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DFA vs NFADeterministic Finite Automata

Characterized as 5-tuple < S, A, T, s0, F >• S is the set of states• A is the alphabet• T is the transition function: S x A S• s0 is the start state• F is the set of accepting states.

Characterized as 5-tuple < S, A, T, s0, F >• S is the set of states• A is the alphabet• T is the transition function: S x (A{}) PPS• s0 is the start state• F is the set of accepting states.

Non-Deterministic Finite Automata

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From NFA To DFA

0

1

2

43a

a b

a

T a b

0 {1} ø {3}

1 ø {2} ø

2 ø ø ø

3 {3} ø {4}

4 {2} ø ø

Transition function for NFA

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Calculating The Transition function for DFA

T a b

0 {1} ø {3}

1 ø {2} ø

2 ø ø ø

3 {3} ø {4}

4 {2} ø ø

T a b

{0,3,4} ? ?

Transition function for DFA

TD({0,3,4},a) = ({0},a) ({3},a) ({4},a) = {1,3} {3,4} {2} = {1,2,3,4}

TD({0,3,4},b) = ({0},b) ({3},b) ({4},b)

= ø ø ø = ø

Transition function for NFA

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Calculating The Transition function for DFA (forts)

T a b

0 {1} ø {3}

1 ø {2} ø

2 ø ø ø

3 {3} ø {4}

4 {2} ø ø

T a b

{0,3,4} {1,2,3,4} ø

Transition function for DFA

ø ? ?

{1,2,3,4} ? ?

ø ø

TD({1,2,3,4},a) = ({1},a) ({2},a) ({3},a) ({2},a) = ø ø {3,4} {2} = {2,3,4}Similarly, TD({1,2,3,4},b) = {2}

{2,3,4} {2}

Transition function for NFA

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Calculating The Transition function for DFA (forts)

T a b

0 {1} ø {3}

1 ø {2} ø

2 ø ø ø

3 {3} ø {4}

4 {2} ø ø

T a b

{0,3,4} {1,2,3,4} ø

Transition function for DFA

ø ø ø

{1,2,3,4} {2,3,4} {2}

{2,3,4} {2,3,4} {2}

{2} ø ø

Transition function for NFA

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Finally

T a b

{0,3,4} {1,2,3,4} ø

Transition function for DFA

ø ? ?

{1,2,3,4} {2,3,4} {2}

ø ø

{2,3,4} {2,3,4} {2}

{2} ø ø

01234

0 2

4

3

1

b

a,bb

a

ab

a

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Exercise 2, solution 2. Construct directly (that is don’t use the transformation procedure given in the lecture) a DFA for each of the following REs.

a) a | b b) a | b* c) ab* | bc*

a

b

a,b

a,b

a,b

a

b

a,b

a

a,b

b

a

b

a,c

a,b

a,b,c

c

b

a)

b)

c)

c

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Exercise 4, solution4. Construct directly (that is don’t use the transformation procedure given in the lecture) a NFA for each of the following REs. a) a*bc*| bc b) a* | ab

a c

ab

c

ab

a

a) b)

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Exercise 5

5. Construct a DFA table for the following NFA transition table. The start state is 0 and there is one accepting state, 2.

a b 0 Ø {1,2} {1} 1 {2} Ø Ø 2 Ø {2} {1}

10

2

a

b

b

b

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Solution to exercise 5

a b

0 Ø {1,2} {1} 1 {2} Ø Ø 2 Ø {2} {1}

T a b

{0,1} ? ?{1,2} {1,2}

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

NFA transition tableDFA transition table

0

1

0 1a,b a,b

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Exercise 6

6. Consider the following NFA over the alphabet {a,b}:a) Find a regular expression for the language accepted by the NFA.b) Write down the transition table for the NFA.c) Transform the NFA into a DFAd) Draw a picture of the resulting DFA.

0 0 0

ab

a

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Solution to exercise 6 (cont’d)

0 1 2

ab

a

Transition table for NFA

T a b

0 {1} Ø {1,2}1 Ø {2} Ø2 {2} Ø Ø

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Solution to exercise 6 (cont’d)

T a b

0 {1} Ø {1,2}1 Ø {2} Ø2 {2} Ø Ø

Transition table for NFA

T a b

{0,1,2} {1,2} {2} {1,2} {2} {2} {2} {2} Ø Ø Ø Ø

Transition table for DFA

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Solution to exercise 6(cont’d)

T a b

{0,1,2} {1,2} {2} {1,2} {2} {2} {2} {2} Ø Ø Ø Ø

Transition table for DFA{0,1,2}

{1,2}

{2}

a

ba,b

a

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Exercise 77. Transform each of the following REs into a NFA, then into a DFA

a) a*b* b) a* | b* c) (a|b)*

Solution to a)

0 12 3

4 5

6

a

b

610

a b

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Solution, exercise 6-a

3210

a b

T a b

{0,1,2,3} {1,2,3} {2,3} {1,2,3} {1,2,3} {2,3} {2,3} Ø {2,3} Ø Ø Ø

0

1

2 3

a

bb

a

b

aa,b

0

2 3

ba

a,b

a

b

Can be reduced to:

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What to do next?

It is your turn to solve the rest of the exercises