Pumping lemma | Theory of Computation - Computer Science Engineering (CSE) PDF Download

Pumping Lemma

There are two Pumping Lemmas, which are defined for
1. Regular Languages, and
2. Context – Free Languages
 
Pumping Lemma for Regular Languages

For any regular language L, there exists an integer n, such that for all x ∈ L with |x| ≥ n, there exists u, v, w ∈ Σ∗, such that x = uvw, and
(1) |uv| ≤ n
(2) |v| ≥ 1
(3) for all i ≥ 0: uvⁱw ∈ L

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. If there exists at least one string made from pumping which is not in L, then L is surely not regular.
The opposite of this may not always be true. That is, if Pumping Lemma holds, it does not mean that the language is regular.

For example, let us prove L₀₁ = {0ⁿ1ⁿ | n ≥ 0} is irregular.
Let us assume that L is regular, then by Pumping Lemma the above given rules follow.
Now, let x ∈ L and |x| ≥ n. So, by Pumping Lemma, there exists u, v, w such that (1) – (3) hold.
 
We show that for all u, v, w, (1) – (3) does not hold.
If (1) and (2) hold then x = 0ⁿ1ⁿ = uvw with |uv| ≤ n and |v| ≥ 1.
So, u = 0^a, v = 0^b, w = 0^c1ⁿ where : a + b ≤ n, b ≥ 1, c ≥ 0, a + b + c = n
But, then (3) fails for i = 0
uv⁰w = uw = 0^a0^c1ⁿ = 0^{a + c}1ⁿ ∉ L, since a + c ≠ n.

For above example, 0ⁿ1ⁿ is CFL, as any string can be the result of pumping at two places, one for 0 and other for 1.
Let us prove, L₀₁₂ = {0ⁿ1ⁿ2ⁿ | n ≥ 0} is not Context-free.
Let us assume that L is Context-free, then by Pumping Lemma, the above given rules follow.
Now, let x ∈ L and |x| ≥ n. So, by Pumping Lemma, there exists u, v, w, x, y such that (1) – (3) hold.
We show that for all u, v, w, x, y (1) – (3) do not hold.

If (1) and (2) hold then x = 0ⁿ1ⁿ2ⁿ = uvwxy with |vwx| ≤ n and |vx| ≥ 1.
(1) tells us that vwx does not contain both 0 and 2. Thus, either vwx has no 0’s, or vwx has no 2’s. Thus, we have two cases to consider.
Suppose vwx has no 0’s. By (2), vx contains a 1 or a 2. Thus uwy has ‘n’ 0’s and uwy either has less than ‘n’ 1’s or has less than ‘n’ 2’s.
But (3) tells us that uwy = uv⁰wx⁰y ∈ L.
So, uwy has an equal number of 0’s, 1’s and 2’s gives us a contradiction. The case where vwx has no 2’s is similar and also gives us a contradiction. Thus L is not context-free.