March 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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[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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Re: [math-fun] Fast factoring ?? (really denesting)
by rwg＠sdf.lonestar.org 08 May '09

08 May '09

> On Fri, Dec 19, 2008 at 12:20 PM, <rwg(a)sdf.lonestar.org> wrote: > >> Amen. I've now tried ~ 10^8 cases of (a^(1/5)-b^(1/5))^(1/3) >> without further success, and for sqrt, only >> >> 3/5 4/5 3/5 2/5 2/5 4/5 1/5 1/5 >> 1/5 1/5 - 2 3 + 3 + 2 2 3 - 2 3 + 2 >> sqrt(4 - 3 ) = ---------------------------------------------------. >> 5 >> >> And, of course, these can only give pentanomials. >> --rwg >> ALGORISMIC MICROGLIAS > > I have no technical knowledge on denesting, but here is how I look at > the problem: > > Since sqrt( a^(1/5) + b^(1/5) ) = a^(-2/5) * sqrt( a + (a^4*b)^(1/5) > ), I'll consider only the form: sqrt( a + b^(1/5) ). If sqrt( a + > b^(1/5) ) could be denested, I'd expect it to be denested into the > form: > > x0 + x1*b^(1/5) + x2*b^(2/5) + x3*b^(3/5) + x4*b^(4/5). Notice that in the (comparatively easy) 4th root case, your conjecture catches 1/4 3/4 1/4 sqrt(161 - 12 5 ) = 2 5 - 3 sqrt(5) + 4 5 + 6 but not 1/4 1/4 1/4 3/4 3/4 1/4 sqrt(2) sqrt(31 - 4 3 5 ) = 3 5 - 2 sqrt(5) + 3 5 + 2 sqrt(3). > That's why I don't think you could find anything longer than > pentanomial. OTOH, sqrt( a^(1/7) + b^(1/7) ) is a better bet for > finding hexanomials or longer. Indeed, but the only one I've found so far is perversely pentanomial: 1/7 Sqrt(7 (2 - 2 )) 1/7 3/7 5/7 6/7 ------------------ = -1 + 2 2 + 2 + 2 - 2 . 1/14 2 > Just my 2 cents. > > Warut > You didn't propose 6th roots. Did you have reason to suspect that they are as infertile as they seem to be? --rwg PRECHRISTMAS PETRARCHISMS ALGORISMIC MICROGLIAS

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Re: [math-fun] If We Have Free Will, Then So Do Electrons
by Dan Asimov 17 Apr '09

17 Apr '09

Mike Stay wrote: << Obligatory reference: http://www.overcomingbias.com/2008/04/zombie-movie.html >> That's utterly hilarious!!! The funniest thing I've read in months. --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] Heterodigital triplets
by Eric Angelini 31 Mar '09

31 Mar '09

> my april math magic problem ... Nice, Erich ! How do you do to be always a step ahead of my thoughts ;-) ? Best, É. -----Message d'origine----- De : math-fun-bounces(a)mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Erich Friedman Envoyé : mardi 31 mars 2009 15:18 À : math-fun Objet : Re: [math-fun] Heterodigital triplets i'm working on something similar: finding integers that can be factored both pandigitally (with or without 0) and unidigitally, such as 5476 × 198 × 32 = 2 × 2 × 2 × 2 × 2 × 22 × 222 × 222 results and variations on the theme are my april math magic problem: http://www.stetson.edu/~efriedma/mathmagic/0409.html erich > Find all heterodigital triplets (a;b;c) such that > a, b, c, s, and p is an heterodigital quintuplet > > There is at least one such quintuplet which is > heteropandigital (all digits from 0-->9 are used) _______________________________________________ math-fun mailing list math-fun(a)mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Heterodigital triplets
by Eric Angelini 31 Mar '09

31 Mar '09

Hello Math-Fun, Let's call 'heterodigital' an integer that doesn't show two identical digits, like 1, 2, 83, 2340, ... An heterodigital doublet is made of two integers whose concatenation is also heterodigital : (123;46) is an heterodigital doublet (123;41) is not. Find all heterodigital triplets (a;b;c) such that a, b, c, s, and p is an heterodigital quintuplet (s = sum = a+b+c and p = product = a*b*c ) A few such examples: a b c s p 1;2;4 7 8 2;4;7 13 56 5;6;8 19 240 There is at least one such quintuplet which is heteropandigital (all digits from 0-->9 are used) Best, É. --- P.-S. I would love to see an equivalent search for all heterodigital quadruplets (a;b;s;p) A sub-list of this being (a;b;s;p;e) where e = exp = a^b examples: (2;3;5;6;8) and (3;2;5;6;9)

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[math-fun] plouffe inverter tables
by Simon Plouffe 31 Mar '09

31 Mar '09

Hello everybody, I managed to make all the tables available (at last!) here http://pictor.math.uqam.ca/~plouffe/pi/ip/ there are 2.68 billion entries to 64 digits in a normalized format. These tables will be updated once in a while. Also, there is a subset of these tables here http://pictor.math.uqam.ca/~plouffe/seq2real/ these tables contain 52 millions entries, they are made from the 150000 sequences from the OEIS, each sequence generates 320 reals as listed in the 140000 tables below, for example , for the sequence A000045 (fibonacci) http://pictor.math.uqam.ca/~plouffe/seq2real/a00/A000045.txt that table contains all the variants, note : in some occasions the signs wehere taken into account. NOTE : there is a total of approximately 350 gigabytes of data in there, beware. Simon Plouffe

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Re: [math-fun] Moby Spacecube
by Dan Asimov 31 Mar '09

31 Mar '09

Interesting. But my browser could not find the second URL, < http://gosper.org/spacecube.jpg >. --Dan << When I first got the mechanical assembly puzzle (http://www.spacecubes.com/more.php?FORMAT=2d) I doubted its (unassisted) feasibility. Having lost all my copies, I sank about $65 into some plumbing supplies and a circular saw, with marginal success: (http://gosper.org/spacecube.jpg) >> _____________________________________________________________________ "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] If We Have Free Will, Then So Do Electrons
by Dan Asimov 31 Mar '09

31 Mar '09

Gareth wrote: << . . . "I don't deny the reality of conscious experience", [Dennett] says. Do you think he's lying? Or so stupid that he simply can't understand the obvious truth that "qualia" == "conscious experience"? . . . >> No, I don't think he's lying, at least not deliberately. When I originally posted about this, I had not, by the way, grasped that Dennett claims he is not denying the existence of conscious experience. But I still maintain that no matter how many words one uses to describe what qualia means, it still boils down to conscious experience. (Qualia is the plural of quale.) People talking about qualia are trying to decompose conscious experience into little pieces that typically make sense (the commonest example being the experience of a color). Of course, conscious experience has subtleties that transcend language's ability to describe things, but I would not disqualify qualia from existence on that basis. So what if people experience colors differently? I know they must, since even my two eyes experience the same color a bit differently. That just means you can't necessarily predict one's experience purely on the basis of the apparent stimulus. I would like to know just what Dennett claims he is disproving, however. --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] If We Have Free Will, Then So Do Electrons
by Dan Asimov 30 Mar '09

30 Mar '09

Yes, I thought I remembered his title, but apparently misremembered it. He is not, in my opinion, "at pains" to say he defines qualia in a new and different way from how virtually all other philosophers define it. And if he does use a standard word in a different from the standard way, he deserves every bit of misunderstanding he may get from it. It's not a "common idea" as you say below or any kind of idea; it is the standard definition of the term. --Dan << On Wednesday 25 March 2009, Dan Asimov wrote: > Ever since I learned from Dennett's "Understanding Consciousness" that > he does not believe in the existence of qualia (aka conscious experience*), > I have seriously wondered whether it's worth the trouble to read anything > else he wrote. 1. Do you mean "Consciousness Explained"? 2. Dennett (in that book, and in the much shorter piece "Quining qualia" which you cite) is at pains to say that he doesn't disbelieve in conscious experience even though he doesn't believe in qualia. (He thinks that the commyon idea that conscious experience == qualia is wrong.) Do you think he's lying, or that the identification conscious experience == qualia is so obviously right that someone who says otherwise isn't worth paying any attention to, even if (say) he happens to be a professional philosopher specializing in consciousness? (For the avoidance of doubt: I do not profess to know whether Dennett is right about qualia, and in particular I am not claiming that the fact that he's a professional philosopher specializing in consciousness guarantees that he is. There are p.p.s.i.c. who disagree with him strongly.) >> _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele

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