Step by Step solution about How to resolve the algorithm Hilbert curve step by step in the C programming language

This C program generates an ASCII art representation of a Hilbert curve of order N (defined as 32 in the code). Here's a detailed explanation of the code:

• Constants and Data Structures:

  • N and K are used to define the size of the Hilbert curve and the spacing between points.
  • MAX is calculated as N * K, which determines the size of the grid to hold the curve's points.
  • point is a struct that represents a point in the 2D grid, containing x and y coordinates.

• rot Function:

  • This function rotates a point around a given center by 90 degrees clockwise or counterclockwise.
  • The rx and ry parameters specify the direction of rotation.
  • t is used as a temporary variable for swapping x and y coordinates.

• d2pt Function:

  • This function takes a direction d and calculates the corresponding point in the Hilbert curve.
  • It applies rotations and updates the point's coordinates based on the direction.
  • s and t are used for tracking the size and direction of the current segment.

• main Function:

  • Initializes a 2D character array pts of size MAX to represent the grid.
  • Initializes a curr and prev point to track the current and previous points on the curve.
  • Loops through all directions (d) from 1 to N * N.
  • For each direction:
    • Calculates the current point curr using the d2pt function.
    • Converts the point coordinates to grid coordinates cx and cy.
    • Draws a dot at the current point.
    • Based on the direction, draws either a vertical line or horizontal line connecting the current and previous points.
  • Finally, prints the pts grid to display the Hilbert curve.

This program generates a Hilbert curve of order N (in this case, 32). The curve resembles a space-filling curve that visits every point in an N x N grid without crossing itself. The ASCII art representation in the grid shows the path of the curve with dots and connecting lines.

#include <stdio.h>

#define N 32
#define K 3
#define MAX N * K

typedef struct { int x; int y; } point;

void rot(int n, point *p, int rx, int ry) {
    int t;
    if (!ry) {
        if (rx == 1) {
            p->x = n - 1 - p->x;
            p->y = n - 1 - p->y;
        }
        t = p->x;
        p->x = p->y;
        p->y = t;
    }
}

void d2pt(int n, int d, point *p) {
    int s = 1, t = d, rx, ry;
    p->x = 0;
    p->y = 0;
    while (s < n) {
        rx = 1 & (t / 2);
        ry = 1 & (t ^ rx);
        rot(s, p, rx, ry);
        p->x += s * rx;
        p->y += s * ry;
        t /= 4;
        s *= 2;
    }
}

int main() {
    int d, x, y, cx, cy, px, py;
    char pts[MAX][MAX];
    point curr, prev;
    for (x = 0; x < MAX; ++x)
        for (y = 0; y < MAX; ++y) pts[x][y] = ' ';
    prev.x = prev.y = 0;
    pts[0][0] = '.';
    for (d = 1; d < N * N; ++d) {
        d2pt(N, d, &curr);
        cx = curr.x * K;
        cy = curr.y * K;
        px = prev.x * K;
        py = prev.y * K;
        pts[cx][cy] = '.';
        if (cx == px ) {
            if (py < cy)
                for (y = py + 1; y < cy; ++y) pts[cx][y] = '|';
            else
                for (y = cy + 1; y < py; ++y) pts[cx][y] = '|';
        }
        else {
            if (px < cx)
                for (x = px + 1; x < cx; ++x) pts[x][cy] = '_';
            else
                for (x = cx + 1; x < px; ++x) pts[x][cy] = '_';
        }
        prev = curr;
    }
    for (x = 0; x < MAX; ++x) {
        for (y = 0; y < MAX; ++y) printf("%c", pts[y][x]);
        printf("\n");
    }
    return 0;
}