The fact that DFAs are limited, it allows us to optimize a DFA. Specifically, we ask: “does this DFA have a minimal number of states?” Formally: can we accept the same language with a particular DFA with another DFA with less states? For every language L, there is a unique (up to state relabeling) minimal-state DFA M^{*} such that L(M^{*}) = L. Furthermore, there exists an efficient algorithm which, given DFA M, will produce this unique M^{*}. (btw; there isn’t a uniquely minimal NFA) Now let’s show it; in many parts. extending \delta Given a DFA, we want to extend its transition function to \Delta: Q \times \Sigma \to Q (i.e. string transitions instead of character transitions)

Meaning, \Delta(q,w) is the state M reached after reading w, starting from state q. Meaning, \Delta(q_0, w) \in F means IFF that w \in L(M). So: w \in \Sigma^{*} distinguishes states q_1, q_2 if Δ(q_1, w) ∈ F ⇔ Δ(q_2, w) ¬ ∈ F—this is because if the right is true, passing w at each of those states do not allow you to forget where you started with (as transitions of both after q_1 and q_2 are the same). Hence, those states are distinguished by w. That is, w allows you to tell which one of q_1 or q_2 you are in. distinguishable states We define state p as distinguishable from state q if there exists some w \in \Sigma^{*} such that exactly one of \Delta(p,w) or \Delta(q,w) is an accept state. That is, w distinguishes p and q. This means that these two states have a different function: you can’t merge them. Conversely, state p and q are indistinguishable if they are not distinguishable; that is, for all w \in \Sigma^{*}, \Delta(p,w) \in F \Leftrightarrow \Delta(q,w) \in F. Indistinguishable states are redundant. ~ Let us define a binary relation ~ on the states of M. Specifically, p \sim q IFF p is indistinguishable from q. That is, p \not\sim q if p is distinguishable from q. \sim is an equivalence relation. state partitioning since ~ is an equivalence relation, we can partition states Q into disjoint equivalence classes. that is, each of the equivalence relation for q means that [q] := \left\{p | p \sim q\right\}; because of transitivity, naming any element of this group will give you the whole group. DFA minimization algorithm input: DFA M output: DFA M_{\min} such that… L(M) = L\left(M_{MIN}\right) M_{MIN} has no inaccessible states (remove any states not accessible from the start state) M_{MIN} is irreducible: for all states p,q\neq p \in Q_{M_{MIN}}, p \not \sim q steps remove inaccessible states lol table-filling algorithm we now want to obtain all of our non distinguishable equivalent state classes. Scratch-work: States of M_{MIN} will be the equivalence classes of the states of M; we cover these equivalence classes with a DP algorithm. Build a DP table: D_{M} =\left\{(p,q) | p,q \in Q, p \sim q\right\}. Cluster them into equivalence classes: EQUIV_{M} = \left\{[q] | q \in Q\right\} To build this table: we know how to find the pairs that \varepsilon distinguishes; then, we iterate to lengthen string length and find those pairs distinguishable with longer strings. The pairs of states then left over would be indistinguishable. base case: (p,q) such that p accepts and q rejects, because they are distinguished by the empty string. iteration: for each grid p,q and see if any symbol \sigma \in \Sigma satisfy—for \delta (p, \sigma) = p’, \delta (q, \sigma) = q’, we have p’ \not \sim q’; if that’s the case, then p \not \sim q we repeat until no more slots can be marked as non-distinguishable by going through an entire iteration without change Time complexity: O(|\Sigma| |q|^{2}) (we go through each vocab item for each state). This algorithm iterates for at most q^{2} times. theorem: M_{MIN} is the unique minimal DFA that is equivalent to M. define our new DFA \begin{equation} M_{MIN} = \left(Q_{MIN}, \Sigma, \delta_{MIN}, q_0_{MIN}, F_{MIN}\right) \end{equation} we will define these vis a vi our new equivalence classes:

the transition function just needs to apply that of m on one member of your equivalence class:

this accepts the same language due to the definition of indistinguishable states (i.e. these equivalence classes). properties as a by product, this algorithm bounds the shortest distinguishing string between two distinguishable states by q^{2} because at each iteration of the algorithm it add at most one char to the end of the distinguishing string, and our system runs at most q^{2} times proof that table-filling algorithm works if (p,q) is marked, then p \not\sim q. Proof: We do induction. Invariant: whatever happened before is correct. Base case: if (p,q) is marked D at the start, then one state is in F and the other isn’t, and so \varepsilon distinguishes p distinguishes q. Suppose (p,q) is marked D at a later point, then there exists states p’, q’ such that p’ \not \sim q’ and also p’ = \delta (p, \sigma ) and q’ = \delta (q, \sigma). Given p’ \not \sim q’, there’s a string w such that \Delta(p’, w) \in F \Leftrightarrow \Delta (q’, w) \not \in F; since we know there exists a symbol \sigma to get from p,q to p, q, we can concatenate that to w like so (\sigma w, processing our new symbol before the rest of the string) to show that \Delta(p, w) \in F \Leftrightarrow \Delta (q, w) \not \in F, so p \not \sim q. if (p,q) is not marked, then p \sim q. Let’s prove this by contradiction. Suppose for contradiction the pair (p,q) is not by D, yet p \not \sim q. Since p \not \sim q, then there exists a string w such that |w| > 0 (this is true because if w was the empty string it would fall into our base case already), \Delta(p, w) \in F \Leftrightarrow \Delta (q, w) \not \in F. If there are many such bad pairs, we choose the one with the shortest distinguishing string w that marks them. Since w’ is non-empty, let’s write w = \sigma w’. Let p’ = \delta (p, \sigma) and q’ = \delta(q, \sigma). So, this means that p’ \not \sim q’ (because we just play the rest of w’ to distinguish them)—yet, they wouldn’t have been marked because if they did our algorithm would have then marked p \not \sim q as well. This means that also p’ and q’ is also a bad pair. Yet, we chose p, q to be the bad pair with the shortest distinguishing string; p’, q’ have shorter distinguishing strings (namely, w’). So, we reach contradiction. DFA minimization is unique theorem: M_{MIN} is the unique minimal DFA that is equivalent to M. theorem proof if M’ is a another minimal DFA for M, then M’ has no inaccessible states and is irreducable; then, lemma 1 applies so there is an isomorphism between M’ and M_{MIN}. lemma 1 Lem. 1: Suppose there’s some irreducible M’ where L(M’) = L(M_{\min}) with no inaccessible states; we have that there exist an isomorphism between M’ and M_{\min}. Proof: we need to construct such an isomorphism. q_0_{\min} \to q_0’ recursion: if p_{\min} \to p’ and \delta_{\min}(p_{\min}, \sigma) =q_{\min } and \delta’(p’, \sigma) =q’ then we assign q_{\min } \to q’ to show isomorphism defined everywhere well defined bijective preserve all transitions (for all p \to p’, then \delta_{\min}(p, \sigma) \to \delta’(p’, \sigma) defined everywhere for all states q of M_{\min }, there exists q’ of m’ such that q \to q’ if q \in M_{\min}, then there is a string w such that \Delta_{\min}(q_0_{\min}, w) = q. let q’ = \Delta’(q’_{0}, w), then q \to q’. we show this by inducing the fact that we can define the mapping of these stages one at a time. one-to-one (injective) we show this as a contradiction. suppose by contradiction p \neq q such that p \to q’ and q \to q’; again, this means that we would (like in the well-defined case, but this time in M_{\min } now), tack on a string which sends us to accept and reject cases surjective for all states q’ of M’, there exists a state q of M_{\min} such that q \to q’. we can show this by noticing that for every state q’ in M’, we have some string w which M’ reaches to get there. we define q as the state of M_{\min} after applying w. We claim q \to q’. well-defined this needs to be a function, so we can’t have cases where its q \to q’ yet also q \to q’’ where q’ \neq q’’. suppose for contradiction q’ and q’’ are distinguishable in M note that this is absurd because the same exact string w caused accept and reject by pulling back the map in M_{\min}.