# 99 questions/Solutions/14

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< 99 questions | Solutions(Difference between revisions)

## Revision as of 19:33, 18 January 2014

(*) Duplicate the elements of a list.

dupli [] = [] dupli (x:xs) = x:x:dupli xs

or, using list comprehension syntax:

dupli list = concat [[x,x] | x <- list]

or, using the list monad:

dupli xs = xs >>= (\x -> [x,x])

concatMap

dupli = concatMap (\x -> [x,x])

concatMap

dupli = concatMap (replicate 2)

foldl

dupli = foldl (\acc x -> acc ++ [x,x]) []

foldr

dupli = foldr (\ x xs -> x : x : xs) []

or, using silliness:

dupli = foldr (\x -> (x:) . (x:)) []

or, even sillier:

dupli = foldr ((.) <$> (:) <*> (:)) []

((.) <$> (:) <*> (:)) = (\y z -> y:y:z)

(.) <$> (:) <*> (:) = ((.) <$> (:)) <*> (:) = -- (<$>) is infixl 4, (<*>) is infixl 4 ((.) . (:)) <*> (:) = -- (<$>) == (.) for functions (\x -> (.) (x:)) <*> (:) = -- definition of (.) \y -> (\x -> (.) (x:)) y (y:) = -- definition of (<*>) for functions \y -> ((.) (y:)) (y:) = -- beta reduction (applying y to (\x -> (.) (x:))) \y -> (y:) . (y:) = -- changing (.) to its prefix form \y -> (\z -> (y:) ((y:) z)) = -- definition of (.) \y z -> y:y:z -- making it look nicer