【译】用 Rust 实现 Advent of Code 2020 第11天开启掘金成长之旅！这是我参与「掘金日新计划 ·

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Advent of Code 2020 Day11 译文（用 Rust 实现 Advent of Code 2020 第11天）

原文链接：fasterthanli.me/series/adve…

原文作者：Amos

译文来自：github.com/suhanyujie/…

译者：suhanyujie

ps：水平有限，如有不当之处，欢迎指正

标签：Rust，advent of code, algo

新的一天，新的问题。

这个问题看起来有点像“康威生命游戏”（Game of Life），或者是一种老式的细胞生命自动机。

我们有一张这样的地图:

L.LL.LL.LL
LLLLLLL.LL
L.L.L..L..
LLLL.LL.LL
L.LL.LL.LL
L.LLLLL.LL
..L.L.....
LLLLLLLLLL
L.LLLLLL.L
L.LLLLL.LL

对于每一次迭代：

如果是 L 符号，会变成 #，如果没有变成 #，则在其相邻的 8 个位置一定有一个可以转换成 #
# 符号，则变成 L，前提是相邻位置中有 4 个以上的单元格是 #
. 符号，则不会改变

这是一个令人感兴趣的问题。我们甚至可以回想一些我们在第 3 天做的东西！

首先是 2D vector 类型：

#[derive(Debug, Clone, Copy, PartialEq)]
struct Vec2 {
    x: i64,
    y: i64,
}

然后是 tile 枚举：

#[derive(Clone, Copy, PartialEq)]
enum Tile {
    Floor,
    EmptySeat,
    OccupiedSeat,
}

impl Default for Tile {
    fn default() -> Self {
        Self::Floor
    }
}

为 Tile 实现一个简洁的 Debug trait：

use std::fmt;

impl fmt::Debug for Tile {
    fn fmt(&self, f: &mut fmt::Formatter<'_>) -> fmt::Result {
        let c = match self {
            Tile::Floor => '.',
            Tile::EmptySeat => 'L',
            Tile::OccupiedSeat => '#',
        };
        write!(f, "{}", c)
    }
}

然后，Map —— 这一次，让它通用一些。只是为了好玩！

struct Map<T> {
    size: Vec2,
    tiles: Vec<T>,
}

如果 T 实现了 Default trait，可以很简单地实现 new:

impl<T> Map<T>
where
    T: Default,
{
    fn new(size: Vec2) -> Self {
        let num_tiles = size.x * size.y;
        Self {
            size,
            tiles: (0..num_tiles)
                .into_iter()
                .map(|_| Default::default())
                .collect(),
        }
    }
}

然后是 Map::index，从 2D 空间到 1D 空间的映射/索引：

impl<T> Map<T> {
    fn index(&self, pos: Vec2) -> Option<usize> {
        if (0..self.size.x).contains(&pos.x) && (0..self.size.y).contains(&pos.y) {
            Some((pos.x + pos.y * self.size.x) as _)
        } else {
            None
        }
    }
}

从这里我们可以很容易地实现 set:

impl<T> Map<T> {
    fn set(&mut self, pos: Vec2, tile: T) {
        if let Some(index) = self.index(pos) {
            self.tiles[index] = tile;
        }
    }
}

对于 get，我们需要 T 是可 Copy 的，这样我们就不需要使用引用：

impl<T> Map<T>
where
    T: Copy,
{
    fn get(&self, pos: Vec2) -> Option<T> {
        self.index(pos).map(|index| self.tiles[index])
    }
}

接下来 —— 我们要处理邻近位置，所以如果我们能够穷举出指定单元格的所有邻近位置，那将会非常简洁:

impl<T> Map<T> {
    fn neighbor_positions(&self, pos: Vec2) -> impl Iterator<Item = Vec2> {
        (-1..=1)
            .map(|dx| (-1..=1).map(move |dy| (dx, dy)))
            .flatten()
            .filter(|&(dx, dy)| !(dx == 0 && dy == 0))
            .map(move |(dx, dy)| Vec2 {
                x: pos.x + dx,
                y: pos.y + dy,
            })
    }
}

酷熊：这看起来有点不太靠谱，测试一下怎么样？

#[test]
fn test_neighbor_positions() {
    use std::collections::HashSet;

    let map = Map::<()>::new(Vec2 { x: 3, y: 3 });
    let positions: HashSet<_> = map
        .neighbor_positions(Vec2 { x: 1, y: 1 })
        .map(|v| (v.x, v.y))
        .collect();
    for p in &[
        (0, 0),
        (0, 1),
        (0, 2),
        (1, 0),
        (2, 0),
        (1, 2),
        (2, 2),
        (2, 1),
    ] {
        assert!(positions.contains(p));
    }
}

$ cargo t -q

running 1 test
.
test result: ok. 1 passed; 0 failed; 0 ignored; 0 measured; 0 filtered out

酷熊：好吧，继续。

接下来，为了更好地连接上下文，我们可以使用 .neighbor_tiles 方法返回 Tile 的迭代器！

impl<T> Map<T>
where
    T: Copy,
{
    fn neighbor_tiles(&self, pos: Vec2) -> impl Iterator<Item = T> + '_ {
        self.neighbor_positions(pos)
            .filter_map(move |pos| self.get(pos))
    }
}

酷熊：+ '_ 是什么意思？

Amos：em，迭代器在借用时，生命周期和 &self 一样长，因为它是从它那里获取的！

酷熊：啊，那么如果是 impl Iterator<Item = T> (没有任何附加注释)，生命周期又是多少呢？

Amos：就像 Box<T> 一样，它默认是 'static 的，这只适用于拥有所有权的类型（比如我们的 neighbor_positions 迭代器，它拥有产生数据项所需的所有权），或者，我猜测，'static 参数本质类似于一个 const，或者形如 Box::leak 或 include_str 中持续存在的东西一样！

好了，现在我们可以开始实现一些“细胞自动机”逻辑了，我们可以基于 Tile 做一些处理：

impl Tile {
    fn next<I>(self, neighbors: I) -> Self
    where
        I: Iterator<Item = Self>,
    {
        match self {
            Self::Floor => Self::Floor,
            Self::EmptySeat => match neighbors
                .filter(|t| matches!(t, Self::OccupiedSeat))
                .count()
            {
                // no one around? we can sit here!
                0 => Self::OccupiedSeat,
                // social distancing please
                _ => Self::EmptySeat,
            },
            Self::OccupiedSeat => {
                match neighbors.filter(|t| matches!(t, Self::OccupiedSeat)).count() {
                    // up to 3 neighbors: still okay for now
                    0..=3 => Self::OccupiedSeat,
                    // that's too many folks!
                    _ => Self::EmptySeat,
                }
            }
        }
    }
}

看看，我们还需要什么...啊，对了，为 Map 实现 Debug trait —— 当且仅当 T 本身满足 Debug 约束的时候。哦，还有 Copy 约束，因为我们的 get 方法中需要用到：

impl<T> fmt::Debug for Map<T>
where
    T: fmt::Debug + Copy,
{
    fn fmt(&self, f: &mut fmt::Formatter<'_>) -> fmt::Result {
        for y in 0..self.size.y {
            for x in 0..self.size.x {
                write!(f, "{:?}", self.get(Vec2 { x, y }).unwrap())?;
            }
            writeln!(f)?;
        }
        Ok(())
    }
}

然后... 你知道怎么做会更好吗？由于这些规则同时适用于所有的 tile，所以如果我们能够迭代所有 tile，那将是非常简洁的。

impl<T> Map<T>
where
    T: Copy,
{
    fn iter(&self) -> impl Iterator<Item = T> + '_ {
        self.tiles.iter().copied()
    }
}

酷熊：但是... 我们还需要知道他们的位置，对吧？这样我们就能推断出他们的邻近位置坐标了？

对! 为了做到这一点，我们引入另一种类型: Positioned:

#[derive(Debug)]
struct Positioned<T>(Vec2, T);

并使 iter 方法返回 position<T> 类型，并将类型作为 Item 的类型:

impl<T> Map<T>
where
    T: Copy,
{
    // 替换前面的 `iter` 方法
    fn iter(&self) -> impl Iterator<Item = Positioned<T>> + '_ {
        (0..self.size.y)
            .map(move |y| {
                (0..self.size.x).map(move |x| {
                    let pos = Vec2 { x, y };
                    Positioned(pos, self.get(pos).unwrap())
                })
            })
            .flatten()
    }
}

我们试试看：

fn main() {
    let mut m = Map::new(Vec2 { x: 3, y: 3 });
    m.set(Vec2 { x: 1, y: 1 }, Tile::OccupiedSeat);

    for tile in m.iter() {
        println!("{:?}", tile);
    }
}

$ cargo run --quiet
Positioned(Vec2 { x: 0, y: 0 }, .)
Positioned(Vec2 { x: 1, y: 0 }, .)
Positioned(Vec2 { x: 2, y: 0 }, .)
Positioned(Vec2 { x: 0, y: 1 }, .)
Positioned(Vec2 { x: 1, y: 1 }, #)
Positioned(Vec2 { x: 2, y: 1 }, .)
Positioned(Vec2 { x: 0, y: 2 }, .)
Positioned(Vec2 { x: 1, y: 2 }, .)
Positioned(Vec2 { x: 2, y: 2 }, .)

酷熊：看起来不错！

然后，最后，作为下一步 —— 我们可以将其收集到 Map 中。

酷熊：这有些不同! 我们该怎么做？

我们可以先看看 Iterator::collect 的签名：

fn collect<B>(self) -> B
where
    B: FromIterator<Self::Item>,

我们只需要实现 FromIterator！

use std::iter::FromIterator;

impl<A> FromIterator<Positioned<A>> for Map<A> {
    fn from_iter<T: IntoIterator<Item = Positioned<A>>>(iter: T) -> Self {
        todo!()
    }
}

还有一个问题，我们的 Map 构造函数需要确定 size:

fn new(size: Vec2) -> Self

我们只是得到已知位置的 tile 流 —— 我们不知道最后的地图有多大！所以我想我们的问题不太适合用 collect。

也许还有另一种方法/trait？一种可以让我们获取现有 Map 的方法？

酷熊：有的，看我找到了什么:

fn extend<T>(&mut self, iter: T)
where
    T: IntoIterator<Item = A>,

啊，看起来有用。

use std::iter::Extend;

impl<A> Extend<Positioned<A>> for Map<A> {
    fn extend<T: IntoIterator<Item = Positioned<A>>>(&mut self, iter: T) {
        for Positioned(pos, tile) in iter {
            self.set(pos, tile)
        }
    }
}

太棒了！

好吧，我想现在我们需要解析我们的地图：

impl Map<Tile> {
    fn parse(input: &[u8]) -> Self {
        let mut columns = 0;
        let mut rows = 1;
        for &c in input.iter() {
            if c == b'\n' {
                rows += 1;
                columns = 0;
            } else {
                columns += 1;
            }
        }

        let mut iter = input.iter().copied();
        let mut map = Self::new(Vec2 {
            x: columns,
            y: rows,
        });
        for row in 0..map.size.y {
            for col in 0..map.size.x {
                let tile = match iter.next() {
                    Some(b'.') => Tile::Floor,
                    Some(b'L') => Tile::EmptySeat,
                    Some(b'#') => Tile::OccupiedSeat,
                    c => panic!("Expected '.', 'L' or '#', but got: {:?}", c),
                };
                map.set(Vec2 { x: col, y: row }, tile);
            }
            iter.next();
        }
        map
    }
}

让我们来做一个快速的测试检查，我们的输入例子是:

L.LL.LL.LL
LLLLLLL.LL
L.L.L..L..
LLLL.LL.LL
L.LL.LL.LL
L.LLLLL.LL
..L.L.....
LLLLLLLLLL
L.LLLLLL.L
L.LLLLL.LL

我们的解析版本是:

fn main() {
    let m = Map::<Tile>::parse(include_bytes!("input.txt"));
    dbg!(&m.size);
    println!("{:?}", m);
}

$ cargo run --quiet
[src/main.rs:230] &m.size = Vec2 {
    x: 10,
    y: 10,
}
L.LL.LL.LL
LLLLLLL.LL
L.L.L..L..
LLLL.LL.LL
L.LL.LL.LL
L.LLLLL.LL
..L.L.....
LLLLLLLLLL
L.LLLLLL.L
L.LLLLL.LL

看起来不错！

现在，我想我们还需要在 Map 上实现一个 next 方法，对吗？但当 T = Tile 时:

impl Map<Tile> {
    fn next(&self) -> Self {
        let mut res = Self::new(self.size);
        res.extend(
            self.iter()
                .map(|Positioned(pos, tile)| Positioned(pos, tile.next(self.neighbor_tiles(pos)))),
        );
        res
    }
}

酷熊：这真是太棒了，我们所有的基础代码都有了回报！

让我们打印一些示例地图的迭代:

$ cargo add itertools
      Adding itertools v0.9.0 to dependencies

fn main() {
    let maps = itertools::iterate(Map::<Tile>::parse(include_bytes!("input.txt")), Map::next);
    for map in maps.take(5) {
        println!("{:?}", map);
    }
}

你真的需要把所有东西都变成迭代器吗？

Amos：现在是寒假，熊，我们还有时间！

$ cargo run --quiet
L.LL.LL.LL
LLLLLLL.LL
L.L.L..L..
LLLL.LL.LL
L.LL.LL.LL
L.LLLLL.LL
..L.L.....
LLLLLLLLLL
L.LLLLLL.L
L.LLLLL.LL

#.##.##.##
#######.##
#.#.#..#..
####.##.##
#.##.##.##
#.#####.##
..#.#.....
##########
#.######.#
#.#####.##

#.LL.L#.##
#LLLLLL.L#
L.L.L..L..
#LLL.LL.L#
#.LL.LL.LL
#.LLLL#.##
..L.L.....
#LLLLLLLL#
#.LLLLLL.L
#.#LLLL.##

#.##.L#.##
#L###LL.L#
L.#.#..#..
#L##.##.L#
#.##.LL.LL
#.###L#.##
..#.#.....
#L######L#
#.LL###L.L
#.#L###.##

#.#L.L#.##
#LLL#LL.L#
L.L.L..#..
#LLL.##.L#
#.LL.LL.LL
#.LL#L#.##
..L.L.....
#L#LLLL#L#
#.LLLLLL.L
#.#L#L#.##

现在回答问题，题目如下：

通过重复应用座位规则来模拟你的座位区域，直到座位状态不再改变。此时，有多少位置被占了？

酷熊：嗯，我们不得不将当前的状态与之前的状态对比... 听起来像是 tuple_windows 的工作！

Amos：确实是! 我不想一个一个地比较所有的 tile...

酷熊：我们可以在 Map 上派生 PartialEq trait？

Amos：这主意不错！

#[derive(PartialEq)]
struct Map<T> {
    size: Vec2,
    tiles: Vec<T>,
}

酷熊热辣小贴士注意: Vec2 已经派生了 PartialEq。至于 T，可能实现了，也可能没有实现。Map<T> 则只有在 T 本身实现/派生 PartialEq 时，才会实现 PartialEq。

Amos：我们怎么确定这些？

酷熊：为什么，我们可以使用 cargo-expand 检查派生宏展开的结果！

#[automatically_derived]
#[allow(unused_qualifications)]
impl<T: ::core::cmp::PartialEq> ::core::cmp::PartialEq for Map<T> {
    #[inline]
    fn eq(&self, other: &Map<T>) -> bool {
        match *other {
            Map {
                size: ref __self_1_0,
                tiles: ref __self_1_1,
            } => match *self {
                Map {
                    size: ref __self_0_0,
                    tiles: ref __self_0_1,
                } => (*__self_0_0) == (*__self_1_0) && (*__self_0_1) == (*__self_1_1),
            },
        }
    }
    #[inline]
    fn ne(&self, other: &Map<T>) -> bool {
        match *other {
            Map {
                size: ref __self_1_0,
                tiles: ref __self_1_1,
            } => match *self {
                Map {
                    size: ref __self_0_0,
                    tiles: ref __self_0_1,
                } => (*__self_0_0) != (*__self_1_0) || (*__self_0_1) != (*__self_1_1),
            },
        }
    }
}

Amos：啊! 看... 生成了。这是合理的。而且它使用的是 core:: 而不是 std:: 中的类型，我想这也是有道理的。

继续 —— 让我们在 Map 上实现 last 方法，它模拟达到一个最终的的状态:

impl Map<Tile> {
    fn last(self) -> Self {
        use itertools::Itertools;
        itertools::iterate(self, Map::next)
            .tuple_windows()
            .find_map(|(prev, next)| if prev == next { Some(next) } else { None });

        todo!();
    }
}

酷熊：啊哦，这不管用:

$ cargo check --quiet
error[E0277]: the trait bound `Map<Tile>: Clone` is not satisfied
   --> src/main.rs:210:14
    |
210 |             .tuple_windows()
    |              ^^^^^^^^^^^^^ the trait `Clone` is not implemented for `Map<Tile>`

error[E0277]: the trait bound `Map<Tile>: Clone` is not satisfied
   --> src/main.rs:211:14
    |
211 |             .find_map(|(prev, next)| if prev == next { Some(next) } else { None });
    |              ^^^^^^^^ the trait `Clone` is not implemented for `Map<Tile>`
    |
    = note: required because of the requirements on the impl of `Iterator` for `TupleWindows<Iterate<Map<Tile>, for<'r> fn(&'r Map<Tile>) -> Map<Tile> {Map::<Tile>::next}>, (Map<Tile>, Map<Tile>)>`

error: aborting due to 2 previous errors

噢！我们也需要在 Map 上派生 Copy。但是如果我们要克隆 Map，我们不希望有太多的分配和拷贝。

酷熊：我们可以直接用 im crate 吗？

让我们看看它有多大程度上是一种“临时替代品”:

$ cargo add im
      Adding im v15.0.0 to dependencies

use im::Vector;

#[derive(PartialEq)]
struct Map<T> {
    size: Vec2,
    tiles: Vector<T>,
}

我们已经有一个问题了：

$ cargo check --quiet
error[E0369]: binary operation `==` cannot be applied to type `Vector<T>`
  --> src/main.rs:71:5
   |
71 |     tiles: Vector<T>,
   |     ^^^^^^^^^^^^^^^^
   |
   = note: this error originates in a derive macro (in Nightly builds, run with -Z macro-backtrace for more info)
help: consider further restricting this bound
   |
68 | #[derive(PartialEq + std::cmp::PartialEq)]
   |                    ^^^^^^^^^^^^^^^^^^^^^

让我们看看 im::Vector 是否实现了 PartialEq。

impl<A: Clone + PartialEq> PartialEq<Vector<A>> for Vector<A>

但前提是 A 实现了 Clone trait

好吧，Tile 绝对实现了 Clone，我们可以这么做:

#[derive(PartialEq)]
struct Map<T>
where
    T: Clone,
{
    size: Vec2,
    tiles: Vector<T>,
}

现在我们需要将这个约束添加到我们的其他 impl 块中 —— 这留给读者作为练习。

酷熊热辣小贴士无论在哪里，我们有 T: Copy 的约束，我们都不需要添加 +Clone，因为实现了 Copy 意味着实现了 Clone：

pub trait Copy: Clone { }

现在我们还可以在 Map 上派生 Clone trait:

#[derive(PartialEq, Clone)]
struct Map<T>
where
    T: Clone,
{
    size: Vec2,
    tiles: Vector<T>,
}

完善我们 last 的实现:

impl Map<Tile> {
    fn last(self) -> Self {
        use itertools::Itertools;
        itertools::iterate(self, Map::next)
            .tuple_windows()
            .find_map(|(prev, next)| if prev == next { Some(next) } else { None })
            .unwrap()
    }
}

我们试试！

fn main() {
    let last = Map::<Tile>::parse(include_bytes!("input.txt")).last();
    println!("{:?}", last);
}

$ cargo run --quiet
#.#L.L#.##
#LLL#LL.L#
L.#.L..#..
#L##.##.L#
#.#L.LL.LL
#.#L#L#.##
..L.L.....
#L#L##L#L#
#.LLLLLL.L
#.#L#L#.##

酷熊：耶，结束了! 接招，停机问题。

而且，既然我们对这个问题的答案是以数字的形式出现，那么让我们来计算一下最终有多少座位被占:

fn main() {
    let last = Map::<Tile>::parse(include_bytes!("input.txt")).last();
    println!("{:?}", last);
    println!(
        "there are {} occupied seats",
        last.iter()
            //      👇  this is a Positioned<Tile>
            .filter(|p| matches!(p.1, Tile::OccupiedSeat))
            .count()
    );
}

$ cargo run --quiet
#.#L.L#.##
#LLL#LL.L#
L.#.L..#..
#L##.##.L#
#.#L.LL.LL
#.#L#L#.##
..L.L.....
#L#L##L#L#
#.LLLLLL.L
#.#L#L#.##

there are 37 occupied seats

这个例子就是这样! 我们用题目的输入来试试：

$ cargo run --quiet
#L#L#.#LL#L##L#.#LL##L##L#.#L#L#L#.#L#L#L#L#L#L#L#L#.#L#L#L#L#.#L##L#.#L#L#L#.#L#L#L#L#LL#L#L#L#L#
LLLLL.LL.L.LLLL.LLL..LLLLL.LLLLLL#LLLLLLLLLLLLLLLLLL.LLLLLLLLL.LLLLLL.LLLLLLL.LLLLLLL.LLLLLLLLLLLL
#L#L#.L#L#.#L#L#L#L#L#L#L#L#L#L.L..#L#L#L#L#L#L#L#L#L#L#L#L#L#.#L#L#L.L#L#L##L#L#L#L#.#.L#L#L#L#L#
LLLLLLLLLL.LLLL.LLLLLLLLLL.LLLL#L#.LLLLLLLLLLLLLLLLL.LLLLLLLLL.LLLLLL#LLLLLLLLLLLLLLL.LLLLLLLLLLLL
#L#L#L#L#L#L#L#.#L##.#L#L#.#L#LLLL.##.L#L#L#..#L#L#L#L#L#L#.L#.#L#L#L#.#L#L#L#L##L#L#.L#L#L#L#L#L#
LLLLLLLLLL.LLLL.LLLL.LLLLL.LLLL#L#LLLLLLLLLL.LLLLLLLLLLLLLLLLL.LLLLLL.LLLLLL.LLLLLLLLLLLLLLLLLLLLL
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there are 2289 occupied seats

就这样吧，我们可以进入下一部分。

第二部分

这是一个新的部分，规则，他们有些改变。

现在我们要找的座位可能不是相邻的，但它是 8 个方位之一。

之前，我们看过这些，我们只看到两个有人的座位:

但是现在，我们朝每个方向前进并观察，直到我们在对应方向上看到一个座位 —— 我们看到 8 个有人坐的座位:

是时候使用一个新方法了! visible_seats 完全可以做到这一点。

impl Map<Tile> {
    fn visible_seats(&self, pos: Vec2) -> impl Iterator<Item = Tile> + '_ {
        use itertools::Itertools;
        (-1..=1)
            .map(|dx| (-1..=1).map(move |dy| (dx, dy)))
            .flatten()
            .filter(|&(dx, dy)| !(dx == 0 && dy == 0))
            // for each direction...
            .map(move |(dx, dy)| {
                // keep moving in set direction
                itertools::iterate(pos, move |v| Vec2 {
                    x: v.x + dx,
                    y: v.y + dy,
                })
                // as long as we're on the map
                .map(move |pos| self.index(pos))
                .while_some()
                // and until we reach a seat
                .filter_map(move |index| match self.tiles[index] {
                    Tile::Floor => None,
                    seat => Some(seat),
                })
                .take(1)
            })
            .flatten()
    }
}

酷熊：好吧，那个肯定需要测试一下。

Amos：同意。

现在是使用 indoc crate 的好时机！

$ cargo add indoc
      Adding indoc v1.0.3 to dependencies

#[test]
fn test_visible_seats() {
    let map = Map::<Tile>::parse(
        indoc::indoc!(
            "
            .......#.
            ...#.....
            .#.......
            .........
            ..#L....#
            ....#....
            .........
            #........
            ...#.....
            "
        )
        .trim()
        .as_bytes(),
    );
    println!("{:?}", map);
    assert_eq!(map.visible_seats(Vec2 { x: 3, y: 4 }).count(), 8);
    assert_eq!(map.visible_seats(Vec2 { x: 8, y: 0 }).count(), 2);
}

$ cargo t -q

running 2 tests
..
test result: ok. 2 passed; 0 failed; 0 ignored; 0 measured; 0 filtered out

看起来不错！

还有第二个测试:

#[test]
fn test_visible_seats2() {
    let map = Map::<Tile>::parse(
        indoc::indoc!(
            "
            .##.##.
            #.#.#.#
            ##...##
            ...L...
            ##...##
            #.#.#.#
            .##.##.
            "
        )
        .trim()
        .as_bytes(),
    );

    assert_eq!(map.visible_seats(Vec2 { x: 3, y: 3 }).count(), 0);
}

$ cargo t -q

running 3 tests
.F.
failures:

---- test_visible_seats2 stdout ----
.##.##.
#.#.#.#
##...##
...L...
##...##
#.#.#.#
.##.##.

thread 'test_visible_seats2' panicked at 'assertion failed: `(left == right)`
  left: `8`,
 right: `0`', src/main.rs:224:5
note: run with `RUST_BACKTRACE=1` environment variable to display a backtrace


failures:
    test_visible_seats2

test result: FAILED. 2 passed; 1 failed; 0 ignored; 0 measured; 0 filtered out

酷熊：看起来很糟糕！

那么，这里发生了什么? 让我们试着找出它看到了哪些 tile:

// in our test
dbg!(map.visible_seats(Vec2 { x: 3, y: 3 }).collect::<Vec<_>>());
assert_eq!(map.visible_seats(Vec2 { x: 3, y: 3 }).count(), 0);

$ cargo t -q seats2

running 1 test
F
failures:

---- test_visible_seats2 stdout ----
.##.##.
#.#.#.#
##...##
...L...
##...##
#.#.#.#
.##.##.

[src/main.rs:224] map.visible_seats(Vec2{x: 3, y: 3,}).collect::<Vec<_>>() = [
    L,
    L,
    L,
    L,
    L,
    L,
    L,
    L,
]
thread 'test_visible_seats2' panicked at 'assertion failed: `(left == right)`
  left: `8`,
 right: `0`', src/main.rs:225:5

酷熊：哦，它只看到它自己! 因为我们是从 pos 开始的，而不是 pos + (dx, dy)。

Amos：是的，看起来我们在调用 itertools::iterate 时把初始值搞错了。

酷熊：或者我们可以跳过这个？

Amos：我也这样觉得。

// keep moving in set direction
itertools::iterate(pos, move |v| Vec2 {
    x: v.x + dx,
    y: v.y + dy,
})
// 👇 new!
.skip(1)

现在我们所有的测试都通过了:

$ cargo t -q

running 3 tests
...
test result: ok. 3 passed; 0 failed; 0 ignored; 0 measured; 0 filtered out

我们调整第二部分问题中指定的 Tile::next 方法:

现在需要 5 个或更多可见的有人坐的座位才能使一个有人坐的座位变成空的(而不是以前规则中的 4 个或更多)

Self::OccupiedSeat => { match neighbors
        .filter(|t| matches!(t, Self::OccupiedSeat))
        .count()
    {
        // 👇 new!
        // up to 4 neighbors: still okay for now
        0..=4 => Self::OccupiedSeat,
        // that's too many folks!
        _ => Self::EmptySeat,
    }
}

然后更改 Map::<Tile>::next 使用 visible_seats 而不是 neighbor_tiles：

impl Map<Tile> {
    fn next(&self) -> Self {
        let mut res = Self::new(self.size);
        res.extend(
            self.iter()
                //                                                          👇👇👇
                .map(|Positioned(pos, tile)| Positioned(pos, tile.next(self.visible_seats(pos)))),
        );
        res
    }
}

我们应该可以开始运行了！

这个程序在 debug 模式时构建需要一点时间:

$ cargo build --quiet
    Finished dev [unoptimized + debuginfo] target(s) in 0.01s
$ time ./target/debug/day11 > /dev/null
./target/debug/day11 > /dev/null  5.68s user 0.00s system 99% cpu 5.680 total

让我们看看 release 构建的版本执行情况:

$ cargo build --release
    Finished release [optimized] target(s) in 0.01s
$ time ./target/release/day11 > /dev/null 
./target/release/day11 > /dev/null  0.24s user 0.00s system 99% cpu 0.240 total

好多了！

酷熊热辣小贴士如果您的 Rust 程序在 crate 运行时比较慢，不要忘记使用 --release 标志来启用优化运行速度！

现在让我们来看看最终答案:

$ cargo run --release
    Finished release [optimized] target(s) in 0.01s
     Running `target/release/day11`
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