aoc/year2019/
day18.rs

1//! # Many-Worlds Interpretation
2//!
3//! Our high-level approach is to simplify the problem into graph path finding. We only
4//! ever need to move directly from key to key, so the maze becomes a graph where the nodes are
5//! keys and the edge weight is the distance between keys. Doors modify which edges
6//! are connected depending on the keys currently possessed.
7//!
8//! We first find the distance between every pair of keys then run
9//! [Dijkstra's algorithm](https://en.wikipedia.org/wiki/Dijkstra%27s_algorithm) to find the
10//! shortest path that visits every node in the graph.
11
12//! The maze is also constructed in such a way to make our life easier:
13//! * There is only ever one possible path to each key. We do not need to consider
14//!   paths of different lengths that need different keys.
15//! * As a corollary, if key `b` lies between `a` and `c` then `|ac| = |ab| + |bc|`. This
16//!   enables a huge optimization that we only need to consider immediate neighbours.
17//!   If we do not possess key `b` then it never makes sense to skip from `a` to `c` since `b` is
18//!   along the way. We can model this by treating keys the same as doors. This optimization
19//!   sped up my solution by a factor of 30.
20//!
21//! On top of this approach we apply some high-level tricks to go faster:
22//! * We store previously seen pairs of `(location, keys collected)` to `total distance` in a map.
23//!   If we are in the same location with the same keys but at a higher cost, then this situation
24//!   can never be optimal so the solution can be discarded.
25//! * When finding the distance between every pair of keys, it's faster to first only find the immediate
26//!   neighbors of each key using a [Breadth first search](https://en.wikipedia.org/wiki/Breadth-first_search)
27//!   then run the [Floyd-Warshall algorithm](https://en.wikipedia.org/wiki/Floyd%E2%80%93Warshall_algorithm)
28//!   to construct the rest of the graph. Even though the Floyd-Warshall asymptotic bound of `O(n³)`
29//!   is higher than the asymptotic bounds of repeated BFS, this was twice as fast in practice
30//!   for my input.
31//!
32//! We also apply some low-level tricks to go even faster:
33//! * The set of remaining keys needed is stored as bits in an `u32`. We can have at most 26 keys
34//!   so this will always fit. For example needing `a`, `b` and `e` is represented as `10011`.
35//! * Robot location is also stored the same way. Robots can only ever be in their initial location
36//!   or at a key, so this gives a max of 26 + 4 = 30 locations. As a nice bonus this allows
37//!   part one and part two to share the same code.
38//! * For fast lookup of distance between keys, the maze is stored as
39//!   [adjacency matrix](https://en.wikipedia.org/wiki/Adjacency_matrix). `a` is index 0, `b` is
40//!   index 1 and robots' initial positions are from 26 to 29 inclusive.
41//!   For example (simplifying by moving robot from index 26 to 2):
42//!
43//!   ```none
44//!       #########    [0 6 2]
45//!       #b.A.@.a# => [6 0 4]
46//!       #########    [2 4 0]
47//!   ```
48use crate::util::bitset::*;
49use crate::util::grid::*;
50use crate::util::hash::*;
51use crate::util::heap::*;
52use std::collections::VecDeque;
53use std::ops::Range;
54
55const RANGE: Range<usize> = 0..30;
56
57/// `position` and `remaining` are both bitfields. For example a robot at key `d` that needs
58/// `b` and `c` would be stored as `position = 1000` and `remaining = 110`.
59#[derive(Clone, Copy, Default, PartialEq, Eq, Hash)]
60struct State {
61    position: u32,
62    remaining: u32,
63}
64
65/// `distance` in the edge weight between nodes. `needed` stores any doors in between as a bitfield.
66#[derive(Clone, Copy)]
67struct Door {
68    distance: u32,
69    needed: u32,
70}
71
72/// `initial` is the complete set of keys that we need to collect. Will always be binary
73/// `11111111111111111111111111` for the real input but fewer for sample data.
74///
75/// `maze` is the adjacency of distances and doors between each pair of keys and the robots
76/// starting locations.
77struct Maze {
78    initial: State,
79    maze: [[Door; 30]; 30],
80}
81
82pub fn parse(input: &str) -> Grid<u8> {
83    Grid::parse(input)
84}
85
86pub fn part1(input: &Grid<u8>) -> u32 {
87    explore(input.width as usize, &input.bytes)
88}
89
90pub fn part2(input: &Grid<u8>) -> u32 {
91    let mut modified = input.bytes.clone();
92    let mut patch = |s: &str, offset: i32| {
93        let middle = (input.width * input.height) / 2;
94        let index = (middle + offset * input.width - 1) as usize;
95        modified[index..index + 3].copy_from_slice(s.as_bytes());
96    };
97
98    patch("@#@", -1);
99    patch("###", 0);
100    patch("@#@", 1);
101
102    explore(input.width as usize, &modified)
103}
104
105fn parse_maze(width: usize, bytes: &[u8]) -> Maze {
106    let mut initial = State::default();
107    let mut found = Vec::new();
108    let mut robots = 26;
109
110    // Find the location of every key and robot in the maze.
111    for (i, &b) in bytes.iter().enumerate() {
112        if let Some(key) = is_key(b) {
113            initial.remaining |= 1 << key;
114            found.push((i, key));
115        }
116        if b == b'@' {
117            initial.position |= 1 << robots;
118            found.push((i, robots));
119            robots += 1;
120        }
121    }
122
123    // Start a BFS from each key and robot's location stopping at the nearest neighbor.
124    // As a minor optimization we re-use the same `todo` and `visited` between each search.
125    let default = Door { distance: u32::MAX, needed: 0 };
126
127    let mut maze = [[default; 30]; 30];
128    let mut visited = vec![usize::MAX; bytes.len()];
129    let mut todo = VecDeque::new();
130
131    for (start, from) in found {
132        todo.push_front((start, 0, 0));
133        visited[start] = from;
134
135        while let Some((index, distance, mut needed)) = todo.pop_front() {
136            if let Some(door) = is_door(bytes[index]) {
137                needed |= 1 << door;
138            }
139
140            if let Some(to) = is_key(bytes[index])
141                && distance > 0
142            {
143                // Store the reciprocal edge weight and doors in the adjacency matrix.
144                maze[from][to] = Door { distance, needed };
145                maze[to][from] = Door { distance, needed };
146                // Faster to stop here and use Floyd-Warshall later.
147                continue;
148            }
149
150            for next in [index + 1, index - 1, index + width, index - width] {
151                if bytes[next] != b'#' && visited[next] != from {
152                    todo.push_back((next, distance + 1, needed));
153                    visited[next] = from;
154                }
155            }
156        }
157    }
158
159    // Fill in the rest of the graph using the Floyd–Warshal algorithm.
160    // As a slight twist we also build the list of intervening doors at the same time.
161    for i in RANGE {
162        maze[i][i].distance = 0;
163    }
164
165    for k in RANGE {
166        for i in RANGE {
167            for j in RANGE {
168                let candidate = maze[i][k].distance.saturating_add(maze[k][j].distance);
169                if maze[i][j].distance > candidate {
170                    maze[i][j].distance = candidate;
171                    // `(1 << k)` is a crucial optimization. By treating intermediate keys like
172                    // doors we speed things up by a factor of 30.
173                    maze[i][j].needed = maze[i][k].needed | (1 << k) | maze[k][j].needed;
174                }
175            }
176        }
177    }
178
179    Maze { initial, maze }
180}
181
182fn explore(width: usize, bytes: &[u8]) -> u32 {
183    let mut todo = MinHeap::with_capacity(5_000);
184    let mut cache = FastMap::with_capacity(5_000);
185
186    let Maze { initial, maze } = parse_maze(width, bytes);
187    todo.push(0, initial);
188
189    while let Some((total, State { position, remaining })) = todo.pop() {
190        // Finish immediately if no keys left.
191        // Since we're using Dijkstra this will always be the optimal solution.
192        if remaining == 0 {
193            return total;
194        }
195
196        // The set of robots is stored as bits in a `u32` shifted by the index of the location.
197        for from in position.biterator() {
198            // The set of keys still needed is also stored as bits in a `u32` similar as robots.
199            for to in remaining.biterator() {
200                let Door { distance, needed } = maze[from][to];
201
202                // u32::MAX indicates that two nodes are not connected. Only possible in part two.
203                if distance != u32::MAX && remaining & needed == 0 {
204                    let next_total = total + distance;
205                    let from_mask = 1 << from;
206                    let to_mask = 1 << to;
207                    let next_state = State {
208                        position: position ^ from_mask ^ to_mask,
209                        remaining: remaining ^ to_mask,
210                    };
211
212                    // Memoize previously seen states to eliminate suboptimal states right away.
213                    cache
214                        .entry(next_state)
215                        .and_modify(|e| {
216                            if next_total < *e {
217                                todo.push(next_total, next_state);
218                                *e = next_total;
219                            }
220                        })
221                        .or_insert_with(|| {
222                            todo.push(next_total, next_state);
223                            next_total
224                        });
225                }
226            }
227        }
228    }
229
230    unreachable!()
231}
232
233// Convenience functions to find keys and robots
234fn is_key(b: u8) -> Option<usize> {
235    b.is_ascii_lowercase().then(|| (b - b'a') as usize)
236}
237
238fn is_door(b: u8) -> Option<usize> {
239    b.is_ascii_uppercase().then(|| (b - b'A') as usize)
240}