# Advent of Code 2018 - Day 6, in Kotlin

Kotlin solutions to parts 1 and 2 of Advent of Code 2018, Day 6: 'Chronal Coordinates'

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Day 6 is here, and with it our first grid problem. Advent of Code usually contains a couple, so this is your opportunity to learn all about Manhattan Distance.

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

Side Note: Today’s code and writeup is way later than usual because I was speaking (about Kotlin!) at the Lead Developer Austin conference. I had a great time presenting and learned so much from the other speakers. It is a very well run conference. If you have an opportunity to attend one of the Lead Developer conferences, I suggest you take the opportunity to do so.

Problem Input

Our input is a list of Strings, which are easily parsed, so I don’t do anything fancy. We will define a `data class` to hold the input below.

#### Day 6, Part 1

The puzzle text can be found here.

First, let’s create a `Point` data class before we do anything else. We’ll probably need it later, so let’s define it in a new place, not associated with this specific challenge. We’ll include the parsing logic for it, hopefully that will be OK going forward, but we can always refactor.

``````data class Point(val x: Int, val y: Int) {
fun distanceTo(otherX: Int, otherY: Int): Int =
abs(x - otherX) + abs(y - otherY)

companion object {
fun of(input: String): Point =
input.split(",")
.map { it.trim().toInt() }
.run { Point(this, this) }
}
}``````

It’s pretty simple - it has an `x` and `y` coordinate, and can determine the distance to another `x` and `y` pair. There is no version of `distanceTo` that takes another `Point` because we don’t actually need it yet. When we do, we’ll add it later. Our parse function splits our string on comma and turns the resulting `String` objects into `Ints`. From there we use `run` as a `map` function to turn our array of `Ints` into a `Point`.

Now let’s look at the challenge. If you think about it, once we have a `List<Point>` we’ll need to know where the bounds are. Meaning: the largest and smallest `x` values, and the largest and smallest `y` values make up a bounding box around our area. We’ll define those as `IntRange` objects, because that seems natural:

``````class Day06(input: List<String>) {

private val points: List<Point> = input.map { Point.of(it) }
private val xRange: IntRange = (points.minBy { it.x }!!.x..points.maxBy { it.x }!!.x)
private val yRange: IntRange = (points.minBy { it.y }!!.y..points.maxBy { it.y }!!.y)

}``````

The next thing we need to think about is how to identify the `Points` that are in infinite areas. In principle, if we are on the edge of a bounding box, the spot closest to us is infinite. So let’s write a function to determine if a `Point` is infinite.

``````private fun isEdge(x: Int, y: Int): Boolean =
x == xRange.first || x == xRange.last || y == yRange.first || y == yRange.last``````

Now that we can do all that, let me show you the solution, and then explain it.

``````fun solvePart1(): Int {
val infinite: MutableSet<Point> = mutableSetOf()
return xRange.asSequence().flatMap { x ->
yRange.asSequence().map { y ->
val closest = points.map { it to it.distanceTo(x, y) }.sortedBy { it.second }.take(2)
if (isEdge(x, y)) {
}
closest.first.takeUnless { closest.second == closest.second }
}
}
.filterNot { it in infinite }
.groupingBy { it }
.eachCount()
.maxBy { it.value }!!
.value
}``````

Put the pitchforks down, I’ll explain!

First, we set up a set for all of the `Points` we discover are infinite. Meaning, those points that are close to an edge. Then we set up a familiar flatmap/map loop over our x and y ranges. This gives us a sequence of x/y pairs representing every spot in our grid. For each of those, we calculate the distance to every `Point`, and take the first two of them. Why two? Because we need to account for places in the grid that are equidistant to two `Points`. We calculate if the current x/y pair is on the edge, and if so put the closest matching `Point` into the `infinite` set.

At this point we take the closest point, only if its distance is not equal to the second closest `Point`. This uses one of my favorite functions - `takeUnless`. There is a `takeIf` as well, which does the opposite. At this point, we have a `List<Point?>`. By filtering out `Point` objects that are in `infinite`, we correctly handle the case where we can’t actually look at infinite spots, as well as filtering out nulls because null is not in the set of infinites (I never took philosophy, but this seems like a question you could really think about for a while).

Once we have that, group and count the `Points`, and find the one with the highest count, returning it.

And that is one hard earned star! Onward!

#### Day 6, Part 2

The puzzle text can be found here.

Don’t worry, this isn’t as hard as it sounds. It took me a bit to understand it. Essentially it says “if you took every spot in the grid, and calculated its distance to every `Point`, how many of them have a total distance of less than 10,000?” That doesn’t seem so bad, let’s code it up!

``````fun solvePart2(range: Int = 10000): Int =
xRange.asSequence().flatMap { x ->
yRange.asSequence().map { y ->
points.map { it.distanceTo(x, y) }.sum()
}
}
.filter { it < range }
.count()``````

As you can see, `range` is an optional argument defaulting to 10,000 so we can unit test it with 32 against the sample input. Next, we do the flatmap/map thing again, and for each point we sum up the distance to all of the `Points`. If we filter out anything whose distance is greater than the `range`, we can count how many results we have for our answer!

Second star earned! I can’t wait for tomorrow!