# ETOOBUSY đźš€ minimal blogging for the impatient

# AoC 2016/11 - Part 1 solution

**TL;DR**

On with Advent of Code puzzle 11 from 2016: implementing graph functions.

In previous post AoC 2016/11 - Initial algorithm: Dijkstra we saw how
to address our problem by leveraging the Dijkstra Algorithm for
finding the *shortest path* from the initial state to the goal. We left
with the need to provide two functions:

`id_of`

to generate a unique and consistent identifier for states, and`successors_for`

, to generate the list of states that can be reached from a given one.

# Identifier

Letâ€™s start with the easier one, which is just a one-liner:

```
sub id_of { join "\n", '---', reverse map { join '', $_->@*} $_[0]->@* }
```

Each floorâ€™s contents are merged together inside the `map`

, with no
spaces; then, all floors are joined together with newlines. This
representation doubles down to provide a visual representation of the
specific state! The three-hyphens string at the beginning is useful to
separate different identifier when printed one after another.

This is really it!

# Successors

My initial idea for finding out successors was based on two halves:

- enumerate all
*potential*moves possible in a given state; - filter out those that lead to frying any microchip.

The enumeration part is easy to doâ€¦ *naĂŻvely*, like this:

- find where the elevator is - only the elevator can determine a move;
- check moves for the elevator that goes up or down one single floor, when possible;
- in each possible move direction, evaluate:
- each item singularly
- each pair of items

I know that this is very crude, but it works (at the expense of some checking). The code below basically implements this approach:

```
sub successors_for ($node) {
my $cidx = elevator_floor_idx($node);
my $cfloor = $node->[$cidx];
my @successors;
for my $tidx ($cidx - 1, $cidx + 1) {
next if $tidx < 0 || $#$node < $tidx;
my $tfloor = $node->[$tidx];
for my $i (1 .. $#$cfloor) {
next unless $cfloor->[$i];
# try to move only this one
if (my @new = move($node, $cidx, $tidx, $i)) {
push @successors, \@new;
}
# now try to move this one with another one
for my $j ($i + 1 .. $#$cfloor) {
next unless $cfloor->[$j];
if (my @new = move($node, $cidx, $tidx, $i, $j)) {
push @successors, \@new;
}
}
}
}
return @successors;
}
```

Finding out where the elevator is does not put any particular problem, itâ€™s a matter of looking for it. Yes, I know that I could track it, but bear with me OK?!?

```
sub elevator_floor_idx ($node) {
for my $candidate (0 .. 4) {
next unless $node->[$candidate][0];
return $candidate;
}
die "wtf?!?";
}
```

The whole logic for *filtering out impossible moves* is encapsulated in
the `move`

sub, which both builds up the landing state and checks for its
validity:

```
sub move ($state, $cidx, $tidx, $i, $j = undef) {
my @new = $state->@*; # shallow copy here
$new[$_] = [$new[$_]->@*] for ($cidx, $tidx); # deep copy here
for my $slot (0, $i, $j) {
next unless defined $slot;
$new[$cidx][$slot] = 0;
$new[$tidx][$slot] = 1;
}
return @new if is_floor_safe($new[$cidx]) && is_floor_safe($new[$tidx]);
return;
}
```

The first part builds up the new landing state. Note that two floors are always copied from the original ones - they are not changed, so it makes sense to reuse them - while two other are created anew to avoid overwriting the previous one. Yes, this might be taxing for the memory, but weâ€™re in developer time saving mode here!

The check about the new state only needs to look at the two floors that
got a change; they are verified through the calls to `is_floor_safe`

:

```
sub is_floor_safe ($floor) {
my $ng = grep {$floor->[2 * $_]} 1 .. $#$floor / 2 or return 1;
for my $midx (1 .. $#$floor / 2) {
next unless $floor->[$midx * 2 - 1];
return 0 unless $floor->[$midx * 2];
}
return 1;
}
```

The algorithm here isâ€¦ working. First we count how many generators we
have on the floor and put the value in `$ng`

. If there is noneâ€¦ the
floor is safe by definition, so we can return immediately.

Otherwise, we have to check that each microchip on the floor is powered by the corresponding generator (a powered microchip is also a protected microchip). For this reason, we iterate over all the microchip slots, and return a false value if we find a microchip without the corresponding generator.

If all these checks are fineâ€¦ we can return a success!

# Time to run it!

The whole code can be found in the local version here.

Running it actually gives us the answer to the first part of the puzzle for day 11. First, a warm-up with the example in the puzzle text:

```
$ cat 11.tmp
The first floor contains a hydrogen-compatible microchip and a lithium-compatible microchip.
The second floor contains a hydrogen generator.
The third floor contains a lithium generator.
The fourth floor contains nothing relevant.
poletti@polebian:2016 (posts-11 *)$ time perl 11.pl 11.tmp
11
real 0m0.078s
user 0m0.072s
sys 0m0.004s
```

Then, on with the real input:

```
$ cat 11.input
The first floor contains a promethium generator and a promethium-compatible microchip.
The second floor contains a cobalt generator, a curium generator, a ruthenium generator, and a plutonium generator.
The third floor contains a cobalt-compatible microchip, a curium-compatible microchip, a ruthenium-compatible microchip, and a plutonium-compatible microchip.
The fourth floor contains nothing relevant.
poletti@polebian:2016 (posts-11 *)$ time perl 11.pl 11.input
33
real 0m31.129s
user 0m30.636s
sys 0m0.292s
```

We completed part 1, yay!