Only Bayes Can Judge Me

  • 14 Posts
  • 579 Comments
Joined 1 year ago
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Cake day: July 4th, 2023

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  • 21!

    Finally managed to beat this one into submission.

    P1

    I created this disgusting mess of a recursive search that happened to work. This problem was really hard to think about due to the levels of indirection. It was also hard because of a bug I introduced into my code that would have been easy to debug with more print statements, but hubris.

    P2

    Recursive solution from P1 was too slow, once I was at 7 robots it was taking minutes to run the code. It didn’t take long to realise that you don’t really care about where the robots other than the keypad robot and the one controlling the keypad robot are since the boundary of each state needs all the previous robots to be on the A button. So with memoisation, you can calculate all the shortest paths for a given robot to each of the directional inputs in constant time, so O(kn) all up where n is the number of robots (25) and k is the complexity of searching for a path over 5 or 11 nodes.

    What helped was looking at the penultimate robot’s button choices when moving the keypad robot. After the first one or two levels, the transitions settle into the table in the appendix. I will not explain the code.

    appendix
      (P(0, 1), P(0, 1)): [],
      (P(0, 1), P(0, 2)): [btn.r],
      (P(0, 1), P(1, 0)): [btn.d, btn.l],
      (P(0, 1), P(1, 1)): [btn.d],
      (P(0, 1), P(1, 2)): [btn.d, btn.r],
      (P(0, 2), P(0, 1)): [btn.l],
      (P(0, 2), P(0, 2)): [],
      (P(0, 2), P(1, 0)): [btn.d, btn.l, btn.l],
      (P(0, 2), P(1, 1)): [btn.l, btn.d],
      (P(0, 2), P(1, 2)): [btn.d],
      (P(1, 0), P(0, 1)): [btn.r, btn.u],
      (P(1, 0), P(0, 2)): [btn.r, btn.r, btn.u],
      (P(1, 0), P(1, 0)): [],
      (P(1, 0), P(1, 1)): [btn.r],
      (P(1, 0), P(1, 2)): [btn.r, btn.r],
      (P(1, 1), P(0, 1)): [btn.u],
      (P(1, 1), P(0, 2)): [btn.u, btn.r],
      (P(1, 1), P(1, 0)): [btn.l],
      (P(1, 1), P(1, 1)): [],
      (P(1, 1), P(1, 2)): [btn.r],
      (P(1, 2), P(0, 1)): [btn.l, btn.u],
      (P(1, 2), P(0, 2)): [btn.u],
      (P(1, 2), P(1, 0)): [btn.l, btn.l],
      (P(1, 2), P(1, 1)): [btn.l],
      (P(1, 2), P(1, 2)): [],
    




  • ok disco

    It took me too long to read the prompt and see that without the shortcuts, it’s a single path. I wasted too much time on search algorithms.

    P1: Here’s what I did: Walk the path. Every time you hit a new grid, check if the two shortcuts you can take will save you 100 ps.

    To calculate the shortcut saving:

    If you index every grid position on the main path from 0, then it takes X ps to reach position X, The time it takes to travel from start to X, then a shortcut to Y, then from Y to the end, is X + 1 + (main path length - Y). The time saved is then just Y - X - 1, modulo maybe like 5 fence post errors.

    P2. The prompt wasn’t really clear about whether or not cheating meant you can only travel through one set of walls before your cheat ends, or if it meant you could just move through walls for 20ps to wherever you could reach. Turns out, it’s the latter.

    The above formula is then a special case of Y - X - manhattan distance(X, Y).



  • Day 19! (the cuervo gold…)

    disc and code

    Ok so my path to this answer was circuitous and I now hate myself a little.

    P1: Ok, a repeated dfs on suffixes. that shouldn’t be too hard. (it was not hard)

    P2: Ok, a repeated dfs is a little too slow for me, I wonder how I can speed it up?

    forgets about memoisation, a thing that you can do to speed this sort of thing up

    I guess the problem is I’m doing an O(mn) match (where m is the number of towels, n is the max towel length) when I can do O(n). I’ll build a prefix tree!

    one prefix tree later

    Ok that still seems to be quite slow. What am I doing wrong?

    remembers that memoisation exists

    Oh I just need to memoise my dp from part 1. Oops.

    Anyway posting the code because I shrunk it down to like two semicolons worth of lines.

    (

    List<String> input = getLines();
    Set<String> ts = Set.from(input.first.split(', '));
    Map<String, int> dp = {};
    
    int dpm(String s) => dp.putIfAbsent(
        s,
        () => s.isNotEmpty
            ? ts
                .where((t) => t.matchAsPrefix(s) != null)
                .map((t) => dpm(s.substring(t.length)))
                .fold(0, (a, b) => a + b)
            : 1);
    
    void d19(bool sub) {
      print(input
          .skip(2)
          .map((s) => dpm(s))
          .map((i) => sub
              ? i
              : i > 0
                  ? 1
                  : 0)
          .reduce((a, b) => a + b));
    }
    





  • 17!

    p1 discussion

    Simultaneously very fun and also the fucking worst.

    Fun: Ooooh, I get to simulate a computer, exciting!

    Worst: Literally 8 edge cases where fucking up even just one can fuck up your hour.

    p2 discussion

    I did this by hand. sort of. I mean I didn’t code up something that found the answer.

    Basically I looked at the program in the input and wrote it out, and realised that A was essentially a loop variable, where the number of iterations was the number of octal digits A would take to represent. The most significant octal digits (octits?) would determine the tail end of the output sequence, so to find the smallest A you can do a DFS starting from the MS octit. I did this by hand.

    EDIT: code. Not gonna explain any of it.
    class Comp {
      List<int> reg;
      List<int> prog;
      int ip = 0;
    
      List<int> output = [];
      late List<(int, bool) Function()> ops;
    
      int get combo => prog[ip + 1] < 4 ? prog[ip + 1] : reg[prog[ip + 1] - 4];
    
      Comp(this.reg, this.prog) {
        ops = [
          () => (reg[0] = (reg[0] >> combo), false),
          () => (reg[1] ^= prog[ip + 1], false),
          () => (reg[1] = combo % 8, false),
          () => (reg[0] != 0) ? (ip = prog[ip + 1], true) : (0, false),
          () => (reg[1] ^= reg[2], false),
          () {
            output.add(combo % 8);
            return (0, false);
          },
          () => (reg[1] = (reg[0] >> combo), false),
          () => (reg[2] = (reg[0] >> combo), false)
        ];
      }
    
      compute() {
        output.clear();
        while (ip < prog.length) {
          if (!ops[prog[ip]]().$2) {
            ip += 2;
          }
        }
      }
    
      reset(int A) {
        ip = 0;
        reg[0] = A;
        reg[1] = 0;
        reg[2] = 0;
      }
    }
    
    void d17(bool sub) {
      List<String> input = getLines();
      Comp c = Comp(
          input.take(3).map((s) => s.split(" ").last).map(int.parse).toList(),
          input.last.split(" ").last.split(",").map(int.parse).toList())
        ..compute();
      print("Part a: ${c.output.join(",")}");
    
      if (!sub) return;
    
      List<int> sols = [];
      bool dfs(int cur) {
        bool found = false;
        sols.add(cur);
        int sol = sols.reduce((a, b) => 8 * a + b);
        c..reset(sol)..compute();
        if (c.prog
            .whereIndexed((i, e) => i >= c.prog.length - c.output.length)
            .foldIndexed(true, (i, p, e) => p && c.output[i] == e)) {
          if (found = c.output.length == c.prog.length) {
            print("Part b: $sol");
          } else {
            for (int i = 0; i < 8 && !(found = found || dfs(i)); i++) {}
          }
        }
    
        sols.removeLast();
        return found;
      }
    
      for (int a = 0; a < 8 && !dfs(a); a++) {}
    }