ETOOBUSY 🚀 minimal blogging for the impatient
AoC 2016/11 - Optimization
TL;DR
On with Advent of Code puzzle 11 from 2016: reading some advice, implementing it… and wow.
In previous post AoC 2016/11 - Part 2 solution we left with… something to read.
Well, folks… it was indeed a very interesting reading.
First… let’s try the code!
The first thing I did was to try the code, of course. In particular, this code.
And something weird happened with my own puzzle input:
$ time ruby 11.rb 11.input
35
57
real 0m0.497s
user 0m0.440s
sys 0m0.012s
First of all, it didn’t take 2 seconds… it took less than a half!
The real weird fact, anyway, was that the solution to the second puzzle (57) was correct… but the solution to the first one (35) was not! Being that thorn in the side that I am… I opened an issue 🤓
Anyway it seemed clear that the direction was correct, and probably the author had been a bit too enthusiastic in optimizing.
Then… let’s read the spoilers
The post on Reddit includes a few hints in the form of spoilers. By default they are not shown, you have to click on them to read.
SPOILER ALERT: I’ll put the most effective hint right below!
THE MOST IMPORTANT, ABSOLUTELY ESSENTIAL: ALL PAIRS ARE INTERCHANGEABLE - The following two states are EQUIVALENT: (HGen@floor0, HChip@floor1, LGen@floor2, LChip@floor2), (LGen@floor0, LChip@floor1, HGen@floor2, HChip@floor2) - prune any state EQUIVALENT TO (not just exactly equal to) a state you have already seen!
This is so true!
And humbling 🙇
If there was a thing that I was sure of… was that the identifier function was correct. Remember what I stated in New identifier:
[The identifier] has to be a string that can represent a state in a one-to-one mapping, so that equivalent states (i.e. states where the elevator, the generators and the microchips are all in the same place) map onto the same identifier, and different states (for either the elevator, the generators or the microchips) map onto different ones.
It turns out that equivalent does not mean what I wrote at all. I only had in mind that two different hashes with the same values inside were equivalent, but this hint changed everything.
So thanks p_tseng for shedding a light!
The new implementation for the id_of
function takes into account the
new insight:
sub id_of ($state) {
state $floor_idx_of = {
map {
my $mask = 0x01 << $_;
map { (($mask << (8 * $_)) => $_) } 0 .. 3;
} 0 .. 7
};
my ($generators, $microchips) = $state->@{qw< generators microchips >};
return join ',', $state->{elevator},
map { $_->@* }
sort { ($a->[0] <=> $b->[0]) || ($a->[1] <=> $b->[1]) }
map {
my $mask = 0x01010101 << $_;
[
$floor_idx_of->{$generators & $mask},
$floor_idx_of->{$microchips & $mask},
];
} 0 .. ($state->{n_elements} - 1);
}
It is quite more complex than before, but not necessarily more complicated. The gist of it is the following:
- for each element generate a pair with the floor id of the generator, followed by the floor id of the microchip;
- sort all these pairs by the generator’s floor first, then by the microchip’s floor;
- flatten all these pairs into a single list;
- pre-pend the elevator floor;
- join everything together with a separator (in our case, this separator might even be the empty string).
Why does this work? Simply put, the only thing that matters is the relation of each generator with respect to its corresponding microchip. Hence we form the pairs to make sure that these two halves stick together, then do the sorting to make sure that different permutations of the same underlying situation are squashed onto the same identifier.
The A* implementation will then do the rest: whenever it encounters the same identifier, it will mark that specific state as a duplicate, because at the end of the day… it is!
You can get the final code from the local version here. Here’s how it goes:
$ time perl 11.pl 11.input2
57
real 0m1.019s
user 0m0.988s
sys 0m0.024s
We are still about half a second off with respect to the solution by p_tseng… but I guess we can declare ourselves very, very satisfied!
Today I Learned…
… that taking things for granted can hide a lot of interesting and useful things.
It might have its merit - after all, we shouldn’t always be reinventing the wheel - but it’s probably also good to get into the habit of questioning also things that seem carved in the stone.
Just in case.