GATE CSE 2015 Set 1 | Question: 52

Question

Consider the DFAs $M$ and $N$ given above. The number of states in a minimal DFA that accept the language $L(M) \cap L(N)$ is_____________.

Comments

Simply the first DFA will accept strings ending with a  and Coming to second DFA it will accept Strings ending with b nowhere in DFA  there could be strings possible ending with both a and b so the language that is under intersection would be null so just simply construct a dfa  with single state but should not accept any string so no final state

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@Kabir5454 @gatecse

Number of states in NFA???

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Isn't every dfa is also a nfa 🤷

Answer 1

Different Approach to solve such questions
1)identify meaning of both DFA and determine intersection of both.
2)make compound table for both then make dfa. 

for more - Good Read
3)determine Regular expression by dissolving all states one by one in both DFA & then find intersection of both.

Answer 2

M accepts the strings which end with a and N accepts the strings which end with B. Their intersection should accept empty language.

Answer 3

A picture is worth a thousand words

Another masterpiece answer

Thanks

Shashank

 

Answer 4

 

For this only 1 state is possible.