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The official awful.systems Advent of Code 2023 thread (awful.systems)
submitted 11 months ago* (last edited 11 months ago) by gerikson@awful.systems to c/notawfultech@awful.systems
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Rules: no spoilers.

The other rules are made up as we go along.

Share code by link to a forge, home page, pastebin (Eric Wastl has one here) or code section in a comment.

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[-] gerikson@awful.systems 3 points 11 months ago

Day 9: Mirage Maintenance

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

discussionWhat 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"

[-] swlabr@awful.systems 3 points 11 months ago

spoilerDP 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?

[-] zogwarg@awful.systems 3 points 11 months ago

spoilerPart 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 ^^.

[-] swlabr@awful.systems 3 points 11 months ago

Now this is content

[-] swlabr@awful.systems 3 points 11 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);
  }));
}

this post was submitted on 01 Dec 2023
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