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Consider the following "board":

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 \end{bmatrix}$$

We have an operation such that flipping any of the $1$'s in the $a_{i,j}$ position will also flip the state of the $a_{i+1,j}, a_{i-1,j}, a_{i,j+1}, a_{i,j-1}$ from a 0 to a 1 or vice versa for $i,j \in\{1,\dots,5\}$ (We may not choose an $a_{i,j}$ if $a_{i,j} = 0$). In the case when $i$ or $j = 0$ or $i$ or $j = 6$, the board wraps around such that $0 = 5$ and $6 = 1$.

For example:

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 \end{bmatrix}\underset{\longrightarrow}{a_{3,3}}\begin{bmatrix} 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 \end{bmatrix} \underset{\longrightarrow}{a_{1,1}}\begin{bmatrix} 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}\underset{\longrightarrow}{a_{1,2}}\begin{bmatrix} 1 & 0 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 \end{bmatrix}$$

However, we may not do

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 \end{bmatrix}\underset{\longrightarrow}{a_{1,2}}\begin{bmatrix} 0 & 1 & 1 & 0 & 1\\ 0 & 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 1 & 1 & 0 & 0 & 1 \end{bmatrix}$$

The goal is to choose a sequence of $a_{i,j}$'s such that there is exactly a single $1$ left on the board.

I am interested in knowing what kind of mathematics is used to determine a solution, or know the existence of a solution given any initial board.

In the example I've given here, one solution is $a_{1,1},a_{1,2},a_{1,3}, a_{1,4},a_{1,5},a_{2,2},a_{2,4},a_{5,3}$. That is,

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 1 \end{bmatrix}\underset{\longrightarrow}{a_{1,1}} \begin{bmatrix} 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}\underset{\longrightarrow}{a_{1,2}} \begin{bmatrix} 1 & 0 & 1 & 0 & 0\\ 1 & 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 \end{bmatrix}\underset{\longrightarrow}{a_{1,3}}\\\begin{bmatrix} 1 & 1 & 0 & 1 & 0\\ 1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 0 & 1 \end{bmatrix}\underset{\longrightarrow}{a_{1,4}}\begin{bmatrix} 1 & 1 & 1 & 0 & 1\\ 1 & 1 & 0 & 1 & 0\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 1 \end{bmatrix}\underset{\longrightarrow}{a_{1,5}}\begin{bmatrix} 0 & 1 & 1 & 1 & 0\\ 1 & 1 & 0 & 1 & 1\\ 0 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0 \end{bmatrix}\underset{\longrightarrow}{a_{2,2}}\\\begin{bmatrix} 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 1 & 1\\ 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0 \end{bmatrix}\underset{\longrightarrow}{a_{2,4}}\begin{bmatrix} 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 1 & 1 & 0 \end{bmatrix}\underset{\longrightarrow}{a_{5,3}}\begin{bmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

There are other solutions that do not seem equivalent up to some reasonable definition of symmetry such as $a_{1,1},a_{5,5},a_{5,1},a_{1,5},a_{2,3},a_{4,3},a_{3,2},a_{3,4},a_{1,4},a_{5,2},a_{2,5},a_{4,1}$.

Thoughts I wrote up the "board" as an adjacency matrix hoping the solution can be solved by linear algebra, but I don't think there is a way to use elementary row operations, or things such as the rank of the matrix to glean any insight into the problem.

Aside from that, both of these solutions end with a $1$ in the middle, however, since the board wraps around, I am pretty certain that the final $1$'s position doesn't matter and it's sufficient to consider only solutions that leave the $1$ in the middle.

What area of mathematics has the best tool to solve this problem? I assume this is already equivalent to something well known in algebra or graph theory, but I have no idea what it might be.

  • 3
    This is a linear algebra problem, but not in the way you're thinking. The set of possible states of the game forms a $25$-dimensional vector space over the finite field $\mathbb{F}_2$ with two elements, and each of your possible moves is a different vector in this vector space. The question is to find out whether or not the original board is in the span of all of these vectors, which can be answered using Gaussian elimination over $\mathbb{F}_2$.2017-01-14
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    @QiaochuYuan Sorry, after some additional thought, I can't convince myself that is true. In particular, if we have say a vector $v_{1}$ with say, $7$ 1's, there are 7 allowable moves, but if we do one of those moves, call it $v_{2}$ the number of $1$s in the new vector $v = v_{1} + v_{2}$ will have a different set of allowable moves. I'm not sure how the span of all possible moves will properly encode whether a move was allowable at each step since addition of the vectors commute, but the order seems to matter in the game.2017-01-15
  • 0
    Oh, I see, I misread that part.2017-01-15
  • 0
    Combinatorial game theory?2017-01-15

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