Pumping lemma for regular language

Question

I have a little confusion in checking whether the given language is regular or not using pumping lemma.

Suppose we have to check whether:

L. The language accepting even number of 0's in regular or not?

We know that it is regular because we can construct a DFA for L. But I want to prove this with pumping lemma.

Now suppose, I take a String w= "0000":

Now will divide the string as x = 0, y = 0, and z = 00. Now on applying pumping lemma for i = 2, I will get the string "00000", which is not present in my language so by pumping lemma its prove that the language is not regular. But it is accepted by DFA ?

Any help will be greatly appreciated
Thank you

Answer 1

You are not completely clear about pumping lemma.

What pumping lemma say:

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 (i.e., w can be divided into three substrings), satisfying the following conditions:

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

But what this statement says is that:

If a language is really a regular language then there must be some way to generate(pump) new strings from all sufficiently large strings.

1. Sufficiently large string means, a string in language that is of the length ≥ P.
   _{So it may not be possible to generate new string from small strings even if language is Regular Language}

2. Some way means, if language is really a regular and our choice of w is correct. Then there should be at lest one way to break w in three parts xyz such that by repeating(pumping) y for any number of times we can generate new strings in the language.
   _{correct choice of w means: w in language and sufficiently large ≥ P}

note: in second point, there may be a chance that even if you breaks w correctly into xyz according to formal definition still some new generated strings are not in language. As you did.

And in this situation you are to retry with some other possible choice of y.

In you chosen string w = "0000" you can break w such that y = 00. And with this choice of y you would always find a new generated string in in Language that is "even number of zeros"

_{One mistake you are doing in your proof that you are doing for a specific string 0000. You should proof for all w ≥ P. So still your proof is incomplete}

Read my this answer IN CONTEXT OF PUMPING LEMMA FOR REGULAR LANGUAGES

In that answer, I have explained that breaking w into xyz and pumping y means finding looping part and repeating looping part to generate new strings in language.
When we proof that some language is regular; then actually we don't know where is the looping part so we try with all possible choices that satisfies pumping lemma's rule 1,2 & 3.

And Pumping lemma says that if language is regular and infinite them there must be a loop in the DFA and every sufficiently large string in language passes through looping part (according to pigeonhole principle) of DFA (and hence y can't be null. That's rule-1 in above formal definition).

Think, loop can be at initial position or at end and so x and z can be null strings.

But actually we don't know where loop falls in DFA so we try with all possible ways.

To proof a language is regular: You are to proof that for all sufficiently long strings(w) in language there is at-least one way(y) to generate new strings in the language by repeating looping part any number (i) of times.

To proof a language is not regular:You are to find at least one sufficiently long strings (w) in language such that there no choice for any way 'y' so that its possible to generate new strings with all possible repetition (i).

To proof using Pumping Lemma: 
+-------------------------+--------------------------+----------------+--------------+
|                         | Sufficient large W in L  |     y          |    i >=0     |
+-------------------------+--------------------------+----------------+--------------+
| language is regular     | For all W (all W can use | At-least one   | For all i>=0 |  
|                         | to generate new W' in L) |                |              |
+-------------------------+--------------------------+----------------+--------------+
| language is NOT regular | Find Any W (at-least 1   | With all (Show | At-least one |   
|                         | W that can't generates   | no possible Y  | i            | 
|                         | new W' in L              | exists)        |              |
+-------------------------+--------------------------+----------------+--------------+

CAUTION:: The Rule always not works to proof 'Weather a Language is Regular?'

Pumping Lemma necessary but not sufficient condition for a language to be regular. A language possible that satisfies these conditions may still be non-regular.
Reference

To proof a language is regular you have some necessary and sufficient conditions for a language to be regular.

Answer 2

Although the accepted answer is complete in its own way, I had to add a few things. I have a very playful way to exploit the pumping lemma to be able to prove that a given language is not a Regular language. Just to have a context to talk about, let me state the lemma:

Pumping Lemma for Regular Languages:

For any regular language L, there exists an integer k.

Such that for all x ∈ L with |x| ≥ k, there exists u, v, w ∈ Σ∗, such that x = uvw, and

(1) |uv| ≤ k

(2) |v| ≥ 1

(3) for all i ≥ 0: u(v^i)w ∈ L

The k is called the Pumping lemma constant. Let me come straight to the point and show you how to go about proving a language L is not regular.

Now to start the game you need two players here. One is the Reader(R) and the other is the Adversary(A).

Input: A language L

The Goal of R: Somehow prove the language L is not regular by some contradiction.

The Goal of A: Somehow be prepared enough to face the arguments of R and do not let him/her create a contradiction.

Now let us start the argument.

A: The language L is not Irregular because none could show the contradiction using pumping lemma with a certain pumping constant k. (Each language is mapped to only one integer k)

R: Let me assume what you say. If language L is regular then it must satisfy the conditions of the pumping lemma. So, let me choose a suitable string x ∈ L (|x| >= k), such that it helps me create a contradiction later.

A: Challenged by the R, A tries its best to find at least one suitable partitioning u, v and w of the string x, such that

x = uvw  and |uv| <= k and |v| > 0

R: With any possible partition given by A, wins the the argument if able to show an integer i >= 0 such that

u(v^i)w ∉ L

Because now the R has shown that the Language L has at least one string x which doesn't have any partition(u, v, w) such that it satisfies the pumping lemma. The contradiction happened because our assumption that L is regular is FALSE. Therefore the language L is proven to be not regular.

If the R is not able to show the above, this is not a Proof of the language being Regular. It just means that L can be a Regular or Irregular language which just happens to satisfy the pumping lemma conditions.

Always remember, the pumping lemma is if(L is regular), then Statements. The vice-versa is not necessarily TRUE. Although it might be TRUE in some cases.

Therefore the pumping lemma is useful only if you want to prove that a language is not regular.

(Source: Theory of Computation(NPTEL): Prof. Somenath Biswas(IIT Kanpur)