TL;DR

On with Advent of Code puzzle 21 from 2021: Many-Worlds, it seems, instead of the Copenhagen interpretation!

This day’s puzzle starts relatively simple with a game of counting, which can be easily (and clearly, I daresay) addressed with some object-oriented programming:

class Player {
has $!position is required is built; has$!score                is built = 0;

method TWEAK (:$!position) { --$!position }

method advance ($amount) {$!position = ($!position +$amount) % 10;
$!score += 1 +$!position;
}

method score () { $!score } } class DeterministicDie { has$!current is built = 99;
has $!count = 0; method roll () { ++$!count;
$!current = ($!current + 1) % 100;
return $!current + 1; } method roll3 () { (1..3).map({self.roll}).sum } method count () {$!count }
}

sub part1 ($inputs) { my @players =$inputs.map: { Player.new(position => $_) }; my$die = DeterministicDie.new();
my $target = 1000; my$current = 0;
loop {
@players[$current].advance($die.roll3);
last if @players[$current].score >=$target;
$current = 1 -$current;
}
return @players[1 - $current].score *$die.count;
}


Class Player takes care to track the position… carefully, offsetting it by one and re-adding the lost position when necessary (i.e. when scoring).

The DeterministicDie class is just the implementation of the description in the puzzle. The roll method takes one value out, while roll3… takes 3 and sums them.

I hope I didn’t mess with the private member variables.

The second part of the puzzle is a totally different beast because it asks us to play all possible different games and count the winning/losing outcome for the two players. Just good, ol’ combinatorics, right?

It’s worth noting that the puzzle tells us explicitly that the universe splits at every roll of the dice, so it’s clearly favoring the Many-Worlds interpretation as opposed to the Copenhagen interpretation. Whoever built this submarine probably knew one thing or two.

Anyway, simulating all possible matches with the brute force is a bit too much here. My solution is in the order of $10^{14}$, which is a tad too much for my limited resources.

In my solution, I assume I can compute the “moves to win” for the players and use it to do the calculation:

sub part2 ($inputs) { my @mtws =$inputs.map: {moves-to-win($_)}; my @wins; for 0, 1 ->$pid {
my $player = @mtws[$pid];
my $other = @mtws[1 -$pid];
my $n-wins = 0; for$player.kv -> $n-moves,$outcome {
my $wins =$outcome<wins> or next;
my $other-go-on =$other{$n-moves - 1 +$pid}<go-on>;
$n-wins +=$wins * $other-go-on; } push @wins,$n-wins;
}

return @wins.max;
}


The moves-to-win() function takes the starting score of a player and provides back a statistic of how many moves are needed to win under the different circumstances, i.e. possible outcomes of the splitting die.

The two starting positions will lead to different statistics, stored in array @mtws.

At this point, we can compare the two, remembering that the first player… moves first, but otherwise comparing the moves to win for the two players and calculating the number of wins along the way (for a player to win with $n$ moves, the other player must be in the situation where it has not won yet).

So we’re left with implementing moves-to-win(), right?

sub moves-to-win ($start) { my %factor-for; for (1..3) X (1..3) X (1..3) ->$tuple {
%factor-for{$tuple.sum}++; } my %mtw; my$target = 21;
my @stack = {position => $start - 1, score => 0, factor => 1, rolls => [3 .. 9]},; while (@stack) { my$top = @stack[* - 1];
if ($top<rolls>.elems == 0) { @stack.pop; next; } my$roll = $top<rolls>.pop; my$position = ($top<position> +$roll) % 10;
my $score =$top<score> + 1 + $position; my$factor = $top<factor> * %factor-for{$roll};
if $score >=$target {
%mtw{@stack.elems}<wins> += $factor; next; # no "recursion" } push @stack, { position =>$position,
score    => $score, factor =>$factor,
rolls    => [3 .. 9];
};
}

my $n = 1; my$residuals = 1;
my $max-moves = %mtw.keys».Int.max; while$residuals > 0 {
die if $n >$max-moves;
my $wins = %mtw{$n}<wins> //= 0;
$residuals = %mtw{$n}<go-on> = $residuals * 27 -$wins;
++$n; } return %mtw; }  In the first part we calculate the statistic of the outcomes of rolling three dice. As an example, a$3$can come out only when the three dice are all equal to 1, while e.g.$7$can come out in many more different ways (e.g.$1$,$3$, and$3$, among all of them). From a single player’s point of view, it’s worth calculating the number of wins with exactly$n$moves only if the$n - 1$moves before did not lead to a victory (i.e. scoring the target value). We adopt a stack-based approach to calculate all cases for the given number of rolls, which is actually the explicit implementation of a recursion. After data collection, it’s counting time. The $residuals variable tracks how many residual ways of going on are left, and it will eventually go to 0 because the target is finite and the game always moves pawns ahead anyway.

Just for fun, I also implemented a recursive solution that I read among other solutions:

sub part2-play ($inputs) { return play2($inputs[0], $inputs[1], 0, 0).max } sub play2 (*@args) { state %cache; state @die = (3,1), (4,3), (5,6), (6,7), (7,6), (8,3), (9,1); state &calc = sub ($p1, $p2,$s1, $s2) { return 0, 1 if$s2 >= 21;
my ($w1,$w2) X= 0;
for @die -> ($d,$n) {
my $np1 = ($p1 + $d) % 10 || 10; my ($v2, $v1) = play2($p2, $np1,$s2, $s1 +$np1);
($w1,$w2) = $w1 +$v1 * $n,$w2 + $v2 *$n;
}
return $w1,$w2;
};
my $key = @args.join(','); return %cache{$key} //= &calc(|@args);
}


This time it’s a real recursion… that takes a bit more time, but it’s probably easier on the reader. Here we are also compacting things a bit, like defining the @die state variable to store the same dice outcome statistic that we previously computed dynamically.

Calling play2 recursively requires some attention and is extremely easy to get wrong. Well, for me at least! Anyway, the principle is the same as before: move on playing as long as the target hasn’t been reached by either player. At each stage, the number of wins is multiplied by the number of occurrences of each roll outcome and the result of the previous recursion.

This alternative solution has the merit of being extremely compact, although it took me some time to implement it right. So it’s not been quick for me!

OK, enough for this day… stay safe folks!