RE: [math-fun] Multiplicative Magic Squares
Sorry Michael, I have not completely read one of your previous messages, you mentioned already the same square:
(p,q,pq; pp,qq,r) [4,4,1] 6480
-----Message d'origine----- De : Christian Boyer [mailto:cboyer@club-internet.fr] Envoyé : lundi 26 septembre 2005 16:14 À : 'math-fun' Objet : RE: [math-fun] Multiplicative Magic Squares
From: "Michael Kleber" <michael.kleber@gmail.com>
So abcdABCD makes a pretty small fraction of all multiplicative magic squares, as the magic product (and the exponents on each prime in its factorization) get large.
Here is an interesting family of 4x4 multiplicative squares, using only three variables a, b, c: 1 abbb bc aaa ac aab ab bb aaab c abb b bbb a aa abc Magic product = a^4 * b^4 * c, for the 4 rows, 4 columns, 2 diagonals and 2 broken diagonals. And also for other sets of four cells (the corners for example, the 4 numbers of each quarter, ...). Apply a=2 and b=3, and you will get one of the various possible examples with magic product 6480. 6480 is the second smallest product after 5040, as listed in my previous email. Christian.
Please, Christian, tell us how you created that list. --Michael Kleber On 9/26/05, Christian Boyer <cboyer@club-internet.fr> wrote:
Sorry Michael, I have not completely read one of your previous messages, you mentioned already the same square:
(p,q,pq; pp,qq,r) [4,4,1] 6480
-----Message d'origine----- De: Christian Boyer [mailto:cboyer@club-internet.fr] Envoyé: lundi 26 septembre 2005 16:14 À: 'math-fun' Objet: RE: [math-fun] Multiplicative Magic Squares
From: "Michael Kleber" <michael.kleber@gmail.com>
So abcdABCD makes a pretty small fraction of all multiplicative magic squares, as the magic product (and the exponents on each prime in its factorization) get large.
Here is an interesting family of 4x4 multiplicative squares, using only three variables a, b, c:
1 abbb bc aaa ac aab ab bb aaab c abb b bbb a aa abc
Magic product = a^4 * b^4 * c, for the 4 rows, 4 columns, 2 diagonals and 2 broken diagonals. And also for other sets of four cells (the corners for example, the 4 numbers of each quarter, ...).
Apply a=2 and b=3, and you will get one of the various possible examples with magic product 6480. 6480 is the second smallest product after 5040, as listed in my previous email.
Christian.
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-- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
3x3 and 4x4 lists created by computing. Of course a bug is always possible, but the lists should be correct. 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 Michael Kleber Envoyé : lundi 26 septembre 2005 17:12 À : math-fun Objet : Re: [math-fun] Multiplicative Magic Squares Please, Christian, tell us how you created that list. --Michael Kleber
Kleber:
Please, Christian, tell us how you created that list. Christian: 3x3 and 4x4 lists created by computing. Of course a bug is always possible, but the lists should be correct.
He means, "Please tell us your algorithm." -- Mike Stay metaweta@gmail.com http://math.ucr.edu/~mike
Michael Kleber:
Please, Christian, tell us how you created that list.
Christian:
3x3 and 4x4 lists created by computing. Of course a bug is always possible, but the lists should be correct.
Mike Stay:
He means, "Please tell us your algorithm."
Yes, Mike, hmmm, hmmm... I pretended to have not well understood the Michael Kleber's question ;-) Before to explain the method, I will try also to answer at least to the more difficult question: what is the minimum 5x5 multiplicative square? I promise to explain my "secret" algorithm (but there is no great secret...) when it will be proved enough good and optimized to successfully answer to the 5x5 question. I agree with Ed, when he said after my sending of the Dudeney's example: "I'm fairly certain Dudeney's 60,466,176 for the 5x5 can be beaten." But the minimum square is difficult to locate... I hope to have enough time to work on the question in the next days. Christian.
Christian Boyer wrote:
Yes, Mike, hmmm, hmmm... I pretended to have not well understood the Michael Kleber's question ;-)
Yes, I had a funny feeling...
Before to explain the method, I will try also to answer at least to the more difficult question:
what is the minimum 5x5 multiplicative square?
I was reluctant to open that can of worms. But roughly the same idea which produced the 7! 4x4 magic square also produces a 9! 5x5: [ 10 24 4 42 9 ] [ 28 54 5 16 3 ] [ 8 2 21 36 30 ] [ 27 20 48 1 14 ] [ 6 7 18 15 32 ] magic product 9! = 362880 = 2^7.3^4.5.7 This time it's built on the permutation matrix [ 1 0 0 0 0 ] [ 0 0 0 1 0 ] [ 0 1 0 0 0 ] [ 0 0 0 0 1 ] [ 0 0 1 0 0 ] Jessica, aka "my wife the weaver", would probably be annoyed if I didn't call this the "satin" matrix, though readers here might prefer to think of it as the order in which you visit vertices of a pentagon when inscribing a star. Anyway, this permutation matrix is a 5x5 magic square with sum 1 -- one of 20 such; see A007016 for further counts. Better yet, it's what David Wilson just reminded us is called "diabolic": every broken diagonal likewise has sum 1. This means that we can take it and four more copies with rows shifted cyclically, and tile a 5x5 matrix, getting each row, column, and broken diagonal to have one entry from each of the five tiling matrices. Laying the transpose on top of that matrix, you get the beauty: [ Aa Bc Ce Db Ed ] [ Cb Dd Ea Ac Be ] [ Ec Ae Bb Cd Da ] [ Bd Ca Dc Ee Ab ] [ De Eb Ad Ba Cc ] The 25 entries in the matrix are all distinct, and it's diabolically magic, with product, if you're thinking multiplicatively, ABCDEabcde. Now we just need to assign those ten variables numerical values such that all 25 entries remain distinct. The 4x4 5040 square came from setting two variables to 1 and the other six to 2-7 in some fortuitious order, and we can do the parallel thing here: set (A,B,C,D)=(2,3,4,6) and (a,b,c,d)=(5,7,8,9), and E=e=1, to get [ 10 24 4 42 9 ] [ 28 54 5 16 3 ] [ 8 2 21 36 30 ] [ 27 20 48 1 14 ] [ 6 7 18 15 32 ] diabolical magic product 9! = 362880 = 2^7.3^4.5.7 Q: Is it always possible to partition the integers from 2 to 2n-1 into two sets of size n-1, such that the products of two elements, one from each set, all distinct integers >=2n? Seems like the answer is "yes with high probability", but can someone change that to just "yes"? Once again, I feel like one ought to be able to argue from elementary principles that this must be the minimum possible, but I'm unable to do so (though I do think the Hilbert basis for the 4x4 case should now make that argument easy). So we can wait for Christian's algorithm to get fast enough to check everything smaller, or hope to use David Wilson's logic to limit the smaller possible signatures, etc. All right, back to real work... --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
From Michael Kleber [ 10 24 4 42 9 ] [ 28 54 5 16 3 ] [ 8 2 21 36 30 ] [ 27 20 48 1 14 ] [ 6 7 18 15 32 ] diabolical magic product 9! = 362880 = 2^7.3^4.5.7
Once again, I feel like one ought to be able to argue from elementary principles that this must be the minimum possible, but I'm unable to do so (though I do think the Hilbert basis for the 4x4 case should now make that argument easy). So we can wait for Christian's algorithm to get fast enough to check everything smaller, or hope to use David Wilson's logic to limit the smaller possible signatures, etc.
Michael, very nice square! However I have a different feeling: I think (of course not sure) that it should be possible to find smaller 5x5 examples. But 362880 should be the smallest possible product of 5x5 pandiagonal magic square (or diabolic). Trying to find publications on multiplicative squares, I have the surprise to find... most of our squares... in the old book "Magic squares and cubes" by W.S. Andrews. Pages 283-294, there is a reprint of an excellent article -with construction methods- written by H.A. Sayles one century ago, date between 1905 and 1916. For example, we can find in the book: - 3x3 (216, 1000, 4096) - 4x4 (5040, 7560, pandiag 14400, ...) - 5x5 (pandiag 362880, pandiag 720720, ...) Good to see that different minds, even separated by one century, find same objects. 362880 seems to be currently the smallest known 5x5. Is it the smallest possible? I will keep you informed of my results (computing not yet started, free time is missing). Christian.
Smallest 5x5 multiplicative magic squares: magic product 302,400 (< 362,880) More info tomorrow! Christian.
participants (3)
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Christian Boyer -
Michael Kleber -
Mike Stay