DEV Community

Cover image for AoC Day 2 - Rock Paper Scissors
Gal Elmalah
Gal Elmalah

Posted on • Updated on • Originally published at galelmalah.com

AoC Day 2 - Rock Paper Scissors

Rock Paper Scissors πŸͺ¨πŸ§»βœ‚

Question

The Elves begin to set up camp on the beach. To decide whose tent gets to be closest to the snack storage, a giant Rock Paper Scissors tournament is already in progress.

God the question's description is always pure gold...now let's jump right in.

We are given a string, our "strategy guide" that represents a rock, paper, scissors turn decision.
The string is divided into two columns, the first represents the first player's decision and the second ours.
For example:

A Y
B X
C Z
Enter fullscreen mode Exit fullscreen mode

We are also told the following:
For player 1 - A = Rock, B = Paper, C = Scissors
For player 2 - X = Rock, Y = Paper, Z = Scissors
We are also given a score table for the outcome and the decision we made, for example, a win is 6 points and choosing paper is 2 points, etc...

Like previous questions, we'll start by parsing our input

Parsing

We can go (see what I did there?) with several options to represent our game but in my opinion, an array of tuples is the simplest option

func parse(raw string) [][]string {
    chunks := strings.Split(string(raw), "\n")
    pairs := make([][]string, len(chunks))
    for i := range pairs {
        pairs[i] = strings.Split(chunks[i], " ")
    }

    return pairs
}
// example output
// [ [A, Y], [B, X], [C, Z] ]
Enter fullscreen mode Exit fullscreen mode

The make function allocates a piece of memory in a certain, specified size for our array

Part 1

We are asked to provide our total score if we play exactly as instructed in the strategy guide.
Let's think about this for a bit, there are several ways we can solve this, we can use a bunch of if statements or some fancy pattern matching, since go does not have pattern matching and I don't want to write a ton of if statements we will go with a hybrid approach.
We will create 3 different mappings:

  1. Represents the points we get for our choice e.g rock, paper, or scissors
  2. Winning state, meaning If we choose X what does the other player need to choose for us to Win
  3. Tie state, essentially the same as .2
scores := map[string]int{
    "X": 1,
    "Y": 2,
    "Z": 3,
}

// If I choose X(Rock) I need him to choose C(scissors) in order to win
win := map[string]string{
    "X": "C", 
    "Y": "A",
    "Z": "B",
}

tie := map[string]string{
    "X": "A",
    "Y": "B",
    "Z": "C",
}
Enter fullscreen mode Exit fullscreen mode

We don't take into account the losing state since its essentially a no-op (0 points)

Building on top of these maps and our parsing logic, we can now solve the first part with the following code

func part1(raw []byte) int {

    pairs := parse(string(raw))

    // X Rock, Y Paper, Z Scissors
    scores := map[string]int{
        "X": 1,
        "Y": 2,
        "Z": 3,
    }

    win := map[string]string{
        "X": "C",
        "Y": "A",
        "Z": "B",
    }

    tie := map[string]string{
        "X": "A",
        "Y": "B",
        "Z": "C",
    }

    score := 0
    for _, pair := range pairs {
        his := pair[0]
        my := pair[1]
        score += scores[my]
        if win[my] == his {
            score += WINNING_POINTS
        }

        if tie[my] == his {
            score += TIE_POINTS
        }

    }
    return score

}
// output for part 1 based on the example is
// 15 -> (8 + 1 + 6)
Enter fullscreen mode Exit fullscreen mode

At each loop iteration we first add the points based on our choice score += scores[my] then we check if his move is what we need based on our player choice, to win or get a tie, and if it is we add the necessary points to our total score.

Part 2

In part two the sneaky elves switch things up a bit.
Instead of our column representing our moves, it represents the turn outcome where X = lose, Y = tie, and Z = win and we need to choose our choice accordingly.
For example, let's look at the first turn A Y, the new meaning of this pair is "player one chose Rock, and the game ended in a tie" building on this information we can create new mappings, the new mappings will be between player 1 choice and the choice player 2 need to make to get to a certain state e.g winning, losing, tie, etc...
Since it's pretty similar to part 1, we will jump right ahead and look at part 2 as a whole

func part2(raw []byte) int {

    var pairs = parse(string(raw))

    // X Lose, Y Tie, Z Win
    scores := map[string]int{
        "X": 1,
        "Y": 2,
        "Z": 3,
    }

    win := map[string]string{
        "C": "X",
        "A": "Y",
        "B": "Z",
    }

    tie := map[string]string{
        "A": "X",
        "B": "Y",
        "C": "Z",
    }

    lose := map[string]string{
        "A": "Z",
        "B": "X",
        "C": "Y",
    }
    score := 0

    for _, pair := range pairs {
        hisMove := pair[0]
        myMove := pair[1]

                // we lose
        if myMove == "X" {
            score += scores[lose[hisMove]]
        }
                // we end in a tie
        if myMove == "Y" {
            score += TIE_POINTS
            score += scores[tie[hisMove]]
        }
                // we win
        if myMove == "Z" {
            score += WINNING_POINTS
            score += scores[win[hisMove]]
        }
    }
    return score
}
Enter fullscreen mode Exit fullscreen mode

For each desired state we check what move we need to do based on player 2 choice and pass it down to the scores map.

That's it we are all done with paper, rock, scissors and I must admit that I didn't think it can be so confusing 🀣

You can find the complete code here
Thanks for reading!

Top comments (0)