RE: [math-fun] Non-commutative magic squares
Rich wrote: << . . . trying out non-commutative squares. This would let us return to the idea of using all the members of a well-defined set. The two most interesting cases to try seem to be using one of the non-commutative groups with 16 elements to make a 4x4, or using 2x2 matrices with elements 0,1 or maybe +-1, also to make a 4x4 square. The 0,1 matrix case would need a magic constant of [00,00]. You need some rules on the order of combining things, and whether the up-going diagonal reads forward or backward.
Am trying to think of a nice non-commutative group of order 16. Here's one: G = { az + b | a in {1,i,-1,-i}, b in Z[i]/2Z[i] }. Here b is uniquely = one of 0,1,i,1+i mod 2Z[i]. (Imagine a square torus T tiled by 4 congruent subsquares. Then G is the group of oriented rigid motions of T that preserve subsquares.) Then define g o h via (az+b)o(cz+d) = (ac)z + (ad+b) where ad+b is represented by its remainder mod 2Z[i] = one of 0,1,i,1+i. Thus 1/(az+b) is given by (1/a)z + -b/a. Questions: 1. Does this G admit any 4x4 magic squares? 2.If so, is the magic constant uniquely determined? [A priori we can just assume group multiplication of a row is left to right, and of a column is top to bottom.] (I'd guess the most promising magic constant is the identity. Also, if it's the identity then any cyclic permutation of a given row or column will multiply to the same thing: rstu = 1 <=> stur = 1 avoiding the sticky problem of how to define the magic product.) Hmm, another nice non-commutative group of order 16 is the isometries of a regular octagon (= D_8). This seems less interesting than G, but the same questions apply.) --Dan ----------------------------------------------------- P.S. (I say: diagonals, schmiagonals. Requiring the two diagonals to have the same magic constant just strikes me as a random rule with little redeeming recreational math value.)
Dan and Rich, excellent idea. And it should be easy to check if your G group admits a 4x4 magic square. Here is a 4x4 multiplicative square using complex numbers: [2 - sqrt(3)*i] [-5*sqrt(3) - 10*i] [4*sqrt(7) - 2*sqrt(21)*i] [-7] [(1/2)*i] [5/2] [sqrt(7)*i] [sqrt(3)/2 - i] [sqrt(7)} [-5*sqrt(7)*i] [14] [-2*sqrt(7) - sqrt(21)*i] |-8] [40*i] [-16*sqrt(7)] [16 + 8*sqrt(3)*i] Magic product = 1960. By Schwartzman, in 1987. Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de dasimov@earthlink.net Envoyé : mardi 11 octobre 2005 08:25 À : math-fun Objet : RE: [math-fun] Non-commutative magic squares Rich wrote: << .. . . trying out non-commutative squares. This would let us return to the idea of using all the members of a well-defined set. The two most interesting cases to try seem to be using one of the non-commutative groups with 16 elements to make a 4x4, or using 2x2 matrices with elements 0,1 or maybe +-1, also to make a 4x4 square. The 0,1 matrix case would need a magic constant of [00,00]. You need some rules on the order of combining things, and whether the up-going diagonal reads forward or backward.
Am trying to think of a nice non-commutative group of order 16. Here's one: G = { az + b | a in {1,i,-1,-i}, b in Z[i]/2Z[i] }. Here b is uniquely = one of 0,1,i,1+i mod 2Z[i]. (Imagine a square torus T tiled by 4 congruent subsquares. Then G is the group of oriented rigid motions of T that preserve subsquares.) Then define g o h via (az+b)o(cz+d) = (ac)z + (ad+b) where ad+b is represented by its remainder mod 2Z[i] = one of 0,1,i,1+i. Thus 1/(az+b) is given by (1/a)z + -b/a. Questions: 1. Does this G admit any 4x4 magic squares? 2.If so, is the magic constant uniquely determined? [A priori we can just assume group multiplication of a row is left to right, and of a column is top to bottom.] (I'd guess the most promising magic constant is the identity. Also, if it's the identity then any cyclic permutation of a given row or column will multiply to the same thing: rstu = 1 <=> stur = 1 avoiding the sticky problem of how to define the magic product.) Hmm, another nice non-commutative group of order 16 is the isometries of a regular octagon (= D_8). This seems less interesting than G, but the same questions apply.) --Dan ----------------------------------------------------- P.S. (I say: diagonals, schmiagonals. Requiring the two diagonals to have the same magic constant just strikes me as a random rule with little redeeming recreational math value.) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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