aoc/year2025/day09.rs
1//! # Movie Theater
2use crate::util::iter::*;
3use crate::util::parse::*;
4
5type Tile = [u32; 2];
6
7struct Candidate {
8 x: u32,
9 y: u32,
10 interval: Interval,
11}
12
13/// The set { x in u32 | l <= x <= r }.
14#[derive(Clone, Copy)]
15struct Interval {
16 l: u32,
17 r: u32,
18}
19
20impl Interval {
21 fn new(l: u32, r: u32) -> Self {
22 debug_assert!(l <= r);
23
24 Interval { l, r }
25 }
26
27 fn intersects(self, other: Self) -> bool {
28 other.l <= self.r && self.l <= other.r
29 }
30
31 fn intersection(self, other: Self) -> Self {
32 debug_assert!(self.intersects(other));
33
34 Interval::new(self.l.max(other.l), self.r.min(other.r))
35 }
36
37 fn contains(self, x: u32) -> bool {
38 self.l <= x && x <= self.r
39 }
40}
41
42pub fn parse(input: &str) -> Vec<Tile> {
43 let mut tiles: Vec<_> = input.iter_unsigned::<u32>().chunk::<2>().collect();
44 tiles.sort_unstable_by_key(|&[x, y]| (y, x));
45 tiles
46}
47
48pub fn part1(tiles: &[Tile]) -> u64 {
49 let (top_left_tiles, top_right_tiles) = get_potential_left_corner_tiles(tiles.iter().copied());
50 let (bottom_left_tiles, bottom_right_tiles) =
51 get_potential_left_corner_tiles(tiles.iter().copied().rev());
52
53 find_largest_from_all_corners(&top_left_tiles, &bottom_right_tiles)
54 .max(find_largest_from_all_corners(&bottom_left_tiles, &top_right_tiles))
55}
56
57/// This function filters `sorted_tiles` into two lists, one containing all tiles that could be the top left
58/// corner of the largest rectangle (assuming the largest rectangle has a top left corner), and the second
59/// containing all tiles that could be the top right corner.
60///
61/// It assumes `sorted_tiles` is sorted in ascending "y" values, or, to get the top right and bottom right corners,
62/// that `sorted_tiles` is sorted in descending "y" order.
63///
64/// It works (for the top left corners, for illustration) by only returning tiles (from the set of all tiles, "T") within
65/// the region:
66///
67/// R = { (x, y) ∈ ℝ² : ∀ (tx, ty) ∈ T, tx ≤ x ⇒ ty ≥ y }
68///
69/// Tiles outside of this region cannot possibly be a corner of the largest rectangle. Assume, for proof by contradiction,
70/// that the top left corner of the largest rectangle is in the complement of the set "R":
71///
72/// R' = { (x, y) ∈ ℝ² : ¬ (∀ (tx, ty) ∈ T, tx ≤ x ⇒ ty ≥ y) }
73/// = { (x, y) ∈ ℝ² : ∃ (tx, ty) ∈ T, tx ≤ x ∧ ty < y }
74///
75/// That is, for the corner (x, y), there exists another tile (tx, ty) that is to the left and above the corner tile, which
76/// means the tile isn't the corner of the largest possible rectangle, completing the proof by contradiction.
77///
78/// The `top_tiles` and `bottom_tiles` are the corner points of this region `R`, built up by scanning through tiles
79/// in either left to right or right to left order.
80fn get_potential_left_corner_tiles(
81 sorted_tiles: impl Iterator<Item = [u32; 2]>,
82) -> (Vec<[u32; 2]>, Vec<[u32; 2]>) {
83 let mut left_tiles = Vec::new();
84 let mut left_tiles_last_x = u32::MAX;
85
86 let mut right_tiles = Vec::new();
87 let mut right_tiles_last_x = u32::MIN;
88
89 let mut it = sorted_tiles.peekable();
90
91 while let Some(first_in_row) = it.next() {
92 let mut last_in_row = first_in_row;
93
94 while let Some(p) = it.next_if(|p| p[1] == first_in_row[1]) {
95 last_in_row = p;
96 }
97
98 let (y, left_x, right_x) = (
99 first_in_row[1],
100 first_in_row[0].min(last_in_row[0]),
101 first_in_row[0].max(last_in_row[0]),
102 );
103
104 if left_x < left_tiles_last_x {
105 left_tiles.push([left_x, y]);
106 left_tiles_last_x = left_x;
107 }
108
109 if right_x > right_tiles_last_x {
110 right_tiles.push([right_x, y]);
111 right_tiles_last_x = right_x;
112 }
113 }
114
115 (left_tiles, right_tiles)
116}
117
118#[inline]
119fn find_largest_from_all_corners(corner: &[[u32; 2]], opposite_corner: &[[u32; 2]]) -> u64 {
120 let mut largest = 0_u64;
121
122 for &p in corner {
123 for &q in opposite_corner {
124 largest =
125 largest.max((p[0].abs_diff(q[0]) + 1) as u64 * (p[1].abs_diff(q[1]) + 1) as u64);
126 }
127 }
128
129 largest
130}
131
132pub fn part2(tiles: &[Tile]) -> u64 {
133 // Track the largest area so far during scanning:
134 let mut largest_area: u64 = 0;
135
136 // Each red tile (`x`, `y`) becomes a candidate for being a top corner of the largest area, and during the
137 // scan, the `interval` containing the maximum possible width is updated:
138 let mut candidates: Vec<Candidate> = Vec::with_capacity(512);
139
140 // Maintain an ordered list of descending edges, i.e. [begin_interval_0, end_interval_0, begin_interval_1, end_interval_1, ...]:
141 let mut descending_edges: Vec<u32> = vec![];
142 let mut intervals_from_descending_edges = vec![];
143
144 // Invariants on the input data (defined by the puzzle) result in points arriving in pairs on the same y line:
145 for [&[x0, y], &[x1, y1]] in tiles.iter().chunk::<2>() {
146 debug_assert_eq!(y, y1);
147
148 // Update the descending edges; since we are scanning from top to bottom, and within each line left to right,
149 // when we, starting from outside of the region, hit a corner tile it is either:
150 //
151 // - The corner of two edges, one going right and one going down. In this case, the `descending_edges` won't contain
152 // the `x` coordinate, and we should "toggle" it on to denote that there is a new descending edge.
153 // - The corner of two edges, one going right and one going up. The `descending_edges` will contain an `x` coordinate
154 // that should be "toggled" off.
155 //
156 // Similar arguments work for when we are scanning inside the edge and we hit the corner that ends the edge; this is also
157 // why corners always arrive in pairs.
158 //
159 // Do the update:
160 for x in [x0, x1] {
161 toggle_value_membership_in_ordered_list(&mut descending_edges, x);
162 }
163
164 // Every pair of descending edges in the ordered list defines a region; find the resulting intervals on this line:
165 update_intervals_from_descending_edges(
166 &descending_edges,
167 &mut intervals_from_descending_edges,
168 );
169
170 // Check the rectangles this red tile could be a bottom tile for, with the current candidates:
171 for candidate in &candidates {
172 for x in [x0, x1] {
173 if candidate.interval.contains(x) {
174 largest_area = largest_area.max(
175 (candidate.x.abs_diff(x) + 1) as u64 * (candidate.y.abs_diff(y) + 1) as u64,
176 );
177 }
178 }
179 }
180
181 // Update candidates when their interval shrinks due to descending edge changes, and drop them when their interval becomes empty:
182 candidates.retain_mut(|candidate| {
183 if let Some(intersection_containing_x) =
184 intervals_from_descending_edges.iter().find(|i| i.contains(candidate.x))
185 {
186 candidate.interval = intersection_containing_x.intersection(candidate.interval);
187
188 true
189 } else {
190 false
191 }
192 });
193
194 // Add any new candidates:
195 for x in [x0, x1] {
196 if let Some(&containing) =
197 intervals_from_descending_edges.iter().find(|i| i.contains(x))
198 {
199 candidates.push(Candidate { x, y, interval: containing });
200 }
201 }
202 }
203
204 largest_area
205}
206
207// Adds `value` if it isn't in `ordered_list`, removes it if it is, maintaining the order.
208fn toggle_value_membership_in_ordered_list(ordered_list: &mut Vec<u32>, value: u32) {
209 match ordered_list.binary_search(&value) {
210 Ok(i) => {
211 ordered_list.remove(i);
212 }
213 Err(i) => {
214 ordered_list.insert(i, value);
215 }
216 }
217}
218
219// Changes the list of descending edges, [begin_interval_0, end_interval_0, begin_interval_1, end_interval_1, ...],
220// into a vector containing the intervals.
221#[inline]
222fn update_intervals_from_descending_edges(descending_edges: &[u32], to_update: &mut Vec<Interval>) {
223 to_update.clear();
224 to_update.extend(descending_edges.chunks_exact(2).map(|c| Interval::new(c[0], c[1])));
225}