CSL and Regular language

Question

if L1 = { aⁿbⁿcⁿ   | n>= 0 } 

and L2 = { aⁿb^mc^k | k,n,m>=0}

L1 is CSL and L2 is regular.

Now  L3 = L1.(L2)*.

Is L3 is regualar or CSL?

Comments

how L₂ could be regular?

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L2 should be dcfl ?

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@Purvi of course it is DCFL

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Then what should be the answer mam ?

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See this closure propertieshttp://gatecse.in/closure-property-of-language-families/

Now, L₂ is DCFL. But DCFL not closed under kleen star.

So, L₂ may or may not be DCFL.

Now, L₁.(L₂)* = it is a concatenation operation. And, CSL closed under concatenation . So, result will be CSL

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Edited the question.. Sorry for my mistake

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Mam if dcfl does not satisfy closure property then we will look for cfl or csl and how do we decide ? I thought we see the language just of the higher type..why are we not considering cfl ?

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yes..

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@srestha,after concatenation,does'nt the language become regular??

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sorry..my mistake..:-)it will not be regular

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Akriti, question edited here :)

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tell me one thing,if it was not a keen closure then would L3 be regular?

i mean like this

L3 =L1XL2

CROSS PEODUCT instead of concatenation

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It's I think L3 is regular

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@Akriti how do u find cross product in TOC?

It is only in DBMS. symbols are different in each subject :)

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Yes there exists concatenation but no cross product in toc

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so sorry,i mixed them..:-P

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Is L2 is regular?

Answer 1

Regular

L1 = { $\epsilon$, abc, aabbcc, ... }

L2 = a*b*c*

L3 is L1.(L2)*, means $a^nb^nc^n(a^*b^*c^*)^*$

The important thing to notice is that n can be 0. So L3 will be  $(a^*b^*c^*)^*$. Which is regular.

Comments

@Archies09 Oh I see it now. Sorry. They are same :)

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So , shouldn't the answer be both ?

Every Regular Language is a CSL right.

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@Harsh That's correct. Every regular language is CSL. So both are correct. Though regular language answer is more specific.

Answer 2

R1(R2)* is simply (a+b+c)*....and hence regular

Answer 3

L1 is CSL

and L2 is regular.

then (L₂)* will be reguler (by using closure properties)

and now L₁.(L₂)*=CSL.Reguler(push up) 

CSL.CSL=CSL

Comments

correct i have also same explanation..