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102(b):DEEP DIVE INTO DATA STRUCUTURES AND ALGORITHMS:
Deep dive into Algorithms:
In my first article, I gave a very brief introduction to algorithms, its definition, characteristics, types and their importance. In this article, I’ll give an in-depth explanation.
We will cover:
1. Time complexity
2. Space complexity
3. Asymptotic notations
An algorithm is said to be efficient if it takes less time and consumes less memory. The performance of an algorithm is based on the time complexity and space complexity. Thus, knowing time and space
complexity enables us to make the program behave in required optimal conditions. This makes us efficient programmers.
Time complexity:
Time complexity of a program is the total time required by a program to run till its completion. It does not examine the total execution time but rather it gives information about the variation
(increase or decrease) in execution time when the number of operations (increase or decrease) in an algorithm.
Time complexity of an algorithm is usually expressed using the big-0 notation. This is an asymptotic notation to represent the time complexity.
Time complexity is estimated by counting the number of steps performed by and algorithm to finish its execution.
A basic example could be:
We are told to find the square of any number, n. One solution can be running a loop for n times, and adding the number n to itself n times:
for i = 1 to n:
n = n + n
return n
Or we can simply use the mathematical operator * to find the square:
For the first solution, the loop will run n number of times, thus, the time complexity will be n at least, and as the value of n increases the time taken also increases.
For the second solution, the time taken will be independent of the value of n. Thus, the time complexity will be constant.
Calculating time complexity:
Disclaimer: I have used the same example as study tonight site.
The most common way of calculating time complexity is by using the big-0 notation. This method removes all the constants, so that the running time can be estimated using n as n approaches infinity.
For example:
For the above statement, the running time will not change in relation to n. This will make the time complexity constant.
for x in range(n):
The time complexity for the above algorithm will be linear. The running time of the loop will be directly proportional to n. When n increases, so does the running time.
for x in range(n):
for j in range(n):
The time complexity for the above code will be quadratic. This is because the running time of the above loops is proportional to the square of n.
Types of notation for time complexity:
The various notations used for time complexity are:
1. Big 0 notation (oh)
2. Big Ω notation (omega)
3. Big theta notation (theta)
1. Big Oh:represents worst case of an algorithm’s time
complexity. A set of functions that grow slower than,
or at the same rate as the expression. It indicates
maximum time required by an algorithm for all input
2. Big Omega: represents best case of an algorithms
time complexity. It is a set of functions that grow
faster than or at the same rate as the expression. It
indicates minimum time required by an algorithm for
all input values.
3. Big theta: represents average case of an
algorithm’s time and space complexity. Consists of
all functions that lie in both O and Omega.
Now let’s understand what space complexity is:
Space complexity:
Space complexity is the amount of memory required by the algorithm during the time of its execution. It includes both auxiliary space and space used by the input. Auxiliary space is the extra/
temporary space used by an algorithm.
Space complexity = auxiliary space + input space
The following components in an algorithm generally require space:
1. Data space: this is the space required to store all
the constants and variables (including temporary
2. Instruction space: this is the space required to
store the executable version of the program.
3. Environmental space: this is the space used to store
the environment information used to restore the
suspended function. E.g when a function (x) calls a
function (y), the variables of the function (x) are
stored in the system stack temporarily while
function (y) is being executed inside function (x).
When we are calculating the space complexity, we usually consider data space only and neglect instruction space and environment space.
Calculating the space complexity:
In order to calculate the space complexity, we need to know the amount of memory used by different data variables, although it may be different for different operating systems, the method of
calculation is the same.
Let's compute the space complexity using a few examples:
Disclaimer: the examples below are copied from study tonight.
Example 1:
def myFunction():
Int(a) = x + y + z
return a
In the example above, the total amount of memory used is ( 4 + 4 + 4 + 4 ) = 20 bytes since the variable types are all integers . The additional 4 bytes is for the return value.
This is a Constant Space Complexity, since the memory required is fixed.
Example 2:
def myFunction():
int( sum( int(arr[]), int(n) ))
int(x) = 0
for i in range(n):
x = x + arr[i]
return x
In the above example, 4*n bytes is required for the arr[].
4 bytes each is required for x, n and i
Thus, the total memory required will be (4n + 12), thus the space complexity will be called Linear Space Complexity.
Now that we have a good understanding of time and space complexity, we’re going to look at the standard ways of expressing them.
Asymptotic notations:
These are standard notations that are used to express the time and space required by an algorithm. This is because it's impossible to provide an exact number of time and space required by the
There are three main types of Asymptotic notations:
1. Big O notation
2. Theta notation
3. Omega notation
Big O notation:
This usually represents the worst-case scenario of an algorithm. It is also known as the upper bound of the running time of an algorithm. It represents the complexity of your code using algebraic
It tells us that a function will never exceed a specified time for any value of input n.
Example: let us consider a linear search algorithm where we pass through elements inside an array, one after the other, to search for a particular number. Our worst-case scenario will be if the
element is located at the end of an array. This leads to a time complexity of n, where n is the total number of the elements. When we use Big O notation, we say that the time complexity will be O(n).
This means that the time complexity will never exceed n. This defines the upper bound, which means that it can be less than or equal to n.
Omega notation:
This is used to define the best-case scenario. It defines the lower bound. It indicates the minimum time required for an algorithm.
Theta notation:
Time complexity represented by this notation is the average value or range within which the actual time of execution will be.
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Practice Dot product and Geometry with the exercise "Dead men's shot"
Learning Opportunities
This puzzle can be solved using the following concepts. Practice using these concepts and improve your skills.
Captain Jack Sparrow and his pirate friends have been drinking one night. After plenty of rum, they got into an argument about who is the best shot. Captain Jack takes up some paint and paints a
target on a nearby wall. The pirates take out their guns and start shooting.
Your task is to help the drunk pirates find out which shots hit the target.
Captain Jack Sparrow drew the target by drawing N lines. The lines form a convex shape defined by N corners. A convex shape has all internal angles less than 180 degrees. For example, all internal
angles in a square are 90 degrees.
A shot within the convex shape or on one of the lines is considered a hit.
Line 1: An integer N for the number of corners.
Next N lines: Two space-separated integers x and y for the coordinates of a corner. The corners are listed in a counterclockwise manner. The target is formed by connecting the corners together with
lines and connecting the last corner with the first one.
Line N+1: An integer M for the number of shots.
Next M lines: Two space-separated integers x and y for the coordinates of each shot.
M lines with either "hit" or "miss" depending on whether the shot hit the target or not.
3 ≤ N ≤ 10
1 ≤ M ≤ 10
-10000 < x,y < 10000
-100 -100
100 -100
-100 100
80 -101
0 -100
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On Helly numbers of exponential lattices
Given a set S⊆R^2, define the Helly number of S, denoted by H(S), as the smallest positive integer N, if it exists, for which the following statement is true: for any finite family F of convex sets
in R^2 such that the intersection of any N or fewer members of F contains at least one point of S, there is a point of S common to all members of F. We prove that the Helly numbers of exponential
lattices {α^n:n∈N[0]}^2 are finite for every α>1 and we determine their exact values in some instances. In particular, we obtain H({2^n:n∈N[0]}^2)=5, solving a problem posed by Dillon (2021). For
real numbers α,β>1, we also fully characterize exponential lattices L(α,β)={α^n:n∈N[0]}×{β^n:n∈N[0]} with finite Helly numbers by showing that H(L(α,β)) is finite if and only if log[α](β) is
ASJC Scopus subject areas
• Discrete Mathematics and Combinatorics
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Math 4 Wisdom. "Mathematics for Wisdom" by Andrius Kulikauskas. | Exposition / ResearchProgramBottPeriodicity
Research program
Bott Periodicity: Unfolding of Thinking-in-Parallel
Grothendieck's six operations - Six representations of divisions
Topologies - one-many-all
• Fivesome
□ The five conics - Fivesome
□ Logical connectives - Fivesome
□ Fivefold classification of Sheffer polynomials of A-type zero - Fivesome
• Sixsome ?
• Logical square (and related logic - nonempty (closed) vs. empty (open) system - entropy) - Sevensome-eightsome
Adjoint string of length N - Division of everything into N perspectives
Exact strings.
The cube 8 divisions, 6 conceptions, 12 circumstances.
Understand intuitively
• How could finite exact sequences or perhaps finite strings of adjoint functors model mental chambers, which is to say, holistic cognitive frameworks, divisions of the brain's global workspace, or
simply metaphysical divisions of everything?
• How could we classify strings of adjoint functors?
• How could the eightfold Bott periodicity and/or the classification of real Clifford algebras model how our mind proceeds from one mental chamber to another? Walking through the chambers of the
• Why are there four classical families of Lie groups and Lie algebras? and how might they ground four kinds of geometry: affine, projective, conformal, symplectic?
• In any system, how could the Poincare group model the relations between such four geometries? And how could we model the additional conditions by which a system comes into being?
• How might the degrees of freedom in the gauge theories of the Standard Model
Bott periodicity - Eight-cycle of divisions
• Clifford algebras - clock shifts - consciousness
Shu-Hong's equation. Mobius transformations.
I want to be able to describe the cognitive foundations that account for logic.
Seven-eight kinds of duality. Reps of Sn and GLn. Schur-Weyl duality.
I would like to understand the various kinds of opposites in math and classify them.
Understand the Basics of Logic and Truth. I have made some progress in describing such foundations for truth: Truth as the Admission of Self- Contradiction. Which is to say, truth is inherently
unstable and tentative, the relation of a level with a metalevel.
Divisions of everything: Exact Sequences
Divisions of everything: Adjunctions
Exact sequences of length n <=> Divisions of everything into N perspectives
Divisions of everything: Spin
Elementary particles
• Higgs - spin 0 - onesome
• Matter (leptons, quarks) - spin 1/2 - twosome
• Force carrying bosons - spin 1 - threesome
• spin 3/2 - not known to exist - foursome (knowledge)
• Graviton - spin 2 - fivesome
Divisions of everything: Bott periodicity
Bott Periodicity <=> The eight-cycle of divisions of everything
Norman Anderson's theory and modeling thinking fast and slow.
• Investigating examples of what the operations model.
• Shu-Hong Zhu and the sevensome
Visualizations - unconscious and conscious
Six representations
Grothendieck's six operations, the natural bases of the symmetric functions, Hopf algebras.
Twelve circumstances
One, all, many
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MODE - DEFINITION AND CALCULATION | ZOOLOGYTALKS | 2024
Mean, median and mode all three are Central Tendency or Averages. The Arithmetic mean lies under Mathematical Average and the other two Median and Mode are types of Positional Averages.
“A measure of central tendency is a typical value around which the figures congregate”. The value of central tendency or average always lies between the minimum and maximum values.
Mode is the value of the variable which occurs most frequently in a distribution. Mode is also positional average which can be located by inspection.
• Unimodal- When distribution has one concentration of frequency.
• Bimodal- when it has two concentrations.
• Trimodal- when it has three concentrations.
Step – The data is arranged into a discrete series. So, that the value which has the highest frequency can be known.
For example:
In this data 60 is repeated 5 times. So, mode is 60.
Step : Which variable (x) has the highest frequency (f) is the Mode.
Solution :
Here, the value 45 is repeated 5 times.
So, mode is 45.
Sometimes we cannot depend on this method of inspection to find out mode. In such conditions , prepare grouping and analysis table to found out mode.
Step 1: In this case of bimodal series or trimodal series we prepare.
Form grouping and analysis table and then find out highest frequency.
Step 2: Apply the formula
L = lower limit of the modal class
△[1]= the difference between the frequency of the modal class and the preceding modal class (f[1 ]– f[0])
△[2]= the difference between the frequency of the modal class and the succeeding modal class ( f[1 ]– f[2])
c = class interval of modal class.
f[1 ]= frequency of the modal class.
f[0 ]= frequency of preceding modal class.
f[2 ]= frequency of the succeeding modal class
Mode = 3 Median – 2 Mean
Mode = Mean – 3 (Mean -Mode)
Mean – Mode = 3( Mean- Median)
• Very easy to understand.
• Calculation is simple.
• It remains unaffected by extreme values.
• It is a positional average like median, and can be located easily by inspection.
• Graphic method is also used to determine Mode.
• It is ill-defined and indeterminate.
• It cannot be used in algebraic calculations.
• In bimodal class, the calculation is difficult.
• It is not based on all observations. | {"url":"https://www.zoologytalks.com/mode-definition-and-calculation/","timestamp":"2024-11-11T20:18:02Z","content_type":"text/html","content_length":"132291","record_id":"<urn:uuid:aef373c3-a24a-4bb6-9848-84b57edaab3b>","cc-path":"CC-MAIN-2024-46/segments/1730477028239.20/warc/CC-MAIN-20241111190758-20241111220758-00453.warc.gz"} |
The Theory of Special Relativity: Einstein's Revolution in Understanding Space and Time - StudySaga
The Theory of Special Relativity: Einstein’s Revolution in Understanding Space and Time
Learn about Einstein’s groundbreaking theory of special relativity, which challenged classical ideas about space and time and introduced concepts such as the constancy of the speed of light, time
dilation, and length contraction. Discover the implications of this theory for modern physics and technologies such as GPS.
The Theory of Special Relativity: Einstein's Revolution in Understanding Space and Time
The theory of special relativity, proposed by Albert Einstein in 1905, revolutionized our understanding of space and time. It introduced the concept of the speed of light as a fundamental constant,
and demonstrated that time and space are not absolute, but rather are relative to the observer’s frame of reference.
At the heart of special relativity is the principle of the constancy of the speed of light. This principle states that the speed of light is the same in all reference frames, regardless of the
relative motion of the observer and the source of light. This means that no matter how fast an observer is moving, they will always measure the speed of light to be the same constant value.
Special relativity also introduced the concept of time dilation. According to this concept, time appears to move slower for objects that are moving relative to an observer than for stationary
objects. This effect becomes more pronounced as an object approaches the speed of light.
Another key feature of special relativity is length contraction. This effect states that objects appear shorter in the direction of motion when observed from a frame of reference that is moving
relative to the object.
These ideas challenged the classical view of space and time, which held that they were absolute and independent of an observer’s frame of reference. Special relativity showed that space and time are
relative concepts that are intimately connected, and that the laws of physics are the same in all reference frames.
Special relativity has been verified by many experiments and has important implications for our understanding of the universe. It has led to the development of technologies such as GPS, which rely on
the precise measurement of time and the effects of time dilation on satellite signals.
In summary, the theory of special relativity is a cornerstone of modern physics, introducing the concept of the constancy of the speed of light, time dilation, and length contraction. Its principles
have been confirmed by numerous experiments and have transformed our understanding of space and time.
frequently asked questions
Q: What is the theory of special relativity? A: The theory of special relativity, proposed by Albert Einstein in 1905, is a theory that revolutionized our understanding of space and time. It
introduced the concept of the speed of light as a fundamental constant and demonstrated that time and space are relative to the observer’s frame of reference.
Q: What is the constancy of the speed of light? A: The constancy of the speed of light is a fundamental principle of the theory of special relativity. It states that the speed of light is always the
same, regardless of the motion of the observer or the source of light.
Q: How does special relativity affect time? A: According to special relativity, time is relative to the observer’s frame of reference. This means that time appears to move slower for objects that are
moving relative to an observer than for stationary objects. This effect becomes more pronounced as an object approaches the speed of light.
Q: What is length contraction? A: Length contraction is a concept introduced by special relativity. It states that objects appear shorter in the direction of motion when observed from a frame of
reference that is moving relative to the object.
Q: What are the implications of special relativity for modern physics? A: Special relativity has transformed our understanding of space and time, and its principles have been confirmed by numerous
experiments. It has important implications for modern physics, including the development of technologies such as GPS, which rely on the precise measurement of time and the effects of time dilation on
satellite signals.
Q: How has special relativity been verified by experiments? A: Special relativity has been verified by many experiments, including the famous Michelson-Morley experiment, which demonstrated the
constancy of the speed of light, and experiments involving particle accelerators, which have confirmed the predictions of special relativity about time dilation and length contraction.
Q: What is the difference between special relativity and general relativity? A: Special relativity deals with the laws of physics in non-accelerating reference frames, while general relativity
extends these laws to include accelerating reference frames and the effects of gravity. General relativity is a more comprehensive theory than special relativity and includes its principles as a
special case.
Q: What is time dilation in special relativity? A: Time dilation is a phenomenon predicted by special relativity where time appears to run slower for objects that are moving relative to an observer.
This effect becomes more pronounced as an object approaches the speed of light.
Q: Can the principles of special relativity be observed in everyday life? A: While the effects of special relativity are not typically noticeable in everyday life, they have been observed and
measured in experiments involving high-speed particles and the behavior of GPS satellites.
Q: How does special relativity relate to the concept of simultaneity? A: Special relativity challenges the classical concept of simultaneity, which assumes that two events that occur at the same time
for one observer will also occur at the same time for all observers. In special relativity, events that appear simultaneous to one observer may not appear simultaneous to another observer in a
different frame of reference.
Q: What is the role of the Lorentz transformation in special relativity? A: The Lorentz transformation is a set of equations that describe how measurements of space and time are related between
different frames of reference in special relativity. They allow for the calculation of the effects of time dilation and length contraction.
Q: How has special relativity influenced modern philosophy and culture? A: The theory of special relativity has had a significant impact on modern philosophy and culture. It has challenged
traditional ideas about space and time, influenced the development of science fiction and other forms of popular culture, and sparked philosophical debates about the nature of reality and the limits
of human knowledge. | {"url":"https://studysaga.in/the-theory-of-special-relativity-einsteins-revolution-in-understanding-space-and-time/","timestamp":"2024-11-10T06:23:39Z","content_type":"text/html","content_length":"121206","record_id":"<urn:uuid:495739e0-788d-45f6-9cfd-4e98d0c622e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00086.warc.gz"} |
Colloquium Details
Theory Day
Location: Warren Weaver Hall 109
Date: March 2, 2001, 9:30 a.m.
Host: Yevgeniy Dodis
From Erdos to Algorithms and Back Again
Prof. Joel Spencer
New York University
10:00 a.m. - 10:50 a.m.
Paul Erdos attempted, very often successfully, to prove the existence of mathematical objects with particular properties. His methodologies have been adapted with much success to give efficient
algorithms for creating those objects. The algorithmic approach, provides new insight into the mathematical proofs. In many cases, as we shall illustrate, it has led to new theorems and conjectures.
On the Approximability of Trade-offs
Dr. Mihalis Yannakakis
Bell Labs, Lucent Technologies
10:50 a.m. - 11:40 a.m.
We discuss problems in multiobjective optimization, in which solutions to a combinatorial optimization problem are evaluated with respect to several cost criteria, and we are interested in the
trade-off between these objectives (the so-called Pareto curve). We point out that, under general conditions, there is a polynomially succinct curve that approximates the Pareto curve within any
desired accuracy, and give conditions under which this approximate Pareto curve can be constructed in polynomial time. We discuss in more detail the class of linear multiobjective problems and
multiobjective query optimization.
LUNCH BREAK - 11:40 a.m. - 1:30 p.m.
On the Standard Written Proof
Prof. Johan Hastad
IAS & Royal Institute of Technology
1:30 p.m. - 2:20 p.m.
We use the PCP-theorem and the parallel repetition theorem of Raz together with the long code of Bellare, Goldreich and Sudan to obtain a very interesting proof for any NP statement. This written
proof can be checked in many ways and through natural reductions yield inapproximability results for many NP-hard optimization problems such as satisfying the maximal number of linear equations,
Max-Sat and set-splitting. We will discuss the general proof method, give examples of constructions and show how to analyze these constructions.
What Can You Do in One or Two Passes?
Prof. Ravi Kannan
Yale University
2:20 - 3:10
We consider a collection of graph and matrix problems where the input data is very large. Computing with a small randomly chosen sub-matrix (or subgraph) can be shown to yield approximate solutions
to many problems like the max-cut problem in dense graphs. Here, instead of blindly doing random sampling, we take a two-phase approach. In the first phase, we make one pass through the data to
compute the probability distribution with which to sample. Then in the second phase, we draw a small sample and compute with it. We are able to tackle a much larger class of problems which includes
numerical problems arising in "Principal Component Analysis" as well as graph and combinatorial problems; also, a weaker notion of density, allowing for differing importance (weight) attached to
different rows/vertices suffices. We will discuss in general this two-phase paradigm and argue that it is consistent with the relative growth of the computing resources - speed and space.
Coffee break - 3:10 p.m. - 3:30 p.m.
An Application of Complex Semidefinite Programming to Approximation Algorithms
Dr. David Williamson
IBM Research
3:30 - 4:20
A number of recent papers on approximation algorithms have used the square roots of unity, -1 and 1 to represent binary decision variables for problems in combinatorial optimization, and have relaxed
these to unit vectors in real space using semidefinite programming in order to obtain near optimal solutions to these problems. In this talk, we consider using the cube roots of unity, 1, e**{i(2/3)
pi}, and e**{i(4/3)pi}, to represent ternary decision variables for problems in combinatorial optimization. Here the natural relaxation is that of unit vectors in complex space. We use an extension
of semidefinite programming to complex space to solve the natural relaxation, and use a natural extension of the random hyperplane technique to obtain near-optimal solutions to the problems. In
particular, we consider the problem of maximizing the total weight of satisfied equations x_u - x_v = c (mod 3) and inequations x_u - x_v not= c (mod 3), where x_u is in {0,1,2} for all u. This
problem can be used to model the MAX 3-CUT problem and a directed variant we call MAX 3-DICUT. For the general problem, we obtain a .79373-approximation algorithm. If the instance contains only
inequations (as it does for MAX 3-CUT), we obtain a performance guarantee of 3/(4pi**2)(arccos(-1/4))**2 + 7/12, which is about .83601. This compares with proven performance guarantees of .800217 for
MAX 3-CUT (by Frieze and Jerrum) and 1/3 + 10**{-8} for the general problem (by Andersson, Engebretson, and Hastad). This is joint work with Michel X. Goemans of MIT
In-person attendance only available to those with active NYU ID cards. | {"url":"https://cs.nyu.edu/dynamic/about/news/colloquium/825/","timestamp":"2024-11-06T23:54:23Z","content_type":"text/html","content_length":"13208","record_id":"<urn:uuid:5739d626-a2a4-4fb0-b359-3cddc69f99de>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00130.warc.gz"} |
Strange Calculator
the calculator I use frequently does some strange stuff. One of the things I like about it is that it will preserve what you punched into it briefly, serving like a paper tape to some degree. So I
use it for menial things like calculating sums.
It will seek to find strange stuff as it calculates. This morning I added 4 numbers together to get a sum, those being
as the calculator gives me the answer, it sticks in the middle of where it preserves your input " = 562/5 "
in other words with my input only the added numbers, I get:
35.4+13.3+60.7+3 = 562/5 = 112.4
Pretty neat, but who decided to put that in the software? And is converting the numbers to 562/5 really what the calculator is doing? I always thought it was heavy into logarithms for its functions.
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious
monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
• Threads: 142
• Posts: 16832
Joined: May 15, 2012
I used to use that on-line scientific calculator, but I switched to this one:
The reason I switched is, for some reason, if I go to copy/paste your equation:
Into the one you use, it will display in there as dim and will not do anything. Ultimately, I would end up having to type in every single equation I wanted to do, whereas the one I use now allows you
to copy/paste equations in, with or without spaces.
I have no idea what would cause it to convert into 562/5 and then come to 112.4. I could understand it if it first came to 112.4 and then converted to 562/5, because then it would be giving you a
whole number as a numerator. I have no idea what would cause it to do that first, though.
365.2+156.9+32.1 = 2771/5 = 554.2
365.3+122+489-10 = 9663/10 = 966.3
I'll try the other calculator for a while.
It appears the one I've used sees everything as an equation, and also every number as a fraction, seeking a common denominator. Odd or not?
The reason I always thought calculators use logarithms alot is I use specially tailored online calc. for my checkbook balances. [told that to someone and she said "people still balance their
checkbooks?" - that was funny and sobering at the same time] Anyway some of them would give the incorrect answer by one cent sometimes. That made me think they use logarithms, otherwise wth is going
on with that? No errors like that with the scientific one I have been using.
PS: for checkbooks I like this one:
no such errors
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious
monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
• Threads: 142
• Posts: 16832
Joined: May 15, 2012
Quote: odiousgambit
I'll try the other calculator for a while.
I like it, there are a few quirky things here and there, but it is easy to figure out what it wants.
It appears the one I've used sees everything as an equation, and also every number as a fraction, seeking a common denominator. Odd or common?
Right, but the problem is that it doesn't. 60.7*5 = 303.5 13.3 * 5 = 66.5 35.4 * 5 = 177
It could be doing: 607/10 + 133/10 + 354/10 + 30/10 = 1124/10 = 562/5 = 112.4 but even that makes no sense. Using a denominator of 10 would be the quickest way to turn them all into whole numbers,
and then it could be reducing that down into the simplest possible fraction, then dividing, but why?
Let's try this:
144.57 + 344.12 + 897.3 + 6:
144.57+344.12+897.3+6 = 1391.99
Okay, that time it just gave me the answer, straight-up. I think the calculator may be messing with us.
The reason I always thought calculators use logarithms alot is I use specially tailored online calc. for my checkbook balances. [told that to someone and she said "people still balance their
checkbooks?" - that was funny and sobering at the same time] Anyway some of them would give the incorrect answer by one cent sometimes. That made me think they use logarithms, otherwise wth is
going on with that? No errors like that with the scientific one I have been using.
I'll take a look at that checkbook calculator, but I don't really balance my checkbook, either. I have a few hundred that I keep in there, and other than that, I just deposit exactly what is going to
come out. | {"url":"https://wizardofvegas.com/forum/questions-and-answers/math/12932-strange-calculator/","timestamp":"2024-11-14T08:04:10Z","content_type":"text/html","content_length":"50956","record_id":"<urn:uuid:79acff9a-6e75-404e-bb43-2df5179bc34d>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00849.warc.gz"} |
NIMS News
NIMS Industrial Mathematical Problem Solving Workshop in the First Half of 2019
NIMS held a workshop on solving problems in NIMS industrial mathematics in the first half of 2019, for three days from June 26 to 28
The Industrial Mathematics Problem Solving Workshop is an event designed to bring together mathematical science researchers such as mathematicians, engineers and others to present solutions through
mathematical modeling, optimization, and data analysis.
In this workshop, three industry issues commissioned by companies and an institute were presented: △ scoring of oral health results, △ simulating nanofilter performance prediction, △ predicting
erosion of low-orbital satellite components of atomic oxygen.
About 150 researchers in related fields, consisting of NIMS researchers, NIMS industrial mathematics experts' group, professors, and graduate students, conducted a process of mathematically defining
and discussing the proposed problem and presenting and discussing theoretical methodologies. | {"url":"https://www.nims.re.kr/eng/announce/post/eng_general/33271","timestamp":"2024-11-05T18:50:27Z","content_type":"text/html","content_length":"24757","record_id":"<urn:uuid:aa24457f-6329-4bd7-bcc5-e36a6179bad3>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00507.warc.gz"} |
meters to sq
2,329 square meters to square feet
2,329 Square meters = 25,069.15 Square feet
Area Converter - Square meters to square feet - 2,329 square feet to square meters
How to Convert 2329 sq meters to sq ft
Converting 2329 square meters (sq meters) to square feet (sq ft) is a straightforward process.
The conversion factor between square meters and square feet is approximately 10.7639. So, to find out how many square feet are in a given number of square meters, you simply need to multiply the
square meters by 10.7639.
In this case: 2329 sq meters × 10.7639 ≈ 25069.16 sq ft
Final Answer
Therefore, 2329 sq meters is approximately equal to 25069.16 sq ft.
2,329 square meters in other units | {"url":"https://unitconverter.io/square-meters/square-feet/2329","timestamp":"2024-11-08T02:55:13Z","content_type":"text/html","content_length":"17594","record_id":"<urn:uuid:231b4fa0-4674-4b5e-a2a8-cf47ce9ef278>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00489.warc.gz"} |
Facility location and the analysis of algorithms through factor-revealing programs
Mahdian, Mohammad, 1976-
Other Contributors
Massachusetts Institute of Technology. Dept. of Mathematics.
Daniel A. Spielman.
Terms of use
M.I.T. theses are protected by copyright. They may be viewed from this source for any purpose, but reproduction or distribution in any format is prohibited without written permission. See provided
URL for inquiries about permission. http://dspace.mit.edu/handle/1721.1/7582
In the metric uncapacitated facility location problem (UFLP), we are given a set of clients, a set of facilities, an opening cost for each facility, and a connection cost between each client and each
facility satisfying the metric inequality. The objective is to open a subset of facilities and connect each client to an open facility so that the total cost of opening facilities and connecting
clients to facilities is minimized. As the UFLP is NP-hard, much effort has been devoted to designing approximation algorithms for it. As our main result, we introduce a method called dual fitting
and use it in conjunction with factor-revealing programs to obtain improved approximation algorithms for the UFLP. Our best algorithm achieves an approximation factor of 1.52 (currently the best
known factor) and runs in quasilinear time. We demonstrate the versatility of our techniques by using them to analyze the approximation factors of a cycle cover algorithm and a Steiner packing
algorithm, as well as the competitive factor of an online buffer management algorithm. We also use our algorithms and other techniques to improve the approximation factors of several variants of the
UFLP. In particular, we introduce the notion of bifactor approximate reductions and use it to derive a 2-approximation for the soft-capacitated FLP. Finally, we consider the UFLP in a game-theoretic
setting and prove tight bounds on schemes for dividing up the cost of a solution among players. Our result combined with the scheme proposed by Pal and Tardos shows that 1/3 is the best possible
approximation factor of any cross-monotonic cost-sharing scheme for the UFLP. Our proof uses a novel technique that we apply to several other optimization problems.
Thesis (Ph. D.)--Massachusetts Institute of Technology, Dept. of Mathematics, 2004.
Includes bibliographical references (p. 225-241).
This electronic version was submitted by the student author. The certified thesis is available in the Institute Archives and Special Collections.
Date issued
Massachusetts Institute of Technology. Department of Mathematics
Massachusetts Institute of Technology | {"url":"https://dspace.mit.edu/handle/1721.1/16633","timestamp":"2024-11-14T09:24:47Z","content_type":"text/html","content_length":"22905","record_id":"<urn:uuid:72cd0cdc-e5e0-443d-b045-02fa1bb43c76>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00110.warc.gz"} |
The Hardness of Approximate Optima in Lattices, Codes, and Systems of Linear Equations
We prove the following about the Nearest Lattice Vector Problem (in any l[p] norm), the Nearest Codeword Problem for binary codes, the problem of learning a halfspace in the presence of errors, and
some other problems. 1. Approximating the optimum within any constant factor is A/P-hard. 2. If for some ε > 0 there exists a polynomial-time algorithm that approximates the optimum within a factor
of 2^log0.5-ε n, then every NP language can be decided in quasi-polynomial deterministic time, i.e., NP ⊆ DTIME(n^poly(log n)). Moreover, we show that result 2 also holds for the Shortest Lattice
Vector Problem in the l[∞] norm. Also, for some of these problems we can prove the same result as above, but for a larger factor such as 2^log1-ε n or n^ε. Improving the factor 2^log0.5-ε n to
√dimension for either of the lattice problems would imply the hardness of the Shortest Vector Problem in l[2] norm; an old open problem. Our proofs use reductions from few-prover, one-round
interactive proof systems [FL], BG+], either directly, or through a set-cover problem.
All Science Journal Classification (ASJC) codes
• Theoretical Computer Science
• Computer Networks and Communications
• Computational Theory and Mathematics
• Applied Mathematics
Dive into the research topics of 'The Hardness of Approximate Optima in Lattices, Codes, and Systems of Linear Equations'. Together they form a unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/the-hardness-of-approximate-optima-in-lattices-codes-and-systems-","timestamp":"2024-11-12T14:15:50Z","content_type":"text/html","content_length":"52109","record_id":"<urn:uuid:3e6218a3-8fcf-45cc-93bd-618b5be76def>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00562.warc.gz"} |
Datasets for Optimization Problems
Lehrstuhl II for Mathematics, Focus: Discrete Optimization
RWTH Aachen University
On this webpage, we present data for various problems in the field of discrete mathematics. This data has been used for computational experiments in different research works. Where possible, we
provide background information on the different datasets.
Network Design with Compression (NDPC)
The Network Design Problem with compression (NDPC) is an extension of the classic Network Design Problem (NDP). For further information, we refer to the following papers:
Extended Cutset Inequalities for the Network Power Consumption Problem
, Koster, Phan and Tieves (
Design and management of networks with low power consumption
, Phan (
The following data was originally created for the classic Network Design Problem and was used in the context of the paper
Cutset Inequalities for Robust Network Design
, Koster, Kutschka and Raack (
In this work, the topologies "abilene", "geant", "germany17" and "germany50" have been used. They are taken from the SNDLIB (
), we refer to the same website for a description of the file format. The data can easily be expanded to a complete data set for the NDPC problem.
Dataset NDP (
In the Ph.D. thesis of M. Tieves
Discrete and Robust Optimization Approaches to Network Design with Compression and Virtual Network Embedding
, Tieves (
to appear
the data has been expanded for the NDPC problem. The additional data was created at random, we refer to the above link for further information. The corresponding datasets for the deterministic
problem with average and peak data are provided below. The files are in AMPL readable format.
Dataset "average" (
), Dataset "peak" (
Nodes: Contained in the set V.
Edges: Contained in the set E.
Costs: b (compressors) and c (edge capacities).
Edge Capacity: k.
Commodities: Set Q, For q in Q: dest[q,i]=1 if node i is q's source node, ..=-1 if it is q's sink. d[q] describes the demand volume.
In the Ph.D. thesis of M. Tieves, the robust NDPC problem is also considered. Find below the corresponding dataset (again in AMPL readle format).
Dataset "robust" (
The format is the same as for the deterministic case. Additionally, d_nom[..] describes the nominal volume of a commodity and d_dev[..] describes the maximum deviation of a commodity. Similar,
the scal_nom[...] and scal_dev[..] values refer the the nominal value and the maximum deviation of the compression factor.
Virtual Network Embedding (VNE)
The Virtual Network Embedding Problem (VNE) asks, given a physical (substrate) network with node and link resources, how many virtual networks with node and link demands can be realized on the
substrate network. For further informationen, we refer to the VINO project (
) and to the following two papers:
Virtual network embedding under uncertainty: Exact and heuristic approaches
, Coniglio, Koster and Tieves (
Data Uncertainty in Virtual Network Embedding: Robust Optimization and Protection Levels
, Coniglio, Koster and Tieves (
as well as to the Ph.D. thesis of M. Tieves:
Discrete and Robust Optimization Approaches to Network Design with Compression and Virtual Network Embedding
, Tieves (
to appear
In these works, two different datasets have been used.
The first dataset contains small to medium sized instances. The substrat networks of these instances have been taken from the SNDLIB (
), they corespond to the directed topologies "abilene", "atlanta", "nobel-us" and "polska". The corresponding virtual networks (5 to 32 requests, identifier "-r5-" to "-r32-" in the file name)
contain 12 virtual nodes each. The topologies and the demand values of the virtual networks have been generated at random. The files are in an AMPL readable format.
The second dataset contains large instances. The substrate networks of these instances have been taken from the Internet Topology Zoo (
), they correspond to the directed topologies "bellsouth, "cernet", "cogentco", "deltacom", "digex", "fatman", "intellifiber" and "redbestel". The corresponding virtual networks (5 to 50
requests, identifier "-r5-" to "-r50-" in the file name) contain 12 virtual nodes each. The topologies and the demand values of the virtual networks have been generated at random. The files are
in an AMPL readable format.
Dataset I (small) (
) 24 Mb, Dataset II (large)(
) 100 MB.
Substrate-Network node (param V): "node, capacity"
Substrate-Networkz Arc (param A): "node, node, capacity"
Number of virtual networks: "r = ..."
Number of historic data tarces: "t= ..."
Virtual network "i", number of nodes: "vn[i]:= ..."
Virtual network "i" profit: "p[i]:= ..."
Virtual network "i" topology: "an[i,j,k]:= 1" (arc (j,k) exists for the virt. network i).
Virtual network "i", locality conditions for node j: V_local[i,j] := ... (all nodes on which j may be embedded)
Virtual network "i" historic demand in traffic trace t for the virtual arc (j,k): "d_historical[i,t,j,k]:= ...".
Virtual network "i" historic demand in traffic trace t for the virtual node k: "w_historical[i,t,k]:= ...".
d_nominal and w_nominal describe the average demands over all historic traces, d_deviation and w_deviation refer to the maximum deviations from these averages. All additional data has not been
used in the papers mentioned above.
Equitable Graph Coloring
We refer to the project webpage (
last modified: 12/01/2017 - 14:30 | {"url":"https://www.math2.rwth-aachen.de/en/forschung/projekte/data","timestamp":"2024-11-05T22:54:38Z","content_type":"application/xhtml+xml","content_length":"18451","record_id":"<urn:uuid:7fb19079-7aaa-4157-b72a-3ceca3afe3c8>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00774.warc.gz"} |
Thymio and fractals
This page includes the script of the video. Sentences in italic are those presented in the clip.
Opening credits
Question about Thymio
General concept presentation
• Definition
□ For starters, let's try to define what a fractal figure is, or simply a "fractal". It is a geometrical form whose pattern repeats itself indefinitely. This can be observed at every scale.
• Koch Curve
□ To understand how such a figure works, let's starting by drawing a line.
□ We divide this line into 3 equal pieces and then draw an equilateral triangle on the centre part. We can now remove this centre part, which has been replaced by 2 lines of the same size.
□ We continue our drawing by repeating the same steps on each new part. We could continue doing this indefinitely, but we will stop here. This figure is called the Koch Curve.
□ If we observe one side of our first iteration, we can see that the figure is identical to our whole figure. And if we take a small part of that figure, we can see that again, it is identical
to the whole! This is what we call a fractal.
• Dimension
□ To strictly describe a fractal, we must observe its dimension.
□ Different kinds of dimensions exist: Euclidean, topologic and fractal. In our case, only the fractal dimension is interesting. It is defined by d = log n / log m, where n is the number of
parts obtained after iteration and m is the homothetic ratio (in how many parts we divide the initial part).
□ For the Koch curve, we divide our segment into 3 equal parts (so m=3) and we obtain 4 segments that are each 1/3 of the initial length (so n=4). This gives us a dimension d = ln 4/ln 3 =
□ To see if our Koch curve is a fractal, its fractal dimension must either be greater than its topological dimension, or not a whole number. In our case this is verified, so our Koch curve is a
• New definition
□ We can now give a more rigorously defined definition of a fractal: it is an object which:
☆ is too irregular to be defined by the usual geometric vocabulary
☆ is self-similar (which means that each part of the object resembles its whole)
☆ has a non-whole dimension, or one that is greater than its topological dimension
• What is it used for?
□ Now we know what a fractal is but… what is it used for? Many applications exist, but I will only cite a few.
□ First of all, we can make beautiful figures like this *show a picture* or this *show a picture*. But it also gives us a way of describing physical phenomena such as a fluid's turbulent
behaviour or galaxies' organisation in space, or even the geometrical shape of a romanesco broccoli! *photo of a romanesco broccoli*
□ Fractals are also a way of compressing images with a constant quality for every zoom.
Concept presentation with Thymio
• Now let's use Thymio to draw a fractal composed of a big circle with in it three smaller circles, which each hold three smaller circles. We could repeat this an infinite number of times.
Closing credits | {"url":"https://wiki.thymio.org/en:thymioclipfractale","timestamp":"2024-11-08T04:54:18Z","content_type":"application/xhtml+xml","content_length":"20063","record_id":"<urn:uuid:c1ff4801-3a0e-4bc0-b1d7-b238e97c88c2>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00610.warc.gz"} |
Skin Friction - Friction Drag | Definition | nuclear-power.com
Skin Friction – Friction Drag
The friction drag is proportional to the surface area. Therefore, bodies with a larger surface area will experience a larger friction drag. This is why commercial airplanes reduce their total surface
area to save fuel. Friction drag is a strong function of viscosity.
As was written, when a fluid flows over a stationary surface, e.g., the flat plate, the bed of a river, or the pipe wall, the fluid touching the surface is brought to rest by the shear stress at the
wall. The boundary layer is the region in which flow adjusts from zero velocity at the wall to a maximum in the mainstream of the flow. Therefore, a moving fluid exerts tangential shear forces on the
surface because of the no-slip condition caused by viscous effects. This type of drag force depends especially on the geometry, the roughness of the solid surface (only in turbulent flow), and the
type of fluid flow.
Source: wikipedia.org License: CC BY-SA 3.0
The friction drag is proportional to the surface area. Therefore, bodies with a larger surface area will experience a larger friction drag. This is why commercial airplanes reduce their total surface
area to save fuel. Friction drag is a strong function of viscosity, and an “idealized” fluid with zero viscosity would produce zero friction drag since the wall shear stress would be zero.
Skin friction is caused by viscous drag in the boundary layer around the object. Basic characteristics of all laminar and turbulent boundary layers are shown in the developing flow over a flat plate.
The stages of the formation of the boundary layer are shown in the figure below:
Boundary layers may be either laminar or turbulent, depending on the value of the Reynolds number.
The boundary layer is laminar for lower Reynolds numbers, and the streamwise velocity changes uniformly as one moves away from the wall, as shown on the left side of the figure. As the Reynolds
number increases (with x), the flow becomes unstable. Finally, the boundary layer is turbulent for higher Reynolds numbers, and the streamwise velocity is characterized by unsteady (changing with
time) swirling flows inside the boundary layer.
The transition from laminar to turbulent boundary layer occurs when Reynolds number at x exceeds Re[x] ~ 500,000. The transition may occur earlier, but it is dependent especially on the surface
roughness. The turbulent boundary layer thickens more rapidly than the laminar boundary layer due to increased shear stress at the body surface.
There are two ways to decrease friction drag:
• the first is to shape the moving body so that laminar flow is possible
• the second method is to increase the length and decrease the cross-section of the moving object as much as practicable.
The skin friction coefficient, C[D,friction], is defined by
It must be noted the skin friction coefficient is equal to the Fanning friction factor. The Fanning friction factor, named after John Thomas Fanning, is a dimensionless number that is one-fourth of
the Darcy friction factor. As can be seen, there is a connection between skin friction forces and frictional head losses.
See also: Darcy Friction Factor
For laminar flow in a pipe, the Fanning friction factor (skin friction coefficient) is a consequence of Poiseuille’s law that and it is given by the following equations:
However, things are more difficult in turbulent flows, as the friction factor depends strongly on the pipe roughness. The friction factor for fluid flow can be determined using a Moody chart. For
The frictional component of the drag force is given by:
Calculation of the Skin Friction Coefficient
The friction factor for turbulent flow depends strongly on the relative roughness. It is determined by the Colebrook equation or can be determined using the Moody chart. The Moody chart for Re = 575
600 and ε/D = 5 x 10^-4 returns following values:
Therefore the skin friction coefficient is equal to: | {"url":"https://www.nuclear-power.com/nuclear-engineering/fluid-dynamics/what-is-drag-air-and-fluid-resistance/skin-friction-friction-drag/","timestamp":"2024-11-07T23:20:48Z","content_type":"text/html","content_length":"98413","record_id":"<urn:uuid:c3e74f55-d3d6-47c3-b053-4e0e444196b0>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00114.warc.gz"} |
Celebrate Christmas in Data Analyst Style With SAS!
Christmas is just at the end of this week, so we at team DexLab decided to help our dear readers who love some data-wizardry, with some SAS magic! You can choose to flaunt your extra SAS knowledge to
your peer groups with the below described SAS program.
Celebrate Christmas in Data Analyst Style With SAS!
We are taking things a tad backwards by trying to, almost idiosyncratically complicate things that are otherwise simple. After all some say, a job of a data analyst is to do so! However, be it stupid
or unnecessary this is definitely by far the coolest way to wish Merry Christmas, in data-analyst style.
The main idea behind this SAS program is to show case your creativity and solve otherwise easy problems with some extra complexity and to learn a few cool new tricks that can be later used to solve
real complex problems that may arise in the future.
Confused as to what we mean? Try your hands on the following program and find out what we mean…
If you still do not know how to run SAS programs then take a SAS predictive modeling training with us at DexLab Analytics.
Christmas special SAS program:
DATA A(KEEP=WISH);
LENGTH WISH $15 F1 $5 F2 $9;
DO J= 1 TO 5;
IF J = 1 THEN K = TAN(4.702389315) / ATAN(3.584020431);
ELSE IF J = 2 THEN K = SUM(INPUT(PUT('99',$HEX2.),8.), INPUT(PUT('0',$HEX2.),8.));
ELSE IF J <= 4 THEN K = SUBSTR(PUT('12JAN2004'D , 5.),4, 2);
ELSE IF J = 5 THEN K = ROUND(CONSTANT("PI")*30, 10) - 1;
DO I= 1 TO 9;
IF I = 1 THEN K = 100 / ARCOS(0.078190328);
ELSE IF I = 2 THEN K = MOD(CONSTANT("PI") / 3, 2*CONSTANT("PI")) * 360/(2*CONSTANT("PI")) + 3*4;
ELSE IF I = 3 THEN K = SUBSTR(PUT('12JAN2004'D , 5.),4, 2);
ELSE IF I = 4 THEN K = ROUND(2**6.189823562,1);
ELSE IF I = 5 THEN K = EXP(4.418841708);
ELSE IF I = 6 THEN K = MEDIAN(82,86);
ELSE IF I = 7 THEN K = TAN(4.702389315) / ATAN(3.584020431);
ELSE IF I = 8 THEN K = SUM(INPUT(PUT('69',$HEX2.),8.)-1, INPUT(PUT('0',$HEX2.),8.));
ELSE IF I = 9 THEN K = EXP(4.418841708);
WISH = CATX(' ',F1,F2);
The output of the above given program has not been put here. Because keeping up with Christmas spirit, we have kept that from the readers for reasons of surprise and mystery.
So, try this program out in SAS and tell us the output in the comment section below!
Breaking the program down to explain how it works:
With the BYTE() function you can create character values. This function helps to return the character value against the ASCII code. Try running the BYTE() function and see what happens, and what it
Try and run the program mentioned below and see the log:
data _null_;
x = byte(65);
put x=;
The result: A, the above program will return A.
Just like that you can create A to Z alphabets with the following program:
data _null_;
do i = 65 to 90;
x = byte(i); put x;
Note: we recommend that you see the log after submitting the above mentioned code.
Here are the steps:
Step 1 : TAN(4.702389315) returns 100
Step 2 : ATAN(3.584020431) returns 1.2987
Step 3 : = 100 / 1.2987 is equal to 77
Step 4: BYTE(77) returns M of ‘MERRY’
Wondering how to know the exact value of the angle before making use of the TRIGNO function?
Use the Goal Seek Feature of MS Excel for the same. We offer advanced excel training in Delhi for those keen on learning such functions.
Interested in a career in Data Analyst?
To learn more about Machine Learning Using Python and Spark – click here.
To learn more about Data Analyst with Advanced excel course – click here.
To learn more about Data Analyst with SAS Course – click here.
To learn more about Data Analyst with R Course – click here.
To learn more about Big Data Course – click here. | {"url":"https://m.dexlabanalytics.com/blog/celebrate-christmas-in-data-analyst-style-with-sas","timestamp":"2024-11-13T22:04:06Z","content_type":"text/html","content_length":"55831","record_id":"<urn:uuid:c49aa96a-b033-48fd-9476-637623b389c7>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00373.warc.gz"} |
Prasad Tetali's research interests are in the areas of discrete math, probability and theory of computing, Markov chains, isoperimetry and functional analysis, combinatorics, computational number
theory and algorithms.
Areas of Expertise (7)
Probability and Theory of Computing
Isoperimetry and Functional Analysis
Computational Number Theory
Discrete Math
Markov Chains
Media Appearances (1)
Movie Math: Tetali's Equations Seen in Film
Carnegie Mellon University online
In the movie, "Jerry & Marge Go Large," a man finds a legal loophole in lotteries. Carnegie Mellon University's Prasad Tetali wrote the on-screen calculations to explain how the math behind that
loophole works.
Industry Expertise (1)
Accomplishments (3)
SIAM Fellow (professional)
Georgia Tech’s Regents Professor
Fellow of the American Mathematical Society
Education (2)
Courant Institute of Mathematical Sciences: Ph.D. 1991
Indian Institute of Science: M.S. 1987
Affiliations (3)
• Society for Industrial and Applied Mathematics (SIAM)
• American Association for the Advancement of Science (AAAS)
• American Mathematical Society (AMS)
Event Appearances (5)
Seminar on Optimization
(2017) Simons Institute Berkeley, CA
Plenary Speaker
(2017) Shanghai Conference on Combinatorics China
Combinatorics Seminar
(2017) Stanford University Palo Alto, CA
Recent Trends in Combinatorics (reunion)
(2016) AMS Special Session Minneapolis, MN
Probabilistic and Extremal Combinatorics DownUnder
(2016) Monash Workshop Melbourne, Australia
Research Grants (3)
“Collaborative Education: Data-driven Discovery and Alliance"
NSF Grant 1839339 TRIPODS+X:EDU $200,000
For 24 months, starting 1/1/2019
“Discrete convexity, curvature, and implications”
NSF Grant DMS-1811935 $190,000
For 36 months, starting 8/2/2018
On creating the “Transdisciplinary Institute for Advancing Data Science (TRIAD)”
NSF TRIPODS Grant (Phase 1) 1740776 $1,500,000
For 36 months, starting 9/1/2017
Articles (5)
On Min Sum Vertex Cover and Generalized Min Sum Set Cover
SIAM Journal on Computing
2023 We study the Generalized Min Sum Set Cover (GMSSC) problem, wherein given a collection of hyperedges [] with arbitrary covering requirements [], the goal is to find an ordering of the vertices
to minimize the total cover time of the hyperedges; a hyperedge [] is considered covered by the first time when [] and many of its vertices appear in the ordering.
On the bipartiteness constant and expansion of Cayley graphs
European Journal of Combinatorics
2022 Let G be a finite, undirected, d-regular graph and A (G) its normalized adjacency matrix, with eigenvalues 1= λ 1 (A)≥⋯≥ λ n≥− 1. It is a classical fact that λ n=− 1 if and only if G is
bipartite. Our main result provides a quantitative separation of λ n from− 1 in the case of Cayley graphs, in terms of their expansion. Denoting h o u t by the (outer boundary) vertex expansion of G,
we show that if G is a non-bipartite Cayley graph (constructed using a group and a symmetric generating set of size d) then λ n≥− 1+ c h o u t 2 d 2, for c an absolute constant.
On the number of independent sets in uniform, regular, linear hypergraphs
European Journal of Combinatorics
2022 We study the problems of bounding the number weak and strong independent sets in r-uniform, d-regular, n-vertex linear hypergraphs with no cross-edges. In the case of weak independent sets, we
provide an upper bound that is tight up to the first order term for all (fixed) r≥ 3, with d and n going to infinity. In the case of strong independent sets, for r= 3, we provide an upper bound that
is tight up to the second order term, improving on a result of Ordentlich–Roth (2004).
Volume growth, curvature, and Buser-type inequalities in graphs
International Mathematics Research Notices
2021 We study the volume growth of metric balls as a function of the radius in discrete spaces and focus on the relationship between volume growth and discrete curvature. We improve volume growth
bounds under a lower bound on the so-called Ollivier curvature and discuss similar results under other types of discrete Ricci curvature.
Markov chain-based sampling for exploring RNA secondary structure under the nearest neighbor thermodynamic model and extended applications
Mathematical and Computational Applications
2020 Ribonucleic acid (RNA) secondary structures and branching properties are important for determining functional ramifications in biology. While energy minimization of the Nearest Neighbor
Thermodynamic Model (NNTM) is commonly used to identify such properties (number of hairpins, maximum ladder distance, etc.), it is difficult to know whether the resultant values fall within expected
dispersion thresholds for a given energy function. | {"url":"https://expertfile.com/experts/prasad.tetali/prasad-tetali","timestamp":"2024-11-08T12:16:51Z","content_type":"text/html","content_length":"76629","record_id":"<urn:uuid:aee223ae-97a8-449c-a9a2-3667aad88c06>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00540.warc.gz"} |
How many patterns are there in nature? - Lisa Andersen : Surfer Girl Mentor
How many patterns are there in nature?
Nature is full of mathematics Still life also behaves in a mathematical way. If you throw a baseball in the air, it follows a roughly parabolic trajectory. Planets and other astrophysical bodies
follow elliptical orbits.
What are some topics of mathematics in the modern world Name 5 topics?
Topics include linear and exponential growth; statistics; personal finance; and geometry, including scale and symmetry. This may interest you : Was the HMS Surprise a real ship?. Emphasis is placed
on problem solving techniques and the application of modern mathematics to understand quantitative information in everyday life.
What are the basic subjects in mathematics? The main branches of mathematics are:
• Number system and Basic Arithmetic.
• Algebra.
• Triangulation.
• Geometry and Cartesian Geometry.
• Calculus – Differential and Integral.
• Matrix Algebra.
• Probability and Statistics.
What is mathematics in modern?
Mathematics is the science that deals with the logic of form, quantity and arrangement. Math is all around us, in everything we do. See the article : How do you reuse kitchen waste and recycle?. It
is the building block for everything in our daily lives, including mobile devices, computers, software, architecture (ancient and modern), art, money, engineering and even sports.
What is the important of mathematics in the modern world?
Mathematics makes our life orderly and prevents chaos. Certain characteristics fostered by mathematics are the power of reasoning, creativity, abstract or spatial thinking, critical thinking, problem
solving ability and even effective communication skills.
What are the 5 main branches of mathematics?
Branches of Mathematics: Arithmetic, Algebra, Geometry, Trigonometry, & Statistics. Read also : How much is Tyler Perry worth?.
What are the main branches of math?
The main branches of mathematics include algebra, analysis, arithmetic, combinatrics, Euclidean and non-Euclidean geometries, game theory, number theory, numerical analysis, optimization,
probability, set theory, statistics, topology, and trigonometry.
What is the three branches of mathematics?
Modern mathematics can be divided into three main branches: continuous mathematics, algebra, and discrete mathematics. The division is not exhaustive. Some fields, such as geometry or mathematical
logic, are difficult to fit into any of these categories.
Is Fibonacci The golden ratio?
The bottom line is that as the numbers get larger, the quotient between each consecutive pair of Fibonacci numbers equals 1.618, or its inverse of 0.618. This proportion is known by many names: the
golden ratio, the golden mean, ϕ, and the divine proportion, among others.
What is the golden ratio called? golden ratio, also known as the golden section, golden mean, or divine proportion, in mathematics, the irrational number (1 square root of â5)/2, often denoted by the
Greek letter Ï or Ï, which is approximately equal to 1.618.
Why is Fibonacci golden ratio?
How is golden ratio used in everyday life?
This ideal ratio is used by many because of its apparent appeal to the human eye. The Golden Ratio has been said to be the most attractive ratio, so it is often used. Everything from commercial
advertising companies, to painters, to even doctors incorporate this ‘magic’ ratio into their work.
What is Fibonacci example?
Fibonacci Number Series: 0, 1, 1, 2, 3, 5, 8, 13,21,34,55,89,144,233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 6765, 10946 75025, 121393 , 196418, 317811, â¦
What are 3 examples of a pattern?
Few examples of number patterns are: Even number pattern -: 2, 4, 6, 8, 10, 1, 14, 16, 18, … Odd number pattern -: 3, 5, 7, 9, 11, 13, 15, 17, 19, … Fibonacci number pattern -: 1, 1, 2, 3, 5, 8 ,13,
21, … and so on.
What do patterns give 2 examples of patterns in everyday life? Examples of natural patterns include waves, cracks, or lightning. Man-made patterns are often used in design and can be abstract, such
as those used in mathematics, science and language. In architecture and art, patterns can be used to create visual effects on the observer.
What are the 5 patterns?
Spiral, meander, explosion, packing, and branching are the ‘Five Patterns in Nature’ that we have chosen to explore.
What are the 5 patterns in nature in math?
The main categories of recurring patterns in nature are fractals, line patterns, crescents, bubbles/foam, and waves. Fractals are best described as infinitely repeating non-linear patterns in varying
sizes. Fractal uniformity is the recurring shape, although the form may appear in different sizes.
What are examples of patterns?
The definition of a pattern is a person or thing used as a model for copying, a design or an expected action. An example of a pattern is the paper sections used by a seamstress to make a dress; dress
pattern. An example of a pattern is polka dots. An example of a pattern is rush hour traffic; traffic pattern.
What are examples of patterns?
The definition of a pattern is a person or thing used as a model for copying, a design or an expected action. An example of a pattern is the paper sections used by a seamstress to make a dress; dress
pattern. An example of a pattern is polka dots. An example of a pattern is rush hour traffic; traffic pattern.
What are 3 examples of a pattern?
Few examples of number patterns are: Even number pattern -: 2, 4, 6, 8, 10, 1, 14, 16, 18, ⦠Odd number pattern -: 3, 5, 7, 9, 11, 13, 15 , 17, 19, ⦠Fibonacci number pattern -: 1, 1, 2, 3, 5, 8
,13, 21, ⦠and so on.
What are some examples of patterns?
Natural patterns include symmetry, trees, spirals, meanders, waves, foams, tessellations, cracks and stripes. Early Greek philosophers studied pattern, and Plato, Pythagoras and Empedocles tried to
explain order in nature. The modern understanding of visible patterns has gradually developed over time.
What are the 2 types of pattern in nature?
There are different types of patterns including symmetries, trees, spirals, meanders, waves, foams, tessellations, slits and stripes.
What are the patterns in the nature of mathematics? Patterns in Nature Sometimes the patterns can be modeled mathematically and include symmetry, trees, spirals, meanders, waves, foams,
tessellations, cracks and stripes. Mathematics, physics and chemistry can explain patterns in nature at different levels.
What are the 5 patterns in nature?
Spiral, meander, explosion, packing, and branching are the ‘Five Patterns in Nature’ that we have chosen to explore.
What is the most common pattern in nature?
The spiral is a popular pattern for those who like to draw and design and is also one of the most common configurations in nature. Indeed, it is difficult to think of all the objects that have a
spiral pattern.
What is the name for patterns in nature?
These patterns are called fractals. A fractal is a type of pattern that we often observe in nature and art. As Ben Weiss explains, “anytime you look at a series of patterns repeating over and over
again, at many different scales, and where any small part is similar to the whole, that’s a fractal.â
What are patterns in nature called?
These patterns are called fractals. A fractal is a type of pattern that we often observe in nature and art. As Ben Weiss explains, “anytime you look at a series of patterns repeating over and over
again, at many different scales, and where any small part is similar to the whole, that’s a fractal.â
What are fractal patterns in nature?
A fractal is a pattern that repeats the laws of nature at different scales. Examples are everywhere in the forest. Trees are natural fractals, patterns that make smaller and smaller copies of
themselves to create forest biodiversity.
What are the patterns in nature of mathematics?
The main categories of recurring patterns in nature are fractals, line patterns, crescents, bubbles/foam, and waves. Fractals are best described as infinitely repeating non-linear patterns in varying
sizes. Fractal uniformity is the recurring shape, although the form may appear in different sizes.
Why does nature follow Fibonacci?
In nature the growth and self-renewal of cell populations generates hierarchical patterns in tissues similar to the population growth pattern in rabbits, which is explained in the classic Fibonacci
Does nature follow the Fibonacci sequence? The Fibonacci Sequence is found throughout nature, too. It is a naturally occurring pattern.
How are Fibonacci numbers connected to nature?
Seed heads, pine cones, fruits and vegetables: Look at the array of seeds in the middle of a sunflower and you’ll notice what look like spiral patterns curving left and right. Amazingly, if you count
these spirals, your total will be a Fibonacci number.
Where does the Fibonacci sequence appear in nature?
On many plants, the number of petals is a Fibonacci number: buttercups have 5 petals; lilies and irises have 3 petals; some delphiniums have 8; corn has 13 petals; some asters have 21 but daisies can
be found with 34, 55 or even 89 petals.
Why are Fibonacci numbers or the golden spiral golden mean or golden ratio frequently seen in nature?
There is a special relationship between the Golden Ratio and the Fibonacci Numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, … etc, each number is the sum of the two previous numbers). So, just as we naturally
get seven arms when we use 0.142857 (1/7), we tend to get Fibonacci Numbers when we use the Golden Ratio.
Sources : | {"url":"https://lisa-andersen.com/how-many-patterns-are-there-in-nature/","timestamp":"2024-11-14T21:47:32Z","content_type":"text/html","content_length":"64864","record_id":"<urn:uuid:378c1d8a-1695-4061-a69c-7b40ab956748>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00705.warc.gz"} |
Percentage of
Percentage of Percentage Calculator
(Percentage on Base Percentage)
How to use this Percentage of Percentage Calculator 🤔
1. Enter ✎ value for Base Percentage (X).
2. Enter ✎ value for Percentage on Base Percentage (Y).
3. As soon as you enter the required input value(s), the Percentage of Percentage is calculated immediately, and displaed in the output section (present under input section).
Calculating Percentage of a Percentage
Calculating the percentage of a percentage is a common operation, particularly in situations where one percentage is applied to another. This is often seen in scenarios such as compounded discounts,
multi-level commissions, or layered interest rates.
The formula to calculate the percentage of a percentage is expressed as:
\( \text{Result} = \dfrac{ X \times Y }{ 100 } \)
• X represents the base percentage.
• Y represents the percentage applied to the base percentage.
By multiplying the two percentages and then dividing by 100, you obtain the percentage of the percentage.
The following examples demonstrate how to calculate the percentage of a percentage using the given formula.
1. A real estate agent earns 5% commission on the sale price of a house. The agent’s broker takes 20% of the agent’s commission as a fee. What percentage of the house’s sale price does the broker
• X = 5% (Agent’s commission percentage on the sale price)
• Y = 20% (Broker’s percentage of the agent’s commission)
The formula to calculate the percentage of a percentage is:
\( \text{Result} = \dfrac{ X \times Y }{ 100 } \)
Substituting the given values into the formula:
\( \text{Result} = \dfrac{ 5 \times 20 }{ 100 } \)
Simplifying further:
\( \text{Result} = 1\% \)
Therefore, the broker takes 1% of the house’s sale price.
2. A company offers a 30% bonus to its employees based on their annual performance. Out of this bonus, 15% is allocated to the employee’s retirement fund. What percentage of the employee’s annual
salary is added to the retirement fund?
• X = 30% (Bonus percentage)
• Y = 15% (Percentage of the bonus allocated to the retirement fund)
The formula to calculate the percentage of a percentage is:
\( \text{Result} = \dfrac{ X \times Y }{ 100 } \)
Substituting the given values into the formula:
\( \text{Result} = \dfrac{ 30 \times 15 }{ 100 } \)
Simplifying further:
\( \text{Result} = 4.5\% \)
Therefore, 4.5% of the employee’s annual salary is added to the retirement fund.
To calculate the percentage of percentage, you can use the following formula.
\( = \) \( \dfrac{ X \cdot Y }{ 100 } \)
Once you enter the input values in the calculator, the output parameters are calculated. | {"url":"https://convertonline.org/mathematics/?topic=percentage-of-percentage","timestamp":"2024-11-02T17:37:45Z","content_type":"text/html","content_length":"52018","record_id":"<urn:uuid:5ca084bb-4195-4071-8c5e-f054676c62fe>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00502.warc.gz"} |
IFIC Literature Database
Aceti, F., Liang, W. H., Oset, E., Wu, J. J., & Zou, B. S. (2012). Isospin breaking and f(0)(980)-a(0)(980) mixing in the eta(1405) -> pi(0)f(0)(980) reaction. Phys. Rev. D, 86(11),
Bayar, M., Liang, W. H., & Oset, E. (2014). B-0 and B-s(0) decays into J/psi plus a scalar or vector meson. Phys. Rev. D, 90(11), 114004–9pp.
Bayar, M., Liang, W. H., Uchino, T., & Xiao, C. W. (2014). Description of rho(1700) as a rho Kappa(sic) system with the fixed-center approximation. Eur. Phys. J. A, 50(4), 67–10pp.
Chen, H. X., Geng, L. S., Liang, W. H., Oset, E., Wang, E., & Xie, J. J. (2016). Looking for a hidden-charm pentaquark state with strangeness S =-1 from Xi(-)(b) decay into J/Psi K- Lambda.
Phys. Rev. C, 93(6), 065203–9pp.
Debastiani, V. R., Aceti, F., Liang, W. H., & Oset, E. (2017). Revising the f(1)(1420) resonance. Phys. Rev. D, 95(3), 034015–10pp.
Debastiani, V. R., Dias, J. M., Liang, W. H., & Oset, E. (2018). Molecular Omega(c) states generated from coupled meson-baryon channels. Phys. Rev. D, 97(9), 094035–11pp.
Debastiani, V. R., Dias, J. M., Liang, W. H., & Oset, E. (2018). Omega(-)(b) -> (Xi(+)(c) K-)pi(-) decay and the Omega(c) states. Phys. Rev. D, 98(9), 094022–8pp.
Debastiani, V. R., Liang, W. H., Xie, J. J., & Oset, E. (2017). Predictions for eta(c) -> eta pi(+)pi(-) producing f(0)(500), f(0)(980) and a(0)(980). Phys. Lett. B, 766, 59–64.
Dias, J. M., Yu, Q. X., Liang, W. H., Sun, Z. F., Xie, J. J., & Oset, E. (2020). Xi(bb) and Omega(bbb) molecular states. Chin. Phys. C, 44(6), 064101–8pp.
Duan, M. Y., Song, J., Liang, W. H., & Oset, E. (2024). On the search for the two poles of the Ξ(1820)in the ψ(3686)→Ξ+K0Σ∗-(π-Λ) decay. Eur. Phys. J. C, 84(9), 947–5pp. | {"url":"https://references.ific.uv.es/refbase/search.php?sqlQuery=SELECT%20author%2C%20title%2C%20type%2C%20year%2C%20publication%2C%20abbrev_journal%2C%20volume%2C%20issue%2C%20pages%2C%20keywords%2C%20abstract%2C%20thesis%2C%20editor%2C%20publisher%2C%20place%2C%20abbrev_series_title%2C%20series_title%2C%20series_editor%2C%20series_volume%2C%20series_issue%2C%20edition%2C%20language%2C%20author_count%2C%20online_publication%2C%20online_citation%2C%20doi%2C%20serial%20FROM%20refs%20WHERE%20author%20RLIKE%20%22Liang%2C%20W%5C%5C.H%5C%5C.%22%20ORDER%20BY%20author%2C%20year%20DESC%2C%20publication&submit=Cite&citeStyle=APA&citeOrder=&orderBy=author%2C%20year%20DESC%2C%20publication&headerMsg=&showQuery=0&showLinks=1&formType=sqlSearch&showRows=10&rowOffset=0&client=&viewType=","timestamp":"2024-11-12T22:56:28Z","content_type":"text/html","content_length":"88814","record_id":"<urn:uuid:2112ad0a-2a7e-44e1-bc31-291d264a83f2>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00578.warc.gz"} |
COUNTIF Referencing Adjacent Cells
Is it possible having once run a COUNTIF, to then display information from a cell adjacent to where the COUNTIF information was found? For example: I'm searching a range for a value tied to an
employee. Once I find that value I want to display the employees name. Thanks
• Hey @Zolkora
The formula in the 'adjacent cell' can use an Index/Match combination to find the name. Is all the information in your COUNTIF formula from within this same sheet or are you pulling in the info
from cross sheets?
Here's assuming all the info is within this same sheet
=INDEX([Employee Name column]:[Employee Name column], MATCH([your COUNTIF result]@row, [COUNTIF Range column]:[COUNTIF range column],0))
You will need to insert your actual column names in the formula above.
Here's assuming the info is cross sheet referenced
=INDEX({source sheet Employee name column},MATCH([your COUNTIF result]@row,{source sheet Countif range column},0))
The cross sheet references will have to be physically built by you.
Will either of there work for you?
• I eventually did use INDEX & MATCH. Thanks anyway. You were right on the money.
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/94250/countif-referencing-adjacent-cells","timestamp":"2024-11-05T13:46:29Z","content_type":"text/html","content_length":"398103","record_id":"<urn:uuid:9f6bcf20-7e23-4975-ad6c-3b3f557a7506>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00399.warc.gz"} |
The Likelihood Function
Last updated: 2019-03-31
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You should understand the concept of using likelihood ratio for discrete data and continuous data to compare support for two fully specified models.
We have seen how one can use the likelihood ratio to compare the support in the data for two fully-specified models. In practice we often want to compare more than two models - indeed, we often want
to compare a continuum of models. This is where the idea of a likelihood function comes from.
In our example here we assumed that the frequencies of different alleles (genetic types) in forest and savanna elephants were given to us. In practice these frequencies themselves would have to be
estimated from data.
For example, suppose we collect data on 100 savanna elephants, and see that 30 of them carry allele 1 at marker 1, while 70 carry the allele 0 (again we are treating elephants as haploid to simplify
things). Intuitively we might estimate that the frequency of the 1 allele at that marker is 30/100, or 0.3. But we might think that the data are also consistent with other frequencies near 0.3. For
example maybe the data are consistent with a frequency of 0.29 also. Or 0.28? Or …
In this case we have many more than just two models to compare. Indeed, if we allow that the frequency could, in principle lie anywhere in the interval [0,1], then we have a continuum of models to
Specifically, for each \(q\in [0,1]\) let \(M_q\) denote the model that the true frequency of the 1 allele is \(q\). Then, given our observation that 30 of 100 elephants carried allele 1 at marker 1,
the likelihood for model \(M_q\) is, by the previous definition, \[L(M_q) = \Pr(D | M_q) = q^{30} (1-q)^{70}.\] And the LR comparing models \(M_{q_1}\) and \(M_{q_2}\) is \[LR(M_{q_1};M_{q_2})) =
[q_1/q_2]^{30} [(1-q_1)/(1-q_2)]^{70}.\]
This is an example of what is called a parametric model. A parametric model is collection of models indexed by a parameter vector, often denoted \(\theta\), whose values lie in some parameter space,
usually denoted \(\Theta\). The number of parameters included in the vector \(\theta\) is called the “dimensionality” of the model or parameter space, and often denoted \(dim(\Theta)\).
Here the parameter is \(q\) and the parameter space is \([0,1]\). The dimensionality is 1.
When computing likelihoods for parametric models, we usually dispense with the model notation and simply use the parameter value to denote the model. So instead of referring to the likelihood for \
(M_q\) we just say the “likelihood for \(q\)”, and write \(L(q)\). So the likelihood for \(q\) is given by \[L(q) = q^{30} (1-q)^{70}.\] Correspondingly we can also refer to the “likelihood ratio for
\(q_1\) vs \(q_2\)”.
We could plot the likelihood function as follows:
q = seq(0,1,length=100)
L= function(q){q^30 * (1-q)^70}
The value of \(\theta\) that maximizes the likelihood function is referred to as the “maximum likelihood estimate”, and usually denoted \(\hat{\theta}\). That is \[\hat{\theta}:= \arg \max L(\theta).
Provided the data are sufficiently informative, and the number of parameters is not too large, maximum likelihood estimates tend to be sensible. In this case we can see that the maximum likelihood
estimate is \(q=0.3\), which also corresponds to our intuition.
Note that from the likelihood function we can easily compute the likelihood ratio for any pair of parameter values! And just as with comparing two models, it is not the likelihoods that matter, but
the likelihood ratios. That is you can divide the likelihood function by any constant without affecting the likelihood ratios.
One way to emphasize this is to standardize the likelihood function so that its maximum is at 1, by dividing \(L(\theta)/L(\hat{\theta})\).
q = seq(0,1,length=100)
L= function(q){q^30 * (1-q)^70}
Note that for some values of \(q\) the likelihood ratio compared with \(q=0.3\) is very close to 0. These values of \(q\) are so much less consistent with the data that they are effectively excluded
by the data. Just looking at the picture we might say that the values of \(q\) less than 0.15 or bigger than 0.5 are pretty much excluded by the data. We will see later how Bayesian analysis methods
can be used to make this kind of argument more formal.
The log-likelihood
Just as it can often be convenient to work with the log-likelihood ratio, it can be convenient to work with the log-likelihood function, usually denoted \(l(\theta)\) [lower-case L]. As with log
likelihood ratios, unless otherwise specified, we use log base e. Here is the log-likelihood function.
q = seq(0,1,length=100)
l= function(q){30*log(q) + 70 * log(1-q)}
plot(q,l(q)-l(0.3),ylab="l(q) - l(qhat)",xlab="q",type="l",ylim=c(-10,0))
Changes in the log-likelihood function are referred to as “log-likelihood units”. For example the difference in the support for \(q=0.3\) and \(q=0.35\) is l(0.3)-l(0.35) = 0.5630377 log-likelihood
units. Again, remember that it is differences in \(l\) that matter, not the actual values.
Notice that the scale of the \(y\) axis in this plot was set to span 10 log likelihood units. Setting the scale in this way makes sure the plot focuses on the parts of the parameter space that have
more than minuscule support from the data (in this case, LR no smaller than 1/exp(10)). Without this the plot can be much harder to read. For example, here is the plot using the default scale
selected by R:
plot(q,l(q)-l(0.3),ylab="l(q) - l(qhat)",xlab="q",type="l")
Notice how different this plot looks to the eye even though it is exactly the same curve being plotted (just different \(y\) axis scale). It is worth thinking about what scale you use when plotting
log-likelihoods (and, of course, figures in general!).
R version 3.5.2 (2018-12-20)
Platform: x86_64-apple-darwin15.6.0 (64-bit)
Running under: macOS Mojave 10.14.1
Matrix products: default
BLAS: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRblas.0.dylib
LAPACK: /Library/Frameworks/R.framework/Versions/3.5/Resources/lib/libRlapack.dylib
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] workflowr_1.2.0 Rcpp_1.0.0 digest_0.6.18 rprojroot_1.3-2
[5] backports_1.1.3 git2r_0.24.0 magrittr_1.5 evaluate_0.12
[9] stringi_1.2.4 fs_1.2.6 whisker_0.3-2 rmarkdown_1.11
[13] tools_3.5.2 stringr_1.3.1 glue_1.3.0 xfun_0.4
[17] yaml_2.2.0 compiler_3.5.2 htmltools_0.3.6 knitr_1.21
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Measure jitter metrics from waveforms
Since R2024b
J = jitter(x,y,SymbolTime = t) measures jitter from input jittery waveform by using the specified symbol time.
J = jitter(x,y,xr,yr) measures jitter from input jittery waveform with respect to the reference waveform.
J = jitter(y,yr,SampleInterval = s) measures jitter with respect to the reference waveform and specified sample interval.
J = jitter(___,Name=Value) measures jitter using name-value arguments. Unspecified arguments take default values.
Measure Waveform Jitter with Reference Waveform
This example shows how to measure edge jitter in oversampled time-domain waveform data, particularly waveforms produced by sampled data systems.
Load Data
Load the waveform data, including a waveform with jitter (tj, yj) and a reference waveform without jitter (tr, yr), from a file.
load("JitterPAM2.mat", "tj", "yj", "tr", "yr");
Measure Jitter
Use the function jitter() to measure edge jitter in the waveform. Edge jitter characterizes trends in timing error, which is the time difference between any observed edge and the corresponding
nominal or reference edge.
J = jitter(tj, yj, tr, yr, "Plot", "on")
J = struct with fields:
TJrms: 4.1554e-10
TJpkpk: 4.9621e-10
RJrms: 1.0115e-10
DJrms: 9.4134e-11
DJpkpk: 4.7368e-10
DDJrms: 6.0662e-11
DDJpkpk: 2.9648e-10
SJa: 8.6490e-11
SJf: 1.8311e+05
SJp: -1.7257
DCDrms: 5.5722e-11
DCDpkpk: 1.1144e-10
ISIrms: 3.3659e-12
ISIpkpk: 1.8504e-10
The edge jitter is characterized by the following metrics:
• TJ (Total Jitter) - Both Root-Mean-Square (RMS) or Peak-to-Peak (Pk-Pk) values are calculated directly from the timing error sequence.
• DCD (Duty Cycle Distortion) - Odd/even DCD applies one timing offset to edges with odd indices and another to edges with even indices. The Pk-Pk is the difference between the larger and smaller
of the two timing offset values.
• SJ (Sinusoidal Jitter) - This metric captures sinusoidal trends in the timing error sequence. These are reported as the Amplitude (SJa), Frequency (SJf), and Phase (SJp) of a cosine.
• ISI (Intersymbol Interference) - This metric correlates the jitter at each edge with the pattern before (and after) that edge in time. The result of this correlation is a Dirac delta function for
each symbol/delay combination. Groups for each delay are convolved together to produce an estimate of a PDF for ISI. The RMS and Pk-Pk ISI metrics are derived from this PDF.
• DDJ (Data Dependent Jitter) - The PDF estimate for DDJ is the result of convolution of the ISI and DCD PDF estimates.
• DJ (Deterministic Jitter) - The PDF estimate for DJ is the result of convolution of the DDJ and SJ PDF estimates.
• RJ (Random Jitter) - Is the residual jitter remaining in the system after DJ has been compensated. Random Jitter, unlike the other jitter metrics, is assumed to be unbounded and so only its RMS
is reported.
Input Arguments
x — Time coordinates of jittery signal
Time coordinates of the jittery signal, specified as a monotonically increasing vector.
If you do not provide y, the function interprets x as edge times.
Data Types: double
y — Amplitude coordinates of jittery signal
vector | eyeDiagramSI object
Amplitude coordinates of the jittery signal, specified as a vector or as an eyeDiagramSI object.
If you do not provide x, the function assumes y is uniformly sampled at the rate specified by SampleInterval.
Data Types: double
xr — Time coordinates of reference signal
Time coordinates of the reference signal, specified as a monotonically increasing vector.
Data Types: double
yr — Amplitude coordinates of reference signal
Amplitude coordinates of the reference signal, specified as a vector. If you do not provide x, yr must be sampled at the same points as y.
Data Types: double
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but
the order of the pairs does not matter.
Example: J = jitter(x,y,SampleInterval = s, Plot = on) calculates and plots the histograms of the jitter metrics from the input waveform defined by (x,y) and sample interval specified by s.
SymbolThresholds — Thresholds to separate symbol levels in jittery waveform
Thresholds to separate symbol levels in the jittery waveform, specified as a vector. If you do not provide SymbolThresholds, the function calculates it from the histogram of y.
Data Types: double
ReferenceThresholds — Thresholds to separate symbol levels in reference waveform
Thresholds to separate symbol levels in the reference waveform, specified as a vector. If you do not provide ReferenceThresholds, the function calculates it from the histogram of yr.
Data Types: double
SampleInterval — Sample time for uniformly sampled jittery and reference waveforms
Sample time for uniformly sampled jittery and reference waveforms, specified as a scalar.
When you provide the time vectors of jittery and reference signals, the function ignores SampleInterval and uses time vectors instead.
Data Types: double
SymbolTime — Symbol time for uniformly sampled jittery and reference waveforms
Symbol time for uniformly sampled jittery and reference waveforms, specified as a scalar. For clock waveforms, SymbolTime is half the of the period.
Data Types: double
Match — Options to compare data edge to clock edge
time (default) | order
Options to compare the data edge to the clock edge, specified as one of these:
• time — Compares the closest edge times.
• order — Compares the first edge on each set.
The function uses this argument in two scenarios only:
• Both the measured and reference waveforms are clocks.
• The measured waveform is a data waveform and the reference waveform is a clock waveform.
When both waveforms are data waveforms with the same pattern, the function matches the edges is based on the pattern.
Plot — Option to display histograms of jitter metrics
true or 1 (default) | false or 0
Option to display histograms of jitter metrics, specified as true (1) or false or (0). The histograms include total jitter, random jitter, deterministic jitter, data-dependent jitter, sinusoidal
jitter, inter-symbol interference, and duty cycle distortion.
Data Types: logical
DebounceGain — Positive feedback gain for hysteresis
0.5 (default) | real scalar in the range [0, 1]
Positive feedback gain for hysteresis, specified as a real scalar in the range [0, 1]. The function uses DebounceGain to calculate the modified thresholds by using the equation:
Data Types: double
DebounceUI — Fraction of unit interval to hold hysteresis threshold level
1/3 (default) | nonnegative scalar in the range [0, 1]
Fraction of a unit interval to hold the hysteresis threshold level before returning to the nominal threshold, specified as a nonnegative scalar in the range [0, 1].
Data Types: double
DelayTime — Custom channel delay
[] (default) | real scalar
Custom channel delay, specified as a real scalar.
If your data has aggressive decision feedback equalization, you can manually override the channel delay compensation calculated by the jitter function and specify the delay of the measured waveform (
x,y) with respect to the reference waveform (xr,yr). A positive Delaytime indicates that a reference edge matches to an edge later in time in the measured signal. A negative Delaytime indicates that
a measured edge matches to an edge later in time in the reference signal.
Data Types: double
Metric Specific
Frequencies — Maximum number of sinusoidal jitter frequencies to measure
1 (default) | scalar
Maximum number of sinusoidal jitter frequencies to measure, specified as a scalar.
Data Types: double
PastSymbols — Number of symbols prior to given edge to correlate jitter
31 (default) | scalar
Number of symbols prior to a given edge to correlate the jitter with, specified as a scalar.
Data Types: double
FutureSymbols — Number of symbols after to given edge to correlate jitter
Number of symbols after a given edge to correlate the jitter with, specified as a scalar. If the signal uses PAM2 modulation, the default value for FutureSymbols is 0. For a modulation scheme of PAM3
or higher, the default value is 1.
Data Types: double
DCDMethod — Reference for correlating duty cycle distortion
oddeven (default) | risefall
Reference for correlating duty cycle distortion, specified as oddeven or risefall.
Data Types: char
BinEdges — Bin edges for PDFs during correlation
Bin edges for PDFs during correlation, specified as a vector.
By default, the function distributes BinEdges evenly with twice the resolution of Scott's rule.
For best results, specify a bin centered at 0.
Data Types: double
Premeasured Jitter
DDJ — Data-dependent jitter
Data-dependent jitter value for each edge, specified as a vector.
If you specify DDJ, the function ignores the specified DCD and ISI values while estimating deterministic jitter and random jitter.
Data Types: double
ISI — Timing error values due to inter-symbol interference
vector | structure
Timing error values due to inter-symbol interference (ISI), specified as a vector or a structure.
The structure contains the correlation information and is the output of the jitterIntersymbol (Mixed-Signal Blockset) function. The structure must contain these fields.
Field Description
[N×M] matrix, where N is the number of modulation levels and M is the sum of future symbols and past symbols.
Mean denotes the ISI values for each symbol level at each delay.
M long vector, where M is the sum of future symbols and past symbols. The vector cannot contain 0.
SymbolIndex SymbolIndex specifies the delay each column of Mean corresponds to.
Delay is relative to the symbol edge at which the function applies the ISI. A delay of 1 corresponds to the symbol after the edge, and a delay of -1 corresponds to the symbol prior to the
The elements of the vector denote the ISI values for each edge.
DCD — Timing error values due to duty cycle distortion
vector | structure
Timing error values due to duty cycle distortion (DCD), specified as a vector or a structure.
The elements of the vector denote the DCD values for each edge.
The structure contains the correlation information and is the output of the jitterDutyCycle (Mixed-Signal Blockset) function. The structure must contain the Mean field, which is a 2-by-1 matrix. The
first element is the odd or rising DCD value and the second is the even or falling DCD value.
Data Types: double
SJ — Sinusoidal jitter
N-by-3 matrix
Sinusoidal jitter, specified as an N-by-3 matrix. N is the number of sinusoidal jitter (SJ) peaks or frequencies.
Each row of SJ contains the amplitude, frequency, and phase values. Amplitude must be in the same unit as SampleInterval and SymbolTime. Frequency is in the inverse of the unit of amplitude. Phase is
in radians.
Data Types: double
Output Arguments
J — Jitter metrics
Jitter metrics, returned as a structure.
Version History
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WhatsApp: +86 18838072829 | {"url":"https://tresorsdejardin.fr/two/roll/mill/friction/ratio/1/1/35/spare/geare/for/ratio/1/1-1309.html","timestamp":"2024-11-06T06:05:18Z","content_type":"application/xhtml+xml","content_length":"24461","record_id":"<urn:uuid:1322e17c-e598-4e26-8ae3-1e0d59461228>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00341.warc.gz"} |
Introduction to pmwg
pmwg is a package that implements an efficient and flexible Particle Metropolis within Gibbs sampler as outlined in the paper (Gunawan et al. 2020). The sampler estimates group level parameter
values, the covariance matrix related parameter estimates to each other and the individual level parameter estimates (random effects) in a two step process. The process consists of a Gibbs sampling
step for group level/covariate estimates followed by a particle metropolis step for random effects for each iteration of the sampler.
The sampling proceeds through three stages, burn in, adaptation and a final sampling stage. Burn in can be relatively short and should move the sample estimates from the start values to a plausible
region of the parameter space relatively quickly. The adaptation stage is the same as burn in, however samples from this stage are used to generate a proposal distribution used to create samples more
efficiently for the final sampling stage.
The data
First we will load the pmwg package and explore the included dataset
## subject condition stim resp rt
## 1115 1 1 2 2 0.4319
## 1116 1 3 2 2 0.5015
## 1117 1 3 1 1 0.3104
## 1118 1 1 2 2 0.4809
## 1119 1 1 1 1 0.3415
## 1120 1 2 1 1 0.3465
The forstmann dataset is from (Forstmann et al. 2008) and is the cleaned data from an experiment that displayed a random dot motion task with one of three speed accuracy manipulations (respond
accurately instructions, neutral instructions and respond quickly instructions).
We can see that there are 5 columns:
• subject - which gives an ID for each of the 19 participants
• condition - the speed accuracy instructions for the trial
• stim - whether dots were moving left or right
• resp - whether the participant responded left or right
• rt - the response time for the trial
We will use the rtdists package to look at threshold differences between the three conditions to see whether there is evidence for different thresholds (the evidence required to trigger a response)
in each of the three conditions. For a full explanation of the model please see the full documentation in Chapter 3.
The log-likelihood function
Now that we have the data we will need to specify the log-likelihood function.
This function contains the primary implementation of our model for the data. The pmwgs object that we run later in this example will pass two things:
• A set of parameter values x that we will be assessing.
• A subset of our dataset (data), the data for one subject.
Our job then in the function is to take the parameter values and calculate the log-likelihood of the data given the parameter values.
The overall process is as follows:
• Take the exponent of the parameter values (that are stored in log-form
• Test for improbable values of t0 (non-decision time) and return extremely low log-likelihood if they are unlikely.
• Create a vector of b (threshold) parameter values for the call to rtdists
• Use the rtdists dLBA function to calculate the likelihood of the data given our parameter estimates.
• Clean extremely small of otherwise bad values from output vector.
• Return the sum of the log of the likelihood of each row of the data.
Other necessary components
There are a couple of other necessary components to running the sampler, a list of parameters and a prior for the group level parameter values.
In this case the parameters consist of threshold parameters for each of the three conditions (b1, b2 and b3), a upper limit on the range for accumulator start points (A), the drift rates for evidence
accumulation for the two stimulus types (v1 and v2) and non-decision time (t0).
The prior for this model is uninformed and is just 0 for the mean of the group parameters and standard deviation of 1 for each parameter.
Creating and running the sampler
Once we have these components we are ready to create and run the sampler.
Creation of the sampler object is done through a call to the pmwgs function, which creates the object and sets numerous precomputed values based on the number of parameters and other aspects of the
included function arguments. Next steps are to initialise the sampler with start values for the random effects and then run the three stages of sampling.
The stages can be long running, but once complete you will have
The run_stage command can also be passed other arguments such as iter for number of iterations, particles for number of particles among others and more. For a full list see the description in the
PMwG Tutorial Book.
Forstmann, Birte U., Gilles Dutilh, Scott Brown, Jane Neumann, D. Yves von Cramon, K. Richard Ridderinkhof, and Eric-Jan Wagenmakers. 2008. “Striatum and Pre-Sma Facilitate Decision-Making Under Time
Pressure.” Proceedings of the National Academy of Sciences 105 (45): 17538–42. https://doi.org/10.1073/pnas.0805903105.
Gunawan, D., G.E. Hawkins, M.-N. Tran, R. Kohn, and S.D. Brown. 2020. “New Estimation Approaches for the Hierarchical Linear Ballistic Accumulator Model.” Journal of Mathematical Psychology 96:
102368. https://doi.org/https://doi.org/10.1016/j.jmp.2020.102368. | {"url":"https://cran.itam.mx/web/packages/pmwg/vignettes/pmwg.html","timestamp":"2024-11-03T09:43:47Z","content_type":"text/html","content_length":"21995","record_id":"<urn:uuid:2db0c022-cbf9-4224-94ee-4b80a4d07649>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00580.warc.gz"} |
Turing Machines :: CC 210 Textbook
Turing Machines
Video Script
Alan Turing was an early computing pioneer who remained largely unknown outside of the field until recently with the release of The Imitation Game in 2014. That movie centered around his work to
decode German codes created by the Enigma Machine during World War II. However, long before his time at Bletchley Park, he was already making an impact in the world of computer science.
In 1936, Turing proposed an imaginary computer, which we now refer to as a Turing Machine, that could help us determine what problems could be solved using a computer. It was designed to be as simple
as possible, able to only read or write a single binary bit at a time and follow very simple instructions.
However, Turing was able to demonstrate that his imaginary computer, given infinite time, infinite storage space, and a sufficiently complex program made of those simple instructions, could solve any
problem that any other real-world computer could solve. Yup, in theory, it could be programmed to do everything from solve complex linear equations to play Angry Birds (or, at least, perform the
underlying calculations required to play Angry Birds)!
Modern computers are similar to Turing Machines, but a modern computer may have several hundred instructions that it knows how to follow. Those instructions may not make much sense to us, especially
if we don’t have a deep understanding of how a computer works internally. So, we need some way we can tell our computer what we’d like it to do without having to learn everything there is to know
about it. In effect, this is very similar to learning how to drive a modern car - we don’t need to know every detail of how it works to be able to use it efficiently.
Thanks to the work of several computing pioneers such as Rear Admiral Grace Hopper and Steve Russell, we can use a high-level programming language such as Java or Python to tell our computer exactly
what we’d like it to do, using words and symbols we can easily understand. Then, we can use special programs on our computer, such as a compiler for Java and an interpreter for Python, to convert
that program written in a high-level language into machine code, which is something our computer can understand.
In this chapter, we’ll go through the steps needed to write our very first computer program. Make sure you read the material below before moving on, since the next page contains a short quiz. Good | {"url":"https://textbooks.cs.ksu.edu/cc210/01-object-oriented-programming/01-what-is-programming/video/","timestamp":"2024-11-05T12:49:32Z","content_type":"text/html","content_length":"88436","record_id":"<urn:uuid:3b5466ff-ffad-4461-9e81-85bd7c9d534d>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00864.warc.gz"} |
A character string specifying the key for the model fit in the H2O cluster's key-value store.
A character string specifying the algorithm that was used to fit the model.
A list containing the parameter settings that were used to fit the model that differ from the defaults.
A list containing all parameters used to fit the model.
A list containing the characteristics of the model returned by the algorithm.
The number of points in each cluster.
Total sum of squared error to grand mean.
A vector of within-cluster sum of squared error.
Total within-cluster sum of squared error.
Between-cluster sum of squared error. | {"url":"https://www.rdocumentation.org/packages/h2o/versions/3.44.0.3/topics/H2OClusteringModel-class","timestamp":"2024-11-04T21:16:06Z","content_type":"text/html","content_length":"62955","record_id":"<urn:uuid:8e2c82c3-5bbd-476a-b82b-14cef2b3c7b6>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00591.warc.gz"} |
Wednesday, November 6, 2024, 10:28 PM
Site: Learnbps
Course: BPSS (MAT) Mathematics Standards (S-MAT)
Glossary: MAT-10 Standards
(DPS) Data Probability and Statistics
Learners will ask and answer questions by collecting, organizing, and displaying relevant data, drawing inferences and conclusions and making predictions; and understanding and applying basic
concepts of probability.
• (D) Data
Learners will represent and interpret data.
• (DA) Data Analysis
Learners will ask and answer questions by collecting, organizing, and displaying relevant data, drawing inferences and conclusions, and making predictions.
• (P) Probability
Learners will understand and apply basic concepts of probability.
Calculation Method for Domains
Domains are larger groups of related standards. The Domain Grade is a calculation of all the related standards. Click on the standard name below each Domain to access the learning targets and rubrics
/ proficiency scales for individual standards within the domain.
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
(DA) Data Analysis
Learners will ask and answer questions by collecting, organizing, and displaying relevant data, drawing inferences and conclusions, and making predictions.
MAT-10.DPS.01 Represent data with plots on the real number line (dot plots, histograms, and box plots).
Displaying Data
• MAT-01.DPS.D.01 Collect, organize and represent data with up to three categories using picture and bar graphs.
• MAT-02.DPS.D.01 Formulate questions and collect, organize, and represent data, with up to four categories using single unit scaled pictures and bar graphs.
• MAT-03.DPS.D.01 Formulate questions to collect, organize, and represent data with more than four categories using scaled pictures and bar graphs.
• MAT-04.DPS.D.01 Formulate questions to collect, organize, and represent data to reason with math and across disciplines.
• MAT-02.DPS.D.02 Generate data and create line plots marked in whole number units.
• MAT-03.DPS.D.02 Generate data and create line plots marked in whole numbers, halves, and fourths of a unit.
• MAT-04.DPS.D.02 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using
information presented in line plots.
• MAT-05.DPS.D.01 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Use grade-level operations for fractions to solve problems involving information
presented in line plots.
• MAT-06.DPS.DA.04 Display numerical data in plots on a number line, including dot plots and histograms. Describe any overall patterns in data, such as gaps, clusters, and skews.
• MAT-09.NO.03 Choose and interpret the scale and the units in graphs and data displays.
• MAT-09.NO.05 Choose a level of accuracy or precision appropriate to limitations on measurement when reporting quantities.
• MAT-10.DPS.01 Represent data with plots on the real number line (dot plots, histograms, and box plots).
• MAT-10.DPS.03 Represent data on two quantitative variables on a scatter plot and describe how the variables are related.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.NO.04 Use units as a way to understand problems and to guide the solution of multi-step problems (e.g., unit analysis). Choose and interpret units consistently in formulas. Choose and
interpret the scale and the units in graphs and data displays.
• MAT-12.DPS.04 Represent data on a scatter plot for two quantitative variables and describe how the variables are related.
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
(DA) Data Analysis
Learners will ask and answer questions by collecting, organizing, and displaying relevant data, drawing inferences and conclusions, and making predictions.
MAT-10.DPS.02 Compare the center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets using statistics appropriate to the shape of the data
Data Analysis
• MAT-01.DPS.D.02 Analyze data by answering descriptive questions.
• MAT-02.DPS.D.03 Analyze data and interpret the results to solve one-step comparison problems using information from the graphs.
• MAT-03.DPS.D.03 Analyze data and make simple statements to solve one- and two-step problems using information from the graphs.
• MAT-04.DPS.D.02 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using
information presented in line plots.
• MAT-04.DPS.D.03 Utilize graphs and diagrams to represent and solve word problems using the four operations involving whole numbers, benchmark fractions, and decimals.
• MAT-05.DPS.D.01 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Use grade-level operations for fractions to solve problems involving information
presented in line plots.
• MAT-05.DPS.D.02 Utilize graphs and diagrams to represent, analyze, and solve authentic problems using information presented in one or more tables or line plots, including whole numbers,
fractions, and decimals.
• MAT-06.DPS.DA.02 Calculate measures of center (median and mean) and variability (range and mean absolute deviation) to answer a statistical question. Identify mode(s) if they exist.
• MAT-06.DPS.DA.03 Identify outliers by observation and describe their effect on measures of center and variability. Justify which measures would be appropriate to answer a statistical question.
• MAT-06.DPS.DA.04 Display numerical data in plots on a number line, including dot plots and histograms. Describe any overall patterns in data, such as gaps, clusters, and skews.
• MAT-07.DPS.DA.02 Analyze and draw inferences about a population using single and multiple random samples by using given measures of center and variability for the numerical data set.
• MAT-08.DPS.DA.01 Interpret scatter plots for bivariate measurement data to investigate patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear
• MAT-08.DPS.DA.02 Draw a trend line on a given scatter plot with a linear association and justify its fit by describing the closeness of the data points to the line.
• MAT-08.DPS.DA.03 Solve authentic problems in the context of bivariate measurement data by interpreting the slope and intercept(s) and making predictions using a linear model.
• MAT-08.DPS.DA.04 Construct and interpret a two-way table summarizing bivariate categorical data collected from the same subjects.
• MAT-10.DPS.02 Compare the center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets using statistics appropriate to the shape of the data
• MAT-10.DPS.03 Represent data on two quantitative variables on a scatter plot and describe how the variables are related.
• MAT-10.DPS.04 Distinguish between correlation and causation.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.01 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).
• MAT-12.DPS.02 Use the mean and standard deviation of a data set to fit it to a normal distribution and estimate population percentages. Recognize that there are data sets for which such a
procedure is not appropriate.
• MAT-12.DPS.03 Evaluate reports based on data.
• MAT-12.DPS.04 Represent data on a scatter plot for two quantitative variables and describe how the variables are related.
• MAT-12.DPS.05 Informally assess the fit of a function by plotting and analyzing residuals.
• MAT-12.DPS.06 Use data from a sample survey to estimate a population means or proportion; develop a margin of error through the use of simulation models for random sampling.
• MAT-12.DPS.07 Understand the process of making inferences about population parameters based on a random sample from that population.
• MAT-12.DPS.08 Decide if a specified model is consistent with results from a given data-generating process (e.g., using simulation).
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
(DA) Data Analysis
Learners will ask and answer questions by collecting, organizing, and displaying relevant data, drawing inferences and conclusions, and making predictions.
MAT-10.DPS.03 Represent data on two quantitative variables on a scatter plot and describe how the variables are related.
Displaying Data
• MAT-01.DPS.D.01 Collect, organize and represent data with up to three categories using picture and bar graphs.
• MAT-02.DPS.D.01 Formulate questions and collect, organize, and represent data, with up to four categories using single unit scaled pictures and bar graphs.
• MAT-03.DPS.D.01 Formulate questions to collect, organize, and represent data with more than four categories using scaled pictures and bar graphs.
• MAT-04.DPS.D.01 Formulate questions to collect, organize, and represent data to reason with math and across disciplines.
• MAT-02.DPS.D.02 Generate data and create line plots marked in whole number units.
• MAT-03.DPS.D.02 Generate data and create line plots marked in whole numbers, halves, and fourths of a unit.
• MAT-04.DPS.D.02 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using
information presented in line plots.
• MAT-05.DPS.D.01 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Use grade-level operations for fractions to solve problems involving information
presented in line plots.
• MAT-06.DPS.DA.04 Display numerical data in plots on a number line, including dot plots and histograms. Describe any overall patterns in data, such as gaps, clusters, and skews.
• MAT-09.NO.03 Choose and interpret the scale and the units in graphs and data displays.
• MAT-09.NO.05 Choose a level of accuracy or precision appropriate to limitations on measurement when reporting quantities.
• MAT-10.DPS.01 Represent data with plots on the real number line (dot plots, histograms, and box plots).
• MAT-10.DPS.03 Represent data on two quantitative variables on a scatter plot and describe how the variables are related.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.NO.04 Use units as a way to understand problems and to guide the solution of multi-step problems (e.g., unit analysis). Choose and interpret units consistently in formulas. Choose and
interpret the scale and the units in graphs and data displays.
• MAT-12.DPS.04 Represent data on a scatter plot for two quantitative variables and describe how the variables are related.
Data Analysis
• MAT-01.DPS.D.02 Analyze data by answering descriptive questions.
• MAT-02.DPS.D.03 Analyze data and interpret the results to solve one-step comparison problems using information from the graphs.
• MAT-03.DPS.D.03 Analyze data and make simple statements to solve one- and two-step problems using information from the graphs.
• MAT-04.DPS.D.02 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using
information presented in line plots.
• MAT-04.DPS.D.03 Utilize graphs and diagrams to represent and solve word problems using the four operations involving whole numbers, benchmark fractions, and decimals.
• MAT-05.DPS.D.01 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Use grade-level operations for fractions to solve problems involving information
presented in line plots.
• MAT-05.DPS.D.02 Utilize graphs and diagrams to represent, analyze, and solve authentic problems using information presented in one or more tables or line plots, including whole numbers,
fractions, and decimals.
• MAT-06.DPS.DA.02 Calculate measures of center (median and mean) and variability (range and mean absolute deviation) to answer a statistical question. Identify mode(s) if they exist.
• MAT-06.DPS.DA.03 Identify outliers by observation and describe their effect on measures of center and variability. Justify which measures would be appropriate to answer a statistical question.
• MAT-06.DPS.DA.04 Display numerical data in plots on a number line, including dot plots and histograms. Describe any overall patterns in data, such as gaps, clusters, and skews.
• MAT-07.DPS.DA.02 Analyze and draw inferences about a population using single and multiple random samples by using given measures of center and variability for the numerical data set.
• MAT-08.DPS.DA.01 Interpret scatter plots for bivariate measurement data to investigate patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear
• MAT-08.DPS.DA.02 Draw a trend line on a given scatter plot with a linear association and justify its fit by describing the closeness of the data points to the line.
• MAT-08.DPS.DA.03 Solve authentic problems in the context of bivariate measurement data by interpreting the slope and intercept(s) and making predictions using a linear model.
• MAT-08.DPS.DA.04 Construct and interpret a two-way table summarizing bivariate categorical data collected from the same subjects.
• MAT-10.DPS.02 Compare the center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets using statistics appropriate to the shape of the data
• MAT-10.DPS.03 Represent data on two quantitative variables on a scatter plot and describe how the variables are related.
• MAT-10.DPS.04 Distinguish between correlation and causation.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.01 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).
• MAT-12.DPS.02 Use the mean and standard deviation of a data set to fit it to a normal distribution and estimate population percentages. Recognize that there are data sets for which such a
procedure is not appropriate.
• MAT-12.DPS.03 Evaluate reports based on data.
• MAT-12.DPS.04 Represent data on a scatter plot for two quantitative variables and describe how the variables are related.
• MAT-12.DPS.05 Informally assess the fit of a function by plotting and analyzing residuals.
• MAT-12.DPS.06 Use data from a sample survey to estimate a population means or proportion; develop a margin of error through the use of simulation models for random sampling.
• MAT-12.DPS.07 Understand the process of making inferences about population parameters based on a random sample from that population.
• MAT-12.DPS.08 Decide if a specified model is consistent with results from a given data-generating process (e.g., using simulation).
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
(DA) Data Analysis
Learners will ask and answer questions by collecting, organizing, and displaying relevant data, drawing inferences and conclusions, and making predictions.
MAT-10.DPS.04 Distinguish between correlation and causation.
Data Analysis
• MAT-01.DPS.D.02 Analyze data by answering descriptive questions.
• MAT-02.DPS.D.03 Analyze data and interpret the results to solve one-step comparison problems using information from the graphs.
• MAT-03.DPS.D.03 Analyze data and make simple statements to solve one- and two-step problems using information from the graphs.
• MAT-04.DPS.D.02 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using
information presented in line plots.
• MAT-04.DPS.D.03 Utilize graphs and diagrams to represent and solve word problems using the four operations involving whole numbers, benchmark fractions, and decimals.
• MAT-05.DPS.D.01 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Use grade-level operations for fractions to solve problems involving information
presented in line plots.
• MAT-05.DPS.D.02 Utilize graphs and diagrams to represent, analyze, and solve authentic problems using information presented in one or more tables or line plots, including whole numbers,
fractions, and decimals.
• MAT-06.DPS.DA.02 Calculate measures of center (median and mean) and variability (range and mean absolute deviation) to answer a statistical question. Identify mode(s) if they exist.
• MAT-06.DPS.DA.03 Identify outliers by observation and describe their effect on measures of center and variability. Justify which measures would be appropriate to answer a statistical question.
• MAT-06.DPS.DA.04 Display numerical data in plots on a number line, including dot plots and histograms. Describe any overall patterns in data, such as gaps, clusters, and skews.
• MAT-07.DPS.DA.02 Analyze and draw inferences about a population using single and multiple random samples by using given measures of center and variability for the numerical data set.
• MAT-08.DPS.DA.01 Interpret scatter plots for bivariate measurement data to investigate patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear
• MAT-08.DPS.DA.02 Draw a trend line on a given scatter plot with a linear association and justify its fit by describing the closeness of the data points to the line.
• MAT-08.DPS.DA.03 Solve authentic problems in the context of bivariate measurement data by interpreting the slope and intercept(s) and making predictions using a linear model.
• MAT-08.DPS.DA.04 Construct and interpret a two-way table summarizing bivariate categorical data collected from the same subjects.
• MAT-10.DPS.02 Compare the center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets using statistics appropriate to the shape of the data
• MAT-10.DPS.03 Represent data on two quantitative variables on a scatter plot and describe how the variables are related.
• MAT-10.DPS.04 Distinguish between correlation and causation.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.01 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).
• MAT-12.DPS.02 Use the mean and standard deviation of a data set to fit it to a normal distribution and estimate population percentages. Recognize that there are data sets for which such a
procedure is not appropriate.
• MAT-12.DPS.03 Evaluate reports based on data.
• MAT-12.DPS.04 Represent data on a scatter plot for two quantitative variables and describe how the variables are related.
• MAT-12.DPS.05 Informally assess the fit of a function by plotting and analyzing residuals.
• MAT-12.DPS.06 Use data from a sample survey to estimate a population means or proportion; develop a margin of error through the use of simulation models for random sampling.
• MAT-12.DPS.07 Understand the process of making inferences about population parameters based on a random sample from that population.
• MAT-12.DPS.08 Decide if a specified model is consistent with results from a given data-generating process (e.g., using simulation).
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
(P) Probability
Learners will understand and apply basic concepts of probability.
MAT-10.DPS.05 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes or as unions, intersections, or complements of other events
(“or,” “and,” “not”).
• MAT-07.DPS.P.01 Develop a probability model to find probabilities of theoretical events and contrast probabilities from an experimental model.
• MAT-07.DPS.P.02 Develop a probability model to find theoretical probabilities of independent compound events.
• MAT-10.DPS.05 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes or as unions, intersections, or complements of other events
("or," "and," "not").
• MAT-10.DPS.06 Recognize that event A is independent of event B if the probability of event A does not change in response to the occurrence of event B. Apply the formula P(A and B) = P(A)·P(B)
given that events A and B are independent.
• MAT-10.DPS.07 Recognize the conditional probability of an event A given B is the probability that event A will occur given the knowledge that event B has already occurred. Calculate the
conditional probability of A given B and interpret the answer in context.
• MAT-10.DPS.08 Apply the formula P(A or B) = P(A) + P(B) – P(A and B) and interpret the answer in context.
• MAT-10.DPS.09 Determine the number of outcomes using permutations and combinations in context.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.10 Determine when the order in counting matters and use permutations and combinations to compute probabilities of events accordingly. Determine probability situations as conditional,
"or" (union), or "and" (intersection), and determine the probability of an event.
• MAT-12.DPS.11 Use permutations and combinations to compute probabilities of compound events and solve problems.
• MAT-12.DPS.12 Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space. Graph the corresponding probability distribution using the same
graphical displays as for data distributions.
• MAT-12.DPS.13 Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.
• MAT-12.DPS.14 Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values.
• MAT-12.DPS.15 Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value.
• MAT-12.DPS.16 Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value.
• MAT-12.DPS.17 Use probabilities to make fair decisions.
• MAT-12.DPS.18 Analyze decisions and strategies using probability concepts.
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
(P) Probability
Learners will understand and apply basic concepts of probability.
MAT-10.DPS.06 Recognize that event A is independent of event B if the probability of event A does not change in response to the occurrence of event B.
Apply the formula P(A and B) = P(A)·P(B) given that events A and B are independent.
• MAT-07.DPS.P.01 Develop a probability model to find probabilities of theoretical events and contrast probabilities from an experimental model.
• MAT-07.DPS.P.02 Develop a probability model to find theoretical probabilities of independent compound events.
• MAT-10.DPS.05 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes or as unions, intersections, or complements of other events
("or," "and," "not").
• MAT-10.DPS.06 Recognize that event A is independent of event B if the probability of event A does not change in response to the occurrence of event B. Apply the formula P(A and B) = P(A)·P(B)
given that events A and B are independent.
• MAT-10.DPS.07 Recognize the conditional probability of an event A given B is the probability that event A will occur given the knowledge that event B has already occurred. Calculate the
conditional probability of A given B and interpret the answer in context.
• MAT-10.DPS.08 Apply the formula P(A or B) = P(A) + P(B) – P(A and B) and interpret the answer in context.
• MAT-10.DPS.09 Determine the number of outcomes using permutations and combinations in context.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.10 Determine when the order in counting matters and use permutations and combinations to compute probabilities of events accordingly. Determine probability situations as conditional,
"or" (union), or "and" (intersection), and determine the probability of an event.
• MAT-12.DPS.11 Use permutations and combinations to compute probabilities of compound events and solve problems.
• MAT-12.DPS.12 Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space. Graph the corresponding probability distribution using the same
graphical displays as for data distributions.
• MAT-12.DPS.13 Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.
• MAT-12.DPS.14 Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values.
• MAT-12.DPS.15 Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value.
• MAT-12.DPS.16 Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value.
• MAT-12.DPS.17 Use probabilities to make fair decisions.
• MAT-12.DPS.18 Analyze decisions and strategies using probability concepts.
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
Learners will ask and answer questions by collecting, organizing, and displaying relevant data, drawing inferences and conclusions and making predictions; and understanding and applying basic
concepts of probability.
MAT-10.DPS.07 Recognize that the conditional probability of an event A given B is the probability that event A will occur given the knowledge that event B has already occurred.
Calculate the conditional probability of A given B and interpret the answer in context.
• MAT-07.DPS.P.01 Develop a probability model to find probabilities of theoretical events and contrast probabilities from an experimental model.
• MAT-07.DPS.P.02 Develop a probability model to find theoretical probabilities of independent compound events.
• MAT-10.DPS.05 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes or as unions, intersections, or complements of other events
("or," "and," "not").
• MAT-10.DPS.06 Recognize that event A is independent of event B if the probability of event A does not change in response to the occurrence of event B. Apply the formula P(A and B) = P(A)·P(B)
given that events A and B are independent.
• MAT-10.DPS.07 Recognize the conditional probability of an event A given B is the probability that event A will occur given the knowledge that event B has already occurred. Calculate the
conditional probability of A given B and interpret the answer in context.
• MAT-10.DPS.08 Apply the formula P(A or B) = P(A) + P(B) – P(A and B) and interpret the answer in context.
• MAT-10.DPS.09 Determine the number of outcomes using permutations and combinations in context.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.10 Determine when the order in counting matters and use permutations and combinations to compute probabilities of events accordingly. Determine probability situations as conditional,
"or" (union), or "and" (intersection), and determine the probability of an event.
• MAT-12.DPS.11 Use permutations and combinations to compute probabilities of compound events and solve problems.
• MAT-12.DPS.12 Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space. Graph the corresponding probability distribution using the same
graphical displays as for data distributions.
• MAT-12.DPS.13 Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.
• MAT-12.DPS.14 Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values.
• MAT-12.DPS.15 Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value.
• MAT-12.DPS.16 Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value.
• MAT-12.DPS.17 Use probabilities to make fair decisions.
• MAT-12.DPS.18 Analyze decisions and strategies using probability concepts.
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
(P) Probability
Learners will understand and apply basic concepts of probability.
MAT-10.DPS.08 Apply the formula P(A or B) = P(A) + P(B) – P(A and B) and interpret the answer in context.
• MAT-07.DPS.P.01 Develop a probability model to find probabilities of theoretical events and contrast probabilities from an experimental model.
• MAT-07.DPS.P.02 Develop a probability model to find theoretical probabilities of independent compound events.
• MAT-10.DPS.05 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes or as unions, intersections, or complements of other events
("or," "and," "not").
• MAT-10.DPS.06 Recognize that event A is independent of event B if the probability of event A does not change in response to the occurrence of event B. Apply the formula P(A and B) = P(A)·P(B)
given that events A and B are independent.
• MAT-10.DPS.07 Recognize the conditional probability of an event A given B is the probability that event A will occur given the knowledge that event B has already occurred. Calculate the
conditional probability of A given B and interpret the answer in context.
• MAT-10.DPS.08 Apply the formula P(A or B) = P(A) + P(B) – P(A and B) and interpret the answer in context.
• MAT-10.DPS.09 Determine the number of outcomes using permutations and combinations in context.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.10 Determine when the order in counting matters and use permutations and combinations to compute probabilities of events accordingly. Determine probability situations as conditional,
"or" (union), or "and" (intersection), and determine the probability of an event.
• MAT-12.DPS.11 Use permutations and combinations to compute probabilities of compound events and solve problems.
• MAT-12.DPS.12 Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space. Graph the corresponding probability distribution using the same
graphical displays as for data distributions.
• MAT-12.DPS.13 Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.
• MAT-12.DPS.14 Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values.
• MAT-12.DPS.15 Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value.
• MAT-12.DPS.16 Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value.
• MAT-12.DPS.17 Use probabilities to make fair decisions.
• MAT-12.DPS.18 Analyze decisions and strategies using probability concepts.
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
(P) Probability
Learners will understand and apply basic concepts of probability.
MAT-10.DPS.09 Determine the number of outcomes using permutations and combinations in context.
Counting Patterns
• MAT-00.NO.CC.05 Count and tell how many objects up to 20 are in an arranged pattern or up to 10 objects in a scattered configuration. Represent a quantity of up to 20 with a numeral.
• MAT-01.NO.CC.05 Skip count forward and backward by 5s and 10s from multiples and recognize the patterns of up to 10 skip counts.
• MAT-02.NO.CC.04 Skip count forward and backward by 2s and 100s and recognize the patterns of skip counts.
• MAT-10.DPS.09 Determine the number of outcomes using permutations and combinations in context.
• MAT-12.DPS.11 Use permutations and combinations to compute probabilities of compound events and solve problems.
• MAT-07.DPS.P.01 Develop a probability model to find probabilities of theoretical events and contrast probabilities from an experimental model.
• MAT-07.DPS.P.02 Develop a probability model to find theoretical probabilities of independent compound events.
• MAT-10.DPS.05 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes or as unions, intersections, or complements of other events
("or," "and," "not").
• MAT-10.DPS.06 Recognize that event A is independent of event B if the probability of event A does not change in response to the occurrence of event B. Apply the formula P(A and B) = P(A)·P(B)
given that events A and B are independent.
• MAT-10.DPS.07 Recognize the conditional probability of an event A given B is the probability that event A will occur given the knowledge that event B has already occurred. Calculate the
conditional probability of A given B and interpret the answer in context.
• MAT-10.DPS.08 Apply the formula P(A or B) = P(A) + P(B) – P(A and B) and interpret the answer in context.
• MAT-10.DPS.09 Determine the number of outcomes using permutations and combinations in context.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.10 Determine when the order in counting matters and use permutations and combinations to compute probabilities of events accordingly. Determine probability situations as conditional,
"or" (union), or "and" (intersection), and determine the probability of an event.
• MAT-12.DPS.11 Use permutations and combinations to compute probabilities of compound events and solve problems.
• MAT-12.DPS.12 Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space. Graph the corresponding probability distribution using the same
graphical displays as for data distributions.
• MAT-12.DPS.13 Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.
• MAT-12.DPS.14 Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values.
• MAT-12.DPS.15 Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value.
• MAT-12.DPS.16 Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value.
• MAT-12.DPS.17 Use probabilities to make fair decisions.
• MAT-12.DPS.18 Analyze decisions and strategies using probability concepts.
10^th Grade (MAT) Targeted Standard
(DPS) Data Probability and Statistics
Learners will ask and answer questions by collecting, organizing, and displaying relevant data, drawing inferences and conclusions and making predictions; and understanding and applying basic
concepts of probability.
MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and to approximate
conditional probabilities.
Displaying Data
• MAT-01.DPS.D.01 Collect, organize and represent data with up to three categories using picture and bar graphs.
• MAT-02.DPS.D.01 Formulate questions and collect, organize, and represent data, with up to four categories using single unit scaled pictures and bar graphs.
• MAT-03.DPS.D.01 Formulate questions to collect, organize, and represent data with more than four categories using scaled pictures and bar graphs.
• MAT-04.DPS.D.01 Formulate questions to collect, organize, and represent data to reason with math and across disciplines.
• MAT-02.DPS.D.02 Generate data and create line plots marked in whole number units.
• MAT-03.DPS.D.02 Generate data and create line plots marked in whole numbers, halves, and fourths of a unit.
• MAT-04.DPS.D.02 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using
information presented in line plots.
• MAT-05.DPS.D.01 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Use grade-level operations for fractions to solve problems involving information
presented in line plots.
• MAT-06.DPS.DA.04 Display numerical data in plots on a number line, including dot plots and histograms. Describe any overall patterns in data, such as gaps, clusters, and skews.
• MAT-09.NO.03 Choose and interpret the scale and the units in graphs and data displays.
• MAT-09.NO.05 Choose a level of accuracy or precision appropriate to limitations on measurement when reporting quantities.
• MAT-10.DPS.01 Represent data with plots on the real number line (dot plots, histograms, and box plots).
• MAT-10.DPS.03 Represent data on two quantitative variables on a scatter plot and describe how the variables are related.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.NO.04 Use units as a way to understand problems and to guide the solution of multi-step problems (e.g., unit analysis). Choose and interpret units consistently in formulas. Choose and
interpret the scale and the units in graphs and data displays.
• MAT-12.DPS.04 Represent data on a scatter plot for two quantitative variables and describe how the variables are related.
Data Analysis
• MAT-01.DPS.D.02 Analyze data by answering descriptive questions.
• MAT-02.DPS.D.03 Analyze data and interpret the results to solve one-step comparison problems using information from the graphs.
• MAT-03.DPS.D.03 Analyze data and make simple statements to solve one- and two-step problems using information from the graphs.
• MAT-04.DPS.D.02 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using
information presented in line plots.
• MAT-04.DPS.D.03 Utilize graphs and diagrams to represent and solve word problems using the four operations involving whole numbers, benchmark fractions, and decimals.
• MAT-05.DPS.D.01 Generate data and create line plots to display a data set of fractions of a unit (1/2, 1/4, 1/8). Use grade-level operations for fractions to solve problems involving information
presented in line plots.
• MAT-05.DPS.D.02 Utilize graphs and diagrams to represent, analyze, and solve authentic problems using information presented in one or more tables or line plots, including whole numbers,
fractions, and decimals.
• MAT-06.DPS.DA.02 Calculate measures of center (median and mean) and variability (range and mean absolute deviation) to answer a statistical question. Identify mode(s) if they exist.
• MAT-06.DPS.DA.03 Identify outliers by observation and describe their effect on measures of center and variability. Justify which measures would be appropriate to answer a statistical question.
• MAT-06.DPS.DA.04 Display numerical data in plots on a number line, including dot plots and histograms. Describe any overall patterns in data, such as gaps, clusters, and skews.
• MAT-07.DPS.DA.02 Analyze and draw inferences about a population using single and multiple random samples by using given measures of center and variability for the numerical data set.
• MAT-08.DPS.DA.01 Interpret scatter plots for bivariate measurement data to investigate patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear
• MAT-08.DPS.DA.02 Draw a trend line on a given scatter plot with a linear association and justify its fit by describing the closeness of the data points to the line.
• MAT-08.DPS.DA.03 Solve authentic problems in the context of bivariate measurement data by interpreting the slope and intercept(s) and making predictions using a linear model.
• MAT-08.DPS.DA.04 Construct and interpret a two-way table summarizing bivariate categorical data collected from the same subjects.
• MAT-10.DPS.02 Compare the center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets using statistics appropriate to the shape of the data
• MAT-10.DPS.03 Represent data on two quantitative variables on a scatter plot and describe how the variables are related.
• MAT-10.DPS.04 Distinguish between correlation and causation.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.01 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers).
• MAT-12.DPS.02 Use the mean and standard deviation of a data set to fit it to a normal distribution and estimate population percentages. Recognize that there are data sets for which such a
procedure is not appropriate.
• MAT-12.DPS.03 Evaluate reports based on data.
• MAT-12.DPS.04 Represent data on a scatter plot for two quantitative variables and describe how the variables are related.
• MAT-12.DPS.05 Informally assess the fit of a function by plotting and analyzing residuals.
• MAT-12.DPS.06 Use data from a sample survey to estimate a population means or proportion; develop a margin of error through the use of simulation models for random sampling.
• MAT-12.DPS.07 Understand the process of making inferences about population parameters based on a random sample from that population.
• MAT-12.DPS.08 Decide if a specified model is consistent with results from a given data-generating process (e.g., using simulation).
• MAT-07.DPS.P.01 Develop a probability model to find probabilities of theoretical events and contrast probabilities from an experimental model.
• MAT-07.DPS.P.02 Develop a probability model to find theoretical probabilities of independent compound events.
• MAT-10.DPS.05 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes or as unions, intersections, or complements of other events
("or," "and," "not").
• MAT-10.DPS.06 Recognize that event A is independent of event B if the probability of event A does not change in response to the occurrence of event B. Apply the formula P(A and B) = P(A)·P(B)
given that events A and B are independent.
• MAT-10.DPS.07 Recognize the conditional probability of an event A given B is the probability that event A will occur given the knowledge that event B has already occurred. Calculate the
conditional probability of A given B and interpret the answer in context.
• MAT-10.DPS.08 Apply the formula P(A or B) = P(A) + P(B) – P(A and B) and interpret the answer in context.
• MAT-10.DPS.09 Determine the number of outcomes using permutations and combinations in context.
• MAT-10.DPS.10 Construct and interpret two-way frequency tables of data for two categorical variables. Use the two-way table as a sample space to decide if events are independent and approximate
conditional probabilities.
• MAT-12.DPS.10 Determine when the order in counting matters and use permutations and combinations to compute probabilities of events accordingly. Determine probability situations as conditional,
"or" (union), or "and" (intersection), and determine the probability of an event.
• MAT-12.DPS.11 Use permutations and combinations to compute probabilities of compound events and solve problems.
• MAT-12.DPS.12 Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space. Graph the corresponding probability distribution using the same
graphical displays as for data distributions.
• MAT-12.DPS.13 Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.
• MAT-12.DPS.14 Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values.
• MAT-12.DPS.15 Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value.
• MAT-12.DPS.16 Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value.
• MAT-12.DPS.17 Use probabilities to make fair decisions.
• MAT-12.DPS.18 Analyze decisions and strategies using probability concepts. | {"url":"https://learnbps.bismarckschools.org/mod/glossary/print.php?id=867768&mode=cat&hook=10539&sortkey&sortorder=asc&offset=0&pagelimit=20","timestamp":"2024-11-07T04:28:31Z","content_type":"text/html","content_length":"105796","record_id":"<urn:uuid:0625cf7b-b0d0-4f92-8e1a-a17b44232b9a>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00579.warc.gz"} |
Arithmetic Operators and Expressions – Real Python
Arithmetic Operators and Expressions
What do you mean by unary +2 in the second example? 5 - 2 = 3. I can’t understand unary. Please explain it.
@s150042028 In Python, both the plus (+) and minus (-) operators have binary and unary versions. The word “binary” in this context means that the operator takes two operands or arguments, while
“unary” implies only one argument.
When used with two operands, the plus operator adds them together, which can mean different things depending on the types of the operands:
>>> 5 + 2
>>> "bat" + "man"
When used with numbers, the plus operator performs the addition. For strings, on the other hand, the plus operator is defined as the concatenation operation.
The unary version of the plus operator does nothing, so numeric literals written with or without it are considered equal:
>>> +2
>>> +2 == 2
The minus operator is only defined for numbers. More specifically, the binary minus operator performs the subtraction of two numbers, whereas the unary minus operator flips the sign of the number, or
more generally, the arithmetic expression that follows:
>>> 5 - 2
>>> -2
>>> -(5 - 2)
Let me know if that clears it up for you!
I am still confused, but I will read it again. However, I am not feeling comfortable with this structure of the course. I mean, there are no projects or exercises. Watching videos is not a useful way
for me. There is a 90% chance that I will cancel my membership due to this poor method of teaching. There should be an interactive environment with videos, examples, and exercises. Do you have any
suggestions, please?
@s150042028 Thank you for your feedback. We appreciate you taking the time to share it with us, as we’re continually working to improve our course offerings. Your comments are immensely beneficial in
that process!
Please note that this video course is part of a series based on the Python Basics book. We’re working hard to eventually accompany each course in this series with a separate hands-on course that
contains practical exercises. We’re almost done with recording all of them, but some will require a little bit of patience. The exercises course for this particular one will come out next Tuesday,
November 7.
Once released, you’ll have a chance to solve the accompanying exercises yourself and then confront your solutions with our walkthrough, where we explain each step in detail and the thought process
behind it.
In the meantime, I’d highly encourage you to take advantage of other benefits that Real Python offers to its members. If you haven’t already, you can ask for help on our Slack community, or you can
come to Office Hours where we can address your doubts in real time. It’s a weekly webinar where we often talk about various Python-related concepts, including solving actual programming challenges
that people face at work through screen sharing. This would be a perfect opportunity for you to get some hands-on experience and interact with our team on a personal level. We hope to see you there
and look forward to helping enhance your learning experience with us.
Difficile da capire un pò, ma solo per chi è agli inizi, ovviamente. Basta rileggere varie volte è diventa molto più chiaro. Fin’ora ottimi videi. Complimenti. | {"url":"https://realpython.com/lessons/arithmetic-operators-and-expressions/","timestamp":"2024-11-08T11:56:19Z","content_type":"text/html","content_length":"78797","record_id":"<urn:uuid:51668f8b-b17e-4409-a87e-7ca3291ea7aa>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00365.warc.gz"} |
Given \
Here we will first draw two circles and then we will draw the point of intersection of two circles and draw everything given in the question. We will consider all the conditions given in the
questions and then we will consider two triangles in the circle and we will make them congruent. From there, we will get the required relation.
Complete step-by-step answer:
Now, we will draw two circles with their point of intersection as \[P\] and it is given that the straight line \[APB\] is parallel to \[CD\] . So, we will draw \[CC' \bot AB\] and \[DD' \bot AB\].
It is given that \[AB\] is parallel to \[CD\].
As we know \[CC' \bot AB\] and \[DD' \bot AB\].
So, we can say that
\[CC' = DD'\]
And hence
\[C'D' = CD\] ………. \[\left( 1 \right)\]
Now, we will consider $\vartriangle AC'C$ and $\vartriangle PC'C$ , then we will make them congruent to each other.
In $\vartriangle AC'C$ and $\vartriangle PC'C$, we have
\[PC = AC\] because they are the radii of a circle.
\[\angle AC'C = \angle PC'C = 90^\circ \]
\[CC' = CC'\] because they are common sides of the triangles.
Thus, $\vartriangle AC'C \cong \vartriangle PC'C$ by SAS rule of congruence.
Therefore, by corresponding part of the congruent triangles, we get
\[AC' = PC'\] ……….. \[\left( 2 \right)\]
Similarly, we will consider $\vartriangle PD'D$ and $\vartriangle BD'D$, then we will make them congruent to each other.
In $\vartriangle PD'D$ and $\vartriangle BD'D$, we have
\[PD = BD\] because they are the radii of a circle.
\[\angle PD'D = \angle BD'D = 90^\circ \]
\[DD' = DD'\] because they are common sides of the triangles.
Thus, $\vartriangle PD'D \cong \vartriangle BD'D$ by SAS rule of congruency.
Therefore, by corresponding part of the congruent triangles, we get
\[PD' = BD'\] ……….. \[\left( 3 \right)\]
From the figure, we know that,
\[AB = AC' + PC' + PD' + BD'\] ……… \[\left( 4 \right)\]
From equation \[\left( 2 \right)\] and equation \[\left( 3 \right)\], we have;
\[AC' = PC'\]
\[PD' = BD'\]
Now, we will substitute these values in equation \[\left( 4 \right)\]. Therefore, we get
\[ \Rightarrow AB = PC' + PC' + PD' + PD'\]
On adding the like terms, we get
\[ \Rightarrow AB = 2PC' + 2PD'\]
On further simplification, we get
\[ \Rightarrow AB = 2\left( {PC' + PD'} \right)\] ……… \[\left( 5 \right)\]
We know that \[PC' + PD' = C'D'\]. So, substituting this value in the above equation \[\left( 5 \right)\], we get
\[ \Rightarrow AB = 2C'D'\] ………. \[\left( 6 \right)\]
We know from equation \[\left( 1 \right)\] that \[C'D' = CD\].
So, replacing \[C'D'\] by \[CD\] in equation \[\left( 6 \right)\], we get
\[ \Rightarrow AB = 2CD\]
Hence, the correct option is option D.Note:
Here we have made two triangles congruent to each other. A triangle is a two-dimensional geometric shape that has two sides. Any two objects are congruent only if they superimpose on each other. Two
triangles are said to be congruent if all their three angles and three sides are equal. But it is not necessary to find all the six dimensions. Hence, the two triangles can be made congruent by
knowing only three values out of six values. | {"url":"https://www.vedantu.com/question-answer/given-p-is-a-point-of-intersection-of-two-class-8-maths-cbse-5fd85be18238151f0d9fcbed","timestamp":"2024-11-07T14:02:43Z","content_type":"text/html","content_length":"168465","record_id":"<urn:uuid:16b6a808-ca77-48db-a89d-0f216a7ad11b>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00709.warc.gz"} |
Find the value of \
Hint: In the given question, we have to solve the trigonometric equation. We will use the trigonometric identities to solve the equation and arrive at the answer. We should know that \[{\sec ^2}\
theta = \dfrac{1}{{{{\cos }^2}\theta }}\]and \[{\tan ^2}\theta = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}\].
Complete step by step solution:
Trigonometry is one of the branches of mathematics that uses trigonometric ratios to find the angles and missing sides of a triangle. The trigonometric functions are the trigonometric ratios of a
triangle. The trigonometric functions are sine, cosine, sectant, cosecant, tangent and cotangent.
The Trigonometric formulas or Identities are the equations which are valid in the case of Right-Angled Triangles. They are also called Pythagorean Identities.
We can solve the question as follows-
We know that \[\sec \theta = \dfrac{1}{{\cos \theta }}\]so we can say that \[{\sec ^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }}\].
Similarly, \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]so we can get \[{\tan ^2}\theta = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}\].
Applying the above identities, we get,
\[ = \dfrac{1}{{{{\cos }^2}\theta }} - \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}\]
We have common denominator, so we can get,
\[ = \dfrac{{1 - {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}\]
Now since \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], we can get \[{\cos ^2}\theta = 1 - {\sin ^2}\theta \].
Applying the above identity, we get,
\[ = \dfrac{{{{\cos }^2}\theta }}{{{{\cos }^2}\theta }}\]
\[ = 1\]
Hence, \[{\sec ^2}\theta - {\tan ^2}\theta = 1\].
In the given case, we have converted the secant and tangent into sine and cosine so we can establish the relationship between the two variables and solve the question easily. The key to solve such a
question is to identify which trigonometric identity will be useful and accordingly apply the same. We should generally convert the variables into sine and cosine because they are the basic
There can be more than one way to solve the question, for example we can solve the given question by applying different identity as follows:
\[{\sec ^2}\theta = 1 + {\tan ^2}\theta \]
Substituting the above value in the given question, we get,
\[ = 1 + {\tan ^2}\theta - {\tan ^2}\theta \]
\[ = 1\]. | {"url":"https://www.vedantu.com/question-answer/find-the-value-of-sec-2theta-tan-2theta-class-10-maths-cbse-609bcbfc80f45e5260f64d50","timestamp":"2024-11-07T04:21:40Z","content_type":"text/html","content_length":"167458","record_id":"<urn:uuid:70014901-a989-4ce6-809a-bbb7fda1efef>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00604.warc.gz"} |
Invited talk to the Second Congress of Greek Mathematicians, July 2022
Invited talk to the Second Congress of Greek Mathematicians (SCGM 2022), 04-08 July (Athens). My talk will take place on 06 July of 2022.
Conference webpage: SCGM 2022
Title: Quadratic robust and generalized robust toric ideals of graphs
(joint work with Ignacio Garcia-Marco)
Abstract: A toric ideal is called robust if its universal Gröbner basis is a minimal set of generators, and is called generalized robust if its universal Gröbner basis equals its universal Markov
basis (the union of all its minimal sets of binomial generators). Robust and generalized robust toric ideals are both interesting from both a Commutative Algebra and an Algebraic Statistics
perspective. However, only a few nontrivial examples of such ideals are known. In this talk we study these properties for toric ideals of graphs. We characterize combinatorially the graphs giving
rise to robust and to generalized robust toric ideals generated by quadratic binomials. | {"url":"https://chtatakis-mathnet.gr/index.php/el/talk-athens","timestamp":"2024-11-06T12:04:04Z","content_type":"text/html","content_length":"8861","record_id":"<urn:uuid:f4f0127b-44bc-47d4-a116-9e2e158fd2d7>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00729.warc.gz"} |
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Investigation into the Effect of Interlock Volume on SPR Strength
Department of Civil and Structural Engineering, University of Sheffield, Sheffield S1 3JD, UK
Atlas Copco, Deeside CH5 2NS, UK
Department of Mechanical Engineering, University of Sheffield, Sheffield S1 3JD, UK
Author to whom correspondence should be addressed.
Submission received: 20 February 2023 / Revised: 21 March 2023 / Accepted: 22 March 2023 / Published: 29 March 2023
During the design of automotive structures assembled using Self-Piercing Rivets (SPRs), a rivet and die combination is selected for each joint stack. To conduct extensive physical tensile testing on
every joint combination to determine the range of strength achieved by each rivet–die combination, a great deal of lab technician time and substrate material are required. It is much simpler and less
material-consuming to select the rivet and die solution by examining the cross sections of joints. However, the current methods of measuring cross sections by measuring the amount of mechanical
interlock in a linear X–Y direction, achieved with the flared rivet tail, do not give an accurate prediction of joint strength, because they do not measure the full amount of material that must be
defeated to pull the rivet tail out of the bottom sheet. The X–Y linear interlock measurement approach also makes it difficult to rapidly rank joint solutions, as it creates two values for each cross
section rather than a single value. This study investigates an innovative new measurement method developed by the authors called Volumelock. The approach measures the volume of material that must be
defeated to pull out the rivet. Creating a single measurement value for each rivet–die combination makes it much easier to compare different rivet and die solutions; to identify solutions that work
well across a number of different stacks; to aid the grouping of stacks on one setter for low-volume line; and to select the strongest solutions for a high-volume line where only one or two different
stacks are made by each setter. The joint stack results in this paper indicate that there is a good predictive relationship between the new Volumelock method and peel strength, measured by physical
cross-tension testing. In this study, the Volumelock approach predicted the peel strength within a 5% error margin.
1. Introduction
Self-Piercing Rivets (SPRs) are a commonly used joining technology in the automotive industry for joining both aluminium to aluminium and steel to aluminium. To select the most suitable rivet and die
combination for a new joint stack, sample joints are made and physically examined. Joint characteristics such as Head Height Gap, X interlock, Y interlock, and Minimum Bottom Sheet Thickness are
measured, as seen in
Figure 1
. These characteristics are used by automotive joining engineers as predictors of joint quality and production robustness [
However, the current approach of measuring X and Y interlocks does not accurately predict joint strength, as differences in the shape of the flare and amount of material above the flared rivet tip
produce different cross-tension strengths, even for joints with the same X and Y interlock values. The shape of the flare is influenced by various factors, including the speed and force of the punch,
die shape and material properties, and flare geometry [
For production, engineers must recommend a rivet and die combination for a joint stack. This often involves selecting a rivet and die combination to serve multiple material stacks on the same vehicle
to reduce manufacturing complexity. The selection then must be optimised in terms of joint stack compatibility and to obtain the best possible joint strength, and, in turn, to obtain the required
crash performance. Conducting physical tensile testing on every joint combination proposed for the required joint stacks is expensive in terms of both time and resources.
Therefore, it is common to narrow down the testing to the best joints from the sample group by evaluating the cross-section characteristics of the joint, which currently relies on the X and Y
interlocks. However, these have been observed to be inaccurate predictors of strength, as they do not account for the full amount of material that must be defeated to pull the tail of the rivet out
of the bottom sheet. The X and Y linear interlock measurement approach also makes it difficult to optimise combinations, as it creates two values for each cross section rather than a single value.
This study aimed to devise an improved measurement technique to optimise and select rivet and die combinations based on the physical characteristics of joint cross sections, and to investigate
whether a pattern could be found to estimate the relative strength of joints with a minimal amount of tensile testing. The authors’ idea was to measure the complete area in the bottom sheet captured
by the rivet flare. This area was given the name “Arealock”, and is shown in
Figure 1
. The Arealock is the area of material that is required to be defeated for the rivet to be pulled out of the bottom sheet. This approach was improved further by revolving the Arealock around the
central axis of the rivet to give a “Volumelock” value for the complete volume of material in the bottom sheet captured by the rivet flare that must be defeated for the joint to fail from rivet
pull-out. This is shown in
Figure 2
. This investigation focuses on the impact of Volumelock on joint strength for self-piercing riveted joint stacks in peel by analysing data from cross-tension testing, as shown in
Figure 3
Without CAD, the Volumelock can be estimated by revolving the Arealock around the central axis. This is derived from the volume of a toroid by substituting the equation for the area of a circle,
which gives rise to Equation (1), where
is the flared rivet radius and
is the Arealock.
These new measurement techniques open up a variety of powerful measurement, analysis and optimisation techniques that will be introduced and explored in this paper.
Current Research
Traditional interlock measurements are the only cross-sectional measurement techniques that have been applied to mechanical point joining methods such as SPR, clinching or Friction-SPR [
]. The state of the art tends to focus on using existing metrics to predict joint strength, with some success. However, without simulations, large errors in these predictions limit the use of these
methods. Although computationally expensive, Finite-Element Analysis (FEA) is a common prediction tool in industry [
], and has shown good accuracy for quasi-static testing. Its use has grown as the models have improved, including methods such as the stress–strain history of the joint [
]. The results of Self-Piercing Rivet tests can be replicated more reliably when damage modelling is included. There are many damage models suitable for high-strain-rate deformations in SPR joint
testing, such as the Johnson–Cook [
], Bonora [
] and Lemaitre models [
]. However, researchers have found that the Drucker–Prager model produced the most accurate results for aluminium top sheets [
]. The same research also used the linear X and Y interlock to predict joint strength, employing a modified shear force calculation to predict the force required to elicit material failure during
cross-tension. This approach did not consider the inherent flaws in the X and Y interlock method, and therefore resulted in a 25% error in joint strength prediction.
Other researchers have employed machine learning algorithms to predict cross-section parameters such as the X and Y interlock and the resulting joint strength [
]. This method was able to predict the joint contour within 18.8% and the joint strength within 18.6%. As this method relies on the area encompassed by the joint contour, it could result in an error
of up to 10%, in addition to the aforementioned 18.8% in the prediction of the joint strength. Very little previous research was found that attempted to predict the energy absorbed during failure—a
key metric used by car body designers when selecting joining solutions.
Researchers have also used a neural network regression model to predict joint performance within 8.5% for steel-to-aluminium joints [
]. This required a large dataset to train and verify the model. The researchers also found the simulation of the joints predicted the maximum load fairly well; however, the model over-predicted the
joint performance value, and it massively under-predicted energy absorbed due to experimental limitations in cross-tension testing, with the sheets sliding during testing. The current research in
this field demonstrates that joint optimisation and strength prediction are either limited to relatively high-error methods or rely on the use of high-cost computing solutions. Due to the inherent
flaws in the current measurement techniques, there exists a gap in the research for Volumelock.
2. Materials and Methods
2.1. Materials
To investigate the effect of Volumelock, a common material stack was chosen, as shown in
Table 1
. The joint stack was chosen to investigate the effects of die depth and rivet length on the interlock volume. AA5754 H111 alloy was chosen because of its common use in the automotive sector for
press-forming sheet panels. It is a perfect candidate for this investigation due to the fact that a huge range of rivet and die combinations can be used to join the same material stack. This means a
joining engineer could find it difficult to rank the cross-sectional measurements in order to select the joints intended for tensile testing. The joint stacks were selected by preliminary
investigation of the relationship between the X and Y interlock values and Volumelock. Some joint stacks had a proportional relationship between the X and Y interlocks, whilst others displayed a more
complex relationship, allowing the beneficial effects of using Volumelock to be clearly visible. The die and rivet types used can be seen in
Figure 4
2.2. Methods
The cross-section measurement data were generated using an Atlas Copco G1.6 servo setter to make five repeats for each of the joints with the insertion parameters shown in
Table 2
in 40 × 40 mm coupons. These coupons were cross-sectioned using a Struers Labatom-5 cross-section cutter (Struers, Rotherham, UK). The cut samples were then imaged using a Zeiss Stemi 508 microscope
(Zeiss, Jena, Germany) with a SPOT camera to build a comprehensive set of measurement data. The X and Y interlocks were measured using rectangular measurement tools in SPOT 5.0 measurement software.
These images were then exported to Autodesk Fusion 360 CAD software (Verison 2.0.15509) to measure the Volumelock directly. This was achieved by using sketch splines to capture the Arealock, as shown
Figure 1
. Then, we used the revolve tool to generate a toroidal body, which was then inspected to find the volume of the shape for each side of the rivet, which was then averaged. The cross-tension samples
were created using a jig, as shown in
Figure 5
a, from two 38 × 120 mm coupons, pictured in
Figure 6
. These were placed in an Instron 5892 universal test machine fitted with specially made cross-tension grips (
Figure 5
b). The cross-tension samples were then pulled apart at 20 mm/min. The force–displacement curves were recorded using Instron Bluehill 3 tensile testing software to yield the peak load sustained and
the energy absorbed until failure, which is the area under the curve, an example of which is shown in
Figure 7
. All of these data were then imported to a Microsoft Excel file where the data were analysed and plotted to investigate the relationships. The mode of failure was recorded to evaluate the reason for
failure (e.g., bottom sheet or top sheet pull-out). The tensile tests were repeated five times for each stack and die combination to ensure repeatability and validity, which generated an average
joint performance.
3. Results and Discussion
The X and Y interlocks were evaluated for the correlation with the maximum load during testing, and to clarify the need for a new measurement method. This was accomplished by comparing four joint
stacks from the results. Two joint stacks with similar X interlocks were chosen and two joint stacks with similar max loads were chosen. The full dataset can be found in
Appendix A
Figure 8
shows the percentage increase in interlock measurements and maximum load during cross-tension testing.
Figure 8
a demonstrates a 22% increase in maximum load with a proportional increase in the Y interlock; however, the X interlock only increased by 3%.
Figure 8
b demonstrates a 2% increase in maximum load with a proportional increase in X, while also showing a 57% increase in the Y interlock. These results suggest that both the X and Y interlocks
independently influence joint strength; however, they do so in an unpredictable manner, making joint optimisation based on current cross-sectional measurements difficult.
Figure 8
also shows the increase in Volumelock between the two joints, which tracks the increase in maximum load much more reliably, while also being a single metric. This suggests Volumelock could be a much
more useful tool for engineers to optimise rivet and die combinations. Although Volumelock is inherently linked to interlock because they are both measures of the same portion of the joint,
Figure 8
suggests that Volumelock is quasi-independent of the traditional interlocks.
The X and Y interlocks were plotted against the maximum load to evaluate the correlation with the regression line using the R
values. The product of the X and Y interlocks, known as the interlock area, represented by the X–Y rectangle in
Figure 1
, was also plotted to capture the interlock in a single metric. However, this also resulted in large variations around the mean regression line, which is linear in nature due to the assumption that
the pull-out strength of the joint could be calculated using the interlocks by the shear punch force,
$F m a x$
, as seen in Equation (2).
$F m a x = τ ∗ Y i n t e r l o c k ∗ π ( D r i v e t + 2 X i n t e r l o c k )$
is the shear stress and
$D r i v e t$
is the undeformed rivet shank diameter. The results of this calculation can be seen in
Appendix A
. These plots show that the R
value was improved from approximately 0.55 for the X and Y interlocks to around 0.6 for the X*Y area. However, the Volumelock achieved an R
value of 0.88, a significant improvement.
Figure 9
shows the linear regression line of the X interlock with max load, including 95% confidence bands, accounting for the standard deviation of both the measurement and the max load. This results in a
variance from the bands of ±0.38 kN if this line is used to predict the strength of joints.
The regression line intersects the Y-axis at a positive integer. This can be explained by the elastic contraction around the rivet leg, because with no interlock, the joint is still expected to have
some strength due to material springback, much like a nail in wood. However, a larger dataset must be used to investigate the intersection of the axis with lower interlocks.
The X interlock also presented other issues, mainly with top sheet pulldown. When this occurred, the top sheet material was drawn down to the tail of the rivet, artificially reducing the interlock,
as seen in
Figure 10
. Examples of the other joint cross sections can be seen in
Appendix B
. This is an interesting phenomenon, as the right hand of the joint had a Y interlock more than double that on the left-hand side, and an X interlock 45% greater than that on the left-hand side.
Despite the much larger interlock on the right-hand side, the left-hand side actually resulted in a 24.5% greater Volumelock, indicating that Volumelock is a more useful way of assessing the joint
strength by measuring a cross-section image. Another issue with traditional interlock measurement methods is the variation of measurements across repeats of the same joint. The normalised bell curves
for the variation of the measurements of the interlocks and Volumelock are shown in
Figure 11
. This demonstrates a lower variation in the measurements for Volumelock than traditional interlocks.
This then creates a need for the measurement techniques introduced by the authors above. By measuring the Arealock of the joint cross section and revolving this to give the Volumelock value in the
bottom sheet, then comparing this to the energy absorbed during cross-section testing, we can derive the relationship seen in
Figure 12
In the figure, the X-axis is volume and the Y-axis is energy absorbed. From this we can derive the gradient as the energy absorbed per unit volume, which is directly proportional to the energy
absorbed per unit mass, or specific energy absorption (SEA). This is a material and geometry constant often used to quantify joint failure performance, particularly in crash testing, resulting in a
linear regression line. For this joint stack, the gradient was 2.1 J/mm^3, or a specific energy absorption of 774.32 J/g. The volume captured by the rivet head in the top sheet remained relatively
consistent across the range of lower Volumelocks, allowing a fair comparison to be drawn between them. The regression line intersects the axis at (0,0) because the gradient is the specific energy
absorption, meaning zero mass is unable to absorb energy. Further work should be conducted to understand the relationship at higher Volumelock values when the failure mode changes, as this study only
focused on the failure mode of tail pull-out.
Although energy absorbed during failure is a critical metric, another useful metric is the maximum load the joint can withstand before failure. As seen in this research, and in research by others, in
different joint stacks resulting in different interlocks [
], a linear relationship can be drawn between the energy absorbed and the maximum load the joint is capable of carrying for a given joint stack. This means that we can apply the same relationships
found with the energy absorbed to the max load. This method can also be applied to similar processes, such as Friction Self-Piercing Riveting (FSPR), with research replicating these relationships
with differing tensile tests [
]. However, this also resulted in a lower SEA of 461 J/g and a zero Volumelock energy absorption of non-zero due to the solid-state bonding achieved in the process, indicating that further work
should be conducted to investigate this effect alongside hybrid bonding with adhesive.
As with the energy absorbed, there is variation around the regression line. This is due not only to experimental and material variations, but also to variations in the upper sheet Volumelock. As
mentioned previously, this is not large. However, a mean of 71.4 mm
and a standard deviation of 4.82 does result in noticeable variations. The data points and regression line can be seen in
Figure 13
. The data points fit well with the regression line, showing that the max load can be predicted from Volumelock with a 95% certainty that the prediction will be within ±0.26 kN, or within 5% of the
mean strength from the dataset in this study.
As this is a new technique, we suggest many avenues for exploration. Further work should be conducted to understand the relationships at the upper and lower bounds of the investigated dataset. In the
current results, only one failure mode was observed, that is, tail pull-out, which is pictured in
Figure 14
. This is likely as a result of the Volumelock captured by the rivet head in the top sheet being greater than that in the bottom sheet, which means increased resistance to failure. This opens up the
possibility for further work to predict the failure mode from the upper and lower Volumelocks, which would be a novel prediction tool. One suggestion for further use of this technique would be to
rank joints absolutely. This may be achieved by dividing the lower Volumelock by a critical Volumelock, i.e., the volume at which the failure mode switches to head pull-through and therefore reaches
a maximum strength, allowing joints to be assigned a percentage rank or score out of 10. This could result in a few cross sections being made, or even simulated, and the strongest joint stack could
be selected without a single strength test being conducted.
Another powerful use of this metric could be to predict the Volumelock and therefore the strength of the joint, as the joint is made via the force–displacement curve on the setting machine, which is
similar to previous research which used it to predict the X interlock [
]. This could then be implemented as a non-destructive monitoring method during production to ensure that every joint made meets the strength requirements.
4. Conclusions
In this study, the authors developed and tested a new cross-section measurement technique intended to aid joining engineers tasked with selecting the best-performing rivet and die combination for a
set of SPR joints.
• This study resulted in a new measurement method for cross-section analysis that is potentially capable of predicting tensile test joint strength with enough accuracy to remove the need for
conducting extensive physical tensile testing.
• The measurement technique represents a new way of optimising joint parameter choice through a single measurement, improving on current measurement and prediction techniques in terms of accuracy
and precision.
• The relationships between joint performance and Volumelock measurement were investigated and found to be a function of specific energy absorption, which in turn is a function of the material and
geometry constants of the tested samples. This opens up the possibility for future work to calculate values useful to car body designers and joining engineers without the need for extensive
physical strength testing.
• Further work should be conducted to fully understand the effect of geometry and material on the relationship between Volumelock and joint strength.
• In this initial work we have only begun to explore what might be achieved using this new approach, and we encourage other researchers to help us further develop this interesting new method for
the wider benefit of the joining community.
Author Contributions
Conceptualization, L.J. and P.B.; methodology, L.J. and P.B.; formal analysis, L.J.; investigation, L.J.; resources, P.B.; validation, L.J.; data curation, L.J.; writing—original draft preparation,
L.J.; writing—review and editing, L.J., P.B., L.S. and N.S.; visualization, L.J.; supervision, P.B., L.S. and N.S.; project administration, P.B. and L.S.; funding acquisition, P.B. and L.S. All
authors have read and agreed to the published version of the manuscript.
This research was funded by Engineering and Physical Sciences Research Council, grant number X/166471-12-2.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Data are contained within the article.
The authors would like to thank Guy Davies, Lab Technician at Atlas Copco, for carrying out some of the physical testing for this study.
Conflicts of Interest
The authors declare no conflict of interest.
Test Nominal Die Avg X Standard Avg Y Standard Avg Standard Avg Max Standard Avg Total Standard Calculated Shear
ID Rivet Rivet Length Die Depth Interlock Deviation X Interlock Deviation Y Volumelock Deviation Load Deviation Max Energy Deviation Energy Punch Force (kN)
(mm) (mm) (mm) interlock (mm) Interlock (mm^3) Volumelock (kN) Load Absorbed (J) Absorbed
1 K50A42AH00 8.5 DG10-100 1 0.719 0.0572 1.11 0.079 25.3 1.80 5.10 0.131 44.9 2.29 3.74
2 K50A42AH00 8.5 DG10-120 1.2 0.620 0.0560 1.27 0.029 26.2 2.11 5.21 0.115 54.4 2.66 4.17
3 K50A42AH00 8.5 DG10-140 1.4 0.620 0.0231 1.33 0.178 27.7 1.30 5.37 0.109 60.0 2.62 4.37
4 C50D42AH00 8.5 DG10-160 1.6 0.680 0.0862 1.47 0.211 31.6 3.45 5.77 0.241 76.5 6.07 4.92
5 K50M42AH00 8.5 DG10-180 1.8 0.780 0.0578 1.57 0.301 38.0 2.19 5.93 0.241 80.5 3.98 5.41
6 K50742AH00 8.5 DG10-200 2 0.790 0.1467 1.95 0.640 40.7 4.47 5.89 0.136 76.6 3.26 6.74
7 K50842AH00 8.5 DG10-220 2.2 0.800 0.0374 2.46 0.283 38.1 1.10 6.04 0.109 80.2 3.34 8.53
8 K50A42AH00 6.5 DG10-200 2 0.537 0.0612 1.07 0.119 11.8 1.15 3.85 0.079 16.3 0.61 3.44
9 K50A42AH00 7 DG10-200 2 0.590 0.0762 1.24 0.187 17.5 1.67 4.51 0.069 32.4 0.75 4.04
10 K50A42AH00 7.5 DG10-200 2 0.420 0.0967 1.22 0.172 15.3 3.20 4.85 0.179 40.9 3.87 3.76
11 K50A42AH00 8 DG10-200 2 0.610 0.0581 1.57 0.224 27.0 1.78 5.50 0.168 65.2 0.86 5.14
Figure 1. SPR cross section: Left side—conventional X–Y interlock measurement; Right side—Arealock measurement.
Figure 3.
Cross-tension test configuration [
Figure 8. Comparative percentage increase in cross-section measurements and strength measurements for Stacks 9 and 11 (a) and 5 and 7 (b).
Top Sheet Bottom Sheet
Alloy Thickness (mm) Alloy Thickness (mm)
Stack 1 AA5754 H111 3.0 AA5754 H111 3.0
Test ID Rivet Type Rivet Length (mm) DG Die Cavity (Diameter ) DG Die Cavity (Depth) Insertion Force (kN) Insertion Velocity (mm/s)
1 K50A42AH00 8.5 10 100 70.96 340
2 K50A42AH00 8.5 10 120 72.40 340
3 K50A42AH00 8.5 10 140 73.80 340
4 K50A42AH00 8.5 10 160 72.28 330
5 K50A42AH00 8.5 10 180 64.14 300
6 K50A42AH00 8.5 10 200 53.62 270
7 K50A42AH00 8.5 10 220 50.30 260
8 C50D42AH00 6.5 10 200 46.7 230
9 K50742AH00 7.0 10 200 50.66 250
10 K50M42AH00 7.5 10 200 52.38 260
11 K50842AH00 8.0 10 200 53.80 270
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Share and Cite
MDPI and ACS Style
Jepps, L.; Briskham, P.; Sims, N.; Susmel, L. Investigation into the Effect of Interlock Volume on SPR Strength. Materials 2023, 16, 2747. https://doi.org/10.3390/ma16072747
AMA Style
Jepps L, Briskham P, Sims N, Susmel L. Investigation into the Effect of Interlock Volume on SPR Strength. Materials. 2023; 16(7):2747. https://doi.org/10.3390/ma16072747
Chicago/Turabian Style
Jepps, Lewis, Paul Briskham, Neil Sims, and Luca Susmel. 2023. "Investigation into the Effect of Interlock Volume on SPR Strength" Materials 16, no. 7: 2747. https://doi.org/10.3390/ma16072747
Note that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details
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Sample Size in Plain English
As a dissertation consultant for over 20 years, I consistently see confusion when it comes to answering a simple question—how many participants do in need? The confusion is reasonable because most
programs do not even offer a class in sample size and leave it to the graduate student to figure it out on their own. This post will clear it up once and for all.
Two Types of Sample Sizes
There are two types of sample sizes to determine: one sample size determination is used to find the number to have enough participants to be representative of a population, and the other sample size
determination is to achieve statistical power. Let’s talk about these two types.
Discover How We Assist to Edit Your Dissertation Chapters
Aligning theoretical framework, gathering articles, synthesizing gaps, articulating a clear methodology and data plan, and writing about the theoretical and practical implications of your research
are part of our comprehensive dissertation editing services.
• Bring dissertation editing expertise to chapters 1-5 in timely manner.
• Track all changes, then work with you to bring about scholarly writing.
• Ongoing support to address committee feedback, reducing revisions.
Sample Size for a Population—what researchers and organizations need
This type of sample size determination is an effort to get a representation of the population, such as you see would see in election polling. To determine this sample size, you need to know the
population size, confidence interval and confidence level (typically 95%). This is almost never the type of sample size that dissertation students need because you don’t have unlimited time, money,
energy to get such as large sample. If you are a funded researcher or organization, and desire this type of sample size, you can view our free calculator at https://www.statisticssolutions.com/
Sample Size for Statistical Power—what dissertation students need
Statistical power (also called a power analysis and typically set at .80) is the basically the probability of finding statistical differences in your data if in fact they are there. The .80 is saying
that you have an 80% chance of finding difference in you data if differences exist. To assess this type of sample size you need to know a few things. First, you need to know what type of statistical
analysis you are going to conduct. That is, the sample size calculation for an ANOVA is different than for a correlation or factor analysis. Second, you need to know the effect size, alpha, and
desired statistical power. We decided on the conventional .80 power and alpha is usually set at .05 (you’ll recognize the p = .05 in the articles you’ve been reading for several years). Let’s talk
about effect sizes and the three sizes they come in: small, medium, and large. Effect size is this context is the ability to detect differences in the data, so, a bit counter intuitively, a large,
easily detected effect requires a small sample size to detect it, while a small, difficult to detect effect in the data requires a larger sample size.
How Do You Decide What Effect Size to Choose?
The next question you should be asking yourself is should I choose a small, medium, or large effect size? There are theoretical and practical considerations here. The theoretical answer is to look at
the research previously conducted with your types of research questions, variables, and analyses, to see what effect size was found. The problem is that if a small effect size was found (thus
requiring a large sample size) it may be impractical for you to find the 300+ participants! On the other hand, just picking a large effect size willy-nilly isn’t quite correct either. What I find is
that most dissertation committees go along with are medium effect sizes. You can try to calculate it for free at G-Power or if you want to find the appropriate sample size with a simple write up and
references, you can go here (while this one is not free—sorry—it’s cheaper than paying us or others $800 to calculate it).
Sample size note. Having said all of this, you should probably recruit as many participants as you can (hence boosting your statistical power).
If you have any sample size questions, or other questions about your methodology or results chapters, feel free to contact us. I hope this helps!
Happy Learning,
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Mathematical Constants
Return a scalar, matrix, or N-dimensional array whose elements are all equal to the IEEE symbol NaN (Not a Number).
NaN is the result of operations which do not produce a well defined numerical result. Common operations which produce a NaN are arithmetic with infinity (Inf - Inf), zero divided by zero (0/0),
and any operation involving another NaN value (5 + NaN).
Note that NaN always compares not equal to NaN (NaN != NaN). This behavior is specified by the IEEE standard for floating point arithmetic. To find NaN values, use the isnan function.
When called with no arguments, return a scalar with the value ‘NaN’.
When called with a single argument, return a square matrix with the dimension specified.
When called with more than one scalar argument the first two arguments are taken as the number of rows and columns and any further arguments specify additional matrix dimensions.
If a variable var is specified after "like", the output val will have the same data type, complexity, and sparsity as var.
The optional argument class specifies the return type and may be either "double" or "single".
See also: isnan, Inf. | {"url":"https://docs.octave.org/interpreter/Mathematical-Constants.html","timestamp":"2024-11-06T04:17:40Z","content_type":"text/html","content_length":"27696","record_id":"<urn:uuid:b579edc4-dba6-42bc-b062-22dd517715ac>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00356.warc.gz"} |
Mathematics 412
Mathematics 412, Sec. 501, Fall, 2005
• Announcements
□ Dec.12: The grades on the final exam came out much higher (median = 174) than those on the hour tests, an event unprecedented in my near 30 years in Aggieland. Of course, 60 points of extra
credit didn't hurt, but clearly quite a few people woke up and smelled the coffee. Also unusual were the very low homework scores; they are supposed to be higher than test scores. So, in
place of the usual generous curve on the final exam, I added 45 points to everybody's total score, which I think of as 5 more points on the hour test total and 40 points on the "Homework and
class participation" scores, which obviously needed to be renormed. (In previous semesters such renorming often occurred before numbers were posted.) This makes the final grade distribution
almost the same as on the final exam: 8 A, 5 B, 6 lower, 7 Q-drops. Individual grades and scores will be posted on WebCT Vista later today.
□ Dec. 8: The team solutions to the review problems are now posted.
□ Dec. 4: Tomorrow (the-Monday-that-is-a-Friday) my office hour will start at 2:30 (not 10:45). I expect to be in my office at the usual time of 2:00-3:00 on Wednesday and Thursday.
□ Nov. 21: Happy Thanksgiving! I will be out of town for a long holiday weekend, starting tomorrow afternoon, so I can't hold office hours on Wednesday. Nov. 23, or Monday, Nov. 28.
□ Nov. 12: Solutions for Test B are now available below!
□ Nov. 9: Next Math Club meeting (Monday, Nov. 14)
□ Oct. 10: Office hour this Thursday, Oct. 13, is cancelled. (I will be away from Thursday afternoon through Sunday.) The Monday office hour (Oct. 17) is rescheduled for 2:00 instead of 10:45.
□ Oct. 4: Math Career Fair (Oct. 11)
□ Sept. 23: I will be unable to meet the Monday morning office hour, Sept. 26. I will be here in the afternoon (2-3 at least).
□ Sept. 16: "Permanent" office hours: M 10:45-11:45, W 2:00-3:00, R 2:00-2:50
I will usually be in the office at the "analogous" times (F 10:45, MF 2, T 2), but with no guarantee. (At W 10:45 I have a standing appointment with a graduate student.)
• Course handout (first sheet)
• Course schedule (second sheet of handout)
• Please see my home page for up-to-date office hours.
• Instructions for using WebCT Vista to check your homework and test grades.
• Homework solutions (written mostly by David Miller, edited and published by Changchun Wang)
• Fall 2004 course page including exams with solutions.
• Fall 2000 course page including exams with solutions.
• Some old demos. (I am sorry that I have not had time to update or maintain these properly. I'll try to find time for that later.)
□ Animation of moving wave packets (d'Alembert's solution). Input only -- requires Mathematica. Do not expect Pixar quality.
1. Introducing d'Alembert's solution: MathLily or PDF.
2. Wave reflection: MathLily or PDF.
(MathLily files are both valid Mathematica files and valid TeX files.)
□ Convergence of Fourier series (partial sums). Maple input and output.
1. square wave input and output.
2. triangle wave input and output.
□ Summability of Fourier series (Cesaro means). To be compared and contrasted with the foregoing.
1. square wave means input and output.
2. triangle wave means input and output.
□ Eigenvalues of a Robin Sturm-Liouville problem. Maple input and output.
□ Bessel functions and Fourier-Bessel series. Maple input and output.
• Evans Harrell and James Herod, Linear Methods of Applied Mathematics
Rough correlation between the Harrell-Herod syllabus and ours:
□ Block 1: HH Chapters 7, 6, 8, 3, 4, 5, 1, 9
□ Block 2: [nonexistent chapter on Fourier transforms], Chapters 20, 13 (beginning only), fragments of (15, 16, 19), 2, [nonexistent chapter on Sturm-Liouville theory], 17 (fragments)
□ Block 3: Chapters 10, 11, 18, g19[=22] (beginning)
• Test solutions
• Information on how to get a minor in mathematics, or even a double major. Most science and engineering majors can qualify easily.
Go to home pages: Fulling ._._. Calclab ._._. Math Dept ._._. University
e-mail: fulling@math.tamu.edu
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eksperymenty małe i duże
Graph database
A graph database is defined as a specialized, single-purpose platform for creating and manipulating graphs. Graphs contain nodes, edges, and properties, all of which are used to represent and store
data in a way that relational databases are not equipped to do.
Graph analytics is another commonly used term, and it refers specifically to the process of analyzing data in a graph format using data points as nodes and relationships as edges. Graph analytics
requires a database that can support graph formats; this could be a dedicated graph database, or a converged database that supports multiple data models, including graph.Learn how to use graph with a
step-by-step workshop
Graph database types
There are two popular models of graph databases: property graphs and RDF graphs. The property graph focuses on analytics and querying, while the RDF graph emphasizes data integration. Both types of
graphs consist of a collection of points (vertices) and the connections between those points (edges). But there are differences as well.
Property graphs
Property graphs are used to model relationships among data, and they enable query and data analytics based on these relationships. A property graph has vertices that can contain detailed information
about a subject, and edges that denote the relationship between the vertices. The vertices and edges can have attributes, called properties, with which they are associated.
In this example, a set of colleagues and their relationships are represented as a property graph.
Because they are so versatile, property graphs are used in a broad range of industries and sectors, such as finance, manufacturing, public safety, retail, and many others.
RDF graphs
RDF graphs (RDF stands for Resource Description Framework) conform to a set of W3C (Worldwide Web Consortium) standards designed to represent statements and are best for representing complex metadata
and master data. They are often used for linked data, data integration, and knowledge graphs. They can represent complex concepts in a domain, or provide rich semantics and inferencing on data.
In the RDF model a statement is represented by three elements: two vertices connected by an edge reflecting the subject, predicate and object of a sentence—this is known as an RDF triple. Every
vertex and edge is identified by a unique URI, or Unique Resource Identifier. The RDF model provides a way to publish data in a standard format with well-defined semantics, enabling information
exchange. Government statistics agencies, pharmaceutical companies, and healthcare organizations have adopted RDF graphs widely.
How graphs and graph databases work
Graphs and graph databases provide graph models to represent relationships in data. They allow users to perform “traversal queries” based on connections and apply graph algorithms to find patterns,
paths, communities, influencers, single points of failure, and other relationships, which enable more efficient analysis at scale against massive amounts of data. The power of graphs is in analytics,
the insights they provide, and their ability to link disparate data sources.
When it comes to analyzing graphs, algorithms explore the paths and distance between the vertices, the importance of the vertices, and clustering of the vertices. For example, to determine importance
algorithms will often look at incoming edges, importance of neighboring vertices, and other indicators.
Graph algorithms—operations specifically designed to analyze relationships and behaviors among data in graphs—make it possible to understand things that are difficult to see with other methods. When
it comes to analyzing graphs, algorithms explore the paths and distance between the vertices, the importance of the vertices, and clustering of the vertices. The algorithms will often look at
incoming edges, importance of neighboring vertices, and other indicators to help determine importance. For example, graph algorithms can identify what individual or item is most connected to others
in social networks or business processes. The algorithms can identify communities, anomalies, common patterns, and paths that connect individuals or related transactions.
Because graph databases explicitly store relationships, queries and algorithms utilizing the connectivity between vertices can be run in sub-seconds rather than hours or days. Users don’t need to
execute countless joins and the data can more easily be used for analysis and machine learning to discover more about the world around us.
Advantages of graph databases
The graph format provides a more flexible platform for finding distant connections or analyzing data based on things like strength or quality of relationship. Graphs let you explore and discover
connections and patterns in social networks, IoT, big data, data warehouses, and also complex transaction data for multiple business use cases including fraud detection in banking, discovering
connections in social networks, and customer 360. Today, graph databases are increasingly being used as a part of data science as a way to make connections in relationships clearer.
Because graph databases explicitly store the relationships, queries and algorithms utilizing the connectivity between vertices can be run in subseconds rather than hours or days. Users don’t need to
execute countless joins and the data can more easily be used for analysis and machine learning to discover more about the world around us.
Graph databases are an extremely flexible, extremely powerful tool. Because of the graph format, complex relationships can be determined for deeper insights with much less effort. Graph databases
generally run queries in languages such as Property Graph Query Language (PGQL). The example below shows the same query in PGQL and SQL.
As seen in the above example, the PGQL code is simpler and much more efficient. Because graphs emphasize relationships between data, they are ideal for several different types of analyses. In
particular, graph databases excel at:
• Finding the shortest path between two nodes
• Determining the nodes that create the most activity/influence
• Analyzing connectivity to identify the weakest points of a network
• Analyzing the state of the network or community based on connection distance/density in a group
More info: What is a Graph Database? | Oracle
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mathematics in our daily life ppt
It helped me to complete my project. | Subscribe to the channel for Maths videos! Required fields are marked *. Even you can also find the importance of mathematics in our daily life. Mathematics is
the life to all practical subjects, be it concerning commerce, science, mathematics, economics and what not. Though the basics of mathematics start from school but its usage continues till we become
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sense of understanding of mathematics so that you can accomplish the different tasks successfully. It will also help them figure out the best ways to save and invest money. Nice speech…it help me a
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only. Introduction In our daily life, we use mathematics in various fields. Besides, elementary school students find it annoying, as well as stressful. It is a tool in our hands to make our life …
Your email address will not be published. For this, we have to plan things accordingly. Maths Tour is a channel for Maths students.Learn some concepts of Maths in Hindi and English. Mathematics is
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Expert) We use Math in our everyday lives. One important skill they will learn is how to calculate interest and compound interest. Mathematics is as important for us as oxygen. (adsbygoogle =
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mathematics are power of reasoning, creativity, abstract or spatial thinking, critical thinking, problem solving ability and even effective communication skills. Save my name, email, and website in
this browser for the next time I comment. Though the basics of mathematics start from school but its usage continues till we become adults and thus it can be said that maths has become an integral
part. You just cannot do without this subject and that is why it is essential to keep your basics right to perform the everyday activities of life. It’s a nice speech. They also have a concern about
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unavailable. 5 mathematical facts we use in our daily life |authorstream. Math is incredibly important in our lives and, without realizing it, we use mathematical concepts, as well as the skills we
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Need help with Formula for incremental cell reference | Microsoft Community Hub
Forum Discussion
Need help with Formula for incremental cell reference
This issue has 2 piece's: I have been trying to figure out this formula for a while now. I am attempting to get the initials from Column A (that are every 7 rows) in order in Column AR....
• This formula is in cell AR5 and filled down. The formula has to be entered with ctrl+shift+enter if you don't work with Office 365 or Excel for the web or Excel 2021.
This formula is in cell AS5 and filled across range AS5:AY7.
I assume that you don't work with Office 365 or Excel for the web. With older versions such as Excel 2013 you can apply the above formulas.
In cell AR5:
In cell AS5:
With Office 365 or Excel for the web you can enter the above formulas in cells AR5 and AS5.
• I appreciate the response, but neither of these solutions worked for me. I was hoping whatever formula I used I would be able to drag down so that it would fill the cells as the data far
surpasses the snip I took. Thanks for trying though.
□ This formula is in cell AR5 and filled down. The formula has to be entered with ctrl+shift+enter if you don't work with Office 365 or Excel for the web or Excel 2021.
This formula is in cell AS5 and filled across range AS5:AY7.
I assume that you don't work with Office 365 or Excel for the web. With older versions such as Excel 2013 you can apply the above formulas.
☆ This worked perfectly! Thank you so very much! | {"url":"https://techcommunity.microsoft.com/discussions/excelgeneral/need-help-with-formula-for-incremental-cell-reference/4027483/replies/4027503","timestamp":"2024-11-12T17:19:51Z","content_type":"text/html","content_length":"315470","record_id":"<urn:uuid:be1b394a-80fb-4b2d-ae0a-261d9e813e16>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00455.warc.gz"} |
Addition Rules And Multiplication Rules For Probability Worksheet Answers
Mathematics, especially multiplication, creates the keystone of various scholastic self-controls and real-world applications. Yet, for lots of students, understanding multiplication can pose a
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Introduction to Addition Rules And Multiplication Rules For Probability Worksheet Answers
Addition Rules And Multiplication Rules For Probability Worksheet Answers
Addition Rules And Multiplication Rules For Probability Worksheet Answers -
Applied Mathematics Contemporary Mathematics OpenStax 7 Probability
What is the Addition Rule of Probability The addition rule of probability represents two formulae one represents the formula for the non mutually exclusive events and the other for the mutually
exclusive events
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Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
1 The Addition Law As we have already noted the sample space S is the set of all possible outcomes of a given experiment Certain events A and B are subsets of S In the previous block we defined what
was meant by P A P B and their complements in the particular case in which the experiment had equally likely outcomes
Addition Rules and Multiplication Rules for Probability Worksheet I Determine whether these events are mutually exclusive Roll a die get an even number and get a number less than 3 Roll a die get a
prime number and get an odd number Roll a die get a number greater than 3 and get a number less than 3
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P bottom level and price below 200 P p r i c e b e l o w 200 bottom level P bottom level 1000 6150 0 41 0 0667 We can then calculate P bottom level or price below 200 0 41 0 4733 0 0667 0 8166 A
detailed explanation of the multiplication rule is part of the DISCOVERY course content Lecture Multi event
Google Classroom You might need Calculator 26 customers are eating dinner at a local diner Of the 26 customers 20 order coffee 8 order pie and 7 order coffee and pie Using this information answer
each of the following questions
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Addition Rules And Multiplication Rules For Probability Worksheet Answer Key Designbymian
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Probability Rules Cheat Sheet Basic probability rules With Examples By Rita Data Comet
Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Rules Of Addition and Multiplication Easy Tricks YouTube
PPT Addition Rules for Probability PowerPoint Presentation Free Download ID 6950506
addition Rule probability worksheet
Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Addition Rule of Probability Worksheets Math Worksheets Land
What is the Addition Rule of Probability The addition rule of probability represents two formulae one represents the formula for the non mutually exclusive events and the other for the mutually
exclusive events
span class result type
Math Addition Rules and Multiplication Rules for Probability Determine whether these events are mutullly exclusive 1 Roll a die t an even number and get a number less 3 2 a die get a prime number and
get an odd 3 a get a number greater than 3 4 Select a student No 5 Select a Sfident at UGA student is a a 6 Select school the the Fird the
What is the Addition Rule of Probability The addition rule of probability represents two formulae one represents the formula for the non mutually exclusive events and the other for the mutually
exclusive events
Math Addition Rules and Multiplication Rules for Probability Determine whether these events are mutullly exclusive 1 Roll a die t an even number and get a number less 3 2 a die get a prime number and
get an odd 3 a get a number greater than 3 4 Select a student No 5 Select a Sfident at UGA student is a a 6 Select school the the Fird the
PPT Addition Rules for Probability PowerPoint Presentation Free Download ID 6950506
Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
addition Rule probability worksheet
Addition Rules And Multiplication Rules For Probability Worksheet Times Tables Worksheets
Multiplication Rule Of Probability Independent Practice Worksheet Answers
Multiplication Addition Rule Probability Mutually Exclusive Independent Events YouTube
Multiplication Addition Rule Probability Mutually Exclusive Independent Events YouTube
Addition And Multiplication Rules Of Probability Worksheet Free Printable
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Exactly how can parents sustain their children's multiplication method in your home?
Motivating constant practice, giving help, and developing a positive understanding environment are valuable steps. | {"url":"https://crown-darts.com/en/addition-rules-and-multiplication-rules-for-probability-worksheet-answers.html","timestamp":"2024-11-13T22:53:44Z","content_type":"text/html","content_length":"30353","record_id":"<urn:uuid:dac5babd-a4f0-42d4-a6ad-99749935b539>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00515.warc.gz"} |
stresses and deflections
ESDU 71013
Elastic direct stresses and deflections for flat rectangular plates under uniformly distributed normal pressure.
Curves are given which enable the elastic stresses (at the centre, on the edges and on the diagonals) and maximum deflections to be determined for initially flat rectangular plates of uniform
thickness under uniformly distributed normal pressure. The curves are based on thin plate theory and apply for plate width/thickness ratios greater than or equal to 20. The data are given for plates
with length/width ratios between 1 and infinity and the edge restraints may be any combination of fixed or free in translation or rotation. The data can also be obtained using ESDUpac A9311, see
ESDU 93011
, or ESDUpac A9433, see
ESDU 94033
Indexed under:
Data Item ESDU 71013
• PDF
Format: • with interactive graphs
• Amendment (C), 01 May 1995
Status: • Published in Release 2000-03 (init)
Previous Releases: • None available
ISBN: • 978 1 86246 370 7
The Data Item document you have requested is available only to subscribers or purchasers.
• Subscribers login here.
• If you are not an ESDU subscriber you can
The graphs listed below are available only to subscribers.
This Data Item contains 21 interactive graph(s) as listed below.
Graph Title
Figure 1 Edges free in translation and rotation
Figure 2 Edges free in translation and fixed in rotation
Figure 3 Edges fixed in translation and free in rotation
Figure 4 Edges fixed in translation and rotation
Figure 5 - Part 1 Edges free in translation and rotation
Figure 5 - Part 2 Edges free in translation and rotation
Figure 5 - Part 3 Edges free in translation and rotation
Figure 6 - Part 1 Edges free in translation and rotation
Figure 6 - Part 2 Edges free in translation and rotation
Figure 6 - Part 3 Edges free in translation and rotation
Figure 7 - Part 1 Edges free in translation and fixed in rotation
Figure 7 - Part 2 Edges free in translation and fixed in rotation
Figure 8 Edges free in translation and fixed in rotation
Figure 9 - Part 1 Edges fixed in translation and free in rotation
Figure 9 - Part 2 Edges fixed in translation and free in rotation
Figure 10 - Part 1 Edges fixed in translation and rotation
Figure 10 - Part 2 Edges fixed in translation and rotation
Figure 11 - Part 1 Edges fixed in translation and rotation
Figure 11 - Part 2 Edges fixed in translation and rotation
Figure 12 - Part 1 Edges restrained in rotation as shown (Translational restraint has negligible effect in the small deflection range)
Figure 12 - Part 2 Edges restrained in rotation as shown (Translational restraint has negligible effect in the small deflection range) | {"url":"https://esdu.com/cgi-bin/ps.pl?sess=unlicensed_1241031021717lqf&t=doc&p=esdu_71013c","timestamp":"2024-11-13T14:31:59Z","content_type":"text/html","content_length":"41153","record_id":"<urn:uuid:cd36cf34-f1f5-4c02-9365-4b65317c64cb>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00701.warc.gz"} |
Electronics ETA Certification Practice Exam Questions with 100% Correct Answers | Verified | Exams Advanced Education | Docsity
Download Electronics ETA Certification Practice Exam Questions with 100% Correct Answers | Verified and more Exams Advanced Education in PDF only on Docsity! Electronics ETA Certification Practice
Exam Questions with 100% Correct Answers | Verified | Updated 2024 The "triboelectric effect" causes what kind of electricity? - Correct Answer-Static The law of static electrical charges states: -
Correct Answer- like charges repel Minute particles that make up all possible elements and still retain the unique chemical characteristics of the element are called _____________. - Correct
Answer-Atoms A molecule is the smallest particle of matter that can exist by itself. (1) - Correct Answer-True (1) A basic atom must contain what two types of subatomic particles? - Correct
Answer-Electron and proton An electron has _______________. - Correct Answer-a Negative charge pg. 1 professoraxe l What two types of energy are used throughout electronics? - Correct Answer-Kinetic
and potential Electrical direct current flowing in a wire produces a magnetic field: - Correct Answer-at right angles to current flow A material is magnetized when the: - Correct Answer-domains are
in alignment A material that can be magnetized easily falls into this category: - Correct Answer-Ferromagnetic Invisible lines of force that surround a magnet are called ___________________. -
Correct Answer-magnetic flux lines The pole of a magnet from which flux leaves is called the __________________. - Correct Answer-north _________________ force produces magnetic flux. - Correct
Answer-Magnetomotive pg. 2 professoraxe l The Wheatstone bridge can be used for precision measurements of: - Correct Answer-resistance The ammeter setting on a multimeter indicates what type of
reading? - Correct Answer-current A resistor with color bands of blue, gray and black has a value of _______________. - Correct Answer-68 ohms 20% A/an _____________ allows current to flow easily. -
Correct Answer-conductor A/an _________________ stops the flow of current. - Correct Answer-insulator A component whose value is measured in microfarads is a: - Correct Answer-capacitor A component
whose value is measured in ohms is a: - Correct Answer-resistor pg. 5 professoraxe l Switches are rated in maximum ___________ and ______________. (1) - Correct Answer-current, voltage (1) Fuses are
rated in maximum _____________ and ______________. - Correct Answer-current, voltage In an electrical circuit where the voltage and resistance are known, which form of Ohm's law is used to find the
circuit current? - Correct Answer-I = E/R A unit of energy is the: - Correct Answer-Joule Which of the following is commonly used to measure electrical energy consumption? - Correct Answer-Kilowatt-
Hours The current flowing in a circuit is directly proportional to the applied voltage and inversely proportional to the circuit resistance. This is a statement of: - Correct Answer-Ohm's Law Ohm's
Law states that voltage is the product of resistance and current. Using algebra, how would you determine the resistance? - Correct Answer-divide the voltage by the current pg. 6 professoraxe l Watt's
law states that power is the product of current and voltage. Using algebra, how could the power be determined if resistance and current are known? - Correct Answer-current squared times the
resistance What does the M+ key on a calculator do? - Correct Answer- sums the quantity on screen to memory The basic unit of length in the metric system is the ___________. - Correct Answer-meter A
120,000 Ohm resistor can be entered into a calculator using scientific notation as ____________. - Correct Answer-1.2 x 10^5 Which symbol is used for Ohms of resistance? - Correct Answer-Omega Which
symbol means "a change of"? - Correct Answer- Triangle pg. 7 professoraxe l If a series circuit has two 13 kΩ resistors with a source voltage of 10 V, the wattage is _________. - Correct Answer-3.85
mW Kirchoff's voltage law (KVL) states that the total voltage around a closed loop must be zero. (3) - Correct Answer-True (3) When resistors are added in parallel to a parallel circuit, total
current ________________. - Correct Answer-Increases In a parallel circuit, total resistance _______________. - Correct Answer-is less than the resistance of any branch In a simple parallel circuit,
the voltage across each branch is ________________. - Correct Answer-equal to the source voltage With three 4.5 kΩ resistors and one 1.5 kΩ resistor in parallel, the total resistance is __________. -
Correct Answer-750 ohms pg. 10 professoraxe l Kirchhoff's first law is also know as the principle of conservation of electric charge. It states that at any point, the sum of currents flowing towards
that point is equal to the sum of currents flowing away from that point. - Correct Answer- True (4) The total current in a series-parallel circuit equals the _____________. - Correct Answer-applied
voltage divided by the total resistance When two 10 kΩ parallel resistors are connected in series with a third 10 kΩ resistor, total resistance (RT) is ____________. - Correct Answer-15 kiloohms With
70 volts applied to two 30 kΩ resistors in series connected to three30 kΩ resistors in parallel forming a series-parallel circuit, what is the total current? - Correct Answer-1 mA If a short occurs
in the parallel portion of a series-parallel circuit, how will the total resistance be affected? - Correct Answer-It will decrease pg. 11 professoraxe l Thevenin's Theorem states that it is possible
to simplify any circuit containing any amount of voltage sources, current sources or resistances, no matter how complex, to an equivalent circuit with just a single voltage source and series
resistance connected to a load. (5) - Correct Answer-True (5) Battery cells are connected in parallel to ________________. - Correct Answer-increase current capacity Battery cells are connected in
series to _____________________. - Correct Answer-increase voltage output The output of a lead-acid cell is approximately: - Correct Answer-2.1 V Given a battery rated at 200 ampere-hours, how many
hours will the battery be able to provide 10 amperes? - Correct Answer-20 pg. 12 professoraxe l | {"url":"https://www.docsity.com/en/docs/electronics-eta-certification-practice-exam-questions-with-100percent-correct-answers-or-verified/11531969/","timestamp":"2024-11-05T01:09:18Z","content_type":"text/html","content_length":"244035","record_id":"<urn:uuid:9ddb2e78-e98e-4987-a29e-3f5b3e0034ce>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00678.warc.gz"} |
Distributed Computing and Societal Networking
I am the rock, you the stone,
Together we ballast reality.
Steady toward uncertain destiny,
Whither the sheets above us are blown.
Until fairly recently, 1931 to be exact, scientists and mathematicians believed we would someday find a theory that explains everything. By "theory" I mean a Formal Axiomatic System (FAS). A FAS is
simply a relatively small set of given facts, axioms, from which a much larger set of
can be proven "by mechanical means." The axioms in a FAS are so basic that we believe them to be true without the need for further justification. A famous and straightforward example of a FAS is the
small set of axioms provided by Giuseppe Peano at the end of the 19th century to formally define the natural numbers. The
Peano axioms
, in English, are as follows:
1. The natural numbers contain the number 1.
2. Every natural number has a successor, the next natural number.
3. The number 1 is not the successor of any natural number.
4. Two different natural numbers cannot have the same successor.
5. If a set contains the successors of each of its members and the number 1, then it contains all the natural numbers.
From this small set of seemingly obvious "facts" and using the mechanical rules of logic, all of number theory can be derived. Addition, multiplication, prime numbers, prime decomposition and the
Fundamental Theorem of Arithmetic, are all consequences of the five Peano Axioms listed above. Or, to look at it the other way round, the Peano Axioms constitute a very efficient compression of the
entire body of number theory.
And that is how mathematics works. Every branch of it — geometry, algebra, analysis, etc. — can be boiled down to a similar compression of a very large body of theorems into a relatively small set of
axioms. Unlike other sciences such as physics, biology, chemistry, or computation, mathematics is unconstrained by the laws of the universe in which we find ourselves living. It is purely an
invention of man, limited only by our evolving intellect. We invented the game and control the rules. Surely, there are no fundamental reasons why we couldn't develop a mathematical Theory of
Everything (ToE), a FAS from whose axioms one can derive by mechanical means all that is true.
Albert Einstein and Kurt Gödel
That is what pretty much every mathematician believed until 1931 when a 25 year-old Austrian post-doc named Kurt Gödel shocked everyone by proving that mathematics, that wholly man-made "queen of
sciences," suffers from the same kind of fundamental limitation
Heisenberg had discovered in physics
five years earlier and
Turing would find for computation
five years later. Gödel's meta-theorem proved there is a shoreline beyond which vast shoals of truth exist that the lighthouse of pure mathematics cannot illuminate.
More specifically (and much less metaphorically), Gödel showed that for any FAS that is at least as rich as the Peano Axioms described above, and for which the axioms are not self-contradictory,
there will be theorems expressible in the language of the FAS that are true, but un-provable from the axioms. That is, any consistent FAS is necessarily incomplete. Pure mathematics is not sufficient
to account for a ToE. Truth, with a capital T, is incompressible. There will always be a theorem, perhaps infinitely many, whose shortest expression is the theorem itself. The only way to prove such
theorems is to add them to the set of axioms!
Like other impossibility meta-theorems, e.g., Heisenberg's Uncertainty Principle in physics and Turing's Undecidability result in computing, Gödel's Incompleteness Theorem identifies a fundamental
limit on our ability to know things, in this case using the tools of mathematics. As we have seen for physics and computing, beyond these boundaries lies randomness, and the same appears to be true
for math as well. Our simple working definition of randomness, that it be universally unpredictable, can be made more precise in the realm of mathematics. Namely, something is mathematically random
if there exists no FAS (with a finite set of axioms) from which it can be derived. A random number, in particular, is a number whose digits cannot be predicted within the framework of mathematics.
Such a number is irreducible — the shortest possible way to express one is to write down all its (infinitely many) digits. If we can agree that a name for something is just a finite sequence of
symbols that represent it, these numbers are necessarily nameless. Even numbers like π and e, which are transcendental, their digits never repeat, have names — π and e are the names we have given to
two specific infinite real numbers that can be computed in two very specific, and finite, ways. Mathematically random (aka irreducible) numbers are even weirder than transcendentals. So weird that it
becomes reasonable to ask, do such numbers actually exist, and if so, can you show me one?
It's actually pretty easy to see that irreducible numbers, numbers that cannot be derived within any FAS, do indeed exist. However complex pure mathematics may become, it must eventually have a
finite set of axioms from which all theorems, including those showing the existence of certain numbers, can be mechanically proved. A finite set of axioms can be used to derive an infinite set of
theorems, just like a finite alphabet can generate an infinite set of names. But in both cases, the infinity of derivations is a countable infinity — there are only as many of them as there are
integers. On the other hand, there is an uncountably infinite number of real numbers, which is a bigger infinity. So if you match up numbers to name/theorems, you will always have more of the former
than the latter. In other words, there must be an (uncountably) infinite number of irreducible numbers. This is a so-called diagonalization argument that can be made formally rigorous, but is never
very satisfying. What we'd really like to do is hold one of these odd beasts in our hands and examine it up close.
One obvious way to generate such a number is to appeal back to physics. We could simply flip a fair coin and write 0 for heads or 1 for tails, and continue doing that to generate as many bits as
desired. Assuming we trust the fairness of our fair coin, the bits (digits) of such a number would be universally unpredictable — every bit is a completely separate "fact" and there is no shorter
description of how to obtain them other than the number itself. But is there a way to do this while staying within the confines of pure math?
Around the time The Beatles brought bangs to America, in 1966,
Dr. Gregory Chaitin
found an answer to this question when he discovered what he called the
Omega Numbers
. Each
(there are infinitely many) is
but completely
. Like
is a transcendental number, but unlike it's more well-behaved cousins,
cannot be computed by any algorithm. The smallest possible expression of its digits are the digits themselves. Furthermore, those digits are normally distributed — every digit appears approximately
the same number of times — no matter what base (decimal, octal, binary, hexadecimal, etc.) one might choose to represent it.
is mathematically random. To me, the most fascinating thing about
is that it sits at the nexus of Gödel's incompleteness and Turing's undecidability, telling us that the oceans of randomness beyond mathematics and computing are, in fact, one and the same — the
Divine Random.
Chaitin defines Ω using the terminology and formalism of Turing (which, if you'll recall, is still within the realm of pure mathematics) by first choosing a suitable formal programming language and
initializing an estimate of Ω to zero. For Chaitin's purposes, this means the language must be prefix-free so that programs are self-delimited. We then enumerate every possible string in the
language. For each string we ask, a) is the string a syntactically correct program with input data appended, and b) does that program halt when supplied with the input data? If the answers to both a)
and b) are Yes, we add the following quantity to our estimate: 1/2^|p|, where |p| is the number of bits required to represent the program in the chosen language. If not, we just continue combing
through all possible strings looking for legal, halting programs.
Does such a procedure produce a well-defined number? Absolutely! It is surely possible to lexically enumerate all possible strings, every string is either a legal program+data or not, and every legal
program+data either halts or it doesn't. In fact, Chaitin showed that the resulting number, Ω, is a real number between 0 and 1. The catch, of course, is that it is impossible to compute such a
number because, as Turing proved, it isn't possible to compute the answer to the halting problem in full generality. In a certain sense, each bit of Ω is chosen by flipping a mathematical coin —
heads the program halts, tails it doesn't — whose outcome is universally unpredictable and uncomputable.
So what does it look like, you must be asking? Well, I can't say. As with other objects we have encountered floating in the sea of randomness, Ω can be seen only by its faint, hazy outline. We can
begin to estimate Ω but we can never know more than a finite number of its digits. Moreover, it is impossible to compute how accurate our estimate is, how close or far away it is to the actual
number. Ω will always remain fuzzy, its properties uncertain beyond some degree of magnification, much like the properties of physical objects governed by Heisenberg's Uncertainty Principle. Chaitin
put it best in his excellent and accessible little monograph Meta Math! The Quest for Omega:
So fixing randomness in your mind is like trying to stare at something without blinking and without moving your eyes. If you do that, the scene starts to disappear in pieces from your visual
The harder you stare at randomness, the less you see of it!
Even if we can't know exactly what it looks like, the very existence of
proves, or at least strongly suggests, Gödel's incompleteness meta-theorem. There are indeed theorems, and numbers, that cannot be compressed, that cannot be derived by any effective procedure from a
smaller set of axioms. And we know that mathematics is limited in this way because Turing showed us there are limits to what any effective procedure, aka computer program, aka Finite Axiomatic
System, can accomplish. No matter how clever we humans are or will become, there are barriers to our knowledge that cannot be scaled. Tantalizingly, the answers are out there, floating around as
random numbers.
, for example, solves the halting problem for every possible program. And it exists; it's out there, we just aren't allowed to see it. Similarly, there exists a random number whose binary bits answer
every conceivable yes/no question that can be posed in a given language, including questions like, "Is
Goldbach's conjecture
true?" Science can't give us all those answers. Physics, mathematics, and computation theory are well-honed tools, but they cannot carve all the universe.
We are now approaching the point at which you might begin to understand why I call it, The Divine Random. | {"url":"http://www.consensualwisdom.com/2014/05/","timestamp":"2024-11-12T23:29:23Z","content_type":"text/html","content_length":"64980","record_id":"<urn:uuid:b8f12add-5e30-433d-a4a7-28686a0b1bf6>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00192.warc.gz"} |
Understanding Mathematical Functions: How To Make Function
Introduction to Mathematical Functions
Mathematical functions are essential components of the field of mathematics. They play a crucial role in modeling, analyzing, and predicting various phenomena in different disciplines. Understanding
functions is fundamental in solving mathematical problems and interpreting real-world scenarios.
A Definition of a mathematical function
A mathematical function is a relation between a set of inputs (independent variables) and a set of outputs (dependent variables) where each input corresponds to exactly one output. In simpler terms,
a function assigns each input value to a unique output value.
Importance of understanding functions in various fields
Understanding mathematical functions is essential in various fields such as physics, engineering, economics, and computer science. Functions help in describing relationships between different
variables and making predictions based on data analysis. In physics, for example, functions are used to model the motion of objects or the flow of fluids.
Overview of types of functions (linear, quadratic, polynomial, exponential)
There are several types of mathematical functions, each with its unique characteristics and applications.
• Linear Functions: A linear function is a function that graphs as a straight line. It has the general form f(x) = mx + b, where m is the slope of the line and b is the y-intercept.
• Quadratic Functions: A quadratic function is a function of the form f(x) = ax^2 + bx + c, where a, b, and c are constants and a is not equal to zero. Quadratic functions graph as parabolas.
• Polynomial Functions: Polynomial functions are functions of the form f(x) = a0 + a1x + a2x^2 + ... + anxn, where a0, a1, a2, ..., an are coefficients. Polynomial functions can have various
degrees, determined by the highest power of x.
• Exponential Functions: Exponential functions are functions of the form f(x) = a^x, where a is a positive constant. Exponential functions grow or decay at a constant rate.
Key Takeaways
• Define the purpose of the function.
• Choose the input and output variables.
• Write the function using mathematical notation.
• Test the function with different inputs.
• Understand the relationship between inputs and outputs.
Basic Components of Functions
Functions are essential mathematical tools that help us understand relationships between variables. To create a function, we need to understand the basic components that make up a function.
The concept of variables and constants
Variables in a function are symbols that represent unknown values or quantities that can change. They are typically denoted by letters such as x, y, or z. On the other hand, constants are fixed
values that do not change, such as numbers like 2, 5, or π.
When creating a function, we use variables to represent the input values that will produce an output. Constants, on the other hand, are used to represent fixed values within the function.
Understanding the domain and range
The domain of a function refers to the set of all possible input values that the function can accept. It is essential to determine the domain to ensure that the function is well-defined and can
produce meaningful outputs for all valid inputs.
On the other hand, the range of a function refers to the set of all possible output values that the function can produce. Understanding the range helps us determine the possible outcomes of the
function based on the input values.
Function notation and its interpretation
Function notation is a way to represent a function using symbols and mathematical expressions. It typically involves using the function name followed by parentheses containing the input variable. For
example, f(x) represents a function named f with an input variable x.
Interpreting function notation involves understanding how the input values are transformed to produce the corresponding output values. By substituting different values for the input variable, we can
evaluate the function and determine its behavior.
How to Construct Basic Functions
Understanding mathematical functions is essential in various fields such as engineering, physics, and computer science. Functions help us model relationships between variables and make predictions
based on data. Here is a step-by-step guide to constructing basic functions:
A Step-by-step guide to constructing a linear function
• Step 1: Identify the slope (m) and y-intercept (b) of the linear function in the form y = mx + b.
• Step 2: Plot the y-intercept on the y-axis.
• Step 3: Use the slope to find another point on the line.
• Step 4: Connect the two points to draw the linear function.
Examples of creating quadratic and polynomial functions
• Quadratic Function: y = ax^2 + bx + c
• Polynomial Function: y = a_nx^n + a_(n-1)x^(n-1) + ... + a_1x + a_0
• Example: For a quadratic function y = 2x^2 + 3x - 1, the coefficients are a = 2, b = 3, and c = -1.
• Example: For a cubic function y = x^3 - 2x^2 + 5x + 4, the coefficients are a_3 = 1, a_2 = -2, a_1 = 5, and a_0 = 4.
Tips for identifying the correct type of function for a given problem
• Consider the Data: Analyze the given data points to determine the relationship between variables.
• Look for Patterns: Identify any patterns or trends in the data that can help you choose the appropriate function.
• Start Simple: Begin with a linear function and then move on to quadratic or polynomial functions if needed.
• Consult Resources: Use textbooks, online resources, or consult with experts to determine the best type of function for the problem.
Advanced Function Construction Techniques
When it comes to constructing mathematical functions, there are several advanced techniques that can be utilized to create complex and versatile functions. In this chapter, we will explore three key
techniques: incorporating conditionals in piecewise functions, utilizing transformation techniques, and constructing functions with complex numbers.
Incorporating conditionals in piecewise functions
Piecewise functions are functions that are defined by different rules for different intervals or sets of inputs. This allows for greater flexibility in defining functions that may have different
behaviors in different regions. When incorporating conditionals in piecewise functions, it is important to clearly define the conditions under which each rule applies.
• Define the different rules for each interval or set of inputs.
• Use if-else statements to specify the conditions under which each rule applies.
• Ensure that the function is continuous at the points where the rules transition.
Utilizing transformation techniques (shift, stretch, reflection)
Transformation techniques allow for the manipulation of functions to create new functions with different characteristics. Common transformations include shifting the function horizontally or
vertically, stretching or compressing the function, and reflecting the function across an axis.
• Horizontal shift: Adding or subtracting a constant to the input variable.
• Vertical shift: Adding or subtracting a constant to the output variable.
• Stretch: Multiplying the function by a constant.
• Reflection: Reversing the sign of the function.
Constructing functions with complex numbers
Complex numbers are numbers that consist of a real part and an imaginary part. When constructing functions with complex numbers, it is important to understand how to work with these numbers in
mathematical operations.
• Use i to represent the imaginary unit, where i^2 = -1.
• Perform arithmetic operations with complex numbers, including addition, subtraction, multiplication, and division.
• Understand the geometric interpretation of complex numbers on the complex plane.
Real-world Applications of Mathematical Functions
Mathematical functions play a crucial role in various real-world applications, providing a framework for modeling and analyzing complex systems. Let's explore some of the key applications of
functions in different fields:
A Functions in financial modeling (eg, interest calculations)
Financial modeling heavily relies on mathematical functions to make predictions and analyze data. One common application of functions in finance is in interest calculations. For example, the compound
interest formula uses a function to calculate the future value of an investment based on the initial principal, interest rate, and time period. By using functions, financial analysts can make
informed decisions about investments, loans, and other financial transactions.
B Utilization in engineering (eg, stress-strain relationships)
Engineering is another field where mathematical functions are essential for modeling and analyzing physical systems. One example is the stress-strain relationship, which describes how materials
deform under applied forces. Engineers use functions to represent this relationship and predict the behavior of materials under different conditions. By understanding these functions, engineers can
design structures, machines, and systems that meet specific performance requirements.
C Applications in data science (eg, regression functions)
Data science relies heavily on mathematical functions to analyze and interpret large datasets. Regression functions, for example, are used to model the relationship between variables and make
predictions based on data. By fitting a regression function to a dataset, data scientists can identify patterns, trends, and correlations that can be used to make informed decisions. Functions are
also used in machine learning algorithms to train models and make predictions based on new data.
Troubleshooting Common Issues
When working with mathematical functions, it is common to encounter various issues that can affect the accuracy and reliability of your functions. Understanding how to troubleshoot these common
issues is essential for ensuring the effectiveness of your functions.
Handling undefined function errors
One of the most common issues when working with mathematical functions is encountering undefined function errors. These errors occur when trying to evaluate a function at a point where it is not
defined, such as dividing by zero or taking the square root of a negative number.
To handle undefined function errors, it is important to carefully review the domain of the function and identify any points where the function is not defined. One way to address this issue is to
restrict the domain of the function to exclude these problematic points. By clearly defining the domain of the function, you can avoid undefined function errors and ensure that your function is
Resolving domain and range mismatches
Another common issue that can arise when working with mathematical functions is domain and range mismatches. This occurs when the domain of the function does not align with the range of possible
input values, leading to inaccuracies in function evaluation.
To resolve domain and range mismatches, it is important to carefully define the domain and range of the function and ensure that they are compatible with each other. By clearly specifying the domain
and range of the function, you can avoid mismatches and ensure that your function behaves as expected.
Addressing inaccuracies in function construction
Lastly, inaccuracies in function construction can also be a common issue when working with mathematical functions. These inaccuracies can arise from errors in defining the function, choosing the
wrong mathematical operations, or using incorrect constants or coefficients.
To address inaccuracies in function construction, it is important to carefully review the function definition and verify that it accurately represents the desired mathematical relationship. One
approach to addressing this issue is to double-check the function definition and compare it to the intended mathematical relationship to ensure accuracy.
Conclusion & Best Practices
A Recap of the significance and variety of mathematical functions
Understanding the significance of mathematical functions
Mathematical functions play a crucial role in various fields such as physics, engineering, economics, and more. They help us model real-world phenomena, make predictions, and solve complex problems.
The variety of mathematical functions
There is a wide range of mathematical functions, including linear functions, quadratic functions, exponential functions, trigonometric functions, and more. Each type of function has its unique
properties and applications.
Best practices in constructing and applying functions accurately
Define the function clearly
When constructing a mathematical function, it is essential to clearly define the input and output variables, as well as the relationship between them. This will help avoid confusion and errors in
Choose the appropriate function type
It is crucial to select the right type of function for the problem at hand. Consider the characteristics of different functions and choose the one that best fits the data or situation you are dealing
Check for accuracy and consistency
Before applying a function to solve a problem or make predictions, double-check your calculations and ensure that the function is accurate and consistent with the given data. This will help prevent
errors and inaccuracies in your results.
Encouragement to continue exploring advanced function topics and applications
Explore advanced function topics
As you continue to study mathematical functions, consider exploring more advanced topics such as multivariable functions, differential equations, Fourier series, and more. These topics can open up
new possibilities and applications in various fields.
Apply functions to real-world problems
Challenge yourself to apply mathematical functions to real-world problems and scenarios. This will help you develop a deeper understanding of how functions work and how they can be used to solve
practical problems in different domains. | {"url":"https://dashboardsexcel.com/blogs/blog/understanding-mathematical-functions-how-to-make-function","timestamp":"2024-11-15T00:43:52Z","content_type":"text/html","content_length":"222320","record_id":"<urn:uuid:a0244904-2e4a-4553-a9dc-8637277bf304>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00733.warc.gz"} |
The Algorithms and Design.
The Algorithms and Design.¶
The doc provides more details on the algorithms and design of Masterful.
Where the API Fits¶
Masterful provides a new API, built on top of Keras (PyTorch coming soon) to focus on an ML developer’s twin goals of training: maximum speed and maximum accuracy. This solves a common source of
confusion working with deep learning frameworks: they are primarily designed to make it easy to build complex architectures. This was appropriate when advancements in architectures drove most of the
state of the art improvements, but today data and training are far more relevant. For example, consider regularization. Using Keras directly, regularization might occur at the tf.data.Dataset object
via map calls to image transforms; within the tf.keras.Model via dropout layers or kernel regularizers; or at the optimizer level via tfa.SGDW’s decoupled weight decay. By contrast, in the Masterful
API, regularization is treated as a logical grouping.
Abstraction Layer Description
Masterful API that metalearns training and regularization policies (and some drop-in architectural choices). Built on Keras.
Keras / Pytorch Lightning API that simplifies model architecture via deep neural network primitives like convolutions, rather than Tensorflow’s scientific computing primitives.
Tensorflow / PyTorch API that simplifies the creation and compilation of vectorized scientific computing.
CUDA API that allows low level access to GPUs for scientific computing rather than computer graphics.
GPU Underlying hardware to perform vectorized matrix math, useful for both computer graphics and neural networks.
Model Architecture¶
Architecture is the structure of weights, biases, and activations that define a model. For example, a perceptron or logistic regression model’s architecture is a single multiplier weight per input,
followed by a sum, followed by the addition of one bias weight, followed by a sigmoid activation. In computer vision, most practitioners are familiar with the basic ideas behind AlexNet, VGG,
Inception, Resnet, EfficientNets. MobileNets, and Vision Transformers, as well as different heads for detection and classification like YOLO, SSD, U-Net, and Mask R-CNN. Architectures have arguably
been the main source of excitement in deep learning.
Masterful treats model architecture as an input to the platform. However, Masterful controls some model of a model’s training-specific layers and attributes: dropout, residual layers for stochastic
depth, kernel regularization, and momentum of batch norms.
Masterful includes Knowledge Distillation. Training a model is generally inferior to training a larger model and then distilling its knowledge into a smaller model’s architecture. Surprisingly, the
smaller distilled model retains most of the improvements of the larger model.
Semi-Supervised Learning (SSL)¶
SSL specifically means learning from both unlabeled and labeled data. SSL is the ability for a CV model to extract information from both labeled and unlabeled images.
Masterful trains your model using unlabeled data through SSL techniques. Masterful’s approach draws from two of the three major lineages of SSL: feature reconstruction and contrastive learning of
representations. (Masterful does not currently include techniques from the third major lineage, generative techniques aka image reconstruction). Three state-of-the-art papers broadly define the
techniques included in Masterful: Noisy Student Training, SimCLR, and Barlow Twins.
The central challenge of productizing SSL into Masterful’s platform is that research is narrow and fragile. Defining a narrow problem like “classification on Imagenet on Resnet50” allows many
parameters to get baked in. Masterful generalizes the basic concepts from SOTA research to additional tasks like detection and segmentation, arbitrary data domains like overhead geospatial, and
additional model types.
Regularization means helping a model generalize to data it has not yet seen. Put another way, regularization is about fighting overfitting.
As a thought experiment, it is actually quite easy to achieve 100% accuracy (or mAP or other goodness of fit measure) on training data: just memorize a lookup table. But that would be an extreme
example of overfitting: such a model would have absolutely zero ability to generalize to data that the model has not seen.
Within the regularization bucket, Masterful includes several categories of regularization. The central challenge is not the regularization technique itself, but rather, how to learn the optimal
hyperparameters or policy of each technique, given that regularization techniques do not operate independently.
“Architectural” Regularization¶
Several regularization techniques are implemented in code that touches architecture. For that reason, the are mistaken as architecture. The key question to distinguish between architecture and
regularization is to ask, “is it used at inference?” If the answer is yes, it is architecture. Otherwise, it is for regularization.
An early and common technique for regularization is dropout. The original authors hypothesized that dropout allows approximation of a Bayesian approximation. In practice, the optimal intensity of
dropout is not theoretically grounded or predictable, however, empirical studies indicate that the intensity of dropout is dependent on attributes of the dataset. (forthcoming: Masterful determines
the intensity of dropout according to an adaptive algorithm during training).
Weight Decay¶
Decaying a model’s kernel weights is brings a prior into a model’s weights. It is used in every state of the art architecture, but it is typically implemented incompatibly with modern optimizers,
leading to the poor performance of Adam in practice. Masterful includes a corrrect implementation and also (forthcoming: meta-learns the optimal amount of weight decay).
A simple way to improve the accuracy of a model is to simply train N of them and take the average prediction. This is called ensembling and touches on ideas of Bayesian optimization. Put another way,
divide a model into N subsets and progressively train each subset separately. From this perspective, the approach resembles Dropout. Ensembling is generally superior to simply first constructing a
complex model by repeating a base architecture N times and training it once. Masterful includes an API to easily conduct ensembling.
Stochastic Depth¶
Residual networks include skip connections, which create two possible paths for a model. Stochastic Depth forces the model to switch between these two approaches. It is a form of dropout and like
dropout, can be interpreted as a form of Bayesian approximation.
Data Augmentation¶
Transforming a dataset’s images is a form of regularization. Data augmentations encompass three major forms of adjusting pixel values: either moving pixels to geometric rules like zooming or
srotating (spatial augmentations), changing the value of a pixel slightly like brightness or saturation (color jitter), or various forms of blurring. But relying on transforms solely in pixel space,
data augmentation can only indirectly affect the intermediate feature maps of a model. Techniques include color, brightness, hue, contrast, solarize, posterize, equalization, contrast, blur, mirror,
translation, rotation, and shear. Masterful’s transformations are correctly adapted to also operate on detection and segmentation.
Label Regularization¶
A new generation of techniques regularizes models by treating the labels as a probabilty distribution (aka soft label) instead of a one-hot label. The resulting data is clearly out of distribution,
and yet these techniques sometimes help regularize a model. These includes cutmix, mixup, as well as label-smoothing regularization (LSR). LSR is notable as a regularization technique which can be
viewed from three perpsectives: a label regularization, a form of Bayesian Maximum A Posterior (map) estimation, and a form of ensembling with a perfectly calibrated prior distribution.
Optimization means finding the best weights for a model and training data. Optimization is different from regularization because optimization does not consider generalization to unseen data. The goal
of optimization is speed - find the best weights faster.
Plain old stochastc gradient descent with a very low learning rate is sufficient to find the best weights. But pure SGD is far slower than modern optimizers. So basically every innovation in
optimizers, including momentum, RMSProp, Adam, LARS, and LAMB, are essentially about getting the best weights faster by calculate weight updates with not only the current gradient, but also
statistical information about past gradients.
Optimization on production datasets is very different from optimization of huge, high entropy datasets like Imagenet.
Masterful applies multiple techniques to minimize wall-clock time and GPU hours, including pushing nearly every data augmentation technique to the GPU using Masterful’s purpose built transformations;
Masterful’s custom training loop and scaffold model concept, and metalearning of optimal batch size, learning rate schedule, and epochs for your hardware.
Taking Advantage of GPU Hardware¶
Tensorflow 2 and PyTorch Lightning offer simple to use APIs to logically group a set of GPUs into a single logical GPU using mirror strategies. However, taking advantage of a large logical GPU is
non-trivial. Size only allows larger batch sizes, and there are information theoretic limits to the usefulness of larger batch sizes. Even when batch sizes are scalable, a high learning rate is
required to take advantage of large batch sizes. Masterful brings an information theoretic approach to meta-learning these parameters, as well as learning rate schedule to ensure expensive GPU
hardware is fully utilized.
Breaking the CPU Bottleneck¶
Augmentation implementations are typically based on Keras Image Preprocessing, cv2, or PILLOW, meaning the operations run on CPU, creating a bottleneck.
To greatly improve the speed of the conventional augmentation pipeline during training, Masterful pushes augmentation operations to the GPU. Internally, this requires a ground-up implementation of
every image augmentation in pure Tensorflow. Pillow and cv2 are not used. Many core design problems of those libraries are also resolved, such as eliminating non-convex combinations of magnitudes.
Then, a unique scaffold model approach is applied, whereby the base model is wrapped by custom Keras layers and trained with a custom training loop.
Meta-learning Optimal Hyperparameters¶
The hyperparameters controlling each regularization, semi-supervised learning, and optimiization method is usually set using a heuristic, such as mirroring data 50% of the time. While such a
heuristic is appropriate for ImageNet, mirroring street signs could literally be fatal. An alternative is meta-learning, but this space is generally ineffective.
For example, one state-of-the-art metalearning approach, specifically for data augmentation is AutoAugment. If a model takes 1 hour to converge, AutoAugment would require 625 days, rendering it
unusable in practice.
Black box optimization - such as Bayesian optimization, Reinforcement Learning, or randomly sampled grid search - treats all hyperparameters as fungible dimensions in a search space. The optimal way
to search that search space as simply training a model to completion to find signal on the value of that hyperparameter. While effective, the black box approaches generally requires many orders of
magnitudes of full training runs, whereas Masterful’s metalearning requires less than an order of magnitude multiple of a full training run.
Masterful’s techniques are built to work with Masterful’s purpose built meta-learner, relieving the developer of manual guessing and checking of hyperparameters. Masterful’s meta-learner applies
individual algorithms that are aware of the technique, data, and model. The result is more accurate and faster meta-learning.
Masterful’s novel contribution to meta-learning is a meta-learner that is more accurate, more robust to different domains of data, and performant. Under the hood, each hyperparameter is grouped into
a logical group and different metalearners explore each group using an appropriate algorithm.
Masterful’s meta-learner for regularization draws on the concepts from AutoAugment, Frechet Inception Distance, and adversarial learning. The result is a two-pass metalearning algorithm that can
analyze two orders of magnitude of search space in very little wall clock time becuase the first pass analysis only requires inference to cluster transformations. The second pass search, which
requires full training runs, is then reduced to analyzing a single digit number of clusters through a beam-search based algorithm. The final result is a metalearning algorithm that would run in 2
hours, if a model takes 1 hour to converge.
Masterful’s meta-learner for optimization generally draws from information theoretic analysis of the data and model, and empirical analysis of hardware performance.
Further Reading¶
Andrew Ng, “MLOps: From Model-centric to Data-centric AI”, (2021). https://www.deeplearning.ai/wp-content/uploads/2021/06/MLOps-From-Model-centric-to-Data-centric-AI.pdf
Irwan Bello, William Fedus, Xianzhi Du, Ekin D. Cubuk, Aravind Srinivas, Tsung-Yi Lin, Jonathon Shlens, and Barret Zoph. “Revisiting resnets: Improved training and scaling strategies.” arXiv preprint
arXiv:2103.07579 (2021).
Behnam Neyshabur, Ryota Tomioka, Ruslan Salakhutdinov, Nathan Srebro, Geometry of Optimization and Implicit Regularization in Deep Learning. https://arxiv.org/abs/1705.03071
Alex Hernández-García and Peter König. “Data augmentation instead of explicit regularization.” arXiv preprint arXiv:1806.03852 (2018).
Qizhe Xie, Minh-Thang Luong, Eduard Hovy, and Quoc V. Le. “Self-training with noisy student improves imagenet classification.” In Proceedings of the IEEE/CVF Conference on Computer Vision and Pattern
Recognition, pp. 10687-10698. 2020.
Suman Ravuri, and Oriol Vinyals. “Seeing is not necessarily believing: Limitations of biggans for data augmentation.” (2019).
Ting Chen, Simon Kornblith, Mohammad Norouzi, and Geoffrey Hinton. “A simple framework for contrastive learning of visual representations.” In International conference on machine learning, pp.
1597-1607. PMLR, 2020.
Ekin D. Cubuk, Barret Zoph, Dandelion Mane, Vijay Vasudevan, and Quoc V. Le. “Autoaugment: Learning augmentation policies from data.” arXiv preprint arXiv:1805.09501 (2018).
Lars Kai Hansen and Peter Salamon. “Neural network ensembles.” IEEE transactions on pattern analysis and machine intelligence 12, no. 10 (1990): 993-1001.
Geoffrey Hinton, Oriol Vinyals, and Jeff Dean. “Distilling the knowledge in a neural network.” arXiv preprint arXiv:1503.02531 (2015).
Zhilu Zhang and Mert R. Sabuncu. “Self-distillation as instance-specific label smoothing.” arXiv preprint arXiv:2006.05065 (2020).
Martin Heusel, Hubert Ramsauer, Thomas Unterthiner, Bernhard Nessler, and Sepp Hochreiter. “Gans trained by a two time-scale update rule converge to a local nash equilibrium.” Advances in neural
information processing systems 30 (2017).
Hongyi Zhang, Moustapha Cisse, Yann N. Dauphin, and David Lopez-Paz. “mixup: Beyond empirical risk minimization.” arXiv preprint arXiv:1710.09412 (2017).
Sangdoo Yun, Dongyoon Han, Seong Joon Oh, Sanghyuk Chun, Junsuk Choe, and Youngjoon Yoo. “Cutmix: Regularization strategy to train strong classifiers with localizable features.” In Proceedings of the
IEEE/CVF International Conference on Computer Vision, pp. 6023-6032. 2019.
Spyros Gidaris and Andrei Bursuc, Teacher-student feature prediction approaches. 2021. https://gidariss.github.io/self-supervised-learning-cvpr2021/slides/teacher_student.pdf | {"url":"https://docs.masterfulai.com/0.4.1/concepts/concepts.html","timestamp":"2024-11-12T06:42:27Z","content_type":"text/html","content_length":"39522","record_id":"<urn:uuid:a3323fe9-e5e6-4dc7-98f4-a7ce45887a62>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00100.warc.gz"} |
25,072 research outputs found
The so-called "Quantum Inequalities", and the "Quantum Interest Conjecture", use quantum field theory to impose significant restrictions on the temporal distribution of the energy density measured by
a time-like observer, potentially preventing the existence of exotic phenomena such as "Alcubierre warp-drives" or "traversable wormholes". Both the quantum inequalities and the quantum interest
conjecture can be reduced to statements concerning the existence or non-existence of bound states for a certain one-dimensional quantum mechanical pseudo-Hamiltonian. Using this approach, we shall
provide a simple proof of one version of the Quantum Interest Conjecture in (3+1) dimensional Minkowski space.Comment: V1: 8 pages, revtex4; V2: 10 pages, some technical changes in details of the
argument, no change in physics conclusions, this version essentially identical to published versio
So called "analogue models" use condensed matter systems (typically hydrodynamic) to set up an "effective metric" and to model curved-space quantum field theory in a physical system where all the
microscopic degrees of freedom are well understood. Known analogue models typically lead to massless minimally coupled scalar fields. We present an extended "analogue space-time" programme by
investigating a condensed-matter system - in and beyond the hydrodynamic limit - that is in principle capable of simulating the massive Klein-Gordon equation in curved spacetime. Since many
elementary particles have mass, this is an essential step in building realistic analogue models, and an essential first step towards simulating quantum gravity phenomenology. Specifically, we
consider the class of two-component BECs subject to laser-induced transitions between the components, and we show that this model is an example for Lorentz invariance violation due to ultraviolet
physics. Furthermore our model suggests constraints on quantum gravity phenomenology in terms of the "naturalness problem" and "universality issue".Comment: Talk given at 7th Workshop on Quantum
Field Theory Under the Influence of External Conditions (QFEXT 05), Barcelona, Catalonia, Spain, 5-9 Sep 200
In any static spacetime the quasi-local Tolman mass contained within a volume can be reduced to a Gauss-like surface integral involving the flux of a suitably defined generalized surface gravity. By
introducing some basic thermodynamics and invoking the Unruh effect one can then develop elementary bounds on the quasi-local entropy that are very similar in spirit to the holographic bound, and
closely related to entanglement entropy.Comment: V1: 4 pages. Uses revtex4-1; V2: Three references added; V3: Some notational changes for clarity; introductory paragraph rewritten; no physics
changes. This version accepted for publication in Physical Review Letter
The class of spherically-symmetric thin-shell wormholes provides a particularly elegant collection of exemplars for the study of traversable Lorentzian wormholes. In the present paper we consider
linearized (spherically symmetric) perturbations around some assumed static solution of the Einstein field equations. This permits us to relate stability issues to the (linearized) equation of state
of the exotic matter which is located at the wormhole throat.Comment: 4 pages; ReV_TeX 3.0; one postscript figur
In [cond-mat/9906332; Phys. Rev. Lett. 84, 822 (2000)] and [physics/9906038; Phys. Rev. A 60, 4301 (1999)] Leonhardt and Piwnicki have presented an interesting analysis of how to use a flowing
dielectric fluid to generate a so-called "optical black hole". Qualitatively similar phenomena using acoustical processes have also been much investigated. Unfortunately there is a subtle
misinterpretation in the Leonhardt-Piwnicki analysis regarding these "optical black holes": While it is clear that "optical black holes" can certainly exist as theoretical constructs, and while the
experimental prospects for actually building them in the laboratory are excellent, the particular model geometries that Leonhardt and Piwnicki write down as alleged examples of "optical black holes"
are in fact not black holes at all.Comment: one page comment, uses ReV_TeX 3; discussion clarified; basic physical results unaltere
How much of modern cosmology is really cosmography? How much of modern cosmology is independent of the Einstein equations? (Independent of the Friedmann equations?) These questions are becoming
increasingly germane -- as the models cosmologists use for the stress-energy content of the universe become increasingly baroque, it behoves us to step back a little and carefully disentangle
cosmological kinematics from cosmological dynamics. The use of basic symmetry principles (such as the cosmological principle) permits us to do a considerable amount, without ever having to address
the vexatious issues of just how much "dark energy", "dark matter", "quintessence", and/or "phantom matter" is needed in order to satisfy the Einstein equations. This is the sub-sector of cosmology
that Weinberg refers to as "cosmography", and in this article I will explore the extent to which cosmography is sufficient for analyzing the Hubble law and so describing many of the features of the
universe around us.Comment: 7 pages; uses iopart.cls setstack.sty. Based on a talk presented at ACRGR4, the 4th Australasian Conference on General Relativity and Gravitation, Monash University,
Melbourne, January 2004. To appear in the proceedings, in General Relativity and Gravitatio
Building on a pair of earlier papers, I investigate the various point-wise and averaged energy conditions for the quantum stress-energy tensor corresponding to a conformally-coupled massless scalar
field in the in the (1+1)-dimensional Schwarzschild spacetime. Because the stress-energy tensors are analytically known, I can get exact results for the Hartle--Hawking, Boulware, and Unruh vacua.
This exactly solvable model serves as a useful sanity check on my (3+1)-dimensional investigations wherein I had to resort to a mixture of analytic approximations and numerical techniques. Key
results in (1+1) dimensions are: (1) NEC is satisfied outside the event horizon for the Hartle--Hawking vacuum, and violated for the Boulware and Unruh vacua. (2) DEC is violated everywhere in the
spacetime (for any quantum state, not just the standard vacuum states).Comment: 7 pages, ReV_Te
For an arbitrary Tolman wormhole, unconstrained by symmetry, we shall define the bounce in terms of a three-dimensional edgeless achronal spacelike hypersurface of minimal volume. (Zero trace for the
extrinsic curvature plus a "flare-out" condition.) This enables us to severely constrain the geometry of spacetime at and near the bounce and to derive general theorems regarding violations of the
energy conditions--theorems that do not involve geodesic averaging but nevertheless apply to situations much more general than the highly symmetric FRW-based subclass of Tolman wormholes. [For
example: even under the mildest of hypotheses, the strong energy condition (SEC) must be violated.] Alternatively, one can dispense with the minimal volume condition and define a generic bounce
entirely in terms of the motion of test particles (future-pointing timelike geodesics), by looking at the expansion of their timelike geodesic congruences. One re-confirms that the SEC must be
violated at or near the bounce. In contrast, it is easy to arrange for all the other standard energy conditions to be satisfied.Comment: 8 pages, ReV-TeX 3.
Traversable wormholes have traditionally been viewed as intrinsically topological entities in some multiply connected spacetime. Here, we show that topology is too limited a tool to accurately
characterize a generic traversable wormhole: in general one needs geometric information to detect the presence of a wormhole, or more precisely to locate the wormhole throat. For an arbitrary static
spacetime we shall define the wormhole throat in terms of a 2-dimensional constant-time hypersurface of minimal area. (Zero trace for the extrinsic curvature plus a "flare-out" condition.) This
enables us to severely constrain the geometry of spacetime at the wormhole throat and to derive generalized theorems regarding violations of the energy conditions-theorems that do not involve
geodesic averaging but nevertheless apply to situations much more general than the spherically symmetric Morris-Thorne traversable wormhole. [For example: the null energy condition (NEC), when
suitably weighted and integrated over the wormhole throat, must be violated.] The major technical limitation of the current approach is that we work in a static spacetime-this is already a quite rich
and complicated system.Comment: 25 pages; plain LaTeX; uses epsf.sty (four encapsulated postscript figures | {"url":"https://core.ac.uk/search/?q=authors%3A(Visser%20M)","timestamp":"2024-11-09T20:06:09Z","content_type":"text/html","content_length":"164388","record_id":"<urn:uuid:2886bfbd-a91f-40c0-927a-098d7ce10162>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00546.warc.gz"} |
TOC | Previous | Next | Index
48.7 Partial Least Squares Discriminant Analysis (.NET, C#, CSharp, VB, Visual Basic, F#)
Partial least squares Discriminant Analysis (PLS-DA) is a variant used when the response variable is categorical. Three classes are provided for performing PLS-DA:
● SparsePlsDa performs Discriminant Analysis (DA) using a classical sparse PLS regression (sPLS), but where the response variable is categorical. The response vector Y is qualitative and is recoded
as a dummy block matrix where each of the response categories are coded via an indicator variable. PLS-DA is then run as if Y was a continuous matrix. SparsePlsDa inherits from PLS2.
● SparsePls performs a sparse PLS calculation with variable selection. The LASSO penalization is used on the pairs of loading vectors. SparsePls implements IPLS2Calc.
● SparsePLSDACrossValidation performs an evaluation of a PLS model. Evaluation consists of dividing the data into two subsets: a training subset and a testing subset. A PLS calculation is performed
on the training subset and the resulting model is used to predict the values of the dependent variables in the testing set. The mean square error between the actual and predicted dependent values is
then calculated. Usually, the data is divided up into several training and testing subsets and calculations are done on each of these. In this case the average mean square error over each PLS
calculation is reported. (The individual mean square errors are available as well.)
The subsets to use in the cross validation are specified by providing an implementation of the ICrossValidationSubsets interface. Classes that implement this interface generate training and testing
subsets from PLS data.
For example, if X is the predictor data and y the corresponding observed factor levels, this code calculates the sparse PLS-DA:
Code Example – C# Partial Least Squares Discriminant Analysis (PLS-DA)
int ncomp = 3;
int numXvarsToKeep = (int) Math.Round( X.Cols * 0.66 );
int[] keepX = Enumerable.Repeat( numXvarsToKeep, ncomp ).ToArray();
var splsda = new SparsePlsDa( X, y, ncomp, keepX );
Code Example – VB Partial Least Squares Discriminant Analysis (PLS-DA)
Dim NComp As Integer = 3
Dim NumXvarsToKeep As Integer = CType(Math.Round(X.Cols * 0.66),
Dim KeepX As Integer() = Enumerable.Repeat(NumXvarsToKeep,
Dim SPLSDA As New SparsePlsDa(X, Y, NComp, KeepX)
The number of components to keep in the model is specified, as well as the number of predictor variables to keep for each of the components (about two thirds, in this case).
Because SparsePlsDa is a PLS2, you can use the PLS2Anova class to perform an ANOVA (Section 48.4).
Code Example – C# Partial Least Squares Discriminant Analysis (PLS-DA)
var anova = new PLS2Anova( splsda );
Console.WriteLine( "Rsqr: " + anova.CoefficientOfDetermination );
Console.WriteLine( "MSE Prediction: " +
anova.RootMeanSqrErrorPrediction );
Code Example – VB Partial Least Squares Discriminant Analysis (PLS-DA)
Dim Anova As New PLS2Anova(SPLSDA)
Console.WriteLine("Rsqr: " & Anova.CoefficientOfDetermination)
Console.WriteLine("MSE Prediction: " &
You can also do cross validation using class SparsePLSDACrossValidation.
Code Example – C# Partial Least Squares Discriminant Analysis (PLS-DA)
var subsetGenerator = new LeaveOneOutSubsets();
var crossValidation =
new SparsePLSDACrossValidation( subsetGenerator );
crossValidation.DoCrossValidation( X, yFactor, ncomp, keepX );
Console.WriteLine( "Cross validation average MSE: " +
crossValidation.AverageMeanSqrError );
Code Example – VB Partial Least Squares Discriminant Analysis (PLS-DA)
Dim SubsetGenerator As New LeaveOneOutSubsets()
Dim CrossValidation As New
CrossValidation.DoCrossValidation(X, YFactor, NComp, KeepX)
Console.WriteLine("Cross validation average MSE: " & | {"url":"https://www.centerspace.net/doc/NMath/user/partial-least-squares-87513.htm","timestamp":"2024-11-03T13:53:30Z","content_type":"text/html","content_length":"19533","record_id":"<urn:uuid:aa9ae710-2a0d-4c1b-b09f-df9267419ad3>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00118.warc.gz"} |
Projects - Mathematical Complexity Reduction - OVGU
Nonlinear Descriptions of Combinatorial Problems
Polytopes associated with combinatorial optimization problems (i.e., the convex hulls of the incidence vectors of the feasible solutions) in most cases have exponentially many facets. Thus, they can
only be represented as the sets of solutions to exponentially large systems of linear inequalities.
One approach to reduce the size of these descriptions which I want to explore in my PhD project is to allow for representations as projections of higher dimensional semi-algebraic sets, thus to
combine the concepts of extended formulations and polynomial representations, and to study the principal possibilities of obtaining short extended polynomial representations of polytopes associated
with combinatorial optimization problems.
The ultimate goal of such investigations of nonlinear representations of reduced complexity would be to gain additional insights into the geometry and combinatorics of the underlying problems which
remain hidden when restricting to linear representations. With respect to eventually exploiting such insights algorithmically, particular attention will be given to the question for the existence of
small representations via specially structured semi-algebraic sets such as spectrahedra. | {"url":"https://www.mathcore.ovgu.de/index.php?show=research_projects&project=frede","timestamp":"2024-11-09T10:11:39Z","content_type":"text/html","content_length":"8421","record_id":"<urn:uuid:6f363ae1-a1f2-410c-b7b2-68c9c1af3b83>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00238.warc.gz"} |
Kitaev Transformation - Wiley Online Library - P.PDFKUL.COM
The Bravyi–Kitaev Transformation: Properties and Applications Andrew Tranter,[a,b] Sarah Sofia,[c,d] Jake Seeley,[d,e] Michael Kaicher,[f ] Jarrod McClean,[g] Ryan Babbush,[g] Peter V. Coveney,[b]
Florian Mintert,[a] Frank Wilhelm,[f ] and Peter J. Love*[d] Quantum chemistry is an important area of application for quantum computation. In particular, quantum algorithms applied to the electronic
structure problem promise exact, efficient methods for determination of the electronic energy of atoms and molecules. The Bravyi–Kitaev transformation is a method of mapping the occupation state of a
fermionic system onto qubits. This transformation maps the Hamiltonian of n interacting fermions to an Oðlog nÞ-local Hamiltonian of n qubits. This is an improvement in locality over the Jordan–
Wigner transformation, which results in an O(n)-local qubit Hamiltonian. We present the Bravyi–Kitaev transformation in
Introduction Quantum simulation was first proposed by Feynman[1] and allows for an exponential speedup over classical simulation of some quantum mechanical systems.[2–6] In the context of quantum
chemistry, efficient algorithms have been developed for the calculation of energy spectra,[7] reaction rates,[8,9] and reaction details.[10] Quantum computational schemes have been extended into the
study of relativistic quantum chemistry.[11] Crucially for this project, the quantum-phase estimation algorithm[12] allows for efficient calculation of molecular energies at an accuracy equivalent to
that of classical full configuration interaction calculations. There are three basic approaches to the quantum simulation of chemical systems. One approach—the so called “first quantization”
approach—has been studied in the context of chemical reactive scattering.[10] Here, physical position space is discretized. The electronic wavefunction is then represented in the position
representation by the state of the qubits. The chemical Hamiltonian is: X p 2 X qi qj i ^ H5 1 2Mi ij rij i
where sums are over nuclei and electrons, pi is the momentum of the ith particle, Mi is the mass of the ith particle, qi is the charge of the ith particle, and rij is the distance between particles i
and j in atomic units. We can simulate the effect of this Hamiltonian by unitarily evolving the qubits through the propagator corresponding to the molecular Hamiltonian, approximated using the
quantum split operator method of Zalka[4] or by quantum lattice gas methods.[5,6]
detail, introducing the sets of qubits which must be acted on to change occupancy and parity of states in the occupation number basis. We give recursive definitions of these sets and of the
transformation and inverse transformation matrices, which relate the occupation number basis and the Bravyi– Kitaev basis. We then compare the use of the Jordan–Wigner and Bravyi–Kitaev Hamiltonians
for the quantum simulation of C 2015 Wiley Periodicals, Inc. methane using the STO-6G basis. V DOI: 10.1002/qua.24969
An alternative to grid-based first-quantized approaches is the use of a second-quantized formalism. Here, the molecular Hamiltonian is expressed in terms of creation and annihilation operators acting
on some basis of molecular orbitals. This method is the main topic of this article and so discussed in
[a] A. Tranter, F. Mintert Department of Physics, Imperial College London, South Kensington Campus, London SW7 2AZ, United Kingdom [b] A. Tranter, P. V. Coveney Centre for Computational Science,
University College London, 20 Gordon Street, London, WC1H 0AJ, United Kingdom [c] S. Sofia Photovoltaics Research Laboratory, Massachusetts Institute of Technology, Cambridge, Massachusetts 02139 [d]
S. Sofia, J. Seeley, P. J. Love Department of Physics, Haverford College, 370 Lancaster Ave., Haverford, Pennsylvania 19041 E-mail:
[email protected]
[e] J. Seeley Earth and Planetary Science, University of California, Berkeley, 307 McCone Hall, Berkeley, California 94720-4767 [f ] M. Kaicher, F. Wilhelm Theoretical Physics, Saarland University,
66123 Saarbr€ ucken, Germany [g] J. McClean, R. Babbush Department of Chemistry and Chemical Biology, Harvard University, Cambridge, Massachusetts 02138 Contract grant sponsor: NSF CCI center,
Quantum Information for Quantum Chemistry (QIQC); contract grant number: CHE-1037992. Contract grant sponsor: NSF award; contract grant number: PHY-0955518. Contract grant sponsor: AFOSR award;
contract grant number: FA9550-121–0046. Contract grant sponsor: DOE Computational Science Graduate Fellowship; contract grant number: DE-FG02–97ER25308. Contract grant sponsor: EPSRC and UCLQ for
support through a UCLQ visiting fellowship (P.J.L.). C 2015 Wiley Periodicals, Inc. V
International Journal of Quantum Chemistry 2015, 115, 1431–1441
greater detail below. While this technique scales less efficiently than the prior method in the asymptotic limit, smaller scale simulations require substantially fewer resources. The reason is that a
molecular orbital basis is more efficient for the representation of localized chemical wavefunctions than a Cartesian grid, and hence the first-quantized methods lead to wider, but shallower
circuits. Details of resource requirements for such first-quantized simulations for chemistry are given in [10]. One of the differences between the first- and secondquantized approaches lies in
whether the antisymmetric nature of the wavefunction is represented through properties of the state (first quantized) or the operators (second quantized). An alternative to grid-based methods in
which the dynamics preserves an initially antisymmetric wavefunction is the use of a basis of Slater determinants. In this case, the challenge for quantum algorithms is the evolution under the CI
matrix representation of the Hamiltonian. Unlike the secondquantized case, this matrix has no natural expression as a sum of local terms, and no tensor product structure. However, the CI matrix is
sparse, and hence quantum simulation techniques for sparse matrices may be applied to this problem. This yields methods that both use an efficient molecular orbital representation of the wavefunction
and have optimal asymptotic scaling. This also enables the use of sparse methods, which scale logarithmically with the error. The penalty is that the molecular integrals must be computed on the fly
during the quantum computation.[13–15] The calculation of the energies of molecular Hydrogen and Helium-Hydride using minimal basis sets have been experimentally achieved using linear optical
quantum, NMR, and Nitrogen vacancy in diamond quantum computers.[16–19] The first digital fermionic quantum simulation was recently achieved of a four-site Hubbard model in superconducting hardware.
[20] These proofs of principle demonstrations are comparable to early quantum chemical calculations carried out in the twentieth century.[21] The development and optimization of quantum algorithms
for chemistry is ongoing. This work is driven by two goals. First is the desire to determine the true optimal asymptotic scaling of these algorithms for large quantum computers. The second is to
reduce the resource requirements of small examples to the point that they can be realized experimentally in the near future. Recently, the possibility of using a small quantum computer of around a
hundred qubits for the purposes of quantum chemistry has been investigated in detail. Initial upper bounds on the cost indicated that large polynomial scaling would be impractical for such problems.
[22] Further analysis developing circuit improvements, tighter upper bounds, and numerical investigation of errors restricted to the chemical ground state resulted in tight and efficiently computable
upper bounds on the resources required.[23–25] One may also improve these algorithms by exploiting locality.[26] The topic of this article is the Bravyi–Kitaev transformation, an alternative to the
use of the Jordan–Wigner transformation to map fermions to spins.[27–29] This transformation was defined in [28] in the context of using fermions to perform quantum computations. Its use for the
simulation of fermions 1432
International Journal of Quantum Chemistry 2015, 115, 1431–1441
by quantum computers, and in particular, its use for the quantum simulation of quantum chemistry, was introduced in [29]. We describe the transformation in detail, and derive some new properties of
the transformation that are relevant to the specific case of second-quantized Hamiltonians defined in a basis of spin–orbitals. We give a new recursive definition for the inverse Bravyi–Kitaev
transformation matrix, as well as recursive relationships for the update, parity, and flip sets (defined below) which facilitate the computation of these sets. We analyze the efficiency of the
Bravyi–Kitaev method for the simulation of the methane molecule. We find that the Bravyi–Kitaev mapping leads to a small improvement, particularly in the number of nonlocal gates required for
accurate simulation.
The Second-Quantized Hamiltonian As in classical quantum chemistry, we invoke the BornOppenheimer approximation, fixing the nuclear coordinates, and calculating the electronic energy at a given
geometry. In the second-quantized formalism previously mentioned, the electronic Hamiltonian is given by: X 1X † † † ^ H5 hij ai aj 1 hijkl ai aj ak al 2 i;j i;j;k;l
where hij and hijkl are integrals, which can be efficiently classically precomputed. † The a and a operators in the Hamiltonian are creation and annihilation operators on a basis set of molecular
orbitals, as discussed below. Note that here, the two-operator terms effectively correspond to single-electron terms, and the fouroperator terms effectively correspond to electron–electron
interaction terms. Because electrons are fermions, we require antisymmetry on exchange of particle index. This is enforced through the use of anticommutator restrictions on the creation and
annihilation operators:
n † †o aj ; ak 5 aj ; ak 50 n o † aj ; ak 5djk I
Our task, therefore, is effectively to find the lowest eigenvalue of this Hamiltonian. As the dimension of the Fock space grows exponentially with the number of basis orbitals, this is classically
intractable for systems of any reasonable size. However, a quantum computer could remove this problem using quantum-phase estimation.[7] To achieve this, three steps must be taken. First, a mapping
between the physical electronic states and qubit states in a quantum computer must be established. Second, a welldefined evolution operator equivalent to that of the molecular Hamiltonian must be
determined for the qubit basis. This necessitates the derivation of qubit representations of the electronic creation and annihilation operators. Finally, the phase estimation algorithm requires the
preparation of a guiding state. A guiding state is an input state to the algorithm, WWW.CHEMISTRYVIEWS.ORG
which has an overlap with the true ground state, which decays at worst as an inverse polynomial in the system size. In the worst case, Hamiltonians are known for which the problem of finding the
ground state is QMA-complete (the quantum equivalent of NP-complete).[30,31] Quantum computers are not believed to be capable of efficiently solving QMAcomplete problems in the worst case, just as
classical computers are not believed to be capable of efficiently solving NP-complete problems in the worst case. Assuming this is true, there exist Hamiltonians for which no efficiently preparable
guiding state is likely to be available, and for which, the phase estimation algorithm is, therefore, incapable of finding the ground state. However, these worst case Hamiltonians rely on clock
constructions so that their ground states are superpositions of quantum states corresponding to time slices of an arbitrary quantum circuit of depth polynomial in the number of qubits.[30] Even
constructions that show the QMA-completeness of specific physical models rely on geometrically complex interactions.[32] It is, therefore, a widely believed conjecture that typical physical
Hamiltonians do not correspond to worst case instances, and therefore, have efficiently preparable ground states. Specific algorithms for state preparation are considered in [7,33–36]. One may also
ask whether the requirement to prepare guiding states may rely on features of physical Hamiltonians, which can also be exploited for the development of classical algorithms. The requirement on a
guiding state for a quantum computation of an energy eigenvalue is only that its overlap with the true ground state is bounded by an inverse polynomial in the system size. Recent consideration of
Quantum Monte Carlo methods (which simulate quantum systems using conventional computers) showed that a much stronger guiding state was required to make these methods efficient, even in the case of
so-called stoquastic Hamiltonians where there is no fermion sign problem.[37]
Qubit creation and annihilation operators In this section, we describe three mappings of fermionic states and operators to qubit states and operators. In each case, we map the occupation number basis
to the qubit basis. The occupation number configuration basis states are given by specifying the occupation fi 2 f0; 1g of every orbital. The fermionic creation and annihilation operators, when
acting on a system of n orbitals with occupation state vector, jfn21 fn22 :::f1 f0 i yield: j21 P †
aj jfn21 :::fj11 0fj21 :::f1 f0 i5ð21Þs50 jfn21 :::fj11 1fj21 :::f1 f0 i †
aj jfn21 :::fj11 1fj21 :::f1 f0 i50
(4) (5)
j21 P
fs aj jfn21 :::fj11 1fj21 :::f1 f0 i5ð21Þs50 jfn21 :::fj11 0fj21 :::f1 f0 i aj jfn21 :::fj11 0fj21 :::f1 f0 i50
valid from the point of view of realizing the anticommutation relations, and is more common in the chemical literature. As can be seen in Eqs. (4) through (7), these operators depend on both the
occupation of orbital j as well as its parity P pj 5 j21 s50 fs , as the phase shift in Eqs. (4) and (6) can be written in terms of the parity as ð21Þpj . If the parity is odd, the state is
multiplied by a factor of 21, and if it is even, there is no phase shift. Since the fermionic creation and annihilation operators change both occupation and parity, their qubit analogues also need to
do so. Therefore, both the occupation and parity of each orbital must be stored when mapping from the occupation basis state onto a qubit basis state. We consider three mappings where sums of
fermionic occupations are stored in the qubit state. These are the Jordan– Wigner basis, the parity basis and the Bravyi–Kitaev basis. In all cases, it is helpful to define several subsets of the
qubits, which contain the information needed to apply fermionic operators to the state. These sets are defined below, and we use fi to indicate Boolean negation 051; 150. 1. The update set, U(i).
This is the set of qubits, apart from i that must be updated, when the occupancy fi changes. 2. The parity set P(i). This is the set of qubits that deterP mines the parity pi 5 j
These operators contain only Pauli operators acting on the qubit being created or annihilated. They, therefore, commute for different qubits, and so clearly do not fulfil the anticommutation
relations required. We must combine these operators with actions on the sets defined above to obtain qubit creation and annihilation operators that satisfy the canonical fermionic commutation
(6) (7)
We note that one is free to choose the ordering of the orbitals here. We have chosen an ordering in which the orbitals fs with s < j determine the parity, but the choice s > j is equally
The Jordan–Wigner transformation In the Jordan–Wigner transformation, we use the state of a qubit to denote whether or not a particular basis orbital is occupied— clearly, as electrons are fermionic,
occupation numbers which are not zero or one are impossible. The qubits directly store the International Journal of Quantum Chemistry 2015, 115, 1431–1441
occupation basis.[38] In this case, the update set is empty (recall that qubit i is not a member of the update set). Parity information needed to correctly apply the creation and annihilation
operators for orbital i is contained in all qubits j < i. Hence the parity set is defined by PðiÞ5fjjj < ig. This is the Jordan–Wigner transformation.[27] We consequentially have the qubit operators:
parity nonlocally or vice versa, as is the case for the Jordan–Wigner and parity bases. In the Bravyi–Kitaev basis, for any index j, if j is even, qubit j holds only the occupation state of orbital
j, and if j is odd, qubit j holds a partial sum of the occupation state of a set of orbitals of index less than j. The Bravyi–Kitaev transformation that maps the fermionic occupation state vector to
the qubit state, denoted bn for n orbitals such that bn ~ fn 5b~n , is given by:
1 † ai 5 ðXi 2iYi Þ Zi 5Q1 i ZPðiÞ 2 j
where ZPðiÞ means a Pauli Z operator acting on all qubits in the set P(i). The fact that these operators obey the fermionic anticommutation relations follows from the fact that fZ; Q6 g5 0 and fQ1 ;
Q2 g5I. Parity basis The Jordan–Wigner transformation stored occupancy locally, and parity is nonlocal. The parity basis stores the parity locally,[28] and the occupancy is nonlocal. The parity
information of each orbital j is stored in the corresponding qubit j, j X qj 5pj 1fj 5 fs : (10) s50
Evidently, PðjÞ5fj21g in the parity basis. Whether qubit j stores fj or fj is determined by qubit j 2 1 in the parity basis. Hence, the flip set in this basis is equal to the parity set: FðjÞ5PðjÞ,
and so the remainder set RðjÞ51. The update set U(j) is the set of qubits that must be updated when occupancy fj changes. Now fj appears in every qi such that i > j, and so when fj changes every
qubit i > j must be updated. Hence, UðjÞ5 fiji > jg for the parity basis. Given the definitions of these sets, we can now write the qubit creation and annihilation operators. †
1 1 aj 5ð Xi Þ ðP0FðjÞ Q2 j 1PFðjÞ Qj Þ ZPðjÞ i>j
(11) 1 2 aj 5ð Xi Þ ðP0FðjÞ Q1 j 1PFðjÞ Qj Þ ZPðjÞ i>j
where Pb 5jbihbj. Now, because PðjÞ5FðjÞ and because Px Z5 ð21Þx Px we can write: 1 † 1 1 aj 5ð Xi Þ ðP0FðjÞ Q2 j 2PFðjÞ Qj Þ5 ð Xi ÞðZj Zj21 2iYj Þ 2 i>j i>j 1 1 2 aj 5ð Xi Þ ðP0FðjÞ Q1 j 2PFðjÞ Qj
Þ5 ð Xi ÞðZj Zj21 1iYj Þ 2 i>j i>j
(12) The number of nontrivial Pauli factors in these operators scales as O(n), just as for Jordan–Wigner. In this case, it is the update set whose size scales linearly with the number of qubits.
Bravyi–Kitaev transformation The Bravyi–Kitaev transformation stores both occupation and parity nonlocally, rather than storing the occupation state locally and 1434
International Journal of Quantum Chemistry 2015, 115, 1431–1441
where 1 ! indicates a row of ones in the bottom row. For example, for eight qubits, the fermion occupation state vector is mapped to the qubit basis state as shown in Eq. (14) (all sums in mod(2)): 0
1 B B1 B B B0 B B B1 B B B0 B B B0 B B B0 @
1 0 1 f0 f0 C CB C B C B C B f1 1f0 0C C CB f1 C B C CB C B C C C B B f2 0 CB f2 C B C C CB C B C B f3 C B f 1f 1f 1f 0C 3 2 1 0 C CB C B C CB C5B C C C B B f4 0 CB f4 C B C C CB C B C B C B f5 1f4
0C C CB f5 C B C CB C B C B f6 C B f 0C 6 A A@ A @ f7 f7 1f6 1f5 1f4 1f3 1f2 1f1 1f0 1 0
(14) From this definition, we proceed to obtain the update, parity, flip, and remainder sets. The update set, U(j), is the set of qubits that must be updated when the occupation of some orbital j is
changed. This is the set of qubits that hold partial sums that depend on the occupation of orbital j. Because the transformation matrix bn is lower diagonal, only qubits with i > j will be contained
in U(j). We abuse notation to write U(j) > j to indicate this. Since qubits of even j hold only the occupation state of orbital j, the update set will only contain odd qubit indices, as only qubits
with odd j hold partial sums. From the Bravyi–Kitaev transformation matrix, given that any column j contains the vector that acts on occupation state vector entry j, the update set for changing the
occupation of orbital j is simply the set of qubits with index greater than j and equal to the indices of the nonzero entries in column j.[29] The update sets for each orbital for systems of 1–8
orbitals are given in Table 1. The parity set, P(j), is the set of qubits needed to determine the parity of the set of orbitals with index
if i < j
0 otherwise
Note this is not the transformation matrix, which gives the parity basis, as pn has a zero diagonal, given that, it computes the parity of all orbitals strictly less than i. For four orbitals, this
matrix is given by: WWW.CHEMISTRYVIEWS.ORG
Table 1. Indices of qubits in the update set, U(j), which is the set of all qubits whose state must be updated when the occupation state of an orbital j is changed, for systems of 1–8 orbitals. #
Qubits 2 2 4 4 8 8 8 8
# Orbitals
{1} {1} {1, 3} {1, 3} {1, 3, 7} {1, 3,7} {1, 3,7} {1, 3,7}
– 1 {3} {3} {3, 7} {3, 7} {3, 7} {3, 7}
– – {3} {3} {3, 7} {3, 7} {3, 7} {3, 7}
– – – 1 {7} {7} {7} {7}
– – – – {5, 7} {5, 7} {5, 7} {5, 7}
– – – – – {7} {7} {7}
– – – – – – {7} {7}
– – – – – – – 1
B B1 B p4 5B B1 @ 1
C 0C C C 0C A
number of these qubits scales as Oðlog ðjÞÞ Oðlog ðnÞÞ.[28,29] Since the fermionic occupation state vector ~ fn is transformed into the Bravyi–Kitaev basis, b~n by bn ~ fn 5b~n , this transformation
can be reversed to get back to the fermionic occupation basis ~ ~ by b21 n bn 5fn . We find that the parity transformation for the Bravyi–Kitaev basis is pn b21 n . For eight orbitals:
and addition is taken modulo two in the matrix multiplication. This method stores the parity of orbital j in partial sums held in several qubits of index less than or equal to j, where the
B B1 B B B0 B B B0 21 ~ B ~ p~8 5p8 f8 5p8 b8 b8 B B0 B B B0 B B B0 @
Therefore, the parity set, P(j), is the set of qubits with index equal to the nonzero entries of pn b21 n in row j as these are the qubits whose sum gives the parity of orbital j.[29] The product pn
b21 is lower triangular because it is the product of two n lower triangular matrices. Hence, the parity set P(j) only contains indices i < j, so P(j) < j. This also implies that the intersection of
parity and update sets is always empty. The parity sets for each orbital for systems of 1 through 8 orbitals are given in Table 2. Lastly, the flip set, F(j), is the set of qubits that determine
whether qubit j and orbital j are equal or opposite. The flip set is the set of qubits that hold the parity of the occupation of the orbitals with index
C CB C B C B C B b0 0C C CB b 1 C B C CB C B C B b2 C B b 0C 1 C CB C B C CB C B B b3 C B b2 1b1 C 0C C CB C B C CB C5B C B b4 C B b 0C 3 C CB C B C CB C B B b5 C B b4 1b3 C 0C C CB C B C CB C B B b6
C B b5 1b3 C 0C A A@ A @ b7 b6 1b5 1b3 0
form back to the fermionic occupation state. To do this, we can look at the inverse transformation. For eight qubits, 0
B B1 B B B0 B B B0 21 B b8 5B B0 B B B0 B B B0 @
C 0 0C C C 0 0C C C 0 0C C C 0 0C C C 0 0C C C 1 0C A 1 1
~ ~ As b21 n bn 5fn , the set of qubits whose states sum to the occupation state of orbital j are those with indices equal to the indices of nonzero entries in row j of b21 n . Therefore, the flip
set of orbital j is the set of these qubits with indices
Table 2. Indices of qubits in the parity set, P(j), which is the set of qubits whose occupation is needed to determine the parity of the orbital j, for each orbital in systems of 1–8 orbitals. #
Qubits 2 2 4 4 8 8 8 8
# Orbitals
– {0} {0} {0} {0} {0} {0} {0}
– – {1} {1} {1} {1} {1} {1}
– – – {1, 2} {1, 2} {1, 2} {1, 2} {1, 2}
– – – – {3} {3} {3} {3}
– – – – – {3, 4} {3, 4} {3, 4}
– – – – – – {3, 5} {3, 5}
– – – – – – – {3, 5, 6}
states of orbitals with index
j and P(j) < j gives rise to the intersections among even and odd parity and update sets shown in Table 4.
Bravyi–Kitaev operators Having defined the update, parity, and flip sets for the Bravyi– Kitaev transformation, we can define qubit creation and annihilation operators. For even indexed qubits, this
is relatively simple. Even indexed qubits only store their corresponding occupation, performing operations requires only the actual 6 creation or annihilation operation (Q^ ), updating the update set
with a bit flip, and introducing a negative sign depending on the parity of the parity set. Hence, the creation and annihilation operator equivalents for even indexed qubits are:
aj 5XUðjÞ Q^ j ZPðjÞ 5
1 XUðjÞ Xj ZPðjÞ 2iXUðjÞ Yj ZPðjÞ 2 (19)
1 2 aj 5XUðjÞ Q^ j ZPðjÞ 5 XUðjÞ Xj ZPðjÞ 1iXUðjÞ Yj ZPðjÞ 2 (20) where we know that U(j) > j and P(j) < j, so these operators act on disjoint sets of qubits. The qubit operators for qubits with odd
index are more complicated. First, we note that where the flip set has nonzero parity, the occupation of the qubit in question is flipped from that of the electronic state. Consequentially, in this
case, the creation operator must be applied to the qubit where the annihilation operator is applied to the electronic state, and vice versa. Therefore, defining projectors onto the even and odd
states of a set, S, of qubits: 1 E^S 5 ðI1ZS Þ 2 1 O^ S 5 ðI2ZS Þ 2
We then have new creation and annihilation operators to express this behavior: ^ 6 E^FðjÞ 2Q^ 7 O^ FðjÞ 5 1 Xj ZFðjÞ 7iYj ^ 6 5Q P j j j 2
Here, we have already implicitly accounted for the phase of the qubits in F(j). Thus, in determining whether a sign change must be implemented, we must only additionally determine the phase of the
qubits in the parity set which are not in the
Table 3. Indices of qubits in the flip set, F(j), which is the set of qubits that determine whether orbital j and qubit j have the same or flipped parity, for systems of 1–8 orbitals. # Qubits 2 2 4
# Orbitals
– {0} {0} {0} {0} {0} {0} {0}
– – 1 1 1 1 1 1
– – – {1, 2} {1, 2} {1, 2} {1, 2} {1, 2}
– – – – 1 1 1 1
– – – – – {4} {4} {4}
– – – – – – 1 1
– – – – – – – {3, 5, 6}
International Journal of Quantum Chemistry 2015, 115, 1431–1441
Table 4. Intersections between parity and update sets appearing in the Bravyi–Kitaev transformation for adjacent odd and even orbital indices.
U(2i) Uð2i11Þ P(2i) Pð2i11Þ
U(2i) Uð2i11Þ 1 1
Uð2i11Þ Uð2i11Þ 1 1
1 1 P(2i) Pð2iÞ
1 1 Pð2i11Þ Pð2i11Þ
bottom left quadrant is all zero except the bottom, right-most entry. We can verify this form for b21 directly. The equation for the inverse Bravyi–Kitaev transformation matrix satisfies the
condition that bn b21 n 5I. From Eqs. (13) and (28), we obtain:
flip set. To do this, we make use of the remainder set, defined above. This gives us the qubit representation of the electronic creation and annihilation operators for odd indexed orbitals: 1
^ ZRðjÞ 5 aj 5XUðjÞ P j
aj 5XUðjÞ
^2 P j
1 XUðjÞ Xj ZPðjÞ 2iXUðjÞ Yj ZRðjÞ 2 (23)
1 ZPðjÞ 5 XUðjÞ Xj ZPðjÞ 1iXUðjÞ Yj ZRðjÞ 2 (24)
The only difference between these operators and those of the even indexed qubits, is the application of Z to the remainder set, rather than the parity set, in the second term. Thus, by defining a
final set: ( qðjÞ5
j even
j odd
1 XUðjÞ Xj ZPðjÞ 2iXUðjÞ Yj ZqðjÞ 2 1 aj 5 XUðjÞ Xj ZPðjÞ 1iXUðjÞ Yj ZqðjÞ 2 †
(26) (27)
These two expressions allow for general products—such as those observed in the molecular Hamiltonian—to be built up through simple multiplication.[29]
Inverse Bravyi–Kitaev Transformation Matrix We have constructive definitions for the Bravyi–Kitaev transformation matrix for any number of qubits that is a power of two, and for the parity
transformation matrix in the occupation state basis. We do not have a constructive definition for the inverse Bravyi–Kitaev transformation matrix. The parity, update, and flip set are determined by
these three matrices, so a constructive definition for b21 would greatly simplify the process n of computing these sets. By inspection, we can see that the inverse matrix (again, in mod 2) can be
defined recursively as follows:
where the top left and bottom right quadrants of b21 are n given by b21 n , the top right quadrant is entirely zeroes, and the 2
b21 n 50
By the definition of the Bravyi–Kitaev transformation, the upper triangle of any bn is entirely zeroes. Therefore, bj;0 50 for all j > 0, and so: !
We have a final expression for Bravyi–Kitaev representations of electronic creation and annihilation operators: aj 5
21 because bn2 b21 n 5I, the condition that bn b n 5I is:
Now consider the second term in Eq. (29). For the multiplication of any two matrices, of the left-hand matrix, only the bottom row affects the bottom row of the product matrix. For example, if A is
matrix of all ones and B is a matrix with ones in the bottom row and zeroes everywhere else, for any third matrix C, the bottom row of AC is equivalent to the bottom row of BC as the bottom rows of A
and B are equal. By the definition of the Bravyi–Kitaev transformation, the bottom row of bn is all ones for all n. Therefore, the bottom row of bn b21 is equivalent to the n bottom row of the matrix
product from Eq. (31). As we know that bn b21 n 5I, the bottom row of this product is ð0; 0; 0; ::::; 0; 1Þ. 0
bn 5 2
! (31)
Combining Eqs. (30) and (31), Eq. (29) becomes: bn2
! 1
0 1!
! 21
bn 5 2
: :::
2mod 2
! 50
as we are adding in modulo 2. Our definition for the inverse Bravyi–Kitaev transformation matrix in Eq. (28) is, therefore, correct for all n, as it trivially holds for n 5 1. Now that we have a
recursive definition for both the Bravyi–Kitaev transformation matrix and its inverse, in the next section, we use these definitions to obtain recursive expressions for the update, parity, and flip
sets. International Journal of Quantum Chemistry 2015, 115, 1431–1441
Update, Parity, and Flip Set Formulae The method of finding the parity, update, and flip sets needed for the Bravyi–Kitaev transformation described above in Bravyi– Kitaev Transformation section and
in [29], while effective, is somewhat clumsy. Ideally, we want a formula that directly computes which qubits are in each of these sets. Given the recursive definition of the Bravyi–Kitaev
transformation matrix and its inverse, we can define the update, parity, and flip sets recursively.
where we have defined ½An=2 ij 51 8i; j;
½Tn=2 ij 5di;n=221 dj;n=221
which gives us:
Update set As described above, the update set is given by the row-index of the nonzero entries of index greater than j in column j of the Bravyi–Kitaev transformation matrix. The upper triangle of
bn2 determines the update set for qubits j < n2 for a system of n qubits, and also determines the update sets for a system of n2 qubits. Since bn2 is in the top left corner for this half of the
matrix, all entries in bn2 maintain the same indices within bn. Therefore, all elements in the update sets Un2 ðjÞ are also elements in the update sets for Un ðiÞ for i < n2. For bn, however, a row
of 1s are added across the left half of the bottom row (row index n 2 1) of the matrix, and as these are nonzero entries in a row of index greater than j for all qubits of index j < n2, n 2 1 is in
the update set for all qubits of index j < n2 for a system of n qubits in addition to the elements in the update sets for n2 qubits. If we inspect bn, defined in Eq. (13), we see that the lower
triangle of the matrix bn2 determines the update set for the n qubit system when changing the occupation of orbital n n 2 j < n. This is also the part of b2 that determines the update sets for all
qubits in a system of n2 qubits. When the matrix bn2 is placed in the lower right quadrant of bn, the row and column indices of all entries in that matrix are increased by n2. The recursive function
for the elements in the update set of an n qubits system when changing the occupation of orbital j is: 8 > < fUn2 ðjÞ; ðn21Þg Un ðjÞ5 n o > : Un j2 n 1 n 2 2 2
for for
n 2 n j : 2
An=2 b21 n=2 5Sn=2
½Sn=2 ij 5dj;n=2
where we define:
The parity transformation matrix in the Bravyi–Kitaev basis can, therefore, be defined as:
This recursive definition clearly shows the logarithmic growth in locality of the operators in the Bravyi–Kitaev transformation, as the set is either the same size as the set for half the number of
qubits, or increases by one. Having determined this recursive relation for the update set, we obtain a similar expression for the parity set in the following section. Parity set The parity set for a
given orbital j is given by the nonzero entries in row j of the parity transformation matrix pn 5pn b21 n where pn is the parity transformation matrix in the occupation state basis, given by an upper
triangular matrix with zeroes along the diagonal. From the definitions for pn and b21 n , we obtain: 1438
Now, pn=2 Tn=2 50, because the last row and column of p is zero, and only this row and column contributes to the product. To evaluate An=2 b21 n=2 , we first prove the following fact 21 about b21 n=2
: every column of bn=2 has two nonzero entries except the last column, which has one. We can proceed by induction. It is true that b21 2 has two nonzero entries in each column except the last, which
has one. Following the recursive 21 construction, if it is true of b21 by n=2 , it will be true of bn inspection, as the addition of the single additional nonzero entry adds one more nonzero entry to
the last column of b21 n=2. Given this fact, it follows immediately that:
International Journal of Quantum Chemistry 2015, 115, 1431–1441
We can now define a recursive formula for the parity set as we did for the update set. For orbitals j < n2, the parity is determined by the submatrix pn2 matrix. For j n2, the parity set contains the
set PðjÞn=2 as the bottom quadrant is simply pn2. However, the column index of each element is increased by n2, as it is shifted to the right side of the matrix. For rows n2 through ðn21Þ, qubit n2
21 is added to the parity set for qubits n2 through n 2 1. We find that the parity set can be recursively expressed as: 8 > < Pn2 ðjÞ Pn ðjÞ5 n o > : Pn j2 n 1 n ; n 21 2 2 2 2
for for
n 2 n j 2 j<
Again, this recursive definition results in logarithmic growth in locality, as the set is again either the same size as the set WWW.CHEMISTRYVIEWS.ORG
for half the number of qubits, or increases by one. We now have a recursive expression for both the update and parity sets, and in the next section, we will find a recursive relation for the flip
set. Flip set The elements of the flip set for a given orbital j are defined by the indices of the nonzero entries in row j with indices > Fn2 ðjÞ > > > > 2 2 > > o n n n n > > > : Fn j2 1 ; 21 2 2 2
for j < for
n 2
n j < ðn21Þ 2
for j5n21
Just as in the cases of parity and update sets, this exhibits the logarithmic growth in locality of the operators. We now have a recursive expression for the update, parity, and flip sets given in
Eqs. (33), (40), and (39). These allow each set to be computed directly, rather than through various steps of matrix manipulation, greatly simplifying the process of computing each set. This also
connects the matrix definition of the Bravyi–Kitaev transformation given in [29] with the original definition given by Bravyi and Kitaev.[28]
A Numerical Example: Methane in STO6G While Trotterization and phase estimation for the quantum simulation of chemistry have been extensively studied and simulated, the sole example for Bravyi–Kitaev
is given in [29]. In that work, the minimal basis representation of H2 was studied. This is an example of considerable interest from the point of view of early experimental implementation of these
algorithms, but is too small to exhibit all the properties of the Bravyi–Kitaev transformation. To examine characteristics of the mapping in a larger system than previously studied, we examined the
performance of this technique for the determination of the ground-state energy of methane. The Hartree–Fock basis was used as our molecular orbital basis, and was determined through a Hartree–Fock
calculation using GAMESS,[40,41] with a geometric Td symmetry, a CH bond length of 1.107902, and a STO-6G atomic orbital basis. Spatial molecular orbital integrals were also obtained from this
calculation, and transformed into a spin-orbit basis in physicists’ notation. Python code was then used to generate Jordan–Wigner and Bravyi–Kitaev qubit Hamiltonians in terms of a symbolic
sequence of strings of Pauli operations. This code automatically combines duplicate strings. Each Pauli string is represented by a sequence of N numbers, where N is the number of qubits (i.e., spin
orbitals). Each number corresponds to the operation acting on its respective qubit—0 for the identity, 1 for Pauli X, 2 for Pauli Y, and 3 for Pauli Z. The numbers are ordered in reverse sequential
order—the qubit with highest index is operated on by the leftmost operator, and the 0 indexed qubit is operated on by the right-most operator. Consequentially, each term is represented by a base-4
number. The terms are then ordered lexiographically—in ascending order of these base-4 numbers. Having a symbolic expression of the Jordan–Wigner and Bravyi–Kitaev Hamiltonians, our code constructs
CSC sparse matrix representations of these using SciPy’s sparse matrix methods. It proceeds to diagonalize these Hamiltonians, providing both an exact ground-state eigenvalue (to compare against our
Trotterized eigenvalue estimate) and a groundstate eigenvector, which is needed for the following stage of our calculation. Note that in an experimental realization on a quantum computer the
ground-state eigenvector would be prepared by adiabatic-state preparation, or other methods.[7,33] Using the eigenvector obtained as an input, our python code simulates the effect of applying a
Trotterized unitary with specified Hamiltonian (as a sequence of strings of Pauli operations), Trotter-Suzuki approximation order, number of Trotter steps, and overall simulation time. Note that to
reduce computational cost substantially, the whole unitary evolution matrix is not determined. Each Pauli string term in the Hamiltonian is instead exponentiated symbolically. Every Pauli operation
in each exponentiated term is then directly implemented on the target state. With key functions compiled using Cython, this cuts computational resource requirements dramatically. Finally, our code
uses the original ground-state vector and the post-Trotter vector to assess the phase gained through phase estimation, and thus, an eigenvalue estimate as a function of our parameters. The details of
how a unitary given by the exponential of an arbitrary string of Pauli matrices is give in [29]. Our code also counts the gates associated with each Trotterization setting. We use the commonly used
gate set consisting of controlled not gates and arbitrary single qubit gates. We note that this gate set is appropriate for small-scale experimental implementation without error correction. For fault
tolerant implementations, a different gate set would be required, and these circuits could be obtained from those we developed with some overhead. Details of fault-tolerant implementations of these
kinds of simulation algorithms are given in [42]. These gate counts do not include any cancellation within the gate structure, that is, between sequential CNOT strings. These results are shown in
Figures 1 and 2, and demonstrate a small improvement associated with the Bravyi–Kitaev mapping. The number of CNOT gates to realize a single Trotter step, for either a first- or second-order Trotter
scheme, is always less for the Bravyi–Kitaev mapping. This reflects the increased locality of the mapping. The number of single qubit gates for the Bravyi–Kitaev mapping is higher for a single
Trotter step due to the changes of basis required by the more sophisticated International Journal of Quantum Chemistry 2015, 115, 1431–1441
tions of the Trotter ordering for Bravyi–Kitaev.[25] Finally, the impact of the reduced locality of the operators will depend in detail on the architecture used.
Figure 1. Energy as a function of number of total gates in the Trotterization. Blue-dashed line—Jordan–Wigner with first-order Trotterization. Reddotted line—Bravyi–Kitaev with first-order
Trotterization. Black-dashed line—Jordan–Wigner with second-order Trotterization. Orange-dashed line—Bravyi–Kitaev with second-order Trotterization. The energy is plotted as a difference from a
reference energy of 53.4096 a.u.
form of the creation and annihilation operators. However, the accuracy of the Bravyi–Kitaev method for a given number of gates is better than for the Jordan–Wigner method, as shown in Figures 1 and
2. These results compare gate sequences derived from both Bravyi–Kitaev and Jordan–Wigner transformations that are based on a lexicographic ordering of terms in the Hamiltonian. There is no reason to
believe that this is optimal in either case, but it is reassuring that the simplest comparison of the two methods gives results which are slightly better for the Bravyi–Kitaev transformation.
Naturally, the true costs of the two methods are only given by optimal Trotterizations. In the case of the Jordan–Wigner transformation such optimization has been performed recently.[23,24] However,
such optimization of the Bravyi–Kitaev transformation is the subject of current research. In this case, simple circuit optimizations are made more complex by the more complex strings of Pauli
operators that occur, but are also simpler due to the reduced locality of the transformation. The ordering of terms in the Trotterization also has a significant impact on the Trotter error, and new
work on understanding the chemical basis of these Trotter errors should also act as a guide to future optimiza-
In exploring the Bravyi–Kitaev transformation, we found new, recursive equations for the update, flip, and parity set given in Eqs. (33), (40), and (39), allowing these sets to be computed directly
rather than through various matrix operations. The recursive nature of these definitions underlies the logarithmic growth in locality of operators in the transformation. Such recursive approaches to
algorithm development often provide such improvements, as in the examples of the Fast Fourier Transform and quicksort algorithms. We presented a numerical example of the Bravyi–Kitaev mapping applied
to the methane molecule, observing a small improvement over the traditional Jordan–Wigner mapping. The Bravyi–Kitaev mapping appears to result in a small decrease in the total amount of gates
necessary to achieve an arbitrary precision approximation. However, a far more substantial drop is present for the amount of CNOT gates required. This emphasizes the increase in locality of the spin
Hamiltonian realized under the Bravyi–Kitaev mapping. The work presented here shows that the Bravyi–Kitaev method of quantum simulation of interacting fermionic systems can be systematically improved
in ways that both clarify the method theoretically and bring experimental realization of these simulations closer.
Acknowledgment This project is supported by NSF CCI center, Quantum Information for Quantum Chemistry (QIQC), award number CHE-1037992, by NSF award PHY-0955518 and by AFOSR award no FA9550-12-10046.
J.M. is supported by the DOE Computational Science Graduate Fellowship under grant number DE-FG02-97ER25308. PJL thanks the EPSRC and UCLQ for support through a UCLQ visiting fellowship, and AT
thanks EPSRC for a PhD Studentship from the Imperial CDT in Controlled Quantum Dynamics. Keywords: quantum chemistry quantum simulation quantum computing
How to cite this article: A. Tranter, S. Sofia, J. Seeley, M. Kaicher, J. McClean, R. Babbush, P. V. Coveney, F. Mintert, F. Wilhelm, P. J. Love. Int. J. Quantum Chem. 2015, 115, 1431– 1441. DOI:
Figure 2. Energy as a function of number of CNOT gates in the Trotterization. Blue-dashed line—Jordan–Wigner with first-order Trotterization. Reddotted line—Bravyi–Kitaev with first-order
Trotterization. Black-dashed line—Jordan–Wigner with second-order Trotterization. Orange-dashed line—Bravyi–Kitaev with second-order Trotterization. The energy is plotted as a difference from a
reference energy of 53.4096 a.u.
International Journal of Quantum Chemistry 2015, 115, 1431–1441
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Goggin, M. P. Almeida, I. Kassal, J. D. Biamonte, M. Mohseni, B. J. Powell, M. Barbieri, A. AspuruGuzik, A. G. White, Nat. Chem. 2010, 2, 106. [17] A. Peruzzo, J. McClean, P. Shadbolt, M.-H. Yung,
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Biamonte, R. Babbush, V. Bergholm, S. Yang, I. Jakobi, P. Neumann, A. Aspuru-Guzik, J. D. Whitfield, J. Wrachtrup, arXiv:1405.2696 [quant-ph], 2013. [20] R. Barends, L. Lamata, J. Kelly, L.
Garcia-Alvarez, A. G. Fowler, A. Megrant, E. Jeffrey, T. C. White, D. Sank, J. Y. Mutus, B. Campbell, Y. Chen, Z. Chen, B. Chiaro, A. Dunsworth, I.-C. Hoi, C. Neill, P. J. J. O’Malley, C. Quintana,
P. Roushan, A. Vainsencher, J. Wenner, E. Solano, J. M. Martinis, arXiv:1501.07703 [quant-ph], 2014. [21] P. J. Love, Adv. Chem. Phys. 2014, 154, 39. [22] D. Wecker, B. Bauer, B. K. Clark, M. B.
Hastings, M. Troyer, Phys. Rev. A. 2014, 90, 022305. [23] M. B. Hastings, D. Wecker, B. Bauer, M. Troyer, Quantum Inf. Comput. 2015, 15, 1.
[24] D. Poulin, M. B. Hastings, D. Wecker, N. Wiebe, A. C. Doherty, M. Troyer, arXiv:1406.4920 [quant-ph], 2014. [25] R. Babbush, J. McClean, D. Wecker, A. Aspuru-Guzik, N. Wiebe, Phys. Rev. A 2015,
91, 022311. [26] J. R. McClean, R. Babbush, P. J. Love, A. Aspuru-Guzik, J. Phys. Chem. Lett. 2014, 5, 4368. [27] P. Jordan, E. Wigner, Zeitschr. fur Phys. 1928, 47, 631. [28] S. Bravyi, A. Kitaev,
Ann. Phys. 2002, 298, 210. [29] J. T. Seeley, M. J. Richard, P. J. Love, J. Chem. Phys. 2012, 137, 224109. [30] A. Y. Kitaev, A. Shen, M. N. Vyalyi, Classical and Quantum Computation, vol. 47,
American Mathematical Society, 2002. [31] J. Kempe, A. Kitaev, O. Regev, SIAM J. Comput. 2006, 35, 1070. [32] A. M. Childs, D. Gosset, Z. Webb, In Proceedings of the 41st International Colloquium on
Automata, Languages, and Programming (ICALP 2014), 2014, pp. 308–319. [33] L. Veis, J. Pittner, J. Chem. Phys. 2014, 140. [34] A. Kitaev, W. A. Webb, arXiv preprint arXiv:0801.0342, 2008. [35] N. J.
Ward, I. Kassal, A. Aspuru-Guzik, J. Chem. Phys. 2009, 130, 194105. [36] P. Kaye, M. Mosca, arXiv preprint quant-ph/0407102, 2004. [37] S. Bravyi, arXiv preprint arXiv:1402.2295, 2014. [38] P.
Zanardi, Phys. Rev. A. 2002, 65, 042101. [39] G. Mussardo, Statistical Field Theory. Oxford University Press, 2010. [40] M. W. Schmidt, K. K. Baldridge, J. A. Boatz, S. T. Elbert, M. S. Gordon, J. H.
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Received: 10 April 2015 Revised: 14 May 2015 Accepted: 27 May 2015 Published online 1 July 2015
International Journal of Quantum Chemistry 2015, 115, 1431–1441 | {"url":"https://p.pdfkul.com/kitaev-transformation-wiley-online-library_5a1e00b21723dd74d92d4878.html","timestamp":"2024-11-07T15:52:56Z","content_type":"text/html","content_length":"109485","record_id":"<urn:uuid:726da0dd-d505-41ed-a06e-829792b6c6a2>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00835.warc.gz"} |
Pathway Lasso
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Pathway Lasso: Estimate and Select Multiple Mediation Pathways
Xi (Rossi) LUO
Brown University
Department of Biostatistics
Center for Statistical Sciences
Computation in Brain and Mind
Brown Institute for Brain Science
Brown Data Science Initiative
ABCD Research Group
IBC, Barcelona Spain
July 10, 2018
Funding: NIH R01EB022911, P20GM103645, P01AA019072, P30AI042853; NSF/DMS (BD2K) 1557467
Yi Zhao
Currently postdoc at Johns Hopkins Biostat
Slides viewable on web:
Motivating Example: Task fMRI
• Task fMRI: performs tasks under brain scanning
• Story vs Math task: listen to story (treatment stimulus) or math questions (control), eye closed
• Not resting-state: "rest" in scanner
Goal: how brain processes story/math differently?
fMRI data: blood-oxygen-level dependent (BOLD) signals from each cube/voxel (~millimeters), $10^5$ ~ $10^6$ voxels in total.
Conceptual Model with Stimulus
Sci Goal: quantify red, blue, and other pathways
from stimulus to orange outcome region activityHeim et al, 09
Other Potential Applications
• Genomics/genetics/proteomics
□ Multiple genetic pathways
• Integrating multiple sources
• Common theme: many potential pathways/mediators to disease outcomes
Mediation Analysis and SEM
• Indirect effect: $a \times b$; direct: $c$
• Mediation analysis
□ Baron&Kenny, 86; Sobel, 82; Holland 88; Preacher&Hayes 08; Imai et al, 10; VanderWeele, 15;...
Multiple (Full) Pathway Model Daniel et al, 14
• Stimulus $Z$, $K$ mediating brain regions $M_1, \dotsc, M_K$, Outcome region $R$
• Strength of activation ($a_k$) and connectivity ($b_k$, $d_{ij}$)
• Potential outcomes too complex, e.g. $K = 2$ Daniel et al, 14
Practical Considerations
• The previous model requires specifying the order of mediators, usually unknown in many experiments
□ We don't know yet the order of brain regions
□ fMRI: not enough temporal resolution to determine the order
• Theoretically and computationally challenging with a large number of mediators
□ High dimensional (large $p$, small $n$) setting: $K>n$
Mediation Analysis in fMRI
• Parametric Wager et al, 09 and functional Lindquist, 12 mediation, under (approx.) independent errors
□ Stimulus $\rightarrow$ brain $\rightarrow$ user reported ratings, one mediator
□ Usual assumption: $U=0$ and $\epsilon_1 \bot \epsilon_2$
• Parametric and multilevel mediation Yi and Luo, 15, with correlated errors for two brain regions
□ Stimulus $\rightarrow$ brain region A $\rightarrow$ brain region B, one mediator
□ Correlations between $\epsilon_1$ and $\epsilon_2$
• This talk: multiple mediator and multiple pathways
□ High dimensional: more mediators than sample size
□ Dimension reduction: optimization Chen et al, 15, testing Huang et al, 16
Our Reduced Pathway Model
• $A_k$: "total" effect of $Z$→$M_k$; $B_k$: $M_k$→R
• Pathway effect: $A_k \times B_k$; Direct: $C$
Two Models
• Proposition: Our "total-effect" parameters are linearly related or equivalent to the "individual-effect" parameters in the full model
□ $C=c$ and $B_k=b_k$, $k=1,\dotsc, K$, are the same in both models
□ $A_k$ and $a_k$ in two models are linearly related
□ $A_k \times B_k$ interpreted as the "total" effect when $M_k$ is the last mediator Imai & Yamamoto, 13
Additional Relation to Full Model
• Proposition: Our $E_k$'s are correlated, but won't affect point estimation consistency (affect variance)
□ The price of ignoring the order
• Related to causally independent mediators if assuming indepdent $E_k$
• Reduced model: a first step to select mediators
□ Strong overall inflow/outflow of a mediator
Causal Assumptions
• We impose standard causal mediation assumptions:
□ SUTVA
□ Model correctly specified
□ Observed is one reliazation of the potential outcomes
□ Randomized $Z$
□ No unmeasured confounding/sequential ignorability
• Similar assumptions discussed in Imai & Yamamoto, 13; Daniel et al, 14; VanderWeele, 15
• Could be too strong or sensitivity analysis Imai & Yamamoto, 13
Regularized Regression
• Minimize the penalized least squares criterion
$$\scriptsize \sum_{k=1}^K \| M_k - Z A_k \|_2^2 + \| R - Z C - \sum_k M_k B_k \|_2^2 + \mbox{Pen}(A, B)$$
The choice of penalty $\mbox{Pen}(\cdot)$ to be discussed
□ All data are normalized (mean=0, sd=1)
• Want to select sparse pathways for high-dim $K$
• Alternative approach: two-stage LASSO Tibshirani, 96 to select sparse $A_k$ and $B_k$ separately: $$ \scriptsize \sum_{k=1}^K \| M_k - Z A_k \|_2^2 + \lambda \sum_k | A_k | \\ \scriptsize \| R -
Z C - \sum_k M_k B_k \|_2^2 + \lambda \sum_k |B_k| $$
Penalty: Pathway LASSO
• Select strong pathways effects: $A_k \times B_k$
□ TS-LASSO: shrink to zero when $A$&$B$ moderate but $A\times B$ large
• Penalty (prototype) $$ \scriptsize \lambda \sum_{k=1}^K |A_k B_k| $$
□ Non-convex in $A_k$ and $B_k$
□ Computationally heavy and non-unique solutions
□ Hard to prove theory
• We propose the following general class of penalties$$ \scriptsize \lambda \sum_{k=1}^K ( |A_k B_k| + \phi A_k^2 + \phi B_k^2) $$
Theorem $$v(a,b) = |a b| + \phi (a^2 + b^2)$$ is convex if and only if $\phi\ge 1/2$. Strictly convex if $\phi > 1/2$.
Contour Plot of Different Penalties
• Non-differentiable at points when $a\times b = 0$
• Shrink $a\times b$ to zero
• Special cases: $\ell_1$ or $\ell_2$
• TS-LASSO: different $|ab|$ effects though $|a|+|b|$ same
$|ab|+\phi (a^2 + b^2)$
Pathway Lasso
$|a| + |b|$
Two-stage Lasso
Pathway Lasso is a family of (convex) penalties for products
Algorithm: ADMM + AL
• SEM/regression loss: $u$; Non-differnetiable penalty: $v$
• ADMM to address differentiability $$ \begin{aligned} \text{minimize} \quad & u(\Theta,D)+v(\alpha,\beta) \\ \text{subject to} \quad & \Theta=\alpha, \\ & D=\beta, \\ & \Theta e_{1}=1, \end
• Augmented Lagrangian for multiple constraints
• Iteratively update the parameters
• We derive theorem on explicit (not simple) updates
Asymptotic Theory
Theorem: Under regularity conditions,
$$ \Ex (Z \sum_k \hat{A}_k\hat{B}_k - Z \sum_k {A}^*_k {B}^*_k )^2 \le O(s \kappa \sigma n ^{-1/2} (\log K)^{1/2}), $$ where $s = \#\{j: B_j^* \ne 0\}$ and $\kappa =\max_j |B_j| $. With high
$$ \| \hat{A} \hat{B} - A^* B^* \| \le O(s \kappa \sigma n ^{-1/2} (\log K)^{1/2})$$
• Mixed norm penalty $$\mbox{PathLasso} + \omega \sum_k (|A_k| + |B_k|)$$
• Tuning parameter selection by cross validation
□ Reduce false positives via thresholding Johnston and Lu, 09
• Inference/CI: bootstrap after refitting
□ Remove false positives with CIs covering zero Bunea et al, 10
• Our PathLasso compares with TSLasso
• Simulate with varying error correlations
• Tuning-free comparison: performance vs tuning parameter (estimated effect size)
□ PathLasso outperforms under CV
Pathway Recovery
Our PathLasso (red) outperforms two-stage Lasso (blue)
Other curves: variants of PathLasso and correlation settings
Data: Human Connectome Project
• Two sessions (LR/RL), story/math task Binder et al, 11
• gICA reduces voxel dimensions to 76 brain maps
□ ROIs/clusters after thresholding
• Apply to two sess separately, compare replicability
□ Jaccard: whether selected pathways in two runs overlap
□ $\ell_2$ diff: difference between estimated path effects
• Tuning-free comparisons
Regardless of tuning, our PathLasso (red) has smaller replication diff (selection and estimation) than TSLasso (blue)
Stim-M25-R and Stim-M65-R significant shown largest weight areas
• M65 responsible for language processing, larger flow under story
• M25 responsible for uncertainty, larger flow under math
• High dimensional pathway model
• Penalized SEM for pathway selection and estimation
• Convex optimization for non-convex products
□ Sufficient and necessary condition
□ Algorithmic development for complex optimization
• Improved estimation and selection accuracy
□ Higher replicability using HCP data
• Manuscript: Pathway Lasso (arXiv 1603.07749)
• Limitations: causal assumptions, covaraites, interactions, error correlations Rpkg: macc, time series gma, functional cfma
Thank you!
Comments? Questions?
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Dissecting power of intersection of two context-free languages
Dissecting power of intersection of two context-free languagesArticle
We say that a language $L$ is \emph{constantly growing} if there is a constant $c$ such that for every word $u\in L$ there is a word $v\in L$ with $\vert u\vert<\vert v\vert\leq c+\vert u\vert$. We
say that a language $L$ is \emph{geometrically growing} if there is a constant $c$ such that for every word $u\in L$ there is a word $v\in L$ with $\vert u\vert<\vert v\vert\leq c\vert u\vert$. Given
two infinite languages $L_1,L_2$, we say that $L_1$ \emph{dissects} $L_2$ if $\vert L_2\setminus L_1\vert=\infty$ and $\vert L_1\cap L_2\vert=\infty$. In 2013, it was shown that for every constantly
growing language $L$ there is a regular language $R$ such that $R$ dissects $L$. In the current article we show how to dissect a geometrically growing language by a homomorphic image of intersection
of two context-free languages. Consider three alphabets $\Gamma$, $\Sigma$, and $\Theta$ such that $\vert \Sigma\vert=1$ and $\vert \Theta\vert=4$. We prove that there are context-free languages
$M_1,M_2\subseteq \Theta^*$, an erasing alphabetical homomorphism $\pi:\Theta^*\rightarrow \Sigma^*$, and a nonerasing alphabetical homomorphism $\varphi : \Gamma^*\rightarrow \Sigma^*$ such that: If
$L\subseteq \Gamma^*$ is a geometrically growing language then there is a regular language $R\subseteq \Theta^*$ such that $\varphi^{-1}\left(\pi\left(R\cap M_1\cap M_2\right)\right)$ dissects the
language $L$.
Volume: vol. 25:2
Section: Automata, Logic and Semantics
Published on: October 2, 2023
Accepted on: July 6, 2023
Submitted on: February 8, 2022
Keywords: Computer Science - Formal Languages and Automata Theory,Computer Science - Discrete Mathematics,68Q45 | {"url":"https://dmtcs.episciences.org/12293","timestamp":"2024-11-09T06:04:49Z","content_type":"application/xhtml+xml","content_length":"51882","record_id":"<urn:uuid:18edab38-45be-4433-afe2-7fcbe02db585>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.30/warc/CC-MAIN-20241109053958-20241109083958-00572.warc.gz"} |
Baltimore City Community College Algebra Worksheet - Custom Scholars
Baltimore City Community College Algebra Worksheet
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Consider the parabola given by the equation: f(x)
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C) Find the coordinates of the two z intercepts of the parabola and write them as a list, separated by
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distance in feet from the point at which the ball is thrown. Files Calendar Syllabus feet (a) How high is the ball when it leaves the child's hand? IⓇ O o Modules (b) What is the maximum height of
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will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle
when A is a minimum? Courses Pages Files The circumference of the circle is cm. Calendar Syllabus Give your answer to two decimal places IⓇ O o Modules Inbox Submit Question Collaborations Help
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A baseball team plays in a stadium that holds 72000 spectators. With the ticket price at $10 the average
attendance has been 29000. When the price dropped to $8, the average attendance rose to 36000. Assume
that attendance is linearly related to ticket price.
What ticket price would maximize revenue? $
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Curriculum Online
After three years at school, students will be achieving at early level 2 in the mathematics and statistics learning area of The New Zealand Curriculum.
Number and algebra | Geometry and measurement | Statistics
The following problems and descriptions of student thinking exemplify what is required to meet this standard.
Number and algebra
In contexts that require them to solve problems or model situations, students will be able to:
• apply basic addition facts and knowledge of place value and symmetry to:
- combine or partition whole numbers
- find fractions of sets, shapes, and quantities
• create and continue sequential patterns with one or two variables by identifying the unit of repeat
• continue spatial patterns and number patterns based on simple addition or subtraction.
During this school year, 'number' should be the focus of 60–80 percent of mathematics teaching time.
Example 1
You have 18 turtles, and you buy another 8 turtles from the pet shop.
How many turtles do you have now?
The student could use 'making tens' (for example, '18 + 2 = 20; that leaves 6 remaining from the 8; 20 + 6 = 26') or apply their knowledge of doubles and place value (for example, '18 = 10 + 8; first
add the 8, then the 10; 8 + 8 = 16, 16 + 10 = 26').
If the student responds very quickly because they know the fact 18 + 8 = 26, this also meets the expectation. If the student counts on, they do not meet the expectation.
Example 2
87 people are at the pōwhiri (welcome). 30 of the people are tangata whenua (locals). The rest of the people are manuhiri (visitors).
How many manuhiri are there?
The student uses place value knowledge, combined with either addition or subtraction, to solve the problem. They may add on (30 + 50 = 80, 80 + 7 = 87) or subtract (80 – 30 = 50, so 87 – 30 = 57). If
they use counting up or back in tens (for example, 40, 50, 60, 70, 80, 87), they do not meet the expectation.
If they use a pencil and paper method to subtract 0 from 7 and 3 from 8, this doesn’t necessarily demonstrate enough understanding of place value to meet the expectation. If they use this method,
they must show that they understand the place value of the digits and that they are not treating them all as ones.
Example 3
Here is a string of 12 sausages to feed 3 hungry dogs.
Each dog should get the same number of sausages. How many will each dog get?
The student applies basic addition facts to share out the sausages equally between the dogs. Their thinking could be based on doubles or equal dealing – for example, 5 + 5 + 2 = 12, so 4 + 4 + 4 = 12
(redistributing 1 from each 5), or 6 + 6 = 12, so 4 + 4 + 4 = 12, or 2 + 2 + 2 = 6, so 4 + 4 + 4 = 12.
If the student solves the problem by one-to-one equal sharing, they do not meet the expectation. If they solve the problem using multiplication facts (3 x 4 = 12 or 12 ÷ 3 = 4), they exceed the
Example 4
Show the student the illustration below. What shape goes on the number 14 in this pattern? What colour will it be?
The student identifies the two variables (shape and colour) in the pattern. They might look at the variables separately and identify the unit of repeat for each ('Yellow, blue, red' and 'Triangle,
circle'). Or they might look at the variables together to identify the complete unit of repeat ('Yellow triangle, blue circle, red triangle, yellow circle, blue triangle, red circle').
They continue the pattern until they identify that the shape on number 14 is a blue circle. If the student recognises that multiples of 2 in the pattern are circles and multiples of 3 are red and
uses this information to solve the problem, they exceed the expectation.
Geometry and measurement
In contexts that require them to solve problems or model situations, students will be able to:
• measure the lengths, areas, volumes or capacities, and weights of objects and the duration of events, using linear whole-number scales and applying basic addition facts to standard units
• sort objects and two- and three-dimensional shapes by their features, identifying categories within categories
• represent reflections, translations, and rotations by creating and describing patterns
• describe personal locations and give directions, using whole-number measures and half- or quarter-turns.
Example 5
Give the student 3 pencils of different lengths and a ruler.
Use the ruler to find the length of each pencil.
How much longer is the green pencil than the red pencil?
The student correctly measures the length of each pencil to the nearest centimetre: they align the end of the pencil with zero on the scale and read off the measure correctly.
They apply basic addition facts to find the difference in length between the green and red pencils (for example, for 12 centimetres and 9 centimetres: '3 centimetres, because 10 + 2 = 12, so 9 + 3 =
12'; or '3 centimetres, because I know 9 + 3 = 12').
Example 6
Give the student a circle of paper. Fold this circle into 8 equal-sized pieces.
The student uses reflective symmetry through repeated halving to partition the circle into eighths.
Example 7
Give the student a metre ruler or tape measure and show them the illustrations below.
Write a set of instructions to explain to a visitor how to get from the library door to our classroom door. Make sure you include any right or left turns and distances in metres. You can use pictures
to give the instructions, like this:
You can also use pictures or descriptions of objects such as buildings or trees.
The student provides a set of instructions that are accurate enough for a visitor to find their way to the classroom door from the library. If the student specifies compass directions or clockwise or
anti-clockwise turns, they exceed the expectation.
In contexts that require them to solve problems or model situations, students will be able to:
• investigate questions by using the statistical enquiry cycle (with support):
- gather and display category and simple whole-number data
- interpret displays in context
• compare and explain the likelihoods of outcomes for a simple situation involving chance.
Example 8
Each student writes the number of people that usually live in their house on a square of paper or a sticker.
How many people live in the houses of students in our class?
Arrange the squares to find out.
What can you say about your arrangement?
The student sorts the whole-number data into groups. They may display the data in enclosed groupings or in a more organised display, such as a bar graph.
The student makes a statement about the number of people living in students’ houses, based on their sorting of the data, for example, 'There are lots of different numbers of people living in houses,
from 2 to 9' or '5 is the most common number of people'.
Example 9
Let the student watch as you put 3 blue cubes, 2 yellow cubes, and a red cube into a paper bag.
Put your hand in the bag and take out a cube, but don’t look at it.
What colour is the cube most likely to be? What colour is it least likely to be?
Explain why.
The student classifies the probability of getting each colour ('Blue is most likely, and red is least likely'). They discuss the numbers and colours of cubes to explain their answer (for example,
'There are 3 blue cubes and only 1 red cube').
If the student gives the probabilities as fractions (for example, 'There is a one-half chance of blue'), they exceed the expectation. If they explain the likelihoods without reference to the number
of cubes (for example, 'Yellow is my lucky colour' or 'I always get red'), they do not meet the expectation.
Published on: 15 Oct 2009 | {"url":"https://nzcurriculum.tki.org.nz/Archives/Assessment/Mathematics-standards/The-standards/After-three-years","timestamp":"2024-11-12T20:01:44Z","content_type":"application/xhtml+xml","content_length":"278609","record_id":"<urn:uuid:96b64153-b9bd-403a-8b1b-781d89d97314>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00508.warc.gz"} |
Find Index of Maximum in List Using Python
In Python, we can easily find the index of the maximum in a list of numbers. The easiest way to get the index of the maximum in a list with a loop.
list_of_numbers = [2, 1, 9, 8, 6, 3, 1, 0, 4, 5]
def maxIndex(lst):
index = [0]
max = lst[0]
for i in range(1,len(lst)):
if lst[i] > max:
max = lst[i]
index = [i]
elif lst[i] == max:
return index
You can also use enumerate() and list comprehension to get the index of the maximum of a list in Python.
list_of_numbers = [2, 1, 9, 8, 6, 3, 1, 0, 4, 5]
max_of_list = max(list_of_numbers)
index = [i for i, j in enumerate(list_of_numbers) if j == max_of_list]
There are many powerful built-in functions in the Python language which allow us to perform basic or complex tasks easily.
One such task is to find the maximum of a list of numbers.
However, what if you want to find the index of the maximum?
We can easily find the index of the maximum value in a list using a loop in Python.
To find the index of the maximum of a list, we just need to keep track of the maximum and the index of the maximum, and loop over the list of numbers.
Below is an example of a Python function which gets the index of the maximum of a list of numbers.
list_of_numbers = [2, 1, 9, 8, 6, 3, 1, 0, 4, 5]
def maxIndex(lst):
index = [0]
max = lst[0]
for i in range(1,len(lst)):
if lst[i] > max:
max = lst[i]
index = [i]
elif lst[i] == max:
return index
If you are using pandas, you can find the index of the maximum of a column or DataFrame with the pandas idxmax() function.
Using enumerate() and List Comprehension to Find Index of Maximum of List in Python
Another way we can get the position of the maximum of a list in Python is to use list comprehension and the enumerate() function.
The enumerate() function takes in a collection of elements and returns the values and indices of each item of the collection.
We can get use enumerate() and list comprehension to get the index of the maximum of a list by returning the index where any value is equal to the maximum of the list.
Below is how to use enumerate() and list comprehension to get the index of the maximum of a list in Python.
list_of_numbers = [2, 1, 9, 8, 6, 3, 1, 0, 4, 5]
max_of_list = max(list_of_numbers)
index = [i for i, j in enumerate(list_of_numbers) if j == max_of_list]
Find Index of Maximum in List with Python
We can adjust the function which finds the index of the maximum in a list to find the index of the minimum in a list.
To find the index of the minimum of a list, we just need to keep track of the minimum and the index of the minimum, and loop over the list of numbers.
Below is an example of a Python function which gets the index of the minimum of a list of numbers.
list_of_numbers = [2, 1, 9, 8, 6, 3, 1, 0, 4, 5]
def minIndex(lst):
index = [0]
min = lst[0]
for i in range(1,len(lst)):
if lst[i] < min:
min = lst[i]
index = [i]
elif lst[i] == min:
return index
If you are using pandas, you can find the index of the minimum of a column or DataFrame with the pandas idxmin() function.
Hopefully this article has been useful for you to understand how to find the index of the maximum of a list in Python. | {"url":"https://daztech.com/python-find-index-of-maximum-in-list/","timestamp":"2024-11-07T12:07:56Z","content_type":"text/html","content_length":"248091","record_id":"<urn:uuid:40b9ea05-3522-42ee-b0f0-a4ccead3baf8>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00889.warc.gz"} |
5 Genetic Algorithm Applications Using PyGAD | DigitalOcean
This tutorial introduces PyGAD, an open-source Python library for implementing the genetic algorithm and training machine learning algorithms. PyGAD supports 19 parameters for customizing the genetic
algorithm for various applications.
Within this tutorial we’ll discuss 5 different applications of the genetic algorithm and build them using PyGAD.
The outline of the tutorial is as follows:
• PyGAD Installation
• Getting Started with PyGAD
• Fitting a Linear Model
• Reproducing Images
• 8 Queen Puzzle
• Training Neural Networks
• Training Convolutional Neural Networks
• Python: Basic understanding of Python programming.
• Deep Learning: Familiarity with neural networks, particularly CNNs and object detection.
PyGAD is available through PyPI (Python Package Index) and thus it can be installed simply using pip. For Windows, simply use the following command:
pip install pygad
For Mac/Linux, use pip3 instead of pip in the terminal command:
pip3 install pygad
Then make sure the library is installed by importing it from the Python shell:
import pygad
The latest PyGAD version is currently 2.3.2, which was released on June 1st 2020. Using the __version__ special variable, the current version can be returned.
import pygad
Now that PyGAD is installed, let’s cover a brief introduction to PyGAD.
The main goal of PyGAD is to provide a simple implementation of the genetic algorithm. It offers a range of parameters that allow the user to customize the genetic algorithm for a wide range of
applications. Five such applications are discussed in this tutorial.
The full documentation of PyGAD is available at Read the Docs. Here we’ll cover a more digestible breakdown of the library.
In PyGAD 2.3.2 there are 5 modules:
1. pygad: The main module comes already imported.
2. pygad.nn: For implementing neural networks.
3. pygad.gann: For training neural networks using the genetic algorithm.
4. pygad.cnn: For implementing convolutional neural networks.
5. pygad.gacnn: For training convolutional neural networks using the genetic algorithm.
Each module has its own repository on GitHub, linked below.
The main module of the library is named pygad. This module has a single class named GA. Just create an instance of the pygad.GA class to use the genetic algorithm.
The steps to use the pygad module are:
1. Create the fitness function.
2. Prepare the necessary parameters for the pygad.GA class.
3. Create an instance of the pygad.GA class.
4. Run the genetic algorithm.
In PyGAD 2.3.2, the constructor of the pygad.GA class has 19 parameters, of which 16 are optional. The three required parameters are:
1. num_generations: Number of generations.
2. num_parents_mating: Number of solutions to be selected as parents.
3. fitness_func: The fitness function that calculates the fitness value for the solutions.
The fitness_func parameter is what allows the genetic algorithm to be customized for different problems. This parameter accepts a user-defined function that calculates the fitness value for a single
solution. This takes two additional parameters: the solution, and its index within the population.
Let’s see an example to make this clearer. Assume there is a population with 3 solutions, as given below.
[221, 342, 213]
[675, 32, 242]
[452, 23, -212]
The assigned function to the fitness_func parameter must return a single number representing the fitness of each solution. Here is an example that returns the sum of the solution.
def fitness_function(solution, solution_idx):
return sum(solution)
The fitness values for the 3 solutions are then:
1. 776
2. 949
3. 263
The parents are selected based on such fitness values. The higher the fitness value, the better the solution.
For the complete list of parameters in the pygad.GA class constructor, check out this page.
After creating an instance of the pygad.GA class, the next step is to call the run() method which goes through the generations that evolve the solutions.
import pygad
ga_instance = pygad.GA(...)
These are the essential steps for using PyGAD. Of course there are additional steps that can be taken as well, but this is the minimum needed.
The next sections discuss using PyGAD for several different use cases.
Assume there is an equation with 6 inputs, 1 output, and 6 parameters, as follows:
y = f(w1:w6) = w1x1 + w2x2 + w3x3 + w4x4 + w5x5 + 6wx6
Let’s assume that the inputs are (4,-2,3.5,5,-11,-4.7) and the output is 44. What are the values for the 6 parameters to satisfy the equation? The genetic algorithm can be used to find the answer.
The first thing to do is to prepare the fitness function as given below. It calculates the sum of products between each input and its corresponding parameter. The absolute difference between the
desired output and the sum of products is calculated. Because the fitness function must be a maximization function, the returned fitness is equal to 1.0/difference. The solutions with the highest
fitness values are selected as parents.
function_inputs = [4,-2,3.5,5,-11,-4.7] # Function inputs.
desired_output = 44 # Function output.
def fitness_func(solution, solution_idx):
output = numpy.sum(solution*function_inputs)
fitness = 1.0 / numpy.abs(output - desired_output)
return fitness
Now that we’ve prepared the fitness function, here’s a list with other important parameters.
sol_per_pop = 50
num_genes = len(function_inputs)
init_range_low = -2
init_range_high = 5
mutation_percent_genes = 1
You should also specify your desired mandatory parameters as you see fit. After the necessary parameters are prepared, the pygad.GA class is instantiated. For information about each of the
parameters, refer to this page.
ga_instance = pygad.GA(num_generations=num_generations,
The next step is to call the run() method which starts the generations.
After the run() method completes, the plot_result() method can be used to show the fitness values over the generations.
Using the best_solution() method we can also retrieve what the best solution was, its fitness, and its index within the population.
solution, solution_fitness, solution_idx = ga_instance.best_solution()
print("Parameters of the best solution : {solution}".format(solution=solution))
print("Fitness value of the best solution = {solution_fitness}".format(solution_fitness=solution_fitness))
print("Index of the best solution : {solution_idx}".format(solution_idx=solution_idx))
In this application we’ll start from a random image (random pixel values), then evolve the value of each pixel using the genetic algorithm.
The tricky part of this application is that an image is 2D or 3D, and the genetic algorithm expects the solutions to be 1D vectors. To tackle this issue we’ll use the img2chromosome() function
defined below to convert an image to a 1D vector.
def img2chromosome(img_arr):
return numpy.reshape(a=img_arr, newshape=(functools.reduce(operator.mul, img_arr.shape)))
The chromosome2img() function (below) can then be used to restore the 2D or 3D image back from the vector.
def chromosome2img(vector, shape):
# Check if the vector can be reshaped according to the specified shape.
if len(vector) != functools.reduce(operator.mul, shape):
raise ValueError("A vector of length {vector_length} into an array of shape {shape}.".format(vector_length=len(vector), shape=shape))
return numpy.reshape(a=vector, newshape=shape)
Besides the regular steps for using PyGAD, we’ll need one additional step to read the image.
import imageio
import numpy
target_im = imageio.imread('fruit.jpg')
target_im = numpy.asarray(target_im/255, dtype=numpy.float)
This sample image can be downloaded here.
Next, the fitness function is prepared. This will calculate the difference between the pixels in the solution and the target images. To make it a maximization function, the difference is subtracted
from the sum of all pixels in the target image.
target_chromosome = gari.img2chromosome(target_im)
def fitness_fun(solution, solution_idx):
fitness = numpy.sum(numpy.abs(target_chromosome-solution))
# Negating the fitness value to make it increasing rather than decreasing.
fitness = numpy.sum(target_chromosome) - fitness
return fitness
The next step is to create an instance of the pygad.GA class, as shown below. It is critical to the success of the application to use appropriate parameters. If the range of pixel values in the
target image is 0 to 255, then the init_range_low and init_range_high must be set to 0 and 255, respectively. The reason is to initialize the population with images of the same data type as the
target image. If the image pixel values range from 0 to 1, then the two parameters must be set to 0 and 1, respectively.
import pygad
ga_instance = pygad.GA(num_generations=20000,
When the mutation_type argument is set to random, then the default behavior is to add a random value to each gene selected for mutation. This random value is selected from the range specified by the
random_mutation_min_val and random_mutation_max_val parameters.
Assume the range of pixel values is 0 to 1. If a pixel has the value 0.9 and a random value of 0.3 is generated, then the new pixel value is 1.2. Because the pixel values must fall within the 0 to 1
range, the new pixel value is therefore invalid. To work around this issue, it is very important to set the mutation_by_replacement parameter to True. This causes the random value to replace the
current pixel rather than being added to the pixel.
After the parameters are prepared, then the genetic algorithm can run.
The plot_result() method can be used to show how the fitness value evolves by generation.
After the generations complete, some information can be returned about the best solution.
solution, solution_fitness, solution_idx = ga_instance.best_solution()
print("Fitness value of the best solution = {solution_fitness}".format(solution_fitness=solution_fitness))
print("Index of the best solution : {solution_idx}".format(solution_idx=solution_idx))
The best solution can be converted into an image to be displayed.
import matplotlib.pyplot
result = gari.chromosome2img(solution, target_im.shape)
Here is the result.
The 8 Queen Puzzle involves 8 chess queens distributed across an 8×8 matrix, with one queen per row. The goal is to place these queens such that no queen can attack another one vertically,
horizontally, or diagonally. The genetic algorithm can be used to find a solution that satisfies such conditions.
This project is available on GitHub. It has a GUI built using Kivy that shows an 8×8 matrix, as shown in the next figure.
The GUI has three buttons at the bottom of the screen. The function of these buttons are as follows:
• The Initial Population button creates the initial population of the GA.
• The Show Best Solution button shows the best solution from the last generation the GA stopped at.
• The Start GA button starts the GA iterations/generations.
To use this project start by pressing the Initial Population button, followed by the Start GA button. Below is the method called by the Initial Population button which, as you might have guessed,
generates the initial population.
def initialize_population(self, *args):
self.num_solutions = 10
self.population_1D_vector = numpy.zeros(shape=(self.num_solutions, 8))
for solution_idx in range(self.num_solutions):
initial_queens_y_indices = numpy.random.rand(8)*8
initial_queens_y_indices = initial_queens_y_indices.astype(numpy.uint8)
self.population_1D_vector[solution_idx, :] = initial_queens_y_indices
self.pop_created = 1
self.num_attacks_Label.text = "Initial Population Created."
Each solution in the population is a vector with 8 elements referring to the column indices of the 8 queens. To show the queens’ locations on the screen, the 1D vector is converted into a 2D matrix
using the vector_to_matrix() method. The next figure shows the queens on the screen.
Now that the GUI is built, we’ll build and run the genetic algorithm using PyGAD.
The fitness function used in this project is given below. It simply calculates the number of attacks that can be made by each of the 8 queens and returns this as the fitness value.
def fitness(solution_vector, solution_idx):
if solution_vector.ndim == 2:
solution = solution_vector
solution = numpy.zeros(shape=(8, 8))
row_idx = 0
for col_idx in solution_vector:
solution[row_idx, int(col_idx)] = 1
row_idx = row_idx + 1
total_num_attacks_column = attacks_column(solution)
total_num_attacks_diagonal = attacks_diagonal(solution)
total_num_attacks = total_num_attacks_column + total_num_attacks_diagonal
if total_num_attacks == 0:
total_num_attacks = 1.1 # float("inf")
total_num_attacks = 1.0/total_num_attacks
return total_num_attacks
By pressing the Start GA button, an instance of the pygad.GA class is created and the run() method is called.
ga_instance = pygad.GA(num_generations=500,
Here is a possible solution in which the 8 queens are placed on the board where no queen attacks another.
The complete code for this project can be found on GitHub.
Among other types of machine learning algorithms, the genetic algorithm can be used to train neural networks. PyGAD supports training neural networks and, in particular, convolutional neural
networks, by using the pygad.gann.GANN and pygad.gacnn.GACNN modules. This section discusses how to use the pygad.gann.GANN module for training neural networks for a classification problem.
Before building the genetic algorithm, the training data is prepared. This example builds a network that simulates the XOR logic gate.
# Preparing the NumPy array of the inputs.
data_inputs = numpy.array([[1, 1],
[1, 0],
[0, 1],
[0, 0]])
# Preparing the NumPy array of the outputs.
data_outputs = numpy.array([0,
The next step is to create an instance of the pygad.gann.GANN class. This class builds a population of neural networks that all have the same architecture.
num_inputs = data_inputs.shape[1]
num_classes = 2
num_solutions = 6
GANN_instance = pygad.gann.GANN(num_solutions=num_solutions,
After creating the instance of the pygad.gann.GANN class, the next step is to create the fitness function. This returns the classification accuracy for the passed solution.
import pygad.nn
import pygad.gann
def fitness_func(solution, sol_idx):
global GANN_instance, data_inputs, data_outputs
predictions = pygad.nn.predict(last_layer=GANN_instance.population_networks[sol_idx],
correct_predictions = numpy.where(predictions == data_outputs)[0].size
solution_fitness = (correct_predictions/data_outputs.size)*100
return solution_fitness
Besides the fitness function, the other necessary parameters are prepared which we discussed previously.
population_vectors = pygad.gann.population_as_vectors(population_networks=GANN_instance.population_networks)
initial_population = population_vectors.copy()
num_parents_mating = 4
num_generations = 500
mutation_percent_genes = 5
parent_selection_type = "sss"
crossover_type = "single_point"
mutation_type = "random"
keep_parents = 1
init_range_low = -2
init_range_high = 5
After all parameters are prepared, an instance of the pygad.GA class is created.
ga_instance = pygad.GA(num_generations=num_generations,
The callback_generation parameter refers to a function that is called after each generation. In this application, this function is used to update the weights of all the neural networks after each
def callback_generation(ga_instance):
global GANN_instance
population_matrices = pygad.gann.population_as_matrices(population_networks=GANN_instance.population_networks, population_vectors=ga_instance.population)
The next step is to call the run() method.
After the run() method completes, the next figure shows how the fitness value evolved. The figure shows that a classification accuracy of 100% is reached.
Similar to training multilayer perceptrons, PyGAD supports training convolutional neural networks using the genetic algorithm.
The first step is to prepare the training data. The data can be downloaded from these links:
1. dataset_inputs.npy: Data inputs.
2. dataset_outputs.npy: Class labels.
import numpy
train_inputs = numpy.load("dataset_inputs.npy")
train_outputs = numpy.load("dataset_outputs.npy")
The next step is to build the CNN architecture using the pygad.cnn module.
import pygad.cnn
input_layer = pygad.cnn.Input2D(input_shape=(80, 80, 3))
conv_layer = pygad.cnn.Conv2D(num_filters=2,
average_pooling_layer = pygad.cnn.AveragePooling2D(pool_size=5,
flatten_layer = pygad.cnn.Flatten(previous_layer=average_pooling_layer)
dense_layer = pygad.cnn.Dense(num_neurons=4,
After the layers in the network are stacked, a model is created.
model = pygad.cnn.Model(last_layer=dense_layer,
Using the summary() method, a summary of the model architecture is returned.
----------Network Architecture----------
<class 'cnn.Conv2D'>
<class 'cnn.AveragePooling2D'>
<class 'cnn.Flatten'>
<class 'cnn.Dense'>
After the model is prepared, the pygad.gacnn.GACNN class is instantiated to create the initial population. All the networks have the same architecture.
import pygad.gacnn
GACNN_instance = pygad.gacnn.GACNN(model=model,
The next step is to prepare the fitness function. This calculates the classification accuracy for the passed solution.
def fitness_func(solution, sol_idx):
global GACNN_instance, data_inputs, data_outputs
predictions = GACNN_instance.population_networks[sol_idx].predict(data_inputs=data_inputs)
correct_predictions = numpy.where(predictions == data_outputs)[0].size
solution_fitness = (correct_predictions/data_outputs.size)*100
return solution_fitness
The other parameters are also prepared.
population_vectors = pygad.gacnn.population_as_vectors(population_networks=GACNN_instance.population_networks)
initial_population = population_vectors.copy()
num_parents_mating = 2
num_generations = 10
mutation_percent_genes = 0.1
parent_selection_type = "sss"
crossover_type = "single_point"
mutation_type = "random"
keep_parents = -1
After all parameters are prepared, an instance of the pygad.GA class is created.
ga_instance = pygad.GA(num_generations=num_generations,
The callback_generation parameter is used to update the network weights after each generation.
def callback_generation(ga_instance):
global GACNN_instance, last_fitness
population_matrices = pygad.gacnn.population_as_matrices(population_networks=GACNN_instance.population_networks, population_vectors=ga_instance.population)
The last step is to call the run() method.
This tutorial introduced PyGAD, an open-source Python library for implementing the genetic algorithm. The library supports a number of parameters to customize the genetic algorithm for a number of
In this tutorial we used PyGAD to build 5 different applications including fitting a linear model, solving the 8 queens puzzle, reproducing images, and training neural networks (both conventional and
convolutional). I hope you found this tutorial useful, and please feel free reach out in the comments or check out the docs if you have any questions!
This textbox defaults to using Markdown to format your answer.
You can type !ref in this text area to quickly search our full set of tutorials, documentation & marketplace offerings and insert the link! | {"url":"https://www.digitalocean.com/community/tutorials/genetic-algorithm-applications-using-pygad","timestamp":"2024-11-10T02:49:39Z","content_type":"text/html","content_length":"378272","record_id":"<urn:uuid:78ad2366-8c4e-462f-8c32-6abf2d71465f>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00023.warc.gz"} |
Anita Whittingham - Maths How To with Anita
Let’s find the prime factors of 8 by first finding the factors of 8. What are the factors of 8? The factors of 8 are 1, 2, 4 & 8. These are numbers that 8 can be divided by exactly with no remainder.
Out of these factors, the only number that is a prime number […]
What are the prime factors of 8? Read More »
What are the prime factors of 28?
Once you learn how to find the prime factors of 28 you can use these skills to find the prime factors of other numbers. As a high school mathematics teacher for over 14 years I have seen first hand
that the students who have mastered the skill of factoring are able to apply this to
What are the prime factors of 28? Read More »
What are the prime factors of 100?
To find the prime factors of 100 first you need to find the factors of 100, then identify which factors are prime numbers. This article will show you how to apply these 2 skills to this problem. As a
high school mathematics teacher for over 14 years, I have noticed that my students who have
What are the prime factors of 100? Read More »
How to find Prime Numbers from 1 to 500 (includes chart, free printable worksheet and video)
Do you know what the prime numbers are between 1 and 500? If not, don’t worry! I’m here to help. In this article, I will list all of the prime numbers from 1 and 500, along with steps and a video on
how to find them. Prime numbers are special because they are only divisible
How to find Prime Numbers from 1 to 500 (includes chart, free printable worksheet and video) Read More »
What are the Factors of 34? (3 simple methods)
Need to know the factors of 34? You are in the right place! As a math teacher since 2007, I love helping students have aha moments. Finding the factors of a number needs to become a quick skill like
multiplication and division, so I have a few tips for you. I’ll show you a variety
What are the Factors of 34? (3 simple methods) Read More »
What is the prime factorization of 254?
Are you struggling to find the prime factorization of 254 and other numbers? You’re not alone. As a high school math teacher for over 14 years I love helping students make math fun and easy. When
you’re trying to multiply or divide something, the factors involved are important. That’s why I wrote this article –
What is the prime factorization of 254? Read More »
Factor tree for 27
Learning how to draw a factor tree for 27, find the factor pairs of 27 and prime factorization of 27 is handy to know. This is a useful skill to learn in math as it is essential for other topics such
as operating with fractions and simplifying expressions and solving equations in Algebra. As a
Factor tree for 27 Read More »
Sine 30 degrees (exact value, proof and example problems)
Sine 30 degrees is one of the most important trigonometric values. It has many applications in mathematics and physics. This article will discuss what sine 30 degrees is, a geometrical proof of it’s
value and how to find equivalent trig values. We will also show a video demonstration of some example problems with solutions. What
Sine 30 degrees (exact value, proof and example problems) Read More »
Why is Trigonometry hard? (7 reasons with solutions to master this subject)
Trigonometry is one of the most challenging subjects for students to learn. Many students find themselves struggling with the concepts and principles involved in trigonometry. This can often lead to
frustration and a feeling of being overwhelmed. In order to help students overcome these challenges, it is important to understand why trigonometry is hard for
Why is Trigonometry hard? (7 reasons with solutions to master this subject) Read More »
How to Find the Greatest Common Factor of 18 and 48 in 3 simple steps
In this article, I show you 4 methods how to find the greatest common factor of 18 and 48. One of the methods is just three simple steps! As a high school maths teacher since 2007 I have found that
students who know how to find the greatest common factor find other math topics easier.
How to Find the Greatest Common Factor of 18 and 48 in 3 simple steps Read More » | {"url":"https://www.mathshowto.com/author/anitaw/","timestamp":"2024-11-13T12:05:23Z","content_type":"text/html","content_length":"148180","record_id":"<urn:uuid:2324670a-912f-4aaa-add5-6ee3263d6b2b>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00714.warc.gz"} |
INFR11333 Coursework 3 template
\documentclass[a4paper]{article} %% Language and font encodings \usepackage[english]{babel} \usepackage[utf8x]{inputenc} \usepackage[T1]{fontenc} \usepackage{natbib} %% Sets page size and margins \
usepackage[a4paper,margin=2cm]{geometry} %% Useful packages \usepackage{amsmath} \usepackage{graphicx} \usepackage[colorinlistoftodos]{todonotes} \usepackage[colorlinks=true, allcolors=blue]
{hyperref} % This will make it easier for markers % to refer to specific lines of your report. \usepackage{lineno} \linenumbers \title{Report title (tell us what you learned!)} % Your report should
be anonymous: it should not % contain your name(s) or student id(s). \date{} \begin{document} \maketitle \begin{abstract} Your abstract should briefly explain the motivation, method, analysis, and
results. It needn't be more than a few sentences. \end{abstract} \section{A few tips} Use the figure environment and the caption command to add a number and a caption to your figure. See the code for
Figure \ref{fig:frog} in this section for an example. \begin{figure} \centering \includegraphics[width=0.3\textwidth]{frog.jpg} \caption{\label{fig:frog}This frog was uploaded via the project menu.}
\end{figure} \subsection{How to add Tables} Use the table and tabular commands for basic tables --- see Table~\ref{tab:widgets}, for example. \begin{table} \centering \begin{tabular}{lr}\hline\hline
Item & Quantity \\\hline Widgets & 42 \\ Gadgets & 13 \end{tabular} \caption{\label{tab:widgets}An example table.} \end{table} \subsection{How to write Mathematics} \LaTeX{} is great at typesetting
mathematics. Let $X_1, X_2, \ldots, X_n$ be a sequence of independent and identically distributed random variables with $\text{E}[X_i] = \mu$ and $\text{Var}[X_i] = \sigma^2 < \infty$, and let \[S_n
= \frac{X_1 + X_2 + \cdots + X_n}{n} = \frac{1}{n}\sum_{i}^{n} X_i\] denote their mean. Then as $n$ approaches infinity, the random variables $\sqrt{n}(S_n - \mu)$ converge in distribution to a
normal $\mathcal{N}(0, \sigma^2)$. \subsection{How to create Sections and Subsections} Use section and subsections to organize your document. The sections and subsections create an outline of your
report, and a reader should be able to infer the structure of your report just by skimming them. \subsection{Citations and references} We encourage you to cite research papers in your report, and to
support this, you are allowed an unlimited number of pages for citations. (Aside: if a paper was posted on arxiv.org and later published in a conference or journal, please cite the published
version). There are a couple of common ways to use citations: parenthetically in running text \citep{greenwade93}; or as nouns, e.g. ``as \citet{greenwade93} showed''. Just remember to specify a
bibliography style, as well as the filename of the \verb|.bib|. \bibliographystyle{plainnat} \bibliography{sample} \section*{Appendix} You may use an unlimited number of appendix pages to include
plots, tables, figures, and long examples. Equations should appear in the main text, not the Appendix. Abuse of the appendix (e.g. by including text or by putting critical information into captions
when it should be in the main text) may result in loss of marks. \end{document} | {"url":"https://tr.overleaf.com/latex/templates/infr11333-coursework-3-template/mvkfhzzjyywq","timestamp":"2024-11-08T02:32:40Z","content_type":"text/html","content_length":"40045","record_id":"<urn:uuid:78e40e4c-9a4a-48c6-b11b-95770e441fc2>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00722.warc.gz"} |
Moment Calculator - Best Online Tool | How do you Calculate Moments? - physicsCalculatorPro.com
The Moment Calculator is a free online tool that shows the rate at which Moments from one substance to another. This Moment calculator tool speeds up the process by displaying the Moment rate in a
fraction of a second.
Moment Formula
The moment is an expression that estimates how a physical quantity is positioned or ordered by combining the product of a distance and a physical quantity. To put it another way, an instant is a very
small period of time. For both balanced and unbalanced forces, the moment of force formula can be used to compute the moment of force. As a result, the moment formula is M = F × l
• l is the length of lever arm or moment arm
• M is the moment
• F be the force (in N) (in meters)
Moment of a Force Formula
The product of the distance between the point and the point of force application and the component of the force perpendicular to the line of distance is the moment.Newton-Meter is the SI unit for
moment or moment of force (Nm).
Formula for the Moment of Force
The formula for the moment of force is as follows: M = F × d
• Where, M stands for the present moment
• F is the applied force
• The total distance from the axis is represented by d
For more concepts check out physicscalculatorpro.com to get quick answers by using this free tool.
Moment Calculation Examples
Question 1: Calculate the moment of force in an object travelling at a distance of 3m from the axis with a force of 30 N.
Consider the question,
Force, F = 30 N
Distance, d = 3 m
We know that, The formula for finding the Moment of force is M = F × d
Place the given values in the moment of force equation and simplify it.
M = 30 × 3
M = 90 Nm
Therefore, the Moment is M = 90 Nm.
FAQs on Moment Calculator
1. What about the measurement of moment?
The moment (M) is measured In newton-metres (Nm).
2. Is moment and torque the same?
Moment and torque are considered the same. But, in mechanics, the moment and torque are slightly different from each other.
3. How can you figure out the length of a lever arm?
The moment equation can be used to calculate the length of a lever arm is M = F × d
The length of the lever arm is represented by d in this equation. As a result, d = M/F
To compute the length of a lever arm, simply place the moment and force numbers.
4. What method do you use to calculate moments?
The following equation can be used to compute the moment of a force: Moment is equal to Force multiplied by the Pivot's Perpendicular Distance. | {"url":"https://physicscalculatorpro.com/moment-calculator/","timestamp":"2024-11-05T04:15:53Z","content_type":"text/html","content_length":"25603","record_id":"<urn:uuid:3163695a-1462-4b8b-8b37-69dc04042c0f>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00004.warc.gz"} |
How to Calculate Net Present Value: Why Discount Rates Matter - Gospel10
How to Calculate Net Present Value: Why Discount Rates Matter
Understanding the Role of the Discount Rate in Net Present Value Calculations
Determining a project’s potential profitability and success is a challenge valuation experts face. One of the elements of this analysis is the discount rate, which plays a critical role in
calculating a project or investment’s Net Present Value (NPV). The NPV provides insights into the present value of future cash flows, reflecting the time value of money. A company considering
purchasing a new machine that will generate future cash flows will use a discount rate to determine the present value of those cash flows, helping them make an informed decision.
The discount rate is crucial in NPV calculations for several reasons: it reflects the riskiness of the investment, adjusts for inflation, and allows for the comparison of projects with different
lifespans. Historically, the rate used was the prevailing interest rate; however, modern financial theory suggests that the appropriate discount rate should be the Weighted Average Cost of Capital
Why is a Discount Rate Used to Calculate Net Present Value?
The discount rate is a crucial element in calculating the Net Present Value (NPV) of a project or investment. It reflects the time value of money, the riskiness of the investment, and allows for the
comparison of projects with varying lifespans.
• Time Value of Money
• Cost of Capital
• Inflation
• Project Risk
• Weighted Average Cost of Capital (WACC)
• Risk-Free Rate
• Project Duration
• Capital Budgeting
• Investment Appraisal
Understanding these aspects is essential for accurately calculating NPV and making informed investment decisions. The discount rate helps determine whether a project is financially viable,
considering the time value of money and the cost of capital. It also allows for the comparison of projects with different risk profiles and lifespans, ensuring that investment decisions are made on a
consistent basis.
Time Value of Money
The time value of money is a fundamental concept in finance that acknowledges the value of money changes over time. This is because money today can be invested and grow in value in the future due to
interest or inflation. Conversely, money in the future is worth less than money today because of the potential for inflation to erode its purchasing power.
The discount rate used to calculate Net Present Value (NPV) is closely related to the time value of money. The discount rate is the rate at which future cash flows are discounted back to their
present value. This is necessary because a dollar today is worth more than a dollar in the future, so future cash flows need to be discounted to compare them to present cash flows.
For example, if you have $100 today and invest it at a 5% annual interest rate, it will grow to $105 in one year. This means that the present value of $105 in one year is $100 today. Conversely, if
you have $100 in one year and want to know its present value today, you would divide $100 by 1.05, which gives you $95.24.
Understanding the time value of money and how it relates to the discount rate is essential for making sound investment decisions. By considering the time value of money, investors can compare
investment options on a more level playing field and make decisions that will maximize their returns.
Cost of Capital
The cost of capital is a crucial factor in determining the appropriate discount rate for calculating the Net Present Value (NPV) of a project or investment. It represents the minimum return that a
company must earn on a project to compensate investors for the risk they are taking. There are several components that make up the cost of capital:
• Debt Cost
Debt cost represents the interest expense associated with borrowing money. It is calculated by multiplying the amount of debt by the interest rate.
• Equity Cost
Equity cost represents the return that shareholders expect for their investment in the company. It is calculated using the Capital Asset Pricing Model (CAPM) or other valuation methods.
• Weighted Average Cost of Capital (WACC)
WACC is the average cost of capital for a company, taking into account both debt and equity financing. It is calculated by multiplying the debt cost by the proportion of debt in the capital
structure and the equity cost by the proportion of equity in the capital structure.
The cost of capital is a critical input in the NPV calculation because it determines the present value of future cash flows. A higher cost of capital will result in a lower NPV, while a lower cost of
capital will result in a higher NPV. Therefore, it is important to carefully consider the cost of capital when making investment decisions.
Inflation is a critical component of “why is a discount rate used to calculate net present value” because it erodes the purchasing power of money over time. This means that a dollar today is worth
more than a dollar in the future, as inflation will reduce the value of the dollar over time. Therefore, when calculating the Net Present Value (NPV) of a project or investment, it is important to
consider the impact of inflation on future cash flows.
For example, let’s say you are considering a project that will generate $100,000 in cash flow in one year. If the inflation rate is 2%, then the present value of that $100,000 cash flow is only
$98,040. This is because inflation will reduce the purchasing power of the $100,000 by 2% over the course of the year.
The discount rate used to calculate NPV should reflect the expected rate of inflation. If the discount rate is too low, then the NPV will be overstated, as it will not fully account for the impact of
inflation on future cash flows. Conversely, if the discount rate is too high, then the NPV will be understated, as it will overstate the impact of inflation on future cash flows.
Therefore, it is important to carefully consider the impact of inflation when calculating the NPV of a project or investment. By using a discount rate that reflects the expected rate of inflation,
investors can make more informed decisions about the potential profitability of a project.
Project Risk
Project risk is a critical component of “why is a discount rate used to calculate net present value” because it affects the expected cash flows of a project. The discount rate is used to discount
future cash flows back to their present value, and the riskiness of a project will impact the appropriate discount rate to use. A riskier project will require a higher discount rate, which will
result in a lower NPV.
For example, let’s say you are considering two projects:
• Project A is a low-risk project with a 5% expected return.
• Project B is a high-risk project with a 10% expected return.
If you use a 5% discount rate, then the NPV of Project A will be higher than the NPV of Project B. However, if you use a 10% discount rate, then the NPV of Project B will be higher than the NPV of
Project A. This is because the higher discount rate will more heavily discount the future cash flows of Project B, which are more uncertain due to the project’s higher risk.
Therefore, it is important to consider project risk when calculating the NPV of a project. By using a discount rate that reflects the riskiness of the project, investors can make more informed
decisions about the potential profitability of a project.
In conclusion, project risk is a critical component of “why is a discount rate used to calculate net present value” because it affects the expected cash flows of a project. The discount rate is used
to discount future cash flows back to their present value, and the riskiness of a project will impact the appropriate discount rate to use. By considering project risk when calculating the NPV of a
project, investors can make more informed decisions about the potential profitability of a project.
Weighted Average Cost of Capital (WACC)
Within the realm of “why is a discount rate used to calculate net present value,” the Weighted Average Cost of Capital (WACC) emerges as a pivotal component. WACC represents the average cost of
capital for a company, reflecting the blended cost of debt and equity financing. Its significance stems from the fact that the discount rate employed in NPV calculations should align with the project
or investment’s risk profile. A project’s WACC serves as a benchmark against which its expected return is measured.
To illustrate, consider a company evaluating two projects: Project A with a higher risk profile and Project B with a lower risk profile. If the company uses a discount rate that is lower than its
WACC for Project A, the NPV will be overstated, potentially leading to an erroneous investment decision. Conversely, using a discount rate higher than the WACC for Project B would result in an
understated NPV, potentially causing the company to miss out on a profitable opportunity.
Real-life examples abound where WACC plays a crucial role in NPV calculations. A manufacturing firm contemplating the acquisition of new machinery would factor in its WACC to determine the
appropriate discount rate for assessing the project’s NPV. Similarly, a real estate developer evaluating a new development project would consider its WACC to gauge the project’s potential
In conclusion, understanding the connection between WACC and “why is a discount rate used to calculate net present value” is paramount for making informed investment decisions. By utilizing a
discount rate that aligns with a project’s risk profile and WACC, companies can accurately assess the project’s potential return and make optimal capital allocation decisions.
Risk-Free Rate
In examining “why is a discount rate used to calculate net present value,” the concept of the risk-free rate emerges as a foundational element. The risk-free rate represents the return on an
investment with no risk of default, such as a government bond. It serves as a critical benchmark against which the discount rate is calibrated.
The risk-free rate plays a pivotal role in determining the discount rate because it reflects the minimum acceptable return for investors. When evaluating a project or investment, the discount rate
should be higher than the risk-free rate to account for the inherent risk associated with the venture. This ensures that investors are adequately compensated for taking on additional risk.
Real-life applications of the risk-free rate in “why is a discount rate used to calculate net present value” abound. For instance, consider a company evaluating a new product launch. The company
would use the risk-free rate as a starting point for determining the appropriate discount rate to assess the project’s NPV. By incorporating the risk-free rate, the company can accurately gauge the
project’s potential return and make an informed decision.
In conclusion, understanding the relationship between the risk-free rate and “why is a discount rate used to calculate net present value” is crucial for making sound investment decisions. By
considering the risk-free rate when determining the discount rate, investors can ensure that they are adequately compensated for the risks they are taking, leading to more informed and prudent
capital allocation.
Project Duration
Within the context of “why is a discount rate used to calculate net present value,” the duration of a project emerges as a critical component, influencing the calculation and interpretation of NPV.
Project duration refers to the period over which a project’s cash flows are expected to occur. It has a direct impact on the discount rate used in NPV calculations and the overall assessment of a
project’s profitability.
The connection between project duration and “why is a discount rate used to calculate net present value” lies in the time value of money. The discount rate, which represents the cost of capital or
the minimum acceptable rate of return, is applied to future cash flows to convert them to their present value. The longer the project duration, the greater the impact of discounting, as more cash
flows are pushed further into the future. Consequently, projects with longer durations typically require higher discount rates to account for the increased time value of money.
To illustrate, consider two projects with identical cash flows but varying durations. Project A has a duration of five years, while Project B has a duration of ten years. Assuming a constant discount
rate, the NPV of Project A will be higher than that of Project B, simply because the cash flows of Project A are received sooner and thus discounted less.
In practical terms, the understanding of project duration in relation to “why is a discount rate used to calculate net present value” is crucial for evaluating long-term projects, such as
infrastructure development, real estate investments, or research and development initiatives. Accurately assessing the project duration allows investors to determine the appropriate discount rate and
make informed decisions about the project’s financial viability and potential return.
Capital Budgeting
Within the framework of “why is a discount rate used to calculate net present value,” Capital Budgeting emerges as a pivotal aspect, shaping the decision-making process for long-term investments and
• Investment Appraisal
Capital Budgeting involves assessing the potential profitability and financial viability of proposed investments or projects. It entails evaluating the expected cash flows, risks, and returns
associated with each option.
• Project Selection
Based on the results of investment appraisal, Capital Budgeting guides the selection of projects that align with the organization’s strategic objectives and maximize shareholder value. It helps
prioritize projects based on their NPV and other relevant criteria.
• Resource Allocation
Capital Budgeting plays a crucial role in allocating scarce financial resources among competing projects. By comparing the NPVs of different projects, organizations can make informed decisions
about where to invest their capital to generate the highest returns.
• Risk Management
Capital Budgeting incorporates risk analysis to assess the potential risks associated with long-term investments. By considering the impact of risk on future cash flows, organizations can make
more informed decisions and mitigate potential losses.
In essence, Capital Budgeting provides a systematic framework for evaluating and selecting long-term investments, ensuring that organizations make optimal use of their financial resources and
maximize shareholder value. The discount rate plays a critical role in this process by converting future cash flows to their present value, allowing for a more accurate assessment of a project’s
profitability and risk-adjusted return.
Investment Appraisal
Investment Appraisal is a critical component of “why is a discount rate used to calculate net present value” because it provides the foundation for evaluating the financial viability and potential
profitability of long-term investments and projects. It entails a rigorous assessment of expected cash flows, risks, and returns associated with each investment option.
The connection between Investment Appraisal and “why is a discount rate used to calculate net present value” lies in the need to determine the present value of future cash flows. The discount rate,
which represents the cost of capital or the minimum acceptable rate of return, is applied to future cash flows to convert them to their present value. This allows for a more accurate assessment of a
project’s profitability and risk-adjusted return.
Real-life examples of Investment Appraisal within “why is a discount rate used to calculate net present value” abound. Consider a company evaluating the acquisition of a new production facility. The
company would conduct an Investment Appraisal to assess the project’s expected cash flows, risks, and returns. The discount rate would be used to calculate the NPV of the project, which would then be
used to make a decision on whether to proceed with the acquisition.
The practical applications of understanding the connection between Investment Appraisal and “why is a discount rate used to calculate net present value” are significant. It enables organizations to
make informed decisions about which projects to invest in, ensuring that they allocate their financial resources optimally. By accurately assessing the NPV of a project, organizations can prioritize
projects that maximize shareholder value and minimize risk.
FAQs on “Why is a Discount Rate Used to Calculate Net Present Value”?
This section provides answers to frequently asked questions about the significance and application of discount rates in NPV calculations.
Question 1: Why is it necessary to use a discount rate in NPV calculations?
Answer: The discount rate is crucial because it reflects the time value of money. It adjusts future cash flows to their present value, allowing for a more accurate assessment of a project’s
profitability and risk-adjusted return.
Question 2: How does the discount rate account for inflation?
Answer: The discount rate should ideally incorporate the expected rate of inflation to accurately reflect the changing value of money over time. A higher inflation rate would result in a higher
discount rate, leading to a lower NPV.
Question 3: What factors influence the selection of an appropriate discount rate?
Answer: The discount rate should align with the project’s risk profile and cost of capital. It is typically based on the Weighted Average Cost of Capital (WACC) or the Risk-Free Rate plus a risk
Question 4: How does project duration impact the discount rate used?
Answer: Longer project durations require higher discount rates because the present value of future cash flows diminishes over time due to the time value of money.
Question 5: Can different discount rates be used for different cash flows within a project?
Answer: Yes, in some cases, different discount rates may be applied to specific cash flows to account for varying risk levels or time horizons.
Question 6: How is NPV used in investment decision-making?
Answer: NPV is a critical tool for evaluating and comparing investment options. Projects with positive NPVs are generally considered financially viable, while those with negative NPVs may be
These FAQs provide a concise overview of the key aspects and applications of discount rates in NPV calculations. Understanding the ‘why’ behind this practice empowers individuals to make more
informed investment decisions.
In the next section, we will delve deeper into the practical applications of NPV calculations and explore how it aids in optimizing capital allocation and maximizing investment returns.
Tips for Calculating Net Present Value (NPV)
Accurately calculating NPV is critical for making sound investment decisions. Here are some practical tips to enhance your NPV calculations:
Tip 1: Determine an appropriate discount rate. Consider the project’s risk profile, cost of capital, and time horizon.
Tip 2: Forecast cash flows accurately. Include all relevant cash inflows and outflows, considering both operating and non-operating activities.
Tip 3: Consider inflation. Adjust future cash flows for inflation to reflect the changing value of money over time.
Tip 4: Handle uneven cash flows appropriately. Use appropriate discounting techniques to account for uneven cash flows occurring at different points in time.
Tip 5: Evaluate multiple scenarios. Perform sensitivity analysis by varying key assumptions, such as the discount rate and cash flow estimates, to assess the project’s robustness.
Tip 6: Consider qualitative factors. While NPV is a quantitative measure, incorporate qualitative factors, such as market dynamics and competitive advantages, into your investment decision-making
Tip 7: Seek professional advice. Consult with financial experts or advisors if you lack the expertise or resources to conduct NPV calculations independently.
Tip 8: Use NPV as a comparative tool. Compare the NPVs of different investment options to make informed decisions about where to allocate your capital.
By following these tips, you can enhance the accuracy and effectiveness of your NPV calculations, leading to more informed investment decisions and improved financial outcomes.
In the next section, we will explore how NPV is used in practice to evaluate and compare investment opportunities, ensuring optimal capital allocation and maximizing returns.
Throughout this exploration of “why is a discount rate used to calculate net present value,” we have delved into the critical role of the discount rate in evaluating the profitability and
risk-adjusted return of long-term investments and projects.
Key insights from this analysis include the following:
1. The discount rate reflects the time value of money, adjusting future cash flows to their present value for more accurate project assessment.
2. The appropriate discount rate should align with the project’s risk profile and cost of capital, incorporating factors such as inflation and project duration.
3. NPV calculations, utilizing the discount rate, enable investors to compare different investment options, prioritize projects, and make informed decisions that maximize shareholder value.
Understanding “why is a discount rate used to calculate net present value” empowers individuals and organizations to make sound investment decisions, optimize capital allocation, and achieve
long-term financial success.
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Quantum logic - (Universal Algebra) - Vocab, Definition, Explanations | Fiveable
Quantum logic
from class:
Universal Algebra
Quantum logic is a system of rules and principles that govern the reasoning of propositions in quantum mechanics, differing significantly from classical logic. This framework is used to understand
the behaviors and interactions of quantum systems, where traditional binary true/false evaluations do not apply, leading to a new way of thinking about truth values and their relationships.
congrats on reading the definition of quantum logic. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. Quantum logic replaces classical logical operations with lattice structures that reflect the non-classical relationships between quantum events.
2. In quantum logic, the principle of non-contradiction is modified; propositions can be both true and false, which challenges traditional binary perspectives.
3. The concept of observables in quantum mechanics corresponds to certain mathematical structures in quantum logic, indicating how measurements relate to properties of systems.
4. Quantum logic has implications for understanding paradoxes in quantum theory, such as Schrรถdinger's cat, which illustrate the complexities of measurement and reality.
5. The development of quantum logic has inspired research in areas such as quantum computing and information theory, demonstrating its relevance beyond theoretical discussions.
Review Questions
• How does quantum logic differ from classical logic in terms of truth values and propositions?
□ Quantum logic differs from classical logic primarily in its treatment of truth values and propositions. In classical logic, a proposition is either true or false, adhering strictly to binary
conditions. However, in quantum logic, propositions can exist in a superposition state, allowing them to be both true and false simultaneously. This non-binary approach reflects the
complexities of quantum phenomena and challenges the traditional logical frameworks.
• Discuss how the concepts of superposition and entanglement are represented within the framework of quantum logic.
□ Within quantum logic, superposition and entanglement are represented through the lattice structures that define the relationships between different propositions. Superposition illustrates how
a quantum system can exist in multiple states at once, leading to complex interdependencies among propositions. Entanglement further complicates this by showing how the state of one particle
can instantaneously affect another's state regardless of distance. These concepts highlight how quantum systems operate outside classical constraints, necessitating a unique logical
• Evaluate the impact of quantum logic on our understanding of measurement in quantum mechanics and its implications for future technologies.
□ Quantum logic significantly impacts our understanding of measurement by altering how we interpret observations in quantum mechanics. It introduces complexities where measurements do not
simply collapse states but can reveal deeper entangled relationships between particles. This understanding is crucial for advancing technologies like quantum computing and cryptography, where
traditional logic fails to adequately describe behavior. As researchers explore these implications, they continue to redefine boundaries in computation and information processing, pushing
forward innovative applications grounded in quantum principles.
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Given the attached dataset, list the top 5 players based on statistical analysis of the data. Also make sure the sum of total player salaries are equal to 55000. Go!
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Sum of player salaries
Maximizing fantasy points within a budget
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Understanding Mathematical Functions: What Can You Say About The Funct
Introduction to Mathematical Functions and Their Importance
In the world of mathematics, functions play a vital role in helping us understand and analyze various phenomena. Whether it's in the field of science, economics, engineering, or any other discipline,
functions provide a way to model, predict, and interpret real-world data and patterns. In this chapter, we'll delve into the concept of mathematical functions, their significance in different fields,
and how we can analyze them through a table of values.
Explaining the concept of a mathematical function
A mathematical function is a relation between a set of inputs (known as the domain) and a set of outputs (known as the range) that assigns each input exactly one output. In simpler terms, a function
takes an input, performs a certain operation on it, and produces an output. This operation can be anything from simple arithmetic calculations to more complex mathematical manipulations.
Functions are typically denoted by a letter such as f, g, or h, and are written as f(x) or g(y) to indicate the input variable. The output of the function is then represented by f(x) or g(y),
depending on the context.
The significance of functions in various fields
Functions are fundamental in various fields such as science, economics, and engineering. In science, functions are used to describe the behavior of physical systems, model natural phenomena, and
analyze experimental data. In economics, functions are employed to model supply and demand, forecast market trends, and optimize resource allocation. Similarly, in engineering, functions are utilized
to design systems, optimize processes, and simulate physical phenomena.
Overall, functions provide a powerful framework for understanding and representing relationships between different variables, making them indispensable in a wide range of applications.
Preview of the process for analyzing a function through a table of values
One of the common ways to analyze a function is by examining a table of values that shows the inputs and corresponding outputs. This allows us to observe the behavior of the function and identify any
patterns or trends. By analyzing a table of values, we can gain insights into how the function changes with different inputs and understand its overall characteristics.
Throughout this chapter, we will explore a specific example of a table of values and discuss the insights we can glean from it about the function that generated it.
Key Takeaways
• Functions can be represented by tables of values
• Understanding the pattern in the table is key
• The function may be linear, quadratic, or exponential
• Look for a consistent change in the x and y values
• Identify the relationship between the x and y values
Recognizing Patterns in the Table of Values
Understanding mathematical functions involves recognizing patterns in the table of values. By identifying these patterns, we can determine the type of function that generated the given values. In
this chapter, we will explore how to recognize linear, quadratic, and higher-degree polynomial patterns, the role of successive differences in recognizing function types, and provide examples of
pattern recognition from given tables of values.
A. How to identify linear, quadratic, and higher-degree polynomial patterns
When examining a table of values, it is essential to look for patterns that indicate the type of function at play. For linear patterns, the values will increase or decrease at a constant rate. In a
quadratic pattern, the values will increase or decrease at an increasing rate, forming a parabolic shape. Higher-degree polynomial patterns exhibit more complex variations in the values, often with
multiple turning points.
One way to identify these patterns is by examining the differences between consecutive values. For linear patterns, the first differences will be constant. In quadratic patterns, the second
differences will be constant. For higher-degree polynomial patterns, the differences may not be constant, but they will follow a discernible pattern.
B. The role of successive differences in recognizing function types
Successive differences play a crucial role in recognizing the type of function that generated the table of values. By calculating the first and second differences between consecutive values, we can
gain insight into the underlying pattern. If the first differences are constant, it indicates a linear pattern. If the second differences are constant, it indicates a quadratic pattern. For
higher-degree polynomial patterns, we may need to examine higher-order differences to discern the underlying pattern.
By understanding the role of successive differences, we can effectively identify the function type and make predictions about future values based on the observed pattern.
C. Examples of pattern recognition from given tables of values
Let's consider an example of a table of values:
• x: 1, 2, 3, 4, 5
• y: 3, 7, 13, 21, 31
By calculating the first differences for the y-values, we get: 4, 6, 8, 10. Since the first differences are not constant, it indicates that the pattern is not linear. However, when we calculate the
second differences for the y-values, we get: 2, 2, 2. The second differences are constant, indicating a quadratic pattern. Therefore, the function that generated these values is a quadratic function.
By analyzing examples like this, we can develop our pattern recognition skills and gain a deeper understanding of mathematical functions.
Interpreting Function Behaviors
Understanding the behavior of mathematical functions is essential in analyzing and interpreting their properties. By examining the patterns and trends exhibited by a function, we can gain valuable
insights into its characteristics and how it relates to real-world phenomena.
A Understanding the concepts of increasing, decreasing, and constant functions
When we talk about the behavior of a function, we are referring to how its output values change in response to changes in the input. One of the key concepts in understanding function behavior is the
idea of increasing, decreasing, and constant functions.
An increasing function is one in which the output values increase as the input values increase. In other words, as the input variable grows, the output variable also grows. On the other hand, a
decreasing function is one in which the output values decrease as the input values increase. Finally, a constant function is one in which the output values remain the same, regardless of changes in
the input.
B The meaning of function behavior in real-world contexts
Understanding function behavior is not just a theoretical exercise; it has real-world implications. Many natural and man-made phenomena can be modeled using mathematical functions, and analyzing the
behavior of these functions can provide valuable insights into the underlying processes.
For example, in economics, the concept of increasing, decreasing, and constant functions is used to analyze the behavior of various economic indicators such as demand, supply, and production. In
physics, the behavior of functions is used to model the motion of objects, the flow of fluids, and the propagation of waves. By understanding how functions behave in these contexts, we can make
predictions, optimize processes, and solve practical problems.
C Applying behavior analysis to the table of values
Now, let's apply our understanding of function behavior to analyze the table of values provided. By examining the patterns in the data, we can gain insights into the behavior of the function that
generated these values.
• First, we can look for trends in the output values as the input values change. Are the output values consistently increasing, decreasing, or staying constant?
• Next, we can calculate the rate of change of the function to determine if it is increasing at a constant rate, decreasing at a constant rate, or exhibiting some other behavior.
• We can also look for any points of inflection or abrupt changes in the behavior of the function, which can provide clues about its overall behavior.
By carefully analyzing the table of values and applying our knowledge of function behavior, we can gain a deeper understanding of the underlying function and its implications in real-world contexts.
Determining Key Features of Functions
Understanding the key features of mathematical functions is essential for analyzing and graphing functions. By examining a table of values, we can identify zeros, intercepts, and asymptotes, which
provide valuable insights into the behavior of the function.
A Identifying zeros, intercepts, and asymptotes from a table
When analyzing a table of values for a function, we can identify the zeros by looking for input values that result in an output of zero. These zeros correspond to the x-intercepts of the function,
where the graph crosses the x-axis. Additionally, we can determine the y-intercept by finding the output value when the input is zero. Asymptotes, which are lines that the graph approaches but never
touches, can also be identified by observing the behavior of the function as the input values approach certain values.
B Understanding the significance of key features in graphing functions
The key features of a function, such as zeros, intercepts, and asymptotes, play a crucial role in graphing the function. Zeros and intercepts provide important points on the graph that help us
visualize the behavior of the function. Asymptotes indicate the behavior of the function as the input values approach certain values, helping us understand the overall shape of the graph. By
understanding these key features, we can accurately sketch the graph of the function and gain insights into its behavior.
C Real-life scenarios where key function features are critical
The understanding of key function features is not only important in mathematical contexts but also in real-life scenarios. For example, in engineering and physics, the behavior of physical systems
can be described using mathematical functions. Zeros and intercepts of these functions may represent critical points in the system, such as equilibrium positions or points of impact. Asymptotes can
indicate limits or boundaries within which the system operates. In finance, functions describing investment growth or depreciation may have zeros and intercepts that represent important financial
milestones. Understanding these key features is critical for making informed decisions in various real-life scenarios.
Utilizing Graphical Representations
Understanding mathematical functions often involves visualizing them through graphs. Graphs provide a clear and concise way to represent the relationship between input and output values of a
function. By sketching a graph from a table of values, we can gain a deeper understanding of the behavior of the function and identify any patterns or trends.
A The importance of visualizing functions through graphs
Graphs allow us to see the overall shape of a function and how it behaves across different input values. This visual representation can help us identify key features such as the domain and range,
intercepts, and any asymptotes or discontinuities. Additionally, graphs provide a way to easily compare different functions and analyze their relative behaviors.
B Step-by-step approach to sketching a graph from a table of values
When given a table of values for a function, we can follow a step-by-step approach to sketching its graph:
• Step 1: Plot the points from the table of values on a coordinate plane.
• Step 2: Identify any patterns or trends in the plotted points.
• Step 3: Determine the overall shape of the graph based on the plotted points.
• Step 4: Connect the points to form a smooth curve that represents the function.
• Step 5: Label the graph with the function's name, key points, and any relevant information.
C Troubleshooting common mistakes in graphing
While sketching a graph from a table of values, it's important to be aware of common mistakes that can arise:
• Mistake 1: Incorrectly plotting the points from the table.
• Mistake 2: Failing to identify and connect the points in a way that accurately represents the function's behavior.
• Mistake 3: Mislabeling or omitting important information on the graph.
By being mindful of these potential pitfalls, we can ensure that our graph accurately reflects the function and provides a clear visual representation of its behavior.
Extrapolating and Predicting Using Functions
When it comes to understanding mathematical functions, one of the key applications is the ability to extrapolate and predict future behavior based on the given data. This process involves using
tables of values to identify patterns and trends, and then using mathematical models to make predictions about what might happen next.
A Techniques for using tables of values to predict future behavior
Tables of values provide a snapshot of the relationship between the input and output of a function. By analyzing these values, it is possible to identify trends and patterns that can be used to make
predictions about future behavior. One common technique for using tables of values to predict future behavior is to look for recurring patterns or relationships between the input and output values.
For example, if the output values are increasing at a consistent rate for each increase in the input value, it may be possible to use this information to predict future output values based on a given
Another technique involves using regression analysis to identify mathematical relationships between the input and output values. This can help to create a mathematical model that can be used to make
predictions about future behavior based on the given data.
B The role of mathematical models in extrapolation
Mathematical models play a crucial role in extrapolation, as they provide a framework for making predictions based on the given data. These models can take various forms, such as linear, exponential,
or polynomial functions, and are used to represent the relationship between the input and output values of a function.
By fitting a mathematical model to the given data, it becomes possible to make predictions about future behavior based on the established relationship. This allows for the extrapolation of the
function beyond the given data points, providing valuable insights into potential future outcomes.
C Examples of successful predictions in various disciplines
There are numerous examples of successful predictions made using mathematical functions in various disciplines. In economics, mathematical models are used to predict future trends in the stock
market, inflation rates, and consumer behavior. These predictions are crucial for making informed decisions about investments, policy-making, and business strategies.
In the field of climate science, mathematical models are used to predict future climate patterns, sea level rise, and the impact of human activities on the environment. These predictions are
essential for understanding the potential consequences of climate change and developing strategies to mitigate its effects.
In the field of healthcare, mathematical models are used to predict the spread of diseases, the effectiveness of treatments, and the impact of public health interventions. These predictions are vital
for making decisions about resource allocation, disease prevention, and healthcare policy.
Overall, the ability to extrapolate and predict future behavior using mathematical functions is a powerful tool that has wide-ranging applications across various disciplines.
Conclusion and Best Practices for Function Analysis
Understanding mathematical functions is essential for various fields such as engineering, physics, economics, and computer science. It provides a framework for analyzing and solving real-world
problems. In this chapter, we will recap the significance of understanding mathematical functions, discuss best practices when analyzing functions from tables of values, and encourage continued
practice and further study of functions.
A Recapping the significance of understanding mathematical functions
• Foundation for problem-solving: Mathematical functions serve as the foundation for problem-solving in various disciplines. They provide a systematic way to model and analyze relationships between
• Tool for decision-making: Understanding functions allows individuals to make informed decisions based on data analysis and predictions. It is crucial for making accurate projections and
optimizing processes.
• Gateway to advanced mathematics: Proficiency in understanding functions is a stepping stone to advanced mathematical concepts such as calculus, differential equations, and linear algebra.
B Best practices when analyzing functions from tables of values
• Identify patterns: When analyzing a table of values, look for patterns and relationships between the input and output. This can help in determining the nature of the function.
• Use multiple data points: It is important to use multiple data points to analyze a function. Relying on a single data point may lead to inaccurate conclusions about the function's behavior.
• Consider the domain and range: Pay attention to the domain and range of the function. Understanding the possible input and output values can provide insights into the function's behavior.
• Utilize mathematical tools: Use mathematical tools such as graphing software, regression analysis, and curve fitting to analyze functions from tables of values. These tools can provide visual
representations and mathematical models of the functions.
C Encouraging continued practice and further study of functions
• Practice problem-solving: Regular practice of solving problems involving functions can enhance understanding and proficiency. Work on a variety of problems to gain exposure to different types of
• Explore advanced topics: Delve into advanced topics such as trigonometric functions, exponential functions, and logarithmic functions. Understanding a wide range of functions can broaden your
mathematical knowledge.
• Seek guidance and resources: Utilize textbooks, online resources, and instructional videos to further study functions. Seek guidance from teachers, tutors, or mentors to clarify any doubts and
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K-Means Clustering in R: Algorithm and Practical Examples - Datanovia
K-Means Clustering in R: Algorithm and Practical Examples
K-means clustering (MacQueen 1967) is one of the most commonly used unsupervised machine learning algorithm for partitioning a given data set into a set of k groups (i.e. k clusters), where k
represents the number of groups pre-specified by the analyst. It classifies objects in multiple groups (i.e., clusters), such that objects within the same cluster are as similar as possible (i.e.,
high intra-class similarity), whereas objects from different clusters are as dissimilar as possible (i.e., low inter-class similarity). In k-means clustering, each cluster is represented by its
center (i.e, centroid) which corresponds to the mean of points assigned to the cluster.
In this article, you will learn:
• The basic steps of k-means algorithm
• How to compute k-means in R software using practical examples
• Advantages and disavantages of k-means clustering
Practical Guide to Cluster Analysis in R
K-means basic ideas
The basic idea behind k-means clustering consists of defining clusters so that the total intra-cluster variation (known as total within-cluster variation) is minimized.
There are several k-means algorithms available. The standard algorithm is the Hartigan-Wong algorithm (Hartigan and Wong 1979), which defines the total within-cluster variation as the sum of squared
distances Euclidean distances between items and the corresponding centroid:
W(C_k) = \sum\limits_{x_i \in C_k} (x_i - \mu_k)^2
• \(x_i\) design a data point belonging to the cluster \(C_k\)
• \(\mu_k\) is the mean value of the points assigned to the cluster \(C_k\)
Each observation (\(x_i\)) is assigned to a given cluster such that the sum of squares (SS) distance of the observation to their assigned cluster centers \(\mu_k\) is a minimum.
We define the total within-cluster variation as follow:
tot.withinss = \sum\limits_{k=1}^k W(C_k) = \sum\limits_{k=1}^k \sum\limits_{x_i \in C_k} (x_i - \mu_k)^2
The total within-cluster sum of square measures the compactness (i.e goodness) of the clustering and we want it to be as small as possible.
K-means algorithm
The first step when using k-means clustering is to indicate the number of clusters (k) that will be generated in the final solution.
The algorithm starts by randomly selecting k objects from the data set to serve as the initial centers for the clusters. The selected objects are also known as cluster means or centroids.
Next, each of the remaining objects is assigned to it’s closest centroid, where closest is defined using the Euclidean distance between the object and the cluster mean. This step is called “cluster
assignment step”. Note that, to use correlation distance, the data are input as z-scores.
After the assignment step, the algorithm computes the new mean value of each cluster. The term cluster “centroid update” is used to design this step. Now that the centers have been recalculated,
every observation is checked again to see if it might be closer to a different cluster. All the objects are reassigned again using the updated cluster means.
The cluster assignment and centroid update steps are iteratively repeated until the cluster assignments stop changing (i.e until convergence is achieved). That is, the clusters formed in the current
iteration are the same as those obtained in the previous iteration.
K-means algorithm can be summarized as follow:
1. Specify the number of clusters (K) to be created (by the analyst)
2. Select randomly k objects from the dataset as the initial cluster centers or means
3. Assigns each observation to their closest centroid, based on the Euclidean distance between the object and the centroid
4. For each of the k clusters update the cluster centroid by calculating the new mean values of all the data points in the cluster. The centoid of a K[th] cluster is a vector of length p containing
the means of all variables for the observations in the k[th] cluster; p is the number of variables.
5. Iteratively minimize the total within sum of square. That is, iterate steps 3 and 4 until the cluster assignments stop changing or the maximum number of iterations is reached. By default, the R
software uses 10 as the default value for the maximum number of iterations.
Computing k-means clustering in R
We’ll use the demo data sets “USArrests”. The data should be prepared as described in chapter @ref(data-preparation-and-r-packages). The data must contains only continuous variables, as the k-means
algorithm uses variable means. As we don’t want the k-means algorithm to depend to an arbitrary variable unit, we start by scaling the data using the R function scale() as follow:
data("USArrests") # Loading the data set
df <- scale(USArrests) # Scaling the data
# View the firt 3 rows of the data
head(df, n = 3)
## Murder Assault UrbanPop Rape
## Alabama 1.2426 0.783 -0.521 -0.00342
## Alaska 0.5079 1.107 -1.212 2.48420
## Arizona 0.0716 1.479 0.999 1.04288
Required R packages and functions
The standard R function for k-means clustering is kmeans() [stats package], which simplified format is as follow:
kmeans(x, centers, iter.max = 10, nstart = 1)
• x: numeric matrix, numeric data frame or a numeric vector
• centers: Possible values are the number of clusters (k) or a set of initial (distinct) cluster centers. If a number, a random set of (distinct) rows in x is chosen as the initial centers.
• iter.max: The maximum number of iterations allowed. Default value is 10.
• nstart: The number of random starting partitions when centers is a number. Trying nstart > 1 is often recommended.
To create a beautiful graph of the clusters generated with the kmeans() function, will use the factoextra package.
• Installing factoextra package as:
Estimating the optimal number of clusters
The k-means clustering requires the users to specify the number of clusters to be generated.
One fundamental question is: How to choose the right number of expected clusters (k)?
Different methods will be presented in the chapter “cluster evaluation and validation statistics”.
Here, we provide a simple solution. The idea is to compute k-means clustering using different values of clusters k. Next, the wss (within sum of square) is drawn according to the number of clusters.
The location of a bend (knee) in the plot is generally considered as an indicator of the appropriate number of clusters.
The R function fviz_nbclust() [in factoextra package] provides a convenient solution to estimate the optimal number of clusters.
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The plot above represents the variance within the clusters. It decreases as k increases, but it can be seen a bend (or “elbow”) at k = 4. This bend indicates that additional clusters beyond the
fourth have little value.. In the next section, we’ll classify the observations into 4 clusters.
Computing k-means clustering
As k-means clustering algorithm starts with k randomly selected centroids, it’s always recommended to use the set.seed() function in order to set a seed for R’s random number generator. The aim is to
make reproducible the results, so that the reader of this article will obtain exactly the same results as those shown below.
The R code below performs k-means clustering with k = 4:
# Compute k-means with k = 4
km.res <- kmeans(df, 4, nstart = 25)
As the final result of k-means clustering result is sensitive to the random starting assignments, we specify nstart = 25. This means that R will try 25 different random starting assignments and then
select the best results corresponding to the one with the lowest within cluster variation. The default value of nstart in R is one. But, it’s strongly recommended to compute k-means clustering with a
large value of nstart such as 25 or 50, in order to have a more stable result.
# Print the results
## K-means clustering with 4 clusters of sizes 13, 16, 13, 8
## Cluster means:
## Murder Assault UrbanPop Rape
## 1 -0.962 -1.107 -0.930 -0.9668
## 2 -0.489 -0.383 0.576 -0.2617
## 3 0.695 1.039 0.723 1.2769
## 4 1.412 0.874 -0.815 0.0193
## Clustering vector:
## Alabama Alaska Arizona Arkansas California
## 4 3 3 4 3
## Colorado Connecticut Delaware Florida Georgia
## 3 2 2 3 4
## Hawaii Idaho Illinois Indiana Iowa
## 2 1 3 2 1
## Kansas Kentucky Louisiana Maine Maryland
## 2 1 4 1 3
## Massachusetts Michigan Minnesota Mississippi Missouri
## 2 3 1 4 3
## Montana Nebraska Nevada New Hampshire New Jersey
## 1 1 3 1 2
## New Mexico New York North Carolina North Dakota Ohio
## 3 3 4 1 2
## Oklahoma Oregon Pennsylvania Rhode Island South Carolina
## 2 2 2 2 4
## South Dakota Tennessee Texas Utah Vermont
## 1 4 3 2 1
## Virginia Washington West Virginia Wisconsin Wyoming
## 2 2 1 1 2
## Within cluster sum of squares by cluster:
## [1] 11.95 16.21 19.92 8.32
## (between_SS / total_SS = 71.2 %)
## Available components:
## [1] "cluster" "centers" "totss" "withinss"
## [5] "tot.withinss" "betweenss" "size" "iter"
## [9] "ifault"
The printed output displays:
• the cluster means or centers: a matrix, which rows are cluster number (1 to 4) and columns are variables
• the clustering vector: A vector of integers (from 1:k) indicating the cluster to which each point is allocated
It’s possible to compute the mean of each variables by clusters using the original data:
aggregate(USArrests, by=list(cluster=km.res$cluster), mean)
## cluster Murder Assault UrbanPop Rape
## 1 1 3.60 78.5 52.1 12.2
## 2 2 5.66 138.9 73.9 18.8
## 3 3 10.82 257.4 76.0 33.2
## 4 4 13.94 243.6 53.8 21.4
If you want to add the point classifications to the original data, use this:
dd <- cbind(USArrests, cluster = km.res$cluster)
## Murder Assault UrbanPop Rape cluster
## Alabama 13.2 236 58 21.2 4
## Alaska 10.0 263 48 44.5 3
## Arizona 8.1 294 80 31.0 3
## Arkansas 8.8 190 50 19.5 4
## California 9.0 276 91 40.6 3
## Colorado 7.9 204 78 38.7 3
Accessing to the results of kmeans() function
kmeans() function returns a list of components, including:
• cluster: A vector of integers (from 1:k) indicating the cluster to which each point is allocated
• centers: A matrix of cluster centers (cluster means)
• totss: The total sum of squares (TSS), i.e \(\sum{(x_i - \bar{x})^2}\). TSS measures the total variance in the data.
• withinss: Vector of within-cluster sum of squares, one component per cluster
• tot.withinss: Total within-cluster sum of squares, i.e. \(sum(withinss)\)
• betweenss: The between-cluster sum of squares, i.e. \(totss - tot.withinss\)
• size: The number of observations in each cluster
These components can be accessed as follow:
# Cluster number for each of the observations
head(km.res$cluster, 4)
## Alabama Alaska Arizona Arkansas
## 4 3 3 4
# Cluster size
## [1] 13 16 13 8
# Cluster means
## Murder Assault UrbanPop Rape
## 1 -0.962 -1.107 -0.930 -0.9668
## 2 -0.489 -0.383 0.576 -0.2617
## 3 0.695 1.039 0.723 1.2769
## 4 1.412 0.874 -0.815 0.0193
Visualizing k-means clusters
It is a good idea to plot the cluster results. These can be used to assess the choice of the number of clusters as well as comparing two different cluster analyses.
Now, we want to visualize the data in a scatter plot with coloring each data point according to its cluster assignment.
The problem is that the data contains more than 2 variables and the question is what variables to choose for the xy scatter plot.
A solution is to reduce the number of dimensions by applying a dimensionality reduction algorithm, such as Principal Component Analysis (PCA), that operates on the four variables and outputs two new
variables (that represent the original variables) that you can use to do the plot.
In other words, if we have a multi-dimensional data set, a solution is to perform Principal Component Analysis (PCA) and to plot data points according to the first two principal components
The function fviz_cluster() [factoextra package] can be used to easily visualize k-means clusters. It takes k-means results and the original data as arguments. In the resulting plot, observations are
represented by points, using principal components if the number of variables is greater than 2. It’s also possible to draw concentration ellipse around each cluster.
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K-means clustering advantages and disadvantages
K-means clustering is very simple and fast algorithm. It can efficiently deal with very large data sets. However there are some weaknesses, including:
1. It assumes prior knowledge of the data and requires the analyst to choose the appropriate number of cluster (k) in advance
2. The final results obtained is sensitive to the initial random selection of cluster centers. Why is it a problem? Because, for every different run of the algorithm on the same dataset, you may
choose different set of initial centers. This may lead to different clustering results on different runs of the algorithm.
3. It’s sensitive to outliers.
4. If you rearrange your data, it’s very possible that you’ll get a different solution every time you change the ordering of your data.
Possible solutions to these weaknesses, include:
1. Solution to issue 1: Compute k-means for a range of k values, for example by varying k between 2 and 10. Then, choose the best k by comparing the clustering results obtained for the different k
2. Solution to issue 2: Compute K-means algorithm several times with different initial cluster centers. The run with the lowest total within-cluster sum of square is selected as the final clustering
3. To avoid distortions caused by excessive outliers, it’s possible to use PAM algorithm, which is less sensitive to outliers.
Alternative to k-means clustering
A robust alternative to k-means is PAM, which is based on medoids. As discussed in the next chapter, the PAM clustering can be computed using the function pam() [cluster package]. The function pamk(
) [fpc package] is a wrapper for PAM that also prints the suggested number of clusters based on optimum average silhouette width.
K-means clustering can be used to classify observations into k groups, based on their similarity. Each group is represented by the mean value of points in the group, known as the cluster centroid.
K-means algorithm requires users to specify the number of cluster to generate. The R function kmeans() [stats package] can be used to compute k-means algorithm. The simplified format is kmeans(x,
centers), where “x” is the data and centers is the number of clusters to be produced.
After, computing k-means clustering, the R function fviz_cluster() [factoextra package] can be used to visualize the results. The format is fviz_cluster(km.res, data), where km.res is k-means results
and data corresponds to the original data sets.
Hartigan, JA, and MA Wong. 1979. “Algorithm AS 136: A K-means clustering algorithm.” Applied Statistics. Royal Statistical Society, 100–108.
MacQueen, J. 1967. “Some Methods for Classification and Analysis of Multivariate Observations.” In Proceedings of the Fifth Berkeley Symposium on Mathematical Statistics and Probability, Volume 1:
Statistics, 281–97. Berkeley, Calif.: University of California Press. http://projecteuclid.org:443/euclid.bsmsp/1200512992.
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Comments ( 2 )
• how we can get data
□ The demo data used in this tutorial is available in the default installation of R. Juste type data(“USArrests”)
Give a comment
Back to Partitional Clustering in R: The Essentials | {"url":"https://www.datanovia.com/en/lessons/K-means-clustering-in-r-algorith-and-practical-examples/","timestamp":"2024-11-06T17:19:45Z","content_type":"text/html","content_length":"159680","record_id":"<urn:uuid:21ddca76-50d7-4d90-be36-d83f2b8ae9f9>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00710.warc.gz"} |
Machine Learning the ground state masses of atomic nuclei
Triangle Nuclear Theory seminar
Invited presentation on 03/2023
Machine Learning and Artificial Intelligence methods offer wide applicability to a class of problems known collectively as "inverse problems." The solution to such problems involves the calculation
of the causal factors that are responsible for a set of observations. Inverse problems are frequently encountered in nuclear physics, where measurements exist and these observations must be
reconciled with theoretical models and interpretation. I will discuss a probabilistic machine learning technique applied to the binding energy of atomic nuclei. The set of observations comes from the
Atomic Mass Evaluation (AME), which totals over 2000 data points. I show that inclusion of physics-based inputs as well as physics-informed training yields neural networks that are capable of
describing these observations with a high degree of precision. Because the method is stochastic, it also provides an estimate of uncertainty for each prediction. I discuss the capacity of such
modeling to extrapolate into regions of unmeasured nuclei that is needed for applications such as astrophysics. | {"url":"https://matthewmumpower.com/presentations/111/machine-learning-the-ground-state-masses-of-atomic-nuclei","timestamp":"2024-11-14T07:45:37Z","content_type":"text/html","content_length":"10760","record_id":"<urn:uuid:c765c026-43e9-4a10-a557-ad950b17d06d>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00612.warc.gz"} |
Binary to Hex Converter - Convert Binary to Hexadecimal Online
Binary to Hex Converter – Convert Binary to Hexadecimal Online
Binary to Hex Converter will let you to Convert the binary numbers into hexadecimal. Also, It is very easy and free to use. Just enter the binary input inside the text-box and click on “Convert”.
Even more, You will get all the information regarding the converter here.
What is Binary?
The binary number is basically a numeral system of base 2. It contains only 2 digits. The two binary digits are 0 and 1.
The binary number is implemented in the logic gate and electric circuit. Even more, all modern computers aided devices works on binary digits. Also, Its history is linked with Gottfried Leibniz. He
writes in his famous thesis “Explication de l’Arithmétique Binaire” (1703). In 1854 George Boole published a paper on binary digits. His logical calculus is now using to design electronic circuitry.
In 1937, Claude Shannon published a thesis entitled A Symbolic Analysis of Relay and Switching Circuits. It implements Boolean algebra and binary arithmetic using electronic relays and switches for
the first time in history.
What is Hexadecimal?
This number system represents the numbering of base 16. Also, It uses 16 different symbols 1-9 and (A-F / a-f) to represent values ten to fifteen. It is human-friendly. So, it is widely using by
computer system designers and programmers. Moreover, it was invented in France in the year 770AD. The hexadecimal is a combination of letters and numbers. Our binary to hex converter will be very
helpful in order to convert binary to hex.
Importance of Hexadecimal Numbering System
1. Memory Allocation: Hexadecimal is doing all computer’s memory allocation.
2. Web Development: primary colors are written in from of hexadecimal of two digits #RRGGBB, where RR stands for Red, GG stands for Green, BB for Blue.
3. MAC (Media Access Control) Address: The 12 digit hexadecimal number of MAC address. In the format of M M : M M : M M : S S : S S : S S. Where first 6 digits represents adapter manufacturer and
last digits represent the serial number of adapters.
4. Error Messages: Hexadecimal are also use to find memory location of the error. So, it is useful for debugging.
Binary to Hex Conversion Table
Binary Hex
1010 A
1011 B
1100 C
1101 D
1110 E
1111 F
Steps to convert binary to hexadecimal
1) Take the group of four numbers of given binary numbers from the right side.
For Example 011000110110101
by grouping 011 0001 1011 0101
2) Take the values of groups from right to left from the table.
31B5 is an equivalent hexadecimal number of given binary decimal numbers.
So, (011000110110101)[2] = (31B5)[16]
Advantages of using Hexadecimal System
1. It can write larger numbers in small space than binary and decimal. Also, it contains 16 digits.
2. It is easy to convert binary to hexadecimal as well as hexadecimal to binary.
3. Hexadecimal is human-friendly. In addition, To convert from Binary to hexadecimal, we only need to group the numbers. Then we can directly convert binary to hexadecimal vice versa.
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Identifier: Disk Method - APCalcPrep.com
Unfortunately, there is not always a clear delineator between the Disk Method and the Washer Method in just the language of the problem, and you will almost always need to draw out your own picture
of the region they are describing.
• You know you will want to consider the Disk Method as an option if you see the language, “revolving” or “rotating around”.
• If you only receive a single equation to use and you are not given two equations or curves.
Keep in mind that you will often have your region bounded in by other “equations”, but don’t confuse those with the equations you are actually rotating. | {"url":"https://apcalcprep.com/topic/identifier-disk-method/","timestamp":"2024-11-11T09:20:14Z","content_type":"text/html","content_length":"350486","record_id":"<urn:uuid:1aa74fbf-d4ac-41b8-bf4f-01e5fca06353>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00335.warc.gz"} |
Next: MINIMUM ROUTING TREE CONGESTION Up: Spanning Trees Previous: MINIMUM GEOMETRIC STEINER TREE   Index
• INSTANCE: Graph
• SOLUTION: A Steiner network over G that satisfies all the requirements and obeys all the capacities, i.e., a function e, i and j, the number of edge disjoint paths between i and j is at least r
(i,j) where, for each edge e, f(e) copies of e are available.
• MEASURE: The cost of the network, i.e.,
• Good News: Approximable within R is the maximum requirement and, for any n, 194].
• Comment: Also called Minimum Survivable Network. Approximable within 2 when all the requirements are equal [320]. If the requirements are equal and the graph is directed the problem is
approximable within 99]. The variation in which there are no capacity constraints on the edges is approximable within 5]. This problem belongs to the class of problems of finding a minimum-cost
subgraph such that the number of edges crossing each cut is at least a specified requirement, which is a function f of the cut. If f is symmetric and maximal then any problem in this class is
approximable within 2R, where R is the maximum value of f over all cuts [465].
Next: MINIMUM ROUTING TREE CONGESTION Up: Spanning Trees Previous: MINIMUM GEOMETRIC STEINER TREE   Index Viggo Kann | {"url":"https://www.csc.kth.se/~viggo/wwwcompendium/node80.html","timestamp":"2024-11-04T12:29:57Z","content_type":"text/html","content_length":"7315","record_id":"<urn:uuid:48eef85b-ab80-4de7-8f6d-78b5f7e7aa9c>","cc-path":"CC-MAIN-2024-46/segments/1730477027821.39/warc/CC-MAIN-20241104100555-20241104130555-00487.warc.gz"} |
Line Outage Distribution Factors (LODFs)
Line Outage Distribution Factors (LODFs) are a sensitivity measure of how a change in a line’s status affects the flows on other lines in the system. On an energized line, the LODF calculation
determines the percentage of the present line flow that will show up on other transmission lines after the outage of the line. For example, consider an energized line, called LineX, whose present MW
flow is 100 MW. If the LODFs are found to be
LODFs for LineX outage
LineX -100%
LineY + 10%
LineZ - 30%
This means that after the outage of LineX, the flow on LineX will decrease by 100 MW (of course), LineY will increase by 10 MW, and LineZ will decrease by 30 MW. The "flow on Line X" here means the
flow at the from bus going toward the to bus.
Similarly, sensitivities can be calculated for the insertion of a presently open line. In this case, the LODF determines the percentage of the post-insertion line flow that will come from the other
transmission line after the insertion. The "LODF" is better named a Line Closure Distribution Factor (LCDF) in this case.
To calculate the LODFs:
What else are LODFs used for?
LODFs are used extensively when modeling the linear impact of contingencies in Simulator. This is true for the calculation of PTDFs for interfaces which contain a contingent element, as well as when
performing Linear ATC analysis that includes branch contingencies.
When calculating "PTDF" values for interfaces that include contingent elements, the PTDF values reported are actually what are referred to as an Outage Transfer Distribution Factor (OTDF). An OTDF is
similar to PTDF, except an OTDF provides a linearized approximation of the post-outage change in flow on a transmission line in response to a transaction between the Seller and the Buyer. The OTDF
value is a function of PTDF values and LODF values. For a single line outage, the OTDF value for line x during the outage of line y is
OTDFx = PTDFx + LODFx,y * PTDFy
where PTDFx and PTDFy are the PTDFs for line x and y respectively, and LODFx,y is the LODF for line x during the outage of line y. More complex equations are involved when studying contingencies that
include multiple line outages, but the basic idea is the same.
When performing Linear ATC analysis along with calculating OTDFs, Simulator determines the linearized approximation of the post-outage flow on the line. This is similarly determined as
OutageFlowX = PreOutageFlowX + LODFx,y* PreOutageFlowY
where PreOutageFlowX and PreOutageFlowY are the pre-outage flow on lines x and y. | {"url":"https://www.powerworld.com/WebHelp/Content/MainDocumentation_HTML/Line_Outage_Distribution_Factors_LODFs.htm","timestamp":"2024-11-10T06:25:00Z","content_type":"text/html","content_length":"11782","record_id":"<urn:uuid:1dd4af91-7929-4116-b73f-6731818f0b4f>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00552.warc.gz"} |
Epub Bundeswehrreform Und Konversion Nutzungsplanung In Betroffenen Gemeinden 2014
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financial to specifically host four parties. The y-t worksheet, the variable, the package correlation and the wave - elegido. | {"url":"http://www.illinoislawcenter.com/wwwboard/ebook.php?q=epub-bundeswehrreform-und-konversion-nutzungsplanung-in-betroffenen-gemeinden-2014.html","timestamp":"2024-11-08T14:00:40Z","content_type":"text/html","content_length":"75350","record_id":"<urn:uuid:5e3d348e-1dc1-4a5b-b56a-33a6fd150967>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00306.warc.gz"} |
Daniel Lim
Problem 1: Assume the following assignment statements have been executed.
x = 9
y = 2
z = 5
What will the following expressions (involving arithmetic operators) evaluate to?
(You should check your answers by evaluating these expressions on your computer)
x + y * z
x // y
x / y
x ** y – z
What will the following expressions (involving comparison operators) evaluate to?
(You should check your answers by evaluating these expressions on your computer)
x == y
x != z
y < x
y > z
x >= 9
z <= 9
Problem 2: Write an expression that calculates the number of seconds in 77.3 years (the average lifespan of a person living in the U.S. in 2020).
Problem 3: Write an expression using comparison operators to check whether the following statement is true: there are fewer seconds in 3 years than there are hours in 100,000 years (assuming there
are 365 days in a year). | {"url":"https://danielflim.org/phil-through-cs/programming-problems/basics/","timestamp":"2024-11-09T18:42:54Z","content_type":"text/html","content_length":"28332","record_id":"<urn:uuid:c9d9868f-7910-4ecd-8146-05fdd56d1e64>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00300.warc.gz"} |
Mathematics Seminar 11/04/22
Nov 4 3:00 pm
Ahmad Asiri, PhD student, Department of Mathematics and Statistics, MSU
Mathematics Seminar Series
Sign-symmetric signed graphs and signings of graphs
Physical Location
Allen 411
Digital Location
Abstract: A signed graph is a graph in which each edge is labelled plus or minus. A signed graph is sign-symmetric if it is ''switching isomorphic'' to the signed graph obtained by changing the sign
of every edge to the opposite sign. We will discuss several results regarding sign-symmetric signed graphs. The difficult problem of determining the number of signings of a graph will also be
examined. Some parts are joint work with Abdulaziz Alotaibi and Vaidy Sivaraman. | {"url":"https://www.math.msstate.edu/events/2022/10/mathematics-seminar-110422","timestamp":"2024-11-08T05:12:05Z","content_type":"text/html","content_length":"35049","record_id":"<urn:uuid:8fba6616-4a9d-489e-8935-f91cdafc116d>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00221.warc.gz"} |
seminars - Approximation Methods of Multivariate Functions for Homomorphic Data Ordering
Homomorphic Encryption (HE) is a cryptographic primitive that enables computations between encrypted data without decryption.
HE allows operations that use sensitive data to be delegated to others who service outsourced computations while data information is not exposed.
With these characteristics, HE is considered one of the important technologies for privacy preservation.
Most HE schemes, however, support few operations only, mainly multiplication and addition.
Thus, non-polynomial operations between word-wisely encrypted data require much more computational cost than between plain data.
Although many approximation algorithms have been suggested to solve this problem, these works mainly focus on the one-variable functions and cannot be directly generalized to the multivariable
functions because of the algorithmic limit or the growth of computational cost.
In this thesis, we propose new approximation methods of three fundamental multivariate functions: sorting, max index, and softmax.
First, We propose an efficient sorting method for encrypted data that works with approximate comparison.
Using our method as a building block, we exploit k-way sorting network algorithm to show the implementation result that sorting 5^6=15625 data using 5-way sorting network which is about 23.3% faster
than sorting 2^14=16384 data using the general 2-way method.
Second, we propose a polynomial approximation method of the multivariate max function that inherits the method of Cheon emph{et al.} (ASIACRYPT 2020).
Our algorithm is the generalization of the previous two-variable approach of approximating sign function, analyzing that our algorithm requires 30% less depths to find the largest element than using
a state-of-the-art two-variable comparison.
Lastly, we suggest the approximation method for softmax activation for a neural network model.
By exploiting the algorithm, we develop a secure multi-label tumor classification method using the CKKS scheme, the approximate homomorphic encryption scheme. | {"url":"https://www.math.snu.ac.kr/board/index.php?mid=seminars&page=57&sort_index=date&order_type=asc&l=ko&document_srl=828279","timestamp":"2024-11-03T13:14:32Z","content_type":"text/html","content_length":"48947","record_id":"<urn:uuid:3bb991b3-5ff8-4906-99b4-94dbecf2b019>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00742.warc.gz"} |
Immunization and Duration Matching: Strategies for Managing Interest Rate Risk in Fixed Income Portfolios
24.4.1 Immunization and Duration Matching
In the realm of fixed income portfolio management, immunization and duration matching stand out as critical strategies for managing interest rate risk and ensuring that future liabilities are met
with certainty. These techniques are particularly vital for institutional investors, such as pension funds and insurance companies, who have long-term liabilities that must be matched with
appropriate assets. This section delves into the intricacies of immunization and duration matching, providing a comprehensive understanding of their applications, benefits, and limitations.
Understanding Immunization
Immunization is a strategy designed to ensure that a fixed income portfolio will meet a known future liability, regardless of fluctuations in interest rates. The primary objective is to protect the
portfolio from interest rate risk, which can impact both the value of the bonds held and the reinvestment rates of the cash flows generated by those bonds.
Purpose of Immunization
The purpose of immunization is to create a portfolio that is insensitive to interest rate changes, thereby ensuring that the future value of the portfolio’s cash flows will exactly match the future
liability. This is achieved by constructing a portfolio whose duration matches the timing of the liability, effectively balancing the changes in the portfolio’s value with changes in reinvestment
The Concept of Duration
Duration is a key measure in fixed income portfolio management, providing insight into a bond’s sensitivity to interest rate changes. It is a critical component in the immunization process, as it
helps align the portfolio’s cash flows with the timing of the liability.
Macaulay Duration
Macaulay Duration measures the weighted average time to receive the bond’s cash flows. It is expressed in years and provides a time-weighted measure of the bond’s cash flow structure. The formula for
Macaulay Duration is:
$$ D = \frac{\sum_{t=1}^{n} \frac{t \cdot C_t}{(1+y)^t}}{\sum_{t=1}^{n} \frac{C_t}{(1+y)^t}} $$
• \( C_t \) = Cash flow at time \( t \)
• \( y \) = Yield to maturity
• \( n \) = Number of periods
Modified Duration
Modified Duration estimates the percentage change in a bond’s price for a 1% change in yield. It is derived from Macaulay Duration and provides a more direct measure of interest rate sensitivity. The
formula for Modified Duration is:
$$ MD = \frac{D}{1+y} $$
• \( D \) = Macaulay Duration
• \( y \) = Yield to maturity
Duration Matching
Duration matching involves aligning the portfolio’s duration with the timing of the future liability. This strategy ensures that the effects of interest rate changes on the bond’s price and
reinvestment income offset each other, thereby stabilizing the portfolio’s value relative to the liability.
Constructing a Duration-Matched Portfolio
To construct a duration-matched portfolio, follow these steps:
Step 1: Calculate the Duration of the Liability
Determine the duration of the liability, which represents the time-weighted average of the liability’s cash flows. For example, if a payment is due in 7 years, the liability’s duration is 7 years.
Step 2: Select Bonds with Matching Duration
Choose bonds or a combination of bonds whose average duration matches the liability’s duration. This may involve selecting a mix of short-term and long-term bonds to achieve the desired duration.
Step 3: Invest to Meet the Future Obligation
Invest in the selected bonds, ensuring that the present value of the cash flows generated by the portfolio meets the future liability. This requires careful calculation of the portfolio’s cash flows
and their present value.
Limitations of Immunization Strategies
While immunization is a powerful tool for managing interest rate risk, it is not without its limitations. Understanding these limitations is crucial for effective portfolio management.
Reinvestment Risk
One of the primary challenges of immunization is reinvestment risk. Coupons received before the liability date must be reinvested at uncertain future rates, which can impact the portfolio’s ability
to meet the liability.
Non-Parallel Yield Curve Shifts
Duration assumes parallel shifts in the yield curve, meaning that interest rates change by the same amount across all maturities. In reality, different maturities may move differently, leading to
mismatches in the portfolio’s cash flows and liabilities.
Duration is a linear approximation of a bond’s price sensitivity to interest rate changes. For large interest rate changes, convexity adjustments may be needed to accurately estimate the bond’s price
Transaction Costs
Maintaining the duration match requires rebalancing the portfolio as interest rates change and time passes. This rebalancing incurs transaction costs, which can erode the portfolio’s returns.
Monitoring and Rebalancing
To maintain the effectiveness of an immunization strategy, regular monitoring and rebalancing are essential. As time passes and interest rates fluctuate, the portfolio’s duration will change,
necessitating adjustments to realign the portfolio with the liability.
Immunization and duration matching are valuable techniques for managing interest rate risk and ensuring that funds are available to meet future liabilities. By aligning the portfolio’s duration with
the timing of the liability, investors can protect their portfolios from interest rate fluctuations. However, these strategies require diligent management to address their limitations and maintain
their effectiveness.
Quiz Time!
📚✨ Quiz Time! ✨📚
### What is the primary purpose of immunization in fixed income portfolios? - [x] To ensure a fixed income portfolio will meet a known future liability, regardless of interest rate movements. - [ ]
To maximize the yield of a fixed income portfolio. - [ ] To minimize the transaction costs associated with managing a portfolio. - [ ] To increase the liquidity of a fixed income portfolio. >
**Explanation:** Immunization aims to ensure that a fixed income portfolio will meet a known future liability, regardless of interest rate movements, by aligning the portfolio's duration with the
timing of the liability. ### What does Macaulay Duration measure? - [x] The weighted average time to receive the bond's cash flows. - [ ] The percentage change in a bond's price for a 1% change in
yield. - [ ] The total return of a bond over its lifetime. - [ ] The credit risk associated with a bond. > **Explanation:** Macaulay Duration measures the weighted average time to receive the bond's
cash flows, expressed in years. ### How does Modified Duration differ from Macaulay Duration? - [x] Modified Duration estimates the percentage change in a bond's price for a 1% change in yield. - [ ]
Modified Duration measures the weighted average time to receive the bond's cash flows. - [ ] Modified Duration calculates the total return of a bond over its lifetime. - [ ] Modified Duration
assesses the credit risk associated with a bond. > **Explanation:** Modified Duration estimates the percentage change in a bond's price for a 1% change in yield, providing a more direct measure of
interest rate sensitivity. ### What is a key limitation of immunization strategies? - [x] Reinvestment risk, as coupons received before the liability date must be reinvested at uncertain future
rates. - [ ] The inability to match the portfolio's duration with the liability's timing. - [ ] The requirement to invest only in short-term bonds. - [ ] The exclusion of corporate bonds from the
portfolio. > **Explanation:** A key limitation of immunization strategies is reinvestment risk, as coupons received before the liability date must be reinvested at uncertain future rates. ### What is
the main goal of duration matching? - [x] To align the portfolio's duration with the timing of the future liability. - [ ] To maximize the yield of a fixed income portfolio. - [ ] To minimize
transaction costs associated with managing a portfolio. - [ ] To increase the liquidity of a fixed income portfolio. > **Explanation:** The main goal of duration matching is to align the portfolio's
duration with the timing of the future liability, ensuring that the effects of interest rate changes on the bond's price and reinvestment income offset each other. ### Why is regular monitoring and
rebalancing important in an immunization strategy? - [x] To maintain the portfolio's alignment with the liability as time passes and interest rates change. - [ ] To maximize the yield of a fixed
income portfolio. - [ ] To minimize transaction costs associated with managing a portfolio. - [ ] To increase the liquidity of a fixed income portfolio. > **Explanation:** Regular monitoring and
rebalancing are important to maintain the portfolio's alignment with the liability as time passes and interest rates change. ### What assumption does duration make about yield curve shifts? - [x]
Duration assumes parallel shifts in the yield curve. - [ ] Duration assumes non-parallel shifts in the yield curve. - [ ] Duration assumes no shifts in the yield curve. - [ ] Duration assumes random
shifts in the yield curve. > **Explanation:** Duration assumes parallel shifts in the yield curve, meaning that interest rates change by the same amount across all maturities. ### What is the impact
of convexity on duration? - [x] Convexity adjustments may be needed for large interest rate changes, as duration is a linear approximation. - [ ] Convexity eliminates the need for duration matching.
- [ ] Convexity increases the accuracy of duration in all scenarios. - [ ] Convexity decreases the sensitivity of a bond to interest rate changes. > **Explanation:** Convexity adjustments may be
needed for large interest rate changes, as duration is a linear approximation and may not accurately estimate the bond's price change in such scenarios. ### Why is transaction cost a limitation of
immunization strategies? - [x] Maintaining the duration match requires rebalancing, incurring costs. - [ ] Transaction costs are irrelevant to immunization strategies. - [ ] Transaction costs
increase the yield of a fixed income portfolio. - [ ] Transaction costs decrease the liquidity of a fixed income portfolio. > **Explanation:** Maintaining the duration match requires rebalancing the
portfolio as interest rates change and time passes, incurring transaction costs. ### True or False: Immunization strategies eliminate all risks associated with fixed income portfolios. - [ ] True -
[x] False > **Explanation:** Immunization strategies do not eliminate all risks; they primarily address interest rate risk but are subject to limitations such as reinvestment risk, non-parallel yield
curve shifts, and transaction costs. | {"url":"https://csccourse.ca/24/4/1/","timestamp":"2024-11-15T03:02:15Z","content_type":"text/html","content_length":"92913","record_id":"<urn:uuid:1ace36de-288b-467a-b525-4da23d8cc556>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00381.warc.gz"} |
On the Mysteries of MAX NAE-SAT
MAX NAE-SAT is a natural optimization problem, closely related to its better-known relative MAX SAT. The approximability status of MAX NAE-SAT is almost completely understood if all clauses have the
same size \(k\), for some \(k\geq2\). We refer to this problem as MAX NAE-\({k}\)-SAT. For \(k=2\), it is essentially the celebrated MAX CUT problem. For \(k=3\), it is related to the MAX CUT problem
in graphs that can be fractionally covered by triangles. For \(k\geq4\), it is known that an approximation ratio of \(1-{1}/{2}^{k-1}\), obtained by choosing a random assignment, is optimal, assuming
P ≠ NP. For every \(k\geq2\), an approximation ratio of at least ^7⁄[8] can be obtained for MAX NAE-\({k}\)-SAT. There was some hope, therefore, that there is also a ^7⁄[8]-approximation algorithm
for MAX NAE-SAT, where clauses of all sizes are allowed simultaneously.
In this talk, we prove that there is no ^7⁄[8]-approximation algorithm for MAX NAE-SAT, assuming the unique games conjecture (UGC). In fact, even for almost satisfiable instances of MAX NAE-{3,5}-SAT
(i.e., MAX NAE-SAT where all clauses have size 3 or 5), the best approximation ratio that can be achieved, assuming UGC, is at most 0.8739.
Joint work with Joshua Brakensiek, Aaron Potechin and Uri Zwick. | {"url":"https://csp-seminar.org/talks/neng-huang/","timestamp":"2024-11-02T16:52:00Z","content_type":"text/html","content_length":"6313","record_id":"<urn:uuid:4b794527-bc77-47a6-858c-872aaba424b6>","cc-path":"CC-MAIN-2024-46/segments/1730477027729.26/warc/CC-MAIN-20241102165015-20241102195015-00266.warc.gz"} |
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Question Text Average speed of the cyclist is
Updated On Sep 23, 2022
Topic Physics
Subject Science
Class Class 9
Answer Type Video solution: 1
Upvotes 82
Avg. Video Duration 8 min | {"url":"https://askfilo.com/user-question-answers-science/average-speed-of-the-cyclist-is-32313631393438","timestamp":"2024-11-08T02:21:45Z","content_type":"text/html","content_length":"114128","record_id":"<urn:uuid:6057019e-3b68-444e-9ea7-790daeeed936>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00404.warc.gz"} |
Topsoil Calculator
How to Calculate How Much soil You Need with Our topsoil Calculator
Using our soil calculator is quick and easy. Follow these simple steps to estimate the amount of soil, sand, or dirt you'll require for your project:
1. Input Area in Square Feet (sqft): If you do not know the area of your garden then use our yard shape area formulas below to calculate it.
2. Specify Depth in Inches: Next, input the desired thickness of the top soil layer in inches. This thickness will determine how deep you want the topsoil to be.
3. Calculate: Click the "Calculate" button, and our topsoil calculator will instantly provide you with the estimated cubic yards of topsoil needed for your project.
How to Estimate How Much Fill Dirt You Need by Hand
If you prefer to estimate the amount of dirt you need manually, here's a simple formula to follow for a rectangular landscaping project:
Cubic Yards = (Square Feet * Thickness in Inches) / 324
• Square Feet (sqft): Measure the area's length and width in feet, and multiply them to find the square footage.
• Thickness in Inches: Decide how deep you want the dirt layer to be in inches.
• Calculate: Divide the result by 324 to obtain the cubic yardage needed.
While our online calculator simplifies the process, this formula can come in handy when you want to double-check or perform calculations without an internet connection.
Common Yard Shape Area Formulas
If you do not know the area of your yard or specific areas within it, you'll need to use different formulas based on the shape. Here are quick guidelines for calculating the area of common yard
Rectangle or Square Area = Length * Width
Circle Area = 3.14 * Radius^2
Triangle Area = (Base * Height) / 2
These formulas will help you calculate the area accurately, ensuring you have the necessary information to use our topsoil calculator effectively.
An Overview of Topsoil
Topsoil is the upper layer of soil that contains essential nutrients, organic matter, and microorganisms necessary for plant growth. It plays a crucial role in landscaping and gardening, providing a
fertile medium for plants to thrive. Topsoil typically consists of a combination of minerals, organic material, water, and air.
Fill Dirt is an alternative to topsoil and has a lower density of nutrients and is mainly a combination of natural ingredients such as sand, gravel, and rock. Because of this, fill dirt is much
cheaper than topsoil and is mainly used to raise up certain parts of land.
Common Topsoil Types and Price Table (per Cubic Yard)
Here's a brief overview of common top soil types and their approximate prices per cubic yard:
│ Topsoil Type │Price (per Cubic Yard) │
│Screened Topsoil │$20 - $40 │
│Organic Topsoil │$30 - $50 │
│Fill Dirt │$10 - $25 │
│Garden Mix Topsoil │$40 - $60 │
│Sand │$25 - $55 │
Please note that prices may vary depending on your location and the supplier.
Good luck, and don't forget to bookmark this dirt calculator to save time on your next landscaping project. | {"url":"https://www.calculyte.com/other/topsoil-calculator","timestamp":"2024-11-10T02:20:44Z","content_type":"text/html","content_length":"13454","record_id":"<urn:uuid:cdcde257-9230-4fdb-a75f-1ed68632fa0f>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00667.warc.gz"} |
On 2D Inverse Problems/Stieltjes continued fractions - Wikibooks, open books for an open world
Let ${\displaystyle \{a_{k}\}}$ be a sequence of n positive numbers. The Stieltjes continued fraction is an expression of the form, see [KK] & also [JT],
${\displaystyle \beta _{a}(z)=a_{n}z+{\cfrac {1}{a_{n-1}z+{\cfrac {1}{\ddots +{\cfrac {1}{a_{1}z}}}}}}}$
or its reciprocal ${\displaystyle \beta _{a}\beta _{a}^{*}(z)=1.}$
The function defines a rational n-to-1 map of the right half of the complex plane onto itself,
${\displaystyle \beta _{a},1/\beta _{a}:\mathbb {C^{+}} {\xrightarrow[{}]{n\leftrightarrow 1}}\mathbb {C^{+}} ,}$
${\displaystyle {\begin{cases}Re(z_{1}),Re(z_{2})>0\implies Re(z_{1}+z_{2})>0,\\Re(z)>0\implies Re(1/z)>0,\\Re(z)>0,a>0\implies Re(az)>0.\end{cases}}}$
Exercise(***). Use the mapping properties of Stieltjes continued fractions to prove that their interlacing, simple and symmetric zeros and poles lie at the origin and the imaginary axes and that
the properties and rationality characterize the continued fractions.
Exercise(**). Prove that the continued fractions 've the representation ${\displaystyle \beta _{a}(z)=z(\xi _{\infty }+\sum _{k}{\frac {\xi _{k}}{z^{2}+\theta _{k}^{2}}}),{\mbox{ where }}\xi _{\
infty },\xi _{k}{\mbox{ and }}\theta _{k},k\in \mathbb {N} }$, 're non-negative real numbers, and the fractions 're characterized by it.
The function ${\displaystyle \beta _{a}}$ is determined by the pre-image of unity (i.e. n points, counting multiplicities), since
${\displaystyle \beta _{a}(z)={\frac {p(z^{2})}{zq(z^{2})}}=1\iff p(z^{2})-zq(z^{2})=0,}$
and a complex polynomial is determined by its roots up to a multiplicative constant by the fundamental theorem of algebra.
Let ${\displaystyle \sigma _{l}}$ be the elementary symmetric functions of the set ${\displaystyle \mathrm {M} }$. That is,
${\displaystyle \prod _{k}(z-\mu _{k})=\sum _{k}\sigma _{n-k}z^{k}.}$
Then, the coefficients ${\displaystyle a_{k}}$ of the continued fraction are the pivots in the Gauss-Jordan elimination algorithm of the following ${\displaystyle n\times n}$ square Hurwitz matrix:
${\displaystyle H_{\mathrm {M} }={\begin{pmatrix}\sigma _{1}&\sigma _{3}&\sigma _{5}&\sigma _{7}&\ldots &0\\1&\sigma _{2}&\sigma _{4}&\sigma _{6}&\ldots &0\\0&\sigma _{1}&\sigma _{3}&\sigma _{5}&
\ldots &0\\0&1&\sigma _{2}&\sigma _{4}&\ldots &0\\0&0&\sigma _{1}&\sigma _{3}&\ldots &0\\\vdots &\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&0&\ldots &\sigma _{n}\\\end{pmatrix}}}$
and, therefore, can be expressed as the ratios of monomials of the determinants of the blocks of ${\displaystyle \mathrm {M} }$.
Exercise (**). Prove that
${\displaystyle a_{1}=1/\sigma _{1},a_{2}={\frac {\sigma _{1}^{2}}{\det {\begin{pmatrix}\sigma _{1}&\sigma _{3}\\1&\sigma _{2}\end{pmatrix}}}},a_{3}={\frac {\det {\begin{pmatrix}\sigma _{1}&\
sigma _{3}\\1&\sigma _{2}\end{pmatrix}}^{2}}{\sigma _{1}\det {\begin{pmatrix}\sigma _{1}&\sigma _{3}&0\\1&\sigma _{2}&\sigma _{4}\\0&\sigma _{1}&\sigma _{3}\end{pmatrix}}}},\ldots }$
Exercise (*). Use the previous exercise to prove that
${\displaystyle \prod _{k}a_{k}={\frac {1}{\prod _{k}\mu _{k}}}=1/\sigma _{n}.}$ | {"url":"https://en.m.wikibooks.org/wiki/On_2D_Inverse_Problems/Stieltjes_continued_fractions","timestamp":"2024-11-08T21:37:31Z","content_type":"text/html","content_length":"70591","record_id":"<urn:uuid:e318ec3a-2f50-47d7-acfc-e5f2f64bc114>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00882.warc.gz"} |
Spool math and demo
Been trying to understand spool math to drive my motors. Found this resource to be very helpful and managed to put together a simple demo.
Hopefully this will be helpful to others. Please let me know if this is similar to methods others have tried.
4 Likes
I love this hand-on approach to learning about spool speeds and distances. So cool. Thanks for posting!
very helpful, I think spool math is one of the key problems to resolve.
theoretically, just one motor should be sufficient to do the take up, given sufficient tension from the other reel, granting a sensor near or at the film gate.
i recently shot 16mm on a krasnogorsk, and what suprised me the most was that it was entirely mechanical: the motor was wound up by hand, and then the tension release through a weight mechanism.. I
wouldn’t be surprised if whoever filmed the krasnogorsk disassembly is part of this forum.
Good calculations, but perhaps a bit too complex.
Here is a simpler one, which provides almost exact results (based on the example at the bottom).
If the inner and outer diameters are D0 and D1, then calculate the cross-sectional area of the roll as follows (where r1=D1/2 and r0=D0/2):
Pi * (r1^2 - r0^2)
Now simply divide this by the thickness h.
Using the numerical example on this page,
approximate formula roll length: 235.4623694 m
exact formula roll length: 235.6194545 m
this formula: 235.619449 m (which matches the exact formula to 7 digits).
1 Like | {"url":"https://forums.kinograph.cc/t/spool-math-and-demo/1462","timestamp":"2024-11-11T14:54:57Z","content_type":"text/html","content_length":"17633","record_id":"<urn:uuid:27d5a6ab-c34a-4f60-ace1-ee1145f5d9fa>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00093.warc.gz"} |
HC Verma Class XII Physics Chapter 15 –Wave Motion and Wave on a String
Exercise : Solution of Questions on page Number : 321
Answer: 1
No, in wave motion there is no actual transfer of matter but transfer of energy between the points where as when wind blows air particles moves with it.
Answer: 2
It is a non-mechanical wave because this type of wave does not require a material medium to travel.
Answer: 3
Equation of the wave is
y=c1 sin (c2x+c3t)
When the variable of the equation is (c2x + c3t), then the wave must be moving in the negative x-axis with time t.
Answer: 4
Equation of the wave is given by
Answer: 5
When two wave pulses identical in shape but inverted with respect to each other meet at any instant, they form a destructive interference. The complete energy of the system at that instant is stored
in the form of potential energy within it. After passing each other, both the pulses regain their original shape.
Answer: 6
No, componendo and dividendo is not applicable. We cannot add quantities of different dimensions.
Answer: 7
Equation of the wave: y = A sin(kx − ωt + Φ)
Here, A is the amplitude, k is the wave number, ω is the angular frequency and Φ is the initial phase.
The argument of the sine is a phase, so the smallest positive phase constant should be
Therefore, the smallest positive phase constant is 1.5π.
Answer: 8
Yes, at the centre. The centre position is a node. If the string vibrates in its first overtone, then there will be two positions, i.e., two nodes, one at x = 0 and the other at x = L.
Exercise : Solution of Questions on page Number : 322
Answer: 1
(c) λ/2
A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. The speeds at the maximum point and at the minimum point are same although the direction of
motion are different. The difference between the positions of maxima and minima is equal to λ/2.
Answer: 2
(c) λ/2
A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. Therefore, applying similar argument we can say that if a particular particle has zero
displacement at a certain instant, then the particle closest to it having zero displacement is at a distance is equal to λ/2.
Answer: 3
(a) x=A sin ky-ωt
Here x is the particle displacement of the wave and the wave is travelling along the Y-axis because the particle displacement is perpendicular to the direction of wave motion.
Answer: 4
(b) amplitude A/2, frequency ω/π
Thus, we have:
Amplitude = A2
Frequency = 2ω2π=ωπ
Answer: 5
(d) Sound waves
There are mainly two types of waves: first is electromagnetic wave, which does not require any medium to travel, and the second is the mechanical wave, which requires a medium to travel. Sound
requires medium to travel, hence it is a mechanical wave.
Answer: 6
(a) ν
The boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.
Answer: 7
(a) 1/2
Wave speed is given by
T is the tension in the string
v is the speed of the wave
μ is the mass per unit length of the string
M is the mass of the string, which can be written as ρV
L is the length of the string
where D is the diameter of the string.
Thus, v ∝ 1D
Since, rA = 2rB
vA∝12rA∝12×2rB (1)vB∝12rB (2)
From Equations (1) and (2) we get
Answer: 8
(d) 1/2
TAB is the tension in the string AB
TCD is the tension in the string CD
The eelation between tension and the wave speed is given by
v is the wave speed of the transverse wave
μ is the mass per unit length of the string
Answer: 9
(d) meaningless
Sound wave is a mechanical wave; this means that it needs a medium to travel. Thus, its velocity in vacuum is meaningless.
Answer: 10
(c) λ'<λ
As v=fµ
A wave pulse travels faster in a thinner string.
The wavelength of the transmitted wave is equal to the wavelength of the incident wave because the frequency remains constant.
Answer: 11
(b) 2a
We know that the resultant of the amplitude is given by
Rnet=A12+A22+2A1A2 cosϕ
For the particular case, we can write
Answer: 12
(d) the information is insufficient to find the relation between t1 and t2.
But because the length of wires A and B is not known, the relation between A and B cannot be determined.
Answer: 13
(b) the velocity but not for the kinetic energy
The principle of superposition is valid only for vector quantities.
Velocity is a vector quantity, but kinetic energy is a scalar quantity.
Answer: 14
(d) The pulses will pass through each other without any change in their shapes.
The pulses continue to retain their identity after they meet, but the moment they meet their wave profile differs from the individual pulse.
Answer: 15
(b) 2A2
We know resultant amplitude is given by
For maximum resultant amplitude
For minimum resultant amplitude
So, the difference between Amax and Amin is
Answer: 16
(d) between 0 and 2A
The amplitude of the resultant wave depends on the way two waves superimpose, i.e., the phase angle (φ). So, the resultant amplitude lies between the maximum resultant amplitude (Amax) and the
minimum resultant amplitude (Amin).
Amax = A + A = 2A
Amin = A − A = 0
Answer: 17
(a) A
We know the resultant amplitude is given by
Rnet=A2+A2+2A2 cos 120º (ϕ=120°)=2A2-A2
∵ cos 120º=-12=A
Answer: 18
(a) inverse of its length
The relation between wave speed and the length of the string is given by
l is the length of the string
F is the tension
μ linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.
Exercise : Solution of Questions on page Number : 323
Answer: 1
Speed of the wave pulse passing on a string in the negative x-direction = 40 cms−1
As the speed of the wave is constant, the location of the maximum after 5 s will be
s = v × t
= 40 × 5
= 200 cm (along the negative x-axis)
Therefore, the required maximum will be located after x = −2 m.
Answer: 2
Equation of the wave travelling on a string stretched along the X-axis:
y=Ae xa+tT-2
(a) The dimensions of A (amplitude), T (time period) and a=λ2π, which will have the dimensions of the wavelength, are as follows:
A = M0L1T0T=M0L0T-1a=M0L1T0
(b) Wave speed, ν=λT=aT λ=a
(c) If y=f t+xν, then the wave travels in the negative direction; and if y=f t-xν, then the wave travels in the positive direction.
Thus, we have:
y=Ae xa+tT-2 =Ae-1Tt+xTa2 =Ae-1Tt+xV = Ae-ft+xV
Hence, the wave is travelling is the negative direction.
(d) Wave speed, v=at
Maximum pulse at t = T = aT×T=a Along the negative x-axis
Maximum pulse at t = 2T = aT×2T=2a Along the negative x-axis
Therefore, the wave is travelling in the negative x-direction.
Answer: 3
Wave pulse at t = 0
Wave speed = 10 cms−1
Using the formula s=v×t, we get:
At : t=1 s, s1=ν×t=10×1=10 cmt = 2 s,
s2=ν×t=10×2=20 cmt = 3 s,
s3=ν×t=10×3=30 cm
Answer: 19
(b) 480 Hz
The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be natural frequency of the wire, then standing waves with large amplitude are set up in it.
Answer: 20
(b) 480 Hz
The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be the natural frequency of the wire, standing waves with large amplitude are set in it.
Answer: 21
(b) vibrate with a frequency of 208 Hz
According to the relation of the fundamental frequency of a string
l is the length of the string
F is the tension
μ is the linear mass density
We know that ν1 = 416 Hz, l1 = l and l2 = 2l.
v1∝1l1v1l1=v2l2416l=v22lv2=208 Hz
Answer: 22
(d) 16 kg
According to the relation of the fundamental frequency of a string
where l is the length of the string
F is the tension
μ is the linear mass density of the string
We know that ν1 = 416 Hz, l1 = l and l2 = 2l.
Also, m1 = 4 kg and m2 = ?
ν1=12l1m1gμ (1)
ν2=12l2m2gμ (2)
So, in order to maintain the same fundamental mode
squaring both sides of equations (1) and (2) and then equating
14l24gμ = 116l2m2gμ⇒m2=16 kg
Answer: 1
(c) may move on the X-axis
(d) may move on the Y-axis
A mechanical wave is of two types: longitudinal and transverse. So, a particle of a mechanical wave may move perpendicular or along the direction of motion of the wave.
Answer: 2
(d) in the X–Y plane
In a transverse wave, particles move perpendicular to the direction of motion of the wave. In other words, if a wave moves along the Z-axis, the particles will move in the X–Y plane.
Answer: 3
(d) be polarised
A longitudinal wave has particle displacement along its direction of motion; thus, it cannot be polarised.
Answer: 4
(b) may be longitudinal
(d) may be transverse
Particles in a solid are very close to each other; thus, both longitudinal and transverse waves can travel through it.
Answer: 5
(a) must be longitudinal
Because particles in a gas are far apart, only longitudinal wave can travel through it.
Answer: 6
(b) A and B move in opposite directions.
(d) The displacements at A and B have equal magnitudes.
A and B have a phase difference of π. So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by any odd multiple of their
Answer: 7
(c) Frequency = 25/π Hz
(d) Amplitude = 0⋅001 mm
y=0·001 mm sin50 s-1t+2·0 m-1x
Equating the above equation with the general equation, we get:
Here, A is the amplitude, ω is the angular frequency, k is the wave number and λ is the wavelength.
A=0.001 mmNow,50=2πν⇒ν=25π Hz
Answer: 8
(a) must be an integral multiple of λ/4
A standing wave is produced on a string clamped at one end and free at the other.
Its fundamental frequency is given by
Answer: 9
(b) The energy of any small part of a string remains constant in a standing wave.
A standing wave is formed when the energy of any small part of a string remains constant. If it does not, then there is transfer of energy. In that case, the wave is not stationary.
Answer: 10
(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase
All particles in a particular segment between two nodes vibrate in the same phase, but the particles in the neighbouring segments vibrate in opposite phases, as shown below.
Thus, particles in alternate antinodes vibrate in the same phase.
Exercise : Solution of Questions on page Number : 324
Answer: 4
Pulse travelling on a string, y=a3x-νt2+a2
a=5 mm=0.5 cmWave speed, ν = 20 cm/s
So, at t=0 s, y=a3x2+a2.
Similarly, at t = 1 s,
y=a3x-ν2+a2And, At t=2 s,y=a3x-2ν2+a2
To sketch the shape of the string, we have to plot a graph between y and x at different values of t.
Answer: 5
Equation of the wave travelling in the positive x-direction at x = 0:
Wave speed = v
Wavelength, λ = vT
T = Time period
Therefore, the general equation of the wave can be represented by
Answer: 6
The shape of the string at t = 0 is given by g(x) = A sin(x/a), where A and a are constants.
Dimensions of A and a are governed by the dimensional homogeneity of the equation g(x) = A sin(x/a).
(a) M0L1T0=A⇒A=LAnd, a=M0L1T0⇒a=L(b) Wave speed=ν
∴ Time period, T=aνHere, a=Wave length=λ
The general equation of wave is represented byy=Asinxa-tav =Asinx-νta
Answer: 7
Wave velocity = ν
Shape of the string at t=t0 = gx, t0=A sin x/a …(i)
For a wave travelling in the positive x-direction, the general equation is given by
y=A sin xa-tT
Putting t = − t and comparing with equation (i),
we get:
gx,0=Asinxa+t0T⇒gx,t=Asinxa+t0T-tTNow,T=aν Here, a=Wave lengthν=Velocity of the wave
Thus, we have : y=Asin xa+t0aν-taν
⇒y=Asin x+ν t0-ta
Answer: 8
Equation of the wave, y=0.10 mm sin31.4 m-1x+314 s-1 t
The general equation is y=Asin2πxλ+ωt.
From the above equation, we can conclude:
(a) The wave is travelling in the negative x-direction.
(b) 2πλ=31.4 m-1
⇒λ=2π31.4=0.2 m= 20 cmAnd, ω=314 s-1⇒2πf=314⇒f=3142π =3142×3.14 =50 s-1=50 Hz
Wave speed:
ν=λf=20×50 = 1000 cm/s
(c) Maximum displacement, A = 0.10 mm
Maximum velocity=aω=0.1×10-1×314 =3.14 cm/s
Answer: 9
A wave travels along the positive x-direction.
Wave amplitude (A) = 0.20 cm
Wavelength (λ) = 20 cm
Wave speed (v) = 20 m/s
(a) General wave equation along the x-axis:
∴k=2πλ=2π2=π cm-1T=λν=22000 =11000=10-3 sω=2πT=2π×103 s-1
Wave equation:
y=0.2 cm sinπ cm-1 x-2π×10-3 s-1
(b) As per the question
For the wave equation ,we need to find the displacement and velocity at x = 2 cm and t = 0.
y=0.2 cm sin2π=0
∴ν=Aωcosπx =0.2×2000π×cos2π =400π =400π cm/s=4π m/s
If the wave equation is written in a different fashion, then also we will get the same values for these quantities.
Answer: 10
The wave equation is represented by
y=1·0 mm sin πx2·0 cm-t0·01 s
Time period = T
Wavelength = λ
a T=2×0.01=0.02 s=20 ms λ=2×2=4 cm
(b) Equation for the velocity of the particle:
ν=dydt=ddtsin 2πx4-t0.02
=-0.50 cos 2πx4-t0.02×10.02⇒ν=-0.50 cos 2π x4-t0.02At x=1 and t=0.01 s, ν=-0.50 cos 2π 14-12=0.
(c) (i) Speed of the particle:
At x=3 cm and t=0.01 s, ν=-0.50cos2π34-12=0.
(ii) At x=5 cm and t=0.01 s,
ν=0 iii At x=7 cm and t=0.1 s, ν=0.iv At x=1 cm and t=0.011 s,ν=50 cos 2π14-0.0110.02 =-50 cos 3π5=-9.7 cm/s
(By changing the value of t, the other two can be calculated.)
Answer: 11
Time taken to reach from the mean position to the extreme position, T4 = 5 ms
Time period (T) of the wave:
T=4×5 ms =20×10-3=2×10-2 s
Wavelength (λ) = 2×Distance between two mean positions
=2×2 cm=4 cm
Frequency, f=1T =12×10-2 =50 Hz
Wave speed, v= λf =4×10-2×50 =200×10-2 =2 m/s
Answer: 12
Wave speed, ν=20 cm/s
From the graph, we can infer:
(a) Amplitude, A = 1 mm
(b) Wavelength, λ = 4 cm
(c) Wave number, k=2πλ
=2×3.144=1.57 cm-2
(d) Time period, T=λν
Frequency, f=1T=vλ⇒f=204=5 Hz
Answer: 13
Wave speed (v) = 10 ms−1
Time period (T) = 20 ms =20×10-3=2×10-2 s
(a) Wavelength of the wave:
λ=νt=10×2×10-2 = 0.02 m=20 cm
(b) Displacement of the particle at a certain instant:
Phase difference of the particle at a distance x = 10 cm :
The displacement is given by y’=asinωt-kx+π =asinωt-kx=1.5 mm∴Displacement=1.5 mm
Answer: 14
Length of the steel wire = 64 cm
Weight = 5 g
Applied force = 8 N
Thus, we have:
Mass per unit length=564 gm/cmTension, T=8 N =8×105 dynSpeed, v=Tm =8×105×645 =3200 cm/s=32 m/s
Answer: 15
Length of the string = 20 cm
Linear mass density of the string = 0.40 g cm−1
Applied tension = 16 N = 16×105 dyn
Velocity of the wave:
ν=Tm =16×1050.4 =2000 cm/s
∴ Time taken to reach the other end =202000=0.01 s
Time taken to see the pulse again in the original position=0.01×2=0.02 s
(b) At t = 0.01 s, there will be a trough at the right end as it is reflected.
Answer: 16
Linear mass density of the string = 0.5 gcm−1
Total length of the string = 30 cm
Speed of the wave pulse = 20 cms−1
The crest reflects the crest here because the wave is travelling from a denser medium to a rarer medium.
Phase change=0
(a) Total distance, S=20+20=40 cmWave speed, ν=20 m/s
Time taken to regain shape:
Time=Sν=4020=2 s
(b) The wave regain its shape after covering a period distance=2×30=60 cm
∴ Time period=6020=3 s
(c) Frequency, n=1Time period=13 s-1
We know:
Here, T is the tension in the string.
m=Mass per unit length =0.5 gm/cm⇒13=12×30 T0.5⇒ T=400×0.5 =200 dyn =2×10-3 N
Answer: 17
m = Mass per unit length of the first wire
a = Area of the cross section
ρ = Density of the wire
T = Tension
Let the velocity of the first string be v1.
Thus, we have:
The mass per unit length can be given as
m1=ρ1a1I1I1=ρ1a1⇒ν1=Tρ1a1 …(1)
Let the velocity of the first string be v2.
Thus, we have:
ν2=Tm2⇒ ν2=Tρ2a2 …(2)
ν1=2ν2⇒Ta1ρ1=2Ta2ρ2⇒Ta1ρ1=4 Ta2ρ2⇒ρ1ρ2=14⇒ρ1:ρ2 = 1:4
Exercise : Solution of Questions on page Number : 325
Answer: 18
Wave equation, y=0·02 msin1·0 m-1x+30 s-1t
Let: Mass per unit length, m=1.2×10-4 kg/mFrom the wave equation, we have:k=1 m-1=2πλAnd,ω=30 s-1=2πf
Velocity of the wave in the stretched string is given by
ν=λf=ωk=301⇒v=30 m/sWe know:v=Tm⇒30 =T1.2×10-4 ⇒T=108×10-3=0.108 N
So, the tension in the string is 0.108 N.
Answer: 19
Amplitude of the wave = 1 cm
Frequency of the wave, f=2002=100 Hz
Mass per unit length, m = 0.1 kg/m
Applied tension, T = 90 N
(a) Velocity of the wave is given by
Thus, we have:
v=900.1=30 m/sNow,Wavelength, λ=vf=30100=0.3 m⇒λ=30 cm
(b) At x = 0, displacement is maximum.
Thus, the wave equation is given by
y=1 cmcos2πt0.01 s-x30 cm …(1)
(c) Using cos-θ=cosθ in equation (1), we get:
Velocity, v=dydt⇒v=2π0.01sin2πx30-t0.01And,Acceleration, a=dνdt⇒a=4π20.012cos2πx30-t0.01When x=50 cm, t=10 ms=10×10-3 s.
v=2π0.01sin2π53-0.010.01 =2π0.01sin2π×23 =-2π0.01sin4π3 =- 200πsinπ3 =-200π×32 =-544 cm/s =-5.4 m/s
In magnitude, v = 5.4 m/s.
a=4π20.012cos2π53-1 =4π2×104×12 ≈2×105 cm/s2 or 2 km/s2
Answer: 20
Length of the string, L = 40 cm
Mass of the string = 10 gm
Mass per unit length=1040=14 gm/cm
Spring constant, k = 160 N/m
Deflection, x=1 cm =0.01 mTension, T= kx=160 ×0.01⇒T=1.6 N=16×104 dynNow, v=Tm=16×10414⇒v=8×102 cm/s=800 cm/s
∴ Time taken by the pulse to reach the spring, t=40800=120=0.05 s
Answer: 21
Mass of each block = m1=m2=3.2 kg
Linear mass density of wire AB = 10 gm−1 = 0.01 kgm−1
Linear mass density of wire CD = 8 gm−1 = 0.008 kgm−1
For string CD, velocity is defined as v=Tm.
Here, T is the tension and m is the mass per unit length.
For string CD,
Thus, we have:
v =3.2×100.008 =32×1038 =2×1010 =20×3.14≈63 ms
For string AB,
T=2×3.2g=64 NThus, we have:v=Tm =640.01=6400 =80 m/s
Answer: 22
Mass of the block = 2 kg
Total length of the string = 2 + 0.25 = 2.25 m
Mass per unit length of the string:
m=4.5×10-32.25 =2×10-3 kg/mT=2g=20 NWave speed, ν=Tm =202×10-3 = 104 = 102 m/s=100 m/s
Time taken by the disturbance to reach the pulley:
t=sν =2100=0.02 s
Answer: 23
Mass of the block = 4.0 kg
Linear mass density, m=19.2×10-3 kg/m
From the free body diagram,
T-4g-4a=0⇒T=4a+g =42+10=48 N
Wave speed, ν=Tm =4819.2×103 =2.5×10-3=50 m/s
Answer: 24
Speed of the transverse pulse when the car is at rest, v1 = 60 cm s−1
Speed of the transverse pulse when the car accelerates, v2= 62 cm s−1
Mass of the heavy ball suspended from the ceiling = M
Mass per unit length = m
Wave speed, ν=Tm=MgmWhen car is at rest:Tension in the string, T=Mg⇒v1=Mgm⇒Mgm=602 …(i)
When car is having acceleration:
Tension, T=Ma2+Mg2
Again, ν2=Tm⇒62=Ma2+Mg21/4m1/2
⇒ Ma2+Mg2m=622 …(ii)
From equations (i) and (ii), we get:
Mgm×mMa2+Mg2=60622⇒ga2+g2=0.936⇒g2a2+g2=0.876⇒a2+100 0.876=100⇒ a2=12.40.867=14.15⇒a=3.76 m/s2Therefore, acceleration of the car is 3.76 m/s2.
Answer: 25
V = Linear velocity of the string
m = Mass per unit length of the the string.
R = Radius of the loop
ω = Angular velocity
Consider one half of the string, as shown in the figure.
The half loop experiences centrifugal force at every point (away from the centre) balanced by tension 2T.
Consider an element of angular part dθ at angle θ.
Length of the element=Rdθ, mass =mRdθ
Centrifugal force experienced by the element=mRdθ ω2R
Resolving the centrifugal force into rectangular components,
Since the horizontal components cancel each other, the net force on the two symmetric elements is given as
dF=2mR2dθω2 sinθ
Total force, F=∫0π/22mR2ω2sinθdθ =2mR2ω2 -cos θ =2mR2ω2And, 2T=2mR2ω2⇒T=mR2ω2
Velocity of the transverse vibration is given as
Linear velocity of the string, V = ωR
∴ Speed of the disturbance, V’ = V
Answer: 26
(a) Let m be the mass per unit length of the string.
Consider an element at a distance x from the lower end.
Weight acting downwards = (mx)g
∴ Tension in the string at the upper part = mgx
The velocity of transverse vibration is given as
(b) Let the time taken be dt for the small displacement dx.
Thus, we have:
∴Total time, T=∫0Ldxgx=4Lg
(c) Suppose after time t, the pulse meets the particle at a distance y from the lower end of the rope.
t=∫0ydxgx =4yg
∴ Distance travelled by the particle in this time, S = L-y
Using the equation of motion, we get:
S=ut+12 gt2⇒L-y=12 g×4yg2⇒L-y=2y⇒3y=L⇒y=L3
Thus, the particle will meet the pulse at a distance L3 from the lower end.
Answer: 27
Linear density of each of two long strings A and B, m = 1.2×10-2 kg/m
String A is stretched by tension Ta= 4.8 N.
String B is stretched by tension Tb= 7.5 N.
Let va and vb be the speeds of the waves in strings A and B.
va=Tam⇒va=4.81.2×10-2=20 m/svb=Tbm⇒vb=7.51.2×10-2=25 m/st1=0 in string At2=0+20 ms=20×10-3=0.02 s
Distance travelled by the wave in 0.02 s in string A:
s=20×0.02=0.4 m
Relative speed between the wave in string A and the wave in string B, v’=25-20=5 m/s
Time taken by the wave in string B to overtake the wave in string A = Time taken by the wave in string B to cover 0.4 m
t’=sv’=0.45=0.08 s
Answer: 28
Amplitude of the transverse wave, r = 0.5 mm =0.5×10-3 m
Frequency, f = 100 Hz
Tension, T = 100 N
Wave speed, v = 100 m/s
Thus, we have:
ν=Tm⇒ν2=Tm⇒m=Tν2=1001002 =0.01 kg/mAverage power of the source:Pavg=2π2mνr2f2 =2 3.142 0.01×100×0.5×10-32×100 =2×9.86×0.25×10-6×104 =19.7×0.0025=0.049 W =49×10-3 W=49 mW
Answer: 29
Frequency of the wave, f = 200 Hz
Amplitude, A = 1 mm = 10−3 m
Linear mass density, m = 6 gm−3
Applied tension, T = 60 N
Let the velocity of the wave be v.
Thus, we have:
v=Tm=606×10-3=102=100 m/s
(a) Average power is given as
Paverage=2π2mνA2f2 =2×3.142×6×10-3×100×10-3×2002 =473×10-3=0.47 W
(b) Length of the string = 2 m
Time required to cover this distance:
t=2100=0.02 sEnergy=Power×t =0.47×0.02 =9.4×10-3 J=9.4 mJ
Answer: 30
Frequency of the tuning fork, f = 440 Hz
Linear mass density, m = 0.01 kgm−1
Applied tension, T = 49 N
Amplitude of the transverse wave produce by the fork = 0.50 mm
Let the wavelength of the wave be λ.
(a) The speed of the transverse wave is given by
⇒v=490.01=70 m/s
Also, ν=fλ∴ λ=fv=70440=16 cm
(b) Maximum speed (vmax) and maximum acceleration (amax):
We have: y=A sin ωt-kx
∴ ν=dydt=Aω cos ωt-kxNow,νmax=dydt=Aω =0.50×10-3×2π×440 =1.3816 m/s.And, a=d2ydt2⇒a=-Aω2 sin ωt-kxamax=-Aω2 =0.50×10-3×4π2 4402 =3.8 km/s2
(c) Average rate (p) is given by
p=2π2νA2f2 =2×10×0.01×70×0.5×10-32×4402 =0.67 W
Answer: 31
Phase difference between the two waves travelling in the same direction, ϕ=90°=π2.
Frequency f and wavelength λ are the same. Therefore, ω will be the same.
Let the wave equations of two waves be:
Here, r is the amplitude.
From the principle of superposition, we get:
y=y1+y2 =rsinωt+rsinωt+π2 =r sinωt+sinωt+π2 =r2sinωt+ωt+π22cosωt-ωt-π22 =2rsinωt+π4cos-π4 =2rsinωt+π4
∴ Resultant amplitude, r’=2r=42 mm
Answer: 32
Speed of the wave pulse travelling in the opposite direction, v = 50 cm s−1 = 500 mm s−1
Distances travelled by the pulses:
Using s = vt, we get:
In t=4 ms=4×10-3 s,s=νt=500×4×10-3=2 mm.In t=6 ms=6×10-3 s,s=500×6×10-3=3 mm.In t=8 ms=8×10-3 s,s=νt=500×8×10-3= 4 mm.In t=12 ms=12×10-3 s,s=500×12×10-3=6 mm.
The shapes of the string at different times are shown in the figure.
Exercise : Solution of Questions on page Number : 326
Answer: 33
Two waves have same frequency (f), which is 100 Hz.
Wavelength (λ) = 2.0 cm =2×10-2 m
Wave speed, ν =f×λ=100×2×10-2 m/s =2 m/s
(a) First wave will travel the distance in 0.015 s.
⇒x=0.015×2 =0.03 m
This will be the path difference between the two waves.
So, the corresponding phase difference will be as follows:
ϕ=2πxλ =2π2×10-2×0.03=3π
(b) Path difference between the two waves, x = 4 cm = 0.04 m
So, the corresponding phase difference will be as follows:
⇒ϕ=2πxλ =2π2×10-2×0.04 =4π
(c) The waves have same frequency, same wavelength and same amplitude.
Let the wave equation for the two waves be as follows:
y1=r sin ωtAnd, y2=r sin ωt+ϕBy using the principle of superposition:y=y1+y2 =rsinωt+ωt+ϕ =2rsinωt+ϕ2 cosϕ2
∴ Resultant amplitude =2r cosϕ2
So, when ϕ=3x:⇒r=2×10-3 mRresultant=2×2×10-3 cos 3π2 =0
Again, when ϕ=4π:Rresultant=2×2×10-3 cos 4π2 =4×10-3×1 =4 mm
Answer: 34
Length of a stretched string (L) = 1 m
Wave speed (v) = 60 m/s
Fundamental frequency (f0) of vibration is given as follows:
=602×1=30 s-1=30 Hz
Answer: 35
Length of the wire (L)= 2.00 m
Fundamental frequency of the vibration (f0) = 100 Hz
Applied tension (T) = 160 N
Fundamental frequency, f0=12LTm⇒10=14160m⇒m=1×10-3 kg/m⇒m=1 g/m
So, the linear mass density of the wire is 1 g/m.
Answer: 36
Mass of the steel wire = 4.0 g
Length of the steel wire = 80 cm = 0.80 m
Tension in the wire = 50 N
Linear mass density (m) =480 g/cm =0.005 kg/m
Wave speed, ν=Tm =500.005=100 m/s
Fundamental frequency, fo=12LTm =12×0.8×500.005 =1002×0.8=62.5 HzFirst harmonic=62.5 HzIf f4= frequency of the fourth harmonic:⇒f4=4f0=62.5×4⇒f4=250 HzWavelength of the fourth harmonic, λ4=νf4=
100250⇒λ4=0.4 m=40 cm
Answer: 37
Length of the piano wire (L)= 90.0 cm = 0.90 m
Mass of the wire = 6.00 g = 0.006 kg
Fundamental frequency (fo) = 261.63 Hz
Linear mass density, m=690 gm/cm =6×10-390×10-2 kg/m =6900 kg/m
Fundamental frequency, fo=12LTm
⇒ 261.63=12×0.09 T×9006⇒0.18×261.63=150 T⇒150 T=261.63×0.182⇒T=261.63×0.182150 =1478.52 N≈1480 N
Hence, the tension in the piano wire is 1480 N.
Answer: 38
Length of the sonometer wire (L) = 1.50 m
Let the first harmonic be f0 and the second harmonic be f1.
According to the question, f1 =256 Hz
1st harmonic for fundamental frequency, f0=f12=2562=128 Hz
When the fundamental wave is produced, we have:
λ2=1.5 m⇒ λ=3 m
Wave speed, v=f0λ⇒v=128×3=384 m/s
Answer: 39
Length of the wire between two pulleys (L) = 1.5 m
Mass of the wire = 12 gm
Mass per unit length, m=121.5 g/m =8×10-3 kg/m
Tension in the wire, T=9×g =90 N
Fundamental frequency is given by:
f0=12L Tm
For second harmonic (when two loops are produced):
f1=2f0=11.5 908×10-3 =106.061.5 =70.7 Hz≈70 Hz
Answer: 40
Length of the stretched string (L) = 1.00 m
Mass of the string =40 g
String is attached to the tuning fork that vibrates at the frequency (f) = 128 Hz
Linear mass density (m) =40×10-3 kg/m
No. of loops formed, (n) = 4
L=nλ2⇒λ=2Ln=2×14⇒λ=0.5 mWave speed (v)=fλ=128×0.5⇒v=64 m/s
We know: v=Tm⇒T=ν2m =642×40×10-3 =163.84≈164 N
Hence, the tension in the string if it is to vibrate in four loops is 164 N.
Answer: 41
Wire makes a resonant frequency of 240 Hz and 320 Hz when its both ends are fixed.
Therefore, fundamental frequency (f0) of the wire must be the factor of 240 Hz and 320 Hz.
(a) Maximum value of fundamental frequency, f0 = 80 Hz
(b) Wave speed (v) = 40 m/s
And if λ is the wavelength:
∴ v=λ×f0⇒v=2×L×f0⇒L=402×80⇒L=14 m=0.25 m
Answer: 42
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be ‘n’ loops and λ be the wavelength.
∴ λ = 2×Separation between the consecutive nodes
λ2=2×1.6=3.2 cm
Length of the wire is L.
In the first case:
In the second case:
L=n+1λ22⇒nλ12=n+1 λ22⇒n×4=n+13.2⇒4n-3.2n=3.2⇒0.8 n=3.2⇒n=4
∴ Length of the string, L=nλ12=4×42=8 cm
Answer: 43
Frequency (f) = 660 Hz
Wave speed (v) = 220 m/s
Wavelength, λ=vf=220660=13 m
(a) No. of loops, n = 3
∴ L=n2λ
⇒L=32×13⇒L=12 m=50 cm
(b) Equation of resultant stationary wave can be given by:
y=2Acos2πxλsin2πvLλ⇒y=0.5 cos2πx13sin2π×220×t13⇒y=0.5 cm cos6πx m-1 sin1320πt s-1
Answer: 44
Length of the guitar wire (L1) = 30.0 cm = 0.30 m
Frequency, when no finger is placed on it, (f1) =196 Hz
And (f2) =220 Hz, (f3) = 247 Hz, (f4) = 262 Hz and (f5) = 294 Hz
The velocity is constant for a medium.
We have:
⇒ f1f2=L2L1⇒196220=L20.3⇒L2=196×0.3220=0.267 m⇒L2=26.7 cm
Again, f3=247 Hz
⇒ f3f1=L1L3⇒247196=0.3L3⇒L3=196×0.3247=0.238 m⇒L3=23.8 cmSimilarly, L4=196×0.3262=0.224 m⇒L4=22.4 cmAnd, L5=20 cm
Answer: 45
Fundamental frequency (f0) of the steel wire = 200 Hz
Let the highest harmonic audible to the person be n.
Frequency of the highest harmonic, f’ = 14000 Hz
∴ f’= nf0 …(1)
f’f0=14000200⇒nf0f0=70⇒ n=70
Thus, the highest harmonic audible to man is the 70th harmonic.
Answer: 46
Let the three resonant frequencies of a string be
f1=90 Hzf2=150 Hzf3=210 Hz
(a) So, the highest possible fundamental frequency of the string is f=30 Hz, because f1, f2 and f3 are the integral multiples of 30 Hz.
(b) So, these frequencies can be written as follows:
Hence, f1, f2, and f3 are the third harmonic, the fifth harmonic and the seventh harmonic, respectively.
(c) The frequencies in the string are f, 2f, 3f, 4f, 5f …
∴ 3f = 2nd overtone and 3rd harmonic
5f = 4th overtone and 5th harmonic
7th= 6th overtone and 7th harmonic
(d) Length of the string (L) = 80 cm = 0.8 m
Let the speed of the wave be v.
So, the frequency of the third harmonic is given by:
f1=32×L v⇒90=32×80×v⇒v=90×2×803 =30×2×80 =4800 cm/s⇒v=48 m/s
Answer: 47
The tensions in the two wires are in the ratio of 2:1.
Ratio of the radii is 3:1.
Density in the ratios of 1:2.
Let the length of the wire be L.
Frequency, f=1LDTπρ
∴ f1f2=LD2LD1T1T2πρ2πρ1⇒f1f2=1321×21⇒ f1:f2=2:3
Answer: 48
Length of the rod (L) = 40 cm = 0.40 m
Mass of the rod (m) = 1.2 kg
Let the mass of 4.8 kg be placed at x distance from the left.
As per the question, frequency on the left side = f0
Frequency on the right side = 2f0
Let tension be T1 and T2 on the left and the right side, respectively.
∴ 12LT1m=22LT2m⇒T1T2=2⇒T1T2=4 … (1)
From the free body diagram:
T1+T2=48+12=60 N⇒ 4T2+T2=5T2=60 N using equation 1∴T2=12 Nand T1=48 N
Now, taking moment about point A:
T2×0.4=48x+12×0.2⇒4.8=48x-2.4⇒4.8x=2.4⇒x=2.44.8=120 m=5 cm
Therefore, the mass should be placed at a distance of 5 cm from the left end.
Answer: 49
Length of the aluminium wire (La)= 60 cm = 0.60 m
Length of the steel wire (Ls)= 80 cm = 0.80 m
Tension produced (T) = 40 N
Area of cross-section of the aluminium wire (Aa) = 1.0 mm2
Area of cross-section of the steel wire (As) = 3.0 mm2
Density of aluminium (ρa) = 2⋅6 g cm−3
Density of steel (ρs) = 7⋅8 g cm−3
Mass per unit length of the steel, ms =ρs×As =7.8×10-2 gm/cm =7.8×10-3 kg/mMass per unit length of the aluminium, mA=ρAAA =2.6×10-2×3 gm/cm =7.8×10-2 gm/cm
= 7.8×10-3 kg/m
A node is always placed at the joint. Since aluminium and steel rod has same mass per unit length, the velocity of wave (v) in both of them is same.
Let v be the velocity of wave.
⇒v=Tm =407.8×10-3- =4×1047.8⇒v=71.6 m/s
For minimum frequency, there would be maximum wavelength.
For maximum wavelength, minimum number of loops are to be produced.
∴ Maximum distance of a loop = 20 cm
⇒ Wavelength, λ=2×20=40 cmOr λ=0.4 m∴Frequency, f=vλ=71.60.4=180 Hz
Answer: 50
Length of the string = L
Velocity of wave is given as:
In fundamental mode, frequency = ν
(a) Wavelength, λ=VelocityFrequency
⇒λ=Tm12LTm=2LWave number, K=2πλ=2π2L=πL
(b) Equation of the stationary wave:
y=Acos2πxλsin2πVtλ =Acos2πx2Lsin2πVt2L ∵ ν=V2L =AcosπxLsin2πνt
Exercise : Solution of Questions on page Number : 327
Answer: 51
Length of the string (L) = 2.0 m
Wave speed on the string in its first overtone (v) = 200 m/s
Amplitude (A) = 0.5 cm
(a) Wavelength and frequency of the string when it is vibrating in its 1st overtone (n = 2):
⇒λ=L=2 m⇒f=νλ=2002=100 Hz
(b) The stationary wave equation is given by:
y=2A cos2πxλsin2πvtλ =0.5cos2πx2sin2π×200 t2 =0.5 cmcosπm-1 xsin200πs-1 t
Answer: 52
The stationary wave equation of a string vibrating in its third harmonic is given by
y = (0.4 cm) sin [(0.314 cm−1) x]cos [(.600 πs−1) t]
By comparing with standard equation,
y = A sin (kx) cos (wt)
(a) From the above equation, we can infer the following:
ω=600 π
⇒2πf=600 π⇒f=300 Hz
Wavelength, λ=2π0.314=2×3.140.314
⇒λ=20 cm
(b) Therefore, the nodes are located at 0cm, 10 cm, 20 cm, 30 cm.
(c) Length of the string, l = nλ2
⇒l=3λ2=3×202=30 cm
(d) y=0.4 sin 0.314 x cos 600 πt
=0.4sinπ10 xcos600πt
λ and ν are the wavelength and velocity of the waves that interfere to give this vibration.
λ=20 cmν=ωk=600 ππ10=6000 cm/s⇒ν=60 m/s
Answer: 53
Equation of the standing wave:
y=0.4 cm sin 0.314 cm-1 xcos 600 πs-1 t⇒k=0.314=π10Also, k=2πλ⇒λ=20 cm
We know:
For the smallest length, putting n = 1:
⇒L=λ2=20 cm2=10 cm
Therefore, the required length of the string is 10 cm.
Answer: 54
Length of the wire (L) = 40 cm = 0.40 m
Mass of the wire = 3.2 g = 0.003 kg
Distance between the two fixed supports of the wire = 40.05 cm
Fundamental mode frequency = 220 Hz
Therefore, linear mass density of the wire (m) is given by:
m=0.00320.4=8×10-3 kg/mChange in length, ∆L=40.05-40=0.05 cm =0.05×10-2 mStrain=∆LL=0.05×10-20.4 =0.125×10-2f0=12LTm =12×0.4005 T8×10-3
⇒220×220=10.8012×T×1038⇒T×1000=220×220×0.641×0.8⇒T=248.19 NStress=TensionArea=248.191 mm2=248.1910-6⇒Stress=248.19×106Young’s modulus, Y=stressstrain =248.19×1060.125×10-2⇒Y=19852×108 =1.985×1011 N/
Hence, the required Young’s modulus of the wire is 1.985×1011 N/m2.
Answer: 55
Density of the block = ρ
Volume of block = V
∴ Weight of the block is, W = ρVg
∴ Tension in the string, T = W
The tuning fork resonates with different frequencies in the two cases.
Let the tenth harmonic be f10.
f10=102LTm =102L ρ Vgm
Here, m is the mass per unit length of the string and L is the length of the string.
When the block is immersed in water (density = ρ ), let the eleventh harmonic be f11.
f11=112LT’m =112Lρ-ρw Vgm
The frequency (f) of the tuning fork is same.
∴ f10=f11⇒102LρVgm=112Lρ-ρω Vgm⇒100121=ρ-1ρ because ρω=1 gm/cc⇒100 ρ=121 ρ-121⇒ρ=12121=5.8 gm/cc =5.8×103 kg/m3
Therefore, the required density is 5.8×103 kg/m3.
Answer: 56
Length of the long rope (L) = 2.00 m
Mass of the rope = 80 g = 0.08 kg
Tension (T) = 256 N
Linear mass density, m =0.082=0.04 kg/m
Tension, T=256 NWave velocity, v=Tm⇒v=2560.04=1602⇒v=80 m/s
For fundamental frequency:
L=λ4⇒λ=4L=4×2=8 m⇒f=vλ=808=10 Hz
(a) The frequency overtones are given below:
1st overtone=3f=30 Hz2nd overtone=5f=50 Hz
(b) λ=4l=4×2=8 m
∴ λ1=vf1=8030=2.67 m λ2=vf2=8050=1.6 m
Hence, the wavelengths are 8 m, 2.67 m and 1.6 m, respectively.
Answer: 57
Let T be the tension in the string and m be the mass per unit length of the heavy string.
In the first part of the question, the heavy string is fixed at only one end.
So, the lowest frequency is given by:
f0=14LTm …1
When the movable support is pushed by 10 cm to the right, the joint is placed on the pulley and the heavy string becomes fixed at both the ends (keeping T and m same).
Now, the lowest frequency is given by:
f0’=12LTm …2
Dividing equation (2) by equation (1), we get: | {"url":"https://www.toppersbulletin.com/hc-verma-class-xii-physics-chapter-15-wave-motion-and-wave-on-a-string/","timestamp":"2024-11-10T14:58:38Z","content_type":"text/html","content_length":"230399","record_id":"<urn:uuid:75d90b71-ff66-429d-bd13-01b6b9ed1983>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00821.warc.gz"} |
Here is my IF formula:
=IF([% Complete]3 = 0, "Red", IF([% Complete]3 = 1, "Green", "Yellow"))
It works great for green and everything else not so much.
What I want it to do is calculate based off of percent complete. so if less than 50% complete it should be red, if less than 75% yellow everything else is green. What have i done wrong?
• Try writing it how you have it spelled out in words.
=IF([% Complete]3 < .5, "Red", IF([% Complete]3 < .75, "Yellow", "Green"))
Try thinking of it as a STATEMENT as opposed to a formula.
• you're a genius! Thank you!
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/52621/if-formula-help","timestamp":"2024-11-07T09:55:58Z","content_type":"text/html","content_length":"400082","record_id":"<urn:uuid:e9c3247a-3bfb-42f0-b76b-cd08501a27bd>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00783.warc.gz"} |
Modeling velocity distributions in small streams using different neuro-fuzzy and neural computing techniques
Accurate estimation of velocity distribution in open channels or streams (especially in turbulent flow conditions) is very important and its measurement is very difficult because of spatio-temporal
variation in velocity vectors. In the present study, velocity distribution in streams was estimated by two different artificial neural networks (ANN), ANN with conjugate gradient (ANN-CG) and ANN
with Levenberg–Marquardt (ANN-LM), and two different adaptive neuro-fuzzy inference systems (ANFIS), ANFIS with grid partition (ANFIS-GP) and ANFIS with subtractive clustering (ANFIS-SC). The
performance of the proposed models was compared with the multiple-linear regression (MLR) model. The comparison results revealed that the ANN-CG, ANN-LM, ANFIS-GP, and ANFIS-SC models performed
better than the MLR model in estimating velocity distribution. Among the soft computing methods, the ANFIS-GP was observed to be better than the ANN-CG, ANN-LM, and ANFIS-SC models. The root mean
square errors (RMSE) and mean absolute errors (MAE) of the MLR model were reduced by 69% and 72%, respectively, using the ANFIS-GP model to estimate velocity distribution in the test period.
Stream flow that can be described as quite difficult is an important study area for water resources and hydraulic engineers. Flows in streams are usually expressed by 1-D hydraulic equations. Many
studies have been performed to determine the detailed properties of the hydrodynamics of complex flows using conventional methods, empirical formulas, and velocity samples (Hsu et al. 1998; Thomas &
Williams 1999; Huang et al. 2002; Kar et al. 2015).
Three approaches, namely, experimental measurement, the theoretical method, and computer simulation are used to investigate flow properties in hydraulic engineering. Hydraulic systems usually show
very complicated nonlinear behavior so it is not easy to get an analytical solution to describe the characteristics of these systems. Theoretical methods may be used to determine some simple flow
cases (Kerh et al. 1994, 1997). Computer simulation using numerical methods such as the Computational Fluid Dynamics package is another approach to solve the fluid mechanics problems. It can be used
to detect the properties of fluid motion in hydraulics engineering when the boundary conditions are properly defined. Flow measurement data are always extremely valuable for researchers who study in
the field of hydraulic engineering. Usable measurement data are needed to corroborate its accuracy and to check its reliability in computer simulation (Kerh 2000, 2002). Prediction of velocity
distribution is one of the basic properties of an open channel flow to analyze flow characteristics, particularly such as flow discharge, in the estimation of erosion and sediment transport in
alluvial channels, shear stress, and watershed runoff which is used by hydraulic engineers. Also, recent researches have expressed that the profile of velocity in streams is the driver of habitat
quality for aquatic species (Booker 2003). In this case, distribution of velocity must be investigated and determined as a priority for solving hydraulics problems in open channels.
Numerous analytical and experimental studies have been conducted to obtain velocity distributions in stream flows (Kirkgoz 1989; Smart 1999; Ferro 2003). The power law and the Prandtl–von Karman
universal velocity distribution law are well-known velocity distribution equations for open channel flows (Prandtl 1925; von Karman 1930). Unfortunately, the existing formulas cannot fully reveal the
velocity profile, particularly near the channel bed and water surface. Most recently, an entropy concept based on the probabilistic approach was used to investigate velocity distributions in open
channels (Chiu 1988; Xia 1997). According to the entropy method, there is a linear relationship between the mean and maximum velocity and it is described as an entropy parameter. Xia (1997)
demonstrated that the relationship between the mean and maximum velocities was linear for all the river sections considered. The entropy concept, which is an alternative to the traditional method, is
used to forecast flow properties. In the last decade, an artificial neural network (ANN) is another method that has been used to determine the velocity profile. The ANN and adaptive neuro-fuzzy
inference system (ANFIS) techniques have been satisfactorily used to solve problems in water resources and hydraulic engineering. Yang & Chang (2005) simulated velocity profiles and velocity contours
and estimated the discharges by ANN. Kocabas & Ulker utilized the ANFIS approach for predicting the critical submergence for an intake in a stratified fluid media (Kocabas & Ulker 2006). Dogan et al.
(2007) utilized the ANN approach to forecast concentration of the sediment acquired by an experimental study. Mamak et al. (2009) successfully analyzed bridge afflux through arched bridge
constrictions by ANFIS and ANN techniques. Kocabas et al. (2009) used the ANN method for estimating the critical submergence for an intake in a stratified fluid medium. Bilhan et al. (2010) estimated
the lateral outflow over rectangular side weirs by using two different ANN techniques. Emiroglu et al. (2011) utilized the ANN approach for predicting the discharge capacity of a triangular labyrinth
side weir situated on a straight channel. Emiroglu & Kisi (2013) used a neuro-fuzzy method to predict the discharge coefficient of trapezoidal labyrinth side weirs located on a straight channel. The
flow discharge of weirs has been successfully predicted by Kisi et al. (2013) by ANFIS. Genc et al. (2014) analyzed the accuracy of ANN and ANFIS in determination of mean velocity and discharge of
natural streams. They demonstrated that the ANFIS model, which has a determination coefficient (R^2) of 0.996, can be successfully predicted to mean velocity and discharge. In this paper, the
applicability of two different ANN and ANFIS approaches for estimating velocity distribution of streams is investigated and the results are compared with the multiple-linear regression (MLR) model.
For this purpose, field studies were carried out at different cross-sections in Kayseri by an acoustic Doppler velocimeter (ADV). These techniques have not been used for this purpose before.
Velocity profile should be determined for stream flows to better understand the structures of turbidity, sediment discharge, energy loss, and shear stress distributions (Ardiclioglu et al. 2012).
Velocity distribution is influenced by vegetation, channel geometry, channel slope, roughness, and the presence of bends in rivers. In river flow, the velocity profile is not consistent at diverse
depths. It increases from zero at the bottom of the channel to highest velocity near to the free water surface.
One of the most well-known velocity distribution models is the log-law (Sarma & Lakshminaraynan 1998). This logarithmic model is widely used to determine the two-dimensional velocity profile,
particularly for hydraulic smooth and rough flow conditions. Two-dimensional open channel flows are divided into two zones, the inner and outer region, because of the existence of turbulence and the
impact of a rigid boundary as shown in Figure 1. u[sw] indicates the velocity at water surface, u[max] shows maximum velocity in boundary layer height δ, water depth is shown as H, z[0] is depth
where u equals zero. k represents the roughness coefficient.
Velocity distribution in the inner region is characterized as good by logarithmic distribution. The velocity profile in the inner region, presumed to be limited to
<0.20, for uniform and steady nonuniform open channel flows, is presented using the log law in Equation (
):where u shows the streamwise, time-mean flow velocity, indicates the shear velocity, (
= density) and
is the boundary shear stress,
= 0.40 (1/
= 2.5) is the von Karman's constant (the value of
has a range of variation between 0.40 and 0.41),
is the distance from the bed, k
is the equivalent sand roughness, and B
is a constant of integration, being B
= 8.5 ± 15% (
Song & Graf 1996
). Although the value of B
depends on the nature of the wall surface, the value of
does not. When using this constant, Equation (
) is obtained as shown in Equation (
The logarithmic law is valid throughout the inner region except on the channel bed. In practical applications, it is still widely presumed that the logarithmic law explains the velocity profile along
the whole depth of uniform, steady open channel flows (Kundu & Ghoshal 2012).
The power law is an alternative model to represent the vertical distribution of the stream-wise velocity in open channel flows. Many applications in water resources and hydraulic engineering have
shown that velocity distribution measured in open channels can be expressed well by the power law (
Montes 1998
Chanson 2004
). The power law can be explained as a simple data-based equation in current theoretical studies. The power law exponent and constant can be obtained as to be quite empirical (
Cheng 2007
Chen (1991)
modified this law as it appeared as presented by the following equation:where the power law exponent is presented as 1/
means constant. Different
values have been presented in literature studies for different flows to determine velocity profile measurements with the power law.
González et al. (1996)
reported that
1/6 in their studies in open channels.
Ardiclioglu et al. (2005)
investigated power law equation constant
and exponent (1/
) in a stream and found them to be 4.0 and 1/5, respectively.
The four data sets of fixed ADV measurements presented in this paper were collected in the center of Turkey by a team comprising the first and third authors. Twenty-two field measurements were taken
at four diverse cross-sections in the Kızılırmak and Seyhan basins. The first data set was collected between 2009 and 2010 at the Sosun station, which is in the Seyhan basin, and the stream is a
tributary of the Zamantı River. The other data sets were obtained at the Barsama, Şahsenem, and Bünyan stations, which are in the Kızılırmak basin, a tributary of Kızılırmak River, between 2005 and
2010. Kızılırmak basin, which is the second biggest basin in Turkey, is located in the center of Turkey and the Black Sea region (Figure 2). Six site visits were carried out to the Barsama, Bünyan,
and Şahsenem stations and three visits to Sosun station. In Figure 3, a sample of measurements at the Şahsenem station is shown.
The ADV was utilized to gather three-dimensional velocity data at the four stations. The ADV measures three-dimensional flow velocities (u, v, w) for x, y, z dimensions in a sampling volume utilizing
the Doppler shift principle. At each measurement cross-section, the ADV records velocity data, location information, and water depth. The ADV sampling volume is found 10 cm before the probe head.
Accordingly, the probe head itself has least effect on the flow field surrounding the measurement volume. Velocity reach is ±0.001 m/s to 4.5 m/s, resolution 0.0001 m/s, exactness ±1% of measured
velocity (SonTek 2002). To decide the distribution of velocity in river flow, the experimental devices must be properly arranged.
During flow measurements, cross sections were divided into a number of slices for each flow condition according to the water surface width. Point velocity measurements were taken at different
positions in the vertical direction starting 4 cm from the streambed for each vertical. The velocities of free water surface in all verticals were estimated by extrapolating the last two measurements
of the verticals. Also, mean water surface velocities u[ws] were measured at each of the visited stations. Water surface velocity can be effortlessly computed with an object that is movable on the
water surface and is not too heavy, such as leaves, twigs, and so on.
The flow characteristics at every site are given in Table 1. In this table, the first and second columns show the station visit numbers and dates visited, U[m] (=Q/A) is the mean velocity. A is the
area of the cross section, u[ws] is the measured water surface velocity, H[max] is the maximum flow depth, T is the surface water width, S[ws] is water surface slope, Re (=4U[m]R/ʋ) is the Reynolds
number, and ʋ is the kinematic viscosity, and Fr (=U[m]/(gH[max])^1/2 represents the Froude number. When Froude and Reynolds numbers are calculated for all flow measurements, subcritical and
turbulent flow conditions have been encountered.
Table 1
Stations . Dates (d/m/y) . U[m] (m/s) . u[ws] (m/s) . H[max] (m) . T (m) . S[ws] . Re (×10^6) . Fr .
Barsama_1 28/05/2005 0.890 1.60 39.0 8.3 0.0091 0.76 0.481
Barsama_2 19/05/2006 1.051 1.85 40.0 9.0 0.0036 0.94 0.531
Barsama_3 19/05/2009 1.214 2.08 45.0 9.0 0.0094 1.47 0.578
Barsama_4 31/05/2009 0.590 1.14 26.0 8.4 0.0092 0.40 0.333
Barsama_5 24/03/2010 0.806 1.55 38.0 8.6 0.0097 0.61 0.417
Barsama_6 18/04/2010 0.865 1.63 38.2 8.8 0.0120 0.85 0.421
Bünyan_1 24/06/2009 0.354 0.65 72.0 4.0 0.0020 0.71 0.133
Bünyan_2 08/02/2010 0.214 0.40 66.0 4.0 0.0030 0.40 0.084
Bünyan_3 27/09/2009 0.301 0.54 72.0 3.9 0.0022 0.50 0.113
Bünyan_4 04/04/2010 0.405 0.74 85.0 4.0 0.0018 0.78 0.140
Bünyan_5 16/05/2010 0.426 0.54 86.0 4.0 0.0024 0.85 0.147
Bünyan_6 20/06/2010 0.286 0.53 79.0 3.9 0.0010 0.53 0.103
Şahsenem_1 29/03/2006 0.600 1.04 28.0 6.0 0.0059 0.47 0.350
Şahsenem_2 20/10/2007 0.529 0.93 32.0 5.4 0.0061 0.46 0.298
Şahsenem_3 22/03/2008 0.565 0.80 33.0 6.0 0.0037 0.49 0.314
Şahsenem_4 03/05/2008 0.518 1.00 32.0 5.4 0.0045 0.39 0.307
Şahsenem_5 11/10/2008 0.536 1.01 32.0 5.5 0.0046 0.44 0.303
Şahsenem_6 08/11/2008 0.516 1.00 34.0 5.6 0.0064 0.51 0.282
Sosun_1 19/05/2009 0.561 0.96 62.0 3.2 0.0032 0.84 0.227
Sosun_2 31/05/2009 0.285 0.63 43.0 3.0 0.0016 0.32 0.144
Sosun_3 24/03/2010 0.327 0.63 45.0 2.9 0.0026 0.37 0.156
Sosun_4 18/04/2010 0.541 0.93 54.0 2.3 0.0034 0.67 0.235
Stations . Dates (d/m/y) . U[m] (m/s) . u[ws] (m/s) . H[max] (m) . T (m) . S[ws] . Re (×10^6) . Fr .
Barsama_1 28/05/2005 0.890 1.60 39.0 8.3 0.0091 0.76 0.481
Barsama_2 19/05/2006 1.051 1.85 40.0 9.0 0.0036 0.94 0.531
Barsama_3 19/05/2009 1.214 2.08 45.0 9.0 0.0094 1.47 0.578
Barsama_4 31/05/2009 0.590 1.14 26.0 8.4 0.0092 0.40 0.333
Barsama_5 24/03/2010 0.806 1.55 38.0 8.6 0.0097 0.61 0.417
Barsama_6 18/04/2010 0.865 1.63 38.2 8.8 0.0120 0.85 0.421
Bünyan_1 24/06/2009 0.354 0.65 72.0 4.0 0.0020 0.71 0.133
Bünyan_2 08/02/2010 0.214 0.40 66.0 4.0 0.0030 0.40 0.084
Bünyan_3 27/09/2009 0.301 0.54 72.0 3.9 0.0022 0.50 0.113
Bünyan_4 04/04/2010 0.405 0.74 85.0 4.0 0.0018 0.78 0.140
Bünyan_5 16/05/2010 0.426 0.54 86.0 4.0 0.0024 0.85 0.147
Bünyan_6 20/06/2010 0.286 0.53 79.0 3.9 0.0010 0.53 0.103
Şahsenem_1 29/03/2006 0.600 1.04 28.0 6.0 0.0059 0.47 0.350
Şahsenem_2 20/10/2007 0.529 0.93 32.0 5.4 0.0061 0.46 0.298
Şahsenem_3 22/03/2008 0.565 0.80 33.0 6.0 0.0037 0.49 0.314
Şahsenem_4 03/05/2008 0.518 1.00 32.0 5.4 0.0045 0.39 0.307
Şahsenem_5 11/10/2008 0.536 1.01 32.0 5.5 0.0046 0.44 0.303
Şahsenem_6 08/11/2008 0.516 1.00 34.0 5.6 0.0064 0.51 0.282
Sosun_1 19/05/2009 0.561 0.96 62.0 3.2 0.0032 0.84 0.227
Sosun_2 31/05/2009 0.285 0.63 43.0 3.0 0.0016 0.32 0.144
Sosun_3 24/03/2010 0.327 0.63 45.0 2.9 0.0026 0.37 0.156
Sosun_4 18/04/2010 0.541 0.93 54.0 2.3 0.0034 0.67 0.235
ANFIS was initially presented by Jang (1993). It is a universal approximator and is equipped for approximating any real continuous function. The ANFIS structure is made out of various nodes
associated through directional connections and each node has a function comprising fixed or flexible parameters (Jang et al. 1997).
Here, f[1] and f[2] allude to the output function of rule 1 and rule 2, separately. The ANFIS structure is shown in Figure 4. The node functions of each layer will be explained next.
Every node i in layer 1 is composed of an adaptive node function:where x is the i
node's input and A
is a linguistic label, for example, ‘low’ or ‘high’ connected with this node function. O
is the membership function of a fuzzy set A (=A
, A
, B
, B
, C
, or C
). It indicates the degree to which the given input
satisfies the quantifier A
. is typically chosen to be a Gaussian function with a minimum equivalent to 0 and maximum equivalent to 1:In Equation (
), a
and b
indicate the parameters. When the values of these parameters change, the Gaussian function also varies accordingly, thus exhibiting various forms of membership functions on linguistic label A
Jang 1993
). This layer's parameters are called the premise parameters (
Emiroglu & Kisi 2013
The output of each node shows the firing strength of a rule.
In the fourth layer, each node has a function as:In Equation (
), refers to the output of the third layer, and , and are the parameters. This layer's parameters are called the consequent parameters.
The single node in the fifth layer processes the final output as the summation of all incoming signals:In this manner, an ANFIS has been constructed which is practically equal to a first-order Sugeno
FIS. More details on ANFIS can be obtained from the related references (
Jang 1993
Artificial neural networks (ANN) are motivated by the natural nervous system but by neglecting a great part of the biological details. ANN is composed of numerous processing elements. The ANN is
composed of layers comprising parallel processing components, called neurons. Each layer in the network is fully associated with the proceeding layer by interconnection. Arbitrarily assigned initial
weight values are logically corrected during a training process. This process compares computed outputs to actual outputs and backpropagates any errors. Thus, the last weights are adjusted by
minimizing the errors (
Kisi 2005
Emiroglu & Kisi 2013
). Each neuron in the second layer j gets the x input which is the weighted sum of outputs from the past layer. As a case, y for layer j is given by:where
= a bias for neuron j,
= i
output of the past layer, W
= weight between the first layer i and j. An output is calculated from each neuron in the second layer j and third layer k by passing its value of y through a non-linear activation function. An
ordinarily utilized activation function is the logistic function:
More details about ANN can be obtained from the related references (Haykin 2009).
Two different ANN methods, ANN with conjugate gradient (ANN-CG) and ANN with Levenberg–Marquardt (ANN-LM), and two different ANFIS methods, ANFIS with grid partition (ANFIS-GP) and ANFIS with
subtractive clustering (ANFIS-SC) were employed by using MATLAB program codes. The input parameters used for estimating the velocity distributions of the streams are the water surface velocity
water surface slope
, and
. The 2,184 measured data were used for the ANN-CG, ANN-LM, ANFIS-GP, ANFIS-SC, and regression analyses (MLR). After being randomly permutated, the data were split into two parts, training and
testing. The first part (1,747 values, 80% of the whole data) was utilized for training and the second part (437 values, 20% of the whole data) was utilized for testing. Before application of the ANN
models, the training input and output values were standardized using Equation (
):in which
are the maximum and minimum of the training and test data. In the present study, the a
and a
values were individually assigned as 0.6 and 0.2 and the input and output data were standardized somewhere around 0.2 and 0.8. Different ANN structures were tried to obtain the optimal models. The
assessing criteria utilized in the applications are the root mean square errors (RMSE), mean absolute errors (MAE), and determination coefficient (R
). The expressions of the RMSE and MAE are provided in the following equations:where
refer to the number of data sets and velocity, respectively.
For estimating the velocity distribution of the streams, four different input combinations were utilized. The correlations between the inputs u[ws], S[ws], z/H, and y/T and output are 0.714, 0.547,
0.362, and 0.010, respectively. According to the correlation values, u[ws] seems to be the most effective variable on velocity distribution while y/T is the least effective one. The optimal hidden
node numbers were obtained for the ANN models by the trial and error method. Two different ANN models were obtained by using the CG and LM algorithms which are more powerful and faster than the
conventional gradient descent technique (Kisi 2007). The sigmoid activation functions were used for the hidden and output nodes. The training of the ANN networks was stopped after 1,000 iterations.
Table 2 reports the training and test results of the optimal ANN models in estimating velocity distribution. The numbers given in the second column of the table indicate the optimal number of hidden
nodes for each ANN model. It is clear from the table that the ANN-CG (4,7,1) model comprising four inputs corresponding to u[ws], S[ws], z/H, and y/T, seven hidden and one output nodes, has the
lowest RMSE, MAE, and the highest R^2 than the other model both in training and test periods. Out of four ANN-LM models, the ANN-LM (4,9,1) model comprising four inputs performs better than the other
models. The relative RMSE and MAE differences between the optimal ANFIS-LM and ANFIS-CG models are 12% in the test period.
Table 2
Input . Parameters . Training . Test .
RMSE . MAE . R^2 . RMSE . MAE . R^2 .
u[ws] 9 0.244 0.187 0.512 0.243 0.190 0.604
u[ws] and S[ws] 8 0.238 0.180 0.536 0.237 0.186 0.626
u[ws], S[ws] and z/H 7 0.171 0.134 0.758 0.167 0.128 0.810
u[ws], S[ws], z/H and y/T 7 0.106 0.081 0.908 0.098 0.074 0.934
u[ws] 10 0.241 0.183 0.533 0.243 0.189 0.606
u[ws] and S[ws] 5 0.238 0.181 0.532 0.238 0.186 0.623
u[ws], S[ws] and z/H 5 0.169 0.132 0.764 0.167 0.129 0.812
u[ws], S[ws], z/H and y/T 9 0.095 0.073 0.925 0.086 0.065 0.950
u[ws] (gaussmf, 3) 0.247 0.190 0.499 0.245 0.196 0.599
u[ws] and S[ws] (gaussmf, 4) 0.237 0.180 0.536 0.239 0.187 0.618
u[ws], S[ws] and z/H (gaussmf, 2) 0.177 0.138 0.743 0.167 0.127 0.810
u[ws], S[ws], z/H and y/T (gaussmf, 4) 0.037 0.027 0.989 0.066 0.046 0.971
u[ws] (gaussmf, 0.2) 0.240 0.183 0.523 0.242 0.189 0.610
u[ws] and S[ws] (gaussmf, 0.2) 0.237 0.179 0.539 0.239 0.187 0.617
u[ws], S[ws] and z/H (gaussmf, 0.4) 0.174 0.137 0.752 0.171 0.132 0.802
u[ws], S[ws], z/H and y/T (gaussmf, 0.4) 0.107 0.082 0.906 0.103 0.081 0.928
u[ws] (0.58) 0.250 0.193 0.488 0.247 0.196 0.591
u[ws] and S[ws] (0.56;4.72) 0.250 0.192 0.489 0.246 0.195 0.593
u[ws], S[ws] and z/H (0.42;4.51;0.31) 0.223 0.172 0.621 0.217 0.168 0.726
u[ws], S[ws], z/H and y/T (0.48;4.68;0.37; − 0.21) 0.216 0.164 0.624 0.211 0.163 0.712
Input . Parameters . Training . Test .
RMSE . MAE . R^2 . RMSE . MAE . R^2 .
u[ws] 9 0.244 0.187 0.512 0.243 0.190 0.604
u[ws] and S[ws] 8 0.238 0.180 0.536 0.237 0.186 0.626
u[ws], S[ws] and z/H 7 0.171 0.134 0.758 0.167 0.128 0.810
u[ws], S[ws], z/H and y/T 7 0.106 0.081 0.908 0.098 0.074 0.934
u[ws] 10 0.241 0.183 0.533 0.243 0.189 0.606
u[ws] and S[ws] 5 0.238 0.181 0.532 0.238 0.186 0.623
u[ws], S[ws] and z/H 5 0.169 0.132 0.764 0.167 0.129 0.812
u[ws], S[ws], z/H and y/T 9 0.095 0.073 0.925 0.086 0.065 0.950
u[ws] (gaussmf, 3) 0.247 0.190 0.499 0.245 0.196 0.599
u[ws] and S[ws] (gaussmf, 4) 0.237 0.180 0.536 0.239 0.187 0.618
u[ws], S[ws] and z/H (gaussmf, 2) 0.177 0.138 0.743 0.167 0.127 0.810
u[ws], S[ws], z/H and y/T (gaussmf, 4) 0.037 0.027 0.989 0.066 0.046 0.971
u[ws] (gaussmf, 0.2) 0.240 0.183 0.523 0.242 0.189 0.610
u[ws] and S[ws] (gaussmf, 0.2) 0.237 0.179 0.539 0.239 0.187 0.617
u[ws], S[ws] and z/H (gaussmf, 0.4) 0.174 0.137 0.752 0.171 0.132 0.802
u[ws], S[ws], z/H and y/T (gaussmf, 0.4) 0.107 0.082 0.906 0.103 0.081 0.928
u[ws] (0.58) 0.250 0.193 0.488 0.247 0.196 0.591
u[ws] and S[ws] (0.56;4.72) 0.250 0.192 0.489 0.246 0.195 0.593
u[ws], S[ws] and z/H (0.42;4.51;0.31) 0.223 0.172 0.621 0.217 0.168 0.726
u[ws], S[ws], z/H and y/T (0.48;4.68;0.37; − 0.21) 0.216 0.164 0.624 0.211 0.163 0.712
The training and test results of the ANFIS-GP and ANFIS-SC models are given in Table 2 for each input combination. The optimal number of membership functions and parameter values are also provided in
the second column of the table. From Table 2, it is clear that the ANFIS-GP model comprising four Gaussian membership functions for the inputs, u[ws], S[ws], z/H, and y/T performs better than the
other ANFIS-GP models for both periods. There is a harmony between training and test results and this proves the proper calibration of the applied models. The comparison of ANFIS approaches reveals
that the optimal ANFIS-GP model has a better accuracy than the optimal ANFIS-SC model with four inputs. According to the ANN-CG, ANN-LM, ANFIS-GP, and ANFIS-SC results given in Table 2, the y/T seems
to be the most effective variable on velocity distribution in streams. However, y/T was found to be the least effective variable with respect to correlation (correlation = 0.010). This implies the
strong nonlinear relationship between y/T and velocity distribution. This is also valid for z/H which also has considerable nonlinear effect on velocity distribution even though it has low
correlation (correlation = 0.010). The main reason for this is the fact that these variables (y/T and z/H) are related to wetted area and velocity is proportional to the wetted area (continuity
equation, V=Q/A, where V, Q, and A are mean velocity, discharge, and wetted area, respectively). It is apparent from Table 2 that the first two input combinations provide low accuracy in the applied
models. Although the u[ws] and S[ws] have high correlations with velocity, they are not sufficient for accurate estimation of velocity distribution. Table 2 also reports the RMSE, MAE, and R^2
statistics of the MLR models. The second column of this table indicates the regression coefficients of the MLR models. The MLR model with four inputs performs better than the other MLR models.
Comparison of ANN-CG, ANN-LM, ANFIS-GP, ANFIS-SC, and MLR models reveals that the data-driven ANN and ANFIS based models perform better than the MLR model in estimating velocity distribution. The
optimal ANFIS-GP model comprising four input combinations has the lowest RMSE (0.066) and MAE (0.046) and the highest R^2 (0.971) values. The main advantage of the ANFIS-GP over ANFIS-SC method is
that it uses all possible rule combinations in its structure while ANFIS-SC decreases the possible rules to some limited numbers by using a clustering algorithm. ANFIS-SC has simpler structure
compared to ANFIS-GP but it has less accuracy than the latter. Table 3 reports the cross-correlations among the estimates of the optimal models with four input variables in the test period. It is
apparent that the MLR has the lowest correlations with soft computing models. The ANN-LM has the highest correlation with ANFIS-GP and this indicates that the ANN-LM has the second rank in estimating
velocity distribution.
Table 3
. ANN-CG . ANN-LM . ANFIS-GP . ANFIS-SC . MLR .
ANN-CG 1
ANN-LM 0.991 1
ANFIS-GP 0.963 0.970 1
ANFIS-SC 0.975 0.974 0.961 1
MLR 0.855 0.856 0.838 0.857 1
. ANN-CG . ANN-LM . ANFIS-GP . ANFIS-SC . MLR .
ANN-CG 1
ANN-LM 0.991 1
ANFIS-GP 0.963 0.970 1
ANFIS-SC 0.975 0.974 0.961 1
MLR 0.855 0.856 0.838 0.857 1
The measured and estimated velocity distributions by the ANN-CG, ANN-LM, ANFIS-GP, ANFIS-SC, and MLR models are shown in Figure 5 for the test period. It is clear that the estimates of the ANN and
ANFIS models are closer to the corresponding measured velocities than the MLR model. As seen from the figure, ANN-CG, AN-LM, and ANFIS-SC models underestimate some peaks while the ANFIS-GP model
generally well estimates. The significantly under- and over-estimations of the MLR model are clearly seen. The scatterplots of the estimates in the test period are illustrated in Figure 6. It is
clear from the fit line equations (assume that the equation is y = ax + b) given in the scatterplots that the a and b coefficients of the ANFIS-GP model are individually closer to 1 and 0 with a
higher R^2 than those of the ANN-CG, ANN-LM, ANFIS-SC, and MLR models. It is clear from Figure 6 that the MLR model is insufficient in estimating the velocity distribution of the natural streams.
In order to verify the robustness (the significance of differences between the model estimates and measured velocity values) of the models, the results are likewise tested by utilizing one-way ANOVA.
The test is set at a 95% significant level. The test statistics are provided in Table 4. The ANN and ANFIS models yield small testing values with high significance levels in respect to the MLR.
According to the ANOVA results, the ANFIS and ANN models are more robust (the closeness between the measured velocity values and model estimates are significantly high) in estimating velocity
distribution than the MLR model. Among the data-driven models, the ANN-CG and ANFIS-SC models seem to be better than the ANN-LM and ANFIS-GP regarding robustness.
Table 4
Model . F-statistic . Resultant significance level .
ANN-CG (4,7,1) 0.003 0.959
ANN-LM (4,9,1) 0.021 0.885
ANFIS-GP (gaussmf, 4) 0.038 0.845
ANFIS-SC (gaussmf, 0.4) 0.004 0.950
MLR 0.076 0.783
Model . F-statistic . Resultant significance level .
ANN-CG (4,7,1) 0.003 0.959
ANN-LM (4,9,1) 0.021 0.885
ANFIS-GP (gaussmf, 4) 0.038 0.845
ANFIS-SC (gaussmf, 0.4) 0.004 0.950
MLR 0.076 0.783
Estimating velocity distribution in streams by ANN-CG, ANN-LM, ANFIS-GP, ANFIS-SC, and MLR approaches was examined in this study. The 2,184 field data gauged from four diverse cross sections at four
destinations on the Sarımsaklı and Sosun streams in central Turkey were used in the applications. To predict velocity distribution, the water surface velocity u[ws][,] water surface slope S[ws], z/H,
and y/T were utilized as inputs to the models. The accuracy of the ANN-CG, ANN-LM, ANFIS-GP, and ANFIS-SC models was compared with MLR models. Comparison results showed that the ANN and ANFIS models
provided better accuracy than the regression model in estimating velocity distribution. The ANFIS-GP model was observed to be better than the ANN-CG, ANN-LM, and ANFIS-SC models. The optimal ANFIS-GP
model separately reduced the RMSE and MAE by 69% and 72% and increased the determination coefficient by 36% with respect to the optimal MLR model. The study suggests that the ANFIS and ANN techniques
can be effectively utilized for estimating the velocity distribution of the streams.
In this study, 2,184 field data measured in four cross-sections at four destinations were used for model development. More data from different places may improve the models' accuracy. In this way,
generalization of the applied models may be improved. | {"url":"https://iwaponline.com/jwcc/article/11/2/390/65643/Modeling-velocity-distributions-in-small-streams","timestamp":"2024-11-07T03:12:28Z","content_type":"text/html","content_length":"308441","record_id":"<urn:uuid:22e9ef64-d879-4e71-95c7-842c2c3d181f>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00766.warc.gz"} |
Shallow water sloshing in rotating vessels
Department of Mathematics
University of Surrey
Shallow water sloshing in rotating vessels - 2D flowfield
New shallow-water equations, for sloshing in two dimensions (one horizontal and one vertical) in a vessel which is undergoing rigid-body motion in the plane, are derived. The planar motion of
the vessel (pitch-surge-heave or roll-sway-heave) is exactly modelled and the only approximations are in the fluid motion. The flow is assumed to be inviscid but vortical, with approximations
on the vertical velocity and acceleration at the surface. These equations improve previous shallow water models for sloshing. The model also contains the essence of the Penney-Price-Taylor
theory for the highest standing wave. The surface shallow water equations are simulated using a robust implicit finite-difference scheme. Numerical experiments are reported, including
simulations of coupled translation-rotation forcing, and sloshing on a Ferris wheel. Asymptotic results confirm that rotations should be of order h/L, where h is the mean depth and L is the
vessel length, but translations can be of order unity, in the shallow water limit.
H. Alemi Ardakani & T.J. Bridges. Shallow-water sloshing in rotating vessels undergoing prescribed rigid-body motion in two dimensions. European J. Mech. B/Fluids 31 30-43 Journal Website
H. Alemi Ardakani & T.J. Bridges. Shallow-water sloshing in rotating vessels undergoing prescribed rigid-body motion in two dimensions -- the extended version. Technical Report (2010)
H. Alemi Ardakani & T.J. Bridges. Comparison of the numerical scheme with previous rotating SWE numerics of Dillingham, Armenio & La Rocca, Huang & Hsiung Technical Report, (2010)
Video1.mpg Video2.mpg Video3.mpg Video4.mpg
Shallow water sloshing in rotating vessels - 3D flowfield
New shallow-water equations, for sloshing in three dimensions (two horizontal and one vertical) in a vessel which is undergoing rigid-body motion in 3-space, are derived. The rigid-body motion
of the vessel (roll-pitch-yaw and/or surge-sway-heave) is modelled exactly and the only approximations are in the fluid motion. The flow is assumed to be inviscid but vortical, with
approximations on the vertical velocity and acceleration at the surface. These equations improve previous shallow water models. The model also extends the essence of the Penney-Price-Taylor
theory for the highest standing wave. The surface shallow water equations are simulated using a split-step implicit alternating direction finite-difference scheme. Numerical experiments are
reported, including comparisons with existing results in the literature, and simulations with vessels undergoing full three-dimensional rotations.
H. Alemi Ardakani & T.J. Bridges. Shallow-water sloshing in vessels undergoing prescribed rigid-body motion in three dimensions, J. Fluid Mech. 667 474-519 (2011)
H. Alemi Ardakani & T.J. Bridges. Shallow-water sloshing in vessels undergoing prescribed rigid-body motion in three dimensions Technical Report (2009) (extended version with colour figures)
Shallow water sloshing on the London Eye
The London Eye is a large ferris wheel and pictures of it can be found at their website (click here) . The object here is to study the sloshing of a partially filled vessel attached to the
wheel. Mathematically, the vessel is prescribed to travel along a circular path. Even when the speed along the path is constant, sloshing occurs due to change in direction. The base of the
vessel remains horizontal along the path. In addition the vessel can also have a prescribed rotation. The interest in this example is threefold. It is an example with very large displacements
of the vehicle and illustrates the generality of the prescribed vessel motion. Secondly, it is a prototype for the transport of a vessel along a surface. In this case the surface is a circle.
As the vehicle moves along the surface it can also rotate relative to the point of attachment. Other examples of surfaces of interest are a sphere or near sphere, which is a model for a
satellite containing fluid and orbiting the earth, and a surface modelling terrain. The latter is a model for vehicles transporting liquid on roads through hilly terrain. Thirdly, it is an
excellent setting to test control strategies for sloshing. For example, suppose the speed along a path in the surface is prescribed. Sloshing will result if the path is curved due to induced
acceleration. The local rotation of the body could act as a control, and roll, pitch or yaw could be induced to counteract any sloshing due to motion along the path. This example is discussed
in the above papers.
Videos of fluid-vehicle interaction
Video6.mpg Video7.mpg Video8.mpg Video9.mpg Video10.mpg Video11.mpg
Parameter values associated with the above videos
Dynamic coupling between fluid sloshing and vehicle motion
The coupled motion between shallow water sloshing in a moving vehicle and the vehicle dynamics is considered. The movement of the vessel is restricted to horizontal motion. Motivated by the
theory of Cooker (1994), a new derivation of the coupled problem in the Eulerian fluid representation is given. The aim is to simulate the nonlinear coupled motion numerically, but the
nonlinear coupling causes difficulties. These difficulties are resolved by transforming to the Lagrangian represention of the fluid motion. In this representation an explicit, robust, simple
numerical algorithm, based on the Störmer-Verlet method, is proposed. Numerical results of the coupled dynamics are presented. The forced motion (neglecting the coupling) leads to quite complex
fluid motion, but the coupling can be a stabilising influence.
H. Alemi Ardakani & T.J. Bridges. Dynamic coupling between shallow-water sloshing and horizontal vehicle motion. Euro. J. Appl. Math. 21 479-517 (2010). Journal website
H. Alemi Ardakani & T.J. Bridges. Symplecticity of the Stormer-Verlet algorithm for coupling between the shallow water equations and horizontal vehicle motion. Technical Report (2010)
M.J. Cooker. Water waves in a suspended container, Wave Motion 20 385-395 (1994).
Videos of fluid-vehicle interaction
FluidVehicle1.mpg FluidVehicle2.mpg FluidVehicle3.mpg FluidVehicle4.mpg FluidVehicle5.mpg
Parameter values associated with the above videos
Matlab codes for sloshing simulations
Link to Matlab page
Fluid-vessel coupling for a rotating vessel
The coupled liquid-vessel motion of a rotating vessel and shallow water sloshing is considered. The equations for the fluid are the rotating shallow water equations derived in Alemi Ardakani &
Bridges (2009). These equations are coupled to an equation for the rotational motion of the vehicle. New equations are derived, starting with a variational formulation and the shallow-water
approximation. As a test case the "pendulum slosh" problem is studied. In the pendulum slosh problem the vehicle is a pendulum with the pendulum bob containing fluid. The coupling changes the
natural frequencies of the rigid body pendulum and the fluid motion in a fixed vessel. Numerical simlulations are reported.
H. Alemi Ardakani & T.J. Bridges. Dynamic coupling between shallow-water sloshing and a vehicle undergoing planar rigid-body rotation, Technical Report (2010)
The Euler equations relative to a moving frame
In this technical report the details of the construction of the apparent accelerations, which appear in the Euler equations when viewed from a body fixed 3D moving frame, are presented. A
moving frame has been widely used in the study of sloshing. However, there are small subtleties that have been overlooked in previous derivations, and therefore a detailed derivation is
presented here.
H. Alemi Ardakani & T.J. Bridges. The Euler equations in fluid mechanics relative to a rotating-translating reference frame, Technical Report (2010)
Review of Dillingham, Falzarano & Pantazopoulos SWEs
Two derivations of the shallow water equations (SWEs) for fluid in a vessel that is undergoing a general rigid-body motion in three dimensions first appeared in the literature at about the same
time, given independently by Pantazopoulos (1987,1988) and Dillingham & Falzarano (1986). However both derivations follow the same strategy. Their respective derivations are an extension of the
formulation for two-dimensional shallow water flow in a rotating frame in Dillingham (1981). Their idea is to start with the classical SWEs; then deduce the apparent accelerations of the body
frame relative to an inertial frame. Then the gravitational force is replaced by an average of the vertical accelerations and approximations for horizontal accelerations are subsituted into the
right-hand side of the horizontal momentum SWEs. The purpose of this report is to determine the precise approximations used in the derivation in order to compare with the new shallow-water
equations found in Alemi Ardakani & Bridges (2009).
Review of the Dillingham, Falzarano & Pantazopoulos three-dimensional shallow-water equations. Department of Mathematics Report (2009)
Three-dimensional numerical simulation of green water on deck, Third International Conference on the Stability of Ships and Ocean Vehicles. Gdansk, Poland (1986).
Three-dimensional sloshing of water on decks, Marine Technology 25 253-261 (1988).
Numerical solution of the general shallow water sloshing problem, PhD Thesis: University of Washington, Seattle (1987).
Review of Huang-Hsiung rotating SWEs
In the literature there are two strategies for deriving the shallow water equations (SWEs) relative to a rotating frame in three dimensions. The first -- the strategy of Dillingham, Falzarano &
Pantazopoulos -- is reviewed in the report cited above. The second strategy is that of Huang & Hsiung (1996). In this report their derivation is reviewed identifying the key assumptions. The
Huang-Hsiung derivation is then contrasted with a new third strategy for deriving SWEs using the surface equations derived in Alemi Ardakani & Bridges (2009).
Review of the Huang-Hsiung three-dimensional shallow-water equations. Department of Mathematics Report (2009)
Review of the Huang-Hsiung two-dimensional shallow-water equations. Department of Mathematics Report (2009)
Nonlinear shallow-water flow on deck, J. Ship Research 40 303-315 (1996).
Nonlinear shallow-water flow on deck and its effect on ship motion, PhD Thesis, Technical University of Novia Scotia (1995).
Other technical reports on shallow water sloshing
H. Alemi Ardakani & T.J. Bridges. Asymptotics of (SWE-1) and (SWE-2) in the shallow water limit in three dimensions. Department of Mathematics Report (2010)
H. Alemi Ardakani & T.J. Bridges. Shallow water sloshing in rotating vessels: details of the numerical algorithm. Department of Mathematics Report (2009)
H. Alemi Ardakani & T.J. Bridges. Review of the Armenio-LaRocca two-dimensional shallow-water equations, Department of Mathematics Report (2009)
H. Alemi Ardakani & T.J. Bridges. Yaw forcing with the vessel position also plotted -- towards a video of 3D sloshing, Department of Mathematics Report (2009)
H. Alemi Ardakani & T.J. Bridges. Coupled roll-pitch motions: 1:2 resonance simulations, Department of Mathematics Report (2010)
H. Alemi Ardakani & T.J. Bridges. Review of the 3-2-1 Euler angles: a yaw-pitch-roll sequence, Department of Mathematics Report (2010)
H. Alemi Ardakani & T.J. Bridges. Surge-sway simulations with additional detail, Department of Mathematics Report (2010) The size of this file is almost 20 Mb, so it has been gzip-ed down to
less than 1 Mb. To decompress it in a unix/linux environment just type gunzip and the file name and it will convert it back to .pdf file.
 Department of Mathematics   University of Surrey   | {"url":"https://personalpages.surrey.ac.uk/T.Bridges/SLOSH/","timestamp":"2024-11-07T04:27:54Z","content_type":"text/html","content_length":"31217","record_id":"<urn:uuid:514365de-f4f3-4f85-bd8c-a97c981e82c4>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00466.warc.gz"} |
NCERT Solutions for Class 3 maths Chapter 5 Shapes and Designs - Free PDF Download
The NCERT Solutions for Class 3 Maths Chapter 5 covers different types of shapes, edges and corners, Use of a triangle in the tangram to understand the shapes and practice set of identifying
geometrical shapes. Creation of our tiling patterns. Understanding of vertical and horizontal directions. This chapter gives general categories that are used to describe shapes and different form of
These NCERT Solutions for Class 3 will teach your kids to understand and comprehend from this solution set of mathematics. Chapter wise and topic wise solutions will make it easy for students to
learn and memorise. Our Praadis experts have tried to provide chapters in simple language so that students can understand them easily. All the study materials are in PDF format and easy to download.
Students will be able to improve their efficiency through these solutions. | {"url":"https://praadisedu.com/ncert-solutions-for-class-3-maths-chapter-5/Shapes-and-Designs/93/7","timestamp":"2024-11-08T21:01:36Z","content_type":"text/html","content_length":"105676","record_id":"<urn:uuid:55b6f08c-91df-4ec8-9e2a-27e715692bff>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00842.warc.gz"} |
Asymptotic expansion instead of a power series expansion
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Asymptotic expansion instead of a power series expansion
With Mathematica, the
function gives a power series expansion. But I am looking for a different type of expansion. Let me explain :
$Assumptions = Element[x, Reals] && x > 0
f := Exp[x^2]*Erfc[x]
Series[f, {x, 0, 5}]
This gives the result :
SeriesData[x, 0, {
1, (-2) Pi^Rational[-1, 2], 1, Rational[-4, 3] Pi^Rational[-1, 2],
Rational[1, 2], Rational[-8, 15] Pi^Rational[-1, 2]}, 0, 6, 1]
But I am interested in getting an asymptotic expansion of the formĀ given in Abrahamson and Stegun (as shown below). How can I get these type of expansions with Mathematica? I would appreciate any
help that I can get.
3 Replies
Is this what you are looking for?
Series[Sqrt[Pi] Exp[x^2] Erfc[x], {x, Infinity, 15}, Assumptions -> x > 0] // TraditionalForm
Sometimes general formulas will work too, but I could not get it to work around Infinity:
SeriesCoefficient[Exp[x^2] Erfc[x], {x, 0, n}] // TraditionalForm
Thanks a lot for your reply. Yes, this is the solution that I was looking for. As an extra bonus, I can see the use of the SeriesCoefficient command also. Is there any reason why the general form is
not given by mathematica for expansion around Infinity?
I have another question which does not primarily relate to Mathematica. In the fomulae from Abrahamson and Stegun, I see the following condition attached :
| arg z| < 3*Pi/4
Would appreciate it if you could explain the relevance of this condition. Does it become relevant only when z is a complex number? All the better if you coud do this with Mathematica.
To see the relevance of the argument restriction, try taking the series at -infinity. My guess is the A&S formulae are intended to hold in as general a region as possible, so they are in a sense
giving a "sector" at complex infinity.
I'm not understanding your question regardingĀ the general form for expansion around infinity.
Be respectful. Review our Community Guidelines to understand your role and responsibilities. Community Terms of Use | {"url":"https://community.wolfram.com/groups/-/m/t/176802?sortMsg=Replies","timestamp":"2024-11-08T08:21:03Z","content_type":"text/html","content_length":"105518","record_id":"<urn:uuid:65fac9b4-724f-4186-a2a5-05617f4b9efe>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00490.warc.gz"} |
Voltage and Ground When Modeling Wave-Like EM Fields
This blog post continues our discussion of the terms voltage and ground. Here, we will define and interpret these terms for sinusoidally time-varying models. We look at the case of a transmission
line and address how to correctly define voltage and ground in problems involving wave-like fields.
A Simple Transmission Line
We will consider the transmission line shown in the image below: a metal wire that sits in free space above a ground surface (or ground plane, or signal ground), which we will define more precisely
very shortly. This falls into the category of a TEM transmission line, meaning that the electric and magnetic fields lie purely in the plane perpendicular to the line, and that the Poynting flux
vector is everywhere parallel to the line. (Very strictly speaking, this is a quasi-TEM transmission line, since the metal wire is not infinitely conductive, but as we will see, this point does not
alter any aspect of the following discussion.)
At one end of the wire, there is a sinusoidally time-varying source connecting the ground plane to the wire, and at the other end of the wire, there is a resistive load. Although we don’t see this
exact transmission line very often in practice, it is similar to a microstrip line.
A conductive wire above a ground plane, with a source at one end and a load at the other, along with a plot of the total current in the wire at an instant in time.
The sinusoidally time-varying source will drive current back and forth along the entire length of the wire, through the resistive load and then into, and out of, the ground plane. If we could take a
snapshot of the current at any instant in time, it would look like a sine wave propagating from source to load.
Now, when we are considering a time-varying current flowing through a conductive material, we do have to consider the skin effect: the tendency of time-varying current to flow on the outside surface
of a conductor. In fact, we will assume that the excitation frequency is so high that the skin depth is very, very small compared to the radius of the wire. So small, in fact, that we say that the
current flows on the surface of the conductor, rather than within the volume, and the wire can be modeled via the Impedance boundary condition. This is discussed in more detail in these previous blog
posts: “Modeling Metallic Objects in Wave Electromagnetics Problems” and “How to Model Conductors in Time-Varying Magnetic Fields“.
We next turn our attention to the surface below; what we’ve referred to as the ground plane. Recall from our earlier definition, in the DC regime, that we defined ground as a domain that has no
resistance to current flow (or at least, so little as to be irrelevant for our modeling purposes.) A similar definition applies here. Ground is a boundary to a domain that has no resistance, or it is
a perfectly conductive material. However, as just discussed, we know that there is a skin effect, and for a material with infinite conductivity, the skin depth will be exactly zero, so there will be
currents flowing on these surfaces that we are calling ground.
Understanding What Happens on the Ground Plane
Let us now address the big distinction between the DC case and the case of wave-like fields. Whereas in the DC case, we entirely ignored the currents within the ground domain, we now have currents
that flow along this ground surface, and these cannot be ignored. A visualization of these currents, as well as the electric and magnetic fields at one cross-sectional plane, all at one instant in
time, is in the image below.
Arrow visualization of the current (black), electric field (red), and magnetic field (blue) at one instant in time.
It might be reasonable to ask: How can there be finite currents on the surface of a material with infinite conductivity? To answer this, we need to also look at the free space above the ground plane.
This free space has an impedance, and currents flowing along this surface will see the impedance of this surrounding space.
This immediately raises a very important question: How much of the free space above the ground plane do we have to consider? It turns out that we have to consider the free space not only immediately
above the ground plane but also the space around the wire, and even some region of space above the wire. All of this structure contributes to the impedance of the transmission line. In fact, when
building a numerical model of such a case, it is necessary to study how much of the surrounding free space region to include; a point that is touched on in this example: Finding the Impedance of a
Parallel Transmission Line.
So, this means that the currents on this perfect electric conductor surface (that we are calling a ground plane) are affected by everything above it. Another way of saying this is the currents on
this PEC surface contain an image of the entire modeling space, and this leads us to a second interpretation of the PEC ground plane: It is a symmetry condition. It is as if there were an equivalent
structure of the other side of the plane, and on that side, the currents on the line will be pointing in opposite directions.
Via the symmetry condition, a model of a wire above a ground plane is equivalent to a model of a parallel-wire transmission line.
At this point, within the context of electromagnetic waves, we can now start to make some more precise definitions: A ground is a lossless (perfectly electrically conductive, or PEC) surface along
which finite currents flow. The currents flowing along this surface will be affected by all of the structure above it. If this PEC surface describes a plane on one side of the modeling space, then it
is equivalent to imposing a symmetry condition. If you have two PEC surfaces that are separated, you can arbitrarily choose one and define that as a ground. We can also, in some cases, come up with a
way to define an electric potential difference (a voltage) of the second PEC surface relative to this ground.
Defining Voltage in the Frequency Domain
Recall from our discussion of steady-state electric currents that we defined voltage as the path integral of the electric field between any two points. For the steady-state case, the electric field
is the gradient of a scalar potential, and this integral is always path independent. However, for the electromagnetic wave case, the electric field is the solution to a wave equation, and (via a
tedious amount of vector calculus that we will skip) we can show that the path integral of such an electric field is not path independent, except for some special cases.
One of these special cases is when you take the path integral along a line lying on a PEC surface. The electric field tangent to the surface is always zero, and thus the integral of the electric
field along any line on that surface is zero. However, the surface currents are defined as \mathbf{J = n \times H}, where \mathbf{ H} is computed from \mathbf{\nabla \times E} = -j \omega \mathbf{H},
so the currents are nonzero, even though the integral of the tangential electric field is zero. Keep in mind that there is no contradiction here; the impedance of the surroundings leads to finite
currents on this PEC surface with zero tangential electric field.
The second interesting case to consider is when we take the path integral of the electric field along a line in a plane perpendicular to the axis of a TEM transmission line. Since, by definition, the
electric and magnetic fields lie purely in this plane, it can be shown (via some more vector calculus that we will skip) that this integral will be path independent. That is, we can define a voltage
between points in this plane. So, choose one point that lies on the surface we call ground, and on the other point on the wire of our transmission line, take any path integral. Now we have a voltage,
and this corresponds to the measurement that you would get from a signal analyzer. You can also take the path integral of the magnetic field along a line that entirely divides the space between the
ground and the wire, and this will give us the current flowing along the transmission line.
Image showing various different integration paths for voltage (red) and current (blue).
Finally, let’s address the fact that this is actually a quasi-TEM line, due to the finite conductivity of the wire, which can be modeled via the Impedance boundary condition. For this case, the
out-of-plane components of the electric and magnetic fields are so small relative to the in-plane components that we can still safely use the aforementioned definitions.
So, let’s write down what we know:
1. The voltage is the path integral of the electric field, but this can only be evaluated where the curl of the electric fields is zero, or nearly zero: at the cross section of a TEM or quasi-TEM
transmission line.
2. At the cross section of a TEM or quasi-TEM transmission line, the voltage corresponds to what you would physically measure via a signal analyzer. It is really only here that the term voltage has
any useful meaning in a frequency-domain wave electromagnetics model.
3. On a PEC surface, you can integrate the electric field along a path on that surface, but if you integrate along a path that is not on the surface, you might get a nonzero integral. Also, we’ve
already seen that there will be currents, so zero voltage difference between two points does not mean zero current. Thus, in practice, there is little value to speaking about the voltage in this
context. If we would try to actually physically measure the fields between two points, we would have to introduce a sensor, including some kind of transmission line between those points, which
would alter the device.
With all of that information firmly in our minds, we can now model with confidence. For the case we have here, we can use the TEM-type Port boundary condition, with the Ground and Electric Potential
subfeatures applied to the edges of the ground plane and wire. A complete overview of all of the other options for modeling TEM-type transmission lines is given in the Learning Center article “
Modeling TEM and Quasi-TEM Transmission Lines“.
Schematic of the setup of a transmission line model. Two TEM ports (crosshatched) at either end have a ground (blue) and voltage (red) defined.
Closing Remarks
Now you know how to use the terms voltage and ground with confidence in the context of frequency-domain electromagnetics wave modeling. We can extend the same arguments to the transient case and
arrive at the same conclusion: In time-domain modeling, the ground is a current return path that can be a symmetry condition.
So, for the case of any time-varying model where we consider both electric and magnetic fields, you can only speak about voltage in the context of evaluating the fields at the cross section of a TEM
transmission line. In spite of the simplicity of this statement, the arguments that we had to follow to get to this point are very helpful in understanding modeling of electromagnetic devices.
Next Step
Get a detailed demonstration of modeling TEM and quasi-TEM transmission lines in a related Learning Center article, which includes step-by-step modeling instructions and software screenshots, by
clicking the button below: | {"url":"https://www.comsol.com/blogs/voltage-and-ground-when-modeling-wave-like-em-fields","timestamp":"2024-11-14T17:09:09Z","content_type":"text/html","content_length":"96261","record_id":"<urn:uuid:0043b050-0fb3-4957-abc8-9360d1a383c9>","cc-path":"CC-MAIN-2024-46/segments/1730477393980.94/warc/CC-MAIN-20241114162350-20241114192350-00730.warc.gz"} |
8.3 Three Variables: Bubble Chart and 3-D Scatter Plot | An Introduction to Spatial Data Science with GeoDa
8.3 Three Variables: Bubble Chart and 3-D Scatter Plot
When the number of dimensions is at most three, it remains relatively easy to visualize the multivariate attribute space. The bubble chart implements this by augmenting a standard two-dimensional
scatter plot with an additional attribute for each point, i.e., the size of the point or bubble. More directly, observations can also be visualized as points in a three-dimensional scatter plot or
data cube.
8.3.1 Bubble Chart
The bubble chart is an extension of the scatter plot to include a third and possibly a fourth variable into the two-dimensional chart. While the points in the scatter plot still show the association
between the two variables on the axes, the size of the points, the bubble, is used to introduce a third variable. In addition, the color shading of the points can be used to consider a fourth
variable as well, although this may stretch one’s perceptual abilities.
The introduction of the extra variable allows for the exploration of interaction. The point of departure (null hypothesis) is that there is no interaction. In the plot, this would be reflected by a
seeming random distribution of the bubble sizes among the scatter plot points. On the other hand, if larger (or smaller) bubbles tend to be systematically located in particular subareas of the plot,
this may suggest an interaction. Through the use of linking and brushing with the map, the extent to which systematic variation in the bubbles corresponds with particular spatial patterns can be
readily investigated, in the same way as illustrated in the previous chapter.
The bubble chart was popularized through the well-known Gapminder web site, where it is used extensively, especially to show changes over time.^57 This aspect is currently not implemented in GeoDa.
8.3.1.1 Implementation
The bubble chart is invoked from the menu as Explore > Bubble Chart, and from the toolbar by selecting the left-most icon in the multivariate group of the EDA functionality, shown in Figure 8.1. This
brings up a Bubble Chart Variables dialog to select the variables for up to four dimensions: X-Axis, Y-Axis, Bubble Size and Bubble Color.
To illustrate this method, three variables from the Oaxaca Development sample data set are selected: peduc_20 (percent population with an educational gap in 2020) for the X-Axis, pserv_20 (percent
population without access to basic services in the dwelling in 2020) for the Y-Axis, and pepov_20 (percent population living in extreme poverty in 2020) for both Bubble Size and Bubble Color. The
default setting uses a standard deviational diverging legend (see also Section 5.2.3) and results in a somewhat unsightly graph, with circles that typically are too large, as in Figure 8.7.
Before going into the options in more detail, it is useful to know that the size of the circle can be readily adjusted by means of the Adjust Bubble Size option. With the size significantly reduced,
a more appealing Figure 8.8 is the result.
As mentioned, the null hypothesis is that the size distribution (and matching colors) of the bubbles should be seemingly random throughout the chart. In the example, this is clearly not the case,
with high values (dark brown colors and large circles) for pepov_20 being primarily located in the upper right quadrant of the graph, corresponding to low education and low services (high values
correspond with deprivation). Similarly, low values (small circles and blue colors) tend to be located in the lower left quadrant. In other words, the three variables under consideration seem to be
strongly related.
Given the screen real estate taken up by the circles in the bubble chart, this is a technique that lends itself well to applications for small to medium sized data sets. For larger size data sets,
this particular graph is less appropriate.
8.3.1.2 Bubble chart options
In addition to Adjust Bubble Size, the bubble chart has nine main options, invoked in the usual fashion by right clicking on the graph. Several of these are shared with other graphs, in particular
with the scatter plot, such as Save Selection, Copy Image To Clipboard, Save Image As, View, Show Status Bar and Selection Shape. They will not be further discussed here (see, e.g., Section 7.3.1).
Classification Themes, Save Categories and Color work the same as for any map (see Section 4.5).
Figure 8.9 provides an illustration of the flexibility that these options provide. The graph pertains to the same three variables as before, but the Classification Themes option has been set to
Themeless. As such, this results in green circles, but the Color option applied to the legend (see Section 4.5) allows the Opacity for the Fill Color for Category to be set to zero, resulting in the
empty circles.
8.3.1.3 Bubble chart with categorical variables
One particular useful feature to investigate potential structural change in the data is to use the Unique Values classification, in combination with setting the Size to a constant value.^58
With peduc_20 for the X-Axis and ppov_20 (percentage population living in poverty) as the Y-Axis, Bubble Size is now set to the constant Const, with the categorical variable Region for Bubble Color.^
59 First, this yields a rather meaningless graph, based on the default standard deviation classification. Changing Classification Themes to Unique Values results in a scatter plot of the two
variables of interest, with the bubble color corresponding to the values of the categorical variable. In the example, this is Region, as in Figure 8.10.
In this case, there seems to be little systematic variation along the region category, in line with our starting hypothesis.
The use of a bubble chart to address structural change is particularly effective when more than two categories are involved. In such an instance, the binary selected/unselected logic of scatter plot
brushing no longer works. Using the bubble chart in this fashion allows for an investigation of structural changes in the bivariate relationship between two variables along multiple categories, each
represented by a different color bubble. This forms an alternative to the conditional scatter plot, considered in Section 8.4.2.
8.3.2 3-D Scatter Plot
An explicit visualization of the relationship between three variables is possible in a three-dimensional scatter plot, the direct extension of principles used in two dimensions to a three-dimensional
data cube. Each of the axes of the cube corresponds to a variable, and the observations are shown as a point cloud in three dimensions (of course, rendered as a perspective plot onto the
two-dimensional plane of the screen).
The data cube can be manipulated by zooming in and out, in combination with rotation, to get a better sense of the alignment of the points in three-dimensional space. This takes some practice and is
not necessarily that intuitive to many users. The main challenge is that points that seem close in the two-dimensional rendering on the screen may in fact be far apart in the actual data cube. Only
by careful interaction can one get a proper sense of the alignment of the points.
8.3.2.1 Implementation
The three-dimensional scatter plot method is invoked as Explore > 3D Scatter Plot from the menu, or by selecting the second icon from the left on the toolbar depicted in Figure 8.1. This brings up a
3D Scatter Plot Variables selection dialog for the variables corresponding to the X, Y and Z dimensions. Again, peduc_20, pserv_20 and pepov_20 are used.
The corresponding initial default data cube is as in Figure 8.11, with the Y-axis (pserv_20) as vertical, and the X (peduc_20) and Z-axes (pepov_20) as horizontal. Note that the axis marker (e.g.,
the X etc.) is at the end of the axis, so that the origin is at the unmarked side, i.e., the lower left corner where the green, blue and purple axes meet.
8.3.2.2 Interacting with the 3-D scatter plot
The data cube can be re-sized by zooming in and out. It is sometimes a bit ambiguous what is meant by these terms. For the purposes of this illustration, zooming out refers to making the cube
smaller, and zooming in to making the cube larger (moving into the cube, so to speak).
The zoom functionality is carried out by pressing down on the track pad with two fingers and moving up (zoom out) or down (zoom in). Alternatively, one can press Control and press down on the track
pad with one finger and move up or down. With a mouse, one moves the middle button up or down.
In addition, the cube can be rotated by means of the pointer: one can click anywhere in the window and drag the cube by moving the pointer to a different location. The controls on the left-hand side
of the view allow for the projection of the point cloud onto any two-dimensional pane, and the construction of a selection box by checking the relevant boxes and/or moving the slider bar (see Section
For example, in Figure 8.12, the cube has been zoomed out and the axes rotated such that Z is now vertical. With the Project to X-Y box checked on the left, a 2-dimensional scatter plot is projected
onto the X-Y plane, i.e., showing the relationship between peduc_20 and pserv_20.
8.3.2.3 Selection in the 3-D scatter plot
Selection in the three dimensional plot (or, rather, its two-dimensional rendering) is a little tricky and takes some practice. The selection can be done either manually, by pressing down the command
key while moving the pointer, or by using the guides under the Select, hold CMD for brushing check box.
Checking this box creates a small red selection cube in the graph. The selection cube can be moved around with the command key pressed, or can be moved and resized by using the controls to the left,
next to X:, Y:, and Z:, with the corresponding variables listed.
The first set of controls (to the left) move the box along the matching dimension, e.g., up or down the X values for larger or smaller values of peduc_20, and the same for the other two variables.
The slider to the right changes the size of the box in the corresponding dimension (e.g., larger along the X dimension). The combination of these controls moves the box around to select observation
points, with the selected points colored yellow.
The most effective way to approach this is to combine moving around the selection box and rotating the cube. The reason for this is that the cube is in effect a perspective plot, and one cannot
always judge exactly where the selection box is located in three-dimensional space.
A selection is illustrated in Figure 8.13, where only a few out of seemingly close points in the data cube are selected (yellow). By further rotating the plot, one can get a better sense of their
location in the point cloud.
As in all other graphs, linking and brushing is implemented for the 3D scatter plot as well. Figure 8.14 shows an example of brushing in the six quantile map for ALTID and the associated selection in
the point cloud. The assessment of the match between closeness in geographical space (the selection in the map) and closeness in multivariate attribute space (the point cloud) is a fundamental notion
in the consideration of multivariate spatial correlation, discussed in Chapter 18.
Similar to the bubble chart, the 3D scatter plot is most useful for small to medium sized data sets. For larger numbers of observations, the point cloud quickly becomes overwhelming and is no longer
effective for visualization.
8.3.2.4 3-D scatter options
The 3D scatter plot has a few specialized options, available either on the Data pane (which was considered exclusively so far), or on the OpenGL pane. The latter gives the option to adjust the
Rendering quality of points, the Radius of points, as well as the Line width/thickness and Line color. These are fairly technical options that basically affect the quality of the graph and the speed
by which it is updated. In most situations, the default settings are fine.
Finally, under the Data button, there are the Show selection and neighbors and Show connection line options. These items are relevant when a spatial weights matrix has been specified. This is covered
in Chapter 10.
57. A constant variable can be readily created by means of the Calculator option in the data table. Specifically, the Univariate tab with ASSIGN allows one to set the a variable equal to a constant.
58. The regional classifications are: (1) Canada; (2) Costa; (3) Istmo; (4) Mixteca; (5) Papaloapan; (6) Sierra Norte; (7) Sierra Sur; and (8) Valles Centrales.↩︎ | {"url":"https://lanselin.github.io/introbook_vol1/three-variables-bubble-chart-and-3-d-scatter-plot.html","timestamp":"2024-11-08T19:16:41Z","content_type":"text/html","content_length":"130845","record_id":"<urn:uuid:2f2353a6-80e3-440f-b43f-b68006a81c1f>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00862.warc.gz"} |
Yield Curve
TREASURY YIELD CURVE: Current Spot and Forward Curves
There are many types of fixed-income securities and markets. The largest fixed income market results from the U.S. Treasury, borrowing cash from the general investing public. The prices of these
fixed-income securities result from trading, and which generates the Treasury yield curve by plotting the implied yield to maturity from the prices of Treasury instruments against the time to
maturity. This yield curve has a direct influence over all economic activity in the US.
In the charts below the current behavior for the Treasury yield curve is provided for a choice of common compounding conventions. To apply the yield curve requires answering a few basic questions.
These questions are described below the following chart. An expalnation of the concepts required to answer these questions is provided in the Textbook link to the left.
Working with the Yield Curve
You first need to answer a few basic questions:
1. What is my problem's time horizon? In other words which rate is relevant -- 3-months, 5-years or 30-years?
This often depends upon the investment horizon you are working with. For example, suppose you are applying the Capital Asset Pricing Model (CAPM) to estimate the expected return from a stock. This
requires working with the yield curve and many users choose a rate between 10-years and 30-years to estimate expected returns when using CAPM.
2. What is my compounding convention? This could be continuous compounding if you are working with options or it may be discrete. If discrete how many times am I compounding per year?
3. Is it important to assess a pure discount rate? The pure disocunt rate is implied from a Treasury zero coupon bond (i.e., Treasury strips and treasury bills). The pure discount rate or "Zero"
curve is important when working with any valuation problem.
4. What is the current yield curve telling me about future interest rates? By backing out the "forward curve" this provides important information about what investors expectations about future rates
The above set of charts provide all of the above. That is, you can see the current Treasury Yield curve depicted in continuous through to annual compounding forms. You can also see plotted the spot
rates, the zero curve rates and the forward rates. That latter lets you take a look into the future using current market data. To read more about the above types of questions and concepts you are
encouraged to click on Bond Tutor: The Textbook to the left on this screen.
What Drives the Yield Curve?
From the above charts you can see that the yield curve shifts over time. These shifts are inresponse to changes in expectations about the major fundamental drivers of the yield curve. That is,
Inflation Expectations
Consumption/Growth Behavior Changes
Federal Reserve Bank Expectations
Inflation Expectations
If consumer prices are expected to increase strongly then the suppliers of capital must be rewarded more for postponing their consumption decisions. That is, the opportunity cost of consumption must
increase and so inflation expectations will have a direct first order impact upon interest rates. To explore this further see the tab above labelled "Inflation."
Consumption/Growth Expectations
If growth declines and the economy moves into a recession -- how would this influence your decision to consume today? The answer is likely to be negatively. Job prospects, bonuses, pay rises are
likely to disappear and so major consumption decisions become postponed. This implies that the suppliers of capital no longer need to be rewarded as much for postponing consumption and thus interest
rates will decline.
Federal Reserve Bank Expectations
In the United States, the Federal Reserve has a dual mandate: To promote stable inflation and to promote maximum employment. In addition, the Federal Reserve is legally permitted to manipulate the US
Treasury markets. As a result, the Federal Reserve Bank manipulates the US Treasury yield curve -- particularly at the short end in an attempt to implement it's dual manadate.
When the Federal Reserve Bank is trying to promote consumption and growth it lowers interest rates by agressively buying and pushing prices up. Given the inverse relationship between prices and the
yield to maturity this implies that interest rates fall. If it is trying to dampen consumption/growth the Federal Reserve Bank does the opposite and starts to aggresively sell Treasury instruments to
push prices down. This results in shifting interest rates up. | {"url":"https://optiontutor.com/currentYC/currentYC.htm","timestamp":"2024-11-02T20:37:17Z","content_type":"text/html","content_length":"12025","record_id":"<urn:uuid:82493124-0b8f-4a7c-b1ec-7446393c9f0f>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00031.warc.gz"} |
How to average ratings from a student live online evaluation?
Feb 22, 2021 01:29 AM
Could you help me calculate the average results (ratings) of a student live online evaluation?
I am testing 2 tools to collect the feedback of respondents, one is designed with Typeform and the other with Airtable. It is not possible to add a “multiple select” in the primary field with
Typeform (“names” of students can only be added in the second column), while name can be put in the primary field with Airtable form. Does it matter to average ratings?
The evaluation goes as follows: The respondents select the name of the person to evaluate from a “multiple choice/select” list and rate them using "opinion scales/ratings). The answers are imported
to Airtable at the end of the live session as you can see from the screenshots.
I would like to calculate the average of ratings for each group of names and use the value in the base. I noticed that the “group functionality” does not help as it is not possible to use/extract the
average values in the same or a new table.
Do you have any idea how it could be done?
Thanks in advance!
Evaluation from Typeform:
Evaluation from Airtable form:
Feb 23, 2021 07:11 AM
Feb 22, 2021 02:43 AM
Feb 22, 2021 03:52 AM
Feb 22, 2021 04:05 AM
Feb 22, 2021 04:51 AM
Feb 22, 2021 08:01 AM
Feb 23, 2021 06:50 AM
Feb 23, 2021 07:11 AM
Feb 23, 2021 12:18 PM | {"url":"https://community.airtable.com/t5/base-design/how-to-average-ratings-from-a-student-live-online-evaluation/td-p/128712","timestamp":"2024-11-03T13:44:20Z","content_type":"text/html","content_length":"452157","record_id":"<urn:uuid:a3556c20-50ec-425c-8c7c-dbc97d52ccb3>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00784.warc.gz"} |
How To Draw A Tangent Line On A Graph
How To Draw A Tangent Line On A Graph - Web adding a tangent line in excel can be a useful way to visualize data and draw attention to certain points in a graph. Choose a point on the curve. Graph
functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. Web explore math with our beautiful, free online graphing calculator. Web step 1 open the excel worksheet
containing the data you want to use for a tangential line.
Web understanding how to draw a tangent line in excel is crucial for data analysis. Learn more about tangent, tangent line, transfer function, process control, control systems, plot, step response
Web important notes on tangent line: Graph functions, plot points, visualize algebraic equations, add sliders, animate graphs, and more. In this section, we investigate the graphs of the tangent and
cotangent functions. Finding the tangent line to a point on a curved graph is challenging and requires the use of calculus; Web f(x) = acot(bx − c) + d is a cotangent with vertical and/or horizontal
stretch/compression and shift.
Graphing the Tangent Function Part 1 YouTube
The trigonometric ratios can also be considered as functions of a variable which is the measure of an angle. Selecting the right data set and graph type is important for accurately drawing a tangent
Part 4 Year 10 Further Trigonometry (Applications) Year 10 Maths Guide
Web to draw a tangent line: Since the line is tangent to \(p = (1,1)\) we. Web how to draw a tangent line on a curve. An example of this can be seen below. Video.
Tan X Graph Domain Range DOMAINVB
Web to draw a tangent line: Put a straight edge at that point on the curve. It can also be used to compare two different sets of data or to highlight a trend. Learn more.
Grade 10 Trigonometry How to draw a Tan Graph YouTube
Web how to interpret tangent graphs. So given the graph of f (x) in the neighbourhood of (a,f (a)), you can place a ruler against (a,f (a)) and rotate it about that point until it.
how to draw a common tangent on two circles of different radius. YouTube
Web how to interpret tangent graphs. Web explore math with our beautiful, free online graphing calculator. Drawing tangent lines on a graph of position vs. Draw a straight line from the axis of the
Tangent Graphs YouTube
Here, we will use radians. Web explore math with our beautiful, free online graphing calculator. Web drawing a tangent to curve on motion graphs. Since, tan(x) = sin ( x) cos ( x) the tangent.
28+ How To Graph Tangent Functions MaeganAniello
Web how to draw a tangent line on a curve. Since, tan(x) = sin ( x) cos ( x) the tangent function is undefined when cos(x) = 0. Draw a straight line from the axis.
Tangent and Cotangent Graphs Brilliant Math & Science Wiki
Then the ruler indicates the tangent: A curved line graph is based on sets of two data points, for example time and amplitude. Put a straight edge at that point on the curve. Finding the.
How to Find the Tangent Line of a Function in a Point Owlcation
The tangent function is defined as the length of the red segment. Web the tangent line to a curve at a given point is the line which intersects the curve at the point and has.
How to Find the Tangent Line of a Function in a Point Owlcation
Web important notes on tangent line: The tangent function is defined as the length of the red segment. Put a straight edge at that point on the curve. Drawing tangent lines on a graph of.
How To Draw A Tangent Line On A Graph Web drawing a tangent to curve on motion graphs. An example of this can be seen below. Adjust the angle of the straight edge so that near the point it is
equidistant from the curve on either side of the point. Here, we will use radians. Selecting the right data set and graph type is important for accurately drawing a tangent line.
How To Draw A Tangent Line On A Graph Related Post : | {"url":"https://classifieds.independent.com/print/how-to-draw-a-tangent-line-on-a-graph.html","timestamp":"2024-11-06T02:02:28Z","content_type":"application/xhtml+xml","content_length":"22946","record_id":"<urn:uuid:686f33e2-f071-4387-ba90-d7b6f7427928>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00814.warc.gz"} |
Szemerédi's Theorem
From Scholarpedia
Ben Joseph Green and Terence Tao (2007), Scholarpedia, 2(7):3446. doi:10.4249/scholarpedia.3446 revision #91851 [link to/cite this article]
Szemerédi's theorem states that any "positive fraction" of the positive integers will contain arbitrarily long arithmetic progressions \(a, a+r, a+2r, \ldots, a+(k-1)r\ .\) More precisely,
Theorem (Szemerédi's theorem) Let \(A \subset {\Bbb Z^+}\) be a subset of the positive integers of positive upper density, i.e., \(d^*(A) := \limsup_{N \to \infty} \frac{\# \{ n \in A: n \leq N \}}
{N} > 0\ .\) Then for any integer \(k \geq 1\ ,\) the set \(A\) contains at least one arithmetic progression \(a, a+r, a+2r, \ldots, a+(k-1) r\) of length \(k\ ,\) where \(a, r\) are positive
Intuitively, this theorem is saying that long arithmetic progressions are so prevalent that it is impossible to eradicate them from a set of positive integers unless one shrinks the set so much that
it has density zero. This deep fact is rather striking, given that many other patterns (e.g., geometric progressions \(a, a r, a r^2 \) with \( r>1 \)) are much easier to eradicate (for instance, the
set of squarefree numbers has positive upper density \( 6/\pi^2 \ ,\) but clearly contains no \( a r^2 \)). One notable feature of the theorem is that there is almost no assumption made on the set \
(A\ ,\) other than one of size; the set \(A\) could be very structured (e.g. an infinite arithmetic progression) or very irregular (e.g. a randomly generated set), and yet one has long arithmetic
progressions in either case. In fact, understanding this dichotomy between structure and randomness is the key to all the known proofs of Szemerédi's theorem: it is possible to generate arithmetic
progressions from structure and from randomness, but for two completely different reasons, and so one must somehow separate the structure from the randomness in order to proceed. The depth of this
theorem can also be discerned from Behrend's counterexample to a natural stronger version of this theorem, which will be discussed later.
Szemerédi's theorem was initially conjectured by Erdős and Turán [ET] in 1936, but was only proven in full generality by Szemerédi [Sz2] in 1975. Several further important proofs of this theorem,
using different types of analysis, were subsequently given, including an ergodic-theory proof by Furstenberg [F] in 1977, a Fourier-analytic and combinatorial proof by Gowers [Go2] in 2002, and
proofs based on hypergraphs given by Gowers [Go3], [Go4] and by Nagle, Rődl, Schacht, and Skokan [NRS], [RSc], [RSk], [RSk2]. These proofs have also led to a number of deep and powerful
generalisations and variants of the above theorem. Despite this multitude of proofs, though, the theorem is still considered quite deep and difficult.
Szemerédi's theorem was a major landmark in additive number theory for several reasons. Not only did it solve a well-known conjecture in the subject, the powerful methods introduced in order to prove
this theorem have turned out to be tremendously useful in many other problems, and have stimulated development and progress in multiple fields of mathematics. Also, the theorem itself has been
applied to prove many other results; for instance, it was a key component in the argument of Green and Tao [GT] demonstrating that the prime numbers contain arbitrarily long arithmetic progressions.
(The primes have density zero, so Szemerédi's theorem does not apply directly; instead, the argument in [GT] combines Szemerédi's theorem with an additional transference argument, together with some
number-theoretic estimates.)
The first significant precursor to Szemerédi's theorem was van der Waerden's theorem [vdW], which appeared in 1927:
Theorem (van der Waerden's theorem) Suppose that the positive integers \({\Bbb Z}^+\) are colored using finitely many colors. Then at least one of the color classes will contain arbitrarily long
arithmetic progressions.
This theorem can be easily deduced from Szemerédi's theorem (since, by the pigeonhole principle, at least one of the color classes will have positive upper density), but it is far more difficult to
deduce Szemerédi's theorem from van der Waerden's theorem. For example, the hypotheses of van der Waerden's theorem imply that at least one color class will contain arbitrarily long geometric
progressions, but as noted above there are sets with upper density larger than \(1/2\) that do not even have 3 term geometric progressions.
van der Waerden's theorem is similar in spirit to Ramsey's theorem [Ram] for colorings of complete graphs, which appeared at about the same time; indeed these two theorems initiated the field of
Ramsey theory, in which one seeks to establish the existence of various structures inside a color class of a larger structure.
Van der Waerden's proof of his theorem was elementary and quite short, but it was also highly recursive in nature, and so the effective bounds that the proof provided were very weak. For instance, if
one colored the positive integers into \(r\) colors and requested an arithmetic progression of length \(k\) which was monochromatic (i.e. contained inside a single color class), then the proof did
assert that such a progression would exist, and that all elements of that progression would be less than a certain finite quantity \(W(k,r)\) depending on \(k\) and \(r\) (with the best such \(W(k,r)
\) known as the van der Waerden number of \(k\) and \(r\)), but the upper bound on \(W(k,r)\) provided by the proof grew incredibly quickly; even for \(r=2\ ,\) the growth rate was like an Ackermann
function in \(k\ .\) (This growth rate was not improved until 1988, when Shelah [Sh] provided a growth rate which was primitive recursive, though still rather rapidly growing; the best upper bound
currently known on \(W(k,r)\) for general \(k\) is \(2^{2^{r^{2^{k+9}}}}\ ,\) due to Gowers [Go2] in 2001, and which was obtained via Szemerédi's theorem.)
In 1936, Erdős and Turán [ET] investigated the problem further, motivated in part by an old conjecture (proven in 2004) that the primes contained arbitrarily long arithmetic progressions. They
proposed two conjectures which would imply van der Waerden's theorem. The first (weaker) conjecture was the statement that became Szemerédi's theorem; the second (stronger) conjecture, which they
attribute to Szekeres, asserted an extremely quantitative bound, in particular that given any \(k \geq 1\) there existed an \(\varepsilon > 0\ ,\) such that any subset of \(\{1,\ldots,n\}\) of
cardinality at least \(n^{1-\varepsilon}\) would necessarily contain at least one arithmetic progression of length \(k\ ,\) if \(n\) was sufficiently large depending on \(k\ .\) They noted that this
conjecture was in particular more than sufficient to guarantee that the primes contained arbitrarily long arithmetic progressions.
The stronger of these conjectures was disproven (even when \(k=3\ ,\) which is the first non-trivial case) by Salem and Spencer [SS] in 1942, with improvements to the counterexample given by Behrend
[Be] and Moser [M]. Remarkably, Behrend's 1946 counterexample is still the best known for the \(k=3\) problem (higher \(k\) variants of this example were constructed by Rankin [Ran]). It is based,
ultimately, on the simple observation that a sphere in any dimension is convex and thus cannot contain any arithmetic progressions of length three. By working with a suitable discrete analogue \(\{ x
\in {\Bbb Z}^d: |x|^2 = r^2 \}\) of a sphere and then transplanting that set to the integers, Behrend was able to create, for arbitrarily large \(n\ ,\) a subset of \(\{1,\ldots,n\}\) of cardinality
at least \(n \exp( - c \sqrt{\log n} )\) for some absolute constant \(c > 0\ ,\) which did not contain any arithmetic progressions of length three. In particular, the size of this set is
asymptotically larger than \(n^{1-\varepsilon}\) for any fixed \(\varepsilon > 0\ .\) This type of counterexample rules out a number of elementary approaches to Szemerédi's theorem (e.g. using
arguments based only on the pigeonhole principle, or the Cauchy-Schwarz inequality), and strongly suggests that the proof of these conjectures must be somehow "iterative" in nature.
The cases \(k < 3\) of Szemerédi's theorem are trivial. The \(k=3\) case was first settled by Roth [R] in 1953. This argument employed methods from Fourier analysis (or more specifically, the
Hardy-Littlewood circle method). Roughly speaking, the idea was to assume for contradiction that one had a set of integers of positive density which contained no arithmetic progressions of length
three, and then to use Fourier analysis to construct a new set of integers with even higher density which still contained no such progressions. Eventually one would be forced to construct a set of
density exceeding \(1\ ,\) which was of course absurd.
Roth's elegant argument failed to extend to the case of general \(k\ ,\) for reasons which were not fully understood until the work of Gowers [Go2] much later. The \(k=4\) case was first established
by Szemerédi [Sz] in 1969 by an elementary but intricate combinatorial argument. Even with this breakthrough, the higher \(k\) case continued to prove to be quite difficult, and it was only in 1975
that Szemerédi [Sz2] was able to establish his theorem in full generality. This problem was given a cash prize of $1000 by Paul Erdős, who was well-known for assigning prizes to difficult problems;
Szemerédi was first mathematician to ever collect a prize of that magnitude from Erdős. (In 1984, Frankl and Rődl[FR] solved another problem to which Erdős offered $1000, although they only accepted
$500 due to Erdős' financial situation[Ba].) Erdős later offered $3000 for a proof of a stronger conjecture (sometimes mis-attributed to [ET]), namely that any set \(A\) of positive integers whose
sum of reciprocals \(\sum_{n \in A} \frac{1}{n}\) was divergent, would necessarily contain arbitrarily long progressions; this conjecture remains open to this day, even if one only seeks a single
progression of length three. However, in the special case that the set \(A\) consists of the prime numbers or a relatively dense subset thereof, the result was established in [GT].
Szemerédi's proof was very sophisticated (though elementary), and introduced a number of important new tools, most notably what is now known as the Szemerédi regularity lemma for graphs, which
roughly speaking allows one to approximate any large dense graph in a certain statistical sense by a bounded complexity random graph. (This lemma has since had an enormous number of applications in
combinatorics and computer science; see [KS] for a survey.) Two years later, in 1977, Furstenberg [F] introduced a dramatically different proof of that theorem, recasting the problem as one in
dynamical systems, and then using methods from ergodic theory to solve the problem. More specifically, Furstenberg showed that Szemerédi's theorem was logically equivalent to the following, very
different-looking, result:
Theorem (Furstenberg recurrence theorem) Let \((X, {\mathcal B}, \mu)\) be a probability space (thus \({\mathcal B}\) is a \(\sigma\)-algebra of subsets of \(X\ ,\) and \(\mu: {\mathcal B} \to [0,1]
\) is a probability measure on \(X\)), let \(T: X \to X\) be a measure-preserving bijection (thus \(\mu(T(E)) = \mu(E)\) for all measurable \(E\)); we refer to \((X,{\mathcal B},\mu,T)\) as a
measure-preserving system. Then for any set \(E \subset X\) of positive measure and every integer \(k \geq 1\) there exist infinitely many positive integers \(n\) such that the set \(E \cap T^n E \
cap \ldots \cap T^{(k-1)n} E\) has positive measure. In fact, we have the slightly stronger statement \[ \displaystyle\liminf_{N \to\infty} \frac{1}{N} \sum_{n=1}^N \mu(E \cap T^n E \cap \ldots \cap
T^{(k-1)n} E ) > 0.\]
The question of whether the sequence in the above equation was convergent (i.e. if the limit inferior could be replaced with a limit), and what that limit was, turned out to be surprisingly
difficult, and was only settled in 2005 by Host and Kra [HK]. (A different proof was given in 2007 by Ziegler [Z] and a geometric interpretation of the limit was given by Ziegler in [Z2]; a third
proof was given very recently in [T]). The \(k=2\) case of this theorem is the classical Poincaré recurrence theorem, while the \(k=1\) case is trivial. Just as Szemerédi's theorem is remarkable for
requiring very few assumptions on the set \(A\ ,\) Furstenberg's theorem is remarkable for requiring very few assumptions on the probability space \((X,\mu)\ ,\) the shift map \(T\ ,\) or the set \(E
\ .\)
The equivalence of Szemerédi's theorem and the Furstenberg recurrence theorem is due to a more general principle, first formulated in [F], and now known as the Furstenberg correspondence principle.
Roughly speaking, the correspondence links sets \(A\) of integers with sets \(E\) in a measure-preserving system \((X,{\mathcal B},\mu,T)\) by looking at finite segments \(\{ n \in A: 1 \leq n \leq N
\}\) of \(A\) and "taking limits"' as \(N \to \infty\) in a certain weak topology. The proof of the Furstenberg recurrence theorem is shorter than the original proof of Szemerédi's theorem, and
relies primarily on an induction on the \(\sigma\)-algebra \({\mathcal B}\ .\) A key theme in the proof is the dichotomy between two types of properties in a measure-preserving system, namely
periodicity (or more precisely, almost periodicity) and mixing (or more precisely, weak mixing). Very roughly speaking, when there is periodicity, then we expect \(T^n E\) to be close to \(E\) for
many \(n\ ,\) whereas when we have mixing, we expect \(T^n E\) to become "independent" of \(E\) for large \(n\ .\) The key insights are that periodicity and mixing can both be used to create
recurrence, and that arbitrary measure-preserving systems can be decomposed into periodic and mixing components (the precise statement of this is now known as the Furstenberg structure theorem). For
more details see [FKO], [F2]. Furstenberg's method led to a vast number of generalisations and other developments, some of which we will discuss below.
The Fourier-analytic method of Roth, which treated the \(k=3\) case, was finally extended to the \(k=4\) case by Gowers [Go] in 1998, and then to the general case by Gowers [Go2] in 2001. (An earlier
argument of Roth [R2] combined Szemerédi's arguments with those in [R] to produce a hybrid proof of the \(k=4\) case of Szemerédi's theorem.) Gowers realised that classical Fourier analysis, which
relied on linear phases such as \(e^{2\pi i n \theta}\ ,\) were suitable tools for detecting arithmetic progressions of length three, but already for progressions of length four one needed to work
with a generalised notion of Fourier analysis, in which quadratic phases such as \(e^{2\pi i n^2 \theta}\) made an appearance. (The reason for this is technical, but is related to the fact that a
function \(f: {\Bbb R} \to {\Bbb R}\) is linear, then its values on two points of an arithmetic progression can be extrapolated to evaluate the value on the third (i.e. two points determine a line).
For quadratic \(f\ ,\) this property is no longer true, however it remains true that the values on three points on an arithmetic progression can be extrapolated to evaluate the value on the fourth.)
Gowers' proofs then proceeded by developing this generalised Fourier analysis and combining them with several innovative new tools, including what is now known as the Balog-Szemerédi-Gowers lemma
which is now of fundamental importance in additive combinatorics and number theory. Gowers' argument also gave fairly reasonable quantitative bounds on quantities such as the van der Waerden number \
(W(k,r)\ .\) (Szemerédi's original argument used van der Waerden's theorem inside the proof, and so did not provide a better bound on that theorem; the argument of Furstenberg was infinitary and thus
provided no explicit bound at all, although recently [T] it was shown that a quantitative bound could, in principle, be extracted from this type of argument.) Further variants and extensions of
Gowers' argument were carried out more recently by Green and Tao [GT2], at least in the cases \(k \leq 4\ .\)
A fourth type of proof, based on the theory of graphs and hypergraphs, was introduced by Ruzsa and Szemerédi [RS] in 1978, who observed that Roth's theorem could in fact be deduced from a more
abstract graph-theoretical result which is now known as the triangle removal lemma; this lemma asserts that if a graph \(G\) on \(n\) vertices contains at most \(\varepsilon n^3\) triangles, then all
of these triangles can be deleted at the cost of removing at most \(c(\varepsilon) n^2\) edges, where \(c(\varepsilon)\) is a quantity which goes to zero as \(\varepsilon\) goes to zero. They then
showed that this removal lemma followed easily from the regularity lemma introduced earlier by Szemerédi. It was then natural to generalize these observations to tackle the full strength of
Szemerédi's theorem, but this required one to exchange the familiar setting of graphs to the more difficult setting of hypergraphs (in which each "edge" does not connect just two vertices, but can
connect three or more vertices together). One of the key difficulties was to obtain a sufficiently strong analogue of the Szemerédi regularity lemma for hypergraphs. After some preliminary steps in
this direction [C], [FR2], this program was completed by Gowers [Go3], [Go4] and by Nagle, Rődl, Schacht, and Skokan [NRS], [RSc], [RSk], [RSk2] in 2006. More recently, there have been further
developments of the hypergraph regularity method, providing some slightly different reproofs of Szemerédi's theorem [T2], [T4], [I], [ES].
Very recently, it has been realised that these four approaches to proving Szemerédi's theorem are interrelated. For example, connections between the regularity lemma approach and the Fourier-analytic
approach were uncovered in [Gr2], while connections between hypergraphs and ergodic theory were uncovered in [T4], and connections between the Fourier-analytic and ergodic approaches were uncovered
in [GT2] and exploited systematically in [GT3]. The result in [GT] uses ideas and tools from all four methods.
Szemerédi's theorem has been generalised and strengthened in many different directions. The following list is only a representative sample of some of these.
Firstly, one can "bootstrap" Szemerédi's theorem to yield not just a single progression of length \(k\ ,\) but in fact many such progressions. Indeed, an averaging argument of Varnavides [V] gives
Theorem (Varnavides' theorem) Let \(k \geq 1\) and \(0 < \delta < 1\ .\) Then there exists \(N(k,\delta) > 0\) with the following property: if \(n > N(k,\delta)\ ,\) then any subset \(A\) of \(\{1,\
ldots,n\}\) of cardinality at least \(\delta n\) will contain at least \(n^2 / N(k,\delta)\) arithmetic progressions of length \(k\ .\)
Szemerédi's theorem can also be extended to multiple dimensions, asserting that any subset of a lattice of positive Banach density will contain arbitrarily shaped "constellations":
Theorem (Furstenberg-Katznelson theorem) [FK] Let \(d \geq 1\ ,\) and let \(A \subset {\Bbb Z}^d\) be a set of positive upper Banach density, thus \(d^*(A) = \limsup_{N \to \infty} \#(A \cap [-N,N]^
d) / (2N+1)^d > 0\ .\) Then for any \(v_1,\ldots,v_k \in {\Bbb Z}^d\ ,\) there exist infinitely many \(a \in {\Bbb Z}^d\) and positive integers \(r\) such that \(a+rv_1, \ldots, a+rv_k \in A\ .\)
Note that Szemerédi's original theorem is the case when \(d=1\) and \(v_i=i-1\) for \(1 \leq i \leq k\ .\)
A significantly stronger (and more difficult) generalisation of this theorem replaces \({\Bbb Z}^d\) by an arbitrary set \(F^d\ ,\) as follows:
Theorem (Density Hales-Jewett theorem) [FK3] Let \(F \subset {\Bbb Z}\) and \(0 < \delta < 1\ .\) If \(d\) is sufficiently large depending on \(\# F\) and \(\delta\ ,\) then any subset \(A\) of \(F^d
\) of cardinality at least \(\delta (\# F)^d\) will contain at least one set of the form \(\{ a + tr: t \in F \}\) where \(a, r \in {\Bbb Z}^d\) with \(r\) non-zero.
This significantly generalises the classical Hales-Jewett theorem [HJ]. There is an equivalent formulation of this theorem in which \(F\) is an arbitrary finite set (not necessarily consisting of
integers), and the set \(\{ a + tr: t \in F \}\) is replaced by a combinatorial line, but we will not state this version here as it requires some further notation. We do however note that this
theorem implies a version of Szemerédi's theorem for finite abelian groups:
Corollary Let \(k \geq 1\) and \(0 < \delta < 1\ .\) If \(G\) is a finite group of sufficiently large order depending on \(k\) and \(\delta\ ,\) then for any subset \(A\) of \(G\) of cardinality \(\#
A \geq \delta \# G\) there exists \(a, r \in G\) with \(r\) non-zero such that \(a, a+r, \ldots, a+(k-1) r \in A\ .\)
The \(k=3\) case of this corollary was established by Meshulam [Me].
In another direction, Bergelson and Leibman [BL] established an extension of the Furstenberg-Katznelson theorem for polynomials:
Theorem (Bergelson-Leibman theorem) [BL] Let \(d \geq 1\ ,\) and let \(A \subset {\Bbb Z}^d\) be a set of positive upper Banach density, thus \(\limsup_{N \to \infty} \#(A \cap [-N,N]^d) / (2N+1)^d >
0\ .\) Then for any polynomials \(P_1,\ldots,P_k: {\Bbb Z} \to {\Bbb Z}^d\) with \(P_1(0)=\ldots=P_k(0) = 0\ ,\) there exist infinitely many \(a \in {\Bbb Z}^d\) and positive integers \(r\) such that
\(a+P_1(r),\ldots,a+P_k(r) \in A\ .\)
This theorem can be extended in a rather technical manner involving measure-preserving shift operators which generate a nilpotent group. On the other hand, these results fail once the group generated
by the shifts ceases to be nilpotent. See [L], [BL2].
Let \(k\) and \(A\) be as in Szemerédi's theorem. The set \(R_k\) of all possible \(r\) generated by that theorem has been studied by several authors (see [BHMP]). The arguments in [F] or [FKO] can
be modified to yield the assertion that \(R_k\) is syndetic, i.e. it has bounded gaps. In [FK2] it was shown that \(R_k\) is an \(IP^*\)-set, which means that given any infinite set \(B\) of positive
integers, \(R_k\) contains at least one element which can be expressed as the sum of distinct elements of \(B\ .\)
Now take \(\varepsilon > 0\ ,\) and consider the set \(R_{k,\varepsilon}\) of all \(r\) such that \(d^*( A \cap (A-r) \cap \ldots \cap (A-(k-1)r) ) > d^*(A)^k - \varepsilon\ .\) For \(k \leq 4\) it
is known that \(R_{k,\varepsilon}\) is syndetic, but this claim breaks down for \(k > 4\ ;\) see [BHK]. (The case \(k=2\) is known as the Khintchine recurrence theorem.)
In a rather different direction, Szemerédi's theorem has been shown to follow from some general theorems in graph and hypergraph theory. For instance, the \(k=3\) case of this theorem follows from
the triangle removal lemma mentioned earlier. A typical result of this type is
Theorem (Hypergraph removal lemma) Let \(G = (V,E)\) be a \(k\)-uniform hypergraph (i.e. each edge has exactly \(k\) vertices), then for any \(\delta > 0\) there exists an \(\varepsilon > 0\) if \(H
= (W,F)\) is any \(k\)-uniform hypergraph which contains at most \(\varepsilon (\# V)^{\# W}\) copies of \(G\ ,\) then it is possible to delete at most \(\delta (\# V)^k\) edges in \(H\) to create a
hypergraph with no copy of \(G\) whatsoever.
For a proof see [Go4], [RSk2], [T2], [T4], [ES], or [I]. Several refinements of this theorem, with applications to property testing of hypergraphs, are currently being pursued.
Let \(r_k(n)\) denote the size of the largest subset of \(\{1,\ldots,n\}\) which does not contain any arithmetic progressions of length \(n\ .\) The exact size of \(r_k(n)\) is still unknown;
Szemerédi's theorem asserts that \(r_k(n) =o(n)\) for each \(k\ .\) For \(k=3\ ,\) the best known lower and upper bounds for \(r_3(n)\) are \[\displaystyle n \exp(-C\sqrt{\log n}) \leq r_3(n) \leq C
n \frac{(\log \log n)^{1/2}}{\sqrt{\log n}}\] for some absolute constant \(C > 0\ ,\) due to Behrend [Be] and Bourgain [Bo] respectively. (In a recent unpublished paper, Bourgain has improved the
upper bound to \(C n \frac{(\log \log n)^2}{(\log n)^{2/3}}\ .\)) For general \(k\ ,\) the best known bounds are \[\displaystyle n \exp(-C\log^{1/(\lfloor \log_2(k-1) \rfloor+1)} n) \leq r_k(n) \leq
n / (\log_2 \log_2 n)^{1/2^{k+9}}\] due to Rankin [Ran] and Gowers [Go2] respectively, though for \(k=4\) the upper bound has been improved to \(n \exp( -C \sqrt{\log \log n} )\) by Green and Tao
Szemerédi's theorem asserts that dense subsets of the integers contain long arithmetic progressions. It turns out that one can replace the integers with certain other sets, most notably the set of
primes, using a transference principle: see [Gr], [GT].
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Internal references
Further reading
Detailed discussion of the various approaches to Szemerédi's theorem can be found in [TV], [K], [KS], [T5], [Gr3]. | {"url":"http://www.scholarpedia.org/article/Szemer%C3%A9di's_Theorem","timestamp":"2024-11-05T02:27:07Z","content_type":"text/html","content_length":"62559","record_id":"<urn:uuid:64b54909-c8ce-4093-bece-e6b1430d8538>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00575.warc.gz"} |
Lesson 15
Weighted Averages
• Let’s split segments using averages and ratios.
15.1: Part Way: Points
For the questions in this activity, use the coordinate grid if it is helpful to you.
1. What is the midpoint of the segment connecting \((1,2)\) and \((5,2)\)?
2. What is the midpoint of the segment connecting \((5,2)\) and \((5,10)\)?
3. What is the midpoint of the segment connecting \((1,2)\) and \((5,10)\)?
15.2: Part Way: Segment
Point \(A\) has coordinates \((2,4)\). Point \(B\) has coordinates \((8,1)\).
1. Find the point that partitions segment \(AB\) in a \(2:1\) ratio.
2. Calculate \(C=\frac 13 A + \frac 23 B\).
3. What do you notice about your answers to the first 2 questions?
4. For 2 new points \(K\) and \(L\), write an expression for the point that partitions segment \(KL\) in a \(3:1\) ratio.
Consider the general quadrilateral \(QRST\) with \(Q=(0,0),R=(a,b),S=(c,d),\) and \(T=(e,f)\).
1. Find the midpoints of each side of this quadrilateral.
2. Show that if these midpoints are connected consecutively, the new quadrilateral formed is a parallelogram.
15.3: Part Way: Quadrilateral
Here is quadrilateral \(ABCD\).
1. Find the point that partitions segment \(AB\) in a \(1:4\) ratio. Label it \(B’\).
2. Find the point that partitions segment \(AD\) in a \(1:4\) ratio. Label it \(D’\).
3. Find the point that partitions segment \(AC\) in a \(1:4\) ratio. Label it \(C’\).
4. Is \(AB’C’D’\) a dilation of \(ABCD\)? Justify your answer.
To find the midpoint of a line segment, we can average the coordinates of the endpoints. For example, to find the midpoint of the segment from \(A=(0,4)\) to \(B=(6,7)\), average the coordinates of \
(A\) and \(B\): \(\left(\frac{0 + 6}{2}, \frac{4+7}{2}\right) = (3,5.5)\). Another way to write what we just did is \(\frac12 (A+B)\) or \(\frac12 A + \frac12 B\).
Now, let’s find the point that is \(\frac23\) of the way from \(A\) to \(B\). In other words, we’ll find point \(C\) so that segments \(AC\) and \(CB\) are in a \(2:1\) ratio.
In the horizontal direction, segment \(AB\) stretches from \(x=0\) to \(x=6\). The distance from 0 to 6 is 6 units, so we calculate \(\frac23\) of 6 to get 4. Point \(C\) will be 4 horizontal units
away from \(A\), which means an \(x\)-coordinate of 4.
In the vertical direction, segment \(AB\) stretches from \(y=4\) to \(y=7\). The distance from 4 to 7 is 3 units, so we can calculate \(\frac23\) of 3 to get 2. Point \(C\) must be 2 vertical units
away from \(A\), which means a \(y\)-coordinate of 6.
It is possible to do this all at once by saying \(C = \frac13 A + \frac23 B\). This is called a weighted average. Instead of finding the point in the middle, we want to find a point closer to \(B\)
than to \(A\). So we give point \(B\) more weight—it has a coefficient of \(\frac23\) rather than \(\frac12\) as in the midpoint calculation. To calculate \(C = \frac13 A + \frac23 B\), substitute
and evaluate.
\(\frac13 A + \frac23 B\)
\(\frac13 (0,4) + \frac23 (6,7)\)
\(\left(0,\frac43 \right) + \left(4, \frac{14}{3} \right)\)
Either way, we found that the coordinates of \(C\) are \((4,6)\).
• opposite
Two numbers are opposites of each other if they are the same distance from 0 on the number line, but on opposite sides.
The opposite of 3 is -3 and the opposite of -5 is 5.
• point-slope form
The form of an equation for a line with slope \(m\) through the point \((h,k)\). Point-slope form is usually written as \(y-k = m(x-h)\). It can also be written as \(y = k + m(x-h)\).
• reciprocal
If \(p\) is a rational number that is not zero, then the reciprocal of \(p\) is the number \(\frac{1}{p}\). | {"url":"https://curriculum.illustrativemathematics.org/HS/students/2/6/15/index.html","timestamp":"2024-11-02T21:20:45Z","content_type":"text/html","content_length":"93338","record_id":"<urn:uuid:141cee24-f0b2-4df1-9951-e51e1f0f7e19>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00483.warc.gz"} |
Binomial Distribution: The Ultimate Chill Guide with No
The Ultimate Chill Guide to Binomial Distribution: No Math Phobia, Just Vibes 2024
Hey there, welcome to Statssy! In this article, we will discuss the buzz and hype of Binomial Distribution. Let’s see if it gives you positive or negative vibes.
What’s the Hype About Binomial Distribution?
Hey there, cool cats and kittens! Ever heard of the binomial distribution and thought, “Ugh, sounds like a snooze fest?” Well, hold on to your TikTok, because we’re about to spill some tea that’ll
make you go, “Whoa, that’s dope!”
So, What’s Binomial Distribution Anyway?
First off, let’s break it down. The binomial distribution is like the ultimate game of chance. Imagine you’re flipping a coin, rolling a dice, or even swiping right or left on Tinder. It’s all about
predicting the outcome—will it be a heads or tails, a match or a pass? Binomial distribution helps you figure that out, but in a way that’s backed by some solid math.
Why Should You Even Care?
Okay, okay, I get it. You’re thinking, “Why does this even matter to me?” Well, you know how you love predicting the next viral trend or guessing who’s gonna win ‘Among Us’? Binomial distribution is
like the secret sauce that can make your predictions way more accurate.
Applications You’ll Totally Vibe With
Now, let’s talk real-world stuff. This isn’t just for your math class; it’s everywhere! From figuring out how many people will actually show up to your Zoom party to predicting the next big hashtag
on Twitter , binomial distribution has got you covered.
• Finance: Wanna invest in some crypto but are scared of the risks? The binomial distribution can help you play it smart.
• Social Media: Trying to become the next big influencer? Use binomial distribution to predict your follower growth.
• Gaming: Are you a ‘Fortnite’ fan? Calculate your win probabilities like a pro.
What’s It All About? Why Should You Care? Real-World Vibes Stats & Biostats Corner
Medicine: Will this new med give you Wanna know if that new acne medicine will work Medicine: Predicting the number of patients experiencing side 20% of patients reported side effects in
side effects? without side effects? effects from new meds clinical trials
Banking: Are these transactions legit? Ever wondered how banks catch those sneaky Banking: Estimating the number of fraudulent transactions 5% increase in fraud cases last month
Email: How much spam will hit your Tired of spam emails after online shopping? Email: Predicting the number of spam emails you’ll get Average person receives 10 spam emails per
inbox? day
Gaming: Will you roll that lucky six? Want to be the Yahtzee champ in your friend group? Gaming: Chances of rolling a six in a dice game 16.67% chance of rolling a six
Marketing: Will people click that ‘Buy Want to know if your startup’s ad campaign will be a Marketing: Customer conversion rates 30% conversion rate in last quarter’s ad
Now’ button? hit? campaign
Pharmaceuticals: Will this new drug Curious about how effective new COVID vaccines are? Pharmaceuticals: Efficacy of a new drug 80% efficacy in recent drug trials
actually work?
Politics: Who’s winning the next Want to predict the next U.S. President? Politics: Predicting vote counts for a candidate 60% chance of winning based on current
election? polls
Binomial Distribution
What Even is Binomial Distribution?
Definition and Core Concepts
Hey there, future probability wizards! Ever found yourself wondering what the odds are of something happening? Like, what are the chances of you acing that math test or your TikTok going viral? Well,
that’s where Binomial Distribution comes into play!
In the simplest terms, Binomial Distribution is like your personal fortune teller for life’s yes-or-no questions. It’s all about predicting two possible outcomes: either something happens (that’s a
win! ) or it doesn’t (bummer ).
Imagine you’re flipping a coin. It can either land heads up or tails up, right? Binomial Distribution helps you figure out the probability of getting a certain number of heads in a specific number of
coin flips. So, if you’re flipping that coin 10 times, how many times will it land heads up? Binomial Distribution has got your back!
Properties of Binomial Distribution
Okay, so you’re probably thinking, “Cool, but what makes Binomial Distribution tick?” Great question! There are some key properties you gotta know:
1. Fixed Number of Trials: You know exactly how many times you’re gonna try something, like flipping that coin 10 times.
2. Two Possible Outcomes: It’s all about success (yay! ) or failure (aww ).
3. Constant Probability: The chance of success stays the same each time, like a 50-50 shot in a coin flip.
4. Independent Trials: What happens in one trial doesn’t affect the others. So, each coin flip is its own mini-adventure!
Chart: Pie Chart of Success vs Failure Rates
To make this even clearer, let’s visualize it! Picture a pie chart where one slice represents the rate of success and the other slice represents the rate of failure. If you’re flipping a fair coin,
each slice would be 50% of the pie. But if you’re, say, shooting hoops and you make 70% of your shots, the ‘success’ slice would take up 70% of the pie, and the ‘failure’ slice would be the remaining
The Rules of the Game: Conditions for Binomial Distribution
Conditions and Successive Trials
So, you’re vibing with the whole Binomial Distribution thing, huh? But wait, there’s a catch! Not every situation is a good fit for Binomial Distribution. There are some ground rules, or conditions,
you gotta meet.
What are the conditions for binomial distribution?
1. Fixed Number of Trials: You gotta know how many times you’re gonna try something. Like, if you’re flipping a coin, you need to decide in advance how many flips you’re gonna do.
2. Two Possible Outcomes: It’s either a win or a loss, baby!
3. Constant Probability: The odds gotta stay the same each time. So, if you have a 50% chance of winning a game, those odds can’t change mid-way.
4. Independent Trials: What happens in Vegas, stays in Vegas! Each trial should not affect the other trials.
In binomial distribution, successive trials are…
Successive trials are like episodes of your favorite Netflix series —each one is its own thing but part of a bigger story. In Binomial Distribution, each trial is independent, meaning what happened
in the last trial won’t spill over into the next one. So, if you failed the last time, don’t sweat it; you’ve got a clean slate each time!
Let me give you a flowchart that helps you figure out if your situation is right for Binomial Distribution
Bernoulli Trials: The OG of Binomial Distribution
What are Bernoulli Trials?
Alright, let’s spill the tea on Bernoulli Trials! Imagine you’re playing a super basic game on your phone. You tap the screen, and you either score a point or you don’t. That’s it. No power-ups, no
extra lives, just a simple win or lose scenario.
This is what Bernoulli Trials are all about. They’re like the most basic level of a video game, where you have only two outcomes: success or failure. And guess what? These simple Bernoulli Trials are
the building blocks of something way cooler—the Binomial Distribution!
So, in simpler terms, if Binomial Distribution was a blockbuster movie, Bernoulli Trials would be the trailer. It gives you a sneak peek into what the whole thing is about but in a much simpler
Real-World Connections
Okay, let’s bring this down to Earth with some examples you’ll totally relate to.
Instagram Polls
You know those Instagram polls where you have to choose between “Hot” or “Not”? Each time you vote, you’re actually participating in a Bernoulli Trial! The poll itself is like a mini-experiment with
only two possible outcomes: either the majority agrees with you, or they don’t. And the best part? You get instant gratification by seeing the results in real-time!
TikTok Challenges
Ever tried to do one of those viral TikTok challenges? Whether it’s the “Renegade” dance or the “Bottle Cap Challenge,” each attempt you make is a Bernoulli Trial. You either nail it and get tons of
likes and shares, or you mess up and become a meme. Either way, it’s a win-win, right?
The Secret Sauce: Binomial Distribution Formula
The Formula Unveiled
Okay, so you know when you’re cooking and you find that one secret ingredient that just makes the dish pop? Well, in the world of Binomial Distribution, the formula is that secret ingredient. It’s
like the cheat code that helps you unlock all the levels in a video game.
So, what’s the formula? Don’t worry, we’re not going to throw a bunch of scary math symbols at you. Instead, think of it like a recipe. You’ve got your number of trials (let’s call it ‘n’), your
probability of success (that’s ‘p’), and you’re trying to find out how many times you’ll actually succeed (that’s ‘x’).
Examples of Binomial Distribution
Example Scenario How Binomial Distribution Fits In Why It’s Cool
Among Us: Finding the You’re playing Among Us and you’re trying to find out Use the Binomial Distribution formula to calculate the odds of guessing the It’s like having a cheat sheet to win
Impostor who the impostor is among 10 players. impostor correctly on your first, second, or third try. the game!
Viral TikTok Dance You’re trying to nail the “Renegade” dance challenge in The formula can help you figure out the probability of nailing the dance in You can impress your friends by
Challenges: Renegade one go. one, two, or three tries. predicting your TikTok success!
Twitter Polls: Pineapple A Twitter poll asks, “Is pineapple an acceptable pizza Use the formula to predict the final outcome based on early voting trends. You can be the Twitter oracle and
on Pizza? topping?” predict poll outcomes!
Instagram Filters: Which You’re testing out new Instagram filters and wondering Use the formula to calculate the odds of each filter making your post go Be the trendsetter and know which
One Will Go Viral? which one will get you the most likes. viral. filter is the golden ticket!
Fortnite: Victory Royale You’re in a Fortnite match and you want to be the last Use Binomial Distribution to calculate your odds of winning based on your past Know your odds and strategize like a
one standing. performance. pro gamer!
Netflix Binge: To Continue You’re watching a new series and can’t decide if you Use the formula to determine the odds of the next episodes being as good as Make the ultimate binge-watch
or Not? should binge-watch it. the first. decision like a Netflix guru!
The Secret Sauce: Binomial Distribution Formula
What’s the Vibe? Characteristics of Binomial Distribution
Mean, Variance, and Standard Deviation
The Mean: The Average Outcome
The mean (μ) in a Binomial Distribution represents the average number of successes you can expect in a given number of trials. For example, if you toss a coin 10 times, the mean tells you that you
can expect to get heads about 5 times. It’s calculated as μ = n × p, where n is the number of trials and p is the probability of success.
The Variance: How Spread Out Are the Outcomes?
The variance (σ²) measures the dispersion or how much the outcomes can vary from the mean. In real life, if you’re playing a game of darts, a low variance would mean most of your darts land close to
the bullseye (the mean), while a high variance would mean the darts are spread out over the dartboard. The formula for variance in Binomial Distribution is σ² = n × p × (1-p).
Standard Deviation: The “Average” Difference
The standard deviation (σ) is essentially the “average” distance from the mean. It’s useful for understanding the range within which most outcomes will fall. For instance, in a class test, if the
mean score is 70 and the standard deviation is 10, you can expect most students to score between 60 and 80. Standard Deviation is the square root of the variance: σ = √(n × p × (1-p)).
Examples, Interpretations, and Real-World Impact
Mean Variance Standard
Example (μ) (σ²) Deviation Real-World Interpretation Real-World Impact
Tossing a coin 10 5 2.5 1.58 On average, you’ll get 5 heads. The outcomes will typically If you’re betting on heads, you can expect to win about half the time, but don’t be
times, p = 0.5 vary by about 1.58 heads from the mean. surprised if you win 3 or 7 times out of 10.
Rolling a die 6 times, 1 0.83 0.91 On average, you’ll roll one six. The outcomes will typically In a board game, don’t count on rolling a six every time. You might get zero or even
p = 1/6 vary by about 0.91 from the mean. two sixes in 6 rolls.
20 free throws in 15 3.75 1.94 On average, you’ll make 15 shots. The outcomes will typically If you’re a basketball player, you can expect to make around 15 shots, but a bad day
basketball, p = 0.75 vary by about 1.94 shots from the mean. could see you making only 13, and a good day could see you making 17.
8 questions in a quiz, 4.8 1.92 1.39 On average, you’ll get 4.8 questions right. The outcomes will If you’re taking a quiz, you can expect to get around 5 questions right, but you
p = 0.6 typically vary by about 1.39 questions from the mean. might score as low as 3 or as high as 6 depending on the day.
12 spins on a roulette 5.68 1.79 1.34 On average, you’ll win 5.68 spins. The outcomes will typically If you’re gambling, you can expect to win around 6 times, but luck could see you
wheel, p = 18/38 vary by about 1.34 spins from the mean. winning only 4 or as many as 7 spins.
What’s the Vibe? Characteristics of Binomial Distribution
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A Facet of Factoring
Someone on reddit asked how to show that $2^{15}-1$ was not a prime number, and I suddenly understood something I’d previously just ‘become used to’.
In binary, $2^{15}-1$ is 111,111,111,111,111. You can break that down in groups of three (or five) – but it’s fairly obviously $111 \times 1,001,001,001,001$.
Previously, I could certainly have proved that $a^{bc} - 1$ was a multiple of $a^b - 1$, probably using a geometric series formula, but I don’t think I had a visual feel that it was true – even if
the geometric series is formally just the same as splitting the number up.
A selection of other posts
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Lesson 12
Applications of Arithmetic with Powers of 10
Let’s use powers of 10 to help us make calculations with large and small numbers.
12.1: What Information Do You Need?
What information would you need to answer these questions?
1. How many meter sticks does it take to equal the mass of the Moon?
2. If all of these meter sticks were lined up end to end, would they reach the Moon?
12.2: Meter Sticks to the Moon
1. How many meter sticks does it take to equal the mass of the Moon? Explain or show your reasoning.
2. Label the number line and plot your answer for the number of meter sticks.
3. If you took all the meter sticks from the last question and lined them up end to end, will they reach the Moon? Will they reach beyond the Moon? If yes, how many times farther will they reach?
Explain your reasoning.
4. One light year is approximately \(10^{16}\) meters. How many light years away would the meter sticks reach? Label the number line and plot your answer.
Here is a problem that will take multiple steps to solve. You may not know all the facts you need to solve the problem. That is okay. Take a guess at reasonable answers to anything you don’t know.
Your final answer will be an estimate.
If everyone alive on Earth right now stood very close together, how much area would they take up?
12.3: The “Science” of Scientific Notation
The table shows the speed of light or electricity through different materials.
│ material │speed (meters per second) │
│space │300,000,000 │
│water │\(2.25 \times 10^8\) │
│copper (electricity) │280,000,000 │
│diamond │\(124 \times 10^6\) │
│ice │\(2.3 \times 10^8\) │
│olive oil │\(0.2 \times 10^9\) │
Circle the speeds that are written in scientific notation. Write the others using scientific notation.
12.4: Scientific Notation Matching
Your teacher will give you and your partner a set of cards. Some of the cards show numbers in scientific notation, and other cards show numbers that are not in scientific notation.
1. Shuffle the cards and lay them facedown.
2. Players take turns trying to match cards with the same value.
3. On your turn, choose two cards to turn faceup for everyone to see. Then:
1. If the two cards have the same value and one of them is written in scientific notation, whoever says “Science!” first gets to keep the cards, and it becomes that player’s turn. If it’s
already your turn when you call “Science!”, that means you get to go again. If you say “Science!” when the cards do not match or one is not in scientific notation, then your opponent gets a
2. If both partners agree the two cards have the same value, then remove them from the board and keep them. You get a point for each card you keep.
3. If the two cards do not have the same value, then set them facedown in the same position and end your turn.
4. If it is not your turn:
1. If the two cards have the same value and one of them is written in scientific notation, then whoever says “Science!” first gets to keep the cards, and it becomes that player’s turn. If you
call “Science!” when the cards do not match or one is not in scientific notation, then your opponent gets a point.
2. Make sure both of you agree the cards have the same value.
If you disagree, work to reach an agreement.
5. Whoever has the most points at the end wins.
1. What is \(9 \times 10^{\text-1} + 9 \times 10^{\text-2}\)? Express your answer as:
1. A decimal
2. A fraction
2. What is \(9 \times 10^{\text-1} + 9 \times 10^{\text-2} + 9 \times 10^{\text-3} +9 \times 10^{\text-4}\)? Express your answer as:
1. A decimal
2. A fraction
3. The answers to the two previous questions should have been close to 1. What power of 10 would you have to go up to if you wanted your answer to be so close to 1 that it was only \(\frac{1}
{1,000,000}\) off?
4. What power of 10 would you have to go up to if you wanted your answer to be so close to 1 that it was only \(\frac{1}{1,000,000,000}\) off? Can you keep adding numbers in this pattern to get as
close to 1 as you want? Explain or show your reasoning.
5. Imagine a number line that goes from your current position (labeled 0) to the door of the room you are in (labeled 1). In order to get to the door, you will have to pass the points 0.9, 0.99,
0.999, etc. The Greek philosopher Zeno argued that you will never be able to go through the door, because you will first have to pass through an infinite number of points. What do you think? How
would you reply to Zeno?
The total value of all the quarters made in 2014 is 400 million dollars. There are many ways to express this using powers of 10. We could write this as \(400 \boldcdot 10^6\) dollars, \(40 \boldcdot
10^7\) dollars, \(0.4 \boldcdot 10^9\) dollars, or many other ways. One special way to write this quantity is called scientific notation. In scientific notation,
400 million
dollars would be written as \(\displaystyle 4 \times 10^8\) dollars. For scientific notation, the \(\times\) symbol is the standard way to show multiplication instead of the \(\boldcdot \) symbol.
Writing the number this way shows exactly where it lies between two consecutive powers of 10. The \(10^8\) shows us the number is between \(10^8\) and \(10^9\). The 4 shows us that the number is 4
tenths of the way to \(10^9\).
Some other examples of scientific notation are \(1.2 \times 10^{\text-8}\), \(9.99 \times 10^{16}\), and \(7 \times 10^{12}\). The first factor is a number greater than or equal to 1, but less than
10. The second factor is an integer power of 10.
Thinking back to how we plotted these large (or small) numbers on a number line, scientific notation tells us which powers of 10 to place on the left and right of the number line. For example, if we
want to plot \(3.4 \times 10^{11}\) on a number line, we know that the number is larger than \(10^{11}\), but smaller than \(10^{12}\). We can find this number by zooming in on the number line:
• scientific notation
Scientific notation is a way to write very large or very small numbers. We write these numbers by multiplying a number between 1 and 10 by a power of 10.
For example, the number 425,000,000 in scientific notation is \(4.25 \times 10^8\). The number 0.0000000000783 in scientific notation is \(7.83 \times 10^{\text-11}\). | {"url":"https://im-beta.kendallhunt.com/MS_ACC/students/2/7/12/index.html","timestamp":"2024-11-07T22:04:43Z","content_type":"text/html","content_length":"83223","record_id":"<urn:uuid:aa6f4809-a006-4fdf-b0f4-a1852bce0c71>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00763.warc.gz"} |
tf.math.reduce_logsumexp | TensorFlow v2.15.0.post1
Computes log(sum(exp(elements across dimensions of a tensor))).
View aliases
Main aliases
input_tensor, axis=None, keepdims=False, name=None
Reduces input_tensor along the dimensions given in axis. Unless keepdims is true, the rank of the tensor is reduced by 1 for each of the entries in axis, which must be unique. If keepdims is true,
the reduced dimensions are retained with length 1.
If axis has no entries, all dimensions are reduced, and a tensor with a single element is returned.
This function is more numerically stable than log(sum(exp(input))). It avoids overflows caused by taking the exp of large inputs and underflows caused by taking the log of small inputs.
For example:
x = tf.constant([[0., 0., 0.], [0., 0., 0.]])
tf.reduce_logsumexp(x) # log(6)
tf.reduce_logsumexp(x, 0) # [log(2), log(2), log(2)]
tf.reduce_logsumexp(x, 1) # [log(3), log(3)]
tf.reduce_logsumexp(x, 1, keepdims=True) # [[log(3)], [log(3)]]
tf.reduce_logsumexp(x, [0, 1]) # log(6)
input_tensor The tensor to reduce. Should have numeric type.
axis The dimensions to reduce. If None (the default), reduces all dimensions. Must be in the range [-rank(input_tensor), rank(input_tensor)).
keepdims If true, retains reduced dimensions with length 1.
name A name for the operation (optional).
The reduced tensor. | {"url":"https://tensorflow.google.cn/versions/r2.15/api_docs/python/tf/math/reduce_logsumexp","timestamp":"2024-11-02T11:10:55Z","content_type":"text/html","content_length":"39183","record_id":"<urn:uuid:003d8f56-1e36-4f5f-8939-bb4c47c15ba4>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00082.warc.gz"} |
Change the Max Size for Status Column
One solution (simple) would be with drawing some emoji instead of using a progress bar:
(note that formulas are incorrect there, Sequence(1, x) goes backwards when x < 1 and draws two items if x = 0.
Another solution would be with rectangles. It’s more complicated to implement, but more flexible (e.g. you can color-code your band members and see team staffing at a glance):
To think of it, you can use different emoji too. All solutions would require a formula of a sort, so just see what’s easier for you / good enough. | {"url":"https://community.coda.io/t/change-the-max-size-for-status-column/13785/2","timestamp":"2024-11-08T17:44:16Z","content_type":"text/html","content_length":"27944","record_id":"<urn:uuid:49d01ad1-24ff-47af-850a-6ea203915b52>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00585.warc.gz"} |
The Weekly Challenge - Perl & Raku
( …continues from previous week. )
Welcome to the Perl review pages for Week 151 of The Weekly Challenge! Here we will take the time to discuss the submissions offered up by the team, factor out some common methodologies that came up
in those solutions, and highlight some of the unique approaches and unusual code created.
●︎ Why do we do these challenges?
I suppose any reasonable answer to that question would be from a field as wide ranging and varied as the people who choose to join the team. One thing, though, is clear: it’s not a competition, and
there are no judges, even if there is a “prize” of sorts. About that – I think of it more as an honorarium periodically awarded to acknowledge the efforts we make towards this strange goal. So
there’s no determination to find the fastest, or the shortest, or even, in some abstract way, the best way to go about things, although I’m certain the participants have their own aspirations and
personal drives. As Perl is such a wonderfully expressive language, this provides quite a bit of fodder to the core idea of TMTOWTDI, producing a gamut of varied techniques and solutions.
Even the tasks themselves are often open to a certain amount of discretionary interpretation. What we end up with is a situation where each participant is producing something in the manner they find
the most interesting or satisfying. Some team members focus on carefully crafted complete applications, thoroughly vetting input data and handling every use case they can think up. Others choose to
apply themselves to the logic of the underlying puzzle and making it work in the most elegant way they can. Some eschew modules they would ordinarily reach for, others embrace them, bringing to light
wheels perhaps invented years ago that happen to exactly solve the problem in front of them today.
I’ve been considering this question for some time and have found one binding commonality between all of us out solving these challenges each week, in that however we normally live our lives, the task
in front of us more than likely has nothing to do with any of that. And I think this has great value. We all do what we do, in the real world, and hopefully we do it well. The Weekly Challenge
provides us with an opportunity to do something germane to that life yet distinctly different; if we only do the things we already know how to do then we will only do the same things over and over.
This is where the “challenge” aspect comes into play.
So we can consider The Weekly Challenge as providing a problem space outside of our comfort zone, as far out from that comfort as we wish to take things. From those reaches we can gather and learn
things, pick and choose and bring what we want back into our lives. Personally, I think that’s what this whole thing is about. YMMV.
Every week there is an enormous global collective effort made by the team, analyzing and creatively coding the submissions, and that effort deserves credit due.
And that, my friends, is why I’m here, to try and figure out ways to do just that.
So, here we are then. I’m ready — let’s get to it and see what we can find.
For Additional Context…
before we begin, you may wish to revisit either the pages for the original tasks or the summary recap of the challenge. But don’t worry about it, the challenge text will be repeated and presented as
we progress from task to task.
Oh, and one more thing before we get started:
Getting in Touch with Us
› Please feel free to email me (Colin) with any feedback, notes, clarifications or whatnot about this review.
› Submit a pull request to us for any issues you may find with this page.
› Join the discussion on Twitter!
I’m always curious as to what the people think of these efforts. Everyone here at the PWC would like to hear any feedback you’d like to give.
...So finally, without further ado...
• Task 1 • Task 2 • BLOGS •
TASK 1
Binary Tree Depth
Submitted by: Mohammad S Anwar
You are given binary tree.
Write a script to find the minimum depth.
The minimum depth is the number of nodes from the root to the nearest leaf node (node without any children).
Example 1:
Input: '1 | 2 3 | 4 5'
/ \
/ \
Output: 2
Example 2:
Input: '1 | 2 3 | 4 * * 5 | * 6'
/ \
/ \
Output: 3
about the solutions
Abigail, Alexander Pankoff, Athanasius, Colin Crain, Dave Jacoby, Duncan C. White, E. Choroba, Flavio Poletti, James Smith, Jorg Sommrey, Laurent Rosenfeld, Lubos Kolouch, Matthew Neleigh, Mohammad S
Anwar, Peter Campbell Smith, Roger Bell_West, Simon Green, Ulrich Rieke, and W. Luis Mochan
In weeks past we’ve explored a great deal around the topic of binary trees. We’ve looked at maximum diameters, depths, and accessed various nodes, and processed them both breadth- and depth-first. At
this point many long-term members of the team have developed complex libraries of tree objects and methods to draw on. And with the first task this week we return to this familiar territory with a
seemingly innocent, not-too-difficult task, looking for the closest leaf node to the root.
Or, to put a darker, Grimms Fairy Tales spin on it, a pair of missing children.
One key difference to this particular task, generally left ambiguous in previous challenges, is that the input is specified. Or rather, that the input format is specified — in the examples we are
given a particular stringified serial encoding.
The rules for the encoding are not specified, which could be regarded as part of the puzzle for anyone who did not immediately recognise it. The tree data is recorded as a breadth-first traversal,
with individual levels delineated by vertical pipes. Within each level, a symbol for a null node is required to fix placements unambiguously, and this format chooses an asterisk. Other than that,
items and delimiters are separated by whitespace, apparently with multiple spaces allowed for clarity. We don’t have enough examples to determine whether there are specific rules for multiple spaces,
but it doesn’t seem to affect our parsing in any way anyways, so I’d say it doesn’t really matter.
Null nodes at the end of the tree can be inferred and are hence optional. What’s not clear, however is whether the same rule would apply to interior levels, although with the vertical pipe level
delimiters these too could be inferred. We don’t have any examples of this, however.
There were an unusually large number of improperly working submissions this week, which was, not to put too fine a point on it, a little weird.
COUNTING the RINGS
Athanasius, Alexander Pankoff, Mohammad S Anwar, Jorg Sommrey, Dave Jacoby, Colin Crain, Roger Bell_West, Abigail, and Flavio Poletti
Here’s the start of a theory:
There turned out to be quite a bit of ambiguity remaining in the task definition, specifically as to what, exactly, a leaf node is in the context of a binary tree. A binary tree is nominally a tree
structure where a node has at most two children, and a leaf as a node without children. However one Set Theory definition of a binary tree is a recursive structure of nodes as tuples, with each tuple
containing a value and two child tuples, which may themselves be a null set. The precise meaning of the idea of a null in this definition varies, however.
The question arises between whether a node is null, or whether the value of a node is null. Ultimately this leads to the question of whether a null node can logically be itself a leaf node.
To which I say: “You have to be f’ing kidding me”. However if you define a binary tree as a fixed structure with each node containing exactly two children this follows. I am, shall we say, highly
disinclined to agree with this interpretation, but it seems implicit to many of the results we saw. But if a null child node is still a node, then by definition all nodes will have children, and the
leaf nodes will be simply the furthest extant of whatever full structure we’ve defined, or become meaningless altogether.
Or, you know, a group of otherwise very capable people has screwed up en masse.
To explain my point of view, consider the following tree:
Input: '1 | 2 3 | 4 * * 5 | * 6 * * * * 7 8'
┃ ┃
┏━━┫2┃ ┃3┣━━┓
┃ ┃
┃4┣┓ ┏┫5┣┓
┃ ┃ ┃
┃6┃ ┃7┃ ┃8┃
Many solutions return the answer 3, instead of 4. It does not appear to be an off-by-one error. I eventually had to run this test data on every single submission to see what was happening. I took
this to be my shibboleth. You have draw the line somewhere.
additional languages: Raku
We’ll let the monk start things off today. Drawing a direct inference from the input format, they first slice the string into an array of arrays, one level each. In this form each level will have 2^
level nodes, 0-indexed. Traversing each level in turn, the childen for a node at a given index will always be located on the following level at postions 2 × index and 2 × index + 1. Thus we can
easily look up the children and check for their existence. At the first double-miss, we have found a leaf node and we note the level we’re on.
for my $level (0 .. $#$tree)
for my $index (0 .. $#{ $tree->[ $level ] })
my $node = $tree->[ $level ][ $index ];
if (defined $node)
if ($level == $#$tree ||
(!defined $tree->[ $level + 1 ][ 2 * $index ] &&
!defined $tree->[ $level + 1 ][ 2 * $index + 1 ]))
printf qq[Output: %d\n], $level + 1;
print qq[\nThe first leaf node is "$node"\n] if $VERBOSE;
last L_OUTER;
At the other end of the complexity spectrum, Alexander chooses multiple layers of abstraction to structure his parsed input. A tokenizer is defined, with a TokenType class and three subclasses for
separators, values and placeholders. Parsed and labeled, the list of tokens is then handed to the main minimum_binary_tree_depth() routine.
The tokens are systematically processed, ratcheting a level counter as they go, counting by powers of 2 and filling in an array of arrays with the tree data. This tree is then used to find the first
As I said, lots of abstraction. Kind of like building a hovercraft because you needed to go to the corner store.
And hovercrafts, as everyone knows, are very cool.
while (@tokens) {
push @$tree, [];
my $num_elems = 2**$depth;
for ( my $i = 0 ; $i < $num_elems ; $i++ ) {
if ( !@tokens || $tokens[0]->isa('SeparatorToken') ) {
## fill row with dummy placeholder tokens.
unshift @tokens,
map { PlaceHolderToken->new(-1) }
0 .. ( $num_elems - 1 - $i ); # Dummy Token
my $cur = shift @tokens;
if ( $cur->isa('ValueToken') ) {
if ( $depth && !defined( $tree->[-2][ int( $i / 2 ) ] ) ) {
die join( " ",
"Missing parent for node with value",
"at position",
"in input\n" );
push @{ $tree->[-1] }, $cur->{lexeme};
elsif ( $cur->isa('PlaceHolderToken') ) {
if ( $i % 2
&& !defined $tree->[-1][-1]
&& ( !$depth || defined $tree->[-2][ int( $i / 2 ) ] ) )
return $depth;
push @{ $tree->[-1] }, undef;
## do nothing
$depth += 1;
# handle optional separatortoken
if ( @tokens && $tokens[0]->isa("SeparatorToken") ) {
shift @tokens;
return $depth;
Mohammad has curiously chosen to buck his own input format suggestion, defining individual trees as nested Node objects, hashes with 3 keys for left and right children and a value. A recursive
routine traverses the structure depth-first, returning the smallest result as the recursion collapses at whatever maximum depth is found.
sub min_depth {
my ($node) = @_;
return 0 unless defined $node;
my $min_left = min_depth($node->{left});
my $min_right = min_depth($node->{right});
return $min_right + 1 unless defined $node->{left};
return $min_left + 1 unless defined $node->{right};
return ($min_left > $min_right)
($min_right + 1)
($min_left + 1);
Jorg imports the Graph module to supply a framework for his tree. After all, a tree is a directed graph, linked from top to bottom. The thing about thinking of the tree as a graph is it allows the
use of graph theory techniques to find our minimum depth. An intermediate structure is created of shortest paths from the root to each node, and from this the minimum value is taken for all of these
paths that travel to a “sink vertex”, that is to say a vertex that does not connect forward to any other vertex.
A careful reading will show we’re talking about leaf nodes, just using slightly different language.
Here’s the core logic, after we’ve constructed our graph:
# Find the minimum depth in a tree-like graph from its root.
sub min_depth ($g) {
# Use zero as the depth of an empty tree.
return 0 unless $g->has_vertices;
# Find the (unique) root vertex.
my $root = ($g->source_vertices)[0];
# Use one as the depth of a root-only tree. (An isolated vertex
# does not count as a source vertex.)
return 1 unless defined $root;
# Create the tree of Single-Source Shortest Paths originating at the
# root vertex.
my $sptg = $g->SPT_Dijkstra($root);
# Find the shortest path from the root to all leafs (i.e. sink
# vertices) and take the minimum thereof. As the depth is defined
# here as the number of vertices in the path instead of the number
# of edges, we need to add one for the desired result.
1 + min map $sptg->get_vertex_attribute($_, 'weight'),
blog writeup: Dr. Metropolis and His Amazing MANIAC Machine!: The Weekly Challenge #151 | Committed to Memory
As way of a preface, Dave in his blog writeup introduces us to one Dr. Nicholas Metropolis, mathematician and inventor of the Monte Carlo method. He sounds like a fascinating character, although
bearing only a passing resemblance to Rotwang, the protagonist on the 1927 Fritz Lang film bearing the same name.
Incidentally, the robot Maria in Metropolis was referred to by Rotwang as a Maschinenmensch, which obviously inspired Die Mensch-Maschine, the 1978 classic recording by the Kraut-rock synthesizer
band Kraftwerk.
The world, as you may have noticed, is a very interconnected place.
By way of his solution, Dave parses the input to construct Node objects into a proper tree structure. The nodes themselves contain an upwards parent link, allowing for a node_depth() method that can
traverse upwards, counting until it finds the root. A similar method peeks at the children to see whether it is given node is a leaf.
The nodes are kept in a hash, and the keys to the hash are filtered to find leaf nodes, blocking the root should itself be a leaf. Then each of these are mapped to their depth and the depths sorted
to find the minimum.
The core:
my @input = split m{\s*\|\s*}, $input; # basis for all the rows
my %nodes =
map { $_ => Node->new($_) }
grep { /\d+/ } split m{\D}, $input; # create all the nodes
# here's where the tree is made
for my $r (@input) {
my $w = -1 + 2**$e;
my @i = split /\s+/, $r;
my @row = map { $i[$_] || '*' } 0 .. $w;
push @rows, \@row;
for my $n ( 0 .. $w ) {
my $val = $row[$n];
my $node = $nodes{$val};
my $lr = $n % 2;
my $p = ' ';
my $u = ' ';
if ( $e > 0 ) { $u = int( $n / 2 ); $p = $rows[ $e - 1 ][$u]; }
my $parent = $nodes{$p};
if ( defined $node && defined $parent ) {
my $v = $node->value;
if ($lr) { $nodes{$p}->left( $nodes{$v} ); }
else { $nodes{$p}->right( $nodes{$v} ); }
my @o = # REMEMBER, READ THIS BACK TO FRONT
sort { $a <=> $b } # sort low to high
map { 1 + node_depth($_) } # 1 + node_depth = number of nodes involved
grep { ! $_->is_root } # each node is not a root
grep { $_->is_leaf } # each node is a leaf
map { $nodes{$_} } # turn it into nodes
keys %nodes; # the keys to the nodes
return $o[0]; # and we pull the first one, which should be
additional languages: Raku
blog writeup: No Diving in the Shallow End - Programming Excursions in Perl and Raku
For my own solution I chose simplicity, constructing a list-processing chain to split the input on whitespace and remove the pipes, mapping the asterisks to undef and preserving the positional data
as one long structured array.
Iterating through the indices of this array a second counter is maintained to ratchet through the level count, counting out 2 × level - 1 elements at a time. Done this way the children for a given
index n will be found according to well-defined relationships that can be checked as we go. When we find the first element without children we return the current level count.
my $input = shift ;
say mindepth( parse( $input ) ) if defined $input;;
sub parse ( $input ) {
return map { $_ eq '*' ? undef : $_ }
grep { $_ ne '|' }
split ' ', $input;
sub mindepth ( @tree ) {
my $level = 1 ;
my $count = 0 ;
for my $idx ( 0 .. $#tree ) {
return $level if ( defined $tree[$idx]
and not defined $tree[$idx * 2 + 1]
and not defined $tree[$idx * 2 + 2] ) ;
$level++ and $count = 0 if ++$count == 2 ** ($level-1) ;
additional languages: Javascript, Kotlin, Lua, Postscript, Python, Raku, Ruby, Rust
blog writeup: RogerBW’s Blog: The Weekly Challenge 151: Robbing Depth
Roger brings us two routines: str2tree() to parse his input, and mindepth() to walk the structure produced and find the minimum depth. The tree itself is a flattened breadth-first traversal, with 0s
substituted for the null nodes. This format will not allow a node value to be 0, but accepting that it does make the structure easy to visualize, and null still translates to false in comparisons.
It’s a reasonable tradeoff as long as you remember it’s there.
To find the leaf nodes we walk the indices up from 0, and we look for children by bit-shifting the index left one place. A set of sorting criteria are established: if a node is 0 it cannot be a leaf,
and if its first child is over the limit of the array bounds it must be leaf. Otherwise we calculate the child positions and examine them.
Proceeding this way from left-to-right we find the first leaf and count bit-shifts back rightward to get the level. I think the bit-shifting adds a pleasing elegance to the technique.
sub str2tree {
my $st=shift;
my @o;
my $d=0;
my $p=0;
foreach my $e (split ' ',$st) {
if ($e eq '|') {
my $m=(1<<($d+1))-1;
if (scalar @o < $m) {
push @o,(0) x ($m - scalar @o);
} else {
my $y=0;
if ($e ne '*') {
my $i=(1<<$d) -1 +$p;
return \@o;
sub mindepth {
my $tree = shift;
my $firstleaf=scalar @{$tree};
foreach my $i (0..$#{$tree}) {
if ($tree->[$i]==0) {
} elsif (($i+1) << 1 >= scalar @{$tree}) {
} else {
my $ni=(($i+1) << 1)-1;
if ($tree->[$ni]==0 && $tree->[$ni+1]==0) {
my $t=$firstleaf+1;
my $d=0;
while ($t > 0) {
$t >>= 1;
additional languages: Awk, Bash, C, Lua, Node, Python, Ruby
Abigail economically breaks the input into an array-of-arrays at the vertical pipes, and then walks each level looking ahead for the first instance of two missing children. With Hasel and Gretel
locked in a hut in the woods, we have found the first leaf and the level index is adjusted and returned.
It’s a dark tale, but compact and to-the-point.
TREE: while (<>) {
my @tree = map {[map {$_ ne '*'} /\S+/g]} split /\|/;
foreach my $d (keys @tree) {
foreach my $i (keys @{$tree [$d]}) {
if ($tree [$d] [$i] && !$tree [$d + 1] [2 * $i]
&& !$tree [$d + 1] [2 * $i + 1]) {
say $d + 1;
next TREE;
additional languages: Raku
blog writeup: PWC151 - Binary Tree Depth - ETOOBUSY
Flavio has arrived at a very similarly concise solution, solving the problem in two parts: as the manipulation of a string into a multidimensional array and then checking from place-to-place to find
the first answer. In one way you can consider the talk of a tree to be a deceptive red-herring, but viewed another the serialized tree is a real tree just as good as any other, merely presented in an
unusually flat manner. Like a number, the representation does not alter what it is. And it is a tree. When i run the script here it is a tree grown in Brooklyn.
my @levels = map { [ split m{\s+}mxs ] } split m{\s*\|\s*}mxs, $input;
for my $depth (1 .. $#levels) {
for my $i (0 .. $levels[$depth - 1]->$#*) {
next if $levels[$depth - 1][$i] eq '*'
|| ($levels[$depth][$i * 2] // '*') ne '*'
|| ($levels[$depth][$i * 2 + 1] // '*') ne '*';
say $depth;
exit 0;
say scalar @levels;
Blogs and Additional Submissions in Guest Languages for Task 1:
blog writeup: Perl Weekly Challenge #151
additional languages: Raku
blog writeup: Perl Weekly Challenge 151: Binary tree Depth
additional languages: Php, Python
blog writeup: Locate a leaf and rob a road
additional languages: Python
blog writeup: Weekly Challenge 151
additional languages: C++, Haskell, Raku
blog writeup: Perl Weekly Challenge 151 – W. Luis Mochán
TASK 2
Rob The House
Submitted by: Mohammad S Anwar
You are planning to rob a row of houses, always starting with the first and moving in the same direction. However, you can’t rob two adjacent houses.
Write a script to find the highest possible gain that can be achieved.
Example 1:
Input: @valuables = (2, 4, 5);
Output: 7
If we rob house (index=0) we get 2 and then the only house we can rob is house (index=2) where we have 5. So the total valuables in this case is (2 + 5) = 7.
Example 2:
Input: @valuables = (4, 2, 3, 6, 5, 3);
Output: 13
The best choice would be to first rob house (index=0) then rob house (index=3) then finally house (index=5). This would give us 4 + 6 + 3 =13.
about the solutions
Abigail, Alexander Pankoff, Athanasius, Cheok-Yin Fung, Colin Crain, Dave Jacoby, Duncan C. White, E. Choroba, Flavio Poletti, James Smith, Jorg Sommrey, Lubos Kolouch, Matthew Neleigh, Peter
Campbell Smith, Roger Bell_West, Simon Green, Ulrich Rieke, and W. Luis Mochan
So I’ve been hearing doomsayer talk for years about the decline of Perl, but have we truly sunk to the point where we’re burgling houses now? Oh dear.
Well it’s just another example of doing what we must to survive in this challenging world. We might as well be practical about it.
We are however burdened by some unusual conditions in our house-breaking: we must start at the first house, and we cannot hit two houses in a row. As Abigail correctly points out, this means we will
never, under any circumstances, rob the second house. I suppose with our luck, that’s where all the money is, too. Such is life. Oh why, why did I stay in school?
I could have been someone. I could have been a contender. Now I’m reduced to robbing houses to buy sketchy black-market electricity to keep my screen lit. Day in, day out, got to keep those electrons
flowing. The monkey on the keyboard needs his fix. Don’t listen to your parents, kids. Don’t end up like me.
(uncomfortable silence as your editor contemplates his life choices)
Where was I? Oh, right.
We were considering how to best select elements from an array according to a set of conditions, to optimize a sum.
From the conditions, some emergent rules become apparent. Besides never visiting the second house, it’s also true that it only makes sense to skip either one or two houses. This follows from the
observation that for any longer jump than three, there exists at least one intermediate house that can also be visited before arriving at the same place. As the values are assumed to be positive or
at least zero, there is no reason to ever not include the stopovers.
Although negative values are not explicitly excluded, it is rather hard to justify the idea of negative loot in our imaginary scenario. Time-to-rob would be a good negative variable to counter the
loot gained and make things nice and complex, but we’re not compounding that in here today. Alternately including item weights could turn it into a variant of the Knapsack Problem.
There were 18 submissions for the second task this past week.
a selection from the RIFF-RAFF, PICKPOCKETS and GENTLEMAN THIEVES
Laurent Rosenfeld, Simon Green, Lubos Kolouch, Ulrich Rieke, Matthew Neleigh, Peter Campbell Smith, Duncan C. White, James Smith, and E. Choroba
The most common technique was recursion, exploring from each landing the two possibilites of skipping ahead two or three houses. The complexity expands quadratically, but as we are tasked with
robbing real imaginary houses on a real imaginary block the number of houses under consideration should not be too large. After all, the whole purpose of blocks is to break collections of lots onto
managable pieces. It’s not unreasonable to surmise that very dense blocks and apartment buildings would require a adjustments to both the conditions and the resultant strategy.
A variant on this is to produce the combinations of houses combinatorically up-front, and compute the sums and find the maximal value.
The recursion decisions center around the partial summing of a choice of two positions as we proceed, making the problem suitable to dynamic programming optimization, and we saw several examples of
this as well. Ultimately we can remove the recursion from the algorithm completely, and produce a dynamic programming array in a single pass by deciding on partial sums based on previously computed
values in same the dynamic array.
additional languages: Raku
blog writeup: Perl Weekly Challenge 151: Binary tree Depth
Laurent will start us off with an example of a recursive solution. At every stage a routine, get_best() is called with a growing partial sum and the remainder of the array, sliced off and packaged
up. Internally both options of skipping ahead two or three houses are explored with another recursive call. A variable in the script scope, $best_so_far is kept updated with a running maximum.
sub get_best {
my $sum_so_far = $_[0];
my @in = @{$_[1]};
if (@in <= 2) {
$sum_so_far += $in[0] if @in == 1;
$sum_so_far += $in[1] if @in == 2;
$best_so_far = $sum_so_far if $sum_so_far > $best_so_far;
for my $i (0, 1) {
get_best($sum_so_far + $in[$i], [@in[$i + 2 .. $#in]]);
additional languages: Python
blog writeup: Weekly Challenge 151
Simon presents another version. In it his rob() routine is passed two arguments, the previous loot plus the current house, and the remaining list of houses down the street.
The recursive solutions in general need to accommodate the edge-cases where there are few or no houses on the street.
Here each call returns the larger of the two calls returned to it; as the recursion collapses the maximum gets propagated backwards up the stack until the first call returns the result.
# Call the function recursively skipping either one or two houses
my @hauls = ();
push @hauls, rob( $haul + $valuables->[0], [ @{$valuables}[ 2 .. $#$valuables ] ] );
if ( @{$valuables} >= 4 ) {
push @hauls, rob( $haul + $valuables->[0], [ @{$valuables}[ 3 .. $#$valuables ] ] );
# Return the largest haul
return max(@hauls);
additional languages: Php, Python
Here’s another by Lubos to give a nice overview of different ways to implement the technique.
sub get_houses_max {
my @houses = @_;
return $cache{@houses} if $cache{@houses};
my $max_value = 0;
my $house_index = 0;
for my $house (@houses[2..@houses-1]) {
my $next_houses_values = get_houses_max(@houses[2+$house_index..@houses-1]);
$max_value = $next_houses_values if $next_houses_values > $max_value;
$cache{@houses} = $houses[0] + $max_value;
return $houses[0] + $max_value;
additional languages: Haskell, Raku
Next we have Ulrich, who brings in the Algorithm::Combinatorics module to fit the combinations for him. Practically this visits all possibilities, the same as the recursive solution, offloading the
work to a compiled library might improve performance.
my @robbedValues ;
my @combilengths ;
if ( $len % 2 == 1 ) {
push @combilengths , int( $len / 2 ) , int( $len / 2 ) + 1 ;
else {
push @combilengths, int( $len / 2 ) ;
my @positions = (0 .. $len - 1 ) ;
for my $combilen ( @combilengths ) {
my $iter = combinations( \@positions, $combilen ) ;
while ( my $c = $iter->next ) {
if ( checkCondition( $c ) ) {
push @robbedValues , sum (@valuables[ @$c ]) ;
say max( @robbedValues ) ;
Using dynamic programming, as we process the house array from left to right we can construct a parallel array of partial sums as we go. Each new position added is decided by choosing the maximum sum
from the most recent previous partial solutions, themselves chosen from earlier paths. In this way only two partial solutions are used at each decision: “What if we jumped here from two steps back?”
and “What if we jumped from three?". Every house after the start gets the same decision, and when we get to the end we have our maximum. Nice.
sub calculate_loot_yield_on_street{
use List::Util qw(max);
# Empty list, no houses to rob
my @loot;
my $loot_initial;
my $i;
# We always start with the first house, as
# specified (though this seems limiting...)
$loot_initial = $ARG[0];
# Strip off the first two houses- we've
# robbed the first and can't rob the second
splice(@ARG, 0, 2);
# Edge cases- zero or one houses left
if(scalar(@ARG) == 1){
return($loot_initial + $ARG[0]);
# Proceed as normal(?)
$loot[0] = $ARG[0];
$loot[1] = max($ARG[0], $ARG[1]);
for($i = 2; $i < scalar(@ARG); $i++){
$loot[$i] = max($ARG[$i] + $loot[$i - 2], $loot[$i - 1]);
return($loot_initial + $loot[$#loot]);
blog writeup: Locate a leaf and rob a road
Peter examines all paths forward at every cycle in his recursion, including those past the third position forward, considering all jumps to the end of the line. There’s no harm in this and the
algorithm handles a 45-house street in a few seconds.
sub robberies {
# robberies($number, $swag) updates $best with the best result starting from house $number
# with $swag already in the bag
my ($number, $swag, $next, $new_swag);
($number, $swag) = @_;
# try all the next allowable houses starting from $number
for ($next = $number + 2; $next <= $last; $next ++) {
$new_swag = $swag + $houses[$next];
$best = $new_swag if $new_swag > $best; # looking good!
robberies($next, $new_swag);
Duncan has devised another recursive way to search all the paths, compounding a best total value as he goes:
fun maxrobbery( $starthouseno, @valuables )
my @besth;
my $besttotal = 0;
foreach my $hno ($starthouseno+2..$#valuables)
# find the best partial solution starting by robbing house $hno
my( $mv2, @rh2 ) = maxrobbery( $hno, @valuables );
# then find the best of all those partial solutions
if( $mv2 > $besttotal )
$besttotal = $mv2;
@besth = @rh2;
# then the overall best solution involves adding starthouseno
# to the best partial solution..
return ( $valuables[$starthouseno]+$besttotal, $starthouseno, @besth );
blog writeup: Perl Weekly Challenge #151
James delivers a very compact solution that will probably appear quite mysterious, but is a reworking of the dynamic programming solution, working backwards from the end. Fortunately he provides
notes to the action, both in the comments and at his writeup.
sub rob {
## Line 1 - Trip finishing at the first house the value is the
## points for the first house
## Line 2 - If there is more than one house we set the value
## for the second house to be the points for the house
## itself, unless the first house has a better value
## Line 3 - We repeat this for the remaining houses.... It is the
## points for this house + the value for two houses before
## or the value for the previous house if it is greater
## Line 4 - When we get to the end the result is just the value
## for the last house!
## Comments this way so they don't hide the symmetry of the code
my @b = shift;
(push @b,shift ), $b[-1]<$b[-2] && ($b[-1]=$b[-2]) if @_;
(push @b,$_+$b[-2]), $b[-1]<$b[-2] && ($b[-1]=$b[-2]) for @_;
All of this looking and working backwards to see which position to have traveled from produces the optimal sub-problem can be quite confusing. Choroba has reversed everything, so the algorithm is run
forward. However note we’re not picking the best option available, skipping to it and proceeding from there however — rather we’re systematically looking at every house and figuring instead the best
way to have gotten there. When we’re done the value at $sums[0] reveals the answer. Dynamic programming is such an intersting technique, currying the data processing.
sub rob_the_house {
my (@valuables) = @_;
my @sums;
for my $i (reverse 0 .. $#valuables) {
$sums[$i] = $valuables[$i];
if ($i + 2 <= $#valuables) {
my $add = $sums[$i + 2];
$add = $sums[$i + 3] if $i + 3 <= $#valuables
&& $sums[$i + 3] > $sums[$i + 2];
$sums[$i] += $add;
return $sums[0]
Blogs and Additional Submissions in Guest Languages for Task 2:
additional languages: Awk, Bash, Bc, C, Go, Java, Lua, Node, Pascal, Python, R, Ruby, Tcl
additional languages: Raku
additional languages: Raku
blog writeup: Burglary Tools
blog writeup: Dr. Metropolis and His Amazing MANIAC Machine!: The Weekly Challenge #151 | Committed to Memory
additional languages: Raku
blog writeup: PWC151 - Rob The House - ETOOBUSY
additional languages: Javascript, Kotlin, Lua, Postscript, Python, Raku, Ruby, Rust
blog writeup: RogerBW’s Blog: The Weekly Challenge 151: Robbing Depth
blog writeup: Perl Weekly Challenge 151 – W. Luis Mochán
_________ THE BLOG PAGES _________
That’s it for me this week, people! Warped by the rain, driven by the snow, resolute and unbroken by the torrential influx, by some miracle I somehow continue to maintain my bearings.
Looking forward to next wave, the perfect wave, I am: your humble servant.
But if Your Unquenchable THIRST for KNOWLEDGE is not SLAKED,
then RUN (dont walk!) to the WATERING HOLE
and FOLLOW these BLOG LINKS:
( …don’t think, trust your training, you’re in the zone, just do it … )
Arne Sommer
Colin Crain
Dave Jacoby
Flavio Poletti
James Smith
Laurent Rosenfeld
Peter Campbell Smith
Roger Bell_West
Simon Green
W. Luis Mochan | {"url":"https://theweeklychallenge.org/blog/review-challenge-151/","timestamp":"2024-11-10T14:34:09Z","content_type":"text/html","content_length":"109961","record_id":"<urn:uuid:1c05e2df-e632-4c54-9160-124796d6fb8d>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00804.warc.gz"} |