June 2008 - mathfun@mailman.xmission.com

Re: [math-fun] no-4-in-plane problem
by Dan Asimov 24 Jun '08

24 Jun '08

Ed wrote: << >From a 3x3x3 grid, select 8 points such that no 4 lie in a plane. The solution is unique. >> I'm not sure what this is quoted from . . . but in any case this is reminiscent of the question (arising from the game "Set"): What's the size of the largest subset S of the group G = (Z_3)^4 such that S does not contain any "affine line". (Where an affine line means any subset g + H of G, where H is any subgroup of G isomorphic to Z_3, and g is any element of G.) By replacing 3 with any integer n >= 2, and replacing the direct sum of 4 copies of Z_n with any number k >= 2 of copies, and replacing "line" with a d-dimensional affine plane for some d, 1 <= d < k, we have a family of analogous questions Q(n; k; d), where the Set question is Q(3; 4; 1). (As Rich once proved, the answer to Q(3; 4; 1) is 20.) [Definition I]: (Where an affine d-plane is just g + H, where H is any subgroup of G = (Z_n)^k isomorphic to (Z_n}^d, and g is any element of G.) I understand these tend to be quite difficult, and only a few answers are currently known (computer searches quickly becoming out of range). * * * For n a prime power p^e, e >= 2, there's an alternative definition of affine plane: [Definition II]: Let V be the vector space of dimension k over the finite field of order p^e. Then an affine d-plane in V is just and subset of V of the form v+P where P is any d-dimensional subspace of V, and v is an arbitrary vector of V. For n = p^e, denote the version of Q(n; k; d) that uses this alternative definition of an affine d-plane by Q'(n; k; d). (possibly easy) QUESTION: For n = p^e, are Q(n; k; d) and Q'(n; k; d) combinatorily equivalent? More precisely, for n = p^e, does there always exist a bijection between (Z_n)^k and (Z_n)^k that induces a bijection between all affine d-place (Def. I) and all affine d-planes (Def. II) ? --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] Rubik's Cube: 23 moves max
by Christian Boyer 24 Jun '08

24 Jun '08

Tomas Rokicki: "The number of moves necessary to solve an arbitrary Rubik's cube configuration has been cut down to 23 moves". http://science.slashdot.org/article.pl?sid=08/06/05/2054249 Christian.

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Re: [math-fun] crop circles
by Steve Witham 23 Jun '08

23 Jun '08

Okay, why, exactly, is the hypothesis that it's humans so good that no other hypothesis is worth bothering with, even though we only have a few people claiming responsibility for a few hundred crop circles out of thousands, plus demos of how easy it is? What's the principle or breakdown here? The things I wonder are like, does every group invent it from first principles, or do they pass it on from group to group? Has anybody studied them like you would crime scenes, looking for patterns of scuff marks, bits of paint, urine spots, wood splinters, etc? Anyway, here's an interesting (though credulous) article that includes a few construction details: http://science.howstuffworks.com/crop-circle.htm It doesn't mention that ropes can be used as rulers as well as compasses. Maybe flashlights pointing up into the sky make good location markers. It does mention one related flow of money, not that I think that's an important factor. Some favorite quotes: "Astronomer and former Boston University professor Gerald S. Hawkins studied several crop circles and found that the positions of the circles, triangles and other shapes were placed based on specific numerical relationships. In one crop circle that had an outer and an inner circle, the area of the outer circle was exactly four times that of the inner circle. The specific placement of the shapes indicates that, whoever the circlemakers are, they have an intricate knowledge of Euclidean geometry (the geometry of a flat surface introduced by the mathematician Euclid of Alexandria)." They know of Ratios! And Geometry! And the ancient art of Making One Circle Twice As Wide As Another, thought to be lost since the days of Pythagoras!! 'This formation was discovered in Eastfield, England, in June 2004. An article in the Western Daily Press called the design "uncannily similar to plans for one of Nikola Tesla's early pieces of equipment."' (Someone must have--uncannily--copied a picture from a book.) Here's my favorite crop circle in the gallery of pictures for the article: http://science.howstuffworks.com/enlarge-image.htm?terms=crop+circles&galle… or http://tinyurl.com/5u2qhs --Steve

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Re: [math-fun] method of construction : crop circles
by James Propp 23 Jun '08

23 Jun '08

Maybe someone could host a crop-circle-making competition with a big cash prize. *That*'d bring the hoaxsters out of hiding --- and maybe convince some people to drop their belief that crop-circles "must" be the work of extraterrestrials. I even thought of proposing such a contest at a Gathering for Gardner (some of the folks who attend have real money!), but the people I spoke to about my proposal convinced me that the payoff to society would be too small. Not many people would change their minds; and frankly, believing that crop-circles are of non-human origins is a fairly harmless human belief (in contrast, say, with the belief that certain humans are vermin who should be eradicated). Personally, I believe that many of the crop-circles in England are the work of aliens. Specifically, legal aliens from across the channel. :-) Jim Propp

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Re: [math-fun] method of construction : crop circles
by Dan Asimov 22 Jun '08

22 Jun '08

Steve wrote: << Not to mention what their motive might be. I'm not saying I'm a convinced believer in an extraordinary origin, but if the earthly constructors are after publicity, why haven't they ever stepped forward and claimed credit (except for a few exceptions which don't account for more than a fraction)? And why have they persisted for 20 years? There's no money in it. >> For what it's worth, Wikipedia says that nowadays it's very hard to get any publicity for a crop circle unless it's extraordinary. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] a certain crop circle and 10 digits of Pi in uk
by Dan Asimov 22 Jun '08

22 Jun '08

The article praised the fact that the E.T.'s had even rounded the last digit (up) correctly. They *also* used an ellipsis at the end to show that "pi is infinite" -- OK, that its decimal expansion is infinite. But these are not very bright E.T.'s, since one should EITHER round the last digit (and indicate that this is only an approximation), OR use ellipsis to show the digits continue . . . but NOT both! --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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Re: [math-fun] a certain crop circle and 10 digits of Pi in uk
by Dan Asimov 22 Jun '08

22 Jun '08

David wrote: << Think about it. If this crop circle were made by aliens, then [lots of reasons they probably weren't made by aliens] >> David, that is totally ridiculous. Everyone knows crop circles are made by aliens. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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[math-fun] a certain crop circle and 10 digits of Pi in uk
by Simon Plouffe 22 Jun '08

22 Jun '08

There is an article about a certain crop circle in uk, The digits are coded in a strange manner, ... I doubt that e.t. is using base 10 to represent pi, this must be done on earth. http://www.dailymail.co.uk/news/article-1027178/Easy-pi-Astrophysicist-solv… Simon Plouffe

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[math-fun] The Chow-Longer sequence
by Fred lunnon 20 Jun '08

20 Jun '08

What OEIS refers to as the "Chow-Long" sequence A078588 --- following the long-established tradition of naming mathematical concepts after people who didn't discover them and quite possibly never even mentioned them --- arises in a problem apparently originating with D. L. Silverman, J. Recr. Math. 9 (4) 208, problem 567 (1976-77). Define a binary sequence of (initially) positive integer n by CL(n) == floor(2*phi*n) - 2*floor(phi*n) where phi denotes the golden section number (1 + sqrt5)/2; the first few values for n > 0 are 1010010110 1001011010 1101001011 0100101001 0110100101 1010110100 1011010010 1001011010 01011010 ... Now partition the positive integers into two sets A = A_0, B = A_1 defined by A_k == { n | CL(n) = k }; so A = { 2, 4, 5, 7, 10, 12, 13, 15, 18, 20, ... }, B = { 1, 3, 6, 8, 9, 11, 14, 16, 17, 19, 21, ... }; then form the sets of sums of pairs of distinct(?) elements from each set; then take the complement of their union, the set S "avoided" by the partition: S = { 1, 2, 3, 5, 8, 13, 21, 34, 55, ... }. This turns out to be the Fibonacci numbers. See for example T. Chow, A new characterization of the Fibonacci-free partition, Fibonacci Q. 29 (1991), 174-180; online at http://www-math.mit.edu/~tchow/cv.html [but really, it is actually much easier to prove than that!] While investigating Knuth's "circle" product, I encountered CL(n) in a different setting, which suggested generalising it to the m-ary sequence CL_m(n) == floor(m*phi*n) - m*floor(phi*n); for instance, taking m = 3, the first few values of CL_3(n) for n > 0 are 1021020210 2101021021 2102102021 0210102102 1210210102 1021210210 2021021010 2102121021 0202102101 0210202102 1010210212 1021020210 2101021021 2102102 ... [I posted this sequence a while back, but nobody hacked it ...] Feeding this back into the original context, we define m sets as before A_k == { n | CL(n) = k }, for m = 3, A = {2, 5, 7, 10, 13, 15, 18, 23, 26, 28, 31, ... }, B = {1, 4, 9, 12, 14, 17, 20, 22, 25, 30, 33, ... }, C = {3, 6, 8, 11, 16, 19, 21, 24, 27, 29, 32, ... }; the sets of sums of m-plets of each set; and finally the set S avoided by their union, for m = 3, S = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 18, 21, 23, 26, 29, 31, 34, 39, 42, 47, 55, 60, 68, 76, 81, 89, 102, 110, 123, 144, 157, 178, 199, 212, 233, 267, 288, 322, 377, 411, 466, 521, 555, 610, 699, 754, 843, 987, ... }. First problem: conjecture the explicit form of this set for m = 3 (easy); second problem: prove it (hmmm...). Fred Lunnon [20/06/08]

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Re: [math-fun] Tiling the Plane with 5-, 6- and 7-gons only
by Dan Asimov 18 Jun '08

18 Jun '08

Actually, I hadn't noticed Michael's solution, which is periodic with all polygons convex, as well as having perpendicular reflection symmetries in each fundamental domain. I prefer his by large margin. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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