Day 18: Ram Run

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
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FAQ

  • sjmulder@lemmy.sdf.org
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    8 days ago

    I think I saw the same! At first I thought it requires pathfinding to see what nodes are connected to the wall, but then someone pointed at disjoint sets and just a glance at Wikipedia made it click right away. What an ingeniously simple but useful data structure! Maybe I’ll reimplement my solution with that - mostly as an exercise for disjoint sets and finding a convenient representation for that in C.

    • Acters@lemmy.world
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      8 days ago

      That would be cool af to see in C, let me know if you do. In python, we can built the two sets, and have the convenient function call of set( [iterate-able object/list/set] ).intersection( [iterate-able object/list/set] ) to see if the two sets touches/intersects as the block that connects the two sets would be in both sets/lists.

      The way I would build the two sets would be to start at the final state with all blocks placed and just union-find all the blocks. When we find that a block appears in both sets, then we stop the union and proceed with the other unions until we find all the blocks that would appear in both sets. then we iteratively find the first block that would appear in both sets. In python the intersection call returns a set, so you can stack the intersect call. like so: set( [top right union set] ).intersection( [bottom left union set] ).intersection( [ one item list with the current block we are checking ] ) technically you can just save the intersections of the first two sets to save a little time because they would not change.

      I didn’t think of this until recently, but I also think it is such a simple and elegant solution. Live and learn! 😄

      hope you are having a good holiday season!