The Art Gallery Guardian

Shortest string distinguishing two regular languages


Let D_0 and D_1 be two DFAs with n and m states, such that L(D_0)\neq L(D_1).

There exist a string of length at most n+m in the symmetric difference of L(D_0) and L(D_1).

We construct the following DFA D from D_0 and D_1. The start state s has a transition \delta(s,0) = s_0 and \delta(s,1) = s_1, where s_0 and s_1 are start state of D_0 and D_1.

Now D represents the language \set{0w|w \in L(D_0)} \cup \set{1w|w\in L(D_1)}. This language has at most n+m+1 equivalent classes in Myhill–Nerode theorem. There exist a string of length less than n+m+1 that differentiate state s_0 and s_1, since s_0 and s_1 can't correspond to the same equivalent class.

Posted by Chao Xu on 2016-03-10.
Tags: DFA.