Day 8 (#8)
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Blown way through my time budget and there's a bunch of low hanging fruit (data dependencies in LCM, for example) but I'm late for work.

Co-authored-by: Smaug123 <patrick+github@patrickstevens.co.uk>
Reviewed-on: #8
This commit is contained in:
2023-12-08 09:07:04 +00:00
parent bcd2bb6349
commit d94663ae0e
8 changed files with 279 additions and 0 deletions

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@@ -9,6 +9,7 @@
<Compile Include="Arr2D.fs" />
<Compile Include="ResizeArray.fs" />
<Compile Include="EfficientString.fs" />
<Compile Include="Arithmetic.fs" />
<Compile Include="Day1.fs" />
<Compile Include="Day2.fs" />
<Compile Include="Day3.fs" />
@@ -16,6 +17,7 @@
<Compile Include="Day5.fs" />
<Compile Include="Day6.fs" />
<Compile Include="Day7.fs" />
<Compile Include="Day8.fs" />
</ItemGroup>
</Project>

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namespace AdventOfCode2023
[<RequireQualifiedAccess>]
module Arithmetic =
[<Struct>]
type EuclidResult<'a> =
{
Hcf : 'a
A : 'a
B : 'a
}
/// Compute floor(sqrt(i)).
let inline sqrt (i : ^a) =
if i <= LanguagePrimitives.GenericOne then
i
else
let rec go start =
let next = start + LanguagePrimitives.GenericOne
let sqr = next * next
if sqr < LanguagePrimitives.GenericZero then
// Overflow attempted, so the sqrt is between start and next
start
elif i < sqr then
start
elif i = sqr then
next
else
go next
go LanguagePrimitives.GenericOne
/// Find Hcf, A, B s.t. A * a + B * b = Hcf, and Hcf is the highest common factor of a and b.
let inline euclideanAlgorithm (a : ^a) (b : ^a) : EuclidResult< ^a > =
let rec go rMin1 r sMin1 s tMin1 t =
if r = LanguagePrimitives.GenericZero then
{
Hcf = rMin1
A = sMin1
B = tMin1
}
else
let newQ = rMin1 / r
go r (rMin1 - newQ * r) s (sMin1 - newQ * s) t (tMin1 - newQ * t)
let maxA = max a b
let minB = min a b
let result =
go
maxA
minB
LanguagePrimitives.GenericOne
LanguagePrimitives.GenericZero
LanguagePrimitives.GenericZero
LanguagePrimitives.GenericOne
if a = maxA then
result
else
{
Hcf = result.Hcf
A = result.B
B = result.A
}

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namespace AdventOfCode2023
open System
open System.Collections.Generic
[<RequireQualifiedAccess>]
module Day8 =
type Instructions =
{
/// "true" is 'R'
Steps : bool array
Nodes : Dictionary<string, string * string>
}
let parse (s : string) =
use mutable lines = StringSplitEnumerator.make '\n' s
lines.MoveNext () |> ignore
let stepsLine = lines.Current.TrimEnd ()
let steps = Array.zeroCreate stepsLine.Length
for i = 0 to stepsLine.Length - 1 do
steps.[i] <- (stepsLine.[i] = 'R')
let dict = Dictionary ()
while lines.MoveNext () do
if not lines.Current.IsEmpty then
use mutable line = StringSplitEnumerator.make' ' ' lines.Current
line.MoveNext () |> ignore
let key = line.Current.ToString ()
line.MoveNext () |> ignore
line.MoveNext () |> ignore
let v1 = line.Current.Slice(1, line.Current.Length - 2).ToString ()
line.MoveNext () |> ignore
let v2 = line.Current.Slice(0, line.Current.Length - 1).ToString ()
dict.[key] <- (v1, v2)
{
Steps = steps
Nodes = dict
}
let part1 (s : string) =
let data = parse s
let mutable i = 0
let mutable currentNode = "AAA"
let mutable answer = 0
while currentNode <> "ZZZ" do
let instruction = data.Nodes.[currentNode]
if data.Steps.[i] then
// "true" is R
currentNode <- snd instruction
else
currentNode <- fst instruction
i <- (i + 1) % data.Steps.Length
answer <- answer + 1
answer
let inline lcm (periods : ^T[]) =
let mutable lcm = periods.[0]
let mutable i = 1
while i < periods.Length do
let euclid = Arithmetic.euclideanAlgorithm lcm periods.[i]
lcm <- (lcm * periods.[i]) / euclid.Hcf
i <- i + 1
lcm
let part2 (s : string) =
let data = parse s
let startingNodes = ResizeArray ()
for key in data.Nodes.Keys do
if key.[key.Length - 1] = 'A' then
startingNodes.Add key
let periods =
Array.init
startingNodes.Count
(fun startNode ->
let mutable i = 0
let mutable currentNode = startingNodes.[startNode]
let mutable answer = 0ul
while currentNode.[currentNode.Length - 1] <> 'Z' do
let instruction = data.Nodes.[currentNode]
if data.Steps.[i] then
// "true" is R
currentNode <- snd instruction
else
currentNode <- fst instruction
i <- (i + 1) % data.Steps.Length
answer <- answer + 1ul
uint64 answer
)
lcm periods