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L={a^n \ n>=0}

M={b^n \ n>=0}

L.M is a regular language and the DFA for this is going to be ending with b and epsilon and it will have two states

Am I correct or not
asked in Theory of Computation by (275 points) | 56 views
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It is a DFA we need a dead state .Total 3 states
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plzzz upload ur DFA which u hav created and plzzz check the concatination properly this language does not going to dead state it is the starting state will also be the final state and the language is

L={epsilon,b,bb,bbb,bbbb        ab,abb,aaab,aaaab,aab,,,,,,,,,,}
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It is a DFA we need a dead state .Total 3 states.Please see this :

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the language which is getting accepted is 

L={epsilon,b,bb,bbb,bbbb        ab,abb,aaab,aaaab,aab,,,,,,,,,,}

ur language is also accepting this language

L(a,aa,aaa,aaaaa,aaaaa,,aaaaa,,,,,,,,,,,,) in the above question this language is not getting accepted

plzz see this DFA this will the DFA of the language  concatination of Language L and language M

it is not necessary that every DFA should always contain Dead state

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L(a,aa,aaa,aaaaa,aaaaa,,aaaaa,,,,,,,,,,,,) in the above question this language is getting accepted.

Because M can but null.

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