Lecture2 - FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 } = {0,1} : Q Q transition function * q Q is start state F = {q 1 , q 2 } Q accept states M = (Q, , , q , F) where 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) is the alphabet (finite) : Q 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, , , q , F) deterministic DFA M accepts a string w if the process ends in a double circle Q is the set of states (finite) is the alphabet (finite) : Q 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, , , q , F) deterministic DFA Let w 1 , ... , w n and w = w 1 ... w n * Then M accepts w if there are r , r 1 , ..., r n Q, s.t . 1. r = q 2. (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) is the alphabet (finite) : Q 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, , , 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 * | 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 * | 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...
View Full Document

This note was uploaded on 02/29/2012 for the course CS 15-453 taught by Professor Edmundm.clarke during the Spring '09 term at Carnegie Mellon.

Page1 / 14

Lecture2 - FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online