Warnsdorff’s algorithm for Knight’s tour problem

31 May, 2022
Problem : A knight is placed on the first block of an empty board and, moving according to the rules of chess, must visit each square exactly once. 
Following is an example path followed by Knight to cover all the cells. The below grid represents a chessboard with 8 x 8 cells. Numbers in cells indicate move number of Knight.  
We have discussed Backtracking Algorithm for solution of Knight’s tour. In this post Warnsdorff’s heuristic is discussed. Warnsdorff’s Rule: 
We can start from any initial position of the knight on the board.We always move to an adjacent, unvisited square with minimal degree (minimum number of unvisited adjacent).
We can start from any initial position of the knight on the board.
We always move to an adjacent, unvisited square with minimal degree (minimum number of unvisited adjacent).
This algorithm may also more generally be applied to any graph. 
Some definitions:  
A position Q is accessible from a position P if P can move to Q by a single Knight’s move, and Q has not yet been visited.
The accessibility of a position P is the number of positions accessible from P.
Algorithm:  
Set P to be a random initial position on the boardMark the board at P with the move number “1”Do following for each move number from 2 to the number of squares on the board: let S be the set of positions accessible from P.Set P to be the position in S with minimum accessibilityMark the board at P with the current move numberReturn the marked board — each square will be marked with the move number on which it is visited.
Set P to be a random initial position on the board
Mark the board at P with the move number “1”
Do following for each move number from 2 to the number of squares on the board: let S be the set of positions accessible from P.Set P to be the position in S with minimum accessibilityMark the board at P with the current move number
let S be the set of positions accessible from P.
Set P to be the position in S with minimum accessibility
Mark the board at P with the current move number
Return the marked board — each square will be marked with the move number on which it is visited.
Below is implementation of above algorithm.  
C++
Java
C#
// C++ program to for Knight's tour problem using// Warnsdorff's algorithm#include <bits/stdc++.h>#define N 8 // Move pattern on basis of the change of// x coordinates and y coordinates respectivelystatic 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 chessboardbool 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("%d\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;} // Driver codeint 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;}
// Java program to for Knight's tour problem using// Warnsdorff's algorithmimport java.util.concurrent.ThreadLocalRandom; class GFG{    public static final int N = 8;     // Move pattern on basis of the change of    // x coordinates and y coordinates respectively    public static final int cx[] = {1, 1, 2, 2, -1, -1, -2, -2};    public static final int cy[] = {2, -2, 1, -1, 2, -2, 1, -1};     // function restricts the knight to remain within    // the 8x8 chessboard    boolean limits(int x, int y)    {        return ((x >= 0 && y >= 0) &&                 (x < N && y < N));    }     /* Checks whether a square is valid and    empty or not */    boolean 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.    Cell nextMove(int a[], Cell cell)    {        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 = ThreadLocalRandom.current().nextInt(1000) % N;        for (int count = 0; count < N; ++count)        {            int i = (start + count) % N;            nx = cell.x + cx[i];            ny = cell.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 null;         // Store coordinates of next point        nx = cell.x + cx[min_deg_idx];        ny = cell.y + cy[min_deg_idx];         // Mark next move        a[ny * N + nx] = a[(cell.y) * N +                           (cell.x)] + 1;         // Update next point        cell.x = nx;        cell.y = ny;         return cell;    }     /* 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)                System.out.printf("%d\t", a[j * N + i]);            System.out.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 */    boolean 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 */    boolean findClosedTour()    {        // Filling up the chessboard matrix with -1's        int a[] = new int[N * N];        for (int i = 0; i < N * N; ++i)            a[i] = -1;         // initial position        int sx = 3;        int sy = 2;         // Current points are same as initial points        Cell cell = new Cell(sx, sy);         a[cell.y * N + cell.x] = 1; // Mark first move.         // Keep picking next points using        // Warnsdorff's heuristic        Cell ret = null;        for (int i = 0; i < N * N - 1; ++i)        {            ret = nextMove(a, cell);            if (ret == null)                return false;        }         // Check if tour is closed (Can end        // at starting point)        if (!neighbour(ret.x, ret.y, sx, sy))            return false;         print(a);        return true;    }     // Driver Code    public static void main(String[] args)    {        // While we don't get a solution        while (!new GFG().findClosedTour())        {            ;        }    }} class Cell{    int x;    int y;     public Cell(int x, int y)    {        this.x = x;        this.y = y;    }} // This code is contributed by SaeedZarinfam
//C# program for Knight’s tour //problem using Warnsdorff’salgorithmusing System;using System.Collections;using System.Collections.Generic; public class GFG{         public static int N = 8;      // Move pattern on basis of the change of    // x coordinates and y coordinates respectively    public int[] cx = new int[] {1, 1, 2, 2, -1, -1, -2, -2};    public int[] cy = new int[] {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.    Cell nextMove(int[] a, Cell cell)    {        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.        Random random = new Random();        int start=random.Next(0, 1000);        for (int count = 0; count < N; ++count)        {            int i = (start + count) % N;            nx = cell.x + cx[i];            ny = cell.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 null;          // Store coordinates of next point        nx = cell.x + cx[min_deg_idx];        ny = cell.y + cy[min_deg_idx];          // Mark next move        a[ny * N + nx] = a[(cell.y) * N +                           (cell.x)] + 1;          // Update next point        cell.x = nx;        cell.y = ny;          return cell;    }      /* 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)                Console.Write(a[j * N + i]+"\t");            Console.Write("\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 = new int[N * N];        for (int i = 0; i < N * N; ++i)            a[i] = -1;          // initial position        int sx = 3;        int sy = 2;          // Current points are same as initial points        Cell cell = new Cell(sx, sy);          a[cell.y * N + cell.x] = 1; // Mark first move.          // Keep picking next points using        // Warnsdorff's heuristic        Cell ret = null;        for (int i = 0; i < N * N - 1; ++i)        {            ret = nextMove(a, cell);            if (ret == null)                return false;        }          // Check if tour is closed (Can end        // at starting point)        if (!neighbour(ret.x, ret.y, sx, sy))            return false;          print(a);        return true;    }         static public void Main (){        // While we don't get a solution        while (!new GFG().findClosedTour())        {            ;        }    }} class Cell{    public int x;    public int y;      public Cell(int x, int y)    {        this.x = x;        this.y = y;    }} //This code is contributed by shruti456rawal
Output: 
59    14    63    32    1    16    19    34    
62    31    60    15    56    33    2    17    
13    58    55    64    49    18    35    20    
30    61    42    57    54    51    40    3    
43    12    53    50    41    48    21    36    
26    29    44    47    52    39    4    7    
11    46    27    24    9    6    37    22    
28    25    10    45    38    23    8    5    
The Hamiltonian path problem is NP-hard in general. In practice, Warnsdorff’s heuristic successfully finds a solution in linear time.
Do you know? “On an 8 × 8 board, there are exactly 26,534,728,821,064 directed closed tours (i.e. two tours along the same path that travel in opposite directions are counted separately, as are rotations and reflections). The number of undirected closed tours is half this number, since every tour can be traced in reverse!”
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SaeedZarinfam
rajeev0719singh
anikakapoor
varshagumber28
simmytarika5
shruti456rawal
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