Eric Lloyd

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# Advent of Code in Kotlin: Day 5, 2021

## Day 5

Alright, today is day 5 of advent of code, we have an interesting challenge involving points and lines (or vectors) in a 2D plane.
Below I will share and explain my proposed solution in Kotlin.

### Looking back on Day 3 and Day 4

Before we jump into day 5's puzzle, I would to share links to my previous days' proposed solutions:

• Day 3 β a few key concepts used are:
• Day 4 β a few key concepts used are:
• data classes for the board, rows and tiles
• chunked method to partition a iterable in groups of a given size n

### Day 5 Problem

You can find the description for this puzzle here, I would summarize the problem as the following:

Given a list of lines or vectors, count the number of points which overlap

### Solution

In my opinion, today's puzzle could be broken down into 3 challenges:

1. modelling the input data
2. converting lines to a list of points
3. counting the points to find overlapping ones

I will share how I approached each of these challenges by sharing parts of my solution.

#### Modelling the input

So here we need to convert each line from the input file to two points on a 2D plane.
The line between two points can be called a vector. I chose to simply call it Line.
Here is how I modelled my data:

data class Point(val x: Int, val y: Int)

data class Line(val from: Point, val to: Point) {
// [...]
}

Then you need to convert the List<String> to a List<Line>. Here is how I did it:

.map { lineFrom(it) }

fun lineFrom(line: String) =
line.split(" -> ")
.map { subLine ->
subLine.split(",")
.map { it.toInt() }
.let { (x, y) -> Point(x, y) }
}
.let { (from, to) -> Line(from, to) }

Once we have modelled the data, we can focus on solving the problem at hand.

#### Converting a line to a list of points

As stated before, we need to count occurrences of each point to find when two points overlap.
Before we can do this, we need to get all given points on a Line.

In my opinion, this part was the most difficult β especially for diagonal lines.
For horizontal and vertical lines, we can leverage the fact that y or x stay consistent for all points.
For diagonal lines, there are 4 cases:

• from.x < to.x and from.y < to.y
• from.x < to.x and from.y > to.y
• from.x > to.x and from.y < to.y
• from.x > to.x and from.y > to.y

In order to traverse the line, we need to either increment or decrement x and y.

• if from.x > to.x we should decrement x
• if from.x < to.x we should increment x

The same thing can be said for y. Instead of handling these 4 cases, we can abstract out the increment/decrement action in a direction function.
Then, we can update x and y using this direction.

data class Line(val from: Point, val to: Point) {
val isHorizontal: Boolean = from.y == to.y
val isVertical: Boolean = from.x == to.x
val isDiagonal: Boolean = abs(from.x - to.x) == abs(from.y - to.y)

/*
* returns all the points along the line.
*/
fun points(): List<Point> {
val points = mutableListOf<Point>()
when {
isHorizontal -> {
val min = minOf(from.x, to.x)
val max = maxOf(from.x, to.x)
for (i in min until max + 1) {
}
return points
}
isVertical -> {
val min = minOf(from.y, to.y)
val max = maxOf(from.y, to.y)
for (i in min until max + 1) {
}
return points
}
isDiagonal -> {
val xDirection = direction(from.x, to.x) // whether to increment or decrement x when traversing the line
val yDirection = direction(from.y, to.y) // whether to increment or decrement y when traversing the line
var currentX = xDirection(from.x)
var currentY = yDirection(from.y)
while (currentX != to.x) {
currentX = xDirection(currentX)
currentY = yDirection(currentY)
}
return points
}
else -> {
throw NotImplementedError("We only support horizontal, vertical or diagonal lines.")
}
}
}
}

fun direction(a1: Int, a2: Int): (Int) -> Int {
if (a1 > a2) return { a -> a - 1 } else return { a -> a + 1 }
}

Once we have points for a given Line, we simply need to count occurences of points for the lines (List<Line>).

#### Counting the points

Counting the points can be done in a very idiomatic way leveraging some built-in function of the Kotlin List class.
Here is the one function for it:

fun overlappingPoints(lines: List<Line>, threshold: Int) =
lines.flatMap { it.points() }
.groupBy { it }
.count { it.value.size >= threshold }

Let me break it down for you:

• flatMap allows getting all the points for all the lines, if we would have simply mapped, we would have had List<List<Point>>, using flatMap allows to flatten the result to a List<Point>.
• groupBy allows transforming the List<Point> to a Map<Point, List<Point>> with the list being a list of duplicates of the same point (one for each time when this point is on a given line).
• count allows counting how many times a given predicate is true for the key/value pair. Here the predicate is { it.value.size >= threshold } which checks whether the number of duplicates for this point is bigger than the given threshold.

You may have noticed how I did not distinguish my solution between Part 1 and Part 2 so far, this is because the only difference is which points to count.
For both parts you need to add a filter based on the type of line (whether it is horizontal, vertical or diagonal).
Here is how I did it:

fun count(lines: List<Line>, filter: (Line) -> Boolean) =
overlappingPoints(lines.filter { filter(it) }, 2)

val count1 = count(lines) { line -> line.isVertical || line.isHorizontal }
val count2 = count(lines) { line -> line.isVertical || line.isHorizontal || line.isDiagonal }

Caveat: for today it seems all lines in the input are one of horizontal, vertical or diagonal (there are no lines at 60 degrees for example).
This means if a line is not horizontal nor vertical you could consider it diagonal without explicit check.

### Kotlin Concepts

Alright, I hope you liked my proposed solution and learned a few things along the way.
The key Kotlin concepts that we have covered are:

Feel free to share with me any feedback / concerns / questions you may have.

### Source code

You can find the source code for my solution here.

Thank you very much for reading!