**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<stdio.h> /*Returns the square root of n. Note that the function */ float squareRoot(float n) { /*We are using n itself as initial approximation This can definitely be improved */ float x = n; float y = 1; float e = 0.000001; /* e decides the accuracy level*/ while(x - y > e) { x = (x + y)/2; y = n/x; } return x; } /* Driver program to test above function*/ int main() { int n = 50; printf ("Square root of %d is %f", n, squareRoot(n)); getchar(); }

## Java

class GFG { /*Returns the square root of n. Note that the function */ static float squareRoot(float n) { /*We are using n itself as initial approximation This can definitely be improved */ float x = n; float y = 1; // e decides the accuracy level double e = 0.000001; while(x - y > e) { x = (x + y)/2; y = n/x; } return x; } /* Driver program to test above function*/ public static void main(String[] args) { int n = 50; System.out.printf ("Square root of " + n + " is " + squareRoot(n)); } } // This code is contriubted by // Smitha DInesh Semwal

## Python 3

# Returns the square root of n. # Note that the function def squareRoot(n): # We are using n itself as # initial approximation This # can definitely be improved x = n y = 1 # e decides the accuracy level e = 0.000001 while(x - y > e): x = (x + y)/2 y = n/x return x # Driver program to test # above function n = 50 print("Square root of", n, "is", round(squareRoot(n), 6)) # This code is contributed by # Smitha Dinesh Semwal

## C#

// C# Porgram for Babylonian // method of square root using System; class GFG { // Returns the square root of n. // Note that the function static float squareRoot(float n) { // We are using n itself as // initial approximation This // can definitely be improved float x = n; float y = 1; // e decides the // accuracy level double e = 0.000001; while(x - y > e) { x = (x + y)/2; y = n/x; } return x; } // Driver Code public static void Main() { int n = 50; Console.Write ("Square root of " + n + " is " + squareRoot(n)); } } // This code is contriubted by nitin mittal.

## PHP

<?php // Returns the square root of n. // Note that the function function squareRoot($n) { // We are using n itself // as initial approximation // This can definitely be // improved $x = $n; $y = 1; /* e decides the accuracy level */ $e = 0.000001; while($x - $y > $e) { $x = ($x + $y)/2; $y = $n / $x; } return $x; } // Driver Code { $n = 50; echo "Square root of $n is ", squareRoot($n); } // This code is contributed by nitin mittal. ?>

Output :

Square root of 50 is 7.071068

**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

// C program for Babylonian // method for square root #include<stdio.h> /* Returns the square root of n. Note that the function will not work for numbers which are not perfect squares*/ unsigned int squareRoot(int n) { int x = n; int y = 1; while(x > y) { x = (x + y) / 2; y = n / x; } return x; } // Driver Code int main() { int n = 49; printf("root of %d is %d", n, squareRoot(n)); getchar(); }

## Java

// Java program for Babylonian // method for square root import java .io.*; public class GFG { /* Returns the square root of n. Note that the function will not work for numbers which are not perfect squares*/ static long squareRoot(int n) { int x = n; int y = 1; while(x > y) { x = (x + y) / 2; y = n / x; } return (long)x; } // Driver Code static public void main (String[] args) { int n = 49; System.out.println( "root of " + n + " is " + squareRoot(n)); } } // This code is contributed by anuj_67.

## Python3

# python3 program for Babylonian # method for square root # Returns the square root of n. # Note that the function # will not work for numbers # which are not perfect squares def squareRoot(n): x = n; y = 1; while(x > y): x = (x + y) / 2; y = n / x; return x; # Driver Code n = 49; print("root of", n, "is", squareRoot(n)); # This code is contributed by mits.

## C#

// C# program for Babylonian // method for square root using System; public class GFG { /* Returns the square root of n. Note that the function will not work for numbers which are not perfect squares*/ static uint squareRoot(int n) { int x = n; int y = 1; while(x > y) { x = (x + y) / 2; y = n / x; } return (uint)x; } // Driver Code static public void Main () { int n = 49; Console.WriteLine( "root of " + n + " is " + squareRoot(n)); } } // This code is contributed by anuj_67.

## PHP

<?php // PHP program for Babylonian // method for square root /* Returns the square root of n. Note that the function will not work for numbers which are not perfect squares */ function squareRoot( $n) { $x = $n; $y = 1; while($x > $y) { $x = ($x + $y) / 2; $y =$n / $x; } return $x; } // Driver Code $n = 49; echo " root of ",$n, " is " , squareRoot($n); // This code is contributed by anuj_67. ?>

Output :

root of 49 is 7

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.

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