Number of Integral Points between Two Points

• Difficulty Level : Medium
• Last Updated : 22 Oct, 2021

Given two points p (x1, y1) and q (x2, y2), calculate the number of integral points lying on the line joining them.
Example : If points are (0, 2) and (4, 0), then the number of integral points lying on it is only one and that is (2, 1).
Similarly, if points are (1, 9) and (8, 16), the integral points lying on it are 6 and they are (2, 10), (3, 11), (4, 12), (5, 13), (6, 14) and (7, 15).

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Simple Approach
Start from any of the given points, reach the other end point by using loops. For every point inside the loop, check if it lies on the line that joins given two points. If yes, then increment the count by 1. Time Complexity for this approach will be O(min(x2-x1, y2-y1)).

Optimal Approach

1. If the edge formed by joining p and q is parallel
to the X-axis, then the number of integral points
between the vertices is :
abs(p.y - q.y)-1

2. Similarly if edge is parallel to the Y-axis, then
the number of integral points in between is :
abs(p.x - q.x)-1

3. Else, we can find the integral points between the
vertices using below formula:
GCD(abs(p.x - q.x), abs(p.y - q.y)) - 1

How does the GCD formula work?
The idea is to find the equation of the line in simplest form, i.e., in equation ax + by +c, coefficients a, b and c become co-prime. We can do this by calculating the GCD (greatest common divisor) of a, b and c and convert a, b and c in the simplest form.
Then, the answer will be (difference of y coordinates) divided by (a) – 1. This is because after calculating ax + by + c = 0, for different y values, x will be number of y values which are exactly divisible by a.
Below is the implementation of above idea.

C++

 // C++ code to find the number of integral points// lying on the line joining the two given points#include #include using namespace std; // Class to represent an Integral point on XY plane.class Point{public:    int x, y;    Point(int a=0, int b=0):x(a),y(b) {}}; // Utility function to find GCD of two numbers// GCD of a and bint gcd(int a, int b){    if (b == 0)       return a;    return gcd(b, a%b);} // Finds the no. of Integral points between// two given points.int getCount(Point p, Point q){    // If line joining p and q is parallel to    // x axis, then count is difference of y    // values    if (p.x==q.x)        return abs(p.y - q.y) - 1;     // If line joining p and q is parallel to    // y axis, then count is difference of x    // values    if (p.y == q.y)        return abs(p.x-q.x) - 1;     return gcd(abs(p.x-q.x), abs(p.y-q.y))-1;} // Driver program to test aboveint main(){    Point p(1, 9);    Point q(8, 16);     cout << "The number of integral points between "         << "(" << p.x << ", " << p.y << ") and ("         << q.x << ", " << q.y << ") is "         << getCount(p, q);     return 0;}

Java

 // Java code to find the number of integral points// lying on the line joining the two given points class GFG{ // Class to represent an Integral point on XY plane.static class Point{    int x, y;    Point(int a, int b)    {        this.x = a;        this.y = b;    }}; // Utility function to find GCD of two numbers// GCD of a and bstatic int gcd(int a, int b){    if (b == 0)    return a;    return gcd(b, a % b);} // Finds the no. of Integral points between// two given points.static int getCount(Point p, Point q){    // If line joining p and q is parallel to    // x axis, then count is difference of y    // values    if (p.x == q.x)        return Math.abs(p.y - q.y) - 1;     // If line joining p and q is parallel to    // y axis, then count is difference of x    // values    if (p.y == q.y)        return Math.abs(p.x - q.x) - 1;     return gcd(Math.abs(p.x - q.x), Math.abs(p.y - q.y)) - 1;} // Driver program to test abovepublic static void main(String[] args){    Point p = new Point(1, 9);    Point q = new Point(8, 16);     System.out.println("The number of integral points between "        + "(" + p.x + ", " + p.y + ") and ("        + q.x + ", " + q.y + ") is "        + getCount(p, q));}} // This code contributed by Rajput-Ji

Python3

 # Python3 code to find the number of# integral points lying on the line# joining the two given points # Class to represent an Integral point# on XY plane.class Point:         def __init__(self, a, b):        self.x = a        self.y = b # Utility function to find GCD# of two numbers GCD of a and bdef gcd(a, b):     if b == 0:        return a    return gcd(b, a % b) # Finds the no. of Integral points# between two given points.def getCount(p, q):     # If line joining p and q is parallel    # to x axis, then count is difference    # of y values    if p.x == q.x:        return abs(p.y - q.y) - 1     # If line joining p and q is parallel    # to y axis, then count is difference    # of x values    if p.y == q.y:        return abs(p.x - q.x) - 1     return gcd(abs(p.x - q.x),               abs(p.y - q.y)) - 1 # Driver Codeif __name__ == "__main__":     p = Point(1, 9)    q = Point(8, 16)     print("The number of integral points",          "between ({}, {}) and ({}, {}) is {}" .           format(p.x, p.y, q.x, q.y, getCount(p, q))) # This code is contributed by Rituraj Jain

C#

 // C# code to find the number of integral points// lying on the line joining the two given pointsusing System; class GFG{ // Class to represent an Integral point on XY plane.public class Point{    public int x, y;    public Point(int a, int b)    {        this.x = a;        this.y = b;    }}; // Utility function to find GCD of two numbers// GCD of a and bstatic int gcd(int a, int b){    if (b == 0)    return a;    return gcd(b, a % b);} // Finds the no. of Integral points between// two given points.static int getCount(Point p, Point q){    // If line joining p and q is parallel to    // x axis, then count is difference of y    // values    if (p.x == q.x)        return Math.Abs(p.y - q.y) - 1;     // If line joining p and q is parallel to    // y axis, then count is difference of x    // values    if (p.y == q.y)        return Math.Abs(p.x - q.x) - 1;     return gcd(Math.Abs(p.x - q.x), Math.Abs(p.y - q.y)) - 1;} // Driver codepublic static void Main(String[] args){    Point p = new Point(1, 9);    Point q = new Point(8, 16);     Console.WriteLine("The number of integral points between "        + "(" + p.x + ", " + p.y + ") and ("        + q.x + ", " + q.y + ") is "        + getCount(p, q));}} /* This code contributed by PrinciRaj1992 */

Javascript



Output:

The number of integral points between (1, 9) and (8, 16) is 6

Reference :
https://www.geeksforgeeks.org/count-integral-points-inside-a-triangle/
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