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| 98 views
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Is it option C?
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I think it should be C.
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doubt : In L1 , Lhas elements x. The relation is given as y = x? what will L1 have?

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L1 would have been regular if n had a fixed value. But it is not regular here.

L2 is regular because {Regular}* is regular

Option(C)

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Say n has fixed value in L. What strings will Lhave in it?

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Let's say n=4

L1: if after iterating x four times it belongs to the set A (which is regular)

If the condition is true it will be accepted else rejected
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For L1.

Suppose {x = 0i1j  / i==j and i,j>=0}

and {y = (oi1j)*/ i,j>=0} we are taking kleen closure because n>=0. and note that y is a complete language over 0 and 1, hence y contain kleen closure of x's string.

now here x is DCFL and y is regular.

there for L1 is not Regular.

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ohh :o

Okay so it's like .. x4 should also belong to A. Only then it'll be accepted. And only then that string is included in L1 ,right?

Shouldn't the language be

L1 = { y | ..... } ?

@srivivek95
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L1 is clearly mentioned to contain x.

L= { y | ..... } ?

is not true.

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Okay

## 1 Answer

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The answer should be C)

L1 is bot regular but L2 is regular.
by Junior (523 points)
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plz any one explain properly i m not geting how they both language are diffrent

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