RE: [math-fun] Multiplicative Magic Squares
Replacement goes a long way. You may take all the 3s in the square below and turn them into 29s. Or some other square having large primes may be replaced with a square with smaller corresponding primes. Even primes within a particular square may be switched. 55440=2^4*3^2*5*7*11 so you can make a corresponding square with 2*3^2*5*7*11^4. No matter the starting square, this reduction leads to some sort of minimal representation, where the factorization of the product gives 2 with the largest exponent, 3 the next and so on. -----Original Message----- From: math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+mdtorge=sandia.gov@mailman.xmission.com] On Behalf Of Michael Kleber Sent: Wednesday, September 21, 2005 2:32 PM To: math-fun Subject: Re: [math-fun] Multiplicative Magic Squares Just blathering off the top of my head: Ed pegg wrote:
Are multiplicative magic squares well studied? For example, using dots merely as spacers, the following has a multiplicative constant of
55440.
231 .40 ..3 ..2 ..6 ..1 132 .70 ..4 .21 .20 .33 .10 .66 ..7 .12
For any prime p, this must still be multiplicatively magic if you replace each entry with the largest power of p dividing it. And that must be additively magic if you take its log_p. For instance, for p=2, the above becomes 0 3 0 1 1 0 2 1 2 0 2 0 1 1 0 2 with additive constant 4, the power of 2 in 55440. Of course, this projection destroys distinctness of entries. For that you just need to find another one like the above such that the ordered pairs of corresponding entries (a,b) are all distinct, and then make a multiplicative one out of 2^a 3^b. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
If you just use powers of 2, the minimal square with distinct entries has multiplicative constant 1073741824 (2^30). 55440 is much smaller. Is it the smallest? --Ed Pegg Jr --- "Torgerson, Mark D" <mdtorge@sandia.gov> wrote:
Replacement goes a long way. You may take all the 3s in the square below and turn them into 29s. Or some other square having large primes may be replaced with a square with smaller corresponding primes. Even primes within a particular square may be switched. 55440=2^4*3^2*5*7*11 so you can make a corresponding square with 2*3^2*5*7*11^4. No matter the starting square, this reduction leads to some sort of minimal representation, where the factorization of the product gives 2 with the largest exponent, 3 the next and so on.
If you just use powers of 2, the minimal square with distinct entries has multiplicative constant 1073741824 (2^30). 55440 is much smaller. Is it the smallest?
For multiplicative magic squares it seems reasonable to allow negative numbers and ask for the minimum absolute value. So you could replace 11 with -1 in your example, giving magic product -5040 . Magic products with other -1 patterns might allow further reductions. If you look at the projections of primes (or unit -1) and consider patterns needed to ensure all entries are unique, information theory gives some lower bounds. What do they say? Are there cases where the information theoretic bound can't be achieved? I conjecture yes. Is this question solved for the simpler problem of dropping the magic constraint on diagonals? That question sounds interesting too. Best, - Scott
--Ed Pegg Jr
--- "Torgerson, Mark D" <mdtorge@sandia.gov> wrote:
Replacement goes a long way. You may take all the 3s in the square below and turn them into 29s. Or some other square having large primes may be replaced with a square with smaller corresponding primes. Even primes within a particular square may be switched. 55440=2^4*3^2*5*7*11 so you can make a corresponding square with 2*3^2*5*7*11^4. No matter the starting square, this reduction leads to some sort of minimal representation, where the factorization of the product gives 2 with the largest exponent, 3 the next and so on.
Ignore the desire for entries to be distinct for a minute. There is a 4x4 magic square with sum 1: 1 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 It's unique, up to rotation and reflection. You can tile the 4x4 square with four of these: A B C D C D A B D C B A B A D C Looks suspiciously familiar from other 4x4 constructions, doesn't it? It's also unique, up to taking the transpose. Take ordered pairs of this with its transpose, and you get Aa Bc Cd Db Cb Dd Ac Ba Dc Ca Bb Ad Bd Ab Da Cc Hooray, these sixteen are all distinct. Under addition, you can let (a,b,c,d)=(0,1,2,3) and (A,B,C,D)=(00,10,20,30) base 4, and you get the usual 4x4 square made with the numbers 0-15. If you want multiplication instead, you can let (a,b,c,d)=(1,2,4,8) and (A,B,C,D)=(1,3,5,7), and you'll get magic product abcdABCD = 6720. I think this is the best choice -- you can use any values such that the sixteen products are pairwise distinct, and that seems the way to minimize it. Translating that back into numbers, you end up with 1 10 28 24 12 56 5 2 40 4 6 7 14 3 8 20 magic product: 6720 = 2^6.3.5.7 --Michael Kleber On 9/21/05, ed pegg <ed@mathpuzzle.com> wrote:
If you just use powers of 2, the minimal square with distinct entries has multiplicative constant 1073741824 (2^30). 55440 is much smaller. Is it the smallest?
--Ed Pegg Jr -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
Last night I got around to:
Aa Bc Cd Db Cb Dd Ac Ba Dc Ca Bb Ad Bd Ab Da Cc
...and then I suggested
(a,b,c,d)=(1,2,4,8) and (A,B,C,D)=(1,3,5,7), and you'll get magic product abcdABCD = 6720. I think this is the best choice[...]
But it's not; I was seduced by the siren song of powers of two. Much better to use (a,b,c,d) = (1,2,3,6) and (A,B,C,D) = (1,4,5,7), so abcdABCD = 7! : 1 12 30 14 10 42 3 4 21 5 8 6 24 2 7 15 magic product = 5040 = 2^4.3^2.5.7 As I was falling asleep I thought I had an information-theoretic proof that this was optimal. [It started by pointing out that (1) in this version, you can specify each entry using only 5 bits -- whether or not to multiply each of 23457 -- and saying that 5040 was the smallest product you could pack 5 bits into; and (2) Rich already optimized 4 bits and got 14400, while 6 bits or more is again going to be worse.] But things that were clear last night are fuzzy this morning, and the above doesn't seem to prove anything at all. Maybe someone else will un-muddle things. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
In his old book on recreative mathematics, Boris Kordiemski described exactly the Michael's method (using two latin squares) for 4x4 multiplicative squares. He also described a method for 3x3 multiplicative squares: a b² a²b a²b² ab 1 b a² ab² With a=2 and b=3, you get 2 9 12 36 6 1 3 4 18 Magic sum = 216 (*). It is the smallest possible multiplicative square. In 1667 (a long time ago...), Arnauld gave a 3x3 multiplicative square. Perhaps the oldest published multiplicative square? Using powers of 2, its magic sum was bigger : 4096. Christian. (*) 216 is also the smallest cube being also sum of three cubes. 216 = 6^3 = 3^3 + 4^3 + 5^3
Of course replace "magic sum" by "magic product", twice in my previous message... -----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 : jeudi 22 septembre 2005 16:30 À : 'math-fun' Objet : RE: [math-fun] Multiplicative Magic Squares In his old book on recreative mathematics, Boris Kordiemski described exactly the Michael's method (using two latin squares) for 4x4 multiplicative squares. He also described a method for 3x3 multiplicative squares: a b² a²b a²b² ab 1 b a² ab² With a=2 and b=3, you get 2 9 12 36 6 1 3 4 18 Magic sum = 216 (*). It is the smallest possible multiplicative square. In 1667 (a long time ago...), Arnauld gave a 3x3 multiplicative square. Perhaps the oldest published multiplicative square? Using powers of 2, its magic sum was bigger : 4096. Christian. (*) 216 is also the smallest cube being also sum of three cubes. 216 = 6^3 = 3^3 + 4^3 + 5^3 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
In our discussion, we have seen 3x3 and 4x4 examples. I have found a 5x5 example in Dudeney, with an interesting and easy-to-use method for even orders. Its first row: 54 / 648 / 1 / 12 / 144 giving the magic product 60,466,176. Because the Dudeney's square is associative, this magic product = 36^5 (the center cell is 36). Christian.
participants (5)
-
Christian Boyer -
ed pegg -
Michael Kleber -
Scott Huddleston -
Torgerson, Mark D