There is a $5 \times 5$ square of lights, of which some number are on and off. "Selecting" a light changes the state of the selected light and all four lights adjacent to it, in effect, changing the state of the lights in a plus centered on the selected light. The ultimate goal is to have every light in the puzzle in its "off" state. Is there a surefire algorithm to solve this using something like a binary matrix? If no, what is the most efficient method to solve this puzzle with only one light on, in the center?
[Math] Method for solving a 5×5 light puzzle
matricesnumber theory
Related Solutions
Here, all work will be done in $\Bbb F_2$, that is, nothing but zeroes and ones.
Construct a $25\times 25$ matrix $A$ such that $A_{i,j} = \begin{cases}1&\text{if button}~j~\text{toggles light}~i\\ 0&\text{otherwise}\end{cases}$
Let $x$ be a $25\times 1$ matrix corresponding to what our button selections were.
Then given an initial configuration of lights, $v$, one has $v+Ax$ is the resulting configuration.
If we have a desired final configuration, $r$, then we wish to solve the matrix equation $v+Ax=r$ for the vector $x$. If it so happens that $A$ is invertible, then this will simply be $x=A^{-1}(r-v)$
Instead of going through the effort of explicitly writing out the $25\times 25$ matrix (it is incredibly tedious), I will show with an example how it works for the $9\times 9$ case corresponding to a $3\times 3$ grid of rooms with lightswitches.
$A=\begin{bmatrix}1&1&0&1&0&0&0&0&0\\ 1&1&1&0&1&0&0&0&0\\ 0&1&1&0&0&1&0&0&0\\ 1&0&0&1&1&0&1&0&0\\ 0&1&0&1&1&1&0&1&0\\ 0&0&1&0&1&1&0&0&1\\ 0&0&0&1&0&0&1&1&0\\ 0&0&0&0&1&0&1&1&1\\ 0&0&0&0&0&1&0&1&1\end{bmatrix}$
As it so happens, is in fact invertible over $\Bbb F_2$ in this case, but that is not always the case. Here we have $A^{-1}=\left[\begin{smallmatrix}1&0&1&0&0&1&1&1&0\\0&0&0&0&1&0&1&1&1\\1&0&1&1&0&0&0&1&1\\0&0&1&0&1&1&0&0&1\\0&1&0&1&1&1&0&1&0\\1&0&0&1&1&0&1&0&0\\1&1&0&0&0&1&1&0&1\\1&1&1&0&1&0&0&0&0\\0&1&1&1&0&0&1&0&1\end{smallmatrix}\right]$
Using this matrix, a starting configuration (say, all lights on) and a desired final configuration (say, all lights off), we can compute $A^{-1}(r-v) = x$ which in this case would be $\left[\begin{smallmatrix}1\\0\\1\\0\\0\\1\\0\\1\\0\\1\end{smallmatrix}\right]$, telling us that to switch the lights from all on to all off in the $3\times 3$ grid, we can flip the switches in rooms $1,3,5,7,9$.
In the case of the $5\times 5$ grid, requiring a $25\times 25$ matrix, we approach similarly.
Our matrix is:
1100010000000000000000000
1110001000000000000000000
0111000100000000000000000
0011100010000000000000000
0001100001000000000000000
1000011000100000000000000
0100011100010000000000000
0010001110001000000000000
0001000111000100000000000
0000100011000010000000000
0000010000110001000000000
0000001000111000100000000
0000000100011100010000000
0000000010001110001000000
0000000001000110000100000
0000000000100001100010000
0000000000010001110001000
0000000000001000111000100
0000000000000100011100010
0000000000000010001100001
0000000000000001000011000
0000000000000000100011100
0000000000000000010001110
0000000000000000001000111
0000000000000000000100011
Unfortunately, as alluded to earlier, this matrix happens to not be invertible. That is to say, given a starting position, there are some ending positions which are impossible to reach. There are also multiple ways to reach the same ending position given a starting position if possible to reach in the first place. Regardless, that is not to say that the desired start and end that we are looking for are impossible. We find that the matrix above happens to be of rank $23$.
Trying to solve the matrix equation then, $Ax=r-v$, we can try to row reduce the augmented matrix $[A~|~(r-v)]$. If the system is inconsistent, then no solution exists. If it is consistent, then you should be able to interpret the results of the reduced form in such a way to get one of the possible vectors $x$ which gets you the desired outcome.
Note: this method can be generalized for similar related problems given arbitrarily shaped grids and lightswitches which may operate differently than those given in the above problem. Say for example where flipping a light in a room in the top row will toggle the lights in all adjacent rooms and itself, flipping a light in the second row will toggle all diagonal rooms and itself, flipping the light in room 20 toggles itself and room 1, etc...
row reducing one finds that by flipping the switches in rooms 2, 3, 5, 7,8, 9, 13, 14,15, 16, 17,19, 20, 21, 22, we will change all lights from on to off or vice versa. I will not bother explicitly solving the second question and leave that as an exercise to the reader.
The column you augment your neighbourhood matrix with needs to be the negative (mod $4$) of the puzzle state. The solution for your example is the negative of the one you've found—namely $$ \begin{matrix} 0&0&3\ \ \\ 3&3&0\ \ \\ 0&3&0\ \ , \end{matrix} $$ which worked when I tried it.
Best Answer
Represent a configuration as a $25$-dimensional vector over $\mathbb{F}_2$, the field with two elements. Each component of the vector corresponds to one of the lights. $1$ represents on, and $0$ represents off.
You can also represent a sequence of moves as a $25$-dimensional vector over $\mathbb{F}_2$ because of two properties:
So the component corresponding to a certain light represents the number times you select it $\pmod{2}$.
Now, let $A$ be the adjacency matrix of the $5\times 5$ rectangular grid graph with $1$'s in the diagonal. Let $\vec{x}$ be the unknown solution to a given initial configuration $\vec{b}$. Then you need to solve the system of $25$ equations $\vec{b}+A\vec{x}=\vec{0}$ which is equivalent to $A\vec{x}=\vec{b}$ over $\mathbb{F}_2$. In your specific problem, you need to set $\vec{b}$ to the vector where the component corresponding to the light in the center is a $1$ and the rest are $0$'s. Solve the system of equations. You may get no solution, a unique solution, or many solutions. If you get many solutions, compare each of the solutions directly to figure out which solution involves pressing the fewest lights.
It so happens that there are $4$ solutions to your problem, and they are all equivalent to each other by rotation. Here's one of these $4$ equally optimal solutions, which toggles $11$ lights: $$\begin{matrix}1&1&0&0&0\\0&0&1&0&0\\1&0&1&1&0\\1&0&0&0&1\\0&1&1&0&1\end{matrix}$$
P.S. I've actually been studying the mathematics of Lights Out for a few years. The topic fascinates me and I'm trying to publish a paper about fractal patterns in the solutions to larger boards of size $2^n\times 2^n$. There are many different classes of board sizes whose solutions give fractal patterns, and my profile picture is the pattern that emerges from boards of size $7(2^n)-1\times 7(2^n)-1$.