Here is a Python version (http://pastebin.com/reW5eaCy):
def valid(a,b,c,d):
return set(w*a + x*b + y*c + z*d
for w in [-1,0,1]
for x in [-1,0,1]
for y in [-1,0,1]
for z in [-1,0,1]) > set(range(1,41))
ws = [(a,b,c,d) for a in range(1,41-1-1-1)
for b in range(a,41-a-1-1)
for c in range(b,41-a-b-1)
for d in range(c,41-a-b-c)
if valid(a,b,c,d)]
I wonder if you can make the cKanren version just as declarative as
this one (cKanren's purpose being declarative).
Jules
On 1 nov, 04:48, David Nolen <dnolen.li...@gmail.com> wrote:
> A blog post explaining the solution step by
> stephttp://dosync.posterous.com/another-taste-of-ckanren.
>
>
>
>
>
>
>
> On Mon, Oct 31, 2011 at 9:36 PM, David Nolen <dnolen.li...@gmail.com> wrote:
> > Here's a correct version that solves the puzzle in ~12ms,
> >https://gist.github.com/1329580. A bit longer but it fun to combine
> > constraints w/ search.
>
> > Will try to find some time to write a more detailed explanation.
>
> > On Sun, Oct 30, 2011 at 9:00 PM, David Nolen <dnolen.li...@gmail.com>wrote:
>
> >> Heh, as someone pointed out this doesn't actually solve the puzzle since
> >> I'm not considering putting stones on either side of the scale. Still I
> >> think the idea of what cKanren can do is communicated :)
>
> >> On Sun, Oct 30, 2011 at 8:34 PM, Brent Millare
> >> <brent.mill...@gmail.com>wrote:
>
> >>> Looks really cool. Can't wait to see the talk.
>
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