1 votes 1 votes How to simplify a regular expression ? Please share a useful resource . Show that a+aa+b+bb* equal to $\phi$* (a+b*) Theory of Computation theory-of-computation regular-expression + – pC asked Jul 21, 2016 pC 2.5k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply LeenSharma commented Jul 21, 2016 reply Follow Share both are not equivalent. Regular expression a+aa+b+bb can generate string aa but regular expression $\phi$*(a+b*)can't able to generate to string aa.ϕ 0 votes 0 votes pC commented Jul 21, 2016 reply Follow Share Could u share some resources for simplifying RE ? as i dont know how to simplify them i cant understand if it were $ \phi $* (a+b)* would they be same ? 0 votes 0 votes pC commented Jul 22, 2016 reply Follow Share @LeenSharma SIr, can u pls show me how to simplify REa+aa+b+bb* 0 votes 0 votes LeenSharma commented Jul 22, 2016 reply Follow Share You can minimize b + bb* into b+ because both generating same strings. here b*={∈,b,bb,bbb.....} and b+={b,bb,bbb,.......} final regular expression will be a + aa + b+ or a + aa + bb* Here bb* ≡ b+ = b + bb* 4 votes 4 votes mcjoshi commented Oct 16, 2016 reply Follow Share @pC I think you can only minimize it to few steps like : $\color{olive}{a( \epsilon + a) + b( \epsilon +b^*)}$, not more. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes {a+aa+b+bb*} this generate the language like L={a,aa,b,bb,bbb,bbbb.....} so on but in next Ø*(a+b*), we know that Ø*=ε so language generate like{a,b,bb,bbb....} aa is not generated so it's not equal. Hira Thakur answered Aug 31, 2016 Hira Thakur comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes both the RE's are not equivallent L2 can produce a or b's or e and l1 can produce a,aa, b or any o of b's owais2580 answered Oct 9, 2016 owais2580 comment Share Follow See all 0 reply Please log in or register to add a comment.