It is only going to be downhill from here

  • Avicenna@programming.devOP
    1·
    2 days ago

    Based on the star distribution today (everyone who has solved part 1 also solved part2) I suspect at least part 2 is easy even if I can’t make a dent on part1 yet

  • NominatedNemesis@reddthat.com
    1·
    3 days ago

    It was easy staight forward brtue force:

    Spoiler 
    For each possible_rectange Find Max:
   If polygon.contains(rectangle) Then rectangle.area() Else 0

    For speed the loop can be run paralelley.

    I had no time to implent the contains myself, but pretty much all languages have a library already.

    • Avicenna@programming.devOP
      2·
      3 days ago

      yea becomes trivial if one uses a polygon library (I did too) but looking at the actual polygon, one can cook up some clever heuristics to do it quite quickly with some basic checks I think.

      • NominatedNemesis@reddthat.com
        1·
        3 days ago

        I’ll check if someone created a guide for the clever solution, cos I’m interested. But I don’t have nearly enough knowledge about geometry to came up with one myshelf. (nor energy after the 10 hour workdays in the last two weeks…)

  • Deebster@programming.dev
    2·
    5 days ago

    I can’t code today because of travel, just as the one that might run slow enough to geek out on benchmarking the various options. Does seem like it’s a bit of a jump in difficulty from previous days.