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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