1 / 14

# Nonregular Languages - PowerPoint PPT Presentation

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 ?

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
Download Presentation

## PowerPoint Slideshow about 'Nonregular Languages' - zavad

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Nonregular Languages

Section 2.4

Wed, Oct 5, 2005

• 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 .

• 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, …

• 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 Lis not equal to any Ln.

• So L is not accepted by any DFA.

• 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!).

• 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 onto2*.

• So, what is an example of a nonregular language?

• 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.

• 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.

• 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.

• 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.

• 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.

• 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.

• 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.

• {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.