Deterministic automata

In the last lecture, we have shown that, for each regular expression $ e$ , we can construct an NFA $ M$ , such that $ L(e) = L(M)$ .

While NFAs are a computational instrument for establishing acceptance, they are not an ideal one. Nondeterminism is difficult to translate per. se. in a programming language. It is also quite inefficient. We illustrate this via our previous example, below. Recall our NFA representation in Haskell:

data NFA = NFA {delta :: [(Int,Char,Int)], fin :: [Int]}

The following code checks if a word $ w$ and NFA $ nfa$ , whether $ w\in L(nfa)$ .

check :: String -> NFA -> Bool
check w nfa = chk (w++"!") nfa [0]
              where chk (x:xs) nfa set  
                          | (x == '!') && ([s | s <- set, s 'elem' fin nfa] /= []) = True
                          | set == [] = False
                          | otherwise = chk xs nfa [s' | s <- set, (s,x,s') <- delta nfa]  

At each step, the procedure builds the set of all successor states $ s'$ which can be reached from some current state $ s$ , on current input $ x$ .

This procedure can be optimised, if we make the following observations:

• we can pre-compute all possible state-combinations (and rule-out those which are inaccessible);
• in so doing, we will have only one transition from one state-combination to the other;
• the number of state combinations may be exponential in the general case, but we will build them only once, not for each word, on each test $ w\in L(nfa)$ , as it is done in the above example.

These observations lead us to Deterministic Finite Automata (DFA). More formally, we can easily obtain the definition for DFAs by enforcing the following restriction on the transition relation $ \Delta$ :

• $ \Delta : K \times \Sigma \rightarrow K$ ,
• the function $ \Delta$ is total (i.e. it is defined for all possible values in its input).

Thus:

• each transition must occur on a symbol,
• exactly one transition is possible for each state-symbol combination

Definition (DFA):

A Deterministic Finite Automata is an NFA, where $ \Delta : K \times \Sigma \rightarrow K$ is a total function. In what follows, we write $ \delta$ instead of $ \Delta$ to refer to the transition function of a DFA.

Definition (Configuration):

The configuration of a DFA is an element of $ K\times\Sigma^*$ .

Definition (One-step move):

We call $ \vdash_M \subseteq (K\times \Sigma^*) \times (K\times \Sigma^*)$ a one-step move relation, over configurations. The relation is defined as follows:

• $ (q,w) \vdash_M (q',w')$ iff $ w=cw'$ for some symbol $ c$ of the alphabet, and $ \delta(q,c)=q'$

We write $ \vdash_M^*$ to refer to the reflexive and transitive closure of $ \vdash_M$ . Hence, $ (q,w)\vdash_M^*(q',w')$ means that DFA $ M$ can reach configuration $ (q',w')$ from configuration $ (q,w)$ is zero or more steps.

Definition (Acceptance):

We say a word $ w$ is accepted by a DFA $ M$ iff $ (q_0,w)\vdash_M^*(q,\epsilon)$ and $ q\in F$ ($ q$ is a final state).

Example(s)

Let $ M=(K,\Sigma, \Delta, q_0, F)$ be an NFA. We assume $ M$ does not contain transitions on words of length larger than 1. If $ (q,w,q')\in\Delta$ for some $ w=c_1\ldots c_n$ of size 2 or more, we construct intermediary states $ q^1, \ldots, q^{n+1}$ as well as transitions $ (q^1,c_1,q^2), \ldots (q^n,c_n,q^{n+1})$ , where $ q=q^1$ and $ q'=q^n$

We denote by $ E_M(q) = \{p\in K\mid (q,\epsilon)\vdash^*_M (p,\epsilon)\}$ , the $ \epsilon$ -closure of state $ q$ . In effect $ E_M(q)$ contains all states reachable from $ q$ via $ \epsilon$ -transitions. When the automaton $ M$ is understood from the context, we omit the subscript and simply write $ E(q)$ .

We build the DFA $ M'=(K',\Sigma, \delta, q_0', F')$ as follows:

• $ K'=2^{K}$ - each state of $ M'$ is a subset of states from the NFA. It may be the case that some such states are not reachable, hence we shall ignore them from our construction;
• $ q_0' = E(q_0)$
• $ \delta(Q,c) = \displaystyle \cup_{q\in Q, (q,c,q')\in\Delta} E(q')$ - a transition from $ Q$ on symbol $ c$ ends in the reunion of all $ \epsilon$ -closures of states in $ M$ reachable from some member of $ Q$ on symbol $ c$ .
• $ F'=\{Q\subseteq K'\mid Q \cap F \neq\emptyset\}$ - a state is final in $ M'$ iff it contains

some final state in $ M$ .

Proposition (1):

For all $ q,p\in K$ , $ (q,w)\vdash^*_M (p,\epsilon)$ iff $ (E(q),w)\vdash^*_{M'}(P,\epsilon)$ , for some $ P$ such that $ p\in P$ .

The proposition states that, for each path in NFA $ M$ starting on $ q$ which consumes word $ w$ (hence ends up in configuration $ (p,\epsilon)$ ), there is an equivalent path in the DFA $ M'$ , which starts in the $ \epsilon$ -closure of $ q$ and ends in some state $ P$ which contains $ p$ — and vice-versa.

Proposition 1 is essential for proving the following result:

Let $ M'$ be a DFA constructed from NFA $ M$ according to the above rules. Then $ L(M)=L(M')$ . </blockquote>

Proof:

The languages of the two machines coincide if, for all words: $ w\in L(M)$ iff $ w\in L(M')$ , thus:

• $ (q_0,w)\vdash^*_M (p,\epsilon)$ with $ p\in F$ iff $ (E(q_0),w)\vdash^*_{M'}(P,\epsilon)$ with $ p\in P$ .

The above statement follows immediately from Proposition 1, where:

• $ q$ is the initial state of $ M$
• $ p$ is some final state of $ M$

as well as from the definition of $ F'$ .

We now turn to the proof of Proposition 1:

Proof:

The proof is by induction over the length of the word $ w$ .

Basis step: $ \mid w\mid=0$ that is $ w=\epsilon$

• direction $ \implies$ :
  1. Suppose $ (q,\epsilon) \vdash^*_M(p,\epsilon)$ .
  2. From the definition of $ E$ and 1., we have that $ p\in E(q)$ .
  3. Since $ \vdash^*_{M'}$ is reflexive, we have $ (E(q),\epsilon)\vdash^*_{M'}(E(q),\epsilon)$ .
  4. Therefore, we have $ E(q),\epsilon)\vdash^*_{M'}(P,\epsilon)$ with $ p\in P$ : $ P$ is actually $ E(q)$ .

• direction $ \impliedby$ :
  1. Suppose $ (E(q),\epsilon) \vdash^*_{M'} (P,\epsilon)$
  2. Since $ \delta$ does not allow $ \epsilon$ -transitions (and $ \vdash^*_{M'}$ is reflexive), it follows that $ E(q)=P$ .
  3. By the definition of $ E$ , we have that $ (q,\epsilon)\vdash^*_{M}(p,\epsilon)$ for any $ p\in E(q)$ .

Induction hypothesis: suppose that the claim is true for all strings w such that $ \mid w\mid\leq k$ for $ k\geq0$

Induction step: we prove for any string $ w$ of length $ k+1$ ; let $ w'=wa$ (hence $ a$ is the last symbol of $ w'$ ).

• direction $ \implies$ :
  1. Suppose $ (q,wa)\vdash^*_{M} (p,\epsilon)$ .
  2. By the definition of $ \vdash^*$ , we have: $ (q,w)\vdash^*_{M} (r_1,a) \vdash_{M} (r_2,\epsilon)\vdash^*_{M} (p,\epsilon)$ . In other words, there is a path from $ q$ which takes us to $ r_1$ by consuming $ w$ , then to $ r_2$ via a one-step transition, then to $ p$ in zero or more $ \epsilon$ -transitions. Notice that $ p$ may be equal to $ r_2$ , which is taken into account since $ \vdash^*_{M}$ is reflexive.
  3. By the construction of $ \vdash^*_{M}$ , we also have $ (q,w)\vdash^*_{M}(r_1,\epsilon)$
  4. From 3. by induction hypothesis, we have $ (E(q),w)\vdash^*_{M'}(R_1,\epsilon)$ with $ r_1 \in R_1$
  5. By construction of $ \vdash^*_{M'}$ , we have $ (E(q),wa)\vdash^*_{M'}(R_1,a)$
  6. Since $ (r_1,a) \vdash_{M} (r_2,\epsilon)$ , by the definition of $ \delta$ , we have $ E(R_2) \subseteq \delta(R_1,a)$ .
  7. Since $ (r_2,\epsilon) \vdash_{M} (p,\epsilon)$ it follows that $ p \in E(r_2)$ , and therefore $ p \in \delta(R_1,a)$ .
  8. In effect, from 5. and 7. we have shown that $ (E(q),wa)\vdash^*_{M'}(R_1,a)\vdash_{M'}(R_2,\epsilon)$ with $ p\in R_2$ , which concludes our proof.

• direction $ \impliedby$ :
  1. Suppose $ (E(q),wa)\vdash^*_{M'}(P,\epsilon)$ .
  2. Since no $ \epsilon$ -transitions are allowed in a DFA, we must have: $ (E(q),wa)\vdash^*_{M'}(R,a)\vdash_{M'}(P,\epsilon)$ .
  3. By construction of $ \vdash^*_{M'}$ : $ (E(q),w)\vdash^*_{M'}(R,\epsilon)$ .
  4. By induction hypothesis, $ (q,w)\vdash^*_{M}(r,\epsilon)$ , with $ r\in R$ .
  5. By construction of $ \vdash^*_{M}$ : $ (q,wa)\vdash^*_{M}(r,a)$ .
  6. Since $ (R,a)\vdash_{M'}(P,\epsilon)$ , by the definition of $ \delta$ , for some $ r\in R$ , $ (r,a,x)\in\Delta$ and $ E(x)\subseteq P$ .
  7. From 6. $ (q,wa)\vdash^*_{M}(r,a)\vdash_{M}(x,\epsilon)$ with $ x\in E(x)\subseteq P$ which completes our proof.