2 votes 2 votes Vïšhàl Kúmãr 3 asked Oct 27, 2018 Vïšhàl Kúmãr 3 1.1k views answer comment Share Follow See all 30 Comments See all 30 30 Comments reply SHUBHAM SHASTRI commented Oct 27, 2018 reply Follow Share no it will not.. Our language is (a^nb^n)^* So, put * =2 so we have (a^nb^n)(a^nb^n) Now we cant define N different for each term ...N can have only one value so , if N=2 we will get "aabbaabb" while if N=3 we will get "aaabbbaaabbb" ...but we will never get "aabbaaabbb". 1 votes 1 votes Verma Ashish commented Oct 27, 2018 reply Follow Share Yes it will be accepted by (a^n b^n)* See entire discussion of this question- https://gateoverflow.in/254165/toc-regular-langauges 0 votes 0 votes Gurdeep Saini commented Oct 27, 2018 i reshown by Gurdeep Saini Oct 27, 2018 reply Follow Share shubham shastri 'solution is correct 0 votes 0 votes Verma Ashish commented Oct 27, 2018 reply Follow Share a^n b^n is accepted by dpda and language is dcfl. And dcfls are not closed under kleens closure But the expression (a^n b^n)* definitely generate aabbaaabbb :) 0 votes 0 votes Swapnil Naik commented Oct 27, 2018 reply Follow Share for L= a+b taking Kleene closure L'= {$\epsilon$,(a+b),(a+b)(a+b),(a+b)(a+b)(a+b),(a+b)(a+b)(a+b)(a+b),.....} L' ={$\epsilon$,a,b,aa,bb,ab,ba,aab,aaa,........} so likewise L = anbn n>0 L = ab+aabb+aaabb+........ its kleene closure will be L' = {$\epsilon$,(ab,aabb,aaabb,....),((ab,aabb,aaabbb,....)(ab,aabb,aaabbb,....)),((ab,aabb,aaabb,....)(ab,aabb,aaabb,....)(ab,aabb,aaabb,....)),.........} so you can get here aabbaaabbb Is this approach correct? 0 votes 0 votes Gurdeep Saini commented Oct 27, 2018 reply Follow Share let suppose if L = anbn ,n>=0 and L2 = kleene closure of L then aabbaaabbb is accepted by L2 will be yes but in the given question language is (anbn)* which is not same as the kleene closure of L so the answer will be no think about it both case is different 1 votes 1 votes Verma Ashish commented Oct 27, 2018 reply Follow Share Gurdeep saini, can you explain a little more. I'm not getting what are u saying..both cases seems to be same 0 votes 0 votes Gurdeep Saini commented Oct 27, 2018 reply Follow Share first consider this example if L= 0n1| n>=0 then kleene closure of this language is (0*1)* not the (0n1)* both are different @verma if L = anbn ,n>=0 and L2 = kleene closure of L then i ask you write the expression for L ? now you can get me 0 votes 0 votes Shaik Masthan commented Oct 27, 2018 reply Follow Share @Gurdeep Saini what you are saying is if it is L and then L* ===> it is correct but if it is directly the language is kleene closure then it is not correct.. right ? from the first line, just assume L* = Lnew. then Lnew it self have kleene closure then it is also not correct by your statement 0 votes 0 votes Shaik Masthan commented Oct 27, 2018 reply Follow Share @srestha and @MiNiPanda and @Magma i am inviting you, to join in the discussion 0 votes 0 votes akash.dinkar12 commented Oct 27, 2018 reply Follow Share Shaik Masthan what r u trying to say in above comment??? 0 votes 0 votes Shaik Masthan commented Oct 27, 2018 reply Follow Share from my first comment... i am trying to say, what gurdeep said is wrong. from my second comment, i am pinging those people to join in this discussion 0 votes 0 votes MiNiPanda commented Oct 27, 2018 reply Follow Share It's hard to find mistake in Swapnil's approach.. But I feel the only thing that can make any difference b/w these two cases: Case 1: L= anbn ,n>=0 L1={∊,ab,aabb,aaabbb...} L2=L1*= {∊,ab,aabb,aaabbb...}.{∊,ab,aabb,aaabbb...}.{∊,ab,aabb,aaabbb...}.... from here we can get aabbaaabbb Case 2: L3=(anbn )*, n>=0 The difference b/w L2 and L3 is : L2=(anbn , n>=0)* and L3=(anbn )*, n>=0 L2 includes the all the possible n's in one instance and that is concatenated with other instances. In L3, it is like (∊)* when n=0 (ab)* when n=1 (aabb)* when n=2 .... 1 votes 1 votes Gurdeep Saini commented Oct 27, 2018 reply Follow Share yes @minipanda is right 0 votes 0 votes Manas Mishra commented Oct 27, 2018 reply Follow Share The definition of Kleene star on V is that means for (anbn )* ,n>=0 it should be (a^0b^0)U(a^1b^1)U(a^2b^2)........ = {∊,ab,aabb,aaabbb.....} so i think aabbaaabbb can't be generated using this 0 votes 0 votes Gurdeep Saini commented Oct 27, 2018 reply Follow Share @manas (anbn )* ,n>=0 is {∊,ab,abab,ababab.......,aabb,aabbaabb,aabbaabbaabb...........,aaabbb,aaabbbaaabbb,aaabbbaaabbbaaabbb.....} 1 votes 1 votes Magma commented Oct 27, 2018 reply Follow Share MiNiPanda great explanation ! 0 votes 0 votes Manas Mishra commented Oct 27, 2018 reply Follow Share @gurdeep yes u r right 0 votes 0 votes MiNiPanda commented Oct 27, 2018 reply Follow Share Thank you @Magma :P 0 votes 0 votes Shaik Masthan commented Oct 27, 2018 reply Follow Share So the conclusion is, i will write L2 = (L1)* where L1 = { an.bn | n ≥ 0 } ---------> accepted aabbaaabbb i can write L3 = { ( an . bn )m | n ≥ 0, m ≥ 0 } ---------> can't accepted aabbaaabbb 2 votes 2 votes MiNiPanda commented Oct 27, 2018 reply Follow Share Yes according to me 0 votes 0 votes Gurdeep Saini commented Oct 27, 2018 reply Follow Share @ shaik yes this is the final answer 0 votes 0 votes srestha commented Oct 27, 2018 reply Follow Share I think 1st ans by Shubham Sastri is correct, because n value can never be changed u can relate this question https://gateoverflow.in/219274/find-the-language 0 votes 0 votes Swapnil Naik commented Oct 27, 2018 reply Follow Share Thank you @MINIPanda, @shaik, @srestha @shubham 1 votes 1 votes srestha commented Oct 29, 2018 reply Follow Share what is answer given? 0 votes 0 votes Mk Utkarsh commented Oct 29, 2018 reply Follow Share It will be accepted if kleen star is on definition on language https://math.stackexchange.com/questions/1812064/possible-strings-of-kleene-star-of-l-anbnn%E2%89%A51 if not then not accepted 0 votes 0 votes srestha commented Oct 29, 2018 reply Follow Share Actually string will be like this $\left \{ \epsilon +a^{n}b^{n}+\left ( a^{n}b^{n} \right )\left ( a^{n}b^{n} \right )+\left ( a^{n}b^{n} \right )\left ( a^{n}b^{n} \right ) \left ( a^{n}b^{n} \right )+....\right \}$ now take any value of n like $n=3$ , $n$ cannot generate above string. And also in one string we cannot take different different value of $n$ right? 0 votes 0 votes Verma Ashish commented Oct 29, 2018 reply Follow Share It all depend on where n is defined... Means whether it is (${a^n}$ ${b^n}$ ; n>0)* or it is (${a^n}$ ${b^n}$)* ;n>0 As it is mentioned in one of the comment by MiniPanda 0 votes 0 votes Gurdeep Saini commented Oct 29, 2018 reply Follow Share comment written by @srestha mam is right and language taken by mam is (an bn)* ;n>=0 0 votes 0 votes Verma Ashish commented Oct 29, 2018 reply Follow Share Yes you are right but constraints on n is not mentioned by questioner... It is already flagged as 'edit necessary' 0 votes 0 votes Please log in or register to add a comment.