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
}

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)
}

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 .
;

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 
#include 
#include 
#include 

std::vector parse_line()
{
    std::string line;
    std::size_t index;
    int num;
    std::vector 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 d) {
            count++;
        }
    }
    return count;
}

int main()
{
    std::vector time;
    std::vector 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 
#include 
#include 
#include 
#include 
#include 

std::vector parse_line()
{
    std::string line;
    std::size_t index;
    long long num;
    std::vector 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 d) {
            count++;
        }
    }
    return count;
}

int main()
{
    std::vector time;
    std::vector 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;
}

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())

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