Equivalence of DFA and NFA

Equivalence of DFA and NFA • In this section to convert the DFA to NFA. • Theorem: Let ‘L’ be the set accepted by a NFA then there exists a DFA that accepts the same the language ‘L’.

Proof: let M={Q,∑,δ,q0,F) be a NFA accepting the language ‘L’ and to define the DFA M1={Q1,∑,δ1,q01,F1) • The final states of ‘M1’are equal to the subsets of sets of states of ‘M’. • Q1=2Q. • Q has a finite set of states. i.e Q={q1,q2…qn} • Q1 has finite set of states accepted by a DFA. • Q={q1,q2.q3….qi} • Q1=[q1,q2,q3….qi]

i.e equivalent to the single state of DFA corresponds to the set of states in NFA. • δ1= ([q1,q2…qi], a) = [p1,p2…pi] if and only if δ = ({q1,q2…qi}, a) = {p1,p2…pi} It is easy to show by an induction method on the length of the string ‘x’ δ1(q01,x) = [q1,q2…qi] if and only if δ = ({q1,q2…qi}, a) = {p1,p2…pi}

Basis step: The result id trivial for |x|=0 • Induction step: suppose length of the string should be length M or less. • δ1(q01,xa) = δ1(δ1(q01,x),a) • By inductive hypothesis of δ1 • δ1(q01,x)= [p1,p2…pi] if and only if δ(q0,x) = {p1,p2…pi}

δ1([p1,p2…pi], a) = [r1,r2…rk] • If and only if δ({p1,p2…pi},a) = {r1,r2….rk} • To complete the proof we have only to add that δ1(q01,x)is in final state F1. • Hence the theorem is proved. i.e L(M) =L(M1).

problem1 • Construct a DFA from an NFA

To construct a DFA M1={q1,∑,δ1,q01,F1) and accepting the language L(M1). • Q1=2Q. Q=2 • 22=4 • δ1( Ø,0) =Ø. • δ1( Ø,1) =Ø. • δ1( q0,0) =[q0q1]. • δ1( q0,1) =[q1]. • δ1( q1,0) =Ø. • δ1( q1,1) =[q0q1] • δ1( [q0q1],0) =δ{q0q1},0) =δ( q0,0) U δ( q1,0) ={q0q1} U Ø => [q0q1]

δ1( [q0q1],1) =δ{q0q1},1) =δ( q0,1) U δ( q1,1) ={q0q1} U {q0q1} => [q0q1] Set of final states {[q1],[q0q1]}

solution

Problem 2 • Construct a DFA from an NFA

To construct a DFA M1={q1,∑,δ1,q01,F1) and accepting the language L(M1). • Q1=2Q. Q=4 • 24=16. • δ1( Ø,0) =Ø. • δ1( Ø,1) =Ø. • δ1( p,0) =[pq]. • δ1( p,1) =[p]. • δ1( q,0) =[r]. • δ1( q,1) =[r] • δ1( r,0) = [s] • δ1( r,1) = Ø • δ1( s,0) = [s] • δ1( s,1) = [s]

δ1( [pq],0) =δ{pq},0) =δ( p,0) U δ( q,0) ={pq} U {r} => {pqr} => [pqr] • δ1( [pq],1) =δ{pq},1) =δ( p,1) U δ( q,1) ={p} U {r} => {pr} =>[pr] • δ1( [pr],0) = [pqs] • δ1( [pr],1) = [p] • δ1( [ps],0) = [pqs] • δ1( [ps],1) = [ps] • δ1( [qr],0) = [rs] • δ1( [qr],1) = [r] • δ1( [qs],0) = [rs]

δ1( [qs],1) = [rs] • δ1( [rs],0) = [s] • δ1( [rs],1) = [s] • δ1( [pqr],0) = [pqrs] • δ1( [pqr],1) = [pr] • δ1( [pqs],0) = [pqrs] • δ1( [pqs],1) = [prs] • δ1( [prs],0) = [pqs] • δ1( [prs],1) = [ps] • δ1( [qrs],0) = [rs] • δ1( [qrs],1) = [rs] • δ1( [pqrs],0) = [pqrs] • δ1( [pqrs],1) = [prs] • Set of finite states { [s],[ps],[rs],[pqs],[prs],[pqrs],[qs],[qrs]}

problem 3 • Construct a DFA from an NFA

problem 4 • Construct a DFA from an NFA

Finite Automata with є-closure • We shall introduce the another extension of FA. • NFA is allowed to make a transition with spontaneously without receiving or reading many input symbols.

Contd..

Example

Transition table

Finite Automata with є-closure

Example

Formal definition of NFA - є closure. • To represent the ∑with NFA A={Q,∑,δ,q0,F) where all components have a same interpretations as for a NFA. • A state in Q. • A members of Є U {є} either an input symbol is empty or a set of strings.

E-closure : Definition

Example: Find the є-closure for all the states of given figure

solution • ECLOSE(1) ={1,2,3,4,6} • ECLOSE(2) ={2,3,6} • ECLOSE(3) ={3,6} • ECLOSE(4) ={4} • ECLOSE(5) ={5} • ECLOSE(6) ={6} • ECLOSE(7) ={7}

Inductive definition on δ

Problem 1: consider the following epsilon with NFA and compute the empty closure of each states and find its equivalent to the DFA.

Let A be a NFA with є-closure which has A={Q,∑,δ,q0,F) • Q={p,q.r} • ∑ = {є,a,b,c} • To find ECLOSE for all states. • ECLOSE(p) = {p,q,r} • ECLOSE(q) = {q,r} • ECLOSE(r) = {r}

δ( p,a) =δ(δ( p,є),a) =δ(δ( p,q,r),a) =δ( p,a) U δ(q,a) U (δ(r,a) =p U Ø U Ø = p = e-closure(p) ={pqr} • δ( p,b) =δ(δ( p,є),b) =δ(δ( p,q,r),b) =δ( p,b) U δ(q,b) U (δ(r,b) = Ø U q U Ø = q = e-closure(q) ={qr}

δ( p,c) =δ(δ( p,є),c) =δ(δ( p,q,r),c) =δ( p,c) U δ(q,c) U (δ(r,c) = Ø U Ø U r = r = e-closure(r) ={r} • δ( q,a) =δ(δ( q,є),a) =δ(δ( q,r),a) =δ(q,a) U (δ(r,a) = Ø U Ø = Ø

δ( q,b) =δ(δ( q,є),b) =δ(δ( q,r),b) =δ(q,b) U (δ(r,b) = q U Ø = q = e-closure (q) ={ q,r} • δ( q,c) =δ(δ( q,є),c) =δ(δ( q,r),c) =δ(q,c) U (δ(r,c) = Ø U r = r = e-closure(r) = {r}

δ( r,a) =δ(δ( r,є),a) =δ(r,a) = Ø • δ( r,b) =δ(δ( r,є),b) =δ(r,b) = Ø • δ( r,c) =δ(δ( r,є),c) =δ(r,c) = r =e-closure(r) = {r}

Equivalence of DFA and NFA