AoC 2016/11 - Initial data structure


On with Advent of Code puzzle 11 from 2016: the initial data structure.

In last post on input parsing we took a quick look at how to turn the very-verbose input into information that we can fit insie a data structure. Now it’s the time to look at the data structure.

As I already wrote, this puzzle was challenging for me. This meant that I had to change strategy, and the underlying data structure, a few times. We will probably see only a couple of them here, so this is the initial take.

My approach is to try and go with what I think it’s the simpler for me as a human to keep track of. So it’s usually nothing optimized, but at least I find it understandable. At that time, at least. So please don’t complain too much, future me!

Representing the state

My initial goal was to keep things simple. I have to represent a four-floors section? Let’s use a four-slots array! Each slot in the array will contain a reference to an array that holds the actual contents of the floor.

I also decided to adopt the following convention inside each floor:

  • each floor size is the same, independently of the actual contents. The size is the maximum number of objects that it can contain, i.e. the elevator, all generators, all microchips;
  • each type of item has a fixed position in the floor’s array:
    • the first position (i.e. index 0) holds the elevator;
    • the second and third positions (i.e. indexes 1 and 2) hold the first element’s microchip and generator, respectively;
    • as an extension, the $k$-th element’s microchip and generator are kept at indexes $2 \cdot k - 1$ and $2 \cdot k$ respectively. But wait! We’re dealing with the Perl language where array indexes start at 0, so assuming that $k$ starts there too our formula becomes $2 \cdot k + 1$, where $k = 0$ represents the first element.

There’s no particular imagination in this representation, actually, because it mimics very closely how the whole thing is pictured in the puzzle input:

F4 .  .  .  .  .  
F3 .  .  .  LG .  
F2 .  HG .  .  .  
F1 E  .  HM .  LM 

So here’s how the initial state state is built:

my @start = map { [(0) x $n_slots] } 1 .. 4;
$start[0][0] = 1; # elevator
while (my ($element, $floor) = each $floor_for{m}->%*) {
   $start[$floor][$slot_idx_of{$element}] = 1;
while (my ($element, $floor) = each $floor_for{g}->%*) {
   $start[$floor][$slot_idx_of{$element} + 1] = 1;

As anticipated, the elevator is put in slot 0 of the first floor (which has index 0). Then we fit microchips in the slot identifiers provided by a helper data structure %slot_idx_for (more on this shortly) and the corresponding generator in the slot immediately after.

The final goal is to move everything on the fourth floor, i.e. on the array at index 3. For this reason, this final state is built as follows:

my @goal = map { [(0) x $n_slots] } 1 .. 3;
push @goal, [(1) x $n_slots];

It’s now time to take a closer look at all these variables and data structures that helped us set up these two states.

The supporting data structures

There are a few supporting data and data structures that are helping in this case:

  • %slot_idx_for keeps track of the slot index that a microchip will end up in;
  • %floor_for keeps track of which floor a microchip or a generator are put in. It actually has two sub-hashes, one for microchips (with the m key) and one for generators (with the g key).
  • $n_slots is the total number of slots.

This last variable is the easier to tackle because it can be derived directly from %slot_idx_for:

my $n_slots = scalar(keys %slot_idx_of) * 2 + 1;

Each floor contains two slots for each element (one for the microchip, one for the generator), plus an additional one for the generator (in index 0).

Let’s not move to the two hashes, which are populated while reading the inputs:

 1 my %floor_idx_of = (
 2    first => 0,
 3    second => 1,
 4    third => 2,
 5    fourth => 3,
 6 );
 7 my %slot_idx_of = ();
 8 my %floor_for;
 9 my $filename = shift || basename(__FILE__) =~ s{\.pl\z}{.tmp}rmxs;
10 open my $fh, '<', $filename;
11 while (<$fh>) {
12    my ($floor) = m{\A The \s+ (\S+) \s+ floor \s+ contains \s+}mxs;
13    my $floor_idx = $floor_idx_of{$floor};
14    while (s{(\S+)-compatible}{}mxs) {
15       my $id = scalar keys %slot_idx_of;
16       $slot_idx_of{$1} //= 2 * $id + 1;
17       $floor_for{m}{$1} =  $floor_idx;
18    }
19    while (s{(\S+) \s+ generator}{}mxs) {
20       $floor_for{g}{$1} = $floor_idx;
21    }
22 }
23 close $fh;

The helper %floor_idx_of (lines 1 through 6) is a mapping that turns the name of the floor into an index inside the main array of the state, hence we find the indexes from 0 to 3 included.

During the main loop for reading the inputs (lines 11 through 22) we leverage the regular expressions in Input parsing to extract the floor name (line 12) to get the floor index (line 13), then iterate over the line to get microchips (lines 14 through 18) and generators (lines 19 through 21).

In these sub-loops we record the specific floor for each element’s component (microchip at line 17, generator at line 20).

Last, we populate %slot_idx_of, that is the slot index for microchips, by potentially assigning a new identifier for the element in $1 (lines 15 and 16). The //= operator makes sure that we only allocate a new identifier when it’s not already present, although this is a paranoid measure because there’s always one single microchip for each element. The new element’s index is found according to the formula we already saw before, remembering that the index of the first element is $k = 0$.


I think I abused your patience enough for today. With probably much more code tha we needed, we managed to populate the initial state from the puzzle input, and the goal state based on the puzzle description. Now… we will concentrate on understanding what’s the minimum number of steps to go from one to the other!

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