My current code (https://github.com/Fluxward/aoc2023/blob/main/21.dart) has a memoisation based approach but my current ailments are preventing me from really thinking about this properly so I am bowing out until I have the wherewithal.

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"

So following from yesterday where I was punished by not going full OO, I decided, hey, this is a problem in which I can do some OOP, so I did. This took very long to do but I regret nothing. If you look at my code, feel free to click your tongue and shake your head at every opportunity I missed to use a design pattern.

Anyway, after a slight snafu with misunderstanding the FF logic and not spotting that some modules can be dummy modules, it all just worked, and I got my answer.

B

This was a bit of a headscratcher, but the answer was surprisingly simple.

First, the answer. Here's how to do it:

• Look for the "rx" module in your input.
• If the module that outputs to rx is a conjunction, keep track of how many button presses it takes for each input of the conjunction to change. The answer is the LCM of all those numbers.
• If the module is a FF, you can also work backwards to get it, but this was not my input so I didn't need to try this.

Getting here was a bit weird. I hoped that I could just run the code from A and spit out the answer when rx went low, but as of time of writing I've been running it now on a separate machine for about an hour and still no result.

My next instinct was to simply work it out from pen and paper. I thought it might be possible (it probably is) but decided to at least experimentally see if the states of the modules connected to rx were cyclic first. I did, and that was enough for me to get to the answer.

My answer was about 230 trillion BPs, which, extrapolating on how long execution is taking on my other machine, might take just under 137 years to calculate naively. Fun!

Further cleaned up version: https://github.com/zogwarg/advent-of-code/blob/main/2023/jq/12-b.jq

Also lost a fair amount of time not not noticing that the sequence should be joined with "?" not with "". (that'll teach me to always run on the example before the full input, when execution time is super long).

Execution time: 17m10s (without memoization a single row was taking multiple minutes, and there's 1000 rows ^^...)

EDIT: see massive improvement by running in parallel in reply.

Satisfyingly short (in lines, not in time writing) some of the longer part is hexadecimal parsing, that doesn't come natively in JQ, I started doing polygon math from part 1, and what took me the longest was properly handling the area contributed by the perimeter. (I toyed with trying very annoying things like computing the outmost vertex at each turn, which is complicated by the fact that you don't initially know which way the digger is turning, and needing previous and next point to disambiguate).

#!/usr/bin/env jq -n -R -f

reduce (
  # Produce stream of the vertices, for the position of the center
  foreach (
    # From hexadecimal representation
    # Get inputs as stream of directions = ["R", 5]
    inputs | scan("#(.+)\\)") | .[0] / ""
    | map(
        if tonumber? // false then tonumber
        else {"a":10,"b":11,"c":12,"d":13,"e":14,"f":15}[.] end
      )
    | [["R","D","L","U"][.[-1]], .[:-1]]
    | .[1] |= (
      # Convert base-16 array to numeric value.
      .[0] * pow(16;4) +
      .[1] * pow(16;3) +
      .[2] * pow(16;2) +
      .[3] * 16 +
      .[4]
    )
  ) as $dir ([0,0];
    if   $dir[0] == "R" then .[0] += $dir[1]
    elif $dir[0] == "D" then .[1] += $dir[1]
    elif $dir[0] == "L" then .[0] -= $dir[1]
    elif $dir[0] == "U" then .[1] -= $dir[1]
    end
  )
  # Add up total area enclosed by path of center
  # And up the are of the perimeter, perimeter * 1/2 + 1
) as [$x, $y] ( #
  {prev: [0,0], area: 0, perimeter_area: 1  };

  # Adds positve rectangles
  # Removes negative rectangles
  .area += ( $x - .prev[0] ) * $y |

  # Either Δx or Δy is 0, so this is safe
  .perimeter_area += (($x - .prev[0]) + ($y - .prev[1]) | abs) / 2 |

  # Keep current position for next vertex
  .prev = [$x, $y]
)

# Output total area
| ( .area | abs ) + .perimeter_area

Satisfyingly very well suited to JQ once you are used to the stream, foreach(init; mod; extract) and recurse(exp) [where every output item of exp as a stream is fed back into recurse] operators. It's a different way of coding but has a certain elegance IMO. This was actually quick to implement, along with re-using the treating a range as a primitive approach of the seeds-to-soil day.

#!/usr/bin/env jq -n -sR -f

inputs / "\n\n"

# Parse rules
| .[0] / "\n"
| .[] |= (
  scan("(.+){(.+)}")
  | .[1] |= (. / ",")
  | .[1][] |= capture("^((?＜reg＞.)(?＜op＞[^\\d]+)(?＜num＞\\d+):)?(?＜to＞[a-zA-Z]+)$")
  | ( .[1][].num | strings ) |= tonumber
  | {key: .[0], value: (.[1]) }
) | from_entries as $rules |

# Split part ranges into new ranges
def split_parts($part; $rule_seq):
  # For each rule in the sequence
  foreach $rule_seq[] as $r (
    # INIT = full range
    {f:$part};

    # OPERATE =
    # Adjust parts being sent forward to next rule
    if $r.reg == null then
      .out = [ .f , $r.to ]
    elif $r.op == "＜" and .f[$r.reg][0] ＜ $r.num then
      ([ .f[$r.reg][1], $r.num - 1] | min ) as $split |
      .out = [(.f | .[$r.reg][1] |= $split ), $r.to ] |
      .f[$r.reg][0] |= ($split + 1)
    elif $r.op == "＞" and .f[$r.reg][1] ＞ $r.num then
      ([ .f[$r.reg][0], $r.num + 1] | max ) as $split |
      .out = [(.f | .[$r.reg][0] |= $split), $r.to ]  |
      .f[$r.reg][1] |= ($split - 1)
    end;

    # EXTRACT = parts sent to other nodes
    # for recursion call
    .out | select(all(.[0][]; .[0] ＜ .[1]))
  )
;

[ # Start with full range of possible sings in input = "in"
  [ {x:[1,4000],m:[1,4000],a:[1,4000],s:[1,4000]} , "in" ] |

  # Recusively split musical parts, into new ranges objects
  recurse(
    if .[1] == "R" or .[1] == "A" then
      # Stop recursion if "Rejected" or "Accepted"
      empty
    else
      # Recursively split
      split_parts(.[0];$rules[.[1]])
    end
    # Keep only part ranges in "Accepted" state
  ) | select(.[1] == "A") | .[0]

  # Total number if parts in each object is the product of the ranges
  | ( 1 + .x[1] - .x[0] ) *
    ( 1 + .m[1] - .m[0] ) *
    ( 1 + .a[1] - .a[0] ) *
    ( 1 + .s[1] - .s[0] )
  # Sum total number of possibly accepted musical parts
] | add

EDIT: Less-thans and greater-thans replaced by fullwidth version, because lemmy is a hungry little goblin.