[racket] Implementation of Simpson's Rule (SICP Exercise 1.29)

The competition does it better, see Python's fsum:

import math

janus = [31.0, 2e+34, -1.2345678901235e+80, 2749.0, -2939234.0, -2e+33,
         3.2e+270, 17.0, -2.4e+270, 4.2344294738446e+170, 1.0, -8e+269,
0.0, 99.0]

print(math.fsum(janus))
# 4.2344294738446e+170

The fsum algorithm is interesting, it essentially emulates
unlimited-precision FP
by using a list of limited-precision FP numbers.

Stephan


2013/11/7 Todd O'Bryan <toddobryan at gmail.com>

> I just found a lovely Java expression to emphasize the inexactness of
> doubles to my AP students. The problem--which I think is from
> HtDP/1e--is to find the value of a bag of coins given the number of
> pennies, nickels, dimes, and quarters. In BlueJ's code pad (or similar
> in DrJava, jGrasp, etc.)
>
> > 0.01 + 0.05 + 0.10 + 0.25
> 0.410000000000000000003
>
> (my number of zeroes may be off)
>
> As one of my students said--"You can do that in your head. What's the
> computer's problem?"
>
> Todd
>
> On Wed, Nov 6, 2013 at 1:30 PM, Neil Toronto <neil.toronto at gmail.com>
> wrote:
> > On 11/06/2013 09:24 AM, Matthias Felleisen wrote:
> >>
> >>
> >> On Nov 6, 2013, at 7:13 AM, Ben Duan <yfefyf at gmail.com> wrote:
> >>
> >>> Thank you, Jens. I didn't know that the inexactness of floating point
> >>> numbers could make such a big difference.
> >>
> >>
> >>
> >>  From HtDP/1e:
> >>
> >> (define JANUS
> >>    (list #i31
> >>          #i2e+34
> >>          #i-1.2345678901235e+80
> >>          #i2749
> >>          #i-2939234
> >>          #i-2e+33
> >>          #i3.2e+270
> >>          #i17
> >>          #i-2.4e+270
> >>          #i4.2344294738446e+170
> >>          #i1
> >>          #i-8e+269
> >>          #i0
> >>          #i99))
> >>
> >>
> >>
> >> ;; [List-of Number] -> Number
> >> ;; add numbers from left to right
> >> (check-expect (sumlr '(1 2 3)) 6)
> >> (define (sumlr l)
> >>    (foldl + 0 l))
> >>
> >> ;; [List-of Number] -> Number
> >> ;; add numbers from right to left
> >> (check-expect (sumrl '(1 2 3)) 6)
> >> (define (sumrl l) (foldr + 0 l))
> >>
> >> Then apply the two functions to JANUS. Enjoy -- Matthias
> >
> >
> > Nice example!
> >
> > You could also (require math) and apply its `sum' or `flsum' to JANUS.
> Then
> > *really* enjoy. :D
> >
> >> (sumlr JANUS)
> > 99.0
> >
> >> (sumrl JANUS)
> > -1.2345678901235e+80
> >
> >> (sum JANUS)
> > 4.2344294738446e+170
> >
> >> (exact->inexact (sumlr (map inexact->exact JANUS)))
> > 4.2344294738446e+170
> >
> > On my computer, using `sum' is about 20x faster than converting JANUS to
> > exact numbers.
> >
> > You can also sort by absolute value before summing, which is a little
> faster
> > still but loses some precision. Do not trust Teh Internets on this one.
> > Popular Q-and-A sites say to sort ascending, which makes intuitive sense:
> > adding a big number to two small numbers in turn might do nothing, but
> > adding a big number to their *sum* might result in something larger.
> >
> >> (expt 2 53.0)
> > 9007199254740992.0
> >
> >> (sumlr (list (expt 2 53.0) 1.0 1.0))
> > 9007199254740992.0
> >
> >> (sumlr (list 1.0 1.0 (expt 2 53.0)))
> > 9007199254740994.0
> >
> > But JANUS shows that sorting ascending doesn't work when summing huge
> > numbers with alternating signs:
> >
> >> (sumlr (sort JANUS (λ (x y) (< (abs x) (abs y)))))
> > 0.0
> >
> >> (sumlr (sort JANUS (λ (x y) (> (abs x) (abs y)))))
> > 4.2344294738446e+170
> >
> > All the research papers on summation by sorting sort descending,
> contrary to
> > the wisdom of Teh Internets. So either do that, or use `sum' or `flsum'
> when
> > you want an accurate sum of flonums.
> >
> > Neil ⊥
> >
> >
> > ____________________
> >  Racket Users list:
> >  http://lists.racket-lang.org/users
>
> ____________________
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>   http://lists.racket-lang.org/users
>
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