[Haskell-cafe] Re: Function to detect duplicates
Daniel Fischer
daniel.is.fischer at web.de
Fri Feb 26 16:03:08 EST 2010
Am Freitag 26 Februar 2010 21:34:28 schrieb Ketil Malde:
> Daniel Fischer <daniel.is.fischer at web.de> skrev:
> > Am Freitag 26 Februar 2010 16:50:42 schrieb Ketil Malde:
> >>> solutions = [[x1,x2,x3,x4,x5,x6,x7,x8,x9,x10]
> >>>
> >>> | x1 <- [0..9]
> >
> > First digit can't be 0, so make it [1 .. 9].
> > Since you use the fact that the last digit must be the 0, pull all
> > others from [1 .. 9].
>
> Originally, I pulled from alternating odds (x1 <- [1,3..9] etc) and
> evens, since this is fairly easy to deduce... I reverted this since the
> point was to use brute force.
Yes, but did you forget x10 or did you think that one was too obvious?
>
> >>> solve :: [Int] -> [[Int]]
> >
> > Not on a 32-bit system. Word would suffice there, but you don't know
> > that in advance, so it'd be Int64 or Integer
>
> Hm? The Ints are just individual digits here.
>
Yup. I didn't realise that you don't call val for the 10-digit number(s).
If you also did x10 <- [0 .. 9] and checked
val [x1, x2, ..., x10] `mod` 10 == 0, it would overflow, that's what I was
thinking of.
> > I would make the length of the prefix a parameter of solve.
>
> I thought about generating a list with solutions for increasing lenghts,
> so that e.g. 'solve [] !! 10' would solve this particular problem.
>
That's nice, but I think it'd be ugly with a DFS, much nicer with a BFS,
like Rafael did.
> > solve prefix =
> > case length prefix of
> > 10 -> return prefix
> > l -> do
> > x <- [0 .. 9]
> > ...
> >
> > over the if-then-else.
>
> Yes, much nicer. Thanks for the feedback!
>
> -k
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