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+1 vote

but what if it is a DFA accepting suppose (ab)* but max such that the number of a nd b's is 10

then how come it accepts sigma*

then how come it accepts sigma*

In case of (ab)* all states are not final state. In question they have mentioned if all states are final state.

okay then assume the dfa is accepting atmost 2 a's

in that case all states are final ststes right

but it isnt accepting sigma*

in that case all states are final ststes right

but it isnt accepting sigma*

0 votes

yes it is true because in dfa we show transition for every symbol on each state .if we have a dfa which accepts all strings (0+1)* for a given input symbols {0,1,2} then for any transition with 2 as the input symbol will lead to a dead state / trap state.when we make all the states as accepting states we include dead state also otherwise it will not be (sigma)^(*).

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