# Day 07 - Handy Haversacks

## Puzzle

You land at the regional airport in time for your next flight. In fact, it looks like you'll even have time to grab some food: all flights are currently delayed due to issues in luggage processing.

Due to recent aviation regulations, many rules (your puzzle input) are being enforced about bags and their contents; bags must be color-coded and must contain specific quantities of other color-coded bags. Apparently, nobody responsible for these regulations considered how long they would take to enforce!

For example, consider the following rules:

```light red bags contain 1 bright white bag, 2 muted yellow bags.
dark orange bags contain 3 bright white bags, 4 muted yellow bags.
bright white bags contain 1 shiny gold bag.
muted yellow bags contain 2 shiny gold bags, 9 faded blue bags.
shiny gold bags contain 1 dark olive bag, 2 vibrant plum bags.
dark olive bags contain 3 faded blue bags, 4 dotted black bags.
vibrant plum bags contain 5 faded blue bags, 6 dotted black bags.
faded blue bags contain no other bags.
dotted black bags contain no other bags.
```

These rules specify the required contents for 9 bag types. In this example, every faded blue bag is empty, every vibrant plum bag contains 11 bags (5 faded blue and 6 dotted black), and so on.

You have a shiny gold bag. If you wanted to carry it in at least one other bag, how many different bag colors would be valid for the outermost bag? (In other words: how many colors can, eventually, contain at least one shiny gold bag?)

In the above rules, the following options would be available to you:

• A bright white bag, which can hold your shiny gold bag directly.
• A muted yellow bag, which can hold your shiny gold bag directly, plus some other bags.
• A dark orange bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
• A light red bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.

So, in this example, the number of bag colors that can eventually contain at least one shiny gold bag is 4.

How many bag colors can eventually contain at least one shiny gold bag? (The list of rules is quite long; make sure you get all of it.)

### Part 2

It's getting pretty expensive to fly these days - not because of ticket prices, but because of the ridiculous number of bags you need to buy!

Consider again your shiny gold bag and the rules from the above example:

• faded blue bags contain 0 other bags.
• dotted black bags contain 0 other bags.
• vibrant plum bags contain 11 other bags: 5 faded blue bags and 6 dotted black bags.
• dark olive bags contain 7 other bags: 3 faded blue bags and 4 dotted black bags.

So, a single shiny gold bag must contain 1 dark olive bag (and the 7 bags within it) plus 2 vibrant plum bags (and the 11 bags within each of those): 1 + 1*7 + 2 + 2*11 = 32 bags!

Of course, the actual rules have a small chance of going several levels deeper than this example; be sure to count all of the bags, even if the nesting becomes topologically impractical!

Here's another example:

```shiny gold bags contain 2 dark red bags.
dark red bags contain 2 dark orange bags.
dark orange bags contain 2 dark yellow bags.
dark yellow bags contain 2 dark green bags.
dark green bags contain 2 dark blue bags.
dark blue bags contain 2 dark violet bags.
dark violet bags contain no other bags.
```

In this example, a single shiny gold bag must contain 126 other bags.

How many individual bags are required inside your single shiny gold bag?

## Solution

`%bag` holds the information for each bag specified. It holds the number of bags a bag holds along with their names. `\$count` holds the count, it holds the solution to the puzzle & will be printed in the end. `@valid_bags` holds the name of bags, this is used later in part 1.

```my %bag;
my Int \$count = 0;
my @valid_bags = "shiny gold",;
```

Iterate over each line of the file & add the bag information to `%bag`.

```for "input".IO.lines -> \$entry {
if \$entry ~~ /^(.*) \s bags \s contain (.*)./ -> \$match {
my \$bag = \$match[0];
for \$match[1].split(", ") {
if \$_ ~~ /(\d) \s (.*) \s bag/ -> \$contain_bag {
push %bag{\$bag}, {
count => \$contain_bag[0].Int,
bag => \$contain_bag[1].Str
};
}
}
}
}
```

For part 1, we wrap the whole thing in a while loop that runs until `\$previous_count` is equal to `\$count`. The first `for` loop within the `MAIN` loop is pushing bags that are to valid to `@valid_bags`. Valid bags are the bags that can contain at least one `shiny gold` bag.

Take the case where `bag 1` holds `shiny gold` bag & `bag 2` holds `bag 1`. This means `bag 2` can also hold a `shiny gold` bag by holding `bag 1` so it should be included in the count. Now the reason we have the `MAIN` loop is that if the statement that says `bag 2 holds bag 1` comes before the statement that says `bag 1 holds shiny gold` then `bag 2` is not counted in `@valid_bags`.

This causes the `\$count` to be less than the actual count, to account for this we just loop indefinitely until `\$previous_count` is equal to `\$count`. `\$previous_count` holds the value of `\$count` at the start of `MAIN` loop. If the value of `\$count` isn't affected then it means that we've accounted for all valid bags & `\$count` holds the final solution.

```if \$part == 1 {
MAIN: while True {
my \$previous_count = \$count;
for keys %bag {
for @(%bag{\$_}) -> \$contain {
push @valid_bags, \$_ if @valid_bags ∋ \$contain<bag>;
}
}

\$count = 0;
COUNT: for keys %bag {
for @(%bag{\$_}) -> \$contain {
if @valid_bags ∋ \$contain<bag> {
\$count++;
next COUNT;
}
}
}
last MAIN if \$previous_count == \$count;
}
} elsif \$part == 2 {
...
}
say "Part \$part: ", \$count;
```

### Part 2

For part 2, we will use `%bag` which was previous defined. And we define a subroutine: `count-bags`. `count-bags` counts the number of bags that some `\$bag` holds. We have to pass `%bag` & `\$bag` to `count-bags`. It'll also account for sub bags, it's an recursive function.

The subroutine is easy to understand, if the bag doesn't hold any other bags then we return 0. If it holds other bags then we count the number of bags it's holding & add it to `\$count`. Later we count the number of bags these sub bags are holding, if it's 0 then the loop exits, if not then we multiply the number of sub bags to the number of bags those sub bags are holding and add it to `\$count`.

This will be easier to understand with an example so take a sample input & follow the code manually.

```# count-bags takes %bag & the bag for which we have to count the bags
# & returns the count. Count here means the number of bags that will
# be inside \$bag.
sub count-bags (
%bag, Str \$bag --> Int # count will be an integer.
) {
return 0 unless %bag{\$bag};
my Int \$count = 0;

for @(%bag{\$bag}) {
my \$sub_bags = count-bags(%bag, \$_<bag>);
\$count += \$_<count>;
unless \$sub_bags == 0 {
\$count += \$_<count> * \$sub_bags;
}
}
return \$count;
}
```

We just run `count-bags` on 'shiny gold' & print the `\$count`.

```\$count = count-bags(%bag, 'shiny gold');
```

Andinus / / Modified: 2020-12-07 Mon 17:11 Emacs 27.2 (Org mode 9.4.4)