**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**:

**Approach:**

- Create a solution matrix of the same structure as a 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 the 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

**Code:**

**Output:**

**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

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;

}

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!

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?