Pumping lemma for regular language


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

Best Solution

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 wP. So still your proof is incomplete


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.

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

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