Sometimes it is interesting to approach things from a different perspective. The goofyness below will have been useful if it interests you in point-free programming. And anyway, I sure had a good time writing it... I am composing several common Haskell idioms: 1) When adjacent list elements are paired is (duplicate,offset one of them, zip together): Prelude> let z = [2,3,5] in zipWith (*) z (tail z) [6,15] 2) Uninterleaving (slicing) a list: Prelude> let slice = fst . foldr (\a ~(x,y) -> (a:y,x)) ([],[]) Prelude> slice [1..10] [1,3,5,7,9] For fun, I will do the slicing by changing the odd-indexed elements to Just x, and the even ones to Nothing, then filter only the Justed elements. There is a twist in your problem: do something special with the last odd (orphaned) element, i.e. append it. One thing to watch out for is that functions like head, tail, last are partial functions that don't play well with the empty list. In my case, I forstall disaster by prepending an unused Nothing value to my list before calling last. For comparison with your concise solution, below is my much more obfuscated one. Believe it or not, this algorithm was the first one that popped into my head. I have encoded it in point-free notation so that it reads along with its algorithm description (and because it's so much fun!) If the use of &&& and >>> is new to you, just read through to get a taste of how it can be used. Note that there is no variable name anywhere in the code, only functions! Solving the same problem several different ways will help build your toolbox. I am not recommending my solution for your exam (it's longer and less efficient than yours), only pointing out that there is always more than one way to do something in Haskell. :) import Control.Arrow((&&&),(>>>)) import Data.Maybe(catMaybes,maybeToList) mapPair = (id &&& tail >>> -- offset list by one uncurry (zipWith (*)) >>> -- multiply adjacent alternate >>> -- mark even elements catMaybes) -- delete even elements &&& -- Tuple this up with... (alternate >>> -- keep odd indices (Nothing:) >>> -- make sure there is a last last >>> -- get last maybeToList) -- keep if it had odd index >>> -- and then... uncurry (++) -- append pair of lists where alternate = zipWith ($) (cycle [Just,const Nothing]) -- Mark even-indexed elements for deletion -- cycle goes on forever, but zipWith stops at -- the end of the shorter list, so no worries. In this formulation, you can see that the whole second half above is there just to handle the orphaned last element, suggesting that there is something unnatural in the problem specification (probably put there to keep you from just using zipWith!). Here is a worked example for [2,3,5,7,11] in case you're not used to point-free programming: (id &&& tail >>> -- ([2,3,5,7,11],[3,5,7,11]) uncurry (zipWith (*)) >>> -- [6,15,35,77] alternate >>> -- [Just 6,Nothing,Just 35,Nothing] catMaybes) -- [6,35] &&& (alternate >>> -- [Just 2,Nothing,Just 5,Nothing,Just 11] (Nothing:) >>> -- [Nothing,Just 2,Nothing,Just 5, -- Nothing,Just 11] last >>> -- Just 11 maybeToList) -- [11] >>> -- ([6,35],[11]) uncurry (++) -- [6,35,11] Dan Weston Alexteslin wrote:
Alexteslin wrote:
Hello,
I just came across with this question on the exam and can not think of implementing it.
mapPair :: (a -> a -> a) -> [a] -> [a]
such that mapPairs f [x1, x2, x3, x4...] = [f x1 x2, f x3 x4,...]
and if the list contains an odd number of elements, the last one is kept unchanged, for example
mapPairs f [x1, x2, x3] = [f x1 x2, x3]
Any ideas will be appreciated, thanks
Oh, I think i just defined it - seems to work. I spent some time on the exam and made silly mistakes:
mapPairs :: (a -> a -> a) -> [a] -> [a] mapPairs f [x] = [x] mapPairs f [] = [] mapPairs f (x:xs) = f x (head xs) : mapPairs f (tail xs)
I was concing f x to (head xs) as well and mapPairs f [x] = x - this is very silly mistake.
Does anyone think this is the right answer?