1.2 Мінімізація булевої функції методом Квайна МакКласкі для одиниць
№ |
X1 |
X2 |
X3 |
X4 |
X5 |
Y |
0 |
0 |
0 |
0 |
0 |
0 |
x |
1 |
0 |
0 |
0 |
0 |
1 |
1 |
2 |
0 |
0 |
0 |
1 |
0 |
0 |
3 |
0 |
0 |
0 |
1 |
1 |
0 |
4 |
0 |
0 |
1 |
0 |
0 |
1 |
5 |
0 |
0 |
1 |
0 |
1 |
1 |
6 |
0 |
0 |
1 |
1 |
0 |
0 |
7 |
0 |
0 |
1 |
1 |
1 |
x |
8 |
0 |
1 |
0 |
0 |
0 |
1 |
9 |
0 |
1 |
0 |
0 |
1 |
1 |
10 |
0 |
1 |
0 |
1 |
0 |
1 |
11 |
0 |
1 |
0 |
1 |
1 |
0 |
12 |
0 |
1 |
1 |
0 |
0 |
1 |
13 |
0 |
1 |
1 |
0 |
1 |
1 |
14 |
0 |
1 |
1 |
1 |
0 |
x |
15 |
0 |
1 |
1 |
1 |
1 |
x |
16 |
1 |
0 |
0 |
0 |
0 |
x |
17 |
1 |
0 |
0 |
0 |
1 |
1 |
18 |
1 |
0 |
0 |
1 |
0 |
1 |
19 |
1 |
0 |
0 |
1 |
1 |
0 |
20 |
1 |
0 |
1 |
0 |
0 |
0 |
21 |
1 |
0 |
1 |
0 |
1 |
x |
22 |
1 |
0 |
1 |
1 |
0 |
1 |
23 |
1 |
0 |
1 |
1 |
1 |
1 |
24 |
1 |
1 |
0 |
0 |
0 |
1 |
25 |
1 |
1 |
0 |
0 |
1 |
0 |
26 |
1 |
1 |
0 |
1 |
0 |
0 |
27 |
1 |
1 |
0 |
1 |
1 |
1 |
28 |
1 |
1 |
1 |
0 |
0 |
x |
29 |
1 |
1 |
1 |
0 |
1 |
x |
30 |
1 |
1 |
1 |
1 |
0 |
x |
31 |
1 |
1 |
1 |
1 |
1 |
x |
K0= K1= K2= K3=
0 0 0 0 0 0 0 0 0 x ü x 0 0 0 x - 0 x x 0 x
0 0 0 0 1 0 0 x 0 0 ü 0 0 x 0 x ü x 1 1 x x
0 0 1 0 0 0 x 0 0 0 ü 0 x x 0 0 ü x x 1 x 1
0 1 0 0 0 x 0 0 0 0 ü 0 x 0 0 x -
1 0 0 0 0 0 0 x 0 1 ü x x 0 0 0 -
0 0 1 0 1 0 x 0 0 1 ü 0 x x 0 1 ü
0 1 0 0 1 0 1 0 x 0 ü x 0 x 0 1 -
0 1 0 1 0 0 1 x 0 0 ü 0 1 x x 0 -
0 1 1 0 0 0 x 1 0 0 ü x 1 x 0 0 ü K=
1 0 0 0 1 x 0 0 0 1 ü 0 1 x 0 x ü 1 0 0 x 0
1 0 0 1 0 1 0 0 0 x ü 0 x 1 0 x ü 1 1 x 1 1
1 1 0 0 0 1 0 0 x 0 - 0 1 1 x x ü 0 x 0 0 x
0 0 1 1 1 1 x 0 0 0 ü 0 x 1 x 1 ü x 0 0 0 x
0 1 1 0 1 x 1 0 0 0 ü x 0 1 x 1 ü x x 0 0 0
0 1 1 1 0 0 0 1 x 1 ü x x 1 0 1 ü x 0 x 0 1
1 0 1 1 0 0 0 1 1 x ü x 1 1 0 x ü 0 1 x x 0
1 0 1 0 1 0 1 x 0 1 ü x x 1 1 1 ü x 1 x 0 0
1 1 1 0 0 0 x 1 0 1 ü x 1 1 1 x ü 1 x 1 1 x
0 1 1 1 1 0 1 x 1 0 ü x 1 1 x 1 ü 0 x x 0 x
1 0 1 1 1 0 1 1 x 0 ü 1 x 1 x 1 ü x 1 1 x x
1 1 0 1 1 1 0 x 0 1 ü 1 x 1 1 x - x x 1 x 1
1 1 1 0 1 x 0 1 0 1 ü 1 1 1 x x ü
1 1 1 1 0 x 1 1 0 0 ü
1 1 1 1 1 1 1 x 0 0 ü
0 x 1 1 1 ü
0 1 1 1 x ü
x 0 1 1 1 ü
1 0 1 x 1 ü
1 0 1 1 x ü
1 1 1 0 x ü
x 1 1 0 1 ü
x 1 1 1 1 ü
1 x 1 1 1 ü
1 1 x 1 1 -
1 1 1 x 1 ü
1 1 1 1 x ü
Після склеювання, будуємо таблицю покриттів. В ній ми знаходимо ядерні строки. Виділяємо ядро.
Таблиця покриттів Табл.2
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0 0 0 0 0 |
0 0 0 0 1 |
0 0 1 0 0 |
0 1 0 0 0 |
1 0 0 0 0 |
0 0 1 0 1 |
0 1 0 0 1 |
0 1 0 1 0 |
0 1 1 0 0 |
1 0 0 0 1 |
1 0 0 1 0 |
1 1 0 0 0 |
0 0 1 1 1 |
0 1 1 0 1 |
0 1 1 1 0 |
1 0 1 1 0 |
1 0 1 0 1 |
1 1 1 0 0 |
0 1 1 1 1 |
1 0 1 1 1 |
1 1 0 1 1 |
1 1 1 0 1 |
1 1 1 1 0 |
1 1 1 1 1 |
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1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
16 |
17 |
18 |
19 |
20 |
21 |
22 |
23 |
24 |
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100x0 |
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ü |
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ü |
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A |
11x11 |
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ü |
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ü |
B |
0x00x |
ü |
ü |
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ü |
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ü |
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C |
x000x |
ü |
ü |
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ü |
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ü |
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D |
xx000 |
ü |
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ü |
ü |
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ü |
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E |
x0x01 |
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ü |
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ü |
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ü |
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ü |
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F |
01xx0 |
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ü |
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ü |
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ü |
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G |
x1x00 |
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ü |
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ü |
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ü |
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H |
1x11x |
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ü |
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ü |
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ü |
ü |
I |
0xx0x |
ü |
ü |
ü |
ü |
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ü |
ü |
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ü |
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ü |
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J |
x11xx |
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ü |
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ü |
ü |
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ü |
ü |
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ü |
ü |
ü |
K |
xx1x1 |
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ü |
ü |
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ü |
ü |
ü |
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ü |
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ü |
ü |
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ü |
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ü |
L |
Сокращенная ДНФ=x1x2x3x5 + x1x2x4x5 + x1x3x4 + x2x3x4 +x3x4x5 +x2x4x5+x1x2x5+x2x4x5+x1x3x4+x1x4+x2x3+x3x4
Ядро: 0xx0x, 01xx0, x0x01, 100x0, xx1x1, 1x11x, 11x11.Ядро покриває все конституанти, тому не будeмо скорочену таблицю покриттів.
Тоді МДНФ = X2X3X4+X3X4X5+X1X3X4+X1X2+X1X2+X1X2X3X4X5+X3X5+X1X4X5