lecture 5: finite automata

Lecture 5: Finite Automata

虞台文大同大學資工所智慧型多媒體研究室

Content Alphabets and Languages Deterministic Finite Automata Nondeterministic Finite Automata Equivalence of DFA &NFA Properties of Languages Accepted by FA’s Finite Automata & Regular Expressions Languages Not Accepted by FA’s

Alphabetsand

Languages

大同大學資工所智慧型多媒體研究室

Alphabets

Alphabet a finite set of symbols Examples:

1 0,1 binary alphabet

3 , , ,a b z Roman alphabet

2 0,1, ,9 decimal alphabet

4 , , , Greek alphabet

5 , , , , , , , , a b c d x y z Z my alphabet

Strings String over an alphabet () a finite

sequence of symbols from * the set of all strings over Examples:

3 , , ,a b z

2 0,1, ,9

4 , , ,

*1001100, 1000011110

*200799, 123789000

*3 ,taxi ixat

More on Strings

Substring Suffix Prefix

" "w HelloWorld

" "u World

" "t Hello

" "v oWo

a prefix of w and a substring of w.

a suffix of w and a substring of w.

a substring of w.

Countability of *

* is countably infinite for any finite alphabet .

Example: {0,1}

00,01,10,11

000,001,010,011,100,101,110,111

* can be enumerated by:

Languages

Language L over any subset of * i.e., L *

How many possible languages over ?

How to specify a language?

* : L w w P has property

uncountably infinite

Operations of Languages

Complement

Intersection

*{ : , }L w w w L

1 2L L L

1 2: or L w w L w L

1 2L L L

1 2: and L w w L w L

Operations of Languages

Complement, Union and Intersection Concatenation

Closure or Kleene Star

Positive Closure

1 2 1 2L L L L L

1 2: ,L w xy x L y L

*1 2 1 2: 0, , ,k kL w w w w k w w w L

*1 2 1 2: 1, , ,k kL LL w w w w k w w w L

Example: Kleene Star

01,1,100L

110001110011w

100011000010v

1909 – 1994

01,1,100L

110001110011w

100011000010v

1909 – 1994

01,1,100L

110001110011w *L ? *

1 2 1 2: 1, , ,k kL LL w w w w k w w w L

1909 – 1994

Language Recognition Device

A machine or algorithm to answer the question

Deterministic Finite Automata

Finite State Machine

Reading Head

Inputtape

Finitecontrol

aa bb aa aa bb bb aa bb aa aa

Definition Deterministic Finite Automaton

A deterministic finite automaton is a 5-tuple

(K, , , s, F)K : a finite set of states : a finite set of symbols

: the transition function K Ks K : the initial state

F K : the set of final states

Reading Head

I nputtape

Finitecontrol

aa bb aa aa bb bb aa bb aa aaaa bb aa aa bb bb aa bb aa aa

Example

K={q0, q1, q2, q3} ={0, 1}

0 1q2 q1

symbol

s=q0 F= {q0}q0

(K, , , s, F)

Operations of a DFA

> 0 1 1 0 0 1 0 1

Finitecontrol

Reading Head

Inputtape

Operations of a DFA

> 0 1 1 0 0 1 0 1

Finitecontrol

Reading Head

Inputtape

Strings are fed into the device by means of an input tape.

Main part of the machine• Initially,

• putting reading head at the leftmost• in state s = q0

• Sense input symbol• Change state based on • Move head right one cell• Give answer when head reaches end

Main part of the machine• Initially,

• putting reading head at the leftmost• in state s = q0

• Sense input symbol• Change state based on • Move head right one cell• Give answer when head reaches end

DFA As a Language Acceptor

> 0 1 1 0 0 1 0 1

Finitecontrol

Reading Head

Inputtape

The input string is accepted by the machine if the reading head ends at a final state.

Otherwise, it is rejected.

Accepted

Rejected JFLAP

> 0 1 1 0 0 1 0 1

Finitecontrol

Reading Head

Inputtape

What language the

machine accepts?What language the

machine accepts?

> 0 1 1 0 0 1 0 1

Finitecontrol

Reading Head

Inputtape

What language the

machine accepts?What language the

machine accepts?

Memory Configuration of a DFA

( , )q uaw

M = (K, , , s, F)

Reading Head

Inputtape

Finitecontrol

( , )q uaw

M = (K, , , s, F)

Reading Head

Inputtape

Finitecontrol

u wDoes not effect the result.

Does not effect the result.

( , )q uaw

M = (K, , , s, F)

( , )q aw

( , )q uaw

M = (K, , , s, F)

( , )q aw

Reading Head

Inputtape

Finitecontrol

2( , )q abbabaa

Yields in One Step M M

( , )q aw M M ( , )q w

if ( , )q a q

Yields

( , )q w ( , )q w

if M M

( , )q w M M M M ( , )q w

in some (including zero) steps

String Acceptance by a DFA

M = (K, , , s, F)

w* is said to be accepted by M iff

( , )s w ( , )q M M* M M

such that q F

Language Defined by a DFA

M = (K, , , s, F)

*( ) : , ( , ) ( , ),L M w w s w q q F M M*

Language L(M) denotes the set of all

strings accepted by M, i.e.,

Example

w=0100100110

(q0, 0100100110)

M M(q2, 100100110)

M M(q3, 00100110)

M M(q1, 0100110)

M M(q3, 100110)

M M(q2, 00110)

M M(q0, 0110)

M M(q2, 110)

M M (q3, 10)

M M (q2, 0)

M M(q0, )

wL(M)?

Example

w=0100100110

(q0, 0100100110)

M M(q2, 100100110)

M M(q3, 00100110)

M M(q1, 0100110)

M M(q3, 100110)

M M(q2, 00110)

M M(q0, 0110)

M M(q2, 110)

M M (q3, 10)

M M (q2, 0)

M M(q0, )

wL(M)?

( , )s w 0( , )q M M*

Example

M *{0,1} and has even( )

numbers of 0's and 1's.

w wL M w

Example

q0 q1 q2 q3> b b b

*{ , } and does not( )

contain three consecutive 's.

w a b wL M w

( ) ?L M

Exercises

Describe informally the languages accepted by the deterministic finite automata shown below:

(a) (b)

Exercises

Construct deterministic finite automata accepting each of the following languages

a) {w{a, b}*: each a in w is immediately preceded and immediately followed by a b}.

b) {w{a, b}*: w has abab as a substring}.c) {w{a, b}*: w has neither aa nor bb as a substrin

g}.d) {w{a, b}*: w has both ab and ba as a substring}.

Nondeterministic

Finite Automata

Review Deterministic Finite Automaton

(K, , , s, F)K : a finite set of states : a finite set of symbols

: the transition function K Ks K : the initial state

: the transition function K K2K : the transition function K 2K

(K, , , s, F)

A nondeterministic finite automaton is a 5-tuple

(K, , , s, F)

Definition Nondeterministic Finite Automaton

K : a finite set of states : a finite set of symbols

s K : the initial state

Transition Functions of FA’s

: K 2K: K K

Transition Functions of FA’s

: K 2K: K K

Change to state q

deterministically.

Change to state q

deterministically.

Change to one of state in Q nondeterministically.

( , )q aw M M ( , )q w

if ( , )q q a

M = (K, , , s, F)

Yields

( , )q w ( , )q w

if is possible.( , )q w M M M M ( , )q w ( , )q w M M M M ( , )q w

M = (K, , , s, F)

String Acceptance by a NFA

M = (K, , , s, F)

( , )s w ( , )q M M* M M

such that q F

Language Defined by an NFA

M = (K, , , s, F)

*( ) : , ( , ) ( , ),L M w w s w q q F M M*

Example: L=(ab+aba)*

q0 q1 q2 q3a b a

DFANFA

Extension: -Move & Multi-Move

q1aq0>

L=(ab+aba)*

Extension: -Move & Multi-Move

q1aq0>

L=(ab+aba)*

: the transition function K* 2K

A nondeterministic finite automaton is a 5-tuple

(K, , , s, F)

Definition Nondeterministic Finite Automaton

K : a finite set of states : a finite set of symbols

s K : the initial state

( , )q w M M ( , )q w

* , ( , )u w uw q q u iff st. and

M = (K, , , s, F)

Yields

( , )q w ( , )q w

if is possible.( , )q w M M M M ( , )q w ( , )q w M M M M ( , )q w

M = (K, , , s, F)

String Acceptance by a NFA

M = (K, , , s, F)

( , )s w ( , )q M M* M M

such that q F

Language Defined by an NFA

M = (K, , , s, F)

*( ) : , ( , ) ( , ),L M w w s w q q F M M*

Equivalence of

DFA &NFA

Equivalence of Finite Automata

Finite automata M1 and M2 are said to be equivalent iff

1 2( ) ( )L M L M

Machine Simulation

NFA DFA

Of CourseOf Course

To be discussedTo be discussed

For each NFA, there is an equivalent NFA

finite automaton without multi-move on

the transition function.

q q’1 k

q p11 p2 pk1 q’2 k

Example

In the following, we will assume the NFA being discussed is one without multi-move unless otherwise stated.

-Closure

M = (K, , , s, F)

In the following, we will assume the NFA being discussed is one without multi-move unless otherwise stated.

( ) : ( , ) ( , )q p K q w p w M M* -Closure of state q

without consumeany input

Example

M ( ) ?L M * *( )a a b bb a

0 ?( )q

1 ?( )q

2 ?( )q

3 ?( )q

4 ?( )q

0 1 2 3, , ,q q q q

1 2 3, ,q q q

3 4,q q

To be used in the next example.

Theorem

For each nondeterministic finite

automaton, there is an equivalent

deterministic finite automaton.

Theorem

Pf) Let M = (K, , , s, F) be an NFA.

To prove the theorem, we must be able to build a DFA M’ = (K’, ’, , s’, F’) such that L(M)=L(M’).

Theorem

Pf) Let M = (K, , , s, F) be an NFA.

To prove the theorem, we must be able to build a DFA M’ = (K’, ’, , s’, F’) such that L(M)=L(M’).

( )s 2 :KQ Q F

Theorem

2KK ( )s s 2 :KF Q Q F

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

Next, we have to show that ( ) ( ).L M L M

Theorem

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

Next, we have to show that ( ) ( ).L M L M

Exercises:

1.Referring to the textbook to complete the proof in detail.

2.To present the proof using PowerPoint.Exercises:

1.Referring to the textbook to complete the proof in detail.

2.To present the proof using PowerPoint.

Example ?

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

M ( ) ?L M * *( )a a b bb a

0 ?( )q

1 ?( )q

2 ?( )q

3 ?( )q

4 ?( )q

0 1 2 3, , ,q q q q

1 2 3, ,q q q

3 4,q q

Example ?

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

M 0 0 1 2 3( ) , , ,q q q q q 1 1 2 3( ) , ,q q q q

2 2( )q q 3 3( )q q 4 3 4( ) ,q q q

Example ?

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

M 0 0 1 2 3( ) , , ,q q q q q 1 1 2 3( ) , ,q q q q

2 2( )q q 3 3( )q q 4 3 4( ) ,q q q

? ,a b0 4{ , , }2 q q

0 1 2 3, , ,q q q qThere are 32 possible subsets. However, most of them are useless.

There are 32 possible subsets. However, most of them are useless.

Example ?

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

M 0 0 1 2 3( ) , , ,q q q q q 1 1 2 3( ) , ,q q q q

2 2( )q q 3 3( )q q 4 3 4( ) ,q q q

? ,a b0 4{ , , }2 q q

0 1 2 3, , ,q q q q 4: :Q K q Q

Example ?

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

M 0 0 1 2 3( ) , , ,q q q q q 1 1 2 3( ) , ,q q q q

2 2( )q q 3 3( )q q 4 3 4( ) ,q q q

?s 0 1 2 3, , ,q q q q

0 1 2 3, , ,q q q q>M’

Example ?

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

M 0 0 1 2 3( ) , , ,q q q q q 1 1 2 3( ) , ,q q q q

2 2( )q q 3 3( )q q 4 3 4( ) ,q q q

0 1 2 3, , ,q q q q> 0 4( ) ( )q q a

2 4( ) ( )q q

Example ?

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

( , )( , ) ( ) ,

q Q p qQ p Q K

( , , , , ) ( , , , , )M K s F M K s F

M 0 0 1 2 3( ) , , ,q q q q q 1 1 2 3( ) , ,q q q q

2 2( )q q 3 3( )q q 4 3 4( ) ,q q q

0 1 2 3, , ,q q q q> 0 4( ) ( )q q a

2 4( ) ( )q q

0 1 2 3 4, , , ,q q q q q

2 3 4, ,q q q

3 4,q qa, b

Example

0 1 2 3, , ,q q q q> 0 4( ) ( )q q a

2 4( ) ( )q q

0 1 2 3 4, , , ,q q q q q

2 3 4, ,q q q

3 4,q qa, b

* *( ) ( )L M a a b bb a ( )L M

Exercises Construct deterministic finite automata equivalent

to the nondeterministic automata shown below:

(a) (b)

Properties of

Languages Accepted by

Finite Automata

The Chomsky Hierarchy

Chomsky Hierarchy

Languages Grammars Automaton

Type 0 Recursively enumerable

unrestricted Turing Machine

Recursive unrestricted Decider

Type 1 Context-Sensitive Context-Sensitive

Linear-Bounded Automaton

Type 2 Context-Free Context-Free Push-Down Automaton

Type 3 Regular Regular NFA or DFA

Non-recursively enumerable

Recursively-enumerable

Recursive

Context-sensitive

Context-free

Regular

Properties of Languages Accepted by FA’s

The class of languages accepted by finite automata

closed under:

a) union;

b) concatenation;

c) Keene star;

d) complementation;

e) intersection.

2L1 2L L

1 2L L*1L

1 2L L

2 1L L

closed under:

a) union;

b) concatenation;

c) Keene star;

d) complementation;

e) intersection.

>M1 >M2

accepts

1 2( ) ( )L M L M

closed under:

a) union;

b) concatenation;

c) Keene star;

d) complementation;

e) intersection.

accepts 1 2( ) ( )L M L M

closed under:

a) union;

b) concatenation;

c) Keene star;

d) complementation;

e) intersection.

accepts *1( )L M

closed under:

a) union;

b) concatenation;

c) Keene star;

d) complementation;

e) intersection.

( , , , , )

M K s F

M K s K F

( )M L M accepts

closed under:

a) union;

b) concatenation;

c) Keene star;

d) complementation;

e) intersection.

1 2L L 1 2L L

1 2L L

Theorem

There are algorithms to answer the following

questions about finite automata:

a) w L(M)?

b) L(M) = ?

c) L(M) = *?

d) L(M1) L(M2) ?

e) L(M1) = L(M2) ?

Theorem

There are algorithms to answer the following

questions about finite automata:

a) w L(M)?

b) L(M) = ?

c) L(M) = *?

d) L(M1) L(M2) ?

e) L(M1) = L(M2) ?

Feed w into M.

Check any final state is reachable Check L(M) = ?

Check L(M1) L(M2) = ? Check L(M1) L(M2) ? L(M2) L(M1) ?

Exercises

Draw state diagrams for nondeterministic finite automata accepting these languages:

a) (ab)*(ba)*aa*

b) ((abaab)*a*)*

c) ((a*b*a*)*b)*

d) (bab)*(bba)*

Finite Automata&

Regular Expressions

The Languages Accepted by FA’s

The languages accepted by FA’s are called regular sets.

How to describe a regular set?– Regular Expression

Definition Regular Expressions

A regular expression is defined inductively over the alphabet { (, ), , , +, *} as follows:

, , and each is a regular expression If and are regular expressions, then

– ( + )

– ( )

are regular expressions

inductive

Definition Regular Expressions

A regular expression is defined inductively over the alphabet { (, ), , , +, *} as follows:

, , and each is a regular expression If and are regular expressions, then

– ( + )

– ( )

are regular expressions

inductive

union, or concatenation

Kleene closure

Writing a Regular Expression

For omitting parentheses, the precedence of ‘operators’ are assigned as follows:

* > • > +((0(1*))+0) 01*+0

What is this language?

Examples

2. (0+1)*

3. (0+1)* 00 (0+1)*

4. (0+) (1+10)*

5. (0+1)*011

6. 0*1*2*

7. 00*11*22*

What do the following regular expressions represent?

More ExamplesIs the following language L an r.e. over =[a-z]?

L={what, which, when, where}.

Write L as an r.e.

L = what + which + when + where

L = (wh)(at + ich + en + ere)

L = (wh)(at + ich) + (whe) (n + re)

Theorem

A language is regular if and only if it is accepted by a finite automaton.

“”

Given a regular expression, how to construct an FA to accept the language described by it?

Given an FA, what is its corresponding regular expression?

Theorem

“”Pf) Suppose that we are given a regular expression r, we can construct an NFA to accept the language described it inductively based on the number of `operators' it possesses as follows.

Basis (zero operators)Case 1: r =

Case 2: r =

Case 3: r = a

qfq0a>

Theorem

Basis (zero operators)

trueHypothesis

Assumed true for r.e.’s few than m operatorsInduction

To be shown true for r.e.’s with m+1 operators

Case 1: r = r1+r2

Case 2: r = r1r2

Case 3: r = r1*

r : m+1 operatorsr1: m operatorsr2: m operators

Theorem

Basis (zero operators)

trueHypothesis

Assumed true for r.e.’s few than m operatorsInduction

To be shown true for r.e.’s with m+1 operators

Case 1: r = r1+r2

Case 2: r = r1r2

Case 3: r = r1*

r : m+1 operatorsr1: m operatorsr2: m operators

The class of languages accepted by fi nite automata is

closed under:

a) union;

b) concatenation;

c) Keene star;

d) complementation;

e) intersection.

2L1 2L L

1 2L L*1L

1 2L L

2 1L L

The class of languages accepted by fi nite automata is

closed under:

a) union;

b) concatenation;

c) Keene star;

d) complementation;

e) intersection.

2L1 2L L

1 2L L*1L

1 2L L

2 1L L

Theorem

“”Pf) Suppose that we are given a DFA, we will determine its corresponding regular expression also inductively.

Consider DFA M = ({q1, q2,…,qn}, , , q1, F).

1 1 1 1( )*k k k k kij ik kk kj ijR R R R R

0 { ( , ) }

{ ( , ) } { }i j

a | δ q a q i jR

a | δ q a q ε i j

*( , ) ( , ) ( , )

i l jkij

q x δ q y qR x

l k y x y

satisfi es if and

M M* M M

Define

Theorem

0 { ( , ) }

{ ( , ) } { }i j

a | δ q a q i jR

a | δ q a q ε i j

*( , ) ( , ) ( , )

i l jkij

q x δ q y qR x

l k y x y

satisfi es if and

M M* M M

Define

kijR is regular for

any i, j, and k.

Theorem

0 { ( , ) }

{ ( , ) } { }i j

a | δ q a q i jR

a | δ q a q ε i j

*( , ) ( , ) ( , )

i l jkij

q x δ q y qR x

l k y x y

satisfi es if and

M M* M M

Define

kijR is regular for

any i, j, and k.1( ) n

ff FL M R

( )L M is reguar.

Equivalence FA & RE

Chomsky Hierarchy

Languages Grammars Automaton

Type 0Recursively enumerable

unrestricted Turing Machine

Recursive unrestricted Decider

Type 1 Context-SensitiveContext-Sensitive

Linear-Bounded Automaton

Type 2 Context-Free Context-Free Push-Down Automaton

Type 3 Regular Regular NFA or DFA

Recursive

Context-sensitive

Context-free

Regular

Example

Find an r.e. to represent the following FA.

M: q1 q2 q3

0 0,1>

3 312 13( )L M R R 3 3

12 13r r

ExampleM: q1 q2 q3

0 0,1> q1 q2 q3

0 0,1q1 q2 q3

0 0,1>

3 312 13( )L M R R 3 3

12 13r r

k = 0 k = 11 0 0 0 0

11 11 11 11 11( )*r r r r r 1 0 0 0 0

12 11 11 12 12( )* 0r r r r r 1 0 0 0 0

13 11 11 13 13( )* 1r r r r r 1 0 0 0 021 21 11 11 21( )*r r r r r 01 0 0 0 022 21 11 12 22( )* 00r r r r r 1 0 0 0 023 21 11 13 23( )* 1 01r r r r r 1 0 0 0 031 31 11 11 31( )*r r r r r 1 0 0 0 032 31 11 12 32( )* 0 1r r r r r 1 0 0 0 033 31 11 13 33( )*r r r r r

0 { ( , ) }

{ ( , ) } { }i j

a | δ q a q i jR

a | δ q a q ε i j

0( , )l i l jl

a δ q a qr

otherwise

1 1 1 1( )*k k k k kij ik kk kj ijr r r r r

ExampleM: q1 q2 q3

0 0,1> q1 q2 q3

0 0,1q1 q2 q3

0 0,1>

3 312 13( )L M R R 3 3

12 13r r

k = 0 k = 10

2 1 1 1 111 12 22 21 11( )* 0( 00)*0r r r r r 2 1 1 1 1

12 12 22 22 12( )* 0( 00)*( 00) 0r r r r r 2 1 1 1 1

13 12 22 23 13( )* 0( 00)*(1 01) 1r r r r r 2 1 1 1 1

21 22 22 21 21( )* ( 00)( 00)*0 0r r r r r 2 1 1 1 1

22 22 22 22 22( )* ( 00)( 00)*( 00) ( 00)r r r r r 2 1 1 1 1

23 22 22 23 23( )* ( 00)( 00)*(1 01) (1 01)r r r r r 2 1 1 1 1

31 32 22 21 31( )* (0 1)( 00)*0r r r r r 2 1 1 1 1

32 32 22 22 32( )* (0 1)( 00)*( 00) (0 1)r r r r r 2 1 1 1 1

33 32 22 23 33( )* (0 1)( 00)*(1 01)r r r r r

k = 2(00)*

0(00)*

(0+1)(00)*0

(0+1)(00)*

+(0+1)0*1

ExampleM: q1 q2 q3

0 0,1> q1 q2 q3

0 0,1q1 q2 q3

0 0,1>

3 312 13( )L M R R 3 3

12 13r r

k = 0 k = 10

k = 2(00)*

0(00)*

(0+1)(00)*0

(0+1)(00)*

+(0+1)0*1

3 2 2 2 212 13 33 32 12( )*

0*1( (0 1)0*1)*(0 1)(00)* 0(00)*

0*1((0 1)0*1)*(0 1)(00)* 0(00)*

r r r r r

3 2 2 2 213 13 33 33 13( )*

0*1( (0 1)0*1)*( (0 1)0*1) 0*1

0*1((0 1)0*1)*

r r r r r

k = 0 k = 10

k = 2(00)*

0(00)*

(0+1)(00)*0

(0+1)(00)*

+(0+1)0*1

ExampleM: q1 q2 q3

0 0,1> q1 q2 q3

0 0,1q1 q2 q3

0 0,1>

3 312 13( )L M R R 3 3

12 13r r

3 2 2 2 212 13 33 32 12( )*

0*1( (0 1)0*1)*(0 1)(00)* 0(00)*

0*1((0 1)0*1)*(0 1)(00)* 0(00)*

r r r r r

3 2 2 2 213 13 33 33 13( )*

0*1( (0 1)0*1)*( (0 1)0*1) 0*1

0*1((0 1)0*1)*

r r r r r

3 312 13 0*1((0 1)0*1)*( (0 1)(00)*) 0(00)*r r 3 3

12 13 0*1((0 1)0*1)*( (0 1)(00)*) 0(00)*r r

Languages Not Accepted by

Finite Automata

What FA’s can? What FA’s cannot?

1 { | {0, ,9}* 2 | ( ) 3 | ( )}L w w val w val w st. and

{ , }* | |w a b wL w

st. is and

contains number of

even 's

3 0n nL a b n

4 nL a n is prime

Which of the following languages are regular?

The Pumping Theory

Let L be an infinite regular set. Then there are strings u,v, and w st. |v|>0 and uvnwL for all n 0.

Pf)Let L is accepted by a DFA M = (K, , , q0 , F) with m states.

Consider input w = 1 2… l, l m.

1 20( , , , , )lq M M 21( , , , )lq M M M M 1( , )llq M M ( , )lq . . .

By pigeonhole, there exists j, k with 0 j < k l st. qj = qk.

qj = qk qmu=1… j

v= j+1… k

w= k+1… mq0>

The Pumping Theory

Let L be an infinite regular set. Then there are strings u,v, and w st. |v|>0 and uvnwL for all n 0.

Pf)Let L is accepted by a DFA M = (K, , , q0 , F) with m states.

Consider input w = 1 2… l, l m.

1 20( , , , , )lq M M 21( , , , )lq M M M M 1( , )llq M M ( , )lq . . .

By pigeonhole, there exists j, k with 0 j < k l st. qj = qk.

v= j+1… k

qj = qk qmu=1… j w= k+1… mq0>

It is then seen that if qmF, then

uw L, uvw L, uv2w L, . . ., uvnwL for all n 0.

ExampleShow that the language L={anbn | n0} is not regular.

Pf) Suppose that L is regular.By pumping theorem,

u v wCase 1: z = a a … a b b … b

uvmwL for all m 0

z = anbn =uvwL st. uvmwL for all m 0.

Recursive

Context-sensitive

Context-free

Regular

L={anbn | n0}L={anbn | n0}

Implementation of DFA

Example Input a 0/1 sting. If the numbers of 0 and 1 in the

string are both even, then it is legal; otherwise, it is illegal.– 0011001001(legal)– 11001001001(illegal)

Writing a C program to do so.

1 1 1 1

Examples:

00100011<CR>

01101011<CR>

001010101<CR>

0010010101<CR>

Finite State Machine

1 1 1 1

symbol(event)

0 1 <CR>q1

The ParserThe Parser

Implementation (I)

symbol(event)

0 1 <CR>q1

#define q0 0#define q1 1#define q2 2#define q3 3#define fini4

Implementation (I)

symbol(event)

0 1 <CR>q1

int parser[4][3]={q1, q2, fini,q0, q3, fini,q3, q0, fini,q2, q1, fini

int state=q0;int event;char* str;

int parser[4][3]={q1, q2, fini,q0, q3, fini,q3, q0, fini,q2, q1, fini

int state=q0;int event;char* str;

Implementation (I)

void ToEvent(char c){

if(c == ’0’) event = 0;else if(c == ’1’) event = 1;else event = 2;

void ToEvent(char c){

if(c == ’0’) event = 0;else if(c == ’1’) event = 1;else event = 2;

Event (or Message) EncodingEvent (or Message) Encoding

Implementation (I)

void main(){

// Ask user to input a 0/1 string// Store the string into strstate = q0; //initializationwhile(state!=fini){

ToEvent(*str++);EventHandler(event);

void main(){

ToEvent(*str++);EventHandler(event);

}} Event (or Message) LoopEvent (or Message) Loop

Implementation (I)

void EventHandler(int event){

int next_state = parser[state][event];switch(next_state){ case fini:

printf(”%s\n”, state==q0 ? ”ok” : ”err”); default:

state = next_state; //change state}

void EventHandler(int event){

int next_state = parser[state][event];switch(next_state){ case fini:

printf(”%s\n”, state==q0 ? ”ok” : ”err”); default:

state = next_state; //change state}

Event HandlerEvent Handler

Implementation (II)

1 1 1 1

symbol(event)

0 1 <CR>pq1

Implementation (II)

symbol(event)

0 1 <CR>pq1

#define q0 0#define q1 1#define q2 2#define q3 3#define fini4

void pq0(), pq1(), pq2(), pq3();void ok(), err();void pq0(), pq1(), pq2(), pq3();void ok(), err();

Implementation (II)

symbol(event)

0 1 <CR>pq1

void pq0(){

state = q0;}

void pq1(){

state = q1;}

void pq0(){

state = q0;}

void pq1(){

state = q1;}

void pq2(){

state = q2;}

void pq3(){

state = q3;}

void pq2(){

state = q2;}

void pq3(){

state = q3;}

Implementation (II)

symbol(event)

0 1 <CR>pq1

void ok(){

printf(”ok\n”);state = fini;

void err(){

printf(”error\n”);state = fini;

void ok(){

printf(”ok\n”);state = fini;

void err(){

printf(”error\n”);state = fini;

Implementation (II)

symbol(event)

0 1 <CR>pq1

typedef void (*FUNCTION)();

FUNCTION parser[4][3]={pq1, pq2, ok,pq0, pq3, err,pq3, pq0, err,pq2, pq1, err

typedef void (*FUNCTION)();

FUNCTION parser[4][3]={pq1, pq2, ok,pq0, pq3, err,pq3, pq0, err,pq2, pq1, err

Implementation (II)

void main(){

ToEvent(*str++);(*parser[state][event])();

void main(){

ToEvent(*str++);(*parser[state][event])();

}} Event (or Message) LoopEvent (or Message) Loop

lecture 5: finite automata

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