Lecture2

# Lecture2 - 15-453 FORMAL LANGUAGES AUTOMATA AND...

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FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY 15-453

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FIRST HOMEWORK IS DUE Thursday, January 22
NON-DETERMINISM THURSDAY JAN 18

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NON-DETERMINISM AND THE PUMPING LEMMA THURSDAY JAN 18
Q = {q 0 , q 1 , q 2 , q 3 } Σ = {0,1} δ : Q × Σ → Q transition function * q 0 Q is start state F = {q 1 , q 2 } Q accept states M = (Q, Σ, δ , q 0 , F) where δ 0 1 q 0 q 0 q 1 q 1 q 2 q 2 q 2 q 3 q 2 q 3 q 0 q 2 * q 2 0 0,1 0 0 1 1 1 q 0 q 1 q 3 M

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Q is the set of states (finite) Σ is the alphabet (finite) δ : Q × Σ → Q is the transition function q 0 Q is the start state F Q is the set of accept states A ^ finite automaton ^ is a 5-tuple M = (Q, Σ, δ , q 0 , F) deterministic DFA M accepts a string w if the process ends in a double circle
Q is the set of states (finite) Σ is the alphabet (finite) δ : Q × Σ → Q is the transition function q 0 Q is the start state F Q is the set of accept states A ^ finite automaton ^ is a 5-tuple M = (Q, Σ, δ , q 0 , F) deterministic DFA Let w 1 , . .. , w n Σ and w = w 1 ... w n Σ* Then M accepts w if there are r 0 , r 1 , . .., r n Q , s.t . 1. r 0 = q 0 2. δ ( r i , w i+1 ) = r i+1 , for i = 0, . .., n-1, and 3. r n F

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Q is the set of states (finite) Σ is the alphabet (finite) δ : Q × Σ → Q is the transition function q 0 Q is the start state F Q is the set of accept states A ^ finite automaton ^ is a 5-tuple M = (Q, Σ, δ , q 0 , F) deterministic DFA A language L is regular if it is recognized by a deterministic finite automaton, i.e. if there is a DFA M such that L = L ( M ). L( M ) = set of all strings machine M accepts
UNION THEOREM The union of two regular languages is also a regular language Intersection THEOREM The intersection of two regular languages is also a regular language

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Complement THEOREM The complement of a regular languages is also a regular language In other words, if L is regular than so is ¬ L, where ¬ L= { w Σ* | w L } Proof ?
THE REGULAR OPERATIONS Union: A B = { w | w A or w B } Intersection: A B = { w | w A and w B } Negation: ¬ A = { w Σ* | w A } Reverse: A R = { w 1 …w k | w k …w 1 A } Concatenation: A B = { vw | v A and w B } Star: A* = { w 1 …w k | k ≥ 0 and each w i A }

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Reverse THEOREM The reverse of a regular languages is also a regular language
REVERSE CLOSURE Regular languages are closed under reverse Assume L is a regular language and M recognizes L We build M R that accepts L R If M accepts w then w describes a directed path in M from start to an accept state Define M R as M with the arrows reversed

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Lecture2 - 15-453 FORMAL LANGUAGES AUTOMATA AND...

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