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 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> |
Output
Integer Point : 6 -3
Time complexity: O(logn)
Auxiliary Space: O(logn)
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