# What is pumping lemma for regular language explain with example?

## What is pumping lemma for regular language explain with example?

Pumping Lemma for Regular Languages In simple terms, this means that if a string v is ‘pumped’, i.e., if v is inserted any number of times, the resultant string still remains in L. Pumping Lemma is used as a proof for irregularity of a language. Thus, if a language is regular, it always satisfies pumping lemma.

## How do you prove a language is regular using pumping lemma?

Here is the Pumping Lemma. If L is a regular language, then there is an integer n > 0 with the property that: (*) for any string x ∈ L where |x| ≥ n, there are strings u, v, w such that (i) x = uvw, (ii) v = ǫ, (iii) |uv| ≤ n, (iv) uvkw ∈ L for all k ∈ N.

How do you demonstrate a language that is regular?

To prove if a language is a regular language, one can simply provide the finite state machine that generates it. If the finite state machine for a given language is not obvious (and this might certainly be the case if a language is, in fact, non-regular), the pumping lemma for regular languages is a useful tool.

### What is pumping lemma write its applications?

Pumping Lemma is to be applied to show that certain languages are not regular. It should never be used to show a language is regular. If L is regular, it satisfies Pumping Lemma. If L does not satisfy Pumping Lemma, it is non-regular.

### What is the main application of pumping lemma in CFLS?

– The pumping lemma for CFL’s states that for sufficiently long strings in a CFL, we can find two, short, nearby substrings that we can “pump” in tandem and the resulting string must also be in the language. Let L be a CFL.

Is a * b * a regular language?

Yes, a*b* represents a regular language. Language description: Any number of a followed by any numbers of b (by any number I mean zero (including null ^ ) or more times). Some example strings are: {^, a, b, aab, abbb, aabbb.}

#### How do you pick a pump lemma?

You pick a string s of length at least p. Mr. Pumping Lemma divides s into three parts uvw, subject to the restrictions that |uv|≤p, |v|≥1. You now “pump” the v part by picking an integer i≠1 to select a word uviw.

#### Is English a regular language?

Back to English. For English, we use the technique of intersection with a regular language to find a subset of English to which we can apply the pumping lemma: then since that language is not regular, and it’s the intersection of English with a regular language, we can conclude that English is not a regular language.

Which kind of proof is used to prove the regularity of a language?

Which kind of proof is used to prove the regularity of a language? Explanation: We use the method of proof by contradiction in pumping lemma to prove that a language is regular or not.

## What is XYZ in pumping lemma?

Let s = xyz where x is the part of the string accepted appearing before the equal states, y is the part of the string appearing between the equal states, and z is the part of the string appearing after the equal states. Now the following should be clear: xyz can be pumped. | y| > 0.

## What is the use of pumping lemma Mcq?

The pumping lemma is often used to prove that a particular language is non-regular. Pumping lemma for regular language is generally used for proving a given grammar is not regular. Hence the correct answer is a given grammar is not regular.