Given a set of finite binary strings S = {s1,...,sk}, we say that a string u is a concatenation over S if it is equal to si1si2 ... sit for some indices
it ε{1, ...,k}.
A friend of yours is considering the following problem: Given two sets of finite binary strings, A = {a1,..., am} and B = {b1,..., bn}, does there exist any string u so that u is both a concatenation over A and a concatenation over B?
Your friend announces, "At least the problem is in NP, since I would just have to exhibit such a string u in order to prove the answer is yes." You point out (politely, of course) that this is a completely inadequate explanation; how do we know that the shortest such string u doesn t have length exponential in the size of the input, in which case it would not be a polynomial-size certificate?
However, it turns out that this claim can be turned into a proof of membership in NP. Specifically, prove the following statement.
If there is a string u that is a concatenation over both A and B, then there is such a string whose length is bounded by a polynomial in the sum of the lengths of the strings in A U B.
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