Step by Step solution about How to resolve the algorithm Knight's tour step by step in the C programming language

The C code you provided implements an algorithm to solve a "Knight's Tour" problem. A Knight's Tour is a path for a knight in a chess game that visits all squares of a given board exactly once.

The code defines a few helper functions and a main function:

• init_board initializes the game board. It creates a 2D array of cells, with some padding around the edges. The padding is used to represent the "outside" of the board, which the knight cannot move to.

• walk_board implements the actual Knight's Tour algorithm. It takes a starting position and a game board as input, and tries to find a path that visits all squares of the board exactly once. The algorithm uses a depth-first search approach, and it backtracks if it reaches a dead end.

• solve is the main function that calls init_board and walk_board to solve the Knight's Tour problem for a given board size. It also handles user input and provides a visual representation of the solution.

The main function takes two command-line arguments: the width and height of the game board. If no arguments are provided, the default board size is 8x8.

The solve function initializes the game board and then calls walk_board to find a solution. If a solution is found, it prints a message to the console and returns 1. If no solution is found, it prints a message to the console and returns 0.

The walk_board function takes a starting position and a game board as input, and tries to find a path that visits all squares of the board exactly once. It uses a depth-first search approach, and it backtracks if it reaches a dead end.

The algorithm works by first occupying the current cell on the board. Then, it reduces the neighbor count of all of the current cell's neighbors. Next, it finds the neighbor with the lowest neighbor count, and moves to that neighbor. The algorithm repeats this process until it has visited all of the squares on the board.

If the algorithm reaches a dead end, it backtracks to the previous cell and tries a different neighbor. If it backtracks to the starting cell, then there is no solution to the Knight's Tour problem for the given board size.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>

typedef unsigned char cell;
int dx[] = { -2, -2, -1, 1, 2,  2,  1, -1 };
int dy[] = { -1,  1,  2, 2, 1, -1, -2, -2 };

void init_board(int w, int h, cell **a, cell **b)
{
	int i, j, k, x, y, p = w + 4, q = h + 4;
	/* b is board; a is board with 2 rows padded at each side */
	a[0] = (cell*)(a + q);
	b[0] = a[0] + 2;

	for (i = 1; i < q; i++) {
		a[i] = a[i-1] + p;
		b[i] = a[i] + 2;
	}

	memset(a[0], 255, p * q);
	for (i = 0; i < h; i++) {
		for (j = 0; j < w; j++) {
			for (k = 0; k < 8; k++) {
				x = j + dx[k], y = i + dy[k];
				if (b[i+2][j] == 255) b[i+2][j] = 0;
				b[i+2][j] += x >= 0 && x < w && y >= 0 && y < h;
			}
		}
	}
}

#define E "\033["
int walk_board(int w, int h, int x, int y, cell **b)
{
	int i, nx, ny, least;
	int steps = 0;
	printf(E"H"E"J"E"%d;%dH"E"32m[]"E"m", y + 1, 1 + 2 * x);

	while (1) {
		/* occupy cell */
		b[y][x] = 255;

		/* reduce all neighbors' neighbor count */
		for (i = 0; i < 8; i++)
			b[ y + dy[i] ][ x + dx[i] ]--;

		/* find neighbor with lowest neighbor count */
		least = 255;
		for (i = 0; i < 8; i++) {
			if (b[ y + dy[i] ][ x + dx[i] ] < least) {
				nx = x + dx[i];
				ny = y + dy[i];
				least = b[ny][nx];
			}
		}

		if (least > 7) {
			printf(E"%dH", h + 2);
			return steps == w * h - 1;
		}

		if (steps++) printf(E"%d;%dH[]", y + 1, 1 + 2 * x);
		x = nx, y = ny;
		printf(E"%d;%dH"E"31m[]"E"m", y + 1, 1 + 2 * x);
		fflush(stdout);
		usleep(120000);
	}
}

int solve(int w, int h)
{
	int x = 0, y = 0;
	cell **a, **b;
	a = malloc((w + 4) * (h + 4) + sizeof(cell*) * (h + 4));
	b = malloc((h + 4) * sizeof(cell*));

	while (1) {
		init_board(w, h, a, b);
		if (walk_board(w, h, x, y, b + 2)) {
			printf("Success!\n");
			return 1;
		}
		if (++x >= w) x = 0, y++;
		if (y >= h) {
			printf("Failed to find a solution\n");
			return 0;
		}
		printf("Any key to try next start position");
		getchar();
	}
}

int main(int c, char **v)
{
	int w, h;
	if (c < 2 || (w = atoi(v[1])) <= 0) w = 8;
	if (c < 3 || (h = atoi(v[2])) <= 0) h = w;
	solve(w, h);

	return 0;
}