是L = {a ^ nb ^ m | n> m}常规或不规则的语言？

Question

I have troubles in solving/proving this problem. Any ideas please?

Answer 1

L = {a ⁿ b ^m | n > m} is not regular language.

Yes, the problem is tricky at first few try and deserve vote-up.

Pumping Lemma a necessary property of regular language is tool for formal proof that language is not regular language.

Formal definition: Pumping lemma for regular languages

Let L be a regular language. Then there exists an integer p ≥ 1 depending only on L such that every string w in L of length at least p ( p is called the "pumping length") can be written as w = xyz (ie, w can be divided into three sub-strings), satisfying the following conditions:

1. | y | ≥ 1
2. | xy | ≤ p
3. for all i ≥ 0, xy ⁱ z ∈ L

Suppose, if you choose string W = a ⁿ b ^m where (n + m) ≥ p and n > m + 1 . Choice of W is valid but this choice you are not able to show that language is not regular language. Because with this W you always have at-least one choice of y=a to pump new strings in language by repeating a for all values of i (for i =0 and i > 1) .

Before I writing my solution to proof the language is not regular. Please understand following points and notice: I made bold every string w and all i in formal definition of pumping lemma above.

• Although with Some Sufficiently large W in language you are able to generate new string in Language but NOT possible WITH ALL ! There are many possible choices for W (below in my proof) with that you can't find any choice of y to generate new string in language for all i >=0 . So because every Sufficiently large W not able to generate new string in language hence language is NOT regular.

read: what pumping lemma formal definition says

Proof: using pumping lemma

Step (1): Choose string W = a ⁿ b ^m where (n + m) ≥ p and n = m + 1 .

Is this choice of W is valid according to pumping lemma?

Yes, such W is in language because number of a = n > number of b =m . W is in language and sufficiently large >= p .

Step (2): Now chose a y to generate new string for all i >= 0 .

And no choice is possible for y this time! Why?

First , it is essay to understand that we can't have b symbol in y because it will either generate new strings out of pattern or in resultant string total number of b will be more than total number of a symbols.

Second , we can't chose y = some a 's because with i=0 you would get a new string in which number of a s will be less then number b s that is not possible in language.( remember number of a in W was just one more then b so removing any a means in resultant string N(a)=N(b) that is not acceptable because n>m )

So in we could find some W those are sufficiently large but using that we can't generate new string in language that contradict with pumping lemma property of regular language hence then language {a ⁿ b ^m | n > m} is not a regular language indeed.