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--- minimization_of_state_machines [2008/09/24 09:58]
vkuncak
+++ minimization_of_state_machines [2015/04/21 17:32]
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-====== Minimization of Deterministic Finite State Machines ======
-We consider **deterministic** [[finite state machine]] $M = (\Sigma,Q,\delta,q_0,F)$.
-**Goal:** build a state machine $M'$ with the least number of states that accepts the language $L(M)$.
-  * we obtain a space-efficient, executable representation of a regular language
-This is the process of //minimization// of $M$.
-  * an easy case of minimizing size of 'generated code' in compiler
-We say that state machine $M$ distinguishes strings $w$ and $w'$ iff it is not the case that ($w \in L(M)$ iff $w' \in L(M)$).
-==== Minimization Algorithm ====
-=== Step 1: Remove unreachable states ===
-We first discard states that are not reachable from the initial state--such states are useless.  In resulting machine, for each state $q$ there exists a string $s$ such that $\delta(q_0,s)=q$, let $s_q$ one such string of minimal length.
-=== (Main) Step 2: Compute Non-Equivalent States ===
-We wish to merge states $q$ and $q'$ into same group as long as they "behave the same" on all future strings $w$, i.e.
-\[
-   \delta(q,w) \in F \mbox{ iff } \delta(q',w) \in F  \ \ \ (*)
-\]
-for all $w$.
-If the condition $(*)$ above holds, we called states **equivalent**.  If the condition does **not** hold, we call states $q$,$q'$ **non-equivalent**.
-States $q$ and $q'$ are $w$-non-equivalent if it is not the case that ($\delta(q,w) \in F \mbox{ iff } \delta(q',w) \in F$).
-Two states are non-equivalent iff they are $w$-non-equivalent for some string $w$.
-Observe that
-  - if $q \in F$ and $q' \notin F$ then $q$ and $q'$ are $\epsilon$-non-equivalent
-  - if $q$ and $q'$ are $w$-non-equivalent and we have $\delta(r,a)=q$, $\delta(r',a)=q'$ for some symbol $a \in \Sigma$, then $r$ and $r'$ are $aw$-non-equivalent
-  - conversely, if $r$ and $r'$ are $w'$-non-equivalent and $w$ is not an empty string, then for $w'=aw$ the states $\delta(r,a)$ and $\delta(r',a)$ are $w$-non-equivalent
-These observations lead to an iterative algorithm for computing non-equivalence relation $\nu$
-  - initially put $\nu = (Q \cap F) \times (Q \setminus F)$  (only final and non-final states are initially non-equivalent)
-  - repeat until no more changes: if $(r,r') \notin \nu$ and there is $a \in \Sigma$ such that $(\delta(r,a),\delta(r',a)) \in \nu$, then
-\[
-   \nu := \nu \cup \{(r,r')\}
-\]
-=== Step 3: Merge States that are not non-equivalent ===
-Relation $Q^2 \setminus \nu$ is an equivalence relation $\sim$.   We define the 'factor automaton' by merging equivalent states:
-  * the initial state is ${q_0}/_{\sim}$
-  * $Q/_{\sim} = \{ \{ y \mid x \sim y \} \mid x \in Q \}$
-  * $F/_{\sim} = \{ \{ y \mid x \sim y \} \mid x \in F \}$
-  * relation $r = \{ ([x],[y]) \mid (x,y) \in \delta \}$ is a function, and we can use it to define a new deterministic automaton (there is a transition in the resulting automaton iff there is a transition between two states in the original automaton)
-This is the minimal automaton.
-==== Correctness of Constructed Automaton ====
-Clearly, this algorithm terminates because in worst case all states become non-equivalent.  We will prove below that the resulting value $\nu$ is the non-equivalence relation, i.e. the complement of relation given by $(*)$ above.
-By induction, we can easily prove that if $(q,q') \in \nu$, then $q$ and $q'$ are non-equivalent.  Similarly we can show that if $q$ and $q'$ are $w$-non-equivalent for $w$ of length $k$, then $(q,q') \in \nu$ by step $k$ of the algorithm.  Because the algorithm terminates, this completes the proof that $\nu$ is the non-equivalence relation.
-Consequently, $Q^2 \setminus \nu$ is the equivalence relation.  From the definition of this equivalence it follows that if two states are equivalent, then so is the result of applying $\delta$ to them.  Therefore, we have obtained a well-defined deterministic automaton.
-==== Minimality of Constructed Automaton ====
-Note that if two distinct states are non-equivalent, there is $w$ such that states $\delta(q_0,s_q w)$ and $\delta(q_0,s_{q'} w)$ have different acceptance, so $M$ distinguishes $s_q w$ and $s_{q'}w$.  Now, if we take any other state machine  $M' = (\Sigma,Q',\delta',q'_0,F')$ with $L(M')=L(M)$, it means that $\delta'(q'_0,s_q) \neq \delta'(q'_0,s_{q'})$, otherwise $M'$ would not distinguish $s_q w$ and $s_{q'} w$.  So, if there are $K$ pairwise non-equivalent states in $M$, then a minimal finite state machine for $L(M)$ must have at least $K$ states.  Note that if the algorithm constructs a state machine with $K$ states, it means that $Q^2 \setminus \tau$ had $K$ equivalence relations, which means that there exist $K$ non-equivalent states.  Therefore, any other deterministic machine will have at least $K$ states, proving that the constructed machine is minimal.
-=== Example ===
-Construct automaton recognizing
-  * language {=,<nowiki><=</nowiki>}
-  * language {=,<=,<nowiki>==</nowiki>}
-Minimize the automaton.