It can be checked that the unique shortest safe-opening word for $2$ buttons is $\mathcal{S}(2) = \sa{11}\cdot \sa{01} \cdot \sa{11}$.
For $u=\sa{0111}$ and $A=\sa{1010}$ there are four nodes $v$ conjugate of $u$, $\sa{0111}$, $\sa{1110}$, $\sa{1101}$, $\sa{1011}$, and consequently edges: $$u\Arrow{A}\sa{1101},\; u\Arrow{A}\sa{0100},\; u\Arrow{A}\sa{0111},\; u\Arrow{A}\sa{0001}.$$
Problem 124: Safe-Opening Words |
The problem addresses a special non-deterministic version of synchronising words. We are given a graph $G$ with edges labelled by symbols and with a unique sink node $s$ on which all edges loop. We are to find a synchronising word $\mathcal{S}$ for which each non-deterministically chosen path labelled by $\mathcal{S}$ goes to $s$ independently of the starting node.
The problem is generally difficult but we consider a very special case called safe-opening words . It shows some surprising operations on words over the alphabet $\Bi_n=\{\sa{0},\sa{1}\}^n$.
The goal is to find a sequence called a safe-opening word $$\mathcal{S}(n) = {A}_1\cdot {A}_2\cdots {A}_{2^n-1}$$ of moves $A_i\in \Bi_n$ having a prefix that opens the safe.
Assuming button positions are numbered $1$ to $n$ from the top position clockwise, a move is described by an $n$-bit word $b_1b_2\cdots b_n$ with the meaning that button at position $i$ is pressed if and only if $b_i=1$. Positions are fixed though buttons can move, i.e. change positions.
The aim is to find a word $\mathcal{S} = {A}_1\cdot {A}_2\cdots {A}_{2^n-1}$ in $\Bi_n^*$ for which each non-deterministically chosen path in $G_n$ labelled by $\mathcal{S}$ leads to the sink $\sa{1}^n$ independently of the starting node.