Equivalence of DFA and NFA n In this
![Equivalence of DFA and NFA n In this section to convert the DFA to Equivalence of DFA and NFA n In this section to convert the DFA to](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-1.jpg)
![n n n n Proof: let M={Q, ∑, δ, q 0, F) be a n n n n Proof: let M={Q, ∑, δ, q 0, F) be a](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-2.jpg)
![i. e equivalent to the single state of DFA corresponds to the set of i. e equivalent to the single state of DFA corresponds to the set of](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-3.jpg)
![n n n Basis step: The result id trivial for |x|=0 Induction step: suppose n n n Basis step: The result id trivial for |x|=0 Induction step: suppose](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-4.jpg)
![n n δ 1([p 1, p 2…pi], a) = [r 1, r 2…rk] If n n δ 1([p 1, p 2…pi], a) = [r 1, r 2…rk] If](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-5.jpg)
![problem 1 n Construct a DFA from an NFA 0 q 0 *q 1 problem 1 n Construct a DFA from an NFA 0 q 0 *q 1](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-6.jpg)
![n n n n n To construct a DFA M 1={q 1, ∑, δ n n n n n To construct a DFA M 1={q 1, ∑, δ](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-7.jpg)
![n δ 1( [q 0 q 1], 1) =δ{q 0 q 1}, 1) =δ( n δ 1( [q 0 q 1], 1) =δ{q 0 q 1}, 1) =δ(](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-8.jpg)
![solution 0 Ø [q 0] [q 1] [q 0 q 1] 1 Ø [q solution 0 Ø [q 0] [q 1] [q 0 q 1] 1 Ø [q](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-9.jpg)
![Problem 2 n Construct a DFA from an NFA 0 1 p {p, q} Problem 2 n Construct a DFA from an NFA 0 1 p {p, q}](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-10.jpg)
![n n n n To construct a DFA M 1={q 1, ∑, δ 1, n n n n To construct a DFA M 1={q 1, ∑, δ 1,](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-11.jpg)
![](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-12.jpg)
![n n n n δ 1( [qs], 1) = [rs] δ 1( [rs], 0) n n n n δ 1( [qs], 1) = [rs] δ 1( [rs], 0)](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-13.jpg)
![problem 3 n Construct a DFA from an NFA 0 1 p {q, s} problem 3 n Construct a DFA from an NFA 0 1 p {q, s}](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-14.jpg)
![problem 4 n Construct a DFA from an NFA 0 1 p {p, q} problem 4 n Construct a DFA from an NFA 0 1 p {p, q}](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-15.jpg)
![Finite Automata with є-closure n n We shall introduce the another extension of FA. Finite Automata with є-closure n n We shall introduce the another extension of FA.](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-16.jpg)
![Contd. . Contd. .](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-17.jpg)
![Example Example](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-18.jpg)
![Transition table Transition table](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-19.jpg)
![Finite Automata with є-closure Finite Automata with є-closure](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-20.jpg)
![Example Example](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-21.jpg)
![Formal definition of NFA - є closure. n To represent the ∑with NFA A={Q, Formal definition of NFA - є closure. n To represent the ∑with NFA A={Q,](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-22.jpg)
![E-closure : Definition E-closure : Definition](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-23.jpg)
![Example: Find the є-closure for all the states of given figure Example: Find the є-closure for all the states of given figure](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-24.jpg)
![solution n n n ECLOSE(1) ={1, 2, 3, 4, 6} ECLOSE(2) ={2, 3, 6} solution n n n ECLOSE(1) ={1, 2, 3, 4, 6} ECLOSE(2) ={2, 3, 6}](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-25.jpg)
![Inductive definition on δ Inductive definition on δ](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-26.jpg)
![Problem 1: consider the following epsilon with NFA and compute the empty closure of Problem 1: consider the following epsilon with NFA and compute the empty closure of](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-27.jpg)
![n n n n Let A be a NFA with є-closure which has A={Q, n n n n Let A be a NFA with є-closure which has A={Q,](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-28.jpg)
![](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-29.jpg)
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![](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-33.jpg)
![Problem 2: consider the following epsilon with NFA and compute the empty closure of Problem 2: consider the following epsilon with NFA and compute the empty closure of](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-34.jpg)
![Problem 3: consider the following epsilon with NFA and compute the empty closure of Problem 3: consider the following epsilon with NFA and compute the empty closure of](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-35.jpg)
- Slides: 35
![Equivalence of DFA and NFA n In this section to convert the DFA to Equivalence of DFA and NFA n In this section to convert the DFA to](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-1.jpg)
Equivalence of DFA and NFA n In this section to convert the DFA to NFA. n Theorem: Let ‘L’ be the set accepted by a NFA then there exists a DFA that accepts the same the language ‘L’.
![n n n n Proof let MQ δ q 0 F be a n n n n Proof: let M={Q, ∑, δ, q 0, F) be a](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-2.jpg)
n n n n Proof: let M={Q, ∑, δ, q 0, F) be a NFA accepting the language ‘L’ and to define the DFA M 1={Q 1, ∑, δ 1, q 01, F 1) The final states of ‘M 1’are equal to the subsets of states of ‘M’. Q 1=2 Q. Q has a finite set of states. i. e Q={q 1, q 2…qn} Q 1 has finite set of states accepted by a DFA. Q={q 1, q 2. q 3…. qi} Q 1=[q 1, q 2, q 3…. qi]
![i e equivalent to the single state of DFA corresponds to the set of i. e equivalent to the single state of DFA corresponds to the set of](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-3.jpg)
i. e equivalent to the single state of DFA corresponds to the set of states in NFA. n δ 1= ([q 1, q 2…qi], a) = [p 1, p 2…pi] if and only if δ = ({q 1, q 2…qi}, a) = {p 1, p 2…pi} It is easy to show by an induction method on the length of the string ‘x’ δ 1(q 01, x) = [q 1, q 2…qi] if and only if δ = ({q 1, q 2…qi}, a) = {p 1, p 2…pi} n
![n n n Basis step The result id trivial for x0 Induction step suppose n n n Basis step: The result id trivial for |x|=0 Induction step: suppose](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-4.jpg)
n n n Basis step: The result id trivial for |x|=0 Induction step: suppose length of the string should be length M or less. δ 1(q 01, xa) = δ 1(q 01, x), a) By inductive hypothesis of δ 1(q 01, x) = [p 1, p 2…pi] if and only if δ(q 0, x) = {p 1, p 2…pi}
![n n δ 1p 1 p 2pi a r 1 r 2rk If n n δ 1([p 1, p 2…pi], a) = [r 1, r 2…rk] If](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-5.jpg)
n n δ 1([p 1, p 2…pi], a) = [r 1, r 2…rk] If and only if δ({p 1, p 2…pi}, a) = {r 1, r 2…. rk} To complete the proof we have only to add that δ 1(q 01, x) is in final state F 1. Hence theorem is proved. i. e L(M) =L(M 1).
![problem 1 n Construct a DFA from an NFA 0 q 0 q 1 problem 1 n Construct a DFA from an NFA 0 q 0 *q 1](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-6.jpg)
problem 1 n Construct a DFA from an NFA 0 q 0 *q 1 1 {q 0, q 1} {q 1} Ø {q 0, q 1}
![n n n n n To construct a DFA M 1q 1 δ n n n n n To construct a DFA M 1={q 1, ∑, δ](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-7.jpg)
n n n n n To construct a DFA M 1={q 1, ∑, δ 1, q 01, F 1) and accepting the language L(M 1). Q 1=2 Q. Q=2 22=4 δ 1( Ø, 0) =Ø. δ 1( Ø, 1) =Ø. δ 1( q 0, 0) =[q 0 q 1]. δ 1( q 0, 1) =[q 1]. δ 1( q 1, 0) =Ø. δ 1( q 1, 1) =[q 0 q 1] δ 1( [q 0 q 1], 0) =δ{q 0 q 1}, 0) =δ( q 0, 0) U δ( q 1, 0) ={q 0 q 1} U Ø => [q 0 q 1]
![n δ 1 q 0 q 1 1 δq 0 q 1 1 δ n δ 1( [q 0 q 1], 1) =δ{q 0 q 1}, 1) =δ(](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-8.jpg)
n δ 1( [q 0 q 1], 1) =δ{q 0 q 1}, 1) =δ( q 0, 1) U δ( q 1, 1) ={q 0 q 1} U {q 0 q 1} => [q 0 q 1] Set of final states {[q 1], [q 0 q 1]}
![solution 0 Ø q 0 q 1 q 0 q 1 1 Ø q solution 0 Ø [q 0] [q 1] [q 0 q 1] 1 Ø [q](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-9.jpg)
solution 0 Ø [q 0] [q 1] [q 0 q 1] 1 Ø [q 0 q 1] [q 0 q 1]
![Problem 2 n Construct a DFA from an NFA 0 1 p p q Problem 2 n Construct a DFA from an NFA 0 1 p {p, q}](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-10.jpg)
Problem 2 n Construct a DFA from an NFA 0 1 p {p, q} {p} q {r} r {s} Ø *s {s}
![n n n n To construct a DFA M 1q 1 δ 1 n n n n To construct a DFA M 1={q 1, ∑, δ 1,](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-11.jpg)
n n n n To construct a DFA M 1={q 1, ∑, δ 1, q 01, F 1) and accepting the language L(M 1). Q 1=2 Q. Q=4 24=16. δ 1( Ø, 0) =Ø. δ 1( Ø, 1) =Ø. δ 1( p, 0) =[pq]. δ 1( p, 1) =[p]. δ 1( q, 0) =[r]. δ 1( q, 1) =[r] δ 1( r, 0) = [s] δ 1( r, 1) = Ø δ 1( s, 0) = [s] δ 1( s, 1) = [s]
![](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-12.jpg)
![n n n n δ 1 qs 1 rs δ 1 rs 0 n n n n δ 1( [qs], 1) = [rs] δ 1( [rs], 0)](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-13.jpg)
n n n n δ 1( [qs], 1) = [rs] δ 1( [rs], 0) = [s] δ 1( [rs], 1) = [s] δ 1( [pqr], 0) = [pqrs] δ 1( [pqr], 1) = [pr] δ 1( [pqs], 0) = [pqrs] δ 1( [pqs], 1) = [prs] δ 1( [prs], 0) = [pqs] δ 1( [prs], 1) = [ps] δ 1( [qrs], 0) = [rs] δ 1( [qrs], 1) = [rs] δ 1( [pqrs], 0) = [pqrs] δ 1( [pqrs], 1) = [prs] Set of finite states { [s], [ps], [rs], [pqs], [prs], [pqrs], [qrs]}
![problem 3 n Construct a DFA from an NFA 0 1 p q s problem 3 n Construct a DFA from an NFA 0 1 p {q, s}](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-14.jpg)
problem 3 n Construct a DFA from an NFA 0 1 p {q, s} {q} q {r} {q, r} r {s} {p} *s Ø {p}
![problem 4 n Construct a DFA from an NFA 0 1 p p q problem 4 n Construct a DFA from an NFA 0 1 p {p, q}](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-15.jpg)
problem 4 n Construct a DFA from an NFA 0 1 p {p, q} {p} q {r} {q} r {s} Ø *s {s}
![Finite Automata with єclosure n n We shall introduce the another extension of FA Finite Automata with є-closure n n We shall introduce the another extension of FA.](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-16.jpg)
Finite Automata with є-closure n n 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 Contd. .](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-17.jpg)
Contd. .
![Example Example](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-18.jpg)
Example
![Transition table Transition table](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-19.jpg)
Transition table
![Finite Automata with єclosure Finite Automata with є-closure](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-20.jpg)
Finite Automata with є-closure
![Example Example](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-21.jpg)
Example
![Formal definition of NFA є closure n To represent the with NFA AQ Formal definition of NFA - є closure. n To represent the ∑with NFA A={Q,](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-22.jpg)
Formal definition of NFA - є closure. n To represent the ∑with NFA A={Q, ∑, δ, q 0, F) where all components have a same interpretations as for a NFA. q q A state in Q. A members of Є U {є} either an input symbol is empty or a set of strings.
![Eclosure Definition E-closure : Definition](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-23.jpg)
E-closure : Definition
![Example Find the єclosure for all the states of given figure Example: Find the є-closure for all the states of given figure](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-24.jpg)
Example: Find the є-closure for all the states of given figure
![solution n n n ECLOSE1 1 2 3 4 6 ECLOSE2 2 3 6 solution n n n ECLOSE(1) ={1, 2, 3, 4, 6} ECLOSE(2) ={2, 3, 6}](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-25.jpg)
solution n n n 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 Inductive definition on δ](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-26.jpg)
Inductive definition on δ
![Problem 1 consider the following epsilon with NFA and compute the empty closure of Problem 1: consider the following epsilon with NFA and compute the empty closure of](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-27.jpg)
Problem 1: consider the following epsilon with NFA and compute the empty closure of each states and find its equivalent to the DFA. є a b c p {q} {p} Ø Ø q {r} Ø {q} Ø *r Ø Ø Ø {r}
![n n n n Let A be a NFA with єclosure which has AQ n n n n Let A be a NFA with є-closure which has A={Q,](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-28.jpg)
n n n n Let A be a NFA with є-closure which has A={Q, ∑, δ, q 0, 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}
![](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-29.jpg)
![](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-30.jpg)
![](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-31.jpg)
![](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-32.jpg)
![](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-33.jpg)
![Problem 2 consider the following epsilon with NFA and compute the empty closure of Problem 2: consider the following epsilon with NFA and compute the empty closure of](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-34.jpg)
Problem 2: consider the following epsilon with NFA and compute the empty closure of each states and find its equivalent to the DFA. є a b c p Ø {p} {q} {r} q {p} {q} {r} Ø *r {q} {r} Ø {p}
![Problem 3 consider the following epsilon with NFA and compute the empty closure of Problem 3: consider the following epsilon with NFA and compute the empty closure of](https://slidetodoc.com/presentation_image_h/05a68cb6075e18806b3a70ad4a79ef89/image-35.jpg)
Problem 3: consider the following epsilon with NFA and compute the empty closure of each states and find its equivalent to the DFA. є a b c p {q, r} Ø {q} {r} q Ø {p} {r} {p, q} *r Ø Ø
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