Advent of Code 2018 - Day 17, in Kotlin • Todd Ginsberg

Today we get to play with extension functions and operators, as well as write a recursive solution.

If you’d rather just view code, the GitHub Repository is here .

Problem Input

Our input is a file with individual lines that are similar, but represent the veins of clay. The trick here is that they vary between x and y axis. We’ll use some of this code in Part 1, but we can talk about the main parsing here where we turn our input into a List<Point>, representing every place where there is clay.

class Day17(rawInput: List<String>) {
    private fun claySpotsFromInput(input: List<String>): List<Point> =
	    input.flatMap { row ->
	        val digits = row.toIntArray()
	        if (row.startsWith("y")) {
	            (digits[1]..digits[2]).map { Point(it, digits[0]) }
	        } else {
	            (digits[1]..digits[2]).map { Point(digits[0], it) }
	        }
	    }

    companion object {
        private val nonDigits = """[xy=,]""".toRegex()

        private fun String.toIntArray(): IntArray =
            this.replace(nonDigits, "").replace("..", " ").split(" ").map { it.toInt() }.toIntArray()
    }	    
}

Instead of coming up with a funky regex, I opted to go for something simple - remove anything that isn’t a number, replace the .. with a space, and split on space. From there we can turn each row into an IntArray of 3 digits each. To turn those numbers into clay, we flatmap over the individual lines, figure out if they are x-focused, or y-focused, and use our IntArray to create a set of Points for the entire x or y range.

We’ll use these below.

⭐ Day 17, Part 1

The puzzle text can be found here.

Now that we know the problem, let’s turn our List<Point> (clay spots) into an Array<CharArray>, representing our grid. We will mutate this grid in order to generate a picture of where the water flows.

Operator Overloading

Because we’ll be getting and setting values of our Array<CharArray>, and testing to see if a Point is in the grid, let’s override some operators as extension functions!

// In Day17

private operator fun Array<CharArray>.get(point: Point): Char =
    this[point.y][point.x]

private operator fun Array<CharArray>.set(point: Point, to: Char) {
    this[point.y][point.x] = to
}

private operator fun Array<CharArray>.contains(point: Point): Boolean =
    point.x >= 0 && point.x < this[0].size && point.y >= 0 && point.y < this.size

**Note** I've moved these extensions from the `Day17` class to the `Extensions.kt` file, because they were reused as-is for [Day 18](../day18/)

This is going to make our code a lot more appealing to look at.

Create The Grid

There are two approaches we could go for here. There is a lot of the map that we just won’t need, and we could trim it to make it take up less space. I didn’t opt for this because I just felt that the input wasn’t all that big, and it complicates the math and isn’t straight forward to explain. You might be wondering why we return the grid and the minimum Y value, and that’s because the instructions state we are to ignore anything before the minimum Y value of our input. And again, sure, we could just not return it, but I felt it’s just easier to explain this than trim it all to fit in a minimal X/Y box.

// In Day17

private val parsedData = createMap(rawInput)
private val grid: Array<CharArray> = parsedData.first
private val minY: Int = parsedData.second
private val fountain: Point = Point(500, 0)

private fun createMap(input: List<String>): Pair<Array<CharArray>, Int> {
    val spots = claySpotsFromInput(input)
    val minY = spots.minBy { it.y }!!.y
    val maxX = spots.maxBy { it.x }!!.x
    val maxY = spots.maxBy { it.y }!!.y

    // Generate based off of maximum sizes
    val grid: Array<CharArray> = (0..maxY).map {
        // Account for zero indexing and flowing off the right side of the map!
        CharArray(maxX + 2).apply { fill('.') }
    }.toTypedArray()

    // Add all clay spots to the grid
    spots.forEach { spot ->
        grid[spot] = '#'
    }
    // Add the fountain
    grid[0][500] = '+'

    return Pair(grid, minY)
}

Here we see the first use of our set operator (grid[spot] = '#')! I think that is one of the most useful features of Kotlin - the ability to extend even generic types with operators, chosen from a set (I dislike some languages ability to define any operator). Another thing to watch out for when creating our grid is to account for zero indexing (so, +1 on the x axis), and the fact that water can flow out the right side of a container (so again, +1 on the x axis).

More Support Functions

We need a way to get the Point to the left, right, and down of a given Point, so we’ll define those as lazy vals in point. We’ll also define up for the sake of completeness.

// In Point

val up by lazy { Point(x, y - 1) }
val down by lazy { Point(x, y + 1) }
val left by lazy { Point(x - 1, y) }
val right by lazy { Point(x + 1, y) }

We also have wallOrStill and flowOrStill defined in our companion, in addition to our parsing logic, which we will use in our solution.

// In Day17 companion:

private val wallOrStill = setOf('#', '~')
private val flowOrStill = setOf('|', '~')

Recursive Flow

I felt that a recursive solution might be easier to reason about than an iterative solution. We’ll start with our main driver function, which makes heavy use of our extension/operators:

// In Day17

private fun flow(source: Point) {
    if (source.down !in grid) {
        return
    }
    if (grid[source.down] == '.') {
        grid[source.down] = '|'
        flow(source.down)
    }
    if (grid[source.down] in wallOrStill && source.right in grid && grid[source.right] == '.') {
        grid[source.right] = '|'
        flow(source.right)
    }
    if (grid[source.down] in wallOrStill && source.left in grid && grid[source.left] == '.') {
        grid[source.left] = '|'
        flow(source.left)
    }
    if (hasWalls(source)) {
        fillLeftAndRight(source)
    }
}

As with all recursive solutions, we start by testing the end condition - that we’ve fallen off the bottom edge of the grid. Assuming that didn’t happen, we continue the flow down (since we know it’s valid) if it is sand (.), calling ourselves. Once we have a downward flow drawn as far as it will go, we scan right and then left. We need to make sure we don’t overflow or underflow our x axis, so we use our in/contains operator here as well. Assuming left or right is sand, and below is is either a wall or a flow, we can draw a flow to our right or left and recurse, to our right and left.

Once all that is done we see if we’re walled in, and if so, fill our row:

// In Day17

private fun hasWalls(source: Point): Boolean =
    hasWall(source, Point::right) && hasWall(source, Point::left)

private fun hasWall(source: Point, nextPoint: (Point) -> Point): Boolean {
    var point = source
    while (point in grid) {
        when (grid[point]) {
            '#' -> return true
            '.' -> return false
            else -> point = nextPoint(point)
        }
    }
    return false
}

private fun fillLeftAndRight(source: Point) {
    fillUntilWall(source, Point::right)
    fillUntilWall(source, Point::left)
}

private fun fillUntilWall(source: Point, nextPoint: (Point) -> Point) {
    var point = source
    while (grid[point] != '#') {
        grid[point] = '~'
        point = nextPoint(point)
    }
}

In these functions we make use of the fact that we can pass functions to functions (“higher order functions”). This saves us a bit of code because we can now just pass Point::left or Point::right instead of having left and right versions of each function.

Now that we have all this, we can start the flow of water from the fountain, calculate what our grid looks like, and count the number of standing or flowing water spots:

// In Day17

fun solvePart1(): Int {
    flow(fountain)
    return grid.filterIndexed { idx, _ -> idx >= minY }.sumBy { row ->
        row.filter { it in flowOrStill }.count()
    }
}

Star earned! Onward!

⭐ Day 17, Part 2

The puzzle text can be found here.

We already have all we need. We could just run the flow once and answer both problems, but I like having part 1 and 2 run in isolation for my tests. Run the flow, and this time just count the standing water:

// In Day17

fun solvePart2(): Int {
    flow(fountain)
    return grid.filterIndexed { idx, _ -> idx >= minY }.sumBy { row ->
        row.filter { it == '~' }.count()
    }
}

Star earned!

⭐ Day 17, Part 1

⭐ Day 17, Part 2

Further Reading