The Rock-Paper-Scissor logic

December 19, 2022 — #AdventOfCode #Java

How would you go about solving this problem?

You have write a method to check wether you win, lose or tie in a game of Rock-Paper-Scissor, based on the choice of you and your opponent.
The method should return 2 if you win, 1 if you tie and 0 if you lose.
And we can use only two if statements and no 'or' or 'and' operators

Lets use some characters and values to represent the choices and outcomes:
Rock = A | Paper = B | Scissor = C
Win = 6 | Tie = 3 | Loss = 0

Here is the program you need to make changes to, replace the places where there's ... with only a single line of your code.

RockPaperScissor.java

public class RockPaperScissor {
    ...
    public static int checkWinner(char player, char opponent) {
        if(...) return 3;
        ...
        if(...) return 6;
        return 0;
    }
}

Try to find the solution yourself before you read the next section.

Need a hint?

RockPaperScissor.java

public class RockPaperScissor {
    //Use an array of characters to represent the choices    static char[] choices = {'C', 'A', 'B', 'C'};    public static int checkWinner(char player, char opponent) {
        if(...) return 3;
        ...
        if(...) return 6;
        return 0;
    }
}

Need one more hint?

Look at this pattern
Whenever the player wins,
these are the plays:
player: A beats C :opponent
player: B beats A :opponent
player: C beats B :opponent

If we organize this a bit differently, we can find a pattern
player choice
A B C A
C A B C <- the array we created
opponent choice

Notice how whenever we win, the opponent's choice is always one value behind of the player's choice in the array.
ie.
when the player choice is B, the opponent's choice is A
when the player choice is A, the opponent's choice is C, because C is one value behind A in the array.
this is the extra element at the start.

Here's my solution to the problem-

RockPaperScissor.java

public class RockPaperScissor {
    static char[] choices = {'C', 'A', 'B', 'C'};
    public static int checkWinner(char player, char opponent) {
        //In case of a tie
        if(player == opponent) return 3;
        //In case of a win
        if(choices[player - 'A'] == opponent)  return 6;
        //In case of a loss
        return 0;
    }
}

Okay, so what did I do here?
I used the choices array to store the choices.
I used the player - 'A' + 1 to get the index of the player's choice in the choices array.
How does that work you ask?
Well, the ASCII value of A is 65 and the ASCII value of C is 67.
So, when the player gives a character input from A to C, we actually have a value from 65 to 67.
In order to find this value in the choices array, we need to subtract 65 from the player input.
So we get a range from 0-2.
But the choices array starts from 1 because of the extra elements on both the ends.
So we add 1 to the value.

Now since we know that we can only win if the opponent's choice is one value behind the player's choice, we can use the choices array to check if the player wins or not.
So to check for the win, we can simply use choices[player - 'A' + 1 - 1] == opponent (simplified to choices[player - 'A'] == opponent)

And that's how you solve this in just three lines of code./ Thanks for reading :))))))

This is from one of the problems in the Advent of Code 2022.