Calculate PI To Arbitrary Precision (Sample Java Code)
By Angsuman Chakraborty, Gaea News NetworkTuesday, November 13, 2007
Sample Java code to calculate PI to arbitrary precision. This uses Machin’s formula: pi/4 = 4*arctan(1/5) - arctan(1/239)
.
Notes:
1. Run Pi with a positive integer precision value. For example to calculate PI to 10 decimal places run:
java Pi 10
2. There are other algorithms and even better ones to calculate PI too. This is just an example. I use it to load test my server.
3. Try running with 1 million (1000000) to give your server a good workout
import java.math.BigDecimal;
public class Pi {
public static void main(String args[]) throws NumberFormatException {
int digits = Integer.parseInt(args[0]);
String pi = computePi(digits).toString();
int freq[] = new int[10];
for(int i = 0;i < 10;i++) freq[i] = 0;
int c;
for(int i = 0;i < pi.length();i++) {
c = pi.charAt(i);
if(c == '.') continue;
c -= '0';
freq[c]++;
}
for(int i = 0;i < 10;i++) {
System.out.println("" + i + " " + freq[i]);
}
}
/** constants used in pi computation */
private static final BigDecimal FOUR = BigDecimal.valueOf(4);
/** rounding mode to use during pi computation */
private static final int roundingMode = BigDecimal.ROUND_HALF_EVEN;
/**
* 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;
}
}
Note: The code has been adapted from a java.sun example.
Discussion
david bandel |
January 30, 2009: 6:26 am
gives array out of bounds exception. not worth my time. probably less efficient than my own pi-calculator anyways. learn some java before you go posting code on the interweb. lolz. |
YOUR VIEW POINT
Shahzad hasan