15 251 Let me show you a machine so simple that you can understand it in less than two minutes Great Theoretical Ideas in Computer Science Deterministic Finite Automata Lecture 20 October 30 2008 11 0 1 0 1 1 0111 111 1 0 0 1 The machine accepts a string if the process ends in a double circle Anatomy of a Deterministic Finite 1 0 Automaton states 1 q1 0 1 accept states F 0 0 0 1 strings of 0s and 1s L M All q12 q0 q0 The Language of Machine M The machine accepts a string if the start state q0 ends in q3a double circle states process Anatomy of a Deterministic Finite Automaton q1 0 1 0 1 q0 1 1 1 q0 0 0 q2 0 0 The alphabet of a finite automaton is the set where the symbols come from 1 0 1 q3 The language of a finite automaton is the set of strings that it accepts q1 L M w w has an even number of 1s Notation M Q q0 F where Q q0 q1 q2 q3 0 1 An alphabet is a finite set e g 0 1 q0 Q is start state F q1 q2 Q accept states A string over is a finite length sequence of elements of The set of all strings over is denoted by For x a string x is the length of x The unique string of length 0 will be denoted by and will be called the empty or null string Q Q transition function q1 0 1 q0 0 1 M q2 0 0 A language over is a set of strings over 1 q1 q1 q2 q2 q2 q3 q2 q3 q0 q2 ABA The Automaton a Q is the set of states is the alphabet Q Q is the transition function q0 Q is the start state L M the language of machine M set of all strings machine M accepts q0 0 q0 1 q3 A finite automaton is a 5 tuple M Q q0 F F Q is the set of accept states 1 b a a b a b b b a Input String Result aba Accept Reject Accept Accept aabb aabba What machine accepts this language What is the language accepted by this machine L all strings in a b that contain at least one a b a b b b a a a L any string ending with a b What machine accepts this language L strings with an odd number of b s and any number of a s What is the language accepted by this machine b a b b a a a a b b L M any string with at least two a s What machine accepts this language Build an automaton that accepts all and only those strings that contain 001 L any string with an a and a b 1 0 1 0 a q a b 0 1 b 0 q0 1 q00 q001 a b b a What machine accepts this language L strings with an even number of ab pairs L all strings containing ababb as a consecutive substring b a b a a q b a qa b qab a qaba b qabab b qababb b b b a a a a b a Invariant I am state s exactly when s is the longest suffix of the input so far forming a prefix of ababb The Grep Problem Input Text T of length t string S of length n Real life Uses of DFAs Grep Problem Does string S appear inside text T Coke Machines Na ve method Thermostats fridge a1 a2 a3 a4 a5 at Elevators Train Track Switches Cost Roughly nt comparisons Automata Solution Build a machine M that accepts any string with S as a consecutive substring Feed the text to M Cost t comparisons time to build M As luck would have it the Knuth Morris Pratt algorithm builds M quickly Lexical Analyzers for Parsers A language is regular if it is recognized by a deterministic finite automaton L w w contains 001 is regular L w w has an even number of 1s is regular Union Theorem Given two languages L1 and L2 define the union of L1 and L2 as L1 L2 w w L1 or w L2 Idea Run both M1 and M2 at the same time Q pairs of states one from M1 and one from M2 q1 q2 q1 Q1 and q2 Q2 Q1 Q2 Theorem The union of two regular languages is also a regular language Theorem The union of two regular languages is also a regular language Proof Sketch Let M1 Q1 1 q10 F1 be finite automaton for L1 and M2 Q2 2 q0 F2 be finite automaton for L2 Theorem The union of two regular languages is also a regular language 0 2 We want to construct a finite automaton M Q q0 F that recognizes L L1 L2 q0 0 1 1 q1 1 1 p0 0 0 p1 Automaton for Union 1 q0 p0 1 0 0 q1 p0 0 0 Theorem The union of two regular languages is also a regular language Corollary Any finite language is regular 1 q0 p1 1 q1 p1 Automaton for Intersection The Regular Operations Union A B w w A or w B 1 q0 p0 1 0 0 q1 p0 Reverse AR w1 wk wk w1 A 0 0 1 q0 p1 1 Intersection A B w w A and w B q1 p1 Negation A w w A Concatenation A B vw v A and w B Star A w1 wk k 0 and each wi A Regular Languages Are Closed Under The Regular Operations We have seen part of the proof for Union The proof for intersection is very similar The proof for negation is easy Are all languages regular Consider the language L anbn n 0 i e a bunch of a s followed by an equal number of b s No finite automaton accepts this language Can you prove this anbn is not regular No machine has enough states to keep track of the number of a s it might encounter That is a fairly weak argument Consider the following example a b a a b a b b a b M accepts only the strings with an equal number of ab s and ba s L strings where the of occurrences of the pattern ab is equal to the number of occurrences of the pattern ba Can t be regular No machine has enough states to keep track of the number of occurrences of ab Let me show you a professional strength proof that anbn is not regular Pigeonhole principle Given n boxes and m n objects at least one box must contain more than one object Letterbox principle If the average number of letters per box is x then some box will have at least x letters similarly some box has at most x Advertisement You can learn much more about these creatures in the FLAC course Formal Languages Automata and Computation There is …
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