这是语法 {a^2n b^n | n >= 0} 是否上下文无关？

Question

这个语法上下文是否免费？

{a^2n b^n | n >= 0}

Answer 1

The object you're asking about is not a grammar but a language: a set of strings. A set of strings can be potentially described by a context-free grammar, in the sense that a some grammar exists whose rules generate exactly that set of strings.

In this case, the expansion of your set looks like this:

{ "", "aab", "aaaabb", "aaaaaabbb", ... }

It's the set of all strings consisting of an even number of a -s, followed by half as many b -s.

This set can be generated by repetitions of the following rules:

1. The empty string is in the set.
2. If a string s is in the set, then the string "aa" + s + "b" is in the set.

That is, from any string which is in the set, we can form a new string which is also in the set, simply by adding two a -s on the left and a b on the right. Together with the base case rule that the empty string is in the set, these rules describe the entire set.

And these rules are equivalent to, and can be written in the form of, a context-free grammar:

s -> ""
  -> "aa" s "b"

The grammar is context free because it only has rules of the form "a -> symbols ...": the left hand side only has a single non-terminal symbol. So none of the generations depend on context: a single symbol is taken, and replaced in isolation according to some available rule.