Thursday, July 2, 2020

July 2020 1HaskellADay Problems and Solutions


  • 2020-07-13: Now that we've selected pairings for today, let's store those pairings into an historical context for today's #haskell problem. 
  • 2020-07-08: Pairing and ... 'Mobbing' (?) ... okay, really??? is our game for today's #haskell problem. A solution to the simple pairing-problem.
  • 2020-07-07: For today's #haskell problem, we lay a foundation for building a team-pairing app. This day's problem addresses a History-type to provide context to the pairing-algorithm. 
  • 2020-07-06: For today's #haskell problem, we look at finding the spanning trees of a graph
  • 2020-07-02: Today's #haskell exercise finds us (acyclic) pathing though a simpler, yet-not-fully connected, graph. 
  • 2020-07-01: For today's #haskell problem we find cycles in graphs, ... MOTORcycles in graphs! AHA! AND MAKE MINE A DUCATI! ON FYE-YARRRR! 🔥 ... no ... wait ... Oh, well. #GraphTheory 

June 2020 1HaskellADay Problems and Solutions

  • YAY! HELLO! Our first #haskell exercise in a while!... and this exercise is about ... wait for it ... exercise
  • For today's #haskell exercise we convert a set of arcs to a graph. #GraphTheory 
  • Thursday, October 17, 2019

    February 2019 1HaskellADay Problems and Solutions

    Monday, February 18, 2019

    April/May 2019 1HaskellADay 1Liners

    • May 27th, 2018:
      data F = F { a, b, c :: String }
      data Stamped a = { time :: Day, stamped :: a }

      f :: Stamped F -> String

      output of f x is  "time a b c" for the respective values of time, a, b, c

      Is there some monadic / applicative elegant definition that does this?
      • Nickolay Kudasov @crazy_fizruk
        f = intercalate “ “ . sequence [ show.time, a.stamped, b.stamped, c.stamped ]
      • Nickolay Kudasov @crazy_fizruk
        f = intercalate " " <$> http://pure.show .time <> (sequence [a, b, c]).stamped
        A bit trickier, but shorter and uses stamped once.
    • April 13th, 2018: given f :: [a] -> b -> [c]
      where c is a derived from g :: a -> b -> c

      You have [a] and [b]

      Write a function h :: [a] -> [b] -> [c] from f

    January 2019 1HaskellADay Problems and Solutions

    Monday, January 28, 2019

    December 2018 1HaskellADay Problems and Solutions

    Monday, December 3, 2018

    November 2018 1HaskellADay Problems and Solutions