This one wasn't too bad. I had a false start trying to flood-fill with multiple generators, and that got silly really quickly.

Then I realized I could just build a big array, chunk through item by item, remove any symbol on an edge, and then just count the places they reached.

fun String.parsePoint() = this.split(""", """).let { found -> Point(found[0].toInt(), found[1].toInt()) } fun Point.manhattanDistance(b: Point): Int = (x - b.x).absoluteValue + (y - b.y).absoluteValue val alpha = ('a'..'z') + ('A'..'Z') fun answer1(input: List<Point>): Int? { val minX = input.minBy { (x, y) -> x }!!.x val minY = input.minBy { (x, y) -> y }!!.y val maxX = input.maxBy { (x, y) -> x }!!.x val maxY = input.maxBy { (x, y) -> y }!!.y val map = input.mapIndexed { i, point -> point to alpha[i].toString() }.toMap().toMutableMap() val output = (minY..maxY).map { y -> (minX..maxX).map { x -> map.toList().groupBy { (k, _) -> (x to y).manhattanDistance(k) }.minByKey()!!.value.let { closest -> when { closest.count() > 1 -> "." else -> closest.first().second } } } } return map.values.filter { v -> v !in output.first() && v !in output.map { it.first() } && v !in output.last() && v !in output.map { it.last() } }.map { v -> output.flatten().count { it == v } }.max() }

This is where I got stuck a bit trying to optimize instead of just letting it finish this code's 30 second cycle.

I'm not happy I couldn't figure out the math, but I'll post some pictures I've generated from my output in penance.

fun answer2(input: List<Point>): Int? { val dist = 10000 val minX = input.minBy { (x, y) -> x }!!.x val minY = input.minBy { (x, y) -> y }!!.y val maxX = input.maxBy { (x, y) -> x }!!.x val maxY = input.maxBy { (x, y) -> y }!!.y val ptc = (maxX - dist..minX + dist).map { x -> println("row $x") (maxY - dist..minY + dist).map { y -> x to y } .filter { point -> input.sumBy(point::manhattanDistance) < dist } }.fold(emptySet<Point>()) { a, b -> a + b } return ptc.count() }

You don't have to construct an array; you can just iterate over min(x)..max(x) and min(y)..max(y) and add + to the nearest input point.

The promised images!

Part 1 - Answer

Part 2 - Answer

Annnnnd, by posting these pictures, I see that the optimization for part 2 is to bound yourself by the points. 10000 is a bit of a red herring.

EDIT: 2018-12-06T11:01:40-08:00 Found some problems with my image processing and fixed them.

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## re: AoC Day 6: Chronal Coordinates VIEW POST

VIEW FULL DISCUSSION## Kotlin Solution

## Part 1

This one wasn't too bad. I had a false start trying to flood-fill with multiple generators, and that got silly really quickly.

Then I realized I could just build a big array, chunk through item by item, remove any symbol on an edge, and then just count the places they reached.

## Part 2

This is where I got stuck a bit trying to optimize instead of just letting it finish this code's 30 second cycle.

I'm not happy I couldn't figure out the math, but I'll post some pictures I've generated from my output in penance.

You don't have to construct an array; you can just iterate over min(x)..max(x) and min(y)..max(y) and add + to the nearest input point.

The promised images!

## Sample Data

Part 1 - Answer

Part 2 - Answer

## Answers

Part 1 - Answer

Part 2 - Answer

Annnnnd, by posting these pictures, I see that the optimization for part 2 is to bound yourself by the points. 10000 is a bit of a red herring.

EDIT: 2018-12-06T11:01:40-08:00

Found some problems with my image processing and fixed them.