Estimating Pi number with Monte Carlo Simulation in Java and Kdb+ q

package com.denizstij.montecarlo.sim;
 
import java.util.Random;
 
/**
*  This class estimates Pi with monte carlo simulation.
* 
 *  We assume we have a square with dimension 1x1 which contains a disk in middle with radious of 0.5
*  where it touches the square on 4 edges. We generate N number of pairs (x,y) between 0 and 1.
 *  Then we check if each pair is in the circle or not. 4 times ratio of total pairs in circle
*  to total simulation is an estimation of Pi. 
 * 
 *  pi=4*(total pairs in circle)/total pairs
* 
 * @author Deniz Turan, http://denizstij.blogspot.co.uk, denizstij AT gmail.com
*
 */
public class PIEstimationWithMonteCarloSimulation {
       // number of total simulation
       private static final long TOTAL_SIMULATION=10000000L;
      
       private double estimatePI(){            
              double pi=0;
              Random randomGen= new Random();
              long inCircleCounter=0;
              for (int i=0;i<TOTAL_SIMULATION;i++){
                     // generate pairs
                     double x = randomGen.nextDouble();
                     double y = randomGen.nextDouble();
                     // check if the pair  in the circle
                     inCircleCounter+=isInCircle(x,y);
              }
              double ratio=(double)inCircleCounter/TOTAL_SIMULATION;
              pi=ratio*4;
              return pi;
       }
      
       /*
       *  Check given two points in a circle: x^2+y^2<=1
       */
       private int isInCircle(double x, double y) {
              double z=Math.sqrt(x*x+y*y);            
              return z<=1?1:0;
       }
 
       public static void main(String[] args) {
              PIEstimationWithMonteCarloSimulation est= new PIEstimationWithMonteCarloSimulation();
              double estimatedPI = est.estimatePI();
              System.out.println("Estimated PI:"+estimatedPI);
       }
}

And now in Kdb+ q :

piEst:{[N]
        sqrArea:4;                                  / area of square with length 2 and containing a circle with radius 1
        xy:flip {2?1.0} each til N;                 / xy points
        p:sqrt sum xy*xy;                           / sqrt(x^2+y^2) <=1 to be in a circle
        totalIncircle: sum p<=1;
        piEst:sqrArea*totalIncircle%N;                    / that’s the estimation of PI
        :piEst
}
q)piEst[1000000]
3.141476

Above code can be reduced to one line such as:

q)N:10000000
q)(1%N)*4*sum 1>=sqrt sum ({x*x}') each flip {2?1.0} each til N
3.141842