15-251Great Theoretical Ideas in Computer ScienceDeterministic Finite AutomataLecture 20 (October 30, 2008)Let me show you a machine so simple that you can understand it in less than two minutes00,1001110111 111111 The machine accepts a string if the process ends in a double circle00,100111 The machine accepts a string if the process ends in a double circleAnatomy of a Deterministic Finite Automatonstatesstatesq0q1q2q3start state (q0)accept states (F)Anatomy of a Deterministic Finite Automaton00,100111q0q1q2q3The alphabet of a finite automaton is the set where the symbols come from:The language of a finite automaton is the set of strings that it accepts{0,1}0,1q0L(M) =All strings of 0s and 1s!The Language of Machine Mq0q10011L(M) ={ w | w has an even number of 1s}An alphabet ! is a finite set (e.g., ! = {0,1})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 xThe unique string of length 0 will be denoted by " and will be called the empty or null stringNotationA language over ! is a set of strings over ! Q is the set of states! is the alphabet" : Q # ! # Q is the transition functionq0 $ Q is the start stateF % Q is the set of accept statesA finite automaton is a 5-tuple M = (Q, !, ", q0, F) L(M) = the language of machine M = set of all strings machine M acceptsQ = {q0, q1, q2, q3}! = {0,1}" : Q # ! " Q transition functionq0 $ Q is start stateF = {q1, q2} % Q accept statesM = (Q, !, ", q0, F) where"01q0q0q1q1q2q2q2q3q2q3q0q2q2 00,100111q0q1q3MInput StringResultabaaabbaabba&bbabaaababAcceptRejectAcceptAccept“ABA” The AutomatonWhat machine accepts this language?L = all strings in {a,b}* that contain at least one aba,baWhat machine accepts this language?L = strings with an odd number of b’s and any number of a’saabbWhat is the language accepted by this machine?L = any string ending with a babbaWhat is the language accepted by this machine?ba,babaL(M) = any string with at least two a’sWhat machine accepts this language?L = any string with an a and a babbaaa,bbWhat machine accepts this language?L = strings with an even number of ab pairsabbaababq q00101q0q0010010,1Build an automaton that accepts all and only those strings that contain 001L = all strings containing ababb as a consecutive substringqqabbbqaqabaabaaqababbqababbba,baaInvariant: I am state s exactly when s is the longest suffix of the input (so far) forming a prefix of ababb.Input: Text T of length t, string S of length nThe “Grep” ProblemProblem: Does string S appear inside text T?a1, a2, a3, a4, a5, …, atNaïve method: Cost: Roughly nt comparisonsAutomata SolutionBuild a machine M that accepts any string with S as a consecutive substringFeed the text to MCost:As luck would have it, the Knuth, Morris, Pratt algorithm builds M quicklyt comparisons + time to build MGrepCoke MachinesThermostats (fridge)ElevatorsTrain Track SwitchesLexical Analyzers for ParsersReal-life Uses of DFAsA language is regular if it is recognized by a deterministic finite automatonL = { w | w contains 001} is regularL = { w | w has an even number of 1s} is regularUnion TheoremGiven two languages, L1 and L2, define the union of L1 and L2 as L1 ' L2 = { w | w $ L1 or w $ L2 } Theorem: The union of two regular languages is also a regular languageTheorem: The union of two regular languages is also a regular languageProof Sketch: Let M1 = (Q1, !, "1, q0, F1) be finite automaton for L1 and M2 = (Q2, !, "2, q0, F2) be finite automaton for L2We want to construct a finite automaton M = (Q, !, ", q0, F) that recognizes L = L1 ' L2 12Idea: 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 # Q2Theorem: The union of two regular languages is also a regular languageq0q10011p0p11100q0,p0q1,p011q0,p1q1,p1110000Automaton for Unionq0,p0q1,p011q0,p1q1,p1110000Automaton for IntersectionTheorem: The union of two regular languages is also a regular languageCorollary: Any finite language is regularThe Regular OperationsUnion: A ' B = { w | w $ A or w $ B } Intersection: A ( B = { w | w $ A and w $ B } Negation: ¬A = { w | w ) A } Reverse: AR = { w1 …wk | wk …w1 $ 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 OperationsWe 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?i.e., a bunch of a’s followed by an equal number of b’sConsider the language L = { anbn | n > 0 }No finite automaton accepts this languageCan you prove this?anbn is not regular. No machine has enough states to keep track of the number of a’s it might encounterThat is a fairly weak argument Consider the following example…L = strings where the # of occurrences of the pattern ab is equal to the number of occurrences of the pattern baCan’t be regular. No machine has enough states to keep track of the number of occurrences of abM accepts only the strings with an equal number of ab’s and ba’s!bbabaaababLet 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 objectLetterbox 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)Theorem: L= {anbn | n > 0 } is not regularProof (by contradiction):Assume that L is regularThen there exists a machine M with k states that accepts LFor each 0 + i + k, let Si be the state M is in after reading ai,i,j + k such that Si = Sj, but i - jM will do the same thing on aibi and ajbi But a valid M must reject ajbi and accept aibiAdvertisementYou can learn much more about these creatures in the FLAC course.Formal Languages, Automata, and Computation• There is a unique smallest automaton for any regular language• It can be found by a fast algorithm.Deterministic Finite Automata• Definition• Testing if they accept a string• Building automataRegular Languages• Definition• Closed Under Union, Intersection, Negation• Using Pigeonhole Principle to show language not regularHere’s What You Need to
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