Question

In a simple example, I’m confused about how to turn this grammar into a LL one by removing the left recursion. Any hints are welcome.

G = {
      A -> A a | A B | a
      B -> b
    }

I get the following by applying this algorithm:

G = {
      A -> a X
      X -> e | A | B X
      B -> b
    }

This seem to work to generate a C pseudo-code for the parser:

void A() {
    switch (token) {
        case 'a' : next(); X(); break;
    }
}

void X() {
    switch (token) {
        case 'e' : finish(); break;
        case 'a' : A(); break;
        case 'b' : B(); X(); break;
    }
}

void B() {
    next();
}

And to generate a parsing tree for the word: aabab:

A ---+
|    |
a    X
     |
     A ---+
     |    |
     a    X ---+
          |    |
          B    X
          |    |
          b    A ---+
               |    |
               a    X ---+
                    |    |
                    B    X
                    |    |
                    b    e

Well, I’m just not sure if it’s right…

Answer 1

First, your grammar seems to accept sequences of as separated by single bs, so that no 2 b come together. (aaa...abaaaaa...abaaa...a) This should be equivalent to something like:

Q -> P | P b P
P -> a | a P