Number of blocks in a chessboard a knight can move to in exactly k moves

Given integers i, j, k and n where (i, j) is the initial position of the Knight on a n * n chessboard, the task is to find the number of positions the Knight can move to in exactly k moves.

Examples:

Input: i = 5, j = 5, k = 1, n = 10
Output: 8

Input: i = 0, j = 0, k = 2, n = 10
Output: 10
The knight can see total 10 different positions in 2nd move.



Approach: Use a recursive approach to solve the problem.
First find all the possible positions where the knight can move to so if the initial position is i, j. Get to all valid locations in single move and recursively find all the possible positions where knight can move to in k – 1 steps from there. The base case of this recursion is when k == 0 (no move to make) then we will mark the position of the chessboard as visited if it is unmarked and increase the count. Finally, display the count .

Below is the implementation of the above approach:

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// function that will be called recursively
int recursive_solve(int i, int j, int steps, int n, 
                      map<pair<int, int>, int> &m)
{
    // If there's no more move to make and
    // this position hasn't been visited before
    if (steps == 0 && m[make_pair(i, j)] == 0) {
  
        // mark the position
        m[make_pair(i, j)] = 1;
  
        // increase the count        
        return 1;
    }
      
    int res = 0;
    if (steps > 0) {
  
        // valid movements for the knight
        int dx[] = { -2, -1, 1, 2, -2, -1, 1, 2 };
        int dy[] = { -1, -2, -2, -1, 1, 2, 2, 1 };
  
        // find all the possible positions
        // where knight can move from i, j
        for (int k = 0; k < 8; k++) {
  
            // if the positions lies within the
            // chessboard
            if ((dx[k] + i) >= 0
                && (dx[k] + i) <= n - 1
                && (dy[k] + j) >= 0
                && (dy[k] + j) <= n - 1) {
  
                // call the function with k-1 moves left
                res += recursive_solve(dx[k] + i, dy[k] + j,
                                       steps - 1, n, m);
            }
        }
    }
    return res;
}
  
// find all the positions where the knight can
// move after k steps
int solve(int i, int j, int steps, int n)
{
    map<pair<int, int>, int> m;
    return recursive_solve(i, j, steps, n, m);
}
  
// driver code
int main()
{
    int i = 0, j = 0, k = 2, n = 10;
  
    cout << solve(i, j, k, n);
  
    return 0;
}

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Python3





# Python3 implementation of above approach

from collections import defaultdict


# Function that will be called recursively

def recursive_solve(i, j, steps, n, m): 


	# If there’s no more move to make and

	# this position hasn’t been visited before

	if steps == 0 and m[(i, j)] == 0: 


		# mark the position

		m[(i, j)] = 1


		# increase the count

		return 1


	res = 0

	if steps > 0: 


		# valid movements for the knight

		dx = [-2, -1, 1, 2, -2, -1, 1, 2]

		dy = [-1, -2, -2, -1, 1, 2, 2, 1] 


		# find all the possible positions

		# where knight can move from i, j

		for k in range(0, 8): 


			# If the positions lies

			# within the chessboard

			if (dx[k] + i >= 0 and

				dx[k] + i <= n - 1 and
				dy[k] + j >= 0 and

				dy[k] + j <= n - 1): 

				# call the function with k-1 moves left 
				res += recursive_solve(dx[k] + i, dy[k] + j, 
									   steps - 1, n, m) 
	
	return res 

# Find all the positions where the 
# knight can move after k steps 
def solve(i, j, steps, n): 

	m = defaultdict(lambda:0) 
	return recursive_solve(i, j, steps, n, m) 

# Driver code 
if __name__ == "__main__": 

	i, j, k, n = 0, 0, 2, 10
	
	print(solve(i, j, k, n)) 

# This code is contributed by Rituraj Jain


[tabbyending]





Output:


10






          
          
          
          

            

    

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