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.Â
CPP
// C++ program to get Integer point on a line#include <bits/stdc++.h>using namespace std;Â
// Utility method for extended Euclidean Algorithmint gcdExtended(int a, int b, int *x, int *y){    // Base Case    if (a == 0)    {        *x = 0;        *y = 1;        return b;    }Â
    int x1, y1; // To store results of recursive call    int gcd = gcdExtended(b%a, a, &x1, &y1);Â
    // Update x and y using results of recursive    // call    *x = y1 - (b/a) * x1;    *y = x1;Â
    return gcd;}Â
// 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]));Â
    int x, y; // To be assigned a value by gcdExtended()    int g = gcdExtended(A, B, &x, &y);Â
    // if C is not divisible by g, then no solution    // is available    if (C % g != 0)        cout << "No possible integer point\n";Â
    elseÂ
        // scaling up x and y to satisfy actual answer        cout << "Integer Point : " << (x * C/g) << " "            << (y * C/g) << endl;}Â
// Driver code to test above methodsint main(){Â Â Â Â int pointU[] = {1, -1};Â Â Â Â int pointV[] = {-4, 1};Â
    printIntegerPoint(pointU, pointV);    return 0;} |
Java
// Java program to get Integer point on a line// Utility method for extended Euclidean Algorithmclass GFG {Â
  public static int x;  public static int y;Â
  // Function for extended Euclidean Algorithm  static int gcdExtended(int a, int b)  {Â
    // Base Case    if (a == 0) {      x = 0;      y = 1;      return b;    }Â
    // To store results of recursive call    int gcd = gcdExtended(b % a, a);    int x1 = x;    int y1 = y;Â
    // Update x and y using results of recursive    // call    int tmp = b / a;    x = y1 - tmp * x1;    y = x1;Â
    return gcd;  }Â
  // 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]));Â
    x = 0; // To be assigned a value by gcdExtended()    y = 0;    int g = gcdExtended(A, B);Â
    // if C is not divisible by g, then no solution    // is available    if (C % g != 0) {      System.out.print("No possible integer point\n");    }Â
    else {Â
      // 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");    }  }Â
  // Driver code to test above methods  public static void main(String[] args)  {    int[] pointU = { 1, -1 };    int[] pointV = { -4, 1 };Â
    printIntegerPoint(pointU, pointV);  }}Â
// The code is contributed by phasing17 |
C#
using System;Â
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;        }Â
        int x1 = 0; // To store results of recursive call        int y1 = 0;        int gcd = gcdExtended(b % a, a, ref x1, ref y1);Â
        // Update x and y using results of recursive        // call        x = y1 - (b / a) * x1;        y = x1;Â
        return gcd;    }Â
    // 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]));Â
        int x            = 0; // To be assigned a value by gcdExtended()        int y = 0;        int g = gcdExtended(A, B, ref x, ref y);Â
        // if C is not divisible by g, then no solution        // is available        if (C % g != 0) {            Console.Write("No possible integer point\n");        }Â
        else {Â
            // 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");        }    }Â
    // Driver code to test above methods    public static void Main()    {        int[] pointU = { 1, -1 };        int[] pointV = { -4, 1 };Â
        printIntegerPoint(pointU, pointV);    }}Â
// The code is contributed by Aarti_Rathi |
Javascript
<script>    // Javascript program to get Integer point on a line         // Function for extended Euclidean Algorithm      function gcdExtended(a,b)      {          // Base Case    if (a == 0) {      x = 0;      y = 1;      return b;    }          // To store results of recursive call    let gcd = gcdExtended(b % a, a);    let x1 = x;    let y1 = y;          // Update x and y using results of recursive    // call    let tmp = Math.floor(b/a);    x = y1 - tmp * x1;    y = x1;          return gcd;      }            // method prints integer point on a line with two      // points U and V.      function printIntegerPoint(pointU,pointV)      {           // Getting coefficient of line    let A = (pointU[1] - pointV[1]);    let B = (pointV[0] - pointU[0]);    let C = (pointU[0] * (pointU[1] - pointV[1])             + pointU[1] * (pointV[0] - pointU[0]));          x = 0; // To be assigned a value by gcdExtended()    y = 0;    let g = gcdExtended(A, B);          // if C is not divisible by g, then no solution    // is available    if (C % g != 0) {      document.write("No possible integer point\n");    }          else {            // scaling up x and y to satisfy actual answer      document.write("Integer Point : ");      document.write((x * C / g));      document.write(" ");      document.write((y * C / g));    }      }         // Driver code to test above methods    let x,y;    let pointU = [1, -1];    let pointV = [-4, 1];    printIntegerPoint(pointU, pointV);         // This Code is contributed by Pushpesh Raj.</script> |
Python3
class GFG:    x = None    y = NoneÂ
    @staticmethod    def gcdExtended(a: int, b: int) -> int:        global x, yÂ
        # Base Case        if a == 0:            x = 0            y = 1            return bÂ
        # To store results of recursive call        gcd = GFG.gcdExtended(b % a, a)        x1 = x        y1 = yÂ
        # Update x and y using results of recursive call        tmp = b // a        x = y1 - tmp * x1        y = x1Â
        return gcdÂ
    @staticmethod    def printIntegerPoint(pointU, pointV):        global x, yÂ
        # Getting coefficient of line        A = (pointU[1] - pointV[1])        B = (pointV[0] - pointU[0])        C = (pointU[0] * (pointU[1] - pointV[1]) + pointU[1] * (pointV[0] - pointU[0]))Â
        x = 0 # To be assigned a value by gcdExtended()        y = 0        g = GFG.gcdExtended(A, B)Â
        # if C is not divisible by g, then no solution is available        if C % g != 0:            print("No possible integer point")        else:            # scaling up x and y to satisfy actual answer            print("Integer Point : ", end="")            print((x * C // g), end=" ")            print((y * C // g))Â
    @staticmethod    def main() -> None:        pointU = [1, -1]        pointV = [-4, 1]Â
        GFG.printIntegerPoint(pointU, pointV)Â
if __name__ == "__main__":Â Â Â Â GFG.main() |
Output
Integer Point : 6 -3
Time complexity: O(logn)
Auxiliary Space: O(logn)
This article is contributed by Aarti_Rathi and Utkarsh Trivedi. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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