Regular Expressions & Pumping Lemma

🎯 Learning Objectives

• Write regular expressions for regular languages
• Convert between regex and DFA/NFA
• Apply the pumping lemma to prove non-regularity
• Use closure properties to construct regular languages

1. Regular Expressions

1.1 Intuition

1.2 Operations

1.3 Examples

1.4 Algebraic Laws

2. Equivalence of Regex and FA

2.1 Regex → NFA (Thompson's Construction)

2.2 DFA → Regex (State Elimination)

1. Add a new start state with ε to old start
2. Add a new accept state with ε from old accepts
3. Eliminate states one by one, updating edge labels
4. The remaining edge from start to accept is the regex

3. Pumping Lemma for Regular Languages

3.1 Intuition

3.2 Formal Statement

1. $|xy| \leq p$ (the pumpable part is near the start)
2. $|y| \geq 1$ (y is non-empty)
3. $xy^iz \in L$ for all $i \geq 0$ (pumping works)

3.3 Proof Template

pseudo
1. Assume L is regular (for contradiction)
2. Let p be the pumping length
3. Choose w ∈ L with |w| ≥ p (be strategic!)
4. For ALL ways to split w = xyz with |xy| ≤ p, |y| ≥ 1:
   Show there exists i such that xy^iz ∉ L
5. Contradiction → L is not regular

3.4 Worked Example 1: $L = \{a^n b^n \mid n \geq 0\}$

1. Assume L is regular
2. Let p be pumping length
3. Choose $w = a^p b^p$
4. Since $|xy| \leq p$, $y$ consists only of a's: $y = a^k$ for $k \geq 1$
5. Pump $i=2$: $xy^2z = a^{p+k}b^p$
   • More a's than b's → not in L
6. Contradiction → L is not regular ✓

3.5 Worked Example 2: $L = \{ww \mid w \in \{0,1\}^*\}$

1. Choose $w = 0^p 1 0^p 1$
2. $|xy| \leq p$, so $y$ is within the first $0^p$ block
3. $y = 0^k$, pump $i=2$: $xy^2z = 0^{p+k} 1 0^p 1$
4. The string before the middle is $0^{p+k}1$ and after is $0^p1$ — not equal
5. Result: not regular ✓

3.6 When Pumping Lemma Fails

4. Common Pitfalls

Pitfall 1: Pumping the wrong string

Pitfall 2: Thinking pumping lemma can prove regularity

Pitfall 3: Forgetting to consider all splits

5. 📝 Practice Questions

Q1: Write a regex for binary strings that do NOT contain "00".

Answer: $(1|01)^*(0|ε)$ — strings where every 0 is followed by 1 (or at end). Q2: Prove $L = \{0^{n^2} \mid n \geq 0\}$ is not regular.

Answer: Assume regular with pumping length p. Choose $w = 0^{p^2}$. Then $y = 0^k$ for $1 \leq k \leq p$. Pump $i=2$: $xy^2z = 0^{p^2+k}$. But $(p+1)^2 = p^2 + 2p + 1 > p^2 + k \geq p^2 + 1$, so $p^2 < p^2+k < (p+1)^2$. Thus $p^2+k$ is not a perfect square → not in L. Contradiction. Q3: Convert regex $(0|1)^*00$ to an NFA.

Answer: NFA with states q0 (start), q1 (after first 0), q2 (accept, after "00"). Transitions: q0—0,1→q0; q0—0→q1; q1—0→q2; q2—0→q2; q2—1→q0. Q4: Is the language L = {w ∈ {0,1} | w has equal number of 0s and 1s} regular?*

Answer: No. Proof using pumping lemma: choose $w = 0^p 1^p$. $y = 0^k$. Pump $i=2$: $0^{p+k}1^p$ has more 0s than 1s → not in L. Not regular. Q5: Show that regular languages are closed under intersection.

Answer: Given DFAs for L₁ and L₂, construct a product DFA where states are pairs (q₁, q₂), transitions are δ((q₁,q₂), a) = (δ₁(q₁,a), δ₂(q₂,a)), and accept states are (q₁,q₂) where q₁ ∈ F₁ AND q₂ ∈ F₂. This DFA accepts exactly L₁ ∩ L₂.

6. 🔗 Cross-References

• Week 1 - DFA/NFA: Equivalence with regex
• Week 3 - DFA Minimization: Minimization uses equivalence classes
• BSCS4032 (Compiler Design): Lexical analysis uses regex → NFA → DFA Join Discord PreviousNFA to DFANextRegex to DFA