CSC 3130: Automata theory and formal languages

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Fall 2009
The Chinese University of Hong Kong
CSC 3130 Automata theory and formal languages
Closure propertiesLimitations of regular
languages
Andrej Bogdanov http//www.cse.cuhk.edu.hk/andrej
b/csc3130
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Operations that preserve regularity

• We saw three operations that preserve regularity
• Union If L, L are regular languages, so is L ?
  L
• Concatenation If L, L are regular languages, so
  is LL
• Star If L is a regular language, so is L
• Exercise If L is regular, is L4 also regular?
• Answer Yes, because

L4 ((LL)L)L
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Example

• The language L of strings that end in 101 is
  regular
• How about the language L of strings that do not
  end in 101?

(01)101
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Example

• Hint A string does not end in 101 if and only if
  it ends in one of the following patterns(or
  it has length 0, 1, or 2)
• So L can be described by the regular expression

000, 001, 010, 011, 100, 110, 111
(01)(000001010010100110111) e (0
1) (0 1)(0 1)
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Complement

• The complement L of a language L is the set of
  all strings that are not in L
• Examples (S 0, 1)
• L1 all strings that end in 101
• L1 all strings that do not end in 101 all
  strings end in 000, , 111 or have length 0, 1,
  or 2
• L2 1 e, 1, 11, 111,
• L2 all strings that contain at least one 0
  (0 1)0(0 1)

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Closure under complement

• If L is a regular language, is L also regular?
• Previous examples indicate answer should be yes
• Theorem

If L is a regular language, so is L.
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Proof of closure under complement

• To argue this, we can use any of the equivalent
  definitions for regular languages
• The DFA definition will be most convenient
• We will assume L is accepted by a DFA, and show
  the same for L

regularexpression
DFA
NFA
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Proof of closure under complement

• Suppose L is regular, then it is accepted by a
  DFA M
• Now consider the DFA M with the accepting and
  rejecting states of M reversed

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Proof of closure under complement

• Now for every input x ? S

M accepts x
After processing x, M ends in an accepting state
Language of M is L
After processing x, M ends in an rejecting state
L is regular
M rejects x
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Intersection

• The intersection L ? L is the set of strings
  that are in both L and L
• Examples
• If L, L are regular, is L ? L also regular?

L ? L
L (0 1)111
L 1
1111
L ? L
Ø
L (0 1)101
L 1
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Closure under intersection

• Theorem
• Proof

If L and L are regular languages, so is L ? L.
L regular
L regular
L regular
L regular
L ? L regular
L ? L regular
L ? L regular.
But
L ? L L ? L
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Reversal

• The reversal wR of a string w is w written
  backwards
• The reversal LR of a language L is the language
  obtained by reversing all its strings

w cave
wR evac
L cat, dog
LR tac, god
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Reversal of regular languages

• L all strings that end in 101 is regular
• How about LR?
• This is the language of all strings beginning in
  101
• Yes, because it is represented by

(01)101
101(01)
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Closure under reversal

• Theorem
• Proof
• We will use the representation of regular
  languages by regular expressions

If L is a regular language, so is LR.
regularexpression
DFA
NFA
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Proof of closure under reversal

• If L is regular, then there is a regular
  expression E that describes it
• We will give a systematic way of reversing E
• Recall that a regular expression can be of the
  following types
• Special expressions ? and e
• Alphabet symbols a, b,
• The union, concatenation, or star of simpler
  expressions
• In each of these cases we show how to do a
  reversal

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Proof of closure under reversal
regular expression E
reversal ER
Æ
Æ
e
e
a (alphabet symbol)
a
E1 E2
E1R E2R
E1E2
E2RE1R
E1
(E1R)
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A question
LDUP ww w ? L
L cat, dog
Ex.
LDUP catcat, dogdog
If L is regular, is LDUP also regular?
?
regularexpression
DFA
NFA
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A question

• Lets try with regular expression
• Lets try with NFA

LDUP LL
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An example
L 01 is regular
LDUP 1, 01, 001, 0001, ...
LDUP 11, 0101, 001001, 00010001, ...
0n10n1 n 0

• Lets try to design an NFA for LDUP

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An example
LDUP 11, 0101, 001001, 00010001, ...
0n10n1 n 0
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Non-regular languages
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A non-regular language

• Another example
• We reason by contradiction
• Suppose we have managed to construct a DFA M for
  L
• We argue something must be wrong with this DFA
• In particular, M must accept some strings outside
  L

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

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

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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 ? L
• 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

• Pumping lemma 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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Arguing 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
L 0n1n n 0
adversary
you
choose n
z 0n11n1
1
write z uvw
i 2
2
uv2w 0j2kl1n1 0n1k1n1 ? L
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Example
LDUP 0n10n1 n 0
adversary
you
choose n
z 0n110n11
1
write z uvw
i 2
2
uv2w 0j2kl10n11 0n1k10n11 ? L
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Which of these are regular?
L1 1n n is divisible by 3 L2 1n n is
prime
S 1
S 0, 1
L3 x x has same number of 0s and 1s L4 x
x has same number of patterns 01 and 10 L5 x
x has more 0s than 1s L6 x x has different
number of 0s and 1s
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Example
L3 x x has same number of 0s and 1s
adversary
you
choose n
z 0n11n1
1
write z uvw
i 2
2
uv2w 0j2kl1n1 0n1k1n1 ? L3
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Example
L4 x x has same number of 01s and 10s
is regular!
adversary
you
choose n
z (01)n1(10)n1
1
write z uvw
i 2
2
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Example
more 10s
1
0
0
r1
r0
1
1
q0
more 01s
0
1
0
1
s0
s1
0
L4 x x has same number of 01s and 10s
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Example
L5 x x has more 0s than 1s
adversary
you
choose n
z 0n11n
1
write z uvw
i 0
2
uv0w 0jl1n1 ? L5
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Example
L6 x x has different number of 0s than 1s
adversary
you
choose n
z ?
1
there is an easier way!
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Example
L2 1p p is prime
adversary
you
choose n
z 1n n gt p is prime
1
write z uvw 1a1b1c
i a c
2
uviw 1a1ib1c 1aibc 1a(ac)bc
1(ac)(b1) 1composite ? L2