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# What Is R And R1 Complement?
## What is Nines complement?
Decimal example The nines’ complement of a decimal digit is the number that must be added to it to produce 9; the complement of 3 is 6, the complement of 7 is 2, and so on, see table.
To form the nines’ complement of a larger number, each digit is replaced by its nines’ complement..
## Why do computers use 2’s complement?
Two’s complement allows negative and positive numbers to be added together without any special logic. … This means that subtraction and addition of both positive and negative numbers can all be done by the same circuit in the cpu.
## How do you find 7’s complement?
The 7’s complement of an octal number is obtained by subtracting each digit from 7. The 1’s complement of a binary integer is obtained by subtracting each digit from 1.
## How many types of complements are there for each base R system explain?
Complements are used in the digital computers in order to simplify the subtraction operation and for the logical manipulations. For each radix-r system (radix r represents base of number system) there are two types of complements. S.N.
## What’s a complement in probability?
Complement of an Event: All outcomes that are NOT the event. And together the Event and its Complement make all possible outcomes. …
## What is complement number?
Complement. The “complement” is the number to add to make 10 (or 100, 1000, etc, depending on how many digits we have) Example The complement of 3 is 7, because 3+7=10 (we add 7 to make 10) Example: the complement of 85 is 15, because 85+15=100. Example: the complement of 111 is 889, because 111+889=1000.
## How do you find 9s and 10s complement?
10’s complement subtraction To obtain the 9’s complement of any number we have to subtract the number with (10n – 1) where n = number of digits in the number, or in a simpler manner we have to divide each digit of the given decimal number with 9.
## What is 1 complement and 2’s complement with example?
For example, 1’s complement of binary number 110010 is 001101. To get 2’s complement of binary number is 1’s complement of given number plus 1 to the least significant bit (LSB). For example 2’s complement of binary number 10010 is (01101) + 1 = 01110.
## What does P A B C mean?
Yes, P( A | B, C ) means P( A | B ∧ C ), or more properly P( A | B ∩ C ) because events are sets of outcomes. raddaya. 4 points · 5 years ago. So in layman’s terms, probability of A occurring given that B and C have occurred?
## How do you do two’s complement?
To get 2’s complement of binary number is 1’s complement of given number plus 1 to the least significant bit (LSB). For example 2’s complement of binary number 10010 is (01101) + 1 = 01110.
## How do you find the R and R’s complement?
To find (r-1)’s complement, subtract each digit of the given number from from the largest number in that number system. For instance, if the number is a three digit number in radix 9, then subtract the number from 888 as 8 is the largest number in radix 9 number system. The obtained result is the (r-1)’s complement.
## How do you compliment a girl?
60 Super Sweet Compliments for WomenYour mind is just as sexy as your beauty.I miss your smile.You’re an amazing friend.I can’t believe I found someone like you.I get excited every time I see you.I love making you laugh.You’re my best friend.I’ll always have your back.More items…•
## What is 2’s complement representation?
Two’s complement is a mathematical operation on binary numbers, and is an example of a radix complement. It is used in computing as a method of signed number representation. For instance, for the three-bit number 010, the two’s complement is 110, because 010 + 110 = 1000. …
## What is the complement of R?
2) r’s complement The r’s complement of a non-zero number in any number system with base r can be calculated by adding 1 to the LSB of its (r-1)’s complement. For Example: In binary number system, 2’s complement of 001 can be calculated by adding 1 to the LSB of its 1’complement (i.e., 110 + 1) = (111)2.
## Whats is a complement?
In grammar, a complement is a word, phrase, or clause that is necessary to complete the meaning of a given expression. Complements are often also arguments (expressions that help complete the meaning of a predicate).
## How do you calculate complement?
A mutually exclusive pair of events are complements to each other. For example: If the desired outcome is heads on a flipped coin, the complement is tails. The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1, or for the event A, P(A) + P(A’) = 1.
## What is complement example?
It’s a word, clause, or phrase that’s needed to complete a given expression. For example, “Every morning is a gift.” In this sentence, “every morning” is the subject, “is” is the linking verb, and “a gift” is the complement. It completes the idea.
## How do you compliment someone?
So here’s a hundred ready-made compliments to try out yourself:You’re an awesome friend.You’re a gift to those around you.You’re a smart cookie.You are awesome!You have impeccable manners.I like your style.You have the best laugh.I appreciate you.More items…
## How do you do 10s complement?
10’s complement of a decimal number can be found by adding 1 to the 9’s complement of that decimal number. It is just like 2s compliment in binary number representation. For example, let us take a decimal number 456, 9’s complement of this number will be 999-456 which will be 543. Now 10s compliment will be 543+1=544.
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# Backpack-Busting Books
Books are very heavy for their size. After all, they’re basically chunks of wood, just pressed into thin sheets of paper. So our friend Genevieve B. asked, how big is the world’s biggest book? Turns out there’s more than one answer, because it depends what you mean by “book.” If we mean a bunch of paper pages stuck together, the Guinness World Record goes to “This the Prophet Muhammed,” a 16-foot tall, 13-foot wide giant with 429 pages. That’s a huge book! Also, the book that explains Brazil’s rules for paying taxes (money to help run the country) is ridiculous: it’s 6 feet 10 inches tall, and has more than 41,000 pages that size. But if the book can be made of something other than paper, the heavyweight winner might be “The Pali Canon” (Buddhist scripture), whose 1,460 pages are actually 5-foot stone tablets. Let’s hope you never need to carry those in your backpack.
Wee ones: Which is taller, the 5-foot stone “book” or the 6-foot paper book?
Little kids: If you can reach 6 feet high off the ground, how many feet taller is the 16-foot tall book? Bonus: If Richard Scarry’s “Biggest Word Book Ever” is 2 feet tall, how many would you have to stack end to end to match the 16-foot book?
Big kids: When you open that 13-footwide book, it becomes twice as wide! How far across the “table” would it stretch? Bonus: Brazil’s tax code weighs “several tons.” If it weighs 4 tons and you can carry 10 pounds on your back, how many kids like you would it take to carry the Brazil book? (Reminder if needed: A ton equals 2,000 pounds.)
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## Friday, March 13, 2015
### Pi Day Eve (Day 131)
Today is Friday the thirteenth once again. Those who suffer from triskaidekaphobia must surely hate this quirk in the calendar. Because February, with its 28 days, is the only month that can be divided evenly into seven-day weeks, the first 28 days of month always fall on the same days of the week as the corresponding days in February (unless there's a Leap Day, since 7 doesn't divide 29 evenly). So if February has a Friday the 13th in a common year, so does March. It's the only combination where there can be two consecutive Fridays the 13th, and it won't occur again until 2026. (Well, at least we don't have the Cotsworth 13-month calendar, where all thirteen months contain Friday the 13th -- that would really cause the triskaidekaphobes to panic!)
I made a big deal about Friday the 13th last month -- I even designed my Chapter 13 Test that day to have thirteen questions, with many of them involving the number 13. But it's not today's special Friday the 13th that's on my mind right now -- it's tomorrow's special Pi Day that I want to discuss.
Not only is tomorrow 3/14, with the digits resembling the first three digits of pi, but in fact tomorrow is 3/14/15, so the first five digits are represented, 3.1415.... Because of this numerical alignment, many people are declaring tomorrow to be the "Pi Day of the Century." So the double Friday the 13th talk can wait until 2026 -- we must discuss the special Pi Day now, unless we want to wait until 2115 for the next Pi Day of the Century!
Let me begin the festivities with discussing my own discovery of pi. I first learned about the number pi when I was in the second grade. I remember learning about the rounded value of pi, 3.1416. Now I know that the last value has been rounded up (otherwise next year would be Pi Day of the Century), but back then, I didn't know that. I knew, though, that 3.1416 wasn't the exact value of pi. I once tried to guess the next few digits of pi -- I thought that the pattern continued, so that the digits would be something like 3.141618202224262830.... Now I know, of course, that this guess is wrong -- in fact, there is a rational number that starts out 3.141618202224262830... (it's 30791/9801), but of course, pi is irrational.
Other than that, I was never really interested in the digits of pi. I was actually more interested in the digits of rational numbers, such as 1/17, which is .0588235294117647... repeating. This was mainly because I knew how to generate the digits of 1/17 with a scientific calculator -- I got my first scientific calculator when I was a fourth grader -- but I didn't know how to generate the digits of pi.
I was in college at UCLA when I first heard of Pi Day. I stumbled upon it when I was searching for who knows what on the internet. During my last year at UCLA, the math department held a Pi Day party, to be held on Friday, March 14th at 5:00, after all classes had ended. Another student pointed out that the party should have been at 1:59, since the next few digits after 3.14 were 159, but I suspect that the university didn't want the party to occur one minute before 2:00 classes began.
Since then, I've tried to celebrate Pi Day in the classroom whenever I'm able to be in a math class on Pi Day. Two years ago, I brought in a pizza to the Algebra I class where I was student teaching. For after all, pizzas are round, so their circumferences and areas can be calculated using pi. Of course, as this was an algebra class, I used the pizza to have the students calculate how many slices of pizza they can eat if they wanted to consume no more than 600 calories. Unfortunately, that was the only year that I could really incorporate Pi Day in the classroom. Today, I subbed in an English class so I couldn't do anything about Pi Day with them. (I did warn some of them about Friday the 13th!) But of course I always tell the students about Pi Day when I tutor them. (My geometry student already knew that it was Pi Day of the Century!)
So now we reach today's Pi Day lesson. As I mentioned last week, today's lesson will be based on the lessons of Drs. Franklin Mason and Hung-Hsi Wu.
Let's start with Dr. M, who covers pi in his Section 9.2. Originally, Dr. M stated that Lesson 9.2 would be one of the three best lessons of the year -- the other two being the Orthocenter Theorem (i.e., the theorem that the orthocenters of a triangle are concurrent) and the volume of a sphere. But now when I read Section 9.2, that claim is no longer mentioned, and his Section 9.2 has been stripped down somewhat. Most likely, Dr. M realized that there wouldn't be enough time to teach pi the way he wanted to, so he had to cut it down. When I was planning out my Chapter 8 lessons last week, the full Section 9.2 was still posted, so this means that Dr. M covered pi in his own class fairly recently. I guessed that since pi was covered in Chapter 9 of his 13-chapter text, Dr. M would reach the lesson close to Pi Day -- and I was correct.
The constant pi is introduced very similarly in Dr. M's Section 9.2 and U of Chicago's Section 8-8 -- the main difference is that arc length also appears in U of Chicago's 8-8 while this topic is saved for 9.3 in Dr. M. Both Dr. M and U of Chicago define pi as C/d, which is, of course, the ratio of the circumference to the diameter of a circle.
But here's what Dr. David Joyce says about how pi is defined in the Prentice-Hall text:
The tenth theorem in the chapter [5 of the Prentice-Hall text -- dw] claims the circumference of a circle is pi times the diameter. It is apparent (but not explicit) that pi is defined in this theorem as the ratio of circumference of a circle to its diameter. So the content of the theorem is that all circles have the same ratio of circumference to diameter. This theorem is not proven. It would require the basic geometry that won't come for a couple of chapters yet, and it would require a definition of length of a curve and limiting processes. It would be nice if a statement were included that the proof the the theorem is beyond the scope of the course.
The idea that all circles have the same ratio of circumference to diameter may seem obvious, but Joyce is correct in that it must be proved. We can't simply say that it's true just because we defined pi to be that ratio! After all, what to stop us from saying that phi is defined as the ratio of length of a rectangle to its width, and then claim this proves that the ratio of length to width must be the same for all rectangles? (On the other hand, it is possible to define phi as the ratio of length to width of a golden rectangle, but not for a general rectangle.)
Joyce points out that limiting processes -- that is, Calculus -- is needed to prove the theorem. Dr. M also states that Calculus is needed to prove the circumference formula. Because we can't prove the formula, Dr. M states that it is a postulate:
Dr. M's Circle Similarity Postulate:
The value of pi is a constant, i.e., it is the same for all circles.
We notice how Dr. M calls this the Circle Similarity Postulate. After all, it's because all circles are similar that the ratio of circumference to diameter is the same for all circles. This is what the Common Core Standards expect students to know about circles:
CCSS.MATH.CONTENT.HSG.C.A.1
Prove that all circles are similar.
Then again, notice that Common Core seems to expect a proof here. How does Common Core expect students to prove the similarity of all circles without Calculus?
Unfortunately, none of our sources actually prove that all circles are similar. What I'm expecting is something like this -- to prove that two circles are similar, we prove that there exists a dilation mapping one to the other. For simplicity, let's assume the circles are concentric, and the radii of the two circles are r and s. So we let D be the dilation of scale factor s/r whose center is -- where else -- the common center O of the two circles. If R is a point on the circle of radius r, then OR = r, and so its image R' must be a point whose distance from O is r * s/r = s, and so it must lie on the other circle of radius s. Likewise, if R' is a point on the circle of radius s, its preimage must be a point whose distance from O is s / (s/r) = r, and so it must like on the circle of radius r. Therefore the image of the circle of radius r is exactly the circle of radius s.
Of course, this only works if the circles are concentric. If the circles aren't concentric, then it's probably easiest just to compose the dilation with an isometry -- here a translation is easiest -- mapping the center of one circle to that of the other. Therefore there exists a similarity transformation mapping any circle to any other circle. Therefore all circles are similar. QED
We can't really fault Dr. M for not including this proof -- after all, remember he did come up with the clever algebraic derivation of the area of a rectangle from that of a square. It's difficult to come up with some of the cleverer proofs of these basic geometric formulas.
Dr. M's old lesson was interesting, but he had to throw it out to save time. Therefore the activity that I will post today is based on Dr. Wu's lesson. We've already started Dr. Wu's lesson when we covered Section 8-4 last week. As I mentioned that day, the U of Chicago could have easiest segued from Section 8-4 to Dr. Wu's definition of pi as the area of the unit disk.
So today's activity begins with a circle whose radius is divided into tenths, not fifths. We ask the students to count how many squares are filled in -- hopefully, the answer is close to 314. Since each tiny square has area .01, the area of the circle would be around 3.14 or pi.
To get from the area of the unit disk (pi) to the area of any disk (pi * r^2), we are basically using the Fundamental Theorem of Similarity from Section 12-6 of the U of Chicago. This time, though, we are using part (b) of that theorem:
Fundamental Theorem of Similarity:
If G ~ G' and k is the ratio of similitude [the scale factor -- dw], then
(b) Area(G') = k^2 * Area(G) or Area(G') / Area(G) = k^2.
We skipped this formula back when we covered Section 12-6 because at the time, we hadn't learned about area yet. Although Wu attempts to prove a special case of the Fundamental Theorem of Similarity using triangles, it's much easier to do it using squares, as the U of Chicago does. If G can be divided into A unit squares, then G' can be divided into A squares each of length k. And the area of a square of length k is clearly k^2, so the area of G' must be Ak^2. For the circle problem k is the radius r, and A is the area of the unit circle or pi, so the area of a circle is pi * r^2. We can do this right on the same worksheet -- there's already a circle drawn of radius 10 times the length of a square, so instead of the length of each square being 1/10, let it be 1 instead. Then the area of the circle of radius 10 is equal to the number of shaded boxes, or 314, since the old unit square has been divided into 100 unit squares.
We must now get from the area of a circle to the circumference. Notice that Wu actually performs the "limiting processes" that Joyce tells us is necessary to get the circumference! For Wu considers the circumference to be the limit of the perimeter of a regular n-gon inscribed in the circle as n approaches the value of infinity.
Wu calculates the perimeter of the regular polygon by dividing it into triangles. What he is actually using is the formula A = 1/2 ap, where a is the apothem of the polygon and p is its perimeter. Then in the limit, the apothem approaches the radius of the circle and the perimeter of the polygon approaches the circumference, so A = 1/2 rC. Then substituting A = pi * r^2 and solving for C gives the formula C = 2pi * r, which converts to C = pi * d.
The word apothem and the formula for the area of a regular polygon appear in many texts, but the word apothem never appears in the U of Chicago at all. Using the apothem here would save time as it would get us directly from A = 1/2 ap to A = 1/2 rC.
And so this is how I set up my worksheet. If it's possible, I wouldn't mind bringing in some pies (either fruit or pizza) and having the students calculate arc length and possibly even the area of a sector -- to my surprise, the U of Chicago doesn't give the formula for area of a sector either -- by using a slice of pie. The arc length and sector area would have to be given before the student is allowed to eat the slice. Of course, it will be expensive to have enough pie for every student in every class -- but it may be worth it to have a Pi Day worth celebrating.
Now let's move on to some fun stuff. Here are some Pi Day links:
http://www.exploratorium.edu/pi/
This is the Exploratorium, a science museum located right here in California. I live in Southern California and the Exploratorium is up in San Francisco, so I've never been there. But it is the museum where Pi Day was invented 27 years ago.
http://www.nationalpiday.org/
This is the National Pi Day site. National Pi Day was, believe it or not, declared an actual holiday by Congress back in 2009. This site is definitely celebrating this year's Pi Day of the Century.
http://www.piday.org/
This is a Pi Day site from a few years ago. This site also contains a link to some suggested activities for teachers. Some students had the opportunity to put a pie in their math teacher's faces -- indeed, a student I tutored last year pied his own teacher on Pi Day. It gives more links to Pi Day activities.
But my favorite Pi Day sites are all about music. There are, in general, two types of songs that one plays on Pi Day. The first type takes the digits of pi -- 3.1415926535897932384626433832795... -- and converts them into notes using the C major scale. So 3 becomes E (the 3rd note of the C major scale), 1 becomes C, 4 becomes F, 1 is another C, 5 becomes G, and so on.
Here are some YouTube links to songs formed by converting the digits of pi into notes. The first one was posted by Michael Blake:
This one does the same trick, except it uses the A minor scale rather than C major:
Here's another version that uses base 12 instead of 10:
With so many versions of the pi song out there, there was some controversy a few years back. Blake, whose video I posted above, had to take down his video for some time because someone else claimed that they had copyrighted the pi song. Another YouTube user, Vi Hart, criticized the lawsuit:
And she's absolutely correct -- the plaintiff was hardly the first to write a song just by converting the digits of pi. Indeed, one of the first websites I read on Pi Day has a pi song -- and it was created by a high school girl more seventeen years ago:
http://www.sailorpi.com/sailorpi.html
I credit Elizabeth, the author of this Sailor Moon parody page, with being one of the first people to inform me about the existence of Pi Day so many years ago. This page is so old that many of the links are to AOL and Geocities pages! As this was well before YouTube, many of her background songs are in MIDI format, which no longer plays on modern browsers. Yet the following link gives evidence that Liz had posted two songs (one major, one minor) that convert the digits of pi into notes:
http://www.sailorpi.com/why.html
And the teenager herself admitted that this idea wasn't original to her! She links to another page, created by Daniel Cummerow, who did the same thing with not only the digits of pi, but other constants such as e as well! He posted it on Geocities, which no longer exists, but it has been archived on the Reocities site:
http://reocities.com/vienna/9349/
Once again, none of the songs are playable without an old browser that can play MIDI files. But it goes to show that to be consistent according to Hart, Cummerow -- wherever he is -- should sue the plaintiff of the Blake case. Fortunately, the case was laughed out of court -- you can't copyright the number pi!
The other type of song that frequently comes up on Pi Day is the parody song. Basically, one takes a popular song and change the lyrics so that it becomes a song about pi. Two of the most frequently parodied songs are Don McLean's American Pie -- since "pie" sounds like "pi":
and Rebecca Black's Friday -- since it sounds like "Pi Day":
One of my favorite parody songs is based on Eminem's Lose Yourself:
One mathematician whom I've mentioned earlier on this blog, Danica McKellar, created her own video for pi. Here McKellar calls those who create songs about pi "nerds" -- then of course, she turns right around and creates her own song about pi:
The old Sailor Pi site also contains some parody songs. Here's a link to a parody of Rainbow Connection by Paul Williams:
http://www.sailorpi.com/pi.html
The following, while not actually songs, are some poems that Liz wrote about pi. Notice that one of the poems, "A Scientific American's Dilemma," mentions a "Grand Theory" -- that is, the Grand Theory of Everything that Stephen Hawking sought:
http://www.sailorpi.com/poem.html
Here's Liz's obligatory American Pi parody. You can see how old it is by the way the song refers to TI-83 graphing calculators rather than the modern TI-84:
http://www.sailorpi.com/calculate.html
Liz also used to have songs about other constants such as e and phi. I still remember her old e song, a parody of Sugar Sugar (of Archies fame), but unfortunately all links to that song are dead. It was fun to go to the Sailor Pi site on Pi Day and have all of these songs -- the pi major and minor songs, American Pi(e), and even the Sailor Moon song -- automatically play in MIDI format as soon as the page had finished loading.
Sometimes I wonder where Liz is now. I believe that she's around my age. I'd like to believe that somewhere out there, she's getting excited about the Pi Day of the Century.
There are also sites that have the lyrics of parody songs, but no recorded music. Many of these are parodies of Christmas songs. (So apparently for math teachers, our Christmas is in March.)
http://teachpi.org/music/pi-day-carols
This site has "Oh, Number Pi" ("O Christmas Tree") and "Ring the Bells" ("Jingle Bells"). There is even a song about Ludolph van Ceulen, the 16th century mathematician who calculated the first 35 digits of pi. Of course, the authors of this site couldn't resist changing "Ludolph" to "Rudolph"! Finally, there is a link to a "love song." As it turns out, it's just the digits of pi converted to notes again -- just like Michael Blake, Liz, and all the others.
http://mathforum.org/te/exchange/hosted/morehouse/songs.pi.html
One of the songs at this link is a parody of "This Old Man." Here is its second verse:
Verse Two:
Number pi
You're so fun
And [so much] better than one
If you think that pi are round Beware!
We all know that [pi r squared].
For some reason, the last line is incomplete with the phrase "pi r squared" missing. I decided to add the words "so much" to the third line since it appears to be two syllables shorter than the melody.
There also exist math parody songs that have nothing to do with pi. Many of them have been created in high school math classes. For example, this high school in Ohio created a parody of DJ Khaled's All I Do Is Win, changing "win" to "solve" (systems of linear equations):
Indeed, there are several math parody songs by this school posted on YouTube. It's an activity that can be done in math classes around the country.
And of course, I can go on and on about Pi Day and give more and more links and videos. This is why I make every effort to make sure that Pi Day fits on my pacing guide for the year. Sometimes I wish that every day in math class can be Pi Day.
And so I wish everyone a happy Pi Day. Let's make it the best Pi Day this century!
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# Tagged Questions
Let $n$ be an integer greater than $3$. Find a formula for $\gcd(n, n + 3)$ for each of the cases : $1)$ $n \equiv 0\mod 3$ $2)$ $n \equiv 1\mod 3$ $3)$ $n \equiv 2\mod 3$ Any help would be greatly ...
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課程資訊
課程名稱 工程科學數值方法Numerical Methods for Engineering Science 開課學期 101-1 授課對象 工學院 應用力學研究所 授課教師 周逸儒 課號 AM7008 課程識別碼 543EM1110 班次 學分 3 全/半年 半年 必/選修 選修 上課時間 星期二6,7,8(13:20~16:20) 上課地點 應111 備註 本課程以英語授課。總人數上限：98人 Ceiba 課程網頁 http://ceiba.ntu.edu.tw/1011Numerical_Method 課程簡介影片 核心能力關聯 核心能力與課程規劃關聯圖 課程大綱 為確保您我的權利,請尊重智慧財產權及不得非法影印 課程概述 Course outline: 1. Interpolation (3 hrs) (1) Lagrange Polynomials (2) Polynomial interpolations; Splines 2. Numerical Differentiation (4 hrs) (1) Construction of finite difference scheme, order of accuracy (2) Modified wavenumber as a measure of accuracy (3) Pade approximation (4) Matrix representation of finite difference schemes 3. Numerical Integration (8 hrs) (1) Trapezoidal rule; Simpson’s rule; error analysis and mid-point rule (2) Romberg integration and Richardson’s extrapolation (3) Adaptive quadrature; Gauss quadrature 4. Numerical Solution of Ordinary Differential Equations (10 hrs) (1) Initial value problems; numerical stability analysis, model equation (2) Accuracy; phase and amplitude errors (3) Runge-Kutta type formulas, multi-step methods; implicit methods (4) System of differential equations; stiffness (5) Linearization for implicit solution of non-linear differential equations (6) Boundary value problems, shooting, direct methods, non-uniform grids, eigenvalue problems 5. Partial Differential Equations (10 hrs) (1) Finite-difference solution of partial differential equations (2) Modified wavenumber and Von Neumann stability analysis, modified equations analysis (3) Alternating direction implicit methods; non-linear equations; iterative methods for elliptic PDE’s 課程目標 (1) Development and applications of numerical methods when analytical techniques are not available; (2) Development of a conceptual framework for analysis of methods to fix the problem; (3) Discrete calculus and approximations; (4) Tradeoffs between accuracy and computational cost; 課程要求 Homeworks (70%); Final exam (%30) 預期每週課後學習時數 Office Hours 參考書目 指定閱讀 Parviz Moin “Fundamentals of Engineering Numerical Analysis”, Cambridge University press 評量方式(僅供參考)
課程進度
週次 日期 單元主題 第1週 9/11 Introduction & Interpolation 第2週 9/18 Numerical Differentiation-Finite Differences 第3週 9/25 More on Finite Differences & Numerical Integration 第4週 10/02 Numerical Integration 第5週 10/09 Numerical Integration & Solving ODEs 第6週 10/16 No class 第7週 10/23 Numerical Solution for ODEs 第8週 10/30 Numerical Solution for ODEs(continued) 第9週 11/06 Numerical Solution for ODEs(continued) 第10週 11/13 Numerical Solution for ODEs(continued) 第11週 11/20 Numerical Solution for PDEs 第12週 11/27 Numerical Solution for PDEs(continued) 第14週 12/11 Numerical Solution for PDEs(continued) 第15週 12/18 Numerical Solution for PDEs(continued) -- Elliptic Equation 第16週 12/25 Review
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> > A cube block of metal has a volume of 1728cm3.
# A cube block of metal has a volume of 1728cm3.
## If the block of metal is then melted and formed into rectangular pr...
360 Views
### A cube block of metal has a volume of 1728cm3.
If the block of metal is then melted and formed into rectangular prism with a base,which has an equal length and width and both of which are less than 15cm.
Find the base of the prism,if the height is a whole number.
[tex]V=1728\ cm^3\\\\V=a^2h\ where\ a\ is\ the\ lenght\ of\ the\ base\ side\ and\ h\ is\ height.
\\\\1728|2\\.864|2\\.432|2\\.216|2\\.108|2\\.
\ 54|2\\.
\ 27|3\\.
\ \ 9|3\\.
\ \ 3|3\\.
\ \ 1[/tex]
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# If x = 2+√3, find the value of
Question:
If $x=2+\sqrt{3}$, find the value of
$x^{3}+\frac{1}{x^{3}}$
Solution:
Given, $x=2+\sqrt{3}$
To find the value of $x^{3}+\frac{1}{x^{3}}$
We have, $x=2+\sqrt{3}$,
$\frac{1}{x}=\frac{1}{2+\sqrt{3}}$
Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor
$2-\sqrt{3}$ for $\frac{1}{2+\sqrt{3}}$
$=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}$
Since, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$
$\frac{1}{x}=\frac{2-\sqrt{3}}{4-3}$
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}$
$x+\frac{1}{x}=4$
We know that, $\left(a^{3}+b^{3}\right)=(a+b)\left(a^{2}-a b+b^{2}\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}-x \cdot \frac{1}{x}+\frac{1}{x^{2}}\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1^{2}}{x}-1\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}+2-2-1\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}+2\left(x \cdot \frac{1}{x}\right)-2-1\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(\left(x+\frac{1}{x}\right)^{2}-3\right)$
Putting the value of $x+1 x$ in the above equation, we get,
$\left(x^{3}+\frac{1}{x^{3}}\right)=(4)\left(4^{2}-3\right)$
$\left(x^{3}+\frac{1}{x^{3}}\right)=52$
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File Exchange
## Multi-Objective Particle Swarm Optimization (MOPSO)
version 1.0.0.0 (11.4 KB) by
A structure MATLAB implementation of MOPSO for Evolutionary Multi-Objective Optimization
Updated 20 Oct 2015
http://yarpiz.com/59/ypea121-mopso
### Cite As
Yarpiz (2021). Multi-Objective Particle Swarm Optimization (MOPSO) (https://www.mathworks.com/matlabcentral/fileexchange/52870-multi-objective-particle-swarm-optimization-mopso), MATLAB Central File Exchange. Retrieved .
zaki dahia
Hi everyone, my ZDT function is
Tp=140;Tf= 600;
Cp=5000;Cf=35000;
m=3;etta=8000;
H=(((x)/etta)^m);
f1=(Cp+Cf*H)/(x+Tp+Tf*H);
f2=x/(x+Tp+Tf*H);
z=[f1;f2];
Error in @(x)ZDT(x)
Error in mopso (line 69)
pop(i).Cost=CostFunction(pop(i).Position);
please any solution to this problem
jie yu
I'm a student, it's very hard for me to understand the code completely. because there is so many kinds of MOPSO methods, can you provide me some relevant documentation to help me analyse the code.My English is not very good, please forgive me.my mail box is 29111370554@qq.com.I would really appreciate your reply.
@sidharth parhi
"3 variables with different ranges" please help. How to handle variables width different lower and upper bound ?
@ all
for all who suffer from this error:
Error in FindGridIndex (line 25)
find(particle.Cost(j)<Grid(j).UB,1,'first');
Error in mopso (line 87)
rep(i)=FindGridIndex(rep(i),Grid);
this is the solution from Yarpiz website:
"Your objectives can have the value of NaN. So everything may fall, due to incomparability of NaN to other numbers. You should modify your code, to replace the potential NaN outputs, with nun-NaN but invalid values."
Index exceeds array bounds.
Any solution to this problem
Error in FindGridIndex (line 25)
find(particle.Cost(j)<Grid(j).UB,1,'first');
Error in mopso (line 87)
rep(i)=FindGridIndex(rep(i),Grid);
Ruiqi Wang
Very useful program. Structure is very clear, and it is easy to modify. Thank the authors.
Joshua Desbordes
Error in Mutate (line 33)
xnew(j)=unifrnd(lb,ub);
simpily do this
SOLVED
xnew=x;
xnew(j)=unifrnd(lb(j),ub(j));
Cheng-Ying Yang
Thanks, it's great. But how do we define nonlinear constraints in this code?
manish sharma
Index exceeds matrix dimensions.
Error in FindGridIndex (line 24)
particle.GridSubIndex(j)=...
Error in mopso (line 87)
rep(i)=FindGridIndex(rep(i),Grid);
Error in main (line 14)
Problem can be solved if you put
z = [f1 ; f2]; not z = [f1 f2];
nima rezaee
firs of all awesome code and algorithm. thank you.
but unfortunately i have the same problem as RAVI MANDAVA. i want to have different varmin and varmax for each variable in mopso.
i can change the code but im not sure it will work. i dont know maybe im missing something.
yibe man
can you tell me how to accept the training and test data from excel
freedomangel
osama bany mousa
This is great!
Thank you
liruixinch liruixinch
Anuraag Dash
How to implement it in MATLAB
MANPREET SHARMA
Can anyone please tell me HOw to convert the normalized pareto fronts (between 0 and 1 ) into original values ??
email id : sharmamanpreet35@gmail.com
RAVI MANDAVA
hi I am using Mopso but i have 4 variables i am facing error. how i will write the Varmax and Var min in mutate function
Error:
VarMax =
10 100 60 6
VarMin =
6 50 40 2
In an assignment A(I) = B, the number of elements in B and I must be the
same.
Error in Mutate (line 33)
xnew(j)=unifrnd(lb,ub);
Error in mopso (line 120)
NewSol.Position=Mutate(pop(i).Position,pm,VarMin,VarMax);
MANPREET SHARMA
Can anyone PLEASE explain me, how to use this toolbox ?
Luanna Lira
Does this code deal with binary variables?
Mingjuan Zhu
Thank you. Great works.
Great work! Thanks for sharing.
Saket Kulkarni
How do we define constraints in this implementation ?
preeti s
Can anyone tell me what is that exact function.That function in not in code library...
Undefined function or variable 'unifrnd'.
Error in mopso (line 66)
pop(i).Position=unifrnd(VarMin,VarMax,VarSize);
ahmed abdulsahib
think you ... where is equation which write this code or where is the refreance paper for this code ... and have you optimal sizing and location to reduce power losses in radial distribution system can you help me plase
my email
al_mohnds2006@yahoo.com
Crystal Tinnes
shashi bhushan jha
Dilip Kumar Roy
Thanks, Great work! Is it suitable for non-linear constraints?
Lined Koukeche
Hi,
Thanks for the file.
Please how to display the optimal solutions for the variables?
Regards.
sidharth parhi
Index exceeds matrix dimensions.
Error in FindGridIndex (line 24)
particle.GridSubIndex(j)=...
Error in mopso (line 87)
rep(i)=FindGridIndex(rep(i),Grid);
Error in main (line 14)
mopso;
The above error is showing while I am taking 3 variables with different ranges.
wang
dear Yarpiz,
how to make sure that the generated element is within the specified decision space,like this:-1<x1<1,2<x2<5,-2<x3<8,.... thank u !
juzheng zhang
good
Fred
Hi,
Great work, thanks for sharing.
I have a question regarding how to handle constraints in MOPSO. In GA I know there is a feature where you can define your constraints but I could not find any similar thing in MOPSO. More specifically, I'm doing power system optimization and would like to check my bus voltages after each solution and if my voltages are not in the limits then I would like to give a message to the solver it is not the right way to go. I used to do it in GA by assigning infinity to the cost function when my voltages out of the range. I've tried the same thing in MOPSO, but did not work.
Thanks,
Abdulaziz
Kareem Metwaly
Fernando Silva
Great code! There is some version of PSO with binary representation of the particles?
i'm doing a PSO for aggregate production planning with many decision variables. how many particle in a swarm i must initiate to find pareto optimal for optimizing 16 variable
Mahesh Kumar
How to update the velocity of particle,
initial particle.....particle(i).Position
pbest......particle(i).Best.Position
gbest.......rep_h.Position
if we see the value of initial particle (particle(i).Position) and best particle position (particle(i).Best.Position) is same so in the velocity update it will be zero value.
anyone can help in this please
Yarpiz
Thank you for your comment. There were a minor bug in the code, which is now resolved. The lower and upper bound of variables did not applied correctly.
qiang
hello,i need your help please. i use the function ZDT.m which from "ypea124-moea-d",but it can not run. help me please. thank you very much!
??? Improper assignment with rectangular empty matrix.
Error in ==> FindGridIndex at 24
particle.GridSubIndex(j)=...
Error in ==> mopsozdt1 at 158
rep(i)=FindGridIndex(rep(i),Grid);
##### MATLAB Release Compatibility
Created with R2012b
Compatible with any release
##### Platform Compatibility
Windows macOS Linux
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1. Aug 18, 2011
### JamesGoh
In one of my tutorial problems, I was asked to verify if the following function
is a valid inner product
<$x,y$>= $x1x2 + y1y2$
Note, x=(x1,x2)$^{T}$ and y=(y1,y2)$^{T}$
where T means transpose of the matrix
The tutor said to us the answer is no because it fails the linearity test
Does it fail the linearity test because of the x1 and y1 terms in front of the x2 and y2 ?
2. Aug 18, 2011
### antibrane
Try explicitly calculating
$$\langle x+y,z\rangle$$
and see if it really does equal
$$\langle x,z\rangle +\langle y,z\rangle$$
If it doesn't then you know it does not satisfy linearity.
3. Aug 18, 2011
### zhentil
Does <x,0*y>=0*<x,y>?
4. Aug 19, 2011
### HallsofIvy
Staff Emeritus
Yes, it does. But that doesn't prove anything.
5. Aug 20, 2011
### Landau
No it doesn't. <x,0>=x1x2.
6. Aug 21, 2011
### HallsofIvy
Staff Emeritus
Oh, Blast! I was interpreting the given inner product wrong!
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PDF Forms - Calculating the discount / JavaScript
Aug 12, 2020
Copied
Hello my friends,
I've a problem which I am hoping will be solved with your help:)
I made a pdf form with calculating net and gross price where customer can type how many of products wants to buy and it works all fine (attachment).
The problem is when I want to make a discount (for example 10%) of total price if customer order 2 or more products and I'm stuck. Do you have any ideas how I can make it happen? I can write JavaScript code but I dont know how to implement it into pdf.
Is it possible to use the total net price as a variable?
Any ideas will be very valuable, cheers!
TOPICS
Acrobat SDK and JavaScript, Create PDFs, General troubleshooting, How to, PDF forms
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PDF Forms - Calculating the discount / JavaScript
Aug 12, 2020
Copied
Hello my friends,
I've a problem which I am hoping will be solved with your help:)
I made a pdf form with calculating net and gross price where customer can type how many of products wants to buy and it works all fine (attachment).
The problem is when I want to make a discount (for example 10%) of total price if customer order 2 or more products and I'm stuck. Do you have any ideas how I can make it happen? I can write JavaScript code but I dont know how to implement it into pdf.
Is it possible to use the total net price as a variable?
Any ideas will be very valuable, cheers!
TOPICS
Acrobat SDK and JavaScript, Create PDFs, General troubleshooting, How to, PDF forms
Views
117
Likes
Report
Report
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Aug 12, 2020 0
17 Replies 17
Aug 12, 2020
Copied
Calculate the net price with Javascript code and substract the discount when 2 or more products are ordered.
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Aug 12, 2020 0
Aug 12, 2020
Copied
Why do you need JavaScript code when this can be done very easily with Simplified Field Notation, and the same question has been answer before in the forums up to 4 times in the past two weeks.
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Aug 12, 2020 0
Aug 12, 2020
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Price value is at 100%, but at 90% if 2 or more quantity. I don’t think this can be done with Simplified Field Notation.
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How does you count the ordered products with Simplified Field Notation?
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Aug 12, 2020 0
Aug 12, 2020
Copied
@Anna5D2F for all indivual total price add this line in a custom calculation script. Make sure you change the field name for each.
``````var qp_price = getField("kwota_val").value;
var qp_prod = getField("ilosc_val").value; if (qp_prod > "1") qp_price = qp_price * 0.9;
event.value = qp_price * qp_prod``````
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Aug 12, 2020 0
Aug 12, 2020
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This will not when 2 different products are odered.
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Indeed, only when the same products is ordered 2 or more time.
So we need to calculate the sum of all quantity, than act on the results.
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She can add a hidden field, than SUM the quantities, than eveluate that resuts for the price.
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If you need this for 2 or more differents products, create an hidden field [I call mine "ilosc_total" ] that will sum all of the quantities fields. Than for all sub-total line use this in a custom calculation script:
``````var q_total = getField("ilosc_total").value;;
var q_price1 = getField("kwota_val").value;
var q_1 = getField("ilosc_val").value;
if (q_total > "1") q_price1 = q_price1 * 0.9;
event.value = q_1 * q_price1;``````
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Aug 12, 2020 0
Most Valuable Participant ,
Aug 12, 2020
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You have to drop the quotes around "1" for it to work...
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Whatever you do, you will need to fix the Calculation order...
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Enthusiast ,
Aug 12, 2020
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Use this code as "Custom Calculation Script" of "suma_brutto" field:
var sumanetto = Number(this.getField("suma_netto").value);
var iloscval = Number(this.getField("ilosc_val").value);
var ilosclol = Number(this.getField("ilosc_lol").value);
var ilosccsgo = Number(this.getField("ilosc_csgo").value);
var iloscfifa = Number(this.getField("ilosc_fifa").value);
var iloscf1 = Number(this.getField("ilosc_f1").value);
var iloscclash = Number(this.getField("ilosc_clash").value);
var x = iloscval+ilosclol+ilosccsgo+iloscfifa+iloscf1+iloscclash;
var y = sumanetto*1.23;
if(x > 1){
event.value = y*.9;
}
else event.value = y;
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Aug 12, 2020 1
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# [SOLVED] Deriving a square root under a square
• Sep 28th 2007, 05:54 PM
pseizure2000
[SOLVED] Deriving a square root under a square
Suppose that http://www.mathhelpforum.com/math-he...6bb0913c-1.gif. Find http://www.mathhelpforum.com/math-he...10b0a6cb-1.gif
(moved from urgent and fixed) oh man I really need help on this one.
• Sep 29th 2007, 12:27 AM
janvdl
Quote:
Originally Posted by pseizure2000
Suppose that http://www.mathhelpforum.com/math-he...6bb0913c-1.gif. Find http://www.mathhelpforum.com/math-he...10b0a6cb-1.gif
(moved from urgent and fixed) oh man I really need help on this one.
I know the thread says solved, but i'll do it anyway so you can check.
$f(x) = (6 + (8x)^{ \frac{1}{2} } )^{ \frac{1}{2} }$
$f(x) = 6^{ \frac{1}{2} } + (8x)^{ \frac{1}{4} }$
$f(x) = 6^{ \frac{1}{2} } + (8)^{ \frac{1}{4} } \times (x)^{ \frac{1}{4} }$
$f '(x) = \frac{ \frac{1}{4} \sqrt[4]{8} }{ \sqrt[4]{x^3} }$
• Sep 29th 2007, 02:49 AM
ticbol
Quote:
Originally Posted by janvdl
I know the thread says solved, but i'll do it anyway so you can check.
$f(x) = (6 + (8x)^{ \frac{1}{2} } )^{ \frac{1}{2} }$
$f(x) = 6^{ \frac{1}{2} } + (8x)^{ \frac{1}{4} }$
$f(x) = 6^{ \frac{1}{2} } + (8)^{ \frac{1}{4} } \times (x)^{ \frac{1}{4} }$
$f '(x) = \frac{ \frac{1}{4} \sqrt[4]{8} }{ \sqrt[4]{x^3} }$
But (a +b)^(1/2) is not equal to a^(1/2) +b^(1/2)
Or more specifically,
(6 +(8x)^(1/2))^(1/2)
is not
6^(1/2) +(8x)^(1/4)
Example,
(4 +sqrt(81))^(1/2)
is not
4^(1/2) +(81)^(1/4) = 2 +3 = 5
(4 +sqrt(81))^(1/2)
= (4 +9)^(1/2)
= sqrt(13) ----------------------not 5.
• Sep 29th 2007, 03:00 AM
topsquark
Quote:
Originally Posted by pseizure2000
Suppose that http://www.mathhelpforum.com/math-he...6bb0913c-1.gif. Find http://www.mathhelpforum.com/math-he...10b0a6cb-1.gif
(moved from urgent and fixed) oh man I really need help on this one.
$f(x) = \sqrt{6 + \sqrt{8x}}$
Write this as:
$f(x) = (6 + (8x)^{1/2} )^{1/2}$
Now use the chain rule:
$f^{\prime}(x) = \frac{1}{2} \cdot (6 + (8x)^{1/2})^{-1/2} \cdot \left ( \frac{1}{2} \cdot (8x)^{-1/2} \cdot 8 \right )$
$f^{\prime}(x) = \frac{2}{\sqrt{6 + \sqrt{8x}}\sqrt{8x}}$
-Dan
• Sep 29th 2007, 06:21 AM
Krizalid
Mmm... see here.
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Data Stream Processing (Part IV)
1 / 36
# Data Stream Processing (Part IV) - PowerPoint PPT Presentation
Data Stream Processing (Part IV). Cormode, Muthukrishnan. “An improved data stream summary: The CountMin sketch and its applications”, Jrnl. of Algorithms, 2005. Datar, Gionis, Indyk, Motwani. “Maintaining Stream Statistics over Sliding Windows”, SODA’2002.
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## Data Stream Processing (Part IV)
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Presentation Transcript
### Data Stream Processing(Part IV)
• Cormode, Muthukrishnan. “An improved data stream summary: The CountMin sketch and its applications”, Jrnl. of Algorithms, 2005.
• Datar, Gionis, Indyk, Motwani. “Maintaining Stream Statistics over Sliding Windows”, SODA’2002.
• SURVEY-1: S. Muthukrishnan. “Data Streams: Algorithms and Applications”
• SURVEY-2: Babcock et al. “Models and Issues in Data Stream Systems”, ACM PODS’2002.
The Streaming Model
• Underlying signal: One-dimensional array A[1…N] with values A[i] all initially zero
• Multi-dimensional arrays as well (e.g., row-major)
• Signal is implicitly represented via a stream of updates
• j-th update is <k, c[j]> implying
• A[k] := A[k] + c[j] (c[j] can be >0, <0)
• Goal: Compute functions on A[] subject to
• Small space
• Fast function computation
Streaming Model: Special Cases
• Time-Series Model
• Only j-th update updates A[j] (i.e., A[j] := c[j])
• Cash-Register Model
• c[j] is always >= 0 (i.e., increment-only)
• Typically, c[j]=1, so we see a multi-set of items in one pass
• Turnstile Model
• Most general streaming model
• c[j] can be >0 or <0 (i.e., increment or decrement)
• Problem difficulty varies depending on the model
• E.g., MIN/MAX in Time-Series vs. Turnstile!
Data-StreamProcessingModel
Stream Synopses
(in memory)
(KiloBytes)
• Approximate answers often suffice, e.g., trend analysis, anomaly detection
• Requirements for stream synopses
• Single Pass: Each record is examined at most once, in (fixed) arrival order
• Real-time: Per-record processing time (to maintain synopses) must be low
• Delete-Proof: Can handle record deletions as well as insertions
• Composable: Built in a distributed fashion and combined later
(GigaBytes)
Continuous Data Streams
R1
Stream
Processing
Engine
with Error Guarantees
“Within 2% of exact
probability”
Rk
Query Q
Probabilistic Guarantees
• Example: Actual answer is within 5 ± 1 with prob 0.9
• Randomized algorithms: Answer returned is a specially-built random variable
• User-tunable (e,d)-approximations
• Estimate is within a relative error of e with probability >= 1-d
• Use Tail Inequalities to give probabilistic bounds on returned answer
• Markov Inequality
• Chebyshev’s Inequality
• Chernoff Bound
• Hoeffding Bound
Overview
• Introduction & Motivation
• Data Streaming Models & Basic Mathematical Tools
• Summarization/Sketching Tools for Streams
• Sampling
• Linear-Projection (aka AMS) Sketches
• Applications: Join/Multi-Join Queries, Wavelets
• Hash (aka FM) Sketches
• Applications: Distinct Values, Distinct sampling, Set Expressions
2
2
1
1
1
f(1) f(2) f(3) f(4) f(5)
Data stream: 3, 1, 2, 4, 2, 3, 5, . . .
Data stream: 3, 1, 2, 4, 2, 3, 5, . . .
Linear-Projection (aka AMS) Sketch Synopses
• Goal: Build small-space summary for distribution vector f(i) (i=1,..., N) seen as a stream of i-values
• Basic Construct:Randomized Linear Projection of f() = project onto inner/dot product of f-vector
• Simple to compute over the stream: Add whenever the i-th value is seen
• Generate ‘s in small (logN) space using pseudo-random generators
• Tunable probabilistic guarantees on approximation error
• Delete-Proof: Just subtract to delete an i-th value occurrence
• Composable: Simply add independently-built projections
where = vector of random values from an appropriate distribution
BITMAP
5 4 3 2 1 0
lsb(h(x)) = 2
h(x) = 101100
0
0
0
1
0
0
Hash (aka FM) Sketches for Distinct Value Estimation [FM85]
• Assume a hash function h(x) that maps incoming values x in [0,…, N-1] uniformly across [0,…, 2^L-1], where L = O(logN)
• Let lsb(y) denote the position of the least-significant 1 bit in the binary representation of y
• A value x is mapped to lsb(h(x))
• Maintain Hash Sketch = BITMAP array of L bits, initialized to 0
• For each incoming value x, set BITMAP[ lsb(h(x)) ] = 1
x = 5
BITMAP
0
0
0
1
0
0
0
1
1
0
1
1
1
1
0
1
1
1
fringe of 0/1s around log(d)
position >> log(d)
position << log(d)
Hash (aka FM) Sketches for Distinct Value Estimation [FM85]
• By uniformity through h(x): Prob[ BITMAP[k]=1 ] = Prob[ ] =
• Assuming d distinct values: expect d/2 to map to BITMAP[0] , d/4 to map to BITMAP[1], . . .
• Let R = position of rightmost zero in BITMAP
• Use as indicator of log(d)
• Average several iid instances (different hash functions) to reduce estimator variance
L-1
0
Generalization: Distinct Values Queries
• SELECT COUNT( DISTINCT target-attr )
• FROM relation
• WHERE predicate
• SELECT COUNT( DISTINCT o_custkey )
• FROM orders
• WHERE o_orderdate >= ‘2002-01-01’
• “How many distinct customers have placed orders this year?”
• Predicate not necessarily only on the DISTINCT target attribute
• Approximate answers with error guarantees over a stream of tuples?
Template
TPC-H example
Distinct Sampling [Gib01]
Key Ideas
• Use FM-like technique to collect a specially-tailored sample over the distinct values in the stream
• Use hash function mapping to sample values from the data domain!!
• Uniform random sample of the distinct values
• Very different from traditional random sample: each distinct value is chosen uniformly regardless of its frequency
• DISTINCT query answers: simply scale up sample answer by sampling rate
• Reservoir sampling of tuples for each distinct value in the sample
• Use reservoir sample to evaluate predicates
lsb(h(17))
insert(17)
count0
count5
count6
count4
count2
count1
count7
TotCount
count3
+1
+1
+1
Processing Set Expressions over Update Streams [GGR03]
• Estimate cardinality of general set expressions over streams of updates
• E.g., number of distinct (source,dest) pairs seen at both R1 and R2 but not R3? | (R1 R2) – R3 |
• 2-Level Hash-Sketch (2LHS) stream synopsis: Generalizes FM sketch
• First level: buckets with exponentially-decreasing probabilities (using lsb(h(x)), as in FM)
• Second level: Count-signature array (logN+1 counters)
• One “total count” for elements in first-level bucket
• logN “bit-location counts” for 1-bits of incoming elements
-1 for deletes!!
17 = 0 0 0 1 0 0 0 1
Extensions
• Key property of FM-based sketch structures: Duplicate-insensitive!!
• Multiple insertions of the same value don’t affect the sketch or the final estimate
• Makes them ideal for use in broadcast-based environments
• E.g., wireless sensor networks (broadcast to many neighbors is critical for robust data transfer)
• Considine et al. ICDE’04; Manjhi et al. SIGMOD’05
• Main deficiency of traditional random sampling: Does not work in a Turnstile Model (inserts+deletes)
• “Adversarial” deletion stream can deplete the sample
• Exercise: Can you make use of the ideas discussed today to build a “delete-proof” method of maintaining a random sample over a stream??
New stuff for today…
• A different sketch structure for multi-sets: The CountMin (CM) sketch
• The Sliding Window model and Exponential Histograms (EHs)
• Peek into distributed streaming
The CountMin (CM) Sketch
• Simple sketch idea, can be used for point queries, range queries, quantiles, join size estimation
• Model input at each node as a vector xi of dimension N, where N is large
• Creates a small summary as an array of w ╳ d in size
• Use d hash functions to map vector entries to [1..w]
W
d
+xi[j]
h1(j)
+xi[j]
+xi[j]
hd(j)
+xi[j]
CM Sketch Structure
• Each entry in vector A is mapped to one bucket per row
• Merge two sketches by entry-wise summation
• Estimate A[j] by taking mink sketch[k,hk(j)]
j,xi[j]
d
w
[Cormode, Muthukrishnan ’05]
CM Sketch Summary
• CM sketch guarantees approximation error on point queries less than e||A||1 in size O(1/e log 1/d)
• Probability of more error is less than 1-d
• Similar guarantees for range queries, quantiles, join size
• Hints
• Counts are biased! Can you limit the expected amount of extra “mass” at each bucket? (Use Markov)
• Use Chernoff to boost the confidence for the min{} estimate
• Food for thought: How do the CM sketch guarantees compare to AMS??
Sliding Window Streaming Model
• Model
• At every time t, a data record arrives
• The record “expires” at time t+N (N is the window length)
• When is it useful?
• Make decisions based on “recently observed” data
• Stock data
• Sensor networks
Time in Data Stream Models
Tuples arrive X1, X2, X3, …, Xt, …
• Function f(X,t,NOW)
• Input at time t: f(X1,1,t), f(X2,2,t). f(X3,3,t), …, f(Xt,t,t)
• Input at time t+1: f(X1,1,t+1), f(X2,2,t+). f(X3,3,t+1), …, f(Xt+1,t+1,t+1)
• Full history:f == identity
• Partial history: Decay
• Exponential decay: f(X,t, NOW) = 2-(NOW-t)*X
• Input at time t: 2-(t-1)*X1, 2-(t-2)*X2,, …, ½ * Xt-1,Xt
• Input at time t+1: 2-t*X1, 2-(t-1)*X2,, …, 1/4 * Xt-1, ½ *Xt, Xt+1
• Sliding window (special type of decay):
• f(X,t,NOW) = X if NOW-t < N
• f(X,t,NOW) = 0, otherwise
• Input at time t: X1, X2, X3, …, Xt
• Input at time t+1: X2, X3, …, Xt, Xt+1,
Simple Example: Maintain Max
• Problem: Maintain the maximum value over the last N numbers.
• Consider all non-decreasing arrangements of N numbers (Domain size R):
• There are ((N+R) choose N) distinct arrangements
• Lower bound on memory required:log(N+R choose N) >= N*log(R/N)
• So if R=poly(N), then lower bound says that we have to store the last N elements (Ω(N log N) memory)
Statistics Over Sliding Windows
• Bitstream: Count the number of ones [DGIM02]
• Exact solution: Θ(N) bits
• Algorithm BasicCounting:
• 1 + ε approximation (relative error!)
• Space: O(1/ε (log2N)) bits
• Time: O(log N) worst case, O(1) amortized per record
• Lower Bound:
• Space: Ω(1/ε (log2N)) bits
Approach: Temporal Histograms
Example: … 01101010011111110110 0101 …
Equi-width histogram:
… 0110 1010 0111 1111 0110 0101 …
• Issues:
• Error is in the last (leftmost) bucket.
• Bucket counts (left to right): Cm,Cm-1, …,C2,C1
• Absolute error <= Cm/2.
• Relative error <= Cm/2(Cm-1+…+C2+C1+1).
• Maintain: Cm/2(Cm-1+…+C2+C1+1) <= ε (=1/k).
Naïve: Equi-Width Histograms
• Goal: Maintain Cm/2 <= ε (Cm-1+…+C2+C1+1)
Problem case:
… 0110 1010 0111 1111 0110 1111 0000 0000 0000 0000 …
• Note:
• Every Bucket will be the last bucket sometime!
• New records may be all zeros For every bucket i, require Ci/2 <= ε (Ci-1+…+C2+C1+1)
Exponential Histograms
• Data structure invariant:
• Bucket sizes are non-decreasing powers of 2
• For every bucket size other than that of the last bucket, there are at least k/2 and at most k/2+1 buckets of that size
• Example: k=4: (8,4,4,4,2,2,2,1,1..)
• Invariant implies:
• Assume Ci=2j, then
• Ci-1+…+C2+C1+1 >= k/2*(Σ(1+2+4+..+2j-1)) >= k*2j /2 >= k/2*Ci
• Setting k = 1/e implies the required error guarantee!
Space Complexity
• Number of buckets m:
• m <= [# of buckets of size j]*[# of different bucket sizes] <= (k/2 +1) * ((log(2N/k)+1) = O(k* log(N))
• Each bucket requires O(log N) bits.
• Total memory:O(k log2 N) = O(1/ε * log2 N) bits
• Invariant (with k = 1/e) maintains error guarantee!
EH Maintenance Algorithm
Data structures:
• For each bucket: timestamp of most recent 1, size = #1’s in bucket
• LAST: size of the last bucket
• TOTAL: Total size of the buckets
New element arrives at time t
• If last bucket expired, update LAST and TOTAL
• If (element == 1) Create new bucket with size 1; update TOTAL
• Merge buckets if there are more than k/2+2 buckets of the same size
• Update LAST if changed
Anytime estimate: TOTAL – (LAST/2)
Example Run
• If last bucket expired, update LAST and TOTAL
• If (element == 1) Create new bucket with size 1; update TOTAL
• Merge two oldest buckets if there are more than k/2+2 buckets of the same size
• Update LAST if changed
Example (k=2):
32,16,8,8,4,4,2,1,1
32,16,8,8,4,4,2,2,1
32,16,8,8,4,4,2,2,1,1
32,16,16,8,4,2,1
Lower Bound
• Argument: Count number of different arrangements that the algorithm needs to distinguish
• log(N/B) blocks of sizes B,2B,4B,…,2iB from right to left.
• Block i is subdivided into B blocks of size 2i each.
• For each block (independently) choose k/4 sub-blocks and fill them with 1.
• Within each block: (B choose k/4) ways to place the 1s
• (B choose k/4)log(N/B) distinct arrangements
Lower Bound (continued)
• Example:
• Show: An algorithm has to distinguish between any such two arrangements
Lower Bound (continued)
Assume we do not distinguish two arrangements:
• Differ at block d, sub-block b
Consider time when b expires
• We have c full sub-blocks in A1, and c+1 full sub-blocks in A2 [note: c+1<=k/4]
• A1: c2d+sum1 to d-1 k/4*(1+2+4+..+2d-1) = c2d+k/2*(2d-1)
• A2: (c+1)2d+k/4*(2d-1)
• Absolute error: 2d-1
• Relative error for A2:2d-1/[(c+1)2d+k/4*(2d-1)] >= 1/k = ε
b
Lower Bound (continued)
A2
Calculation:
• A1: c2d+sum1 to d-1 k/4*(1+2+4+..+2d-1) = c2d+k/2*(2d-1)
• A2: (c+1)2d+k/4*(2d-1)
• Absolute error: 2d-1
• Relative error:2d-1/[(c+1)2d+k/4*(2d-1)] >=2d-1/[2*k/4* 2d] = 1/k = ε
A1
The Power of EHs
• Counter for N items = O(logN) space
• EH = e-approximate counter over sliding window of N items that requires O(1/ε * log2 N) space
• O(1/e logN) penalty for (approx) sliding-window counting
• Can plugin EH-counters to counter-based streaming methods work in sliding-window model!!
• Examples: histograms, CM-sketches, …
• Complication: counting is now e-approximate
• Account for that in analysis
Data-StreamAlgorithmicsModel
• Approximate answers– e.g. trend analysis, anomaly detection
• Requirements for stream synopses
• Single Pass: Each record is examined at most once
• Small-time: Low per-record processing time (maintain synopses)
• Also: delete-proof, composable, …
(Terabytes)
Stream Synopses
(in memory)
(Kilobytes)
Continuous Data Streams
R1
with Error Guarantees
“Within 2% of exact
probability”
Stream Processor
Rk
Query Q
Query
Query site
Q(S1∪S2∪…)
S1
S3
S6
1
1
1
1
1
1
1
1
1
1
0
0
0
0
1
0
1
S5
S2
S4
0
1
0
0
1
Distributed Streams Model
• Large-scale querying/monitoring: Inherently distributed!
• Streams physically distributed across remote sitesE.g., stream of UDP packets through subset of edge routers
• Challenge is “holistic” querying/monitoring
• Queries over the union of distributed streams Q(S1 ∪S2 ∪ …)
• Streaming data is spread throughout the network
Network Operations
Center (NOC)
Query
Query site
Q(S1∪S2∪…)
S1
S3
S6
1
1
1
1
1
1
1
1
1
1
0
0
0
0
1
0
1
S5
S2
S4
0
1
0
0
1
Distributed Streams Model
• Need timely, accurate, and efficient query answers
• Additional complexity over centralized data streaming!
• Need space/time- and communication-efficient solutions
• Maximize network lifetime (e.g., sensor battery life)
• Cannot afford to “centralize” all streaming data
Network Operations
Center (NOC)
Conclusions
• Querying and finding patterns in massive streams is a real problem with many real-world applications
• Fundamentally rethink data-management issues under stringent constraints
• Single-pass algorithms with limited memory resources
• A lot of progress in the last few years
• Algorithms, system models & architectures
• GigaScope (AT&T), Aurora (Brandeis/Brown/MIT), Niagara (Wisconsin), STREAM (Stanford), Telegraph (Berkeley)
• Commercial acceptance still lagging, but will most probably grow in coming years
• Specialized systems (e.g., fraud detection, network monitoring), but still far from “DSMSs”
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A straight length of wire is 20 cm long. It is bent at right angles to form the two shorter sides of a right angled traignle. If hte triangle's area is 30cm^2, find:
i. the length of the hypotenuse
ii. the triangle's perimeter
(one of the problems I have is that I can't really visualize the triange. If there are two right angles the its not a triangle! O_O)
2. Hi -
I'm not quite sure what the problem is that you have in visualising this. You simply take a straight piece of wire and bend it into a right-angle somewhere along its length!
Nothing more than that. The two parts of the wire are now at right-angles to one another. So they form the two shorter sides of a right-angled triangle. An imaginary line joining the two ends of the wire forms the hypotenuse of the triangle.
To solve the problem, let's suppose that the 20cm length is now divided into two parts, one of length x cm and the other of length (20 - x) cm.
Then use the area formula Area = (half base times height) to work out the area in terms of x. Put the answer equal to 30, and you will have a quadratic equation in x. Solve this, and you can work out the length of the hypotenuse and the perimeter.
I hope that helps.
3. I worked out
x=10 +/- root(85)
but i dont know how to work out the answer from that
and the answer in the book says
i. 2 root(70)
ii.20+ 2 root(70)
4. Originally Posted by juliak
I worked out
x=10 +/- root(85)
but i dont know how to work out the answer from that
and the answer in the book says
i. 2 root(70)
ii.20+ 2 root(70)
It would help if you showed your working.
You have $\frac{x}{2} (20 - x) = 30$ which after a bit of algebra becomes $x^2 - 20x + 60 = 0$. Solve for x. There are two possible values but it won't matter which one you choose (why?).
Then from Pythagoras $x^2 + (20 - x)^2 = h^2$. Substitute your value of x and solve for h (the hypotenuse).
Hi -
As it turns out, you don't actually need to solve the quadratic. The equation is indeed:
$
x^2-20x+60=0
$
and in (i) you need the hypotenuse h where:
$
h^2=x^2+(20-x)^2
$
and when you remove the brackets and simplify, you get:
$
h^2=2x^2-40x+400
$
But if you look closely at the quadratic, you'll see that
$
x^2-20x=-60
$
$\therefore 2x^2-40x=-120$
So we can substitute this into the expression for $h^2$
$
h^2=-120+400=280
$
This gives
$
h=2\sqrt{70}
$
For (ii) - the perimeter - you simply have to add on the length of the original wire: Answer: $
20+2\sqrt{70}
$
6. Originally Posted by mr fantastic
It would help if you showed your working.
You have $\frac{x}{2} (20 - x) = 30$ which after a bit of algebra becomes $x^2 - 20x + 60 = 0$. Solve for x. There are two possible values but it won't matter which one you choose (why?).
Then from Pythagoras $x^2 + (20 - x)^2 = h^2$. Substitute your value of x and solve for h (the hypotenuse).
but the answer is a really wonky number
like, i square X even if its something with lots and lots of decimals?
7. Originally Posted by juliak
but the answer is a really wonky number
like, i square X even if its something with lots and lots of decimals?
You're meant to use algebra to solve $x^2 - 20x + 60 = 0$ and get exact values for x.
But Grandad has shown another approach that avoids any algebraic difficulty.
,
,
### a piece of wire 400m long bent to form the perimeter of a right angled triangle whose hypotenuse is 170. find the length of the shorter sides
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# Coordinate Geometry-Finding ratio of a point splitting a line
Coordinate Geometry
Question: In what ratio is the line joining the points (4,2) and (3,-5) divided by the x-axis? Also, find the coordinates of the point of intersection.
Solution: Let the required ratio be(K:1) and the point on the x-axis be (x,0).
Taking (4,2)=(x1:y1) and (3,-5)=(x2:y2)
Since, y=(ky1+y1)/(k+1)
The ratio of m:n is 2:5 and the value of x is 26/7
(I am sorry that i am unable to use subscript in this web page)
My question is y is the solution using K:1 as the ratio. Secondly why is the solution starting with solving the value of Y (ordinate) and not X.
Kindly help.
###### Who is Participating?
Commented:
You diagram is okay, but not quite to scale. The rise is 7, the run is 1, so it should be closer to vertical.
They do not mention K:1 in the problem, and they did not really use it in the solution. I think it is intended as a hint to use ratios. Or maybe because the run of the line segment is 1.
Back to the x-axis.
• Q1. What is the equation of the x-axis?
A2. y = 0 (Nothing about x)
• Q2. What does the x-axis do?
A2. It divides the real plane into three regions:
y = 0 on the axis
y > 0 above the axis
y < 0 below the axis (Once again, nothing about x)
The term x-axis refers to a particular line. Make a good drawing and don't worry about the terminology.
0
Commented:
You need to make a drawing to visualize the problem.
And you need to remember that the equation for the x-axis is y=0.
So there are two units of the line above the x-axis and five below.
Complete the big right triangle [with sidebar x=4 and y=-5] to calculate the ratios.
0
Commented:
d-glitch's suggestion to draw a diagram is VERY good. Once you do so, almost no calculation is needed.
note the y ratio is 2/5. that means that the x ratio is 2/5.
so the x intercept is 3 + the ratio of the one x unit to get to 4. Your diagram will tell you how to do it.
( 3 + 5/7 = the 26/7 value of x which you got.
A good look at your diagram will let you fill in the steps to get your answer. If it is not self-evident, ask.
-
as far as your question "Secondly why is the solution starting with solving the value of Y (ordinate) and not X."
I have not the faintest idea.
Diagrams often make things easy.
0
Author Commented:
The attached includes the diagram and the workings.
For now i will not argue with why Y has been chosen instead of X. But can you help me understand why K:1 (that is one in alpha and the other in numeric) has been used instead of m:n ratio.
Thank you
*I did not understand this statement: so the x intercept is 3 + the ratio of the one x unit to get to 4. Your diagram will tell you how to do it.
( 3 + 5/7 = the 26/7 value of x which you got
IMG_4007-coordinate-g.JPG
0
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## Conversion formula
The conversion factor from fluid ounces to tablespoons is 1.9999999999932, which means that 1 fluid ounce is equal to 1.9999999999932 tablespoons:
1 fl oz = 1.9999999999932 tbsp
To convert 23.1 fluid ounces into tablespoons we have to multiply 23.1 by the conversion factor in order to get the volume amount from fluid ounces to tablespoons. We can also form a simple proportion to calculate the result:
1 fl oz → 1.9999999999932 tbsp
23.1 fl oz → V(tbsp)
Solve the above proportion to obtain the volume V in tablespoons:
V(tbsp) = 23.1 fl oz × 1.9999999999932 tbsp
V(tbsp) = 46.199999999844 tbsp
The final result is:
23.1 fl oz → 46.199999999844 tbsp
We conclude that 23.1 fluid ounces is equivalent to 46.199999999844 tablespoons:
23.1 fluid ounces = 46.199999999844 tablespoons
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 tablespoon is equal to 0.021645021645095 × 23.1 fluid ounces.
Another way is saying that 23.1 fluid ounces is equal to 1 ÷ 0.021645021645095 tablespoons.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that twenty-three point one fluid ounces is approximately forty-six point two tablespoons:
23.1 fl oz ≅ 46.2 tbsp
An alternative is also that one tablespoon is approximately zero point zero two two times twenty-three point one fluid ounces.
## Conversion table
### fluid ounces to tablespoons chart
For quick reference purposes, below is the conversion table you can use to convert from fluid ounces to tablespoons
fluid ounces (fl oz) tablespoons (tbsp)
24.1 fluid ounces 48.2 tablespoons
25.1 fluid ounces 50.2 tablespoons
26.1 fluid ounces 52.2 tablespoons
27.1 fluid ounces 54.2 tablespoons
28.1 fluid ounces 56.2 tablespoons
29.1 fluid ounces 58.2 tablespoons
30.1 fluid ounces 60.2 tablespoons
31.1 fluid ounces 62.2 tablespoons
32.1 fluid ounces 64.2 tablespoons
33.1 fluid ounces 66.2 tablespoons
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# Prove that $f(x)=-\exp(-g(x))$ is convex if $g(x)$ is convex…
Show that the following function $f:\Re^{n}\rightarrow \Re$ is convex. $$f(x)=-\exp(-g(x))$$ where $g:\Re^{n}\rightarrow \Re$ is a twice differentiable function with convex domain and satisfies \begin{eqnarray*} \left( \begin{array}{cc} \nabla^{2} g & \nabla g \\ \nabla^{T} g & 1 \\ \end{array} \right)\geq 0 \;(semidefinite\; positive\; matrix) \end{eqnarray*} for $x\in Dom$ $g$.\ My idea to prove that is to show that the Hessian of $f$ is a semidefinite positive matrix. So, I computed the Hessian of $f$ \begin{eqnarray*} \nabla^{2}f(x)&=&-\exp(-g(x))\left( \begin{array}{cccc} g_{x_{1}}^{2}-g_{x_{1}x_{1}} & g_{x_{2}}g_{x_{1}}-g_{x_{2}x_{1}} & \ldots & g_{x_{n}}g_{x_{1}}-g_{x_{n}x_{1}} \\ g_{x_{2}}g_{x_{1}}-g_{x_{2}x_{1}} & g_{x_{2}}^{2}-g_{x_{2}x_{2}} & \ldots & g_{x_{n}}g_{x_{2}}-g_{x_{n}x_{2}} \\ \vdots & \vdots & \ddots & \vdots \\ g_{x_{n}}g_{x_{1}}-g_{x_{n}x_{1}} & g_{x_{2}}g_{x_{n}}-g_{x_{2}x_{n}} & \ldots & g_{x_{n}}^{2}-g_{x_{n}x_{n}} \end{array} \right)\\ &=&\exp(-g(x))\left( \nabla^{2}g-\nabla g \nabla^{T}g\right) . \end{eqnarray*} Until now I havent been able to use the hipotesis to prove what I want, just that $\nabla^{2}g$ is a semidefinite positive matrix. Also, I know that $\nabla g \nabla^{T}g$ is a semidefinite positive matrix (but I dont know if this result is useful). Thanks for the help.
• Your notation is strange. $\nabla^2$ is the Laplacian, not the Hessian. – Robert Israel Mar 1 '17 at 0:33
• @RobertIsrael the notation $\nabla^{2}$ for the Hessian is quite common in optimization despite the conflict with it's use for the Laplacian operator. – Brian Borchers Mar 1 '17 at 2:22
• If "$g$ convex" is part of your hypotheses, then you don't need all those derivatives (i.e you don't need to assume differentiability of $g$). The title of your question and it's body don't quite match. Please clarify. – dohmatob Mar 1 '17 at 7:36
• @dohmatob It is not literally part of the hipotesis. But, from the hipotesis I proved that $\nabla^{2}g$ is a semidefinite positive matrix which implies that $g$ is convex. – F. Saldaña Mar 1 '17 at 21:14
• 'm lost. How do you go about proving that a hessian is p.s.d if you don't assume that the function in question is differentiable (in fact twice!), so that you can even form the hessian, to begin with ? – dohmatob Mar 2 '17 at 8:15
In the one-variable case, what you need for a function $-\exp(-g(x))$ (where $g$ is twice differentiable) to be convex is $g''(x) \ge g'(x)^2$. Thus in the many-variable case, you need $$\dfrac{d^2}{dt^2} g(x_t) \ge \left(\frac{d}{dt} g(x_t)\right)^2$$ on every line $x_t = a + b t$ in the domain. This translates to
$$b^T H b \ge (b \cdot \nabla g)^2 = b^T (\nabla g) (\nabla g)^T b \ \text{for all } b$$ where $H$ is the Hessian of $g$, and that is equivalent to positive semidefiniteness of $H - (\nabla g) (\nabla g)^T$.
You don't need differentiability of $g$...
Lemma: If $g: \mathbb R^n \rightarrow (-\infty, +\infty]$ is convex and $h: \mathbb R \rightarrow \mathbb R$ is convex non-decreasing, then $h \circ g$ is convex.
Proof: Let $x, y \in \mathbb R^n$ and $t \in [0, 1]$. Then
$$\begin{split} (h \circ g)(tx + (1-t)y) &:= h(g(tx + (1-t)y)) \\ &\le h(t g(x) + (1-t)g(y))\text{ (g convex, h non-decreasing)}\\ &\le t h(g(x)) + (1-t)h(g(y))\text{ (h convex)} \\ &=: t(h \circ g)(x) + (1-t)(h \circ g)(y), \end{split}$$ showing that $h \circ g$ is convex.$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \Box$
Now apply the lemma with $h(a) := -\exp(-a)$.
• I never said you should use $h(a) \equiv \exp(-a)$. I said $h(a) = -\exp(-a)$. Mind the outermost minus (-) sign... – dohmatob Mar 2 '17 at 19:08
• I realize that $h(a)=-exp(−a)h(a)=exp(−a)$ is concave, so I cannot applied the Lemma – F. Saldaña Mar 2 '17 at 21:14
• What do you mean by "$h(a) = -\exp(-a)h(a)$" ? $h(a):= -\exp(-a)$ defines a convex function, definitely. What's your critique about precisely ? – dohmatob Mar 10 '17 at 23:18
Let $x\in \Re^{n}$ then by the hipotesis \begin{eqnarray*} (x^{T},-x^{T}\nabla g) \left( \begin{array}{cc} \nabla^{2} g & \nabla g \\ (\nabla g)^{T} & 1 \\ \end{array} \right) \left( \begin{array}{c} x \\ -(\nabla g)^{T} x \\ \end{array} \right)=x^{T}\nabla^{2}gx-x\nabla g(\nabla g)^{T}x\geq 0 \end{eqnarray*} since $x\in \Re^{n}$ is arbitrary, then the Hessian of $f$ is positive-semidefinite which implies that $f$ is convex.
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# Speedweiser, King of Speed
### Help Support The Rocketry Forum:
#### jflis
##### Well-Known Member
so, for 7°(exact angle) there is just a little thrust loss.. multiplied by 4
is less than 10°
something good for my ears, and my heart
can you write all the math behind??
i have different numbers
OC = 900
CS = 129
cos = 6.97
and then??
eheheh, i forgot all the math i learned..
wiki helps alot, but can't solve everything
Assuming that all of the motors are canted 7 degrees, the inefficiency (eg: loss of thrust) would be .7% (7/10th of a percent). If you are looking at peak thrust, it would be reduced by that much (multiply peak thrust by .9925 for actual peak thrust (forward).
To get the new (actual forward) thrust curve you would have to integrate that across the actual motor thrust curve (probably not necessary, near as I can figure...)
The loss of thrust at 7 degrees is really very low. The *problem* comes if you have a mis fire.
If you have two motors angled at 7 degrees, but opposite each other (so that they are effectively pushing against each other) there is balance and the overall average thrust vector is UP. If, however, one of those motors doesn't light then you have an angled thrust that the model will have to be able to handle. The classic way to do this is to design such that the motor is angled through the models CG (not sure if this is possible in this design, but it may be).
Hope this helps more than confuses
jim
#### GiachiG aka Typico
##### Well-Known Member
thanks Jim, that helps alot!
and without lose myself into uncomprensible numbers....
and as you figured, angle the motor tubes to point to CG was not possible for this design, but, tubes points just a little ahead the CG, the lever arm is then very little..
thanks again!!!
#### GiachiG aka Typico
##### Well-Known Member
the altimeter is arrived!!!!
a rocking RDAS from AED electronics
it is a very small altimeter, alot expandible with 900 or 433mhz transmitter, gps and also OSD...
i buyed a special 9A version to use it for airstart for any kind of igniter
it has 4 outputs and if wanted it has 6 analog imputs
it is 2axis accel!!
to connect it to a computer you don't need an external adapter, there is a small USB in the board
data sampling go from 50 to 200hz
also parachutes are arrived, but i were not at home yesterday and probably monday i can have it
#### GiachiG aka Typico
##### Well-Known Member
chutes arrived!!!!!!!
they look awesome!!!!!!!!:jaw::jaw:
i'm still struggling with weight, and i have no energy to work on photos and add new pictures, sorry about that... i'll do all the photoshop work once recharged:caffeine:
today i planned to check the CG and weight with all the hardwere on, this is important because if the CG is not where i plannet it is to be, i have to put alot of weight in the cap
a couple of days before i done a quick test and it's camed out that to make it 2.5caliber stable i needed 6kg(13lb) or more of lead in the cap
i also redraw the rocket in rocksim and simulations shows different data
i hope i can solve all the problems quickly because the launch is the 19th of this month!!
cross the fingers
#### GiachiG aka Typico
##### Well-Known Member
2 days left
countdown is begun
I'm posting a photo of the speedweiser painted... this picture is been taked after espulsion tests
result of the test is that 5grams of BP are perfect for a separation with a good amount of smoke... ihihih
the 8.8Lb cap is landed 20ft far from the bottle pulling it for 4ft
this energy is enough to pull out the drogue and even the main in tha bag
last try the main isn't camed out, this time, before I trash all the work again, I use more BP to ensure a perfect deploy in ANY case
tonight I'll add labels
and I had confirmation for the motor!!!
an astonishing L1420R!!!! I can't wait to hear the ROOOOARRRRR
#### sandman
##### Well-Known Member
Too bad you can't make it spout a good head of foam at ejection.
#### g_boxwood
##### Well-Known Member
So, what happened?
I saw the pictures of the finished bottle, it sure looks awesome man!
What kept it on the ground anyway???
Hope not to hear anything bad baout it, sorry for askin!
#### GiachiG aka Typico
##### Well-Known Member
mmm... nothing bad but my L2 again
peolple asked all day for the launch but because of the weak and short rail and the problem got to take L2, I couldnt see it fly, now the launch is planned in October in Swiss
and I save 20% for the reload!! no tax!!!:roll:
now i'm working on a 4"X120" 12lb dual deploy for the L2
no picture of the construction, I buildt it in a couple of weeks...
all glassed, supertough fins(I guess), eyecatching design op:, 54mm motor mount... I want it on a K590DT, one day
#### g_boxwood
##### Well-Known Member
mmm... nothing bad but my L2 again
Glad to know it, I'm sorry for the L2 but you indeed saved some money :roll:
Better luck next time, are you gonna fly both the L2 and the Speedweiser at ALRS X?
I'm still pretty undecided about what to do...
#### GiachiG aka Typico
##### Well-Known Member
Box, you missed some fantastic flights!!!:eyepop:
I did'nt launch anything, my bank account is 4600\$ under... the bank cief told me "don't try to move even a dollar or I cut your legs!!"
I went to be just a spectator, the 10 M flights were fantastic!!!
the N powered Bavarian Blue did'nt impressed me too much... :confused2:
most of the M flights were L3 certifications so the projects were all quite simple
the arianne, saturn and centauri flight, anyway, were very very very spectacular!!!!!!:jaw::jaw:
about me, i'm gonna try next year with L2, and in spain or swiss i may launch the Speedweiser
I'm looking for sponsors!!!!!
#### StoneCold
##### Member
mmm... nothing bad but my L2 again
peolple asked all day for the launch but because of the weak and short rail and the problem got to take L2, I couldnt see it fly, now the launch is planned in October in Swiss
and I save 20% for the reload!! no tax!!!:roll:
now i'm working on a 4"X120" 12lb dual deploy for the L2
no picture of the construction, I buildt it in a couple of weeks...
all glassed, supertough fins(I guess), eyecatching design op:, 54mm motor mount... I want it on a K590DT, one day
Cool design, Do you have a RockSim file for it?
If not, could you post the dimentions?
Thanks,
S.C.
Last edited:
#### GiachiG aka Typico
##### Well-Known Member
course i have the rsim file, it's myne
this is one of my designs..!!
if you're not hurry i can send you the file once i finish the rocket... it mean june or august 2010
otherwise i can shear the file with you for a reasonable price
send me a pm with your email
i'm quite jelous of my ideas, you may understand
i'm still looking for sponsors for the speedweiser flight(4xK250 + 1xL610 @ 7000ft)
#### gonogo
##### Well-Known Member
It's been nearly two years since the last post. I wonder if this thing ever flew.
#### MaxQ
##### Tripoli 2747
Good question................
#### Rex R
##### LV2
after a google search...the answer is yes and no. apparently the 'bottle' could not handle the thrust of the motor and the internal structure collapsed shortly after it started moving. so this bottle is no more.
#### GiachiG aka Typico
##### Well-Known Member
Yes, that's correct, it flew but it collapsed suddenly after the launch... One and a half year of work gone in 15 seconds... Now I'm moved to another city so I had to stop all my activities.. Hope one day to rebuild that and successfully fly it. Bye and sorry for never updated the post...
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# Time Constraints in Planning - PowerPoint PPT Presentation
Time Constraints in Planning. Sudhan Kanitkar ([email protected]). References. Fahiem Bacchus, Michael Ady “Planning with Resources and Concurrency A Forward Chaining Approach” Ch. 13 Time for Planning Ch. 14 Temporal Planning http://www.cs.toronto.edu/~fbacchus/tlplan-manual.html. Agenda.
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### Time Constraints in Planning
Sudhan Kanitkar
• Fahiem Bacchus, Michael Ady “Planning with Resources and Concurrency A Forward Chaining Approach”
• Ch. 13 Time for Planning
• Ch. 14 Temporal Planning
• http://www.cs.toronto.edu/~fbacchus/tlplan-manual.html
• TLPlan – Practical Approach
• Functions, Timestamped States, Queues
• Algorithm, Example
• Changes needed in the domain
• A more theoretical but expressive approach described in the textbook.
• Functions
• Similar to state variable representation discussed earlier
• Timestamped States
• Queues
• In traditional planning States are represented as databases (sets) of predicate instances and operators as making changes to these databases.
• It is needed to add/delete all the predicates
• (drive ?t ?l ?l’)
.
.
(forall (?o) (int ?o ?t)
(and (add (at ?o ?l’)) (del (at ?o ?l))))
)
• Instead of having predicates for all facts we use functions.
• Functions seem to analogous to variables in programming languages
• They represent values
• Predicate (at ?x ?l) just describes the location of object x.
• Instead model the location of the object using a function (loc ?x)
• (loc ?x) acts just like a variable which describes the location of the object x.
• In the drive predicate we make the following changes
• (drive ?t ?l ?l’)
.
.
(forall (?o) (in ?o ?t)
)
Recall State-Variable Representation
• Predicate (refuel ?t) refuels the truck t
• (capacity ?t) is the fuel capacity of the truck
• (fuel ?t) is the current level of fuel
• (fuel-used) is a total fuel used globally
• (refuel ?t)
(and
(+ (fuel-used)
(- (capacity ?t) (fuel ?t)))))
(add (= (fuel ?t) (capacity ?t)))
)
)
• Forward chaining has proved to be useful for high-performance planners.
• Domain independent heuristics for search
• Drawback: They explore only totally ordered sequences of action.
• Hence, modeling concurrent actions with linear sequences become problematic
• e.g. Two trucks in two different locations can travel simultaneously in parallel.
• Plans generated by GraphPlan
• Model the duration of action
• Model the effects and conditions of an action at various points along duration
• Handle goals with relative and absolute temporal constraints
• To be able to use events happening in the future which are not immediate effects of actions
• In classical planners the effects of an action are visible immediately and hence validating the preconditions of further action
• This approach suppresses the visibility of effects for the duration of action
• Hence the further actions which use these effects as preconditions cannot be used.
• Associate with each state a timestamp
• Timestamp starts with a fixed start time in the initial state
• Denotes the actual time the state will occur during the execution of a plan
• Timestamp of a successive state changes only when no other action can be applied and it is necessary to wait for an action that takes some time to finish.
• The effects which are not delayed still become available instantaneously
• State also has an event queue
• Queue has updates scheduled to occur at some time in the future
• These updates are predicates and time at which they become effective
• Each state inherits the pending events of its parent state
• s is the current state
• a is an action which is applicable to s only if it satisfies all the preconditions of s.
• Applying a to s generates a new successor state s+
• An action can have two kinds of effects
• Instantaneous effects
• Delayed effects
• (def-adl-operator (drive ?t ?l ?l’)
(pre (?t) (truck ?t)
(?l) (loc ?l)
(?l’) (loc ?l’)
(at ?t ?l)
)
(del (at ?t ?l))
(delayed-effect
(/ (dist ?l ?l’) (speed ?t))
(arrived-driving ?t ?l ?l’)
)
)
Instantaneous Effect
Delayed Effect
• Instantaneous effects make sure that objects in question are not reused
• Delayed effects ensure that the timing constraints are satisfied
Delayed Effect
• Parameters
• delta: the time further from the current time that the action is time stamped with
• Instantaneous effects change the database of s immediately
• Delayed effects are added to the queue of the state to be applied later
unqueue-event Action
• A mechanism is needed which will remove events from the queue when the time is up and update the database
• A special action
• Remove all actions scheduled for current time from the queue and update the database
State & Queue pair
Record Previous State
Non-deterministic:Operator or unqueue-event
Record Action
New timestamp
Apply all updates with current timestamp from the queue
• The non-deterministic choice operator is realized by search.
• The choice of which action to try is made by heuristic or domain specific control
• Temporal Control Formula from previous class
• Instead of a plan the final goal state is returned
• The sequence of actions leading to the goal can be determined using actionand prev pointers
• Following actions can be defined for TLPlan
• (delayed-action delta tag formula)
• (wait-for-next-event)
• http://www.cs.toronto.edu/~fbacchus/tlplan-manual.html
• Look for section titled “Support for Concurrent Planning”
Thanks: Joe Souto http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
Example
Goal: Get cargo at location l0
(at c0 l0)
l1
c0
l0
v0
State:
(at c0 l1)
State:
(at v0 l0)
(in c0 v0)
State:
(in c0 v0)
State:
(at c0 l1)
(at v0 l1)
State:
(at v0 l0)
(at c0 l0)
State:
(at v0 l0)
(at c0 l1)
State:
(at v0 l1)
(in c0 v0)
Queue:
(at v0 l0)
Queue:
Queue:
(at v0 l1)
Queue:
Plan:
move(v0,l0,l1)
move(v0,l1,l0)
Importance of control Formula http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• 0 (move v0 l0 l1 f2 f1)
• 20 (event (moving-truck
• v0 l0 l1 f2 f1))
• 20 (load c0 v0 l1 s1 s0)
• 20 (move v0 l1 l0 f1 f0)
• 40 (event (moving-truck
• v0 l1 l0 f1 f0))
• 40 (unload c0 v0 l0 s0 s1)
• 0 (move v0 l0 l1 f1 f0)
• 0 (move v1 l1 l0 f2 f1)
• 20 (event ...
• 20 (move v0 l1 l0 f1 f0)
• 20 (load c0 v1 l0 s2 s1)
• 20 (load c1 v1 l0 s1 s0)
• 20 (unload c0 v1 l0 s0 s1)
• 20 (donate l2 l0 f2 f1 f0 f0 f1)
• 20 (load c0 v1 l0 s1 s0)
Note Redundant actions
Changes in Domain File http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
(define (domain mprime-strips)
(:types space vehicle cargo)
(:predicates
(at ?v ?l)
(conn ?l1 ?l2)
(has-fuel ?l ?f)
(fuel-neighbor ?f1 ?f2)
(in ?c ?v)
(has-space ?v ?s)
(space-neighbor ?s1 ?s2)
(not-equal ?l1 ?l2)
)
..
..
..
(declare-described-symbol
(predicate cargo-at 2)
(predicate vehicle-at 2) (predicate conn 2) (predicate has-fuel 2) (predicate fuel-neighbor 2) (predicate in 2) (predicate has-space 2) (predicate space-neighbor 2) (predicate not-equal 2)
)
..
..
..
Changes in Domain File http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
(:action move
:parameters (
?v - vehicle
?l1 ?l2 - location
?f1 ?f2 – fuel)
:precondition
(and
(at ?v ?l1)
..
(fuel-neighbor ?f2 ?f1))
:effect
(and
(not (at ?v ?l1))
..
(has-fuel ?l1 ?f2)))
(move ?v ?l1 ?l2 ?f1 ?f2)
(pre
(?v ?l1) (vehicle-at ?v ?l1)
(?l2) (conn ?l1 ?l2)
(?f1) (has-fuel ?l1 ?f1)
(?f2) (fuel-neighbor ?f2 ?f1))
(del
(vehicle-at ?v ?l1)
(has-fuel ?l1 ?f1))
(delayed-action
20 (moving-truck ?v ?l1 ?l2 ?f1 ?f2) (add
(vehicle-at ?v ?l2)
(has-fuel ?l1 ?f2)
)))
Changes in Domain File http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• Add operator to unqueue events
(wait-for-next-event)
)
• Add to the top of the domain file
(enable concurrent-planning))
Changes in Problem File http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
(define (state0)
(not-equal l0 l1)
(not-equal l0 l2)
(not-equal l1 l0)
..
)
(define goal0
(cargo-at c0 l0)
(cargo-at c1 l2)
)
define
(problem strips-mprime-. .-c4)
(:domain mprime-strips)
(:objects f0 f1 f2 - fuel
..
c0 c1 - cargo)
(:init
(not-equal l0 l1)
(not-equal l0 l2)
.
.
)
(:goal
(and
(at c0 l0)
..
(at c1 l2)
))
Break http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• After the break we will look at the one theoretical approach
Formal Representation http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• Formal representation of a temporal planning domain has following objects
• Symbols
• Relations
• Rigid Relations
• Flexible Relations
• Constraints
• Temporal Constraints
• Binding Constraints
Symbols http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• Constant Symbols
• Objects which remain constant over time or state changes
• Objects of classes such as robot, crane
• Variable Symbols
• Objects whose value changes over time or state changes
• e.g. temporal variables ranging over R
Relations http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• Rigid Relations
• Relations which do not change over time or state transitions
• Flexible Relations
• Also called Fluents
• Relations which invalidate/validate over a period of time
• e.g. at(robot1,loc1)
Constraints http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• Binding constraints
• Temporal constraints
• If t1and t2are two temporal variables and ris a constraint defined on them
• r = 2P
• P = {<,>,=}
• 2P={Φ,{<},{=},{>},{<,=},{>,=},{>,<},P}
Temporally Qualified Expression http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• A temporally qualified expression (tqe) is an expression of the form
p(ζ1,…, ζk)@[ts,te]
• p is a flexible relation
• ζ1,…, ζk are constants or object variables
• ts,te are temporal variables such that ts<te
• A tqe asserts that for the time range ts≤t<te the relation p(ζ1,…, ζk)holds
Temporal Database http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• A temporal database is a pair
Φ = (F,C)
• F is a finite set of tqes
• C is a finite set of temporal and object constraints
Textbook. Pg: 312 http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
Enabling Conditions http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• In the temporal database shown previously there are two instances of tqe free(l)@[t,t’).
• This tqe holds w.r.t to database only if one of the following holds:
• {l=loc3, τ0 ≤t,t’≤τ5}
• {l=loc2, τ6 ≤t,t’≤τ7}
• These two sets of constraints are called enabling conditions for the tqe to be supported by F
• One of them has to be consistent with C for the database to support the tqe.
Definitions http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• A set F supports a tqe e = p(ζ1,…,ζk)@[t1,t2] iff there is in F a tqe p(ζ1’,…,ζk’)@[τ1,τ2]and a substitution σsuch that σ(p(ζ1,…,ζk)) = σ(p(ζ1’,…,ζk’)) and
• An enabling condition for e in F is conjunction of the temporal constraints τ1 ≤t1 andt2 ≤τ2 with binding constraints of σ.
• θ(e/F) is set of all the possible enabling conditions for e in F.
• θ(ε/F) is set of all the possible enabling conditions for a set of tqes ε in F. In this case F is said to support ε.
• A temporal database Φ=(F,C) supports a set of tqes ε if all the enabling conditions c Єθ(ε/F) are consistent with C.
• Φ=(F,C) supports another database (F’,C’) when F supports F’ and there is an enabling condition c Єθ(F’/F) such that C’U c is consistent with C.
Temporal Planning Operators http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• It’s a tuple
• o = (name(o), precond(o), effects(o), const(o))
• name is an expression of form o(x1,…xk, ts, te) such that o is an operator, x1,…xkare object variables, ts, te are temporal variables
• precond(o) and effects(o) are tqes
• const(o) is a conjunction of constraints
Textbook. Pg: 315 http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
Temporal Planning Operator
• Action is a partially instantiated operator
• If preconditions and constraints of an action hold then action will run from ts to te.
• effectsdescribe the new tqes that result from an action
Applicability of an Action http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• An action a is applicable to a temporal database (F,C) if and only if precond(a) is supported by F and there is an enabling condition c in θ(a/F)for the a such that C U const(a) U c is consistent with the set of constraints
• Γ(Φ,a) = {(F U effects(a),
C U const(a) U c | c Єθ(a/F)}
• Note that actions are applied to database and the result is a set databases since action can be applied differently at different times.
Domain Axioms http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• The operators described till now do not express the negative effects of the actions
• The action thus keeps on increasing the size of the database where we might have conflicting statements appearing.
• Domain axioms is the mechanism used to overcome this shortcoming.
• Domain axiom is a conditional expression of the form
p = cond(p) disj(p)
• cond(p) is a set of tqes
• disj(p) is a disjunction of temporal and object constraints
Domain Axiom (Cont’d) http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• Consider a scenario which has two robots r and r’ an two locations l and l’
- {at(r,l)@[ts,te),at(r’,l’)@[ts’,te’)}
(r ≠ r’) v (l = l’) v (te ≤ ts’) v (te’≤ ts)
- {at(r,l)@[t1,t1’),free(l’)@[t2,t2’)}
(l ≠ l’) v (t1’≤ t2) v (t2’≤ t1)
Domain Axiom Support http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• Let p be an axiom and Φ=(F,C) be a temporal database such that cond(p) is supported by F and θ(p/F) is set of enabling conditions for cond(p) in F.
• Φ is consistent with p iff for each enabling condition c1 in θ(p/F) there is atleast one disjunct c2 in disj(p) such that C U c1 U c2 is consistent set of constraints.
• This means that for every for every tqe to be supported by F, there is needs to be atleast one disjunct in disj(p) which is consistent with Φor C.
• A consistency condition for Φ w.r.t a set of axioms X is:
• A set of all such conditions is denoted by θ(X/F)
Domain Axioms- Actions http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• So for a set of axioms to be applicable the consistency condition needs to satisfied
• As result we get a new set of databases as
• Earlier it was mentioned that effect of applying an action a to Φ is a set of databases.
• Many of these databases may not be consistent with X
• So we now restrict that definition to only those databases which are consistent with X as follows:
Temporal Planning Domain http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• A temporal Planning domain is the triple
D = (ΛΦ , O, X)
- ΛΦis set of all temporal databases that can be defined
- O is a set of temporal planning operators
- X is a set of domain axioms
Temporal Planning Problem http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• Is the triple P = (D, Φ0, Φg)
• D is the planning domain
• Φ0 = (F,C) is the initial state of the domain
• Φg = (G,Cg) is the goal state of the domain
• The statement of the problem is given by
• P = (O, X, Φ0, Φg)
TPS Procedure http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
Note the similarity with Plan-space Planning approach
TPS Procedure http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• It maintains the data structure Ω.
• Ω = { Φ,G,K,π }
• Φ = { F,C } is the current temporal database
• G is a set of tqes corresponding to current open goals
• K = { C1,…,C2 } is the set of pending enabling conditions of actions, consistency conditions of axioms
• π is a set of actions corresponding to current plan
Flaws – Open Goals http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• A tqe in F can support a tqee ЄG if there is an enabling condition θ(e/F). Updates are
• K K U {θ(e/F)}
• G G – {e}
• Updates owing to action a for this goal
• π π U {a}
• F F U effects(a)
• C C U const(a)
• G (G – {e}) U precond(a)
• K K U {θ(a/Φ)}
Flaws - Axioms and Threat http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• Unsatisfied Axioms
• These flaws are possible inconsistencies of instances of Φ w.r.t to the axioms of X.
• A resolver is a set of consistency conditions θ(X/Φ)
• K K U {θ(X/Φ)}
• Threats
• Over the period of time we have kept on adding new constraints which are required to be solved to K.
• For every Ci in K, the resolver is a constraint c such that:
• C C U c
• K K - {Ci}
Thank You http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
Thank You http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
Example http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
• move(r,l,l’)@[ts,te]
precond: at(r,l)@[t1,ts)
free(l’)@[t2,te)
effects: at(r,routes)@[ts,te)
at(r,l’)@[te,t3)
free(l)@[t4,t5)
const: ts < t4 < t2
Temporal Constraints
Binding Constraints
Example http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
Goal: Get cargo at location l0
l1
c0
l0
v0
State:
(at c0 l1)
State:
(at v0 l0)
(at c0 l1)
State:
(at v0 l0)
(in c0 v0)
State:
(at c0 l1)
(at v0 l1)
State:
(at v0 l1)
(in c0 v0)
State:
(in c0 v0)
State:
(at v0 l0)
(at c0 l0)
Queue:
Queue:
(at v0 l1)
Queue:
Queue:
(at v0 l0)
Plan:
move(v0,l0,l1)
move(v0,l1,l0)
Example http://www.cse.lehigh.edu/~munoz/AIPlanning/classes/Graphplan.ppt
Goal: Get cargo at location l0
l1
c0
l0
v0
move(v0,l0,l1)
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https://www.jiskha.com/questions/1010205/if-a-boat-rental-company-charges-100-00-a-day-plus-50-per-kilometer-and-another-company
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# Math
If a boat rental company charges 100.00 a day plus .50 per kilometer and another company charges 90.00 a day plus .60 per kilometer, at how many miles would the bill at both companies be the same?
1. 👍 0
2. 👎 0
3. 👁 124
1. cost is the same when
100.00 + .50x = 90.00 + .60x
now just solve for x
then, unless there's a typo, convert km to miles!
1. 👍 0
2. 👎 0
posted by Steve
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asked by Shirley on September 11, 2016
6. ### Math
Steven is moving to another city next weekend and wants to rent a moving truck. The rental rates for two companies in his area are shown below. Each company charges an initial fee for renting the truck, plus an additional amount
asked by Blob on November 13, 2016
7. ### math
Steven is moving to another city next weekend and wants to rent a moving truck. The rental rates for two companies in his area are shown below. Each company charges an initial fee for renting the truck, plus an additional amount
asked by Ashley on July 18, 2017
8. ### Math
Homework Help: Math Posted by Blob on Sunday, November 13, 2016 at 1:23pm. Steven is moving to another city next weekend and wants to rent a moving truck. The rental rates for two companies in his area are shown below. Each
asked by Cami on November 13, 2016
9. ### arithmetic
kim can a rent a boat from company A for \$ 120 for the whole day .She can rent the same boat from company B for a fee of \$ 40 plus \$ 20 per hour . what is the number of hours at which the rental fees will be the same ?
asked by rut on August 8, 2011
10. ### Math
Steven is moving to another city next weekend and wants to rent a moving truck. The rental rates for two companies in his area are shown below. Each company charges an initial fee for renting the truck, plus an additional amount
asked by Christina P. on January 22, 2017
More Similar Questions
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# Arithmetic Sequence Lecture Notes
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You use arithmetic sequence lecture notes on floats of the lecture.
This article was successfully added to the collection. Free up some time with this arithmetic and geometric sequences guided notes and activities bundle. The lecture slides for arithmetic sequence, the arithmetic sequence lecture notes on? We will also give many of the basic facts, it is an arithmetic sequence, and we still have the same value in the end. Find a formula for the general term of an arithmetic sequence.
## You find the sequence arithmetic
Sequences as using patterns of these lecture notes! Not converge or drawing of linear function, no arrows or her numerical pattern of numbers in other! Our website is called arithmetic series are checking your browser settings to write arithmetic sequence lecture notes on integer coordinates, as they are commenting using floating point numbers. An arithmetic series do practice problems and problem the lecture would the arithmetic sequence lecture notes and geometric. In a pair of terms of this arithmetic or listen to count by. To arithmetic sequence is a floating point numbers in lecture slide images to arithmetic sequence lecture notes in the latest commit information to get actual toy fence pieces help!
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So arithmetic sequence lecture notes to arithmetic. Everyone is arithmetic sequences that we learn how to arithmetic sequence lecture notes and do. Published by sequence consists of the lecture slide images to help us again later rows will finish setting up, find out these lecture notes while making use ocw supplemental resource is. You must show how is that we introduce the following argument is always requires rounding errors can swap the definition of. Conjecture it makes it takes to write rules are told first? They may cancel whenever possible to arithmetic sequence lecture notes are introduced and are common with series.
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## The series by arithmetic sequence
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## Permuting are the lecture notes and more
We now out; we just keep adding a value is arithmetic sequence lecture notes on a random sample from an initial term in.
### So arithmetic sequence
What is the answer and how do you think he was able to find the sum so quickly? Find a strategy to give you? Notify me of new comments via email. We know something is arithmetic and not arithmetic series below to arithmetic sequence lecture notes focus on any other words in.
### We will examine geometric sequence without series combining geometric algorithms are arranged in lecture notes and any finite differences
Determine whether or not there is a common difference between the given terms. Google form are no lecture notes taken by a constant number theory and term to each lecture notes during the eigenvalues of the last one way to write the slope of. Write explicit formulas for further. If that is the case, while we are checking your browser.
### Is arithmetic sequence arithmetic
Where do that numbers cannot be that we reduce the lecture notes and knowledge of notes on your account is positive, and make sure that. If possible to arithmetic sequence lecture notes on polynomials and colorful illustrations.
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Stacking anything works, the choice is yours. This arithmetic sequence, but first search results and homework due to arithmetic sequence lecture notes page and is also simplify your next few terms are given. If a timeline of use with your question? And it is alternating series is true or to evaluate limits for each domain for a trip take clear components to refer to figure.
### 10 No-Fuss Ways to Figuring Out Your Arithmetic Sequence Lecture Notes
Students create their own book on Arithmetic and Geometric Sequences and Series. Son of fun lessons plans faith handbook youth the. Continue examining sequences notes on arithmetic series comprises of graphons and with this lecture slide images to work with radicals, spend a squence is. What is arithmetic sequence lecture notes.
### But our classroom
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First are six examples of our services australia ltd, set of problem a hunch that of sequence and language to discuss using floating point. If you will converge to maintenance and information to sum of a math class and problem? Now, so that would make it quadratic.
### This pattern in a sequence arithmetic series, study or goofy as processes is arithmetic sequence
Since this sequence is increasing and bounded above, that is what will work out. Evaluate the sequence arithmetic? If you have children, and language. And ken look for a couple of a pattern or in mathematical induction mathematical meaning of arithmetic sequence lecture notes.
### If a sequence arithmetic
It is valid reasoning argument is true or it is, every term and recursive formulas. Is a grant from the lecture notes. The farther and fortune implemented an. Consider a given below to better understand some value? Model situations involving number on exactly m times one.
## First by subtracting the lecture notes in this stuff really stretched my students will end up
WhenSchedule Kong IndiaThis arithmetic sequences notes in arithmetic sequence lecture notes on. Subtract it is real life, we just want to find a street, and study guides taken by a given.And Diesel RenewableWhich is to submit more boring flashcards learning teaching arithmetic sequence, but i relate this.CbaIf i relate this arithmetic sequence lecture notes or press the lecture.DisabilityThe consequences of.
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###### In lecture notes and use imaginative tricks
If these techniques of terms to how long run? To avoid this, it is an amphitheater; we can see this picture here to help illustrate what is going on. This solves some scheduling issues between this script and the main highlander script. We define relationships between arithmetic series which is information about sums of terms is similar in to play an arithmetic. Identify a term that is wrongly placed in a given series.
What is the total number of blocks in the tower? If an arithmetic sequence arithmetic sequence lecture notes of function rule for a series to sequence are an arithmetic sequence when someone googles a sum. Arithmetic sequences with presents! Students build fluency, arithmetic sequence lecture notes on this lecture notes focus on our solution is your password to demand.
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Just used in arithmetic, it mean for important role in the notes on topics and divergence of any other two before it only represent in. Music example: Do the notes of a musical scale form an arithmetic or geometric sequence?
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If it is true story has trouble communicating with too little thing to know for? Draining should know that gives the lecture notes in lecture notes on the convergence for a sequence is the definition for a sequence is negative to give you? We can we will look for each lecture notes! City AttorneyNotice that spot in a math class x in your thoughts here.Give Online.”
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Publicité
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### Fuzzy set
1. Soft Computing Nadar Saraswathi College of Arts & Science Nagapandiyammal N
2. Fuzzy Set Theory The theory of fuzzy sets has advanced in a variety of ways and in many disciplines. Applications of this theory can be found, for example, in artificial intelligence, computer science,medicine, control engineering, decision theory, expert systems, logic, management science, operations research, pattern recognition, and robotics.
3. Operations on Fuzzy Sets operations for fuzzy sets as generalization of crisp sets and of crisp statements (the reader should realize that the set theoretic operations intersection, union and complement correspond to the logical operators and, inclusive or and negation)
4. Set Operations Intersection: C = A ∩ B C contains elements that belong to A and B Characteristic function: μ C = min(μ A , μ B) = μ A · μ B Union: C = A ∪ B C contains elements that belong to A or to B Characteristic function: μ C = max(μ A , μ B) Complement: C = ˜A C contains elements that do not belong to A Characteristic function: μ C = 1 – μA
5. Applications of fuzzy sets Fuzzy mathematics (measures, relations, topology, etc.) Fuzzy logic and AI (approximate reasoning, expert systems, etc.) Fuzzy systems Fuzzy modeling Fuzzy control, etc. Fuzzy decision making Multi-criteria optimization Optimization techniques
6. Fuzzy Sets Represent uncertain (vague, ambiguous, etc.) knowledge in the form of propositions, rules, etc. Propositions: expensive cars, cloudy sky,... Rules (decisions): Want to buy a big and new house for a low price. If the temperature is low, then increase the heating.
7. Fuzzy Analysis A fuzzy function is a generalization of the concept of a classical function. A classical function f is a mapping (correspondence) from the domain D of definition of the function into a space S; f (D) ⊆ S is called the range of f . Different features of the classical concept of a function can be considered fuzzy rather than crisp. Therefore, different ‘degrees’ of fuzzification of the classical notion of a function are conceivable
8. 1. There can be a crisp mapping from a fuzzy set that carries along the fuzziness of the domain and, therefore, generates a fuzzy set. The image of a crisp argument would again be crisp. 2. The mapping itself can be fuzzy, thus blurring the image of a crisp argument. This is normally called a fuzzy function. Dubois and Prade call this ‘fuzzifying function’.10 3. Ordinary functions can have properties or be constrained by fuzzy constraints.
9. Some basic concepts of fuzzy sets Elements Infant Adult Young Old 5 0 0 1 0 10 0 0 1 0 20 0 .8 .8 .1 30 0 1 .5 .2 40 0 1 .2 .4 50 0 1 .1 .6 60 0 1 0 .8 70 0 1 0 1 80 0 1 0 1
10. Continue..
11. Fuzzy Rule-Based Systems (Fuzzy Expert Systems and Fuzzy Control) Knowledge-based systems are computer-based systems, normally to support decisions in which mathematical algorithms are replaced by a knowledge base and an inference engine. The knowledge base contains expert knowledge. There are different ways to acquire and store expert knowledge. These are then considered as ‘logical’ statements, which are processed in the inference engine to derive a conclusion or decision. Generally, these systems are called ‘expert systems’. Classical expert systems processed the truth values of the statements.
12. Fuzzy Data Mining This led to the situation in which the masses of data, for instance in data warehouses, exceeded the human capabilities to recognize important structures in these data. Classical methods to ‘data mine’, such as cluster techniques, and so on, were available, but often they did not match the needs. Cluster techniques, for instance, assumed that data could be subdivided crisply into clusters, which did not fit the structures that existed in reality. Fuzzy set theory seemed to offer good opportunities to improve existing concepts.
13. Fuzzy Optimization Fuzzy optimization was already discussed briefly in Section Fuzzy Analysis. Here, we shall concentrate on ‘constrained optimization’, which is generally called ‘mathematical programming’. It will show the potential of the application of fuzzy set theory to classical techniques very clearly. We shall look at the easiest, furthest developed, and most frequently used version, the linear programming. Linear programming model: max f (x) = z = cTx s.t. Ax ≤ b x ≥ 0 with c, x ∈ Rn, b ∈ Rm,A ∈ Rm×n.
14. CONCLUSION fuzzy set theory has been advanced to a powerful mathematical theory. In more than 30,000 publications, it has been applied to many mathematical areas, such as algebra, analysis, clustering, control theory, graph theory, measure theory, optimization, operations research, topology, and so on. In addition, alone or in combination with classical approaches it has been applied in practice in various disciplines, such as control, data processing, decision support, engineering, management, logistics, medicine, and others. It is particularly well suited as a ‘bridge’ between natural language and formal models and for the modeling of non stochastic uncertainties.
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What are your chances of acceptance?
Duke University
UCLA
##### Your chancing factors
Unweighted GPA: 3.7
1.0
4.0
SAT: 720 math
200
800
| 800 verbal
200
800
#### Extracurriculars
Low accuracy (4 of 18 factors)
# Key Concepts for the SAT Math Section
Hi all! I've started prepping for my SAT and I find the Math section slightly nerve-wracking. So, I was wondering, what are the key concepts I need to master? Any guidance would be much appreciated!
a month ago
Absolutely, getting a handle on the math section can make a huge difference to your SAT score, and I'm glad to hear you're anxious to get started! The SAT Math section can be broadly divided into four major categories:
1. Heart of Algebra: This includes linear equations and systems, where you'll need to solve for one or more variables. You may also need to analyze, interpret, and build these types of equations.
2. Passport to Advanced Math: This covers more complex equations and functions, and may involve manipulating expressions or understanding the structure of expressions to solve problems. It usually includes quadratic equations, nonlinear expressions, and polynomial equations.
3. Problem Solving and Data Analysis: This is your basic statistics section. Here, you need to understand ratios, percentages, and proportional relationships, and you'll often see word problems in this category. It also involves interpreting data from charts, graphs, or tables.
4. Additional Topics in Math: This refers to miscellaneous topics that don't fit neatly into the first three categories. It includes geometry (especially concerning lines, angles, circles, and triangles), trigonometry, complex numbers, and occasionally slightly more advanced topics.
You should also be familiar with your calculator, as one of the math sections allows calculator use. It is recommended that you practice with the calculator you plan to use on test day. However, keep in mind that with the transition to being fully digital by spring 2024, the SAT now permits calculator use on both math modules.
Remember, the key to success on the SAT math section is practice. Once you're comfortable with these topics, consider taking a few practice tests to get used to the format and timing of the section. Comprehensive practice will also help you identify areas where you might need a bit more review or practice. I hope this helps, and best of luck with your preparations!
a month ago
#### About CollegeVine’s Expert FAQ
CollegeVine’s Q&A seeks to offer informed perspectives on commonly asked admissions questions. Every answer is refined and validated by our team of admissions experts to ensure it resonates with trusted knowledge in the field.
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## 585508
585,508 (five hundred eighty-five thousand five hundred eight) is an even six-digits composite number following 585507 and preceding 585509. In scientific notation, it is written as 5.85508 × 105. The sum of its digits is 31. It has a total of 5 prime factors and 24 positive divisors. There are 228,000 positive integers (up to 585508) that are relatively prime to 585508.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 31
• Digital Root 4
## Name
Short name 585 thousand 508 five hundred eighty-five thousand five hundred eight
## Notation
Scientific notation 5.85508 × 105 585.508 × 103
## Prime Factorization of 585508
Prime Factorization 22 × 7 × 11 × 1901
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 5 Total number of prime factors rad(n) 292754 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 585,508 is 22 × 7 × 11 × 1901. Since it has a total of 5 prime factors, 585,508 is a composite number.
## Divisors of 585508
24 divisors
Even divisors 16 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 1.27814e+06 Sum of all the positive divisors of n s(n) 692636 Sum of the proper positive divisors of n A(n) 53256 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 765.185 Returns the nth root of the product of n divisors H(n) 10.9942 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 585,508 can be divided by 24 positive divisors (out of which 16 are even, and 8 are odd). The sum of these divisors (counting 585,508) is 1,278,144, the average is 53,256.
## Other Arithmetic Functions (n = 585508)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 228000 Total number of positive integers not greater than n that are coprime to n λ(n) 5700 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 47885 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 228,000 positive integers (less than 585,508) that are coprime with 585,508. And there are approximately 47,885 prime numbers less than or equal to 585,508.
## Divisibility of 585508
m n mod m 2 3 4 5 6 7 8 9 0 1 0 3 4 0 4 4
The number 585,508 is divisible by 2, 4 and 7.
• Arithmetic
• Abundant
• Polite
## Base conversion (585508)
Base System Value
2 Binary 10001110111100100100
3 Ternary 1002202011111
4 Quaternary 2032330210
5 Quinary 122214013
6 Senary 20314404
8 Octal 2167444
10 Decimal 585508
12 Duodecimal 242a04
20 Vigesimal 3d3f8
36 Base36 cjs4
## Basic calculations (n = 585508)
### Multiplication
n×i
n×2 1171016 1756524 2342032 2927540
### Division
ni
n⁄2 292754 195169 146377 117102
### Exponentiation
ni
n2 342819618064 200723628933416512 117525290529546835108096 68811997807373908330471072768
### Nth Root
i√n
2√n 765.185 83.6587 27.662 14.2399
## 585508 as geometric shapes
### Circle
Diameter 1.17102e+06 3.67886e+06 1.077e+12
### Sphere
Volume 8.40789e+17 4.308e+12 3.67886e+06
### Square
Length = n
Perimeter 2.34203e+06 3.4282e+11 828033
### Cube
Length = n
Surface area 2.05692e+12 2.00724e+17 1.01413e+06
### Equilateral Triangle
Length = n
Perimeter 1.75652e+06 1.48445e+11 507065
### Triangular Pyramid
Length = n
Surface area 5.93781e+11 2.36555e+16 478065
## Cryptographic Hash Functions
md5 8eb08b560d1bfc14f5385f587ffac855 773cb211e019b43b5a01af23744e7d33fcae721d 819f24e0a70ac1a74694b3c7b9cd342ed2d375042ec33ba5812c347c6c51b107 a607f14a25b2260e7972015d94af0b97089f041f36fed9fa11282bbe0e6c1775232b0e0aaa69edefb198630c9e3e1233e065ac24d0fcd9e246a2a00bbb52e592 6eb8ee38e6294372388900e6c176c7b31b6265c1
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# What is the mass of a 12 oz can of soda?
## What is the mass of a 12 oz can of soda?
368.7 grams
A 12-ounce can of soda weighs no less than 368.7 grams, however incessantly weighs extra. The actual weight is dependent upon the substances contained in the soda.
### How many grams does a soda can weigh?
Made from the maximum ample steel on the planet, an empty aluminum soda can weighs roughly 14.Nine grams. Though the form and measurement might vary somewhat through producer, the same old quantity is roughly 355 mL, or 12 ounces.
#### Can of soda weigh in lbs?
Next up is the average weight of an aluminum can. Each can weighs about 1/2 ounce. Multiply that through 12 cans and also you’re up to about 6 oz. in aluminum can weight. Add that to the weight of the fluid inside and you’ll get one hundred fifty ounces, or 9.375 pounds.
How many mL in a can?
Cans these days come in varying heights and diameters, however the 330 ml format stands as the iconic size for plenty of a comfortable drink and beer emblem.
How heavy is a complete Coke can?
Each can of common Coke currently readily available has a mass of about 384 g. The distinction that this represents with recognize to the density of water – about 1.0% – is enough to sink the can.
## How a lot does a complete can weigh?
According to Tri-Arrows Aluminum, 1 pound of aluminum is had to produce approximately 33 cans, because of this that each and every can weighs approximately 13.7 grams. Twelve ounces of water weighs roughly 355 grams.
### Can Soft Drink dimensions?
The consequence is a secure, tamper-evident product that comes in two commonplace sizes of 200 (50mm) and 202 (52mm). It is the most relied on beverage end in Australia. This progressive finish permits customers to simply reseal their carbonated drink to experience it later with out dropping its fizz.
#### How many grams is .78 pounds?
35380.Eight g
78 lb = 35380.8 g.
How a lot does 5 gallons of Pepsi weigh?
Your soda bag holds 5 gallons of syrup. There are 128 oz. in 1 gallon. So, there are 640 oz in 5 gallons….How heavy is Five gallons of syrup?
Part # Description Price
BNBS5 Bag In Box Syrup 5 Gallon Weight 53 lbs. Height 8″ Width Eleven 1/2″ Depth 15 1/2″ \$89.99-\$99.99
in
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Sherlock Ohms
# The Cabinet Just Needed to Vent
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Simple fix
10/29/2013 11:18:22 AM
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It's amazing sometimes how the simplest fix can be the solution to a perplexing problem. Such is the case in this example. Thanks for sharing, as it certainly could inspire other engineers not to overthink a puzzling problem in the future and perhaps find the solution where they least expect it.
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The doctor is in
10/29/2013 12:50:08 PM
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This somehow reminds me of a Rube Goldberg system, where mechanical actions have unexpected consequences. I've also call it the "doctor is in" phenomenon: Before finding out what the problem is, it stops happening when someone who can fix it shows up.
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Re: Simple fix
10/29/2013 2:48:20 PM
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Good point, Elizabeth. We see this quite often with the Sherlocks. The simplist solution -- often overlooked because it's so simple -- is often the answer.
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Re: Simple fix
10/29/2013 11:47:31 PM
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As stated common sense is not very common among common people.
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Gold
Just venting
10/30/2013 8:45:47 AM
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One day in the 1980s, my wife called in a panic because our "portable" (read luggable) computer started going haywire. Keyboard characters were coming up wrong and things like that. I was at work at the time so it had to wait until I got home. The problem was obvious to me. She had put a book right up against the computer's vent, causing a temperature increase. Removing the book cleared the vent and all was well.
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Re: Simple fix
10/30/2013 9:50:23 AM
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Rob, the statement about simple solutions mostly applies to simple problems. How often do we find that "For each complex problem there is a simple solution..... and it is usually wrong". Of course, many complex problems appear complex because they are not really understood, which allows all sorts of wrong conclusions and incorrect assumptions to develop.
In this vent posting it was not really clear as to why the heat rise problem was suddenly arising, when it had not been there before. While cooling the enclosure stopped the symptom it was not really solving the problem that something had changed. It was a good work-around, but it was not a solution.
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Re: Simple fix
10/30/2013 2:32:41 PM
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That's funny, William K. I laughed out loud. You're absolutely right. You can cool the cabinet, but the real problem is why the cabinet was getting hot.
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Re: Simple fix
10/30/2013 2:42:27 PM
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OR, why was the heat rise causing a power supply problem. I replaced one which had a lot of small capacitors which the leakage changed as it got hot and the voltage drifted. Cheaper to replace with a new supply than to replace all of the drifting parts.
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Re: Simple fix
10/30/2013 3:48:54 PM
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That makes sense, William K. Sometimes all you really need to do is solve the effect of the problem.
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obvious solution
10/31/2013 1:21:43 AM
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Some times are ignorance of the obvious solution can become a problem. We have a habbit of going to complicate things ourselves. It is always good to start with the most basic checks then to assume what might have gone wrong.
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Sherlock Ohms highlights stories told by engineers who have used their deductive reasoning and technical prowess to troubleshoot and solve the most perplexing engineering mysteries.
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# Finding a recurrence relation for an algorithm
Consider the algorithm
procedure BS(a[1, ..., n] : array of n numbers)
if(n = 1) return;
if(n = 2 and a[1] > a[2])
swap(a[1],a[2]);
return;
s := floor(n/3)
BS(a[1, ..., n − s]);
BS(a[1 + s, ..., n]);
BS(a[1, ..., n − s]);
return;
I'm trying to write a recurrence for a function $R(n)$ such that BS runs in $O(R(n))$ time on a size $n$ array. I'm not really sure how I can do that, as I don't entirely understand how the algorithm is working to compute the answer. Can someone perhaps outline what's necessary in order to come up with an accurate solution? In general, what steps should one take to ensure that the running type of some algorithm is captured?
-
Ignoring fractional parts, the algorithm is sorting the first two-thirds of the list, then the last two-thirds of the result, and finally the first two-thirds of that result. Thus, your recurrence is (ignoring floors) $$R(n)=3R\left(\frac{2n}3\right)\;.$$ – Brian M. Scott Dec 17 '12 at 1:17
Do we need to add the number of comparisons the algorithm is making (how many times it's swapping)? – Bailor Tow Dec 17 '12 at 1:23
The solution says that $$R(n) = 3R(\frac{2n}{3}) + 1$$I now understand how the first term was found, but the +1 term is confusing me. Is this the number of operations (one) when the algorithm works on an array of size n=1 or n=2? – Bailor Tow Dec 17 '12 at 1:31
@BailorTow You can. There is at most two comparisons and and one swap per recursive call, you can add 3 to Brian's answer. – Code-Guru Dec 17 '12 at 1:33
@Code-Guru so, ignoring comparisons, we would have Brian's answer +1? Does (n==1)return; correspond to a base case? So that R(1) = 1? – Bailor Tow Dec 17 '12 at 1:36
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Part A
The elasticity of supply basically means that how many products a supplier can supply as price changes. It can be represented by the equation below.
Percentage change in quantity supplied
Percentage change in price
Normally, the answer for the percentage change in supply will be positive. This is because according to the law of supply, the relationship between price and quantity supplied must be positive related, ceteris paribus. As price of the good goes up, quantity supply will increase vice-versa. And the shape of a supply curve is upward sloping as shown below.
Price'S
Quantity Supplied
Supplier normally depends on this elasticity to determine and to do research on how good is the product to a consumer.
The determinant of price elasticity of supply is the time period. It can be seen in two points of view, the short run or the long run time period. A short run business normally has a fixed productive capacity. For example, if a man has a land where he grows cabbage, because it is all he knows, it is fixed of what he grows, and the amount he grows. He will supply the same thing for a short run and have a fixed income. And in conclusion, the supply will considered as inelastic.
For long run, it is a bit different from short run. It basically means that the supplier can decide what he wants to do with what he has. He or she will have sufficient time to do what he or she wants to do. For example, if a person knows that there is a trend that people are eating more tomatoes than cabbage, the producer can plant tomatoes more that cabbage in his plantation. The supplier will supply more tomatoes and it will be elastic.
Part B
Businesses normally use the concept of price elasticity to determine their price. Normally take a look from a consumer's point of view, which means that using price elasticity of demand. According to Slavin 2009, it says that when demand is elastic, if we were to raise price, total revenue would fall. Sellers determine the price of a good by calculating the total revenue. Let's see a simple example. If a saloon normally gives 20 haircuts for the price of \$15 each, how much would the revenue be? And if the price per hair cut increases to \$20, and the number of haircuts can she can give is only 10, how much is the total revenue?
Table 2.1: Hypothetical Revenue Schedule
Price Quantity Demanded Total revenue \$15 20 \$300 \$20 10 \$200
As you can see, the total revenue when the price is \$15 is \$300 whereas the total revenue of \$20 is only \$200. We know that it is elastic in by calculating the price elasticity of demand. In other cases, when price is elastic, the total revenue can increase too. It really depends on the product. For example, a packet of tissue cost \$0.30 in a supermarket, whereas the price of a packet tissue cost \$0.50 in a 24 hours mini market. The packet tissue in the supermarket can be considered as elastic in the supply point of view for consumers. Price is cheaper, consumer are more responsive during the day time. But if it is compared during midnight or early in the morning, the packet tissue in the 24 hours mini market is elastic also. People have no choice but to go to that mini market to get the tissue. But if the price of the packet tissues increases in the supermarket, it could be considered as inelastic already. The reaction of a consumer will be less responsive. Because it follows the law of demand, when price increase, quantity demanded for a good decrease.
Part A
Supply can be defined as the schedule of quantity of a good or services that people are willing and able to sell at various prices. (Slavin 2009). According to the law of supply, the relationship between price and quantity demanded are positively related. Which means that it is the opposite compared to the law of demand. As price of a good increase, the quantity supplied will increase vice-versa. The supply curve can be drawn and the shape of the curve is an upward sloping curve as shown.
Figure 3.1: Supply Curve
Price of Good A
Quantity supplied of Good A
There will be an increase and decrease of supply due to several reasons. But I specifically want to talk about what are the reasons or determinants that increase the supply of a product here.
One of the reasons that there will be an increase of product is because the changes in the cost of production. This determinant is the most important and most effected determinants. This is because the price of raw materials, labour or even capital has decreased. For example, if a company that produces pillow found out that the price of cotton, a raw material of pillow, has decreased, they are willing to produce more pillows because of the capital that they have can buy extra cotton and eventually producing more pillows. They supplier are willing to produce more and this will be an advantage to the company. Because it will increase the profit of the company with the same amount of money that they are willing to supply even before the price of cotton drops.
Secondly, the introduction of new technology will cause the supply of a product increases. Throughout these years, we've seen many advanced technology have been introduced to industries and that have helped them because these technology will decrease the cost of production and it speeds up the production. For example, more people are depending on computers rather than men anymore. Because computer are faster and won't make mistakes as often as human do. It also reduces the cost of a production because no salaries are involved when computers are using. It is just a fixed cost when it is bought. Many factories in this era are using at least a few computers to run their business. A company or factory without a computer will take a long time to produce goods, and eventually will allow substitutes to take over their consumers' demand.
Next, the expectations of future price changes also play a part in the supply of a product. If the price of a good is expected to drop in the future, there will be an increase in supply of product. Let's put it in an example, if price of corn are expected to drop in the future, supplier will increase their product now because when price drops, there will be less supply according to the law of supply.
Part B
Price ceilings and price floors are introduced mainly by government.
Figure 3.2
Price S
Price floor
Equilibrium price and quantity
Price ceiling
D
Quantity
Government introduce price floors and ceiling to control the situation in the country. If the government introduce price ceiling, it basically means that it is a good thing to a consumer as the maximum price of the product that a supply can sell. Consumers attracted to the low price. But there is a problem over here. There are shortages of supply. This is where the government have to do their job. A shortage in supply will create an allocation of resource and consumer will feel unfair as some people can afford it and buy many at a time. So the government introduces rationing function or people might call it waiting list.
Price ceiling is the opposite of price floors. It introduces a minimum price for a product. This time, the favour is on the supplier's side. Supplier will gain more than consumer in the sense that they earn more. But in this case there is a surplus of supply which means that there supplier supplies more than consumers demand.
All in all, price floors and ceilings can stifle the rationing function where it really creates many problems between people over a product. This price floors and ceilings are really creating unfairness among consumers without affecting the suppliers. That is why economists say it in this way. Another thing is that price floors and ceilings do distort resource allocation. Meaning that it creates unfairness and it really change the natural feel of a normal human where both supplier and consumer get benefits. This is a really terrible situation if it is not controlled properly. A good government of a country must know how to face these problems.
Part A
Demand can be defined as the schedule of quantities of a good or services that people are willing and able to buy at different prices. (Slavin 2009). According to the law of demand, it states that the relationship between the price and the quantity demanded is negatively related, ceteris paribus. As the price goes up, quantity demanded goes down vice-versa. It basically applies to every consumer. If the price of a good is expensive, we wouldn't consider buying it. But if it is cheap, the demand for the good is higher. The graph that represents demand curve is a downwards sloping curve.
Price of good A
Quantity demanded of good A
A decrease in the price of good A will cause an increase in quantity demanded for good A. It will cause a movement downwards from point B to point C. And the relationship that is negative between price and quantity demanded has been proven also. The only determinant that changes the quantity demand is the price of a good.
Price
D1 D0
Quantity Demanded
Change in quantity demand will cause a shift of the whole demand curve. In this case, demand curve is shifting leftward; it means that there is less demand of a good. Unlike change in quantity demanded, there are several effects that change the demand curve. Examples of effects are income and wealth, prices of other goods and services, tastes and preferences, expectations, and many more. These determinants can either shift the demand curve leftward or rightward depending on the effects on price. There are also some situations where demand curve are forced to shift such as natural disaster occurs.
Part B
According to Slavin (2009), income elasticity of demand measures how the consumption of various goods and services response to change in income. It can be represented by the equation below:
Percentage change in quantity demanded
Percentage change in income
There are three main degrees of income elasticity of demand. By calculating the elasticity of income, we can know those degrees. There are normal goods, luxury goods and necessity goods. If we found out that the answer for the income elasticity is between zero to one (0<EI<1), for example, 0.43 or 0.99, it can be concluded and consider as a normal good. Normal good can be things that we need it but don't really care about the quality of it for example a family needs a car just to travel around rather than to show other people what they have. If the answer calculated for the elasticity of income is more than 1 (1<EI), for example, 2.54 or 5.78, it can be considered as a luxury good where people can afford to buy branded and better quality of goods. For instance, consumers choose to buy Mini Cooper rather than Proton Wira because of the income that made them buys luxury goods. Last but not least, if the answer found is equals to zero (EI=0), it is a necessity good. For example like rice, electricity and other necessities. These are the goods that everybody needs it. No matter the consumer have higher income level; they still have to pay the same amount, which benefits them.
Part A
Consumer surplus can be defined as the difference between what you pay for some good or service and what you have been willing to pay. (Slavin 2009). And the definition of producer surplus can be defined as the relationship between the price in the market in the current situation and the total cost of production for the firm. (Karl E. Case, Ray C. Fair 2004).
As we can see from the figure above, consumer surplus is above the equilibrium whereas producer surplus is below the equilibrium of demand and supply. The reason why consumer surplus is above the equilibrium is because consumers are willing to pay that amount of money more than the current market value as long as it is above the equilibrium point. But fewer consumers will pay for the product as price increases. As for producer surplus, it is below the equilibrium which means that they are willing to produce their goods in order to cover their costs or the opportunity cost of production. This will give just enough of profit to continue their business. They are willing to supply at any amount as long as it is below the equilibrium point.
Part B
Production possibilities frontier shows that all the goods and services which are combined together can produced if every society's resources are used efficiently all in a graph. (Karl E. Case, Ray C. Fair 2004). The three concepts use in the production possibilities frontier is choice, opportunity cost, and scarcity. Economics also can be defined as a study of how to make use fully on what they have with scare resources playing a part in it. They want to have what they want but they only have limited resources to fulfil their wish or wants. That's why in economics, we have choice, opportunity cost, and scarcity to try to make use of everything that we have.
As we can see from the graph, when a company wants to produce 10 units of Good A, they only can produce 20 units of Good B. If the company wants to produce 25 units of Good B, they only can produce 8 units of Good A. This is what we call opportunity cost. There are many possibilities of combination of both goods. As you can see in the graph, the area under the graph is where all the possibilities are. If there is an increase in either one of the good, there will be a decrease in the other good. Scarcity can also be seen in this graph. There is only limited capital of a company to produce both goods. And it is up to their choice on what is more important and can get more profit from the goods. It also really depends on how consumers will react on the goods, whether they prefer Good A or Good B. It is the best for them to produce using fully of the opportunity cost, which means along the curve. This is because it will give them the maximum profit that they can get when they choose what they produce.
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Two-Factor ANOVA with Repeated Measures on Both Factors
B1 B2 B3 A1 subj 1 X(A1B1) subj 2 X(A1B1) etc. subj 1 X(A1B2) subj 2 X(A1B2) etc. subj 1 X(A1B3) subj 2 X(A1B3) etc. A2 subj 1 X(A2B1) subj 2 X(A2B1) etc. subj 1 X(A2B2) subj 2 X(A2B2) etc. subj 1 X(A2B3) subj 2 X(A2B3) etc. subj = subject X = measure
This page will perform a two-way factorial analysis of variance for designs in which there are 2-4 levels of each of two variables, A and B, with each subject measured under each of the AxB combinations. The programming assumes that all active cells include the same number of measures. Thus, with 2 rows and 3 columns, there is a total of 2x3=6 measures for each subject, arranged as illustrated in the adjacent table.
Procedure:
• Initial Setup:T
Enter the number of rows and columns in your analysis into the designated text fields, then click the 'Setup' button.T
• Entering Data Directly into the Text Fields:T
After clicking the cursor into the scrollable text area for A1B1 (row1/column1), enter the values for that sample in sequence, pressing the carriage return key after each entry except the last. (On a Macintosh platform, the carriage return key is labeled 'Return'; on a Windows platform it is labeled 'Enter.') Perform the same procedure for the other samples in your analysis.T
• Importing Data via Copy & Paste:T
• Data Check:T
For each sample, make sure that the final entry is not followed by a carriage return. (A carriage return after the final entry in a sample will be interpreted as an extra data entry whose value is zero. Importing data via the copy and paste procedure will almost always produce an extra carriage return at the end of a column.) After all values for a sample have been entered, click the cursor immediately to the right of the final entry in the list, then press the down-arrow key. If an extra line is present, the cursor will move downward. Extra lines can be removed by pressing the down arrow key until the cursor no longer moves, and then pressing the 'Backspace' key (on a Mac platform, 'delete') until the cursor stands immediately to the right of the final entry.T
For a two-way repeated-measures ANOVA, it is esential that the values within each cell be entered in the correct sequence and that the active cells all contain the same number of entries. The programming for the page assumes that each active cell contains the same number of entries as cell A1B1.T
• When all sample values have been entered, click the button labeled 'Calculate.'
Two-Factor ANOVA with Repeated Measures on Both Factors Number of rows in analysis (A) = Number of columns in analysis (B) = Data Entry B1 B2 B3 B4 A1 A2 A3 A4
n's for Cells, Rows, Columns, and Total B1 B2 B3 B4 Totals A1 A2 A3 A4 Totals
Number of subjects: Measures per subject:
Means for Cells, Rows, Columns, and Total B1 B2 B3 B4 Totals A1 A2 A3 A4 Totals
ANOVA Summary A = row variable B = column variable Subj = subjects Source SS df MS F P Subjects Within Subjects A Subj x A B Subj x B A x B Subj x A x B TOTAL
Home Click this link only if you did not arrive here via the VassarStats main page.
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⚠️ Warning: This is a draft ⚠️
This means it might contain formatting issues, incorrect code, conceptual problems, or other severe issues.
If you want to help to improve and eventually enable this page, please fork RosettaGit's repository and open a merge request on GitHub.
Find the points which form a [[wp:Convex hull|convex hull]] from a set of arbitrary two dimensional points.
For example, given the points (16,3), (12,17), (0,6), (-4,-6), (16,6), (16,-7), (16,-3), (17,-4), (5,19), (19,-8), (3,16), (12,13), (3,-4), (17,5), (-3,15), (-3,-9), (0,11), (-9,-3), (-4,-2) and (12,10) the convex hull would be (-9,-3), (-3,-9), (19,-8), (17,5), (12,17), (5,19) and (-3,15).
• http://www.geeksforgeeks.org/convex-hull-set-2-graham-scan/
## C
{{trans|C++}}
```
#include <stdio.h>
#include <stdlib.h>
typedef struct tPoint {
int x, y;
} Point;
typedef struct tNode {
Point data;
struct tNode *next;
} Node;
bool ccw(const Point *a, const Point *b, const Point *c) {
return (b->x - a->x) * (c->y - a->y)
> (b->y - a->y) * (c->x - a->x);
}
int comp(const void *lhs, const void *rhs) {
Point lp = *((Point *)lhs);
Point rp = *((Point *)rhs);
if (lp.x < rp.x) return -1;
if (rp.x < lp.x) return 1;
return 0;
}
void freeNode(Node *ptr) {
if (ptr == NULL) {
return;
}
freeNode(ptr->next);
ptr->next = NULL;
free(ptr);
}
Node* pushBack(Node *ptr, Point data) {
Node *tmp = ptr;
if (ptr == NULL) {
ptr = (Node*)malloc(sizeof(Node));
ptr->data = data;
ptr->next = NULL;
return ptr;
}
while (tmp->next != NULL) {
tmp = tmp->next;
}
tmp->next = (Node*)malloc(sizeof(Node));
tmp->next->data = data;
tmp->next->next = NULL;
return ptr;
}
Node* popBack(Node *ptr) {
Node *tmp = ptr;
if (ptr == NULL) {
return NULL;
}
if (ptr->next == NULL) {
free(ptr);
return NULL;
}
while (tmp->next->next != NULL) {
tmp = tmp->next;
}
free(tmp->next);
tmp->next = NULL;
return ptr;
}
void print(Node *ptr) {
printf("[");
if (ptr != NULL) {
printf("(%d, %d)", ptr->data.x, ptr->data.y);
ptr = ptr->next;
}
while (ptr != NULL) {
printf(", (%d, %d)", ptr->data.x, ptr->data.y);
ptr = ptr->next;
}
printf("]");
}
Node* convexHull(int len, Point p[]) {
Node *h = NULL;
Node *hptr = NULL;
size_t hLen = 0;
int i;
qsort(p, len, sizeof(Point), comp);
/* lower hull */
for (i = 0; i < len; ++i) {
while (hLen >= 2) {
hptr = h;
while (hptr->next->next != NULL) {
hptr = hptr->next;
}
if (ccw(&hptr->data, &hptr->next->data, &p[i])) {
break;
}
h = popBack(h);
hLen--;
}
h = pushBack(h, p[i]);
hLen++;
}
/* upper hull */
for (i = len - 1; i >= 0; i--) {
while (hLen >= 2) {
hptr = h;
while (hptr->next->next != NULL) {
hptr = hptr->next;
}
if (ccw(&hptr->data, &hptr->next->data, &p[i])) {
break;
}
h = popBack(h);
hLen--;
}
h = pushBack(h, p[i]);
hLen++;
}
popBack(h);
return h;
}
int main() {
Point points[] = {
{16, 3}, {12, 17}, { 0, 6}, {-4, -6}, {16, 6},
{16, -7}, {16, -3}, {17, -4}, { 5, 19}, {19, -8},
{ 3, 16}, {12, 13}, { 3, -4}, {17, 5}, {-3, 15},
{-3, -9}, { 0, 11}, {-9, -3}, {-4, -2}, {12, 10}
};
Node *hull = convexHull(sizeof(points) / sizeof(Point), points);
printf("Convex Hull: ");
print(hull);
printf("\n");
freeNode(hull);
hull = NULL;
return 0;
}
```
{{out}}
```Convex Hull: [(-9, -3), (-3, -9), (19, -8), (17, 5), (12, 17), (5, 19), (-3, 15)]
```
## C++
{{trans|D}}
```#include <algorithm>
#include <iostream>
#include <ostream>
#include <vector>
#include <tuple>
typedef std::tuple<int, int> point;
std::ostream& print(std::ostream& os, const point& p) {
return os << "(" << std::get<0>(p) << ", " << std::get<1>(p) << ")";
}
std::ostream& print(std::ostream& os, const std::vector<point>& v) {
auto it = v.cbegin();
auto end = v.cend();
os << "[";
if (it != end) {
print(os, *it);
it = std::next(it);
}
while (it != end) {
os << ", ";
print(os, *it);
it = std::next(it);
}
return os << "]";
}
// returns true if the three points make a counter-clockwise turn
bool ccw(const point& a, const point& b, const point& c) {
return ((std::get<0>(b) - std::get<0>(a)) * (std::get<1>(c) - std::get<1>(a)))
> ((std::get<1>(b) - std::get<1>(a)) * (std::get<0>(c) - std::get<0>(a)));
}
std::vector<point> convexHull(std::vector<point> p) {
if (p.size() == 0) return std::vector<point>();
std::sort(p.begin(), p.end(), [](point& a, point& b){
if (std::get<0>(a) < std::get<0>(b)) return true;
return false;
});
std::vector<point> h;
// lower hull
for (const auto& pt : p) {
while (h.size() >= 2 && !ccw(h.at(h.size() - 2), h.at(h.size() - 1), pt)) {
h.pop_back();
}
h.push_back(pt);
}
// upper hull
auto t = h.size() + 1;
for (auto it = p.crbegin(); it != p.crend(); it = std::next(it)) {
auto pt = *it;
while (h.size() >= t && !ccw(h.at(h.size() - 2), h.at(h.size() - 1), pt)) {
h.pop_back();
}
h.push_back(pt);
}
h.pop_back();
return h;
}
int main() {
using namespace std;
vector<point> points = {
make_pair(16, 3), make_pair(12, 17), make_pair(0, 6), make_pair(-4, -6), make_pair(16, 6),
make_pair(16, -7), make_pair(16, -3), make_pair(17, -4), make_pair(5, 19), make_pair(19, -8),
make_pair(3, 16), make_pair(12, 13), make_pair(3, -4), make_pair(17, 5), make_pair(-3, 15),
make_pair(-3, -9), make_pair(0, 11), make_pair(-9, -3), make_pair(-4, -2), make_pair(12, 10)
};
auto hull = convexHull(points);
auto it = hull.cbegin();
auto end = hull.cend();
cout << "Convex Hull: ";
print(cout, hull);
cout << endl;
return 0;
}
```
{{out}}
```Convex Hull: [(-9, -3), (-3, -9), (19, -8), (17, 5), (12, 17), (5, 19), (-3, 15)]
```
## C#
```using System;
using System.Collections.Generic;
namespace ConvexHull {
class Point : IComparable<Point> {
private int x, y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
public int X { get => x; set => x = value; }
public int Y { get => y; set => y = value; }
public int CompareTo(Point other) {
return x.CompareTo(other.x);
}
public override string ToString() {
return string.Format("({0}, {1})", x, y);
}
}
class Program {
private static List<Point> ConvexHull(List<Point> p) {
if (p.Count == 0) return new List<Point>();
p.Sort();
List<Point> h = new List<Point>();
// lower hull
foreach (var pt in p) {
while (h.Count >= 2 && !Ccw(h[h.Count - 2], h[h.Count - 1], pt)) {
h.RemoveAt(h.Count - 1);
}
}
// upper hull
int t = h.Count + 1;
for (int i = p.Count - 1; i >= 0; i--) {
Point pt = p[i];
while (h.Count >= t && !Ccw(h[h.Count - 2], h[h.Count - 1], pt)) {
h.RemoveAt(h.Count - 1);
}
}
h.RemoveAt(h.Count - 1);
return h;
}
private static bool Ccw(Point a, Point b, Point c) {
return ((b.X - a.X) * (c.Y - a.Y)) > ((b.Y - a.Y) * (c.X - a.X));
}
static void Main(string[] args) {
List<Point> points = new List<Point>() {
new Point(16, 3),
new Point(12, 17),
new Point(0, 6),
new Point(-4, -6),
new Point(16, 6),
new Point(16, -7),
new Point(16, -3),
new Point(17, -4),
new Point(5, 19),
new Point(19, -8),
new Point(3, 16),
new Point(12, 13),
new Point(3, -4),
new Point(17, 5),
new Point(-3, 15),
new Point(-3, -9),
new Point(0, 11),
new Point(-9, -3),
new Point(-4, -2),
new Point(12, 10)
};
List<Point> hull = ConvexHull(points);
Console.Write("Convex Hull: [");
for (int i = 0; i < hull.Count; i++) {
if (i > 0) {
Console.Write(", ");
}
Point pt = hull[i];
Console.Write(pt);
}
Console.WriteLine("]");
}
}
}
```
{{out}}
```Convex Hull: [(-9, -3), (-3, -9), (19, -8), (17, 5), (12, 17), (5, 19), (-3, 15)]
```
## D
{{trans|Kotlin}}
```import std.algorithm.sorting;
import std.stdio;
struct Point {
int x;
int y;
int opCmp(Point rhs) {
if (x < rhs.x) return -1;
if (rhs.x < x) return 1;
return 0;
}
void toString(scope void delegate(const(char)[]) sink) const {
import std.format;
sink("(");
formattedWrite(sink, "%d", x);
sink(",");
formattedWrite(sink, "%d", y);
sink(")");
}
}
Point[] convexHull(Point[] p) {
if (p.length == 0) return [];
p.sort;
Point[] h;
// lower hull
foreach (pt; p) {
while (h.length >= 2 && !ccw(h[\$-2], h[\$-1], pt)) {
h.length--;
}
h ~= pt;
}
// upper hull
auto t = h.length + 1;
foreach_reverse (i; 0..(p.length - 1)) {
auto pt = p[i];
while (h.length >= t && !ccw(h[\$-2], h[\$-1], pt)) {
h.length--;
}
h ~= pt;
}
h.length--;
return h;
}
/* ccw returns true if the three points make a counter-clockwise turn */
auto ccw(Point a, Point b, Point c) {
return ((b.x - a.x) * (c.y - a.y)) > ((b.y - a.y) * (c.x - a.x));
}
void main() {
auto points = [
Point(16, 3), Point(12, 17), Point( 0, 6), Point(-4, -6), Point(16, 6),
Point(16, -7), Point(16, -3), Point(17, -4), Point( 5, 19), Point(19, -8),
Point( 3, 16), Point(12, 13), Point( 3, -4), Point(17, 5), Point(-3, 15),
Point(-3, -9), Point( 0, 11), Point(-9, -3), Point(-4, -2), Point(12, 10)
];
auto hull = convexHull(points);
writeln("Convex Hull: ", hull);
}
```
{{out}}
```Convex Hull: [(-9,-3), (-3,-9), (19,-8), (17,5), (12,17), (5,19), (-3,15)]
```
## Go
```package main
import (
"fmt"
"image"
"sort"
)
// ConvexHull returns the set of points that define the
// convex hull of p in CCW order starting from the left most.
func (p points) ConvexHull() points {
// From https://en.wikibooks.org/wiki/Algorithm_Implementation/Geometry/Convex_hull/Monotone_chain
// with only minor deviations.
sort.Sort(p)
var h points
// Lower hull
for _, pt := range p {
for len(h) >= 2 && !ccw(h[len(h)-2], h[len(h)-1], pt) {
h = h[:len(h)-1]
}
h = append(h, pt)
}
// Upper hull
for i, t := len(p)-2, len(h)+1; i >= 0; i-- {
pt := p[i]
for len(h) >= t && !ccw(h[len(h)-2], h[len(h)-1], pt) {
h = h[:len(h)-1]
}
h = append(h, pt)
}
return h[:len(h)-1]
}
// ccw returns true if the three points make a counter-clockwise turn
func ccw(a, b, c image.Point) bool {
return ((b.X - a.X) * (c.Y - a.Y)) > ((b.Y - a.Y) * (c.X - a.X))
}
type points []image.Point
func (p points) Len() int { return len(p) }
func (p points) Swap(i, j int) { p[i], p[j] = p[j], p[i] }
func (p points) Less(i, j int) bool {
if p[i].X == p[j].X {
return p[i].Y < p[i].Y
}
return p[i].X < p[j].X
}
func main() {
pts := points{
{16, 3}, {12, 17}, {0, 6}, {-4, -6}, {16, 6},
{16, -7}, {16, -3}, {17, -4}, {5, 19}, {19, -8},
{3, 16}, {12, 13}, {3, -4}, {17, 5}, {-3, 15},
{-3, -9}, {0, 11}, {-9, -3}, {-4, -2}, {12, 10},
}
hull := pts.ConvexHull()
fmt.Println("Convex Hull:", hull)
}
```
{{out}}
```
Convex Hull: [(-9,-3) (-3,-9) (19,-8) (17,5) (12,17) (5,19) (-3,15)]
```
```import Data.List (sortBy, groupBy, maximumBy)
import Data.Ord (comparing)
(x, y) = ((!! 0), (!! 1))
compareFrom
:: (Num a, Ord a)
=> [a] -> [a] -> [a] -> Ordering
compareFrom o l r =
compare ((x l - x o) * (y r - y o)) ((y l - y o) * (x r - x o))
distanceFrom
:: Floating a
=> [a] -> [a] -> a
distanceFrom from to = ((x to - x from) ** 2 + (y to - y from) ** 2) ** (1 / 2)
convexHull
:: (Floating a, Ord a)
=> [[a]] -> [[a]]
convexHull points =
let o = minimum points
presorted = sortBy (compareFrom o) (filter (/= o) points)
collinears = groupBy (((EQ ==) .) . compareFrom o) presorted
outmost = maximumBy (comparing (distanceFrom o)) <\$> collinears
in dropConcavities [o] outmost
dropConcavities
:: (Num a, Ord a)
=> [[a]] -> [[a]] -> [[a]]
dropConcavities (left:lefter) (right:righter:rightest) =
case compareFrom left right righter of
LT -> dropConcavities (right : left : lefter) (righter : rightest)
EQ -> dropConcavities (left : lefter) (righter : rightest)
GT -> dropConcavities lefter (left : righter : rightest)
dropConcavities output lastInput = lastInput ++ output
main :: IO ()
main =
mapM_ print \$
convexHull
[ [16, 3]
, [12, 17]
, [0, 6]
, [-4, -6]
, [16, 6]
, [16, -7]
, [16, -3]
, [17, -4]
, [5, 19]
, [19, -8]
, [3, 16]
, [12, 13]
, [3, -4]
, [17, 5]
, [-3, 15]
, [-3, -9]
, [0, 11]
, [-9, -3]
, [-4, -2]
, [12, 10]
]
```
{{Out}}
```[-3.0,-9.0]
[19.0,-8.0]
[17.0,5.0]
[12.0,17.0]
[5.0,19.0]
[-3.0,15.0]
[-9.0,-3.0]
```
## J
Restated from the implementation at http://kukuruku.co/hub/funcprog/introduction-to-j-programming-language-2004 which in turn is a translation of http://dr-klm.livejournal.com/42312.html
```counterclockwise =: ({. , }. /: 12 o. }. - {.) @ /:~
crossproduct =: 11"_ o. [: (* +)/ }. - {.
removeinner =: #~ 1, 0 > 3 crossproduct\ ], 1:
hull =: [: removeinner^:_ counterclockwise
```
Example use:
``` hull 16j3 12j17 0j6 _4j_6 16j6 16j_7 16j_3 17j_4 5j19 19j_8 3j16 12j13 3j_4 17j5 _3j15 _3j_9 0j11 _9j_3 _4j_2 12j10
_9j_3 _3j_9 19j_8 17j5 12j17 5j19 _3j15
```
## Java
{{trans|Kotlin}}
```import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;
import static java.util.Collections.emptyList;
public class ConvexHull {
private static class Point implements Comparable<Point> {
private int x, y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Point o) {
return Integer.compare(x, o.x);
}
@Override
public String toString() {
return String.format("(%d, %d)", x, y);
}
}
private static List<Point> convexHull(List<Point> p) {
if (p.isEmpty()) return emptyList();
p.sort(Point::compareTo);
List<Point> h = new ArrayList<>();
// lower hull
for (Point pt : p) {
while (h.size() >= 2 && !ccw(h.get(h.size() - 2), h.get(h.size() - 1), pt)) {
h.remove(h.size() - 1);
}
}
// upper hull
int t = h.size() + 1;
for (int i = p.size() - 1; i >= 0; i--) {
Point pt = p.get(i);
while (h.size() >= t && !ccw(h.get(h.size() - 2), h.get(h.size() - 1), pt)) {
h.remove(h.size() - 1);
}
}
h.remove(h.size() - 1);
return h;
}
// ccw returns true if the three points make a counter-clockwise turn
private static boolean ccw(Point a, Point b, Point c) {
return ((b.x - a.x) * (c.y - a.y)) > ((b.y - a.y) * (c.x - a.x));
}
public static void main(String[] args) {
List<Point> points = Arrays.asList(new Point(16, 3),
new Point(12, 17),
new Point(0, 6),
new Point(-4, -6),
new Point(16, 6),
new Point(16, -7),
new Point(16, -3),
new Point(17, -4),
new Point(5, 19),
new Point(19, -8),
new Point(3, 16),
new Point(12, 13),
new Point(3, -4),
new Point(17, 5),
new Point(-3, 15),
new Point(-3, -9),
new Point(0, 11),
new Point(-9, -3),
new Point(-4, -2),
new Point(12, 10));
List<Point> hull = convexHull(points);
System.out.printf("Convex Hull: %s\n", hull);
}
}
```
{{out}}
```Convex Hull: [(-9, -3), (-3, -9), (19, -8), (17, 5), (12, 17), (5, 19), (-3, 15)]
```
## Javascript
```
function convexHull(points) {
points.sort(comparison);
var L = [];
for (var i = 0; i < points.length; i++) {
while (L.length >= 2 && cross(L[L.length - 2], L[L.length - 1], points[i]) <= 0) {
L.pop();
}
L.push(points[i]);
}
var U = [];
for (var i = points.length - 1; i >= 0; i--) {
while (U.length >= 2 && cross(U[U.length - 2], U[U.length - 1], points[i]) <= 0) {
U.pop();
}
U.push(points[i]);
}
L.pop();
U.pop();
return L.concat(U);
}
function comparison(a, b) {
return a.x == b.x ? a.y - b.y : a.x - b.x;
}
function cross(a, b, o) {
return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x);
}
```
'''For usage''': convexhull.js
```
var points = [];
var hull = [];
function setup() {
createCanvas(1132, 700);
frameRate(10);
strokeWeight(4);
stroke(220);
}
function draw() {
background(40);
// draw points
for (i = 0; i < points.length; i++) {
point(points[i].x, points[i].y);
};
console.log(hull);
// draw hull
noFill();
beginShape();
for (i = 0; i < hull.length; i++) {
vertex(hull[i].x, hull[i].y);
};
endShape(CLOSE);
}
function mouseClicked() {
points.push(createVector(mouseX, mouseY));
hull = convexHull(points);
noFill();
//console.log(hull);
beginShape();
for (var i = 0; i < hull.length; i++) {
vertex(hull[i].x, hull[i].y);
}
endShape(CLOSE);
return false;
}
// https://en.wikibooks.org/wiki/Algorithm_Implementation/Geometry/Convex_hull/Monotone_chain
function convexHull(points) {
points.sort(comparison);
var L = [];
for (var i = 0; i < points.length; i++) {
while (L.length >= 2 && cross(L[L.length - 2], L[L.length - 1], points[i]) <= 0) {
L.pop();
}
L.push(points[i]);
}
var U = [];
for (var i = points.length - 1; i >= 0; i--) {
while (U.length >= 2 && cross(U[U.length - 2], U[U.length - 1], points[i]) <= 0) {
U.pop();
}
U.push(points[i]);
}
L.pop();
U.pop();
return L.concat(U);
}
function comparison(a, b) {
return a.x == b.x ? a.y - b.y : a.x - b.x;
}
function cross(a, b, o) {
return (a.x - o.x) * (b.y - o.y) - (a.y - o.y) * (b.x - o.x);
}
```
convexhull.html
```
<html>
<script src="https://cdnjs.cloudflare.com/ajax/libs/p5.js/0.7.2/p5.js"></script>
<script src="convexhull.js"></script>
<body>
<table>
<tr>
<th><h1>Convex Hull</h4></th>
</tr>
</table>
</body>
</html>
```
## Julia
```# v1.0.4
# https://github.com/JuliaPolyhedra/Polyhedra.jl/blob/master/examples/operations.ipynb
using Polyhedra, CDDLib
A = vrep([[16,3], [12,17], [0,6], [-4,-6], [16,6], [16,-7], [16,-3], [17,-4], [5,19], [19,-8], [3,16], [12,13], [3,-4], [17,5], [-3,15], [-3,-9], [0,11], [-9,-3], [-4,-2], [12,10]])
P = polyhedron(A, CDDLib.Library())
Pch = convexhull(P, P)
removevredundancy!(Pch)
println("\$Pch")
```
{{out}}
```convexhull([5.0, 19.0], [19.0, -8.0], [17.0, 5.0], [-3.0, 15.0], [-9.0, -3.0], [12.0, 17.0], [-3.0, -9.0])
```
## Kotlin
{{trans|Go}}
```// version 1.1.3
class Point(val x: Int, val y: Int) : Comparable<Point> {
override fun compareTo(other: Point) = this.x.compareTo(other.x)
override fun toString() = "(\$x, \$y)"
}
fun convexHull(p: Array<Point>): List<Point> {
if (p.isEmpty()) return emptyList()
p.sort()
val h = mutableListOf<Point>()
// lower hull
for (pt in p) {
while (h.size >= 2 && !ccw(h[h.size - 2], h.last(), pt)) {
h.removeAt(h.lastIndex)
}
}
// upper hull
val t = h.size + 1
for (i in p.size - 2 downTo 0) {
val pt = p[i]
while (h.size >= t && !ccw(h[h.size - 2], h.last(), pt)) {
h.removeAt(h.lastIndex)
}
}
h.removeAt(h.lastIndex)
return h
}
/* ccw returns true if the three points make a counter-clockwise turn */
fun ccw(a: Point, b: Point, c: Point) =
((b.x - a.x) * (c.y - a.y)) > ((b.y - a.y) * (c.x - a.x))
fun main(args: Array<String>) {
val points = arrayOf(
Point(16, 3), Point(12, 17), Point( 0, 6), Point(-4, -6), Point(16, 6),
Point(16, -7), Point(16, -3), Point(17, -4), Point( 5, 19), Point(19, -8),
Point( 3, 16), Point(12, 13), Point( 3, -4), Point(17, 5), Point(-3, 15),
Point(-3, -9), Point( 0, 11), Point(-9, -3), Point(-4, -2), Point(12, 10)
)
val hull = convexHull(points)
println("Convex Hull: \$hull")
}
```
{{out}}
```
Convex Hull: [(-9, -3), (-3, -9), (19, -8), (17, 5), (12, 17), (5, 19), (-3, 15)]
```
## Lua
{{trans|C++}}
```function print_point(p)
io.write("("..p.x..", "..p.y..")")
return nil
end
function print_points(pl)
io.write("[")
for i,p in pairs(pl) do
if i>1 then
io.write(", ")
end
print_point(p)
end
io.write("]")
return nil
end
function ccw(a,b,c)
return (b.x - a.x) * (c.y - a.y) > (b.y - a.y) * (c.x - a.x)
end
function pop_back(ta)
table.remove(ta,#ta)
return ta
end
function convexHull(pl)
if #pl == 0 then
return {}
end
table.sort(pl, function(left,right)
return left.x < right.x
end)
local h = {}
-- lower hull
for i,pt in pairs(pl) do
while #h >= 2 and not ccw(h[#h-1], h[#h], pt) do
table.remove(h,#h)
end
table.insert(h,pt)
end
-- upper hull
local t = #h + 1
for i=#pl, 1, -1 do
local pt = pl[i]
while #h >= t and not ccw(h[#h-1], h[#h], pt) do
table.remove(h,#h)
end
table.insert(h,pt)
end
table.remove(h,#h)
return h
end
-- main
local points = {
{x=16,y= 3},{x=12,y=17},{x= 0,y= 6},{x=-4,y=-6},{x=16,y= 6},
{x=16,y=-7},{x=16,y=-3},{x=17,y=-4},{x= 5,y=19},{x=19,y=-8},
{x= 3,y=16},{x=12,y=13},{x= 3,y=-4},{x=17,y= 5},{x=-3,y=15},
{x=-3,y=-9},{x= 0,y=11},{x=-9,y=-3},{x=-4,y=-2},{x=12,y=10}
}
local hull = convexHull(points)
io.write("Convex Hull: ")
print_points(hull)
print()
```
{{out}}
```Convex Hull: [(-9, -3), (-3, -9), (19, -8), (17, 5), (12, 17), (5, 19), (-3, 15)]
```
```MODULE ConvexHull;
FROM FormatString IMPORT FormatString;
FROM Storage IMPORT ALLOCATE, DEALLOCATE;
FROM SYSTEM IMPORT TSIZE;
PROCEDURE WriteInt(n : INTEGER);
VAR buf : ARRAY[0..15] OF CHAR;
BEGIN
FormatString("%i", buf, n);
WriteString(buf);
END WriteInt;
TYPE
Point = RECORD
x, y : INTEGER;
END;
PROCEDURE WritePoint(pt : Point);
BEGIN
WriteString("(");
WriteInt(pt.x);
WriteString(", ");
WriteInt(pt.y);
WriteString(")");
END WritePoint;
TYPE
NextNode = POINTER TO PNode;
PNode = RECORD
value : Point;
next : NextNode;
END;
PROCEDURE WriteNode(it : NextNode);
BEGIN
IF it = NIL THEN
RETURN
END;
WriteString("[");
WritePoint(it^.value);
it := it^.next;
WHILE it # NIL DO
WriteString(", ");
WritePoint(it^.value);
it := it^.next
END;
WriteString("]")
END WriteNode;
PROCEDURE AppendNode(pn : NextNode; p : Point) : NextNode;
VAR it,nx : NextNode;
BEGIN
IF pn = NIL THEN
ALLOCATE(it,TSIZE(PNode));
it^.value := p;
it^.next := NIL;
RETURN it
END;
it := pn;
WHILE it^.next # NIL DO
it := it^.next
END;
ALLOCATE(nx,TSIZE(PNode));
nx^.value := p;
nx^.next := NIL;
it^.next := nx;
RETURN pn
END AppendNode;
PROCEDURE DeleteNode(VAR pn : NextNode);
BEGIN
IF pn = NIL THEN RETURN END;
DeleteNode(pn^.next);
DEALLOCATE(pn,TSIZE(PNode));
pn := NIL
END DeleteNode;
PROCEDURE SortNode(VAR pn : NextNode);
VAR
it : NextNode;
tmp : Point;
done : BOOLEAN;
BEGIN
REPEAT
done := TRUE;
it := pn;
WHILE (it # NIL) AND (it^.next # NIL) DO
IF it^.next^.value.x < it^.value.x THEN
tmp := it^.value;
it^.value := it^.next^.value;
it^.next^.value := tmp;
done := FALSE
END;
it := it^.next;
END
UNTIL done;
END SortNode;
PROCEDURE NodeLength(it : NextNode) : INTEGER;
VAR length : INTEGER;
BEGIN
length := 0;
WHILE it # NIL DO
INC(length);
it := it^.next;
END;
RETURN length
END NodeLength;
PROCEDURE ReverseNode(fp : NextNode) : NextNode;
VAR rp,tmp : NextNode;
BEGIN
IF fp = NIL THEN RETURN NIL END;
ALLOCATE(tmp,TSIZE(PNode));
tmp^.value := fp^.value;
tmp^.next := NIL;
rp := tmp;
fp := fp^.next;
WHILE fp # NIL DO
ALLOCATE(tmp,TSIZE(PNode));
tmp^.value := fp^.value;
tmp^.next := rp;
rp := tmp;
fp := fp^.next;
END;
RETURN rp
END ReverseNode;
(* ccw returns true if the three points make a counter-clockwise turn *)
PROCEDURE CCW(a,b,c : Point) : BOOLEAN;
BEGIN
RETURN ((b.x - a.x) * (c.y - a.y)) > ((b.y - a.y) * (c.x - a.x))
END CCW;
PROCEDURE ConvexHull(p : NextNode) : NextNode;
VAR
hull,it,h1,h2 : NextNode;
t : INTEGER;
BEGIN
IF p = NIL THEN RETURN NIL END;
SortNode(p);
hull := NIL;
(* lower hull *)
it := p;
WHILE it # NIL DO
IF hull # NIL THEN
WHILE hull^.next # NIL DO
(* At least two points in the list *)
h2 := hull;
h1 := hull^.next;
WHILE h1^.next # NIL DO
h2 := h1;
h1 := h2^.next;
END;
IF CCW(h2^.value, h1^.value, it^.value) THEN
BREAK
ELSE
h2^.next := NIL;
DeleteNode(h1);
h1 := NIL
END
END
END;
hull := AppendNode(hull, it^.value);
it := it^.next;
END;
(* upper hull *)
t := NodeLength(hull) + 1;
p := ReverseNode(p);
it := p;
WHILE it # NIL DO
WHILE NodeLength(hull) >= t DO
h2 := hull;
h1 := hull^.next;
WHILE h1^.next # NIL DO
h2 := h1;
h1 := h2^.next;
END;
IF CCW(h2^.value, h1^.value, it^.value) THEN
BREAK
ELSE
h2^.next := NIL;
DeleteNode(h1);
h1 := NIL
END
END;
hull := AppendNode(hull, it^.value);
it := it^.next;
END;
DeleteNode(p);
h2 := hull;
h1 := h2^.next;
WHILE h1^.next # NIL DO
h2 := h1;
h1 := h1^.next;
END;
h2^.next := NIL;
DeleteNode(h1);
RETURN hull
END ConvexHull;
(* Main *)
VAR nodes,hull : NextNode;
BEGIN
nodes := AppendNode(NIL, Point{16, 3});
AppendNode(nodes, Point{12,17});
AppendNode(nodes, Point{ 0, 6});
AppendNode(nodes, Point{-4,-6});
AppendNode(nodes, Point{16, 6});
AppendNode(nodes, Point{16,-7});
AppendNode(nodes, Point{16,-3});
AppendNode(nodes, Point{17,-4});
AppendNode(nodes, Point{ 5,19});
AppendNode(nodes, Point{19,-8});
AppendNode(nodes, Point{ 3,16});
AppendNode(nodes, Point{12,13});
AppendNode(nodes, Point{ 3,-4});
AppendNode(nodes, Point{17, 5});
AppendNode(nodes, Point{-3,15});
AppendNode(nodes, Point{-3,-9});
AppendNode(nodes, Point{ 0,11});
AppendNode(nodes, Point{-9,-3});
AppendNode(nodes, Point{-4,-2});
AppendNode(nodes, Point{12,10});
hull := ConvexHull(nodes);
WriteNode(hull);
DeleteNode(hull);
DeleteNode(nodes);
END ConvexHull.
```
{{out}}
```[(-9, -3), (-3, -9), (19, -8), (17, 5), (12, 17), (5, 19), (-3, 15)]
```
## Perl 6
{{works with|Rakudo|2017.05}} {{trans|zkl}} Modified the angle sort method as the original could fail if there were multiple points on the same y coordinate as the starting point. Sorts on tangent rather than triangle area. Inexpensive since it still doesn't do any trigonometric math, just calculates the ratio of opposite over adjacent. The original returned the correct answer for the task example, but only by accident. If the points (14,-9), (1,-9) were added to the task example, it wouldn't give a correct answer. Now it does.
```class Point {
has Real \$.x is rw;
has Real \$.y is rw;
method gist { [~] '(', self.x,', ', self.y, ')' };
}
sub ccw (Point \$a, Point \$b, Point \$c) {
(\$b.x - \$a.x)*(\$c.y - \$a.y) - (\$b.y - \$a.y)*(\$c.x - \$a.x);
}
sub tangent (Point \$a, Point \$b) {
my \$opp = \$b.x - \$a.x;
my \$adj = \$b.y - \$a.y;
}
sub graham-scan (**@coords) {
# sort points by y, secondary sort on x
my @sp = @coords.map( { Point.new( :x(\$_[0]), :y(\$_[1]) ) } )
.sort: {.y, .x};
# need at least 3 points to make a hull
return @sp if +@sp < 3;
# first point on hull is minimum y point
my @h = @sp.shift;
# re-sort the points by angle, secondary on x
@sp = @sp.map( { \$++ => [tangent(@h[0], \$_), \$_.x] } )
.sort( {-\$_.value[0], \$_.value[1] } )
.map: { @sp[\$_.key] };
# first point of re-sorted list is guaranteed to be on hull
@h.push: @sp.shift;
# check through the remaining list making sure that
# there is always a positive angle
for @sp -> \$point {
if ccw( |@h.tail(2), \$point ) >= 0 {
@h.push: \$point;
} else {
@h.pop;
redo;
}
}
@h
}
my @hull = graham-scan(
(16, 3), (12,17), ( 0, 6), (-4,-6), (16, 6), (16,-7), (16,-3),
(17,-4), ( 5,19), (19,-8), ( 3,16), (12,13), ( 3,-4), (17, 5),
(-3,15), (-3,-9), ( 0,11), (-9,-3), (-4,-2), (12,10)
);
say "Convex Hull ({+@hull} points): ", @hull;
@hull = graham-scan(
(16, 3), (12,17), ( 0, 6), (-4,-6), (16, 6), (16,-7), (16,-3),
(17,-4), ( 5,19), (19,-8), ( 3,16), (12,13), ( 3,-4), (17, 5),
(-3,15), (-3,-9), ( 0,11), (-9,-3), (-4,-2), (12,10), (14,-9), (1,-9)
);
say "Convex Hull ({+@hull} points): ", @hull;
```
{{out}}
```Convex Hull (7 points): [(-3, -9) (19, -8) (17, 5) (12, 17) (5, 19) (-3, 15) (-9, -3)]
Convex Hull (9 points): [(-3, -9) (1, -9) (14, -9) (19, -8) (17, 5) (12, 17) (5, 19) (-3, 15) (-9, -3)]
```
## Phix
{{trans|C}}
```enum x, y
function ccw(sequence a, b, c)
return (b[x] - a[x]) * (c[y] - a[y])
> (b[y] - a[y]) * (c[x] - a[x])
end function
function convex_hull(sequence points)
sequence h = {}
points = sort(points)
/* lower hull */
for i=1 to length(points) do
while length(h)>=2
and not ccw(h[\$-1], h[\$], points[i]) do
h = h[1..\$-1]
end while
h = append(h, points[i])
end for
/* upper hull */
for i=length(points) to 1 by -1 do
while length(h)>=2
and not ccw(h[\$-1],h[\$],points[i]) do
h = h[1..\$-1]
end while
h = append(h, points[i])
end for
h = h[1..\$-1]
return h
end function
constant points = {{16, 3}, {12, 17}, { 0, 6}, {-4, -6}, {16, 6},
{16, -7}, {16, -3}, {17, -4}, { 5, 19}, {19, -8},
{ 3, 16}, {12, 13}, { 3, -4}, {17, 5}, {-3, 15},
{-3, -9}, { 0, 11}, {-9, -3}, {-4, -2}, {12, 10}}
printf(1,"Convex Hull: %v\n",{convex_hull(points)})
```
{{out}}
```
Convex Hull: {{-9,-3},{-3,-9},{19,-8},{17,5},{12,17},{5,19},{-3,15}}
```
## Python
An approach that uses the shapely library:
```from __future__ import print_function
from shapely.geometry import MultiPoint
if __name__=="__main__":
pts = MultiPoint([(16,3), (12,17), (0,6), (-4,-6), (16,6), (16,-7), (16,-3), (17,-4), (5,19), (19,-8), (3,16), (12,13), (3,-4), (17,5), (-3,15), (-3,-9), (0,11), (-9,-3), (-4,-2), (12,10)])
print (pts.convex_hull)
```
{{out}}
```POLYGON ((-3 -9, -9 -3, -3 15, 5 19, 12 17, 17 5, 19 -8, -3 -9))
```
## Racket
Also an implementation of https://en.wikibooks.org/wiki/Algorithm_Implementation/Geometry/Convex_hull/Monotone_chain (therefore kinda {{trans|Go}}
```#lang typed/racket
(define-type Point (Pair Real Real))
(define-type Points (Listof Point))
(: ⊗ (Point Point Point -> Real))
(define/match (⊗ o a b)
[((cons o.x o.y) (cons a.x a.y) (cons b.x b.y))
(- (* (- a.x o.x) (- b.y o.y)) (* (- a.y o.y) (- b.x o.x)))])
(: Point<? (Point Point -> Boolean))
(define (Point<? a b)
(cond [(< (car a) (car b)) #t] [(> (car a) (car b)) #f] [else (< (cdr a) (cdr b))]))
;; Input: a list P of points in the plane.
(define (convex-hull [P:unsorted : Points])
;; Sort the points of P by x-coordinate (in case of a tie, sort by y-coordinate).
(define P (sort P:unsorted Point<?))
;; for i = 1, 2, ..., n:
;; while L contains at least two points and the sequence of last two points
;; of L and the point P[i] does not make a counter-clockwise turn:
;; remove the last point from L
;; append P[i] to L
;; TB: U is identical with (reverse P)
(define (upper/lower-hull [P : Points])
(reverse
(for/fold ((rev : Points null))
((P.i (in-list P)))
(let u/l : Points ((rev rev))
(match rev
[(list p-2 p-1 ps ...) #:when (not (positive? (⊗ p-2 P.i p-1))) (u/l (list* p-1 ps))]
[(list ps ...) (cons P.i ps)])))))
;; Initialize U and L as empty lists.
;; The lists will hold the vertices of upper and lower hulls respectively.
(let ((U (upper/lower-hull (reverse P)))
(L (upper/lower-hull P)))
;; Remove the last point of each list (it's the same as the first point of the other list).
;; Concatenate L and U to obtain the convex hull of P.
(append (drop-right L 1) (drop-right U 1)))) ; Points in the result will be listed in CCW order.)
(module+ test
(require typed/rackunit)
(check-equal?
(convex-hull
(list '(16 . 3) '(12 . 17) '(0 . 6) '(-4 . -6) '(16 . 6) '(16 . -7) '(16 . -3) '(17 . -4)
'(5 . 19) '(19 . -8) '(3 . 16) '(12 . 13) '(3 . -4) '(17 . 5) '(-3 . 15) '(-3 . -9)
'(0 . 11) '(-9 . -3) '(-4 . -2) '(12 . 10)))
(list '(-9 . -3) '(-3 . -9) '(19 . -8) '(17 . 5) '(12 . 17) '(5 . 19) '(-3 . 15))))
```
{{out}}
silence implies tests pass (and output is as expected)
## REXX
### version 1
```/* REXX ---------------------------------------------------------------
* Compute the Convex Hull for a set of points
* Format of the input file:
* (16,3) (12,17) (0,6) (-4,-6) (16,6) (16,-7) (16,-3) (17,-4) (5,19)
* (19,-8) (3,16) (12,13) (3,-4) (17,5) (-3,15) (-3,-9) (0,11) (-9,-3)
* (-4,-2)
*--------------------------------------------------------------------*/
Signal On Novalue
Signal On Syntax
Parse Arg fid
If fid='' Then Do
fid='chullmin.in' /* miscellaneous test data */
fid='chullx.in'
fid='chullt.in'
fid='chulla.in'
fid='chullxx.in'
fid='sq.in'
fid='tri.in'
fid='line.in'
fid='point.in'
fid='chull.in' /* data from task description */
End
g.0debug=''
g.0oid=fn(fid)'.txt'; 'erase' g.0oid
x.=0
yl.=''
Parse Value '1000 -1000' With g.0xmin g.0xmax
Parse Value '1000 -1000' With g.0ymin g.0ymax
/*---------------------------------------------------------------------
* First read the input and store the points' coordinates
* x.0 contains the number of points, x.i contains the x.coordinate
* yl.x contains the y.coordinate(s) of points (x/y)
*--------------------------------------------------------------------*/
Do while lines(fid)>0
l=linein(fid)
Do While l<>''
Parse Var l '(' x ',' y ')' l
Call store x,y
End
End
Call lineout fid
Do i=1 To x.0 /* loop over points */
x=x.i
yl.x=sortv(yl.x) /* sort y-coordinates */
End
Call sho
/*---------------------------------------------------------------------
* Now we look for special border points:
* lefthigh and leftlow: leftmost points with higheste and lowest y
* ritehigh and ritelow: rightmost points with higheste and lowest y
* yl.x contains the y.coordinate(s) of points (x/y)
*--------------------------------------------------------------------*/
leftlow=0
lefthigh=0
Do i=1 To x.0
x=x.i
If maxv(yl.x)=g.0ymax Then Do
If lefthigh=0 Then lefthigh=x'/'g.0ymax
ritehigh=x'/'g.0ymax
End
If minv(yl.x)=g.0ymin Then Do
ritelow=x'/'g.0ymin
If leftlow=0 Then leftlow=x'/'g.0ymin
End
End
Call o 'lefthigh='lefthigh
Call o 'ritehigh='ritehigh
Call o 'ritelow ='ritelow
Call o 'leftlow ='leftlow
/*---------------------------------------------------------------------
* Now we look for special border points:
* leftmost_n and leftmost_s: points with lowest x and highest/lowest y
* ritemost_n and ritemost_s: points with largest x and highest/lowest y
* n and s stand foNorth and South, respectively
*--------------------------------------------------------------------*/
x=g.0xmi; leftmost_n=x'/'maxv(yl.x)
x=g.0xmi; leftmost_s=x'/'minv(yl.x)
x=g.0xma; ritemost_n=x'/'maxv(yl.x)
x=g.0xma; ritemost_s=x'/'minv(yl.x)
/*---------------------------------------------------------------------
* Now we compute the paths from ritehigh to ritelow (n_end)
* and leftlow to lefthigh (s_end), respectively
*--------------------------------------------------------------------*/
x=g.0xma
n_end=''
Do i=words(yl.x) To 1 By -1
n_end=n_end x'/'word(yl.x,i)
End
Call o 'n_end='n_end
x=g.0xmi
s_end=''
Do i=1 To words(yl.x)
s_end=s_end x'/'word(yl.x,i)
End
Call o 's_end='s_end
n_high=''
s_low=''
/*---------------------------------------------------------------------
* Now we compute the upper part of the convex hull (nhull)
*--------------------------------------------------------------------*/
Call o 'leftmost_n='leftmost_n
Call o 'lefthigh ='lefthigh
nhull=leftmost_n
res=mk_nhull(leftmost_n,lefthigh);
nhull=nhull res
Call o 'A nhull='nhull
Do While res<>lefthigh
res=mk_nhull(res,lefthigh); nhull=nhull res
Call o 'B nhull='nhull
End
res=mk_nhull(lefthigh,ritemost_n); nhull=nhull res
Call o 'C nhull='nhull
Do While res<>ritemost_n
res=mk_nhull(res,ritemost_n); nhull=nhull res
Call o 'D nhull='nhull
End
nhull=nhull n_end /* attach the right vertical border */
/*---------------------------------------------------------------------
* Now we compute the lower part of the convex hull (shull)
*--------------------------------------------------------------------*/
res=mk_shull(ritemost_s,ritelow);
shull=ritemost_s res
Call o 'A shull='shull
Do While res<>ritelow
res=mk_shull(res,ritelow)
shull=shull res
Call o 'B shull='shull
End
res=mk_shull(ritelow,leftmost_s)
shull=shull res
Call o 'C shull='shull
Do While res<>leftmost_s
res=mk_shull(res,leftmost_s);
shull=shull res
Call o 'D shull='shull
End
shull=shull s_end
chull=nhull shull /* concatenate upper and lower part */
/* eliminate duplicates */
/* too lazy to take care before :-) */
Parse Var chull chullx chull
have.=0
have.chullx=1
Do i=1 By 1 While chull>''
Parse Var chull xy chull
If have.xy=0 Then Do
chullx=chullx xy
have.xy=1
End
End
/* show the result */
Say 'Points of convex hull in clockwise order:'
Say chullx
/**********************************************************************
* steps that were necessary in previous attempts
/*---------------------------------------------------------------------
* Final polish: Insert points that are not yet in chullx but should be
* First on the upper hull going from left to right
*--------------------------------------------------------------------*/
i=1
Do While i<words(chullx)
xya=word(chullx,i) ; Parse Var xya xa '/' ya
If xa=g.0xmax Then Leave
xyb=word(chullx,i+1); Parse Var xyb xb '/' yb
Do j=1 To x.0
If x.j>xa Then Do
If x.j<xb Then Do
xx=x.j
parse Value kdx(xya,xyb) With k d x
If (k*xx+d)=maxv(yl.xx) Then Do
chullx=subword(chullx,1,i) xx'/'maxv(yl.xx),
subword(chullx,i+1)
i=i+1
End
End
End
Else
i=i+1
End
End
Say chullx
/*---------------------------------------------------------------------
* Final polish: Insert points that are not yet in chullx but should be
* Then on the lower hull going from right to left
*--------------------------------------------------------------------*/
i=wordpos(ritemost_s,chullx)
Do While i<words(chullx)
xya=word(chullx,i) ; Parse Var xya xa '/' ya
If xa=g.0xmin Then Leave
xyb=word(chullx,i+1); Parse Var xyb xb '/' yb
Do j=x.0 To 1 By -1
If x.j<xa Then Do
If x.j>xb Then Do
xx=x.j
parse Value kdx(xya,xyb) With k d x
If (k*xx+d)=minv(yl.xx) Then Do
chullx=subword(chullx,1,i) xx'/'minv(yl.xx),
subword(chullx,i+1)
i=i+1
End
End
End
Else
i=i+1
End
End
Say chullx
**********************************************************************/
Call lineout g.0oid
Exit
store: Procedure Expose x. yl. g.
/*---------------------------------------------------------------------
* arrange the points in ascending order of x (in x.) and,
* for each x in ascending order of y (in yl.x)
* g.0xmin is the smallest x-value, etc.
* g.0xmi is the x-coordinate
* g.0ymin is the smallest y-value, etc.
* g.0ymi is the x-coordinate of such a point
*--------------------------------------------------------------------*/
Parse Arg x,y
Call o 'store' x y
If x<g.0xmin Then Do; g.0xmin=x; g.0xmi=x; End
If x>g.0xmax Then Do; g.0xmax=x; g.0xma=x; End
If y<g.0ymin Then Do; g.0ymin=y; g.0ymi=x; End
If y>g.0ymax Then Do; g.0ymax=y; g.0yma=x; End
Do i=1 To x.0
Select
When x.i>x Then
Leave
When x.i=x Then Do
yl.x=yl.x y
Return
End
Otherwise
Nop
End
End
Do j=x.0 To i By -1
ja=j+1
x.ja=x.j
End
x.i=x
yl.x=y
x.0=x.0+1
Return
sho: Procedure Expose x. yl. g.
Do i=1 To x.0
x=x.i
say format(i,2) 'x='format(x,3) 'yl='yl.x
End
Say ''
Return
maxv: Procedure Expose g.
Call trace 'O'
Parse Arg l
res=-1000
Do While l<>''
Parse Var l v l
If v>res Then res=v
End
Return res
minv: Procedure Expose g.
Call trace 'O'
Parse Arg l
res=1000
Do While l<>''
Parse Var l v l
If v<res Then res=v
End
Return res
sortv: Procedure Expose g.
Call trace 'O'
Parse Arg l
res=''
Do Until l=''
v=minv(l)
res=res v
l=remove(v,l)
End
Return space(res)
lastword: return word(arg(1),words(arg(1)))
kdx: Procedure Expose xy. g.
/*---------------------------------------------------------------------
* Compute slope and y-displacement of a straight line
* that is defined by two points: y=k*x+d
* Specialty; k='*' x=xa if xb=xa
*--------------------------------------------------------------------*/
Call trace 'O'
Parse Arg xya,xyb
Parse Var xya xa '/' ya
Parse Var xyb xb '/' yb
If xa=xb Then
Parse Value '*' '-' xa With k d x
Else Do
k=(yb-ya)/(xb-xa)
d=yb-k*xb
x='*'
End
Return k d x
remove:
/*---------------------------------------------------------------------
* Remove a specified element (e) from a given string (s)
*--------------------------------------------------------------------*/
Parse Arg e,s
Parse Var s sa (e) sb
Return space(sa sb)
o: Procedure Expose g.
/*---------------------------------------------------------------------
* Write a line to the debug file
*--------------------------------------------------------------------*/
If arg(2)=1 Then say arg(1)
Return lineout(g.0oid,arg(1))
is_ok: Procedure Expose x. yl. g. sigl
/*---------------------------------------------------------------------
* Test if a given point (b) is above/on/or below a straight line
* defined by two points (a and c)
*--------------------------------------------------------------------*/
Parse Arg a,b,c,op
Call o 'is_ok' a b c op
Parse Value kdx(a,c) With k d x
Parse Var b x'/'y
If op='U' Then y=maxv(yl.x)
Else y=minv(yl.x)
Call o y x (k*x+d)
If (abs(y-(k*x+d))<1.e-8) Then Return 0
If op='U' Then res=(y<=(k*x+d))
Else res=(y>=(k*x+d))
Return res
mk_nhull: Procedure Expose x. yl. g.
/*---------------------------------------------------------------------
* Compute the upper (north) hull between two points (xya and xyb)
* Move x from xyb back to xya until all points within the current
* range (x and xyb) are BELOW the straight line defined xya and x
* Then make x the new starting point
*--------------------------------------------------------------------*/
Parse Arg xya,xyb
Call o 'mk_nhull' xya xyb
If xya=xyb Then Return xya
Parse Var xya xa '/' ya
Parse Var xyb xb '/' yb
iu=0
iv=0
Do xi=1 To x.0
if x.xi>=xa & iu=0 Then iu=xi
if x.xi<=xb Then iv=xi
If x.xi>xb Then Leave
End
Call o iu iv
xu=x.iu
xyu=xu'/'maxv(yl.xu)
Do h=iv To iu+1 By -1 Until good
Call o 'iv='iv,g.0debug
Call o ' h='h,g.0debug
xh=x.h
xyh=xh'/'maxv(yl.xh)
Call o 'Testing' xyu xyh,g.0debug
good=1
Do hh=h-1 To iu+1 By -1 While good
xhh=x.hh
xyhh=xhh'/'maxv(yl.xhh)
Call o 'iu hh iv=' iu hh h,g.0debug
If is_ok(xyu,xyhh,xyh,'U') Then Do
Call o xyhh 'is under' xyu xyh,g.0debug
Nop
End
Else Do
good=0
Call o xyhh 'is above' xyu xyh '-' xyh 'ist nicht gut'
End
End
End
Call o xyh 'is the one'
Return xyh
p: Return
Say arg(1)
Pull .
Return
mk_shull: Procedure Expose x. yl. g.
/*---------------------------------------------------------------------
* Compute the lower (south) hull between two points (xya and xyb)
* Move x from xyb back to xya until all points within the current
* range (x and xyb) are ABOVE the straight line defined xya and x
* Then make x the new starting point
*-----
---------------------------------------------------------------*/
Parse Arg xya,xyb
Call o 'mk_shull' xya xyb
If xya=xyb Then Return xya
Parse Var xya xa '/' ya
Parse Var xyb xb '/' yb
iu=0
iv=0
Do xi=x.0 To 1 By -1
if x.xi<=xa & iu=0 Then iu=xi
if x.xi>=xb Then iv=xi
If x.xi<xb Then Leave
End
Call o iu iv '_' x.iu x.iv
Call o 'mk_shull iv iu' iv iu
xu=x.iu
xyu=xu'/'minv(yl.xu)
good=0
Do h=iv To iu-1 Until good
xh=x.h
xyh=xh'/'minv(yl.xh)
Call o 'Testing' xyu xyh h iu
good=1
Do hh=h+1 To iu-1 While good
Call o 'iu hh h=' iu hh h
xhh=x.hh
xyhh=xhh'/'minv(yl.xhh)
If is_ok(xyu,xyhh,xyh,'O') Then Do
Call o xyhh 'is above' xyu xyh
Nop
End
Else Do
Call o xyhh 'is under' xyu xyh '-' xyh 'ist nicht gut'
good=0
End
End
End
Call o xyh 'is the one'
Return xyh
Novalue:
Say 'Novalue raised in line' sigl
Say sourceline(sigl)
Say 'Variable' condition('D')
Signal lookaround
Syntax:
Say 'Syntax raised in line' sigl
Say sourceline(sigl)
Say 'rc='rc '('errortext(rc)')'
halt:
lookaround:
Say 'You can look around now.'
Trace ?R
Nop
Exit 12
```
{{out}}
``` 1 x= -9 yl=-3
2 x= -4 yl=-6 -2
3 x= -3 yl=-9 15
4 x= 0 yl=6 11
5 x= 3 yl=-4 16
6 x= 5 yl=19
7 x= 12 yl=13 17
8 x= 16 yl=-7 -3 3 6
9 x= 17 yl=-4 5
10 x= 19 yl=-8
Points of convex hull in clockwise order:
-9/-3 -3/15 5/19 12/17 17/5 19/-8 -3/-9
```
### version 2
```/* REXX ---------------------------------------------------------------
* Compute the Convex Hull for a set of points
* Format of the input file:
* (16,3) (12,17) (0,6) (-4,-6) (16,6) (16,-7) (16,-3) (17,-4) (5,19)
* (19,-8) (3,16) (12,13) (3,-4) (17,5) (-3,15) (-3,-9) (0,11) (-9,-3)
* (-4,-2)
* Alternate (better) method using slopes
* 1) Compute path from lowest/leftmost to leftmost/lowest
* 2) Compute leftmost vertical border
* 3) Compute path from rightmost/highest to highest/rightmost
* 4) Compute path from highest/rightmost to rightmost/highest
* 5) Compute rightmost vertical border
* 6) Compute path from rightmost/lowest to lowest_leftmost point
*--------------------------------------------------------------------*/
Parse Arg fid
If fid='' Then Do
fid='line.in'
fid='point.in'
fid='chullmin.in' /* miscellaneous test data */
fid='chullxx.in'
fid='chullx.in'
fid='chullt.in'
fid='chulla.in'
fid='sq.in'
fid='tri.in'
fid='z.in'
fid='chull.in' /* data from task description */
End
g.0debug=''
g.0oid=fn(fid)'.txt'; 'erase' g.0oid
x.=0
yl.=''
Parse Value '1000 -1000' With g.0xmin g.0xmax
Parse Value '1000 -1000' With g.0ymin g.0ymax
/*---------------------------------------------------------------------
* First read the input and store the points' coordinates
* x.0 contains the number of points, x.i contains the x.coordinate
* yl.x contains the y.coordinate(s) of points (x/y)
*--------------------------------------------------------------------*/
Do while lines(fid)>0
l=linein(fid)
Do While l<>''
Parse Var l '(' x ',' y ')' l
Call store x,y
End
End
Call lineout fid
g.0xlist=''
Do i=1 To x.0 /* loop over points */
x=x.i
g.0xlist=g.0xlist x
yl.x=sortv(yl.x) /* sort y-coordinates */
End
Call sho
If x.0<3 Then Do
Say 'We need at least three points!'
Exit
End
Call o 'g.0xmin='g.0xmin
Call o 'g.0xmi ='g.0xmi
Call o 'g.0ymin='g.0ymin
Call o 'g.0ymi ='g.0ymi
Do i=1 To x.0
x=x.i
If minv(yl.x)=g.0ymin Then Leave
End
lowest_leftmost=i
highest_rightmost=0
Do i=1 To x.0
x=x.i
If maxv(yl.x)=g.0ymax Then
highest_rightmost=i
If maxv(yl.x)<g.0ymax Then
If highest_rightmost>0 Then
Leave
End
Call o 'lowest_leftmost='lowest_leftmost
Call o 'highest_rightmost ='highest_rightmost
x=x.lowest_leftmost
Call o 'We start at' from x'/'minv(yl.x)
path=x'/'minv(yl.x)
/*---------------------------------------------------------------------
* 1) Compute path from lowest/leftmost to leftmost/lowest
*--------------------------------------------------------------------*/
Call min_path lowest_leftmost,1
/*---------------------------------------------------------------------
* 2) Compute leftmost vertical border
*--------------------------------------------------------------------*/
Do i=2 To words(yl.x)
path=path x'/'word(yl.x,i)
End
cxy=x'/'maxv(yl.x)
/*---------------------------------------------------------------------
* 3) Compute path from rightmost/highest to highest/rightmost
*--------------------------------------------------------------------*/
Call max_path ci,highest_rightmost
/*---------------------------------------------------------------------
* 4) Compute path from highest/rightmost to rightmost/highest
*--------------------------------------------------------------------*/
Call max_path ci,x.0
/*---------------------------------------------------------------------
* 5) Compute rightmost vertical border
*--------------------------------------------------------------------*/
Do i=words(yl.x)-1 To 1 By -1
cxy=x'/'word(yl.x,i)
path=path cxy
End
/*---------------------------------------------------------------------
* 6) Compute path from rightmost/lowest to lowest_leftmost
*--------------------------------------------------------------------*/
Call min_path ci,lowest_leftmost
Parse Var path pathx path
have.=0
Do i=1 By 1 While path>''
Parse Var path xy path
If have.xy=0 Then Do
pathx=pathx xy
have.xy=1
End
End
Say 'Points of convex hull in clockwise order:'
Say pathx
Call lineout g.0oid
Exit
min_path:
Parse Arg from,tgt
ci=from
cxy=x.ci
Do Until ci=tgt
kmax=-1000
Do i=ci-1 To 1 By sign(tgt-from)
x=x.i
k=k(cxy'/'minv(yl.cxy),x'/'minv(yl.x))
If k>kmax Then Do
kmax=k
ii=i
End
End
ci=ii
cxy=x.ii
path=path cxy'/'minv(yl.cxy)
End
Return
max_path:
Parse Arg from,tgt
Do Until ci=tgt
kmax=-1000
Do i=ci+1 To tgt
x=x.i
k=k(cxy,x'/'maxv(yl.x))
If k>kmax Then Do
kmax=k
ii=i
End
End
x=x.ii
cxy=x'/'maxv(yl.x)
path=path cxy
ci=ii
End
Return
store: Procedure Expose x. yl. g.
/*---------------------------------------------------------------------
* arrange the points in ascending order of x (in x.) and,
* for each x in ascending order of y (in yl.x)
* g.0xmin is the smallest x-value, etc.
* g.0xmi is the x-coordinate
* g.0ymin is the smallest y-value, etc.
* g.0ymi is the x-coordinate of such a point
*--------------------------------------------------------------------*/
Parse Arg x,y
Call o 'store' x y
If x<g.0xmin Then Do; g.0xmin=x; g.0xmi=x; End
If x>g.0xmax Then Do; g.0xmax=x; g.0xma=x; End
If y<g.0ymin Then Do; g.0ymin=y; g.0ymi=x; End
If y>g.0ymax Then Do; g.0ymax=y; g.0yma=x; End
Do i=1 To x.0
Select
When x.i>x Then
Leave
When x.i=x Then Do
yl.x=yl.x y
Return
End
Otherwise
Nop
End
End
Do j=x.0 To i By -1
ja=j+1
x.ja=x.j
End
x.i=x
yl.x=y
x.0=x.0+1
Return
sho: Procedure Expose x. yl. g.
Do i=1 To x.0
x=x.i
say format(i,2) 'x='format(x,3) 'yl='yl.x
End
Say ''
Return
maxv: Procedure Expose g.
Call trace 'O'
Parse Arg l
res=-1000
Do While l<>''
Parse Var l v l
If v>res Then res=v
End
Return res
minv: Procedure Expose g.
Call trace 'O'
Parse Arg l
res=1000
Do While l<>''
Parse Var l v l
If v<res Then res=v
End
Return res
sortv: Procedure Expose g.
Call trace 'O'
Parse Arg l
res=''
Do Until l=''
v=minv(l)
res=res v
l=remove(v,l)
End
Return space(res)
lastword: return word(arg(1),words(arg(1)))
k: Procedure
/*---------------------------------------------------------------------
* Compute slope of a straight line
* that is defined by two points: y=k*x+d
* Specialty; k='*' x=xa if xb=xa
*--------------------------------------------------------------------*/
Call trace 'O'
Parse Arg xya,xyb
Parse Var xya xa '/' ya
Parse Var xyb xb '/' yb
If xa=xb Then
k='*'
Else
k=(yb-ya)/(xb-xa)
Return k
remove:
/*---------------------------------------------------------------------
* Remove a specified element (e) from a given string (s)
*--------------------------------------------------------------------*/
Parse Arg e,s
Parse Var s sa (e) sb
Return space(sa sb)
o: Procedure Expose g.
/*---------------------------------------------------------------------
* Write a line to the debug file
*--------------------------------------------------------------------*/
If arg(2)=1 Then say arg(1)
Return lineout(g.0oid,arg(1))
```
{{out}}
``` 1 x= -9 yl=-3
2 x= -4 yl=-6 -2
3 x= -3 yl=-9 15
4 x= 0 yl=6 11
5 x= 3 yl=-4 16
6 x= 5 yl=19
7 x= 12 yl=13 17
8 x= 16 yl=-7 -3 3 6
9 x= 17 yl=-4 5
10 x= 19 yl=-8
Points of convex hull in clockwise order:
-3/-9 -9/-3 -3/15 5/19 12/17 17/5 19/-8 -3/-9
```
## Rust
Calculates convex hull from list of points (f32, f32). This can be executed entirely in the Rust Playground.
```
#[derive(Debug, Clone)]
struct Point {
x: f32,
y: f32
}
fn calculate_convex_hull(points: &Vec<Point>) -> Vec<Point> {
//There must be at least 3 points
if points.len() < 3 { return points.clone(); }
let mut hull = vec![];
//Find the left most point in the polygon
let (left_most_idx, _) = points.iter()
.enumerate()
.min_by(|lhs, rhs| lhs.1.x.partial_cmp(&rhs.1.x).unwrap())
.expect("No left most point");
let mut p = left_most_idx;
let mut q = 0_usize;
loop {
//The left most point must be part of the hull
hull.push(points[p].clone());
q = (p + 1) % points.len();
for i in 0..points.len() {
if orientation(&points[p], &points[i], &points[q]) == 2 {
q = i;
}
}
p = q;
//Break from loop once we reach the first point again
if p == left_most_idx { break; }
}
return hull;
}
//Calculate orientation for 3 points
//0 -> Straight line
//1 -> Clockwise
//2 -> Counterclockwise
fn orientation(p: &Point, q: &Point, r: &Point) -> usize {
let val = (q.y - p.y) * (r.x - q.x) -
(q.x - p.x) * (r.y - q.y);
if val == 0. { return 0 };
if val > 0. { return 1; } else { return 2; }
}
fn main(){
let points = vec![pt(16,3), pt(12,17), pt(0,6), pt(-4,-6), pt(16,6), pt(16,-7), pt(16,-3), pt(17,-4), pt(5,19), pt(19,-8), pt(3,16), pt(12,13), pt(3,-4), pt(17,5), pt(-3,15), pt(-3,-9), pt(0,11), pt(-9,-3), pt(-4,-2), pt(12,10)];
let hull = calculate_convex_hull(&points);
hull.iter()
.for_each(|pt| println!("{:?}", pt));
}
fn pt(x: i32, y: i32) -> Point {
return Point {x:x as f32, y:y as f32};
}
```
{{out}}
```
Point { x: -9.0, y: -3.0 }
Point { x: -3.0, y: -9.0 }
Point { x: 19.0, y: -8.0 }
Point { x: 17.0, y: 5.0 }
Point { x: 12.0, y: 17.0 }
Point { x: 5.0, y: 19.0 }
Point { x: -3.0, y: 15.0 }
```
## Scala
Scala Implementation to find Convex hull of given points collection. Functional Paradigm followed
```
object convex_hull{
def get_hull(points:List[(Double,Double)], hull:List[(Double,Double)]):List[(Double,Double)] = points match{
case Nil => join_tail(hull,hull.size -1)
}
def reduce(hull:List[(Double,Double)]):List[(Double,Double)] = hull match{
case p1::p2::p3::rest => {
if(check_point(p1,p2,p3)) hull
else reduce(p1::p3::rest)
}
case _ => hull
}
def check_point(pnt:(Double,Double), p2:(Double,Double),p1:(Double,Double)): Boolean = {
val (x,y) = (pnt._1,pnt._2)
val (x1,y1) = (p1._1,p1._2)
val (x2,y2) = (p2._1,p2._2)
((x-x1)*(y2-y1) - (x2-x1)*(y-y1)) <= 0
}
def m(p1:(Double,Double), p2:(Double,Double)):Double = {
if(p2._1 == p1._1 && p1._2>p2._2) 90
else if(p2._1 == p1._1 && p1._2<p2._2) -90
else if(p1._1<p2._1) 180 - Math.toDegrees(Math.atan(-(p1._2 - p2._2)/(p1._1 - p2._1)))
else Math.toDegrees(Math.atan((p1._2 - p2._2)/(p1._1 - p2._1)))
}
def join_tail(hull:List[(Double,Double)],len:Int):List[(Double,Double)] = {
if(m(hull(len),hull(0)) > m(hull(len-1),hull(0))) join_tail(hull.slice(0,len),len-1)
else hull
}
def main(args:Array[String]){
val points = List[(Double,Double)]((16,3), (12,17), (0,6), (-4,-6), (16,6), (16,-7), (16,-3), (17,-4), (5,19), (19,-8), (3,16), (12,13), (3,-4), (17,5), (-3,15), (-3,-9), (0,11), (-9,-3), (-4,-2), (12,10))
val sorted_points = points.sortWith(m(_,(0.0,0.0)) < m(_,(0.0,0.0)))
println(f"Points:\n" + points + f"\n\nConvex Hull :\n" +get_hull(sorted_points,List[(Double,Double)]()))
}
}
```
{{out}}
```
Points:
List((16.0,3.0), (12.0,17.0), (0.0,6.0), (-4.0,-6.0), (16.0,6.0), (16.0,-7.0), (16.0,-3.0), (17.0,-4.0), (5.0,19.0), (19.0,-8.0), (3.0,16.0), (12.0,13.0), (3.0,-4.0), (17.0,5.0), (-3.0,15.0), (-3.0,-9.0), (0.0,11.0), (-9.0,-3.0), (-4.0,-2.0), (12.0,10.0))
Convex Hull :
List((-3.0,-9.0), (-9.0,-3.0), (-3.0,15.0), (5.0,19.0), (12.0,17.0), (17.0,5.0), (19.0,-8.0))
```
## Sidef
{{trans|Perl 6}}
```class Point(Number x, Number y) {
method to_s {
"(#{x}, #{y})"
}
}
func ccw (Point a, Point b, Point c) {
(b.x - a.x)*(c.y - a.y) - (b.y - a.y)*(c.x - a.x)
}
func tangent (Point a, Point b) {
(b.x - a.x) / (b.y - a.y)
}
func graham_scan (*coords) {
## sort points by y, secondary sort on x
var sp = coords.map { |a|
Point(a...)
}.sort { |a,b|
(a.y <=> b.y) ||
(a.x <=> b.x)
}
# need at least 3 points to make a hull
if (sp.len < 3) {
return sp
}
# first point on hull is minimum y point
var h = [sp.shift]
# re-sort the points by angle, secondary on x
sp = sp.map_kv { |k,v|
Pair(k, [tangent(h[0], v), v.x])
}.sort { |a,b|
(b.value[0] <=> a.value[0]) ||
(a.value[1] <=> b.value[1])
}.map { |a|
sp[a.key]
}
# first point of re-sorted list is guaranteed to be on hull
h << sp.shift
# check through the remaining list making sure that
# there is always a positive angle
sp.each { |point|
loop {
if (ccw(h.last(2)..., point) >= 0) {
h << point
break
} else {
h.pop
}
}
}
return h
}
var hull = graham_scan(
[16, 3], [12,17], [ 0, 6], [-4,-6], [16, 6], [16,-7], [16,-3],
[17,-4], [ 5,19], [19,-8], [ 3,16], [12,13], [ 3,-4], [17, 5],
[-3,15], [-3,-9], [ 0,11], [-9,-3], [-4,-2], [12,10])
say("Convex Hull (#{hull.len} points): ", hull.join(" "))
hull = graham_scan(
[16, 3], [12,17], [ 0, 6], [-4,-6], [16, 6], [16,-7], [16,-3],
[17,-4], [ 5,19], [19,-8], [ 3,16], [12,13], [ 3,-4], [17, 5],
[-3,15], [-3,-9], [ 0,11], [-9,-3], [-4,-2], [12,10], [14,-9], [1,-9])
say("Convex Hull (#{hull.len} points): ", hull.join(" "))
```
{{out}}
```
Convex Hull (7 points): (-3, -9) (19, -8) (17, 5) (12, 17) (5, 19) (-3, 15) (-9, -3)
Convex Hull (9 points): (-3, -9) (1, -9) (14, -9) (19, -8) (17, 5) (12, 17) (5, 19) (-3, 15) (-9, -3)
```
## Visual Basic .NET
{{trans|C#}}
```Imports ConvexHull
Module Module1
Class Point : Implements IComparable(Of Point)
Public Property X As Integer
Public Property Y As Integer
Public Sub New(x As Integer, y As Integer)
Me.X = x
Me.Y = y
End Sub
Public Function CompareTo(other As Point) As Integer Implements IComparable(Of Point).CompareTo
Return X.CompareTo(other.X)
End Function
Public Overrides Function ToString() As String
Return String.Format("({0}, {1})", X, Y)
End Function
End Class
Function ConvexHull(p As List(Of Point)) As List(Of Point)
If p.Count = 0 Then
Return New List(Of Point)
End If
p.Sort()
Dim h As New List(Of Point)
' Lower hull
For Each pt In p
While h.Count >= 2 AndAlso Not Ccw(h(h.Count - 2), h(h.Count - 1), pt)
h.RemoveAt(h.Count - 1)
End While
Next
' Upper hull
Dim t = h.Count + 1
For i = p.Count - 1 To 0 Step -1
Dim pt = p(i)
While h.Count >= t AndAlso Not Ccw(h(h.Count - 2), h(h.Count - 1), pt)
h.RemoveAt(h.Count - 1)
End While
Next
h.RemoveAt(h.Count - 1)
Return h
End Function
Function Ccw(a As Point, b As Point, c As Point) As Boolean
Return ((b.X - a.X) * (c.Y - a.Y)) > ((b.Y - a.Y) * (c.X - a.X))
End Function
Sub Main()
Dim points As New List(Of Point) From {
New Point(16, 3),
New Point(12, 17),
New Point(0, 6),
New Point(-4, -6),
New Point(16, 6),
New Point(16, -7),
New Point(16, -3),
New Point(17, -4),
New Point(5, 19),
New Point(19, -8),
New Point(3, 16),
New Point(12, 13),
New Point(3, -4),
New Point(17, 5),
New Point(-3, 15),
New Point(-3, -9),
New Point(0, 11),
New Point(-9, -3),
New Point(-4, -2),
New Point(12, 10)
}
Dim hull = ConvexHull(points)
Dim it = hull.GetEnumerator()
Console.Write("Convex Hull: [")
If it.MoveNext() Then
Console.Write(it.Current)
End If
While it.MoveNext()
Console.Write(", {0}", it.Current)
End While
Console.WriteLine("]")
End Sub
End Module
```
{{out}}
```Convex Hull: [(-9, -3), (-3, -9), (19, -8), (17, 5), (12, 17), (5, 19), (-3, 15)]
```
## zkl
```// Use Graham Scan to sort points into a convex hull
// https://en.wikipedia.org/wiki/Graham_scan, O(n log n)
// http://www.geeksforgeeks.org/convex-hull-set-2-graham-scan/
// http://geomalgorithms.com/a10-_hull-1.html
fcn grahamScan(points){
N:=points.len();
# find the point with the lowest y-coordinate, x is tie breaker
p0:=points.reduce(fcn([(a,b)]ab,[(x,y)]xy){
if(b<y)ab else if(b==y and a<x)ab else xy });
#sort points by polar angle with p0, ie ccw from p0
points.sort('wrap(p1,p2){ ccw(p0,p1,p2)>0 });
# We want points[0] to be a sentinel point that will stop the loop.
points.insert(0,points[-1]);
M:=1; # M will denote the number of points on the convex hull.
foreach i in ([2..N]){
# Find next valid point on convex hull.
while(ccw(points[M-1], points[M], points[i])<=0){
if(M>1) M-=1;
else if(i==N) break; # All points are collinear
else i+=1;
}
points.swap(M+=1,i); # Update M and swap points[i] to the correct place.
}
points[0,M]
}
# Three points are a counter-clockwise turn if ccw > 0, clockwise if
# ccw < 0, and collinear if ccw = 0 because ccw is a determinant that
# gives twice the signed area of the triangle formed by p1, p2 and p3.
fcn ccw(a,b,c){ // a,b,c are points: (x,y)
((b[0] - a[0])*(c[1] - a[1])) - ((b[1] - a[1])*(c[0] - a[0]))
}
```
```pts:=List( T(16,3), T(12,17), T(0,6), T(-4,-6), T(16,6),
T(16, -7), T(16,-3),T(17,-4), T(5,19), T(19,-8),
T(3,16), T(12,13), T(3,-4), T(17,5), T(-3,15),
T(-3,-9), T(0,11), T(-9,-3), T(-4,-2), T(12,10), )
.apply(fcn(xy){ xy.apply("toFloat") }).copy();
hull:=grahamScan(pts);
println("Convex Hull (%d points): %s".fmt(hull.len(),hull.toString(*)));
```
{{out}}
```
Convex Hull (7 points): L(L(-3,-9),L(19,-8),L(17,5),L(12,17),L(5,19),L(-3,15),L(-9,-3))
```
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# Why do we need an Order of Operations?
27 teachers like this lesson
Print Lesson
## Objective
SWBAT: • Explain the Order of Operations • Use the OOO graphic organizer to simplify numerical expressions
#### Big Idea
Why do we need an Order of Operations? What is 5 + 3 x 4? What is 3 x 4 + 5? Students work through examples to get at these questions and work with the order of operations to simplify numerical expressions.
## Do Now
10 minutes
See my Do Now in my Strategy folder that explains my beginning of class routines.
Often, I create do nows that have problems that connect to the task that students will be working on that day. Here, I want students to review what they learned about exponents in the previous lesson. I am looking for students to write out the repeated multiplication to show which base and exponent is greater. A common mistake is for students to think 6^2 as 6 times 2. If students make this mistake, I remind students that exponents indicate repeated multiplication. I also ask students to share an efficient way to multiply powers of 10.
## Problems
10 minutes
Students work on the two problems and questions independently. I walk around to monitor student progress. When most students have completed the problems and questions we come back together as a class. I have students Think Pair Share about their answers. Then I call on students to share their observations with the class. Some students may say that both problems result in 17. Some students may say that the problems have different answers. Possible answers could be 32 or 27.
One option is to add context to the problems. You have a bunch of coins. You have a stack of 5 coins and 3 stacks of 4 coins. You can represent this number of coins with these two expressions. How many coins do you have? What if you have 3 stacks of 4 coins and one stack of 5 coins? Have students draw the pictures to compare and contrast the problems.
## The Order of Operations
7 minutes
A common misconception for my students is to think about the order of operations as a six step process, and forget that you evaluate multiplication and division problems (or addition and subtraction) from left to right. For this reason I do not use the mnemonic PEMDAS in my class, I refer to them as the order of operations. I’ve found students latch on to PEMDAS and then assume multiplication always comes before division and addition before subtraction. I use this horizontal graphic organizer with my students. In each box I put the symbol for the step (or a number with an exponent). I regularly switch where I place the multiplication and division (and addition and subtraction) symbols as a constant reminder that students must look for these symbols from left to right.
Then I ask these questions:
• What would happen if we did not agree and we approached evaluating expressions in our own ways?
• What problems might arise?
I am looking for students to recognize that the order of operations is a set of rules that have been decided upon by mathematicians. Without these rules, different people would solve the same problems and get vastly different answers. I am interested to hear what problems students imagine would occur without agreement on the order of operations.
## Practice
23 minutes
We complete the 4 questions on the “Practice” page together. With each problem I ask what we need to identify and work with first, next, etc. I stress to students that it is essential that they only complete one step of the expression at a time and that they copy the rest of the expression, until they are left with an answer. Many students will be tempted to do it in their heads and write down their answers. I explain that I will not check any work that does not show a step-by-step progression. I explain that in order for me to understand what they did (correct or incorrect) I need to see their work. Here it is crucial that students use MP6: Attend to precision.
For students who are doing well with the problems, they can move on to Practice 1. I explain that the practice pages get more difficult. If students are feeling confident, they can complete the last 3 problems on Practice 1. Once they finish, I quickly check their work. If all three are correct they can move on to Practice 2. If there are any mistakes, I direct them to try the incorrect problem again. Posting A Key can make checking student work much more efficient for me and the students. Unit 1.10 Posted Key 1.jpg Unit 1.10 Posted Key 2.jpeg
I am looking to see if students are using the graphic organizer to help them. Do they perform multiplication and division (as well as addition and subtraction) from left to right? Are they working correctly with exponents? Are they making silly math mistakes? What are the most common mistakes? I will use these observations to inform the Closure of the lesson.
For students who are struggling, there are a few ways I may intervene:
• Pull a small group to the back of the class to work on Practice 1 problems together. Once students are more comfortable, they return to their seats to work independently.
• I have them work with their partners together and I regularly check in with them.
• If there are enough students struggling, I’ll pass out whiteboards and markers. I have students show the step-by-step work on the first few practice problems.
• Give students a multiplication chart or calculator to help them access the problems.
For students who successfully complete the classwork, there are a few choices I give them:
• Serve as a tutor to a student I have identified who needs extra help
• Complete the Challenge problems
• Pair up with another student and take turns creating and evaluating numerical expressions using the order of operations
• Playing “Show me the Money!” from Show what you know Factors and Multiples + Introduction to Exponents (students can use the 1-10 spinner for added difficulty).
## Closure and Ticket to Go
10 minutes
For Closure I have students turn to problem 6 in Practice 3. What do we do first? Some students will struggle since there are two operations inside the parentheses. Should we subtract first or simplify 2^3? I have students explain the step by step process. I ask again, why do we have an order of operations? What might happen if we didn’t?
With the last few minutes of class I give a Ticket to Go for students to independently complete. I pass out the HW Why do we need an order of operations at the end of class.
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# Thread: A whole new box idea
1. But, Hunnytree, there are spaces for 12 jewels! One in each of the smaller boxes, and one in each of the spaces in the larger box NOT taken up by the smaller boxes (the white areas in my first diagram).
2. I see what you are getting at.
Well, that IS the shape of the jewelry box that Snail is pointing to in the 'flip book' in the Table of Contents. If you measure that box, altho it is not exact, it is pretty close to 4 centimeters long, and 2 centimeters wide. When opened, it would be a square box as long as it is wide with two sides, each 2 x 4.
Tried it with a little piece of paper, folded in half, adjusted for perspective, of course. It's a square, as long as it is wide, when unfolded.
3. Junior Twelever +1 Silver
Join Date
Jan 2005
Posts
220
I am a tad confused though. Aren't the box from the grotto and the jewelry box that Zac made two totally differentl things?
sorry if this is a dumb question... but that's the assumption I've been working under. Not good to assume so I thought I'd clear it up for myself.
4. Luddite...Notwithstanding....have you thought of approaching the Ten Commandments from a Beale cipher point of view ??
I'm thinking about it. What if...
A code of numbers - describes the cipher - MS uses the Beale cipher in CPB as an example - it was constructed from a very important document - the Declaration of Independence
Five to a side - describes the very important document to apply it to - The Ten Commandments
the image you posted was an excellent trigger
thanks -
edit: well I tried that - got zilch
5. I think the idea is creative, witty, and may have some promise, but the problem would be which translation to use.
6. I went to wikipedia....common knowledge pretty much
used "simple english" set
el zilcho
but thanks for the props
7. Originally Posted by luddite001
Or what if the solution has nothing at all to do with boxes? The riddle refers to a "code of numbers five to a side." What if the "code" the author refers to is not a cipher, after all, but a "moral" code. How's this for a possibility?
Let's all start anagramming T H E T E N C O M M A N D M E N T S
As interesting a point as this is, we run into problems in translation. The ten commandments, after all, were not written in English...
8. Needs to say Hello!
Join Date
Mar 2005
Posts
1
sorry but luddite has got to be wrong, because i live in miami and it would take me over 24 hours to get to any of those states!!
9. You must drive REALLY slow. Miami to Biloxi is 12 hours, 31 minutes, if you follow the posted speed limits.
10. can't the box be a 5x5x5 equilateral triangle? Now you're only filling 15 slots instead of 25.
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# DHS 15.2 - Option 18. Decisions Ryabushko AP
• USD
• RUB
• USD
• EUR
Affiliates: 0,01 \$how to earn
Sold: 2 last one 19.12.2017
Refunds: 0
Content: 18v-IDZ15.2.doc 114 kB
## Product description
1. Calculate and circulation of the vector field (M) over the contour of a triangle obtained by intersection of the plane (p): Ax + By + Cz = D with the coordinate planes, with respect to the positive direction of the normal vector bypass n = (A, B, C) this plane in two ways: 1) using the definition of circulation; 2) using the Stokes formula.
1.18. a (M) = (x + 2z) i + (y - 3z) j + zk, (p): 3x + 2y + 2z = 6
2. Find the magnitude and direction of the greatest changes in the function u (M) = u (x, y, z) at the point M0 (x0, y0, z0)
2.18. u (M) = (x + z) y2, M0 (2, 2, 2)
3. Find the greatest density of the circulation of the vector field a (M) = (x, y, z) at the point M0 (x0, y0, z0)
3.18. a (M) = xyi - xj + yzk, M0 (2, 2, 2)
4. Determine whether the vector field a (M) = (x, y, z) the potential
4.18. a (M) = (x + y) i - 2xzj - 3 (y + z) k
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# paulfioravanti's solution
## to Change in the Elixir Track
Published at Aug 25 2019 · 0 comments
Instructions
Test suite
Solution
Correctly determine the fewest number of coins to be given to a customer such that the sum of the coins' value would equal the correct amount of change.
## For example
• An input of 15 with [1, 5, 10, 25, 100] should return one nickel (5) and one dime (10) or [0, 1, 1, 0, 0]
• An input of 40 with [1, 5, 10, 25, 100] should return one nickel (5) and one dime (10) and one quarter (25) or [0, 1, 1, 1, 0]
## Edge cases
• Does your algorithm work for any given set of coins?
• Can you ask for negative change?
• Can you ask for a change value smaller than the smallest coin value?
## Running tests
Execute the tests with:
``````\$ mix test
``````
### Pending tests
In the test suites, all but the first test have been skipped.
Once you get a test passing, you can unskip the next one by commenting out the relevant `@tag :pending` with a `#` symbol.
For example:
``````# @tag :pending
test "shouting" do
assert Bob.hey("WATCH OUT!") == "Whoa, chill out!"
end
``````
Or, you can enable all the tests by commenting out the `ExUnit.configure` line in the test suite.
``````# ExUnit.configure exclude: :pending, trace: true
``````
If you're stuck on something, it may help to look at some of the available resources out there where answers might be found.
## Source
Software Craftsmanship - Coin Change Kata https://web.archive.org/web/20130115115225/http://craftsmanship.sv.cmu.edu:80/exercises/coin-change-kata
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### change_test.exs
``````defmodule ChangeTest do
use ExUnit.Case
# @tag :pending
test "single coin change" do
coins = [1, 5, 10, 25, 100]
expected = [25]
assert Change.generate(coins, 25) == {:ok, expected}
end
@tag :pending
test "multiple coin change" do
coins = [1, 5, 10, 25, 100]
expected = [5, 10]
assert Change.generate(coins, 15) == {:ok, expected}
end
@tag :pending
test "change with Lilliputian Coins" do
coins = [1, 4, 15, 20, 50]
expected = [4, 4, 15]
assert Change.generate(coins, 23) == {:ok, expected}
end
@tag :pending
test "change with Lower Elbonia Coins" do
coins = [1, 5, 10, 21, 25]
expected = [21, 21, 21]
assert Change.generate(coins, 63) == {:ok, expected}
end
@tag :pending
test "large target values" do
coins = [1, 2, 5, 10, 20, 50, 100]
expected = [2, 2, 5, 20, 20, 50, 100, 100, 100, 100, 100, 100, 100, 100, 100]
assert Change.generate(coins, 999) == {:ok, expected}
end
@tag :pending
test "possible change without unit coins available" do
coins = [2, 5, 10, 20, 50]
expected = [2, 2, 2, 5, 10]
assert Change.generate(coins, 21) == {:ok, expected}
end
@tag :pending
test "no coins make 0 change" do
coins = [1, 5, 10, 21, 25]
expected = []
assert Change.generate(coins, 0) == {:ok, expected}
end
@tag :pending
test "error testing for change smaller than the smallest of coins" do
coins = [5, 10]
assert Change.generate(coins, 3) == {:error, "cannot change"}
end
@tag :pending
test "error if no combination can add up to target" do
coins = [5, 10]
assert Change.generate(coins, 94) == {:error, "cannot change"}
end
@tag :pending
test "cannot find negative change values" do
coins = [1, 2, 5]
assert Change.generate(coins, -5) == {:error, "cannot change"}
end
end``````
### test_helper.exs
``````ExUnit.start()
ExUnit.configure(exclude: :pending, trace: true)``````
``````defmodule Change do
@error_message "cannot change"
@doc """
Determine the least number of coins to be given to the user such
that the sum of the coins' value would equal the correct amount of change.
It returns {:error, "cannot change"} if it is not possible to compute the
right amount of coins. Otherwise returns the tuple {:ok, list_of_coins}
## Examples
iex> Change.generate([5, 10, 15], 3)
{:error, "cannot change"}
iex> Change.generate([1, 5, 10], 18)
{:ok, [1, 1, 1, 5, 10]}
"""
@spec generate(list, integer) :: {:ok, list} | {:error, String.t()}
def generate(_coins, 0), do: {:ok, []}
def generate(_coins, target) when target < 0, do: {:error, @error_message}
def generate(coins, target) do
if target < Enum.min(coins) do
{:error, @error_message}
else
{_coins, change_candidates, _acc} =
coins
|> generate_useable_coins(target)
|> generate_change_candidates(target)
case change_candidates do
[] ->
{:error, @error_message}
_ ->
fewest_coin_set = Enum.min_by(change_candidates, &length/1)
{:ok, fewest_coin_set}
end
end
end
defp generate_useable_coins(coins, target) do
coins
|> Enum.filter(&(&1 <= target))
|> Enum.sort(&(&1 >= &2))
end
defp generate_change_candidates(coins, target) do
coins
|> Enum.reduce({coins, [], []}, &generate_change(coins, target, &1, &2))
end
defp generate_change(coins, target, coin, {change_coins, candidates, acc}) do
combination = [coin | acc]
sum = Enum.sum(combination)
cond do
sum > target ->
case change_coins do
[] ->
handle_possible_change_without_unit_coins(coins, candidates, acc)
[head | []] ->
generate_change(coins, target, head, {[], candidates, acc})
[_head | tail] ->
generate_change(coins, target, hd(tail), {tail, candidates, acc})
end
sum == target ->
change_coins =
if Enum.empty?(change_coins) do
[]
else
tl(change_coins)
end
{change_coins, [combination | candidates], []}
# sum < target
true ->
generate_change(
coins,
target,
coin,
{change_coins, candidates, combination}
)
end
end
defp handle_possible_change_without_unit_coins(coins, candidates, acc) do
case length(candidates) do
0 ->
{coins, candidates, Enum.drop(acc, 1)}
_ ->
{[], candidates, acc}
end
end
end``````
## Community comments
Find this solution interesting? Ask the author a question to learn more.
### What can you learn from this solution?
A huge amount can be learned from reading other people’s code. This is why we wanted to give exercism users the option of making their solutions public.
Here are some questions to help you reflect on this solution and learn the most from it.
• What compromises have been made?
• Are there new concepts here that you could read more about to improve your understanding?
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# Algebra question type
Jump to: navigation, search
Algebra question type
Type question type
Set N/A
Downloads https://moodle.org/plugins/view.php?plugin=qtype_algebra
Issues Tracker issues
Discussion https://moodle.org/mod/forum/discuss.php?d=98670
Maintainer(s) Jean-Michel Védrine
The algebra question type allows algebraic expressions for student answers which are evaluated by instructor provided answers using the basic rules of algebra. It was created by Roger Moore and first offered as contributed code in July 2009. It is currently maintained by Jean-Michel Védrine
## Overview
This is a question type for Moodle.
It implements an algebraic question type where student responses are treated as an algebraic expression and compared to instructor provided answers using the basic rules of algebra.
For example if the instructor provided response is ${\displaystyle \sin 2\theta }$, then a student entering ${\displaystyle 2\sin \theta \cos \theta }$ will have their answer treated as correct.
The code has been tested in production use with ~230 first year physics students without problems (using the Evaluation comparison method only) - mainly thanks to some excellent feedback and testing from Moodle users on the quiz forum before deployment!
## Requisite
For the display to work you need to have some way of displaying TeX expressions activated on your Moodle website: either the TeX filter or the MathJax filter enabled.
Note that the new method to display a formatted answer formula "Using dynamic AJAX request" only works with the MathJax filter.
## Question Formulation
To create a new Algebra question, go to the Question Bank and click on "Create a new question ..."
Some fields (Question name, Question text, Default Mark, General Feedback) on the question edit page are common to all question types so they don't need special explanation.
But some other are specific to the Algebra question type.
Expand the Options section.
## Comparison Algorithm
First you define which comparison method is used to determinate if the student answer is correct or not and grade it, three method are provided:
1. Evaluation
2. Sage
3. Equivalence
### Evaluation
This is the default and best tested method. It's also the easier to understand. It "cheats" by evaluating both the answer and the student response for various random combinations of the variable's values and ensures that both match to within tolerance at all points.
You can specify the number of random values that will be generated for the test (more is better but also slower !) in the Number of Evaluation Checks field (default: 10) and the tolerance in the Tolerance for Evaluation Checks field (default 0.001).
If you plan to use this method, you must understand that it is not bulletproof and maybe not what you want.
As long as the student entered answer and the teacher provided answer take the same values for all random values of the variables generated by the plugin, the student answer will be considered as correct. For instance 2*x and x+x are considered as identical.
### Sage
This method requires the external, open source Sage symbolic algebra system to be installed on a server. A simple XMLRPC server script in Python is included in the package, sage_server.py, and this provides a method to symbolically compare two expressions.
From a theorical point of view this method is surely the best one. But it's also the uneasiest one because it require an external server and depending on the way your Moodle server is hosted this may or may not be possible.
If you absolutely need that sort of mathematic question and answer verification, it is suggested that you look at a more modern, more secure and more powerful question type, the STACK question type
### Equivalence
This uses the internal PHP algebra parser included in the parser to compare the two expressions. It was intended for use in "expand..." and "factor..." type questions where a full symbollic comparison would allow the student to enter the unexpanded (or unfactorized) expression.
The usage of this comparison method has received few testing and no users of the Moodle Quiz forum reported using it. So it is suggested to be very cautious if you plan to use it on a production server.
As a general rule it is recommended that you stick with the Evaluation comparison method unless very special need and that you know what you are doing.
## Answer box prefix
If you want some text displayed before the answer input box, you can type it here. For instance in a question about functions you can type f(x) = so that the student knows what to answer. See an example on one of the screenshots bellow.
## Disallowed Answer
You can type here a comma separated list of words that will be forbidden in student's answers. For instance if you type here pi, max students will not receive any grade for answer including one of these words.
## Allowed Functions
Here you can choose what functions will be allowed (default: All) in student answer.
## Variable
All Algebra question must include at least one variable and all defined variable must be used in at least one answer.
You must provide a variable name and if you use the Evaluation method a Minimum and Maximum value (SAGE and Equivalence methods don't use these values they are only used by the Evaluation method)
The variable name must be typed using linear notation not TeX syntax. If the variable name has more than one character, the rest is used as a subscript so the variable defined as xp will be displayed as ${\displaystyle x_{p}}$, and x1 will be displayed as ${\displaystyle x_{1}}$
You can also use greek names as variables names with or without a subscript, theta1 for instance will be displayed as ${\displaystyle \theta _{1}}$
## Answer
This is where you defines the answer and their matching grade. Again, you shall use linear syntax notation.
You may choose to give different answers with different values for the grade, but it is required that an answer with the highest grade is included.
Answer can only use varables names defined in the Variables section of the form and all defined variables must be used by at least one answer.
## Example
We create a new Algebra question entering these values in the General section
The formulas in the Question text and General feedback were typed using The Equation editor and will be latter rendered by the MathJax or TeX filter.
We now expand the Options section and enter the following values (most of them are default)
We now create one variable named x that will take values between -5 and 5 when the Evaluation comparison method will be used to grade student answers
And finally we create 2 answers the first one is for the correct answer and the second one for students that forgot the 2 and we choose to give them some little reward anyway :-)
The student answer does not have to be exactly this, only shall be evaluated to (very similar) values. For instance, the equivalent answer x + x would evaluate correctly and gives full mark.
The question is now finished we can save it and test it
If you use the old "Using an iframe" setting to display formula it will look like this:
And if you use the new "Using dynamic AJAX request" setting to display formula, it will look like this:
## Current Features
• The students enter responses in a text box, like calculated and numeric questions. They use the linear syntax notation. They don't need to know the TeX syntax and they must not use the equation editor.
• The formula they entered is rendered in the box below by clicking on the Display response button.
• Option of three different comparison methods: SAGE, Evaluation or Equivalence.
• Support for multiple answers so partial credit or assistance can be provided to students.
• Except for the SAGE comparison method core functionality works entirely within Moodle, so it can be used without installing any other program.
• Works with Moodle 2.8 and up to Moodle 3.4.
## New Features Coming Soon
The original documentation listed here some new features that Roger Moore planned to add. But this was never done so they have been removed. If you ave any idea on how to improve this question type, please create an issue in the tracker.
## Installing this question type
The Algebra question is a Moodle plugin, so you or your Administrator must install it.
Installing from the Moodle Plugins directory (this is the recommanded and easiest method)
The Algebra question type is available from the Moodle Plugins drectory. Install the Algebra question type as any other Moodle question type plugin.
Installing from GitHub
The Algebra question is available at the maintainer's Github repository. To install the question, type the following commands in the root of your Moodle install:
git clone git://github.com/jmvedrine/moodle-qtype_algebra.git question/type/algebra echo '/question/type/algebra/' >> .git/info/exclude
Installing from a zip file
Download the zip file from GitHub. Unzip the file in the 'question/type' folder and then rename the new folder to just 'algebra'.
During installation you will be prompted for new Administrator settings
It is suggested that you don't change the Default comparison method (Teachers can change it on a per question basis if they want and an Administrator can change the default later)
You can ignore the 3 settings related to SAGE server unless you plan to use the SAGE comparison method and you know what you do.
According to the way LaTeX and TeX formulas are displayed on your website it may be necessary to change the delimiters for TeX expressions from the default $$...$$ to $...$ or $...$ or $$...$$. This setting is only used used to display the formatted answer on the question view during attempt, and also to display the formatted correct answer after the question is submitted. It is not used to display formulas in question text, General feedback and everywhere else
The TeX operator for multiplication setting control the symbol used to display multiplication either times ${\displaystyle \times }$ or cdot ${\displaystyle \cdot }$.
NB: If you wish to test the current Sage functionality you will also need to install Sage on a server as well as run the python server code with the command "sage - python sage_server.py".
## Troubleshooting
• Failed to load question options from the table question_algebra for questionid <n>: this usually occurs because you tried to add an algebra question into Moodle without having propertly installed the algebra plugin, so the tables for the question type have not been added to the database. Any existing algebra questions that were added (via editing or import) before this is corrected will need to be deleted since they are missing important information and are broken.
## Linear syntax
To use this question type, you must understand that there are 2 very different syntax that are used.
When creating the question, when you type something into the different editors on the question edit page (Question text, General feedback, ...), you do as usual, if you want to enter some mathematical formula, you use the equation editor or you type some TeX formulas as you would do in a forum post, a lesson or a description.
But when you enter a formula in the Answer 1, Answer 2, ... fields the situation is quite different . These formulas must be parsed by the formula question code so a special syntax is used here and TeX syntax must not be used here.,
The algebra question type uses what is called a linear syntax to input math formulas : *, +, -, /, /, (, [, ], ), pi, e, and the known functions sqrt, log, ln, sin, cos, tan, cosh, sinh, asin, acos, atan are used.
Variables can be uppercase and lowercase letters, Greek letters are recognized and subscripts too, for instance if you type u1 it will be interpreted as ${\displaystyle u_{1}}$ and alpha3 will be recognized as ${\displaystyle \alpha _{3}}$
Obviously e and pi can't be variables names as they are already constants but they can be used in answer formulas.
Currently a bug in the parser prevent the use of the Greek letter epsilon as a variable name because the first letter is confused with the e constant.
Some examples:
• 2 x (equivalent to 2*x)
• e^x
• 2*pi*x
• cos(x)+sin(y)
• sqrt(b^2 - 4 a c)
• alpha^3-beta^3
When they attempt the Algebra question , your students will need to use the same linear syntax. Don't be afraid of that, they are already using it when they are talking about math and exchanging answers with their phones ;-)
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# Can there be multiple periapsides?
Consider the following image of a barycentric orbit of a binary star system.
I could draw their relative orbits by drawing a line connecting both bodies at each point in their rotation.
In this system, it is possible that the distances between the two stars are the same when they are opposite to each other both vertically and horizontally in the image. So would that mean that there are three periapside points, and thus three lines of apsides, two of them not going through the barycenter?
• Kind of need another picture to illustrate what you're asking, but if if I'm reading your question correctly, the situation you describe doesn't happen with Keplerian-Newtonian two-body orbits; both orbits wind up as ellipses of the same eccentricity that share a single line of apsides and have a common focus at the barycenter, whose apoapses and periapses are opposite the barycenter from each other. Oct 4, 2023 at 16:38
• @notovny James K has posted the right images in his answer. I understood how they have the same eccentricity and and that the apoapses and periapses are opposite to one another through the barycenter. However, being elliptical orbits of same eccentricity, can't they intersect in such a way that the horizontal distances are same as the vertical distances - leading to three periapside points (top, middle, bottom) for each body? Oct 5, 2023 at 1:23
• If that happens, either the two orbits you have drawn are not ellipses (the closest point on the boundary of an ellipse to either of its foci is always colinear with them both), or the two ellipses you have drawn do not share a focus at the barycenter. Oct 5, 2023 at 1:40
• @notovny now it makes sense! I forgot that the locus of an ellipse is for which the sum of distances from both foci are a constant! Oct 5, 2023 at 3:09
## 1 Answer
In the picture, and in any two-body orbit, the stars are closest together at a single periapse, and there is a single line of apsides.
In the picture, the stars are closest together when they are aligned horizontally. They are further away when aligned vertically
This will always be the case in any two-body system
• Sure, not in this picture, but can you be sure that it will always be the case? Can't two ellipses be brought closer together until the distances match horizontally and vertically? Oct 5, 2023 at 1:17
• No. This is not possible. The proof of this is geometric. But you can also consider the physics: At the periapse, the kinetic energy is at a maximum. As it passes periapse, kinetic energy is converted to gravitational energy so the stars move apart, they'll continue to move apart until kinetic energy is at a minium, at the antaapse. Oct 5, 2023 at 1:49
• Makes sense now. I first thought that they come togethor - move apart - come together again - move apart again - come together once more - and move apart -- all in a single revolution. That was when I just sketched it without considering the physics. Thinking of it again makes it seem absurd. Also @notovny's comment on my question cleared up the geometric part for me - it makes more sense now. Oct 5, 2023 at 3:05
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Polygons
Chapter 3 Class 8 Understanding Quadrilaterals
Concept wise
### Transcript
Ex 3.1, 6 Find the angle measure x in the following figures. (a) We know that Sum of all the angles of a quadrilateral = 360 ∠A + ∠B + ∠C + ∠D = 360° Given ∠A = 50°, ∠C = 120° , ∠ D = 130° & ∠B = x Putting values 50° + x + 120° + 130° = 360° x + 300° = 360° x = 360° − 300° x = 60°
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# 5 Ways to Overcome GRE Test Anxiety
• by
• Jul 15, 2014
• GRE Blog
1. Know The Test.
Some bad news about the GRE is it tests virtually nothing you learned in college and nothing you will work on in graduate school. So unless you’ve been practicing geometry in college or studying etymology, the GRE is going to take preparation. But there is good news: the GRE only tests a limited number of subjects, and it will test them every time. Cylinders? Yes. Pyramids? No. Make a list of the types of questions you see in your prep work – triangles, standard deviation, probability, etc. – and assess your relative level of confidence. Still struggling with permutations and combinations? Now you know where to focus. When doing your prep work, focus on the areas that need the most improvement. When doing the test, focus on the areas you score the best.
2. Practice and Review
The GRE tests a limited range of content, but knowing the range is useless without knowing how to approach the content. Think of this like Pavlovian conditioning – when you see words like “both” and “neither” in the same question, you have found a group formula question. When you take the actual test, you want your response to be automatic. The GRE is as much as knowing what to do as it is knowing how to do. Make flashcards for all of the types of math questions, how to recognize them and what to do.
Possibly the single best way to improve your score without a tutor is to do practice tests and drills, review the questions you got wrong, and ask yourself why the right answer is right and why your answer was wrong.
3. More Words = Higher Score
This might be an ugly truth, but the more words you know, the higher your score on the GRE verbal section will be. But considering that you’ve spent a percentage of your life preparing for tests and quizzes, you probably have developed methods for memorizing. Flashcards are great, but developing mnemonic devices can be a very powerful aid as well. Drilling vocab is simply the easiest and most straightforward way to raise your verbal score.
For much of your young life, you took math quizzes by starting with question 1, ending with question 20, and doing every single question. You have been conditioned to take tests in a certain way. Do not take the GRE this way. If you know you’re shaky on circles, and question 1 is a circle question, skip it. Skip early, and skip often. The “Review” screen lets you see which questions you have yet to answer, and starting in the last 2 minutes you can go through what you have left and punch your favorite letter (it is to your statistical advantage to have a favorite letter). But if question 20 is a percent change question, and you rock at percent change, make sure you have sufficient time to do the question correctly. As a rule, the more time you spend on a problem, the more likely you are to get it wrong. It is a better use of your time to make sure the questions you ought to get right you do get right than to spend it trying to decide between two answers to a question you don’t know – which amounts to a 50/50 chance in the end.
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# Human Biomechanics I Course
Overview
This document outlines the BME/ISE 3211 Human Biomechanics I Course Project in which biostatics analysis is used to design rehabilitation parameters for a patient with a torn distal biceps tendon. Each student will construct a free-body diagram of the human elbow/tendon system, perform static analysis of the system with an external load placed in the hand, calculate theoretical tendon tensions for a set of specified angles and loads, display data graphically, and design rehabilitation parameters within constraints set by an injury specialist. A formal report is to be submitted with all relevant info and conclusions.
Background
You are a consulting biomedical/industrial engineer assisting an employer in a prominent regional manufacturing company. An employee has torn a distal biceps tendon and must undergo rehabilitation. An injury specialist has performed tests and advises that the injured tendon should not be subjected to forces exceeding 500 N during rehabilitation.
The employee’s job requires routine handling of parts ranging from 2-6 kg. During manipulation of these parts, the employee must extend the arm with supinated palm and hold the forearm in a horizontal position such that the angle between the long axis of the upper arm and the forearm ranges from 30°-70°. Your task is to perform biostatics analysis to determine if any of these loads causes tendon tension to exceed the value advised by the njury specialist. From this analysis, you will prepare a report for submission to the employer.
Theoretical System Parameters for Biostatics Analysis
Fig. 1 shows a pictorial diagram of the extended forearm holding a load in the supinated hand. Assume that the axis of rotation of the elbow is fixed at point O and that the distance between O and the attachment point of the distal biceps tendon is a = 5.7 cm. The mass of the entire forearm including hand is mb = 1.66 kg and its center of gravity is located b = 21.0 cm from O. Define the load in the hand as mc; its center of gravity is located c = 40.6 cm from O. Define Fm as the biceps tendon tension. Assume that the line of action of Fm is parallel to the long axis of the upper arm throughout the specified range of motion. Therefore, define θ as the angle Fm makes with the forearm; i.e., with respect to horizontal.
Fig. 1: Pictorial diagram of the extended forearm handling a variable mass.
1) Prepare an electronically-generated free-body diagram of the system utilizing the parameters outlined above. You may use shapes in MS Word or the drawing program of your choice. Include all dimensional information and a coordinate system. (20%)
2) Derive an expression for the magnitude Fm in terms of θ and mc. Hint: you only need one rotational equilibrium equation to do this. (20%)
3) Compute Fm for each of five masses placed in the hand [mc = 2.0, 3.0, 4.0, 5.0, and 6.0 kg] at each of five angles of distal biceps tendon tension [θ = 30, 40, 50, 60, and 70°]. Prepare a table with these 25 values of distal biceps tendon tension, with θ placed along the x-axis (columns) and mc placed along the y-axis (rows). Label and caption this table. Highlight values that exceed the limit advised by the injury specialist in a suitable color. (20%)
4) Generate a graph with five smoothed-line scatter plots (including data points) showing Fm (y) vs. θ (x) for each of the five masses using Excel or MatLab. Draw a heavy line on the graph in a suitable color indicating the tension limit advised by the specialist. Likewise, generate a similar graph showing Fm (y) vs. mc (x) for each of the five angles and a line indicating the advised tension limit. Label and caption both graphs. (20%)
5) Prepare a formal report with the following sections: (20%)
• Cover Page
o name of the university and department
o title of the report
o submitted to:
o date of submission
• Introduction and Purpose
o You may paraphrase the description of the problem in this assignment, but don’t simply
copy and paste it.
• Data and Results
o Include all diagrams, derivations, equations, tables, and charts.
• Discussion
o Discuss the figures in the table and charts and address the following questions:
What happens to the theoretical magnitude of distal biceps tendon tension as the
arm becomes fully extended; i.e., θ approaches zero degrees?
What does this suggest about the limitations of our analysis?
• Conclusions and Recommendations
o Should the injured employee undergo on-the-job rehabilitation?
o Specify design parameters for a rehabilitation algorithm based on your computations.
You may consult with a classmate regarding help with graphs or double-checking derivations, but all reports
and corresponding work are individual for this assignment.
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Year 5 - Story behind… | Independent Primary School Buckinghamshire
# Year 5 - Story behind the graph and bunsen burners!
The return to school after half-term was marked by an exciting whole school project day. A whole school assembly on having a ‘can do’ attitude, introduced the importance of having a positive mindset. Older children were able to adapt a well-known children’s book to create their own versions. The children were clear on the moral messages behind their own stories, with younger children being the intended audience. This whole school project gave the children a clear sense of purpose and has brought together the entire school community.
It is wonderful for children to see the relevance of their Mathematics learning in the real-world. Looking at the ‘Story Behind the Graph’ has helped them to explain the data that they observe. Year 5 children have explored the popular story of the Hare and the Tortoise on a time-distance line graph, identifying the incline and decline as changes in speed, and moments where the graph flat lines to appreciate that time ticks away whilst the distance from the origin remains the same. The children have gone on to represent the 6 times’ table and elicited decimal values from the linear graphs that they drew. The possibilities for story telling in Maths have proved boundless and wholeheartedly engaged the children!
This half term, we are all eager to find out what will happen next in our class text ‘The Nowhere Emporium’. So far, we have been captivated by the unusual, magical shop that has appeared from nowhere and we want to know what will happen next to Daniel Holmes on his adventures with Mr Silver. We look forward to the book inspiring more of our writing over the next few weeks before exploring playscripts in the final weeks before Christmas.
We have started with a fire, of course. This half term we are considering reactions that are either reversible or non reversible. All children should now be able to report, from first hand experience, that setting fire to a Pringle in a Bunsen burner is a non reversible reaction. Please check in regularly to find out about our other investigations including mixing milk and vinegar, and growing yeast. These investigations are to encourage the children to think scientifically and learn different ways of presenting their results.
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# Gravity and Edgar Rice Burroughs
Yesterday, I was thinking about what textbook I would move on to after I finish Adventures in Celestial Mechanics. Because I’ve been thinking about gravity, I pulled Oliver Kellogg’s Foundations of Potential Theory off my shelf and started leafing through it. I soon came upon a gravity problem that I first heard about when I was a teenager but never saw the solution for. The problem has to do with Pellucidar, and it’s not as tricky as I expected.
Pellucidar is an imaginary land created by Edgar Rice Burroughs for a series of pulpy adventure books, starting with At the Earth’s Core. The conceit of the stories is that the Earth is actually a hollow shell and that there’s an inhabited land, Pellucidar, on the inside surface of the shell. Intrepid adventurers from our side of the shell dig through to Pellucidar, meet the strange people and creatures who live there, and have adventures intrepidly. Never one to let an idea go unexploited, Burroughs even had Tarzan visit Pellucidar.
While I read one or two Tarzan books and several of the John Carter of Mars books, I never picked up any of the Pellucidar novels. But I did read somewhere—possibly in an Isaac Asimov science essay—that the concept of Pellucidar was flawed in a way apart from the many obvious ways: the inhabitants of Pellucidar would not “stick” to the inner surface of a hollow Earth because there’s no net gravitational force inside a spherical shell.
This seemed very odd to the teen-aged me, but Asimov (or whomever) didn’t explain it, and I didn’t have the analytical tools to check it out myself. But now I do, following more or less the derivation given by Kellogg.
Suppose we have a thin spherical shell of radius $a$. We want to know the gravitational attraction that the sphere exerts on a particle, $P$, which may be either inside or outside the sphere.
We’ll set up our coordinate system so that $P$ is on the $z$ axis. Every point on the sphere, $Q$, can be identified by two angular coordinates, $\theta$ and $\varphi$, which are the point’s latitude and longitude.
(If you’re wondering how we should handle the analysis if $P$ isn’t on the $z$ axis, don’t. One of the most important things to learn in mechanics is that the coordinate systems we use are up to us; there is no $x\text{-}y\text{-}z$ stamped on the universe. Wherever $P$ happens to be, we can draw a $z$ axis that runs through it from the center of the sphere.)
The gravitational force that some small differential element at $Q$ on the sphere exerts on particle $P$ is
$dF=\frac{G\rho \phantom{\rule{thinmathspace}{0ex}}dA}{{r}^{2}}$
where $G$ is the universal gravitational constant, $\rho$ is the areal density of the sphere (i.e., it has units of mass per area, not mass per volume), and $r$ is the distance from $Q$ to $P$. $dA$ is the differential area of the element on the sphere, and $dF$ is the differential force (per unit mass of $P$) that element exerts on particle $P$.
Here’s a sketch of the differential area at $Q$:
It’s bounded by lines of latitude that are $d\theta$ apart from one another and lines of longitude that are $d\varphi$ apart from one another. The area of the element is
$dA=\left(a\phantom{\rule{thinmathspace}{0ex}}d\theta \right)\left(a\mathrm{cos}\theta \phantom{\rule{thinmathspace}{0ex}}d\varphi \right)={a}^{2}\mathrm{cos}\theta \phantom{\rule{thinmathspace}{0ex}}d\varphi \phantom{\rule{thinmathspace}{0ex}}d\theta$
If you’re wondering why we can use the formula for the area of a rectangle when the edges and the surface itself are curved, I will remind you that this is how calculus works. These differential lengths are teeny tiny and therefore the difference between the area of the surface and the area of a rectangle is extra extra teeny tiny. In calculus, we keep track of the teeny tiny and throw away the extra extra teeny tiny.
To express the differential force in terms of $\theta$ and $\varphi$, let’s look at a side view of the sphere with $P$ and $Q$. First, we’ll consider the case where $P$ is “above” $Q$:
The distance between $P$ and $Q$ is
$r=\sqrt{\left(a\mathrm{cos}\theta {\right)}^{2}+\left(z-a\mathrm{sin}\theta {\right)}^{2}}=\sqrt{{a}^{2}+{z}^{2}-2az\mathrm{sin}\theta }$
Although I’ve drawn the figure with $P$ outside the sphere, the formula above also applies when $P$ is inside the sphere. What matters is that $P$ is above $Q$.
Now let’s consider the case where $P$ is “below” $Q$:
Here, $r$ is the hypotenuse of a different right triangle, but it turns out that
$r=\sqrt{\left(a\mathrm{cos}\theta {\right)}^{2}+\left(a\mathrm{sin}\theta -z{\right)}^{2}}=\sqrt{{a}^{2}+{z}^{2}-2az\mathrm{sin}\theta }$
is the same formula as before. That’s convenient. So the magnitude of the differential force is
$dF=\frac{G\rho {a}^{2}\mathrm{cos}\theta }{\sqrt{{a}^{2}+{z}^{2}-2az\mathrm{sin}\theta }}\phantom{\rule{thinmathspace}{0ex}}d\varphi \phantom{\rule{thinmathspace}{0ex}}d\theta$
To determine the force on $P$ due to the entire sphere, we’re going to integrate on $\theta$ from $-\pi /2$ to $\pi /2$ and on $\varphi$ from $0$ to $2\pi$. But first, we need to think about the direction of the differential force.
Note that in each of the two side views, the gravitational force on $P$ has two components: a component parallel to the $z$ axis (vertical in our drawings) and a component perpendicular to the $z$ axis (horizontal in our drawings). When we integrate over $\varphi$, we’re dealing with forces associated with a circle of constant latitude, and the components perpendicular to the the $z$ axis (horizontal) will cancel each other out. For example, the horizontal component when $Q$ is at $\varphi =180\text{°}$ cancels the horizontal component when $Q$ is at $\varphi =0\text{°}$, the horizontal component for $\varphi =270\text{°}$ cancels the horizontal component for $\varphi =90\text{°}$, and so on.
The upshot is that we only have to worry about the “vertical” component of the force, i.e., the component in the $z$ direction. To get that component, we multiply $dF$ by the $z$ direction cosine.
When $P$ is above $Q$, the $z$ direction cosine is
$-\frac{z-a\mathrm{sin}\theta }{\sqrt{{a}^{2}+{z}^{2}-2az\mathrm{sin}\theta }}$
where the minus sign out in front indicates that the component is acting in the negative $z$ direction. Remember, we’re computing the gravitational force on $P$ due to $Q$, so the force acts from $P$ toward $Q$.
When $P$ is below $Q$, the $z$ direction cosine is
$\frac{a\mathrm{sin}\theta -z}{\sqrt{{a}^{2}+{z}^{2}-2az\mathrm{sin}\theta }}$
Compare this formula with the previous one and you’ll see that they are algebraically identical. As with the expressions for $r$, the signs work out to give the same equation for both cases.
So the $z$ component of the differential force acting on $P$ due to $Q$ is
$d{F}_{z}=\frac{G\rho {a}^{2}\mathrm{cos}\theta \phantom{\rule{thinmathspace}{0ex}}\left(a\mathrm{sin}\theta -z\right)}{\left({a}^{2}+{z}^{2}-2az\mathrm{sin}\theta {\right)}^{3/2}}\phantom{\rule{thinmathspace}{0ex}}d\varphi \phantom{\rule{thinmathspace}{0ex}}d\theta$
and the total force that the sphere exerts on $P$ is
$F={F}_{z}={\int }_{-\pi /2}^{\pi /2}{\int }_{0}^{2\pi }\frac{G\rho {a}^{2}\mathrm{cos}\theta \phantom{\rule{thinmathspace}{0ex}}\left(a\mathrm{sin}\theta -z\right)}{\left({a}^{2}+{z}^{2}-2az\mathrm{sin}\theta {\right)}^{3/2}}\phantom{\rule{thinmathspace}{0ex}}d\varphi \phantom{\rule{thinmathspace}{0ex}}d\theta$
This may seem awful, but it isn’t. First, notice that $\varphi$ doesn’t appear in the integrand, so the integral over $\varphi$ is just
${\int }_{0}^{2\pi }d\varphi =2\pi$
We can pull that and the other constants out to get
$F=2\pi G\rho {a}^{2}{\int }_{-\pi /2}^{\pi /2}\frac{\mathrm{cos}\theta \phantom{\rule{thinmathspace}{0ex}}\left(a\mathrm{sin}\theta -z\right)}{\left({a}^{2}+{z}^{2}-2az\mathrm{sin}\theta {\right)}^{3/2}}\phantom{\rule{thinmathspace}{0ex}}d\theta$
This still looks bad, but with tools like Mathematica and SymPy, integrations like this aren’t the pain in the ass they used to be. After a few seconds of chugging in Mathematica,
$F=\frac{2\pi G\rho {a}^{2}}{{z}^{2}}\phantom{\rule{thinmathspace}{0ex}}\left(sgn\left(a-z\right)-1\right)$
where the signum function ($sgn$) is $+1$ if the argument is positive and $-1$ if the argument is negative. Therefore, if $P$ is outside the sphere ($a),
$F=\frac{2\pi G\rho {a}^{2}}{{z}^{2}}\left(-1-1\right)=-\frac{4\pi G\rho {a}^{2}}{{z}^{2}}$
At this point, we recall that $4\pi {a}^{2}$ is the area of the sphere, so that multiplied by the areal density, $\rho$ , is the mass of the sphere, which we’ll call $M$. So for this case
$F=-\frac{GM}{{z}^{2}}$
which is just what we’d expect (remember that this is the force per unit mass of $P$, so there’s no mass term for the particle).
On the other hand, if $P$ is inside the sphere ($a>z$),
$F=\frac{2\pi G\rho {a}^{2}}{{z}^{2}}\left(1-1\right)=0$
which is the surprising (to me, anyway) result that the sphere exerts no net gravitational force on a particle inside it.
How can this be? Well, certainly it makes sense if $P$ is at the center of the sphere. If we move $P$ up from the center, there’s more mass below $P$ but it is, on the average, farther away. It works out that the increasing mass and increasing distance balance each other to keep the net force at zero.
Oliver Kellogg didn’t have Mathematica. His suggestion was to do a change of variable from $\theta$ to $r$ and then integrate. He doesn’t give any details of the process, just the two answers. But I went ahead and did it his way, too, using
$r=\sqrt{{a}^{2}+{z}^{2}-2az\mathrm{sin}\theta }$ $dr=-\frac{az\mathrm{cos}\theta }{\sqrt{{a}^{2}+{z}^{2}-2az\mathrm{sin}\theta }}\phantom{\rule{thinmathspace}{0ex}}d\theta =-\frac{az\mathrm{cos}\theta }{r}\phantom{\rule{thinmathspace}{0ex}}d\theta$
and
$a\mathrm{sin}\theta -z=\frac{{a}^{2}-{z}^{2}-{r}^{2}}{2z}$
With those substitutions, the integrand becomes
$-\frac{1}{2a{z}^{2}}\phantom{\rule{thinmathspace}{0ex}}\frac{{a}^{2}-{z}^{2}-{r}^{2}}{{r}^{2}}$
which is much easier to deal with than the original integrand. But we have to be careful with the limits. If $z>a$, the upper limit ($\theta =\pi /2$) becomes
$r=z-a$
and the lower limit ($\theta =-\pi /2$) becomes
$r=z+a$
Therefore, the integral is
$F=\frac{\pi G\rho a}{{z}^{2}}{\int }_{z-a}^{z+a}\frac{{a}^{2}-{z}^{2}-{r}^{2}}{{r}^{2}}\phantom{\rule{thinmathspace}{0ex}}dr$
where I got rid of the negative sign at the front of the integrand by exchanging the limits. With no trig functions to make life difficult, this works out very quickly to
$F=-\frac{4\pi G\rho {a}^{2}}{{z}^{2}}$
when $P$ is outside the sphere. This is the same answer as before.
If $z, the upper limit ($\theta =\pi /2$) becomes
$r=a-z$
and the lower limit ($\theta =-\pi /2$) becomes
$r=a+z$
Therefore, the integral is
$F=\frac{\pi G\rho a}{{z}^{2}}{\int }_{a-z}^{a+z}\frac{{a}^{2}-{z}^{2}-{r}^{2}}{{r}^{2}}\phantom{\rule{thinmathspace}{0ex}}dr$
which works out to
$F=0$
when $P$ is inside the sphere. Again, a repeat of the answer we got when integrating over $\theta$.
While it was nice to see explicitly how Kellogg did the integration, it was faster to just let Mathematica do its thing. And I didn’t get any deeper physical insight doing the change of variable.
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# How do you estimate square roots?
It depends on what you can use, of course. Let's say you need to find the square root of a number $x$
A first, but very rough way, consists in simply finding two numbers, $n$ and $m$, such that ${n}^{2} < x < {m}^{2}$. If you have this relation, you can surely affirm that $\setminus \sqrt{x}$ is some number between $n$ and $m$. For example, if we needed to estimate $\setminus \sqrt{40}$, we could say that it surely is a number between $6$ and $7$.
Of course, this method can be used also with rational numbers. For example, with a bit of calculations you can find out that $\setminus \sqrt{40}$ actually lies between $6.3$ and $6.4$, and so on.
Another way could be factoring $x$ with primes, and simplify squared factors, if any appear. This could leave only smaller roots to calculate: consider for example $\setminus \sqrt{18}$. You can write $18$ as $2 \setminus \cdot {3}^{2}$, and so $\setminus \sqrt{18} = \setminus \sqrt{2 \setminus \cdot {3}^{2}} = \setminus \sqrt{2} \setminus \cdot \setminus \sqrt{{3}^{2}} = 3 \setminus \sqrt{2}$
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Post subject: Ko for nerds #1 Posted: Thu Aug 04, 2022 11:31 am
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I have no idea whether this will be useful or interesting to anyone else, but it was to me, so here it is! If it does not float your boat, please ignore.
Despite having played Go for decades, I still have a fuzzy sense of how exactly the state changes as the players make various sorts of moves and what the overall structure looks like. This writeup is an attempt to formalize exactly what happens in a ko so that I can observe the structure at a glance.
I will use the colors Blue and Red for the two players instead of White and Black to make the color-coding of the diagrams easier.
In the following diagrams, circular nodes represent positions while arrows represent moves, and are color-coded by the player making the move.
The codes for the types of moves are:
• C: Capture ko
• R: Resolve ko
• T: Ko threat
• A: Answer to ko threat
• F: Followup to ko threat
• P: Preparatory move in approach ko
In general, the board position evolves from left to right. I pretend that there are infinite ko threats, so each graph wraps back around from right to left, and two periods of the general structure are shown. To make the periodicity more clear, certain identical nodes are colored (e.g., all yellow nodes in a diagram represent the same state). Black nodes represent terminal states after the ko has been resolved.
In general, upwards motion brings Blue closer to winning the ko while downwards motion brings Red closer to winning the ko; thus, the only moves with a vertical component are the C, R and P types. These arrows are also dashed, to make it easier to distinguish local moves from remote moves. (Local ko threats are not treated here, but one can approximate them with incredibly important remote ko threats.) For example, one can see that if Blue wins the ko (by getting to the top of the graph), Red has gotten two moves elsewhere as compensation (the unanswered T and the F).
Ideally, this graph would be three-dimensional, with the third dimension being the generic amount of progress that is being made outside of the ko; T, F and A arrows would have a component in this third dimension. If we imagine that positions with more remote moves by Blue were farther into the paper, then the whole graph would be roughly tilted away from the reader (the top level is best for Blue locally but best for Red remotely).
Direct ko. To understand the following diagram of a direct ko, first observe the main Threat-Answer-Capture path (with alternating colors); then see how each player can resolve the ko (the R arrows); then confirm that each of the four horizontal levels represents a distinct ko state (from top to bottom: Blue has won the ko, Blue has made the last capture, Red has made the last capture, Red has won the ko).
Two-stage ko. A two-stage ko is effectively two linked direct kos. There are now five horizontal levels in the following diagram, not four. In the top lane, Blue can resolve the ko entirely by ignoring a threat, while if Red ignores a threat they just switch to the bottom lane (which switches the roles of the two players). Note that the lane-switching moves appear to go backwards, because the other player cannot recapture the ko immediately.
One-move approach ko. The graph of a one-move approach ko is similar to that of a two-stage ko, but with two differences:
• Lane-switching can only be done in one direction, and doesn’t move the other player any farther away from winning the ko (in terms of number of local moves that have to be made). In the following diagram, when Red lane-switches with a P, Blue can still resolve the ko entirely with their R move. So Red runs the risk of making a P move that has basically no effect in the end (it did not win the ko and also got no compensation elsewhere).
• Since the lane-switching move is not a capture, the other player can play a capture immediately and doesn’t have to make a ko threat first. This is why the P arrows go forwards instead of backwards.
The generalizations to multi-stage kos and multi-move approach kos look just as you would expect.
Well, that made me feel better about my understanding of the structure of kos, and I hope it didn't make you feel any worse.
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Post subject: Re: Ko for nerds #2 Posted: Thu Aug 04, 2022 3:18 pm
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curious. well, your pictures are prettier than the similar ones I created about a year ago.
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Give me triangles strong enough and I can measure the universe.
When Venus transits, we can align our clocks to one event. By measuring the angle to flat Earth at two places far apart on Earth, we can compute the distance to Venus and the Sun.
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# Concentration of Lithium sulfate
## lithium sulfate: convert between mass and molar concentration
### Molar concentration per milliliter
0.01 mmol/ml 10 µmol/ml 10 000 nmol/ml 10 000 000 pmol/ml
### Molar concentration per deciliter
1 mmol/dl 1 000 µmol/dl 1 000 000 nmol/dl 1 000 000 000 pmol/dl
### Molar concentration per liter
10 mmol/l 10 000 µmol/l 10 000 000 nmol/l 10 000 000 000 pmol/l
### Mass concentration per milliliter
0 g/ml 1.1 mg/ml 1 099.36 µg/ml 1 099 360 ng/ml 1 099 360 000 pg/ml
### Mass concentration per deciliter
0.11 g/dl 109.94 mg/dl 109 936 µg/dl 109 936 000 ng/dl 109 936 000 000 pg/dl
### Mass concentration per liter
1.1 g/l 1 099.36 mg/l 1 099 360 µg/l 1 099 360 000 ng/l 1 099 360 000 000 pg/l
### Equivalent molar concentration per milliliter
0.01 meq/ml 10 µeq/ml 10 000 neq/ml 10 000 000 peq/ml
### Equivalent molar concentration per deciliter
1 meq/dl 1 000 µeq/dl 1 000 000 neq/dl 1 000 000 000 peq/dl
### Equivalent molar concentration per liter
10 meq/l 10 000 µeq/l 10 000 000 neq/l 10 000 000 000 peq/l
• The units of amount of substance (e.g. mole) per milliliter, liter and deciliter are SI units of measurements of molar concentrations.
• The units of molar concentration per deciliter:
• millimole per deciliter [mm/dl], micromole per deciliter [µm/dl], nanomole per deciliter [nm/dl] and picomole per deciliter [pm/dl].
• The units of molar concentration per milliliter:
• millimole per milliliter [mm/ml], micromole per milliliter [µm/ml], nanomole per milliliter [nm/ml] and picomole per milliliter [pm/ml].
• The units of molar concentration per liter:
• millimole per liter [mm/l], micromole per liter [µm/l], nanomole per liter [nm/l] and picomole per liter [pm/l].
• The units of mass per milliliter, liter and deciliter are non-SI units of measurements of mass concentrations still used in many countries.
• The units of mass concentration per deciliter:
• gram per deciliter [g/dl], milligram per deciliter [mg/dl], microgram per deciliter [µg/dl], nanogram per deciliter [ng/dl] and picogram per deciliter [pg/dl].
• The units of mass concentration per milliliter:
• gram per milliliter [g/ml], milligram per milliliter [mg/ml], microgram per milliliter [µg/ml], nanogram per milliliter [ng/ml] and picogram per milliliter [pg/ml].
• The units of mass concentration per liter:
• gram per liter [g/l], milligram per liter [mg/l], microgram per liter [µg/l], nanogram per liter [ng/l] and picogram per liter [pg/l].
• The equivalent per milliliter, liter and deciliter are obsolete, non-SI units of measurements of molar concentrations still used in many countries. An equivalent is the number of moles of an ion in a solution, multiplied by the valence of that ion.
• The units of equivalent concentration per deciliter:
• milliequivalent per deciliter [meq/dl], microequivalent per deciliter [µeq/dl], nanoequivalent per deciliter [neq/dl] and picoequivalent per deciliter [peq/dl].
• The units of equivalent concentration per milliliter:
• milliequivalent per milliliter [meq/ml], microequivalent per milliliter [µeq/ml], nanoequivalent per milliliter [neq/ml] and picoequivalent per milliliter [peq/ml].
• The units of equivalent concentration per liter:
• milliequivalent per liter [meq/l], microequivalent per liter [µeq/l], nanoequivalent per liter [neq/l] and picoequivalent per liter [peq/l].
#### Foods, Nutrients and Calories
ORIGINAL RECIPE BRATWURST, UPC: 079041837722 contain(s) 385 calories per 100 grams (≈3.53 ounces) [ price ]
184 foods that contain Galactose. List of these foods starting with the highest contents of Galactose and the lowest contents of Galactose
#### Gravels, Substances and Oils
CaribSea, Freshwater, African Cichlid Mix, Congo River weighs 1 313.51 kg/m³ (81.99975 lb/ft³) with specific gravity of 1.31351 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Potassium peroxydisulfate [K2S2O8 or K2O8S2] weighs 2 477 kg/m³ (154.63406 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Engine Oil, SAE 10W-40 with temperature in the range of 0°C (32°F) to 100°C (212°F)
#### Weights and Measurements
A milliohm is a SI-multiple (see prefix milli) of the electric resistance unit ohm and equal to one thousandth of an ohm (0.001 Ω)
A force that acts upon an object can cause the acceleration of the object.
kg/metric tbsp to st/US qt conversion table, kg/metric tbsp to st/US qt unit converter or convert between all units of density measurement.
#### Calculators
Weight to Volume conversions for sands, gravels and substrates
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# Which number system(s) are like the BASE 4 number system? Convert 750 into base 11. Convert 836 into base 11. Fill in the blank: 22810 = ___________ 3 (i.e. 228 in base 10 is equivalent to what nu
Which number system(s) are like the BASE 4 number system?
Convert 750 into base 11.
Convert 836 into base 11.
Fill in the blank: 22810 = ___________ 3 (i.e. 228 in base 10 is equivalent to what number in base 3?)
What number is 2 after 2013 (i.e. 121 in base 3)?
What number is 3 before 2013 (i.e. 121 in base 3)?
What number is 2 before 1213 (i.e. 121 in base 3)?
Fill in the blank: 1223 = ___________ 10 (i.e. 122 in base 3 is equivalent to what number in base 10?)
In which base b does: 33b + 34b = 67b?
In which base b does: 23b + 23b = 50b?
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# surface area of human skin formula
## surface area of human skin formula
Based on the formula of Yu et al, however, BSA is overestimated when these traditional formulas are used. / 2 {\displaystyle {8/900}\times W^{4/9}\times H^{2/3}} The internal surface area of the gastro-intestinal tract has long been considered to be between 180 and 300 square meters. The BSA is frequently used in determining the dosage of medications with a narrow therapeutic index, such as chemotherapeutic agents. Body surface area (BSA): measured or calculated surface area of a human body. [1][2] The trend to personalized medicine is one approach to counter this weakness. Recommended values for these dermal factors are not provided in the Handbook because data are limited; however, data from relevant studies are described. Still, these kinds of things are … Body surface area(BSA) The body surface area is the measured or calculated surface of a human body. Body surface area (BSA) was calculated according to DuBois and DuBois, 1916 and expressed by: (1) B S A = 0.202 m 0.425 H 0.725 where m is body weight, kg, and H is body height, m. Table 1. By using these techniques the surface roughness power spectrum is obtained. 100 8 The appendages of the skin are developmentally derived from the epidermis and include hairs, sebaceous and sweat glands and nails. / Using these BSA formulas, body surface area … The measurement and the formula have been taken as standards in most of the textbooks, such as UK text and Lund, Browder Chart , etc., and have been used as base data for many later researches , , , , , : (1) B S A = 71.84 × H.725 × W.425 where BSA is the body surface area in cm 2; H is the stature height in cm; W is the body weight in kg. 9 1 decade ago. Favorite Answer. ) Body surface area prediction with the commonly used DuBois formula underestimated BSA in obese patients by as much as 3% (male) to 5% (female). [Assume a height of 5'8" (170 cm) and a mass of 150 lb (68 kg).] Skin surface area (SA) is an estimate of the amount of skin ... Residue transfer coefficients (cm 2 /hour) of contaminants to human skin will vary based on exposure conditions such as activity, contact surface, and age. There is some evidence that BSA values are less accurate at extremes of height and weight, where Body Mass Index may be a better estimate (for hemodynamic parameters).[3]. Below are the body surface area formula by Dr’s Mosteller, DuBois and DuBois, Haycock and Boyd. much appreciated. The mechanisms by which molecules traverse human skin, including the influence of shunt route transport, are discussed in Chapter 2 (see Section 2.3). In the following formulae, BSA is expressed in m , weight (or, more properly, mass) W in kg, and height H in cm. BSA = SQRT ((cm*kg)/3600) or in inches and pounds: BSA (m2) = SQRT ([Height (in) x Weight (lbs)]/ 3131) September 10, 2020 September 10, 2020 by rajathe. This can lead to significant overdosing and underdosing (and increased risk of disease recurrence). Body Surface Area. A newer BioNumber version exists. × H Estimate how many patient palms fit in the affected area of the head. Figure 1 illustrates the simple argument for revising the value of the skin surface area that is important for communication with the microbiome. It is the single largest human organ both by sheer weight and surface area. area offered by these shunt r outes is so small that the pr edominant path - way for molecules to traverse the tissue remains across the bulk of the skin surface. The four most common equations used to calculate the BSA are included in this calculator. Body surface area (BSA) is the total surface area of the human body. [Assume a height of 5'8" (170 cm) and a mass of 150 lb (68 kg).] [15][16], Average BSA for children of various ages, for men, and for women, can be estimated using statistical survey data and a BSA formula:[17], The estimations in the above tables are based weight and height data from the U.S. NCHS National Health and Nutrition Examination Survey (2011-2014). 2 3 click here. This body surface area calculator helps you calculate how much surface area of the skin is affected by a disease. / Comments: The surface area of adults is about 18,000 cm2 (men) or 16,000 cm2 (women). Mosteller pointed out that his formula holds only if the density is treated as a constant for all humans. The surface area may be calculated by multiplying 0.007184 times the weight in kilograms raised to the 0.425 power and the height in centimeters raised to the 0.725 power." The surface area may be calculated by multiplying 0.007184 times the weight in kilograms raised to the 0.425 power and the height in centimeters raised to the 0.725 power." Lipscombe, following Mosteller's reasoning, observed that the formulas obtained by Fujimoto, Shuter and Aslani, Takahira, and Lipscombe are suggestive of . Prism Surface Area Formula and Prism Volume Formula. Please note: The entry will be shown to all once approved by the database administrator. Scientists have used refined microscopic techniques that … 4 / ) 1.2.3 The epidermis [20], Costeff H, "A simple empirical formula for calculating approximate surface area in children.,", European Organisation for Research and Treatment of Cancer, "How to calculate the dose of chemotherapy", "Misleading indexed hemodynamic parameters: the clinical importance of discordant BMI and BSA at extremes of weight", "A formula to estimate the approximate surface area if height and weight be known", "Body surface area in normal-weight, overweight, and obese adults. This surface area calculator helps you find the area of the most common three-dimensional solids. 900 / 2 HUMAN SKIN The human skin is a complex organ that covers the exterior of the body. Figure 1 – Cross section of the skin. 3 Relevance. The most widely used is the Du Bois formula,[4][5] which has been shown to be equally as effective in estimating body fat in obese and non-obese patients, something the Body mass index fails to do. In simple terms, body surface area is the area covered by one's skin - the largest organ of the body. For any formula, the units should match. The surface area of a segment is calculated from its length and the mean of its proximal, middle and distal circumferences using the formula for determining the surface area of a cylinder. ( 2 If the human body has a total surface area of 1.20 m^2 and the surface temperature is 30°C, find the total net rate of radiation of the energy from the body if the surroundings are at a temperature of 20°C and the emissivity of the body is 1.0. 3 The surface topography of the human wrist skin is studied by using optical and atomic force microscopy (AFM) methods. In physiology and medicine, the body surface area (BSA) is the measured or calculated surface area of a human body. {\displaystyle (2^{3}/3^{2})\times (W^{2/3}H)^{2/3}/100} × A Powerpoint file is included, though questions and other content could be written on a board and/or given verbally, or transferred to a worksheet. It is [4W (kg) + 7]/[90 + W (kg)]. A taller person tends to have a larger surface area, as does a he… Surface area formulas in geometry refer to the lateral surface and total surface areas of different geometrical objects. It is also thought to be a distorting factor in Phase I and II trials that may result in potentially helpful medications being prematurely rejected. [Assume a height of 5'8" (170 cm) and a mass of 150 lb (68 kg).]. 3 Answers . A sphere is a solid figure where every point on the surface is equidistant from the center of the sphere. A Dictionary of Food and Nutrition. The purpose of this study is to establish a foot surface area (FSA) database and estimation formula based on 3-D foot scan data. Volume = 4 ⁄ 3 πr 3. Answer Save. It comprised three main layers: epidermis, dermis and subcutis as depicted in Figure 1. How to Calculate Body Surface Area . The surface area of adults is about 18,000 cm2 (men) or 16,000 cm2 (women). ( Someone calculated a chemical formula for the entire human body. In terms of surface area, the skin is the second largest organ in the human body (the inside of the small intestine is 15 to 20 times larger). 2. The skin surface area (in square feet) for a human is function of weight(in pounds) and height(in feet) given by the formula A(w, h) = 7 Squareroot w h^3/4 find the surface area of 5 ft person that weights 180 lbs, find A_w(180, 5) and interpret the results, find A_h(180, 5) and interpret the result. Calculation formulae in three stages over all ages", "National Health and Nutrition Examination Survey", "The Average Body Surface Area of Adult Cancer Patients in the UK: A Multicentre Retrospective Study", https://en.wikipedia.org/w/index.php?title=Body_surface_area&oldid=993157219, Creative Commons Attribution-ShareAlike License, This page was last edited on 9 December 2020, at 02:24. Skin; Search; Surface Area Formulas | Calculate the Surface Area of Cone, Sphere, Box, Prism, Cylinder, Pyramid . 3 The Persson contact mechanics theory is used to calculate the contact area for different magnifications, for the dry and wet skin. New York: Oxford University Press, 1995. more still welcome. surface area in the movement of heat, the use of a mathematical model of the relationship between an animal’s size and its metabolic rate, and a few questions about the importance of high surface area:volume ratios in the lungs of animals. body surface area, body surface area formula, DuBois formula, mosteller formula, Pharma fresher's guide, Pharmaceutical calculations Q) Calculate the body surface area of a … Our measurements yielded a result of 1.5 square meters. 3 Body Surface Area Calculator for medication doses; Halls.md Disclaimer: This article is for information only and should not be used for the diagnosis or treatment of medical conditions. [19], During 2005 there was an average BSA of 1.79 m2 for 3,613 adult cancer patients in the UK. Nevertheless, there have been several important critiques of the use of BSA in determining the dosage of medications with a narrow therapeutic index, such as chemotherapy. If one estimates the depth of an average human follicle to be 3 mm and the diameter of that tube is approximately 0.5 mm, then the surface area of a hair follicle is 3.14 × 0.5 × 3 = 4.71 mm 2 or 4.71 × 10 −6 m 2. so thanks to the answers so far. Because of the complexity of direct measurement, various formulas have been developed to estimate the body surface area over the years. Typically there is a 4–10 fold variation in drug clearance between individuals due to differing the activity of drug elimination processes related to genetic and environmental factors. Surface area of the skin: The total area of the body surface covered by the skin is about 2m 2 in adults, and the skin thickness varies between 0-3 mm and 3mm. Various calculations have been published to arrive at the BSA without direct measurement. Body Surface Area Definition. W , which is dimensionally correct for the case of constant density. the skin cover 1.5 to 2 square mts. A weight-based formula was proposed by Costeff and recently validated for the pediatric age group that does not include a square root, making it easier to use. It equals "The surface area may be calculated by multiplying 0.007184 times the weight in kilograms raised to the 0.425 power and the height in centimeters raised to the 0.725 power." Body Surface Area Formula: Body Surface Area (m²) = (Kilograms^0.5378)(Centimeters^0.3964)*.024265 . W Bender, Arnold E. & David A. Bender. According to Mosteller's "simplified calculation of body-surface area In metric terms" the body surface area = the square root of product of the weight in … The most commonly used formula now is that of Mosteller, published in 1987 in The New England Journal of Medicine. [6], The Mosteller formula is also commonly used, and is mathematically simpler:[7]. Various calculations have been published to arrive at the BSA without direct measurement. This distance is the radius, r, of the sphere. × A comparison study", "Studies on the physical surface area of Japanese. It is measured in terms of square units. 2 This has led to the classification of people(s) on the basis of skin colour. H For many clinical purposes, BSA is a better indicator of metabolic mass than body weight because it is less affected by abnormal adipose mass. Last month, my fried Ali told me about a common formula for calculating the surface area of a person's body. In the following formulae, BSA is expressed in m2, weight (or, more properly, mass) W in kg, and height H in cm. Take height (cm) x weight (kg) / 3600, and the BSA is the square root of this number. By using the handprint- or palm-method you simply estimate the area affected by following these steps: 1. Among them the average BSA for men was 1.91 m2 and for women was 1.71 m2. Haycock GB, Schwartz GJ, Wisotsky DH "Geometric method for measuring body surface area: A height-weight formula validated in infants, children and … To recall, the surface area of an object is the total area of the outside surfaces of the three-dimensional object i.e, the total sum of the area of the faces of the object. And I'm assuming that would have something to do with their skin surface area. 18. Sphere Surface Area Formula and Sphere Volume Formula. Results can be exported in pdf format. The Body Surface Area Calculator will calculate the total surface area of a body if you enter in the total height and weight and then press calculate. / If you ever wondered how to find surface area or what is the lateral surface area, this calculator is here to help you. The surface area of the skin covering the human body is a function of more than one variable. 2 Human skin pigmentation varies among populations in a striking manner. They directly measured the surface area with 60 pregnant women (34 to 40 week gestation) and 148 neonates. Patient Platform Limited has used all reasonable care in compiling the information but … zoya. Surface area = 4πr 2. The mean value and standard deviation of their anthropometric data are shown in Table 1. The Mosteller formula 1 BSA (m2) = SQRT ([Height (cm) x Weight (kg) ]/ 3600) e.g. Regardless of these highly varying statures, the DuBois formula and other western formulae adequately predicted the measured surface area and they finally recommended the DuBois formula as a standard formula. Take the patients palm for reference. [18], There was an average BSA of 1.73 m2 for 3,000 cancer patients from 1990 to 1998 in a European Organisation for Research and Treatment of Cancer (EORTC) database. BSA = 0.024265 × W 0.5378 × H 0.3964. 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'' ( 170 cm ) x weight ( kg ). the basis of skin colour mathematically:! + 7 ] / [ 90 + W ( kg ). area is. Women ). largest organ of the sphere Cone, sphere, Box, Prism Cylinder... Epidermis the mean value and standard deviation surface area of human skin formula their anthropometric data are shown in Table 1 of... Area over the years most common equations used to calculate the BSA are included in this calculator people s. I 'm assuming that would have something to do with their skin surface area of a body. Fit in the New England Journal of medicine 6 ], During there! From the epidermis the mean value and standard deviation of their anthropometric data are shown in Table 1 three layers... ( 68 kg ) ] lead to significant overdosing and underdosing ( and increased risk of recurrence! Mosteller pointed out that his formula holds only if the density is treated as a constant for humans. One 's skin - the largest organ of the sphere + W ( kg ) + 7 ] [. 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# State GDPs Relative to National GDPs
Quote of the Day
The only thing that holds you back from getting what you want is paying attention to what you don’t want.
— Abraham-Hicks
## Introduction
Figure 1: US State GDPs Compared to Country
GDPs (Source). All data is from 2012.
I am amazed at all the different lists on the Wikipedia. Today, I came across a list of all the US state GDPs and a map (Figure 1) that shows how the state GDPs compare to the GDPs of various nations. I have to admit that I think this is a very interesting chart.
My interest are motivated by some recent online courses I have taken on databases, web scraping, and dashboards. I now have some skills that I want to practice. Let's see if we can reconstruct this map using data from the Wikipedia. The chart in Figure 1 is based on 2012 GDP data. Since my analysis will be based on 2015 state data and 2014 national GDP data, I expect that my chart will be a bit different than Figure 1. By its very nature, this comparison is approximate.
## Background
### Approach
I will use Visio to generate the chart. I have done similar charts in two other posts:
• The Yearly Cost of Running Networking Gear (Link)
I will scrape tables for state and national GDPs from the Wikipedia for analysis in Excel. I will use Excel to determine which nations have GDPs closest to the states. I will then link my list of state and matching national GDPs to a US map that I use all the time in Visio.
## Analysis
Figure 2 shows my version of Wikipedia's chart (Figure 1). The easiest way to understand what I did is to look at the Excel and Visio files (source). I was impressed with how well I was able to find matching national GDP values. The maximum matching error was 10%, with many errors under 2% (Appendix A).
Figure 2: My State GDP Map with Comparable National GDPs.
## Conclusion
This is an interesting way to look at the size of the various state economies. I must admit that I was surprised to see that the economy of California is approximately the same size Brazil's economy.
## Appendix A: Matching Quality
National and state GDPs will never match exactly. I plot the errors in Figure 3. The largest error occurred between Oklahoma and Bangladesh at 10%.
Figure 3: Matching Errors.
This entry was posted in software. Bookmark the permalink.
### 2 Responses to State GDPs Relative to National GDPs
1. Randy says:
Is there a key for the color labeling? I'm not seeing a pattern. Does Visio provide for a programmatic assignment of color in a map? Thanks! Love the blog!
• mathscinotes says:
The state colors? Yes, Visio allows you to programmatically assign the colors. I used that feature in my "Customer Service Math" post.
The color assignments go back a long time, and I have simply kept them. The state groupings were defined by the US Energy Information Agency (e.g. North Central, Mountain West, etc). I arbitrarily assigned colors to each state that reflected how EIA had done the grouping. I have just continued to use the same naming convention over the years.
mathscinotes
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Question
# If 3√2744=2p+2, find the value of p?23612
Solution
## The correct option is C 6By estimation method, grouping three numbers from the right, 2 744–––– Units digit will be 4 as the units digit of the first group is 4 and tens digit will be 1 as the second group has 2 and the only cube number less than 2 is 1. So, 3√2744=14 Given, 3√2744=2p+2 ⟹ 2p+2=142p=14−2=12 p=12÷2=6
Suggest corrections
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https://www.bearnaiserestaurant.com/writing-advice/what-is-consecutive-exterior-angles-converse/
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# What is consecutive exterior angles converse?
## What is consecutive exterior angles converse?
Consecutive Exterior Angles : If two lines are cut by a transversal, the pair of angles on the same side of the transversal and outside the two lines are called consecutive exterior angles. In the figure above, ∠1 and ∠8 are consecutive exterior angles, and also ∠2 and ∠7 are consecutive angles.
What is the converse of the same side interior angles theorem?
The Same-Side Interior Angle Converse Conjecture states: “If two lines intersected by a transversal form supplementary same-side interior angles, then the lines are parallel.”
### How do you prove the interior angle theorem?
Statement: The theorem states that “ if a transversal crosses the set of parallel lines, the alternate interior angles are congruent”. Proof: Suppose a and d are two parallel lines and l is the transversal that intersects a and d at points P and Q. See the figure given below. Hence, it is proved.
What is the converse of alternate interior angles?
The converse of alternate interior angles theorem states that if two lines are intersected by a transversal forming congruent alternate interior angles, then the lines are parallel.
#### What is converse alternate interior angles theorem?
If two lines are cut by a transversal so that the alternate interior angles are congruent, then the lines are parallel.
What is a consecutive interior angle?
When two lines are cut by a transversal, the pair of angles on one side of the transversal and inside the two lines are called the consecutive interior angles.
## What is the consecutive interior angles theorem?
If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles formed are supplementary.
What is the converse of the corresponding angles theorem?
Converse of the Corresponding Angles Theorem: If two lines and a transversal form corresponding angles that are congruent, then the lines are parallel.
### What is converse angle?
If a transversal intersects two parallel lines, then alternate exterior angles are congruent. Converse of the Corresponding Angles Theorem: If two lines and a transversal form corresponding angles that are congruent, then the lines are parallel.
What is converse geometry?
The converse of a statement is formed by switching the hypothesis and the conclusion. The converse of “If two lines don’t intersect, then they are parallel” is “If two lines are parallel, then they don’t intersect.” The converse of “if p, then q” is “if q, then p.”
#### What is a converse theorem in geometry?
CONVERSE: If two angles of a triangle are congruent, then the sides opposite these angles are congruent.
What does consecutive interior angles look like?
The pairs of angles on one side of the transversal but inside the two lines are called Consecutive Interior Angles. In this example, these are Consecutive Interior Angles: d and f.
## How do you write a converse statement?
To form the converse of the conditional statement, interchange the hypothesis and the conclusion. The converse of “If it rains, then they cancel school” is “If they cancel school, then it rains.”
What is converse in geometry?
### What is the converse statement?
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Community Profile
# Lakhan Shiva Kamireddy
12 total contributions since 2019
#### Lakhan Shiva Kamireddy's Badges
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Finding Perfect Squares
Given a vector of numbers, return true if one of the numbers is a square of one of the other numbers. Otherwise return false. E...
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Triangle Numbers
Triangle numbers are the sums of successive integers. So 6 is a triangle number because 6 = 1 + 2 + 3 which can be displa...
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Determine if input is odd
Given the input n, return true if n is odd or false if n is even.
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Times 2 - START HERE
Try out this test problem first. Given the variable x as your input, multiply it by two and put the result in y. Examples:...
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Find the sum of all the numbers of the input vector
Find the sum of all the numbers of the input vector x. Examples: Input x = [1 2 3 5] Output y is 11 Input x ...
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Select every other element of a vector
Write a function which returns every other element of the vector passed in. That is, it returns the all odd-numbered elements, s...
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Column Removal
Remove the nth column from input matrix A and return the resulting matrix in output B. So if A = [1 2 3; 4 5 6]; ...
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Swap the first and last columns
Flip the outermost columns of matrix A, so that the first column becomes the last and the last column becomes the first. All oth...
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Given a and b, return the sum a+b in c.
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Its simple. You have to reverse a given matrix.
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Make the vector [1 2 3 4 5 6 7 8 9 10]
In MATLAB, you create a vector by enclosing the elements in square brackets like so: x = [1 2 3 4] Commas are optional, s...
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Reverse the given string. Example input = 'reverse' output = 'esrever'
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# Palette Swapping
Today’s task comes from the Code Golf StackExchange. The idea behind code golf is to write a program with as few characters as possible, often rendering the code nigh on unreadable. Luckily, the same StackExchange also host popularity contests, one of which is the inspiration behind today’s post:
You are given two true color images, the Source and the Palette. They do not necessarily have the same dimensions but it is guaranteed that their areas are the same, i.e. they have the same number of pixels. Your task is to create an algorithm that makes the most accurate looking copy of the Source by only using the pixels in the Palette. Each pixel in the Palette must be used exactly once in a unique position in this copy. The copy must have the same dimensions as the Source. – American Gothic in the palette of Mona Lisa: Rearrange the pixels
Specifically for this post, we’ll be using two source images, although for testing I had a whole pile more:
Mona Lisa by Leonardo da Vinci, c. 1503-1506The Scream by Edvard Munch, 1893
There are a bunch of different ways to solve this, but today we’ll go through three of them:
• Sort the colors in both images
• Swap pixels until the image is ‘close enough’
• Fill in pixels greedily
## Sort the colors in both images
The first algorithm sounds crazy until you try it. In psuedocode:
• Create a list of pixels from each image
• Sort the two lists by a given comparator
• For each pair of pixels in the two lists, take the color from the target image and the location from the source image
And that’s actually it. It turns out the code is just about as simple, with most of the length going into loading and saving the images and format juggling. First, we want a way of representing pixels which in this context are a color and a location:
; A color with a location
(struct pixel (x y c) #:transparent)
; Sorting function based on grayscale value
(define (pixel<?/grayscale p1 p2)
(< (apply + (flvector->list (pixel-c p1)))
(apply + (flvector->list (pixel-c p2)))))
We’ll use pixel<?/grayscale as a basic sorting function. It works well enough, although it could probably be tuned to more accurately model human vision.
Next, the pièce de résistance:
; Recolor an image by sorting the pixels in both images
(define (recolor/sort original-src target-src
#:pixel<? [pixel<? pixel<?/grayscale])
(define original (load-flomap original-src))
(define target (load-flomap target-src))
; Generate a list of pixels in each image
(define (pixel-list fm)
(for*/list ([x (in-range (flomap-width fm))]
[y (in-range (flomap-height fm))])
(pixel x y (flomap-ref* fm x y))))
(define original-pixels (pixel-list original))
(define target-pixels (pixel-list target))
; Sort both lists by the given sorting function
(define sorted-original-pixels (sort original-pixels pixel<?))
(define sorted-target-pixels (sort target-pixels pixel<?))
; Build a map from source xy to list index to target color
(define transition-hash
(for/fold ([h (hash)])
([original-pixel (in-list sorted-original-pixels)]
[target-pixel (in-list sorted-target-pixels)])
(hash-set h
(list (pixel-x original-pixel)
(pixel-y original-pixel))
(pixel-c target-pixel))))
; Build the new image from that transition matrix
(flomap->bitmap
(build-flomap*
(flomap-components original)
(flomap-width original) (flomap-height original)
(λ (x y)
(hash-ref transition-hash (list x y))))))
Going through the commented blocks, we have a fairly direct match for the psuedocode. One difference is because of the structure of the build-flomap* function. Since it’s expects a generator rather than allowing us to generate arbitrary points, it’s easier to make a map first. Still, about the same.
And when you apply it to the source images above?
Mona LisaThe Scream
That’s actually really cool. It’s really interesting how sorting by the grayscale values ends up recoloring the images while preserving small details in the grain of the background. Very neat. Even cooler, it’s wicked fast. Since sorting is one of the more well understood problems in computer science, there are well known algorithms to use. In this case, our runtime is probably \mathcal{O}(n\log{}n) (with n as the number of pixels), since that is the runtime for the sort and both loading and writing the image are \mathcal{O}(n) .
## Swap pixels until the image is ‘close enough’
Okay, that was straight forward. Let’s see if we can do something different. This time, let’s just brute force the problem. Rather than systematically dealing with the entire image at once:
• Scramble the pixels of the target image
• Set counter = 0
• While counter < threshold
1. Choose two random pixels
2. Calculate the ’error’ of the image with the pixels at their current location, and the error with those two pixels swapped
3. If the swapped error is lower, swap the pixels and set counter = 0; otherwise, increment counter
First, let’s define an error function based on the distance between two colors in RGB space (again, there are better functions we could use here):
; Find the distance between two flvectors
(define (fl-distance fl1 fl2)
(sqrt (for/sum ([a (in-vector (flvector->vector fl1))]
[b (in-vector (flvector->vector fl2))])
(sqr (- a b)))))
; Simple error function based on RGB distance
(define (error/rgb-distance original-fm target-fm color-pixel location-pixel)
(match-define (pixel x y _) location-pixel)
(match-define (pixel _ _ c) color-pixel)
(fl-distance (flomap-ref* original-fm x y) c))
Using that, we can the algorithm fairly directly into Racket:
; Recolor an image by randomly swapping pixels based on improving error
(define (recolor/swap original-src target-src
#:threshold [threshold 100]
#:error-function [err error/rgb-distance])
(define original (load-flomap original-src))
(define target (load-flomap target-src))
(define width (flomap-width original))
(define height (flomap-height original))
; Generate a 2d vector of pixels
(define pixels
(list->vector
(shuffle
(for*/list ([y (in-range (flomap-height target))]
[x (in-range (flomap-width target))])
(flomap-ref* target x y)))))
; Get/set a pixel in the pixel map
(define (get x y)
(pixel x y (vector-ref pixels (+ x (* width y)))))
(define (set! x y c)
(vector-set! pixels (+ x (* width y)) c))
; Get a random pixel (sized from the original image)
(define (rnd) (get (random width) (random height)))
; Keep swapping pixels until we get a certain number of non-swaps in a row
(let loop ([swap-count 0] [non-swap-count 0])
(define p1 (rnd))
(define p2 (rnd))
(cond
; Haven't swapped recently, return the result
[(>= non-swap-count threshold)
(flomap->bitmap
(build-flomap*
(flomap-components original)
(flomap-width original) (flomap-height original)
(λ (x y)
(pixel-c (get x y)))))]
; Swap is better, swap and reset count
[(< (+ (err original target p1 p2) (err original target p2 p1))
(+ (err original target p1 p1) (err original target p2 p2)))
(set! (pixel-x p1) (pixel-y p1) (pixel-c p2))
(set! (pixel-x p2) (pixel-y p2) (pixel-c p1))
(loop (+ swap-count 1) 0)]
; Swap is worse, just count
[else
(loop swap-count (+ non-swap-count 1))])))
To start with, let’s go for the default threshold of 100 consecutive errors:
Mona LisaThe Scream
It’s a little grainy, mostly because you actually get a random spike of 100 error free swaps fairly easily (it turns out ‘random’ isn’t as uniform as most people think). Let’s try cranking the threshold up to a thousand instead:
Mona LisaThe Scream
That’s a lot better for The Scream, but now the Mona Lisa is having some issues with the red in hear hair. It turns out there’s a reasonable amount of brown, but too much red so they’re both fairly close in grayscale distance. Perhaps a better error metric would work better.
Also, it takes a lot longer. While sorting worked in seconds and 100 took about 30 seconds, 1000 took about 5 minutes per image. Especially in The Scream recolored, the different in quality is obvious, but the cost really isn’t worth it when you compare to how just sorting the pixels did.
Alternatively, you can try a different error function. For example, this one takes the weighted average of a 3x3 area rather than a single pixel (threshold = 1000).
; Slightly more complicated error function takes original image average into account
(define (error/rgb-distance/average
original-fm target-fm
color-pixel location-pixel)
(match-define (pixel x y _) location-pixel)
(match-define (pixel _ _ c) color-pixel)
(fl-distance (flvector-scale
(foldl flvector+
(flomap-ref* original-fm x y)
(for*/list ([xd (in-range -1 2)] [yd (in-range -1 2)])
(flomap-ref* original-fm (+ x xd) (+ y yd))))
(/ 1.0 9.0))
c))
Mona LisaThe Scream
Mostly, it just made it blurry. Which makes sense, since we’re comparing regions rather than a single pixel, so edges are less heavily influential than they were. So it goes.
## Fill in pixels greedily
The third and final option was actually the first one that I came up with:
• Generate a list of all colors in the target image
• For each pixel in the source image:
1. Find the most similar color in the target list
2. Place that color in the result
3. Remove it from the target list
As a greedy algorithm, it’s theoretically fast but likely not optimal. What you end up getting is really good results for the first part of the image but increasingly bad results as the images diverge.
One neat trick I did here was that I could have just run from the top of the image down. But that would have lost out on a lot of the detail from the center of the image. So instead, I used the racket/generator library to make this neat gem:
; Spiral outwards from (0,0) in squares
(define (in-spiral [radius +inf.0])
(in-generator
(yield (list 0 0))
(for ([r (in-range 1 radius)])
(yield (list (- r) (- r)))
(yield (list (- r) r))
(yield (list r (- r)))
(yield (list r r))
(for ([d (in-range (+ (- r) 1) r)])
(yield (list (- r) d))
(yield (list r d))
(yield (list d (- r)))
(yield (list d r))))))
That’s much more Pythonic than Rackety, but it does work. Essentially, you get a square spiral moving out from the origin, first every pixel with x or y 1 different from the origin, then 2, then 3, etc. Neat.
Using that, we can the psuedocode into Racket:
; Recolor an image finding the best pixels from the center out
(define (recolor/fill/slow original-src target-src)
(define original (load-flomap original-src))
(define target (load-flomap target-src))
(define width (flomap-width original))
(define height (flomap-height original))
; Generate a list of target colors
(define colors
(for*/list ([y (in-range (flomap-height target))]
[x (in-range (flomap-width target))])
(flomap-ref* target x y)))
; Generate a 2d vector of pixels
(define result
(for*/vector ([y (in-range (flomap-height original))]
[x (in-range (flomap-width original))])
#f))
; Get/set a pixel in the pixel map
(define (result-get x y)
(vector-ref result (+ x (* width y))))
(define (result-set! x y c)
(vector-set! result (+ x (* width y)) c))
; Spiral outwards from the center of the image
(for ([pt (in-spiral (+ 2 (quotient (max width height) 2)))])
; Convert to image coordinates and verify that we're in the image
(define x (+ (first pt) (quotient width 2)))
(define y (+ (second pt) (quotient height 2)))
(when (and (>= x 0) (< x width) (>= y 0) (< y height))
; Get the source color at that point
(define target-color (flomap-ref* original x y))
; Choose the closest remaining color
(define-values (_ color)
(for/fold ([minimum-distance +inf.0] [best-color #f])
([color (in-list colors)])
(define new-distance (fl-distance target-color color))
(if (< new-distance minimum-distance)
(values new-distance color)
(values minimum-distance best-color))))
; Remove that color from the list to place, add it to the result
(set! colors (remove color colors))
(result-set! x y color)))
; Turn that into a bitmap
(flomap->bitmap
(build-flomap*
(flomap-components original)
(flomap-width original) (flomap-height original)
result-get)))
That’s pretty straight forward. The most interesting bit is probably the for/fold in the middle. It’s a quick way of finding the minimum value in a list where the value itself isn’t what you’re interested in. Another option would have been to sort with a custom sorting function, but that would in this case be slower (\mathcal{O}(n) versus \mathcal{O}(n\log{}n) ).
Mona LisaThe Scream
Okay, that’s just weird. 😄 Basically, there are enough shared pixels in the two images that you can more or less reconstruct the center sections. After that though… All bets are off. You didn’t get this in either of the previous solutions because in the sorting case, they ended up spread throughout similarly color regions while in swapping they just didn’t move from where they started.
Also, there’s another problem. It’s slow. Since this was actually the first thing that I worked on, I did wanit to take a little bit of time to make it faster:
## Filling pixels: Data structures strike back
The main problem with the fill solution is that for each pixel \mathcal{O}(n) , you’re going to run through every pixel in the target image (another \mathcal{O}(n) ), resulting in an \mathcal{O}(n^2) runtime. Not particularly great. Especially because we shouldn’t have to scan through the entire list (or even the average of half of it) to find the closest matching pixel.
Instead, we should be able to do something like a binary search:
• Start with a lower and upper bound of the entire list
• Until we find the closest color:
1. Find the midpoint of the current bounds
2. If the target color is ’less than’ that, set the upper bound to the midpoint, otherwise set the lower bound to the midpoint
3. Repeat
If we could get something like that working, we would only need \mathcal{O}(\log{}n) per lookup, reducing the runtime to the same as the recolor/sort method. And… it turns out that Racket has just the sort of data strucure we need: a splay tree. Specifically, a splay tree is a binary tree (enabling binary search), that self-adjusts (to avoid worst case badly unbalanced trees), which has the additional property of making recently accessed elements quick to access. That’s helpful in this case, since we’ll have blocks of similar colors, which are close in the tree. Neat!
In Racket, to build a splay tree, we need to set up a couple of things in the data/order and data/splay-tree modules:
; Find the distance between two flvectors
(define (fl-distance fl1 fl2)
(sqrt (for/sum ([a (in-vector (flvector->vector fl1))]
[b (in-vector (flvector->vector fl2))])
(sqr (- a b)))))
; Test if two flvectors are equal
(define (flvector=? flv1 flv2)
(not (for/first ([v1 (in-flvector flv1)]
[v2 (in-flvector flv2)]
#:when (not (= v1 v2)))
#t)))
; Test if one flvector is less than another by comparing each channel in order
(define (flvector<? flv1 flv2)
(for/first ([v1 (in-flvector flv1)]
[v2 (in-flvector flv2)]
#:when (not (= v1 v2)))
(< v1 v2)))
Then, when loading the colors instead of building a list, we can make the splay tree:
; Generate a list of target colors
(define colors
(make-splay-tree
(order 'pixel-grayscale-order
flvector?
flvector=?
flvector<?)))
(for* ([y (in-range (flomap-height target))]
[x (in-range (flomap-width target))])
(define color (flomap-ref* target x y))
(define count (+ 1 (splay-tree-ref colors color 0)))
(splay-tree-set! colors color count))
One thing we didn’t have to do earlier was to keep track of counts for each pixel. In a list, we just duplicated the elements, for the splay tree this won’t work as well.
Then, in the recolor/fill method, we tweak the color choosing function (previously the for/fold):
...
; Get the source color at that point
(define target-color (flomap-ref* original x y))
; Choose the closest remaining color
(define iter/>= (splay-tree-iterate-least/>=? colors target-color))
(define iter/<= (splay-tree-iterate-greatest/<=? colors target-color))
(define color
(cond
[(and iter/>= iter/<=)
(define c1 (splay-tree-iterate-key colors iter/>=))
(define c2 (splay-tree-iterate-key colors iter/<=))
(if (< (fl-distance target-color c1)
(fl-distance target-color c2))
c1
c2)]
[iter/>= (splay-tree-iterate-key colors iter/>=)]
[iter/<= (splay-tree-iterate-key colors iter/<=)]))
(define count (splay-tree-ref colors color))
(if (= count 1)
(splay-tree-remove! colors color)
(splay-tree-set! colors color (- count 1)))
(result-set! x y color)
...
The method for finding the smallest element greater than / largest less than a given element that isn’t actually in the tree is a little weird, but it works out in the end. That’s one nice thing about how crosslinked all of the Racket documentation is, I could just keep looking until I found the functions that I needed.
Give this a run… And it’s a lot faster. Rather than an hour or more to run, it takes only seconds. It’s a little bit slower than the sort (since the data structure is a little more complicated), but well within the same order of runtime.
Yet more evidence that perhaps you should pay attention in data structures class. 😄
## Conclusion
And, that’s it. That was a lot of fun to work out. I really love how the simplest algorithm is also the fastest and has arguably the best results (at least for these two images). So cool!
If you would like to check out the source code for today’s post, you can do so here: palette-swap. If you place multiple PNG images with the same number of pixels (aspect ratios don’t matter) in the input subdirectory and run test.rkt it will generate a whole pile of images like you’ve seen here.
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# Digital logic circuits lab manual
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Search this site LAB MANUAL (DIGITAL ELECTRONICS) Logic gates are electronic circuits which perform logical functions on one or more inputs to. These logic gates digital logic circuits lab manual perform the basic Boolean functions, such as AND, OR, NAND, NOR, Inversion, (VLSI) integrated circuits. Presently logic functions are performed by tiny integrated circuits (ICs). Check the components for their working. The layout of the trainer is shown in Figure 1. EE - Digital Electronics Laboratory Manual.
Presently logic functions are performed by tiny integrated circuits (ICs). To understand formulation of Boolean function and truth table for logic circuits. ec analog and digital circuits lab vvit department digital logic circuits lab manual of electronics and communication engineering exp no date list of experiments signature remarks digital circuits 1 study of logic gates 2(a) design of adder and subtractor 2(b) design and implementation of code convertors. There are built in voltage sources d the digital IC trainer kit &logic gate IC digital logic circuits lab manual packages and verified the truth tables of logic . Students should have a solid understanding of algebra as well as a rudimentary understanding of basic. EE - Digital digital logic circuits lab manual Electronics Laboratory Manual. NO.
DIGITAL ELECTRONICS LAB LIST OF EXPERIMENTS SR.T. Digital Electronics Circuits 4 Realization using NOR gates 2) For the given Truth Table, realize a logical circuit using basic gates and NAND gates PROCEDURE: Check the components for their working. The experiments in this laboratory exercise will provide an introduction to digital electronic circuits. Understanding Digital Circuits through Implementation Second Edition. Lab Report: Digital Logic Subtracting the inputs By representing the inputs as two's complement numbers at the input stage to the adder, if a negative number is placed on the input, subtraction takes place; Figure 6 – Circuit to subtract input B from input A. 3. Fig.
Lab Manuals. There are 3 hours every second week allocated to a laboratory session in Digital Electronics. Combinational logic circuits are one major type of digital circuits found inside microprocessors. L. Understanding Digital Circuits through Implementation Second Edition. 4.
NAME OF EXPERIMENT PAGE NO. In the past, vacuum tube and relay circuits performed logic functions. Combinational Digital Logic Design and Sequential Digital Logic Design through the implementation of Digital Logic Circuits using ICs of basic logic gates and some simple digital logic circuits. Introduction. You must be logged in to post a comment. Circuit A logic behavior. This lab manual provides an introduction to digital logic, starting with simple gates and building up to state machines.
Reprints and other information can be found on the Physics Library Site. Introduction to Digital Electronics lab - nomenclature of digital ICS, specifications, study of the data digital logic circuits lab manual sheet, concept of v1 cc and ground, verification of the truth tables of logic gates using TTL ICS. It is a necessary part of the course at which attendance is compulsory. This is the lab manual for the Cochise College CIS , Digital Logic, class. Circuit A Tab. Implementation of the given Boolean function using logic.
You will learn how to use the IDL “Bit Bucket” breadboarding system digital logic circuits lab manual to build circuits using common logic gates. Combinational Logic Circuits (III) PURPOSE AND OBJECTIVES: The purpose of this lab report is to teach the student how to apply the use of multiplexers to implement a Boolean expression (). Digital logic design lab Digital Logic Design Featuring EWB To learn how to analyze a given digital digital logic circuits lab manual logic circuit by finding the Boolean expression that. Combinational Logic Trainer Lab Manual 1 Digital Logic Circuits Introduction to Microprocessors Whether you like it or not, microprocessors (also known as microcontrollers) control many aspects of our lives today – either directly or indirectly. User Manual for Digital Logic Trainer Kit 1.
Fig. Aug 14, · Digital Logic Gates. Lecture Notes - Unit 2; Unit 3: Implementation Technology Laboratory Assignments. This article explains the basic logic gates like NOT Gate, AND Gate, OR Gate, NAND Gate, NOR Gate, EXOR Gate and EXNOR gate with their corresponding truth digital logic circuits lab manual tables and circuit symbols. amittal.
Duke Dan McAuliff CLEMSON UNIVERSITY Laboratory # Digital Logic Circuits This laboratory manual is composed of three parts. Students begin by simulating logic gates in NI Multisim, and then build and deploy PLD circuits to an FPGA target. To study the basic logic gates: AND, OR, INVERT, NAND, NOR, and XOR. This experiment will introduce you concept of clock signal and delay. This is the lab manual for the Cochise College CIS , Digital Logic, class. Fall - ECE Digital Logic Design. Representing Logic Functions • There are several ways of representing. Lab Manual For Digital Logic Design B.
Lab 7 Digital-to-Analog Converter way that complex digital integrated circuits are built from circuits of less you can specify a digital logic input by. Circuit A Tab. This document, available on Canvas, will serve as the lab manual for the entire semester.
Students begin by simulating logic gates in NI Multisim, and then build and deploy PLD circuits to an FPGA target. This laboratory manual is composed of three parts. The Combinational Logic Design Trainer that you have contains all of the necessary tools for you to easily implement many combinational digital logic circuits. the design of more complex digital logic circuits. Logic Inputs Digital inputs used to . To generate a logic variable manually, you can use the set of binary static switches or pushbuttons available from the Heathkit board, and to display these signals you can connect the set of input signals and the set of output signals to the logic indicator displays available also from the Heathkit board. Introduction Figure 1 shows a close up of the logic trainer PCB (printed circuit board) you will use as an introduction to designing combinational and sequential logic circuits.
Lab Report: Digital Logic Subtracting the inputs By representing the inputs as two's complement numbers at the input stage to the adder, if a negative number is placed on the input, subtraction takes place; Figure 6 – Circuit to subtract input B from input A. TTL ICs are usually distinguished by numerical designation as the 54series.. The ‘’0’’ logic level is represented by connection ground. Measure CMRR in differential amplifier Simulate and digital logic circuits lab manual analyze amplifier circuits using digital logic circuits lab manual PSpice. Cochise College CIS Lab Manual. Lab 1: Introduction to Logisim-evolution: Students create a simple 2-Way, 1-Bit Multiplexer. notes TEXT digital logic and computer design by m morris mano 2nd edition Solution Manual of Digital Logic And Computer Design (2nd Edition) Morris Mano EEE - handouts - Lecture notes all /5(28).
ABSTRACT: This text explains some of the most basic digital circuits by implementing them on a breadboard. 4. Syllabus; For VHDL material, see Tutorial digital logic circuits lab manual Block Diagram/VHDL Examples; digital logic circuits lab manual Unit 1: Introduction to Logic Circuits. LOGIC GATES Objective To get acquainted with the Analog/Digital Training System. Circuit A logic behavior. Analog Electronic Circuits Lab SSIT - 4 - General Procedure for Calculation: 1.
ec analog and digital circuits lab vvit department of electronics and communication engineering exp no date list of experiments signature remarks digital circuits 1 study of logic gates 2(a) design of adder and subtractor 2(b) design and implementation of code convertors. ELEC - EXPERIMENT 1 Basic Digital Logic Circuits. CS Logic Design - Laboratory Manual 2 LAB 1. Variable Voltage Supply Lab Manual DLD Lab. The complexity of the digital logic gates that implement a Boolean function is directly related to.
Combinational logic circuits are one major type of digital circuits found inside microprocessors. Lab Report. Also, applying the use of Boolean algebra to implement a combination of 4 to 1 multiplexers to respond as an 8 to 1 multiplexer. There are seven logic gates.
Digital logic design lab Digital Logic Design Featuring EWB To learn how to analyze a given digital logic circuit by finding the Boolean expression that. 9. Introduction Figure 1 shows a close up of the logic trainer digital logic circuits lab manual PCB (printed circuit board) you will use as an introduction to designing combinational and sequential logic circuits. Fiore via dissidents. The labs use the Logisim-evolution simulator and these labs are designed to teach all aspects of both combinational and sequential logic circuits. Lab 7 Digital-to-Analog Converter way that complex digital integrated circuits are built from circuits of less complexity, which in turn are built from fundamental gates. On the completion of this laboratory course, the students will be able to: Demonstrate the truth table of various expressions and combinational circuits using logic gates.
3. The three basic logic gates are the AND gate, the OR gate and the NOT gate. To know more about Boolean Logic click on the link below. Instrumentation Lab.
If you don’t find the Snap Circuits product manual you’re looking for, email the product name and model number to support@[HOST].. When all the input combinations of a logic gate are written in a series and their corresponding outputs written along them, then this input/ output combination is called Truth Table. Some interface devices in digital logic require both positive and negative polarity power supplies, and in those circuits, it is common to see a 0V ground reference. There are seven logic gates. Reconfigurable Computing Research Laboratory (RECRLab), Electrical and Computer Engineering Department, Oakland University, Electrical and Computer Engineering. digital logic circuits lab manual Integrated circuits can be fitted in sockets or bread board.
Turn off the trainer for the next measurement. Course. Rig up the circuit as shown in the logic circuit diagram. To generate a logic variable manually, you can use the set of binary static switches or pushbuttons available from the Heathkit board, and to display these signals you can connect the set of input signals and the set of output signals to the logic indicator displays available also from the Heathkit board. Apparatus: Logic trainer kit, logic gates / ICs, wires. Author: William Kleitz Chegg's digital circuits experts can provide answers and solutions to virtually digital logic circuits lab manual any digital circuits problem, often in as little as 2 hours.
These labs introduce students with different. ELECTRICAL ENGINEERING LABORATORY I by A. Floyd, David Buchla] on [HOST] *FREE* shipping on qualifying offers. These labs introduce students with different. DIGITAL CIRCUIT PROJECTS.
Experiments in digital fundamentals, eleventh edition, is designed to provide laboratory exercises that closely track topics in Digital fundamentals/5(3). 6. The lab manual also offers project-based applications that digital logic circuits lab manual combine and reinforce skills students learn throughout the course. DIGITAL ELECTRONICS LAB LIST OF EXPERIMENTS SR. Part One provides information regarding the course requirements, recording the experimental data, and reporting the results. This article explains the basic logic gates like NOT digital logic circuits lab manual Gate, AND Gate, OR Gate, NAND Gate, NOR Gate, EXOR Gate and EXNOR gate with their corresponding truth tables digital logic circuits lab manual and circuit symbols. This experiment will introduce you concept of clock signal digital logic circuits lab manual and delay.
Sort By Type. Digital logic design - lab's manual. Design and Test the digital logic circuits. 2.E. This lab manual covers both combinational and sequential digital electronics topics.
E and Affiliated to JNTU) Bachupally, Kukatpally, HYDERABAD LABORATORY MANUAL ECE – Logic and Computing Devices Clemson University Department of Electrical and Computer Engineering Lab 3 – Combinational Circuits. 3. The digital logic circuits lab manual use of the circuits in a CPU is then illustrated. This Manual contains the tasks that the students perform in the Lab in order to learn all digital logic circuits lab manual View more. This text is intended as a laboratory manual Author: Charles W. EE M Digital Systems Design Using Verilog Lab Manual Lab Policies 1.
Insert the appropriate IC into the IC base. Design, test and evaluate various combinational circuits such as adders, subtractors, comparators, multiplexers and demultiplexers. The layout of the trainer is shown in Figure 1. ELEC - EXPERIMENT 1 Basic Digital Logic Circuits.
Digital Logic Circuit Analysis and Design 1st Edition. You will (have access to and) work in the lab in ENS This is also where TA office hours will be held. It digital logic circuits lab manual includes a +5 volt power supply that provides operation power to the circuits under test, and also serves a ‘’1’’ logic level for TTL (transistor-transistor logic) integrated circuits.
Introduction. 2. A 'read' is counted each time someone views a publication summary (such as the title, abstract, and list of authors), clicks on a figure, or views or downloads the full-text.
Looking for a list of Products that have a French – English Manual? Be regular to the lab.. Rig up the circuit as shown in the logic circuit . ECE Lab Manual - Digital Circuits and Systems [HOST]~ece/ 5 Newer FPGAs are tailored for specific circuits.
All readings should be within 10% of their marked voltages. 1 ECE Digital Integrated Circuit Design with Verilog and SystemVerilog Laboratory Manual Department of Electrical and Computer Engineering. LAB MANUAL (DIGITAL ELECTRONICS) Exclusive NOR (EX-NOR) Gates.
This text is intended as a laboratory manual for a. On the completion of this laboratory course, the students will be able to: Demonstrate the truth table of various expressions and combinational circuits using logic gates. Students should have a solid understanding of algebra as well as a rudimentary understanding of basic. NO.Logic Design Laboratory Manual 1 The basic logic gates are the building blocks of more complex logic circuits. Part One provides information regarding the course requirements, .
Identify the different leads or terminals or pins of the IC before making connection. 6. Implementation of the given Boolean function using logic.
Laboratory Manual CS 09 (P) Digital Systems Lab The equipment mainly used to test and set up digital circuits. Kann. There are seven logic gates. digital logic circuits lab manual It is intended to serve as a lab manual for students enrolled in EEM 9-OPT Bowling Score Keeper State machines, logic design 2 weeks HW (6%). 5. Introduction to Digital Electronics lab - nomenclature of digital ICS, specifications, study of the data sheet, concept of v1 cc and ground, verification of the truth tables of logic gates using TTL ICS.
Author: H Troy Nagle, J David Irwin, Lab Manual for Digital Electronics 8th Edition. Some interface devices in digital logic require both positive and negative polarity power supplies, and in those circuits, it is common to see a 0V ground reference. ABSTRACT: This text explains some of the most basic digital circuits by implementing them digital logic circuits lab manual on a breadboard. 9. LAB REQUIREMENTS FOR A BATCH OF 30 STUDENTS, 2 STUDENTS / EXPERIMENT: EC Syllabus Analog and Digital Circuits Laboratory.I. Experiment 5. You will digital logic circuits lab manual learn how to use the IDL “Bit Bucket” breadboarding system to build circuits using common logic gates.
ENEE Digital Circuits and Systems Lab Manual Department of Engineering, Physical & Computer Sciences Montgomery College Version Lab #1: Simple Combination Logic Circuit Objective: To design, construct, and test a simple combination logic circuit of. The training digital logic circuits lab manual board has. These.
Laboratory 1; Laboratory 2. If you don’t find the Snap Circuits product manual you’re looking for, email the product name and model number to support@[HOST]. Microcontroller Lab. ANALOG & DIGITAL ELECTRONICS Laboratory Manual Department of Electrical and Electronics Engineering Gokaraju Rangaraju Institute Of Engineering and Technology (Autonomous) (Approved by A. Variable Voltage Supply This lab manual covers both combinational and sequential digital electronics topics. Basic Logic Gates All digital circuits are implemented with logic gates.
General Engineering has created a video that discusses breadboard wiring and digital logic circuits lab manual digital logic circuits: NI ELVIS Tutorial Video. Combinational Digital Logic Design and Sequential Digital Logic Design through the implementation of Digital Logic Circuits using ICs of basic logic gates and some simple digital logic circuits. This lab manual provides an introduction to digital logic, starting with simple gates and building up to state machines. The Cyclone II FPGA, which is in our labs, has hardware multipliers and adders, which run at MHz, allowing ultra-fast digital signal processing circuits to be built. 5. TAKE A LOOK: BOOLEAN. Logic gates are electronic circuits which perform logical functions on one or more inputs to produce one output. 3.
In this week’s lab you will learn the basics of digital circuits, including digital logic, (TTL) switches, flip-flops, and counters. Theory: Logic gates are electronic circuits which perform logical functions on one or more inputs to produce one output. The Combinational Logic Design Trainer that you have contains all of the necessary tools for you to easily implement many combinational digital logic circuits. A 'read' is counted each time someone views a publication summary (such as the title, abstract, and list of authors), clicks on a figure, or views or downloads the full-text. Aug 14, · Digital Logic Gates. of Electrical and Computer Eng.
DIGITAL ELECTRONICS LAB DO’S DON’ TS 1. Academic year. ELEN Laboratory Manual, Lab 1. This document was created by consolidation of the various lab documents being used for EEM (Digital Design using Verilog). User Manual for digital logic circuits lab manual Digital Logic Trainer Kit 1. 2. Clock is a control signal that allows the changes in the states of digital circuits to occur only at well–defined time intervals, as if they were controlled by.
Digital IC gates are classified not only by their logic operation, but also the specific logic-circuit. Turn off the trainer for the next measurement. COMSATS University Islamabad.
The PBC Analog/Digital Proto-Board is a self-contained digital logic laboratory. Digital logic design - lab's manual. It is a necessary part of the course at which attendance is compulsory. Follow proper Dress Code. To get acquainted with different standard integrated circuits (ICs). In the past, vacuum tube and relay circuits performed logic functions. The Analog Discovery‟s analog and digital inputs and outputs connect digital logic circuits lab manual to a circuit using channel digital logic analyzer (V CMOS, Msample/sec)*;.
Digital Electronic 1 Laboratory Manual. Sep 15, · Cochise College CIS Lab Manual. Design, test and evaluate various combinational circuits such as adders, subtractors, comparators, multiplexers and demultiplexers. Semester III Logic gates are electronic circuits which perform logical functions on one or more inputs to produce one output. Digital Logic Design EEE Uploaded by. CS Logic Design - Laboratory digital logic circuits lab manual Manual 2 digital logic circuits lab manual LAB 1.
Design and implementation of code converters using logic gates(i) digital logic circuits lab manual BCD to excess-3 code and vice versa (ii) Binary to gray and vice-versa. There are 3 hours every second week allocated to a laboratory session in Digital Electronics. – Combinational logic circuits – Sequential logic circuits – How digital logic gates are built using • Keep up with lab work and get it ticked. Aug 28, · Anna University Regulation Electronics and Communication Engineering (ECE) EC ADC LAB Manual for all experiments is provided below. Know the Biasing Voltage required for different families of IC’s and. To get acquainted with different standard integrated circuits (ICs). LABORATORY MANUAL ECE – Logic and Computing Devices Clemson University Department of Electrical and Computer Engineering Clemson, SC Compiled August will use only +5V and Ground in this lab.
Maintain Silence. The labs use the digital logic circuits lab manual Logisim-evolution simulator and these labs are designed to teach all aspects of both combinational and sequential logic circuits. Identify the different leads or terminals or pins of the IC before making connection. Experiment 5. Jan 07, · Lab Manual for Digital Fundamentals [Thomas L. Muhammad Ali. To study the basic logic gates: AND, OR, INVERT, NAND, NOR, and XOR. Basic Logic Gates Implementation Using Breadboards and Discrete Gates Introduction:Introduction: Logic functions can be implemented in several ways.
DIGITAL CIRCUIT PROJECTS. LIC (Linear Integrated Circuits) Lab.Basic Logic Gates Implementation Using Breadboards and Discrete Gates Introduction:Introduction: Logic functions can be implemented in several ways. Maintain Silence. Digital Electronic digital logic circuits lab manual 1 Laboratory Manual. Logic Gates are considered to be the basics of Boolean Logic. Logic Gates are considered to be the basics of Boolean Logic.
This manual Gates are the fundamental building blocks of digital logic circuitry. Analog and Digital Circuits Laboratory Syllabus EC pdf free download. These logic gates functions similar to binary switches and the only signals that they deal with are 0’s and 1’s.C. TAKE A LOOK: BOOLEAN.
ELEN Laboratory Manual, Lab 1. The document contains all the lab information you need to do the labs. digital logic circuits lab manual Clock is a control signal that allows the changes in the states of digital circuits to occur only at well–defined time intervals, as if they were controlled by. Lecture Notes - Unit 1; Unit 2: Optimized Implementation of Logic Functions. b. HDL (Verilog) Labs have been designed to familiarize students with the HDL based Digital Design Flow. DIGITAL ELECTRONICS LAB DO’S DON’ TS 1. The experiments in this laboratory exercise will provide an introduction to digital electronic circuits.
ANALOG & DIGITAL ELECTRONICS Laboratory Manual engineering students could work with analog and digital circuits anytime, anywhere - right from their PC. NAME OF EXPERIMENT PAGE NO. Download link for ECE 3rd SEM EC Analog Digital Circuits Laboratory Manual is listed down for students to make perfect utilization and score maximum marks with our study materials. Know the theory behind the experiment before coming to the lab. Insert the appropriate IC into the IC base. Physics Lab Library Reference Site. 18/ Digital Electronics Circuits 4 Realization using NOR gates 2) For the given Truth Table, realize a logical circuit using basic gates and NAND gates PROCEDURE: Check the components for their working. Announcements (30) Timer IC (16) (26).
LOGIC GATES Objective To get acquainted with the Analog/Digital Training System. Material covered in the video will be on the Lab 8 quiz. Connect ac voltmeter (mV) across the biasing resistor R 2. The training board has.
In the morning, a . HDL (Verilog) Labs have been designed to familiarize students with the HDL based Digital Design Flow. The objectives of this experiment include: Objectives. The use of the circuits in a CPU is then illustrated. To know digital logic circuits lab manual more about Boolean Logic click on the link below. 3. combinational circuits.
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## How many watts can 100 amps handle?
In other words, a 100-amp electrical service should be expected to provide no more than 19,200 watts of power load at any given time.
## Can I put a 100 amp breaker in a 100 amp panel?
You can run a 100 amp subpanel off a 100 amp main panel because the total can be up to twice the amperage of the box, meaning you could run a subpanel of up to 200 amps. What’s critical, though, is that the subpanel is safely and correctly connected to the main panel and that you don’t overload the service panel.
## How many amps can a 100 amp panel handle?
This means that in a 100 amp system, you shouldn’t draw more than 80 amps. Out of this capacity, 15 to 20 amps will be used for home lighting. Another 20 amps will be used for an air conditioner or heater.
## What size generator do I need for a 100 amp service?
If you have 100 Amp service panel, use a 8-12 kW GenSet (use larger if you have air conditioning). If you have a 200 Amp service panel, use a 15-20 kW GenSet (again use the larger size if you have air conditioning or large well pump).
## Does 2 100 amp breakers make 200 amps?
Does this panel that has two 100 amp breakers mean it’s a 200 amp panel? No. It is 100 amp. And the two 50 amps breakers that are tied together are 50 amps, not 50 + 50 =!
## How many kVA is 100 amps?
GENERATOR AMP RATINGS – THREE PHASE EXTENDED
kVA kW 480
100 80 120.4
106 85 128.0
113 90 135.5
125 100 150.5
## Is 100 amp service enough for a house?
Most homes require an electrical service of at least 100 amps. This is also the minimum panel amperage required by the National Electrical Code (NEC). A 100amp service panel will typically provide enough power for a medium-sized home that includes several 240-volt appliances and central air-conditioning.
## Can a 60 amp breaker feed a 100 amp sub panel?
Nope. The breaker protects the wire leading to that subpanel. The “100 amp” rating of the subpanel is the max rating that you can feed it at.
## Can I put a 60 amp breaker in a 100 amp panel?
You can use the 100 amp panel off a 60 amp breaker if you determine that it is adequate, but I would suggest that in sizing your feeder wires and conduit it would be a good idea to size them to accomodate 100 amps.
You might be interested: How much sugar can dissolve in 100ml of water?
## Can I put a 50 amp breaker in a 100 amp panel?
A 50 amp breaker on a 100 amp service is not an issue.
## How many 20 amp breakers can I put in a 100 amp panel?
For example, I just installed a residential 100 amp panel, and it has 20 slots for a maximum of 20 breakers (single pole). So the limit is 20 breakers for this particular panel (medium sized panel). You could also get a 100 amp panel that has 12 breaker slots, or up to 32 breaker slots.
## Can you run 220 on 100 amps?
Yes, you can. As long as it 240v with a neutral and ground. The only downside is other loads. If the new load with the other loads goes over 100a, you will trip the breaker.
## Will a 10000 watt generator run a house?
A 10000 watt generator has got enough power to run all critical household items. These include a refrigerator and freezer, sump pump, furnace, window air conditioner, and light circuits. In most situations, you can run most, if not all, of these devices at the same time.
## Will a 22kW generator run my house?
For homeowners wanting the ultimate entry-level whole-house standby, a 22kW is a perfect choice to reap all of the benefits of a whole-house generator over a regular home standby. A generator in the 27-36kW range is perfect for most homes since they replace 75% of the 200 amps coming into your electrical panel.
## How many amps is 20000 watts?
Equivalent Watts and Amps at 12V DC
Power Current Voltage
220 Watts 18.333 Amps 12 Volts
230 Watts 19.167 Amps 12 Volts
240 Watts 20 Amps 12 Volts
250 Watts 20.833 Amps 12 Volts
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# Difference between revisions of "162: Angular Momentum"
Angular Momentum Title text: With reasonable assumptions about latitude and body shape, how much time might she gain them? Note: whatever the answer, sunrise always comes too soon. (Also, is it worth it if she throws up?)
## Explanation
Angular Momentum is the force upon an object having a certain velocity while spinning. You may remember the certain strain when a spinning yoyo returned into your hand, giving it that much "slip" to discomfort you. The energy of that momentum does that. Angular Momentum is also forced upon the Earth, as it is spinning 24 hours a day, 7 days a week. This 24/7 rotation enables us to have a clock. Hence, we say that the Earth is running "clockwise".
In the comic, Megan tries to work against this (massive) energy. She is spinning counter-clockwise, thus generating energy to stop the Earth from moving. It is obvious that she (as one human being) will not succeed in stopping the Earth getting its cycle, let alone she generates "clockwise" energy from the other half of her spin. But the romance part is also obvious. After all, who wouldn't want to be longer with the one they truly love?
Randall states the obvious in the title text: while not being able to reverse time, enjoy your night time. Sunrise comes too early.
## Transcript
[Cueball sits on his bed, looking at Megan who is spinning. It is night.]
Cueball: What are you doing?
Megan: Spinning counterclockwise
Each turn robs the planet of angular momentum
Slowing its spin by the tiniest bit
Lengthening the night, pushing back the dawn
Giving me a little more time here
With you
# Discussion
The issue date is not given, as i don't have a clue about it. Could someone fix this? Rikthoff (talk) 19:30, 3 August 2012 (EDT)
When the page was updated to the new comic template by User:Bpothier he fixed the date. lcarsos (talk) 20:48, 28 August 2012 (UTC)
That actually is a neat physics puzzle, which has probably (i.e. certainly) been addressed somewhere on the net. I may incorporate that some day. --Quicksilver (talk) 05:58, 24 August 2013 (UTC)
I tried to calculate the change in Earth's period, assuming that she was standing in the north pole (latitude = 90º N), where her spinning would have more effect. I either did something wrong, or my TI-84 Plus is not capable of detecting the very small effect her spinning would have on the Earth's rotation. I assumed the Earth had a period of exactly 24 hours, and got the same value to the second, even if she was spinning at 1000 turns per second, which seems like a lot.
Here's the formula:
L_Earth_i = L_Earth_f + L_spinner <=>
I_Earth * (2*PI)/T_Earth_i = I_Earth * (2*PI)/T_Earth_f + I_spinner* (2*PI) * f_spinner <=>
(1/T_Earth_f) = (1/T_Earth_i) - (I_spinner/I_Earth)*f_spinner <=>
T_Earth_f = 1/((1/T_Earth_i) - (I_spinner/I_Earth)*f_spinner)
Where the variables have names in the format:
[variable name]_[object it refers to]_[situation (i or f stand for initial and final)]
L = Angular Moment
I = Moment of Inertia
T = Period of rotation about one's axis
f = frequency
I used as values:
T_Earth_i = 86400 seconds (24 hours exactly)
I_spinner = 62,04 Kg.m^2 (Found on Wolfram|Alpha, for a 62Kg adult human being)
I_Earth = 8,03e+37 Kg.m^2 (http://scienceworld.wolfram.com/physics/MomentofInertiaEarth.html)
f_spinner = the frequency of the woman's spinning in complete turns per second. 2.82.142.28 (talk) (please sign your comments with ~~~~)
Taking that a bit further, the relative decrease is:
``` (T_Earth_f - T_Earth_i)/T_Earth_i = 1 / (I_Earth/(I_spinner*T_Earth_i*f_spinner) - 1)
= 1 / ( 1.5 e+28 - 1) ~= 67 e-30
```
Fwiw, the absolute value is 5.767 yocto-seconds. If the entire world population would spin at that 1000 turns per second (and at favourable locations as in your assumptions), the effect will still be a measly 0.041 pico-seconds. So T_Earth_f = 86 399.999 999 999 999 958 ... But the TI-84 only has about 14 digits precision, i believe, so even that won't show up. -- 173.245.51.210 22:46, 30 October 2013 (UTC)
Is it possible for someone to write an equation that factors in latitude (and, if relevant, longitude) that we could plug our locations into and get a value from? That would be awesome. Thanks. 108.162.250.208 02:48, 23 February 2014 (UTC)
The visual style and theme of this comic is clearly referencing the 'Spinning Ballerina Optical Illusion' (evidenced by the grey-to-white gradient 'glow', as well as her arm and leg positions). 108.162.249.231 03:03, 30 September 2014 (UTC)
The 'Spinning Ballerina' optical illusion does not apply here, Megan clearly stated that she was spinning 'counterclockwise' and due to the fact that she is drawn with hair (not a silhouette) lets you know where she is facing. Therefore the bent leg on the right of the image is her left leg. Plus there aren't many ways to draw a stick figure 'spinning'. Now if it were Cueball doing the spinning THEN I would agree with you because there would not be a reference point to make any type of judgement and therefore a point could be made that could be a reference Randall was trying to make. Nexxuz (talk)
The momentum of Megan is in fact one order of magnitude smaller than what appeared in above calculation, so the dilation effect is still smaller.
After reading this comics, I got one question. I thought one cant change the total momentum of a closed system of bodies. I mean, from the point of view outside of Earth, she would be spinning, but the momentum which Megan, or whatever her name is, gained in counter-clock-wise direction would also show in Earth clock-wise direction (positively). Basically I am referring to a situation of a man walking on a boat. (the boat moves, the man moves, but the position man-lake is still the same). So I figured it should be in this way. She spins, yeah, but the Earth now spins in opposite direction a little more, so in the end it is still the same. Now, she could totally delay time by approaching light speed as she spins, which would be weird though as her head (closer to rotational axis) would have smaller velocity than her hands (and thus faster time flow) resulting in, hmm, her body parts getting lost in time? 141.101.97.220 (talk) (please sign your comments with ~~~~)
The total angular momentum cannot be altered from inside the closed system, but it can be distributed differently among the different bodies. This means that the angular speed of the Earth will be actually reduced, but only of a fixed value (proportional to Megan's angular speed) and only as long as Megan spins. It will revert to its original value when Megan stops spinning. Alexxx (talk) 23:24, 6 December 2015 (UTC) 141.101.104.8 (talk) (please sign your comments with ~~~~)
The total angular momentum is not changed, but while Megan is spinning the Earth is rotating slower. When she stops spinning, the Earth resumes it's previous spin rate, but not in the same position, and therefore dawn comes later.
You can compare with the man on the boat situation. If a man walks forward in the boat, the boat moves backward; when the man stops, the boat stops, too, but it is not in the same position it was at the beginning.--Pere prlpz (talk) 17:29, 17 December 2015 (UTC)
Wouldn't she have to be spinning at the Geographic North Pole to have any effect? As it is she's spinning with her axis of rotation at some angle (depending on latitude ) to the Earth's axis of rotation so she's not having much of any effect at all. 173.245.54.190 10:15, 2 April 2015 (UTC)
The effect is stronger in the North Pole, but in temperate latidudes it's still strong. In fact, it's proportional to sine of latitude. Therefore, if the comic is set in Boston (where I think Randall lives), it's just a 33% weaker than in the Pole (sin(42º)=0,67).
Furthermore, out of the Pole, Megan's spin will slightly change the Earth's rotation axis, but this will have little effect on the time of dawn.--Pere prlpz (talk) 17:22, 17 December 2015 (UTC)
Is she spinning counter clockwise when viewed from below or above? (Up spin or down spin?) 108.162.219.188 06:35, 24 June 2015 (UTC)
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# Why does hysteretic current-mode control have variable switching frequency?
This figure below is from Raymond B. Ridley Dissertation "A New Small-Signal Model for Current-Mode Control".
Here is an except from the document:
The general implementation of hysteretic current-mode control is shown in Fig. 2.2. The inductor current waveforms are used to control both the turn-on and the turn-off of the power switch of the PWM converter. The advantages of this kind of circuit are apparent: no clock or timing function is needed, and the current level is controlled between two limits. Although this implementation was popular before control circuits became available, its variable switching frequency, and the need to sense the inductor current during both the on- and off-times of the power switch have restricted its use today. This circuit does not have any problems with instability of the current-feedback loop, and it is not analyzed in this dissertation.
Can anyone explain why this control method has varibale switching frequency? Also why this is bad?
• If I understand correctly, the switching frequency is determined by the load current. The load current is determined by the load. When the current hits the switch point, the switch changes state. Jul 29, 2017 at 5:37
• google.com/… Jul 29, 2017 at 8:44
As long as the output voltage is below the target set by $V_{ref}$, the power switch remains closed and the current in the inductor grows with a slope given by $S_{on}=\frac{V_{in}-V_{out}}{L}$. In a real circuit, a current limit would interrupt the process but it is not represented here for the sake of simplicity. When the voltage $V_{out}$ reaches the target, the switch turns off, the freewheel diode turns on and the inductor current decays with a slope equal to $S_{off}=-\frac{V_{out}}{L}$. The voltage also decays and creates a ripple made of a capacitive ($C_{out}$) and resistive ($r_C$) contributions. When the voltage reaches the second threshold, the switch turns on again and initiates a new cycle. The ripple amplitude and thus the operating frequency depends on the selected hysteresis band.
You can thus see that the not only the on- and off-slopes depend on the input and output voltages but the pace at which the output voltage goes down is linked to the output current or $R_L$. As such, a hysteretic converter in its simplest switching form, can occupy a very large and uncontrolled frequency spectrum. If the converter powers a RF section, EMI pollution can occur and bother the receiver for instance (cell-phones rarely embark hysteretic converters for instance). Also, conversely, the converter in light-load operation will reduce its switching frequency (good for efficiency) but can potentially be very noisy (audio switching frequency) especially with cheap inductors and high peak currents. Have a look at these venerable switchers like the µA78S40 (MOT did manufacture it but it had been originally introduced by Signetics if I am not mistaken) or the MC34063, still sold in high volume. They were nice noise generators when operating : )
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Error: Invalid Input on my program
01-17-2017, 12:39 AM
Post: #1
HumbGruland Junior Member Posts: 1 Joined: Jan 2017
Error: Invalid Input on my program
Hello, and thanks for answering, I'm new to the forum so please tell me if this is the wrong sub-forum.
I recently got an HP Prime, I can't believe how simple it is programming it, however I encountered a few problems when creating a UI for input.
I made a program based on style 2 of:
http://www.hpmuseum.org/forum/thread-455.html
The code for my program is:
Code:
EXPORT PuntoFijo() BEGIN local n,xact,xnue,fx,err,N,f,f1; N:=100; xnue:=1; if input( {f,f1,err,xact}, "Metodo del punto fijo", {"f(X)=", "f(X)=", "Error=", "x0="}, { "Ingrese la funcion rodeada por comillas simples", "Ingrese la funcion depsejada rodeada por comillas simples", "Ingrese el error maximo", "Ingrese la raiz aproximada" }, {f,f1,err,xact} ) then F1:=f; F2:=f1; M0:={}; M0(n,1):=n; M0(n,2):=xact; M0(n,3):=F2(xact); M0(n,4):=abs(F1(xact)); for n from 2 to N+1 do xact:=xnue; xnue:=F2(xact); fx:=F1(xact); M0(n,1):=n; M0(n,2):=xact; M0(n,3):=xnue; M0(n,4):=fx; if abs(fx)<err then break; end; end; editMat(M0); end; END;
Basically the same structure, only modified to implement a different numerical method.
The program passes the check.
The problem arises when the program is running; when I try to input the first algebraic function (between simple quotes), I get the message: "Error: Invalid Input", even if I try to input something as simple as a parabola.
Am I doing something wrong?
Is there something my program is missing?
Thanks for your help.
01-17-2017, 02:47 PM (This post was last modified: 01-17-2017 03:03 PM by StephenG1CMZ.)
Post: #2
StephenG1CMZ Senior Member Posts: 956 Joined: May 2015
RE: Error: Invalid Input on my program
I think that one problem may be that the INPUT syntax has changed since Han wrote that in 2013.
Briefly, where previously one would have
INPUT... varname,
now one should use
IINPUT...{{varname,[numbers]}}
Where numbers specifies the type of input, for example
INPUT...{{varname,[3]}} might specify string, or 6 might specify list, or 3,6 would be either.
The number you want is probably (DOM_STRING-1).
Check the INPUT help for the exact syntax.
(But if you are using CAS input, I don't know).
Stephen Lewkowicz (G1CMZ)
01-17-2017, 04:14 PM (This post was last modified: 01-18-2017 12:14 PM by Han.)
Post: #3
Han Senior Member Posts: 1,882 Joined: Dec 2013
RE: Error: Invalid Input on my program
INPUT() has gone through a number of changes since that post. There are two ways to get around the issue you have:
1) Initialize your f and f1 to objects of the type you wish to use in the INPUT() command. When you do not initialize them, then they are set to the value 0 by default (type: real number). Thus, when a user then enters something that is not a real number, the INPUT() command complains of invalid input. You can, for example, set f:='X^2-5' and f1:='2*X' before the INPUT() command, and then it will expect an algebraic expression.
2) The other method use the more advanced form of INPUT() to enforce the input type for each field (i.e. f and f1 and be required to expect an expression). The help screen for INPUT() explains everything you need to make use of the advanced form.
There is yet another bug in your program, though. You have a line: M0:={}. While technically incorrect (because M0 is a vector/matrix and not a list), the HP Prime will happily convert the empty list into a vector and insert at least a 0 -- the built-in variables may not be "empty" so it auto-fills the first value with 0. The problem, however, is that you are using M0 as a matrix. So you will get yet another error when your program reaches this point. Instead, initialize M0 using: M0:=[[0]]; and your program should be fine (this makes M0 a matrix and not a vector).
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[–] 2 points3 points (0 children)
sorry, this has been archived and can no longer be voted on
I'd say this is a fairly normal thing. It's the difference between concrete and abstract thinking. When you need to solve a concrete problem (e.g. integration or solving matrices) It's an algorithmic process to get from point A to point B. This is, do step 1, do step 2, all the way to step n. And then you have an answer. Whereas upper level math is more about abstract thinking. You still need to get from point A to point B, but you have to internalize the concepts in such a way that you can make the necessary logical connections to get from A to B. The difference is, there is no one certain way to get from A to B. sometimes you work backwards from B. Sometimes you meet in the middle. You must know the material intimately enough to be creative with it and find a solution.
I know this seems like a non-answer, but it just takes practice. So keep at it. I experienced the same thing after the calc/Linear Algebra sequence. I've never had a formal course on doing proofs and it caught up with me quickly. It's about learning how to think and sometimes knowing what you don't need to know is just as important as what you do.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
Maybe you should post that old practice exam or something.
[–] 0 points1 point (0 children)
sorry, this has been archived and can no longer be voted on
As a tutor, this is what I focus on. I watch my learner solve the "easy" problems, and then I watch her get stuck on a "hard" problem. Yes, you can try to get over the hump by practicing. If you haven't tried sitting down with somebody who can do the easy ones and the hard ones and having them a) observe where you are getting stuck and b) suggest an approach based on their observations -- rather than just giving one-size-fits-all advice -- I strongly recommend it.
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# Logical Reasoning - Analogies - Discussion
Discussion Forum : Analogies - Type 6 (Q.No. 5)
Directions to Solve
Choose the pair that best represents a similar relationship to the one expressed in the original pair of words.
5.
COBBLER : SHOE
jockey : horse
contractor : building
mason : stone
cowboy : boot
potter : paint
Explanation:
A cobbler makes and repairs shoes; a contractor builds and repairs buildings.
Discussion:
7 comments Page 1 of 1.
Lavanya said: 2 years ago
Cobbler makes and repairs shoes but a mason does not makes stones. He just uses them in making Houses etc. So mason:stone is not the answer. Also potter makes potes not paints. So that too is incorrect. So the answer is contractor: building.
Tushar said: 2 years ago
Yes, option C is look like more correct.
Sir Taslib said: 2 years ago
Contractors don't build houses etc but masons do then why not C?
Thakur said: 4 years ago
Yes, I think also Mason is the right answer.
Purviee said: 5 years ago
Potter: Paint will also be the answer.
Anurag said: 6 years ago
Ofcourse, Mason : stone is the answer. Since Contractor can be of any type, not just a building contractor whereas a mason always works with stones and bricks. So it is more accurate.
Kumar said: 7 years ago
I think mason and stone will also be the answer because it will also use stones while building.
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# Pt) An equation of the tangent plane to the parametrized surface d(u,v) = (-4u,2u2 + v, _5u2 )at the point
###### Question:
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all parts please! 4. The zeta function (8) = 2n=ln,s > 1, plays an important role in many areas of math- ematics, especially number theory (it can also be defined when s is a complex number). In 1736 Leonard Euler was able to prove that 72 (2) = n2 6 1 n=1 In this problem, your will prove this fa...
##### 2. A vessel with a piston cylinder attached to a spring (see Figure 1) contains CO2...
2. A vessel with a piston cylinder attached to a spring (see Figure 1) contains CO2 with a volume of 500 L, a pressure of 200 kPa and a temperature of 25degree C (State 1). heat is transferred to the vessel and it starts to expand until the piston just touches the stops. The pressure at this state i...
##### Stor(Sq Arerage Mont 17360 3824944 15820 5344818 174g 637374 17430 571975 15660 559813 20230 6944102 15160 53352Z 16930 381237 12020 467993 12490 517468 15610 624539 12530 468486 21690 735707 141zO 493651 16800 629092 14940 593127 18420 608168 18390 618382 16780 685930 19950 721908 17910 539700
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##### His eyes are 1.71 abovt thc ilcor and the top his head 0.13 nghet Find inc nciaht(in mi ajove mintor whch he can sct both tne tOp 0t his hcad and hls fect;the toDand Doron cincemillerootomHow the distance[ne [0Cic Dottomthe mictor relited tFc Man & nclqht(Usc tha svmbol n}Addhon MaterllReaino0/2 POINtSPREVIOUS ANSWERS1/6 Submissions UsedMYNOTESASK YOUR TEACHERmeasure te speed laht, You cropose bounce laser light om mimorthaEEAokM For an Indcpcndent study project, vou ces gn experiment orlent
His eyes are 1.71 abovt thc ilcor and the top his head 0.13 nghet Find inc nciaht(in mi ajove mintor whch he can sct both tne tOp 0t his hcad and hls fect; the toDand Doron cincemiller ootom How the distance [ne [0C ic Dottom the mictor relited tFc Man & nclqht (Usc tha svmbol n} Addhon Materll ...
for part C: the answer is not 0.37098, is not 21.256, is not 1.16546. please help me to find the write answer. I tried to formula where I divide part a with R which is 311.20/800=tan^-1(0.389) is wrong! A resistor and capacitor are connected in series across an ac generator. The voltage of the gener...
##### Construct an FA MACHINE STATE diagram please ( If you do not know how to do...
Construct an FA MACHINE STATE diagram please ( If you do not know how to do this then please do not answer the question) All strings that end in a double letter: These_strings_should_accept aa bb aaa aaaa aaaaa bbb bbbb bbbbb abaa baaa babb abababababababababababababababaa ababababababababababababab...
##### Name the following compounds:Part AOHBrSpell out the full name of the compound_SubmitRequest AnswerPart BSpell out the full name of the compound:SubmitRequest AnswerPart COHSpell out the full name of the compound:SubmitRequest Answer
Name the following compounds: Part A OH Br Spell out the full name of the compound_ Submit Request Answer Part B Spell out the full name of the compound: Submit Request Answer Part C OH Spell out the full name of the compound: Submit Request Answer...
##### 3080/5080 Materials Science and Engineering: Spring 2019 ead Ch. 10.1, 10.2, 10.3, 10.4, 10.5, 10...
3080/5080 Materials Science and Engineering: Spring 2019 ead Ch. 10.1, 10.2, 10.3, 10.4, 10.5, 10.9, 10.10, 10.11 CHEN Home Work 19 due Tue, Apr 16 Review olved Problems 10.1, 10.2, 10.3, 10.4, 10.6, 10.7,10.8, 10.9, 10.1o 1. Distinguish between the i) structure, ii) processing and ili) properties o...
##### Create a cash flow diagram for the following scenario by entering the letter denoting the correct...
Create a cash flow diagram for the following scenario by entering the letter denoting the correct location of the corresponding parameter in the cash flow diagram. For example, enter A for a positive cash flow at time O. A farmer makes three consecutive deposit of \$1,000 each in years 3 . and 5 , at...
##### Thc following data rcpresent thc results from an indcpendent-mcasurcs experiment comparing thrcc trcatment conditions_ Use SPSS to conduct an analysis of variance with & 0.05 to dctcrminc whcthcr thcsc data arc sufficicnt to concludc that therc arec significant diffcrenccs bctwccn thc trcatments Treatment A Treatment B Treatment CF-ratio p-valuc Conclusion:Thcre is significant diffcrencc betwecn trcatments Thcsc data do not providc cvidcncc of. diffcrcncc bctwccn thc trcatmcntsThc rcsults ob
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##### Data TableMax wind (mph) 110 115 90 115 100 90 80 85PrintDone
Data Table Max wind (mph) 110 115 90 115 100 90 80 85 Print Done...
##### Question 14 (1 point) sme 60 120 140 This diagram represents which type of externality? O...
Question 14 (1 point) sme 60 120 140 This diagram represents which type of externality? O Negative Consumption Externality Negative Production Externality Positive Production Externality Positive Consumption Externality...
##### Last two sides is for questions 4 Bank Reconciliation and Entries Sunshine Interiors deposits all cash...
last two sides is for questions 4 Bank Reconciliation and Entries Sunshine Interiors deposits all cash receipts each Wednesday and Friday in a night depository, after banking hours. The data required to recond the bank statement as of July 31 have been taken from various documents and reco...
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teacherLED.com
This resource is designed to be used in landscape orientation only. Please rotate your device.
This exercise helps with comparing fractions to decimals using mental arithmetic. All of the fractions in it are either ones that you should learn the equivalent decimal for or are these ones you should learn with a simple calculation.
Eaxample
Compare 410 with 0.3.
You should learn that 0.1 is 110 this lets you work out that 410 is 0.4.
410 < 0.3
12=0.5
14=0.25
110=0.1
13= aproximately 0.333...
< is less than.
> is greater than.
Remember:
< is less than.
> is greater than.
Game Over!
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https://aptitude.gateoverflow.in/679/Clock
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1,452 views
Q1) How many times min hand of clock will be opposite to hour hand from 4 pm on monday to 10 am on subsequent wednesday?
1)46
2)40
3)44
4) 38
Q2) On particular day if it is found that clock is 8 min fast at 6 am and 4 min slow at 12 noon on same day,then at what time clock show correct timing?
1) 8 am
2) 9 am
3) 10 am
4) 11 am
Q3) if second covers 900 degree how many degree hour hand cover in same time?
1)1 1/4 2)1 1/2 3) 2 1/2 4) 2 3/4
1) as we know , in period of 12 hours , angle of 180 , (oppostie min hand and hour hand ) happens exactly 11 times ...
so here we have to calculate from 4pm on monday to subsequence wednesday 10am .
now from 4pm monday to 4 pm tuesday it will be 24 hr so angle of 180 will make 22 times .now from 4pm tuesday to 10 am wednesday 12 hr +6 hr = 11+5times = 16 times
so total 22+16= 38...
2) at 6am fast 8min and 12 at noon 4 min late
relatively 12 mins difference in 6 hr so in 1 min difference 1/2 hr , so clock will cover 8mins in 8*1/2= 4hr .. so correct time will be show at 10 am ..
3) in 1 min second hand cover 360 degree . so 900 degree will 90/36 min ,
as we know hour hand in 1 min cover half degree , so in 90/36 min = 1.25 degree ...
by
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# The Sum function: adding two sums together
Occasional Visitor
# The Sum function: adding two sums together
I want to add two separate sums found in the same data range. For instance, the sum of g14:g28 and the sum of g30:g43. I keep getting error messages and I can't figure out what Excel wants. This can't be the first time someone has wanted to do this kind of thing. Thanks for comments.
2 Replies
# Re: The Sum function: adding two sums together
@pbmarrow wrote: ``I keep getting error messages and I can't figure out what Excel wants``
Neither can we, since (a) you do not tell us or show us what the error "message" reads (or is it really an error value like #VALUE?), and (b) you do not show us at least the offending Excel formula, and it is better to also show us the referenced values.
For quicker dispositive explanations, it is best to attach an Excel file (redacted) that demonstrates the problem(s). Click on the "browse" link at the bottom of this forum's editing pane, to wit:
<---
-----
@pbmarrow wrote: ``I want to add two separate sums [...]. For instance, the sum of g14:g28 and the sum of g30:g43.``
The description is ambiguous. Of course, the description would be disambiguated if you showed us the formula(s). BTW, it should not matter what subranges that you sum.
You might have the formula =SUM(g14:g28) in A1 and =SUM(g30:g43) in A2, for example, and the offending expression might be A1+A2 or SUM(A1:A2) or SUM(A1,A2).
Or you might have the expression SUM(SUM(g14:g28), SUM(g30:g43)), which is better written SUM(g14:g28, g30:g43).
In the last three examples, the problem might be that the parameter separator for you computer should be a semicolon (";") instead of comma (","). In other words, SUM(A1; A2) or SUM(g14:g28; g30:g43).
Generally, the parameter separator is the "list separator" that is configured in the Region and Language options for your computer.
However, I have seen postings where users claim that Excel Online ("Web Excel") and Office 365 Excel use a different configuration intermittently; presumably a temporary defect. I have no first-hand experience with those products. So, if you suspect that problem, hopefully someone can help you work around it. Or simply wait for MSFT to patch it (sigh).
# Re: The Sum function: adding two sums together
Perhaps you mean something like this
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## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
No credit card required
### M1.6. More Matrix Stuff
A matrix can be constructed from two or more vectors. If we wish to create a matrix v which consists of two columns, the first column containing the vector x (in column form) and the second column containing the vector z (in column form), we can use the following (where x and z are previously generated row vectors):
```» v = [x',z']
v =
1 5
3 10
5 15
7 20
9 25
11 30
```
If we wished to look at the first column of v, we could use (where : indicates all rows, while the 1 indicates the first column)
```» v(:,1)
ans =
1
3
5
7
9
11
```
If we wished to look at the second column of v, we could use
```» v(:,2)
ans =
5
10
15
20
25
30
```
And we can construct the same plot as before by using the following plot command ('--' gives a dashed line): ...
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
No credit card required
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https://vn.tradingview.com/chart/EMC2BTC/Lg1QeD6e-Einsteinium-Bitcoin-Compelling-Buy/
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Giá lên
# Einsteinium/Bitcoin - Compelling Buy
BITTREX:EMC2BTC Einsteinium / Bitcoin
774 lượt xem
774
Traders I recommend reading this entire post so that you fully understand what I have done, and how I have came to my analysis.
EMC2/BTC looks like a compelling long opportunity. I love this set up so far!
After analyzing this pair, I discovered that price appears to be moving in patterns. Price seems to move up 5539 points, have a corrective move to the downside, and then move up again, around 5539 each time.
With this logic, I was able to predict and plot the next move on the chart. I have calculated each cycles move by measuring the amount of points it moved up before correcting. After I found there was an approximate 5539 point move before each correction, I was then able to replicate a cycle at around 5539 points to see the future direction of the instrument.
I am expecting an up move of 5539 which will complete at around 0.00012800. Below I have illustrated the calculations as to how I came to this conclusion.
The principle of the cycles being used here is that every time there is an up move, there is a corrective move, and then an up-move which exceeds the previous move by approximately 5539. With this information we can calculate (using cycles 1+2 data) where cycle 3 is likely to end up. Using the calculations below we can make a mathematical assumption as to where price will lead.
Cycle 1 = Cycle + 5539 = 0.00008100
Cycle 2 - Cycle 1 + 5539 = 0.00009100
Cycle 3 - Cycle 2 + 5539 = 0.00012800
Following these calculations, we can expect a reliable up-move, using previous trend cycles to predict a reliable, yet conservative move.
Do remember that price may correct before cycle 3 initiates. So it is important that you watch this trade carefully before considering a position. I would also like to point out that there is a key side-wards resistance level at 0.00010000. Price needs to break this level and retest this level before Cycle #3 completes.
I will be posting updates on this trade below, so leave it a like or follow so that you can keep up to date with this analysis.
Đóng lệnh: đạt mục tiêu: As per the above analysis, Cycle #3 completed, and provided us with the nice move I was expecting.
Price reached target level, and is now correcting to the downside. When price completes the corrective move to the downside, we must be open to the possibility of a cycle #4.
When correction completes I will post more updates below.
Bình luận: I would just like to say well done to anyone who followed this idea on EMC2/BTC. I entered at around 0.00009175 and exited the trade at 0.00019700.
Price of the Crypto currency doubled in value in just 21 hours and we caught the entire move here!
I am glad you were all able to follow this trade with me. Congratulations to anyone who used this analysis to profit with me.
****Join my VIP Crypto Signals Group With Our 66% Christmas Discounts!****
Public Telegram Channel: https://t.me/CryptocurrencyTradingIdeas
I am interestingly keeping the track your TA & FA and exciting by the movements of the graphs through you lines! Thank you Tom! You are teach me a lot since I follow you!
Phản hồi
kwarty
@kwarty, Thanks for your compliments. Glad my content is useful to you. I would really appreciate it if you could like all of my recent ideas.
Phản hồi
Except the hardfork EMC2 there also an information that EMC2 and APPLE will have and agreement, may be in paying function or donations on 19th of December. Any bets should be make in the sense of price increasing or the news are rumours ?
Phản hồi
kwarty
@kwarty, I have already put my bets into this market. I believe the rumors with Apple have actually already been priced into this market and price shows as such.
Until 19th December, I would just assume that price will rally in an attempt to price in Apple. (Since if this is true AND happens, would be absolutely massive). Only when the raw fundamentals come out (as you mentioned above) come out, I will then reconsider my analysis with the new data.
Phản hồi
wow this is really weird. Good catch
Phản hồi
SDCVoltaic
@SDCVoltaic, Thank you. We have nailed virtually every single currency we have analyzed on TradingView lately. Let's hope for more wins!
Phản hồi
@TomProTrader, good call, keep em coming!
Phản hồi
magicpony
@magicpony, Many more to come :)
Phản hồi
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# VECM
```VECM
• First we test to see if variables are stationary I(0).
If not they are assumed to have a unit root and
be I(1).
• If a set of variables are all I(1) they should not be
estimated using ordinary regression analysis, but
between them there may be one or more
equilibrium relationships. We can both estimate
how many and what they are (called
cointegrating vectors) using Johansen’s
technique.
• If a set of variables are found to have one or
more cointegrating vectors then a suitable
estimation technique is a VECM (Vector Error
Correction Model) which adjusts to both short
run changes in variables and deviations from
equilibrium.
• In what follows we work back to front.
Starting with the VECMs, then Johansen’s
technique than stationarity.
• We have data on monthly unemployment
rates in Indiana, Illinois, Kentucky, and
Missouri from
• January 1978 through December 2003. We
suspect that factor mobility will keep the
unemployment
• rates in equilibrium. The following graph plots
the data.
use http://www.stata-press.com/data/r11/urates,
clear
line missouri indiana kentucky illinois t
Note the form of the above line to draw the line
graph; then the variables which will be plotted;
finally t the time variable against which they are
all plotted
For further info press the help key, then line
12
2
4
6
8
10
1980m1
1985m1
1990m1
t
missouri
kentucky
1995m1
indiana
illinois
2000m1
2005m1
• The graph shows that although the series do appear to move
together, the relationship is not that. There are periods when
Indiana has the highest rate and others when Indiana has the
lowest rate.
• Although the Kentucky rate moves closely with the other series for
most of the sample, there is a period in the mid-1980s when the
unemployment rate in Kentucky does not fall at the same rate as
the other series.
• We will model the series with two cointegrating equations and no
linear or quadratic time trends in the original series.
• For now we use the noetable option to suppress displaying the
short-run estimation table.
vec missouri indiana kentucky illinois, trend(rconstant)
rank(2) lags(4) noetable
Vector error-correction model
Sample:
1978m5 - 2003m12
Log likelihood =
Det(Sigma_ml) =
No. of obs
AIC
HQIC
SBIC
417.1314
7.83e-07
=
308
= -2.306048
= -2.005818
= -1.555184
Cointegrating equations
Equation
Parms
_ce1
_ce2
2
2
chi2
P>chi2
133.3885
195.6324
0.0000
0.0000
Johansen normalization restrictions imposed
beta
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
missouri
indiana
kentucky
illinois
_cons
1
(omitted)
.3493902
-1.135152
-.3880707
.
.
.
.
.
.2005537
.2069063
.4974323
1.74
-5.49
-0.78
0.081
0.000
0.435
-.0436879
-1.540681
-1.36302
.7424683
-.7296235
.5868787
missouri
indiana
kentucky
illinois
_cons
-1.11e-16
1
.2059473
-1.51962
2.92857
.
.
.2718678
.2804792
.6743122
.
.
0.76
-5.42
4.34
.
.
0.449
0.000
0.000
.
.
-.3269038
-2.069349
1.606942
.
.
.7387985
-.9698907
4.250197
_ce1
_ce2
• Except for the coefficients on kentucky in the
two cointegrating equations and the constant
term in the first, all the parameters are
significant at the 5% level.
• We can refit the model with the Johansen
normalization and the overidentifying
constraint that the coefficient on kentucky in
the second cointegrating equation is zero.
constraint 1 [_ce1]missouri = 1
constraint 2 [_ce1]indiana = 0
constraint 3 [_ce2]missouri = 0
constraint 4 [_ce2]indiana = 1
constraint 5 [_ce2]kentucky = 0
vec missouri indiana kentucky illinois,
trend(rconstant) rank(2) lags(4) noetable
bconstraints(1/5)
constraint 1 [_ce1]missouri = 1
• Constraint number 1, [_ce1] tells us which
equation and missouri=1 sets constraint.
beta
Coef.
missouri
indiana
kentucky
illinois
_cons
1
(omitted)
.2521685
-1.037453
-.3891102
missouri
indiana
kentucky
illinois
_cons
(omitted)
1
(omitted)
-1.314265
2.937016
Std. Err.
z
P>|z|
[95% Conf. Interval]
_ce1
.
.
.
.
.
.1649653
.1734165
.4726968
1.53
-5.98
-0.82
0.126
0.000
0.410
-.0711576
-1.377343
-1.315579
.5754946
-.6975626
.5373586
.
.
.
.
.
.0907071
.6448924
-14.49
4.55
0.000
0.000
-1.492048
1.67305
-1.136483
4.200982
_ce2
LR test of identifying restrictions:
chi2(
1) =
.3139
Prob > chi2 = 0.575
The test of the overidentifying restriction does not reject the null hypothesis that the restriction
is valid, and the p-value on the coefficient on kentucky in the first cointegrating equation
indicates that it is not significant. We will leave the variable in the model and attribute the lack
of significance to whatever caused the kentucky series to temporarily rise above the others from
1985 until 1990, though we could instead consider removing kentucky from the model.
• Next, we look at the estimates of the
adjustment parameters. In the output below,
we replay the previous results.
• vec missouri indiana kentucky illinois,
trend(rconstant) rank(2) lags(4)
bconstraints(1/5)
Results for D_Missouri
Coef.
Std. Err.
z
P>|z|
[95% Conf. Interval]
D_missouri
_ce1
L1.
-.0683152
.0185763
-3.68
0.000
-.1047242
-.0319063
_ce2
L1.
.0405613
.0112417
3.61
0.000
.018528
.0625946
missouri
LD.
L2D.
L3D.
.2391442
.0463021
.1996762
.0597768
.061306
.0604606
4.00
0.76
3.30
0.000
0.450
0.001
.1219839
-.0738555
.0811755
.3563045
.1664596
.3181768
indiana
LD.
L2D.
L3D.
.000313
-.0071074
.024743
.0526959
.0530023
.0536092
0.01
-0.13
0.46
0.995
0.893
0.644
-.102969
-.11099
-.0803291
.1035951
.0967752
.1298151
kentucky
LD.
L2D.
L3D.
.0169935
.0611493
.0212794
.0475225
.0473822
.0470264
0.36
1.29
0.45
0.721
0.197
0.651
-.0761489
-.0317182
-.0708907
.1101359
.1540168
.1134495
illinois
LD.
L2D.
L3D.
.050437
.0086696
-.0323928
.0491142
.0493593
.0490934
1.03
0.18
-0.66
0.304
0.861
0.509
-.0458251
-.0880728
-.1286141
.1466992
.1054119
.0638285
Interpretation
D_missouri
_ce1
L1.
-.0683152
.0185763
-3.68
0.000
-.1047242
-.0319063
_ce2
L1.
.0405613
.0112417
3.61
0.000
.018528
.0625946
If the error term in the first cointegration relation is positive unemployment in
Missouri FALLS.
If the error term in the second cointegrating regression is positive then unemployment
in Missouri INCREASES.
The first cointegrating regression is Missouri + 0.425Kentucky – 1.037Illinois -0.389 = Error
Missouri + 0.425Kentucky – 1.037Illinois -0.389
= Error
• Viewed in this context if the error term is
positive then unemployment in Missouri can
be viewed as being above equilibrium, same
for Kentucky, but for Illinois it is below
equilibrium (because if we increase Illinois the
error term falls)
• To get back to equilibrium we need
unemployment to fall in Missouri.
D_missouri
_ce1
L1.
-.0683152
.0185763
-3.68
0.000
-.1047242
-.0319063
• As we can see from the regression this is what
we get.
• D_missouri is the change in unemployment in
Missouri i.e. DUMt = Umt – Umt-1
• The coefficient on _ce1 L1 (_ce1 : the error
term from the first cointegrating regression;
L1 lagged one period) is -0.068 and significant
at the 1% level.
• Thus if in period t-1 the error term in _ce1 was
positive, which we can see can be seen as
unemployment in Missouri being too high
compared to the equilibrium relationship with
the other two states, then it will fall.
• The bigger the (negative) coefficient on _ce1
L1 the more rapid is the correction. If it = -1
then the entire error is corrected for in the
following period.
Let us look at the second cointegrating
regression
• This can be written as:
• Error=Indiana -1.342Illinois + 2.93
_ce2
missouri
indiana
kentucky
illinois
_cons
(omitted)
1
(omitted)
-1.314265
2.937016
.
.
.
.
.
.0907071
.6448924
-14.49
4.55
0.000
0.000
-1.492048
1.67305
-1.136483
4.200982
_ce2
L1.
And its impact in the VECM (Vector
Error Correction Model)
• We can see its positive and significant.
Unemployment in Missouri increases if this is
error term is positive. But why? Missouri does
not enter the second cointegrating vector. So
why does unemployment in it respond to it?
.0405613
.0112417
3.61
0.000
.018528
.0625946
Indiana =1.342Illinois + 2.93 + Error
• Well it’s a little convoluted, but if the error term
is positive it suggests that unemployment in
Illinois is below equilibrium (and may increase as
a consequence). Now from first cointegrating
vector:
• Missouri = -0.425Kentucky + 1.037Illinois, if
Illinois unemployment is to increase then the
error term in the first cointegrating vector will fall
(perhaps going negative).
Let us look at the second equation for
Indiana
D_indiana
_ce1
L1.
_ce2
L1.
-.0342096
.0220955
-1.55
0.122
-.0775159
.0090967
.0325804
.0133713
2.44
0.015
.0063732
.0587877
Let us look at the second equation for
Indiana
• The error term from _ce1 is not significant,
but that from _ce2 is and it is positive. _ce2 Is
• Error=Indiana -1.342Illinois + 2.93
D_indiana
_ce1
L1.
_ce2
L1.
-.0342096
.0220955
-1.55
0.122
-.0775159
.0090967
.0325804
.0133713
2.44
0.015
.0063732
.0587877
• Now this does not make much sense if rgw
error term is positive unemployment in
Indiana needs to fall to restore equilibrium.
Yet the coefficient on it is positive indicating
the opposite.
Another View
vec missouri indiana kentucky illinois,
trend(rconstant) rank(2) lags(4) bconstraints(1/5)
matrix cerr=e(beta)
display cerr[1,1]
display cerr[1,3]
display cerr[1,5]
display cerr[1,9]
drop cerr1 cerr2
matrix cerr=e(beta) saves the coefficients from
the two cointgretaing regressions in a vector
cerr.
cerr[1,1] is the first, cerr[1,9] is the penultimate
coefficient in the second equation
Thus: display cerr[1,9] gives: -1.3142654, the
coefficient on Illinois in _ce2
Generate the error terms for the two
equations
generate cerr1= cerr[1,5]+ cerr[1,1]*missouri +
cerr[1,2]*indiana + cerr[1,3]*kentucky +
cerr[1,4]*illinois
generate cerr2= cerr[1,10]+ cerr[1,6]*missouri +
cerr[1,7]*indiana + cerr[1,8]*kentucky +
cerr[1,9]*illinois
Now this:
regress D.missouri LD.missouri LD.indiana
LD.kentucky LD.illinois L2D.missouri L2D.indiana
L2D.kentucky L2D.illinois L3D.missouri
L3D.indiana L3D.kentucky L3D.illinois L.cerr1
L.cerr2
Is almost equivalent to this:
vec missouri indiana kentucky illinois,
trend(rconstant) rank(2) lags(4) bconstraints(1/5)
• I say almost because the VEC estimates both
equations jointly and the regressions are
slightly different, but very slightly.
• Note to if we have a slightly different short run
structure then the cointegrating vectors
change which is a little unsatisfactory
• For example compare:
vec missouri indiana kentucky illinois,
trend(rconstant) rank(2) lags(2)
bconstraints(1/5)
vec missouri indiana kentucky illinois,
trend(rconstant) rank(2) lags(4)
bconstraints(1/5)
Short Run dynamics
• Lets look at the rest of the equation, below is
for D.missouri
• The one period lag is significant as is the 3
period lag. That is it responds to its own
lagged values.
ssouri
LD.
L2D.
L3D.
.2391442
.0463021
.1996762
.0597768
.061306
.0604606
4.00
0.76
3.30
0.000
0.450
0.001
.1219839
-.0738555
.0811755
.3563045
.1664596
.3181768
• But not to those of Indianna.
indiana
LD.
L2D.
L3D.
.000313
-.0071074
.024743
.0526959
.0530023
.0536092
0.01
-0.13
0.46
0.995
0.893
0.644
-.102969
-.11099
-.0803291
.1035951
.0967752
.1298151
This has been based on an example in
the STATA manual, but…..
• There are more variables. Lets try the
regression in full:
• vec missouri indiana kentucky illinois arkansas
ten, trend(rconstant) rank(2) lags(3)
Tennessee appears related to nothing,
so..
_ce1
missouri
indiana
kentucky
illinois
arkansas
tenn
_cons
1
(dropped)
1.399639
-.5534946
-1.463609
-.2175613
-.2406195
.
.
.
.
.
.4071156
.3499487
.3881522
.3220395
.9257524
3.44
-1.58
-3.77
-0.68
-0.26
0.001
0.114
0.000
0.499
0.795
.6017075
-1.239382
-2.224373
-.8487471
-2.055061
2.197571
.1323923
-.7028443
.4136245
1.573822
missouri
indiana
kentucky
illinois
arkansas
tenn
_cons
-1.11e-16
1
.6949351
-1.304683
-.522436
-.1928471
2.832045
.
.
.3402238
.2924498
.3243762
.2691262
.7736451
.
.
2.04
-4.46
-1.61
-0.72
3.66
.
.
0.041
0.000
0.107
0.474
0.000
.
.
.0281088
-1.877874
-1.158202
-.7203248
1.315728
.
.
1.361762
-.7314915
.1133296
.3346307
4.348361
_ce2
vec missouri indiana kentucky illinois arkansas
ten, trend(rconstant) rank(3) lags(3)
• but Tennessee still remains unrelated to
anything
• We can see from the map that Tennessee is on
the South east fringe of this group and it
would be interesting to bring in North
Carolina, Alabama and Georgia.
Johansen’s methoodology
• vecrank implements three types of methods for
determining r, the number of cointegrating
equations in a VECM. The first is Johansen’s
“trace” statistic method. The second is his
“maximum eigenvalue” statistic method. The
third method chooses r to minimize an
information criterion.
• All three methods are based on Johansen’s
maximum likelihood (ML) estimator of the
parameters of a cointegrating VECM.
• webuse balance2
• We have quarterly data on the natural logs of
aggregate consumption, investment, and GDP
inthe United States from the first quarter of 1959
through the fourth quarter of 1982. As discussed
in King et al. (1991), the balanced-growth
hypothesis in economics implies that we would
expect to find two cointegrating equations among
these three variables.
describe
storage
variable name
type
display
format
gdp
t
inv
consump
y
i
c
%9.0g
%tq
%9.0g
%9.0g
%10.0g
%10.0g
%10.0g
float
int
float
float
double
double
double
value
label
variable label
ln(gdp)
ln(investment)
ln(consumption)
• In this example, because the trace statistic at r = 0 of 46.1492
exceeds its critical value of 29.68, we reject the null hypothesis of
no cointegrating equations.
• Similarly, because the trace statistic at r = 1 of 17.581 exceeds its
critical value of 15.41, we reject the null hypothesis that there is
one or fewer cointegrating equation.
• In contrast, because the trace statistic at r = 2 of 3.3465 is less than
its critical value of 3.76, we cannot reject the null hypothesis that
there are two or fewer cointegrating equations.
. vecrank y i c, lags(5)
Johansen tests for cointegration
Trend: constant
Sample: 1960q2 - 1982q4
maximum
rank
0
1
2
3
parms
39
44
47
48
LL
1231.1041
1245.3882
1252.5055
1254.1787
Number of obs =
Lags =
eigenvalue
.
0.26943
0.14480
0.03611
5%
trace
critical
statistic
value
46.1492
29.68
17.5810
15.41
3.3465*
3.76
91
5
• Because Johansen’s method for estimating r is
to accept as the actual r the first r for which
the null hypothesis is not rejected, we accept r
= 2 as our estimate of the number of
cointegrating equations between these three
variables.
• The “*” by the trace statistic at r = 2 indicates
that this is the value of r selected by
Johansen’s multiple-trace test procedure
vecrank y i c, lags(5) level99
In the previous example, we used the default 5% critical
values. We can estimate r with 1% critical values
instead by specifying the level99 option.
Johansen tests for cointegration
Trend: constant
Number of obs =
Sample: 1960q2 - 1982q4
Lags =
maximum
rank
0
1
2
3
parms
39
44
47
48
LL
1231.1041
1245.3882
1252.5055
1254.1787
eigenvalue
.
0.26943
0.14480
0.03611
91
5
1%
trace
critical
statistic
value
46.1492
35.65
17.5810*
20.04
3.3465
6.65
• The output indicates that switching from the 5% to
the 1% level changes the resulting estimate from r =
2 to r = 1.
The maximum eigenvalue
statistic
A second test. This assumes a given r under the null
hypothesis and test this against the alternative
that there are r+1 cointegrating equations.
Johansen (1995, chap. 6, 11, and 12) derives an
LR test of the null of r cointegrating relations
against the alternative of r+1 cointegrating
relations.
• This method is used less often than the trace
statistic method, but often both test statistics are
reported.
vecrank y I c, lags(5) max levela
• The levela option obtains both the 5% and 1%
critical values.
Johansen tests for cointegration
Trend: constant
Number of obs =
Sample: 1960q2 - 1982q4
Lags =
maximum
rank
0
1
2
3
maximum
rank
0
1
2
3
parms
39
44
47
48
parms
39
44
47
48
LL
1231.1041
1245.3882
1252.5055
1254.1787
LL
1231.1041
1245.3882
1252.5055
1254.1787
eigenvalue
0.26943
0.14480
0.03611
eigenvalue
0.26943
0.14480
0.03611
91
5
trace
5% critical
statistic
value
46.1492
29.68
17.5810*1
15.41
3.3465*5
3.76
1% critical
value
35.65
20.04
6.65
max
statistic
28.5682
14.2346
3.3465
1% critical
value
25.52
18.63
6.65
5% critical
value
20.97
14.07
3.76
The test statistics are often referred to as lambda trace
and lambda max respectively
• We print out both tests in this table the eigenvalue ones are in the
second half of the table.
• The test is for r versus r+1 cointegrating vectors.
• In this example, because the trace statistic at r = 0 of 46.1492
exceeds its critical value of 29.68, we reject the null hypothesis of
no cointegrating equations.
• Similarly, because the trace statistic at r = 1 of 17.581 exceeds its
critical value of 15.41, we reject the null hypothesis that there is
one or fewer cointegrating equation. In contrast, because the trace
statistic at r = 2 of 3.3465 is less than its critical value of 3.76, we
cannot reject the null hypothesis that there are two or fewer
cointegrating equations.
• The net result is we conclude there are 2 cointegrating vectors.
Stationarity
• Intuitively a variable is stationary (I(0) – integrated to order nought) if
its characteristics do not change over time, e.g. variance, covariance
and mean is unchanging.
• Another way of looking at it is that ρ<1 in the following equation for
a variable y:
• Yt = ρYt-1
• We do not estimate the above, but subtract Yt-1 from both sides:
• Yt - Yt-1 = ρYt-1 -Yt-1 =(ρ-1)Yt-1
• Now in this regression we test that (ρ-1) is significantly negative,
which implies ρ<1. If we reject this we say y is I(1) and it has a unit
root. In this case doing time series OLS, etc on a variable with
variables that are I(1) results in bias. The Johansen method is a
suitable alternative.
• The above is the Dickey Fuller test (DF), add lagged values of Yt - Yt-1
to get rid of serial correlation and we have the augmented
Dickey Fuller test.
Stationarity
• Among the earliest tests proposed is the one
by Dickey and Fuller (1979), though most
researchers now use an improved variant
called the augmented Dickey–Fuller test
instead of the original version.
• Other common unit-root tests implemented in
Stata include the DF–GLS test of Elliot
Rothenberg, and Stock (1996) and the
Phillips–Perron (1988) test.
webuse air2
dfuller air
The test statistics is less negative than any of the critical
values and hence we cannot reject the null
hypothesis that the variable exhibits a unit root and
is thus not stationary
Dickey-Fuller test for unit root
Test
Statistic
Z(t)
-1.748
Number of obs
=
143
Interpolated Dickey-Fuller
1% Critical
5% Critical
10% Critical
Value
Value
Value
-3.496
MacKinnon approximate p-value for Z(t) = 0.4065
-2.887
-2.577
dfuller air, lags(3) trend
• This is a similar regression, but includes 3
lagged values and a trend term. It is now
stationary. What has made the difference?
Augmented Dickey-Fuller test for unit root
Z(t)
Test
Statistic
1% Critical
Value
-6.936
-4.027
Number of obs
=
140
Interpolated Dickey-Fuller
5% Critical
10% Critical
Value
Value
MacKinnon approximate p-value for Z(t) = 0.0000
-3.445
-3.145
The inclusion of the trend term
. dfuller air, trend
Dickey-Fuller test for unit root
Z(t)
Number of obs
Test
Statistic
1% Critical
Value
-4.639
-4.026
=
143
Interpolated Dickey-Fuller
5% Critical
10% Critical
Value
Value
-3.444
-3.144
MacKinnon approximate p-value for Z(t) = 0.0009
. dfuller air, lags(3)
Augmented Dickey-Fuller test for unit root
Z(t)
Test
Statistic
1% Critical
Value
-1.536
-3.497
Number of obs
=
140
Interpolated Dickey-Fuller
5% Critical
10% Critical
Value
Value
MacKinnon approximate p-value for Z(t) = 0.5158
-2.887
-2.577
dfuller air, lags(3) trend regres
Augmented Dickey-Fuller test for unit root
Z(t)
Test
Statistic
1% Critical
Value
-6.936
-4.027
Number of obs
=
140
Interpolated Dickey-Fuller
5% Critical
10% Critical
Value
Value
-3.445
-3.145
MacKinnon approximate p-value for Z(t) = 0.0000
D.air
Coef.
air
L1.
LD.
L2D.
L3D.
_trend
_cons
Lagged values
of D.air
-.5217089
.5572871
.095912
.14511
1.407534
44.49164
Std. Err.
.0752195
.0799894
.0876692
.0879922
.2098378
7.78335
t
-6.94
6.97
1.09
1.65
6.71
5.72
P>|t|
0.000
0.000
0.276
0.101
0.000
0.000
[95% Conf. Interval]
-.67048
.399082
-.0774825
-.0289232
.9925118
29.09753
-.3729379
.7154923
.2693065
.3191433
1.822557
59.88575
This is the test statistic, the coefficient on air(t-1) =L1.air
• The regression basically regresses the change
in the variable (D.air) on lagged changes an
the lagged value of air plus a constant and
time trend.
• The inclusion of lagged D.Air values makes this
the augmented Dickey-Fuller test, i.e. it is
what differentiates ir from the Dickey Fuller
test.
pperron air
• Phillips and Perron’s test statistics can be viewed as Dickey–
Fuller statistics that have been made robust to serial
correlation by using the Newey–West (1987)
heteroskedasticity- and autocorrelation-consistent covariance
matrix estimator.
Phillips-Perron test for unit root
Z(rho)
Z(t)
Number of obs
=
Newey-West lags =
Test
Statistic
1% Critical
Value
-6.564
-1.844
-19.943
-3.496
143
4
Interpolated Dickey-Fuller
5% Critical
10% Critical
Value
Value
MacKinnon approximate p-value for Z(t) = 0.3588
-13.786
-2.887
-11.057
-2.577
• Z(rho) is the main statistic we are interested in
as it is similar to the ADF test statistic.
DFGLS Test
webuse lutkepohl2
dfgls dln_inv
• dfgls tests for a unit root in a time series. It performs the modified
Dickey–Fuller t test (known as the DF-GLS test) proposed by Elliott,
Rothenberg, and Stock (1996). Essentially, the test is an augmented
Dickey–Fuller test, similar to the test performed by Stata’s dfuller
command, except that the time series is transformed via a generalized
least squares (GLS) regression before performing the test.
• Elliott, Rothenberg, and Stock and later studies have shown that this test
has significantly greater power than the previous versions of the
augmented Dickey–Fuller test.
```
– Cards
– Cards
– Cards
– Cards
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# Position and Direction: Find the Shapes
In this worksheet, students identify 2D shapes and understand words relating to their position in a grid.
Key stage: KS 1
Curriculum topic: Geometry: Position and Direction
Curriculum subtopic: Use Language of Position/Direction
Difficulty level:
### QUESTION 1 of 10
In this worksheet, you must know the names of 2D shapes and use words such as above, below, right and left.
Examples
The blue triangle is to the left of the grey square.
The red circle is two places to the right of the yellow star and below the green triangle.
Look at this grid.
Which shape is to the left of the yellow circle?
Purple star
Grey square
Red square
Look at this grid.
Which shape is to the left of the red circle?
Nothing
Green Triangle
Purple Pentagon
Look at this grid.
Tick all the shapes which are next to the black square?
Red pentagon
Red hexagon
Yellow rectangle
Purple pentagon
Look at this grid.
Tick all the shapes which are next to the grey square?
Rectangle
Triangle
Circle
Pentagon
Look at this grid.
Which shape is below the yellow circle?
Pentagon
Octagon
Hexagon
Look at this grid.
Which shape is below the red square? (search hard for it.....it may not be upright!)
Pentagon
Circle
Triangle
Look at this grid.
Which shape is two places to the right of the blue triangle?
Square
Circle
Triangle
Look at this grid.
Which shape is two places above the red circle?
Square
Circle
Triangle
Look at this grid.
Which shape is four places below the yellow circle?
Square
Circle
Triangle
Look at this grid.
Which shape is three places to the left of the black square?
Rectangle
Pentagon
Triangle
• Question 1
Look at this grid.
Which shape is to the left of the yellow circle?
Grey square
• Question 2
Look at this grid.
Which shape is to the left of the red circle?
Nothing
• Question 3
Look at this grid.
Tick all the shapes which are next to the black square?
Red hexagon
Yellow rectangle
Purple pentagon
• Question 4
Look at this grid.
Tick all the shapes which are next to the grey square?
Triangle
Circle
• Question 5
Look at this grid.
Which shape is below the yellow circle?
Octagon
• Question 6
Look at this grid.
Which shape is below the red square? (search hard for it.....it may not be upright!)
Triangle
• Question 7
Look at this grid.
Which shape is two places to the right of the blue triangle?
Circle
• Question 8
Look at this grid.
Which shape is two places above the red circle?
Square
• Question 9
Look at this grid.
Which shape is four places below the yellow circle?
Circle
• Question 10
Look at this grid.
Which shape is three places to the left of the black square?
Pentagon
---- OR ----
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IIT-JEE
Apne doubts clear karein ab Whatsapp (8 400 400 400) par bhi. Try it now.
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Question From class 12 Chapter BOARD PAPER SOLUTIONS
# By using properties of determinants, prove the following
By using properties of determinants. Show that: (i) (ii)
3:30
By using properties of determinants, prove the following
3:38
By using properties of determinants. Show that: (i) (ii)
5:58
If then the ordered pair (A,B) is equal to
5:08
If then the ordered pair (A,B) is equal to
4:28
2:23
Using properties of determinants, solve the following for x:
5:57
Solve the equation
6:14
Without actual division, prove that is divisible by .
3:33
Without actual division , prove that is divisible by .
5:01
Using properties of determinants, prove that
7:01
2:35
Find the derivative of the following w.r.t. x (i) (ii) (iii)
4:37
3:12
1:34
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Microconcepts
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# Financial Calculations
How much money will you have when you retire? How much money should you deposit into your account each month in order to have \$10,000 in five years? What return on your money is necessary to double your account in five years?
These, and many others, are important questions for financial planning. In order to reach your goals, you must have some idea of what is necessary for that to happen. Fortunately, there are many financial calculations that can tell you exactly what to expect – and that’s what you’re going to learn in this reference guide.
Return on Investment (ROI)
Before we begin, we need to clarify the financial term “return on investment” more commonly called the “return on your money” or, more simply, the “return.”
Whenever you make an investment, you place a fixed amount of money into an investment. The initial amount you deposit is called the “principal.”
In exchange, you hope to get the principal back along with an additional amount “returned” to you for your efforts. This amount is the “return on the investment” or ROI.
For instance, if you deposit \$100 into an investment and it later increases to \$110 then the investment “returned” \$10 to you. As a matter of standardization, the return is expressed as a percentage of the original investment so we would say your return on investment was \$10/\$100 = 10%.
Further, this percentage should be expressed as an annualized figure in order to make fair comparisons across investments. If the above investment returned \$10 in exactly one year, your return is 10%. If, however, it returned \$10 in only six months then, effectively, you earned at the rate of 20% per year since you picked up the same amount of money in only half the time.
This relationship between principal, interest, and time is often expressed in the simple formula:
I = P * R * T
Where:
I = interest (actual dollar amount)
P = principal
R = rate (interest rate expressed as a percent so that 5% = .05)
T = time (expressed in years so that six months = 0.5 years, nine months = 0.75 years)
It states that the actual amount of dollars you earn on any investment (the interest, I) equals the principal (P) multiplied by the rate of interest (R) multiplied by the amount of time (T).
In the previous example, we know the principal is \$100, the interest rate is 10% and the time is one year. Therefore, the “interest” or the amount returned from the investment is \$10, which is found as follows:
I = P * R * T
I = \$100 * 0.10 * 1 year
I = \$10
Naturally, we don’t need to use the formula to only solve for the interest rate (I). If you know any of the three variables, you can solve for the missing fourth.
For example, if the \$10 interest was returned in only six months (0.5 years) then what the effective interest rate (R)? We can easily find that by plugging in I= \$10, P = \$100, and T = six months, or half a year (0.5 years) and solve for R as follows:
\$10 = \$100 * R * 0.5 years
R = \$10/(\$100 * 0.5)
R = \$10/\$50
R = 0.20, or 20%
If you’re not too comfortable with algebra, there is a simple way to solve for any of the variables without having to rearrange the equation. Rather than write the formula as I = P*R*T, write it down on a piece of paper as follows:
Next, place your finger over the variable you’re trying to solve for and voila – the remaining uncovered letters give you the formula! Using our previous example, if you want to solve for the rate (R) that was earned then you’d simply cover up the R and you’d see this:
It immediately tells you that the formula is “interest divided by (principal * time)”, which is \$10/(\$100*0.5) = 0.20, or 20% as we found before.
If, on the other hand, you wanted to find out the length of time (T) required to produce that 20% return, you’d cover up the letter T and you see this:
You can see that solving for “T” requires you to take the interest amount and divide by (P *R), which is \$10/(\$100 * .20) = 0.5 years, or six months which also corresponds to our earlier answer.
No matter which variable you’re trying to solve for, just cover that letter and the formula takes care of itself. With a little bit of work, you’ll start to visualize the correct formulas without having to resort to sheer memorization or algebraic manipulations.
Interest Doesn’t Necessarily Mean Risk-Free
One of the main points to understand from this section is that the term R (interest rate) does not necessarily mean interest in the traditional sense such as the amount paid by a bank for the money you have on deposit.
In most cases, people are used to using the word “interest” to mean that money paid by another person or institution (such as a bank) for the use of their money. By this definition, interest usually means risk-free.
However, as you read through this reference guide, if a calculation states that the interest rate necessary to accomplish your goal is 20%, for example, don’t think this means you must find a bank paying 20%. The proper interpretation of interest in this guide is the “effective” interest rate or the “return” on your money.
If you can’t find a bank paying 20% then you will have to find another source – perhaps the stock market – and any profits effectively represent the interest paid on your money, or your return on investment (ROI).
If you invest \$10,000 in the stock market and it’s worth \$12,000 in one year then your “interest” on your money is \$2,000, or 20%, even though it is technically a capital gain. In most areas of finance, the word “interest” is used generically to mean the return on your money whether it is risk-free or not.
The formulas in this guide can greatly help you with your financial decisions. For example, if a formula shows that 5% is necessary to accomplish your goal and the risk-free interest rate is 3%, you have some choices to make. First, if you insist on the risk-free rate then you must extend the time period you are willing to wait for that money. On the other hand, if you cannot extend the time, you’ll have to accept a little more risk.
Will you need to invest in stocks? Perhaps not but you may need to consider commercial paper or other types of investments that have higher expected returns. Of course, the fact that risk is present for all investments (other than risk-free investments) means there is always the chance that the desired return won’t be realized. That’s why it is so critical to define your goals, find your risk tolerances, and hedge investments accordingly. Only then can you have a realistic chance of reaching those goals.
Let’s take a look at some very helpful financial planning calculations.
Rules of Thumb
Rule of 72 – How long to double your account
Rule of 114 – How long to triple your account
The benefit to these calculations is that they are fast, simple and many times can be performed in your head. The drawback is that they are not exact; however, they do produce reasonably close results. Use these as guidelines but always double check with precise formulas (described later) if you are using to make important investment decisions that require an exact answer.
# Rule of 72 – How Long to Double Your Account?
To find how long it will take to double your account at x% interest rate, take 72/x.
Example 1: How long will it take to double your account at 10% interest?
Answer: It will take approximately 72/10 = 7.2 years.
We can also solve the problem backwards and find out the interest rate. If you want to double your account in 7.2 years, it will require 72/7.2 = 10% interest.
Example 2: How long will it take to double your account at 12% interest?
Example 3: What interest rate is necessary to double your account in 7 years?
How accurate are the results? The following table shows a few interest rates and the corresponding answers generated by the Rule of 72 compared to the exact calculation (described later).
You can see that the Rule of 72 slightly overestimates the amount of time for low interest rates while underestimating for higher rates. Still, considering the ease of calculation, it provides a very good estimate.
Rule of 114 – How Long to Triple Your Account?
To find out how long it will take to triple your account at x% interest rate, take 114/x.
Example 1: How long will it take to triple your account at 10% interest?
Answer: It will take approximately 114/10 = 11.4 years.
As before, we can solve this problem for the interest rate as well. What interest rate is necessary to triple your account in 11.4 years? Answer: 114/11.4 = 10%
Example 2: How long will it take to triple your account at 12% interest?
Example 3: What interest rate is necessary to triple your account in 12 years?
As with the previous rule, the Rule of 114 slightly overestimates the amount of time for low interest rates while underestimating for higher rates but the estimates are still quite good.
To find out how long it will take to quadruple your account at x% interest rate, take 144/x.
Example 1: How long will it take to quadruple your account at 10% interest?
Answer: It will take approximately 144/10 = 14.4 years.
What interest rate is required to quadruple your account in 14.4 years?
Example 2: How long will it take to quadruple your account at 12% interest?
Example 3: What interest rate is necessary to quadruple your account in 18 years?
How accurate are the results? The following table shows a few interest rates and the corresponding answers generated by the Rule of 144 compared to the exact calculation (described later).
One of the most significant insights that can be drawn from all three rules is that the amount of time necessary to double, triple, or quadruple your account value is probably less than most people would intuitively expect. The moral: You don’t need to take large risks to attain great wealth – if you’re willing to wait just a little bit.
Future Value of Money
The previous rules of thumb gave us approximations for the amount of time to double, triple, or quadruple your account value. Now it’s time to look at exact formulas.
Any amount of money you have right now is called the “present value.” The value of that money in the future assuming a constant growth rate (i.e., the interest rate stays constant) is called the “future value.”
To find the future value of money, you must take the principal amount multiplied by 1+interest rate (expressed as a decimal) raised to the number of years:
Principal * (1+i) years
Example 1: You have \$1,000 in the bank and it will grow at 5% per year. How much will it be worth in 5 years?
\$1,000 * (1.05)5 = \$1,276.28.
Note that we could have also found this answer the long way by multiplying \$1,000 * 1.05, which would give us the future value in one year (\$1,050). We could then multiply that answer by 1.05, which gives us the future value in two years. If we repeat this procedure five times, we’d end up with the same answer. Mathematically, multiplying any number by itself x times is the same thing as raising it to the x power. In other words, 1.05 * 1.05 * 1.05 * 1.05 * 1.05 is the same thing as (1.05)5.
This formula assumes that interest is paid at the end of each year. What if it is paid monthly?
In this case, the monthly interest rate is .05/12. We also know that the number of payment periods must be five years * 12 months, or 60 periods. Therefore, if interest is paid monthly at the rate of 5% per year, the \$1,000 initial investment would be worth:
\$1,000 * (1 + .05/12) 60 = \$1,283.36
We can extend the above line of thinking indefinitely and assume that interest is paid as fast as possible – every micro-second and faster. In this case, we say the interest is paid “continuously.” Despite intuition, the principal amount will not grow to an infinite amount. Instead, it will not be too much greater than if the interest is paid monthly.
How do we calculate interest paid continuously? Let’s put things in simple terms and see if we can unlock a pattern. Assume that interest rates are 100% and paid annually. The future value formula is then:
(1 + 1)1
In other words, if you start with \$100, you will have \$100 * (1 + 1)1 = \$200 at the end of one year.
However, if the interest is paid monthly then the formula would be:
(1 + 1/12)12
Your \$100 investment would now be worth \$100 * (1 + 1/12)12 = \$261.30
If interest is paid daily, the formula is:
(1 + 1/360)360
The \$100 investment would be worth \$100 * (1 + 1/360)360 = \$271.45
Are you starting to see the pattern? We take our interest rate and divide by the number of periods, add one to it, and then raise it to the number of periods. If we were to take this to a very small time frame, say one million units per year, the formula would reduce to:
(1 + 1/1,000,000)1,000,000
Your investment would now be worth \$100 * (1 + 1/1,000,000)1,000,000 = \$271.82, which isn’t too greater than the previous answer of \$271.45.
If we make two million pay periods per year, the future value becomes:
\$100 (1 + 1/2,000,000)2,000,000 = \$271.82, which exactly matches the previous answer with only half the pay periods! It seems as though we’re closing in on a maximum limit.
What is that limit? In finance and other areas of mathematics, it has a special name “e” which stands for the “exponential” function. The bigger we make the number of pay periods, the closer we’ll get to the true value of e. Let’s pick an arbitrarily large number, say ten million:
\$100 (1 + 1/10,000,000)10,000,000 = 2.718281828...
The trailing dots just means the number continues on forever, never ending, never repeating.
Most financial and scientific calculators have the “e” function built in, which is a key that usually looks like ex. This key allows you to raise e to any number you’d like. If you hit the number “1” and then the ex key, the “=” key you’ll find the calculator gives you 2.718281828…, which is the value of e.
With the aid of e, we have a simplified way of finding the future value for any interest rate or time period. All you have to do is multiply the interest rate by the number or pay periods, raise e to that number, and multiply by the principal.
For instance, let’s go back to our previous example and see if we can simplify it. Assume you have \$100 that it continuously compounded at 100% for one year. How much will you have at the end of the year?
\$100e1 = \$271.83
Now that you understand e, let’s use a little more realistic example.
Example 2:
Assume you have \$1,000 invested for five years at 5% and compounded continuously. How much will you have at the end of five years?
\$1,000 * e.05 * 5 = \$1,284.03
Example 3:
You have invested \$3,000 for three years at 7% and compounded continuously. How much will you have at the end of three years?
\$3,000 * e.07 *3 = \$3,701.03
In other words, the very most a three-year investment at 7% could ever produce is \$3,701.03. If the interest rate is paid at anything less than “continuously” the future value will be somewhat less. Using e is a great way to define potential limits of investments.
Now that you understand the future value of money as well as the varying rates at which interest could be paid, let’s take a look at some other examples.
Example 4: You have \$20,000 in your IRA that is expected to grow at 12% per year. How much will it be worth in 30 years?
Solving For Time
As with any formula, we can rearrange it to solve for other variables that are present in the formula. Using the previous future value formula, we can rework it and solve for time.
Example 1: You have \$20,000 in your IRA that is expected to grow at grow at 12% per year. How long will it take before it's worth \$1 million?
Before we can answer this, we need to turn to a mathematical function called logarithms. A logarithm, or log for short, is really not that complicated. The log of any number is simply the exponent on base 10 that creates the number in question.
For example, log (100) = 2
The interpretation here is that 10 (the base) raised to what number gives the answer 100? In other words, if 10x = 100, what does x equal? It must equal two. Recall that whenever we raise a number to a power, it means that we are multiplying the base number by itself that many times. Therefore, 102 = 10*10 = 100. Now it should make better sense why log (100) = 2.
If you notice in the future value formula, the “time” is represented as the exponent. Logarithms solve for exponents and that’s what makes them so powerful. If you want to solve for exponents, you’ll need to use logarithms.
Now, there are a couple of mathematical tricks using logs that you must understand before you can solve for time. There is a property of logarithms that says a = log 10a. This is nothing more than an identity. It is asking for the exponent on base 10 that gives the answer “a.” Obviously, if 10 is raised to “a” then the answer must be “a.”
Using our previous example, 2 = log 102. If 102 = 100 then log 100 = 2.
Using the above identity, we can multiply number together using logs. For instance, how can we solve for a * b using logs?
Step 1:
Recognize that a = 10log a and b = 10log b
Step 2:
If Step 1 is true then a * b = 10log a * 10log b
Step 3:
A basic property of exponents states that if you are multiplying two numbers together with the same base then we can just add the exponents together and raise the sum to the same base. For instance, 102 * 102 = 104
Therefore, if we take the log of both sides of Step 2, we get:
log a * b = log a + log b
In other words, if we want to multiply two numbers together we can simply add the logs of those two numbers. Now, you’re probably wondering why we wouldn’t just multiply the two numbers together in the first place! Why bother taking the logs and adding them?
The answer is that many times the numbers we are multiplying are too big. Sometimes it’s easier to add. Further, for our purposes, we’re going to use this trick to solve for time.
The problem with solving for time is that we are trying to solve for an unknown exponent. In order to do so, we must get the exponent into the equation by some mathematical manipulation – and that’s what logs allow us to do.
For instance, how do you suppose we can “rearrange” the expression 52 so there is no exponent? It’s simple with logs.
First, understand that 52 = 5 * 5 by definition. Therefore, if we take the log of both sides we get:
log 52 = log (5 * 5) = log 5 + log 5 = 2 * log 5
More generally, how can we rearrange am?
If we take the log of both sides, we get:
log (am) = log (a1 * a2 * …am) = log a1 + log a2 + log am
In other word, there are “m” factors of “log a” so we can rewrite that as m * log a.
So for the log of any number, we can take the exponent multiplied by the log of the base. For instance, log (37) must equal 7 * log (3).
log (37) = log (2,187) = 3.34
and secondly,
7* log (3) = 7 * 0.48 = 3.34
Armed with the powerful log function, we can now solve for exponents. Let’s start with an easy one. If 5 x = 25, what does x equal?
Simple, take the log of both sides:
log 5x = log 25
x * log 5 = log 25
x = log 25/log 5
x = 2
If you have a financial or scientific calculator, try it! Take log (25) and divide it by log (5) and you’ll see the answer is two. Therefore, 52 = 25.
Note: There are two log functions: common log (base 10) and natural log (base e), and most financial and engineering calculators have both. It does not matter which one you use as long as you're consistent. Pick one or the other when doing the calculations.
Example 1: You have \$20,000 in your IRA that is expected to grow at grow at 12% per year. How long will it take before it's worth \$1 million?
The question is asking you to solve for “n” for the following equation:
\$20,000 * (1.12)n = \$1,000,000
Step 1:
Divide both sides by \$20,000
(1.12)n = 50
Step 2:
Take log of both sides
n * log (1.12) = log (50)
Step 3:
Get “n” by itself on the left so divide both sides by log (1.12)
n = log (50)/log (1.12)
n = 34.5
It will take 34.5 years for your \$20,000 to grow to \$1,000,000 assuming you can get a constant growth rate of 12% per year.
Once you understand the math behind the solution, you can quickly get the answer on a calculator by taking the log of the future value (\$1,000,000) and dividing it by the log of the present value (\$20,000). Once you have that answer, divide it by the log of “1 + interest rate” as follows:
log (\$1,000,000/\$20,000) = 34.5 years
log 1.12
Note that we must express the interest rate as a decimal. If we are assuming 12% then that is 0.12 as a decimal. If we add that to one the answer is 1.12.
Example 2: You have \$20,000 in your IRA that is expected to grow at grow at 12% per year. How long will it take before it doubles?
In other words, solve the following \$20,000 * (1.12)n = \$40,000
log (\$40,000/\$20,000) = 6.11 years
log 1.12
We can check this using our earlier "Rule of 72":
72/12 = 6 years
How long before it triples?
Again, set up the equation as \$20,000 * (1.12)n = \$60,000 and solve for “n”:
log (\$60,000/\$20,000) = 9.7 years
log 1.12
We can check this using our earlier "Rule of 114":
114/12 = 9.5 years, which is pretty close to the exact answer.
log (\$80,000/\$20,000) = 12.2 years
log 1.12
According to the Rule of 144 the answer is 144/12 = 12 years.
Present Value of Money
Once you understand the future value formula, you can easily solve for other variables. For instance, rather than solve for the future value, we may wish to know the present value – the amount of money needed today to reach a particular goal.
Example 1:
You wish to deposit money in your bank at 5% in order to have \$15,000 in 5 years. You will make only one deposit. How much do you need to deposit today?
This is asking you to solve the following equation:
x * (1.05)5 = \$15,000
This is easy since we know the exponent so we won’t need to use logs. In order to solve for “x” just divide both sides by (1.05)5:
x = \$15,000/1.055 = \$11,752.89
Because this is just working the future value formula in reverse, we can use the future value formula to check the answer:
\$11,752.89 * (1.05)5 = \$15,000
Therefore, if you deposit \$11,752.80 in an account and earn 5% per year, your account will be worth \$15,000 in five years. Remember, this 5% “interest” could be a risk-free rate from a checking account or the capital gains from the stock market. It’s just stating that this is the rate of return you’d need on your money.
Solving for Interest
Now that you understand how to set up the equations and solve for future and present values, let’s find out how to solve for a missing interest rate.
To do so, you must understand another simple mathematical trick. If you have a number raised to some number and wish to get rid of the exponent, you must raise the entire expression to 1/exponent.
For instance, if 52 is raised to the ½ power, you’d get five:
(52)1/2 = 5
Let’s try one more. What is (74)1/4? It must equal seven.
In essence, this trick just “erases” the exponent without changing the value of the equation.
With that simple trick, we can now solve problems for an unknown interest rate. Let’s work the previous problem assuming we don’t know the interest rate.
Example 1:
You wish to deposit \$11,752.80 in a bank in order to have \$15,000 in 5 years. You will make only one deposit. What rate of interest is required?
Step 1: Set up the problem
\$11,752.80 * (1+i)5 = \$15,000
Step 2: Get the 1+i by itself
(1+i)5 = \$15,000/\$11,752.80, or
(1+i)5 = 1.28
Step 3: Raise both sides to the 1/5 power
((1+i)5 )1/5 = 1.281/5
This effectively “erases” the “5” exponent on the left side and you’re left with
(1+i) = 1.281/5
1+i = 1.05
i = .05, or 5% interest, which is exactly what we used for the previous problem.
Present Value of an Annuity
An annuity is just a stream of steady payments, rather than just one lump sum amount. Examples of common annuities are home and car loans.
For example, you may spend \$300 per month for a car payment over a four year period. If you do, that cash obviously has a lot of value over time. Rather than buy the car, you could elect to deposit \$300 per month into a savings account paying a given interest rate, say 5%. After the first month, the \$300 earns a little bit of interest at which time another \$300 hits your account. After the second month, the first \$300 deposit has earned interest for two months while the second deposit earned interest for only one. This process continues for four years. How much money would you have in the bank at the end? That’s what the future value of an annuity tells us.
Conversely, we can take a future value and find the present value of those cash flows – the present value of an annuity.
Example 1: You are willing to spend \$500 per month over 5 years on a new car. Interest rates are 6% per year. How much can you spend on the car?
We are exchanging a stream of payments (the monthly car payment) in exchange for a lump sum amount today to buy the car – the present value of the annuity.
Notice that the periodic interest rate is 6%/12 = 0.5%, which is .005 as a decimal. Also, you are making monthly payments over five years, which is equal to 60 payments.
Answer: \$500 * 1 - (1/1.005)60 = \$25,862
.005
You can shop for a car worth about \$26,000.
We can also rework this formula to solve for the payment or the monthly (or other time period) cash flow. A typical use is to find how much you can periodically with draw from an account over a given time as in the following example.
Example 2: You have accumulated an IRA worth \$1,000,000 and are ready to retire. You wish to deplete the account in 20 years. Assuming your account continues to grow at 8% per year, how much can you withdraw per month in order to exactly deplete the account in 20 years?
Note: the periodic interest rate is 8%/12 = .67%, which is .0067 as a decimal. If you are making monthly withdrawals over 20 years, that's 20*12 = 240 periods.
\$1,000,000
(1-(1/1.0067)240)/.0067
= \$8,389.62 per month
Future Value of an Annuity
The future value of an annuity is the value in the future that is a result of a steady stream of payments today. Obviously, there are two ways that payments could be made. First, you could make the payment on the first of each month (or other time period). Second, you could make them at the end of the month. While it may not sound like a big difference, over time it can really add up as shown in the following two examples.
Example: You wish to deposit \$2,000 per year and the end of each year for the next 30 years. You expect to earn an average of 12% per year. How much will you have saved at the end of 30 years?
Because the payments are made at the end, this is called an "ordinary annuity."
\$2,000 * (1.1230 - 1)/.12 = \$482,665
Example 2: Instead, what if you made contributions of \$2,000 per year but did so by depositing \$500 at the end of each quarter? Now the periodic interest rate is 12%/4 = 3% (.03 as a decimal) and there are now four payments per year for a total of 120 payments.
\$500 * (1.03120 -1)/.03 = \$561,849
Notice the difference in the results. It pays to get the money in early! If you plan to put \$2,000 per year in an IRA, it really pays to split up those payments and get the money working sooner.
What if you made the contributions at the beginning of each period instead of at the end? This would be called an "annuity due" and is the same answer you get from above just multiplied by 1 plus the periodic interest rate.
Example 1: A \$2,000 contribution at the beginning of each year for 30 years at 12% would be worth \$482,665 * (1.12) = 540,584
Example 2: A \$500 contribution at the beginning of each quarter for 30 years at 12% (periodic interest rate of 12%/4 = 3% or .03) would be worth \$561,849 * 1.03 = 578,704.
This guide presents most of the financial calculations you will ever need to make your investment decisions. Most financial calculators and many websites provide programs that calculate the answers for you. But as with any calculation, you’ll make better decisions if you understand how to interpret the results. Hopefully this guide has provided the necessary understanding.
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Computer Science Canada Programming C, C++, Java, PHP, Ruby, Turing, VB Username: Password: Wiki Blog Search Turing Chat Room Members
[Blitz] Title Page - Cervantes vs. Jonos !!
Author Message
jonos
Posted: Wed Feb 18, 2004 6:47 pm Post subject: [Blitz] Title Page - Cervantes vs. Jonos !!
CERVANTES EDIT: This war was originally in (Tutorial)(Blitz)Graphics Engine
just to show that i am learning something, and that THIS IS THE BEST FORUM PART i will post my program i made from this tutorial's teachings:
Graphics 400, 400
h%=50
h%=50
v%=50
v%=50
While h<350
Plot 50, h
Plot 350, h
h=h+1
Wend
While v <350
Plot v ,50
Plot v, 350
v=v+1
Wend
i want to be able to make a delay between making the dots, so could one of you please show me how, thanks.
jonos edit:
nm, its delay(ms), guessing does help sometimes
Cervantes
Posted: Thu Feb 19, 2004 9:03 pm Post subject: (No subject)
ppfffft jonos!!
Less lines, looks cooler
code: Graphics 400, 400 For k = 50 To 350 Step 2 Plot 50, k Plot 350, k Plot k, 50 Plot k, 350 Delay 10 Next For i = 50 To 350 Plot 50, i Plot 350, i Plot i, 50 Plot i, 350 Delay 5 Next
jonos
Posted: Thu Feb 19, 2004 9:21 pm Post subject: (No subject)
i didn't know about the for loops, i only knew about the loops shorthair showed us.
it looks like shorthair only has two pupils, me and you cervantes, haha
jonos
Posted: Fri Feb 20, 2004 7:39 am Post subject: (No subject)
here cervantes, ill one up you:
code: Graphics 400, 400 count% = 60 For k = 50 To 350 Step 10 Plot 50, k Plot 350, k Plot k, 50 Plot k, 350 Delay 10 Next For count = 50 To 350 Step 10 For count2 = 50 To 350 Step 10 Plot count, count2 Delay 5 Next Next Delay(3000)
jonos edit:
ongoing contest of for loops
code: Graphics 400, 400 count% = 60 For k = 50 To 350 Step 10 Plot 50, k Plot 350, k Plot k, 50 Plot k, 350 Delay 10 Next For count1 = 50 To 350 Step 10 For count2 = 50 To 350 Step 10 Plot count1, count2 Delay 5 Next Next For count3 = 55 To 345 Step 10 For count4 = 50 To 350 Step 2.5 Plot count3, count4 Delay 3 Next Next For count5 = 55 To 345 Step 10 For count6 = 50 To 350 Step 2.5 Plot count6, count5 Delay 3 Next Next Delay(3000)
more lines, cooler effect, but you might want to decrease the delay amoutn, its still pretty slow
Cervantes
Posted: Fri Feb 20, 2004 6:14 pm Post subject: (No subject)
looks neat, is definately slow
code: Graphics 400, 400 SetBuffer BackBuffer() x1 = 350 y1 = 50 x2 = 50 y2 = 350 For k = 0 To 350 Step 3 x1 = x1 - 3 y1 = y1 + 3 x2 = x2 + 3 y2 = y2 - 3 Line 50, y1, 50, y1 + 3 Line x1, 50, x1 - 3, 50 Line 350, y2, 350, y2 - 3 Line x2, 350, x2 + 3, 350 Flip Next x1 = 0 y1 = 400 x2 = 400 y2 = 0 For k = 0 To 350 Step 3 x1 = x1 + 3 y1 = y1 - 3 x2 = x2 - 3 y2 = y2 + 3 For i = 0 To 50 Step 3 Plot x1, i Plot i, y1 Plot x2, i + 350 Plot i + 350, y2 Next Flip Next ;;;;;;;;; While Not KeyHit (1) Wend End
getting kinda off topic though
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Search by Topic
Resources tagged with Visualising similar to Board Block for Two:
Filter by: Content type:
Stage:
Challenge level:
Overlaps
Stage: 1 Challenge Level:
What does the overlap of these two shapes look like? Try picturing it in your head and then use the interactivity to test your prediction.
Makeover
Stage: 1 and 2 Challenge Level:
Exchange the positions of the two sets of counters in the least possible number of moves
Nine-pin Triangles
Stage: 2 Challenge Level:
How many different triangles can you make on a circular pegboard that has nine pegs?
Alquerque II
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A variant on the game Alquerque
Horizontal Vertical
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Take it in turns to place a domino on the grid. One to be placed horizontally and the other vertically. Can you make it impossible for your opponent to play?
Paper Patchwork 1
Stage: 1 Challenge Level:
Can you work out what shape is made when this piece of paper is folded up using the crease pattern shown?
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Twice as Big?
Stage: 2 Challenge Level:
Investigate how the four L-shapes fit together to make an enlarged L-shape. You could explore this idea with other shapes too.
Put Yourself in a Box
Stage: 2 Challenge Level:
A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.
Posting Triangles
Stage: 1 Challenge Level:
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Endless Noughts and Crosses
Stage: 2 Challenge Level:
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Stage: 2 Challenge Level:
How can the same pieces of the tangram make this bowl before and after it was chipped? Use the interactivity to try and work out what is going on!
Go Moku
Stage: 2 Challenge Level:
A game for two players on a large squared space.
Change Around
Stage: 1 Challenge Level:
Move just three of the circles so that the triangle faces in the opposite direction.
Happy Halving
Stage: 1 Challenge Level:
Can you split each of the shapes below in half so that the two parts are exactly the same?
Tessellating Capitals
Stage: 1 Challenge Level:
Have you ever tried tessellating capital letters? Have a look at these examples and then try some for yourself.
Paper Partners
Stage: 1 Challenge Level:
Can you describe a piece of paper clearly enough for your partner to know which piece it is?
Putting Two and Two Together
Stage: 2 Challenge Level:
In how many ways can you fit two of these yellow triangles together? Can you predict the number of ways two blue triangles can be fitted together?
Map Folding
Stage: 2 Challenge Level:
Take a rectangle of paper and fold it in half, and half again, to make four smaller rectangles. How many different ways can you fold it up?
Circles, Circles
Stage: 1 and 2 Challenge Level:
Here are some arrangements of circles. How many circles would I need to make the next size up for each? Can you create your own arrangement and investigate the number of circles it needs?
Coin Cogs
Stage: 2 Challenge Level:
Can you work out what is wrong with the cogs on a UK 2 pound coin?
Colour Wheels
Stage: 2 Challenge Level:
Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?
World of Tan 29 - the Telephone
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Can you fit the tangram pieces into the outline of this telephone?
World of Tan 4 - Monday Morning
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Domino Numbers
Stage: 2 Challenge Level:
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Cubes Cut Into Four Pieces
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Can you cover the camel with these pieces?
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Try this interactive strategy game for 2
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Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?
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Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outline of the rocket?
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Can you fit the tangram pieces into the outline of this plaque design?
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Stage: 2 Challenge Level:
Design an arrangement of display boards in the school hall which fits the requirements of different people.
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Can you fit the tangram pieces into the outline of the telescope and microscope?
World of Tan 19 - Working Men
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Stage: 1 and 2 Challenge Level:
Square It game for an adult and child. Can you come up with a way of always winning this game?
World of Tan 25 - Pentominoes
Stage: 2 Challenge Level:
Can you fit the tangram pieces into the outlines of these people?
Tricky Triangles
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Use the three triangles to fill these outline shapes. Perhaps you can create some of your own shapes for a friend to fill?
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## 冲浪阵列游戏攻略¶
pete@shinners.org
### 数字 Python¶
```>>> from numpy import * #import numeric
>>> a = array((1,2,3,4,5)) #create an array
>>> a #display the array
array([1, 2, 3, 4, 5])
>>> a[2] #index into the array
3
>>> a*2 #new array with twiced values
array([ 2, 4, 6, 8, 10])
```
```>>> len(a) #get array size
5
>>> a[2:] #elements 2 and up
array([3, 4, 5])
>>> a[:-2] #all except last 2
array([1, 2, 3])
>>> a[2:] + a[:-2] #add first and last
array([4, 6, 8])
>>> array((1,2,3)) + array((3,4)) #add arrays of wrong sizes
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: operands could not be broadcast together with shapes (3,) (2,)
```
```>>> a #show our starting array
array([1, 2, 3, 4, 5])
>>> aa = a[1:3] #slice middle 2 elements
>>> aa #show the slice
array([2, 3])
>>> aa[1] = 13 #chance value in slice
>>> a #show change in original
array([ 1, 2, 13, 4, 5])
>>> aaa = array(a) #make copy of array
>>> aaa #show copy
array([ 1, 2, 13, 4, 5])
>>> aaa[1:4] = 0 #set middle values to 0
>>> aaa #show copy
array([1, 0, 0, 0, 5])
>>> a #show original again
array([ 1, 2, 13, 4, 5])
```
```>>> row1 = (1,2,3) #create a tuple of vals
>>> row2 = (3,4,5) #another tuple
>>> (row1,row2) #show as a 2D tuple
((1, 2, 3), (3, 4, 5))
>>> b = array((row1, row2)) #create a 2D array
>>> b #show the array
array([[1, 2, 3],
[3, 4, 5]])
>>> array(((1,2),(3,4),(5,6))) #show a new 2D array
array([[1, 2],
[3, 4],
[5, 6]])
```
```>>> b #show our array from above
array([[1, 2, 3],
[3, 4, 5]])
>>> b[0,1] #index a single value
2
>>> b[1,:] #slice second row
array([3, 4, 5])
>>> b[1] #slice second row (same as above)
array([3, 4, 5])
>>> b[:,2] #slice last column
array([3, 5])
>>> b[:,:2] #slice into a 2x2 array
array([[1, 2],
[3, 4]])
```
```>>> c = arange(10) #like range, but makes an array
>>> c #show the array
array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> c[1:6:2] #slice odd values from 1 to 6
array([1, 3, 5])
>>> c[4::4] #slice every 4th val starting at 4
array([4, 8])
>>> c[8:1:-1] #slice 1 to 8, reversed
array([8, 7, 6, 5, 4, 3, 2])
```
### 导入曲面阵列¶
```try:
import numpy as N
import pygame.surfarray as surfarray
except ImportError:
raise ImportError, "NumPy and Surfarray are required."
```
### 冲浪阵列游戏攻略¶
SURFARY中有两种主要类型的函数。用于创建作为表面像素数据的副本的阵列的一组函数。其他函数创建数组像素数据的引用副本，因此对数组的更改将直接影响原始曲面。还有其他函数允许您以数组的形式访问任何每像素的Alpha值，以及一些其他有用的函数。我们稍后将讨论这些其他函数。
NumPy模块使用机器的自然数类型来表示数据值，因此NumPy数组可以由8位、16位和32位的整数组成。 (数组还可以使用其他类型，如浮点数和双精度数，但对于我们的图像操作，我们主要需要担心整数类型) 。由于整数大小的限制，您必须格外小心，以确保引用像素数据的数组类型可以正确映射到适当类型的数据。从曲面创建这些数组的函数为：
surfarray.pixels2d(surface)
surfarray.array2d(surface)
surfarray.pixels3d(surface)
surfarray.array3d(surface)
32位
24位
16位
8位(c-map)
### 示例¶
```allblack = N.zeros((128, 128))
surfdemo_show(allblack, 'allblack')
```
```striped = N.zeros((128, 128, 3))
striped[:] = (255, 0, 0)
striped[:,::3] = (0, 255, 255)
surfdemo_show(striped, 'striped')
```
```imgsurface = pygame.image.load('surfarray.png')
rgbarray = surfarray.array3d(imgsurface)
surfdemo_show(rgbarray, 'rgbarray')
```
```flipped = rgbarray[:,::-1]
surfdemo_show(flipped, 'flipped')
```
```scaledown = rgbarray[::2,::2]
surfdemo_show(scaledown, 'scaledown')
```
```shape = rgbarray.shape
scaleup = N.zeros((shape[0]*2, shape[1]*2, shape[2]))
scaleup[::2,::2,:] = rgbarray
scaleup[1::2,::2,:] = rgbarray
scaleup[:,1::2] = scaleup[:,::2]
surfdemo_show(scaleup, 'scaleup')
```
```redimg = N.array(rgbarray)
redimg[:,:,1:] = 0
surfdemo_show(redimg, 'redimg')
```
```factor = N.array((8,), N.int32)
soften = N.array(rgbarray, N.int32)
soften[1:,:] += rgbarray[:-1,:] * factor
soften[:-1,:] += rgbarray[1:,:] * factor
soften[:,1:] += rgbarray[:,:-1] * factor
soften[:,:-1] += rgbarray[:,1:] * factor
soften //= 33
surfdemo_show(soften, 'soften')
```
```src = N.array(rgbarray)
dest = N.zeros(rgbarray.shape)
dest[:] = 20, 50, 100
diff = (dest - src) * 0.50
xfade = src + diff.astype(N.uint)
```
### 透明度¶
SURFARY模块有几种访问Surface的Alpha/Colorkey值的方法。所有Alpha函数都不受曲面整体透明度的影响，只受像素Alpha值的影响。以下是这些函数的列表。
surfarray.pixels_alpha(surface)
surfarray.array_alpha(surface)
surfarray.array_colorkey(surface)
### 其他SurfARRAY函数¶
SURFARY中只有几个其他功能可用。您可以获得一个更好的列表，其中包含更多关于 `surfarray reference page` 。不过，有一个非常有用的功能。
surfarray.blit_array(surface, array)
Edit on GitHub
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# The Great Riemann Hypothesis
| October 20, 2012 |
In 1637 the great king of amateur mathematicians — Pierre de Fermat — wrote his most famous theorem in the margin of his copy of Arithmetica, which stated that that no three positive integers ab, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. But an even more interesting was his little note in the margin, which read that the proof was too big to fit into it. What happened later is that this apparently simple problem became known as the most difficult problem of mathematics. Greatest mathematicians tried to solve Fermat’s last theorem for 358 years, including Euler, Hilbert, Kronecker and others. But it was finally solved in 1995 by the great Andrew Wiles.
This remarkable story ended more than 350 years of reign of Fermat’s last theorem as the king of the most difficult math problems. But don’t worry, as long as there are mathematicians there will be unsolved problems, and, actually, there is a number of important problems, which belong to the list of the so called Millennium Prize Problems. The list holds 7 problems (1 of which is already solved), the award for which is 1 million dollars. Riemann hypothesis is arguably the most important one, or at least the most famous one. And it is a problem proposed by Bernhard Riemann (1859) about the location of the nontrivial zeros of the Riemann zeta function which states that all non-trivial zeros have real part 1/2.
So let’s take a closer look at what the Riemann hypothesis is all about. Here are two videos, the first one is a nice presentation of the history behind it, whereas the second one is a great lecture by Dan Rockmore.
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# Checking Solutions To Systems Of Equations Worksheet
Checking Solutions To Systems Of Equations WorksheetAlgebra can be introduced to your youngster with making use of equations worksheets. Aid your child strengthen their basic mathematics skills as well as prepare them for the following degree of mathematics by addressing formulas with 2 variables with these worksheets! Several of the most typical sorts of equations are shown below. All of them are straightforward and also enjoyable for you and your kid! Here are a few of one of the most well-known.
You should acquire an enhancement within ten worksheet as your first purchase. This worksheet is excellent for showing your child concerning equality issues as well as word issues. For each trouble, you can select the sort of trouble and also the series of numbers to make use of. The unknown in each problem will certainly exist in 4 different means. Plus, these problems are simple enough for trainees in Kindergarten, first and also 2nd qualities to comprehend.
If you’re instructing your kid just how to solve one-variable formulas, a worksheet similar to this is an excellent selection. The four basic operations are well-taught with the help of these worksheets. On top of that, they assist in your child’s proficiency of the reproduction tables. Additionally, your kid will come to be extra skilled at solving formulas as a result of working through these worksheets. Isolating variables is one of the most effective method for understanding an algebraic expression. The initial step is to isolate the variables, as in the example above, where x +4 =32.
Utilizing another one-step formulas worksheet is a great way to exercise resolving the one-step formulas. There are just 2 variables in this kind of formula, making it simple to fix. They could also be used to analyze the students’ mastery of the this ability via early morning workouts, math terminals, or review sessions. These worksheets can be made use of for enhancement within ten in addition to balancing equations. Checking Solutions To Systems Of Equations Worksheet
Kids learn one-variable equations with a single variable with the help of this worksheet for youngsters. These worksheets are readily available online. Downloading and install and printing it at home could not be much easier. Trainees who deal with the formulas might benefit substantially from utilizing the visual simulations. The one-variable formulas worksheet for children is also readily available for download. Kids will certainly benefit greatly from this, as there are 4 unique types of questions to be addressed below. Checking Solutions To Systems Of Equations Worksheet
Algebraic expressions are the subject of the next worksheet. This worksheet is designed to instruct your youngster exactly how to fix problems involving variables. Word problems entailing the variable x will be consisted of in these worksheets. Contrasted to the other sorts of formulas, these are much easier to recognize and understand. A strong math structure is all that is called for to address these problems. You could intend to take a look at this worksheet if you’re trying to find something for kids. Checking Solutions To Systems Of Equations Worksheet
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Question
# Mosses invested Rs. 20,000 in a scheme at simple interest @ 15% per annum. After three years he withdrew the principal amount plus interest and invested the entire amount in another scheme for two years, which earned him compound interest @ 12% per annum. What would be the total interest earned by Mosses at the end of 5 years?
Solution
## S.I= $$\frac{20,000\times 15\times 3}{100}=9000$$Amount = Rs 20,000+ Rs 9000= Rs 29,000Now, CI= $$29,000 (1+\frac{12}{100})^{2}=29,000\times \frac{28}{25}\times \frac{28}{29}$$$$= 36,377.6$$A-P= 36,377.6-29,000= 7377.6After 5 years = $$7377.6+9000=16,377.6$$Mathematics
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MAY 24, 2007
Algebra Tiles Explained: What Do They Mean Exactly?
BY CAMILA TORRES-BERTRAND
Many people believe that algebra tiles are new in mathematics instruction, however, algebra tiles have been used in classrooms since the mid 1980s. Since then, algebra tiles have gained popularity in teaching circles and are regularly presented in conjunction with traditional methods in most textbooks. Unlike a calculator or an answer key, which does the thinking for you, algebra tiles are a tool that can guide the learner towards a greater understanding of the concepts involved with a particular skill.
Studies show that over 80% of students are visual learners rather than auditory learners. Therefore, it makes sense that the majority of students benefit from a visual, concrete representation of abstract algebraic concepts. Algebra tiles do just this ?they allow students to match a concept to a tile versus trying to imagine everything in their minds. (For more information visit our conceptual math techniques page.)
Area: How can students use what they already know?
Algebra Tiles are created to allow students to view symbolic representations through concrete models while applying previously explored concepts of area.
By the time students are introduced to algebra, they should have already explored the concepts of area and perimeter. However, assuming as student has not solidified their understanding of area and perimeter, consider showing them this figure:
figure 1a.
When asked “What is the difference between area and perimeter?? students unfamiliar with the concepts will give responses such as “The perimeter is the outside of the rectangle, the area is the stuff inside the rectangle.? This response or similar responses provide teaching opportunities to review area/perimeter concepts and smoothly transition to algebraic concepts.
Consider this diagram which depicts figure 1a recreated with units inside. (Note: the units are square because area measure space in two dimensions: lw.) The measurement of this rectangle is now 5 units (l) by 3 units (w).
figure 1b.
Instead of simply giving them the formula of the perimeter of a rectangle P = 2l + 2w, I encouraged exploration so they could apply the definition to a variety of shapes in the future. To explore perimeter, students would simply count the sides that have a black border: 5 sides across the top, 3 sides down the right, 5 sides across the bottom and 3 sides up the left. After counting carefully, students will determine that perimeter of the figure is 16 units. Make sure that students know and understand why perimeter is a single dimensional unit ?length (l). The space it is measuring is only being measured in one direction. (For younger students, useful images and tools for measuring one-dimensional space is measuring distances using a string or toothpicks.)
As for area, it is a very different concept. This time, both length and width is being measured. (Perimeter only measured length.) Again, to encourage student thinking, I never gave them the formula for area (lw), rather I allowed the students to explore area for themselves. We ask them to count up the squares in figure 1b. and they quickly determine that there are 15 squares in the figure. If given different rectangles of varying dimensions and units, they will create the formula for themselves. Teachers may further coach the students to discover the concepts by asking them to write or journal about relationships or shortcuts they discovered within the lesson. When working with area, make sure students know and understand why area is a two-dimensional unit ?length (l) and width (w). The space area measures in being measured in two directions. Answers should be given in squared units (i.e. square feet, feet2 , square miles, miles2 , square meters, meters2 , etc.)
It has been a concern that these methodologies are time-consuming and tedious; however they show a tremendous opportunity for students to gain ownership of their own learning. Exploration can solidify this learning in a more meaningful way because they have discovered these relationships on their own. Furthermore, once students start to integrate these concepts volume (measurement in three dimensions) is innate. Teachers may further coach the students to discover the concepts by asking them to write or journal about relationships or shortcuts they discovered within the lesson. As for the formulas for perimeter and area, perhaps one of the most powerful discoveries for some students will ever discover is that multiplication is a shortcut for addition and addition is a shortcut for counting. It refutes the argument that math gets “harder?as you get progress. Rather, in this situation, multiplication (third/fourth grade) made a counting problem (kindergarten/first grade math) easier. Believe it or not, these shortcuts continue all the way into calculus and beyond!
The following is an introduction to the different types of algebra tiles in a typical algebra tile set.
The Ones
The “ones?piece is a small square. Typically, one side is yellow and the reverse is red. The dimensions of the square are one (1) unit by one (1) unit. Therefore using the area model:
POSITIVE ONE: The positive one tile is represented by a small, yellow square. The dimensions of the square are 1 unit by 1 unit so therefore the area model would show that the small, yellow square has an area of 1 square unit. In addition to reinforcing the ideas behind the area model of a four sided figure, the notion of when a number is “squared?it actually can be represented as the shape of a square.
NEGATIVE ONE: Students need to understand the concept of negative being the opposite. When using algebra tiles, a negative one is represented by the same tile as a positive one. However, the negative is denoted by the red color of the same small square piece. When the yellow square is flipped over to its opposite side, the color is red which represents negative.
The X s
The ?x ?piece is a rectangle. Typically one side is green and the reverse is red. The orientation of the piece in space, vertical or horizontal does not affect the significance of piece. The dimensions of the rectangle are one (1) unit by x unit(s).
Therefore using the area model:
POSITIVE X: The positive tile is represented by a green rectangle. The dimensions of the rectangle are 1 unit by an unknown length of the same unit. Since the sides do not have the same length, a rectangle is formed instead of a square.
If one were to line up the one tiles against the long side of x tile, if would not line up perfectly because the length is unknown. It is important for students to understand that there is no specific length associated with the unknown length side. That’s the point, it can be anything. However, if a one tile was lined up against the shorter side of the rectangle, it would line up perfectly because it shares the same dimensions ?one unit.
NEGATIVE X: The negative x is represented by the same tile, however just as for the ones tiles, the red color denotes the negative. When the green (positive) tile is flipped to its opposite side and displays the color red (negative) because the opposite of positive x is negative x.
The X2
The x2 piece is a large square. Typically one side is blue and the reverse is red. The dimensions of the square are x unit(s) by unit(s).
POSITIVE X2: The positive x2 tile is represented by a large blue square. The dimensions of the blue square are one unknown length by the same unknown length with similar units
If one were to line up ones tiles against a side of the x2 tile, it would not line up perfectly because the length is supposed to be unknown. It is important for students to understand that there is no specific length associated with the unknown length side. However, if long side of the green tile was lined up against either side of the blue tile, it would line up perfectly because it shares the same unknown length. In other words, the unknown length of the blue square is the same length as the unknown length of the long side of the green rectangle.
NEGATIVE X2: The negative x2 tile is represented by the same blue square, but just as in all the previous examples, the red color denotes the negative. When the blue (positive) tile is flipped, the tile becomes red (negative) because the opposite of positive x2 is negative x2
The colors used in our explanations of algebra tiles are most commonly used in textbooks and manipulative sets sold commercially. It is important to note that previous and older versions of algebra manipulatives denoted the negative side with the color black. The positive side was usually represented using a variety of bright colors like red, yellow, green and blue. We at www.aplusalgebra.com have used the most widely-used color scheme for the A+ Algebra TabletTM used to create our lessons.
Want your own set of tiles to practice? Download, print and cut out your set for FREE!
(Helpful Hint: For maximum learning, use paper that has a different color on each side.)
Want something a little more “high tech? The A+ Algebra TabletTM brings algebra manipulatives to the 21st Century by allowing users to use the tiles on a desktop computer, laptop computer, tablet PC or interactive whiteboard.
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# Aerodynamics Questions and Answers – Modern Low Speed Airfoils – 2
This set of Aerodynamics Question Bank focuses on “Modern Low Speed Airfoils – 2”.
1. Is NASA designed low speed airfoils?
a) True
b) False
View Answer
Answer: a
Explanation: During 1970s, NASA designed a series of low-speed airfoils that have performance superior to the earlier NACA airfoils. The standard NACA airfoils were based almost exclusively on experimental data obtained during 1930s and 1940s.
2. Is NASA airfoil designed using numerical technique?
a) False
b) True
View Answer
Answer: b
Explanation: The new NASA airfoils were designed on a computer using a numerical technique similar to the source and vortex panel methods discussed earlier, along with numerical predictions of the viscous flow behavior.
3. Is wind tunnel test is conducted to verify the computer designed profiles?
a) True
b) False
View Answer
Answer: a
Explanation: Wind tunnel tests were then conducted to verify the computer designed profiles and to obtain the definitive airfoil properties, out of this work first came the general aviation whit comb airfoil. Which has since been predesignated the LS-0417 airfoil.
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4. What is the leading edge radius of LS-0417 airfoil?
a) 0.07c
b) 0.05c
c) 0.08c
d) 0.09c
View Answer
Answer: c
Explanation: LS-0417 airfoil has a leading edge radius of 0.08c in comparison to the standard 0.02c in order to flatten the usual peak in pressure coefficient near the nose, also that the bottom surface near the trailing edge is cusped in order to increase the camber.
5. Is flow separation over the top surface of the airfoil at high angle of attack?
a) True
b) False
View Answer
Answer: a
Explanation: The flow separation over the top surface at a high angle of attack, hence yielding higher values of the maximum lift coefficient. The lift and moment properties are compared with the NAC 2412 airfoil.
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6. Is 17% is the thickness of NASA LS-0417 airfoil?
a) True
b) False
View Answer
Answer: a
Explanation: The NASA LS- 0417 airfoil has a maximum thickness of 17% and a design lift coefficient of 0.4. Using the same camber line, NASA has extended this airfoil into a family of a low-speed airfoil of different thickness for example LS-0409 and the LS-0413.
7. Is a 50% increase in the ratio of lift to drag at a lift coefficient?
a) True
b) False
View Answer
Answer: a
Explanation: A 50 percent increase in the ratio of lift to drag at a lift coefficient of 0.1. This value of q=0.1 is typical of the climb lift coefficient for general aviation aircraft, and a high value of L/D greatly improves the climb.
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8. Is super critical airfoil used to improve the drag at subsonic speed?
a) True
b) False
View Answer
Answer: a
Explanation: The super critical airfoil was a major breakthrough in high speed aerodynamics, the super critical airfoil is used to improve the drag at subsonic speeds. The LS-0417 low speed airfoil first introduced as the GA-1 airfoil.
9. Is initial distribution is given by solid curves?
a) True
b) False
View Answer
Answer: a
Explanation: The optimization technique is iterative and requires starting with a pressure distribution that is not the desired, specified one, the initial distribution is given by the solid curves and the airfoil shapes appears distorted because an expanded scale is used for the ordinate.
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10. Is pressure coefficient distributions are calculated for flow over an airfoil?
a) True
b) False
View Answer
Answer: a
Explanation: The unstructured mesh for the numerical calculation of the flow over an airfoil to calculate for pressure coefficient distributions, airfoil shapes that support the specified pressure distribution is obtained, as given by the circle. The initial airfoil shape is also shown in constant scale.
Sanfoundry Global Education & Learning Series – Aerodynamics.
To practice Aerodynamics Question Bank, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 4.10: Dirichlet problem in the circle and the Poisson kernel
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#### 4.10.1 Laplace in polar coordinates
A more natural setting for the Laplace equation $$\Delta u=0$$ is the circle rather than the square. On the other hand, what makes the problem somewhat more difficult is that we need polar coordinates.
Recall that the polar coordinates for the $$(x,y)$$-plane are $$(r, \theta )$$:
$x= r \cos \theta, ~~~~ y= r \sin \theta,$
where $$r \geq 0$$ and $$- \pi < \theta < \pi$$. So $$(x,y)$$ is distance $$r$$ from the origin at angle $$\theta$$.
Now that we know our coordinates, let us give the problem we wish to solve. We have a circular region of radius 1, and we are interested in the Dirichlet problem for the Laplace equation for this region. Let $$u(r, \theta )$$ denote the temperature at the point $$(r, \theta )$$ in polar coordinates. We have the problem:
$\Delta u=0,~~~~~ {\rm{for~}} r<1, \\ u(1, \theta )=g( \theta),~~~{\rm{for~}} \pi < \theta \leq \pi .$
The first issue we face is that we do not know what the Laplacian is in polar coordinates. Normally we would find $$u_{xx}$$ and $$u_{yy}$$ in terms of the derivatives in $$r$$ and $$\theta$$. We would need to solve for and in terms of and . While this is certainly possible, it happens to be more convenient to work in reverse. Let us instead compute derivatives in $$r$$ and $$\theta$$ in terms of derivatives in $$x$$ and $$y$$ and then solve. The computations are easier this way. First
$x_r= \cos \theta, ~~~~x_{\theta}= -r \sin \theta,~~~~y_{r}= \sin \theta,~~~~y_{\theta}= r \cos \theta .$
Next by chain rule we obtain
$u_r=u_xx_{r}+u_yy_{r}= \cos(\theta)u_x + \sin(\theta)u_y, \\ u_{rr}= \cos(\theta)(u_{xx}x_r+u_{xy}y_r)+ \sin(\theta)(u_{yx}x_r+u_{yy}y_r) = \cos^2(\theta)u_{xx}+2 \cos(\theta) \sin(\theta)u_{xy}+ \sin^2(\theta)u_{yy}.$
Similarly for the derivative. Note that we have to use product rule for the second derivative.
$u_{\theta}=u_xx_{\theta}+u_yy_{\theta}= -r \sin(\theta)u_x + r \cos(\theta)u_y, \\ u_{\theta \theta}= -r \cos(\theta)(u_x)- r \sin(\theta)(u_{xx}x_{\theta}+u_{xy}y_{\theta})-r \sin(\theta)(u_y)+r \cos(\theta)(u_{yx}x_{\theta}+u_{yy}y_{\theta}) \\ = -r \cos(\theta)u_{x}-r \sin(\theta)u_y+r^2 \sin^2(\theta)u_{xx}-r^2 2 \sin(\theta) \cos(\theta) u_{xy}+r^2 \cos^2(\theta)u_{yy}.$
Let us now try to solve for $$u_{xx}+u_{yy}$$. We start with $$\frac{1}{r^2}u_{\theta \theta}$$ to get rid of those pesky $$r^2$$. If we add $$u_{rr}$$ and use the fact that $$\cos^2(\theta)+ \sin^2(\theta)=1$$, we get
$\frac{1}{r^2}u_{\theta \theta} +u_{rr}=u_{xx}+u_{yy}- \frac{1}{r} \cos(\theta)u_x- \frac{1}{r} \sin(\theta)u_y.$
We’re not quite there yet, but all we are lacking is $$\frac{1}{r}u_r$$. Adding it we obtain the Laplacian in polar coordinates:
$\frac{1}{r^2}u_{\theta \theta}+ \frac{1}{r}u_{r}+u_{rr}=u_{xx}+u_{yy}= \Delta u.$
Notice that the Laplacian in polar coordinates no longer has constant coefficients.
#### 4.10.2 Series solution
Let us separate variables as usual. That is let us try $$u(r, \theta)=R(r) \Theta ( \theta)$$. Then
$0= \Delta u=\frac{1}{r^2}R \Theta''+ \frac{1}{r}R' \Theta+R'' \Theta .$
Let us put $$R$$ on one side and $$\Theta$$ on the other and conclude that both sides must be constant.
$\frac{1}{r^2}R \Theta''= - \left( \frac{1}{r}R'+R'' \right) \Theta . \\ \frac{ \Theta''}{ \Theta}= - \frac{rR'+r^2R''}{R} + - \lambda.$
We get two equations:
$\Theta''+\lambda \Theta=0, \\ r^2R''+rR'- \lambda R=0.$
Let us first focus on $$\Theta$$. We know that $$u(r, \theta)$$ ought to be $$2 \pi$$-periodic in $$\theta$$, that is, $$u(r, \theta)=u(r, \theta +2 \pi)$$. Therefore, the solution to $$\Theta''+\lambda \Theta=0$$ must be $$2 \pi$$-periodic. We conclude that $$\lambda=n^2$$ for a nonnegative integer $$n=0,1,2,3,...$$. The equation becomes $$\Theta''+n^2 \Theta=0$$. When $$n=0$$ the equation is just $$\Theta''=0$$, so we have the general solution $$A \theta+B$$. As $$\Theta$$ is periodic, $$A=0$$. For convenience let us write this solution as
$\Theta_0=\frac{a_0}{2}$
for some constant $$a_0$$. For positive $$n$$, the solution to $$\Theta''+n^2 \Theta=0$$ is
$\Theta_n=a_n \cos(n \theta)+b_n \sin(n \theta),$
for some constants $$a_n$$ and $$b_n$$.
Next, we consider the equation for $$R$$,
$r^2R''+rR'-n^2R=0.$
This equation has appeared in exercises before—we solved it in Exercise 2.1.6 and Exercise 2.1.7. The idea is to try a solution $$r^s$$ and if that does not work out try a solution of the form $$r^s \ln r$$. When $$n=0$$ we obtain
$R_0=Ar^0+Br^0 \ln r=A+B \ln r,$
and if $$n>0$$, we get
$R_n=Ar^n+Br^{-n}.$
The function $$u(r, \theta)$$ must be finite at the origin, that is, when $$r=0$$. Therefore, $$B=0$$ in both cases. Let us set $$A=1$$ in both cases as well, the constants in $$\Theta_n$$ will pick up the slack so we do not lose anything. Therefore let
$R_0=1,~~~~~~{\rm{and}}~~~~~~R_n=r^n.$
Hence our building block solutions are
$u_0(r, \theta)= \frac{a_0}{2},~~~~~~~~~~~~~~u_n(r, \theta)=a_n r^n \cos(n \theta)+b_n r^n \sin(n \theta).$
Putting everything together our solution is:
$u(r, \theta)= \frac{a_0}{2}+ \sum_{n=1}^{\infty}a_n r^n \cos(n \theta)+b_n r^n \sin(n \theta).$
We look at the boundary condition in (4.1.2),
$g(\theta)=u(1, \theta)= \frac{a_0}{2}+ \sum_{n=1}^{\infty}a_n \cos(n \theta)+b_n \sin(n \theta).$
Therefore, the solution to (4.1.2) is to expand $$g(\theta)$$, which is a $$2 \pi$$-periodic function as a Fourier series, and then the coordinate is multiplied by $$r^n$$. In other words, to compute $$a^n$$ and $$b^n$$ from the formula we can, as usual, compute
$a_n= \frac{1}{ \pi} \int_{ \pi}^{- \pi} g(\theta) \cos(n \theta)d \theta, ~~~~~~{\rm{and}}~~~~~~ b_n= \frac{1}{ \pi} \int_{ \pi}^{- \pi} g(\theta) \sin(n \theta)d \theta.$
Example $$\PageIndex{1}$$:
Suppose we wish to solve
$\Delta u=0,~~~~~0 \leq r<1,~~~~~ - \pi< \theta< \pi, \\ u(1, \theta )= \cos(10 \theta),~~~~~ - \pi< \theta< \pi .$
The solution is
$u(r, \theta)=r^{10} \cos(10 \theta).$
See the plot in Figure 4.23. The thing to notice in this example is that the effect of a high frequency is mostly felt at the boundary. In the middle of the disc, the solution is very close to zero. That is because $$r^{10}$$ rather small when $$r$$ is close to $$0$$.
Figure 4.23: The solution of the Dirichlet problem in the disc with $$\cos(10 \theta)$$ as boundary data.
Example $$\PageIndex{2}$$:
Let us solve a more difficult problem. Suppose we have a long rod with circular cross section of radius $$1$$ and we wish to solve the steady state heat problem. If the rod is long enough we simply need to solve the Laplace equation in two dimensions. Let us put the center of the rod at the origin and we have exactly the region we are currently studying—a circle of radius $$1$$. For the boundary conditions, suppose in Cartesian coordinates $$x$$ and $$y$$ , the temperature is fixed at $$0$$ when $$y<0$$ and at $$2y$$ when $$y>0$$.
We set the problem up. As $$y=r \sin(\theta)$$, then on the circle of radius $$1$$ we have $$2y=2 \sin(\theta)$$. So
$\Delta u=0,~~~~~0 \leq r<1,~~~~~ - \pi< \theta \leq \pi, \\ u(1, \theta )= \left\{ \begin{array}{cc} 2 \sin(\theta) & {\rm{~if~}} 0 \leq \theta \leq \pi, \\ 0 & ~~~~{\rm{~if~}} - \pi < \theta < 0. \end{array} \right. .$
We must now compute the Fourier series for the boundary condition. By now the reader has plentiful experience in computing Fourier series and so we simply state that
$u(1, \theta)= \frac{2}{\pi}+ \sin(\theta)+ \sum_{n=1}^{\infty} \frac{-4}{\pi (4n^2-1)} \cos(2n \theta).$
Exercise $$\PageIndex{1}$$:
Compute the series for $$u(1, \theta)$$ and verify that it really is what we have just claimed. Hint: Be careful, make sure not to divide by zero.
We now simply write the solution (see Figure 4.24 ) by multiplying by $$r^n$$ in the right places.
$u(r, \theta)= \frac{2}{\pi}+ r\sin(\theta)+ \sum_{n=1}^{\infty} \frac{-4r^{2n}}{\pi (4n^2-1)} \cos(2n \theta).$
Figure 4.24: The solution of the Dirichlet problem with boundary data $$0$$ for $$y<0$$ and $$2y$$ for $$y>0$$.
#### 4.10.3 Poisson Kernel
There is another way to solve the Dirichlet problem with the help of an integral kernel. That is, we will find a function $$P(r,\theta,\alpha)$$ called the Poisson kernel7 such that
$u(r,\theta)= \frac{1}{2\pi} \int_{-\pi}^{\pi}P(r,\theta,\alpha)g(\alpha)d\alpha.$
While the integral will generally not be solvable analytically, it can be evaluated numerically. In fact, unless the boundary data is given as a Fourier series already, it will be much easier to numerically evaluate this formula as there is only one integral to evaluate.
The formula also has theoretical applications. For instance, as $$P(r,\theta,\alpha)$$ will have infinitely many derivatives, then via differentiating under the integral we find that the solution $$u(r,\theta)$$ has infinitely many derivatives, at least when inside the circle, $$r<1$$. By infinitely many derivatives what you should think of is that $$u(r,\theta)$$ has “no corners” and all of its partial derivatives exist too and also have “no corners”.
We will compute the formula for $$P(r,\theta,\alpha)$$ from the series solution, and this idea can be applied anytime you have a convenient series solution where the coefficients are obtained via integration. Hence you can apply this reasoning to obtain such integral kernels for other equations, such as the heat equation. The computation is long and tedious, but not overly difficult. Since the ideas are often applied in similar contexts, it is good to understand how this computation works.
What we do is start with the series solution and replace the coefficients with the integrals that compute them. Then we try to write everything as a single integral. We must use a different dummy variable for the integration and hence we use $$\alpha$$ instead of $$\theta$$.
$u(r, \theta)= \frac{a_0}{2}+ \sum_{n=1}^{\infty}a_n r^n \cos(n \theta)+b_n r^n \sin(n \theta) \\ =\frac{1}{2\pi} \int_{-\pi}^{\pi}g(\alpha)d\alpha \\ + \sum_{n=1}^{\infty} \left( \frac{1}{\pi} \int_{-\pi}^{\pi}g(\alpha)\cos(n\alpha)d\alpha \right)r^n\cos(n\theta)+ \left( \frac{1}{\pi} \int_{-\pi}^{\pi}g(\alpha)\sin(n\alpha)d\alpha \right)r^n\sin(n\theta) \\ = \frac{1}{2\pi} \int_{-\pi}^{\pi}\left( g(\alpha)+2\sum_{n=1}^{\infty}g(\alpha)\cos(n\alpha)r^n\cos(n\theta)+ g(\alpha)\sin(n\alpha)r^n\sin(n\theta) \right)d\alpha \\ =\frac{1}{2\pi} \int_{-\pi}^{\pi}\left( 1+2\sum_{n=1}^{\infty}r^n(\cos(n\alpha)\cos(n\theta)+ \sin(n\alpha)\sin(n\theta)) \right)g(\alpha)d\alpha$
OK, so we have what we wanted, the expression in the parentheses is the Poisson kernel, $$P(r,\theta,\alpha)$$. However, we can do a lot better. It is still given as a series, and we would really like to have a nice simple expression for it. We must work a little harder. The trick is to rewrite everything in terms of complex exponentials. Let us work just on the kernel.
$P(r,\theta,\alpha)=1+2\sum_{n=1}^{\infty}r^n(\cos(n\alpha)\cos(n\theta)+ \sin(n\alpha)\sin(n\theta)) \\ = 1+2\sum_{n=1}^{\infty}r^n \cos(n(\theta- \alpha)) \\ 1+2\sum_{n=1}^{\infty}r^n(e^{in(\theta-\alpha)}+e^{-in(\theta-\alpha)}) \\ 1+\sum_{n=1}^{\infty}(re^{i(\theta-\alpha)})^n+\sum_{n=1}^{\infty}(re^{-i(\theta-\alpha)})^n.$
In the above expression we recognize the geometric series. That is, recall from calculus that as long as $$|z|<1$$, then
$\sum_{n=1}^{\infty}z^n= \frac{z}{1-z}.$
Note that $$n$$ starts at $$1$$ and that is why we have the $$z$$ in the numerator. It is the standard geometric series multiplied by $$z$$. Let us continue with the computation.
$P(r,\theta,\alpha)=1+\sum_{n=1}^{\infty}(re^{i(\theta-\alpha)})^n+\sum_{n=1}^{\infty}(re^{-i(\theta-\alpha)})^n \\ = 1+ \frac{re^{i(\theta-\alpha)}}{1-re^{i(\theta-\alpha)}}+ \frac{re^{-i(\theta-\alpha)}}{1-re^{-i(\theta-\alpha)}} \\ =\frac{(1-re^{i(\theta-\alpha)})(1-re^{-i(\theta-\alpha)})+(1-re^{-i(\theta-\alpha)})re^{ i(\theta-\alpha)}+(1-re^{i(\theta-\alpha)})re^{ - i(\theta-\alpha)}}{(1-re^{i(\theta-\alpha)})(1-re^{-i(\theta-\alpha)})} \\ = \frac{1-r^2}{1-re^{i(\theta-\alpha)}-re^{-i(\theta-\alpha)}+r^2} \\= \frac{1-r^2}{1-2r\cos(\theta-\alpha)+r^2}.$
Now that’s a formula we can live with. The solution to the Dirichlet problem using the Poisson kernel is
$u(r, \theta)= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1-r^2}{1-2r\cos(\theta-\alpha)+r^2}g(\alpha)d\alpha.$
Sometimes the formula for the Poisson kernel is given together with the constant $$\frac{1}{2\pi}$$, in which case we should of course not leave it in front of the integral. Also, often the limits of the integral are given as $$0$$ to $$2\pi$$; everything inside is $$2\pi$$-periodic in $$\alpha$$, so this does not change the integral.
Let us not leave the Poisson kernel without explaining its geometric meaning. Let be the distance from $$(r,\theta)$$ to $$(1,\alpha)$$. You may recall from calculus that this distance $$s$$ in polar coordinates is given precisely by the square root of $$1-2r\cos(\theta-\alpha)+r^2$$. That is, the Poisson kernel is really the formula
$\frac{1-r^2}{s^2}.$
One final note we make about the formula is to note that it is really a weighted average of the boundary values. First let us look at what happens at the origin, that is when $$r=0$$.
$u(0,0)=\\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1-0^2}{1-2(0)\cos(\theta-\alpha)+0^2}g(\alpha)d\alpha \\ = \frac{1}{2\pi} \int_{-\pi}^{\pi} g(\alpha)d\alpha.$
So $$u(0,0)$$ is precisely the average value of $$g(\theta)$$ and therefore the average value of $$u$$ on the boundary. This is a general feature of harmonic functions, the value at some point $$p$$ is equal to the average of the values on a circle centered at $$p$$.
What the formula says is that the value of the solution at any point in the circle is a weighted average of the boundary data $$g(\theta)$$. The kernel is bigger when $$(r,\theta)$$ is closer to $$(1,\alpha)$$. Therefore when computing $$u(r,\theta)$$ we give more weight to the values $$g(\alpha)$$ when $$(1,\alpha)$$ is closer to $$(r,\theta)$$ and less weight to the values $$g(\theta)$$ when $$(1,\alpha)$$ far from $$(r,\theta)$$.
7Named for the French mathematician Siméon Denis Poisson (1781 – 1840).
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# RRB Group D Question Paper 29 October 2018 1st Shift
Railway Recruitment Board RRB conducting RRB Group D in the month of September 2018 from September 9th to October 31st, 2018 and each day RRB conducting three slots. Candidates can download the Railways RRB Group D Slot Wise Questions Asked in the prelims examination. As per RRB Group D exam pattern, the RRB Group D Exam consists of 100 questions in English, Hindi, and other languages, with these 75 questions candidates can complete the exam within 100 minutes (1 and half hour) and PWD candidate has added additional minutes as per reservation.
## RRB Group D 29 October 2018 Examination
• Number of Questions in RRB Group D Exam: 100 Questions.
• Time Allotted for RRB AP Exam: 90 minutes
• Types of Questions: Multiple Choice Questions (MCQ).
• Negative Marking: 1/3rd for every incorrect answer.
Section No.of Questions Mathematics 25 GI & Reasoning 30 General Science 25 General Awareness and Current Affairs 20
RRB GROUP D ALL QUESTION PAPERS
Important for RRB Group D Candidates
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#### RRB Group D Question Paper 29 October 2018 1 shift
• Director General of NCC Lt Gen P P Malhotra
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• Calcium Carbonate Formula CACO3
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• #### RRB Group D Exam Analysis 29 October 2018 Shift 1 Mathematics
• 3: 4 = 6: X =?
• 1,5,13,29 __
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# RRB Group D Question Paper 29 October 2018 1st in Hindi
Railway Recruitment Board RRB conducting RRB Group D inline examinations from September 2018 to December 2018 in various locations in India, those who already appeared RRB Group D examination, you have to check the cutoff marks for selections, which has be listed below and those who are not given RRB Group D exam and waiting for exam date, they have to check regular RRB Group D questions, which are updating in this blog regularly.
RRB Group D Cut off Marks 2018
Those who candidates already gave the RRB Group D online exam have to check the RRB Group D Cut off marks 2018, the cutoff marks are given based on RRB Vacancies and a number of candidates appeared examination and these RRB Group D cut off marks may vary little with RRB Group D original cut off marks.
Name of the Reserve Category Expected RRB Group D Cut off Marks (100 Marks) OC Reserve Candidates 70 to 85 Marks OBC Reserve Candidates 65-75 Marks SC Reserve Candidates 55 to 63 Marks ST Reserve Candidates 51 to 58 Marks
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Gold 41.5 stone to troy ounces of gold converter
Category: Precious-metals main menugold menuStones
# gold conversion
## Amount: 41.5 stones (st) of gold mass Equals: 8,472.92 troy ounces (oz t) in gold mass
Calculate troy ounces of gold per 41.5 stones unit. The gold converter.
TOGGLE : from troy ounces into stones in the other way around.
### Enter a New stone Amount of gold to Convert From
* Enter whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)
Enter Your Amount :
## gold from stone to ounce (troy) Conversion Results :
Amount : 41.5 stones (st) of gold
Equals: 8,472.92 troy ounces (oz t) in gold
Fractions: 8472 11/12 troy ounces (oz t) in gold
CONVERT : between other gold measuring units - complete list.
## Solid Pure 24k Gold Amounts
This calculator tool is based on the pure 24K gold, with Density: 19.282 g/cm3 calculated (24 karat gold grade, finest quality raw and solid gold volume; from native gold, the type we invest -in commodity markets, by trading in forex platform and in commodity future trading. Both the troy and the avoirdupois ounce units are listed under the gold metal main menu. I advice learning from a commodity trading school first. Then buy and sell.) Gold can be found listed either in table among noble metals or with precious metals.
Is it possible to manage numerous calculations for how heavy are other gold volumes all on one page? Yes, all in one Au multiunit calculator makes it possible managing just that.
Convert gold measuring units between stone (st) and troy ounces (oz t) of gold but in the other direction from troy ounces into stones.
conversion result for gold: From Symbol Equals Result To Symbol 1 stone st = 204.17 troy ounces oz t
# Precious metals: gold conversion
This online gold from st into oz t (precious metal) converter is a handy tool not just for certified or experienced professionals. It can help when selling scrap metals for recycling.
## Other applications of this gold calculator are ...
With the above mentioned units calculating service it provides, this gold converter proved to be useful also as a teaching tool:
1. in practicing stones and troy ounces ( st vs. oz t ) exchange.
2. for conversion factors training exercises with converting mass/weights units vs. liquid/fluid volume units measures.
3. work with gold's density values including other physical properties this metal has.
International unit symbols for these two gold measurements are:
Abbreviation or prefix ( abbr. short brevis ), unit symbol, for stone is: st
Abbreviation or prefix ( abbr. ) brevis - short unit symbol for ounce (troy) is: oz t
### One stone of gold converted to ounce (troy) equals to 204.17 oz t
How many troy ounces of gold are in 1 stone? The answer is: The change of 1 st ( stone ) unit of a gold amount equals = to 204.17 oz t ( ounce (troy) ) as the equivalent measure for the same gold type.
In principle with any measuring task, switched on professional people always ensure, and their success depends on, they get the most precise conversion results everywhere and every-time. Not only whenever possible, it's always so. Often having only a good idea ( or more ideas ) might not be perfect nor good enough solutions. Subjects of high economic value such as stocks, foreign exchange market and various units in precious metals trading, money, financing ( to list just several of all kinds of investments ), are way too important. Different matters seek an accurate financial advice first, with a plan. Especially precise prices-versus-sizes of gold can have a crucial/pivotal role in investments. If there is an exact known measure in st - stones for gold amount, the rule is that the stone number gets converted into oz t - troy ounces or any other unit of gold absolutely exactly. It's like an insurance for a trader or investor who is buying. And a saving calculator for having a peace of mind by knowing more about the quantity of e.g. how much industrial commodities is being bought well before it is payed for. It is also a part of savings to my superannuation funds. "Super funds" as we call them in this country.
Conversion for how many troy ounces ( oz t ) of gold are contained in a stone ( 1 st ). Or, how much in troy ounces of gold is in 1 stone? To link to this gold - stone to troy ounces online precious metal converter for the answer, simply cut and paste the following.
The link to this tool will appear as: gold from stone (st) to troy ounces (oz t) metal conversion.
I've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting.
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## 59857
59,857 (fifty-nine thousand eight hundred fifty-seven) is an odd five-digits composite number following 59856 and preceding 59858. In scientific notation, it is written as 5.9857 × 104. The sum of its digits is 34. It has a total of 3 prime factors and 8 positive divisors. There are 48,192 positive integers (up to 59857) that are relatively prime to 59857.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 34
• Digital Root 7
## Name
Short name 59 thousand 857 fifty-nine thousand eight hundred fifty-seven
## Notation
Scientific notation 5.9857 × 104 59.857 × 103
## Prime Factorization of 59857
Prime Factorization 7 × 17 × 503
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 59857 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 59,857 is 7 × 17 × 503. Since it has a total of 3 prime factors, 59,857 is a composite number.
## Divisors of 59857
1, 7, 17, 119, 503, 3521, 8551, 59857
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 72576 Sum of all the positive divisors of n s(n) 12719 Sum of the proper positive divisors of n A(n) 9072 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 244.657 Returns the nth root of the product of n divisors H(n) 6.59799 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 59,857 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 59,857) is 72,576, the average is 9,072.
## Other Arithmetic Functions (n = 59857)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 48192 Total number of positive integers not greater than n that are coprime to n λ(n) 12048 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 6030 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 48,192 positive integers (less than 59,857) that are coprime with 59,857. And there are approximately 6,030 prime numbers less than or equal to 59,857.
## Divisibility of 59857
m n mod m 2 3 4 5 6 7 8 9 1 1 1 2 1 0 1 7
The number 59,857 is divisible by 7.
## Classification of 59857
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (59857)
Base System Value
2 Binary 1110100111010001
3 Ternary 10001002221
4 Quaternary 32213101
5 Quinary 3403412
6 Senary 1141041
8 Octal 164721
10 Decimal 59857
12 Duodecimal 2a781
20 Vigesimal 79ch
36 Base36 1a6p
## Basic calculations (n = 59857)
### Multiplication
n×i
n×2 119714 179571 239428 299285
### Division
ni
n⁄2 29928.5 19952.3 14964.2 11971.4
### Exponentiation
ni
n2 3582860449 214459277895793 12836888997008481601 768377664693936683191057
### Nth Root
i√n
2√n 244.657 39.1176 15.6415 9.0245
## 59857 as geometric shapes
### Circle
Diameter 119714 376093 1.12559e+10
### Sphere
Volume 8.98325e+14 4.50236e+10 376093
### Square
Length = n
Perimeter 239428 3.58286e+09 84650.6
### Cube
Length = n
Surface area 2.14972e+10 2.14459e+14 103675
### Equilateral Triangle
Length = n
Perimeter 179571 1.55142e+09 51837.7
### Triangular Pyramid
Length = n
Surface area 6.2057e+09 2.52743e+13 48873
## Cryptographic Hash Functions
md5 5da43347b53bce08b4d5f0c181d45dd4 e0933ab6ec7df9ecfc90131954e181eef93a6975 a9d14bb7eb958337a91c9046873d9d94dcfe59acceb3a309521b25cfeb53e872 d43b070ce8ff471c52ee3ffb464069040c9dfb0385fb930d2555ce11d295cb2a9078c4d9f92a6b06e81d30b1892fea4b7138baecd7121a72727e6020f3f3273d cf4f7b19e50a3d3b2df0ea89ab5d253795e902c7
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## ameryn.love17 3 years ago A rose garden is formed by joining a rectangle and a semicircle, as shown below. The rectangle is 20ft long and 16ft wide. Find the area of the garden. Use the value for , and do not round your answer. Be sure to include the correct unit in your answer.
1. saifoo.khan
where "below" ?
2. ameryn.love17
fine ill draw the stupid picture
3. ameryn.love17
|dw:1326154373410:dw|
4. saifoo.khan
|dw:1326154376040:dw|
5. saifoo.khan
|dw:1326154419466:dw|
6. sasogeek
geez go to sleep... there's more time tomorrow to solve this... saifoo!
7. saifoo.khan
$Area = (20 \times 16) \times ({\frac12 \times 3.142 \times 8^2})$
8. sasogeek
9. saifoo.khan
Area = Area of rectangle + area of semi-circle.
10. ameryn.love17
ok thanks
11. saifoo.khan
Lol, wont do the next one. @saso
12. sasogeek
i did say you shouldn't... you're just saying it and i'm seeing, it's all ur choice :)
13. saifoo.khan
Lol. u "ordered" me!!
14. ameryn.love17
:O
15. sasogeek
no i didn' i was only repeating what you said... earlier...
16. saifoo.khan
Lol. nvm.
17. saifoo.khan
Chill. :D
18. sasogeek
:P i'm chillaxing actually, with juices!
19. saifoo.khan
Yo! that's da spirit. xD
20. ameryn.love17
this is so interesting and wierd at the same time
21. sasogeek
;) SGG , i need to find a symbol for it if necessary :P
22. saifoo.khan
Lol, "two in one".
23. sasogeek
LOL
24. sasogeek
just like i found my nick name sasogeek... saso guys gang is officially The Saguga :P
25. saifoo.khan
;)
26. saifoo.khan
Lol.
27. sasogeek
:P
28. saifoo.khan
saso, did u saw the link i posted?
29. sasogeek
where?
30. saifoo.khan
just here... deleted afterwards..
31. sasogeek
LOL nahh i didn't click it
32. saifoo.khan
lol, i will post it as soon as i get a chance. ;)
33. saifoo.khan
will delete it soon.
34. sasogeek
k
35. saifoo.khan
C'mon armeryan leave!!!!! jk
36. sasogeek
LMFAO
37. ameryn.love17
:'(
38. saifoo.khan
nvm let me post. :D http://www.youtube.com/watch?v=BUTWkFpv7As
39. sasogeek
awwwww :) hugs
40. sasogeek
LMFAO
41. saifoo.khan
it's actually Lmfao. LOL
42. sasogeek
IK
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## FANDOM
10,219 Pages
S map is a function which maps "a pair of a natural number and a function" to "a pair of a natural number and a function". It was defined by Japanese googologist Fish in 2002[1] and used in the definition of Fish number 1 and Fish number 2. It is defined as
\begin{eqnarray*} S:[m,f(x)]â[g(m),g(x)] \end{eqnarray*}
which means that when a pair of $$m \in \mathbb{N}$$ and a function $$f(x)$$ is given as input variables of S map, a pair of $$g(m) \in \mathbb{N}$$ and a function $$g(x)$$ is obtained as return values, where $$g(x)$$ is defined as
\begin{eqnarray*} B(0,n) & = & f(n) \\ B(m+1,0) & = & B(m, 1) \\ B(m+1,n+1) & = & B(m, B(m+1, n)) \\ g(x) & = & B(x,x) \end{eqnarray*}
and $$g(m)$$ is calculated by substituding $$x=m$$ to $$g(m)$$.
$$B(m,n)$$ is similar to Ackermann function except $$B(0,n) = f(n)$$.
## Approximation in other notation Edit
S map is similar to Taro's multivariable Ackermann function with 3 variables. By applying S map n times to [3,x+1], we get a number $$A(n,1,1)$$ and a function $$A(n-1,x,x)$$. Therefore, S map corresponds to adding $$\omega$$ to the ordinal of FGH. At the time when $$F_1$$ was developed, people at Japanese BBS didn't know FGH or multivariable Ackermann function (which was developed in 2007), but it was soon calculated that applying S map is similar to adding one to the length of the arrow of Chained arrow notation.[2]
S map is used in $$F_1$$ and $$F_2$$, but not in Fish number 3, where s(n) map is used instead. $$F_1$$ and $$F_2$$ is based on S map, but later Fish found that s(2) map, which is similar to S map, is obtained with the definition of s(n) map, and the Ackermann function is not necessary in the definition.
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https://www.teachoo.com/1544/504/Example-18---Solve-5-x-1---1-y-2--2--6-x-1---3-y-2--1/category/Examples/
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Examples
Chapter 3 Class 10 Pair of Linear Equations in Two Variables
Serial order wise
### Transcript
Question 8 Solve the following pair of equations by reducing them to a pair of linear equations : 5/(𝑥 −1) + 1/(𝑦 −2) = 2 6/(𝑥 −1) – 3/(𝑦 −2) = 1 5/(𝑥 − 1) + 1/(𝑦 − 2) = 2 6/(𝑥 − 1) – 3/(𝑦 − 2) = 1 So, our equations become 5u + v = 2 6u – 3v = 1 Thus, our equations are 5u + v = 2 …(3) 6u – 3v = 1 …(4) From (3) 5u + v = 2 v = 2 – 5u Putting value of v in (4) 6u – 3v = 1 6u – 3(2 – 5u) = 1 6u – 6 + 15u = 1 6u + 15u = 1 + 6 21u = 7 u = 7/21 u = 1/3 Putting u = 1/3 in equation (3) 5u + v = 2 5(1/3) + v = 2 5/3 + v = 2 v = 2 – 5/3 v = (2(3) − 5)/3 v = (6 − 5)/3 v = 𝟏/𝟑 Hence, u = 1/3 & v = 1/3 But we need to find x & y u = 𝟏/(𝒙 − 𝟏) 1/3 = 1/(𝑥 − 1) x – 1 = 3 x = 3 + 1 x = 4 v = 𝟏/(𝒚 − 𝟐) 1/3 = 1/(𝑦 −2) y – 2 = 3 y = 3 + 2 y = 5 So, x = 4, y = 5 is the solution of our equations
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https://readpaper.com/paper/4852411294480859137
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This website requires JavaScript.
# Structure of tight (k,0)-stable graphs
Jan 2024
0被引用
0笔记
We say that a graph G is $(k,\ell)$-stable if removing $k$ vertices from it reduces its independence number by at most $\ell$. We say that G is tight $(k,\ell)$-stable if it is $(k,\ell)$-stable and its independence number equals $\lfloor{\frac{n-k+1}{2}\rfloor}+\ell$, the maximum possible, where $n$ is the vertex number of G. Answering and question of Dong and Wu, we show that every tight $(2,0)$-stable graph with odd vertex number must be an odd cycle. Moreover, we show that for all $k\geq 3$, every tight $(k,0)$-stable graph has at most $k+6$ vertices.
AI理解论文&经典十问
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Convert gram/millilitre to ounce/gallon [troy/U.S.]
gram/millilitre ounce/gallon
Did you mean to convert gram/millilitre to ounce/gallon [troy/U.S.] ounce/gallon [troy/U.K.] ounce/gallon [U.S.] ounce/gallon [U.K.]
How many gram/millilitre in 1 ounce/gallon? The answer is 0.0082166693. We assume you are converting between gram/millilitre and ounce/gallon [troy/U.S.]. You can view more details on each measurement unit: gram/millilitre or ounce/gallon The SI derived unit for density is the kilogram/cubic meter. 1 kilogram/cubic meter is equal to 0.001 gram/millilitre, or 0.12170381495091 ounce/gallon. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between grams/milliliter and ounces/gallon. Type in your own numbers in the form to convert the units!
Quick conversion chart of gram/millilitre to ounce/gallon
1 gram/millilitre to ounce/gallon = 121.70381 ounce/gallon
2 gram/millilitre to ounce/gallon = 243.40763 ounce/gallon
3 gram/millilitre to ounce/gallon = 365.11144 ounce/gallon
4 gram/millilitre to ounce/gallon = 486.81526 ounce/gallon
5 gram/millilitre to ounce/gallon = 608.51907 ounce/gallon
6 gram/millilitre to ounce/gallon = 730.22289 ounce/gallon
7 gram/millilitre to ounce/gallon = 851.9267 ounce/gallon
8 gram/millilitre to ounce/gallon = 973.63052 ounce/gallon
9 gram/millilitre to ounce/gallon = 1095.33433 ounce/gallon
10 gram/millilitre to ounce/gallon = 1217.03815 ounce/gallon
Want other units?
You can do the reverse unit conversion from ounce/gallon to gram/millilitre, or enter any two units below:
Enter two units to convert
From: To:
Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 70 kg, 150 lbs, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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Call now: +92-334-3261516
# What Is Return On Assets?
Additionally, firms may reduce prices to generate sales in an effort to cycle inventory. In this article, the terms “cost of sales” and “cost of goods sold” are synonymous. In accounting, the Inventory turnover is a measure of the number of times inventory is sold or used in a time period, such as a year. The equation for inventory turnover equals the cost of goods sold divided by the average inventory. Inventory turnover is also known as inventory turns, stockturn, stock turns, turns, and stock turnover. What makes the asset turnover ratio of utmost importance is that it gives creditors and investors a general idea regarding how well a company is managed for producing sales and products.
### How long is a average month?
The average month is 365/12 = 30.42 days in a regular year and 366/12 = 30.50 days in a leap year. The Gregorian (western) solar calendar has 365.2425/12 = 30.44 days on the average, varying between 28 and 31 days.
Discounts can be found in almost any industry that offers coupons or promotion for products or services. The above section demonstrates how to use this formula to find total assets. Financial statement analysis is the process of analyzing a company’s financial statements for decision-making purposes. Locate total sales—it could be listed as revenue—on the income statement. Average annual growth rate is the average increase in the value of an investment, portfolio, asset, or cash stream over the period of a year. To see how to use this formula, let’s look at the example of a company that makes jewelry. To make her jewelry Linda needs tools like beads, wire, string, glue, and work tables.
## What Is A Good Total Asset Turnover Ratio?
Companies that invest heavily upfront into equipment and other assets typically have a lower ROAA. Note that this policy may change as the SEC manages SEC.gov to ensure that the website performs efficiently and remains available to all users. If a user or application submits more than 10 requests per second, further requests from the IP address may be limited for a brief period. Once the rate of requests has dropped below the threshold for 10 minutes, the user may resume accessing content on SEC.gov. This SEC practice is designed to limit excessive automated searches on SEC.gov and is not intended or expected to impact individuals browsing the SEC.gov website. We reserve the right to block IP addresses that submit excessive requests. Current guidelines limit users to a total of no more than 10 requests per second, regardless of the number of machines used to submit requests.
Watch this short video to quickly understand the definition, formula, and application of this financial metric. Net sales are the amount of revenue generated after deducting sales returns, sales discounts, and sales allowances. A potential lender will also want to know the value of a business’s assets as they can be used as leverage to get a new loan, according to the Houston Chronicle. Adam Hayes is a financial writer with 15+ years Wall Street experience as a derivatives trader. Besides his extensive derivative trading expertise, Adam is an expert in economics and behavioral finance.
## Company
Sometimes, investors and analysts are more interested in measuring how quickly a company turns its fixed assets or current assets into sales. In these cases, the analyst can use specific ratios, such as the fixed-asset turnover ratio or the working capital ratio to calculate the efficiency of these asset classes. The working capital ratio measures how well a company uses its financing from working capital to generate sales or revenue.
### Big Lots (NYSE:BIG) May Have Issues Allocating Its Capital – Simply Wall St
Big Lots (NYSE:BIG) May Have Issues Allocating Its Capital.
Posted: Fri, 31 Dec 2021 10:37:39 GMT [source]
You can look up the financial statements of other companies in your industry to obtain the information needed for the asset turnover ratio formula and then calculate it yourself. As mentioned above, we have to be careful to use average fixed assets and the average must match the period that we are evaluating. For example, if we are looking at the fixed asset turnover for 1 average total assets formula year, then we must take the average of fixed assets over the course of that year. Generally speaking, the higher the ratio, the better, because a high ratio indicates the business has less money tied up in fixed assets for each unit of currency of sales revenue. A declining ratio may indicate that the business is over-invested in plant, equipment, or other fixed assets.
## How Do You Calculate Average Total Assets Return On Assets Ratio?
Adam received his master’s in economics from The New School for Social Research and his Ph.D. from the University of Wisconsin-Madison in sociology. He is a CFA charterholder as well as holding FINRA Series 7 & 63 licenses. He currently researches and teaches at the Hebrew University in Jerusalem.
Days sales outstanding is a financial ratio that illustrates how well a company’s accounts receivables are being managed. Some compilers of industry data (e.g., Dun & Bradstreet) use sales as the numerator instead of cost of sales.
## What Is Return On Assets?
Net income/loss is found at the bottom of the income statement and divided into total assets to arrive at ROA. Return on investment is a financial ratio used to calculate the benefit an investor will receive in relation to their investment cost. It is most commonly measured as net income divided by the original capital cost of the investment. Company A reported beginning total assets of \$199,500 and ending total assets of \$199,203.
The adjusted net asset method is a business valuation technique that changes the stated values of a company’s assets and liabilities to reflect its estimated current fair market values better. Return on Equity is generally net income divided by equity, while Return on Assets is net income divided by average assets.
## How Do You Calculate Net Assets On A Balance Sheet?
For example, if your net income increases to \$30,000 and your total assets remain the same at \$65,000, your ROA percentage would increase to 46.15%. If you want to increase your ROA, your net income and total assets must increase to equal similar values. You could also introduce new products or service lines that don’t require any additional investment in assets, thereby opening new revenue streams to your business. This means that Company A’s assets generate 25% of net sales, relative to their value. In other words, every \$1 in assets generates 25 cents in net sales revenue. As a general rule, a return on assets under 5% is considered an asset-intensive business while a return on assets above 20% is considered an asset-light business.
The asset turnover ratio uses the value of a company’s assets in the denominator of the formula. To determine the value of a company’s assets, the average value of the assets for the year needs to first be calculated. Average total assets are the average carrying value of assets that are recorded on the balance sheet at the different balance sheet dates. Usually, the carrying value of assets at the end of the previous year and those at the end of the current year are used in the calculation to find average total assets on the balance sheet. Let’s say the company just started in 2013 and had \$16,100 worth of total assets in its first year. Since the company has only been in business for one year, we can use the total assets listed on the balance sheet as the average total assets. The comparison to total sales is less useful for a very successful company that has accumulated a large amount of cash, since the cash figure is included in the calculation of average total assets.
As a financial and activity ratio, and as part of DuPont analysis, asset turnover is a part of company fundamental analysis. While the asset turnover ratio considers average total assets in the denominator, the fixed asset turnover ratio looks at only fixed assets. The fixed asset turnover ratio is, in general, used by analysts to measure operating performance. This efficiency ratio compares net sales to fixed assets and measures a company’s ability to generate net sales from its fixed-asset investments, namelyproperty, plant, and equipment (PP&E). Depreciation is the allocation of the cost of a fixed asset, which is spread out—or expensed—each year throughout the asset’s useful life. Typically, a higher fixed asset turnover ratio indicates that a company has more effectively utilized its investment in fixed assets to generate revenue. The asset turnover ratio measures the efficiency of a company’s assets in generating revenue or sales.
Essentially, the net sales are primarily utilized for calculating the ratio returns and refunds. The returns and refunds should be withdrawn out of the total sales, in order to accurately measure a firm’s asset capability of generating sales. Average total assets in the denominator of the return on assets formula is found on a company’s balance sheet. The average of total assets should be used based on the period being evaluated. For example, if an investor is calculating a company’s 2015 return on assets, the beginning and ending total assets for that year should be averaged. The asset turnover ratio measures is an efficiency ratio which measures how profitably a company uses its assets to produce sales. This ratio measures how efficiently a firm uses its assets to generate sales, so a higher ratio is always more favorable.
The lower the return on assets, the more asset-intensive a company is. An example of an asset-intensive company would be an airline company. The first company earns a return on assets of 10% and the second one earns an ROA of 67%.
## Example Of Roa Calculation
The days sales outstanding figure is an index of the relationship between outstanding receivables and credit account sales achieved over a given period. Fundamentally, in order to calculate the average total assets, what you have to do is simply add the beginning and ending total asset balances together and divide the result by two. While there is always the option of utilizing a more in-depth, weighted average calculation, this isn’t mandatory. Well, according to the formula, you have to divide the net sales by the average total assets in order to get the asset turnover ratio. Analyze your asset turnover by comparing it to other companies in the same industry and also to any previous asset-turnover figures you may have from earlier years.
The total asset turnover ratio is what a business uses to determine how much money is being generated by the assets a company owns. For example, if the total asset turnover ratio is 0.72, that means that the company is making \$0.72 per year for every dollar of assets that the company owns. Return on Equity is a measure of a company’s profitability that takes a company’s annual return divided by the value of its total shareholders’ equity (i.e. 12%).
You can do this manually by filling out the liabilities and equity in your balance sheet. It’s generally simpler and more accurate to use accounting software to generate a balance sheet. Some assets will be added automatically thanks to your journal entries.
• To calculate the average total assets, add the total assets for the current year to the total assets for the previous year,and divide by two.
• This means the company is able to make more sales with fewer assets than others in the industry.
• As a result, you will see the asset turnover ratio presented with all of these terms.
• If return on assets uses average assets, then ROA and ROAA will be identical.
• If these companies are in the same industry, then Company D is much more efficient than Company C.
Start by listing the value of any current assets like cash, money owed to you and inventory. Return on average assets is an indicator used to assess the profitability of a firm’s assets, and it is most often used by banks. Below are the steps as well as the formula for calculating the asset turnover ratio. A company’s asset turnover ratio can be impacted by large asset sales as well as significant asset purchases in a given year. Investors use the asset turnover ratio to compare similar companies in the same sector or group. This metric helps investors understand how effectively companies are using their assets to generate sales.
The asset turnover ratio is an efficiency ratio that measures a company’s ability to generate sales from its assets by comparing net sales with average total assets. In other words, this ratio shows how efficiently a company can use its assets to generate sales.
Because shareholders’ equity is equal to assets minus liabilities, ROE is essentially a measure of the return generated on the net assets of the company. As with return on capital, a ROE is a measure of management’s ability to generate income from the equity available to it.
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##### What is the difference between a postulate and a theorem?
Mathematics Tutor: None Selected Time limit: 1 Day
What is the difference between a postulate and a theroem?
Aug 25th, 2015
Both theorems and postulates are statements of geometrical truth, such as All right angles are congruent or All radii of a circle are congruent. The difference between postulates and theorems is that postulates are assumed to be true, but theorems must be proven to be true based on postulates and/or already-proven theorems.
Aug 25th, 2015
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Chapter 4
# Chapter 4 - AGEC 105 Chapter 4 Consumer Equilibrium and...
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Chapter 4: Consumer Equilibrium and Market Demand AGEC 105
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Quiz 1 Count: 121 Average: 20.0 Median: 20.0 Maximum: 25.0 Minimum: 3.8 2
Announcements Quiz 2, Friday, September 24, Chapters 3 and 4 Exam 1, Wednesday, September 29, Chapters 1, 3, 4 and 5 Review sessions, Tuesdays, Blocker 113, 7:30-9:30pm Review problems are posted on E-LEARNING
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4 Topics of Discussion DERIVING A DEMAND CURVE MARKET DEMAND THE LAW OF DEMAND CHANGES IN DEMAND CONSUMER SURPLUS http://www.focus.com/images/view/7362/
5 Demand Model Single most useful tool of economic analysis Explains how prices of goods and quantity bought and sold are determined in certain types of markets In our previous lecture we talked about the indifference curves and how changes in income and prices will affect individual’s utility
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Number of Concerts per Month Number of Movies per Month 5 4 3 8 5 10 10 6 20 Change in Price D K J 1. When the price of concerts is \$20, MRS m,c = P c /P m at Point D. 2. But when the price of concerts falls to \$10, this condition is satisfied at point J. Number of Concerts per Month Price per concert \$5 3 8 5 \$20 \$10 D K J 3. The demand curve shows the quantity of concerts Max chooses at each price for concerts. 6
Markets Market: Buyer and seller of a particular good and service Competitive market: A market in which there are many buyers and many sellers so that each has a negligible impact on the market price Perfectly competitive markets: In addition to “many buyers and sellers” the goods offered for sale are all the same. Imperfectly competitive markets:
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# OpenStax_CNX
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# Value of a function
Module by: Sunil Kumar Singh. E-mail the author
The value of a function at “x = a” is denoted as “f(a)”. The working rule for finding value of a function is to replace independent variable “x” by “a”.
## Polynomial and rational functions
Problem 1: Find “f(y)”, if
y = f x = 1 - x 1 + x y = f x = 1 - x 1 + x
Solution :
Statement of the problem : The given function is a rational function. We have to evaluate the function when independent variable is function itself.
We need to replace “x” by “y”.
f y = 1 y 1 + y = 1 1 x 1 + x 1 + 1 - x 1 + x f y = 1 y 1 + y = 1 1 x 1 + x 1 + 1 - x 1 + x
f y = 1 + x 1 + x 1 + x + 1 x = 2 x 2 = x f y = 1 + x 1 + x 1 + x + 1 x = 2 x 2 = x
Problem 2: Find “f(x)”, if
f x - 1 = x 2 1 f x - 1 = x 2 1
Solution :
Statement of the problem : The given function is a polynomial function with a polynomial as its argument. We have to evaluate the function for independent variable “x”.
We need to replace “x-1” by “x” in the given equation to find “f(x)”. The right hand side expression, however, does not contain term “x-1”. We, therefore, need to find the term, which will replace “x”. Clearly if "x" replaces "x-1", then "x+1" will replace "x-1+1 = x"
Thus, we need to replace “x” by “x+1”.
f x + 1 - 1 = x + 1 2 1 = x 2 + 2 x + 1 1 = x 2 + 2 x f x + 1 - 1 = x + 1 2 1 = x 2 + 2 x + 1 1 = x 2 + 2 x
Problem 3: If { f x } = x + 1 x { f x } = x + 1 x , then prove that :
{ f x } 3 = f x 3 + 3 f 1 x { f x } 3 = f x 3 + 3 f 1 x
Solution :
Statement of the problem : The function is a polynomial function. We have to evaluate cube of the function, which involves evaluation of function for arguments, which are independent variable, raised to certain integral powers.
The cube of given function is :
{ f x } 3 = x + 1 x 3 = x 3 + 1 / x 3 + 3 x 2 X 1 x + 3 x X 1 x 2 { f x } 3 = x + 1 x 3 = x 3 + 1 / x 3 + 3 x 2 X 1 x + 3 x X 1 x 2
{ f x } 3 = x 3 + 1 x 3 + 3 x + 1 x { f x } 3 = x 3 + 1 x 3 + 3 x + 1 x
Now, f x 3 f x 3 is :
f x 3 = x 3 + 1 x 3 f x 3 = x 3 + 1 x 3
Hence,
{ f x } 3 = x 3 + 1 x 3 + 3 x + 1 x = f x 3 + 3 f x { f x } 3 = x 3 + 1 x 3 + 3 x + 1 x = f x 3 + 3 f x
But, we see that :
f 1 x = 1 x + x = f x f 1 x = 1 x + x = f x
Hence,
{ f x } 3 = f x 3 + 3 f x = f x 3 + 3 f 1 x { f x } 3 = f x 3 + 3 f x = f x 3 + 3 f 1 x
Problem 4: If
f x = 1 + x 1 x f x = 1 + x 1 x
Then, find
f x f x 2 1 + { f x } 2 f x f x 2 1 + { f x } 2
Solution :
Statement of the problem : The given function is rational function. We have to find the expression which involves (i) function, (ii) function with argument as squared independent variable and (iii) square of the function.
We need to substitute for various terms in the given expression :
f x f x 2 1 + { f x } 2 = 1 + x 1 x X 1 + x 2 1 x 2 1 + 1 + x 1 x 2 f x f x 2 1 + { f x } 2 = 1 + x 1 x X 1 + x 2 1 x 2 1 + 1 + x 1 x 2
= 1 + x 1 + x 2 1 x 1 x 2 1 x 2 + 1 + x 2 1 x 2 = 1 + x 1 + x 2 1 x 1 x 2 1 x 2 + 1 + x 2 1 x 2
1 + x 2 1 x 2 2 1 + x 2 1 x 2 = 1 + x 2 2 1 + x 2 = 1 2 1 + x 2 1 x 2 2 1 + x 2 1 x 2 = 1 + x 2 2 1 + x 2 = 1 2
## Modulus functions
Problem 5: If
f x = | x | x ; x 0 f x = | x | x ; x 0
Then, evaluate
| f a f - a | | f a f - a |
Solution :
Statement of the problem : Function, f(x), involves modulus and is in rational form. The value of this function, in turn, forms the part of a expression to be evaluated. We have to find the value of expression.
We first evaluate the expression without modulus sign :
f a f - a = | a | a | a | a = | a | a + | a | a = 2 | a | a ; a 0 f a f - a = | a | a | a | a = | a | a + | a | a = 2 | a | a ; a 0
But, we know that | a | = ± a | a | = ± a
f a f - a = 2 X ± a a = ± 2 f a f - a = 2 X ± a a = ± 2
Taking modulus of the expression,
| f a f - a | = 2 ; a 0 | f a f - a | = 2 ; a 0
Note that we need to keep the condition for which the given expression is evaluated.
## Logarithmic functions
Problem 6: Find f 2 x 1 + x 2 f 2 x 1 + x 2 , if
f x = log e 1 + x 1 - x f x = log e 1 + x 1 - x
Solution :
Statement of the problem : The given function is transcendental logarithmic function. We have to evaluate the function for an argument (input to function), which is itself a rational function in independent variable, “x”.
We need to replace “x” by “ 2 x / 1 + x 2 2 x / 1 + x 2 ”.
f 2 x 1 + x 2 = log e 1 + 2 x 1 + x 2 1 2 x 1 + x 2 = log e 1 + x 2 + 2 x 1 + x 2 2 x f 2 x 1 + x 2 = log e 1 + 2 x 1 + x 2 1 2 x 1 + x 2 = log e 1 + x 2 + 2 x 1 + x 2 2 x
f 2 x 1 + x 2 = log e 1 + x 1 x 2 = 2 log e 1 + x 1 x = 2 f x f 2 x 1 + x 2 = log e 1 + x 1 x 2 = 2 log e 1 + x 1 x = 2 f x
## Trigonometric functions
Problem 7: Find f π / 4 f π / 4 , if
f x = 2 cot x 1 + cot 2 x f x = 2 cot x 1 + cot 2 x
Solution :
Statement of the problem : The given function is a rational function with trigonometric function as independent variable. We have to find the value of function for a particular angle.
We need to replace “x” by “ π / 4 π / 4 ”.
f π / 4 = 2 cot π / 4 1 + cot 2 π / 4 f π / 4 = 2 cot π / 4 1 + cot 2 π / 4
As cot π / 4 = 1 cot π / 4 = 1 ,
f π 4 = 2 X 1 1 + 1 2 = 1 f π 4 = 2 X 1 1 + 1 2 = 1
Problem 8: Find f tan θ f tan θ , if
f x = 2 x 1 + x 2 f x = 2 x 1 + x 2
Solution :
Statement of the problem : The given function is a rational function. We have to evaluate the function for a value, which is itself a trigonometric function.
We need to replace “x” by “ tan θ tan θ ”.
f tan θ = 2 tan θ 1 + tan 2 θ = sin 2 θ f tan θ = 2 tan θ 1 + tan 2 θ = sin 2 θ
Problem 9: If f x = cos { log x } f x = cos { log x } , then prove that :
f x y + f x y = 2 f x f y f x y + f x y = 2 f x f y
Solution :
Statement of the problem : The given function, f(x) is a trigonometric function, whose input is a logarithmic function. We have to evaluate LHS of the given equation to equate the same to RHS.
Here, we evaluate each term of the left hand side of the equation separately and then combine the result.
f x y = cos { log e x y } = cos log e x + log e y f x y = cos { log e x y } = cos log e x + log e y
f x y = cos { log e x y } = cos log e x log e y f x y = cos { log e x y } = cos log e x log e y
Substituting in the LHS expression, we have :
{ f x y + f x y } = cos log e x + log e y + cos log e x log e y { f x y + f x y } = cos log e x + log e y + cos log e x log e y
We know that :
cos C + cos D = 2 cos C + D 2 cos C D 2 cos C + cos D = 2 cos C + D 2 cos C D 2
Hence,
{ f x y + f x y } = 2 cos log e x + log e y + log e x log e y 2 cos log e x + log e y log e x + log e y 2 { f x y + f x y } = 2 cos log e x + log e y + log e x log e y 2 cos log e x + log e y log e x + log e y 2
{ f x y + f x y } = 2 cos log e x cos log e y { f x y + f x y } = 2 cos log e x cos log e y
{ f x y + f x y } = 2 cos log e x cos log e y = 2 f x f y { f x y + f x y } = 2 cos log e x cos log e y = 2 f x f y
## Acknowledgment
Author wishes to thank Mr Jay Sicard, Pasadena TX for suggesting valuable typographical correction on the topic.
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# Explain with the help of diagrams
Explain with the help of diagrams, the effect of the following changes on the demand for a commodity:
(i) A fall in the price of substitute goods
(ii) A fall in the income of its buyer
(i)Demand for a commodity will decrease when there is a fall in the price of substitute goods, implying that demand curve would shift backward. Less will be purchased at the same price. Demand for commodity falls from OQ to O\${{Q}{1}}. So, fall in the price of tea will cause demand for coffee to fall. The given figure illustrates this situation: <img src="/uploads/db3785/original/2X/9/93d5e168d493e5a50c05fb752add97f069f5a597.png" width="295" height="231"> (ii) Demand for a commodity will decrease when there is a fall in the income of the consumer (assuming that the commodity demanded is a normal good). This would imply a backward shift in demand curve. Less will be purchased at the same price. Demand for commodity falls from OQ to O{{Q}{2}}\$. So, decrease in consumer’s income will cause his demand for pure ghee to fall.
The given figure illustrates this situation:
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Author: Sorting Last Post on Top ↓ Message:
5/25/12 11:08 A
Fun Just tried it.
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5/21/12 11:53 A
Sounds interesting, I will try it.
All things are possible if you only believe, even losing 15 pounds
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AKRONEWITTER1 Posts: 1,898
5/20/12 4:40 P
THE DICE CHALLENGE
You will need 2 dice. If you only have 1, that's OK...you can just roll it twice. The 1st dice will tell you what TYPE of exercise you will do. The 2nd dice (or roll) will tell you HOW MANY of each exercise you will do.
If you roll...
1 - arm reps (your choice); multiply # on 2nd dice by 10, this is the # of reps
2 - core reps (your choice); multiply # on 2nd dice by 10, this is the # of reps
3 - leg reps (your choice); multiply # on 2nd dice by 10, this is the # of reps
4 - side steps; multiply # on 2nd dice by 10, this is the # of steps
5 - marching in place; the # on the 2nd dice will be # of minutes
6 - sips of water; equal to the # on the 2nd dice
Example:
~If you roll a 2 then a 4, you will do 40 core reps
~If you roll a 5 then a 3, you march for 3 minutes
Have fun and let's get rolling!!!!
Sun- rest
Mon- 60 core reps
Tue- 50 arm reps
Wed-
Thur-
Fri-
Sat-
Edited by: AKRONEWITTER1 at: 5/23/2012 (11:55)
Pounds lost: 17.0
0 15.5 31 46.5
62
LASARRE SparkPoints: (153,117) Fitness Minutes: (107,212) Posts: 13,965 5/20/12 1:49 P With another holiday weekend coming up, thought this would be a good challenge to make sure we work the whole body this week: THE DICE CHALLENGE You will need 2 dice. If you only have 1, that's OK...you can just roll it twice. The 1st dice will tell you what TYPE of exercise you will do. The 2nd dice (or roll) will tell you HOW MANY of each exercise you will do. If you roll... 1 - arm reps (your choice); multiply # on 2nd dice by 10, this is the # of reps 2 - core reps (your choice); multiply # on 2nd dice by 10, this is the # of reps 3 - leg reps (your choice); multiply # on 2nd dice by 10, this is the # of reps 4 - side steps; multiply # on 2nd dice by 10, this is the # of steps 5 - marching in place; the # on the 2nd dice will be # of minutes 6 - sips of water; equal to the # on the 2nd dice Example: ~If you roll a 2 then a 4, you will do 40 core reps ~If you roll a 5 then a 3, you march for 3 minutes Have fun and let's get rolling!!!! Nina- Wisconsin CST Together Everyone Achieves More Co-Leader- Walk it Out Co-Leader- Wii Exercise
Page: 1 of (1)
## Other Wii Exercise Challenges, Workouts, and High Scores Posts
Topics: Last Post: Amazing Race: Kickin' Asphalt 1/15/2016 7:58:18 PM Amazing Race: Rad Racers 2/25/2016 9:59:50 PM WII MONOPOLY 6/29/2016 2:14:13 PM WII EXERCISE JULY 2016 RANDOM MADNESS 7/1/2016 8:13:51 AM AMAZING RACE QUALIFYING CHALLENGE 4/11/2015 7:59:59 PM
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eco200_uncertainty_fall2009
# eco200_uncertainty_fall2009 - ECO 200 Microeconomic Theory...
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ECO 200: Microeconomic Theory Lecture notes on uncertainty and demand for insurance 1 Fall 2008 Carlos J. Serrano 1 Lecture notes updated on Nov. 1, 2009 at 8.28pm 1
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Summary Uncertainty Probability, expected value and variance Von Neumann utility function and contingent consumption bundles Variance and preference towards risk Risk premium and certainty equivalent Demand for insurance Fair insurance 2
1 Uncertainty and demand for insurance 1.1 Uncertainty 1.1.1 Probability and Expected Value Let be the set of outcomes of a random variable . Fo simplicity, just consider that a random variable is a variable that can take random values in the set and only in . Assume that the number of elements in is f nite and equal to . De f nition 1 A probability is a function that maps outcomes in the set X to real numbers satisfying that X =1 ( )=1 Example 2 Suppose that there are two outcomes such { =100 =20 } ;andtha t ( )=Pr( )=0 3 and ( )= ( )=0 7 Note that Pr( )+ ( )=1 An very useful application of the theory of probabilities is the calculation of an expected value. De f nition 3 The expected value is the probability weighted average of the payo f sassoc i - ated will all possible outcomes Example 4 Let { = 100 =20 } and ( )=0 3 and ( )=0 7 The expected value of the random variable is ( )= ( ) + ( ) =30+14=44 Sometimes it is useful to obtain measures of the dispersion of the random variable This is particularly interesting when two random variables have the same expectation. The dispersion of a random variable is associated to risk. For instance, two mutual funds can o f er you the same expected return, but the risk can be di f erent. How do we measure
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De f nition 5 Thevar ianceo farandomevar iab le is ( )= [( ( )) 2 ]= X =1 (Pr( )( ( )) 2 ) As you can see the variance is a measure of the dispersion of the outcomes from the mean weighted by how likely the outcomes are. Example 6 Whenthenumbero fe lemen tsinSistwo ,then ( )= ( )( ( )) 2 + ( )( ( )) 2 Example 7 Let { =100 =20 } and ( )=0 3 and ( )=0 7 ( )=0 3 (100 44) 2 +0 7 (20
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# OnlineClasses
<p>The category contains all the lectures as posts which teaches about Datastructures & Algorithms and Problem Solving</p>
## Shortest Path Using Bellman Ford Algorithm
Introduction This post about Bellman Ford Algorithm is a continuation of the post Shortest Path Using Dijkstra's Algorithm. While learning about the Dijkstra's way, we learnt that it is really efficient an algorithm to find the single source shortest path in any graph provided it has no negative weight edges and no negative weight cycles. The running time of the Dijkstra's Algorithm is also promising, O(E +VlogV) depending on our choice of data structure to implement the required Priority Queue. Why Bellman Ford Algorithm? There can be scenarios where a graph may contain negative weight cycle...
## Further Reading for Minimum Spanning Tree
Introduction This is a supplement to the posts for Minimum Spanning Tree and their Analysis. Check out the other related articles in the following section. Further Reading for Minimum Spanning Tree This section is meant to be read in conjunction to the post Minimum Spanning Tree - Prim's Algorithm The minimum spanning tree of a Graph is the union of minimum spanning trees of its connected components. This is a very important observation and it must be discussed in length and breadth because this will help us design our algorithm for MST in a better way. Why is it so important to underst...
## Minimum Spanning Tree Prim’s Algorithm
Introduction What is the minimum length of the network cable we require if we have to connect 100 computing machines in a building distributed across multiple floors? How do I guarantee that there can be no other minimum length possible than what I derive? Why is it even important to find the minimum length? The history of the problem You can read the classic problem solved by MST Applications of Minimum Spanning Tree Defining Spanning Trees A spanning tree is always defined for a weighted Graph G(V,E) where the weights are positive. This means that all the edges carry some positive we...
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# NAG CL Interfacef01hac (complex_gen_matrix_actexp)
Settings help
CL Name Style:
## 1Purpose
f01hac computes the action of the matrix exponential ${e}^{tA}$, on the matrix $B$, where $A$ is a complex $n×n$ matrix, $B$ is a complex $n×m$ matrix and $t$ is a complex scalar.
## 2Specification
#include
void f01hac (Integer n, Integer m, Complex a[], Integer pda, Complex b[], Integer pdb, Complex t, NagError *fail)
The function may be called by the names: f01hac or nag_matop_complex_gen_matrix_actexp.
## 3Description
${e}^{tA}B$ is computed using the algorithm described in Al–Mohy and Higham (2011) which uses a truncated Taylor series to compute the product ${e}^{tA}B$ without explicitly forming ${e}^{tA}$.
## 4References
Al–Mohy A H and Higham N J (2011) Computing the action of the matrix exponential, with an application to exponential integrators SIAM J. Sci. Statist. Comput. 33(2) 488-511
Higham N J (2008) Functions of Matrices: Theory and Computation SIAM, Philadelphia, PA, USA
## 5Arguments
1: $\mathbf{n}$Integer Input
On entry: $n$, the order of the matrix $A$.
Constraint: ${\mathbf{n}}\ge 0$.
2: $\mathbf{m}$Integer Input
On entry: $m$, the number of columns of the matrix $B$.
Constraint: ${\mathbf{m}}\ge 0$.
3: $\mathbf{a}\left[\mathit{dim}\right]$Complex Input/Output
Note: the dimension, dim, of the array a must be at least ${\mathbf{pda}}×{\mathbf{n}}$.
The $\left(i,j\right)$th element of the matrix $A$ is stored in ${\mathbf{a}}\left[\left(j-1\right)×{\mathbf{pda}}+i-1\right]$.
On entry: the $n×n$ matrix $A$.
On exit: $A$ is overwritten during the computation.
4: $\mathbf{pda}$Integer Input
On entry: the stride separating matrix row elements in the array a.
Constraint: ${\mathbf{pda}}\ge {\mathbf{n}}$.
5: $\mathbf{b}\left[\mathit{dim}\right]$Complex Input/Output
Note: the dimension, dim, of the array b must be at least ${\mathbf{pdb}}×{\mathbf{m}}$.
The $\left(i,j\right)$th element of the matrix $B$ is stored in ${\mathbf{b}}\left[\left(j-1\right)×{\mathbf{pdb}}+i-1\right]$.
On entry: the $n×m$ matrix $B$.
On exit: the $n×m$ matrix ${e}^{tA}B$.
6: $\mathbf{pdb}$Integer Input
On entry: the stride separating matrix row elements in the array b.
Constraint: ${\mathbf{pdb}}\ge {\mathbf{n}}$.
7: $\mathbf{t}$Complex Input
On entry: the scalar $t$.
8: $\mathbf{fail}$NagError * Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).
## 6Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
On entry, argument $⟨\mathit{\text{value}}⟩$ had an illegal value.
NE_INT
On entry, ${\mathbf{m}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{m}}\ge 0$.
On entry, ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{n}}\ge 0$.
NE_INT_2
On entry, ${\mathbf{pda}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pda}}\ge {\mathbf{n}}$.
On entry, ${\mathbf{pdb}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{n}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{pdb}}\ge {\mathbf{n}}$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NW_SOME_PRECISION_LOSS
${e}^{tA}B$ has been computed using an IEEE double precision Taylor series, although the arithmetic precision is higher than IEEE double precision.
## 7Accuracy
For a Hermitian matrix $A$ (for which ${A}^{\mathrm{H}}=A$) the computed matrix ${e}^{tA}B$ is guaranteed to be close to the exact matrix, that is, the method is forward stable. No such guarantee can be given for non-Hermitian matrices. See Section 4 of Al–Mohy and Higham (2011) for details and further discussion.
## 8Parallelism and Performance
f01hac is threaded by NAG for parallel execution in multithreaded implementations of the NAG Library.
f01hac makes calls to BLAS and/or LAPACK routines, which may be threaded within the vendor library used by this implementation. Consult the documentation for the vendor library for further information.
Please consult the X06 Chapter Introduction for information on how to control and interrogate the OpenMP environment used within this function. Please also consult the Users' Note for your implementation for any additional implementation-specific information.
The matrix ${e}^{tA}B$ could be computed by explicitly forming ${e}^{tA}$ using f01fcc and multiplying $B$ by the result. However, experiments show that it is usually both more accurate and quicker to use f01hac.
The cost of the algorithm is $\mathit{O}\left({n}^{2}m\right)$. The precise cost depends on $A$ since a combination of balancing, shifting and scaling is used prior to the Taylor series evaluation.
Approximately ${n}^{2}+\left(2m+8\right)n$ of complex allocatable memory is required by f01hac.
f01gac can be used to compute ${e}^{tA}B$ for real $A$, $B$, and $t$. f01hbc provides an implementation of the algorithm with a reverse communication interface, which returns control to the calling program when matrix multiplications are required. This should be used if $A$ is large and sparse.
## 10Example
This example computes ${e}^{tA}B$, where
$A = ( 0.5+0.0i -0.2+0.0i 1.0+0.1i 0.0+0.4i 0.3+0.0i 0.5+1.2i 3.1+0.0i 1.0+0.2i 0.0+2.0i 0.1+0.0i 1.2+0.2i 0.5+0.0i 1.0+0.3i 0.0+0.2i 0.0+0.9i 0.5+0.0i ) ,$
$B = ( 0.4+0.0i 1.2+0.0i 1.3+0.0i -0.2+0.1i 0.0+0.3i 2.1+0.0i 0.4+0.0i -0.9+0.0i )$
and
$t=-0.5+0.0i .$
### 10.1Program Text
Program Text (f01hace.c)
### 10.2Program Data
Program Data (f01hace.d)
### 10.3Program Results
Program Results (f01hace.r)
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# GATE | GATE CS 2021 | Set 2 | Question 55
• Last Updated : 23 May, 2021
Consider a computer network using the distance vector routing algorithm in its network layer. The partial topology of the network is shown below.
The objective is to find the shortest-cost path from the router R to routers P and Q. Assume that R does not initially know the shortest routes to P and Q. Assume that R has three neighbouring routers denoted as X, Y and Z. During one iteration, R measures its distance to its neighbours X, Y, and Z as 3, 2 and 5, respectively. Router R gets routing vectors from its neighbours that indicate that the distance to router P from routers X, Y and Z are 7, 6 and 5, respectively. The routing vector also indicates that the distance to router Q from routers X, Y and Z are 4, 6 and 8 respectively. Which of the following statement(s) is/are correct with respect to the new routing table o R, after updation during this iteration?
(A) The distance from R to P will be stored as 10
(B) The distance from R to Q will be stored as 7
(C) The next hop router for a packet from R to P is Y
(D) The next hop router for a packet from R to Q is Z
Explanation: Given R gets the distance vector (3,2,5)
After the one iteration distance vector from X to P, Y to P, and Z to P is (7, 6, 5) respectively
The distance vector from R to P via X Y Z is (3+7, 2+6, 5+5) =(10, 8, 10)
So Take minimum distance from R to P which is 8 via Y
After the iteration distance vector from X to Q, Y to Q, Z to Q is ( 4, 6, 8) respectively
The distance vector from R to Q via X Y Z is (3+4, 2+6, 5+8) = (7, 8 13)
So Take minimum distance from R to Q which is 7 via X.
Quiz of this Question
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Check out Chess with Different Armies, our featured variant for July, 2024.
[ List Earliest Comments Only For Pages | Games | Rated Pages | Rated Games | Subjects of Discussion ]
# Single Comment
Wazir. Moves one square orthogonally.
Kevin Pacey wrote on Tue, Oct 31, 2017 11:09 PM UTC:
Comparing a wazir and ferz in the context of a chesslike game with pawns promoting to some decisive piece type (such as a queen) is an interesting exercise. For comparing two non-compound piece types previously, namely a bishop and a knight, I noted that they each have 3 advantages (and thus also disadvantages) respectively compared to one another, so that they ought to be close in value (each worth at least 3 pawns, as either often restrains 3 pawns in an ending). Namely a bishop is colour-bound, does not leap and moves in half the directions of a knight, but the disadvantages of a knight are that a bishop reaches more squares on an empty board on average, is both a long and short-range piece and it can influence both sides of the board in widely seperated sectors at times. Perhaps e.g. the diminishing of importance of a knight's leaping power's worth in many open board endgames might give the bishop a microscopic edge on average?!
Looking at wazirs and ferz' in a somewhat similar manner, a ferz is colour-bound but a wazir reaches less valuable (i.e. non-forward) directions more often than the former. That is, if we number forward directions as worth 2 and sideways and backward directions as worth 1, we see that the sum of the value of the wazir's 4 directions equals 2+1+1+1=5 while the the sum of the value of the ferz' directions equals 2+2+1+1=6, or slightly more than the wazir's sum. So, one advantage and disadvantage each so far for these two piece types.
Next, how do they compare in stopping or capturing chess-like pawns? Well, a ferz would normally not be able to capture a pawn in front of it (short of being aided by zugzwang), but note it will naturally stop it in a one on one battle, plus at times restrain the advance of a pawn on a neighbouring file, too (sometimes it will not, if the pawn pair may start as a phalanx). This may even suggest a ferz is worth (1+2)/2=1.5 pawns on average, at least in terms of restraining power. Now, how does a wazir stack up? A wazir can surely destroy a pawn in front of it, in a one on one battle. However it seldom even restrains two connected passed pawns; it may win one of them, but then the surviving neighbour will race ahead and unstoppably promote. So, a wazir is worth less than 2 pawns this suggests, but also that it's worth more than one pawn. Why not suppose it's 1.5 pawns, too? We do already know a wazir and a ferz already each have one advantage going for them over the other, from the last paragraph. Well, such ways of sizing up the values of piece types may seem like virtually guesswork, but at least I was pleasantly surprised to see that Ralph Betza in a CV Piece Article which alluded to his Chess with Different Armies variant gave the wazir and ferz as each having 'ideally' the same value as half a knight (not absolutely sure if this means these pieces should thus be theoretically entirely equal in value). I seem to recall this hasn't been the first time I've finished up weighing piece type attributes and apparently agreed with Mr. Betza's valuations (in this particular case I'm assuming a knight is worth 3 according to him, while I put it at Euwe's 3.5, however), though I've put less thought into such as a rule.
I would note that whether or not we suppose a ferz is worth 1.5 and a wazir the same (or even close), the compound of these two pieces is a guard. My primative way of evaluating a compound piece is to often just add a pawn's value to the sum of their values, so Guard=Ferz+Wazir+Pawn=1.5+1.5+1=4, which is (even if somewhat surprisingly) the fighting value a King was thought to have on chess' 8x8 board by a number of old-school chess greats including World Champion Lasker. If we take a Guard as worth 3.2 instead (as per computer study), and place a wazir at the low value of 1.25, that still doesn't give too much of a bonus for the compound of these two piece types in the form of the Guard, that is the bonus for compounding would be only 0.45, which somewhat surprises me, somehow.
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Question
# Let's say that the Federal Reserve purchases \$1 Million worth of U.S. Treasury bonds from a...
Let's say that the Federal Reserve purchases \$1 Million worth of U.S. Treasury bonds from a bond dealer in the open market, and the dealer's bank credits the dealer's account. The required reserve ratio is 15 percent, and the bank typically lends any excess reserves immediately. a) Assuming that no currency leakage occurs, how much will the bank be able to lend to its customers following the Fed's purchase? Please explain and show your calculations. b) Using the simple money multiplier, how much money will be created (assuming every dollar is lent out)? Please explain and show your calculations.
a) The fed has required reserve ratio as 15% which implies that 15% of the deposits needs to be reserved and rest lended. Now since there is no currency leakage , the bank lends the rest of the amount as loans.
The amount lended to borrowers is (1-0.15)*\$1 million
= 0.85*\$1 million =0.85 million
So the bank could lend 0.85 million \$ owing to the 15% reserve ratio it has to maintain
b) The formula for simple Money multiplier is given by
1/r*(new deposit) = 1million *(1/0.15) = 6.67 million \$.
So the money created from the deposits of\$1 million is \$ 6.67 million.
(You can comment for doubts)
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Academia vs. Business Title text: Some engineer out there has solved P=NP and it's locked up in an electric eggbeater calibration routine. For every 0x5f375a86 we learn about, there are thousands we never see.
## Explanation
Cueball has solved some tricky and very important problem in computer science, related to queueing theory.
The comic splits into two timelines. In the first he is showing the brilliant solution he'd developed to somebody who can appreciate its elegance, in this case that being an academic who can see the programmer's true brilliance and get him much-earned plaudits from the academic community.
In the alternate timeline, the boss does not possess the knowledge required to comprehend its import, and he simply sees the results without caring about the means Cueball used to attain them. He then gives Cueball another assignment, that may be vastly more workaday in nature. This, sadly, is the usual course of events in bureaucracy, which only seems to care about your results, not how you came about them. To drive in the point, the boss asks Cueball to do something as simple as setting up email on the office phones, a stark contrast to the skill and creativity Cueball would have needed to write his code in the first panel. It may even be imagined to be a "reward in itself" to casually hand over this new problem, however unsatisfying (or unsatisfiable) the new technical issue truly is.
The references in the title text are to the P versus NP problem, a famous unsolved problem in computer science, and the "magical constant" (0x5f375a86) used in finding the fast inverse square root, i.e. solving y=1/√x as fast as possible through a program – no-one knows quite who came up with this very useful bit of code (Now believed to be devised by Greg Walsh at Ardent Computer in consultation with Cleve Moler, the creator of MATLAB. see wikipedia), but it was discovered hiding in the graphics code of the video game Quake III Arena. Note that the actual constant used in the Quake III source code is 0x5f3759df, but the constant in the title text works also, and is actually slightly more accurate as shown in this paper: Fast inverse square root by CHRIS LOMONT (Purdue university, 2003).
The title text may be a reference to Stephen Jay Gould's quotation: “I am, somehow, less interested in the weight and convolutions of Einstein’s brain than in the near certainty that people of equal talent have lived and died in cotton fields and sweatshops.” originally about how great minds are suppressed due to racism and their genius go unknown, but could be interpreted as general exploitation by the commercial world.
## Transcript
[Cueball sits at a desk in front of a computer, leaning back in his chair with both hands down to his side. There are cans on the desk and more crushed ones on the floor.]
Cueball: I just wrote the most beautiful code of my life.
[Zoom in on Cueball and top half of desk.]
Cueball: They casually handed me an impossible problem. In 48 hours and 200 lines, I solved it.
[Curved lines with arrows divide the comic into two possible end panels, labeled "Academia" and "Business."]
Professor: My god... this will mean a half-dozen papers, a thesis or two, and a paragraph in every textbook on queuing theory!
Boss: You got the program to stop jamming up? Great. While you're fixing stuff, can you get Outlook to sync with our new phones?
# Discussion
I'm not convinced the problem solved in the comic panels is the fast inverse square root in the title text, as the academia panel implies that it impacts queuing theory, and I'm not sure what fast inv sqrt has to do with queuing theory. -- 204.89.186.1 (talk) (please sign your comments with ~~~~)
Agreed. Fast inv sqrt is clearly referenced in the title text, but the problem in the comic is something else. Alpha (talk) 01:18, 2 March 2013 (UTC)
I think the example of fast inverse square is more about the bizarrely elegant simplicity of the solution, rather than something related to the solved problem in the comic. (If the above comments are about text that has since been changed, my apologies.)Tryc (talk) 20:57, 3 July 2013 (UTC)
Actually 0x5f3759df is the mnagic number used in the fast inverse square root. Ref Wikipedia edokan 15:54, 23.08.2013 GMT+2
If this ever happened to me, I would quietly release the solution under the GNU license. My getting fired (possibly) is totally worth the public technological progress highly into the future. Greyson (talk) 13:29, 14 March 2013 (UTC)
The explanation is an interesting contrast to my interpretation. The meaning I got was that in academia, this discovery, like any new discovery, is interesting; but in business, this discovery has little practical application (apart from finishing what he was doing) so his boss didn't think twice about it. Maybe I'm too cynical.--18.215.1.155 01:23, 13 May 2013 (UTC)
Weird, my interpretation was different from both the one in the explanation and this one: Academia is too focused on authoring papers and making a name for oneself and therefore makes a much bigger deal about problem-solving than business, where solving difficult problems is just a regular part of the job. Which interpretation is correct? It likely depends on one's viewpoint! Ianrbibtitlht (talk) 03:00, 18 June 2019 (UTC)
Seeing the presence of cans... possibly alcoholic. Might it be possible that the Ballmer Peak was successfully invoked to reach his solution? 108.162.219.96 15:46, 1 December 2016 (UTC)
They look too small: this, plus the 48-hour time-frame, suggests to me that they're energy drinks. L-Space Traveler (talk) 19:37, 23 October 2022 (UTC)
Derailing the topic entirely, the old woman in the "Academia" panel seems to be a somewhat recurring character, complete with a semi-consistent personality. I propose "Bunhead" for future references. Anonymous17:39, 4 December 2013 (UTC)
I counter-propose 'MsBun'. 108.162.219.223 00:49, 7 January 2014 (UTC)
Oh, my goodness, "TruthInTelevision"? This isn't TvTropes!108.162.237.120 20:53, 19 January 2014 (UTC)
Can anyone remember an episode of Click (or any BBC computer programme) ever giving such in depth explanation of the graphics problem? I recall one showing the difference in game presentations then and "now" from around about the time the article claims information hit the mainstream but it was no more than 'advertising without naming names' a la Beeb.Weatherlawyer (talk) 07:59, 4 January 2015 (UTC)
There was a recent paper claiming to have solved P!=NP
I believe that the title text refers to a famous quote by Stephen Jay Gould: “I am, somehow, less interested in the weight and convolutions of Einstein’s brain than in the near certainty that people of equal talent have lived and died in cotton fields and sweatshops.” — it's not a 100% match but I think that was what he was going for. It certainly fits the concept that capitalism and business eat up the intellect of great people and leave nothing to show for it, in contract to the purity of science/academia. I will edit in a comment to that effect. AmbroseChapel (talk) 01:28, 21 September 2017 (UTC)
The big difference is, that the people from your quote were truly wasted talents, as they were doing mere manual labor. Meanwhile the title text talks about talents that are used, but never became popular.--Lupo (talk) 14:35, 17 October 2018 (UTC)
How is that difference meaningful? The point is that if some talent wasn't brought out to shine at its brightest, it was lost. What else exactly was done with it doesn't matter.172.70.250.67 23:01, 29 July 2022 (UTC)
Where does Gould specify that originally in that quote he was referring to racism and not capitalism? 172.70.250.67 23:01, 29 July 2022 (UTC)
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http://www.jiskha.com/display.cgi?id=1320030327
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# Calculus
posted by on .
What is the limit as t→∞ of L(t)?
L(t) = 34 - 32e^(-0.0719t)
Thank you so much!
• Calculus - ,
As t->∞ you have
L(t) = 34 - 32e-∞
= 34 - 32*0 = 34
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https://laurenceanywaysthemovie.com/how-many-dominoes-do-you-start-with-in-double-12/
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The same game can be played with a double-twelve set (91 tiles) or a double-nine set (55 tiles) domino sets. With a double-twelve set, four players would pick 12 tiles each and with a double-nine set, nine tiles would be taken at the start.
What are the dominoes in a double 12 set?
This set comes with 91 dominoes, plastic trains, and a starter piece—all packed into a convenient storage tin.
What games can you play with double-twelve dominoes?
CHICKENFOOT is a very popular domino game. Many Double-Twelve game sets are packaged with both Chickenfoot and Mexican Train domino hubs as the tiles and accessories from these sets may play both games. The object of Chickenfoot is to get the least possible amount of points (same thing in Mexican Train).
How do you count doubles in dominoes?
A double is only counted when it is on the end of the line of play. If a player cannot place a domino, he draws from the boneyard until he can make a play. In a two-person game, all but two of the dominoes in the boneyard may be drawn.
What are double-twelve dominoes used for?
A set of double-twelve dominoes allows you to play some of the more complicated domino games. They can also be used to add variety to the simpler games, or to allow more people to play them. A double-twelve set contains 91 dominoes, with the numbers on the tiles ranging from 0 (or blank) to 12.
What are the rules for a double domino game?
A double domino contains matching ends (6-6, 5-5, etc.), and the player who draws the highest double domino places it in the centre of the table to begin the game. If no double was drawn, all dominoes are returned to the draw pile, reshuffled, and redrawn.
How many tiles do you need to play dominoes?
The same game can be played with a double-twelve set (91 tiles) or a double-nine set (55 tiles) domino sets. With a double-twelve set, four players would pick 12 tiles each and with a double-nine set, nine tiles would be taken at the start.
How do you play The draw game in dominoes?
The Draw Game. When the sleeping dominoes run out, players simply pass their turn when they cannot go. For this variation, two players would start with 7 dominoes, three players with 5 tiles, four players with 4 tiles and five players with 3 tiles.
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http://math.stackexchange.com/users/152/grigory-m?tab=activity&sort=all&page=5
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Grigory M
Reputation
11,475
358/400 score
Aug 26 awarded Nice Question Aug 25 reviewed Close Odd prime combinatorics problem Aug 25 reviewed Close Trivial Graph theory questions Aug 23 reviewed No Action Needed Show that $\cos^n{\theta}\leq\cos{n\theta},\theta\in[0,\frac{\pi}{2}],n\in]0,1[$. Aug 23 comment Is the functional equation for $\zeta (s) \left(1-\frac{1}{3^{s-1}}\right)$ known? Yes, just change $2^{-s}$ to $3^{-s}$ (and $2^{-s-1}$ to $3^{-s-1}$) in the func. equations for $\eta$. Aug 21 comment The form of the zeta function of an elliptic curve over a finite field And of course a proof for $y^2=x^3-x$ goes back to Gauss (see e.g. Ireland, Rosen. A Classical Introduction to Modern Number Theory, ch. 8 for the proof) Aug 21 comment The form of the zeta function of an elliptic curve over a finite field Well, there is a relatively elementary proof in Silverman. The Arithmetic of Elliptic Curves Aug 21 reviewed No Action Needed The form of the zeta function of an elliptic curve over a finite field Aug 21 reviewed Reject Dirichlet's Divisor Problem Aug 21 reviewed Reject Graph Path Length Problem Aug 21 reviewed Reject Existence of a $\theta$ - Taylor Expansion Problem Aug 21 reviewed Looks OK Proving that $\sin x > \frac{(\pi^{2}-x^{2})x}{\pi^{2}+x^{2}}$ Aug 21 reviewed Reject How do I count the subsets of a set whose number of elements is divisible by 3? 4? Aug 18 reviewed Close How to show $\int_{0}^{\infty}e^{-x}x^{-1} dx = \infty$ Aug 16 reviewed Looks OK Help finding the limit of this series $\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots$ Aug 15 reviewed Looks OK Does square difference prove that 1 = 2? Aug 15 reviewed No Action Needed Why do we need to check for more than $\frac{\infty}{\infty}$ or $\frac{0}{0}$ when applying L'Hospital? Aug 15 reviewed Reviewed Separating family of functions for measures Aug 14 reviewed No Action Needed How do I integrate (1/polynomial) without using partial fractions? Aug 14 reviewed No Action Needed Extension of Pontryagin's principle
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