Let $S=\{1,2,\dots,m+n-1\}$.

An $m\times n$ matrix($\in S^{m\times n}$) is called silver matrix if

(a) There is no same numbers in the row or column. (like latin square)

(b) {$i$ th row}$\cup${$i$ th column}=S for all $1\leq i\leq min(m,n)$

Does silver matrix exist for all $m\neq n$ ?

If this conjecture is true, $d(K_m\times K_n, m+n-1) = mn-min(m,n)$ ($m\neq n$)

($d$ is defining number, $\times$ is cartesian product)

don'tknow the answer? – Yemon Choi Feb 12 '12 at 8:03