A89: Re: Converting (sorry. first e-mail didn't go through or something)

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A89: Re: Converting (sorry. first e-mail didn't go through or something)

sorry. i feel like such an idiot. the first e-mail didn't go through or 
something. Anyway, here's what the origional e-mail said:

"Does anyone know of a program to convert programs written in C (for the 
computer) to programs for the TI-89. I have the source code attached 
(downloaded it from a site). It calculates PI to whatever decimal place you 
want. Thanks."

The file is displayed at the bottom of this e-mail (in Text format). If 
anyone could help me, I would really appreciate it. Thanks.

Sincerely and feeling like an idiot,
Josh

/*  This program employs the recently discovered digit extraction scheme
    to produce hex digits of pi.  This code is valid up to ic = 2^24 on 
    systems with IEEE arithmetic. */

/*  David H. Bailey     960429 */

#include <stdio.h>
#include <math.h>

main()
{
  double pid, s1, s2, s3, s4;
  double series (int m, int n);
  void ihex (double x, int m, char c[]);
  int ic = 1000000;
#define NHX 16
  char chx[NHX];

/*  ic is the hex digit position -- output begins at position ic + 1. */

  s1 = series (1, ic);
  s2 = series (4, ic);
  s3 = series (5, ic);
  s4 = series (6, ic);
  pid = 4. * s1 - 2. * s2 - s3 - s4;
  pid = pid - (int) pid + 1.;
  ihex (pid, NHX, chx);
  printf ("Pi hex digit computation\n");
  printf ("position = %i + 1\n %20.15f\n %12.12s\n", ic, pid, chx);
}

void ihex (double x, int nhx, char chx[])

/*  This returns, in chx, the first nhx hex digits of the fraction of x.
*/

{
  int i;
  double y;
  char hx[] = "0123456789ABCDEF";

  y = fabs (x);

  for (i = 0; i < nhx; i++){
    y = 16. * (y - floor (y));
    chx[i] = hx[(int) y];
  }
}

double series (int m, int ic)

/*  This routine evaluates the series  sum_k 16^(ic-k)/(8*k+m) 
    using the modular exponentiation technique. */

{
  int k;
  double ak, eps, p, s, t;
  double expm (double x, double y);
#define eps 1e-17

  s = 0.;

/*  Sum the series up to ic. */

  for (k = 0; k < ic; k++){
    ak = 8 * k + m;
    p = ic - k;
    t = expm (p, ak);
    s = s + t / ak;
    s = s - (int) s;
  }

/*  Compute a few terms where k >= ic. */

  for (k = ic; k <= ic + 100; k++){
    ak = 8 * k + m;
    t = pow (16., (double) (ic - k)) / ak;
    if (t < eps) break;
    s = s + t;
    s = s - (int) s;
  }
  return s;
}

double expm (double p, double ak)

/*  expm = 16^p mod ak.  This routine uses the left-to-right binary 
    exponentiation scheme.  It is valid for  ak <= 2^24. */

{
  int i, j;
  double p1, pt, r;
#define ntp 25
  static double tp[ntp];
  static int tp1 = 0;

/*  If this is the first call to expm, fill the power of two table tp. */

  if (tp1 == 0) {
    tp1 = 1;
    tp[0] = 1.;

    for (i = 1; i < ntp; i++) tp[i] = 2. * tp[i-1];
  }

  if (ak == 1.) return 0.;

/*  Find the greatest power of two less than or equal to p. */

  for (i = 0; i < ntp; i++) if (tp[i] > p) break;

  pt = tp[i-1];
  p1 = p;
  r = 1.;

/*  Perform binary exponentiation algorithm modulo ak. */

  for (j = 1; j <= i; j++){
    if (p1 >= pt){
      r = 16. * r;
      r = r - (int) (r / ak) * ak;
      p1 = p1 - pt;
    }
    pt = 0.5 * pt;
    if (pt >= 1.){
      r = r * r;
      r = r - (int) (r / ak) * ak;
    }
  }

  return r;
}
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