From starting state q0 , the last state will be q0 only if there is repeated sequence of ab so that on input a, transition from q0 to q1 will happen and from input b, transition from q1 to q0 will happen.
Hence, regular expression R0 for the given DFA such that last state should be q0 is given by R0 = (ab)* which means any possible repetition of sequence ab. Hence
State q1 is reached, only when previous state was q0 and input a is supplied. Hence regular expression for R1 can be obtained from regular expression for R0 by inserting a at the end. Hence regular expression for R1 = (ab)*a i.e.
State q2 is reached either when input b is given to state q0 or input b is given to state q3 . State q3 is reached when either input a is given to q1 or input a is given to q2 .
Regular expression for q2 is R2 = (ab)*b + (ab)*aa(ba)*b .
Here, (ab)*b is the path to reach q2 from q0 and (ab)*aa(ba)*b is the path to reach q2 from q0 to q1 then from q1 to q3 and then q3 to q2 .
Regular expression for q3 is R3 = b(ab)*a + (ab)*aa(ba)*
Please comment for any clarification.
Consider the finite automaton M = (Q,{a, b},8,90,F) defined by the following illustration. -0.00 7 92...
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that h(mn ) h ( m)n, h ( ) and that if m < n then h ( m ) < n ( n ) = . Exercise 2.7.4. [Used in Theorem 2.7.1.] Complete the missing part of Step 3 of the proof of Theorem 2.7.1. That is, prove that k is surjective. Exercise 2.7.5. [Used in Theorem 2.7.1.] Let Ri and R2 be ordered fields that satisf We were unable to transcribe this imageWe were unable to transcribe this...
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