Dec 2012 paper III answer

We can disambiguate by using a similar approach to the dangling else, and
decide that each b should be associated with the nearest a. This means that
the expansion within an ab pair should always be balanced. This leads to
the following grammar:

S -> a S | S1 S | epsilon
S1 -> a S1 S1 b | epsilon

It is easy to verify that this generates the same strings as the original
grammar, and the parse tree is always unique, because one b is always associated
with the most recent a.

Note that the answer is not necessarily unique. If the grammar is ambiguous,
it means that we get to choose between possible parses, and each choice is
in a sense a different language. For example, given the ambiguous grammar
for expressions:

  E -> E + E | E * E | id

We say that the unambiguous grammar we want is:

  E -> E + T | T,  T -> T * T | id

because it gives us the proper precedence between the two operators. But that
choice is in no way mandated by the grammar. We could just as well choose:

  E -> E * T | T , T -> T + T | id

which generates the same strings, but gives the opposite precedence to
operators.