Why is {a^nb^n | n >= 0} not regular?

Question

In a CS course I'm taking there is an example of a language that is not regular:

{a^nb^n | n >= 0}

I can understand that it is not regular since no Finite State Automaton/Machine can be written that validates and accepts this input since it lacks a memory component. (Please correct me if I'm wrong)

The wikipedia entry on Regular Language also lists this example, but does not provide a (mathematical) proof why it is not regular.

Can anyone enlighten me on this and provide proof for this, or point me too a good resource?

Answer 1

What you're looking for is Pumping lemma for regular languages.

Here is an example with your exact problem:

Examples:
Let L = {a^mb^m | m ≥ 1}.
Then L is not regular.
Proof: Let n be as in Pumping Lemma.
Let w = aⁿbⁿ.
Let w = xyz be as in Pumping Lemma.
Thus, xy²z ∈ L, however, xy²z contains more a’s than b’s.

Answer 2

Because you can't write a finite state machine that will 'count' identical sequences of 'a' and 'b' symbols. In a nutshell, FSMs cannot 'count'. Try imagining such a FSM: how many states would you give to symbol 'a'? How many to 'b'? What if your input sequence has more?

Note that if you had n <= X with X an integer value you could prepare such FSM (by having one with a lots of states, but still a finite number); such language would be regular.

Answer 3

Finite State Automaton has no data structure (stack) - memory as in case of push down automaton. Yeah it can give you some 'a's followed by some 'b's but not exact amount of 'a' followed by that no 'b'.