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[Solved]: Undecidability of the PCP problem with bounded width

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Problem Detail: 

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 $w$?

Asked By : Al Learner

Answered By : Shaull

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.

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Question Source : http://cs.stackexchange.com/questions/16357

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