Draw a circle without floating point arithmetic

Last Updated : 21 Jun, 2022

Given a radius of a circle, draw the circle without using floating point arithmetic.
Following program uses a simple concept. Let the radius of the circle be r. Consider a square of size (2r+1)*(2r+1) around the circle to be drawn. Now walk through every point inside the square. For every  point (x,y), if (x, y) lies inside the circle (or  x2+ y2 < r2), then print it, otherwise print space.
 

Approach: 

1. draw the rectangle

2. check if the point lie inside or on the circle using formula  x2+ y2 < r2.

3. If it fulfill the condition print the star else print blank space

Algorithm: 

Step 1: Take the input of radius

Step 2: Calculate the size of rectangle

Step 3: Draw the rectangle using the nested for loop

Step 4: Start coordinates from left most corner

Step 5: Check if the coordinates lie inside the circle, if yes Print the star else print blank space.

C++
// C++ code to demonstrate to draw
// circle without floating 
// point arithmetic
#include <stdio.h>

void drawCircle(int r)
{
    // Consider a rectangle of size N*N
    int N = 2*r+1;

    int x, y;  // Coordinates inside the rectangle

    // Draw a square of size N*N.
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            // Start from the left most corner point
            x = i-r;
            y = j-r;

            // If this point is inside the circle, print it
            if (x*x + y*y <= r*r+1 )
                printf(".");
            else // If outside the circle, print space
                printf(" ");
            printf(" ");
        }
        printf("\n");
    }
}

// Driver Program to test above function
int  main()
{
    drawCircle(8);
    return 0;
}
Java
// Java code to demonstrate to draw
// circle without floating 
// point arithmetic

class GFG
{
static void drawCircle(int r)
{
    // Consider a rectangle of size N*N
    int N = 2*r+1;

    int x, y; // Coordinates inside the rectangle

    // Draw a square of size N*N.
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < N; j++)
        {
            // Start from the left most corner point
            x = i-r;
            y = j-r;

            // If this point is inside the circle, print it
            if (x*x + y*y <= r*r+1 )
                System.out.print(".");
            else 
                // If outside the circle, print space
                System.out.print(" ");

            System.out.print(" ");
        }

        System.out.println();
    }
}

// Driver Program to test above function
public static void main(String arg[])
{
    drawCircle(8);
}
}

// This code is contributed
// by Anant Agarwal.
Python3
# Python3 code to demonstrate to draw
# circle without floating
# point arithmetic

def drawCircle(r):

    # Consider a rectangle of size N*N
    N = 2 * r + 1

    # Draw a square of size N*N.
    for i in range(N):
        for j in range(N):

            # Start from the left most corner point
            x = i - r
            y = j - r

            # If this point is inside the circle, 
            # print it
            if x * x + y * y <= r * r + 1:
                print(".", end = " ")
                
            # If outside the circle, print space
            else:
                print(" ", end = " ")
        print()

# Driver Code
if __name__ == "__main__":
    drawCircle(8)

# This code is contributed 
# by vibhu4agarwal
C#
// C# code to demonstrate to draw 
// circle without floating 
// point arithmetic 
using System;

public class GFG{
    static void drawCircle(int r) 
{ 
    // Consider a rectangle of size N*N 
    int N = 2*r+1; 

    int x, y; // Coordinates inside the rectangle 

    // Draw a square of size N*N. 
    for (int i = 0; i < N; i++) 
    { 
        for (int j = 0; j < N; j++) 
        { 
            // Start from the left most corner point 
            x = i-r; 
            y = j-r; 

            // If this point is inside the circle, print it 
            if (x*x + y*y <= r*r+1 ) 
                Console.Write("."); 
            else
                // If outside the circle, print space 
                Console.Write(" "); 

            Console.Write(" "); 
        } 

        Console.WriteLine(); 
    } 
} 

// Driver Program to test above function
    static public void Main (){
        drawCircle(8); 
    
     
    }
// This code is contributed 
// by ajit. 
}
PHP
<?php
// PHP code to demonstrate to draw circle 
// without floating point arithmetic

function drawCircle($r)
{
    // Consider a rectangle of size N*N
    $N = 2 * $r + 1;

    $y; // Coordinates inside the rectangle

    // Draw a square of size N*N.
    for ($i = 0; $i < $N; $i++)
    {
        for ( $j = 0; $j < $N; $j++)
        {
            // Start from the left most
            // corner point
            $x = $i - $r;
            $y = $j - $r;

            // If this point is inside the
            // circle, print it
            if ($x * $x + $y * $y <= $r * $r + 1)
                echo (".");
            else // If outside the circle, 
                 // print space
                echo (" ");
            echo (" ");
        }
    echo "\n";
    }
}

// Driver Code
drawCircle(8);

// This code is contributed by akt_mit
?>
JavaScript
<script>

function drawCircle(r)
{
    // Consider a rectangle of size N*N
    let N = 2*r+1;
  
    let x, y; // Coordinates inside the rectangle
  
    // Draw a square of size N*N.
    for (let i = 0; i < N; i++)
    {
        for (let j = 0; j < N; j++)
        {
            // Start from the left most corner point
            x = i - r;
            y = j - r;
  
            // If this point is inside the circle, print it
            if (x * x + y * y <= r * r + 1 )
                document.write(".");
            else 
                // If outside the circle, print space
                document.write(" &nbsp");
  
            document.write("&nbsp ");
        }
  
        document.write("<br>");
    }
}

drawCircle(8);
// This code is contributed by avanitrachhadiya2155
</script>

Output: 
 

Time Complexity: O(N2)

Space Complexity: O(N2)
 

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