Question

4. (15) Show that EQDFA E P.

Answer 1

SOLUTION-

EQ_DFA is the decidability problem to check whether two DFAs accept the same language or not and so it can be represented as {<A,B> | A,B are DFA, and L(A) = L(B)}.

Let us consider A, B be two DFAs and start by constructing the Symmetric difference of A and B (let's call it C).

so,$L(C) = (L(A) \cap \overline{L(B)}) \cup (\overline{L(A)} \cap L(B))$

So, we can say thet C is DFA ,as DFA is closed under union, intersection and complement

Construction of C can be finished in polynomial time as all of union, intersection and Complement of a DFA can be constructed in polynomial time.

Verifying L(A) = L(B) is same as verifying L(C) = $\phi$. The below TM decides whether a language is empty or not in polynomial time as follows:

Consider a Turing Machine T = On input <C> where C is a DFA.

1. Mark the start state of C
2. Mark all states that have a transition coming into it from a state that is already marked
3. Repeat step 2 until no new state is marked
4. If no final state of A is marked, accept; otherwise, reject

Therefore we get to decide EQ_DFA in polynomial time. Therefore EQ_DFA is in P.

IF YOU HAVE ANY DOUBT PLEASE COMMENT DOWN BELOW I WILL SOLVE IT FOR YOU:)
----------------PLEASE RATE THE ANSWER-----------THANK YOU!!!!!!!!----------