Algorithm:
This method can be derived from (but predates) Newton–Raphson method.
1 Start with an arbitrary positive start value x (the closer to the
root, the better).
2 Initialize y = 1.
3. Do following until desired approximation is achieved.
a) Get the next approximation for root using average of x and y
b) Set y = n/x
Implementation:
C++
#include <iostream>
using namespace std;
class gfg {
public:
float squareRoot(float n)
{
float x = n;
float y = 1;
float e = 0.000001;
while (x - y > e) {
x = (x + y) / 2;
y = n / x;
}
return x;
}
};
int main()
{
gfg g;
int n = 50;
cout << "Square root of " << n << " is " << g.squareRoot(n);
getchar();
}
|
C
#include <stdio.h>
float squareRoot(float n)
{
float x = n;
float y = 1;
float e = 0.000001;
while (x - y > e) {
x = (x + y) / 2;
y = n / x;
}
return x;
}
int main()
{
int n = 50;
printf("Square root of %d is %f", n, squareRoot(n));
getchar();
}
|
Java
class GFG {
static float squareRoot(float n)
{
float x = n;
float y = 1;
double e = 0.000001;
while (x - y > e) {
x = (x + y) / 2;
y = n / x;
}
return x;
}
public static void main(String[] args)
{
int n = 50;
System.out.printf("Square root of "
+ n + " is " + squareRoot(n));
}
}
|
Python 3
def squareRoot(n):
x = n
y = 1
e = 0.000001
while(x - y > e):
x = (x + y)/2
y = n / x
return x
n = 50
print("Square root of", n, "is",
round(squareRoot(n), 6))
|
C#
using System;
class GFG {
static float squareRoot(float n)
{
float x = n;
float y = 1;
double e = 0.000001;
while (x - y > e) {
x = (x + y) / 2;
y = n / x;
}
return x;
}
public static void Main()
{
int n = 50;
Console.Write("Square root of "
+ n + " is " + squareRoot(n));
}
}
|
PHP
<?php
function squareRoot($n)
{
$x = $n;
$y = 1;
$e = 0.000001;
while($x - $y > $e)
{
$x = ($x + $y)/2;
$y = $n / $x;
}
return $x;
}
{
$n = 50;
echo "Square root of $n is ", squareRoot($n);
}
?>
|
Javascript
<script>
function squareRoot( n)
{
let x = n;
let y = 1;
let e = 0.000001;
while (x - y > e) {
x = (x + y) / 2;
y = n / x;
}
return x;
}
let n = 50;
document.write( "Square root of "+n+" is " + squareRoot(n).toFixed(6));
</script>
|
Output :
Square root of 50 is 7.071068
Time Complexity: O(n1/2)
Auxiliary Space: O(1)
Example:
n = 4 /*n itself is used for initial approximation*/
Initialize x = 4, y = 1
Next Approximation x = (x + y)/2 (= 2.500000),
y = n/x (=1.600000)
Next Approximation x = 2.050000,
y = 1.951220
Next Approximation x = 2.000610,
y = 1.999390
Next Approximation x = 2.000000,
y = 2.000000
Terminate as (x - y) > e now.
If we are sure that n is a perfect square, then we can use following method. The method can go in infinite loop for non-perfect-square numbers. For example, for 3 the below while loop will never terminate.
C++
#include <iostream>
using namespace std;
class gfg {
public:
float squareRoot(float n)
{
float x = n;
float y = 1;
while (x > y) {
x = (x + y) / 2;
y = n / x;
}
return x;
}
};
int main()
{
gfg g;
int n = 49;
cout << "Square root of " << n << " is " << g.squareRoot(n);
getchar();
}
|
C
#include <stdio.h>
unsigned int squareRoot(int n)
{
int x = n;
int y = 1;
while (x > y) {
x = (x + y) / 2;
y = n / x;
}
return x;
}
int main()
{
int n = 49;
printf("root of %d is %d", n, squareRoot(n));
getchar();
}
|
Java
import java.io.*;
public class GFG {
static long squareRoot(int n)
{
int x = n;
int y = 1;
while (x > y) {
x = (x + y) / 2;
y = n / x;
}
return (long)x;
}
static public void main(String[] args)
{
int n = 49;
System.out.println("root of "
+ n + " is " + squareRoot(n));
}
}
|
Python3
def squareRoot(n):
x = n;
y = 1;
while(x > y):
x = (x + y) / 2;
y = n / x;
return x;
n = 49;
print("root of", n, "is", squareRoot(n));
|
C#
using System;
public class GFG {
static uint squareRoot(int n)
{
int x = n;
int y = 1;
while (x > y) {
x = (x + y) / 2;
y = n / x;
}
return (uint)x;
}
static public void Main()
{
int n = 49;
Console.WriteLine("root of "
+ n + " is " + squareRoot(n));
}
}
|
PHP
<?php
function squareRoot( $n)
{
$x = $n;
$y = 1;
while($x > $y)
{
$x = ($x + $y) / 2;
$y =$n / $x;
}
return $x;
}
$n = 49;
echo " root of ", $n, " is ", squareRoot($n);
?>
|
Javascript
<script>
function squareRoot(n) {
var x = n;
var y = 1;
while (x > y) {
x = (x + y) / 2;
y = n / x;
}
return x;
}
var n = 49;
document.write("root of " + n + " is " + squareRoot(n));
</script>
|
Output :
root of 49 is 7
Time Complexity: O(n1/2)
Auxiliary Space: O(1)
References;
http://en.wikipedia.org/wiki/Square_root
http://en.wikipedia.org/wiki/Babylonian_method#Babylonian_method
Asked by Snehal
Please write comments if you find any bug in the above program/algorithm, or if you want to share more information about Babylonian method.