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--- Day 17: Clumsy Crucible ---
The lava starts flowing rapidly once the Lava Production Facility is operational. As you leave, the reindeer offers you a parachute, allowing you to quickly reach Gear Island.
As you descend, your bird's-eye view of Gear Island reveals why you had trouble finding anyone on your way up: half of Gear Island is empty, but the half below you is a giant factory city!
You land near the gradually-filling pool of lava at the base of your new lavafall. Lavaducts will eventually carry the lava throughout the city, but to make use of it immediately, Elves are loading it into large crucibles on wheels.
The crucibles are top-heavy and pushed by hand. Unfortunately, the crucibles become very difficult to steer at high speeds, and so it can be hard to go in a straight line for very long.
To get Desert Island the machine parts it needs as soon as possible, you'll need to find the best way to get the crucible from the lava pool to the machine parts factory. To do this, you need to minimize heat loss while choosing a route that doesn't require the crucible to go in a straight line for too long.
Fortunately, the Elves here have a map (your puzzle input) that uses traffic patterns, ambient temperature, and hundreds of other parameters to calculate exactly how much heat loss can be expected for a crucible entering any particular city block.
For example:
2413432311323
3215453535623
3255245654254
3446585845452
4546657867536
1438598798454
4457876987766
3637877979653
4654967986887
4564679986453
1224686865563
2546548887735
4322674655533
Each city block is marked by a single digit that represents the amount of heat loss if the crucible enters that block. The starting point, the lava pool, is the top-left city block; the destination, the machine parts factory, is the bottom-right city block. (Because you already start in the top-left block, you don't incur that block's heat loss unless you leave that block and then return to it.)
Because it is difficult to keep the top-heavy crucible going in a straight line for very long, it can move at most three blocks in a single direction before it must turn 90 degrees left or right. The crucible also can't reverse direction; after entering each city block, it may only turn left, continue straight, or turn right.
One way to minimize heat loss is this path:
2>>34^>>>1323
32v>>>35v5623
32552456v>>54
3446585845v52
4546657867v>6
14385987984v4
44578769877v6
36378779796v>
465496798688v
456467998645v
12246868655<v
25465488877v5
43226746555v>
This path never moves more than three consecutive blocks in the same direction and incurs a heat loss of only 102.
Directing the crucible from the lava pool to the machine parts factory, but not moving more than three consecutive blocks in the same direction, what is the least heat loss it can incur?
Your puzzle answer was 928.
--- Part Two ---
The crucibles of lava simply aren't large enough to provide an adequate supply of lava to the machine parts factory. Instead, the Elves are going to upgrade to ultra crucibles.
Ultra crucibles are even more difficult to steer than normal crucibles. Not only do they have trouble going in a straight line, but they also have trouble turning!
Once an ultra crucible starts moving in a direction, it needs to move a minimum of four blocks in that direction before it can turn (or even before it can stop at the end). However, it will eventually start to get wobbly: an ultra crucible can move a maximum of ten consecutive blocks without turning.
In the above example, an ultra crucible could follow this path to minimize heat loss:
2>>>>>>>>1323
32154535v5623
32552456v4254
34465858v5452
45466578v>>>>
143859879845v
445787698776v
363787797965v
465496798688v
456467998645v
122468686556v
254654888773v
432267465553v
In the above example, an ultra crucible would incur the minimum possible heat loss of 94.
Here's another example:
111111111111
999999999991
999999999991
999999999991
999999999991
Sadly, an ultra crucible would need to take an unfortunate path like this one:
1>>>>>>>1111
9999999v9991
9999999v9991
9999999v9991
9999999v>>>>
This route causes the ultra crucible to incur the minimum possible heat loss of 71.
Directing the ultra crucible from the lava pool to the machine parts factory, what is the least heat loss it can incur?
| 100
|
--- Day 20: Race Condition ---
The Historians are quite pixelated again. This time, a massive, black building looms over you - you're right outside the CPU!
While The Historians get to work, a nearby program sees that you're idle and challenges you to a race. Apparently, you've arrived just in time for the frequently-held race condition festival!
The race takes place on a particularly long and twisting code path; programs compete to see who can finish in the fewest picoseconds. The winner even gets their very own mutex!
They hand you a map of the racetrack (your puzzle input). For example:
###############
#...#...#.....#
#.#.#.#.#.###.#
#S#...#.#.#...#
#######.#.#.###
#######.#.#...#
#######.#.###.#
###..E#...#...#
###.#######.###
#...###...#...#
#.#####.#.###.#
#.#...#.#.#...#
#.#.#.#.#.#.###
#...#...#...###
###############
The map consists of track (.) - including the start (S) and end (E) positions (both of which also count as track) - and walls (#).
When a program runs through the racetrack, it starts at the start position. Then, it is allowed to move up, down, left, or right; each such move takes 1 picosecond. The goal is to reach the end position as quickly as possible. In this example racetrack, the fastest time is 84 picoseconds.
Because there is only a single path from the start to the end and the programs all go the same speed, the races used to be pretty boring. To make things more interesting, they introduced a new rule to the races: programs are allowed to cheat.
The rules for cheating are very strict. Exactly once during a race, a program may disable collision for up to 2 picoseconds. This allows the program to pass through walls as if they were regular track. At the end of the cheat, the program must be back on normal track again; otherwise, it will receive a segmentation fault and get disqualified.
So, a program could complete the course in 72 picoseconds (saving 12 picoseconds) by cheating for the two moves marked 1 and 2:
###############
#...#...12....#
#.#.#.#.#.###.#
#S#...#.#.#...#
#######.#.#.###
#######.#.#...#
#######.#.###.#
###..E#...#...#
###.#######.###
#...###...#...#
#.#####.#.###.#
#.#...#.#.#...#
#.#.#.#.#.#.###
#...#...#...###
###############
Or, a program could complete the course in 64 picoseconds (saving 20 picoseconds) by cheating for the two moves marked 1 and 2:
###############
#...#...#.....#
#.#.#.#.#.###.#
#S#...#.#.#...#
#######.#.#.###
#######.#.#...#
#######.#.###.#
###..E#...12..#
###.#######.###
#...###...#...#
#.#####.#.###.#
#.#...#.#.#...#
#.#.#.#.#.#.###
#...#...#...###
###############
This cheat saves 38 picoseconds:
###############
#...#...#.....#
#.#.#.#.#.###.#
#S#...#.#.#...#
#######.#.#.###
#######.#.#...#
#######.#.###.#
###..E#...#...#
###.####1##.###
#...###.2.#...#
#.#####.#.###.#
#.#...#.#.#...#
#.#.#.#.#.#.###
#...#...#...###
###############
This cheat saves 64 picoseconds and takes the program directly to the end:
###############
#...#...#.....#
#.#.#.#.#.###.#
#S#...#.#.#...#
#######.#.#.###
#######.#.#...#
#######.#.###.#
###..21...#...#
###.#######.###
#...###...#...#
#.#####.#.###.#
#.#...#.#.#...#
#.#.#.#.#.#.###
#...#...#...###
###############
Each cheat has a distinct start position (the position where the cheat is activated, just before the first move that is allowed to go through walls) and end position; cheats are uniquely identified by their start position and end position.
In this example, the total number of cheats (grouped by the amount of time they save) are as follows:
There are 14 cheats that save 2 picoseconds.
There are 14 cheats that save 4 picoseconds.
There are 2 cheats that save 6 picoseconds.
There are 4 cheats that save 8 picoseconds.
There are 2 cheats that save 10 picoseconds.
There are 3 cheats that save 12 picoseconds.
There is one cheat that saves 20 picoseconds.
There is one cheat that saves 36 picoseconds.
There is one cheat that saves 38 picoseconds.
There is one cheat that saves 40 picoseconds.
There is one cheat that saves 64 picoseconds.
You aren't sure what the conditions of the racetrack will be like, so to give yourself as many options as possible, you'll need a list of the best cheats. How many cheats would save you at least 100 picoseconds?
Your puzzle answer was 1485.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
The programs seem perplexed by your list of cheats. Apparently, the two-picosecond cheating rule was deprecated several milliseconds ago! The latest version of the cheating rule permits a single cheat that instead lasts at most 20 picoseconds.
Now, in addition to all the cheats that were possible in just two picoseconds, many more cheats are possible. This six-picosecond cheat saves 76 picoseconds:
###############
#...#...#.....#
#.#.#.#.#.###.#
#S#...#.#.#...#
#1#####.#.#.###
#2#####.#.#...#
#3#####.#.###.#
#456.E#...#...#
###.#######.###
#...###...#...#
#.#####.#.###.#
#.#...#.#.#...#
#.#.#.#.#.#.###
#...#...#...###
###############
Because this cheat has the same start and end positions as the one above, it's the same cheat, even though the path taken during the cheat is different:
###############
#...#...#.....#
#.#.#.#.#.###.#
#S12..#.#.#...#
###3###.#.#.###
###4###.#.#...#
###5###.#.###.#
###6.E#...#...#
###.#######.###
#...###...#...#
#.#####.#.###.#
#.#...#.#.#...#
#.#.#.#.#.#.###
#...#...#...###
###############
Cheats don't need to use all 20 picoseconds; cheats can last any amount of time up to and including 20 picoseconds (but can still only end when the program is on normal track). Any cheat time not used is lost; it can't be saved for another cheat later.
You'll still need a list of the best cheats, but now there are even more to choose between. Here are the quantities of cheats in this example that save 50 picoseconds or more:
There are 32 cheats that save 50 picoseconds.
There are 31 cheats that save 52 picoseconds.
There are 29 cheats that save 54 picoseconds.
There are 39 cheats that save 56 picoseconds.
There are 25 cheats that save 58 picoseconds.
There are 23 cheats that save 60 picoseconds.
There are 20 cheats that save 62 picoseconds.
There are 19 cheats that save 64 picoseconds.
There are 12 cheats that save 66 picoseconds.
There are 14 cheats that save 68 picoseconds.
There are 12 cheats that save 70 picoseconds.
There are 22 cheats that save 72 picoseconds.
There are 4 cheats that save 74 picoseconds.
There are 3 cheats that save 76 picoseconds.
Find the best cheats using the updated cheating rules. How many cheats would save you at least 100 picoseconds?
| 101
|
--- Day 14: Docking Data ---
As your ferry approaches the sea port, the captain asks for your help again. The computer system that runs this port isn't compatible with the docking program on the ferry, so the docking parameters aren't being correctly initialized in the docking program's memory.
After a brief inspection, you discover that the sea port's computer system uses a strange bitmask system in its initialization program. Although you don't have the correct decoder chip handy, you can emulate it in software!
The initialization program (your puzzle input) can either update the bitmask or write a value to memory. Values and memory addresses are both 36-bit unsigned integers. For example, ignoring bitmasks for a moment, a line like mem[8] = 11 would write the value 11 to memory address 8.
The bitmask is always given as a string of 36 bits, written with the most significant bit (representing 2^35) on the left and the least significant bit (2^0, that is, the 1s bit) on the right. The current bitmask is applied to values immediately before they are written to memory: a 0 or 1 overwrites the corresponding bit in the value, while an X leaves the bit in the value unchanged.
For example, consider the following program:
mask = XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
mem[8] = 11
mem[7] = 101
mem[8] = 0
This program starts by specifying a bitmask (mask = ....). The mask it specifies will overwrite two bits in every written value: the 2s bit is overwritten with 0, and the 64s bit is overwritten with 1.
The program then attempts to write the value 11 to memory address 8. By expanding everything out to individual bits, the mask is applied as follows:
value: 000000000000000000000000000000001011 (decimal 11)
mask: XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
result: 000000000000000000000000000001001001 (decimal 73)
So, because of the mask, the value 73 is written to memory address 8 instead. Then, the program tries to write 101 to address 7:
value: 000000000000000000000000000001100101 (decimal 101)
mask: XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
result: 000000000000000000000000000001100101 (decimal 101)
This time, the mask has no effect, as the bits it overwrote were already the values the mask tried to set. Finally, the program tries to write 0 to address 8:
value: 000000000000000000000000000000000000 (decimal 0)
mask: XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
result: 000000000000000000000000000001000000 (decimal 64)
64 is written to address 8 instead, overwriting the value that was there previously.
To initialize your ferry's docking program, you need the sum of all values left in memory after the initialization program completes. (The entire 36-bit address space begins initialized to the value 0 at every address.) In the above example, only two values in memory are not zero - 101 (at address 7) and 64 (at address 8) - producing a sum of 165.
Execute the initialization program. What is the sum of all values left in memory after it completes? (Do not truncate the sum to 36 bits.)
Your puzzle answer was 8471403462063.
--- Part Two ---
For some reason, the sea port's computer system still can't communicate with your ferry's docking program. It must be using version 2 of the decoder chip!
A version 2 decoder chip doesn't modify the values being written at all. Instead, it acts as a memory address decoder. Immediately before a value is written to memory, each bit in the bitmask modifies the corresponding bit of the destination memory address in the following way:
If the bitmask bit is 0, the corresponding memory address bit is unchanged.
If the bitmask bit is 1, the corresponding memory address bit is overwritten with 1.
If the bitmask bit is X, the corresponding memory address bit is floating.
A floating bit is not connected to anything and instead fluctuates unpredictably. In practice, this means the floating bits will take on all possible values, potentially causing many memory addresses to be written all at once!
For example, consider the following program:
mask = 000000000000000000000000000000X1001X
mem[42] = 100
mask = 00000000000000000000000000000000X0XX
mem[26] = 1
When this program goes to write to memory address 42, it first applies the bitmask:
address: 000000000000000000000000000000101010 (decimal 42)
mask: 000000000000000000000000000000X1001X
result: 000000000000000000000000000000X1101X
After applying the mask, four bits are overwritten, three of which are different, and two of which are floating. Floating bits take on every possible combination of values; with two floating bits, four actual memory addresses are written:
000000000000000000000000000000011010 (decimal 26)
000000000000000000000000000000011011 (decimal 27)
000000000000000000000000000000111010 (decimal 58)
000000000000000000000000000000111011 (decimal 59)
Next, the program is about to write to memory address 26 with a different bitmask:
address: 000000000000000000000000000000011010 (decimal 26)
mask: 00000000000000000000000000000000X0XX
result: 00000000000000000000000000000001X0XX
This results in an address with three floating bits, causing writes to eight memory addresses:
000000000000000000000000000000010000 (decimal 16)
000000000000000000000000000000010001 (decimal 17)
000000000000000000000000000000010010 (decimal 18)
000000000000000000000000000000010011 (decimal 19)
000000000000000000000000000000011000 (decimal 24)
000000000000000000000000000000011001 (decimal 25)
000000000000000000000000000000011010 (decimal 26)
000000000000000000000000000000011011 (decimal 27)
The entire 36-bit address space still begins initialized to the value 0 at every address, and you still need the sum of all values left in memory at the end of the program. In this example, the sum is 208.
Execute the initialization program using an emulator for a version 2 decoder chip. What is the sum of all values left in memory after it completes?
| 102
|
--- Day 7: Handy Haversacks ---
You land at the regional airport in time for your next flight. In fact, it looks like you'll even have time to grab some food: all flights are currently delayed due to issues in luggage processing.
Due to recent aviation regulations, many rules (your puzzle input) are being enforced about bags and their contents; bags must be color-coded and must contain specific quantities of other color-coded bags. Apparently, nobody responsible for these regulations considered how long they would take to enforce!
For example, consider the following rules:
light red bags contain 1 bright white bag, 2 muted yellow bags.
dark orange bags contain 3 bright white bags, 4 muted yellow bags.
bright white bags contain 1 shiny gold bag.
muted yellow bags contain 2 shiny gold bags, 9 faded blue bags.
shiny gold bags contain 1 dark olive bag, 2 vibrant plum bags.
dark olive bags contain 3 faded blue bags, 4 dotted black bags.
vibrant plum bags contain 5 faded blue bags, 6 dotted black bags.
faded blue bags contain no other bags.
dotted black bags contain no other bags.
These rules specify the required contents for 9 bag types. In this example, every faded blue bag is empty, every vibrant plum bag contains 11 bags (5 faded blue and 6 dotted black), and so on.
You have a shiny gold bag. If you wanted to carry it in at least one other bag, how many different bag colors would be valid for the outermost bag? (In other words: how many colors can, eventually, contain at least one shiny gold bag?)
In the above rules, the following options would be available to you:
A bright white bag, which can hold your shiny gold bag directly.
A muted yellow bag, which can hold your shiny gold bag directly, plus some other bags.
A dark orange bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
A light red bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
So, in this example, the number of bag colors that can eventually contain at least one shiny gold bag is 4.
How many bag colors can eventually contain at least one shiny gold bag? (The list of rules is quite long; make sure you get all of it.)
| 103
|
--- Day 23: Category Six ---
The droids have finished repairing as much of the ship as they can. Their report indicates that this was a Category 6 disaster - not because it was that bad, but because it destroyed the stockpile of Category 6 network cables as well as most of the ship's network infrastructure.
You'll need to rebuild the network from scratch.
The computers on the network are standard Intcode computers that communicate by sending packets to each other. There are 50 of them in total, each running a copy of the same Network Interface Controller (NIC) software (your puzzle input). The computers have network addresses 0 through 49; when each computer boots up, it will request its network address via a single input instruction. Be sure to give each computer a unique network address.
Once a computer has received its network address, it will begin doing work and communicating over the network by sending and receiving packets. All packets contain two values named X and Y. Packets sent to a computer are queued by the recipient and read in the order they are received.
To send a packet to another computer, the NIC will use three output instructions that provide the destination address of the packet followed by its X and Y values. For example, three output instructions that provide the values 10, 20, 30 would send a packet with X=20 and Y=30 to the computer with address 10.
To receive a packet from another computer, the NIC will use an input instruction. If the incoming packet queue is empty, provide -1. Otherwise, provide the X value of the next packet; the computer will then use a second input instruction to receive the Y value for the same packet. Once both values of the packet are read in this way, the packet is removed from the queue.
Note that these input and output instructions never block. Specifically, output instructions do not wait for the sent packet to be received - the computer might send multiple packets before receiving any. Similarly, input instructions do not wait for a packet to arrive - if no packet is waiting, input instructions should receive -1.
Boot up all 50 computers and attach them to your network. What is the Y value of the first packet sent to address 255?
Your puzzle answer was 15662.
--- Part Two ---
Packets sent to address 255 are handled by a device called a NAT (Not Always Transmitting). The NAT is responsible for managing power consumption of the network by blocking certain packets and watching for idle periods in the computers.
If a packet would be sent to address 255, the NAT receives it instead. The NAT remembers only the last packet it receives; that is, the data in each packet it receives overwrites the NAT's packet memory with the new packet's X and Y values.
The NAT also monitors all computers on the network. If all computers have empty incoming packet queues and are continuously trying to receive packets without sending packets, the network is considered idle.
Once the network is idle, the NAT sends only the last packet it received to address 0; this will cause the computers on the network to resume activity. In this way, the NAT can throttle power consumption of the network when the ship needs power in other areas.
Monitor packets released to the computer at address 0 by the NAT. What is the first Y value delivered by the NAT to the computer at address 0 twice in a row?
| 104
|
--- Day 16: Reindeer Maze ---
It's time again for the Reindeer Olympics! This year, the big event is the Reindeer Maze, where the Reindeer compete for the lowest score.
You and The Historians arrive to search for the Chief right as the event is about to start. It wouldn't hurt to watch a little, right?
The Reindeer start on the Start Tile (marked S) facing East and need to reach the End Tile (marked E). They can move forward one tile at a time (increasing their score by 1 point), but never into a wall (#). They can also rotate clockwise or counterclockwise 90 degrees at a time (increasing their score by 1000 points).
To figure out the best place to sit, you start by grabbing a map (your puzzle input) from a nearby kiosk. For example:
###############
#.......#....E#
#.#.###.#.###.#
#.....#.#...#.#
#.###.#####.#.#
#.#.#.......#.#
#.#.#####.###.#
#...........#.#
###.#.#####.#.#
#...#.....#.#.#
#.#.#.###.#.#.#
#.....#...#.#.#
#.###.#.#.#.#.#
#S..#.....#...#
###############
There are many paths through this maze, but taking any of the best paths would incur a score of only 7036. This can be achieved by taking a total of 36 steps forward and turning 90 degrees a total of 7 times:
###############
#.......#....E#
#.#.###.#.###^#
#.....#.#...#^#
#.###.#####.#^#
#.#.#.......#^#
#.#.#####.###^#
#..>>>>>>>>v#^#
###^#.#####v#^#
#>>^#.....#v#^#
#^#.#.###.#v#^#
#^....#...#v#^#
#^###.#.#.#v#^#
#S..#.....#>>^#
###############
Here's a second example:
#################
#...#...#...#..E#
#.#.#.#.#.#.#.#.#
#.#.#.#...#...#.#
#.#.#.#.###.#.#.#
#...#.#.#.....#.#
#.#.#.#.#.#####.#
#.#...#.#.#.....#
#.#.#####.#.###.#
#.#.#.......#...#
#.#.###.#####.###
#.#.#...#.....#.#
#.#.#.#####.###.#
#.#.#.........#.#
#.#.#.#########.#
#S#.............#
#################
In this maze, the best paths cost 11048 points; following one such path would look like this:
#################
#...#...#...#..E#
#.#.#.#.#.#.#.#^#
#.#.#.#...#...#^#
#.#.#.#.###.#.#^#
#>>v#.#.#.....#^#
#^#v#.#.#.#####^#
#^#v..#.#.#>>>>^#
#^#v#####.#^###.#
#^#v#..>>>>^#...#
#^#v###^#####.###
#^#v#>>^#.....#.#
#^#v#^#####.###.#
#^#v#^........#.#
#^#v#^#########.#
#S#>>^..........#
#################
Note that the path shown above includes one 90 degree turn as the very first move, rotating the Reindeer from facing East to facing North.
Analyze your map carefully. What is the lowest score a Reindeer could possibly get?
| 105
|
--- Day 4: Repose Record ---
You've sneaked into another supply closet - this time, it's across from the prototype suit manufacturing lab. You need to sneak inside and fix the issues with the suit, but there's a guard stationed outside the lab, so this is as close as you can safely get.
As you search the closet for anything that might help, you discover that you're not the first person to want to sneak in. Covering the walls, someone has spent an hour starting every midnight for the past few months secretly observing this guard post! They've been writing down the ID of the one guard on duty that night - the Elves seem to have decided that one guard was enough for the overnight shift - as well as when they fall asleep or wake up while at their post (your puzzle input).
For example, consider the following records, which have already been organized into chronological order:
[1518-11-01 00:00] Guard #10 begins shift
[1518-11-01 00:05] falls asleep
[1518-11-01 00:25] wakes up
[1518-11-01 00:30] falls asleep
[1518-11-01 00:55] wakes up
[1518-11-01 23:58] Guard #99 begins shift
[1518-11-02 00:40] falls asleep
[1518-11-02 00:50] wakes up
[1518-11-03 00:05] Guard #10 begins shift
[1518-11-03 00:24] falls asleep
[1518-11-03 00:29] wakes up
[1518-11-04 00:02] Guard #99 begins shift
[1518-11-04 00:36] falls asleep
[1518-11-04 00:46] wakes up
[1518-11-05 00:03] Guard #99 begins shift
[1518-11-05 00:45] falls asleep
[1518-11-05 00:55] wakes up
Timestamps are written using year-month-day hour:minute format. The guard falling asleep or waking up is always the one whose shift most recently started. Because all asleep/awake times are during the midnight hour (00:00 - 00:59), only the minute portion (00 - 59) is relevant for those events.
Visually, these records show that the guards are asleep at these times:
Date ID Minute
000000000011111111112222222222333333333344444444445555555555
012345678901234567890123456789012345678901234567890123456789
11-01 #10 .....####################.....#########################.....
11-02 #99 ........................................##########..........
11-03 #10 ........................#####...............................
11-04 #99 ....................................##########..............
11-05 #99 .............................................##########.....
The columns are Date, which shows the month-day portion of the relevant day; ID, which shows the guard on duty that day; and Minute, which shows the minutes during which the guard was asleep within the midnight hour. (The Minute column's header shows the minute's ten's digit in the first row and the one's digit in the second row.) Awake is shown as ., and asleep is shown as #.
Note that guards count as asleep on the minute they fall asleep, and they count as awake on the minute they wake up. For example, because Guard #10 wakes up at 00:25 on 1518-11-01, minute 25 is marked as awake.
If you can figure out the guard most likely to be asleep at a specific time, you might be able to trick that guard into working tonight so you can have the best chance of sneaking in. You have two strategies for choosing the best guard/minute combination.
Strategy 1: Find the guard that has the most minutes asleep. What minute does that guard spend asleep the most?
In the example above, Guard #10 spent the most minutes asleep, a total of 50 minutes (20+25+5), while Guard #99 only slept for a total of 30 minutes (10+10+10). Guard #10 was asleep most during minute 24 (on two days, whereas any other minute the guard was asleep was only seen on one day).
While this example listed the entries in chronological order, your entries are in the order you found them. You'll need to organize them before they can be analyzed.
What is the ID of the guard you chose multiplied by the minute you chose? (In the above example, the answer would be 10 * 24 = 240.)
| 106
|
--- Day 14: One-Time Pad ---
In order to communicate securely with Santa while you're on this mission, you've been using a one-time pad that you generate using a pre-agreed algorithm. Unfortunately, you've run out of keys in your one-time pad, and so you need to generate some more.
To generate keys, you first get a stream of random data by taking the MD5 of a pre-arranged salt (your puzzle input) and an increasing integer index (starting with 0, and represented in decimal); the resulting MD5 hash should be represented as a string of lowercase hexadecimal digits.
However, not all of these MD5 hashes are keys, and you need 64 new keys for your one-time pad. A hash is a key only if:
It contains three of the same character in a row, like 777. Only consider the first such triplet in a hash.
One of the next 1000 hashes in the stream contains that same character five times in a row, like 77777.
Considering future hashes for five-of-a-kind sequences does not cause those hashes to be skipped; instead, regardless of whether the current hash is a key, always resume testing for keys starting with the very next hash.
For example, if the pre-arranged salt is abc:
The first index which produces a triple is 18, because the MD5 hash of abc18 contains ...cc38887a5.... However, index 18 does not count as a key for your one-time pad, because none of the next thousand hashes (index 19 through index 1018) contain 88888.
The next index which produces a triple is 39; the hash of abc39 contains eee. It is also the first key: one of the next thousand hashes (the one at index 816) contains eeeee.
None of the next six triples are keys, but the one after that, at index 92, is: it contains 999 and index 200 contains 99999.
Eventually, index 22728 meets all of the criteria to generate the 64th key.
So, using our example salt of abc, index 22728 produces the 64th key.
Given the actual salt in your puzzle input, what index produces your 64th one-time pad key?
| 107
|
--- Day 21: Monkey Math ---
The monkeys are back! You're worried they're going to try to steal your stuff again, but it seems like they're just holding their ground and making various monkey noises at you.
Eventually, one of the elephants realizes you don't speak monkey and comes over to interpret. As it turns out, they overheard you talking about trying to find the grove; they can show you a shortcut if you answer their riddle.
Each monkey is given a job: either to yell a specific number or to yell the result of a math operation. All of the number-yelling monkeys know their number from the start; however, the math operation monkeys need to wait for two other monkeys to yell a number, and those two other monkeys might also be waiting on other monkeys.
Your job is to work out the number the monkey named root will yell before the monkeys figure it out themselves.
For example:
root: pppw + sjmn
dbpl: 5
cczh: sllz + lgvd
zczc: 2
ptdq: humn - dvpt
dvpt: 3
lfqf: 4
humn: 5
ljgn: 2
sjmn: drzm * dbpl
sllz: 4
pppw: cczh / lfqf
lgvd: ljgn * ptdq
drzm: hmdt - zczc
hmdt: 32
Each line contains the name of a monkey, a colon, and then the job of that monkey:
A lone number means the monkey's job is simply to yell that number.
A job like aaaa + bbbb means the monkey waits for monkeys aaaa and bbbb to yell each of their numbers; the monkey then yells the sum of those two numbers.
aaaa - bbbb means the monkey yells aaaa's number minus bbbb's number.
Job aaaa * bbbb will yell aaaa's number multiplied by bbbb's number.
Job aaaa / bbbb will yell aaaa's number divided by bbbb's number.
So, in the above example, monkey drzm has to wait for monkeys hmdt and zczc to yell their numbers. Fortunately, both hmdt and zczc have jobs that involve simply yelling a single number, so they do this immediately: 32 and 2. Monkey drzm can then yell its number by finding 32 minus 2: 30.
Then, monkey sjmn has one of its numbers (30, from monkey drzm), and already has its other number, 5, from dbpl. This allows it to yell its own number by finding 30 multiplied by 5: 150.
This process continues until root yells a number: 152.
However, your actual situation involves considerably more monkeys. What number will the monkey named root yell?
| 108
|
--- Day 24: Blizzard Basin ---
With everything replanted for next year (and with elephants and monkeys to tend the grove), you and the Elves leave for the extraction point.
Partway up the mountain that shields the grove is a flat, open area that serves as the extraction point. It's a bit of a climb, but nothing the expedition can't handle.
At least, that would normally be true; now that the mountain is covered in snow, things have become more difficult than the Elves are used to.
As the expedition reaches a valley that must be traversed to reach the extraction site, you find that strong, turbulent winds are pushing small blizzards of snow and sharp ice around the valley. It's a good thing everyone packed warm clothes! To make it across safely, you'll need to find a way to avoid them.
Fortunately, it's easy to see all of this from the entrance to the valley, so you make a map of the valley and the blizzards (your puzzle input). For example:
#.#####
#.....#
#>....#
#.....#
#...v.#
#.....#
#####.#
The walls of the valley are drawn as #; everything else is ground. Clear ground - where there is currently no blizzard - is drawn as .. Otherwise, blizzards are drawn with an arrow indicating their direction of motion: up (^), down (v), left (<), or right (>).
The above map includes two blizzards, one moving right (>) and one moving down (v). In one minute, each blizzard moves one position in the direction it is pointing:
#.#####
#.....#
#.>...#
#.....#
#.....#
#...v.#
#####.#
Due to conservation of blizzard energy, as a blizzard reaches the wall of the valley, a new blizzard forms on the opposite side of the valley moving in the same direction. After another minute, the bottom downward-moving blizzard has been replaced with a new downward-moving blizzard at the top of the valley instead:
#.#####
#...v.#
#..>..#
#.....#
#.....#
#.....#
#####.#
Because blizzards are made of tiny snowflakes, they pass right through each other. After another minute, both blizzards temporarily occupy the same position, marked 2:
#.#####
#.....#
#...2.#
#.....#
#.....#
#.....#
#####.#
After another minute, the situation resolves itself, giving each blizzard back its personal space:
#.#####
#.....#
#....>#
#...v.#
#.....#
#.....#
#####.#
Finally, after yet another minute, the rightward-facing blizzard on the right is replaced with a new one on the left facing the same direction:
#.#####
#.....#
#>....#
#.....#
#...v.#
#.....#
#####.#
This process repeats at least as long as you are observing it, but probably forever.
Here is a more complex example:
#.######
#>>.<^<#
#.<..<<#
#>v.><>#
#<^v^^>#
######.#
Your expedition begins in the only non-wall position in the top row and needs to reach the only non-wall position in the bottom row. On each minute, you can move up, down, left, or right, or you can wait in place. You and the blizzards act simultaneously, and you cannot share a position with a blizzard.
In the above example, the fastest way to reach your goal requires 18 steps. Drawing the position of the expedition as E, one way to achieve this is:
Initial state:
#E######
#>>.<^<#
#.<..<<#
#>v.><>#
#<^v^^>#
######.#
Minute 1, move down:
#.######
#E>3.<.#
#<..<<.#
#>2.22.#
#>v..^<#
######.#
Minute 2, move down:
#.######
#.2>2..#
#E^22^<#
#.>2.^>#
#.>..<.#
######.#
Minute 3, wait:
#.######
#<^<22.#
#E2<.2.#
#><2>..#
#..><..#
######.#
Minute 4, move up:
#.######
#E<..22#
#<<.<..#
#<2.>>.#
#.^22^.#
######.#
Minute 5, move right:
#.######
#2Ev.<>#
#<.<..<#
#.^>^22#
#.2..2.#
######.#
Minute 6, move right:
#.######
#>2E<.<#
#.2v^2<#
#>..>2>#
#<....>#
######.#
Minute 7, move down:
#.######
#.22^2.#
#<vE<2.#
#>>v<>.#
#>....<#
######.#
Minute 8, move left:
#.######
#.<>2^.#
#.E<<.<#
#.22..>#
#.2v^2.#
######.#
Minute 9, move up:
#.######
#<E2>>.#
#.<<.<.#
#>2>2^.#
#.v><^.#
######.#
Minute 10, move right:
#.######
#.2E.>2#
#<2v2^.#
#<>.>2.#
#..<>..#
######.#
Minute 11, wait:
#.######
#2^E^2>#
#<v<.^<#
#..2.>2#
#.<..>.#
######.#
Minute 12, move down:
#.######
#>>.<^<#
#.<E.<<#
#>v.><>#
#<^v^^>#
######.#
Minute 13, move down:
#.######
#.>3.<.#
#<..<<.#
#>2E22.#
#>v..^<#
######.#
Minute 14, move right:
#.######
#.2>2..#
#.^22^<#
#.>2E^>#
#.>..<.#
######.#
Minute 15, move right:
#.######
#<^<22.#
#.2<.2.#
#><2>E.#
#..><..#
######.#
Minute 16, move right:
#.######
#.<..22#
#<<.<..#
#<2.>>E#
#.^22^.#
######.#
Minute 17, move down:
#.######
#2.v.<>#
#<.<..<#
#.^>^22#
#.2..2E#
######.#
Minute 18, move down:
#.######
#>2.<.<#
#.2v^2<#
#>..>2>#
#<....>#
######E#
What is the fewest number of minutes required to avoid the blizzards and reach the goal?
| 109
|
--- Day 9: All in a Single Night ---
Every year, Santa manages to deliver all of his presents in a single night.
This year, however, he has some new locations to visit; his elves have provided him the distances between every pair of locations. He can start and end at any two (different) locations he wants, but he must visit each location exactly once. What is the shortest distance he can travel to achieve this?
For example, given the following distances:
London to Dublin = 464
London to Belfast = 518
Dublin to Belfast = 141
The possible routes are therefore:
Dublin -> London -> Belfast = 982
London -> Dublin -> Belfast = 605
London -> Belfast -> Dublin = 659
Dublin -> Belfast -> London = 659
Belfast -> Dublin -> London = 605
Belfast -> London -> Dublin = 982
The shortest of these is London -> Dublin -> Belfast = 605, and so the answer is 605 in this example.
What is the distance of the shortest route?
| 110
|
--- Day 2: Rock Paper Scissors ---
The Elves begin to set up camp on the beach. To decide whose tent gets to be closest to the snack storage, a giant Rock Paper Scissors tournament is already in progress.
Rock Paper Scissors is a game between two players. Each game contains many rounds; in each round, the players each simultaneously choose one of Rock, Paper, or Scissors using a hand shape. Then, a winner for that round is selected: Rock defeats Scissors, Scissors defeats Paper, and Paper defeats Rock. If both players choose the same shape, the round instead ends in a draw.
Appreciative of your help yesterday, one Elf gives you an encrypted strategy guide (your puzzle input) that they say will be sure to help you win. "The first column is what your opponent is going to play: A for Rock, B for Paper, and C for Scissors. The second column--" Suddenly, the Elf is called away to help with someone's tent.
The second column, you reason, must be what you should play in response: X for Rock, Y for Paper, and Z for Scissors. Winning every time would be suspicious, so the responses must have been carefully chosen.
The winner of the whole tournament is the player with the highest score. Your total score is the sum of your scores for each round. The score for a single round is the score for the shape you selected (1 for Rock, 2 for Paper, and 3 for Scissors) plus the score for the outcome of the round (0 if you lost, 3 if the round was a draw, and 6 if you won).
Since you can't be sure if the Elf is trying to help you or trick you, you should calculate the score you would get if you were to follow the strategy guide.
For example, suppose you were given the following strategy guide:
A Y
B X
C Z
This strategy guide predicts and recommends the following:
In the first round, your opponent will choose Rock (A), and you should choose Paper (Y). This ends in a win for you with a score of 8 (2 because you chose Paper + 6 because you won).
In the second round, your opponent will choose Paper (B), and you should choose Rock (X). This ends in a loss for you with a score of 1 (1 + 0).
The third round is a draw with both players choosing Scissors, giving you a score of 3 + 3 = 6.
In this example, if you were to follow the strategy guide, you would get a total score of 15 (8 + 1 + 6).
What would your total score be if everything goes exactly according to your strategy guide?
| 111
|
--- Day 19: An Elephant Named Joseph ---
The Elves contact you over a highly secure emergency channel. Back at the North Pole, the Elves are busy misunderstanding White Elephant parties.
Each Elf brings a present. They all sit in a circle, numbered starting with position 1. Then, starting with the first Elf, they take turns stealing all the presents from the Elf to their left. An Elf with no presents is removed from the circle and does not take turns.
For example, with five Elves (numbered 1 to 5):
1
5 2
4 3
Elf 1 takes Elf 2's present.
Elf 2 has no presents and is skipped.
Elf 3 takes Elf 4's present.
Elf 4 has no presents and is also skipped.
Elf 5 takes Elf 1's two presents.
Neither Elf 1 nor Elf 2 have any presents, so both are skipped.
Elf 3 takes Elf 5's three presents.
So, with five Elves, the Elf that sits starting in position 3 gets all the presents.
With the number of Elves given in your puzzle input, which Elf gets all the presents?
Your puzzle answer was 1815603.
--- Part Two ---
Realizing the folly of their present-exchange rules, the Elves agree to instead steal presents from the Elf directly across the circle. If two Elves are across the circle, the one on the left (from the perspective of the stealer) is stolen from. The other rules remain unchanged: Elves with no presents are removed from the circle entirely, and the other elves move in slightly to keep the circle evenly spaced.
For example, with five Elves (again numbered 1 to 5):
The Elves sit in a circle; Elf 1 goes first:
1
5 2
4 3
Elves 3 and 4 are across the circle; Elf 3's present is stolen, being the one to the left. Elf 3 leaves the circle, and the rest of the Elves move in:
1 1
5 2 --> 5 2
4 - 4
Elf 2 steals from the Elf directly across the circle, Elf 5:
1 1
- 2 --> 2
4 4
Next is Elf 4 who, choosing between Elves 1 and 2, steals from Elf 1:
- 2
2 -->
4 4
Finally, Elf 2 steals from Elf 4:
2
--> 2
-
So, with five Elves, the Elf that sits starting in position 2 gets all the presents.
With the number of Elves given in your puzzle input, which Elf now gets all the presents?
| 112
|
--- Day 15: Lens Library ---
The newly-focused parabolic reflector dish is sending all of the collected light to a point on the side of yet another mountain - the largest mountain on Lava Island. As you approach the mountain, you find that the light is being collected by the wall of a large facility embedded in the mountainside.
You find a door under a large sign that says "Lava Production Facility" and next to a smaller sign that says "Danger - Personal Protective Equipment required beyond this point".
As you step inside, you are immediately greeted by a somewhat panicked reindeer wearing goggles and a loose-fitting hard hat. The reindeer leads you to a shelf of goggles and hard hats (you quickly find some that fit) and then further into the facility. At one point, you pass a button with a faint snout mark and the label "PUSH FOR HELP". No wonder you were loaded into that trebuchet so quickly!
You pass through a final set of doors surrounded with even more warning signs and into what must be the room that collects all of the light from outside. As you admire the large assortment of lenses available to further focus the light, the reindeer brings you a book titled "Initialization Manual".
"Hello!", the book cheerfully begins, apparently unaware of the concerned reindeer reading over your shoulder. "This procedure will let you bring the Lava Production Facility online - all without burning or melting anything unintended!"
"Before you begin, please be prepared to use the Holiday ASCII String Helper algorithm (appendix 1A)." You turn to appendix 1A. The reindeer leans closer with interest.
The HASH algorithm is a way to turn any string of characters into a single number in the range 0 to 255. To run the HASH algorithm on a string, start with a current value of 0. Then, for each character in the string starting from the beginning:
Determine the ASCII code for the current character of the string.
Increase the current value by the ASCII code you just determined.
Set the current value to itself multiplied by 17.
Set the current value to the remainder of dividing itself by 256.
After following these steps for each character in the string in order, the current value is the output of the HASH algorithm.
So, to find the result of running the HASH algorithm on the string HASH:
The current value starts at 0.
The first character is H; its ASCII code is 72.
The current value increases to 72.
The current value is multiplied by 17 to become 1224.
The current value becomes 200 (the remainder of 1224 divided by 256).
The next character is A; its ASCII code is 65.
The current value increases to 265.
The current value is multiplied by 17 to become 4505.
The current value becomes 153 (the remainder of 4505 divided by 256).
The next character is S; its ASCII code is 83.
The current value increases to 236.
The current value is multiplied by 17 to become 4012.
The current value becomes 172 (the remainder of 4012 divided by 256).
The next character is H; its ASCII code is 72.
The current value increases to 244.
The current value is multiplied by 17 to become 4148.
The current value becomes 52 (the remainder of 4148 divided by 256).
So, the result of running the HASH algorithm on the string HASH is 52.
The initialization sequence (your puzzle input) is a comma-separated list of steps to start the Lava Production Facility. Ignore newline characters when parsing the initialization sequence. To verify that your HASH algorithm is working, the book offers the sum of the result of running the HASH algorithm on each step in the initialization sequence.
For example:
rn=1,cm-,qp=3,cm=2,qp-,pc=4,ot=9,ab=5,pc-,pc=6,ot=7
This initialization sequence specifies 11 individual steps; the result of running the HASH algorithm on each of the steps is as follows:
rn=1 becomes 30.
cm- becomes 253.
qp=3 becomes 97.
cm=2 becomes 47.
qp- becomes 14.
pc=4 becomes 180.
ot=9 becomes 9.
ab=5 becomes 197.
pc- becomes 48.
pc=6 becomes 214.
ot=7 becomes 231.
In this example, the sum of these results is 1320. Unfortunately, the reindeer has stolen the page containing the expected verification number and is currently running around the facility with it excitedly.
Run the HASH algorithm on each step in the initialization sequence. What is the sum of the results? (The initialization sequence is one long line; be careful when copy-pasting it.)
Your puzzle answer was 511215.
--- Part Two ---
You convince the reindeer to bring you the page; the page confirms that your HASH algorithm is working.
The book goes on to describe a series of 256 boxes numbered 0 through 255. The boxes are arranged in a line starting from the point where light enters the facility. The boxes have holes that allow light to pass from one box to the next all the way down the line.
+-----+ +-----+ +-----+
Light | Box | | Box | ... | Box |
----------------------------------------->
| 0 | | 1 | ... | 255 |
+-----+ +-----+ +-----+
Inside each box, there are several lens slots that will keep a lens correctly positioned to focus light passing through the box. The side of each box has a panel that opens to allow you to insert or remove lenses as necessary.
Along the wall running parallel to the boxes is a large library containing lenses organized by focal length ranging from 1 through 9. The reindeer also brings you a small handheld label printer.
The book goes on to explain how to perform each step in the initialization sequence, a process it calls the Holiday ASCII String Helper Manual Arrangement Procedure, or HASHMAP for short.
Each step begins with a sequence of letters that indicate the label of the lens on which the step operates. The result of running the HASH algorithm on the label indicates the correct box for that step.
The label will be immediately followed by a character that indicates the operation to perform: either an equals sign (=) or a dash (-).
If the operation character is a dash (-), go to the relevant box and remove the lens with the given label if it is present in the box. Then, move any remaining lenses as far forward in the box as they can go without changing their order, filling any space made by removing the indicated lens. (If no lens in that box has the given label, nothing happens.)
If the operation character is an equals sign (=), it will be followed by a number indicating the focal length of the lens that needs to go into the relevant box; be sure to use the label maker to mark the lens with the label given in the beginning of the step so you can find it later. There are two possible situations:
If there is already a lens in the box with the same label, replace the old lens with the new lens: remove the old lens and put the new lens in its place, not moving any other lenses in the box.
If there is not already a lens in the box with the same label, add the lens to the box immediately behind any lenses already in the box. Don't move any of the other lenses when you do this. If there aren't any lenses in the box, the new lens goes all the way to the front of the box.
Here is the contents of every box after each step in the example initialization sequence above:
After "rn=1":
Box 0: [rn 1]
After "cm-":
Box 0: [rn 1]
After "qp=3":
Box 0: [rn 1]
Box 1: [qp 3]
After "cm=2":
Box 0: [rn 1] [cm 2]
Box 1: [qp 3]
After "qp-":
Box 0: [rn 1] [cm 2]
After "pc=4":
Box 0: [rn 1] [cm 2]
Box 3: [pc 4]
After "ot=9":
Box 0: [rn 1] [cm 2]
Box 3: [pc 4] [ot 9]
After "ab=5":
Box 0: [rn 1] [cm 2]
Box 3: [pc 4] [ot 9] [ab 5]
After "pc-":
Box 0: [rn 1] [cm 2]
Box 3: [ot 9] [ab 5]
After "pc=6":
Box 0: [rn 1] [cm 2]
Box 3: [ot 9] [ab 5] [pc 6]
After "ot=7":
Box 0: [rn 1] [cm 2]
Box 3: [ot 7] [ab 5] [pc 6]
All 256 boxes are always present; only the boxes that contain any lenses are shown here. Within each box, lenses are listed from front to back; each lens is shown as its label and focal length in square brackets.
To confirm that all of the lenses are installed correctly, add up the focusing power of all of the lenses. The focusing power of a single lens is the result of multiplying together:
One plus the box number of the lens in question.
The slot number of the lens within the box: 1 for the first lens, 2 for the second lens, and so on.
The focal length of the lens.
At the end of the above example, the focusing power of each lens is as follows:
rn: 1 (box 0) * 1 (first slot) * 1 (focal length) = 1
cm: 1 (box 0) * 2 (second slot) * 2 (focal length) = 4
ot: 4 (box 3) * 1 (first slot) * 7 (focal length) = 28
ab: 4 (box 3) * 2 (second slot) * 5 (focal length) = 40
pc: 4 (box 3) * 3 (third slot) * 6 (focal length) = 72
So, the above example ends up with a total focusing power of 145.
With the help of an over-enthusiastic reindeer in a hard hat, follow the initialization sequence. What is the focusing power of the resulting lens configuration?
| 113
|
--- Day 3: Crossed Wires ---
The gravity assist was successful, and you're well on your way to the Venus refuelling station. During the rush back on Earth, the fuel management system wasn't completely installed, so that's next on the priority list.
Opening the front panel reveals a jumble of wires. Specifically, two wires are connected to a central port and extend outward on a grid. You trace the path each wire takes as it leaves the central port, one wire per line of text (your puzzle input).
The wires twist and turn, but the two wires occasionally cross paths. To fix the circuit, you need to find the intersection point closest to the central port. Because the wires are on a grid, use the Manhattan distance for this measurement. While the wires do technically cross right at the central port where they both start, this point does not count, nor does a wire count as crossing with itself.
For example, if the first wire's path is R8,U5,L5,D3, then starting from the central port (o), it goes right 8, up 5, left 5, and finally down 3:
...........
...........
...........
....+----+.
....|....|.
....|....|.
....|....|.
.........|.
.o-------+.
...........
Then, if the second wire's path is U7,R6,D4,L4, it goes up 7, right 6, down 4, and left 4:
...........
.+-----+...
.|.....|...
.|..+--X-+.
.|..|..|.|.
.|.-X--+.|.
.|..|....|.
.|.......|.
.o-------+.
...........
These wires cross at two locations (marked X), but the lower-left one is closer to the central port: its distance is 3 + 3 = 6.
Here are a few more examples:
R75,D30,R83,U83,L12,D49,R71,U7,L72
U62,R66,U55,R34,D71,R55,D58,R83 = distance 159
R98,U47,R26,D63,R33,U87,L62,D20,R33,U53,R51
U98,R91,D20,R16,D67,R40,U7,R15,U6,R7 = distance 135
What is the Manhattan distance from the central port to the closest intersection?
Your puzzle answer was 557.
--- Part Two ---
It turns out that this circuit is very timing-sensitive; you actually need to minimize the signal delay.
To do this, calculate the number of steps each wire takes to reach each intersection; choose the intersection where the sum of both wires' steps is lowest. If a wire visits a position on the grid multiple times, use the steps value from the first time it visits that position when calculating the total value of a specific intersection.
The number of steps a wire takes is the total number of grid squares the wire has entered to get to that location, including the intersection being considered. Again consider the example from above:
...........
.+-----+...
.|.....|...
.|..+--X-+.
.|..|..|.|.
.|.-X--+.|.
.|..|....|.
.|.......|.
.o-------+.
...........
In the above example, the intersection closest to the central port is reached after 8+5+5+2 = 20 steps by the first wire and 7+6+4+3 = 20 steps by the second wire for a total of 20+20 = 40 steps.
However, the top-right intersection is better: the first wire takes only 8+5+2 = 15 and the second wire takes only 7+6+2 = 15, a total of 15+15 = 30 steps.
Here are the best steps for the extra examples from above:
R75,D30,R83,U83,L12,D49,R71,U7,L72
U62,R66,U55,R34,D71,R55,D58,R83 = 610 steps
R98,U47,R26,D63,R33,U87,L62,D20,R33,U53,R51
U98,R91,D20,R16,D67,R40,U7,R15,U6,R7 = 410 steps
What is the fewest combined steps the wires must take to reach an intersection?
| 114
|
--- Day 18: Boiling Boulders ---
You and the elephants finally reach fresh air. You've emerged near the base of a large volcano that seems to be actively erupting! Fortunately, the lava seems to be flowing away from you and toward the ocean.
Bits of lava are still being ejected toward you, so you're sheltering in the cavern exit a little longer. Outside the cave, you can see the lava landing in a pond and hear it loudly hissing as it solidifies.
Depending on the specific compounds in the lava and speed at which it cools, it might be forming obsidian! The cooling rate should be based on the surface area of the lava droplets, so you take a quick scan of a droplet as it flies past you (your puzzle input).
Because of how quickly the lava is moving, the scan isn't very good; its resolution is quite low and, as a result, it approximates the shape of the lava droplet with 1x1x1 cubes on a 3D grid, each given as its x,y,z position.
To approximate the surface area, count the number of sides of each cube that are not immediately connected to another cube. So, if your scan were only two adjacent cubes like 1,1,1 and 2,1,1, each cube would have a single side covered and five sides exposed, a total surface area of 10 sides.
Here's a larger example:
2,2,2
1,2,2
3,2,2
2,1,2
2,3,2
2,2,1
2,2,3
2,2,4
2,2,6
1,2,5
3,2,5
2,1,5
2,3,5
In the above example, after counting up all the sides that aren't connected to another cube, the total surface area is 64.
What is the surface area of your scanned lava droplet?
| 115
|
--- Day 21: Chronal Conversion ---
You should have been watching where you were going, because as you wander the new North Pole base, you trip and fall into a very deep hole!
Just kidding. You're falling through time again.
If you keep up your current pace, you should have resolved all of the temporal anomalies by the next time the device activates. Since you have very little interest in browsing history in 500-year increments for the rest of your life, you need to find a way to get back to your present time.
After a little research, you discover two important facts about the behavior of the device:
First, you discover that the device is hard-wired to always send you back in time in 500-year increments. Changing this is probably not feasible.
Second, you discover the activation system (your puzzle input) for the time travel module. Currently, it appears to run forever without halting.
If you can cause the activation system to halt at a specific moment, maybe you can make the device send you so far back in time that you cause an integer underflow in time itself and wrap around back to your current time!
The device executes the program as specified in manual section one and manual section two.
Your goal is to figure out how the program works and cause it to halt. You can only control register 0; every other register begins at 0 as usual.
Because time travel is a dangerous activity, the activation system begins with a few instructions which verify that bitwise AND (via bani) does a numeric operation and not an operation as if the inputs were interpreted as strings. If the test fails, it enters an infinite loop re-running the test instead of allowing the program to execute normally. If the test passes, the program continues, and assumes that all other bitwise operations (banr, bori, and borr) also interpret their inputs as numbers. (Clearly, the Elves who wrote this system were worried that someone might introduce a bug while trying to emulate this system with a scripting language.)
What is the lowest non-negative integer value for register 0 that causes the program to halt after executing the fewest instructions? (Executing the same instruction multiple times counts as multiple instructions executed.)
Your puzzle answer was 11050031.
--- Part Two ---
In order to determine the timing window for your underflow exploit, you also need an upper bound:
What is the lowest non-negative integer value for register 0 that causes the program to halt after executing the most instructions? (The program must actually halt; running forever does not count as halting.)
| 116
|
--- Day 18: Like a GIF For Your Yard ---
After the million lights incident, the fire code has gotten stricter: now, at most ten thousand lights are allowed. You arrange them in a 100x100 grid.
Never one to let you down, Santa again mails you instructions on the ideal lighting configuration. With so few lights, he says, you'll have to resort to animation.
Start by setting your lights to the included initial configuration (your puzzle input). A # means "on", and a . means "off".
Then, animate your grid in steps, where each step decides the next configuration based on the current one. Each light's next state (either on or off) depends on its current state and the current states of the eight lights adjacent to it (including diagonals). Lights on the edge of the grid might have fewer than eight neighbors; the missing ones always count as "off".
For example, in a simplified 6x6 grid, the light marked A has the neighbors numbered 1 through 8, and the light marked B, which is on an edge, only has the neighbors marked 1 through 5:
1B5...
234...
......
..123.
..8A4.
..765.
The state a light should have next is based on its current state (on or off) plus the number of neighbors that are on:
A light which is on stays on when 2 or 3 neighbors are on, and turns off otherwise.
A light which is off turns on if exactly 3 neighbors are on, and stays off otherwise.
All of the lights update simultaneously; they all consider the same current state before moving to the next.
Here's a few steps from an example configuration of another 6x6 grid:
Initial state:
.#.#.#
...##.
#....#
..#...
#.#..#
####..
After 1 step:
..##..
..##.#
...##.
......
#.....
#.##..
After 2 steps:
..###.
......
..###.
......
.#....
.#....
After 3 steps:
...#..
......
...#..
..##..
......
......
After 4 steps:
......
......
..##..
..##..
......
......
After 4 steps, this example has four lights on.
In your grid of 100x100 lights, given your initial configuration, how many lights are on after 100 steps?
Your puzzle answer was 814.
--- Part Two ---
You flip the instructions over; Santa goes on to point out that this is all just an implementation of Conway's Game of Life. At least, it was, until you notice that something's wrong with the grid of lights you bought: four lights, one in each corner, are stuck on and can't be turned off. The example above will actually run like this:
Initial state:
##.#.#
...##.
#....#
..#...
#.#..#
####.#
After 1 step:
#.##.#
####.#
...##.
......
#...#.
#.####
After 2 steps:
#..#.#
#....#
.#.##.
...##.
.#..##
##.###
After 3 steps:
#...##
####.#
..##.#
......
##....
####.#
After 4 steps:
#.####
#....#
...#..
.##...
#.....
#.#..#
After 5 steps:
##.###
.##..#
.##...
.##...
#.#...
##...#
After 5 steps, this example now has 17 lights on.
In your grid of 100x100 lights, given your initial configuration, but with the four corners always in the on state, how many lights are on after 100 steps?
| 117
|
--- Day 13: Distress Signal ---
You climb the hill and again try contacting the Elves. However, you instead receive a signal you weren't expecting: a distress signal.
Your handheld device must still not be working properly; the packets from the distress signal got decoded out of order. You'll need to re-order the list of received packets (your puzzle input) to decode the message.
Your list consists of pairs of packets; pairs are separated by a blank line. You need to identify how many pairs of packets are in the right order.
For example:
[1,1,3,1,1]
[1,1,5,1,1]
[[1],[2,3,4]]
[[1],4]
[9]
[[8,7,6]]
[[4,4],4,4]
[[4,4],4,4,4]
[7,7,7,7]
[7,7,7]
[]
[3]
[[[]]]
[[]]
[1,[2,[3,[4,[5,6,7]]]],8,9]
[1,[2,[3,[4,[5,6,0]]]],8,9]
Packet data consists of lists and integers. Each list starts with [, ends with ], and contains zero or more comma-separated values (either integers or other lists). Each packet is always a list and appears on its own line.
When comparing two values, the first value is called left and the second value is called right. Then:
If both values are integers, the lower integer should come first. If the left integer is lower than the right integer, the inputs are in the right order. If the left integer is higher than the right integer, the inputs are not in the right order. Otherwise, the inputs are the same integer; continue checking the next part of the input.
If both values are lists, compare the first value of each list, then the second value, and so on. If the left list runs out of items first, the inputs are in the right order. If the right list runs out of items first, the inputs are not in the right order. If the lists are the same length and no comparison makes a decision about the order, continue checking the next part of the input.
If exactly one value is an integer, convert the integer to a list which contains that integer as its only value, then retry the comparison. For example, if comparing [0,0,0] and 2, convert the right value to [2] (a list containing 2); the result is then found by instead comparing [0,0,0] and [2].
Using these rules, you can determine which of the pairs in the example are in the right order:
== Pair 1 ==
- Compare [1,1,3,1,1] vs [1,1,5,1,1]
- Compare 1 vs 1
- Compare 1 vs 1
- Compare 3 vs 5
- Left side is smaller, so inputs are in the right order
== Pair 2 ==
- Compare [[1],[2,3,4]] vs [[1],4]
- Compare [1] vs [1]
- Compare 1 vs 1
- Compare [2,3,4] vs 4
- Mixed types; convert right to [4] and retry comparison
- Compare [2,3,4] vs [4]
- Compare 2 vs 4
- Left side is smaller, so inputs are in the right order
== Pair 3 ==
- Compare [9] vs [[8,7,6]]
- Compare 9 vs [8,7,6]
- Mixed types; convert left to [9] and retry comparison
- Compare [9] vs [8,7,6]
- Compare 9 vs 8
- Right side is smaller, so inputs are not in the right order
== Pair 4 ==
- Compare [[4,4],4,4] vs [[4,4],4,4,4]
- Compare [4,4] vs [4,4]
- Compare 4 vs 4
- Compare 4 vs 4
- Compare 4 vs 4
- Compare 4 vs 4
- Left side ran out of items, so inputs are in the right order
== Pair 5 ==
- Compare [7,7,7,7] vs [7,7,7]
- Compare 7 vs 7
- Compare 7 vs 7
- Compare 7 vs 7
- Right side ran out of items, so inputs are not in the right order
== Pair 6 ==
- Compare [] vs [3]
- Left side ran out of items, so inputs are in the right order
== Pair 7 ==
- Compare [[[]]] vs [[]]
- Compare [[]] vs []
- Right side ran out of items, so inputs are not in the right order
== Pair 8 ==
- Compare [1,[2,[3,[4,[5,6,7]]]],8,9] vs [1,[2,[3,[4,[5,6,0]]]],8,9]
- Compare 1 vs 1
- Compare [2,[3,[4,[5,6,7]]]] vs [2,[3,[4,[5,6,0]]]]
- Compare 2 vs 2
- Compare [3,[4,[5,6,7]]] vs [3,[4,[5,6,0]]]
- Compare 3 vs 3
- Compare [4,[5,6,7]] vs [4,[5,6,0]]
- Compare 4 vs 4
- Compare [5,6,7] vs [5,6,0]
- Compare 5 vs 5
- Compare 6 vs 6
- Compare 7 vs 0
- Right side is smaller, so inputs are not in the right order
What are the indices of the pairs that are already in the right order? (The first pair has index 1, the second pair has index 2, and so on.) In the above example, the pairs in the right order are 1, 2, 4, and 6; the sum of these indices is 13.
Determine which pairs of packets are already in the right order. What is the sum of the indices of those pairs?
| 118
|
--- Day 3: Spiral Memory ---
You come across an experimental new kind of memory stored on an infinite two-dimensional grid.
Each square on the grid is allocated in a spiral pattern starting at a location marked 1 and then counting up while spiraling outward. For example, the first few squares are allocated like this:
17 16 15 14 13
18 5 4 3 12
19 6 1 2 11
20 7 8 9 10
21 22 23---> ...
While this is very space-efficient (no squares are skipped), requested data must be carried back to square 1 (the location of the only access port for this memory system) by programs that can only move up, down, left, or right. They always take the shortest path: the Manhattan Distance between the location of the data and square 1.
For example:
Data from square 1 is carried 0 steps, since it's at the access port.
Data from square 12 is carried 3 steps, such as: down, left, left.
Data from square 23 is carried only 2 steps: up twice.
Data from square 1024 must be carried 31 steps.
How many steps are required to carry the data from the square identified in your puzzle input all the way to the access port?
Your puzzle answer was 552.
--- Part Two ---
As a stress test on the system, the programs here clear the grid and then store the value 1 in square 1. Then, in the same allocation order as shown above, they store the sum of the values in all adjacent squares, including diagonals.
So, the first few squares' values are chosen as follows:
Square 1 starts with the value 1.
Square 2 has only one adjacent filled square (with value 1), so it also stores 1.
Square 3 has both of the above squares as neighbors and stores the sum of their values, 2.
Square 4 has all three of the aforementioned squares as neighbors and stores the sum of their values, 4.
Square 5 only has the first and fourth squares as neighbors, so it gets the value 5.
Once a square is written, its value does not change. Therefore, the first few squares would receive the following values:
147 142 133 122 59
304 5 4 2 57
330 10 1 1 54
351 11 23 25 26
362 747 806---> ...
What is the first value written that is larger than your puzzle input?
| 119
|
--- Day 18: Like a GIF For Your Yard ---
After the million lights incident, the fire code has gotten stricter: now, at most ten thousand lights are allowed. You arrange them in a 100x100 grid.
Never one to let you down, Santa again mails you instructions on the ideal lighting configuration. With so few lights, he says, you'll have to resort to animation.
Start by setting your lights to the included initial configuration (your puzzle input). A # means "on", and a . means "off".
Then, animate your grid in steps, where each step decides the next configuration based on the current one. Each light's next state (either on or off) depends on its current state and the current states of the eight lights adjacent to it (including diagonals). Lights on the edge of the grid might have fewer than eight neighbors; the missing ones always count as "off".
For example, in a simplified 6x6 grid, the light marked A has the neighbors numbered 1 through 8, and the light marked B, which is on an edge, only has the neighbors marked 1 through 5:
1B5...
234...
......
..123.
..8A4.
..765.
The state a light should have next is based on its current state (on or off) plus the number of neighbors that are on:
A light which is on stays on when 2 or 3 neighbors are on, and turns off otherwise.
A light which is off turns on if exactly 3 neighbors are on, and stays off otherwise.
All of the lights update simultaneously; they all consider the same current state before moving to the next.
Here's a few steps from an example configuration of another 6x6 grid:
Initial state:
.#.#.#
...##.
#....#
..#...
#.#..#
####..
After 1 step:
..##..
..##.#
...##.
......
#.....
#.##..
After 2 steps:
..###.
......
..###.
......
.#....
.#....
After 3 steps:
...#..
......
...#..
..##..
......
......
After 4 steps:
......
......
..##..
..##..
......
......
After 4 steps, this example has four lights on.
In your grid of 100x100 lights, given your initial configuration, how many lights are on after 100 steps?
| 120
|
--- Day 12: Leonardo's Monorail ---
You finally reach the top floor of this building: a garden with a slanted glass ceiling. Looks like there are no more stars to be had.
While sitting on a nearby bench amidst some tiger lilies, you manage to decrypt some of the files you extracted from the servers downstairs.
According to these documents, Easter Bunny HQ isn't just this building - it's a collection of buildings in the nearby area. They're all connected by a local monorail, and there's another building not far from here! Unfortunately, being night, the monorail is currently not operating.
You remotely connect to the monorail control systems and discover that the boot sequence expects a password. The password-checking logic (your puzzle input) is easy to extract, but the code it uses is strange: it's assembunny code designed for the new computer you just assembled. You'll have to execute the code and get the password.
The assembunny code you've extracted operates on four registers (a, b, c, and d) that start at 0 and can hold any integer. However, it seems to make use of only a few instructions:
cpy x y copies x (either an integer or the value of a register) into register y.
inc x increases the value of register x by one.
dec x decreases the value of register x by one.
jnz x y jumps to an instruction y away (positive means forward; negative means backward), but only if x is not zero.
The jnz instruction moves relative to itself: an offset of -1 would continue at the previous instruction, while an offset of 2 would skip over the next instruction.
For example:
cpy 41 a
inc a
inc a
dec a
jnz a 2
dec a
The above code would set register a to 41, increase its value by 2, decrease its value by 1, and then skip the last dec a (because a is not zero, so the jnz a 2 skips it), leaving register a at 42. When you move past the last instruction, the program halts.
After executing the assembunny code in your puzzle input, what value is left in register a?
| 121
|
--- Day 9: Stream Processing ---
A large stream blocks your path. According to the locals, it's not safe to cross the stream at the moment because it's full of garbage. You look down at the stream; rather than water, you discover that it's a stream of characters.
You sit for a while and record part of the stream (your puzzle input). The characters represent groups - sequences that begin with { and end with }. Within a group, there are zero or more other things, separated by commas: either another group or garbage. Since groups can contain other groups, a } only closes the most-recently-opened unclosed group - that is, they are nestable. Your puzzle input represents a single, large group which itself contains many smaller ones.
Sometimes, instead of a group, you will find garbage. Garbage begins with < and ends with >. Between those angle brackets, almost any character can appear, including { and }. Within garbage, < has no special meaning.
In a futile attempt to clean up the garbage, some program has canceled some of the characters within it using !: inside garbage, any character that comes after ! should be ignored, including <, >, and even another !.
You don't see any characters that deviate from these rules. Outside garbage, you only find well-formed groups, and garbage always terminates according to the rules above.
Here are some self-contained pieces of garbage:
<>, empty garbage.
<random characters>, garbage containing random characters.
<<<<>, because the extra < are ignored.
<{!>}>, because the first > is canceled.
<!!>, because the second ! is canceled, allowing the > to terminate the garbage.
<!!!>>, because the second ! and the first > are canceled.
<{o"i!a,<{i<a>, which ends at the first >.
Here are some examples of whole streams and the number of groups they contain:
{}, 1 group.
{{{}}}, 3 groups.
{{},{}}, also 3 groups.
{{{},{},{{}}}}, 6 groups.
{<{},{},{{}}>}, 1 group (which itself contains garbage).
{<a>,<a>,<a>,<a>}, 1 group.
{{<a>},{<a>},{<a>},{<a>}}, 5 groups.
{{<!>},{<!>},{<!>},{<a>}}, 2 groups (since all but the last > are canceled).
Your goal is to find the total score for all groups in your input. Each group is assigned a score which is one more than the score of the group that immediately contains it. (The outermost group gets a score of 1.)
{}, score of 1.
{{{}}}, score of 1 + 2 + 3 = 6.
{{},{}}, score of 1 + 2 + 2 = 5.
{{{},{},{{}}}}, score of 1 + 2 + 3 + 3 + 3 + 4 = 16.
{<a>,<a>,<a>,<a>}, score of 1.
{{<ab>},{<ab>},{<ab>},{<ab>}}, score of 1 + 2 + 2 + 2 + 2 = 9.
{{<!!>},{<!!>},{<!!>},{<!!>}}, score of 1 + 2 + 2 + 2 + 2 = 9.
{{<a!>},{<a!>},{<a!>},{<ab>}}, score of 1 + 2 = 3.
What is the total score for all groups in your input?
Your puzzle answer was 11846.
--- Part Two ---
Now, you're ready to remove the garbage.
To prove you've removed it, you need to count all of the characters within the garbage. The leading and trailing < and > don't count, nor do any canceled characters or the ! doing the canceling.
<>, 0 characters.
<random characters>, 17 characters.
<<<<>, 3 characters.
<{!>}>, 2 characters.
<!!>, 0 characters.
<!!!>>, 0 characters.
<{o"i!a,<{i<a>, 10 characters.
How many non-canceled characters are within the garbage in your puzzle input?
| 122
|
--- Day 11: Hex Ed ---
Crossing the bridge, you've barely reached the other side of the stream when a program comes up to you, clearly in distress. "It's my child process," she says, "he's gotten lost in an infinite grid!"
Fortunately for her, you have plenty of experience with infinite grids.
Unfortunately for you, it's a hex grid.
The hexagons ("hexes") in this grid are aligned such that adjacent hexes can be found to the north, northeast, southeast, south, southwest, and northwest:
n /
nw +--+ ne
/ -+ +-
/
sw +--+ se
/ s You have the path the child process took. Starting where he started, you need to determine the fewest number of steps required to reach him. (A "step" means to move from the hex you are in to any adjacent hex.)
For example:
ne,ne,ne is 3 steps away.
ne,ne,sw,sw is 0 steps away (back where you started).
ne,ne,s,s is 2 steps away (se,se).
se,sw,se,sw,sw is 3 steps away (s,s,sw).
Your puzzle answer was 818.
--- Part Two ---
How many steps away is the furthest he ever got from his starting position?
| 123
|
--- Day 12: Hot Springs ---
You finally reach the hot springs! You can see steam rising from secluded areas attached to the primary, ornate building.
As you turn to enter, the researcher stops you. "Wait - I thought you were looking for the hot springs, weren't you?" You indicate that this definitely looks like hot springs to you.
"Oh, sorry, common mistake! This is actually the onsen! The hot springs are next door."
You look in the direction the researcher is pointing and suddenly notice the massive metal helixes towering overhead. "This way!"
It only takes you a few more steps to reach the main gate of the massive fenced-off area containing the springs. You go through the gate and into a small administrative building.
"Hello! What brings you to the hot springs today? Sorry they're not very hot right now; we're having a lava shortage at the moment." You ask about the missing machine parts for Desert Island.
"Oh, all of Gear Island is currently offline! Nothing is being manufactured at the moment, not until we get more lava to heat our forges. And our springs. The springs aren't very springy unless they're hot!"
"Say, could you go up and see why the lava stopped flowing? The springs are too cold for normal operation, but we should be able to find one springy enough to launch you up there!"
There's just one problem - many of the springs have fallen into disrepair, so they're not actually sure which springs would even be safe to use! Worse yet, their condition records of which springs are damaged (your puzzle input) are also damaged! You'll need to help them repair the damaged records.
In the giant field just outside, the springs are arranged into rows. For each row, the condition records show every spring and whether it is operational (.) or damaged (#). This is the part of the condition records that is itself damaged; for some springs, it is simply unknown (?) whether the spring is operational or damaged.
However, the engineer that produced the condition records also duplicated some of this information in a different format! After the list of springs for a given row, the size of each contiguous group of damaged springs is listed in the order those groups appear in the row. This list always accounts for every damaged spring, and each number is the entire size of its contiguous group (that is, groups are always separated by at least one operational spring: #### would always be 4, never 2,2).
So, condition records with no unknown spring conditions might look like this:
#.#.### 1,1,3
.#...#....###. 1,1,3
.#.###.#.###### 1,3,1,6
####.#...#... 4,1,1
#....######..#####. 1,6,5
.###.##....# 3,2,1
However, the condition records are partially damaged; some of the springs' conditions are actually unknown (?). For example:
???.### 1,1,3
.??..??...?##. 1,1,3
?#?#?#?#?#?#?#? 1,3,1,6
????.#...#... 4,1,1
????.######..#####. 1,6,5
?###???????? 3,2,1
Equipped with this information, it is your job to figure out how many different arrangements of operational and broken springs fit the given criteria in each row.
In the first line (???.### 1,1,3), there is exactly one way separate groups of one, one, and three broken springs (in that order) can appear in that row: the first three unknown springs must be broken, then operational, then broken (#.#), making the whole row #.#.###.
The second line is more interesting: .??..??...?##. 1,1,3 could be a total of four different arrangements. The last ? must always be broken (to satisfy the final contiguous group of three broken springs), and each ?? must hide exactly one of the two broken springs. (Neither ?? could be both broken springs or they would form a single contiguous group of two; if that were true, the numbers afterward would have been 2,3 instead.) Since each ?? can either be #. or .#, there are four possible arrangements of springs.
The last line is actually consistent with ten different arrangements! Because the first number is 3, the first and second ? must both be . (if either were #, the first number would have to be 4 or higher). However, the remaining run of unknown spring conditions have many different ways they could hold groups of two and one broken springs:
?###???????? 3,2,1
.###.##.#...
.###.##..#..
.###.##...#.
.###.##....#
.###..##.#..
.###..##..#.
.###..##...#
.###...##.#.
.###...##..#
.###....##.#
In this example, the number of possible arrangements for each row is:
???.### 1,1,3 - 1 arrangement
.??..??...?##. 1,1,3 - 4 arrangements
?#?#?#?#?#?#?#? 1,3,1,6 - 1 arrangement
????.#...#... 4,1,1 - 1 arrangement
????.######..#####. 1,6,5 - 4 arrangements
?###???????? 3,2,1 - 10 arrangements
Adding all of the possible arrangement counts together produces a total of 21 arrangements.
For each row, count all of the different arrangements of operational and broken springs that meet the given criteria. What is the sum of those counts?
Your puzzle answer was 7047.
--- Part Two ---
As you look out at the field of springs, you feel like there are way more springs than the condition records list. When you examine the records, you discover that they were actually folded up this whole time!
To unfold the records, on each row, replace the list of spring conditions with five copies of itself (separated by ?) and replace the list of contiguous groups of damaged springs with five copies of itself (separated by ,).
So, this row:
.# 1
Would become:
.#?.#?.#?.#?.# 1,1,1,1,1
The first line of the above example would become:
???.###????.###????.###????.###????.### 1,1,3,1,1,3,1,1,3,1,1,3,1,1,3
In the above example, after unfolding, the number of possible arrangements for some rows is now much larger:
???.### 1,1,3 - 1 arrangement
.??..??...?##. 1,1,3 - 16384 arrangements
?#?#?#?#?#?#?#? 1,3,1,6 - 1 arrangement
????.#...#... 4,1,1 - 16 arrangements
????.######..#####. 1,6,5 - 2500 arrangements
?###???????? 3,2,1 - 506250 arrangements
After unfolding, adding all of the possible arrangement counts together produces 525152.
Unfold your condition records; what is the new sum of possible arrangement counts?
| 124
|
--- Day 16: The Floor Will Be Lava ---
With the beam of light completely focused somewhere, the reindeer leads you deeper still into the Lava Production Facility. At some point, you realize that the steel facility walls have been replaced with cave, and the doorways are just cave, and the floor is cave, and you're pretty sure this is actually just a giant cave.
Finally, as you approach what must be the heart of the mountain, you see a bright light in a cavern up ahead. There, you discover that the beam of light you so carefully focused is emerging from the cavern wall closest to the facility and pouring all of its energy into a contraption on the opposite side.
Upon closer inspection, the contraption appears to be a flat, two-dimensional square grid containing empty space (.), mirrors (/ and ), and splitters (| and -).
The contraption is aligned so that most of the beam bounces around the grid, but each tile on the grid converts some of the beam's light into heat to melt the rock in the cavern.
You note the layout of the contraption (your puzzle input). For example:
.|.......
|.-......
.....|-...
........|.
..........
............./.\..
.-.-/..|..
.|....-|...//.|....
The beam enters in the top-left corner from the left and heading to the right. Then, its behavior depends on what it encounters as it moves:
If the beam encounters empty space (.), it continues in the same direction.
If the beam encounters a mirror (/ or ), the beam is reflected 90 degrees depending on the angle of the mirror. For instance, a rightward-moving beam that encounters a / mirror would continue upward in the mirror's column, while a rightward-moving beam that encounters a mirror would continue downward from the mirror's column.
If the beam encounters the pointy end of a splitter (| or -), the beam passes through the splitter as if the splitter were empty space. For instance, a rightward-moving beam that encounters a - splitter would continue in the same direction.
If the beam encounters the flat side of a splitter (| or -), the beam is split into two beams going in each of the two directions the splitter's pointy ends are pointing. For instance, a rightward-moving beam that encounters a | splitter would split into two beams: one that continues upward from the splitter's column and one that continues downward from the splitter's column.
Beams do not interact with other beams; a tile can have many beams passing through it at the same time. A tile is energized if that tile has at least one beam pass through it, reflect in it, or split in it.
In the above example, here is how the beam of light bounces around the contraption:
>|<<<....
|v-.^....
.v...|->>>
.v...v^.|.
.v...v^...
.v...v^...v../2\..
<->-/vv|..
.|<<<2-|..v//.|.v..
Beams are only shown on empty tiles; arrows indicate the direction of the beams. If a tile contains beams moving in multiple directions, the number of distinct directions is shown instead. Here is the same diagram but instead only showing whether a tile is energized (#) or not (.):
######....
.#...#....
.#...#####
.#...##...
.#...##...
.#...##...
.#..####..
########..
.#######..
.#...#.#..
Ultimately, in this example, 46 tiles become energized.
The light isn't energizing enough tiles to produce lava; to debug the contraption, you need to start by analyzing the current situation. With the beam starting in the top-left heading right, how many tiles end up being energized?
Your puzzle answer was 7884.
--- Part Two ---
As you try to work out what might be wrong, the reindeer tugs on your shirt and leads you to a nearby control panel. There, a collection of buttons lets you align the contraption so that the beam enters from any edge tile and heading away from that edge. (You can choose either of two directions for the beam if it starts on a corner; for instance, if the beam starts in the bottom-right corner, it can start heading either left or upward.)
So, the beam could start on any tile in the top row (heading downward), any tile in the bottom row (heading upward), any tile in the leftmost column (heading right), or any tile in the rightmost column (heading left). To produce lava, you need to find the configuration that energizes as many tiles as possible.
In the above example, this can be achieved by starting the beam in the fourth tile from the left in the top row:
.|<2<....
|v-v^....
.v.v.|->>>
.v.v.v^.|.
.v.v.v^...
.v.v.v^...v.v/2\..
<-2-/vv|..
.|<<<2-|..v//.|.v..
Using this configuration, 51 tiles are energized:
.#####....
.#.#.#....
.#.#.#####
.#.#.##...
.#.#.##...
.#.#.##...
.#.#####..
########..
.#######..
.#...#.#..
Find the initial beam configuration that energizes the largest number of tiles; how many tiles are energized in that configuration?
| 125
|
--- 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 554.
--- 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?
| 126
|
--- Day 7: Handy Haversacks ---
You land at the regional airport in time for your next flight. In fact, it looks like you'll even have time to grab some food: all flights are currently delayed due to issues in luggage processing.
Due to recent aviation regulations, many rules (your puzzle input) are being enforced about bags and their contents; bags must be color-coded and must contain specific quantities of other color-coded bags. Apparently, nobody responsible for these regulations considered how long they would take to enforce!
For example, consider the following rules:
light red bags contain 1 bright white bag, 2 muted yellow bags.
dark orange bags contain 3 bright white bags, 4 muted yellow bags.
bright white bags contain 1 shiny gold bag.
muted yellow bags contain 2 shiny gold bags, 9 faded blue bags.
shiny gold bags contain 1 dark olive bag, 2 vibrant plum bags.
dark olive bags contain 3 faded blue bags, 4 dotted black bags.
vibrant plum bags contain 5 faded blue bags, 6 dotted black bags.
faded blue bags contain no other bags.
dotted black bags contain no other bags.
These rules specify the required contents for 9 bag types. In this example, every faded blue bag is empty, every vibrant plum bag contains 11 bags (5 faded blue and 6 dotted black), and so on.
You have a shiny gold bag. If you wanted to carry it in at least one other bag, how many different bag colors would be valid for the outermost bag? (In other words: how many colors can, eventually, contain at least one shiny gold bag?)
In the above rules, the following options would be available to you:
A bright white bag, which can hold your shiny gold bag directly.
A muted yellow bag, which can hold your shiny gold bag directly, plus some other bags.
A dark orange bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
A light red bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
So, in this example, the number of bag colors that can eventually contain at least one shiny gold bag is 4.
How many bag colors can eventually contain at least one shiny gold bag? (The list of rules is quite long; make sure you get all of it.)
Your puzzle answer was 335.
--- Part Two ---
It's getting pretty expensive to fly these days - not because of ticket prices, but because of the ridiculous number of bags you need to buy!
Consider again your shiny gold bag and the rules from the above example:
faded blue bags contain 0 other bags.
dotted black bags contain 0 other bags.
vibrant plum bags contain 11 other bags: 5 faded blue bags and 6 dotted black bags.
dark olive bags contain 7 other bags: 3 faded blue bags and 4 dotted black bags.
So, a single shiny gold bag must contain 1 dark olive bag (and the 7 bags within it) plus 2 vibrant plum bags (and the 11 bags within each of those): 1 + 1*7 + 2 + 2*11 = 32 bags!
Of course, the actual rules have a small chance of going several levels deeper than this example; be sure to count all of the bags, even if the nesting becomes topologically impractical!
Here's another example:
shiny gold bags contain 2 dark red bags.
dark red bags contain 2 dark orange bags.
dark orange bags contain 2 dark yellow bags.
dark yellow bags contain 2 dark green bags.
dark green bags contain 2 dark blue bags.
dark blue bags contain 2 dark violet bags.
dark violet bags contain no other bags.
In this example, a single shiny gold bag must contain 126 other bags.
How many individual bags are required inside your single shiny gold bag?
| 127
|
--- Day 14: Docking Data ---
As your ferry approaches the sea port, the captain asks for your help again. The computer system that runs this port isn't compatible with the docking program on the ferry, so the docking parameters aren't being correctly initialized in the docking program's memory.
After a brief inspection, you discover that the sea port's computer system uses a strange bitmask system in its initialization program. Although you don't have the correct decoder chip handy, you can emulate it in software!
The initialization program (your puzzle input) can either update the bitmask or write a value to memory. Values and memory addresses are both 36-bit unsigned integers. For example, ignoring bitmasks for a moment, a line like mem[8] = 11 would write the value 11 to memory address 8.
The bitmask is always given as a string of 36 bits, written with the most significant bit (representing 2^35) on the left and the least significant bit (2^0, that is, the 1s bit) on the right. The current bitmask is applied to values immediately before they are written to memory: a 0 or 1 overwrites the corresponding bit in the value, while an X leaves the bit in the value unchanged.
For example, consider the following program:
mask = XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
mem[8] = 11
mem[7] = 101
mem[8] = 0
This program starts by specifying a bitmask (mask = ....). The mask it specifies will overwrite two bits in every written value: the 2s bit is overwritten with 0, and the 64s bit is overwritten with 1.
The program then attempts to write the value 11 to memory address 8. By expanding everything out to individual bits, the mask is applied as follows:
value: 000000000000000000000000000000001011 (decimal 11)
mask: XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
result: 000000000000000000000000000001001001 (decimal 73)
So, because of the mask, the value 73 is written to memory address 8 instead. Then, the program tries to write 101 to address 7:
value: 000000000000000000000000000001100101 (decimal 101)
mask: XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
result: 000000000000000000000000000001100101 (decimal 101)
This time, the mask has no effect, as the bits it overwrote were already the values the mask tried to set. Finally, the program tries to write 0 to address 8:
value: 000000000000000000000000000000000000 (decimal 0)
mask: XXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXX0X
result: 000000000000000000000000000001000000 (decimal 64)
64 is written to address 8 instead, overwriting the value that was there previously.
To initialize your ferry's docking program, you need the sum of all values left in memory after the initialization program completes. (The entire 36-bit address space begins initialized to the value 0 at every address.) In the above example, only two values in memory are not zero - 101 (at address 7) and 64 (at address 8) - producing a sum of 165.
Execute the initialization program. What is the sum of all values left in memory after it completes? (Do not truncate the sum to 36 bits.)
| 128
|
--- Day 25: Code Chronicle ---
Out of ideas and time, The Historians agree that they should go back to check the Chief Historian's office one last time, just in case he went back there without you noticing.
When you get there, you are surprised to discover that the door to his office is locked! You can hear someone inside, but knocking yields no response. The locks on this floor are all fancy, expensive, virtual versions of five-pin tumbler locks, so you contact North Pole security to see if they can help open the door.
Unfortunately, they've lost track of which locks are installed and which keys go with them, so the best they can do is send over schematics of every lock and every key for the floor you're on (your puzzle input).
The schematics are in a cryptic file format, but they do contain manufacturer information, so you look up their support number.
"Our Virtual Five-Pin Tumbler product? That's our most expensive model! Way more secure than--" You explain that you need to open a door and don't have a lot of time.
"Well, you can't know whether a key opens a lock without actually trying the key in the lock (due to quantum hidden variables), but you can rule out some of the key/lock combinations."
"The virtual system is complicated, but part of it really is a crude simulation of a five-pin tumbler lock, mostly for marketing reasons. If you look at the schematics, you can figure out whether a key could possibly fit in a lock."
He transmits you some example schematics:
#####
.####
.####
.####
.#.#.
.#...
.....
#####
##.##
.#.##
...##
...#.
...#.
.....
.....
#....
#....
#...#
#.#.#
#.###
#####
.....
.....
#.#..
###..
###.#
###.#
#####
.....
.....
.....
#....
#.#..
#.#.#
#####
"The locks are schematics that have the top row filled (#) and the bottom row empty (.); the keys have the top row empty and the bottom row filled. If you look closely, you'll see that each schematic is actually a set of columns of various heights, either extending downward from the top (for locks) or upward from the bottom (for keys)."
"For locks, those are the pins themselves; you can convert the pins in schematics to a list of heights, one per column. For keys, the columns make up the shape of the key where it aligns with pins; those can also be converted to a list of heights."
"So, you could say the first lock has pin heights 0,5,3,4,3:"
#####
.####
.####
.####
.#.#.
.#...
.....
"Or, that the first key has heights 5,0,2,1,3:"
.....
#....
#....
#...#
#.#.#
#.###
#####
"These seem like they should fit together; in the first four columns, the pins and key don't overlap. However, this key cannot be for this lock: in the rightmost column, the lock's pin overlaps with the key, which you know because in that column the sum of the lock height and key height is more than the available space."
"So anyway, you can narrow down the keys you'd need to try by just testing each key with each lock, which means you would have to check... wait, you have how many locks? But the only installation that size is at the North--" You disconnect the call.
In this example, converting both locks to pin heights produces:
0,5,3,4,3
1,2,0,5,3
Converting all three keys to heights produces:
5,0,2,1,3
4,3,4,0,2
3,0,2,0,1
Then, you can try every key with every lock:
Lock 0,5,3,4,3 and key 5,0,2,1,3: overlap in the last column.
Lock 0,5,3,4,3 and key 4,3,4,0,2: overlap in the second column.
Lock 0,5,3,4,3 and key 3,0,2,0,1: all columns fit!
Lock 1,2,0,5,3 and key 5,0,2,1,3: overlap in the first column.
Lock 1,2,0,5,3 and key 4,3,4,0,2: all columns fit!
Lock 1,2,0,5,3 and key 3,0,2,0,1: all columns fit!
So, in this example, the number of unique lock/key pairs that fit together without overlapping in any column is 3.
Analyze your lock and key schematics. How many unique lock/key pairs fit together without overlapping in any column?
| 129
|
--- Day 1: The Tyranny of the Rocket Equation ---
Santa has become stranded at the edge of the Solar System while delivering presents to other planets! To accurately calculate his position in space, safely align his warp drive, and return to Earth in time to save Christmas, he needs you to bring him measurements from fifty stars.
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!
The Elves quickly load you into a spacecraft and prepare to launch.
At the first Go / No Go poll, every Elf is Go until the Fuel Counter-Upper. They haven't determined the amount of fuel required yet.
Fuel required to launch a given module is based on its mass. Specifically, to find the fuel required for a module, take its mass, divide by three, round down, and subtract 2.
For example:
For a mass of 12, divide by 3 and round down to get 4, then subtract 2 to get 2.
For a mass of 14, dividing by 3 and rounding down still yields 4, so the fuel required is also 2.
For a mass of 1969, the fuel required is 654.
For a mass of 100756, the fuel required is 33583.
The Fuel Counter-Upper needs to know the total fuel requirement. To find it, individually calculate the fuel needed for the mass of each module (your puzzle input), then add together all the fuel values.
What is the sum of the fuel requirements for all of the modules on your spacecraft?
Your puzzle answer was 3465154.
--- Part Two ---
During the second Go / No Go poll, the Elf in charge of the Rocket Equation Double-Checker stops the launch sequence. Apparently, you forgot to include additional fuel for the fuel you just added.
Fuel itself requires fuel just like a module - take its mass, divide by three, round down, and subtract 2. However, that fuel also requires fuel, and that fuel requires fuel, and so on. Any mass that would require negative fuel should instead be treated as if it requires zero fuel; the remaining mass, if any, is instead handled by wishing really hard, which has no mass and is outside the scope of this calculation.
So, for each module mass, calculate its fuel and add it to the total. Then, treat the fuel amount you just calculated as the input mass and repeat the process, continuing until a fuel requirement is zero or negative. For example:
A module of mass 14 requires 2 fuel. This fuel requires no further fuel (2 divided by 3 and rounded down is 0, which would call for a negative fuel), so the total fuel required is still just 2.
At first, a module of mass 1969 requires 654 fuel. Then, this fuel requires 216 more fuel (654 / 3 - 2). 216 then requires 70 more fuel, which requires 21 fuel, which requires 5 fuel, which requires no further fuel. So, the total fuel required for a module of mass 1969 is 654 + 216 + 70 + 21 + 5 = 966.
The fuel required by a module of mass 100756 and its fuel is: 33583 + 11192 + 3728 + 1240 + 411 + 135 + 43 + 12 + 2 = 50346.
What is the sum of the fuel requirements for all of the modules on your spacecraft when also taking into account the mass of the added fuel? (Calculate the fuel requirements for each module separately, then add them all up at the end.)
| 130
|
--- Day 20: Pulse Propagation ---
With your help, the Elves manage to find the right parts and fix all of the machines. Now, they just need to send the command to boot up the machines and get the sand flowing again.
The machines are far apart and wired together with long cables. The cables don't connect to the machines directly, but rather to communication modules attached to the machines that perform various initialization tasks and also act as communication relays.
Modules communicate using pulses. Each pulse is either a high pulse or a low pulse. When a module sends a pulse, it sends that type of pulse to each module in its list of destination modules.
There are several different types of modules:
Flip-flop modules (prefix %) are either on or off; they are initially off. If a flip-flop module receives a high pulse, it is ignored and nothing happens. However, if a flip-flop module receives a low pulse, it flips between on and off. If it was off, it turns on and sends a high pulse. If it was on, it turns off and sends a low pulse.
Conjunction modules (prefix &) remember the type of the most recent pulse received from each of their connected input modules; they initially default to remembering a low pulse for each input. When a pulse is received, the conjunction module first updates its memory for that input. Then, if it remembers high pulses for all inputs, it sends a low pulse; otherwise, it sends a high pulse.
There is a single broadcast module (named broadcaster). When it receives a pulse, it sends the same pulse to all of its destination modules.
Here at Desert Machine Headquarters, there is a module with a single button on it called, aptly, the button module. When you push the button, a single low pulse is sent directly to the broadcaster module.
After pushing the button, you must wait until all pulses have been delivered and fully handled before pushing it again. Never push the button if modules are still processing pulses.
Pulses are always processed in the order they are sent. So, if a pulse is sent to modules a, b, and c, and then module a processes its pulse and sends more pulses, the pulses sent to modules b and c would have to be handled first.
The module configuration (your puzzle input) lists each module. The name of the module is preceded by a symbol identifying its type, if any. The name is then followed by an arrow and a list of its destination modules. For example:
broadcaster -> a, b, c
%a -> b
%b -> c
%c -> inv
&inv -> a
In this module configuration, the broadcaster has three destination modules named a, b, and c. Each of these modules is a flip-flop module (as indicated by the % prefix). a outputs to b which outputs to c which outputs to another module named inv. inv is a conjunction module (as indicated by the & prefix) which, because it has only one input, acts like an inverter (it sends the opposite of the pulse type it receives); it outputs to a.
By pushing the button once, the following pulses are sent:
button -low-> broadcaster
broadcaster -low-> a
broadcaster -low-> b
broadcaster -low-> c
a -high-> b
b -high-> c
c -high-> inv
inv -low-> a
a -low-> b
b -low-> c
c -low-> inv
inv -high-> a
After this sequence, the flip-flop modules all end up off, so pushing the button again repeats the same sequence.
Here's a more interesting example:
broadcaster -> a
%a -> inv, con
&inv -> b
%b -> con
&con -> output
This module configuration includes the broadcaster, two flip-flops (named a and b), a single-input conjunction module (inv), a multi-input conjunction module (con), and an untyped module named output (for testing purposes). The multi-input conjunction module con watches the two flip-flop modules and, if they're both on, sends a low pulse to the output module.
Here's what happens if you push the button once:
button -low-> broadcaster
broadcaster -low-> a
a -high-> inv
a -high-> con
inv -low-> b
con -high-> output
b -high-> con
con -low-> output
Both flip-flops turn on and a low pulse is sent to output! However, now that both flip-flops are on and con remembers a high pulse from each of its two inputs, pushing the button a second time does something different:
button -low-> broadcaster
broadcaster -low-> a
a -low-> inv
a -low-> con
inv -high-> b
con -high-> output
Flip-flop a turns off! Now, con remembers a low pulse from module a, and so it sends only a high pulse to output.
Push the button a third time:
button -low-> broadcaster
broadcaster -low-> a
a -high-> inv
a -high-> con
inv -low-> b
con -low-> output
b -low-> con
con -high-> output
This time, flip-flop a turns on, then flip-flop b turns off. However, before b can turn off, the pulse sent to con is handled first, so it briefly remembers all high pulses for its inputs and sends a low pulse to output. After that, flip-flop b turns off, which causes con to update its state and send a high pulse to output.
Finally, with a on and b off, push the button a fourth time:
button -low-> broadcaster
broadcaster -low-> a
a -low-> inv
a -low-> con
inv -high-> b
con -high-> output
This completes the cycle: a turns off, causing con to remember only low pulses and restoring all modules to their original states.
To get the cables warmed up, the Elves have pushed the button 1000 times. How many pulses got sent as a result (including the pulses sent by the button itself)?
In the first example, the same thing happens every time the button is pushed: 8 low pulses and 4 high pulses are sent. So, after pushing the button 1000 times, 8000 low pulses and 4000 high pulses are sent. Multiplying these together gives 32000000.
In the second example, after pushing the button 1000 times, 4250 low pulses and 2750 high pulses are sent. Multiplying these together gives 11687500.
Consult your module configuration; determine the number of low pulses and high pulses that would be sent after pushing the button 1000 times, waiting for all pulses to be fully handled after each push of the button. What do you get if you multiply the total number of low pulses sent by the total number of high pulses sent?
| 131
|
--- Day 1: Inverse Captcha ---
The night before Christmas, one of Santa's Elves calls you in a panic. "The printer's broken! We can't print the Naughty or Nice List!" By the time you make it to sub-basement 17, there are only a few minutes until midnight. "We have a big problem," she says; "there must be almost fifty bugs in this system, but nothing else can print The List. Stand in this square, quick! There's no time to explain; if you can convince them to pay you in stars, you'll be able to--" She pulls a lever and the world goes blurry.
When your eyes can focus again, everything seems a lot more pixelated than before. She must have sent you inside the computer! You check the system clock: 25 milliseconds until midnight. With that much time, you should be able to collect all fifty stars by December 25th.
Collect stars by solving puzzles. Two puzzles will be made available on each day millisecond in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!
You're standing in a room with "digitization quarantine" written in LEDs along one wall. The only door is locked, but it includes a small interface. "Restricted Area - Strictly No Digitized Users Allowed."
It goes on to explain that you may only leave by solving a captcha to prove you're not a human. Apparently, you only get one millisecond to solve the captcha: too fast for a normal human, but it feels like hours to you.
The captcha requires you to review a sequence of digits (your puzzle input) and find the sum of all digits that match the next digit in the list. The list is circular, so the digit after the last digit is the first digit in the list.
For example:
1122 produces a sum of 3 (1 + 2) because the first digit (1) matches the second digit and the third digit (2) matches the fourth digit.
1111 produces 4 because each digit (all 1) matches the next.
1234 produces 0 because no digit matches the next.
91212129 produces 9 because the only digit that matches the next one is the last digit, 9.
What is the solution to your captcha?
Your puzzle answer was 1044.
--- Part Two ---
You notice a progress bar that jumps to 50% completion. Apparently, the door isn't yet satisfied, but it did emit a star as encouragement. The instructions change:
Now, instead of considering the next digit, it wants you to consider the digit halfway around the circular list. That is, if your list contains 10 items, only include a digit in your sum if the digit 10/2 = 5 steps forward matches it. Fortunately, your list has an even number of elements.
For example:
1212 produces 6: the list contains 4 items, and all four digits match the digit 2 items ahead.
1221 produces 0, because every comparison is between a 1 and a 2.
123425 produces 4, because both 2s match each other, but no other digit has a match.
123123 produces 12.
12131415 produces 4.
What is the solution to your new captcha?
| 132
|
--- Day 20: Pulse Propagation ---
With your help, the Elves manage to find the right parts and fix all of the machines. Now, they just need to send the command to boot up the machines and get the sand flowing again.
The machines are far apart and wired together with long cables. The cables don't connect to the machines directly, but rather to communication modules attached to the machines that perform various initialization tasks and also act as communication relays.
Modules communicate using pulses. Each pulse is either a high pulse or a low pulse. When a module sends a pulse, it sends that type of pulse to each module in its list of destination modules.
There are several different types of modules:
Flip-flop modules (prefix %) are either on or off; they are initially off. If a flip-flop module receives a high pulse, it is ignored and nothing happens. However, if a flip-flop module receives a low pulse, it flips between on and off. If it was off, it turns on and sends a high pulse. If it was on, it turns off and sends a low pulse.
Conjunction modules (prefix &) remember the type of the most recent pulse received from each of their connected input modules; they initially default to remembering a low pulse for each input. When a pulse is received, the conjunction module first updates its memory for that input. Then, if it remembers high pulses for all inputs, it sends a low pulse; otherwise, it sends a high pulse.
There is a single broadcast module (named broadcaster). When it receives a pulse, it sends the same pulse to all of its destination modules.
Here at Desert Machine Headquarters, there is a module with a single button on it called, aptly, the button module. When you push the button, a single low pulse is sent directly to the broadcaster module.
After pushing the button, you must wait until all pulses have been delivered and fully handled before pushing it again. Never push the button if modules are still processing pulses.
Pulses are always processed in the order they are sent. So, if a pulse is sent to modules a, b, and c, and then module a processes its pulse and sends more pulses, the pulses sent to modules b and c would have to be handled first.
The module configuration (your puzzle input) lists each module. The name of the module is preceded by a symbol identifying its type, if any. The name is then followed by an arrow and a list of its destination modules. For example:
broadcaster -> a, b, c
%a -> b
%b -> c
%c -> inv
&inv -> a
In this module configuration, the broadcaster has three destination modules named a, b, and c. Each of these modules is a flip-flop module (as indicated by the % prefix). a outputs to b which outputs to c which outputs to another module named inv. inv is a conjunction module (as indicated by the & prefix) which, because it has only one input, acts like an inverter (it sends the opposite of the pulse type it receives); it outputs to a.
By pushing the button once, the following pulses are sent:
button -low-> broadcaster
broadcaster -low-> a
broadcaster -low-> b
broadcaster -low-> c
a -high-> b
b -high-> c
c -high-> inv
inv -low-> a
a -low-> b
b -low-> c
c -low-> inv
inv -high-> a
After this sequence, the flip-flop modules all end up off, so pushing the button again repeats the same sequence.
Here's a more interesting example:
broadcaster -> a
%a -> inv, con
&inv -> b
%b -> con
&con -> output
This module configuration includes the broadcaster, two flip-flops (named a and b), a single-input conjunction module (inv), a multi-input conjunction module (con), and an untyped module named output (for testing purposes). The multi-input conjunction module con watches the two flip-flop modules and, if they're both on, sends a low pulse to the output module.
Here's what happens if you push the button once:
button -low-> broadcaster
broadcaster -low-> a
a -high-> inv
a -high-> con
inv -low-> b
con -high-> output
b -high-> con
con -low-> output
Both flip-flops turn on and a low pulse is sent to output! However, now that both flip-flops are on and con remembers a high pulse from each of its two inputs, pushing the button a second time does something different:
button -low-> broadcaster
broadcaster -low-> a
a -low-> inv
a -low-> con
inv -high-> b
con -high-> output
Flip-flop a turns off! Now, con remembers a low pulse from module a, and so it sends only a high pulse to output.
Push the button a third time:
button -low-> broadcaster
broadcaster -low-> a
a -high-> inv
a -high-> con
inv -low-> b
con -low-> output
b -low-> con
con -high-> output
This time, flip-flop a turns on, then flip-flop b turns off. However, before b can turn off, the pulse sent to con is handled first, so it briefly remembers all high pulses for its inputs and sends a low pulse to output. After that, flip-flop b turns off, which causes con to update its state and send a high pulse to output.
Finally, with a on and b off, push the button a fourth time:
button -low-> broadcaster
broadcaster -low-> a
a -low-> inv
a -low-> con
inv -high-> b
con -high-> output
This completes the cycle: a turns off, causing con to remember only low pulses and restoring all modules to their original states.
To get the cables warmed up, the Elves have pushed the button 1000 times. How many pulses got sent as a result (including the pulses sent by the button itself)?
In the first example, the same thing happens every time the button is pushed: 8 low pulses and 4 high pulses are sent. So, after pushing the button 1000 times, 8000 low pulses and 4000 high pulses are sent. Multiplying these together gives 32000000.
In the second example, after pushing the button 1000 times, 4250 low pulses and 2750 high pulses are sent. Multiplying these together gives 11687500.
Consult your module configuration; determine the number of low pulses and high pulses that would be sent after pushing the button 1000 times, waiting for all pulses to be fully handled after each push of the button. What do you get if you multiply the total number of low pulses sent by the total number of high pulses sent?
Your puzzle answer was 681194780.
--- Part Two ---
The final machine responsible for moving the sand down to Island Island has a module attached named rx. The machine turns on when a single low pulse is sent to rx.
Reset all modules to their default states. Waiting for all pulses to be fully handled after each button press, what is the fewest number of button presses required to deliver a single low pulse to the module named rx?
| 133
|
--- Day 17: Reservoir Research ---
You arrive in the year 18. If it weren't for the coat you got in 1018, you would be very cold: the North Pole base hasn't even been constructed.
Rather, it hasn't been constructed yet. The Elves are making a little progress, but there's not a lot of liquid water in this climate, so they're getting very dehydrated. Maybe there's more underground?
You scan a two-dimensional vertical slice of the ground nearby and discover that it is mostly sand with veins of clay. The scan only provides data with a granularity of square meters, but it should be good enough to determine how much water is trapped there. In the scan, x represents the distance to the right, and y represents the distance down. There is also a spring of water near the surface at x=500, y=0. The scan identifies which square meters are clay (your puzzle input).
For example, suppose your scan shows the following veins of clay:
x=495, y=2..7
y=7, x=495..501
x=501, y=3..7
x=498, y=2..4
x=506, y=1..2
x=498, y=10..13
x=504, y=10..13
y=13, x=498..504
Rendering clay as #, sand as ., and the water spring as +, and with x increasing to the right and y increasing downward, this becomes:
44444455555555
99999900000000
45678901234567
0 ......+.......
1 ............#.
2 .#..#.......#.
3 .#..#..#......
4 .#..#..#......
5 .#.....#......
6 .#.....#......
7 .#######......
8 ..............
9 ..............
10 ....#.....#...
11 ....#.....#...
12 ....#.....#...
13 ....#######...
The spring of water will produce water forever. Water can move through sand, but is blocked by clay. Water always moves down when possible, and spreads to the left and right otherwise, filling space that has clay on both sides and falling out otherwise.
For example, if five squares of water are created, they will flow downward until they reach the clay and settle there. Water that has come to rest is shown here as ~, while sand through which water has passed (but which is now dry again) is shown as |:
......+.......
......|.....#.
.#..#.|.....#.
.#..#.|#......
.#..#.|#......
.#....|#......
.#~~~~~#......
.#######......
..............
..............
....#.....#...
....#.....#...
....#.....#...
....#######...
Two squares of water can't occupy the same location. If another five squares of water are created, they will settle on the first five, filling the clay reservoir a little more:
......+.......
......|.....#.
.#..#.|.....#.
.#..#.|#......
.#..#.|#......
.#~~~~~#......
.#~~~~~#......
.#######......
..............
..............
....#.....#...
....#.....#...
....#.....#...
....#######...
Water pressure does not apply in this scenario. If another four squares of water are created, they will stay on the right side of the barrier, and no water will reach the left side:
......+.......
......|.....#.
.#..#.|.....#.
.#..#~~#......
.#..#~~#......
.#~~~~~#......
.#~~~~~#......
.#######......
..............
..............
....#.....#...
....#.....#...
....#.....#...
....#######...
At this point, the top reservoir overflows. While water can reach the tiles above the surface of the water, it cannot settle there, and so the next five squares of water settle like this:
......+.......
......|.....#.
.#..#||||...#.
.#..#~~#|.....
.#..#~~#|.....
.#~~~~~#|.....
.#~~~~~#|.....
.#######|.....
........|.....
........|.....
....#...|.#...
....#...|.#...
....#~~~~~#...
....#######...
Note especially the leftmost |: the new squares of water can reach this tile, but cannot stop there. Instead, eventually, they all fall to the right and settle in the reservoir below.
After 10 more squares of water, the bottom reservoir is also full:
......+.......
......|.....#.
.#..#||||...#.
.#..#~~#|.....
.#..#~~#|.....
.#~~~~~#|.....
.#~~~~~#|.....
.#######|.....
........|.....
........|.....
....#~~~~~#...
....#~~~~~#...
....#~~~~~#...
....#######...
Finally, while there is nowhere left for the water to settle, it can reach a few more tiles before overflowing beyond the bottom of the scanned data:
......+....... (line not counted: above minimum y value)
......|.....#.
.#..#||||...#.
.#..#~~#|.....
.#..#~~#|.....
.#~~~~~#|.....
.#~~~~~#|.....
.#######|.....
........|.....
...|||||||||..
...|#~~~~~#|..
...|#~~~~~#|..
...|#~~~~~#|..
...|#######|..
...|.......|.. (line not counted: below maximum y value)
...|.......|.. (line not counted: below maximum y value)
...|.......|.. (line not counted: below maximum y value)
How many tiles can be reached by the water? To prevent counting forever, ignore tiles with a y coordinate smaller than the smallest y coordinate in your scan data or larger than the largest one. Any x coordinate is valid. In this example, the lowest y coordinate given is 1, and the highest is 13, causing the water spring (in row 0) and the water falling off the bottom of the render (in rows 14 through infinity) to be ignored.
So, in the example above, counting both water at rest (~) and other sand tiles the water can hypothetically reach (|), the total number of tiles the water can reach is 57.
How many tiles can the water reach within the range of y values in your scan?
| 134
|
--- Day 13: Care Package ---
As you ponder the solitude of space and the ever-increasing three-hour roundtrip for messages between you and Earth, you notice that the Space Mail Indicator Light is blinking. To help keep you sane, the Elves have sent you a care package.
It's a new game for the ship's arcade cabinet! Unfortunately, the arcade is all the way on the other end of the ship. Surely, it won't be hard to build your own - the care package even comes with schematics.
The arcade cabinet runs Intcode software like the game the Elves sent (your puzzle input). It has a primitive screen capable of drawing square tiles on a grid. The software draws tiles to the screen with output instructions: every three output instructions specify the x position (distance from the left), y position (distance from the top), and tile id. The tile id is interpreted as follows:
0 is an empty tile. No game object appears in this tile.
1 is a wall tile. Walls are indestructible barriers.
2 is a block tile. Blocks can be broken by the ball.
3 is a horizontal paddle tile. The paddle is indestructible.
4 is a ball tile. The ball moves diagonally and bounces off objects.
For example, a sequence of output values like 1,2,3,6,5,4 would draw a horizontal paddle tile (1 tile from the left and 2 tiles from the top) and a ball tile (6 tiles from the left and 5 tiles from the top).
Start the game. How many block tiles are on the screen when the game exits?
Your puzzle answer was 247.
--- Part Two ---
The game didn't run because you didn't put in any quarters. Unfortunately, you did not bring any quarters. Memory address 0 represents the number of quarters that have been inserted; set it to 2 to play for free.
The arcade cabinet has a joystick that can move left and right. The software reads the position of the joystick with input instructions:
If the joystick is in the neutral position, provide 0.
If the joystick is tilted to the left, provide -1.
If the joystick is tilted to the right, provide 1.
The arcade cabinet also has a segment display capable of showing a single number that represents the player's current score. When three output instructions specify X=-1, Y=0, the third output instruction is not a tile; the value instead specifies the new score to show in the segment display. For example, a sequence of output values like -1,0,12345 would show 12345 as the player's current score.
Beat the game by breaking all the blocks. What is your score after the last block is broken?
| 135
|
--- Day 8: Resonant Collinearity ---
You find yourselves on the roof of a top-secret Easter Bunny installation.
While The Historians do their thing, you take a look at the familiar huge antenna. Much to your surprise, it seems to have been reconfigured to emit a signal that makes people 0.1% more likely to buy Easter Bunny brand Imitation Mediocre Chocolate as a Christmas gift! Unthinkable!
Scanning across the city, you find that there are actually many such antennas. Each antenna is tuned to a specific frequency indicated by a single lowercase letter, uppercase letter, or digit. You create a map (your puzzle input) of these antennas. For example:
............
........0...
.....0......
.......0....
....0.......
......A.....
............
............
........A...
.........A..
............
............
The signal only applies its nefarious effect at specific antinodes based on the resonant frequencies of the antennas. In particular, an antinode occurs at any point that is perfectly in line with two antennas of the same frequency - but only when one of the antennas is twice as far away as the other. This means that for any pair of antennas with the same frequency, there are two antinodes, one on either side of them.
So, for these two antennas with frequency a, they create the two antinodes marked with #:
..........
...#......
..........
....a.....
..........
.....a....
..........
......#...
..........
..........
Adding a third antenna with the same frequency creates several more antinodes. It would ideally add four antinodes, but two are off the right side of the map, so instead it adds only two:
..........
...#......
#.........
....a.....
........a.
.....a....
..#.......
......#...
..........
..........
Antennas with different frequencies don't create antinodes; A and a count as different frequencies. However, antinodes can occur at locations that contain antennas. In this diagram, the lone antenna with frequency capital A creates no antinodes but has a lowercase-a-frequency antinode at its location:
..........
...#......
#.........
....a.....
........a.
.....a....
..#.......
......A...
..........
..........
The first example has antennas with two different frequencies, so the antinodes they create look like this, plus an antinode overlapping the topmost A-frequency antenna:
......#....#
...#....0...
....#0....#.
..#....0....
....0....#..
.#....A.....
...#........
#......#....
........A...
.........A..
..........#.
..........#.
Because the topmost A-frequency antenna overlaps with a 0-frequency antinode, there are 14 total unique locations that contain an antinode within the bounds of the map.
Calculate the impact of the signal. How many unique locations within the bounds of the map contain an antinode?
Your puzzle answer was 220.
--- Part Two ---
Watching over your shoulder as you work, one of The Historians asks if you took the effects of resonant harmonics into your calculations.
Whoops!
After updating your model, it turns out that an antinode occurs at any grid position exactly in line with at least two antennas of the same frequency, regardless of distance. This means that some of the new antinodes will occur at the position of each antenna (unless that antenna is the only one of its frequency).
So, these three T-frequency antennas now create many antinodes:
T....#....
...T......
.T....#...
.........#
..#.......
..........
...#......
..........
....#.....
..........
In fact, the three T-frequency antennas are all exactly in line with two antennas, so they are all also antinodes! This brings the total number of antinodes in the above example to 9.
The original example now has 34 antinodes, including the antinodes that appear on every antenna:
##....#....#
.#.#....0...
..#.#0....#.
..##...0....
....0....#..
.#...#A....#
...#..#.....
#....#.#....
..#.....A...
....#....A..
.#........#.
...#......##
Calculate the impact of the signal using this updated model. How many unique locations within the bounds of the map contain an antinode?
| 136
|
--- Day 11: Monkey in the Middle ---
As you finally start making your way upriver, you realize your pack is much lighter than you remember. Just then, one of the items from your pack goes flying overhead. Monkeys are playing Keep Away with your missing things!
To get your stuff back, you need to be able to predict where the monkeys will throw your items. After some careful observation, you realize the monkeys operate based on how worried you are about each item.
You take some notes (your puzzle input) on the items each monkey currently has, how worried you are about those items, and how the monkey makes decisions based on your worry level. For example:
Monkey 0:
Starting items: 79, 98
Operation: new = old * 19
Test: divisible by 23
If true: throw to monkey 2
If false: throw to monkey 3
Monkey 1:
Starting items: 54, 65, 75, 74
Operation: new = old + 6
Test: divisible by 19
If true: throw to monkey 2
If false: throw to monkey 0
Monkey 2:
Starting items: 79, 60, 97
Operation: new = old * old
Test: divisible by 13
If true: throw to monkey 1
If false: throw to monkey 3
Monkey 3:
Starting items: 74
Operation: new = old + 3
Test: divisible by 17
If true: throw to monkey 0
If false: throw to monkey 1
Each monkey has several attributes:
Starting items lists your worry level for each item the monkey is currently holding in the order they will be inspected.
Operation shows how your worry level changes as that monkey inspects an item. (An operation like new = old * 5 means that your worry level after the monkey inspected the item is five times whatever your worry level was before inspection.)
Test shows how the monkey uses your worry level to decide where to throw an item next.
If true shows what happens with an item if the Test was true.
If false shows what happens with an item if the Test was false.
After each monkey inspects an item but before it tests your worry level, your relief that the monkey's inspection didn't damage the item causes your worry level to be divided by three and rounded down to the nearest integer.
The monkeys take turns inspecting and throwing items. On a single monkey's turn, it inspects and throws all of the items it is holding one at a time and in the order listed. Monkey 0 goes first, then monkey 1, and so on until each monkey has had one turn. The process of each monkey taking a single turn is called a round.
When a monkey throws an item to another monkey, the item goes on the end of the recipient monkey's list. A monkey that starts a round with no items could end up inspecting and throwing many items by the time its turn comes around. If a monkey is holding no items at the start of its turn, its turn ends.
In the above example, the first round proceeds as follows:
Monkey 0:
Monkey inspects an item with a worry level of 79.
Worry level is multiplied by 19 to 1501.
Monkey gets bored with item. Worry level is divided by 3 to 500.
Current worry level is not divisible by 23.
Item with worry level 500 is thrown to monkey 3.
Monkey inspects an item with a worry level of 98.
Worry level is multiplied by 19 to 1862.
Monkey gets bored with item. Worry level is divided by 3 to 620.
Current worry level is not divisible by 23.
Item with worry level 620 is thrown to monkey 3.
Monkey 1:
Monkey inspects an item with a worry level of 54.
Worry level increases by 6 to 60.
Monkey gets bored with item. Worry level is divided by 3 to 20.
Current worry level is not divisible by 19.
Item with worry level 20 is thrown to monkey 0.
Monkey inspects an item with a worry level of 65.
Worry level increases by 6 to 71.
Monkey gets bored with item. Worry level is divided by 3 to 23.
Current worry level is not divisible by 19.
Item with worry level 23 is thrown to monkey 0.
Monkey inspects an item with a worry level of 75.
Worry level increases by 6 to 81.
Monkey gets bored with item. Worry level is divided by 3 to 27.
Current worry level is not divisible by 19.
Item with worry level 27 is thrown to monkey 0.
Monkey inspects an item with a worry level of 74.
Worry level increases by 6 to 80.
Monkey gets bored with item. Worry level is divided by 3 to 26.
Current worry level is not divisible by 19.
Item with worry level 26 is thrown to monkey 0.
Monkey 2:
Monkey inspects an item with a worry level of 79.
Worry level is multiplied by itself to 6241.
Monkey gets bored with item. Worry level is divided by 3 to 2080.
Current worry level is divisible by 13.
Item with worry level 2080 is thrown to monkey 1.
Monkey inspects an item with a worry level of 60.
Worry level is multiplied by itself to 3600.
Monkey gets bored with item. Worry level is divided by 3 to 1200.
Current worry level is not divisible by 13.
Item with worry level 1200 is thrown to monkey 3.
Monkey inspects an item with a worry level of 97.
Worry level is multiplied by itself to 9409.
Monkey gets bored with item. Worry level is divided by 3 to 3136.
Current worry level is not divisible by 13.
Item with worry level 3136 is thrown to monkey 3.
Monkey 3:
Monkey inspects an item with a worry level of 74.
Worry level increases by 3 to 77.
Monkey gets bored with item. Worry level is divided by 3 to 25.
Current worry level is not divisible by 17.
Item with worry level 25 is thrown to monkey 1.
Monkey inspects an item with a worry level of 500.
Worry level increases by 3 to 503.
Monkey gets bored with item. Worry level is divided by 3 to 167.
Current worry level is not divisible by 17.
Item with worry level 167 is thrown to monkey 1.
Monkey inspects an item with a worry level of 620.
Worry level increases by 3 to 623.
Monkey gets bored with item. Worry level is divided by 3 to 207.
Current worry level is not divisible by 17.
Item with worry level 207 is thrown to monkey 1.
Monkey inspects an item with a worry level of 1200.
Worry level increases by 3 to 1203.
Monkey gets bored with item. Worry level is divided by 3 to 401.
Current worry level is not divisible by 17.
Item with worry level 401 is thrown to monkey 1.
Monkey inspects an item with a worry level of 3136.
Worry level increases by 3 to 3139.
Monkey gets bored with item. Worry level is divided by 3 to 1046.
Current worry level is not divisible by 17.
Item with worry level 1046 is thrown to monkey 1.
After round 1, the monkeys are holding items with these worry levels:
Monkey 0: 20, 23, 27, 26
Monkey 1: 2080, 25, 167, 207, 401, 1046
Monkey 2:
Monkey 3:
Monkeys 2 and 3 aren't holding any items at the end of the round; they both inspected items during the round and threw them all before the round ended.
This process continues for a few more rounds:
After round 2, the monkeys are holding items with these worry levels:
Monkey 0: 695, 10, 71, 135, 350
Monkey 1: 43, 49, 58, 55, 362
Monkey 2:
Monkey 3:
After round 3, the monkeys are holding items with these worry levels:
Monkey 0: 16, 18, 21, 20, 122
Monkey 1: 1468, 22, 150, 286, 739
Monkey 2:
Monkey 3:
After round 4, the monkeys are holding items with these worry levels:
Monkey 0: 491, 9, 52, 97, 248, 34
Monkey 1: 39, 45, 43, 258
Monkey 2:
Monkey 3:
After round 5, the monkeys are holding items with these worry levels:
Monkey 0: 15, 17, 16, 88, 1037
Monkey 1: 20, 110, 205, 524, 72
Monkey 2:
Monkey 3:
After round 6, the monkeys are holding items with these worry levels:
Monkey 0: 8, 70, 176, 26, 34
Monkey 1: 481, 32, 36, 186, 2190
Monkey 2:
Monkey 3:
After round 7, the monkeys are holding items with these worry levels:
Monkey 0: 162, 12, 14, 64, 732, 17
Monkey 1: 148, 372, 55, 72
Monkey 2:
Monkey 3:
After round 8, the monkeys are holding items with these worry levels:
Monkey 0: 51, 126, 20, 26, 136
Monkey 1: 343, 26, 30, 1546, 36
Monkey 2:
Monkey 3:
After round 9, the monkeys are holding items with these worry levels:
Monkey 0: 116, 10, 12, 517, 14
Monkey 1: 108, 267, 43, 55, 288
Monkey 2:
Monkey 3:
After round 10, the monkeys are holding items with these worry levels:
Monkey 0: 91, 16, 20, 98
Monkey 1: 481, 245, 22, 26, 1092, 30
Monkey 2:
Monkey 3:
...
After round 15, the monkeys are holding items with these worry levels:
Monkey 0: 83, 44, 8, 184, 9, 20, 26, 102
Monkey 1: 110, 36
Monkey 2:
Monkey 3:
...
After round 20, the monkeys are holding items with these worry levels:
Monkey 0: 10, 12, 14, 26, 34
Monkey 1: 245, 93, 53, 199, 115
Monkey 2:
Monkey 3:
Chasing all of the monkeys at once is impossible; you're going to have to focus on the two most active monkeys if you want any hope of getting your stuff back. Count the total number of times each monkey inspects items over 20 rounds:
Monkey 0 inspected items 101 times.
Monkey 1 inspected items 95 times.
Monkey 2 inspected items 7 times.
Monkey 3 inspected items 105 times.
In this example, the two most active monkeys inspected items 101 and 105 times. The level of monkey business in this situation can be found by multiplying these together: 10605.
Figure out which monkeys to chase by counting how many items they inspect over 20 rounds. What is the level of monkey business after 20 rounds of stuff-slinging simian shenanigans?
Your puzzle answer was 57838.
--- Part Two ---
You're worried you might not ever get your items back. So worried, in fact, that your relief that a monkey's inspection didn't damage an item no longer causes your worry level to be divided by three.
Unfortunately, that relief was all that was keeping your worry levels from reaching ridiculous levels. You'll need to find another way to keep your worry levels manageable.
At this rate, you might be putting up with these monkeys for a very long time - possibly 10000 rounds!
With these new rules, you can still figure out the monkey business after 10000 rounds. Using the same example above:
== After round 1 ==
Monkey 0 inspected items 2 times.
Monkey 1 inspected items 4 times.
Monkey 2 inspected items 3 times.
Monkey 3 inspected items 6 times.
== After round 20 ==
Monkey 0 inspected items 99 times.
Monkey 1 inspected items 97 times.
Monkey 2 inspected items 8 times.
Monkey 3 inspected items 103 times.
== After round 1000 ==
Monkey 0 inspected items 5204 times.
Monkey 1 inspected items 4792 times.
Monkey 2 inspected items 199 times.
Monkey 3 inspected items 5192 times.
== After round 2000 ==
Monkey 0 inspected items 10419 times.
Monkey 1 inspected items 9577 times.
Monkey 2 inspected items 392 times.
Monkey 3 inspected items 10391 times.
== After round 3000 ==
Monkey 0 inspected items 15638 times.
Monkey 1 inspected items 14358 times.
Monkey 2 inspected items 587 times.
Monkey 3 inspected items 15593 times.
== After round 4000 ==
Monkey 0 inspected items 20858 times.
Monkey 1 inspected items 19138 times.
Monkey 2 inspected items 780 times.
Monkey 3 inspected items 20797 times.
== After round 5000 ==
Monkey 0 inspected items 26075 times.
Monkey 1 inspected items 23921 times.
Monkey 2 inspected items 974 times.
Monkey 3 inspected items 26000 times.
== After round 6000 ==
Monkey 0 inspected items 31294 times.
Monkey 1 inspected items 28702 times.
Monkey 2 inspected items 1165 times.
Monkey 3 inspected items 31204 times.
== After round 7000 ==
Monkey 0 inspected items 36508 times.
Monkey 1 inspected items 33488 times.
Monkey 2 inspected items 1360 times.
Monkey 3 inspected items 36400 times.
== After round 8000 ==
Monkey 0 inspected items 41728 times.
Monkey 1 inspected items 38268 times.
Monkey 2 inspected items 1553 times.
Monkey 3 inspected items 41606 times.
== After round 9000 ==
Monkey 0 inspected items 46945 times.
Monkey 1 inspected items 43051 times.
Monkey 2 inspected items 1746 times.
Monkey 3 inspected items 46807 times.
== After round 10000 ==
Monkey 0 inspected items 52166 times.
Monkey 1 inspected items 47830 times.
Monkey 2 inspected items 1938 times.
Monkey 3 inspected items 52013 times.
After 10000 rounds, the two most active monkeys inspected items 52166 and 52013 times. Multiplying these together, the level of monkey business in this situation is now 2713310158.
Worry levels are no longer divided by three after each item is inspected; you'll need to find another way to keep your worry levels manageable. Starting again from the initial state in your puzzle input, what is the level of monkey business after 10000 rounds?
| 137
|
--- Day 24: Immune System Simulator 20XX ---
After a weird buzzing noise, you appear back at the man's cottage. He seems relieved to see his friend, but quickly notices that the little reindeer caught some kind of cold while out exploring.
The portly man explains that this reindeer's immune system isn't similar to regular reindeer immune systems:
The immune system and the infection each have an army made up of several groups; each group consists of one or more identical units. The armies repeatedly fight until only one army has units remaining.
Units within a group all have the same hit points (amount of damage a unit can take before it is destroyed), attack damage (the amount of damage each unit deals), an attack type, an initiative (higher initiative units attack first and win ties), and sometimes weaknesses or immunities. Here is an example group:
18 units each with 729 hit points (weak to fire; immune to cold, slashing)
with an attack that does 8 radiation damage at initiative 10
Each group also has an effective power: the number of units in that group multiplied by their attack damage. The above group has an effective power of 18 * 8 = 144. Groups never have zero or negative units; instead, the group is removed from combat.
Each fight consists of two phases: target selection and attacking.
During the target selection phase, each group attempts to choose one target. In decreasing order of effective power, groups choose their targets; in a tie, the group with the higher initiative chooses first. The attacking group chooses to target the group in the enemy army to which it would deal the most damage (after accounting for weaknesses and immunities, but not accounting for whether the defending group has enough units to actually receive all of that damage).
If an attacking group is considering two defending groups to which it would deal equal damage, it chooses to target the defending group with the largest effective power; if there is still a tie, it chooses the defending group with the highest initiative. If it cannot deal any defending groups damage, it does not choose a target. Defending groups can only be chosen as a target by one attacking group.
At the end of the target selection phase, each group has selected zero or one groups to attack, and each group is being attacked by zero or one groups.
During the attacking phase, each group deals damage to the target it selected, if any. Groups attack in decreasing order of initiative, regardless of whether they are part of the infection or the immune system. (If a group contains no units, it cannot attack.)
The damage an attacking group deals to a defending group depends on the attacking group's attack type and the defending group's immunities and weaknesses. By default, an attacking group would deal damage equal to its effective power to the defending group. However, if the defending group is immune to the attacking group's attack type, the defending group instead takes no damage; if the defending group is weak to the attacking group's attack type, the defending group instead takes double damage.
The defending group only loses whole units from damage; damage is always dealt in such a way that it kills the most units possible, and any remaining damage to a unit that does not immediately kill it is ignored. For example, if a defending group contains 10 units with 10 hit points each and receives 75 damage, it loses exactly 7 units and is left with 3 units at full health.
After the fight is over, if both armies still contain units, a new fight begins; combat only ends once one army has lost all of its units.
For example, consider the following armies:
Immune System:
17 units each with 5390 hit points (weak to radiation, bludgeoning) with
an attack that does 4507 fire damage at initiative 2
989 units each with 1274 hit points (immune to fire; weak to bludgeoning,
slashing) with an attack that does 25 slashing damage at initiative 3
Infection:
801 units each with 4706 hit points (weak to radiation) with an attack
that does 116 bludgeoning damage at initiative 1
4485 units each with 2961 hit points (immune to radiation; weak to fire,
cold) with an attack that does 12 slashing damage at initiative 4
If these armies were to enter combat, the following fights, including details during the target selection and attacking phases, would take place:
Immune System:
Group 1 contains 17 units
Group 2 contains 989 units
Infection:
Group 1 contains 801 units
Group 2 contains 4485 units
Infection group 1 would deal defending group 1 185832 damage
Infection group 1 would deal defending group 2 185832 damage
Infection group 2 would deal defending group 2 107640 damage
Immune System group 1 would deal defending group 1 76619 damage
Immune System group 1 would deal defending group 2 153238 damage
Immune System group 2 would deal defending group 1 24725 damage
Infection group 2 attacks defending group 2, killing 84 units
Immune System group 2 attacks defending group 1, killing 4 units
Immune System group 1 attacks defending group 2, killing 51 units
Infection group 1 attacks defending group 1, killing 17 units
Immune System:
Group 2 contains 905 units
Infection:
Group 1 contains 797 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 184904 damage
Immune System group 2 would deal defending group 1 22625 damage
Immune System group 2 would deal defending group 2 22625 damage
Immune System group 2 attacks defending group 1, killing 4 units
Infection group 1 attacks defending group 2, killing 144 units
Immune System:
Group 2 contains 761 units
Infection:
Group 1 contains 793 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 183976 damage
Immune System group 2 would deal defending group 1 19025 damage
Immune System group 2 would deal defending group 2 19025 damage
Immune System group 2 attacks defending group 1, killing 4 units
Infection group 1 attacks defending group 2, killing 143 units
Immune System:
Group 2 contains 618 units
Infection:
Group 1 contains 789 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 183048 damage
Immune System group 2 would deal defending group 1 15450 damage
Immune System group 2 would deal defending group 2 15450 damage
Immune System group 2 attacks defending group 1, killing 3 units
Infection group 1 attacks defending group 2, killing 143 units
Immune System:
Group 2 contains 475 units
Infection:
Group 1 contains 786 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 182352 damage
Immune System group 2 would deal defending group 1 11875 damage
Immune System group 2 would deal defending group 2 11875 damage
Immune System group 2 attacks defending group 1, killing 2 units
Infection group 1 attacks defending group 2, killing 142 units
Immune System:
Group 2 contains 333 units
Infection:
Group 1 contains 784 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 181888 damage
Immune System group 2 would deal defending group 1 8325 damage
Immune System group 2 would deal defending group 2 8325 damage
Immune System group 2 attacks defending group 1, killing 1 unit
Infection group 1 attacks defending group 2, killing 142 units
Immune System:
Group 2 contains 191 units
Infection:
Group 1 contains 783 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 181656 damage
Immune System group 2 would deal defending group 1 4775 damage
Immune System group 2 would deal defending group 2 4775 damage
Immune System group 2 attacks defending group 1, killing 1 unit
Infection group 1 attacks defending group 2, killing 142 units
Immune System:
Group 2 contains 49 units
Infection:
Group 1 contains 782 units
Group 2 contains 4434 units
Infection group 1 would deal defending group 2 181424 damage
Immune System group 2 would deal defending group 1 1225 damage
Immune System group 2 would deal defending group 2 1225 damage
Immune System group 2 attacks defending group 1, killing 0 units
Infection group 1 attacks defending group 2, killing 49 units
Immune System:
No groups remain.
Infection:
Group 1 contains 782 units
Group 2 contains 4434 units
In the example above, the winning army ends up with 782 + 4434 = 5216 units.
You scan the reindeer's condition (your puzzle input); the white-bearded man looks nervous. As it stands now, how many units would the winning army have?
Your puzzle answer was 38008.
--- Part Two ---
Things aren't looking good for the reindeer. The man asks whether more milk and cookies would help you think.
If only you could give the reindeer's immune system a boost, you might be able to change the outcome of the combat.
A boost is an integer increase in immune system units' attack damage. For example, if you were to boost the above example's immune system's units by 1570, the armies would instead look like this:
Immune System:
17 units each with 5390 hit points (weak to radiation, bludgeoning) with
an attack that does 6077 fire damage at initiative 2
989 units each with 1274 hit points (immune to fire; weak to bludgeoning,
slashing) with an attack that does 1595 slashing damage at initiative 3
Infection:
801 units each with 4706 hit points (weak to radiation) with an attack
that does 116 bludgeoning damage at initiative 1
4485 units each with 2961 hit points (immune to radiation; weak to fire,
cold) with an attack that does 12 slashing damage at initiative 4
With this boost, the combat proceeds differently:
Immune System:
Group 2 contains 989 units
Group 1 contains 17 units
Infection:
Group 1 contains 801 units
Group 2 contains 4485 units
Infection group 1 would deal defending group 2 185832 damage
Infection group 1 would deal defending group 1 185832 damage
Infection group 2 would deal defending group 1 53820 damage
Immune System group 2 would deal defending group 1 1577455 damage
Immune System group 2 would deal defending group 2 1577455 damage
Immune System group 1 would deal defending group 2 206618 damage
Infection group 2 attacks defending group 1, killing 9 units
Immune System group 2 attacks defending group 1, killing 335 units
Immune System group 1 attacks defending group 2, killing 32 units
Infection group 1 attacks defending group 2, killing 84 units
Immune System:
Group 2 contains 905 units
Group 1 contains 8 units
Infection:
Group 1 contains 466 units
Group 2 contains 4453 units
Infection group 1 would deal defending group 2 108112 damage
Infection group 1 would deal defending group 1 108112 damage
Infection group 2 would deal defending group 1 53436 damage
Immune System group 2 would deal defending group 1 1443475 damage
Immune System group 2 would deal defending group 2 1443475 damage
Immune System group 1 would deal defending group 2 97232 damage
Infection group 2 attacks defending group 1, killing 8 units
Immune System group 2 attacks defending group 1, killing 306 units
Infection group 1 attacks defending group 2, killing 29 units
Immune System:
Group 2 contains 876 units
Infection:
Group 2 contains 4453 units
Group 1 contains 160 units
Infection group 2 would deal defending group 2 106872 damage
Immune System group 2 would deal defending group 2 1397220 damage
Immune System group 2 would deal defending group 1 1397220 damage
Infection group 2 attacks defending group 2, killing 83 units
Immune System group 2 attacks defending group 2, killing 427 units
After a few fights...
Immune System:
Group 2 contains 64 units
Infection:
Group 2 contains 214 units
Group 1 contains 19 units
Infection group 2 would deal defending group 2 5136 damage
Immune System group 2 would deal defending group 2 102080 damage
Immune System group 2 would deal defending group 1 102080 damage
Infection group 2 attacks defending group 2, killing 4 units
Immune System group 2 attacks defending group 2, killing 32 units
Immune System:
Group 2 contains 60 units
Infection:
Group 1 contains 19 units
Group 2 contains 182 units
Infection group 1 would deal defending group 2 4408 damage
Immune System group 2 would deal defending group 1 95700 damage
Immune System group 2 would deal defending group 2 95700 damage
Immune System group 2 attacks defending group 1, killing 19 units
Immune System:
Group 2 contains 60 units
Infection:
Group 2 contains 182 units
Infection group 2 would deal defending group 2 4368 damage
Immune System group 2 would deal defending group 2 95700 damage
Infection group 2 attacks defending group 2, killing 3 units
Immune System group 2 attacks defending group 2, killing 30 units
After a few more fights...
Immune System:
Group 2 contains 51 units
Infection:
Group 2 contains 40 units
Infection group 2 would deal defending group 2 960 damage
Immune System group 2 would deal defending group 2 81345 damage
Infection group 2 attacks defending group 2, killing 0 units
Immune System group 2 attacks defending group 2, killing 27 units
Immune System:
Group 2 contains 51 units
Infection:
Group 2 contains 13 units
Infection group 2 would deal defending group 2 312 damage
Immune System group 2 would deal defending group 2 81345 damage
Infection group 2 attacks defending group 2, killing 0 units
Immune System group 2 attacks defending group 2, killing 13 units
Immune System:
Group 2 contains 51 units
Infection:
No groups remain.
This boost would allow the immune system's armies to win! It would be left with 51 units.
You don't even know how you could boost the reindeer's immune system or what effect it might have, so you need to be cautious and find the smallest boost that would allow the immune system to win.
How many units does the immune system have left after getting the smallest boost it needs to win?
| 138
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--- Day 14: Space Stoichiometry ---
As you approach the rings of Saturn, your ship's low fuel indicator turns on. There isn't any fuel here, but the rings have plenty of raw material. Perhaps your ship's Inter-Stellar Refinery Union brand nanofactory can turn these raw materials into fuel.
You ask the nanofactory to produce a list of the reactions it can perform that are relevant to this process (your puzzle input). Every reaction turns some quantities of specific input chemicals into some quantity of an output chemical. Almost every chemical is produced by exactly one reaction; the only exception, ORE, is the raw material input to the entire process and is not produced by a reaction.
You just need to know how much ORE you'll need to collect before you can produce one unit of FUEL.
Each reaction gives specific quantities for its inputs and output; reactions cannot be partially run, so only whole integer multiples of these quantities can be used. (It's okay to have leftover chemicals when you're done, though.) For example, the reaction 1 A, 2 B, 3 C => 2 D means that exactly 2 units of chemical D can be produced by consuming exactly 1 A, 2 B and 3 C. You can run the full reaction as many times as necessary; for example, you could produce 10 D by consuming 5 A, 10 B, and 15 C.
Suppose your nanofactory produces the following list of reactions:
10 ORE => 10 A
1 ORE => 1 B
7 A, 1 B => 1 C
7 A, 1 C => 1 D
7 A, 1 D => 1 E
7 A, 1 E => 1 FUEL
The first two reactions use only ORE as inputs; they indicate that you can produce as much of chemical A as you want (in increments of 10 units, each 10 costing 10 ORE) and as much of chemical B as you want (each costing 1 ORE). To produce 1 FUEL, a total of 31 ORE is required: 1 ORE to produce 1 B, then 30 more ORE to produce the 7 + 7 + 7 + 7 = 28 A (with 2 extra A wasted) required in the reactions to convert the B into C, C into D, D into E, and finally E into FUEL. (30 A is produced because its reaction requires that it is created in increments of 10.)
Or, suppose you have the following list of reactions:
9 ORE => 2 A
8 ORE => 3 B
7 ORE => 5 C
3 A, 4 B => 1 AB
5 B, 7 C => 1 BC
4 C, 1 A => 1 CA
2 AB, 3 BC, 4 CA => 1 FUEL
The above list of reactions requires 165 ORE to produce 1 FUEL:
Consume 45 ORE to produce 10 A.
Consume 64 ORE to produce 24 B.
Consume 56 ORE to produce 40 C.
Consume 6 A, 8 B to produce 2 AB.
Consume 15 B, 21 C to produce 3 BC.
Consume 16 C, 4 A to produce 4 CA.
Consume 2 AB, 3 BC, 4 CA to produce 1 FUEL.
Here are some larger examples:
13312 ORE for 1 FUEL:
157 ORE => 5 NZVS
165 ORE => 6 DCFZ
44 XJWVT, 5 KHKGT, 1 QDVJ, 29 NZVS, 9 GPVTF, 48 HKGWZ => 1 FUEL
12 HKGWZ, 1 GPVTF, 8 PSHF => 9 QDVJ
179 ORE => 7 PSHF
177 ORE => 5 HKGWZ
7 DCFZ, 7 PSHF => 2 XJWVT
165 ORE => 2 GPVTF
3 DCFZ, 7 NZVS, 5 HKGWZ, 10 PSHF => 8 KHKGT
180697 ORE for 1 FUEL:
2 VPVL, 7 FWMGM, 2 CXFTF, 11 MNCFX => 1 STKFG
17 NVRVD, 3 JNWZP => 8 VPVL
53 STKFG, 6 MNCFX, 46 VJHF, 81 HVMC, 68 CXFTF, 25 GNMV => 1 FUEL
22 VJHF, 37 MNCFX => 5 FWMGM
139 ORE => 4 NVRVD
144 ORE => 7 JNWZP
5 MNCFX, 7 RFSQX, 2 FWMGM, 2 VPVL, 19 CXFTF => 3 HVMC
5 VJHF, 7 MNCFX, 9 VPVL, 37 CXFTF => 6 GNMV
145 ORE => 6 MNCFX
1 NVRVD => 8 CXFTF
1 VJHF, 6 MNCFX => 4 RFSQX
176 ORE => 6 VJHF
2210736 ORE for 1 FUEL:
171 ORE => 8 CNZTR
7 ZLQW, 3 BMBT, 9 XCVML, 26 XMNCP, 1 WPTQ, 2 MZWV, 1 RJRHP => 4 PLWSL
114 ORE => 4 BHXH
14 VRPVC => 6 BMBT
6 BHXH, 18 KTJDG, 12 WPTQ, 7 PLWSL, 31 FHTLT, 37 ZDVW => 1 FUEL
6 WPTQ, 2 BMBT, 8 ZLQW, 18 KTJDG, 1 XMNCP, 6 MZWV, 1 RJRHP => 6 FHTLT
15 XDBXC, 2 LTCX, 1 VRPVC => 6 ZLQW
13 WPTQ, 10 LTCX, 3 RJRHP, 14 XMNCP, 2 MZWV, 1 ZLQW => 1 ZDVW
5 BMBT => 4 WPTQ
189 ORE => 9 KTJDG
1 MZWV, 17 XDBXC, 3 XCVML => 2 XMNCP
12 VRPVC, 27 CNZTR => 2 XDBXC
15 KTJDG, 12 BHXH => 5 XCVML
3 BHXH, 2 VRPVC => 7 MZWV
121 ORE => 7 VRPVC
7 XCVML => 6 RJRHP
5 BHXH, 4 VRPVC => 5 LTCX
Given the list of reactions in your puzzle input, what is the minimum amount of ORE required to produce exactly 1 FUEL?
| 139
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--- Day 21: Scrambled Letters and Hash ---
The computer system you're breaking into uses a weird scrambling function to store its passwords. It shouldn't be much trouble to create your own scrambled password so you can add it to the system; you just have to implement the scrambler.
The scrambling function is a series of operations (the exact list is provided in your puzzle input). Starting with the password to be scrambled, apply each operation in succession to the string. The individual operations behave as follows:
swap position X with position Y means that the letters at indexes X and Y (counting from 0) should be swapped.
swap letter X with letter Y means that the letters X and Y should be swapped (regardless of where they appear in the string).
rotate left/right X steps means that the whole string should be rotated; for example, one right rotation would turn abcd into dabc.
rotate based on position of letter X means that the whole string should be rotated to the right based on the index of letter X (counting from 0) as determined before this instruction does any rotations. Once the index is determined, rotate the string to the right one time, plus a number of times equal to that index, plus one additional time if the index was at least 4.
reverse positions X through Y means that the span of letters at indexes X through Y (including the letters at X and Y) should be reversed in order.
move position X to position Y means that the letter which is at index X should be removed from the string, then inserted such that it ends up at index Y.
For example, suppose you start with abcde and perform the following operations:
swap position 4 with position 0 swaps the first and last letters, producing the input for the next step, ebcda.
swap letter d with letter b swaps the positions of d and b: edcba.
reverse positions 0 through 4 causes the entire string to be reversed, producing abcde.
rotate left 1 step shifts all letters left one position, causing the first letter to wrap to the end of the string: bcdea.
move position 1 to position 4 removes the letter at position 1 (c), then inserts it at position 4 (the end of the string): bdeac.
move position 3 to position 0 removes the letter at position 3 (a), then inserts it at position 0 (the front of the string): abdec.
rotate based on position of letter b finds the index of letter b (1), then rotates the string right once plus a number of times equal to that index (2): ecabd.
rotate based on position of letter d finds the index of letter d (4), then rotates the string right once, plus a number of times equal to that index, plus an additional time because the index was at least 4, for a total of 6 right rotations: decab.
After these steps, the resulting scrambled password is decab.
Now, you just need to generate a new scrambled password and you can access the system. Given the list of scrambling operations in your puzzle input, what is the result of scrambling abcdefgh?
Your puzzle answer was cbeghdaf.
--- Part Two ---
You scrambled the password correctly, but you discover that you can't actually modify the password file on the system. You'll need to un-scramble one of the existing passwords by reversing the scrambling process.
What is the un-scrambled version of the scrambled password fbgdceah?
| 140
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--- Day 11: Chronal Charge ---
You watch the Elves and their sleigh fade into the distance as they head toward the North Pole.
Actually, you're the one fading. The falling sensation returns.
The low fuel warning light is illuminated on your wrist-mounted device. Tapping it once causes it to project a hologram of the situation: a 300x300 grid of fuel cells and their current power levels, some negative. You're not sure what negative power means in the context of time travel, but it can't be good.
Each fuel cell has a coordinate ranging from 1 to 300 in both the X (horizontal) and Y (vertical) direction. In X,Y notation, the top-left cell is 1,1, and the top-right cell is 300,1.
The interface lets you select any 3x3 square of fuel cells. To increase your chances of getting to your destination, you decide to choose the 3x3 square with the largest total power.
The power level in a given fuel cell can be found through the following process:
Find the fuel cell's rack ID, which is its X coordinate plus 10.
Begin with a power level of the rack ID times the Y coordinate.
Increase the power level by the value of the grid serial number (your puzzle input).
Set the power level to itself multiplied by the rack ID.
Keep only the hundreds digit of the power level (so 12345 becomes 3; numbers with no hundreds digit become 0).
Subtract 5 from the power level.
For example, to find the power level of the fuel cell at 3,5 in a grid with serial number 8:
The rack ID is 3 + 10 = 13.
The power level starts at 13 * 5 = 65.
Adding the serial number produces 65 + 8 = 73.
Multiplying by the rack ID produces 73 * 13 = 949.
The hundreds digit of 949 is 9.
Subtracting 5 produces 9 - 5 = 4.
So, the power level of this fuel cell is 4.
Here are some more example power levels:
Fuel cell at 122,79, grid serial number 57: power level -5.
Fuel cell at 217,196, grid serial number 39: power level 0.
Fuel cell at 101,153, grid serial number 71: power level 4.
Your goal is to find the 3x3 square which has the largest total power. The square must be entirely within the 300x300 grid. Identify this square using the X,Y coordinate of its top-left fuel cell. For example:
For grid serial number 18, the largest total 3x3 square has a top-left corner of 33,45 (with a total power of 29); these fuel cells appear in the middle of this 5x5 region:
-2 -4 4 4 4
-4 4 4 4 -5
4 3 3 4 -4
1 1 2 4 -3
-1 0 2 -5 -2
For grid serial number 42, the largest 3x3 square's top-left is 21,61 (with a total power of 30); they are in the middle of this region:
-3 4 2 2 2
-4 4 3 3 4
-5 3 3 4 -4
4 3 3 4 -3
3 3 3 -5 -1
What is the X,Y coordinate of the top-left fuel cell of the 3x3 square with the largest total power?
Your puzzle answer was 20,68.
--- Part Two ---
You discover a dial on the side of the device; it seems to let you select a square of any size, not just 3x3. Sizes from 1x1 to 300x300 are supported.
Realizing this, you now must find the square of any size with the largest total power. Identify this square by including its size as a third parameter after the top-left coordinate: a 9x9 square with a top-left corner of 3,5 is identified as 3,5,9.
For example:
For grid serial number 18, the largest total square (with a total power of 113) is 16x16 and has a top-left corner of 90,269, so its identifier is 90,269,16.
For grid serial number 42, the largest total square (with a total power of 119) is 12x12 and has a top-left corner of 232,251, so its identifier is 232,251,12.
What is the X,Y,size identifier of the square with the largest total power?
| 141
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--- Day 15: Science for Hungry People ---
Today, you set out on the task of perfecting your milk-dunking cookie recipe. All you have to do is find the right balance of ingredients.
Your recipe leaves room for exactly 100 teaspoons of ingredients. You make a list of the remaining ingredients you could use to finish the recipe (your puzzle input) and their properties per teaspoon:
capacity (how well it helps the cookie absorb milk)
durability (how well it keeps the cookie intact when full of milk)
flavor (how tasty it makes the cookie)
texture (how it improves the feel of the cookie)
calories (how many calories it adds to the cookie)
You can only measure ingredients in whole-teaspoon amounts accurately, and you have to be accurate so you can reproduce your results in the future. The total score of a cookie can be found by adding up each of the properties (negative totals become 0) and then multiplying together everything except calories.
For instance, suppose you have these two ingredients:
Butterscotch: capacity -1, durability -2, flavor 6, texture 3, calories 8
Cinnamon: capacity 2, durability 3, flavor -2, texture -1, calories 3
Then, choosing to use 44 teaspoons of butterscotch and 56 teaspoons of cinnamon (because the amounts of each ingredient must add up to 100) would result in a cookie with the following properties:
A capacity of 44*-1 + 56*2 = 68
A durability of 44*-2 + 56*3 = 80
A flavor of 44*6 + 56*-2 = 152
A texture of 44*3 + 56*-1 = 76
Multiplying these together (68 * 80 * 152 * 76, ignoring calories for now) results in a total score of 62842880, which happens to be the best score possible given these ingredients. If any properties had produced a negative total, it would have instead become zero, causing the whole score to multiply to zero.
Given the ingredients in your kitchen and their properties, what is the total score of the highest-scoring cookie you can make?
| 142
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--- Day 21: Springdroid Adventure ---
You lift off from Pluto and start flying in the direction of Santa.
While experimenting further with the tractor beam, you accidentally pull an asteroid directly into your ship! It deals significant damage to your hull and causes your ship to begin tumbling violently.
You can send a droid out to investigate, but the tumbling is causing enough artificial gravity that one wrong step could send the droid through a hole in the hull and flying out into space.
The clear choice for this mission is a droid that can jump over the holes in the hull - a springdroid.
You can use an Intcode program (your puzzle input) running on an ASCII-capable computer to program the springdroid. However, springdroids don't run Intcode; instead, they run a simplified assembly language called springscript.
While a springdroid is certainly capable of navigating the artificial gravity and giant holes, it has one downside: it can only remember at most 15 springscript instructions.
The springdroid will move forward automatically, constantly thinking about whether to jump. The springscript program defines the logic for this decision.
Springscript programs only use Boolean values, not numbers or strings. Two registers are available: T, the temporary value register, and J, the jump register. If the jump register is true at the end of the springscript program, the springdroid will try to jump. Both of these registers start with the value false.
Springdroids have a sensor that can detect whether there is ground at various distances in the direction it is facing; these values are provided in read-only registers. Your springdroid can detect ground at four distances: one tile away (A), two tiles away (B), three tiles away (C), and four tiles away (D). If there is ground at the given distance, the register will be true; if there is a hole, the register will be false.
There are only three instructions available in springscript:
AND X Y sets Y to true if both X and Y are true; otherwise, it sets Y to false.
OR X Y sets Y to true if at least one of X or Y is true; otherwise, it sets Y to false.
NOT X Y sets Y to true if X is false; otherwise, it sets Y to false.
In all three instructions, the second argument (Y) needs to be a writable register (either T or J). The first argument (X) can be any register (including A, B, C, or D).
For example, the one-instruction program NOT A J means "if the tile immediately in front of me is not ground, jump".
Or, here is a program that jumps if a three-tile-wide hole (with ground on the other side of the hole) is detected:
NOT A J
NOT B T
AND T J
NOT C T
AND T J
AND D J
The Intcode program expects ASCII inputs and outputs. It will begin by displaying a prompt; then, input the desired instructions one per line. End each line with a newline (ASCII code 10). When you have finished entering your program, provide the command WALK followed by a newline to instruct the springdroid to begin surveying the hull.
If the springdroid falls into space, an ASCII rendering of the last moments of its life will be produced. In these, @ is the springdroid, # is hull, and . is empty space. For example, suppose you program the springdroid like this:
NOT D J
WALK
This one-instruction program sets J to true if and only if there is no ground four tiles away. In other words, it attempts to jump into any hole it finds:
.................
.................
@................
#####.###########
.................
.................
.@...............
#####.###########
.................
..@..............
.................
#####.###########
...@.............
.................
.................
#####.###########
.................
....@............
.................
#####.###########
.................
.................
.....@...........
#####.###########
.................
.................
.................
#####@###########
However, if the springdroid successfully makes it across, it will use an output instruction to indicate the amount of damage to the hull as a single giant integer outside the normal ASCII range.
Program the springdroid with logic that allows it to survey the hull without falling into space. What amount of hull damage does it report?
Your puzzle answer was 19349722.
--- Part Two ---
There are many areas the springdroid can't reach. You flip through the manual and discover a way to increase its sensor range.
Instead of ending your springcode program with WALK, use RUN. Doing this will enable extended sensor mode, capable of sensing ground up to nine tiles away. This data is available in five new read-only registers:
Register E indicates whether there is ground five tiles away.
Register F indicates whether there is ground six tiles away.
Register G indicates whether there is ground seven tiles away.
Register H indicates whether there is ground eight tiles away.
Register I indicates whether there is ground nine tiles away.
All other functions remain the same.
Successfully survey the rest of the hull by ending your program with RUN. What amount of hull damage does the springdroid now report?
| 143
|
--- Day 24: Air Duct Spelunking ---
You've finally met your match; the doors that provide access to the roof are locked tight, and all of the controls and related electronics are inaccessible. You simply can't reach them.
The robot that cleans the air ducts, however, can.
It's not a very fast little robot, but you reconfigure it to be able to interface with some of the exposed wires that have been routed through the HVAC system. If you can direct it to each of those locations, you should be able to bypass the security controls.
You extract the duct layout for this area from some blueprints you acquired and create a map with the relevant locations marked (your puzzle input). 0 is your current location, from which the cleaning robot embarks; the other numbers are (in no particular order) the locations the robot needs to visit at least once each. Walls are marked as #, and open passages are marked as .. Numbers behave like open passages.
For example, suppose you have a map like the following:
###########
#0.1.....2#
#.#######.#
#4.......3#
###########
To reach all of the points of interest as quickly as possible, you would have the robot take the following path:
0 to 4 (2 steps)
4 to 1 (4 steps; it can't move diagonally)
1 to 2 (6 steps)
2 to 3 (2 steps)
Since the robot isn't very fast, you need to find it the shortest route. This path is the fewest steps (in the above example, a total of 14) required to start at 0 and then visit every other location at least once.
Given your actual map, and starting from location 0, what is the fewest number of steps required to visit every non-0 number marked on the map at least once?
Your puzzle answer was 442.
--- Part Two ---
Of course, if you leave the cleaning robot somewhere weird, someone is bound to notice.
What is the fewest number of steps required to start at 0, visit every non-0 number marked on the map at least once, and then return to 0?
| 144
|
--- Day 2: Bathroom Security ---
You arrive at Easter Bunny Headquarters under cover of darkness. However, you left in such a rush that you forgot to use the bathroom! Fancy office buildings like this one usually have keypad locks on their bathrooms, so you search the front desk for the code.
"In order to improve security," the document you find says, "bathroom codes will no longer be written down. Instead, please memorize and follow the procedure below to access the bathrooms."
The document goes on to explain that each button to be pressed can be found by starting on the previous button and moving to adjacent buttons on the keypad: U moves up, D moves down, L moves left, and R moves right. Each line of instructions corresponds to one button, starting at the previous button (or, for the first line, the "5" button); press whatever button you're on at the end of each line. If a move doesn't lead to a button, ignore it.
You can't hold it much longer, so you decide to figure out the code as you walk to the bathroom. You picture a keypad like this:
1 2 3
4 5 6
7 8 9
Suppose your instructions are:
ULL
RRDDD
LURDL
UUUUD
You start at "5" and move up (to "2"), left (to "1"), and left (you can't, and stay on "1"), so the first button is 1.
Starting from the previous button ("1"), you move right twice (to "3") and then down three times (stopping at "9" after two moves and ignoring the third), ending up with 9.
Continuing from "9", you move left, up, right, down, and left, ending with 8.
Finally, you move up four times (stopping at "2"), then down once, ending with 5.
So, in this example, the bathroom code is 1985.
Your puzzle input is the instructions from the document you found at the front desk. What is the bathroom code?
| 145
|
--- Day 19: Aplenty ---
The Elves of Gear Island are thankful for your help and send you on your way. They even have a hang glider that someone stole from Desert Island; since you're already going that direction, it would help them a lot if you would use it to get down there and return it to them.
As you reach the bottom of the relentless avalanche of machine parts, you discover that they're already forming a formidable heap. Don't worry, though - a group of Elves is already here organizing the parts, and they have a system.
To start, each part is rated in each of four categories:
x: Extremely cool looking
m: Musical (it makes a noise when you hit it)
a: Aerodynamic
s: Shiny
Then, each part is sent through a series of workflows that will ultimately accept or reject the part. Each workflow has a name and contains a list of rules; each rule specifies a condition and where to send the part if the condition is true. The first rule that matches the part being considered is applied immediately, and the part moves on to the destination described by the rule. (The last rule in each workflow has no condition and always applies if reached.)
Consider the workflow ex{x>10:one,m<20:two,a>30:R,A}. This workflow is named ex and contains four rules. If workflow ex were considering a specific part, it would perform the following steps in order:
Rule "x>10:one": If the part's x is more than 10, send the part to the workflow named one.
Rule "m<20:two": Otherwise, if the part's m is less than 20, send the part to the workflow named two.
Rule "a>30:R": Otherwise, if the part's a is more than 30, the part is immediately rejected (R).
Rule "A": Otherwise, because no other rules matched the part, the part is immediately accepted (A).
If a part is sent to another workflow, it immediately switches to the start of that workflow instead and never returns. If a part is accepted (sent to A) or rejected (sent to R), the part immediately stops any further processing.
The system works, but it's not keeping up with the torrent of weird metal shapes. The Elves ask if you can help sort a few parts and give you the list of workflows and some part ratings (your puzzle input). For example:
px{a<2006:qkq,m>2090:A,rfg}
pv{a>1716:R,A}
lnx{m>1548:A,A}
rfg{s<537:gd,x>2440:R,A}
qs{s>3448:A,lnx}
qkq{x<1416:A,crn}
crn{x>2662:A,R}
in{s<1351:px,qqz}
qqz{s>2770:qs,m<1801:hdj,R}
gd{a>3333:R,R}
hdj{m>838:A,pv}
{x=787,m=2655,a=1222,s=2876}
{x=1679,m=44,a=2067,s=496}
{x=2036,m=264,a=79,s=2244}
{x=2461,m=1339,a=466,s=291}
{x=2127,m=1623,a=2188,s=1013}
The workflows are listed first, followed by a blank line, then the ratings of the parts the Elves would like you to sort. All parts begin in the workflow named in. In this example, the five listed parts go through the following workflows:
{x=787,m=2655,a=1222,s=2876}: in -> qqz -> qs -> lnx -> A
{x=1679,m=44,a=2067,s=496}: in -> px -> rfg -> gd -> R
{x=2036,m=264,a=79,s=2244}: in -> qqz -> hdj -> pv -> A
{x=2461,m=1339,a=466,s=291}: in -> px -> qkq -> crn -> R
{x=2127,m=1623,a=2188,s=1013}: in -> px -> rfg -> A
Ultimately, three parts are accepted. Adding up the x, m, a, and s rating for each of the accepted parts gives 7540 for the part with x=787, 4623 for the part with x=2036, and 6951 for the part with x=2127. Adding all of the ratings for all of the accepted parts gives the sum total of 19114.
Sort through all of the parts you've been given; what do you get if you add together all of the rating numbers for all of the parts that ultimately get accepted?
Your puzzle answer was 319062.
--- Part Two ---
Even with your help, the sorting process still isn't fast enough.
One of the Elves comes up with a new plan: rather than sort parts individually through all of these workflows, maybe you can figure out in advance which combinations of ratings will be accepted or rejected.
Each of the four ratings (x, m, a, s) can have an integer value ranging from a minimum of 1 to a maximum of 4000. Of all possible distinct combinations of ratings, your job is to figure out which ones will be accepted.
In the above example, there are 167409079868000 distinct combinations of ratings that will be accepted.
Consider only your list of workflows; the list of part ratings that the Elves wanted you to sort is no longer relevant. How many distinct combinations of ratings will be accepted by the Elves' workflows?
| 146
|
--- Day 11: Space Police ---
On the way to Jupiter, you're pulled over by the Space Police.
"Attention, unmarked spacecraft! You are in violation of Space Law! All spacecraft must have a clearly visible registration identifier! You have 24 hours to comply or be sent to Space Jail!"
Not wanting to be sent to Space Jail, you radio back to the Elves on Earth for help. Although it takes almost three hours for their reply signal to reach you, they send instructions for how to power up the emergency hull painting robot and even provide a small Intcode program (your puzzle input) that will cause it to paint your ship appropriately.
There's just one problem: you don't have an emergency hull painting robot.
You'll need to build a new emergency hull painting robot. The robot needs to be able to move around on the grid of square panels on the side of your ship, detect the color of its current panel, and paint its current panel black or white. (All of the panels are currently black.)
The Intcode program will serve as the brain of the robot. The program uses input instructions to access the robot's camera: provide 0 if the robot is over a black panel or 1 if the robot is over a white panel. Then, the program will output two values:
First, it will output a value indicating the color to paint the panel the robot is over: 0 means to paint the panel black, and 1 means to paint the panel white.
Second, it will output a value indicating the direction the robot should turn: 0 means it should turn left 90 degrees, and 1 means it should turn right 90 degrees.
After the robot turns, it should always move forward exactly one panel. The robot starts facing up.
The robot will continue running for a while like this and halt when it is finished drawing. Do not restart the Intcode computer inside the robot during this process.
For example, suppose the robot is about to start running. Drawing black panels as ., white panels as #, and the robot pointing the direction it is facing (< ^ > v), the initial state and region near the robot looks like this:
.....
.....
..^..
.....
.....
The panel under the robot (not visible here because a ^ is shown instead) is also black, and so any input instructions at this point should be provided 0. Suppose the robot eventually outputs 1 (paint white) and then 0 (turn left). After taking these actions and moving forward one panel, the region now looks like this:
.....
.....
.<#..
.....
.....
Input instructions should still be provided 0. Next, the robot might output 0 (paint black) and then 0 (turn left):
.....
.....
..#..
.v...
.....
After more outputs (1,0, 1,0):
.....
.....
..^..
.##..
.....
The robot is now back where it started, but because it is now on a white panel, input instructions should be provided 1. After several more outputs (0,1, 1,0, 1,0), the area looks like this:
.....
..<#.
...#.
.##..
.....
Before you deploy the robot, you should probably have an estimate of the area it will cover: specifically, you need to know the number of panels it paints at least once, regardless of color. In the example above, the robot painted 6 panels at least once. (It painted its starting panel twice, but that panel is still only counted once; it also never painted the panel it ended on.)
Build a new emergency hull painting robot and run the Intcode program on it. How many panels does it paint at least once?
Your puzzle answer was 2021.
--- Part Two ---
You're not sure what it's trying to paint, but it's definitely not a registration identifier. The Space Police are getting impatient.
Checking your external ship cameras again, you notice a white panel marked "emergency hull painting robot starting panel". The rest of the panels are still black, but it looks like the robot was expecting to start on a white panel, not a black one.
Based on the Space Law Space Brochure that the Space Police attached to one of your windows, a valid registration identifier is always eight capital letters. After starting the robot on a single white panel instead, what registration identifier does it paint on your hull?
| 147
|
--- Day 18: Operation Order ---
As you look out the window and notice a heavily-forested continent slowly appear over the horizon, you are interrupted by the child sitting next to you. They're curious if you could help them with their math homework.
Unfortunately, it seems like this "math" follows different rules than you remember.
The homework (your puzzle input) consists of a series of expressions that consist of addition (+), multiplication (*), and parentheses ((...)). Just like normal math, parentheses indicate that the expression inside must be evaluated before it can be used by the surrounding expression. Addition still finds the sum of the numbers on both sides of the operator, and multiplication still finds the product.
However, the rules of operator precedence have changed. Rather than evaluating multiplication before addition, the operators have the same precedence, and are evaluated left-to-right regardless of the order in which they appear.
For example, the steps to evaluate the expression 1 + 2 * 3 + 4 * 5 + 6 are as follows:
1 + 2 * 3 + 4 * 5 + 6
3 * 3 + 4 * 5 + 6
9 + 4 * 5 + 6
13 * 5 + 6
65 + 6
71
Parentheses can override this order; for example, here is what happens if parentheses are added to form 1 + (2 * 3) + (4 * (5 + 6)):
1 + (2 * 3) + (4 * (5 + 6))
1 + 6 + (4 * (5 + 6))
7 + (4 * (5 + 6))
7 + (4 * 11 )
7 + 44
51
Here are a few more examples:
2 * 3 + (4 * 5) becomes 26.
5 + (8 * 3 + 9 + 3 * 4 * 3) becomes 437.
5 * 9 * (7 * 3 * 3 + 9 * 3 + (8 + 6 * 4)) becomes 12240.
((2 + 4 * 9) * (6 + 9 * 8 + 6) + 6) + 2 + 4 * 2 becomes 13632.
Before you can help with the homework, you need to understand it yourself. Evaluate the expression on each line of the homework; what is the sum of the resulting values?
Your puzzle answer was 3885386961962.
--- Part Two ---
You manage to answer the child's questions and they finish part 1 of their homework, but get stuck when they reach the next section: advanced math.
Now, addition and multiplication have different precedence levels, but they're not the ones you're familiar with. Instead, addition is evaluated before multiplication.
For example, the steps to evaluate the expression 1 + 2 * 3 + 4 * 5 + 6 are now as follows:
1 + 2 * 3 + 4 * 5 + 6
3 * 3 + 4 * 5 + 6
3 * 7 * 5 + 6
3 * 7 * 11
21 * 11
231
Here are the other examples from above:
1 + (2 * 3) + (4 * (5 + 6)) still becomes 51.
2 * 3 + (4 * 5) becomes 46.
5 + (8 * 3 + 9 + 3 * 4 * 3) becomes 1445.
5 * 9 * (7 * 3 * 3 + 9 * 3 + (8 + 6 * 4)) becomes 669060.
((2 + 4 * 9) * (6 + 9 * 8 + 6) + 6) + 2 + 4 * 2 becomes 23340.
What do you get if you add up the results of evaluating the homework problems using these new rules?
| 148
|
--- Day 5: Doesn't He Have Intern-Elves For This? ---
Santa needs help figuring out which strings in his text file are naughty or nice.
A nice string is one with all of the following properties:
It contains at least three vowels (aeiou only), like aei, xazegov, or aeiouaeiouaeiou.
It contains at least one letter that appears twice in a row, like xx, abcdde (dd), or aabbccdd (aa, bb, cc, or dd).
It does not contain the strings ab, cd, pq, or xy, even if they are part of one of the other requirements.
For example:
ugknbfddgicrmopn is nice because it has at least three vowels (u...i...o...), a double letter (...dd...), and none of the disallowed substrings.
aaa is nice because it has at least three vowels and a double letter, even though the letters used by different rules overlap.
jchzalrnumimnmhp is naughty because it has no double letter.
haegwjzuvuyypxyu is naughty because it contains the string xy.
dvszwmarrgswjxmb is naughty because it contains only one vowel.
How many strings are nice?
| 149
|
--- Day 11: Corporate Policy ---
Santa's previous password expired, and he needs help choosing a new one.
To help him remember his new password after the old one expires, Santa has devised a method of coming up with a password based on the previous one. Corporate policy dictates that passwords must be exactly eight lowercase letters (for security reasons), so he finds his new password by incrementing his old password string repeatedly until it is valid.
Incrementing is just like counting with numbers: xx, xy, xz, ya, yb, and so on. Increase the rightmost letter one step; if it was z, it wraps around to a, and repeat with the next letter to the left until one doesn't wrap around.
Unfortunately for Santa, a new Security-Elf recently started, and he has imposed some additional password requirements:
Passwords must include one increasing straight of at least three letters, like abc, bcd, cde, and so on, up to xyz. They cannot skip letters; abd doesn't count.
Passwords may not contain the letters i, o, or l, as these letters can be mistaken for other characters and are therefore confusing.
Passwords must contain at least two different, non-overlapping pairs of letters, like aa, bb, or zz.
For example:
hijklmmn meets the first requirement (because it contains the straight hij) but fails the second requirement requirement (because it contains i and l).
abbceffg meets the third requirement (because it repeats bb and ff) but fails the first requirement.
abbcegjk fails the third requirement, because it only has one double letter (bb).
The next password after abcdefgh is abcdffaa.
The next password after ghijklmn is ghjaabcc, because you eventually skip all the passwords that start with ghi..., since i is not allowed.
Given Santa's current password (your puzzle input), what should his next password be?
| 150
|
--- Day 3: Mull It Over ---
"Our computers are having issues, so I have no idea if we have any Chief Historians in stock! You're welcome to check the warehouse, though," says the mildly flustered shopkeeper at the North Pole Toboggan Rental Shop. The Historians head out to take a look.
The shopkeeper turns to you. "Any chance you can see why our computers are having issues again?"
The computer appears to be trying to run a program, but its memory (your puzzle input) is corrupted. All of the instructions have been jumbled up!
It seems like the goal of the program is just to multiply some numbers. It does that with instructions like mul(X,Y), where X and Y are each 1-3 digit numbers. For instance, mul(44,46) multiplies 44 by 46 to get a result of 2024. Similarly, mul(123,4) would multiply 123 by 4.
However, because the program's memory has been corrupted, there are also many invalid characters that should be ignored, even if they look like part of a mul instruction. Sequences like mul(4*, mul(6,9!, ?(12,34), or mul ( 2 , 4 ) do nothing.
For example, consider the following section of corrupted memory:
xmul(2,4)%&mul[3,7]!@^do_not_mul(5,5)+mul(32,64]then(mul(11,8)mul(8,5))
Only the four highlighted sections are real mul instructions. Adding up the result of each instruction produces 161 (2*4 + 5*5 + 11*8 + 8*5).
Scan the corrupted memory for uncorrupted mul instructions. What do you get if you add up all of the results of the multiplications?
Your puzzle answer was 171183089.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
As you scan through the corrupted memory, you notice that some of the conditional statements are also still intact. If you handle some of the uncorrupted conditional statements in the program, you might be able to get an even more accurate result.
There are two new instructions you'll need to handle:
The do() instruction enables future mul instructions.
The don't() instruction disables future mul instructions.
Only the most recent do() or don't() instruction applies. At the beginning of the program, mul instructions are enabled.
For example:
xmul(2,4)&mul[3,7]!^don't()_mul(5,5)+mul(32,64](mul(11,8)undo()?mul(8,5))
This corrupted memory is similar to the example from before, but this time the mul(5,5) and mul(11,8) instructions are disabled because there is a don't() instruction before them. The other mul instructions function normally, including the one at the end that gets re-enabled by a do() instruction.
This time, the sum of the results is 48 (2*4 + 8*5).
Handle the new instructions; what do you get if you add up all of the results of just the enabled multiplications?
| 151
|
--- Day 22: Sporifica Virus ---
Diagnostics indicate that the local grid computing cluster has been contaminated with the Sporifica Virus. The grid computing cluster is a seemingly-infinite two-dimensional grid of compute nodes. Each node is either clean or infected by the virus.
To prevent overloading the nodes (which would render them useless to the virus) or detection by system administrators, exactly one virus carrier moves through the network, infecting or cleaning nodes as it moves. The virus carrier is always located on a single node in the network (the current node) and keeps track of the direction it is facing.
To avoid detection, the virus carrier works in bursts; in each burst, it wakes up, does some work, and goes back to sleep. The following steps are all executed in order one time each burst:
If the current node is infected, it turns to its right. Otherwise, it turns to its left. (Turning is done in-place; the current node does not change.)
If the current node is clean, it becomes infected. Otherwise, it becomes cleaned. (This is done after the node is considered for the purposes of changing direction.)
The virus carrier moves forward one node in the direction it is facing.
Diagnostics have also provided a map of the node infection status (your puzzle input). Clean nodes are shown as .; infected nodes are shown as #. This map only shows the center of the grid; there are many more nodes beyond those shown, but none of them are currently infected.
The virus carrier begins in the middle of the map facing up.
For example, suppose you are given a map like this:
..#
#..
...
Then, the middle of the infinite grid looks like this, with the virus carrier's position marked with [ ]:
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . # . . .
. . . #[.]. . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
The virus carrier is on a clean node, so it turns left, infects the node, and moves left:
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . # . . .
. . .[#]# . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
The virus carrier is on an infected node, so it turns right, cleans the node, and moves up:
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . .[.]. # . . .
. . . . # . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
Four times in a row, the virus carrier finds a clean, infects it, turns left, and moves forward, ending in the same place and still facing up:
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . #[#]. # . . .
. . # # # . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
Now on the same node as before, it sees an infection, which causes it to turn right, clean the node, and move forward:
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
. . # .[.]# . . .
. . # # # . . . .
. . . . . . . . .
. . . . . . . . .
. . . . . . . . .
After the above actions, a total of 7 bursts of activity had taken place. Of them, 5 bursts of activity caused an infection.
After a total of 70, the grid looks like this, with the virus carrier facing up:
. . . . . # # . .
. . . . # . . # .
. . . # . . . . #
. . # . #[.]. . #
. . # . # . . # .
. . . . . # # . .
. . . . . . . . .
. . . . . . . . .
By this time, 41 bursts of activity caused an infection (though most of those nodes have since been cleaned).
After a total of 10000 bursts of activity, 5587 bursts will have caused an infection.
Given your actual map, after 10000 bursts of activity, how many bursts cause a node to become infected? (Do not count nodes that begin infected.)
| 152
|
--- Day 18: Settlers of The North Pole ---
On the outskirts of the North Pole base construction project, many Elves are collecting lumber.
The lumber collection area is 50 acres by 50 acres; each acre can be either open ground (.), trees (|), or a lumberyard (#). You take a scan of the area (your puzzle input).
Strange magic is at work here: each minute, the landscape looks entirely different. In exactly one minute, an open acre can fill with trees, a wooded acre can be converted to a lumberyard, or a lumberyard can be cleared to open ground (the lumber having been sent to other projects).
The change to each acre is based entirely on the contents of that acre as well as the number of open, wooded, or lumberyard acres adjacent to it at the start of each minute. Here, "adjacent" means any of the eight acres surrounding that acre. (Acres on the edges of the lumber collection area might have fewer than eight adjacent acres; the missing acres aren't counted.)
In particular:
An open acre will become filled with trees if three or more adjacent acres contained trees. Otherwise, nothing happens.
An acre filled with trees will become a lumberyard if three or more adjacent acres were lumberyards. Otherwise, nothing happens.
An acre containing a lumberyard will remain a lumberyard if it was adjacent to at least one other lumberyard and at least one acre containing trees. Otherwise, it becomes open.
These changes happen across all acres simultaneously, each of them using the state of all acres at the beginning of the minute and changing to their new form by the end of that same minute. Changes that happen during the minute don't affect each other.
For example, suppose the lumber collection area is instead only 10 by 10 acres with this initial configuration:
Initial state:
.#.#...|#.
.....#|##|
.|..|...#.
..|#.....#
#.#|||#|#|
...#.||...
.|....|...
||...#|.#|
|.||||..|.
...#.|..|.
After 1 minute:
.......##.
......|###
.|..|...#.
..|#||...#
..##||.|#|
...#||||..
||...|||..
|||||.||.|
||||||||||
....||..|.
After 2 minutes:
.......#..
......|#..
.|.|||....
..##|||..#
..###|||#|
...#|||||.
|||||||||.
||||||||||
||||||||||
.|||||||||
After 3 minutes:
.......#..
....|||#..
.|.||||...
..###|||.#
...##|||#|
.||##|||||
||||||||||
||||||||||
||||||||||
||||||||||
After 4 minutes:
.....|.#..
...||||#..
.|.#||||..
..###||||#
...###||#|
|||##|||||
||||||||||
||||||||||
||||||||||
||||||||||
After 5 minutes:
....|||#..
...||||#..
.|.##||||.
..####|||#
.|.###||#|
|||###||||
||||||||||
||||||||||
||||||||||
||||||||||
After 6 minutes:
...||||#..
...||||#..
.|.###|||.
..#.##|||#
|||#.##|#|
|||###||||
||||#|||||
||||||||||
||||||||||
||||||||||
After 7 minutes:
...||||#..
..||#|##..
.|.####||.
||#..##||#
||##.##|#|
|||####|||
|||###||||
||||||||||
||||||||||
||||||||||
After 8 minutes:
..||||##..
..|#####..
|||#####|.
||#...##|#
||##..###|
||##.###||
|||####|||
||||#|||||
||||||||||
||||||||||
After 9 minutes:
..||###...
.||#####..
||##...##.
||#....###
|##....##|
||##..###|
||######||
|||###||||
||||||||||
||||||||||
After 10 minutes:
.||##.....
||###.....
||##......
|##.....##
|##.....##
|##....##|
||##.####|
||#####|||
||||#|||||
||||||||||
After 10 minutes, there are 37 wooded acres and 31 lumberyards. Multiplying the number of wooded acres by the number of lumberyards gives the total resource value after ten minutes: 37 * 31 = 1147.
What will the total resource value of the lumber collection area be after 10 minutes?
| 153
|
--- Day 22: Crab Combat ---
It only takes a few hours of sailing the ocean on a raft for boredom to sink in. Fortunately, you brought a small deck of space cards! You'd like to play a game of Combat, and there's even an opponent available: a small crab that climbed aboard your raft before you left.
Fortunately, it doesn't take long to teach the crab the rules.
Before the game starts, split the cards so each player has their own deck (your puzzle input). Then, the game consists of a series of rounds: both players draw their top card, and the player with the higher-valued card wins the round. The winner keeps both cards, placing them on the bottom of their own deck so that the winner's card is above the other card. If this causes a player to have all of the cards, they win, and the game ends.
For example, consider the following starting decks:
Player 1:
9
2
6
3
1
Player 2:
5
8
4
7
10
This arrangement means that player 1's deck contains 5 cards, with 9 on top and 1 on the bottom; player 2's deck also contains 5 cards, with 5 on top and 10 on the bottom.
The first round begins with both players drawing the top card of their decks: 9 and 5. Player 1 has the higher card, so both cards move to the bottom of player 1's deck such that 9 is above 5. In total, it takes 29 rounds before a player has all of the cards:
-- Round 1 --
Player 1's deck: 9, 2, 6, 3, 1
Player 2's deck: 5, 8, 4, 7, 10
Player 1 plays: 9
Player 2 plays: 5
Player 1 wins the round!
-- Round 2 --
Player 1's deck: 2, 6, 3, 1, 9, 5
Player 2's deck: 8, 4, 7, 10
Player 1 plays: 2
Player 2 plays: 8
Player 2 wins the round!
-- Round 3 --
Player 1's deck: 6, 3, 1, 9, 5
Player 2's deck: 4, 7, 10, 8, 2
Player 1 plays: 6
Player 2 plays: 4
Player 1 wins the round!
-- Round 4 --
Player 1's deck: 3, 1, 9, 5, 6, 4
Player 2's deck: 7, 10, 8, 2
Player 1 plays: 3
Player 2 plays: 7
Player 2 wins the round!
-- Round 5 --
Player 1's deck: 1, 9, 5, 6, 4
Player 2's deck: 10, 8, 2, 7, 3
Player 1 plays: 1
Player 2 plays: 10
Player 2 wins the round!
...several more rounds pass...
-- Round 27 --
Player 1's deck: 5, 4, 1
Player 2's deck: 8, 9, 7, 3, 2, 10, 6
Player 1 plays: 5
Player 2 plays: 8
Player 2 wins the round!
-- Round 28 --
Player 1's deck: 4, 1
Player 2's deck: 9, 7, 3, 2, 10, 6, 8, 5
Player 1 plays: 4
Player 2 plays: 9
Player 2 wins the round!
-- Round 29 --
Player 1's deck: 1
Player 2's deck: 7, 3, 2, 10, 6, 8, 5, 9, 4
Player 1 plays: 1
Player 2 plays: 7
Player 2 wins the round!
== Post-game results ==
Player 1's deck:
Player 2's deck: 3, 2, 10, 6, 8, 5, 9, 4, 7, 1
Once the game ends, you can calculate the winning player's score. The bottom card in their deck is worth the value of the card multiplied by 1, the second-from-the-bottom card is worth the value of the card multiplied by 2, and so on. With 10 cards, the top card is worth the value on the card multiplied by 10. In this example, the winning player's score is:
3 * 10
+ 2 * 9
+ 10 * 8
+ 6 * 7
+ 8 * 6
+ 5 * 5
+ 9 * 4
+ 4 * 3
+ 7 * 2
+ 1 * 1
= 306
So, once the game ends, the winning player's score is 306.
Play the small crab in a game of Combat using the two decks you just dealt. What is the winning player's score?
| 154
|
--- Day 4: Ceres Search ---
"Looks like the Chief's not here. Next!" One of The Historians pulls out a device and pushes the only button on it. After a brief flash, you recognize the interior of the Ceres monitoring station!
As the search for the Chief continues, a small Elf who lives on the station tugs on your shirt; she'd like to know if you could help her with her word search (your puzzle input). She only has to find one word: XMAS.
This word search allows words to be horizontal, vertical, diagonal, written backwards, or even overlapping other words. It's a little unusual, though, as you don't merely need to find one instance of XMAS - you need to find all of them. Here are a few ways XMAS might appear, where irrelevant characters have been replaced with .:
..X...
.SAMX.
.A..A.
XMAS.S
.X....
The actual word search will be full of letters instead. For example:
MMMSXXMASM
MSAMXMSMSA
AMXSXMAAMM
MSAMASMSMX
XMASAMXAMM
XXAMMXXAMA
SMSMSASXSS
SAXAMASAAA
MAMMMXMMMM
MXMXAXMASX
In this word search, XMAS occurs a total of 18 times; here's the same word search again, but where letters not involved in any XMAS have been replaced with .:
....XXMAS.
.SAMXMS...
...S..A...
..A.A.MS.X
XMASAMX.MM
X.....XA.A
S.S.S.S.SS
.A.A.A.A.A
..M.M.M.MM
.X.X.XMASX
Take a look at the little Elf's word search. How many times does XMAS appear?
| 155
|
--- 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?
| 156
|
--- Day 15: Warehouse Woes ---
You appear back inside your own mini submarine! Each Historian drives their mini submarine in a different direction; maybe the Chief has his own submarine down here somewhere as well?
You look up to see a vast school of lanternfish swimming past you. On closer inspection, they seem quite anxious, so you drive your mini submarine over to see if you can help.
Because lanternfish populations grow rapidly, they need a lot of food, and that food needs to be stored somewhere. That's why these lanternfish have built elaborate warehouse complexes operated by robots!
These lanternfish seem so anxious because they have lost control of the robot that operates one of their most important warehouses! It is currently running amok, pushing around boxes in the warehouse with no regard for lanternfish logistics or lanternfish inventory management strategies.
Right now, none of the lanternfish are brave enough to swim up to an unpredictable robot so they could shut it off. However, if you could anticipate the robot's movements, maybe they could find a safe option.
The lanternfish already have a map of the warehouse and a list of movements the robot will attempt to make (your puzzle input). The problem is that the movements will sometimes fail as boxes are shifted around, making the actual movements of the robot difficult to predict.
For example:
##########
#..O..O.O#
#......O.#
#.OO..O.O#
#[email protected].#
#O#..O...#
#O..O..O.#
#.OO.O.OO#
#....O...#
##########
<vv>^<v^>v>^vv^v>v<>v^v<v<^vv<<<^><<><>>v<vvv<>^v^>^<<<><<v<<<v^vv^v>^
vvv<<^>^v^^><<>>><>^<<><^vv^^<>vvv<>><^^v>^>vv<>v<<<<v<^v>^<^^>>>^<v<v
><>vv>v^v^<>><>>>><^^>vv>v<^^^>>v^v^<^^>v^^>v^<^v>v<>>v^v^<v>v^^<^^vv<
<<v<^>>^^^^>>>v^<>vvv^><v<<<>^^^vv^<vvv>^>v<^^^^v<>^>vvvv><>>v^<<^^^^^
^><^><>>><>^^<<^^v>>><^<v>^<vv>>v>>>^v><>^v><<<<v>>v<v<v>vvv>^<><<>^><
^>><>^v<><^vvv<^^<><v<<<<<><^v<<<><<<^^<v<^^^><^>>^<v^><<<^>>^v<v^v<v^
>^>>^v>vv>^<<^v<>><<><<v<<v><>v<^vv<<<>^^v^>^^>>><<^v>>v^v><^^>>^<>vv^
<><^^>^^^<><vvvvv^v<v<<>^v<v>v<<^><<><<><<<^^<<<^<<>><<><^^^>^^<>^>v<>
^^>vv<^v^v<vv>^<><v<^v>^^^>>>^^vvv^>vvv<>>>^<^>>>>>^<<^v>^vvv<>^<><<v>
v^^>>><<^^<>>^v^<v^vv<>v^<<>^<^v^v><^<<<><<^<v><v<>vv>>v><v^<vv<>v^<<^
As the robot (@) attempts to move, if there are any boxes (O) in the way, the robot will also attempt to push those boxes. However, if this action would cause the robot or a box to move into a wall (#), nothing moves instead, including the robot. The initial positions of these are shown on the map at the top of the document the lanternfish gave you.
The rest of the document describes the moves (^ for up, v for down, < for left, > for right) that the robot will attempt to make, in order. (The moves form a single giant sequence; they are broken into multiple lines just to make copy-pasting easier. Newlines within the move sequence should be ignored.)
Here is a smaller example to get started:
########
#..O.O.#
##@.O..#
#...O..#
#.#.O..#
#...O..#
#......#
########
<^^>>>vv<v>>v<<
Were the robot to attempt the given sequence of moves, it would push around the boxes as follows:
Initial state:
########
#..O.O.#
##@.O..#
#...O..#
#.#.O..#
#...O..#
#......#
########
Move <:
########
#..O.O.#
##@.O..#
#...O..#
#.#.O..#
#...O..#
#......#
########
Move ^:
########
#[email protected].#
##..O..#
#...O..#
#.#.O..#
#...O..#
#......#
########
Move ^:
########
#[email protected].#
##..O..#
#...O..#
#.#.O..#
#...O..#
#......#
########
Move >:
########
#..@OO.#
##..O..#
#...O..#
#.#.O..#
#...O..#
#......#
########
Move >:
########
#...@OO#
##..O..#
#...O..#
#.#.O..#
#...O..#
#......#
########
Move >:
########
#...@OO#
##..O..#
#...O..#
#.#.O..#
#...O..#
#......#
########
Move v:
########
#....OO#
##..@..#
#...O..#
#.#.O..#
#...O..#
#...O..#
########
Move v:
########
#....OO#
##..@..#
#...O..#
#.#.O..#
#...O..#
#...O..#
########
Move <:
########
#....OO#
##.@...#
#...O..#
#.#.O..#
#...O..#
#...O..#
########
Move v:
########
#....OO#
##.....#
#..@O..#
#.#.O..#
#...O..#
#...O..#
########
Move >:
########
#....OO#
##.....#
#...@O.#
#.#.O..#
#...O..#
#...O..#
########
Move >:
########
#....OO#
##.....#
#....@O#
#.#.O..#
#...O..#
#...O..#
########
Move v:
########
#....OO#
##.....#
#.....O#
#.#.O@.#
#...O..#
#...O..#
########
Move <:
########
#....OO#
##.....#
#.....O#
#.#O@..#
#...O..#
#...O..#
########
Move <:
########
#....OO#
##.....#
#.....O#
#.#O@..#
#...O..#
#...O..#
########
The larger example has many more moves; after the robot has finished those moves, the warehouse would look like this:
##########
#.O.O.OOO#
#........#
#OO......#
#OO@.....#
#O#.....O#
#O.....OO#
#O.....OO#
#OO....OO#
##########
The lanternfish use their own custom Goods Positioning System (GPS for short) to track the locations of the boxes. The GPS coordinate of a box is equal to 100 times its distance from the top edge of the map plus its distance from the left edge of the map. (This process does not stop at wall tiles; measure all the way to the edges of the map.)
So, the box shown below has a distance of 1 from the top edge of the map and 4 from the left edge of the map, resulting in a GPS coordinate of 100 * 1 + 4 = 104.
#######
#...O..
#......
The lanternfish would like to know the sum of all boxes' GPS coordinates after the robot finishes moving. In the larger example, the sum of all boxes' GPS coordinates is 10092. In the smaller example, the sum is 2028.
Predict the motion of the robot and boxes in the warehouse. After the robot is finished moving, what is the sum of all boxes' GPS coordinates?
| 157
|
--- Day 8: Resonant Collinearity ---
You find yourselves on the roof of a top-secret Easter Bunny installation.
While The Historians do their thing, you take a look at the familiar huge antenna. Much to your surprise, it seems to have been reconfigured to emit a signal that makes people 0.1% more likely to buy Easter Bunny brand Imitation Mediocre Chocolate as a Christmas gift! Unthinkable!
Scanning across the city, you find that there are actually many such antennas. Each antenna is tuned to a specific frequency indicated by a single lowercase letter, uppercase letter, or digit. You create a map (your puzzle input) of these antennas. For example:
............
........0...
.....0......
.......0....
....0.......
......A.....
............
............
........A...
.........A..
............
............
The signal only applies its nefarious effect at specific antinodes based on the resonant frequencies of the antennas. In particular, an antinode occurs at any point that is perfectly in line with two antennas of the same frequency - but only when one of the antennas is twice as far away as the other. This means that for any pair of antennas with the same frequency, there are two antinodes, one on either side of them.
So, for these two antennas with frequency a, they create the two antinodes marked with #:
..........
...#......
..........
....a.....
..........
.....a....
..........
......#...
..........
..........
Adding a third antenna with the same frequency creates several more antinodes. It would ideally add four antinodes, but two are off the right side of the map, so instead it adds only two:
..........
...#......
#.........
....a.....
........a.
.....a....
..#.......
......#...
..........
..........
Antennas with different frequencies don't create antinodes; A and a count as different frequencies. However, antinodes can occur at locations that contain antennas. In this diagram, the lone antenna with frequency capital A creates no antinodes but has a lowercase-a-frequency antinode at its location:
..........
...#......
#.........
....a.....
........a.
.....a....
..#.......
......A...
..........
..........
The first example has antennas with two different frequencies, so the antinodes they create look like this, plus an antinode overlapping the topmost A-frequency antenna:
......#....#
...#....0...
....#0....#.
..#....0....
....0....#..
.#....A.....
...#........
#......#....
........A...
.........A..
..........#.
..........#.
Because the topmost A-frequency antenna overlaps with a 0-frequency antinode, there are 14 total unique locations that contain an antinode within the bounds of the map.
Calculate the impact of the signal. How many unique locations within the bounds of the map contain an antinode?
| 158
|
--- Day 6: Memory Reallocation ---
A debugger program here is having an issue: it is trying to repair a memory reallocation routine, but it keeps getting stuck in an infinite loop.
In this area, there are sixteen memory banks; each memory bank can hold any number of blocks. The goal of the reallocation routine is to balance the blocks between the memory banks.
The reallocation routine operates in cycles. In each cycle, it finds the memory bank with the most blocks (ties won by the lowest-numbered memory bank) and redistributes those blocks among the banks. To do this, it removes all of the blocks from the selected bank, then moves to the next (by index) memory bank and inserts one of the blocks. It continues doing this until it runs out of blocks; if it reaches the last memory bank, it wraps around to the first one.
The debugger would like to know how many redistributions can be done before a blocks-in-banks configuration is produced that has been seen before.
For example, imagine a scenario with only four memory banks:
The banks start with 0, 2, 7, and 0 blocks. The third bank has the most blocks, so it is chosen for redistribution.
Starting with the next bank (the fourth bank) and then continuing to the first bank, the second bank, and so on, the 7 blocks are spread out over the memory banks. The fourth, first, and second banks get two blocks each, and the third bank gets one back. The final result looks like this: 2 4 1 2.
Next, the second bank is chosen because it contains the most blocks (four). Because there are four memory banks, each gets one block. The result is: 3 1 2 3.
Now, there is a tie between the first and fourth memory banks, both of which have three blocks. The first bank wins the tie, and its three blocks are distributed evenly over the other three banks, leaving it with none: 0 2 3 4.
The fourth bank is chosen, and its four blocks are distributed such that each of the four banks receives one: 1 3 4 1.
The third bank is chosen, and the same thing happens: 2 4 1 2.
At this point, we've reached a state we've seen before: 2 4 1 2 was already seen. The infinite loop is detected after the fifth block redistribution cycle, and so the answer in this example is 5.
Given the initial block counts in your puzzle input, how many redistribution cycles must be completed before a configuration is produced that has been seen before?
| 159
|
--- Day 11: Radioisotope Thermoelectric Generators ---
You come upon a column of four floors that have been entirely sealed off from the rest of the building except for a small dedicated lobby. There are some radiation warnings and a big sign which reads "Radioisotope Testing Facility".
According to the project status board, this facility is currently being used to experiment with Radioisotope Thermoelectric Generators (RTGs, or simply "generators") that are designed to be paired with specially-constructed microchips. Basically, an RTG is a highly radioactive rock that generates electricity through heat.
The experimental RTGs have poor radiation containment, so they're dangerously radioactive. The chips are prototypes and don't have normal radiation shielding, but they do have the ability to generate an electromagnetic radiation shield when powered. Unfortunately, they can only be powered by their corresponding RTG. An RTG powering a microchip is still dangerous to other microchips.
In other words, if a chip is ever left in the same area as another RTG, and it's not connected to its own RTG, the chip will be fried. Therefore, it is assumed that you will follow procedure and keep chips connected to their corresponding RTG when they're in the same room, and away from other RTGs otherwise.
These microchips sound very interesting and useful to your current activities, and you'd like to try to retrieve them. The fourth floor of the facility has an assembling machine which can make a self-contained, shielded computer for you to take with you - that is, if you can bring it all of the RTGs and microchips.
Within the radiation-shielded part of the facility (in which it's safe to have these pre-assembly RTGs), there is an elevator that can move between the four floors. Its capacity rating means it can carry at most yourself and two RTGs or microchips in any combination. (They're rigged to some heavy diagnostic equipment - the assembling machine will detach it for you.) As a security measure, the elevator will only function if it contains at least one RTG or microchip. The elevator always stops on each floor to recharge, and this takes long enough that the items within it and the items on that floor can irradiate each other. (You can prevent this if a Microchip and its Generator end up on the same floor in this way, as they can be connected while the elevator is recharging.)
You make some notes of the locations of each component of interest (your puzzle input). Before you don a hazmat suit and start moving things around, you'd like to have an idea of what you need to do.
When you enter the containment area, you and the elevator will start on the first floor.
For example, suppose the isolated area has the following arrangement:
The first floor contains a hydrogen-compatible microchip and a lithium-compatible microchip.
The second floor contains a hydrogen generator.
The third floor contains a lithium generator.
The fourth floor contains nothing relevant.
As a diagram (F# for a Floor number, E for Elevator, H for Hydrogen, L for Lithium, M for Microchip, and G for Generator), the initial state looks like this:
F4 . . . . .
F3 . . . LG .
F2 . HG . . .
F1 E . HM . LM
Then, to get everything up to the assembling machine on the fourth floor, the following steps could be taken:
Bring the Hydrogen-compatible Microchip to the second floor, which is safe because it can get power from the Hydrogen Generator:
F4 . . . . .
F3 . . . LG .
F2 E HG HM . .
F1 . . . . LM
Bring both Hydrogen-related items to the third floor, which is safe because the Hydrogen-compatible microchip is getting power from its generator:
F4 . . . . .
F3 E HG HM LG .
F2 . . . . .
F1 . . . . LM
Leave the Hydrogen Generator on floor three, but bring the Hydrogen-compatible Microchip back down with you so you can still use the elevator:
F4 . . . . .
F3 . HG . LG .
F2 E . HM . .
F1 . . . . LM
At the first floor, grab the Lithium-compatible Microchip, which is safe because Microchips don't affect each other:
F4 . . . . .
F3 . HG . LG .
F2 . . . . .
F1 E . HM . LM
Bring both Microchips up one floor, where there is nothing to fry them:
F4 . . . . .
F3 . HG . LG .
F2 E . HM . LM
F1 . . . . .
Bring both Microchips up again to floor three, where they can be temporarily connected to their corresponding generators while the elevator recharges, preventing either of them from being fried:
F4 . . . . .
F3 E HG HM LG LM
F2 . . . . .
F1 . . . . .
Bring both Microchips to the fourth floor:
F4 E . HM . LM
F3 . HG . LG .
F2 . . . . .
F1 . . . . .
Leave the Lithium-compatible microchip on the fourth floor, but bring the Hydrogen-compatible one so you can still use the elevator; this is safe because although the Lithium Generator is on the destination floor, you can connect Hydrogen-compatible microchip to the Hydrogen Generator there:
F4 . . . . LM
F3 E HG HM LG .
F2 . . . . .
F1 . . . . .
Bring both Generators up to the fourth floor, which is safe because you can connect the Lithium-compatible Microchip to the Lithium Generator upon arrival:
F4 E HG . LG LM
F3 . . HM . .
F2 . . . . .
F1 . . . . .
Bring the Lithium Microchip with you to the third floor so you can use the elevator:
F4 . HG . LG .
F3 E . HM . LM
F2 . . . . .
F1 . . . . .
Bring both Microchips to the fourth floor:
F4 E HG HM LG LM
F3 . . . . .
F2 . . . . .
F1 . . . . .
In this arrangement, it takes 11 steps to collect all of the objects at the fourth floor for assembly. (Each elevator stop counts as one step, even if nothing is added to or removed from it.)
In your situation, what is the minimum number of steps required to bring all of the objects to the fourth floor?
| 160
|
--- Day 21: Step Counter ---
You manage to catch the airship right as it's dropping someone else off on their all-expenses-paid trip to Desert Island! It even helpfully drops you off near the gardener and his massive farm.
"You got the sand flowing again! Great work! Now we just need to wait until we have enough sand to filter the water for Snow Island and we'll have snow again in no time."
While you wait, one of the Elves that works with the gardener heard how good you are at solving problems and would like your help. He needs to get his steps in for the day, and so he'd like to know which garden plots he can reach with exactly his remaining 64 steps.
He gives you an up-to-date map (your puzzle input) of his starting position (S), garden plots (.), and rocks (#). For example:
...........
.....###.#.
.###.##..#.
..#.#...#..
....#.#....
.##..S####.
.##..#...#.
.......##..
.##.#.####.
.##..##.##.
...........
The Elf starts at the starting position (S) which also counts as a garden plot. Then, he can take one step north, south, east, or west, but only onto tiles that are garden plots. This would allow him to reach any of the tiles marked O:
...........
.....###.#.
.###.##..#.
..#.#...#..
....#O#....
.##.OS####.
.##..#...#.
.......##..
.##.#.####.
.##..##.##.
...........
Then, he takes a second step. Since at this point he could be at either tile marked O, his second step would allow him to reach any garden plot that is one step north, south, east, or west of any tile that he could have reached after the first step:
...........
.....###.#.
.###.##..#.
..#.#O..#..
....#.#....
.##O.O####.
.##.O#...#.
.......##..
.##.#.####.
.##..##.##.
...........
After two steps, he could be at any of the tiles marked O above, including the starting position (either by going north-then-south or by going west-then-east).
A single third step leads to even more possibilities:
...........
.....###.#.
.###.##..#.
..#.#.O.#..
...O#O#....
.##.OS####.
.##O.#...#.
....O..##..
.##.#.####.
.##..##.##.
...........
He will continue like this until his steps for the day have been exhausted. After a total of 6 steps, he could reach any of the garden plots marked O:
...........
.....###.#.
.###.##.O#.
.O#O#O.O#..
O.O.#.#.O..
.##O.O####.
.##.O#O..#.
.O.O.O.##..
.##.#.####.
.##O.##.##.
...........
In this example, if the Elf's goal was to get exactly 6 more steps today, he could use them to reach any of 16 garden plots.
However, the Elf actually needs to get 64 steps today, and the map he's handed you is much larger than the example map.
Starting from the garden plot marked S on your map, how many garden plots could the Elf reach in exactly 64 steps?
Your puzzle answer was 3795.
--- Part Two ---
The Elf seems confused by your answer until he realizes his mistake: he was reading from a list of his favorite numbers that are both perfect squares and perfect cubes, not his step counter.
The actual number of steps he needs to get today is exactly 26501365.
He also points out that the garden plots and rocks are set up so that the map repeats infinitely in every direction.
So, if you were to look one additional map-width or map-height out from the edge of the example map above, you would find that it keeps repeating:
.................................
.....###.#......###.#......###.#.
.###.##..#..###.##..#..###.##..#.
..#.#...#....#.#...#....#.#...#..
....#.#........#.#........#.#....
.##...####..##...####..##...####.
.##..#...#..##..#...#..##..#...#.
.......##.........##.........##..
.##.#.####..##.#.####..##.#.####.
.##..##.##..##..##.##..##..##.##.
.................................
.................................
.....###.#......###.#......###.#.
.###.##..#..###.##..#..###.##..#.
..#.#...#....#.#...#....#.#...#..
....#.#........#.#........#.#....
.##...####..##..S####..##...####.
.##..#...#..##..#...#..##..#...#.
.......##.........##.........##..
.##.#.####..##.#.####..##.#.####.
.##..##.##..##..##.##..##..##.##.
.................................
.................................
.....###.#......###.#......###.#.
.###.##..#..###.##..#..###.##..#.
..#.#...#....#.#...#....#.#...#..
....#.#........#.#........#.#....
.##...####..##...####..##...####.
.##..#...#..##..#...#..##..#...#.
.......##.........##.........##..
.##.#.####..##.#.####..##.#.####.
.##..##.##..##..##.##..##..##.##.
.................................
This is just a tiny three-map-by-three-map slice of the inexplicably-infinite farm layout; garden plots and rocks repeat as far as you can see. The Elf still starts on the one middle tile marked S, though - every other repeated S is replaced with a normal garden plot (.).
Here are the number of reachable garden plots in this new infinite version of the example map for different numbers of steps:
In exactly 6 steps, he can still reach 16 garden plots.
In exactly 10 steps, he can reach any of 50 garden plots.
In exactly 50 steps, he can reach 1594 garden plots.
In exactly 100 steps, he can reach 6536 garden plots.
In exactly 500 steps, he can reach 167004 garden plots.
In exactly 1000 steps, he can reach 668697 garden plots.
In exactly 5000 steps, he can reach 16733044 garden plots.
However, the step count the Elf needs is much larger! Starting from the garden plot marked S on your infinite map, how many garden plots could the Elf reach in exactly 26501365 steps?
| 161
|
--- Day 10: Pipe Maze ---
You use the hang glider to ride the hot air from Desert Island all the way up to the floating metal island. This island is surprisingly cold and there definitely aren't any thermals to glide on, so you leave your hang glider behind.
You wander around for a while, but you don't find any people or animals. However, you do occasionally find signposts labeled "Hot Springs" pointing in a seemingly consistent direction; maybe you can find someone at the hot springs and ask them where the desert-machine parts are made.
The landscape here is alien; even the flowers and trees are made of metal. As you stop to admire some metal grass, you notice something metallic scurry away in your peripheral vision and jump into a big pipe! It didn't look like any animal you've ever seen; if you want a better look, you'll need to get ahead of it.
Scanning the area, you discover that the entire field you're standing on is densely packed with pipes; it was hard to tell at first because they're the same metallic silver color as the "ground". You make a quick sketch of all of the surface pipes you can see (your puzzle input).
The pipes are arranged in a two-dimensional grid of tiles:
| is a vertical pipe connecting north and south.
- is a horizontal pipe connecting east and west.
L is a 90-degree bend connecting north and east.
J is a 90-degree bend connecting north and west.
7 is a 90-degree bend connecting south and west.
F is a 90-degree bend connecting south and east.
. is ground; there is no pipe in this tile.
S is the starting position of the animal; there is a pipe on this tile, but your sketch doesn't show what shape the pipe has.
Based on the acoustics of the animal's scurrying, you're confident the pipe that contains the animal is one large, continuous loop.
For example, here is a square loop of pipe:
.....
.F-7.
.|.|.
.L-J.
.....
If the animal had entered this loop in the northwest corner, the sketch would instead look like this:
.....
.S-7.
.|.|.
.L-J.
.....
In the above diagram, the S tile is still a 90-degree F bend: you can tell because of how the adjacent pipes connect to it.
Unfortunately, there are also many pipes that aren't connected to the loop! This sketch shows the same loop as above:
-L|F7
7S-7|
L|7||
-L-J|
L|-JF
In the above diagram, you can still figure out which pipes form the main loop: they're the ones connected to S, pipes those pipes connect to, pipes those pipes connect to, and so on. Every pipe in the main loop connects to its two neighbors (including S, which will have exactly two pipes connecting to it, and which is assumed to connect back to those two pipes).
Here is a sketch that contains a slightly more complex main loop:
..F7.
.FJ|.
SJ.L7
|F--J
LJ...
Here's the same example sketch with the extra, non-main-loop pipe tiles also shown:
7-F7-
.FJ|7
SJLL7
|F--J
LJ.LJ
If you want to get out ahead of the animal, you should find the tile in the loop that is farthest from the starting position. Because the animal is in the pipe, it doesn't make sense to measure this by direct distance. Instead, you need to find the tile that would take the longest number of steps along the loop to reach from the starting point - regardless of which way around the loop the animal went.
In the first example with the square loop:
.....
.S-7.
.|.|.
.L-J.
.....
You can count the distance each tile in the loop is from the starting point like this:
.....
.012.
.1.3.
.234.
.....
In this example, the farthest point from the start is 4 steps away.
Here's the more complex loop again:
..F7.
.FJ|.
SJ.L7
|F--J
LJ...
Here are the distances for each tile on that loop:
..45.
.236.
01.78
14567
23...
Find the single giant loop starting at S. How many steps along the loop does it take to get from the starting position to the point farthest from the starting position?
Your puzzle answer was 6842.
--- Part Two ---
You quickly reach the farthest point of the loop, but the animal never emerges. Maybe its nest is within the area enclosed by the loop?
To determine whether it's even worth taking the time to search for such a nest, you should calculate how many tiles are contained within the loop. For example:
...........
.S-------7.
.|F-----7|.
.||.....||.
.||.....||.
.|L-7.F-J|.
.|..|.|..|.
.L--J.L--J.
...........
The above loop encloses merely four tiles - the two pairs of . in the southwest and southeast (marked I below). The middle . tiles (marked O below) are not in the loop. Here is the same loop again with those regions marked:
...........
.S-------7.
.|F-----7|.
.||OOOOO||.
.||OOOOO||.
.|L-7OF-J|.
.|II|O|II|.
.L--JOL--J.
.....O.....
In fact, there doesn't even need to be a full tile path to the outside for tiles to count as outside the loop - squeezing between pipes is also allowed! Here, I is still within the loop and O is still outside the loop:
..........
.S------7.
.|F----7|.
.||OOOO||.
.||OOOO||.
.|L-7F-J|.
.|II||II|.
.L--JL--J.
..........
In both of the above examples, 4 tiles are enclosed by the loop.
Here's a larger example:
.F----7F7F7F7F-7....
.|F--7||||||||FJ....
.||.FJ||||||||L7....
FJL7L7LJLJ||LJ.L-7..
L--J.L7...LJS7F-7L7.
....F-J..F7FJ|L7L7L7
....L7.F7||L7|.L7L7|
.....|FJLJ|FJ|F7|.LJ
....FJL-7.||.||||...
....L---J.LJ.LJLJ...
The above sketch has many random bits of ground, some of which are in the loop (I) and some of which are outside it (O):
OF----7F7F7F7F-7OOOO
O|F--7||||||||FJOOOO
O||OFJ||||||||L7OOOO
FJL7L7LJLJ||LJIL-7OO
L--JOL7IIILJS7F-7L7O
OOOOF-JIIF7FJ|L7L7L7
OOOOL7IF7||L7|IL7L7|
OOOOO|FJLJ|FJ|F7|OLJ
OOOOFJL-7O||O||||OOO
OOOOL---JOLJOLJLJOOO
In this larger example, 8 tiles are enclosed by the loop.
Any tile that isn't part of the main loop can count as being enclosed by the loop. Here's another example with many bits of junk pipe lying around that aren't connected to the main loop at all:
FF7FSF7F7F7F7F7F---7
L|LJ||||||||||||F--J
FL-7LJLJ||||||LJL-77
F--JF--7||LJLJ7F7FJ-
L---JF-JLJ.||-FJLJJ7
|F|F-JF---7F7-L7L|7|
|FFJF7L7F-JF7|JL---7
7-L-JL7||F7|L7F-7F7|
L.L7LFJ|||||FJL7||LJ
L7JLJL-JLJLJL--JLJ.L
Here are just the tiles that are enclosed by the loop marked with I:
FF7FSF7F7F7F7F7F---7
L|LJ||||||||||||F--J
FL-7LJLJ||||||LJL-77
F--JF--7||LJLJIF7FJ-
L---JF-JLJIIIIFJLJJ7
|F|F-JF---7IIIL7L|7|
|FFJF7L7F-JF7IIL---7
7-L-JL7||F7|L7F-7F7|
L.L7LFJ|||||FJL7||LJ
L7JLJL-JLJLJL--JLJ.L
In this last example, 10 tiles are enclosed by the loop.
Figure out whether you have time to search for the nest by calculating the area within the loop. How many tiles are enclosed by the loop?
| 162
|
--- Day 7: Recursive Circus ---
Wandering further through the circuits of the computer, you come upon a tower of programs that have gotten themselves into a bit of trouble. A recursive algorithm has gotten out of hand, and now they're balanced precariously in a large tower.
One program at the bottom supports the entire tower. It's holding a large disc, and on the disc are balanced several more sub-towers. At the bottom of these sub-towers, standing on the bottom disc, are other programs, each holding their own disc, and so on. At the very tops of these sub-sub-sub-...-towers, many programs stand simply keeping the disc below them balanced but with no disc of their own.
You offer to help, but first you need to understand the structure of these towers. You ask each program to yell out their name, their weight, and (if they're holding a disc) the names of the programs immediately above them balancing on that disc. You write this information down (your puzzle input). Unfortunately, in their panic, they don't do this in an orderly fashion; by the time you're done, you're not sure which program gave which information.
For example, if your list is the following:
pbga (66)
xhth (57)
ebii (61)
havc (66)
ktlj (57)
fwft (72) -> ktlj, cntj, xhth
qoyq (66)
padx (45) -> pbga, havc, qoyq
tknk (41) -> ugml, padx, fwft
jptl (61)
ugml (68) -> gyxo, ebii, jptl
gyxo (61)
cntj (57)
...then you would be able to recreate the structure of the towers that looks like this:
gyxo
/
ugml - ebii
/
| jptl
|
| pbga
/ /
tknk --- padx - havc
| qoyq
|
| ktlj
/
fwft - cntj
xhth
In this example, tknk is at the bottom of the tower (the bottom program), and is holding up ugml, padx, and fwft. Those programs are, in turn, holding up other programs; in this example, none of those programs are holding up any other programs, and are all the tops of their own towers. (The actual tower balancing in front of you is much larger.)
Before you're ready to help them, you need to make sure your information is correct. What is the name of the bottom program?
| 163
|
--- Day 7: The Sum of Its Parts ---
You find yourself standing on a snow-covered coastline; apparently, you landed a little off course. The region is too hilly to see the North Pole from here, but you do spot some Elves that seem to be trying to unpack something that washed ashore. It's quite cold out, so you decide to risk creating a paradox by asking them for directions.
"Oh, are you the search party?" Somehow, you can understand whatever Elves from the year 1018 speak; you assume it's Ancient Nordic Elvish. Could the device on your wrist also be a translator? "Those clothes don't look very warm; take this." They hand you a heavy coat.
"We do need to find our way back to the North Pole, but we have higher priorities at the moment. You see, believe it or not, this box contains something that will solve all of Santa's transportation problems - at least, that's what it looks like from the pictures in the instructions." It doesn't seem like they can read whatever language it's in, but you can: "Sleigh kit. Some assembly required."
"'Sleigh'? What a wonderful name! You must help us assemble this 'sleigh' at once!" They start excitedly pulling more parts out of the box.
The instructions specify a series of steps and requirements about which steps must be finished before others can begin (your puzzle input). Each step is designated by a single letter. For example, suppose you have the following instructions:
Step C must be finished before step A can begin.
Step C must be finished before step F can begin.
Step A must be finished before step B can begin.
Step A must be finished before step D can begin.
Step B must be finished before step E can begin.
Step D must be finished before step E can begin.
Step F must be finished before step E can begin.
Visually, these requirements look like this:
-->A--->B--
/ C -->D----->E
/
---->F-----
Your first goal is to determine the order in which the steps should be completed. If more than one step is ready, choose the step which is first alphabetically. In this example, the steps would be completed as follows:
Only C is available, and so it is done first.
Next, both A and F are available. A is first alphabetically, so it is done next.
Then, even though F was available earlier, steps B and D are now also available, and B is the first alphabetically of the three.
After that, only D and F are available. E is not available because only some of its prerequisites are complete. Therefore, D is completed next.
F is the only choice, so it is done next.
Finally, E is completed.
So, in this example, the correct order is CABDFE.
In what order should the steps in your instructions be completed?
Your puzzle answer was CGKMUWXFAIHSYDNLJQTREOPZBV.
--- Part Two ---
As you're about to begin construction, four of the Elves offer to help. "The sun will set soon; it'll go faster if we work together." Now, you need to account for multiple people working on steps simultaneously. If multiple steps are available, workers should still begin them in alphabetical order.
Each step takes 60 seconds plus an amount corresponding to its letter: A=1, B=2, C=3, and so on. So, step A takes 60+1=61 seconds, while step Z takes 60+26=86 seconds. No time is required between steps.
To simplify things for the example, however, suppose you only have help from one Elf (a total of two workers) and that each step takes 60 fewer seconds (so that step A takes 1 second and step Z takes 26 seconds). Then, using the same instructions as above, this is how each second would be spent:
Second Worker 1 Worker 2 Done
0 C .
1 C .
2 C .
3 A F C
4 B F CA
5 B F CA
6 D F CAB
7 D F CAB
8 D F CAB
9 D . CABF
10 E . CABFD
11 E . CABFD
12 E . CABFD
13 E . CABFD
14 E . CABFD
15 . . CABFDE
Each row represents one second of time. The Second column identifies how many seconds have passed as of the beginning of that second. Each worker column shows the step that worker is currently doing (or . if they are idle). The Done column shows completed steps.
Note that the order of the steps has changed; this is because steps now take time to finish and multiple workers can begin multiple steps simultaneously.
In this example, it would take 15 seconds for two workers to complete these steps.
With 5 workers and the 60+ second step durations described above, how long will it take to complete all of the steps?
| 164
|
--- Day 23: Crab Cups ---
The small crab challenges you to a game! The crab is going to mix up some cups, and you have to predict where they'll end up.
The cups will be arranged in a circle and labeled clockwise (your puzzle input). For example, if your labeling were 32415, there would be five cups in the circle; going clockwise around the circle from the first cup, the cups would be labeled 3, 2, 4, 1, 5, and then back to 3 again.
Before the crab starts, it will designate the first cup in your list as the current cup. The crab is then going to do 100 moves.
Each move, the crab does the following actions:
The crab picks up the three cups that are immediately clockwise of the current cup. They are removed from the circle; cup spacing is adjusted as necessary to maintain the circle.
The crab selects a destination cup: the cup with a label equal to the current cup's label minus one. If this would select one of the cups that was just picked up, the crab will keep subtracting one until it finds a cup that wasn't just picked up. If at any point in this process the value goes below the lowest value on any cup's label, it wraps around to the highest value on any cup's label instead.
The crab places the cups it just picked up so that they are immediately clockwise of the destination cup. They keep the same order as when they were picked up.
The crab selects a new current cup: the cup which is immediately clockwise of the current cup.
For example, suppose your cup labeling were 389125467. If the crab were to do merely 10 moves, the following changes would occur:
-- move 1 --
cups: (3) 8 9 1 2 5 4 6 7
pick up: 8, 9, 1
destination: 2
-- move 2 --
cups: 3 (2) 8 9 1 5 4 6 7
pick up: 8, 9, 1
destination: 7
-- move 3 --
cups: 3 2 (5) 4 6 7 8 9 1
pick up: 4, 6, 7
destination: 3
-- move 4 --
cups: 7 2 5 (8) 9 1 3 4 6
pick up: 9, 1, 3
destination: 7
-- move 5 --
cups: 3 2 5 8 (4) 6 7 9 1
pick up: 6, 7, 9
destination: 3
-- move 6 --
cups: 9 2 5 8 4 (1) 3 6 7
pick up: 3, 6, 7
destination: 9
-- move 7 --
cups: 7 2 5 8 4 1 (9) 3 6
pick up: 3, 6, 7
destination: 8
-- move 8 --
cups: 8 3 6 7 4 1 9 (2) 5
pick up: 5, 8, 3
destination: 1
-- move 9 --
cups: 7 4 1 5 8 3 9 2 (6)
pick up: 7, 4, 1
destination: 5
-- move 10 --
cups: (5) 7 4 1 8 3 9 2 6
pick up: 7, 4, 1
destination: 3
-- final --
cups: 5 (8) 3 7 4 1 9 2 6
In the above example, the cups' values are the labels as they appear moving clockwise around the circle; the current cup is marked with ( ).
After the crab is done, what order will the cups be in? Starting after the cup labeled 1, collect the other cups' labels clockwise into a single string with no extra characters; each number except 1 should appear exactly once. In the above example, after 10 moves, the cups clockwise from 1 are labeled 9, 2, 6, 5, and so on, producing 92658374. If the crab were to complete all 100 moves, the order after cup 1 would be 67384529.
Using your labeling, simulate 100 moves. What are the labels on the cups after cup 1?
Your puzzle answer was 65432978.
--- Part Two ---
Due to what you can only assume is a mistranslation (you're not exactly fluent in Crab), you are quite surprised when the crab starts arranging many cups in a circle on your raft - one million (1000000) in total.
Your labeling is still correct for the first few cups; after that, the remaining cups are just numbered in an increasing fashion starting from the number after the highest number in your list and proceeding one by one until one million is reached. (For example, if your labeling were 54321, the cups would be numbered 5, 4, 3, 2, 1, and then start counting up from 6 until one million is reached.) In this way, every number from one through one million is used exactly once.
After discovering where you made the mistake in translating Crab Numbers, you realize the small crab isn't going to do merely 100 moves; the crab is going to do ten million (10000000) moves!
The crab is going to hide your stars - one each - under the two cups that will end up immediately clockwise of cup 1. You can have them if you predict what the labels on those cups will be when the crab is finished.
In the above example (389125467), this would be 934001 and then 159792; multiplying these together produces 149245887792.
Determine which two cups will end up immediately clockwise of cup 1. What do you get if you multiply their labels together?
| 165
|
--- Day 22: Monkey Market ---
As you're all teleported deep into the jungle, a monkey steals The Historians' device! You'll need get it back while The Historians are looking for the Chief.
The monkey that stole the device seems willing to trade it, but only in exchange for an absurd number of bananas. Your only option is to buy bananas on the Monkey Exchange Market.
You aren't sure how the Monkey Exchange Market works, but one of The Historians senses trouble and comes over to help. Apparently, they've been studying these monkeys for a while and have deciphered their secrets.
Today, the Market is full of monkeys buying good hiding spots. Fortunately, because of the time you recently spent in this jungle, you know lots of good hiding spots you can sell! If you sell enough hiding spots, you should be able to get enough bananas to buy the device back.
On the Market, the buyers seem to use random prices, but their prices are actually only pseudorandom! If you know the secret of how they pick their prices, you can wait for the perfect time to sell.
The part about secrets is literal, the Historian explains. Each buyer produces a pseudorandom sequence of secret numbers where each secret is derived from the previous.
In particular, each buyer's secret number evolves into the next secret number in the sequence via the following process:
Calculate the result of multiplying the secret number by 64. Then, mix this result into the secret number. Finally, prune the secret number.
Calculate the result of dividing the secret number by 32. Round the result down to the nearest integer. Then, mix this result into the secret number. Finally, prune the secret number.
Calculate the result of multiplying the secret number by 2048. Then, mix this result into the secret number. Finally, prune the secret number.
Each step of the above process involves mixing and pruning:
To mix a value into the secret number, calculate the bitwise XOR of the given value and the secret number. Then, the secret number becomes the result of that operation. (If the secret number is 42 and you were to mix 15 into the secret number, the secret number would become 37.)
To prune the secret number, calculate the value of the secret number modulo 16777216. Then, the secret number becomes the result of that operation. (If the secret number is 100000000 and you were to prune the secret number, the secret number would become 16113920.)
After this process completes, the buyer is left with the next secret number in the sequence. The buyer can repeat this process as many times as necessary to produce more secret numbers.
So, if a buyer had a secret number of 123, that buyer's next ten secret numbers would be:
15887950
16495136
527345
704524
1553684
12683156
11100544
12249484
7753432
5908254
Each buyer uses their own secret number when choosing their price, so it's important to be able to predict the sequence of secret numbers for each buyer. Fortunately, the Historian's research has uncovered the initial secret number of each buyer (your puzzle input). For example:
1
10
100
2024
This list describes the initial secret number of four different secret-hiding-spot-buyers on the Monkey Exchange Market. If you can simulate secret numbers from each buyer, you'll be able to predict all of their future prices.
In a single day, buyers each have time to generate 2000 new secret numbers. In this example, for each buyer, their initial secret number and the 2000th new secret number they would generate are:
1: 8685429
10: 4700978
100: 15273692
2024: 8667524
Adding up the 2000th new secret number for each buyer produces 37327623.
For each buyer, simulate the creation of 2000 new secret numbers. What is the sum of the 2000th secret number generated by each buyer?
| 166
|
--- Day 15: Beacon Exclusion Zone ---
You feel the ground rumble again as the distress signal leads you to a large network of subterranean tunnels. You don't have time to search them all, but you don't need to: your pack contains a set of deployable sensors that you imagine were originally built to locate lost Elves.
The sensors aren't very powerful, but that's okay; your handheld device indicates that you're close enough to the source of the distress signal to use them. You pull the emergency sensor system out of your pack, hit the big button on top, and the sensors zoom off down the tunnels.
Once a sensor finds a spot it thinks will give it a good reading, it attaches itself to a hard surface and begins monitoring for the nearest signal source beacon. Sensors and beacons always exist at integer coordinates. Each sensor knows its own position and can determine the position of a beacon precisely; however, sensors can only lock on to the one beacon closest to the sensor as measured by the Manhattan distance. (There is never a tie where two beacons are the same distance to a sensor.)
It doesn't take long for the sensors to report back their positions and closest beacons (your puzzle input). For example:
Sensor at x=2, y=18: closest beacon is at x=-2, y=15
Sensor at x=9, y=16: closest beacon is at x=10, y=16
Sensor at x=13, y=2: closest beacon is at x=15, y=3
Sensor at x=12, y=14: closest beacon is at x=10, y=16
Sensor at x=10, y=20: closest beacon is at x=10, y=16
Sensor at x=14, y=17: closest beacon is at x=10, y=16
Sensor at x=8, y=7: closest beacon is at x=2, y=10
Sensor at x=2, y=0: closest beacon is at x=2, y=10
Sensor at x=0, y=11: closest beacon is at x=2, y=10
Sensor at x=20, y=14: closest beacon is at x=25, y=17
Sensor at x=17, y=20: closest beacon is at x=21, y=22
Sensor at x=16, y=7: closest beacon is at x=15, y=3
Sensor at x=14, y=3: closest beacon is at x=15, y=3
Sensor at x=20, y=1: closest beacon is at x=15, y=3
So, consider the sensor at 2,18; the closest beacon to it is at -2,15. For the sensor at 9,16, the closest beacon to it is at 10,16.
Drawing sensors as S and beacons as B, the above arrangement of sensors and beacons looks like this:
1 1 2 2
0 5 0 5 0 5
0 ....S.......................
1 ......................S.....
2 ...............S............
3 ................SB..........
4 ............................
5 ............................
6 ............................
7 ..........S.......S.........
8 ............................
9 ............................
10 ....B.......................
11 ..S.........................
12 ............................
13 ............................
14 ..............S.......S.....
15 B...........................
16 ...........SB...............
17 ................S..........B
18 ....S.......................
19 ............................
20 ............S......S........
21 ............................
22 .......................B....
This isn't necessarily a comprehensive map of all beacons in the area, though. Because each sensor only identifies its closest beacon, if a sensor detects a beacon, you know there are no other beacons that close or closer to that sensor. There could still be beacons that just happen to not be the closest beacon to any sensor. Consider the sensor at 8,7:
1 1 2 2
0 5 0 5 0 5
-2 ..........#.................
-1 .........###................
0 ....S...#####...............
1 .......#######........S.....
2 ......#########S............
3 .....###########SB..........
4 ....#############...........
5 ...###############..........
6 ..#################.........
7 .#########S#######S#........
8 ..#################.........
9 ...###############..........
10 ....B############...........
11 ..S..###########............
12 ......#########.............
13 .......#######..............
14 ........#####.S.......S.....
15 B........###................
16 ..........#SB...............
17 ................S..........B
18 ....S.......................
19 ............................
20 ............S......S........
21 ............................
22 .......................B....
This sensor's closest beacon is at 2,10, and so you know there are no beacons that close or closer (in any positions marked #).
None of the detected beacons seem to be producing the distress signal, so you'll need to work out where the distress beacon is by working out where it isn't. For now, keep things simple by counting the positions where a beacon cannot possibly be along just a single row.
So, suppose you have an arrangement of beacons and sensors like in the example above and, just in the row where y=10, you'd like to count the number of positions a beacon cannot possibly exist. The coverage from all sensors near that row looks like this:
1 1 2 2
0 5 0 5 0 5
9 ...#########################...
10 ..####B######################..
11 .###S#############.###########.
In this example, in the row where y=10, there are 26 positions where a beacon cannot be present.
Consult the report from the sensors you just deployed. In the row where y=2000000, how many positions cannot contain a beacon?
| 167
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--- Day 18: Operation Order ---
As you look out the window and notice a heavily-forested continent slowly appear over the horizon, you are interrupted by the child sitting next to you. They're curious if you could help them with their math homework.
Unfortunately, it seems like this "math" follows different rules than you remember.
The homework (your puzzle input) consists of a series of expressions that consist of addition (+), multiplication (*), and parentheses ((...)). Just like normal math, parentheses indicate that the expression inside must be evaluated before it can be used by the surrounding expression. Addition still finds the sum of the numbers on both sides of the operator, and multiplication still finds the product.
However, the rules of operator precedence have changed. Rather than evaluating multiplication before addition, the operators have the same precedence, and are evaluated left-to-right regardless of the order in which they appear.
For example, the steps to evaluate the expression 1 + 2 * 3 + 4 * 5 + 6 are as follows:
1 + 2 * 3 + 4 * 5 + 6
3 * 3 + 4 * 5 + 6
9 + 4 * 5 + 6
13 * 5 + 6
65 + 6
71
Parentheses can override this order; for example, here is what happens if parentheses are added to form 1 + (2 * 3) + (4 * (5 + 6)):
1 + (2 * 3) + (4 * (5 + 6))
1 + 6 + (4 * (5 + 6))
7 + (4 * (5 + 6))
7 + (4 * 11 )
7 + 44
51
Here are a few more examples:
2 * 3 + (4 * 5) becomes 26.
5 + (8 * 3 + 9 + 3 * 4 * 3) becomes 437.
5 * 9 * (7 * 3 * 3 + 9 * 3 + (8 + 6 * 4)) becomes 12240.
((2 + 4 * 9) * (6 + 9 * 8 + 6) + 6) + 2 + 4 * 2 becomes 13632.
Before you can help with the homework, you need to understand it yourself. Evaluate the expression on each line of the homework; what is the sum of the resulting values?
| 168
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--- Day 6: Custom Customs ---
As your flight approaches the regional airport where you'll switch to a much larger plane, customs declaration forms are distributed to the passengers.
The form asks a series of 26 yes-or-no questions marked a through z. All you need to do is identify the questions for which anyone in your group answers "yes". Since your group is just you, this doesn't take very long.
However, the person sitting next to you seems to be experiencing a language barrier and asks if you can help. For each of the people in their group, you write down the questions for which they answer "yes", one per line. For example:
abcx
abcy
abcz
In this group, there are 6 questions to which anyone answered "yes": a, b, c, x, y, and z. (Duplicate answers to the same question don't count extra; each question counts at most once.)
Another group asks for your help, then another, and eventually you've collected answers from every group on the plane (your puzzle input). Each group's answers are separated by a blank line, and within each group, each person's answers are on a single line. For example:
abc
a
b
c
ab
ac
a
a
a
a
b
This list represents answers from five groups:
The first group contains one person who answered "yes" to 3 questions: a, b, and c.
The second group contains three people; combined, they answered "yes" to 3 questions: a, b, and c.
The third group contains two people; combined, they answered "yes" to 3 questions: a, b, and c.
The fourth group contains four people; combined, they answered "yes" to only 1 question, a.
The last group contains one person who answered "yes" to only 1 question, b.
In this example, the sum of these counts is 3 + 3 + 3 + 1 + 1 = 11.
For each group, count the number of questions to which anyone answered "yes". What is the sum of those counts?
Your puzzle answer was 6662.
--- Part Two ---
As you finish the last group's customs declaration, you notice that you misread one word in the instructions:
You don't need to identify the questions to which anyone answered "yes"; you need to identify the questions to which everyone answered "yes"!
Using the same example as above:
abc
a
b
c
ab
ac
a
a
a
a
b
This list represents answers from five groups:
In the first group, everyone (all 1 person) answered "yes" to 3 questions: a, b, and c.
In the second group, there is no question to which everyone answered "yes".
In the third group, everyone answered yes to only 1 question, a. Since some people did not answer "yes" to b or c, they don't count.
In the fourth group, everyone answered yes to only 1 question, a.
In the fifth group, everyone (all 1 person) answered "yes" to 1 question, b.
In this example, the sum of these counts is 3 + 0 + 1 + 1 + 1 = 6.
For each group, count the number of questions to which everyone answered "yes". What is the sum of those counts?
| 169
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--- Day 1: Not Quite Lisp ---
Santa was hoping for a white Christmas, but his weather machine's "snow" function is powered by stars, and he's fresh out! To save Christmas, he needs you to collect fifty stars by December 25th.
Collect stars by helping Santa solve 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!
Here's an easy puzzle to warm you up.
Santa is trying to deliver presents in a large apartment building, but he can't find the right floor - the directions he got are a little confusing. He starts on the ground floor (floor 0) and then follows the instructions one character at a time.
An opening parenthesis, (, means he should go up one floor, and a closing parenthesis, ), means he should go down one floor.
The apartment building is very tall, and the basement is very deep; he will never find the top or bottom floors.
For example:
(()) and ()() both result in floor 0.
((( and (()(()( both result in floor 3.
))((((( also results in floor 3.
()) and ))( both result in floor -1 (the first basement level).
))) and )())()) both result in floor -3.
To what floor do the instructions take Santa?
| 170
|
--- Day 1: Report Repair ---
After saving Christmas five years in a row, you've decided to take a vacation at a nice resort on a tropical island. Surely, Christmas will go on without you.
The tropical island has its own currency and is entirely cash-only. The gold coins used there have a little picture of a starfish; the locals just call them stars. None of the currency exchanges seem to have heard of them, but somehow, you'll need to find fifty of these coins by the time you arrive so you can pay the deposit on your room.
To save your vacation, you 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!
Before you leave, the Elves in accounting just need you to fix your expense report (your puzzle input); apparently, something isn't quite adding up.
Specifically, they need you to find the two entries that sum to 2020 and then multiply those two numbers together.
For example, suppose your expense report contained the following:
1721
979
366
299
675
1456
In this list, the two entries that sum to 2020 are 1721 and 299. Multiplying them together produces 1721 * 299 = 514579, so the correct answer is 514579.
Of course, your expense report is much larger. Find the two entries that sum to 2020; what do you get if you multiply them together?
| 171
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--- Day 8: Handheld Halting ---
Your flight to the major airline hub reaches cruising altitude without incident. While you consider checking the in-flight menu for one of those drinks that come with a little umbrella, you are interrupted by the kid sitting next to you.
Their handheld game console won't turn on! They ask if you can take a look.
You narrow the problem down to a strange infinite loop in the boot code (your puzzle input) of the device. You should be able to fix it, but first you need to be able to run the code in isolation.
The boot code is represented as a text file with one instruction per line of text. Each instruction consists of an operation (acc, jmp, or nop) and an argument (a signed number like +4 or -20).
acc increases or decreases a single global value called the accumulator by the value given in the argument. For example, acc +7 would increase the accumulator by 7. The accumulator starts at 0. After an acc instruction, the instruction immediately below it is executed next.
jmp jumps to a new instruction relative to itself. The next instruction to execute is found using the argument as an offset from the jmp instruction; for example, jmp +2 would skip the next instruction, jmp +1 would continue to the instruction immediately below it, and jmp -20 would cause the instruction 20 lines above to be executed next.
nop stands for No OPeration - it does nothing. The instruction immediately below it is executed next.
For example, consider the following program:
nop +0
acc +1
jmp +4
acc +3
jmp -3
acc -99
acc +1
jmp -4
acc +6
These instructions are visited in this order:
nop +0 | 1
acc +1 | 2, 8(!)
jmp +4 | 3
acc +3 | 6
jmp -3 | 7
acc -99 |
acc +1 | 4
jmp -4 | 5
acc +6 |
First, the nop +0 does nothing. Then, the accumulator is increased from 0 to 1 (acc +1) and jmp +4 sets the next instruction to the other acc +1 near the bottom. After it increases the accumulator from 1 to 2, jmp -4 executes, setting the next instruction to the only acc +3. It sets the accumulator to 5, and jmp -3 causes the program to continue back at the first acc +1.
This is an infinite loop: with this sequence of jumps, the program will run forever. The moment the program tries to run any instruction a second time, you know it will never terminate.
Immediately before the program would run an instruction a second time, the value in the accumulator is 5.
Run your copy of the boot code. Immediately before any instruction is executed a second time, what value is in the accumulator?
Your puzzle answer was 1137.
--- Part Two ---
After some careful analysis, you believe that exactly one instruction is corrupted.
Somewhere in the program, either a jmp is supposed to be a nop, or a nop is supposed to be a jmp. (No acc instructions were harmed in the corruption of this boot code.)
The program is supposed to terminate by attempting to execute an instruction immediately after the last instruction in the file. By changing exactly one jmp or nop, you can repair the boot code and make it terminate correctly.
For example, consider the same program from above:
nop +0
acc +1
jmp +4
acc +3
jmp -3
acc -99
acc +1
jmp -4
acc +6
If you change the first instruction from nop +0 to jmp +0, it would create a single-instruction infinite loop, never leaving that instruction. If you change almost any of the jmp instructions, the program will still eventually find another jmp instruction and loop forever.
However, if you change the second-to-last instruction (from jmp -4 to nop -4), the program terminates! The instructions are visited in this order:
nop +0 | 1
acc +1 | 2
jmp +4 | 3
acc +3 |
jmp -3 |
acc -99 |
acc +1 | 4
nop -4 | 5
acc +6 | 6
After the last instruction (acc +6), the program terminates by attempting to run the instruction below the last instruction in the file. With this change, after the program terminates, the accumulator contains the value 8 (acc +1, acc +1, acc +6).
Fix the program so that it terminates normally by changing exactly one jmp (to nop) or nop (to jmp). What is the value of the accumulator after the program terminates?
| 172
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--- Day 9: Mirage Maintenance ---
You ride the camel through the sandstorm and stop where the ghost's maps told you to stop. The sandstorm subsequently subsides, somehow seeing you standing at an oasis!
The camel goes to get some water and you stretch your neck. As you look up, you discover what must be yet another giant floating island, this one made of metal! That must be where the parts to fix the sand machines come from.
There's even a hang glider partially buried in the sand here; once the sun rises and heats up the sand, you might be able to use the glider and the hot air to get all the way up to the metal island!
While you wait for the sun to rise, you admire the oasis hidden here in the middle of Desert Island. It must have a delicate ecosystem; you might as well take some ecological readings while you wait. Maybe you can report any environmental instabilities you find to someone so the oasis can be around for the next sandstorm-worn traveler.
You pull out your handy Oasis And Sand Instability Sensor and analyze your surroundings. The OASIS produces a report of many values and how they are changing over time (your puzzle input). Each line in the report contains the history of a single value. For example:
0 3 6 9 12 15
1 3 6 10 15 21
10 13 16 21 30 45
To best protect the oasis, your environmental report should include a prediction of the next value in each history. To do this, start by making a new sequence from the difference at each step of your history. If that sequence is not all zeroes, repeat this process, using the sequence you just generated as the input sequence. Once all of the values in your latest sequence are zeroes, you can extrapolate what the next value of the original history should be.
In the above dataset, the first history is 0 3 6 9 12 15. Because the values increase by 3 each step, the first sequence of differences that you generate will be 3 3 3 3 3. Note that this sequence has one fewer value than the input sequence because at each step it considers two numbers from the input. Since these values aren't all zero, repeat the process: the values differ by 0 at each step, so the next sequence is 0 0 0 0. This means you have enough information to extrapolate the history! Visually, these sequences can be arranged like this:
0 3 6 9 12 15
3 3 3 3 3
0 0 0 0
To extrapolate, start by adding a new zero to the end of your list of zeroes; because the zeroes represent differences between the two values above them, this also means there is now a placeholder in every sequence above it:
0 3 6 9 12 15 B
3 3 3 3 3 A
0 0 0 0 0
You can then start filling in placeholders from the bottom up. A needs to be the result of increasing 3 (the value to its left) by 0 (the value below it); this means A must be 3:
0 3 6 9 12 15 B
3 3 3 3 3 3
0 0 0 0 0
Finally, you can fill in B, which needs to be the result of increasing 15 (the value to its left) by 3 (the value below it), or 18:
0 3 6 9 12 15 18
3 3 3 3 3 3
0 0 0 0 0
So, the next value of the first history is 18.
Finding all-zero differences for the second history requires an additional sequence:
1 3 6 10 15 21
2 3 4 5 6
1 1 1 1
0 0 0
Then, following the same process as before, work out the next value in each sequence from the bottom up:
1 3 6 10 15 21 28
2 3 4 5 6 7
1 1 1 1 1
0 0 0 0
So, the next value of the second history is 28.
The third history requires even more sequences, but its next value can be found the same way:
10 13 16 21 30 45 68
3 3 5 9 15 23
0 2 4 6 8
2 2 2 2
0 0 0
So, the next value of the third history is 68.
If you find the next value for each history in this example and add them together, you get 114.
Analyze your OASIS report and extrapolate the next value for each history. What is the sum of these extrapolated values?
Your puzzle answer was 2101499000.
--- Part Two ---
Of course, it would be nice to have even more history included in your report. Surely it's safe to just extrapolate backwards as well, right?
For each history, repeat the process of finding differences until the sequence of differences is entirely zero. Then, rather than adding a zero to the end and filling in the next values of each previous sequence, you should instead add a zero to the beginning of your sequence of zeroes, then fill in new first values for each previous sequence.
In particular, here is what the third example history looks like when extrapolating back in time:
5 10 13 16 21 30 45
5 3 3 5 9 15
-2 0 2 4 6
2 2 2 2
0 0 0
Adding the new values on the left side of each sequence from bottom to top eventually reveals the new left-most history value: 5.
Doing this for the remaining example data above results in previous values of -3 for the first history and 0 for the second history. Adding all three new values together produces 2.
Analyze your OASIS report again, this time extrapolating the previous value for each history. What is the sum of these extrapolated values?
| 173
|
--- Day 16: Ticket Translation ---
As you're walking to yet another connecting flight, you realize that one of the legs of your re-routed trip coming up is on a high-speed train. However, the train ticket you were given is in a language you don't understand. You should probably figure out what it says before you get to the train station after the next flight.
Unfortunately, you can't actually read the words on the ticket. You can, however, read the numbers, and so you figure out the fields these tickets must have and the valid ranges for values in those fields.
You collect the rules for ticket fields, the numbers on your ticket, and the numbers on other nearby tickets for the same train service (via the airport security cameras) together into a single document you can reference (your puzzle input).
The rules for ticket fields specify a list of fields that exist somewhere on the ticket and the valid ranges of values for each field. For example, a rule like class: 1-3 or 5-7 means that one of the fields in every ticket is named class and can be any value in the ranges 1-3 or 5-7 (inclusive, such that 3 and 5 are both valid in this field, but 4 is not).
Each ticket is represented by a single line of comma-separated values. The values are the numbers on the ticket in the order they appear; every ticket has the same format. For example, consider this ticket:
.--------------------------------------------------------.
| ????: 101 ?????: 102 ??????????: 103 ???: 104 |
| |
| ??: 301 ??: 302 ???????: 303 ??????? |
| ??: 401 ??: 402 ???? ????: 403 ????????? |
'--------------------------------------------------------'
Here, ? represents text in a language you don't understand. This ticket might be represented as 101,102,103,104,301,302,303,401,402,403; of course, the actual train tickets you're looking at are much more complicated. In any case, you've extracted just the numbers in such a way that the first number is always the same specific field, the second number is always a different specific field, and so on - you just don't know what each position actually means!
Start by determining which tickets are completely invalid; these are tickets that contain values which aren't valid for any field. Ignore your ticket for now.
For example, suppose you have the following notes:
class: 1-3 or 5-7
row: 6-11 or 33-44
seat: 13-40 or 45-50
your ticket:
7,1,14
nearby tickets:
7,3,47
40,4,50
55,2,20
38,6,12
It doesn't matter which position corresponds to which field; you can identify invalid nearby tickets by considering only whether tickets contain values that are not valid for any field. In this example, the values on the first nearby ticket are all valid for at least one field. This is not true of the other three nearby tickets: the values 4, 55, and 12 are are not valid for any field. Adding together all of the invalid values produces your ticket scanning error rate: 4 + 55 + 12 = 71.
Consider the validity of the nearby tickets you scanned. What is your ticket scanning error rate?
Your puzzle answer was 32842.
--- Part Two ---
Now that you've identified which tickets contain invalid values, discard those tickets entirely. Use the remaining valid tickets to determine which field is which.
Using the valid ranges for each field, determine what order the fields appear on the tickets. The order is consistent between all tickets: if seat is the third field, it is the third field on every ticket, including your ticket.
For example, suppose you have the following notes:
class: 0-1 or 4-19
row: 0-5 or 8-19
seat: 0-13 or 16-19
your ticket:
11,12,13
nearby tickets:
3,9,18
15,1,5
5,14,9
Based on the nearby tickets in the above example, the first position must be row, the second position must be class, and the third position must be seat; you can conclude that in your ticket, class is 12, row is 11, and seat is 13.
Once you work out which field is which, look for the six fields on your ticket that start with the word departure. What do you get if you multiply those six values together?
| 174
|
--- Day 4: Scratchcards ---
The gondola takes you up. Strangely, though, the ground doesn't seem to be coming with you; you're not climbing a mountain. As the circle of Snow Island recedes below you, an entire new landmass suddenly appears above you! The gondola carries you to the surface of the new island and lurches into the station.
As you exit the gondola, the first thing you notice is that the air here is much warmer than it was on Snow Island. It's also quite humid. Is this where the water source is?
The next thing you notice is an Elf sitting on the floor across the station in what seems to be a pile of colorful square cards.
"Oh! Hello!" The Elf excitedly runs over to you. "How may I be of service?" You ask about water sources.
"I'm not sure; I just operate the gondola lift. That does sound like something we'd have, though - this is Island Island, after all! I bet the gardener would know. He's on a different island, though - er, the small kind surrounded by water, not the floating kind. We really need to come up with a better naming scheme. Tell you what: if you can help me with something quick, I'll let you borrow my boat and you can go visit the gardener. I got all these scratchcards as a gift, but I can't figure out what I've won."
The Elf leads you over to the pile of colorful cards. There, you discover dozens of scratchcards, all with their opaque covering already scratched off. Picking one up, it looks like each card has two lists of numbers separated by a vertical bar (|): a list of winning numbers and then a list of numbers you have. You organize the information into a table (your puzzle input).
As far as the Elf has been able to figure out, you have to figure out which of the numbers you have appear in the list of winning numbers. The first match makes the card worth one point and each match after the first doubles the point value of that card.
For example:
Card 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53
Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19
Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1
Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83
Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36
Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11
In the above example, card 1 has five winning numbers (41, 48, 83, 86, and 17) and eight numbers you have (83, 86, 6, 31, 17, 9, 48, and 53). Of the numbers you have, four of them (48, 83, 17, and 86) are winning numbers! That means card 1 is worth 8 points (1 for the first match, then doubled three times for each of the three matches after the first).
Card 2 has two winning numbers (32 and 61), so it is worth 2 points.
Card 3 has two winning numbers (1 and 21), so it is worth 2 points.
Card 4 has one winning number (84), so it is worth 1 point.
Card 5 has no winning numbers, so it is worth no points.
Card 6 has no winning numbers, so it is worth no points.
So, in this example, the Elf's pile of scratchcards is worth 13 points.
Take a seat in the large pile of colorful cards. How many points are they worth in total?
Your puzzle answer was 25004.
--- Part Two ---
Just as you're about to report your findings to the Elf, one of you realizes that the rules have actually been printed on the back of every card this whole time.
There's no such thing as "points". Instead, scratchcards only cause you to win more scratchcards equal to the number of winning numbers you have.
Specifically, you win copies of the scratchcards below the winning card equal to the number of matches. So, if card 10 were to have 5 matching numbers, you would win one copy each of cards 11, 12, 13, 14, and 15.
Copies of scratchcards are scored like normal scratchcards and have the same card number as the card they copied. So, if you win a copy of card 10 and it has 5 matching numbers, it would then win a copy of the same cards that the original card 10 won: cards 11, 12, 13, 14, and 15. This process repeats until none of the copies cause you to win any more cards. (Cards will never make you copy a card past the end of the table.)
This time, the above example goes differently:
Card 1: 41 48 83 86 17 | 83 86 6 31 17 9 48 53
Card 2: 13 32 20 16 61 | 61 30 68 82 17 32 24 19
Card 3: 1 21 53 59 44 | 69 82 63 72 16 21 14 1
Card 4: 41 92 73 84 69 | 59 84 76 51 58 5 54 83
Card 5: 87 83 26 28 32 | 88 30 70 12 93 22 82 36
Card 6: 31 18 13 56 72 | 74 77 10 23 35 67 36 11
Card 1 has four matching numbers, so you win one copy each of the next four cards: cards 2, 3, 4, and 5.
Your original card 2 has two matching numbers, so you win one copy each of cards 3 and 4.
Your copy of card 2 also wins one copy each of cards 3 and 4.
Your four instances of card 3 (one original and three copies) have two matching numbers, so you win four copies each of cards 4 and 5.
Your eight instances of card 4 (one original and seven copies) have one matching number, so you win eight copies of card 5.
Your fourteen instances of card 5 (one original and thirteen copies) have no matching numbers and win no more cards.
Your one instance of card 6 (one original) has no matching numbers and wins no more cards.
Once all of the originals and copies have been processed, you end up with 1 instance of card 1, 2 instances of card 2, 4 instances of card 3, 8 instances of card 4, 14 instances of card 5, and 1 instance of card 6. In total, this example pile of scratchcards causes you to ultimately have 30 scratchcards!
Process all of the original and copied scratchcards until no more scratchcards are won. Including the original set of scratchcards, how many total scratchcards do you end up with?
| 175
|
--- Day 15: Rambunctious Recitation ---
You catch the airport shuttle and try to book a new flight to your vacation island. Due to the storm, all direct flights have been cancelled, but a route is available to get around the storm. You take it.
While you wait for your flight, you decide to check in with the Elves back at the North Pole. They're playing a memory game and are ever so excited to explain the rules!
In this game, the players take turns saying numbers. They begin by taking turns reading from a list of starting numbers (your puzzle input). Then, each turn consists of considering the most recently spoken number:
If that was the first time the number has been spoken, the current player says 0.
Otherwise, the number had been spoken before; the current player announces how many turns apart the number is from when it was previously spoken.
So, after the starting numbers, each turn results in that player speaking aloud either 0 (if the last number is new) or an age (if the last number is a repeat).
For example, suppose the starting numbers are 0,3,6:
Turn 1: The 1st number spoken is a starting number, 0.
Turn 2: The 2nd number spoken is a starting number, 3.
Turn 3: The 3rd number spoken is a starting number, 6.
Turn 4: Now, consider the last number spoken, 6. Since that was the first time the number had been spoken, the 4th number spoken is 0.
Turn 5: Next, again consider the last number spoken, 0. Since it had been spoken before, the next number to speak is the difference between the turn number when it was last spoken (the previous turn, 4) and the turn number of the time it was most recently spoken before then (turn 1). Thus, the 5th number spoken is 4 - 1, 3.
Turn 6: The last number spoken, 3 had also been spoken before, most recently on turns 5 and 2. So, the 6th number spoken is 5 - 2, 3.
Turn 7: Since 3 was just spoken twice in a row, and the last two turns are 1 turn apart, the 7th number spoken is 1.
Turn 8: Since 1 is new, the 8th number spoken is 0.
Turn 9: 0 was last spoken on turns 8 and 4, so the 9th number spoken is the difference between them, 4.
Turn 10: 4 is new, so the 10th number spoken is 0.
(The game ends when the Elves get sick of playing or dinner is ready, whichever comes first.)
Their question for you is: what will be the 2020th number spoken? In the example above, the 2020th number spoken will be 436.
Here are a few more examples:
Given the starting numbers 1,3,2, the 2020th number spoken is 1.
Given the starting numbers 2,1,3, the 2020th number spoken is 10.
Given the starting numbers 1,2,3, the 2020th number spoken is 27.
Given the starting numbers 2,3,1, the 2020th number spoken is 78.
Given the starting numbers 3,2,1, the 2020th number spoken is 438.
Given the starting numbers 3,1,2, the 2020th number spoken is 1836.
Given your starting numbers, what will be the 2020th number spoken?
Your puzzle answer was 959.
--- Part Two ---
Impressed, the Elves issue you a challenge: determine the 30000000th number spoken. For example, given the same starting numbers as above:
Given 0,3,6, the 30000000th number spoken is 175594.
Given 1,3,2, the 30000000th number spoken is 2578.
Given 2,1,3, the 30000000th number spoken is 3544142.
Given 1,2,3, the 30000000th number spoken is 261214.
Given 2,3,1, the 30000000th number spoken is 6895259.
Given 3,2,1, the 30000000th number spoken is 18.
Given 3,1,2, the 30000000th number spoken is 362.
Given your starting numbers, what will be the 30000000th number spoken?
| 176
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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?
| 177
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--- Day 12: Hot Springs ---
You finally reach the hot springs! You can see steam rising from secluded areas attached to the primary, ornate building.
As you turn to enter, the researcher stops you. "Wait - I thought you were looking for the hot springs, weren't you?" You indicate that this definitely looks like hot springs to you.
"Oh, sorry, common mistake! This is actually the onsen! The hot springs are next door."
You look in the direction the researcher is pointing and suddenly notice the massive metal helixes towering overhead. "This way!"
It only takes you a few more steps to reach the main gate of the massive fenced-off area containing the springs. You go through the gate and into a small administrative building.
"Hello! What brings you to the hot springs today? Sorry they're not very hot right now; we're having a lava shortage at the moment." You ask about the missing machine parts for Desert Island.
"Oh, all of Gear Island is currently offline! Nothing is being manufactured at the moment, not until we get more lava to heat our forges. And our springs. The springs aren't very springy unless they're hot!"
"Say, could you go up and see why the lava stopped flowing? The springs are too cold for normal operation, but we should be able to find one springy enough to launch you up there!"
There's just one problem - many of the springs have fallen into disrepair, so they're not actually sure which springs would even be safe to use! Worse yet, their condition records of which springs are damaged (your puzzle input) are also damaged! You'll need to help them repair the damaged records.
In the giant field just outside, the springs are arranged into rows. For each row, the condition records show every spring and whether it is operational (.) or damaged (#). This is the part of the condition records that is itself damaged; for some springs, it is simply unknown (?) whether the spring is operational or damaged.
However, the engineer that produced the condition records also duplicated some of this information in a different format! After the list of springs for a given row, the size of each contiguous group of damaged springs is listed in the order those groups appear in the row. This list always accounts for every damaged spring, and each number is the entire size of its contiguous group (that is, groups are always separated by at least one operational spring: #### would always be 4, never 2,2).
So, condition records with no unknown spring conditions might look like this:
#.#.### 1,1,3
.#...#....###. 1,1,3
.#.###.#.###### 1,3,1,6
####.#...#... 4,1,1
#....######..#####. 1,6,5
.###.##....# 3,2,1
However, the condition records are partially damaged; some of the springs' conditions are actually unknown (?). For example:
???.### 1,1,3
.??..??...?##. 1,1,3
?#?#?#?#?#?#?#? 1,3,1,6
????.#...#... 4,1,1
????.######..#####. 1,6,5
?###???????? 3,2,1
Equipped with this information, it is your job to figure out how many different arrangements of operational and broken springs fit the given criteria in each row.
In the first line (???.### 1,1,3), there is exactly one way separate groups of one, one, and three broken springs (in that order) can appear in that row: the first three unknown springs must be broken, then operational, then broken (#.#), making the whole row #.#.###.
The second line is more interesting: .??..??...?##. 1,1,3 could be a total of four different arrangements. The last ? must always be broken (to satisfy the final contiguous group of three broken springs), and each ?? must hide exactly one of the two broken springs. (Neither ?? could be both broken springs or they would form a single contiguous group of two; if that were true, the numbers afterward would have been 2,3 instead.) Since each ?? can either be #. or .#, there are four possible arrangements of springs.
The last line is actually consistent with ten different arrangements! Because the first number is 3, the first and second ? must both be . (if either were #, the first number would have to be 4 or higher). However, the remaining run of unknown spring conditions have many different ways they could hold groups of two and one broken springs:
?###???????? 3,2,1
.###.##.#...
.###.##..#..
.###.##...#.
.###.##....#
.###..##.#..
.###..##..#.
.###..##...#
.###...##.#.
.###...##..#
.###....##.#
In this example, the number of possible arrangements for each row is:
???.### 1,1,3 - 1 arrangement
.??..??...?##. 1,1,3 - 4 arrangements
?#?#?#?#?#?#?#? 1,3,1,6 - 1 arrangement
????.#...#... 4,1,1 - 1 arrangement
????.######..#####. 1,6,5 - 4 arrangements
?###???????? 3,2,1 - 10 arrangements
Adding all of the possible arrangement counts together produces a total of 21 arrangements.
For each row, count all of the different arrangements of operational and broken springs that meet the given criteria. What is the sum of those counts?
| 178
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--- Day 9: Sensor Boost ---
You've just said goodbye to the rebooted rover and left Mars when you receive a faint distress signal coming from the asteroid belt. It must be the Ceres monitoring station!
In order to lock on to the signal, you'll need to boost your sensors. The Elves send up the latest BOOST program - Basic Operation Of System Test.
While BOOST (your puzzle input) is capable of boosting your sensors, for tenuous safety reasons, it refuses to do so until the computer it runs on passes some checks to demonstrate it is a complete Intcode computer.
Your existing Intcode computer is missing one key feature: it needs support for parameters in relative mode.
Parameters in mode 2, relative mode, behave very similarly to parameters in position mode: the parameter is interpreted as a position. Like position mode, parameters in relative mode can be read from or written to.
The important difference is that relative mode parameters don't count from address 0. Instead, they count from a value called the relative base. The relative base starts at 0.
The address a relative mode parameter refers to is itself plus the current relative base. When the relative base is 0, relative mode parameters and position mode parameters with the same value refer to the same address.
For example, given a relative base of 50, a relative mode parameter of -7 refers to memory address 50 + -7 = 43.
The relative base is modified with the relative base offset instruction:
Opcode 9 adjusts the relative base by the value of its only parameter. The relative base increases (or decreases, if the value is negative) by the value of the parameter.
For example, if the relative base is 2000, then after the instruction 109,19, the relative base would be 2019. If the next instruction were 204,-34, then the value at address 1985 would be output.
Your Intcode computer will also need a few other capabilities:
The computer's available memory should be much larger than the initial program. Memory beyond the initial program starts with the value 0 and can be read or written like any other memory. (It is invalid to try to access memory at a negative address, though.)
The computer should have support for large numbers. Some instructions near the beginning of the BOOST program will verify this capability.
Here are some example programs that use these features:
109,1,204,-1,1001,100,1,100,1008,100,16,101,1006,101,0,99 takes no input and produces a copy of itself as output.
1102,34915192,34915192,7,4,7,99,0 should output a 16-digit number.
104,1125899906842624,99 should output the large number in the middle.
The BOOST program will ask for a single input; run it in test mode by providing it the value 1. It will perform a series of checks on each opcode, output any opcodes (and the associated parameter modes) that seem to be functioning incorrectly, and finally output a BOOST keycode.
Once your Intcode computer is fully functional, the BOOST program should report no malfunctioning opcodes when run in test mode; it should only output a single value, the BOOST keycode. What BOOST keycode does it produce?
| 179
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--- Day 4: The Ideal Stocking Stuffer ---
Santa needs help mining some AdventCoins (very similar to bitcoins) to use as gifts for all the economically forward-thinking little girls and boys.
To do this, he needs to find MD5 hashes which, in hexadecimal, start with at least five zeroes. The input to the MD5 hash is some secret key (your puzzle input, given below) followed by a number in decimal. To mine AdventCoins, you must find Santa the lowest positive number (no leading zeroes: 1, 2, 3, ...) that produces such a hash.
For example:
If your secret key is abcdef, the answer is 609043, because the MD5 hash of abcdef609043 starts with five zeroes (000001dbbfa...), and it is the lowest such number to do so.
If your secret key is pqrstuv, the lowest number it combines with to make an MD5 hash starting with five zeroes is 1048970; that is, the MD5 hash of pqrstuv1048970 looks like 000006136ef....
Your puzzle answer was 117946.
--- Part Two ---
Now find one that starts with six zeroes.
| 180
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--- Day 4: High-Entropy Passphrases ---
A new system policy has been put in place that requires all accounts to use a passphrase instead of simply a password. A passphrase consists of a series of words (lowercase letters) separated by spaces.
To ensure security, a valid passphrase must contain no duplicate words.
For example:
aa bb cc dd ee is valid.
aa bb cc dd aa is not valid - the word aa appears more than once.
aa bb cc dd aaa is valid - aa and aaa count as different words.
The system's full passphrase list is available as your puzzle input. How many passphrases are valid?
| 181
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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 380612.
--- 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?
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--- Day 2: Password Philosophy ---
Your flight departs in a few days from the coastal airport; the easiest way down to the coast from here is via toboggan.
The shopkeeper at the North Pole Toboggan Rental Shop is having a bad day. "Something's wrong with our computers; we can't log in!" You ask if you can take a look.
Their password database seems to be a little corrupted: some of the passwords wouldn't have been allowed by the Official Toboggan Corporate Policy that was in effect when they were chosen.
To try to debug the problem, they have created a list (your puzzle input) of passwords (according to the corrupted database) and the corporate policy when that password was set.
For example, suppose you have the following list:
1-3 a: abcde
1-3 b: cdefg
2-9 c: ccccccccc
Each line gives the password policy and then the password. The password policy indicates the lowest and highest number of times a given letter must appear for the password to be valid. For example, 1-3 a means that the password must contain a at least 1 time and at most 3 times.
In the above example, 2 passwords are valid. The middle password, cdefg, is not; it contains no instances of b, but needs at least 1. The first and third passwords are valid: they contain one a or nine c, both within the limits of their respective policies.
How many passwords are valid according to their policies?
Your puzzle answer was 519.
--- Part Two ---
While it appears you validated the passwords correctly, they don't seem to be what the Official Toboggan Corporate Authentication System is expecting.
The shopkeeper suddenly realizes that he just accidentally explained the password policy rules from his old job at the sled rental place down the street! The Official Toboggan Corporate Policy actually works a little differently.
Each policy actually describes two positions in the password, where 1 means the first character, 2 means the second character, and so on. (Be careful; Toboggan Corporate Policies have no concept of "index zero"!) Exactly one of these positions must contain the given letter. Other occurrences of the letter are irrelevant for the purposes of policy enforcement.
Given the same example list from above:
1-3 a: abcde is valid: position 1 contains a and position 3 does not.
1-3 b: cdefg is invalid: neither position 1 nor position 3 contains b.
2-9 c: ccccccccc is invalid: both position 2 and position 9 contain c.
How many passwords are valid according to the new interpretation of the policies?
| 183
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--- Day 19: Linen Layout ---
Today, The Historians take you up to the hot springs on Gear Island! Very suspiciously, absolutely nothing goes wrong as they begin their careful search of the vast field of helixes.
Could this finally be your chance to visit the onsen next door? Only one way to find out.
After a brief conversation with the reception staff at the onsen front desk, you discover that you don't have the right kind of money to pay the admission fee. However, before you can leave, the staff get your attention. Apparently, they've heard about how you helped at the hot springs, and they're willing to make a deal: if you can simply help them arrange their towels, they'll let you in for free!
Every towel at this onsen is marked with a pattern of colored stripes. There are only a few patterns, but for any particular pattern, the staff can get you as many towels with that pattern as you need. Each stripe can be white (w), blue (u), black (b), red (r), or green (g). So, a towel with the pattern ggr would have a green stripe, a green stripe, and then a red stripe, in that order. (You can't reverse a pattern by flipping a towel upside-down, as that would cause the onsen logo to face the wrong way.)
The Official Onsen Branding Expert has produced a list of designs - each a long sequence of stripe colors - that they would like to be able to display. You can use any towels you want, but all of the towels' stripes must exactly match the desired design. So, to display the design rgrgr, you could use two rg towels and then an r towel, an rgr towel and then a gr towel, or even a single massive rgrgr towel (assuming such towel patterns were actually available).
To start, collect together all of the available towel patterns and the list of desired designs (your puzzle input). For example:
r, wr, b, g, bwu, rb, gb, br
brwrr
bggr
gbbr
rrbgbr
ubwu
bwurrg
brgr
bbrgwb
The first line indicates the available towel patterns; in this example, the onsen has unlimited towels with a single red stripe (r), unlimited towels with a white stripe and then a red stripe (wr), and so on.
After the blank line, the remaining lines each describe a design the onsen would like to be able to display. In this example, the first design (brwrr) indicates that the onsen would like to be able to display a black stripe, a red stripe, a white stripe, and then two red stripes, in that order.
Not all designs will be possible with the available towels. In the above example, the designs are possible or impossible as follows:
brwrr can be made with a br towel, then a wr towel, and then finally an r towel.
bggr can be made with a b towel, two g towels, and then an r towel.
gbbr can be made with a gb towel and then a br towel.
rrbgbr can be made with r, rb, g, and br.
ubwu is impossible.
bwurrg can be made with bwu, r, r, and g.
brgr can be made with br, g, and r.
bbrgwb is impossible.
In this example, 6 of the eight designs are possible with the available towel patterns.
To get into the onsen as soon as possible, consult your list of towel patterns and desired designs carefully. How many designs are possible?
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--- Day 14: Regolith Reservoir ---
The distress signal leads you to a giant waterfall! Actually, hang on - the signal seems like it's coming from the waterfall itself, and that doesn't make any sense. However, you do notice a little path that leads behind the waterfall.
Correction: the distress signal leads you behind a giant waterfall! There seems to be a large cave system here, and the signal definitely leads further inside.
As you begin to make your way deeper underground, you feel the ground rumble for a moment. Sand begins pouring into the cave! If you don't quickly figure out where the sand is going, you could quickly become trapped!
Fortunately, your familiarity with analyzing the path of falling material will come in handy here. You scan a two-dimensional vertical slice of the cave above you (your puzzle input) and discover that it is mostly air with structures made of rock.
Your scan traces the path of each solid rock structure and reports the x,y coordinates that form the shape of the path, where x represents distance to the right and y represents distance down. Each path appears as a single line of text in your scan. After the first point of each path, each point indicates the end of a straight horizontal or vertical line to be drawn from the previous point. For example:
498,4 -> 498,6 -> 496,6
503,4 -> 502,4 -> 502,9 -> 494,9
This scan means that there are two paths of rock; the first path consists of two straight lines, and the second path consists of three straight lines. (Specifically, the first path consists of a line of rock from 498,4 through 498,6 and another line of rock from 498,6 through 496,6.)
The sand is pouring into the cave from point 500,0.
Drawing rock as #, air as ., and the source of the sand as +, this becomes:
4 5 5
9 0 0
4 0 3
0 ......+...
1 ..........
2 ..........
3 ..........
4 ....#...##
5 ....#...#.
6 ..###...#.
7 ........#.
8 ........#.
9 #########.
Sand is produced one unit at a time, and the next unit of sand is not produced until the previous unit of sand comes to rest. A unit of sand is large enough to fill one tile of air in your scan.
A unit of sand always falls down one step if possible. If the tile immediately below is blocked (by rock or sand), the unit of sand attempts to instead move diagonally one step down and to the left. If that tile is blocked, the unit of sand attempts to instead move diagonally one step down and to the right. Sand keeps moving as long as it is able to do so, at each step trying to move down, then down-left, then down-right. If all three possible destinations are blocked, the unit of sand comes to rest and no longer moves, at which point the next unit of sand is created back at the source.
So, drawing sand that has come to rest as o, the first unit of sand simply falls straight down and then stops:
......+...
..........
..........
..........
....#...##
....#...#.
..###...#.
........#.
......o.#.
#########.
The second unit of sand then falls straight down, lands on the first one, and then comes to rest to its left:
......+...
..........
..........
..........
....#...##
....#...#.
..###...#.
........#.
.....oo.#.
#########.
After a total of five units of sand have come to rest, they form this pattern:
......+...
..........
..........
..........
....#...##
....#...#.
..###...#.
......o.#.
....oooo#.
#########.
After a total of 22 units of sand:
......+...
..........
......o...
.....ooo..
....#ooo##
....#ooo#.
..###ooo#.
....oooo#.
...ooooo#.
#########.
Finally, only two more units of sand can possibly come to rest:
......+...
..........
......o...
.....ooo..
....#ooo##
...o#ooo#.
..###ooo#.
....oooo#.
.o.ooooo#.
#########.
Once all 24 units of sand shown above have come to rest, all further sand flows out the bottom, falling into the endless void. Just for fun, the path any new sand takes before falling forever is shown here with ~:
.......+...
.......~...
......~o...
.....~ooo..
....~#ooo##
...~o#ooo#.
..~###ooo#.
..~..oooo#.
.~o.ooooo#.
~#########.
~..........
~..........
~..........
Using your scan, simulate the falling sand. How many units of sand come to rest before sand starts flowing into the abyss below?
| 185
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--- Day 9: Marble Mania ---
You talk to the Elves while you wait for your navigation system to initialize. To pass the time, they introduce you to their favorite marble game.
The Elves play this game by taking turns arranging the marbles in a circle according to very particular rules. The marbles are numbered starting with 0 and increasing by 1 until every marble has a number.
First, the marble numbered 0 is placed in the circle. At this point, while it contains only a single marble, it is still a circle: the marble is both clockwise from itself and counter-clockwise from itself. This marble is designated the current marble.
Then, each Elf takes a turn placing the lowest-numbered remaining marble into the circle between the marbles that are 1 and 2 marbles clockwise of the current marble. (When the circle is large enough, this means that there is one marble between the marble that was just placed and the current marble.) The marble that was just placed then becomes the current marble.
However, if the marble that is about to be placed has a number which is a multiple of 23, something entirely different happens. First, the current player keeps the marble they would have placed, adding it to their score. In addition, the marble 7 marbles counter-clockwise from the current marble is removed from the circle and also added to the current player's score. The marble located immediately clockwise of the marble that was removed becomes the new current marble.
For example, suppose there are 9 players. After the marble with value 0 is placed in the middle, each player (shown in square brackets) takes a turn. The result of each of those turns would produce circles of marbles like this, where clockwise is to the right and the resulting current marble is in parentheses:
[-] (0)
[1] 0 (1)
[2] 0 (2) 1
[3] 0 2 1 (3)
[4] 0 (4) 2 1 3
[5] 0 4 2 (5) 1 3
[6] 0 4 2 5 1 (6) 3
[7] 0 4 2 5 1 6 3 (7)
[8] 0 (8) 4 2 5 1 6 3 7
[9] 0 8 4 (9) 2 5 1 6 3 7
[1] 0 8 4 9 2(10) 5 1 6 3 7
[2] 0 8 4 9 2 10 5(11) 1 6 3 7
[3] 0 8 4 9 2 10 5 11 1(12) 6 3 7
[4] 0 8 4 9 2 10 5 11 1 12 6(13) 3 7
[5] 0 8 4 9 2 10 5 11 1 12 6 13 3(14) 7
[6] 0 8 4 9 2 10 5 11 1 12 6 13 3 14 7(15)
[7] 0(16) 8 4 9 2 10 5 11 1 12 6 13 3 14 7 15
[8] 0 16 8(17) 4 9 2 10 5 11 1 12 6 13 3 14 7 15
[9] 0 16 8 17 4(18) 9 2 10 5 11 1 12 6 13 3 14 7 15
[1] 0 16 8 17 4 18 9(19) 2 10 5 11 1 12 6 13 3 14 7 15
[2] 0 16 8 17 4 18 9 19 2(20)10 5 11 1 12 6 13 3 14 7 15
[3] 0 16 8 17 4 18 9 19 2 20 10(21) 5 11 1 12 6 13 3 14 7 15
[4] 0 16 8 17 4 18 9 19 2 20 10 21 5(22)11 1 12 6 13 3 14 7 15
[5] 0 16 8 17 4 18(19) 2 20 10 21 5 22 11 1 12 6 13 3 14 7 15
[6] 0 16 8 17 4 18 19 2(24)20 10 21 5 22 11 1 12 6 13 3 14 7 15
[7] 0 16 8 17 4 18 19 2 24 20(25)10 21 5 22 11 1 12 6 13 3 14 7 15
The goal is to be the player with the highest score after the last marble is used up. Assuming the example above ends after the marble numbered 25, the winning score is 23+9=32 (because player 5 kept marble 23 and removed marble 9, while no other player got any points in this very short example game).
Here are a few more examples:
10 players; last marble is worth 1618 points: high score is 8317
13 players; last marble is worth 7999 points: high score is 146373
17 players; last marble is worth 1104 points: high score is 2764
21 players; last marble is worth 6111 points: high score is 54718
30 players; last marble is worth 5807 points: high score is 37305
What is the winning Elf's score?
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--- Day 8: Two-Factor Authentication ---
You come across a door implementing what you can only assume is an implementation of two-factor authentication after a long game of requirements telephone.
To get past the door, you first swipe a keycard (no problem; there was one on a nearby desk). Then, it displays a code on a little screen, and you type that code on a keypad. Then, presumably, the door unlocks.
Unfortunately, the screen has been smashed. After a few minutes, you've taken everything apart and figured out how it works. Now you just have to work out what the screen would have displayed.
The magnetic strip on the card you swiped encodes a series of instructions for the screen; these instructions are your puzzle input. The screen is 50 pixels wide and 6 pixels tall, all of which start off, and is capable of three somewhat peculiar operations:
rect AxB turns on all of the pixels in a rectangle at the top-left of the screen which is A wide and B tall.
rotate row y=A by B shifts all of the pixels in row A (0 is the top row) right by B pixels. Pixels that would fall off the right end appear at the left end of the row.
rotate column x=A by B shifts all of the pixels in column A (0 is the left column) down by B pixels. Pixels that would fall off the bottom appear at the top of the column.
For example, here is a simple sequence on a smaller screen:
rect 3x2 creates a small rectangle in the top-left corner:
###....
###....
.......
rotate column x=1 by 1 rotates the second column down by one pixel:
#.#....
###....
.#.....
rotate row y=0 by 4 rotates the top row right by four pixels:
....#.#
###....
.#.....
rotate column x=1 by 1 again rotates the second column down by one pixel, causing the bottom pixel to wrap back to the top:
.#..#.#
#.#....
.#.....
As you can see, this display technology is extremely powerful, and will soon dominate the tiny-code-displaying-screen market. That's what the advertisement on the back of the display tries to convince you, anyway.
There seems to be an intermediate check of the voltage used by the display: after you swipe your card, if the screen did work, how many pixels should be lit?
| 187
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--- Day 23: LAN Party ---
As The Historians wander around a secure area at Easter Bunny HQ, you come across posters for a LAN party scheduled for today! Maybe you can find it; you connect to a nearby datalink port and download a map of the local network (your puzzle input).
The network map provides a list of every connection between two computers. For example:
kh-tc
qp-kh
de-cg
ka-co
yn-aq
qp-ub
cg-tb
vc-aq
tb-ka
wh-tc
yn-cg
kh-ub
ta-co
de-co
tc-td
tb-wq
wh-td
ta-ka
td-qp
aq-cg
wq-ub
ub-vc
de-ta
wq-aq
wq-vc
wh-yn
ka-de
kh-ta
co-tc
wh-qp
tb-vc
td-yn
Each line of text in the network map represents a single connection; the line kh-tc represents a connection between the computer named kh and the computer named tc. Connections aren't directional; tc-kh would mean exactly the same thing.
LAN parties typically involve multiplayer games, so maybe you can locate it by finding groups of connected computers. Start by looking for sets of three computers where each computer in the set is connected to the other two computers.
In this example, there are 12 such sets of three inter-connected computers:
aq,cg,yn
aq,vc,wq
co,de,ka
co,de,ta
co,ka,ta
de,ka,ta
kh,qp,ub
qp,td,wh
tb,vc,wq
tc,td,wh
td,wh,yn
ub,vc,wq
If the Chief Historian is here, and he's at the LAN party, it would be best to know that right away. You're pretty sure his computer's name starts with t, so consider only sets of three computers where at least one computer's name starts with t. That narrows the list down to 7 sets of three inter-connected computers:
co,de,ta
co,ka,ta
de,ka,ta
qp,td,wh
tb,vc,wq
tc,td,wh
td,wh,yn
Find all the sets of three inter-connected computers. How many contain at least one computer with a name that starts with t?
Your puzzle answer was 1327.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
There are still way too many results to go through them all. You'll have to find the LAN party another way and go there yourself.
Since it doesn't seem like any employees are around, you figure they must all be at the LAN party. If that's true, the LAN party will be the largest set of computers that are all connected to each other. That is, for each computer at the LAN party, that computer will have a connection to every other computer at the LAN party.
In the above example, the largest set of computers that are all connected to each other is made up of co, de, ka, and ta. Each computer in this set has a connection to every other computer in the set:
ka-co
ta-co
de-co
ta-ka
de-ta
ka-de
The LAN party posters say that the password to get into the LAN party is the name of every computer at the LAN party, sorted alphabetically, then joined together with commas. (The people running the LAN party are clearly a bunch of nerds.) In this example, the password would be co,de,ka,ta.
What is the password to get into the LAN party?
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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?
| 189
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--- Day 19: Not Enough Minerals ---
Your scans show that the lava did indeed form obsidian!
The wind has changed direction enough to stop sending lava droplets toward you, so you and the elephants exit the cave. As you do, you notice a collection of geodes around the pond. Perhaps you could use the obsidian to create some geode-cracking robots and break them open?
To collect the obsidian from the bottom of the pond, you'll need waterproof obsidian-collecting robots. Fortunately, there is an abundant amount of clay nearby that you can use to make them waterproof.
In order to harvest the clay, you'll need special-purpose clay-collecting robots. To make any type of robot, you'll need ore, which is also plentiful but in the opposite direction from the clay.
Collecting ore requires ore-collecting robots with big drills. Fortunately, you have exactly one ore-collecting robot in your pack that you can use to kickstart the whole operation.
Each robot can collect 1 of its resource type per minute. It also takes one minute for the robot factory (also conveniently from your pack) to construct any type of robot, although it consumes the necessary resources available when construction begins.
The robot factory has many blueprints (your puzzle input) you can choose from, but once you've configured it with a blueprint, you can't change it. You'll need to work out which blueprint is best.
For example:
Blueprint 1:
Each ore robot costs 4 ore.
Each clay robot costs 2 ore.
Each obsidian robot costs 3 ore and 14 clay.
Each geode robot costs 2 ore and 7 obsidian.
Blueprint 2:
Each ore robot costs 2 ore.
Each clay robot costs 3 ore.
Each obsidian robot costs 3 ore and 8 clay.
Each geode robot costs 3 ore and 12 obsidian.
(Blueprints have been line-wrapped here for legibility. The robot factory's actual assortment of blueprints are provided one blueprint per line.)
The elephants are starting to look hungry, so you shouldn't take too long; you need to figure out which blueprint would maximize the number of opened geodes after 24 minutes by figuring out which robots to build and when to build them.
Using blueprint 1 in the example above, the largest number of geodes you could open in 24 minutes is 9. One way to achieve that is:
== Minute 1 ==
1 ore-collecting robot collects 1 ore; you now have 1 ore.
== Minute 2 ==
1 ore-collecting robot collects 1 ore; you now have 2 ore.
== Minute 3 ==
Spend 2 ore to start building a clay-collecting robot.
1 ore-collecting robot collects 1 ore; you now have 1 ore.
The new clay-collecting robot is ready; you now have 1 of them.
== Minute 4 ==
1 ore-collecting robot collects 1 ore; you now have 2 ore.
1 clay-collecting robot collects 1 clay; you now have 1 clay.
== Minute 5 ==
Spend 2 ore to start building a clay-collecting robot.
1 ore-collecting robot collects 1 ore; you now have 1 ore.
1 clay-collecting robot collects 1 clay; you now have 2 clay.
The new clay-collecting robot is ready; you now have 2 of them.
== Minute 6 ==
1 ore-collecting robot collects 1 ore; you now have 2 ore.
2 clay-collecting robots collect 2 clay; you now have 4 clay.
== Minute 7 ==
Spend 2 ore to start building a clay-collecting robot.
1 ore-collecting robot collects 1 ore; you now have 1 ore.
2 clay-collecting robots collect 2 clay; you now have 6 clay.
The new clay-collecting robot is ready; you now have 3 of them.
== Minute 8 ==
1 ore-collecting robot collects 1 ore; you now have 2 ore.
3 clay-collecting robots collect 3 clay; you now have 9 clay.
== Minute 9 ==
1 ore-collecting robot collects 1 ore; you now have 3 ore.
3 clay-collecting robots collect 3 clay; you now have 12 clay.
== Minute 10 ==
1 ore-collecting robot collects 1 ore; you now have 4 ore.
3 clay-collecting robots collect 3 clay; you now have 15 clay.
== Minute 11 ==
Spend 3 ore and 14 clay to start building an obsidian-collecting robot.
1 ore-collecting robot collects 1 ore; you now have 2 ore.
3 clay-collecting robots collect 3 clay; you now have 4 clay.
The new obsidian-collecting robot is ready; you now have 1 of them.
== Minute 12 ==
Spend 2 ore to start building a clay-collecting robot.
1 ore-collecting robot collects 1 ore; you now have 1 ore.
3 clay-collecting robots collect 3 clay; you now have 7 clay.
1 obsidian-collecting robot collects 1 obsidian; you now have 1 obsidian.
The new clay-collecting robot is ready; you now have 4 of them.
== Minute 13 ==
1 ore-collecting robot collects 1 ore; you now have 2 ore.
4 clay-collecting robots collect 4 clay; you now have 11 clay.
1 obsidian-collecting robot collects 1 obsidian; you now have 2 obsidian.
== Minute 14 ==
1 ore-collecting robot collects 1 ore; you now have 3 ore.
4 clay-collecting robots collect 4 clay; you now have 15 clay.
1 obsidian-collecting robot collects 1 obsidian; you now have 3 obsidian.
== Minute 15 ==
Spend 3 ore and 14 clay to start building an obsidian-collecting robot.
1 ore-collecting robot collects 1 ore; you now have 1 ore.
4 clay-collecting robots collect 4 clay; you now have 5 clay.
1 obsidian-collecting robot collects 1 obsidian; you now have 4 obsidian.
The new obsidian-collecting robot is ready; you now have 2 of them.
== Minute 16 ==
1 ore-collecting robot collects 1 ore; you now have 2 ore.
4 clay-collecting robots collect 4 clay; you now have 9 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 6 obsidian.
== Minute 17 ==
1 ore-collecting robot collects 1 ore; you now have 3 ore.
4 clay-collecting robots collect 4 clay; you now have 13 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 8 obsidian.
== Minute 18 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
1 ore-collecting robot collects 1 ore; you now have 2 ore.
4 clay-collecting robots collect 4 clay; you now have 17 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 3 obsidian.
The new geode-cracking robot is ready; you now have 1 of them.
== Minute 19 ==
1 ore-collecting robot collects 1 ore; you now have 3 ore.
4 clay-collecting robots collect 4 clay; you now have 21 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 5 obsidian.
1 geode-cracking robot cracks 1 geode; you now have 1 open geode.
== Minute 20 ==
1 ore-collecting robot collects 1 ore; you now have 4 ore.
4 clay-collecting robots collect 4 clay; you now have 25 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 7 obsidian.
1 geode-cracking robot cracks 1 geode; you now have 2 open geodes.
== Minute 21 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
1 ore-collecting robot collects 1 ore; you now have 3 ore.
4 clay-collecting robots collect 4 clay; you now have 29 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 2 obsidian.
1 geode-cracking robot cracks 1 geode; you now have 3 open geodes.
The new geode-cracking robot is ready; you now have 2 of them.
== Minute 22 ==
1 ore-collecting robot collects 1 ore; you now have 4 ore.
4 clay-collecting robots collect 4 clay; you now have 33 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 4 obsidian.
2 geode-cracking robots crack 2 geodes; you now have 5 open geodes.
== Minute 23 ==
1 ore-collecting robot collects 1 ore; you now have 5 ore.
4 clay-collecting robots collect 4 clay; you now have 37 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 6 obsidian.
2 geode-cracking robots crack 2 geodes; you now have 7 open geodes.
== Minute 24 ==
1 ore-collecting robot collects 1 ore; you now have 6 ore.
4 clay-collecting robots collect 4 clay; you now have 41 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 8 obsidian.
2 geode-cracking robots crack 2 geodes; you now have 9 open geodes.
However, by using blueprint 2 in the example above, you could do even better: the largest number of geodes you could open in 24 minutes is 12.
Determine the quality level of each blueprint by multiplying that blueprint's ID number with the largest number of geodes that can be opened in 24 minutes using that blueprint. In this example, the first blueprint has ID 1 and can open 9 geodes, so its quality level is 9. The second blueprint has ID 2 and can open 12 geodes, so its quality level is 24. Finally, if you add up the quality levels of all of the blueprints in the list, you get 33.
Determine the quality level of each blueprint using the largest number of geodes it could produce in 24 minutes. What do you get if you add up the quality level of all of the blueprints in your list?
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--- Day 24: Air Duct Spelunking ---
You've finally met your match; the doors that provide access to the roof are locked tight, and all of the controls and related electronics are inaccessible. You simply can't reach them.
The robot that cleans the air ducts, however, can.
It's not a very fast little robot, but you reconfigure it to be able to interface with some of the exposed wires that have been routed through the HVAC system. If you can direct it to each of those locations, you should be able to bypass the security controls.
You extract the duct layout for this area from some blueprints you acquired and create a map with the relevant locations marked (your puzzle input). 0 is your current location, from which the cleaning robot embarks; the other numbers are (in no particular order) the locations the robot needs to visit at least once each. Walls are marked as #, and open passages are marked as .. Numbers behave like open passages.
For example, suppose you have a map like the following:
###########
#0.1.....2#
#.#######.#
#4.......3#
###########
To reach all of the points of interest as quickly as possible, you would have the robot take the following path:
0 to 4 (2 steps)
4 to 1 (4 steps; it can't move diagonally)
1 to 2 (6 steps)
2 to 3 (2 steps)
Since the robot isn't very fast, you need to find it the shortest route. This path is the fewest steps (in the above example, a total of 14) required to start at 0 and then visit every other location at least once.
Given your actual map, and starting from location 0, what is the fewest number of steps required to visit every non-0 number marked on the map at least once?
| 191
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--- Day 24: It Hangs in the Balance ---
It's Christmas Eve, and Santa is loading up the sleigh for this year's deliveries. However, there's one small problem: he can't get the sleigh to balance. If it isn't balanced, he can't defy physics, and nobody gets presents this year.
No pressure.
Santa has provided you a list of the weights of every package he needs to fit on the sleigh. The packages need to be split into three groups of exactly the same weight, and every package has to fit. The first group goes in the passenger compartment of the sleigh, and the second and third go in containers on either side. Only when all three groups weigh exactly the same amount will the sleigh be able to fly. Defying physics has rules, you know!
Of course, that's not the only problem. The first group - the one going in the passenger compartment - needs as few packages as possible so that Santa has some legroom left over. It doesn't matter how many packages are in either of the other two groups, so long as all of the groups weigh the same.
Furthermore, Santa tells you, if there are multiple ways to arrange the packages such that the fewest possible are in the first group, you need to choose the way where the first group has the smallest quantum entanglement to reduce the chance of any "complications". The quantum entanglement of a group of packages is the product of their weights, that is, the value you get when you multiply their weights together. Only consider quantum entanglement if the first group has the fewest possible number of packages in it and all groups weigh the same amount.
For example, suppose you have ten packages with weights 1 through 5 and 7 through 11. For this situation, some of the unique first groups, their quantum entanglements, and a way to divide the remaining packages are as follows:
Group 1; Group 2; Group 3
11 9 (QE= 99); 10 8 2; 7 5 4 3 1
10 9 1 (QE= 90); 11 7 2; 8 5 4 3
10 8 2 (QE=160); 11 9; 7 5 4 3 1
10 7 3 (QE=210); 11 9; 8 5 4 2 1
10 5 4 1 (QE=200); 11 9; 8 7 3 2
10 5 3 2 (QE=300); 11 9; 8 7 4 1
10 4 3 2 1 (QE=240); 11 9; 8 7 5
9 8 3 (QE=216); 11 7 2; 10 5 4 1
9 7 4 (QE=252); 11 8 1; 10 5 3 2
9 5 4 2 (QE=360); 11 8 1; 10 7 3
8 7 5 (QE=280); 11 9; 10 4 3 2 1
8 5 4 3 (QE=480); 11 9; 10 7 2 1
7 5 4 3 1 (QE=420); 11 9; 10 8 2
Of these, although 10 9 1 has the smallest quantum entanglement (90), the configuration with only two packages, 11 9, in the passenger compartment gives Santa the most legroom and wins. In this situation, the quantum entanglement for the ideal configuration is therefore 99. Had there been two configurations with only two packages in the first group, the one with the smaller quantum entanglement would be chosen.
What is the quantum entanglement of the first group of packages in the ideal configuration?
Your puzzle answer was 10723906903.
--- Part Two ---
That's weird... the sleigh still isn't balancing.
"Ho ho ho", Santa muses to himself. "I forgot the trunk".
Balance the sleigh again, but this time, separate the packages into four groups instead of three. The other constraints still apply.
Given the example packages above, this would be some of the new unique first groups, their quantum entanglements, and one way to divide the remaining packages:
11 4 (QE=44); 10 5; 9 3 2 1; 8 7
10 5 (QE=50); 11 4; 9 3 2 1; 8 7
9 5 1 (QE=45); 11 4; 10 3 2; 8 7
9 4 2 (QE=72); 11 3 1; 10 5; 8 7
9 3 2 1 (QE=54); 11 4; 10 5; 8 7
8 7 (QE=56); 11 4; 10 5; 9 3 2 1
Of these, there are three arrangements that put the minimum (two) number of packages in the first group: 11 4, 10 5, and 8 7. Of these, 11 4 has the lowest quantum entanglement, and so it is selected.
Now, what is the quantum entanglement of the first group of packages in the ideal configuration?
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--- Day 8: I Heard You Like Registers ---
You receive a signal directly from the CPU. Because of your recent assistance with jump instructions, it would like you to compute the result of a series of unusual register instructions.
Each instruction consists of several parts: the register to modify, whether to increase or decrease that register's value, the amount by which to increase or decrease it, and a condition. If the condition fails, skip the instruction without modifying the register. The registers all start at 0. The instructions look like this:
b inc 5 if a > 1
a inc 1 if b < 5
c dec -10 if a >= 1
c inc -20 if c == 10
These instructions would be processed as follows:
Because a starts at 0, it is not greater than 1, and so b is not modified.
a is increased by 1 (to 1) because b is less than 5 (it is 0).
c is decreased by -10 (to 10) because a is now greater than or equal to 1 (it is 1).
c is increased by -20 (to -10) because c is equal to 10.
After this process, the largest value in any register is 1.
You might also encounter <= (less than or equal to) or != (not equal to). However, the CPU doesn't have the bandwidth to tell you what all the registers are named, and leaves that to you to determine.
What is the largest value in any register after completing the instructions in your puzzle input?
| 193
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--- Day 6: Universal Orbit Map ---
You've landed at the Universal Orbit Map facility on Mercury. Because navigation in space often involves transferring between orbits, the orbit maps here are useful for finding efficient routes between, for example, you and Santa. You download a map of the local orbits (your puzzle input).
Except for the universal Center of Mass (COM), every object in space is in orbit around exactly one other object. An orbit looks roughly like this:
|
|
AAA--> o o <--BBB
|
|
/
/
In this diagram, the object BBB is in orbit around AAA. The path that BBB takes around AAA (drawn with lines) is only partly shown. In the map data, this orbital relationship is written AAA)BBB, which means "BBB is in orbit around AAA".
Before you use your map data to plot a course, you need to make sure it wasn't corrupted during the download. To verify maps, the Universal Orbit Map facility uses orbit count checksums - the total number of direct orbits (like the one shown above) and indirect orbits.
Whenever A orbits B and B orbits C, then A indirectly orbits C. This chain can be any number of objects long: if A orbits B, B orbits C, and C orbits D, then A indirectly orbits D.
For example, suppose you have the following map:
COM)B
B)C
C)D
D)E
E)F
B)G
G)H
D)I
E)J
J)K
K)L
Visually, the above map of orbits looks like this:
G - H J - K - L
/ /
COM - B - C - D - E - F
I
In this visual representation, when two objects are connected by a line, the one on the right directly orbits the one on the left.
Here, we can count the total number of orbits as follows:
D directly orbits C and indirectly orbits B and COM, a total of 3 orbits.
L directly orbits K and indirectly orbits J, E, D, C, B, and COM, a total of 7 orbits.
COM orbits nothing.
The total number of direct and indirect orbits in this example is 42.
What is the total number of direct and indirect orbits in your map data?
| 194
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--- Day 9: Explosives in Cyberspace ---
Wandering around a secure area, you come across a datalink port to a new part of the network. After briefly scanning it for interesting files, you find one file in particular that catches your attention. It's compressed with an experimental format, but fortunately, the documentation for the format is nearby.
The format compresses a sequence of characters. Whitespace is ignored. To indicate that some sequence should be repeated, a marker is added to the file, like (10x2). To decompress this marker, take the subsequent 10 characters and repeat them 2 times. Then, continue reading the file after the repeated data. The marker itself is not included in the decompressed output.
If parentheses or other characters appear within the data referenced by a marker, that's okay - treat it like normal data, not a marker, and then resume looking for markers after the decompressed section.
For example:
ADVENT contains no markers and decompresses to itself with no changes, resulting in a decompressed length of 6.
A(1x5)BC repeats only the B a total of 5 times, becoming ABBBBBC for a decompressed length of 7.
(3x3)XYZ becomes XYZXYZXYZ for a decompressed length of 9.
A(2x2)BCD(2x2)EFG doubles the BC and EF, becoming ABCBCDEFEFG for a decompressed length of 11.
(6x1)(1x3)A simply becomes (1x3)A - the (1x3) looks like a marker, but because it's within a data section of another marker, it is not treated any differently from the A that comes after it. It has a decompressed length of 6.
X(8x2)(3x3)ABCY becomes X(3x3)ABC(3x3)ABCY (for a decompressed length of 18), because the decompressed data from the (8x2) marker (the (3x3)ABC) is skipped and not processed further.
What is the decompressed length of the file (your puzzle input)? Don't count whitespace.
Your puzzle answer was 115118.
--- Part Two ---
Apparently, the file actually uses version two of the format.
In version two, the only difference is that markers within decompressed data are decompressed. This, the documentation explains, provides much more substantial compression capabilities, allowing many-gigabyte files to be stored in only a few kilobytes.
For example:
(3x3)XYZ still becomes XYZXYZXYZ, as the decompressed section contains no markers.
X(8x2)(3x3)ABCY becomes XABCABCABCABCABCABCY, because the decompressed data from the (8x2) marker is then further decompressed, thus triggering the (3x3) marker twice for a total of six ABC sequences.
(27x12)(20x12)(13x14)(7x10)(1x12)A decompresses into a string of A repeated 241920 times.
(25x3)(3x3)ABC(2x3)XY(5x2)PQRSTX(18x9)(3x2)TWO(5x7)SEVEN becomes 445 characters long.
Unfortunately, the computer you brought probably doesn't have enough memory to actually decompress the file; you'll have to come up with another way to get its decompressed length.
What is the decompressed length of the file using this improved format?
| 195
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--- Day 6: Signals and Noise ---
Something is jamming your communications with Santa. Fortunately, your signal is only partially jammed, and protocol in situations like this is to switch to a simple repetition code to get the message through.
In this model, the same message is sent repeatedly. You've recorded the repeating message signal (your puzzle input), but the data seems quite corrupted - almost too badly to recover. Almost.
All you need to do is figure out which character is most frequent for each position. For example, suppose you had recorded the following messages:
eedadn
drvtee
eandsr
raavrd
atevrs
tsrnev
sdttsa
rasrtv
nssdts
ntnada
svetve
tesnvt
vntsnd
vrdear
dvrsen
enarar
The most common character in the first column is e; in the second, a; in the third, s, and so on. Combining these characters returns the error-corrected message, easter.
Given the recording in your puzzle input, what is the error-corrected version of the message being sent?
Your puzzle answer was xhnqpqql.
--- Part Two ---
Of course, that would be the message - if you hadn't agreed to use a modified repetition code instead.
In this modified code, the sender instead transmits what looks like random data, but for each character, the character they actually want to send is slightly less likely than the others. Even after signal-jamming noise, you can look at the letter distributions in each column and choose the least common letter to reconstruct the original message.
In the above example, the least common character in the first column is a; in the second, d, and so on. Repeating this process for the remaining characters produces the original message, advent.
Given the recording in your puzzle input and this new decoding methodology, what is the original message that Santa is trying to send?
| 196
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--- Day 25: Combo Breaker ---
You finally reach the check-in desk. Unfortunately, their registration systems are currently offline, and they cannot check you in. Noticing the look on your face, they quickly add that tech support is already on the way! They even created all the room keys this morning; you can take yours now and give them your room deposit once the registration system comes back online.
The room key is a small RFID card. Your room is on the 25th floor and the elevators are also temporarily out of service, so it takes what little energy you have left to even climb the stairs and navigate the halls. You finally reach the door to your room, swipe your card, and - beep - the light turns red.
Examining the card more closely, you discover a phone number for tech support.
"Hello! How can we help you today?" You explain the situation.
"Well, it sounds like the card isn't sending the right command to unlock the door. If you go back to the check-in desk, surely someone there can reset it for you." Still catching your breath, you describe the status of the elevator and the exact number of stairs you just had to climb.
"I see! Well, your only other option would be to reverse-engineer the cryptographic handshake the card does with the door and then inject your own commands into the data stream, but that's definitely impossible." You thank them for their time.
Unfortunately for the door, you know a thing or two about cryptographic handshakes.
The handshake used by the card and the door involves an operation that transforms a subject number. To transform a subject number, start with the value 1. Then, a number of times called the loop size, perform the following steps:
Set the value to itself multiplied by the subject number.
Set the value to the remainder after dividing the value by 20201227.
The card always uses a specific, secret loop size when it transforms a subject number. The door always uses a different, secret loop size.
The cryptographic handshake works like this:
The card transforms the subject number of 7 according to the card's secret loop size. The result is called the card's public key.
The door transforms the subject number of 7 according to the door's secret loop size. The result is called the door's public key.
The card and door use the wireless RFID signal to transmit the two public keys (your puzzle input) to the other device. Now, the card has the door's public key, and the door has the card's public key. Because you can eavesdrop on the signal, you have both public keys, but neither device's loop size.
The card transforms the subject number of the door's public key according to the card's loop size. The result is the encryption key.
The door transforms the subject number of the card's public key according to the door's loop size. The result is the same encryption key as the card calculated.
If you can use the two public keys to determine each device's loop size, you will have enough information to calculate the secret encryption key that the card and door use to communicate; this would let you send the unlock command directly to the door!
For example, suppose you know that the card's public key is 5764801. With a little trial and error, you can work out that the card's loop size must be 8, because transforming the initial subject number of 7 with a loop size of 8 produces 5764801.
Then, suppose you know that the door's public key is 17807724. By the same process, you can determine that the door's loop size is 11, because transforming the initial subject number of 7 with a loop size of 11 produces 17807724.
At this point, you can use either device's loop size with the other device's public key to calculate the encryption key. Transforming the subject number of 17807724 (the door's public key) with a loop size of 8 (the card's loop size) produces the encryption key, 14897079. (Transforming the subject number of 5764801 (the card's public key) with a loop size of 11 (the door's loop size) produces the same encryption key: 14897079.)
What encryption key is the handshake trying to establish?
| 197
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--- Day 21: Springdroid Adventure ---
You lift off from Pluto and start flying in the direction of Santa.
While experimenting further with the tractor beam, you accidentally pull an asteroid directly into your ship! It deals significant damage to your hull and causes your ship to begin tumbling violently.
You can send a droid out to investigate, but the tumbling is causing enough artificial gravity that one wrong step could send the droid through a hole in the hull and flying out into space.
The clear choice for this mission is a droid that can jump over the holes in the hull - a springdroid.
You can use an Intcode program (your puzzle input) running on an ASCII-capable computer to program the springdroid. However, springdroids don't run Intcode; instead, they run a simplified assembly language called springscript.
While a springdroid is certainly capable of navigating the artificial gravity and giant holes, it has one downside: it can only remember at most 15 springscript instructions.
The springdroid will move forward automatically, constantly thinking about whether to jump. The springscript program defines the logic for this decision.
Springscript programs only use Boolean values, not numbers or strings. Two registers are available: T, the temporary value register, and J, the jump register. If the jump register is true at the end of the springscript program, the springdroid will try to jump. Both of these registers start with the value false.
Springdroids have a sensor that can detect whether there is ground at various distances in the direction it is facing; these values are provided in read-only registers. Your springdroid can detect ground at four distances: one tile away (A), two tiles away (B), three tiles away (C), and four tiles away (D). If there is ground at the given distance, the register will be true; if there is a hole, the register will be false.
There are only three instructions available in springscript:
AND X Y sets Y to true if both X and Y are true; otherwise, it sets Y to false.
OR X Y sets Y to true if at least one of X or Y is true; otherwise, it sets Y to false.
NOT X Y sets Y to true if X is false; otherwise, it sets Y to false.
In all three instructions, the second argument (Y) needs to be a writable register (either T or J). The first argument (X) can be any register (including A, B, C, or D).
For example, the one-instruction program NOT A J means "if the tile immediately in front of me is not ground, jump".
Or, here is a program that jumps if a three-tile-wide hole (with ground on the other side of the hole) is detected:
NOT A J
NOT B T
AND T J
NOT C T
AND T J
AND D J
The Intcode program expects ASCII inputs and outputs. It will begin by displaying a prompt; then, input the desired instructions one per line. End each line with a newline (ASCII code 10). When you have finished entering your program, provide the command WALK followed by a newline to instruct the springdroid to begin surveying the hull.
If the springdroid falls into space, an ASCII rendering of the last moments of its life will be produced. In these, @ is the springdroid, # is hull, and . is empty space. For example, suppose you program the springdroid like this:
NOT D J
WALK
This one-instruction program sets J to true if and only if there is no ground four tiles away. In other words, it attempts to jump into any hole it finds:
.................
.................
@................
#####.###########
.................
.................
.@...............
#####.###########
.................
..@..............
.................
#####.###########
...@.............
.................
.................
#####.###########
.................
....@............
.................
#####.###########
.................
.................
.....@...........
#####.###########
.................
.................
.................
#####@###########
However, if the springdroid successfully makes it across, it will use an output instruction to indicate the amount of damage to the hull as a single giant integer outside the normal ASCII range.
Program the springdroid with logic that allows it to survey the hull without falling into space. What amount of hull damage does it report?
| 198
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--- Day 6: Tuning Trouble ---
The preparations are finally complete; you and the Elves leave camp on foot and begin to make your way toward the star fruit grove.
As you move through the dense undergrowth, one of the Elves gives you a handheld device. He says that it has many fancy features, but the most important one to set up right now is the communication system.
However, because he's heard you have significant experience dealing with signal-based systems, he convinced the other Elves that it would be okay to give you their one malfunctioning device - surely you'll have no problem fixing it.
As if inspired by comedic timing, the device emits a few colorful sparks.
To be able to communicate with the Elves, the device needs to lock on to their signal. The signal is a series of seemingly-random characters that the device receives one at a time.
To fix the communication system, you need to add a subroutine to the device that detects a start-of-packet marker in the datastream. In the protocol being used by the Elves, the start of a packet is indicated by a sequence of four characters that are all different.
The device will send your subroutine a datastream buffer (your puzzle input); your subroutine needs to identify the first position where the four most recently received characters were all different. Specifically, it needs to report the number of characters from the beginning of the buffer to the end of the first such four-character marker.
For example, suppose you receive the following datastream buffer:
mjqjpqmgbljsphdztnvjfqwrcgsmlb
After the first three characters (mjq) have been received, there haven't been enough characters received yet to find the marker. The first time a marker could occur is after the fourth character is received, making the most recent four characters mjqj. Because j is repeated, this isn't a marker.
The first time a marker appears is after the seventh character arrives. Once it does, the last four characters received are jpqm, which are all different. In this case, your subroutine should report the value 7, because the first start-of-packet marker is complete after 7 characters have been processed.
Here are a few more examples:
bvwbjplbgvbhsrlpgdmjqwftvncz: first marker after character 5
nppdvjthqldpwncqszvftbrmjlhg: first marker after character 6
nznrnfrfntjfmvfwmzdfjlvtqnbhcprsg: first marker after character 10
zcfzfwzzqfrljwzlrfnpqdbhtmscgvjw: first marker after character 11
How many characters need to be processed before the first start-of-packet marker is detected?
Your puzzle answer was 1757.
--- Part Two ---
Your device's communication system is correctly detecting packets, but still isn't working. It looks like it also needs to look for messages.
A start-of-message marker is just like a start-of-packet marker, except it consists of 14 distinct characters rather than 4.
Here are the first positions of start-of-message markers for all of the above examples:
mjqjpqmgbljsphdztnvjfqwrcgsmlb: first marker after character 19
bvwbjplbgvbhsrlpgdmjqwftvncz: first marker after character 23
nppdvjthqldpwncqszvftbrmjlhg: first marker after character 23
nznrnfrfntjfmvfwmzdfjlvtqnbhcprsg: first marker after character 29
zcfzfwzzqfrljwzlrfnpqdbhtmscgvjw: first marker after character 26
How many characters need to be processed before the first start-of-message marker is detected?
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