This is a solution to Advent of Code 2023 day 22, written in Raku.

https://adventofcode.com/2023/day/22

### Part One

Figure how the blocks will settle based on the snapshot. Once they've settled, consider disintegrating a single brick; how many bricks could be safely chosen as the one to get disintegrated?

```
constant \x = 0; constant \y = 1; constant \z = 2;
my @bricks = '22-test.txt'.IO.lines>>.comb(/\d+/)>>.Int>>.rotor(3);
my @stack[330;10;10];
for @bricks.kv -> $k, @v {
for @v[0][x] .. @v[1][x] -> $x {
for @v[0][y] .. @v[1][y] -> $y {
for @v[0][z] .. @v[1][z] -> $z {
@stack[$z;$y;$x] = $k;
}
}
}
}
say @stack;
```

### Part Two

```
constant \x = 0; constant \y = 1; constant \z = 2;
my @bricks = '22-input.txt'.IO.lines>>.comb(/\d+/)>>.Int>>.rotor(3);
say @bricks.map(
-> @b {
@b[0][z] max @b[1][z]
}).max;
```

327

`say (3,4,5) >>->> 1`

(2 3 4)