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Matrix 1: \begin{matrix} 0&1&1&0\\ 1&0&1&0\\ 1&1&0&1\\ 0&0&1&0 \end{matrix}

Matrix 2: \begin{matrix} 0&1&1&1\\ 1&0&0&0\\ 1&0&0&1\\ 1&0&1&0 \end{matrix}

I've looked on google to find out how to do this, but I can't find an answer that makes sense to me. As far as I can tell, there is no efficient algorithm to do this, so you need to check all the permutations... but I don't even know how to start. Any help would be great, Thanks.

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Draw the graphs... –  draks ... Nov 28 '12 at 14:40

1 Answer 1

up vote 3 down vote accepted

As isomorphims have to preserve degree, there are only 2 possible ones, the first given by $$ \phi_1 \colon 1\mapsto 3, 2 \mapsto 4, 3 \mapsto 1, 4 \mapsto 2 $$ and $$ \phi_2 \colon 1\mapsto 4, 2 \mapsto 3, 3 \mapsto 1, 4 \mapsto 2 $$ $\phi_1$ maps the first matrix to (that means the following matrix has $a_{\phi_1^{-1}(i), \phi_1^{-1}(j)}$ as $(i,j)$-th entry. $$ \begin{pmatrix} 0&1&1&1\\ 1&0&0&0\\ 1&0&0&1\\ 1&0&1&0 \end{pmatrix} $$ and so $\phi_1$ is an isomorphism.

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OK - This helps a lot. I can see now how you applied the algorithm to do this. –  agent154 Nov 28 '12 at 14:49

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