July 2007 - mathfun@mailman.xmission.com

[math-fun] Re: ENIAC and maple : a simple test
by Steve Witham 21 Jul '07

21 Jul '07

Here a Java applet: http://www.tiac.net/~sw/DumbFlops/dumbflops.html It's not the same calculation as in Simon's loop-- a multiplication rather than a division. I wrote it in 1999: int i, reps; double x, y; ... for( i = 0; i < reps; i++ ) { x = i; y += x * x; } My 600MHz PPC Mac gets about 40 Million reps/sec, or about 15 clock cycles per rep. Wikipedia says ENIAC it could multiply two 10-digit fixed-point numbers in 14 clock cycles. --Steve

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[math-fun] semicyclic polygons
by Richard Schroeppel 21 Jul '07

21 Jul '07

The rational equation that MK discusses can step solutions in at least three different ways, along with some simple modifications. A^2 + B^2 + C^2 + 2ABC = 1 With starting solution 1/2, 1/2, 1/2. The simplest mods are to permute A,B,C; and to negate any pair. The linear step is to fix A and B; then C is the root of a quadratic whose two roots sum to -2AB, whence A,B,C' is a solution, C' = -2AB-C. Applying this twice just regenerates C, but you can step away by permuting A,B,C. The quadratic stepper is the fix A, then consider the equation B^2 + 2A BC + C^2 = 1 - A^2. For our starting solution of 1/2,1/2,/1/2, the new equation is B^2 + BC + C^2 = 3/4. We can generate more rational solutions from this by finding rational solutions of X^2 + XY + Y^2 = 1 or integer solutions of S^2 + ST + T^2 = N^2. These come from taking N = U^2 + UV + V^2 = norm(U - wV), and squaring. Take U=2, V=1, N=7, giving S = U^2 - V^2 = 3, T = 2UV + V^2 = 5, and then X = 3/7, Y = 5/7. Multiply (B-wC)*(X-wY) = B'-wC', giving B' = -1/7, C' = 13/14. Together with the unchanged A, the new solution A,B',C' = (1/2,-1/7,13/14). A final stepper can be derived from treating the equation as a cubic surface. Take any two distinct solutions and draw the line connecting them. As long as the delta of the two solutions has no 0 components, the line intersects the cubic surface in one more point. That point can be computed easily: Assume the two solutions are A,B,C and A+a,B+b,C+c. The connecting line is A+ta,B+tb,C+tc. Write the surface equation, 2 abc t^3 + (2abC+2aBc+2Abc+a^2+b^2+c^2) t^2 + xxx t + yyy = 1 The sum of the three roots for t is - coeff(t^2) / 2abc, and two of the roots are t=0 and t=1. This is the standard elliptic curve trick, and it works for cubic surfaces (and cubics with even more variables). This is already too long, so I won't go through the grommy details. I've ignored the unphysicalness of solutions with negative numbers. I'm guessing that half of all our stepped-to solutions will have an even number of negative components, and we can complement the negative pair to get a positive solution. Puzzle: Can all the other solutions be got by starting from 1/2,1/2,1/2 and using the various stepping methods? Given any solution, is there a path back to 1/2,1/2,1/2 with decreasing denominators? Rich --------- copy of MKs message ---------- On 4/16/07, Michael Kleber <michael.kleber(a)gmail.com> wrote: > A friend recently mentioned to me the following ARML > problem from 1989: "A convex hexagon is inscribed in > a circle. If its successive sides are 2, 2, 7, 7, 11, 11, > compute the diameter of the circumscribed circle." > > (Anyone who wants to solve this on their own may go do > so now, and come back later for the rest of my question.) > > Since we can reorder sides of this cyclic hexagon, the > question is the same as asking for the diameter D of a > circle in which you can inscribe a quadrilateral with side > lengths 2,7,11,D. So maybe we should generalze from > Pythagorean triangle, and say that a convex n-gon is > Pythagorean if it has integral "hypotenuse" D and edges > a1, a2,...,a_{n-1}, and can be inscribed in a semi-circle > with diameter D. The obvious question is, what's the > generalization of the Pythagorean theorem? > > In the quadrilateral case, the relation you get -- necessraily > symmetric on the edges a_i -- turns out to be > D * (a1^2 + a2^2 + a3^2) + 2 a1 a2 a3 = D^3 > It's maybe more elegant-looking if you say that you're searching > for rational solutions to the D=1 version, > a^2 + b^2 + c^2 + 2abc = 1 > but maybe not, since the LHS isn't homogeneous. > > Solutions include all Pythagorean triangles, by setting a3=0, > and lots of solutions like 2-2-7-8, whose trigonometric meaning > I'll leave you to ponder. Hmm, I suppose the regular hexagon's > sol'n 1-1-1-2 also falls into this category. > > But the interesting solutions start with the 2-7-11-D from the > ARML, then 2-9-12-16, 6-11-14-21, 1-12-22-26, 3-14-25-30, > and so on. Do these come up anywhere else? > > And what's the relation between the edges and the hypotenuse > for the n-gon case? > > Likely this is all well-known to those who know it, but not to me... > > --Michael

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[math-fun] Emma Lehmer r.i.p., Henri Cohen retirement
by Richard Schroeppel 20 Jul '07

20 Jul '07

John Brillhart has told me about the death of Emma Lehmer, May 7, 2007. He's written an obituary to appear in an AMS publication. Happier news, Henri Cohen's retirement is being celebrated with a meeting in Bordeaux, Oct 15-19, see http://www.math.u-bordeaux.fr/emnt2007/ Rich

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RE: [math-fun] Re: solid conclusion(?)
by Ralph Gosper 20 Jul '07

20 Jul '07

Rich writes>we can freely reorder the sides, so the area >is independent of the order for an inscribed polygon. >What's the radius of the circle? For the quadrilateral, sqrt((a d + b c) (b d + a c) (c d + a b)) r = -----------------------------------------, 4 A (which I overlooked in Mathworld and stupidly rederived) where A is the area sqrt((c + b + a - d) (d - c + b + a) (d + c - b + a) (d + c + b - a)) --------------------------------------------------------------------- 4 (Brahmagupta). (Did he really have the general case with the cos^2(opposite angles)? I.e., he knew from trig?) These conjectures map back to my(?) "Brahmagupta" solid angle formula: is it maximal when the pyramid inscribes in a circular cone? Etc for more sides. The trig for the n=4 case is not, so far, repeatable in polite company. MiKle> Igor Pak's recent "The area of cyclic polygons: Recent progress on Robbins' conjectures", available off his web page ( http://math.mit.edu/~pak ), says just about everything there is to know now. rwg>The formulas must be doozies. Wow, the degrees (in z=16 Area^2) go up exponentially: 1, 2, 7, 14, 38, 76,..., n/2 binom(n-1,floor((n-1)/2)) - 2^(n-2), for n = 3,4,5,... . (Sic 'em, Neil.) But how can n=3 (Heron) and n=4 (Brahmagupta) differ, when their formulas are nearly identical? Because the theorem also assumes that the coeffs are polynomials in the *squares* of the sides, producing, for n=4, the product of Brahmagupta and a conjugate describing the nonconvex case. So possibly the n=6 case also factors? Anyway, the n=5 case is already unsolvable in radicals. The area of a general quadrilateral is very simply half the product of the diagonals times the sine of the angle between them, so there are probably simpler hexagon (e.g.) area formulas involving the chord lengths. --rwg

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[math-fun] Checkers, etc.
by Steve Gray 20 Jul '07

20 Jul '07

Here's the actual press release. Schaeffer is the guy to whom correspondence should be addressed. Jonathan Schaeffer , E-mail: jonathan(a)cs.ualberta.ca <mailto:jonathan@cs.ualberta.ca> The game of checkers has roughly 500 billion billion possible^ positions (5 x 10^20 ). The task of solving the game, determining^ the final result in a game with no mistakes made by either player,^ is daunting. Since 1989, almost continuously, dozens of computers^ have been working on solving checkers, applying state-of-the-art^ artificial intelligence techniques to the proving process. This^ paper announces that checkers is now solved: perfect play by^ both sides leads to a draw. This is the most challenging popular^ game to be solved to date, roughly one million times more complex^ than Connect Four. Artificial intelligence technology has been^ used to generate strong heuristic-based game-playing programs,^ such as DEEP BLUE for chess. Solving a game takes this to the^ next level, by replacing the heuristics with perfection. Does anyone know if a quantum computer, assuming they're perfected, whatever that means, could solve Go or chess? Steve Gray

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[math-fun] All Factors of n
by Eric Angelini 19 Jul '07

19 Jul '07

Hello Math-Fun, does someone know of an on-line factorization applet showing ALL divisors of n (not only the prime ones) Not this applet, for instance : http://www.merriampark.com/factor.htm Neither this famous one : http://www.alpertron.com.ar/ECM.HTM Thanks, Best, É.

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RE : [math-fun] All Factors of n
by Eric Angelini 19 Jul '07

19 Jul '07

Thanks to Edwin, Neil and Mike -- this is beautiful ! Best, É. ________________________________ De: math-fun-bounces+eric.angelini=kntv.be(a)mailman.xmission.com de la part de Edwin Clark Date: jeu. 19/07/2007 21:25 À: math-fun Objet : Re: [math-fun] All Factors of n On Thu, 19 Jul 2007, Eric Angelini wrote: > > does someone know of an on-line factorization applet showing > ALL divisors of n (not only the prime ones) Use for example the command Divisors(12345678); at this site http://magma.maths.usyd.edu.au/calc/ and click on evaluate. You will get [ 1, 2, 3, 6, 9, 18, 47, 94, 141, 282, 423, 846, 14593, 29186, 43779, 87558, 131337, 262674, 685871, 1371742, 2057613, 4115226, 6172839, 12345678 ] Total time: 0.350 seconds, Total memory usage: 6.46MB _______________________________________________ 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] Re: solid conclusion(?)
by Bill Gosper 19 Jul '07

19 Jul '07

> This is analogous to the maximal area of a quadrilateral >with sides a,b,c,d (or a,b,d,c): sqrt((c + b + a - d) (d - c + b + a) (d + c - b + a) (d + c + b - a)) ---------------------------------------------------------------------, 4 >with diagonals (a d + b c) (b d + a c) (b d + a c) (c d + a b) sqrt(-----------------------) and sqrt(-----------------------). c d + a b a d + b c >Whose formula is this? Brahmagupta's (cyclic case). His formula makes it clear that the area is maximized when the vertices lie on a circle. >And what about arbitrary pentagons, etc? Are the maximal areas fixed >w.r.t. permuting the sides? If the area is maximized when the vertices lie on a circle, we can freely swap adjacent sides, preserving the chord (and circumscribing) arc, and thus area. --rwg

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Re: [math-fun] ENIAC and maple : a simple test
by Eugene Salamin 18 Jul '07

18 Jul '07

Bill Gosper asks: Why does everybody ignore Macsyma? Perhaps it's because, thanks to Dick Petti and the other former Macsyma Inc. stockholders, Macsyma has self-destructed. You can't buy the software (macsyma.com is now an entertainment site), and it is being maintained and upgraded only by unofficial programmers, and yet it cannot be open sourced. What would be the legal status of using a Macsyma that was not purchased years ago from Macsyma Inc? If a project discloses its use of Macsyma, will its members be subject to legal action? Could funding be found to create an open source computer algebra system? Extract the best ideas out of Macsyma, Maple, Mathematica, Mathcad, etc., chuck the ancient excess baggage, and make a fresh start. Gene ____________________________________________________________________________________ Be a better Heartthrob. Get better relationship answers from someone who knows. Yahoo! Answers - Check it out. http://answers.yahoo.com/dir/?link=list&sid=396545433

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Re:[math-fun] ENIAC and maple : a simple test
by Bill Gosper 18 Jul '07

18 Jul '07

Simon Plouffe wrote: > I was reading an article on the ENIAC, that monster could add 100,000 10 digits numbers in 1 second. At the time it was impressive in 1946 of course. I made this simple test, a naive test. I took maple 11, with a 3.0 Ghz PC pentium. ##################################### Digits:=10: som:=0: for i from 1 to 1e99 do: som:=som+1.0/i: if i mod 100000 = 0 then lprint(i,som) fi: od: ##################################### It displays the partial sum of the harmonic series every 100000 terms. this is not optimized, a lazy program. >Well, this is the bad news. That loop made on a modern computer with one of the top programs goes at the SAME speed. It takes 1 second to display each step. Mma seems comparable: Timing[Block[{s = 0}, For[{n = 1}, n <= 100000, s += 1/N[n++, 10]]; s]] {0.906 Second, 12.09014613} Macsyma (using 16 digit doublefloats instead of 10 digit bigfloats): (c4) (showtime:true,block([s:0],for n thru 100000 do s:s+1.0d0/n,s)) Time= 1594 msec. (d4) 12.0901461298633d0 (c5) harm(n):=block([s:0.0d0],mode_declare([n,k,s],float),for k thru n do s:s+1.0d0/k,s) Time= 0 msec. (d5) harm(n):=block([s:0.0d0],mode_declare([n,k,s],float), 1.0d0 for k thru n do s : s + -----, s) k (c6) harm(100000.0) Time= 1672 msec. (d6) 12.0901461298633d0 BUT (c7) compile(harm) time= 0 msec. (d7) [harm] (c8) harm(100000.0) Time= 78 msec. (d8) 12.0901461298633d0 Why does everybody ignore Macsyma? It should be mentioned that these were run in conflict with a second, rogue Mma process trying to FactorInteger repunit 71. Pausing continued to hog cpu, and subsession won't enter, so I just whacked the process priority. --rwg

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