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

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 –

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 –**

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#include <bits/stdc++.h> using namespace std; #define N 8 // Below arrays details all 8 possible movements // for a knight int row[] = { 2, 2, -2, -2, 1, 1, -1, -1 }; int col[] = { -1, 1, 1, -1, 2, -2, 2, -2 }; // Check if (x, y) is valid chess board coordinates // Note that a knight cannot go out of the chessboard bool valid(int x, int y) { if (x < 0 || y < 0 || x >= N || y >= N) return false; return true; } // queue node used in BFS struct Node { // (x, y) represents chess board coordinates // dist represent its minimum distance from the source int x, y, dist; // As we are using struct as a key in a std::map, // we need to overload below operators // Alternatively we can use std::pair<int, int> as a key // to store coordinates of the matrix in the map 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); } }; // Find minimum number of steps taken by the knight // from source to reach destination using BFS int BFS(Node src, Node dest) { // map to check if matrix cell is visited before or not map<Node, bool> visited; // create a queue and enqueue first node queue<Node> q; q.push(src); // run till queue is not empty while (!q.empty()) { // pop front node from queue and process it Node node = q.front(); q.pop(); int x = node.x; int y = node.y; int dist = node.dist; // if destination is reached, return distance if (x == dest.x && y == dest.y) return dist; // Skip if location is visited before if (!visited.count(node)) { // mark current node as visited visited[node] = true; // check for all 8 possible movements for a knight // and enqueue each valid movement into the queue for (int i = 0; i < 8; ++i) { // Get the new valid position of Knight from current // position on chessboard and enqueue it in the // queue with +1 distance int x1 = x + row[i]; int y1 = y + col[i]; if (valid(x1, y1)) q.push({x1, y1, dist + 1}); } } } // return INFINITY if path is not possible return INT_MAX; } // main function int main() { // source coordinates Node src = {0, 7}; // destination coordinates Node dest = {7, 0}; cout << BFS(src, dest) << endl; return 0; } |

**Output: **

6

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

**Thanks for reading.**

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## Leave a Reply

great!

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

What is the answer?

We can store complete path from source cell to destination in a vector. Please refer below program for similar implementation –

https://www.techiedelight.com/find-shortest-path-source-destination-matrix-satisfies-given-constraints/

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?