Knight’s Tour Problem in C

// C Program to Solve Knight’s Tour using Backtracking

#include <stdio.h>

#define N 8

// Function to check if (x, y) is a valid move for the
// knight
int isSafe(int x, int y, int sol[N][N])
{
    return (x >= 0 && x < N && y >= 0 && y < N
            && sol[x][y] == -1);
}

// Function to print the solution matrix
void printSolution(int sol[N][N])
{
    for (int x = 0; x < N; x++) {
        for (int y = 0; y < N; y++)
            printf(" %2d ", sol[x][y]);
        printf("\n");
    }
}

// Utility function to solve the Knight's Tour problem
int solveKTUtil(int x, int y, int movei, int sol[N][N],
                int xMove[N], int yMove[N])
{
    int k, next_x, next_y;
    if (movei == N * N)
        return 1;

    // Try all next moves from the current coordinate x, y
    for (k = 0; k < 8; k++) {
        next_x = x + xMove[k];
        next_y = y + yMove[k];
        if (isSafe(next_x, next_y, sol)) {
            sol[next_x][next_y] = movei;
            if (solveKTUtil(next_x, next_y, movei + 1, sol,
                            xMove, yMove)
                == 1)
                return 1;
            else
                // Backtracking
                sol[next_x][next_y] = -1;
        }
    }
    return 0;
}

// This function solves the Knight's Tour problem using
// Backtracking
int solveKT()
{
    int sol[N][N];

    // Initialization of solution matrix
    for (int x = 0; x < N; x++)
        for (int y = 0; y < N; y++)
            sol[x][y] = -1;

    // xMove[] and yMove[] define the next move of the
    // knight xMove[] is for the next value of x coordinate
    // yMove[] is for the next value of y coordinate
    int xMove[8] = { 2, 1, -1, -2, -2, -1, 1, 2 };
    int yMove[8] = { 1, 2, 2, 1, -1, -2, -2, -1 };

    // Starting position of knight
    sol[0][0] = 0;

    // Start from (0, 0) and explore all tours using
    // solveKTUtil()
    if (solveKTUtil(0, 0, 1, sol, xMove, yMove) == 0) {
        printf("Solution does not exist");
        return 0;
    }
    else
        printSolution(sol);

    return 1;
}

int main()
{
    solveKT();
    return 0;
}

// C Program to Solve Knight’s Tour using Warnsdorff’s Rule

#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define N 8

// Move pattern on basis of the change of
// x coordinates and y coordinates respectively
static int cx[N] = { 1, 1, 2, 2, -1, -1, -2, -2 };
static int cy[N] = { 2, -2, 1, -1, 2, -2, 1, -1 };

// function restricts the knight to remain within
// the 8x8 chessboard
bool limits(int x, int y)
{
    return ((x >= 0 && y >= 0) && (x < N && y < N));
}

/* Checks whether a square is valid and empty or not */
bool isempty(int a[], int x, int y)
{
    return (limits(x, y)) && (a[y * N + x] < 0);
}

/* Returns the number of empty squares adjacent to (x, y) */
int getDegree(int a[], int x, int y)
{
    int count = 0;
    for (int i = 0; i < N; ++i)
        if (isempty(a, (x + cx[i]), (y + cy[i])))
            count++;

    return count;
}

// Picks next point using Warnsdorff's heuristic.
// Returns false if it is not possible to pick next point.
bool nextMove(int a[], int* x, int* y)
{
    int min_deg_idx = -1, c, min_deg = (N + 1), nx, ny;

    // Try all N adjacent of (*x, *y) starting from a random
    // adjacent. Find the adjacent with minimum degree.
    int start = rand() % N;
    for (int count = 0; count < N; ++count) {
        int i = (start + count) % N;
        nx = *x + cx[i];
        ny = *y + cy[i];
        if ((isempty(a, nx, ny))
            && (c = getDegree(a, nx, ny)) < min_deg) {
            min_deg_idx = i;
            min_deg = c;
        }
    }

    // If we could not find a next cell
    if (min_deg_idx == -1)
        return false;

    // Store coordinates of next point
    nx = *x + cx[min_deg_idx];
    ny = *y + cy[min_deg_idx];

    // Mark next move
    a[ny * N + nx] = a[(*y) * N + (*x)] + 1;

    // Update next point
    *x = nx;
    *y = ny;

    return true;
}

/* displays the chessboard with all the legal knight's moves
 */
void print(int a[])
{
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j)
            printf("%2d\t", a[j * N + i]);
        printf("\n");
    }
}

/* checks its neighbouring squares */
/* If the knight ends on a square that is one knight's move
 * from the beginning square, then tour is closed */
bool neighbour(int x, int y, int xx, int yy)
{
    for (int i = 0; i < N; ++i)
        if (((x + cx[i]) == xx) && ((y + cy[i]) == yy))
            return true;

    return false;
}

/* Generates the legal moves using Warnsdorff's heuristics.
 * Returns false if not possible */
bool findClosedTour()
{
    // Filling up the chessboard matrix with -1's
    int a[N * N];
    for (int i = 0; i < N * N; ++i)
        a[i] = -1;

    // Random initial position
    int sx = rand() % N;
    int sy = rand() % N;

    // Current points are same as initial points
    int x = sx, y = sy;
    a[y * N + x] = 1; // Mark first move.

    // Keep picking next points using Warnsdorff's heuristic
    for (int i = 0; i < N * N - 1; ++i)
        if (nextMove(a, &x, &y) == 0)
            return false;

    // Check if tour is closed (Can end at starting point)
    if (!neighbour(x, y, sx, sy))
        return false;

    print(a);
    return true;
}

int main()
{
    // To make sure that different random initial positions
    // are picked.
    srand(time(NULL));

    // While we don't get a solution
    while (!findClosedTour()) {
        ;
    }

    return 0;
}