Re: [Haskell-cafe] Re: Function to detect duplicates

26 Feb 2010


      Am Freitag 26 Februar 2010 21:34:28 schrieb Ketil Malde:
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Daniel Fischer  skrev:
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Am Freitag 26 Februar 2010 16:50:42 schrieb Ketil Malde:
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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?
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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.
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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.
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solve prefix =
      case length prefix of
        10 -> return prefix
        l -> do
           x <- [0 .. 9]
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over the if-then-else.
Yes, much nicer.  Thanks for the feedback!
-k