Determinization
Remember that a language \(L\) is finitely recognizable if there is a finite automaton \(\mathcal A = (Q, A, \delta, F)\) and a state \(x \in Q\) such that \(L = \mathcal L(\mathcal A, x)\). Last time, we saw that \(\mathsf{Fin} \subseteq \mathsf{TDFin}\). Today, we are going to show the converse: that \(\mathsf{TDFin} \subseteq \mathsf{Fin}\), which would prove the following theorem:
The difficult part of the proof of the Finite Recognizability Theorem is showing that \(\mathsf{TDFin} \subseteq \mathsf{Fin}\). The idea behind the proof is an automaton construction: this is a kind of function that takes an automaton and a state as input, and produces an automaton and a state as output. The automaton construction we need is
- Input: a finite automaton \(\mathcal A = (Q, A, \delta, F)\) and a state \(x \in Q\)
- Output: a finite automaton \(\mathrm{Det}(\mathcal A) = (Q', A, \delta', F')\) and a state \(S \in Q'\) such that \[\mathcal L(\mathcal A, x) = \mathcal L(\mathrm{Det}(\mathcal A), S)\]
The automaton construction we are going to describe below, \(\mathrm{Det}\), is called determinization. Before we get into it, it helps to see a bit of intuition for where it comes from. Remember how a state of an automaton reads a word:
- Write the word to a tape and place a marker on the starting state.
- If there is no letter under the tape head, stop. Otherwise, Read the letter under the tape head. Call it \(a_i\).
- For each active state \(y\) (a state with a marker on it) with a transition \(y \xrightarrow{a_i} z\), remove the marker from \(y\) and place a marker on \(z\). Now \(z\) is active.
- Move the tape head to the right and go back to 2.
We know that \(x_0\) accepts \(aa\) here because one of the active states is accepting at the end. But how do we turn this into a deterministic path?
At the moment, multiple states are active at each step. The idea behind turning this into a deterministic automaton is to create a state in our determinized automaton for each possible collection of active states. So, the first collection would just be \(\{x_0\}\), since that's where we start. The outgoing transitions from \(x_0\) are now collected into a single state: after reading an \(a\), the next active state would be the collection \(\{x_1, x_2\}\). Now repeat this again: after reading a second \(a\), collect all the outgoing transitions from \(\{x_1, x_2\}\) into a single transition to the collection \(\{x_3, x_4, x_5\}\). The collections \(\{x_0\}\), \(\{x_1, x_2\}\), and \(\{x_3, x_4, x_5\}\) are three of the states that we get from determinizing the automaton.
Above, the collection \(\{x_3, x_4, x_5\}\) is a accepting state because one of the states it contains is accepting: indeed, observe that \(x_0\) accepts \(aa\). This is where the definition of \(\mathrm{Det}(\mathcal A)\) below comes from.
- \(Q' = 2^Q\) is the set of subsets of \(Q\), i.e., the set of all collections of states of \(\mathcal A\),
- \(\delta' = \big\{ (U, a, V) \mid V = \bigcup_{x \in U} \delta(x, a) \big\}\), i.e., \(U \xrightarrow{a} V\) if \(V\) is the set of all states with incoming \(a\) transitions from states in \(U\),
- \(F' = \big\{ U \mid F \cap U \neq \emptyset\big\}\), i.e., a collection of states is accepting if it contains an accepting state of \(\mathcal A\)
The set \(2^Q\), consisting of all the subsets of \(Q\), is called the powerset of \(Q\). You might sometimes hear this determinization construction called the powerset construction.
- \(Q = \{x\}\)
- \(Q = \{x, y\}\)
- \(Q = \{x, y, z\}\)
- \(Q = \{\}\)
A two-state automaton \(\mathcal A_{eg}\) over the alphabet \(A = \{a, b\}\).
Notice that there are no \(b\)-transitions.
Formally, this state diagram represents the automaton \(\mathcal A_{eg} = (Q, A, \delta, F)\) given by \[\begin{aligned} Q &= \{x_0, x_1\} \\ A &= \{a, b\} \\ \delta &= \{(x_0, a, x_0), (x_0, a, x_1), (x_1, a, x_0)\}\\ F &= \{x_1\} \end{aligned}\] This is not a deterministic automaton: there are two \(a\)-transitions leaving \(x_0\).
In the determinization of \(\mathcal A_{eg}\), the states are collections of states from \(\mathcal A_{eg}\) (in other words, subsets of \(Q\)). There are two states in \(\mathcal A_{eg}\), so there are \(2^2 = 4\) states in \(\mathrm{Det}(\mathcal A_{eg})\). Specifically, \[ Q' = 2^Q = \big\{ \{\}, \{x_0\}, \{x_1\}, \{x_0,x_1\} \big\} \] Let's look at an example of a transition: since \(x_0 \xrightarrow{a} x_0\) and \(x_0 \xrightarrow{a} x_1\), the transition \(\{x_0\} \xrightarrow{a} \{x_0, x_1\}\) appears in \(\mathrm{Det}(\mathcal A_{eg})\). Similarly, since \(x_1 \xrightarrow{a} x_0\), there is a self \(a\)-transition \(\{x_0, x_1\} \xrightarrow{a} \{x_0, x_1\}\) in \(\mathrm{Det}(\mathcal A_{eg})\). In total, the state diagram looks like this:
A state diagram for the determinization of the two state automaton, \(\mathrm{Det}(\mathcal A_{eg})\).
A two-state nondeterministic automaton \(\mathcal A = (Q, A, \delta, F)\), where
\[\begin{aligned}
Q &= \{x_0, x_1\} \\
A &= \{a, b\} \\
\delta &= \{
(x_0, a, x_0),
(x_0, a, x_1),
(x_0, a, x_1),
(x_1, a, x_0),
(x_1, b, x_1),
\} \\
F &= \{x_0, x_1\}
\end{aligned}\]
- \(\{x_0\} \xrightarrow{a} \{x_0,x_1\}\) is a transition in \(\mathrm{Det}(\mathcal A)\). What about \(\{x_0\} \xrightarrow{a} \{x_1\}\)?
- What are the accepting states of \(\mathrm{Det}(\mathcal A)\)?
- Draw the complete state diagram for \(\mathrm{Det}(\mathcal A)\).
Determinization Works
The point of the previous few exercises and problems is to show that \(\mathrm{Det}(\mathcal A)\) is a total deterministic finite automaton when \(\mathcal A\) is a finite automaton. Our next goal is to show that this automaton construction is correct, in the sense that accomplishes our goal for this construction. The way we designed \(\mathrm{Det}(\mathcal A)\) is that for any state \(x \in Q\), \(x\) accepts the same language as \(\{x\}\).
Let \(w \in \mathcal L(\mathcal A, x)\). Since \(w\) is a word, there are letters \(a_1, a_2, \dots, a_n \in A\) such that \(w = a_1a_2\cdots a_n\). Now \(x\) accepts \(w\), which means that there is a path through \(\mathcal A\) of the form \[ x \xrightarrow{a_1} x_1 \xrightarrow{a_2} \cdots \xrightarrow{a_n} x_n \] where \(x_n \in F\). Let's build the path from \(\{x\}\) to an accepting state of \(\mathrm{Det}(\mathcal A)\). Recursively, we define \(U_0 = \{x\}\), and for each \(0 \le i < n\), \[ U_{i+1} = \bigcup_{y \in U_i} \delta(y, a_i) \] So, for example, \(U_1\) is the set of states with incoming transitions from \(x\), \(U_2\) is the set of states with incoming transitions from states in \(U_1\), and so on. Notice that \(x_i \in U_i\) for each \(1 \le i \le n\).
By the definition of the transition relation \(\delta'\) of \(\mathrm{Det}(\mathcal A)\), the \(U_i\) form the path \[ \{x_0\} = U_0 \xrightarrow{a_1} U_1 \xrightarrow{a_2} U_2 \xrightarrow{a_3} \cdots \xrightarrow{a_{n}} U_n \] Since \(x_n \in U_n\) and \(x_n \in F\) (is an accepting state of \(\mathcal A\)), \(U_n\) is an accepting state of \(\mathrm{Det}(\mathcal A)\). This means that \(w = a_1a_2\cdots a_n \in \mathcal L(\mathrm{Det}(\mathcal A), \{x\})\).
We have just shown that every element of \(\mathcal L(\mathcal A, x)\) is an element of \(\mathcal L(\mathrm{Det}(\mathcal A), \{x\})\). In other words, \(\mathcal L(\mathcal A, x) \subseteq \mathcal L(\mathrm{Det}(\mathcal A), \{x\})\).
This allows for the following proof of the Finite Recognizability Theorem.
Let \(L \in \mathsf{Fin}\). Then by definition of finite recognizability, there is a finite automaton \(\mathcal A = (Q, A, \delta, F)\) and a state \(x \in Q\) such that \(L = \mathcal L(\mathcal A, x)\). We need to find a total deterministic finite automaton \(\mathcal A'\) and a state \(z\) of \(\mathcal A'\) such that \(\mathcal L(\mathcal A', z) = L\). By the Determinized State Theorem, \(\mathcal A' = \mathrm{Det}(\mathcal A)\) is such an automaton: it is a total deterministic automaton with the state \(\{x\}\), and \[ L = \mathcal L(\mathcal A, x) = \mathcal L(\mathrm{Det}(\mathcal A), \{x\}) \] It follows that \(L\) is a total deterministic finitely recognizable language, i.e., \(L \in \mathsf{TDFin}\).