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

Huh. It looks like they're also using Shewchuk's O(n*log(n)) 
average-case distillation algorithm. Like Racket's math library does. :D

Neil ⊥

On 11/07/2013 04:24 AM, Stephan Houben wrote:
> 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 <mailto: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
>     <mailto: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
>     <mailto: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 ⊥
>      >
>      >
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