Automata

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Non Deterministic Automata

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1q 2q

3q

a

a

a

0q

}{aAlphabet =

Nondeterministic Finite Accepter (NFA)

3

1q 2q

3q

a

a

a

0q

Two choices

}{aAlphabet =


4

No transition

1q 2q

3q

a

a

a

0q

Two choices No transition

}{aAlphabet =


5

a a

0q

1q 2q

3q

a

a

First Choice

a

6

a a

0q

1q 2q

3q

a

a

a

First Choice

7

a a

0q

1q 2q

3q

a

a

First Choice

a

8

a a

0q

1q 2q

3q

a

a

a “accept”

First Choice

9

a a

0q

1q 2q

3q

a

a

Second Choice

a

10

a a

0q

1q 2qa

a

Second Choice

a

3q

11

a a

0q

1q 2qa

a

a

3q

Second Choice

No transition:the automaton hangs

12

a a

0q

1q 2qa

a

a

3q

Second Choice

“reject”

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Observation

An NFA accepts a stringifthere is a computation of the NFAthat accepts the string

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Example

aa is accepted by the NFA:

0q

1q 2q

3q

a

a

a

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Lambda Transitions

1q 3qa0q

1q a

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a a

1q 3qa0q

2q a

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a a

1q 3qa0q

2q a

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a a

1q 3qa0q

2q a

(read head doesn’t move)

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a a

1q 3qa0q

2q a

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a a

1q 3qa0q

2q a

“accept”

String is acceptedaa

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Language accepted: }{aaL

1q 3qa0q

2q a

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Another NFA Example

0q 1q 2qa b

3q

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a b

0q 1q 2qa b

3q

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0q 2qa b

3q

a b

1q

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a b

0q 1qa b

3q2q

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a b

0q 1qa b

3q2q

“accept”

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0qa b

a b

Another String

a b

1q 2q 3q

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0qa b

a b a b

1q 2q 3q

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0qa b

a b a b

1q 2q 3q

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0qa b

a b a b

1q 2q 3q

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0qa b

a b a b

1q 2q 3q

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0qa b

a b a b

1q 2q 3q

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0qa b

a b a b

1q 2q 3q

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a b a b

0qa b

1q 2q 3q

“accept”

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ab

ababababababL ...,,,

Language accepted

0q 1q 2qa b

3q

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Another NFA Example

0q 1q 2q0

11,0

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*10

...,101010,1010,10,

L

0q 1q 2q0

11,0

Language accepted

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Formal Definition of NFAs FqQM ,,,, 0

:Q

::0q

:F

Set of states, i.e. 210 ,, qqq

: Input aplhabet, i.e. ba,Transition function

Initial state

Final states

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10 1, qq

0

11,0

Transition Function

0q 1q 2q

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0q

0

11,0

},{)0,( 201 qqq

1q 2q

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0q

0

11,0

1q 2q

},{),( 200 qqq

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0q

0

11,0

1q 2q

)1,( 2q

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Extended Transition Function

*

0q

5q4q

3q2q1qa

aa

b

10 ,* qaq

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540 ,,* qqaaq

0q

5q4q

3q2q1qa

aa

b

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0320 ,,,* qqqabq

0q

5q4q

3q2q1qa

aa

b

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Formally

wqq ij ,*It holds

if and only if

there is a walk from towith label

iq jqw

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The Language of an NFA

0q

5q4q

3q2q1qa

aa

b

540 ,,* qqaaq

M

)(MLaa

50 ,qqF

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0q

5q4q

3q2q1qa

aa

b

0320 ,,,* qqqabq MLab

50 ,qqF

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0q

5q4q

3q2q1qa

aa

b

50 ,qqF

540 ,,* qqabaaq )(MLaaba

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0q

5q4q

3q2q1qa

aa

b

50 ,qqF

10 ,* qabaq MLaba

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0q

5q4q

3q2q1qa

aa

b

aaababaaML *

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FormallyThe language accepted by NFA is:

where

and there is some

M

,...,, 321 wwwML

,...},{),(* 0 jim qqwq

Fqk (final state)

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0q kq

w

w

w

),(* 0 wq MLw

Fqk

iq

jq

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Equivalence of NFAs and DFAs

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Equivalence of Machines

For DFAs or NFAs:

Machine is equivalent to machine

if

1M 2M

21 MLML

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Example

0q 1q 2q

0

11,0

0q 1q 2q

0

11

0

1,0

NFA

DFA

*}10{1 ML

*}10{2 ML

1M

2M

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Since

machines and are equivalent

*1021 MLML

1M 2M

0q 1q 2q

0

11,0

0q 1q 2q

0

11

0

1,0

DFA

NFA 1M

2M

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Question: NFAs = DFAs ?

Same power?Accept the same languages?

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Question: NFAs = DFAs ?

Same power?Accept the same languages?

YES!

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We will prove:

Languages acceptedby NFAs

Languages acceptedby DFAs

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We will prove:



NFAs and DFAs have the same computation power

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Step 1

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Step 1

Proof: Every DFA is also an NFA

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Step 1

Proof: Every DFA is also an NFA

A language accepted by a DFAis also accepted by an NFA

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Step 2

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Step 2

Proof: Any NFA can be converted to anequivalent DFA

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Step 2

Proof: Any NFA can be converted to anequivalent DFA

A language accepted by an NFAis also accepted by a DFA

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NFA to DFA

a

b

a

0q 1q 2q

NFA

DFA

0q

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NFA to DFA

a

b

a

0q 1q 2q

NFA

DFA

0q 21,qqa

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NFA to DFA

a

b

a

0q 1q 2q

NFA

DFA

0q 21,qqa

b

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NFA to DFA

a

b

a

0q 1q 2q

NFA

DFA

0q 21,qqa

b

a

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NFA to DFA

a

b

a

0q 1q 2q

NFA

DFA

0q 21,qqa

b

ab

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NFA to DFA

a

b

a

0q 1q 2q

NFA

DFA

0q 21,qqa

b

ab

ba,

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NFA to DFA

a

b

a

0q 1q 2q

NFA

DFA

0q 21,qqa

b

ab

ba,

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NFA to DFA: Remarks

We are given an NFA

We want to convert it to an equivalent DFA

With

M

M

)(MLML

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If the NFA has states

the DFA has states in the powerset

,...,, 210 qqq

,....,,,,,,, 7432110 qqqqqqq

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Procedure NFA to DFA

1. Initial state of NFA:

Initial state of DFA:

0q

0q

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Example

a

b

a

0q 1q 2q

NFA

DFA

0q

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2. For every DFA’s state

Compute in the NFA

Add transition

},...,,{ mji qqq

...

,,*

,,*

aq

aq

j

i

},...,,{ mji qqq

},...,,{},,...,,{ mjimji qqqaqqq

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Exampe

a

b

a

0q 1q 2q

NFA

0q 21,qqa

DFA

},{),(* 210 qqaq

210 ,, qqaq

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Repeat Step 2 for all letters in alphabet,

untilno more transitions can be added.

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Example

a

b

a

0q 1q 2q

NFA

DFA

0q 21,qqa

b

ab

ba,

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3. For any DFA state

If some is a final state in the NFA

Then, is a final state in the DFA

},...,,{ mji qqq

jq

},...,,{ mji qqq

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Example

a

b

a

0q 1q 2q

NFA

DFA

0q 21,qqa

b

ab

ba,

Fq 1

Fqq 21,

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Theorem Take NFA M

Apply procedure to obtain DFA M

Then and are equivalent :M M

MLML

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Finally

We have proven



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We have proven

Regular Languages

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We have proven

Regular LanguagesRegular Languages

Automata

Documents

Transcript of Automata