Advent of Code 2020: Day 4

Day 4 presented another text parsing challenge. In the first part, I just needed to validate if each section contained a specific seven strings, which is easy enough to solve in Python. For part two, I need to now look at the text following each of these strings, and apply some validation rules. At first I thought I’d throw out my part 1 work and start processing all the data into a Python dict. But then I realized I could just write a regex for each validation, and use the same pattern.

Challenge

The puzzle can be found here. I’m given a bunch of text lines of “passport” data in key:value format, with eight required keys. The example data is:

ecl:gry pid:860033327 eyr:2020 hcl:#fffffd
byr:1937 iyr:2017 cid:147 hgt:183cm

iyr:2013 ecl:amb cid:350 eyr:2023 pid:028048884
hcl:#cfa07d byr:1929

hcl:#ae17e1 iyr:2013
eyr:2024
ecl:brn pid:760753108 byr:1931
hgt:179cm

hcl:#cfa07d eyr:2025 pid:166559648
iyr:2011 ecl:brn hgt:59in

In the first part, I’m asked to validate passports by if they contain seven (ignoring the eighth for the sake of the story) specific key values.

In the second part, I need to validate the data for each field.

In both cases, I’m asked to return the number of valid passports.

Solution

Part 1

On reading in the data, I’ll split on \n\n to get each passport into it’s own string, newlines and all.

I created a list of the keys I was looking for, and use it in a nested list comprehension. The inner list is a loop over all the keys, checking in a given passport to see if they are in there:

all([k in p for k in keys]

If all seven keys are there, this is true.

Now I’ll loop over the passports and count the numbers of true:

num_keys_valid = sum([all([k in p for k in keys]) for p in passports])

Putting that into a script, it returns the answer:

#!/usr/bin/env python3

import sys


keys = [
    "byr",  # (Birth Year)
    "iyr",  # (Issue Year)
    "eyr",  # (Expiration Year)
    "hgt",  # (Height)
    "hcl",  # (Hair Color)
    "ecl",  # (Eye Color)
    "pid",  # (Passport ID)
    # "cid", # (Country ID)
]

with open(sys.argv[1], "r") as f:
    passports = f.read().split("\n\n")


num_keys_valid = sum([all([k in p for k in keys]) for p in passports])
print(f"Part 1: {num_keys_valid}")

Part 2

At first I thought my part one skim of the data like this would have to completely change structure for part two, reading each passport into a dictionary and applying validity rules. However, as I started thinking about it, each field’s rules could be expressed as a regex, and so I turned the keys from part 1 into a dict with a regex string for each key, and did a similar loop.

What got me really stuck for longer than I’m proud to admit was that I wasn’t checking for the end of the string. The thing that was getting me was my original regex for pid:

r"pid:\s*\d{9}"

This found all the nine-digit pids, but it also found a 10 digit one, which isn’t valid. I’ll add a \b to the end, which matches on any whitespace or a newline, and that fixes it (I should have \b on most of these):

#!/usr/bin/env python3

import re
import sys


keys = {
    "byr": r"byr:\s*(19[2-9]\d|200[0-2])\b",  # (Birth Year) - four digits; at least 1920 and at most 2002.
    "iyr": r"iyr:\s*20(1\d|20)\b",  # (Issue Year) - four digits; at least 2010 and at most 2020.
    "eyr": r"eyr:\s*20(2\d|30)\b",  # (Expiration Year) - four digits; at least 2020 and at most 2030.
    "hgt": r"hgt:\s*(1([5-8]\d|9[0-3])cm|(59|6\d|7[0-6])in)",  # (Height) - a number followed by either cm or in:
    # "If cm, the number must be at least 150 and at most 193.
    # "If in, the number must be at least 59 and at most 76.
    "hcl": r"hcl:\s*#[0-9a-f]{6}\b",  # (Hair Color) - a # followed by exactly six characters 0-9 or a-f.
    "ecl": r"ecl:\s*(amb|blu|brn|gry|grn|hzl|oth)\b",  # (Eye Color) - exactly one of: amb blu brn gry grn hzl oth.
    "pid": r"pid:\s*\d{9}\b",  # (Passport ID) - a nine-digit number, including leading zeroes.
    # "cid", # (Country ID) - ignored, missing or not.
}

with open(sys.argv[1], "r") as f:
    passports = f.read().split("\n\n")

num_keys_valid = sum([all([k in p for k in keys]) for p in passports])
print(f"Part 1: {num_keys_valid}")

num_data_valid = sum([all([re.search(keys[k], p) for k in keys]) for p in passports])
print(f"Part 2: {num_data_valid}")

Running this gives both answers (still instantly):

$ time python3 day4.py 04-puzzle_input.txt
Part 1: 245
Part 2: 133

real    0m0.065s
user    0m0.053s
sys     0m0.008s

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