Pumping Lemma for Context-Free Languages

The Pumping Lemma for Context-Free Languages (CFLs) is a fundamental property used to prove that a language is not context-free. It is similar in working to the pumping lemma for regular languages.

Conditions for Pumping Lemma in Context-Free Languages

If $L$ is a context-free language, then there exists a constant $p$ (called the pumping length) such that every string w ∈ L with ∣w∣≥p can be written as:

w = uvxyzw 

such that the following conditions hold:

Condition 01: Length restriction:

∣vxy∣≤p

The length of “middle” part “vxy” sould always less than or equal to pumping “p” value.

Condition 02: Pumping parts not empty:

∣vy∣ ≥ 1

At least one of “v” or “y” is not empty. So that string can pump.

Condition 03: Closure under pumping

uvⁱxyⁱz ∈ L for all i≥0

The string remains in the language for any number of repetitions (including zero).

Pumping Lemma Algorithm in Context-Free Languages

let explain the pumping lemma algorithm in context free languages

Key Differences in Context-Free vs. Regular Pumping Lemma

Here’s a clear comparison showing the key differences between the Pumping Lemma for Regular Languages and Context-Free Languages:

Pumping Lemma Examples

Example 01: Using Pumping Lemma prove that the language A = { yy | y ∈ {0,1}*is not context free language

Below is a descriptive diagram that explains that why the language A = { yy | y ∈ {0,1}* } is not a context-free language.

Step-by-step proof using the Pumping Lemma for CFLs:

Let explain again why the language A = { yy | y ∈ {0,1}* } is not a context-free language by using different values of pumping “p” and “i”.

1. Assume A is context-free.

2. Let the pumping length be p=4 (or any value ≥ 1).

3. Choose a string S=01010101 ∈ A, because this is of the form yy, where y=0101. So we can select the following string:

   S=0101 0101

4. According to the pumping lemma, S can be split as:

   S=uvxyz 

   with:

   • ∣vxy∣≤p

   • ∣vy∣≥1

   • uvⁱxyⁱz∈A, for all $i\ge 0$

5. Let’s say:

   • u=0

   • v=1

   • x=0

   • y=1

   • z=0101

   Then:

   $$=01010101$$

6. Now pump with i=2

   $$0110110101$$

7. Check the result:

   • The new string is 0110110101, which is not of the form yy (it’s not a repeat of the same string).

8. So, the pumped string is not in language A.

Example 02: Using Pumping Lemma prove that the language A = { aⁿbⁿcⁿ | n ≥ 0 } is not a Context Free Language

Below is a descriptive diagram that explains that why the language A ={ aⁿbⁿcⁿ | n ≥ 0 } is not a context-free language.

Explain: Pumping Lemma for CFLs

Let’s explain that, A = { aⁿ bⁿ cⁿ | n ≥ 1 } is not context-free by using different values of pumping “p” and “i”.

Note: By using different values of “p” and “i” as in the above diagram.

• Assume A is context-free.
• Let the pumping length be $p=3$ (or any value ≥ 1).
• Choose a string S=aaabbbccc ∈ A, i.e., a³b³c³.
• According to the pumping lemma rule, it can be written as:

S=uvxyz

With conditions:

• ∣vxy∣≤p
• ∣vy∣≥1
• uvⁱxyⁱz ∈ A, where $i\ge 0$

Let’s say:

• u=a
• v=a
• x=a
• y=b
• z=bbccc

Then S=uvxyz=aaabbbccc, Now pump with i=2

S = uv²xy²z=aaaabbbbccc 

This gives 4 a’s, 4 b’s, and 3 c’s, which is not in A. Therefore, the pumped string is not in language A.

Example 03: Using Pumping Lemma prove that the language A = { aⁿb²ⁿ | n ≥ 0 } is not a Context Free Language

Below is a descriptive diagram that explains that why the language A ={ aⁿb²ⁿ | n ≥ 0 } is not a context-free language.

Explain: Pumping Lemma for CFLs

Let’s explain that A = { aⁿ b²ⁿ | n ≥ 0 } is not a context-free language by using different values of pumping “p” and “i”.

Note: By using different values of “p” and “i” as in the above diagram.

• Assume A is context-free.
• Let the pumping length be $p=3$ (or any value ≥ 1).
• Choose a string  S=a³b⁶=aaabbbbbb ∈ A.
• According to the pumping lemma rule, it can be written as:

S=uvxyz

With conditions:

• ∣vxy∣≤p
• ∣vy∣≥1
• uvⁱxyⁱz ∈ A, where $i\ge 0$

Let’s say:

• $u=a$
• v=a
• x=a
• y=ε (empty, optional)
• z=bbbbbb

Then

S=uvxyz=aaabbbbbb

Now pump with i=2

S = uv²xy²z=aaaabbbbbb=a⁴b⁶

But in A, for every “a “$n”$, the number of b’s must be twice the number of a’s.

• Here: 4 a’s → should have 8 b’s

• But we have only 6 b’s → not in A