Advent of Code 2022 Day 2

Challenge – Part 1

https://adventofcode.com/2022/day/2

Example

For example, suppose you were given the following strategy guide:

A Y
B X
C Z

This strategy guide predicts and recommends the following:

• In the first round, your opponent will choose Rock (A), and you should choose Paper (Y). This ends in a win for you with a score of 8 (2 because you chose Paper + 6 because you won).
• In the second round, your opponent will choose Paper (B), and you should choose Rock (X). This ends in a loss for you with a score of 1 (1 + 0).
• The third round is a draw with both players choosing Scissors, giving you a score of 3 + 3 = 6.

In this example, if you were to follow the strategy guide, you would get a total score of 15 (8 + 1 + 6).

Solution

The first step is to take in the sample data and split at the newline to get the individual rounds:

data = """A Y
B X
C Z"""

def test_split_data():
    rounds = data.split("\n")
    assert rounds == ['A Y', 'B X', 'C Z']

Than I came up with the idea to create a dictionary with all possible constellations and the appropriate scores:

strategy_dict = {
    'A X': 4,  # 1 + 3
    'A Y': 8,  # 2 + 6
    'A Z': 3,  # 3 + 0
    'B X': 1,  # 1 + 0
    'B Y': 5,  # 2 + 3
    'B Z': 9,  # 3 + 6
    'C X': 7,  # 1 + 6
    'C Y': 2,  # 2 + 0
    'C Z': 6,  # 3 + 3
}

After that we iterate over all rounds and sum up the scores. For the sample data it worked quite nicely like this:

def test_lookup_strategy_dict():
    score = 0
    rounds = data.split("\n")
    for round in rounds:
        score += strategy_dict[round]
    assert score == 15

And now the final conclusion. We read in the data from a file aoc_data_02.txt and apply the code from above:

def test_data_from_file_lookup_strategy_dict():
    with open("aoc_data_02.txt", "r") as f:
        data = f.read()

    score = 0
    rounds = data.split("\n")
    for round in rounds:
        score += strategy_dict[round]
    assert score == 11063

This did the trick! Nice 🙂

Challenge – Part 2

Now we are confronted with a new calculation of the scores

In the first round, your opponent will choose Rock (A), and you need the round to end in a draw (Y), so you also choose Rock. This gives you a score of 1 + 3 = 4.
In the second round, your opponent will choose Paper (B), and you choose Rock so you lose (X) with a score of 1 + 0 = 1.
In the third round, you will defeat your opponent’s Scissors with Rock for a score of 1 + 6 = 7.

Solution

The approach with a dictionary for lookup paid out. By altering the scores for every combination of moves we can reuse the code from part 1

strategy_dict = {
    'A X': 3,  # 3 + 0
    'A Y': 4,  # 1 + 3
    'A Z': 8,  # 2 + 6
    'B X': 1,  # 1 + 0
    'B Y': 5,  # 2 + 3
    'B Z': 9,  # 3 + 6
    'C X': 2,  # 2 + 0
    'C Y': 6,  # 3 + 3
    'C Z': 7,  # 1 + 6
}

def test_data_from_file_lookup_strategy_dict():
    with open("aoc_data_02.txt", "r") as f:
        data = f.read()

    score = 0
    rounds = data.split("\n")
    for round in rounds:
        score += strategy_dict[round]
    assert score == 10349

Et voilá! Day 2 is done