Polynomial bound on sum of strings

I am struggling with the following problem:

Given a set of finite binary strings $S=\{s_1,\ldots,s_k\}$, we say that a string $u$ is a concatenation over $S$ if it is equal to $s_{i_{1}} s_{i_{2}} \cdots s_{i_{t}}$ for some indices $i_1,\ldots, i_t \in \{1,\ldots, k\}.$

Your friend is considering the following problem: given two sets of finite binary strings $A=\{a_1,\ldots,a_m\}$ and $B=\{b_1,\ldots , b_n \}$, 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 $\mathcal{NP}$, since I would just have to exhibit such a string $u$ in order to prove the answer is yes." You point out that this an inadequate explanation. How do we know the shortest string $u$ doesn't have length exponential in the size of the input?

Prove the following: 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 $A\cup B$.

Now, I have actually found a solution and the solution claims that the maximum length of $u$ is at most $n^2L^2$ where we assume $m \leq n$ and $L$ denotes the maximum length of any string in $A \cup B$. The solution then goes on to exhibit a proof by contradiction using the pigeonhole principle to show that if we assume $u$ has length greater than $n^2L^2$ we arrive at a contradiction.

My question is where does the bound $n^2L^2$ come from? I know that $n$ is the number of strings in set $B$ and $L$ is the max length string in $A \cup B$. But I feel as if the largest possible concatenation over both would be more like $2nL$ since we would need to include every string in both $A$ and $B$ and the longest string in either is $L$ and there are at most $n$ strings in each. What am I missing? I think I can handle the remainder of the proof once I understand where the $n^2L^2$ bound is coming from.

• It would be easier to read and understand your post if you used Latex typeset instead of plain text. – fade2black Oct 18 '17 at 21:26
• Presumably it comes from the proof. To answer that question, you'll need to understand the proof. Have you tried understanding the proof? The proof starts by assuming $u$ has length greater than $n^2 L^2$; see where in the proof that assumption is used, and probably you'll have the answer to your question. – D.W. Oct 18 '17 at 22:58

You can also prove this using automata theory. For a set $S = \{s_1,\ldots,s_k\}$ of strings over $\Sigma$, consider the following language over $\Sigma \cup \{1,\ldots,k\}$: $$L_S = (1s_1+2s_2+\cdots+ks_k)^+.$$ This language is accepted by a DFA of size roughly $\|S\| := |s_1|+\cdots+|s_k|$. The exact size depends on the model of DFA - in our case it makes sense to ask for at most one transition out of each state with a given label (rather than exactly one transition, as is usually the case).

Now we want to do the same, but for two sets of strings $A = \{a_1,\ldots,a_m\}$, $B = \{b_1,\ldots,b_n\}$. To this end, we consider languages $L_A,L_B$ which are defined as before, but with two differences:

• $1,\ldots,k$ are replaced with $A_1,\ldots,A_m$ for $L_A$ and $B_1,\ldots,B_n$ for $L_B$.
• The symbols $A_1,\ldots,A_m$ are ignored in $L_B$, and the symbols $B_1,\ldots,B_n$ are ignored in $L_A$.

As an example, suppose that $A = \{a,ba\}$ and that $B = \{ab,a\}$. Then the following word will be in our language: $A_1B_1aA_2bB_2a$.

As before, we can construct DFAs for $L_A,L_B$ having roughly $\|A\|,\|B|$ states, respectively. Using the product construction, we get a DFA of size roughly $\|A\|\cdot\|B\|$ for the intersection. It is well-known (and can be proved using the pigeonhole principle) that if a DFA having $N$ states accepts any word, then it accepts some word of size less than $N$. This implies your claim.

In terms of lower bounds, it is easy to refute your conjecture that $2nL$ is enough. Consider $A = \{ 0^L \}$ and $B = \{ 0^{L-1} \}$. It is not hard to check that the minimal solution has length $L(L-1) \approx L^2$.

It's less clear what is the correct dependence on $n$.