Babylonian Algorithm for Square Roots in Java no comments and JOptionPane

BabylonianSqrt6.java

import javax.swing.JOptionPane;
import java.io.*;
import static java.lang.Math.*;
public class BabylonianSqrt6 {
	public static void main(String[] args) {
	
	String S;
	S = JOptionPane.showInputDialog(null, "Enter a Whole Integer");
	
	int SigFigs = 0;
	for (int i = 0, len = S.length(); i < len; i++) {
		if (Character.isDigit(S.charAt(i))) {
			SigFigs++;
			}
	}
	
	
	if (SigFigs % 2 == 0)  
	{
	
	double V = log(SigFigs)/log(2);  
	double G = Math.round(V);		
	
	double R = Integer.parseInt(S);
	
	double x0 = 6*(Math.pow(10,G));
	double x1 = (0.5)*((x0)+(R/x0));
	double x2 = (0.5)*((x1)+(R/x1));
	double x3 = (0.5)*((x2)+(R/x2));
	double x4 = (0.5)*((x3)+(R/x3));
	double x5 = (0.5)*((x4)+(R/x4));
	double x6 = (0.5)*((x5)+(R/x5));
	double x7 = (0.5)*((x6)+(R/x6));
	double x8 = (0.5)*((x7)+(R/x7));
	double x9 = (0.5)*((x8)+(R/x8));
	double x10 = (0.5)*((x9)+(R/x9));
	String str = Double.toString(x10);
	System.out.println("Square root of " + S + " by the Babylonian Guess/iteration method is approximately " +str);
	}
	else 
	{
	
	double V = log(SigFigs)/log(2);
	double G = Math.round(V);
	
	double R = Integer.parseInt(S);
	
	double x0 = 2*(Math.pow(10,G));
	double x1 = (0.5)*((x0)+(R/x0));
	double x2 = (0.5)*((x1)+(R/x1));
	double x3 = (0.5)*((x2)+(R/x2));
	double x4 = (0.5)*((x3)+(R/x3));
	double x5 = (0.5)*((x4)+(R/x4));
	double x6 = (0.5)*((x5)+(R/x5));
	double x7 = (0.5)*((x6)+(R/x6));
	double x8 = (0.5)*((x7)+(R/x7));
	double x9 = (0.5)*((x8)+(R/x8));
	double x10 =(0.5)*((x9)+(R/x9));
	String str = Double.toString(x10);
	System.out.println("Square root of " + S + " is approximately " +str);
	}
	
	double M = Integer.parseInt(S);
	double B = sqrt(M);
	
	System.out.println("Compare this to the java Sqrt() function result: " + B);
	
	}
}