# Advent of Code 2023 - Day 24

By Eric Burden | January 2, 2024

It’s that time of year again! Just like last year, I’ll be posting my solutions to the Advent of Code puzzles. This year, I’ll be solving the puzzles in Kotlin. I’ll post my solutions and code to GitHub as well. If you haven’t given AoC a try, I encourage you to do so along with me!

# Day 24 - Never Tell Me The Odds

Find the problem description HERE.

## The Input - Hailstones in a Hurry

Today’s input represents a list of the positions and velocities (in the form of change in position per nanosecond) of hailstones, whizzing in surprisingly consistent trajectories through the air. Since we have nice, consistent line separators, parsing is pretty straightforward. Instead of integers, I’ll use Doubles for the numeric type, since we know from the puzzle that we’re going to need to handle decimal numbers.


/**
* This class represents one of the hailstones.
*
* @property x The initial X-coordinate of the stone.
* @property y The initial Y-coordinate of the stone.
* @property z The initial Z-coordinate of the stone.
* @property dx The initial X velocity of the stone.
* @property dy The initial Y velocity of the stone.
* @property dz The initial Z velocity of the stone.
*/
data class Hailstone(
val x: Double,
val y: Double,
val z: Double,
val dx: Double,
val dy: Double,
val dz: Double
) {
companion object {
/**
* Parse a [Hailstone] from a line of the input file
*
* @param str The string representation of a [Hailstone].
* @return The [Hailstone] represented by str.
*/
fun fromString(str: String): Hailstone {
val splitRegex = Regex("""[, @]+""")
val numbers = str.split(splitRegex).map { it.toDouble() }

// Still can't destructure a list with six item all at once.
val (x, y, z) = numbers
val (dx, dy, dz) = numbers.drop(3)

return Hailstone(x, y, z, dx, dy, dz)
}
}
}

class Day24(input: List<String>) {

// Parse each line from the input into a [Hailstone].
private val parsed =
input.filter { it.isNotEmpty() }.map(Hailstone::fromString)
}


With our stones in our pockets (so to speak), let’s get to calculating line intersections.

## Part One - How Many Parsecs?

Looks like we’re producing hail instead of snow, and, just like Mother Nature, we’re contemplating the best way to smash those ice balls into smithereens to produce those lovely floating crystals we all know and love. That is how that works, right? Yes, of course it is. Now, we just need to see whether Mother Nature is already on the job or if she needs some help from us. For part one, “all” we need to do is determine for each pair of hailstones whether their paths moving forward in time will intersect on the X/Y plane.


/**
* This class represents a rectangular region in 2D space.
*
* @property left The X-value of the left edge of the region.
* @property right The X-value of the right edge of the region.
* @property top The Y-value of the top edge of the region.
* @property bottom The Y-value of the bottom edge of the region.
*/
data class Region2D(
val left: Double, val right: Double, val top: Double, val bottom: Double
) {
/**
* Indicates whether an X/Y position is contained within the [Region2D]
*/
fun contains(location: Pair<Double, Double>) =
location.first in left..right && location.second in top..bottom
}

data class Hailstone( ... ) {
// companion object { ... }

/**
* Find the X,Y intersection between the future paths of two hailstones
*
* Given that we can ignore both the time (partially) and the z-axis of
* the paths of these hailstones, we're really looking at a system of
* two equations and two unknowns. We can use the formula for finding the
* intersection of two lines from:
* https://en.wikipedia.org/wiki/Intersection_(geometry)#Two_line_segments.
*
* From what I understand, t (and u) represents the proportion along
* the first line segment where it intersects with the second line segment.
*
* @param h2 The other hailstone whose path may intersect with this one.
* @return The X/Y position of the intersection, if there is one. If the
* lines are parallel, or the intersection occurred prior to the
* hailstones' starting positions, return null.
*/
private fun intersectionXYWith(h2: Hailstone): Pair<Double, Double>? {
// These lines are parallel, they don't intersect
val denominator = (dx * h2.dy) - (dy * h2.dx)
if (denominator == 0.0) return null

val h1 = this
val t = ((h2.x - h1.x) * h2.dy - (h2.y - h1.y) * h2.dx) / denominator
val u = ((h2.x - h1.x) * h1.dy - (h2.y - h1.y) * h1.dx) / denominator

// Intersection is prior to the stones' initial position (negative
// time).
if (t < 0 || u < 0) return null

// This is from the Wikipedia article as well!
return (h1.x + t * h1.dx) to (h1.y + t * h1.dy)
}

fun intersectionIn2DRegion(h2: Hailstone, region2D: Region2D): Boolean {
val intersection = intersectionXYWith(h2) ?: return false
return region2D.contains(intersection)
}
}

class Day24(input: List<String>) {

// private val parsed = ...

// For each unique pair of hailstones, identify whether the X/Y paths of the
// hailstones will cross. Return the count of pairs that cross paths.
fun solvePart1(region2D: Region2D): Int {
var totalIntersectionsInRegion = 0
for (i1 in 0 until parsed.size - 1) for (i2 in i1 until parsed.size) {
val h1 = parsed[i1]
val h2 = parsed[i2]
if (h1.intersectionIn2DRegion(h2, region2D)) {
totalIntersectionsInRegion += 1
}
}
}
}


I’m thinking that’s not going to be enough collisions without our intervention…

## Part Two - There Is No Try

Yeah, let’s be honest, we knew we weren’t going to be able to ignore the Z-axis forever. But, what we couldn’t have expected (although maybe we should have) is that we’d be solving this problem with a rock. And linear algebra. Thankfully, my math is probably a bit better than my ability to throw a rock completely straight and with pinpoint accuracy over long distances. So, let’s go with math.


// data class Region2D( ... ) { ... }

data class Hailstone( ... ) {
// companion object { ... }

// private fun intersectionXYWith(h2: Hailstone): Pair<Double, Double>? { ... }

// fun intersectionIn2DRegion(h2: Hailstone, region2D: Region2D): Boolean { ... }
}

/**
* Solve a system of linear equations using Gaussian Elimination
*
* Yep, this comes from Wikipedia too. Specifically:
* https://en.wikipedia.org/wiki/Gaussian_elimination. As I understand it, this
* works by converting a system of linear equation into a simpler series of
* equations.
*
* For the example, solving for the X, Y, dX, and dY components, the matrix
* of coefficients is simplified like so:
*
*  [[-2, -1, -6, -1,  -44],     >     [[1, 0, 0, 0, 24],   (X)
*   [-1,  1, -6,  2,    9],     >      [0, 1, 0, 0, 13],   (Y)
*   [ 0,  1, -6, -8,   -3],     >      [0, 0, 1, 0, -3],   (dX)
*   [-3, -2, 12,  8, -126]]     >      [0, 0, 0, 1,  1]]   (dY)
*
* This is called the "reduced row echelon" form, and achieving this state means
* that our constants on the right-hand side of the equation _is_ the value for
* the unknown variable, like:
*
*    -2X - 1Y -  6dX - 1dY =  -44  ->  1X + 0Y + 0dX + 0dY = 24  ->  X = 24
*    -1X + 1Y -  6dX + 2dY =    9  ->  0X + 1Y + 0dX + 0dY = 13  ->  Y = 13
*     0X + 1Y -  6dX - 8dY =   -3  ->  0X + 0Y + 1dX + 0dY = -3  -> dX = -3
*    -3X - 2Y + 12dX + 8dY = -126  ->  0X + 0Y + 0dX + 1dY =  1  -> dy =  1
*
* @param coefficients The coefficients of the system of linear equations.
* @return The coefficients on the right-hand side of the simplified system
* of equations.
*/
fun gaussianElimination(coefficients: List<MutableList<Double>>): List<Double> {
val rows = coefficients.size
val cols = coefficients.first().size

// This only works on a square matrix (with one extra column for the
// coefficient on the right-hand side of the equation).
require(rows == cols - 1) {
throw Exception(
"The number of coefficients on the left side of the" +
"equation should be equal to the number of equations."
)
}

// We operate on each row in the matrix of coefficients.
for (row in coefficients.indices) {

// Normalize the row starting with the diagonal value of each row.
val pivot = coefficients[row][row]
for (col in coefficients[row].indices) {
coefficients[row][col] /= pivot
}

// Sweep the other rows with row
for (otherRow in coefficients.indices) {
if (row == otherRow) continue

val factor = coefficients[otherRow][row]
for (col in coefficients[otherRow].indices) {
coefficients[otherRow][col] -= factor * coefficients[row][col]
}
}
}

return coefficients.map { it.last() }.toList()
}

class Day24(input: List<String>) {

// private val parsed = ...

// fun solvePart1(region2D: Region2D): Int { ... }

/**
* Argh! It's geometry! And linear algebra! Yuck. Here's how this goes:
*
* Say the rock we want to throw to smash all the hailstones starts out at
* xR, yR, zR, dxR, dyR, dzR, but we don't actually know what any of those
* values is. We can start "simply" by identifying where on each axis
* and when as time (t) the rock should collide with one hailstone
* (with properties of, say: x, y, z, dx, dy, dz) as:
*
*      xR + (t * dxR) = x + (t * dx)
*      yR + (t * dyR) = y + (t * dy)
*      zR + (t * dzR) = z + (t * dz)
*
* Rearranging to solve for t, we get:
*
*      t = (xR - x)/(dx - dxR) = (yR - y)/(dy - dyR) = (zR - z)/(dz - dzR).
*
* For _just_ two axes (start with X/Y again), we can isolate the
* values related to just the rock (which won't change from one hailstone
* to another in order to solve the puzzle) by rearranging the relationship
* between the X and Y axes like so:
*
*      (xR - x)/(dx - dxR) = (yR - y)/(dy - dyR)
*      (xR - x)(dy - dyR)  = (yR - y)(dx - dxR)
*      xR*dy - x*dy - xR*dyR + x*dyR = yR*dx - yR*dxR - y*dx + y*dxR
*      yR*dxR - xR*dyR = yR*dx - y*dx + y*dxR - xR*dy + x*dy - x*dyR
*
* Because the terms (yR*dxR - xR*dyR) should be the same no matter which
* hailstone we consider (in order to hit all the hailstones), we can
* alternatively consider another hailstone with properties of, say:
* x', y', z', dx', dy', dz', like so:
*
*      yR*dxR - xR*dyR = yR*dx' - y'*dx' + y'*dxR - xR*dy' + x'*dy' - x'*dyR
*
* Because (yR*dxR - xR*dyR) is unchanging, it must be true that:
*
*      yR*dx - y*dx + y*dxR - xR*dy + x*dy - x*dyR = yR*dx' - y'*dx' + y'*dxR - xR*dy' + x'*dy' - x'*dyR
*      (dy'-dy)xR + (dx - dx')yR + (y - y')dxR + (x' - x)dyR = y*dx - x*dy -y'*dx' + x'dy'
*
* Since we need to solve for the properties of the rock, we can substitute
* the actual values from any pair of hailstones into this equation. We'll
* need at least four pairs of hailstones to solve for the four unknowns.
*
* We can perform the same rearrangement for the X and Z axes like so:
*
*      (xR - x)/(dx - dxR) = (zR - z)/(dz - dzR)
*      (xR - x)(dz - dzR)  = (zR - z)(dx - dxR)
*      xR*dz - x*dz - xR*dzR + x*dzR = zR*dx - zR*dxR - z*dx + z*dxR
*      zR*dxR - xR*dzR = zR*dx  -  z*dx  + z*dxR  - xR*dz  + x*dz   - x*dzR
*                      = zR*dx' - z'*dx' + z'*dxR - xR*dz' + x'*dz' - x'*dzR
*      zR*dx - z*dx + z*dxR - xR*dz + x*dz  - x*dzR = zR*dx' - z'*dx' + z'*dxR - xR*dz' + x'*dz' - x'*dzR
*      (dz'-dz)xR + (dx - dx')zR + (z - z')dxR + (x' - x)dzR = z*dx - x*dz -z'*dx' + x'dz'
*
* The neat thing is, if we already know xR and dxR from solving the first set
* of equations, then we only have two unknowns remaining (zR and dzR), for
* which we only need two pairs of hailstones by rearranging the system of
* equations like:
*
*       (dx - dx')zR + (x' - x)dzR = z*dx - x*dz -z'*dx' + x'dz' - (dz'-dz)xR - (z - z')dxR
*
*  Now, let's get solving!
*/
fun solvePart2(): Long {
val (rockX, rockY, rockDX, rockDY) = gaussianElimination(
parsed.take(5).windowed(2).map { (h1, h2) ->
// This comes from:
// (dy'-dy)xR + (dx - dx')yR + (y - y')dxR + (x' - x)dyR = y*dx - x*dy -y'*dx' + x'dy'
mutableListOf(
h2.dy - h1.dy,  // (dy' - dy)
h1.dx - h2.dx,  // (dx - dx')
h1.y - h2.y,    // (y - y')
h2.x - h1.x,    // (x' - x')

// This is the right-hand side of the equation, or
// y*dx - x*dy -y'*dx' + x'dy'
((h1.y * h1.dx) + (-h1.x * h1.dy) + (-h2.y * h2.dx) + (h2.x * h2.dy))
)
}).map { it.roundToLong() } // Prevents issues with precision

val (rockZ, rockDZ) = gaussianElimination(
parsed.take(3).windowed(2).map { (h1, h2) ->
mutableListOf(
// This comes from:
// (dx - dx')zR + (x' - x)dzR = z*dx - x*dz -z'*dx' + x'dz' - (dz'-dz)xR - (z - z')dxR
h1.dx - h2.dx,  // (dx - dx')
h2.x - h1.x,    // (x' - x)

// This is the right-hand side again
//  z*dx - x*dz - z'*dx' + x'dz' - (dz'-dz)xR - (z - z')dxR
// @formatter:off
( (h1.z  * h1.dx)                       // z*dx
- (h1.x  * h1.dz)                       // x*dz
- (h2.z  * h2.dx)                       // z'*dx'
+ (h2.x  * h2.dz)                       // x'*dz'
- ((h2.dz - h1.dz) * rockX.toDouble())  // (dz'-dz)xR
- ((h1.z  - h2.z)  * rockDX))           // (z - z')dxR
//@formatter:on
)
}).map { it.roundToLong() }

// We've solved for all the necessary parameters now, return the sum!
return rockX + rockY + rockZ
}
}


I (re)learned a lot about solving systems of equations today, including learning about Gaussian Elimination for the first time. Neat!

## Wrap Up

Today’s problem really showcased the value of math. I really needed to access a number of resources (articles and videos) to understand exactly how the math worked. I never had the benefit of a linear algebra course in college, which is kind of a shame since working through these kinds of rearrangements is really fun. It took me a while post-holiday travels to get back to Advent of Code and the blog, and this was a nice puzzle to nearly wrap up the year with.