Number Scrunching

[MediaSlayer]

what does it mean

[LeGoMyFnLeg]

Number crunching.

One of the most famous examples.

  program piqp

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

      real*8 pid, s1, s2, s3, s4, series
      parameter (ic = 100000, nhx = 12)
      character*1 chx(nhx)

c  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 = mod (4.d0 * s1 - 2.d0 * s2 - s3 - s4, 1.d0) + 1.d0
      call ihex (pid, nhx, chx)
      write (6, 1) ic, pid, chx
1    format ('Pi hex digit computation'/'position =',i12,' + 1'/
    $  f20.15/12a1)

      stop
      end

      subroutine ihex (x, nhx, chx)

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

      real*8 x, y
      character*1 chx(nhx), hx(0:15)
      data hx/'0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'A',
    $  'B', 'C', 'D', 'E', 'F'/

      y = abs (x)

      do i = 1, nhx
        y = 16.d0 * (y - aint (y))
        chx(i) = hx(int(y))
      enddo

      return
      end

      function series (m, ic)

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

      real*8 ak, eps, expm, p, s, series, t
      parameter (eps = 1d-17)

      s = 0.d0

c  Sum the series up to ic.

      do k = 0, ic - 1
        ak = 8 * k + m
        p = ic - k
        t = expm (p, ak)
        s = mod (s + t / ak, 1.d0)
      enddo

c  Compute a few terms where k >= ic.

      do k = ic, ic + 100
        ak = 8 * k + m
        t = 16.d0 ** (ic - k) / ak
        if (t .lt. eps) goto 100
        s = mod (s + t, 1.d0)
      enddo

100  continue

      series = s
      return
      end

      function expm (p, ak)

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

&nbsp; &nbsp; &nbsp; parameter (ntp = 25)
&nbsp; &nbsp; &nbsp; real*8 ak, expm, p, p1, pt, r, tp(ntp)
&nbsp; &nbsp; &nbsp; save tp
&nbsp; &nbsp; &nbsp; data tp/ntp*0.d0/

c&nbsp; If this is the first call to expm, fill the power of two table tp.

&nbsp; &nbsp; &nbsp; if (tp(1) .eq. 0.d0) then
&nbsp; &nbsp; &nbsp; &nbsp; tp(1) = 1.d0

&nbsp; &nbsp; &nbsp; &nbsp; do i = 2, ntp
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tp(i) = 2.d0 * tp(i-1)
&nbsp; &nbsp; &nbsp; &nbsp; enddo
&nbsp; &nbsp; &nbsp; endif

&nbsp; &nbsp; &nbsp; if (ak .eq. 1.d0) then
&nbsp; &nbsp; &nbsp; &nbsp; expm = 0.d0
&nbsp; &nbsp; &nbsp; &nbsp; return
&nbsp; &nbsp; &nbsp; endif

c&nbsp; Find the greatest power of two less than or equal to p.

&nbsp; &nbsp; &nbsp; do i = 1, ntp
&nbsp; &nbsp; &nbsp; &nbsp; if (tp(i) .gt. p) goto 100
&nbsp; &nbsp; &nbsp; enddo

100&nbsp; pt = tp(i-1)
&nbsp; &nbsp; &nbsp; p1 = p
&nbsp; &nbsp; &nbsp; r = 1.d0

c&nbsp; Perform binary exponentiation algorithm modulo ak.

110&nbsp; continue

&nbsp; &nbsp; &nbsp; if (p1 .ge. pt) then
&nbsp; &nbsp; &nbsp; &nbsp; r = mod (16.d0 * r, ak)
&nbsp; &nbsp; &nbsp; &nbsp; p1 = p1 - pt
&nbsp; &nbsp; &nbsp; endif
&nbsp; &nbsp; &nbsp; pt = 0.5d0 * pt
&nbsp; &nbsp; &nbsp; if (pt .ge. 1.d0) then
&nbsp; &nbsp; &nbsp; &nbsp; r = mod (r * r, ak)
&nbsp; &nbsp; &nbsp; &nbsp; goto 110
&nbsp; &nbsp; &nbsp; endif

&nbsp; &nbsp; &nbsp; expm = mod (r, ak)

&nbsp; &nbsp; &nbsp; return
&nbsp; &nbsp; &nbsp; end



Another example

http://www.activemind.com/Mysterious/Topic...e_equation.html (http://www.activemind.com/Mysterious/Topics/SETI/drake_equation.html)

[noname12]

You could have just used a dictionary defenition... <_<

[LeGoMyFnLeg]

Feel free to add to the discussion, I believe that&#39;s what forums are for?

[noname12]

Nah I prefer to drop a little comment here and there and then walk off again.

[LeGoMyFnLeg]

Should I provide a definition for that?

[noname12]

:lol: