Here is my list-based version. There are redundant calls to get the length of the same list, but I didn't feel like factoring them out (call it an exercise for the reader). The key to its simplicity is that shifting an element is a similarity transform of shifting the first element, with pre- and post list rotation. The shift just maps (h:xs++ys) to xs++(h:ys). Rotating a list is easy with the drop . cycle pattern, and shares the list up to the point of rotation (when brought in from cycle 2 to end the list).

Dan

module RotateList where

import Control.Arrow((&&&))

rotateList :: Int -> [a] -> [a]
rotateList offset = uncurry take
                  . (length &&& uncurry drop .
                                (mod offset  . length &&& cycle))

shiftElem :: Int -> [a] -> [a]
shiftElem  _      []    = []
shiftElem  offset (h:t) = a ++ (h:b)
            where (a,b) = splitAt ((offset-1) `mod` (length t) + 1) t

-- rotateElem is a similarity transform of shiftElem
rotateElem :: Int -> Int -> [a] -> [a]
rotateElem start offset = rotateList (negate start)
                        . shiftElem  offset
                        . rotateList start


kevin birch wrote:
On 火, 2007-6月-05, at 02:54, Greg Fitzgerald wrote:

> rotating the fourth element 2 positions would result in: [1, 2, 4, 3, 5]
Seems odd.  Should that be [4,1,2,3,5]?

Yes, I meant to use the 5 element in my second example. Sorry for the confusion.

> Is there an idomatic way to handle both of these cases in a function?
Generally people like to see your attempt at a solution before giving the idomatic one so that they are sure it's not a homework question. What do you have so far?

Yeah, I only wish I had gone to a school that would be forward thinking enough to each FP. ;-)

Here is my version:

rotate :: Array Integer Card -> Integer -> Integer -> Array Integer Card
rotate a i n
| i <= u - n = a // [(i, a ! (i + 1)), (i + 1, a ! (i + 2)), (i + 2, a ! i)] | otherwise = a // zip [l..u] (h ++ [a ! i] ++ filter (not . (== (a ! i))) t)
    where (l, u) = bounds a
          (h, t) = splitAt (fromInteger ((i - u) + n)) $ elems a

This function is part of my implementation of the Solitaire encryption algorithm, so that is why I have the reference to a Card data type. This does what I want, and seems basically idiomatic, but perhaps it could be better.

Thanks,
Kevin


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