Java Program to Find Square Root of a Number Without sqrt Method

FindSquareRootExample1 .java

import java.util.Scanner;  
public class FindSquareRootExample1   
{  
public static void main(String[] args)    
{   
System.out.print("Enter a number: ");  
//creating object of the Scanner class  
Scanner sc = new Scanner(System.in);  
//reading a number form the user  
int n = sc.nextInt();  
//calling the method and prints the result  
System.out.println("The square root of "+ n+ " is: "+squareRoot(n));  
}  
//user-defined method that contains the logic to find the square root  
public static double squareRoot(int num)   
{  
//temporary variable  
double t;  
double sqrtroot=num/2;  
do   
{  
t=sqrtroot;  
sqrtroot=(t+(num/t))/2;  
}   
while((t-sqrtroot)!= 0);  
return sqrtroot;  
}  
}  

Output 1:

Enter a number: 12
The square root of 12 is: 3.4641016151377544

Output 2:

Enter a number: 25
The square root of 25 is: 5.0

Let's see another logic to find the square root.

In the following example, we have used the following procedure to find the square root.

• We have initialized an iterator variable i=1.
• Check the number that we have entered is a perfect square or not. If the square of i is equal to n, i will be the square root value of n.
• Otherwise, find the lowest value of i. Remember that the square of i must be greater than n. The square root of a number lies between i-1 and i. After performing the steps, we use the binary search algorithm to find the square root of a number up to n decimal places.
• Increment the variable i by 1.

Binary Search Algorithm

• Find the midvalue of i-1 and i.
• Find the square of midvalue and compare it with n.
  • If midvalue * midvalue = n, the midvalue is the square root of the given number. Compare the square of midvalue with n (up to n decimal places) if the difference is minor, the midvalue will be the square root of the number.
  • If midvalue * midvalue > n, the square root belongs to the first half.
  • If midvalue * midvalue < n, the square root belongs to the second half.

Let's implement the algorithm in a Java program and find the square root of a number.

FindSquareRootExample2.java

import java.util.Scanner;  
public class FindSquareRootExample2   
{  
public static void main(String[] args)   
{  
double number = 0, sqrt=0;  
//object of the Scanner class  
Scanner sc = new Scanner(System.in);  
System.out.print("Enter a number: ");  
//reading a double value from the user  
number = sc.nextDouble();  
//method calling  
sqrt = squareRoot(number);  
//prints the result  
System.out.println(The square root of "+number+ " is " +sqrt);  
}  
//user-defined method to find the square root of a number  
private static double squareRoot(double number)   
{  
//iterator variable      
int i = 1;  
while(true)   
{  
//for perfect square numbers  
if(i*i == number)  
return i;  
//for not perfect square numbers  
else if(i*i > number)   
//returns the value calculated by the method decimalSqrt()  
return decimalSqrt(number,i-1,i);  
//increments the variable i by 1  
i++;  
}  
}  
// recursive method to find the square root of a number up to 7 decimal places    
private static double decimalSqrt(double number, double i, double j)   
{  
//calculates the middle of i and j  
double midvalue = (i+j)/2;  
//finds the square of the midvalue  
double square = midvalue * midvalue;  
//compares the midvalue with square up to n decimal places  
if(square==number||Math.abs(square-number)<0.0000001)   
return midvalue;   
//if the square root belongs to second half  
else if(square>number)  
return decimalSqrt(number, i, midvalue);  
//if the square root belongs to first half  
else  
return decimalSqrt(number, midvalue, j);  
}  
}  

Output 1:

Enter a number: 625
The square root of 625.0 is 25.0

Output 2:

Enter a number: 129
The square root of 129.0 is 11.357816688716412