this post was submitted on 10 Dec 2024
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[–] [email protected] 3 points 2 weeks ago (2 children)

Day 11!

discussion p1 + 2I'd say this was pretty easy. My instinct was that 64 bit ints were big enough for this problem, and that memoising would also be a good idea. I didn't experiment with a control though, so I don't know if memoisation was truly necessary. Either way, my initial solution for 1. was performant enough for part 2.

[–] [email protected] 2 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

followupSo memoisation is predictably needed for part 2 to run in time. It's an O(e^n^), so it takes seconds by step 39 and minutes by step 47.

[–] [email protected] 3 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

re:followupIf you somehow wanted your whole final array it would also require over 1 Peta byte ^^, memoization definetely reccomended.

[–] [email protected] 4 points 2 weeks ago

spoilerIt's one AOC problem zogwarg, what could it cost? 10 PB?

[–] [email protected] 2 points 2 weeks ago* (last edited 2 weeks ago) (1 children)

11 discussion with spoliersWell my pt1 solution would require something like at least 1.5 petabytes RAM to hold the fully expanded array, so it was back to the drawing board for pt2 😁

Luckily I noticed the numbers produced in every iteration were incredibly repetitive, so I assigned a separate accumulator to each one, and every iteration I only kept the unique numbers and updated the corresponding accumulators with how many times they had appeared, and finally I summed the accumulators.

The most unique numbers in one iteration were 3777, the 75 step execution was basically instant.

edit: other unhinged attempts included building a cache with how many pebbles resulted from a number after x steps that I would start using after reaching the halfway point, so every time I found a cached number I would replace that branch with the final count according to the remaining steps, but I couldn't think of a way to actually track how many pebbles result downstream from a specific pebble, but at least it got me thinking about tracking something along each pebble.

11 code

// F# as usual
// fst and snd are tuple deconstruction helpers

[<TailCall>]
let rec blink (idx:int) (maxIdx:int) (pebbles : (int64*int64) list) =
    if idx = maxIdx
    then pebbles |> List.sumBy snd
    else
        pebbles
        // Expand array
        |> List.collect (fun (pebbleId, pebbleCount) -> 
            let fpb = float pebbleId
            let digitCount = Math.Ceiling(Math.Log(fpb + 1.0,10))      
            match pebbleId with
            | 0L -> [ 1L, pebbleCount ]
            | x when digitCount % 2.0 = 0.0 -> 
                let factor = Math.Pow(10,digitCount/2.0)
                let right = fpb % factor
                let left = (fpb - right) / factor
                [int64 left, pebbleCount; int64 right,pebbleCount]   
            | x -> [ x * 2024L, pebbleCount ])
        // Compress array
        |> List.groupBy fst
        |> List.map (fun (pebbleId, pebbleGroup) -> pebbleId, pebbleGroup |> List.sumBy snd)
        |> blink (idx+1) maxIdx


"./input.example"
|> Common.parse
|> List.map (fun pebble -> pebble,1L)
|> blink 0 25 
|> Global.shouldBe 55312L

"./input.actual"
|> Common.parse
|> List.map (fun pebble -> pebble,1L)
|> blink 0 75 
|> printfn "Pebble count after 75 blinks is %d" 

[–] [email protected] 3 points 2 weeks ago

re: 11p2 discussion

I was also on my way to building a multilevel memoization cache with "branches", when I just happened to stumble on an incredibly elegant solution in the subreddit. I stole it and awarded myself only 1 point for today. Because I'm worth it.