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Lec02BRE - y ε F2 If M1 has 4 states and M2 has 3 states...

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Regular Operations by Ariel De Prada Regular Operations (let A and B be languages) U – union; A U B = {x | x ε A V x ε B} o – concatenation; A o B = {xy | x ε A ٨ y ε B} Example: 01101 E ε A ٨ 01101 ε B 0 ε A ٨ 1101 ε B 01 ε A ٨ 101 ε B 011 ε A ٨ 01 ε B 0110 ε A ٨ 1 ε B 01101 ε A ٨ E ε B * - star; A* = {x1, x2, x3, …, xk | k >= 0 ٨ xi ε A} for all i; 1 <= I <= k The union of 2 regular languages is in fact a regular language. Example M1 = { Q1, Σ, δ1, q1, F1} M2 = { Q2, Σ, δ2, q2, F2} Build a machine M3 so that the language it accepts is A1 U A2 Q3 = Q1 X Q2 (Cartesian Product) Σ stays the same δ3 = Q3 X Σ => Q3 q3 = (q1, q2) F3 = F1 X F2 = {(x,y) | x ε F1
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Unformatted text preview: ٧ y ε F2} If M1 has 4 states and M2 has 3 states then we will need at most 12 states in M3 to keep track of all combinations of states. We can think of the combinations as ordered pairs. To find a transition of the machine that results from A U B you must run both machines on the input and the resulting state is the concatenation of the two resulting states. Example: Start State (S1, T1) Q3 = {(S1, T1), (S1, T2), (S1, T3), (S2, T1), (S2, T2), (S2, T3)} Σ = (0,1) δ3 = 1 S1,T1 S1,T2 S2,T1 S1,T2 S1,T3 S2,T2 S1,T3 S1,T1 S2,T3 S2,T1 S2,T2 S1,T1 S2,T2 S2,T3 S1,T2 S2,T3 S2,T1 S1T3 Finish State Set {(S1,T1), (S1,T2), (S1,T3), (S2,T1)}...
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