Transcript Finite Automata & Regular Languages

The Pumping Lemma
Hopcroft, Motawi, Ullman, Chap 4,
sections 4.1 and 4.2
A language that is not regular
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Consider
L = {w | w = anbn for all n >= 0}
There is no DFA that accepts L, no regular
expression that describes L
How do we prove this?
Intuition: DFAs cannot “count”, that is, DFAs
cannot remember that n a’s have been
recognized so that n b’s should follow
Proof technique: by contradiction
Proving that a language is
not regular
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Suppose there is a DFA that
recognizes L, and let k be the number of
states in that DFA
Consider recognizing the strings ab, a2b2,
a3b3, …, akbk, ak+1bk+1
Note what state the DFA is in as it reads the
last a. For the k+1 examples, there should
be two examples where they are in the same
state (pigeonhole principle)
Proving that a language is
not regular
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This means there are two strings: arbr,
ar+pbr+p, such that, the prefixes ar and ar+p
brings the DFA to the same state, say qt. That
is,
q0 s qt for both s = ar and s = ar+p
This in turn means that
qt v qtfor v = ap
Note also that
qt w qf for both w = br and w = br+p
Implications: arbr+p and ar+2pbr+p are
acceptable in this DFA. A contradiction.
Summarizing the strategy
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We looked for a state that can be
pumped so that substrings admissible
by starting and ending with that state
can be arbitrarily inserted in the
acceptable string
We formalize this notion through
the Pumping Lemma
Pumping Lemma
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Let L be a regular language
There exists an n such that for all z in L,
|z| >= n, z can be broken down into
three substrings: z = uvw,
where, for all i, uviw is in L
Note: |uv| <= n and |v| >= 1
Proof of this lemma follows the outline
of our earlier argument
Using the pumping lemma in
our example
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L is all strings of the form apbp
We are guaranteed there is an n as
specified in the pumping lemma
We choose anbn in L and express this as
uvw. Since |uv| <=n, v consists
entirely of a’s, say v=ak. This means,
strings like an-kbn and an+kbn should be in
L, which is a contradiction