Theorem 1.39 on Page 55 of textbook [Introduction to the Theory of Computation:Sipser Introduction to the Theory of Computation 3rd Edition] says that every NFA has an equivalent DFA, followed by the construction

procedures of equivalent DFA given a NFA. In this project, will write a program using Using Java Programming to implement part of such construction procedures.

Read part of the transition table of a NFA from a txt file, suppose this table only contains two columns: the first

column lists out the states, while the second column lists out what each state jumps to upon the input e. For

example, given the following transition table in a txt file:

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1,{2, 3}

2,empty 3,{4}

4,empty

We know the corresponding NFA contains the following set of states: 1, 2, 3, 4 . Also state 1 jumps to states 2,

3 upon the input of ε, while state 2 does not have an outgoing arrowhead with ε on, namely it jumps to ∅ upon the input of ε (in the txt file we use “empty” to denote empty set ∅).

Read part of the transition table of a NFA from a txt file, and then print out the set of states the equivalent DFA has and E(q) for each q in the set of states of the NFA. For example, for the above part of the transition table of a NFA, the print out should be

State set of the equivalent DFA = {empty, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2,

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

E(1) = {1,2,3,4}

E(2) = {2}

E(3) = {3,4}

E(4) = {4}

Note that your program must be able to work for any input, not only just for the above given example.

The following are a few more examples you can use to test your program (I highly recommend that you come up with more examples to make sure the correctness of your program):

(1) 1,{3}

2,empty 3,empty

For the above part of the transition table of a NFA, the print out should be

State set of the equivalent DFA = {empty, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

E(1) = {1,3}

E(2) = {2}

E(3) = {3}

(2) 1,{3}

2,{1}

3,empty

For the above part of the transition table of a NFA, the print out should be

State set of the equivalent DFA = {empty, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

E(1) = {1,3}

E(2) = {1,2,3}

E(3) = {3}

(3) 1,{2,3}

2,empty 3,{4,5}

4,empty 5,empty

For the above part of the transition table of a NFA, the print out should be

State set of the equivalent DFA = {empty, {1}, {2}, {3}, {4}, {5}, {1, 2}, {1, 3}, {2, 3}, {1, 4}, {2, 4}, {3,

4}, {1, 5}, {2, 5}, {3, 5}, {4, 5}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 5}, {1, 3, 5}, {2, 3, 5}, {1,

4, 5}, {2, 4, 5}, {3, 4, 5}, {1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5}, {1, 2, 3, 4, 5}}

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

E(2) = {2}

E(3) = {3, 4, 5}

E(4) = {4}

E(5) = {5}

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

Example:

1,{2, 3}

2,empty

3,{4}

4,empty

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

Test 1:

1,{3}

2,empty

3,empty

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

Test 2:

1,{3}

2,{1}

3,empty

– – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

Test 3:

1,{2,3}

2,empty

3,{4,5}

4,empty

5,empty

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