zogwarg

joined 2 years ago
[–] zogwarg@awful.systems 1 points 1 month ago (1 children)

Updated ReasoningOk it probably works because it isn't bang center but a bit up of center, most other steps most be half half noise vertically, and the reason it doesn;t minimize on an earlier horizontal step (where every step is mostly half half), is because the middle points on the trunk, that don't contribute to the overall product therefore minimizing it even lower.

[–] zogwarg@awful.systems 3 points 1 month ago (5 children)

Day 14, got very lucky on this one, but too tired to think about why part 2 still worked.

spoiler

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

#     Board size     # Our list of robots positions and speed #
[101,103] as [$W,$H] | [ inputs | [scan("-?\\d+")|tonumber] ] |

#     Making the assumption that the easter egg occurs when   #
#           When the quandrant product is minimized           #
def sig:
  reduce .[] as [$x,$y] ([];
    if $x < ($W/2|floor) and $y < ($H/2|floor) then
      .[0] += 1
    elif $x < ($W/2|floor) and $y > ($H/2|floor) then
      .[1] += 1
    elif $x > ($W/2|floor) and $y < ($H/2|floor) then
      .[2] += 1
    elif $x > ($W/2|floor) and $y > ($H/2|floor) then
      .[3] += 1
    end
  ) | .[0] * .[1] * .[2] * .[3];

#           Only checking for up to W * H seconds             #
#   There might be more clever things to do, to first check   #
#       vertical and horizontal alignement separately         #
reduce range($W*$H) as $s ({ b: ., bmin: ., min: sig, smin: 0};
  .b |= (map(.[2:4] as $v | .[0:2] |= (
    [.,[$W,$H],$v] | transpose | map(add) 
    | .[0] %= $W | .[1] %= $H
  ))) 
  | (.b|sig) as $sig |
  if $sig < .min then
    .min = $sig | .bmin = .b | .smin = $s 
  end | debug($s)
)

| debug(
  #    Contrary to original hypothesis that the easter egg    #
  #  happens in one of the quandrants, it occurs almost bang  #
  # in the center, but this is still somehow the min product  #       
  reduce .bmin[] as [$x,$y] ([range($H)| [range($W)| " "]];
    .[$y][$x] = "█"
  ) |
  .[] | add
)

| .smin + 1 # Our easter egg step

And a bonus tree:

[–] zogwarg@awful.systems 3 points 1 month ago* (last edited 1 month ago) (1 children)

I liked day 13, a bit easy but in the right way.

Edit:

SpoilersAlthough saying "minimum" was a bit evil when all of the systems had exactly 1 solution (not necessarily in ℕ^2), I wonder if it's puzzle trickiness, anti-LLM (and unfortunate non comp-sci souls) trickiness or if the puzzle was maybe scaled down from a version where there are more solutions.

[–] zogwarg@awful.systems 3 points 1 month ago* (last edited 1 month ago) (1 children)

re:followupIf you somehow wanted your whole final array it would also require over 1 Peta byte ^^, memoization definetely reccomended.

[–] zogwarg@awful.systems 4 points 1 month ago

Day 11

Some hacking required to make JQ work on part 2 for this one.

Part 1, bruteforce blessedly short

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

last(limit(1+25;
  [inputs] | recurse(map(
    if . == 0 then 1 elif (tostring | length%2 == 1) then .*2024 else
      tostring | .[:length/2], .[length/2:] | tonumber
    end
  ))
)|length)

Part 2, some assembly required, batteries not included

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

reduce (inputs|[.,0]) as [$n,$d] ({};     debug({$n,$d,result}) |
  def next($n;$d): # Get next           # n: number, d: depth  #
      if $d == 75                    then          1
    elif $n == 0                     then [1          ,($d+1)]
    elif ($n|tostring|length%2) == 1 then [($n * 2024),($d+1)]
    else #    Two new numbers when number of digits is even    #
      $n|tostring| .[0:length/2], .[length/2:] | [tonumber,$d+1]
    end;

  #         Push onto call stack           #
  .call = [[$n,$d,[next($n;$d)]], "break"] |

  last(label $out | foreach range(1e9) as $_ (.;
    # until/while will blow up recursion #
    # Using last-foreach-break pattern   #
    if .call[0] == "break" then break $out
    elif
      all( #     If all next calls are memoized        #
          .call[0][2][] as $next
        | .memo["\($next)"] or ($next|type=="number"); .
      )
    then
      .memo["\(.call[0][0:2])"] = ([ #                 #
          .call[0][2][] as $next     # Memoize result  #
        | .memo["\($next)"] // $next #                 #
      ] | add ) |  .call = .call[1:] # Pop call stack  #
    else
      #    Push non-memoized results onto call stack   #
      reduce .call[0][2][] as [$n,$d] (.;
        .call = [[$n,$d, [next($n;$d)]]] + .call
      )
    end
  ))
  # Output final sum from items at depth 0
  | .result = .result + .memo["\([$n,0])"]
) | .result

[–] zogwarg@awful.systems 11 points 1 month ago

I remember being quite ticked off by her takes about free will, and specifically severly misrepresenting compatibilism and calling philosphers stupid for coming up with the idea.

[–] zogwarg@awful.systems 2 points 1 month ago (1 children)

One look day 9 and I had to leave it until after work (puzzles unlock at 2PM for me), it wasn't THAT hard, but I had to leave it until later.

[–] zogwarg@awful.systems 1 points 1 month ago

re:10Mwahaha I'm just lazy and did are "unique" (single word dropped for part 2) of start/end pairs.

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

([
     inputs/ "" | map(tonumber? // -1) | to_entries
 ] | to_entries | map( # '.' = -1 for handling examples #
     .key as $y | .value[]
   | .key as $x | .value   | { "\([$x,$y])":[[$x,$y],.] }
)|add) as $grid | #           Get indexed grid          #

[
  ($grid[]|select(last==0)) | [.] |    #   Start from every '0' head
  recurse(                             #
    .[-1][1] as $l |                   # Get altitude of current trail
    (                                  #
      .[-1][0]                         #
      | ( .[0] = (.[0] + (1,-1)) ),    #
        ( .[1] = (.[1] + (1,-1)) )     #
    ) as $np |                         #   Get all possible +1 steps
    if $grid["\($np)"][1] != $l + 1 then
      empty                            #     Drop path if invalid
    else                               #
    . += [ $grid["\($np)"] ]           #     Build path if valid
    end                                #
  ) | select(last[1]==9)               #   Only keep complete trails
    | . |= [first,last]                #      Only Keep start/end
]

# Get score = sum of unique start/end pairs.
| group_by(first) | map(unique|length) | add

[–] zogwarg@awful.systems 1 points 1 month ago* (last edited 1 month ago)

Day 9 discussionPart two for me was also very slow until I, speed up the index search by providing a lower bound for the insertion. for every insertion of size "N", I have an array lower = [null, 12, 36, ...], since from the left any time you find free space for a given size, the next time must be at an index at least one larger, which makes it close to being O(N) [assuming search for the next free space is more or less constant] instead of O(N^2), went from about 30s to 2s. https://github.com/zogwarg/advent-of-code/blob/main/2024/jq/09-b.jq

[–] zogwarg@awful.systems 4 points 1 month ago

Day 8

Al lot of grid index shuffling these past few days! Not too difficult yet though, will this year be gentler or much harsher later?

Part 2 code in JQ#!/usr/bin/env jq -n -R -f

[ inputs / "" ] | [.,.[0]|length] as [$H,$W] |

#----- In bound selectors -----#
def x: select(. >= 0 and . < $W);
def y: select(. >= 0 and . < $H);

reduce (
  [
    to_entries[] | .key as $y | .value |
    to_entries[] | .key as $x | .value |
    [ [$x,$y],. ]  | select(last!=".")
  ] | group_by(last)[] # Every antenna pair #
    | combinations(2)  | select(first < last)
) as [[[$ax,$ay]],[[$bx,$by]]] ({};
  # Assign linear anti-nodes #
  .[ range(-$H;$H) as $i | "\(
    [($ax+$i*($ax-$bx)|x), ($ay+$i*($ay-$by)|y)] | select(length==2)
  )"] = true
) | length

[–] zogwarg@awful.systems 3 points 2 months ago

Day 3 well suited to JQ

Part 2#!/usr/bin/env jq -n -R -f

reduce (
  inputs |   scan("do\\(\\)|don't\\(\\)|mul\\(\\d+,\\d+\\)")
         | [[scan("(do(n't)?)")[0]], [ scan("\\d+") | tonumber]]
) as [[$do], [$a,$b]] (
  { do: true, s: 0 };
    if $do == "do" then .do = true
  elif $do         then .do = false
  elif .do         then .s = .s + $a * $b end
) | .s

[–] zogwarg@awful.systems 6 points 2 months ago (2 children)

I would say it goes further and that they have a (pseudo?)magical trust in their own intuitions, as if they are crystal clear revalations from the platonic realms.

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