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

# Get LR instructions
( input / "" | map(if . == "L" then 0 else 1 end )) as $LR |
( $LR | length ) as $l |

# Make map {"AAA":["BBB","CCC"], ...}
(
  [
    inputs | select(.!= "") | [ scan("[A-Z]{3}") ] | {(.[0]): .[1:]}
  ] | add
) as $map |

# List primes for GCM test / factorization
[
  2, 3, 5, 7, 11, 13, 17, 19,
  23, 29, 31, 37, 41, 43, 47,
  53, 59, 61, 67, 71, 73, 79,
  83, 89, 97
] as $primes |

reduce (
  $map | keys[] | select(test("..A")) | { s: 0, i: 0, c: .} |

  # For each "..A" starting position
  # Produce visited [ "KEY", pos mod $l ], until loop is detected
  until (.i as $i | .[.c] // [] | contains([$i]);
    .s as $s | .i as $i | .c as $c        |
    $map[$c][$LR[$i]] as $next            | # Get next KEY
    .[$c] = (( .[$c] // [ $s ] ) + [$i] ) | # Append ( .s ≡ $l ) to array for KEY (first = .s non mod)
    .s = ( $s + 1 )  | .i = (.s % $l )    | # Update cursors, for next iteration
    .c = $next
  )
  | .[.c][0] as $start_loop_idx | (.s - $start_loop_idx) as $loop_size
  | [ to_entries[] | select(.key[-1:] == "Z") ]
  | if (
      length != 1                                           # Only one ..Z per loop
      or ( .[0].value[0] != $loop_size )                    # First ..Z idx = loop size
      or ( [ .[0].value[0] / $l ] | inside($primes) | not ) # loop_size = ( prime x $l )
      or ( .[0].value[0] / $l  > $l )                       # GCM(loop_sizes) = $l
    ) then "Input does not fit expected pattern" | halt_error else
      # Under these conditions, synched positions of ..Zs happen at:
      # LCM = Π(loop_size_i / GCM) * GCM

      # loop_size_i / GCM
      .[0].value[0] / $l
    end
) as $i (1; . * $i)

# Output LCM = first step where, all ghosts are on "..Z" nodes
| . * $l

I read the problem as follows: for N galaxies, find N/2 pairings such that the sum of distances is minimal.

At this point, I was like, wow, the difficulty has ramped up. A DP? That will probably run out of memory with most approaches, requiring a BitSet. I dove in, eager to implement a janky data structure to solve this humdinger.

I wrote the whole damn thing. The bitset, the DP, everything. I ran the code, and WOAH, that number was much smaller than the sample answer. I reread the prompt and realised I had it all wrong.

It wasn't all for naught, though. A lot of the convenience stuff I'd written was fine. Also, I implemented a sparse parser, which helped for b.

b: I was hoping they were asking for what I had accidentally implemented for a. My hopes were squandered.

Anyway, this was pretty trivial with a sparse representation of the galaxies.

part b: Before diving into the discussion, I timed how long 1000 cycles takes to compute, and apparently, it would take 1643175 seconds or just over 19 days to compute 1 billion cycles naively. How fun!

So, how do you cut down that number? First, that number includes a sparse map representation, so you can tick that box off.

Second is intuiting that the result of performing a cycle is cyclical after a certain point. You can confirm this after you've debugged whatever implementation you have for performing cycles- run it a few times on the sample input and you'll find that the result has a cycle length of 7 after the second interaction.

Once you've got that figured out, it's a matter of implementing some kind of cycle detection and modular arithmetic to get the right answer without having to run 1000000000 cycles. For the record, mine took about 400 cycles to find the loop.

a) I used a regex to do some parsing because I haven't looked at dart regex much and wanted to dip my toes a little.

I considered doing this "properly" with OO classes and subclasses for the different rules. I felt that it would be too difficult and just wrote something janky instead. In hindsight, this was probably the wrong choice, especially since grappling with all the nullable types I had in my single rule class became a little too complex for my melting brain (it is HOT in Australia right now; also my conjunctivae are infected from my sinus infection. So my current IQ is like down 40 IQ points from its normal value of probably -12)

b) There may have been a trick here to simplify the programming (not the processing). Again, I felt that directly implementing the specified algorithm was the only real way forward. In brief:

1. Start with an "open" set containing a part with 4 ranges from [1, 4001) and an "accepted" set that is empty.
2. Start at the workflow "in"
3. For each rule in the current workflow:

• If the rule accepts part of the ranges in the open set, remember those ranges in a closed set and remove them from the open set.
• Remove anything the rule rejects from the open set.
• If the rule redirects to a different workflow W, split off the applicable ranges and recurse at 3 with the current workflow as W.
• Keep in the open set anything the rule doesn't consider.

Because this is AOC, I assumed that the input would be nice and wouldn't have anything problematic like overlapping ranges, and I was right. I had a very stupid off by one error that took me a while to find as well.

b. It's not much of a difference to solve b. The trick here is that if you did the bit stuff I suggested above, you'd quickly realise that a smudge interpreted as a binary number is a power of two. Finding a smudge is equivalent to if the bitwise XOR of two numbers is a power of 2, which can be done with some bitwise magic.

Completed when waiting for the second leg of my Christmas holidays flight. (It was a long wait, can I blame jet-lag?).

Have a more compact implementation of LCM/GCD, something tells me it will come in handy In future editions. (I’ve also progressively been doing past years)

Essentially it reaches a (bi stable?) steady state between two numbers, which makes sense- if you can only make single rook steps, then the reachable squares will alternate every cycle.

Don't know if I'll try solve this again tonight but mentally I have now understood the solution.

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?