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.toro...@gmail.com> wrote: > On 11/06/2013 09:24 AM, Matthias Felleisen wrote: >> >> >> On Nov 6, 2013, at 7:13 AM, Ben Duan <yfe...@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 ____________________ Racket Users list: http://lists.racket-lang.org/users
