Admission in india 2015

Admission in India 2015Admission in India 2015By:

admission.edhole.com

Finite Automata andFinite Automata andNon DeterminismNon Determinism

http://cis.k.hosei.ac.jp/~yukita/


Definition 1.1: Finite AutomatonDefinition 1.1: Finite Automaton

. theis 5.

and , theis 4.

, theis : 3.

, thecalledset finite a is 2.


where),,,,,( tupple-5 a is A

0

0

states accept of set

state start

function transition

alphabet

states

automatonfinite

QF

Qq

QQ

Q

FqQ

3admission.edhole.com

State Diagram for State Diagram for MM11

4

q3q1 q2

0 1

10

0, 1


Data Representation for MData Representation for M11

.}{ 5.

and state,start theis 4.

,

10

as described is 3.

{0,1} 2.

},,{ 1.

2

1

223

232

211

321

qF

q

qqq

qqq

qqq

qqqQ


Task 01Task 01DFADFA

1. Implement M1 with your favorite programming language.

2. GUI• Two buttons for input 0 and 1• State chart with the current state

highlighted


Language of MLanguage of M11

}1last thefollow 0s ofnumber even an

and 1 oneleast at contains |{)(

. say that can We

accepts. machine that strings all ofset theis

),( as written , The

1 wwML

AM

M

MLM

recognizes

machine oflanguage


State Diagram for MState Diagram for M55

8

q2

q1

q0

0

1

12

2, <RESET>

1, <RESET>

2

0, <RESET> 0


Data Representation for MData Representation for M55

.}{ 5.

and state,start theis 4.

,

210

as described is 3.

,0,1,2}RESET{ 2.

},,{ 1.

0

0

10202

02101

21000

210

qF

q

qqqqq

qqqqq

qqqqq

R

qqqQ


Informal Description of Informal Description of MM55

M5 keeps a running count of the sum of the numerical symbols it reads, modulo 3.

Every time it receives the <RESET> symbol it resets the count to 0.

M5 accepts if the sum is 0, modulo 3.


Definition 1.7: Regular Definition 1.7: Regular LanguageLanguage

A language is called a regular language if some finite automaton recognizes it.


Example 1.9: A finite automaton Example 1.9: A finite automaton EE22

E2 recognizes the regular language of all strings that contain the string 001 as a substring.

0010, 1001, 001, and 1111110011110 are all accepted,

but 11 and 0000 are not.


Find a set of states of Find a set of states of EE22

You1. haven’t just seen any symbols of the

pattern,2. have just seen a 0,3. have just seen 00 or,4. have just seen the entire pattern 001.Assign the states q,q0,q00, and q001 to these

possibilities.


Draw a State Diagram for Draw a State Diagram for EE22

14

q00 q001

0 0, 1

1

1

0q0

0

q

1


Regular Operations on Regular Operations on LanguagesLanguages

operation. a onelast theand

,operations are operations first two The

}.each and 0|{ :

}. and |{ :

}.or |{ :

languages. be and Let

21*

unary

binary

AxkxxxA

ByAxxyBA

BxAxxBA

BA

ik

Star

ionConcatenat

Union


Example 1.11Example 1.11

}. ,goodbadbad

d,goodbadgoo goodbad, good od,goodgoodgo

badbad, goodbad, goodgood, bad, good, ,{

and },badgirl badboy, goodgirl, goodboy,{

},girl boy, bad, good,{

have weThen, }.girl boy,{ and }bad good,{Let

}.z,,ba,{ be alphabet Let the

*

A

BA

BA

BA


Theorem 1.12 Closedness for Theorem 1.12 Closedness for UnionUnion

. is so

languages,regular are and if s,other wordIn

operation.union the

under closed is languagesregular of class The

21

21

AA

AA


Proof of Theorem 1.12Proof of Theorem 1.12

on.constructi theof scorrectnes check the Then,

.),,,,( where

, recognize toConstruct

.),,,,( where, recognize and

),,,,,( where, recognize Let

21

2222222

1111111

FqQM

AAM

FqQMAM

FqQMAM

Proof of Th 1.12 18admission.edhole.com

Construction of Construction of MM

}.or |),{()()( 5.

).,( 4.

)),(),,(()),,((;,),( 3.

(Why?) .alphabet

same thehave and that assumecan We2.

}. and |),{( 1.

2211212121

210

22112121

21

22112121

FrFrrrFQQFF

qqq

arararraQrr

MM

QrQrrrQQQ


CorrectnessCorrectness

You should check the following.1. For any string recognized by M1 is

recognized by M.2. For any string recognized by M2 is

recognized by M.3. For any string recognized by M is

recognized by M1 or M2.


Theorem 1.13 Closedness for Theorem 1.13 Closedness for concatenationconcatenation

. is so

languages,regular are and if s,other wordIn operation.

ionconcatenatunder closed is languagesregular of class The

21

21

AA

AA


NondeterminismNondeterminism

To prove Theorem 1.13, we need nondeterminism.

Nondeterminism is a generalization of determinism. So, every deterministic automaton is automatically a nondeterministic automaton.


Nondetermistic Finite AutomataNondetermistic Finite Automata

A nondeterministic finite automaton can be different from a deterministic one in that◦for any input symbol, nondeterministic one can

transit to more than one states.◦epsilon transition

NFA and DFA stand for nondeterministic finite automaton and deterministic finite automaton, respectively.


NFA NFA NN11

24

q3q1 q2

0,1

1 0, q4

0,1

1


Parallel world and NFAParallel world and NFA25

... ...accept


Example 1.14 NFA Example 1.14 NFA NN22

26

q3q1 q2

0,1

1 0, q4

0,1

Let language A consist of all strings over {0,1} containing a 1 in the third position from the end. N2 recognizes A.


A DFA equivalent to A DFA equivalent to NN22

27

q010q000 q100

0

0 0 q110

q011q001 q101

q111

1

0

1

00

10

11

01



28

0

Let language A consist of all strings 0k , where k is a multiple of 2 or 3. N3 recognizes A.

0

0

0

0


A DFA equivalent to A DFA equivalent to NN33

29

0q ２q １ q3

q50

q0

0 0 0 q4

0

q-1

11

1 1 1

1

0, 1



30

q1

q2q3

a

b

a,b

a

N4 accepts , a, baba, and baa. N4 does not accept b, nor babba.


Definition 1.17: NFADefinition 1.17: NFA

. theis 5.

and , theis 4.

, theis 2: 3.



where),,,,,( tupple-5 a is A

0

0

states accept of set

state start

function transition

alphabet

states

automatonfinite nistic nondetermi

QF

Qq

Q

Q

FqQ

Q


Example 1.18 NFA Example 1.18 NFA NN11}.{ 5.

state.start theis 4.

}{}{

}{

}{}{

},{}{

10

asgiven is 3.

}1,0{ 2.

},,,,{ 1.

4

1

444

43

332

2111

4321

qF

q

qqq

qq

qqq

qqqq

qqqqQ

32

q3q1 q2

0,1

1 0,q4

0,1

1


In what situation is Non In what situation is Non Determinism relevant?Determinism relevant?

Von Neumann machines are deterministic.However, there are many cases where

machine specification is all we need.


Theorem 1.19Theorem 1.19

Every nondeterministic finite automaton has an equivalent deterministic finite automaton.◦Def. The two machines are equivalent is they

recognize the same language.


Proof of Th. 1.19Proof of Th. 1.19

}.|{ 4.

and },{ 3.

,),(),( 2.

,2 1.

such that ),,,,(Construct

arrows. no has first that assume usLet

. language some grecognizinNFA thebe ),,,,(Let

00

0

0

FRQRF

qq

araR

Q

FqQM

N

AFqQN

Rr

Q


Incorporate Incorporate arrows arrows

0

0 0

( ) { | can be reached from by traveling along zero or

more arrows.}

We modify as follows.

( , ) { | ( ( , )) for some }.

We modify as follows.

({ }).

We omit the correctness p

E R q q R

R a q Q q E r a r R

q

q E q

roof.


Corollary 1.20Corollary 1.20

A language is regular if and only if some nondeterministic finite automaton recognizes it.


Example 1.21 NFA Example 1.21 NFA NN44 to DFA to DFA

}}.3,2,1{},3,2{},3,1{},2,1{},3{},2{},1{,{2

as takesbemay set state s' The .DFA equivalentan

construct want to we,}}3{,1,},,{},3,2,1{{Given

}3,2,1{

4

DD

baN

38

1

2 3a

b

a,b

a


Task 02Task 02Parallel WorldParallel World

1. Write a program that simulates N4. 2. GUI

◦ Three buttons for input 0, 1, and epsion.◦ State chart that reflect the branching of the

world.


Start and Accept statesStart and Accept states

follows. asgiven is diagram state The

}}.3,2,1{},3,2{},3,1{},3{{ is statesaccept ofset The

}.3,1{})1({ is statestart The E


The state diagram of The state diagram of DD41

{1} {2} {1,2}

{3} {1,3} {2,3} {1,2,3}

a,b

a,b

a

aa

a

a

a

b

b bb

b

b


Theorem 1.22Theorem 1.22 The class of regular The class of regular languages is closed under the union languages is closed under the union operation.operation.

42

N1

N2

N



. and for

and for },{

for ),(

for ),(

),( 4.

3.

of statestart theis state The 2.

}{ 1.

. recognize to),,,,{ Construct

. recognize ),,,,(

and , recognize ),,,,(Let

0

021

22

11

21

0

210

210

222222

111111

aqq

aqqqq

Qqaq

Qqaq

aq

FFF

N.q

QQqQ

AAFqQN

AFqQN

AFqQN


Theorem 1.23Theorem 1.23 The class of regular The class of regular languages is closed under the languages is closed under the concatenation operation.concatenation operation.

44

N1N2

N



.for ),(

and for }{),(

and for ),(

and for ),(

),( 4.

of that as same theis statesaccept ofset The 3.

. of that as same theis statestart The 2.

1.


. recognize ),,,,(

and , recognize ),,,,(Let

22

121

11

111

2

11

21

2121

222222

111111

Qqaq

aFqqaq

aFqaq

FqQqaq

aq

N

Nq

QQQ

AAFqQN

AFqQN

AFqQN


Theorem 1.24Theorem 1.24 The class of regular The class of regular languages is closed under the star languages is closed under the star operation.operation.

46

N1

N



. and for

and for }{

and for }{),(

and for ),(

and for ),(

),( 4.

}{ 3.

state.start new theis statestart The 2.

}{ 1.


. recognize ),,,,(Let

0

01

111

11

111

10

0

10

*10

111111

aqq

aqqq

aFqqaq

aFqaq

FqQqaq

aq

FqF

q

QqQ

AFqQN

AFqQN


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