chap2+finiteautomataStud

8/10/2019 chap2+finiteautomataStud

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SSK5204

CHAPTER2: FINITEAUTOMATA

Dr. Nor Fazlida Mohd Sani

Dept. of Computer Science

Fac. of Computer Science and Information

Technology, UPM.

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Finite Automata

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EXAMPLEOFAFINITEAUTOMATON

There are statesevenand odd, evenis the startstate

The automaton takes inputsfrom alphabet {L, R}

The state evenis an accepting stateThere are transitionssaying what to do for every

state and every alphabet symbol

even odd

L, R

L, R

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DETERMINISTICFINITEAUTOMATA

A finite automaton(DFA) is a 5-tuple (Q, S, d, q0,

F)where

Qis a finite set of states

Sis an alphabet d:QSQis a transition function

q0Qis the initial state

F Q is a set of accepting states(or final states).

In diagrams, the accepting states will be

denoted by double loops 4


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EXAMPLE

5

q0 q1 q21 0

0 0,11

alphabet S= {0, 1}

statesQ= {q0, q1, q2}

initial state q0

accepting states F= {q0, q1}

states

inputs

0 1q0q1q2

q0 q1q2

q2q2

q1

table of

transition function d:


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LANGUAGEOFADFA

Language ofMis {x {L, R}*: xhas even length} 6

The language of a DFA (Q, S, d, q0, F)is the set of

all strings over Sthat, starting from q0and

following the transitions as the string is read left

to right, will reach some accepting state.

M: even odd

L, R

L, R


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q0 q1

b a

a

b

What are the languages of these automata?

EXAMPLES

q0

q1

q2

q3

q4

a

a

a a

a

b

b

bb

b

q0 q1

0 1

1 0 q2

0, 1

S= {a, b} S= {a, b}

S= {0, 1}

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EXAMPLES

Construct a DFA over alphabet {0, 1} that acceptsall strings with at most three 1s

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EXAMPLES

Construct a DFA over alphabet {0, 1} that acceptsall strings with at most three 1s

Answer

q0 q1

0

1 1 q2

0

q31 q4+

0, 10

1

0

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EXAMPLES

Construct a DFA that accepts the language

L= {010, 1} ( S= {0, 1} )

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EXAMPLES

Construct a DFA over alphabet {0, 1}that

accepts all strings that end in 01

Hint: The DFA must remember the last 2bits of the string it is reading

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EXAMPLES

Construct a DFA over alphabet {0, 1}that accepts

all strings that end in 101

Sketch of answer:

0

1

qe

q0

q1

q00

q10

q01

q11

q000

q001

q101

q111

0

1

0

1

0

1

1

1

1

1

0

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Nondeterminism

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EXAMPLE

0

1

qe

q0

q1

q00

q10

q01

q11

q000

q001

q101

q111

0

1

0

1

0

1

1

1

1

1

0

Construct a DFA over alphabet {0, 1}thataccepts those strings that end in 101

Sketch of answer:

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WOULDBEEASIERIF

Suppose we could guesswhen the string we are

reading has only 3 symbols left

Then we could simply look for the sequence 101

and accept if we see it

qdie

1

0 13 symbols left

This is nota DFA!

0

1

0

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NONDETERMINISM

Nondeterminismis the ability to make guesses,

which we can later verify

How a nondeterministicautomaton for strings that

end in 101works:

1. Guessif you are approaching end of input

2. If guess is yes, look for 101and accept if you see it

3. If guess is no, read one more symbol and go to step1

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NONDETERMINISTICFINITEAUTOMATON

This is a kind of automaton that allows you to

make guesses

Each state can have zero, one, or moreoutgoingtransitions labeled by the same symbol

1 0 1

0, 1

q0 q1 q2 q3

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THEABILITYTOMAKECHOICES

1 0 1

0, 1

q0 q1 q2 q3

State q0has two transitions labeled 1

Upon reading 1, we have the choiceof staying in

q0or moving to q1

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1 0 1

0, 1

q0 q1 q2 q3

State q1has no transition labeled 1

Upon reading 1 in q1, we die; upon reading 0, we

continue to q2

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1 0 1

0, 1

q0 q1 q2 q3

State q3has no transition going out

Upon reading 0or 1in q3, we die

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MEANINGOFNFA

1 0 1

0, 1

q0 q1 q2 q3

Guess if you are 3

symbols away from end of

input

If so, guess you

will see the pattern101

Check that you are at

the end of input

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HOWTORUNANNFA

1 0 1

0

q0 q1 q2 q3

input: 0 1 1 0 1

,1

The NFA can have several active statesat the same time

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EXAMPLE

Construct an NFA over alphabet {0, 1} that accepts

those strings that contain the pattern001

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EXAMPLE

Construct an NFA over alphabet {0, 1} that accepts thosestrings that contain the pattern 001 somewhere

Answer

0 0 1

0, 1

q0 q1 q2 q3

0, 1

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DEFINITION

A nondeterministic finite automaton(NFA) is a5-tuple (Q, S, d, q0, F)where

Qis a finite set of states

Sis an alphabet

d:Q(S {e}) subsets of Qis a transition function q0Qis the initial state

F Q is a set of accepting states (or final states).

Differences from DFA: transition functiondcan go into several states

It allows e-transitions

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LANGUAGEOFANNFA

The language of an NFA is the set of allstrings for which there is somepath that,starting from q0, leads to an accepting state as

the string is read left to right (and e-transitions

are taken for free). Example

e, 00, 001, 101are accepted, but 11, 0110are not

q0e, 1 0

0

eq1 q2

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EXAMPLE

alphabet S = {0, 1}

states Q= {q0, q1, q2}

initial state q0

accepting states F= {q2}

states

inputs0 1

q0

q1

q2

{q1}

table of

transition function d :

{q0, q1}

q0e, 1 0

0

eq1 q2

e

{q1}

{q2}

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EXAMPLES

q0e, 1 e

0

q1 q2

q0e, 1

e

0

q1 q2

e1 2

00

0

01

2

3

4

5

q0

e, 1e

0q1 q2

01

2 3

4

q0e, 1

e

0

q1 q2

01

2

3

54

or

or

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EXAMPLES

q0e , 1 e

0

q1 q21 5

001

0

q0

e, 1e

0q1 q2

01

4

q0e, 1

e

0

q1 q2

01

2

3

4

1012

3

11

q0e, 1

e

0

q1 q2

01

2

STOP

STOP30


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EXAMPLEOFe-TRANSITIONS

Construct an NFA that accepts all strings with an evennumber of 0s oran odd number of 1s

r0 r1

1 1

0

0

even number of 0s

s0 s1

0 0

1

1

odd number of 1s

q0

e

e

e-transitions can be taken for free (without reading input)31


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NFA to DFA conversion

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NFASAREASPOWERFULASDFAS

Obviously, an NFA can do everything a DFA can do

How about the other way?

If Lis the language of some NFA, thenit is also the language of some DFA.

YES

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CONVERTINGNFASINTODFAS

We will show a general way to convert every NFA into an

equivalent DFA

Step 1:Simplify NFA by eliminating e-transitions

Step 2: Convert simplified NFA (without es)

We do this first

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NFA TODFA CONVERSIONINTUITION

1 0

0, 1

q0 q1 q2NFA:

DFA: 1q0 q0or q1

1

q0or q2

1

0 0

0

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NFA TODFA CONVERSIONINTUITION

1 0

0, 1

q0 q1 q2NFA:

DFA: 1q0 {q0, q1}

1

{q0, q2}

1

0 0

0

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GENERALMETHOD

NFA DFAstates q0, q1, , qn {q0}, {q1}, {q0,q1}, , {q0,,qn}

one for each subset of statesin the NFA

initial state q0 {q0}

transitions d d({qi1,,qik}, a) =

d(qi1

, a) d(qik

, a)

accepting

states

F Q F = {S: Scontains some statein F}

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CONVERTINGVIATHEGENERALMETHOD

1 0

0, 1

q0 q1 q2NFA:

DFA:

{q0, q1}

{q0, q2} {q0, q1, q2}

{q1, q2}

{q0}

{q1}

{q2}

0, 1

0

1

0

1

0

1

0 1

0

1

0

1

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CONVERTINGVIATHEGENERALMETHOD

{q0, q1}

{q0, q2} {q0, q1, q2}

{q1, q2}

{q0}

{q1}

{q2}

0, 1

0

1

0

1

0

1

0 1

0

1

0

1

After eliminating the dead states andtransitions, we end up with the same picture

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WHYTHEMETHODWORKS

At the end, the DFA accepts when it is in a state thatcontains some accepting state of NFA

So the DFA accepts only when the NFA accepts too

After reading nsymbols, the DFA is in state{qi1,,qik}if and only if the NFA is in one of the

states qi1,,qik

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CONVERTINGNFASINTODFAS

We will show a general way to convert every NFA into an

equivalent DFA

Step 1:Simplify NFA by eliminating e-transitions

Step 2: Convert simplified NFA (without es)

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ELIMINATINGe-TRANSITIONS

q0 q1 q2e, 1 0

0

eeNFA:

Transition table of corresponding NFA:

states

inputs0 1

q0

q1

q2

{q1, q2}{q0, q1, q2}

{q0, q1, q2}

Accepting states of NFA: q0, q1, q242


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q0 q1 q2e, 1

0

0

eNFA:

NFA without es: q0 q1 q20, 1

0

0

0

0, 1

0

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ELIMINATINGe-TRANSITIONS: GENERAL

METHOD

For every pair of states qi, qj and every symbol a S,replace every pathlike

For every accept state qf, make accepting all states

connected to it via es:

qi q5 q2 q0 qje e ea

q4

qi

q5

a

e

e

qi qja qi

a

e e

e

q9 q7 q3 qf44


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q0 q1 q2e, 1

0

0

eNFA:

NFA without es: q0 q1 q20

0

0

00

, 1

0, 1

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