/**
 * Do a friends of one of FHK's problems.  Sum(n=0..infinity) 1/(n+pi).
 * First forwards (finding out where no futher change happens, and then
 * backwards.  Note the maths says the answer is infinite...
 */

/**
 * @author silviabreu
 *
 */
public class SumFwdBack {

	private static double g(int i) { 
		return 1.0/(i + Math.PI); 
	}


	/**
	 * @param args
	 */
	public static void main(String[] args) {
		double d, dd;
	    int n = 0, i;
	    float f = 0, ff;
	    do { ff = f;
	         f += 1.0/(n + Math.PI);
	         n++;
	    } while (f != ff);
	    System.out.printf("Forward (float) sum until no change: %11.7f at n=%d\n", f, n);
	    f = 0;
	    for (i = n-1; i>=0; i--)
	        f += 1.0/(i + Math.PI);
	    System.out.printf("Backward (float) sum:                %11.7f\n", f);
	    d = 0;
	    for (i = 0; i<n; i++)
	        d += g(i);
	    System.out.printf("Forward (double) sum:                %20.16f\n", d);
	    d = 0;
	    for (i = n-1; i>=0; i--)
	        d += g(i);
	    System.out.printf("Backward (double) sum:               %20.16f\n", d);

	    /* now use Kahan float summation, using ff as guard digits...           */
	    ff = 0; f = (float) g(0);
	    for (i = 1; i<n; i++)
	    {   float y  = (float) (g(i) - ff); 
	        float t = f + y;
	        ff = (t - f) - y;
	        f = t;
	    }
	    System.out.printf("Forward (float) Kahan sum:           %11.7f\n", f);
	    ff = 0; f = (float) g(n-1);
	    for (i = n-2; i>=0; i--)
	    {   float y  = (float) (g(i) - ff); 
	        float t = f + y;
	        ff = (t - f) - y;
	        f = t;
	    }
	    System.out.printf("Backward (float) Kahan sum:          %11.7f\n", f);

	    /* now use Kahan summation, using dd as guard digits...                 */
	    dd = 0; d = g(0);
	    for (i = 1; i<n; i++)
	    {   double y  = g(i) - dd; 
	        double t = d + y;
	        dd = (t - d) - y;
	        d = t;
	    }
	    System.out.printf("Forward (double) Kahan sum:          %20.16f\n", d);
	    dd = 0; d = g(n-1);
	    for (i = n-2; i>=0; i--)
	    {   double y  = g(i) - dd; 
	        double t = d + y;
	        dd = (t - d) - y;
	        d = t;
	    }
	    System.out.printf("Backward (double) Kahan sum:         %20.16f\n", d);
	}

}