May 2010 - 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] parametrizing the Koch snowflake curve
by Bill Gosper 24 Apr '15

24 Apr '15

Fourier epicycles thru nth: (d16) sum(((product((1 - (( - 1)^j * tan(((%pi * (2 * k + (1/3)))/(2^j)))/(sqrt(3)))),j,1,inf)) * (%e)^(%i * (6 * k + 1) * t)/((2 * k + (1/3))^2)),k, - n,n) See http://www.tweedledum.com/rwg/cog.htm Also http://gosper.org/fst.dvi Pix: http://gosper.org/fst1.png ,http://gosper.org/fst2.png , http://gosper.org/fst3.png ,http://gosper.org/fst4.png . (For parametric, take Re and Im.) Rich's base 4 recursion loops for rational arg. Find the fixed point of the pending stack ops to simulate infinite recursion, and get an exact val on the curve. Equate to this Fourier series for a *non-obvious* identity. --rwg > If you view the starting triangle as running from 0 to 3, > and write the fraction base 4, you can use the > fraction digits to select which part of the (sub)edge > you are on. The orientation of the edge can be calculated > from the 1s and 2s of the high order portion of the fraction, > and that should give a somewhat kludgy sum for the (x,y) > coordinates of the image point. > RWG likely has some nice FFT clock sum expression. > > Rich > > > -----Original Message----- > From: math-fun-bounces(a)mailman.xmission.com [mailto: math-fun-bounces(a)mailman.xmission.com] On > Behalf Of N. J. A. Sloane > Sent: Thursday, February 25, 2010 12:43 PM > To: math-fun(a)mailman.xmission.com > Cc: njas(a)research.att.com > Subject: Re: [math-fun] parametrizing the Koch snowflake curve > > Jim Propp asked: > >> Does anyone (Bill Gosper, maybe?) know an explicit parametrization > of the Koch snowflake curve? > > The Cantor set consists of the points 0 <= x <= 1 which have a base 3 "decimal" expansion that > does not contain a 1. Is there a similar characterization of the Koch snowflake? > > Neil Sloane > >

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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] Slanted coverage
by Bill Gosper 08 Jul '11

08 Jul '11

In those ubiquitous little tables of area and volume formulae in handbooks and textbooks, there's often a picture of a skewed cone next to the right cone to emphasize that they have the same volume formula. Did you ever wonder why there's never the skewed cone area formula? For the special case with the apex directly above a point on the circumference, I get pi h r 3F2[-1/2,1/4,3/4; 1/2, 1; -4 r^2/h^2], which doesn't even converge for h < 2 r. The area element for the general x,y,z parametric surface is just Out[85]= (1,0) (0,1) Norm[Cross[({x, y, z}) , ({x, y, z}) ]] (I don't recall this from 18.03.) E.g., for the torus with radii r and R, In[116]:= % /. {x :> (Cos[#2]*(R + r*Sin[#1]) &), y :> (Sin[#2]*(R + r*Sin[#1]) &), z :> (r*Cos[#1] &)} Out[116]= Norm[Cross[({Cos[#2] (R + r Sin[#1]) & , Sin[#2] (R + r Sin[#1]) & , (1,0) r Cos[#1] & }) , ({Cos[#2] (R + r Sin[#1]) & , (0,1) Sin[#2] (R + r Sin[#1]) & , r Cos[#1] & }) ]] In[117]:= % /. f_[L_List] :> f /@ L /. Abs[w_] -> w Out[117]= Sqrt[Power[-(-r Sin[#1] & ) (Cos[#2] (R + r Sin[#1]) & ) + (0 & ) (r Cos[#1] Sin[#2] & ), 2] + Power[-(0 & ) (r Cos[#1] Cos[#2] & ) + (-r Sin[#1] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2] + Power[(r Cos[#1] Cos[#2] & ) (Cos[#2] (R + r Sin[#1]) & ) - (r Cos[#1] Sin[#2] & ) (-(R + r Sin[#1]) Sin[#2] & ), 2]] In[119]:= % /. (f_ &) :> (f &)[t, u] Out[119]= 2 2 2 2 Sqrt[r Cos[u] Sin[t] (R + r Sin[t]) + 2 2 2 2 r Sin[t] (R + r Sin[t]) Sin[u] + 2 2 2 (r Cos[t] Cos[u] (R + r Sin[t]) + r Cos[t] (R + r Sin[t]) Sin[u] ) ] In[120]:= FullSimplify[%] Out[120]= 2 2 Sqrt[r (R + r Sin[t]) ] In[125]:= Assuming[R > r > 0, Integrate[%120, {t, 0, 2*Pi}, {u, 0, 2*Pi}]] Out[125]= 2 4 Pi r R Did Archimedes know this? --rwg

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[math-fun] €8,000 (~$10,000) prizes on magic squares
by Christian Boyer 11 Oct '10

11 Oct '10

While magic squares have been known and studied for many centuries, it is surprising that for certain types of magic squares we still do not know today which are the SMALLEST possible! In an effort to make progress on these unsolved problems, twelve prizes totaling €8,000 (~$10,000) and twelve bottles of champagne are offered for the solutions to twelve enigmas. Details here: www.multimagie.com/English/MagicSquaresEnigmasE.pdf Christian Boyer. www.multimagie.com/indexengl.htm

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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] Fibonacci Somos
by Bill Gosper 31 May '10

31 May '10

On Thu, May 27, 2010 at 9:00 PM, Bill Gosper <billgosper(a)gmail.com> wrote: > Let's call sequences growing like x^n^2 "supergeometric". > The Somos recurrence s[n]:=(s[n-1]*s[n-3]+s[n-2]^2)/s[n-4], for > s[1],... = 1,1,1,1,..., grows supergeometrically. But if we initialize > with 1,1,-1,-2/3,..., we get a supergeometrically *shrinking* sequence > = a period 8 sequence times 3^-(n^2/16): > > s[n] = -2*%i^(n^2+1)*q^(n^2/64)*theta[1](%i*n*log(q)/8+n*%pi/2,q)/ > theta[4](0,q)*3^(n^2/16) > > 2 2 > n + 1 n /64 i n log(q) n pi > 2 i q theta (---------- + ----, q) > 1 8 2 > s = - ------------------------------------------ > n 2 > n /16 > theta (0, q) 3 > 4 > > = 1,1,-1,-2/3,1/3,1/9,-1/27,0,1/243,-1/729,-1/2187,2/3^9,... > > where the simplest equation I've found for q is > > 3^(1/4)*theta[4](0,q) = -theta[2](0,%i*q^(1/4))*%e^(7*i*%pi/8) > > 7 i pi > ------ > 1/4 1/4 8 > 3 theta (0, q) = - theta (0, i q ) e > 4 2 > > Can we do better? > > Similarly, the period 8 sequence > (1,1,-1,0,1,-1,-1,0)* > comes from > > s[n] = theta[1](n*%pi/4,q)/(theta[2](0,q)*%i^(n^2/4-1)) > > n pi > theta (----, q) > 1 4 > s = -------------------- > n 2 > n > -- - 1 > 4 > theta (0, q) i > 2 > > with q a root of > > theta[2](0,q) = theta[2](0,%i*q)*sqrt(%i)/sqrt(2) > > theta (0, i q) sqrt(i) > 2 > theta (0, q) = ---------------------- > 2 sqrt(2) > > And then there's the even simpler period 5 sequence > (1,1,-1,-1,0)* > from > s[n] = theta[1](2*n*%pi/5,q)/theta[1](%pi/5,q) > > 2 n pi > theta (------, q) > 1 5 > s = ----------------- > n pi > theta (--, q) > 1 5 > > with q defined by > > theta[1](%pi/5,q) = theta[1](2*n*%pi/5,q) > > pi 2 n pi > theta (--, q) = theta (------, q). > 1 5 1 5 > > This theta is presumably algebraic? That would bode ill for closedform q. > The > only thetas I know how to invert are algebraic times gamma(s). > > Of course, all of these can be expressed as a pile of trigs But not when > we generalize to > 1,1,-1,x,... . Can anyone say why this makes polynomials in x? > With x = phi and -1/phi we get period 6 sequences times supergeometrics. Dividing their difference by rt5 gives 6 interlaced fibonacci(qudratic)s: http://gosper.org/fibsom.pdf . There must be a nice relation between the p and q nomes. > --rwg >

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Re: [math-fun] iter_exp()
by Christian G. Bower 29 May '10

29 May '10

------ Original Message ------ Received: Mon, 24 May 2010 03:59:20 PM PDT From: Steve Witham <sw(a)tiac.net> To: math-fun(a)mailman.xmission.com Subject: [math-fun] iter_exp() > I'm (morbidly?) fascinated by iterated exponentials and I like > the idea of extrapolating a continuous extension of a function > from simple behavior at a limit. The stuff I was doing with > sinh and tanh, seemed a bit arbitrary, though. > > ... > The current place I've wandered to is to define > > lim > n -> oo iter_exp( z ) = exp^n( z0 + z0^(z-n) ) > > ... > > iter_exp( z + 1 ) = exp( iter_exp( z ) ) > > Which is nice, but there are no z's where iter_exp(z) = 0 or 1 or e > or e^e... I may have to learn some complex analysis. Or maybe > category theory or LaTeX. I don't think there are z's that map to 0, but remember exp is not 1-1 in the complex plane. Find a place where iter_exp(z) = 2 pi i and that yields a place where iter_exp(z) = 1 I find one around 1.511 - .217i Giving us 1 at around 2.511 - .217i Christian > ... > > --Steve >

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[math-fun] Gardner, an other imprint
by wouter meeussen 28 May '10

28 May '10

what did it for me: Gardner. "Mathematical Games." Scientific American, Vol. 241, No 2, pp. 18-24; June, 1976; all about Catalans, trees and Bell numbers. Readable, intriguing, enticing, correct, elegant. Cute pictures too. It was in the personal-computer-stone-age, mind you. W.

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