CSC 3130: Automata theory and formal languages

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Fall 2008
The Chinese University of Hong Kong
CSC 3130 Automata theory and formal languages
Limitations of finite automata
Andrej Bogdanov http//www.cse.cuhk.edu.hk/andrej
b/csc3130
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Many examples of regular languages
((0 1)0 (0 1)(0 1)(0 1))
all strings not containing pattern 010 followed
by 101
Is every language regular?
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Which are regular?
L1 0n1m n, m 0 L2 0n1n n 0
S 0, 1
S 1
L3 1n n is divisible by 3 L4 1n n is
prime
S 0, 1
L5 x x has same number of 0s and 1s L6 x
x has same number of patterns 01 and 10
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Which are regular?
01, so regular
L1 0n1m n, m 0

• How about
• Lets try to design a DFA for it it appears
  that we need infinitely many states!

L2 0n1n n 0 e, 01, 0011, 000111,
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A non-regular language

• Theorem
• To prove this, we argue by contradiction
• Suppose we have managed to construct a DFA M for
  L2
• We show something must be wrong with this DFA
• In particular, M must accept some strings not in
  L2

The language L2 0n1n n 0 is not regular.
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A non-regular language
imaginary DFA for L2 with n states
M
x

• What happens when we run M on input x 0n11n1?
• M better accept, because x ? L2

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A non-regular language
0
M
x
0
0
0
0
0r1n1

• What happens when we run M on input x 0n11n1?
• M better accept, because x ? L2
• But since M has n states, it must revisit at
  least one of its states while reading 0n1

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Pigeonhole principle
Suppose you are tossing n 1 balls into n bins.
Then two balls end up in the same bin.

• Here, balls are 0s, bins are states

If you have a DFA with n states and it reads n
1 consecutive 0s, then it must end up in the same
state twice.
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A non-regular language
0
M
x
0
0
0
0
0r1n1

• What happens when we run M on input x 0n11n1?
• M better accept, because x ? L2
• But since M has n states, it must revisit at
  least one of its states while reading 0n1
• But then the DFA must contain a loop with 0s

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A non-regular language
0
M
0
0
0
0
0r1n1

• The DFA will then also accept strings that go
  around the loop multiple times
• But such strings have more 0s than 1s, so they
  are not in L2!

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General method for showing non-regularity

• Every regular language L has a property
• For every sufficiently long input z in L, there
  is a middle part in z that, even if repeated
  several times, keeps the input inside L

an
z
a1
ak1
an-1
ak


an1am
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Pumping lemma for regular languages

• Theorem For every regular language L

There exists a number n such that for every
string z in L, we can write z u v w where ?
uv n ? v 1 ? For every i 0, the
string u vi w is in L.
v
z


w
u
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Proving non-regularity

• If L is regular, then
• So to prove L is not regular, it is enough to
  show

There exists n such that for every z in L, we can
write z u v w where ?uv n, ?v 1 and ?
For every i 0, the string u vi w is in L.
For every n there exists z in L, such that for
every way of writing z u v w where?uv n
and ?v 1, the string u vi w is not in L for
some i 0.
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Proving non regularity
For every n there exists z in L, such that for
every way of writing z u v w where?uv n
and ?v 1, the string u vi w is not in L for
some i 0.

• This is a game between you and an imagined
  adversary

adversary choose nwrite z uvw (uv n,v
1)
you choose z ? Lchoose iyou win if uviw ? L
1
2
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Proving non-regularity

• You need to give a strategy that, regardless of
  what the adversary does, always wins you the game

adversary choose nwrite z uvw (uv n,v
1)
you choose z ? Lchoose iyou win if uviw ? L
1
2
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Example
adversary choose nwrite z uvw (uv n,v
1)
you choose z ? Lchoose iyou win if uviw ? L
1
2

• L2 0n1n n 0 S 0, 1

adversary choose nwrite z uvw (uv n,v
1)
you z 0n11n1i 2 uviw 0j2kl1n1
0n1k1n1 ? L
1
2
u 0j, v 0k, w 0l1n1
j k l n 1
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More examples
L3 1n n is divisible by 3 L4 1n n is
prime
S 1
S 0, 1
L5 x x has same number of 0s and 1s L6 x
x has same number of patterns 01 and 10 L7 x
x has more 0s than 1s L8 x x has different
number of 0s and 1s