Notes for Comp 497 (Comp 454) Week 5 2/22/05. Today we will look at some of the rest of the material in Part 1 of the book.

Transcription

1 Notes for Comp 497 (Comp 454) Week 5 2/22/05 Today we will look at some of the rest of the material in Part 1 of the book Errata (Chapter 9) Chapter 9 p. 177, line 11, replace "for L 1 " by "for L 2 ". p. 179, 3rd from the end, replace "L 1 " and "r 1 " by "L 2 " and "r 2 ". In Chapter 9 we look at regular languages those that can be defined by a regular expression. Not all languages are regular (a simple example of a non-regular language is the set of well-formed parenthesis strings). In Chapter 9, Cohen gives some results about languages that are regular. THEOREM 10 If L 1 and L 2 are regular languages then so are the languages L 1 L 2 L 1 +L 2 L 1 * This seems obvious given what we know about the construction of regular expressions. There must be expressions that define L 1 and L 2 (call these regular expressions R 1 and R 2 ) so we can make new expressions R 1 R 2 R 1 +R 2 (R 1 )* that define the three languages above making them regular languages too. We can also show it using TG Complements and intersections. The complement of a language is simply the set of strings that it does not contain. Note that we need to be careful to define the alphabet. For example, if our language is a(a+b)* with alphabet (a,b) the language is the set of strings starting with a. The complement is the set of strings that do not start with a. We intend this to be b(a+b)*. We assume that the complement is defined over the same alphabet so we don t worry about strings like xyz and pq999. We will follow Cohen and use L to signify the complement of L.

2 THEOREM 11 If L is a regular language then so is L We know that there is a FA that accepts L so we just modify it to accept L instead and that makes L regular. Modification means making the final states non-final states and vice versa. See the example pp THEOREM 12 If L 1 and L 2 are regular then so is L 1 L 2 (intersection) Proof uses the results of Theorems 10 and 11 (union and complement) because L 1 L 2 = (L 1 + L 2 ) There is an extensive example of this in action beginning on Page 174 and going all the way to page 180. Cohen also shows a proof by construction we can create a machine based on the FAs for L 1 and L 2 that accepts only those strings that are simultaneously accepted by the FA for L 1 and the FA for L 2 in other words the strings that are in both L 1 and L 2. Errata (Chapter 10) Chapter 10 p. 200, 3 lines from the end and 7 lines from end, replace "R/Q" by "R/P". We now look at non-regular languages and introduce the pumping lemma. See links on web site for pumping lemma pages; you will see that it has wider applications than regular languages and we will see it again in Part 2 of the book. We know a n b n 1 < n < 100 is regular because it is finite We know a n b m m,n >1 is regular because it is equivalent to a*b* But what about a n b n n > 1? (Some a s followed by the same number of b s) A language for which no FA can be devised is nonregular (from Kleene s theorem it follows that no TG exists and no RE). We can show that a n b n is not regular by showing that no FA can exist. Proof is by contradiction. Suppose that there is such a recognizer and it has 500 states It must be capable of recognizing the string a 501 b 501 because this is in a n b n There must be a cycle in the machine because 501 is greater than the number of states in the machine.

3 Suppose the cycle is length T Machine must also recognize a 501+T b 501 But this is not in the language a n b n Contradiction So our assumption that an FA for a n b n exists is wrong and a n b n is nonregular. This cycle idea is a tool that lets us prove that certain other languages are also nonregular. Look at the pumping lemma. THEOREM 13 If L is a regular infinite language then it must contain all the strings of the form xy n z where x and z are (possible null) strings, y is a non null string and n>0 If w is a string with more letters than our FA recognizer has states then we can divide w into three parts x, y and z x is that part of the string processed pre-loop y is that part of the string processed in a loop z is that part of the string processed post-loop Example (p. 192) is a 6-state machine therefore strings longer than 6 characters must be processed with a loop. Consider the acceptance of bbbabbababa w = b b b a b b a b a b a from state to state x y y z Let us revisit a n b n using the pumping lemma. Pumping lemma says that there must be strings x, y, z (y not null) such that all words of the form xy n z are in L. Is there any way we can identify x, y and z for the language a n b n? A typical string in a n b n is aaaaaaabbbbbbb Question revolves round the identity of y. It cannot be null (by definition)

4 If it contains only a s then when we pump it, the string will have more a s than b s If it contains only b s then when we pump it, the string will have more b s than a s It cannot contain both a s and b s because the only part of the string having both a and b is the part where the a s stop and the b s begin. If we set y = ab for example then when we pump it we would get abab So no strings x,y,z with the appropriate properties exist So a n b n is nonregular. We can use this result in proving others. For example the language EQUAL (equal numbers of a s and b s in each string) is nonregular because if it were regular then a n b n would be regular [ follows from identity a n b n = a*b* EQUAL ] and Theorem 12 : the intersection of two regular languages is regulars And we know a n b n is nonregular. What about a language such as a n ba n? { b aba aabaa } If it is regular, the pumping lemma tells us that there must be strings of the form xy n z in the language. But if there are, what is in string y? It cannot contain a b because when we pump it, we would get more than one b It cannot contain only a s to the left of the b because when we pump it we would get more a s at the beginning of the string than at the end. Similarly it cannot contains the a s that follow the b. So there is no match for y so the language is not regular. Similarly page 194 proof of nonregularity of a n b n ab n+1 THEOREM 14 If L is an infinite language accepted by FA with N states then for all words (w) in L longer than N characters, there are strings x, y and z (y not null) and length(x) + length(y) N such that w = xy n z (n = 1,2,3 ) are in L So those strings longer than the number of states, must involve some looping. Note how Cohen uses this to show that PALINDROME is non-regular. The Pumping Lemma of Theorem 13 doesn t work because

5 Strings of the form a b n a (x=a y=b z=a are in PALINDROME, i.e. aba, abba, abbba, abbbba and so on.) Proof of nonregularity of PALINDROME is by contradiction. Assume there is a FA that recognizes PALINDROME and that it has 213 states. It must accept a 220 ba 220 because it is a palindrome Which, by Theorem 14 must be of the form xy n z where length(x)+length(y) 213 But it cannot be because x and y would have to be entirely a s and then pumping y would make string unbalanced MyHill-Nerode Theorem We can think of each state in a FA as being associated with a society of strings {S 1, S 2, } all of which bring the machine to that state from a start state. Suppose that strings X and Y are in the same society. Now consider strings Xz and Yz. It must be the case that either both Xz and Yz are accepted or both Xz and Yz are rejected by the FA THEOREM 15 Given a language L, two strings x,y are in the same class if, for all possible strings z, either both xz and yz are in L or both are not. Thus The language L divides the set of possible strings into separate classes If L is regular, the number of classes L creates is finite If the number of classes L creates is finite, then L is regular Proof (pages ) looks at the three parts above Part 1: a string cannot be in two separate classes. Since it must be in at least one, it must be in exactly one. Part 2: If L is regular there is an FA that recognizes it. With each state is set of strings that cause transition to that state. Some of these classes of strings could be identical. The number of classes cannot exceed the number of states so it is finite. Part 3: If we can identify the classes, we can construct an FA which will prove that L is regular. Cohen sketches a method for arriving at a set of connected states that satisfies the definition of FA.

6 Consider the examples on Pp in particular the ones that show that we can use this latest result to prove that a n b n and a n ba n are nonregular. Quotient Languages We are happy with the idea of a product of two languages X = WY meaning that every string in X is a string from W followed by a string from Y. We can define division in some sense by saying that Q = R/P If R = PQ. However, given R and P there may not be a unique Q Note p. 201 that if we have R = PQ and we know that P and R are regular. Q may or may not be regular. Reading Assignment Read Chapters 9 and 10; we skip the last part of chapter 10 (mid page 201 onwards). Next week we will finish Part I of the book and begin Part II Homework #2 is due today. Program 1 Homework Write the shortest high-level language program you can to test to see if a string is in EVEN-EVEN (p.49). It should be possible to write the program in well under a page even with input/output. Due date: March 15