Question

How to prove that the grammar G is not LL(1)

The grammar is:

L => LE | E
E => (C) | (F) | V | T
C => ifEE | ifEEE
F => +L | -L | *L | printL
V => a | b | c | d
T => 0 | 1 | 2 | 3

Answer 1

The given grammar

L => LE | E
E => (C) | (F) | V | T
C => ifEE | ifEEE
F => +L | -L | *L | printL
V => a | b | c | d
T => 0 | 1 | 2 | 3

In this grammar there is left recursion in production

L => LE | E

Since there is left recursion the grammar G is not LL(1)

To prove the grammar G is not LL(1) we need to check for above two conditions

In Grammar G

First(E)={(,a,b,c,d,0,1,2,3}

First(LE)={First(E)}

={(,a,b,c,d,0,1,2,3}

It is clearly visible that condition 1 is violated

ie)

There for all terminals 0,1,2,3,a,b,c,d

(L,0) , (L,1), (L,2),....,(L,d)

there will be two entries in predicative parsing table

L->LE and L->E

So the given grammar G is not LL(1)