Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision | Next revision Both sides next revision | ||
|
closure_properties_of_finite_state_machines [2007/05/11 17:11] ghid.maatouk |
closure_properties_of_finite_state_machines [2007/05/24 00:42] vkuncak |
||
|---|---|---|---|
| Line 13: | Line 13: | ||
| Let $A = (\Sigma, Q_A, q_0^A, \Delta_A, F_A)$ and $B = (\Sigma, Q_B, q_0^B, \Delta_B, F_B)$.\\ | Let $A = (\Sigma, Q_A, q_0^A, \Delta_A, F_A)$ and $B = (\Sigma, Q_B, q_0^B, \Delta_B, F_B)$.\\ | ||
| Then the union of the two automata will be $C=(\Sigma, Q_A \cup Q_B, q_0, \Delta_A \cup \Delta_B \cup \{(q_0,\epsilon, q_0^A), (q_0,\epsilon, q_0^B)\}, F_A \cup F_B)$. | Then the union of the two automata will be $C=(\Sigma, Q_A \cup Q_B, q_0, \Delta_A \cup \Delta_B \cup \{(q_0,\epsilon, q_0^A), (q_0,\epsilon, q_0^B)\}, F_A \cup F_B)$. | ||
| + | |||
| ==== Intersection ==== | ==== Intersection ==== | ||
| - | Product of state spaces for deterministic machines.\\ | + | Product of state spaces, even for non-deterministic machines.\\ |
| - | For nondeterministic machines, must use DeMorgan's laws with complementation and union. Note that every time we want to complement an machine, we must determinize it first. | + | |
| ==== Concatenation ==== | ==== Concatenation ==== | ||