September 2014 - mathfun@mailman.xmission.com

Re: [math-fun] iterating the inverse of a spacefill converges quadratically near its fixed pt.
by Bill Gosper 06 Sep '14

06 Sep '14

* Fred Lunnon <fred.lunnon(a)gmail.com> [Sep 06. 2014 15:22]: On 9/6/14, Bill Gosper <billgosper(a)gmail.com> wrote: [...] In[32]:= RealDigits[%, 16] Out[32]= {{6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 8}, 0} ... >> That isn't the Thue-Morse sequence! Jörg>It is the TM seq. in hexadecimal. For a start, it contains both consecutive 0's and consecutive 1's ... ... as the TM seq. indeed does: https://oeis.org/A010060 WFL Thanks for speaking up, Jörg. I would have spent several days trying to get Fred's joke. Shifting the integer one bit and converting to base 4: In[46]:= IntegerDigits[ FromDigits[{6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6}, 16]/2, 4] gives a "squarefree" (wasn't it once stutterfree?) sequence on four letters: Out[46]= {3, 1, 0, 3, 0, 2, 3, 1, 0, 2, 3, 0, 3, 1, 0, 3, 0, 2, 3, 0, 3, 1, 0, 2, 3, 1, 0, 3, 0, 2, 3, 1, 0, 2, 3, 0, 3, 1, 0, 2, 3, 1, 0, 3, 0, 2, 3, 0, 3, 1, 0, 3, 0, 2, 3, 1, 0, 2, 3, 0, 3, 1, 0, 3, 0, 2, 3, 0, 3, 1, 0, 2, 3, 1, 0, 3, 0, 2, 3, 0, 3, 1, 0, 3, 0, 2, 3, 1, 0, 2, 3, 0, 3, 1, 0, 2, 3, 1, 0, 3, 0, 2, 3, 1, 0, 2, 3, 0, 3, 1, 0, 3, 0, 2, 3, 0, 3, 1, 0, 2, 3, 1, 0, 3, 0, 2, 3} which retains its virtue after conflating 2&1: In[47]:= % /. {2 -> 1, 3 -> 2} Out[47]= {2, 1, 0, 2, 0, 1, 2, 1, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2, 1, 0, 1, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 2, 1, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 2, 1, 0, 1, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2, 1, 0, 1, 2, 0, 2, 1, 0, 2, 0, 1, 2, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 1, 2} Not obviously related to http://oeis.org/A086713 ? It might be nice to add this and a few variants to OEIS, but what is the likelihood of a match when a seeqr finds one independently? --rwg

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[math-fun] iterating the inverse of a spacefill converges quadratically near its fixed pt.
by Bill Gosper 06 Sep '14

06 Sep '14

E.g., the leading octal digits of the middle fixed point of the Dragon: In[36]:= FromDigits[#, 8]/8^Length[#] &@{6, 4, 1, 7, 6, 4, 0, 0, 0, 3, 2, 0, 0, 0, 0, 0, 0, 0, 1, 5, 0} Out[36]= 941199549051961357/1152921504606846976 The inverse In[37]:= undrag[%] Out[37]= {1085129200228273162089124938402233613/ 1329227995784915872903807060280344576} is unique. In[38]:= RealDigits[%[[1]], 8] Out[38]= {{6, 4, 1, 7, 6, 4, 0, 0, 0, 3, 2, 0, 0, 0, 0, 0, 0, 0, 1, 5, 0, 3, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 4, 1, 5}, 0} And *all* these digits are good! Another e.g.: The leading hex digits of Thue-Morse (HAKMEM's "parity constant"): (1/2)*(2 - Product[1 - 2^(-2^k), {k, 0, Infinity}]) In[31]:= %/2 /.∞-> 6 Out[31]= 280701669626288379716180548005699333331/ 680564733841876926926749214863536422912 In[32]:= RealDigits[%, 16] Out[32]= {{6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 8}, 0} The exact constant satisfies hilbert[x]==I+x . Inverting I+x, In[33]:= unbert[%% + I] Out[33]= { 573106971234556259792196936825953480449973086345124920770960412939461034308365/ 1389505070847794345082851820104254894239239815987686768473491008094957555679232, 573106971234556259792196936825953480449973086345124920770960412939461034308367/ 1389505070847794345082851820104254894239239815987686768473491008094957555679232} Two very close values In[34]:= RealDigits[%[[1]], 16] Out[34]= {{6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 6, 9, 9, 6, 9, 6, 6, 9, 6, 9, 9, 6, 9, 6, 6, 9, 9, 6, 6, 9, 6, 9, 9, 6, 6, {10}}, 0} In[39]:= Length[Out[#][[1]]] & /@ {32, 34} Out[39]= {33, 66} Digits doubled. Intuitively, you need 2n bits of t to get n bits of each of the imaginary and real parts of spacefill(t). When we take the inverse of x+0i or x+I, we are actually supplying n bits of x and n bits of 0 or 1, so we've cranked in 2n bits, and we get them back. --rwg Note that 1-Thue also satisfies hilbert[x]==I+x due to the left-right symmetry of the function: hilbert[1-t] = 1-Conjugate[hilbert[t]] . "That can't be right." But it is.

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[math-fun] antiprism oddity
by Bill Gosper 05 Sep '14

05 Sep '14

Twist "top" face half a "click" wrt to its opponent. This makes an ordinary prism with rectangular faces diagonally slashed. If you twist a whole click the other way, the whole figure will mirror-image. I.e., the slashes flip to the opposite diagonal. You could make a prank animation of the top face continuously rotating wrt the bottom, by repeatedly fading out the slash and fading up its mirror image, or have both slashes present at the "prism points". --rwg

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[math-fun] Gaussian v. Logistic pdf's, cdf's
by Henry Baker 05 Sep '14

05 Sep '14

Warren raised some interesting points about the various distributions and scoring methods. Since we're talking about _models_ here, let's assume that: 1. The Elo chess system "works" & is "faithful", i.e., a. the eventually stationary scores actually imply the correct probabilities for wins; and b. the logistic pdf so produced actually matches the statistics of real chess players. (Some of the literature seems to indicate that the standard chess updating rule is slow to converge, but I only care that it does converge for my 1a. Warren will quibble with 1b, but here I'm simply assuming it.) 2. We further assume that the population size is large enough to be able to approximate well with continuous pdf's & cdf's. Let pdf_normal(x) and cdf_normal(x) be the gaussian pdf and its integral function, respectively. Let quantile_normal(q) be the inverse function to cdf_normal(x); i.e., quantile_normal(cdf_normal(x))=x. Let pdf_logistic(y) and cdf_logistic(y) by the logistic pdf and its integral function, respectively. Let quantile_logistic(q) be the inverse function to cdf_logistic(y); i.e., quantile_logistic(cdf_logistic(y))=y. Since cdf_normal and cdf_logistic are both monotonic with image [0,1), we can relate x,y by a 1-1 coorespondence: cdf_normal(x)=cdf_logistic(y) Thus, we can "transform" from "logistic coordinates" to "Gaussian coordinates" via the function quantile_normal(cdf_normal(x))=x=quantile_normal(cdr_logistic(y)) Furthermore, this "transform" preserves ordering, and enables us to relate more traditional statistical terminology to this problem. We can also scale our Gaussian coordinates so that the mean is 100 and the standard deviation is 15 (i.e., IQ-type scaling). We can also scale our logistic scoring so that the mean is 100 and the logistic score 115 is the same quantile as the Gaussian value 115 (= 1 standard deviation for the Gaussian distribution). Of course, none of the other integral standard deviations will match, but a least we tried. Perhaps Warren has already gone through this thought process?

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[math-fun] 1/2+i/4 is a quadruple point of the Hilbert curve.
by Bill Gosper 05 Sep '14

05 Sep '14

(Which is obvious if you watch the sweep.) The preimages are {5/48, 7/48, 41/48, 43/48}. "Proof": In[19]:= Select[Tally[hilbert /@ Range[0, 1, 1/60/3/5/4]], #[[2]] > 2 &] Out[19]= {{{1/4 + I/4}, 3}, {{1/2 + I/4}, 4}, {{1/2 + I/2}, 3}, {{1/4 + I/2}, 3}, {{1/4 + (3 I)/4}, 3}, {{3/4 + I/2}, 3}, {{3/4 + (3 I)/4}, 3}, {{3/4 + I/4}, 3}} In[20]:= unbert[#[[1, 1]]] & /@ % Out[20]= {{1/24, 1/8, 5/24}, {5/48, 7/48, 41/48, 43/48}, {1/6, 1/2, 5/ 6}, {3/16, 13/48, 23/48}, {7/24, 3/8, 11/24}, {25/48, 35/48, 13/ 16}, {13/24, 5/8, 17/24}, {19/24, 7/8, 23/24}} unbert? hilbert, not Hilbert? Big news: Julian has delivered an astonishing little Mma function that subsumes *all* of my exact rational fractal evaluators, plus their (multivalued) inverses! E.g., 1/6+i/3 is a triple point of the Dragon[0,1] that joins 0+0i to 1+0i: In[5]:= undrag[1/6 + I/3] Out[5]= {13/60, 7/20, 23/60} In[35]:= dragun /@ %5 Out[35]= {{1/6 + I/3}, {1/6 + I/3}, {1/6 + I/3}} where dragun[t_] := piecewiserecursivefractal[t, Identity, Piecewise[{{{1}, 0 <= # <= 1/2}, {{2}, 1/2 <= # <= 1}}, {}] &, {2*# &, 2*(1 - #) &}, {(1 + I)*#/2 &, (I - 1)*#/2 + 1 &}] undrag[z_] := piecewiserecursivefractal[z, Identity, If[-(1/3) <= Re[#] <= 7/6 && -(1/3) <= Im[#] <= 2/3, {1, 2}, {}] &, {#*(1 - I) &, (1 - #)*(1 + I) &}, {#/2 &, 1 - #/2 &}] and the underlying little bombshell, ClearAll[piecewiserecursivefractal]; piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := piecewiserecursivefractal[x, g_, which, iters, fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := Block[{y}, y /. Solve[f[y] == h[y], y]]); Union @@ ((fns[[#]] /@ piecewiserecursivefractal[iters[[#]][x], Composition[f, fns[[#]]], which, iters, fns]) & /@ which[x])); (*[Julian's comment]: For example, here's a dragon (prf:=piecewiserecursivefractal):In[290]:=prf[7/22,Identity,Piecewise[{{{1},0\[LessEqual]#\[LessEqual]1/2},{{2},1/2\[LessEqual]#\[LessEqual]1}},{}]&,{2*#&,2*(1-#)&},{(1+I)*#/2&,(I-1)*#/2+1&}] Out[290]={23/82+(39 I)/82} It's written to allow multi-valued functions,and in particular can be used for the inverse dragon:In[289]:=prf[13/12+I/48,Identity,If[-(1/3)\[LessEqual]Re[#]\[LessEqual]7/6&&-(1/3)\[LessEqual]Im[#]\[LessEqual]2/3,{1,2},{}]&,{#*(1-I)&,(1-#)*(1+I)&},{#/2&,1-#/2&}] Out[289]={3793/3840,1267/1280,3803/3840} The parameters are x,the input;f,the tracking variable;which,which takes x and gives which (if any) of the iters and fns should be applied;iters,a list of functions (which specifies which ones to use) that take x and give a new x to feed back in;and fns,the functions (again specified by which) that take the whole fractal into part of itself.In order to convert a which,iters,fns triplet into the triplet corresponding to the inverse function,you replace iters and fns with each other's inverses,and think carefully about what which should look like for the inverse (in the case of a dragon or Hilbert curve*or another closed bounded fractal whose fns are contraction mappings you can just use a bounding box).Note that the implementation can be improved.In particular,if you have a particular kind of function in mind you could get a speedup by replacing the Composition and Solve with something which works on the relevant kinds of functions (in particular with dragon,Hilbert,inverse dragon,etc.you can use a pair of constants (as in the curve-specific implementations) and a difference ratio because the functions are linear).The syntax can also be improved if you have some idea what you want to do.*) Further e.g.: In[6]:= hilbert[t_] := piecewiserecursivefractal[t, Identity, {Min[4, 1 + Floor[4*#]]} &, {1 - 4*# &, 4*# - 1 &, 4*# - 2 &, 4 - 4*# &}, {I*(1 - #)/2 &, (I + #)/2 &, (I + 1 + #)/2 &, 1 + #*I/2 &}] In[16]:= unbert[z_] := piecewiserecursivefractal[z, Identity, If[0 <= Re[#] <= 1 && 0 <= Im[#] <= 1, Range[4], {}] &, {1 - 2*#/I &, 2*# - I &, 2*# - I - 1 &, (# - 1)*2/I &}, {(1 - #)/4 &, (# + 1)/4 &, (# + 2)/4 &, 1 - #/4 &}] In[17]:= unbert[1/2 + I/2] Out[17]= {1/6, 1/2, 5/6} In[18]:= hilbert /@ % Out[18]= {{1/2 + I/2}, {1/2 + I/2}, {1/2 + I/2}} This should make it easy for someone to collect Jörg's T-shirt. --rwg

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[math-fun] Chess ratings
by Warren D Smith 05 Sep '14

05 Sep '14

> Henry Baker: > My results from several hours of Googling: > > 1. Huge amount of research on chess ratings. scholar.google.com "chess" > "Elo" > > 2. Elo model was invented by Zermelo 1928 and Bradley&Terry 1952, > rediscovered by Ford 1957. > > Bradley, Ralph A. and Terry, Milton E. (1952). "The rank analysis of > incomplete block > designs. I. The method of paired comparisons." Biometrika, 39, 324-45. > > Elo, Arpad E. (1978). The rating of chessplayers, past and present. Arco > Publishing: New York. > > https://en.wikipedia.org/wiki/Elo_rating_system > > Ford, Lester R. Jr. (1957). "Solution of a ranking problem from binary > comparisons." > American Mathematical Monthly, 64(8), 28-33. > > As best I can tell, the Elo model is based on the "logistic" distribution > rather than the Gaussian distribution. They look very similar, but > the logistic has slightly fatter tails. --not so. Elo's system in his book (which I read prior to writing my 1984 BS thesis) based on Gaussian, the others based on logistic. But these functions employed in the rating formulas, not as probability distributions of rating distribution. My 1984 bachelor's thesis at MIT, which my paper #20 was based on, shows Elo was wrong to do what he did, and logistic is superior to Gaussian where he used it, at least in the sense I proved "minimatch self-consistency" results for logistic (my "WM^n model") while Gaussian has no such. If Glickman claims otherwise it is perhaps because he never read Elo's book, or for some other reason, I have no idea. However, it appears in the meantime that the world in general has somehow realized that Gaussians suck (presumably due to experiments, since few members of the world showed any signs of reading my BS thesis), and most (all?) rating agencies have switched away from Elo's original system and to a logistic-base system. > sech((x-mu)/(2s))^2/(4s) > > Just remember "sech-mate" ;-) > > Its cdf is the hyperbolic tangent function! > > https://en.wikipedia.org/wiki/Logistic_distribution > > The best discussion of ZermElo I was able to find is: > > "Introductory note to 1928" > Mark E. Glickman 10 pages > > "Zermelo's 1928 paper on measuring participants' playing strengths in chess > tournaments > is a remarkable work in the history of paired comparison modeling." > > http://www.glicko.net/research/preface-z28.pdf

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Re: [math-fun] Chess ratings probability distribution function
by Henry Baker 05 Sep '14

05 Sep '14

One of Warren's papers talked of the ICS ("Internet Chess Server"), which had (1993) 2800 registered players of which perhaps 30 were computers. At 10:04 AM 9/5/2014, Eric Angelini wrote: >Is it possible to rate humans when they play against chess computers? >Best, >Ã.

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Re: [math-fun] chess ratings
by Henry Baker 05 Sep '14

05 Sep '14

Thanks, Warren! These papers are very relevant & interesting. I'm already quite familiar with your "Time-sharing model", as my father used this model when teaching me tennis! At 08:28 AM 9/5/2014, Warren D Smith wrote: >Amazingly enough, my Bachelor's thesis was on this very subject -- rating systems. >The internet chess club rating system is, or at least was, based on my work. >See > >http://rangevoting.org/WarrenSmithPages/homepage/works.html >paper #20 >and >paper #88. > >Paper 20 discusses among other things, a "fluid dynamics" model of >learning as reflected in ratings, and discusses the rating >distribution, and how to optimize rating system parameters. The >rating distribution has features of a "extreme value distribution." >Paper 88 discusses a new approach. > >Glickman kind of annoyed me by a combination of ignoring and appropriating my work, >which I told him all about before he got going on the topic. But I do not know >what Glickman's latest stuff is, haven't looked at this in years.

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Re: [math-fun] Chess ratings probability distribution function
by Dave Dyer 05 Sep '14

05 Sep '14

ELO ratings are not bad if all the games under consideration are played in similar settings and with similar degree of seriousness, and against the whole pool of players (ie; tournament settings) They're extremely unreliable if the games are casual, the player pool is variable, or if the players get to select their opponents.

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[math-fun] Chess ratings probability distribution function
by Henry Baker 05 Sep '14

05 Sep '14

I found this (somewhat old) picture of the distribution of chess ratings: http://zwim.free.fr/ics/rating_distribution.gif It isn't exactly Gaussian, but it seems to have Gaussian qualities. 1. Has anyone done research on the distribution of chess ratings? 2. What is the _interpretation_ of chess ratings? I.e., if A has chess rating Ar and B has chess rating Br, what is the probability that A beats B?

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