More efficient choice comparison for Rock Paper Scissors

Question

This is an ongoing school project that I would like to improve. The point is to make the code as efficient (or short) as possible. I would like to reduce it by finding an al

Answer 1

Quite late to the party, but here's a slightly different way to the other answers in here. I've included the answer in TypeScript, but you can obviously remove the types if you like.

type Weapon = "Rock" | "Paper" | "Scissors";
type Result = "Player" | "Computer" | "Draw";
type GameLogic = Record>;

// Map that tracks what each weapon beats.
const outcomes: GameLogic = {
  Rock: ["Scissors"],
  Paper: ["Rock"],
  Scissors: ["Paper"]
};

const determineWinner = (playerOneWeapon: Weapon, playerTwoWeapon: Weapon): Result => {
  if (playerOneWeapon === playerTwoWeapon) {
    return "Draw";
  }

  return outcomes[playerOneWeapon].includes(playerTwoWeapon)
    ? "Player One"
    : "Player Two";
}

This implementation has the ability to scale nicely, too. As you can add extra weapons into the mix and the implementation of determineWinner doesn't change - by adding to the Weapon type and the outcomes map:

type Weapon = "Rock" | "Paper" | "Scissors" | "Lizard" | "Spock";
const outcomes: GameLogic = {
  Rock: ["Scissors", "Lizard"],
  Paper: ["Rock", "Spock"],
  Scissors: ["Paper", "Lizard"],
  Lizard: ["Spock", "Paper"],
  Spock: ["Rock", "Scissors"]
};

We can now support a game where each weapon beats each weapon exactly twice and each weapon loses to a weapon exactly twice (as opposed to the classic variation where everything beats one weapon and loses to one weapon).