CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 3 Mälardalen University

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CD5560

FABER

Formal Languages, Automata and Models of Computation

Lecture 3

Mälardalen University2010

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Content

Finite Automata, FADeterministic Finite Automata, DFANondeterministic Automata NFANFA DFA Equivalence

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Finite Automata FA

(Finite State Machines)

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There is no formal general definition for "automaton". Instead, there are various kinds of automata, each with it's own formal definition.

• has some form of input

• has some form of output

• has internal states,

• may or may not have some form of storage

• is hard-wired rather than programmable

Generally, an automaton

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Finite Automaton

Input

String

Output

String

Finite

Automaton

6

Finite Accepter

Input

“Accept”

or

“Reject”

String

Finite

Automaton

Output

7

Nodes = States Edges = Transitions

An edge with several symbols is a short-hand for several edges:

FA as Directed Graph

1qa

0q

1qba,0q1q

a

0qb

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Deterministic Finite Automata DFA

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• Deterministic there is no element of choice• Finite only a finite number of states and arcs • Acceptors produce only a yes/no answer

DFA

10

Transition Graph

initial

state

final

state

“accept”statetransition

abba -Finite Acceptor

0q 1q 2q 3q 4qa b b a

5q

a a bb

ba,

ba,

},{ baAlphabet =

11

Formally

For a DFA

Language accepted by :

FqQM ,,,, 0

M

FwqwML ,*:* 0

alphabet transition

function

initial

state

final

states

12

Observation

Language accepted by

FwqwML ,*:* 0

M

FwqwML ,*:* 0

MLanguage rejected by

13

Regular Languages

All regular languages form a language family

LM MLL

A language is regular if there is

a DFA such that

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Formal definitions

Deterministic Finite Accepter (DFA)

FqQM ,,,, 0

Q

0q

F

: set of states

: input alphabet

: transition function

: initial state

: set of final states

15

Input Aplhabet


5q

a a bb

ba,

ba,

ba,

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Set of States

Q


5q

a a bb

ba,

543210 ,,,,, qqqqqqQ

ba,

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Initial State

0q

1q 2q 3q 4qa b b a

5q

a a bb

ba,

ba,

0q

18

Set of Final States

F

0q 1q 2q 3qa b b a

5q

a a bb

ba,

4qF

ba,

4q

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Transition Function


5q

a a bb

ba,

QQ :

ba,

20

10 , qaq

2q 3q 4qa b b a

5q

a a bb

ba,

ba,

0q 1q

21

50 , qbq

1q 2q 3q 4qa b b a

5q

a a bb

ba,

ba,

0q

22


5q

a a bb

ba,

ba,

32 , qbq

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Transition Function


5q

a a bb

ba,

ba,

a b0q

2q

5q

1q 5q

5q5q

3q

4q

1q 5q 2q

5q 3q

4q5q

5q 5q

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Extended Transition Function

*

QQ *:*


5q

a a bb

ba,

ba,

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20 ,* qabq

3q 4qa b b a

5q

a a bb

ba,

ba,

0q 1q 2q

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40 ,* qabbaq


5q

a a bb

ba,

ba,

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50 ,* qabbbaaq

1q 2q 3q 4qa b b a

5q

a a bb

ba,

ba,

0q

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50 ,* qabbbaaq

1q 2q 3q 4qa b b a

5q

a a bb

ba,

ba,

0q

Observation: There is a walk from to

with label

0q 5qabbbaa

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Recursive Definition

)),,(*(,*

,*

awqwaq

qq


5q

a a bb

ba,

ba,

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5q

a a bb

ba,

ba,

21

0

0

0

0

,

,,

,,,*

),,(*

,*

qbq

baq

baq

baq

abq

31

Languages Accepted by DFAs

Take DFA

Definition:

The language contains

all input strings accepted by

= { strings that drive to a final state}

M

MLM

ML M

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Example


5q

a a bb

ba,

ba,

abbaML M

accept

},{ baAlphabet =

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Another Example


5q

a a bb

ba,

ba,

abbaabML ,, M

acceptacceptaccept

},{ baAlphabet =

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Formally

For a DFA

Language accepted by :

FqQM ,,,, 0

M

FwqwML ,*:* 0

alphabet transition

function

initial

state

final

states

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Observation

Language accepted by

FwqwML ,*:* 0

M

FwqwML ,*:* 0

MLanguage rejected by

36

More Examples

a

b ba,

ba,

0q 1q 2q

}0:{ nbaML n

accept trap state},{ baAlphabet =

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ML = { all strings with prefix }ab

a b

ba,

0q 1q 2q

accept

ba,3q

ab

},{ baAlphabet =

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ML = { all strings without substring }001

0 00 001

1

0

1

10

0 1,0

}1,0{Alphabet =

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Regular Languages

All regular languages form a language family

LM MLL

A language is regular if there is

a DFA such that

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Example

*,: bawawaL

a

b

ba,

a

b

ba

0q 2q 3q

4q

is regular

The language

},{ baAlphabet =

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Nondeterministic Automata NFA

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• Nondeterministic there is an element of choice: in a given

state NFA can act on a given string in different ways. Several start/final states are allowed. -transitions are allowed.

• Finite only a finite number of states and arcs • Acceptors produce only a yes/no answer

NFA

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1q 2q

3q

a

a

a

0q

Two choices

}{aAlphabet =

Nondeterministic Finite Accepter (NFA)

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a a

0q

1q 2q

3q

a

a

First Choice

a

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a a

0q

1q 2q

3q

a

a

a

First Choice

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a a

0q

1q 2q

3q

a

a

First Choice

a

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a a

0q

1q 2q

3q

a

a

a “accept”

First Choice

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a a

0q

1q 2q

3q

a

a

Second Choice

a

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a a

0q

1q 2qa

a

Second Choice

a

3q

50

a a

0q

1q 2qa

a

a

3q

Second Choice

No transition:

the automaton hangs

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a a

0q

1q 2qa

a

a

3q

Second Choice

“reject”

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Observation

An NFA accepts a string if

there is a computation of the NFA

that 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

2q 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

},{ baAlphabet =

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a b

0q 1q 2qa b

3q

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a b

0q 2qa b

3q1q

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a b

0q 1qa b

3q2q

65

a b

0q 1qa b

3q2q

“accept”

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a b

Another String

a b

0q a b

1q 2q 3q

},{ baAlphabet =

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a b a b

0q a b

1q 2q 3q

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a b a b

0q a b

1q 2q 3q

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a b a b

0q a b

1q 2q 3q

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a b a b

0q a b

1q 2q 3q

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a b a b

0q a b

1q 2q 3q

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a b a b

0q a b

1q 2q 3q

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a b a b

0q a b

1q 2q 3q

“accept”

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ab

ababababababL ...,,,

Language accepted

0q 1q 2qa b

3q

},{ baAlphabet =

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Another NFA Example

0q 1q 2q0

11,0

}1,0{Alphabet =

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*10

...,101010,1010,10,

L

0q 1q 2q0

11,0

Language accepted

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Formal Definition of NFA

FqQM ,,,, 0

:Q

::0q

:F

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

: Input alphabet, i.e. ba,

Transition function

Initial state

Final states

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10 1, qq

Transition Function

0

11,0

0q 1q 2q

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},{)0,( 201 qqq

0q

0

11,0

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

*

10 ,* qaq

0q

5q4q

3q2q1qaaa

b

(Utvidgad övergångsfunktion)

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540 ,,* qqaaq

0q

5q4q

3q2q1qaaa

b

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0320 ,,,* qqqabq

0q

5q4q

3q2q1qaaa

b

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Formally

wqq ij ,*

if and only if

there is a walk from to

with label iq jq

w

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The Language of an NFA

540 ,,* qqaaq

0q

5q4q

3q2q1qaaa

b

M

)(MLaa

50 ,qqF

},{ baAlphabet =

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0320 ,,,* qqqabq

0q

5q4q

3q2q1qaaa

b

MLab

50 ,qqF

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50 ,qqF

540 ,,* qqabaaq )(MLabaa

0q

5q4q

3q2q1qaaa

b

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10 ,* qabaq MLaba

0q

5q4q

3q2q1qaaa

b

50 ,qqF

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0q

5q4q

3q2q1qaaa

b

aaababaaML *

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Formally

The language accepted by NFA M

,...,, 321 wwwML

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

Fqk (final state)

where

and there is some (at least one)

is:

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MLw

0q kq

w

w

w

),(* 0 wq

Fqk

iq

jq

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NFA DFA Equivalence

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Equivalence of NFAs and DFAs

Accept the same languages?

YES!

NFAs DFAs ?

Same power?

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We will prove:

Languages

accepted

by NFAs

Languages

accepted

by DFAsNFAs and DFAs have the same

computation power!

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Languages

accepted

by NFAs

Languages

accepted

by DFAs

Step 1

Proof Every DFA is also an NFA

A language accepted by a DFA

is also accepted by an NFA

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Languages

accepted

by NFAs

Languages

accepted

by DFAs

Step 2

Proof Any NFA can be converted to an

equivalent DFA

A language accepted by an NFA

is also accepted by a DFA

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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

Step 1

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Procedure NFA to DFA 2. For every DFA’s state

Compute in the NFA

},...,,{ mji qqq

...

,,*

,,*

aq

aq

j

i

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

},...,,{ mji qqq

Add transition

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Example

NFA

DFA

a

b

a

0q 1q 2q

},{),(* 210 qqaq

0q 21,qqa

210 ,, qqaq

Step 2

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Repeat Step 2 for all letters in alphabet,

until

no more transitions can be added.

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Example

a

b

a

0q 1q 2q

NFA

DFA

0q 21,qqa

b

ab

ba,

Step 3

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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,

Step 4

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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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Languages

accepted

by NFAs

Languages

accepted

by DFAsWe have proven (proof by construction):

Regular Languages

END OF PROOF

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Nondeterministic

vs.

Deterministic Automata

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Formal Definition of NFA

FqQM ,,,, 0

:Q Set of states, i.e. 210 ,, qqq: Input alphabet, i.e. ba,: Transition function

:0q Initial state

:F Final (accepting) states

NFA is a mathematical model defined as a quintuple:

110

Deterministic Finite Automata

A deterministic finite automaton (DFA) is a special case of a nondeterministic finite automaton in which

1. no state has an -transition, i.e. a transition on input , and

2. for each state q and input symbol a, there is at most one edge labeled a leaving q.

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STATEINPUT SYMBOL

a b

0

1

2

{0, 1}

--

{0}

{2}

{3}

Transition table for the finite automaton above

A nondeterministic finite automaton

b

0start

1a

2b b

3

a

Example

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NFA accepting aa* + bb*

0start

1

a2

a

3

b4

b

Example

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NFA accepting (a+b)*abb

0start

1a

2b b

b

a a

a

b

3

a

Example

114

NFA recognizing three different patterns.

(a) NFA for a, abb, and a*b+.

(b) Combined NFA.

Example

4

1start a

2

3start a

65b b

7start b

8

ba

4

1

start

a2

3a

65b b

7b

8

ba

0

115

Ways to think of nondeterminism

• always make the correct guess• “backtracking” (systematically try all possibilities)

For a particular string, imagine a tree of possiblestate transitions:

q0

q3

q0

q4

q2

q1a

a

a

a

b

a

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Advantages of nondeterminism

• an NFA can be smaller, easier to construct and easier to understand than a DFA that accepts the same language

• useful for proving some theorems• good introduction to nondeterminism in more

powerful computational models, where nondeterminism plays an important role

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Space and time taken to recognize regular expressions:- NFA more compact but take time to backtrack all choices- DFA take place, but save time

AUTOMATON SPACE TIME

NFA

DFA

O(|r|)

O(2|r|)

O(|r||x|)

O(|x|)

Determinism vs. nondeterminism

(Where r is regular expression, and x is input string)

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Equivalent automata

Two finite automata M1 and M2 are equivalent if

L(M1) = L(M2)

that is, if they both accept the same language.

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Equivalence of NFAs and DFAs

To show that NFAs and DFAs accept the same class of languages, we show two things:

– Any language accepted by a DFA can also be accepted by some NFA (As DFA is a special case of NFA)

– Any language accepted by a NFA can also be accepted by some (corresponding, specially constructed) DFA

120

Proof strategy

To show that any language accepted by a NFA is also accepted by some DFA, we describe an algorithm that takes any NFA and converts it into a DFA that accepts the same language.

The algorithm is called the “subset construction algorithm”.

We can use mathematical induction (on the length of a string accepted by the automaton) to prove the DFA that is constructed accepts the same language as the NFA.

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Converting NFA to DFA Subset Construction

http://www.math.uu.se/~salling/Movies/SubsetConstruction.mov

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Subset construction

Given a NFA constructs a DFA that accepts the same language

The equivalent DFA simulates the NFA by keeping track of the possible states it could be in. Each state of the DFA is a subset of the set of states of the NFA -hence, the name of the algorithm.

If the NFA has n states, the DFA can have as many as 2n states, although it usually has many less.

123

Steps of subset construction

The initial state of the DFA is the set of all states the NFA can be in without reading any input.

For any state {qi,qj,…,qk} of the DFA and any input a, the next state of the DFA is the set of all states of the NFA that can result as next states if the NFA is in any of the states qi,qj,…,qk when it reads a. This includes states that can be reached by reading a, followed by any number of -moves. Use this rule to keep adding new states and transitions until it is no longer possible to do so.

The accepting states of the DFA are those states that contain an accepting state of the NFA.

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Example

Here is a NFA that we want to convert to an equivalent DFA.

a

b

b

01

2

b

a

a

b

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{0,1}

The start state of the DFA is the set of states the NFA can be in before reading any input. This includes the start state of the NFA and any states that can be reached by a -transition.

a

b

b

b

a

a

01

2

b

NFA

DFA

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{0,1}

a

b

{2}

For start state {0,1}, make transitions for each possible input, here a and b. Reading b from start {0,1}, we reach state {2}. Means from either {0}, or {1} we reach {2}.

a

b

b

b

a

a

01

2

b

NFADFA

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For state {2}, we create a transition for each possible input, a and b. From {2}, with b we are either back to {2} (loop) or we reach {1}- see the little framed original NFA. So from {2}, with b we end in state {1, 2}. Reading a leads us from {2} to {0} in the original NFA, which means state {0, 1} in the new DFA.

ba

a

b{0,1}{1,2}

{2}

a

b

b

b

a

a

01

2

b

NFADFA

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For state {1, 2}, we make again transition for eachpossible input, a and b. From {2} a leads us to {0}. From {1} with a we are back to {1}. So, we reach {0, 1} with a from {1,2}. With b we are back to {1,2}.

At this point, a transition is defined for every state-input pair.

ba

a

b{0,1}

{1,2}

{2}

b

a

DFA a

b

b

b

a

a

01

2

b

NFA

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The last step is to mark the final states of the DFA.As {1} was the accepting state in NFA, all states containing {1} in DFA will be accepting states: ({0, 1} and {1, 2}).

ba

a

b{0,1}

{1,2}

{2}

b

a

DFA a

b

b

b

a

a

01

2

b

NFA

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Subset Construction Algorithm

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Subset Construction

States of nondeterministic M´ will correspond to sets of states of deterministic M

Where q0 is start state of M, use {q0} as start state of M´.

Accepting states of M´ will be those state-sets containing at least one accepting state of M.

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Subset Construction (cont.)

For each state-set S and for each s in alphabet of M, we draw an arc labeled s from state S to that state-set consisting of all and only the s-successors of members of S.

Eliminate any state-set, as well as all arcs incident upon it, such that there is no path leading to it from {q0}.

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The power set of a finite set, Q, consists of 2|Q| elements

The DFA corresponding to a given NFA with Q states have a finite number of states, 2|Q|.

If |Q| = 0 then Q is the empty set, | P(Q)| = 1 = 20.

If |Q| = N and N 1, we construct subset of a given set so that for each element of the initial set there are two alternatives, either is the element member of a subset or not. So we have

2 · 2 · 2 · 2 · 2 · 2 · 2…. ·2 = 2N

N times

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From an NFA to a DFA

Subset Construction

Operation Description

- closure(s)

- closure(T)

Move(T,a)

Set of NFA states reachable from an NFA state s on -transitions along

Set of NFA states reachable from some NFA state s in T on -transitions along

Set of NFA states reachable from some NFA state set with a transition on input symbol a

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From an NFA to a DFA

Subset Construction

Initially, -closure (s0) is the only states in D and it is unmarked

while there is an unmarked state T in D do mark T; for each input symbol a do U:= e-closure(move(T,a)); if U is not in D then add U as an unmarked state to D Dtran[T,a]:=U; end(for) end(while)

CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 3 Mälardalen University

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Transcript of CD5560 FABER Formal Languages, Automata and Models of Computation Lecture 3 Mälardalen University