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aoc/year2019/
day17.rs

1//! # Set and Forget
2//!
3//! The key insight is that this is not a pathfinding problem but a *compression*
4//! problem. We need to reduce the robot's path into repetitions of three patterns.
5//! This is essentially a very simple version of the well-known
6//! [LZW](https://en.wikipedia.org/wiki/Lempel-Ziv-Welch)
7//! algorithm used by the `GIF` and `ZIP` file formats.
8//!
9//! First we find the complete path with a simple heuristic:
10//! * Rotate left or right to face the current path segment (a horizontal or vertical line).
11//! * Go forward until we hit the end of the current path segment.
12//! * If it's a dead end then finish.
13//!
14//! Then we look for three patterns that can be repeated in any order to form the whole path.
15//! Without loss of generality the first pattern anchored at the start is always `A`,
16//! the next `B` and the last `C`.
17//!
18//! A good chunk of the Intcode runtime is spent on unpacking compressed memory into the grid
19//! that is first displayed to the user; this effort is the same between both parts. Running the
20//! entire solution in parse thus reduces the overall runtime.
21use super::intcode::*;
22use crate::util::hash::*;
23use crate::util::parse::*;
24use crate::util::point::*;
25use std::fmt::Write as _;
26use std::iter::once;
27use std::ops::ControlFlow;
28
29type Input = (FastSet<Point>, i64);
30
31struct Movement<'a> {
32    routine: String,
33    functions: [Option<&'a str>; 3],
34}
35
36/// The camera output points from left to right, top to bottom.
37pub fn parse(input: &str) -> Input {
38    // Only run the part two program; its initial output matches part one, and avoiding the
39    // startup costs of a second machine results in a faster solution.
40    let code: Vec<_> = once(2).chain(input.iter_signed().skip(1)).collect();
41    let mut computer = Computer::new(&code);
42
43    let mut x = 0;
44    let mut y = 0;
45    let mut scaffold = FastSet::new();
46    let mut position = ORIGIN;
47    let mut direction = ORIGIN;
48
49    while let State::Output(next) = computer.run() {
50        let next = next as u8;
51        let point = Point::new(x, y);
52
53        match next {
54            b'\n' => {
55                y += 1;
56                x = 0;
57                continue;
58            }
59            b'#' => {
60                scaffold.insert(point);
61            }
62            b'<' | b'>' | b'^' | b'v' => {
63                scaffold.insert(point);
64                position = point;
65                direction = Point::from(next);
66            }
67            _ => (),
68        }
69
70        x += 1;
71    }
72
73    // With the scaffold now available, construct the compressed path.
74    let path = build_path(&scaffold, position, direction);
75    let mut movement = Movement { routine: String::new(), functions: [None; 3] };
76
77    let _unused = compress(&path, &mut movement);
78
79    // Convert trailing comma ',' into a trailing newline '\n'
80    let mut rules = String::new();
81    let parts = once(movement.routine.as_str()).chain(movement.functions.into_iter().flatten());
82    for s in parts {
83        rules.push_str(s);
84        rules.pop();
85        rules.push('\n');
86    }
87
88    computer.input_ascii(&rules);
89    let score = visit(computer);
90
91    (scaffold, score)
92}
93
94pub fn part1(input: &Input) -> i32 {
95    let (scaffold, _) = input;
96    scaffold
97        .iter()
98        .filter(|&point| ORTHOGONAL.iter().all(|&delta| scaffold.contains(&(*point + delta))))
99        .map(|point| point.x * point.y)
100        .sum()
101}
102
103pub fn part2(input: &Input) -> i64 {
104    input.1
105}
106
107/// Use a simple heuristic to build a path that visits every part of the scaffold at least once.
108/// This string will be too long to use directly in the robot's movement functions, so we'll
109/// need to compress it first.
110fn build_path(scaffold: &FastSet<Point>, mut position: Point, mut direction: Point) -> String {
111    let mut path = String::new();
112
113    loop {
114        let left = direction.counter_clockwise();
115        let right = direction.clockwise();
116
117        if scaffold.contains(&(position + left)) {
118            direction = left;
119        } else if scaffold.contains(&(position + right)) {
120            direction = right;
121        } else {
122            break path;
123        }
124
125        let mut next = position + direction;
126        let mut magnitude = 0;
127
128        while scaffold.contains(&next) {
129            position = next;
130            next += direction;
131            magnitude += 1;
132        }
133
134        let direction = if direction == left { 'L' } else { 'R' };
135        let _ = write!(path, "{direction},{magnitude},");
136    }
137}
138
139/// Find three patterns that can be repeated in any order to build the whole path.
140///
141/// Uses a greedy backtracking algorithm that attempts to match as much of the remaining string
142/// as possible with known patterns, before trying combinations of a new pattern (up to the maximum
143/// movement function length of 20 characters).
144fn compress<'a>(path: &'a str, movement: &mut Movement<'a>) -> ControlFlow<()> {
145    // Nothing left to match, we've finished successfully.
146    if path.is_empty() {
147        return ControlFlow::Break(());
148    }
149    // Safety check just in case very short sequences can match the entire input.
150    if movement.routine.len() > 21 {
151        return ControlFlow::Continue(());
152    }
153
154    for (i, &name) in ['A', 'B', 'C'].iter().enumerate() {
155        movement.routine.push(name);
156        movement.routine.push(',');
157
158        if let Some(needle) = movement.functions[i] {
159            // Try known patterns first.
160            if let Some(remaining) = path.strip_prefix(needle) {
161                compress(remaining, movement)?;
162            }
163        } else {
164            // Then combinations up to length 20 characters.
165            for (needle, remaining) in segments(path) {
166                movement.functions[i] = Some(needle);
167                compress(remaining, movement)?;
168                movement.functions[i] = None;
169            }
170        }
171
172        movement.routine.pop();
173        movement.routine.pop();
174    }
175
176    ControlFlow::Continue(())
177}
178
179/// Fun with iterators.
180fn segments(path: &str) -> impl Iterator<Item = (&str, &str)> {
181    path.bytes()
182        .enumerate()
183        // Index of every comma ',' in the string.
184        .filter_map(|(i, b)| (b == b',').then_some(i))
185        // Maximum length for movement function is 20 characters.
186        .take_while(|&i| i < 21)
187        // Include trailing comma in "needle" to make matching easier.
188        .map(|i| path.split_at(i + 1))
189        // Movement is always pairs of (rotation, magnitude) so return every second comma.
190        .skip(1)
191        .step_by(2)
192}
193
194#[cfg(not(feature = "frivolity"))]
195fn visit(mut computer: Computer) -> i64 {
196    // Disable continuous video feed.
197    computer.input_ascii("n\n");
198
199    let mut result = 0;
200    while let State::Output(next) = computer.run() {
201        result = next;
202    }
203    result
204}
205
206/// Non essential but fun. Animates the robot traversing the scaffold.
207#[cfg(feature = "frivolity")]
208fn visit(mut computer: Computer) -> i64 {
209    use crate::util::ansi::*;
210    use std::thread::sleep;
211    use std::time::Duration;
212
213    let mut result = 0;
214    let mut previous = ' ';
215    let mut buffer = String::new();
216
217    // Enable continuous video feed.
218    computer.input_ascii("y\n");
219
220    while let State::Output(next) = computer.run() {
221        result = next;
222        let ascii = (next as u8) as char;
223
224        // Highlight the robot's position.
225        match ascii {
226            '^' | 'v' | '<' | '>' => {
227                let _ = write!(&mut buffer, "{BOLD}{YELLOW}{ascii}{RESET}");
228            }
229            _ => buffer.push(ascii),
230        }
231
232        // Each frame is separated by a blank line.
233        if ascii == '\n' && previous == '\n' {
234            print!("{HOME}{CLEAR}{buffer}");
235            sleep(Duration::from_millis(25));
236            buffer.clear();
237        }
238
239        previous = ascii;
240    }
241
242    result
243}