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

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
}

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

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())

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

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()
    }
}

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 }

Python

Questions and feedback welcome!

import re

from functools import reduce
from operator import mul

from .solver import Solver

def upper_bound(start: int, stop: int, predicate: Callable[[int], bool]) -> int:
  """Find the smallest integer in [start, stop) for which the predicate is
   false, or stop if the predicate is always true.

   The predicate must be monotonic, i.e. predicate(x + 1) implies predicate(x).
   """
  assert start < stop
  if not predicate(start):
    return start
  if predicate(stop - 1):
    return stop
  while start + 1 < stop:
    mid = (start + stop) // 2
    if predicate(mid):
      start = mid
    else:
      stop = mid
  return stop

def travel_distance(hold: int, limit: int) -> int:
  dist = hold * (limit - hold)
  return dist

def ways_to_win(time: int, record: int) -> int:
  definitely_winning_hold = time // 2
  assert travel_distance(definitely_winning_hold, time) > record
  minimum_hold_to_win = upper_bound(
      1, definitely_winning_hold, lambda hold: travel_distance(hold, time) <= record)
  minimum_hold_to_lose = upper_bound(
      definitely_winning_hold, time, lambda hold: travel_distance(hold, time) > record)
  return minimum_hold_to_lose - minimum_hold_to_win

class Day06(Solver):

  def __init__(self):
    super().__init__(6)
    self.times = []
    self.distances = []

  def presolve(self, input: str):
    times, distances = input.rstrip().split('\n')
    self.times = [int(time) for time in re.split(r'\s+', times)[1:]]
    self.distances = [int(distance) for distance in re.split(r'\s+', distances)[1:]]

  def solve_first_star(self):
    ways= []
    for time, record in zip(self.times, self.distances):
      ways.append(ways_to_win(time, record))
    return reduce(mul, ways)

  def solve_second_star(self):
    time = int(''.join(map(str, self.times)))
    distance = int(''.join(map(str, self.distances)))
    return ways_to_win(time, distance)

Crystal

# part 1
times = input[0][5..].split.map &.to_i
dists = input[1][9..].split.map &.to_i

prod = 1
times.each_with_index do |time, i|
	start, last = find_poss(time, dists[i])
	prod *= last - start + 1
end
puts prod

# part 2
time = input[0][5..].chars.reject!(' ').join.to_i64
dist = input[1][9..].chars.reject!(' ').join.to_i64

start, last = find_poss(time, dist)
puts last - start + 1

def find_poss(time, dist)
	start = 0
	last  = 0
	(1...time).each do |acc_time|
		if (time-acc_time)*acc_time > dist
			start = acc_time
			break
	end     end
	(1...time).reverse_each do |acc_time|
		if (time-acc_time)*acc_time > dist
			last = acc_time
			break
	end     end
	{start, last}
end

Not sure if it’s the most optimal, but I figured it’s probably quicker to calculate the first point when you start winning, and then reverse it to get the last point when you’ll last win. Subtracting the two to get the total number of ways to win.

Takes about 3 seconds to run on the real input

class Race:
    def __init__(self, time, distance):
        self.time = time
        self.distance = distance

    def get_win(self, start, stop, step):
        for i in range(start, stop, step):
            if (self.time - i) * i > self.distance:
                return i

    def get_winners(self):
        return (
            self.get_win(0, self.time, 1),
            self.get_win(self.time, 0, -1),
        )

race = Race(71530, 940200)
winners = race.get_winners()
print(winners[1] - winners[0] + 1)