Day 2

The input

Each line is a different game. Each game contains draws separated by semicolon. Each draw contains number of cubes that were selected in each color: red, green, or blue.

input_file = open("../../DATA/2023/input_day02.txt", "r");
#input_file = open("DATA/2023/example_input_day02.txt", "r");
peek(input_file, String)

"Game 1: 4 blue, 16 green, 2 red; 5 red, 11 blue, 16 green; 9 green, 11 blue; 10 blue, 6 green, 4 red\nGame 2: 15 green, 20 red, 8 blue; 12 green, 7 red; 10 green, 2 blue, 15 red; 13 blue, 15 red\nGame 3: 8 red, 2 blue; 3 green, 10 blue, 10 red; 7 green, 4 blue, 7 red; 8 r" ⋯ 10303 bytes ⋯ "98: 4 blue, 9 red; 10 red, 1 green, 11 blue; 7 blue, 1 red; 1 red, 6 blue, 1 green\nGame 99: 7 red, 6 green, 2 blue; 8 red; 16 green, 7 red, 4 blue\nGame 100: 1 red, 1 green, 9 blue; 6 blue, 4 green, 3 red; 4 red, 2 green; 3 green, 2 red, 11 blue; 6 green, 5 blue, 1 red\n"

Part 1

The problem

Determine which games would have been possible if the bag had been loaded with only 12 red cubes, 13 green cubes, and 14 blue cubes.

First, let’s create a dataset with all draws.

all_games = DataFrame(
    game_no = 0,
    colors = "",
    draws = 0
);

We will also need a function to find these different numbers and colors.

function find_colors_numbers(draws::String, color::String)
    out_num_cubes = Int32[];
    idx = 1;
    while idx < lastindex(draws)
        m = match(
            Regex(" (?<num_cubes>\\d+) $color"),
            draws,
            idx
        );
        if isnothing(m)
            break
        end
        push!(out_num_cubes, parse(Int32, m["num_cubes"]));
        idx = m.offset + length(m.match);
    end
    return out_num_cubes
end

find_colors_numbers (generic function with 1 method)

all_colors = ["red", "green", "blue"];

line = readline(input_file);
while line != ""
    split_first = split(line, ":");
    game_no = split(
        first(split_first), " "
    ) |>
    x -> parse(Int32, last(x));
    draws = map(
        find_colors_numbers,
        [line, line, line],
        all_colors
    );
    cur_games = DataFrame(
        game_no = game_no,
        colors = all_colors,
        draws = draws
    ) |>
        x -> flatten(x, :draws);
    append!(all_games, cur_games);

    # read the next line:
    global line = readline(input_file);
end

close(input_file)

The solution

Let’s check which draws gave more than 12 red cubes, 13 green cubes, and 14 blue cubes.

all_no_draws = [12, 13, 14];

function get_games_not_possible(color, no_draws)
    cur_subset = subset(
        all_games,
        :colors => col -> col .== color,
        :draws => dr -> dr .> no_draws
    );
    return unique(cur_subset.game_no)
end

get_games_not_possible (generic function with 1 method)

all_games_ids_not_possible = map(
        get_games_not_possible,
        all_colors,
        all_no_draws
    ) |>
        x -> unique(vcat(x...));
all_games_ids_possible = setdiff(unique(all_games.game_no), all_games_ids_not_possible);

The sum of these numbers is 2685.

Part 2

The problem

What is the minimum number of cubes that makes each game possible?

This means, that I need to find the minimum number in each game, for each color.

min_cubes_all_games = combine(
    groupby(all_games, [:game_no, :colors]),
    :draws => (x -> max(x...)) => :max_draw
)

301×3 DataFrame

276 rows omitted

Row	game_no	colors	max_draw
	Int64	String	Int64
1	0		0
2	1	red	5
3	1	green	16
4	1	blue	11
5	2	red	20
6	2	green	15
7	2	blue	13
8	3	red	10
9	3	green	7
10	3	blue	13
11	4	red	14
12	4	green	13
13	4	blue	18
⋮	⋮	⋮	⋮
290	97	red	4
291	97	green	1
292	97	blue	15
293	98	red	10
294	98	green	1
295	98	blue	11
296	99	red	8
297	99	green	16
298	99	blue	4
299	100	red	4
300	100	green	6
301	100	blue	11

The solution

Then, for each game, multiply the minimal numbers and sum up.

powers_per_game = combine(
    groupby(min_cubes_all_games, :game_no),
    :max_draw => (x -> reduce(*, x)) => :power
)

101×2 DataFrame

76 rows omitted

Row	game_no	power
	Int64	Int64
1	0	0
2	1	880
3	2	3900
4	3	910
5	4	3276
6	5	216
7	6	198
8	7	384
9	8	40
10	9	728
11	10	288
12	11	1980
13	12	70
⋮	⋮	⋮
90	89	2250
91	90	51
92	91	864
93	92	1848
94	93	18
95	94	378
96	95	840
97	96	810
98	97	60
99	98	110
100	99	512
101	100	264

The sum of these numbers is 83707.