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minimization_of_state_machines [2008/09/20 18:07] 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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| 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. | 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 {=,<=} | ||
| + | * language {=,<=,==} | ||
| + | Minimize the automaton. | ||