TDDD65 Introduction to the Theory of Computation

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1 TDDD65 Introduction to the Theory of Computation Lecture 2 Gustav Nordh, IDA gustav.nordh@liu.se

2 Outline - Lecture 2 Closure properties of regular languages Regular expressions Equivalence of regular expressions and finite automata Pumping lemma for regular languages

3 Closure properties of regular languages The natural numbers N = {0, 1, 2, 3,...} are closed under multiplication in the sense that for any natural numbers x and y, x y is again a natural number The natural numbers are not closed under subtraction (3 5 = 2 which is not a natural number) Definition We say that a class of languagesc is closed under an operation op if applying op to any languages from C results in a language in C

4 Closure properties of regular languages Understanding under which operations a class of languagesc is closed is important!

5 Closure properties of regular languages Theorem The class of regular languages is closed under union (if L 1 and L 2 are regular languages, then so is L 1 L 2 ) Proof.

6 Closure properties of regular languages Theorem The class of regular languages is closed under concatenation (if L 1 and L 2 are regular languages, then so is L 1 L 2 ) Proof.

7 Closure properties of regular languages Theorem The class of regular languages is closed under star (if L 1 is a regular language, then so is L 1 ) Proof.

8 Regular expressions

9 Regular expressions

10 Regular expressions

11 Definition of regular expressions Definition L(R) denotes the language described by the regular expression R. R is a regular expression if R is 1 a for a Σ, L(a) = {a} 2, L() = {} 3, L( ) = 4 R 1 + R 2 where R 1 and R 2 are regular expressions, L(R 1 + R 2 ) = L(R 1 ) L(R 2 ) 5 R 1 R 2 where R 1 and R 2 are regular expressions, L(R 1 R 2 ) = L(R 1 )L(R 2 ) 6 R 1 where R 1 is a regular expression, L(R 1 ) = L(R 1) has higher precedence than concatenation and +, concatenation has higher precedence than +

12 Examples of regular expressions Example (0+1) 0 binary strings ending with 0 (0+1) 00(0+1) binary strings with at least two consecutive 0 s ( ) 1234( )

13 Equivalence with finite automata Theorem A language is regular if and only if some regular expression describes it

14 Equivalence with finite automata If a language is described by a regular expression then it is recognized by a NFA

15 Equivalence with finite automata If a language is described by a regular expression then it is recognized by a NFA Proof. R = a for a Σ, L(R) = {a} R =, L(R) = {} a R =, L(R) =

16 Equivalence with finite automata If a language is described by a regular expression then it is recognized by a NFA Proof. R = R 1 + R 2, L(R) = L(R 1 ) L(R 2 )

17 Equivalence with finite automata If a language is described by a regular expression then it is recognized by a NFA Proof. R = R 1 R 2, L(R) = L(R 1 )L(R 2 )

18 Equivalence with finite automata If a language is described by a regular expression then it is recognized by a NFA Proof. R = R 1, L(R) = L(R 1)

19 Equivalence with finite automata: Example Example (0+1) to NFA (0+1) 0 1

20 Equivalence with finite automata RegExp Closure Properties NFA Subset Construction DFA

21 Equivalence with finite automata RegExp Closure Properties GNFA Construction NFA Subset Construction DFA

22 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Idea: Use a generalized NFA (GNFA) where the transition arrows can be labeled by regular expressions (aa) a ab q 1 q 2 ab+ba

23 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Idea: Use a generalized NFA (GNFA) where the transition arrows can be labeled by regular expressions Given a DFA Add a new start state with an transition to the old start state Add a new accept state with transitions from all old accept states Replace transitions of the form a, b, c by a+b + c

24 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Add a new start state with an transition to the old start state Add a new accept state with transitions from all old accept states Replace transitions of the form a, b, c by a+b + c q 1 q 2 0

25 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Add a new start state with an transition to the old start state Add a new accept state with transitions from all old accept states Replace transitions of the form a, b, c by a+b + c 0 1 q s 1 q 1 q 2 q F 0

26 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Add a new start state with an transition to the old start state Add a new accept state with transitions from all old accept states Replace transitions of the form a, b, c by a+b + c Eliminate a state different from the start and accept state (reducing the number of states by 1)

27 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate a state different from the start and accept state (reducing the number of states by 1) R 2 R 1 R 3 q 1 q e q 2 R 4

28 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate a state different from the start and accept state (reducing the number of states by 1) R 2 R 1 R 3 q 1 q e q 2 R 4 q 1 R 1 R 2 R 3 + R 4 q 2

29 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate a state different from the start and accept state (reducing the number of states by 1) 0 1 q s 1 q 1 q 2 q F 0

30 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate q q s 1 q 1 q 2 q F 0

31 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate q q s 1 q 1 q 2 q F 0 Using the rule R 1 R 2 R 3 + R 4 the new transition from q 1 to q F is labeled 11 +

32 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate q q s 1 q 1 q 2 q F 0 Using the rule R 1 R 2 R 3 + R 4 the new transition from q 1 to q F is labeled 11 + Using the rule R 1 R 2 R 3 + R 4 the new transition from q 1 to q 1 is labeled

33 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate q q s 1 q 1 q 2 q F q s q q F

34 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate q q s q q F

35 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate q q s q q F Using the rule R 1 R 2 R 3 + R 4 the new transition from q s to q F is labeled (11 0+0) (11 + )+

36 Equivalence with finite automata If a language is recognized by a DFA then it is described by a regular expression Eliminate q q s q q F q s (11 0+0) (11 + )+ q F

37 Equivalence with finite automata RegExp Closure Properties GNFA Construction NFA Subset Construction DFA

38 Nonregular languages

39 Nonregular languages Example L = {a n b n n 0} is not a regular language Why?

40 Nonregular languages Example L = {a n b n n 0} is not a regular language Why? Imagine a DFA that recognize L When reading a string the DFA needs to keep track of the number of a s seen so far so that it can check that the same number of b s follow The number of a s is unbounded so the DFA needs to remember an unbounded number This is not possible since the memory of the DFA is bounded (finite number of states)

41 Nonregular languages, pumping Given a sufficiently long string s in a regular language L Then the DFA that recognizes L will be in the same state q t several times s = s 1 s j 1 s j s k 1 s k s n L }{{}}{{}}{{} q t q t q accept

42 Nonregular languages, pumping Given a sufficiently long string s in a regular language L Then the DFA that recognizes L will be in the same state q t several times So for any i 0 s = s 1 s j 1 s j s k 1 s k s n L }{{}}{{}}{{} q t q t q accept s = s 1 s j 1 (s j s k 1 ) i s k s n L }{{}}{{}}{{} q t q t q accept

43 The Pumping for Regular Languages If L is a regular language, then there exists a positive integer p (the pumping length) such that every string s L, s p, can be partitioned into three pieces, s = xyz, such that the following conditions hold: y > 0, xy p, and for each i 0, xy i z L

44 Proof of the Pumping If L is a regular language, then there exists a positive integer p (the pumping length) such that every string s L, s p, can be partitioned into three pieces, s = xyz, such that the following conditions hold: y > 0, xy p, and for each i 0, xy i z L Proof. If L is regular, let M be a DFA recognizing L. Let p be the number of states of M and let s = s 1 s 2 s n L with n p. Let r 1 r n+1 be the sequence of states that M enters while reading s. Among the first p+1 states in this sequence, two must be the same. Let r j = r k (j < k) be such two. Let x = s 1 s j 1, y = s j s k 1, and z = s k s n.

45 Proof of the Pumping If L is a regular language, then there exists a positive integer p (the pumping length) such that every string s L, s p, can be partitioned into three pieces, s = xyz, such that the following conditions hold: y > 0, xy p, and for each i 0, xy i z L Proof. If L is regular, let M be a DFA recognizing L. Let p be the number of states of M and let s = s 1 s 2 s n L with n p. Let r 1 r n+1 be the sequence of states that M enters while reading s. Among the first p+1 states in this sequence, two must be the same. Let r j = r k (j < k) be such two. Let x = s 1 s j 1, y = s j s k 1, and z = s k s n. y = s j s k 1 x = s 1 s j 1 z = sk s n r 1 r j = r k r n+1

46 Proof of the Pumping If L is a regular language, then there exists a positive integer p (the pumping length) such that every string s L, s p, can be partitioned into three pieces, s = xyz, such that the following conditions hold: y > 0, xy p, and for each i 0, xy i z L Proof. If L is regular, let M be a DFA recognizing L. Let p be the number of states of M and let s = s 1 s 2 s n L with n p. Let r 1 r n+1 be the sequence of states that M enters while reading s. Among the first p+1 states in this sequence, two must be the same. Let r j = r k (j < k) be such two. Let x = s 1 s j 1, y = s j s k 1, and z = s k s n. y i, i 0 x = s 1 s j 1 z = sk s n r 1 r j = r k r n+1

47 Proof of the Pumping If L is a regular language, then there exists a positive integer p (the pumping length) such that every string s L, s p, can be partitioned into three pieces, s = xyz, such that the following conditions hold: y > 0, xy p, and for each i 0, xy i z L Proof. If L is regular, let M be a DFA recognizing L. Let p be the number of states of M and let s = s 1 s 2 s n L with n p. Let r 1 r n+1 be the sequence of states that M enters while reading s. Among the first p+1 states in this sequence, two must be the same. Let r j = r k (j < k) be such two. Let x = s 1 s j 1, y = s j s k 1, and z = s k s n. We have xy i z L for each i 0.

48 Proof of the Pumping If L is a regular language, then there exists a positive integer p (the pumping length) such that every string s L, s p, can be partitioned into three pieces, s = xyz, such that the following conditions hold: y > 0, xy p, and for each i 0, xy i z L Proof. If L is regular, let M be a DFA recognizing L. Let p be the number of states of M and let s = s 1 s 2 s n L with n p. Let r 1 r n+1 be the sequence of states that M enters while reading s. Among the first p+1 states in this sequence, two must be the same. Let r j = r k (j < k) be such two. Let x = s 1 s j 1, y = s j s k 1, and z = s k s n. We have y > 0 (since j k). We have xy = s 1 s k 1 p since r k occurs among the first p + 1 places in the sequence of states (so k p+1).

49 Strategy for showing that a language L is not regular 1 Assume that L is regular (with the aim of reaching a contradiction). 2 Choose a string s L, such that s p where p is the pumping length given by the Pumping. The Pumping then says that s can be partitioned into three pieces s = xyz where y > 0 and xy p, such that xy i z L for all i 0. 3 Show that for ALL POSSIBLE partitions of s = xyz satisfying y > 0 and xy p there exists an i such that xy i z / L. If we succeed to show this, then we have a contradiction with the Pumping and our assumption that L is regular is wrong.

50 The standard example Example L = {a n b n n 0} is not regular. Proof. 1 Assume that L is regular 2 Choose s = a p b p where p is the pumping length given by the Pumping. s L and s p, so the Pumping says that s can be partitioned into three pieces s = xyz where y > 0 and xy p, such that xy i z L for all i 0. 3 s = a p b p = xyz where y > 0 and xy p. For all such partitions of s, y is a string of a s of length at least 1. Choose i = 2, xy 2 z contains more a s than b s, thus, xy 2 z / L. This contradicts the Pumping, hence, our assumption that L is regular is wrong.

51 Regular languages: Summary Regular languages are the languages recognized by DFAs For every NFA there is an equivalent DFA The subset construction A language can be described by a regular expression if and only if it can be recognized by a DFA GNFA construction, closure properties There are simple non-regular languages Pumping lemma

FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

15-453 FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY THE PUMPING LEMMA FOR REGULAR LANGUAGES and REGULAR EXPRESSIONS TUESDAY Jan 21 WHICH OF THESE ARE REGULAR? B = {0 n 1 n n 0} C = { w w has equal number