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 Algorithmint 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 methodsint 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 Algorithmclass 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 |
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 = None @staticmethod def gcdExtended(a: int, b: int) -> int: global x, y # Base Case if a == 0: x = 0 y = 1 return b # To store results of recursive call gcd = GFG.gcdExtended(b % a, a) x1 = x y1 = y # Update x and y using results of recursive call tmp = b // a x = y1 - tmp * x1 y = x1 return gcd @staticmethod def printIntegerPoint(pointU, pointV): global x, y # 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])) x = 0 # To be assigned a value by gcdExtended() y = 0 g = GFG.gcdExtended(A, B) # 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)) @staticmethod def main() -> None: pointU = [1, -1] pointV = [-4, 1] GFG.printIntegerPoint(pointU, pointV)if __name__ == "__main__": 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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