Question

Can anyone tell whether "aabbaaabbb" is accepted by (a^n b^n)^* or not??

Comments

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".

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Yes it will be accepted by (a^n b^n)*

See entire discussion of this question-

https://gateoverflow.in/254165/toc-regular-langauges

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shubham shastri 'solution  is correct

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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 :)

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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?

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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

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Gurdeep saini, can you explain a little more.

I'm not getting what are u saying..both cases seems to be same

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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

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@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* = L_new.  then

L_new it self have kleene closure then it is also not correct by your statement

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@srestha   and @MiNiPanda  and  @Magma

i am inviting you, to join in the discussion

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Shaik Masthan

what r u trying to say in above comment???

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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

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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 ....

 

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yes  @minipanda is right

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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

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@manas 

(aⁿbⁿ )* ,n>=0   is {∊,ab,abab,ababab.......,aabb,aabbaabb,aabbaabbaabb...........,aaabbb,aaabbbaaabbb,aaabbbaaabbbaaabbb.....}

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MiNiPanda great explanation !

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@gurdeep yes u r right

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Thank you @Magma :P

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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

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Yes according to me

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@ shaik

yes this is the final answer

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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

 

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Thank you @MINIPanda, @shaik, @srestha @shubham

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what is answer given?

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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

 

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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?

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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

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comment written by @srestha mam is right 

and language taken by mam is (aⁿ bⁿ)* ;n>=0

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Yes you are right but constraints on n is not mentioned by questioner...

It is already flagged as 'edit necessary'