# Regular Expression

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## Can I write a* + b* = (a + b)* ????

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NO  you cannot write
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To check equivalence of two regex create minimal DFA for both regex and if they are same then yes they are equivalent.
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sumit goyal 1 please explain the difference

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a* + b *  wil give { a , aa ,aaa ,aaaa ,......  , b,bb,bbb,bbbb,bbbb  , epsilon }

(a+b)* will give all possible combinations
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so how to create a DFA for this? a* + b *
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It is drawn like this

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sumit goyal 1 You forgot dead state.$4$ states will be there moreover first state is final

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i cretaed a nfa , so i didnot take dead state ,

NO,

we can write (a+b)*=(a*+b*)*

=(a*+b)*

=(a+b*)*

=(a*b*)*

=a*(ba*)*

=b*(ab*) *

or

make minimal dfa for both the RE and check LHS != RHS ,infact  language both are different

edited by
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$(a+b)^* \neq (ab^*)^*$

And last one should be b*(ab*)*

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yes..by mistake correct one is

(a+b)*=(a*b*)*

=b*(ab*)*

=a*(ba*)*

1 vote

$1^{st}$ one is RHS.
$2^{nd}$ one is LHS.
LHS $\neq$ RHS

No ,because both are not same as u can see by the 1st REstring produce like ab,aabb.a,b,aa,bbetc but u can't get string like aba,aaabbabab,ababa etc .

whereas by the 2nt RE u will get all string produce by a and b.

## Related questions

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486 views
Given answer is option c. Can anyone tell me how?
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.