Find an Integer point on a line segment with given two ends

Given two points pointU and pointV in XY-space, we need to find a point which has integer coordinates and lies on a line going through points pointU and pointV. Examples:

If  pointU = (1, -1 and pointV = (-4, 1)
then equation of line which goes 
through these two points is,
2X + 5Y = -3
One point with integer co-ordinate which
satisfies above equation is (6, -3)

We can see that once we found the equation of line, this problem can be treated as Extended Euclid algorithm problem, where we know A, B, C in AX + BY = C and we want to find out the value of X and Y from the equation. In above Extended Euclid equation, C is gcd of A and B, so after finding out the line equation from given two points if C is not a multiple of gcd(A, B) then we can conclude that there is no possible integer coordinate on the specified line. If C is a multiple of g, then we can scale up the founded X and Y coefficients to satisfy the actual equation, which will be our final answer. 

CPP

// C++ program to get Integer point on a line
#include <bits/stdc++.h>
using namespace std;

// Utility method for extended Euclidean Algorithm
int gcdExtended(int a, int b, int *x, int *y)
{
    // Base Case
    if (a == 0)
    {
        *x = 0;
        *y = 1;
        return b;
    }

    int x1, y1; // To store results of recursive call
    int gcd = gcdExtended(b%a, a, &x1, &y1);

    // Update x and y using results of recursive
    // call
    *x = y1 - (b/a) * x1;
    *y = x1;

    return gcd;
}

// method prints integer point on a line with two
// points U and V.
void printIntegerPoint(int pointU[], int pointV[])
{
    // Getting coefficient of line
    int A = (pointU[1] - pointV[1]);
    int B = (pointV[0] - pointU[0]);
    int C = (pointU[0] * (pointU[1] - pointV[1]) +
            pointU[1] * (pointV[0] - pointU[0]));

    int x, y; // To be assigned a value by gcdExtended()
    int g = gcdExtended(A, B, &x, &y);

    // if C is not divisible by g, then no solution
    // is available
    if (C % g != 0)
        cout << "No possible integer point\n";

    else

        // scaling up x and y to satisfy actual answer
        cout << "Integer Point : " << (x * C/g) << " "
            << (y * C/g) << endl;
}

// Driver code to test above methods
int main()
{
    int pointU[] = {1, -1};
    int pointV[] = {-4, 1};

    printIntegerPoint(pointU, pointV);
    return 0;
}

Java

// Java program to get Integer point on a line
// Utility method for extended Euclidean Algorithm
class GFG {

  public static int x;
  public static int y;

  // Function for extended Euclidean Algorithm
  static int gcdExtended(int a, int b)
  {

    // Base Case
    if (a == 0) {
      x = 0;
      y = 1;
      return b;
    }

    // To store results of recursive call
    int gcd = gcdExtended(b % a, a);
    int x1 = x;
    int y1 = y;

    // Update x and y using results of recursive
    // call
    int tmp = b / a;
    x = y1 - tmp * x1;
    y = x1;

    return gcd;
  }

  // method prints integer point on a line with two
  // points U and V.
  public static void printIntegerPoint(int[] pointU,
                                       int[] pointV)
  {
    // Getting coefficient of line
    int A = (pointU[1] - pointV[1]);
    int B = (pointV[0] - pointU[0]);
    int C = (pointU[0] * (pointU[1] - pointV[1])
             + pointU[1] * (pointV[0] - pointU[0]));

    x = 0; // To be assigned a value by gcdExtended()
    y = 0;
    int g = gcdExtended(A, B);

    // if C is not divisible by g, then no solution
    // is available
    if (C % g != 0) {
      System.out.print("No possible integer point\n");
    }

    else {

      // scaling up x and y to satisfy actual answer
      System.out.print("Integer Point : ");
      System.out.print((x * C / g));
      System.out.print(" ");
      System.out.print((y * C / g));
      System.out.print("\n");
    }
  }

  // Driver code to test above methods
  public static void main(String[] args)
  {
    int[] pointU = { 1, -1 };
    int[] pointV = { -4, 1 };

    printIntegerPoint(pointU, pointV);
  }
}

// The code is contributed by phasing17

C#

using System;

public static class GFG {
    // C++ program to get Integer point on a line
    // Utility method for extended Euclidean Algorithm
    public static int gcdExtended(int a, int b, ref int x,
                                  ref int y)
    {
        // Base Case
        if (a == 0) {
            x = 0;
            y = 1;
            return b;
        }

        int x1 = 0; // To store results of recursive call
        int y1 = 0;
        int gcd = gcdExtended(b % a, a, ref x1, ref y1);

        // Update x and y using results of recursive
        // call
        x = y1 - (b / a) * x1;
        y = x1;

        return gcd;
    }

    // method prints integer point on a line with two
    // points U and V.
    public static void printIntegerPoint(int[] pointU,
                                         int[] pointV)
    {
        // Getting coefficient of line
        int A = (pointU[1] - pointV[1]);
        int B = (pointV[0] - pointU[0]);
        int C = (pointU[0] * (pointU[1] - pointV[1])
                 + pointU[1] * (pointV[0] - pointU[0]));

        int x
            = 0; // To be assigned a value by gcdExtended()
        int y = 0;
        int g = gcdExtended(A, B, ref x, ref y);

        // if C is not divisible by g, then no solution
        // is available
        if (C % g != 0) {
            Console.Write("No possible integer point\n");
        }

        else {

            // scaling up x and y to satisfy actual answer
            Console.Write("Integer Point : ");
            Console.Write((x * C / g));
            Console.Write(" ");
            Console.Write((y * C / g));
            Console.Write("\n");
        }
    }

    // Driver code to test above methods
    public static void Main()
    {
        int[] pointU = { 1, -1 };
        int[] pointV = { -4, 1 };

        printIntegerPoint(pointU, pointV);
    }
}

// The code is contributed by Aarti_Rathi

Output

Integer Point : 6 -3

Time complexity: O(logn)
Auxiliary Space: O(logn)

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