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[math-fun] record computation of Pi ,
by Bill Gosper 12 Dec '16

12 Dec '16

Digits. Feh. What a waste of chips and neurons. The only excuse: Now that they've done it, converting to the continued fraction might be slightly easier than extracting the cf directly. But if you want both, it might be easier to extract the decimal from the cf. Or maybe not, since they probably wrote their own multiprecision routines to use base 10^19 vs 2^64 to dodge the big radix conversion cost. Or is floating point still so much faster than fixed that they sacrificed several digits per "word" to redundant exponents, like the old Mersenne finders? --rwg Date: 2016-12-10 17:53 From: Simon Plouffe <simon.plouffe(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> They made a new computation record of Pi, http://www.numberworld.org/y-cruncher/records.html 22 459 157 718 361 digits. This is big. the digits seems to be rather normal in base 10 and 16. https://arxiv.org/ftp/arxiv/papers/1612/1612.00489.pdf also : 22.459157718361 is Pi^E. Some explanations here : http://pi2e.ch/ best regards, Simon Plouffe

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[math-fun] Dissecting a triangle into polygons with 180-degree rotational symmetry
by James Propp 12 Dec '16

12 Dec '16

Fun puzzle (maybe an oldie): Find a dissection of a triangle (any triangle you like!) into polygons with 180-degree rotational symmetry, or prove that no such dissection is possible. Jim Propp

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Re: [math-fun] Dodecahedron dissection
by Dan Asimov 12 Dec '16

12 Dec '16

Nice question! Cleverly avoiding addressing it, I wonder: * If there is a good way of ordering the f-vectors F = (f_0, f_1, f2, f_3) (where f_k is the count of k-dimensional faces of a polyhedron, so Fred's dissection would have F = (20, 61, 66, 24)) so that one could seek a polyhedron for which F is optimal with respect to this ordering. * I've seen similar questions for polygons where the dissection sought is one into acute triangles. E.g., the minimal number of these for a square is 7. So: What is the smallest number of acute tetrahedra that a regular solid can be dissected into? An acute tetrahedron is one whose dihedral angles are all acute.* —Dan _______________ P.S. Just learned two things on the subject: 1) The smallest known number of acute tetrahedra that the cube can be dissected into (as of 2010) was 1370. 2) It is known that 3-space can be dissected into acute simplices, but 5-space cannot be. ----- Fred <fred.lunnon(a)gmail.com> wrote on Dec 12, 2016 at 10:48 AM: ----- Problem: dissect into tetrahedra the regular dodecahedron with interior, so that the resulting complex is symmetric under reflection in its centre. I have a solution with 20 vertices, 61 edges, 66 triangles, 24 tetrahedra: can any of these counts be reduced? [ Further details concerning this solution, and the motivation behind the problem, available on demand. ] -----

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[math-fun] Dodecahedron dissection
by Fred Lunnon 12 Dec '16

12 Dec '16

Problem: dissect into tetrahedra the regular dodecahedron with interior, so that the resulting complex is symmetric under reflection in its centre. I have a solution with 20 vertices, 61 edges, 66 triangles, 24 tetrahedra: can any of these counts be reduced? [ Further details concerning this solution, and the motivation behind the problem, available on demand. ] Fred Lunnon

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[math-fun] record computation of Pi ,
by Bill Gosper 12 Dec '16

12 Dec '16

I should expect floating point to have been equally useless for the mere kilodigit efforts of the early Mersenne prime chasers. I now realize I have no idea how they did it. Ah, I forgot about C.R.T.. The Chudnovskys used it for a gigadigit on their Porkroast 1. It enabled them to punt the huge common divisors that hampered my effort. --rwg Date: 2016-12-12 00:07 From: Joerg Arndt <arndt(a)jjj.de> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> * Bill Gosper <billgosper(a)gmail.com> [Dec 12. 2016 08:38]: > [...] > Or is floating > point still so much faster than fixed that they sacrificed several digits > per "word" to redundant exponents, like the old Mersenne finders? --rwg For this (giga-digits) precision floating point is useless. They certainly used (a variant of) the "three primes" NTTs, doing CRT at the very end. Sadly, the information given by the program author is rather sketchy. E.g., I'd like to know whether Montgomery multiplication is used. The computation IS an impressive feat, even though the guy had a machine with a droolworthy 1.2 TB of RAM. Best regards, jj

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[math-fun] record computation of Pi ,
by Bill Gosper 11 Dec '16

11 Dec '16

Counting repeated additions to estimate an integer quotient is silly. Knuth Vol 2 has bignum algorithms for both Euclidean and binary GCD. In both cases, you run a toy algorithm using just a few bytes of the operands. This will behave initially like the full precision process. Record the steps in a matrix of smallnums. About when the toy process runs out of significance, or when the smallnums threaten largeness, discard its state and apply the matrix to the two bignums. This will make them smaller, and in the binary case, much "evener". It doesn't matter if the last few terms or bits were wrong--the process is self-correcting. It will simply cough up a few nonpositive terms which can be smoothed out in (sub?)linear time. --rwg Date: 2016-12-11 10:53 From: Simon Plouffe <simon.plouffe(a)gmail.com> Hello, yes, speaking of wich, To go from a decimal number to continued fraction there is one simple algorithm that uses only additions. You just need a for-next loop, you start with x0 (supposedly fractional) Start with S0= 0. do Sn+1 = Sn+Xn, until Xn+1 is greater than 1. Output the index n (the partial quotient). X0 = Xn+1 od: In principle this is clean, no inverse to compute. The only drawback of this method is that when the partial quotient is really big then the procedures hangs. But if the number is believed to be non-exotic, well, that's a living. The big advantage is that the computation can be carried out to billions of digits wide with not much overhead, it is a bit slow perhaps but very economical in memory. Does this have ever been attempted on a big number like Pi ?? Question : what is the cost of operations to inverse a number of let's say 22459 billion digits ?? a lot ? it could be improved with this trick. When you have a certain X0 in the process - Evaluate roughly the ratio of 1/X0 = ratio , (let's say to 50 digits) then multiply the big number by that [ ratio ] + 1 , [ ] being the integer part. For all practical purpose the ratio won't be bigger than 20-30 digits. so this is simple in fact, only 1 multiplication by 1 number of limited size. Too simple ? Best regards, Simon Plouffe Le 2016-12-11 à 18:04, Bill Gosper a écrit : > Digits. Feh. What a waste of chips and neurons. The only excuse: > Now that they've done it, converting to the continued fraction might > be slightly easier than extracting the cf directly. But if you want both, > it might be easier to extract the decimal from the cf. Or maybe not, > since they probably wrote their own multiprecision routines to use base > 10^19 vs 2^64 to dodge the big radix conversion cost. Or is floating > point still so much faster than fixed that they sacrificed several digits > per "word" to redundant exponents, like the old Mersenne finders? --rwg > > Date: 2016-12-10 17:53 > From: Simon Plouffe <simon.plouffe(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > They made a new computation record of Pi, > > http://www.numberworld.org/y-cruncher/records.html > > 22 459 157 718 361 digits. > > This is big. > > the digits seems to be rather normal in base 10 and 16. > > https://arxiv.org/ftp/arxiv/papers/1612/1612.00489.pdf > > also : 22.459157718361 is Pi^E. > > Some explanations here : http://pi2e.ch/ > > best regards, > Simon Plouffe

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Re: [math-fun] record computation of Pi ,
by Dan Asimov 11 Dec '16

11 Dec '16

<< Digits. Feh. What a waste of chips and neurons. Sure seems that way to me. If anyone can please explain what the big deal is, I'm listening. —Dan >From: Bill Gosper <billgosper(a)gmail.com> >Sent: Dec 11, 2016 9:04 AM >Digits. Feh. What a waste of chips and neurons. The only excuse: >Now that they've done it, converting to the continued fraction might >be slightly easier than extracting the cf directly. But if you want both, >it might be easier to extract the decimal from the cf. Or maybe not, >since they probably wrote their own multiprecision routines to use base >10^19 vs 2^64 to dodge the big radix conversion cost. Or is floating >point still so much faster than fixed that they sacrificed several digits >per "word" to redundant exponents, like the old Mersenne finders? --rwg > >Date: 2016-12-10 17:53 >From: Simon Plouffe <simon.plouffe(a)gmail.com> >To: math-fun <math-fun(a)mailman.xmission.com> >Reply-To: math-fun <math-fun(a)mailman.xmission.com> > >They made a new computation record of Pi, > >http://www.numberworld.org/y-cruncher/records.html > >22 459 157 718 361 digits. > >This is big. > > the digits seems to be rather normal in base 10 and 16. > >https://arxiv.org/ftp/arxiv/papers/1612/1612.00489.pdf > >also : 22.459157718361 is Pi^E. > >Some explanations here : http://pi2e.ch/ > >best regards, > Simon Plouffe >_______________________________________________ >math-fun mailing list >math-fun(a)mailman.xmission.com >https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] record computation of Pi ,
by Simon Plouffe 10 Dec '16

10 Dec '16

They made a new computation record of Pi, http://www.numberworld.org/y-cruncher/records.html 22 459 157 718 361 digits. This is big. the digits seems to be rather normal in base 10 and 16. https://arxiv.org/ftp/arxiv/papers/1612/1612.00489.pdf also : 22.459157718361 is Pi^E. Some explanations here : http://pi2e.ch/ best regards, Simon Plouffe

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[math-fun] Grochow/Moore: "Matrix multiplication algorithms from group orbits"
by Joerg Arndt 09 Dec '16

09 Dec '16

Some of us may like the preprint at https://arxiv.org/abs/1612.01527 Abstract: We show how to construct highly symmetric algorithms for matrix multiplication. In particular, we consider algorithms which decompose the matrix multiplication tensor into a sum of rank-1 tensors, where the decomposition itself consists of orbits under some finite group action. We show how to use the representation theory of the corresponding group to derive simple constraints on the decomposition, which we solve by hand for n=2,3,4,5, recovering Strassen's algorithm (in a particularly symmetric form) and new algorithms for larger n. While these new algorithms do not improve the known upper bounds on tensor rank or the matrix multiplication exponent, they are beautiful in their own right, and we point out modifications of this idea that could plausibly lead to further improvements. Our constructions also suggest further patterns that could be mined for new algorithms, including a tantalizing connection with lattices. Best regards, jj

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Re: [math-fun] overqualified typesetters
by Bill Gosper 09 Dec '16

09 Dec '16

Date: 2016-12-09 06:45 From: James Propp <jamespropp(a)gmail.com> I just posted to MathOverflow (tag: "terminology") to see if anyone can provide info on the radiculus function. Maybe G.H. Hardy collaborated with Alfred Haar on a two-author paper about the radiculus function, with Hardy getting priority in the author-list despite Haar being alphabetically prior? :-) Jim Propp On Friday, December 9, 2016, Dan Asimov <asimov(a)msri.org> wrote: > Do you call this utterly miscellaneous because why choose that particular > product, or because of how it's displayed (instead of gathering the powers > of 3 — or better, sqrt(3) — in one place) ? > > —Dan The former. I wouldn't consider eccentric notation "miscellaneous". But Gauss gave us a formula for Γ(x+1/n) Γ(x+2/n) ... Γ(x+1). Why exhibit this nebbish special case of it? --rwg > > > > On Dec 9, 2016, at 12:50 AM, Bill Gosper <billgosper(a)gmail.com > <javascript:;>> wrote: > > > > the utterly > > miscellaneous > > Product[Gamma[k/3], {k, 8}] == 640/3^6 (\[Pi]/Sqrt[3])^3 >

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