Day 1: Secret Entrance

Megathread guidelines

  • Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
  • You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL

FAQ

all 28 comments

sorted by: hot top controversial new old
[–] 8 points 9 months ago* (last edited 9 months ago) (1 child)

DOS + BIOS boot (hybrid binary)

Repo | day01.asm | day01.c (prototype) | .COM download

Written in x86-16 assembly. Works as a DOS program but also as disk image for older PCs with BIOS support (or older VMs). Getting this setup to work was tricky, especially since when I started this, I had only basic x86-16 experience! Needless to say I've spent much time staring at hex numbers.

The very start of the file determines if it's running in DOS or as a bootloader, in which case it'll have to load the remainder of the file from disk and rearrange the memory layout to emulate the DOS situation.

Right now this is using just one segment of memory (64K). Input data is compressed using a custom run-length encoding scheme. The compressed background image is directly decoded into the VGA framebuffer, after which it is overwritten with the decompressed input file. Space is tight!

My main goal is to finish a few days at least with one or more game consoles supported (GameBoy Advance would be cool) and to do some cool palette tricks, like a day/night transition or animated water and stars.

  • source
  • hideshow 1 child comment
  • [–] 6 points 9 months ago (6 children)

    Rust

    #[derive(Default)]
    pub struct Day1Solver {
        input: Vec<i64>,
    }
    
    impl Solver for Day1Solver {
        fn presolve(&mut self, input: &str) {
            self.input = input
                .trim()
                .split("\n")
                .map(|line| {
                    if let Some(n) = line.strip_prefix('L') {
                        -n.parse::<i64>().unwrap()
                    } else if let Some(n) = line.strip_prefix('R') {
                        n.parse().unwrap()
                    } else {
                        panic!("what: {line}");
                    }
                })
                .collect();
        }
    
        fn solve_part_one(&mut self) -> String {
            let mut p = 50;
            let mut count = 0;
            for n in self.input.clone() {
                p += n;
                if p % 100 == 0 {
                    count += 1;
                }
            }
            count.to_string()
        }
    
        fn solve_part_two(&mut self) -> String {
            let mut count = 0;
            let mut p = 1000000000050;
            for i in self.input.clone() {
                if p % 100 == 0 {
                    count += (i / 100).abs();
                } else {
                    count += ((p + i) / 100 - p / 100).abs();
                    if i < 0 && (p + i) % 100 == 0 {
                        count += 1;
                    }
                }
                p += i;
            }
            count.to_string()
        }
    }
    
  • source
  • hideshow 6 child comments
  • [–] 3 points 9 months ago (1 child)

    Nice solution. Just a little Rust tip if you don't mind: In this case you can avoid cloning the input Vec in the loops by instead looping over references into the list with for n in self.input.iter() or simpler for n in &self.input. The only difference is that n will be of type &i64 instead of i64.

  • source
  • parent
  • hideshow 1 child comment
  • [–] 1 point 9 months ago (1 child)

    I'm far from sure I understand what you're doing in part 2, but I think maybe you hit the same logic bug I did (I solved it with a shameful if statement that I will have to fix later).

  • source
  • parent
  • hideshow 1 child comment
  • [–] 1 point 9 months ago

    the idea is that crossing the zero is easy to detect by dividing the total dial movement by 100. I.e. if you cross from 120 to 90 you will detect that 120/100=1 changed to 90/100=0. The only case when this doesn’t work is when you stop at zero going in the negative direction, hence the extra if

  • source
  • parent
  • [–] 4 points 9 months ago*

    Golang

    func part1() {
    	// file, _ := os.Open("sample.txt")
    	file, _ := os.Open("input.txt")
    	defer file.Close()
    
    	scanner := bufio.NewScanner(file)
    	n := 100
    	current := 50
    	pointingAt0 := 0
    
    	for scanner.Scan() {
    		line := scanner.Text()
    		num, _ := strconv.Atoi(line[1:])
    		if line[0] == 'L' {
    			current = ((current-num)%n + n) % n
    		} else {
    			current = (current + num) % n
    		}
    		if current == 0 {
    			pointingAt0++
    		}
    	}
    
    	fmt.Println(pointingAt0)
    }
    
    func part2() {
    	// file, _ := os.Open("sample.txt")
    	file, _ := os.Open("input.txt")
    	defer file.Close()
    
    	scanner := bufio.NewScanner(file)
    	n := 100
    	current := 50
    	pointingAt0 := 0
    
    	for scanner.Scan() {
    		line := scanner.Text()
    		num, _ := strconv.Atoi(line[1:])
    
    		rounds := num / n
    		pointingAt0 += rounds
    		num = num % n
    		new := -1
    
    		if line[0] == 'L' {
    			new = ((current-num)%n + n) % n
    			if current != 0 && (new > current || new == 0) {
    				pointingAt0++
    			}
    		} else {
    			new = (current + num) % n
    			if current != 0 && (new < current || new == 0) {
    				pointingAt0++
    			}
    		}
    
    		current = new
    	}
    
    	fmt.Println(pointingAt0)
    }
    
  • source
  • [–] 4 points 9 months ago (1 child)

    A straightforward day 1 for a Javascript / NodeJS solution.

    Poorly Optimized JS Solution

    let position = 50;
    let answer1 = 0;
    let answer2 = 0;
    
    const sequence = require('fs').readFileSync('input-day1.txt', 'utf-8');
    
    sequence.split('\n').forEach(instruction => {
        const turn = instruction.charAt(0);
        let distance = parseInt(instruction.slice(1), 10);
    
        while (distance > 0) {
            distance -= 1;
    
            if (turn === 'L') {
                position -= 1;
            } else if (turn === 'R') {
                position += 1;
            }
    
            if (position >= 100) {
                position -= 100;
            } else if (position < 0) {
                position += 100;
            }
    
            if (position === 0) {
                answer2 += 1;
            }
        }
    
        if (position === 0) {
            answer1 += 1;
        }
    });
    
    console.log(`Part 1 Answer: ${answer1}`);
    console.log(`Part 2 Answer: ${answer2}`);
    

  • source
  • hideshow 1 child comment
  • [–] 3 points 9 months ago* (last edited 9 months ago)

    Elixir

    Oh wow I'm late but that's because I decided to do this year in Elixir which I am not really experienced enough yet as it seems. Part 1 was ok but I didn't really put enough thought into Part 2 and got stuck for a while. However I got it done after a few failed tries and a few questions to AI on how to do things in Elixir.

    #!/usr/bin/elixir
    
    defmodule SafeDial do
    
    def readInstructions(filename) do
    	filename
    	|> File.stream!()
    	|> Stream.map(&String.trim/1)
    	|> Stream.reject(&(&1 == ""))
    	|> Stream.map(fn line ->
    		case Regex.run(~r/^([A-Za-z])(\d+)$/, line) do
    			[_, letter, number] -> {String.to_atom(letter), String.to_integer(number)}
    			_ -> nil
    		end
    	end)
    	|> Enum.reject(&(&1 == nil))
    end
    
    def evalInstructions(instructions) do
    	initialPos = 50
    	zeroCount = 0
    	checkInstruction(instructions, initialPos, zeroCount)
    end
    
    defp checkInstruction(instructions, currPos, zeroCount) do
    	case instructions do
    		[] -> {currPos, zeroCount}
    		[instruct | rest] ->
    			nextPos =
    				case instruct do
    					{dir, step} when dir == :R and currPos + rem(step,100) > 99 ->
    						currPos + rem(step,100) - 100
    					{dir, step} when dir == :R and currPos + rem(step,100) > 99 ->
    						currPos + rem(step,100) - 100
    					{dir, step} when dir == :R ->
    						currPos + rem(step,100)
    					{dir, step} when dir == :L and currPos - rem(step,100) < 0 ->
    						currPos - rem(step,100) + 100
    					{dir, step} when dir == :L ->
    						currPos - rem(step,100)
    					_ -> raise "[ ERROR ] unkown instruction: #{inspect(instruct)}"
    				end
    			newZeroCount = if nextPos == 0, do: zeroCount + 1, else: zeroCount
    			checkInstruction(rest, nextPos, newZeroCount)
    		other -> raise "[ ERROR ] unknown instruction: #{inspect(other)}"
    	end
    end
    
    def evalInstructionsPart2(instructions) do
    	initialPos = 50
    	zeroCount = 0
    	checkInstructionPart2(instructions, initialPos, zeroCount)
    end
    
    defp checkInstructionPart2(instructions, currPos, zeroCount) do
    	case instructions do
    		[] -> {currPos, zeroCount}
    		[instruct | rest] ->
    			{nextPos, zeroCount1} =
    				case instruct do
    					{dir, step} when dir == :R and currPos + rem(step,100) == 100 ->
    						{currPos + rem(step,100) - 100, zeroCount+floor(step/100)}
    					{dir, step} when dir == :R and currPos + rem(step,100) > 99 ->
    						{currPos + rem(step,100) - 100, zeroCount+floor(step/100)+1}
    					{dir, step} when dir == :R ->
    						{currPos + rem(step,100), zeroCount+floor(step/100)}
    					{dir, step} when dir == :L and currPos == 0 and currPos - rem(step,100) < 0 ->
    						{currPos - rem(step,100) + 100, zeroCount+floor(step/100)}
    					{dir, step} when dir == :L and currPos - rem(step,100) < 0 ->
    						{currPos - rem(step,100) + 100, zeroCount+floor(step/100)+1}
    					{dir, step} when dir == :L ->
    						{currPos - rem(step,100), zeroCount+floor(step/100)}
    					_ -> raise "[ ERROR ] unkown instruction: #{inspect(instruct)}"
    				end
    			newZeroCount = if nextPos == 0, do: zeroCount1 + 1, else: zeroCount1
    			checkInstructionPart2(rest, nextPos, newZeroCount)
    		other -> raise "[ ERROR ] unknown instruction: #{inspect(other)}"
    	end
    end
    
    end #module
    
    IO.puts "### PART 1 ###"
    instructions = SafeDial.readInstructions("input/day01Input.txt")
    {finalPos, zeroCount} = SafeDial.evalInstructions(instructions)
    IO.puts "final position:#{finalPos} zero count:#{zeroCount}"
    
    IO.puts "### PART 2 ###"
    instructions = SafeDial.readInstructions("input/day01Input.txt")
    {finalPos, zeroCount} = SafeDial.evalInstructionsPart2(instructions)
    IO.puts "final position:#{finalPos} zero count:#{zeroCount}"
    

    Edit: removed debug output

  • source
  • [–] 3 points 9 months ago* (last edited 9 months ago)

    c

    #include "aoc.h"
    #include <stdio.h>
    #include <string.h>
    
    constexpr usize LINE_BUFSZ = (1 << 3);
    constexpr i32 START = 50;
    constexpr i32 TOP = 100;
    
    static void
    solve(Mode mode) {
      FILE* input = fopen("input", "r");
      c8 line[LINE_BUFSZ] = {};
      u32 zs = 0;
      i32 idx = START;
      while (fgets(line, sizeof(line), input)) {
        line[strcspn(line, "\n")] = 0;
        i32 val = 0;
        sscanf(&line[1], "%d", &val);
        if (mode == MODE_ONE) {
          i32 d = line[0] == 'L' ? -val : val;
          idx = (idx + d) % TOP;
          idx += idx < 0 ? TOP : 0;
          idx == 0 ? zs++ : 0;
        } else {
          for (i32 i = 0; i < val; i++) {
            idx = line[0] == 'L' ? idx - 1 : idx + 1;
            idx = idx == -1 ? TOP - 1 : idx;
            idx = idx == TOP ? 0 : idx;
            idx == 0 ? zs++ : 0;
          }
        }
      }
      fclose(input);
      printf("%u\n", zs);
    }
    
    i32
    main(void) {
      solve(MODE_ONE);
      solve(MODE_TWO);
    }
    
  • source
  • [–] 3 points 9 months ago

    Nim

    That was the rough first day for me. Part 1 was ok. For part 2 I didn't want to go the easy route, so I was trying to find simple formulaic solution, but my answer was always off by some amount. And debugging was hard, because I I was getting the right answer for example input.
    After 40 minutes I wiped everything clean and wrote a bruteforce.

    Later that day I returned and solved this one properly. I had to draw many schemes and consider all the edge cases carefully to come up with code below.

    type
      AOCSolution[T,U] = tuple[part1: T, part2: U]
    
    proc solve(input: string): AOCSolution[int, int] =
      var dial = 50
      for line in input.splitLines():
        let value = parseInt(line[1..^1])
        let sign = if line[0] == 'L': -1 else: 1
        let offset = value mod 100
        result.part2 += value div 100
    
        if dial != 0:
          if sign < 0 and offset >= dial or
             sign > 0 and offset >= (100-dial): inc result.part2
    
        dial = (dial + offset * sign).euclmod(100)
        if dial == 0: inc result.part1
    

    Full solution at Codeberg: solution.nim

  • source
  • [–] 3 points 9 months ago (1 child)

    Haskell

    import Control.Arrow
    import Control.Monad
    import Control.Monad.Writer.Strict
    import Data.Char
    import Data.Functor
    import Data.Functor.Identity
    import Text.ParserCombinators.ReadP
    
    n = 100
    start = 50
    
    parse = fst . last . readP_to_S (endBy rotation (char '\n'))
      where
        rotation = (*) <$> ((char 'L' $> (-1)) <++ (char 'R' $> 1)) <*> (read <$> munch isDigit)
    
    part1 = length . filter (== 0) . fmap (`mod` n) . scanl (+) start
    
    spins :: Int -> Int -> Writer [Int] Int
    spins acc x = do
        when (abs x >= n) . tell . pure $ abs x `div` n -- full loops
        let res = acc + (x `rem` n)
            res' = res `mod` n
    
        when (res /= res') . tell . pure $ 1
    
        return res'
    
    part2 = runIdentity . fmap sum . execWriterT . foldM spins start
    
    main = getContents >>= (print . (part1 &&& part2) . parse)
    
  • source
  • hideshow 1 child comment
  • [–] 2 points 9 months ago

    Rust

    Almost missed, that "the dial starts by pointing at 50".

    View on github

    const N: i32 = 100;
    
    fn parse_line(l: &str) -> (i32, i32) {
        let dir = match l.chars().next().unwrap() {
            'L' => -1,
            'R' => 1,
            _ => panic!(),
        };
        let dist = l[1..].parse::<i32>().unwrap();
        (dir, dist)
    }
    
    fn part1(input: String) {
        let mut pos = 50;
        let mut count0 = 0;
        for l in input.lines() {
            let (dir, dist) = parse_line(l);
            pos = (pos + dir * dist) % N;
            if pos == 0 {
                count0 += 1;
            }
        }
        println!("{count0}");
    }
    
    fn part2(input: String) {
        let mut pos = 50;
        let mut count0 = 0;
        for l in input.lines() {
            let (dir, dist) = parse_line(l);
            if dir == 1 {
                count0 += (pos + dist) / N;
            } else {
                count0 += ((N - pos) % N + dist) / N;
            }
            pos = (pos + dir * dist).rem_euclid(N);
        }
        println!("{count0}");
    }
    
    util::aoc_main!();
    
  • source
  • [–] 2 points 9 months ago

    So, obviously the bot didnt work, thanks Ategon for covering :D

    Here is the dumb solution for pt2, everyone can share in my shame:

    #[test]
        fn test_2025_1_part2() {
            let input = std::fs::read_to_string("input/2025/day_1.txt").unwrap();
            let turns = input
                .lines()
                .map(|line| {
                    let value = line[1..].parse::<i32>().unwrap();
                    if line[0..1] == *"L" {
                        -value
                    } else {
                        value
                    }
                })
                .collect::<Vec<i32>>();
            let mut zero_ctr = 0;
            let mut pos = 50;
            for turn in turns {
                print!("{pos}->");
    
                let mut zero_passes = 0;
                if turn > 0 {
                    for _ in 0..turn {
                        pos += 1;
                        if pos == 100 {
                            pos = 0;
                            zero_passes += 1;
                        }
                    }
                } else {
                    for _ in 0..-turn {
                        pos -= 1;
                        if pos == 0 {
                            zero_passes += 1;
                        }
                        if pos == -1 {
                            pos = 99;
                        }
                    }
                };
                println!("{turn}->{pos}: {zero_passes}");
                zero_ctr += zero_passes;
            }
            println!("zero ctr: {}", zero_ctr);
        }
    
  • source
  • [–] 2 points 9 months ago

    Python

    Solved part 1 in a matter of minutes, spent literal days tinkering with part 2 before resorting to brute force, just to finally get a correct result, which allowed me to fix the more optimised approach. Brute force still turned out surprisingly performant, scaling linearly in time with the rotation distance.

    from itertools import accumulate
    from math import copysign
    from pathlib import Path
    from typing import List
    
    
    def parse_input(input: str) -> List[int]:
        return list(map(lambda s: int(f"{'-' if s[0] == 'L' else ''}{s[1:]}"), input.splitlines()))
    
    
    def part_one(input: str) -> int:
        return len(list(filter(lambda x: x == 0, accumulate(parse_input(input), lambda v, i: (i + v) % 100, initial=50))))
    
    
    def part_two(input: str) -> int:
        v, c = 50, 0
        for i in parse_input(input):
            c += i // (d := int(copysign(100, i)))  # full rotations
            count_underflow = v != 0  # was counted in previous iteration
            v += (i % d)  # remainder
            if count_underflow:
                c += abs(v // 100)  # under-/overflows
            if v == 0:
                c += 1
            v %= 100
        return c
    
    
    if __name__ == "__main__":
        input = Path("_2025/_1/input").read_text("utf-8")
        print(part_one(input))
        print(part_two(input))
    

    Brute force part 2

    def part_two(input: str) -> int:
        from itertools import chain
        return len(list(filter(
            lambda x: x == 0,
            accumulate(chain(*[[int(copysign(1, i))] * abs(i) for i in parse_input(input)]), lambda v, i: (i + v) % 100, initial = 50)
        )))
    
  • source
  • [–] 2 points 9 months ago

    Haskell

    I was late to the part on this one and forgot to post my solution :3

    import Data.List  
    
    readInput = map readMove . lines  
      where  
        readMove (d : ds) =  
          let n = read ds :: Int  
           in case d of  
                'L' -> -n  
                'R' -> n  
    
    part1 = length . filter ((== 0) . (`mod` 100)) . scanl' (+) 50  
    
    part2 = fst . foldl' count (0, 50)  
      where  
        count (z, p) d =  
          let (q, r) = (p + d) `divMod` 100  
              a = if p == 0 && d < 0 then -1 else 0  
              b = if r == 0 && d < 0 then 1 else 0  
           in (z + abs q + a + b, r)  
    
    main = do  
      input <- readInput <$> readFile "input01"  
      print $ part1 input  
      print $ part2 input  
    
  • source
  • [–] 2 points 9 months ago (1 child)

    Uiua

    Today's lesson: Never think that scanning the first 100 lines of input will give you a good understanding of it.

    Part 2 is really messy and could probably be much simpler, but I couldn't get the logic straight in my head otherwise.

    "L68 L30 R48 L5 R60 L55 L1 L99 R14 L82"
    βŠœβ‹•βŠΈβ‰ @\sβˆ§βœβŠ‘β‹…@Β―βŠšβŠΈβŒ•"L"βˆ§βœβŠ‘β‹…@+βŠšβŠΈβŒ•"R"
    P₁ ← ⧻⊚=0\(β—Ώ100+)βŠ‚50
    P ← (
      βŠƒ(-Γ—100Γ—|/+)⌊÷100⌡⟜(⊸±) # Count and remove all over-rotations
      β‰βŠŸβ†˜Β―1⊸(\(β—Ώ100+)βŠ‚50)     # Take all positions and next moves.
      β–½βŠΈβ‰‘(β‰ 0⊒)                # Ignore any starting from zero
      +/+β†₯βŠƒ(<0|>100)≑/+       # Sum the pairs, check for passing zero.
    )
    Pβ‚‚ ← +βŠƒP P₁
    βŠƒ(P₁|Pβ‚‚)
    
  • source
  • hideshow 1 child comment
  • [–] 2 points 9 months ago

    Just had a look at the Uiua Discord, and Part 2 can be simplified a little...

    "L68 L30 R48 L5 R60 L55 L1 L99 R14 L82"
    βŠœβ‹•βŠΈβ‰ @\sβˆ§βœβŠ‘β‹…@Β―βŠšβŠΈβŒ•"L"βˆ§βœβŠ‘β‹…@+βŠšβŠΈβŒ•"R"
    P₁ ← ⧻⊚=0\(β—Ώ100+)βŠ‚50
    Pβ‚‚ ← /+=0β—Ώ100\+β–½βŒ΅βŸœΒ±βŠ‚50
    βŠƒP₁ Pβ‚‚
    

    Sometimes I could just cry.

  • source
  • parent
  • C# ( c sharp )

    using System.Collections;
    using System.Collections.Generic;
    
    namespace ConsoleApp1
    {
        public static class Program
        {
            public static void Part1()
            {
                var lines = File.ReadAllLines("C:\\Users\\aman\\RiderProjects\\ConsoleApp1\\ConsoleApp1\\input.txt");
                var dialReading = 50;
                int result = 0;
                foreach (var line in lines)
                {
                    if (dialReading == 0)
                    {
                        result += 1;
                    }
                    char dir = line[0];
                    int rotation =  int.Parse(line.Substring(1));
                    if (dir == 'R')
                    {
                        dialReading += rotation;
                        dialReading %= 100;
                    }
                    else
                    {
                        int diff = dialReading - rotation;
                        if (diff > 0)
                        {
                            dialReading -= rotation;
                            dialReading %= 100;
                        }
                        else
                        {
                            dialReading = dialReading + 100 - rotation;
                            dialReading %= 100;
                        }
                    }
                }
    
                Console.WriteLine(result);
            }
    
            public static void Part2()
            {
                var lines = File.ReadAllLines("C:\\Users\\aman\\RiderProjects\\ConsoleApp1\\ConsoleApp1\\input.txt");
                var dialReading = 50;
                int result = 0;
                foreach (var line in lines)
                {
                    char dir = line[0];
                    int rotation =  int.Parse(line.Substring(1));
                    if (dir == 'R')
                    {
                        while (rotation > 0)
                        {
                            if (dialReading == 0)
                                result += 1;
                            dialReading += 1;
                            dialReading %= 100;
                            rotation -= 1;
                        }
                    }
                    else
                    {
                        while (rotation > 0)
                        {
                            if (dialReading == 0)
                                result += 1;
                            dialReading -= 1;
                            if ( dialReading < 0)
                                dialReading += 100;
                            dialReading %= 100;
                            rotation -= 1;
                        }
                    }
                }
    
                Console.WriteLine(result);
            }
            public static void Main(string[] args)
            {
                Part1();
                Part2();
            }
        }
    }
    
  • source
  • [–] 1 point 9 months ago

    The struggled with a counting solution for a long time. I submitted with a simple enumerative solution in the end but managed to get it right after some pause time:

    Haskell

    Fast to Run, Stepwise Solution

    {-# LANGUAGE LambdaCase #-}
    {-# LANGUAGE OrPatterns #-}
    module Main (main) where
    
    import Control.Monad ( (<$!>) )
    import qualified Data.List as List
    
    main :: IO ()
    main = do
      rotations <- (fmap parseRotation . init . lines) <$!> getContents
      print $ part1 rotations
      print $ part2 rotations
    
    part2 :: [Either Int Int] -> Int
    part2 rotations = let
    
        foldRotation (position, zeroCount) operation = case operation of
          Left y -> let
            (zeroPasses, y') = y `divMod` 100
            position' = (position - y') `mod` 100
            zeroCount' = zeroPasses + zeroCount + if position <= y' then fromEnum $ position /= 0 else 0
            in (position', zeroCount')
          Right y -> let
            (zeroPasses, y') = y `divMod` 100
            position' = (position + y') `mod` 100
            zeroCount' = zeroPasses + zeroCount + if y' + position >= 100 then 1 else 0
            in (position', zeroCount')
    
      in snd $ List.foldl' foldRotation (50, 0) rotations
    
    part1 :: [Either Int Int] -> Int
    part1 rotations = let
        positions = List.scanl applyRotation 50 rotations
      in List.length . filter (== 0) $ positions
    
    applyRotation :: Int -> Either Int Int -> Int
    applyRotation x = \case
      Left y -> (x - y) `mod` 100
      Right y -> (x + y) `mod` 100
    
    parseRotation :: String -> Either Int Int
    parseRotation = \case
      'R':rest -> Right $ read rest
      'L':rest -> Left $ read rest
      bad -> error $ "invalid rotation operation: " ++ bad
    

    Fast to Code, Exhaustively Enumerating Solution

    -- | Old solution enumerating all the numbers
    
    part2' :: [Either Int Int] -> Int
    part2' rotations = let
      intermediatePositions _ [] = []
      intermediatePositions x (op:ops) = case op of
        Left 0; Right 0 -> intermediatePositions x ops
        Left y -> let x' = pred x `mod` 100 in x' : intermediatePositions x' (Left (pred y) : ops)
        Right y -> let x' = succ x `mod` 100 in x' : intermediatePositions x' (Right (pred y) : ops)
      in List.length . List.filter (== 0) . intermediatePositions 50 $ rotations
    

  • source
  • [–] 1 point 9 months ago* (last edited 9 months ago)
    [–] 1 point 9 months ago

    I agree with strlcpy -- computing was better in the 1980s. Let's try the version of the 1980s where Lisp Machines were going to power the AI future. It can't be any worse than the AI future we've got right now.

    (This is SBCL, not a Lisp Machine emulator, because I'm not that hardcore.)

    (defun parse-line (line)
      (let ((sign (if (eql (char line 0) #\R) 1 -1))
            (number (parse-integer (subseq line 1))))
        (* sign number)))
    
    (defun read-inputs (filename)
      (let ((input-lines (uiop:read-file-lines filename)))
        (mapcar #'parse-line input-lines)))
    
    (defun rotate (pos rotation)
      (mod (+ pos rotation) 100))
    
    (defun main-1 (filename)
      (let ((rotations (read-inputs filename))
            (pos 50))
        (loop for rotation in rotations
              do (setf pos (rotate pos rotation))
              sum (if (= pos 0) 1 0))))
    
    (defun zero-crossings (pos rotation)
      (if (> rotation 0)
          (floor (+ rotation pos) 100)
          (let ((neg-pos (if (zerop pos) pos (- pos 100))))
            (- (ceiling (+ rotation neg-pos) 100)))))
    
    (defun main-2 (filename)
      (let ((rotations (read-inputs filename))
            (pos 50))
        (loop for rotation in rotations
              sum (zero-crossings pos rotation) into crossings
              do (setf pos (rotate pos rotation))
              finally (return crossings))))
    
  • source
  • [–] 1 point 9 months ago* (last edited 9 months ago)

    Rust

    use std::{cmp::Ordering, fs, str::FromStr};
    
    use color_eyre::eyre::{Result, bail};
    
    const DIAL_NUMBERS: isize = 100;
    
    #[derive(Clone, Copy)]
    struct Dial(usize);
    
    impl Dial {
        fn new() -> Self {
            Self(50)
        }
    
        fn rotate(&mut self, rot: isize) -> usize {
            let pass_0s;
    
            let total = rot.wrapping_add_unsigned(self.0);
            match total.cmp(&0) {
                Ordering::Equal => {
                    pass_0s = 1;
                    self.0 = 0;
                }
                Ordering::Less => {
                    // Starting on 0 means we don't cross it
                    let started_0 = if self.0 == 0 { 1 } else { 0 };
                    pass_0s = 1 + (-total / DIAL_NUMBERS) as usize - started_0;
                    self.0 = (self.0 as isize + rot).rem_euclid(DIAL_NUMBERS) as usize;
                }
                Ordering::Greater => {
                    let full_turns = total / DIAL_NUMBERS;
                    pass_0s = full_turns as usize;
                    self.0 = (total - DIAL_NUMBERS * full_turns) as usize;
                }
            };
    
            pass_0s
        }
    
        fn sequence(&mut self, s: &str) -> Result<(usize, usize)> {
            let mut end_0s = 0;
            let mut pass_0s = 0;
    
            for l in s.lines() {
                let num = isize::from_str(&l[1..]).unwrap();
                let dir = l.bytes().next().unwrap();
                pass_0s += match dir {
                    b'L' => self.rotate(-num),
                    b'R' => self.rotate(num),
                    _ => bail!("b{dir} is an invalid rotation direction"),
                };
                if self.0 == 0 {
                    end_0s += 1;
                }
            }
    
            Ok((end_0s, pass_0s))
        }
    }
    
    fn parts(filepath: &str) -> Result<(usize, usize)> {
        let input = fs::read_to_string(filepath)?;
        let mut dial = Dial::new();
        let res = dial.sequence(&input)?;
        Ok(res)
    }
    
    fn main() -> Result<()> {
        color_eyre::install()?;
    
        let (p1, p2) = parts("d01/input.txt")?;
        println!("Part 1: {p1}");
        println!("Part 2: {p2}");
        Ok(())
    }
    

    I lost a lot of time struggling with edge cases in part two since I thought it was wasteful to run rem_euclid() (i.e. modulus) when I'd already got the information to do it more efficiently. While I got there in the end my final code was a bit ugly. This prettier, "unoptimised" version runs in 1.6ms, so I was being an idiot.

  • source