Advent of Code Week 3 - you're lost in a maze of twisty mazes, all alike

What is this, day 16?

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 #

I literally created different test inputs for all the examples given and that found a lot of bugs for me. Specifically the difference between literal and combo operators.

DFS (it's all dfs all the time now, this is my life now, thanks AOC) pruned by unless-I-ever-passed-through-here-with-a-smaller-score-before worked well enough for Pt1. In Pt2 in order to get all the paths I only had to loosen the filter by a) not pruning for equal scores and b) only prune if the direction also matched.

Pt2 was easier for me because while at first it took me a bit to land on lifting stuff from Djikstra's algo to solve the challenge maze before the sun turns supernova, as I tend to store the paths for debugging anyway it was trivial to group them by score and count by distinct tiles.