Nonregular Languages Section 2.4 Wed, Oct 5, 2005
Countability of the Set of DFAs Theorem: The set of all DFAs (over an alphabet ) is countable. Proof: For a given n > 0, let S n be the set of all DFAs with exactly n states. How many DFAs are in S n ? There are n choices for the initial state. For each state, there are n | | choices for the transitions coming out of that state. Therefore, there are (n | | ) n = n | |n choices for .
Countability of the Set of DFAs There are 2 n choices for the final states. Therefore, the number of DFAs with exactly n states is n n | |n 2 n. The set of all DFAs is S 1 S 2 S 3 … This is a countable set since it is the union of a countable number of finite sets. Thus, we can enumerate the DFAs as M 0, M 1, M 2, M 3, …
The Existence of a Non-Regular Language There exists a language that is not accepted by any DFA (provided ). Proof: Let L n = L(M n ). Let x be any symbol in . Let s n = x n, for all n 0. Define a new language L by the rule that s n L if and only if s n L n. Then L is not equal to any L n. So L is not accepted by any DFA.
The Existence of a Non-Regular Language This is another example of a diagonalization argument. It is a non-constructive proof. It does not provide us with an example (unless we actually figure out what each M n is!).
The Existence of a Non-Regular Language Another non-constructive proof is based on a cardinality argument. The set of all languages is 2 *, which is uncountable since its cardinality is equal to the cardinality of 2 N, which we know to be uncountably infinite. The set of DFAs is countable. Therefore, the function f(M) = L(M) cannot be onto 2 *. So, what is an example of a nonregular language?
The Pumping Lemma The Pumping Lemma: Let L be an infinite regular language. There exists an integer n 1 such that any string w L, with |w| n, can be represented as the concatenation xyz such that y is non-empty, |xy| n, and xy i z L for every i 0.
Proof of the Pumping Lemma Proof: Let n be the number of states. Let w be any string in L with at least n symbols. After processing n symbols, we must have returned to a previously visited state (the Pigeonhole Principle). Let q be the first revisited state. Let x be the string processed from s to q. Let y be the string processed around the loop from q back to q. Let z be the string from q to the end, a final state f.
Proof, continued Then clearly |y| > 0 and |xy| n. It is also clear that xy i z L for all i 0, since we may travel the loop as many times as we like, including 0 times.
The Pumping Lemma The Pumping Lemma says that if L is regular, then certain properties hold. The contrapositive of the Pumping Lemma says that if certain properties do not hold, then L is not regular. Therefore, you cannot use the Pumping Lemma to conclude that a language is regular, but only that it is not regular. That’s good, because that is exactly what we want to do.
The Standard Example of a Nonregular Language Let L = {a i b i | i 0}. Suppose that L is regular. “Let n be the n of the Pumping Lemma” and consider the string w = a n b n. Then w can be decomposed as xyz where |y| > 0 and |xy| n. Therefore, xy consists only of a’s. It follows that y = a k, for some k > 0.
Standard Example of a Nonregular Language According to the Pumping Lemma, xy 2 z is in L. However, xy 2 z =a n + k b n, which is not in L. since n + k n. This is a contradiction. Therefore, L is not regular.
A Second Example of a Nonregular Language Let L = {w * | w contains an equal number of a’s and b’s}. Suppose L is regular. Let L 1 = L(a * b * ). Then L L 1 = {a i b i | i 0} would also be regular, which is a contradiction. Therefore, L is not regular.
More Examples {w * | w contains an unequal number of a’s and b’s}. {w * | w contains more a’s than b’s}. {w * | w = zz for some z * }. Consider w = a n ba n b and use the Pumping Lemma. {w * | w zz for some z * }. Notice that we use the Pumping Lemma only when necessary; other arguments are often simpler.