# GATE | GATE-CS-2003 | Question 55

Consider the NFA M shown below.

Let the language accepted by M be L. Let L1 be the language accepted by the NFA M1, obtained by changing the accepting state of M to a non-accepting state and by changing the non-accepting state of M to accepting states. Which of the following statements is true ?**(A)** L1 = {0, 1}* – L**(B)** L1 = {0, 1}***(C)** L1 ⊆ L**(D)** L1 = L**Answer:** **(B)****Explanation:**

In case of a Deterministic Finite Automata (DFA) when we change the accepting states into non-accepting states and non-accepting states into accepting states, the new DFA obtained accepts the complement of the language accepted by the initial DFA. It is because we have one single movement for a particular input alphabet from one state so the strings accepted by the transformed DFA will be all those which are not accepted by the actual DFA.

But it is not the case with the NFA’s (Non-Deterministic Finite Automata). In case of NFA we need to have a check on the language accepted by the NFA. The NFA obtained by changing the accepting states to non-accepting states and non-accepting states to accepting states is as follows:-

Here we can see that as i. The initial state is an accepting state hence null string is always accepted by the NFA. ii. There is a movement from state 1 to state 2 on both {0, 1} input alphabets and further any number of 1’s and 0’s or even none in the string lets the string be at an accepting state(state 2). Hence the language accepted by the NFA can be any string with any combination of 0’s and 1’s including a null string i.e. {null, 0, 1, 00, 01, 10, 11,……………..} so L1= {0, 1}*.

This Explanation has been contributed by **Yashika Arora.**

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