Advent of Code F# - Day 9 | theburningmonk.com

Yan Cui

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ps. look out for all my other solutions for Advent of Code challenges here.

Day 9

See details of the challenge here.

The input for today’s challenge is a very long string like this:

(6×9)JUORKH(10×13)LNWIKDMACM(126×14)(21×8)QLKUJNVVZIQGGFCJZMPHK(2×1)ZH(59×3)(38×14)KELEPIDYLCGJUBCXACRSOCEZYXLO…

First, let’s see how we’re gonna parse this input.

The approach I went with is to recursively split the input with input.Split([| ‘(‘; ‘)’ |], 3) which will return either:

the whole string if there’s no (n x count) section, eg. “ADVENT” => “ADVENT”
or an array of 3 elements
- for “X(1×3)YZUVW”, it’ll return [| “X”; “1×3”; “YZUVW” |]
- for “(1×3)YZUVW”, it’ll return [| “”; “1×3”; “YZUVW” |]

To parse the number of characters (n) and number of times to repeat (count) from “1×3” we can introduce an active pattern (the Repeat pattern below) to do the job.

	open System
	open System.IO

	let input = File.ReadAllText(__SOURCE_DIRECTORY__ + "/Day09Input.txt")

	let (\|Int\|_\|) x =
	match Int32.TryParse x with
	\| true, x -> Some x
	\| _ -> None

	let (\|Repeat\|_\|) (input : string) =
	match input.Split('x') with
	\| [\| Int n; Int count \|] -> Some (n, count)
	\| _ -> None

view raw Day09_parsing.fsx hosted with

by GitHub

Now we can recursively split the string and count the length of the decompressed string.

	let decompressV1 input =
	let rec loop (input : string) acc =
	match input.Split([\| '('; ')' \|], 3) with
	\| [\| text \|] -> acc + text.Length
	\| [\| head; Repeat(n, count); rest \|] ->
	let acc = acc + head.Length + n * count
	loop (rest.Substring n) acc

	loop input 0

	let part1 = decompressV1 input

view raw Day09_Part1.fsx hosted with

by GitHub

Part 2

Apparently, the file actually uses version two of the format.

In version two, the only difference is that markers within decompressed data are
decompressed. This, the documentation explains, provides much more substantial
compression capabilities, allowing many-gigabyte files to be stored in only a
few kilobytes.

For example:

(3×3)XYZ still becomes XYZXYZXYZ, as the decompressed section contains no
markers.
X(8×2)(3×3)ABCY becomes XABCABCABCABCABCABCY, because the decompressed data from
the (8×2) marker is then further decompressed, thus triggering the (3×3) marker
twice for a total of six ABC sequences.
(27×12)(20×12)(13×14)(7×10)(1×12)A decompresses into a string of A repeated
241920 times.
(25×3)(3×3)ABC(2×3)XY(5×2)PQRSTX(18×9)(3×2)TWO(5×7)SEVEN becomes 445 characters
long.
Unfortunately, the computer you brought probably doesn’t have enough memory to
actually decompress the file; you’ll have to come up with another way to get its
decompressed length.

What is the decompressed length of the file using this improved format?

Part 2 requires slightly more work and, as the challenge hinted, we might overflow on int so let’s use int64 instead.

The main difference to part 1 is that we need to make the decompressV2 function itself recursive, and call into it from inside the loop function to work out the decompressed length of the repeated section.

	let rec decompressV2 input =
	let rec loop (input : string) acc =
	match input.Split([\| '('; ')' \|], 3) with
	\| [\| text \|] -> acc + int64 text.Length
	\| [\| head; Repeat(n, count); rest \|] ->
	let repeatLen = rest.Substring(0, n) \|> decompressV2
	let acc = acc + int64 head.Length + int64 count * repeatLen
	loop (rest.Substring n) acc

	loop input 0L

	let part2 = decompressV2 input

view raw Day09_Part2.fsx hosted with

by GitHub

Advent of Code F# – Day 9

Day 9

Part 2

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