Difference between revisions of "99 questions/Solutions/14"

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or, using silliness:
 
or, using silliness:
 
<haskell>
 
<haskell>
dupli = foldr (\x -> ((x:) . (x:)) []
+
dupli = foldr (\x -> (x:) . (x:)) []
 
</haskell>
 
</haskell>
  +
  +
or, even sillier:
  +
<haskell>
  +
dupli = foldr ((.) <$> (:) <*> (:)) []
  +
</haskell>
  +
  +
and here is the proof that <hask>((.) <$> (:) <*> (:)) = (\y z -> y:y:z)</hask>:
  +
  +
<haskell>
  +
(.) <$> (:) <*> (:) =
  +
((.) <$> (:)) <*> (:) = -- (<$>) is infixl 4, (<*>) is infixl 4
  +
((.) . (:)) <*> (:) = -- (<$>) == (.) 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
  +
</haskell>
  +
  +
  +
[[Category:Programming exercise spoilers]]

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])

or, using concatMap:

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

also using concatMap:

dupli = concatMap (replicate 2)

or, using foldl:

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

or, using foldr:

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

or, using silliness:

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

or, even sillier:

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

and here is the proof that ((.) <$> (:) <*> (:)) = (\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