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Ryan Palo
Ryan Palo

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Advent of Code 2020 Solution Megathread - Day 21: Allergen Assessment

Whew. Finals weekend this weekend put me way off my pace. I'm still trying to finish up the math one from a few days ago. But I'm not giving up! Four more days! It looks like I'm not the only one either, with only two submissions yesterday! It definitely was a tough one.

The Puzzle

In today’s puzzle, we're simply trying to feed ourselves, but it's not going to be easy. The foods are listed with some of the allergens that are present. But not all, necessarily. So. Deal with that uncertainty.

The Leaderboards

As always, this is the spot where I’ll plug any leaderboard codes shared from the community.

Ryan's Leaderboard: 224198-25048a19
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If you want to generate your own leaderboard and signal boost it a little bit, send it to me either in a DEV message or in a comment on one of these posts and I'll add it to the list above.

Yesterday’s Languages

Updated 09:23AM 12/21/2020 PST.

Language Count
Python 1
JavaScript 1

Merry Coding!

Top comments (6)

meseta profile image
Yuan Gao

Lots and lots of set theory

Part 2 result is at the end; Part 1 result is somewhere halfway down.

import re
from collections import defaultdict
rule = re.compile("^((?:\w+ )+)\(contains (.*?)\)$")

data = open("demo.txt").read().splitlines()
all_candidates = defaultdict(list)
all_ingredients = []
for line in data:
    ing_str, all_str = rule.match(line).groups()

    ingredients = ing_str.strip().split(" ")
    allergens = all_str.split(", ")

    for allergen in allergens:


candidates = {}
for allergen_name, ingredient_list in all_candidates.items():
    candidates[allergen_name] = set.intersection(*ingredient_list)

print("appearances", sum(all_ingredients.count(ingredient) for ingredient in set(all_ingredients).difference(set.union(*candidates.values()))))

confirmed = {}

while len(confirmed) < len(candidates):
    for allergen in candidates:
        new_set = candidates[allergen].difference(confirmed.values())

        if len(new_set) == 1:
            confirmed[allergen] = new_set.pop()

        candidates[allergen] = new_set

print(",".join(v for k, v in sorted(confirmed.items(), key=lambda item: item[0])))
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bgaster profile image
Benedict Gaster • Edited

YEsterday was not fun Haskell, I did get it done, but it is horrible. Mostly my solution, some people on Reddit did it with mutable arrays, however, the whole point of doing AoC with Haskell, was to spend sometime programming in a purely functional language, for fun, away from my normal language of choice Rust and C++ if I have too. But in truth I think yesterday would have been a lot easier and more fun in Rust, at least for me :-)

Anyway today was a lot easier and got it done pretty quickly.

split :: String -> Char -> String -> [String]
split [] _ ss | null ss = []
              | otherwise = [ss]
split (x:xs) c ss | c == x = ss : split xs c ""
                  | otherwise = split xs c (ss ++ [x])

parse :: [String] -> [(S.Set String, S.Set String)]
parse = map parse'
        parse' xs = let [fs, as] = split xs '(' ""
                        fs'      = split fs ' ' ""
                        (a:as')  = map (takeWhile (/=')') . dropWhile (==' ')) (split as ',' "")
                    in (S.fromList fs', S.fromList $ drop 9 a:as')

generate foods = valid <$> 
                 M.fromListWith (++) (do (is, as) <- foods
                                         a       <- S.toList as
                                         return (a, [is]))
        valid is = S.filter (\g -> all (g `S.member`) is) $ S.unions is

task1 foods = length $ filter (`S.notMember` allValids) $ concatMap (S.toList . fst) foods
  isValid ing = any (\a -> ing `S.member` (generate foods M.! a))
  allValids = S.unions $ map (\(ings, a) -> S.filter (`isValid` a) ings) foods

task2 = intercalate "," . map (S.elemAt 0 . snd) . M.toAscList . go . generate
    -- Just like day 16 find foods which contain a particular allergen and repeat
    -- though a process of elimination
    go :: M.Map String (S.Set String) -> M.Map String (S.Set String)
    go v
      | all (\s -> S.size s == 1) (M.elems v) = v
      | otherwise = let found = S.unions $ M.elems $ M.filter (\s -> S.size s == 1) v
                    in go $ (\s -> if S.size s == 1 then s else s S.\\ found) v

main = do
    is <- readFile "day21_input" <&> lines <&> parse
    print (task1 is)
    print (task2 is)
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rpalo profile image
Ryan Palo

Finishing this one up a little late, but it's my second one completed today, so I'm catching back up! Lots of sets, but pretty straightforward once I figured out how you could possibly even rule things in or out for sure. The code came out simpler than I expected.

"""Day 21: Allergen Assessment

Find out through process of elimination which foreign ingredients
contain which allergens.

from collections import Counter

def parse(filename):
    """Parse the input file into Python objects.

    Input file has lines like this:
        sdfsjdfk sjdkf lfke ljrl (contains dairy, soy)

        filename (str): the name of the input file to open

        candidates (Dict[str, Set[str]]): a dict with keys that are the
            allergens (dairy, soy, etc.) and values that are the sets
            of all possible ingredients that could contain those
        counts (Counter): a count of how many lines each ingredient
            appears in.

    candidates = dict()
    counts = Counter()
    with open(filename, "r") as f:
        for line in f:
            ingredients, _contains, allergens = line.partition(" (contains ")
            ingredients = set(ingredients.split())
            counts += Counter(ingredients)
            for allergen in allergens.strip(")\n").split(", "):
                if allergen in candidates:
                    candidates[allergen] &= ingredients
                    candidates[allergen] = ingredients.copy()

    return candidates, counts

def find_allergens(candidates):
    """Uses allergens that only have one possible ingredient that could
    contain them to rule that ingredient out for other allergens until
    all allergens have exactly one ingredient that could contain it.

    Returns a dict of allergen name to ingredient name to simplify things.
    while not all(len(ingredients) == 1 for ingredients in candidates.values()):
        for allergen, ingredients in candidates.items():
            if len(ingredients) == 1:
                for other_al, other_ingredients in candidates.items():
                    if other_al != allergen:
                        other_ingredients -= ingredients
    return {allergen: ingredients.pop() for allergen, ingredients in candidates.items()}

def discard_unsafe_ingredient_counts(allergens, counts):
    """Given a set of decided allergens, removes each of these unsafe
    ingredients from a dict of counts so that only the counts of totally
    safe ingredients are retained.
    for ingredient in allergens.values():
        del counts[ingredient]

def part1(counts):
    """Part 1 asks us to add up the number of occurrences of any totally
    safe ingredient.

    Assumes `counts` has already had its unsafe ingredients discarded.
    return sum(counts.values())

def part2(allergens):
    """Part 2 wants the list of unsafe ingredients, sorted by their
    allergen alphabetically, joined together into one string by commas.
    return ",".join(
        ingredient for (allergen, ingredient) in sorted(allergens.items())

if __name__ == "__main__":
    test_candidates, test_counts = parse("data/day21_test.txt")
    test_allergens = find_allergens(test_candidates)
    discard_unsafe_ingredient_counts(test_allergens, test_counts)
    assert 5 == part1(test_counts)
    assert "mxmxvkd,sqjhc,fvjkl" == part2(test_allergens)
    print("All tests passed!")

    candidates, counts = parse("data/day21.txt")
    allergens = find_allergens(candidates)
    discard_unsafe_ingredient_counts(allergens, counts)
    print("Part 1:", part1(counts))
    print("Part 2:", part2(allergens))
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neilgall profile image
Neil Gall • Edited

Felt like a replay of day 16.

My first solution used lots and lots of string copies. Then I refactored it to use slices of the original parsed string all the way through, just to prove to myself I understand Rust lifetimes and ownership. Works!

use std::collections::{HashMap, HashSet};
use parser::*;

// -- model

#[derive(Debug, PartialEq)]
struct Food<'a> {
    ingredients: HashSet<&'a str>,
    allergens: HashSet<&'a str>

impl<'a> Food<'a> {
    fn new(ingredients: &[&'a str], allergens: &[&'a str]) -> Self {
        Food {
            ingredients: ingredients.iter().cloned().collect(),
            allergens: allergens.iter().cloned().collect()

struct Model<'a> {
    foods: Vec<Food<'a>>,
    ingredients_by_allergen: HashMap<&'a str, HashSet<&'a str>>,

impl<'a> Model<'a> {
    fn new(input: &'a str) -> Self {
        let ingredients = one_or_more(whitespace_wrap(word_ref));
        let allergens = word_ref
            .sep_by(match_literal(", "))
            .between(match_literal("(contains "), match_literal(")"));

        let food = pair(ingredients, allergens, |ingredients, allergens| Food {
            ingredients: ingredients.into_iter().collect(),
            allergens: allergens.into_iter().collect()

        let foods = one_or_more(whitespace_wrap(food)).parse(input).unwrap().1;

        Model {
            ingredients_by_allergen: HashMap::new()

    fn determine_allergens(&mut self) {
        while !self.is_fully_determined() {

    fn associate_ingredients_with_allergens(&mut self) {
        for food in self.foods.iter() {
            for allergen in food.allergens.iter() {
                match self.ingredients_by_allergen.get_mut(allergen) {
                    Some(existing) => {
                        *existing = existing.intersection(&food.ingredients).cloned().collect();
                    None => {
                        self.ingredients_by_allergen.insert(allergen, food.ingredients.clone());

    fn is_fully_determined(&self) -> bool {
        self.ingredients_by_allergen.values().all(|ingredients| ingredients.len() < 2)

    fn eliminate_duplicate_matches(&mut self) {
        let determined: HashSet<&'a str> = self.ingredients_by_allergen.values()
                if ingredients.len() == 1 { ingredients.iter().next() } else { None }

        for ingredients in self.ingredients_by_allergen.values_mut().filter(|ings| ings.len() > 1) {
            *ingredients = ingredients.difference(&determined).cloned().collect();

    fn ingredients_with_allergen(&self) -> HashSet<&'a str> {
        self.ingredients_by_allergen.values().flat_map(|values| values.iter().cloned()).collect()

    fn all_ingredients(&self) -> HashSet<&'a str> {
        self.foods.iter().flat_map(|food| food.ingredients.iter().cloned()).collect()

    fn ingredients_with_no_allergen(&self) -> HashSet<&'a str> {

    fn ingredients_alphabetically_by_allergen(&self) -> Vec<&'a str> {
        let mut allergens: Vec<&'a str> = self.ingredients_by_allergen.keys().cloned().collect();
            self.ingredients_by_allergen.get(a).and_then(|ings| ings.iter().next())

// -- problems 

fn part1(model: &Model) -> usize {

fn part2(model: &Model) -> String {

fn main() {
    let input = std::fs::read_to_string("./input.txt").unwrap();
    let mut model = Model::new(&input);

    println!("part 1 {}", part1(&model));
    println!("part 2 {}", part2(&model));

mod tests {
    use super::*;

    fn test_input() -> &'static str {
        "mxmxvkd kfcds sqjhc nhms (contains dairy, fish)
         trh fvjkl sbzzf mxmxvkd (contains dairy)
         sqjhc fvjkl (contains soy)
         sqjhc mxmxvkd sbzzf (contains fish)"        

    fn test_parser() {
        let model = Model::new(test_input());
        assert_eq!(model.foods, vec![
                &["mxmxvkd", "kfcds", "sqjhc", "nhms"],
                &["dairy", "fish"]
                &["trh", "fvjkl", "sbzzf", "mxmxvkd"],
                &["sqjhc", "fvjkl"],
                &["sqjhc", "mxmxvkd", "sbzzf"],

    fn test_part1() {
        let mut model = Model::new(test_input());
        assert_eq!(part1(&model), 5);

    fn test_part2() {
        let mut model = Model::new(test_input());
        assert_eq!(part2(&model), "mxmxvkd,sqjhc,fvjkl");
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mellen profile image
Matt Ellen • Edited

Another javascript gist. It's been a few days since i could solve both parts.

Today's solution took a "by eye" shortcut, as I could see I only needed three iterations, rather than figuring out the stopping condition.

thibpat profile image
Thibaut Patel

Today was a lot more enjoyable than yesterday! Here is my JavaScript video explanation of the day.