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<ba...@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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