Thursday, March 1, 2018

February 2018 1HaskellADay problems and solutions

Tuesday, February 6, 2018

January 2018 1Liner 1HaskellADay problems and solutions

  • January 8th, 2018: from Nicoλas‏ @BeRewt
    A small @1HaskellADay, old-school. Define foo:

    > foo 3 [1..5]
    [([1,2,3], 4), ([2,3,4], 5)]

    > foo 2 [1..4]
    [([1,2], 3), ([2,3], 4)]

    > foo 2 [1..20]
    [([1,2],3), ([2,3],4), ..., ([18,19],20)]

    > foo 20 [1..2]
    []
    • Demiurge With a Teletype @mrkgrnao
      foo n
        = tails
        # filter (length # (> n))
        # map (splitAt n # second head)

      (#) = flip (.)
    • Andreas Källberg @Anka213
      I haven't tested it, but this should work:
      foo n xs = [ (hd,x) | (hd , x:_) <- n="" splitat=""> tails xs ]
    • <- n="" splitat="">Nicoλas @BeRewt foo n = zip <$> fmap (take n) . tails <*> drop n
  • January 5th, 2018: You have the following DAG-paths:

    a -> b -> c -> e
    a -> b -> d -> e
    q -> r -> s
    w -> x
    y -> z

    and many more.

    From a path, provide a bi-directional encoding* given maximum graph depth is, say, 7, max number of roots is, say, 10, and max number of nodes is, say, 1000.
    • *bi-directional encoding of a graph path:

      DAG path -> enc is unique for an unique DAG path
      enc -> DAG path yields the same DAG path that created the unique enc.

      *DAG: "Directed, acyclic graph."
  • January 5th, 2018: given s :: Ord k => a -> (k,[v])

    define f using s

    f :: Ord k => [a] -> Map k [v]

    with no duplicate k in [a]
    • Christian Bay @the_greenbourne f = foldr (\e acc -> uncurry M.insert (s e) acc) M.empty
      • me: you can curry away the acc variable easily
      • Christian Bay @the_greenbourne You're right :)
        f = foldr (uncurry M.insert . s) M.empty
    • Bazzargh @bazzargh fromList.(map s) ?
      • me: Yuppers

Wednesday, January 31, 2018

January 2018 1HaskellADay Problems and Solutions

Friday, January 5, 2018

December 2017 1HaskellADay 1Liners problems and solutions

  • December 29th, 2017:
    given f :: Monad m => n -> a -> m (Maybe b)
    define g :: Monad m => n -> a -> m (a, Maybe b)
    using f and ... arrows? Kleisli category?
    • Bazzargh @bazzargh (\n a->liftM ((,) a) (f n a)) ... according to pointfree.io, that's `liftM2 fmap (,) . f` but I can't pretend to get the transformation
  • December 29th, 2017:
    given f :: a -> b
    define g :: [a] -> [Maybe c] -> [(b, c)]

    >>> g [1,2,3] [Just 7, Nothing, Just 10]
    [("1",7),("3",10)]

    when f = show
    • matt @themattchan
      g = catMaybes ... zipWith (fmap . (,) . f)
      where (...) = (.).(.)
    • garrison @GarrisonLJ g a b = map (f.id***fromJust) . filter (isJust . snd) $ zip a b
    • TJ Takei @karoyakani g = (catMaybes .) . zipWith ((<$>) . (,) . f)
  • December 29th, 2017: define f :: [(a,b)] -> ([a], [b])
    • Андреев Кирилл @nonaem00 and matt @themattchan unzip
    • Victoria C @ToriconPrime f = fmap fst &&& fmap snd
      • (in a vacuum, a more general type signature would be inferred, but the compiler limits itself as instruct)

Tuesday, January 2, 2018

December 2017 1HaskellADay problems and solutions

Friday, December 29, 2017

November 2017 1HaskellADay 1Liner problem and solutions

  • November 5th, 2017: f :: Map Int [a] -> [b] - > [(Int, b)]
    for, e.g.: f mapping bs
    length bs == length (concat (Map.elems mapping))
    define f
    • Andreas Källberg @Anka213 Using parallel list comprehensions:
      f mp bs = [ (k,b) | (k,as) <-assocs mp, a <- as | b <- bs]
    • Steve Trout @strout f = zip . foldMapWithKey (fmap . const)

Thursday, November 30, 2017

November 2017 1HaskellADay problems and solutions