Pumping Lemma for context-free Languages

Pumping Lemma for context-free Languages Visually

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Visual Simulation Step Execution Pumping Lemma for CFL Constructor Performance Analysis

What is the Pumping Lemma for CFLs?

The Pumping Lemma for context-free Languages is a more complex version of the pumping lemma that applies to context-free languages. It provides a necessary condition for a language to be context-free and is used to prove that certain languages are not context-free.

Unlike the regular pumping lemma which uses a 3-part lhG (xyz), the CFL pumping lemma uses a 5-part lhG (uvxyz) with more sophisticated pumping conditions.

Key Difference: The CFL pumping lemma allows for two separate parts (v and y) to be pumped simultaneously, reflecting the nested structure of context-free languages.

Interactive CFL Pumping Lemma Tester

Test strings against the pumping lemma for L = @{a^n b^n c^n | n ≥ 1@}

Formal Statement of the CFL Pumping Lemma

Pumping Lemma for context-free Languages
If $L$ is a context-free language, then there exists a pumping length $p \geq 1$ such that for any string $s \in L$ with $|s| \geq p$, we can write $s = uvxyz$ where:
  1. $|vxy| \leq p$ (the middle part is bounded)
  2. $|vy| \geq 1$ (at least one of v or y is non-empty)
  3. $uv^ixy^iz \in L$ for all $i \geq 0$ (pumping property)
String lhG (5 parts):
u v x y z
where v and y are pumped together: uv^i xy^i z

Interactive String lhG Visualizer

Visualize how strings are decomposed according to the CFL pumping lemma

String lhG: s = uvxyz

2
2
2
2
Conditions Check:
Pumping iterations (i):
0

How to Use the CFL Pumping Lemma

To prove a language L is not context-free:

1
Assume L is context-free (proof by contradiction)
2
Choose a string s ∈ L with |s| ≥ p (where p is the pumping length)
3
Consider all possible ways to decompose s = uvxyz satisfying |vxy| ≤ p and |vy| ≥ 1
4
Show that for some i ≥ 0, uv^i xy^i z ∉ L (contradiction)
5
Conclude L is not context-free

Classic Example: L = @{a^n b^n c^n | n ≥ 1@}

Let's prove that L = @{a^n b^n c^n | n ≥ 1@} is not context-free.

Language L: @{abc, aabbcc, aaabbbccc, aaaabbbbcccc, ...@}
Step 1: Assume L is context-free. Then there exists a pumping length p.
Step 2: Choose string s = a^p b^p c^p ∈ L with |s| = 3p ≥ p.
Step 3: By the pumping lemma, s = uvxyz where |vxy| ≤ p, |vy| ≥ 1. Since |vxy| ≤ p, the substring vxy can contain at most 2 different symbols.
Step 4: Consider cases:
  • If vxy contains only a's and b's: pumping changes ratio of a's and b's to c's
  • If vxy contains only b's and c's: pumping changes ratio of a's to b's and c's
  • If vxy contains only one symbol: pumping changes ratios
Contradiction: In all cases, uv^i xy^i z has unequal counts of a's, b's, and c's for some i ≠ 1, so it's not in L. This violates the pumping lemma condition.
Conclusion: L = @{a^n b^n c^n | n ≥ 1@} is not context-free.