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

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
}

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(''))]
    ]);

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;
}

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)

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()
    }
}

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 change of pace from the previous puzzles, more maths and less parsing :::spoiler Python

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())

Language: Python

def race(charge_time: int, total_time: int) -> int:
    return charge_time * (total_time - charge_time)

def main(stream=sys.stdin) -> None:
    times     = [int(t) for t in stream.readline().split(':')[-1].split()]
    distances = [int(d) for d in stream.readline().split(':')[-1].split()]
    product   = 1

    for time, distance in zip(times, distances):
        ways     = [c for c in range(1, time + 1) if race(c, time) > distance]
        product *= len(ways)

    print(product)

def race(charge_time: int, total_time: int):
    return charge_time * (total_time - charge_time)

def main(stream=sys.stdin) -> None:
    time     = int(''.join(stream.readline().split(':')[-1].split()))
    distance = int(''.join(stream.readline().split(':')[-1].split()))
    ways     = [c for c in range(1, time + 1) if race(c, time) > distance]
    print(len(ways))

GitHub Repo

[Language: Lean4]

This one was straightforward, especially since Lean's Floats are 64bits. There is one interesting piece in the solution though, and that's the function that combines two integers, which I wrote because I want to use the same parse function for both parts. This combineNumbers function is interesting, because it needs a proof of termination to make the Lean4 compiler happy. Or, in other words, the compiler needs to be told that if n is larger than 0, n/10 is a strictly smaller integer than n. That proof actually exists in Lean's standard library, but the compiler doesn't find it by itself. Supplying it is as easy as invoking the simp tactic with that proof, and a proof that n is larger than 0.

As with the previous days, I won't post the full source here, just the relevant parts. The full solution is on github, including the main function of the program, that loads the input file and runs the solution.

structure Race where
  timeLimit : Nat
  recordDistance : Nat
  deriving Repr

private def parseLine (header : String) (input : String) : Except String (List Nat) := do
  if not $ input.startsWith header then
    throw s!"Unexpected line header: {header}, {input}"
  let input := input.drop header.length |> String.trim
  let numbers := input.split Char.isWhitespace
    |> List.map String.trim
    |> List.filter (not ∘ String.isEmpty)
  numbers.mapM $ Option.toExcept s!"Failed to parse input line: Not a number {input}" ∘  String.toNat?

def parse (input : String) : Except String (List Race) := do
  let lines := input.splitOn "\n"
    |> List.map String.trim
    |> List.filter (not ∘ String.isEmpty)
  let (times, distances) ← match lines with
    | [times, distances] =>
      let times ← parseLine "Time:" times
      let distances ← parseLine "Distance:" distances
      pure (times, distances)
    | _ => throw "Failed to parse: there should be exactly 2 lines of input"
  if times.length != distances.length then
    throw "Input lines need to have the same number of, well, numbers."
  let pairs := times.zip distances
  if pairs = [] then
    throw "Input does not have at least one race."
  return pairs.map $ uncurry Race.mk

-- okay, part 1 is a quadratic equation. Simple as can be
-- s = v * tMoving
-- s = tPressed * (tLimit - tPressed)
-- (tPressed - tLimit) * tPressed + s = 0
-- tPressed² - tPressed * tLimit + s = 0
-- tPressed := tLimit / 2 ± √(tLimit² / 4 - s)
-- beware: We need to _beat_ the record, so s here is the record + 1

-- Inclusive! This is the smallest number that can win, and the largest number that can win
private def Race.timeRangeToWin (input : Race) : (Nat × Nat) :=
  let tLimit  := input.timeLimit.toFloat
  let sRecord := input.recordDistance.toFloat
  let tlimitHalf := 0.5 * tLimit
  let theRoot := (tlimitHalf^2 - sRecord - 1.0).sqrt
  let lowerBound := tlimitHalf - theRoot
  let upperBound := tlimitHalf + theRoot
  let lowerBound := lowerBound.ceil.toUInt64.toNat
  let upperBound := upperBound.floor.toUInt64.toNat
  (lowerBound,upperBound)

def part1 (input : List Race) : Nat :=
  let limits := input.map Race.timeRangeToWin
  let counts := limits.map $ λ p ↦ p.snd - p.fst + 1 -- inclusive range
  counts.foldl (· * ·) 1

-- part2 is the same thing, but here we need to be careful.
-- namely, careful about the precision of Float. Which luckily is enough, as confirmed by pen&paper
-- but _barely_ enough.
-- If Lean's Float were an actual C float and not a C double, this would not work.

-- we need to concatenate the numbers again (because I don't want to make a separate parse for part2)
private def combineNumbers (left : Nat) (right : Nat) : Nat :=
  let rec countDigits := λ (s : Nat) (n : Nat) ↦
    if p : n > 0 then
      have : n > n / 10 := by simp[p, Nat.div_lt_self]
      countDigits (s+1) (n/10)
    else
      s
  let d := if right = 0 then 1 else countDigits 0 right
  left * (10^d) + right

def part2 (input : List Race) : Nat :=
  let timeLimits := input.map Race.timeLimit
  let timeLimit := timeLimits.foldl combineNumbers 0
  let records := input.map Race.recordDistance
  let record := records.foldl combineNumbers 0
  let limits := Race.timeRangeToWin $ {timeLimit := timeLimit, recordDistance := record}
  limits.snd - limits.fst + 1 -- inclusive range

open DayPart
instance : Parse ⟨6, by simp⟩ (ι := List Race) where
  parse := parse

instance : Part ⟨6, _⟩ Parts.One (ι := List Race) (ρ := Nat) where
  run := some ∘ part1

instance : Part ⟨6, _⟩ Parts.Two (ι := List Race) (ρ := Nat) where
  run := some ∘ part2

Factor on github (with comments and imports):

I didn't use any math smarts.

: input>data ( -- races )
  "vocab:aoc-2023/day06/input.txt" utf8 file-lines     
  [ ": " split harvest rest [ string>number ] map ] map
  first2 zip                                           
;

: go ( press-ms total-time -- distance )
  over - *
;

: beats-record? ( press-ms race -- ? )
  [ first go ] [ last ] bi >
;

: ways-to-beat ( race -- n )
  dup first [1..b)          
  [                         
    over beats-record?      
  ] map [ ] count nip       
;

: part1 ( -- )
  input>data [ ways-to-beat ] map-product .
;

: input>big-race ( -- race )
  "vocab:aoc-2023/day06/input.txt" utf8 file-lines             
  [ ":" split1 nip " " without string>number ] map
;

: part2 ( -- )
  input>big-race ways-to-beat .
;

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 }

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)
}

[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)

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)

That was so much better than yesterday. Went with algebra but looks like brute force would have worked.

import re
import argparse
import math

# i feel doing this with equations is probably the
# "fast" way.

# we can re-arange stuff so we only need to find the point
# the line crosses the 0 line

# distance is speed * (time less time holding the button (which is equal to speed)):
# -> d = v * (t - v)
# -> v^2 -vt +d = 0

# -> y=0 @ v = t +- sqrt( t^2 - 4d) / 2

def get_cross_points(time:int, distance:int) -> list | None:
    pre_root = time**2 - (4 * distance)
    if pre_root < 0:
        # no solutions
        return None
    if pre_root == 0:
        # one solution
        return [(float(time)/2)]
    sqroot = math.sqrt(pre_root)
    v1 = (float(time) + sqroot)/2
    v2 = (float(time) - sqroot)/2
    return [v1,v2]

def float_pair_to_int_pair(a:float,b:float):
    # if floats are equal to int value, then we need to add one to value
    # as we are looking for values above 0 point
    
    if a > b:
        # lower a and up b
        if a == int(a):
            a -= 1
        if b == int(b):
            b += 1

        return [math.floor(a),math.ceil(b)]
    if a < b:
        if a == int(a):
            a += 1
        if b == int(b):
            b -= 1
        return [math.floor(b),math.ceil(a)]

def main(line_list: list):
    time_section,distance_section = line_list
    if (args.part == 1):
        time_list = filter(None , re.split(' +',time_section.split(':')[1]))
        distance_list = filter(None ,  re.split(' +',distance_section.split(':')[1]))
        games = list(zip(time_list,distance_list))
    if (args.part == 2):
        games = [ [time_section.replace(' ','').split(':')[1],distance_section.replace(' ','').split(':')[1]] ]
    print (games)
    total = 1
    for t,d in games:
        cross = get_cross_points(int(t),int(d))
        cross_int = float_pair_to_int_pair(*cross)
        print (cross_int)
        total *= cross_int[0] - cross_int[1] +1
    
    print(f"total: {total}")

if __name__ == "__main__":
    parser = argparse.ArgumentParser(description="day 6 solver")
    parser.add_argument("-input",type=str)
    parser.add_argument("-part",type=int)
    args = parser.parse_args()
    filename = args.input
    if filename == None:
        parser.print_help()
        exit(1)
    file = open(filename,'r')
    main([line.rstrip('\n') for line in file.readlines()])
    file.close()

Today's problems felt really refreshing after yesterday.

Solution in Rust 🦀

View formatted code on GitLab

use std::{
    collections::HashSet,
    env, fs,
    io::{self, BufRead, BufReader, Read},
};

fn main() -> io::Result<()> {
    let args: Vec = env::args().collect();
    let filename = &args[1];
    let file1 = fs::File::open(filename)?;
    let file2 = fs::File::open(filename)?;
    let reader1 = BufReader::new(file1);
    let reader2 = BufReader::new(file2);

    println!("Part one: {}", process_part_one(reader1));
    println!("Part two: {}", process_part_two(reader2));
    Ok(())
}

fn parse_data(reader: BufReader) -> Vec> {
    let lines = reader.lines().flatten();
    let data: Vec<_> = lines
        .map(|line| {
            line.split(':')
                .last()
                .expect("text after colon")
                .split_whitespace()
                .map(|s| s.parse::().expect("numbers"))
                .collect::>()
        })
        .collect();
    data
}

fn calculate_ways_to_win(time: u64, dist: u64) -> HashSet {
    let mut wins = HashSet

:::

Golang

Pretty straightforward. The only optimization I did is that the pairs are symmetric (3ms hold and 4ms travel is the same as 4ms hold and 3ms travel).

file, _ := os.Open("input.txt")
defer file.Close()
scanner := bufio.NewScanner(file)

scanner.Scan()
times := strings.Fields(strings.Split(scanner.Text(), ":")[1])
scanner.Scan()
distances := strings.Fields(strings.Split(scanner.Text(), ":")[1])
n := len(times)
countProduct := 1

for i := 0; i < n; i++ {
	t, _ := strconv.Atoi(times[i])
	d, _ := strconv.Atoi(distances[i])
	count := 0
	for j := 0; j <= t/2; j++ {
		if j*(t-j) > d {
			if t%2 == 0 && j == t/2 {
				count++
			} else {
				count += 2
			}
		}
	}

	countProduct *= count
}

fmt.Println(countProduct)

file, _ := os.Open("input.txt")
defer file.Close()
scanner := bufio.NewScanner(file)

scanner.Scan()
time := strings.ReplaceAll(strings.Split(scanner.Text(), ":")[1], " ", "")
scanner.Scan()
distance := strings.ReplaceAll(strings.Split(scanner.Text(), ":")[1], " ", "")

t, _ := strconv.Atoi(time)
d, _ := strconv.Atoi(distance)
count := 0

for j := 0; j <= t/2; j++ {
	if j*(t-j) > d {
		if t%2 == 0 && j == t/2 {
			count++
		} else {
			count += 2
		}
	}
}

fmt.Println(count)

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

C++

Yesterday, I decided to code in Tcl. That program is still running, i will go back to the day 5 post once it finishes :)

Today was super simple. My first attempt worked in both cases, where the hardest part was really switching my ints to long longs. Part 1 worked on first compile and part 2 I had to compile twice after I realized the data type needs. Still, that change was made by search and replace.

I guess today was meant to be a real time race to get first answer? This is like day 1 stuff! Still, I have kids and a job so I did not get to stay up until the problem was posted.

I used C++ because I thought something intense may be coming on the part 2 problem, and I was burned yesterday. It looks like I spent another fast language on nothing! I think I'll keep zig in the hole for the next number cruncher.

Oh, and yes my TCL program is still running...

My solutions can be found here:

• https://github.com/pngwen/advent-2023/blob/main/day-6a.cpp

// File: day-6a.cpp
// Purpose: Solution to part of day 6 of advent of code in C++
//          https://adventofcode.com/2023/day/6
// Author: Robert Lowe
// Date: 6 December 2023
#include <iostream>
#include <vector>
#include <string>
#include <sstream>

std::vector<int> parse_line()
{
    std::string line;
    std::size_t index;
    int num;
    std::vector<int> result;
    
    // set up the stream
    std::getline(std::cin, line);
    index = line.find(':');
    std::istringstream is(line.substr(index+1));

    while(is>>num) {
        result.push_back(num);
    }

    return result;
}

int count_wins(int t, int d) 
{
    int count=0;
    for(int i=1; i<t; i++) {
        if(t*i-i*i > d) {
            count++;
        }
    }
    return count;
}

int main()
{
    std::vector<int> time;
    std::vector<int> dist;
    int product=1;

    // get the times and distances
    time = parse_line();
    dist = parse_line();

    // count the total number of wins
    for(auto titr=time.begin(), ditr=dist.begin(); titr!=time.end(); titr++, ditr++) {
        product *= count_wins(*titr, *ditr);
    }

    std::cout << product << std::endl;
}

• https://github.com/pngwen/advent-2023/blob/main/day-6b.cpp

// File: day-6b.cpp
// Purpose: Solution to part 2 of day 6 of advent of code in C++
//          https://adventofcode.com/2023/day/6
// Author: Robert Lowe
// Date: 6 December 2023
#include <iostream>
#include <vector>
#include <string>
#include <sstream>
#include <algorithm>
#include <cctype>

std::vector<long long> parse_line()
{
    std::string line;
    std::size_t index;
    long long num;
    std::vector<long long> result;
    
    // set up the stream
    std::getline(std::cin, line);
    line.erase(std::remove_if(line.begin(), line.end(), isspace), line.end());
    index = line.find(':');
    std::istringstream is(line.substr(index+1));

    while(is>>num) {
        result.push_back(num);
    }

    return result;
}

long long count_wins(long long t, long long d) 
{
    long long count=0;
    for(long long i=1; i<t; i++) {
        if(t*i-i*i > d) {
            count++;
        }
    }
    return count;
}

int main()
{
    std::vector<long long> time;
    std::vector<long long> dist;
    long long product=1;

    // get the times and distances
    time = parse_line();
    dist = parse_line();

    // count the total number of wins
    for(auto titr=time.begin(), ditr=dist.begin(); titr!=time.end(); titr++, ditr++) {
        product *= count_wins(*titr, *ditr);
    }

    std::cout << product << std::endl;
}

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

    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)

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

My solutions, as always, in C: https://git.sr.ht/~aidenisik/aoc23/tree/master/item/day6

It's nice that the bruteforce method doesn't take HOURS for this one. My day 5 bruteforce is still running :(