# Warnsdorff’s algorithm for Knight’s tour problem

• Difficulty Level : Expert
• Last Updated : 09 Jul, 2021

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.

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In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students. We have discussed Backtracking Algorithm for solution of Knight’s tour. In this post Warnsdorff’s heuristic is discussed.
Warnsdorff’s Rule:

1. We can start from any initial position of the knight on the board.
2. 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:

1. Set P to be a random initial position on the board
2. Mark the board at P with the move number “1”
3. 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 accessibility
• Mark the board at P with the current move number
4. 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++

 `// C++ program to for Kinight's tour problem using``// Warnsdorff's algorithm``#include ``#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``(``"%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;` `    ``// Randome 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 code``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;``}`

## Java

 `// Java program to for Kinight's tour problem using``// Warnsdorff's algorithm``import` `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`

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, Warnsdorf’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!”