April 2014 - mathfun@mailman.xmission.com

Re: [math-fun] Aknother knight's surprise?
by Adam P. Goucher 15 Apr '14

15 Apr '14

> [...] > > Whichever, is there now some way to reverse the previous identification, > and lift the embedding back to 36 vertices? Yes! You implied the existence of such a procedure yourself, when you said "vertices opposite across the torus have all neighbours in common". Specifically, for every vertex v in the 18-vertex graph, create vertices v_1 and v_2 in the new graph. Then connect u_i with v_j in the 36-vertex graph iff u is connected to v in the 18-vertex graph. The 36-vertex graph should have 2^18 times more symmetries than the 18-vertex graph, given by the wreath product C_2 wr G, where G is the automorphism group of the 18-vertex graph (as a subgroup of S_18) -- the one you claim (probably correctly) has order 144. Sincerely, Adam P. Goucher > Fred Lunnon > > > On 4/15/14, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > > On 4/15/14, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > >> I have posted a diagram, showing as much of the symmetry of this graph > >> as > >> planar representation allows, at > >> https://www.dropbox.com/s/aeea6bqlrt23tck/semi_6x6.png > >> Note the 4-fold diagonal coincidences --- whereby hangs another tale > >> entirely! > > > > Ugh --- grotesquely aliased lines --- see instead > > https://www.dropbox.com/s/8b0stotr8343mq2/semi_6x6.pdf > > > > The external edges obviously form a tour (closed Hamiltonian path); > > the remaining diagonals also form a tour, isomorphic to the first. > > > > Fred Lunnon [15/04/14] > > > >> On 4/14/14, Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > >>> Since you dispatched 4x4 so readily, perhaps you can manage something > >>> about the graph genus 6x6 knight moves on a torus instead? > >>> > >>> In the 4x4 case the vertices had valency 4 instead of 8 . > >>> The 6x6 case degenerates as well, less predictably: vertices opposite > >>> across the torus have all neighbours in common; identification yields > >>> 18 vertices and 36 edges and valency 4 immersed on a Klein bottle. > >>> The symmetry group has order 144 . > >>> > >>> Fred Lunnon > > > > _______________________________________________ > 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] Aknother knight's surprise?
by Fred Lunnon 15 Apr '14

15 Apr '14

Since you dispatched 4x4 so readily, perhaps you can manage something about the graph genus 6x6 knight moves on a torus instead? In the 4x4 case the vertices had valency 4 instead of 8 . The 6x6 case degenerates as well, less predictably: vertices opposite across the torus have all neighbours in common; identification yields 18 vertices and 36 edges and valency 4 immersed on a Klein bottle. The symmetry group has order 144 . Fred Lunnon

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Re: [math-fun] knight moves on 6x6 torus chessboard
by Adam P. Goucher 15 Apr '14

15 Apr '14

> --so if I understand aright, you took the 6x6 torus chessboard, and > identified antipodal > squares, thus getting 18 = (6*6/2) vertices, then used knight moves > to define edges? > Thus getting a graph of valency 4? > Wrong. This would yield valency 8. No, it wouldn't. (+1, +2) = (-2, -1) modulo (3, 3), so pairs of knight moves would land on the same square (after `projectivisation' (quotienting out by identifying `antipodal' squares)). Have I really outwitted Warren whilst under the influence of three glasses of Cotes-du-Rhone? :) Sincerely, Adam P. Goucher

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[math-fun] knight moves on 6x6 torus chessboard
by Warren D Smith 15 Apr '14

15 Apr '14

Knight moves on 6x6 torus board define a 36-vertex bipartite (18 dark, 18 light squares) graph with all valencies=8. Note antipodal squares always have the opposite square-color. >Ugh --- grotesquely aliased lines --- see instead https://www.dropbox.com/s/8b0stotr8343mq2/semi_6x6.pdf The external edges obviously form a tour (closed Hamiltonian path); the remaining diagonals also form a tour, isomorphic to the first. Fred Lunnon [15/04/14] --so if I understand aright, you took the 6x6 torus chessboard, and identified antipodal squares, thus getting 18 = (6*6/2) vertices, then used knight moves to define edges? Thus getting a graph of valency 4? Wrong. This would yield valency 8. So I do not understand what you did to create your graph.

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[math-fun] Very nice lecture on quasicrystals
by Henry Baker 15 Apr '14

15 Apr '14

Paul Steinhardt of Princeton on impossible crystals 54 minutes, 132 MBytes https://www.youtube.com/watch?v=a0wo_yAh0Ps

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Re: [math-fun] [CASE:721791] Re: BarnesG and Hyperfactorial of n/4 and n/6
by Bill Gosper 15 Apr '14

15 Apr '14

Hi Jason, All you need is the relation In[270]:= FunctionExpand[%266, x > 0] Out[270]= BarnesG[x] == E^(-(1/2) (-1 + x) x - PolyGamma[-2, x]) (2 \[Pi])^(x/2) Gamma[x]^(-1 + x) Then In[271]:= Table[ FunctionExpand[%], {x, {1/2, 1/3, 1/4, 1/6, 2/3, 3/4, 5/6}}] Out[271]= {True, BarnesG[1/3] == ( 3^(1/72) E^( 1/9 + \[Pi]/(18 Sqrt[3]) - PolyGamma[1, 1/3]/(12 Sqrt[3] \[Pi])))/( Glaisher^(4/3) Gamma[1/3]^(2/3)), BarnesG[1/4] == E^(3/32 - Catalan/(4 \[Pi]))/( Glaisher^(9/8) Gamma[1/4]^(3/4)), BarnesG[1/6] == E^( 5/72 + \[Pi]/(12 Sqrt[3]) - PolyGamma[1, 1/3]/(8 Sqrt[3] \[Pi]))/( 2^(1/72) 3^(1/144) (Glaisher Gamma[1/6])^(5/6)), BarnesG[2/3] == ( 3^(1/72) E^( 1/9 - \[Pi]/(18 Sqrt[3]) + PolyGamma[1, 1/3]/(12 Sqrt[3] \[Pi])))/( Glaisher^(4/3) Gamma[2/3]^(1/3)), BarnesG[3/4] == E^(3/32 + Catalan/(4 \[Pi]))/( Glaisher^(9/8) Gamma[3/4]^(1/4)), BarnesG[5/6] == E^( 5/72 - \[Pi]/(12 Sqrt[3]) + PolyGamma[1, 1/3]/(8 Sqrt[3] \[Pi]))/( 2^(1/72) 3^(1/144) Glaisher^(5/6) Gamma[5/6]^(1/6))} In[272]:= N[%] Out[272]= {True, True, True, True, True, True, True} --Bill Gosper On Mon, Apr 14, 2014 at 10:51 AM, Wolfram Technical Support < support(a)wolfram.com> wrote: > > Hello Bill Gosper, > > Thank you for the further information. > > I have passed those notes along to our development team. I will let you > know if I hear anything back about this issue. > > We are always interested in improving Mathematica and I want to thank you > again for taking the time to contact us about this issue. If you find any > further issues, please feel free to contact us again. > > Sincerely, > > Jason Grigsby > Wolfram Technology Group > Wolfram Research, Inc > www.wolfram.com > > > > > > On: 04/13/2014 10:56pm billgosper(a)gmail.com submitted the following Case: > On Tue, Mar 11, 2014 at 2:53 AM, Bill Gosper wrote: > > > Papers of mine and your own Victor Adamchik give > > Integrate[Log[Gamma[t]],{t,0,r}] > > for various rational r. We also give this integral in terms of BarnesG. > > But > > FunctionExpand[BarnesG[r]] only "works" for r=n/2, which it gives in > terms > > of > > the goddam useless Glaisher symbol. Zeta'[-1] spontaneously turns into > > 1/12 - Log[Glaisher]. Why?? Why not then arbitrarily smash every Zeta[3] > > to > > be Apery? Or Apéry, to make it harder to type. > > > > Here are G[1/3] and -1/3 with z1 for Zeta[-1], to avoid the Glaisher > > braindamage. > > > This should be Zeta'[-1], but the following are wrong! > > > > > Out[67]= {BarnesG[1/3] -> 3^(25/72)* > > E^((1/72)*(96*z1 + > > (Sqrt[3]*(-PolyGamma[1, 1/3] + > > PolyGamma[1, 2/3]))/Pi))* > > Gamma[1/3]^(1/Gamma[4/3])*Gamma[4/3]^(4/3), > > BarnesG[-(1/3)] -> 3^(1/72)* > > E^((1/72)*(96*z1 + > > (Sqrt[3]*(PolyGamma[1, 1/3] - > > PolyGamma[1, 2/3]))/Pi))* > > Gamma[-(1/3)]^(1/Gamma[2/3])* > > Gamma[2/3]^(2/3)} > > > > Profuse apologies to all for inadequate testing! These should read > {BarnesG[1/3]==E^(1/72 (8+(Sqrt[3] > (-PolyGamma[1,1/3]+PolyGamma[1,2/3]))/\[Pi]))/(3^(47/72) > (E^(1/12-(Zeta^\[Prime])[-1]))^(4/3) ((1/3)!)^(2/3)), > BarnesG[-(1/3)]==-((3 3^(1/72) E^(1/72 (8+(Sqrt[3] > (PolyGamma[1,1/3]-PolyGamma[1,2/3]))/\[Pi])) ((1/3)!/\[Pi])^(4/3))/(2 > 2^(1/3) (E^(1/12-(Zeta^\[Prime])[-1]))^(4/3)))} > > My (inadequate) excuse: FunctionExpand just flat out returned a wrong > answer! Nonreproducible, of course, but > I have the .nb recording the event. My guess is that I damaged Mathematica > somehow while trying to banish > the vexatious Glaisher symbol. > > Numerical test: > In[175]:= N[ReleaseHold[%]] > > Out[175]= {True, False} > > HAH! So much for testing. They're both true, dammit. > (There's an invisible HoldForm around the Zeta'[-1]s) > --rwg > > In[68]:= Zeta'[-1] > > > > Out[68]= 1/12 - Log[Glaisher] > > > > > > FunctionExpand also appears to disunderstand the tuplication formula > > Product[Hyperfactorial[-(i/n) + z], {i, 0, -1 + n}] == > > (E^(z1/n - n*z1)*Hyperfactorial[n*z]^(1/n))/ > > n^((1 + 6*n*z + 6*n^2*z^2)/(12*n)) > > > > whose need it artificially creates for itself when given the Alternating > > Sign > > Matrix formula > > In[171]:= FunctionExpand[Product[(3 k + 1)!/(n + k)!, {k, 0, n - 1}], > > n > 0 && n \[Element] Integers] > > > > Out[171]= (3^(2 - 5 n + 3/2 n (1 + n)) > > 8^(-1 + n) (BarnesG[5/3] BarnesG[7/3])^(-2 + > > n) (BarnesG[8/3] BarnesG[10/3])^(1 - n) > > BarnesG[2/3 + n] BarnesG[1 + n] BarnesG[4/3 + n] BarnesG[2 + n]^( > > 2 - n) BarnesG[3 + n]^(-1 + n) Gamma[2 + n]^( > > 1 - n))/(BarnesG[1 + 2 n] Gamma[1 + n]) > > > > In[175]:= FunctionExpand[Table[%171, {n, 5}]] > > > > Out[175]= {1, 2, ( > > 9 BarnesG[5/3] BarnesG[7/3] BarnesG[11/3] BarnesG[13/3])/( > > 5 BarnesG[8/3]^2 BarnesG[10/3]^2), ( > > 2187 BarnesG[5/3]^2 BarnesG[7/3]^2 BarnesG[14/3] BarnesG[16/3])/( > > 7000 BarnesG[8/3]^3 BarnesG[10/3]^3), ( > > 1594323 BarnesG[5/3]^3 BarnesG[7/3]^3 BarnesG[17/3] BarnesG[19/3])/( > > 274400000 BarnesG[8/3]^4 BarnesG[10/3]^4)} > > > > This is an integer sequence! (A005130, 1,2,7,42,...). > > --Bill Gosper > > In[71]:= $LicenseID > > > > Out[71]= "L3290-7570" > > > > >

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Re: [math-fun] A knight's surprise?
by Bill Gosper 14 Apr '14

14 Apr '14

If you draw the 16 ways to slide, vs the 8 ways to hop, you get something like http://www.log24.com/log/pix10A/100619-FourDiamondTesseract.gif --rwg Bill Gosper wrote: Years ago Scott Kim told me that when shown the rules of chess as a child, he exclaimed "Ooh, a tesseract!" when they came to the knight. --rwg Adam>Why, exactly? As far as I can see, the tesseract only appears when one imposes that the knight inhabits a 4-by-4 toroidal grid. [Massive crop]

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[math-fun] BarnesG and Hyperfactorial of n/4 and n/6
by Bill Gosper 14 Apr '14

14 Apr '14

Papers of mine and your own Victor Adamchik give Integrate[Log[Gamma[t]],{t,0,r}] for various rational r. We also give this integral in terms of BarnesG. But FunctionExpand[BarnesG[r]] only "works" for r=n/2, which it gives in terms of the goddam useless Glaisher symbol. Zeta'[-1] spontaneously turns into 1/12 - Log[Glaisher]. Why?? Why not then arbitrarily smash every Zeta[3] to be Apery? Or Apéry, to make it harder to type. Here are G[1/3] and -1/3 with z1 for Zeta[-1], to avoid the Glaisher braindamage. Out[67]= {BarnesG[1/3] -> 3^(25/72)* E^((1/72)*(96*z1 + (Sqrt[3]*(-PolyGamma[1, 1/3] + PolyGamma[1, 2/3]))/Pi))* Gamma[1/3]^(1/Gamma[4/3])*Gamma[4/3]^(4/3), BarnesG[-(1/3)] -> 3^(1/72)* E^((1/72)*(96*z1 + (Sqrt[3]*(PolyGamma[1, 1/3] - PolyGamma[1, 2/3]))/Pi))* Gamma[-(1/3)]^(1/Gamma[2/3])* Gamma[2/3]^(2/3)} In[68]:= Zeta'[-1] Out[68]= 1/12 - Log[Glaisher] FunctionExpand also appears to disunderstand the tuplication formula Product[Hyperfactorial[-(i/n) + z], {i, 0, -1 + n}] == (E^(z1/n - n*z1)*Hyperfactorial[n*z]^(1/n))/ n^((1 + 6*n*z + 6*n^2*z^2)/(12*n)) whose need it artificially creates for itself when given the Alternating Sign Matrix formula In[171]:= FunctionExpand[Product[(3 k + 1)!/(n + k)!, {k, 0, n - 1}], n > 0 && n \[Element] Integers] Out[171]= (3^(2 - 5 n + 3/2 n (1 + n)) 8^(-1 + n) (BarnesG[5/3] BarnesG[7/3])^(-2 + n) (BarnesG[8/3] BarnesG[10/3])^(1 - n) BarnesG[2/3 + n] BarnesG[1 + n] BarnesG[4/3 + n] BarnesG[2 + n]^( 2 - n) BarnesG[3 + n]^(-1 + n) Gamma[2 + n]^( 1 - n))/(BarnesG[1 + 2 n] Gamma[1 + n]) In[175]:= FunctionExpand[Table[%171, {n, 5}]] Out[175]= {1, 2, ( 9 BarnesG[5/3] BarnesG[7/3] BarnesG[11/3] BarnesG[13/3])/( 5 BarnesG[8/3]^2 BarnesG[10/3]^2), ( 2187 BarnesG[5/3]^2 BarnesG[7/3]^2 BarnesG[14/3] BarnesG[16/3])/( 7000 BarnesG[8/3]^3 BarnesG[10/3]^3), ( 1594323 BarnesG[5/3]^3 BarnesG[7/3]^3 BarnesG[17/3] BarnesG[19/3])/( 274400000 BarnesG[8/3]^4 BarnesG[10/3]^4)} This is an integer sequence! (A005130, 1,2,7,42,...). --Bill Gosper In[71]:= $LicenseID Out[71]= "L3290-7570"

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Re: [math-fun] A knight's surprise?
by Adam P. Goucher 13 Apr '14

13 Apr '14

> Bill Gosper wrote: > > Years ago Scott Kim told me that when shown the rules of chess as a > child, he exclaimed "Ooh, a tesseract!" when they came to the knight. > --rwg Why, exactly? As far as I can see, the tesseract only appears when one imposes that the knight inhabits a 4-by-4 toroidal grid. > Fred Lunnon wrote: > > Yeah, that's much more direct than my argument! WFL Thanks. What was your intended proof? Sincerely, Adam P. Goucher > > On 4/13/14, Adam P. Goucher <apgoucher(a)gmx.com> wrote: > >> Very nice discovery! > >> > >> > >> Lemma: The graph is isomorphic to the 4-by-4 toroidal grid graph. > >> > >> Proof: draw the graph in the way you defined (which is on the 4-by-4 > >> torus, but has lots of edges crossing). Then, translate all of the black > >> vertices by (2,2) whilst preserving the positions of the white vertices. > >> Then every vertex is connected to its four neighbours and nothing else. > >> > >> -------- > >> > >> Corollary: (2) > >> > >> Proof: Trivial. > >> > >> -------- > >> > >> Corollary: (1) > >> > >> Proof: The graph is isomorphic to the product C4 x C4 (where C4 is > >> the cycle graph on four vertices), so is isomorphic to the product > >> K2 x K2 x K2 x K2 (since C4 is isomorphic to K2 x K2). Consequently, the > >> graph is isomorphic to the skeleton of the tesseract (four-dimensional > >> hypercube). > >> > >> -------- > >> > >> The last isomorphism I used (C4 x C4 == K2 x K2 x K2 x K2) may be > familiar > >> to any electronic engineers amongst you, on the basis that it allows > >> Karnaugh maps to work. > >> > >> > >> Sincerely, > >> > >> > >> Adam P. Goucher > >> > >> > >>> ----- Original Message ----- > >>> From: Fred Lunnon > >>> Sent: 04/13/14 12:04 PM > >>> To: math-fun > >>> Subject: [math-fun] A knight's surprise? > >>> > >>> Consider the graph whose vertices & edges represent squares & knight > >>> moves on a 4x4 chessboard tiling the torus. > >>> > >>> (1) Find a spatial embedding with the full symmetry of the graph. > >>> > >>> (2) Deduce that the graph can be embedded (without crossings) on the > >>> torus. > >>> > >>> Fred Lunnon > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] A knight's surprise?
by Adam P. Goucher 13 Apr '14

13 Apr '14

Very nice discovery! Lemma: The graph is isomorphic to the 4-by-4 toroidal grid graph. Proof: draw the graph in the way you defined (which is on the 4-by-4 torus, but has lots of edges crossing). Then, translate all of the black vertices by (2,2) whilst preserving the positions of the white vertices. Then every vertex is connected to its four neighbours and nothing else. -------- Corollary: (2) Proof: Trivial. -------- Corollary: (1) Proof: The graph is isomorphic to the product C4 x C4 (where C4 is the cycle graph on four vertices), so is isomorphic to the product K2 x K2 x K2 x K2 (since C4 is isomorphic to K2 x K2). Consequently, the graph is isomorphic to the skeleton of the tesseract (four-dimensional hypercube). -------- The last isomorphism I used (C4 x C4 == K2 x K2 x K2 x K2) may be familiar to any electronic engineers amongst you, on the basis that it allows Karnaugh maps to work. Sincerely, Adam P. Goucher > ----- Original Message ----- > From: Fred Lunnon > Sent: 04/13/14 12:04 PM > To: math-fun > Subject: [math-fun] A knight's surprise? > > Consider the graph whose vertices & edges represent squares & knight > moves on a 4x4 chessboard tiling the torus. > > (1) Find a spatial embedding with the full symmetry of the graph. > > (2) Deduce that the graph can be embedded (without crossings) on the torus. > > Fred Lunnon > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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