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Class Discussion
Can you draw a DFA that accepts the language
{akbk| k = 0,1,2,} over the alphabet ={a,b}?
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Limitations of FA
Many languages are non-regular:
{anbn| n = 0,1,2,}
{0i2
| i 0,1,2,} {0
p| p is a prime number }
the set of all well-formed parentheses overthe alphabet ={(,
)}
the set of all palindromes over the alphabet={a, b}
...
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Why Impossible?
We want to prove that L = {aibi| i = 0,1,2,} isnon-regular.
Prove by contradiction:
Assume that there is a DFA M that recognizes
L. Let n be the total number of states in M.
Consider the path followed by the input string
anbnin M:
q(0) q(1) q(n) q(n+1) q(2n)a b b b......
a a
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Why Impossible?
Since M has only n states, there must be at least onestate
visited twice in the first n transitions. Let this
state be visited at the ithand the jthsteps, where j > i.
By skipping the loop, an-(j-i)bnshould also be accepted,
but this is contradictory since an-(j-i)bnL
q(0)
q(i) = q(j)
q(2n)F
The path in M when reading anbn
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Another Example
We want to prove that L = {1k2| k > 0} is non-
regular. Prove by contradiction:
Assume that there is a DFA M that recognizesL. Let n be the
total number of states in M.
M should also accept the string 1n2
q(0) q(1) q(2) q(n2)1 1
...1 1
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Another Example Since n2n and M has only n states, there must
be
at least two equal states from q(0)to q (n2). Let q(i)= q(j)be
the first repeated state where j-i = m n.
By repeating the loop one more time, 1(n2+m) is also
accepted by M, which is a contradiction, since
(n2+m) cannot be a square (the next square after n2
is (n+1)2
but n2
+m < (n+1)2
).
q(0)
q(i) = q(j)
q(n2)FThe path in M
when reading 1n2
a loop of
length m
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Pumping Lemma for Regular Set
If L be a regular language, there is a constant n such
that if z is any word in L and |z| n, we can write z
= uvw in such a way that |uv| n, |v| 1, and for alli 0, uviw is
in L. (This constant n can be the
number of states of a DFA accepting L.)
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Proof of Pumping Lemma
If L is a regular language, there are DFAs thatrecognize L. Let
n be the number of states in onesuch DFA (called M) recognizing
L.
If z is a word in L with |z| = k n:
where z(i) is the ithsymbol of the string z.
Since k n and M has only n states, there must be
at least one repeated states from q(0)to q(k). Letq(i)be the
first such repeated state.
q(0) q(1) q(2) q(k)z(2) z(k)
...z(1) z(3)
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Proof of Pumping Lemma
Let u be the string obtained by traversing from q(0)to q(i),
and v be the string obtained by traversing the loop once
(|v|1). In the traversal from q(0)
to q(i)
and then through
the loop once back to q(i), nothing except q(i)repeats, so
|uv|n. By traversing the loop 0 or more times, we obtain
uviw where i 0. These strings should all be accepted by M
since they can all reach q(k)which is a final state in M.
a path in Mq(0)
q(k)F
q(i)
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