CSCI 301 L26 Notes

Lecture 26 - Notes

Goals

• Be able to prove that regular languages are closed under regular operations including union, concatenation, and star.
• Know the definition of regular expressions
• Be able to construct regular expressions for a given language, and be able to describe the language represented by a regular expression.

Announcements

• Week 6 Survey due tonight
• A6 due Friday night
• Midterm wrapper also due Friday night
• If you haven’t gotten your exam back, you can collect it after class or in office hours.

Closure of Regular Languages under Regular Operations

We’re finally ready to do the (actually quite straightforward, now!) proof of the closure of regular operations.

Theorem: The set of regular languages is closed under the regular operations.

Regular Operations

• The union of two languages \(A\) and \(B\) is defined as \(A \cup B = \{w : w \in A \text{ or } w \in B\}\).
• The concatenation or product of two languages \(A\) and \(B\) is defined as \(AB = \{ww' : w \in A \text{ and } w' \in B\}\).
• The closure or star (or Kleene closure) of a language \(A\) is defined as: \(A^* = \{u_1 u_2 \ldots u_k : k \ge 0 \text{ and } u_i \in A \text{ for all } i = 1, 2, \ldots, k\}\)
  • In other words, it’s the language containing all possible concatenations of any choice of 0 or more strings over \(A\). Notice that 0 is an option here, so \(\epsilon \in A^*\) for any language \(A\).
  • Another equivalent recursive definition is:
    • Let \(A^0 = \{\epsilon\}\)
    • Let \(A^k = AA^{k-1}\)
    • \(A^* = \bigcup_{k=0}^{\infty} A^k\)

Proof:

Case: \(A \cup B\) - proof by construction:

Let \(M_A = (Q_A, \Sigma, \delta_A, q_A, F_A)\) be a DFA accepting \(A\) and \(M_B = (Q_B, \Sigma, \delta_B, q_B, F_B)\) be a DFA accepting \(B\).

Construct an NFA \(N = (Q, \Sigma', \delta, q, F)\) as follows:

1. Let \(Q_B'\) be a set containing \(Q_B\)’s elements, but renamed to ensure that \(Q_A \cap Q_B = \varnothing\); also let \(s\) be a new state that is not a member of \(Q_A\) or \(Q_B\).
   • The NFA’s set of states is \(Q = Q_A \cup Q_B' \cup \{s\}\).
2. The NFA’s alphabet is \(\Sigma' = \Sigma \cup \{\epsilon\}\).
3. The NFA’s transition function \(\delta = \delta \cup \{((s, \epsilon), q_A), ((s, \epsilon), q_B)\}\).
4. The NFA’s start state is \(q = s\).
5. The NFA’s accept states are \(F = F_A \cup F_B\).

This machine accepts \(A \cup B\) because at the start of processing, it may nondeterministically decide to make an \(\epsilon\)-transition to either \(A\)’s start state or \(B\)’s start state, then proceed to process the string exactly as \(M_A\) or \(M_B\) would. Therefore, if the string is in \(A\) or \(B\), then there is a way that \(N\) can process the string and end in an accept state.

• Remaining cases: Exercises!

Regular Expressions

Regular expressions are a meta-language for describing regular languages. They are defined based on the properties of regular languages that we’ve now explored thoroughly, so a lot of what follows will probably be unsurprising.

Definition: A regular expression over an alphabet \(\Sigma\) is defined as follows:

• \(\epsilon\) is a regular expression, describing the language \(\{\epsilon\}\)
• \(\varnothing\) is a regular expression describing the language \(\varnothing\)
• For each \(a \in \Sigma\), \(a\) is a regular expression describing the language \(\{a\}\)
• If \(R_1\) and \(R_2\) are regular expressions, then \(R_1 \cup R_2\) is a regular expression describing the language \(L(R_1) \cup L(R_2)\)
• If \(R_1\) and \(R_2\) are regular expressions, then \(R_1R_2\) is a regular expression describing the language \(L(R_1) L(R_2)\)
• If \(R\) is a regular expression, then \(R^*\) is a regular expression describing the language \(L^*\)

Examples:

What language does \((0 \cup 1)01^*\) describe?

​ \(\{00, 001, 0011, \ldots, 10, 101, 1011, \ldots\}\)

What regular expression describes \(\{w: w \text{ has $1011$ as a substring}\}\)?

​ \((0 \cup 1)^*1011(0 \cup 1)^*\)