Advent of Code 2022 Day 2
Day 2: Rock Paper Scissors
https://adventofcode.com/2022/day/2
We’re playing a Rock, Paper, Scissors tournament with the elves. The input represents an encryped strategy guide.
Each line has 2 letters seperated by a space.
- The first letter is
A
,B
, orC
. - The second letter is
X
,Y
, orZ
.
An example input looks like this:
A YB XC Z
Each round is worth some points, all scores get summed up, and whoever has the highest total at the end of the tournament wins.
Your score for a single round is the sum of the score for the shape you played, and the score for the outcome of the round.
Shape scores:
- Rock: 1
- Paper: 2
- Scissors: 3
Outcome scores:
- Loss: 0
- Draw: 3
- Win: 6
The first letter in the input is what your opponent is going to play.
A
for Rock, B
for Paper, and C
for Scissors.
Before the elf can tell you what the second letter means, they leave.
Part 1
Winning every round would be suspicious, so whatever that second letter is, it has to be important.
You assume the second letter is the shape you should play.
X
for Rock, Y
for Paper, and Z
for Scissors.
If “Rock”, “Paper”, and “Scissors” have positions in a list.
- To win, move 1 position to the right (and wrap around from “Scissors” to “Rock”!)
- To draw, keep the same position
- To lose, move 1 position to the left (and wrap around from “Rock” to “Scissors”!)
So I translated both A
, B
, C
, and X
, Y
, Z
to 0
, 1
, and 2
respectively.
Thankfully, both ABC and XYZ are sequences where the ASCII value of a letter increases by 1 each step.
- The value for the shape the opponent plays is known.
- The value for the shape I play is known.
- To calculate the score for this round, we need to know the outcome of the round.
With those two pieces of information a game of Rock, Paper, Scissors can be expressed as the following equation:
outcome = my_shape - opponent_shape + 1 (mod 3)
This expresses outcome
as a number from 0 to 2:
- 0 for loss
- 1 for draw
- 2 for win
pub fn part_1() -> String {let input = std::fs::read_to_string("src/day02.txt").unwrap();input.lines()// map every line to the score for that round.map(|line| {// transform A B C and X Y Z to 0 1 2 respectively by using their ASCII orderlet bytes = line.as_bytes();let left = (bytes[0] - b'A') as i8;let right = (bytes[2] - b'X') as i8;// 0 for rock, 1 for paper, 2 for scissors// 0 for loss, 1 for draw, 2 for winlet opponent_shape = left;let my_shape = right;let outcome = (my_shape - opponent_shape + 1).rem_euclid(3);let shape_score = my_shape + 1;let outcome_score = 3 * outcome;(shape_score + outcome_score) as u32}).sum::<u32>().to_string()}
Part 2
The elf returns and tells you that the second letter in the input is the desired outcome.
X
for loss, Y
for draw, Z
for win.
- The value for the shape the opponent plays is known.
- The value for the outcome of the round is known.
- To calculate the score for this round, we need to know the shape we need to play.
We rearrange the equation from part1 to solve for my_shape
instead of outcome
:
my_shape = opponent_shape - 1 + outcome (mod 3)
pub fn part_2() -> String {let input = std::fs::read_to_string("src/day02.txt").unwrap();input.lines()// map every line to the score for that round.map(|line| {// transform A B C and X Y Z to 0 1 2 respectively by using their ASCII orderlet bytes = line.as_bytes();let left = (bytes[0] - b'A') as i8;let right = (bytes[2] - b'X') as i8;// 0 for rock, 1 for paper, 2 for scissors// 0 for loss, 1 for draw, 2 for winlet opponent_shape = left;let outcome = right;let my_shape = (opponent_shape - 1 + outcome).rem_euclid(3);let shape_score = my_shape + 1;let outcome_score = 3 * outcome;(shape_score + outcome_score) as u32}).sum::<u32>().to_string()}
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