Day 04 - Passport Processing
Puzzle
- This puzzle is taken from: https://adventofcode.com/2020/day/4
You arrive at the airport only to realize that you grabbed your North Pole Credentials instead of your passport. While these documents are extremely similar, North Pole Credentials aren't issued by a country and therefore aren't actually valid documentation for travel in most of the world.
It seems like you're not the only one having problems, though; a very long line has formed for the automatic passport scanners, and the delay could upset your travel itinerary.
Due to some questionable network security, you realize you might be able to solve both of these problems at the same time.
The automatic passport scanners are slow because they're having trouble detecting which passports have all required fields. The expected fields are as follows:
byr
(Birth Year)iyr
(Issue Year)eyr
(Expiration Year)hgt
(Height)hcl
(Hair Color)ecl
(Eye Color)pid
(Passport ID)cid
(Country ID)
Passport data is validated in batch files (your puzzle input). Each passport is represented as a sequence of key:value pairs separated by spaces or newlines. Passports are separated by blank lines.
Here is an example batch file containing four passports:
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
The first passport is valid - all eight fields are present. The second passport is invalid - it is missing hgt (the Height field).
The third passport is interesting; the only missing field is cid, so it looks like data from North Pole Credentials, not a passport at all! Surely, nobody would mind if you made the system temporarily ignore missing cid fields. Treat this "passport" as valid.
The fourth passport is missing two fields, cid and byr. Missing cid is fine, but missing any other field is not, so this passport is invalid.
According to the above rules, your improved system would report 2 valid passports.
Count the number of valid passports - those that have all required fields. Treat cid as optional. In your batch file, how many passports are valid?
Part 2
The line is moving more quickly now, but you overhear airport security talking about how passports with invalid data are getting through. Better add some data validation, quick!
You can continue to ignore the cid field, but each other field has strict rules about what values are valid for automatic validation:
byr
(Birth Year) - four digits; at least 1920 and at most 2002.iyr
(Issue Year) - four digits; at least 2010 and at most 2020.eyr
(Expiration Year) - four digits; at least 2020 and at most 2030.hgt
(Height) - a number followed by eithercm
orin
: Ifcm
, the number must be at least150
and at most193
. Ifin
, the number must be at least59
and at most76
.hcl
(Hair Color) - a#
followed by exactly six characters0-9
ora-f
.ecl
(Eye Color) - exactly one of:amb
blu
brn
gry
grn
hzl
oth
.pid
(Passport ID) - a nine-digit number, including leading zeroes.cid
(Country ID) - ignored, missing or not.
Your job is to count the passports where all required fields are both present and valid according to the above rules. Here are some example values:
byr valid: 2002 byr invalid: 2003 hgt valid: 60in hgt valid: 190cm hgt invalid: 190in hgt invalid: 190 hcl valid: #123abc hcl invalid: #123abz hcl invalid: 123abc ecl valid: brn ecl invalid: wat pid valid: 000000001 pid invalid: 0123456789
Here are some invalid passports:
eyr:1972 cid:100 hcl:#18171d ecl:amb hgt:170 pid:186cm iyr:2018 byr:1926 iyr:2019 hcl:#602927 eyr:1967 hgt:170cm ecl:grn pid:012533040 byr:1946 hcl:dab227 iyr:2012 ecl:brn hgt:182cm pid:021572410 eyr:2020 byr:1992 cid:277 hgt:59cm ecl:zzz eyr:2038 hcl:74454a iyr:2023 pid:3556412378 byr:2007
Here are some valid passports:
pid:087499704 hgt:74in ecl:grn iyr:2012 eyr:2030 byr:1980 hcl:#623a2f eyr:2029 ecl:blu cid:129 byr:1989 iyr:2014 pid:896056539 hcl:#a97842 hgt:165cm hcl:#888785 hgt:164cm byr:2001 iyr:2015 cid:88 pid:545766238 ecl:hzl eyr:2022 iyr:2010 hgt:158cm hcl:#b6652a ecl:blu byr:1944 eyr:2021 pid:093154719
Count the number of valid passports - those that have all required fields and valid values. Continue to treat cid as optional. In your batch file, how many passports are valid?
Solution
The input file is split by \n\n
& stored in @passports
. We iterate over
@passports
& seperate each field in %field
. Then we check if all
required fields are present, if not then skip the passport. If true then
increment $valid_passports
which holds the number of valid passports.
my $input = slurp "input"; my @passports = $input.split("\n\n"); my $valid_passports = 0; MAIN: for @passports -> $passport { my %fields; for $passport.words -> $field { my ($key, $value) = $field.split(":"); %fields{$key} = $value; } # Check for fields that are strictly required. `cid' can be # skipped. Skip this passport if it's not valid. for <byr iyr eyr hgt hcl ecl pid> -> $field { next MAIN unless %fields{$field}; } # Do validation in part 2. if $part == 2 { ... } $valid_passports++; } say "Part $part: " ~ $valid_passports;
Part 2
For part 2, more validation is performed. The rules are simple & well-defined in Puzzle section.
I had forgotten the default
block in given/when
& that meant it was
counting passports which don't have hgt
set to either cm
or in
as valid.
In my case it was a single passport.
next MAIN unless ( (1920 ≤ %fields{<byr>} ≤ 2002) and (2010 ≤ %fields{<iyr>} ≤ 2020) and (2020 ≤ %fields{<eyr>} ≤ 2030) and (<amb blu brn gry grn hzl oth> ∋ %fields{<ecl>}) and (%fields{<pid>} ~~ /^\d ** 9$/) and (%fields{<hcl>} ~~ /^'#' <[\d a..f]> ** 6/) ); given substr(%fields{<hgt>}, *-2) { when 'cm' { next MAIN unless 150 ≤ substr(%fields{<hgt>}, 0, *-2) ≤ 193; } when 'in' { next MAIN unless 59 ≤ substr(%fields{<hgt>}, 0, *-2) ≤ 76; } default { next MAIN; } }