Pumping Lemma and Nonregular Languages

🌱 What Is the Pumping Lemma? (In simple words)

Think of a DFA like a small machine with a fixed number of states.

Now imagine you feed the machine a very long string from some language.
Since the machine has only limited states but the string is very long,
the machine must reuse some state again.

This repeated state creates a loop.
And because of the loop, some part of the string can be “pumped” —
meaning we can repeat it many times, and the machine will still accept the string.

This idea is the heart of the Pumping Lemma.

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🎈 The Intuition Behind Pumping

Let’s use a simple analogy.

Imagine you walk inside a small circular park.
The park has only 10 stepping stones placed in a ring.
But you take 40 steps.

Since there are more steps than stones,
you must step on the same stone more than once.

That’s the DFA:

• stones = states
• your steps = characters in the input string

Whenever you loop over the same stone again, you have found a repeating path.

This repeating path can be done again and again.

This is the “pumping” idea.

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📌 What the Pumping Lemma Says

The pumping lemma states:

If a language is regular, then every long enough string in the language
can be broken into 3 parts:

s = xyz

such that

• The middle part y is the part we can “pump”.
• Repeating y any number of times (0, 1, 2, …) keeps the string inside the language.

In short: xyⁱz is always in the language for all i ≥ 0.

If you pump the string and it still stays in the language, the language might be regular.

If pumping breaks the string —
the language is definitely NOT regular.

This is how we prove nonregularity.

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🧩 Simple Diagram Idea

Below is a simple ASCII-style diagram to visualize the idea.

        +-------------------+
        |       Loop        |
   --->(A)----y---->(B)----+
        ^                  |
        |                  |
        +--------x-------->+
                 |
                 v
                (C)

• A → B is the “y” loop (the part that can be repeated).
• x is what happens before the loop begins.
• z is what comes after the loop ends.

So the string becomes:

s = x y z

And because of the loop, we can do:

x y² z
x y³ z
x y⁰ z (removing y)

All must still be valid strings in the language if the language is regular.

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🧨 How the Pumping Lemma Helps Us Prove a Language Is Not Regular

We use a simple strategy:

1. Assume the language is regular.
2. Then it must follow the pumping lemma.
3. Choose a special long string from the language.
4. Try to pump it.
5. If pumping creates a string not in the language,
   → contradiction!
   → the language is NOT regular.

This method is like checking if someone is telling the truth by following their logic
until it breaks.

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🌉 Example: L = { aⁿbⁿ | n ≥ 0 }

This language contains strings like:

ab, aabb, aaabbb, aaaabbbb, ...

Same number of a’s followed by the same number of b’s.

Let’s prove this language is NOT regular.

Step 1 — Assume L is regular

Then the pumping lemma must hold.

Step 2 — Pick a long string

Let’s choose

s = a^p b^p

where p is the pumping length.

Step 3 — Break it into xyz

The pumping lemma says:

• |xy| ≤ p
• |y| > 0

This means y only contains a’s, because the first p letters are all a’s.

Step 4 — Pump y

Let’s remove y (pump i = 0):

We get a string with fewer a’s but the same number of b’s:

a^(p - |y|) b^p

This is NOT in the language L, because the counts don’t match anymore.

Step 5 — Contradiction!

So the pumping lemma fails.
Therefore:

🔴 L is NOT regular.

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🎯 Why the Pumping Lemma Is So Important

The pumping lemma is one of the most important tools in theory of computation because:

• It helps us identify languages that DFAs cannot handle.
• It separates the world of regular languages from more powerful ones.
• It prepares your mind for the need of more complex models like
  context-free grammars, pushdown automata, and Turing machines.

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