1. Nonregular Languages Section 2.4 Wed, Oct 5, 2005

2. 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 .

3. 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, …

4. 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.

5. 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!).

6. 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?

7. 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.

8. 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.

9. 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.

10. 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.

11. 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.

12. 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.

13. 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.

14. 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.