July 2020 - mathfun@mailman.xmission.com

[math-fun] Rolling polyhedra
by James Propp 26 Dec '20

26 Dec '20

I want to think about rolling polyhedra, and the polyhedra I want to think about rolling are polyhedra in which the sum of the angles at each vertex is a submultiple of 360 degrees. For instance, given any triangle T, we can create a tetrahedron whose four faces are all congruent to T, with angle-sum 180 degrees at each vertex (there's a name for such tetrahedra but I forget what it is). Are there other polyhedra in which the angles at each vertex sum to a submultiple of 360 degrees (not necessarily the same one at each vertex)? Jim Propp

4 10

[math-fun] Chess variant
by David Wilson 13 Nov '20

13 Nov '20

In Paul Morphy's "Opera House" chess game, Morphy never moves a piece backward. I propose Opera House Chess, with initial position and moves as in chess, except pieces cannot move backward. Is it playable? With victory conditions modified?

4 4

[math-fun] One parameter quintic solution
by Bill Gosper 01 Nov '20

01 Nov '20

This is probably ho-hum, but while trying to simplify a Chowla-surdberg, I found that 0 = a + 80 x + 20 x^3 + x^5 has solutions {((-a - Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((2 I \[Pi])/5))/2^( 1/5), ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), ((-a - Sqrt[4096 + a^2])^(1/5) E^(-((2 I \[Pi])/5)))/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^(-((4 I \[Pi])/5)))/2^( 1/5), (-a + Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a - Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^( 1/5), (-a - Sqrt[4096 + a^2])^(1/5)/2^( 1/5) + ((-a + Sqrt[4096 + a^2])^(1/5) E^((4 I \[Pi])/5))/2^(1/5)} Is it from a two-parameter family? C.f. https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics --rwg

1 3

[math-fun] Yet another Heighway Dragon theorem. And some Terdragon bits.
by Bill Gosper 29 Aug '20

29 Aug '20

A couple of months ago, Julian found the first half of this identity for the (self avoiding) "median curve" <http://gosper.org/dragmed.png> of the usual dyadic rational sweep over 2^k+1 points. 1/2 (dragun[2^-k n] + dragun[2^-k (1 + n)]) == dragun[2^-k (1/2 - 1/4 (-1)^(1 + n) + n)] == dragun[2^(-3 - k) (-4 + 8 (1 + n) + Sqrt[2] Cos[1/4 (1 + n) π] - 2 Cos[1/2 (1 + n) π] - Sqrt[2] Cos[3/4 (1 + n)π] - (2 + Sqrt[2]) Sin[1/4 (1 + n) π] + 2 Sin[1/2 (1 + n) π] - (-2 + Sqrt[2]) Sin[3/4 (1 + n) π])] I just "conjectured" this 2nd half. The Heighway function never sees the trigs and radicals, which are artifacts of a period 8 perturbation of the sweep parameter. Mandelbrot seemed slightly embarrassed by his kludgy averaging|smoothing recipe for self avoidance. I'll bet he would have been pleased to see it fall out of these simple formulas. Though not nearly as pleased as just being able to exactly evaluate continuous fractal curves anywhere on the continuum. It turns out that the (self-avoidng) median curve of the Terdragon <http://gosper.org/Terdmed.png> is identical to merely phase shifting it by half a step. It's possible that this is a new theorem, since it's not obvious how to take half a step on a triadic fractal. But it's hard to find a nice sampling pattern for the median <http://gosper.org/Terdshftmed.png> curve of the median curve <http://gosper.org/Terdshftmed.png>, because all its points are double! Leaving you with an exponential search (over the two-valued inverses) for a simple pattern. —rwg

2 3

[math-fun] Exploiting uniqueness
by James Propp 22 Aug '20

22 Aug '20

Does anyone know of a name attached to this problem-solving tactic? Jim Propp ---------- Forwarded message --------- From: David Will <daw5de(a)virginia.edu> Date: Tue, Jun 30, 2020 at 8:17 PM Subject: Question about the self-referential aptitude test To: <jamespropp(a)gmail.com> Dr. Propp: Good evening. I know I'm a bit late to the party, but I just wanted to say I'm a big fan of your self-referential aptitude test, which I just discovered today. In the instructions, you mentioned this idea that the foreknowledge of the puzzle having a unique solution can, in and of itself, be used as a clue when solving the puzzle. I enjoy Japanese-style puzzles, and I've previously encountered this same phenomenon in games such as slitherlink and masyu. I'm also a graduate student in mathematics at the University of Virginia, so I've been curious about this idea from a mathematical perspective. I was hoping you'd be able to answer a quick question. Do you know of a name for this phenomenon? I'm lacking in knowledge about game theory (is that even the right subject?), so I was unsuccessful in my searches online. I'd like to see if this phenomenon has been studied with any sort of rigor. In particular, I want to know if it would be possible to construct a puzzle that requires this logic in order to be solved (in some loose sense, ignoring brute-force guessing). Thank you for your time. I'd greatly appreciate it if you'd be willing to entertain my curiosity. Best, Andrew Will

5 5

Re: [math-fun] FactorInteger
by Bill Gosper 20 Aug '20

20 Aug '20

Slight improvement: Mathematica's front end by default displays a centerdot for implicit multiplication. Strangely, this doesn't work in old the FI function below, which displays blanks. Blanks also (by default) separate digits in blocks of 3. So the display is sometimes ambiguous. But Wolfram CE Harry Calkins showed me this workaround: FI⨉[n : _Rational | _Integer] := Defer @@ {FactorInteger@n} /. {{a_Integer, 1} -> a, {a_Integer, b_} -> a^b} /. {a_} -> a /. List -> Times E.g. In[144]:= FI⨉@BernoulliB@50]@// tim During evaluation of In[144]:= 0.006522,1 (seconds) Out[144]= (5^2 417202699 47464429777438199)/(2 3 11) (The first Bernoulli number with a square factor!) Dirty trick. I used a tiny ⨉ symbol instead of x, so you need Front End completion to input FI⨉. You may want to edit the definition. Instead of little centerdots between the factors, there's a Preferences variable Appearances|Numbers|Multiplication|Automatically display multiplication symbol between multiplied numbers Use symbol: which you can bash to be a mini ⨉. Sorry, I don't know how to bash it under program control, as in an init file. Unfactor is way nicer than with FactorInteger: UnFI⨉@Defer[x_] := x In[153]:= UnFI⨉@%144 Out[153]= 495057205241079648212477525/66 —rwg On Wed, Jan 4, 2017 at 12:44 AM Bill Gosper <billgosper(a)gmail.com> wrote: > There are > > In[14]:= 52! > Out[14]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > ways to shuffle a deck. > That's just a "little" over a googol ... > > In[366]:= Log[10.^100, %] > Out[366]= 0.679066483922048 > > to the 2/3 power. The square of the cube root. And vice versa. So if > anybody asks how many > ways to shuffle a deck, "Oh, about the cube root of a googol squared." > But this is actually low > by a fairly substantial factor of > In[372]:= (10.^100)^(.679 - 2/3) > > Out[372]= 17.1132830416181 > > The actual number of shufflings is enormously composite: > > In[15]:= FactorInteger@%14 > Out[15]= > {{2,49},{3,23},{5,12},{7,8},{11,4},{13,4},{17,3},{19,2},{23,2},{29,1},{31,1},{37,1},{41,1},{43,1},{47,1}} > > But this is ugly and hard to read. The doc suggests > > In[16]:= CenterDot@@(Superscript@@@%) > Out[16]= > 2^49\[CenterDot]3^23\[CenterDot]5^12\[CenterDot]7^8\[CenterDot]11^4\[CenterDot]13^4\[CenterDot]17^3\[CenterDot]19^2\[CenterDot]23^2\[CenterDot]29^1\[CenterDot]31^1\[CenterDot]37^1\[CenterDot]41^1\[CenterDot]43^1\[CenterDot]47^1 > > This actually looks neat. Except for the ^1s. Try it. But it's unclear > (actually, a nice little Mathematica exercise) how to get your original > number back. > So I wrote something called FI > > In[364]:= FI[52!] > Out[364]= 2^49 3^23 5^12 7^8 11^4 13^4 17^3 19^2 23^2 29 31 37 41 43 47 > > In[365]:= ReleaseHold@% > Out[365]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Here is the definition: > FI[n : _Rational | _Integer] := > Times @@ ((HoldForm @@ {#1[[1]]})^#1[[2]] &) /@ FactorInteger[n] > > And here's a spoiler for how to invert FactorInteger: > > In[367]:= Times @@ (Power @@@ FactorInteger[73 137]) > Out[367]= 10001 > > which also works on that CenterDot expression: > > In[368]:= Times @@ (Power @@@ %16) > > Out[368]= > 80658175170943878571660636856403766975289505440883277824000000000000 > > Finally, you might enjoy finding a fairly nice formula for how many 0s are > on the end of n!, and, > more generally, what are exponents of the various prime factors. > --Bill >

1 1

[math-fun] 1 spiral + 10 diagonals = 1 sequence
by Eric ANGELINI 30 Jul '20

30 Jul '20

Hello Math-Fun (this is an attempt to send a message to the mailing list via my new gmail address I've subscribed in order to avoid warning messages that people sometimes get from me; the hereunder message will be more readableif you click this link: http://cinquantesignes.blogspot.com/2020/07/1-spiral-10-diagonals-1-sequenc… ) > Pick any digit zero and start a clockwise spiral on this infinite square lattice (streaked by the same succession of ten different diagonals, one digit per square); the resulting sequence *S* is the lexicographically earliest of distinct positive terms: 012569874301256987430125698743012569874301256987430125698743 125698743012569874301256987430125698743012569874301256987430 256987430125698743012569874301256987430125698743012569874301 569874301256987430125698743012569874301256987430125698743012 698743012569874301256987430125698743012569874301256987430125 987430125698743012569874301*2**5*6987430125698743012569874301256 874301256987430125698743012*569874301*256987430125698743012569 743012569874301256987430125*698743012*569874301256987430125698 430125698743012569874301256*987430125*698743012569874301256987 301256987430125698743012569*8743**0**1256*987430125698743012569873 012569874301256987430125698*74301*2*569*874301256987430125698743 125698743012569874301256987*430125698*743012569874301256987430 256987430125698743012569874*301256987*430125698743012569874301 569874301256987430125698743*012569874*301256987430125698743012 698743012569874301256987430125698743012569874301256987430125 987430125698743012569874301256987430125698743012569874301256 874301256987430125698743012569874301256987430125698743012569 743012569874301256987430125698743012569874301256987430125698 430125698743012569874301256987430125698743012569874301256987 301256987430125698743012569874301256987430125698743012569873 ... *S* = *0*, *1*, *2*, *10*, *3*, *4*, *30*, *12*, *5*, *6*, *52*, *103*, *47* , *8*, *7*, *430*, *125*, *69*, *89*, *65*, *210*, *34*, *78*, *9*, *698*, *74*, *301*, *25*, *6987*, *478*, *96*, *5210*, *347*, *896*, *525*, ... Any other order of the ten diagonals will not produce a lexicographically earliest sequence of distinct positive terms (I guess). Best, É.

1 0

Re: [math-fun] High schoolers: simplify this expression
by Henry Baker 27 Jul '20

27 Jul '20

This method isn't for high schoolers, but should be doable for undergrads: Find the cube root of 10+sqrt(108). (This method works for all(?) fields Q(sqrt(D)), not just D=3.) 1. Find the minimal polynomial for 10+sqrt(108). (%i1) (x-(10+sqrt(108)))*(x-(10-sqrt(108))),expand; 2 (%o1) x - 20 x - 8 2. Produce the companion matrix for this minimal polynomial. (%i2) polytocompanion(%,x); [ 0 8 ] (%o2) [ ] [ 1 20 ] Its determinant is the 'norm' of 10+sqrt(108), which by inspection is -8. Note that -8 = (-2)^3; i.e., -8 is a perfect cube. The general 2x2 companion matrix looks like: [ 0 - Determinant ] (%o4) [ ] [ 1 Trace ] Let T=Trace and N=Determinant: (%i5) %,Trace=T,Determinant=N; [ 0 - N ] (%o5) [ ] [ 1 T ] Cube this matrix: (%i6) %^^3,factor; [ 2 ] [ - N T - N (T - N) ] (%o6) [ ] [ 2 2 ] [ T - N T (T - 2 N) ] 'Companionize' this matrix by producing the companion matrix with the same characteristic polynomial as our cubed matrix: (%i7) polytocompanion(charpoly(%,z),z),factor; [ 3 ] [ 0 - N ] (%o7) [ ] [ 2 ] [ 1 T (T - 3 N) ] Equate this matrix with our integer matrix above, and extract equations in N,T: (%i8) list_matrix_entries(col(%-matrix([0,8],[1,20]),2)); 3 2 (%o8) [- N - 8, T (T - 3 N) - 20] Eliminate N and factor: (%i9) eliminate(%,[N]),factor; 2 2 (%o9) [- (T - 2) (T + 2 T + 4) (T + 2 T + 10) 4 3 2 (T - 2 T - 6 T - 20 T + 100)] The simplest integer answer is T=2 and N=-2. So (10+sqrt(108))^(1/3) = (T/2) + sqrt((T/2)^2 - N) = 1 + sqrt( 1^2 - -2) = 1 + sqrt(3) At 10:43 AM 7/10/2020, Henry Baker wrote: >True. > >Now do it like a high schooler... > >At 09:17 AM 7/10/2020, Simon Plouffe wrote: >>Hello, >> >>the expression is simplified in a fraction of a second (sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3); to 2 by Maple. >> >>Best regards, >> >>Simon Plouffe >> >>Le ven. 10 juil. 2020 Ã 17:47, Henry Baker <hbaker1(a)pipeline.com> a ÃÂ©crit : >>> (sqrt(108)+10)^(1/3)-(sqrt(108)-10)^(1/3) >>> >>> [I thought this was a pretty cool problem that came up on the Maxima email list.]

2 1

Re: [math-fun] Small chunks of letters
by Éric Angelini 26 Jul '20

26 Jul '20

Thanks Allan! (and this is the first time I encounter the « single token » concept — very instructive!-) Here is, I guess, the longest (though short) seq of terms counting the consonants used so far: THRee FiVe eiGHT TWeLVe SiXTeeN TWeNTy TWeNTySiX THiRTy THiRTySeVeN FoRTyoNe FoRTySeVeN FiFTyoNe FiFTySeVeN SiXTyTWo SiXTyeiGHT SeVeNTyFouR stop Best, É. > Le 26 juil. 2020 à 16:04, Allan Wechsler <acwacw(a)gmail.com> a écrit : > > This sequence is the fixed point of the morphism 1 -> 12; 2 -> 3; 3 -> > "41"; "41" -> 2122. Here, "41" must be treated as a single token. > > Limits of such morphisms are not usually periodic. The Thue-Morse sequence > is a famous example. > >>> On Sat, Jul 25, 2020, 4:24 PM Éric Angelini <eric.angelini(a)skynet.be> wrote: >> ONE TWO THREE FOUR ONE TWO ONE TWO TWO THREE ONE TWO THREE THREE FOUR ONE >> ONE TWO ... >> Insert a dot after each vowel: >> O.NE. TWO. THRE.E. FO.U.R O.NE. TWO. O.NE. TWO. TWO. THRE.E. O.NE.TWO. >> THRE.E. THRE.E. FO.U.R O.NE. O.NE. TW.O ... >> The size of the successive chunks is given >> by the sequence itself. >> Question: does the said sequence enter into >> a loop? >> Best, >> É. >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0

[math-fun] Small chunks of letters
by Éric Angelini 26 Jul '20

26 Jul '20

ONE TWO THREE FOUR ONE TWO ONE TWO TWO THREE ONE TWO THREE THREE FOUR ONE ONE TWO ... Insert a dot after each vowel: O.NE. TWO. THRE.E. FO.U.R O.NE. TWO. O.NE. TWO. TWO. THRE.E. O.NE.TWO. THRE.E. THRE.E. FO.U.R O.NE. O.NE. TW.O ... The size of the successive chunks is given by the sequence itself. Question: does the said sequence enter into a loop? Best, É.

2 1