Lecture 4: More on Regexps, Non-Regular Languages

Transcription

1 6.045 Lecture 4: More on Regexps, Non-Regular Languages

2 6.045 Announcements: - Pset 1 is on piazza (as of last night) - If you don t have piazza access but are registered for 6.045, send to TAs with your mit.edu address!

3 Deterministic Finite Automata Computation with finite memory

4 Non-Deterministic Finite Automata Computation with finite memory and guessing

5 Regular Expressions: Computation as Description A different way of thinking about computation: What is the complexity of describing the strings in the language?

6 Syntax Inductive Definition of Regexp Let Σ be an alphabet. We define the regular expressions over Σ inductively: For all Σ, is a regexp is a regexp is a regexp If R 1 and R 2 are both regexps, then (R 1 R 2 ), (R 1 + R 2 ), and (R 1 )* are regexps Examples:, 0, (1)*, (0+1)*, ((((0)*1)*1) + (10))

7 Semantics Definition: Regexps Represent Languages The regexp Σ represents the language { } The regexp represents {} The regexp represents If R 1 and R 2 are regular expressions representing L 1 and L 2 then: (R 1 R 2 ) represents L 1 L 2 (R 1 + R 2 ) represents L 1 L 2 (R 1 )* represents L 1 * Example: (10 + 0*1) represents {10} {0 k 1 k 0}

8 Regexps Represent Languages For every regexp R, define L(R) to be the language that R represents A string w Σ* is accepted by R (or, w matches R) if w L(R) Examples: 0, 010, and match (01)* matches (0+1)*0

9 DFAs NFAs Regular Expressions! L can be represented by some regexp L is regular We saw: L can be represented by some regexp L is regular Every regexp can be converted into an NFA Now we ll show: L is regular L can be represented by some regexp Every DFA can be converted into a regexp

10 Generalized NFAs (GNFA) Idea: Transform an DFA for L into a regular expression by removing states and re-labeling the arcs connected to those states with regular expressions Rather than reading in just 0 or 1 letters from the string on an arc, we can read in entire substrings

11 Generalized NFA (GNFA) cb a*b a q 0 q 1 q 2 Accept string x there is some path of regexps R 1,, R k from start state to final such that x matches R 1 R k This GNFA recognizes L(a*b(cb)*a), the set of strings matched by a*b(cb)*a

12 DFA Add unique start and accept states Goal: Replace DFA with a single regexp R Then, L(R) = L(DFA)

13 DFA While the machine has more than 2 states: Pick an internal state, rip it out and re-label the arrows with regexps, to account for paths through the missing state a b c

14 DFA While the machine has more than 2 states: Pick an internal state, rip it out and re-label the arrows with regexps, to account for paths through the missing state ab*c

15 GNFA While the machine has more than 2 states: In general: R(q 1,q 3 ) R(q 1,q 2 ) R(q 2,q 3 ) q 1 q 2 q 3 R(q 2,q 2 )

16 GNFA While the machine has more than 2 states: In general: R(q 1,q 2 )R(q 2,q 2 )*R(q 2,q 3 ) + R(q 1,q 3 ) q 1 q 3

17 a a,b b q 0 q 1 q 2 q 3 R(q 0,q 3 ) = (a*b)(a+b)* represents L(N)

18 a,b a*b q 0 q 2 q 3 R(q 0,q 3 ) = (a*b)(a+b)* represents L(N)

19 (a*b)(a+b)* q 0 q 3 R(q 0,q 3 ) = (a*b)(a+b)* represents L(N)

20 Formally: CONVERT(G): Given a DFA, add q start and q acc to create G For all q, q, define R(q,q ) = σ σ k s.t. δ(q,σ i ) = q (Takes a GNFA, outputs a regexp) If #states = 2 return R(q start, q acc ) If #states > 2 pick q rip Q different from q start and q acc define Q = Q {q rip } define R on Q -{q acc } x Q -{q start } as: defines a new GNFA G R (q i,q j ) = R(q i,q rip )R(q rip,q rip )*R(q rip,q j ) + R(q i,q j ) return CONVERT(G ) Theorem: Let R = CONVERT(G). Then L(R) = L(G). Claim: L(G ) = L(G) [Sipser, p.73-74]

21 Theorem: Let R = CONVERT(G). Then L(R) = L(G). Proof by induction on k, the number of states in G Base Case: k = 2 CONVERT outputs R(q start, q acc ) Inductive Step: Assume theorem is true for k-1 state GNFAs Let G have k states. Let G be the k-1 state GNFA. First show that L(G) = L(G ) [Sipser, p ] G has k-1 states, so by induction, L(G ) = L(CONVERT(G )) = L(CONVERT(G)) = L(R) by I.H. Therefore L(R)=L(G). QED

22 b a q 1 b a q 2 b a q 3 Convert the DFA to a regular expression

23 b a q 1 b b a a q 2 q 3

24 bb b q 1 a a + ba b q 2

25 bb + (a + ba)b*a q 1 b + (a + ba)b* (bb + (a + ba)b*a)* (b + (a + ba)b*)

26 Convert the NFA to a regular expression q 1 a, b q 2 b a b b q 3

27 Convert the NFA to a regular expression q 1 a, b q 2 b b a b q 3

28 Convert the NFA to a regular expression q 1 a (a + b)b*b bb*b q 3

29 Convert the NFA to a regular expression q 1 (a + b)b*b(bb*b)* (a + b)b*b(bb*b)*a ((a + b)b*b(bb*b)*a)*( + (a + b)b*b(bb*b)*)

30 DFAs NFAs DEFINITION GNFAs Regular Languages Regular Expressions

31 Some Languages Are Not Regular: Limitations on DFAs/NFAs a.k.a. Lower Bounds on DFAs/NFAs

32 Σ = {0,1} Regular or Not? C = { w w has equal number of 1s and 0s} NOT REGULAR! D = { w w has equal number of occurrences of 01 and 10 } REGULAR!

33 Σ = {0,1} D = { w w has equal number of occurrences of 01 and 10} = { w w = 1, w = 0, or w =, or w starts with a 0 and ends with a 0, or w starts with a 1 and ends with a 1 } (0+1)*0 + 1(0+1)*1 Claim: A string w has equal occurrences of 01 and 10 w starts and ends with the same bit.

34 A Language That Has No DFA Theorem: A = {0 n 1 n n 0} is not regular Proof Idea: No DFA can remember the number of 0 s, if we feed it more 0 s than its number of states. In that case, the DFA can t accurately compare the number of 0 s to the number of 1s!

35 A Language That Has No DFA Theorem: A = {0 n 1 n n 0} is not regular Proof: By contradiction. Assume A is regular. Then A has a DFA D with P states, for some P > 0. Consider feeding input w = 0 P+1 to D. By pigeonhole principle, some state q of D is visited more than once while reading in w. Therefore, D is in state q after reading 0 S, and is also in q after reading 0 R, for some R < S P+1. What happens when D reads 1 S starting from state q? D must accept, because 0 S 1 S in A. D must reject, because 0 R 1 S is not in A. Contradiction!

36 Another Language That Has No DFA Thm: EQ = {w w has an equal number of 0s and 1s} is not regular Proof: By contradiction. Assume EQ is regular. Observation: EQ L(0*1*) = {0 n 1 n n 0} EQ is regular and L(0*1*) is regular EQ L(0*1*) is regular. (Regular Languages are closed under intersection!) But {0 n 1 n n 0} is not regular! Contradiction!

37 The Pumping Lemma: Structure in Regular Languages Let L be any regular language There is a positive integer P s.t. for all strings w L with w P there are x, y, z where w = xyz, and: 1. y > 0 (that is, y ) 2. xy P 3. xz L and xyyz L (in fact, xy yz L) Why is it called the pumping lemma? The word xyz can be pumped into a longer string xyyz

38 Proof: Let M be a DFA that recognizes L Let P be the number of states in M Let w be a string where w L and w P We show: w = xyz y 1. y > 0 2. xy P 3. xz L and xyyz L x w z q 0 q j q k q w Claim: There must exist j and k such that 0 j < k P, and q j = q k

39 Proving a Language is NOT Regular Let L be any language. Suppose for every integer P there is a w L with w P so that for all ways to write w = xyz, where: 1. y > 0 (that is, y ) 2. xy P Either xz L or xyyz L THEN L is not regular!

40 Applying the Pumping Lemma Theorem: EQ = { w w has equal number of 1s and 0s} is not regular. By contradiction. Assume EQ is regular. Let P be as in pumping lemma. Let w = 0 P 1 P and note that w EQ. If EQ is regular, then there is a way to write w as w = xyz, y > 0, xy P, and xyyz, xz is also in EQ Claim: The string y must be all zeroes. Why? Because xy P and w = xyz = 0 P 1 P But then xyyz has more 0s than 1s! Contradiction!

41 Applying the Pumping Lemma Let s prove that SQ = {0 n2 n 0} is not regular Assume SQ is regular. Let w = 0 P2 If SQ is regular, then we can write w = xyz, y > 0, xy P, such that xz and xyyz are also in SQ We have: xyyz = 0 P2 + y Observe that 0 < y P So xyyz = P 2 + y P 2 + P < P 2 + 2P + 1 = (P+1) 2 and P 2 < xyyz < (P+1) 2 Therefore xyyz is not a perfect square! Hence 0 P2 + y = xyyz SQ, so our assumption must be false. That is, SQ is not regular!