## Advent of Code 2022 Day 2

## Part 1

- It was bound to happen eventually
- Learning my ABCs and 123s
- Going from A to Z, so to speak
- My algorithm in JavaScript

### It was bound to happen eventually

"Want to learn programming?" they say.

"Why not build a simple game?" they say.

"How about Rock Paper Scissors?" they say.

AoC has avoided it seven years. Until now.

How exciting!

### Learning my ABCs and 123s

- A and X and 1 for Rock
- B and Y and 2 for Paper
- C and Z and 3 for Scissors

Opponent's point of view:

- A beats Z
- C beats Y
- B beats X

Player's point of view:

- X beats C
- Y beats A
- Z beats B

And the formula for a round's score:

- Amount above for what player chose, plus...
- 0 for player loss
- 3 for tie
- 6 for player win

### Going from A to Z, so to speak

I need to turn this:

```
A Y
B X
C Z
```

Into this:

```
8
1
6
```

In order to get `15`

.

Starting with the first round:

```
A Y
```

I could use a few dictionaries to map winners and amounts:

```
const elf = { A: 'Z', C: 'Y', B: 'X' }
const player = { X: 'C', Y: 'A', Z: 'B' }
const score = { X: 1, Y: 2, Z: 3 }
```

Three conditions:

```
if (player[col2] == col1) {
// player wins
} else if (elf[col1] == col2) {
// elf wins
} else {
// 'twas a draw
}
```

Would this work on the first round?

```
A Y
player[Y] == 'A' // true
player wins: 6
Y = 2
6 + 2 = 8
Success!
```

Great! Second round?

```
B X
player[X] == 'C' // false
elf[B] == 'X' // true
player loses: 0
X = 1
1 + 0 = 1
Success!
```

Great! Third round?

```
C Z
player[Z] == 'C' // false
elf[C] == 'Z' // false
draw: 3
Z = 3
3 + 3 = 6
Success!
```

It seems this algorithm and data structure will help me solve Part 1!

### My algorithm in JavaScript

```
const elf = { A: 'Z', C: 'Y', B: 'X' }
const player = { X: 'C', Y: 'A', Z: 'B' }
const score = { X: 1, Y: 2, Z: 3 }
return input
.split('\n')
.reduce((total, round) => {
let [col1, col2] = round.split(' ')
if (player[col2] == col1) {
return total += score[col2] + 6
} else if (elf[col1] == col2) {
return total += score[col2]
} else {
return total += score[col2] + 3
}
}, 0)
```

## Part 2

- Just what I need
- My algorithm in JavaScript

### Just what I need

An expectedly clever switch of the rules!

- X needs me to lose
- Y needs me to draw
- Z needs me to win

How will I decide what I play?

I'll reference example round one again:

```
A Y
```

`Y`

needs me to draw (pick the same thing as my opponent).

What draw's to an `A`

? `X`

How might I represent that - and losing and winning - as a data structure?

```
{
A: { X: 'Z', Y: 'X', Z: 'Y' },
B: { X: 'X', Y: 'Y', Z: 'Z' },
C: { X: 'Y', Y: 'Z', Z: 'X' },
}
```

Here I've mapped the first column to the second column to the player's necessary choice.

Trying round one again:

```
A Y
A: Y: X
Y is a draw = 3
X = 1
3 + 1 = 4
Success!
```

Trying round two:

```
B X
B: X: X
X is a loss = 0
X = 1
0 + 1 = 1
Success!
```

```
C Z
C: Z: X
Z is a win = 6
X = 1
6 + 1 = 7
Success!
```

It seems this algorithm and data structure will help me solve Part 2!

### My algorithm in JavaScript

```
const move = {
A: { X: 'Z', Y: 'X', Z: 'Y' },
B: { X: 'X', Y: 'Y', Z: 'Z' },
C: { X: 'Y', Y: 'Z', Z: 'X' },
}
const score = { X: 1, Y: 2, Z: 3 }
const outcome = { X: 0, Y: 3, Z: 6 }
return input
.split('\n')
.reduce((total, round) => {
let [col1, col2] = round.split(' ')
return total += score[move[col1][col2]] + outcome[col2]
}, 0)
```

## I did it!!

- I solved both parts!
- I worked my way toward solutions by writing what I observed and thought, correcting myself along the way!
- I was surprised that my code for Part 2 is shorter than my code for Part 1! I didn't need all the conditions!

This was my funnest time playing Rock Paper Scissors!

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