322 CSE - Introduction to Formal Methods in Computer Science The Myhill-Nerode Theorem
Dave Bacon
Department of Computer Science & Engineering, University of Washington
The pumping lemma for regular languages is nice, but it has one fatal drawback, and that is that it is not a complete characterization of whether a language is regular. In other words this means that there are languages which are not regular, but which obey the pumping lemma. Here we will describe a condition which, unlike the pumping lemma, can be used as a complete test for whether a language is regular or not. It is called the Myhill-Nerode theorem.
I.
TWO EQUIVALENCE RELATIONS
To begin let us start with two useful equivalence relations. The first equivalence relation is defined over any language. Let A be a language in . Then we say that the strings x and y are indistinguishable by A if and only if, for every string z , either both x and y are in the language or both x and y are not in the language. When two strings are indistinguishable with respect to A, we write x A y. Lets give an example. Consider the language A = {01k |k 0} {00}. The two strings 01 and 011 are indistinguishable in A. Why? Because 01z is in A iff z 1 . Similarly 011z is in A iff z 1 . Thus the strings 01z and 011z are both in A at the same time, or both not in A at the same time. Now, A is an equivalence relation. Recall that a relation is an equivalence relation if it is reflexive, symmetric, and transitive. The first of these means that x A x which is clearly true. The second of these means that x A y = y A x, which is also clearly true since our definition was symmetric. The final of these means that if x A y and y A z, then x A z which is the only slightly tricky one, but if you give it a little thought you will see it is true. Now given that A is an equivalence relation what can we do? Recall that if we have an equivalence relation we can divide the set the relation is defined over into equivalence classes. These equivalence classes are subsets of the set we are working with which are disjoint (the intersection of any two is empty) and which are complete (the union of all of the equivalence classes is the entire set.) In other words equivalence relations give us a way to define equivalence classes, and these equivalence classes fully partition the set we are working on. Lets work out an example for the language we considered above A = {01k |k 0} {00} (we work with the alphabet = {0, 1}. ) It is easy to see, by generalizing our example above, the {01k |k > 0 are equivalent to each other. Note that 0 is not equivalent over A to 01, since, 00 is in A, but 010 is not in A. Further, using similar reasoning 00 is not equivalent to 0. Further is not equivalent to any other string in A, since there is no portion of A which ends with arbitrary strings. But is equivalent to strings which are not in A. Indeed, all of the elements that are not in A are equivalent to each other (you should check this.) Thus the equivalence classes of A are {0}, {00}, {01k |k > 0}, and - A. Next lets give another equivalence relation, but this time defined over a DFA M . To do this we will need to define a concept called delta-star. The transition function of a DFA, tell us, given a current state, and a symbol, what new state to transition to. tell us, given a current state, and a string, what new state would the DFA end up, starting at the current state and then reading that particular string. We can define inductively. First note that : Q Q. Then define this function as Delta-star definition: 1. Base case: (q, ) = q 2. Inductive step: For x and a , (q, xa) = ( (q, x), a) Okay, given the definition of we can now begin to define our new equivalence relation. Given a DFA M = (Q, , , q0 , F ), we say that two string x and y are indistinguishable, if and only if (q0 , x) = (q0 , y). In other words the two strings are indistinguishable if, starting at the initial state, one ends up at the same state upon reading either x or y. If this is true, we write x M y. Again, it is easy to check that M is and equivalence relation. Thus we can use this equivalence relation to partition up into equivalence classes as before. Further note that M only has a finite number of equivalence classes, since the equivalence is defined by ending up in the same state, and there are only a finite number of states for a DFA. Further note that if the DFA is made up of reachable states, then the number of equivalence classes will just be equal to the number of states of the machine M .
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II. A LEMMA AND A NEW METHOD FOR PROVING A LANGUAGE NOT REGULAR
Now lets prove a useful lemma relating the equivalence class of a machine M and the language accepted by this machine, L(M ). The lemma is as follows: If A = L(M ) for a DFA M , then for any x, y , if x M y then x A y. Notice that at this point this is only an if statement, not an if and only if (yet!). So why is this true? Lets prove it! Let A = L(M ) and M = (Q, , , q0 , F ). Therefore w A if and only if (q0 , w) F . Suppose also that x M y. Then (q0 , x) = (q0 , y). Let z . Then (q0 , xz) = (yz) (since after we get to the state reached by both x and y the DFA will proceed deterministically.) Therefore xz A if and only if (q0 , xz) F . Further since (q0 , xz) = (yz) this means that xz A if and only if (q0 , yz) F . But (q0 , yz) F if and only if yz A. Therefore xz A if and only if yz A. It thus follows that x A y. What good is this lemma? Well, it tells us something nice. It says that whenever two elements arrive at the same state in M , they are in the same equivalence class of A . This means that each equivalence class of A is a union of equivalence classes of M . One nice consequence of this is, since M has equivalence which classes are finite in number, that we can now show that If A is regular, then A has a finite number of equivalence classes. Why is this true? If A is regular, then there is a DFA M which recognizes the language A. Our lemma from above tells us that A has at most the number of equivalence classes of M (this largest case comes about when A equivalence classes are identical to the equivalence classes of M .) Since the number of equivalence classes of M is finite, this implies that the number of equivalence classes of A is finite. Okay well now that we've proved this, what good is it? Well now we have a new method for proving that languages are not regular! How? Well, again, by contradiction. We proceed by assuming that the language, A, we are considering is regular. Then if we can show that the number of equivalence classes under A is not finite, then we have proved a contradiction to our deduction above. To prove that the number of equivalence classes under A is not finite, we don't really need to identify all of the equivalence classes: all we really need to do is to provide an infinite sequence of strings and prove that they are not equivalent to each other. Lets do an example! Consider the language A = {0n 1n |n 0}. Lets prove that this language is not regular, using the corollary above. Assume that A is regular and therefore, by the above corollary, it has a finite number of equivalence classes. Lets examine the sequence of strings x1 , x2 , . . . where xi = 0i , where i 1. Let us show that no two of these are equivalent to each other with respect to A. Consider 0i and 0j where i = j. Let z = 1i . Now 0i z = 0i 1i which is in A. But 0j z = 0j 1i is not in A. Thus these two strings 0i and 0j are not equivalent to each other. Thus, since the sequence of string x1 , x2 , . . . is infinite, we have shown that the number of equivalence classes of A is infinite. This contradicts our assumption that the number of equivalence classes of A is finite. Thus A must not be regular. Wah-lah. Okay, so we have a brand spanking new method for showing that a language is not regular. But does this method provide us anything more than the pumping lemma did? Well the answer to that is the reason I'm blabbing on about this. This is the fact that, unlike the pumping lemma, this method is always guaranteed to work. The statement of this is the Myhill-Nerode theorem.
III.
MYHILL-NERODE THEOREM
Myhill-Nerode Theorem A is regular if and only if A has a finite number of equivalence classes. In addition there is a DFA M with A = L(M ) having precisely one state for each equivalence class of A . We have already proved one direction of this theorem (that if A is regular, then it has a finite number of equivalence classes under A .) Now lets prove the opposite direction: that if A has a finite number of equivalence classes, then A is regular. We will do this by our favorite trick, by showing that if A has a finite number of equivalence classes, then we can construct a DFA M which recognizes A. Let M = (Q, , , q0 , F ) be the DFA we are going to construct. Further let A1 , A2 , . . . , Ar denote the equivalence classes of A . The Ai are disjoint and partition . Define the states of the DFA as Q = {q1 , . . . , qr }. Our goal will be to define the transition function and q0 and F such that (q0 , x) = qr if and only if x Ar . Lets begin by defining q0 . Let q0 = qi such that Ai . In other words, the start state corresponds to the equivalence class that contain...