# ETOOBUSY 🚀 minimal blogging for the impatient

# AoC 2016/15 - Chinese Reminder Theorem - again!

**TL;DR**

The Advent of Code puzzle 15 from 2016 has more Chinese Remainder Theorem!

As you might have noticed, I’ve been taking a look at the 2016 edition of the puzzles in Advent of Code.

No, this will not be another series!

It so happens that puzzle 15 has a long and involved description about capsules, discs, alignments, exact timings… I had to read it twice and put my brain in full imaginative mode.

Anyway, as I read through it, I started suspecting that it had to do with the Chinese Remainder Theorem. Which, indeed, it does.

First thought was: *again?!?* Then I had that *Back to the Future*
moment when I realized that this puzzle came *before* the ones I
discussed last December 🙄

Second thought was: *I don’t want to understand this problem, I want to
COOOODE!*.

# He who its first…

… hits twice, right?

So I went ahead, took the relevant functions from cglib’s Numbers.pm, put some parsing, a bit of this, a bit of that… and ended up with the following:

```
#!/usr/bin/env perl
use 5.024;
use warnings;
use English qw< -no_match_vars >;
use autodie;
use experimental qw< postderef signatures >;
no warnings qw< experimental::postderef experimental::signatures >;
use File::Basename qw< basename >;
use Data::Dumper; $Data::Dumper::Indent = 1;
use Storable 'dclone';
$|++;
my @stuff;
my $filename = shift || basename(__FILE__) =~ s{\.pl\z}{.tmp}rmxs;
open my $fh, '<', $filename;
while (<$fh>) {
my ($delay, $n, $position) = m{
\A Disc \s+ \#(\d+) \s+
has \s+ (\d+) \s+ positions .*?
at \s+ position \s+ (\d+)
}mxs or die $_;
push @stuff, $n, ($delay + $position) % $n;
}
close $fh;
say((chinese_remainder_theorem(@stuff))[1]);
# chinese_remainder_theorem and egcd below... nothing new
```

I have to admit that I was *a bit unsure* about the ```
($delay +
$position) % $n
```

- it was somehow a shot in the dark.

Anyway, I run it over the example input and *presto!* - it worked! Right
off the bat!

```
$ cat 15.tmp
Disc #1 has 5 positions; at time=0, it is at position 4.
Disc #2 has 2 positions; at time=0, it is at position 1.
$ perl 15-1.pl 15.tmp
5
```

OK, on with my puzzle input then:

```
$ cat 15.input
Disc #1 has 13 positions; at time=0, it is at position 1.
Disc #2 has 19 positions; at time=0, it is at position 10.
Disc #3 has 3 positions; at time=0, it is at position 2.
Disc #4 has 7 positions; at time=0, it is at position 1.
Disc #5 has 5 positions; at time=0, it is at position 3.
Disc #6 has 17 positions; at time=0, it is at position 5.
$ perl 15-1.pl 15.input
64118
```

Only that… **NO, it does not work!!!**

I hit first… but I hit wrong!

# Back to the paper

This must be my humbling year, because I’m reminded so many times of how many ways I have to fail!

Well, I meant *learn* 👨🎓

Let’s see… if we start at time $T$, the first disk is reached after a delay of $1$ at time $T + 1$, and if its starting position (at $t = 0$) is $P_{1, 0}$ then its position at $T + 1$ will be $T + 1 + p_1 \pmod {n_1}$. We have similar relations for the other discs:

\[P_{1, T + 1} \equiv T + 1 + P_{1, 0} \pmod {n_1} \\ P_{2, T + 2} \equiv T + 2 + P_{2, 0} \pmod {n_2} \\ ... \\ P_{i, T + i} \equiv T + i + P_{i, 0} \pmod {n_i}\]If we really need that capsule, each of the left-hand sides MUST be $0$, which brings us to:

\[T \equiv -1 - P_{1, 0} \pmod {n_1} \\ T \equiv -2 - P_{2, 0} \pmod {n_2} \\ ... \\ T \equiv -i - P_{i, 0} \pmod {n_i}\]This can also be rewritten as:

\[r_1 = T \pmod {n_1} = n_1 - (1 + P_{1, 0} \pmod {n_1}) \\ r_2 = T \pmod {n_2} = n_2 - (2 + P_{2, 0} \pmod {n_2}) \\ ... \\ r_i = T \pmod {n_i} = n_i - (i + P_{i, 0} \pmod {n_i})\]*I knew it!*

It’s also funny that it worked for the example input: simply put, $1 \equiv -1 \pmod 2$, so the sign flip didn’t matter!

So the right code is actually this… *a small change for a program, but
a big step ahead for a puzzle solver*!

```
#!/usr/bin/env perl
use 5.024;
use warnings;
use English qw< -no_match_vars >;
use autodie;
use experimental qw< postderef signatures >;
no warnings qw< experimental::postderef experimental::signatures >;
use File::Basename qw< basename >;
use Data::Dumper; $Data::Dumper::Indent = 1;
use Storable 'dclone';
$|++;
my @stuff;
my $filename = shift || basename(__FILE__) =~ s{\.pl\z}{.tmp}rmxs;
open my $fh, '<', $filename;
while (<$fh>) {
my ($delay, $n, $position) = m{
\A Disc \s+ \#(\d+) \s+
has \s+ (\d+) \s+ positions .*?
at \s+ position \s+ (\d+)
}mxs or die $_;
push @stuff, $n, $n - ($delay + $position) % $n;
}
close $fh;
say((chinese_remainder_theorem(@stuff))[1]);
# ...
```

Let’s run it…

```
$ perl 15.pl 15.input
376777
```

Yay, this is correct now!

# A final thought

I work in the telecommunications industry and most of my… *coding
occasions* come in relation to relatively small integrations, so I
definitely have a biased view.

This said… I wonder if the Chineses discovered this theorem just for the fun of puzzle builders and solvers in the 21st century!

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