0 0 votes Can I write $a^* + b^* = (a + b)^*$ ???? Theory of Computation regular-expression theory-of-computation + – mrinmoyh 2.1k views answer comment Share Follow Print See all 8 Comments 8 8 Comments reply sumit goyal 1 commented Jan 31, 2018 reply Follow flag NO you cannot write 0 0 replyShare Mk Utkarsh commented Jan 31, 2018 reply Follow flag To check equivalence of two regex create minimal DFA for both regex and if they are same then yes they are equivalent. 0 0 replyShare Mk Utkarsh commented Jan 31, 2018 reply Follow flag sumit goyal 1 please explain the difference 0 0 replyShare sumit goyal 1 commented Jan 31, 2018 reply Follow flag a* + b * wil give { a , aa ,aaa ,aaaa ,...... , b,bb,bbb,bbbb,bbbb , epsilon } (a+b)* will give all possible combinations 1 1 replyShare Mk Utkarsh commented Jan 31, 2018 reply Follow flag so how to create a DFA for this? a* + b * 0 0 replyShare sumit goyal 1 commented Jan 31, 2018 i edited by sumit goyal 1 Jan 31, 2018 reply Follow flag It is drawn like this 0 0 replyShare Inspiron commented Jan 31, 2018 reply Follow flag @ sumit goyal 1 You forgot dead state.$4$ states will be there moreover first state is final 0 0 replyShare sumit goyal 1 commented Jan 31, 2018 reply Follow flag i cretaed a nfa , so i didnot take dead state , 0 0 replyShare Please log in or register to add a comment.
2 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 2 2 Comments reply Verma Ashish commented Jun 21, 2019 i edited by Verma Ashish Jun 21, 2019 reply Follow flag $(a+b)^* \neq (ab^*)^*$ And last one should be b*(ab*)* 0 0 replyShare Ravijha commented Jun 21, 2019 reply Follow flag yes..by mistake correct one is (a+b)*=(a*b*)* =b*(ab*)* =a*(ba*)* 0 0 replyShare Please log in or register to add a comment.
1 1 vote $1^{st}$ one is RHS. $2^{nd}$ one is LHS. LHS $\neq$ RHS Inspiron answered Jan 31, 2018 Inspiron comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes @MRINMOY_HALDER check this https://gateoverflow.in/169116/regular-expression Sahil1994 answered Jan 31, 2018 Sahil1994 comment Share Follow 0 reply Please log in or register to add a comment.
0 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 0 reply Please log in or register to add a comment.