Transcript Properties of Regular Languages Fall 2006 Costas Busch - RPI For regular languages we will prove that: Union: Concatenation: L1 and L1  L2 L1L2 Star: L1 * Reversal: R L1 Complement: L1 Intersection: L1  L2 Fall 2006 L2 Costas Busch.

Properties of
Regular Languages
Fall 2006
Costas Busch - RPI
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For regular languages
we will prove that:
Union:
Concatenation:
L1
and
L1  L2
L1L2
Star:
L1 *
Reversal:
R
L1
Complement:
L1
Intersection:
L1  L2
Fall 2006
L2
Costas Busch - RPI
Are regular
Languages
2
We say: Regular languages are closed under
Union:
Concatenation:
L1  L2
L1L2
Star:
L1 *
Reversal:
R
L1
Complement:
L1
Intersection:
L1  L2
Fall 2006
Costas Busch - RPI
3
A useful transformation: use one accept state
NFA
a
2 accept states
b
a
b
Equivalent
NFA
a
b
Fall 2006
1 accept state
a
b


Costas Busch - RPI
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NFA
In General
Equivalent NFA



Fall 2006
Costas Busch - RPI
Single
accepting
state
5
Extreme case
NFA without accepting state
Add an accepting state
without transitions
Fall 2006
Costas Busch - RPI
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Take two languages
Regular language
L1
Regular language
LM1   L1
NFA
Fall 2006
LM 2   L2
NFA
M1
Single accepting state
L2
M2
Single accepting state
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Example
n0
L1  {a b}
n
M1
a
b
M2
L2  ba
Fall 2006
b
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a
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NFA for
L1  L2
Union
M1


Fall 2006
M2
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NFA for
Example
n
L1  L2  {a b}  {ba}
L1  {a b}
n
a


b
L2  {ba}
b
Fall 2006
a
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Concatenation
NFA for
L1L2
M1
M2

Fall 2006
Costas Busch - RPI
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Example
NFA for
L1L2  {a b}{ba}  {a bba}
n
L1  {a b}
n
n
L2  {ba}
a
b
Fall 2006

b
Costas Busch - RPI
a
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Star Operation
NFA for L1 *
w  w1w2  wk

wi  L1
M1


  L1 *

Fall 2006
Costas Busch - RPI
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Example
NFA for
L1*  {a b} *
n

L1  {a b}
n
a


b

Fall 2006
Costas Busch - RPI
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Reverse
R
NFA for L1
L1

M1
M1
1. Reverse all transitions
2. Make initial state accepting state
and vice versa
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Costas Busch - RPI
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Example
M1
a
L1  {a b}
n
b
a
R
L1
Fall 2006
 {ba }
n

M1
b
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Complement
L1
M1
L1
1. Take the DFA that accepts

M1
L1
2. Make accepting states non-final,
and vice-versa
Fall 2006
Costas Busch - RPI
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Example
M1
a, b
a
L1  {a b}
n
b
L1  {a, b} * {a b}
n

M1
a, b
a
b
Fall 2006
a, b
Costas Busch - RPI
a, b
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Intersection
L1
regular
We show
L2
Fall 2006
regular
L1  L2
regular
Costas Busch - RPI
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DeMorgan’s Law:
Fall 2006
L1  L2  L1  L2
L1 , L2
regular
L1 , L2
regular
L1  L2
regular
L1  L2
regular
L1  L2
regular
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Example
L1  {a b} regular
n
L2  {ab, ba} regular
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L1  L2  {ab}
regular
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Another Proof for Intersection Closure
Machine
M1
Machine
M2
DFA for
L1
DFA for
L2
Construct a new DFA
M that accepts L1  L2
M simulates in parallel M1 and M 2
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Costas Busch - RPI
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States in M
qi , p j
State in
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M1
State in
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M2
23
DFA
q1
DFA
M1
a
q2
a
p1
transition
M2
p2
transition
DFA
q1, p1
M
a
q2 , p2
New transition
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Costas Busch - RPI
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DFA
DFA
M1
M2
q0
p0
initial state
initial state
DFA
M
q0 , p0
New initial state
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DFA M1
DFA
pj
qi
accept state
M2
pk
accept states
DFA
M
qi , p j
qi , pk
New accept states
Both constituents must be accepting states
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Example:
n0
m0
L1  {a b}
L2  {ab }
M1
M2
n
a
q0
b
q2
m
q1
p0
a, b
b
a, b
Fall 2006
a
b
p1
a
p2
a, b
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Automaton for intersection
L  {a nb}  {ab n }  {ab}
a, b
q0 , p0
a
b
q1, p2
q0 , p1
b
q1, p1
a
b
a
a
b
q0 , p2
q2 , p2
q2 , p1
a
b
a, b
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Costas Busch - RPI
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M
simulates in parallel
M1
and
M2
M accepts string w if and only if:
and
M1
accepts string
w
M2
accepts string
w
L( M )  L( M1)  L( M 2 )
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