GATE | GATE CS 2010 | Question 65

The grammar S → aSa | bS | c is
(A) LL(1) but not LR(1)
(B) LR(1)but not LR(1)
(C) Both LL(1)and LR(1)
(D) Neither LL(1)nor LR(1)


Answer: (C)

Explanation:

First(aSa) = a
First(bS) = b
First(c) = c
All are mutually disjoint i.e no common terminal 
between them, the given grammar is LL(1).

As the grammar is LL(1) so it will also be LR(1) as LR parsers are
more powerful then LL(1) parsers. and all LL(1) grammar are also LR(1)
So option C is correct. 

Below are more details.

A grammar is LL(1) if it is possible to choose the next production by looking at only the next token in the input string.

Formally, grammar G is LL(1) if and only if 
   For all productions  A → α1 |  α2 | ... | αn, 
      First(αi) ∩ First(αj) =  ∅, 1 ≤ i,j  ≤ n,  i ≠ j.
   For every non-terminal A such that First(A) contains ε, 
      First(A) ∩ Follow(A) = ∅ 

Source: https://s3-ap-southeast-1.amazonaws.com/erbuc/files/4147_870334aa-9922-4f78-9c2c-713e7a7f0d53.pdf

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