Properties of Regular Languages. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 1

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1 Properties of Regular Languages Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 1

2 Properties of Regular Languages Pumping Lemma. Every regular language satisfies the pumping lemma. If somebody presents you with language, use the pumping lemma to show a whether it is regular or not. Closure properties. Building automata from components through operations, e.g. given L and M we can build an automaton for L M. Decision properties. Computational analysis of automata, e.g. are two automata equivalent. Minimization techniques. We can save money since we can build smaller machines. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 2

3 Pumping Lemma Informally Consider the language L = {0 n 1 n n 1}. If L were regular then it would be recognized by some DFA A, with, say, k states. Let A read 0 k. On the way it will travel as follows: start p 0 0 p 1 00 p 2 0 k p k Since there are only k states, the pigeon hole principle tells us that after reading two different prefixes 0 i and 0 j, A must be in the same state (i.e., i<j and p i = p j = q). Suppose instead that after i or j 0 s, the automaton A starts receiving 1 s. After receiving i 1 s, it must accept if it previously received i 0 s; but not if it received j 0 s. Since it was in state q when 1 s started, it cannot remember whether it received i or j 0 s, so we can fool A, make wrong decisions. So L is not regular. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 3

4 The Pumping Lemma for Regular Languages Theorem: Let L be regular language. n, ω L : ω n ω = xyz such that 1. y ε 2. xy n 3. k 0, xy k z L That is, we can always find a nonempty string y not too far from the beginning of ω that can be pumped; that is repeating it any number of times, or deleting it (k=0 case), keeps the resulting string in the language L. Proof: Suppose L is regular. Then L=L(A) for some DFA A, with say n states. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 4

5 PL proof Now, break ω = xyz as follows: 1. x = a 1 a 2 a i. 2. y = a i+1 a i+2 a j. 3. z = a j+1 a j+2 a m That is Pigeon hole principle Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 5

6 PL Proof Now, consider what happens if A receives the input xy k z for k 0. If k = 0, automaton goes from p 0 to p i and accepting state. If k > 0, automaton goes from p 0 to p i, circles from p i to p i k times, and then goes to accepting state. That is xy k z is in L. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 6

7 PL Example 4.2 Let L eq be the language of strings with equal number of zero's and one's. Consider the {0 n 1 n n 1} L eq. Pumping lemma n, ω L : ω n, ω = xyz such that y ε, xy n, k 0, xy k z L xz has n 1 s but it has fewer than n 0 s. Thus xz L eq. We proved that L eq is not regular. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 7

8 PL example 4.3 Show that L pr = { 1 p : p is prime} is not regular. Proof: If it was, it d satisfy the pumping lemma: n, ω L : ω n, ω = xyz such that y ε, xy n, k 0, xy k z L Let n be given by pumping lemma. Chose a prime p n + 2. Is xy p-m z L pr? xy p-m z = xz + (p-m) y = (p-m) + (p-m)m=(p-m)(1+m) y ε m 1 m+1 2, m = y xy n, p n + 2 (p-m) (n+2) n 2 (p-m)(1+m) 4 which is not (always) prime. So, language is not regular. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 8

9 Closure Properties of regular languages Now, we shall discuss several theorems of the form If certain languages are regular, and a language L is formed from them by certain operations, then L is also regular. The simplest is below. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 9

10 Theorem: If L is a regular language over Σ, then is also a regular language. L = Σ * L Proof: Let L = L(A) for some DFA A = (Q, Σ, δ, q 0, F). Then complement is L(B), where B = (Q, Σ, δ, q 0, Q-F). That is, B is exactly like A but accepting states A have become nonaccepting states of B, and vice versa. Example 4.6: An automaton that accepts all and only strings ending in 01. Regular Expression: (0+1)*01. NFA diagram Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 10

11 Example 4.6 Complement of L(A) is therefore all strings of 0 s and 1 s that do not end in 01. What is the regular expression for complement L(A)? Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 11

12 We have already showed that if L and M are regular, their complements and union are also regular. So it follows. Remember that concatenation of languages L and M L M = {w : w = xy, x L, y M} Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 12

13 Example 4.9 Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 13

14 Closure Under Difference Theorem 4.10: If L and M are regular languages, then so is L-M. Proof: Observe that L M = L I M. Since complement of regular language and intersection of regular languages are regular, if follows. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 14

15 Reversal The reversal of a string a 1 a 2 a n is the string written backwards, that is, a n a n-1 a 1. We use ω R for the reversal of string ω. Thus, 0010 R is 0100, and ε R =ε. Reversal of a language L, written L R, is the language consisting of the all strings. For instance, if L = {001, 10, 111}, then L R = {100, 01, 111}. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 15

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17 Theorem 4.11: If L is a regular language, so is L R. Assume that L is defined by regular expression E. The proof is a structural induction on the size of E. We show that there is another regular expression E R such that L(E) R = L(E R ). Basis: If E is ε, Ø or a, for some symbol a, then E R is the same as E. That is {ε} R = {ε}, Ø R = Ø, and {a} R = {a}. Thus L(E) R = L(E R ) = L(E). Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 17

18 Induction: There are three cases, depending on the form of E. 1. E = F + G. Then E R = F R + G R. Reversal of the union of two languages is union of reversal of the languages. 2. E = FG. Then E R = G R F R. Example: L(F) = {01, 111} and L(G) = {00,10} then L(FG) = { 0100, 0110, 11100, 11110} = L(F)L(G). L(FG) R = {0010, 0110, 00111, 01111} or FR = , GR = E R = G R F R = L(G R F R ) = {0010, 00111, 0110, 01111} 3. E = F*. Then E R = (F R )* Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 18

19 Induction: (continue) 3. E = F*. Then E R = (F R )* L(F*) = { w Σ*: w 1 w 2 w k for some k 0 and w 1,, w k L(F) } Then w R =w kr w 2R w 1R. Each w ir is in L(F R ), and thus w R is in L(F R )* = L(E) R. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 19

20 Homomorphism A string homomorphism is a function on strings that works by substituting a particular string for each symbol. That is, a homomorphism on Σ is a function h: Σ* Θ* where Σ and Θ are alphabets. Let ω = a 1 a 2 a n Σ*. Then h(ω) = h(a 1 )h(a 2 ) h(a n ). Further we can apply homomorphism to a language by applying it to each of the strings in the language. That is h(l) = {h(ω) ω L}. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 20

21 Example Let h: {0,1}* {a,b}* such that h(0) = ab and h(1) = ε. Then for example, h(0011) = abab. h(l(10*1)) = L((ab)*). Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 21

22 Theorem 4.14 If L is a regular language over alphabet Σ, and h is homomorphism on Σ, then h(l) is also regular. Proof: In general, ω = a 1 a 2 a n Σ*. Then h(ω) = h(a 1 )h(a 2 ) h(a n ). Thus if E is a regex let h(e) be the expression we obtained by replacing each symbol a of Σ in E by h(a). Assume L is defined by regular expression R. We claim that h(r) defines the language h(l(r)) Proof is an easy structural induction. Whenever we take a subexpression E of R and apply h to it to get h(e), h(l(e)) = L(h(E)). Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 22

23 Basis: if E is ε or Ø, then h(e) = E and L(E) is either {ε} or Ø. Thus L(h(E)) = L(E) = h(l(e)). If E = a Σ. Then L(E) = {a}, so h(l(e)) = {h(a)}. L(h(a)) = {h(a)}. Thus we conclude L(h(E)) = h(l(e)) Since E is regular, so is h(e) and h(l(e)). Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 23

24 Induction: There are three cases: union, concatenation, and closure. L = L(E+F). L = L(E.F). L = L(E*). Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 24

25 Inverse Homomorphism Homomorphism may also be applied backwards. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 25

28 It is an easy induction on ω to show that Since the accepting states of A and B are the same, B accepts ω if and only if A accepts h(ω). Put another way, B accepts exactly those strings ω that are in h -1 (L). Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 28

29 Decision Properties We consider the following: 1. Converting among representations for regular languages. In particular, observe the time complexity of the algorithms that perform the conversions. 2. Is L = Ø? 3. Is ω L? 4. Do two descriptions define the same language? i.e., equivalence of languages. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 29

30 Converting Among Representations Converting NFA s to DFA s Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 30

31 O(8 n 4 2 n ) Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 31

36 Equivalence and Minimization of Automata Two descriptions of two regular languages actually define the same language involves considerable intellectual mechanics. Is it possible to minimize a DFA? That is, we can take any DFA and find an equivalent DFA that has the minimum number of states. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 36

37 Equivalence and Minimization of Automata Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 37

38 Equivalence and Minimization of Automata In other words p and q are distinguishable if and only if Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 38

39 Example 4.18 Certain pair of states are obviously not equivalent. For example, C and G are not equivalent because one is accepting state and the other one is not. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 39

40 Example 4.18 A and G are not equivalent because 01 distinguishes A from G. State C is accepting but E is not. Any input string that takes only one of the state to accepting state is sufficient to show the states are not equivalent. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 40

41 Example 4.18 Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 41

42 Table Filling (TF) Algorithm o o o o o o o o o o o o o o o o o o o o o o o o o Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 42

43 Theorem 4.20: If two states are not distinguished by the Table Filling algorithm, then the states are equivalent. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 43

44 Testing Equivalence of Regular Languages Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 44

45 Consider the DFA s each accepts the empty strings and all strings that end in 0; that is the language of regular expression ε+(0+1)*0. Distinguishing the accepting states of different DFA s is different. We have to check their transition properties. For example {A,D} pair is not distinguishable because on 0 they go to themselves, and on 1 they go to the pair {B,E} which has not yet been distinguished. Since start states are found equivalent by this test, DFA s do accept the same language. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 45

46 Minimization of DFA s Another important consequence of the test is that we can minimize DFA s. That is, we can find an equivalent DFA that has few states as any DFA accepting the same language. The central idea behind the minimization of DFA s is the notion of state equivalence. Let s partition the states into blocks such that: 1. All the states in a block are equivalent. 2. No two states chosen from two different blocks are equivalent. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 46

47 Minimization Example: The partition of the states into equivalent blocks is: ({A,E},{B,H},{C},{D,F},{G}) Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 47

48 Theorem 4.23 The equivalence of states is transitive. That is in some DFA A = (Q, Σ, δ, q 0, F), p q and q r then p r. Suppose Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 48

49 Minimization Algorithm of DFA A = (Q, Σ, δ, q 0, F) 1. Use table-filling algorithm to find all the pair of equivalent states. 2. Partition the set of states Q into blocks of mutually equivalent states. 3. Construct the minimum-state equivalent DFA B = (Q, Σ, γ, q 0, F ) where γ (p, a) = δ(p,a) by using the blocks as its states. a) The start state of B is the block containing the start state of A. b) The set of accepting states of B is the set of blocks containing accepting states of A. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 49

50 Example 4.25 ({A,E},{B,H},{C},{D,F},{G}) Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 50

51 Theorem 4.26 If A is a DFA, and M the DFA constructed from A by the algorithm described in minimization algorithm, then M has as few states as any DFA equivalent to A. Mart 2006 Ankara Üniversitesi Bilgisayar Mühendisliği - TY 51