Title
Review: DFA Union
To gain practice with designing constructions, let’s prove the following
Theorem: the class of DFA-decidable languages is closed under union.
In other words, if \(L_1 = L(D_1)\) and \(L_2 = L(D_2)\), then we must prove the existence of a DFA \(D\) such that \(L_1 \cup L_2 = L(D)\).
Idea: simulate \(D_1\) and \(D_2\) in parallel, and accept if either \(D_1\) or \(D_2\) accepts.
\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) 
\(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) 
\(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string:
\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) 
\(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: a
\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) 
\(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) 
\(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) 
\(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaaaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaaaaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaaaaaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaaaaaaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaaaaaaaaaa\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)
Input string: aaaaaaaaaaaaaaaWhat do we need to keep track of to execute this simulation?
Which pair of states are the machines in?
Closure Under Intersection
We already showed that the DFA-decidable languages are closed under union:
If \(L_1 = L(D_1)\) and \(L_2 = L(D_2)\), then we can construct \(D\) such that \(L(D) = L_1 \cup L_2\).
Formally:
- Let \(D_1 = (Q_1, \Sigma, \delta_1, s_1, F_1)\) and
\(\phantom{\text{Let }} D_2 = (Q_2, \Sigma, \delta_2, s_2, F_2)\) - Construct \(D = (Q, \Sigma, \delta, s, F)\):
- \(Q = Q_1 \times Q_2 = \setbuild{(p_1, p_2)}{p_1 \in Q_1 \text{ and } p_2 \in Q_2}\)
- \(s = (s_1, s_2)\)
- For all \((p_1, p_2) \in Q\) and \(a \in \Sigma\): \[\delta((p_1, p_2), a) = (\delta_1(p_1, a), \delta_2(p_2, a))\]
- \[F = \setbuild{(p_1, p_2)}{p_1 \in F_1 \text{ or } p_2 \in F_2}\]
What about intersection?
Closure Under Intersection
Theorem: the DFA-decidable languages are closed under intersection:
If \(L_1 = L(D_1)\) and \(L_2 = L(D_2)\), then we can construct \(D\) such that \(L(D) = L_1 \color{red} \cap \color{black} L_2\).
Formally:
- Let \(D_1 = (Q_1, \Sigma, \delta_1, s_1, F_1)\) and
\(\phantom{\text{Let }} D_2 = (Q_2, \Sigma, \delta_2, s_2, F_2)\) - Construct \(D = (Q, \Sigma, \delta, s, F)\):
- \(Q = Q_1 \times Q_2 = \setbuild{(p_1, p_2)}{p_1 \in Q_1 \text{ and } p_2 \in Q_2}\)
- \(s = (s_1, s_2)\)
- For all \((p_1, p_2) \in Q\) and \(a \in \Sigma\): \[\delta((p_1, p_2), a) = (\delta_1(p_1, a), \delta_2(p_2, a))\]
- \[F = \setbuild{(p_1, p_2)}{p_1 \in F_1 {\color{red} \text{ and }} p_2 \in F_2}\]
Just update the accept states!
DFA Concatenation Attempt
Let’s consider the concatenation of
\(L_1 = \setbuild{w \in \{0, 1\}^*}{ \#_1(w) \equiv 1 \mod 2}\), and \(L_2 = \setbuild{w \in \{0, 1\}^*}{ \#_1(w) \equiv 1 \mod 3}\)
If we immediately “jump” from odd to \(q_0\) by redirecting the transition \(\texttt{even} \stackrel{1}{\to} \texttt{odd}\) into \(\texttt{even} \stackrel{1}{\to} \texttt{q0}\), then the string \(1111 = 111 \circ 1\) is rejected despite \(111 \in L_1\) and \(1 \in L_2\).
The same “where to split?” difficulty is encountered for Kleene star.
NFAs to the Rescue
The more convenient and flexible NFA model makes it simpler to construct a machine deciding the concatenation and Kleene star of languages.
Once we demonstrate how to convert an NFA to a DFA, this will prove that DFA-decidable languages are closed under concatenation and Kleene star.
First: how can we simplify the union construction with NFAs?
\(L(D_1) = \setbuild{a^n}{n \equiv 2 \mod 3}\) \(L(D_2) = \setbuild{a^n}{n \equiv 1 \mod 5 \; \text{ or } \; n \equiv 4 \mod 5}\)Just create a new start state and add \(\emptystring\)-transitions to the start states of \(D_1\) and \(D_2\).
NFAs to the Rescue
The more convenient and flexible NFA model makes it simpler to construct a machine deciding the concatenation and Kleene star of languages.
Once we demonstrate how to convert an NFA to a DFA, this will prove that DFA-decidable languages are closed under concatenation and Kleene star.
First: how can we simplify the union construction with NFAs?
Just create a new start state and add \(\emptystring\)-transitions to the start states of \(D_1\) and \(D_2\).
NFAs to the rescue: Kleene star example
Consider the star of: \(L(N_1) = \setbuild{w \in \{\string{a, b}\}^*}{ w \text{ has an odd number of } \string{b} \text{'s followed by an } \string{a}}\)
This changes the behavior of the original NFA \(N_1\)! … accepts \(\string{bb}\) even though \(\string{bb} \notin L(N_1)^*\)
NFAs to the rescue: Kleene star example
Consider the star of: \(L(N_1) = \setbuild{w \in \{\string{a, b}\}^*}{ w \text{ has an odd number of } \string{b} \text{'s followed by an } \string{a}}\)
