August 2009 - mathfun@mailman.xmission.com

[math-fun] muffin problem -- clusters and bounds
by Huddleston, Scott 01 Dec '17

01 Dec '17

I enclose examples of some of the bounds in my 28 Feb. post, and a derivation of the most involved example. S(12/7) = 3/7 = r/4 = (3-r)/3 Muffins: 1/7 * { [3,4]*12 } Persons: 1/7 * { [4*3]*4, [3*4]*3 } S(7/4) = 5/12 = (3-r)/3 = (2r-1)/6 Muffins: 1/12 * { [5,7]*6, [6*2] } Persons: 1/12 * { [5*3,6]*2, [7*3]*2 } S(40/23) = 19/46 = (2r-1)/6 = (12-5r)/8 Muffins: 1/46 * { [19,27]*24, [20,26]*12, [23*2]*4 } Persons: 1/46 * { [19*3,23]*8, [20*4]*3, [27*2,26]*12 } S(228/131) = 54/131 = (12-5r)/8 = (7r-6)/15 Muffins: 1/131 * { [54,77]*120, [55,76]*24, [57,74]*60, [65,66]*24 } Persons: 1/131 * { [54*3,66]*24, [54*2,55,65]*24, [57*4]*15, [77*2,74]*60, [76*3]*8 } S(449/258) = 319/774 = (7r-6)/15 = (73-32r)/42 Muffins: 1/774 * { [319,455]*210, [320,454]*30, [325,449]*42, [336,438]*30, [337,437]*90, [384,398]*42, [387*2]*5 } Persons: 1/774 * { [319*3,390]*42, [319*2,325,384]*42, [320*3,387]*10, [336,337*3]*30, [455*2,437]*90, [455,454,438]*30, [449*3]*14 } Here's a derivation of 449/258 example. 449 muffins -> 898 pieces, used in 124 4-parts with 496 pieces and 134 3-parts with 402 pieces. We split muffins between a 3-part and a 4-part when possible, but 4-parts contain 94 more pieces than 3-parts. We divide these 94 pieces (47 muffins) as evenly as possible among the 124 4-parts. 47 pairs of 4-parts contain both halves of a muffin, which I'll call an A^1 cluster and write A^1 = [a1*3,m]*2 where m denotes the two muffin halves and the other pieces are denoted a1, indicating a set of pieces of possibly different sizes that occure in A^1 clusters. 34 4-parts contain only muffin pieces whose mates appear in 3-parts. These I call A^0 clusters and write A^0 = [a0*4] The 449/258 example uses 30 A^0 clusters ([a0*4])*30 (120 a0 pieces, 30 4-parts) 47 A^1 clusters ([a1*3,m]*2)*47 (282 a1 pieces, 94 4-parts) Note that the A^1 cluster by itself sets an upper bound on S -- the cluster is 2 persons (size 2r), it cntains 1 complete muffin (size 1, pieces m*2), and it has 6 unmatched muffin pieces (a1*6), giving S(449/258) <= avg(a1) = (2r-1)/6. A^0 clusters also bound S by S(449/258) <= avg(a0) = r/4, but the A^1 cluster bound is tighter. Note that this (2r-1)/6 bound holds for examples 7/4 and 40/23. The muffin mates of a0 pieces (resp. a1 pieces) are z0 pieces (resp. z1 pieces). We distribute z0 and z1 pieces as evenly as possible across the 134 3-parts (402 pieces). With 282 z1 pieces, at least 14 3-parts must consist of only z1 pieces. These are Z^0 clusters, Z^0 = [z1*3] = [z10*3] where I refine piece labels so that z1 pieces appearing in Z^0 clusters are labelled z10. I create general Z^k clusters by finding minimal sets of persons that contain complete a0/z0 muffins. Maximally even distribution of the 120 z0 pieces places one per 3-part. Then a Z^1 cluster, closed under a0/z0 muffins, is Z^1 = [a0*4],[z0,z1*2]*4 = A^0,[z0,z11*2]*4 where z1 pieces appearing in Z^1 clusters are relabelled z11. Z^0 clusters bound S by S(449/258) <= avg(a10) = 1-avg(z10) = 1-r/3 = (3-r)/3 and Z^1 clusters provide an even tighter bound S(449/258) <= avg(a11) = 1-avg(z11) = 1-(5r-4)/8 = (12-5r)/8. The 449/258 example uses 14 Z^0 clusters = ([z10*3])*14 ( 42 z10 pieces, 14 3-parts) 30 Z^1 clusters = (A^0,[z0,z11*2]*4)*30 (240 z11 pieces, 120 3-parts) As before, we refine the labelling of a1 pieces into a10 and a11 pieces, according to z10 or z11 complements. The next level of clustering is constructed by taking closure under a10/z10 muffins. The 282 a1 pieces in the 47 A^1 clusters are divided into 240 a11 pieces and 42 a10 pieces. Thus at least 5 A^1 clusters must contain only a11 pieces, and with maximally even distribution we have 42 A^1 clusters with 1 a10 piece and 5 a11 pieces. Continuing to refine pieces according to their appearance in A^10 or A^11 clusters, we have A^10 = A^1 = {a110*6, ...} (6 a110 pieces and some other pieces) A^11 = Z^0, (A^1)*3 = (a111*15, ...) with 449/258 consisting of 5 A^10 clusters ( 5 A^1 clusters, 30 a110 pieces) 14 A^11 clusters (42 A^1 clusters, 252 a111 pieces) A^11 clusters provide bound S(449/258) <= avg(a111) = (7r-6)/15 The 449/258 example lies on the intersection of this bound and the next level cluster bound. The next cluster level takes a110/z110 muffin closure and induces clusters Z^10 = Z^1 Z^11 = A^10, (Z^1)*6 with 449/258 consisting of 0 Z^10 clusters 5 Z^11 clusters (i.e., only Z^11 clusters) and bound S(449/258) <= avg(a1111) = 1-avg(z1111) = (73-32r)/42. I conclude with some computer generated additional information on these bounds. If anyone sort of understands this post and would like to use my program as is (i.e., no manual and documentation is mostly my posts), drop me a note and I'll send the C++ source (300 lines). B<4>(r) = (1r-0)/4; S'(r) <= B<4>(r) on (3/2,2/1] tight bound on (3/2,8/5] and at r = (4+8k)/(2+5k) (3/2,8/5] contains mediant(3/2,8/5) = 11/7 S'( (3/2,8/5] ) = (3/8,2/5]; S'(11/7) = 11/28 B<>(r) = (1-0r)/2 B<4>(r) = B<>(r) at r = 2/1 = 2.0000000 S'(2/1) = 1/2 = 0.50000000 B<4,1>(r) = (3-1r)/3; S'(r) <= B<4,1>(r) on [12/7,2/1) tight bound on [9/5,2/1) and at r = (12+9k)/(7+5k) [9/5,2/1) contains mediant(9/5,2/1) = 11/6 S'( [9/5,2/1) ) = [2/5,1/3); S'(11/6) = 7/18 B<4>(r) = (1r-0)/4 B<4,1>(r) = B<4>(r) at r = 12/7 = 1.7142857 S'(12/7) = 3/7 = 0.42857143 B<4,1,1>(r) = (2r-1)/6; S'(r) <= B<4,1,1>(r) on (12/7,7/4] tight bound on (12/7,59/34] and at r = (21+59k)/(12+34k) (12/7,59/34] contains mediant(12/7,7/4) = 19/11 S'( (12/7,59/34] ) = (17/42,7/17]; S'(19/11) = 9/22 B<4,1>(r) = (3-1r)/3 B<4,1,1>(r) = B<4,1>(r) at r = 7/4 = 1.7500000 S'(7/4) = 5/12 = 0.41666667 B<4,1,1,1>(r) = (12-5r)/8; S'(r) <= B<4,1,1,1>(r) on [40/23,7/4) tight bound on [148/85,7/4) and at r = (80+148k)/(46+85k) [148/85,7/4) contains mediant(47/27,7/4) = 54/31 S'( [148/85,7/4) ) = [7/17,13/32); S'(54/31) = 51/124 B<4,1,1>(r) = (2r-1)/6 B<4,1,1,1>(r) = B<4,1,1>(r) at r = 40/23 = 1.7391304 S'(40/23) = 19/46 = 0.41304348 B<4,1,1,1,1>(r) = (7r-6)/15; S'(r) <= B<4,1,1,1,1>(r) on (40/23,228/131] tight bound on (40/23,1119/643] and at r = (228+1119k)/(131+643k) (40/23,1119/643] contains mediant(40/23,47/27) = 87/50 S'( (40/23,1119/643] ) = (142/345,265/643]; S'(87/50) = 103/250 B<4,1,1,1>(r) = (12-5r)/8 B<4,1,1,1,1>(r) = B<4,1,1,1>(r) at r = 228/131 = 1.7404580 S'(228/131) = 54/131 = 0.41221374 B<4,1,1,1,1,1>(r) = (73-32r)/42; S'(r) <= B<4,1,1,1,1,1>(r) on [449/258,228/131) tight bound on [9349/5372,228/131) and at r = (1347+9349k)/(774+5372k) [9349/5372,228/131) contains mediant(134/77,47/27) = 181/104 S'( [9349/5372,228/131) ) = [1107/2686,2267/5502); S'(181/104) = 75/182 B<4,1,1,1,1>(r) = (7r-6)/15 B<4,1,1,1,1,1>(r) = B<4,1,1,1,1>(r) at r = 449/258 = 1.7403101 S'(449/258) = 319/774 = 0.41214470 B<4,1,1,1,1,1,1>(r) = (61r-60)/112; S'(r) <= B<4,1,1,1,1,1,1>(r) on (449/258,764/439] tight bound on (449/258,149516/85913] and at r = (10696+149516k)/(6146+85913k) (449/258,149516/85913] contains mediant(449/258,764/439) = 1213/697 S'( (449/258,149516/85913] ) = (11909/28896,35408/85913]; S'(1213/697) = 32173/78064 B<4,1,1,1,1,1>(r) = (73-32r)/42 B<4,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1>(r) at r = 764/439 = 1.7403189 S'(764/439) = 2533/6146 = 0.41213798 B<4,1,1,1,1,1,1,1>(r) = (1101-487r)/615; S'(r) <= B<4,1,1,1,1,1,1,1>(r) on [160212/92059,764/439) tight bound on [1313469/754729,764/439) and at r = (160212+6567345k)/(92059+3773645k) [1313469/754729,764/439) contains mediant(19549/11233,764/439) = 20313/11672 S'( [1313469/754729,764/439) ) = [1555262/3773645,111271/269985); S'(20313/11672) = 986147/2392760 B<4,1,1,1,1,1,1>(r) = (61r-60)/112 B<4,1,1,1,1,1,1,1>(r) = B<4,1,1,1,1,1,1>(r) at r = 160212/92059 = 1.7403187 S'(160212/92059) = 37941/92059 = 0.41213787

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Re: [math-fun] Disk polyominoes
by Allan Wechsler 13 Apr '12

13 Apr '12

A funster who might or might not wish to remain anonymous has corrected me in private. A3 does not equal 2; it equals 1. The corrected sequence is: 1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 3, 3 ... As before, I'm not entirely confident of the last couple of values. Thank you for the correction.

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[math-fun] Machine Go
by rcs＠xmission.com 12 Oct '11

12 Oct '11

Science Daily has a report of a recent success for a Go program. http://www.sciencedaily.com/releases/2009/05/090514083931.htm The program Huygens defeated two professional Go players. The program had handicaps of 6 or 7 stones. Question for the Go players among us: Does a handicap of 6 stones mean that the starting board has 6 stones added for the weaker player, or that 6 is added to the final score? Rich

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[math-fun] Isodiametric problem
by victor miller 16 Sep '10

16 Sep '10

A few days ago, at the AMS meeting, I saw a neat talk by Michael Mossinghoff about "isodiametic" problems -- e.g. Fix the diameter of a conv ex n-gon and ask for the ones with maximum perimeter. He reduced the problem to finding univariate polynomials whose non-zero coefficients alternate between +/- 1 and are divisible by a particular cyclotomic polynomial.. He was able to calculate all such up to about 500. Each such polynomial is compactly encoded by th length of the gaps between the non-zero terms. For small n almost all were periodic. He showed how a certain family of periodic constrctions give solutions, but as n got bigger the "sporadic" ones swamped them. He conjectued that these are eventually all of them. He had lots of pictures. The paper is not available yet. Victor

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[math-fun] Systems of continued fractions
by Mike Stay 02 Sep '09

02 Sep '09

Given a finite continued fraction, you can start at the last term and build up the cf by | b d c ab + dc | - -> a + --- = ------- | c b b where 'b/c' is the current fraction, 'a' is the current term, and 'd' is usually 1. This operation is nicely represented with a 2x2 matrix: | [ a d ] [ b ] = [ ab + dc ] | [ 1 0 ] [ c ] [ b ] I'm looking at a situation where I have multiple fractions: | b0 b1 b2 | --, --, -- | c0 c1 c2 and an update rule that takes these to | d00 c0 d01 c1 d02 c2 | a0 + ------ + ------ + ------, | b0 b1 b2 | | d10 c0 d11 c1 d12 c2 | a1 + ------ + ------ + ------, | b0 b1 b2 | | d20 c0 d21 c1 d22 c2 | a2 + ------ + ------ + ------, | b0 b1 b2 Is there a way to do this case with matrices? -- Mike Stay - metaweta(a)gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com

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[math-fun] Origami torus
by Fred lunnon 02 Sep '09

02 Sep '09

The original problem was to construct a polyhedron, homeomorphic to a torus, with angular defect zero at every corner (and everywhere else): properly embedded in 3-space, so that the interiors are homeomorphic. I found a 2-parameter family of "origami polytore" solutions with 20 faces (4 square + 16 triangular), 12 vertices, 32 edges, each comprising: a square cuboid of depth c with corners (+/-1,+/-1,+/-c), deprived of its top and bottom faces, with 8 edges and 8 corners connected by triangle faces to the skew quadrangle with corners (+q,0,+h), (0,-q,-h), (-q,0,+h), (0,+q,-h). Given the depth c, what radii q and heights h are possible? Theorem: when the polytore is origami, either h^2 = -(q^2-2)*( 1/2 + c^2/(q-2)^2 ) --- the "oval" branch, or h^2 = -(q^2)*( 1/2 + c^2/((q-2)^2-2) ) --- the "cusp" branch. There are just two congruency classes of corner: cuboid, and quadrangle. By the Gauss-Bonnet theorem, the sum of the angular defects vanishes; so we need ensure only that they vanish at cuboid corners. The latter having two square faces, it remains to ensure that the other three face angles A,B,C sum to \pi. >From the coordinates of the corners, the squares of the edge lengths can be found by Pythagoras; then in turn by the cosine rule, the squared sines SA,SB,SC of the face angles are expressible as _rational_ functions of c,q,h. The squared sine SABC of the sum angle satisfies a quartic polynomial equation with coefficients polynomial in SA,SB,SC [of which I am inordinately proud --- see "Horrid trig identity"], which we need to have root zero, so constant coefficient zero. The latter turns out to factorise as the square of the product of the polynomials implicit in the theorem; so we have a necessary condition for origami-city. For sufficiency, in practice, all we need to do is check that the sum is correct at some convenient point, such as q = 1, then rely on continuity of the root as a function of q. In theory, I'm not entirely happy about this: just exactly _how_ can we be sure that some other root of the quartic does not intrude? That caveat aside, the interesting upshot of this frivolous investigation was that there is another branch of solutions (cusp) besides those originally found (oval) by trial-and-lots-of-error. Both branches are plotted for c = 1 at <hq_plot.jpg>; those solutions giving proper embeddings (rather than mere immersions) lie between 0 < q < 1. The new branch is less well house-trained than the old: at q = 2-sqrt2, h roars off to oo. At one point the branches actually cross: although somehow I keep feeling this solution should have some interesting property, in fact it looks just like most of the others <polytore.jpg>. There is available a Maple worksheet <polytore.mw> on this topic, incorporating the above still in rotatable form, and an animation of the polyhedron as q varies. Further projects suggested by this investigation include the extension to regular base polygons with any even number edges (rather than just squares, as above); and an animation of the polyhedron as folding up from a planar map. The latter poses a nontrivial robotics question: is it possible to fold the polyhedron from a sheet on which opposite edges have already been joined [forcing it to be folded flat, but four layers thick]? My attempts to find somewhere public to put all this stuff having been so far cruelly stymied, I must again offer to send the files to anybody expressing an interest. Fred Lunnon [18/08/09]

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[math-fun] Shoe Polish
by Hans Havermann 28 Aug '09

28 Aug '09

I was writing today about my once-upon-a-time grade 13 mathematics teacher who, every time a student carelessly referred to something on the blackboard with the pronoun "it", would shout out: "It! Shoe polish." Now, I believe It! was a brand of shoe polish at one time, but I can't seem to find an online mention of it. Anybody?

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[math-fun] Horrid trig identity
by Fred lunnon 25 Aug '09

25 Aug '09

On 8/14/09, Fred lunnon <fred.lunnon(a)gmail.com> wrote: > ... > What's more, using Gaussian curvature (in the form of a "2-D cosine") > gives a far easier way to compute the face angle sums than employed > by my existing kludgy program (or my dodgy sine formula). WFL Well, no as it turns out --- computing face angles from polyhedron edges turns out less troublesome. But any algebraic proof of Fred's identity [with appropriate ambiguity concerning which Fred is thus immortalised], to avoid introducing arcsines, must deal with cosines and sines, which are not additive; and to avoid square roots, must deal with squares of sines. So for two angles [my earlier dodgy identity should have read] SAB^2 - 2*(SA-2*SA*SB+SB)*SAB + (SA-SB)^2 = 0, with SA -> sin^2(A), SB -> sin^2(B), SAB -> sin^2(A+B). But the polytore proof needs three angles, for which the equivalent identity (according to my Maple investigations) has degree 4 in SABC, 8796 terms in SA,SB,SC --- and anyway refuses to check out numerically, let alone formally. So where does one go from here? WFL

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[math-fun] Life imitates art
by rwg＠sdf.lonestar.org 21 Aug '09

21 Aug '09

Exploring a recent Life discovery by Mitchell Riley, I found the little green inset, which, after ~1.25G, produces the strangely varied textures shown in http://gosper.org/diparaa125G.png . (Scale: 2^-18) --rwg

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[math-fun] Fw: Re: Usain Bolt
by hbaker1 19 Aug '09

19 Aug '09

The early 30mph reports were wrong; see excellent analysis at the URL mentioned below. -----Forwarded Message----- >From: Shane Skowron <sskowron(a)fordham.edu> >Sent: Aug 19, 2009 9:52 AM > >Right. He never ran 30mph, unless we're rounding up? >His peak speed from 60m to 80m was 27.78 mph. > >Here's an analysis: >http://www.sportsscientists.com/2009/08/analysis-of-bolts-958-wr.html > >Shane

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