Automata Theory. Lecture on Discussion Course of CS120. Runzhe SJTU ACM CLASS

Transcription

1 Automata Theory Lecture on Discussion Course of CS2

2 This Lecture is about Mathematical Models of Computation.

3 Why Should I Care? - Ways of thinking. - Theory can drive practice. - Don t be an Instrumentalist.

4 Mathematical Models of Computation (predated computers as we know them) Automata and Languages: (94 s) finite automata, regular languages, pushdown automata, context-free languages, pumping lemmas. 2 Computability Theory: (93-4 s) Turing Machines, decidability, reducibility, the arithmetic hierarchy, the recursion theorem, the Post correspondence problem. 3 Complexity Theory and Applications: (96-7 s) time complexity, classes P and NP, NP-completeness, space complexity PSPACE, PSPACE-completeness, the polynomial hierarchy, randomized complexity, classes RP and BPP.

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6 Let me emphasize Proofs a bit more. A good proof:. Correct 2. Easy to understand

7 Suppose with. True or False: There are always two numbers in one divides the other. such that

8 LEVEL HINT : THE PIGEONHOLE PRINCIPLE If you put 6 pigeons in 5 holes then at least one hole will have more than one pigeon

9 LEVEL HINT : THE PIGEONHOLE PRINCIPLE If you put 6 pigeons in 5 holes then at least one hole will have more than one pigeon

10 LEVEL HINT : THE PIGEONHOLE PRINCIPLE If you put 6 pigeons in 5 holes then at least one hole will have more than one pigeon HINT 2: Every integer can be written as, where is an odd number

11 LEVEL 2 Proof Idea: Given: with. Show: There is an integer and and. such that

12 LEVEL 3 Proof: Suppose with. Write every number in A as a = 2 k m, where m is an odd number. How many odd numbers in {,,2n-}? n Since A = n+, there must be two numbers in A with the same odd part. Say a and a 2 have the same odd part m. Then a = 2 i m and a 2 = 2 j m, so one must divide the other.

13 We expect your proofs to have three levels: The first level should be a one-word or one-phrase HINT of the proof (e.g. Proof by contradiction, Proof by induction, Follows from the pigeonhole principle ) The second level should be a short oneparagraph description or KEY IDEA The third level should be the FULL PROOF

14 Double standards :-) HINT + KEY IDEA (+ FULL PROOF)

15 Deterministic Finite Automata (DFA)

16 NOTATION An alphabet Σ is a finite set (e.g., Σ = {,}) A string over Σ is a finite-length sequence of elements of Σ Σ* denotes the set of finite length sequences of elements of Σ For x a string, x is the length of x The unique string of length will be denoted by ε and will be called the empty or null string A language over Σ is a set of strings over Σ, ie, a subset of Σ*

17 Finite Automata finite # of internal states Input (String) -> -> Output (Yes/No) Divice

18 A Deterministic Finite Automata, Read string left to right The machine accepts a string if the process ends in a double circle.

19 A Deterministic Finite Automata states q, q q 3 q 2 states The machine accepts a string if the process ends in a double circle.

20 A Deterministic Finite Automata states final states (F) q, q q 3 start state (q ) q 2 states The machine accepts a string if the process ends in a double circle.

21 A Deterministic Finite Automata is represented by a 5-tuple M = (Q, Σ, δ, q, 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 Let w,..., w n Σ and w = w... w n Σ* Then M accepts w if there are r, r,..., r n Q, s.t. r =q δ(r i, w i+ ) = r i+, for i =,..., n-, and r n F

22 A Deterministic Finite Automata is represented by a 5-tuple M = (Q, Σ, δ, q, 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 L(M) = the language of machine M = set of all strings machine M accepts

23 Warming-up q, L(M) ={,}*

24 Warming-up L(M) ={,}*

25 Warming-up L(M) = { w w has an even number of s}

26 Warming-up q q L(M) = { w w has an even number of s}

27 Warming-up Build an automaton that accepts all and only those strings that contain

28 Warming-up Build an automaton that accepts all and only those strings that contain, q q q q

29 Regular Language A language L is regular if it is recognized by a deterministic finite automaton (DFA), i.e. if there is a DFA M such that L = L (M). L = { w w contains } is regular L = { w w has an even number of s} is regular

30 UNION THEOREM Given two languages, L and L 2, define the union of L and L 2 as L L 2 = { w w L or w L 2 } Theorem: The union of two regular languages is also a regular language.

31 Theorem: The union of two regular languages is also a regular language. Proof: Let M = (Q, Σ, δ, q, F ) be finite automaton for L and 2 M 2 = (Q 2, Σ, δ 2, q, F 2 ) be finite automaton for L 2 We want to construct a finite automaton M = (Q, Σ, δ, q, F) that recognizes L = L L 2

32 Theorem: The union of two regular languages is also a regular language. Idea: Run both M and M 2 at the same time! Q = pairs of states, one from M and one from M 2 = { (q, q 2 ) q Q and q 2 Q 2 } = Q Q 2 2 q = (q, q ) F = { (q, q 2 ) q F or q 2 F 2 } δ( (q,q 2 ), σ) = (δ (q, σ), δ 2 (q 2, σ))

33 Theorem: The union of two regular languages is also a regular language. M = q q M 2 = p p

34 Theorem: The union of two regular languages is also a regular language. q,p q,p M = q,p q,p

35 INTERSECTION THEOREM Given two languages, L and L 2, define the intersection of L and L 2 as L L 2 = { w w L and w L 2 } Theorem: The intersection of two regular languages is also a regular language.

36 COMPLEMENT THEOREM Given language L, define the complement of L as L= { w Σ* w L } Theorem: The complement of a regular languages is also a regular language.

37 THE REGULAR OPERATIONS Union: L L 2 = { w w L or w L 2 } Intersection: L L 2 = { w w L and w L 2 } Negation: L = { w Σ* w L } Reverse: L R = { w w k w k w L } Concatenation: L L 2 = { vw v L and w L 2 } Star: L* = { w w k k and each w i L }

38 REVERSE THEOREM Given language L, define the reverse of L R as L R = { w w k w k w L } Theorem: The reverse of a regular languages is also a regular language.

39 Theorem: The reverse of a regular languages is also a regular language. Proof: Let M = (Q, Σ, δ, q, F) that recognizes L. If M accepts w then w describes a directed path in M from start to an accept state We want to construct a finite automaton M R as M with the arrows reversed.

40 M R IS NOT ALWAYS A DFA! - It may have many start states - Some states may have too many outgoing edges, or none

41 M R IS NOT ALWAYS A DFA!,

42 M R IS NOT ALWAYS A DFA!,

43 Non-Determinism, We will say that the machine accepts if there is some way to make it reach an accept state

44 Example,,,ε At each state, possibly zero, one or many out arrows for each σ Σ or with label ε

45 Example, ε Possibly many start states

46 Example L(M)={,}

47 A Non-Deterministic Finite Automata (NFA) is represented by a 5-tuple N = (Q, Σ, δ, Q, F) : Q is the set of states (finite) Σ is the alphabet (finite) δ : Q Σ {ε} 2 Q is the transition function Q Q is the set of start state F Q is the set of accept states Let w,..., w n Σ {ε} and w = w... w n Σ ε * Then N accepts w if there are r, r,..., r n Q, s.t. r =Q r i+ δ(r i, w i+ ), for i =,..., n-, and r n F

48 A Non-Deterministic Finite Automata (NFA) is represented by a 5-tuple N = (Q, Σ, δ, Q, F) : Q is the set of states (finite) Σ is the alphabet (finite) δ : Q Σ {ε} 2 Q is the transition function Q Q is the set of start state F Q is the set of accept states L(N) = the language of machine N = set of all strings machine N accepts

49 Deterministic Computation Non-Deterministic Computation reject accept or reject accept

50 NFAs ARE SIMPLER THAN DFAs An NFA that recognizes the language {}: A DFA that recognizes the language {}:,,

51 FROM NFA TO DFA Theorem: Every NFA has an equivalent DFA N is equivalent to M if L(N) = L (M) Corollary: A language is regular iff it is recognized by an NFA Corollary: L is regular iff L R is regular

52 FROM NFA TO DFA Input: N = (Q, Σ, δ, Q, F) Output: M = (Q, Σ, δ, q, F ) reject To learn if NFA accepts, we could do the computation in parallel, maintaining the set of states where all threads are. Idea: Q = 2 Q accept

53 FROM NFA TO DFA Input: N = (Q, Σ, δ, Q, F) Output: M = (Q, Σ, δ, q, F ) Q = 2 Q δ : Q Σ Q ε ε ε δ (R,σ) = ε( δ(r,σ) ), r R q = ε(q ) F = { R Q f R for some f F } For R Q, the ε-closure of R, ε(r) = {q that can be reached from some r R by traveling along zero or more ε arrows},

54 FROM NFA TO DFA () + () +

55 Regular Languages Closure Under Concatenation DFA -> DFA2 NFA Given DFAs M and M 2, construct NFA by connecting all accept states in M to start states in M 2.

56 Regular Languages Closure Under Star Let L be a regular language and M be a DFA for L We construct an NFA N that recognizes L* ε, ε ε

57 THE REGULAR OPERATIONS Union: L L 2 = { w w L or w L 2 } Intersection: L L 2 = { w w L and w L 2 } Negation: L = { w Σ* w L } Reverse: L R = { w w k w k w L } Concatenation: L L 2 = { vw v L and w L 2 } Star: L* = { w w k k and each w i L }

58 Are all languages regular?

59 Consider the language L = { n n n > } No finite automaton accepts this language. Can you prove this?

60 Idea? n n is not regular. No machine has enough states to keep track of the number of s it might encounter

61 That is a fairly weak argument Consider the following example

62 L = { w w has equal number of occurrences of and } No machine has enough states to keep track of the number of s it might encounter.

63 THE PUMPING LEMMA Let L be a regular language with L = Then there exists a positive integer P such that Any x L, x P can be written as x = uvw where:. v > 2. uv P 3. uv i w L for any i

64 THE PUMPING LEMMA Assume x L is such that x P Let P be the number of states in M We show x = uvw where:. v > 2. uv P 3. uv i w L for any i PIGEONHOLE: There must be j > i such that q i = q j u v w q q i q j q x

65 THE PUMPING LEMMA Assume x L is such that x P Let P be the number of states in M We show x = uvw where:. v > 2. uv P 3. uv i w L for any i PIGEONHOLE: There must be j > i such that q i = q j u v w q q i q j q x

66 THE PUMPING LEMMA L = { n n n > } HINT: Assume L is regular, and try pumping. u v w q q i q j q x

67 THE PUMPING LEMMA L = { n n n > } HINT: Assume L is regular, and try pumping. u v w q q i q j q x

68 References: CMU FLAC: Formal Languages, Automata and Computation, Spring 24 Stanford CS54: Introduction to Automata and Complexity Theory, Spring 29 John E. Hopcroft, etc., Introduction to Automata Theory, Languages, and Computation(Third Edition). Pearson; 26