC ADE
CAE DE AD kAB BC AC C
，（边长比等于相似比，这里 k 称作相似比） （2） 2
1212
AED ABC
AE DE S kS AB BC
（面积比等于相似比的平方） ④直角三角形射影定理：
6.三角形的四心五线
重心：中线的交点，将中线分成 2 :1 两部分. 内心：内切圆圆心，角平分线交点，到三角形三边距离相等. 外心：外接圆圆心，中垂线交点，到三角形三个定点距离相等. 垂心：高线交点. 【注意】等边三角形四心合一
三、四边形
1.平行四边形
平行四边形两边长为 a 、 b ，以 a 边为底边的高为 h ，面积为 S ah ，周长 2C a b
2.矩形和正方形
矩形两边长为 a 、 b ，面积为 S ab ，周长 2C a b ，对角线 2 2
l a b
3.菱形
菱形四边长均为 a ，以 a 为底边的高为 h ，两条对角线为 1l 、 2l ，
222
AB BC AB BD BC BD AB AC BC AC DC BC DC AC AD CD AD BD DC DB DA
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面积为 1 2
12
S ah l l ，周长 4C a
4.梯形
上底为 a ，下底为 b ，高为 h ，中位线
12
l a b ，面积
12
S a b h
梯形的中位线平行于梯形的两底边，并且等于两底和的一半． 特殊四边形的有关判定：
图形 性质 判定 对称性
平行 四边形
①对边平行且相等； ②对角相等； ③对角线互相平分．
①两组对边分别平行的四边形； ②两组对边分别相等的四边形； ③一组对边平行且相等的四边形； ④两组对角分别相等的四边形； ⑤对角线互相平分的四边形．
中心 对称
矩形 ①对边平行且相等； ②四个角都相等都是直角； ③对角线互相平分且相等．
①有一个角是直角的平行四边形； ②有三个角是直角的四边形； ③对角线相等的平行四边形．
轴对称 中心对 称
菱形 ①对边平行且四条边都相 等； ②对角相等； ③对角线互相垂直平分， 并且每一条对角线平分一组 对角．
①有一组邻边相等的平行四边形； ②四条边相等的四边形； ③对角线互相垂直的平行四边形．
轴对称 中心对 称
正方形 ①对边平行且四条边都相 等； ②四个角都相等都是直角； ③两条对角线互相垂直平 分且相等， 每一条对角线平 分一组对角．
①有一个角是直角的菱形； ②有一组邻边相等的矩形； ③两条对角线垂直的矩形； ④两条对角线相等的菱形．
轴对称 中心对 称
等腰 梯形
①一组对边平行而另一组 对边不平行，两腰相等；
①两腰相等的梯形； ②同一条底边上的两个角相等的梯
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四、圆和扇形
1.角的弧度
把圆弧长度和半径的比值称为一个圆周角的弧度，弧度就是用
的算式来表示角度，其中
180
， 90 2
o
， 60 3
o
， 45 4
o
， 30 6
o
.
2.圆
圆的圆心为 O ，半径为 r ，则周长为 2C r
，面积是 2
S r
.
3.扇形
扇形半径为 r ，圆心角角度为 o
n ,弧度为
，扇形的弧长 2360
nl r r
弧长 ，面积 2 1360 2
nS r l r
弧长 扇形 .
4.常用定理 圆周角定理： 一条弧所对圆周角等于它所对圆心角的一半. 推论：直径所对的圆周角是直角.
②同一条底边上两个角相 等； ③对角线相等．
形； ③两条对角线相等的梯形． 24 小时热线电话：4008-219-288
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第二节 基础过关题型
【题型 1】三角形基本概念 【例 6.1-1】（2014-10）三条长度分别为 a ， b ， c 的线段能构成一个三角形. （1） a b c （2） b c a
【例 6.1-2】（2014-1） 如图， O 是半圆圆心， C 是半圆上的一点， OD AC ，则能确
定 OD 的长.
（1）已知 BC 的长 （2）已知 AO 的长
【思路电拨】 1.利用平行线、三角形等关于角度的基本概念，求某些夹角的角度. 2.利用特殊三角形的特点，包括直角三角形、等腰三角形、等边三角形等，求出相应三 角形的面积.
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【例 6.1-3】 （2014-1） 如图， 已知 3AE AB ， 2BF BC ，若 ABC 的面积是 2， 则 AEF
的面积为（ ）. (A) 14 (B) 12 (C) 10 (D) 8 (E) 6 【题型 2】三角形全等和相似 【例 6.2-1】（2009-1）直角三角形 ABC 的斜边 13 AB 厘米，直角边 5AC 厘米，把 AC
对折到 AB 上去与斜边相重合，点 C 与点 E 重合，折痕为 AD ，则图中阴影部分的面积为
（ ）. （A） 20 （B） 40 3 （C） 38 3
（D） 14 （E） 12
【例 6.2-2】 （2018） 如图， 在矩形 ABCD 中， AE FC ，则三角形 AED 与四边形 BCFE
能拼接成一个直角三角形. （1） 2EB FC （2） ED EF
【思路电拨】 利用相似三角形，全等三角形或相邻三角形的关系，结合比例的思想， 解决三角形与三角形之间边与边，面积与面积的关系问题. 24 小时热线电话：4008-219-288
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【题型 3】三角形内切圆与外接圆 【例 6.3-1】已知 ABC ， 8AB AC ， 30 o
ABC 的外接圆面积为（ ）. （A） 8
（B） 16
（B） 32
（D） 60
（E） 64
【例 6.3-2】（2018）如图，圆 O 是三角形 ABC 的内切圆，若三角形 ABC 的面积与周长的 大小之比为 1:2，则圆 O 的面积为( ). （A）
（B） 2
（C） 3
（C） 4
（E） 5
【题型 4】判断三角形形状 【例 6.4】（2009-10） ABC 是等边三角形. （1） ABC 的三边满足 ac bc ab cba 222
（2） ABC 的三边满足 0222223 bc bac ab baa
【思路电拨】 1.基本解题思路是因式分解 2.判断三角形形状时，一般为特殊三角形，即等边三角形、等腰三角形、直角三角形、 等腰直角三角形 3.易错点：“等腰直角三角形”与“等腰或直角三角形”概念不同.
【思路电拨】 1.三角形外心：三角形的三条边的垂直平分线的交点. 2.三角形内心：三角形的三条角平分线的交点. 24 小时热线电话：4008-219-288
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【题型 5】 四边形及相关性质 【例 6.5-1】 （2014-10）如图所示，在平行四边形 ABCD 中， ABC 的平分线交 AD 于 E ，
150 o
BED ，则 A 的大小为（ ）. （A） 100 o （B） 110 o （C） 120 o
（D） 130 o （E） 150 o
【例 6.5-2】（2007-10）如图，正方形 ABCD 四条边与圆 O 相切，而正方形 EFGH 是圆 O 的 内接正方形.已知正方形 ABCD 面积为 1，则正方形 EFGH 面积是（ ）. （A） 23 （B） 12 （C） 22 （D） 23 （E） 14
【例 6.5-3】（2015）如图，梯形 ABCD 的上底与下底分别为 5,7， E 为 AC 与 BD 的交 点， MN 过点 E 且平行与 AD ，则 MN （ ）
（A） 26 5 （B） 11 2 （C） 35 6
（D） 36 7 （E） 40 7
【思路电拨】 掌握四边形的基本概念及特殊性质， 并利用这些性质图形进行变形等处理， 从而达到快速解题的目的.
D C
BA
E F
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【题型 6】与圆弧相关的应用 【例 7.6-1】 (2014-10).一个长为 8cm， 宽为 6cm 的长方形木板在桌面上做无滑动的滚动 （顺 时针方向），如右图所示，第二次滚动中被一小木块垫住而停止，使木板边沿 AB 与桌面 成 30 o 角，则木板滚动中，点 A 经过的路径长为（ ）. （A） 4 （B） 5 （C） 6 （D） 7 （E） 8
【例 7.6-2】 （2010-10） 图中， 阴影甲的面积比阴影乙的面积多 2
28 cm ， 40 AB cm ，CB
垂直 AB ，则 BC 的长为（ ） cm .（
取到小数点后两位.）
（A）30 （B）32 （C）34 （D）36 （E）40
【思路电拨】掌握圆和扇形相关知识，如圆弧，圆心角，圆周角等，利用圆和扇形的性 质解题. 24 小时热线电话：4008-219-288
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【例 7.6-3】(1999-10)如图，半圆 ADB 以 C 为半径为 1，且 CD AB ，延长 BD 和 AD ，分别与以 B ， A 为圆心半径为 2 的圆弧交于 E ， F 两点，则图中阴影部分的面积是（ ）. （A） 12 2
（B） (1 2)
（C） 12
（D） ( 3 1)
（E） (2 3)
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第七章 解析几何 第一节 考试要点剖析
一、平面直角坐标系
1.点
点在平面直角坐标系中的表示： ,P x y
两点 1 1,A x y 与 2 2,B x y 之间的距离：
222121
d x x y y
2.线段的定比分点
设 A B、 是两个不同点， 它们的坐标依次为 1 1 2 2, , , ,A x y B x y P 是线段 AB 上的一个点， 且分 AB 的定比为 AH HB
,则 P 的坐标为 1 2 1 2
,1 1
x x y y
.
若 P 为 AB 中点，则中点坐标为 1 2 1 2
,2 2
x x y y
y
22(,)B x y
x
11(,)A x y
B
y
A
( , )P x y
x 第三象限
o x
y
第一象限 第二象限
第四象限
【大纲考点】1.平面直角坐标系；2.直线方程与圆的方程；3.两点间距离公式与点到直 线距离公式. 【命题剖析】解析几何主要考察的是在平面直角坐标系中直线、圆、直线与直线的位置 关系、直线与圆的位置关系.一般在考试中考察 3 个题目，命题方向主要体现在三个方 面：一是对于基本方程表达式的求解与计算；二是确定图形之间的位置关系；三是对于 对称的考察. 24 小时热线电话：4008-219-288
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二、直线
1.直线的倾斜角和斜率
（1）倾斜角：直线向上的方向与 x 轴正半轴形成的夹角称为倾斜角.倾斜角的取值范围 为 o
[0,180 )
（2）斜率：倾斜角的正切值称为斜率.特别地，当倾斜角是 90°时，斜率不存在. （3）两点斜率公式：设直线 l 上有两个点 1 1,A x y 、 2 2,B x y ，则 2 12 1
y yk x x
2.直线方程的几种形式
（1） 斜截式： 斜率为 k ，在 y 轴上截距为 b（即过点 0, P b ）,的直线方程是 y kx b .（2）点斜式：过点 0 0,P x y ，斜率为 k 的直线方程为 0 0y y k x x .（3）两点式：过两个点 1 1,A x y 、 2 2,B x y 的直线方程为 1 12 1 2 1
y y x xy y x x
，其中 1 2 1 2,x x y y .（4）截距式：在 x 轴上的截距为 a （即过点 1 , 0 P a ），在 y 轴上的截距为 b
（即过点 2 0, P b ）的直线方程为 1
x ya b
， 0, 0a b .（5）一般式： 0ax by c ， ,a b 不能同时为零.
3.点到直线的距离
点 0 0,x y 到直线 0ax by c 的距离 0 02 2
ax by cd a b
.注：点与直线位置关系具体分以下情况，即将点代入直线方程 ①若代入后直线等于 0 时，则点在直线上. ②若代入后直线大于 0 或小于 0 时，则点在直线两侧. 24 小时热线电话：4008-219-288
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4.两条直线的位置关系
直线形式 位置关系
斜截式
11
:l y k x b
22
:l y k x b
一般式
1111
: 0l a x b y c
2222
: 0l a x b y c
平行： 1 2/ / l l 1 2 1 2,k k b b 1 1 12 2 2
a b ca b c
相交 1 2k k 1 12 2
a ba b
垂直: 1 2l l
（相交特殊情况）
12
1k k 或
1
0k 且 2k 不存在
12121212
= 1 0
a a a a b b b b
【评注】求两条相交直线交点坐标的方法为，联立直线方程，求解二元一次方程的根，即 为交点的横纵坐标.
三、圆
1.圆的方程 （1）标准方程 当圆心为 0 0,x y ，半径为 r 时，圆的标准方程为
22200
x x y y r .特别地，当圆 心在原点 0, 0 时，圆的标准方程为 2 2 2
x y r .（2）一般方程
22
0x y ax by c .配方后得到： 2 2 2 2 42 2 4
a b a b cx y
，要求
22
4 0a b c .圆心 ,2 2
a b ，半径 2 2 4 02
a b cr .【评注】若给定一个方程 2 2 0x y ax by c ，可以通过 2 2 4a b c 来判断是否是圆，
若 2 2 4 0a b c ， 2 2 0x y ax by c 表示圆. 24 小时热线电话：4008-219-288
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2.点与圆的关系
点 ,p pP x y ，圆
22200
x x y y r
点与圆的关系
2222002
,,
pp
rx x y y rr
点在圆内 ,点在圆上 点在圆外
3.直线与圆的位置关系
直线 l : 0ax by c ,圆 O ：
22200
x x y y r ，d 为圆心 0 0,x y 到直线 l 的距 离. （1）相离：圆心到直线的距离大于圆的半径，即 d r
此时图像没有交点，即方程组
22200
0ax by cx x y y r
无实根， 0 .（2）相切：圆心到直线的距离等于圆的半径，即 d r
此时 图像只有 唯一交 点，即 方程组
22200
0ax by cx x y y r
有两 个相等的 实数根 ，
0 .（3）相交：圆心到直线的距离小于圆的半径，即 d r
此时图像有两个不同的交点，即方程组
22200
0ax by cx x y y r
，有两个不相等的实数 根， 0 .24 小时热线电话：4008-219-288
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4.圆与圆的位置关系
圆 1O ：
222111
x x y y r ；圆 2O ：
222222
x x y y r ；设 1 2r r ； d 为两 圆圆心 1 1,x y 和 2 2,x y 的距离，即圆心距.
位置关系 图形 成立条件 内公切 线条数
外公切 线条数
外离 1 2d r r 2 2
外切 1 2d r r 1 2
相交 1 2 1 2r r d r r 0 2
内切 1 2d r r 0 1
内含 1 2d r r 0 024 小时热线电话：4008-219-288
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第二节 基础过关题
一、点、线、圆基本概念题
【题型 1】点和点的位置关系 【例 7.1-1】 直线 l 经过 (2,1) A ， 2
(1, )B m （ m R 两点） ，那么直线 l 的斜率取值范围 （ ）.（A） [1, ) （B） ( , ) （C） ( ,1) （D） ( ,1] （E）以上均不对 【例 7.1-2】直线 y x 上的两点 ,P Q 的横坐标分别是 1、5，则 PQ （ ）. （A）4 （B） 4 2 （C） 2 （D） 2 2 （E） 3 2
【思路电拨】若两点坐标为 1 1( , )x y ， 2 2( , )x y 则有： 1.中点公式： 1 2 1 2
,2 2
x x y y
2.斜率公式： 2 12 1
y yk x x
；3.两点之间的距离公式：
222121
d x x y y
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【题型 2】直线 【 例 8.2-1 】 （ 1999-1 ） 已 知 直 线 1l ： 03)1()2( yaxa 与 直 线 2l ：
( 1) (2 3) 2 0a x a y 互相垂直，则 a 等于（ ）. （A） 1 （B）1 （C） 1 （D） 32
（E）0 【例 8.2-2】直线 ( 2) 3 0m x y 与 (3 2) 1 0m x y 平行，则实数 m 的值是（ ）. （A）1 （B）2 （C）3 （D）4 （E）不存在
【思路电拨】 1.直线方程的五种形式： （1）点斜式（2）斜截式（3）两点式（4）截距式（5）一般形式 2.判断直线经过哪些象限 一般将直线化为斜截式，然后画草图分析. 3.两直线的位置关系 （1）平行（2）相交（3）垂直 24 小时热线电话：4008-219-288
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【题型 3】 圆的方程 【例 7.3-1】 （1997-1）圆方程 0142 22 yyxx 的圆心是（ ）. （A） ( 1, 2) （B） )2,1( （C） )2,2( （D） )2,2( （E） )2,1(
【例 7.3-2】（2008-1）动点 ( , )x y 的轨迹是圆. （1） 1 4x y
（2） 2 2
3( ) 6 9 1 0x y x y
【例 7.4-3】（1998-1）设 AB 为圆 C 的直径，点 A，B 的坐标分别是 )1,5()5,3( 、 ，则圆 C 的方程是（ ）. （A） 80 )6()2( 22 yx （B） 20 )3()1( 22 yx
（C） 80 )4()2( 22 yx （D） 2 2
( 2) ( 6) 20 x y
（E） 20 22 yx
【思路点拨】熟练掌握圆的标准方程的概念和形式，并结合画图解决问题. 24 小时热线电话：4008-219-288
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二、点、线、圆之间的位置关系（重点）
【题型 4】点到直线的距离 【例 7.4-1】点 3, 4 A , 2, 1B 到直线 y kx 的距离之比为 1: 2 .（1） 94
k （2） 78
k
【题型 5】点和圆的位置关系 【例 7.5】曲线 2 2 2
: 2 2 ( 1) 0C x y ax y a 与原点位置关系是点在圆外. （1） 0 1a （2） 1a
【思路点拨】点 0 0,x y 到直线 0ax by c 的距离 0 02 2
ax by cd a b
.
【思路电拨】 点与圆的位置关系， 点 0 0( , )P x y ，圆： 2 2 2
( ) ( )x a y b r
（1）点在圆内 d r 2 2 20 0( ) ( )x a y b r
（2）点在圆上 d r 2 2 20 0( ) ( )x a y b r
（3）点在圆外 d r 2 2 20 0( ) ( )x a y b r 24 小时热线电话：4008-219-288
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变式思维训练
若连续掷 2 次骰子，第一次掷得的点数为 m ，第二次掷得的点数为 n ，则点 ( , )P m n 落在圆 2 2 16 x y 内的概率为（ ）. （A） 19 （B） 29 （C） 736 （D） 10 36 （E）以上均不对 【题型 6】直线和圆的位置关系 【例 7.6-1】（2014-10）直线 ( 2) y k x 与圆 2 2 1x y 相切. （1） 12
k （2） 33
k
【例 7.6-2】 （2011-10）已知直线 y kx 与圆 2 2 2x y y 有两个交点 A ，B . 若 AB 的长度大于 2 ，则 k 的取值范围是（ ）. （A） ( , 1) （B） ( 1, 0) （C） (0,1)
（D） (1, ) （E） ( , 1) (1, )
直线的圆的位置关系 （1）相交： d r
（2）相切： d r
（3）相离： d r
【注意】遇到相交问题已知弦长时，常利用勾股定理将弦长转化为圆心到直线的距离 .24 小时热线电话：4008-219-288
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【例 7.6-3】（2014-1） 已知直线 l 是圆 2 2 5x y 在点（1,2）处的切线，则 l 在 y 轴上 截距为（ ）. (A) 25 (B) 23 (C) 32 (D) 52 (E) 5
【题型 7】圆与圆的位置关系 【例 7.7-2】（2008-1）圆 1C ： 2 2 23( ) ( 2) 2
x y r 与圆 2C ： 2 2
6 8 0x x y y 有交 点（1） 50 2
r （2） 15 2
r
三、综合题
【题型 8】 解析几何与图像有关问题 【例 7.8-1】（2008-10）过点 )0,2(A 向圆 122 yx 作两条切线 AM 和 AN （见下图）， 则两切线和弧 MN 所围成的面积（图中阴影部分）为（ ）.
（A） 31
（B） 61
（ C ） 623
（D） 63
（E） 33
【思路电拨】圆与圆的位置关系包括外离、外切、相交、内切、内含共五种.
【思路电拨】解析几何要先根据所给的方程或表达式画出图像，然后借助平面几何的知 识来求解面积 .24 小时热线电话：4008-219-288
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【例 7.8-2】（2012-1）在直角坐标系中，若平面区域 D 中所有点的坐标（ ,x y ）均满足：
0 6x ， 0 6y ， 3y x ， 2 2
x y 9 ，则 D 的面积是（ ）. （A） )41(49
（B） )44(9
（C） )43(9
（D） )2(49
（E） )1(49
【题型 9】过定点的曲线系 【例 7.9-1】（2008-10）曲线 2 2 1ax by 通过 4 个定点. （1） 1a b
（2） 2a b
【例 7.9-2】（2014-10）圆盘 2 2 2( )x y x y 被直线 L 分成面积相等的两部分. （1） : 2L x y
（2） 12: yxL
【思路电拨】恒过定点问题的解法：先将参数看作未知数，其它字母看作已知数，将 方程整理为 0a b
，然后再令 0, 0a b ，最后解方程组求出定点即可. 24 小时热线电话：4008-219-288
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【题型 10】最值问题 【例 7.10-1】动点 ( , )P x y 在圆 2 2 1 0x y 则 12
yx
的最大值是（ ）. （A） 2 （B） 2 （C） 12 （D） 12
（E） 43
【例 7.10-2】曲线 2 2
2 0x x y 上的点到直线 3 4 12 0x y 的最短距离是（ ）. （A） 35 （B） 45 （C） 1 （D） 43 （E） 2
【思路电拨】与圆有关的最值问题策略： 1. 形如 y bx a
或 ax by 的式子求最值，一般转化为线性规划问题，直接令式子等于 k. 2. 形如 2 2
( ) ( )x a x b 的式子一般转化为两点之间的距离公式求解. 3. 点、线、圆之间涉及距离的最值问题常常画图分析，一般先求到圆心的距离，然后 根据题意加或减半径. 24 小时热线电话：4008-219-288
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【题型 11】对称问题 （1）关于点对称 【例 7.11-1】点 (6, 3) A 关于点 (1, 2) P 的对称点 'A 的坐标为（ ）. （A） ( 4, 1) （B） (1, 2) （C） (1, 4) （D） (6, 3) （E）以上均不是 【例 7.11-2】直线 : 2 3 6 0l x y 关于点 (1, 1) P 对称的直线方程为（ ）. （A） 2 3 8 0x y （B） 3 2 8 0x y （C） 2 8 0x y
（D） 2 3 0x y （E）以上均不对 （1）关于直线对称 【例 7.11-3】（2007-10）点 (2,3) P 关于直线 0x y 的对称点是（ ）. （A）（4，3） （B）（－2，3） （C）（－3，－2） （D）（－2，－3） （E）（－4，－3）
【思路电拨】1.点关于直线对称，常规方法是采用方程组解决. 已知直线 l ： 0Ax By C ，求点 1P 关于直线 l 的对称点 2P ，对称关系中包括有两个 等量关系：①线段 1 2P P 的中点在对称轴上② 1 2P P 与对称轴 l 互相垂直. 2.关于特殊直线对称（坐标轴、斜率为 1 ， x a ， y b 等特殊直线可以快速求解） 【注意】对于选择题来说，一般作图求解速度最佳. 24 小时热线电话：4008-219-288
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【例 7.11-4】（2013-1）点（0,4）关于直线 2 1 0x y 的对称点为（ ）. （A）（2,0） （B）（-3,0） （C）（-6,1） （D）（4,2） （E）（-4,2） 【例 7.11-5】 （2019）设圆 C 与圆 2 2
5 2x y 关于 2y x 对称，则圆 C 方程为（ ）. （A）
22
3 4 2x y （B）
22
4 3 2x y （C）
22
3 4 2x y
（D）
22
3 4 2x y （E）
22
3 4 2x y
【题型 12】线性规划问题 【例 7.12-1】（2018）已知点 ( ,0), (1,3), (2,1), ( , )P m A B x y 点 在三角形 PAB 上，则 x y 的最小值与最大值分别为-2 和 1. (1) 1m (2) 2m 24 小时热线电话：4008-219-288
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第八章 立体几何 第一节 考试要点剖析
一、长方体
设三条相邻的棱长度分别为 , ,a b c .
1.全面积： 2F ab bc ac
2.体积： V abc
3.体对角线： 2 2 2
d a b c
4.所有棱长之和： 4l a b c
特别地，当 a b c 时的长方体称作正方体，且有 2
6F a ， 3
V a ， 3d a
二、柱体
1.柱体的分类
圆柱：底面为圆的柱体称为圆柱 棱柱：底面为多边形的柱体称为棱柱
2.柱体的一般公式
无论圆柱还是棱柱，侧面展开图都是一个矩形，其中一边长为底面周长，另一边长为柱体 的高. 侧面积： S 底面周长 高
体积： V 底面积 高
【大纲考点】空间立体几何：长方体、柱体、球体 【命题剖析】主要考察长方体、柱体、球体等立体几何图形的表面积、体积以及和提及 相关问题的求解，重点考察体积和表面积的计算和运用.
'A 'B
'C'D
A B
C
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3.圆柱体的公式
设高为 h ，底面半径为 r
体积： 2
V r h
侧面积： 2S rh
全面积： 2
2 2 2F S S rh r
侧 底
三、球
设球的半径为 r
1.球的表面积： 2
4S r
2.球的体积： 343
V r
四、长方体、正方体、圆柱体和球的关系
设圆柱体底面半径为 r ，球半径为 R ，圆柱体的高为 h
内切球 外接球
长方体 无，只有正方体才有 体对角线 2l R
正方体 棱长 2a R 体对角线 2 3l R a
圆柱
只有轴截面是正方形的圆柱（等边圆 柱）才有，此时有 2 2r h R 22
2 2R h r
【评注】1.在这些关系中，一定要注意寻找几何关系时要利用几何体的轴截面； 2.关系是相互的，可以说正方体的外接球，也可以说球的内接正方体，其性质是一样的.
h
C
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第二节 基础过关题型
【题型 1】长方体和正方体 【例 8.1-1】（1997-10）一个长方体，长与宽之比为 2:1，宽与高之比是 3:2，若长方体 的全部棱长之和是 220 厘米，则长方体的体积是（ ）. （A）2880 立方厘米 （B）7200 立方厘米 （C）4600 立方厘米 （D）4500 立方厘米 （E）3600 立方厘米 【例 8.1-2】(2016-1)现有长方形木板 340 张，正方形木板 160 张（图 1）这些木板恰好可 以装配成若干竖式和横式的无盖箱子（图 2），装配成的竖式和横式箱子的个数为（ ）.
图 1 图 2 （A）25，80 （B）60，50 （C）20，70 （D）64，40 （E）40，60
【思路电拨】熟练掌握长方体和正方体的概念以及体积、表面积、体对角线等求法.
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【例 8.1-3】（2019）如图，六边形 ABCDEF 是平面与棱长为 2 的正方体所截得到的， 若 , , ,A B D E 分别为相应棱的中点，则六边形 ABCDEF 的面积是（ ）.
（A） 32 （B） 3 （C）2 3
（D）3 3 （E）4 3
【题型 2】圆柱体 【例 8.2-1】（1999-1）一个两头密封的圆柱形水桶，水平横放时桶内有水部分占水桶一 头原周长的 14 ，则水桶直立时水的高度和桶的高度之比是（ ）.
(A) 14 (B) 1 14
(C) 1 14 2
(D) 18 (E) 4
【思路电拨】熟练掌握圆柱体的概念以及体积、表面积、截面的求法. 24 小时热线电话：4008-219-288
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【例 8.2-2】(2012-1)如图，一个储物罐的下半部分是底面直径与高均是 20m 的圆柱形、 上半部分（顶部）是半球形，已知底面与顶部的造价是 400 元/m 2
，侧面的造价是 300 元/
m 2
,该储物罐的造价是（ ）.（ 3.14）
（A）56.52 万元 （B）62.8 万元 （C）75.36 万元 （D）87.92 万元 （E）100.48 万元 【题型 3】球体
【例 8.3-1】球的体积增加到原来的 27 倍. （1）半径增加到原来 3 倍 （2）表面积增加到原来的 9 倍 【例 8.3-2】(2017)如图，一个铁球沉入水池中，则能确定铁球的体积.
（1）已知铁球露出水面的高度 （2）已知水深及铁球与水面交线的周长
【思路电拨】 熟练掌握球体的概念以及体积、表面积的求法.设球的半径为 R 则1. 球的表面积 2
4S R
.2. 球的体积 343
V R
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【题型 4】几何体的“切”与“接” 【例 8.4-1】现有一个半径为 R 的球体，拟用刨床将其加工成正方体，则能加工成的最大 的正方体的体积是（ ）. （A） 3
38 R (B) 3
938 R (C) 3
34 R (D) 3
31 R （E） 3
93 R
【例 8.4-2】如图，在半径为 10 厘米的球体上开一个底面半径是 6 厘米的圆柱形洞，则洞
的内壁面积为（ ）.（单位：平方厘米） (A) 48π (B) 288π (C) 96π (D) 576π （E） 192π
【思路电拨】组合体的关键是找等量关系，常见的等量关系有：
长方体的外接球：球的直径 =长方体体对角线 .2. 圆柱体的外接球：球的直径 =圆柱体体对角线 .3. 正方体内切球：内切球的直径 =正方体棱长 .
【评注】以上关系均源自勾股定理关系，如遇到复杂题可自行找到勾股定理关系分析 .24 小时热线电话：4008-219-288
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第九章 排列组合 第一节 考试要点剖析
一、加法原理（分类计数原理）
1.定义
做一件事情，完成它有 n 类方法.第一类方法中有 1M 种不同的方法，第二类方法中有
2
M 种不同的方法，……，第 n 类方法中有 nM 种不同的方法，那么完成这件事情共有
123n
M M M M+ + + 种不同的方法. 例如，从武汉到上海有火车、飞机、轮船 3 种交通 方式可供选择，一天中火车、飞机、轮船分别有 1 2 3, ,k k k 个班次，那么一天中某人从武汉 到上海共有 1 2 3N k k k= + + 种不同的方法.
2.本质
分类完成.类与类之间相互独立，即：任何一类中的任何一种方法都能够单独完成此 事件.
二、乘法原理（分步计数原理）
1.定义
做一件事情， 完成它有 n 个步骤， 第一个步骤有 1M 种不同的方法， 第二个步骤有 2M
种 不 同 的 方 法 ， … … ， 第 n 个 步 骤 有 nM 种 不 同 的 方 法 ， 那 么 完 成 这 件 事 共 有
123n
M M M M 种不同的方法. 例如：某人从三亚到哈尔滨游玩，途中必须换乘.
【大纲考点】1.加法原理、乘法原理；2.排列与排列数；3.组合与组合数. 【命题剖析】对于排列组合，不仅要掌握加法原理与乘法原理的本质，更要从定义出发 对各种排列组合类型题做一个综合归纳.本章命题主要体现在两个方面：1.考察概念型 的题目，主要围绕加法原理、乘法原理、排列数、组合数来展开；2.考察排列组合类型 题的做题技巧. 24 小时热线电话：4008-219-288
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换乘的经过如下：从三亚到武汉，有 1k 种航班可选；再从武汉到沈阳，有 2k 种航班可选； 最后从沈阳到哈尔滨，有 3k 趟高铁可选，那么某人从三亚到哈尔滨共有 1 2 3N k k k 种不同的方法.
2.本质
分步完成. 步与步之间相互连贯，即：每一步都完成了才能完成此事件.
三、排列
1.定义
从 n 个不同元素中，任取出 ( )m m n 个元素，并按照一定的顺序排成一列，称为从
n 个不同元素中取出 m 个元素的一个排列，排列的总个数记为 mnP （或 mnA ）.
2.排列数计算公式
!( 1)( 2) ( 1) ( )!
mn
nP n n n n m n m
.说明： （i)公式特征：第一个因数是 n ，后面每一个因数比前面一个少 1，最后一个因数是
1n m ，共有 m 个因数； （ii）全排列数： ( 1)( 2) 2 1 !nnP n n n n （称为 n 的阶乘）. 含义： n 个不同的元素和 n 个不同的位置一一对应的方法总数. （iii）规定 0 1nP ， 0! 1 ．
四、组合
1.定义
从 n 个不同的元素中，任取出 ( )m m n 个元素，并成一组，称为从 n 个不同元素中取 出 m 个元素的一个组合，组合的总个数记为 mnC .24 小时热线电话：4008-219-288
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2.组合数计算公式
① !!( )! !
mmnn
PnC m n m m
.② m n mn nC C
说明： （i)公式特征：等式两边下标相同，上标之和等于下标； （ii)当 2
nm 时，计算 mnC 时，可用 mnnC 简化计算； （iii） x yn nC C x y x y n 或 ；（iV) 规定： 10 nC .3.常用组合恒等式: ① nnnnnn CCCC 2210 ;② )(2 1420 为偶数 nCCCC nnnnnn
;③ )(2 1531 为奇数 nCCCC nnnnnn
;由②、③两恒等式说明，①式中奇数项和与偶数项和的值相等均为 1
2 n .24 小时热线电话：4008-219-288
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第三节 基础过关题型
一、两个公式（排列与组合）
【题型 1】排列组合公式运用 【例 9.1-1】（2008-1）公路上各站之间共有 90 种不同的车票. （1）公路 AB 上有 10 个车站，每两站之间都有往返车票 （2）公路 AB 上有 9 个车站，每两站之间都有往返车票 【例 9.1-2】（2012-1）某商店经营 15 种商品，每次在橱窗内陈列 5 种，若每两次陈列的 商品不完全相同，则最多可陈列（ ）. （A）3000 次 （B） 3003 次 （C）4000 次 （D） 4003 次 （E）4300 次 【例 9.1-3】（2001-10）若 2 21 1
31
mnmn
C Cn
，则（ ）. （A） 2m n （B） 2m n （C） 1
nk
m k
（D） 1
1 nk
m k
（E） 1m n
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二、四大符号
【题型 2】四大符号基本原理 【例 9.2-1】（2002-10）某办公室有男职工 5 人，女职工 4 人，欲从中抽取 3 人支援其它 工作，但至少有 2 位男士，问抽调方案有（ ）. （A）50 种 （B）40 种 （C）30 种 （D）20 种 （E）10 种 【例 9.2-2】（1999-10） 从 0,1,2，3,5,7，11 七个数字中每次取两个相乘，不同的积有 （ ）. （A）15 种 （B）16 种 （C）19 种 （D）23 种 （E）21 种 【例 9.2-3】（2001-10）一个班里有 5 名男工和 4 名女工，若要安排 3 名男工和 2 名女工 分别担任不同的工作，则不同的安排方法数有（ ）. （A）300 种 （B）720 种 （C）1440 种 (D)7200 种 （E）3600 种
【思路电拨】表现形式：排列组合的最基本应用.处理方法：先分类后分步.
1.加法：分类原理. 2.减法：剔除法. 3.乘法：分步原理. 4.除法：消序. 24 小时热线电话：4008-219-288
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【例 9.2-4】（2017）将 6 人分成 3 组，每组 2 人，则不同的分组方式共有（ ）种. （A）12 （B）15 （C）30 （D）45 （E）90
三、八大基本策略
【题型 3】捆绑法和插空法 【例 9.3-1】（2011-1）3 个三口之家一起观看演出，他们买了同一排的 9 张连座票，则每 一家的人都坐在一起的不同方案共有（ ）. （A） 2
(3!) 种 （B） 3
(3!) 种 （C） 3
3(3!) 种 （D） 4
(3!) 种 （E）9! 种【例 9.3-2】7 名运动员接连出场，其中 3 名美国选手必须连续出场，2 名俄罗斯选手不能 接连出场的排法有（ ）种. （A）72 （B）216 （C）144 （D）432 （E）864
【思路电拨】1.相邻问题用捆绑法. 2.不相邻问题用插空法. 24 小时热线电话：4008-219-288
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【题型 4】分房问题（翻译成“每”字句） 【例 9.4】（2007-10）有 5 人报名参加 3 项不同的培训，每人都只报一项，则不同的报法 有（ ）. （A）243 种 （B）125 种 （C）81 种 （D）60 种 （E）以上结论均不正确 【题型 5】特殊优先法 【例 8.5】（1999-1）加工某产品需要经过 5 个工种，其中某一工种不能最后加工，试问 可安排（ ）种工序. （A）96 （B）102 （C）112 （D）92 （E）86 【题型 6】分组分配问题 【例 8.6-1】（2010-1）某大学派出 5 名志愿者到西部 4 所中学支教，若每所中学至少有 一名志愿者，则不同的分配方案共有（ ）. （A）240 种 （B）144 种 （C）120 种 （D）60 种 （E）24 种
【思路电拨】表现形式：元素排队. 处理方法：分步进行：先将特殊的元素排好，再将 余下的元素进行排列.基本原理：两个原理的应用.
【思路电拨】解决不同元素的分配，且每部分的元素至少一个的问题.
【思路电拨】1.表现形式：不同的元素分给不同的元素，无限制. 2.处理方法：翻译成“每”字句或利用乘法原理. 24 小时热线电话：4008-219-288
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【例 8.6-2】（2000-1）甲、乙、丙三位教师分配到 6 个班级，若其中甲教一个班，乙教 两个班，丙教三个班，则共有分配方法（ ）. (A)720 种 (B)360 种 (C)120 种 (D)60 种 (E)20 种 【题型 7】错排问题 【例 8.7】（2014-1）某单位决定对 4 个部门的经理进行轮岗，要求 4 个部门的经理必须 到其它部门任职，则不同的轮岗方案有（ ）种. (A)3 (B)6 (C)8 (D)9 （E）10 【题型 8】隔板法 【例 9.8-1】（2009-10）将 10 个相同的球随机放入编号为 1，2，3，4 的四个盒子中，则 每个盒子不空的投放方法有（ ）种. (A)72 (B)84 (C)96 (D)108 （E）120
【思路电拨】此类问题必须满足两个条件：（1）所分物品规格必须完全相同；（2） 参与分物品的每个成员至少分到 1 个，若满足这两个条件，则当物品有 n 个，参与分 物品的成员有 m 个时，分法有 11
mn
C 种.
【思路电拨】错排速记：两两错排共有 1 种分法，三三错排共有 2 种分法，四四错排共 有 9 种分法，五五错排共有 44 种分法. 24 小时热线电话：4008-219-288
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【例 9.8-2】 20 名三好学生名额分给 3 个班， 每个班至少 2 个名额， 则总的分法数是 （ ）.（A） 39C （B） 216 C （C） 317 C （D） 318 C （E） 59C
【题型 9】全能元素问题 【例 9.9】（2011-10）在 8 名志愿者中，只能做英语翻译的有 4 人，只能做法语翻译的有 3 人,既能做英语翻译又能做法语翻译的有 1 人，现从这些志愿者中选取 3 人做翻译工作， 确保英语和法语都有翻译的不同选法共有（ ）种. (A)12 (B)18 (C)21 (D)30 （E）51 【题型 10】成双成对问题 【例 9.10】（2002-10）从 6 双不同的鞋子中任意取 4 只，则其中没有成双鞋子的概率是 （ ）. （A） 411 （B） 511 （C） 16 33 （D） 23 （E）以上均不对
【思路电拨】全能元素问题一般按是否选全能元素进行分类.
【思路电拨】成双成对问题，无论题目是否要求取双，第一步一律取双，第二步再按 照题意再取即可. 24 小时热线电话：4008-219-288
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三、综合运用
【题型 11】数字问题 【例 9.11】（2014-10）.用 0,1,2,3,4,5 组成没有重复数字的四位数,其中千位数字大于 百位数字且百位数字大于十位数字的四位数的个数是（ ）. （A）36 （B）40 （C）48 （D）60 （E）72 【题型 12】涂色问题 【例 9.12-1】（2000-1）用五种不同的颜色涂在图中四个区域里，每一个区域涂一种颜色 且相邻的区域涂不同颜色，则共有（ ）种不同的染色方法. （A）120 （B）180 （C）210 （D）300 （E）510 【例 9.12-2】如图，一环形花坛分成四块，现有 4 种不同的花供选种，要求每块里种一种 花，且相邻的 2 块种不同的花，则不同的种类总数为（ ）种. （A）96 （B）84 （C）60 （D）48 （E）36
【思路电拨】数字排队问题应注意以下几点： 1.处理方法：先分好类，再分步进行：先将特殊的数字排好，再将余下的数字进行排列. 2.基本原理：两个原理的应用.
【思路电拨】常见的涂色问题有以下两种形式： 类型 1：分步原理涂色 ；类型 2：环形涂色可套公式 ( 1) ( 1)( 1) k k
N s s .24 小时热线电话：4008-219-288
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【题型 13】环形排列问题 【例 9.13】把八人围成圆桌就餐，有（ ）种排法. （A）6！ （B）7！ （C）8！ （D）3！ （E）以上均不对 【题型 14】排座位问题 【例 9.14】（2008-1）有两排座位，前排 6 个座，后排 7 个座.若安排 2 人就坐，规定前 排中间两个座位不能坐，且此 2 人始终不能相邻而坐，则不同的坐法种数为（ ）. （A）92 （B）93 （C）94 （D）95 （E）96 【题型 15】多项式展开系数 【例 9.15】（2013-1）在 2 5
( 3 1) x x 的展开式中， 2
x 系数为（ ）. （A）5 （B）10 （C）45 （D）90 （E）95
【思路电拨】环形排列问题： n 个人的环形排列的方法数 ( 1)! n 种.
【思路电拨】将多项式展开系数问题看作排列组合问题进行分析. 24 小时热线电话：4008-219-288
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第十章 概率初步 第一节 考试要点剖析
一、基本概念
1.随机试验
设 E 为一试验,如果 E 满足以下条件，则称试验 E 是一个随机试验. （1）试验可以在相同的条件下重复地进行； （2）每次试验的可能结果不止一个,并且能事先明确试验的所有可能结果； （3）进行一次试验之前,不能确定哪一个结果会出现.
2.事件
随机试验的每一个可能结果,称为基本事件,而一个随机试验的所有基本事件所构成 的集合,称为样本空间,通常用字母 来表示,即 1 2, ,..., ,... nw w w 样本空间 中的 点 iw ,即基本事件（也称做样本点）. 常用大写英文字母 , ,A B C …表示事件,如设”掷两 枚均匀的硬币,至少一个正面向上”为事件 B .事件可分为以下三类： （1）必然事件：在一定条件下必然发生的事件叫做必然事件,如”在标准大气压下,水的温 度达到 100 o C 时沸腾”,就是必然事件. （2）不可能事件：在一定的条件下不可能发生的事件叫做不可能事件,如”在常温下,铁熔 化”, 就是不可能事件.记作“ ”（3）随机事件：在一定的条件下可能发生也可能不发生的事件叫做随机事件,如”掷一枚 硬币,正面向上”,就是随机事件.
【大纲考点】1.事件及其简单运算；2.古典概型；3.独立概型. 【命题剖析】1.掌握事件之间的关系及运算；2.根据概率的定义运用排列组合的基本方 法处理古典概型；3.理解独立事件的含义，掌握常考的独立概型，并能熟练应用伯努利 公式. 24 小时热线电话：4008-219-288
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随机事件的概率
在重复同一实验时,若进行了 n 次试验,事件 A 发生了 m 次,则称 mn 为事件 A 发生的 频率.在大量重复同一试验时,事件 A 发生的频率 mn 总是接近于某个常数,在它附近摆动, 就把这个常数叫做事件 A 的概率,记做 ( )P A ,显然 0 1P A .
4.事件的基本关系
在一个样本空间中,可以定义多个事件,在概率论中常要讨论不同事件之间的关系和运 算,为了直观,常用几何图形来表示事件,一般用某一个矩形区域表示样本空间,该区域中的 一个子区域表示某个事件.这样,便可以用集合的文氏图来表示事件的各种关系和运算法 则. 事件之间常见的有四种关系和三种运算: （1）包含关系: 如果事件 A 发生必然导致事件 B 发生,即事件 A 的所有样本点都包含在事件 B 中,则 称事件 B 包含是事件 A ,或称事件 A 包含于事件 B 中,记为 A B .（2）相等关系 如果 A B 与 B A 同时成立,即事件 A 与事件 B 具有完全相同的样本点,则称事件
A 与事件 B 相等,记做 A B .（3）对立关系 事件“非 A ”,表示事件 A 不发生,称为事件 A 的对立事件或逆事件,它是由样本空间 中所有不属于 A 的样本点构成的集合,记做 A ,显然 A A .（4）互斥关系 如果两事件 A 与 B 不可能同时发生,即事件 A 与 B 不包含公共样本点,则称事件 A
与 B 是互斥的.若 n 个事件是两两互斥的,则称这 n 个事件是互斥的.显然,对立事件必然是 互斥的,而互斥的事件不一定是对立. （5）事件的和(或并)运算 两个事件 A 和 B 中至少一个发生 ,称为事件 A 与 B 的和(或并),它是由属于 A 或属 于 B 的所有样本点构成的集合,记做 A B 或者 A B ,“ n 个事件 1 2, ,... nA A A 同时发生” 这一事件,称为 1 2, ,... nA A A 的和(或并),记做 1 2 ... nA A A (或简记为 1
nii
A
)（6）事件的积(或交)运算
两个事件 A 与 B 同时发生,称为事件 A 与 B 的积(或交),它是由既属
于 A 又属于 B 的所 24 小时热线电话：4008-219-288
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有公共样本点构成的集合,记做 A B (或简记为 AB ),“ n 个事件 1 2, ,... nA A A 同时发生” 这一事件,称为 1 2, ,... nA A A 的积(或交),记做 1 2 ... nA A A (或简记为 1 2, ,... nA A A 或
1
nii
A
). 显然,事件 A 与 B 互斥,即为 AB .（7） 事件的差运算 事件 A 发生而事件 B 不发生,称为事件 A 与事件 B 的差,它是由属于 A 而不属于 B 的那些样本点构成的集合,记做 A B ,显然 A A , A B A AB AB .
5.事件之间的运算规律:
（1） 交换律: A B B A ; A B B A
（2） 结合律: A B C A B C ; A B C A B C
（3） 分配律: A B C A B A C ; A B C A B A C
（4） 吸收律: A A B A ; A A B A
（5） 补元律: A A ; A A
（6） 德·摩根定律 A B A B , A B A B ,更一般的 1 1
nniiii
A A
,或者 1 1
nniiii
A A
.讨论事件之间的关系和运算是学习概率论的一个基本手段,要熟练掌握.在遇到实际问 题时,要认清问题所涉及的事件是什么,要善于用文字正确地写出所讨论的事件.另外,事件 的关系与运算跟集合的关系与运算十分相像,要学会用集合中的文氏图来理解事件之间的 关系.
事件的概率性质
（1） 0P ;（2） 对于任意的事件 A , 有 0 1P A ;
（3）有限可加性: 若 1 2, ,..., nA A A 两两互斥,则 24 小时热线电话：4008-219-288
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1 2 1 2... ... n nP A A A P A P A P A ;（4）逆事件的概率: 1P A P A ;（5）减法公式: 设 ,A B 是两个事件,则有 P A B P A P AB ;特别地, 当 B A 时, 有 P A B P A P B ,从而有 P B P A ,即概率具有单调性; （6）加法公式: ①设 ,A B 是两个事件,则有 P A B P A P B P AB ;②设 , ,A B C 是三个事件,则有
P A B C P A P B P C P AB P AC P BC P ABC
还可以推广到有限情况.
二、独立事件
设 A，B 是两个事件，如果事件 A 的发生与事件 B 的发生互不影响，则称这两个事件互 相独立. （1）独立事件 A、B 同时发生的概率： ( ) ( ) ( )P AB P A P B
（2）独立事件 A、B 至少有一个发生的概率： ( ) 1 ( ) ( )P A B P A P B
（3）独立事件 A、B 至多有一个发生的概率： ( ) 1 ( ) ( )P A B P A P B 24 小时热线电话：4008-219-288
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三、古典概型
一个随机实验 ,如果它满足以下的三个条件， 则称此随机实验 属于古典概率模型, 简称古典概型. (1)样本空间只含有限个基本事件； (2)每个基本事件发生的可能性都相同. 对于古典事件概型，如果样本空间 中的基本事件的总数是 n ，而事件 A 包含的基本 事件数为 m ，那么事件 A 的概率为
( ) mP A n
四、伯努利概型
伯努利试验（Bernoulli experiment）是在同样的条件下重复地、相互独立地进行的 一种随机试验，其特点是该随机试验只有两种可能结果：发生或者不发生。我们假设该项 试验独立重复地进行了 n 次，那么就称这一系列重复独立的随机试验为 n 重伯努利试验， 或称为伯努利概型. 在伯努利实验中， 设事件 A 发生的概率为 p ，则 n 重伯努利实验中事件 A 恰好发生 k 次
(0 )k n 的概率为 (1 ) 0,1, 2,..., k k n kn nP k C P P k n
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第二节 基础过关题型
一、独立事件求概率
【题型 1】独立事件求概率 【例 10.1-1】（2018）甲、乙两人进行围棋比赛，约定先胜 2 盘者赢得比赛.已知每盘棋 甲获胜的概率是 0.6，乙获胜的概率是 0.4，若乙在第一盘获胜，则甲赢得比赛的概率为 ( ). （A）0.144 （B）0.288 （C）0.36 （D）0.4 （E）0.6 【例 10.1-2】（2012-1）经统计，某机场的一个安检口每天中午办理安检手续的乘客人数 及相应的概率如下表：
乘客人数 0~5 6~10 11~15 16~20 21~25 25 以上
概率 0.1 0.2 0.2 0.25 0.2 0.05
该安检口 2 天中至少有 1 天中午办理安检手续的乘客人数超过 15 的概率是（ ）. （A）0.2 （B）0.25 （C）0.4 （D）0.5 （E）0.75 【例 10.1-3】（2000-1）某人忘记三位号码锁（每位均有 0 ～9 十个数码中的一个）的最 后一个数码，因此在正确拨出前两个数码后，只能随机地试拨最后一位数码，每拨一次算 做一次试开，则他在第 4 次试开时才将锁打开的概率是（ ）. （A） 14 （B） 15 （C） 25 （D） 110 （E）以上均不对
【思路电拨】1.独立事件同时发生的概率公式： ( ) ( ) ( )P AB P A P B
2.互斥事件有一个发生的概率公式： ( ) ( ) ( )P A B P A P B
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二、古典概型
【题型 2】古典概型之穷举法 【例 10.2-1】（2012-1）在一次商品促销活动中，主持人出示一个 9 位数，让顾客猜测商 品的价格，商品的价格是该 9 位数中从左到右相邻的 3 个数字组成的 3 位数，若主持人出 示的是 513535319，则顾客一次猜中价格的概率是（ ）. （A） 71 （B） 61 （C） 51 （D） 72 （E） 31
【例 10.2-2】.（2016）在分别标记了数字 1, 2,3, 4,5,6 的 6 张卡片中随机抽取 3 张，其中数 字之和等于 10 的概率是（ ）. （A）0.05 （B）0.1 （C）0.15 （D）0.2 （E）0.25 【例 10.2-3】（2001-10）从集合 0,1,3,5,7 中先任取一个数记为 a ，放回集合后再任取 一个数记为 b .若 0ax by 能表示一条直线，则该直线的斜率-1 的概率是（ ）. （A） 425 （B） 16 （C） 14 （D） 115 （E）以上结论均不对
【思路电拨】简单的古典概型计算分子时只需用穷举法即可，此外还要注意： 1.古典概型公式： ( ) mP A n
.2.常用正难则反策略. 24 小时热线电话：4008-219-288
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【题型 3】古典概型之排列组合公式 【例 10.3-1】（2007-10） 从含有 2 件次品、 2( 2) n n ＞ 件正品的 n 件产品中随机抽查 2 件，其中恰有 1 件次品的概率为 0.6. （1） 5n （2） 6n
【例 10.3-2】（2010-1）某商店举行店庆活动，顾客消费达到一定的数量后，可以在 4 种 赠品中随机选取 2 件不同的赠品，任意两位顾客所选的赠品中，恰有 1 件赠品相同的概 率是（ ）. （A） 16 （B） 14 （C） 13 （D） 12 （E） 23
【例 10.3-3】（2010-10） 在 10 道备选试题中，甲能答对 8 题，乙能答对 6 题.若某次考 试从这 10 道备选题中随机抽出 3 道作为考题，至少答对 2 题才算合格，则甲乙两人考 试都合格的概率是（ ）. （A） 28 45 （B） 23 （C） 14 15 （D） 26 45 （E） 815
【例 10.3-4】（2014-1）已知袋中装有红、黑、白三种颜色的球若干只，则红球最多. （1）随机取出的一球是白球的概率为 25
（2）随机取出的两球中至少有一只黑球的概率小于 1524 小时热线电话：4008-219-288
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【题型 4】古典概型之骰子问题 【例 10.4】（2008-10）若以连续掷两枚骰子分别得到的点数 a 和 b 作为点 M 的坐标，则 点 M 落入圆 2 2 18 x y 内（不含圆周）的概率是（ ）. （A） 736 （B） 29 （C） 14 （D） 518 （E） 11 36
【题型 5】古典概型之正方体涂漆问题 【例 10.5】（2001-1）将一块各面均涂有红漆的正方体锯成 125 个大小相同的小正方体， 从这些小正立方体中随机抽取一个，所取到的小正方体至少两个面涂有红漆的概率是 （ ）. （A）0.064 （B）0.216 （C）0.288 （D）0.352 （E）0.235
【思路电拨】将一个正方体六个面涂成红色，然后切成 3
n 个小正方体，则 1.3 面红色的小正方体有 8 个，位于原正方体的顶点上. 2.2 面红色的小正方体有 12( 2) n ，位于原正方体的 12 条棱上. 3.1 面红色的小正方体有 2
6( 2) n ，位于原正方体的 6 个面上. 4.没有红色的小正方体有 3
(2) n
，位于原正方体的内部. 24 小时热线电话：4008-219-288
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三、伯努利概型
【题型 6】伯努利概型 【例 10.6-1】（1998-10）掷一枚不均匀的硬币，正面向上的概率为 23 ，若将次硬币投掷 4 次，则正面向上 3 次的概率是（ ）. （A） 881 （B） 827 （C） 32 81 （D） 12 （E） 26 27
【例 10.6-2】（1999-1）进行一系列独立的实验，每次实验成功的概率为 p ，则在成功 2 次前已经失败 3 次的概率为（ ）（A） 2 3
4 (1 )p p （B） 3
4 (1 )p p （C） 2 3
10 (1 )p p
（D） 2 3
(1 )p p (E) 3
(1 )p
四、几何概型
【题型 7】几何概型 【例 10.7】(2014-10)在矩形 ABCD 的边 CD 上随机取一点 P ,使得 AB 是 APB 的最大 边的概率大于 .21
（1） .47
AB AD （2） .21
AB AD
【思路电拨】 n 次独立重复实验恰好发生 k 次的概率可用 ( ) (1 )k k kn nP k C p p 24 小时热线电话：4008-219-288
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第十一章 数据描述 第一节 考试要点剖析 一、基本概念
1.平均值 设一组样本数据 1 2, , ,nx x x 称 1 2 nx x xx n
为这 n 个数据的平均值. 特别地：样本均值也可以记为 ( )x E X ，并且满足线性关系，即
( ) ( )E aX b aE X b .平均值反应的是这组数据的整体水平. 2.方差 设一组样本数据 1 2, , ,nx x x 其平均值为 x ，则称
222212
( ) ( ) ( )nx x x x x xs n
为这 n 个样本的方差. 特别地：样本方差也可以记为 2 ( )s D X ，并且满足 2
( ) ( )D aX b a D X .3.标准差 因为方差与原始数据的单位不同，且平方后可能扩大了离差的程度，将方差的算术平 方根分称为这组数据的标准差，即
22212
( ) ( ) ( )nx x x x x xs n
.特别地：样本方差与标准差反应的是这组数据的波动程度，也称为离散程度.方差越 大说明波动越大数据越不稳定，反之方差越小说明波动越小数据越稳定.
4.众数：出现次数最多的数称之为众数.
【大纲考点】1.平均值； 2.方差和标准差； 3.数据的图表表示 （直方图， 饼形图， 数表） .【命题剖析】1.掌握平均值、方差和标准差的计算公式并理解他们的实际意义；2、掌 握直方图等在数据描述中的含义. 24 小时热线电话：4008-219-288
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5.中位数 将数据由小到大排列，若有奇数个数据，则正中间的数为中位数;若有偶数个数据， 则中间两个数的平均值为中位数. 6.极差 一组数据中最大值与最小值的差.
二、图表
1.直方图 把数据分为若干个小组，每组的组距保持一致，并在直角坐标系的横轴上标出每组 的位置（以组距作为底），计算每组所包含的数据个数（频数），以该组的“频率/组距” 为高作矩形，这样得出若干个矩形构成的图叫做直方图. 特别地：众数是最高矩形底边中点的横坐标；中位数左边和右边的直方图的面积相 等；平均数是直方图的重心，它等于每个小矩形的面积乘以小矩形底边中点横坐标之和. 2.饼形图 饼形图是一个划分为几个扇形的圆形统计表，用于描述量、频率或者百分比之间的 相对关系.一般应用某部分所占的百分比等于对应扇形所占整个圆的比例加以计算. 24 小时热线电话：4008-219-288
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第二节 基础过关题型
【题型 1】平均数与方差计算 【例 11.1-1】（2010-10） 某学生在军训时进行打靶测试，共射击 10 次.他的第 6、7、 8、9 次射击分别射中 9.0 环、8.4 环、8.1 环、9.3 环，他的前 9 次射击的平均环数高 于前 5 次的平均环数.若要使 10 次射击的平均环数超过 8.8 环， 则他第 10 次射击至少 应该射中（ ）环.（报靶成绩精确到 0.1 环） （A）9.0 （B）9.2 （C）9.4 （D）9.5 （E）9.9 【例 11.1-2】（2019）10 名同学的语文和数学成绩如表： 语文和数学均值分别为 1E 和 2E ，标准差分别是 1
和 2
，则（ ）. （A） 1 2E E＞ ， 1 2
＞ （B） 1 2E E＞ ， 1 2
＜ （C） 1 2E E＞ ， 1 2
=
（D） 1 2E E＜ ， 1 2
＞ （E） 1 2E E＜ ， 1 2
＜
语文成绩 90 92 94 88 86 95 87 89 91 93
数学成绩 94 88 96 93 90 85 84 80 82 98
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【思路点拨】若有 n 个数据，分别为 1 2, 3, , , , nx x x x ，则 1. 平均数 1 2 3 nx x x xx n
，是表示一组数据集中趋势的量数，显然一组 数据不能整体都大于平均数，亦不可能整体都小于平均数 .
方差：在计算完平均数 x 的基础上， 方差
22222123
( ) ( ) ( ) ( )nx x x x x x x xS n
，方差反应数据的波动情 况，方差越大，数据波动性大（越不稳定）；反之，方差越小，数据波动性小（越 稳定）.基于上述原因，在比较两组数据方差大小时，可以利用直接观察法. 24 小时热线电话：4008-219-288
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【例 11.1-3】（2014-1）已知 , , , ,M a b c d e 是一个整数集合 .则能确定集合 M .
（1） , , , ,a b c d e 的平均值为 10
（2） , , , ,a b c d e 的方差为 2
【例 11.1-4】（2017）甲、乙、丙三人每轮各投篮 10 次，投了三轮，投中数如下表：
第一轮 第二轮 第三轮
甲 2 5 8
乙 5 2 5
丙 8 4 9
记 1 2 3, ,
分别为甲、乙、丙投中数的方差，则（ ）. （A） 1 2 3
（B） 1 3 2
（C） 2 1 3
（D） 2 3 1
（E） 3 2 1
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【题型 2】平均值与方差的变化规律 【例 11.2-1】 若样本 12 1 x ， 12 2 x ，……， 12 nx 的平均数为 11，方差为 16，则样本
21 x ， 22 x ，……， 2nx 的平均数及方差分别为（ ）. （A）10,2 （B）11,3 （C）11,2 （D）12,4 （E）7,4 【例 11.2-2】 设两组数据 1 : 3, 4,5,6,7 S 和 2 : 4,5,6,7, S a ，则能确定 a 的值. （1） 1S 与 2S 的均值相等 （2） 1S 与 2S 的方差相等
【思路点拨】若原有 n 个数据 1 2, 3, , , , nx x x x 的平均数为为 x ，方差为 2
S ， 则 1.平均数： ①新数据： 1 2, 3, , , , nax ax ax ax 的平均数为 ax ；②新数据： 1 2 3 ,, , , , nax b ax b ax b ax b 的平均数为 ax b ；2.方差： ①新数据： 1 2, 3, , , , nax ax ax ax 的方差为 2 2
a S ；②新数据： 1 2 3 ,, , , , nax b ax b ax b ax b 的方差为 2 2
a S ；【注意】若一组数据同步增长得到新数据，则新数据方差等于原数据方差，即同 步增长，方差不变. 24 小时热线电话：4008-219-288
交大安泰天使基金定向扶持企业（代码：Z17157681）
153
【题型 3】饼图 【例 11.3】某小区退休工人业余爱好的统计图如图所示，则 25 m .（1）共有 60 人，喜欢各项体育项目的人数极差是 m 名（2）喜欢太极拳的有 75 人，喜欢羽毛球的有 m 人【题型 4】直方图 【例 11.4】某 棉 纺 厂 为 了 了 解 一 批 棉 花 的 质 量 ，从 中 随 机 抽 取 了 100 根 棉 花 纤 维 的 长 度（ 棉 花 纤 维 的 长 度 是 棉 花 质 量 的 重 要 指 标 ），所 得 数 据 都 在 区 间 [5，40] 之 间 ， 其 频 率 分 布 直 方 图 如 图 所 示 ， 则 抽 取 的 100 根 中 ， 有 （ ） 根 在 棉 花 纤
维 长 度 小 于 20 mm .
(A)20 （ B ） 25 （ C ） 26 （ D） 28 （ E） 30
【题型 5】其它统计图 【例 11.5】（2019）某影城统计了一季度的观众人数，如图，则一季度的男士观众人数与
女士观众之比为（ ）. （A）3:4 （B） 5:6 （C） 12:13 （D） 13:12 （E） 4:3
太极拳
50%
其它
羽毛球
60 °
90 °乒乓球 |
188195 | https://www.youtube.com/watch?v=MuX7T4PM1Mc | Can you evaluate a log for a negative number
Brian McLogan
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👉 Learn all about the properties of logarithms. The logarithm of a number say a to the base of another number say b is a number say n which when raised as a power of b gives a. (i.e. log [base b] (a) = n means that b^n = a). The logarithm of a negative number is not defined. (i.e. it is not possible to find the logarithm of a negative number). There are some properties of logarithm which are helpful in simplifying and evaluating logarithm problems. Knowledge of these properties is a valuable tool in solving logarithm problems.
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welcome so what I like to do is just show you how to evaluate log base 3 of negative you're not going to get a lot of these examples coming up um because once you once you kind of go through this example you'll be able to see that uh um it's very kind of basic to be able to under U understand how to solve any of these when you're going to have try to evaluate a logarithm for a negative number so the best thing I can do is actually I'm going to take this expression and actually rewrite it as an equation so I can rewrite this in exponential form and what this says is 3 to x = 9 so what we want to do is we're trying to find we're trying to find a value that I can raise my base 32 to equal a negative number and the thing is this is going to be impossible so there is no solution for this so let's just go and take a look at see you know why is that going to be the case well we know that 3^2 equal POS 9 right so we could say all right well what else what about if we did -2 then so if I did 3 to the2 power well remember by using your negative um negative exponent that just represents 3 or 1 over 3^2 which is just going to be 1 nth so again that's not going to produce a negative number and then maybe you might want to think of you know what about 3 to the 1/2 what if I made it a fraction well remember by using your rational exponent that just represents the square root of three so either way if you're thinking about this when you're trying to produce multiplication and you have your base is positive POS it's never going to produce multip multiplying a number by itself is never going to produce itself into a negative number therefore this example or this expression all right is going to have no solution there you go |
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188197 | https://www.nature.com/articles/s41467-022-33110-5 | Tubular cell polyploidy protects from lethal acute kidney injury but promotes consequent chronic kidney disease | Nature Communications
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Tubular cell polyploidy protects from lethal acute kidney injury but promotes consequent chronic kidney disease
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Published: 04 October 2022
Tubular cell polyploidy protects from lethal acute kidney injury but promotes consequent chronic kidney disease
Letizia De Chiara1,
Carolina Conte1,
Roberto Semeraro2,
Paula Diaz-Bulnes3,
Maria Lucia Angelotti1,
Benedetta Mazzinghi4,
Alice Molli1,4,
Giulia Antonelli1,
Samuela LandiniORCID: orcid.org/0000-0001-7423-03255,
Maria Elena Melica1,
Anna Julie PeiredORCID: orcid.org/0000-0001-9732-13691,
Laura MaggiORCID: orcid.org/0000-0003-2862-95912,
Marta Donati1,
Gilda La Regina1,
Marco Allinovi6,
Fiammetta RavagliaORCID: orcid.org/0000-0001-7121-23247,
Daniele Guasti8,
Daniele BaniORCID: orcid.org/0000-0001-6302-901X8,
Luigi CirilloORCID: orcid.org/0000-0002-0440-73561,4,
Francesca Becherucci4,
Francesco GuzziORCID: orcid.org/0000-0003-0788-50787,
Alberto Magi9,
Francesco AnnunziatoORCID: orcid.org/0000-0001-8798-75892,10,
Laura Lasagni1,
Hans-Joachim Anders11,
Elena LazzeriORCID: orcid.org/0000-0002-9505-21151na1&
…
Paola RomagnaniORCID: orcid.org/0000-0002-1774-80881,4na1
Show authors
Nature Communicationsvolume 13, Article number:5805 (2022) Cite this article
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Abstract
Acute kidney injury (AKI) is frequent, often fatal and, for lack of specific therapies, can leave survivors with chronic kidney disease (CKD). We characterize the distribution of tubular cells (TC) undergoing polyploidy along AKI by DNA content analysis and single cell RNA-sequencing. Furthermore, we study the functional roles of polyploidization using transgenic models and drug interventions. We identify YAP1-driven TC polyploidization outside the site of injury as a rapid way to sustain residual kidney function early during AKI. This survival mechanism comes at the cost of senescence of polyploid TC promoting interstitial fibrosis and CKD in AKI survivors. However, targeting TC polyploidization after the early AKI phase can prevent AKI-CKD transition without influencing AKI lethality. Senolytic treatment prevents CKD by blocking repeated TC polyploidization cycles. These results revise the current pathophysiological concept of how the kidney responds to acute injury and identify a novel druggable target to improve prognosis in AKI survivors.
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Acute kidney injury
Article 15 July 2021
Introduction
Acute kidney injury (AKI) is a syndrome characterized by an acute deterioration of kidney function impacting 20.0-31.7% of hospitalized patients and 8.3% of outpatients1. With no specific medical treatment available, AKI is potentially lethal and hence a global health concern1. In addition, AKI survivors face a high risk of chronic kidney disease (CKD)2, referred to as “AKI-CKD transition” that is associated with a high cardiovascular risk and the likelihood of progression toward end stage kidney disease (ESKD)3,4,5. The traditional concept of kidney function recovery after AKI assumes a widespread proliferative capacity and regeneration of injured tubular epithelial cells (TC) based on diffuse positivity for cell cycle markers in the kidney, which is, however, inconsistent with the high incidence of CKD after AKI6,7. We recently reported that these cell cycle markers do not faithfully indicate cell proliferation but rather cell cycle entry to initiate polyploidization in TC in response to AKI8. Polyploidization is the process that leads a normally diploid cell to acquire additional set(s) of chromosomes9. Most mammalian cells are diploid, but hepatocytes and cardiomyocytes become progressively polyploid during lifetime via an alternative cell cycle named endoreplication10,11,12,13. In contrast to hepatocytes and cardiomyocytes10,11,12,13, TC remain mostly mononuclear after AKI, explaining why this phenomenon had remained undiscovered8. In this study, we aimed to investigate the biological relevance and the structural consequences of TC polyploidization in the kidney, which are currently unknown. We hypothesized that polyploidy in TC is an evolutionary conserved mechanism developed to permit a quick recovery of kidney function after damage assuring survival. However, life-preserving stress responses frequently drive long-term organ failure, as shown in other organs14,15,16. Therefore, we carefully assessed long-term outcomes. To do so, we employed numerous conditional transgenic mouse lines under the control of the Pax8 promoter, which is consistently and exclusively8,17 expressed by renal TC, to achieve targeted recombination in all the tubular segments.
In this study, we demonstrate that during AKI, YAP1-driven polyploidization of TC acts as an immediate compensatory mechanism to augment residual kidney function and to avoid early death for kidney failure. However, as a trade-off, TC polyploidy promotes TC senescence and progressive interstitial fibrosis, i.e., AKI-CKD transition. Importantly, delayed blockade of YAP1-driven polyploidization can avoid the development of CKD and improve long-term outcome in those that survive the early injury phase. These findings identify a previously unrecognized life-preserving type of cellular response, a Janus-faced role of cell polyploidization in AKI and a novel drug target to improve long-term prognosis in survivors of the acute phase.
Results
After AKI, most TC enter the cell cycle and either die or become polyploid
To identify ploidy in mononuclear TC, we combined measurement of the DNA content with assessment of cell cycle phase in Pax8/FUCCI2aR mice by flow cytometry8 (Fig.1a, b and Supplementary Fig.1a–e). Ischemic AKI induced a strong but short-lasting entry into the cell cycle in 39.8 ± 12.5% of TC at day 2 that became mVenus+ or mCherry+mVenus+ (Fig.1a, b). At this stage, it was impossible to establish how many of these cycling TC would complete division, endoreplication or eventually die. At day 3 after AKI, the percentage of cells in the G1 phase (i.e., mCherry+ with a 2C DNA content, defined as non-cycling TC) had remained stable in comparison to day 2, whereas the percentage of cycling cells (i.e., mCherry+mVenus+ with DNA content = 2C and mVenus+ cells up to 4C DNA content) had decreased from 39.8 ± 12.5% to 8.8 ± 4.7%. This suggested that those TC that enter the S phase of the cell cycle at day 2, at day 3 either: (1) keep cycling (2) die or (3) become polyploid via endoreplication. Indeed, considering this distribution as 100%, we found that at day 3, 17 ± 2.4% of TC (i.e., 38.8 ± 9.2% of the cells that were cycling at day 2) had acquired ≥4C DNA content (i.e., had become mCherry+, mCherry+mVenus+ with a 4C DNA content or mVenus+ with a DNA content ≥8C), suggesting one or even multiple endoreplication cycles (Fig.1a–c). By day 5, the percentage of cycling cells was reduced to 1.1 ± 0.6% remaining consistently low till day 30 (10.7 ± 8.7%) (Fig.1b). Polyploid TC decreased from 17 ± 2.4% at day 3 to 10.9 ± 1.6% of total TC at day 5, remaining stable thereafter and persisting until day 30 (12.6 ± 2.3%) (Fig.1a, b). Of note, virtually all dying cells at the different time points were cycling cells (i.e., mVenus+ or mCherry+mVenus+) based on the FUCCI2aR reporter, suggesting that TC death occurred during the S or G2/M phase of the cell cycle (Fig.1d). Consistently, 19.6 ± 13.7% of TC (i.e., 44.4 ± 15.8% of the TC that were cycling at day 2), died between day 2 and 3 (Fig.1a, b) and were detectable as dead TC in S or G2/M phase by flow cytometry and confocal microscopy in kidney tissue as well as lost TC in the urine (Fig.1d–h). Taken together these results show that, in the early phase of AKI, most TC that start DNA synthesis die and are lost into the urine, while most surviving TC undergo polyploidization within the remaining tubules.
Fig. 1: The majority of tubular cells (TC) entering cell cycle becomes polyploid or dies following AKI.
a FACS plots of TC in Pax8/FUCCI2aR mice (n = 6) after IRI, showing dead (<2C), diploid (2C), tetraploid (4C) and octaploid or greater (≥8C) TC. Colours match those of the FUCCI2aR reporter. b FUCCI2aR TC distribution after IRI. Dead TC: t0 vs t3 p = 0.015, t0 vs t30 p = 0.008; t0 vs t5 p = 0.015, cycling TC: t0 vs t2 p = 0.002, t2 vs t3 p = 0.002, t2 vs t5 p = 0.002, t2 vs t30 p = 0.015; Polyploid TC: t0 vs t3 p = 0.002, t0 vs t5 p = 0.026, t0 vs t30 p = 0.008, t2 vs t3 p = 0.002, t3 vs t5 p = 0.004, t3 vs t30 p = 0.026 (n = 6 each time point). c Distribution of polyploid TC across cell cycle after IRI (n = 6). d Distribution of FUCCI2aR dead TC analysed by FACS after IRI (n = 6). Representative pictures of e healthy and f two days after IRI Pax8/FUCCI2aR mice showing necrotic tubu les () with mVenus+ TC. DAPI counterstains nuclei. Bars 75 µm. The inlet is a higher magnification of the tubule indicated by the arrow. g FACS analysis of cells from the urine of Pax8/FUCCI2aR mice after IRI. h Distribution across cell cycle of FUCCI2aR TC from the urine of healthy and after IRI mice (multiple urine samples pulled together, n = 6 mice for each time point). i, j Heterozygous Pax8/Confetti mice 30 days after IRI. Bars 20 µm. k 3D reconstruction of cortical tubules. Bar 20 µm. Arrows indicate bi-coloured TC. AQP1 staining in heterozygous Pax8/Confetti mice 30 days after IRI, arrows show bi-coloured TC in a representative l S1 segment and m S2 segment. Bars 20 µm. n Transmission electron microscopy of a binucleated polyploid TC 30 days after IRI. Bar 2 µm. o Bi-coloured TC distribution in tubular segments of heterozygous Pax8/Confetti mice after IRI (n = 9). t0: healthy, t2: day 2 after IRI, t3: day 3 after IRI, t5: day 5 after IRI, t30: day 30 after IRI. IRI ischemia reperfusion injury. S1: S1 segment of proximal tubule, S2: S2 segment of proximal tubule, S3: S3 segment of proximal tubule, DST distal straight tubule, TALH thick ascending limb of loop of Henle, CD collecting duct. AQP1 Aquaporin-1. Statistical significance was calculated by two-sided Mann-Whitney test; numbers on graphs represent exact p values. Data are expressed as mean ± SEM in graph 1b and 1h. Box-and-whisker plots: line = median, box = 25–75%, whiskers = outlier (coef. 1.5).
Full size image
TC polyploidization occurs distant from the site of injury
The FUCCI2aR reporter is a ubiquitin-based system that permits detection of the cell cycle state of polyploid TC at a specific moment but neither their permanent tracing, underestimating the phenomenon, nor their localization. Indeed, ischemic necrosis affects specifically S3 segments of the proximal tubules in the outer medulla raising the question why AKI involves widespread cell cycle marker positivity all over the cortex (Supplementary Fig.2a, b)18,19. To effectively localize polyploid TC to different tubular segments, we induced AKI in heterozygous Pax8/Confetti mice. As shown in the Supplementary Fig.2c–g, in these mice the diploid cells recombine only one fluorochrome appearing as mono-coloured cells, while polyploid cells with 4C DNA content recombine two fluorochromes, appearing as bi-coloured or mono-coloured cells20. Indeed, when analysing the three fluorochromes that have the same probability of recombination (RFP, CFP and YFP), we can calculate that 2/3 of polyploid cells will recombine two different colours appearing bi-coloured and 1/3 will recombine the same colour twice, appearing mono-coloured, resulting undistinguishable from diploid cells (see Supplementary Fig.2c–g). Bi-coloured polyploid TC localized mostly in the cortex (13.3 ± 2.5% over total coloured TC in the cortex vs 2.7 ± 0.4% in the outer stripe of outer medulla, Fig.1i–m and Supplementary Fig.2h–j), representing a 2.2-fold increase upon AKI in comparison to sham control (13.3 ± 2.5% vs 5.9 ± 0.7%). Bi-coloured polyploid TC in the cortex were not only mononucleated but also sporadically binucleated, as confirmed by electron microscopy (Fig.1n).
To better dissect polyploid TC distribution in the various portions of the nephron, we then performed immunofluorescence staining for: 1. Aquaporin-1 (AQP1) to identify the various segments of the proximal tubule (namely S1, S2 and S3) and the thin descending limb of the Henle’s loop; 2. Tamm Horsfall protein (THP) to identify the thick ascending limb of the Henle’s loop and the distal straight tubule; 3. Aquaporin-2 (AQP2) to identify the collecting ducts (Fig.1l–o, Supplementary Fig.2k–m). S1, S2 and S3 segments were distinguished based on their morphology and localization in distinct kidney areas21. Bi-coloured polyploid TC mostly localized to S1 and S2 segments of the cortex sparing the injured outer medulla (Fig.1l–o, Supplementary Fig.2k–m). Collectively, diffuse TC polyploidization distant from the injury site suggests a functional role beyond structural replacement of TC lost in the injured outer medulla.
Polyploid TC arise in a YAP1-related manner
To analyse TC polyploidization, we performed single cell RNA-sequencing (scRNA-seq) on primary cultures of human proximal tubular epithelial cells (hPTC), which we have found to contain a fraction of polyploid hPTC (Supplementary Fig.3a–d). Unsupervised clustering revealed eleven clusters with distinct expression patterns (Supplementary Fig.3e, f and Supplementary Table1). Five clusters were enriched in genes previously associated with different phases of endoreplication-mediated polyploidy: 1. Ribosome biogenesis (cluster 10)22,23,24; 2. Hypertrophy, DNA synthesis and chromatin decondensation (clusters 4, 5, 7)25,26; 3. Hippo pathway effector YAP1 targets (cluster 9)9,27,28 (Supplementary Table1). In particular, clusters 10, 5, 4, 7 displayed a pattern reported in other organs as characteristic of early, and cluster 9 of late, hypertrophy phases (Fig.2a)25,26. A trajectory analysis along pseudotime suggested that hPTC polyploidization started with increased RNA synthesis and ribosome biogenesis, followed by entry and progression through the cell cycle (Fig.2b, c). Subsequently, the cells transcribed YAP1 mRNA followed by expression of the endoreplication-specific regulators E2F1, E2F7 and E2F89 and YAP1 target genes, particularly CTGF28,29 (Fig.2c), while TAZ mRNA expression was not altered (Supplementary Fig.3g). Sorting of mCherry-hPTC (i.e., hPTC transduced with a mCherry-G1 vector) with ≥4C DNA content (polyploid hPTC) showed enrichment for the YAP1 downstream target CTGF and the SEMA5A30 gene, characteristic of hypertrophic cluster 9 (Fig.2d–f and Supplementary Fig.3h–j). Blocking YAP1 nuclear translocation with verteporfin31 significantly decreased the percentage of polyploid mCherry-hPTC further suggesting that YAP1 controls hPTC polyploidy in vitro (Fig.2g, h and Supplementary Fig.3k). However, as verteporfin blocks nuclear translocation also of the YAP1 paralog TAZ to some extent (Supplementary Fig.3l), we silenced YAP1 and TAZ independently in mCherry-hPTC. These experiments confirmed that only YAP1 significantly reduced mCherry-hPTC polyploidy (Supplementary Fig.3m–q). Next, we performed scRNA-seq analysis on healthy mouse kidney and on kidneys at days 2 and 30 after ischemic injury (Supplementary Fig.4a, b). We identified and further analysed proximal (P) TC on day 2 and 30 after injury, when polyploidization occurs (Fig.3a, b, Supplementary Fig.4c–e and Supplementary Table2). In vivo, scRNA-seq analysis on PTC at day 2 after AKI revealed high levels of ribosome biogenesis and enrichment in hypertrophy genes, culminating in cluster 8 (Fig.3c, d). More importantly, in vivo cluster 8 closely resembled the signature already observed in polyploid hPTC in vitro (Supplementary Tables1, 2, highlighted genes) and was enriched with endoreplication-specific regulators E2f1, E2f7 and E2f8 and YAP1 target genes9,28,29 (Fig.3e–g). A trajectory analysis along pseudotime showed similar results to those obtained in vitro with increased ribosome biogenesis, followed by entry and progression through the cell cycle, culminating in the expression of endoreplication-specific regulators E2f1, E2f7, E2f8 and YAP1 target genes (Fig.3h, i). The closely related cluster 9 from kidneys at day 30 after AKI was also enriched in hypertrophy genes (Fig.3g). Interestingly, increased ribosomal synthesis was accompanied by a progressive increase of transcriptome abundance along the trajectory in endoreplicating clusters, confirming that polyploid cells are hypertrophic cells (Fig.3j, k). Protein analysis showed that nuclear translocation of YAP1 started immediately after AKI and persisted until day 30 (Supplementary Fig.5a, b). Conversely, TAZ expression was not significantly altered following AKI (Supplementary Fig.5c, d). These results demonstrate that a subset of PTC after AKI shows polyploidization-related hypertrophy signature and that PTC undergo polyploidization in a YAP1-related manner.
Fig. 2: Polyploidization of human proximal tubular cells (hPTC) is controlled by YAP1.
a Matrix plot of hPTC clusters showing hypertrophy genes. b Monocle2 pseudotime trajectory of hPTC clusters and “facet” trajectory (right) showing cluster distribution in order of appearance. c Monocle2 trajectory of hPTC coloured by pseudotime, mean ribosome gene expression difference, cell cycle phases, YAP1 mRNA accumulation, E2F-markers (E2F1, E2F7, E2F8) and YAP1-targets (ANKDR1, BIRC5, CTGF, CDK1, CCNB1, AKT1). d Cell cycle distribution of mCherry-hPTC. e Gene expression of sorted polyploid mCherry-hPTC over diploid mCherry-hPTC (n = 4). f Matrix plot of cluster 9 characteristic genes. g Cell cycle distribution of mCherry-hPTC treated with DMSO or with verteporfin (VP). A representative experiment out of 6 is shown. h Percentage of polyploid mCherry-hPTC in DMSO-treated or VP-treated culture (n = 6). Statistical significance was calculated by two-sided Mann-Whitney test; numbers on graphs represent exact p values. Box-and-whisker plots: line = median, box = 25–75%, whiskers = outlier (coef. 1.5).
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Fig. 3: Polyploidization of proximal tubular cells (PTC) after AKI in mice is controlled by YAP1.
a UMAP of experimental time distribution and b cluster distribution of mouse PTC at day 2 (t2) and 30 (t30) after IRI. c UMAP showing ribosome gene expression difference and d hypertrophy genes (Top2a, Mcm7, Ccna2, Ccnb2, Tmsb10, S100a11). e UMAP showing E2F-markers (E2f1, E2f7, E2f8) and f YAP1-targets (Ankdr1, Birc5, Ctgf, Cdk1, Ccnb1, Akt1) in PTC. g Matrix plot of PTC at 2 and 30 days after IRI, showing hypertrophy genes. h Monocle2 trajectory of PTC clusters and trajectory of PTC coloured by pseudotime, mean ribosome gene expression difference, cell cycle phases, E2F-markers (E2f1, E2f7, E2f8) and YAP1-targets (Ankdr1, Birc5, Ctgf, Cdk1, Ccnb1, Akt1). i Monocle2 “facet” trajectory showing cluster distribution in order of appearance. j Violin plots showing ribosome distribution in clusters grouped by ribosome gene expression. Median distribution is represented by the white dot; the black bar in the centre of the violin represents the interquartile range between the first and third quartile; the black lines stretched from the bar represent the lower/upper adjacent values defined as first interquartile −1.5 and third interquartile +1.5, respectively. k Bar plot showing the binned normalized counts distribution in each cluster group. IRI ischemia reperfusion injury.
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YAP1 controls TC polyploidization through E2F7, E2F8 and AKT1
We next tried to establish the mechanism through which YAP1 triggers polyploidy in TC. Firstly, we screened the genes previously shown to be involved in endoreplication9,27 and enriched in the polyploid clusters that emerged from the scRNA-seq analysis in vitro and in vivo. Upon verteporfin treatment, E2F7, E2F8 and AKT1 were shown to be the most strongly downregulated in mCherry-hPTC (Fig.4a), suggesting they may be directly regulated by YAP1 activation. As verteporfin partially affects TAZ localization along with YAP1, we then confirmed that only YAP1 knock-down and not TAZ knock-down successfully downregulated E2F7, E2F8 and AKT1 in mCherry-hPTC (Fig.4b). Importantly, chromatin immunoprecipitation (ChIP) assay proved that YAP1 directly binds to the promoters and activates all of them, with E2F8 showing the strongest enrichment, as well as CTGF which was used as an internal control29 (Fig.4c–f). Accordingly, silencing of E2F7, E2F8 and AKT1 resulted in a significant downregulation of polyploidy in mCherry-hPTC, replicating the results obtained with YAP1 silencing (Fig.4g–i). Collectively, these results demonstrate that YAP1 controls polyploidization of TC via E2F7, E2F8 and AKT1.
Fig. 4: YAP1 controls polyploidization via AKT1, E2F7 and E2F8 activation.
a qRT-PCR analysis of YAP1 targets in DMSO and verteporfin (VP) treated mCherry-hPTC (n = 4). CTGF was used as a positive control. YAP1 was used as a negative control. # Significance between DMSO and VP treated mCherry-hPTC for each gene analysed, p = 0.028. b qRT-PCR quantification of E2F7, E2F8 and AKT1 in scramble, YAP1, and TAZ knock-down (KD) mCherry-hPTC (n = 4). # Significance between scramble and YAP1-KD mCherry-hPTC for each gene analysed, p = 0.028. Scramble and TAZ-KD conditions are not significantly different. Chromatin immunoprecipitation assay showing YAP1 binding on c CTGF, d E2F7, e E2F8 and f AKT1 promoters in DMSO and VP treated mCherry-hPTC (n = 4). #Significance between DMSO and VP treated mCherry-hPTC, p = 0.028. g KD efficiency of E2F7, E2F8 and AKT1 GapmeRs (n = 4). #Significance between scramble and KD mCherry-hPTC, p = 0.028. h Cell cycle distribution of mCherry-hPTC transfected with scramble, E2F7, E2F8 and AKT1 GapmeRs. A representative experiment out of 4 is shown. i Percentage of polyploid mCherry-hPTC in scramble-treated, E2F7-KD, E2F8-KD and AKT1-KD cultures (n = 4). # Significance between scramble and KD mCherry-hPTC, p = 0.028. hPTC: human proximal tubular cells. Statistical significance was calculated by two-sided Mann-Whitney test. Box-and-whisker plots: line = median, box = 25–75%, whiskers = outlier (coef. 1.5). Bar plots: line = mean, whisker = outlier (coef. 1.5).
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TC polyploidization preserves kidney function assuring survival after AKI
To determine the functional role of YAP1-dependent polyploidization, we used mice with a TC-specific YAP1 depletion upon exposure to doxycycline (Pax8/YAP1 ko,Supplementary Fig.6a–h). YAP1 was successfully knocked-out in TC upon doxycycline-induced recombination (Supplementary Fig.6i, j) while TAZ expression was not affected (Supplementary Fig.6k, l). These mice developed less cycling TC two days after unilateral ischemia reperfusion injury (ischemic injury) and less polyploid TC at day 3 in comparison to Pax8/WT mice (Fig.5a–c) and showed a more pronounced decline of kidney function at day 2 (Fig.5d). However, the unilateral ischemic model cannot truly evaluate the role of polyploidization on kidney function recovery due to the confounding effect of the contralateral kidney during compensation. To address this point, we used a model of nephrotoxic AKI with bilateral kidney involvement. Pax8/YAP1 ko mice showed a lower percentage of cycling TC and of polyploid TC after AKI (Fig.5e–g), while TC death and number were comparable (Fig.5h, i). We confirmed that only YAP1 upregulation and not TAZ was impaired in Pax8/YAP1 ko mice 2 days after nephrotoxic injury in comparison to Pax8/WT mice (Supplementary Fig.6m–p). Importantly, this model identified YAP1-driven TC polyploidy as a survival mechanism protecting from hyperkalemia and uremic death (Fig.5j–l). Indeed, at 48 h, the injured kidneys of Pax8/YAP1 ko mice, as well as the TC, were smaller compared to WT mice (Fig.5m–o), consistent with the concept that hypertrophy of cortical TC enlarges kidney dimensions during AKI, a phenomenon well-known clinically. Considering that Pax8 is selectively expressed in TC and that TC death and number were comparable between WT and KO mice, the reduced survival of Pax8/YAP1 ko mice can only be related to a defect in polyploidization-mediated hypertrophy of TC. Altogether, these data show that YAP1-driven polyploidization-dependent hypertrophy of cortical TC is a previously unknown protecting mechanism of immediate cellular adaptation that sustains residual kidney function during the early and potentially fatal phase of AKI.
Fig. 5: Early tubular cell (TC) polyploidization preserves residual kidney function and assures survival during AKI.
a FACS plots of Pax8/FUCCI2aR (Pax8/WT) (n = 6) and Pax8/FUCCI2aR/YAP1ko (Pax8/YAP1 ko) (n = 6) showing diploid (2C), tetraploid (4C) and octaploid or greater (≥8C) TC. Colours match the FUCCI2aR reporter. b Percentage of cycling TC in healthy (t0) mice and after IRI (n = 6). c Percentage of polyploid TC in healthy (t0) mice and after IRI (n = 6). (§) Significance within Pax8/WT mice: t0 vs t2 p = 0.015, t0 vs t3 p = 0.002, t0 vs t5 p = 0.002, t2 vs t3 p = 0.002, t2 vs t5 p = 0.026, t3 vs t5 p = 0.002. (†) Significance within Pax8/YAP1 ko mice t0 vs t2 p = 0.002, t0 vs t3 p = 0.002, t2 vs t3 p = 0.04. d Glomerular filtration rate (GFR) measurement (n = 8). (t0 = healthy, t2 = day 2 after IRI, t3 = day 3 after IRI, t5 = day 5 after IRI). e FACS plots of Pax8/WT (n = 6) and Pax8/YAP1 ko (n = 6) TC after nephrotoxic injury, showing diploid (2C), tetraploid (4C) and octaploid or greater (≥8C) TC. f Percentage of cycling TC in healthy (t0) mice and after nephrotoxic injury (n = 6). g Percentage of polyploid TC in healthy mice (t0) and after nephrotoxic injury (n = 6). h Percentage of dead TC after nephrotoxic injury (n = 6). i Total FUCCI2aR TC number after nephrotoxic injury (n = 4). j Survival analysis of mice after nephrotoxic injury. Kaplan-Meier analysis showed a significant difference at Log rank comparison X2 = 17.663, p = 0.0004 (n = 24 Pax8/WT, n = 14 Pax8/YAP1 ko, none censored). k Blood urea nitrogen measurement (n = 10). l Potassium level in the serum (n = 10). m Kidney weight (n = 18), p = 0.0000016. n Picture of Pax8/WT and Pax8/YAP1 ko kidneys 2 days after nephrotoxic injury. o Cell surface area of mCherry+ TC at day 1 and 2 after nephrotoxic injury. 30 TC for each mouse (t1 n = 2; t2 n = 2) were counted. (t0 = healthy, t1 = day 1 after nephrotoxic AKI, t2 = day 2 after nephrotoxic AKI, p = 5.4 × 10−10). Statistical significance was calculated by two-sided Mann-Whitney test; numbers on graphs represent exact p values or are provided in the legend. Box-and-whisker plots: line = median, box = 25–75%, whiskers = outlier (coef. 1.5).
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YAP1-driven TC polyploidization leads to CKD
Next, we studied the consequences of sustained TC polyploidization in a conditional transgenic mouse model of continuous YAP1 activation, referred to as Pax8/SAV1 ko mice (Supplementary Fig.7a–e)32. Persistent YAP1 activation in TC increased kidney size (Fig.6a–c) without affecting the absolute number of FUCCI2aR TC (Supplementary Fig.7f) whereas TAZ was not upregulated (Supplementary Fig.7g, h). By contrast, persistent YAP1 activation in TC increased the percentage of polyploid TC already in healthy conditions (Fig.6d–f). This was associated with prominent interstitial fibrosis compared to WT controls (Fig.6g–j). Enhanced TC polyploidy in Pax8/SAV1 ko mice associated with expression of β-galactosidase, a marker of cell senescence (Fig.6k, l). All these changes were accompanied by a progressive glomerular filtration rate (GFR) decline (Fig.6m). Thus, induced overexpression of YAP1 in TC of healthy mice drives polyploidization, interstitial fibrosis and senescence of TC, i.e., progressive CKD.
Fig. 6: Pax8/SAV1 ko healthy mice show increased tubular cell (TC) polyploidization and spontaneously progress toward CKD.
a Representative pictures for immunohistochemistry of active-YAP1 staining. Bars 100 µm. A representative experiment out of 4 is shown. b Picture of Pax8/FUCCI2aR/SAV1 ko (Pax8/SAV1 ko) and Pax8/WT kidneys. c Kidney weight in Pax8/WT (n = 30) and Pax8/SAV1 ko mice (n = 16), p = 7.2 × 10−6. FACS plots of TC in d Pax8/WT and in e Pax8/SAV1 ko mice, showing diploid (2C), tetraploid (4C) and octaploid or greater (≥8C) TC. Colours match the FUCCI2aR reporter (n = 8). f Percentage of polyploid TC (n = 8). g Representative pictures of Masson’s trichrome staining (n = 8). Bars 100 µm. h Tubular score evaluated on Masson’s trichrome staining (n = 8). i Sequential scanning of kidney section stained for fibronectin. DAPI counterstains nuclei. Bars 500 µm. j Quantification of fibronectin deposition by digital morphometry (n = 8). k Senescence-associated β-galactosidase assay (n = 8). Bars 100 µm. l Percentage of β-galactosidase+ TC (n = 8). m GFR measurement for 30 days (n = 5). Statistical significance was calculated by two-sided Mann-Whitney test; numbers on graphs represent exact p values or are provided in the legend. Two-way ANOVA test of significance followed by Bonferroni post-test was employed for graph m. Data are expressed as mean ± SEM in graph m. Box-and-whisker plots: line = median, box = 25–75%, whiskers = outlier (coef. 1.5).
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TC polyploidization aggravates AKI-CKD transition but attenuates AKI
We then evaluated how enhanced polyploidization of TC in Pax8/SAV1 ko mice affects the different phases of AKI. At day 30 after unilateral ischemic injury we did not observe any difference in TC number in Pax8/SAV1 ko compared to kidneys of Pax8/WT mice (Supplementary Fig.8a). However, polyploidization, interstitial fibrosis and TC senescence were enhanced in ischemic Pax8/SAV1 ko mice in comparison to ischemic Pax8/WT mice, indicating that YAP1-activated polyploidization promotes AKI-CKD transition (Supplementary Fig.8b–j). Of note, in ischemic Pax8/SAV1 ko mice polyploidization, interstitial fibrosis and TC senescence were enhanced also in comparison to healthy Pax8/SAV1 ko mice, indicating that also AKI significantly contributes to CKD (Supplementary Fig.8b–j). In addition, we observed binucleated TC at electron and confocal microscopy, as already observed in Pax8/Confetti mice with normal YAP1 function (Fig.7a and Supplementary Fig.8c, d). Polyploid TC kept cycling and produced TC with 8C or more of DNA content at 30 days prominently in Pax8/SAV1 ko mice (Fig.7b, c). Re-analysis of scRNA-seq data showed that polyploid TC cluster(s) displayed a profibrotic and senescent signature, both in vitro and in vivo (Fig.7d and Supplementary Figs.8k, 9a–f). In addition, we observed enhanced TC polyploidization, interstitial fibrosis, and TC senescence in Pax8/SAV1 ko mice also in the nephrotoxic AKI (Fig.7e–k and Supplementary Fig.8l). Consistently, Pax8/SAV1 ko mice showed an attenuated loss of kidney function at day 2 after AKI but developed a more severe CKD post-AKI, in comparison to both Pax8/WT mice that underwent nephrotoxic AKI and healthy Pax8/SAV1 ko mice as well as an increased survival rate (Fig.7l, m). Thus, enhanced TC polyploidization sustains residual kidney function and improves survival during early AKI but subsequently promotes AKI to CKD transition.
Fig. 7: YAP1-driven tubular cell (TC) polyploidization attenuates AKI but aggravates AKI-CKD transition.
a Transmission electron microscopy of a binucleated TC in Pax8/SAV1 ko mice at day 30 after IRI. Bar 5 µm. b FACS plots of TC in Pax8/WT (n = 6) and in Pax8/SAV1 ko mice (n = 6) at day 30 after IRI, showing diploid (2C), tetraploid (4C) and octaploid or greater (≥8C) TC. Colours match the FUCCI2aR reporter. c Cell cycle distribution of polyploid TC at day 30 after IRI (n = 6). d Matrix plot showing epithelial, fibrotic and senescence genes in mouse proximal tubular cells at day 2 and 30 after IRI. e FACS plots of TC at day 30 (t30) after nephrotoxic injury in Pax8/WT (n = 8) and in Pax8/SAV1 ko mice (n = 8). f Percentage of polyploid TC after nephrotoxic injury (n = 8). g Sequential scanning of whole kidney sections stained for fibronectin at day 30 after nephrotoxic injury. DAPI counterstains nuclei. Bars 500 µm. h Quantification of fibronectin deposition by digital morphometry after nephrotoxic injury (n = 8). i Tubular score evaluated on Masson’s trichrome staining (n = 8). j Senescence-associated β-galactosidase assay at day 30 after nephrotoxic injury (n = 8). Bars 100 µm. k Percentage of β-galactosidase+ TC after nephrotoxic injury (n = 8). l Blood urea nitrogen measurement after nephrotoxic injury (n = 10 nephrotoxic Pax8/WT and Pax8/SAV1 ko, n = 6 healthy Pax8/SAV1 ko). #statistical significance calculated between nephrotoxic Pax8/WT and Pax8/SAV1 ko, statistical significance calculated between nephrotoxic and healthy Pax8/SAV1 ko. m Survival analysis after nephrotoxic injury. Kaplan-Meier analysis showed a significant difference at Log rank comparison X2 = 5.07, p = 0.024 (n = 24 Pax8/WT, n = 11 Pax8/SAV1 ko, none censored). Statistical significance was calculated by two-sided Mann-Whitney test; numbers on graphs represent exact p values or are provided in the legend. Data are expressed as mean ± SEM in graph l. Box-and-whisker plots: line = median, box = 25–75%, whiskers = outlier (coef. 1.5).
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TC polyploidy in human biopsies correlates with kidney fibrosis after AKI
To validate these findings in humans, we investigated the presence of TC polyploidization in 45 patients who underwent diagnostic kidney biopsy after one or more AKI episodes (Supplementary Table3) and categorized them into “early” and “late” according to the AKI-to-biopsy interval (see Methods). Co-expression of nuclear p-H3 and perinuclear CDK4 identified cells in G1 with chromatin remodeling, i.e., polyploid cells (Fig.8a, b)8,33. Their quantification in the tubule (stained with phalloidin) showed a substantial increase of polyploid TC in kidney biopsies of AKI patients in comparison to healthy kidneys (Fig.8c). The percentage of polyploid TC in biopsies obtained early after AKI was significantly higher in comparison to the percentage in biopsies taken late after AKI (Fig.8d), in line with the decrease of polyploidization observed overtime after AKI in mice. In addition, in early biopsies, DNA content was significantly increased in YAP1+ nuclei in comparison to YAP1− nuclei indicating that where YAP1 is active TC have an increased DNA content (i.e., they are polyploid) (Fig.8e, f). Importantly, level of polyploid TC correlated with fibronectin expression in kidney biopsies of patients of the “late” group (Fig.8g–i). Taken together, these results demonstrate that the extent of TC polyploidization associates with kidney fibrosis after AKI also in humans.
Fig. 8: YAP1-related tubular cell (TC) polyploidization correlates with fibrosis in human kidney biopsies.
a Representative sequential scanning of kidney biopsy from a CKD after AKI patient showing CDK4, p-H3 and Phalloidin staining (n = 45). DAPI counterstains nuclei. Bar 250 µm. b, b’ Higher magnification of kidney section with split images. Bar 50 µm. c Percentage of polyploid TC in CKD after AKI (n = 45) vs healthy patients (n = 18). d Percentage of polyploid TC in “early” (n = 14) and “late” groups (n = 31). e Representative sequential scanning of kidney biopsy from a CKD after AKI patient showing YAP1 and Phalloidin staining(n = 8). PicoGreen counterstains nuclei. Bar 250 µm. e’ Higher magnification of kidney section with split images. Bar 25 µm. f DNA content quantification of YAP1+ nuclei vs YAP1− nuclei of TC over diploid TC(n = 8, for each biopsy 25 YAP1+/Phalloidin+ and 25 YAP1−/Phalloidin+ nuclei respectively were quantified), p = 2.19 × 1011. g Scoring of glomerular sclerosis (GS) and interstitial fibrosis and tubular atrophy (IFTA) in kidney biopsies of “late” group (n = 16). h Representative sequential scanning of biopsies stained for fibronectin (n = 16). DAPI counterstains nuclei. Bars 250 µm. i Linear correlation between polyploid TC and fibronectin deposition quantified by digital morphometry in the “late” group (n = 16) was assessed by Pearson correlation coefficient. Statistical significance was calculated by two-sided Mann-Whitney test; numbers on graphs represent exact p values or are provided in the legend. Box-and-whisker plots: line = median, box = 25–75%, whiskers = outlier (coef. 1.5).
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Delayed onset of YAP1-inhibition attenuates AKI-CKD transition
We then speculated that delayed YAP1 inhibition could prevent AKI-CKD transition without affecting AKI survival. We employed the selective inhibitor of YAP1/TEAD transcriptional activity CA334,35 and started treatment of Pax8/FUCCI2aR and non-induced (WT) mice 4 days after nephrotoxic AKI, i.e., after recovery of function loss had occurred (Fig.9a). Delayed treatment with CA3 prevented AKI-CKD transition, as defined by kidney function at day 30 (Fig.9a, b). Percentages of non-cycling polyploid TC (i.e., polyploid TC in G1) compared to vehicle controls were similar at day 30 (Fig.9c), suggesting that they had been generated before the start of CA3 treatment and were growth-arrested. By contrast, cycling polyploid TC were reduced, proving that inhibiting YAP1 blocked further endoreplication cycles of polyploid TC (Fig.9c, d). The total number of FUCCI2aR TC did not differ between the two groups, proving that polyploid TC do not proliferate (Fig.9e). In addition, mice treated with the YAP1 inhibitor CA3 displayed a better-preserved kidney tissue (Fig.9f) as well as a significant reduction of senescent TC and fibronectin staining compared to control mice (Fig.9g–j). We further validated these results in the ischemic injury model (Supplementary Fig.10a–g).
Fig. 9: Time-dependent inhibition of tubular cells (TC) polyploidization attenuates AKI-CKD transition.
a Blood urea nitrogen measurement in healthy WT mice (n = 5) and after nephrotoxic injury (n = 7). b qRT-PCR for YAP1-downstream target CTGF in nephrotoxic injury after CA3 treatment (n = 7). c FACS plots of TC in Pax8/FUCCI2aR mice after nephrotoxic injury, showing diploid (2C), tetraploid (4C) and octaploid or greater (≥8C) TC (n = 8). Colours match the FUCCI2aR reporter. d Percentage of total and ≥8C cycling polyploid TC in Pax8/FUCCI2aR mice after nephrotoxic injury (n = 8). e Total number of FUCCI2aR TC after nephrotoxic injury (n = 8). f PAS staining in WT mice at day 30 after nephrotoxic injury (n = 7). Bars 50 µm. g Senescence-associated β-galactosidase assay in WT mice at day 30 after nephrotoxic injury (n = 7). Bars 100 µm. h Percentage of β-galactosidase+ TC (n = 7). i Sequential scanning of kidney sections stained for fibronectin in WT mice at day 30 after nephrotoxic injury (n = 7). Bars 500 µm. DAPI counterstains nuclei. j Quantification of fibronectin deposition by digital morphometry (n = 7). k Blood urea nitrogen measurement in Pax8/WT and Pax8/YAP1 ko mice after nephrotoxic injury (n = 8). l FACS plots of TC after nephrotoxic injury showing diploid (2C), tetraploid (4C) and octaploid or greater (≥8C) TC (n = 8). Colours match the FUCCI2aR reporter. m Percentage of total and ≥8C cycling polyploid TC (n = 8). n Percentage of β-galactosidase+ TC (n = 8). o Quantification of fibronectin deposition by digital morphometry (n = 8). p Blood urea nitrogen measurement in Pax8/WT mice after nephrotoxic injury (n = 7). q FACS plots of TC in Pax8/FUCCI2aR mice after nephrotoxic injury, showing diploid (2C), tetraploid (4C) and octaploid or greater (≥8C) TC (n = 7). Colours match the FUCCI2aR reporter. r Percentage of total and ≥8C cycling polyploid TC (n = 7). s Percentage of β-galactosidase+ TC (n = 7). t Quantification of fibronectin deposition by digital morphometry (n = 7). t30: day 30 after nephrotoxic AKI. t4 recombination: doxycycline administered 4 days after nephrotoxic AKI. Senolytic treatment: quercetin+dasatinib. PAS: Periodic-acid schiff.Statistical significance was calculated by two-sided Mann-Whitney test; numbers on graphs represent exact p values. Data are expressed as mean ± SEM in graph a, k, and p. Box-and-whisker plots: line = median, box = 25–75%, whiskers = outlier (coef. 1.5).
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To exclude any systemic effect of the CA3 administration, we repeated the same experimental design in Pax8/YAP1 ko mice in which we triggered recombination by doxycycline administration 4 days after nephrotoxic as well as ischemic injury (Fig.9k–o and Supplementary Fig.10h–p). Pax8/WT mice recombined at day 4 were used as controls. Delayed recombination of YAP1 prevented CKD progression (Fig.9k) and reduced further endoreplication cycles of polyploid TC (Fig.9l, m). Accordingly, mice displayed less fibrosis and senescent TC compared to control mice (Fig.9n, o and Supplementary Fig.10h, i). Animal studies have shown that senescence occurs upon AKI, drives CKD and senolytic treatment can avoid AKI to CKD transition36,37. By administering senolytic therapy (i.e., dasatinib in combination with quercetin)36,38 at day 4 after nephrotoxic AKI we observed a significant decrease in the percentage of cycling polyploid TC, in association with a significantly preserved kidney function at day 30, decreased percentage of senescent TC and fibrosis development (Fig.9p–t and Supplementary Fig.10q, r). These results indicate that cycling polyploid TC were responsible for the senescent profibrotic phenotype and were the primary trigger of CKD progression. Thus, YAP1-inhibition initiated right after the early injury phase of AKI, can attenuate ongoing polyploidization of TC, which is sufficient to prevent the long-term trade-off of this survival mechanism, i.e., AKI-CKD transition.
Discussion
AKI can be induced by dehydration, toxins, drug exposure, trauma or infections, representing one of the ancestral challenges to individual and species survival since the early phases of evolution. Nowadays, dialysis can save patients’ life in the acute phase. However, without artificial replacement of kidney function, AKI can be lethal in few hours because of liquid and electrolyte misbalance. Investigating the temporal and spatial distribution and biological significance of TC polyploidy in response to AKI revealed that: 1. in the early phase after AKI, TC polyploidization outside the site of injury is an adaptive stress response to TC loss at the site of injury in order to assure survival by sustaining residual kidney function; 2. during the late phase of AKI, cycling polyploid TC become senescent and develop a profibrotic phenotype promoting CKD development; 3. blocking polyploidization by YAP1 inhibition in the right window of opportunity avoids CKD after AKI without interfering with the protective activity of TC polyploidy in early AKI; 4. senolytic therapy prevents CKD after AKI by blocking continuous TC polyploidization. Collectively, these results have three important implications:
First, they revise the existing concept on how the kidney responds to AKI which is currently based on staining with “proliferation markers”, such as Ki67, PCNA, BrdU that cannot distinguish endoreplicating from proliferating cells6,7,8. Widespread TC proliferation was considered to mediate tissue reconstitution, driving functional organ recovery after AKI. By contrast, our data dissociate the recovery of kidney function from structural repair at the site of kidney injury. We demonstrate that polyploidization-mediated hypertrophy of TC distant from the injury site provides a life support in the early phase of AKI.
Second, they solve the apparent inconsistency of why even while complete functional recovery upon AKI is frequent, AKI survivors often end up in CKD1. The current paradigm refers this phenomenon to a process defined as ‘maladaptive repair’39. In contrast, our data demonstrate that interstitial fibrosis and CKD represent trade-offs of an acute response required to prevent death in the early AKI phase. Other life-preserving acute stress responses such as the neurohumoral response, the renin-angiotensin-system or mineralocorticoid receptor signaling, also come with the long-term trade-off of promoting chronic organ degeneration and failure40,41. Consistently, the role of TC polyploidization in preventing AKI fatality ultimately promotes AKI-CKD transition via TC senescence and interstitial fibrosis.
Third, our study identifies TC polyploidization as a target for therapeutic intervention to attenuate the transition of AKI to CKD with all its consequences, such as progression of CKD and possibly the related development of cardiovascular disease2,3,4,5. The paucity of valid therapeutic interventions keeps resulting in high mortality rates during the acute phase and a poor long-term prognosis1,2. Current treatment of AKI is still confined to supportive strategies such as dialysis. Promising trials of agents, including diuretics, dopamine and atrial natriuretic peptide, were effective in animal experiments but failed when translated into clinical settings42. The results of our study provide an explanation for these observations, showing that the same polyploid TC population that is responsible for hypertrophy and recovery of kidney function immediately after AKI is also responsible for senescence, interstitial fibrosis and CKD in the long run. Indeed, we proved that YAP1 controls TC polyploidization offering a druggable target to block CKD development. However, systemic inhibition of YAP1 does not exclude that cells other than TC could contribute to the observed effect. Nevertheless, the comparable results obtained after triggering YAP1 inactivation selectively in TC after AKI demonstrates that the drug-mediated YAP1-inhibition prevents CKD after AKI by blocking TC polyploidization and subsequent TC senescence and interstitial fibrosis. Interestingly, Zheng et al. reported that YAP1 inhibition is associated with in vivo amelioration of kidney fibrosis and suggested that YAP1 activation after AKI is not compatible with a G2/M cell cycle arrest43. Consistently, Gerhardt et al. observed no sign of G2/M cell cycle arrest after AKI using scRNA-seq44. Intriguingly, the results of our study suggest that the previously reported cells in G2/M45,46 rather represent polyploid cells in G1. Indeed, the method usually applied of measuring DNA content alone cannot distinguish polyploid cells in G1 from diploid cells in G2/M without a simultaneous assessment of the cell cycle phase45,46. Nevertheless, even the sophisticated methods applied in this study have a sensitivity limitation and still underestimate the percentage of polyploid TC after AKI for the following reasons: 1. The Confetti reporter misses the polyploid TC that recombine twice or multiple times the same colour; 2. The FUCCI2aR reporter can detect as polyploid only those TC that have completed the cell cycle and are in G1 with a 4C DNA content in a certain moment or as cycling polyploid cells only if their DNA content is >of 4C. Taken together, these results show that polyploidy after AKI is a diffuse phenomenon, explaining its crucial role in the recovery of kidney function in the early phase, as well as in promoting senescence and fibrosis in the late phase after AKI. This implies that treatments for AKI must consider the right window of opportunity because targeting polyploid TC too early would be detrimental for patient survival. Conversely, polyploid TC may become a right target for treatment once the early phase of AKI is bypassed but profibrotic polyploid TC have not yet induced CKD development. We propose that blocking polyploidization at least can protect from CKD development after AKI. Whether other mechanisms in addition to inhibition of TC polyploidization contribute to the beneficial effects of YAP1 inhibition after AKI remains to be established and will require further studies.
In summary, our findings revise the current concept of the biological response to AKI, providing a unified model of kidney adaptation to acute tubular necrosis. During injury, TC polyploidization is the fastest way to increase the residual function of differentiated TC. Increasing function without delay is required to avoid AKI fatality during the early phase, that is why this mechanism likely was positively selected during evolution, at the cost of tissue remodeling and chronic dysfunction later in life47,48. Nevertheless, YAP1-driven TC polyploidization is an attractive therapeutic target to prevent CKD when blocked during the recovery phase of AKI. This concept may apply also to other organs with a kidney-like low turnover of parenchymal cells.
Methods
Mice
To visualize the cell cycle progression of Pax8+ TC, the Pax8.rtTA;TetO.Cre;R26.FUCCI2aR (Pax8/FUCCI2aR) mouse model was employed. This model was obtained by crossing Pax8.rtTA;TetO.Cre mice8 with mice harboring the Fluorescent Ubiquitin-based Cell cycle Indicator (FUCCI2aR) Cre-dependent reporter (European Mouse Mutant Archive (EMMA), INFRAFRONTIER-I3, Neuherberg-München, Germany), which consists of a bicistronic Cre-activable reporter of two fluorescent proteins whose expression alternates based on cell cycle phase: mCherry-hCdt1 (30/120) (red), expressed by nuclei of cells in G1 phase, and mVenus-hGem (1/110) (green), expressed by nuclei of cells in S/G2/M. Cells can also appear as yellow at the G1/S boundary49. Mice were developed on a full C57Bl/6 background8. Triple transgenic mice were crossed with either SAV1 knock-out (Sav1 tm1.1Dupa, JAX:027933) to obtain Pax8/SAV1 ko mice or with YAP1 knock-out mice (Yap1 tm1.1Dupa, JAX:027929) to obtain Pax8/YAP1 ko mice both from the Jackson Laboratory. To induce reporter expression, at 5 weeks of age mice were given 4 mg/ml doxycycline hyclate (Merck) in drinking water supplemented with 2.5% sucrose (Merck) for 10 days, followed by a 7-day washout. Un-induced animals showed no leakage nor non-specific transgene expression of the transgenic systems. At the end of the washout period, male mice underwent a unilateral ischemia reperfusion injury (IRI, ischemic injury) of 30 min, and were then sacrificed at day 2, 3, 5 and 30 days post-IRI. Sham operated mice were used as controls. As IRI tolerance is profoundly increased in female mice compared with that observed in male mice, female mice underwent glycerol-induced nephrotoxic AKI (nephrotoxic-AKI, nephrotoxic injury). Mice were sacrificed at day 1, 2 and 30 days after nephrotoxic-AKI. Healthy mice were used as controls. Animals with identical genotype and similar age were assigned to experimental groups in a blinded manner. Successful transgene recombination was assessed at the time of sacrifice by DNA and RNA expression as described below.
For the treatment with YAP1 inhibitor CA3 (CIL56), Pax8/FUCCI2aR mice and non-induced (WT) littermates underwent nephrotoxic-AKI (female mice) or ischemic-AKI (male mice) at 7 weeks of age. Following evaluation of the degree of kidney damage by BUN or GFR measurement, mice were assigned to the treated or control group ensuring evenly distribution in the experimental groups. CA3 (CIL56) was dissolved in DMSO and prior to injection diluted in PBS and sonicated. At day 4 after AKI mice were divided in two groups receiving CA3 (1.5 mg/kg, Selleck, S8661) or vehicle (PBS with 5% DMSO) injection via i.p. every other day until 30 days and then sacrificed34,35. To trigger recombination of Pax8/YAP1 ko and Pax8/WT mice after ischemic (male) and nephrotoxic (female) AKI, mice at 7 weeks of age were given 4 mg/ml doxycycline hyclate (Merck) in drinking water supplemented with 2.5% sucrose (Merck) at 4 days after AKI for 10 days, followed by a 7-day washout. For the senolytic treatment, Pax8/FUCCI2aR mice underwent nephrotoxic-AKI (female mice) at 7 weeks of age. Following evaluation of the degree of kidney damage by BUN measurement, mice were assigned to the treated or control group ensuring evenly distribution in the experimental groups. At day 4 after AKI mice were divided in two groups receiving Quercetin (50 mg/Kg, Q4951, Merck) + Dasatinib (5 mg/Kg, CDS023389, Merck)36 or vehicle (PBS with 5% DMSO) via gavage every other day until 30 days and then sacrificed.
The number of mice used, numbers of replicates, and statistical values (where applicable) are provided in the figure legends. Animal experiments were approved by the Institutional Review Board and by the Italian Ministry of Health and performed in accordance with institutional, regional, and state guidelines and in adherence to the National Institutes of Health Guide for the Care and Use of Laboratory Animals. Mice were housed in a specific pathogen-free facility with free access to chow and water and a 12-hour day/night cycle. The references to the ethics approvals are the following: 809/2017-PR; 272/2018-PR; 689/2019-PR; 864/2021-PR and 239/2022-PR.
Confetti reporter for polyploidy evaluation
The Confetti reporter, upon induction, allowed the stochastic labeling of cells by permanent recombination of a single colour-encoding gene (red, yellow, green, or blue fluorescent proteins, RFP, YFP, GFP, or CFP), with GFP cells occurring at lower frequency than other colours and being extremely rare8,20. For this reason, GFP+ TC were not counted and excluded from the calculation. At 7 weeks of age heterozygous male Pax8.rtTA;TetO.Cre;Rosa.Confetti (Pax8/Confetti)mice8 underwent unilateral ischemia reperfusion injury (IRI) to trigger polyploid TC formation and 13 days after damage they were given 4 mg/ml doxycycline hyclate to activate reporter recombination. After washout, diploid Pax8+ TC (i.e., cells with one set of chromosomes, DNA content 2C) can stochastically express only one of the four colour of the Confetti reporter. In contrast, cells carrying two (DNA content 4C) sets of chromosomes (i.e., polyploid TC) can activate two fluorochromes at the same time, resulting in bi-coloured or mono-coloured TC. Mice were then sacrificed at the end of the 7-day washout. Age-matched control mice received the same induction without undergoing IRI. As GFP+ TC were not included in the analysis, quantitative analysis of polyploid TC was performed by counting TC with the following bi-coloured combinations of fluorescence: RFP + YFP, RFP + CFP, YFP + CFP. The count was normalized on the total number of Confetti TC (GFP+ TC and their combinations were excluded accordingly). Bi-coloured TC were counted by two independent blinded observers. We additionally calculated the efficiency of reporter recombination in our mice, by counting the number of Confetti+ TC over the total TC which is 84.7 ± 1.7%. The percentage of observed bi-coloured TC is 13.3 ± 2.5%. When normalized for the efficiency of reporter recombination, the percentage of bi-coloured TC correspond to 15.7%.
Based on the likelihood of event occurrence, we estimated that 1/3 of polyploid TC with 4C DNA content will recombine twice the same colour and cannot be distinguished from the mono-coloured diploid ones (Supplementary Fig.2c–g). The acquisition of Pax8/Confetti mice was performed with Leica SP8 STED 3X confocal microscope (Leica Microsystems) with the following set for wavelength excitation: cyan 470 nm, green 488 nm, yellow 514 nm, and red 543 nm. The number of mice used, numbers of replicates, and statistical values (where applicable) are provided in the figure legends.
Genotyping
Tail biopsies were incubated overnight at 55 °C in lysis reagent (1 M TrisHCl pH 8.5, 0.5 M EDTA, 20% SDS, 4 M NaCl, 0.1 mg/mL proteinase K neutralized with 40 mM TrisHCl, all from Merck), centrifuged and DNA extracted using isopropanol (Merck). Primers and PCR parameters were obtained from Jackson Laboratory online resources of the relative strain purchased or previously reported experimental procedure8. To distinguish transgene homozygosity from heterozygosity, qRT-PCR were performed by using 5 ng/µl of genomic DNA with LightCycler® 480 SYBR Green I Master (Roche Diagnostics). The reactions were performed using a LightCycler® 480 (Roche Diagnostics) with a program consisting of 40 cycles each constituted of an initiation phase at 95 °C for 5 s, annealing phase at 60 °C for 10 s and amplification phase at 95 °C for 10 s, followed by 65 °C for 60 s. The following primers were used:
Pax8.rtTA forward 5′-AACGCACTGTACGCTCTGTC-3′ and reverse 5′-GAATCGGTGGTAGGTGTCTC-3′; TetO.Cre forward 5′-TCGCTGCATTACCGGTCGATGC-3′ and reverse 5′-CCATGAGTGAACGAACCTGGTCG-3′. TCRα genomic DNA was used as gene housekeeping for relative quantification and was amplified by using the forward 5′-CAAATGTTGCTTGTCTGGTG-3′ and the reverse 5′-GTCAGTCGAGTGCACAGTTT-3′ primers.
To distinguish FUCCI2aR homozygosity from heterozygosity, PCR was performed with the following primers and parameters: FUCCI-P3 5′-TCCCTCGTGATCTGCAACTCCAGTC-3′, FUCCI-P4 5′-AACCCCAGATGACTACCTATCCTCC-3′, and FUCCI-P7 5′-GGGGGATTGGGAAGACAATAGC-3′; 3 min 94 °C, 35 cycles of 94 °C 30 s, 58 °C 30 s and, finally, 72 °C for 30 s. SAV1 and YAP1 genotyping was performed with the following primers and parameters: SAV1 forward 5′-CATAGAAGCAGGGGATTCTGA-3′ and reverse 5′-ACATGCTGACCACAAGCAGA-3′; YAP1 forward 5′- AGG ACA GCC AGG ACT ACA CAG-3′ and reverse 5′- CAC CAG CCT TTA AAT TGA GAA C-3′; 3 min 94 °C, 35 cycles of 94 °C 30 s, 58 °C 30 s and finally, 72 °C for 30 s.
To confirm successful recombination of the alleles (YAP1 and SAV1), a small piece of kidney cortex was excised and RNA and DNA were extracted according to manufacturer’ s instruction (Qiagen). Primers and PCR parameters were obtained from previously published procedures50,51. PCR was performed with the following primers and parameters:
SAV1-S1 5′-AGGGGATTCTGACATTTCAGTCAGTT-3′, SAV1-S2 5′-AGTCACATGCTGACCACAAGCAGAA-3′ SAV1-S3; 5′-TGCCATTAAGTGTAATCACTGG-3′; 3 min 94 °C, 35 cycles of 94 °C 30 s, 56 °C 30 s and finally, 72 °C for 1 min; YAP1-P1 5′-CCATTTGTCCTCATCTCTTACTAAC-3′, YAP1-P2 5′-GATTGGGCACTGTCAATTAATGGGCTT-3′, YAP1-P3: 5′-CAGTCTGTAACAACCAGTCAGGGATAC-3′; 3 min 94 °C, 35 cycles of 94 °C 30 s, 63 °C 30 s and finally, 72 °C for 1 min.
Quantitative Real-Time PCR
Total RNA was extracted from small kidney cortex pieces at the time of sacrifice in all Pax8/SAV1 ko and Pax8/YAP1 ko mice to further confirm recombination using the RNeasy Mini Kit (Qiagen). Pax8/WT were used as controls. cDNA was synthesized using High Capacity cDNA Reverse Transcription Kit (Applied Biosystems). Data was normalized on HPRT1 expression. For sorted cells and total mCherry-hPTC, RNA was extracted using RNeasy Microkit (Qiagen) and retrotranscribed using TaqMan Reverse Transcription Reagents (Thermo Fisher Scientific). TaqMan RT-PCR for 18S, CTGF, SEMA5A, CCL2, COLα1, CDKN1A, SERPINE1, YAP1, TAZ, E2F7, E2F8, AKT1, CDK1 and CCNB1 was performed using customized TaqMan assays (Thermo Fisher Scientific) on a 7900HT Fast Real-Time (Applied Biosystem).
Mouse tissue collection and analysis
All animals were weighted prior to sacrifice by CO 2 chamber; kidneys were collected, measured and weighted. Kidney weight was normalized on total body weight. For histological analysis, kidneys were fixed in 10% buffered formalin (Bio-Optica) overnight at room temperature, dehydrated then embedded in paraffin, and 5 µm sections were prepared for Periodic-Acid Schiff staining (PAS, Bio-Optica) and Masson’s Trichrome staining (Bio-Optica). The scoring of tissue damage was carried out by two independent blinded observers through a semiquantitative evaluation, based on arbitrary score on Masson’s trichrome sections under the optical microscope Leica DM750 (Leica Microsystems, Mannheim, Germany). Whole sections were analysed. The score ranged from 0 to 4+ as follows 0: no changes; 1: damage to <25% of the interstitial area; 2: damage to 25–50% of the interstitial area; 3: damage to 50–75% and 4: damage to >75% of the interstitial area52. For confocal analysis of the FUCCI2aR reporter and tubular markers, kidneys were incubated in 4% paraformaldehyde in PBS (both from Merck) for 2 h at 4 °C followed by immersion in a 15% sucrose solution in PBS for 2 h at 4 °C and, subsequently, in a 30% sucrose solution in PBS overnight at 4 °C, then frozen8. 10 µm sections of renal tissues were analysed on Leica SP8 STED 3X confocal microscope (Leica Microsystems). For β-galactosidase, fibronectin (1:100, Abcam, ab2413) and Ki67 (1:50, Abcam, ab15580) staining, kidneys were snap-frozen. Only cortical fibrosis was analysed. For β-galactosidase staining, snap-frozen tissues were cut in 10 µm sections and fixed with 2% formalin (Bio-Optica) and 0.2% glutaraldehyde (Merck)in pH6 PBS for 20 min at RT then incubated O/N with X-GAL (Thermo Fisher Scientific, B1690) at a final concentration of 1 mg/ml. The nuclei were counterstained with Nuclear Fast Red (Merck). Sections were then dehydrated and mounted with SafeMount (Bio-Optica) and analysed under the optical microscope Leica DM750 (Leica Microsystems). A total of 5 fields for each conditions were counted. For fibronectin and Ki67 staining, snap-frozen tissues were cut in 10 µm sections and fixed in 4% PFA for 30 min.
Renal ischemia reperfusion injury
Renal ischemia was performed on male mice as previously described8,53. Briefly, mice were anesthetized by intraperitoneal injection of Ketamine (100 mg/kg)/Xylazine (10 mg/kg, Bio98 S.r.L, Milan, Italia). The mouse was placed on a thermostatic station laying on the right side and the body temperature was kept at 37 °C, shaved, and disinfected with Povidone-iodine and an incision of 1–1.5 cm on the skin on the left side was performed. The left kidney was then externalized and the renal artery was clamped for 30 min. After clap removal, the muscle layer was sutured, followed by the closure of the skin wound with metal clips. Immediately after the wound closure, 0.5 ml warm sterile saline (0.9% NaCl) was given subcutaneously to each mouse to rehydrate it. The right contralateral kidney was maintained untouched. Sham-operated mice underwent the same surgical procedure without left renal artery clamping.
Nephrotoxic AKI
Rhabdomyolysis-induced AKI was performed on female mice, by intramuscular injection with hypertonic glycerol (8 ml/kg body weight of a 50% glycerol solution; Merck) into the inferior hind limbs8.
Transcutaneous measurement of glomerular filtration rate
Measurement of the glomerular filtration rate (GFR) was done as previously described8,53. Mice where anesthetized with isoflurane and a miniaturized imager device built from two light-emitting diodes, a photodiode, and a battery (Mannheim Pharma and Diagnostics GmbH, Mannheim, Germany) were mounted via a double-sided adhesive tape onto the shaved animals’ neck. For the duration of recording (~1.5 h) each animal was conscious and kept in a single cage. FITC-sinistrin (Mannheim Pharma and Diagnostics GmbH,) was resuspended in a sterile saline solution to a final concertation of 30 mg/ml. Injection volume (150 mg/kg) was calculated per each mouse according to the body weight. Prior to the intravenous injection of FITC-sinistrin the skin’s background signal was recorded for 5 min. After removing the imager device, the data were analysed using MPD Studio software ver.RC6 (MediBeacon GmbH Cubex41, Mannheim, Germany)54. The GFR [µl/min] was calculated from the decrease of fluorescence intensity over time (i.e., plasma half-life of FITC-sinistrin) using a two-compartment model, the animals body weight and an empirical conversion factor, as previously described8, using the following formula:
$${{{{{\rm{GFR}}}}}}\,[{{{{{\rm{\mu }}}}}}{{{{{\rm{l}}}}}}/{{{{{\rm{min }}}}}}]=\frac{14616.8\,[{{{{{\rm{\mu }}}}}}{{{{{\rm{l}}}}}}]}{{{{{{{\rm{t}}}}}}}_{1/2}({{{{{\rm{FITC}}}}}}-{{{{{\rm{sinistrin}}}}}})[{{{{{\rm{min }}}}}}]}\,{{{{{\rm{\times }}}}}}\frac{{{{{{\rm{bw}}}}}}\,[{{{{{\rm{g}}}}}}]}{100\,[{{{{{\rm{g}}}}}}]}$$
For each time point, GFR value was normalized on the value at baseline and on the sham value except for healthy Pax8/WT, Pax8/YAP1 ko and Pax8/SAV1 ko mice and the measurement at day 2 after AKI in Pax8/WT and Pax8/YAP1 ko.
Blood urea nitrogen and Potassium quantification
Kidney function was assessed at different time points by collecting a small amount of blood from mice with a metal lancet from the submandibular plexus in order to measure BUN levels or potassium levels. Blood parameters were measured in EDTA anticoagulated plasma samples using Reflotron (Roche Diagnostics), according to the manufacturer’s protocols.
FACS analysis on kidney tissue
Cell cycle analysis was performed on total FUCCI2aR cells (mCherry+ and mVenus+ cells) in Pax8/FUCCI2aR, Pax8/SAV1 ko, and Pax8/YAP1 ko mice. Kidneys were processed to obtain a single cell suspension as previously reported8. Briefly, kidneys were minced using a scalpel and then incubated at 37 °C for 20 min in 1.5 ml of digestion buffer (300 U/ml Collagenase II and 1 mg/ml Pronase E, Merck). The solution was pipetted up and down with a cut 1000 µl pipette tip every 5 min. The digested kidneys were gently pressed through a graded mesh screen (150 mesh, Merck) and the flow through was washed extensively with HBSS (Thermo Fisher Scientific). Following centrifugation, the pellets was digested again with digestion buffer at 37 °C for 20 min, the suspensions were sheared with a 27-G needle every 10 min. Erythrocytes were lysed with NH4Cl 0.8%. Single-cell suspensions were fixed with 1% PFA for 1 h at RT and with 70% ethanol overnight at 4 °C. Incubation with anti-DsRed (1:25, Clontech, 632496) or isotype control (normal rabbit IgG, 1:250, Thermo Fisher Scientific, 02-6102) was followed by Alexa Fluor 647 goat anti-rabbit (1:100, Thermo Fisher Scientific, A-21245) as secondary antibody to detect mCherry+ TC, whereas for detection of mVenus+ TC was used an anti-GFP-488 (1:100, Termo Fisher Scientific, A21311). Cells were then incubated with DAPI (4′,6-diamidino-2-phenylindole, 1:2000, Thermo Fisher Scientific), to perform the DNA content analysis. The assessment of polyploid TC was performed using a MacsQuant instrument (Miltenyi Biotec). Alexa Fluor 647 secondary antibody was excited by a 633 nm laser line, GFP was excited by a 488 nm laser line, DAPI was excited by a laser at 405 nm. Polyploid TC were defined as mCherry+ or mCherry+mVenus+ cells with a DNA content ≥4C and mVenus+ with a DNA content ≥8C. Polyploid TC were further divided in non-cycling polyploid TC (mCherry+ DNA content ≥4C) and cycling polyploid TC (mCherry+ mVenus+ cells with a DNA content ≥4C and mVenus+ with a DNA content ≥8C). Cycling cells were defined as mCherry+ mVenus+ with DNA content = 2C and mVenus+ cells up to 4C DNA content. Dying cells were defined as FUCCI2aR+ cells with DNA content <2C. To analyse the redistribution of cycling cells at day 2 after AKI, we calculated the redistribution of those cycling cells per each mouse in category (1) cycling, (2) dying, and (3) polyploid as 100%. Cell doublets were excluded from the analysis, as previously reported by us8 and shown in Supplementary Fig.1. For urine analysis, multiple urine samples were manually collected from each mouse at t0, t2, and t3 after IRI and pulled together. Staining was performed as described above. Number of FUCCI2aR cells was quantified by using a MacsQuant instrument (Miltenyi Biotec) and normalized on the percentage of induction of each mouse calculated by counting the number of mCherry+ and mVenus+ TC on the total TC number on confocal microscope. Data were analysed by FlowLogic software (FlowLogic 7.2.1, Inivai Technology).
Immunofluorescence
Confocal microscopy was performed on 10 µm sections of murine renal tissues, on hPTC and 10 µm sections of human kidney biopsies on Leica SP8 STED 3X confocal microscope (Leica Microsystems). For fibronectin acquisition and CDK4/p-H3 staining sequential scanning of whole kidney sections was performed using Leica SP8 STED 3X confocal microscope (Leica Microsystems). Fibronectin deposition was quantified using Image J software (RRID:SCR_003070) and normalized on the section area. The following antibodies were used: anti-aquaporin-1 (AQP1, 1:100, Millipore, AB2219), anti-Tamm-Horsfall (THP, 1:20, Cederlane, CL1032A), anti-aquaporin-2 (AQP2, 1:500, Abcam, Ab62628), anti-GFP polyclonal antibody (1:100, Thermo Fisher Scientific, A-11122), anti-phosphorylated Histone 3 (p-H3, 1:2000, Abcam, ab14955), anti-cyclin dependent kinase 4 (CDK4, 1:50, Santa Cruz Biotechnology, SC-601), Phalloidin-633 and Phalloidin-488 (1:40, Thermo Fisher Scientific, A22284 and A12379), anti-TAZ (1:100, Abcam, ab110239), and anti-active YAP1 (1:500, Abcam, ab205270) staining. Alexa-Fluor secondary antibodies were obtained from Thermo Fisher Scientific. Nuclei were counterstained with DAPI (1:1000, Thermo Fisher Scientific) excited with UV laser. Cell surface area based on Phalloidin-488 staining on mCherry+ TC of Pax8/FUCCI2aR and Pax8/FUCCI2aR/YAP1 ko mice were measured using Image J software (RRID:SCR_003070)8. The analysis was restricted to PTC.
Immunohistochemistry
Immunohistochemistry was performed following previously published protocols53. Briefly, paraffin-embedded tissues were cut in 5 µm sections, deparaffinized, and rehydrated. The sections were then incubated in 3% H2O2 (Merck) for 10 min to block the endogenous peroxidase activity followed by blocking of endogenous Avidin/Biotin according to manufacturer’ s instruction (Vector Laboratories, Thermo Fisher Scientific, SP-2001). Antigen retrieval was performed in sodium citrate buffer (10 mM, pH 6) al 95 °C for 40 min followed by 30 min of cooling at RT in deionized water. The sections were then blocked and incubated with the primary antibody anti-active YAP1 (1:500, Abcam, ab205270) or anti-TAZ (1:100, Abcam, ab110239). All the buffers were prepared in PBS + 0.1% Triton X-100 reduced (Merck). 3 3,3′-Diaminobenzidine (DAB, Merck) was used as a peroxidase substrate. Sections were dehydrated and mounted using SafeMount (Bio-Optica). As negative control, primary antibody was substituted with an isotype-matched antibody with irrelevant specificity
Transmission Electron Microscopy
Kidney tissue samples of about 2 mm 3 were obtained from Pax8/WT or Pax8/SAV1 ko 30 days after IRI, embedded in epoxy resin, and routinely processed for transmission electron microscopy observation. Ultrathin sections were cut using a LKB‐Nova ultramicrotome (LKB, Bromma, Sweden, www.lkb.com), counterstained with uranyl acetate and alkaline bismuth subnitrate, and examined under a JEM 1010 transmission electron microscope (Jeol, Tokyo, Japan, www.jeol.com) at 80 kV.
Kidney biopsy evaluation in humans
All consecutive patients with different nephropathies who underwent renal biopsy in our Nephrology Units (Nephrology, Dialysis and Transplantation, Azienda Ospedaliero- Universitaria Careggi and Nephrology and Dialysis, Meyer Children’s University Hospital, Florence, Italy) between 2012 and 2021 were screened in this study. Exclusion criteria were: (1) Positive immunofluorescence staining suggestive of immune-mediated nephropathy; (2) Less than two creatinine values within 48 h in the period before renal biopsy which does not allow the diagnosis of AKI; (3) Insufficient clinical information and no or lacking laboratory findings. Following these criteria, 45 renal biopsies from patients (mean age 52 ± 17) that developed CKD after AKI were analysed. 18 healthy kidneys (mean age 59.22 ± 3.65;gender (male) 12/18, 66.6%) were used as controls.Authorization and approval to this study was granted by the Ethical Committee on human experimentation of the Azienda Ospedaliero- Universitaria Careggi (Clinical Trial Center (CTC) AOU Careggi, authorizations: n. OSS_10243) and by the Meyer Children’s University Hospital, Florence, Italy (Clinical Trial Office, Meyer, authorizations: n.150/2016). According to the Kidney Disease Improving Global Outcomes (KDIGO) guidelines, patients were classified as CKD if they had preoperative GFR less than 60 mL/min/1.73m255. The estimated glomerular filtration rate (eGFR) was calculated using Chronic Kidney Disease Epidemiology Collaboration equations (CKD-EPI creatinine formula)56. CKD after AKI biopsies were divided in “early” and “late” group according to the time elapsed from the AKI episode and the biopsy (2-15 days between the AKI episode and the biopsy, >15 days between the AKI episode and the biopsy). Normal kidney fragments were obtained from the pole opposite to the tumour of patients who underwent radical nephrectomy for localized kidney tumours. Formal consents were obtained by the donors or relatives. The AKI stage was classified according to KDIGO Guidelines57. Demographic, clinical and histopathological features of patients who developed CKD after AKI that we analysed in this study are reported in Supplementary Table3. Polyploid TC were quantified as cells positive for CDK4 and p-H3 (see immunofluorescence section). The scoring of CKD after AKI biopsies was carried out by two independent blinded observers through a semiquantitative evaluation, based on arbitrary score on fibronectin sections. Global glomerulosclerosis (GS) ranged from 1+ to 3+ as follows 1+: GS < 25% of total glomeruli; 2+: GS 25–50% of total glomeruli; and 3+: GS 50−70% of total glomeruli. Cortical interstitial fibrosis and tubular atrophy (IFTA) ranged from 0 to 3 as follows 0: involving <10% of cortical tubulo-interstitial area; 1+: involving 10–25% of cortical tubulo-interstitial area; 2+: involving 26–50% of cortical tubulo-interstitial area; and 3+: involving >50% of cortical tubulo-interstitial area. Fibronectin staining was carried out in biopsies belonging to the “late” group and only where kidney cortical tissue was available.
Ploidy quantification on kidney biopsies
DNA nuclear content was measured as previously described58 with some modifications. Biopsies were fixed 20 min in 4% PFA and then incubated with 0.75 µl/mL PicoGreen (Thermo Fisher Scientific) for 15 min at room temperature which was determined to be the best incubation times to obtain a non-saturated signals. Importantly, PicoGreen allows the precise and accurate dsDNA concentration measurements. The sections were then stained as with anti-active YAP1 (1:500, Abcam, ab205270) and Phalloidin-633 (1:40, Thermo Fisher Scientific, A22284) antibody as described in the immunofluorescence section. Biopsies were then imaged using Leica SP8 STED 3X confocal microscope (Leica Microsystems) on sequential scanning of whole kidney biopsies. A healthy biopsy was used as a diploid internal control for ploidy measurements by imaging the biopsy at the same gain and settings. Using Image J software (RRID:SCR_003070) regions were drawn around each nucleus and PicoGreen intensity was measured within each outlined nuclear region. The average background was calculated and subtracted from the measured intensities. The average PicoGreen intensity of diploid healthy tubular cells (2C) were analysed and calculated. The ploidy of YAP1+ /Phalloidin+ and YAP1−/Phalloidin+ nuclei was calculated by normalizing the PicoGreen intensity to average value of the diploid cells. For each biopsy 25 YAP1+/Phalloidin+ and YAP1−/Phalloidin+ nuclei respectively were quantified.
Cell culture, virus transduction, and GapmeRs transfection
Primary hPTC cultures (ATCC-PCS-400-010) were maintained in REGM (Lonza, CC-3190). hPTC were seeded at a density 10 5 cells/6-well. The following day cells were transduced with a pRetroX-G1-Red (Clontech, 631436) to allow the identification of cells in G1 phase. A MOI of 10 was used (Retro-X™ qRT-PCR Titration Kit, 631453) according to manufacturer’s instruction. In this plasmid the cell cycle indicator hCdt1 (30-120) is tagged with the red fluorescent protein mCherry. After transduction cells are referred to as hPTC-mCherry. 48 h after transduction, the media was changed and the cells were split within a week. For confocal imaging, hPTC were seeded in Lab-Tek™ II Chamber Slide™ System (Nunc, Thermo Fisher Scientific) and stained O/N with anti-active YAP1 (1:500, Abcam, ab205270) or anti-TAZ (1:100, Abcam ab110239). Nuclei were counterstained with DAPI (1:1000, Thermo Fisher Scientific).
For FACS analysis, cells were seeded and analysed after 48 h. Cell cycle analysis and gating strategy to exclude cell doublets was performed on total hPTC as previously published with some modifications8. Cells were trypsinized (Euroclone) and single-cell suspensions were fixed with 0.25% PFA for 30 min on ice and with 70% ethanol overnight. Incubation with anti-DsRed (1:25, Clontech, 632496) or isotype control was followed by incubation with Alexa Fluor 647 goat anti-rabbit (1:100, Thermo Fisher Scientific, A-21245) as secondary antibody to detect mCherry+ hPTC. In the verteporfin experiment,Alexa Fluor 488 goat anti-rabbit (1:100, Thermo Fisher Scientific, A-11008) was employed.hPTC were then incubated with DAPI (1:2000, Thermo Fisher Scientific) to perform the DNA content analysis and analysed on MacsQuant instrument (Miltenyi Biotec). Polyploid hPTC were defined as mCherry+ cells with a DNA content ≥4C. For verteporfin treatment, a dose response curve was performed and the concentration of 0.6 µM for 48 h was chosen. DMSO (Merck) was used as vehicle control. After 48 h treatment, cells were harvested and analyses by FACS or mRNA was extracted to evaluate YAP1, CTGF, E2F7, E2F8, AKT1, CDK1 and CCNB1 expression. For knock-down experiments, YAP1 (LG00806288-DDA), TAZ (LG00806278-DDA), E2F7 (LG00804017-DDA), E2F8 (LG00804007-DDA) and AKT1 (LG00803998-DDA) antisense LNA-GapmeRs were designed by the Qiagen selection tool (Qiagen, Hilden, Germany) and administered to mCherry-hPTC using Xfect transfection reagent (Takara Bio, Kusatsu, Japan). In detail, 10 5 cells/6-well were seeded and transfected for 6 h with 250 nM antisense LNA-GapmeRs for YAP1, TAZ, E2F7, E2F8 and AKT1 or Negative control (scramble, LG00248643-DFA). The efficacy of mRNA knock-down was evaluated 48 h after transfection by qRT-PCR. Primers are described in the quantitative Real-Time PCR section. After 48 h mCherry-hPTC were harvested and analysed by FACS.
Chromatin immunoprecipitation assay
Primary hPTC (10 7) treated with DMSO or verteporfin for 48 h were fixed with 1% formaldehyde (Merck) in REGM (Lonza, CC-3190) for 20 min at 4 °C, following by quenching with 0.125 M glycine for 10 min at 4 °C. Further, chromatin was shared into DNA fragment of 500-1000 pb long. 100 μg of chromatin was diluted into ChIP dilution buffer containing protease and phosphatase inhibitor (Merck) and incubated with specific antibodies against YAP1 (10 µg/IP, Novus Biologicals, NB110-58358) or normal rabbit IgG (7 µg/IP Thermo Fisher Scientific, 02-6102) overnight at 4 °C. Antibody-chromatin complexes were recovered with Salmon Sperm DNA/Protein A-Agarose beads (Merck) for 1 h at 4 °C, washed and eluted from the beads with elution buffer. After reverse crosslinking and proteinase K treatment, DNA was extracted and analysed by quantitative qRT-PCR on a 7900HT Fast Real-Time (Applied Biosystem) using specific primers: CTGF (F: TGGTGCGAAGAGGATAGGG; R: CGGATTGATCCTGACCCCTTG), AKT1 (F: GAATGGTTGACTCCCCTCGG; R: GCGGCCAAGAGTGACCTAAA) E2F7 (F: CCTCCTAATTATTTGCAATTTGCCG; R: GACTGGAAGCCAAACCAGAA) E2F8: (F: ATTCCCCAACTTTGAGCAAGGG; R: GTTGGGGGAAAAGTTCAGCAAC). Enrichment was calculated using the formula: ΔCt= Ct (bound) – [Ct (input) - log2 (Input dilution factor)], where chromatin obtained before immunoprecipitation was used as the input control. Data are expressed as fold enrichment (FR) of each specific antibody over a negative control antibody relative to negative control; FR = 2 exp - [ΔCt (specific antibody) - ΔCt (normal IgG)].
Fluorescence-activated cell sorting
hPTC transduced with pRetroX-G1-Red (mCherry-G1) retrovirus were trypsinized (Euroclone) at passage 2 after transduction. The cells were then fixed with PFA 0.25%, 0.5% saponin (Merck) with the addition of 1:25 RNAase inhibitor (Promega, N2615). Then anti-DsRed (1:25, Clontech, 632496) or isotype control was incubated for 1 h at RT followed by 1 h incubation with secondary antibody Alexa Fluor 647 goat anti-rabbit (1:100, Thermo Fisher Scientific, A-21245) to detect mCherry+ hPTC. All the antibodies were diluted in 0,5% saponin (Merck) with the addition of 1:100 RNAase inhibitor (Applied Biosystems, N8080119). All the solutions were diluted in RNAase-free PBS prepared with DEPC water (Merck). The procedure was carried out on ice. Finally, hPTC were incubated with DAPI (1:1000, Thermo Fisher Scientific) to perform the DNA content analysis and sorted on the FACSAria III BD (Bioscience). Alexa Fluor 647 secondary antibody was excited by a 633 nm laser line, DAPI was excited by a 405 nm laser line. Data were analysed by FacsDiva software (Beckman Coulter).
Single cell RNA-sequencing
hPTC were harvested and filtered using the FlowmiTM Tip strainer (Miltenyi Biotec) to remove clumps and debris, followed by viability assessment with trypan blue (Merck) staining and Propidium Iodide (Miltenyi Biotec)/Calcein AM (Thermo Fisher Scientific) staining. The single cell suspension with at least 98% viability was run on a 10x Chromium Single Cell instrument (10x Genomics) following Manufacturer’s instructions, as previously described53. 3′ gene expression libraries were constructed as previously described53 and were sequenced on an Illumina NextSeq550 (Illumina Inc., RRID:SCR_020138).
Wild-type C57Bl/6 mice underwent IRI at 7 weeks of age and were sacrificed 2 (n = 2) and 30 days (n = 3) after AKI. Healthy mouse kidney (n = 1) was used as control. Kidneys were minced into 1-mm pieces with a razor blade and incubated at 37 °C in enzyme dissociation buffer containing 250 U/ml Liberase (Roche) and 40 U/ml DNase I (Merck). After 10 min, the solution was transferred to a Miltenyi C-tube, and the gentleMACS D1 program was run (Miltenyi Biotec). These steps were repeated twice. The reaction was stopped by adding 10% FBS. The solution was then passed through a 40 µm cell strainer. The dissociated cells were then incubated with RBC lysis buffer. We then proceeded to remove the dead cells employing the dead cell removal kit (Miltenyi Biotec, 130-090-101) according to manufacturer’s instruction. Cells were then filtered using the FlowmiTM Tip strainer and run on a 10x Chromium Single Cell instrument (10x Genomics). This method generated single cell suspension with greater than 95% viability.
Bioinformatic analysis
Preprocessing (hPTC & Mice)
Raw sequencing data were processed using the 10x Genomics Cell Ranger pipeline (version 3.0.1). First, cellranger mkfastq demultiplexed libraries based on sample indices and converted the barcode and read data to FASTQ files. Second, cellranger count took FASTQ files and performed alignment to the human GRCh38 and mouse mm10 reference genome59 respectively, and then proceeded with filtering and unique molecular identifier (UMI) counting.
Quality check and normalization (hPTC & Mice)
Next, by means of scanpy toolbox (v1.7.2)60 we performed the data analysis starting from quality control on both datasets to remove poor-quality cells and badly detected genes. For hPTC we filtered out cells with a mitochondrial read rate >20% and expressed <350 genes, and for mice, >40% and <500 respectively. Cell-specific biases were normalized by dividing the measured counts by the size factor obtained through the scran computeSumFactors method, which implements the deconvolution strategy for scaling normalization61. Finally, all counts were log-transformed after addition of a pseudocount of 1. After quality control and filtering, we obtained 11,054 hPTC, 3,625 cells from healthy mouse kidney, 2,690 cells and 2,937 cells from IRI at day 2 and 30 respectively.
Batch correction, dimensional reduction, and visualization (hPTC & Mice)
Next, we mitigated the batch effect in both datasets through the ComBat method62 to later proceed with feature selection to keep “informative” genes only used for dimensional reduction through principal component analysis (PCA). The first 50 principle components (PCs) were used to construct a neighborhood graph of observations63 through the pp.neighbors function, which relies on the Uniform Manifold Approximation and Projection (UMAP) algorithm to estimate connectivity of data points.
Downstream analysis of hPTC
As first step, we clustered data by tl.louvain function at different resolutions (0.5, 1, 1.5, 2) and 0.5 proved to be the best, producing eleven clusters of hPTC in vitro. To annotate the clusters, we obtained and defined the marker genes for each cluster, we ran the tl.rank_gene_groups function using the Wilcoxon rank-sum method. Cell cycle analysis was performed by creating two lists of genes associated to the S and G2/M phases based on cell cycle genes previously defined64. Next, we performed cell cycle scoring through the tl.score_cell_cycle_genes function to score S and G2/M phases. To score a gene list, the algorithm calculates the difference of mean expression of the given list and the mean expression of reference genes. To build the reference, the function randomly chooses a bunch of genes matching the distribution of the expression of the given list. Finally, we used Monocle v. 2.14.0 to generate a trajectory with “selected” clusters to infer the consequentiality of events65. For this analysis, we excluded clusters 1 and 6 based on the expression of solute carrier family (Slc) genes, which demonstrated that they were not proximal tubular cells but cells belonging to other portions of nephron. As input to Monocle’s Reversed Graph Embedding algorithm, we selected a set of 2188 highly variable genes contributing to the second principle component defined via the PCA, which captures variation in cellular maturation stages.
Downstream analysis of mice
At first, we clustered data at different resolutions as for hPTCs and obtained the marker genes for produced clusters. Based on results we kept data at resolution 1, the clusters that either did not express specific cell type genes or expressed marker genes of different cell types have been iteratively subclustered (louvain function with resolution equal to 0.5). Next, we proceeded with the analysis by isolating proximal tubule clusters and recalculated the neighborhood graph on the latent space in order to cluster and annotate the data. After clustering, we evaluated the normalized count distribution in order to check the presence of polyploid cells. To this end, we binned the normalized counts and we made a barplot showing the distribution of the 5 obtained bins (2000, 2500, 3000, 3500, 4000) in clusters grouped according to the ribosome distribution. Finally, we used Monocle2, with default parameters and 2939 highly variable genes driving the maturation from t2 to t30, to perform a trajectory analysis and infer the consequentiality of events.
Statistics and reproducibility
Comparison between groups was performed by two-sided Mann-Whitney test or through the analysis of variance for multiple comparisons (ANOVA for repeated measures) with Bonferroni post hoc analysis or with fisher’s exact test (two-tailed p-value). A p-value <0.05 was considered statistically significant. Statistical analysis was performed using SPSS (RRID:SCR_002865) and OriginPro (RRID:SCR_015636) statistical software. Correlation between the percentage of polyploid TC and the fibronectin deposition was tested using Pearson correlation coefficient. Kaplan–Meier estimates were used to generate an overall survival curve for Pax8/WT and Pax8/SAV1 ko mice after nephrotoxic injury, and for Pax8/WT and Pax8/YAP1 ko mice after nephrotoxic injury. Differences among groups were assessed by log-rank test.
Figures1e, f, a representative experiment out of 4 is shown. Figures1n, 7a, a representative experiment out of 2 is shown.
Reporting summary
Further information on research design is available in theNature Research Reporting Summary linked to this article.
Data availability
The authors declare that all data supporting the findings of this study are available within the article and its Supplementary Information Files. Processed data for all human and mouse scRNA-seq libraries generated in this study have been deposited in the Gene Expression Omnibus (GEO) database under accession code GSE212275 for the human dataset and GSE212273 for the mouse dataset.Source data are provided with this paper.
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Acknowledgements
This research was funded by the European Union’s Marie Sklodowska-Curie fellowship program to L.D.C., grant agreement No. 845774. This study was also funded by the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation program (grant agreement no. 101019891) to P.R and by PRIN: progetti di ricerca di interesse nazionale (2017T95E9X) to P.R. L.D.C. is also the recipient of a L’Oreal-Unesco for women in science award. H.J.A. is supported by the Deutsche Forschungsgemeinschaft (AN372/314-4, 27-1, 30-1).
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1. These authors contributed equally: Elena Lazzeri, Paola Romagnani.
Authors and Affiliations
Department of Experimental and Clinical Biomedical Sciences “Mario Serio”, University of Florence, Florence, 50139, Italy
Letizia De Chiara,Carolina Conte,Maria Lucia Angelotti,Alice Molli,Giulia Antonelli,Maria Elena Melica,Anna Julie Peired,Marta Donati,Gilda La Regina,Luigi Cirillo,Laura Lasagni,Elena Lazzeri&Paola Romagnani
Department of Experimental and Clinical Medicine, University of Florence, Florence, 50139, Italy
Roberto Semeraro,Laura Maggi&Francesco Annunziato
Translational immunology, Instituto de Investigación Sanitaria del Principado de Asturias ISPA, 33011, Oviedo, Asturias, España
Paula Diaz-Bulnes
Nephrology and Dialysis Unit, Meyer Children’s University Hospital, Florence, 50139, Italy
Benedetta Mazzinghi,Alice Molli,Luigi Cirillo,Francesca Becherucci&Paola Romagnani
Medical Genetics Unit, Meyer Children’s University Hospital, Florence, 50139, Italy
Samuela Landini
Nephrology, Dialysis and Transplantation Unit, Careggi University Hospital, Florence, 50134, Italy
Marco Allinovi
Nephrology and Dialysis Unit, Santo Stefano Hospital, Prato, 59100, Italy
Fiammetta Ravaglia&Francesco Guzzi
Department of Experimental & Clinical Medicine, Imaging Platform, University of Florence, Florence, 50139, Italy
Daniele Guasti&Daniele Bani
Department of Information Engineering, University of Florence, Florence, 50139, Italy
Alberto Magi
Flow Cytometry Diagnostic Center and Immunotherapy (CDCI), Careggi University Hospital, Florence, 50134, Italy
Francesco Annunziato
Division of Nephrology, Department of Internal Medicine IV, LMU Hospital, Munich, 80336, Germany
Hans-Joachim Anders
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Contributions
L.D.C., E.L., and P.R. designed the study and interpreted the data. L.D.C. performed or supervised all the experiments. E.L. performed the analysis of biopsies and Confetti mice. C.C. and M.D. performed flow cytometry and cell cycle analysis and mouse experiments. R.S. analysed the data from the scRNA-seq analysis. P.D.B. performed the knock-down and ChIP assays and helped with mouse experiments. M.L.A and G.A. designed and performed immunofluorescence and confocal microscopy. B.M. carried out all scRNA-seq and assisted with data analysis. S.L. validated and sequenced the single-cell libraries. A.J.P. and A.M. carried out mouse experiments. L.M. performed cell sorting experiments. M.E.M performed GFR measurement and mouse experiments. M.A., F.G., and F.R. organized patient tissue collection, assisted with statistical analysis, and scored the human samples blinded. G.L.R. performed mouse genotyping and assisted with mouse experiments. D.G. and D.B. performed electron microscopy experiments. L.C. and F.B. assisted with statistical analysis and critically revised the manuscript. A.M. helped with scRNA-seq analysis. F.A. assisted and advised on flow cytometry data interpretation. L.L. and H.J.A. critically revised and edited the manuscript and advised on data interpretation. P.R., E.L., H.J.A., and L.D.C. wrote the manuscript and organized the figures. All authors read and approved the final manuscript.
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Correspondence to Elena Lazzeri or Paola Romagnani.
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De Chiara, L., Conte, C., Semeraro, R. et al. Tubular cell polyploidy protects from lethal acute kidney injury but promotes consequent chronic kidney disease. Nat Commun13, 5805 (2022).
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Fig. 1: The majority of tubular cells (TC) entering cell cycle becomes polyploid or dies following AKI.
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Fig. 2: Polyploidization of human proximal tubular cells (hPTC) is controlled by YAP1.
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Fig. 3: Polyploidization of proximal tubular cells (PTC) after AKI in mice is controlled by YAP1.
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Fig. 4: YAP1 controls polyploidization via AKT1, E2F7 and E2F8 activation.
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Fig. 5: Early tubular cell (TC) polyploidization preserves residual kidney function and assures survival during AKI.
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Fig. 6: Pax8/SAV1 ko healthy mice show increased tubular cell (TC) polyploidization and spontaneously progress toward CKD.
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Fig. 7: YAP1-driven tubular cell (TC) polyploidization attenuates AKI but aggravates AKI-CKD transition.
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Fig. 8: YAP1-related tubular cell (TC) polyploidization correlates with fibrosis in human kidney biopsies.
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Fig. 9: Time-dependent inhibition of tubular cells (TC) polyploidization attenuates AKI-CKD transition.
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188198 | http://spiff.rit.edu/classes/phys283/lectures/impedance/impedance.html | Copyright © Michael Richmond. This work is licensed under a Creative Commons License.
Impedance of mechanical waves
What is "impedance", in general?
How can we calculate the impedance of a string?
Check the limiting cases
Impedance at the boundary
Impedance and the flow of energy
Impedance depends on tension, too
Impedance in electrical devices
For more information
What is "impedance", in general?
In ordinary English, the word impede has the following definition:
To stop in progress; obstruct; hinder
The word's origin, in Latin, refers to "entangling the feet."
This meaning transfers broadly into the jargon of physics. Impedance is a measure of the degree to which something is prevented from moving forward. In the world of electrical circuits, current is the quantity which is being prevented from progressing (and we'll come back to that later). In the world of strings and waves, however, the important item is the transverse velocity of the string.
In particular, physicists define the impedance of a system which sends waves down a string in the following manner.
Or, in short,
Imagine that we start with a long string, stretched under some tension. We can create waves by attaching a motor to one end of the string, pulling and pushing it with some frequency. Those oscillations will propagate down the string, carrying some of the energy of the motor to the other end of the string.
A system with large impedance will require us to pull the string with a large force in order to give the string some particular transverse velocity; a system with small impedance, on the other hand, will require only a small force from the motor to create the vertical motions of the same speed.
In this context, impedance sounds a bit like inertia. So we might expect that the linear mass density of the string will be involved in the calculation of impedance.
It might also remind you a bit of this equation from the past:
In a damped harmonic oscillator, there is also some connection between the force applied to a system, and the resulting speed of oscillation.
How can we calculate the impedance of a string?
Okay, now that we know how impedance is defined, let's try to calculate its value for some simple string system. We'll begin with a string of linear mass density μ stretched with tension T. Near one end of the string, we place a motor with a small arm that attaches to the string. When the motor turns, the arm moves up and down, imparting some force F at a fixed frequency to the string. As a result, the string moves up and down with some particular vertical velocity v y; in addition, a wave moves to the right along the string, away from the motor.
The diagram below shows a little section of the string next to the arm which provides the driving force. Consider the point P on the string.
At point P, the tension in the string can be broken into horizontal and vertical components. We need to find the vertical force F y in order to calculate the impedance.
The magnitude of the vertical force is simply the vertical component of the tension.
Now, as long as the string deflects by only a small angle (meaning that all vertical motions are much smaller than the length of the string), then the ratio of the vertical to horizontal components is approximately
Therefore, we can express the magnitude of the vertical force as
If one looks at the picture again, we can see that at this moment, the vertical force exerted on the string by the motor is downwards, or in the negative y direction; but the slope of the string is positive, so
Let's assume that the position of the string is given by our old friend
Q: Write an expression for the force in the y-direction.
Right.
Okay, so that's the force.
In order to compute the impedance, we also need to know the speed of the string in the transverse, or vertical direction.
Q: What is the speed of the string in the vertical direction?
Yes, the vertical speed is the derivative of position with respect to time.
We can put these two expressions together to compute the impedance of this section (Region I) of the string.
Some of the terms cancel, leaving us with tension times wave number, divided by angular frequency.
But that combination of wave number and angular frequency is related to the velocity of the wave ALONG the string (not transverse to it), so
The velocity of waves travelling along this section of the string also depends on tension
and it turns out that the mechanical impedance of a long string depends on two things: its tension, and its linear mass density.
Check the limiting cases
Let's see if this makes any sense by considering the limiting cases.
very dense string As the density of the string increases, the impedance increases. That means that if we keep applying the same force, but increase the density, the vertical speed of the string should decrease toward zero.
Yes, that make sense.
very tenuous string On the other hand, if we keep applying the same force, but DEcrease the density, the vertical speed of the string should increase; as the density approaches zero, the transverse speed of the string should approach infinity.
That makes sense, too.
very high tension The tension of the string is important: the higher the tension, the smaller the ratio of the motor's force to the tension -- which means the smaller the angle of the string, and the smaller the vertical motion.
More tension leads to smaller vertical motions, which means higher impedance. Yes, that sounds correct.
very small tension If the tension in the string is initially very small, then the force we apply through the motor will dominate; this will eventually invalidate our assumption that the angle of the string away from the horizontal is small, but it seems reasonable that the smaller the tension, the faster the string will move vertically, and thus the smaller the impedance.
Impedance at the boundary
The really interesting aspects of impedance occur when we consider the boundary between two different pieces of string. Suppose we have two pieces of string with identical tensions, but different linear mass densities.
What will happen if a travelling wave from Region I
reaches the boundary?
As you may recall, we concluded last time that the coefficients of the reflected and transmitted waves are given by the expressions
But since the impedance is directly related to the wave number,
and since the tension and angular frequency are the same in both regions, we can simply replace the wave number k in each region with the impedance Z for that region:
Impedance and the flow of energy
So, now we know how the change in impedance at a boundary between two strings is related to the coefficients of the reflected and transmitted waves at that boundary.
Suppose we send a travelling wave
moving along a string of impedance Z I, so that it approaches a boundary with a string of impedance Z II.
Q: If we want to transfer as much energy as possible to the new string, in Region II, what value should the impedance Z II have? a) Z II should be (nearly) zero b) Z II should be (nearly) infinite c) Z II should be somewhere in between
The answer is not a), but c). In fact, the best choice for the impedance in Region II is
Why? Let's consider several cases.
Z II close to zero In this case, the coefficient of the transmitted wave will be as large as it can possibly be: TWICE the size of the incident wave. That's the good news, and it might make one think that the transmitted wave will carry the maximum amount of energy into the new region. But consider two things.
First, the coefficient of the REFLECTED wave is not zero, but negative one times the coefficient of the incident wave. The amplitude of the reflected wave is the same size as that of the incident wave --- so it carries all the energy of the incident wave back, away from Region II.
Remember that the energy carried by a wave depends on several factors. One expression is
Yes, if the impedance in Region II is close to zero, then the coefficient A will be large; but a very small impedance also means (for our fixed tension) that the linear mass density must be very small. As the linear mass density approaches zero, the power carried by any wave through this medium must also approach zero.
Z II close to infinity This is simple: the amplitude of the transmitted wave goes to zero, so the amount of energy it can carry will also go to zero. In the limiting case, the incident wave hits a "brick wall," and the end of string I is fixed at zero.
No energy will move into Region II in this case.
Z II equal to Z I If the impedance on the right side of the boundary is exactly the same as the impedance on the left side of the boundary -- and the tension on both sides is the same -- then it must mean that the linear mass density is the same on both sides, too.
Qualitatively, the string doesn't change at all at the boundary, so ... there is no boundary. The incident wave just keeps travelling to the right, into Region II, carrying all its energy.
Quantitatively, the expression for the reflected coefficient goes to zero, and the expression for the transmitted coefficient goes to one.
The maximum transfer of energy occurs NOT when the impedance in Region II is zero, but when the impedance values are the same on both sides of the boundary, a condition called impedance matching.
One can find analogous situations in other sorts of systems. There's a good list in Transverse Waves by David Morin. I'll list just a couple.
Ultrasound technicians send sound waves from a device into the body, then listen for echoes to create images of internal structures. If any of the sound waves are reflected off the boundaries between device and air, or air and skin, on the way in, or on the way out, then the images lose contrast and resolution.
Would it help to place the patient and device into a vacuum chamber? In a gas of very low density, the speed of the sound waves would be faster than in air -- but the energy transferred by those waves to the skin and tissue would be very much smaller. So decreasing the "resistance" of the air to sound waves would not help.
In order to minimize reflections, technicians spread a special gel on the skin, then push the device into the gel. This removes the device-air interface, and replaces it with device-gel and gel-skin boundaries. The gel is specially designed to have impedance which is similar to both the device and the skin.
The gears on a bicycle sit between the legs of the cyclist and the wheels of the bike. If the rider pushes with some particular force and frequency, the a gear setting which is very low will provide little resistance (that's good), but also move the wheels only a small distance forward for each revolution of the pedals (that's bad). A gear setting which is very high will turn the wheels a large distance for every cycle of the pedals (good), but do so very slowly for the given force (bad).
In between these two extremes will be some particular gear setting which maximizes the speed of the bicycle for the rider's leg motions. Choosing the right gear is equivalent to matching the impedance of the rider and the wheels.
Impedance depends on tension, too
The examples so far have assumed that the tension in the two strings is identical ... but that doesn't necessarily have to be true. Although it seems a bit awkward, one can imagine setting up equipment which would allow two strings to have different tensions, yet remain connected in a physically meaningful manner. All it takes is a ring sliding without friction along a vertical pole.
If we are allowed to change both the linear mass density μ and the tension T, then we can devise two strings with quite different properties which end up yielding the same impedance; all it takes is
In fact, it is even possible to create a boundary with no reflection between a string and ... a pool of liquid. Huh?
No, it's true. As mentioned earlier, one can view the impedance of a mechanical system as the ratio of the tranverse force on a medium to the transverse velocity of the medium. Suppose that we place the end of a string into a pool of water, or oil, or some other liquid. When a travelling wave reaches the end of the string, it causes that end to oscillate by exerting a force in the transverse direction on the string; that force leads to some particular tranverse velocity.
Objects which move through any liquid will experience a force opposing their motion, the liquid equivalent to air resistance. In many cases, this resistive force is linearly proportional to the speed of the object.
So, if we choose a liquid which happens to have a resistive coefficient b which is exactly the same as the impedance of the string
then the wave will transfer all its energy into the pool of liquid.
Impedance in electrical devices
We've been discussing mechanical impedance, but the concept is a very important one in the area of electric circuits as well. In that case, we are still interested in a ratio of a "driving force" to some sort of "velocity", but the quantities are a bit different.
driving force becomes voltage
velocity becomes the motion of charge, or current
So, in the case of a direct-current circuit, the impedence is simply
Q: Is there a more common name for impedance in a direct-current circuit?
Sure -- we just call it "resistance."
But when the voltage varies with time, as in an AC circuit, this relationship can become much more complicated. Electrical devices such as inductors and capacitors interact with time-varying voltage in particular ways; yes, the current usually does change over time, with the same frequency as the voltage; but not always at the same time.
In other words, in AC circuits, one must pay attention to the phase of the quantities, as well as to their amplitudes. The topic becomes rather complex, and it's one in which I have very little experience or knowledge.
But one thing that IS the same as in our mechanical system is that the flow of energy between circuits can be maximized by matching the impedance of the circuits at their junction.
For more information
David Morin at Harvard University has written a nice chapter describing transverse waves; the material on impedance is a good reference.
Transverse Waves by David Morin (PDF)
Copyright © Michael Richmond. This work is licensed under a Creative Commons License. |
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Measuring Cups – Definition with Examples
Home » Math Vocabulary » Measuring Cups – Definition with Examples
What Is a Cup?
Cup Definition
How to Measure Using a Cup?
Solved Examples
Practice Problems
Frequently Asked Questions
What Is a Cup?
A cup is a unit of volume measurement of volume equal to 16 tablespoons, pint, quart, or 8 fluid ounces.
It is used in cooking to measure liquids and powdery substances.
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Cup Definition
A cup is a customary unit used to measure volume. The U.S. measurement units for capacity or volume includes ounces, cups, pints, quarts, and gallons.
1 cup tablespoons pint quart fluid ounces.
Cup measurements are commonly associated with cooking or serving. A measuring cup generally measures liquids such as milk, water, oil, etc., and powdered substances such as sugar, flour, baking soda, etc. Based on these two types of measurements, it is respectively divided into:
Liquid measuring cup
Dry measuring cup
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How to Measure Using a Cup?
There are various scale markings on a measuring cup at different heights. They help in cup measurement. The substance to be measured is poured into the cup until the required level is reached. There are different measuring cup sizes. A standard set includes cup, cup, cup and 1 cup.
Generally, fluid ounces and pints are considered the two standard units of measurement for measuring cups. If the liquid reaches the height of 6, then there are 6 oz or cup of liquid. We can find fractions of a cup between two numbers.
Cups, Tablespoons, Teaspoons
Cups can also be measured with the help of tablespoons and teaspoons, both are much smaller units of volume. 1 tablespoon is equal to 3 teaspoons.
Note that
1 tablespoon teaspoons
Let’s understand the cup-teaspoon-tablespoon relationship using the following image.
Cups, Milliliters, Ounces, Pints
A metric cup is a bit different than a US cup. 1 Cup is equal to 8 fluid ounces in US Standard Volume.1 metric cup is 250 milliliters (which is about 8.5 fluid ounces). A US cup is about 237-240 mL. The following table shows the relationship between cups, milliliters, fluid ounces, pints, and tablespoons.
How to Make Cup with Measuring Cups?
Usually, the measuring cup sizes which are common to find are 1 cup, cup, cup, and cup. It is challenging to find a cup since measuring sets rarely have a cup that is th of a cup in size.
So, to make cups, you will need to add cups three times, as shown below:
If you are also wondering how to measure cup, cup and cup using tablespoons, let’s revise again using a well-illustrated image!
How to Measure a Cup of Water without Using a Measuring Cup?
You can measure a cup of water even if you don’t have a measuring cup. Using an object as a reference point is very helpful to measure water in the absence of a measuring cup. Having some visual images in our mind can help us remember the water amount to be measured.
Here are some instances typically used as a guide when the required tools or measuring cups are not available:
A teaspoon: size of a finger’s tip from the joint, approximately
A tablespoon: an ice cube’s size, approximately
cup: one large egg’s size, approximately
cup: a tennis ball’s size approximately
1 cup: an apple’s size, approximately
Fun Facts!
When translating recipes from American to British, variations are present in liquid measurement values.
If you use a cup to measure all the ingredients, you will notice that every time the cup weighs differently depending on the ingredient’s density.
A variety of measuring cups and spoons are required when baking and cooking. The culinary industry does not, however, have a set size requirement.
Conclusion
A measuring cup is of two types, namely, liquid and dry measuring cup. The former helps measure the volume of liquid-based things like water, milk, oil, etc., with the help of scale markings. The latter helps measure the quantity of powdered materials like sugar, baking soda, flour, etc.
Knowing how to measure a cup is imperative as it helps in measuring several things in the kitchen for cooking and outside the kitchen as well.
Solved Examples
1. Lilly wants to bake a cake. She knows she will have to add 1 cup of sugar, but she only has tablespoons available. How many tablespoons of sugar equal 1 cup?
Solution:
We know that 1 cup tablespoons.
So, Lilly will have to pour in the sugar 16 times using the tablespoons to equal 1 cup.
2. How many teaspoons will make cups of oil?
Solution:
We know that cups teaspoons.
So, Chris will need to pour in 36 teaspoons of oil to equal cups of oil.
3. How to fill a 2 liter water bottle using a measuring cup?
Solution:
1 cup
2 cups
So, 4 cups
Thus, to fill a 2 liter bottle, we will need to pour 8 cups of water.
4. A recipe requires 1 cup oil. What’s the quantity of oil in fluid ounces?
Solution:
1 cup equals 8 fluid ounces. So, the recipe requires 8 fluid ounces of oil.
5. Nick needs to use cup of washing powder to do his laundry. Which type will he use to measure its quantity: a dry or liquid measuring cup? How should he pour the powder using tablespoons?
Solution:
Nick will use a dry measuring cup as this type of cup is usually used for adding or pouring powdery substances like sugar, baking soda, washing powder, etc.
Also, cup tablespoons. So, he will need to pour 4 tablespoons of powder.
Practice Problems
Measuring Cups - Definition with Examples
Attend this quiz & Test your knowledge.
1
A recipe needs 24 tablespoons of flour. How to add the flour using measuring cups?
1 cup and cup
1 cup and cup
cup and cup
and
CorrectIncorrect
Correct answer is: 1 cup and cup
1 cup tablespoons, and 12 cup = 8 tablespoons.
.
So, 1 cup and another cup will equal 24 tablespoons of flour.
2
Which of the following cannot be measured with the help of measuring cups?
Water and sugar
Milk and flour
Oil and rice
Brick and rocks
CorrectIncorrect
Correct answer is: Brick and rocks
Brick and rocks (neither liquid nor powder) can’t be measured with the help of measuring cups, while all the other items can.
3
How many teaspoons of milk make two and a half cups of milk?
96
106
120
126
CorrectIncorrect
Correct answer is: 120
1 cup teaspoons.
So, 2 cups teaspoons - (i)
And, cup teaspoons - (ii)
Upon adding (i) and (ii), you will be able to reach the final answer as,
.
4
2 cups ___ oz
8
16
32
64
CorrectIncorrect
Correct answer is: 16
1 cup oz
So, 2 cups oz
5
Find the correct option.
1 cup and cup equals 10 ounces.
1 cup and cup equals 12 ounces.
1 cup and cup equals 8 ounces.
1 cup and cup equals 6 ounces.
CorrectIncorrect
Correct answer is: 1 cup and cup equals 10 ounces.
1 cup ounces
cups ounces.
Upon adding the two values, you will get
1 cup and cup .
Frequently Asked Questions
Name two other units of measurement on a measuring cup that can be present besides fluid ounces and pints.
What is the symbol of fluid ounces in math?
What is the equivalent value of pints for 1 cup?
How is a pint expressed in math?
How much is a cup of water in fluid ounces?
What is cup to tbsp conversion? What is 1 cup in tbsp? |
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