AOC 22 Day 2 🪨📄✂️

Problem🤔

Rock Paper Scissors: 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.

Rock Paper Scissors is a game between two players. Each game contains many rounds; in each round, the players each simultaneously choose one of Rock, Paper, or Scissors using a hand shape. Then, a winner for that round is selected: Rock defeats Scissors, Scissors defeats Paper, and Paper defeats Rock. If both players choose the same shape, the round instead ends in a draw.

Appreciative of your help yesterday, one Elf gives you an encrypted strategy guide (your puzzle input) that they say will be sure to help you win. “The first column is what your opponent is going to play: A for Rock, B for Paper, and C for Scissors. The second column—” Suddenly, the Elf is called away to help with someone’s tent.

The second column, you reason, must be what you should play in response: X for Rock, Y for Paper, and Z for Scissors. Winning every time would be suspicious, so the responses must have been carefully chosen.

The winner of the whole tournament is the player with the highest score. Your total score is the sum of your scores for each round. The score for a single round is the score for the shape you selected (1 for Rock, 2 for Paper, and 3 for Scissors) plus the score for the outcome of the round (0 if you lost, 3 if the round was a draw, and 6 if you won).

Sample Data

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

A Y
B X
C Z

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).

Part 1

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

What would your total score be if everything goes exactly according to your strategy guide?

Solution 2 1💡

import { input } from './input.ts';
 
type opKey = keyof typeof opp;
type meKey = keyof typeof me;
const opp = {
  A: 1, //🪨
  B: 2, //📄
  C: 3, //✂️
};
const me = {
  X: 1, //🪨
  Y: 2, //📄
  Z: 3, //✂️
};
 
const winScore = (oKey: opKey, mKey: meKey) => {
  if (opp[oKey] === me[mKey]) return 3;
 
  if (opp[oKey] < me[mKey]) {
    // 🪨 🆚 ✂️
    if (opp[oKey] === 1 && me[mKey] === 3) return 0;
    return 6;
  }
  if (opp[oKey] > me[mKey]) {
    // 🪨 🆚 ✂️
    if (me[mKey] === 1 && opp[oKey] === 3) return 6;
    return 0;
  }
  throw new Error(oKey + ' | ' + mKey);
};
 
const roundScore = (oKey: opKey, mKey: meKey) => {
  const baseScore = me[mKey];
  const winPoints = winScore(oKey, mKey);
  return baseScore + winPoints;
};
 
const rounds = input.split('\n').map((r) => {
  return r.split(' ') as [opKey, meKey];
});
 
const sum = rounds.reduce((ttl, [op, my]) => {
  const rs = roundScore(op, my);
  return ttl + rs;
}, 0);
console.log(sum);

import { input } from './input.ts';
 
type opKey = keyof typeof opp;
type meKey = keyof typeof me;
const opp = {
  A: 1, //🪨
  B: 2, //📄
  C: 3, //✂️
};
const me = {
  X: 1, //🪨
  Y: 2, //📄
  Z: 3, //✂️
};
 
const winScore = (oKey: opKey, mKey: meKey) => {
  if (opp[oKey] === me[mKey]) return 3;
 
  if (opp[oKey] < me[mKey]) {
    // 🪨 🆚 ✂️
    if (opp[oKey] === 1 && me[mKey] === 3) return 0;
    return 6;
  }
  if (opp[oKey] > me[mKey]) {
    // 🪨 🆚 ✂️
    if (me[mKey] === 1 && opp[oKey] === 3) return 6;
    return 0;
  }
  throw new Error(oKey + ' | ' + mKey);
};
 
const roundScore = (oKey: opKey, mKey: meKey) => {
  const baseScore = me[mKey];
  const winPoints = winScore(oKey, mKey);
  return baseScore + winPoints;
};
 
const rounds = input.split('\n').map((r) => {
  return r.split(' ') as [opKey, meKey];
});
 
const sum = rounds.reduce((ttl, [op, my]) => {
  const rs = roundScore(op, my);
  return ttl + rs;
}, 0);
console.log(sum);

Each option has a numerical value: rock is 1, paper is 2, and scissors is 3. The code defines two objects, opp and me, that map the player’s choices to their numerical values. For example, opp['A'] is 1, and me['Y'] is 2.

The winScore function takes the choices made by the opponent and the player (represented by the opKey and mKey parameters, respectively), and returns the number of points that the player gets for winning the round. If the player and the opponent choose the same option, the player gets 3 points. If the player’s choice beats the opponent’s choice, the player gets 6 points. For example, if the player chooses rock and the opponent chooses scissors, the player wins and gets 6 points. If the player’s choice is weaker than the opponent’s choice, the player gets 0 points.

The roundScore function takes as input the choices made by the opponent and the player, and returns the total number of points that the player gets for the round. This is equal to the number of points the player gets for their choice plus the number of points they get for winning the round.

The rounds variable is an array that contains the choices made by the opponent and the player in each round of the game. For example, if the input is using the sample data, then rounds will be [["A", "Y"], ["B", "X"], ["C", "Z"]].

Finally, the code uses the reduce method to compute the total number of points the player gets over all rounds of the game. It does this by iterating over the rounds array and adding up the points the player gets in each round, using the roundScore function. The result is logged to the console with the console.log function.

Part 2

The Elf finishes helping with the tent and sneaks back over to you. “Anyway, the second column says how the round needs to end: X means you need to lose, Y means you need to end the round in a draw, and Z means you need to win. Good luck!”

The total score is still calculated in the same way, but now you need to figure out what shape to choose so the round ends as indicated. The example above now goes like this:

• 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.
• Now that you’re correctly decrypting the ultra top secret strategy guide, you would get a total score of 12.

What would your total score be if everything goes exactly according to your strategy guide?

Solution 2💡

import { input } from './input.ts';
 
type rKey = keyof typeof roundOutcome;
type meKey = keyof typeof me;
const rounds = input.split('\n').map((r) => {
  return r.split(' ') as [rKey, meKey];
});
 
const roundOutcome = {
  //Rock
  A: {
    X: 'Scissors',
    Y: 'Rock',
    Z: 'Paper',
  },
 
  //Paper
  B: {
    X: 'Rock',
    Y: 'Paper',
    Z: 'Scissors',
  },
 
  //Scissors
  C: {
    X: 'Paper',
    Y: 'Scissors',
    Z: 'Rock',
  },
} as const;
const me = {
  X: 1, //🪨
  Y: 2, //📄
  Z: 3, //✂️
};
const winPoints = {
  X: 0, //❌
  Y: 3, //➖
  Z: 6, //✅
};
const Points = {
  Rock: 1, //🪨
  Paper: 2, //📄
  Scissors: 3, //✂️
};
 
const roundScore = (rKey: rKey, wKey: meKey) => {
  const myAction = roundOutcome[rKey][wKey];
  const basePoint = Points[myAction];
  const wp = winPoints[wKey];
  return basePoint + wp;
};
 
const sum = rounds.reduce((accumulator, [op, my]) => {
  const rs = roundScore(op, my);
  return accumulator + rs;
}, 0);
console.log(sum);

import { input } from './input.ts';
 
type rKey = keyof typeof roundOutcome;
type meKey = keyof typeof me;
const rounds = input.split('\n').map((r) => {
  return r.split(' ') as [rKey, meKey];
});
 
const roundOutcome = {
  //Rock
  A: {
    X: 'Scissors',
    Y: 'Rock',
    Z: 'Paper',
  },
 
  //Paper
  B: {
    X: 'Rock',
    Y: 'Paper',
    Z: 'Scissors',
  },
 
  //Scissors
  C: {
    X: 'Paper',
    Y: 'Scissors',
    Z: 'Rock',
  },
} as const;
const me = {
  X: 1, //🪨
  Y: 2, //📄
  Z: 3, //✂️
};
const winPoints = {
  X: 0, //❌
  Y: 3, //➖
  Z: 6, //✅
};
const Points = {
  Rock: 1, //🪨
  Paper: 2, //📄
  Scissors: 3, //✂️
};
 
const roundScore = (rKey: rKey, wKey: meKey) => {
  const myAction = roundOutcome[rKey][wKey];
  const basePoint = Points[myAction];
  const wp = winPoints[wKey];
  return basePoint + wp;
};
 
const sum = rounds.reduce((accumulator, [op, my]) => {
  const rs = roundScore(op, my);
  return accumulator + rs;
}, 0);
console.log(sum);

Lets define an object called roundOutcome that maps each player’s choice to a string representing the outcome of the round: Rock, Paper, or Scissors. For example, roundOutcome['A']['X'] is Scissors, because if the first player chooses rock and the second player chooses rock, the outcome is that the second player wins with scissors.

We can also define an object called me that maps each player’s choice to a numerical value: rock is 1, paper is 2, and scissors is 3. This is similar to the opp and me objects in the previous example.

The rounds array stays the same.

The roundScore function takes the choices made by the two players represented by the rKey and wKey parameters, respectively, and returns the number of points that the player gets for the round. It does this by looking up the outcome of the round in the roundOutcome object and using the Points object to get the number of points associated with that outcome. It then adds the number of points the player gets for winning the round, which is looked up in the winPoints object.

Finally, we can use the reduce method to compute the total number of points.