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Lecture2

# Lecture2 - FORMAL LANGUAGES AUTOMATA AND COMPUTABILITY...

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Unformatted text preview: FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY 15-453 FIRST HOMEWORK IS DUE Thursday, January 22 NON-DETERMINISM THURSDAY JAN 18 NON-DETERMINISM AND THE PUMPING LEMMA THURSDAY JAN 18 Q = {q , q 1 , q 2 , q 3 } Σ = {0,1} δ : Q × Σ → Q transition function * q ∈ Q is start state F = {q 1 , q 2 } ⊆ Q accept states M = (Q, Σ, δ , q , F) where δ 1 q q q 1 q 1 q 2 q 2 q 2 q 3 q 2 q 3 q q 2 * q 2 0,1 1 1 1 q q 1 q 3 M Q is the set of states (finite) Σ is the alphabet (finite) δ : Q × Σ → Q is the transition function q ∈ 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 , 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 ∈ 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 , F) deterministic DFA Let w 1 , ... , w n ∈ Σ and w = w 1 ... w n ∈ Σ* Then M accepts w if there are r , r 1 , ..., r n ∈ Q, s.t . 1. r = q 2. δ (r i , w i+1 ) = r i+1 , for i = 0, ..., n-1, and 3. r n ∈ F Q is the set of states (finite) Σ is the alphabet (finite) δ : Q × Σ → Q is the transition function q ∈ 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 , 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 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 } 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...
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Lecture2 - FORMAL LANGUAGES AUTOMATA AND COMPUTABILITY...

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