Question

1. Using the given details of a PDA with Q = {90, 91, 92, 93}; { = {a, b}; Z = 0; F = {q3} S = {90} and the following transitions construct a PDA. Show all the stack operations for the string "aaaabbbba" and tell whether the string is acceptable or not. (90, a, 0) = (91, 1) (q0, b, 0) = (q3, 1) (91, a, 1) = (92, 1) 5 (91, b, 1) = (92, 1) 5 (92, a, 1) = (92, 1) (92, b, 1) = (q3, 1) (q3, b, 1) = (q3, A) 5 (93, a, 0) = (92, 1) (5 (90, b, 0) = (93,1 ) ► top of the stack is 0, then pop) 2. Consider the given PDA and check if it accepts 0001110. 0, €/ q1 q2 0, € /0 1, 1/6 94 93 1,1/ € 0, € /o

Answer 1

Solution:

The solution is explained in the screenshot attached below.

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PDA is ansbructed bollows. (ao/1) (511) 23 az (5.1) mes String is aaaabbbba 10-12 D bbbb and then 0 final state, The state is a. It is not So the string is not accepted. no but not a The final state is alz

string i 000 illo Move from ايه ar 000 -> move bom al 2 When comes mes the from to 3 ar at and moves pop stack from Move a3 to a3 The final move from alui state. The string is accepted. au is final