Newsgroups: sci.math.research From: Hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 1998/01/05 Subject: Re: Simon Singh's "Fermat's Last Theorem" > Lieven Marchand wrote: > >[Aczel] ... accuses Weil of an elaborate plot to steal the credit > >for the Taniyama-Shimura conjecture from its authors. [ One of the > >subchapters has as title Intrigues and Treachery, and another one > >The lie. ] In light of the dubious research on other material in > >the book, I would very much like to hear some knowledgeable people > >comment on these allegations. Aczel is extracting and popularizing Lang's notice "Some History of the Shimura-Taniyama Conjecture", from the November 1995 _Notices of the AMS_, also at http://www.ams.org/notices/199511/lang.html. As in other parts of the book, Aczel has added his own witty comments to make the narrative more exciting--and less trustworthy. Even with the authoritative version, though, it is unclear to me how much of the controversy is about lies and treachery and how much is about exaggeration and sensationalism. richard.gr...@ox.compsoc.org.uk (Richard Green) writes: > ... Not only is the historical material somewhat off-topic, but > the mathematics is often badly explained.... The more technical > material on elliptic curves is presented in a similarly > unenlightening way. Actually, the elliptic curve stuff is even more broken than most of the rest. For instance, Aczel's graphs on page 93 are footnoted 13. Adapted from Kenneth A. Ribet and Brian Hayes, "Fermat's Last Theorem and Modern Arithmetic," _American Scientist_, Vol. 82, March-April 1994, pp. 144-156. The figures seem to have been "adapted" by (1) reducing them by half, so that they are nearly illegible, and (2) removing the captions that explain that the first two are examples of elliptic curves, but the second two are not elliptic curves, because they are associated with singular cubic equations. Aczel's book has a large collection of disappointing figures. Page 6 has a picture of the wrong _Arithmetica_, lacking Fermat's comment. Page 25 was drawn by someone who seems to have lacked both T-square and protractor (or their computational equivalent). Pages 82 and 83 are trying to say something, but nothing comes out. And the top of page 86 needs just a little more low-res ray-tracing. An inverse problem is in the caption on Gerd Faltings's photograph on page 137. Perhaps "many feared that Faltings would now beat [Wiles] to the true proof" in 1993, but nothing of this little bit of suspense made it to the narrative (or even the index). > I haven't read Singh's book, but it seems to be by far the better work > from what I've heard other people say about it. Better, but not by so very far. Singh doesn't explain the math, either. And he misstates the Frey equation--it's supposed to be y^2 = x (x - A^N) (x + B^N). The whole point of the Frey equation (as I understand it) is that the differences between its (y=0) roots are A^N, B^N, and A^N + B^N, and it's the differences being Nth powers that makes the curve non-modular. Also, as mentioned previously, Singh _never_ mentions that Wiles proved only the semistable case of Taniyama-Shimura. If you want to learn some relevant mathematics, I would recommend the Ribet and Hayes article cited above. It isn't perfect--particularly confusing are the parts from p.147 column 3 to p.148 column 2, where they use the equivalent formulation A^N + B^N + C^N = 0, (A B C not zero) but don't bother to mention the change. When you read "C" as "-C", the argument suddenly starts making sense. There's another error on page 155, where they justify a property by congruence that should be justified by relative primality, but that's relatively benign. And Figure 1 is misleading but unrelated to the article--someone wanted to show off their Mathematica. What all three of these lack is a good description of modular (or automorphic?) forms. Here Aczel gets the most explicit, but I'm not sure whether to trust him. Is it really just any function f:C->C that has a nontrivial symmetry f(z) = f( (az+b)/(cz+d) ) ? Does f have to be continuous? If I find out what a modular form is, I might--mind you, I said 'might'--want to look up L-series. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: Hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 1998/01/11 Subject: Re: n-rooks problem , n-bishops problem qs...@aol.com (QSCGZ) writes: > >>>: How many solutions are there, to place n chess-rooks (2*n-2 bishops) > >>>: on an n x n chessboard, such that no rook (bishop) attacks another one ? > >>>: Symmetric solutions are counted only once. Assume n>1. For bishops, consider a family of 2n-1 parallel diagonals. The endmost diagonals (of length 1) cannot both have bishops on them, and the remaining 2n-3 diagonals can have at most one bishop each. This establishes the bound of 2n-2 bishops. For k > 1, suppose we populate all but one of the diagonals of length less than k, and consider the diagonal(s) of length k. Each such diagonal will have k-2 squares forbidden by previously placed bishops, so the only places that bishops can be placed are at one of the endpoints. This establishes that all bishops will be on the edges of the board. So any arrangement of 2n-2 non-attacking bishops will have two bishops at the ends of the long diagonals (i.e., in adjacent corners) and two bishops at opposite corners of each of n-2 45-degree rectangles: . . . . 2 . . . . . x' `x . . . x'. . .`1 Either there are bishops at both 2s . x'. . . x'. or bishops at both 1s. 1'. . . x'. . .`x . x'. . . . .`2'. . . . So there are 2^n possible ways of placing the bishops (before reducing by symmetry). Because exactly two adjacent corners are populated, no such pattern can be symmetric under rotation or diagonal reflection. For orthogonal reflection, the symmetric patterns will have an axis determined by the populated corners, and then the floor((n-2)/2) pairs of rectangles must agree: . . 4 . 2 . b . x'.`x' `x . If this is symmetric through the reflection 3'. x'.`x .`1 determined by the bs, then either there are .`x'. . .`x'. bishops at the 2s and 4s or there are bishops 1'.`x . x'.`3 at the 1s and 3s.. .`x .`x'. x'. . .`2'.`4'. b If n is odd, the equilateral rectangle will always be symmetric under orthogonal reflection. So there are 2^floor((n+3)/2) symmetric positions and 2^n-2^floor((n+3)/2) asymmetric positions; the former appear in 4 orientations, and the latter in 8. So the total number of arrangements is (2^floor((n+3)/2))/4 + (2^n-2^floor((n+3)/2))/8 = 2^(floor((n-3)/2) + 2^(n-3). This works out to: n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 positions 1 1 1 2 3 6 10 20 36 72 136 272 528 1056 2080 4160 and for n=100, there are 158456325028528956662064611328 positions. The analysis for rooks is harder, since there are five kinds of symmetry possible (reflection through two diagonals, reflection through one diagonal, four-fold rotation, two-fold rotation, and asymmetry). Still, it's not impossible to carry out by hand for small boards. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: Hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 1998/01/14 Subject: Re: n-rooks problem , n-bishops problem (rook spoiler) qs...@aol.com (QSCGZ) asked: > How many solutions are there, to place n chess-rooks... > on an n x n chessboard, such that no rook attacks another one ? > Symmetric solutions are counted only once. There are eight symmetry operations of the chessboard: three rotations, four reflections, and the identity. Counting the number of positions that are left unchanged by the various symmetry operations is a tedious but not terribly difficult combinatorial exercise. For n>1 there are [n/2]! R_Q(n) = ------ if [n/2] is even, 0 otherwise [n/4]! positions unchanged by 90-degree rotations, R_S(n) = [n/2]! 2^[n/2] positions unchanged by 180-degree rotations, [n/2] n! R_L(n) = Sum -------------- i=0 (n-2i)! i! 2^i positions unchanged by each diagonal reflection, none unchanged by an orthogonal reflection, and n! unchanged by the identity. From the Cauchy-Frobenius-Polya-(not)Burnside theorem we can compute the number of positions up to symmetry by counting the average, over all symmetry operations, of the number of positions left unchanged: R(n) = (2 R_Q(n) + R_S(n) + 2 R_L(n) + 0 + n!)/8. n 0 1 2 3 4 5 6 7 8 9 10 11 12 R(n) 1 1 1 2 7 23 115 694 5,282 46,066 456,454 4,999,004 59,916,028 and R(100)=11,665,776,930,493,019,085,212,404,857,033,337,561,339,- 496,033,047,702,683,574,120,486,902,199,999,159,757,- 068,380,354,180,755,001,424,896,581,199,329,890,890,- 062,330,104,725,878,634,060,229,093,973,813,100,544. This is sequence A000903 in Sloane and Plouffe (see http://www.research.att.com/~njas/sequences). Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: Hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 1998/01/16 Subject: Re: (Oh no!) Still more dice... fifty_degrees_c@yahoo.com writes: > Assuming anyone is even vaguely interested, I have had more > pea-brained thoughts on the 3x3 dice cube. First off, it's a 3x3x3 dice cube. Or a 3^3 cube. But a cube is not a square (unless it's a sixth power, of course). > With the condition that a die can only be placed on another die if > the same value appears on the faces in contact i.e. 3 against 3, 1 > against 1 etc. the sum of the face values on the cube exterior (V) > appears always to be 168. This is obvious: Any line parallel to the axes through cube faces passes through faces n,7-n,n,7-n. Opposing facelets sum to 7. This makes a more interesting puzzle on a cube of even side, provided none of the dice are mirror images of each other. I don't know the answer to that. Incidentally, everyone knows that opposite faces of a die add to seven. John H. Conway notes another rule: "chaos clockwise-counters counter." In other words, three numbers around a corner increase counterclockwise just if they form an arithmetic sequence. It's rare you'll find a mirror-image die, apart from things like fuzzy dice and other toys. He has a third rule, which tells you the orientation of the 2, 3, and 6. The 2 and 3 faces each define a diagonal, and the 6 can define a diagonal by mentally extending it to an N or Z. The faces whose diagonals go through the 2-3-6 corner are called "homeward" and the faces that don't are called "wayward". The "waywardness" of a die is the sum of its wayward faces if at most one face is wayward, and seven minus the sum of its homeward faces if at most one face is homeward. I think he said he had collected normal dice of all 8 waywardnesses, though homeward dice are the most common. Dan Hoey@AIC.NRL.Navy.Mil --- Newsgroups: rec.puzzles From: Hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 1998/01/23 Subject: Nontransitive dice [ repost ] [ I posted this last week but it doesn't seem to have gotten out. I hope you didn't get duplicate copies. ] fifty_degrees_c@yahoo.com asked about a set of four nontransitive "fantasy" dice (where die two usually beats die one, die three beats die two, die four beats die three, and die one beats die four). He askes if there is a limit to the number of dice in the set. There is no upper limit, because you can always add a new die by cloning. For example, the given nontransitive dice {004444, 333333, 222266, 111555} can be extended to form {005555, 333333, 444444, 222277, 111666}. The minimum number of dice is three. Unfortunately, neither the January 1998 _Esquire_ article you refer to, nor the rec.puzzles archive entry that Matthew Daly mentions in a followup (http://xraysgi.ims.uconn.edu/rpa-output/probability/transitivity.s), nor Martin Gardner's book _Wheels, Life, and other mathematical amusements_ (which reprints his October 1974 _Scientific American_ column on Efron's dice) really makes it clear that you have a set of only three dice that exhibit nontransitivity. It can even be done with dice numbered from 1 to 5 (no number appearing on two different dice, to prevent ties) in essentially one way: {222255, 333333, 114444}. Unfortunately, the amounts by which die two beats die one, three beats two, and one beats three are not equal. However, if we allow seven distinct values among the faces, we can have a set of three nontransitive dice that beat each other equally. There are four sets of three nontransitive dice with this property. Two of the sets are kzormdromes*, like the five-pip set above. The other two sets are hivevied** versions of each other. Can you find these sets of dice? Remember that they must be three six-sided dice numbered from three disjoint subsets of {0,1,2,3,4,5,6}. [notes] *Kzormdrome: A string that when written in a reversed alphabet yields the reverse of the string. "Wizard" and "hovels" are kzormdromes. **Hivevied: Written backwards in a reversed alphabet. When you hivevy "shrilly" you get "boorish". A kzormdrome hivevies to itself. Dan Hoey Hoey@AIC.NRL.navy.Mil --- Newsgroups: sci.math From: Hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 1998/02/14 Subject: Re: Graph Theory help req. jons5...@telcel.net.ve (Jos'e H. Nieto) writes: > sebarn...@aol.com (SEBarnett) wrote: > > (1) Use induction on k to prove that the edges of a connected > > simple graph with 2k edges can be partitioned into paths of length > > 2.... While Sr. Nieto's proof is gratifyingly concise, I rather like the one I came up with, on the basis of its easy construction or possibly on the basis of its algorithmic efficiency. If the graph has any edges, it must have a vertex of degree two or greater. Let V be such a vertex and let U and W be two of its neighboring vertices. Removing the edges UV and VW will separate the graph into at most three components, which we name by the vertices of {U,V,W} they contain: either {UVW}, {U,VW}, {V,UW}, {W,UV}, or {U,V,W}. If there is no component with an odd number of edges, then removing the path UVW leaves us with connected components of at most 2k-2 edges, that can have their edges partitioned into 2-paths by the inductive hypothesis. If G - {UV,VW} has a component with an odd number of edges, then there must be at least two such components (because there are an even number of edges in all). Examining the five possible cases, we can be sure that at least one of the components with an odd number of edges is U, W, or UW. If U, then take G1 = component U + edge UV. If W, then take G1 = component W + edge VW. If UW, then take G1 = component UW + edge UV. In each case, take G2 = G - G1. ___ .' ___ .' ________ .' /G1 \ .' /G1 \ .' /G1 \ .' \__U/ .' \__W/ .' \__U__W__/ .' \ .' ____ \ .' ____ \ .' ____ V .' V___/ V .' V___/ V .' V G2 \ .' / G2 : .' / G2 : .' / \___/ .' W_____: .' U_____: .' W .' \___/ .' \___/ .' In either case, we have made G1 a connected graph with an even number of edges between 2 and 2k-2. G2, being the rest of G, must have an even number of edges and can be seen to be connected. So we can inductively partition the edges of G1 and G2 into 2-paths. QED. I wonder if this theorem may be a contender for having as many different proofs as the problem of tiling a rectangle with rectangles that each have an integer edge (see Stan Wagon's "Fourteen proofs of a result about tiling a rectangle" [Amer Math Monthly 94(1987) 601-617] and Alexander Shen's "Math Entertainments" [Math Intell 19.1(1997)12-13 and 19.4(1997)49-50]). Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: news.admin.net-abuse.email From: Hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 1998/03/17 Subject: Re: Wallace's new rant. Sam wrote: : In case you haven't noticed (it's a tiny addition), Wallace doesn't like : being called a moron. _Wallace_ doesn't like it? Think of how morons feel--they're the ones being insulted. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Date: Tue, 7 Apr 98 17:23:41 EDT From: Dan Hoey To: "Andy Spragg" Subject: Re: What is it with Andy Spragg? Andy "What is it with" Spragg writes: > So there you have it folks. I rest my case. Anyone wanting to side with > Guido, I'd be interested to hear from you. Sorry, I can't see her side at all. "Completely barking mad" seems the most likely assessment. Still, "venal, vapid, and vain" is a possibility--sort of a Marilyn vos Savant without the engaging prose style. You remarked that Guido's earlier answer didn't introduce "too many" errors. I wonder if you have a theory on what > It is easily seen that the rational numbers, Q, also may be > uniquely factored. is supposed to mean? Other than the trivial observation that all nonzero elements of a field are units, in which case the statement about R not being a UFD is nonsense. Or were you just allowing her a generous error budget? Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Letter to the WSFA Journal (June 1998 issue) Dear WSFA I'm sorry to have to take my e-mail address off the WSFA web page. While I am happy to receive e-mail direct to me from WSFAns, I do not want to be on any list that is used to send e-mail to everyone in WSFA. The reason for this is that such e-mail lists tend to devolve from infrequent emergency announcements, to frequent announcements that duplicate what I find out at WSFA meetings, and finally to matters that are not expected to be of interest to everyone on the list but which the sender hopes to be applicable to someone on the list. The ease of sending e-mail to a large list of people means that more time is spent by the people reading such a note than was spent writing it. And the ease of copying such a list of addresses from one e-mail message to another means that one message sent to such a list of addresses begets others. Those of you who have your e-mail addresses on the WSFA web page will know that it was used for two general announcements this week, and I do not envision that they will cease. So if you want to send e-mail to everyone in WSFA, please treat me as having no e-mail address. I do welcome person-to-person communication, and I hope that I may make this easier by listing my e-mail address in the off-line WSFA directory without having it used as a distribution list. Dan Hoey Hoey@AIC.NRL.Navy.Mil --- Newsgroups: sci.math.research From: Hoey@AIC.NRL.Navy.Mil (Dan Hoey) Date: 1998/11/03 Subject: Re: Finite Groups Generated by 2 Elements jheck...@my-dejanews.com writes: > What a coincidence! A friend of mine who's a professor of > mathematics at Cal State Fullerton just recently challenged me with > a closely related problem - specifically, what subgroup of S_20 is > generated by the 2 elements X = (1,2,3,...,20), a 20-cycle, and > Y = (1,20)(2,19), a product of 2 disjoint 2-cycles? > I was able to convince myself that, in fact, these 2 elements > generate the entire symmetric group S_20. But my proof relies on > non-elementary permutation group theory.... A simple demonstration is to note that T_0 = Y X^-1 (Y X)^17 X Y = (3,20). Taking T_k = X^-k T_0 X^k for k=1,2,...,18 yields the conjugates T_1=(1,4), T_2=(2,5), ..., T_18=(18,1). It is then straightforward to construct any element of S_20 from products of the transpositions {T_0, T_1, ..., T_18}. But there are other products of disjoint 2-cycles that will not work with X to generate S_20 -- (1,3)(2,4), for example. I suspect that the product of two disjoint transpositions (a,b)(c,d) will, with X, generate S_20 just when gcd(a-b,c-d,20)=1; it shouldn't be hard to test this with some software such as GAP. > In regards to your question, I could have sworn I saw someone post here > recently that all (finite?) symmetric groups can be generated by 2 elements, > but I don't remember s/he giving a reference. I haven't yet found such a > theorem in the reference I cited above, but I haven't worked all the way > through it yet, either. A transposition (1,2) and its conjugates by (1,2,...,n) generate S_n by a construction similar to the above. This doesn't work for infinite symmetric groups, though you can generate the group of nigh-translations {s in S_Z : s : x |-> x+c with finitely many exceptions}. and perhaps other interesting countable subgroups of S_Z. See HSM Coxeter, WOJ Moser, _Generators and relations for discrete groups_ Erg. Math., Springer, 1957 for a lot of information on this topic. Dan Hoey Posted and e-mailed Hoey@AIC.NRL.Navy.Mil ---