ace test series

Question

L1=0*1*0*

L2=0^n1*0^n

L=L1 INTERSECTION L2

THEN L IS

A.DCFL BUT NOT CFL

B.CFL BUT NOT CSL

C.DCFL BUT NOT CSL

D.NONE OF THE ABOVE

Comments

L = L1 intersection L2 = 0ⁿ1*0ⁿ , which is DCFL

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THE DOUBT IS  THAT .......I DID  LIKE L1 IS REGULAR AND L2 IS DCFL SO THEIR INTERSECTION WILL BE DCFL (i.e language is closed under regular intersection) hence dcfl so also cfl so also csl ....so answer should be D but they answered option A ...........so how to choose according to their intention.

Answer 1

L1 = {epsilon, 0,00,000.....,1,11,111......01,001......010,00100,000100......}

L2=0^n1*0^n | n>=0
 

L2={epsilon, 00, 010,00100......}

L=L1 INTERSECTION L2 = {0^n1*0^n |n>=0}

L is DCFL

Option D is correct.

Comments

HOW L2 IS DCFL