Consider the two functions: $g_{\sa{a}}(i)=(i+1) \bmod n$, $g_{\sa{b}}(i) = \min\{i,g_{\sa{a}}(i)\}$. For $n=3$ they are illustrated by the automaton. As shown on the table below the word $w=\sa{baab}$ is synchronising since the image of the set $\{0,1,2\}$ by $g_w$ is the singleton $\{0\}$. The word obviously synchronises every pair but the table shows additional synchonisers like $\sa{b}$ for the pair $(0,2)$, $\sa{baa}$ for the pair $(0,1)$ and $\sa{ba}$ for the pair $(1,2)$.
$w$ | $\sa{b}$ | $\sa{a}$ | $\sa{a}$ | $\sa{b}$ | ||||||
$g_w(0)$ | = | 0 | $\leftarrow$ | 2 | $\leftarrow$ | 1 | $\leftarrow$ | 0 | $\leftarrow$ | 0 |
$g_w(1)$ | = | 0 | $\leftarrow$ | 0 | $\leftarrow$ | 2 | $\leftarrow$ | 1 | $\leftarrow$ | 1 |
$g_w(2)$ | = | 0 | $\leftarrow$ | 2 | $\leftarrow$ | 1 | $\leftarrow$ | 0 | $\leftarrow$ | 2 |
Problem 123: Synchronising Words |
The problem deals with the synchronisation of the composition of functions from $I_n=\{0,1,\dots,n-1\}$ to itself for a positive integer $n$. For two such functions $f_{\sa{a}}$ and $f_{\sa{b}}$, a word $w\in\{\sa{a},\sa{b}\}^+$ encodes their composition $f_w = h_0\circ h_1\circ\cdots\circ h_{|w|-1}$ when $h_i=f_{\sa{a}}$ if $w[i]=\sa{a}$ and $h_i=f_{\sa{b}}$ if $w[i]=\sa{b}$. (Note functions $h_i$ are applied to elements of $I_n$ in decreasing order of $i$ as usual, i.e. from right to left according to $w$.) The word $w$ is said to be {\it synchronising} if the set $f_w(I_n)$ contains a single element.
A useful notion is that of a pair synchroniser: a word $u\in\{\sa{a},\sa{b}\}^*$ is a synchroniser of the pair $(i,j)$, $i,j\in I_n$, if $f_u(i)=f_u(j)$.
For any positive integer $n$ the word $w=\sa{b}(\sa{a}^{n-1}\sa{b})^{n-2}$ is a synchronising word of the functions $g_{\sa{a}}$ and $g_{\sa{b}}$. It is more difficult to see that it is a shortest such word in this particular case. This yields a quadratic lower bound on the length of a synchronising words.