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Problem 57: Avoidability of Binary Words

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Some patterns occur in all long enough words. They are said to be unavoidable. The notion obviously depends on the alphabet size and in the problem we consider binary patterns.

A word $w$ is said to avoid a set $X$ of words if no factor of $w$ belongs to $X$. The set $X$ is said to be avoidable if there is an infinite word avoiding it, or equivalently on a finite alphabet, if there are infinitely any words avoiding it. The goal is to test if a set of words drawn from the binary alphabet $\Bi=\{\sa{a},\sa{b}\}$ is avoidable.

To design the test we define two reductions on a set $X\subseteq\Bi^+$.

reduce1 (remove super-word):
If $x,y\in X$ and $x$ is a factor of $y$ remove $y$ from $X$.
reduce2 (chop last letter):
If $x$ is a suffix of $y\neq\varepsilon$ and $x\bar{a},ya\in X$ substitute $y$ for $ya$ in $X$ (the bar morphism exchanges $\sa{a}$ and $\sa{b}$).
Avoidable$(X \textrm{ non-empty set of binary words})$ 
    \begin{algorithmic}
  \WHILE{$\mbox{reduce1 or reduce2 are applicable to } X$}
    \STATE $X\leftarrow \mbox{reduce1}(X)$ \OR $\mbox{reduce2}(X)$
  \ENDWHILE
  \IF{$X\ne \{\tt{a},\tt{b}\}$}
    \RETURN{True}
  \ELSE
    \RETURN{False}
  \ENDIF
    \end{algorithmic}

Show that a set $X$ of binary words is avoidable if and only if Avoidable$(X)$=True.

Show that a set $X\subseteq\Bi^{\leq n}$ is avoidable if and only if it is avoided by a word of length larger than $2^{n-1}+n-2$ and that the bound is optimal.

References

  • M. Crochemore, M. Lerest, and P. Wender. An optimal test on finite unavoidable sets of words. Inf. Process. Lett., 16(4):179-180, 1983.
  • M. Lothaire. Combinatorics on Words. Addison-Wesley, 1983. Reprinted in 1997.