[math-fun] 6x6 multiplicative magic squares
After 3x3, 4x4, 5x5, what about 6x6? My algorithm needs to be better optimized to attack the question. I will work on it in the next days. The smallest published 6x6 seems to be an example constructed by Borkovitz & Hwang, with magic product = 2,000,376,000. My question: can somebody produce an example with a smaller product? Michael Kleber? Rich? Unfortunately the latin squares method, used in previous messages for 4x4 and 5x5, does not work for 6x6: the famous "36-officers problem" of Euler has no solution! Christian.
Christian's message reminded me of a chat I had with Dan Asimov a couple of weeks ago. We were thinking about other kinds of magic squares, and had both thought about 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. Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Christian Boyer Sent: Mon 10/10/2005 4:07 AM To: 'math-fun' Subject: [math-fun] 6x6 multiplicative magic squares After 3x3, 4x4, 5x5, what about 6x6? My algorithm needs to be better optimized to attack the question. I will work on it in the next days. The smallest published 6x6 seems to be an example constructed by Borkovitz & Hwang, with magic product = 2,000,376,000. My question: can somebody produce an example with a smaller product? Michael Kleber? Rich? Unfortunately the latin squares method, used in previous messages for 4x4 and 5x5, does not work for 6x6: the famous "36-officers problem" of Euler has no solution! Christian. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Latest (and good) news! I can construct a lot of 6x6 multiplicative squares. My current smallest 6x6 multiplicative magic square has a magic product more than 70 times smaller than the Borkovitz & Hwang square. The B&H square used 36 distinct integers, the biggest being 1,260. My current best 6x6 example uses also 36 integers (of course...), but the biggest integer is 78: an extremely compact set of integers! But I need more checking before to be sure that it will be impossible to get better 6x6 results, and will keep you informed. 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 Christian Boyer Envoyé : lundi 10 octobre 2005 12:07 À : 'math-fun' Objet : [math-fun] 6x6 multiplicative magic squares After 3x3, 4x4, 5x5, what about 6x6? My algorithm needs to be better optimized to attack the question. I will work on it in the next days. The smallest published 6x6 seems to be an example constructed by Borkovitz & Hwang, with magic product = 2,000,376,000. My question: can somebody produce an example with a smaller product? Michael Kleber? Rich? Unfortunately the latin squares method, used in previous messages for 4x4 and 5x5, does not work for 6x6: the famous "36-officers problem" of Euler has no solution! Christian. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (2)
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Christian Boyer -
Schroeppel, Richard