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I tried applying a method where we write equation as state with incoming transition on given dfa -

q1 = $\epsilon$ + bq1+bq2

q2 = aq1+aq2

but then able to reach till this conclusion only:

q2 = ab*+aq2+abq2b*

How to solve further?

Here q1 is a start state.

| 153 views
+2

the way of writing eqn is wrong...

the eqn R = Q+RP, ===> R = QP* when P is free from ϵ

Q1 = ϵ + Q1 b + Q2 b  ====> Q1 = ( ϵ + Q2 b ) b*

Q2 = Q1.a + Q2.a = ( ϵ + Q2 b ) b* a + Q2 . a = b* a + Q2 . b . b* . a + Q2 . a

= b* a + Q2 . ( b . b* . a + a ) = b* a + Q2 . ( b+ a + a ) = b* a + Q2 . ( b+ + ϵ ) a = b* a + Q2 . ( b* ) a .

∴ Q2 = b* a + Q2 . ( b* a ) ====> Q2 = b* a ( b*  a )*

only Q2 is the final state, RE = b* a ( b*  a )*

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if it is multiple choice question, just check the each RE can deployed by the given FA or not?
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yeah that was the reason I wasn't able to take common Q2 outside. Thanks Shaik
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Kindly Tell me where I went wrong . Thanks

+2

@Pawan Kumar 2

you got b* a ( a + $b^{+}$ a)* = b* a ( a + $b^{+}$ a)* = b* a (  ( ∈ + $b^{+}$ ) a )* = b* a ( $b^{*}$a )*

Both are equivalent

Note that, For a language may be more than one RE exist !

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Thanks for confirming

brother

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That also given in Peter Linz book