Place K-knights such that they do not attack each other in C++

C++Server Side ProgrammingProgramming

In this problem, we are given three integer value K, N, M. our task is to place K knights in an NxM chessboard such that no two knights attack each other. There can be cases with 0 valid ways and also cases with multiple valid ways. You need to print all valid cases.

Knight is a chess piece that moves two moves ahead and then one move to the left of right. It can move in any direction in the chessboard.

Attack is the position when one piece can be in the same place as other pieces in one chance of its valid moves.

Let’s take an example to understand the problem,

Input − M = 3, N = 3, K = 5

Output

K A K
A K A
K A K

A K A
K K K
A K A

To solve this problem, we will start placing knights one by one in each row, column by column. And check the position than are attacking after each placement. On placing knights we will check if it is safe or not. If it is safe, then we will place it and then move to the next position. We will use backtracking so, that all possible ways can be obtained and for this, we will create a new board after each placement of knight for backtracking. This is how we will obtain all possible solutions using backtracking.

Example

Program to show the implementation of our solution,

 Live Demo

#include <iostream>
using namespace std;
int m, n, k, count = 0;
void displayPositions(char** board){
   cout<<endl;
   for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
         cout<<board[i][j]<<"\t";
      }
      cout<<endl;
   }
}
void canattack(int i, int j, char a,
char** board){
   if ((i + 2) < m && (j - 1) >= 0) {
      board[i + 2][j - 1] = a;
   }
   if ((i - 2) >= 0 && (j - 1) >= 0) {
      board[i - 2][j - 1] = a;
   }
   if ((i + 2) < m && (j + 1)< n) {
      board[i + 2][j + 1] = a;
   }
   if ((i - 2) >= 0 && (j + 1) < n) {
      board[i - 2][j + 1] = a;
   }
   if ((i + 1) < m && (j + 2) <n) {
      board[i + 1][j + 2] = a;
   }
   if ((i - 1) >= 0 && (j + 2) < n) {
      board[i - 1][j + 2] = a;
   }
   if ((i + 1) < m && (j - 2) >= 0) {
      board[i + 1][j - 2] = a;
   }
   if ((i - 1) >= 0 && (j - 2) >= 0) {
      board[i - 1][j - 2] = a;
   }
}
bool canPlace(int i, int j, char** board){
   if (board[i][j] == '_')
      return true;
   else
      return false;
}
void place(int i, int j, char k, char a,
char** board, char** new_board){
   for (int y = 0; y < m; y++) {
      for (int z = 0; z < n; z++) {
         new_board[y][z] = board[y][z];
      }
   }
   new_board[i][j] = k;
   canattack(i, j, a, new_board);
}
void placeKnights(int k, int sti, int stj, char** board){
   if (k == 0) {
      displayPositions(board);
      count++;
   } else {
      for (int i = sti; i < m; i++) {
         for (int j = stj; j < n; j++) {
            if (canPlace(i, j, board)) {
               char** new_board = new char*[m];
               for (int x = 0; x < m; x++) {
                  new_board[x] = new char[n];
               }
               place(i, j, 'K', 'A', board, new_board);
               placeKnights(k - 1, i, j, new_board);
            }
         }
         stj = 0;
      }
   }
}
int main() {
   m = 3, n = 3, k = 5;
   char** board = new char*[m];
   for (int i = 0; i < m; i++)
   board[i] = new char[n];
   for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++)
      board[i][j] = '_';
   }
   cout<<"The ways in which "<<k<<" knights can be placed in "<<m<<"x"<<n<<" chessboard are :\n";
   placeKnights(k, 0, 0, board);
   return 0;
}

Output

The ways in which 5 knights can be placed in 3x3 chessboard are :
K A K
A K A
K A K

A K A
K K K
A K A

Here, we have marked positions of knights by K and positions where they are attacked by A.

raja
Published on 17-Apr-2020 10:42:13
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