![]() I think the only way is to establish what the linear dependencies are, to show that the only patterns that have no effect on the lights are those where all the squares in an even number of rows are pressed. This is obvious when $ageneral this can allow for shorter solutions to those patterns that can be achieved. This pattern clearly does not have a unique solution. So we can get from all-off to all-on by pressing either the first row, or the second row, etc. This is easy to see because pressing the squares of a single row toggles everything. If $b$ (the length of a row) is odd and $a$ even, then we are not in this ideal situation. We know that tapping every square will change the all-off state to all-on (everything toggles $a b-1$ times) and from the above we know that this is the unique solution so $a*b$ moves are necessary. This means that there are only $2^$ move sequences, every combination of states has a unique solution.įor $a\times b$ boards with both $a$ and $b$ even, we are in this ideal situation, because tapping a square together with all the squares in the same row and those in the same column changes only that single square (it toggles $a b-1$ times, the column toggles $a$ times, the row $b$ times, the rest $2$ times). But tapping a square twice has no net effect, so such repeated moves can be skipped. ![]() If you were to tap a square more than once during a solution, you can reorder the moves to make those two taps consecutive. It does not matter to that square what order the moves are performed, as all the moves that affect it will affect it in the same way.Įach square need not be tapped more than once. If you look at the effect of a move sequence on a single square, all that matters to that square is how often it is toggled. Like all these games, they have the following properties: ![]()
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