+1 vote
129 views
$L=\{x^n:n \in N\} \cup\{x^ny^n|n \in N\}$

This language does not have LL(k) parser while being deterministic context free.Why?
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0
is this DCFL ?
0
yes. If we assume end delimiter is there. i.e., all strings end with a delimiter like '$'. 0 If we assume end delimiter is there then it is DCFL, in that case it should have LL(k) for some k right sir ? then there is no meaning in this question ! +2 No. DCFL is guaranteed to have only LR(1) not necessarily LL(k) 0 if it is LL(k) for some k, then it is LR(m) for some m. but converse not true ! Forgot that basic point, thanks sir !! +1 @Arjun sir and @Shaik Masthan-I am trying to build a proof point out if anywhere I go wrong. The grammar can be$S->A|B\\A->xA|x\\B->xBy|xy$For string let's say$w_1=x^3$we need to use production$S->A$and for string$w_2=x^3y^3$we need to use the production$S->B$So here we are in a dilemma which one to use. Similarly, for a given string of length say$n\$, we are not sure how much of lookahead we need for each string so that production to be used at each step can be uniquely determined.

So, the given language does not have an LL(K) parser for any fixed k.

Am I going in the correct direction?

0
yes, intuitively that is correct 👍 Similar to why LR(k) is more powerful than LL(k).
0
sir, can you show in formal way, why it doesn't have any LL(k) parser?
+1

Till DCFL there exist a LR(K) parser.

by Boss (36.9k points)

+1 vote