can removing left recursion introduce ambiguity?

Question

Let's assume we have the following CFG G:

A -> A b A
A -> a

Which should produce the strings a, aba, ababa, abababa, and so on. Now I want to remove the left recursion to make it suitable for predictive parsing. The dragon book gives the following rule to remove immediate left recursions. Given

A -> Aa | b

rewrite as

A  -> b A'
A' -> a A'
    | ε

If we simply apply the rule to the grammar from above, we get grammar G':

A  -> a A'
A' -> b A A'
    | ε

Which looks good to me, but apparently this grammar is not LL(1), because of some ambiguity. I get the following First/Follow sets:

First(A) = { "a" }
First(A') = { ε, "b" }
Follow(A) = { $, "b" }
Follow(A') = { $, "b" }

From which I construct the parsing table

    |   a          |   b            |  $           |
----------------------------------------------------
A   | A -> a A'    |                |              |
A'  |              | A' -> b A A'   | A' -> ε      |
    |              | A' -> ε        |              |

and there is a conflict in T[A',b], so the grammar isn't LL(1), although there are no left recursions any more and there are also no common prefixes of the productions so that it would require left factoring.

I'm not completely sure where the ambiguity comes from. I guess that during parsing the stack would fill with S'. And you can basically remove it (reduce to epsilon), if it isn't needed any more. I think this is the case if another S' is below on on the stack.

I think the LL(1) grammar G'' that I try to get from the original one would be:

A  -> a A'
A' -> b A
    | ε

Am I missing something? Did I do anything wrong?

Is there a more general procedure for removing left recursion that considers this edge case? If I want to automatically remove left recursions I should be able to handle this, right?

Is the second grammar G' a LL(k) grammar for some k > 1?

Answer 1

The original grammar was ambiguous, so it is not surprising that the new grammar is also ambiguous.

Consider the string a b a b a. We can derive this in two ways from the original grammar:

A ⇒ A b A
  ⇒ A b a
  ⇒ A b A b a
  ⇒ A b a b a
  ⇒ a b a b a

A ⇒ A b A
  ⇒ A b A b A
  ⇒ A b A b a
  ⇒ A b a b a
  ⇒ a b a b a

Unambiguous grammars are, of course possible. Here are right- and left-recursive versions:

A ⇒ a              A ⇒ a
A ⇒ a b A          A ⇒ A b a

(Although these represent the same language, they have different parses: the right-recursive version associates to the right, while the left-recursive one associates to the left.)

Answer 2

Removing left recursion cannot introduce ambiguity. This kind of transformation preserves ambiguity. If the CFG is already ambiguous, the result will be ambiguous too, and if the original is not, the resulting neither.