TL;DR

On with Advent of Code puzzle 17 from 2022: Merry Christmas everybody, and a nice puzzle.

Day 17th was a good day puzzle-wise, because I solved it and I enjoyed doing so.

This time we have falling stuff resembling a tetris game. Only it’s not tetris at all: different pieces, no rotations, potentially infinite height… I mean, there’s only the falling part, and the accumulation on the bottom.

I represented the falling pieces through ASCII-art sprites, stored as arrays of arrays:

sub sprites { # each is reversed
return (['****'], < .*. *** .*. >, < ..* ..* *** >, < * * * * >, < ** ** >)
.map( { $_».comb».Array.Array } ).Array; }  These sprites fall from the ceiling and shifted around according to some inputs, each represented by a character. It makes sense to read and store them independently: sub get-inputs ($filename) { [ $filename.IO.comb(/\S/) ] }  Both the pieces and the inputs are supposed to appear periodically, according to their respective amounts. It makes sense to create a little iterator-helper class, working on the array of elements: class ArrayIterator { has @!items is built; has$!i = 0;
method get {
my $j =$!i++;
$!i %= @!items; return @!items[$j];
}
method at-start { $!i == 0 } method idx { return$!i }
method dump { say @!items.elems }
}


Fun fact is that I’m probably supposed to do iterators differently (and idiomatically) in Raku. Whatever. This class gets the items during object creation, and allows getting items, figure out which item we got last, if we’re at the start, and so on.

Both parts of the puzzle can be addressed with a single function, properly driven. It’s OK to discuss it at this point, even though we still don’t know what class Field does:

sub part12 ($inputs,$max) {
my $field = Field.new; my$dit = ArrayIterator.new(items => @$inputs); my$sit = ArrayIterator.new(items => sprites());

my %last-seen-indexes;
for 0 ..^ $max ->$i {
my $ip =$sit.idx ~ '/' ~ $dit.idx; if %last-seen-indexes{$ip}:exists { # look for period
my $last = %last-seen-indexes{$ip};
my $period = (($i, $field.top) «-»$last).Array;
if $field.check-period($period) {
my $delta =$max - $i;$field.drop($sit,$dit, '#') for ^ ($delta %$period);
return 1 + $field.top + ($delta / $period).Int *$period;
}
}
%last-seen-indexes{$ip} = ($i, $field.top);$field.drop($sit,$dit, '#');
}

return 1 + $field.top; }  The two iterators $sit and $dit allow us to generate elements as needed, and $field helps us with the actual business logic.

As observed by many, dropping pieces in these conditions leads to a periodic arrangement of pieces, i.e. it’s possible to detect (from a certain point on) an arrangement that is then repeated indefintely.

From a certain point underlines the fact that the very first pieces fall onto the ground and not on top of the previous iteration of the periodic shape. This might make the first pieces arrangements different from the periodic section.

For this reason, there are two “modes” of operation: one looking for a reliable shape that repeats periodically, and another one using it to calculate all the remaining parts to fill in with big chunks.

The discovery phase is just dropping one piece at a time with line:

$field.drop($sit, $dit, '#');  We will see how Field does this shortly. Looking for the period is more interesting. To really be on the safe side, we must ensure that looking back at the dropped pieces, we find exact replicas that are going to repeat themselves. The period is driven by the joint positioning of the two iterators on the same spot, over and over on the same pair. This is why we’re using $ip as a marker, formed by both indexes as provided by the iterator class.

The actual check is performed inside the $field; if it is successful, though, we take the lazy approach and move on dropping a few more pieces, until we are only left with stacking an integer amount of periodical aggregates. It’s time for class Field, at last: class Field { has @!data; has$!top = -1;
has $!offset = 0; method fits ($sprite, $x,$y is copy) {
@!data.push: [< . . . . . . . >] while @!data <= $y; for @$sprite -> $row { for 0 ..$row.end -> $dx { return False if$row[$dx] eq '*' && @!data[$y][$x +$dx] ne '.';
}
--$y; } return True; } method overlay ($sprite, $x,$y is copy, $char = '*') {$!top = max($!top,$y);
for @$sprite ->$row {
@!data[$y][$x + $_] =$row[$_] eq '.' ?? @!data[$y][$x +$_] !! $char for 0 ..$row.end;
--$y; } } method landing-position ($sprite, $dit) { my$x = 2;
my $y =$!top + $sprite.elems + 3; loop { my$movement = $dit.get; my$nx = $movement eq '<' ??$x - 1 !! $x + 1;$x = $nx if 0 <=$nx <= 7 - $sprite.elems && self.fits($sprite, $nx,$y);
#say "$movement$x $y"; my$ny = $y - 1;$y = $ny if$ny >= 0 && self.fits($sprite,$x, $ny); return$x, $y if$y != $ny; #say "v$x $y"; } } method drop ($sit, $dit,$c = '*') {
my $sprite =$sit.get;
my ($x,$y) = self.landing-position($sprite,$dit);
self.overlay($sprite,$x, $y,$c);
return self;
}

method check-period ($period) { my ($n, $height) =$period.Slip;
return False unless 4 * $height + 10 <=$!top;
for 0 ..^ $height ->$offset {
my $closer = @!data[$!top - 1 * $height -$offset].join('');
my $farther = @!data[$!top - 2 * $height -$offset].join('');
return False if $closer ne$farther;
}
return True;
}

method print {
for @!data.reverse -> $row { put '|',$row.join(''), '|';
}
put '+-------+';
}

method top { \$!top }
}


It’s a lot of code, and shows something that I’m actually happy about: attempting to do my future self a favor and make it readable, without (too much) cleverness.

Dropping a piece with the drop method involves a first phase where we figure out where the piece is going to land (i.e. finding the landing-position), then fixing the piece in place with overlay.

As mentioned, checking for the period is the interesting part (as well as the key to solving part 2). One tricky part to keep in mind is that there are two periods to deal with in this puzzle: one is the amount of drawing from both iterators before we arrive to the same exact arrangement that we already saw in the past, which we already saw in the driving loop; another period is the length of the stack of pieces that accumulate through the first period.

To some extent, the first one is a time-based period, while this second one is a space-based period, driven by a candidate height.

The magic constants in the method check-period are there to ensure that the effects of the bottom floor (which is identically full) have been brushed off. In my specific puzzle input this does not really make a difference, because of how the first pieces settle down; we’re aiming for general solutions though.

The check is straightforward: just compare two consecutive stacks of the candidate height period and check that they’re the same. When this happens… it’s a True.

Well, this has been an interesting ride, so cheers and Merry Christmas!

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