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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