Thanks Richard! What you described about symmetries has been used during the construction, which was based on a central auxiliary cube of order 3. About the size 6, a magic cube has been found only 2 months ago by Walter Trump. You can download it from my web site www.multimagie.com/indexengl.htm, and click on "Perfect magic cubes" in the left menu. But this page has not yet been updated with the new magic cube of order 5. Look also at the Walter's web site, www.trump.de/magic-squares I think that the smallest unknown perfect cubes are now of order 10. Kind regards. Christian.
-----Message d'origine----- De: Richard Schroeppel [mailto:rcs@CS.Arizona.EDU] Date: vendredi 14 novembre 2003 19:32 À: math-fun@mailman.xmission.com Cc: cboyer@club-internet.fr; TrumpNbg@aol.com Objet: magic cube, size 5
Congratulations to Christian Boyer and Walter Trump, who have found a magic cube of size 5!
They sent me a .gif picture of the cube. I've transcribed it below. K=315. I spot checked a couple of rows but haven't done a full check of all 109 lines. [109 = (7^3-5^3)/2.] There is some amount of symmetry, opposite cells adding to 126, but it's irregular. Sometimes the symmetry is reflection through a center plane rather than through the center cell. Hopefully we'll hear more about the technique for finding the cube soon.
I think this leaves size 6 as the only magic cube size of undetermined existence. Gardner's columns c. 1974 had a magic cube of size 7, and I think 8+ had already been done. Size 7 is a straightforward generalization of the stepping method for the magic 3^2 and 5^2. I proved there's no magic 4^3 (published in MGs column), and the nonex. of 3^3 was already well known.
Curiously, the magic 6^2 is harder to construct than squares of size 4,5,7,8. 4K+2 seems to be a hard size. I have some simple results on magic 6^3 properties: various groups of cells must have the expected average value. (Make a semihemidemi-magic 3^3 by folding the 6^3 along the three center planes, adding the cells. Various rows & agonals must have the average value, and algebra gives a little bit more.) The nonexistence of magic 5^4 follows from each 5^3 having the same required center value. There should be enough line types to prove that a magic 7^4 must have center 1201, but I haven't done the algebra. I've tried to show the impossibility of the magic 6^4 but have only partial results, which don't lean either way.
This magic 5 cube supports my conjecture that existence of a magic object (of size >1) is equivalent to the corresponding tictactoe game being a draw. (Since 4^3 TTT is a long battle for the win, 5^3 seems certain to be a draw.)
Let the analysis begin!
Rich rcs@cs.arizona.edu
----- Walter Trump & Christian Boyer Magic 5^3 (<= Nov 14, 2003)
25 16 80 104 90 115 98 4 1 97 42 111 85 2 75 66 72 27 102 48 67 18 119 106 5
91 77 71 6 70 52 64 117 69 13 30 118 21 123 23 26 39 92 44 114 116 17 14 73 95
47 61 45 76 86 107 43 38 33 94 89 68 63 58 37 32 93 88 83 19 40 50 81 65 79
31 53 112 109 10 12 82 34 87 100 103 3 105 8 96 113 57 9 62 74 56 120 55 49 35
121 108 7 20 59 29 28 122 125 11 51 15 41 124 84 78 54 99 24 60 36 110 46 22 101
----- From: "Christian Boyer" <cboyer@club-internet.fr> To: <rcs@CS.Arizona.EDU> Cc: "Walter Trump" <TrumpNbg@aol.com> Subject: "The center of an order 5 magic cube must be 63" Date: Fri, 14 Nov 2003 18:05:49 +0100
Dear Richard, You will probably be happy by these news. As reported by Martin Gardner, and as reported at http://www.inwap.com/pdp10/hbaker/hakmem/number.html#item50, you have proved in 1972 that a perfect magic cube of order 5, if it exits, has a central value = 63. But nobody had found such a cube. We, my German friend Walter Trump and me, have computed this week the first perfect cube of order 5. 109 ways to sum 315. Of course, as you proved, its central value is 63: look at the attached file! Best regards. Christian Boyer. www.multimagie.com/indexengl.htm