search
Log In
0 votes
327 views

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

in Theory of Computation
retagged by
327 views
0
NO  you cannot write
0
To check equivalence of two regex create minimal DFA for both regex and if they are same then yes they are equivalent.
0

sumit goyal 1 please explain the difference 

1
a* + b *  wil give { a , aa ,aaa ,aaaa ,......  , b,bb,bbb,bbbb,bbbb  , epsilon }

(a+b)* will give all possible combinations
0
so how to create a DFA for this? a* + b *
0

It is drawn like this

0

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

0
i cretaed a nfa , so i didnot take dead state ,

4 Answers

2 votes

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
0

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

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

 

0

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

0 votes
0 votes

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

0 votes
1 answer
2
125 views
The answer given is none of these !! I think 1’st statement is the correct one, but still want to confirm
asked Jan 16, 2019 in Theory of Computation Nandkishor3939 125 views
0 votes
1 answer
3
1.1k views
which one of the following regular expression describe the language over {a,b} consist of no pair of consecutive a’s? a. (b*abb*) (a+€) b. (b+ab)* (a+€) c. (b*abb*)*(a+€)+b* d. (b*ab*)*(a+€)+b*(a+€)
asked Dec 28, 2018 in Theory of Computation Ram Swaroop 1.1k views
0 votes
0 answers
4
296 views
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.
asked Oct 24, 2018 in Theory of Computation Swapnil Naik 296 views
...