Usando ~ para NOT y N para NAND, una búsqueda por computadora (sin compartir términos entre salidas) encuentra una solución con 82 NAND sin compartir. La búsqueda manual de términos de uso compartido lo reduce a 54 NAND, y una búsqueda por computadora que incluye el uso compartido lo reduce aún más a 37 NAND. El mínimo podría ser incluso más bajo, ya que el método ciertamente no es exhaustivo.
Aquí está el programa que recrea la tabla anterior. Cada línea está etiquetada con las NANDS para esa salida.
#include <stdio.h>
int N(int x, int y) { return 1 & ~(x & y); }
void main(void)
{
int i0, i1, i2, i3;
for (i3 = 0; i3 <= 1; i3++) {
for (i2 = 0; i2 <= 1; i2++) {
for (i1 = 0; i1 <= 1; i1++) {
for (i0 = 0; i0 <= 1; i0++) {
printf("%d %d %d %d : %d %d %d %d %d %d %d\n", i3, i2, i1, i0,
/* 14 */ N(N(N(i3, N(i2, i1)), N(N(i2, i0), ~i1)), N(N(i2, N(i3, N(i3, i0))), N(i0, N(i3, N(i3, i1))))),
/* 12 */ N(N(N(i3, i0), N(i2, N(i1, i0))), N(~i1, N(N(i3, i0), N(~i3, N(i2, i0))))),
/* 10 */ N(N(i0, N(i3, i1)), N(N(i3, i2), N(i1, N(i1, N(~i3, ~i2))))),
/* 16 */ N(N(N(i2, i1), N(N(i3, i0), N(i2, i0))), N(N(i0, N(i1, ~i2)), N(N(i3, i2), N(N(i3, i1), N(i2, N(i2, i1)))))),
/* 7 */ N(N(i3, i2), N(N(i2, N(i2, i1)), N(i0, N(i3, i1)))),
/* 11 */ N(N(i3, N(i2, N(i3, i1))), N(N(i1, N(i2, N(i2, i0))), N(i0, N(i2, ~i3)))),
/* 12 */ N(N(i3, i0), ~N(N(i1, N(i2, i0)), N(N(i3, i2), N(~i3, N(i2, N(i2, i1)))))) );
} } } }
}
Y aquí está el resultado:
0 0 0 0 : 1 1 1 1 1 1 0
0 0 0 1 : 0 1 1 0 0 0 0
0 0 1 0 : 1 1 0 1 1 0 1
0 0 1 1 : 1 1 1 1 0 0 1
0 1 0 0 : 0 1 1 0 0 1 1
0 1 0 1 : 1 0 1 1 0 1 1
0 1 1 0 : 1 0 1 1 1 1 1
0 1 1 1 : 1 1 1 0 0 0 0
1 0 0 0 : 1 1 1 1 1 1 1
1 0 0 1 : 1 1 1 1 0 1 1
1 0 1 0 : 1 1 1 0 1 1 1
1 0 1 1 : 0 0 1 1 1 1 1
1 1 0 0 : 1 0 0 1 1 1 0
1 1 0 1 : 0 1 1 1 1 0 1
1 1 1 0 : 1 0 0 1 1 1 1
1 1 1 1 : 1 0 0 0 1 1 1
Y aquí están las ecuaciones equivalentes, compartiendo términos que lo reducen a 54 NAND:
/* 1 */ int z1 = 1 - i1;
/* 1 */ int z2 = 1 - i2;
/* 1 */ int z3 = 1 - i3;
/* 1 */ int n_i2_i0 = N(i2, i0);
/* 1 */ int n_i2_i1 = N(i2, i1);
/* 1 */ int n_i3_i0 = N(i3, i0);
/* 1 */ int n_i3_i1 = N(i3, i1);
/* 1 */ int n_i3_i2 = N(i3, i2);
/* 1 */ int n_i0_n_i3_i1 = N(i0, n_i3_i1);
/* 1 */ int n_i2_n_i2_i1 = N(i2, n_i2_i1);
printf("%d %d %d %d : %d %d %d %d %d %d %d\n", i3, i2, i1, i0,
/* 9 */ N(N(N(i3, n_i2_i1), N(n_i2_i0, z1)), N(N(i2, N(i3, n_i3_i0)), N(i0, N(i3, n_i3_i1)))),
/* 7 */ N(N(n_i3_i0, N(i2, N(i1, i0))), N(z1, N(n_i3_i0, N(z3, n_i2_i0)))),
/* 5 */ N(n_i0_n_i3_i1, N(n_i3_i2, N(i1, N(i1, N(z3, z2))))),
/* 8 */ N(N(n_i2_i1, N(n_i3_i0, n_i2_i0)), N(N(i0, N(i1, z2)), N(n_i3_i2, N(n_i3_i1, n_i2_n_i2_i1)))),
/* 2 */ N(n_i3_i2, N(n_i2_n_i2_i1, n_i0_n_i3_i1)),
/* 8 */ N(N(i3, N(i2, n_i3_i1)), N(N(i1, N(i2, n_i2_i0)), N(i0, N(i2, z3)))),
/* 6 */ N(n_i3_i0, ~N(N(i1, n_i2_i0), N(n_i3_i2, N(z3, n_i2_n_i2_i1)))) );
Y aquí está la solución 37 NAND:
x0fff = N(i3, i2);
x33ff = N(i3, i1);
x55ff = N(i3, i0);
x0f0f = not(i2);
x3f3f = N(i2, i1);
x5f5f = N(i2, i0);
xcfcf = N(i2, x3f3f);
xf3f3 = N(i1, x0f0f);
x5d5d = N(i0, xf3f3);
xaaa0 = N(x55ff, x5f5f);
xfc30 = N(x33ff, xcfcf);
xd5df = N(x3f3f, xaaa0);
xf3cf = N(x0fff, xfc30);
xaeb2 = N(x5d5d, xf3cf);
x7b6d = N(xd5df, xaeb2);
xb7b3 = N(i1, x7b6d);
xf55f = N(x0fff, xaaa0);
xcea0 = N(x33ff, xf55f);
x795f = N(xb7b3, xcea0);
xd7ed = N(xaeb2, x795f);
xfaf0 = N(x55ff, x0f0f);
xae92 = N(x55ff, x7b6d);
xdd6d = N(x33ff, xae92);
x279f = N(xfaf0, xdd6d);
xaf0f = N(i2, x55ff);
x50ff = N(i3, xaf0f);
xef4c = N(xb7b3, x50ff);
x1cb3 = N(xf3cf, xef4c);
xef7c = N(xf3cf, x1cb3);
xfb73 = N(i1, x279f);
x2c9e = N(xd7ed, xfb73);
xdf71 = N(xf3cf, x2c9e);
xdd55 = N(i0, x33ff);
xf08e = N(x0fff, xdf71);
x2ffb = N(xdd55, xf08e);
x32ba = N(xcfcf, xdd55);
xfd45 = N(x0fff, x32ba);
printf("%d %d %d %d : %d %d %d %d %d %d %d\n", i3, i2, i1, i0,
xd7ed, x279f, x2ffb, x7b6d, xfd45, xdf71, xef7c);
} } } }
