Day 21 was welcome relief after day 20. In this one, I’ll parse a list of foods, each with an ingredients list and a listing of some (not necessarily all) of the allergies. I’ll use that list to pair up allergens to ingredients.

## Challenge

The puzzle can be found here. I’m given list, where each line is the list of ingredients in the food and at least some of the allergens the food may contain (absence of a label doesn’t mean that food doesn’t contain that allergen). An example list of four foods is:

mxmxvkd kfcds sqjhc nhms (contains dairy, fish)
trh fvjkl sbzzf mxmxvkd (contains dairy)
sqjhc fvjkl (contains soy)
sqjhc mxmxvkd sbzzf (contains fish)


In part one, I’m asked to find all the ingredients that can’t contain be an allergen.

In part two, I solve the system and find which ingredient contains each allergen.

## Solution

### Part 1

I’ll keep a list of all the ingredients (originally I used a set, but the challenge for part one is to count the number of times the safe ingredients appear, not the number of ingredients) and a dictionary by allergen. I’ll parse the input saving the ingredients to the big list, and keeping track of which ingredients could be associated with each allergen.

all_ingredients = []
all_allergens = {}

for line in data:
ilist, alist = line.strip('\n)').split(' (contains ')
ilist = ilist.split(' ')
alist = alist.split(', ')
all_ingredients += ilist
for allergen in alist:
if allergen in all_allergens:
all_allergens[allergen] &= set(ilist)
else:
all_allergens[allergen] = set(ilist)


If an allergen is already in the dictionary, then the potential ingredients that could contain that allergen are ingredients that are in both sets, or the intersection of the two sets, which is defined in Python as &. If the allergen isn’t already in the dictionary, then the potential ingredients are the list of ingredients in this food.

Once that’s processed, to find the list of safe ingredients, I’ll get a list of all the foods in the allergens list, and then get the ingredients list without the foods in that list:

allergen_foods = set([i for v in all_allergens.values() for i in v])
safe_foods = [i for i in all_ingredients if i not in allergen_foods]
print(f'Part 1: {len(safe_foods)}')


### Part 2

Now I just need to run through the allergens dict looking for any with only one potential ingredient. I’ll save that pair, and then remove the allergen from the dict and the ingredient from all of the other allergens’ lists.

canonical = {}
while all_allergens:
known = [(k,list(v)[0]) for k,v in all_allergens.items() if len(v) == 1]
for k,v in known:
canonical[k] = v
del all_allergens[k]
for a in all_allergens:
if v in all_allergens[a]:
all_allergens[a].remove(v)


When that’s done, the sorted list from canonical is used to make the string:

can_str = ','.join([v for k,v in sorted(canonical.items())])
print(f'Part 2: {can_str}')


### Final Code

This one is basically instant:

\$ time python3 day21.py 21-puz.txt
Part 1: 2573
Part 2: bjpkhx,nsnqf,snhph,zmfqpn,qrbnjtj,dbhfd,thn,sthnsg

real    0m0.037s
user    0m0.032s
sys     0m0.000s



This code is pretty ugly, but it works:

#!/usr/bin/env python3

import sys

with open(sys.argv[1], "r") as f:

all_ingredients = []
all_allergens = {}

for line in data:
ilist, alist = line.strip("\n)").split(" (contains ")
ilist = ilist.split(" ")
alist = alist.split(", ")
all_ingredients += ilist
for allergen in alist:
if allergen in all_allergens:
all_allergens[allergen] &= set(ilist)
else:
all_allergens[allergen] = set(ilist)

allergen_foods = set([i for v in all_allergens.values() for i in v])
safe_foods = [i for i in all_ingredients if i not in allergen_foods]
print(f"Part 1: {len(safe_foods)}")

canonical = {}
while all_allergens:
known = [(k, list(v)[0]) for k, v in all_allergens.items() if len(v) == 1]
for k, v in known:
canonical[k] = v
del all_allergens[k]
for a in all_allergens:
if v in all_allergens[a]:
all_allergens[a].remove(v)

can_str = ",".join([v for k, v in sorted(canonical.items())])
print(f"Part 2: {can_str}")