22-b.jq

Massive time gains with parallelization + optimized next function (2x speedup) by doing the 3 xor operation in “one operation”, Maybe I prefer the grids ^^:

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

#────────────────── Big-endian to_bits ───────────────────#
def to_bits:
  if . == 0 then [0] else { a: ., b: [] } | until (.a == 0;
      .a /= 2 |
      if .a == (.a|floor) then .b += [0]
                          else .b += [1] end | .a |= floor
  ) | .b end;
#────────────────── Big-endian from_bits ────────────────────────#
def from_bits: [ range(length) as $i | .[$i] * pow(2; $i) ] | add;

( # Get index that contribute to next xor operation.
  def xor_index(a;b): [a, b] | transpose | map(add);
  [ range(24) | [.] ]
  | xor_index([range(6) | [-1]] + .[0:18] ; .[0:24])
  | xor_index(.[5:29] ; .[0:24])
  | xor_index([range(11) | [-1]] + .[0:13]; .[0:24])
  | map(
      sort | . as $indices | map(
        select( . as $i |
          $i >= 0 and ($indices|indices($i)|length) % 2 == 1
        )
      )
    )
) as $next_ind |

# Optimized Next, doing XOR of indices simultaneously a 2x speedup #
def next: . as $in | $next_ind | map( [ $in[.[]] // 0 ] | add % 2 );

#  Still slow, because of from_bits  #
def to_price($p): $p | from_bits % 10;

# Option to run in parallel using xargs, Eg:
#
# seq 0 9 | \
# xargs -P 10 -n 1 -I {} bash -c './2024/jq/22-b.jq input.txt \
# --argjson s 10 --argjson i {} > out-{}.json'
# cat out-*.json | ./2024/jq/22-b.jq --argjson group true
# rm out-*.json
#
# Speedup from naive ~50m -> ~1m
def parallel: if $ARGS.named.s and $ARGS.named.i  then
   select(.key % $ARGS.named.s == $ARGS.named.i)  else . end;

#════════════════════════════ X-GROUP ═══════════════════════════════#
if $ARGS.named.group then

# Group results from parallel run #
reduce inputs as $dic ({}; reduce (
      $dic|to_entries[]
  ) as {key: $k, value: $v} (.; .[$k] += $v )
)

else

#════════════════════════════ X-BATCH ═══════════════════════════════#
reduce (
  [ inputs ] | to_entries[] | parallel
) as { value: $in } ({};  debug($in) |
  reduce range(2000) as $_ (
    .curr = ($in|to_bits) | .p = to_price(.curr) | .d = [];
    .curr |= next | to_price(.curr) as $p
    | .d = (.d+[$p-.p])[-4:]  | .p = $p # Four differences to price
    | if .a["\($in)"]["\(.d)"]|not then # Record first price
         .a["\($in)"]["\(.d)"] = $p end # For input x 4_diff
  )
)

# Summarize expected bananas per 4_diff sequence #
| [ .a[] | to_entries[] ]
| group_by(.key)
| map({key: .[0].key, value: ([.[].value]|add)})
| from_entries

end |

#═══════════════════════════ X-FINALLY ══════════════════════════════#
if $ARGS.named.s | not then

#     Output maximum expexted bananas      #
to_entries | max_by(.value) | debug | .value

end

Well it’s not a grid! My chosen language does not have bitwise operators so it’s a bit slow. Have to resort to manual parallelization.

Also did this by hand to get my precious gold star, but then actually went back and implemented it Some JQ extension required:

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

#─────────── Big-endian to_bits and from_bits ────────────#
def to_bits:
  if . == 0 then [0] else { a: ., b: [] } | until (.a == 0;
      .a /= 2 |
      if .a == (.a|floor) then .b += [0]
                          else .b += [1] end | .a |= floor
  ) | .b end;
def from_bits:
  { a: 0, b: ., l: length, i: 0 } | until (.i == .l;
    .a += .b[.i] * pow(2;.i) | .i += 1
  ) | .a;
#──────────── Big-endian xor returns integer ─────────────#
def xor(a;b): [a, b] | transpose | map(add%2) | from_bits ;

[ inputs | scan("\\d+") | tonumber ] | .[3:] |= [.]
| . as [$A,$B,$C,$pgrm] |


# Assert  #
if  [first(
        range(8) as $x |
        range(8) as $y |
        range(8) as $_ |
        [
          [2,4],  # B = A mod 8            # Zi
          [1,$x], # B = B xor x            # = A[i*3:][0:3] xor x
          [7,5],  # C = A << B (w/ B < 8)  # = A(i*3;3) xor x
          [1,$y], # B = B xor y            # Out[i]
          [0,3],  # A << 3                 # = A(i*3+Zi;3) xor y
          [4,$_], # B = B xor C            #               xor Zi
          [5,5],  # Output B mod 8         #
          [3,0]   # Loop while A > 0       # A(i*3;3) = Out[i]
        ] | select(flatten == $pgrm)       #         xor A(i*3+Zi;3)
      )] == []                             #         xor constant
then "Reverse-engineering doesn't neccessarily apply!" | halt_error
 end |

#  When minimizing higher bits first, which should always produce   #
# the final part of the program, we can recursively add lower bits  #
#          Since they are always stricly dependent on a             #
#                  xor of Output x high bits                        #

def run($A):
  # $A is now always a bit array                    #
  #                 ┌──i is our shift offset for A  #
  { p: 0, $A,$B,$C, i: 0} | until ($pgrm[.p] == null;

    $pgrm[.p:.p+2] as [$op, $x]       | # Op & literal operand
    [0,1,2,3,.A,.B,.C,null][$x] as $y | # Op &  combo  operand

    # From analysis all XOR operations can be limited to 3 bits  #
    # Op == 2 B is only read from A                              #
    # Op == 5 Output is only from B (mod should not be required) #
      if $op == 0 then .i += $y
    elif $op == 1 then .B = xor(.B|to_bits[0:3]; $x|to_bits[0:3])
    elif $op == 2
     and $x == 4  then .B = (.A[.i:.i+3] | from_bits)
    elif $op == 3
     and (.A[.i:]|from_bits) != 0
                  then .p = ($x - 2)
    elif $op == 3 then .
    elif $op == 4 then .B = xor(.B|to_bits[0:3]; .C|to_bits[0:3])
    elif $op == 5 then .out += [ $y % 8 ]
    elif $op == 6 then .B = (.A[.i+$y:][0:3] | from_bits)
    elif $op == 7 then .C = (.A[.i+$y:][0:3] | from_bits)
    else "Unexpected op and x: \({$op,$x})" | halt_error
    end | .p += 2
  ) | .out;

[ { A: [], i: 0 } | recurse (
    #  Keep all candidate A that produce the end of the program,  #
    #  since not all will have valid low-bits for earlier parts.  #
    .A = ([0,1]|combinations(6)) + .A | # Prepend all 6bit combos #
    select(run(.A) == $pgrm[-.i*2-2:] ) # Match pgrm from end 2x2 #
    | .i += 1
    # Keep only the full program matches, and convert back to int #
  ) | select(.i == ($pgrm|length/2)) | .A | from_bits
]

| min # From all valid self-replicating intputs output the lowest #

Definitely a bit tedious, I had to “play” a whole session to spot bugs that I had. It took me far longer than average. I had buggy dissepearing boxes because of update order, I would reccomend a basic test case of pushing a line/pyramid of boxes in every direction.

Ok 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.

#!/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:

Although 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.

If you somehow wanted your whole final array it would also required over 1 Peta byte ^^, memoization definetely reccomended.

#!/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)

#!/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

#!/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

#!/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