Transcript Single Final State for NFAs and DFAs Observation Any Finite Automaton (NFA or DFA) can be converted to an equivalent NFA with a single.

Single Final State
for NFAs and DFAs
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Observation
Any Finite Automaton (NFA or DFA)
can be converted to an equivalent NFA
with a single final state
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a
Example
NFA
b
a
b
Equivalent NFA
a
a
b
b


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NFA
In General
Equivalent NFA



Single
final state
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Extreme Case
NFA without final state
Add a final state
Without transitions
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Some Properties of
Regular Languages
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Properties
For regular languages
we will prove that:
Union:
Concatenation:
Star:
L1
and
L2
L1  L2
L1L2
Are regular
Languages
L1 *
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We Say:
Regular languages are closed under
Union:
Concatenation:
Star:
L1  L2
L1L2
L1 *
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Regular language
L1
LM1   L1
NFA
M1
Single final state
Regular language
L2
LM 2   L2
NFA
M2
Single final state
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Example
M1
a
L1  {a b}
n
b
M2
L2  ba
b
a
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NFA for
L1  L2
Union
M1



M2

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Example
NFA for
L1  L2  {a b}  {ba}
n
L1  {a b}
n
a


b

L2  {ba}
b

a
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Concatenation
NFA for
L1L2
M1
M2


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Example
NFA for
L1L2  {a b}{ba}  {a bba}
n
L1  {a b}
n
n
L2  {ba}
a
b

b
a

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Star Operation
NFA for L1 *

  L1 *
M1



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Example
NFA for
L1*  {a b} *
n

L1  {a b}
n
a


b

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Regular Expressions
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Regular Expressions
Regular expressions
describe regular languages
Example:
( a  b  c) *
describes the language
a, bc*   , a, bc, aa, abc, bca,...
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Recursive Definition
Primitive regular expressions:
Given regular expressions
r1
,  , 
and r2
r1  r2
r1  r2
r1 *
Are regular expressions
r1 
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Examples
A regular expression:
a  b  c  * (c  )
Not a regular expression:
a  b  
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Languages of Regular Expressions
L r  :
language of regular expression
r
Example
L(a  b  c) *   , a, bc, aa, abc, bca,...
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Definition
For primitive regular expressions:
L   
L    
La   a
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Definition (continued)
For regular expressions r1 and r2
Lr1  r2   Lr1   Lr2 
Lr1  r2   Lr1  Lr2 
Lr1 *  Lr1  *
Lr1   Lr1 
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Example
Regular expression: a  b   a *
La  b   a *  La  b La *
 La  b La *
 La   Lb La  *
 a b a *
 a, b , a, aa, aaa,...
 a, aa, aaa,..., b, ba, baa,...
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Example
Regular expression
r  a  b  * a  bb
Lr   a, bb, aa, abb, ba, bbb,...
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Example
Regular expression
Lr   {a b
r  aa  * bb * b
2n 2m
b : n, m  0}
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Example
Regular expression
r  (0  1) * 00 (0  1) *
L(r ) = { all strings with at least
two consecutive 0 }
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Example
Regular expression
r  (1  01) * (0   )
L(r ) = { all strings without
two consecutive 0 }
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Equivalent Regular Expressions
Definition:
Regular expressions
are equivalent if
r1
and
r2
L(r1)  L(r2 )
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Example
L = { all strings with at least
two consecutive 0 }
r1  (1  01) * (0   )
r2  (1* 011*) * (0   )  1* (0   )
L(r1)  L(r2 )  L
and r2
are equivalent
regular expr.
r1
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Regular Expressions
and
Regular Languages
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Theorem
Languages
Generated by
Regular Expressions

Regular
Languages
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Theorem - Part 1
Languages
Generated by
Regular Expressions

Regular
Languages
1. For any regular expression r
the language L(r ) is regular
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Theorem - Part 2
Languages
Generated by
Regular Expressions

Regular
Languages
2. For any regular language L there is
a regular expression r with L(r )  L
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Proof - Part 1
1. For any regular expression r
the language L(r ) is regular
Proof by induction on the size of
r
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Induction Basis
Primitive Regular Expressions:
,  , 
NFAs
L( M1 )    L()
L( M 2 )  {}  L( )
regular
languages
L( M 3 )  {a}  L(a)
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Inductive Hypothesis
Assume
for regular expressions r1 and r2
that
L(r1) and L(r2 ) are regular languages
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Inductive Step
We will prove:
Lr1  r2 
Lr1  r2 
Lr1 *
Lr1 
Are regular
Languages
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By definition of regular expressions:
Lr1  r2   Lr1   Lr2 
Lr1  r2   Lr1  Lr2 
Lr1 *   Lr1  *
Lr1   Lr1 
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By inductive hypothesis we know:
L(r1) and L(r2 ) are regular languages
We also know:
Regular languages are closed under
union
Lr1   Lr2 
concatenation Lr1  Lr2 
star
 Lr1  *
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Therefore:
Lr1  r2   Lr1   Lr2 
Lr1  r2   Lr1  Lr2 
Are regular
languages
Lr1 *   Lr1  *
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And trivially:
L((r1))
is a regular language
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Proof – Part 2
2. For any regular language L there is
a regular expression r with L(r )  L
Proof by construction of regular expression
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Since L is regular take the
NFA M that accepts it
L( M )  L
Single final state
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From M construct the equivalent
Generalized Transition Graph
transition labels
are regular expressions
Example:
M
a
c
a, b
a
c
ab
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Another Example:
a
q0
b
b
q1 a, b
q2
b
b
b
a
q0
q1 a  b q2
b
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Reducing the states:
b
a
q0
b
q1 a  b q2
b
bb * a
q0
b
bb * (a  b)
q2
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Resulting Regular Expression:
bb * a
q0
b
bb * (a  b)
q2
r  (bb * a) * bb * (a  b)b *
L( r )  L( M )  L
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In General
e
Removing states:
d
qi
c
qj
q
a
b
ae * d
ce * b
ce * d
qj
qi
ae * b
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Obtaining the final regular expression:
r4
r1
r3
q0
r2
qf
r  r1 * r2 (r4  r3r1 * r2 ) *
L( r )  L( M )  L
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