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[math-fun] Several years ago I wrote a terse quintic solver-in-radicals-where-possible.
by Bill Gosper 14 Nov '20

14 Nov '20

Unfortunately, I no longer have enough brain cells to remember or figure out how it works. It somehow gets exact expressions by taking the inverse Fourier transform of numerically approximated roots, without all the Galois group and resolvent rigamarole. I.e., it claims to find the radicals when they exist, subject to adjustable limits on coefficient size. Two examples (tim is just a timing function): In[451]:= squint[x^5 - 5 x x - 3] // tim During evaluation of In[451]:= 0.319225 seconds, 5 roots Out[451]= {-((-1)^(3/5)/(2^(2/5) (5/(5 - Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5))) + ((-1)^(2/5) (1/5 (5 - Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^(1/5))/2^( 2/5) - (-1 - 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5)/2^( 2/5) - ((-1)^(3/5) (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^(2/5), 1/(2^(2/5) (5/(5 - Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5)) + (1/5 (5 - Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^(1/5)/2^( 2/5) + ((-1)^(4/5) (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^( 2/5) + (1 + 1/Sqrt[5] + 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5)/2^(2/5), ((-1)^(4/5) (1/5 (5 - Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5))/2^(2/5) - (1/5 (-5 + Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5)/2^(2/5) + ((-1)^(2/5) (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^( 2/5) + ((-1)^(2/5) (1 + 1/Sqrt[5] + 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^(2/5), (-1)^(4/5)/(2^(2/5) (5/(5 - Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5)) - (1/5 (-5 + Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^(1/5)/2^( 2/5) - ((-1)^(1/5) (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^( 2/5) - ((-1)^(3/5) (1 + 1/Sqrt[5] + 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^(2/5), (-1)^(2/5)/( 2^(2/5) (5/(5 - Sqrt[5] - Sqrt[6 (5 - 11/Sqrt[5])]))^( 1/5)) - ((-1)^(3/5) (1/5 (5 - Sqrt[5] + Sqrt[6 (5 - 11/Sqrt[5])]))^(1/5))/2^( 2/5) + (1 + 1/Sqrt[5] - 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5)/2^( 2/5) + ((-1)^(4/5) (1 + 1/Sqrt[5] + 2 Sqrt[3/10 + 33/(50 Sqrt[5])])^(1/5))/2^(2/5)} In[445]:= squint[-1 + 13755 x + 3740 x^2 + 1260 x^3 - 5 x^4 + x^5] Out[445]= {1 + 5 2^(1/5) - 10 2^(2/5) + 10 2^(3/5) - 5 2^(4/5), 1 + 5 (-1)^(4/5) 2^(1/5) + 10 (-1)^(3/5) 2^(2/5) + 10 (-1)^(2/5) 2^(3/5) + 5 (-1)^(1/5) 2^(4/5), 1 - 5 (-1)^(3/5) 2^(1/5) + 10 (-1)^(1/5) 2^(2/5) + 10 (-1)^(4/5) 2^(3/5) - 5 (-1)^(2/5) 2^(4/5), 1 + 5 (-1)^(2/5) 2^(1/5) - 10 (-1)^(4/5) 2^(2/5) - 10 (-1)^(1/5) 2^(3/5) + 5 (-1)^(3/5) 2^(4/5), 1 - 5 (-2)^(1/5) - 10 (-2)^(3/5) - 10 (-1)^(2/5) 2^(2/5) - 5 (-1)^(4/5) 2^(4/5)} (It can take Mathematica several minutes to check one of these symbolically.) Here is the definition: squint[q_] := ToRadicals[ Block[{rts = (#1[[1, 2]] & ) /@ NSolve[q == 0, WorkingPrecision -> 99], boa, foos}, foos = (InverseFourier[rts[[#1]]/Sqrt[5]]^5 & ) /@ Permutations[Range[5]]; foos = (Together[#1[[1, 2]]]^(1/5) & ) /@ Solve[0 == Rationalize[Expand[Times @@ (#1 - Rest[MinimalBy[foos, Denominator@Rationalize[Plus @@ #, 9.^-69] &,1][[1]]])]]]; boa = Rationalize@Mean@rts; boa + Union[(foos . #1 & ) /@ Select[Tuples[I^(4 Range@5/5), 4], MemberQ[Chop[#1 . foos + boa - rts], 0] & ]]]] One thing I remember is boa is "b over a", as in a x^5 + b x^4 + . . . note boa = Rationalize@Mean@rts; Something I don't remember is feeling any surprise that this works. It would be funny if it comes as a surprise to algebraists. Actually, it wouldn't be funny because then I'd have to publish something. That I don't remember. Replacing 5 by 7, 4 by 6, and Solve[0==. . .] with a sextic solver, Julian and I found the analogous septic solver. He expressed mistrust, but it has yet to fail. —rwg

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[math-fun] Chess variant
by James Propp 13 Nov '20

13 Nov '20

Ordinary chess forbids moving into check (and hence from moving into checkmate), which means you automatically lose if there's no way to avoid moving into check. Is there a variant of chess in which you're not allowed to make a move that would cause you to lose 2 moves in the future? (Replacing 2 by larger integers, we get a game in which the legality of a move can become rather murky!) Jim

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[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?

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[math-fun] Fwd: Chess variant
by M F Hasler 13 Nov '20

13 Nov '20

On Fri Nov 13 2020, 15:10, James Propp <jamespropp(a)gmail.com> wrote : > Ordinary chess forbids moving into check (and hence from moving into > checkmate), which means you automatically lose if there's no way to avoid > moving into check. > Is there a variant of chess in which you're not allowed to make a move > that would cause you to lose 2 moves in the future? > In Blitz chess, according to FIDE rules until 2017, a player could make an illegal move and the opponent would win if he noticed this on his turn (otherwise the referee should call the game a draw). So in particular one could ignore a checkmate situation (the rules do not say that you have to announce check or checkmate, many high class players consider it is even very impolite to announce check) and instead make a move which leaves your king under attack. The opponent could then take the king and thus win the game. This was actually very popular, but a somewhat ancient way of proceeding which was officially deprecated. The opponent should, instead of taking the king, rather claim his win because the other player had made an illegal move. However, e.g., the German Chess Federation had ruled that taking the king was a way of signalling the opponent's illegal move. (see https://www.schachbund.de/news/hau-weg-den-koenig-wirklich.html) The player might get a blame for doing so, but would nonetheless win the game. So with these older rules one player could make a move that left his king in check and then his king could've been taken. However, the next move should be either the king being taken, or the game ended as a draw by the referee. (When there's no referee, I assume it often happened that both of the players didn't notice the check situation and continued playing otherwise. Probably the guy who was in check would remark it soon after and move out of it, and the game would thereafter continue "legally".) - Maximilian (Replacing 2 by larger integers, we get a game in which the legality of > a move can become rather murky!) > > Jim

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Re: [math-fun] Chess variant
by Éric Angelini 13 Nov '20

13 Nov '20

> > which means you automatically lose if there's no way to avoid moving into check. ... hmm, rather 1/2-1/2 if the King or another chess unit of the same color cannot move. à+ É. Catapulté de mon aPhone > Le 13 nov. 2020 à 20:11, James Propp <jamespropp(a)gmail.com> a écrit : > > which means you automatically lose if there's no way to avoid > moving into check.

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Re: [math-fun] Cyclic Difference Sets
by Dan Asimov 11 Nov '20

11 Nov '20

Are we supposed to guess what you're talking about, or would you be so kind as to tell us? —Dan ----- A perfect cyclic difference set with n values has length L = (n-1)^2 + n and works mod L. All differences up to L-1 exist uniquely. 3 values is structurally unique, 1 solution. 4 values has 2 solutions. 5 values has 1 solution. 6 values has 5 solutions. 7 values has 0 solutions. 8 values has 1 solution. Do any subsequent values have more than 1 solution? {{{7,3},{0,1,3}}, {{13,4},{0,1,3,9}},{{13,4},{0,1,4,6}}, {{21,5},{0,1,4,14,16}}, {{31,6},{0,1,3,8,12,18}},{{31,6},{0,1,3,10,14,26}},{{31,6},{0,1,4,6,13,21}},{{31,6},{0,1,4,10,12,17}},{{31,6},{0,1,8,11,13,17}}, {{57,8},{0,1,3,13,32,36,43,52}}, {{73,9},{0,1,12,20,26,30,33,35,57}},{{91,10},{0,1,5,7,27,35,44,67,77,80}},{{133,12},{0,1,3,17,21,58,65,73,100,105,111,124}},{{183,14},{0,1,3,24,41,52,57,66,70,96,102,149,164,176}},{{273,17},{0,1,22,33,83,122,135,141,145,159,175,200,226,229,231,238,246}},{{307,18},{0,1,3,30,37,50,55,76,98,117,129,133,157,189,199,222,293,299}},{{381,20},{0,1,13,16,21,45,75,82,86,92,128,142,180,206,229,231,264,286,354,363}}} -----

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[math-fun] Another quaternion eigenvalue pun
by Henry Baker 11 Nov '20

11 Nov '20

I've been fiddling with Sylvester's Equation in quaternions: ax-xb=c, for a,b, s.t. aa'=bb' and a+a'=b+b'. Consider the linear function f(x)=ax-xb. Note that (a-b)+(a-b)' = a - b + a' - b' = a+a' - (b+b') = a+a' - (a+a') = 0 So (a-b) = -(a'-b'). I contend that (a-b) is an *eigenvector* of f(x), i.e., f(a-b) = lambda (a-b) for some 'eigenvalue' lambda. Let's try (a-b) out: f(a-b) = a(a-b) - (a-b)b = a(a-b) - (-(a'-b'))b = a(a-b) + (a'-b')b = a(a-b) + a'b - b'b = a(a-b) + a'b - a'a = a(a-b) + a'(b-a) = a(a-b) - a'(a-b) = (a-a')(a-b) ; lambda = (a-a') So, (a-b) is an eigenvector for f(x) for the 'eigenvalue' (a-a'). But (a-a') is a (pure) quaternion *vector* -- i.e., it isn't a real or complex number at all ! Pure vectors play the part in quaternions of pure imaginaries in complex numbers. For example, the square of a pure vector v is -N(v), so pure vectors are the square roots of negative real numbers. We notice that if this Sylvester expression ax-xb is represented in *matrix* form, it is a skew symmetric matrix whose eigenvalues are (surprise, surprise) *pure imaginaries*. Indeed, the absolute value of these imaginaries are identical to the absolute value of (a-a'). So if we 'unlift' (a-a') and -(a-a') back into the complex numbers, they are indeed the eigenvalues of the matrix form of the Sylvester equation. What is the other eigenvector for ax-xb ? We consider f(f(x)): a(a-a')(a-b) - (a-a')(a-b)b = (a-a')a(a-b) - (a-a')(a-b)b = (a-a') (a(a-b) - (a-b)b) = (a-a') (a-a')(a-b) = -N(a-a') (a-b) So -N(a-a') is indeed a *real number* eigenvalue for f(f(x)) with eigenvector (a-b), which follows from the corresponding theorem that the *square* of a skew symmetric matrix is a symmetric negative semidefinite matrix. We have also proved that (a-a')(a-b) is also an eigenvector, so we can span our 2D range space. As a bonus, our eigenvectors are *orthogonal*. dot(q,r) = (qr'+rq')/2, so q,r are orthogonal iff qr'+rq' = 0. dot((a-a')(a-b),(a-b)) = (a-b)[(a-a')(a-b)]'+(a-a')(a-b)(a-b)' = (a-b)(a-b)'(a-a')' + (a-a')(a-b)(a-b)' = N(a-b)(a-a')' + (a-a')N(a-b) = N(a-b) ((a-a')' + (a-a')) = N(a-b) (a' - a + a - a') = N(a-b) 0 = 0 So, (a-b) and (a-a')(a-b) form our orthogonal eigenbasis. Finally, to solve ax-xb=c, we have to express c as a linear combination of our eigenbase vectors, which task is simplified due to their orthogonality.

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[math-fun] Group theory question suggested by "Infinite products of Hurwitz quaternions"
by Dan Asimov 10 Nov '20

10 Nov '20

The complex numbers {x + iy} (x,y, real) lie on the unit circle when x^2 + y^2 = 1. These form the circle group under multiplication. If in addition x and y are rational numbers, this is a countably infinite subgroup of the circle group. What is its structure as a group? That problem was solved in this paper: <https://www.maa.org/sites/default/files/pdf/upload_library/22/Allendoerfer/…>, which shows (Theorem 1) that the group is isomorphic to the direct sum of the group Z/4Z and countably many infinite cyclic groups. Likewise, the set of quaternions {x + iy + zj + wk} (x,y,z,w real) with x^2 + y^2 + z^2 + w^2 = 1 forms a group under multiplication: it is the versatile group S^3 of unit quaternions that is isomorphic also to the groups Spin(3) and SU(2).* Once again, if we restrict x, y, z, w to be rational numbers, this becomes a countable subgroup of S^3. Question: What is the structure of this group of rational unit quaternions? —Dan ————— * Spin(3) is the double cover of the rotation group SO(3) or R^3, and SU(2) is the group of 2x2 complex unitary matrices** having determinant = 1. ** The conjugate transpose is the inverse matrix.

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[math-fun] Infinite products of Hurwitz quaternions;
by Henry Baker 10 Nov '20

10 Nov '20

'Euclid tries to deal with noncommutativity' A real number can be approximated by a single rational number which is a ratio of integers. The algorithm to produce this rational number is the oldest non-trivial algorithm: Euclid's algorithm which is a greedy algorithm for removing the largest integer part; inverting the remainder, removing the largest integer part, etc. The efficiency of the algorithm is ever-so- slightly improved by using a 'rounding' division instead of the greediest division, but the greediest algorithm still produces the same rational approximation to the real number. So let's move to the complex numbers and their approximation by a ratio of Gaussian integers. Presumably the same arguments continue to work, but I've not seen any discussion of the exact number of steps, or whether the sequence of convergents is always the same. But now we want to consider the approximation of quaternions having real coefficients by an infinite product of 'integer' quaternions, where by 'product' we include both quaternion multiplication and quaternion division. Due to noncommutativity, we may have to extend our product on either the left OR the right. Note that the *product* itself is important; we can't simply produce 4 rational numbers which approximate the 4 real coefficients of the given real quaternion, and *add* them together. We want an actual sequence of integer quaternions representing a quaternion *product*. Here's a potential algorithm: 1. Start with the approximation a=1. 2. If we pick a quaternion c to extend our current approximation, we have *four* choices to evaluate: extending on the left -- e.g. a:=ca or a:=(1/c)a -- or extending on the right a:=ac or a:=a(1/c). We somehow choose the 'greediest' 'integer' quaternion c, and the greediest extension method of the four, which produces the largest reduction in the difference between the norm of the given real quaternion and the norm of the approximating product of integer quaternions. 3. Rinse and repeat. There is a small question about whether to use quaternions with actual integer coefficients or the Hurwitz quaternions with half-integer coefficients; presumably one or the other will work 'better' and produce more elegant results. ------ So now comes the question: how to make the 'right' choice at each step? Also, IS there a method of 'cheating', whereby we can build up our approximating quaternion by multiplying and dividing by rational integers and then putting the results together with a fixed number of full quaternion multiplications and divisions? I.e., given 4 rational numbers q0,q1,a2,q3 which approximate the coefficients of the real quaternion, is there a easy way to construct a quaternion *product* of 'integer' quaternions to produce the final value q0+q1*I+q2*J+q3*K ? --- Finally, assuming that one has such a quaternion product sequence, how does one show that it is the *shortest* such sequence? I.e., is there any way to do 'collapsing' moves?

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Re: [math-fun] Brownie problem
by Dan Asimov 10 Nov '20

10 Nov '20

Of course, the equation could be set up to take account solely of the ratio of volumes (assuming all pieces are the same volume), without counting anything. —Dan ----- Alan confirms that he means that 1/n of the pieces are in the interior for some positive integer n. -----

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