So, it's known that PCP is undecidable even when we fix the number of tiles to $n \geq 7$.

I'm wondering, can anything similar be said for when there is a fixed word length?

To be precise, here's the problem:

Given fixed $m$ and $n$, with $n \geq 7$, and words $u_1, \ldots u_n$ and $v_1 \ldots v_n$ such that $|u_i| \leq m$ and $|v_i| \leq m$, is there an index sequence $i_1, \ldots i_k$ such that $u_{i_1} \cdots u_{i_k} = v_{i_1} \cdots v_{i_k}$.

For what values of $m$, if any, is this known to be undecidable?

Note that this is similar to this question, but none of the 8 linked papers seemed by their titles to answer my question, and I haven't fully read all 8 of them yet.

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    $\begingroup$ There is another parameter, the target alphabet size. This has to be unbounded since otherwise your problem is trivially decidable. $\endgroup$ Jul 29, 2014 at 22:47

1 Answer 1


For all $m \geq 3$, the problem is undecidable.

Proof by reduction from the word problem of unrestricted grammars:

  1. Take an arbitrary formal grammar. W.l.o.g. all left and right sides of rules have length at most $3$.

    This can be seen by translating any grammar into an equivalent TM and then converting back.

  2. Map the resulting grammar to PCP instances; no tile is longer than the longest left or right side of a rule.

    That is, with step 1, all tiles have length $\geq 3$.

  • $\begingroup$ That is: bound on world length $m$, but no bound on number of words $n$. The question seems to look for bounds on both, but see comment by Yuval. (nevertheless +1) $\endgroup$ Jul 30, 2014 at 9:39
  • $\begingroup$ @HendrikJan Hm, true, missed that ("fixed $n$"). I wonder how to go about that; maybe via a small universal TM? $\endgroup$
    – Raphael
    Jul 30, 2014 at 9:42
  • $\begingroup$ Originally I was looking for fixed $m$ and $n$, but this is still a very interesting result regardless! $\endgroup$
    – jmite
    Jul 30, 2014 at 17:53
  • $\begingroup$ Actually, now that I think about it, it's really weird if you have fixed $m$ and $n$ since technically then you're dealing with a finite, therefore regular, language. There probably is no algorithm which can generate a description of that language (RE, DFA etc.) since you'd need to solve each PCP instance, but there is a finite number of instances with fixed $m$ and $n$. $\endgroup$
    – jmite
    Jul 30, 2014 at 20:12
  • 1
    $\begingroup$ @jmite If you bound $\Sigma$, yes. (As Yuval noted.) $\endgroup$
    – Raphael
    Jul 30, 2014 at 20:39

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