UP / HOME

Day 02 - Password Philosophy

Table of Contents

Puzzle

Your flight departs in a few days from the coastal airport; the easiest way down to the coast from here is via toboggan.

The shopkeeper at the North Pole Toboggan Rental Shop is having a bad day. "Something's wrong with our computers; we can't log in!" You ask if you can take a look.

Their password database seems to be a little corrupted: some of the passwords wouldn't have been allowed by the Official Toboggan Corporate Policy that was in effect when they were chosen.

To try to debug the problem, they have created a list (your puzzle input) of passwords (according to the corrupted database) and the corporate policy when that password was set.

For example, suppose you have the following list:

1-3 a: abcde
1-3 b: cdefg
2-9 c: ccccccccc

Each line gives the password policy and then the password. The password policy indicates the lowest and highest number of times a given letter must appear for the password to be valid. For example, 1-3 a means that the password must contain a at least 1 time and at most 3 times.

In the above example, 2 passwords are valid. The middle password, cdefg, is not; it contains no instances of b, but needs at least 1. The first and third passwords are valid: they contain one a or nine c, both within the limits of their respective policies.

How many passwords are valid according to their policies?

Part 2

While it appears you validated the passwords correctly, they don't seem to be what the Official Toboggan Corporate Authentication System is expecting.

The shopkeeper suddenly realizes that he just accidentally explained the password policy rules from his old job at the sled rental place down the street! The Official Toboggan Corporate Policy actually works a little differently.

Each policy actually describes two positions in the password, where 1 means the first character, 2 means the second character, and so on. (Be careful; Toboggan Corporate Policies have no concept of "index zero"!) Exactly one of these positions must contain the given letter. Other occurrences of the letter are irrelevant for the purposes of policy enforcement.

Given the same example list from above:

  • 1-3 a: abcde is valid: position 1 contains a and position 3 does not.
  • 1-3 b: cdefg is invalid: neither position 1 nor position 3 contains b.
  • 2-9 c: ccccccccc is invalid: both position 2 and position 9 contain c.

How many passwords are valid according to the new interpretation of the policies?

Solution

Raku grammar is used to parse the file. I took the grammar from tyil's solution.

# Password grammar was taken from tyil's solution.
grammar Password {
    rule TOP { <num> '-' <num> <letter> ':' <password> }
    token num { \d+ }
    token letter { \w }
    token password { \w+ }
}

The program takes a positional input $part which instructs it on which part it should solve. The default is to solve for part 1.

$valid_passwords stores the number of passwords that are considered valid. The checks are made inside the if/elsif loop & it continues from there only if the password is considered valid.

sub MAIN (
    # Part to run.
    Int $part where * == 1|2 = 1
) {
    my $valid_passwords = 0;

    for "input".IO.lines -> $entry {
        if Password.parse($entry) -> $match {
            if $part == 1 {
                ...
            } elsif $part == 2 {
                ...
            }
            # All checks were completed & this is a valid password.
            $valid_passwords++;
        }
    }
    say "Part $part: " ~ $valid_passwords;
}

Nothing fancy is done, we just count the number of times each character is used & then perform the check.

my %chars;
$match<password>.comb.map({ %chars{$_}++ });

# Check if the letter exists in the password.
next unless %chars{$match<letter>};

# Check for length.
next unless $match<num>[0].Int <= %chars{$match<letter>};
next unless %chars{$match<letter>} <= $match<num>[1].Int;

Part 2

This part too just performs the check.

elsif $part == 2 {
    my $combed = $match<password>.comb;
    my $first = $match<num>[0].Int - 1;
    my $second = $match<num>[1].Int - 1;

    next if $combed[$first] eq $combed[$second];
    next unless $combed[$first] eq $match<letter> || $combed[$second] eq $match<letter>;
}

Andinus / / Modified: 2020-12-04 Fri 20:42 Emacs 27.1 (Org mode 9.3)