# Lights out game on a grid

Problem1

Let G be a graph, let A be the adjacency matrix of G. Solve the equation Ax=b in \F_2.

The problem is equivalent to the lights out game. Each vertex has state 0 or 1. Activate a vertex flips the state of itself and all its neighbors. Find a set of activations that turns all state into 0. Originally I thought this problem can be solved in O(n^{\omega/2}) when G is planar graph on n vertices by nested dissection. However, only recently I found out the matrix must be non-singular. Therefore nested dissection does not apply.

Recently I saw an algorithm that shows if the graph is a n\times n grid, then it can be solved in O(n^3) time. The solution in Chinese and can be seen here.

Given a n\times n grid graph. Let v_{i,j} be the node on the ith row and jth column. Let b_{i,j} be the state of the vertex v_{i,j}. The state is in \F_2 If we activates a node, the state of the node and its neighbors change by 1. The set of activated node is called the activation set.

We are interested in finding an activation set S, such the state of all nodes after activate S is 0.

Let S be the activation set, and S_1 to be the activation set of the first row.

Theorem2

S_1 uniquely determines S. Moreover, One can compute S from S_1 in O(n^2) time.

Proof

Indeed, consider apply activation to the nodes in S_1. Consider any vertex in row 1. If it 0, then the remaining neighbor (on the second row) cannot be activated. If it is 1, then the remaining neighbor has to be activated.

Let D[i,j] indicates if we activate v_{i,j} or not. We create formal variables Z=\set{z_1,\ldots,z_n}. Here z_i is an indicator variable that represents if v_{1,i} is activated or not. The base case D[1,j] = z_j. The observation shows that for each i and j, D[i+1,j] = 1 + D[i,j-1]+D[i,j]+D[i,j+1]+D[i-1,j]+b_{i,j}. We can express D[i,j] as a sum of elements in Z and a constant, and we are summing a constant number of previous states. So it has size O(n). We can compute the expression of D[i,j] in O(n) time. So computing all D[i,j] for i\geq 2 takes O(n^3) time.

We are interested in D[n,1],\ldots,D[n,n]. We can see it is of the following form.

\displaystyle \begin{aligned} D[n,1] &= c_{1,1} z_1+\ldots +c_{1,n} z_{n} + u_{1}\\ D[n,2] &= c_{2,1} z_1+\ldots +c_{2,n} z_{n} + u_{2}\\ \vdots &\qquad \vdots\\ D[n,n] &= c_{m,1} z_1+\ldots + c_{m,n} z_n + u_{n} \end{aligned}

We solve the equation Cz=u. Note here C is just a n\times n matrix. We finds z_1,\ldots,z_n. So now we have found the activation set restricted on the first row. We can use it to find the entire activation set.

The total running time is O(n^3). Building the table D and solving Cz=u. One can generalize this a bit further. We can obtain O(m^2n) running time for a m\times n grid, where m\leq n. Also, there is no reason we have to work in \F_2, any arbitrary field is fine.

Theorem3

Let G be a m\times n grid and A is a matrix where the non-zero entires are precisely the position of 1s in the adjacency matrix of A. Finding Ax=b can be done in O(m^2n) time.

I did not think too much into it, but maybe it works for all integral domains too. Interestingly, this algorithm is so special, that we have no idea how to extend it to other graphs. Maybe it works for directed graph, maybe it works for subgraph of the grid graphs.

It would be really interesting to see an algorithm with running time O(n^{3/2}) for a planar graph of n vertices.

Posted by Chao Xu on 2019-01-12.
Tags: algorithm, algebra.