Chess Knight Problem | Find Shortest path from source to destination

Given a chess board, find the shortest distance (minimum number of steps) taken by a Knight to reach given destination from given source.

 

For example,

Input: N = 8 (8 x 8 board), Source = (7, 0) Destination  = (0, 7)

Output: Minimum number of steps required is 6

Explanation: The Knight’s movement can be demonstrated in figure below

  chess-board

 


 

The idea is to use BFS as it is a Shortest Path problem. Below is the complete algorithm.


1. Create an empty queue and enqueue source cell having
   distance 0 from source (itself)

2. do till queue is not empty

   a) Pop next unvisited node from queue

   b) If the popped node is destination node, return its distance

   c) else we mark current node as visited and for each of 8 possible
      movements for a knight, we enqueue each valid movement into the
      queue with +1 distance (min distance of given node from source
      = min distance of parent from source + 1)

A knight can move in 8 possible directions from a given cell as illustrated in below figure –

  knight-movements

We can find all the possible locations the Knight can move to from the given location by using the array that stores the relative position of Knight movement from any location. For example, if the current location is (x, y), we can move to (x + row[k], y + col[k]) for 0 <= k <=7 using below array.

row[] = [ 2, 2, -2, -2, 1, 1, -1, -1 ]
col[] = [ -1, 1, 1, -1, 2, -2, 2, -2 ]

So, from position (x, y) Knight’s can move to:

(x + 2, y – 1)
(x + 2, y + 1)
(x – 2, y + 1)
(x – 2, y – 1)
(x + 1, y + 2)
(x + 1, y – 2)
(x – 1, y + 2)
(x – 1, y – 2)

Note that in BFS, all cells having shortest path as 1 are visited first, followed by their adjacent cells having shortest path as 1 + 1 = 2 and so on.. so if we reach any node in BFS, its shortest path = shortest path of parent + 1. So, the first occurrence of the destination cell gives us the result and we can stop our search there. It is not possible that the shortest path exists from some other cell for which we haven’t reached the given node yet. If any such path was possible, we would have already explored it.

 
C++ implementation –
 

Download   Run Code

Output:

6

 
Exercise: Extend the solution to print the paths as well.

 
Thanks for reading.




Please use ideone or C++ Shell or any other online compiler link to post code in comments.
Like us? Please spread the word and help us grow. Happy coding 🙂
 





Leave a Reply

Notify of
avatar
Sort by:   newest | oldest | most voted
bihn
Guest
bihn

Great stuff! There is also a variation of this problem where you calculate the minimum number of moves to get to a position in an infinite coordinate plane using a formula in constant time (there is some hard coding involved for values within 4 spaces of the origin or so. Any plans to attack that one?

Oleg Abrazhaev
Guest

great!

iko
Guest

“Exercise: Extend the solution to print the paths as well.”
What is the answer?

pratik Sharma
Guest
pratik Sharma

bool const operator==(const Node& o) const
    {
        return x == o.x && y == o.y;
    }
 
    bool operator<(const Node& o) const
    {
        return x < o.x || (x == o.x && y < o.y);
    }
//someone please explain this bunch

nilesh
Guest
nilesh

Thanks, I have found your website to be very useful… especially the codes are neat and well explained.

Sovit
Guest
Sovit

Interesting that you kept track of visited/unvisited marks. I had used similar approach of using the closest point approach for moving the knight in the beginning. But as the knight reaches a proximity distance (which is very close to the target), I used a different set of algorithm.

Please look into my solution and let me know what you think:

http://sovitpoudel.com.np/2016/06/03/a-knights-watch/

Thanks!

wpDiscuz