Added Solution for 2021 day 21

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Burnus 2023-05-05 14:42:10 +02:00
parent 912bb4b2d9
commit 4d4fd2de5a
5 changed files with 227 additions and 0 deletions

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[package]
name = "day21_dirac_dice"
version = "0.1.0"
edition = "2021"
# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
[dependencies]

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There's not much to do as you slowly descend to the bottom of the ocean. The submarine computer challenges you to a nice game of *Dirac Dice*.
This game consists of a single [die](https://en.wikipedia.org/wiki/Dice), two [pawns](https://en.wikipedia.org/wiki/Glossary_of_board_games#piece), and a game board with a circular track containing ten spaces marked `1` through `10` clockwise. Each player's *starting space* is chosen randomly (your puzzle input). Player 1 goes first.
Players take turns moving. On each player's turn, the player rolls the die *three times* and adds up the results. Then, the player moves their pawn that many times *forward* around the track (that is, moving clockwise on spaces in order of increasing value, wrapping back around to `1` after `10`). So, if a player is on space `7` and they roll `2`, `2`, and `1`, they would move forward 5 times, to spaces `8`, `9`, `10`, `1`, and finally stopping on `2`.
After each player moves, they increase their *score* by the value of the space their pawn stopped on. Players' scores start at `0`. So, if the first player starts on space `7` and rolls a total of `5`, they would stop on space `2` and add `2` to their score (for a total score of `2`). The game immediately ends as a win for any player whose score reaches *at least `1000`*.
Since the first game is a practice game, the submarine opens a compartment labeled *deterministic dice* and a 100-sided die falls out. This die always rolls `1` first, then `2`, then `3`, and so on up to `100`, after which it starts over at `1` again. Play using this die.
For example, given these starting positions:
```
Player 1 starting position: 4
Player 2 starting position: 8
```
This is how the game would go:
* Player 1 rolls `1`+`2`+`3` and moves to space `10` for a total score of `10`.
* Player 2 rolls `4`+`5`+`6` and moves to space `3` for a total score of `3`.
* Player 1 rolls `7`+`8`+`9` and moves to space `4` for a total score of `14`.
* Player 2 rolls `10`+`11`+`12` and moves to space `6` for a total score of `9`.
* Player 1 rolls `13`+`14`+`15` and moves to space `6` for a total score of `20`.
* Player 2 rolls `16`+`17`+`18` and moves to space `7` for a total score of `16`.
* Player 1 rolls `19`+`20`+`21` and moves to space `6` for a total score of `26`.
* Player 2 rolls `22`+`23`+`24` and moves to space `6` for a total score of `22`.
...after many turns...
* Player 2 rolls `82`+`83`+`84` and moves to space `6` for a total score of `742`.
* Player 1 rolls `85`+`86`+`87` and moves to space `4` for a total score of `990`.
* Player 2 rolls `88`+`89`+`90` and moves to space `3` for a total score of `745`.
* Player 1 rolls `91`+`92`+`93` and moves to space `10` for a final score, `1000`.
Since player 1 has at least `1000` points, player 1 wins and the game ends. At this point, the losing player had `745` points and the die had been rolled a total of `993` times; `745 * 993 = *739785*`.
Play a practice game using the deterministic 100-sided die. The moment either player wins, *what do you get if you multiply the score of the losing player by the number of times the die was rolled during the game?*
Your puzzle answer was `518418`.
\--- Part Two ---
----------
Now that you're warmed up, it's time to play the real game.
A second compartment opens, this time labeled *Dirac dice*. Out of it falls a single three-sided die.
As you experiment with the die, you feel a little strange. An informational brochure in the compartment explains that this is a *quantum die*: when you roll it, the universe *splits into multiple copies*, one copy for each possible outcome of the die. In this case, rolling the die always splits the universe into *three copies*: one where the outcome of the roll was `1`, one where it was `2`, and one where it was `3`.
The game is played the same as before, although to prevent things from getting too far out of hand, the game now ends when either player's score reaches at least `*21*`.
Using the same starting positions as in the example above, player 1 wins in `*444356092776315*` universes, while player 2 merely wins in `341960390180808` universes.
Using your given starting positions, determine every possible outcome. *Find the player that wins in more universes; in how many universes does that player win?*
Your puzzle answer was `116741133558209`.
Both parts of this puzzle are complete! They provide two gold stars: \*\*
At this point, you should [return to your Advent calendar](/2021) and try another puzzle.
If you still want to see it, you can [get your puzzle input](21/input).

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use core::fmt::Display;
use std::{num::ParseIntError, collections::BTreeMap};
#[derive(Debug, PartialEq, Eq)]
pub enum ParseError {
ParseIntError(std::num::ParseIntError),
LineMalformed(String),
}
impl From<ParseIntError> for ParseError {
fn from(value: ParseIntError) -> Self {
Self::ParseIntError(value)
}
}
impl Display for ParseError {
fn fmt(&self, f: &mut std::fmt::Formatter<'_>) -> std::fmt::Result {
match self {
Self::ParseIntError(e) => write!(f, "Unable to parse into integer: {e}"),
Self::LineMalformed(v) => write!(f, "Line is malformed: {v}"),
}
}
}
const WINNING_SCORE_V1: usize = 1000;
const WINNING_SCORE_V2: usize = 21;
#[derive(Clone, PartialOrd, Ord, PartialEq, Eq)]
struct Player {
score: usize,
position: usize,
won: bool,
}
impl TryFrom<&str> for Player {
type Error = ParseError;
fn try_from(value: &str) -> Result<Self, Self::Error> {
let (_, pos) = value.rsplit_once(' ').ok_or(Self::Error::LineMalformed(value.to_string()))?;
Ok(Self {
position: pos.parse()?,
score: 0,
won: false,
})
}
}
impl Player {
fn play(&mut self, die: &mut DeterministicDie) -> bool {
self.roll(die.roll() + die.roll() + die.roll());
if self.score >= WINNING_SCORE_V1 {
self.won = true;
true
} else {
false
}
}
fn roll(&mut self, face: usize) {
self.position = (self.position + face - 1) % 10 + 1;
self.score += self.position;
}
}
#[derive(Default)]
struct DeterministicDie {
roll_counter: usize,
}
impl DeterministicDie {
fn roll(&mut self) -> usize {
self.roll_counter += 1;
(self.roll_counter - 1) % 100 + 1
}
}
pub fn run(input: &str) -> Result<(usize, usize), ParseError> {
let mut players: Vec<_> = input.lines().map(Player::try_from).collect::<Result<Vec<_>, _>>()?;
let second = max_wins(&players);
let mut die = DeterministicDie::default();
'outer: loop {
for player in players.iter_mut() {
if player.play(&mut die) {
break 'outer;
}
}
}
let first = players.iter().find(|p| !p.won).map(|p| p.score).unwrap() * die.roll_counter;
Ok((first, second))
}
fn max_wins(players: &[Player]) -> usize {
let mut wins = [0, 0];
let mut open_set = BTreeMap::from([(players.to_vec(), 1)]);
let roll_results = [
(3, 1),
(4, 3),
(5, 6),
(6, 7),
(7, 6),
(8, 3),
(9, 1),
];
while let Some(current) = open_set.pop_first() {
let players = current.0;
let count = current.1;
for d1 in roll_results {
let mut player_1 = players[0].clone();
player_1.roll(d1.0);
if player_1.score >= WINNING_SCORE_V2 {
wins[0] += d1.1 * count;
} else {
for d2 in roll_results {
let mut player_2 = players[1].clone();
player_2.roll(d2.0);
if player_2.score >= WINNING_SCORE_V2 {
wins[1] += d1.1 * d2.1 * count;
} else {
open_set.entry(vec![player_1.clone(), player_2]).and_modify(|ct| *ct += count * d1.1 * d2.1).or_insert(count * d1.1 * d2.1);
}
}
}
}
}
*wins.iter().max().unwrap()
}
#[cfg(test)]
mod tests {
use super::*;
use std::fs::read_to_string;
fn read_file(name: &str) -> String {
read_to_string(name).expect(&format!("Unable to read file: {name}")[..]).trim().to_string()
}
#[test]
fn test_sample() {
let sample_input = read_file("tests/sample_input");
assert_eq!(run(&sample_input), Ok((739785, 444356092776315)));
}
#[test]
fn test_challenge() {
let challenge_input = read_file("tests/challenge_input");
assert_eq!(run(&challenge_input), Ok((518418, 116741133558209)));
}
}

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Player 1 starting position: 8
Player 2 starting position: 1

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Player 1 starting position: 4
Player 2 starting position: 8