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CS 208: Computing Theory Assoc. Prof. Dr. Brahim Hnich Faculty of Computer Sciences Izmir University of Economics

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Automata Theory Finite Sate Machines

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Motivation Question: What is a computer? Too complex to allow a precise mathematical model We introduce an abstract idealized computer Computational model One such a model is called finite automaton

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Key ideas Formal definition of a finite automaton Deterministic vs. nondeterministic finite automaton Regular languages Operations on regular languages Regular expressions Pumping Lemma

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Automata Theory Deterministic Finite Automaton

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Example: Automatic door Front pad Rear pad door Opens when someone approaches Holds open until person clears Doesn’t open when someone is standing behind the door

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Automatic door Sensors The door has 4 sensors FRONT: someone on rear pad BEHIND: someone on the rear pad BOTH: someone on both pads NEITHER: no one on either pad

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Automatic door states The door has two states OPEN CLOSE

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Automatic door: state diagram CLOSEDOPEN

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Automatic door: state diagram CLOSEDOPEN FRONT REAR BOTH

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Automatic door: state diagram CLOSEDOPEN REAR BOTH NEITHER FRONT REAR BOTH

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Automatic door: state diagram CLOSEDOPEN FRONT REAR BOTH NEITHER FRONT REAR BOTH

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Automatic door: state diagram CLOSEDOPEN FRONT NEITHER REAR BOTH NEITHER FRONT REAR BOTH

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Automatic door: state transition table CLOSEDOPEN FRONT NEITHER REAR BOTH NEITHER FRONT REAR BOTH NEITHERFRONT REARBOTH CLOSED OPEN CLOSED OPEN CLOSED CLOSED CLOSED OPEN OPEN OPEN

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Informal definition The machine M has States: q1,q2, and q3 Start state: q1 (incoming arrow from nowhere) Accept state: q2 (thick circle) State transitions: arrows q1 q2 q3 0110 0,1

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Informal definition The machine M on input string begins in start state q1 after reading each symbol, 1 takes transition with matching symbol q1 q2 q3 0110 0,1

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Informal definition After processing all the input string machine M produces output accept if M is in an accepting state reject otherwise q1 q2 q3 0110 0,1

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Informal definition What happens on input string? 1101 0010 01100 q1 q2 q3 0110 0,1

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Informal definition What happens on input string? 1101 q1 q2 q3 0110 0,1

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Informal definition What happens on input string? 1101 q1 q2 q3 0110 0,1

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Informal definition What happens on input string? 1101 q1 q2 q3 0110 0,1

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Informal definition What happens on input string? 1101 Accept q1 q2 q3 0110 0,1

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Informal definition What happens on input string? 0010 q1 q2 q3 0110 0,1

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Informal definition What happens on input string? 0010 q1 q2 q3 0110 0,1

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Informal definition What happens on input string? 0010 q1 q2 q3 0110 0,1

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Informal definition What happens on input string? 0010 Reject q1 q2 q3 0110 0,1

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Informal definition Can you describe the language consisting of all string that M actually accepts? q1 q2 q3 0110 0,1

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Informal definition M actually accepts any string that ends with a 1 1,01,11,01010101010101,… any string the ends with an even number of 0’s after the last 1 100, 101010000,… q1 q2 q3 0110 0,1

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Formal Definitions A finite automaton is a 5-tuple (Q,∑,ξ,q0,F), where Q is a finite set of states ∑ is a finite set of symbols called the alphabet ξ : Q x ∑ Q is the transition function q0 Є Q is the start state F ⊆ Q is the set of accept states or final states

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Back to M M is a 5-tuple (Q,∑,ξ,q1,F), where Q is a finite set of states {q1,q2,q3} ∑ is {0,1} ξ : Q x ∑ Q is the transition function q1 Є Q is the start state F ⊆ Q is {q2} q1 q2 q3 0110 0,1 0101 q1 q2 q3 q1q2 q3q2q2 ξ

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Another example: M1 M1 is a 5-tuple (Q,∑,ξ,q1,F), where Q is a finite set of states {q1,q2} ∑ is {0,1} ξ : Q x ∑ Q is the transition function q1 Є Q is the start state F ⊆ Q is {q2} q1 q2 011 0101 q1 q2 q1q2 ξ 0

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Another example: M1 What language does M1 accept? q1 q2 011 0

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Another example:M1 What language does M1 accept? M1 accepts all the strings that end with a 1 q1 q2 011 0

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Another example:M2 q1r1 ab q2r2 s a a b b a ab b

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Another example:M2 What language does M2 accept? q1r1 ab q2r2 s a a b b a ab b

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Another example:M2 What language does M2 accept? M1 accepts all the strings that start and end with the same symbol! q1r1 ab q2r2 s a a b b a ab b

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Languages The language of machine M is the set of strings A that M accepts L(M) = A Note that M can accept many strings … but M accepts only one language We say that M recognizes language A if A= {w | M accepts w}

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Regular language A language is called a regular language if some finite automaton recognizes it

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Formal model of computation Let M= (Q,∑,ξ,q 0,F) be a Deterministic Finite Automaton (DFA) Let w=w 1 w 2 …w n be a string over ∑ M accepts w if a sequence of states r 0,…,r n exist such that: r 0 =q 0 ξ(r i,w i+1 )=r i+1, i<n r n Є F

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Formal model of computation Let M= (Q,∑,ξ,q 0,F) be a Deterministic Finite Automaton (DFA) Let w=w 1 w 2 …w n be a string over ∑ M accepts w if a sequence of states r 0,…,r n exist such that: r 0 =q 0 : the machine starts in the start state ξ(r i,w i+1 )=r i+1, i<n: the machines goes from state to state according to the transition function r n Є F: machine accepts its input if it ends up in an accept state

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The regular operations Union operation A U B = {x | x Є A or x Є B} It takes all string and A and B and puts them together into one language

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The regular operations Union operation A U B = {x | x Є A or x Є B} It takes all string in A and B and puts them together into one language Let A={good, bad} and B ={boy, girl} A U B ={good, bad, boy, girl}

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The regular operations Concatenation operation A o B = {xy| x Є A and y Є B} It attaches a string from B to the end of a string from A in the new language

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The regular operations Concatenation operation A o B = {xy| x Є A and y Є B} It attaches a string from B to the end of a string from A in the new language Let A={good, bad} and B ={boy, girl} A o B = {goodboy, goodgirl, badboy, badgirl}

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The regular operations Star operation A*={x 1 x 2 …x k | k is positive and each x i Є A} It applies to a single language Attaches any number of string from A to get a string in the new language

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The regular operations Star operation A*={x 1 x 2 …x k | k is positive and each x i Є A} It applies to a single language Attaches any number of string from A to get a string in the new language Let A={good, bad} A* = {Ø, good,bad, goodgood, goodbad, badbad, badgood, gooodgooodgood, goodgoodbad, goodbadgood,…}

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“Closed under” The set of natural number N={1,2,3,….} is closed under multiplication For any a and b in N, a x b is also in N 3 x 5 = 15: 3, 5, and 15 are all naturals In general, a collection of objects is closed under some operation if applying that operation to members of the collection returns and object still in the collection Note; the set of natural numbers is not closed under division 3 / 5 is not a natural number

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Regular languages closed under union Theorem: The class of regular languages is closed under the union operation If A1 and A2 are regular languages, so is A1 U A2

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Proof idea How to prove such a theorem? Remember that the only result we have seen so far is “A language is called a regular language if some finite automaton recognizes it”

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Proof idea “A language is called a regular language if some finite automaton recognizes it” So, let us look at our theorem from this perspective: If A1 and A2 are regular languages, so is A1 U A2 M1

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Proof idea “A language is called a regular language if some finite automaton recognizes it” So, let us look at our theorem from this perspective: If A1 and A2 are regular languages, so is A1 U A2 M1M2

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Proof idea “A language is called a regular language if some finite automaton recognizes it” So, let us look at our theorem from this perspective: If A1 and A2 are regular languages, so is A1 U A2 M1M2M Can we construct?

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Proof: approach Approach: proof by construction Some M1 accepts A1 Some M2 accepts A2 Construct M that accepts A1 U A2

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Proof: first attempt Given M1 and M2 First simulate M1 If M1 doesn’t accept then simulate M2

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Proof: first attempt Given M1 and M2 First simulate M1 If M1 doesn’t accept then simulate M2 What is wrong with this approach? Once the symbols are read by M1, they are lost! We cannot simulate M2! We need another way

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Proof: a second attempt We need another way: Simulate both machines simultaneously! That way only one pass thru the input is necessary But, can we keep of both simulations with finite memory?

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Proof: a second attempt But, can we keep of both simulations with finite memory? Answer: All you need to remember is the sate that each machine would be in if it had read up to this point in the input You need to remember a pair of states!

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Proof: a second attempt q1 q2 r1 r2 q1,r1 M1 M2 M

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Proof: a second attempt q1 q2 r1 r2 q1,r1 q1,r2 M1 M2 M

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Proof: a second attempt q1 q2 r1 r2 q1,r1 q1,r2 q2,r1 M1 M2 M

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Proof: a second attempt q1 q2 r1 r2 q1,r1 q1,r2 q2,r1 q2,r2 M1 M2 M

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Proof: a second attempt q1 q2 r1 r2 q1,r1 q1,r2 q2,r1 q2,r2 M1 M2 M

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Proof: a second attempt: Accept sates q1 q2 r1 r2 q1,r1 q1,r2 q2,r1q2,r2 M1 M2 M

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Proof: a second attempt: Accept sates q1 q2 r1 r2 q1,r1 q1,r2 q2,r1q2,r2 M1 M2 M

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Proof: formal details Let M1 = (Q 1,∑,ξ 1,q 1,F 1 ) accept A1 Let M2 = (Q 2,∑,ξ 2,q 2,F 2 ) accept A2 Define M = (Q,∑,ξ,q,F) as follows Q = {(r 1,r 2 ) | r 1 Є Q 1 and r 2 Є Q 2 } This set is the Cartesian product of sets Q 1 and Q 2 Denoted by Q 1 x Q 2

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Proof: formal details Let M1 = (Q 1,∑,ξ 1,q 1,F 1 ) accept A1 Let M2 = (Q 2,∑,ξ 2,q 2,F 2 ) accept A2 Define M = (Q,∑,ξ,q,F) as follows … ∑ is the same

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Proof: formal details Let M1 = (Q 1,∑,ξ 1,q 1,F 1 ) accept A1 Let M2 = (Q 2,∑,ξ 2,q 2,F 2 ) accept A2 Define M = (Q,∑,ξ,q,F) as follows … … ξ, the transition function is defined as follows ξ((r 1,r 2 ),a) = (ξ1(r 1,a), ξ2(r 2,a))

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Proof: formal details Let M1 = (Q 1,∑,ξ 1,q 1,F 1 ) accept A1 Let M2 = (Q 2,∑,ξ 2,q 2,F 2 ) accept A2 Define M = (Q,∑,ξ,q,F) as follows … … … q is the pair (q 1,q 2 )

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Proof: formal details Let M1 = (Q 1,∑,ξ 1,q 1,F 1 ) accept A1 Let M2 = (Q 2,∑,ξ 2,q 2,F 2 ) accept A2 Define M = (Q,∑,ξ,q,F) as follows … … … … F is the set of pairs in which either member is an accept state of M1 or M2 F = {(r 1,r 2 ) | r 1 Є F 1 or r 2 Є F 2 }

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Proof: formal details Let M1 = (Q 1,∑,ξ 1,q 1,F 1 ) accept A1 Let M2 = (Q 2,∑,ξ 2,q 2,F 2 ) accept A2 Define M = (Q,∑,ξ,q,F) as follows … … … … Why not choose F as follows F = {(r 1,r 2 ) | r 1 Є F 1 and r 2 Є F 2 }=F 1 x F 2 ?

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What about concatenation? Theorem: The class of regular languages is closed under the concatenation operation If A1 and A2 are regular languages, so is A1 o A2

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Proof Idea Simulate M1 for a while, then switch to M2 BUT: when to switch? … leads us into nondeterminism

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Conclusions Deterministic Finite State Machines Regular Languages Regular Operations Union Concatenation Star Next time: Nondeterministic FA

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