Difference between revisions of "99 questions/Solutions/14"
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</haskell> |
</haskell> |
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| + | or, using the list instance of <hask>Applicative</hask>: |
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| ⚫ | |||
| + | <haskell> |
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| + | dupli = (<**> [id,id]) |
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| + | </haskell> |
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| + | |||
| ⚫ | |||
<haskell> |
<haskell> |
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dupli = concatMap (\x -> [x,x]) |
dupli = concatMap (\x -> [x,x]) |
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</haskell> |
</haskell> |
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| − | also using concatMap: |
+ | also using <hask>concatMap</hask>: |
<haskell> |
<haskell> |
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dupli = concatMap (replicate 2) |
dupli = concatMap (replicate 2) |
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| + | </haskell> |
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| + | |||
| + | or, using <hask>foldl</hask>: |
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| + | <haskell> |
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| + | dupli = foldl (\acc x -> acc ++ [x,x]) [] |
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</haskell> |
</haskell> |
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| − | or, using foldr: |
+ | or, using <hask>foldr</hask>: |
<haskell> |
<haskell> |
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dupli = foldr (\ x xs -> x : x : xs) [] |
dupli = foldr (\ x xs -> x : x : xs) [] |
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</haskell> |
</haskell> |
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| + | |||
| + | or, using silliness: |
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| + | <haskell> |
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| + | dupli = foldr (\x -> (x:) . (x:)) [] |
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| + | </haskell> |
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| + | |||
| + | or, even sillier: |
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| + | <haskell> |
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| + | dupli = foldr ((.) <$> (:) <*> (:)) [] |
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| + | </haskell> |
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| + | |||
| + | and here is the proof that <hask>((.) <$> (:) <*> (:)) = (\y z -> y:y:z)</hask>: |
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| + | |||
| + | <haskell> |
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| + | (.) <$> (:) <*> (:) = |
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| + | ((.) <$> (:)) <*> (:) = -- (<$>) is infixl 4, (<*>) is infixl 4 |
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| + | ((.) . (:)) <*> (:) = -- (<$>) == (.) for functions |
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| + | (\x -> (.) (x:)) <*> (:) = -- definition of (.) |
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| + | \y -> (\x -> (.) (x:)) y (y:) = -- definition of (<*>) for functions |
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| + | \y -> ((.) (y:)) (y:) = -- beta reduction (applying y to (\x -> (.) (x:))) |
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| + | \y -> (y:) . (y:) = -- changing (.) to its prefix form |
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| + | \y -> (\z -> (y:) ((y:) z)) = -- definition of (.) |
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| + | \y z -> y:y:z -- making it look nicer |
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| + | </haskell> |
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| + | |||
| + | |||
| + | [[Category:Programming exercise spoilers]] |
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Latest revision as of 20:45, 28 December 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 the list instance of Applicative:
dupli = (<**> [id,id])
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