#Game of 5 Lights
Rules
A row of 5 lights is controlled by 5 buttons. Lights can be on/off Each push-button changes the state of the lights (on or off) of the light directly above it and the states of the lights immediately adjacent to the left and the right.
On/Off suggest binary notation, we should work with $\mathbb{Z}_2$
States of 5 lights are represented by a vector in $\mathbb{Z}_2^5$, for example:
$$ \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 1 \end{bmatrix} $$lights 1, 3, and 5 are on.
Actions of 5 buttons are given by:
$$ a = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \quad b = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} \quad c = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{bmatrix} \quad d = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \\ 1 \end{bmatrix} \quad e = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 1 \end{bmatrix} $$$s$ is the initial configuration: $s = \begin{bmatrix} 0 \ 0 \ 0 \ 0 \ 0 \end{bmatrix}$ so find some combination of $s + a + b + c + d + e = t$ where $t = \begin{bmatrix} 1 \ 0 \ 1 \ 0 \ 1 \end{bmatrix}$
$t$ is the target configuration from an initial configuration $s$.
Actions of 5 buttons are given by $s + x_1 a + x_2 b + x_3 c + x_4 d + x_5 e = t \quad \text{where} \quad x_1, x_2, x_3, x_4, x_5 \in \mathbb{Z}_2$ and $a, b, c, d, e \in \mathbb{R}^5$
$$ x_1 a + x_2 b + x_3 c + x_4 d + x_5 e = t - s $$ $$ \begin{bmatrix} 1 & 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} t_1 - s_1 \\ t_2 - s_2 \\ t_3 - s_3 \\ t_4 - s_4 \\ t_5 - s_5 \end{bmatrix} $$ Solution $$ s = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \quad t = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 1 \end{bmatrix} \quad \rightarrow \quad t - s = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 1 \end{bmatrix} \text{ over } \mathbb{Z}_2 $$ $$ \text{Augmented Matrix} = \begin{pmatrix} 1 & 1 & 0 & 0 & 0 & \big| & 1 \\ 1 & 1 & 1 & 0 & 0 & \big| & 0 \\ 0 & 1 & 1 & 1 & 0 & \big| & 1 \\ 0 & 0 & 1 & 1 & 1 & \big| & 0 \\ 0 & 0 & 0 & 1 & 1 & \big| & 1 \end{pmatrix} \rightarrow \text{Reduced Row Echelon Form} \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & \big| & 1 \\ 0 & 1 & 0 & 0 & -1 & \big| & 1 \\ 0 & 0 & 1 & 0 & 0 & \big| & -1 \\ 0 & 0 & 0 & 1 & 1 & \big| & 1 \\ 0 & 0 & 0 & 0 & 0 & \big| & 0 \end{pmatrix} $$ $$ \rightarrow \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & \big| & 1 \\ 0 & 1 & 0 & 0 & 1 & \big| & 1 \\ 0 & 0 & 1 & 0 & 0 & \big| & 1 \\ 0 & 0 & 0 & 1 & 1 & \big| & 1 \\ 0 & 0 & 0 & 0 & 0 & \big| & 0 \end{pmatrix} \text{ Reduced over } \mathbb{Z}_2 \, (-1 \equiv 1) $$ $$ \begin{pmatrix} 1 & 0 & 0 & 0 & 1 & \big| & 0 \\ 0 & 1 & 0 & 0 & 1 & \big| & 1 \\ 0 & 0 & 1 & 0 & 0 & \big| & 1 \\ 0 & 0 & 0 & 1 & 1 & \big| & 1 \\ 0 & 0 & 0 & 0 & 0 & \big| & 0 \end{pmatrix} $$ $$ x_1 = -x_5 \\ x_2 = 1 - x_5 \\ x_3 = 1 \\ x_4 = 1 - x_5 \\ x_5 \text{ is free variable} $$ $$ \rightarrow \text{Solutions when } x_5 = 0 \text{ or } x_5 = 1 $$Operations in $\mathbb{Z}_2$ is modulo(2), so values are only 0, 1 and $2 \equiv 0$ and $-1 \equiv 1$.
Solutions $(x_5 = 0) \rightarrow$
$$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 1 \\ 0 \end{bmatrix} \quad \text{and } (x_5 = 1) \rightarrow \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} - 1 \\ 0 \\ 1 \\ 0 \\ 1 \end{bmatrix} \equiv \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \\ 1 \end{bmatrix} $$ Solutions: Press 1B + 1C + 1D or Press 1A + 1C + 1E
Eigenvectors and Eigenvalues