Backtracking – Knight’s Tour Problem

Objective : A knight’s tour is a sequence of moves of a knight on a chessboard such that the knight visits every square only once. If the knight ends on a square that is one knight’s move from the beginning square (so that it could tour the board again immediately, following the same path), the tour is closed, otherwise it is open. (Source : http://en.wikipedia.org/wiki/Knight%27s_tour)

Example:

Path-follwed-by-Knight-to-cover-all-the-cells

Path-foll0wed-by-Knight-to-cover-all-the-cells

Approach:

  • Create a solution matrix of the same structure as chessboard.
  • Start from 0,0 and index = 0. (index will represent the no of cells has been covered by the knight)
  • Check current cell is not already used if not then mark that cell (start with 0 and keep incrementing it, it will show us the path for the knight).
  • Check if index = N*N-1, means Knight has covered all the cells. return true and print the solution matrix.
    • Now try to solve rest of the problem recursively by making index +1. Check all 8 directions. (Knight can move to 8 cells from its current position.) Check the boundary conditions as well
    • 8-moves-of-a-Knight

      8-moves-of-a-Knight

    • If none of the 8 recursive calls return true, BACKTRACK and undo the changes ( put 0 to corresponding cell in solution matrix) and return false.
  • See the code for better understanding.

 

Code:


Output:
   00   59   38   33   30   17   08   63
   37   34   31   60   09   62   29   16
   58   01   36   39   32   27   18   07
   35   48   41   26   61   10   15   28
   42   57   02   49   40   23   06   19
   47   50   45   54   25   20   11   14
   56   43   52   03   22   13   24   05
   51   46   55   44   53   04   21   12

__________________________________________________
Top Companies Interview Questions..-

Google Microsoft Amazon Facebook more..

If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
__________________________________________________

  • Jo Smith

    The solution presented works quickly (milliseconds). However, note that choice of ordering of the possible moves affects how quickly the result comes back.

    For example, this alternative ordering of moves doesn’t seem to ever return a result:

    // go left and up
    if (canMove(row – 1, column – 2, N)
    && findPath(row – 1, column – 2, index + 1, N)) {
    return true;
    }
    // go left and down
    if (canMove(row + 1, column – 2, N)
    && findPath(row + 1, column – 2, index + 1, N)) {
    return true;
    }
    // go right and up
    if (canMove(row – 1, column + 2, N)
    && findPath(row – 1, column + 2, index + 1, N)) {
    return true;
    }
    // go down and right
    if (canMove(row + 2, column + 1, N)
    && findPath(row + 2, column + 1, index + 1, N)) {
    return true;
    }
    // go up and left
    if (canMove(row – 2, column – 1, N)
    && findPath(row – 2, column – 1, index + 1, N)) {
    return true;
    }
    // go down and left
    if (canMove(row + 2, column – 1, N)
    && findPath(row + 2, column – 1, index + 1, N)) {
    return true;
    }
    // go right and down
    if (canMove(row + 1, column + 2, N)
    && findPath(row + 1, column + 2, index + 1, N)) {
    return true;
    }
    // go up and right
    if (canMove(row – 2, column + 1, N)
    && findPath(row – 2, column + 1, index + 1, N)) {
    return true;
    }

    • Roxana Sarbu

      What is the explanation for this , because , theoretically , it should “enter ” in one of the cases above;
      It depends on the start point also?
      Thanks!

  • Hore Hores

    I noticed that if you changed the initial starting point in this code, the result gets a little weird. The 00 spot is somewhere random, and not connected with 01. Any thoughts?

%d bloggers like this: