Nonregular Languages

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# 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 &gt; 0, let S n be the set of all DFAs with exactly n states. How many DFAs are in S n ?

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### 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 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 .
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, …
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 Lis not equal to any Ln.
• 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 Mn 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 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
• 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 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 xyiz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 = {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.
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.
• 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 L1 = L(a*b*).
• Then LL1 = {aibi | 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 = 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.