Explore topic-wise MCQs in Automata Theory.

This section includes 1 Mcqs, each offering curated multiple-choice questions to sharpen your Automata Theory knowledge and support exam preparation. Choose a topic below to get started.

1.

Let u=’1101’, v=’0001’, then uv=11010001 and vu= 00011101.Using the given information what is the identity element for the string?a) u-1b) v-1c) u-1v-1d) εView Answer a) 0101011b) 0101010c) 010100d) 100001View Answer 4.Predict the following step in the given bunch of steps which accepts a strings which is of even length and has a prefix=’01’δ (q0, ε) =q0 < δ(q0,0) =δ (δ (q0, ε),0) =δ(q0,0) =q1 < _______________

A. u-1b) v-1c) u-1v-1d) εView Answer a) 0101011b) 0101010c) 010100d) 100001View Answer 4.Predict the following step in the given bunch of steps which accepts a strings which is of even length and has a prefix=’01’δ (q0, ε) =q0 < δ(q0,0) =δ (δ (q0, ε),0) =δ(q0,0) =q1 < _______________a) δ (q0, 011) =δ (δ (q0,1), 1) =δ (q2, 1) =q3
B. v-1c) u-1v-1d) εView Answer a) 0101011b) 0101010c) 010100d) 100001View Answer 4.Predict the following step in the given bunch of steps which accepts a strings which is of even length and has a prefix=’01’δ (q0, ε) =q0 < δ(q0,0) =δ (δ (q0, ε),0) =δ(q0,0) =q1 < _______________a) δ (q0, 011) =δ (δ (q0,1), 1) =δ (q2, 1) =q3b) δ (q0, 01) =δ (δ (q0, 0), 1) = δ (q1, 1) =q2
C. u-1v-1d) εView Answer a) 0101011b) 0101010c) 010100d) 100001View Answer 4.Predict the following step in the given bunch of steps which accepts a strings which is of even length and has a prefix=’01’δ (q0, ε) =q0 < δ(q0,0) =δ (δ (q0, ε),0) =δ(q0,0) =q1 < _______________a) δ (q0, 011) =δ (δ (q0,1), 1) =δ (q2, 1) =q3b) δ (q0, 01) =δ (δ (q0, 0), 1) = δ (q1, 1) =q2c) δ (q0, 011) =δ (δ (q01, 1), 1) =δ (q2, 0) =q3
D. εView Answer a) 0101011b) 0101010c) 010100d) 100001View Answer 4.Predict the following step in the given bunch of steps which accepts a strings which is of even length and has a prefix=’01’δ (q0, ε) =q0 < δ(q0,0) =δ (δ (q0, ε),0) =δ(q0,0) =q1 < _______________a) δ (q0, 011) =δ (δ (q0,1), 1) =δ (q2, 1) =q3b) δ (q0, 01) =δ (δ (q0, 0), 1) = δ (q1, 1) =q2c) δ (q0, 011) =δ (δ (q01, 1), 1) =δ (q2, 0) =q3d) δ (q0, 0111) =δ (δ (q0, 011), 0) = δ (q3, 1) =q2View Answer
Answer» E.
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