arithmos
MAKE THE MATHEMATICS YOURSELF

Hilbert’s binary labyrinth in C

A complete program. Standard C, a compiler, and the rule behind the shape.

Output generated by this C program: Hilbert’s binary labyrinth
Generated by the C program below Download image ↗
FROM SOURCE TO SHAPE

Run it in three steps.

  1. Save the source.

    Download geo-hilbert.c into a folder on your computer.

  2. Compile it.

    In that folder, run this with GCC or Clang:

    cc -std=c11 -O2 geo-hilbert.c -lm -o geo-hilbert
  3. Make the image.
    ./geo-hilbert

    Open geo-hilbert.svg in a browser to see the result.

On Windows with GCC, name the executable geo-hilbert.exe and run it from the same folder.

THE RULE IN THE PROGRAM

How the picture is built

Decode base-four digits with recursive rotations and reflections to visit every cell of a power-of-two square grid with adjacent steps.

Make it your own

ORDER = 5, allowed 1..7. Number of vertices is 4^ORDER.

A finite polyline approximates the limiting Hilbert curve; only the infinite limit fills the square. Each run produces one image; use Graphic mode for the interactive animation.

The complete source

geo-hilbert.c · 101 lines · no graphics libraries
Download .c ↓
/* Arithmos: geo-hilbert
 * Compile: cc -std=c11 -O2 geo-hilbert.c -lm -o geo-hilbert
 * Run:     ./geo-hilbert
 * Output:  geo-hilbert.svg (open this file in a browser)
 * Optional output path: ./geo-hilbert my-image.svg
 * Edit the constants in draw() to explore another case.
 */
#include <math.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

static FILE *nt_out;
#define NT_PI 3.14159265358979323846

/* The small SVG writer keeps this program free of graphics dependencies.
 * Coordinates are pixels on a 1000 x 700 drawing surface.
 * t runs from 0 to 1 through mint, blue, rose, and gold.
 */
static inline void nt_color(double t, char hex[8]) {
    const double stops[4][3] = {
        {91,227,201}, {128,146,240}, {218,138,220}, {244,200,127}
    };
    t = fmax(0.0, fmin(1.0,t)) * 3.0;
    int band = (int)fmin(2.0,floor(t));
    double blend = t - band;
    int r[3];
    for (int k=0;k<3;k++) r[k]=(int)lround(stops[band][k]*(1.0-blend)+stops[band+1][k]*blend);
    snprintf(hex,8,"#%02x%02x%02x",r[0],r[1],r[2]);
}
static inline void nt_line(double x,double y,double X,double Y,double t,double alpha,double width) {
    char color[8];nt_color(t,color);
    fprintf(nt_out,"<path d=\"M%.3f %.3f L%.3f %.3f\" fill=\"none\" stroke=\"%s\" stroke-opacity=\"%.3f\" stroke-width=\"%.3f\"/>\n",x,y,X,Y,color,alpha,width);
}
static inline void nt_dot(double x,double y,double r,double t,double alpha) {
    char color[8];nt_color(t,color);
    fprintf(nt_out,"<circle cx=\"%.3f\" cy=\"%.3f\" r=\"%.3f\" fill=\"%s\" fill-opacity=\"%.3f\"/>\n",x,y,r,color,alpha);
}
static inline void nt_circle(double x,double y,double r,double t,double alpha,double width) {
    char color[8];nt_color(t,color);
    fprintf(nt_out,"<circle cx=\"%.3f\" cy=\"%.3f\" r=\"%.3f\" fill=\"none\" stroke=\"%s\" stroke-opacity=\"%.3f\" stroke-width=\"%.3f\"/>\n",x,y,r,color,alpha,width);
}
static inline void nt_rect(double x,double y,double w,double h,double t,double alpha) {
    char color[8];nt_color(t,color);
    fprintf(nt_out,"<rect x=\"%.3f\" y=\"%.3f\" width=\"%.3f\" height=\"%.3f\" fill=\"%s\" fill-opacity=\"%.3f\"/>\n",x,y,w,h,color,alpha);
}
static inline void nt_text(double x,double y,const char *text) {
    fprintf(nt_out,"<text x=\"%.3f\" y=\"%.3f\" fill=\"#ededf3\" font-family=\"monospace\" font-size=\"16\">",x,y);
    for (;*text;text++) {
        if (*text=='&') fputs("&amp;",nt_out);
        else if (*text=='<') fputs("&lt;",nt_out);
        else if (*text=='>') fputs("&gt;",nt_out);
        else fputc(*text,nt_out);
    }
    fputs("</text>\n",nt_out);
}
static inline int nt_gcd(int a,int b) {a=abs(a);b=abs(b);while(b){int r=a%b;a=b;b=r;}return a;}
static inline int nt_prime(int n) {if(n<2)return 0;for(int d=2;d<=n/d;d++)if(n%d==0)return 0;return 1;}

/* Edit ORDER (1..7). The grid has 2^ORDER cells along each side. */
static void hilbert_xy(int side,int d,int *xx,int *yy) {
    int x=0,y=0,t=d;
    for(int s=1;s<side;s*=2) {
        int rx=1&(t>>1),ry=1&(t^rx);
        if(!ry){if(rx){x=s-1-x;y=s-1-y;}int u=x;x=y;y=u;}
        x+=s*rx;y+=s*ry;t>>=2;
    }
    *xx=x;*yy=y;
}
static void draw(void) {
    const int ORDER=5;
    int side=1<<ORDER,count=side*side,px=0,py=0;
    double s=560.0/(side-1);
    for(int d=0;d<count;d++) {
        int x,y;hilbert_xy(side,d,&x,&y);
        if(d)nt_line(220+s*px,630-s*py,220+s*x,630-s*y,(double)d/(count-1),.95,2);
        px=x;py=y;
    }
    nt_text(30,35,"Hilbert curve: every grid cell visited once, with adjacent steps");
}

int main(int argc, char **argv) {
    if (argc > 2) {
        fprintf(stderr, "Usage: %s [OUTPUT.svg]\n", argv[0]);
        return EXIT_FAILURE;
    }
    const char *filename = argc == 2 ? argv[1] : "geo-hilbert.svg";
    nt_out = fopen(filename, "wb");
    if (!nt_out) { perror(filename); return EXIT_FAILURE; }
    fputs("<svg xmlns=\"http://www.w3.org/2000/svg\" width=\"1000\" height=\"700\" viewBox=\"0 0 1000 700\">\n"
          "<rect width=\"1000\" height=\"700\" fill=\"#171721\"/>\n", nt_out);
    draw();
    fputs("</svg>\n", nt_out);
    int failed = ferror(nt_out);
    if (fclose(nt_out) != 0) failed = 1;
    if (failed) { fputs("Could not finish writing the image.\n", stderr); return EXIT_FAILURE; }
    printf("Wrote %s\n", filename);
    return EXIT_SUCCESS;
}
All 40 explorations, ready to compile.Download all C examples ↓