TL;DR

On with Advent of Code puzzle 23 from 2021: another tough puzzle!

It was about after solving day 19 that I thought OK, now it should get a little easier. Hear hear, I was so much wrong, and the puzzle from day 22 should have warned me.

This time weâ€™re introduced to amphipods, little creatures that have probably been generated in Hanoi (maybe in one of its towers).

At this point in december I was kind of depleted of puzzle energies and completely dazzled, like a boxer who took way too many hits. I just wanted to arrive to that bell sound which would give me some additional rest.

So I coded a solution that was relying on depth-first searching and some search-tree cutting techniques to bring down the amount of computation. Except that it was still taking ages.

So I cheated. I mean, sort of. I printed new, better solutions as long as I found them, and then tried them after a while that no new one was printed. And, eventually, I got it.

Yes, I am ashamed of me now. Well, even then.

Stolen or not, anyway, I had the keys that unlock my mental block to look at other playerâ€™s solution before having solved the puzzle myself, and the thread was full of Dijkstra, Dijkstra!

Well, of course it already occurred to me to adopt a best-first approach, but remember the boxer?!? Anyway, it would have been the right thing to do, and I did it afterwards (along with implementing Dijkstra.rakumod in cglib-raku, at last!).

I had the luck to recover a lot from my previous implementation, though, because also in that case I was visiting a graph and I was already thinking in terms of â€śfinding all successors from a given nodeâ€ť. Time and again, though, I learned that a good algorithm makes all the difference.

``````sub successors-factory (\$graph) {
return sub (\$state) {
my \$nodes = \$state<nodes>;
my (@ok, @target);
for (7 .. \$nodes.end).reverse -> \$j {
if @ok[\$j + 4] // 1 { @ok[\$j] = \$nodes[\$j] == \$j % 4 }
else                { @ok[\$j] = 0 }
my \$class = \$j % 4;
next if defined(@target[\$class]) || @ok[\$j];
@target[\$class] = \$nodes[\$j] == 4 ?? \$j !! 0; # real target > 0
}

my @letter_for = < B C D A >;
my \$positions = \$state<positions>;
my @succs;
for ^\$positions -> \$apod {
my \$p = \$positions[\$apod];
next if @ok[\$p];
my \$class = (\$apod + 3) % 4;
if (\$p <= 6) { # in the corridor
my \$t = @target[\$class] or next;
my \$cost = cost(\$graph, \$state, \$p, \$t) or next;
@succs.push: new-state(\$state, \$apod, \$t, \$cost);
}
else { # in a "room"
my \$t = @target[\$class];
if (\$t && (my \$cost = cost(\$graph, \$state, \$p, \$t))) {
@succs.push: new-state(\$state, \$apod, \$t, \$cost);
next;
}
for 0 .. 6 -> \$t {
my \$cost = cost(\$graph, \$state, \$p, \$t) or next;
@succs.push: new-state(\$state, \$apod, \$t, \$cost);
}
}
}
return @succs;
}
}
``````

In my particular representation, the whole thing is kept in a single array with all positions, and the â€śroomsâ€ť where amphipods have to go start from position 7. This is why I have the fancy `@letter_for` array that places `A` at the end: the position of a letter is the remainder of the division of the associated â€śroomâ€ť modulo 4.

We iterate over all amphipods, skipping those that are already in place (`@ok[\$p]`) and looking at the position of the other ones:

• in the corridor? Then they can only go in their â€śroomâ€ť
• in a room? Then we first check if they can go in their â€śroomâ€ť, and as a fallback we consider sending them in the corridor.

This puzzle drove me a little crazy, but was worth the effort, because I had a small bug in my PriorityQueue.rakumod (and its counterpart in Perl) that prevented me from updating the saved items as they changed priority on the way. Solving puzzles saves code!

OK, enough rambling for todayâ€¦ stay safe!

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