Uiua

It's been a while since I posted one of these, but I thought this would be straightforward in Uiua. Turns out that bitwise operations are a bit (haha) of a pain, so the Rng operation is very slow at 4sec for live data.

I took this as an opportunity to play with the ⧈(stencil) operator which probably slowed things down too.

Data ← 1_10_100_2024
Xor  ← °⋯◿2⬚0+∩⋯ # Bitwise xor of two numbers.
Rng  ← ⊙◌◿,Xor×2048.◿,Xor⌊÷32.◿,Xor×64.⊙16777216
Runs ← &p⍜now((⇌[⍥(Rng.)])2000 Data) # len seed
First ← (
  ⊟⊂0⧈₂/-.◿10 ↘¯1                      # Build run, gen pair diffs
  ⍣⊢0▽⊸(>0)⊢⧈(⨬(⋅0|⊣)≍¯2_1_¯1_3°⊟) 2_4 # Take first value for matching diffs.
)
&p /+≡⊣Runs
&p /+≡First Runs

Rust

Not too hard today, apart from yesterday's visit to a cocktail bar leaving me a little hazy in the mind.

use std::{fs, str::FromStr};

use color_eyre::eyre::{Report, Result};
use gxhash::{HashMap, HashMapExt};

const SECRETS_PER_DAY: usize = 2000;
const SEQ_LEN: usize = 4;

type Sequence = [i8; SEQ_LEN];

fn produce(n: usize) -> usize {
    let n = (n ^ (n * 64)) % 16777216;
    let n = (n ^ (n / 32)) % 16777216;
    (n ^ (n * 2048)) % 16777216
}

#[derive(Debug)]
struct Buyer {
    prices: [u8; SECRETS_PER_DAY + 1],
    changes: [i8; SECRETS_PER_DAY],
}

impl Buyer {
    fn price_at_seq(&self, seq: &Sequence) -> Option<u8> {
        self.changes
            .windows(SEQ_LEN)
            .position(|win| win == *seq)
            .and_then(|i| self.price_for_window(i))
    }

    fn price_for_window(&self, i: usize) -> Option<u8> {
        self.prices.get(i + SEQ_LEN).copied()
    }
}

struct BananaMarket {
    buyers: Vec<Buyer>,
}

impl FromStr for BananaMarket {
    type Err = Report;

    fn from_str(s: &str) -> Result<Self, Self::Err> {
        let buyer_seeds = s
            .lines()
            .map(|s| s.parse::<usize>())
            .collect::<Result<Vec<_>, _>>()?;

        let buyers = buyer_seeds
            .into_iter()
            .map(|seed| {
                let mut prices = [0; SECRETS_PER_DAY + 1];
                let mut changes = [0; SECRETS_PER_DAY];

                let mut secret = seed;
                let mut price = (seed % 10) as u8;
                prices[0] = price;
                for i in 0..SECRETS_PER_DAY {
                    let last_price = price;
                    secret = produce(secret);
                    price = (secret % 10) as u8;
                    prices[i + 1] = price;
                    changes[i] = price as i8 - last_price as i8;
                }
                Buyer { prices, changes }
            })
            .collect();
        Ok(Self { buyers })
    }
}

impl BananaMarket {
    fn sell_with_seq(&self, seq: &Sequence) -> usize {
        self.buyers
            .iter()
            .map(|b| b.price_at_seq(seq).unwrap_or(0) as usize)
            .sum()
    }

    fn maximise_bananas(&self) -> usize {
        let mut cache: HashMap<Sequence, usize> = HashMap::new();

        for seq in self
            .buyers
            .iter()
            .flat_map(|buyer| buyer.changes.windows(SEQ_LEN))
        {
            let seq = seq.try_into().unwrap();
            cache.entry(seq).or_insert_with(|| self.sell_with_seq(&seq));
        }

        cache.into_values().max().unwrap_or(0)
    }
}

fn part1(filepath: &str) -> Result<usize> {
    let input = fs::read_to_string(filepath)?
        .lines()
        .map(|s| s.parse::<usize>())
        .collect::<Result<Vec<_>, _>>()?;
    let res = input
        .into_iter()
        .map(|n| (0..SECRETS_PER_DAY).fold(n, |acc, _| produce(acc)))
        .sum();
    Ok(res)
}

fn part2(filepath: &str) -> Result<usize> {
    let input = fs::read_to_string(filepath)?;
    let market = BananaMarket::from_str(&input)?;
    Ok(market.maximise_bananas())
}

fn main() -> Result<()> {
    color_eyre::install()?;

    println!("Part 1: {}", part1("d22/input.txt")?);
    println!("Part 2: {}", part2("d22/input.txt")?);
    Ok(())
}

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.

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

import Control.Arrow
import Data.Bits
import Data.List
import qualified Data.Map as M

parse = fmap (secretNums . read) . lines

secretNums :: Int -> [Int]
secretNums = take 2001 . iterate (step1 >>> step2 >>> step3)
 where
  step1 n = ((n `shiftL` 06) `xor` n) .&. 0xFFFFFF
  step2 n = ((n `shiftR` 05) `xor` n) .&. 0xFFFFFF
  step3 n = ((n `shiftL` 11) `xor` n) .&. 0xFFFFFF

part1 = sum . fmap last
part2 = maximum . M.elems . M.unionsWith (+) . fmap (deltas . fmap (`mod` 10))

deltas l = M.fromListWith (\n p -> p) $ flip zip (drop 4 l) $ zip4 diffs (tail diffs) (drop 2 diffs) (drop 3 diffs)
 where
  diffs = zipWith (-) (tail l) l

main = getContents >>= print . (part1 &&& part2) . parse

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)
}

Dart

Well, that was certainly a bit easier than yesterday...

I know running a window over each full list of 2000 prices rather than looking for cycles etc means I'm doing a lot of unnecessary work, but it only takes a couple of seconds, so that'll do.

import 'package:collection/collection.dart';
import 'package:more/more.dart';

rng(int i) {
  i = ((i << 6) ^ i) % 16777216;
  i = ((i >> 5) ^ i) % 16777216;
  i = ((i << 11) ^ i) % 16777216;
  return i;
}

Iterable<int> getPrices(int val, int rounds) {
  var ret = [val];
  for (var _ in 1.to(rounds)) {
    ret.add(val = rng(val));
  }
  return ret.map((e) => e % 10);
}

int run(int val, int rounds) => 0.to(rounds).fold(val, (s, t) => s = rng(s));
part1(lines) => [for (var i in lines.map(int.parse)) run(i, 2000)].sum;

part2(List<String> lines) {
  var market = <int, int>{}.withDefault(0);
  for (var seed in lines.map(int.parse)) {
    var seen = <int>{};
    for (var w in getPrices(seed, 2000).window(5)) {
      var key = // Can't use lists as keys, so make cheap hash.
          w.window(2).map((e) => e[1] - e[0]).reduce((s, t) => (s << 4) + t);
      if (seen.contains(key)) continue;
      seen.add(key);
      market[key] += w.last;
    }
  }
  return market.values.max;
}

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

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