Day 22: Monkey Market
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Go
Re-familiarizing myself with Go. The solution to Part 2 is fairly simply, the whole packing of the sequence into a single integer to save on memory was an optimization I did afterwards based on looking at other solutions. I thought it was cool.
package main import ( "bufio" "fmt" "os" "strconv" ) type SequenceMap struct { Data map[int32]int } func PackSeq(numbers [4]int8) int32 { var packed int32 for i, num := range numbers { packed |= int32(num+9) << (i * 5) } return packed } func UnpackSeq(packed int32) [4]int8 { var numbers [4]int8 for i := range numbers { numbers[i] = int8((packed>>(i*5))&0x1F) - 9 } return numbers } func NewSequenceMap() SequenceMap { return SequenceMap{make(map[int32]int)} } func (m *SequenceMap) Increment(seq [4]int8, val int) { pSeq := PackSeq(seq) acc, ok := m.Data[pSeq] if ok { m.Data[pSeq] = acc + val } else { m.Data[pSeq] = val } } func (m *SequenceMap) Has(seq [4]int8) bool { pSeq := PackSeq(seq) _, ok := m.Data[pSeq] return ok } type Generator struct { Secret int64 LastPrice int8 ChangeSequence []int8 } func NewGenerator(Secret int64) Generator { var ChangeSequence []int8 return Generator{Secret, int8(Secret % 10), ChangeSequence} } func (g *Generator) Mix(value int64) *Generator { g.Secret = g.Secret ^ value return g } func (g *Generator) Prune() *Generator { g.Secret = g.Secret % 16777216 return g } func (g *Generator) Next() { g.Mix(g.Secret * 64).Prune().Mix(g.Secret / 32).Prune().Mix(g.Secret * 2048).Prune() Price := int8(g.Secret % 10) g.ChangeSequence = append(g.ChangeSequence, Price-g.LastPrice) g.LastPrice = Price if len(g.ChangeSequence) > 4 { g.ChangeSequence = g.ChangeSequence[1:] } } func ParseInput() []int64 { if fileInfo, _ := os.Stdin.Stat(); (fileInfo.Mode() & os.ModeCharDevice) != 0 { fmt.Println("This program expects input from stdin.") os.Exit(1) } scanner := bufio.NewScanner(os.Stdin) var numbers []int64 for scanner.Scan() { line := scanner.Text() num, err := strconv.ParseInt(line, 10, 64) if err != nil { fmt.Printf("ERROR PARSING VALUE: %s\n", line) os.Exit(1) } numbers = append(numbers, num) } return numbers } func main() { numbers := ParseInput() m := NewSequenceMap() sum := int64(0) for i := 0; i < len(numbers); i += 1 { g := NewGenerator(numbers[i]) tM := NewSequenceMap() for j := 0; j < 2000; j += 1 { g.Next() if len(g.ChangeSequence) == 4 { if !tM.Has([4]int8(g.ChangeSequence)) { tM.Increment([4]int8(g.ChangeSequence), 1) if g.LastPrice > 0 { m.Increment([4]int8(g.ChangeSequence), int(g.LastPrice)) } } } } sum += g.Secret } fmt.Printf("Part One: %d\n", sum) var bestSeq [4]int8 bestPrice := 0 for pSeq, price := range m.Data { if price > bestPrice { bestPrice = price bestSeq = UnpackSeq(pSeq) } } fmt.Printf("Part Two: %d\n", bestPrice) fmt.Printf("Best Sequence: %d\n", bestSeq) }
Rust
Nice breather today (still traumatized from the robots). At some point I thought you had to do some magic for predicting special properties of the pseudorandom function, but no, just collect all values, have a big table for all sequences and in the end take the maximum value in that table. Part 1 takes 6.7ms, part 2 19.2ms.
Solution
fn step(n: u32) -> u32 { let a = (n ^ (n << 6)) % (1 << 24); let b = a ^ (a >> 5); (b ^ (b << 11)) % (1 << 24) } fn part1(input: String) { let sum = input .lines() .map(|l| { let n = l.parse().unwrap(); (0..2000).fold(n, |acc, _| step(acc)) as u64 }) // More than 2¹⁰ 24-bit numbers requires 35 bits .sum::<u64>(); println!("{sum}"); } const N_SEQUENCES: usize = 19usize.pow(4); fn sequence_key(sequence: &[i8]) -> usize { sequence .iter() .enumerate() .map(|(i, x)| (x + 9) as usize * 19usize.pow(i as u32)) .sum() } fn part2(input: String) { // Table for collecting the amount of bananas for every possible sequence let mut table = vec![0; N_SEQUENCES]; // Mark the sequences we encountered in a round to ensure that only the first occurence is used let mut seen = vec![false; N_SEQUENCES]; for l in input.lines() { let n = l.parse().unwrap(); let (diffs, prices): (Vec<i8>, Vec<u8>) = (0..2000) .scan(n, |acc, _| { let next = step(*acc); let diff = (next % 10) as i8 - (*acc % 10) as i8; *acc = next; Some((diff, (next % 10) as u8)) }) .unzip(); for (window, price) in diffs.windows(4).zip(prices.iter().skip(3)) { let key = sequence_key(window); if !seen[key] { seen[key] = true; table[key] += *price as u32; } } // Reset seen sequences for next round seen.fill(false); } let bananas = table.iter().max().unwrap(); println!("{bananas}"); } util::aoc_main!();
Also on github
Haskell
A nice easy one today; shame I couldn’t start on time. I had a go at refactoring to reduce the peak memory usage, but it just ended up a mess. Here’s a tidy version.
import Data.Bits import Data.List import Data.Map (Map) import Data.Map qualified as Map next :: Int -> Int next = flip (foldl' (\x n -> (x `xor` shift x n) .&. 0xFFFFFF)) [6, -5, 11] bananaCounts :: Int -> Map [Int] Int bananaCounts seed = let secrets = iterate next seed prices = map (`mod` 10) secrets changes = zipWith (-) (drop 1 prices) prices sequences = map (take 4) $ tails changes in Map.fromListWith (const id) $ take 2000 (zip sequences (drop 4 prices)) main = do input <- map read . lines <$> readFile "input22" print . sum $ map ((!! 2000) . iterate next) input print . maximum $ Map.unionsWith (+) $ map bananaCounts input
Haskell
I have no Idea how to optimize this and am looking forward to the other solutions that probably run in sub-single-second times. I like my solution because it was simple to write which I hadn’t managed in the previous days, runs in 17 seconds with no less than 100MB of RAM.
import Control.Arrow import Data.Bits (xor) import Data.Ord (comparing) import qualified Data.List as List import qualified Data.Map as Map parse :: String -> [Int] parse = map read . filter (/= "") . lines mix = xor prune = flip mod 16777216 priceof = flip mod 10 nextSecret step0 = do let step1 = prune . mix step0 $ step0 * 64 let step2 = prune . mix step1 $ step1 `div` 32 let step3 = prune . mix step2 $ step2 * 2048 step3 part1 = sum . map (head . drop 2000 . iterate nextSecret) part2 = map (iterate nextSecret >>> take 2001 >>> map priceof >>> (id &&& tail) >>> uncurry (zipWith (curry (uncurry (flip (-)) &&& snd))) >>> map (take 4) . List.tails >>> filter ((==4) . length) >>> map (List.map fst &&& snd . List.last) >>> List.foldl (\ m (s, p) -> Map.insertWith (flip const) s p m) Map.empty ) >>> Map.unionsWith (+) >>> Map.assocs >>> List.maximumBy (comparing snd) main = getContents >>= print . (part1 &&& part2) . parse
Haha, same! Mine runs in a bit under 4s compiled, but uses a similar 100M-ish peak. Looks like we used the same method.
Maybe iterate all the secrets in parallel, and keep a running note of the best sequences so far? I’m not sure how you’d decide when to throw away old candidates, though. Sequences might match one buyer early and another really late.