November 2012 - mathfun@mailman.xmission.com

[math-fun] Fwd: harder 7x7x7?
by Bill Gosper 03 Jun '15

03 Jun '15

I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf ---------- Forwarded message ---------- From: Bill Gosper <billgosper(a)gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314(a)gmail.com> Cc: Julian Ziegler Hunts <julianj.zh(a)gmail.com>, Jon Ziegler < jonathan.zh(a)gmail.com>, Corey Ziegler Hunts <corwin.zh(a)gmail.com>, Michael Beeler <mikebeeler(a)verizon.net>, rcs(a)xmission.com, stan(a)isaacs.com Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill

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[math-fun] What is elliptic K and E closed form?
by Bill Gosper 25 May '15

25 May '15

I think the answer has changed in the last 50 yrs. A query from constantologist Steve Finch revived my interest in whether special value identities like EllipticK[(1 + Sqrt[2])^4] == ((1/4 - I/2)π^(3/2))/ ((2 + Sqrt[2]) Gamma[3/4]^2) and EllipticK[8 2^(1/4) (-1 + Sqrt[2])^3* E^(I ArcTan[(3 (3 + Sqrt[2]))/(4 2^(1/4))])] == (I (1 + Sqrt[2]) E^(-2 I ArcCot[2^(1/4)]) π^(3/2))/ (4 Sqrt[2] Gamma[3/4]^2) are already known, and where one might find tables much more extensive than, e.g., http://functions.wolfram.com/EllipticIntegrals/EllipticK/03/01/ or http://dlmf.nist.gov/19.6 . DLMF says "Exact values of K(k) and E(k) for various special values of k are given in Byrd and Friedman (1971, 111.10 and 111.11)" (1st English edition, 1954) Neil and his Mom were kind enough to fetch me this comprehensive (*Handbook* of *Elliptic* Integrals for Engineers and Physicists<http://www.amazon.com/Handbook-Elliptic-Integrals-Engineers-Physicists/dp/B…>), and to my surprise and disappointment, there was not a single identity of the above form, i.e. algebraics times Gammas(rationals). How could this be? My guess: In the 1950s K and E were considered closed forms while Gamma was not! It makes sense: K and E are of lower computational complexity. We Funsters once concluded that Gamma[k/24] was rapidly computable thanks to elliptic K identities. But now, in papers like "EVALUATION OF COMPLETE ELLIPTIC INTEGRALS OF THE FIRST KIND AT SINGULAR MODULI" (people.math.carleton.ca/~*williams*/papers/*pdf*/*299*.*pdf*) "evaluation" means "determin[ation] explicitly in terms of Gamma values". Which is what I always thought. See, e.g., http://en.wikipedia.org/wiki/Elliptic_integral#Special_values --rwg

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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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[math-fun] That fractal pentagram again
by Bill Gosper 09 Apr '15

09 Apr '15

For some pictorial. mildly trick questions: http://gosper.org/foupent.pdf (1.6M) --rwg

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[math-fun] why can a line of slope 10^n determine n decimal digits of π?
by Gary Antonick 18 Jun '14

18 Jun '14

Just noticed this yesterday and can't figure out why it works. Draw a circle with radius 1 around 0,0. Draw a line with slope -10 that hits the circle at 1,0. This line will also hit the circle again and make a little arc. The length of the arc will be about 1/31 the perimeter of the circle. Now try again with a line of slope -100. The line will create an arc 1/314 of the circle. Starting to look a bit like pi. Am wondering why this is. In case I made a mistake here's what I'm trying to describe.. if the circle were a clock the first line would cut an arc from 15 minutes (3:00) to almost the 13 minute mark.

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[math-fun] sqrt(pi)
by Bill Gosper 24 Mar '13

24 Mar '13

Pairwise combining the terms of A&S 6.5.31 (upper Gamma CF), then adding last week's lower gamma CF: In[185]:= Equal[Gamma[ z], ((y^z*((1/(cfk[n*(z - n), -z + y + 2*n + 1, List[n, Infinity]] - z + y + 1)) + (1/(z + cfk[n*y, z - y + n, List[n, Infinity]] - y))))/(E^y))] z 1 Gamma[z] == (y (------------------------------------------- + -y + z + cfk[n y, n - y + z, {n, Infinity}] 1 y -----------------------------------------------------------)) / E 1 + y - z + cfk[n (-n + z), 1 + 2 n + y - z, {n, Infinity}] where cfk:=ContinuedFractionK and y>0 can be chosen to optimize convergence. Specializing, In[186]:= % /. z -> 1/2 /. y -> 1 Out[186]= 1 Sqrt[Pi] == (-------------------------------------- + 1 1 -(-) + cfk[n, -(-) + n, {n, Infinity}] 2 2 1 ------------------------------------------) / E 3 1 3 - + cfk[(- - n) n, - + 2 n, {n, Infinity}] 2 2 2 In[192]:= N[List @@ %186 /. cfk[num_, den_, _] :> ContinuedFractionK[num, den, {n, 9}]] Out[192]= {1.77245, 1.77244} This is the first non-asymptotic sum or CF I've seen for sqrt(pi). But the sum behind the lowergamma cf has been there all along, so I must not have been paying attention. --rwg

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[math-fun] poisoned wine bottles
by Warren Smith 02 Jan '13

02 Jan '13

> From: Michael Kleber <michael.kleber(a)gmail.com> > You have 1000 bottles of wine, which you need for a party that starts in an > hour. Two of the bottles are poisoned. > You have ten rats, and you can cause each rat to drink from any subset of > the bottles, which will cause it to die if the bottle is poisoned. But the > poison takes most of an hour to act, so you only get one round of giving > wine to rats, after which you must throw away any bottle which you do not > know is safe. > What do you do? I'd be happy to hear either worst-case or average-case > results. > Obviously if only one bottle were poisoned, you'd number the bottles and > assign one rat to each bit of the binary representation, and the dead rats > would spell out the binary of the unique poisoned bottle number. But two > poisoned bottles seems to make this much more subtle. --Interesting puzzle. Let W be the number of wine bottles, P the number poisoned, 1<=P<=W, and R the number of rats. The task is to identify the poisoned wine bottles with a single round of feeding each rat a wine-mixture from a selected bottle-subset. Say rat j gets fed subset Sj. Consider this WxR boolean matrix M: M_ij = 1 if wine bottle i is in set Sj. Otherwise M_ij=0. The essential property demanded (for a worst-case result allowing us to identify the P poisoned bottles) is that for any P-element subset of the rows of the matrix, its logical-OR (as an R-bit binary word) is DISTINCT from that arising from any other P-element subset. (This logical OR's locations for its 1-bits give the names of the dead rats.) So evidently for a solution to exist we need that binomial(W, P) <= 2^R. It was proposed that W=1000, P=2. Hence the left hand side is 499500=1000*999/2. Hence a solution cannot exist unless there are at least 19 rats, R>=19. Since Kleber had R=10, we conclude there is no solution guaranteed to identify the two poisoned bottles. We can however adopt the weaker goal of trying to identify some subset of bottles we can guarantee are safe. In that case, just partition the bottles into R equal-size subsets each with W/R bottles. Then P rats die whereupon we've identified (R-P)*W/R safe wine bottles. With R=10, P=2, W=1000, we identify 800 safe ones this way. Returning to the original problem of identifying the poisoned bottles, Kleber's binary method shows the inequality I gave is tight (up to integer roundoff) if P=1. But what if P>=2? I will now attempt to sketch an argument that my trivial inequality is not too far off being tight, for any FIXED P. To prove this, consider selecting the matrix M at random, each bit chosen by an i.i.d. coin toss, where the coin used is biased with a particular well-chosen constant bias. Specifically, I choose the bias in such a way that the expected number of 1-bits in the R-vector got by ORing P rows of the matrix, comes out to be R/2. I claim this argument will show that R = C * lg( binomial(W,P) ) rats suffice, for some positive constant C. To prove this (i.e. to supply the details of the proof, which I have not provided since I am lazy), you need to argue that with positive probability, all those P-row-ORs will indeed be distinct. This can be done by arguing the expected number of conflicting P-row-subset pairs is <1. Then since probability>0, that proves some set of such coin tosses must exist, which have this property, hence a suitable experimental-design exists. QED. Hence, with P=2, since 2 is fixed, there is a solution with R=O(logW) rats and somebody less lazy than I could even work out a good bound on the constant C (1<C<5 sounds safe).

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[math-fun] antikythera device: an ancient computer
by Robert Baillie 06 Dec '12

06 Dec '12

a while back, someone posted a link to this youtube video of the antikythera device made out of legos: http://www.youtube.com/watch?v=RLPVCJjTNgk here is an interesting (58 minute) bbc documentary that explains in more detail how the device actually works, as far as we know today: http://www.youtube.com/watch?v=-rUsgL2VGeU many of the gears had a prime number of teeth. this was very helpful in figuring out what the device was used for, and how it worked. here are several shorter videos that might also be of interest: The Antikythera Mechanism - 2D (8 minutes) http://www.youtube.com/watch?v=UpLcnAIpVRA The Antikythera Mechanism - 3D (4 minutes) http://www.youtube.com/watch?v=L1CuR29OajI Antikythera mechanism working model.mov (4 minutes) http://www.youtube.com/watch?v=4eUibFQKJqI in case anyone passes through bozeman, montana, there is a great computer museum that has a replica of the antikythera device. the museum also has an old ibm 1620 computer, a curta calculator, and many other neat old things: http://www.compustory.com/ note: their rather primitive website doesn't do the place justice! bob baillie

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Re: [math-fun] checkerboard version of an e-related dating problem--a graph theoretic approach
by W. Edwin Clark 02 Dec '12

02 Dec '12

Concerning Allan Wechsler's approach: Note that the number of valid (jammed) configurations for n vertices on a line is the number of maximal matchings of the path graph with n vertices. And the number of singletons for a given configurations is the number of vertices not in the matching. [This same characterization holds for a grid graph or indeed any simple graph.] I don't know if there is a good algorithm in general for enumerating the set of all maximal matchings for a graph, but I wrote a crude one for this case and was able to find directly Allan's sequences a[n],b[n] for n from 0 to 20: a = 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200 b = 0, 1, 0, 2, 2, 3, 6, 7, 12, 17, 24, 36, 50, 72, 102, 143, 202, 282, 394, 549, 762 I cannot find b in the OEIS. My value for b[4] is 2 which differs from Allan's b[4] = 3. I didn't work out the recurrences, but Maple's gfun:-listtorec guessed that the recurrences for a and b are given by: {a(n+3)-a(n+1)-a(n) =0, a(0) = 1, a(1) = 1, a(2) = 1} {(n+3)*b(n+4)+(-5-n)*b(n+2)+(-6-n)*b(n+1) = 0, b(0) = 0, b(1) = 1, b(2) = 0, b(3) = 2} --Edwin On Wed, Nov 28, 2012 at 3:42 PM, Allan Wechsler <acwacw(a)gmail.com> wrote: > Going back to trying to make sense of the 1D problem, the number of valid > configurations for N people satisfies a[0]=a[1]=a[2]=1, a[n+3]=a[n]+a[n+1]. > There are about half a dozen sequences in OEIS with this "tribonacci" > rule, with minor differences in offset and initial conditions. > > Let b[N] be the aggregate number of singletons in all valid N-person > configurations. I'm getting a recurrence b[n+3] = a[n] + b[n] + b[n+1]. > Now b[0] = 0, b[1] = 1, b[2] = 0, and then the recurrence gives 2, 3, 3, > 6, 7, 10, 15, 19 ... The last entry diverges from A121833, but the rule is > so simple that I can't believe it's not in OEIS, so I must have > miscalculated. > > Anyway, the wanted expected number of singletons is lim b[n]/a[n]. Peering > at the small entries I realize I must have blundered somewhere. Can > somebody with more attention to give it please repeat this reasoning but do > it correctly? > > On Wed, Nov 28, 2012 at 3:09 PM, Dan Asimov <dasimov(a)earthlink.net> wrote: > > > Good point, but it wasn't intentional -- I was just careless. > > > > So, I should've said "then restrict attention to only those outcomes > > satisfying the conditions of the problem: That each person holds > > at most one hand and no two singles are adjacent." > > > > Or, put another way: After giving each pair of adjacent hands an > > independent ½ probability of holding each other, restrict attention > > to the configurations having a maximal set of adjacent pairs of > > gridpoints each connected by the line segment between them, > > such that no line segment touches another one. > > > > (This almost suggests a way to obtain the results for 1D.) > > > > --Dan > > > > > > On 2012-11-28, at 11:44 AM, Andy Latto wrote: > > > > > On Wed, Nov 28, 2012 at 1:49 PM, Dan Asimov <dasimov(a)earthlink.net> > > wrote: > > >> Whether in the 1D or 2D grid case (or for that matter in nD), > > >> one way to give this problem well-defined probabilities > > >> (that I think is in some sense the most natural way) > > >> is to first give each pair of adjacent hands an independent > > >> probability of ½ as to whether they or not they hold each other -- > > >> and then restrict attention to only those outcomes > > >> satisfying the conditions of the problem: That each > > >> person holds at most one hand. > > > > > > That's a different problem, because you're ignoring the second > > > condition of the problem; no two adjacent single people. If you > > > restrict attention to those, you have a formulation equivalent to > > > Allan's, I think, but different from both of mine. > > > > > > Andy > > > > > >> > > >> I'm not sure if this agrees with the probabilities > > >> in one of Andy's two methods. > > >> > > >> The problem becomes more symmetrical if the 1D case is a circle > > >> instead of a line (and in nD a K^n grid on an n-dimensional torus) -- > > >> but this won't affect the asymptotic expected fraction of singles. > > >> > > >> --Dan > > >> > > >> > > >> On 2012-11-28, at 6:49 AM, Andy Latto wrote: > > >> > > >>> On Wed, Nov 28, 2012 at 12:17 AM, Gary Antonick > > >>> <gantonick(a)post.harvard.edu> wrote: > > >>>> Cornell mathematician Steven Strogatz thinks this is an unsolved > > problem > > >>>> but am wondering if there's something being overlooked. > > >>>> > > >>>> Here's the basic -- and solved -- problem. There's a long line of > > people > > >>>> standing near each other. They're told to hold hands with someone - > > either > > >>>> the person to their right or to their left, if available, but not > > both. > > >>>> The ratio of the resulting singles to all the people in line > > approaches > > >>>> 1/e^2 as the line increases. > > > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > _______________________________________________ > 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] end-of-semester riddle
by Veit Elser 30 Nov '12

30 Nov '12

Which fundamental physical constant prefers its martinis shaken, not stirred? And if you were curious about the probability a "random" complex NxN Hermitian matrix has only positive eigenvalues, the answer (for large N) is sqrt(3)^(-N^2) -Veit

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