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A1 - Assignment 1 solution set Jalaj Upadhyay 1 Question 1...

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Assignment 1: solution set Jalaj Upadhyay 1 Question 1 Let M := (Σ , Q, S, F, δ ) be an automaton defined over the alphabet set Σ using the transition table δ , where S is the set of the starting states, Q is the set of all states, and F is the set of accepting states. For this question, we assume Σ = { 0 , 1 } . (a) Q := { q 0 , q 1 } , S := { q 0 } and F = Q . δ ( q 0 , 0) = q 0 , δ ( q 0 , 1) = q 1 , and δ ( q 1 , 1) = q 1 . In order to prove the correctness of the above NFA, we prove both the direction. First, to see that all words in L are accepted by L ( M ) , note that we can break L as a union of 0 + 1 + 0 + 1 + . The first part of this union is trivial, the second part is accepted by F = q 0 , the third part by a transition from q 0 to q 1 and then staying thereafter in q 1 . For the last part, we first loop in q 0 and then make a transition of the previous form on seeing the first 1 . To show the converse, we prove the contrapositive. Any words in the complement of L has a substring of the form 10 . Now, notice that once the NFA is fed in 1 , it moves to q 1 , hence forth any encounter with 0 will result in crashing of NFA. Hence, if w ¯ L , it is not accepted by M , proving the result. (b) Q := { q 0 , q 1 , q 2 , q 3 , q 4 , q 5 } , S := { q 0 } and F = { q 2 , q 5 } . The transition table is as follows: δ ( q 0 , 0) = q 1 δ ( q 0 , 1) = q 3 δ ( q 1 , 0) = q 2 δ ( q 1 , 1) = q 4 δ ( q 2 , 0) = q 2 δ ( q 2 , 0) = q 5 δ ( q 3 , 0) = q 4 δ ( q 4 , 0) = q 5 δ ( q 5 , 0) = q 5 . Note that there is encountering a 0 , we make a transition to the next state, except when we are in the accepting states. Also, when there is a 1 , we make a transition from q i to q i +3 for i = { 0 , 1 } . This will be useful in proving the correctness.

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A1 - Assignment 1 solution set Jalaj Upadhyay 1 Question 1...

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