# Shifting the Stars: Advent of Code with Galilean Optimization

by Justin Le ♦

(TL;DR: scroll down to the very bottom for a summary and the closed form solution)

Another short Advent of Code post! Advent of Code 2018 is in full swing; we’re 40% of the way through. Every once in a while, if I find a fun way to solve a problem, I’ll make a short post about it. You can check out my other ones here on the series page, and you can also find my daily reflections here, as well. And, again, if you’re following along in Haskell, why not hop on glguy’s semi-official Haskell Leaderboard (join code `43100-84040706`

)! There are also Haskellers on freenode ##adventofcode, and also #adventofcode on the Functional Programming slack. You might also find my advent of code api haskell bindings helpful too!

Today, we’re going to be using linear algebra, calculus, and galilian transformations to solve the *Day 10* challenge. (That’s right, this isn’t just a Haskell blog, I do have math posts on occasion too :) )

## Part 1

It’s no use; your navigation system simply isn’t capable of providing walking directions in the arctic circle, and certainly not in 1018.

The Elves suggest an alternative. In times like these, North Pole rescue operations will arrange points of light in the sky to guide missing Elves back to base. Unfortunately, the message is easy to miss: the points move slowly enough that it takes hours to align them, but have so much momentum that they only stay aligned for a second. If you blink at the wrong time, it might be hours before another message appears.

You can see these points of light floating in the distance, and record their position in the sky and their velocity, the relative change in position per second (your puzzle input). The coordinates are all given from your perspective; given enough time, those positions and velocities will move the points into a cohesive message!

Rather than wait, you decide to fast-forward the process and calculate what the points will eventually spell.

For example, suppose you note the following points:

`position=< 9, 1> velocity=< 0, 2> position=< 7, 0> velocity=<-1, 0> position=< 3, -2> velocity=<-1, 1> position=< 6, 10> velocity=<-2, -1> position=< 2, -4> velocity=< 2, 2> position=<-6, 10> velocity=< 2, -2> position=< 1, 8> velocity=< 1, -1> position=< 1, 7> velocity=< 1, 0> position=<-3, 11> velocity=< 1, -2> position=< 7, 6> velocity=<-1, -1> position=<-2, 3> velocity=< 1, 0> position=<-4, 3> velocity=< 2, 0> position=<10, -3> velocity=<-1, 1> position=< 5, 11> velocity=< 1, -2> position=< 4, 7> velocity=< 0, -1> position=< 8, -2> velocity=< 0, 1> position=<15, 0> velocity=<-2, 0> position=< 1, 6> velocity=< 1, 0> position=< 8, 9> velocity=< 0, -1> position=< 3, 3> velocity=<-1, 1> position=< 0, 5> velocity=< 0, -1> position=<-2, 2> velocity=< 2, 0> position=< 5, -2> velocity=< 1, 2> position=< 1, 4> velocity=< 2, 1> position=<-2, 7> velocity=< 2, -2> position=< 3, 6> velocity=<-1, -1> position=< 5, 0> velocity=< 1, 0> position=<-6, 0> velocity=< 2, 0> position=< 5, 9> velocity=< 1, -2> position=<14, 7> velocity=<-2, 0> position=<-3, 6> velocity=< 2, -1>`

Each line represents one point. Positions are given as

`<X, Y>`

pairs: X represents how far left (negative) or right (positive) the point appears, while Y represents how far up (negative) or down (positive) the point appears.At

`0`

seconds, each point has the position given. Each second, each point’s velocity is added to its position. So, a point with velocity`<1, -2>`

is moving to the right, but is moving upward twice as quickly. If this point’s initial position were`<3, 9>`

, after`3`

seconds, its position would become`<6, 3>`

.Over time, the points listed above would move like this:

`Initially: ........#............. ................#..... .........#.#..#....... ...................... #..........#.#.......# ...............#...... ....#................. ..#.#....#............ .......#.............. ......#............... ...#...#.#...#........ ....#..#..#.........#. .......#.............. ...........#..#....... #...........#......... ...#.......#.......... After 1 second: ...................... ...................... ..........#....#...... ........#.....#....... ..#.........#......#.. ...................... ......#............... ....##.........#...... ......#.#............. .....##.##..#......... ........#.#........... ........#...#.....#... ..#...........#....... ....#.....#.#......... ...................... ...................... After 2 seconds: ...................... ...................... ...................... ..............#....... ....#..#...####..#.... ...................... ........#....#........ ......#.#............. .......#...#.......... .......#..#..#.#...... ....#....#.#.......... .....#...#...##.#..... ........#............. ...................... ...................... ...................... After 3 seconds: ...................... ...................... ...................... ...................... ......#...#..###...... ......#...#...#....... ......#...#...#....... ......#####...#....... ......#...#...#....... ......#...#...#....... ......#...#...#....... ......#...#..###...... ...................... ...................... ...................... ...................... After 4 seconds: ...................... ...................... ...................... ............#......... ........##...#.#...... ......#.....#..#...... .....#..##.##.#....... .......##.#....#...... ...........#....#..... ..............#....... ....#......#...#...... .....#.....##......... ...............#...... ...............#...... ...................... ......................`

After 3 seconds, the message appeared briefly:

`HI`

. Of course, your message will be much longer and will take many more seconds to appear.

What message will eventually appear in the sky?

The problem tells us to talk about a system of N particles, each moving at constant velocity. From this, we can see that the position of particle \(i\) at time \(t\) is \(\mathbf{r}_i + \mathbf{v}_i t\), where \(\mathbf{r}_i\) is the *initial position* vector, and \(\mathbf{v}_i\) is the *velocity* vector.

More generally, we can express this in terms of matrices. If we talk talk about \(R\) as the \(N \times 2\) matrix of initial positions, and \(V\) as the \(N \times 2\) matrix of initial velocities:

\[ R = \begin{bmatrix} x_1 & y_1 \\ x_2 & y_2 \\ \vdots & \vdots \\ x_N & y_N \end{bmatrix} \]

\[ V = \begin{bmatrix} v_{x1} & v_{y1} \\ v_{x2} & v_{y2} \\ \vdots & \vdots \\ v_{xN} & v_{yN} \end{bmatrix} \]

Then we can say that the state of the total system at time \(t\) is given by \(R + V t\)

Now, how can we find the time when all of the letters are aligned?

For this, we can make a *somewhat justified guess*: looking at the input data, we see that things start out as “scattered”, and end up in a clean clustered arrangement. We know that the ending arrangement must be clustered fairly close together because we only have a few hundred points in the input data set, whereas the start times are all in the thousands or higher. And, once things get clustered, they will also get un-clustered right away, because of the randomness of the directions of motion.

This gives us a clue: if we can find the \(t\) that will give us the \(R + V t\) with the *least variance*, we are good to go! All of a sudden, this is now an optimization problem. Find the \(t\) that minimizes the variance of x plus the variance of y. We can find this by finding the formula for the sum of variances, taking the first derivative, and setting it to zero.

The typical formula for finding the sum of variances of a matrix \(M\) is to take the trace of the covariance matrix, \(\mathrm{Tr} \left[ \left(M - \mu_M \right)^T \left(M - \mu_M \right) \right]\). However, in this form, it’s not too fun to work with. That’s because we have to re-compute the mean of of the positions at every point, and things will get messy before they get clean.

Conceptually, however, we have a powerful tool: the Center of Mass frame. Essentially, because our system has no external forces (and no net acceleration), we can *perform a Galilean transform* into a frame of reference where the center of mass is *fixed at the origin*, and *never changes*. If we can do this, then we only need to compute \(\mathrm{Tr} \left(M^T M \right)\) (since we guarantee that the mean of \(M\) is 0), which is relatively easy peasy.

Because our system has points of all equal “mass”, we can shift \(R\) into \(\hat{R}\) (\(R\) shifted into the center of mass frame) and \(V\) into \(\hat{V}\) by just subtracting by the *initial* mean:

\[ \begin{aligned} \hat{R} & = R - \mu_R \\ \hat{V} & = V - \mu_V \end{aligned} \]

This means that our formula for variance at time T is now simple to manipulate. Because we now know that center of mass *is always zero*, we can compute the sum of variance as:

\[ \lvert \Sigma(t) \rvert = \mathrm{Tr} \left[ \left( \hat{R} + \hat{V} t \right)^T \left( \hat{R} + \hat{V} t \right) \right] \]

We can do some simplification, remembering that the trace distributes over addition, and that \(\mathrm{Tr} \left( X^T Y \right) = \mathrm{Tr} \left (Y^T X \right)\):

\[ \lvert \Sigma(t) \rvert = \mathrm{Tr} (\hat{R}^T \hat{R}) + 2 \mathrm{Tr} (\hat{R}^T \hat{V}) t + \mathrm{Tr} (\hat{V}^T \hat{V}) t^2 \]

Now, we want to minimize the sum of variances. So to do that, we can take the first derivative with respect to \(t\), and set it to be zero:

\[ \begin{aligned} \frac{d}{d t} \lvert \Sigma(t) \rvert & = 2 \mathrm{Tr} (\hat{R}^T \hat{V}) + 2 \mathrm{Tr} (\hat{V}^T \hat{V}) t \\ 0 & = 2 \mathrm{Tr} (\hat{R}^T \hat{V}) + 2 \mathrm{Tr} (\hat{V}^T \hat{V}) t_f \\ t_f & = - \frac{\mathrm{Tr} (\hat{R}^T \hat{V})}{\mathrm{Tr}(\hat{V}^T \hat{V})} \end{aligned} \]

And just like that, we have a formula for \(t_f\)!

We can simplify this a little more by remembering that the trace of a matrix multiplication is the sum of the dot products of the columns of the first matrix by the rows of the second. That means we can write:

\[ t_f = - \frac{\sum_i \hat{\mathbf{r}}_i \cdot \hat{\mathbf{v}}_i}{\sum_i \hat{\mathbf{v}}_i \cdot \hat{\mathbf{v}}_i} \]

Once we find this, we can plug into our original form, to find that our final points are, in our un-shifted coordinates, \(R + V t_f\). This is because Galilean transformations leave time unchanged, unlike other frame transformations, like the Lorentz transform. However, we have a simpler option: we could just leave our answer in shifted coordinates as well, since we only care about the shape of the result, and not the absolute position.

We can write this as a Haskell function, assuming we take in a list of `V2 Double`

for velocities and `V2 Double`

for positions, from the *linear* library:

```
-- | Shift so that centroid is at zero
centralize :: [V2 Double] -> [V2 Double]
centralize ps = map (subtract mean) ps
where
mean = sum ps L.^/ fromIntegral (length ps)
-- ^ component-wise division
-- | Sum of dot products
sumOfDots :: [V2 Double] -> [V2 Double] -> Double
sumOfDots xs ys = sum (zipWith L.dot xs ys)
findWord
:: [V2 Double] -- ^ velocities
-> [V2 Double] -- ^ initial positions
-> ([V2 Double], Double) -- ^ points in word, and final time t
findWord (centralize->vs) (centralize->xs) = (final, t)
where
t = negate (sumOfDots xs vs / sumOfDots vs vs)
final = zipWith (\v x -> x + t L.*^ v) vs xs
```

We don’t even need to round the answer — we can directly make a scatter plot of these points and read off what they look like :)

## Part 2

Good thing you didn’t have to wait, because that would have taken a long time - much longer than the

`3`

seconds in the example above.Impressed by your sub-hour communication capabilities, the Elves are curious:

exactly how many seconds would they have needed to wait for that message to appear?

This one is just \(t\), which we solved for in the last part! This time, we do need to remember to `round`

it before we submit.

## Message in the Stars

Optimization by finding the first derivative is a common tool in math that is definitely under-utilized! In practice, unless we have a really clean system, we won’t be able to analytically “solve for zero” in most situations. However, this system shows all of the signs of being well-behaved: the thing we are minimizing is quadratic on our variable, so the first derivative will be linear on our variable, making “solving for zero” very simple.

To do this in a clean way we:

- Represented our system as a matrix formula, giving us key linear algebra insights we could exploit.
- Saw that our problem is feasible, because our thing we are minimizing is quadratic in our variable, meaning the derivative is linear in our variable.
- Made this feasible by using a Galilean transform to shift things into the center-of-mass frame, so that the mean is
*fixed*over the entire time span, and*set to zero*. This made the final solution simple enough to work out on a small sheet of notebook paper.

And isn’t it cute that we use the *Galilean* transform, named after someone who is famous for having studied the motion of astronomical bodies? Maybe that was a subtle hint from the author of the challenges ;)

Anyway, I thought this was a fun twist on the typical Advent of Code challenges. It’s always fun when something that you might think can only be solved by simulation turns out to have a closed-form solution…but even more fun when the closed-form solution turns out to just be simple linear algebra:

\[ t_f = - \frac{\sum_i \hat{\mathbf{r}}_i \cdot \hat{\mathbf{v}}_i}{\sum_i \hat{\mathbf{v}}_i \cdot \hat{\mathbf{v}}_i} \]

“It’s just dot products all the way down.”