Circle and Lattice Points

3

Given a circle of radius r in 2-D with origin or (0, 0) as center. The task is to find the total lattice points on circumference. Lattice Points are points with coordinates as integers in 2-D space.

Example:

Input  : r = 5.
Output : 12
Below are lattice points on a circle with
radius 5 and origin as (0, 0).
(0,5), (0,-5), (5,0), (-5,0),
(3,4), (-3,4), (-3,-4), (3,-4),
(4,3), (-4,3), (-4,-3), (4,-3).
are 12 lattice point.

To find lattice points, we basically need to find values of (x, y) which satisfy the equation x2 + y2 = r2.
For any value of (x, y) that satisfies the above equation we actually have total 4 different combination which that satisfy the equation. For example if r = 5 and (3, 4) is a pair which satisfies the equation, there are actually 4 combinations (3, 4) , (-3,4) , (-3,-4) , (3,-4). There is an exception though, in case of (0, r) or (r, 0) there are actually 2 points as there is no negative 0.

// Initialize result as 4 for (r, 0), (-r. 0),
// (0, r) and (0, -r)
result = 4

Loop for x = 1 to r-1 and do following for every x.
    If r*r - x*x is a perfect square, then add 4 
    tor result.  

Below is C++ implementation of above idea.

CPP

// C++ program to find countLattice points on a circle
#include<bits/stdc++.h>
using namespace std;

// Function to count Lattice points on a circle
int countLattice(int r)
{
    if (r <= 0)
        return 0; 

    // Initialize result as 4 for (r, 0), (-r. 0),
    // (0, r) and (0, -r)
    int result = 4;

    // Check every value that can be potential x
    for (int x=1; x<r; x++)
    {
        // Find a potential y
        int ySquare = r*r - x*x;
        int y = sqrt(ySquare);

        // checking whether square root is an integer
        // or not. Count increments by 4 for four 
        // different quadrant values
        if (y*y == ySquare)
            result += 4;
    }

    return result;
}

// Driver program
int main()
{
    int r = 5;
    cout << countLattice(r);
    return 0;
}

Java

// Java program to find
// countLattice points on a circle

class GFG
{

// Function to count
// Lattice points on a circle
static int countLattice(int r)
{
    if (r <= 0)
        return 0; 
 
    // Initialize result as 4 for (r, 0), (-r. 0),
    // (0, r) and (0, -r)
    int result = 4;
 
    // Check every value that can be potential x
    for (int x=1; x<r; x++)
    {
        // Find a potential y
        int ySquare = r*r - x*x;
        int y = (int)Math.sqrt(ySquare);
 
        // checking whether square root is an integer
        // or not. Count increments by 4 for four 
        // different quadrant values
        if (y*y == ySquare)
            result += 4;
    }
 
    return result;
}

// Driver code
public static void main(String arg[]) 
{
    int r = 5;
    System.out.println(countLattice(r));
}
}
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find
# countLattice podefs on a circle

import math

# Function to count Lattice
# podefs on a circle
def countLattice(r):

    if (r <= 0):
        return 0  

    # Initialize result as 4 for (r, 0), (-r. 0),
    # (0, r) and (0, -r)
    result = 4 

    # Check every value that can be potential x
    for x in range(1, r):
    
        # Find a potential y
        ySquare = r*r - x*x 
        y = int(math.sqrt(ySquare)) 

        # checking whether square root is an defeger
        # or not. Count increments by 4 for four 
        # different quadrant values
        if (y*y == ySquare):
            result += 4 
     

    return result 
 

# Driver program
r = 5 
print(countLattice(r)) 

# This code is contributed by
# Smitha Dinesh Semwal


Output:

12

Reference:
http://mathworld.wolfram.com/CircleLatticePoints.html

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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