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    <p>I think brute force is not necessary.</p> <p>The problem can be rephrased in terms of a group. The matrices over the field with 2 elements constitute an commutative group with respect to addition.</p> <p>As pointed out before, the question whether A can be toggled into B is equivalent to see if A-B can be toggled into 0. Note that toggling of row i is done by adding a matrix with only ones in the row i and zeros otherwise, while the toggling of column j is done by adding a matrix with only ones in column j and zeros otherwise.</p> <p>This means that A-B can be toggled to the zero matrix if and only if A-B is contained in the subgroup generated by the toggling matrices.</p> <p>Since addition is commutative, the toggling of columns takes place first, and we can apply the approach of Marius first to the columns and then to the rows.</p> <p>In particular the toggling of the columns must make any row either all ones or all zeros. there are two possibilites:</p> <ol> <li><p>Toggle columns such that every 1 in the first row becomes zero. If after this there is a row in which both ones and zeros occur, there is no solution. Otherwise apply the same approach for the rows (see below).</p></li> <li><p>Toggle columns such that every 0 in the first row becomes 1. If after this there is a row in which both ones and zeros occur, there is no solution. Otherwise apply the same approach for the rows (see below).</p></li> </ol> <p>Since the columns have been toggled successfully in the sense that in each row contains only ones or zeros, there are two possibilities:</p> <ol> <li><p>Toggle rows such that every 1 in the first column becomes zero.</p></li> <li><p>Toggle rows such that every 0 in the first row becomes zero.</p></li> </ol> <p>Of course in the step for the rows, we take the possibility which results in less toggles, i.e. we count the ones in the first column and then decide how to toggle.</p> <p>In total, only 2 cases have to be considered, namely how the columns are toggled; for the row step, the toggling can be decided by counting to minimuze the number of toggles in the second step.</p>
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