# ETOOBUSY đźš€ minimal blogging for the impatient

# AoC 2016/11 - New algorithm: A*

**TL;DR**

On with Advent of Code puzzle 11 from 2016: moving to the A* algorithm.

One drawback of using Dijkstraâ€™s algorithm for our search is that it makes
a lot of expansions to get from the start node to the goal one. This was
somehow acceptable with the shorter inputs, but became prohibitive with
the longer one in **part 2**.

Thereâ€™s a more complex, but also more efficient, algorithm that will yield
the optimal path by generally expanding less nodes, taking less time in
the process. This is the A* Algorithm. On the good side, we
*already* have a self-contained implementation for the algorithm in
cglib, i.e. AstarX.pm.

This is the call we can adopt:

```
my $outcome2 = astar(
start => [@start],
goal => [@goal],
distance => sub { return 1 },
heuristic => sub ($v, $goal) {
my $d = 0;
for my $fid (0 .. 2) {
my $weight = 3 - $fid;
$d += $weight * scalar grep {$_} $v->[$fid]->@*;
}
return $d;
},
identifier => \&id_of,
successors => \&successors_for,
);
say scalar($outcome2->@*) - 1;
```

One difference is that the returned value is a list of nodes from the
start to the goal, both included. For this reason, we print the listâ€™s
lenght *minus one*, so that we know how many *steps* we have to take.

The second difference is the presence of the `heuristic`

parameter. This
is an estimation of the distance between two nodes, but in the algorithm
is only used to establish the *distance of a node to the goal*. If we
set this identically to `0`

we would fall back to Dijkstraâ€™s Algorithm,
and in general we need to provide either a correct value, or an
*underestimated* one.

Of course we donâ€™t know the correct value at this stage, otherwise we
might use that value to solve our initial problem. We can provide an
*underestimation* though, i.e. the bare minimum number of moves for each
item to the target fourth floor. Itâ€™s a bit crude but itâ€™s a start.

Running the new code (local version here) tells us weâ€™re heading in the right direction:

```
$ time perl 11.pl 11.tmp
11
real 0m0.053s
user 0m0.048s
sys 0m0.008s
$ time perl 11.pl 11.input
33
real 0m10.165s
user 0m10.036s
sys 0m0.116s
```

It now takes one third of the previous time to solve part 1 - nothing terribly better, but still an improvement!

Alas, this does not suffice to address all our concerns for this puzzle.
When provided the new *extended* input with *elerium* and *dilithium*â€¦
it still eats a lot of memory and provides no answer in *reasonable
time*, so we have to do MOAR!

*Comments? Octodon, , GitHub, Reddit, or drop me a line!*