Day 22: Monkey Market

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FAQ

  • Zikeji@programming.dev
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    59 minutes ago

    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)
    }
    
  • Gobbel2000@programming.dev
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    1 hour ago

    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

  • lwhjp@lemmy.sdf.org
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    3 hours ago

    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
    
  • VegOwOtenks@lemmy.world
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    5 hours ago

    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
    
    • lwhjp@lemmy.sdf.org
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      3 hours ago

      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.