Gaussian distribution of NBA scores


Joseph Nebus:

The Prior Probability blog points out an interesting graph, showing the most common scores in basketball teams, based on the final scores of every NBA game. It’s actually got three sets of data there, one for all basketball games, one for games this decade, and one for basketball games of the 1950s. Unsurprisingly there’s many more results for this decade — the seasons are longer, and there are thirty teams in the league today, as opposed to eight or nine in 1954. (The Baltimore Bullets played fourteen games before folding, and the games were expunged from the record. The league dropped from eleven teams in 1950 to eight for 1954-1959.)

I’m fascinated by this just as a depiction of probability distributions: any team can, in principle, reach most any non-negative score in a game, but it’s most likely to be around 102. Surely there’s a maximum possible score, based on the fact a team has to get the ball and get into position before it can score; I’m a little curious what that would be.

Prior Probability itself links to another blog which reviews the distribution of scores for other major sports, and the interesting result of what the most common basketball score has been, per decade. It’s increased from the 1940s and 1950s, but it’s considerably down from the 1960s.

Originally posted on prior probability:

You can see the most common scores in such sports as basketball, football, and baseball in Philip Bump’s fun Wonkblog post here. Mr Bump writes: “Each sport follows a rough bell curve … Teams that regularly fall on the left side of that curve do poorly. Teams that land on the right side do well.” Read more about Gaussian distributions here.

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Reading the Comics, December 14, 2014: Pictures Gone Again? Edition


I’ve got enough comics to do a mathematics-comics roundup post again, but none of them are the King Features or Creators or other miscellaneous sources that demand they be included here in pictures. I could wait a little over three hours and give the King Features Syndicate comics another chance to say anything on point, or I could shrug and go with what I’ve got. It’s a tough call. Ah, what the heck; besides, it’s been over a week since I did the last one of these.

Bill Amend’s FoxTrot (December 7) bids to get posted on mathematics teachers’ walls with a bit of play on two common uses of the term “degree”. It’s also natural to wonder why the same word “degree” should be used to represent the units of temperature and the size of an angle, to the point that they even use the same symbol of a tiny circle elevated from the baseline as a shorthand representation. As best I can make out, the use of the word degree traces back to Old French, and “degré”, meaning a step, as in a stair. In Middle English this got expanded to the notion of one of a hierarchy of steps, and if you consider the temperature of a thing, or the width of an angle, as something that can be grown or shrunk then … I’m left wondering if the Middle English folks who extended “degree” to temperatures and angles thought there were discrete steps by which either quantity could change.

As for the little degree symbol, Florian Cajori notes in A History Of Mathematical Notations that while the symbol (and the ‘ and ” for minutes and seconds) can be found in Ptolemy (!), in describing Babylonian sexagesimal fractions, this doesn’t directly lead to the modern symbols. Medieval manuscripts and early printed books would use abbreviations of Latin words describing what the numbers represented. Cajori rates as the first modern appearance of the degree symbol an appendix, composed by one Jacques Peletier, to the 1569 edition of the text Arithmeticae practicae methods facilis by Gemma Frisius (you remember him; the guy who made triangulation into something that could be used for surveying territories). Peletier was describing astronomical fractions, and used the symbol to denote that the thing before it was a whole number. By 1571 Erasmus Reinhold (whom you remember from working out the “Prutenic Tables”, updated astronomical charts that helped convince people of the use of the Copernican model of the solar system and advance the cause of calendar reform) was using the little circle to represent degrees, and Tycho Brahe followed his example, and soon … well, it took a century or so of competing symbols, including “Grad” or “Gr” or “G” to represent degree, but the little circle eventually won out. (Assume the story is more complicated than this. It always is.)

Mark Litzer’s Joe Vanilla (December 7) uses a panel of calculus to suggest something particularly deep or intellectually challenging. As it happens, the problem isn’t quite defined well enough to solve, but if you make a reasonable assumption about what’s meant, then it becomes easy to say: this expression is “some infinitely large number”. Here’s why.

The numerator is the integral \int_{0}^{\infty} e^{\pi} + \sin^2\left(x\right) dx . You can think of the integral of a positive-valued expression as the area underneath that expression and between the lines marked by, on the left, x = 0 (the number on the bottom of the integral sign), and on the right, x = \infty (the number on the top of the integral sign). (You know that it’s x because the integral symbol ends with “dx”; if it ended “dy” then the integral would tell you the left and the right bounds for the variable y instead.) Now, e^{\pi} + \sin^2\left(x\right) is a number that depends on x, yes, but which is never smaller than e^{\pi} (about 23.14) and e^{\pi} + 1 (about 24.14). So the area underneath this expression has to be at least as big as the area within a rectangle that’s got a bottom edge at y = 0, a top edge at y = 23, a left edge at x = 0, and a right edge at x infinitely far off to the right. That rectangle’s got an infinitely large area. The area underneath this expression has to be no smaller than that.

Just because the numerator’s infinitely large doesn’t mean that the fraction is, though. It’s imaginable that the denominator is also infinitely large, and more wondrously, is large in a way that makes the ratio some more familiar number like “3”. Spoiler: it isn’t.

Actually, as it is, the denominator isn’t quite much of anything. It’s a summation; that’s what the capital sigma designates there. By convention, the summation symbol means to evaluate whatever expression there is to the right of it — in this case, it’s x^{\frac{1}{e}} + \cos\left(x\right) — for each of a series of values of some index variable. That variable is normally identified underneath the sigma, with a line such as x = 1, and (again by convention) for x = 2, x = 3, x = 4, and so on, until x equals whatever the number on top of the sigma is. In this case, the bottom doesn’t actually say what the index should be, although since “x” is the only thing that makes sense as a variable within the expression — “cos” means the cosine function, and “e” means the number that’s about 2.71828 unless it’s otherwise made explicit — we can suppose that this is a normal bit of shorthand like you use when context is clear.

With that assumption about what’s meant, then, we know the denominator is whatever number is represented by \left(1^{\frac{1}{e}} + \cos\left(1\right)\right) + \left(2^{\frac{1}{e}} + \cos\left(2\right)\right) + \left(3^{\frac{1}{e}} + \cos\left(3\right)\right) + \left(4^{\frac{1}{e}} + \cos\left(4\right)\right) +  \cdots + \left(10^{\frac{1}{e}} + \cos\left(10\right)\right) (and 1/e is about 0.368). That’s a number about 16.549, which falls short of being infinitely large by an infinitely large amount.

So, the original fraction shown represents an infinitely large number.

Mark Tatulli’s Lio (December 7) is another “anthropomorphic numbers” genre comic, and since it’s Lio the numbers naturally act a bit mischievously.

Greg Evans’s Luann Againn (December 7, I suppose technically a rerun) only has a bit of mathematical content, as it’s really playing more on short- and long-term memories. Normal people, it seems, have a buffer of something around eight numbers that they can remember without losing track of them, and it’s surprisingly easy to overload that. I recall reading, I think in Joseph T Hallinan’s Why We Make Mistakes: How We Look Without Seeing, Forget Things In Seconds, And Are All Pretty Sure We are Way Above Average, and don’t think I’m not aware of how funny it would be if I were getting this source wrong, that it’s possible to cheat a little bit on the size of one’s number-buffer.

Hallinan (?) gave the example of a runner who was able to remember strings of dozens of numbers, well past the norm, but apparently by the trick of parsing numbers into plausible running times. That is, the person would remember “834126120820” perfectly because it could be expressed as four numbers, “8:34, 1:26, 1:20, 8:20”, that might be credible running times for something or other and the runner was used to remembering such times. Supporting the idea that this trick was based on turning a lot of digits into a few small numbers was that the runner would be lost if the digits could not be parsed into a meaningful time, like, “489162693077”. So, in short, people are really weird in how they remember and don’t remember things.

Harley Schwadron’s 9 to 5 (December 8) is a “reluctant student” question who, in the tradition of kids in comic strips, tosses out the word “app” in the hopes of upgrading the action into a joke. I’m sympathetic to the kid not wanting to do long division. In arithmetic the way I was taught it, this was the first kind of problem where you pretty much had to approximate and make a guess what the answer might be and improve your guess from that starting point, and that’s a terrifying thing when, up to that point, arithmetic has been a series of predictable, discrete, universally applicable rules not requiring you to make a guess. It feels wasteful of effort to work out, say, what seven times your divisor is when it turns out it’ll go into the dividend eight times. I am glad that teaching approaches to arithmetic seem to be turning towards “make approximate or estimated answers, and try to improve those” as a general rule, since often taking your best guess and then improving it is the best way to get a good answer, not just in long division, and the less terrifying that move is, the better.

Justin Boyd’s Invisible Bread (December 12) reveals the joy and the potential menace of charts and graphs. It’s a reassuring red dot at the end of this graph of relevant-graph-probabilities.

Several comics chose to mention the coincidence of the 13th of December being (in the United States standard for shorthand dating) 12-13-14. Chip Sansom’s The Born Loser does the joke about how yes, this sequence won’t recur in (most of our) lives, but neither will any other. Stuart Carlson and Jerry Resler’s Gray Matters takes a little imprecision in calling it “the last date this century to have a consecutive pattern”, something the Grays, if the strip is still running, will realize on 1/2/34 at the latest. And Francesco Marciuliano’s Medium Large uses the neat pattern of the dates as a dip into numerology and the kinds of manias that staring too closely into neat patterns can encourage.

The Short, Unhappy Life Of A Doomed Conjecture


So last month amongst the talk about the radius of a circle inscribed in a Pythagorean right triangle I mentioned that I had, briefly, floated a conjecture that might have spun off it. It didn’t, though I promised to describe the chain of thought I had while exploring it, on the grounds that the process of coming up with mathematical ideas doesn’t get described much, and certainly doesn’t get described for the sorts of fiddling little things that make up a trifle like this.

A triangle with sides a, b, and c, and an inscribed circle. From the center of the circle are lines going to the vertices of the triangle, dividing the circle into three smaller triangles, with bases of lengths, a, b, and c respectively and all with the same height, r, the radius of the inscribed circle.

A triangle (meant to be a right triangle) with an inscribed circle of radius r. The triangle is divided into three smaller triangles meeting at the center of the inscribed circle.

The point from which I started was a question about the radius of a circle inscribed in the right triangle with legs of length 5, 12, and 13. This turns out to have a radius of 2, which is interesting because it’s a whole number. It turns out to be simple to show that for a Pythagorean right triangle, that is, a right triangle whose legs are a Pythagorean triple — like (3, 4, 5), or (5, 12, 13), any where the square of the biggest number is the same as what you get adding together the squares of the two smaller numbers — the inscribed circle has a radius that’s a whole number. For example, the circle you could inscribe in a triangle of sides 3, 4, and 5 would have radius 1. The circle inscribed in a triangle of sides 8, 15, and 17 would have radius 3; so does the circle inscribed in a triangle of sides 7, 24, and 25.

Since I now knew that (and in multiple ways: HowardAt58 had his own geometric solution, and you can also do this algebraically) I started to wonder about the converse. If a Pythagorean right triangle’s inscribed circle has a whole number for a radius, can does knowing a circle has a whole number for a radius tell us anything about the triangle it’s inscribed in? This is an easy way to build new conjectures: given that “if A is true, then B must be true”, can it also be that “if B is true, then A must be true”? Only rarely will that be so — it’s neat when it is — but we might be able to patch something up, like, “if B, C, and D are all simultaneously true, then A must be true”, or perhaps, “if B is true, then at least E must be true”, where E resembles A but maybe doesn’t make such a strong claim. Thus are tiny little advances in mathematics created.

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What Do I Need To Pass This Class? (December 2014 Edition)


It’s finals season, at least for colleges that run on a semesterly schedule, and a couple of my posts are turning up in search query results again. So I thought it worth drawing a little more attention to them and hopefully getting people what they need sooner.

The answer: you need to study a steady but not excessive bit every night from now to before the exam; you need to get a full night of sleep before the exam; and you really needed to pay attention in class and do the fiddly little assignments all semester, so, sorry it’s too late for that. Also you need to not pointlessly antagonize your professor; even if you don’t like this class, you could have taken others to meet your academic requirement, so don’t act like you were dragged into Topics in Civilization: Death against your will even if it does satisfy three general-education requirements at the cost of being a 7:50 am section.

Anyway, that doesn’t help figuring out whether you can relax as soon as you get 82.3 percent of the final right or if you have to strain to get that 82.6. So let me point to those: What Do I Need To Pass This Class? (December 2013 Edition) gives an expression for working out the score you need, and shows how to develop that formula, based on things like the pre-final grade, the weight given the exam, and extra credit (or demerits) that you’ve received, and is therefore good for absolutely any weighted-average based course grade you might have.

That also involves formulas, though, and I know that makes people nervous, so What Do I Need To Get An A In This Class? simplifies matters a bit by working out a couple common cases: for finals worth 40, 33, 30, 25, and 20 percent of the class, based on pre-final averages, what final exam grade do you need to get to at least a given level. Good luck, but you really shouldn’t be scrounging for points. Study because it’s fun to learn things and the grades will be good of their own accord.

Reading the Comics, December 5, 2014: Good Questions Edition


This week’s bundle of mathematics-themed comic strips has a pretty nice blend, to my tastes: about half of them get at good and juicy topics, and about half are pretty quick and easy things to describe. So, my thanks to Comic Strip Master Command for the distribution.

Bill Watterson’s Calvin and Hobbes (December 1, rerun) slips in a pretty good probability question, although the good part is figuring out how to word it: what are the chances Calvin’s Dad was thinking of 92,376,051 of all the possible numbers out there? Given that there’s infinitely many possible choices, if every one of them is equally likely to be drawn, then the chance he was thinking of that particular number is zero. But Calvin’s Dad couldn’t be picking from every possible number; all humanity, working for its entire existence, will only ever think of finitely many numbers, which is the kind of fact that humbles me when I stare too hard at it. And people, asked to pick a number, have ones they prefer: 37, for example, or 17. Christopher Miller’s American Cornball: A Laffopedic Guide To The Formerly Funny (a fine guide to jokes that you see lingering around long after they were actually funny) notes that what number people tend to pick seems to vary in time, and in the early 20th century 23 was just as funny a number as you could think of on a moment’s notice.

Zach Weinersmith’s Saturday Morning Breakfast Cereal (December 1) is entitled “how introductory physics problems are written”, and yeah, that’s about the way that a fair enough problem gets rewritten so as to technically qualify as a word problem. I think I’ve mentioned my favorite example of quietly out-of-touch word problems, a 1970s-era probability book which asked the probability of two out of five transistors in a radio failing. That was blatantly a rewrite of a problem about a five-vacuum-tube radio (my understanding is many radios in that era used five tubes) and each would have a non-negligible chance of failing on any given day. But that’s a slightly different problem, as the original question would have made good sense when it was composed, and it only in the updating became ridiculous.

Julie Larson’s The Dinette Set (December 2) illustrates one of the classic sampling difficulties: how can something be generally true if, in your experience, it isn’t? If you make the reasonable assumption that there’s nothing exceptional about you, then, shouldn’t your experience of, say, fraction of people who exercise, or average length of commute, or neighborhood crime rate be tolerably close to what’s really going on? You could probably build an entire philosophy-of-mathematics course around this panel before even starting the question of how do you get a fair survey of a population.

Scott Hilburn’s The Argyle Sweater (December 3) tells a Roman numeral joke that actually I don’t remember encountering before. Huh.

Samson’s Dark Side Of The Horse (December 3) does some play with mathematical symbols and of course I got distracted by thinking what kind of problem Horace was working on in the first panel; it looks obvious to me that it’s something about the orbit of one body around another. In principle, it might be anything, since the great discovery of algebra is that you can replace numbers with symbols like “a” and work out relations without having to know anything about them. “G”, for example, tends to mean the gravitational constant of the universe, and “GM” makes this identification almost certain: gravitation problems need the masses of a main body, like a planet, and a smaller body, like a satellite, and that’s usually represented as either m1 and m2 or as M and m.

In orbital mechanics problems, “a” often refers to the semimajor axis — the long diameter of the ellipse the orbiting body makes — and “e” the eccentricity — a measure of how close to a circle the ellipse is (an eccentricity of zero means it’s a circle — but the fact that there’s subscripts of k makes that identification suspect: subscripts are often used to distinguish which of multiple similar things you mean to talk about, and if it’s just one body orbiting the other there’s no need for that. So what is Horace working on?

The answer is: Horace is working on an orbital perturbation problem, describing how far from the true circular orbit a satellite will drift when you consider things like atmospheric drag and the slightly non-spherical shape of the Earth. ak is still a semi-major axis and ek the eccentricity, but of the little changes from the original orbit, rather than the original orbit themselves. And now I wonder if Samson plucked the original symbol just because it looked so graphically pleasant, or if Samson was slipping in a further joke about the way an attractive body will alter another body’s course.

Jenny Campbell’s Flo and Friends (December 4) offers a less exciting joke: it’s a simple word problem joke, playing on the ambiguity of “calculate how many seconds there are in the year”. Now, the dull way to work this out is to multiply 60 seconds per minute times 60 minutes per hour times 24 hours per day times 365 (or 365.25, or 365.2422 if you want to start that argument) days per year. But we can do almost as well and purely in our heads, if we remember that a million seconds is almost twelve days long. How many twelve-day stretches are there in a year? Well, right about 31 — after all, the year is (nearly) 12 groups of 31 days, and therefore it’s also 31 groups of 12 days. Therefore the year is about 31 million seconds long. If we pull out the calculator we find that a 365-day year is 31,536,000 seconds, but isn’t it more satisfying to say “about 31 million seconds” like we just did?

John Deering’s Strange Brew (December 4) took me the longest time to work out what the joke was supposed to be. I’m still not positive but I think it’s just one colleague sneering at the higher mathematics of another.

Todd the Dinosaur's abacus only goes up to 2.

Patrick Roberts’s Todd the Dinosaur (December 5) discovers there are numbers bigger than 2.

Patrick Roberts’s Todd the Dinosaur (December 5) discovers that some numbers are quite big ones, actually. There is a challenge in working with really big numbers, even if they’re usually bigger than 2. Usually we’re not interested in a number by itself, and would rather do some kind of calculation with it, and that’s boring to do too much of, but a computer can only work with so many digits at once. The average computer uses floating point arithmetic schemes which will track, at most, about 19 decimal digits, on the reasonable guess that twenty decimal digits is the difference between 3.1415926535897932384 and 3.1415926535897932385 and how often is that difference — a millionth of a millionth of a millionth of a percent — going to matter? If it does, then, you do the kind of work that gets numerical mathematicians their big paydays: using schemes that work with more digits, or chopping up a problem so that you never have to use all 86 relevant digits at once, or rewriting your calculation so that you don’t need so many digits of accuracy all at once.

David Bloomier, former math teacher, has the number 4 car, identified by addition, division, square root, and finger count.

Daniel Beyer’s Offbeat Comics (December 5) gives four ways to represent the number 4. Five, if you count the caption.

Daniel Beyer’s Offbeat Comics (December 5) gives a couple of ways to express the number 4 — including, look closely, holding up fingers — as part of a joke about the driver being a former mathematics teacher.

Greg Cravens’s The Buckets (December 5) is the old, old, old joke about never using algebra in real life. Do English teachers get this same gag about never using the knowledge of how to diagram sentences? In any case, I did use my knowledge of sentence-diagramming, and for the best possible application: I made fun of a guy on the Internet with it.

I advise against reading the comments — I mean, that’s normally good advice, but comic strips attract readers who want to complain about how stupid kids are anymore and strips that mention education give plenty of grounds for it — but I noticed one of the early comments said “try to do any repair at home without some understanding of it”. I like the claim, but, I can’t think of any home repair I’ve done that’s needed algebra. The most I’ve needed has been working out the area of a piece of plywood I needed, but if multiplying length by width is algebra than we’ve badly debased the term. Even my really ambitious project, building a PVC-frame pond cover, isn’t going to be one that uses algebra unless we take an extremely generous view of the subject.

Advanced November 2014 Statistics


So that little bit I added in my last statistics post, tracking how many days went between the first and the last reading of an article according to WordPress’s figures? I was curious, and went through my posts from mid-October through mid-November to see how long the readership lifespan of an average post was. I figured stuff after mid-November may not have quite had long enough for people to gradually be done with it.

I’d expected the typical post to have what’s called a Poisson distribution, in number of page views per day, with a major peak in the first couple days after it’s published and then, maybe, a long stretch of exceedingly minor popularity. I think that’s what’s happening, although the problem of small numbers means it’s a pretty spotty pattern. Also confounding things is that a post can sometimes get a flurry of publicity long after its main lifespan has passed. So I decided to count both how long each post had between its first and last-viewed days, and also the “first span”, how many days it was until the first day without page views, to use as proxy for separating out late revivals.

Post Days Read First Span
How To Numerically Integrate Like A Mathematician 45 8
Reading the Comics, October 14, 2014: Not Talking About Fourier Transforms Edition 25 7
How Richard Feynman Got From The Square Root of 2 to e 41 4
Reading The Comics, October 20, 2014: No Images This Edition 5 5
Calculus without limits 5: log and exp 25 3
Reading the Comics, October 25, 2014: No Images Again Edition 28 2
How To Hear Drums 14 6
My Math Blog Statistics, October 2014 30 4
Reading The Comics, November 4, 2014: Will Pictures Ever Reappear Edition 9 6
Echoing “Fourier Echoes Euler” 12 5
Some Stuff About Edmond Halley 11 2
Reading The Comics, November 9, 2014: Finally, A Picture Edition 11 4
About An Inscribed Circle 13 5
Reading The Comics, November 14, 2014: Rectangular States Edition 15 1
Radius of the inscribed circle of a right angled triangle 12 5

For what it’s worth, the mean lifespan of a post is 19.7 days, with standard deviation of 12.0 days. The mean lifespan of the first flush of popularity is 4.5 days, with a standard deviation of 1.9 days.

I suspect the thing that brings out these late rushes of popularity are things like the monthly roundup posts, which send people back to articles whose lifespans had expired weeks before; or when there’s a running thread as in the circle-inscribed-in-a-triangle theme that encourages people to go back again and again. And I’m curious how long articles would last without this sort of threading between them.

My Math Blog Statistics, November 2014


October 2014 was my fourth-best month in the mathematics blog here, if by “best” we mean “has a number of page views”, and second-best if by “best” we mean “has a number of unique visitors”. And now November 2014 has taken October’s place on both counts, by having bigger numbers for both page views and visitors, as WordPress reveals such things to me. Don’t tell October; that’d just hurt its feelings. Plus, I got to the 19,000th page view, and as of right now I’m sitting at 19,181; it’s conceivable I might reach my 20,000th viewer this month, though that would be a slight stretch.

But the total number of page views grew from 625 up to 674, and the total number of visitors from 323 to 366. The number of page views is the highest since May 2014 (751), although this is the greatest number of visitors since January 2014 (473), the second month when WordPress started revealing those numbers to us mere bloggers. I like the trends, though; since June the number of visitors has been growing at a pretty steady rate, although steadily enough I can’t say whether it’s an arithmetic or geometric progression. (In an arithmetic progression, the difference between two successive numbers is about constant, for example: 10, 15, 20, 25, 30, 35, 40. In a geometric progression, the ratio between two successive numbers is about constant, for example: 10, 15, 23, 35, 53, 80, 120.) Views per visitor dropped from 1.93 to 1.84, although I’m not sure even that is a really significant difference.

The countries sending me the most readers were just about the same set as last month: the United States at 458; Canada recovering from a weak October with 27 viewers; Argentina at 20; Austria and the United Kingdom tied at 19; Australia at 17; Germany at 16 and Puerto Rico at 14.

Sending only one reader this month were: Belgium, Bermuda, Croatia, Estonia, Guatemala, Hong Kong, Italy, Lebanon, Malaysia, the Netherlands, Norway, Oman, the Philippines, Romania, Singapore, South Korea, and Sweden. (Really, Singapore? I’m a little hurt. I used to live there.) The countries repeating that from October were Estonia, the Netherlands, Norway, and Sweden; Sweden’s going on three months with just a single reader each. I don’t know what’s got me only slightly read in Scandinavia and the Balkans.

My most-read articles for November were pretty heavily biased towards the comics, with a side interest in that Pythagorean triangle problem with an inscribed circle. Elke Stangl had wondered about the longevity of my most popular posts, and I was curious too, so I’m including in brackets a note about the number of days between the first and the last view which WordPress has on record. This isn’t a perfect measure of longevity, especially for the most recent posts, but it’s a start.

As ever there’s no good search term poetry, but among the things that brought people here were:

  • trapezoid
  • how many grooves are on one side of an lp record?
  • origin is the gateway to your entire gaming universe.
  • cauchy funny things done
  • trapezoid funny
  • yet another day with no plans to use algebra

Won’t lie; that last one feels a little personal. But the “origin is the gateway” thing keeps turning up and I don’t know why. I’d try to search for it but that’d just bring me back here, leaving me no more knowledgeable, wouldn’t it?

Reading the Comics, November 28, 2014: Greatest Hits Edition?


I don’t ever try speaking for Comic Strip Master Command, and it almost never speaks to me, but it does seem like this week’s strips mentioning mathematical themes was trying to stick to the classic subjects: anthropomorphized numbers, word problems, ways to measure time and space, under-defined probability questions, and sudoku. It feels almost like a reunion weekend to have all these topics come together.

Dan Thompson’s Brevity (November 23) is a return to the world-of-anthropomorphic-numbers kind of joke, and a pun on the arithmetic mean, which is after all the statistic which most lends itself to puns, just edging out the “range” and the “single-factor ANOVA F-Test”.

Phil Frank Joe Troise’s The Elderberries (November 23, rerun) brings out word problem humor, using train-leaves-the-station humor as a representative of the kinds of thinking academics do. Nagging slightly at me is that I think the strip had established the Professor as one of philosophy and while it’s certainly not unreasonable for a philosopher to be interested in mathematics I wouldn’t expect this kind of mathematics to strike him as very interesting. But then there is the need to get the idea across in two panels, too.

Jonathan Lemon’s Rabbits Against Magic (November 25) brings up a way of identifying the time — “half seven” — which recalls one of my earliest essays around here, “How Many Numbers Have We Named?”, because the construction is one that I find charming and that was glad to hear was still current. “Half seven” strikes me as similar in construction to saying a number as “five and twenty” instead of “twenty-five”, although I’m ignorant as to whether the actually is any similarity.

Scott Hilburn’s The Argyle Sweater (November 26) brings out a joke that I thought had faded out back around, oh, 1978, when the United States decided it wasn’t going to try converting to metric after all, now that we had two-liter bottles of soda. The curious thing about this sort of hyperconversion (it’s surely a satiric cousin to the hypercorrection that makes people mangle a sentence in the misguided hope of perfecting it) — besides that the “yard” in Scotland Yard is obviously not a unit of measure — is the notion that it’d be necessary to update idiomatic references that contain old-fashioned units of measurement. Part of what makes idioms anything interesting is that they can be old-fashioned while still making as much sense as possible; “in for a penny, in for a pound” is a sensible thing to say in the United States, where the pound hasn’t been legal tender since 1857; why would (say) “an ounce of prevention is worth a pound of cure” be any different? Other than that it’s about the only joke easily found on the ground once you’ve decided to look for jokes in the “systems of measurement” field.

Mark Heath’s Spot the Frog (November 26, rerun) I’m not sure actually counts as a mathematics joke, although it’s got me intrigued: Surly Toad claims to have a stick in his mouth to use to give the impression of a smile, or 37 (“Sorry, 38”) other facial expressions. The stick’s shown as a bundle of maple twigs, wound tightly together and designed to take shapes easily. This seems to me the kind of thing that’s grown as an application of knot theory, the study of, well, it’s almost right there in the name. Knots, the study of how strings of things can curl over and around and cross themselves (or other strings), seemed for a very long time to be a purely theoretical playground, not least because, to be addressable by theory, the knots had to be made of an imaginary material that could be stretched arbitrarily finely, and could be pushed frictionlessly through it, which allows for good theoretical work but doesn’t act a thing like a shoelace. Then I think everyone was caught by surprise when it turned out the mathematics of these very abstract knots also describe the way proteins and other long molecules fold, and unfold; and from there it’s not too far to discovering wonderful structures that can change almost by magic with slight bits of pressure. (For my money, the most astounding thing about knots is that you can describe thermodynamics — the way heat works — on them, but I’m inclined towards thermodynamic problems.)

Veronica was out of town for a week; Archie's test scores improved. This demonstrates that test scores aren't everything.

Henry Scarpelli and Crag Boldman’s Archie for the 28th of November, 2014. Clearly we should subject this phenomenon to scientific inquiry!

Henry Scarpelli and Crag Boldman’s Archie (November 28, rerun) offers an interesting problem: when Veronica was out of town for a week, Archie’s test scores improved. Is there a link? This kind of thing is awfully interesting to study, and awfully difficult to: there’s no way to run a truly controlled experiment to see whether Veronica’s presence affects Archie’s test scores. After all, he never takes the same test twice, even if he re-takes a test on the same subject (and even if the re-test were the exact same questions, he would go into it the second time with relevant experience that he didn’t have the first time). And a couple good test scores might be relevant, or might just be luck, or it might be that something else happened to change that week that we haven’t noticed yet. How can you trace down plausible causal links in a complicated system?

One approach is an experimental design that, at least in the psychology textbooks I’ve read, gets called A-B-A, or A-B-A-B, experiment design: measure whatever it is you’re interested in during a normal time, “A”, before whatever it is whose influence you want to see has taken hold. Then measure it for a time “B” where something has changed, like, Veronica being out of town. Then go back as best as possible to the normal situation, “A” again; and, if your time and research budget allow, going back to another stretch of “B” (and, hey, maybe even “A” again) helps. If there is an influence, it ought to appear sometime after “B” starts, and fade out again after the return to “A”. The more you’re able to replicate this the sounder the evidence for a link is.

(We’re actually in the midst of something like this around home: our pet rabbit was diagnosed with a touch of arthritis in his last checkup, but mildly enough and in a strange place, so we couldn’t tell whether it’s worth putting him on medication. So we got a ten-day prescription and let that run its course and have tried to evaluate whether it’s affected his behavior. This has proved difficult to say because we don’t really have a clear way of measuring his behavior, although we can say that the arthritis medicine is apparently his favorite thing in the world, based on his racing up to take the liquid and his trying to grab it if we don’t feed it to him fast enough.)

Ralph Hagen’s The Barn (November 28) has Rory the sheep wonder about the chances he and Stan the bull should be together in the pasture, given how incredibly vast the universe is. That’s a subtly tricky question to ask, though. If you want to show that everything that ever existed is impossibly unlikely you can work out, say, how many pastures there are on Earth multiply it by an estimate of how many Earth-like planets there likely are in the universe, and take one divided by that number and marvel at Rory’s incredible luck. But that number’s fairly meaningless: among other obvious objections, wouldn’t Rory wonder the same thing if he were in a pasture with Dan the bull instead? And Rory wouldn’t be wondering anything at all if it weren’t for the accident by which he happened to be born; how impossibly unlikely was that? And that Stan was born too? (And, obviously, that all Rory and Stan’s ancestors were born and survived to the age of reproducing?)

Except that in this sort of question we seem to take it for granted, for instance, that all Stan’s ancestors would have done their part by existing and doing their part to bringing Stan around. And we’d take it for granted that the pasture should exist, rather than be a farmhouse or an outlet mall or a rocket base. To come up with odds that mean anything we have to work out what the probability space of all possible relevant outcomes is, and what the set of all conditions that satisfy the concept of “we’re stuck here together in this pasture” is.

Mark Pett’s Lucky Cow (November 28) brings up sudoku puzzles and the mystery of where they come from, exactly. This prompted me to wonder about the mechanics of making sudoku puzzles and while it certainly seems they could be automated pretty well, making your own amounts to just writing the digits one through nine nine times over, and then blanking out squares until the puzzle is hard. A casual search of the net suggests the most popular way of making sure you haven’t blanking out squares so that the puzzle becomes unsolvable (in this case, that there’s two or more puzzles that fit the revealed information) is to let an automated sudoku solver tell you. That’s true enough but I don’t see any mention of any algorithms by which one could check if you’re blanking out a solution-foiling set of squares. I don’t know whether that reflects there being no algorithm for this that’s more efficient than “try out possible solutions”, or just no algorithm being more practical. It’s relatively easy to make a computer try out possible solutions, after all.

A paper published by Mária Ercsey-Ravasz and Zoltán Toroczkai in Nature Scientific Reports in 2012 describes the recasting of the problem of solving sudoku into a deterministic, dynamical system, and matches the difficulty of a sudoku puzzle to chaotic behavior of that system. (If you’re looking at the article and despairing, don’t worry. Go to the ‘Puzzle hardness as transient chaotic dynamics’ section, and read the parts of the sentence that aren’t technical terms.) Ercsey-Ravasz and Toroczkai point out their chaos-theory-based definition of hardness matches pretty well, though not perfectly, the estimates of difficulty provided by sudoku editors and solvers. The most interesting (to me) result they report is that sudoku puzzles which give you the minimum information — 17 or 18 non-blank numbers to start — are generally not the hardest puzzles. 21 or 22 non-blank numbers seem to match the hardest of puzzles, though they point out that difficulty has got to depend on the positioning of the non-blank numbers and not just how many there are.

No Conjecture For 19,000


I failed to notice when it happened but my little blog here reached its 19,000th page view sometime on Saturday the 22nd. I’m sorry to have missed it; I like keeping track of these little milestones and I expect to be pretty self-confident if and when I hit 20,000. I know these aren’t enormous numbers, as even mathematics blogging goes, but I do feel like I’m getting a larger audience, that’s sometimes also more engaged, and that’s gratifying.

Since I don’t want to just seem to be bragging about a really minor accomplishment I hoped to include a conjecture that would be a nice little puzzle, following up on the right-triangle stuff done so recently here. But in writing out the problem exactly, I realized that the conjecture I had in mind was (a) false, and (b) really obviously false. Which is a shame, but so many attempts at figuring out something turn out that way. I’m deciding whether to swallow my pride and lay out the line of thought that ended up in disappointment, on the grounds that you never get to see a mathematician go through the stages of discovery only to come up with a flop; that’s not the sort of thing that gets into papers, much less textbooks, unless there’s a good juicy scandal behind it. We’ll see.

Another Reason Why It’s Got To Be 2


To circle back around that inscribed circle problem, about what the radius of the circle that just fits inside a right triangle with sides of length 5, 12, and 13: I’d had an approach for solving it different from HowardAt58’s geometric answer. This isn’t to imply that his answer’s wrong, I should point out: problems can often be solved by several different yet equally valid approaches. (It might almost be the definition of cutting-edge research if it’s a problem there’s only one approach for.)

A triangle with sides a, b, and c, with an inscribed circle, and three radial lines, one reaching to each side.

Figure 1. A triangle (meant to be a right triangle) with an inscribed circle of radius r. Three radial lines, perpendicular to the bases they touch, are included.

So here’s another geometry-based approach to finding what the radius of the circle that just fits inside the triangle has to be. We started off with the right triangle, and sides a and b and c; and there’s a circle inscribed in it. This is the biggest circle that’ll fit within the triangle. The circle has some radius, and we’ll just be a little daring and original and use the symbol “r” to stand for that radius. We can draw a line from the center of the circle to the point where the circle touches each of the legs, and that line is going to be of length r, because that’s the way circles work. My drawing, Figure 1, looks a little bit off because I was sketching this out on my iPad and being more exact about all this was just so, so much work.

The next step is to add three more lines to the figure, and this is going to make it easier to see what we want. What we’re adding are liens that go from the center of the circle to each of the corners of the original triangle. This divides the original triangle into three smaller ones, which I’ve lightly colored in as amber (on the upper left), green (on the upper right), and blue (on the bottom). The coloring is just to highlight the new triangles. I know the figure is looking even sketchier; take it up with how there’s no good mathematics-diagram sketching programs for a first-generation iPad, okay?

A triangle with sides a, b, and c, and an inscribed circle. From the center of the circle are lines going to the vertices of the triangle, dividing the circle into three smaller triangles, with bases of lengths, a, b, and c respectively and all with the same height, r, the radius of the inscribed circle.

Figure 2. A triangle (meant to be a right triangle) with an inscribed circle of radius r. The triangle is divided into three smaller triangles meeting at the center of the inscribed circle.

If we can accept my drawings for what they are already, then, there’s the question of why I did all this subdividing, anyway? The good answer is: looking at this Figure 2, do you see what the areas of the amber, green, and blue triangles have to be? Well, the area of a triangle generally is half its base times its height. A base is the line connecting two of the vertices, and the height is the perpendicular distane between the third vertex and that base. So, for the amber triangle, “a” is obviously a base, and … say, now, isn’t “r” the height?

It is: the radius line is perpendicular to the triangle leg. That’s how inscribed circles work. You can prove this, although you might convince yourself of it more quickly by taking the lid of, say, a mayonnaise jar and a couple of straws. Try laying down the straws so they just touch the jar at one point, and so they cross one another (forming a triangle), and try to form a triangle where the straw isn’t perpendicular to the lid’s radius. That’s not proof, but, it’ll probably leave you confident it could be proven.

So coming back to this: the area of the amber triangle has to be one-half times a times r. And the area of the green triangle has to be one-half times b times r. The area of the blue triangle, yeah, one-half times c times r. This is great except that we have no idea what r is.

But we do know this: the amber triangle, green triangle, and blue triangle together make up the original triangle we started with. So the areas of the amber, green, and blue triangles added together have to equal the area of the original triangle, and we know that. Well, we can calculate that anyway. Call that area “A”. So we have this equation:

\frac12 ar + \frac12 br + \frac12 cr = A

Where a, b, and c we know because those are the legs of the triangle, and A we may not have offhand but we can calculate it right away. The radius has to be twice the area of the original triangle divided by the sum of a, b, and c. If it strikes you that this is twice the area of the circle divided by its perimeter, yeah, that it is.

Incidentally, we haven’t actually used the fact that this is a right triangle. All the reasoning done would work if the original triangle were anything — equilateral, isosceles, scalene, whatever you like. If the triangle is a right angle, the area is easy to work out — it’s one-half times a times b — but Heron’s Formula tells us the area of a triangle knowing nothing but the lengths of its three legs. So we have this:

(Right triangle)

r = \frac{1}{a + b + c} \cdot \left(a\cdot b\right)

(Arbitrary triangle)

r = \frac{1}{a + b + c} \cdot 2 \sqrt{p\cdot(p - a)\cdot(p - b)\cdot(p - c)} \mbox{ where }  p = \frac12\left(a + b + c\right) .

Since we started out with a Pythagorean right triangle, with sides 5, 12, and 13, then: a = 5, b = 12, c = 13; a times b is 60; a plus b plus c is 30; and therefore the radius of the inscribed circle is 60 divided by 30, or, 2.