TL;DR

On with Advent of Code puzzle 19 from 2021: GPS is smarter.

This day’s puzzle is a definite increase in difficulty for this year. Or is it really? Looking back at it, the solution I found is devoid of cleverness but it still manages to run well below the solution from the previous day.

Anyway, it was hard to stitch all parts together, with the constant fear that something might be buggy as well the constant time pressure. I mean, the next day approaching.

Before moving on, let’s rant a bit and also explain why GPS is smarter. I mean… how on Earth is it possible that a scanner senses a bunch of beacons with an exact measurement of their relative position, but each beacon does not emit a unique identifier?!?

At the highest abstraction level, my algorithm for placing all scanners is the following:

my @unbound = get-inputs($filename); # [1] my @bound = @unbound.shift; # [2] my$alice-id = 0;                                     # [3]
while @unbound.elems > 0 {                            # [4]
my $alice = @bound[$alice-id];                    # [5]
my @left;
for @unbound -> $umberto { # [6] if my$new = match-unbound($alice,$berto) {  # [7]
@bound.push: $new; } else { @left.push:$umberto;
}
}
die 'disconnected' if ++$alice-id > @bound.end; # [8] @unbound = @left; # [9] }  I started laying this out as pseudocode, then I figured that the Raku implementation was clearer! Here’s a few comments on the algorithm: • we keep two arrays, one with the unbound inputs ([1]) and one with the bound ones, i.e. the scanners that we have placed in the same coordinate systems. • The @bound array is initialized with the first scanner ([2]), which will be our reference one for the coordinate system. • Integer index variable $alice-id will track our analysis to bind more and more elements, so it starts from the first (and only) element in @bound ([3]) and it MUST always be valid to index something in @bound ([8]).
• We loop until all scanners have been bound and set in our chosen coordinate systems ([4]).
• Variable $alice ([5]) is our reference scanner for finding and binding new neighbors. It is taken from the @bound array, so it’s already in the right coordinate system. Alas, in each iteration not all unbound scanners will be properly bound, so @left will keep track of those that will have to wait some more time. • Variable $umberto ([6]) iterates over the unbound elements to see if it can be matched against $alice, i.e. if they are neighbors. • All the magic of the match between $alice and $umberto is encapsulated within function match-unbound ([7]). This returns the representation of $umberto in the target coordinates system if the match is successful, leading to its addition to @bound, or nothing if the match fails, in which case $umberto is added to @left for future consideration. • After sweeping through all @unbound scanners, we prepare for the next iteration by setting the @left scanners as those @unbound elements to analyze. If any. Now this is a rather verbose description for a quite boring algorithm, whose complexity is not even that great. If there are$N$scanners, the worst case would require$N * (N - 1)$matches for an overall complexity of$O(N^2)$. Anyway, it’s at least simple: just keep onboarding new elements from the pool of unknowns until all of them have been placed. Like doing a puzzle in a very systematic way. Both parts of the puzzle can be solved easily once we have all our scanners in place, with their absolute positions (and that of all beacons too). In the first part we are required to understand how many different beacons are there. My go-to solution in these cases is to populate a hash with the unique identifiers of all beacons (in this case, their position represented as a string) and then count how many elements are in the hash: my %beacon-at; for @bound ->$scanner {
for $scanner<coords>.List ->$p {
my $key =$p.join(',');
%beacon-at{$key}++; } } my$part1 = %beacon-at.elems;


The second part is conceptually simpler: take any pair of scanners, calculate their Manhattan distance, then take the maximum value. Again not a fantastic complexity (still $O(N^2)$) but quite effective.

my $part2 = (@bound X @bound).map( -> ($foo, $bar) { ($foo<origin> «-» \$bar<origin>)».abs.sum }
).max;


In this case Raku is helping us a lot:

• hyperoperator «-» allows us to take the difference of the positions of a pair of scanners, coordinate by coordinate, getting back the difference as a vector;
• hyperoperator ».abs takes the absolute value through all the dimensions, which is also the Manhattan distance component for each dimension;
• .sum does the final sum to get the Manhattan distance;
• .max takes the maximum value out of a list of Manhattan distances.

I guess it’s everything for today, right?

What? Well, ehr… yes! We’re missing the details for match-unbound(), of course!

But… I guess this will be for another post. And another year!

Stay safe, stay tuned and enjoy the last hours of 2021!