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

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