Лабораторная работа №3. Определить свойства ка. Построить ндка. Реализовать преобразование ндка в дка.

3.1. Сгенерировать цепочку символов по заданному языку L.

00101010+111

3.2. По заданному языку построить НДКА. Представить в виде диаграмм.

KA = {Q, ∑, δ, q0, F}

Q = { A, B, C, S₀, q_f } – состояния автомата. Q = V

∑ = {1, 0, +} – алфавит. ∑ = T

δ – правила перехода. δ = P

q₀ – начальное состояние. q₀ = S₀

F – множество заключительных состояний. F = {q_f}

НДКА с ε-переходами.

δ(S0, 0) = {1}

δ(1, 0) = {2}

δ(2, ε ) = {3}

δ(3, ε) = {4}

δ(3, ε) = {6}

δ(4, 1 ) = {5}

δ(6, 0) = {7}

δ(5, ε) = {8}

δ(7, ε ) = {8}

δ(8, ε) = {9}

δ(9, +) = {10}

δ(10, ε ) = {11}

δ(11, 1 ) = {12}

δ(12, 1) = {11}

δ(11, ε) = {q_f}

δ(12, ε ) = { q_f }

Диаграмма переходов.

ε

ε

5

4

1

ε ε

9

8

3

2

1

0

0 0 ε ε

7

6

0 ε

ε +

q_f

12

11

10

ε 1 ε

1

3.3. Реализовать преобразование НДКА в ДКА.

Шаг 1.

ε-closure(0) = {0} = A

ε-closure(move(A, 1)) = Ø

Dtran[A, 1] = Ø

ε-closure(move(A, +)) = Ø

Dtran[A, +] = Ø

ε-closure(move(A, 0)) = ε-closure({1}) = {1} = B

Dtran[A, 0] = B

ε-closure(move(B, 1)) = Ø

Dtran[B, 1] = Ø

ε-closure(move(B, 0)) = ε-closure({2}) = {2, 3, 4, 6, 9} = C

Dtran[B, 0] = C

ε-closure(move(B, +)) = Ø

Dtran[B, +] = Ø

ε-closure(move(C, 1)) = ε-closure({5}) = {3, 4, 5, 6, 8, 9} = D

Dtran[C, 1] = D

ε-closure(move(C, 0)) = ε-closure({7}) = {3, 4, 6, 7, 8, 9} = E

Dtran[C, 0] = E

ε-closure(move(C, +)) = ε-closure({10}) = {11} = F

Dtran[C, +] = F

ε-closure(move(D, 1)) = Ø

Dtran[D, 1] = Ø

ε-closure(move(D, 0)) = ε-closure({7}) = {3, 4, 6, 7, 8, 9} = E

Dtran[D, 0] = E

ε-closure(move(D, +)) = ε-closure({10}) = {11} = F

Dtran[D, +] = F

ε-closure(move(E, 0)) = Ø

Dtran[E, 0] = Ø

ε-closure(move(E, 1)) = ε-closure({5}) = {3, 4, 5, 6, 8, 9} = D

Dtran[E, 1] = D

ε-closure(move(E, +)) = ε-closure({10}) = {11} = F

Dtran[E, +] = F

ε-closure(move(F, 0)) = Ø

Dtran[F, 0] = Ø

ε-closure(move(F, 1)) = ε-closure({11, 12}) = {qf} = K

Dtran[F, 1] = K

ε-closure(move(F, +)) = Ø

Dtran[F, +] = Ø

Шаг 2. Построение таблицы переходов Dtran.

Состояние	Входной символ
	1	0	+
A	Ø	B	Ø
B	Ø	C	Ø
C	D	E	F
D	Ø	E	F
E	D	Ø	F
F	K	Ø	Ø

Диаграмма переходов ДКА.

Текст программы.

myAutomate ka = new myAutomate(new ArrayList() { "S0", "A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "qf" },

new ArrayList() { "0", "1", "+", ""}, new ArrayList() { "qf" },"S0");

ka.AddRule("S0", "0", "A");

ka.AddRule("A", "0", "B");

ka.AddRule("B", "1", "C");

ka.AddRule("B", "0", "C");

ka.AddRule("C", "+", "D");

ka.AddRule("D", "1", "qf");

ka.Execute(Console.ReadLine()); // lab 1-3 KA

myAutomate ndka = new myAutomate(new ArrayList() {"S0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18", "19", "20", "qf"}, new ArrayList() { "1", "0","+", "-"}, new ArrayList() { "qf" }, "S0");

ndka.AddRule("S0", "0", "1");

ndka.AddRule("1", "0", "2");

ndka.AddRule("2", "", "3");

ndka.AddRule("3", "", "4");

ndka.AddRule("3", "", "6");

ndka.AddRule("4", "1", "5");

ndka.AddRule("6", "0", "7");

ndka.AddRule("5", "", "8");

ndka.AddRule("7", "", "8");

ndka.AddRule("8", "", "9");

ndka.AddRule("9", "+", "10");

ndka.AddRule("10", "", "11");

ndka.AddRule("11", "1", "12");

ndka.AddRule("12", "1", "11");

ndka.AddRule("11", "", "qf");

ndka.AddRule("12", "", "qf");

myAutomate dka = new myAutomate();

dka.BuildDeltaDKAutomate(ndka);

dka.DebugAuto();

dka.Execute(Console.ReadLine());