Day 2: Cube Conundrum
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Found a C per-char solution, that is, no lines, splitting, lookahead, etc. It wasn’t even necessary to keep match lengths for the color names because they all have unique characters, e.g. ‘b’ only occurs in “blue” so then you can attribute the count to that color.
int main() { int p1=0,p2=0, id=1,num=0, r=0,g=0,b=0, c; while ((c = getchar()) != EOF) if (c==',' || c==';' || c==':') num = 0; else if (c>='0' && c<='9') num = num*10 + c-'0'; else if (c=='d') r = MAX(r, num); else if (c=='g') g = MAX(g, num); else if (c=='b') b = MAX(b, num); else if (c=='\n') { p1 += (r<=12 && g<=13 && b<=14) * id; p2 += r*g*b; r=g=b=0; id++; } printf("%d %d\n", p1, p2); return 0; }
Golfed:
c,p,P,i,n,r,g,b;main(){while(~ (c=getchar()))c==44|c==58|59== c?n=0:c>47&c<58?n=n*10+c-48:98 ==c?b=b>n?b:n:c=='d'?r=r>n?r:n :c=='g'?g=g>n?g:n:10==c?p+=++i *(r<13&g<14&b<15),P+=r*g*b,r=g =b=0:0;printf("%d %d\n",p,P);}
Rust
Pretty straightforward this time, the bulk of the work was clearly in parsing the input.
Not too tricky today. Part 2 wasn’t as big of a curveball as yesterday thankfully. I don’t think it’s the cleanest code I’ve ever written, but hey - the whole point of this is to get better at Rust, so I’ll definitely be learning as I go, and coming back at the end to clean a lot of these up. I think for this one I’d like to look into a parsing crate like nom to clean up all the spliting and unwrapping in the two from() methods.
https://github.com/capitalpb/advent_of_code_2023/blob/main/src/solvers/day02.rs
#[derive(Debug)] struct Hand { blue: usize, green: usize, red: usize, } impl Hand { fn from(input: &str) -> Hand { let mut hand = Hand { blue: 0, green: 0, red: 0, }; for color in input.split(", ") { let color = color.split_once(' ').unwrap(); match color.1 { "blue" => hand.blue = color.0.parse::().unwrap(), "green" => hand.green = color.0.parse::().unwrap(), "red" => hand.red = color.0.parse::().unwrap(), _ => unreachable!("malformed input"), } } hand } } #[derive(Debug)] struct Game { id: usize, hands: Vec, } impl Game { fn from(input: &str) -> Game { let (id, hands) = input.split_once(": ").unwrap(); let id = id.split_once(" ").unwrap().1.parse::().unwrap(); let hands = hands.split("; ").map(Hand::from).collect(); Game { id, hands } } } pub struct Day02; impl Solver for Day02 { fn star_one(&self, input: &str) -> String { input .lines() .map(Game::from) .filter(|game| { game.hands .iter() .all(|hand| hand.blue <= 14 && hand.green <= 13 && hand.red <= 12) }) .map(|game| game.id) .sum::() .to_string() } fn star_two(&self, input: &str) -> String { input .lines() .map(Game::from) .map(|game| { let max_blue = game.hands.iter().map(|hand| hand.blue).max().unwrap(); let max_green = game.hands.iter().map(|hand| hand.green).max().unwrap(); let max_red = game.hands.iter().map(|hand| hand.red).max().unwrap(); max_blue * max_green * max_red }) .sum::() .to_string() } }
Ruby
https://github.com/snowe2010/advent-of-code/blob/master/ruby_aoc/2023/day02/day02.rb
Second part was soooo much easier today than yesterday. Helps I solved it exactly how he wanted you to solve it I think.
I’m going to work on some code golf now.
Golfed P2 down to 133 characters:
p g.map{_1.sub!(/.*:/,'') m=Hash.new(0) _1.split(?;){|r|r.split(?,){|a|b,c=a.split m[c]=[m[c],b.to_i].max}} m.values.reduce(&:*)}.sum
That’s a nice golf! Clever use of the hash and nice compact reduce. I got my C both-parts solution down to 210 but it’s not half as nice.
Thanks! Your C solution includes main, whereas I do some stuff to parse the lines before hand. I think it would only be 1 extra character if I wrote it to parse the input manually, but I just care for ease of use with these AoC problems so I don’t like counting that, makes it harder to read for me lol. Your solution is really inventive. I was looking for something like that, but didn’t ever get to your conclusion. I wonder if that would be longer in my solution or shorter 🤔
I had some time, so here’s a terrible solution in Uiua (Run it here) :
Lim ← [14 13 12] {"Game 1: 3 blue, 4 red; 1 red, 2 green, 6 blue; 2 green" "Game 2: 1 blue, 2 green; 3 green, 4 blue, 1 red; 1 green, 1 blue" "Game 3: 8 green, 6 blue, 20 red; 5 blue, 4 red, 13 green; 5 green, 1 red" "Game 4: 1 green, 3 red, 6 blue; 3 green, 6 red; 3 green, 15 blue, 14 red" "Game 5: 6 red, 1 blue, 3 green; 2 blue, 1 red, 2 green"} LtoDec ← ∧(+ ×10:) :0 StoDec ← LtoDec▽≥0. ▽≤9. -@0 FilterMax! ← /↥≡(StoDec⊢↙ ¯1)⊔⊏⊚≡(/×^1⊢).⊔ # Build 'map' of draws for each game ∵(□≡(∵(⬚@\s↙2 ⊔) ⇌) ↯¯1_2 ↘ 2⊜□≠@\s . ⊔) # Only need the max for each colour ≡(⊂⊂⊃⊃(FilterMax!(="bl")) (FilterMax!(="gr")) (FilterMax!(="re"))) # part 1 - Compare against limits, and sum game numbers /+▽:+1⇡⧻. ≡(/×≤0-Lim). # part 2 - Multiply the maxes in each game and then sum. /+/×⍉:
This was mostly straightforward… basically just parsing input. Here are my condensed solutions in Python
Part 1
Game = dict[str, int] RED_MAX = 12 GREEN_MAX = 13 BLUE_MAX = 14 def read_game(stream=sys.stdin) -> Game: try: game_string, cubes_string = stream.readline().split(':') except ValueError: return {} game: Game = defaultdict(int) game['id'] = int(game_string.split()[-1]) for cubes in cubes_string.split(';'): for cube in cubes.split(','): count, color = cube.split() game[color] = max(game[color], int(count)) return game def read_games(stream=sys.stdin) -> Iterator[Game]: while game := read_game(stream): yield game def is_valid_game(game: Game) -> bool: return all([ game['red'] <= RED_MAX, game['green'] <= GREEN_MAX, game['blue'] <= BLUE_MAX, ]) def main(stream=sys.stdin) -> None: valid_games = filter(is_valid_game, read_games(stream)) sum_of_ids = sum(game['id'] for game in valid_games) print(sum_of_ids)
Part 2
For the second part, the main parsing remainded the same. I just had to change what I did with the games I read.
def power(game: Game) -> int: return game['red'] * game['green'] * game['blue'] def main(stream=sys.stdin) -> None: sum_of_sets = sum(power(game) for game in read_games(stream)) print(sum_of_sets)
Factor on github (with comments and imports):
: known-color ( color-phrases regexp -- n ) all-matching-subseqs [ 0 ] [ [ split-words first string>number ] map-supremum ] if-empty ; : line>known-rgb ( str -- game-id known-rgb ) ": " split1 [ split-words last string>number ] dip R/ \d+ red/ R/ \d+ green/ R/ \d+ blue/ [ known-color ] tri-curry@ tri 3array ; : possible? ( known-rgb test-rgb -- ? ) v<= [ ] all? ; : part1 ( -- ) "vocab:aoc-2023/day02/input.txt" utf8 file-lines [ line>known-rgb 2array ] [ last { 12 13 14 } possible? ] map-filter [ first ] map-sum . ; : part2 ( -- ) "vocab:aoc-2023/day02/input.txt" utf8 file-lines [ line>known-rgb nip product ] map-sum . ;
Getting my head around parsing tricks for python, maybe abusing dicts as a replacement for a types, but appears to be working: https://gist.github.com/purplemonkeymad/983eec7ff0629e8834163b17ec673958
Rust (Rank 7421/6311) (Time after start 00:32:27/00:35:35)
Extremely easy part 2 today, I would say easier than part 1 but they share the same sort of framework
Code Block
(Note lemmy removed some characters, code link shows them all)
use std::fs; fn part1(input: String) -> i32 { const RED: i32 = 12; const GREEN: i32 = 13; const BLUE: i32 = 14; let mut sum = 0; for line in input.lines() { let [id, content] = line.split(": ").collect::>()[0..2] else { continue }; let id = id.split(" ").collect::>()[1].parse::().unwrap(); let marbles = content.split("; ").map(|x| { x.split(", ").collect::>() }).collect::>>(); let mut valid = true; for selection in marbles { for marble in selection { let marble_split = marble.split(" ").collect::>(); let marble_amount = marble_split[0].parse::().unwrap(); let marble_color = marble_split[1]; if marble_color == "red" && marble_amount > RED { valid = false; break; } if marble_color == "green" && marble_amount > GREEN { valid = false; break; } if marble_color == "blue" && marble_amount > BLUE { valid = false; break; } } } if !valid { continue; } sum += id; } return sum; } fn part2(input: String) -> i32 { let mut sum = 0; for line in input.lines() { let [id, content] = line.split(": ").collect::>()[0..2] else { continue }; let id = id.split(" ").collect::>()[1].parse::().unwrap(); let marbles = content.split("; ").map(|x| { x.split(", ").collect::>() }).collect::>>(); let mut red = 0; let mut green = 0; let mut blue = 0; for selection in marbles { for marble in selection { let marble_split = marble.split(" ").collect::>(); let marble_amount = marble_split[0].parse::().unwrap(); let marble_color = marble_split[1]; if marble_color == "red" && marble_amount > red { red = marble_amount; } if marble_color == "green" && marble_amount > green { green = marble_amount; } if marble_color == "blue" && marble_amount > blue { blue = marble_amount; } } } sum += red * green * blue; } return sum; } fn main() { let input = fs::read_to_string("data/input.txt").unwrap(); println!("{}", part1(input.clone())); println!("{}", part2(input.clone())); }
String parsing! Always fun in C!
https://github.com/sjmulder/aoc/blob/master/2023/c/day02.c
int main(int argc, char **argv) { char ln[256], *sr,*srd,*s; int p1=0,p2=0, id, r,g,b; for (id=1; (sr = fgets(ln, sizeof(ln), stdin)); id++) { strsep(&sr, ":"); r = g = b = 0; while ((srd = strsep(&sr, ";"))) while ((s = strsep(&srd, ","))) if (strchr(s, 'd')) r = MAX(r, atoi(s)); else if (strchr(s, 'g')) g = MAX(g, atoi(s)); else if (strchr(s, 'b')) b = MAX(b, atoi(s)); p1 += (r <= 12 && g <= 13 && b <= 14) * id; p2 += r * g * b; } printf("%d %d\n", p1, p2); return 0; }
Dart solution
Quite straightforward, though there’s a sneaky trap in the test data for those of us who don’t read the rules carefully enough.
Read, run and edit this solution in your browser: https://dartpad.dev/?id=203b3f0a9a1ad7a51daf14a1aeb6cf67
parseLine(String s) { var game = s.split(': '); var num = int.parse(game.first.split(' ').last); var rounds = game.last.split('; '); var cubes = [ for (var (e) in rounds) { for (var ee in e.split(', ')) ee.split(' ').last: int.parse(ee.split(' ').first) } ]; return MapEntry(num, cubes); } /// collects the max of the counts from both maps. Map merge2(Map a, Map b) => { for (var k in {...a.keys, ...b.keys}) k: max(a[k] ?? 0, b[k] ?? 0) }; var limit = {"red": 12, "green": 13, "blue": 14}; bool isGood(Map test) => limit.entries.every((e) => (test[e.key] ?? 0) <= e.value); part1(List lines) => lines .map(parseLine) .where((e) => e.value.every(isGood)) .map((e) => e.key) .sum; part2(List lines) => lines .map(parseLine) .map((e) => e.value.reduce(merge2)) .map((e) => e.values.reduce((s, t) => s * t)) .sum;
Quite straightforward, though there’s a sneaky trap in the test data for those of us who don’t read the rules carefully enough.
What’s that? I didn’t notice anything, perhaps I was lucky.
Oh, I misread the rules as each game having rounds of draws without replacement and the test data gave the same result for that reading, so when I confidently submitted my answer I got a bit of a surprise.
Same here, yesterday felt like a trap, but didn’t run into anything today?
My (awful) Python solves. Much easier than day 1’s, although I did run into an issue with trimming whitespace characters with my approach (Game 96 wouldn’t flag properly).
Part 1
with open('02A_input.txt', 'r') as file: data = file.readlines() possibleGames=[] for game in data: # Find Game number game = game.removeprefix("Game ") gameNumber = int(game[0:game.find(":")]) # Break Game into rounds (split using semicolons) game=game[game.find(":")+1:] rounds=game.split(";") # For each round, determine the maximum number of Red, Blue, Green items shown at a time rgb=[0,0,0] for round in rounds: combos=round.split(",") for combo in combos: combo=combo.strip() number=int(combo[0:combo.find(" ")]) if combo.endswith("red"): if number>rgb[0]: rgb[0]=number elif combo.endswith("green"): if number>rgb[1]: rgb[1]=number elif combo.endswith("blue"): if number>rgb[2]: rgb[2]=number # If Red>12, Green>13, Blue>14, append Game number to possibleGames if not (rgb[0]>12 or rgb[1]>13 or rgb[2]>14): possibleGames.append(gameNumber) print(sum(possibleGames))
Part 2
with open('02A_input.txt', 'r') as file: data = file.readlines() powers=[] for game in data: # Find Game number game = game.removeprefix("Game ") # Break Game into rounds (split using semicolons) game=game[game.find(":")+1:] rounds=game.split(";") # For each round, determine the maximum number of Red, Blue, Green items shown at a time # Note: This could be faster, since we don't need to worry about actual rounds rgb=[0,0,0] for round in rounds: combos=round.split(",") for combo in combos: combo=combo.strip() number=int(combo[0:combo.find(" ")]) if combo.endswith("red"): if number>rgb[0]: rgb[0]=number elif combo.endswith("green"): if number>rgb[1]: rgb[1]=number elif combo.endswith("blue"): if number>rgb[2]: rgb[2]=number # Multiple R, G, B to find the "power" of the game # Append Power to the list powers.append(rgb[0]*rgb[1]*rgb[2]) print(sum(powers))
Mostly an input parsing problem this time, but it was fun to use Hares tokenizer functions:
lua
-- SPDX-FileCopyrightText: 2023 Jummit -- -- SPDX-License-Identifier: GPL-3.0-or-later local colors = {"blue", "red", "green"} local available = {red = 12, blue = 14, green = 13} local possible = 0 local id = 0 local min = 0 for game in io.open("2.input"):lines() do id = id + 1 game = game:gsub("Game %d+: ", "").."; " local max = {red = 0, blue = 0, green = 0} for show in game:gmatch(".-; ") do for _, color in ipairs(colors) do local num = tonumber(show:match("(%d+) "..color)) if num then max[color] = math.max(max[color], num) end end end min = min + max.red * max.blue * max.green local thisPossible = true for _, color in ipairs(colors) do if max[color] > available[color] then thisPossible = false break end end if thisPossible then possible = possible + id end end print(possible) print(min)
hare
// SPDX-FileCopyrightText: 2023 Jummit // // SPDX-License-Identifier: GPL-3.0-or-later use strconv; use types; use strings; use io; use bufio; use os; use fmt; const available: []uint = [12, 13, 14]; fn color_id(color: str) const uint = { switch (color) { case "red" => return 0; case "green" => return 1; case "blue" => return 2; case => abort(); }; }; export fn main() void = { const file = os::open("2.input")!; defer io::close(file)!; const scan = bufio::newscanner(file, types::SIZE_MAX); let possible: uint = 0; let min: uint = 0; for (let id = 1u; true; id += 1) { const line = match(bufio::scan_line(&scan)!) { case io::EOF => break; case let line: const str => yield strings::sub( line, strings::index(line, ": ") as size + 2, strings::end); }; let max: []uint = [0, 0, 0]; let tok = strings::rtokenize(line, "; "); for (true) { const show = match(strings::next_token(&tok)) { case void => break; case let show: str => yield show; }; const pairs = strings::tokenize(show, ", "); for (true) { const pair: (str, str) = match(strings::next_token(&pairs)) { case void => break; case let pair: str => let tok = strings::tokenize(pair, " "); yield ( strings::next_token(&tok) as str, strings::next_token(&tok) as str ); }; let color = color_id(pair.1); let amount = strconv::stou(pair.0)!; if (amount > max[color]) max[color] = amount; }; }; if (max[0] <= available[0] && max[1] <= available[1] && max[2] <= available[2]) { fmt::printfln("{}", id)!; possible += id; }; min += max[0] * max[1] * max[2]; }; fmt::printfln("{}", possible)!; fmt::printfln("{}", min)!; };
Was pretty simple in Python with a regex to get the game number, and then the count of color. for part 2 instead of returning true/false whether the game is valid, you just max the count per color. No traps like in the first one, that I’ve seen, so it was surprisingly easy
def process_game(line: str): game_id = int(re.findall(r'game (\d+)*', line)[0]) colon_idx = line.index(":") draws = line[colon_idx+1:].split(";") # print(draws) if is_game_valid(draws): # print("Game %d is possible"%game_id) return game_id return 0 def is_game_valid(draws: list): for draw in draws: red = get_nr_of_in_draw(draw, 'red') if red > MAX_RED: return False green = get_nr_of_in_draw(draw, 'green') if green > MAX_GREEN: return False blue = get_nr_of_in_draw(draw, 'blue') if blue > MAX_BLUE: return False return True def get_nr_of_in_draw(draw: str, color: str): if color in draw: nr = re.findall(r'(\d+) '+color, draw) return int(nr[0]) return 0 # f = open("input.txt", "r") f = open("input_real.txt", "r") lines = f.readlines() sum = 0 for line in lines: sum += process_game(line.strip().lower()) print("Answer: %d"%sum)
C# - a tad bit overkill for this. I was worried that we’d need more statistical analysis in part 2, so I designed my solution around that. I ended up cutting everything back when
Max()
was enough for both parts.https://github.com/warriordog/advent-of-code-2023/tree/main/Solutions/Day02