-
Notifications
You must be signed in to change notification settings - Fork 0
/
18_many_worlds_interpretation.rb
207 lines (181 loc) · 6.74 KB
/
18_many_worlds_interpretation.rb
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
require_relative 'lib/search'
def bitfield(chars, range)
base = range.begin.ord
chars.select { |c| range.cover?(c) }.map { |c| 1 << (c.ord - base) }.reduce(0, :|)
end
def key_to_key(flat_input, width, sources)
# AoC-specific optimisation:
# For all paths between key -> key that contain doors,
# the doors block the ONLY path to the key.
# Given this property is true, all key -> key paths are precomputed.
# This property is false for these inputs, with sources:
# https://www.reddit.com/r/adventofcode/comments/ecj4e7/2019_day_18_challenging_input/
# ##########
# #.a###.Ab#
# #.B..@.###
# #...######
# ##########
# https://www.reddit.com/r/adventofcode/comments/ecgyey/2019_day_18_part_1_im_not_seeing_how_to_optimize/fbc3iih/
# #######
# #....@#
# #.###A#
# #.###b#
# #.aBCc#
# #######
idx = sources.each_with_index.to_h
# BFS from {start, each key} to all other single keys.
# Paths picking up multiple keys will be computed after, using this info.
# Positions will be renumbered to be their index in the list.
sources.map { |src|
have_new_key = ->pos { pos != src && (?a..?z).cover?(flat_input[pos]) }
keys = Search.bfs(
src, num_goals: Float::INFINITY,
neighbours: ->pos {
next [] if have_new_key[pos]
[pos - width, pos + width, pos - 1, pos + 1].select { |npos|
flat_input[npos] != ?#
}
},
goal: have_new_key,
)
keys[:goals].to_h { |pos, dist|
path = Search.path_of(keys[:prev], pos)
things_on_path = path.map { |path_pos| flat_input[path_pos] }
[idx[pos], {
pos: idx[pos],
dist: dist,
# Represent keys and doors as bitfields so set intersections become cheap
keys: bitfield([flat_input[pos], flat_input[src]], ?a..?z),
doors: bitfield(things_on_path, ?A..?Z),
}.freeze]
}
}
end
def all_pairs(keys_from)
# https://en.wikipedia.org/wiki/Floyd%E2%80%93Warshall_algorithm
# Using this is much faster than traveling the entire map for each key.
keys_from.each_index { |k|
keys_from.each_index { |i|
next if k == i
next unless (ik = keys_from.dig(i, k))
keys_from.each_index { |j|
next if i == j || k == j
next unless (kj = keys_from.dig(k, j))
new_dist = ik[:dist] + kj[:dist]
ij = keys_from.dig(i, j)
if !ij || ij[:dist] > new_dist
keys_from[i][j] = {
pos: kj[:pos],
dist: new_dist,
keys: ik[:keys] | kj[:keys],
doors: ik[:doors] | kj[:doors],
}.freeze
end
}
}
}
keys_from.map { |vs| vs.values.sort_by { |v| -v[:dist] }.freeze }.freeze
end
def all_keys_time(keys_from, num_keys, robots)
all_keys = (1 << num_keys) - 1
# Pack all robot positions into one int.
# Now that positions are renumbered to max 31 (4 robots + 26 keys + 1 dummy start),
# they fit in 5 bits.
# With the key bitfield taking 26 bits, the entire state fits within 46 bits.
bits_per_robot = keys_from.size.bit_length
robot_mask = (1 << bits_per_robot) - 1
robot_base = (0...robots.size).map { |i| bits_per_robot * i + num_keys }
cost, _junk = Search.astar(
# Assumption: Renumbering done by keys_from put the robots first.
robots.zip(robot_base).map { |bot, base| bot << base }.reduce(0, :|),
neighbours: ->(robots_and_keys) {
keys = robots_and_keys & all_keys
robot_base.flat_map { |base|
robot = (robots_and_keys >> base) & robot_mask
keys_from[robot].filter_map { |key|
# Have these keys already.
next if key[:keys] | keys == keys
# Don't have all keys needed.
next unless key[:doors] | keys == keys
[(robots_and_keys & ~(robot_mask << base)) | (key[:pos] << base) | key[:keys], key[:dist]]
}
}
},
# heuristic - max dist to remaining keys is at most harmless,
# but does help for certain inputs, it seems.
# Also tried unsuccessfully:
# * MST of remaining keys
# * number remaining keys * minimum distance between two keys
# * Dijkstra's:
# heuristic: Hash.new(0),
heuristic: ->(robots_and_keys) {
keys = robots_and_keys & all_keys
robot_base.sum { |base|
robot = (robots_and_keys >> base) & robot_mask
# since keys_from is sorted in descending order of dist:
not_picked_up = keys_from[robot].find { |key|
key[:keys] | keys != keys
}
not_picked_up&.[](:dist) || 0
}
},
goal: ->(robots_and_keys) { robots_and_keys & all_keys == all_keys },
)
cost
end
input = ARGF.map(&:chomp).map(&:freeze).freeze
# Represent position as y * width + x, indexing into flattened grid.
# The edge of the grid is all walls, so this is fine.
width = input.map(&:size).max
flat_input = input.map { |l| l.ljust(width, ' ') }.join.freeze
keys = []
robots = []
input.each_with_index { |row, y|
row.chars.each_with_index { |cell, x|
pos = y * width + x
keys << pos if (?a..?z).cover?(cell)
robots << pos if cell == ?@
}
}
# If this is a part 1 that can be converted to a part 2, then do both.
# If not, it's fine, just do what's given, since tests use maps that only do part 1 or only do part 2.
can_part_2 = robots.size == 1 && begin
bot = robots[0]
diagonal = [-width, width].product([-1, 1]).map(&:sum)
orthogonal = [-width, width, -1, 1]
surrounding = diagonal + orthogonal
surrounding.all? { |s| flat_input[bot + s] == ?. }
end
if can_part_2
# Calculate the key-to-key map for part 2,
# then transform it into one for part 1.
# (This is faster because of traveling the map fewer times)
robots = diagonal.map { |diag| bot + diag }
flat_input = flat_input.dup
orthogonal.each { |orth| flat_input[bot + orth] = ?# }
flat_input.freeze
k2k = key_to_key(flat_input, width, [bot] + robots + keys)
k2k1 = k2k.map(&:dup)
add_pair = ->(i, j, dist) {
k2k1[i][j] = {pos: j, dist: dist, keys: 0, doors: 0}.freeze
k2k1[j][i] = {pos: i, dist: dist, keys: 0, doors: 0}.freeze
}
(1..4).each { |i|
# Allow each key to go back to the corner (part 2 entrances).
# Normally, they would not try to because the corner has no keys.
k2k1[i].each { |k, v| k2k1[k][i] = v.merge(pos: i) }
# Centre (part 1 entrance) is 2 away from each corner (part 2 entrances)
add_pair[0, i, 2]
}
(1..4).to_a.combination(2) { |i, j|
# Allow each corner (part 2 entrances) to reach each other.
y1, x1 = robots[i - 1].divmod(width)
y2, x2 = robots[j - 1].divmod(width)
add_pair[i, j, (y1 - y2).abs + (x1 - x2).abs]
}
puts all_keys_time(all_pairs(k2k1), keys.size, [0])
puts all_keys_time(all_pairs(k2k), keys.size, (1..robots.size).to_a)
else
k2k = key_to_key(flat_input, width, robots + keys)
puts all_keys_time(all_pairs(k2k), keys.size, (0...robots.size).to_a)
end