For a given number find the square root using log function. Number may be int, float or double.
Examples:
Input : n = 9
Output : 3Input : n = 2.93
Output : 1.711724
We can find square root of a number using sqrt() method:
C++
// C++ program to demonstrate finding // square root of a number using sqrt() #include<bits/stdc++.h> int main( void )
{ double n = 12;
printf ( "%lf " , sqrt (n));
return 0;
} |
Java
// Java program to demonstrate finding // square root of a number using sqrt() import java.io.*;
class GFG {
public static void main (String[] args) {
double n = 12 ;
System.out.println(Math.sqrt(n));
// This code is contributed by akt_mit }
} |
Python3
# Python3 program to demonstrate finding # square root of a number using sqrt() import math
if __name__ = = '__main__' :
n = 12
print (math.sqrt(n))
# This code is contributed by # Sanjit_Prasad |
C#
// C# program to demonstrate finding // square root of a number using sqrt() using System;
class GFG
{ public static void Main()
{ double n = 12;
Console.Write(Math.Sqrt(n));
} } // This code is contributed // by Akanksha Rai |
PHP
<?php // PHP program to demonstrate finding // square root of a number using sqrt() $n = 12;
echo sqrt( $n );
// This code is contributed by jit_t ?> |
Javascript
<script> // Javascript program to demonstrate finding // square root of a number using sqrt() var n = 12;
document.write(Math.sqrt(n).toFixed(6)); // This code is contributed by aashish1995 </script> |
Output
3.464102
Time complexity: O(log2n), for using sqrt() function.
Auxiliary space: O(1)
We can also find square root using log2() library function:
C++
// C++ program to demonstrate finding // square root of a number using log2() #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
// Java program to demonstrate finding // square root of a number using log2() import java.io.*;
class GFG
{ static double squareRoot( double n)
{ return Math.pow( 2 , 0.5 * (Math.log(n) /
Math.log( 2 )));
} // Driver Code public static void main (String[] args)
{ double n = 12 ;
System.out.println(squareRoot(n));
} } // This code is contributed by akt_mit |
Python
# Python program to demonstrate finding # square root of a number using sqrt() import math
# function to return squareroot def squareRoot(n):
return pow ( 2 , 0.5 * math.log2(n))
# Driver program n = 12
print (squareRoot(n))
# This code is contributed by # Sanjit_Prasad |
C#
// C# program to demonstrate finding // square root of a number using log2() 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));
}
//This code is contributed by akt_mit } |
PHP
<?php // PHP program to demonstrate finding // square root of a number using log2() function squareRoot( $n )
{ return pow(2, 0.5 * log( $n , 2));
} // Driver Code $n = 12;
echo squareRoot( $n );
// This code is contributed by ajit ?> |
Javascript
<script> // Javascript program to demonstrate finding
// square root of a number using log2()
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> |
Output
3.464102
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))
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