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Maya Herrscher e6293fe077 Add solution for day 13 2021-12-13 07:13:07 +01:00
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.gitignore
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AoC2021/Bene/ AoC2021/Bene/
__pycache__
2021/Bene/input.txt
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end-MY
MY-xc
ho-NF
start-ho
NF-xc
NF-yf
end-yf
xc-TP
MY-qo
yf-TP
dc-NF
dc-xc
start-dc
yf-MY
MY-ho
EM-uh
xc-yf
ho-dc
uh-NF
yf-ho
end-uh
start-NF
2021/Bene/main.py
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#!/bin/env python3
import sys
import string
from collections import Counter
def isSmall(node): return all(c in string.ascii_lowercase for c in node)
def ext(path, ns, visits=1):
if path[-1] == 'end': return (yield path)
for n in ns[path[-1]]:
if n == 'start': continue
ext = path + (n,)
m = list(map(lambda x: x[1], Counter(n for n in ext if isSmall(n)).most_common(2)))
if m[0] > visits or len(m) > 1 and m[1] > 1: continue
yield ext
if __name__ == '__main__':
edges = [line.strip().split('-') for line in open(sys.argv[1])]
ns = { n: set() for e in edges for n in e }
for a,b in edges: ns[a].add(b); ns[b].add(a)
ps = set([('start',)])
while (n := set(np for p in ps for np in ext(p, ns))) != ps: ps = n
print(len(ps))
ps = set([('start',)])
while (n := set(np for p in ps for np in ext(p, ns, visits=2))) != ps: ps = n
print(len(ps))
2021/challenges/challenge14
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--- Day 14: Extended Polymerization ---
The incredible pressures at this depth are starting to put a strain on your submarine. The submarine has polymerization equipment that would produce suitable materials to reinforce the submarine, and the nearby volcanically-active caves should even have the necessary input elements in sufficient quantities.
The submarine manual contains instructions for finding the optimal polymer formula; specifically, it offers a polymer template and a list of pair insertion rules (your puzzle input). You just need to work out what polymer would result after repeating the pair insertion process a few times.
For example:
NNCB
CH -> B
HH -> N
CB -> H
NH -> C
HB -> C
HC -> B
HN -> C
NN -> C
BH -> H
NC -> B
NB -> B
BN -> B
BB -> N
BC -> B
CC -> N
CN -> C
The first line is the polymer template - this is the starting point of the process.
The following section defines the pair insertion rules. A rule like AB -> C means that when elements A and B are immediately adjacent, element C should be inserted between them. These insertions all happen simultaneously.
So, starting with the polymer template NNCB, the first step simultaneously considers all three pairs:
The first pair (NN) matches the rule NN -> C, so element C is inserted between the first N and the second N.
The second pair (NC) matches the rule NC -> B, so element B is inserted between the N and the C.
The third pair (CB) matches the rule CB -> H, so element H is inserted between the C and the B.
Note that these pairs overlap: the second element of one pair is the first element of the next pair. Also, because all pairs are considered simultaneously, inserted elements are not considered to be part of a pair until the next step.
After the first step of this process, the polymer becomes NCNBCHB.
Here are the results of a few steps using the above rules:
Template: NNCB
After step 1: NCNBCHB
After step 2: NBCCNBBBCBHCB
After step 3: NBBBCNCCNBBNBNBBCHBHHBCHB
After step 4: NBBNBNBBCCNBCNCCNBBNBBNBBBNBBNBBCBHCBHHNHCBBCBHCB
This polymer grows quickly. After step 5, it has length 97; After step 10, it has length 3073. After step 10, B occurs 1749 times, C occurs 298 times, H occurs 161 times, and N occurs 865 times; taking the quantity of the most common element (B, 1749) and subtracting the quantity of the least common element (H, 161) produces 1749 - 161 = 1588.
Apply 10 steps of pair insertion to the polymer template and find the most and least common elements in the result. What do you get if you take the quantity of the most common element and subtract the quantity of the least common element?
Your puzzle answer was 3697.
--- Part Two ---
The resulting polymer isn't nearly strong enough to reinforce the submarine. You'll need to run more steps of the pair insertion process; a total of 40 steps should do it.
In the above example, the most common element is B (occurring 2192039569602 times) and the least common element is H (occurring 3849876073 times); subtracting these produces 2188189693529.
Apply 40 steps of pair insertion to the polymer template and find the most and least common elements in the result. What do you get if you take the quantity of the most common element and subtract the quantity of the least common element?
Your puzzle answer was 4371307836157.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/challenges/challenge15
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--- Day 15: Chiton ---
You've almost reached the exit of the cave, but the walls are getting closer together. Your submarine can barely still fit, though; the main problem is that the walls of the cave are covered in chitons, and it would be best not to bump any of them.
The cavern is large, but has a very low ceiling, restricting your motion to two dimensions. The shape of the cavern resembles a square; a quick scan of chiton density produces a map of risk level throughout the cave (your puzzle input). For example:
1163751742
1381373672
2136511328
3694931569
7463417111
1319128137
1359912421
3125421639
1293138521
2311944581
You start in the top left position, your destination is the bottom right position, and you cannot move diagonally. The number at each position is its risk level; to determine the total risk of an entire path, add up the risk levels of each position you enter (that is, don't count the risk level of your starting position unless you enter it; leaving it adds no risk to your total).
Your goal is to find a path with the lowest total risk. In this example, a path with the lowest total risk is highlighted here:
1163751742
1381373672
2136511328
3694931569
7463417111
1319128137
1359912421
3125421639
1293138521
2311944581
The total risk of this path is 40 (the starting position is never entered, so its risk is not counted).
What is the lowest total risk of any path from the top left to the bottom right?
Your puzzle answer was 487.
--- Part Two ---
Now that you know how to find low-risk paths in the cave, you can try to find your way out.
The entire cave is actually five times larger in both dimensions than you thought; the area you originally scanned is just one tile in a 5x5 tile area that forms the full map. Your original map tile repeats to the right and downward; each time the tile repeats to the right or downward, all of its risk levels are 1 higher than the tile immediately up or left of it. However, risk levels above 9 wrap back around to 1. So, if your original map had some position with a risk level of 8, then that same position on each of the 25 total tiles would be as follows:
8 9 1 2 3
9 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
Each single digit above corresponds to the example position with a value of 8 on the top-left tile. Because the full map is actually five times larger in both dimensions, that position appears a total of 25 times, once in each duplicated tile, with the values shown above.
Here is the full five-times-as-large version of the first example above, with the original map in the top left corner highlighted:
11637517422274862853338597396444961841755517295286
13813736722492484783351359589446246169155735727126
21365113283247622439435873354154698446526571955763
36949315694715142671582625378269373648937148475914
74634171118574528222968563933317967414442817852555
13191281372421239248353234135946434524615754563572
13599124212461123532357223464346833457545794456865
31254216394236532741534764385264587549637569865174
12931385212314249632342535174345364628545647573965
23119445813422155692453326671356443778246755488935
22748628533385973964449618417555172952866628316397
24924847833513595894462461691557357271266846838237
32476224394358733541546984465265719557637682166874
47151426715826253782693736489371484759148259586125
85745282229685639333179674144428178525553928963666
24212392483532341359464345246157545635726865674683
24611235323572234643468334575457944568656815567976
42365327415347643852645875496375698651748671976285
23142496323425351743453646285456475739656758684176
34221556924533266713564437782467554889357866599146
33859739644496184175551729528666283163977739427418
35135958944624616915573572712668468382377957949348
43587335415469844652657195576376821668748793277985
58262537826937364893714847591482595861259361697236
96856393331796741444281785255539289636664139174777
35323413594643452461575456357268656746837976785794
35722346434683345754579445686568155679767926678187
53476438526458754963756986517486719762859782187396
34253517434536462854564757396567586841767869795287
45332667135644377824675548893578665991468977611257
44961841755517295286662831639777394274188841538529
46246169155735727126684683823779579493488168151459
54698446526571955763768216687487932779859814388196
69373648937148475914825958612593616972361472718347
17967414442817852555392896366641391747775241285888
46434524615754563572686567468379767857948187896815
46833457545794456865681556797679266781878137789298
64587549637569865174867197628597821873961893298417
45364628545647573965675868417678697952878971816398
56443778246755488935786659914689776112579188722368
55172952866628316397773942741888415385299952649631
57357271266846838237795794934881681514599279262561
65719557637682166874879327798598143881961925499217
71484759148259586125936169723614727183472583829458
28178525553928963666413917477752412858886352396999
57545635726865674683797678579481878968159298917926
57944568656815567976792667818781377892989248891319
75698651748671976285978218739618932984172914319528
56475739656758684176786979528789718163989182927419
67554889357866599146897761125791887223681299833479
Equipped with the full map, you can now find a path from the top left corner to the bottom right corner with the lowest total risk:
11637517422274862853338597396444961841755517295286
13813736722492484783351359589446246169155735727126
21365113283247622439435873354154698446526571955763
36949315694715142671582625378269373648937148475914
74634171118574528222968563933317967414442817852555
13191281372421239248353234135946434524615754563572
13599124212461123532357223464346833457545794456865
31254216394236532741534764385264587549637569865174
12931385212314249632342535174345364628545647573965
23119445813422155692453326671356443778246755488935
22748628533385973964449618417555172952866628316397
24924847833513595894462461691557357271266846838237
32476224394358733541546984465265719557637682166874
47151426715826253782693736489371484759148259586125
85745282229685639333179674144428178525553928963666
24212392483532341359464345246157545635726865674683
24611235323572234643468334575457944568656815567976
42365327415347643852645875496375698651748671976285
23142496323425351743453646285456475739656758684176
34221556924533266713564437782467554889357866599146
33859739644496184175551729528666283163977739427418
35135958944624616915573572712668468382377957949348
43587335415469844652657195576376821668748793277985
58262537826937364893714847591482595861259361697236
96856393331796741444281785255539289636664139174777
35323413594643452461575456357268656746837976785794
35722346434683345754579445686568155679767926678187
53476438526458754963756986517486719762859782187396
34253517434536462854564757396567586841767869795287
45332667135644377824675548893578665991468977611257
44961841755517295286662831639777394274188841538529
46246169155735727126684683823779579493488168151459
54698446526571955763768216687487932779859814388196
69373648937148475914825958612593616972361472718347
17967414442817852555392896366641391747775241285888
46434524615754563572686567468379767857948187896815
46833457545794456865681556797679266781878137789298
64587549637569865174867197628597821873961893298417
45364628545647573965675868417678697952878971816398
56443778246755488935786659914689776112579188722368
55172952866628316397773942741888415385299952649631
57357271266846838237795794934881681514599279262561
65719557637682166874879327798598143881961925499217
71484759148259586125936169723614727183472583829458
28178525553928963666413917477752412858886352396999
57545635726865674683797678579481878968159298917926
57944568656815567976792667818781377892989248891319
75698651748671976285978218739618932984172914319528
56475739656758684176786979528789718163989182927419
67554889357866599146897761125791887223681299833479
The total risk of this path is 315 (the starting position is still never entered, so its risk is not counted).
Using the full map, what is the lowest total risk of any path from the top left to the bottom right?
Your puzzle answer was 2821.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/challenges/challenge16
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--- Day 16: Packet Decoder ---
As you leave the cave and reach open waters, you receive a transmission from the Elves back on the ship.
The transmission was sent using the Buoyancy Interchange Transmission System (BITS), a method of packing numeric expressions into a binary sequence. Your submarine's computer has saved the transmission in hexadecimal (your puzzle input).
The first step of decoding the message is to convert the hexadecimal representation into binary. Each character of hexadecimal corresponds to four bits of binary data:
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001
A = 1010
B = 1011
C = 1100
D = 1101
E = 1110
F = 1111
The BITS transmission contains a single packet at its outermost layer which itself contains many other packets. The hexadecimal representation of this packet might encode a few extra 0 bits at the end; these are not part of the transmission and should be ignored.
Every packet begins with a standard header: the first three bits encode the packet version, and the next three bits encode the packet type ID. These two values are numbers; all numbers encoded in any packet are represented as binary with the most significant bit first. For example, a version encoded as the binary sequence 100 represents the number 4.
Packets with type ID 4 represent a literal value. Literal value packets encode a single binary number. To do this, the binary number is padded with leading zeroes until its length is a multiple of four bits, and then it is broken into groups of four bits. Each group is prefixed by a 1 bit except the last group, which is prefixed by a 0 bit. These groups of five bits immediately follow the packet header. For example, the hexadecimal string D2FE28 becomes:
110100101111111000101000
VVVTTTAAAAABBBBBCCCCC
Below each bit is a label indicating its purpose:
The three bits labeled V (110) are the packet version, 6.
The three bits labeled T (100) are the packet type ID, 4, which means the packet is a literal value.
The five bits labeled A (10111) start with a 1 (not the last group, keep reading) and contain the first four bits of the number, 0111.
The five bits labeled B (11110) start with a 1 (not the last group, keep reading) and contain four more bits of the number, 1110.
The five bits labeled C (00101) start with a 0 (last group, end of packet) and contain the last four bits of the number, 0101.
The three unlabeled 0 bits at the end are extra due to the hexadecimal representation and should be ignored.
So, this packet represents a literal value with binary representation 011111100101, which is 2021 in decimal.
Every other type of packet (any packet with a type ID other than 4) represent an operator that performs some calculation on one or more sub-packets contained within. Right now, the specific operations aren't important; focus on parsing the hierarchy of sub-packets.
An operator packet contains one or more packets. To indicate which subsequent binary data represents its sub-packets, an operator packet can use one of two modes indicated by the bit immediately after the packet header; this is called the length type ID:
If the length type ID is 0, then the next 15 bits are a number that represents the total length in bits of the sub-packets contained by this packet.
If the length type ID is 1, then the next 11 bits are a number that represents the number of sub-packets immediately contained by this packet.
Finally, after the length type ID bit and the 15-bit or 11-bit field, the sub-packets appear.
For example, here is an operator packet (hexadecimal string 38006F45291200) with length type ID 0 that contains two sub-packets:
00111000000000000110111101000101001010010001001000000000
VVVTTTILLLLLLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBBBBBB
The three bits labeled V (001) are the packet version, 1.
The three bits labeled T (110) are the packet type ID, 6, which means the packet is an operator.
The bit labeled I (0) is the length type ID, which indicates that the length is a 15-bit number representing the number of bits in the sub-packets.
The 15 bits labeled L (000000000011011) contain the length of the sub-packets in bits, 27.
The 11 bits labeled A contain the first sub-packet, a literal value representing the number 10.
The 16 bits labeled B contain the second sub-packet, a literal value representing the number 20.
After reading 11 and 16 bits of sub-packet data, the total length indicated in L (27) is reached, and so parsing of this packet stops.
As another example, here is an operator packet (hexadecimal string EE00D40C823060) with length type ID 1 that contains three sub-packets:
11101110000000001101010000001100100000100011000001100000
VVVTTTILLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBCCCCCCCCCCC
The three bits labeled V (111) are the packet version, 7.
The three bits labeled T (011) are the packet type ID, 3, which means the packet is an operator.
The bit labeled I (1) is the length type ID, which indicates that the length is a 11-bit number representing the number of sub-packets.
The 11 bits labeled L (00000000011) contain the number of sub-packets, 3.
The 11 bits labeled A contain the first sub-packet, a literal value representing the number 1.
The 11 bits labeled B contain the second sub-packet, a literal value representing the number 2.
The 11 bits labeled C contain the third sub-packet, a literal value representing the number 3.
After reading 3 complete sub-packets, the number of sub-packets indicated in L (3) is reached, and so parsing of this packet stops.
For now, parse the hierarchy of the packets throughout the transmission and add up all of the version numbers.
Here are a few more examples of hexadecimal-encoded transmissions:
8A004A801A8002F478 represents an operator packet (version 4) which contains an operator packet (version 1) which contains an operator packet (version 5) which contains a literal value (version 6); this packet has a version sum of 16.
620080001611562C8802118E34 represents an operator packet (version 3) which contains two sub-packets; each sub-packet is an operator packet that contains two literal values. This packet has a version sum of 12.
C0015000016115A2E0802F182340 has the same structure as the previous example, but the outermost packet uses a different length type ID. This packet has a version sum of 23.
A0016C880162017C3686B18A3D4780 is an operator packet that contains an operator packet that contains an operator packet that contains five literal values; it has a version sum of 31.
Decode the structure of your hexadecimal-encoded BITS transmission; what do you get if you add up the version numbers in all packets?
Your puzzle answer was 971.
--- Part Two ---
Now that you have the structure of your transmission decoded, you can calculate the value of the expression it represents.
Literal values (type ID 4) represent a single number as described above. The remaining type IDs are more interesting:
Packets with type ID 0 are sum packets - their value is the sum of the values of their sub-packets. If they only have a single sub-packet, their value is the value of the sub-packet.
Packets with type ID 1 are product packets - their value is the result of multiplying together the values of their sub-packets. If they only have a single sub-packet, their value is the value of the sub-packet.
Packets with type ID 2 are minimum packets - their value is the minimum of the values of their sub-packets.
Packets with type ID 3 are maximum packets - their value is the maximum of the values of their sub-packets.
Packets with type ID 5 are greater than packets - their value is 1 if the value of the first sub-packet is greater than the value of the second sub-packet; otherwise, their value is 0. These packets always have exactly two sub-packets.
Packets with type ID 6 are less than packets - their value is 1 if the value of the first sub-packet is less than the value of the second sub-packet; otherwise, their value is 0. These packets always have exactly two sub-packets.
Packets with type ID 7 are equal to packets - their value is 1 if the value of the first sub-packet is equal to the value of the second sub-packet; otherwise, their value is 0. These packets always have exactly two sub-packets.
Using these rules, you can now work out the value of the outermost packet in your BITS transmission.
For example:
C200B40A82 finds the sum of 1 and 2, resulting in the value 3.
04005AC33890 finds the product of 6 and 9, resulting in the value 54.
880086C3E88112 finds the minimum of 7, 8, and 9, resulting in the value 7.
CE00C43D881120 finds the maximum of 7, 8, and 9, resulting in the value 9.
D8005AC2A8F0 produces 1, because 5 is less than 15.
F600BC2D8F produces 0, because 5 is not greater than 15.
9C005AC2F8F0 produces 0, because 5 is not equal to 15.
9C0141080250320F1802104A08 produces 1, because 1 + 3 = 2 * 2.
What do you get if you evaluate the expression represented by your hexadecimal-encoded BITS transmission?
Your puzzle answer was 831996589851.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/challenges/challenge17
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--- Day 17: Trick Shot ---
You finally decode the Elves' message. HI, the message says. You continue searching for the sleigh keys.
Ahead of you is what appears to be a large ocean trench. Could the keys have fallen into it? You'd better send a probe to investigate.
The probe launcher on your submarine can fire the probe with any integer velocity in the x (forward) and y (upward, or downward if negative) directions. For example, an initial x,y velocity like 0,10 would fire the probe straight up, while an initial velocity like 10,-1 would fire the probe forward at a slight downward angle.
The probe's x,y position starts at 0,0. Then, it will follow some trajectory by moving in steps. On each step, these changes occur in the following order:
The probe's x position increases by its x velocity.
The probe's y position increases by its y velocity.
Due to drag, the probe's x velocity changes by 1 toward the value 0; that is, it decreases by 1 if it is greater than 0, increases by 1 if it is less than 0, or does not change if it is already 0.
Due to gravity, the probe's y velocity decreases by 1.
For the probe to successfully make it into the trench, the probe must be on some trajectory that causes it to be within a target area after any step. The submarine computer has already calculated this target area (your puzzle input). For example:
target area: x=20..30, y=-10..-5
This target area means that you need to find initial x,y velocity values such that after any step, the probe's x position is at least 20 and at most 30, and the probe's y position is at least -10 and at most -5.
Given this target area, one initial velocity that causes the probe to be within the target area after any step is 7,2:
.............#....#............
.......#..............#........
...............................
S........................#.....
...............................
...............................
...........................#...
...............................
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTT#TT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
In this diagram, S is the probe's initial position, 0,0. The x coordinate increases to the right, and the y coordinate increases upward. In the bottom right, positions that are within the target area are shown as T. After each step (until the target area is reached), the position of the probe is marked with #. (The bottom-right # is both a position the probe reaches and a position in the target area.)
Another initial velocity that causes the probe to be within the target area after any step is 6,3:
...............#..#............
...........#........#..........
...............................
......#..............#.........
...............................
...............................
S....................#.........
...............................
...............................
...............................
.....................#.........
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................T#TTTTTTTTT
....................TTTTTTTTTTT
Another one is 9,0:
S........#.....................
.................#.............
...............................
........................#......
...............................
....................TTTTTTTTTTT
....................TTTTTTTTTT#
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
One initial velocity that doesn't cause the probe to be within the target area after any step is 17,-4:
S..............................................................
...............................................................
...............................................................
...............................................................
.................#.............................................
....................TTTTTTTTTTT................................
....................TTTTTTTTTTT................................
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The probe appears to pass through the target area, but is never within it after any step. Instead, it continues down and to the right - only the first few steps are shown.
If you're going to fire a highly scientific probe out of a super cool probe launcher, you might as well do it with style. How high can you make the probe go while still reaching the target area?
In the above example, using an initial velocity of 6,9 is the best you can do, causing the probe to reach a maximum y position of 45. (Any higher initial y velocity causes the probe to overshoot the target area entirely.)
Find the initial velocity that causes the probe to reach the highest y position and still eventually be within the target area after any step. What is the highest y position it reaches on this trajectory?
Your puzzle answer was 11781.
--- Part Two ---
Maybe a fancy trick shot isn't the best idea; after all, you only have one probe, so you had better not miss.
To get the best idea of what your options are for launching the probe, you need to find every initial velocity that causes the probe to eventually be within the target area after any step.
In the above example, there are 112 different initial velocity values that meet these criteria:
23,-10 25,-9 27,-5 29,-6 22,-6 21,-7 9,0 27,-7 24,-5
25,-7 26,-6 25,-5 6,8 11,-2 20,-5 29,-10 6,3 28,-7
8,0 30,-6 29,-8 20,-10 6,7 6,4 6,1 14,-4 21,-6
26,-10 7,-1 7,7 8,-1 21,-9 6,2 20,-7 30,-10 14,-3
20,-8 13,-2 7,3 28,-8 29,-9 15,-3 22,-5 26,-8 25,-8
25,-6 15,-4 9,-2 15,-2 12,-2 28,-9 12,-3 24,-6 23,-7
25,-10 7,8 11,-3 26,-7 7,1 23,-9 6,0 22,-10 27,-6
8,1 22,-8 13,-4 7,6 28,-6 11,-4 12,-4 26,-9 7,4
24,-10 23,-8 30,-8 7,0 9,-1 10,-1 26,-5 22,-9 6,5
7,5 23,-6 28,-10 10,-2 11,-1 20,-9 14,-2 29,-7 13,-3
23,-5 24,-8 27,-9 30,-7 28,-5 21,-10 7,9 6,6 21,-5
27,-10 7,2 30,-9 21,-8 22,-7 24,-9 20,-6 6,9 29,-5
8,-2 27,-8 30,-5 24,-7
How many distinct initial velocity values cause the probe to be within the target area after any step?
Your puzzle answer was 4531.
Both parts of this puzzle are complete! They provide two gold stars: **
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--- Day 18: Snailfish ---
You descend into the ocean trench and encounter some snailfish. They say they saw the sleigh keys! They'll even tell you which direction the keys went if you help one of the smaller snailfish with his math homework.
Snailfish numbers aren't like regular numbers. Instead, every snailfish number is a pair - an ordered list of two elements. Each element of the pair can be either a regular number or another pair.
Pairs are written as [x,y], where x and y are the elements within the pair. Here are some example snailfish numbers, one snailfish number per line:
[1,2]
[[1,2],3]
[9,[8,7]]
[[1,9],[8,5]]
[[[[1,2],[3,4]],[[5,6],[7,8]]],9]
[[[9,[3,8]],[[0,9],6]],[[[3,7],[4,9]],3]]
[[[[1,3],[5,3]],[[1,3],[8,7]]],[[[4,9],[6,9]],[[8,2],[7,3]]]]
This snailfish homework is about addition. To add two snailfish numbers, form a pair from the left and right parameters of the addition operator. For example, [1,2] + [[3,4],5] becomes [[1,2],[[3,4],5]].
There's only one problem: snailfish numbers must always be reduced, and the process of adding two snailfish numbers can result in snailfish numbers that need to be reduced.
To reduce a snailfish number, you must repeatedly do the first action in this list that applies to the snailfish number:
If any pair is nested inside four pairs, the leftmost such pair explodes.
If any regular number is 10 or greater, the leftmost such regular number splits.
Once no action in the above list applies, the snailfish number is reduced.
During reduction, at most one action applies, after which the process returns to the top of the list of actions. For example, if split produces a pair that meets the explode criteria, that pair explodes before other splits occur.
To explode a pair, the pair's left value is added to the first regular number to the left of the exploding pair (if any), and the pair's right value is added to the first regular number to the right of the exploding pair (if any). Exploding pairs will always consist of two regular numbers. Then, the entire exploding pair is replaced with the regular number 0.
Here are some examples of a single explode action:
[[[[[9,8],1],2],3],4] becomes [[[[0,9],2],3],4] (the 9 has no regular number to its left, so it is not added to any regular number).
[7,[6,[5,[4,[3,2]]]]] becomes [7,[6,[5,[7,0]]]] (the 2 has no regular number to its right, and so it is not added to any regular number).
[[6,[5,[4,[3,2]]]],1] becomes [[6,[5,[7,0]]],3].
[[3,[2,[1,[7,3]]]],[6,[5,[4,[3,2]]]]] becomes [[3,[2,[8,0]]],[9,[5,[4,[3,2]]]]] (the pair [3,2] is unaffected because the pair [7,3] is further to the left; [3,2] would explode on the next action).
[[3,[2,[8,0]]],[9,[5,[4,[3,2]]]]] becomes [[3,[2,[8,0]]],[9,[5,[7,0]]]].
To split a regular number, replace it with a pair; the left element of the pair should be the regular number divided by two and rounded down, while the right element of the pair should be the regular number divided by two and rounded up. For example, 10 becomes [5,5], 11 becomes [5,6], 12 becomes [6,6], and so on.
Here is the process of finding the reduced result of [[[[4,3],4],4],[7,[[8,4],9]]] + [1,1]:
after addition: [[[[[4,3],4],4],[7,[[8,4],9]]],[1,1]]
after explode: [[[[0,7],4],[7,[[8,4],9]]],[1,1]]
after explode: [[[[0,7],4],[15,[0,13]]],[1,1]]
after split: [[[[0,7],4],[[7,8],[0,13]]],[1,1]]
after split: [[[[0,7],4],[[7,8],[0,[6,7]]]],[1,1]]
after explode: [[[[0,7],4],[[7,8],[6,0]]],[8,1]]
Once no reduce actions apply, the snailfish number that remains is the actual result of the addition operation: [[[[0,7],4],[[7,8],[6,0]]],[8,1]].
The homework assignment involves adding up a list of snailfish numbers (your puzzle input). The snailfish numbers are each listed on a separate line. Add the first snailfish number and the second, then add that result and the third, then add that result and the fourth, and so on until all numbers in the list have been used once.
For example, the final sum of this list is [[[[1,1],[2,2]],[3,3]],[4,4]]:
[1,1]
[2,2]
[3,3]
[4,4]
The final sum of this list is [[[[3,0],[5,3]],[4,4]],[5,5]]:
[1,1]
[2,2]
[3,3]
[4,4]
[5,5]
The final sum of this list is [[[[5,0],[7,4]],[5,5]],[6,6]]:
[1,1]
[2,2]
[3,3]
[4,4]
[5,5]
[6,6]
Here's a slightly larger example:
[[[0,[4,5]],[0,0]],[[[4,5],[2,6]],[9,5]]]
[7,[[[3,7],[4,3]],[[6,3],[8,8]]]]
[[2,[[0,8],[3,4]]],[[[6,7],1],[7,[1,6]]]]
[[[[2,4],7],[6,[0,5]]],[[[6,8],[2,8]],[[2,1],[4,5]]]]
[7,[5,[[3,8],[1,4]]]]
[[2,[2,2]],[8,[8,1]]]
[2,9]
[1,[[[9,3],9],[[9,0],[0,7]]]]
[[[5,[7,4]],7],1]
[[[[4,2],2],6],[8,7]]
The final sum [[[[8,7],[7,7]],[[8,6],[7,7]]],[[[0,7],[6,6]],[8,7]]] is found after adding up the above snailfish numbers:
[[[0,[4,5]],[0,0]],[[[4,5],[2,6]],[9,5]]]
+ [7,[[[3,7],[4,3]],[[6,3],[8,8]]]]
= [[[[4,0],[5,4]],[[7,7],[6,0]]],[[8,[7,7]],[[7,9],[5,0]]]]
[[[[4,0],[5,4]],[[7,7],[6,0]]],[[8,[7,7]],[[7,9],[5,0]]]]
+ [[2,[[0,8],[3,4]]],[[[6,7],1],[7,[1,6]]]]
= [[[[6,7],[6,7]],[[7,7],[0,7]]],[[[8,7],[7,7]],[[8,8],[8,0]]]]
[[[[6,7],[6,7]],[[7,7],[0,7]]],[[[8,7],[7,7]],[[8,8],[8,0]]]]
+ [[[[2,4],7],[6,[0,5]]],[[[6,8],[2,8]],[[2,1],[4,5]]]]
= [[[[7,0],[7,7]],[[7,7],[7,8]]],[[[7,7],[8,8]],[[7,7],[8,7]]]]
[[[[7,0],[7,7]],[[7,7],[7,8]]],[[[7,7],[8,8]],[[7,7],[8,7]]]]
+ [7,[5,[[3,8],[1,4]]]]
= [[[[7,7],[7,8]],[[9,5],[8,7]]],[[[6,8],[0,8]],[[9,9],[9,0]]]]
[[[[7,7],[7,8]],[[9,5],[8,7]]],[[[6,8],[0,8]],[[9,9],[9,0]]]]
+ [[2,[2,2]],[8,[8,1]]]
= [[[[6,6],[6,6]],[[6,0],[6,7]]],[[[7,7],[8,9]],[8,[8,1]]]]
[[[[6,6],[6,6]],[[6,0],[6,7]]],[[[7,7],[8,9]],[8,[8,1]]]]
+ [2,9]
= [[[[6,6],[7,7]],[[0,7],[7,7]]],[[[5,5],[5,6]],9]]
[[[[6,6],[7,7]],[[0,7],[7,7]]],[[[5,5],[5,6]],9]]
+ [1,[[[9,3],9],[[9,0],[0,7]]]]
= [[[[7,8],[6,7]],[[6,8],[0,8]]],[[[7,7],[5,0]],[[5,5],[5,6]]]]
[[[[7,8],[6,7]],[[6,8],[0,8]]],[[[7,7],[5,0]],[[5,5],[5,6]]]]
+ [[[5,[7,4]],7],1]
= [[[[7,7],[7,7]],[[8,7],[8,7]]],[[[7,0],[7,7]],9]]
[[[[7,7],[7,7]],[[8,7],[8,7]]],[[[7,0],[7,7]],9]]
+ [[[[4,2],2],6],[8,7]]
= [[[[8,7],[7,7]],[[8,6],[7,7]]],[[[0,7],[6,6]],[8,7]]]
To check whether it's the right answer, the snailfish teacher only checks the magnitude of the final sum. The magnitude of a pair is 3 times the magnitude of its left element plus 2 times the magnitude of its right element. The magnitude of a regular number is just that number.
For example, the magnitude of [9,1] is 3*9 + 2*1 = 29; the magnitude of [1,9] is 3*1 + 2*9 = 21. Magnitude calculations are recursive: the magnitude of [[9,1],[1,9]] is 3*29 + 2*21 = 129.
Here are a few more magnitude examples:
[[1,2],[[3,4],5]] becomes 143.
[[[[0,7],4],[[7,8],[6,0]]],[8,1]] becomes 1384.
[[[[1,1],[2,2]],[3,3]],[4,4]] becomes 445.
[[[[3,0],[5,3]],[4,4]],[5,5]] becomes 791.
[[[[5,0],[7,4]],[5,5]],[6,6]] becomes 1137.
[[[[8,7],[7,7]],[[8,6],[7,7]]],[[[0,7],[6,6]],[8,7]]] becomes 3488.
So, given this example homework assignment:
[[[0,[5,8]],[[1,7],[9,6]]],[[4,[1,2]],[[1,4],2]]]
[[[5,[2,8]],4],[5,[[9,9],0]]]
[6,[[[6,2],[5,6]],[[7,6],[4,7]]]]
[[[6,[0,7]],[0,9]],[4,[9,[9,0]]]]
[[[7,[6,4]],[3,[1,3]]],[[[5,5],1],9]]
[[6,[[7,3],[3,2]]],[[[3,8],[5,7]],4]]
[[[[5,4],[7,7]],8],[[8,3],8]]
[[9,3],[[9,9],[6,[4,9]]]]
[[2,[[7,7],7]],[[5,8],[[9,3],[0,2]]]]
[[[[5,2],5],[8,[3,7]]],[[5,[7,5]],[4,4]]]
The final sum is:
[[[[6,6],[7,6]],[[7,7],[7,0]]],[[[7,7],[7,7]],[[7,8],[9,9]]]]
The magnitude of this final sum is 4140.
Add up all of the snailfish numbers from the homework assignment in the order they appear. What is the magnitude of the final sum?
Your puzzle answer was 3305.
--- Part Two ---
You notice a second question on the back of the homework assignment:
What is the largest magnitude you can get from adding only two of the snailfish numbers?
Note that snailfish addition is not commutative - that is, x + y and y + x can produce different results.
Again considering the last example homework assignment above:
[[[0,[5,8]],[[1,7],[9,6]]],[[4,[1,2]],[[1,4],2]]]
[[[5,[2,8]],4],[5,[[9,9],0]]]
[6,[[[6,2],[5,6]],[[7,6],[4,7]]]]
[[[6,[0,7]],[0,9]],[4,[9,[9,0]]]]
[[[7,[6,4]],[3,[1,3]]],[[[5,5],1],9]]
[[6,[[7,3],[3,2]]],[[[3,8],[5,7]],4]]
[[[[5,4],[7,7]],8],[[8,3],8]]
[[9,3],[[9,9],[6,[4,9]]]]
[[2,[[7,7],7]],[[5,8],[[9,3],[0,2]]]]
[[[[5,2],5],[8,[3,7]]],[[5,[7,5]],[4,4]]]
The largest magnitude of the sum of any two snailfish numbers in this list is 3993. This is the magnitude of [[2,[[7,7],7]],[[5,8],[[9,3],[0,2]]]] + [[[0,[5,8]],[[1,7],[9,6]]],[[4,[1,2]],[[1,4],2]]], which reduces to [[[[7,8],[6,6]],[[6,0],[7,7]]],[[[7,8],[8,8]],[[7,9],[0,6]]]].
What is the largest magnitude of any sum of two different snailfish numbers from the homework assignment?
Your puzzle answer was 4563.
Both parts of this puzzle are complete! They provide two gold stars: **
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--- Day 19: Beacon Scanner ---
As your probe drifted down through this area, it released an assortment of beacons and scanners into the water. It's difficult to navigate in the pitch black open waters of the ocean trench, but if you can build a map of the trench using data from the scanners, you should be able to safely reach the bottom.
The beacons and scanners float motionless in the water; they're designed to maintain the same position for long periods of time. Each scanner is capable of detecting all beacons in a large cube centered on the scanner; beacons that are at most 1000 units away from the scanner in each of the three axes (x, y, and z) have their precise position determined relative to the scanner. However, scanners cannot detect other scanners. The submarine has automatically summarized the relative positions of beacons detected by each scanner (your puzzle input).
For example, if a scanner is at x,y,z coordinates 500,0,-500 and there are beacons at -500,1000,-1500 and 1501,0,-500, the scanner could report that the first beacon is at -1000,1000,-1000 (relative to the scanner) but would not detect the second beacon at all.
Unfortunately, while each scanner can report the positions of all detected beacons relative to itself, the scanners do not know their own position. You'll need to determine the positions of the beacons and scanners yourself.
The scanners and beacons map a single contiguous 3d region. This region can be reconstructed by finding pairs of scanners that have overlapping detection regions such that there are at least 12 beacons that both scanners detect within the overlap. By establishing 12 common beacons, you can precisely determine where the scanners are relative to each other, allowing you to reconstruct the beacon map one scanner at a time.
For a moment, consider only two dimensions. Suppose you have the following scanner reports:
--- scanner 0 ---
0,2
4,1
3,3
--- scanner 1 ---
-1,-1
-5,0
-2,1
Drawing x increasing rightward, y increasing upward, scanners as S, and beacons as B, scanner 0 detects this:
...B.
B....
....B
S....
Scanner 1 detects this:
...B..
B....S
....B.
For this example, assume scanners only need 3 overlapping beacons. Then, the beacons visible to both scanners overlap to produce the following complete map:
...B..
B....S
....B.
S.....
Unfortunately, there's a second problem: the scanners also don't know their rotation or facing direction. Due to magnetic alignment, each scanner is rotated some integer number of 90-degree turns around all of the x, y, and z axes. That is, one scanner might call a direction positive x, while another scanner might call that direction negative y. Or, two scanners might agree on which direction is positive x, but one scanner might be upside-down from the perspective of the other scanner. In total, each scanner could be in any of 24 different orientations: facing positive or negative x, y, or z, and considering any of four directions "up" from that facing.
For example, here is an arrangement of beacons as seen from a scanner in the same position but in different orientations:
--- scanner 0 ---
-1,-1,1
-2,-2,2
-3,-3,3
-2,-3,1
5,6,-4
8,0,7
--- scanner 0 ---
1,-1,1
2,-2,2
3,-3,3
2,-1,3
-5,4,-6
-8,-7,0
--- scanner 0 ---
-1,-1,-1
-2,-2,-2
-3,-3,-3
-1,-3,-2
4,6,5
-7,0,8
--- scanner 0 ---
1,1,-1
2,2,-2
3,3,-3
1,3,-2
-4,-6,5
7,0,8
--- scanner 0 ---
1,1,1
2,2,2
3,3,3
3,1,2
-6,-4,-5
0,7,-8
By finding pairs of scanners that both see at least 12 of the same beacons, you can assemble the entire map. For example, consider the following report:
--- scanner 0 ---
404,-588,-901
528,-643,409
-838,591,734
390,-675,-793
-537,-823,-458
-485,-357,347
-345,-311,381
-661,-816,-575
-876,649,763
-618,-824,-621
553,345,-567
474,580,667
-447,-329,318
-584,868,-557
544,-627,-890
564,392,-477
455,729,728
-892,524,684
-689,845,-530
423,-701,434
7,-33,-71
630,319,-379
443,580,662
-789,900,-551
459,-707,401
--- scanner 1 ---
686,422,578
605,423,415
515,917,-361
-336,658,858
95,138,22
-476,619,847
-340,-569,-846
567,-361,727
-460,603,-452
669,-402,600
729,430,532
-500,-761,534
-322,571,750
-466,-666,-811
-429,-592,574
-355,545,-477
703,-491,-529
-328,-685,520
413,935,-424
-391,539,-444
586,-435,557
-364,-763,-893
807,-499,-711
755,-354,-619
553,889,-390
--- scanner 2 ---
649,640,665
682,-795,504
-784,533,-524
-644,584,-595
-588,-843,648
-30,6,44
-674,560,763
500,723,-460
609,671,-379
-555,-800,653
-675,-892,-343
697,-426,-610
578,704,681
493,664,-388
-671,-858,530
-667,343,800
571,-461,-707
-138,-166,112
-889,563,-600
646,-828,498
640,759,510
-630,509,768
-681,-892,-333
673,-379,-804
-742,-814,-386
577,-820,562
--- scanner 3 ---
-589,542,597
605,-692,669
-500,565,-823
-660,373,557
-458,-679,-417
-488,449,543
-626,468,-788
338,-750,-386
528,-832,-391
562,-778,733
-938,-730,414
543,643,-506
-524,371,-870
407,773,750
-104,29,83
378,-903,-323
-778,-728,485
426,699,580
-438,-605,-362
-469,-447,-387
509,732,623
647,635,-688
-868,-804,481
614,-800,639
595,780,-596
--- scanner 4 ---
727,592,562
-293,-554,779
441,611,-461
-714,465,-776
-743,427,-804
-660,-479,-426
832,-632,460
927,-485,-438
408,393,-506
466,436,-512
110,16,151
-258,-428,682
-393,719,612
-211,-452,876
808,-476,-593
-575,615,604
-485,667,467
-680,325,-822
-627,-443,-432
872,-547,-609
833,512,582
807,604,487
839,-516,451
891,-625,532
-652,-548,-490
30,-46,-14
Because all coordinates are relative, in this example, all "absolute" positions will be expressed relative to scanner 0 (using the orientation of scanner 0 and as if scanner 0 is at coordinates 0,0,0).
Scanners 0 and 1 have overlapping detection cubes; the 12 beacons they both detect (relative to scanner 0) are at the following coordinates:
-618,-824,-621
-537,-823,-458
-447,-329,318
404,-588,-901
544,-627,-890
528,-643,409
-661,-816,-575
390,-675,-793
423,-701,434
-345,-311,381
459,-707,401
-485,-357,347
These same 12 beacons (in the same order) but from the perspective of scanner 1 are:
686,422,578
605,423,415
515,917,-361
-336,658,858
-476,619,847
-460,603,-452
729,430,532
-322,571,750
-355,545,-477
413,935,-424
-391,539,-444
553,889,-390
Because of this, scanner 1 must be at 68,-1246,-43 (relative to scanner 0).
Scanner 4 overlaps with scanner 1; the 12 beacons they both detect (relative to scanner 0) are:
459,-707,401
-739,-1745,668
-485,-357,347
432,-2009,850
528,-643,409
423,-701,434
-345,-311,381
408,-1815,803
534,-1912,768
-687,-1600,576
-447,-329,318
-635,-1737,486
So, scanner 4 is at -20,-1133,1061 (relative to scanner 0).
Following this process, scanner 2 must be at 1105,-1205,1229 (relative to scanner 0) and scanner 3 must be at -92,-2380,-20 (relative to scanner 0).
The full list of beacons (relative to scanner 0) is:
-892,524,684
-876,649,763
-838,591,734
-789,900,-551
-739,-1745,668
-706,-3180,-659
-697,-3072,-689
-689,845,-530
-687,-1600,576
-661,-816,-575
-654,-3158,-753
-635,-1737,486
-631,-672,1502
-624,-1620,1868
-620,-3212,371
-618,-824,-621
-612,-1695,1788
-601,-1648,-643
-584,868,-557
-537,-823,-458
-532,-1715,1894
-518,-1681,-600
-499,-1607,-770
-485,-357,347
-470,-3283,303
-456,-621,1527
-447,-329,318
-430,-3130,366
-413,-627,1469
-345,-311,381
-36,-1284,1171
-27,-1108,-65
7,-33,-71
12,-2351,-103
26,-1119,1091
346,-2985,342
366,-3059,397
377,-2827,367
390,-675,-793
396,-1931,-563
404,-588,-901
408,-1815,803
423,-701,434
432,-2009,850
443,580,662
455,729,728
456,-540,1869
459,-707,401
465,-695,1988
474,580,667
496,-1584,1900
497,-1838,-617
527,-524,1933
528,-643,409
534,-1912,768
544,-627,-890
553,345,-567
564,392,-477
568,-2007,-577
605,-1665,1952
612,-1593,1893
630,319,-379
686,-3108,-505
776,-3184,-501
846,-3110,-434
1135,-1161,1235
1243,-1093,1063
1660,-552,429
1693,-557,386
1735,-437,1738
1749,-1800,1813
1772,-405,1572
1776,-675,371
1779,-442,1789
1780,-1548,337
1786,-1538,337
1847,-1591,415
1889,-1729,1762
1994,-1805,1792
In total, there are 79 beacons.
Assemble the full map of beacons. How many beacons are there?
Your puzzle answer was 323.
--- Part Two ---
Sometimes, it's a good idea to appreciate just how big the ocean is. Using the Manhattan distance, how far apart do the scanners get?
In the above example, scanners 2 (1105,-1205,1229) and 3 (-92,-2380,-20) are the largest Manhattan distance apart. In total, they are 1197 + 1175 + 1249 = 3621 units apart.
What is the largest Manhattan distance between any two scanners?
Your puzzle answer was 10685.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/challenges/challenge20
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--- Day 20: Trench Map ---
With the scanners fully deployed, you turn their attention to mapping the floor of the ocean trench.
When you get back the image from the scanners, it seems to just be random noise. Perhaps you can combine an image enhancement algorithm and the input image (your puzzle input) to clean it up a little.
For example:
..#.#..#####.#.#.#.###.##.....###.##.#..###.####..#####..#....#..#..##..##
#..######.###...####..#..#####..##..#.#####...##.#.#..#.##..#.#......#.###
.######.###.####...#.##.##..#..#..#####.....#.#....###..#.##......#.....#.
.#..#..##..#...##.######.####.####.#.#...#.......#..#.#.#...####.##.#.....
.#..#...##.#.##..#...##.#.##..###.#......#.#.......#.#.#.####.###.##...#..
...####.#..#..#.##.#....##..#.####....##...##..#...#......#.#.......#.....
..##..####..#...#.#.#...##..#.#..###..#####........#..####......#..#
#..#.
#....
##..#
..#..
..###
The first section is the image enhancement algorithm. It is normally given on a single line, but it has been wrapped to multiple lines in this example for legibility. The second section is the input image, a two-dimensional grid of light pixels (#) and dark pixels (.).
The image enhancement algorithm describes how to enhance an image by simultaneously converting all pixels in the input image into an output image. Each pixel of the output image is determined by looking at a 3x3 square of pixels centered on the corresponding input image pixel. So, to determine the value of the pixel at (5,10) in the output image, nine pixels from the input image need to be considered: (4,9), (4,10), (4,11), (5,9), (5,10), (5,11), (6,9), (6,10), and (6,11). These nine input pixels are combined into a single binary number that is used as an index in the image enhancement algorithm string.
For example, to determine the output pixel that corresponds to the very middle pixel of the input image, the nine pixels marked by [...] would need to be considered:
# . . # .
#[. . .].
#[# . .]#
.[. # .].
. . # # #
Starting from the top-left and reading across each row, these pixels are ..., then #.., then .#.; combining these forms ...#...#.. By turning dark pixels (.) into 0 and light pixels (#) into 1, the binary number 000100010 can be formed, which is 34 in decimal.
The image enhancement algorithm string is exactly 512 characters long, enough to match every possible 9-bit binary number. The first few characters of the string (numbered starting from zero) are as follows:
0 10 20 30 34 40 50 60 70
| | | | | | | | |
..#.#..#####.#.#.#.###.##.....###.##.#..###.####..#####..#....#..#..##..##
In the middle of this first group of characters, the character at index 34 can be found: #. So, the output pixel in the center of the output image should be #, a light pixel.
This process can then be repeated to calculate every pixel of the output image.
Through advances in imaging technology, the images being operated on here are infinite in size. Every pixel of the infinite output image needs to be calculated exactly based on the relevant pixels of the input image. The small input image you have is only a small region of the actual infinite input image; the rest of the input image consists of dark pixels (.). For the purposes of the example, to save on space, only a portion of the infinite-sized input and output images will be shown.
The starting input image, therefore, looks something like this, with more dark pixels (.) extending forever in every direction not shown here:
...............
...............
...............
...............
...............
.....#..#......
.....#.........
.....##..#.....
.......#.......
.......###.....
...............
...............
...............
...............
...............
By applying the image enhancement algorithm to every pixel simultaneously, the following output image can be obtained:
...............
...............
...............
...............
.....##.##.....
....#..#.#.....
....##.#..#....
....####..#....
.....#..##.....
......##..#....
.......#.#.....
...............
...............
...............
...............
Through further advances in imaging technology, the above output image can also be used as an input image! This allows it to be enhanced a second time:
...............
...............
...............
..........#....
....#..#.#.....
...#.#...###...
...#...##.#....
...#.....#.#...
....#.#####....
.....#.#####...
......##.##....
.......###.....
...............
...............
...............
Truly incredible - now the small details are really starting to come through. After enhancing the original input image twice, 35 pixels are lit.
Start with the original input image and apply the image enhancement algorithm twice, being careful to account for the infinite size of the images. How many pixels are lit in the resulting image?
Your puzzle answer was 5391.
--- Part Two ---
You still can't quite make out the details in the image. Maybe you just didn't enhance it enough.
If you enhance the starting input image in the above example a total of 50 times, 3351 pixels are lit in the final output image.
Start again with the original input image and apply the image enhancement algorithm 50 times. How many pixels are lit in the resulting image?
Your puzzle answer was 16383.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/challenges/challenge21
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--- Day 21: Dirac Dice ---
There's not much to do as you slowly descend to the bottom of the ocean. The submarine computer challenges you to a nice game of Dirac Dice.
This game consists of a single die, two pawns, and a game board with a circular track containing ten spaces marked 1 through 10 clockwise. Each player's starting space is chosen randomly (your puzzle input). Player 1 goes first.
Players take turns moving. On each player's turn, the player rolls the die three times and adds up the results. Then, the player moves their pawn that many times forward around the track (that is, moving clockwise on spaces in order of increasing value, wrapping back around to 1 after 10). So, if a player is on space 7 and they roll 2, 2, and 1, they would move forward 5 times, to spaces 8, 9, 10, 1, and finally stopping on 2.
After each player moves, they increase their score by the value of the space their pawn stopped on. Players' scores start at 0. So, if the first player starts on space 7 and rolls a total of 5, they would stop on space 2 and add 2 to their score (for a total score of 2). The game immediately ends as a win for any player whose score reaches at least 1000.
Since the first game is a practice game, the submarine opens a compartment labeled deterministic dice and a 100-sided die falls out. This die always rolls 1 first, then 2, then 3, and so on up to 100, after which it starts over at 1 again. Play using this die.
For example, given these starting positions:
Player 1 starting position: 4
Player 2 starting position: 8
This is how the game would go:
Player 1 rolls 1+2+3 and moves to space 10 for a total score of 10.
Player 2 rolls 4+5+6 and moves to space 3 for a total score of 3.
Player 1 rolls 7+8+9 and moves to space 4 for a total score of 14.
Player 2 rolls 10+11+12 and moves to space 6 for a total score of 9.
Player 1 rolls 13+14+15 and moves to space 6 for a total score of 20.
Player 2 rolls 16+17+18 and moves to space 7 for a total score of 16.
Player 1 rolls 19+20+21 and moves to space 6 for a total score of 26.
Player 2 rolls 22+23+24 and moves to space 6 for a total score of 22.
...after many turns...
Player 2 rolls 82+83+84 and moves to space 6 for a total score of 742.
Player 1 rolls 85+86+87 and moves to space 4 for a total score of 990.
Player 2 rolls 88+89+90 and moves to space 3 for a total score of 745.
Player 1 rolls 91+92+93 and moves to space 10 for a final score, 1000.
Since player 1 has at least 1000 points, player 1 wins and the game ends. At this point, the losing player had 745 points and the die had been rolled a total of 993 times; 745 * 993 = 739785.
Play a practice game using the deterministic 100-sided die. The moment either player wins, what do you get if you multiply the score of the losing player by the number of times the die was rolled during the game?
Your puzzle answer was 913560.
--- Part Two ---
Now that you're warmed up, it's time to play the real game.
A second compartment opens, this time labeled Dirac dice. Out of it falls a single three-sided die.
As you experiment with the die, you feel a little strange. An informational brochure in the compartment explains that this is a quantum die: when you roll it, the universe splits into multiple copies, one copy for each possible outcome of the die. In this case, rolling the die always splits the universe into three copies: one where the outcome of the roll was 1, one where it was 2, and one where it was 3.
The game is played the same as before, although to prevent things from getting too far out of hand, the game now ends when either player's score reaches at least 21.
Using the same starting positions as in the example above, player 1 wins in 444356092776315 universes, while player 2 merely wins in 341960390180808 universes.
Using your given starting positions, determine every possible outcome. Find the player that wins in more universes; in how many universes does that player win?
Your puzzle answer was 110271560863819.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/challenges/challenge22
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--- Day 22: Reactor Reboot ---
Operating at these extreme ocean depths has overloaded the submarine's reactor; it needs to be rebooted.
The reactor core is made up of a large 3-dimensional grid made up entirely of cubes, one cube per integer 3-dimensional coordinate (x,y,z). Each cube can be either on or off; at the start of the reboot process, they are all off. (Could it be an old model of a reactor you've seen before?)
To reboot the reactor, you just need to set all of the cubes to either on or off by following a list of reboot steps (your puzzle input). Each step specifies a cuboid (the set of all cubes that have coordinates which fall within ranges for x, y, and z) and whether to turn all of the cubes in that cuboid on or off.
For example, given these reboot steps:
on x=10..12,y=10..12,z=10..12
on x=11..13,y=11..13,z=11..13
off x=9..11,y=9..11,z=9..11
on x=10..10,y=10..10,z=10..10
The first step (on x=10..12,y=10..12,z=10..12) turns on a 3x3x3 cuboid consisting of 27 cubes:
10,10,10
10,10,11
10,10,12
10,11,10
10,11,11
10,11,12
10,12,10
10,12,11
10,12,12
11,10,10
11,10,11
11,10,12
11,11,10
11,11,11
11,11,12
11,12,10
11,12,11
11,12,12
12,10,10
12,10,11
12,10,12
12,11,10
12,11,11
12,11,12
12,12,10
12,12,11
12,12,12
The second step (on x=11..13,y=11..13,z=11..13) turns on a 3x3x3 cuboid that overlaps with the first. As a result, only 19 additional cubes turn on; the rest are already on from the previous step:
11,11,13
11,12,13
11,13,11
11,13,12
11,13,13
12,11,13
12,12,13
12,13,11
12,13,12
12,13,13
13,11,11
13,11,12
13,11,13
13,12,11
13,12,12
13,12,13
13,13,11
13,13,12
13,13,13
The third step (off x=9..11,y=9..11,z=9..11) turns off a 3x3x3 cuboid that overlaps partially with some cubes that are on, ultimately turning off 8 cubes:
10,10,10
10,10,11
10,11,10
10,11,11
11,10,10
11,10,11
11,11,10
11,11,11
The final step (on x=10..10,y=10..10,z=10..10) turns on a single cube, 10,10,10. After this last step, 39 cubes are on.
The initialization procedure only uses cubes that have x, y, and z positions of at least -50 and at most 50. For now, ignore cubes outside this region.
Here is a larger example:
on x=-20..26,y=-36..17,z=-47..7
on x=-20..33,y=-21..23,z=-26..28
on x=-22..28,y=-29..23,z=-38..16
on x=-46..7,y=-6..46,z=-50..-1
on x=-49..1,y=-3..46,z=-24..28
on x=2..47,y=-22..22,z=-23..27
on x=-27..23,y=-28..26,z=-21..29
on x=-39..5,y=-6..47,z=-3..44
on x=-30..21,y=-8..43,z=-13..34
on x=-22..26,y=-27..20,z=-29..19
off x=-48..-32,y=26..41,z=-47..-37
on x=-12..35,y=6..50,z=-50..-2
off x=-48..-32,y=-32..-16,z=-15..-5
on x=-18..26,y=-33..15,z=-7..46
off x=-40..-22,y=-38..-28,z=23..41
on x=-16..35,y=-41..10,z=-47..6
off x=-32..-23,y=11..30,z=-14..3
on x=-49..-5,y=-3..45,z=-29..18
off x=18..30,y=-20..-8,z=-3..13
on x=-41..9,y=-7..43,z=-33..15
on x=-54112..-39298,y=-85059..-49293,z=-27449..7877
on x=967..23432,y=45373..81175,z=27513..53682
The last two steps are fully outside the initialization procedure area; all other steps are fully within it. After executing these steps in the initialization procedure region, 590784 cubes are on.
Execute the reboot steps. Afterward, considering only cubes in the region x=-50..50,y=-50..50,z=-50..50, how many cubes are on?
Your puzzle answer was 580810.
--- Part Two ---
Now that the initialization procedure is complete, you can reboot the reactor.
Starting with all cubes off, run all of the reboot steps for all cubes in the reactor.
Consider the following reboot steps:
on x=-5..47,y=-31..22,z=-19..33
on x=-44..5,y=-27..21,z=-14..35
on x=-49..-1,y=-11..42,z=-10..38
on x=-20..34,y=-40..6,z=-44..1
off x=26..39,y=40..50,z=-2..11
on x=-41..5,y=-41..6,z=-36..8
off x=-43..-33,y=-45..-28,z=7..25
on x=-33..15,y=-32..19,z=-34..11
off x=35..47,y=-46..-34,z=-11..5
on x=-14..36,y=-6..44,z=-16..29
on x=-57795..-6158,y=29564..72030,z=20435..90618
on x=36731..105352,y=-21140..28532,z=16094..90401
on x=30999..107136,y=-53464..15513,z=8553..71215
on x=13528..83982,y=-99403..-27377,z=-24141..23996
on x=-72682..-12347,y=18159..111354,z=7391..80950
on x=-1060..80757,y=-65301..-20884,z=-103788..-16709
on x=-83015..-9461,y=-72160..-8347,z=-81239..-26856
on x=-52752..22273,y=-49450..9096,z=54442..119054
on x=-29982..40483,y=-108474..-28371,z=-24328..38471
on x=-4958..62750,y=40422..118853,z=-7672..65583
on x=55694..108686,y=-43367..46958,z=-26781..48729
on x=-98497..-18186,y=-63569..3412,z=1232..88485
on x=-726..56291,y=-62629..13224,z=18033..85226
on x=-110886..-34664,y=-81338..-8658,z=8914..63723
on x=-55829..24974,y=-16897..54165,z=-121762..-28058
on x=-65152..-11147,y=22489..91432,z=-58782..1780
on x=-120100..-32970,y=-46592..27473,z=-11695..61039
on x=-18631..37533,y=-124565..-50804,z=-35667..28308
on x=-57817..18248,y=49321..117703,z=5745..55881
on x=14781..98692,y=-1341..70827,z=15753..70151
on x=-34419..55919,y=-19626..40991,z=39015..114138
on x=-60785..11593,y=-56135..2999,z=-95368..-26915
on x=-32178..58085,y=17647..101866,z=-91405..-8878
on x=-53655..12091,y=50097..105568,z=-75335..-4862
on x=-111166..-40997,y=-71714..2688,z=5609..50954
on x=-16602..70118,y=-98693..-44401,z=5197..76897
on x=16383..101554,y=4615..83635,z=-44907..18747
off x=-95822..-15171,y=-19987..48940,z=10804..104439
on x=-89813..-14614,y=16069..88491,z=-3297..45228
on x=41075..99376,y=-20427..49978,z=-52012..13762
on x=-21330..50085,y=-17944..62733,z=-112280..-30197
on x=-16478..35915,y=36008..118594,z=-7885..47086
off x=-98156..-27851,y=-49952..43171,z=-99005..-8456
off x=2032..69770,y=-71013..4824,z=7471..94418
on x=43670..120875,y=-42068..12382,z=-24787..38892
off x=37514..111226,y=-45862..25743,z=-16714..54663
off x=25699..97951,y=-30668..59918,z=-15349..69697
off x=-44271..17935,y=-9516..60759,z=49131..112598
on x=-61695..-5813,y=40978..94975,z=8655..80240
off x=-101086..-9439,y=-7088..67543,z=33935..83858
off x=18020..114017,y=-48931..32606,z=21474..89843
off x=-77139..10506,y=-89994..-18797,z=-80..59318
off x=8476..79288,y=-75520..11602,z=-96624..-24783
on x=-47488..-1262,y=24338..100707,z=16292..72967
off x=-84341..13987,y=2429..92914,z=-90671..-1318
off x=-37810..49457,y=-71013..-7894,z=-105357..-13188
off x=-27365..46395,y=31009..98017,z=15428..76570
off x=-70369..-16548,y=22648..78696,z=-1892..86821
on x=-53470..21291,y=-120233..-33476,z=-44150..38147
off x=-93533..-4276,y=-16170..68771,z=-104985..-24507
After running the above reboot steps, 2758514936282235 cubes are on. (Just for fun, 474140 of those are also in the initialization procedure region.)
Starting again with all cubes off, execute all reboot steps. Afterward, considering all cubes, how many cubes are on?
Your puzzle answer was 1265621119006734.
Both parts of this puzzle are complete! They provide two gold stars: **
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--- Day 23: Amphipod ---
A group of amphipods notice your fancy submarine and flag you down. "With such an impressive shell," one amphipod says, "surely you can help us with a question that has stumped our best scientists."
They go on to explain that a group of timid, stubborn amphipods live in a nearby burrow. Four types of amphipods live there: Amber (A), Bronze (B), Copper (C), and Desert (D). They live in a burrow that consists of a hallway and four side rooms. The side rooms are initially full of amphipods, and the hallway is initially empty.
They give you a diagram of the situation (your puzzle input), including locations of each amphipod (A, B, C, or D, each of which is occupying an otherwise open space), walls (#), and open space (.).
For example:
#############
#...........#
###B#C#B#D###
#A#D#C#A#
#########
The amphipods would like a method to organize every amphipod into side rooms so that each side room contains one type of amphipod and the types are sorted A-D going left to right, like this:
#############
#...........#
###A#B#C#D###
#A#B#C#D#
#########
Amphipods can move up, down, left, or right so long as they are moving into an unoccupied open space. Each type of amphipod requires a different amount of energy to move one step: Amber amphipods require 1 energy per step, Bronze amphipods require 10 energy, Copper amphipods require 100, and Desert ones require 1000. The amphipods would like you to find a way to organize the amphipods that requires the least total energy.
However, because they are timid and stubborn, the amphipods have some extra rules:
Amphipods will never stop on the space immediately outside any room. They can move into that space so long as they immediately continue moving. (Specifically, this refers to the four open spaces in the hallway that are directly above an amphipod starting position.)
Amphipods will never move from the hallway into a room unless that room is their destination room and that room contains no amphipods which do not also have that room as their own destination. If an amphipod's starting room is not its destination room, it can stay in that room until it leaves the room. (For example, an Amber amphipod will not move from the hallway into the right three rooms, and will only move into the leftmost room if that room is empty or if it only contains other Amber amphipods.)
Once an amphipod stops moving in the hallway, it will stay in that spot until it can move into a room. (That is, once any amphipod starts moving, any other amphipods currently in the hallway are locked in place and will not move again until they can move fully into a room.)
In the above example, the amphipods can be organized using a minimum of 12521 energy. One way to do this is shown below.
Starting configuration:
#############
#...........#
###B#C#B#D###
#A#D#C#A#
#########
One Bronze amphipod moves into the hallway, taking 4 steps and using 40 energy:
#############
#...B.......#
###B#C#.#D###
#A#D#C#A#
#########
The only Copper amphipod not in its side room moves there, taking 4 steps and using 400 energy:
#############
#...B.......#
###B#.#C#D###
#A#D#C#A#
#########
A Desert amphipod moves out of the way, taking 3 steps and using 3000 energy, and then the Bronze amphipod takes its place, taking 3 steps and using 30 energy:
#############
#.....D.....#
###B#.#C#D###
#A#B#C#A#
#########
The leftmost Bronze amphipod moves to its room using 40 energy:
#############
#.....D.....#
###.#B#C#D###
#A#B#C#A#
#########
Both amphipods in the rightmost room move into the hallway, using 2003 energy in total:
#############
#.....D.D.A.#
###.#B#C#.###
#A#B#C#.#
#########
Both Desert amphipods move into the rightmost room using 7000 energy:
#############
#.........A.#
###.#B#C#D###
#A#B#C#D#
#########
Finally, the last Amber amphipod moves into its room, using 8 energy:
#############
#...........#
###A#B#C#D###
#A#B#C#D#
#########
What is the least energy required to organize the amphipods?
Your puzzle answer was 18195.
--- Part Two ---
As you prepare to give the amphipods your solution, you notice that the diagram they handed you was actually folded up. As you unfold it, you discover an extra part of the diagram.
Between the first and second lines of text that contain amphipod starting positions, insert the following lines:
#D#C#B#A#
#D#B#A#C#
So, the above example now becomes:
#############
#...........#
###B#C#B#D###
#D#C#B#A#
#D#B#A#C#
#A#D#C#A#
#########
The amphipods still want to be organized into rooms similar to before:
#############
#...........#
###A#B#C#D###
#A#B#C#D#
#A#B#C#D#
#A#B#C#D#
#########
In this updated example, the least energy required to organize these amphipods is 44169:
#############
#...........#
###B#C#B#D###
#D#C#B#A#
#D#B#A#C#
#A#D#C#A#
#########
#############
#..........D#
###B#C#B#.###
#D#C#B#A#
#D#B#A#C#
#A#D#C#A#
#########
#############
#A.........D#
###B#C#B#.###
#D#C#B#.#
#D#B#A#C#
#A#D#C#A#
#########
#############
#A........BD#
###B#C#.#.###
#D#C#B#.#
#D#B#A#C#
#A#D#C#A#
#########
#############
#A......B.BD#
###B#C#.#.###
#D#C#.#.#
#D#B#A#C#
#A#D#C#A#
#########
#############
#AA.....B.BD#
###B#C#.#.###
#D#C#.#.#
#D#B#.#C#
#A#D#C#A#
#########
#############
#AA.....B.BD#
###B#.#.#.###
#D#C#.#.#
#D#B#C#C#
#A#D#C#A#
#########
#############
#AA.....B.BD#
###B#.#.#.###
#D#.#C#.#
#D#B#C#C#
#A#D#C#A#
#########
#############
#AA...B.B.BD#
###B#.#.#.###
#D#.#C#.#
#D#.#C#C#
#A#D#C#A#
#########
#############
#AA.D.B.B.BD#
###B#.#.#.###
#D#.#C#.#
#D#.#C#C#
#A#.#C#A#
#########
#############
#AA.D...B.BD#
###B#.#.#.###
#D#.#C#.#
#D#.#C#C#
#A#B#C#A#
#########
#############
#AA.D.....BD#
###B#.#.#.###
#D#.#C#.#
#D#B#C#C#
#A#B#C#A#
#########
#############
#AA.D......D#
###B#.#.#.###
#D#B#C#.#
#D#B#C#C#
#A#B#C#A#
#########
#############
#AA.D......D#
###B#.#C#.###
#D#B#C#.#
#D#B#C#.#
#A#B#C#A#
#########
#############
#AA.D.....AD#
###B#.#C#.###
#D#B#C#.#
#D#B#C#.#
#A#B#C#.#
#########
#############
#AA.......AD#
###B#.#C#.###
#D#B#C#.#
#D#B#C#.#
#A#B#C#D#
#########
#############
#AA.......AD#
###.#B#C#.###
#D#B#C#.#
#D#B#C#.#
#A#B#C#D#
#########
#############
#AA.......AD#
###.#B#C#.###
#.#B#C#.#
#D#B#C#D#
#A#B#C#D#
#########
#############
#AA.D.....AD#
###.#B#C#.###
#.#B#C#.#
#.#B#C#D#
#A#B#C#D#
#########
#############
#A..D.....AD#
###.#B#C#.###
#.#B#C#.#
#A#B#C#D#
#A#B#C#D#
#########
#############
#...D.....AD#
###.#B#C#.###
#A#B#C#.#
#A#B#C#D#
#A#B#C#D#
#########
#############
#.........AD#
###.#B#C#.###
#A#B#C#D#
#A#B#C#D#
#A#B#C#D#
#########
#############
#..........D#
###A#B#C#.###
#A#B#C#D#
#A#B#C#D#
#A#B#C#D#
#########
#############
#...........#
###A#B#C#D###
#A#B#C#D#
#A#B#C#D#
#A#B#C#D#
#########
Using the initial configuration from the full diagram, what is the least energy required to organize the amphipods?
Your puzzle answer was 50265.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/challenges/challenge24
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--- Day 24: Arithmetic Logic Unit ---
Magic smoke starts leaking from the submarine's arithmetic logic unit (ALU). Without the ability to perform basic arithmetic and logic functions, the submarine can't produce cool patterns with its Christmas lights!
It also can't navigate. Or run the oxygen system.
Don't worry, though - you probably have enough oxygen left to give you enough time to build a new ALU.
The ALU is a four-dimensional processing unit: it has integer variables w, x, y, and z. These variables all start with the value 0. The ALU also supports six instructions:
inp a - Read an input value and write it to variable a.
add a b - Add the value of a to the value of b, then store the result in variable a.
mul a b - Multiply the value of a by the value of b, then store the result in variable a.
div a b - Divide the value of a by the value of b, truncate the result to an integer, then store the result in variable a. (Here, "truncate" means to round the value toward zero.)
mod a b - Divide the value of a by the value of b, then store the remainder in variable a. (This is also called the modulo operation.)
eql a b - If the value of a and b are equal, then store the value 1 in variable a. Otherwise, store the value 0 in variable a.
In all of these instructions, a and b are placeholders; a will always be the variable where the result of the operation is stored (one of w, x, y, or z), while b can be either a variable or a number. Numbers can be positive or negative, but will always be integers.
The ALU has no jump instructions; in an ALU program, every instruction is run exactly once in order from top to bottom. The program halts after the last instruction has finished executing.
(Program authors should be especially cautious; attempting to execute div with b=0 or attempting to execute mod with a<0 or b<=0 will cause the program to crash and might even damage the ALU. These operations are never intended in any serious ALU program.)
For example, here is an ALU program which takes an input number, negates it, and stores it in x:
inp x
mul x -1
Here is an ALU program which takes two input numbers, then sets z to 1 if the second input number is three times larger than the first input number, or sets z to 0 otherwise:
inp z
inp x
mul z 3
eql z x
Here is an ALU program which takes a non-negative integer as input, converts it into binary, and stores the lowest (1's) bit in z, the second-lowest (2's) bit in y, the third-lowest (4's) bit in x, and the fourth-lowest (8's) bit in w:
inp w
add z w
mod z 2
div w 2
add y w
mod y 2
div w 2
add x w
mod x 2
div w 2
mod w 2
Once you have built a replacement ALU, you can install it in the submarine, which will immediately resume what it was doing when the ALU failed: validating the submarine's model number. To do this, the ALU will run the MOdel Number Automatic Detector program (MONAD, your puzzle input).
Submarine model numbers are always fourteen-digit numbers consisting only of digits 1 through 9. The digit 0 cannot appear in a model number.
When MONAD checks a hypothetical fourteen-digit model number, it uses fourteen separate inp instructions, each expecting a single digit of the model number in order of most to least significant. (So, to check the model number 13579246899999, you would give 1 to the first inp instruction, 3 to the second inp instruction, 5 to the third inp instruction, and so on.) This means that when operating MONAD, each input instruction should only ever be given an integer value of at least 1 and at most 9.
Then, after MONAD has finished running all of its instructions, it will indicate that the model number was valid by leaving a 0 in variable z. However, if the model number was invalid, it will leave some other non-zero value in z.
MONAD imposes additional, mysterious restrictions on model numbers, and legend says the last copy of the MONAD documentation was eaten by a tanuki. You'll need to figure out what MONAD does some other way.
To enable as many submarine features as possible, find the largest valid fourteen-digit model number that contains no 0 digits. What is the largest model number accepted by MONAD?
Your puzzle answer was 99394899891971.
--- Part Two ---
As the submarine starts booting up things like the Retro Encabulator, you realize that maybe you don't need all these submarine features after all.
What is the smallest model number accepted by MONAD?
Your puzzle answer was 92171126131911.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/challenges/challenge25
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>--- Day 25: Sea Cucumber ---
This is it: the bottom of the ocean trench, the last place the sleigh keys could be. Your submarine's experimental antenna still isn't boosted enough to detect the keys, but they must be here. All you need to do is reach the seafloor and find them.
At least, you'd touch down on the seafloor if you could; unfortunately, it's completely covered by two large herds of sea cucumbers, and there isn't an open space large enough for your submarine.
You suspect that the Elves must have done this before, because just then you discover the phone number of a deep-sea marine biologist on a handwritten note taped to the wall of the submarine's cockpit.
"Sea cucumbers? Yeah, they're probably hunting for food. But don't worry, they're predictable critters: they move in perfectly straight lines, only moving forward when there's space to do so. They're actually quite polite!"
You explain that you'd like to predict when you could land your submarine.
"Oh that's easy, they'll eventually pile up and leave enough space for-- wait, did you say submarine? And the only place with that many sea cucumbers would be at the very bottom of the Mariana--" You hang up the phone.
There are two herds of sea cucumbers sharing the same region; one always moves east (>), while the other always moves south (v). Each location can contain at most one sea cucumber; the remaining locations are empty (.). The submarine helpfully generates a map of the situation (your puzzle input). For example:
v...>>.vv>
.vv>>.vv..
>>.>v>...v
>>v>>.>.v.
v>v.vv.v..
>.>>..v...
.vv..>.>v.
v.v..>>v.v
....v..v.>
Every step, the sea cucumbers in the east-facing herd attempt to move forward one location, then the sea cucumbers in the south-facing herd attempt to move forward one location. When a herd moves forward, every sea cucumber in the herd first simultaneously considers whether there is a sea cucumber in the adjacent location it's facing (even another sea cucumber facing the same direction), and then every sea cucumber facing an empty location simultaneously moves into that location.
So, in a situation like this:
...>>>>>...
After one step, only the rightmost sea cucumber would have moved:
...>>>>.>..
After the next step, two sea cucumbers move:
...>>>.>.>.
During a single step, the east-facing herd moves first, then the south-facing herd moves. So, given this situation:
..........
.>v....v..
.......>..
..........
After a single step, of the sea cucumbers on the left, only the south-facing sea cucumber has moved (as it wasn't out of the way in time for the east-facing cucumber on the left to move), but both sea cucumbers on the right have moved (as the east-facing sea cucumber moved out of the way of the south-facing sea cucumber):
..........
.>........
..v....v>.
..........
Due to strong water currents in the area, sea cucumbers that move off the right edge of the map appear on the left edge, and sea cucumbers that move off the bottom edge of the map appear on the top edge. Sea cucumbers always check whether their destination location is empty before moving, even if that destination is on the opposite side of the map:
Initial state:
...>...
.......
......>
v.....>
......>
.......
..vvv..
After 1 step:
..vv>..
.......
>......
v.....>
>......
.......
....v..
After 2 steps:
....v>.
..vv...
.>.....
......>
v>.....
.......
.......
After 3 steps:
......>
..v.v..
..>v...
>......
..>....
v......
.......
After 4 steps:
>......
..v....
..>.v..
.>.v...
...>...
.......
v......
To find a safe place to land your submarine, the sea cucumbers need to stop moving. Again consider the first example:
Initial state:
v...>>.vv>
.vv>>.vv..
>>.>v>...v
>>v>>.>.v.
v>v.vv.v..
>.>>..v...
.vv..>.>v.
v.v..>>v.v
....v..v.>
After 1 step:
....>.>v.>
v.v>.>v.v.
>v>>..>v..
>>v>v>.>.v
.>v.v...v.
v>>.>vvv..
..v...>>..
vv...>>vv.
>.v.v..v.v
After 2 steps:
>.v.v>>..v
v.v.>>vv..
>v>.>.>.v.
>>v>v.>v>.
.>..v....v
.>v>>.v.v.
v....v>v>.
.vv..>>v..
v>.....vv.
After 3 steps:
v>v.v>.>v.
v...>>.v.v
>vv>.>v>..
>>v>v.>.v>
..>....v..
.>.>v>v..v
..v..v>vv>
v.v..>>v..
.v>....v..
After 4 steps:
v>..v.>>..
v.v.>.>.v.
>vv.>>.v>v
>>.>..v>.>
..v>v...v.
..>>.>vv..
>.v.vv>v.v
.....>>vv.
vvv>...v..
After 5 steps:
vv>...>v>.
v.v.v>.>v.
>.v.>.>.>v
>v>.>..v>>
..v>v.v...
..>.>>vvv.
.>...v>v..
..v.v>>v.v
v.v.>...v.
...
After 10 steps:
..>..>>vv.
v.....>>.v
..v.v>>>v>
v>.>v.>>>.
..v>v.vv.v
.v.>>>.v..
v.v..>v>..
..v...>v.>
.vv..v>vv.
...
After 20 steps:
v>.....>>.
>vv>.....v
.>v>v.vv>>
v>>>v.>v.>
....vv>v..
.v.>>>vvv.
..v..>>vv.
v.v...>>.v
..v.....v>
...
After 30 steps:
.vv.v..>>>
v>...v...>
>.v>.>vv.>
>v>.>.>v.>
.>..v.vv..
..v>..>>v.
....v>..>v
v.v...>vv>
v.v...>vvv
...
After 40 steps:
>>v>v..v..
..>>v..vv.
..>>>v.>.v
..>>>>vvv>
v.....>...
v.v...>v>>
>vv.....v>
.>v...v.>v
vvv.v..v.>
...
After 50 steps:
..>>v>vv.v
..v.>>vv..
v.>>v>>v..
..>>>>>vv.
vvv....>vv
..v....>>>
v>.......>
.vv>....v>
.>v.vv.v..
...
After 55 steps:
..>>v>vv..
..v.>>vv..
..>>v>>vv.
..>>>>>vv.
v......>vv
v>v....>>v
vvv...>..>
>vv.....>.
.>v.vv.v..
After 56 steps:
..>>v>vv..
..v.>>vv..
..>>v>>vv.
..>>>>>vv.
v......>vv
v>v....>>v
vvv....>.>
>vv......>
.>v.vv.v..
After 57 steps:
..>>v>vv..
..v.>>vv..
..>>v>>vv.
..>>>>>vv.
v......>vv
v>v....>>v
vvv.....>>
>vv......>
.>v.vv.v..
After 58 steps:
..>>v>vv..
..v.>>vv..
..>>v>>vv.
..>>>>>vv.
v......>vv
v>v....>>v
vvv.....>>
>vv......>
.>v.vv.v..
In this example, the sea cucumbers stop moving after 58 steps.
Find somewhere safe to land your submarine. What is the first step on which no sea cucumbers move?
Your puzzle answer was 520.
--- Part Two ---
Suddenly, the experimental antenna control console lights up:
Sleigh keys detected!
According to the console, the keys are directly under the submarine. You landed right on them! Using a robotic arm on the submarine, you move the sleigh keys into the airlock.
Now, you just need to get them to Santa in time to save Christmas! You check your clock - it is Christmas. There's no way you can get them back to the surface in time.
Just as you start to lose hope, you notice a button on the sleigh keys: remote start. You can start the sleigh from the bottom of the ocean! You just need some way to boost the signal from the keys so it actually reaches the sleigh. Good thing the submarine has that experimental antenna! You'll definitely need 50 stars to boost it that far, though.
The experimental antenna control console lights up again:
Energy source detected.
Integrating energy source from device "sleigh keys"...done.
Installing device drivers...done.
Recalibrating experimental antenna...done.
Boost strength due to matching signal phase: 1 star
Only 49 stars to go.
If you like, you can [Remotely Start The Sleigh Again].
Both parts of this puzzle are complete! They provide two gold stars: **
2021/code/AoC14.py
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#!/usr/bin/env python3
import sys
from collections import Counter
from copy import deepcopy
def step(cons,s):
for _ in range(s):
ncons = {}
for (a,b),v in cons.items():
if (a, rules[a+b]) not in ncons: ncons[(a, rules[a+b])] = 0
if (rules[a+b], b) not in ncons: ncons[(rules[a+b], b)] = 0
ncons[(a, rules[a+b])] += v
ncons[(rules[a+b],b)] += v
cons = deepcopy(ncons)
return cons
def countp(cons):
ex = {}
for k in nex.keys():
ex[k[0]] = 0
ex[k[1]] = 0
ex[start[0]] += 1
ex[start[-1]] += 1
for k,v in nex.items():
ex[k[0]] += v
ex[k[1]] += v
return ex
if __name__ == '__main__':
start, rules = open(sys.argv[1]).read().split('\n\n')
rules = [r.split(' -> ') for r in rules.split('\n')]
rules = {a[0]:a[1] for a in rules[:-1]}
nex = {}
for a,b in zip(start[:-1],start[1:]):
if (a,b) not in nex: nex[(a,b)] = 0
nex[(a,b)] +=1
# challenge 1
nex = step(nex,10)
res = countp(nex)
res1 = str(max(res.values())//2-min(res.values())//2)
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
nex = step(nex,30)
res = countp(nex)
res2 = str(max(res.values())//2-min(res.values())//2)
print("challenge 2:" +"\n" + res2 + "\n")
2021/code/AoC15.py
-75
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@@ -1,75 +0,0 @@
#!/usr/bin/env python3
import sys
def memo(f):
memo = {}
def f2(a,x,y,c,e):
if (x,y) not in memo: memo[(x,y)] = f(a,x,y,c,e)
return memo[(x,y)]
return f2
def nb(sx, sy):
n = [(0,1), (1,0), (0,-1), (-1,0)]
return [(max(0,sx+x), max(0,sy+y)) for x,y in n]
def dijkstra(graph, s):
a = {}
v = {}
Q = initG(graph, s, a, v)
lq = len(Q)
while len(Q) > 0:
u = min([u for u in Q], key=lambda x: a[x])
Q.remove(u)
for vo in nb(*u):
if vo in Q:
distance_update(graph, u,vo,a,v)
if (len(Q)/lq * 100 % 1) == 0:
print("|", end='', flush=True)
print("\n")
return v
def initG(graph, s, a, v):
for vo in graph.keys():
a[vo] = 9999999999999999
v[vo] = 'null'
a[s] = 0
Q = [n for n in graph.keys()]
return Q
def distance_update(g, u,vo,a,v):
alt = a[u] + g[u]
if alt < a[vo]:
a[vo] = alt
v[vo] = u
def shortest_path(g, d, v):
w = [d]
u = d
r = 0
while v[u] != 'null':
r += g[u]
u = v[u]
w.append(u)
return w.reverse(), r
if __name__ == '__main__':
nums = [[int(i) for i in list(line.strip('\n'))] for line in open(sys.argv[1])]
# challenge 1
g1 = {(x,y) : nums[y][x] for x in range(len(nums[0])) for y in range(len(nums))}
v = dijkstra(g1, (0,0))
w, r = shortest_path(g1, (len(nums)-1, len(nums[0])-1), v)
res1 = str(r)
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2 - inefficient
g2 = {(x+i*len(nums[0]),y+j*len(nums)) : (nums[y][x] + i + j if nums[y][x] + i + j < 10 else (nums[y][x] + i + j) % 10 + 1) for x in range(len(nums[0])) for y in range(len(nums)) for i in range(5) for j in range(5)}
v = dijkstra(g2, (0,0))
w, r = shortest_path(g2, (len(nums[0])*5-1, len(nums)*5-1), v)
res2 = str(r)
print("challenge 2:" +"\n" + res2 + "\n")
2021/code/AoC16.py
-68
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@@ -1,68 +0,0 @@
#!/usr/bin/env python3
import sys
from bitstring import BitArray
from operator import mul
from functools import reduce
def parse_packet(packet, s, np=100000000, length=1000000000):
nex = s
end = s+length-1
bv = 0
numbers = []
while np > 0 and nex < min(len(packet)+1,end):
if (len(packet[nex:end+1]) == 0 or int(''.join(packet[nex:end+1]),2) == 0 or int(''.join(packet[nex:]),2) == 0):
if length < 1000000000: nex = end+1
break
v = int(''.join(packet[nex:nex+3]),2)
bv += v
t = int(''.join(packet[nex+3:nex+6]),2)
nex += 6
if t == 4:
num = []
while (n := packet[nex]) == '1':
num.append(list(packet[nex+1:nex+5]))
nex += 5
num.append(list(packet[nex+1:nex+5]))
nex += 5
i = int(''.join([''.join(n) for n in num]),2)
numbers.append(i)
else:
I = packet[nex]
nex += 1
if I == '1':
l = int(''.join(packet[nex:nex+11]),2)
nex += 11
b, nex, numbs = parse_packet(packet, nex, np=l)
bv += b
elif I == '0':
l = int(''.join(packet[nex:nex+15]),2)
nex += 15
b, nex, numbs = parse_packet(packet, nex, length=l)
bv += b
if t == 0: numbers.append(sum(numbs))
elif t == 1: numbers.append(reduce(mul, numbs, 1))
elif t == 2: numbers.append(min(numbs))
elif t == 3: numbers.append(max(numbs))
elif t == 5: numbers.append(int(numbs[0] > numbs[1]))
elif t == 6: numbers.append(int(numbs[0] < numbs[1]))
elif t == 7: numbers.append(int(numbs[0] == numbs[1]))
np -= 1
return bv, nex, numbers
if __name__ == '__main__':
hexa = [list(line.strip('\n')) for line in open(sys.argv[1])]
bits = [list(BitArray(hex=c).bin) for c in hexa[0]]
bits = [b for l in bits for b in l]
bv, _, num = parse_packet(bits,0,np=1)
# challenge 1
res1 = str(bv)
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
res2 = str(num[0])
print("challenge 2:" + "\n" + res2 + "\n")
2021/code/AoC17.py
-40
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@@ -1,40 +0,0 @@
#!/usr/bin/env python3
import sys, re
def parse(l):
return re.compile(r"target area: x=(?P<x1>\d+)..(?P<x2>\d+), y=(?P<y1>-?\d+)..(?P<y2>-?\d+)").search(l)
def step(cx,cy,x,y):
cx += x
cy += y
x = x-1 if x > 0 else (x+1 if x < 0 else 0)
y -= 1
return cx,cy,x,y
def hits_target(x,y,t):
cx, cy, rt, my = 0,0, False, 0
while cx <= max(t[0]) and cy >= min(t[1]):
cx,cy,x,y = step(cx,cy,x,y)
my = max(my,cy)
if (min(t[0]) <= cx <= max(t[0])) and (min(t[1]) <= cy <= max(t[1])): rt = True
return my, rt
if __name__ == '__main__':
x1,x2,y1,y2 = map(int,parse(open(sys.argv[1]).read()).groups())
target = ((x1,x2),(y1,y2))
dic = {}
for x in range(target[0][1]+1):
for y in range(min(target[1]),target[0][1]+max(map(abs,target[1]))):
my,rt = hits_target(x,y,target)
if rt: dic[(x,y)] = my
# challenge 1
res1 = str(max(dic.values()))
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
res2 = str(len(dic))
print("challenge 2:" +"\n" + res2 + "\n")
2021/code/AoC18.py
-100
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@@ -1,100 +0,0 @@
#!/usr/bin/env python3
import sys, math
from functools import reduce
from copy import deepcopy
def magn(n):
if type(n) == int: return n
return 3*magn(n[0]) + 2*magn(n[1])
def explode(osn):
sn = deepcopy(osn)
l = None
r = 0
d = False
for i,n1 in enumerate(sn):
if type(n1) == int:
l = (0,i,n1)
if r != 0:
sn[i] = n1 + r
return sn
else:
for j,n2 in enumerate(n1):
if type(n2) == int:
l = (1,[i,j],n2)
if r != 0:
n1[j] = n2 + r
return sn
else:
for k,n3 in enumerate(n2):
if type(n3) == int:
l = (2,[i,j,k],n3)
if r != 0:
n2[k] = n3 + r
return sn
else:
for m,n4 in enumerate(n3):
if type(n4) == int:
l = (3,[i,j,k,m],n4)
if r != 0:
n3[m] = n4 + r
return sn
elif not d:
d = True
r = n4[1]
if l != None:
if l[0] == 0: sn[l[1]] = l[2] + n4[0]
elif l[0] == 1: sn[l[1][0]][l[1][1]] = l[2] + n4[0]
elif l[0] == 2: sn[l[1][0]][l[1][1]][l[1][2]] = l[2] + n4[0]
elif l[0] == 3: sn[l[1][0]][l[1][1]][l[1][2]][l[1][3]] = l[2] + n4[0]
n3[m] = 0
else:
if type(n4[0]) != int:
print('help!')
n4[0] += r
return sn
return sn
def split(osn):
sn = deepcopy(osn)
if isinstance(sn,list):
for i,n in enumerate(sn):
o = sn[i]
sn[i] = split(n)
if sn[i] != o: return sn
elif sn > 9:
return [math.floor(sn/2),math.ceil(sn/2)]
return sn
def snreduce(sn):
new = 0
old = sn
while old != new:
old = new if new != 0 else old
new = explode(old)
if new != old:
continue
new = split(old)
return new
def snadd(sn1,sn2):
sn = [sn1,sn2]
return snreduce(sn)
if __name__ == '__main__':
nums = [eval(line.strip('\n')) for line in open(sys.argv[1])]
# challenge 1
res1 = str(magn(reduce(snadd,nums)))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
mm = 0
for n in nums:
for m in nums:
mm = max(mm, magn(snadd(n,m)))
res2 = str(mm)
print("challenge 2:" + "\n" + res2 + "\n")
2021/code/AoC19.py
-94
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@@ -1,94 +0,0 @@
#!/usr/bin/env python3
import sys, math
import numpy as np
def rotations(x,y,z):
r = [
# x is facing x
[x, y, z],
[x, -z, y],
[x, -y, -z],
[x, z, -y],
# x is facing -x
[-x, -y, z],
[-x, -z, -y],
[-x, y, -z],
[-x, z, y],
# x is facing y
[-z, x, -y],
[y, x, -z],
[z, x, y],
[-y, x, z],
# x is facing -y
[z, -x, -y],
[y, -x, z],
[-z, -x, y],
[-y, -x, -z],
# x is facing z
[-y, -z, x],
[z, -y, x],
[y, z, x],
[-z, y, x],
# x is facing -z
[z, y, -x],
[-y, z, -x],
[-z, -y, -x],
[y, -z, -x]
]
for point in r:
yield np.array(point)
def overlap(s1,s2,off):
se1 = set([tuple(p) for p in s1])
se2 = set([tuple(p-off) for p in s2])
return len(se1.intersection(se2))
def test(s1,s2):
rot = [[r for r in rotations(*p)] for p in s2]
rot2 = [[r[i] for r in rot] for i in range(24)]
mc = 0
for i in range(24):
for p1 in s1:
for p2 in rot2[i]:
dif = p2 - p1
com = overlap(s1,rot2[i],dif)
if com > mc:
mc = com
if mc >= 12:
return [r - dif for r in rot2[i]], dif
return None, None
if __name__ == '__main__':
nums = [[np.array([i for i in map(int,n.split(','))]) for n in line.split('\n')[1:] if n != ''] for line in open(sys.argv[1]).read().split('\n\n')]
scanners = []
matched = {}
matched[0] = nums[0]
al = set([tuple(i) for i in matched[0]])
while len(matched) != len(nums):
for i in range(len(nums)):
if i in matched: continue
r = None
r, d = test(al,nums[i])
scanners.append(d)
if r != None:
matched[i] = r
al = al.union(set([tuple(i) for i in matched[i]]))
break
m = 0
for a in scanners:
for b in scanners:
n = sum(abs(a[i] - b[i]) for i in range(3))
if n > m:
m = n
# challenge 1
res1 = str(len(al))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
res2 = str(m)
print("challenge 2:" + "\n" + res2 + "\n")
2021/code/AoC20.py
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#!/usr/bin/env python3
import sys
from copy import deepcopy
def imagine(im):
for y in im:
l = ''
for x in y: l += x
print(l)
def n(im,x,y,o):
index = ''
h = [(-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)]
for j,i in h:
if (0 <= x+i < len(im[0]) and 0 <= y+j < len(im)): index += '0' if im[y+j][x+i] == '.' else '1'
else: index += '0' if o == '.' else '1'
return algo[int(index,2)]
def enhance(im, o):
nex = deepcopy(im)
for y in range(len(im)):
for x in range(len(im[0])):
nex[y][x] = n(im,x,y,o)
return nex, '.' if o == '#' or mode == 'k' else '#'
if __name__ == '__main__':
algo, im = open(sys.argv[1]).read().split('\n\n')
algo = list(algo.strip('\n'))
im = [list(i) for i in im.split('\n') if i != '']
print('------------------ start -------------------')
bim = [[im[i-50][j-50] if (49 < i <= len(im)+49 and 49 < j <= len(im[0])+49) else '.' for j in range(len(im[0]) + 100)] for i in range(len(im) + 100)]
imagine(bim)
mode = 's' if algo[0] == '#' else 'k'
nim, o = bim, '.'
for i in range(50):
print('-------------------- ' + str(i) + ' ---------------------')
nim, o = enhance(nim,o)
imagine(nim)
if i == 1: c1 = nim
# challenge 1
res1 = str(sum(sum(1 if i == '#' else 0 for i in j) for j in c1))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
res2 = str(sum(sum(1 if i == '#' else 0 for i in j) for j in nim))
print("challenge 2:" + "\n" + res2 + "\n")
2021/code/AoC21.py
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#!/usr/bin/env python3
import sys, re
def memo(fc):
memo = {}
def f2(a,b,c,d,e,f,g):
if (a,b,c,d,e,f,g) not in memo: memo[(a,b,c,d,e,f,g)] = fc(a,b,c,d,e,f,g)
return memo[(a,b,c,d,e,f,g)]
return f2
def parse(line):
groups = re.compile(r"Player 1 starting position: (?P<p1>\d+)\nPlayer 2 starting position: (?P<p2>\d+)").search(line)
return groups
def step(p1,p2):
d,s1,s2,die,dr = 0,0,0,0,0
t = 1
while s1 < 1000 and s2 < 1000:
mv = d*3 + 6
d = d + 3 % 100
dr += 3
if t == 1:
p1 = (p1 + mv - 1) % 10 + 1
s1 += p1
if t == 2:
p2 = (p2 + mv - 1) % 10 + 1
s2 += p2
t = t%2+1
return dr*s2 if s2 < 1000 else dr*s1
@memo
def qstep(p1,p2,s1,s2,dv,t,dt):
if s1 >= 21:
return {1:1,2:0}
elif s2 >= 21:
return {1:0,2:1}
wins = {1: 0, 2: 0}
if t == 1:
if dt == 3:
nt = t%2+1
np1 = (p1 + dv - 1) % 10 + 1
ns1 = s1+np1
win = qstep(np1,p2,ns1,s2,0,nt,0)
wins[1] += win[1]
wins[2] += win[2]
else:
dt += 1
for d in range(1,4):
ndv = dv+d
win = qstep(p1,p2,s1,s2,ndv,t,dt)
wins[1] += win[1]
wins[2] += win[2]
if t == 2:
if dt == 3:
nt = t%2+1
np2 = (p2 + dv - 1) % 10 + 1
ns2 = s2+np2
win = qstep(p1,np2,s1,ns2,0,nt,0)
wins[1] += win[1]
wins[2] += win[2]
else:
dt += 1
for d in range(1,4):
ndv = dv+d
win = qstep(p1,p2,s1,s2,ndv,t,dt)
wins[1] += win[1]
wins[2] += win[2]
return wins
if __name__ == '__main__':
p1,p2 = map(int,parse(open(sys.argv[1]).read()).groups())
# challenge 1
res1 = str(step(p1,p2))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
results = qstep(p1,p2,0,0,0,1,0)
res2 = str(max(results.values()))
print("challenge 2:" + "\n" + res2 + "\n")
2021/code/AoC22.py
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#!/usr/bin/env python3
import sys,re
from collections import Counter
def parse(l):
return re.compile(r"(?P<mode>\w+) x=(?P<x1>-?\d+)..(?P<x2>-?\d+),y=(?P<y1>-?\d+)..(?P<y2>-?\d+),z=(?P<z1>-?\d+)..(?P<z2>-?\d+)").search(l)
def volume(k):
if k[1]-k[0] < 0 or k[3]-k[2] < 0 or k[5]-k[4] < 0: return 0
return (k[1]-k[0]+1)*(k[3]-k[2]+1)*(k[5]-k[4]+1)
def count_cubes(c):
o = 0
for k,v in c.items():
if v == 'on':
o += volume(k)
return o
def intersect(c1,c2):
x1,x2,y1,y2,z1,z2 = c1
lx,hx,ly,hy,lz,hz = c2
xos = max(lx,x1)
xoe = min(hx,x2)
yos = max(ly,y1)
yoe = min(hy,y2)
zos = max(lz,z1)
zoe = min(hz,z2)
return volume((xos,xoe,yos,yoe,zos,zoe)) > 0
#return (c1[0] in range(c2[0],c2[1]+1) or c1[1] in range(c2[0],c2[1]+1)) and (c1[2] in range(c2[2],c2[3]+1) or c1[3] in range(c2[2],c2[3]+1)) and (c1[4] in range(c2[4],c2[5]+1) or c1[5] in range(c2[4],c2[5]+1))
def split(c1,c2):
x1,x2,y1,y2,z1,z2 = c1
lx,hx,ly,hy,lz,hz = c2
xos = max(lx,x1)
xoe = min(hx,x2)
yos = max(ly,y1)
yoe = min(hy,y2)
zos = max(lz,z1)
zoe = min(hz,z2)
v = 0
for x in [(lx,xos-1),(xos,xoe),(xoe+1,hx)]:
for y in [(ly,yos-1),(yos,yoe),(yoe+1,hy)]:
for z in [(lz,zos-1),(zos,zoe),(zoe+1,hz)]:
v += volume((*x,*y,*z))
if volume((*x,*y,*z)) > 0 and (x,y,z) != ((xos,xoe),(yos,yoe),(zos,zoe)):
yield (*x,*y,*z)
def init(steps):
cubes = {}
for m,x1,x2,y1,y2,z1,z2 in steps:
if not (-50 <= x1 <= 50 and -50 <= x2 <= 50): continue
for x in range(x1,x2+1):
if not (-50 <= y1 <= 50 and -50 <= y2 <= 50): continue
for y in range(y1,y2+1):
if not (-50 <= z1 <= 50 and -50 <= z2 <= 50): continue
for z in range(z1,z2+1):
cubes[(x,y,z)] = m
return cubes
def reboot(steps):
cubes = {}
for c1 in steps:
m,c1 = c1[0],c1[1:]
atm = [k for k in cubes.keys()]
for c2 in atm:
om = cubes[c2]
if intersect(c1,c2) or intersect(c2,c1):
for c in split(c1,c2):
if not (intersect(c,c1) or intersect(c1,c)):
cubes[c] = om
cubes.pop(c2)
cubes[c1] = m
return cubes
if __name__ == '__main__':
nums = map(lambda x: (x[0],int(x[1]),int(x[2]),int(x[3]),int(x[4]),int(x[5]),int(x[6])),[parse(l).groups() for l in open(sys.argv[1])])
nums = [n for n in nums]
# challenge 1
cubes1 = init(nums)
c = Counter([v for k,v in cubes1.items() if -50 <= k[0] <= 50 and -50 <= k[1] <= 50 and -50 <= k[2] <= 50])
res1 = str(c.most_common(2))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
cubes2 = reboot(nums)
res2 = str(count_cubes(cubes2))
print("challenge 2:" + "\n" + res2 + "\n")
2021/code/AoC23.py
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#!/usr/bin/env python3
import sys
from functools import cache
@cache
def path(f1,f2):
path = []
y, x = f1
while y != 0:
y -= 1
path.append((y,x))
while x != f2[1]:
x += 1 if f2[1] > x else -1
path.append((y,x))
while y != f2[0]:
y += 1
path.append((y,x))
assert path[-1] == f2
return path
def pathvalid(path,board):
for p in path:
if p in board: return False
return True
def cost(a,w):
if a == 'A': return len(w) * 1
if a == 'B': return len(w) * 10
if a == 'C': return len(w) * 100
if a == 'D': return len(w) * 1000
def endstate(board):
for f,a in board.items():
if f[1] != target[a]: return False
return True
def target_position(f,b,lowest):
a = b[f]
if f[1] != target[a]: return False
for y in range(f[0],lowest+1):
if (y,target[a]) not in b: return False
if b[(y,target[a])] != a: return False
return True
def totarget(board,lowest):
change = False
costs = 0
nb = {k: v for k,v in board.items()}
for f,a in board.items():
if target_position(f,nb,lowest): continue
t = (lowest,target[a])
while t in nb and nb[t] == a:
t = (t[0]-1,target[a])
if t == (0,target[a]): continue
p = path(f,t)
if pathvalid(p,nb):
change = True
nb.pop(f)
nb[t] = a
costs += cost(a,p)
return nb,costs,change
def dfs(board,lowest,costs=0,curmin=sys.maxsize):
ch = True
nb = {k: v for k,v in board.items()}
while ch:
nb,co,ch = totarget(nb,lowest)
costs += co
if endstate(nb):
return costs
if costs > curmin: return None
board = {k: v for k,v in nb.items()}
for f,a in board.items():
if f[0] != 0 and not(target_position(f,board,lowest)):
for f2 in hallway:
p = path(f,f2)
if pathvalid(p, board):
cb = {k: v for k,v in board.items()}
cb.pop(f)
cb[f2] = a
nc = dfs(cb,lowest,costs+cost(a,p),curmin)
if nc is not None and nc < curmin: curmin = nc
if curmin < sys.maxsize: return curmin
if __name__ == '__main__':
# challenge 1
lines = open(sys.argv[1]).readlines()
start = {}
for y,l in enumerate(lines[1:]):
for x,f in enumerate(l[1:]):
if f in 'ABCD': start[(y,x)] = f
hallway = [(0,0), (0,1), (0,3), (0,5), (0,7), (0,9), (0,10)]
target = {'A': 2, 'B': 4, 'C': 6, 'D': 8}
res1 = str(dfs(start,2))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
lines = open(sys.argv[2]).readlines()
start = {}
for y,l in enumerate(lines[1:]):
for x,f in enumerate(l[1:]):
if f in 'ABCD': start[(y,x)] = f
res2 = str(dfs(start,4))
print("challenge 2:" + "\n" + res2 + "\n")
2021/code/AoC24.py
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#!/usr/bin/env python3
import sys, re, math
from operator import add, mul, truediv, mod
from functools import reduce
def parse(line):
l = line.split(' ')
if l[0] == 'inp': i,o = l
else: i,o = l[0], [l[1],l[2]]
return (i,o)
def replist(l,v,r):
return [r if i == v else i for i in l]
def eql(a,b):
return int(a == b)
def div(a,b):
return math.trunc(reduce(truediv,[a,b]))
def alu(instruction, inp):
reg = {'w': 0, 'x': 0, 'y': 0, 'z': 0}
inp = list(str(inp))
index = 0
for i,ops in instruction:
if i == 'inp':
reg[ops] = inp[index]
index += 1
else:
reg[ops[0]] = reduce(eval(i),[int(reg[o]) if o in 'wxyz' else int(o) for o in ops])
return reg
if __name__ == '__main__':
instr = [parse(line.strip('\n')) for line in open(sys.argv[1])]
for i in range(1,9):
for k in range(11111111111,99999999999):
for j in ['18','27']:
r = alu(instr,str(i)+j+str(k))
print(str(i)+j+str(k),r)
if r['z'] == 0: print(i,j,k)
for i in range(99999999999999,9999999999999,-1):
if i % 10000 == 0: print(i)
if '0' in str(i): continue
r = alu(instr,str(i))
if r['z'] == 0:
print(i)
break
# challenge 1
res1 = ""
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
res2 = ""
print("challenge 2:" + "\n" + res2 + "\n")
2021/code/AoC25.py
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#!/usr/bin/env python3
import sys
from copy import deepcopy
def cprint(cucs):
for l in cucs:
print(''.join(l))
print('\n')
def step(cucs):
nc = deepcopy(cucs)
for y,l in enumerate(cucs):
for x,c in enumerate(l):
if c == '>':
if cucs[y][(x+1)%len(l)] == '.':
nc[y][x] = '.'
nc[y][(x+1)%len(l)] = '>'
cucs = deepcopy(nc)
for y,l in enumerate(cucs):
for x,c in enumerate(l):
if c == 'v':
if cucs[(y+1)%len(cucs)][x] == '.':
nc[y][x] = '.'
nc[(y+1)%len(cucs)][x] = 'v'
return nc
if __name__ == '__main__':
cucs = [list(line.strip('\n')) for line in open(sys.argv[1])]
# challenge 1
nc = []
i = 0
while nc != cucs:
print(i)
i += 1
cprint(nc)
cucs = deepcopy(nc) if nc != [] else cucs
nc = step(cucs)
res1 = str(i)
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
res2 = "doesn't exist :)"
print("challenge 2:" + "\n" + res2 + "\n")
2021/code/aoc24.pl
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:- use_module(library(clpfd)).
% den state kannst du mit nem dictoniary modellieren:
initState(state{ x: 0, y: 0, z:0, w: 0 , in: I}, I).
% Dictonary lookup: initstate(State), R = State.x
% Dictonary update: initstate(State), NewState = State.put(x, 1)
% Convert List of input digits into single number
%input_number(+Inputs, -Number)
input_number(Inputs, Number) :- atomic_list_concat(Inputs, Number).
% read program helper functions
noteof(L) :- \+ L = [end_of_file].
string_term(String,Term) :- term_string(Term, String).
parse_instr(String, Inst) :-
split_string(String, " ", "", Words),
maplist(string_term, Words, Inst).
% Read a list of Instructions from Filename
% read_program(+Filename, -Prog)
read_program(Filename, Prog) :-
open(Filename, read, Str),
read_string(Str, _, S),
split_string(S, "\n", "", Lines),
maplist(parse_instr, Lines, Instrs),
include(noteof, Instrs, Prog),
close(Str).
% Takes input and state and binds O to a number
num(N,S,O) :-
atom(N),
O = S.N.
num(N,S,O) :-
number(N),
O = N.
eq(A,B,R) :-
A = B,
R = 1.
eq(A,B,R) :-
A #\= B,
R = 0.
read_input(In,F,NIn) :-
[F | NIn] = In.
% interprets an instruction
% interpret(+list,+state,R)
interpret([inp, A],InState,OutState) :-
read_input(InState.in,C,NInput),
BS = InState.put(A,C),
OutState = BS.put(in,NInput).
interpret([add, A, B],InState,OutState) :-
num(B,InState,C),
R #= InState.A + C,
OutState = InState.put(A,R).
interpret([mul, A, B],InState,OutState) :-
num(B,InState,C),
R #= InState.A*C,
OutState = InState.put(A,R).
interpret([div, A, B],InState,OutState) :-
num(B,InState,C),
R #= InState.A div C,
OutState = InState.put(A,R).
interpret([mod, A, B],InState,OutState) :-
num(B,InState,C),
R #= InState.A mod C,
OutState = InState.put(A,R).
interpret([eql, A, B],InState,OutState) :-
num(B,InState,C),
eq(InState.A, C, R),
OutState = InState.put(A,R).
rprogram(State,[],O) :-
O = State.z.
rprogram(State,[I|Instr], O) :-
interpret(I,State,Out),
rprogram(Out,Instr,O).
program(Instr,Input) :-
length(Input, 14),
Input ins 1..9,
initState(S,Input),
rprogram(S,Instr, O),
O = 0.
max_monad(M) :-
read_program(input24,Instr),
program(Instr,R),
labeling([max,down],R),
input_number(R,M),
write(M).
min_monad(M) :-
read_program(input24,Instr),
program(Instr,R),
labeling([min],R),
input_number(R,M),
write(M).
2021/day01/AoC01.py
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#!/usr/bin/env python3
import sys
if __name__ == '__main__':
measurements = open(sys.argv[1]).read().strip('\n\n').split('\n')
meas = [int(i) for i in measurements]
# challenge 1
incr = 0
last = 0
for i in meas:
if i > last:
incr += 1
last = i
res1 = str(incr - 1)
print("challenge 1: " + res1 + "\n")
# challenge 2
incr = 0
first = meas[0]
second = meas[1]
third = meas[2]
last = first + second + third
for i in meas[3:]:
first = second
second = third
third = i
nex = first + second + third
if nex > last:
incr += 1
last = nex
res2 = str(incr)
print("challenge 2: " + res2 + "\n")
2021/day01/challenge01
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--- Day 1: Sonar Sweep ---
You're minding your own business on a ship at sea when the overboard alarm goes off! You rush to see if you can help. Apparently, one of the Elves tripped and accidentally sent the sleigh keys flying into the ocean!
Before you know it, you're inside a submarine the Elves keep ready for situations like this. It's covered in Christmas lights (because of course it is), and it even has an experimental antenna that should be able to track the keys if you can boost its signal strength high enough; there's a little meter that indicates the antenna's signal strength by displaying 0-50 stars.
Your instincts tell you that in order to save Christmas, you'll need to get all fifty stars by December 25th.
Collect stars by solving puzzles. Two puzzles will be made available on each day in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!
As the submarine drops below the surface of the ocean, it automatically performs a sonar sweep of the nearby sea floor. On a small screen, the sonar sweep report (your puzzle input) appears: each line is a measurement of the sea floor depth as the sweep looks further and further away from the submarine.
For example, suppose you had the following report:
199
200
208
210
200
207
240
269
260
263
This report indicates that, scanning outward from the submarine, the sonar sweep found depths of 199, 200, 208, 210, and so on.
The first order of business is to figure out how quickly the depth increases, just so you know what you're dealing with - you never know if the keys will get carried into deeper water by an ocean current or a fish or something.
To do this, count the number of times a depth measurement increases from the previous measurement. (There is no measurement before the first measurement.) In the example above, the changes are as follows:
199 (N/A - no previous measurement)
200 (increased)
208 (increased)
210 (increased)
200 (decreased)
207 (increased)
240 (increased)
269 (increased)
260 (decreased)
263 (increased)
In this example, there are 7 measurements that are larger than the previous measurement.
How many measurements are larger than the previous measurement?
Your puzzle answer was 1400.
--- Part Two ---
Considering every single measurement isn't as useful as you expected: there's just too much noise in the data.
Instead, consider sums of a three-measurement sliding window. Again considering the above example:
199 A
200 A B
208 A B C
210 B C D
200 E C D
207 E F D
240 E F G
269 F G H
260 G H
263 H
Start by comparing the first and second three-measurement windows. The measurements in the first window are marked A (199, 200, 208); their sum is 199 + 200 + 208 = 607. The second window is marked B (200, 208, 210); its sum is 618. The sum of measurements in the second window is larger than the sum of the first, so this first comparison increased.
Your goal now is to count the number of times the sum of measurements in this sliding window increases from the previous sum. So, compare A with B, then compare B with C, then C with D, and so on. Stop when there aren't enough measurements left to create a new three-measurement sum.
In the above example, the sum of each three-measurement window is as follows:
A: 607 (N/A - no previous sum)
B: 618 (increased)
C: 618 (no change)
D: 617 (decreased)
E: 647 (increased)
F: 716 (increased)
G: 769 (increased)
H: 792 (increased)
In this example, there are 5 sums that are larger than the previous sum.
Consider sums of a three-measurement sliding window. How many sums are larger than the previous sum?
Your puzzle answer was 1429.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day02/AoC02.py
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#!/usr/bin/env python3
import sys
if __name__ == '__main__':
lines = open(sys.argv[1]).readlines()
dirs = [i.strip('\n').split(" ") for i in lines]
# challenge 1
dep = 0
pos = 0
for d in dirs:
if d[0] == 'forward':
pos += int(d[1])
elif d[0] == 'up':
dep -= int(d[1])
elif d[0] == 'down':
dep += int(d[1])
else:
print(d)
res1 = str(dep * pos)
print("challenge 1:" + res1 + "\n")
# challenge 2
dep = 0
pos = 0
aim = 0
for d in dirs:
if d[0] == 'forward':
pos += int(d[1])
dep += int(d[1]) * aim
elif d[0] == 'up':
aim -= int(d[1])
elif d[0] == 'down':
aim += int(d[1])
else:
print(d)
res2 = str(dep * pos)
print("challenge 2:" + res2 + "\n")
2021/day02/challenge02
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--- Day 2: Dive! ---
Now, you need to figure out how to pilot this thing.
It seems like the submarine can take a series of commands like forward 1, down 2, or up 3:
forward X increases the horizontal position by X units.
down X increases the depth by X units.
up X decreases the depth by X units.
Note that since you're on a submarine, down and up affect your depth, and so they have the opposite result of what you might expect.
The submarine seems to already have a planned course (your puzzle input). You should probably figure out where it's going. For example:
forward 5
down 5
forward 8
up 3
down 8
forward 2
Your horizontal position and depth both start at 0. The steps above would then modify them as follows:
forward 5 adds 5 to your horizontal position, a total of 5.
down 5 adds 5 to your depth, resulting in a value of 5.
forward 8 adds 8 to your horizontal position, a total of 13.
up 3 decreases your depth by 3, resulting in a value of 2.
down 8 adds 8 to your depth, resulting in a value of 10.
forward 2 adds 2 to your horizontal position, a total of 15.
After following these instructions, you would have a horizontal position of 15 and a depth of 10. (Multiplying these together produces 150.)
Calculate the horizontal position and depth you would have after following the planned course. What do you get if you multiply your final horizontal position by your final depth?
Your puzzle answer was 1648020.
--- Part Two ---
Based on your calculations, the planned course doesn't seem to make any sense. You find the submarine manual and discover that the process is actually slightly more complicated.
In addition to horizontal position and depth, you'll also need to track a third value, aim, which also starts at 0. The commands also mean something entirely different than you first thought:
down X increases your aim by X units.
up X decreases your aim by X units.
forward X does two things:
It increases your horizontal position by X units.
It increases your depth by your aim multiplied by X.
Again note that since you're on a submarine, down and up do the opposite of what you might expect: "down" means aiming in the positive direction.
Now, the above example does something different:
forward 5 adds 5 to your horizontal position, a total of 5. Because your aim is 0, your depth does not change.
down 5 adds 5 to your aim, resulting in a value of 5.
forward 8 adds 8 to your horizontal position, a total of 13. Because your aim is 5, your depth increases by 8*5=40.
up 3 decreases your aim by 3, resulting in a value of 2.
down 8 adds 8 to your aim, resulting in a value of 10.
forward 2 adds 2 to your horizontal position, a total of 15. Because your aim is 10, your depth increases by 2*10=20 to a total of 60.
After following these new instructions, you would have a horizontal position of 15 and a depth of 60. (Multiplying these produces 900.)
Using this new interpretation of the commands, calculate the horizontal position and depth you would have after following the planned course. What do you get if you multiply your final horizontal position by your final depth?
Your puzzle answer was 1759818555.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day03/AoC03.py
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#!/usr/bin/env python3
import sys
def inverted(i):
if i == 0:
return 1
else:
return 0
def filter(lis, pos, i):
if len(lis) == 1:
return lis
if sum([b[pos] for b in lis]) >= sum([inverted(b[pos]) for b in lis]):
val = i
else:
val = inverted(i)
return [i for i in lis if i[pos] == val]
if __name__ == '__main__':
lines = open(sys.argv[1]).readlines()
bins = [[int(i) for i in list(l.strip('\n'))] for l in lines]
# challenge 1
lis = [0] * len(bins[0])
for i in range(len(bins[0])):
lis[i] = sum([b[i] for b in bins])
g = ''
e = ''
for i in range(len(bins[0])):
if lis[i] > 0.5 * len(lines):
g += '1'
e += '0'
else:
g += '0'
e += '1'
gamma = int(g, 2)
eps = int(e,2)
res1 = str(gamma * eps)
print("challenge 1:" + res1 + "\n")
ox = bins
co2 = bins
# challenge 2
for i in range(len(bins[0])):
ox = filter(ox, i, 1)
co2 = filter(co2, i, 0)
oxr = int(''.join(str(i) for i in ox[0]), 2)
csr = int(''.join(str(i) for i in co2[0]), 2)
res2 = str(oxr * csr)
print("challenge 2:" + res2 + "\n")
2021/day03/challenge03
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--- Day 3: Binary Diagnostic ---
The submarine has been making some odd creaking noises, so you ask it to produce a diagnostic report just in case.
The diagnostic report (your puzzle input) consists of a list of binary numbers which, when decoded properly, can tell you many useful things about the conditions of the submarine. The first parameter to check is the power consumption.
You need to use the binary numbers in the diagnostic report to generate two new binary numbers (called the gamma rate and the epsilon rate). The power consumption can then be found by multiplying the gamma rate by the epsilon rate.
Each bit in the gamma rate can be determined by finding the most common bit in the corresponding position of all numbers in the diagnostic report. For example, given the following diagnostic report:
00100
11110
10110
10111
10101
01111
00111
11100
10000
11001
00010
01010
Considering only the first bit of each number, there are five 0 bits and seven 1 bits. Since the most common bit is 1, the first bit of the gamma rate is 1.
The most common second bit of the numbers in the diagnostic report is 0, so the second bit of the gamma rate is 0.
The most common value of the third, fourth, and fifth bits are 1, 1, and 0, respectively, and so the final three bits of the gamma rate are 110.
So, the gamma rate is the binary number 10110, or 22 in decimal.
The epsilon rate is calculated in a similar way; rather than use the most common bit, the least common bit from each position is used. So, the epsilon rate is 01001, or 9 in decimal. Multiplying the gamma rate (22) by the epsilon rate (9) produces the power consumption, 198.
Use the binary numbers in your diagnostic report to calculate the gamma rate and epsilon rate, then multiply them together. What is the power consumption of the submarine? (Be sure to represent your answer in decimal, not binary.)
Your puzzle answer was 3242606.
--- Part Two ---
Next, you should verify the life support rating, which can be determined by multiplying the oxygen generator rating by the CO2 scrubber rating.
Both the oxygen generator rating and the CO2 scrubber rating are values that can be found in your diagnostic report - finding them is the tricky part. Both values are located using a similar process that involves filtering out values until only one remains. Before searching for either rating value, start with the full list of binary numbers from your diagnostic report and consider just the first bit of those numbers. Then:
Keep only numbers selected by the bit criteria for the type of rating value for which you are searching. Discard numbers which do not match the bit criteria.
If you only have one number left, stop; this is the rating value for which you are searching.
Otherwise, repeat the process, considering the next bit to the right.
The bit criteria depends on which type of rating value you want to find:
To find oxygen generator rating, determine the most common value (0 or 1) in the current bit position, and keep only numbers with that bit in that position. If 0 and 1 are equally common, keep values with a 1 in the position being considered.
To find CO2 scrubber rating, determine the least common value (0 or 1) in the current bit position, and keep only numbers with that bit in that position. If 0 and 1 are equally common, keep values with a 0 in the position being considered.
For example, to determine the oxygen generator rating value using the same example diagnostic report from above:
Start with all 12 numbers and consider only the first bit of each number. There are more 1 bits (7) than 0 bits (5), so keep only the 7 numbers with a 1 in the first position: 11110, 10110, 10111, 10101, 11100, 10000, and 11001.
Then, consider the second bit of the 7 remaining numbers: there are more 0 bits (4) than 1 bits (3), so keep only the 4 numbers with a 0 in the second position: 10110, 10111, 10101, and 10000.
In the third position, three of the four numbers have a 1, so keep those three: 10110, 10111, and 10101.
In the fourth position, two of the three numbers have a 1, so keep those two: 10110 and 10111.
In the fifth position, there are an equal number of 0 bits and 1 bits (one each). So, to find the oxygen generator rating, keep the number with a 1 in that position: 10111.
As there is only one number left, stop; the oxygen generator rating is 10111, or 23 in decimal.
Then, to determine the CO2 scrubber rating value from the same example above:
Start again with all 12 numbers and consider only the first bit of each number. There are fewer 0 bits (5) than 1 bits (7), so keep only the 5 numbers with a 0 in the first position: 00100, 01111, 00111, 00010, and 01010.
Then, consider the second bit of the 5 remaining numbers: there are fewer 1 bits (2) than 0 bits (3), so keep only the 2 numbers with a 1 in the second position: 01111 and 01010.
In the third position, there are an equal number of 0 bits and 1 bits (one each). So, to find the CO2 scrubber rating, keep the number with a 0 in that position: 01010.
As there is only one number left, stop; the CO2 scrubber rating is 01010, or 10 in decimal.
Finally, to find the life support rating, multiply the oxygen generator rating (23) by the CO2 scrubber rating (10) to get 230.
Use the binary numbers in your diagnostic report to calculate the oxygen generator rating and CO2 scrubber rating, then multiply them together. What is the life support rating of the submarine? (Be sure to represent your answer in decimal, not binary.)
Your puzzle answer was 4856080.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day04/AoC04.py
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#!/usr/bin/env python3
import sys
def printb(board):
for line in board:
print(str(line[0]) + ' ' + str(line[1]) + ' ' + str(line[2]) + ' ' + str(line[3]) + ' ' + str(line[4]))
print("\n")
def check_win(board):
for l in board:
if l == ['x' ,'x', 'x', 'x', 'x']:
return True
hit = 0
for i in range(len(board)):
for l in board:
if l[i] == 'x':
hit += 1
if hit == len(board):
return True
hit = 0
return False
def play(boards):
for num in nums:
boards = [mark_board(board, num) for board in boards]
for board in boards:
if check_win(board):
return num * sum([sum([i for i in l if i != 'x']) for l in board])
def play2(boards):
for num in nums:
boards = [mark_board(board, num) for board in boards]
if sum([1 if check_win(board) else 0 for board in boards]) == (len(boards) - 1):
for i in range(len(boards)):
if not check_win(boards[i]):
last = i
if sum([1 if check_win(board) else 0 for board in boards]) == (len(boards)):
if check_win(boards[last]):
board = boards[last]
return num * sum([sum([i for i in l if i != 'x']) for l in board])
def mark_board(board, num):
return [['x' if col == num else col for col in li] for li in board]
if __name__ == '__main__':
lines = open(sys.argv[1]).read().split('\n\n') # or split \n\n or sth similar
nums = [int(i) for i in lines[0].split(',')]
boards = [[[int(n) for n in l.split(' ') if n != ''] for l in li.split('\n')] for li in lines[1:]]
boards = [[l for l in b if l != []] for b in boards]
# challenge 1
res1 = str(play(boards))
print("challenge 1:" + res1 + "\n")
# challenge 2
res2 = str(play2(boards))
print("challenge 2:" + res2 + "\n")
2021/day04/challenge04
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--- Day 4: Giant Squid ---
You're already almost 1.5km (almost a mile) below the surface of the ocean, already so deep that you can't see any sunlight. What you can see, however, is a giant squid that has attached itself to the outside of your submarine.
Maybe it wants to play bingo?
Bingo is played on a set of boards each consisting of a 5x5 grid of numbers. Numbers are chosen at random, and the chosen number is marked on all boards on which it appears. (Numbers may not appear on all boards.) If all numbers in any row or any column of a board are marked, that board wins. (Diagonals don't count.)
The submarine has a bingo subsystem to help passengers (currently, you and the giant squid) pass the time. It automatically generates a random order in which to draw numbers and a random set of boards (your puzzle input). For example:
7,4,9,5,11,17,23,2,0,14,21,24,10,16,13,6,15,25,12,22,18,20,8,19,3,26,1
22 13 17 11 0
8 2 23 4 24
21 9 14 16 7
6 10 3 18 5
1 12 20 15 19
3 15 0 2 22
9 18 13 17 5
19 8 7 25 23
20 11 10 24 4
14 21 16 12 6
14 21 17 24 4
10 16 15 9 19
18 8 23 26 20
22 11 13 6 5
2 0 12 3 7
After the first five numbers are drawn (7, 4, 9, 5, and 11), there are no winners, but the boards are marked as follows (shown here adjacent to each other to save space):
22 13 17 11 0 3 15 0 2 22 14 21 17 24 4
8 2 23 4 24 9 18 13 17 5 10 16 15 9 19
21 9 14 16 7 19 8 7 25 23 18 8 23 26 20
6 10 3 18 5 20 11 10 24 4 22 11 13 6 5
1 12 20 15 19 14 21 16 12 6 2 0 12 3 7
After the next six numbers are drawn (17, 23, 2, 0, 14, and 21), there are still no winners:
22 13 17 11 0 3 15 0 2 22 14 21 17 24 4
8 2 23 4 24 9 18 13 17 5 10 16 15 9 19
21 9 14 16 7 19 8 7 25 23 18 8 23 26 20
6 10 3 18 5 20 11 10 24 4 22 11 13 6 5
1 12 20 15 19 14 21 16 12 6 2 0 12 3 7
Finally, 24 is drawn:
22 13 17 11 0 3 15 0 2 22 14 21 17 24 4
8 2 23 4 24 9 18 13 17 5 10 16 15 9 19
21 9 14 16 7 19 8 7 25 23 18 8 23 26 20
6 10 3 18 5 20 11 10 24 4 22 11 13 6 5
1 12 20 15 19 14 21 16 12 6 2 0 12 3 7
At this point, the third board wins because it has at least one complete row or column of marked numbers (in this case, the entire top row is marked: 14 21 17 24 4).
The score of the winning board can now be calculated. Start by finding the sum of all unmarked numbers on that board; in this case, the sum is 188. Then, multiply that sum by the number that was just called when the board won, 24, to get the final score, 188 * 24 = 4512.
To guarantee victory against the giant squid, figure out which board will win first. What will your final score be if you choose that board?
Your puzzle answer was 65325.
--- Part Two ---
On the other hand, it might be wise to try a different strategy: let the giant squid win.
You aren't sure how many bingo boards a giant squid could play at once, so rather than waste time counting its arms, the safe thing to do is to figure out which board will win last and choose that one. That way, no matter which boards it picks, it will win for sure.
In the above example, the second board is the last to win, which happens after 13 is eventually called and its middle column is completely marked. If you were to keep playing until this point, the second board would have a sum of unmarked numbers equal to 148 for a final score of 148 * 13 = 1924.
Figure out which board will win last. Once it wins, what would its final score be?
Your puzzle answer was 4624.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day05/AoC05.py
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#!/usr/bin/env python3
import sys
import re
def parse(line):
groups = re.compile(r"(?P<x1>\d+),(?P<y1>\d+) -> (?P<x2>\d+),(?P<y2>\d+)").search(line)
return groups
def calc(x1, y1, x2, y2, mem):
if x1 == x2:
if y1 > y2:
y = y2
y2 = y1
y1 = y
for y in range(y1, y2+1):
if (x1, y) not in mem:
mem[(x1, y)] = 1
else:
mem[(x1, y)] += 1
elif y1 == y2:
if x1 > x2:
x = x2
x2 = x1
x1 = x
for x in range(x1, x2+1):
if (x, y1) not in mem:
mem[(x, y1)] = 1
else:
mem[(x, y1)] += 1
def calc2(x1, y1, x2, y2):
if x1 != x2 and y1 != y2:
if x1 > x2:
dif = x1-x2+1
m1 = -1
else:
dif = x2-x1+1
m1 = 1
if y1 > y2:
m2 = -1
else:
m2 = 1
for i in range(dif):
if (x1+m1*i, y1+m2*i) not in memo2:
memo2[(x1+m1*i, y1+m2*i)] = 1
else:
memo2[(x1+m1*i, y1+m2*i)] += 1
if __name__ == '__main__':
lines = open(sys.argv[1]).readlines()
coords = [tuple(map(int,parse(line).groups())) for line in lines]
# challenge 1
memo = {}
for c in coords:
calc(*c,memo)
res1 = 0
for v in memo.values():
if v > 1:
res1 += 1
print("challenge 1:" + "\n" + str(res1) + "\n")
# challenge 2
memo2 = {}
for c in coords:
calc(*c,memo2)
calc2(*c)
res2 = 0
for v in memo2.values():
if v > 1:
res2 += 1
print("challenge 2:" + "\n" + str(res2) + "\n")
2021/day05/challenge05
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--- Day 5: Hydrothermal Venture ---
You come across a field of hydrothermal vents on the ocean floor! These vents constantly produce large, opaque clouds, so it would be best to avoid them if possible.
They tend to form in lines; the submarine helpfully produces a list of nearby lines of vents (your puzzle input) for you to review. For example:
0,9 -> 5,9
8,0 -> 0,8
9,4 -> 3,4
2,2 -> 2,1
7,0 -> 7,4
6,4 -> 2,0
0,9 -> 2,9
3,4 -> 1,4
0,0 -> 8,8
5,5 -> 8,2
Each line of vents is given as a line segment in the format x1,y1 -> x2,y2 where x1,y1 are the coordinates of one end the line segment and x2,y2 are the coordinates of the other end. These line segments include the points at both ends. In other words:
An entry like 1,1 -> 1,3 covers points 1,1, 1,2, and 1,3.
An entry like 9,7 -> 7,7 covers points 9,7, 8,7, and 7,7.
For now, only consider horizontal and vertical lines: lines where either x1 = x2 or y1 = y2.
So, the horizontal and vertical lines from the above list would produce the following diagram:
.......1..
..1....1..
..1....1..
.......1..
.112111211
..........
..........
..........
..........
222111....
In this diagram, the top left corner is 0,0 and the bottom right corner is 9,9. Each position is shown as the number of lines which cover that point or . if no line covers that point. The top-left pair of 1s, for example, comes from 2,2 -> 2,1; the very bottom row is formed by the overlapping lines 0,9 -> 5,9 and 0,9 -> 2,9.
To avoid the most dangerous areas, you need to determine the number of points where at least two lines overlap. In the above example, this is anywhere in the diagram with a 2 or larger - a total of 5 points.
Consider only horizontal and vertical lines. At how many points do at least two lines overlap?
Your puzzle answer was 5373.
--- Part Two ---
Unfortunately, considering only horizontal and vertical lines doesn't give you the full picture; you need to also consider diagonal lines.
Because of the limits of the hydrothermal vent mapping system, the lines in your list will only ever be horizontal, vertical, or a diagonal line at exactly 45 degrees. In other words:
An entry like 1,1 -> 3,3 covers points 1,1, 2,2, and 3,3.
An entry like 9,7 -> 7,9 covers points 9,7, 8,8, and 7,9.
Considering all lines from the above example would now produce the following diagram:
1.1....11.
.111...2..
..2.1.111.
...1.2.2..
.112313211
...1.2....
..1...1...
.1.....1..
1.......1.
222111....
You still need to determine the number of points where at least two lines overlap. In the above example, this is still anywhere in the diagram with a 2 or larger - now a total of 12 points.
Consider all of the lines. At how many points do at least two lines overlap?
Your puzzle answer was 21514.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day06/AoC06.py
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#!/usr/bin/env python3
import sys
from copy import deepcopy
def sim(lf, days):
nf = {}
for i in range(days):
for k,v in lf.items():
if k == 7:
nf[6] = (nf[6] + v) if 6 in nf else v
elif k == 6:
nf[5] = v
elif k == 0:
nf[8] = v
nf[6] = (nf[6] + v) if 6 in nf else v
else:
nf[k-1] = v
lf = deepcopy(nf)
nf = {}
return sum(lf.values())
if __name__ == '__main__':
lines = open(sys.argv[1]).read().strip('\n\n').split('\n') # or split \n\n or sth similar
nums = [int(i) for i in lines[0].split(",")]
lf1 = {}
lf1[6] = 0
lf1[7] = 0
lf1[8] = 0
for n in nums:
if n not in lf1:
lf1[n] = 1
else:
lf1[n] += 1
lf2 = deepcopy(lf1)
# challenge 1
res1 = str(sim(lf1, 80))
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
res2 = str(sim(lf2, 256))
print("challenge 2:" +"\n" + res2 + "\n")
2021/day06/challenge06
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--- Day 6: Lanternfish ---
The sea floor is getting steeper. Maybe the sleigh keys got carried this way?
A massive school of glowing lanternfish swims past. They must spawn quickly to reach such large numbers - maybe exponentially quickly? You should model their growth rate to be sure.
Although you know nothing about this specific species of lanternfish, you make some guesses about their attributes. Surely, each lanternfish creates a new lanternfish once every 7 days.
However, this process isn't necessarily synchronized between every lanternfish - one lanternfish might have 2 days left until it creates another lanternfish, while another might have 4. So, you can model each fish as a single number that represents the number of days until it creates a new lanternfish.
Furthermore, you reason, a new lanternfish would surely need slightly longer before it's capable of producing more lanternfish: two more days for its first cycle.
So, suppose you have a lanternfish with an internal timer value of 3:
After one day, its internal timer would become 2.
After another day, its internal timer would become 1.
After another day, its internal timer would become 0.
After another day, its internal timer would reset to 6, and it would create a new lanternfish with an internal timer of 8.
After another day, the first lanternfish would have an internal timer of 5, and the second lanternfish would have an internal timer of 7.
A lanternfish that creates a new fish resets its timer to 6, not 7 (because 0 is included as a valid timer value). The new lanternfish starts with an internal timer of 8 and does not start counting down until the next day.
Realizing what you're trying to do, the submarine automatically produces a list of the ages of several hundred nearby lanternfish (your puzzle input). For example, suppose you were given the following list:
3,4,3,1,2
This list means that the first fish has an internal timer of 3, the second fish has an internal timer of 4, and so on until the fifth fish, which has an internal timer of 2. Simulating these fish over several days would proceed as follows:
Initial state: 3,4,3,1,2
After 1 day: 2,3,2,0,1
After 2 days: 1,2,1,6,0,8
After 3 days: 0,1,0,5,6,7,8
After 4 days: 6,0,6,4,5,6,7,8,8
After 5 days: 5,6,5,3,4,5,6,7,7,8
After 6 days: 4,5,4,2,3,4,5,6,6,7
After 7 days: 3,4,3,1,2,3,4,5,5,6
After 8 days: 2,3,2,0,1,2,3,4,4,5
After 9 days: 1,2,1,6,0,1,2,3,3,4,8
After 10 days: 0,1,0,5,6,0,1,2,2,3,7,8
After 11 days: 6,0,6,4,5,6,0,1,1,2,6,7,8,8,8
After 12 days: 5,6,5,3,4,5,6,0,0,1,5,6,7,7,7,8,8
After 13 days: 4,5,4,2,3,4,5,6,6,0,4,5,6,6,6,7,7,8,8
After 14 days: 3,4,3,1,2,3,4,5,5,6,3,4,5,5,5,6,6,7,7,8
After 15 days: 2,3,2,0,1,2,3,4,4,5,2,3,4,4,4,5,5,6,6,7
After 16 days: 1,2,1,6,0,1,2,3,3,4,1,2,3,3,3,4,4,5,5,6,8
After 17 days: 0,1,0,5,6,0,1,2,2,3,0,1,2,2,2,3,3,4,4,5,7,8
After 18 days: 6,0,6,4,5,6,0,1,1,2,6,0,1,1,1,2,2,3,3,4,6,7,8,8,8,8
Each day, a 0 becomes a 6 and adds a new 8 to the end of the list, while each other number decreases by 1 if it was present at the start of the day.
In this example, after 18 days, there are a total of 26 fish. After 80 days, there would be a total of 5934.
Find a way to simulate lanternfish. How many lanternfish would there be after 80 days?
Your puzzle answer was 362666.
--- Part Two ---
Suppose the lanternfish live forever and have unlimited food and space. Would they take over the entire ocean?
After 256 days in the example above, there would be a total of 26984457539 lanternfish!
How many lanternfish would there be after 256 days?
Your puzzle answer was 1640526601595.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day07/AoC07.py
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#!/usr/bin/env python3
import sys
def fuel(crabs, x):
return sum(abs(c-x) for c in crabs)
def fuel_incr(crabs, x):
return sum(sc[abs(c-x)] for c in crabs)
if __name__ == '__main__':
lines = open(sys.argv[1]).read().strip('\n\n').split('\n') # or split \n\n or sth similar
nums = [int(i) for i in lines[0].split(",")]
# challenge 1
res1 = str(min(fuel(nums, i) for i in range(max(nums))))
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
sc = [0] * (max(nums)+1)
for i in range(max(nums)+1):
sc[i] = sc[i-1] + i
res2 = str(min(fuel_incr(nums, i) for i in range(max(nums))))
print("challenge 2:" +"\n" + res2 + "\n")
2021/day07/challenge07
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--- Day 7: The Treachery of Whales ---
A giant whale has decided your submarine is its next meal, and it's much faster than you are. There's nowhere to run!
Suddenly, a swarm of crabs (each in its own tiny submarine - it's too deep for them otherwise) zooms in to rescue you! They seem to be preparing to blast a hole in the ocean floor; sensors indicate a massive underground cave system just beyond where they're aiming!
The crab submarines all need to be aligned before they'll have enough power to blast a large enough hole for your submarine to get through. However, it doesn't look like they'll be aligned before the whale catches you! Maybe you can help?
There's one major catch - crab submarines can only move horizontally.
You quickly make a list of the horizontal position of each crab (your puzzle input). Crab submarines have limited fuel, so you need to find a way to make all of their horizontal positions match while requiring them to spend as little fuel as possible.
For example, consider the following horizontal positions:
16,1,2,0,4,2,7,1,2,14
This means there's a crab with horizontal position 16, a crab with horizontal position 1, and so on.
Each change of 1 step in horizontal position of a single crab costs 1 fuel. You could choose any horizontal position to align them all on, but the one that costs the least fuel is horizontal position 2:
Move from 16 to 2: 14 fuel
Move from 1 to 2: 1 fuel
Move from 2 to 2: 0 fuel
Move from 0 to 2: 2 fuel
Move from 4 to 2: 2 fuel
Move from 2 to 2: 0 fuel
Move from 7 to 2: 5 fuel
Move from 1 to 2: 1 fuel
Move from 2 to 2: 0 fuel
Move from 14 to 2: 12 fuel
This costs a total of 37 fuel. This is the cheapest possible outcome; more expensive outcomes include aligning at position 1 (41 fuel), position 3 (39 fuel), or position 10 (71 fuel).
Determine the horizontal position that the crabs can align to using the least fuel possible. How much fuel must they spend to align to that position?
Your puzzle answer was 357353.
--- Part Two ---
The crabs don't seem interested in your proposed solution. Perhaps you misunderstand crab engineering?
As it turns out, crab submarine engines don't burn fuel at a constant rate. Instead, each change of 1 step in horizontal position costs 1 more unit of fuel than the last: the first step costs 1, the second step costs 2, the third step costs 3, and so on.
As each crab moves, moving further becomes more expensive. This changes the best horizontal position to align them all on; in the example above, this becomes 5:
Move from 16 to 5: 66 fuel
Move from 1 to 5: 10 fuel
Move from 2 to 5: 6 fuel
Move from 0 to 5: 15 fuel
Move from 4 to 5: 1 fuel
Move from 2 to 5: 6 fuel
Move from 7 to 5: 3 fuel
Move from 1 to 5: 10 fuel
Move from 2 to 5: 6 fuel
Move from 14 to 5: 45 fuel
This costs a total of 168 fuel. This is the new cheapest possible outcome; the old alignment position (2) now costs 206 fuel instead.
Determine the horizontal position that the crabs can align to using the least fuel possible so they can make you an escape route! How much fuel must they spend to align to that position?
Your puzzle answer was 104822130.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day08/AoC08.py
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#!/usr/bin/env python3
import sys
def search_num(d, val):
for k,v in d.items():
if val == v:
return str(k)
def decode(inp, out):
nums = {}
for i in sorted(inp, key=len):
if len(i) == 2: nums[1] = set(i)
elif len(i) == 3: nums[7] = set(i)
elif len(i) == 4: nums[4] = set(i)
elif len(i) == 5: #2, 3, 5
if nums[1] <= set(i): nums[3] = set(i)
elif len(i) == 6: #0, 6, 9
if nums[4] <= set(i): nums[9] = set(i)
elif nums[1] <= set(i): nums[0] = set(i)
else: nums[6] = set(i)
elif len(i) == 7: nums[8] = set(i)
for i in inp:
if set(i) not in nums.values():
if set(i) <= nums[6]: nums[5] = set(i)
else: nums[2] = set(i)
return int(''.join([search_num(nums, set(o)) for o in out]))
if __name__ == '__main__':
lines = open(sys.argv[1]).read().strip('\n\n').split('\n') # or split \n\n or sth similar
inp = [i.split(' | ') for i in lines]
inout = [(i[0].split(' '), i[1].split(' ')) for i in inp]
# challenge 1
sums = 0
for _,out in inout:
for o in out:
if len(o) == 2 or len(o) == 3 or len(o) == 4 or len(o) == 7:
sums += 1
res1 = str(sums)
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
res2 = str(sum([decode(i,o) for i,o in inout]))
print("challenge 2:" +"\n" + res2 + "\n")
2021/day08/challenge08
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--- Day 8: Seven Segment Search ---
You barely reach the safety of the cave when the whale smashes into the cave mouth, collapsing it. Sensors indicate another exit to this cave at a much greater depth, so you have no choice but to press on.
As your submarine slowly makes its way through the cave system, you notice that the four-digit seven-segment displays in your submarine are malfunctioning; they must have been damaged during the escape. You'll be in a lot of trouble without them, so you'd better figure out what's wrong.
Each digit of a seven-segment display is rendered by turning on or off any of seven segments named a through g:
0: 1: 2: 3: 4:
aaaa .... aaaa aaaa ....
b c . c . c . c b c
b c . c . c . c b c
.... .... dddd dddd dddd
e f . f e . . f . f
e f . f e . . f . f
gggg .... gggg gggg ....
5: 6: 7: 8: 9:
aaaa aaaa aaaa aaaa aaaa
b . b . . c b c b c
b . b . . c b c b c
dddd dddd .... dddd dddd
. f e f . f e f . f
. f e f . f e f . f
gggg gggg .... gggg gggg
So, to render a 1, only segments c and f would be turned on; the rest would be off. To render a 7, only segments a, c, and f would be turned on.
The problem is that the signals which control the segments have been mixed up on each display. The submarine is still trying to display numbers by producing output on signal wires a through g, but those wires are connected to segments randomly. Worse, the wire/segment connections are mixed up separately for each four-digit display! (All of the digits within a display use the same connections, though.)
So, you might know that only signal wires b and g are turned on, but that doesn't mean segments b and g are turned on: the only digit that uses two segments is 1, so it must mean segments c and f are meant to be on. With just that information, you still can't tell which wire (b/g) goes to which segment (c/f). For that, you'll need to collect more information.
For each display, you watch the changing signals for a while, make a note of all ten unique signal patterns you see, and then write down a single four digit output value (your puzzle input). Using the signal patterns, you should be able to work out which pattern corresponds to which digit.
For example, here is what you might see in a single entry in your notes:
acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab |
cdfeb fcadb cdfeb cdbaf
(The entry is wrapped here to two lines so it fits; in your notes, it will all be on a single line.)
Each entry consists of ten unique signal patterns, a | delimiter, and finally the four digit output value. Within an entry, the same wire/segment connections are used (but you don't know what the connections actually are). The unique signal patterns correspond to the ten different ways the submarine tries to render a digit using the current wire/segment connections. Because 7 is the only digit that uses three segments, dab in the above example means that to render a 7, signal lines d, a, and b are on. Because 4 is the only digit that uses four segments, eafb means that to render a 4, signal lines e, a, f, and b are on.
Using this information, you should be able to work out which combination of signal wires corresponds to each of the ten digits. Then, you can decode the four digit output value. Unfortunately, in the above example, all of the digits in the output value (cdfeb fcadb cdfeb cdbaf) use five segments and are more difficult to deduce.
For now, focus on the easy digits. Consider this larger example:
be cfbegad cbdgef fgaecd cgeb fdcge agebfd fecdb fabcd edb |
fdgacbe cefdb cefbgd gcbe
edbfga begcd cbg gc gcadebf fbgde acbgfd abcde gfcbed gfec |
fcgedb cgb dgebacf gc
fgaebd cg bdaec gdafb agbcfd gdcbef bgcad gfac gcb cdgabef |
cg cg fdcagb cbg
fbegcd cbd adcefb dageb afcb bc aefdc ecdab fgdeca fcdbega |
efabcd cedba gadfec cb
aecbfdg fbg gf bafeg dbefa fcge gcbea fcaegb dgceab fcbdga |
gecf egdcabf bgf bfgea
fgeab ca afcebg bdacfeg cfaedg gcfdb baec bfadeg bafgc acf |
gebdcfa ecba ca fadegcb
dbcfg fgd bdegcaf fgec aegbdf ecdfab fbedc dacgb gdcebf gf |
cefg dcbef fcge gbcadfe
bdfegc cbegaf gecbf dfcage bdacg ed bedf ced adcbefg gebcd |
ed bcgafe cdgba cbgef
egadfb cdbfeg cegd fecab cgb gbdefca cg fgcdab egfdb bfceg |
gbdfcae bgc cg cgb
gcafb gcf dcaebfg ecagb gf abcdeg gaef cafbge fdbac fegbdc |
fgae cfgab fg bagce
Because the digits 1, 4, 7, and 8 each use a unique number of segments, you should be able to tell which combinations of signals correspond to those digits. Counting only digits in the output values (the part after | on each line), in the above example, there are 26 instances of digits that use a unique number of segments (highlighted above).
In the output values, how many times do digits 1, 4, 7, or 8 appear?
Your puzzle answer was 504.
--- Part Two ---
Through a little deduction, you should now be able to determine the remaining digits. Consider again the first example above:
acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab |
cdfeb fcadb cdfeb cdbaf
After some careful analysis, the mapping between signal wires and segments only make sense in the following configuration:
dddd
e a
e a
ffff
g b
g b
cccc
So, the unique signal patterns would correspond to the following digits:
acedgfb: 8
cdfbe: 5
gcdfa: 2
fbcad: 3
dab: 7
cefabd: 9
cdfgeb: 6
eafb: 4
cagedb: 0
ab: 1
Then, the four digits of the output value can be decoded:
cdfeb: 5
fcadb: 3
cdfeb: 5
cdbaf: 3
Therefore, the output value for this entry is 5353.
Following this same process for each entry in the second, larger example above, the output value of each entry can be determined:
fdgacbe cefdb cefbgd gcbe: 8394
fcgedb cgb dgebacf gc: 9781
cg cg fdcagb cbg: 1197
efabcd cedba gadfec cb: 9361
gecf egdcabf bgf bfgea: 4873
gebdcfa ecba ca fadegcb: 8418
cefg dcbef fcge gbcadfe: 4548
ed bcgafe cdgba cbgef: 1625
gbdfcae bgc cg cgb: 8717
fgae cfgab fg bagce: 4315
Adding all of the output values in this larger example produces 61229.
For each entry, determine all of the wire/segment connections and decode the four-digit output values. What do you get if you add up all of the output values?
Your puzzle answer was 1073431.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day09/AoC09.py
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#!/usr/bin/env python3
import sys
def risk_level(m, x, y):
adj = []
if x-1 > -1:
adj.append(m[x-1][y])
try: adj.append(m[x+1][y])
except IndexError: pass
if y-1 > -1:
adj.append(m[x][y-1])
try: adj.append(m[x][y+1])
except IndexError: pass
for a in adj:
if a <= m[x][y]:
return 0
return m[x][y] + 1
def basin(m, x, y):
while (x,y) not in bas:
adj = {}
adj[(0,0)] = m[x][y]
if x-1 > -1:
adj[(-1,0)] = m[x-1][y]
try: adj[(1,0)] = m[x+1][y]
except IndexError: pass
if y-1 > -1:
adj[(0,-1)] = m[x][y-1]
try: adj[(0,1)] = m[x][y+1]
except IndexError: pass
mi = min(adj.items(),key=lambda x: x[1])
x += mi[0][0]
y += mi[0][1]
bas[(x,y)] += 1
if __name__ == '__main__':
lines = open(sys.argv[1]).read().strip('\n\n').split('\n') # or split \n\n or sth similar
mp = [[int(i) for i in l] for l in lines]
bas = {}
risk = 0
for x in range(len(mp)):
for y in range(len(mp[0])):
r = risk_level(mp, x, y)
risk += r
if r > 0:
bas[(x,y)] = 0
for x in range(len(mp)):
for y in range(len(mp[0])):
if mp[x][y] != 9:
basin(mp, x, y)
b = sorted(bas.values(), reverse=True)
# challenge 1
res1 = str(risk)
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
res2 = str(b[0]*b[1]*b[2])
print("challenge 2:" +"\n" + res2 + "\n")
2021/day09/AoC09B.py
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#!/usr/bin/env python3
import sys
from operator import mul
from functools import reduce
def adj(mp, x, y):
adj = {}
if x-1 > -1: adj[(-1,0)] = mp[x-1][y]
if x+1 < len(mp): adj[(1,0)] = mp[x+1][y]
if y-1 > -1: adj[(0,-1)] = mp[x][y-1]
if y+1 < len(mp[0]): adj[(0,1)] = mp[x][y+1]
return adj
def risk(mp, x, y):
ad = adj(mp, x, y).values()
if all(map(lambda i: i > mp[x][y], ad)):
bas[(x,y)] = 0
return mp[x][y] + 1
else: return 0
def basin(mp, x, y):
nex = min(adj(mp, x, y).items(), key=lambda i: i[1])
if (x,y) in bas: (i,j) = (x,y)
elif (x,y) in mem: (i,j) = mem[(x,y)]
else:
(i,j) = basin(mp, x+nex[0][0], y+nex[0][1])
return (i,j)
bas[(i,j)] += 1
return (i,j)
if __name__ == '__main__':
m = [[int(i) for i in l] for l in open(sys.argv[1]).read().strip('\n\n').split('\n')]
bas,mem = {},{}
# challenge 1
res1 = str(sum([sum([risk(m, x, y) for y in range(len(m[0]))]) for x in range(len(m))]))
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
([([basin(m, x, y) for y in range(len(m[0])) if m[x][y] < 9]) for x in range(len(m))])
res2 = str(reduce(mul, sorted(bas.values(), reverse=True)[0:3]))
print("challenge 2:" +"\n" + res2 + "\n")
2021/day09/challenge09
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--- Day 9: Smoke Basin ---
These caves seem to be lava tubes. Parts are even still volcanically active; small hydrothermal vents release smoke into the caves that slowly settles like rain.
If you can model how the smoke flows through the caves, you might be able to avoid it and be that much safer. The submarine generates a heightmap of the floor of the nearby caves for you (your puzzle input).
Smoke flows to the lowest point of the area it's in. For example, consider the following heightmap:
2199943210
3987894921
9856789892
8767896789
9899965678
Each number corresponds to the height of a particular location, where 9 is the highest and 0 is the lowest a location can be.
Your first goal is to find the low points - the locations that are lower than any of its adjacent locations. Most locations have four adjacent locations (up, down, left, and right); locations on the edge or corner of the map have three or two adjacent locations, respectively. (Diagonal locations do not count as adjacent.)
In the above example, there are four low points, all highlighted: two are in the first row (a 1 and a 0), one is in the third row (a 5), and one is in the bottom row (also a 5). All other locations on the heightmap have some lower adjacent location, and so are not low points.
The risk level of a low point is 1 plus its height. In the above example, the risk levels of the low points are 2, 1, 6, and 6. The sum of the risk levels of all low points in the heightmap is therefore 15.
Find all of the low points on your heightmap. What is the sum of the risk levels of all low points on your heightmap?
Your puzzle answer was 506.
--- Part Two ---
Next, you need to find the largest basins so you know what areas are most important to avoid.
A basin is all locations that eventually flow downward to a single low point. Therefore, every low point has a basin, although some basins are very small. Locations of height 9 do not count as being in any basin, and all other locations will always be part of exactly one basin.
The size of a basin is the number of locations within the basin, including the low point. The example above has four basins.
The top-left basin, size 3:
2199943210
3987894921
9856789892
8767896789
9899965678
The top-right basin, size 9:
2199943210
3987894921
9856789892
8767896789
9899965678
The middle basin, size 14:
2199943210
3987894921
9856789892
8767896789
9899965678
The bottom-right basin, size 9:
2199943210
3987894921
9856789892
8767896789
9899965678
Find the three largest basins and multiply their sizes together. In the above example, this is 9 * 14 * 9 = 1134.
What do you get if you multiply together the sizes of the three largest basins?
Your puzzle answer was 931200.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day10/AoC10.py
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#!/usr/bin/env python3
import sys
import statistics
def corr(line):
st = []
for b in line:
if b in match.keys(): st.append(b)
if b in error.keys():
if not (st.pop(), b) in match.items(): return error[b]
return 0
def inc(line):
st, score = [], 0
for b in line:
if b in match.keys(): st.append(b)
if b in comp.keys(): st.pop()
miss = [match[b] for b in st]
for b in reversed(miss):
score *= 5
score += comp[b]
return score
if __name__ == '__main__':
inp = [list(i) for i in open(sys.argv[1]).read().strip('\n\n').split('\n')]
match = {'(': ')', '{': '}', '[': ']', '<': '>'}
# challenge 1
error = {')': 3, ']': 57, '}': 1197, '>': 25137}
res1 = str(sum([corr(line) for line in inp]))
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
comp = {')': 1, ']': 2, '}': 3, '>': 4}
res2 = str(statistics.median([inc(line) for line in inp if corr(line) == 0]))
print("challenge 2:" +"\n" + res2 + "\n")
2021/day10/challenge10
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--- Day 10: Syntax Scoring ---
You ask the submarine to determine the best route out of the deep-sea cave, but it only replies:
Syntax error in navigation subsystem on line: all of them
All of them?! The damage is worse than you thought. You bring up a copy of the navigation subsystem (your puzzle input).
The navigation subsystem syntax is made of several lines containing chunks. There are one or more chunks on each line, and chunks contain zero or more other chunks. Adjacent chunks are not separated by any delimiter; if one chunk stops, the next chunk (if any) can immediately start. Every chunk must open and close with one of four legal pairs of matching characters:
If a chunk opens with (, it must close with ).
If a chunk opens with [, it must close with ].
If a chunk opens with {, it must close with }.
If a chunk opens with <, it must close with >.
So, () is a legal chunk that contains no other chunks, as is []. More complex but valid chunks include ([]), {()()()}, <([{}])>, [<>({}){}[([])<>]], and even (((((((((()))))))))).
Some lines are incomplete, but others are corrupted. Find and discard the corrupted lines first.
A corrupted line is one where a chunk closes with the wrong character - that is, where the characters it opens and closes with do not form one of the four legal pairs listed above.
Examples of corrupted chunks include (], {()()()>, (((()))}, and <([]){()}[{}]). Such a chunk can appear anywhere within a line, and its presence causes the whole line to be considered corrupted.
For example, consider the following navigation subsystem:
[({(<(())[]>[[{[]{<()<>>
[(()[<>])]({[<{<<[]>>(
{([(<{}[<>[]}>{[]{[(<()>
(((({<>}<{<{<>}{[]{[]{}
[[<[([]))<([[{}[[()]]]
[{[{({}]{}}([{[{{{}}([]
{<[[]]>}<{[{[{[]{()[[[]
[<(<(<(<{}))><([]([]()
<{([([[(<>()){}]>(<<{{
<{([{{}}[<[[[<>{}]]]>[]]
Some of the lines aren't corrupted, just incomplete; you can ignore these lines for now. The remaining five lines are corrupted:
{([(<{}[<>[]}>{[]{[(<()> - Expected ], but found } instead.
[[<[([]))<([[{}[[()]]] - Expected ], but found ) instead.
[{[{({}]{}}([{[{{{}}([] - Expected ), but found ] instead.
[<(<(<(<{}))><([]([]() - Expected >, but found ) instead.
<{([([[(<>()){}]>(<<{{ - Expected ], but found > instead.
Stop at the first incorrect closing character on each corrupted line.
Did you know that syntax checkers actually have contests to see who can get the high score for syntax errors in a file? It's true! To calculate the syntax error score for a line, take the first illegal character on the line and look it up in the following table:
): 3 points.
]: 57 points.
}: 1197 points.
>: 25137 points.
In the above example, an illegal ) was found twice (2*3 = 6 points), an illegal ] was found once (57 points), an illegal } was found once (1197 points), and an illegal > was found once (25137 points). So, the total syntax error score for this file is 6+57+1197+25137 = 26397 points!
Find the first illegal character in each corrupted line of the navigation subsystem. What is the total syntax error score for those errors?
Your puzzle answer was 390993.
--- Part Two ---
Now, discard the corrupted lines. The remaining lines are incomplete.
Incomplete lines don't have any incorrect characters - instead, they're missing some closing characters at the end of the line. To repair the navigation subsystem, you just need to figure out the sequence of closing characters that complete all open chunks in the line.
You can only use closing characters (), ], }, or >), and you must add them in the correct order so that only legal pairs are formed and all chunks end up closed.
In the example above, there are five incomplete lines:
[({(<(())[]>[[{[]{<()<>> - Complete by adding }}]])})].
[(()[<>])]({[<{<<[]>>( - Complete by adding )}>]}).
(((({<>}<{<{<>}{[]{[]{} - Complete by adding }}>}>)))).
{<[[]]>}<{[{[{[]{()[[[] - Complete by adding ]]}}]}]}>.
<{([{{}}[<[[[<>{}]]]>[]] - Complete by adding ])}>.
Did you know that autocomplete tools also have contests? It's true! The score is determined by considering the completion string character-by-character. Start with a total score of 0. Then, for each character, multiply the total score by 5 and then increase the total score by the point value given for the character in the following table:
): 1 point.
]: 2 points.
}: 3 points.
>: 4 points.
So, the last completion string above - ])}> - would be scored as follows:
Start with a total score of 0.
Multiply the total score by 5 to get 0, then add the value of ] (2) to get a new total score of 2.
Multiply the total score by 5 to get 10, then add the value of ) (1) to get a new total score of 11.
Multiply the total score by 5 to get 55, then add the value of } (3) to get a new total score of 58.
Multiply the total score by 5 to get 290, then add the value of > (4) to get a new total score of 294.
The five lines' completion strings have total scores as follows:
}}]])})] - 288957 total points.
)}>]}) - 5566 total points.
}}>}>)))) - 1480781 total points.
]]}}]}]}> - 995444 total points.
])}> - 294 total points.
Autocomplete tools are an odd bunch: the winner is found by sorting all of the scores and then taking the middle score. (There will always be an odd number of scores to consider.) In this example, the middle score is 288957 because there are the same number of scores smaller and larger than it.
Find the completion string for each incomplete line, score the completion strings, and sort the scores. What is the middle score?
Your puzzle answer was 2391385187.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day11/AoC11.py
-51
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@@ -1,51 +0,0 @@
#!/usr/bin/env python3
import sys
def flash(octop, x, y):
for i in [(-1,-1), (-1,0), (-1,1), (0,-1), (0,1), (1,-1), (1,0), (1,1)]:
if -1 < y+i[0] < len(octop) and -1 < x+i[1] < len(octop):
if octop[y+i[0]][x+i[1]] > 0: octop[y+i[0]][x+i[1]] += 1
octop[y][x] = 0
return octop
def flashable(octo):
for y in octo:
for x in y:
if x > 9:
return True
return False
def step(octo):
nex = [[x+1 for x in y] for y in octo]
flashs = 0
while flashable(nex):
for x in range(len(octo[0])):
for y in range(len(octo)):
if nex[y][x] > 9:
nex = flash(nex,x,y)
flashs += 1
return nex, flashs
if __name__ == '__main__':
lines = open(sys.argv[1]).read().strip('\n\n').split('\n') # or split \n\n or sth similar
octos = [[int(i) for i in l] for l in lines]
# challenge 1
flashes, i = 0, 0
while True:
i += 1
octos, f = step(octos)
if(i <= 100):
flashes += f
if sum([sum([1 for x in range(len(octos[0])) if octos[y][x] == 0]) for y in range(len(octos))]) == 100:
first = i
break
res1 = str(flashes)
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
res2 = str(first)
print("challenge 2:" +"\n" + res2 + "\n")
2021/day11/challenge11
-348
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@@ -1,348 +0,0 @@
--- Day 11: Dumbo Octopus ---
You enter a large cavern full of rare bioluminescent dumbo octopuses! They seem to not like the Christmas lights on your submarine, so you turn them off for now.
There are 100 octopuses arranged neatly in a 10 by 10 grid. Each octopus slowly gains energy over time and flashes brightly for a moment when its energy is full. Although your lights are off, maybe you could navigate through the cave without disturbing the octopuses if you could predict when the flashes of light will happen.
Each octopus has an energy level - your submarine can remotely measure the energy level of each octopus (your puzzle input). For example:
5483143223
2745854711
5264556173
6141336146
6357385478
4167524645
2176841721
6882881134
4846848554
5283751526
The energy level of each octopus is a value between 0 and 9. Here, the top-left octopus has an energy level of 5, the bottom-right one has an energy level of 6, and so on.
You can model the energy levels and flashes of light in steps. During a single step, the following occurs:
First, the energy level of each octopus increases by 1.
Then, any octopus with an energy level greater than 9 flashes. This increases the energy level of all adjacent octopuses by 1, including octopuses that are diagonally adjacent. If this causes an octopus to have an energy level greater than 9, it also flashes. This process continues as long as new octopuses keep having their energy level increased beyond 9. (An octopus can only flash at most once per step.)
Finally, any octopus that flashed during this step has its energy level set to 0, as it used all of its energy to flash.
Adjacent flashes can cause an octopus to flash on a step even if it begins that step with very little energy. Consider the middle octopus with 1 energy in this situation:
Before any steps:
11111
19991
19191
19991
11111
After step 1:
34543
40004
50005
40004
34543
After step 2:
45654
51115
61116
51115
45654
An octopus is highlighted when it flashed during the given step.
Here is how the larger example above progresses:
Before any steps:
5483143223
2745854711
5264556173
6141336146
6357385478
4167524645
2176841721
6882881134
4846848554
5283751526
After step 1:
6594254334
3856965822
6375667284
7252447257
7468496589
5278635756
3287952832
7993992245
5957959665
6394862637
After step 2:
8807476555
5089087054
8597889608
8485769600
8700908800
6600088989
6800005943
0000007456
9000000876
8700006848
After step 3:
0050900866
8500800575
9900000039
9700000041
9935080063
7712300000
7911250009
2211130000
0421125000
0021119000
After step 4:
2263031977
0923031697
0032221150
0041111163
0076191174
0053411122
0042361120
5532241122
1532247211
1132230211
After step 5:
4484144000
2044144000
2253333493
1152333274
1187303285
1164633233
1153472231
6643352233
2643358322
2243341322
After step 6:
5595255111
3155255222
3364444605
2263444496
2298414396
2275744344
2264583342
7754463344
3754469433
3354452433
After step 7:
6707366222
4377366333
4475555827
3496655709
3500625609
3509955566
3486694453
8865585555
4865580644
4465574644
After step 8:
7818477333
5488477444
5697666949
4608766830
4734946730
4740097688
6900007564
0000009666
8000004755
6800007755
After step 9:
9060000644
7800000976
6900000080
5840000082
5858000093
6962400000
8021250009
2221130009
9111128097
7911119976
After step 10:
0481112976
0031112009
0041112504
0081111406
0099111306
0093511233
0442361130
5532252350
0532250600
0032240000
After step 10, there have been a total of 204 flashes. Fast forwarding, here is the same configuration every 10 steps:
After step 20:
3936556452
5686556806
4496555690
4448655580
4456865570
5680086577
7000009896
0000000344
6000000364
4600009543
After step 30:
0643334118
4253334611
3374333458
2225333337
2229333338
2276733333
2754574565
5544458511
9444447111
7944446119
After step 40:
6211111981
0421111119
0042111115
0003111115
0003111116
0065611111
0532351111
3322234597
2222222976
2222222762
After step 50:
9655556447
4865556805
4486555690
4458655580
4574865570
5700086566
6000009887
8000000533
6800000633
5680000538
After step 60:
2533334200
2743334640
2264333458
2225333337
2225333338
2287833333
3854573455
1854458611
1175447111
1115446111
After step 70:
8211111164
0421111166
0042111114
0004211115
0000211116
0065611111
0532351111
7322235117
5722223475
4572222754
After step 80:
1755555697
5965555609
4486555680
4458655580
4570865570
5700086566
7000008666
0000000990
0000000800
0000000000
After step 90:
7433333522
2643333522
2264333458
2226433337
2222433338
2287833333
2854573333
4854458333
3387779333
3333333333
After step 100:
0397666866
0749766918
0053976933
0004297822
0004229892
0053222877
0532222966
9322228966
7922286866
6789998766
After 100 steps, there have been a total of 1656 flashes.
Given the starting energy levels of the dumbo octopuses in your cavern, simulate 100 steps. How many total flashes are there after 100 steps?
Your puzzle answer was 1705.
--- Part Two ---
It seems like the individual flashes aren't bright enough to navigate. However, you might have a better option: the flashes seem to be synchronizing!
In the example above, the first time all octopuses flash simultaneously is step 195:
After step 193:
5877777777
8877777777
7777777777
7777777777
7777777777
7777777777
7777777777
7777777777
7777777777
7777777777
After step 194:
6988888888
9988888888
8888888888
8888888888
8888888888
8888888888
8888888888
8888888888
8888888888
8888888888
After step 195:
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
If you can calculate the exact moments when the octopuses will all flash simultaneously, you should be able to navigate through the cavern. What is the first step during which all octopuses flash?
Your puzzle answer was 265.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day12/AoC12.py
-48
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@@ -1,48 +0,0 @@
#!/usr/bin/env python3
import sys
import re
from copy import deepcopy
def parse(l):
groups = re.compile(r"(?P<p1>\w+)-(?P<p2>\w+)").search(l)
p1, p2 = groups.groups()
for p in (p1,p2):
if not p in opts.keys(): opts[p] = set()
if not p in opts.keys(): opts[p] = set()
if p.islower() and p not in ('start', 'end'): opts['small'].add(p)
if p.islower() and p not in ('start', 'end'): opts['small'].add(p)
opts[p1].add(p2)
opts[p2].add(p1)
def paths(p1, l, twice=''):
p = deepcopy(p1)
p.append(l)
if p.count(twice) > 1: twice = ''
if l == 'end':
pths.add(tuple(p))
return
for n in opts[l]:
if not ((n != twice and n in p and n in opts['small']) or n == 'start'):
paths(p, n, twice)
if __name__ == '__main__':
lines = open(sys.argv[1]).read().strip('\n\n').split('\n') # or split \n\n or sth similar
opts = {'start': set(), 'end': set(), 'small': set()}
[parse(i) for i in lines]
# challenge 1
pths = set()
paths([], 'start')
res1 = str(len(pths))
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
pths = set()
[paths([], 'start', t) for t in opts['small']]
res2 = str(len(pths))
print("challenge 2:" +"\n" + res2 + "\n")
2021/day12/challenge12
-152
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@@ -1,152 +0,0 @@
--- Day 12: Passage Pathing ---
With your submarine's subterranean subsystems subsisting suboptimally, the only way you're getting out of this cave anytime soon is by finding a path yourself. Not just a path - the only way to know if you've found the best path is to find all of them.
Fortunately, the sensors are still mostly working, and so you build a rough map of the remaining caves (your puzzle input). For example:
start-A
start-b
A-c
A-b
b-d
A-end
b-end
This is a list of how all of the caves are connected. You start in the cave named start, and your destination is the cave named end. An entry like b-d means that cave b is connected to cave d - that is, you can move between them.
So, the above cave system looks roughly like this:
start
/ \
c--A-----b--d
\ /
end
Your goal is to find the number of distinct paths that start at start, end at end, and don't visit small caves more than once. There are two types of caves: big caves (written in uppercase, like A) and small caves (written in lowercase, like b). It would be a waste of time to visit any small cave more than once, but big caves are large enough that it might be worth visiting them multiple times. So, all paths you find should visit small caves at most once, and can visit big caves any number of times.
Given these rules, there are 10 paths through this example cave system:
start,A,b,A,c,A,end
start,A,b,A,end
start,A,b,end
start,A,c,A,b,A,end
start,A,c,A,b,end
start,A,c,A,end
start,A,end
start,b,A,c,A,end
start,b,A,end
start,b,end
(Each line in the above list corresponds to a single path; the caves visited by that path are listed in the order they are visited and separated by commas.)
Note that in this cave system, cave d is never visited by any path: to do so, cave b would need to be visited twice (once on the way to cave d and a second time when returning from cave d), and since cave b is small, this is not allowed.
Here is a slightly larger example:
dc-end
HN-start
start-kj
dc-start
dc-HN
LN-dc
HN-end
kj-sa
kj-HN
kj-dc
The 19 paths through it are as follows:
start,HN,dc,HN,end
start,HN,dc,HN,kj,HN,end
start,HN,dc,end
start,HN,dc,kj,HN,end
start,HN,end
start,HN,kj,HN,dc,HN,end
start,HN,kj,HN,dc,end
start,HN,kj,HN,end
start,HN,kj,dc,HN,end
start,HN,kj,dc,end
start,dc,HN,end
start,dc,HN,kj,HN,end
start,dc,end
start,dc,kj,HN,end
start,kj,HN,dc,HN,end
start,kj,HN,dc,end
start,kj,HN,end
start,kj,dc,HN,end
start,kj,dc,end
Finally, this even larger example has 226 paths through it:
fs-end
he-DX
fs-he
start-DX
pj-DX
end-zg
zg-sl
zg-pj
pj-he
RW-he
fs-DX
pj-RW
zg-RW
start-pj
he-WI
zg-he
pj-fs
start-RW
How many paths through this cave system are there that visit small caves at most once?
Your puzzle answer was 5076.
--- Part Two ---
After reviewing the available paths, you realize you might have time to visit a single small cave twice. Specifically, big caves can be visited any number of times, a single small cave can be visited at most twice, and the remaining small caves can be visited at most once. However, the caves named start and end can only be visited exactly once each: once you leave the start cave, you may not return to it, and once you reach the end cave, the path must end immediately.
Now, the 36 possible paths through the first example above are:
start,A,b,A,b,A,c,A,end
start,A,b,A,b,A,end
start,A,b,A,b,end
start,A,b,A,c,A,b,A,end
start,A,b,A,c,A,b,end
start,A,b,A,c,A,c,A,end
start,A,b,A,c,A,end
start,A,b,A,end
start,A,b,d,b,A,c,A,end
start,A,b,d,b,A,end
start,A,b,d,b,end
start,A,b,end
start,A,c,A,b,A,b,A,end
start,A,c,A,b,A,b,end
start,A,c,A,b,A,c,A,end
start,A,c,A,b,A,end
start,A,c,A,b,d,b,A,end
start,A,c,A,b,d,b,end
start,A,c,A,b,end
start,A,c,A,c,A,b,A,end
start,A,c,A,c,A,b,end
start,A,c,A,c,A,end
start,A,c,A,end
start,A,end
start,b,A,b,A,c,A,end
start,b,A,b,A,end
start,b,A,b,end
start,b,A,c,A,b,A,end
start,b,A,c,A,b,end
start,b,A,c,A,c,A,end
start,b,A,c,A,end
start,b,A,end
start,b,d,b,A,c,A,end
start,b,d,b,A,end
start,b,d,b,end
start,b,end
The slightly larger example above now has 103 paths through it, and the even larger example now has 3509 paths through it.
Given these new rules, how many paths through this cave system are there?
Your puzzle answer was 145643.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day12/test12A
-7
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@@ -1,7 +0,0 @@
start-A
start-b
A-c
A-b
b-d
A-end
b-end
2021/day12/test12B
-10
View File
@@ -1,10 +0,0 @@
dc-end
HN-start
start-kj
dc-start
dc-HN
LN-dc
HN-end
kj-sa
kj-HN
kj-dc
2021/day12/test12C
-18
View File
@@ -1,18 +0,0 @@
fs-end
he-DX
fs-he
start-DX
pj-DX
end-zg
zg-sl
zg-pj
pj-he
RW-he
fs-DX
pj-RW
zg-RW
start-pj
he-WI
zg-he
pj-fs
start-RW
2021/day13/AoC13.py
-38
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@@ -1,38 +0,0 @@
#!/usr/bin/env python3
import sys
def printi(dots):
st = ''
for y in range(max(dots,key=lambda x: x[1])[1]+1):
for x in range(max(dots,key=lambda x: x[0])[0]+1):
if (x,y) in dots: st += '#'
else: st += '.'
print(st)
st = ''
def fold(f, dots):
ndots, c = set(), f[1]
if f[0] == 'x': r = 0
elif f[0] == 'y': r = 1
for dot in dots:
if dot[r] < c: ndots.add(dot)
elif dot[r] > c: ndots.add([(c-(dot[0]-c),dot[1]), (dot[0],c-(dot[1]-c))][r])
return ndots
if __name__ == '__main__':
lines = open(sys.argv[1]).read().split('\n\n')
dots = set([tuple(map(int, line.split(','))) for line in lines[0].split('\n')])
folds = [i for i in map(lambda x: (x[0],int(x[1])), [tuple(line.strip('fold along ').split('=')) for line in lines[1].split('\n')[:-1]])]
# challenge 1
res1 = str(sum(1 for dot in fold(folds[0], dots)))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
for f in folds: dots = fold(f,dots)
printi(dots)
res2 = "read yourself please xD"
print("challenge 2:" + "\n" + res2 + "\n")
2021/day13/challenge13
-121
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@@ -1,121 +0,0 @@
--- Day 13: Transparent Origami ---
You reach another volcanically active part of the cave. It would be nice if you could do some kind of thermal imaging so you could tell ahead of time which caves are too hot to safely enter.
Fortunately, the submarine seems to be equipped with a thermal camera! When you activate it, you are greeted with:
Congratulations on your purchase! To activate this infrared thermal imaging
camera system, please enter the code found on page 1 of the manual.
Apparently, the Elves have never used this feature. To your surprise, you manage to find the manual; as you go to open it, page 1 falls out. It's a large sheet of transparent paper! The transparent paper is marked with random dots and includes instructions on how to fold it up (your puzzle input). For example:
6,10
0,14
9,10
0,3
10,4
4,11
6,0
6,12
4,1
0,13
10,12
3,4
3,0
8,4
1,10
2,14
8,10
9,0
fold along y=7
fold along x=5
The first section is a list of dots on the transparent paper. 0,0 represents the top-left coordinate. The first value, x, increases to the right. The second value, y, increases downward. So, the coordinate 3,0 is to the right of 0,0, and the coordinate 0,7 is below 0,0. The coordinates in this example form the following pattern, where # is a dot on the paper and . is an empty, unmarked position:
...#..#..#.
....#......
...........
#..........
...#....#.#
...........
...........
...........
...........
...........
.#....#.##.
....#......
......#...#
#..........
#.#........
Then, there is a list of fold instructions. Each instruction indicates a line on the transparent paper and wants you to fold the paper up (for horizontal y=... lines) or left (for vertical x=... lines). In this example, the first fold instruction is fold along y=7, which designates the line formed by all of the positions where y is 7 (marked here with -):
...#..#..#.
....#......
...........
#..........
...#....#.#
...........
...........
-----------
...........
...........
.#....#.##.
....#......
......#...#
#..........
#.#........
Because this is a horizontal line, fold the bottom half up. Some of the dots might end up overlapping after the fold is complete, but dots will never appear exactly on a fold line. The result of doing this fold looks like this:
#.##..#..#.
#...#......
......#...#
#...#......
.#.#..#.###
...........
...........
Now, only 17 dots are visible.
Notice, for example, the two dots in the bottom left corner before the transparent paper is folded; after the fold is complete, those dots appear in the top left corner (at 0,0 and 0,1). Because the paper is transparent, the dot just below them in the result (at 0,3) remains visible, as it can be seen through the transparent paper.
Also notice that some dots can end up overlapping; in this case, the dots merge together and become a single dot.
The second fold instruction is fold along x=5, which indicates this line:
#.##.|#..#.
#...#|.....
.....|#...#
#...#|.....
.#.#.|#.###
.....|.....
.....|.....
Because this is a vertical line, fold left:
#####
#...#
#...#
#...#
#####
.....
.....
The instructions made a square!
The transparent paper is pretty big, so for now, focus on just completing the first fold. After the first fold in the example above, 17 dots are visible - dots that end up overlapping after the fold is completed count as a single dot.
How many dots are visible after completing just the first fold instruction on your transparent paper?
Your puzzle answer was 745.
--- Part Two ---
Finish folding the transparent paper according to the instructions. The manual says the code is always eight capital letters.
What code do you use to activate the infrared thermal imaging camera system?
Your puzzle answer was ABKJFBGC.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day14/AoC14.py
-49
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@@ -1,49 +0,0 @@
#!/usr/bin/env python3
import sys
from collections import Counter
from copy import deepcopy
def step(cons,s):
for _ in range(s):
ncons = {}
for (a,b),v in cons.items():
if (a, rules[a+b]) not in ncons: ncons[(a, rules[a+b])] = 0
if (rules[a+b], b) not in ncons: ncons[(rules[a+b], b)] = 0
ncons[(a, rules[a+b])] += v
ncons[(rules[a+b],b)] += v
cons = deepcopy(ncons)
return cons
def countp(cons):
ex = {}
for k in nex.keys():
ex[k[0]] = 0
ex[k[1]] = 0
ex[start[0]] += 1
ex[start[-1]] += 1
for k,v in nex.items():
ex[k[0]] += v
ex[k[1]] += v
return ex
if __name__ == '__main__':
start, rules = open(sys.argv[1]).read().split('\n\n')
rules = [r.split(' -> ') for r in rules.split('\n')]
rules = {a[0]:a[1] for a in rules[:-1]}
nex = {}
for a,b in zip(start[:-1],start[1:]):
if (a,b) not in nex: nex[(a,b)] = 0
nex[(a,b)] +=1
# challenge 1
nex = step(nex,10)
res = countp(nex)
res1 = str(max(res.values())//2-min(res.values())//2)
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
nex = step(nex,30)
res = countp(nex)
res2 = str(max(res.values())//2-min(res.values())//2)
print("challenge 2:" +"\n" + res2 + "\n")
2021/day14/challenge14
-65
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@@ -1,65 +0,0 @@
--- Day 14: Extended Polymerization ---
The incredible pressures at this depth are starting to put a strain on your submarine. The submarine has polymerization equipment that would produce suitable materials to reinforce the submarine, and the nearby volcanically-active caves should even have the necessary input elements in sufficient quantities.
The submarine manual contains instructions for finding the optimal polymer formula; specifically, it offers a polymer template and a list of pair insertion rules (your puzzle input). You just need to work out what polymer would result after repeating the pair insertion process a few times.
For example:
NNCB
CH -> B
HH -> N
CB -> H
NH -> C
HB -> C
HC -> B
HN -> C
NN -> C
BH -> H
NC -> B
NB -> B
BN -> B
BB -> N
BC -> B
CC -> N
CN -> C
The first line is the polymer template - this is the starting point of the process.
The following section defines the pair insertion rules. A rule like AB -> C means that when elements A and B are immediately adjacent, element C should be inserted between them. These insertions all happen simultaneously.
So, starting with the polymer template NNCB, the first step simultaneously considers all three pairs:
The first pair (NN) matches the rule NN -> C, so element C is inserted between the first N and the second N.
The second pair (NC) matches the rule NC -> B, so element B is inserted between the N and the C.
The third pair (CB) matches the rule CB -> H, so element H is inserted between the C and the B.
Note that these pairs overlap: the second element of one pair is the first element of the next pair. Also, because all pairs are considered simultaneously, inserted elements are not considered to be part of a pair until the next step.
After the first step of this process, the polymer becomes NCNBCHB.
Here are the results of a few steps using the above rules:
Template: NNCB
After step 1: NCNBCHB
After step 2: NBCCNBBBCBHCB
After step 3: NBBBCNCCNBBNBNBBCHBHHBCHB
After step 4: NBBNBNBBCCNBCNCCNBBNBBNBBBNBBNBBCBHCBHHNHCBBCBHCB
This polymer grows quickly. After step 5, it has length 97; After step 10, it has length 3073. After step 10, B occurs 1749 times, C occurs 298 times, H occurs 161 times, and N occurs 865 times; taking the quantity of the most common element (B, 1749) and subtracting the quantity of the least common element (H, 161) produces 1749 - 161 = 1588.
Apply 10 steps of pair insertion to the polymer template and find the most and least common elements in the result. What do you get if you take the quantity of the most common element and subtract the quantity of the least common element?
Your puzzle answer was 3697.
--- Part Two ---
The resulting polymer isn't nearly strong enough to reinforce the submarine. You'll need to run more steps of the pair insertion process; a total of 40 steps should do it.
In the above example, the most common element is B (occurring 2192039569602 times) and the least common element is H (occurring 3849876073 times); subtracting these produces 2188189693529.
Apply 40 steps of pair insertion to the polymer template and find the most and least common elements in the result. What do you get if you take the quantity of the most common element and subtract the quantity of the least common element?
Your puzzle answer was 4371307836157.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day14/input14
-102
View File
@@ -1,102 +0,0 @@
CPSSSFCFOFVFNVPKBFVN
NV -> V
CF -> O
BB -> F
SB -> H
KF -> O
SP -> H
CS -> V
VF -> F
PC -> H
PH -> H
SF -> F
CP -> B
BC -> H
PB -> V
KO -> B
CV -> S
ON -> B
PV -> F
OO -> B
VV -> B
NO -> B
SH -> N
FC -> C
VO -> B
NN -> C
HH -> S
CK -> C
PF -> N
SN -> K
OK -> F
FH -> S
BP -> K
HO -> K
FB -> P
HC -> N
FP -> P
NC -> H
PK -> O
BV -> P
HK -> S
PP -> N
VC -> K
CH -> C
KS -> V
KB -> V
FN -> P
BS -> O
PS -> N
NS -> B
PN -> N
NP -> H
CB -> S
SV -> O
OC -> H
BO -> C
HN -> N
HP -> N
OF -> H
FS -> S
KV -> P
HV -> S
VS -> P
BH -> N
CC -> V
VN -> H
NF -> B
NK -> N
CN -> F
FV -> P
NH -> S
OV -> H
KN -> F
SO -> H
OP -> N
KC -> P
HB -> B
BN -> V
VP -> N
HS -> S
VK -> C
VH -> H
OS -> C
FO -> B
NB -> P
KP -> V
SS -> O
BK -> F
SK -> N
HF -> O
PO -> F
OH -> B
KK -> O
FK -> S
VB -> V
OB -> C
KH -> H
SC -> F
FF -> H
CO -> V
BF -> H
2021/day14/test14
-18
View File
@@ -1,18 +0,0 @@
NNCB
CH -> B
HH -> N
CB -> H
NH -> C
HB -> C
HC -> B
HN -> C
NN -> C
BH -> H
NC -> B
NB -> B
BN -> B
BB -> N
BC -> B
CC -> N
CN -> C
2021/day15/AoC15.py
-75
View File
@@ -1,75 +0,0 @@
#!/usr/bin/env python3
import sys
def memo(f):
memo = {}
def f2(a,x,y,c,e):
if (x,y) not in memo: memo[(x,y)] = f(a,x,y,c,e)
return memo[(x,y)]
return f2
def nb(sx, sy):
n = [(0,1), (1,0), (0,-1), (-1,0)]
return [(max(0,sx+x), max(0,sy+y)) for x,y in n]
def dijkstra(graph, s):
a = {}
v = {}
Q = initG(graph, s, a, v)
lq = len(Q)
while len(Q) > 0:
u = min([u for u in Q], key=lambda x: a[x])
Q.remove(u)
for vo in nb(*u):
if vo in Q:
distance_update(graph, u,vo,a,v)
if (len(Q)/lq * 100 % 1) == 0:
print("|", end='', flush=True)
print("\n")
return v
def initG(graph, s, a, v):
for vo in graph.keys():
a[vo] = 9999999999999999
v[vo] = 'null'
a[s] = 0
Q = [n for n in graph.keys()]
return Q
def distance_update(g, u,vo,a,v):
alt = a[u] + g[u]
if alt < a[vo]:
a[vo] = alt
v[vo] = u
def shortest_path(g, d, v):
w = [d]
u = d
r = 0
while v[u] != 'null':
r += g[u]
u = v[u]
w.append(u)
return w.reverse(), r
if __name__ == '__main__':
nums = [[int(i) for i in list(line.strip('\n'))] for line in open(sys.argv[1])]
# challenge 1
g1 = {(x,y) : nums[y][x] for x in range(len(nums[0])) for y in range(len(nums))}
v = dijkstra(g1, (0,0))
w, r = shortest_path(g1, (len(nums)-1, len(nums[0])-1), v)
res1 = str(r)
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2 - inefficient
g2 = {(x+i*len(nums[0]),y+j*len(nums)) : (nums[y][x] + i + j if nums[y][x] + i + j < 10 else (nums[y][x] + i + j) % 10 + 1) for x in range(len(nums[0])) for y in range(len(nums)) for i in range(5) for j in range(5)}
v = dijkstra(g2, (0,0))
w, r = shortest_path(g2, (len(nums[0])*5-1, len(nums)*5-1), v)
res2 = str(r)
print("challenge 2:" +"\n" + res2 + "\n")
2021/day15/challenge15
-164
View File
@@ -1,164 +0,0 @@
--- Day 15: Chiton ---
You've almost reached the exit of the cave, but the walls are getting closer together. Your submarine can barely still fit, though; the main problem is that the walls of the cave are covered in chitons, and it would be best not to bump any of them.
The cavern is large, but has a very low ceiling, restricting your motion to two dimensions. The shape of the cavern resembles a square; a quick scan of chiton density produces a map of risk level throughout the cave (your puzzle input). For example:
1163751742
1381373672
2136511328
3694931569
7463417111
1319128137
1359912421
3125421639
1293138521
2311944581
You start in the top left position, your destination is the bottom right position, and you cannot move diagonally. The number at each position is its risk level; to determine the total risk of an entire path, add up the risk levels of each position you enter (that is, don't count the risk level of your starting position unless you enter it; leaving it adds no risk to your total).
Your goal is to find a path with the lowest total risk. In this example, a path with the lowest total risk is highlighted here:
1163751742
1381373672
2136511328
3694931569
7463417111
1319128137
1359912421
3125421639
1293138521
2311944581
The total risk of this path is 40 (the starting position is never entered, so its risk is not counted).
What is the lowest total risk of any path from the top left to the bottom right?
Your puzzle answer was 487.
--- Part Two ---
Now that you know how to find low-risk paths in the cave, you can try to find your way out.
The entire cave is actually five times larger in both dimensions than you thought; the area you originally scanned is just one tile in a 5x5 tile area that forms the full map. Your original map tile repeats to the right and downward; each time the tile repeats to the right or downward, all of its risk levels are 1 higher than the tile immediately up or left of it. However, risk levels above 9 wrap back around to 1. So, if your original map had some position with a risk level of 8, then that same position on each of the 25 total tiles would be as follows:
8 9 1 2 3
9 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
Each single digit above corresponds to the example position with a value of 8 on the top-left tile. Because the full map is actually five times larger in both dimensions, that position appears a total of 25 times, once in each duplicated tile, with the values shown above.
Here is the full five-times-as-large version of the first example above, with the original map in the top left corner highlighted:
11637517422274862853338597396444961841755517295286
13813736722492484783351359589446246169155735727126
21365113283247622439435873354154698446526571955763
36949315694715142671582625378269373648937148475914
74634171118574528222968563933317967414442817852555
13191281372421239248353234135946434524615754563572
13599124212461123532357223464346833457545794456865
31254216394236532741534764385264587549637569865174
12931385212314249632342535174345364628545647573965
23119445813422155692453326671356443778246755488935
22748628533385973964449618417555172952866628316397
24924847833513595894462461691557357271266846838237
32476224394358733541546984465265719557637682166874
47151426715826253782693736489371484759148259586125
85745282229685639333179674144428178525553928963666
24212392483532341359464345246157545635726865674683
24611235323572234643468334575457944568656815567976
42365327415347643852645875496375698651748671976285
23142496323425351743453646285456475739656758684176
34221556924533266713564437782467554889357866599146
33859739644496184175551729528666283163977739427418
35135958944624616915573572712668468382377957949348
43587335415469844652657195576376821668748793277985
58262537826937364893714847591482595861259361697236
96856393331796741444281785255539289636664139174777
35323413594643452461575456357268656746837976785794
35722346434683345754579445686568155679767926678187
53476438526458754963756986517486719762859782187396
34253517434536462854564757396567586841767869795287
45332667135644377824675548893578665991468977611257
44961841755517295286662831639777394274188841538529
46246169155735727126684683823779579493488168151459
54698446526571955763768216687487932779859814388196
69373648937148475914825958612593616972361472718347
17967414442817852555392896366641391747775241285888
46434524615754563572686567468379767857948187896815
46833457545794456865681556797679266781878137789298
64587549637569865174867197628597821873961893298417
45364628545647573965675868417678697952878971816398
56443778246755488935786659914689776112579188722368
55172952866628316397773942741888415385299952649631
57357271266846838237795794934881681514599279262561
65719557637682166874879327798598143881961925499217
71484759148259586125936169723614727183472583829458
28178525553928963666413917477752412858886352396999
57545635726865674683797678579481878968159298917926
57944568656815567976792667818781377892989248891319
75698651748671976285978218739618932984172914319528
56475739656758684176786979528789718163989182927419
67554889357866599146897761125791887223681299833479
Equipped with the full map, you can now find a path from the top left corner to the bottom right corner with the lowest total risk:
11637517422274862853338597396444961841755517295286
13813736722492484783351359589446246169155735727126
21365113283247622439435873354154698446526571955763
36949315694715142671582625378269373648937148475914
74634171118574528222968563933317967414442817852555
13191281372421239248353234135946434524615754563572
13599124212461123532357223464346833457545794456865
31254216394236532741534764385264587549637569865174
12931385212314249632342535174345364628545647573965
23119445813422155692453326671356443778246755488935
22748628533385973964449618417555172952866628316397
24924847833513595894462461691557357271266846838237
32476224394358733541546984465265719557637682166874
47151426715826253782693736489371484759148259586125
85745282229685639333179674144428178525553928963666
24212392483532341359464345246157545635726865674683
24611235323572234643468334575457944568656815567976
42365327415347643852645875496375698651748671976285
23142496323425351743453646285456475739656758684176
34221556924533266713564437782467554889357866599146
33859739644496184175551729528666283163977739427418
35135958944624616915573572712668468382377957949348
43587335415469844652657195576376821668748793277985
58262537826937364893714847591482595861259361697236
96856393331796741444281785255539289636664139174777
35323413594643452461575456357268656746837976785794
35722346434683345754579445686568155679767926678187
53476438526458754963756986517486719762859782187396
34253517434536462854564757396567586841767869795287
45332667135644377824675548893578665991468977611257
44961841755517295286662831639777394274188841538529
46246169155735727126684683823779579493488168151459
54698446526571955763768216687487932779859814388196
69373648937148475914825958612593616972361472718347
17967414442817852555392896366641391747775241285888
46434524615754563572686567468379767857948187896815
46833457545794456865681556797679266781878137789298
64587549637569865174867197628597821873961893298417
45364628545647573965675868417678697952878971816398
56443778246755488935786659914689776112579188722368
55172952866628316397773942741888415385299952649631
57357271266846838237795794934881681514599279262561
65719557637682166874879327798598143881961925499217
71484759148259586125936169723614727183472583829458
28178525553928963666413917477752412858886352396999
57545635726865674683797678579481878968159298917926
57944568656815567976792667818781377892989248891319
75698651748671976285978218739618932984172914319528
56475739656758684176786979528789718163989182927419
67554889357866599146897761125791887223681299833479
The total risk of this path is 315 (the starting position is still never entered, so its risk is not counted).
Using the full map, what is the lowest total risk of any path from the top left to the bottom right?
Your puzzle answer was 2821.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day15/input15
-100
View File
@@ -1,100 +0,0 @@
3576219475874583191916312133474175459337114195988185136398151631965391991813219974121211251194786128
1713881999231519357799114192443351147195293575386923868711388519361669464234397975938889146414199688
2151922113728581598379349983928489926261239862882888968998791929828316718921297117449631496888879522
3291722311299652548445221254456535112944431135116631433192483199481594768169444158222431199929579291
1472287315918113113959972189351931626141472991992791413146153989481318998519171231313898962992319251
8978798318873934543151823824117132228417869392811216547172919119998739911121199246294815175868992773
4711149989992717728465321161611967173674239298593935117599173279267114155265393397571372124856718732
9843245319692998612781837729915931371619886371128818266322121623171343132131239399743799827972585645
7183211712173272827112268971459424836293613294388761928991542137194981939169315879277494581991465822
2385721996578834579224651431411717198341139751429156637229888849181154922897746418293184413519931124
6188313732929496115412842928273812153715747411177189187726751991191841181119822534511125656211991265
8797112815141319881991983914861912838966446818811271274141461487947171138945321912192198815813211761
6131514331472951494911549373681638637191295997625615885219668259269297171939941498581919718223529141
3916711951831246219991217199412116783411218814949692281987659123812288937983969116847986196936523779
9111121227192813382983442234171148191131821497435387936912532695293912972491131977272231919798477344
9994893855122115213919995636978443773161196511775674783979472217917498285423121574533491643891265199
9245818921321877317663254739619499915821162412924816263648913499811482474392667993739139899758919393
1717618849193152316325119128859116191132672215933282127611134982961918179152189411131259943819997123
3697848795188715264992962617223177221118919863975198881197837936258983111819488622362584888223331999
1798259211962116675922944295117391732295963342131192112612386194749389991199821195162975124619149837
5365696872959269494682128272373619348283964191994187749824841195582221478325111187843285531119934786
6717218734155417511564162235845524121493736962442884479191718924594119911521121422231982184671122116
9346412964321187889663345611132183519588187791111227269112129947122999176114184938494824342939375892
2912811929919891428891223818529115949522464779949526791515797886471246181118183121786524127979998512
7353285281129811915971956929718433284441492484117162562212973157265812816731843791838571691421923924
3239161911151141369392413114665193927469988999219478935992541193184945926828911597158489999651151172
9229716159183212219791136121569512978927999177196111291668152292896495877722611129462617229299962112
5994461796358342348216915883991574569784887978731368749888449599246152414391635925763156831839199496
1569319989929452917219726368583887924179117119284971126181346919789949393972171262535283921926293357
9983673949614595396426594299567925779229751982433631919159114841538978735893759492592481573495963491
3725632421623191323789991288299296415942162988289929992238877153268488127718372211951117951486111926
1219319492614741819142363452974731825929485665978219725131218962626179997611254524524915768686381164
8248145978417221154822663831966124539168641311913211295941167919583189792718599334616827916567313217
3483959252172951953519211189255336234459522397526829492685113849198611624111948187714151963198671919
7812289559812455122297839958719917624841136314322613942214613249662544617928521344873779598816941946
6868958791119815122313931178313637489923479492171741443993619161318781194421239654468391996636923611
6999166111129212624151555925991311538952878946191234511973499736237742491811348911419449161895978178
3532973784281182181552481928144111246711742443831184236962981395142863562121961323169398316921741321
8433727191521455193223914171391447439164791659118411391327413285243717794299999523811137619197999791
8169125649963232439519311131691539694589231314268683691811431259295132344195953951287639889737849189
8928161919934761971442719391119881113318323987227229831917897151155121657624142152884624176461913498
2949199492196424164751892146395174669111921799261527568751283712133148212821261936297739912939743599
9931616797885267969938591912672996183187926259991298429278229232257219428413174972999193796827459799
3138768154311742899419449741617489465993174645228814938145273266941566223933119891379181296548999222
1338961668848636157444719898575174699419997972125194827176897565699991611654214368232592793825617931
9283643143692917314535119897889711191946688911436965194232811624712861576291987716317193114445363435
8718611398299869991788113111214414269554542114329621286592313532782819131919527228417764279755899812
8128249329314753847725634378737239939769913811444529224619715471961999278915341955188871714271145449
1286217894699991715153139268991918988338684391252271254127193342854156438873836258484975151167996316
4191439845728418348368191516746499654147257851256861534634988281199187541199892317145428248531129171
9799283477747387253566811199414256363339122372629545249542373591951156699415998714124115397811359516
3191985767689943186838293175311429527499612824265518522881922917119891424717751985621941995373481629
8127933395192162886259956635985821651844893796541627799276258692298972881471991789861232424926758792
4349683335349673317521761437148933192989961489478395417258212949865876242843242464123471244631449998
3983861771861891251448475182984833811718959521263444111998834975935221115813839514917176912994677265
3361812177831144971531441184671112122322312961859915722393944269211333951428175372918136112814926411
2492891241191991149319731182431211211971267682931551534199328118919751416881458658314218331565282229
1316191719349114251772629411233483317516143684488291792691839198723174287211241498955137331159443961
1295632882814319179937625629218199381191127966181781921829819969815912382191294796979515315471941392
1952318366571121641614137937116326339638859628226299126699118638171813164397332391891669798271559135
1123676641998146639499323488991286491444174358479157916696196841912233773269986812299996975117612192
2496461124219121332118311532239527696198787778122345883331631119147159212262659717289674149798169111
3994915157217763242581922474512992479146377117411981999549977495462427199524334285127723311599337186
2856249932899117429471617195415229923432185189218399415357391954728651191591964713418532917111296399
1916415314958527255299763241981929186681257596912948977397875248189748655173819381269954513499896991
4214987991254389723346911149661118597277889213928551947599177915888233175182921923128853579525754718
3582279578128192231339943295188563998645998149526284997789731787116991199391984141983177825883111821
3318115949152512995142691231324793411691198458654473675135993122216324137859959272731599996923465431
9431374329994955614816239978928993493129129641248882311462522524447519388869238199852941551532223632
6749132793518479891929891686579791123712167784913331559922969149125915827657111724348929978159964476
6899888192198113996189996152913216258469182711411338723293823541574149295918323829819122941288693461
9833728958423343229397292861582211971713299293611778961979819175799893793971822162383179128129741328
9719459171294372288418126215914532995125999274359633129923191213332337513812761149562146497198614329
6323381739518574154814713691189651618965915153972197836392492518147816118266818195856772641337997448
9515391698994467556612815425651153225477392335219196697477512338938131933398177519117437353271396315
2881769116189531115832167881372917543267529333184621316918481786111252281738298936313393416198972192
1262821848371618529985323295192217356152412696271318793198457225899782896767339942896611231999996177
7611115824284211421951282951346181221891123993118231772195952872865982323819415318945742389969468591
6289176957971112617212793915412958239574931721216722823933461763146195116737161463663388274118352919
2739192179312181318764617662499189345995579913463399176128133623751919135855775349195561978131664671
7419179135228196296753972713643556671138381494838116814212892453118963456121883515171981991493112197
2351962484381996196119791327953881225881124837211622273581789216339733276411173141146816794919111532
4828494193941149581788515219884312818182198261739598611663311138255885557141492174934279519487959315
6266441934234532987821537914696156313934795144412166199339199381197916324959558778257894958593689511
1218836684761811498991162549547375891999161535133511817388897491266957819219117988952234747197245136
7714718988198712181711441996373181351671962319416917816121119483959397254815953421521642116185224622
6315878819598412161782318179586237893968466116199183185764132999388411414821117761121199725212571499
9298381112525419214597534611761811137687414291394795245955115116482261788724982532573616634235949295
2513837618149681781571514782119218925111312739891194994857915371836413112952729282168681511999132471
9859244419394114469927618849921646911349163815889492121537711577378978633345419421881312649299234749
6241913851991287776811271361777919519213726771122769111499925516752149976397933753482495291399144132
7549785323789821947513966787125532113931855716249321712383292746672763573114656989299712174939322181
9188852123922512181114117536955617335173328988385462159798591117758397298294919117614163738414981949
2974419341361292817129453536824618155299993142149899966226312193848695911943416589913541194129913422
1672761999113289962322634642122163368424955339864962951462929691975798622972131599198673671183449929
3462285147627397867977869242451116812198183328383813959284449969199819896174618466572298882951459875
6153638634925613615423256118159531429294172496996993184762972841759153392753653191942715379879852345
6516291128536155113712481272992636373774429965994966353494115221979925592128929821611928492143929534
3993993153713717844719492314391992638719111242291783837279215399219822611221331985991484863975594721
5995891457939115884519762228128552168915218299831775259316392719113962218429351553267511523198149462
2021/day15/test15
-10
View File
@@ -1,10 +0,0 @@
1163751742
1381373672
2136511328
3694931569
7463417111
1319128137
1359912421
3125421639
1293138521
2311944581
2021/day16/AoC16.py
-68
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@@ -1,68 +0,0 @@
#!/usr/bin/env python3
import sys
from bitstring import BitArray
from operator import mul
from functools import reduce
def parse_packet(packet, s, np=100000000, length=1000000000):
nex = s
end = s+length-1
bv = 0
numbers = []
while np > 0 and nex < min(len(packet)+1,end):
if (len(packet[nex:end+1]) == 0 or int(''.join(packet[nex:end+1]),2) == 0 or int(''.join(packet[nex:]),2) == 0):
if length < 1000000000: nex = end+1
break
v = int(''.join(packet[nex:nex+3]),2)
bv += v
t = int(''.join(packet[nex+3:nex+6]),2)
nex += 6
if t == 4:
num = []
while (n := packet[nex]) == '1':
num.append(list(packet[nex+1:nex+5]))
nex += 5
num.append(list(packet[nex+1:nex+5]))
nex += 5
i = int(''.join([''.join(n) for n in num]),2)
numbers.append(i)
else:
I = packet[nex]
nex += 1
if I == '1':
l = int(''.join(packet[nex:nex+11]),2)
nex += 11
b, nex, numbs = parse_packet(packet, nex, np=l)
bv += b
elif I == '0':
l = int(''.join(packet[nex:nex+15]),2)
nex += 15
b, nex, numbs = parse_packet(packet, nex, length=l)
bv += b
if t == 0: numbers.append(sum(numbs))
elif t == 1: numbers.append(reduce(mul, numbs, 1))
elif t == 2: numbers.append(min(numbs))
elif t == 3: numbers.append(max(numbs))
elif t == 5: numbers.append(int(numbs[0] > numbs[1]))
elif t == 6: numbers.append(int(numbs[0] < numbs[1]))
elif t == 7: numbers.append(int(numbs[0] == numbs[1]))
np -= 1
return bv, nex, numbers
if __name__ == '__main__':
hexa = [list(line.strip('\n')) for line in open(sys.argv[1])]
bits = [list(BitArray(hex=c).bin) for c in hexa[0]]
bits = [b for l in bits for b in l]
bv, _, num = parse_packet(bits,0,np=1)
# challenge 1
res1 = str(bv)
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
res2 = str(num[0])
print("challenge 2:" + "\n" + res2 + "\n")
2021/day16/challenge16
-127
View File
@@ -1,127 +0,0 @@
--- Day 16: Packet Decoder ---
As you leave the cave and reach open waters, you receive a transmission from the Elves back on the ship.
The transmission was sent using the Buoyancy Interchange Transmission System (BITS), a method of packing numeric expressions into a binary sequence. Your submarine's computer has saved the transmission in hexadecimal (your puzzle input).
The first step of decoding the message is to convert the hexadecimal representation into binary. Each character of hexadecimal corresponds to four bits of binary data:
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001
A = 1010
B = 1011
C = 1100
D = 1101
E = 1110
F = 1111
The BITS transmission contains a single packet at its outermost layer which itself contains many other packets. The hexadecimal representation of this packet might encode a few extra 0 bits at the end; these are not part of the transmission and should be ignored.
Every packet begins with a standard header: the first three bits encode the packet version, and the next three bits encode the packet type ID. These two values are numbers; all numbers encoded in any packet are represented as binary with the most significant bit first. For example, a version encoded as the binary sequence 100 represents the number 4.
Packets with type ID 4 represent a literal value. Literal value packets encode a single binary number. To do this, the binary number is padded with leading zeroes until its length is a multiple of four bits, and then it is broken into groups of four bits. Each group is prefixed by a 1 bit except the last group, which is prefixed by a 0 bit. These groups of five bits immediately follow the packet header. For example, the hexadecimal string D2FE28 becomes:
110100101111111000101000
VVVTTTAAAAABBBBBCCCCC
Below each bit is a label indicating its purpose:
The three bits labeled V (110) are the packet version, 6.
The three bits labeled T (100) are the packet type ID, 4, which means the packet is a literal value.
The five bits labeled A (10111) start with a 1 (not the last group, keep reading) and contain the first four bits of the number, 0111.
The five bits labeled B (11110) start with a 1 (not the last group, keep reading) and contain four more bits of the number, 1110.
The five bits labeled C (00101) start with a 0 (last group, end of packet) and contain the last four bits of the number, 0101.
The three unlabeled 0 bits at the end are extra due to the hexadecimal representation and should be ignored.
So, this packet represents a literal value with binary representation 011111100101, which is 2021 in decimal.
Every other type of packet (any packet with a type ID other than 4) represent an operator that performs some calculation on one or more sub-packets contained within. Right now, the specific operations aren't important; focus on parsing the hierarchy of sub-packets.
An operator packet contains one or more packets. To indicate which subsequent binary data represents its sub-packets, an operator packet can use one of two modes indicated by the bit immediately after the packet header; this is called the length type ID:
If the length type ID is 0, then the next 15 bits are a number that represents the total length in bits of the sub-packets contained by this packet.
If the length type ID is 1, then the next 11 bits are a number that represents the number of sub-packets immediately contained by this packet.
Finally, after the length type ID bit and the 15-bit or 11-bit field, the sub-packets appear.
For example, here is an operator packet (hexadecimal string 38006F45291200) with length type ID 0 that contains two sub-packets:
00111000000000000110111101000101001010010001001000000000
VVVTTTILLLLLLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBBBBBB
The three bits labeled V (001) are the packet version, 1.
The three bits labeled T (110) are the packet type ID, 6, which means the packet is an operator.
The bit labeled I (0) is the length type ID, which indicates that the length is a 15-bit number representing the number of bits in the sub-packets.
The 15 bits labeled L (000000000011011) contain the length of the sub-packets in bits, 27.
The 11 bits labeled A contain the first sub-packet, a literal value representing the number 10.
The 16 bits labeled B contain the second sub-packet, a literal value representing the number 20.
After reading 11 and 16 bits of sub-packet data, the total length indicated in L (27) is reached, and so parsing of this packet stops.
As another example, here is an operator packet (hexadecimal string EE00D40C823060) with length type ID 1 that contains three sub-packets:
11101110000000001101010000001100100000100011000001100000
VVVTTTILLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBCCCCCCCCCCC
The three bits labeled V (111) are the packet version, 7.
The three bits labeled T (011) are the packet type ID, 3, which means the packet is an operator.
The bit labeled I (1) is the length type ID, which indicates that the length is a 11-bit number representing the number of sub-packets.
The 11 bits labeled L (00000000011) contain the number of sub-packets, 3.
The 11 bits labeled A contain the first sub-packet, a literal value representing the number 1.
The 11 bits labeled B contain the second sub-packet, a literal value representing the number 2.
The 11 bits labeled C contain the third sub-packet, a literal value representing the number 3.
After reading 3 complete sub-packets, the number of sub-packets indicated in L (3) is reached, and so parsing of this packet stops.
For now, parse the hierarchy of the packets throughout the transmission and add up all of the version numbers.
Here are a few more examples of hexadecimal-encoded transmissions:
8A004A801A8002F478 represents an operator packet (version 4) which contains an operator packet (version 1) which contains an operator packet (version 5) which contains a literal value (version 6); this packet has a version sum of 16.
620080001611562C8802118E34 represents an operator packet (version 3) which contains two sub-packets; each sub-packet is an operator packet that contains two literal values. This packet has a version sum of 12.
C0015000016115A2E0802F182340 has the same structure as the previous example, but the outermost packet uses a different length type ID. This packet has a version sum of 23.
A0016C880162017C3686B18A3D4780 is an operator packet that contains an operator packet that contains an operator packet that contains five literal values; it has a version sum of 31.
Decode the structure of your hexadecimal-encoded BITS transmission; what do you get if you add up the version numbers in all packets?
Your puzzle answer was 971.
--- Part Two ---
Now that you have the structure of your transmission decoded, you can calculate the value of the expression it represents.
Literal values (type ID 4) represent a single number as described above. The remaining type IDs are more interesting:
Packets with type ID 0 are sum packets - their value is the sum of the values of their sub-packets. If they only have a single sub-packet, their value is the value of the sub-packet.
Packets with type ID 1 are product packets - their value is the result of multiplying together the values of their sub-packets. If they only have a single sub-packet, their value is the value of the sub-packet.
Packets with type ID 2 are minimum packets - their value is the minimum of the values of their sub-packets.
Packets with type ID 3 are maximum packets - their value is the maximum of the values of their sub-packets.
Packets with type ID 5 are greater than packets - their value is 1 if the value of the first sub-packet is greater than the value of the second sub-packet; otherwise, their value is 0. These packets always have exactly two sub-packets.
Packets with type ID 6 are less than packets - their value is 1 if the value of the first sub-packet is less than the value of the second sub-packet; otherwise, their value is 0. These packets always have exactly two sub-packets.
Packets with type ID 7 are equal to packets - their value is 1 if the value of the first sub-packet is equal to the value of the second sub-packet; otherwise, their value is 0. These packets always have exactly two sub-packets.
Using these rules, you can now work out the value of the outermost packet in your BITS transmission.
For example:
C200B40A82 finds the sum of 1 and 2, resulting in the value 3.
04005AC33890 finds the product of 6 and 9, resulting in the value 54.
880086C3E88112 finds the minimum of 7, 8, and 9, resulting in the value 7.
CE00C43D881120 finds the maximum of 7, 8, and 9, resulting in the value 9.
D8005AC2A8F0 produces 1, because 5 is less than 15.
F600BC2D8F produces 0, because 5 is not greater than 15.
9C005AC2F8F0 produces 0, because 5 is not equal to 15.
9C0141080250320F1802104A08 produces 1, because 1 + 3 = 2 * 2.
What do you get if you evaluate the expression represented by your hexadecimal-encoded BITS transmission?
Your puzzle answer was 831996589851.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day16/input16
-1
View File
@@ -1 +0,0 @@
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
2021/day17/AoC17.py
-40
View File
@@ -1,40 +0,0 @@
#!/usr/bin/env python3
import sys, re
def parse(l):
return re.compile(r"target area: x=(?P<x1>\d+)..(?P<x2>\d+), y=(?P<y1>-?\d+)..(?P<y2>-?\d+)").search(l)
def step(cx,cy,x,y):
cx += x
cy += y
x = x-1 if x > 0 else (x+1 if x < 0 else 0)
y -= 1
return cx,cy,x,y
def hits_target(x,y,t):
cx, cy, rt, my = 0,0, False, 0
while cx <= max(t[0]) and cy >= min(t[1]):
cx,cy,x,y = step(cx,cy,x,y)
my = max(my,cy)
if (min(t[0]) <= cx <= max(t[0])) and (min(t[1]) <= cy <= max(t[1])): rt = True
return my, rt
if __name__ == '__main__':
x1,x2,y1,y2 = map(int,parse(open(sys.argv[1]).read()).groups())
target = ((x1,x2),(y1,y2))
dic = {}
for x in range(target[0][1]+1):
for y in range(min(target[1]),target[0][1]+max(map(abs,target[1]))):
my,rt = hits_target(x,y,target)
if rt: dic[(x,y)] = my
# challenge 1
res1 = str(max(dic.values()))
print("challenge 1:" +"\n" + res1 + "\n")
# challenge 2
res2 = str(len(dic))
print("challenge 2:" +"\n" + res2 + "\n")
2021/day17/challenge17
-136
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@@ -1,136 +0,0 @@
--- Day 17: Trick Shot ---
You finally decode the Elves' message. HI, the message says. You continue searching for the sleigh keys.
Ahead of you is what appears to be a large ocean trench. Could the keys have fallen into it? You'd better send a probe to investigate.
The probe launcher on your submarine can fire the probe with any integer velocity in the x (forward) and y (upward, or downward if negative) directions. For example, an initial x,y velocity like 0,10 would fire the probe straight up, while an initial velocity like 10,-1 would fire the probe forward at a slight downward angle.
The probe's x,y position starts at 0,0. Then, it will follow some trajectory by moving in steps. On each step, these changes occur in the following order:
The probe's x position increases by its x velocity.
The probe's y position increases by its y velocity.
Due to drag, the probe's x velocity changes by 1 toward the value 0; that is, it decreases by 1 if it is greater than 0, increases by 1 if it is less than 0, or does not change if it is already 0.
Due to gravity, the probe's y velocity decreases by 1.
For the probe to successfully make it into the trench, the probe must be on some trajectory that causes it to be within a target area after any step. The submarine computer has already calculated this target area (your puzzle input). For example:
target area: x=20..30, y=-10..-5
This target area means that you need to find initial x,y velocity values such that after any step, the probe's x position is at least 20 and at most 30, and the probe's y position is at least -10 and at most -5.
Given this target area, one initial velocity that causes the probe to be within the target area after any step is 7,2:
.............#....#............
.......#..............#........
...............................
S........................#.....
...............................
...............................
...........................#...
...............................
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTT#TT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
In this diagram, S is the probe's initial position, 0,0. The x coordinate increases to the right, and the y coordinate increases upward. In the bottom right, positions that are within the target area are shown as T. After each step (until the target area is reached), the position of the probe is marked with #. (The bottom-right # is both a position the probe reaches and a position in the target area.)
Another initial velocity that causes the probe to be within the target area after any step is 6,3:
...............#..#............
...........#........#..........
...............................
......#..............#.........
...............................
...............................
S....................#.........
...............................
...............................
...............................
.....................#.........
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................T#TTTTTTTTT
....................TTTTTTTTTTT
Another one is 9,0:
S........#.....................
.................#.............
...............................
........................#......
...............................
....................TTTTTTTTTTT
....................TTTTTTTTTT#
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
....................TTTTTTTTTTT
One initial velocity that doesn't cause the probe to be within the target area after any step is 17,-4:
S..............................................................
...............................................................
...............................................................
...............................................................
.................#.............................................
....................TTTTTTTTTTT................................
....................TTTTTTTTTTT................................
....................TTTTTTTTTTT................................
....................TTTTTTTTTTT................................
....................TTTTTTTTTTT..#.............................
....................TTTTTTTTTTT................................
...............................................................
...............................................................
...............................................................
...............................................................
................................................#..............
...............................................................
...............................................................
...............................................................
...............................................................
...............................................................
...............................................................
..............................................................#
The probe appears to pass through the target area, but is never within it after any step. Instead, it continues down and to the right - only the first few steps are shown.
If you're going to fire a highly scientific probe out of a super cool probe launcher, you might as well do it with style. How high can you make the probe go while still reaching the target area?
In the above example, using an initial velocity of 6,9 is the best you can do, causing the probe to reach a maximum y position of 45. (Any higher initial y velocity causes the probe to overshoot the target area entirely.)
Find the initial velocity that causes the probe to reach the highest y position and still eventually be within the target area after any step. What is the highest y position it reaches on this trajectory?
Your puzzle answer was 11781.
--- Part Two ---
Maybe a fancy trick shot isn't the best idea; after all, you only have one probe, so you had better not miss.
To get the best idea of what your options are for launching the probe, you need to find every initial velocity that causes the probe to eventually be within the target area after any step.
In the above example, there are 112 different initial velocity values that meet these criteria:
23,-10 25,-9 27,-5 29,-6 22,-6 21,-7 9,0 27,-7 24,-5
25,-7 26,-6 25,-5 6,8 11,-2 20,-5 29,-10 6,3 28,-7
8,0 30,-6 29,-8 20,-10 6,7 6,4 6,1 14,-4 21,-6
26,-10 7,-1 7,7 8,-1 21,-9 6,2 20,-7 30,-10 14,-3
20,-8 13,-2 7,3 28,-8 29,-9 15,-3 22,-5 26,-8 25,-8
25,-6 15,-4 9,-2 15,-2 12,-2 28,-9 12,-3 24,-6 23,-7
25,-10 7,8 11,-3 26,-7 7,1 23,-9 6,0 22,-10 27,-6
8,1 22,-8 13,-4 7,6 28,-6 11,-4 12,-4 26,-9 7,4
24,-10 23,-8 30,-8 7,0 9,-1 10,-1 26,-5 22,-9 6,5
7,5 23,-6 28,-10 10,-2 11,-1 20,-9 14,-2 29,-7 13,-3
23,-5 24,-8 27,-9 30,-7 28,-5 21,-10 7,9 6,6 21,-5
27,-10 7,2 30,-9 21,-8 22,-7 24,-9 20,-6 6,9 29,-5
8,-2 27,-8 30,-5 24,-7
How many distinct initial velocity values cause the probe to be within the target area after any step?
Your puzzle answer was 4531.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day17/input17
-1
View File
@@ -1 +0,0 @@
target area: x=111..161, y=-154..-101
2021/day17/test17
-1
View File
@@ -1 +0,0 @@
target area: x=20..30, y=-10..-5
2021/day18/AoC18.py
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@@ -1,100 +0,0 @@
#!/usr/bin/env python3
import sys, math
from functools import reduce
from copy import deepcopy
def magn(n):
if type(n) == int: return n
return 3*magn(n[0]) + 2*magn(n[1])
def explode(osn):
sn = deepcopy(osn)
l = None
r = 0
d = False
for i,n1 in enumerate(sn):
if type(n1) == int:
l = (0,i,n1)
if r != 0:
sn[i] = n1 + r
return sn
else:
for j,n2 in enumerate(n1):
if type(n2) == int:
l = (1,[i,j],n2)
if r != 0:
n1[j] = n2 + r
return sn
else:
for k,n3 in enumerate(n2):
if type(n3) == int:
l = (2,[i,j,k],n3)
if r != 0:
n2[k] = n3 + r
return sn
else:
for m,n4 in enumerate(n3):
if type(n4) == int:
l = (3,[i,j,k,m],n4)
if r != 0:
n3[m] = n4 + r
return sn
elif not d:
d = True
r = n4[1]
if l != None:
if l[0] == 0: sn[l[1]] = l[2] + n4[0]
elif l[0] == 1: sn[l[1][0]][l[1][1]] = l[2] + n4[0]
elif l[0] == 2: sn[l[1][0]][l[1][1]][l[1][2]] = l[2] + n4[0]
elif l[0] == 3: sn[l[1][0]][l[1][1]][l[1][2]][l[1][3]] = l[2] + n4[0]
n3[m] = 0
else:
if type(n4[0]) != int:
print('help!')
n4[0] += r
return sn
return sn
def split(osn):
sn = deepcopy(osn)
if isinstance(sn,list):
for i,n in enumerate(sn):
o = sn[i]
sn[i] = split(n)
if sn[i] != o: return sn
elif sn > 9:
return [math.floor(sn/2),math.ceil(sn/2)]
return sn
def snreduce(sn):
new = 0
old = sn
while old != new:
old = new if new != 0 else old
new = explode(old)
if new != old:
continue
new = split(old)
return new
def snadd(sn1,sn2):
sn = [sn1,sn2]
return snreduce(sn)
if __name__ == '__main__':
nums = [eval(line.strip('\n')) for line in open(sys.argv[1])]
# challenge 1
res1 = str(magn(reduce(snadd,nums)))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
mm = 0
for n in nums:
for m in nums:
mm = max(mm, magn(snadd(n,m)))
res2 = str(mm)
print("challenge 2:" + "\n" + res2 + "\n")
2021/day18/challenge18
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--- Day 18: Snailfish ---
You descend into the ocean trench and encounter some snailfish. They say they saw the sleigh keys! They'll even tell you which direction the keys went if you help one of the smaller snailfish with his math homework.
Snailfish numbers aren't like regular numbers. Instead, every snailfish number is a pair - an ordered list of two elements. Each element of the pair can be either a regular number or another pair.
Pairs are written as [x,y], where x and y are the elements within the pair. Here are some example snailfish numbers, one snailfish number per line:
[1,2]
[[1,2],3]
[9,[8,7]]
[[1,9],[8,5]]
[[[[1,2],[3,4]],[[5,6],[7,8]]],9]
[[[9,[3,8]],[[0,9],6]],[[[3,7],[4,9]],3]]
[[[[1,3],[5,3]],[[1,3],[8,7]]],[[[4,9],[6,9]],[[8,2],[7,3]]]]
This snailfish homework is about addition. To add two snailfish numbers, form a pair from the left and right parameters of the addition operator. For example, [1,2] + [[3,4],5] becomes [[1,2],[[3,4],5]].
There's only one problem: snailfish numbers must always be reduced, and the process of adding two snailfish numbers can result in snailfish numbers that need to be reduced.
To reduce a snailfish number, you must repeatedly do the first action in this list that applies to the snailfish number:
If any pair is nested inside four pairs, the leftmost such pair explodes.
If any regular number is 10 or greater, the leftmost such regular number splits.
Once no action in the above list applies, the snailfish number is reduced.
During reduction, at most one action applies, after which the process returns to the top of the list of actions. For example, if split produces a pair that meets the explode criteria, that pair explodes before other splits occur.
To explode a pair, the pair's left value is added to the first regular number to the left of the exploding pair (if any), and the pair's right value is added to the first regular number to the right of the exploding pair (if any). Exploding pairs will always consist of two regular numbers. Then, the entire exploding pair is replaced with the regular number 0.
Here are some examples of a single explode action:
[[[[[9,8],1],2],3],4] becomes [[[[0,9],2],3],4] (the 9 has no regular number to its left, so it is not added to any regular number).
[7,[6,[5,[4,[3,2]]]]] becomes [7,[6,[5,[7,0]]]] (the 2 has no regular number to its right, and so it is not added to any regular number).
[[6,[5,[4,[3,2]]]],1] becomes [[6,[5,[7,0]]],3].
[[3,[2,[1,[7,3]]]],[6,[5,[4,[3,2]]]]] becomes [[3,[2,[8,0]]],[9,[5,[4,[3,2]]]]] (the pair [3,2] is unaffected because the pair [7,3] is further to the left; [3,2] would explode on the next action).
[[3,[2,[8,0]]],[9,[5,[4,[3,2]]]]] becomes [[3,[2,[8,0]]],[9,[5,[7,0]]]].
To split a regular number, replace it with a pair; the left element of the pair should be the regular number divided by two and rounded down, while the right element of the pair should be the regular number divided by two and rounded up. For example, 10 becomes [5,5], 11 becomes [5,6], 12 becomes [6,6], and so on.
Here is the process of finding the reduced result of [[[[4,3],4],4],[7,[[8,4],9]]] + [1,1]:
after addition: [[[[[4,3],4],4],[7,[[8,4],9]]],[1,1]]
after explode: [[[[0,7],4],[7,[[8,4],9]]],[1,1]]
after explode: [[[[0,7],4],[15,[0,13]]],[1,1]]
after split: [[[[0,7],4],[[7,8],[0,13]]],[1,1]]
after split: [[[[0,7],4],[[7,8],[0,[6,7]]]],[1,1]]
after explode: [[[[0,7],4],[[7,8],[6,0]]],[8,1]]
Once no reduce actions apply, the snailfish number that remains is the actual result of the addition operation: [[[[0,7],4],[[7,8],[6,0]]],[8,1]].
The homework assignment involves adding up a list of snailfish numbers (your puzzle input). The snailfish numbers are each listed on a separate line. Add the first snailfish number and the second, then add that result and the third, then add that result and the fourth, and so on until all numbers in the list have been used once.
For example, the final sum of this list is [[[[1,1],[2,2]],[3,3]],[4,4]]:
[1,1]
[2,2]
[3,3]
[4,4]
The final sum of this list is [[[[3,0],[5,3]],[4,4]],[5,5]]:
[1,1]
[2,2]
[3,3]
[4,4]
[5,5]
The final sum of this list is [[[[5,0],[7,4]],[5,5]],[6,6]]:
[1,1]
[2,2]
[3,3]
[4,4]
[5,5]
[6,6]
Here's a slightly larger example:
[[[0,[4,5]],[0,0]],[[[4,5],[2,6]],[9,5]]]
[7,[[[3,7],[4,3]],[[6,3],[8,8]]]]
[[2,[[0,8],[3,4]]],[[[6,7],1],[7,[1,6]]]]
[[[[2,4],7],[6,[0,5]]],[[[6,8],[2,8]],[[2,1],[4,5]]]]
[7,[5,[[3,8],[1,4]]]]
[[2,[2,2]],[8,[8,1]]]
[2,9]
[1,[[[9,3],9],[[9,0],[0,7]]]]
[[[5,[7,4]],7],1]
[[[[4,2],2],6],[8,7]]
The final sum [[[[8,7],[7,7]],[[8,6],[7,7]]],[[[0,7],[6,6]],[8,7]]] is found after adding up the above snailfish numbers:
[[[0,[4,5]],[0,0]],[[[4,5],[2,6]],[9,5]]]
+ [7,[[[3,7],[4,3]],[[6,3],[8,8]]]]
= [[[[4,0],[5,4]],[[7,7],[6,0]]],[[8,[7,7]],[[7,9],[5,0]]]]
[[[[4,0],[5,4]],[[7,7],[6,0]]],[[8,[7,7]],[[7,9],[5,0]]]]
+ [[2,[[0,8],[3,4]]],[[[6,7],1],[7,[1,6]]]]
= [[[[6,7],[6,7]],[[7,7],[0,7]]],[[[8,7],[7,7]],[[8,8],[8,0]]]]
[[[[6,7],[6,7]],[[7,7],[0,7]]],[[[8,7],[7,7]],[[8,8],[8,0]]]]
+ [[[[2,4],7],[6,[0,5]]],[[[6,8],[2,8]],[[2,1],[4,5]]]]
= [[[[7,0],[7,7]],[[7,7],[7,8]]],[[[7,7],[8,8]],[[7,7],[8,7]]]]
[[[[7,0],[7,7]],[[7,7],[7,8]]],[[[7,7],[8,8]],[[7,7],[8,7]]]]
+ [7,[5,[[3,8],[1,4]]]]
= [[[[7,7],[7,8]],[[9,5],[8,7]]],[[[6,8],[0,8]],[[9,9],[9,0]]]]
[[[[7,7],[7,8]],[[9,5],[8,7]]],[[[6,8],[0,8]],[[9,9],[9,0]]]]
+ [[2,[2,2]],[8,[8,1]]]
= [[[[6,6],[6,6]],[[6,0],[6,7]]],[[[7,7],[8,9]],[8,[8,1]]]]
[[[[6,6],[6,6]],[[6,0],[6,7]]],[[[7,7],[8,9]],[8,[8,1]]]]
+ [2,9]
= [[[[6,6],[7,7]],[[0,7],[7,7]]],[[[5,5],[5,6]],9]]
[[[[6,6],[7,7]],[[0,7],[7,7]]],[[[5,5],[5,6]],9]]
+ [1,[[[9,3],9],[[9,0],[0,7]]]]
= [[[[7,8],[6,7]],[[6,8],[0,8]]],[[[7,7],[5,0]],[[5,5],[5,6]]]]
[[[[7,8],[6,7]],[[6,8],[0,8]]],[[[7,7],[5,0]],[[5,5],[5,6]]]]
+ [[[5,[7,4]],7],1]
= [[[[7,7],[7,7]],[[8,7],[8,7]]],[[[7,0],[7,7]],9]]
[[[[7,7],[7,7]],[[8,7],[8,7]]],[[[7,0],[7,7]],9]]
+ [[[[4,2],2],6],[8,7]]
= [[[[8,7],[7,7]],[[8,6],[7,7]]],[[[0,7],[6,6]],[8,7]]]
To check whether it's the right answer, the snailfish teacher only checks the magnitude of the final sum. The magnitude of a pair is 3 times the magnitude of its left element plus 2 times the magnitude of its right element. The magnitude of a regular number is just that number.
For example, the magnitude of [9,1] is 3*9 + 2*1 = 29; the magnitude of [1,9] is 3*1 + 2*9 = 21. Magnitude calculations are recursive: the magnitude of [[9,1],[1,9]] is 3*29 + 2*21 = 129.
Here are a few more magnitude examples:
[[1,2],[[3,4],5]] becomes 143.
[[[[0,7],4],[[7,8],[6,0]]],[8,1]] becomes 1384.
[[[[1,1],[2,2]],[3,3]],[4,4]] becomes 445.
[[[[3,0],[5,3]],[4,4]],[5,5]] becomes 791.
[[[[5,0],[7,4]],[5,5]],[6,6]] becomes 1137.
[[[[8,7],[7,7]],[[8,6],[7,7]]],[[[0,7],[6,6]],[8,7]]] becomes 3488.
So, given this example homework assignment:
[[[0,[5,8]],[[1,7],[9,6]]],[[4,[1,2]],[[1,4],2]]]
[[[5,[2,8]],4],[5,[[9,9],0]]]
[6,[[[6,2],[5,6]],[[7,6],[4,7]]]]
[[[6,[0,7]],[0,9]],[4,[9,[9,0]]]]
[[[7,[6,4]],[3,[1,3]]],[[[5,5],1],9]]
[[6,[[7,3],[3,2]]],[[[3,8],[5,7]],4]]
[[[[5,4],[7,7]],8],[[8,3],8]]
[[9,3],[[9,9],[6,[4,9]]]]
[[2,[[7,7],7]],[[5,8],[[9,3],[0,2]]]]
[[[[5,2],5],[8,[3,7]]],[[5,[7,5]],[4,4]]]
The final sum is:
[[[[6,6],[7,6]],[[7,7],[7,0]]],[[[7,7],[7,7]],[[7,8],[9,9]]]]
The magnitude of this final sum is 4140.
Add up all of the snailfish numbers from the homework assignment in the order they appear. What is the magnitude of the final sum?
Your puzzle answer was 3305.
--- Part Two ---
You notice a second question on the back of the homework assignment:
What is the largest magnitude you can get from adding only two of the snailfish numbers?
Note that snailfish addition is not commutative - that is, x + y and y + x can produce different results.
Again considering the last example homework assignment above:
[[[0,[5,8]],[[1,7],[9,6]]],[[4,[1,2]],[[1,4],2]]]
[[[5,[2,8]],4],[5,[[9,9],0]]]
[6,[[[6,2],[5,6]],[[7,6],[4,7]]]]
[[[6,[0,7]],[0,9]],[4,[9,[9,0]]]]
[[[7,[6,4]],[3,[1,3]]],[[[5,5],1],9]]
[[6,[[7,3],[3,2]]],[[[3,8],[5,7]],4]]
[[[[5,4],[7,7]],8],[[8,3],8]]
[[9,3],[[9,9],[6,[4,9]]]]
[[2,[[7,7],7]],[[5,8],[[9,3],[0,2]]]]
[[[[5,2],5],[8,[3,7]]],[[5,[7,5]],[4,4]]]
The largest magnitude of the sum of any two snailfish numbers in this list is 3993. This is the magnitude of [[2,[[7,7],7]],[[5,8],[[9,3],[0,2]]]] + [[[0,[5,8]],[[1,7],[9,6]]],[[4,[1,2]],[[1,4],2]]], which reduces to [[[[7,8],[6,6]],[[6,0],[7,7]]],[[[7,8],[8,8]],[[7,9],[0,6]]]].
What is the largest magnitude of any sum of two different snailfish numbers from the homework assignment?
Your puzzle answer was 4563.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day18/input18
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[[[[9,5],[9,4]],[[6,5],[7,0]]],4]
[[[5,2],[[7,2],1]],[[[7,5],[0,8]],[[6,9],[7,3]]]]
[[[9,7],[0,1]],9]
[1,[[7,3],[[3,7],[3,2]]]]
[[9,[[0,8],7]],[[3,1],[[6,6],[9,0]]]]
[4,[[4,4],[[7,7],1]]]
[[[[6,2],[5,1]],[[3,3],9]],[7,[[5,7],[5,0]]]]
[[[[4,8],[4,9]],[1,[9,3]]],[1,[1,[6,1]]]]
[[[[4,7],[3,4]],[8,3]],[[3,7],[0,[1,8]]]]
[[[6,[4,8]],[4,5]],[4,[1,3]]]
[[[0,7],0],[[6,[1,8]],[9,[7,9]]]]
[[[[4,8],[3,9]],[4,5]],[1,1]]
[[[4,2],[0,[6,7]]],[[[1,8],2],[8,8]]]
[[[[1,1],7],5],[[6,[5,6]],[6,[7,5]]]]
[[[[3,2],5],[[5,3],1]],[[[0,4],[9,6]],9]]
[[6,[7,6]],9]
[[[[4,0],[0,1]],7],1]
[[[[1,3],4],6],[[1,[4,2]],[1,4]]]
[[[[6,9],[4,1]],[[6,3],[0,8]]],[[4,0],[[3,2],[2,9]]]]
[[[3,6],[[2,0],[3,2]]],[2,5]]
[[[[4,3],5],5],[[4,[4,0]],6]]
[[[[4,0],3],[[3,5],8]],[[8,[4,4]],[[9,9],[4,1]]]]
[[[2,7],6],1]
[[[[5,3],[8,4]],[0,0]],4]
[[[0,[8,1]],0],3]
[[[6,5],[8,2]],[[[6,9],[6,1]],[9,9]]]
[0,[[4,9],6]]
[[9,[[9,9],4]],[[[4,7],1],2]]
[[8,0],[[[0,7],6],[[6,4],2]]]
[[1,[[2,4],8]],1]
[[[[1,3],4],[[1,3],0]],[[[1,2],3],2]]
[[[[2,1],2],[5,[2,8]]],[2,[[6,0],2]]]
[[[8,[1,0]],[[6,7],[9,6]]],[[2,[9,7]],5]]
[[[3,[2,0]],[[3,2],[0,0]]],[[[4,6],[9,4]],[[7,8],[5,1]]]]
[[3,[[9,9],[7,2]]],[[1,3],[2,[3,2]]]]
[4,[4,[[9,5],6]]]
[[[[5,7],7],[[3,4],0]],[[9,[8,2]],[2,3]]]
[[[[2,1],[5,7]],4],[[[6,3],8],[[1,6],[5,1]]]]
[[[4,4],[[0,9],[7,8]]],[[2,[2,5]],5]]
[1,[5,[[3,7],[8,2]]]]
[[[[9,5],[8,6]],[5,5]],[[[9,2],8],[[9,3],[3,8]]]]
[0,[[9,5],[[3,7],7]]]
[[[8,[0,4]],[[2,9],6]],[[6,[8,0]],4]]
[[0,[3,5]],[[5,[0,1]],[[3,6],7]]]
[[2,[7,1]],[[[5,0],[7,7]],[[2,3],9]]]
[[5,[9,[3,9]]],[[8,[3,7]],[[7,6],[3,0]]]]
[[[4,[2,5]],5],[3,1]]
[[[[4,3],1],[[5,7],6]],[0,[3,1]]]
[[8,9],[[[0,7],5],[6,[5,7]]]]
[[6,8],[[5,8],[[8,2],[6,0]]]]
[[1,[5,6]],5]
[[[6,1],[9,[1,2]]],1]
[[5,[7,[4,8]]],[[4,[2,9]],5]]
[[[2,2],[[7,1],3]],[[[9,7],[4,6]],[1,[0,1]]]]
[[3,[6,[4,5]]],2]
[[[0,2],[[8,1],[0,6]]],[[7,[9,6]],0]]
[[[[1,0],[5,1]],[[0,6],5]],[[[1,8],8],[[0,2],5]]]
[[6,[[3,6],6]],[[[9,7],[6,4]],[[9,5],1]]]
[[[0,[5,6]],[9,0]],[[2,9],9]]
[1,[[4,[9,3]],0]]
[[1,0],[[1,9],[4,8]]]
[[[9,3],[7,0]],[[[5,1],[3,8]],9]]
[[[3,9],[[5,9],2]],[[7,2],1]]
[[1,[[3,0],[7,6]]],[7,[8,1]]]
[0,[6,[[7,1],[1,1]]]]
[[4,[[5,0],[2,1]]],[[[8,8],[8,1]],7]]
[[[[9,3],[4,3]],4],[7,5]]
[[9,[[7,4],[8,3]]],[[[1,9],7],[[1,6],[3,1]]]]
[[6,9],[5,[0,[5,1]]]]
[[[8,7],3],[[4,8],[0,7]]]
[[[[3,1],2],[[1,6],[4,3]]],[0,6]]
[[5,[[5,4],3]],[[8,8],9]]
[[5,[3,[4,5]]],[[2,[6,0]],[6,1]]]
[[[[9,5],3],6],[[8,[1,9]],[[5,2],5]]]
[[[7,5],[[3,6],4]],[6,[[5,1],[0,1]]]]
[[1,[[4,8],[1,3]]],7]
[[4,[[4,0],5]],[[[6,2],7],[[4,8],[4,9]]]]
[[[[2,3],[0,9]],[7,2]],[4,5]]
[[[[7,7],[8,0]],[7,7]],[[[6,6],[3,2]],[4,[4,3]]]]
[[[[8,7],6],[[5,5],0]],[[6,[7,3]],[[4,1],[1,7]]]]
[[[2,[2,2]],[[5,2],1]],[[9,[9,2]],6]]
[[[[1,7],6],[[8,8],5]],[6,[1,[1,7]]]]
[[[[8,6],[3,2]],[[5,2],[2,0]]],[[[8,7],2],[[5,5],2]]]
[[[8,[9,0]],[[9,5],[7,5]]],[[5,1],[[1,1],[4,6]]]]
[5,[9,[[0,2],7]]]
[8,[[0,[4,9]],[[7,4],9]]]
[[[[2,9],5],[[0,6],[6,6]]],[[0,6],[[4,2],[9,9]]]]
[7,[[[4,3],3],[[5,4],[6,0]]]]
[[0,[8,[1,1]]],5]
[[[1,8],[[4,6],[9,7]]],[[[6,6],[2,6]],[4,3]]]
[[0,[[7,5],[9,9]]],[[9,7],[6,2]]]
[[[9,[3,0]],[[1,4],0]],[[1,1],1]]
[[[0,7],[[3,0],8]],[[6,[8,0]],[[4,5],[4,0]]]]
[[[[2,9],[4,2]],[5,[9,3]]],[4,[2,[3,4]]]]
[[[1,[7,3]],[[5,7],0]],[6,[[6,5],2]]]
[4,5]
[[7,9],[6,[[6,5],[1,0]]]]
[[4,[[7,5],8]],[[4,0],[[6,6],[0,4]]]]
[[[9,[7,7]],[[4,2],7]],4]
[[0,[0,3]],5]
2021/day18/test18
-10
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[[[0,[4,5]],[0,0]],[[[4,5],[2,6]],[9,5]]]
[7,[[[3,7],[4,3]],[[6,3],[8,8]]]]
[[2,[[0,8],[3,4]]],[[[6,7],1],[7,[1,6]]]]
[[[[2,4],7],[6,[0,5]]],[[[6,8],[2,8]],[[2,1],[4,5]]]]
[7,[5,[[3,8],[1,4]]]]
[[2,[2,2]],[8,[8,1]]]
[2,9]
[1,[[[9,3],9],[[9,0],[0,7]]]]
[[[5,[7,4]],7],1]
[[[[4,2],2],6],[8,7]]
2021/day19/AoC19.py
-94
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#!/usr/bin/env python3
import sys, math
import numpy as np
def rotations(x,y,z):
r = [
# x is facing x
[x, y, z],
[x, -z, y],
[x, -y, -z],
[x, z, -y],
# x is facing -x
[-x, -y, z],
[-x, -z, -y],
[-x, y, -z],
[-x, z, y],
# x is facing y
[-z, x, -y],
[y, x, -z],
[z, x, y],
[-y, x, z],
# x is facing -y
[z, -x, -y],
[y, -x, z],
[-z, -x, y],
[-y, -x, -z],
# x is facing z
[-y, -z, x],
[z, -y, x],
[y, z, x],
[-z, y, x],
# x is facing -z
[z, y, -x],
[-y, z, -x],
[-z, -y, -x],
[y, -z, -x]
]
for point in r:
yield np.array(point)
def overlap(s1,s2,off):
se1 = set([tuple(p) for p in s1])
se2 = set([tuple(p-off) for p in s2])
return len(se1.intersection(se2))
def test(s1,s2):
rot = [[r for r in rotations(*p)] for p in s2]
rot2 = [[r[i] for r in rot] for i in range(24)]
mc = 0
for i in range(24):
for p1 in s1:
for p2 in rot2[i]:
dif = p2 - p1
com = overlap(s1,rot2[i],dif)
if com > mc:
mc = com
if mc >= 12:
return [r - dif for r in rot2[i]], dif
return None, None
if __name__ == '__main__':
nums = [[np.array([i for i in map(int,n.split(','))]) for n in line.split('\n')[1:] if n != ''] for line in open(sys.argv[1]).read().split('\n\n')]
scanners = []
matched = {}
matched[0] = nums[0]
al = set([tuple(i) for i in matched[0]])
while len(matched) != len(nums):
for i in range(len(nums)):
if i in matched: continue
r = None
r, d = test(al,nums[i])
scanners.append(d)
if r != None:
matched[i] = r
al = al.union(set([tuple(i) for i in matched[i]]))
break
m = 0
for a in scanners:
for b in scanners:
n = sum(abs(a[i] - b[i]) for i in range(3))
if n > m:
m = n
# challenge 1
res1 = str(len(al))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
res2 = str(m)
print("challenge 2:" + "\n" + res2 + "\n")
2021/day19/challenge19
-378
View File
@@ -1,378 +0,0 @@
--- Day 19: Beacon Scanner ---
As your probe drifted down through this area, it released an assortment of beacons and scanners into the water. It's difficult to navigate in the pitch black open waters of the ocean trench, but if you can build a map of the trench using data from the scanners, you should be able to safely reach the bottom.
The beacons and scanners float motionless in the water; they're designed to maintain the same position for long periods of time. Each scanner is capable of detecting all beacons in a large cube centered on the scanner; beacons that are at most 1000 units away from the scanner in each of the three axes (x, y, and z) have their precise position determined relative to the scanner. However, scanners cannot detect other scanners. The submarine has automatically summarized the relative positions of beacons detected by each scanner (your puzzle input).
For example, if a scanner is at x,y,z coordinates 500,0,-500 and there are beacons at -500,1000,-1500 and 1501,0,-500, the scanner could report that the first beacon is at -1000,1000,-1000 (relative to the scanner) but would not detect the second beacon at all.
Unfortunately, while each scanner can report the positions of all detected beacons relative to itself, the scanners do not know their own position. You'll need to determine the positions of the beacons and scanners yourself.
The scanners and beacons map a single contiguous 3d region. This region can be reconstructed by finding pairs of scanners that have overlapping detection regions such that there are at least 12 beacons that both scanners detect within the overlap. By establishing 12 common beacons, you can precisely determine where the scanners are relative to each other, allowing you to reconstruct the beacon map one scanner at a time.
For a moment, consider only two dimensions. Suppose you have the following scanner reports:
--- scanner 0 ---
0,2
4,1
3,3
--- scanner 1 ---
-1,-1
-5,0
-2,1
Drawing x increasing rightward, y increasing upward, scanners as S, and beacons as B, scanner 0 detects this:
...B.
B....
....B
S....
Scanner 1 detects this:
...B..
B....S
....B.
For this example, assume scanners only need 3 overlapping beacons. Then, the beacons visible to both scanners overlap to produce the following complete map:
...B..
B....S
....B.
S.....
Unfortunately, there's a second problem: the scanners also don't know their rotation or facing direction. Due to magnetic alignment, each scanner is rotated some integer number of 90-degree turns around all of the x, y, and z axes. That is, one scanner might call a direction positive x, while another scanner might call that direction negative y. Or, two scanners might agree on which direction is positive x, but one scanner might be upside-down from the perspective of the other scanner. In total, each scanner could be in any of 24 different orientations: facing positive or negative x, y, or z, and considering any of four directions "up" from that facing.
For example, here is an arrangement of beacons as seen from a scanner in the same position but in different orientations:
--- scanner 0 ---
-1,-1,1
-2,-2,2
-3,-3,3
-2,-3,1
5,6,-4
8,0,7
--- scanner 0 ---
1,-1,1
2,-2,2
3,-3,3
2,-1,3
-5,4,-6
-8,-7,0
--- scanner 0 ---
-1,-1,-1
-2,-2,-2
-3,-3,-3
-1,-3,-2
4,6,5
-7,0,8
--- scanner 0 ---
1,1,-1
2,2,-2
3,3,-3
1,3,-2
-4,-6,5
7,0,8
--- scanner 0 ---
1,1,1
2,2,2
3,3,3
3,1,2
-6,-4,-5
0,7,-8
By finding pairs of scanners that both see at least 12 of the same beacons, you can assemble the entire map. For example, consider the following report:
--- scanner 0 ---
404,-588,-901
528,-643,409
-838,591,734
390,-675,-793
-537,-823,-458
-485,-357,347
-345,-311,381
-661,-816,-575
-876,649,763
-618,-824,-621
553,345,-567
474,580,667
-447,-329,318
-584,868,-557
544,-627,-890
564,392,-477
455,729,728
-892,524,684
-689,845,-530
423,-701,434
7,-33,-71
630,319,-379
443,580,662
-789,900,-551
459,-707,401
--- scanner 1 ---
686,422,578
605,423,415
515,917,-361
-336,658,858
95,138,22
-476,619,847
-340,-569,-846
567,-361,727
-460,603,-452
669,-402,600
729,430,532
-500,-761,534
-322,571,750
-466,-666,-811
-429,-592,574
-355,545,-477
703,-491,-529
-328,-685,520
413,935,-424
-391,539,-444
586,-435,557
-364,-763,-893
807,-499,-711
755,-354,-619
553,889,-390
--- scanner 2 ---
649,640,665
682,-795,504
-784,533,-524
-644,584,-595
-588,-843,648
-30,6,44
-674,560,763
500,723,-460
609,671,-379
-555,-800,653
-675,-892,-343
697,-426,-610
578,704,681
493,664,-388
-671,-858,530
-667,343,800
571,-461,-707
-138,-166,112
-889,563,-600
646,-828,498
640,759,510
-630,509,768
-681,-892,-333
673,-379,-804
-742,-814,-386
577,-820,562
--- scanner 3 ---
-589,542,597
605,-692,669
-500,565,-823
-660,373,557
-458,-679,-417
-488,449,543
-626,468,-788
338,-750,-386
528,-832,-391
562,-778,733
-938,-730,414
543,643,-506
-524,371,-870
407,773,750
-104,29,83
378,-903,-323
-778,-728,485
426,699,580
-438,-605,-362
-469,-447,-387
509,732,623
647,635,-688
-868,-804,481
614,-800,639
595,780,-596
--- scanner 4 ---
727,592,562
-293,-554,779
441,611,-461
-714,465,-776
-743,427,-804
-660,-479,-426
832,-632,460
927,-485,-438
408,393,-506
466,436,-512
110,16,151
-258,-428,682
-393,719,612
-211,-452,876
808,-476,-593
-575,615,604
-485,667,467
-680,325,-822
-627,-443,-432
872,-547,-609
833,512,582
807,604,487
839,-516,451
891,-625,532
-652,-548,-490
30,-46,-14
Because all coordinates are relative, in this example, all "absolute" positions will be expressed relative to scanner 0 (using the orientation of scanner 0 and as if scanner 0 is at coordinates 0,0,0).
Scanners 0 and 1 have overlapping detection cubes; the 12 beacons they both detect (relative to scanner 0) are at the following coordinates:
-618,-824,-621
-537,-823,-458
-447,-329,318
404,-588,-901
544,-627,-890
528,-643,409
-661,-816,-575
390,-675,-793
423,-701,434
-345,-311,381
459,-707,401
-485,-357,347
These same 12 beacons (in the same order) but from the perspective of scanner 1 are:
686,422,578
605,423,415
515,917,-361
-336,658,858
-476,619,847
-460,603,-452
729,430,532
-322,571,750
-355,545,-477
413,935,-424
-391,539,-444
553,889,-390
Because of this, scanner 1 must be at 68,-1246,-43 (relative to scanner 0).
Scanner 4 overlaps with scanner 1; the 12 beacons they both detect (relative to scanner 0) are:
459,-707,401
-739,-1745,668
-485,-357,347
432,-2009,850
528,-643,409
423,-701,434
-345,-311,381
408,-1815,803
534,-1912,768
-687,-1600,576
-447,-329,318
-635,-1737,486
So, scanner 4 is at -20,-1133,1061 (relative to scanner 0).
Following this process, scanner 2 must be at 1105,-1205,1229 (relative to scanner 0) and scanner 3 must be at -92,-2380,-20 (relative to scanner 0).
The full list of beacons (relative to scanner 0) is:
-892,524,684
-876,649,763
-838,591,734
-789,900,-551
-739,-1745,668
-706,-3180,-659
-697,-3072,-689
-689,845,-530
-687,-1600,576
-661,-816,-575
-654,-3158,-753
-635,-1737,486
-631,-672,1502
-624,-1620,1868
-620,-3212,371
-618,-824,-621
-612,-1695,1788
-601,-1648,-643
-584,868,-557
-537,-823,-458
-532,-1715,1894
-518,-1681,-600
-499,-1607,-770
-485,-357,347
-470,-3283,303
-456,-621,1527
-447,-329,318
-430,-3130,366
-413,-627,1469
-345,-311,381
-36,-1284,1171
-27,-1108,-65
7,-33,-71
12,-2351,-103
26,-1119,1091
346,-2985,342
366,-3059,397
377,-2827,367
390,-675,-793
396,-1931,-563
404,-588,-901
408,-1815,803
423,-701,434
432,-2009,850
443,580,662
455,729,728
456,-540,1869
459,-707,401
465,-695,1988
474,580,667
496,-1584,1900
497,-1838,-617
527,-524,1933
528,-643,409
534,-1912,768
544,-627,-890
553,345,-567
564,392,-477
568,-2007,-577
605,-1665,1952
612,-1593,1893
630,319,-379
686,-3108,-505
776,-3184,-501
846,-3110,-434
1135,-1161,1235
1243,-1093,1063
1660,-552,429
1693,-557,386
1735,-437,1738
1749,-1800,1813
1772,-405,1572
1776,-675,371
1779,-442,1789
1780,-1548,337
1786,-1538,337
1847,-1591,415
1889,-1729,1762
1994,-1805,1792
In total, there are 79 beacons.
Assemble the full map of beacons. How many beacons are there?
Your puzzle answer was 323.
--- Part Two ---
Sometimes, it's a good idea to appreciate just how big the ocean is. Using the Manhattan distance, how far apart do the scanners get?
In the above example, scanners 2 (1105,-1205,1229) and 3 (-92,-2380,-20) are the largest Manhattan distance apart. In total, they are 1197 + 1175 + 1249 = 3621 units apart.
What is the largest Manhattan distance between any two scanners?
Your puzzle answer was 10685.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day19/input19
-780
View File
@@ -1,780 +0,0 @@
--- scanner 0 ---
-757,414,492
-593,762,-478
-608,779,-508
-761,323,468
-583,-536,626
539,660,453
745,-932,544
419,509,439
62,-24,118
338,-793,-584
-775,808,-521
708,771,-798
-90,-99,12
400,-811,-639
850,-816,468
715,652,-760
-820,276,460
737,-883,437
716,659,-718
-619,-458,-521
-551,-451,-643
477,574,399
-653,-464,-508
-596,-505,704
438,-706,-637
-589,-621,598
--- scanner 1 ---
-386,481,-530
568,822,748
56,78,41
-653,837,672
-825,-465,-698
-842,-495,-683
517,-321,717
401,-495,-638
442,717,-695
-661,-765,361
558,611,-702
-871,-508,-516
-339,473,-471
-709,-772,450
441,-376,741
-477,505,-534
388,-630,-740
473,639,-708
-528,845,807
-610,763,777
583,809,645
502,-587,734
395,-508,-672
-37,-6,-116
-623,-735,471
505,810,757
--- scanner 2 ---
-695,364,-810
-642,483,453
745,-960,-441
743,235,-394
15,-139,-14
-452,-639,536
635,652,646
810,-702,549
700,-762,505
-453,-534,569
621,-860,-431
695,501,604
-510,489,384
-706,370,-596
-536,388,405
-628,402,-625
633,540,465
802,-735,390
606,-923,-539
578,271,-459
105,-2,-130
-517,-596,-542
-571,-576,-529
-486,-440,473
654,315,-514
-443,-634,-618
--- scanner 3 ---
812,-526,-776
447,879,691
-2,88,14
-853,-613,-699
868,-612,-712
-746,793,-620
-860,-681,-577
491,-498,660
-529,-701,302
502,814,567
-540,613,383
445,628,-447
355,810,596
403,-351,685
-807,800,-831
-33,-43,-159
-731,776,-886
583,616,-432
-582,553,534
-602,-816,250
471,485,-398
872,-622,-827
-568,-778,372
377,-461,620
-721,615,495
-847,-579,-712
--- scanner 4 ---
-383,-542,-722
703,879,-598
587,-459,579
-424,322,707
-398,371,564
113,1,1
629,-610,639
-400,-767,313
-455,-643,378
777,869,-619
-308,-672,382
793,544,399
-536,356,576
828,-723,-592
-389,-498,-592
768,-571,577
854,755,392
-621,489,-516
-611,512,-432
849,-622,-566
-443,-593,-655
753,753,-624
778,810,416
780,-628,-698
-654,590,-372
--- scanner 5 ---
366,-674,-398
646,574,-495
810,-712,775
-776,-616,-664
709,499,-484
538,832,519
-851,-513,601
330,-736,-449
-832,-477,488
754,549,-531
534,-704,-421
589,756,371
-769,699,-884
742,-698,801
-608,351,427
616,920,401
-821,-527,-723
-149,-42,-32
-775,702,-638
-576,337,535
-750,-560,-579
-717,658,-756
632,-708,837
-18,44,105
-824,-294,581
-538,374,539
--- scanner 6 ---
-670,706,-858
342,376,641
405,-867,-344
-768,-540,-416
-591,518,863
-711,414,845
-776,-509,-579
281,-922,466
-464,-713,728
378,352,-541
288,342,-625
287,-929,-439
-757,-508,-659
-18,-130,-81
-577,685,-840
371,-842,552
-550,433,837
-523,-701,877
303,376,-402
247,-928,631
-659,-686,773
333,388,704
445,389,626
-433,677,-834
358,-840,-351
--- scanner 7 ---
-608,662,-491
-421,395,723
537,400,-654
-801,-707,334
-716,-378,-908
-606,481,-603
-691,-669,364
722,-794,-779
-712,-755,327
-551,-443,-872
-64,-119,-13
50,75,-130
590,472,837
655,-609,-741
-668,620,-642
-445,471,778
569,546,847
-428,456,512
595,496,721
733,-580,-773
563,421,-755
669,469,-660
-627,-498,-828
799,-779,496
860,-721,542
820,-593,474
--- scanner 8 ---
529,-663,-800
519,-548,-804
-429,-739,586
271,298,-446
-718,433,-595
505,549,874
562,-663,915
615,-599,-744
-729,576,-465
351,484,-448
-529,-792,673
305,358,-378
-68,-144,62
498,443,758
528,-676,921
-836,389,447
-828,282,496
-796,-643,-252
-491,-794,568
-652,604,-655
492,471,843
-653,-636,-220
460,-798,928
-941,313,563
-115,1,182
-609,-662,-266
--- scanner 9 ---
819,-614,-852
-394,-498,662
-568,774,-852
-488,737,-781
-672,-417,-703
-664,-337,-815
652,-627,-823
498,457,699
595,-492,407
-568,636,624
645,572,-483
575,387,738
47,-73,82
-546,514,691
697,-627,420
-442,625,-847
-356,-496,767
-671,-375,-696
460,335,665
745,655,-394
663,-614,414
865,584,-431
637,-657,-806
-700,627,697
-294,-551,749
--- scanner 10 ---
12,120,-29
-342,-679,-939
530,-282,501
-809,684,-468
489,-308,416
572,350,621
-403,-504,359
577,584,627
403,-417,-730
-66,14,-156
414,664,-519
598,-372,-703
650,-300,397
-506,-519,372
-884,650,-644
-765,612,-643
-394,-441,371
462,-268,-718
-374,674,418
-308,-514,-945
398,839,-455
-344,592,336
363,708,-493
-479,717,344
623,433,604
-432,-593,-859
--- scanner 11 ---
891,-554,-721
-666,637,-444
-607,-541,-415
29,2,4
162,154,-54
-404,-217,531
532,510,-451
713,-606,830
928,-640,-742
-654,574,-435
-681,-584,-520
-609,441,448
-689,-459,-385
-398,-451,536
743,571,-458
921,-592,-831
534,611,565
477,586,-475
781,-478,790
408,661,539
-653,652,-547
759,-664,769
-524,415,489
418,587,560
-421,-300,582
-628,433,609
--- scanner 12 ---
-377,614,-406
-552,-831,694
527,-699,-733
-754,618,829
-411,-958,-621
745,422,-840
509,-816,-841
-425,703,-413
849,417,-798
605,657,787
-689,453,793
-338,-820,700
772,341,-775
649,784,719
670,-589,575
-539,-918,-553
-353,-897,-575
-463,593,-443
463,-695,-855
130,-179,-79
513,-525,626
-634,529,749
703,610,736
45,-46,33
661,-442,574
-523,-753,662
--- scanner 13 ---
909,550,-668
112,-89,-14
-273,384,725
824,456,386
-373,374,601
-793,-530,-506
-594,-593,-472
787,-689,-321
-657,472,-591
717,-801,-298
821,591,-739
-310,482,586
-637,485,-537
493,-740,530
749,-693,-284
724,419,337
474,-627,566
-531,-606,790
-33,37,38
-471,-562,680
501,-647,347
618,425,401
-699,-626,-473
944,544,-718
-498,-447,782
-599,561,-463
--- scanner 14 ---
551,-821,-732
-578,-311,780
558,-481,467
-688,548,475
743,667,230
628,740,282
-667,-431,700
384,-850,-719
-866,523,-640
608,409,-597
-743,-376,744
-94,25,-89
458,-392,550
-849,538,-906
566,-804,-762
506,422,-732
-493,-640,-462
-603,-602,-492
705,701,352
-582,-688,-455
338,-429,479
-873,462,-851
540,416,-719
-626,687,514
-686,729,363
--- scanner 15 ---
-702,-746,443
739,592,-374
597,814,559
763,-590,-410
-670,658,-388
682,788,518
727,655,-365
-335,568,609
-345,642,630
-858,-671,-565
-459,719,-381
-404,643,-405
438,-883,401
-843,-533,-553
701,796,679
-892,-559,-696
409,-907,638
-725,-756,527
-624,-789,383
430,-889,545
726,577,-477
-80,-67,-39
-461,618,635
773,-674,-320
58,23,81
739,-451,-335
--- scanner 16 ---
-755,578,-603
487,691,-754
593,698,-638
20,82,-38
-923,-807,452
382,-791,415
-559,792,522
-834,-736,364
-836,-723,-539
552,845,568
-922,-847,-618
-610,838,475
-739,456,-638
394,-761,-691
678,891,498
358,-716,554
382,-752,-628
738,925,566
565,704,-723
348,-806,638
-667,922,511
-779,551,-493
-916,-813,-453
-153,145,64
-861,-689,357
395,-649,-613
--- scanner 17 ---
664,-630,-754
805,704,436
-431,686,597
-587,-662,388
-631,-532,382
569,439,-825
400,-738,620
-717,-556,-781
-461,651,805
-984,750,-627
423,-822,610
-711,-688,-850
459,413,-747
674,775,343
780,-635,-778
790,-656,-704
-840,657,-649
-53,56,-150
-812,-662,-829
451,333,-808
-112,-109,-68
-936,553,-605
-682,-599,442
708,813,465
501,-882,653
-541,612,682
--- scanner 18 ---
552,283,-704
606,-679,-568
33,2,166
471,-600,469
15,-169,53
-565,601,731
-501,373,-422
-680,574,833
846,372,668
634,306,-684
-534,478,-421
563,-505,400
703,258,-627
-665,-549,-505
-466,-662,726
618,-629,465
552,-774,-505
-521,541,841
-617,-644,-611
-605,-701,664
804,401,618
-686,-650,-479
-424,-813,666
893,472,557
-597,279,-455
546,-730,-635
--- scanner 19 ---
835,-511,-678
895,455,462
650,-578,-640
-505,-664,948
-359,-696,-553
161,-95,23
790,538,-336
749,472,-391
888,283,439
30,-35,133
-384,431,727
-455,-638,822
-462,405,617
820,-434,597
-380,-670,-654
784,307,460
699,385,-327
621,-500,-672
-401,325,609
-428,-564,-559
-560,649,-682
762,-373,643
-604,545,-778
-431,-643,938
634,-413,575
-582,742,-826
--- scanner 20 ---
-600,-782,533
596,577,-670
620,550,-780
-646,690,-519
286,-347,543
763,-848,-655
-541,-753,-645
413,595,320
321,586,395
-752,697,-635
-75,63,-4
677,-724,-734
-518,-812,451
-554,-650,-660
-526,-677,576
-882,525,309
-724,703,-550
-890,466,262
-24,138,-191
461,-391,471
-175,5,-154
462,-272,509
-553,-762,-582
-858,529,400
373,551,219
580,455,-635
757,-814,-820
--- scanner 21 ---
-377,-514,-910
844,-627,-723
759,-599,-717
-445,-454,-944
-402,821,411
929,642,-755
749,-810,733
734,-797,630
812,-798,742
-294,-810,772
-514,787,542
-376,772,554
875,-705,-710
876,388,726
-447,-640,-864
916,507,-653
800,333,702
-433,-834,697
-803,392,-696
-517,-844,787
922,520,-687
-743,443,-600
22,-64,-168
160,4,-35
815,599,719
-799,507,-686
--- scanner 22 ---
817,529,-684
728,823,688
-473,-503,418
-72,102,143
-547,657,684
559,-544,861
-476,801,651
821,-620,-435
699,781,638
-521,742,740
628,870,577
-465,-410,601
526,-703,853
622,-557,-401
-721,-457,-213
727,-524,-402
-528,962,-575
499,-588,751
-580,834,-534
-689,-349,-227
66,6,64
794,611,-664
842,680,-607
-446,-523,637
-522,716,-578
-759,-461,-250
--- scanner 23 ---
-352,-697,-348
432,569,514
-44,-72,111
-406,627,809
-581,-621,391
-473,788,-390
819,-664,596
877,399,-834
566,-655,-584
782,-784,667
366,523,626
473,-689,-636
-378,729,728
-410,628,711
730,-564,679
-673,698,-393
-334,-673,-384
-432,-525,427
361,649,477
-528,-485,420
-310,-680,-500
151,90,68
-518,655,-370
946,312,-770
592,-590,-584
865,328,-773
--- scanner 24 ---
-519,780,-365
634,802,-649
-674,902,562
105,13,-47
-667,878,446
-662,-634,635
-479,684,-358
465,-735,700
-522,-563,-663
704,539,455
430,-788,740
757,-463,-300
-606,-678,-646
677,-509,-447
794,739,452
-657,-639,463
-706,-723,627
735,834,-666
-595,-597,-832
575,-753,793
661,694,432
-62,53,53
633,806,-747
678,-478,-454
-691,860,410
-615,631,-329
--- scanner 25 ---
183,2,-88
656,638,481
894,-843,430
-325,-646,403
-658,-347,-552
619,806,-682
657,512,589
562,876,-811
5,-20,46
-679,-329,-645
-650,-353,-408
-315,-562,313
536,-608,-751
586,835,-891
-759,588,-782
701,481,532
-639,600,608
867,-836,383
-349,-785,328
567,-771,-743
956,-748,312
535,-528,-732
-702,558,533
-554,511,-793
-765,497,-720
-786,627,617
--- scanner 26 ---
568,-612,-772
-494,-703,-776
689,-640,-900
512,-573,342
765,568,275
-699,525,741
508,448,-580
558,403,-627
-458,-494,437
58,44,-10
-695,623,-517
-620,-556,487
-452,-603,-788
-689,415,705
929,541,305
675,-619,407
479,-562,414
-703,347,679
-665,791,-494
-527,-407,477
128,-52,-173
583,511,-591
698,-517,-736
-727,827,-545
-409,-773,-839
841,619,423
--- scanner 27 ---
743,-494,-529
20,142,-6
374,-668,657
-568,-551,-817
374,719,-731
-538,-500,730
-515,-415,-788
-385,-491,820
-427,-538,747
-393,697,-399
411,815,380
-365,799,-480
530,822,418
420,-486,700
-702,757,526
-419,-455,-781
403,-501,670
-626,716,631
-397,698,-497
157,78,120
379,593,-582
523,823,460
445,639,-694
832,-594,-612
-640,682,457
795,-714,-541
2021/day19/test19
-136
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@@ -1,136 +0,0 @@
--- scanner 0 ---
404,-588,-901
528,-643,409
-838,591,734
390,-675,-793
-537,-823,-458
-485,-357,347
-345,-311,381
-661,-816,-575
-876,649,763
-618,-824,-621
553,345,-567
474,580,667
-447,-329,318
-584,868,-557
544,-627,-890
564,392,-477
455,729,728
-892,524,684
-689,845,-530
423,-701,434
7,-33,-71
630,319,-379
443,580,662
-789,900,-551
459,-707,401
--- scanner 1 ---
686,422,578
605,423,415
515,917,-361
-336,658,858
95,138,22
-476,619,847
-340,-569,-846
567,-361,727
-460,603,-452
669,-402,600
729,430,532
-500,-761,534
-322,571,750
-466,-666,-811
-429,-592,574
-355,545,-477
703,-491,-529
-328,-685,520
413,935,-424
-391,539,-444
586,-435,557
-364,-763,-893
807,-499,-711
755,-354,-619
553,889,-390
--- scanner 2 ---
649,640,665
682,-795,504
-784,533,-524
-644,584,-595
-588,-843,648
-30,6,44
-674,560,763
500,723,-460
609,671,-379
-555,-800,653
-675,-892,-343
697,-426,-610
578,704,681
493,664,-388
-671,-858,530
-667,343,800
571,-461,-707
-138,-166,112
-889,563,-600
646,-828,498
640,759,510
-630,509,768
-681,-892,-333
673,-379,-804
-742,-814,-386
577,-820,562
--- scanner 3 ---
-589,542,597
605,-692,669
-500,565,-823
-660,373,557
-458,-679,-417
-488,449,543
-626,468,-788
338,-750,-386
528,-832,-391
562,-778,733
-938,-730,414
543,643,-506
-524,371,-870
407,773,750
-104,29,83
378,-903,-323
-778,-728,485
426,699,580
-438,-605,-362
-469,-447,-387
509,732,623
647,635,-688
-868,-804,481
614,-800,639
595,780,-596
--- scanner 4 ---
727,592,562
-293,-554,779
441,611,-461
-714,465,-776
-743,427,-804
-660,-479,-426
832,-632,460
927,-485,-438
408,393,-506
466,436,-512
110,16,151
-258,-428,682
-393,719,612
-211,-452,876
808,-476,-593
-575,615,604
-485,667,467
-680,325,-822
-627,-443,-432
872,-547,-609
833,512,582
807,604,487
839,-516,451
891,-625,532
-652,-548,-490
30,-46,-14
2021/day20/AoC20.py
-53
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@@ -1,53 +0,0 @@
#!/usr/bin/env python3
import sys
from copy import deepcopy
def imagine(im):
for y in im:
l = ''
for x in y: l += x
print(l)
def n(im,x,y,o):
index = ''
h = [(-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)]
for j,i in h:
if (0 <= x+i < len(im[0]) and 0 <= y+j < len(im)): index += '0' if im[y+j][x+i] == '.' else '1'
else: index += '0' if o == '.' else '1'
return algo[int(index,2)]
def enhance(im, o):
nex = deepcopy(im)
for y in range(len(im)):
for x in range(len(im[0])):
nex[y][x] = n(im,x,y,o)
return nex, '.' if o == '#' or mode == 'k' else '#'
if __name__ == '__main__':
algo, im = open(sys.argv[1]).read().split('\n\n')
algo = list(algo.strip('\n'))
im = [list(i) for i in im.split('\n') if i != '']
print('------------------ start -------------------')
bim = [[im[i-50][j-50] if (49 < i <= len(im)+49 and 49 < j <= len(im[0])+49) else '.' for j in range(len(im[0]) + 100)] for i in range(len(im) + 100)]
imagine(bim)
mode = 's' if algo[0] == '#' else 'k'
nim, o = bim, '.'
for i in range(50):
print('-------------------- ' + str(i) + ' ---------------------')
nim, o = enhance(nim,o)
imagine(nim)
if i == 1: c1 = nim
# challenge 1
res1 = str(sum(sum(1 if i == '#' else 0 for i in j) for j in c1))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
res2 = str(sum(sum(1 if i == '#' else 0 for i in j) for j in nim))
print("challenge 2:" + "\n" + res2 + "\n")
2021/day20/challenge20
-118
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@@ -1,118 +0,0 @@
--- Day 20: Trench Map ---
With the scanners fully deployed, you turn their attention to mapping the floor of the ocean trench.
When you get back the image from the scanners, it seems to just be random noise. Perhaps you can combine an image enhancement algorithm and the input image (your puzzle input) to clean it up a little.
For example:
..#.#..#####.#.#.#.###.##.....###.##.#..###.####..#####..#....#..#..##..##
#..######.###...####..#..#####..##..#.#####...##.#.#..#.##..#.#......#.###
.######.###.####...#.##.##..#..#..#####.....#.#....###..#.##......#.....#.
.#..#..##..#...##.######.####.####.#.#...#.......#..#.#.#...####.##.#.....
.#..#...##.#.##..#...##.#.##..###.#......#.#.......#.#.#.####.###.##...#..
...####.#..#..#.##.#....##..#.####....##...##..#...#......#.#.......#.....
..##..####..#...#.#.#...##..#.#..###..#####........#..####......#..#
#..#.
#....
##..#
..#..
..###
The first section is the image enhancement algorithm. It is normally given on a single line, but it has been wrapped to multiple lines in this example for legibility. The second section is the input image, a two-dimensional grid of light pixels (#) and dark pixels (.).
The image enhancement algorithm describes how to enhance an image by simultaneously converting all pixels in the input image into an output image. Each pixel of the output image is determined by looking at a 3x3 square of pixels centered on the corresponding input image pixel. So, to determine the value of the pixel at (5,10) in the output image, nine pixels from the input image need to be considered: (4,9), (4,10), (4,11), (5,9), (5,10), (5,11), (6,9), (6,10), and (6,11). These nine input pixels are combined into a single binary number that is used as an index in the image enhancement algorithm string.
For example, to determine the output pixel that corresponds to the very middle pixel of the input image, the nine pixels marked by [...] would need to be considered:
# . . # .
#[. . .].
#[# . .]#
.[. # .].
. . # # #
Starting from the top-left and reading across each row, these pixels are ..., then #.., then .#.; combining these forms ...#...#.. By turning dark pixels (.) into 0 and light pixels (#) into 1, the binary number 000100010 can be formed, which is 34 in decimal.
The image enhancement algorithm string is exactly 512 characters long, enough to match every possible 9-bit binary number. The first few characters of the string (numbered starting from zero) are as follows:
0 10 20 30 34 40 50 60 70
| | | | | | | | |
..#.#..#####.#.#.#.###.##.....###.##.#..###.####..#####..#....#..#..##..##
In the middle of this first group of characters, the character at index 34 can be found: #. So, the output pixel in the center of the output image should be #, a light pixel.
This process can then be repeated to calculate every pixel of the output image.
Through advances in imaging technology, the images being operated on here are infinite in size. Every pixel of the infinite output image needs to be calculated exactly based on the relevant pixels of the input image. The small input image you have is only a small region of the actual infinite input image; the rest of the input image consists of dark pixels (.). For the purposes of the example, to save on space, only a portion of the infinite-sized input and output images will be shown.
The starting input image, therefore, looks something like this, with more dark pixels (.) extending forever in every direction not shown here:
...............
...............
...............
...............
...............
.....#..#......
.....#.........
.....##..#.....
.......#.......
.......###.....
...............
...............
...............
...............
...............
By applying the image enhancement algorithm to every pixel simultaneously, the following output image can be obtained:
...............
...............
...............
...............
.....##.##.....
....#..#.#.....
....##.#..#....
....####..#....
.....#..##.....
......##..#....
.......#.#.....
...............
...............
...............
...............
Through further advances in imaging technology, the above output image can also be used as an input image! This allows it to be enhanced a second time:
...............
...............
...............
..........#....
....#..#.#.....
...#.#...###...
...#...##.#....
...#.....#.#...
....#.#####....
.....#.#####...
......##.##....
.......###.....
...............
...............
...............
Truly incredible - now the small details are really starting to come through. After enhancing the original input image twice, 35 pixels are lit.
Start with the original input image and apply the image enhancement algorithm twice, being careful to account for the infinite size of the images. How many pixels are lit in the resulting image?
Your puzzle answer was 5391.
--- Part Two ---
You still can't quite make out the details in the image. Maybe you just didn't enhance it enough.
If you enhance the starting input image in the above example a total of 50 times, 3351 pixels are lit in the final output image.
Start again with the original input image and apply the image enhancement algorithm 50 times. How many pixels are lit in the resulting image?
Your puzzle answer was 16383.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day20/input20
-102
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@@ -1,102 +0,0 @@
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#####.#.##.###.##...#.##.##.#.#....#.##.#.#..#.#.##...##.########..#.#....#####.#######........#..#.
#...###..####.##.#.#####..##.###...###.##..#..##...##.....####...#.......#....#.#.###..#......#..##.
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2021/day20/test20
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..#.#..#####.#.#.#.###.##.....###.##.#..###.####..#####..#....#..#..##..###..######.###...####..#..#####..##..#.#####...##.#.#..#.##..#.#......#.###.######.###.####...#.##.##..#..#..#####.....#.#....###..#.##......#.....#..#..#..##..#...##.######.####.####.#.#...#.......#..#.#.#...####.##.#......#..#...##.#.##..#...##.#.##..###.#......#.#.......#.#.#.####.###.##...#.....####.#..#..#.##.#....##..#.####....##...##..#...#......#.#.......#.......##..####..#...#.#.#...##..#.#..###..#####........#..####......#..#
#..#.
#....
##..#
..#..
..###
2021/day21/AoC21.py
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#!/usr/bin/env python3
import sys, re
def memo(fc):
memo = {}
def f2(a,b,c,d,e,f,g):
if (a,b,c,d,e,f,g) not in memo: memo[(a,b,c,d,e,f,g)] = fc(a,b,c,d,e,f,g)
return memo[(a,b,c,d,e,f,g)]
return f2
def parse(line):
groups = re.compile(r"Player 1 starting position: (?P<p1>\d+)\nPlayer 2 starting position: (?P<p2>\d+)").search(line)
return groups
def step(p1,p2):
d,s1,s2,die,dr = 0,0,0,0,0
t = 1
while s1 < 1000 and s2 < 1000:
mv = d*3 + 6
d = d + 3 % 100
dr += 3
if t == 1:
p1 = (p1 + mv - 1) % 10 + 1
s1 += p1
if t == 2:
p2 = (p2 + mv - 1) % 10 + 1
s2 += p2
t = t%2+1
return dr*s2 if s2 < 1000 else dr*s1
@memo
def qstep(p1,p2,s1,s2,dv,t,dt):
if s1 >= 21:
return {1:1,2:0}
elif s2 >= 21:
return {1:0,2:1}
wins = {1: 0, 2: 0}
if t == 1:
if dt == 3:
nt = t%2+1
np1 = (p1 + dv - 1) % 10 + 1
ns1 = s1+np1
win = qstep(np1,p2,ns1,s2,0,nt,0)
wins[1] += win[1]
wins[2] += win[2]
else:
dt += 1
for d in range(1,4):
ndv = dv+d
win = qstep(p1,p2,s1,s2,ndv,t,dt)
wins[1] += win[1]
wins[2] += win[2]
if t == 2:
if dt == 3:
nt = t%2+1
np2 = (p2 + dv - 1) % 10 + 1
ns2 = s2+np2
win = qstep(p1,np2,s1,ns2,0,nt,0)
wins[1] += win[1]
wins[2] += win[2]
else:
dt += 1
for d in range(1,4):
ndv = dv+d
win = qstep(p1,p2,s1,s2,ndv,t,dt)
wins[1] += win[1]
wins[2] += win[2]
return wins
if __name__ == '__main__':
p1,p2 = map(int,parse(open(sys.argv[1]).read()).groups())
# challenge 1
res1 = str(step(p1,p2))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
results = qstep(p1,p2,0,0,0,1,0)
res2 = str(max(results.values()))
print("challenge 2:" + "\n" + res2 + "\n")
2021/day21/challenge21
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--- Day 21: Dirac Dice ---
There's not much to do as you slowly descend to the bottom of the ocean. The submarine computer challenges you to a nice game of Dirac Dice.
This game consists of a single die, two pawns, and a game board with a circular track containing ten spaces marked 1 through 10 clockwise. Each player's starting space is chosen randomly (your puzzle input). Player 1 goes first.
Players take turns moving. On each player's turn, the player rolls the die three times and adds up the results. Then, the player moves their pawn that many times forward around the track (that is, moving clockwise on spaces in order of increasing value, wrapping back around to 1 after 10). So, if a player is on space 7 and they roll 2, 2, and 1, they would move forward 5 times, to spaces 8, 9, 10, 1, and finally stopping on 2.
After each player moves, they increase their score by the value of the space their pawn stopped on. Players' scores start at 0. So, if the first player starts on space 7 and rolls a total of 5, they would stop on space 2 and add 2 to their score (for a total score of 2). The game immediately ends as a win for any player whose score reaches at least 1000.
Since the first game is a practice game, the submarine opens a compartment labeled deterministic dice and a 100-sided die falls out. This die always rolls 1 first, then 2, then 3, and so on up to 100, after which it starts over at 1 again. Play using this die.
For example, given these starting positions:
Player 1 starting position: 4
Player 2 starting position: 8
This is how the game would go:
Player 1 rolls 1+2+3 and moves to space 10 for a total score of 10.
Player 2 rolls 4+5+6 and moves to space 3 for a total score of 3.
Player 1 rolls 7+8+9 and moves to space 4 for a total score of 14.
Player 2 rolls 10+11+12 and moves to space 6 for a total score of 9.
Player 1 rolls 13+14+15 and moves to space 6 for a total score of 20.
Player 2 rolls 16+17+18 and moves to space 7 for a total score of 16.
Player 1 rolls 19+20+21 and moves to space 6 for a total score of 26.
Player 2 rolls 22+23+24 and moves to space 6 for a total score of 22.
...after many turns...
Player 2 rolls 82+83+84 and moves to space 6 for a total score of 742.
Player 1 rolls 85+86+87 and moves to space 4 for a total score of 990.
Player 2 rolls 88+89+90 and moves to space 3 for a total score of 745.
Player 1 rolls 91+92+93 and moves to space 10 for a final score, 1000.
Since player 1 has at least 1000 points, player 1 wins and the game ends. At this point, the losing player had 745 points and the die had been rolled a total of 993 times; 745 * 993 = 739785.
Play a practice game using the deterministic 100-sided die. The moment either player wins, what do you get if you multiply the score of the losing player by the number of times the die was rolled during the game?
Your puzzle answer was 913560.
--- Part Two ---
Now that you're warmed up, it's time to play the real game.
A second compartment opens, this time labeled Dirac dice. Out of it falls a single three-sided die.
As you experiment with the die, you feel a little strange. An informational brochure in the compartment explains that this is a quantum die: when you roll it, the universe splits into multiple copies, one copy for each possible outcome of the die. In this case, rolling the die always splits the universe into three copies: one where the outcome of the roll was 1, one where it was 2, and one where it was 3.
The game is played the same as before, although to prevent things from getting too far out of hand, the game now ends when either player's score reaches at least 21.
Using the same starting positions as in the example above, player 1 wins in 444356092776315 universes, while player 2 merely wins in 341960390180808 universes.
Using your given starting positions, determine every possible outcome. Find the player that wins in more universes; in how many universes does that player win?
Your puzzle answer was 110271560863819.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day21/input21
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Player 1 starting position: 4
Player 2 starting position: 1
2021/day21/test21
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Player 1 starting position: 4
Player 2 starting position: 8
2021/day22/AoC22.py
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#!/usr/bin/env python3
import sys,re
from collections import Counter
def parse(l):
return re.compile(r"(?P<mode>\w+) x=(?P<x1>-?\d+)..(?P<x2>-?\d+),y=(?P<y1>-?\d+)..(?P<y2>-?\d+),z=(?P<z1>-?\d+)..(?P<z2>-?\d+)").search(l)
def volume(k):
if k[1]-k[0] < 0 or k[3]-k[2] < 0 or k[5]-k[4] < 0: return 0
return (k[1]-k[0]+1)*(k[3]-k[2]+1)*(k[5]-k[4]+1)
def count_cubes(c):
o = 0
for k,v in c.items():
if v == 'on':
o += volume(k)
return o
def intersect(c1,c2):
x1,x2,y1,y2,z1,z2 = c1
lx,hx,ly,hy,lz,hz = c2
xos = max(lx,x1)
xoe = min(hx,x2)
yos = max(ly,y1)
yoe = min(hy,y2)
zos = max(lz,z1)
zoe = min(hz,z2)
return volume((xos,xoe,yos,yoe,zos,zoe)) > 0
#return (c1[0] in range(c2[0],c2[1]+1) or c1[1] in range(c2[0],c2[1]+1)) and (c1[2] in range(c2[2],c2[3]+1) or c1[3] in range(c2[2],c2[3]+1)) and (c1[4] in range(c2[4],c2[5]+1) or c1[5] in range(c2[4],c2[5]+1))
def split(c1,c2):
x1,x2,y1,y2,z1,z2 = c1
lx,hx,ly,hy,lz,hz = c2
xos = max(lx,x1)
xoe = min(hx,x2)
yos = max(ly,y1)
yoe = min(hy,y2)
zos = max(lz,z1)
zoe = min(hz,z2)
v = 0
for x in [(lx,xos-1),(xos,xoe),(xoe+1,hx)]:
for y in [(ly,yos-1),(yos,yoe),(yoe+1,hy)]:
for z in [(lz,zos-1),(zos,zoe),(zoe+1,hz)]:
v += volume((*x,*y,*z))
if volume((*x,*y,*z)) > 0 and (x,y,z) != ((xos,xoe),(yos,yoe),(zos,zoe)):
yield (*x,*y,*z)
def init(steps):
cubes = {}
for m,x1,x2,y1,y2,z1,z2 in steps:
if not (-50 <= x1 <= 50 and -50 <= x2 <= 50): continue
for x in range(x1,x2+1):
if not (-50 <= y1 <= 50 and -50 <= y2 <= 50): continue
for y in range(y1,y2+1):
if not (-50 <= z1 <= 50 and -50 <= z2 <= 50): continue
for z in range(z1,z2+1):
cubes[(x,y,z)] = m
return cubes
def reboot(steps):
cubes = {}
for c1 in steps:
m,c1 = c1[0],c1[1:]
atm = [k for k in cubes.keys()]
for c2 in atm:
om = cubes[c2]
if intersect(c1,c2) or intersect(c2,c1):
for c in split(c1,c2):
if not (intersect(c,c1) or intersect(c1,c)):
cubes[c] = om
cubes.pop(c2)
cubes[c1] = m
return cubes
if __name__ == '__main__':
nums = map(lambda x: (x[0],int(x[1]),int(x[2]),int(x[3]),int(x[4]),int(x[5]),int(x[6])),[parse(l).groups() for l in open(sys.argv[1])])
nums = [n for n in nums]
# challenge 1
cubes1 = init(nums)
c = Counter([v for k,v in cubes1.items() if -50 <= k[0] <= 50 and -50 <= k[1] <= 50 and -50 <= k[2] <= 50])
res1 = str(c.most_common(2))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
cubes2 = reboot(nums)
res2 = str(count_cubes(cubes2))
print("challenge 2:" + "\n" + res2 + "\n")
2021/day22/challenge22
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--- Day 22: Reactor Reboot ---
Operating at these extreme ocean depths has overloaded the submarine's reactor; it needs to be rebooted.
The reactor core is made up of a large 3-dimensional grid made up entirely of cubes, one cube per integer 3-dimensional coordinate (x,y,z). Each cube can be either on or off; at the start of the reboot process, they are all off. (Could it be an old model of a reactor you've seen before?)
To reboot the reactor, you just need to set all of the cubes to either on or off by following a list of reboot steps (your puzzle input). Each step specifies a cuboid (the set of all cubes that have coordinates which fall within ranges for x, y, and z) and whether to turn all of the cubes in that cuboid on or off.
For example, given these reboot steps:
on x=10..12,y=10..12,z=10..12
on x=11..13,y=11..13,z=11..13
off x=9..11,y=9..11,z=9..11
on x=10..10,y=10..10,z=10..10
The first step (on x=10..12,y=10..12,z=10..12) turns on a 3x3x3 cuboid consisting of 27 cubes:
10,10,10
10,10,11
10,10,12
10,11,10
10,11,11
10,11,12
10,12,10
10,12,11
10,12,12
11,10,10
11,10,11
11,10,12
11,11,10
11,11,11
11,11,12
11,12,10
11,12,11
11,12,12
12,10,10
12,10,11
12,10,12
12,11,10
12,11,11
12,11,12
12,12,10
12,12,11
12,12,12
The second step (on x=11..13,y=11..13,z=11..13) turns on a 3x3x3 cuboid that overlaps with the first. As a result, only 19 additional cubes turn on; the rest are already on from the previous step:
11,11,13
11,12,13
11,13,11
11,13,12
11,13,13
12,11,13
12,12,13
12,13,11
12,13,12
12,13,13
13,11,11
13,11,12
13,11,13
13,12,11
13,12,12
13,12,13
13,13,11
13,13,12
13,13,13
The third step (off x=9..11,y=9..11,z=9..11) turns off a 3x3x3 cuboid that overlaps partially with some cubes that are on, ultimately turning off 8 cubes:
10,10,10
10,10,11
10,11,10
10,11,11
11,10,10
11,10,11
11,11,10
11,11,11
The final step (on x=10..10,y=10..10,z=10..10) turns on a single cube, 10,10,10. After this last step, 39 cubes are on.
The initialization procedure only uses cubes that have x, y, and z positions of at least -50 and at most 50. For now, ignore cubes outside this region.
Here is a larger example:
on x=-20..26,y=-36..17,z=-47..7
on x=-20..33,y=-21..23,z=-26..28
on x=-22..28,y=-29..23,z=-38..16
on x=-46..7,y=-6..46,z=-50..-1
on x=-49..1,y=-3..46,z=-24..28
on x=2..47,y=-22..22,z=-23..27
on x=-27..23,y=-28..26,z=-21..29
on x=-39..5,y=-6..47,z=-3..44
on x=-30..21,y=-8..43,z=-13..34
on x=-22..26,y=-27..20,z=-29..19
off x=-48..-32,y=26..41,z=-47..-37
on x=-12..35,y=6..50,z=-50..-2
off x=-48..-32,y=-32..-16,z=-15..-5
on x=-18..26,y=-33..15,z=-7..46
off x=-40..-22,y=-38..-28,z=23..41
on x=-16..35,y=-41..10,z=-47..6
off x=-32..-23,y=11..30,z=-14..3
on x=-49..-5,y=-3..45,z=-29..18
off x=18..30,y=-20..-8,z=-3..13
on x=-41..9,y=-7..43,z=-33..15
on x=-54112..-39298,y=-85059..-49293,z=-27449..7877
on x=967..23432,y=45373..81175,z=27513..53682
The last two steps are fully outside the initialization procedure area; all other steps are fully within it. After executing these steps in the initialization procedure region, 590784 cubes are on.
Execute the reboot steps. Afterward, considering only cubes in the region x=-50..50,y=-50..50,z=-50..50, how many cubes are on?
Your puzzle answer was 580810.
--- Part Two ---
Now that the initialization procedure is complete, you can reboot the reactor.
Starting with all cubes off, run all of the reboot steps for all cubes in the reactor.
Consider the following reboot steps:
on x=-5..47,y=-31..22,z=-19..33
on x=-44..5,y=-27..21,z=-14..35
on x=-49..-1,y=-11..42,z=-10..38
on x=-20..34,y=-40..6,z=-44..1
off x=26..39,y=40..50,z=-2..11
on x=-41..5,y=-41..6,z=-36..8
off x=-43..-33,y=-45..-28,z=7..25
on x=-33..15,y=-32..19,z=-34..11
off x=35..47,y=-46..-34,z=-11..5
on x=-14..36,y=-6..44,z=-16..29
on x=-57795..-6158,y=29564..72030,z=20435..90618
on x=36731..105352,y=-21140..28532,z=16094..90401
on x=30999..107136,y=-53464..15513,z=8553..71215
on x=13528..83982,y=-99403..-27377,z=-24141..23996
on x=-72682..-12347,y=18159..111354,z=7391..80950
on x=-1060..80757,y=-65301..-20884,z=-103788..-16709
on x=-83015..-9461,y=-72160..-8347,z=-81239..-26856
on x=-52752..22273,y=-49450..9096,z=54442..119054
on x=-29982..40483,y=-108474..-28371,z=-24328..38471
on x=-4958..62750,y=40422..118853,z=-7672..65583
on x=55694..108686,y=-43367..46958,z=-26781..48729
on x=-98497..-18186,y=-63569..3412,z=1232..88485
on x=-726..56291,y=-62629..13224,z=18033..85226
on x=-110886..-34664,y=-81338..-8658,z=8914..63723
on x=-55829..24974,y=-16897..54165,z=-121762..-28058
on x=-65152..-11147,y=22489..91432,z=-58782..1780
on x=-120100..-32970,y=-46592..27473,z=-11695..61039
on x=-18631..37533,y=-124565..-50804,z=-35667..28308
on x=-57817..18248,y=49321..117703,z=5745..55881
on x=14781..98692,y=-1341..70827,z=15753..70151
on x=-34419..55919,y=-19626..40991,z=39015..114138
on x=-60785..11593,y=-56135..2999,z=-95368..-26915
on x=-32178..58085,y=17647..101866,z=-91405..-8878
on x=-53655..12091,y=50097..105568,z=-75335..-4862
on x=-111166..-40997,y=-71714..2688,z=5609..50954
on x=-16602..70118,y=-98693..-44401,z=5197..76897
on x=16383..101554,y=4615..83635,z=-44907..18747
off x=-95822..-15171,y=-19987..48940,z=10804..104439
on x=-89813..-14614,y=16069..88491,z=-3297..45228
on x=41075..99376,y=-20427..49978,z=-52012..13762
on x=-21330..50085,y=-17944..62733,z=-112280..-30197
on x=-16478..35915,y=36008..118594,z=-7885..47086
off x=-98156..-27851,y=-49952..43171,z=-99005..-8456
off x=2032..69770,y=-71013..4824,z=7471..94418
on x=43670..120875,y=-42068..12382,z=-24787..38892
off x=37514..111226,y=-45862..25743,z=-16714..54663
off x=25699..97951,y=-30668..59918,z=-15349..69697
off x=-44271..17935,y=-9516..60759,z=49131..112598
on x=-61695..-5813,y=40978..94975,z=8655..80240
off x=-101086..-9439,y=-7088..67543,z=33935..83858
off x=18020..114017,y=-48931..32606,z=21474..89843
off x=-77139..10506,y=-89994..-18797,z=-80..59318
off x=8476..79288,y=-75520..11602,z=-96624..-24783
on x=-47488..-1262,y=24338..100707,z=16292..72967
off x=-84341..13987,y=2429..92914,z=-90671..-1318
off x=-37810..49457,y=-71013..-7894,z=-105357..-13188
off x=-27365..46395,y=31009..98017,z=15428..76570
off x=-70369..-16548,y=22648..78696,z=-1892..86821
on x=-53470..21291,y=-120233..-33476,z=-44150..38147
off x=-93533..-4276,y=-16170..68771,z=-104985..-24507
After running the above reboot steps, 2758514936282235 cubes are on. (Just for fun, 474140 of those are also in the initialization procedure region.)
Starting again with all cubes off, execute all reboot steps. Afterward, considering all cubes, how many cubes are on?
Your puzzle answer was 1265621119006734.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day22/input22
-420
View File
@@ -1,420 +0,0 @@
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on x=-39178..-2942,y=61606..79222,z=-12921..17911
on x=-27409..-12810,y=-52323..-14724,z=70950..84284
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on x=-21207..4193,y=-7234..-2286,z=-85723..-60361
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on x=-65712..-61563,y=-49387..-24770,z=-44808..-18999
on x=-69404..-48271,y=-68838..-43356,z=-7999..13427
on x=-14554..-6184,y=67577..76141,z=25618..45348
off x=17080..39490,y=54822..76095,z=-33504..-20429
off x=72126..88153,y=-25075..-2924,z=-35655..-21906
off x=-9585..24318,y=-79462..-62167,z=36467..45387
on x=50711..62365,y=8410..24520,z=44319..69714
off x=26163..48176,y=49270..76862,z=25486..40214
off x=-66006..-47007,y=26306..40268,z=-57257..-44008
on x=-18103..978,y=-66402..-53423,z=-58187..-49950
off x=-82231..-53929,y=-30302..-2792,z=-50350..-33862
off x=-80773..-50142,y=26085..50767,z=-37036..-5317
on x=-73412..-50645,y=-37254..-15738,z=27795..47197
on x=53256..56693,y=-27500..-12752,z=39288..60009
on x=9678..25000,y=9814..20035,z=-80762..-69083
off x=75905..77562,y=-35226..-12588,z=-23403..2007
on x=-74228..-51192,y=-60471..-47282,z=-18153..11152
on x=-88144..-65383,y=27947..50190,z=-22007..7068
on x=-57769..-30200,y=-74568..-46266,z=1997..7401
on x=-53231..-25606,y=-1090..21355,z=57688..84914
off x=3894..23908,y=-12364..22961,z=-89732..-71854
on x=-29012..-9560,y=-78082..-62997,z=-29218..-8539
off x=-35760..-28704,y=-69703..-59299,z=26665..59052
on x=-9820..-311,y=-50467..-30267,z=45150..77323
off x=-34746..-29058,y=53504..84718,z=15584..24656
off x=-80685..-48403,y=-19777..4826,z=33281..66712
on x=67395..77602,y=-46450..-19661,z=-31973..-18038
on x=-70033..-64759,y=19552..51911,z=11121..35381
off x=-73590..-42942,y=-46390..-22759,z=40874..56566
off x=-36575..-18344,y=61042..86630,z=6307..26426
off x=-46370..-27813,y=23975..41903,z=54255..72840
2021/day22/test22A
-4
View File
@@ -1,4 +0,0 @@
on x=10..12,y=10..12,z=10..12
on x=11..13,y=11..13,z=11..13
off x=9..11,y=9..11,z=9..11
on x=10..10,y=10..10,z=10..10
2021/day22/test22B
-22
View File
@@ -1,22 +0,0 @@
on x=-20..26,y=-36..17,z=-47..7
on x=-20..33,y=-21..23,z=-26..28
on x=-22..28,y=-29..23,z=-38..16
on x=-46..7,y=-6..46,z=-50..-1
on x=-49..1,y=-3..46,z=-24..28
on x=2..47,y=-22..22,z=-23..27
on x=-27..23,y=-28..26,z=-21..29
on x=-39..5,y=-6..47,z=-3..44
on x=-30..21,y=-8..43,z=-13..34
on x=-22..26,y=-27..20,z=-29..19
off x=-48..-32,y=26..41,z=-47..-37
on x=-12..35,y=6..50,z=-50..-2
off x=-48..-32,y=-32..-16,z=-15..-5
on x=-18..26,y=-33..15,z=-7..46
off x=-40..-22,y=-38..-28,z=23..41
on x=-16..35,y=-41..10,z=-47..6
off x=-32..-23,y=11..30,z=-14..3
on x=-49..-5,y=-3..45,z=-29..18
off x=18..30,y=-20..-8,z=-3..13
on x=-41..9,y=-7..43,z=-33..15
on x=-54112..-39298,y=-85059..-49293,z=-27449..7877
on x=967..23432,y=45373..81175,z=27513..53682
2021/day22/test22C
-60
View File
@@ -1,60 +0,0 @@
on x=-5..47,y=-31..22,z=-19..33
on x=-44..5,y=-27..21,z=-14..35
on x=-49..-1,y=-11..42,z=-10..38
on x=-20..34,y=-40..6,z=-44..1
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off x=-43..-33,y=-45..-28,z=7..25
on x=-33..15,y=-32..19,z=-34..11
off x=35..47,y=-46..-34,z=-11..5
on x=-14..36,y=-6..44,z=-16..29
on x=-57795..-6158,y=29564..72030,z=20435..90618
on x=36731..105352,y=-21140..28532,z=16094..90401
on x=30999..107136,y=-53464..15513,z=8553..71215
on x=13528..83982,y=-99403..-27377,z=-24141..23996
on x=-72682..-12347,y=18159..111354,z=7391..80950
on x=-1060..80757,y=-65301..-20884,z=-103788..-16709
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on x=-29982..40483,y=-108474..-28371,z=-24328..38471
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on x=-98497..-18186,y=-63569..3412,z=1232..88485
on x=-726..56291,y=-62629..13224,z=18033..85226
on x=-110886..-34664,y=-81338..-8658,z=8914..63723
on x=-55829..24974,y=-16897..54165,z=-121762..-28058
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on x=-57817..18248,y=49321..117703,z=5745..55881
on x=14781..98692,y=-1341..70827,z=15753..70151
on x=-34419..55919,y=-19626..40991,z=39015..114138
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on x=-32178..58085,y=17647..101866,z=-91405..-8878
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on x=-111166..-40997,y=-71714..2688,z=5609..50954
on x=-16602..70118,y=-98693..-44401,z=5197..76897
on x=16383..101554,y=4615..83635,z=-44907..18747
off x=-95822..-15171,y=-19987..48940,z=10804..104439
on x=-89813..-14614,y=16069..88491,z=-3297..45228
on x=41075..99376,y=-20427..49978,z=-52012..13762
on x=-21330..50085,y=-17944..62733,z=-112280..-30197
on x=-16478..35915,y=36008..118594,z=-7885..47086
off x=-98156..-27851,y=-49952..43171,z=-99005..-8456
off x=2032..69770,y=-71013..4824,z=7471..94418
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off x=37514..111226,y=-45862..25743,z=-16714..54663
off x=25699..97951,y=-30668..59918,z=-15349..69697
off x=-44271..17935,y=-9516..60759,z=49131..112598
on x=-61695..-5813,y=40978..94975,z=8655..80240
off x=-101086..-9439,y=-7088..67543,z=33935..83858
off x=18020..114017,y=-48931..32606,z=21474..89843
off x=-77139..10506,y=-89994..-18797,z=-80..59318
off x=8476..79288,y=-75520..11602,z=-96624..-24783
on x=-47488..-1262,y=24338..100707,z=16292..72967
off x=-84341..13987,y=2429..92914,z=-90671..-1318
off x=-37810..49457,y=-71013..-7894,z=-105357..-13188
off x=-27365..46395,y=31009..98017,z=15428..76570
off x=-70369..-16548,y=22648..78696,z=-1892..86821
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off x=-93533..-4276,y=-16170..68771,z=-104985..-24507
2021/day23/AoC23.py
-109
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@@ -1,109 +0,0 @@
#!/usr/bin/env python3
import sys
from functools import cache
@cache
def path(f1,f2):
path = []
y, x = f1
while y != 0:
y -= 1
path.append((y,x))
while x != f2[1]:
x += 1 if f2[1] > x else -1
path.append((y,x))
while y != f2[0]:
y += 1
path.append((y,x))
assert path[-1] == f2
return path
def pathvalid(path,board):
for p in path:
if p in board: return False
return True
def cost(a,w):
if a == 'A': return len(w) * 1
if a == 'B': return len(w) * 10
if a == 'C': return len(w) * 100
if a == 'D': return len(w) * 1000
def endstate(board):
for f,a in board.items():
if f[1] != target[a]: return False
return True
def target_position(f,b,lowest):
a = b[f]
if f[1] != target[a]: return False
for y in range(f[0],lowest+1):
if (y,target[a]) not in b: return False
if b[(y,target[a])] != a: return False
return True
def totarget(board,lowest):
change = False
costs = 0
nb = {k: v for k,v in board.items()}
for f,a in board.items():
if target_position(f,nb,lowest): continue
t = (lowest,target[a])
while t in nb and nb[t] == a:
t = (t[0]-1,target[a])
if t == (0,target[a]): continue
p = path(f,t)
if pathvalid(p,nb):
change = True
nb.pop(f)
nb[t] = a
costs += cost(a,p)
return nb,costs,change
def dfs(board,lowest,costs=0,curmin=sys.maxsize):
ch = True
nb = {k: v for k,v in board.items()}
while ch:
nb,co,ch = totarget(nb,lowest)
costs += co
if endstate(nb):
return costs
if costs > curmin: return None
board = {k: v for k,v in nb.items()}
for f,a in board.items():
if f[0] != 0 and not(target_position(f,board,lowest)):
for f2 in hallway:
p = path(f,f2)
if pathvalid(p, board):
cb = {k: v for k,v in board.items()}
cb.pop(f)
cb[f2] = a
nc = dfs(cb,lowest,costs+cost(a,p),curmin)
if nc is not None and nc < curmin: curmin = nc
if curmin < sys.maxsize: return curmin
if __name__ == '__main__':
# challenge 1
lines = open(sys.argv[1]).readlines()
start = {}
for y,l in enumerate(lines[1:]):
for x,f in enumerate(l[1:]):
if f in 'ABCD': start[(y,x)] = f
hallway = [(0,0), (0,1), (0,3), (0,5), (0,7), (0,9), (0,10)]
target = {'A': 2, 'B': 4, 'C': 6, 'D': 8}
res1 = str(dfs(start,2))
print("challenge 1:" + "\n" + res1 + "\n")
# challenge 2
lines = open(sys.argv[2]).readlines()
start = {}
for y,l in enumerate(lines[1:]):
for x,f in enumerate(l[1:]):
if f in 'ABCD': start[(y,x)] = f
res2 = str(dfs(start,4))
print("challenge 2:" + "\n" + res2 + "\n")
2021/day23/challenge23
-329
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@@ -1,329 +0,0 @@
--- Day 23: Amphipod ---
A group of amphipods notice your fancy submarine and flag you down. "With such an impressive shell," one amphipod says, "surely you can help us with a question that has stumped our best scientists."
They go on to explain that a group of timid, stubborn amphipods live in a nearby burrow. Four types of amphipods live there: Amber (A), Bronze (B), Copper (C), and Desert (D). They live in a burrow that consists of a hallway and four side rooms. The side rooms are initially full of amphipods, and the hallway is initially empty.
They give you a diagram of the situation (your puzzle input), including locations of each amphipod (A, B, C, or D, each of which is occupying an otherwise open space), walls (#), and open space (.).
For example:
#############
#...........#
###B#C#B#D###
#A#D#C#A#
#########
The amphipods would like a method to organize every amphipod into side rooms so that each side room contains one type of amphipod and the types are sorted A-D going left to right, like this:
#############
#...........#
###A#B#C#D###
#A#B#C#D#
#########
Amphipods can move up, down, left, or right so long as they are moving into an unoccupied open space. Each type of amphipod requires a different amount of energy to move one step: Amber amphipods require 1 energy per step, Bronze amphipods require 10 energy, Copper amphipods require 100, and Desert ones require 1000. The amphipods would like you to find a way to organize the amphipods that requires the least total energy.
However, because they are timid and stubborn, the amphipods have some extra rules:
Amphipods will never stop on the space immediately outside any room. They can move into that space so long as they immediately continue moving. (Specifically, this refers to the four open spaces in the hallway that are directly above an amphipod starting position.)
Amphipods will never move from the hallway into a room unless that room is their destination room and that room contains no amphipods which do not also have that room as their own destination. If an amphipod's starting room is not its destination room, it can stay in that room until it leaves the room. (For example, an Amber amphipod will not move from the hallway into the right three rooms, and will only move into the leftmost room if that room is empty or if it only contains other Amber amphipods.)
Once an amphipod stops moving in the hallway, it will stay in that spot until it can move into a room. (That is, once any amphipod starts moving, any other amphipods currently in the hallway are locked in place and will not move again until they can move fully into a room.)
In the above example, the amphipods can be organized using a minimum of 12521 energy. One way to do this is shown below.
Starting configuration:
#############
#...........#
###B#C#B#D###
#A#D#C#A#
#########
One Bronze amphipod moves into the hallway, taking 4 steps and using 40 energy:
#############
#...B.......#
###B#C#.#D###
#A#D#C#A#
#########
The only Copper amphipod not in its side room moves there, taking 4 steps and using 400 energy:
#############
#...B.......#
###B#.#C#D###
#A#D#C#A#
#########
A Desert amphipod moves out of the way, taking 3 steps and using 3000 energy, and then the Bronze amphipod takes its place, taking 3 steps and using 30 energy:
#############
#.....D.....#
###B#.#C#D###
#A#B#C#A#
#########
The leftmost Bronze amphipod moves to its room using 40 energy:
#############
#.....D.....#
###.#B#C#D###
#A#B#C#A#
#########
Both amphipods in the rightmost room move into the hallway, using 2003 energy in total:
#############
#.....D.D.A.#
###.#B#C#.###
#A#B#C#.#
#########
Both Desert amphipods move into the rightmost room using 7000 energy:
#############
#.........A.#
###.#B#C#D###
#A#B#C#D#
#########
Finally, the last Amber amphipod moves into its room, using 8 energy:
#############
#...........#
###A#B#C#D###
#A#B#C#D#
#########
What is the least energy required to organize the amphipods?
Your puzzle answer was 18195.
--- Part Two ---
As you prepare to give the amphipods your solution, you notice that the diagram they handed you was actually folded up. As you unfold it, you discover an extra part of the diagram.
Between the first and second lines of text that contain amphipod starting positions, insert the following lines:
#D#C#B#A#
#D#B#A#C#
So, the above example now becomes:
#############
#...........#
###B#C#B#D###
#D#C#B#A#
#D#B#A#C#
#A#D#C#A#
#########
The amphipods still want to be organized into rooms similar to before:
#############
#...........#
###A#B#C#D###
#A#B#C#D#
#A#B#C#D#
#A#B#C#D#
#########
In this updated example, the least energy required to organize these amphipods is 44169:
#############
#...........#
###B#C#B#D###
#D#C#B#A#
#D#B#A#C#
#A#D#C#A#
#########
#############
#..........D#
###B#C#B#.###
#D#C#B#A#
#D#B#A#C#
#A#D#C#A#
#########
#############
#A.........D#
###B#C#B#.###
#D#C#B#.#
#D#B#A#C#
#A#D#C#A#
#########
#############
#A........BD#
###B#C#.#.###
#D#C#B#.#
#D#B#A#C#
#A#D#C#A#
#########
#############
#A......B.BD#
###B#C#.#.###
#D#C#.#.#
#D#B#A#C#
#A#D#C#A#
#########
#############
#AA.....B.BD#
###B#C#.#.###
#D#C#.#.#
#D#B#.#C#
#A#D#C#A#
#########
#############
#AA.....B.BD#
###B#.#.#.###
#D#C#.#.#
#D#B#C#C#
#A#D#C#A#
#########
#############
#AA.....B.BD#
###B#.#.#.###
#D#.#C#.#
#D#B#C#C#
#A#D#C#A#
#########
#############
#AA...B.B.BD#
###B#.#.#.###
#D#.#C#.#
#D#.#C#C#
#A#D#C#A#
#########
#############
#AA.D.B.B.BD#
###B#.#.#.###
#D#.#C#.#
#D#.#C#C#
#A#.#C#A#
#########
#############
#AA.D...B.BD#
###B#.#.#.###
#D#.#C#.#
#D#.#C#C#
#A#B#C#A#
#########
#############
#AA.D.....BD#
###B#.#.#.###
#D#.#C#.#
#D#B#C#C#
#A#B#C#A#
#########
#############
#AA.D......D#
###B#.#.#.###
#D#B#C#.#
#D#B#C#C#
#A#B#C#A#
#########
#############
#AA.D......D#
###B#.#C#.###
#D#B#C#.#
#D#B#C#.#
#A#B#C#A#
#########
#############
#AA.D.....AD#
###B#.#C#.###
#D#B#C#.#
#D#B#C#.#
#A#B#C#.#
#########
#############
#AA.......AD#
###B#.#C#.###
#D#B#C#.#
#D#B#C#.#
#A#B#C#D#
#########
#############
#AA.......AD#
###.#B#C#.###
#D#B#C#.#
#D#B#C#.#
#A#B#C#D#
#########
#############
#AA.......AD#
###.#B#C#.###
#.#B#C#.#
#D#B#C#D#
#A#B#C#D#
#########
#############
#AA.D.....AD#
###.#B#C#.###
#.#B#C#.#
#.#B#C#D#
#A#B#C#D#
#########
#############
#A..D.....AD#
###.#B#C#.###
#.#B#C#.#
#A#B#C#D#
#A#B#C#D#
#########
#############
#...D.....AD#
###.#B#C#.###
#A#B#C#.#
#A#B#C#D#
#A#B#C#D#
#########
#############
#.........AD#
###.#B#C#.###
#A#B#C#D#
#A#B#C#D#
#A#B#C#D#
#########
#############
#..........D#
###A#B#C#.###
#A#B#C#D#
#A#B#C#D#
#A#B#C#D#
#########
#############
#...........#
###A#B#C#D###
#A#B#C#D#
#A#B#C#D#
#A#B#C#D#
#########
Using the initial configuration from the full diagram, what is the least energy required to organize the amphipods?
Your puzzle answer was 50265.
Both parts of this puzzle are complete! They provide two gold stars: **
2021/day23/input23
-5
View File
@@ -1,5 +0,0 @@
#############
#...........#
###A#C#B#A###
#D#D#B#C#
#########
2021/day23/input23-2
-7
View File
@@ -1,7 +0,0 @@
#############
#...........#
###A#C#B#A###
#D#C#B#A#
#D#B#A#C#
#D#D#B#C#
#########
2021/day23/test23
-5
View File
@@ -1,5 +0,0 @@
#############
#...........#
###B#C#B#D###
#A#D#C#A#
#########
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