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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 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

Nonregular Languages

Section 2.4

Wed, Oct 5, 2005

countability of the set of dfas
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 dfas3
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
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 language5
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 language6
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
  • 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 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
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 lemma10
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
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
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.
a second example of a nonregular language
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
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.
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