God Plays Dice

I've been busy, but here's a quick one: The quadratic equation, Dr. Seuss style, by Katie Benedetto. My favorite stanza:

Each value there was, well she renamed each one
Claiming that nice names would make this more fun.
“Variables, well, they’re whatever we wish
So like one, two, red, blue, I’ll call this one “fish”!

This is the antidote to if the IRS had discovered the quadratic formula.

How many solutions does a typical quadratic equation have? Discuss.

(I'm deliberately leaving the problem vague, because it's more interesting that way.)

From everything.com: what are finite fields, anyway? From this I learned that "number sets grow when a mathematician and an equation love each other very much." And it's a pretty good exposition of the construction of a finite field, too.

By Igor Kriz and Paul Siegel, at Scientific American: Rubik's Cube Inspired Puzzles Demonstrate Math's "Simple Groups".

The Rubik's cube can be said to be a physical embodiment of the Rubik group, which is a certain subgroup of S₄₈. (Why 48? There are 54 "facets" of the Rubik's cube, nine on each side; six of these don't move, leaving 48.) This subgroup has an easy presentation with six generators, namely rotations of the six faces. As the authors point out, other "Rubik's-type" puzzles embody groups of the same general sort.

The authors have invented puzzles based on certain sporadic groups, namely theMathieu groups M₁₂ and M₂₄ and the Conway group Co₀. These right now only exist as computer programs, although the authors claim that a physical version of the M₂₄ puzzle could be built.

A possibly interesting, although not-at-all well-defined question -- which groups are "buildable", in the sense that one can build a physical object that represents them?

The programs (which run on Windows machines) can be downloaded from Igor Kriz's home page.

Thanks to John Armstrong for pointing me to the article.

A Computational Introduction to Number Theory and Algebra, online book, by Victor Shoup. (There's also a print edition, but the online PDF-book will remain freely available.) It is what it sounds like; "computational" doesn't mean "non-mathematical" but rather means that a lot of the applications are chosen with regard to their usability in computer science, specifically cryptography.

From reddit, where various commenters have pointed out things like "the author says that book is elementary but it really isn't." Of course, this is fairly common. The author actually says

The mathematical prerequisites are minimal: no particular mathematical concepts beyond what is taught in a typical undergraduate calculus sequence are assumed.
The computer science prerequisites are also quite minimal: it is assumed that the reader is proficient in programming, and has had some exposure to the analysis of algorithms, essentially at the level of an undergraduate course on algorithms and data structures.

As usual, "elementary" means something like "a smart, well-trained undergrad could read and understand it"; it's a term of art like anything else. But in some provinces of the Internet there seems to be an idea that everything should be immediately understandable to all readers at first glance. (I use Reddit for the links it gives me to interesting news, not for the comments.)

I came across the polynomial f(x) = 2x² + 3x - 5 during a calculation I was doing a few days ago. I wanted to factor it. Sure, I could have done it the usual way. But I have a better intuition for factoring numbers than I do for factoring polynomials. So I plug in x = 10; then f(10) = 225.225 factors into 9 times 25. Perhaps this reflects a factorization f(x) = g(x) h(x), where g(10) = 9, h(10) = 25.

Indeed, it does: 2x² + 3x - 5 = (x-1)(2x+5). Of course, this gives a whole family of integer factorizations, plugging in different integers for x.

Of course, this doesn't work in general; consider for example 2x² + 2x + 5, which doesn't factor at all. And when the trick is spelled out explicitly it seems to be irredeemably flawed -- how did I know to take (x-1)(2x+5), say, and not (x-1)(3x-5)? (More importantly, can this be explained without reference to the original polynomial?) One could perhaps point out that, say, 184 = (8)(23), which is just f(9) = g(9) h(9), and so on; from a family of such facts it might be possible to deduce the polynomial factorization, but at that point it's just not worth the trouble. These sorts of tricks, like jokes, rarely stand up to explanation.

If the IRS had discovered the quadratic formula -- the algorithm for solving a quadratic equation, in the style of an IRS tax form. (This has been floating around for a while; I'm just reviving it.)

It seems silly here because everyone knows the quadratic formula. But the cubic formula is quite ugly, and takes up a couple lines; it's better expressed as an algorithm, which basically says to make a few changes of variables that give a quadratic and then solve the resulting quadratic. (Certainly if one were in the business of solving cubics, one would probably memorize the algorithm, not the formula.) And I don't even want to think about writing down an explicit formula for the general solution of the quartic in terms of radicals.

But tax forms are really just walking ordinary people through dealing with calculations involving certain equations. Sometimes I think taxation would make more sense to me if I actually saw the equations.

That is, "all mathematics is divided into three parts". I don't actually know Latin. I fear I may have said that Mathematica, the software, is divided into three parts.

This occurred to me this morning, when I saw the following three-fold characterization of the conic sections, which may amuse. (I'm basically copying this from Leila Schneps' lecture notes.)

Conic sections can be defined geometrically, analytically, or algebraically.

• Geometrically: a conic section is the curve obtained when a plane intersects with a cone. The conic section is an ellipse, parabola, or hyperbola according to whether the plane is less steeply slanted, exactly as steeply slanted, or more steeply slanted than the cone's generating line.


• Analytically: given a point F the focus and a line D the directrix, a conic section is the locus of points P such that d(P,F)/d(P,D) is a constant e, the eccentricity. The conic section is an ellipse, parabola, or hyperbola according to whether the eccentricity is less than, equal to, or greater than 1.


• Algebraically: a conic section is the solution set of Ax² + Bxy + Cy² + Dx + Ey + F = 0. The conic section is an ellipse, parabola, or hyperbola according to whether B² - 4AC, the discriminant, is negative, zero, or positive.

The three definitions are equivalent.

It's fitting that the quote is in Latin, because I don't know Latin, and I also don't believe that mathematics is divided into only these three parts. In particular, you all know that I like probability and combinatorics, which don't naturally fit into this tripartite scheme. But it's a nice division of what one might call "classical" mathematics (it's roughly the content of the semi-standard first-year graduate mathematics curriculum, for example).

Calling the second definition "analytic" is a bit of a stretch, though.

As at least two readers have pointed out, my claim that the triviality of the homophony group was part of the unwritten folklore is false. Take a look at the paper Quotients Homophones des Groupes Libres/Homophonic Quotients of Free Groups, by Mestre, Schoof, Washington, and Zagier. They use a lot of the same relations that I did in my proof; in particular damn = damn, damned = dammed, barred = bard, bass = base, jeans = genes, ruff = rough, phase = faze. (The last three of these seem to me to be the ones that most people will find.) Like my proof, v is the last generator to fall; they use chivvy = chivy or leitmotif = leitmotiv. I'm not sure how I feel about these; in both cases these seem like alternate spellings, not different words. veldt = felt, as suggested by a commenter on the earlier entry, feels "right" to me; these are quite clearly two different words with different meanings, veldt being a certain kind of open space in Africa, felt being the stuff out of which the tips of markers are made. (I think this is the one I came up with the first time I saw this problem.) They don't seem satisfied with their proofs for v, either; they introduce the space as a 27th generator, and use avowers = of ours to show the triviality of v.

The paper in question is bilingual; the proof that the English homophony group is trivial is given in French, and the proof that the French homophony group is trivial is given in English. I particularly like the acknowledgements:

The third and fourth authors were partially supported by NSF and (by the results of this paper) numerous other government agencies.

A related problem is as follows: consider the group on twenty-six letters A, B, ..., Z with the relations that two words are equivalent if they are permutations of each other and each appears as a word in some dictionary of choice. Identify the center of the group. According to Steven E. Landsburg, "The Jimmy's Book", The American Mathematical Monthly, Vol. 93, No. 8 (Oct., 1986), pp. 636-638, much work was done on this problem at Chicago in the seventies.

To get started: post = pots = stop = opts = spot = tops. Since post = pots, s and t commute; since pots = opts, o and p commute. This is clearly much harder than the homophony group problem. By the way, elation = toenail. (I've been playing lots of Scrabble recently; that set of seven letters seems to come up a lot.)

From Justin Roberts at UCSD, an old homework problem:

"B: (Fun for all the family, ages 6-66...)The homophony group (of English) is the group with 26 generators a,b, ..., z and one relation for every pair of English words which sound the same. Example: "knight = night" shows that k=1, "earn=urn" shows that ea=u, and so on. (Scrabble rules apply; no proper names or slang!) Prove that the group is trivial! (I would guess it's also trivial in French but probably non-trivial in Spanish and very non-trivial in Japanese!)."

I was looking for something else, but I came across this problem, which I saw before in my first algebra class when I'd just learned what a group was; it's part of the folklore, apparently, and I feel like it's good to have the folklore written down. (It might be written down somewhere on the Internet under another name, but it's hard to find because "group" is such a common word in non-mathematical circumstances.)

I can't do it. I did it before, but I remember it took a lot of looking at lists of homophones; I can't find the particular lists of homophones I used years ago. Most of the relations are simpler than they look, because generally if two words in English are spelled the same they have a lot of letters in common. Here's a partial solution -- I prove that the English homophony group is a subgroup of the free group on two generators, namely m and v. (Or maybe three.)

First, we deal with the vowels. The first homophones that come to anyone's mind are probably to = too = two. From to = too we get o = 1. Thus to = two becomes t = tw, so w = 1. (Yes, I know, w isn't a vowel...) Next, consider the relations

rain = rein = reign, sign = sine, earn = urn.

From rain = rein we get a = e. From rain = reign we get ai = eig; but since a = e we can reduce this to i = ig, or g = 1. Thus we can eliminate the "g" in sign = sine to get sin = sine, so e = 1; recalling a = e, we have a = 1. Finally, from earn = urn we have ea = u; since e = a = 1 we have u = 1.

Another candidate for "most canonical homophone" is probably there = their (I'll ignore "they're" because I don't want to deal with apostrophes); we can clip the "the" in each of these to get re = ir. But e = 1, so this is r = ir, or i = 1. Next, i = eye, but i = e = 1, so this reduces to y = 1. (Does "i" count as a word? Of course I mean "I", the first-person singular pronoun. Roberts said "Scrabble rules apply", but Scrabble rules have never had to weigh in on whether "I" is playable. Scrabble rules out words which are "always capitalized" but the way in which "I" is always capitalized seems different from the way that a proper name is.)

At this point we've shown that the letters a, e, g, i, o, u, w, and y are trivial; this set of letters shouldn't be too surprising to anyone reasonable familiar with English spelling.

There are a lot of silent k's; we take no = know. Since w = o = 1, this reduces to n = kn, so k = 1. Similarly, we can leverage the silent h with wine = whine. lam = lamb gives us b = 1, and dam = damn gives n = 1.

Somewhere around here the problem seems to get harder. One key insight is that groups don't have nontrivial idempotents; if we can show that a generator is equal to its square, then it must be equal to 1. This gets us a few more letters: from bard = barred we have r = rr (recall e = 1), so r is trivial. Similarly, from medal = meddle we get dal = ddle; but a = e = 1, so dl = ddl; thus d = 1. This actually isn't how I take care of t, although there are enough double t's in the language that it's robably possible to handle t this way; I take packed = pact, from which ked = t; but k, e, d have already been shown trivial, so t = 1. A bit more trickily, thinking in terms of double letters, bawl = ball gives us w = l; but w = 1, so l = 1.


That's sixteen letters shown trivial so far: a, b, d, e, g, h, i, k, l, n, o, r, t, u, w, y. The remaining letters are c, f, j, m, p, q, s, v, x, and z. These seem trickier somehow... to my ear they have pretty well-defined sound. The easy letters to get rid of, you'll remember, are the vowels, which can basically sound however you want them to in English.

Still, a few more fall quickly. base = bass (the musical kind, not the fish), so s = 1; razor = raiser can now take advantage of the times when s sounds like z, since all the other letters involved are trivial, to give us s = z, so z = 1. The letters that are phonetically kind of k-like fall next; key = quay gives k = q, so q = 1, and choir = quire (the pair I'm proudest of finding, by the way) gives c = q, so c = 1.

Six letters to go: f, j, m, p, v, x.

lox = locks, which takes care of x. genes = jeans, killing j. phase = faze; only the p and f are nontrivial here, so p = f. There are two cheap tricks to handle this. The first is hiccup = hiccough, where all the letters involved are trivial except p, so p = 1; but are those separate words, or alternative spellings of the same word? The second is if = iff, so f = 1. (The problem here, of course, is that iff is a bit iffy as a word... is it really a word, or just an abbreviation for "if and only if"?) Two half-solutions add up to one whole solution, though, so I say that p and f are trivial.

That leaves two generators, m and v. If there are relations between them, they're trivializing ones; we're not going to get a relation like "mv = vm" (which would bring us down from the free group on two generators to the free abelian group on two generators), since that would require two homophonous words in English, both containing m and v, where in one the m comes first and in one the v comes first. I've tried to use the double-letter trick; the closest I can get is "clamor" and "clammer" to get rid of m, but these aren't really pronounced the same.

Any ideas on how to finish this up?

My vocabulary in other languages isn't rich enough to do this. But I agree with Roberts' judgement about French, Spanish, and Japanese; French is full of silent letters, even more so than English.

Edited, 12:24 pm: In the comments, dan suggests dammed = damned to show that m is trivial. And ruff = rough shows f is trivial, which gets around the objections I had before. I want to take advantage of the fact that the f in of is pronounced like v usually is, but I can't.