Count all possible position that can be reached by Modified Knight

Given a chessboard of size 8 x 8 and the current position of Mirandote. All the rules of this chess game are same but the knight is modified, we call new knight as “Mirandote”. The moves of Mirandote is given by blue color where its current position is denoted by red color in the following image :

The task is to find how many possible positions exist in Chessboard that can be reached by Mirandote in exactly S steps.

Examples:

Input: row = 4, col = 4, steps = 1
Output: 12
All the 12 moves denoted by the following image by blue color :


Input: row = 4, col = 4, steps = 2
Output: 55



Solution:

We can observe that all the possible position with respect to current position can be written in the form of row and column. This thing is illustrated by the following image :

We can call a function recursively for each possible position and count all the possible position.

Below is the required implementation to find the positions:

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// C++ implementation to find the
// possible positions
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the positions
void findSteps(int current_row, int current_column,
               int curr, int board_size, int steps,
               int* visited)
{
    // Bound checking
    if (current_row >= board_size || current_row < 0
        || current_column >= board_size || current_column < 0
        || curr > steps) {
        return;
    }
  
    // If steps is equal to current steps,
    // that means current position is reached by Mirandote
    if (curr == steps) {
        *((visited + (current_row)*board_size) + current_column) = 1;
        return;
    }
  
    // Recursive calls for each possible position.
    // Position of a, b, c, ..., l given in above image.
    /* a = */ findSteps(current_row - 2, current_column - 1,
                        curr + 1, board_size, steps, visited);
  
    /* b = */ findSteps(current_row - 2, current_column + 1,
                        curr + 1, board_size, steps, visited);
  
    /* c = */ findSteps(current_row - 1, current_column - 2,
                        curr + 1, board_size, steps, visited);
  
    /* d = */ findSteps(current_row - 1, current_column - 1,
                        curr + 1, board_size, steps, visited);
  
    /* e = */ findSteps(current_row - 1, current_column + 1,
                        curr + 1, board_size, steps, visited);
  
    /* f = */ findSteps(current_row - 1, current_column + 2,
                        curr + 1, board_size, steps, visited);
  
    /* g = */ findSteps(current_row + 1, current_column - 2,
                        curr + 1, board_size, steps, visited);
  
    /* h = */ findSteps(current_row + 1, current_column - 1,
                        curr + 1, board_size, steps, visited);
  
    /* i = */ findSteps(current_row + 1, current_column + 1,
                        curr + 1, board_size, steps, visited);
  
    /* j = */ findSteps(current_row + 1, current_column + 2,
                        curr + 1, board_size, steps, visited);
  
    /* k = */ findSteps(current_row + 2, current_column - 1,
                        curr + 1, board_size, steps, visited);
  
    /* l = */ findSteps(current_row + 2, current_column + 1,
                        curr + 1, board_size, steps, visited);
  
    return;
}
  
int countSteps(int current_row, int current_column,
               int board_size, int steps)
{
  
    // Visited array
    int visited[board_size][board_size];
  
    // Initialize visited array to zero
    for (int i = 0; i < board_size; i++) {
        for (int j = 0; j < board_size; j++) {
            visited[i][j] = 0;
        }
    }
  
    int answer = 0;
  
    // Function call where initial step count is 0
    findSteps(current_row, current_column, 0,
              board_size, steps, (int*)visited);
  
    for (int i = 0; i < board_size; i++) {
        for (int j = 0; j < board_size; j++) {
  
            // If value of element is 1, that implies,
            // the position can be reached by Mirandote.
            if (visited[i][j] == 1) {
                answer++;
            }
        }
    }
  
    return answer;
}
  
// Driver code
int main()
{
    int board_size = 8, steps = 1;
    int current_row = 4, current_column = 4;
  
    cout << countSteps(current_row, current_column,
                       board_size, steps);
    return 0;
}

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Output:

12

Time complexity of above algorithm is O(12^S), where S is the number of steps.



My Personal Notes arrow_drop_up

My name is Saurav Chandra I am from Rajasthan I am currently pursuing my B Tech from National Institute of Technology Hamirpur (H P) in Computer Science and Engineering Department My area of interests are Algorithms and Data Structures

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