Talk:Lights Out (video game)

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[edit] Mathematical proof by matrix analysis

Let \mathcal{L} be the set of all 5×5 matrices over the two-element field {0,1}. Then any board setup in Lights Out can be represented by a matrix in \mathcal{L}, where 1 corresponds to a light that is on and 0 to one that is off.

However, if we regard the elements of \mathcal{L} as simply matrices over {0,1}, then we can add them according to the rules of matrix addition. And we know from matrix theory that \mathcal{L} is an Abelian group under matrix addition.

If m,n\in\{1,2,3,4,5\}, let P(m,n) denote the matrix whose (i,j)th entry is 1 if (i,j) = (m,n) or (m±1,n) or (m,n±1) , and 0 otherwise. Then for a board setup L, pressing the (m,n)th light corresponds to adding the matrix P(m,n) to L.

This analysis proves the two statements stated in the main article. The first statement is proved by the fact that matrix addition is commutative and associative, while the second statement stems from the fact that \forall\,L\in\mathcal{L}, L+L=\mathbf{0}, the zero matrix (board with all lights off).

Jane Fairfax 13:38, 12 May 2007 (UTC)

[edit] strategy missing something

using the strategy, what happens when you're left with a light at D or E (single light at the last row in pos E for example)? gujamin 16:52, 24 August 2007 (UTC)

If that were to happen, it would mean that the position you started with was insoluble. Spacepotato 21:47, 18 October 2007 (UTC)