Welcome everyone to the 2023 advent of code! Thank you all for stopping by and participating in it in programming.dev whether youre new to the event or doing it again.
This is an unofficial community for the event as no official spot exists on lemmy but ill be running it as best I can with Sigmatics modding as well. Ill be running a solution megathread every day where you can share solutions with other participants to compare your answers and to see the things other people come up with
Day 1: Trebuchet?!
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)
- Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ or pastebin (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)
FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
🔒This post will be unlocked when there is a decent amount of submissions on the leaderboard to avoid cheating for top spots
🔓 Edit: Post has been unlocked after 6 minutes
Java
My take on a modern Java solution (parts 1 & 2).
spoiler
package thtroyer.day1; import java.util.*; import java.util.stream.IntStream; import java.util.stream.Stream; public class Day1 { record Match(int index, String name, int value) { } Map numbers = Map.of( "one", 1, "two", 2, "three", 3, "four", 4, "five", 5, "six", 6, "seven", 7, "eight", 8, "nine", 9); /** * Takes in all lines, returns summed answer */ public int getCalibrationValue(String... lines) { return Arrays.stream(lines) .map(this::getCalibrationValue) .map(Integer::parseInt) .reduce(0, Integer::sum); } /** * Takes a single line and returns the value for that line, * which is the first and last number (numerical or text). */ protected String getCalibrationValue(String line) { var matches = Stream.concat( findAllNumberStrings(line).stream(), findAllNumerics(line).stream() ).sorted(Comparator.comparingInt(Match::index)) .toList(); return "" + matches.getFirst().value() + matches.getLast().value(); } /** * Find all the strings of written numbers (e.g. "one") * * @return List of Matches */ private List findAllNumberStrings(String line) { return IntStream.range(0, line.length()) .boxed() .map(i -> findAMatchAtIndex(line, i)) .filter(Optional::isPresent) .map(Optional::get) .sorted(Comparator.comparingInt(Match::index)) .toList(); } private Optional findAMatchAtIndex(String line, int index) { return numbers.entrySet().stream() .filter(n -> line.indexOf(n.getKey(), index) == index) .map(n -> new Match(index, n.getKey(), n.getValue())) .findAny(); } /** * Find all the strings of digits (e.g. "1") * * @return List of Matches */ private List findAllNumerics(String line) { return IntStream.range(0, line.length()) .boxed() .filter(i -> Character.isDigit(line.charAt(i))) .map(i -> new Match(i, null, Integer.parseInt(line.substring(i, i + 1)))) .toList(); } public static void main(String[] args) { System.out.println(new Day1().getCalibrationValue(args)); } }
I’m a bit late to the party. I forgot about this.
Anyways, my (lazy) C solutions: https://git.sr.ht/~aidenisik/aoc23/tree/master/item/day1
Uiua solution
I may add solutions in Uiua depending on how easy I find them, so here’s today’s (also available to run online):
Inp ← {"four82nine74" "hlpqrdh3" "7qt" "12" "1one"} # if needle is longer than haystack, return zeros SafeFind ← ((⌕|-.;)< ∩⧻ , ,) FindDigits ← (× +1⇡9 ⊠(□SafeFind∩⊔) : Inp) "123456789" ⊜□ ≠@\s . "one two three four five six seven eight nine" ∩FindDigits BuildNum ← (/+∵(/+⊂⊃(×10↙ 1)(↙ 1⇌) ▽≠0.⊔) /↥) ∩BuildNum+,
or stripping away all the fluff:
Inp ← {"four82nine74" "hlpqrdh3" "7qt" "12" "1one"} ⊜□ ≠@\s."one two three four five six seven eight nine" "123456789" ∩(×+1⇡9⊠(□(⌕|-.;)<⊙:∩(⧻.⊔)):Inp) ∩(/+∵(/+⊂⊃(×10↙1)(↙1⇌)▽≠0.⊔)/↥)+,
I wanted to see if it was possible to do part 1 in a single line of Python:
print(sum([(([int(i) for i in line if i.isdigit()][0]) * 10 + [int(i) for i in line if i.isdigit()][-1]) for line in open("input.txt")]))
Python
Questions and feedback welcome!
import re from .solver import Solver class Day01(Solver): def __init__(self): super().__init__(1) self.lines = [] def presolve(self, input: str): self.lines = input.rstrip().split('\n') def solve_first_star(self): numbers = [] for line in self.lines: digits = [ch for ch in line if ch.isdigit()] numbers.append(int(digits[0] + digits[-1])) return sum(numbers) def solve_second_star(self): numbers = [] digit_map = { "one": 1, "two": 2, "three": 3, "four": 4, "five": 5, "six": 6, "seven": 7, "eight": 8, "nine": 9, "zero": 0, } for i in range(10): digit_map[str(i)] = i for line in self.lines: digits = [digit_map[digit] for digit in re.findall( "(?=(one|two|three|four|five|six|seven|eight|nine|[0-9]))", line)] numbers.append(digits[0]*10 + digits[-1]) return sum(numbers)
[Language: C#]
This isn’t the most performant or elegant, it’s the first one that worked. I have 3 kids and a full time job. If I get through any of these, it’ll be first pass through and first try that gets the correct answer.
Part 1 was very easy, just iterated the string checking if the char was a digit. Ditto for the last, by reversing the string. Part 2 was also not super hard, I settled on re-using the iterative approach, checking each string lookup value first (on a substring of the current char), and if the current char isn’t the start of a word, then checking if the char was a digit. Getting the last number required reversing the string and the lookup map.
Part 1:
var list = new List((await File.ReadAllLinesAsync(@".\Day 1\PuzzleInput.txt"))); int total = 0; foreach (var item in list) { //forward string digit1 = string.Empty; string digit2 = string.Empty; foreach (var c in item) { if ((int)c >= 48 && (int)c <= 57) { digit1 += c; break; } } //reverse foreach (var c in item.Reverse()) { if ((int)c >= 48 && (int)c <= 57) { digit2 += c; break; } } total += Int32.Parse(digit1 +digit2); } Console.WriteLine(total);
Part 2:
var list = new List((await File.ReadAllLinesAsync(@".\Day 1\PuzzleInput.txt"))); var numbers = new Dictionary() { {"one" , 1} ,{"two" , 2} ,{"three" , 3} ,{"four" , 4} ,{"five" , 5} ,{"six" , 6} ,{"seven" , 7} ,{"eight" , 8} , {"nine" , 9 } }; int total = 0; string digit1 = string.Empty; string digit2 = string.Empty; foreach (var item in list) { //forward digit1 = getDigit(item, numbers); digit2 = getDigit(new string(item.Reverse().ToArray()), numbers.ToDictionary(k => new string(k.Key.Reverse().ToArray()), k => k.Value)); total += Int32.Parse(digit1 + digit2); } Console.WriteLine(total); string getDigit(string item, Dictionary numbers) { int index = 0; int digit = 0; foreach (var c in item) { var sub = item.AsSpan(index++); foreach(var n in numbers) { if (sub.StartsWith(n.Key)) { digit = n.Value; goto end; } } if ((int)c >= 48 && (int)c <= 57) { digit = ((int)c) - 48; break; } } end: return digit.ToString(); }
APL
spoiler
args ← {1↓⍵/⍨∨\⍵∊⊂'--'} ⎕ARG inputs ← ⎕FIO[49]¨ args words ← 'one' 'two' 'three' 'four' 'five' 'six' 'seven' 'eight' 'nine' digits ← '123456789' part1 ← {↑↑+/{(10×↑⍵)+¯1↑⍵}¨{⍵~0}¨+⊃(⍳9)+.×digits∘.⍷⍵} "Part 1: ", part1¨ inputs part2 ← {↑↑+/{(10×↑⍵)+¯1↑⍵}¨{⍵~0}¨+⊃(⍳9)+.×(words∘.⍷⍵)+digits∘.⍷⍵} "Part 2: ", part2¨ inputs )OFF
I feel ok about part 1, and just terrible about part 2.
day01.factor
on github (with comments and imports):: part1 ( -- ) "vocab:aoc-2023/day01/input.txt" utf8 file-lines [ [ [ digit? ] find nip ] [ [ digit? ] find-last nip ] bi 2array string>number ] map-sum . ; MEMO: digit-words ( -- name-char-assoc ) [ "123456789" [ dup char>name "-" split1 nip ,, ] each ] H{ } make ; : first-digit-char ( str -- num-char/f i/f ) [ digit? ] find swap ; : last-digit-char ( str -- num-char/f i/f ) [ digit? ] find-last swap ; : first-digit-word ( str -- num-char/f ) [ digit-words keys [ 2dup subseq-index dup [ [ digit-words at ] dip ,, ] [ 2drop ] if ] each drop ! ] H{ } make [ f ] [ sort-keys first last ] if-assoc-empty ; : last-digit-word ( str -- num-char/f ) reverse [ digit-words keys [ reverse 2dup subseq-index dup [ [ reverse digit-words at ] dip ,, ] [ 2drop ] if ] each drop ! ] H{ } make [ f ] [ sort-keys first last ] if-assoc-empty ; : first-digit ( str -- num-char ) dup first-digit-char dup [ pick 2dup swap head nip first-digit-word dup [ [ 2drop ] dip ] [ 2drop ] if nip ] [ 2drop first-digit-word ] if ; : last-digit ( str -- num-char ) dup last-digit-char dup [ pick 2dup swap 1 + tail nip last-digit-word dup [ [ 2drop ] dip ] [ 2drop ] if nip ] [ 2drop last-digit-word ] if ; : part2 ( -- ) "vocab:aoc-2023/day01/input.txt" utf8 file-lines [ [ first-digit ] [ last-digit ] bi 2array string>number ] map-sum . ;
Part 02 in Rust 🦀 :
use std::{ collections::HashMap, env, fs, io::{self, BufRead, BufReader}, }; fn main() -> io::Result<()> { let args: Vec = env::args().collect(); let filename = &args[1]; let file = fs::File::open(filename)?; let reader = BufReader::new(file); let number_map = HashMap::from([ ("one", "1"), ("two", "2"), ("three", "3"), ("four", "4"), ("five", "5"), ("six", "6"), ("seven", "7"), ("eight", "8"), ("nine", "9"), ]); let mut total = 0; for _line in reader.lines() { let digits = get_text_numbers(_line.unwrap(), &number_map); if !digits.is_empty() { let digit_first = digits.first().unwrap(); let digit_last = digits.last().unwrap(); let mut cat = String::new(); cat.push(*digit_first); cat.push(*digit_last); let cat: i32 = cat.parse().unwrap(); total += cat; } } println!("{total}"); Ok(()) } fn get_text_numbers(text: String, number_map: &HashMap<&str, &str>) -> Vec { let mut digits: Vec = Vec::new(); if text.is_empty() { return digits; } let mut sample = String::new(); let chars: Vec = text.chars().collect(); let mut ptr1: usize = 0; let mut ptr2: usize; while ptr1 < chars.len() { sample.clear(); ptr2 = ptr1 + 1; if chars[ptr1].is_digit(10) { digits.push(chars[ptr1]); sample.clear(); ptr1 += 1; continue; } sample.push(chars[ptr1]); while ptr2 < chars.len() { if chars[ptr2].is_digit(10) { sample.clear(); break; } sample.push(chars[ptr2]); if number_map.contains_key(&sample.as_str()) { let str_digit: char = number_map.get(&sample.as_str()).unwrap().parse().unwrap(); digits.push(str_digit); sample.clear(); break; } ptr2 += 1; } ptr1 += 1; } digits }
Thanks, used this as input for reading the Day 2 file and looping the lines, just getting started with rust :)
Trickier than expected! I ran into an issue with Lua patterns, so I had to revert to a more verbose solution, which I then used in Hare as well.
Lua:
lua
-- SPDX-FileCopyrightText: 2023 Jummit -- -- SPDX-License-Identifier: GPL-3.0-or-later local sum = 0 for line in io.open("1.input"):lines() do local a, b = line:match("^.-(%d).*(%d).-$") if not a then a = line:match("%d+") b = a end if a and b then sum = sum + tonumber(a..b) end end print(sum) local names = { ["one"] = 1, ["two"] = 2, ["three"] = 3, ["four"] = 4, ["five"] = 5, ["six"] = 6, ["seven"] = 7, ["eight"] = 8, ["nine"] = 9, ["1"] = 1, ["2"] = 2, ["3"] = 3, ["4"] = 4, ["5"] = 5, ["6"] = 6, ["7"] = 7, ["8"] = 8, ["9"] = 9, } sum = 0 for line in io.open("1.input"):lines() do local firstPos = math.huge local first for name, num in pairs(names) do local left = line:find(name) if left and left < firstPos then firstPos = left first = num end end local last for i = #line, 1, -1 do for name, num in pairs(names) do local right = line:find(name, i) if right then last = num goto found end end end ::found:: sum = sum + tonumber(first * 10 + last) end print(sum)
Hare:
hare
// SPDX-FileCopyrightText: 2023 Jummit // // SPDX-License-Identifier: GPL-3.0-or-later use fmt; use types; use bufio; use strings; use io; use os; const numbers: [](str, int) = [ ("one", 1), ("two", 2), ("three", 3), ("four", 4), ("five", 5), ("six", 6), ("seven", 7), ("eight", 8), ("nine", 9), ("1", 1), ("2", 2), ("3", 3), ("4", 4), ("5", 5), ("6", 6), ("7", 7), ("8", 8), ("9", 9), ]; fn solve(start: size) void = { const file = os::open("1.input")!; defer io::close(file)!; const scan = bufio::newscanner(file, types::SIZE_MAX); let sum = 0; for (let i = 1u; true; i += 1) { const line = match (bufio::scan_line(&scan)!) { case io::EOF => break; case let line: const str => yield line; }; let first: (void | int) = void; let last: (void | int) = void; for (let i = 0z; i < len(line); i += 1) :found { for (let num = start; num < len(numbers); num += 1) { const start = strings::sub(line, i, strings::end); if (first is void && strings::hasprefix(start, numbers[num].0)) { first = numbers[num].1; }; const end = strings::sub(line, len(line) - 1 - i, strings::end); if (last is void && strings::hasprefix(end, numbers[num].0)) { last = numbers[num].1; }; if (first is int && last is int) { break :found; }; }; }; sum += first as int * 10 + last as int; }; fmt::printfln("{}", sum)!; }; export fn main() void = { solve(9); solve(0); };
My solutin in Elixir for both part 1 and part 2 is below. It does use regex and with that there are many different ways to accomplish the goal. I’m no regex master so I made it as simple as possible and relied on the language a bit more. I’m sure there are cooler solutions with no regex too, this is just what I settled on:
https://pastebin.com/u1SYJ4tY
defmodule AdventOfCode.Day01 do def part1(args) do number_regex = ~r/([0-9])/ args |> String.split(~r/\n/, trim: true) |> Enum.map(&first_and_last_number(&1, number_regex)) |> Enum.map(&number_list_to_integer/1) |> Enum.sum() end def part2(args) do number_regex = ~r/(?=(one|two|three|four|five|six|seven|eight|nine|[0-9]))/ args |> String.split(~r/\n/, trim: true) |> Enum.map(&first_and_last_number(&1, number_regex)) |> Enum.map(fn number -> Enum.map(number, &replace_word_with_number/1) end) |> Enum.map(&number_list_to_integer/1) |> Enum.sum() end defp first_and_last_number(string, regex) do matches = Regex.scan(regex, string) [_, first] = List.first(matches) [_, last] = List.last(matches) [first, last] end defp number_list_to_integer(list) do list |> List.to_string() |> String.to_integer() end defp replace_word_with_number(string) do numbers = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"] String.replace(string, numbers, fn x -> (Enum.find_index(numbers, &(&1 == x)) + 1) |> Integer.to_string() end) end end
I think I found a decently short solution for part 2 in python:
DIGITS = { 'one': '1', 'two': '2', 'three': '3', 'four': '4', 'five': '5', 'six': '6', 'seven': '7', 'eight': '8', 'nine': '9', } def find_digit(word: str) -> str: for digit, value in DIGITS.items(): if word.startswith(digit): return value if word.startswith(value): return value return '' total = 0 for line in puzzle.split('\n'): digits = [ digit for i in range(len(line)) if (digit := find_digit(line[i:])) ] total += int(digits[0] + digits[-1]) print(total)
Looks very elegant! I’m having trouble understanding how this finds digit “words” from the end of the line though, as they should be spelled backwards IIRC? I.e.
eno
,owt
,eerht
It simply finds all possible digits and then locates the last one through the reverse indexing. It’s not efficient because there’s no shortcircuiting, but there’s no need to search the strings backwards, which is nice. Python’s startswith method is also hiding a lot of that other implementations have done explicitly.
My solution in rust. I’m sure there’s a lot more clever ways to do it but this is what I came up with.
Code
use std::{io::prelude::*, fs::File, path::Path, io }; fn main() { run_solution(false); println!("\nPress enter to continue"); let mut buffer = String::new(); io::stdin().read_line(&mut buffer).unwrap(); run_solution(true); } fn run_solution(check_for_spelled: bool) { let data = load_data("data/input"); println!("\nProcessing Data..."); let mut sum: u64 = 0; for line in data.lines() { // Doesn't seem like the to_ascii_lower call is needed but... just in case let first = get_digit(line.to_ascii_lowercase().as_bytes(), false, check_for_spelled); let last = get_digit(line.to_ascii_lowercase().as_bytes(), true, check_for_spelled); let num = (first * 10) + last; // println!("\nLine: {} -- First: {}, Second: {}, Num: {}", line, first, last, num); sum += num as u64; } println!("\nFinal Sum: {}", sum); } fn get_digit(line: &[u8], from_back: bool, check_for_spelled: bool) -> u8 { let mut range: Vec = (0..line.len()).collect(); if from_back { range.reverse(); } for i in range { if is_num(line[i]) { return (line[i] - 48) as u8; } if check_for_spelled { if let Some(num) = is_spelled_num(line, i) { return num; } } } return 0; } fn is_num(c: u8) -> bool { c >= 48 && c <= 57 } fn is_spelled_num(line: &[u8], start: usize) -> Option { let words = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine"]; for word_idx in 0..words.len() { let mut i = start; let mut found = true; for c in words[word_idx].as_bytes() { if i < line.len() && *c != line[i] { found = false; break; } i += 1; } if found && i <= line.len() { return Some(word_idx as u8 + 1); } } return None; } fn load_data(file_name: &str) -> String { let mut file = match File::open(Path::new(file_name)) { Ok(file) => file, Err(why) => panic!("Could not open file {}: {}", Path::new(file_name).display(), why), }; let mut s = String::new(); let file_contents = match file.read_to_string(&mut s) { Err(why) => panic!("couldn't read {}: {}", Path::new(file_name).display(), why), Ok(_) => s, }; return file_contents; }
[Language: Lean4]
I’ll only post the actual parsing and solution. I have written some helpers which are in other files, as is the main function. For the full code, please see my github repo.
Part 2 is a bit ugly, but I’m still new to Lean4, and writing it this way (structural recursion) just worked without a proof or termination.
Solution
def parse (input : String) : Option $ List String := some $ input.split Char.isWhitespace |> List.filter (not ∘ String.isEmpty) def part1 (instructions : List String) : Option Nat := let firstLast := λ (o : Option Nat × Option Nat) (c : Char) ↦ let digit := match c with | '0' => some 0 | '1' => some 1 | '2' => some 2 | '3' => some 3 | '4' => some 4 | '5' => some 5 | '6' => some 6 | '7' => some 7 | '8' => some 8 | '9' => some 9 | _ => none if let some digit := digit then match o.fst with | none => (some digit, some digit) | some _ => (o.fst, some digit) else o let scanLine := λ (l : String) ↦ l.foldl firstLast (none, none) let numbers := instructions.mapM ((uncurry Option.zip) ∘ scanLine) let numbers := numbers.map λ l ↦ l.map λ (a, b) ↦ 10*a + b numbers.map (List.foldl (.+.) 0) def part2 (instructions : List String) : Option Nat := -- once I know how to prove stuff propery, I'm going to improve this. Maybe. let instructions := instructions.map String.toList let updateState := λ (o : Option Nat × Option Nat) (n : Nat) ↦ match o.fst with | none => (some n, some n) | some _ => (o.fst, some n) let extract_digit := λ (o : Option Nat × Option Nat) (l : List Char) ↦ match l with | '0' :: _ | 'z' :: 'e' :: 'r' :: 'o' :: _ => (updateState o 0) | '1' :: _ | 'o' :: 'n' :: 'e' :: _ => (updateState o 1) | '2' :: _ | 't' :: 'w' :: 'o' :: _ => (updateState o 2) | '3' :: _ | 't' :: 'h' :: 'r' :: 'e' :: 'e' :: _ => (updateState o 3) | '4' :: _ | 'f' :: 'o' :: 'u' :: 'r' :: _ => (updateState o 4) | '5' :: _ | 'f' :: 'i' :: 'v' :: 'e' :: _ => (updateState o 5) | '6' :: _ | 's' :: 'i' :: 'x' :: _ => (updateState o 6) | '7' :: _ | 's' :: 'e' :: 'v' :: 'e' :: 'n' :: _ => (updateState o 7) | '8' :: _ | 'e' :: 'i' :: 'g' :: 'h' :: 't' :: _ => (updateState o 8) | '9' :: _ | 'n' :: 'i' :: 'n' :: 'e' :: _ => (updateState o 9) | _ => o let rec firstLast := λ (o : Option Nat × Option Nat) (l : List Char) ↦ match l with | [] => o | _ :: cs => firstLast (extract_digit o l) cs let scanLine := λ (l : List Char) ↦ firstLast (none, none) l let numbers := instructions.mapM ((uncurry Option.zip) ∘ scanLine) let numbers := numbers.map λ l ↦ l.map λ (a, b) ↦ 10*a + b numbers.map (List.foldl (.+.) 0)
Ruby
https://github.com/snowe2010/advent-of-code/blob/master/ruby_aoc/2023/day01/day01.rb
Part 1
execute(1, test_file_suffix: "p1") do |lines| lines.inject(0) do |acc, line| d = line.gsub(/\D/,'') acc += (d[0] + d[-1]).to_i end end
Part 2
map = { "one": 1, "two": 2, "three": 3, "four": 4, "five": 5, "six": 6, "seven": 7, "eight": 8, "nine": 9, } execute(2) do |lines| lines.inject(0) do |acc, line| first_num = line.sub(/(one|two|three|four|five|six|seven|eight|nine)/) do |key| map[key.to_sym] end last_num = line.reverse.sub(/(enin|thgie|neves|xis|evif|ruof|eerht|owt|eno)/) do |key| map[key.reverse.to_sym] end d = first_num.chars.select { |num| numeric?(num) } e = last_num.chars.select { |num| numeric?(num) } acc += (d[0] + e[0]).to_i end end
Then of course I also code golfed it, but didn’t try very hard.
P1 Code Golf
execute(1, alternative_text: "Code Golf 60 bytes", test_file_suffix: "p1") do |lines| lines.inject(0){|a,l|d=l.gsub(/\D/,'');a+=(d[0]+d[-1]).to_i} end
P2 Code Golf (ignore the formatting, I just didn’t want to reformat to remove all the spaces, and it’s easier to read this way.)
execute(1, alternative_text: "Code Golf 271 bytes", test_file_suffix: "p1") do |z| z.inject(0) { |a, l| w = %w(one two three four five six seven eight nine) x = w.join(?|) f = l.sub(/(#{x})/) { |k| map[k.to_sym] } g = l.reverse.sub(/(#{x.reverse})/) { |k| map[k.reverse.to_sym] } d = f.chars.select { |n| n.match?(/\d/) } e = g.chars.select { |n| n.match?(/\d/) } a += (d[0] + e[0]).to_i } end
Thank you for sharing this. I also wrote a regular expression with
\d|eno|owt
and so on, and I was not so proud of myself :). Good to know I wasn’t the only one :).I was trying so hard to avoid doing this and landed on a pretty nice solution (for me). It’s funny sering everyone’s approach especially when you have no problem running through that barrier that I didn’t want to 😆