Today was easy enough that I felt confident enough to hammer out a solution in Uiua, so read and enjoy (or try it out live):

{"Time:      7  15   30"
 "Distance:  9  40  200"}
StoInt ← /(+ ×10) ▽×⊃(≥0)(≤9). -@0
Count ← (
  ⊙⊢√-×4:×.⍘⊟.           # Determinant, and time
  +1-⊃(+1↥0⌊÷2-)(-1⌈÷2+) # Diff of sanitised roots
)
≡(↘1⊐⊜∘≠@\s.)
⊃(/×≡Count⍉∵StoInt)(Count⍉≡(StoInt⊐/⊂))

Rust

I went with solving the quadratic equation, so part 2 was just a trivial change in parsing. It was a bit janky to find the integer that is strictly larger than a floating point number, but it all worked out.

Nim

Hey, waitaminute, this isn’t a programming puzzle. This is algebra homework!

Part 2 only required a trivial change to the parsing, the rest of the code still worked. I kept the data as singleton arrays to keep it compatible.

• Day 6, part 1
• Day 6, part 2

Haskell

This problem has a nice closed form solution, but brute force also works.

(My keyboard broke during part two. Yet another day off the bottom of the leaderboard…)

import Control.Monad
import Data.Bifunctor
import Data.List

readInput :: String -> [(Int, Int)]
readInput = map (\[t, d] -> (read t, read d)) . tail . transpose . map words . lines

-- Quadratic formula
wins :: (Int, Int) -> Int
wins (t, d) =
  let c = fromIntegral t / 2 :: Double
      h = sqrt (fromIntegral $ t * t - 4 * d) / 2
   in ceiling (c + h) - floor (c - h) - 1

main = do
  input <- readInput <$> readFile "input06"
  print $ product . map wins $ input
  print $ wins . join bimap (read . concatMap show) . unzip $ input

Dart Solution

I decided to use the quadratic formula to solve part 1 which slowed me down while I struggled to remember how it went, but meant that part 2 was a one line change.

This year really is a roller coaster…

int countGoodDistances(int time, int targetDistance) {
  var det = sqrt(time * time - 4 * targetDistance);
  return (((time + det) / 2).ceil() - 1) -
      (max(((time - det) / 2).floor(), 0) + 1) +
      1;
}

solve(List> data, [param]) {
  var distances = data.first
      .indices()
      .map((ix) => countGoodDistances(data[0][ix], data[1][ix]));
  return distances.reduce((s, t) => s * t);
}

getNums(l) => l.split(RegExp(r'\s+')).skip(1);

part1(List lines) =>
    solve([for (var l in lines) getNums(l).map(int.parse).toList()]);

part2(List lines) => solve([
      for (var l in lines) [int.parse(getNums(l).join(''))]
    ]);

Rust

Feedback welcome! Feel like I’m getting the hand of Rust more and more.

use regex::Regex;
pub fn part_1(input: &str) {
    let lines: Vec<&str> = input.lines().collect();
    let time_data = number_string_to_vec(lines[0]);
    let distance_data = number_string_to_vec(lines[1]);

    // Zip time and distance into a single iterator
    let data_iterator = time_data.iter().zip(distance_data.iter());

    let mut total_possible_wins = 1;
    for (time, dist_req) in data_iterator {
        total_possible_wins *= calc_possible_wins(*time, *dist_req)
    }
    println!("part possible wins: {:?}", total_possible_wins);
}

pub fn part_2(input: &str) {
    let lines: Vec<&str> = input.lines().collect();
    let time_data = number_string_to_vec(&lines[0].replace(" ", ""));
    let distance_data = number_string_to_vec(&lines[1].replace(" ", ""));

    let total_possible_wins = calc_possible_wins(time_data[0], distance_data[0]);
    println!("part 2 possible wins: {:?}", total_possible_wins);
}

pub fn calc_possible_wins(time: u64, dist_req: u64) -> u64 {
    let mut ways_to_win: u64 = 0;

    // Second half is a mirror of the first half, so only calculate first part
    for push_time in 1..=time / 2 {
        // If a push_time crosses threshold the following ones will too so break loop
        if push_time * (time - push_time) > dist_req {
            // There are (time+1) options (including 0).
            // Subtract twice the minimum required push time, also removing the longest push times
            ways_to_win += time + 1 - 2 * push_time;
            break;
        }
    }
    ways_to_win
}

fn number_string_to_vec(input: &str) -> Vec {
    let regex_number = Regex::new(r"\d+").unwrap();
    let numbers: Vec = regex_number
        .find_iter(input)
        .filter_map(|m| m.as_str().parse().ok())
        .collect();
    numbers
}

Raku

I spent a lot more time than necessary optimizing the count-ways-to-beat function, but I’m happy with the result. This is my first time using the | operator to flatten a list into function arguments.

edit: unfortunately, the lemmy web page is unable to properly display the source code in a code block. It doesn’t display text enclosed in pointy brackets <>, perhaps it looks too much like HTML. View code on github.

use v6;

sub MAIN($input) {
    my $file = open $input;

    grammar Records {
        token TOP {  "\n"  "\n"* }
        token times { "Time:" \s* +%\s+ }
        token distances { "Distance:" \s* +%\s+ }
        token num { \d+ }
    }

    my $records = Records.parse($file.slurp);

    my $part-one-solution = 1;
    for $records».Int Z $records».Int -> $record {
        $part-one-solution *= count-ways-to-beat(|$record);
    }
    say "part 1: $part-one-solution";

    my $kerned-time = $records.join.Int;
    my $kerned-distance = $records.join.Int;
    my $part-two-solution = count-ways-to-beat($kerned-time, $kerned-distance);
    say "part 2: $part-two-solution";
}

sub count-ways-to-beat($time, $record-distance) {
    # time = button + go
    # distance = go * button
    # 0 = go^2 - time * go + distance
    # go = (time +/- sqrt(time**2 - 4*distance))/2

    # don't think too hard:
    # if odd t then t/2 = x.5,
    #   so sqrt(t**2-4*d)/2 = 2.3 => result = 4
    #   and sqrt(t**2-4*d)/2 = 2.5 => result = 6
    #   therefore result = 2 * (sqrt(t**2-4*d)/2 + 1/2).floor
    # even t then t/2 = x.0
    #   so sqrt(t^2-4*d)/2 = 2.x => result = 4 + 1(shared) = 5
    #   therefore result = 2 * (sqrt(t^2-4*d)/2).floor + 1
    # therefore result = 2 * ((sqrt(t**2-4*d)+t%2)/2).floor + 1 - t%2
    # Note: sqrt produces a Num, so perhaps the result could be off by 1 or 2,
    #       but it solved my AoC inputs correctly 😃.

    my $required-distance = $record-distance + 1;
    return 2 * ((sqrt($time**2 - 4*$required-distance) + $time%2)/2).floor + 1 - $time%2;
}

A nice change of pace from the previous puzzles, more maths and less parsing

import math
import re


def create_table(filename: str) -> list[tuple[int, int]]:
    with open('day6.txt', 'r', encoding='utf-8') as file:
        times: list[str] = re.findall(r'\d+', file.readline())
        distances: list[str] = re.findall(r'\d+', file.readline())
    table: list[tuple[int, int]] = []
    for t, d in zip(times, distances):
        table.append((int(t), int(d)))
    return table


def get_possible_times_num(table_entry: tuple[int, int]) -> int:
    t, d = table_entry
    l_border: int = math.ceil(0.5 * (t - math.sqrt(t**2 -4 * d)) + 0.0000000000001)  # Add small num to ensure you round up on whole numbers
    r_border: int = math.floor(0.5*(math.sqrt(t**2 - 4 * d) + t) - 0.0000000000001)  # Subtract small num to ensure you round down on whole numbers
    return r_border - l_border + 1


def puzzle1() -> int:
    table: list[tuple[int, int]] = create_table('day6.txt')
    possibilities: int = 1
    for e in table:
        possibilities *= get_possible_times_num(e)
    return possibilities


def create_table_2(filename: str) -> tuple[int, int]:
    with open('day6.txt', 'r', encoding='utf-8') as file:
        t: str = re.search(r'\d+', file.readline().replace(' ', '')).group(0)
        d: str = re.search(r'\d+', file.readline().replace(' ', '')).group(0)
    return int(t), int(d)


def puzzle2() -> int:
    t, d = create_table_2('day6.txt')
    return get_possible_times_num((t, d))


if __name__ == '__main__':
    print(puzzle1())
    print(puzzle2())

A nice simple one today. And only a half second delay for part two instead of half an hour. What a treat. I could probably have nicer input parsing, but that seems to be the theme this year, so that will become a big focus of my next round through these I’m guessing. The algorithm here to get the winning possibilities could also probably be improved upon by figuring out what the number of seconds for the current record is, and only looping from there until hitting a number that doesn’t win, as opposed to brute-forcing the whole loop.

https://github.com/capitalpb/advent_of_code_2023/blob/main/src/solvers/day06.rs

#[derive(Debug)]
struct Race {
    time: u64,
    distance: u64,
}

impl Race {
    fn possible_ways_to_win(&self) -> usize {
        (0..=self.time)
            .filter(|time| time * (self.time - time) > self.distance)
            .count()
    }
}

pub struct Day06;

impl Solver for Day06 {
    fn star_one(&self, input: &str) -> String {
        let mut race_data = input
            .lines()
            .map(|line| {
                line.split_once(':')
                    .unwrap()
                    .1
                    .split_ascii_whitespace()
                    .filter_map(|number| number.parse::().ok())
                    .collect::>()
            })
            .collect::>();

        let times = race_data.pop().unwrap();
        let distances = race_data.pop().unwrap();

        let races = distances
            .into_iter()
            .zip(times)
            .map(|(time, distance)| Race { time, distance })
            .collect::>();

        races
            .iter()
            .map(|race| race.possible_ways_to_win())
            .fold(1, |acc, count| acc * count)
            .to_string()
    }

    fn star_two(&self, input: &str) -> String {
        let race_data = input
            .lines()
            .map(|line| {
                line.split_once(':')
                    .unwrap()
                    .1
                    .replace(" ", "")
                    .parse::()
                    .unwrap()
            })
            .collect::>();

        let race = Race {
            time: race_data[0],
            distance: race_data[1],
        };

        race.possible_ways_to_win().to_string()
    }
}

C

Brute forced it, runs in 60 ms or so. Only shortcut is quitting the loop when the distance drops below the record. I didn’t bother with the closed form solution here because a) it ran so fast and b) I was concerned about floats, rounding and off-by-one errors. Will probably implement it later!

GitHub link

Edit: implemented the closed form solution. Feels dirty copying a formula without really understanding it…

GitHub link (closed form)

Scala3

// math.floor(i) == i if i.isWhole, but we want i-1
def hardFloor(d: Double): Long = (math.floor(math.nextAfter(d, Double.NegativeInfinity))).toLong
def hardCeil(d: Double): Long = (math.ceil(math.nextAfter(d, Double.PositiveInfinity))).toLong

def wins(t: Long, d: Long): Long =
    val det = math.sqrt(t*t/4.0 - d)
    val high = hardFloor(t/2.0 + det)
    val low = hardCeil(t/2.0 - det)
    (low to high).size

def task1(a: List[String]): Long = 
    def readLongs(s: String) = s.split(raw"\s+").drop(1).map(_.toLong)
    a match
        case List(s"Time: $time", s"Distance: $dist") => readLongs(time).zip(readLongs(dist)).map(wins).product
        case _ => 0L

def task2(a: List[String]): Long =
    def readLong(s: String) = s.replaceAll(raw"\s+", "").toLong
    a match
        case List(s"Time: $time", s"Distance: $dist") => wins(readLong(time), readLong(dist))
        case _ => 0L

Nim

Today’s puzzle was too easy. I solved it with bruteforce in 20 minutes, but that’s boring. So here’s the optimized solution with quadratic formula.

Total runtime: 0.008 ms
Puzzle rating: Too Easy 5/10
Code: day_06/solution.nim

Well, this one ended up being a really nice and terse solution when done naïvely, here with a trimmed snippet from my simple solution;

puts "Part 1:", @races.map do |race|
  (0..race[:time]).count { |press_dur| press_dur * (race[:time] - press_dur) > race[:distance] }
end.inject(:*)

full_race_time = @races.map { |r| r[:time].to_s }.join.to_i
full_race_dist = @races.map { |r| r[:distance].to_s }.join.to_i

puts "Part 2:", (0..full_race_time).count { |press_dur| press_dur * (full_race_time - press_dur) > full_race_dist }

[JavaScript] Relatively easy one today

Paste

function part1(input) {
  const split = input.split("\n");
  const times = split[0].match(/\d+/g).map((x) => parseInt(x));
  const distances = split[1].match(/\d+/g).map((x) => parseInt(x));

  let sum = 0;

  for (let i = 0; i < times.length; i++) {
    const time = times[i];
    const recordDistance = distances[i];

    let count = 0;

    for (let j = 0; j < time; j++) {
      const timePressed = j;
      const remainingTime = time - j;

      const travelledDistance = timePressed * remainingTime;

      if (travelledDistance > recordDistance) {
        count++;
      }
    }

    if (sum == 0) {
      sum = count;
    } else {
      sum = sum * count;
    }
  }

  return sum;
}

function part2(input) {
  const split = input.split("\n");
  const time = parseInt(split[0].split(":")[1].replace(/\s/g, ""));
  const recordDistance = parseInt(split[1].split(":")[1].replace(/\s/g, ""));

  let count = 0;

  for (let j = 0; j < time; j++) {
    const timePressed = j;
    const remainingTime = time - j;

    const travelledDistance = timePressed * remainingTime;

    if (travelledDistance > recordDistance) {
      count++;
    }
  }

  return count;
}

Was a bit late with posting the solution thread and solving this since I ended up napping until 2am, if anyone notices theres no solution thread and its after the leaderboard has been filled (can check from the stats page if 100 people are done) feel free to start one up (I just copy paste the text in each of them)

Rust:

For part 2, all I did was edit the input and change i32 to i64.

use std::fs;
use std::path::PathBuf;

use clap::Parser;

#[derive(Parser)]
#[command(author, version, about, long_about = None)]
struct Cli {
    input_file: PathBuf,
}

fn main() {
    // Parse CLI arguments
    let cli = Cli::parse();

    // Read file
    let input_text = fs::read_to_string(&cli.input_file)
        .expect(format!("File \"{}\" not found", cli.input_file.display()).as_str());

    let input_lines: Vec<&str> = input_text.lines().collect();

    let times: Vec = input_lines[0]
        .split_ascii_whitespace()
        .skip(1)
        .map(|s| s.parse().unwrap())
        .collect();

    let distances: Vec = input_lines[1]
        .split_ascii_whitespace()
        .skip(1)
        .map(|s| s.parse().unwrap())
        .collect();

    println!("{:?}", times);
    println!("{:?}", distances);

    let mut product: i64 = 1;

    for (time, distance) in times.iter().zip(distances) {
        let mut n = 0;
        for b in 0..*time {
            if time * b - b * b > distance {
                n += 1;
            }
        }

        product *= n;
    }

    println!("{}", product)
}