Rules: no spoilers.

The other rules are made up as we go along.

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  • @geriksonOPEnglish
    3•10 months ago

    Day 9: Mirage Maintenance

    My solution: https://github.com/gustafe/aoc2023/blob/main/d09-Mirage-Maintenance.pl

    discussion

    What can I say. Shockingly simple.

    I just literally followed the instructions, and got a solution in 20ms. This despite literally creating each intermediate array yet only using the ends. I’m sure I used way to much memory but you know? I’m using a $5/mo VPS for everything and unless I’m barking totally up the wrong tree I’ve never exceeded its memory limits.

    On the subreddit I see people discussing recursion and “dynamic programming” (which is an empty signifier imho) but I really don’t see the need, unless you wanna be “elegant”

    • @swlabrEnglish
      3•9 months ago
      spoiler

      DP to me is when you use memoisation and sometimes recursion and you want to feel smarter about what you did.

      I also struggle to think of the need for DP, even in a more “elegant” approach. Maybe if you wanted to do an O(n) memory solution instead of n^2, or something. Not saying this out of derision. I do like looking at elegant code, sometimes you learn something.

      I feel like there’s an unreadable Perl one line solution to this problem, wanna give that a go, @gerikson?

      • @zogwargEnglish
        3•9 months ago
        spoiler

        Part 2 only, but Part 1 is very similar.

        #!/usr/bin/env jq -n -R -f
[
  # For each line, get numbers eg: [ [1,2,3] ]
  inputs / " " | map(tonumber) | [ . ] |

  # Until latest row is all zeroes
  until (.[-1] | [ .[] == 0] | all;
   . += [
     # Add next row, where for element(i) = prev(i+1) - prev(i)
     [ .[-1][1:] , .[-1][0:-1] ] | transpose | map(.[0] - .[1])
    ]
  )
  # Get extrapolated previous element for first row
  |  [ .[][0] ] | reverse | reduce .[] as $i (0; $i - . )
]

# Output sum of extapolations for all lines
| add

        I’m pretty sure you could make this one line and unreadable ^^.

        • @swlabrEnglish
          3•9 months ago

          Now this is content

        • @swlabrEnglish
          3•9 months ago

          Here’s where I landed in dart

          no comments 
          d9(bool s) {
  print(getLines().fold(0, (p, e) {
    int pre(List h, bool s) {
      return h.every((e) => e == 0)
          ? 0
          : (pre(List.generate(h.length - 1, (i) => h[i + 1] - h[i]), s)) *
                  (s ? -1 : 1) +
              (s ? h.first : h.last);
    }

    return p + pre(stois(e), s);
  }));
}