I'm only just starting to learn about the PCP, and I'm especially interested in the Bounded version. From Wikipedia:

One of the most important variants of PCP is the bounded Post correspondence problem, which asks if we can find a match using no more than k tiles, including repeated tiles.

I am not sure I understand what Wiki means by a 'tile' in this context.

I do understand that a typical PCP problem takes as input two equal-length lists of nonempty strings, each of which contains some number of words. An example is this:

M = (abb, aa, aaa) and N = (bba, aaa, aa)

In terms of the Bounded PCP, what would the integer K be here? Is it the length an individual word can be (which in this case is maxed at 3)? Is it the length of a single list?

Hope this is clear. I appreciate any help profusely in advance.

  • $\begingroup$ You've flipped around the problem. You've given as input a pair of lists, but it takes as input a list of pairs. $\endgroup$ Nov 14, 2016 at 4:47

1 Answer 1


Suppose your list is M=(bba,ab,bba,a) N=(bb,aa,bb,baa) Then its equivalent tile representation is as follows

enter image description here

Now your pcp problem is equivalent to finding a sequence of tiles(repetition allowed) such that concatenating all upper halves and concatenating all lower halves gives you same string. K here is nothing but number of tiles that can be used

Edit : As Yuval Filmus subtly mentions in his comment, there are many other interpretations of K. In an another formulation K can also be the maximum length of concatenated string, in which case the bounded PCP problem is Pspace complete. But usually many books refer to K as being number of tiles that are allowed.Please refer to Sipser book for more.

  • $\begingroup$ There might be several different definitions of bounded, see for example this answer. In particular, the Wikipedia definition seems to be different. $\endgroup$ Nov 14, 2016 at 8:12
  • $\begingroup$ I would say something like "finding a multiset of tiles" instead, to stress you can repeat the same tile. If you can't then the problem becomes decidable. $\endgroup$
    – chi
    Nov 14, 2016 at 8:50
  • $\begingroup$ sequence would be much better representation to insist on the order . Anyways thanks for pointing out :) $\endgroup$ Nov 14, 2016 at 9:29
  • $\begingroup$ Thank you for the answer. For this version of the Bounded PCP, the question is whether we can find a matching solution using no more than k tiles, as opposed to the general PCP which allows an infinite supply of such tiles, right? $\endgroup$ Nov 14, 2016 at 15:19
  • $\begingroup$ @Fiery Phoenix Yes thats true $\endgroup$ Nov 15, 2016 at 11:26

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