For a given number find the square root using log function. Number may be int, float or double.
Examples:
Input : n = 9
Output : 3
Input : n = 2.93
Output : 1.711724
We can find square root of a number using sqrt() method:
C++
#include<bits/stdc++.h>
int main( void )
{
double n = 12;
printf ( "%lf " , sqrt (n));
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main (String[] args) {
double n = 12 ;
System.out.println(Math.sqrt(n));
}
}
|
Python3
import math
if __name__ = = '__main__' :
n = 12
print (math.sqrt(n))
|
C#
using System;
class GFG
{
public static void Main()
{
double n = 12;
Console.Write(Math.Sqrt(n));
}
}
|
PHP
<?php
$n = 12;
echo sqrt( $n );
?>
|
Javascript
<script>
var n = 12;
document.write(Math.sqrt(n).toFixed(6));
</script>
|
Time complexity: O(log2n), for using sqrt() function.
Auxiliary space: O(1)
We can also find square root using log2() library function:
C++
#include<bits/stdc++.h>
double squareRoot( double n)
{
return pow (2, 0.5*log2(n));
}
int main( void )
{
double n = 12;
printf ( "%lf " , squareRoot(n));
return 0;
}
|
Java
import java.io.*;
class GFG
{
static double squareRoot( double n)
{
return Math.pow( 2 , 0.5 * (Math.log(n) /
Math.log( 2 )));
}
public static void main (String[] args)
{
double n = 12 ;
System.out.println(squareRoot(n));
}
}
|
Python
import math
def squareRoot(n):
return pow ( 2 , 0.5 * math.log2(n))
n = 12
print (squareRoot(n))
|
C#
using System;
public class GFG{
static double squareRoot( double n)
{
return Math.Pow(2, 0.5 * (Math.Log(n) /Math.Log(2)));
}
static public void Main (){
double n = 12;
Console.WriteLine(squareRoot(n));
}
}
|
PHP
<?php
function squareRoot( $n )
{
return pow(2, 0.5 * log( $n , 2));
}
$n = 12;
echo squareRoot( $n );
?>
|
Javascript
<script>
function squareRoot(n)
{
return Math.pow(2, 0.5 * (Math.log(n) /Math.log(2)));
}
let n = 12;
document.write(squareRoot(n).toFixed(15));
</script>
|
Time complexity: O(log2log2N), complexity of using log(N) is log(logN), and pow(x,N) is log(N), so pow(2,0.5*log(n)) will be log(logN).
Auxiliary space: O(1)
How does the above program work?
let d be our answer for input number n
then n(1/2) = d
apply log2 on both sides
log2(n(1/2)) = log2(d)
log2(d) = 1/2 * log2(n)
d = 2(1/2 * log2(n))
d = pow(2, 0.5*log2(n))
This article is contributed by Tumma Umamaheswararao from Jntuh College of Engineering . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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