How Can I demonstrate this grammar is not ambiguous?

Question

I know I need to show that there is no string that I can get using only left most operation that will lead to two different parsing trees. But how can I do it? I know there is not a simple way of doing it, but since this exercise is on the compilers Dragon book, then I am pretty sure there is a way of showing (no need to be a formal proof, just justfy why) it. The Gramar is:

S-> SS* | SS+ | a

What this grammar represents is another way of simple arithmetic(I do not remember the name if this technique of anyone knows, please tell me ) : normal sum arithmetic has the form a+a, and this just represents another way of summing and multiplying. So aa+ also mean a+a, aaa*+ is a*a+a and so on

Answer 1

The easiest way to prove that a CFG is unambiguous is to construct an unambiguous parser. If the grammar is LR(k) or LL(k) and you know the value of k , then that is straightforward.

This particular grammar is LR(0), so the parser construction is almost trivial; you should be able to do it on a single sheet of paper (which is worth doing before you try to look up the answer.

The intuition is simple: every production ends with a different terminal symbol, and those terminal symbols appear nowhere else in the grammar. So when you read a symbol, you know precisely which production to use to reduce; there is only one which can apply, and there is no left-hand side you can shift into.

If you invert the grammar to produce Polish (or Łukasiewicz) notation, then you get a trivial LL grammar. Again the parsing algorithm is obvious, since every right hand side starts with a unique terminal, so there is only one prediction which can be made:

S → * S S | + S S | a

So that's also unambiguous. But the infix grammar is ambiguous:

S → S * S | S + S | a

The easiest way to provide ambiguity is to find a sentence which has two parses; one such sentence in this case is:

a + a + a

Answer 2

I think the example string aa actually shows what you need. Can it not be parsed as:

S => SS* => aa OR S => SS+ => aa