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 Sn be the set of all DFAs with exactly n states.
How many DFAs are in Sn?
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 2n choices for the final states.
Therefore, the number of DFAs with exactly n states is
n  n||n  2n.
The set of all DFAs is
S1  S2  S3  …
This is a countable set since it is the union of a countable number of finite sets.
Thus, we can enumerate the DFAs as M0, M1, M2, M3, …

4. The Existence of a Non-Regular Language
There exists a language that is not accepted by any DFA (provided   ).
Proof:
Let Ln = L(Mn).
Let x be any symbol in .
Let sn = xn, for all n  0.
Define a new language L by the rule that sn  L if and only if sn  Ln.
Then L is not equal to any Ln.
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 Mn 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 2N, 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
xyiz  L for every i  0.

8. Proof of the Pumping Lemma
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 xyiz  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 = {aibi | i  0}.
Suppose that L is regular.
“Let n be the n of the Pumping Lemma” and consider the string w = anbn.
Then w can be decomposed as xyz where |y| > 0 and |xy|  n.
Therefore, xy consists only of a’s.
It follows that y = ak, for some k > 0.

12. Standard Example of a Nonregular Language
According to the Pumping Lemma, xy2z is in L.
However, xy2z =an + kbn, 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 L1 = L(a*b*).
Then L  L1 = {aibi | 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 = anbanb 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.