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Construct DFA's for the following languages:

1. $L=\left\{w \mid w \in \{a,b\}^*, \text{ w has baab as a substring } \right\}$
2. $L=\left\{w \mid w \in \{a,b\}^*, \text{ w has an odd number of a's and an odd number of b's } \right\}$

edited | 1k views
+1
what if the remaining input will stand or loop within the state....................ie instead of giving b as back input let its loop there itself...............and same do for all????????

DFA for A:

Part (B):

by Loyal (8.1k points)
edited
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How dfa A will accept baabbaab?? I think there should be transition from final stage to state 2???
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Observe, after "baab" , there is a loop (a+b)*, so after once "baab" occurred, it will accept anything. so yout string "baabbaab" will definitely be accepted.

DFA for (B)

by Boss (41.1k points)
edited
+1
q3 will be final.

q1 is for odd no of a's and even no of b's as it accepting a, bba, etc.
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Updated.

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