这是LL（1）语法吗？

Question

I have derived the following grammar:

S -> a | aT
T -> b | bR
R -> cb | cbR

I understand that in order for a grammar to be LL(1) it has to be non-ambiguous and right-recursive. The problem is that I do not fully understand the concept of left-recursive and right-recursive grammars. I do not know whether or not the following grammar is right recursive. I would really appreciate a simple explanation of the concept of left-recursive and right-recursive grammars, and if my grammar is LL(1).

Many thanks.

Answer 1

This grammar is not LL(1). In an LL(1) parser, it should always be possible to determine which production to use next based on the current nonterminal symbol and the next token of the input.

Let's look at this production, for example:

S → a | aT

Now, suppose that I told you that the current nonterminal symbol is S and the next symbol of input was an a. Could you determine which production to use? Unfortunately, without more context, you couldn't do so: perhaps you're suppose to use S → a, and perhaps you're supposed to use S → aT. Using similar reasoning, you can see that all the other productions have similar problems.

This doesn't have anything to do with left or right recursion, but rather the fact that no two productions for the same nonterminal in an LL(1) grammar can have a nonempty common prefix. In fact, a simple heuristic for checking if a grammar is not LL(1) is to see if you can find two production rules like this.

Hope this helps!

Answer 2

The grammar has only a single recursive rule: the last one where R is the symbol on the left, and also appears on the right. It is right-recursive because in the grammar rule, R is the rightmost symbol. The rule refers to R, and that reference is rightmost.

The language is LL(1). How we know this is that we can easily construct a recursive descent parser that uses no backtracking and at most one token of lookahead.

But such a parser would be based on a slightly modified version of the grammar.

For instance the two productions: S -> a and S -> a T could be merged into a single one that can be expressed by the EBNF S -> a [ T ] . ( S derives a , followed by optional T ). This rule can be handled by a single parsing function for recognizing S .

The function matches a and then looks for the optional T , which would be indicated by the next input symbol being b .

We can write an LL(1) grammar for this, along these lines:

S -> a T_opt
T_opt -> b R_opt
T_opt -> <empty>
... et cetera

The optionality of T is handled explicitly, by making T (which we rename to T_opt ) capable of deriving the empty string, and then condensing to a single rule for S , so that we don't have two phrases that both start with a .

So in summary, the language is LL(1), but the given grammar for it isn't. Since the language is LL(1) it is possible to find another grammar which is LL(1), and that grammar is not far off from the given one.