# DFA of a string in which 3rd symbol from RHS is ‘a’

Prerequisite – Finite Automata Introduction, DFA of a string in which 2nd symbol from RHS is ‘a’

**Problem –** Draw deterministic finite automata (DFA) of the language containing the set of all strings over {a, b} in which 3rd symbol from RHS is ‘a’.

The strings in which 3rd last symbol is “a” are:

aaa, aab, aaab, aaaa, aabbaaa, bbbaba etc

The DFA which accept the language of such strings is similar the DFA which accept the regular expression:

L = (a+b)*.a.(a+b).(a+b)

**For example:**

Input:babaaOutput:NOT ACCEPTEDInput:aaabbOutput:ACCEPTED

Constructing the DFA of the given problem directly is very complicated. So, here we are going to design the non-deterministic finite automata (NFA) and then convert it to the deterministic finite automata (DFA).

The NFA of the language containing all the strings in which 3rd symbol from the RHS is “a” is:

Here, A is the initial state and D is the final state.

Now, we are going to construct the state transition table of the above NFA.

After that we will draw the state transition table of DFA using subset configuration on the state transition table of NFA. We will mention all the possible transition for a and b.

Now it’s become very easy to draw the DFA with the help of its transition table. In this DFA, we have eight different states A, AB, AC, AD, ABC, ABD, ACD and ABCD, where AD, ABD, ACD and ABCD are the final states and A is the initial state of the DFA.

This is our required DFA of the language containing the set of all strings over {a, b} in which 3rd symbol from RHS is ‘a’.