## 8/07/2011

### Example of expensive computation

Sometimes I need to simulate some expensive computation, or long-running task, without using Thread.sleep or Object.wait. A real busy waiting. The following example of computing PI is adapted from this RMI tutorial:

`package test;import java.math.BigDecimal;public class Pi implements java.io.Serializable {   private static final long serialVersionUID = 227L;   private static final BigDecimal FOUR = BigDecimal.valueOf(4);   private static final int roundingMode = BigDecimal.ROUND_HALF_EVEN;   public static void main(String[] args) {       BigDecimal result = computePi(Integer.parseInt(args));       System.out.println("Pi is " + result);   }   /**    * Compute the value of pi to the specified number of digits after the    * decimal point. The value is computed using Machin's formula:    * pi/4 = 4*arctan(1/5) - arctan(1/239)    * and a power series expansion of arctan(x) to sufficient precision.    */   public static BigDecimal computePi(int digits) {       int scale = digits + 5;       BigDecimal arctan1_5 = arctan(5, scale);       BigDecimal arctan1_239 = arctan(239, scale);       BigDecimal pi = arctan1_5.multiply(FOUR).subtract(arctan1_239).multiply(FOUR);       return pi.setScale(digits, BigDecimal.ROUND_HALF_UP);   }   /**    * Compute the value, in radians, of the arctangent of the inverse of the    * supplied integer to the specified number of digits after the decimal    * point. The value is computed using the power series expansion for the arc    * tangent:    * arctan(x) = x - (x^3)/3 + (x^5)/5 - (x^7)/7 + (x^9)/9 ...    */   public static BigDecimal arctan(int inverseX, int scale) {       BigDecimal result, numer, term;       BigDecimal invX = BigDecimal.valueOf(inverseX);       BigDecimal invX2 = BigDecimal.valueOf(inverseX * inverseX);       numer = BigDecimal.ONE.divide(invX, scale, roundingMode);       result = numer;       int i = 1;       do {           numer = numer.divide(invX2, scale, roundingMode);           int denom = 2 * i + 1;           term = numer.divide(BigDecimal.valueOf(denom), scale, roundingMode);           if ((i % 2) != 0) {               result = result.subtract(term);           } else {               result = result.add(term);           }           i++;       } while (term.compareTo(BigDecimal.ZERO) != 0);       return result;   }}`
To run and time the task:
`\$ time java test.Pi 50000Pi is 3.14159265358979323846264338327950288419716939937510 ...9.263u 0.187s 0:09.22 102.3% 0+0k 0+6io 8pf+0w`
It takes about 10 seconds to compute PI with 5000 digits after the decimal point.