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Outline

1 What is Non-Deterministic Finite Automaton(NFA)

Shaifur (CSE, BUET) CSE211: NFA Class 4 2 / 23

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Outline


2 Examples of NFA


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Outline


2 Examples of NFA

3 Equivalence of NFA and DFA


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What is NFA

Non-Deterministic Finite Automaton (NFA)For a single input symbol, transition from the current state to one ormore states

Transition from current state to one or more states withoutconsuming any input symbol (in other word, transition consuming )


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NFA Example

NFA N1 has four states


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NFA Example


For input 1, q1 q1 OR q1 q2

State q2 has only transition for input 0 BUTnot for 1


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NFA Example


For input 1, q1 q1 OR q1 q2

State q2 has only transition for input 0 BUTnot for 1N1 can transit q2 q3 without any input!


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Examples of NFA (Contd.)

NFA Decision Tree


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NFA Decision Tree for N1 for computation of010110


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NFA N1

Equivalent DFA D1


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An NFA which all strings of the form 0

k

wherek

is a multiple of 2 or 3


( )

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An NFA which all strings of the form , a, baba, baa, . . .But does not accept b, bb, babba etc.


F l D fi i f NFA

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Formal Defintion of NFA

NFA

a 5-tuple (Q,, ,q 0, F) where:

1 Q is a finite set of states

2 is a finite alphabet

3 :Q P(Q)

4 q0Q is the start-state

5 F Q is the accept-state set


E l f NFA (C d )

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Give the formal definition of for the following NFA:


E l f NFA (C d )

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Give the formal definition of for the following NFA:

Q= {q1, q2, q3, q4}

={0, 1}start-state,q0=q1

accept-states, F ={q4}

is:0 1

q1 {q1} {q1, q2}

q2 {q3} {q3}q3 {q4} q4 {q4} {q4}


C t ti b NFA

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Computation by NFA

Formal Definition of Computation by NFA

N= (Q,, ,q 0, F) is an NFA, w is a string ofN accepts w ifw=y1 y2 . . . ym where each yi and sequence of

states r0, r1, . . . , rm each exists in Q such that:

1 r0=q02 ri+1(ri, yi+1) form= 0, 1, . . . ,m 1 and

3 rm F


E i l f NFA d DFA

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Equivalence of NFA and DFA

Equivalence of Machines

Two machines are equivalent if they recognize the same languages

Theorem 1.39

Every NFA has an equivalent DFA


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NFA DFA (Contd )

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NFA=DFA (Contd.)

NFA has an equivalent DFA . . . proof contd.

Let N= (Q,, ,q 0, F) be the NFAand M= (Q,, , q0, F

) be the equivalent DFA. Both N and Mrecognizes the same language A.

Case I: There is no arrow in the NFA N.1 Q =P(Q), because every state ofM is aset of statesofN

2 q0={q0}

3 F ={RQ| R contains an accept-state ofN}

4 For RQ and aQ

(R, a) ={qQ| q(r, a) for some rR}(R, a) =

rR

(r, a) set of all states reachable from all states ofRfor input a according to the ofN


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NFA=DFA (Contd )

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NFA=DFA (Contd.)

Corollary 1.40A language is regularif and only ifsome non-deterministic finiteautomaton recognizes it

ProofA language is regularifsome NFA recognizes itIf some NFA recognizes the language, so does an equivalent DFA.Hence, the language is regular.

A language is regularonly ifsome NFA recognizes itIf a language is regular, a DFA recognizes it. A DFA is also an NFA.So, an NFA also recognizes the regular language.


Examples of Conversion from NFA to DFA

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Examples of Conversion from NFA to DFA

DFA= (Q,, ,q 0, F)

Q= {, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}, }

={a, b}

q0=E({1}) ={1, 3}

F ={ All states ofQ that have a final state ofN}={{1}, {1, 2}, {1, 3}, {1, 2, 3}}


Examples of Conversion from NFA to DFA (Contd )

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Examples of Conversion from NFA to DFA (Contd.)

Transition Function =?

On input a: {2} {2, 3}, on input b: {2} {3}

On input a: {1} , on input b: {1} {2}

On input a: {3} {1, 3}, on input b: {3}

On input a: {1, 2} {2, 3}, on input b: {1, 2} {2, 3}

Similarly, for rest of the states . . .Shaifur (CSE, BUET) CSE211: NFA Class 4 18 / 23

Correction to class lecture

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Correction to class lecture

attention: follow after applying (), not before!

({1}, a) = INCORRECT!, (E({1}), a) =({1, 3}, a) ={1}But, ({3}, a) =E({1}) ={1, 3}



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After conversion . . .



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After conversion and simplification . . .

Dropping the states from which there is no outgoing arrow


Try Yourself

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y

Convert the following NFA into the equivalent DFA


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Question?


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