Equivalence, DFA, NDFAEquivalence, DFA, NDFA

Sequential Machine Theory

Prof. K. J. HintzDepartment of Electrical and Computer

Engineering

Lecture 2

Updated and modified by Marek Perkowski

Equivalence Relation on AEquivalence Relation on AEquivalence Relation on AEquivalence Relation on A

• An Equivalence Relation (Not Relationship) Is Not an Equality Relation

• A Relation is an Equivalence Relation if and only if (iff) it is:– Reflexive– Symmetric– Transitive

Equivalence Relation on AEquivalence Relation on A

A

A

A

A

on ~ relation eequivalenc an is

e)(transitiv '',','','',' ',

)(symmetric ',,' ',

)(reflexive ', ,

R

RRR,

RR

R

then

aaaaaaaaa

and

aaaaaa

and

aaaa

if

Non-Algebraic Equivalence Relation Example

Non-Algebraic Equivalence Relation Example

Equivalence Relation on the Set of All Triangles on a Plane

“is congruent to” or “is similar to”

– Reflexive, each triangle is similar to itself,

Equivalence Relation ExampleEquivalence Relation Example

Symmetric, if

is similar to

then

is similar to

Equivalence Relation ExampleEquivalence Relation Example

Transitive, if

is similar to

and

is similar to

then

is similar to

Inclusion RelationInclusion Relation

ji

jiji

jjjj

iiii

ji

iff

then

, also are states all

)" in included is " as (read

,,,

,,,

, states, ofset theon defined partitions 2 Given

321

321

BBB

BBB

BBB

S

Inclusion Relation ExampleInclusion Relation Example

, in" included is"

,

5,4,3,2,1

5,4,3,2,1

5,4,3,2,1

21

21

2

14131211

1

or,

then

and

let

BBBB

Illustrate a bigger lattice graphically

Partition NotationPartition Notation

• Overbar Indicates States Which Are Elements of the Same -block.

• Single States Are Not Normally Listed

1

11 12 13 14

1 2

1 2 3 4 5

1 2 3 4 5

1 2 1 2 3 4 5

, , ,

,

,

B B B B

is written as

and

Relations May Be OrderingsRelations May Be Orderings

• Partial Ordering

• Total Ordering, aka Chain

• Well Ordering (not discussed here)

Partial Ordering (PO)Partial Ordering (PO)Partial Ordering (PO)Partial Ordering (PO)

• Given an Inclusion Relation, R: s s’, Defined on some Elements of the Set S such that s, s’ S, R Is a Partial Ordering If It Is:– Reflexive– Anti-Symmetric (asymmetric)– Transitive– Not all orderings are specified, therefore partial

Illustrate on a lattice

Properties of Partial OrderingsProperties of Partial Orderings

– Reflexive• s s for all s S

– Anti-Symmetric (asymmetric)If

and

then

s s

s s

s s s s

'

, S

e.g., let : “older than”

if Sam is older than Bill,

then Bill cannot be older

than Sam

But if this symbol means older or the same age, then OK

Properties of POProperties of PO

– TransitiveIf

and

then

s s

s s

s s

'

e.g., If the Redskins beat the Patriots

and the Patriots beat the Cowboys

then the Redskins will beat the Cowboys

Total OrderingTotal Ordering

• aka– Chain, simply ordered set, totally ordered set

• A Partial Ordering for Which All Orderings Are Specified

• A Chain Is “Connected” Because

s s s s s s or , S

POSETPOSET

• Partially Ordered SET– A set on which a partial ordering is specified– ( S, ) where is defined– Not a chain since not all elements are

connected

• We Will Revisit This Concept in a later part of the Course

Finite AutomataFinite Automata

A Deterministic semi-automaton*, aka Completely Specified Deterministic Semi-automaton is a Triple

with no Mealy machine output function, Beta ()

* Ginzburg, 1968

,,

or ,,,

ΣK

IS

M

M

FSM Set PropertiesFSM Set Properties

,,,,,,,,,

,

inputs ofset finite A,,,,

states ofset finite A,,,,

1210

1210

ihgfedcba

m

n

sississis

S

iiiiI

ssss

ISSIS

KS

SIsa sc

ib

Language RecognizerLanguage Recognizer

• aka, Rabin-Scott Automata (machine), Automaton, Language Recognizer

• A Recognizer Is a Quintuple of Sets

with S, I, as before

M rs S I F, , , , s0

s

s sp q

0

the single, unique, initial state

A finite set of final statesF S, ,

Kleene StarKleene Star

• a* = e, a, aa, aaa, aaaa, ...

• The Kleene Star, *, means NONE or more occurrences of something

• Star is an overloaded operator so be aware of context

• a+= ONE or more occurrences of something.

• a+ is Kleene Star less the null string, .

Kleene ClosureKleene Closure

• Kleene Closure Is Not Identical to Kleene Star– “*” Symbol is the same (overloaded)

• Kleene Closure/Star Closure– Found in descriptions of formal language– Language consisting of all strings over some

alphabet

StringString

• An Ordered Concatenation of Symbols From an Alphabet

• Used in Place of “Word” to Decouple From Common Concept of Word in Informal Language

• If = { a, 1, 0, b, % } then a “1%0b” is a string.

RecognizerRecognizer

If x I*, i.e., a string of input symbols selected from the set of allowable input symbols,

and the application of x to the recognizer results in a final state F,

then the recognizer “accepts” the string.

StringsStrings

A String, x, Is Accepted by a Recognizer Left-most Letter First, i.e.,

if the input to a recognizer is a string w,

and if w = w’

then is the first letter of the string which causes a state

transition. Subsequent letters from left to right do the same.

State TransitionState Transition

Let There Be Two Configurations for a Machine

somefor

',',

relation, by the related are which',' and ,

1 ww

iff

wqwq

wqwqM

R,

1

q q’S

String ExampleString Example

Let

w = a b b a

then

w = a w’

and

w’ = b b a

Recognizer as Directed GraphRecognizer as Directed GraphRecognizer as Directed GraphRecognizer as Directed Graph

• Arbitrary State

• State Transition

• Start (initial) State

• Final State

1

q

q q’

-

+ or

or

Let I = { a, b }

• Accepts no strings since no final state

• Accepts all strings

• Dead State

Recognizer ExamplesRecognizer Examples

a, b

-

a, b

a, b

Recognizer ExamplesRecognizer Examples

• Accepts only , the null string

a,b

a,b

+/-

This is start and final state

Recognizer ExampleRecognizer Example

This Recognizer Accepts the Language

L= { ab, a (aa) b, a (aa) (aa) b, ...

ab (bb), ab (bb) (bb), ... }

L = a (aa)* b ( b (aa)* b )*

- a

a

b

b

New Example: Rabin-Scott MachineNew Example: Rabin-Scott Machine

S

F

1 2 3 4

1

40

, , ,

, , ;I a

s

M rs S I F, , , , s0

ps\input a + ;1 2 3 32 3 1 43 3 3 34 3 3 3

4

Now we transform this table to a graph

The same Rabin-Scott machine as a graph

The same Rabin-Scott machine as a graph

L (M) = { x I* |* ( 1, x ) = 4 }

L (M) = { a; , a+a; , a+a+a; , ... }

aa

2

3 4

+

+ ;

;

a + ;

a + ;

1

Non-Deterministic FSMNon-Deterministic FSM

A Non-deterministic Finite Automata Is a Quintuple with S, I, s0, F

as in a recognizer, but, M nd S I F, , , , s0

,,,,,,

,,,,,,

functiona not relation,a is

,,* *

jhged

fedcba

sssxs

sxssxs

SISSIS

Non-Deterministic FSMNon-Deterministic FSM

• State May Change– to two (or any) different states in response to

the same input at the same state– in response to a string rather than just a single

element from the set of inputs– in response to a null string input

DFA-NDFA TheoremDFA-NDFA Theorem

• Every NDFA Can Be Replaced by an Equivalent DFA

• Equivalent Means Not Only Accepting All Strings Accepted by the NDFA, but Also NOT Accepting Any Other strings

NDFA ExampleNDFA Example

Non-deterministic Since

( ( 1, a ), 2 ) and ( ( 1, a ), 3 )

1

ab

2

3 4a

b

1

NDFA ExampleNDFA Example

Non-deterministic Since Not Completely Specified

ab

4

1

abb

L = ab abb,

NDFA ExampleNDFA Example

Non-deterministic Since State Changes in Response to a Null String.

ab

2

3 4a

1

bb

NDFA to DFANDFA to DFANDFA to DFANDFA to DFA• Theorem

– For each NDFA there is an equivalent DFA

• Constructive Proof• 4 Difficulties to Resolve

– Missing transitions– Single transitions due to | strings | > 1– Transitions due to strings– Multiple transitions

Problem: Missing TransitionsProblem: Missing Transitions

• I = { a, b }

• In DFA, all i I must be accounted for in each state

a

b ?

Solution: Missing TransitionsSolution: Missing Transitions

Add a “sink” state which is not a final state and terminate all missing transitions there.

a

b

a, b

a, b

Problem: | strings | > 1Problem: | strings | > 1

• Single transition due to string of size > 1

• Add intermediate states and “sink”, other characters in those states go to “sink” state

a

ab

a, b

b

a

Problem: StringsProblem: Strings

Can’t just combine states sincea

ab

a bb

a

b

Solution: Strings & Multiple Transitions

Solution: Strings & Multiple Transitions

• Eliminate by defining the set of next states which occur in response to no input, call this function E( )

• E( ) is called the “equivalents of ( )

NDFA ExampleNDFA Example

> 1

2

3

4

a b

b a

b

a

State EquivalentsState Equivalents

E( 1 ) = {self, explicit alternative} = { 1, 3 }

E( 2 ) = { 2 }

E( 3 ) = { 3 }

E( 4 ) = { 4 }

• Define a new machine based on the old using the E( ) states

> 1

2

3

4

a b

b a

b

a

New MachineNew Machine

E a

E b

E a

E b

E a

E b

E a

E b E E b

1 2 3 4

1 5

2 5

2 3

3 3 4

3 5

4 5

4 1 4 4 1 3 4

, , , ,

, , ,

, ,

, ,

, , ,

, ,

, ,

, , , , , , ,

state 5 is new dead state

> 1

2

3

4

a b

b a

b

a

New Machine Transition TableNew Machine Transition Table

Equivalent

OldState

NewState

Inputa

Inputb

E ( 1 ) { 1, 3 } A { 2,3,4 } = F 5 = EE ( 2 ) 2 B 5 = E 3 = CE ( 3 ) 3 C { 3, 4 } = G 5 = EE ( 4 ) 4 D 5 = E { E( 1 ), 4 }

{ 1, 3, 4 } = H

> 1

2

3

4

a b

b a

b

a

New Machine Transition TableNew Machine Transition Table

Equivalent

Old State

New State

Input a

Input b

E ( 5 ) 5 E 5 = E 5 = E 2, 3, 4 F { 5, 3, 4 } = I { 3, 5, E(1), 4}

{ 1, 3, 4, 5} = J 3, 4 G { 3, 4, 5 } = I { 5, E(1), 4}

{ 1, 3, 4, 5} = J

1, 3, 4 H 2, 3, 4, 5 = L { 5, E(1), 4} { 1, 3, 4, 5} = J

> 1

2

3

4

a b

b a

b

a

New Machine Transition TableNew Machine Transition TableOldState

NewState

Inputa

Inputb

3, 4, 5 I { 3, 4, 5 } = I { 5, E(1), 4}{ 1, 3, 4, 5} = J

1, 3, 4, 5 J { 2, 3, 4, 5 }= L { 5, E(1), 4}{ 1, 3, 4, 5} = J

2, 3, 4, 5 L { 5, 3, 4 } = I { 3, 5, 4, E(1) }{ 1, 3, 4, 5} = J

> 1

2

3

4

a b

b a

b

a

DFA Equivalent of NDFADFA Equivalent of NDFA

> 1

2

3

4

a b

b a

b

a

Reduced DFA EquivalentReduced DFA Equivalent

Comparison of both machinesComparison of both machines

Discuss other methods of doing this transformation

HomeworkHomeworkHomeworkHomework• This is only homework example. Another homework may

be assigned.• 1. Describe any problem from real life, possibly related to

your previous research, experience from other classes or otherwise as a non-deterministic finite automaton NDA. You may use Kleene star and empty symbols, if necessary.

• 2. Convert it to a deterministic automaton DA.• 3. Try to minimize DA to Minimized MDA.• 4. Realize MDA using JK Flip-Flops and logic gates. • 5. Analyze its behavior using the logical diagram. Compare

to the initial description. Draw and write conclusions.