627 views
1 votes
1 votes
An $\epsilon$ free LL(1) grammar is also a SLR(1) and hence LALR(1) and LR(1) too.

Is this statement true ?

Please log in or register to answer this question.

Related questions

0 votes
0 votes
2 answers
1
0 votes
0 votes
1 answer
2
Ana_101 asked 3 days ago
37 views
S - A BA - f S fA - b b B dB - ƐB - cFirst(S) =First(A) =First(B) =Follow(S) =Follow(A) =Follow(B) =
2 votes
2 votes
2 answers
3
Subhrangsu asked Apr 16, 2022
9,410 views
Is the following grammar LL(1) ?S→ aABbCD | ϵA→ ASd | ϵB→ Sac| hC | ϵC→ Sf |CgD→ aBD | ϵDraw the LL(1) parsing table for the given grammar.
0 votes
0 votes
1 answer
4
Subhrangsu asked Apr 15, 2022
2,261 views
Compute FIRST and FOLLOW sets:S→ aAC | bBA→ Abc| Abd | eB→ f | gC→ h | i