Closest-pair problem/C

From Rosetta Code
Closest-pair problem/C is part of Closest pair problem. You may find other members of Closest pair problem at Category:Closest pair problem.
#include <stdio.h>
#include <stdlib.h>
#include <values.h>
#include <math.h>
#include <string.h>
 
typedef struct { double x, y; } point_t, *point;
 
inline double dist(point a, point b)
{
double dx = a->x - b->x, dy = a->y - b->y;
return dx * dx + dy * dy;
}
 
inline int cmp_dbl(double a, double b)
{
return a < b ? -1 : a > b ? 1 : 0;
}
 
int cmp_x(const void *a, const void *b) {
return cmp_dbl( (*(const point*)a)->x, (*(const point*)b)->x );
}
 
int cmp_y(const void *a, const void *b) {
return cmp_dbl( (*(const point*)a)->y, (*(const point*)b)->y );
}
 
double brute_force(point* pts, int max_n, point *a, point *b)
{
int i, j;
double d, min_d = MAXDOUBLE;
 
for (i = 0; i < max_n; i++) {
for (j = i + 1; j < max_n; j++) {
d = dist(pts[i], pts[j]);
if (d >= min_d ) continue;
*a = pts[i];
*b = pts[j];
min_d = d;
}
}
return min_d;
}
 
double closest(point* sx, int nx, point* sy, int ny, point *a, point *b)
{
int left, right, i;
double d, min_d, x0, x1, mid, x;
point a1, b1;
point *s_yy;
 
if (nx <= 8) return brute_force(sx, nx, a, b);
 
s_yy = malloc(sizeof(point) * ny);
mid = sx[nx/2]->x;
 
/* adding points to the y-sorted list; if a point's x is less than mid,
add to the begining; if more, add to the end backwards, hence the
need to reverse it */

left = -1; right = ny;
for (i = 0; i < ny; i++)
if (sy[i]->x < mid) s_yy[++left] = sy[i];
else s_yy[--right]= sy[i];
 
/* reverse the higher part of the list */
for (i = ny - 1; right < i; right ++, i--) {
a1 = s_yy[right]; s_yy[right] = s_yy[i]; s_yy[i] = a1;
}
 
min_d = closest(sx, nx/2, s_yy, left + 1, a, b);
d = closest(sx + nx/2, nx - nx/2, s_yy + left + 1, ny - left - 1, &a1, &b1);
 
if (d < min_d) { min_d = d; *a = a1; *b = b1; }
d = sqrt(min_d);
 
/* get all the points within distance d of the center line */
left = -1; right = ny;
for (i = 0; i < ny; i++) {
x = sy[i]->x - mid;
if (x <= -d || x >= d) continue;
 
if (x < 0) s_yy[++left] = sy[i];
else s_yy[--right] = sy[i];
}
 
/* compare each left point to right point */
while (left >= 0) {
x0 = s_yy[left]->y + d;
 
while (right < ny && s_yy[right]->y > x0) right ++;
if (right >= ny) break;
 
x1 = s_yy[left]->y - d;
for (i = right; i < ny && s_yy[i]->y > x1; i++)
if ((x = dist(s_yy[left], s_yy[i])) < min_d) {
min_d = x;
d = sqrt(min_d);
*a = s_yy[left];
*b = s_yy[i];
}
 
left --;
}
 
free(s_yy);
return min_d;
}
 
#define NP 1000000
int main()
{
int i;
point a, b;
 
point pts = malloc(sizeof(point_t) * NP);
point* s_x = malloc(sizeof(point) * NP);
point* s_y = malloc(sizeof(point) * NP);
 
for(i = 0; i < NP; i++) {
s_x[i] = pts + i;
pts[i].x = 100 * (double) rand()/RAND_MAX;
pts[i].y = 100 * (double) rand()/RAND_MAX;
}
 
/* printf("brute force: %g, ", sqrt(brute_force(s_x, NP, &a, &b)));
printf("between (%f,%f) and (%f,%f)\n", a->x, a->y, b->x, b->y); */

 
memcpy(s_y, s_x, sizeof(point) * NP);
qsort(s_x, NP, sizeof(point), cmp_x);
qsort(s_y, NP, sizeof(point), cmp_y);
 
printf("min: %g; ", sqrt(closest(s_x, NP, s_y, NP, &a, &b)));
printf("point (%f,%f) and (%f,%f)\n", a->x, a->y, b->x, b->y);
 
/* not freeing the memory, let OS deal with it. Habit. */
return 0;
}
Compile and run (1000000 points, bruteforce method not shown because it takes forever):
 
% gcc -Wall -lm -O2 test.c
% time ./a.out
min: 5.6321e-05; point (19.657247,79.900855) and (19.657303,79.900862)
./a.out 7.16s user 0.08s system 96% cpu 7.480 total