语言{0 ^ m 1 ^ m 2 ^ n |的结果是什么？ n> = 0，m> n}

Question

I would like to know how to get the productions for language {0^m 1^m 2^n | n>=0, m > n}.

This is what I had which I'm not sure if it's correct. Please correct me if I'm wrong:

 S -> 01A | 0B1A | 00B11A
 A -> 2A | 2 | λ
 B -> 01

Thank you.

Answer 1

This language is not context free. This can be shown using the pumping lemma for context-free languages. You end up with five cases; three of the cases pump only one of the symbols 0, 1 or 2; and two cases pump adjacent symbols. Note that you can almost do it except that we can choose the string 0^(p+1) 1^(p+1) 2^p and even when you span 0s and 1s and pump them evenly you still fail the m>n test when pumping down.

There are more powerful grammars than context free. We can produce a general grammar for this language.

S -> RT
R -> 0R1X | 01

X1 -> 1X

XT -> T2 | T
1T -> e

The first two productions produce strings of the form 0^n (01) (1X)^n T.

The third production produces strings of the form 0^n (01) 1^n X^n T. It allows the X's to "float" to the right past all the 1s.

The last two productions produce strings of the form 0^n (01) 1^n 2^m, m<=n. These allow the T to "float" left past the Xs, translating each X into either 2 or nothing.