这是语法LL（1）

Question

I have been asked to convert:

S → Sa | bSb | bc 

to LL(1) so far I have:

S → bY
Y → SbF | cF
F → aF | ε

Is this LL(1)? If not would this be LL(1):

S → bY
Y → bYbF | cF
F → aF | ε

if neither of these would somebody please give me the correct answer and why thanks in advance!

Answer 1

This is what I would do:

S → Sa | bSb | bc 

1. Remove left recursion:

    F -> aF | EPSILON``` 

2. Now left factor:

    F -> aF | EPSILON X -> SbF | cF``` 

3. Check the First and Follows:

    S: b X: b, c F: a, EPSILON``` ```Follows(): S: $, b X: $, b F: $, b``` Everything checks out so it is LL(1) parsable.