含糊不清的语法？

Question

hi there is this question in the book that said

Given this grammer

A --> AA | (A) | epsilon

a- what it generates\\

b- show that is ambiguous

now the answers that i think of is

a- adjecent paranthesis

b- it generates diffrent parse tree so its abmbiguous and i did a draw showing two scenarios .

is this right or there is a better answer ?

Answer 1

a is almost correct.
Grammar really generates () , ()() , ()()() , … sequences.
But due to second rule it can generate (()) , ()((())) , etc.

b is not correct.
This grammar is ambiguous due ot immediate left recursion: A → AA .

How to avoid left recursion: one , two .

Answer 2

a) Nearly right...

This grammar generates exactly the set of strings composed of balanced parenthesis. To see why is that so, let's try to make a quick demonstration.

First: Everything that goes out of your grammar is a balanced parenthesis string. Why?, simple induction:

• Epsilon is a balanced (empty) parenthesis string.
• if A is a balanced parenthesis string, the (A) is also balanced.
• if A1 and A2 are balanced, so is A1A2 (I'm using too different identifiers just to make explicit the fact that A -> AA doesn't necessary produces the same for each A).

Second: Every set of balanced string is produced by your grammar. Let's do it by induction on the size of the string.

• If the string is zero-sized, it must be Epsilon.
• If not, then being N the size of the string and M the length of the shortest prefix that is balanced (note that the rest of the string is also balanced):
  • If M = N then you can produce that string with (A).
  • If M < N the you can produce it with A -> AA, the first M characters with the first A and last N - M with the last A. In either case, you have to produce a string shorter than N characters, so by induction you can do that. QED.

For example: (()())(())

We can generate this string using exactly the idea of the demonstration.

A -> AA -> (A)A -> (AA)A -> ((A)(A))A -> (()())A -> (()())(A) -> (()())((A)) -> (()())(())

b) Of course left and right recursion is enough to say it's ambiguous, but to see why specially this grammar is ambiguous, follow the same idea for the demonstration:

It is ambiguous because you don't need to take the shortest balanced prefix. You could take the longest balanced (or in general any balanced prefix) that is not the size of the string and the demonstration (and generation) would follow the same process.

Ex: (())()()

You can chose A -> AA and generate with the first A the (()) substring, or the (())() substring.

Answer 3

Yes you are right.

That is what ambigious grammar means.

the problem with mbigious grammars is that if you are writing a compiler, and you want to identify each token in certain line of code (or something like that), then ambigiouity wil inerrupt you in identifying as you will have "two explainations" to that line of code.

Answer 4

It sounds like your approach for part B is correct, showing two independent derivations for the same string in the languages defined by the grammar.

However, I think your answer to part A needs a little work. Clearly you can use the second clause recursively to obtain strings like (((((epsilon))))), but there are other types of derivations possible using the first clause and second clause together.