I just found a way to generate this kind of pseudo-random number, created by the creator of the Perlin Noise, Ken Perlin.
He left one of his own Talks available online here. Throughout the explanation, it leads to one of your other sites, where it shows the original algorithm.
For completeness, and because the algorithm is not patented/available online (besides I have already made the appropriate references), I will put the algorithm his here:
/* coherent noise function over 1, 2 or 3 dimensions */
/* (copyright Ken Perlin) */
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#define B 0x100
#define BM 0xff
#define N 0x1000
#define NP 12 /* 2^N */
#define NM 0xfff
static p[B + B + 2];
static float g3[B + B + 2][3];
static float g2[B + B + 2][2];
static float g1[B + B + 2];
static start = 1;
static void init(void);
#define s_curve(t) ( t * t * (3. - 2. * t) )
#define lerp(t, a, b) ( a + t * (b - a) )
#define setup(i,b0,b1,r0,r1)\
t = vec[i] + N;\
b0 = ((int)t) & BM;\
b1 = (b0+1) & BM;\
r0 = t - (int)t;\
r1 = r0 - 1.;
double noise1(double arg)
{
int bx0, bx1;
float rx0, rx1, sx, t, u, v, vec[1];
vec[0] = arg;
if (start) {
start = 0;
init();
}
setup(0, bx0,bx1, rx0,rx1);
sx = s_curve(rx0);
u = rx0 * g1[ p[ bx0 ] ];
v = rx1 * g1[ p[ bx1 ] ];
return lerp(sx, u, v);
}
float noise2(float vec[2])
{
int bx0, bx1, by0, by1, b00, b10, b01, b11;
float rx0, rx1, ry0, ry1, *q, sx, sy, a, b, t, u, v;
register i, j;
if (start) {
start = 0;
init();
}
setup(0, bx0,bx1, rx0,rx1);
setup(1, by0,by1, ry0,ry1);
i = p[ bx0 ];
j = p[ bx1 ];
b00 = p[ i + by0 ];
b10 = p[ j + by0 ];
b01 = p[ i + by1 ];
b11 = p[ j + by1 ];
sx = s_curve(rx0);
sy = s_curve(ry0);
#define at2(rx,ry) ( rx * q[0] + ry * q[1] )
q = g2[ b00 ] ; u = at2(rx0,ry0);
q = g2[ b10 ] ; v = at2(rx1,ry0);
a = lerp(sx, u, v);
q = g2[ b01 ] ; u = at2(rx0,ry1);
q = g2[ b11 ] ; v = at2(rx1,ry1);
b = lerp(sx, u, v);
return lerp(sy, a, b);
}
float noise3(float vec[3])
{
int bx0, bx1, by0, by1, bz0, bz1, b00, b10, b01, b11;
float rx0, rx1, ry0, ry1, rz0, rz1, *q, sy, sz, a, b, c, d, t, u, v;
register i, j;
if (start) {
start = 0;
init();
}
setup(0, bx0,bx1, rx0,rx1);
setup(1, by0,by1, ry0,ry1);
setup(2, bz0,bz1, rz0,rz1);
i = p[ bx0 ];
j = p[ bx1 ];
b00 = p[ i + by0 ];
b10 = p[ j + by0 ];
b01 = p[ i + by1 ];
b11 = p[ j + by1 ];
t = s_curve(rx0);
sy = s_curve(ry0);
sz = s_curve(rz0);
#define at3(rx,ry,rz) ( rx * q[0] + ry * q[1] + rz * q[2] )
q = g3[ b00 + bz0 ] ; u = at3(rx0,ry0,rz0);
q = g3[ b10 + bz0 ] ; v = at3(rx1,ry0,rz0);
a = lerp(t, u, v);
q = g3[ b01 + bz0 ] ; u = at3(rx0,ry1,rz0);
q = g3[ b11 + bz0 ] ; v = at3(rx1,ry1,rz0);
b = lerp(t, u, v);
c = lerp(sy, a, b);
q = g3[ b00 + bz1 ] ; u = at3(rx0,ry0,rz1);
q = g3[ b10 + bz1 ] ; v = at3(rx1,ry0,rz1);
a = lerp(t, u, v);
q = g3[ b01 + bz1 ] ; u = at3(rx0,ry1,rz1);
q = g3[ b11 + bz1 ] ; v = at3(rx1,ry1,rz1);
b = lerp(t, u, v);
d = lerp(sy, a, b);
return lerp(sz, c, d);
}
static void normalize2(float v[2])
{
float s;
s = sqrt(v[0] * v[0] + v[1] * v[1]);
v[0] = v[0] / s;
v[1] = v[1] / s;
}
static void normalize3(float v[3])
{
float s;
s = sqrt(v[0] * v[0] + v[1] * v[1] + v[2] * v[2]);
v[0] = v[0] / s;
v[1] = v[1] / s;
v[2] = v[2] / s;
}
static void init(void)
{
int i, j, k;
for (i = 0 ; i < B ; i++) {
p[i] = i;
g1[i] = (float)((random() % (B + B)) - B) / B;
for (j = 0 ; j < 2 ; j++)
g2[i][j] = (float)((random() % (B + B)) - B) / B;
normalize2(g2[i]);
for (j = 0 ; j < 3 ; j++)
g3[i][j] = (float)((random() % (B + B)) - B) / B;
normalize3(g3[i]);
}
while (--i) {
k = p[i];
p[i] = p[j = random() % B];
p[j] = k;
}
for (i = 0 ; i < B + 2 ; i++) {
p[B + i] = p[i];
g1[B + i] = g1[i];
for (j = 0 ; j < 2 ; j++)
g2[B + i][j] = g2[i][j];
for (j = 0 ; j < 3 ; j++)
g3[B + i][j] = g3[i][j];
}
}
Related question (not exactly a duplicate, but some answers there may apply here)
– mgibsonbr
Hi mgibsonbr! Thanks, I had seen this question, but not the first answer. I think we can make a test :)
– carlosrafaelgn
"The return being an integer or floating point number is indifferent." I would suggest always working with integers, so that you have no problem changing processor architecture (if the map is desired to be the same on any machine). Depending on the type of map you make, a rounding or precision difference can give you a voucher on one machine and a hill on another.
– Bacco
I agree, @Bacco, this is dangerous in many applications! But in this particular case, no! Good ;) This time, it is enough that the result is "beautiful" and OK!
– carlosrafaelgn