Dfa to regular expression

Question

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.

Comments

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 

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@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 Shaik Masthan

brother

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That also given in Peter Linz book

Answer 1

(b+aa*b)*