I have no clue how to prove this question.

Consider the language

$L = \{ \langle D_1, D_2, ... ,D_K \rangle : k \in {N},$ the $D_i$ are DFAs and ${\bigcap}_{i=1}^k L(D_i) = \emptyset \}$

Prove that L is NP-hard.

Can someone guide me through this problem? I understand that I will have to reduce it into something, I just don't know what to reduce to.

  • 1
    $\begingroup$ Hint: the intersection operation is very similar to the AND operation... taking the intersection of $k$ DFAs could be paralleled to taking the conjunction of $k$ boolean expressions. I haven't worked out the reduction in full, but it was the first thing I noticed when seeing the problem. I assume you're familiar with 3-SAT? $\endgroup$ – jmite Nov 26 '13 at 0:42
  • $\begingroup$ Yes, that is the direction I am currently pursuing. But I wasn't sure whether I can get the expected proof. $\endgroup$ – vanblaze Nov 26 '13 at 1:38

Here are a few hints (the basic idea was in the comment of jmite). Try to reduce from CNF-SAT

  1. You have to encode a truth assignment as a word. Say the word $\tt 101001001$ encodes $x_1=\text{TRUE}$, $x_2=\text{FALSE}$, and so on
  2. Try to construct a DEA that accepts all words (truth assignments) that satisfy a clause. Yo know that the truth values come in order, so yo can decide check every literal one by one.
  3. The intersection of the DEAs for the clauses accept the truth assignments that satisfy all clauses. So if there exists some truth assignment this language wont be empty.
  4. You are left with a formal proof, and you have to check that the list of DEAs is not too long. In particular you have to check if the reduction is polytime.

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