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 Answer
You are not completely clear about pumping lemma.
What pumping lemma say:
Formal definition: Pumping lemma for regular languages
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.
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
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 breakw
in three partsxyz
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 ≥ Pnote: in second point, there may be a chance that even if you breaks
w
correctly intoxyz
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 breakw
such thaty = 00
. And with this choice ofy
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 incompleteRead my this answer IN CONTEXT OF PUMPING LEMMA FOR REGULAR LANGUAGES
In that answer, I have explained that breaking
w
intoxyz
and pumpingy
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
andz
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).
CAUTION:: The Rule always not works to proof 'Weather a Language is Regular?'
To proof a language is regular you have some necessary and sufficient conditions for a language to be regular.