COMS W3261
Computer Science Theory
Lecture 7: September 30, 2009
Constructing an NFA from a Regular Expression

1. Outline

• Review
• From regular expressions to ε-NFA's
• Algebraic laws for regular expressions
• The pumping lemma for regular languages

2. Review

• From DFA's to regular expressions

3. From RE's to ε-NFA's

• Theorem: If L = L(R) for some regular expression R, then there is an ε-NFA N such that L = L(N).
• Proof: See HMU, Sect. 3.2.1, pp. 93-97.
• The proof is in the form of an algorithm that takes as input a regular expression R of length n and inductively constructs from it an equivalent ε-NFA that has
• exactly one start state and one final state
• at most 2n states
• no arcs coming into its start state
• no arcs leaving its final state
• at most two arcs leaving any nonfinal state

4. Algebraic Laws for Regular Expressions

• Algebraic laws can be used to simplify regular expressions
• Here are some of the most important algebraic identities for regular expressions
• Union is commutative: L + M = M + L
• Union is associative: (L + M) + N = L + (M + N)
• Concatenation is associative: (LM)N = L(MN)
• ∅ is the identity for union: ∅ + L = L + ∅ = L
• ε is the identity for concatenation: εL = Lε = L
• ∅ is the annihilator for concatenation: ∅L = L∅ = ∅
• Concatenation left distributes over union: L(M + N) = LM + LN
• Concatenation right distributes over union: (M + N)L = ML + NL
• Union is idempotent: L + L = L
• L** = L*
• ∅* = ε
• ε* = ε

5. The Pumping Lemma for Regular Languages

• The Pumping Lemma: For every nonfinite regular language L, there exists a constant n that depends on L such that for all w in L with |w| ≥ n, there exists a decomposition of w into xyz such that
1. y ≠ ε,
2. |xy| ≤ n, and
3. for all k ≥ 0, the string xy^kz is in L.
• Proof: See HMU, p. 129.

6. Applications of the Pumping Lemma

• One important use of the pumping lemma is to prove some languages are not regular.
• Example: The language L consisting of all strings of a's and b's of the form aⁱbⁱ, i ≥ 0, is not regular.
• The proof will be by contradiction. Assume L is regular. Then by the pumping lemma there is a constant n associated with L such that for all w in L with |w| ≥ n, w can be written as xyz such that
1. y ≠ ε,
2. |xy| ≤ n, and
3. for all k ≥ 0, the string xy^kz is in L.
• Since |xy| ≤ n, xy = a^m for some 0 < m ≤ n.
• Setting k = 0, condition (3) of the pumping lemma says xz must also be in L.
• But xz is of the form a^pbⁿ, where p < n.
• This contradicts the conclusion that xz must be in L.

7. Practice Problems

1. Convert the regular expression a(a+b)*b into an equivalent ε-NFA.
2. Convert your ε-NFA into an equivalent DFA.
3. Show that the language consisting of strings of a's where the number of a's is a perfect square is not regular.
4. Show that the language consisting of strings of balanced parentheses is not regular.

8. Reading Assignment

• HMU: Sects. 3.2-3.7, 4.1