Find path traveling which bishop traverse each black cell atleast once

Given two integers X and Y such that X+Y is even, Which denotes the initial position of the bishop on a chessboard of 8 x 8 dimension. Then the task is to output a continuous path with not more than 32 coordinates(total black cells on board) such that traveling on that path bishop covers all the cells (i, j) having even sum (i + j = even). These cells need to be covered at least once over the course of travel, return the number of coordinates taken in the path, and the coordinates of the path to be followed.

Note: A single cell can be traversed more than once.

Examples:

Input: X = 5, Y = 3
Output:
19
5 3
4 4
1 1
8 8
7 7
8 6
3 1
1 3
6 8
5 7
4 8
1 5
5 1
8 4
7 3
8 2
7 1
1 7
2 8
Explanation:

Graphical Explanation of input case 1

Initially, the bishop is placed at (5, 3). Then it moves as follows: A (4, 4) -> B (1, 1) -> C (8, 8 ) -> D (7, 7) -> E (8, 6) -> F (3, 1) and so on. So, In total there are 19 moves, Which are shown in the output. Following this output coordinates all the cells (i, j) having even sum, ie. (i + j) = even, will be traversed. Formally, all black cells will be traversed.

Input: X =1, Y = 1
Output:
17
1 1
8 8
7 7
8 6
. . . . and so on.
Explanation: In total there will be 17 moves, and the starting of the path is as: (1, 1) -> (8, 8) -> (7, 7) -> (8, 6) and so on . . . .

Approach: Implement the idea below to solve the problem:

The problem is observation based and can be solved by using those observations. There exists a common path that applies for all the inputs of X and Y. By thinking about this problem deeply, we can observe a common path.

Steps were taken to solve the problem:

• Create a method commonPath() containing the below co-ordinates:
  • (1, 1)
    (8, 8)
    (7, 7)
    (8, 6)
    (3, 1)
    (1, 3)
    (6, 8)
    (5, 7)
    (4, 8)
    (1, 5)
    (5, 1)
    (8, 4)
    (7, 3)
    (8, 2)
    (7, 1)
    (1, 7)
    (2, 8)
• If ( X == Y && X == 1 ) then print all the points of the common_path() method.
• Else if ( X == Y && X != 1 ) then print 18 coordinates, Formally (X, Y) with all the points of the common_path() method.
• Else print 19 coordinates, Formally (X, Y), ( (X+Y)/2, (X+Y)/2 ) all the points of common_path() method.

Below is the code to implement the approach:

C++

#include<iostream>
using namespace std;

// Method for printing path
void pathPrinter(int X, int Y)
{
    // Condition for implementing
    // approach
    if (X == Y && X == 1) {
        cout << 17 << endl;
        // call commonPath
    }
    else if (X == Y && X != 1) {
        cout << 18 << endl;
        cout << X << " " << Y << endl;
        // call commonPath
    }
    else {
        cout << 19 << endl;
        cout << X << " " << Y << endl;
        int d = (X + Y) / 2;
        cout << d << " " << d << endl;
        // call commonPath
    }
}

// Method containing common path
// co-ordinates for all inputs
void commonPath()
{
    cout << "1 1" << endl;
    cout << "8 8" << endl;
    cout << "7 7" << endl;
    cout << "8 6" << endl;
    cout << "3 1" << endl;
    cout << "1 3" << endl;
    cout << "6 8" << endl;
    cout << "5 7" << endl;
    cout << "4 8" << endl;
    cout << "1 5" << endl;
    cout << "5 1" << endl;
    cout << "8 4" << endl;
    cout << "7 3" << endl;
    cout << "8 2" << endl;
    cout << "7 1" << endl;
    cout << "1 7" << endl;
    cout << "2 8" << endl;
}

// Driver Function
int main()
{
    // Inputs
    int X = 6;
    int Y = 2;

    // Function call
    pathPrinter(X, Y);
    commonPath();

}

Java

// Java code to implement the approach
import java.io.*;
import java.lang.*;
import java.util.*;

// Driver Function
class GFG {
    public static void main(String[] args)
    {

        // Inputs
        int X = 6;
        int Y = 2;

        // Function call
        pathPrinter(X, Y);
    }

    // Method for printing path
    static void pathPrinter(int X, int Y)
    {

        // Condition for implementing
        // approach
        if (X == Y && X == 1) {
            System.out.println(17);
            commonPath();
        }
        else if (X == Y && X != 1) {
            System.out.println(18);
            System.out.println(X + " " + Y);
            commonPath();
        }
        else {
            System.out.println(19);
            System.out.println(X + " " + Y);
            int d = (X + Y) / 2;
            System.out.println(d + " " + d);
            commonPath();
        }
    }

    // Method containing common path
    // co-ordinates for all inputs
    public static void commonPath()
    {
        System.out.println(1 + " " + 1);
        System.out.println(8 + " " + 8);
        System.out.println(7 + " " + 7);
        System.out.println(8 + " " + 6);
        System.out.println(3 + " " + 1);
        System.out.println(1 + " " + 3);
        System.out.println(6 + " " + 8);
        System.out.println(5 + " " + 7);
        System.out.println(4 + " " + 8);
        System.out.println(1 + " " + 5);
        System.out.println(5 + " " + 1);
        System.out.println(8 + " " + 4);
        System.out.println(7 + " " + 3);
        System.out.println(8 + " " + 2);
        System.out.println(7 + " " + 1);
        System.out.println(1 + " " + 7);
        System.out.println(2 + " " + 8);
    }
}

Python3

# Method for printing path
def pathPrinter(X, Y):

    # Condition for implementing
    # approach
    if X == Y and X == 1:
        print(17)
        # call commonPath

    elif X == Y and X != 1:
        print(18)
        print(X, Y)
        # call commonPath

    else:
        print(19)
        print(X, Y)
        d = (X + Y) // 2
        print(d, d)
        # call commonPath

# Method containing common path
# co-ordinates for all inputs
def commonPath():
    print("1 1")
    print("8 8")
    print("7 7")
    print("8 6")
    print("3 1")
    print("1 3")
    print("6 8")
    print("5 7")
    print("4 8")
    print("1 5")
    print("5 1")
    print("8 4")
    print("7 3")
    print("8 2")
    print("7 1")
    print("1 7")
    print("2 8")

# Driver Function
# Inputs
X = 6
Y = 2

# Function call
pathPrinter(X, Y)
commonPath()
# This code is contributed by prasad264

C#

// C# code to implement the approach
using System;

public class GFG {
    // Method for printing path
    static void pathPrinter(int X, int Y)
    {

        // Condition for implementing
        // approach
        if (X == Y && X == 1) {
            Console.WriteLine(17);
            commonPath();
        }
        else if (X == Y && X != 1) {
            Console.WriteLine(18);
            Console.WriteLine(X + " " + Y);
            commonPath();
        }
        else {
            Console.WriteLine(19);
            Console.WriteLine(X + " " + Y);
            int d = (X + Y) / 2;
            Console.WriteLine(d + " " + d);
            commonPath();
        }
    }

    // Method containing common path
    // co-ordinates for all inputs
    public static void commonPath()
    {
        Console.WriteLine(1 + " " + 1);
        Console.WriteLine(8 + " " + 8);
        Console.WriteLine(7 + " " + 7);
        Console.WriteLine(8 + " " + 6);
        Console.WriteLine(3 + " " + 1);
        Console.WriteLine(1 + " " + 3);
        Console.WriteLine(6 + " " + 8);
        Console.WriteLine(5 + " " + 7);
        Console.WriteLine(4 + " " + 8);
        Console.WriteLine(1 + " " + 5);
        Console.WriteLine(5 + " " + 1);
        Console.WriteLine(8 + " " + 4);
        Console.WriteLine(7 + " " + 3);
        Console.WriteLine(8 + " " + 2);
        Console.WriteLine(7 + " " + 1);
        Console.WriteLine(1 + " " + 7);
        Console.WriteLine(2 + " " + 8);
    }

    // Driver Code
    static public void Main()
    {
        // Inputs
        int X = 6;
        int Y = 2;

        // Function call
        pathPrinter(X, Y);
    }
}

// This code is contributed by Rohit Pradhan

Javascript

// Method for printing path
function pathPrinter(X, Y) {
  // Condition for implementing approach
  if (X === Y && X === 1) {
    console.log(17);
    // call commonPath
  } else if (X === Y && X !== 1) {
    console.log(18);
    console.log(X + " " + Y);
    // call commonPath
  } else {
    console.log(19);
    console.log(X + " " + Y);
    let d = Math.floor((X + Y) / 2);
    console.log(d + " " + d);
    // call commonPath
  }
}

// Method containing common path coordinates for all inputs
function commonPath() {
  console.log("1 1");
  console.log("8 8");
  console.log("7 7");
  console.log("8 6");
  console.log("3 1");
  console.log("1 3");
  console.log("6 8");
  console.log("5 7");
  console.log("4 8");
  console.log("1 5");
  console.log("5 1");
  console.log("8 4");
  console.log("7 3");
  console.log("8 2");
  console.log("7 1");
  console.log("1 7");
  console.log("2 8");
}

// Driver function
function main() {
  // Inputs
  let X = 6;
  let Y = 2;

  // Function call
  pathPrinter(X, Y);
  commonPath();
}

main();

Output

19
6 2
4 4
1 1
8 8
7 7
8 6
3 1
1 3
6 8
5 7
4 8
1 5
5 1
8 4
7 3
8 2
7 1
1 7
2 8

Time Complexity: O(1)
Auxiliary Space: O(1)