Circle and Lattice Points
Last Updated :
20 Jun, 2022
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 the implementation of above idea.
CPP
#include<bits/stdc++.h>
using namespace std;
int countLattice( int r)
{
if (r <= 0)
return 0;
int result = 4;
for ( int x=1; x<r; x++)
{
int ySquare = r*r - x*x;
int y = sqrt (ySquare);
if (y*y == ySquare)
result += 4;
}
return result;
}
int main()
{
int r = 5;
cout << countLattice(r);
return 0;
}
|
Java
class GFG
{
static int countLattice( int r)
{
if (r <= 0 )
return 0 ;
int result = 4 ;
for ( int x= 1 ; x<r; x++)
{
int ySquare = r*r - x*x;
int y = ( int )Math.sqrt(ySquare);
if (y*y == ySquare)
result += 4 ;
}
return result;
}
public static void main(String arg[])
{
int r = 5 ;
System.out.println(countLattice(r));
}
}
|
Python3
import math
def countLattice(r):
if (r < = 0 ):
return 0
result = 4
for x in range ( 1 , r):
ySquare = r * r - x * x
y = int (math.sqrt(ySquare))
if (y * y = = ySquare):
result + = 4
return result
r = 5
print (countLattice(r))
|
C#
using System;
class GFG {
static int countLattice( int r)
{
if (r <= 0)
return 0;
int result = 4;
for ( int x = 1; x < r; x++)
{
int ySquare = r*r - x*x;
int y = ( int )Math.Sqrt(ySquare);
if (y*y == ySquare)
result += 4;
}
return result;
}
public static void Main()
{
int r = 5;
Console.Write(countLattice(r));
}
}
|
PHP
<?php
function countLattice( $r )
{
if ( $r <= 0)
return 0;
$result = 4;
for ( $x = 1; $x < $r ; $x ++)
{
$ySquare = $r * $r - $x * $x ;
$y = ceil (sqrt( $ySquare ));
if ( $y * $y == $ySquare )
$result += 4;
}
return $result ;
}
$r = 5;
echo countLattice( $r );
?>
|
Javascript
<script>
function countLattice(r) {
if (r <= 0)
return 0;
var result = 4;
for (x = 1; x < r; x++)
{
var ySquare = r * r - x * x;
var y = parseInt( Math.sqrt(ySquare));
if (y * y == ySquare)
result += 4;
}
return result;
}
var r = 5;
document.write(countLattice(r));
</script>
|
Output:
12
Time Complexity: O(1)
Auxiliary Space: O(1)
Reference:
http://mathworld.wolfram.com/CircleLatticePoints.html
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