GO Classes Test Series 2024 | Mock GATE | Test 11 | Question: 62

Answer 1

A short cut method to eliminate the options is to observe only ‘c’ can lead us to final state and also on encountering 1st ‘c’ we are getting into final state 4 from state 3 , but if we again encounter ‘c’ i.e 2 times , we get away from final state ( back to 3).
That means this will accept strings with odd c’s , which is generated by c(cc)*.

Thus option A , is correct solution.

Answer 2

By using a transition table we can draw the given finite automata which look like the below:

The above FSM accepts strings like $abbc,abbbc,abbccc,abbccccc,...$ one occurrence of $a$, followed by two or more numbers of $b’s$, followed by an odd number of c’s.

Option D) it is wrong because it generates strings of ending even number of c’s like $cc,cccc,cccccc,..$

Option C) It can generate $abbbcc$ which is rejected by the above FA.

Option B) It can generate one or more occurrences of c’s same as the above reason.

Option A) it is true. RE is $abb^*bc(cc)^*$. as we know that $r.r^*=r^+$ after simplifying the reg expression we get $abb^+c(cc)^*$

Option $(A)$ is correct.