Consider the NPDA Q q0 q1 q2 0 1 0 1 q0 F q2 gate cse 2015

The Eduladder is a community of students, teachers, and programmers just interested to make you pass any exams. So we solve previous year question papers for you.
See Our team
Wondering how we keep quality?
Got unsolved questions?

Ask Questions
gate-cse-2015-->View question

Consider the NPDA 〈Q = {q0, q1, q2}, Σ = {0, 1}, Γ = {0, 1, ⊥}, δ, q0, ⊥, F = {q2}〉 -gate-cse-2015

 where (as per usual convention) Q is the set of states, Σ is the input alphabet, Γ is stack alphabet, δ is the state transition function, q0 is the initial state, ⊥ is the initial stack symbol, and F is the set of accepting states, The state transition is as follows:

 Which one of the following sequences must follow the string 101100 so that the overall string is accepted by the automaton?


Taged users:

Be first to like this question

Be first to dislike this question

Talk about thisDelete|Like|Dislike|


Answer: (B) Explanation: In q0 state for ‘1’, a ‘1’ is pushed and for a ‘0’ a ‘0’ is pushed. In q1 state, for a ‘0’ a ‘1’ is popped and for a ‘1’ a ‘0’ is popped. So, the given PDA is accepting all strings of of the form x0x’r or x1x’r or xx’r, where x’r is the reverse of the 1’s complement of x. The given string 101100 has 6 letters and we are given 5 letter strings. So, x0 is done, with x = 10110. So, x’r = (01001)r = 10010. Hence option B is correct.

Be first to like this answer

Be first to dislike this answer
Talk about this|Once you have earned teacher badge you can edit this questionDelete|Like|Dislike|

Can you help us to add better answer here? Please see this

Not the answer you're looking for? Browse other questions from this Question paper or ask your own question.

Join eduladder!