0 votes 0 votes Can I write $a^* + b^* = (a + b)^*$ ???? Theory of Computation regular-expression theory-of-computation + – mrinmoyh asked Jan 31, 2018 retagged Jun 20, 2019 by Cristine mrinmoyh 876 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply sumit goyal 1 commented Jan 31, 2018 reply Follow Share NO you cannot write 0 votes 0 votes Mk Utkarsh commented Jan 31, 2018 reply Follow Share To check equivalence of two regex create minimal DFA for both regex and if they are same then yes they are equivalent. 0 votes 0 votes Mk Utkarsh commented Jan 31, 2018 reply Follow Share sumit goyal 1 please explain the difference 0 votes 0 votes sumit goyal 1 commented Jan 31, 2018 reply Follow Share a* + b * wil give { a , aa ,aaa ,aaaa ,...... , b,bb,bbb,bbbb,bbbb , epsilon } (a+b)* will give all possible combinations 1 votes 1 votes Mk Utkarsh commented Jan 31, 2018 reply Follow Share so how to create a DFA for this? a* + b * 0 votes 0 votes sumit goyal 1 commented Jan 31, 2018 i edited by sumit goyal 1 Jan 31, 2018 reply Follow Share It is drawn like this 0 votes 0 votes Inspiron commented Jan 31, 2018 reply Follow Share @ sumit goyal 1 You forgot dead state.$4$ states will be there moreover first state is final 0 votes 0 votes sumit goyal 1 commented Jan 31, 2018 reply Follow Share i cretaed a nfa , so i didnot take dead state , 0 votes 0 votes Please log in or register to add a comment.
2 votes 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 Ravijha answered Jun 21, 2019 edited Jun 21, 2019 by Ravijha Ravijha comment Share Follow See all 2 Comments See all 2 2 Comments reply Verma Ashish commented Jun 21, 2019 i edited by Verma Ashish Jun 21, 2019 reply Follow Share $(a+b)^* \neq (ab^*)^*$ And last one should be b*(ab*)* 0 votes 0 votes Ravijha commented Jun 21, 2019 reply Follow Share yes..by mistake correct one is (a+b)*=(a*b*)* =b*(ab*)* =a*(ba*)* 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes $1^{st}$ one is RHS. $2^{nd}$ one is LHS. LHS $\neq$ RHS Inspiron answered Jan 31, 2018 Inspiron comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes @MRINMOY_HALDER check this https://gateoverflow.in/169116/regular-expression Sahil1994 answered Jan 31, 2018 Sahil1994 comment Share Follow See all 0 reply Please log in or register to add a comment.
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. Poonam Gupta 1 answered Feb 13, 2018 Poonam Gupta 1 comment Share Follow See all 0 reply Please log in or register to add a comment.