Menggunakan ~ untuk NOT dan N untuk NAND, pencarian komputer (tanpa berbagi istilah antara output) menemukan solusi dengan 82 NANDs tanpa berbagi. Mencari istilah berbagi secara manual akan menurunkannya menjadi 54 NAND, dan pencarian komputer yang mencakup pembagian lebih lanjut menguranginya menjadi 37 NAND. Minimal mungkin bahkan lebih rendah, karena metode ini tentu tidak lengkap.
Inilah program yang membuat ulang tabel di atas. Setiap baris diberi label dengan NANDS untuk output itu.
#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)))))) );
} } } }
}
Dan inilah hasilnya:
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
Dan berikut ini persamaan yang setara, istilah berbagi yang membuatnya menjadi 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)))) );
Dan inilah solusi 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);
} } } }
