Given two ordered sets of words $a_1, a_2, ..., a_k$, $b_1, b_2, ..., b_k$ taking values in some discrete alphabet $A$, a solution to the PCP problem is a sequence $i_1, ..., i_n$ taking values in $1, 2,..., k$ such that $a_{i_1}|a_{i_2}|...|a_{i_n}=b_{i_1}|b_{i_2}|...|b_{i_n}$ where $|$ means concatenation. $k$ can be called the length of the problem, $n$ the length of the solution and if we let $w$ be the length of the largest word in $a_1, a_2, ..., a_k, b_1, b_2, ..., b_k$, $w$ is called the width of the problem.

I know that the PCP problem becomes decidable in several scenarios, for instance: for bounded $n$, or if $A$ is unary, etc. On the other hand for $k\geq7$ PCP is still undecidable. My question is, is there any result known for bounded values of $w$?

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    $\begingroup$ You may be interested in this question. $\endgroup$
    – Raphael
    Commented Nov 13, 2013 at 22:10

1 Answer 1


If $w$ is bounded, and the alphabet size is bounded, then the number of possible words is bounded. Thus, there is only a finite number of possible instances, so the problem becomes a finite languages, and hence decidable.

If you don't bound the alphabet, then you can encode any width of $w$ by adding more letters, so the problem is the same as standard PCP, and hence undecidable.

  • $\begingroup$ Yes, maybe my argument is flawed. What I was thinking about is that still with $w$ fixed which limits the length of the ---input--- word and $k$ fixed; the number of words is unlimited since there are infinite ways of combining them. Note that $n$ would be unbounded... $\endgroup$
    – Al Learner
    Commented Oct 23, 2013 at 9:49
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    $\begingroup$ $n$ is unbounded, but that doesn't matter: suppose you only have 1000 possible instances of PCP. Then you have at most a 1000 "yes" instances, and the rest are "no", so the language is finite. $\endgroup$
    – Shaull
    Commented Oct 23, 2013 at 10:00

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