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minimization_of_state_machines [2008/09/20 18:01] vkuncak |
minimization_of_state_machines [2008/09/24 09:57] vkuncak |
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This is the process of //minimization// of $M$. | 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 $w' \in L(M)$). | We say that state machine $M$ distinguishes strings $w$ and $w'$ iff it is not the case that ($w \in L(M)$ iff $w' \in L(M)$). | ||
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==== Minimization Algorithm ==== | ==== Minimization Algorithm ==== | ||
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- | We show how to minimize a deterministic [[finite state automaton]]. | ||
=== Step 1: Remove unreachable states === | === Step 1: Remove unreachable states === | ||
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Observe that | Observe that | ||
- | - if $q \in F$ and $q' \notin F$ then $q$ and $q'$ are $\epsilon$-non-equivalent. | + | - 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. | + | - 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 two observations lead to an iterative algorithm for computing non-equivalence relation $\nu$ | + | 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) | - 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 | - 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 | ||
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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. | 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. | ||
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==== Minimality of Constructed Automaton ==== | ==== Minimality of Constructed Automaton ==== | ||
- | Note that if two 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. | + | 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. |
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+ | === Example === | ||
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+ | Construct automaton recognizing | ||
+ | * language {=,<=} | ||
+ | * language {=,<=,==} | ||