The problem is
  lexOrder s@[_] = s
where you just give back what you receive, i.e. [Char].
But you claim to give back [[Char]].
Try [s] on the right-hand side.

On 05/04/2011 02:41 PM, Barbara Shirtcliff wrote: 
On May 4, 2011, at 7:21 AM, Ivan Lazar Miljenovic wrote:


On 4 May 2011 13:13, Barbara Shirtcliff <barcs@gmx.com> wrote:

Hi,

In the following solution to problem 24, why is nub ignored?
I.e. if you do lexOrder of "0012," you get twice as many permutations as with "012," even though I have used nub.

[snip]

lexOrder :: [Char] -> [[Char]]
lexOrder s
 | length s == 1    = [s]
 | length s == 2    = z : [reverse z]
 | otherwise        = concat $ map (\n -> h n) [0..((length s) - 1)]
                   where z = sort $ nub s -- why is the nub ignored here?
                         h :: Int -> [String]
                         h n = map (z!!n :) $ lexOrder $ filter (\c -> lexI c z /= n) z

As a guess, I think it's from the usage of length on the right-hand size.

Also, note that "lexOrder s@[_] = [s]" is nicer than "lexOrder s |
length s == 1 = [s]".

I agree that that initial version was a little clumsy, but your suggestion doesn't really seem to work:


lexOrder :: [Char] -> [[Char]]
lexOrder s@[_] = s
lexOrder s =
         concat $ map (\n -> h n) [0..((length z) - 1)]
         where z = sort $ nub s 
               h :: Int -> [String]
               h n = map (z!!n :) $ lexOrder $ filter (\c -> lexI c z /= n) z


Euler.hs:8:18:
    Couldn't match expected type `[Char]' with actual type `Char'
    Expected type: [[Char]]
      Actual type: [Char]
    In the expression: s
    In an equation for `lexOrder': lexOrder s@[_] = s




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