Richard, you are perfectly right. With cubes, the situation is more complicated. And with hypercubes, the situation is much more complicated! On your question, as mentioned in my MI paper, a nxnxn "perfect magic cube" has 3n²+6n+4 magic lines: your (a) + (c1) + (b1) + also (b2). With cubes, from each cell, you have 13 possible directions. The best possible cubes are often called "pandiagonal perfect magic cubes". See www.multimagie.com/English/Panperfectcubes.htm, or www.multimagie.com/English/PanperfectMMC.htm all is magic in any direction from any cell! It seems impossible to create a pandiagonal perfect magic cube of an order smaller than 8, but I have never seen such a proof. Tomorrow, I can send you more details on your question if you wish. Directly, details can be too long and boring for [math-fun]. Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Richard Guy Envoyé : lundi 10 mars 2008 20:12 À : math-fun Objet : Re: [math-fun] Magic cubes Thankyou, Christian! R, Could you, or someone else, clarify the definition of `magic cube'? For magic squares, there are (a) n rows & n columns (b) diagonals, 2 main & 2(n-1)broken. I understand that, in a magic square, (a) & main (b) have the magic total, and that a `pandiagonal' (`Nasik'?) square has (a) & all of (b). When we come to cubes, the situation is more complicated. There are (a) n^2 rows in each of 3 orthogonal directions (b) `2-dimensional' diagonals in 6 directions, which may be classified as [don't think I've got the numbers right]: (b1) 12 `face' (`visible') diagonals, (b2) 6(n-2) `edge' (`interior') diags, (b3) 12(n-1) broken `face' diagonals, (b4) 12(n-2)^2? broken interior diags. (c) `3-D' diags in 4 directions: (c1) 4 main body diagonals, (c2) 12(n-2)? with ends on edges, (c3) 12(n-2)^2? with ends on faces [these last two being broken]. Trenkler's cubes have (a) and (c1). Does a `perfect' cube have (a), (c1) & (b1) ?? On Mon, 10 Mar 2008, Christian Boyer wrote:
Richard, supplemental information.
If you plan to update problem D15 of UPINT, read my paper published in The Mathematical Intelligencer, Spring 2005, pages 52-64: I organized this paper around nine quotations from your own text, UPINT 3rd edition 2004. The 5x5x5 magic cube question is one of them (quotation #8, pages 60-62 of the M.I. paper).
Christian.
-----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Richard Guy Envoyé : vendredi 7 mars 2008 17:36 À : Math Fun Objet : [math-fun] Magic cubes
I have a question in UPINT, D15: ``Has anyone constructed a 5 x 5 x 5 magic cube, or proved its impossibility'' (just possibly asked by Rich Schroeppel?). I've recently downloaded a 2-page article by Maria'n Trenkler, An algorithm for making magic cubes, which was evidently published in Pi Mu Epsilon J, 12 #2 (Spring 2005) 105-106. It produces magic cubes of any order > 2. Am I the only person not to know about this? R.
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