Cartesian Magicness of 3-Dimensional Boards
Abstract
A (p,q,r)-board that has pq+pr+qr squares consists of a (p,q)-, a (p,r)-, and a (q,r)-rectangle. Let S be the set of the squares. Consider a bijection f : S [1,pq+pr+qr]. Firstly, for 1 i p, let xi be the sum of all the q+r integers in the i-th row of the (p,q+r)-rectangle. Secondly, for 1 j q, let yj be the sum of all the p+r integers in the j-th row of the (q,p+r)-rectangle. Finally, for 1 k r, let zk be the the sum of all the p+q integers in the k-th row of the (r,p+q)-rectangle. Such an assignment is called a (p,q,r)-design if \xi : 1 i p\=\c1\ for some constant c1, \yj : 1 j q\=\c2\ for some constant c2, and \zk : 1 k r\=\c3\ for some constant c3. A (p,q,r)-board that admits a (p,q,r)-design is called (1) Cartesian tri-magic if c1, c2 and c3 are all distinct; (2) Cartesian bi-magic if c1, c2 and c3 assume exactly 2 distinct values; (3) Cartesian magic if c1 = c2 = c3 (which is equivalent to supermagic labeling of K(p,q,r)). Thus, Cartesian magicness is a generalization of magic rectangles into 3-dimensional space. In this paper, we study the Cartesian magicness of various (p,q,r)-board by matrix approach involving magic squares or rectangles. In Section~2, we obtained various sufficient conditions for (p,q,r)-boards to admit a Cartesian tri-magic design. In Sections~3 and~4, we obtained many necessary and (or) sufficient conditions for various (p,q,r)-boards to admit (or not admit) a Cartesian bi-magic and magic design. In particular, it is known that K(p,p,p) is supermagic and thus every (p,p,p)-board is Cartesian magic. We gave a short and simpler proof that every (p,p,p)-board is Cartesian magic.
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