September 2003 - mathfun@mailman.xmission.com

[math-fun] Simpsons TV show writers
by Thane Plambeck 22 Sep '03

22 Sep '03

http://www.mathsci.appstate.edu/~sjg/simpsonsmath/degrees.html Thane Plambeck 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/ehome.htm

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Re: [math-fun] Simpsons TV show writers
by asimovd＠aol.com 22 Sep '03

22 Sep '03

Henry asked:(in light of Thane's link to the math backgrounds of so many Simpsons writers): << Amazing! But how come math teachers aren't funny/entertaining in class? >> Well, I think we now know where the five funny ones have gone (and Tom Lehrer doesn't teach math anymore)! --Dan

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[math-fun] round robin
by Michael Kleber 20 Sep '03

20 Sep '03

Hi Funsters, Surely this is well known, but somehow not to me. How do you arrange round robin tournament pairings? I actually needed to do it for six teams, and after a little doodling I came up with a pretty symmetric- looking pairing scheme. Numbering points in cyclic order, you pair 1-2 3-4 5-6 and its (one) cyclic shift, and pair 1-4 2-6 3-5 and its (two) cyclic shifts. The first two match each person with the two sitting on either side of them, and the last three match each person with their antipode and antipode's neighbors. Is it always possible to do this? (Where "this" is a set of pairings that's invariant under cyclic shifts.) For an odd number of people, I can certainly come up with one. Number the people from -n to n, and have each k play -k, with 0 sitting out. This plus its cyclic shifts (just adding mod n) do the job, and it's easy to see why -- draw the people in order in a circle, and each round's pairings are a full family of parallel lines. For any even number of players, you can run the above scheme, with one distinguished person playing againt the guy who would have sat out above. But I don't like singling someone out like that. (Though I guess you can still draw asymmetric-looking picture of it, by having the distinguished person sitting in the middle of the circle.) But is there always a way that's as symmetric as the 6-person example? And of course there are the counting questions: in how many different ways (modulo renaming the players, of course!) can you pair people up, both in general and with the symmetry requirement. I looked in the EIS for round-robin and didn't get any answers. Okay, now you can all tell me how well-known all this is. --Michael Kleber kleber(a)brandeis.edu

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[math-fun] spam blocking, not mathematically
by Chris Landauer 19 Sep '03

19 Sep '03

hihi, all - it turns out the the single most effective way to block the kind of spam that i get is to look for words or just letter sequences with embedded html comments, things of the form (this from one i got this morning) REVOLUTIONARY detecting and removing these messages would take care of almost all the spam i get, except for the nigeria scam and its brethren, and the messages with no content except a link (and since an embedded comment is needed only as a way to hide a word from a text search, that automatically makes the containing spam as far as i'm concerned) i suspect that removing these messages first would make any statistical approach far more effective, since many include extra (or only) nonsense words to circumvent language-based discrimination (i got this example also this morning) wpj<...html stuff...>a osnkni tolzbo mxwzys a finally, apparwently you can also encode ordinary ascii letters to disguise them, such as the example below for the letter i i finally, i have not yet seen even one legitimate message from any of the new domains (.biz, .info, etc.), though i imagine that that will eventually happen if anyone wants a good hard test of their spam filter rules or tools, please let me help, since i get enough mail that i do want to make discriminating it from spam interestingly difficult (i run my mail on a unix machine, so windows tools won't help me) more later, cal Dr. Christopher Landauer Aerospace Integration Science Center cal(a)aero.org +1 (310) 336-1361

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[math-fun] non-square products of squares?
by Marc LeBrun 19 Sep '03

19 Sep '03

Let some set be closed under an associative multiplication operator. If it commutes then the product of distinct squares is also square: aabb = abab. Contrarily, can you construct a non-commuting system where this is usually false? That is, where non-trivial products of squares are never square? (Trivializers include null/idem-potent elements, perhaps all self-squares a=aa). For example is there a simple model using 2x2 matrices over some Zn?

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[math-fun] Spam-blocking, Bayesian methods
by Guy Haworth 18 Sep '03

18 Sep '03

I would appreciate a pointer to a readily accessible article or three on how Bayesian methods can be used to block Spam. Pointers to any products implementing these methods also welcomed. Guy

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Re: [math-fun] non-square products of squares?
by asimovd＠aol.com 18 Sep '03

18 Sep '03

Marc asks: << Let some set be closed under an associative multiplication operator. If it commutes then the product of distinct squares is also square: aabb = abab. Contrarily, can you construct a non-commuting system where this is usually false? >> The simplest example I can imagine is the set of all finite strings of a's and b's with the operation of concatenation, aka the free semigroup on 2 generators. --Dan.

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[math-fun] TUP's and TOP's
by Jon Perry 18 Sep '03

18 Sep '03

New paper @; http://www.users.globalnet.co.uk/~perry/maths/threadedpartitions/threadedpar titions.htm Jon Perry perry(a)globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/ http://www.users.globalnet.co.uk/~perry/DIVMenu/ BrainBench MVP for HTML and JavaScript http://www.brainbench.com

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[math-fun] Total Harmony?
by Guy Haworth 15 Sep '03

15 Sep '03

Does the hairy-ball theorem prove that total harmony in the world is impossible - or is it rendered possible by the existence of areas of sea? Guy Haworth

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[math-fun] stalking the wild quaternary game, or misere .3102 with a bounty
by Thane Plambeck 15 Sep '03

15 Sep '03

I'll pay US $200 for a complete analysis of the following impartial game, which (for lack of a better term), I'll call the WILD QUATERNARY GAME. The game is played with a number of heaps of beans of various sizes. There are two players. They take turns making legal moves. The person who makes the last legal move *loses* the game. There are three types of legal moves. Each type of move involves selecting one particular heap, and then removing 1, 2 or 4 beans from that heap, subject to the following additional conditions: 1) A single bean can be removed from any heap, no matter how many beans are in it. 2) Two beans can be removed only if the heap contains exactly two beans. 3) Four beans can be removed from a heap, but only if the heap contains more than 4 beans. In the Winning Ways "octal code" notation, this is the misere game .3102. By a "complete analysis", I mean an explicit mathematical description of the first- or second-player winning positions of the game for arbitrarily-many arbitrarily-sized heaps (ie, 'algorithmic'' solutions that boil down to a recursive search of a game tree don't cut it). For example, if I had changed the problem statement so that the last player to make a legal move *wins* the game, a complete description of the second-player winning positions is All positions with a) An even number of heaps of size (5k+1 or 5k+4) AND b) An even number of heaps of size (5k+2). Thane Plambeck thane(a)best.com 650 321 4884 office 650 323 4928 fax http://www.qxmail.com/ehome.htm

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