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--- Day 23: Coprocessor Conflagration ---
You decide to head directly to the CPU and fix the printer from there. As you get close, you find an experimental coprocessor doing so much work that the local programs are afraid it will halt and catch fire. This would cause serious issues for the rest of the computer, so you head in and see what you can do.
The code it's running seems to be a variant of the kind you saw recently on that tablet. The general functionality seems very similar, but some of the instructions are different:
set X Y sets register X to the value of Y.
sub X Y decreases register X by the value of Y.
mul X Y sets register X to the result of multiplying the value contained in register X by the value of Y.
jnz X Y jumps with an offset of the value of Y, but only if the value of X is not zero. (An offset of 2 skips the next instruction, an offset of -1 jumps to the previous instruction, and so on.)
Only the instructions listed above are used. The eight registers here, named a through h, all start at 0.
The coprocessor is currently set to some kind of debug mode, which allows for testing, but prevents it from doing any meaningful work.
If you run the program (your puzzle input), how many times is the mul instruction invoked?
Your puzzle answer was 3025.
--- Part Two ---
Now, it's time to fix the problem.
The debug mode switch is wired directly to register a. You flip the switch, which makes register a now start at 1 when the program is executed.
Immediately, the coprocessor begins to overheat. Whoever wrote this program obviously didn't choose a very efficient implementation. You'll need to optimize the program if it has any hope of completing before Santa needs that printer working.
The coprocessor's ultimate goal is to determine the final value left in register h once the program completes. Technically, if it had that... it wouldn't even need to run the program.
After setting register a to 1, if the program were to run to completion, what value would be left in register h?
| 0
|
--- Day 10: The Stars Align ---
It's no use; your navigation system simply isn't capable of providing walking directions in the arctic circle, and certainly not in 1018.
The Elves suggest an alternative. In times like these, North Pole rescue operations will arrange points of light in the sky to guide missing Elves back to base. Unfortunately, the message is easy to miss: the points move slowly enough that it takes hours to align them, but have so much momentum that they only stay aligned for a second. If you blink at the wrong time, it might be hours before another message appears.
You can see these points of light floating in the distance, and record their position in the sky and their velocity, the relative change in position per second (your puzzle input). The coordinates are all given from your perspective; given enough time, those positions and velocities will move the points into a cohesive message!
Rather than wait, you decide to fast-forward the process and calculate what the points will eventually spell.
For example, suppose you note the following points:
position=< 9, 1> velocity=< 0, 2>
position=< 7, 0> velocity=<-1, 0>
position=< 3, -2> velocity=<-1, 1>
position=< 6, 10> velocity=<-2, -1>
position=< 2, -4> velocity=< 2, 2>
position=<-6, 10> velocity=< 2, -2>
position=< 1, 8> velocity=< 1, -1>
position=< 1, 7> velocity=< 1, 0>
position=<-3, 11> velocity=< 1, -2>
position=< 7, 6> velocity=<-1, -1>
position=<-2, 3> velocity=< 1, 0>
position=<-4, 3> velocity=< 2, 0>
position=<10, -3> velocity=<-1, 1>
position=< 5, 11> velocity=< 1, -2>
position=< 4, 7> velocity=< 0, -1>
position=< 8, -2> velocity=< 0, 1>
position=<15, 0> velocity=<-2, 0>
position=< 1, 6> velocity=< 1, 0>
position=< 8, 9> velocity=< 0, -1>
position=< 3, 3> velocity=<-1, 1>
position=< 0, 5> velocity=< 0, -1>
position=<-2, 2> velocity=< 2, 0>
position=< 5, -2> velocity=< 1, 2>
position=< 1, 4> velocity=< 2, 1>
position=<-2, 7> velocity=< 2, -2>
position=< 3, 6> velocity=<-1, -1>
position=< 5, 0> velocity=< 1, 0>
position=<-6, 0> velocity=< 2, 0>
position=< 5, 9> velocity=< 1, -2>
position=<14, 7> velocity=<-2, 0>
position=<-3, 6> velocity=< 2, -1>
Each line represents one point. Positions are given as <X, Y> pairs: X represents how far left (negative) or right (positive) the point appears, while Y represents how far up (negative) or down (positive) the point appears.
At 0 seconds, each point has the position given. Each second, each point's velocity is added to its position. So, a point with velocity <1, -2> is moving to the right, but is moving upward twice as quickly. If this point's initial position were <3, 9>, after 3 seconds, its position would become <6, 3>.
Over time, the points listed above would move like this:
Initially:
........#.............
................#.....
.........#.#..#.......
......................
#..........#.#.......#
...............#......
....#.................
..#.#....#............
.......#..............
......#...............
...#...#.#...#........
....#..#..#.........#.
.......#..............
...........#..#.......
#...........#.........
...#.......#..........
After 1 second:
......................
......................
..........#....#......
........#.....#.......
..#.........#......#..
......................
......#...............
....##.........#......
......#.#.............
.....##.##..#.........
........#.#...........
........#...#.....#...
..#...........#.......
....#.....#.#.........
......................
......................
After 2 seconds:
......................
......................
......................
..............#.......
....#..#...####..#....
......................
........#....#........
......#.#.............
.......#...#..........
.......#..#..#.#......
....#....#.#..........
.....#...#...##.#.....
........#.............
......................
......................
......................
After 3 seconds:
......................
......................
......................
......................
......#...#..###......
......#...#...#.......
......#...#...#.......
......#####...#.......
......#...#...#.......
......#...#...#.......
......#...#...#.......
......#...#..###......
......................
......................
......................
......................
After 4 seconds:
......................
......................
......................
............#.........
........##...#.#......
......#.....#..#......
.....#..##.##.#.......
.......##.#....#......
...........#....#.....
..............#.......
....#......#...#......
.....#.....##.........
...............#......
...............#......
......................
......................
After 3 seconds, the message appeared briefly: HI. Of course, your message will be much longer and will take many more seconds to appear.
What message will eventually appear in the sky?
| 1
|
--- 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?
Your puzzle answer was 95444.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
Now that you know what the best paths look like, you can figure out the best spot to sit.
Every non-wall tile (S, ., or E) is equipped with places to sit along the edges of the tile. While determining which of these tiles would be the best spot to sit depends on a whole bunch of factors (how comfortable the seats are, how far away the bathrooms are, whether there's a pillar blocking your view, etc.), the most important factor is whether the tile is on one of the best paths through the maze. If you sit somewhere else, you'd miss all the action!
So, you'll need to determine which tiles are part of any best path through the maze, including the S and E tiles.
In the first example, there are 45 tiles (marked O) that are part of at least one of the various best paths through the maze:
###############
#.......#....O#
#.#.###.#.###O#
#.....#.#...#O#
#.###.#####.#O#
#.#.#.......#O#
#.#.#####.###O#
#..OOOOOOOOO#O#
###O#O#####O#O#
#OOO#O....#O#O#
#O#O#O###.#O#O#
#OOOOO#...#O#O#
#O###.#.#.#O#O#
#O..#.....#OOO#
###############
In the second example, there are 64 tiles that are part of at least one of the best paths:
#################
#...#...#...#..O#
#.#.#.#.#.#.#.#O#
#.#.#.#...#...#O#
#.#.#.#.###.#.#O#
#OOO#.#.#.....#O#
#O#O#.#.#.#####O#
#O#O..#.#.#OOOOO#
#O#O#####.#O###O#
#O#O#..OOOOO#OOO#
#O#O###O#####O###
#O#O#OOO#..OOO#.#
#O#O#O#####O###.#
#O#O#OOOOOOO..#.#
#O#O#O#########.#
#O#OOO..........#
#################
Analyze your map further. How many tiles are part of at least one of the best paths through the maze?
| 2
|
--- 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?
| 3
|
--- Day 18: Like a Rogue ---
As you enter this room, you hear a loud click! Some of the tiles in the floor here seem to be pressure plates for traps, and the trap you just triggered has run out of... whatever it tried to do to you. You doubt you'll be so lucky next time.
Upon closer examination, the traps and safe tiles in this room seem to follow a pattern. The tiles are arranged into rows that are all the same width; you take note of the safe tiles (.) and traps (^) in the first row (your puzzle input).
The type of tile (trapped or safe) in each row is based on the types of the tiles in the same position, and to either side of that position, in the previous row. (If either side is off either end of the row, it counts as "safe" because there isn't a trap embedded in the wall.)
For example, suppose you know the first row (with tiles marked by letters) and want to determine the next row (with tiles marked by numbers):
ABCDE
12345
The type of tile 2 is based on the types of tiles A, B, and C; the type of tile 5 is based on tiles D, E, and an imaginary "safe" tile. Let's call these three tiles from the previous row the left, center, and right tiles, respectively. Then, a new tile is a trap only in one of the following situations:
Its left and center tiles are traps, but its right tile is not.
Its center and right tiles are traps, but its left tile is not.
Only its left tile is a trap.
Only its right tile is a trap.
In any other situation, the new tile is safe.
Then, starting with the row ..^^., you can determine the next row by applying those rules to each new tile:
The leftmost character on the next row considers the left (nonexistent, so we assume "safe"), center (the first ., which means "safe"), and right (the second ., also "safe") tiles on the previous row. Because all of the trap rules require a trap in at least one of the previous three tiles, the first tile on this new row is also safe, ..
The second character on the next row considers its left (.), center (.), and right (^) tiles from the previous row. This matches the fourth rule: only the right tile is a trap. Therefore, the next tile in this new row is a trap, ^.
The third character considers .^^, which matches the second trap rule: its center and right tiles are traps, but its left tile is not. Therefore, this tile is also a trap, ^.
The last two characters in this new row match the first and third rules, respectively, and so they are both also traps, ^.
After these steps, we now know the next row of tiles in the room: .^^^^. Then, we continue on to the next row, using the same rules, and get ^^..^. After determining two new rows, our map looks like this:
..^^.
.^^^^
^^..^
Here's a larger example with ten tiles per row and ten rows:
.^^.^.^^^^
^^^...^..^
^.^^.^.^^.
..^^...^^^
.^^^^.^^.^
^^..^.^^..
^^^^..^^^.
^..^^^^.^^
.^^^..^.^^
^^.^^^..^^
In ten rows, this larger example has 38 safe tiles.
Starting with the map in your puzzle input, in a total of 40 rows (including the starting row), how many safe tiles are there?
| 4
|
--- Day 20: Trench Map ---
With the scanners fully deployed, you turn their attention to mapping the floor of the ocean trench.
When you get back the image from the scanners, it seems to just be random noise. Perhaps you can combine an image enhancement algorithm and the input image (your puzzle input) to clean it up a little.
For example:
..#.#..#####.#.#.#.###.##.....###.##.#..###.####..#####..#....#..#..##..##
#..######.###...####..#..#####..##..#.#####...##.#.#..#.##..#.#......#.###
.######.###.####...#.##.##..#..#..#####.....#.#....###..#.##......#.....#.
.#..#..##..#...##.######.####.####.#.#...#.......#..#.#.#...####.##.#.....
.#..#...##.#.##..#...##.#.##..###.#......#.#.......#.#.#.####.###.##...#..
...####.#..#..#.##.#....##..#.####....##...##..#...#......#.#.......#.....
..##..####..#...#.#.#...##..#.#..###..#####........#..####......#..#
#..#.
#....
##..#
..#..
..###
The first section is the image enhancement algorithm. It is normally given on a single line, but it has been wrapped to multiple lines in this example for legibility. The second section is the input image, a two-dimensional grid of light pixels (#) and dark pixels (.).
The image enhancement algorithm describes how to enhance an image by simultaneously converting all pixels in the input image into an output image. Each pixel of the output image is determined by looking at a 3x3 square of pixels centered on the corresponding input image pixel. So, to determine the value of the pixel at (5,10) in the output image, nine pixels from the input image need to be considered: (4,9), (4,10), (4,11), (5,9), (5,10), (5,11), (6,9), (6,10), and (6,11). These nine input pixels are combined into a single binary number that is used as an index in the image enhancement algorithm string.
For example, to determine the output pixel that corresponds to the very middle pixel of the input image, the nine pixels marked by [...] would need to be considered:
# . . # .
#[. . .].
#[# . .]#
.[. # .].
. . # # #
Starting from the top-left and reading across each row, these pixels are ..., then #.., then .#.; combining these forms ...#...#.. By turning dark pixels (.) into 0 and light pixels (#) into 1, the binary number 000100010 can be formed, which is 34 in decimal.
The image enhancement algorithm string is exactly 512 characters long, enough to match every possible 9-bit binary number. The first few characters of the string (numbered starting from zero) are as follows:
0 10 20 30 34 40 50 60 70
| | | | | | | | |
..#.#..#####.#.#.#.###.##.....###.##.#..###.####..#####..#....#..#..##..##
In the middle of this first group of characters, the character at index 34 can be found: #. So, the output pixel in the center of the output image should be #, a light pixel.
This process can then be repeated to calculate every pixel of the output image.
Through advances in imaging technology, the images being operated on here are infinite in size. Every pixel of the infinite output image needs to be calculated exactly based on the relevant pixels of the input image. The small input image you have is only a small region of the actual infinite input image; the rest of the input image consists of dark pixels (.). For the purposes of the example, to save on space, only a portion of the infinite-sized input and output images will be shown.
The starting input image, therefore, looks something like this, with more dark pixels (.) extending forever in every direction not shown here:
...............
...............
...............
...............
...............
.....#..#......
.....#.........
.....##..#.....
.......#.......
.......###.....
...............
...............
...............
...............
...............
By applying the image enhancement algorithm to every pixel simultaneously, the following output image can be obtained:
...............
...............
...............
...............
.....##.##.....
....#..#.#.....
....##.#..#....
....####..#....
.....#..##.....
......##..#....
.......#.#.....
...............
...............
...............
...............
Through further advances in imaging technology, the above output image can also be used as an input image! This allows it to be enhanced a second time:
...............
...............
...............
..........#....
....#..#.#.....
...#.#...###...
...#...##.#....
...#.....#.#...
....#.#####....
.....#.#####...
......##.##....
.......###.....
...............
...............
...............
Truly incredible - now the small details are really starting to come through. After enhancing the original input image twice, 35 pixels are lit.
Start with the original input image and apply the image enhancement algorithm twice, being careful to account for the infinite size of the images. How many pixels are lit in the resulting image?
| 5
|
--- Day 12: Hill Climbing Algorithm ---
You try contacting the Elves using your handheld device, but the river you're following must be too low to get a decent signal.
You ask the device for a heightmap of the surrounding area (your puzzle input). The heightmap shows the local area from above broken into a grid; the elevation of each square of the grid is given by a single lowercase letter, where a is the lowest elevation, b is the next-lowest, and so on up to the highest elevation, z.
Also included on the heightmap are marks for your current position (S) and the location that should get the best signal (E). Your current position (S) has elevation a, and the location that should get the best signal (E) has elevation z.
You'd like to reach E, but to save energy, you should do it in as few steps as possible. During each step, you can move exactly one square up, down, left, or right. To avoid needing to get out your climbing gear, the elevation of the destination square can be at most one higher than the elevation of your current square; that is, if your current elevation is m, you could step to elevation n, but not to elevation o. (This also means that the elevation of the destination square can be much lower than the elevation of your current square.)
For example:
Sabqponm
abcryxxl
accszExk
acctuvwj
abdefghi
Here, you start in the top-left corner; your goal is near the middle. You could start by moving down or right, but eventually you'll need to head toward the e at the bottom. From there, you can spiral around to the goal:
v..v<<<<
>v.vv<<^
.>vv>E^^
..v>>>^^
..>>>>>^
In the above diagram, the symbols indicate whether the path exits each square moving up (^), down (v), left (<), or right (>). The location that should get the best signal is still E, and . marks unvisited squares.
This path reaches the goal in 31 steps, the fewest possible.
What is the fewest steps required to move from your current position to the location that should get the best signal?
| 6
|
--- Day 19: Medicine for Rudolph ---
Rudolph the Red-Nosed Reindeer is sick! His nose isn't shining very brightly, and he needs medicine.
Red-Nosed Reindeer biology isn't similar to regular reindeer biology; Rudolph is going to need custom-made medicine. Unfortunately, Red-Nosed Reindeer chemistry isn't similar to regular reindeer chemistry, either.
The North Pole is equipped with a Red-Nosed Reindeer nuclear fusion/fission plant, capable of constructing any Red-Nosed Reindeer molecule you need. It works by starting with some input molecule and then doing a series of replacements, one per step, until it has the right molecule.
However, the machine has to be calibrated before it can be used. Calibration involves determining the number of molecules that can be generated in one step from a given starting point.
For example, imagine a simpler machine that supports only the following replacements:
H => HO
H => OH
O => HH
Given the replacements above and starting with HOH, the following molecules could be generated:
HOOH (via H => HO on the first H).
HOHO (via H => HO on the second H).
OHOH (via H => OH on the first H).
HOOH (via H => OH on the second H).
HHHH (via O => HH).
So, in the example above, there are 4 distinct molecules (not five, because HOOH appears twice) after one replacement from HOH. Santa's favorite molecule, HOHOHO, can become 7 distinct molecules (over nine replacements: six from H, and three from O).
The machine replaces without regard for the surrounding characters. For example, given the string H2O, the transition H => OO would result in OO2O.
Your puzzle input describes all of the possible replacements and, at the bottom, the medicine molecule for which you need to calibrate the machine. How many distinct molecules can be created after all the different ways you can do one replacement on the medicine molecule?
Your puzzle answer was 509.
--- Part Two ---
Now that the machine is calibrated, you're ready to begin molecule fabrication.
Molecule fabrication always begins with just a single electron, e, and applying replacements one at a time, just like the ones during calibration.
For example, suppose you have the following replacements:
e => H
e => O
H => HO
H => OH
O => HH
If you'd like to make HOH, you start with e, and then make the following replacements:
e => O to get O
O => HH to get HH
H => OH (on the second H) to get HOH
So, you could make HOH after 3 steps. Santa's favorite molecule, HOHOHO, can be made in 6 steps.
How long will it take to make the medicine? Given the available replacements and the medicine molecule in your puzzle input, what is the fewest number of steps to go from e to the medicine molecule?
| 7
|
--- Day 15: Oxygen System ---
Out here in deep space, many things can go wrong. Fortunately, many of those things have indicator lights. Unfortunately, one of those lights is lit: the oxygen system for part of the ship has failed!
According to the readouts, the oxygen system must have failed days ago after a rupture in oxygen tank two; that section of the ship was automatically sealed once oxygen levels went dangerously low. A single remotely-operated repair droid is your only option for fixing the oxygen system.
The Elves' care package included an Intcode program (your puzzle input) that you can use to remotely control the repair droid. By running that program, you can direct the repair droid to the oxygen system and fix the problem.
The remote control program executes the following steps in a loop forever:
Accept a movement command via an input instruction.
Send the movement command to the repair droid.
Wait for the repair droid to finish the movement operation.
Report on the status of the repair droid via an output instruction.
Only four movement commands are understood: north (1), south (2), west (3), and east (4). Any other command is invalid. The movements differ in direction, but not in distance: in a long enough east-west hallway, a series of commands like 4,4,4,4,3,3,3,3 would leave the repair droid back where it started.
The repair droid can reply with any of the following status codes:
0: The repair droid hit a wall. Its position has not changed.
1: The repair droid has moved one step in the requested direction.
2: The repair droid has moved one step in the requested direction; its new position is the location of the oxygen system.
You don't know anything about the area around the repair droid, but you can figure it out by watching the status codes.
For example, we can draw the area using D for the droid, # for walls, . for locations the droid can traverse, and empty space for unexplored locations. Then, the initial state looks like this:
D
To make the droid go north, send it 1. If it replies with 0, you know that location is a wall and that the droid didn't move:
#
D
To move east, send 4; a reply of 1 means the movement was successful:
#
.D
Then, perhaps attempts to move north (1), south (2), and east (4) are all met with replies of 0:
##
.D#
#
Now, you know the repair droid is in a dead end. Backtrack with 3 (which you already know will get a reply of 1 because you already know that location is open):
##
D.#
#
Then, perhaps west (3) gets a reply of 0, south (2) gets a reply of 1, south again (2) gets a reply of 0, and then west (3) gets a reply of 2:
##
#..#
D.#
#
Now, because of the reply of 2, you know you've found the oxygen system! In this example, it was only 2 moves away from the repair droid's starting position.
What is the fewest number of movement commands required to move the repair droid from its starting position to the location of the oxygen system?
| 8
|
--- Day 5: Supply Stacks ---
The expedition can depart as soon as the final supplies have been unloaded from the ships. Supplies are stored in stacks of marked crates, but because the needed supplies are buried under many other crates, the crates need to be rearranged.
The ship has a giant cargo crane capable of moving crates between stacks. To ensure none of the crates get crushed or fall over, the crane operator will rearrange them in a series of carefully-planned steps. After the crates are rearranged, the desired crates will be at the top of each stack.
The Elves don't want to interrupt the crane operator during this delicate procedure, but they forgot to ask her which crate will end up where, and they want to be ready to unload them as soon as possible so they can embark.
They do, however, have a drawing of the starting stacks of crates and the rearrangement procedure (your puzzle input). For example:
[D]
[N] [C]
[Z] [M] [P]
1 2 3
move 1 from 2 to 1
move 3 from 1 to 3
move 2 from 2 to 1
move 1 from 1 to 2
In this example, there are three stacks of crates. Stack 1 contains two crates: crate Z is on the bottom, and crate N is on top. Stack 2 contains three crates; from bottom to top, they are crates M, C, and D. Finally, stack 3 contains a single crate, P.
Then, the rearrangement procedure is given. In each step of the procedure, a quantity of crates is moved from one stack to a different stack. In the first step of the above rearrangement procedure, one crate is moved from stack 2 to stack 1, resulting in this configuration:
[D]
[N] [C]
[Z] [M] [P]
1 2 3
In the second step, three crates are moved from stack 1 to stack 3. Crates are moved one at a time, so the first crate to be moved (D) ends up below the second and third crates:
[Z]
[N]
[C] [D]
[M] [P]
1 2 3
Then, both crates are moved from stack 2 to stack 1. Again, because crates are moved one at a time, crate C ends up below crate M:
[Z]
[N]
[M] [D]
[C] [P]
1 2 3
Finally, one crate is moved from stack 1 to stack 2:
[Z]
[N]
[D]
[C] [M] [P]
1 2 3
The Elves just need to know which crate will end up on top of each stack; in this example, the top crates are C in stack 1, M in stack 2, and Z in stack 3, so you should combine these together and give the Elves the message CMZ.
After the rearrangement procedure completes, what crate ends up on top of each stack?
Your puzzle answer was ZWHVFWQWW.
--- Part Two ---
As you watch the crane operator expertly rearrange the crates, you notice the process isn't following your prediction.
Some mud was covering the writing on the side of the crane, and you quickly wipe it away. The crane isn't a CrateMover 9000 - it's a CrateMover 9001.
The CrateMover 9001 is notable for many new and exciting features: air conditioning, leather seats, an extra cup holder, and the ability to pick up and move multiple crates at once.
Again considering the example above, the crates begin in the same configuration:
[D]
[N] [C]
[Z] [M] [P]
1 2 3
Moving a single crate from stack 2 to stack 1 behaves the same as before:
[D]
[N] [C]
[Z] [M] [P]
1 2 3
However, the action of moving three crates from stack 1 to stack 3 means that those three moved crates stay in the same order, resulting in this new configuration:
[D]
[N]
[C] [Z]
[M] [P]
1 2 3
Next, as both crates are moved from stack 2 to stack 1, they retain their order as well:
[D]
[N]
[C] [Z]
[M] [P]
1 2 3
Finally, a single crate is still moved from stack 1 to stack 2, but now it's crate C that gets moved:
[D]
[N]
[Z]
[M] [C] [P]
1 2 3
In this example, the CrateMover 9001 has put the crates in a totally different order: MCD.
Before the rearrangement process finishes, update your simulation so that the Elves know where they should stand to be ready to unload the final supplies. After the rearrangement procedure completes, what crate ends up on top of each stack?
| 9
|
--- Day 21: Keypad Conundrum ---
As you teleport onto Santa's Reindeer-class starship, The Historians begin to panic: someone from their search party is missing. A quick life-form scan by the ship's computer reveals that when the missing Historian teleported, he arrived in another part of the ship.
The door to that area is locked, but the computer can't open it; it can only be opened by physically typing the door codes (your puzzle input) on the numeric keypad on the door.
The numeric keypad has four rows of buttons: 789, 456, 123, and finally an empty gap followed by 0A. Visually, they are arranged like this:
+---+---+---+
| 7 | 8 | 9 |
+---+---+---+
| 4 | 5 | 6 |
+---+---+---+
| 1 | 2 | 3 |
+---+---+---+
| 0 | A |
+---+---+
Unfortunately, the area outside the door is currently depressurized and nobody can go near the door. A robot needs to be sent instead.
The robot has no problem navigating the ship and finding the numeric keypad, but it's not designed for button pushing: it can't be told to push a specific button directly. Instead, it has a robotic arm that can be controlled remotely via a directional keypad.
The directional keypad has two rows of buttons: a gap / ^ (up) / A (activate) on the first row and < (left) / v (down) / > (right) on the second row. Visually, they are arranged like this:
+---+---+
| ^ | A |
+---+---+---+
| < | v | > |
+---+---+---+
When the robot arrives at the numeric keypad, its robotic arm is pointed at the A button in the bottom right corner. After that, this directional keypad remote control must be used to maneuver the robotic arm: the up / down / left / right buttons cause it to move its arm one button in that direction, and the A button causes the robot to briefly move forward, pressing the button being aimed at by the robotic arm.
For example, to make the robot type 029A on the numeric keypad, one sequence of inputs on the directional keypad you could use is:
< to move the arm from A (its initial position) to 0.
A to push the 0 button.
^A to move the arm to the 2 button and push it.
>^^A to move the arm to the 9 button and push it.
vvvA to move the arm to the A button and push it.
In total, there are three shortest possible sequences of button presses on this directional keypad that would cause the robot to type 029A: <A^A>^^AvvvA, <A^A^>^AvvvA, and <A^A^^>AvvvA.
Unfortunately, the area containing this directional keypad remote control is currently experiencing high levels of radiation and nobody can go near it. A robot needs to be sent instead.
When the robot arrives at the directional keypad, its robot arm is pointed at the A button in the upper right corner. After that, a second, different directional keypad remote control is used to control this robot (in the same way as the first robot, except that this one is typing on a directional keypad instead of a numeric keypad).
There are multiple shortest possible sequences of directional keypad button presses that would cause this robot to tell the first robot to type 029A on the door. One such sequence is v<<A>>^A<A>AvA<^AA>A<vAAA>^A.
Unfortunately, the area containing this second directional keypad remote control is currently -40 degrees! Another robot will need to be sent to type on that directional keypad, too.
There are many shortest possible sequences of directional keypad button presses that would cause this robot to tell the second robot to tell the first robot to eventually type 029A on the door. One such sequence is <vA<AA>>^AvAA<^A>A<v<A>>^AvA^A<vA>^A<v<A>^A>AAvA^A<v<A>A>^AAAvA<^A>A.
Unfortunately, the area containing this third directional keypad remote control is currently full of Historians, so no robots can find a clear path there. Instead, you will have to type this sequence yourself.
Were you to choose this sequence of button presses, here are all of the buttons that would be pressed on your directional keypad, the two robots' directional keypads, and the numeric keypad:
<vA<AA>>^AvAA<^A>A<v<A>>^AvA^A<vA>^A<v<A>^A>AAvA^A<v<A>A>^AAAvA<^A>A
v<<A>>^A<A>AvA<^AA>A<vAAA>^A
<A^A>^^AvvvA
029A
In summary, there are the following keypads:
One directional keypad that you are using.
Two directional keypads that robots are using.
One numeric keypad (on a door) that a robot is using.
It is important to remember that these robots are not designed for button pushing. In particular, if a robot arm is ever aimed at a gap where no button is present on the keypad, even for an instant, the robot will panic unrecoverably. So, don't do that. All robots will initially aim at the keypad's A key, wherever it is.
To unlock the door, five codes will need to be typed on its numeric keypad. For example:
029A
980A
179A
456A
379A
For each of these, here is a shortest sequence of button presses you could type to cause the desired code to be typed on the numeric keypad:
029A: <vA<AA>>^AvAA<^A>A<v<A>>^AvA^A<vA>^A<v<A>^A>AAvA^A<v<A>A>^AAAvA<^A>A
980A: <v<A>>^AAAvA^A<vA<AA>>^AvAA<^A>A<v<A>A>^AAAvA<^A>A<vA>^A<A>A
179A: <v<A>>^A<vA<A>>^AAvAA<^A>A<v<A>>^AAvA^A<vA>^AA<A>A<v<A>A>^AAAvA<^A>A
456A: <v<A>>^AA<vA<A>>^AAvAA<^A>A<vA>^A<A>A<vA>^A<A>A<v<A>A>^AAvA<^A>A
379A: <v<A>>^AvA^A<vA<AA>>^AAvA<^A>AAvA^A<vA>^AA<A>A<v<A>A>^AAAvA<^A>A
The Historians are getting nervous; the ship computer doesn't remember whether the missing Historian is trapped in the area containing a giant electromagnet or molten lava. You'll need to make sure that for each of the five codes, you find the shortest sequence of button presses necessary.
The complexity of a single code (like 029A) is equal to the result of multiplying these two values:
The length of the shortest sequence of button presses you need to type on your directional keypad in order to cause the code to be typed on the numeric keypad; for 029A, this would be 68.
The numeric part of the code (ignoring leading zeroes); for 029A, this would be 29.
In the above example, complexity of the five codes can be found by calculating 68 * 29, 60 * 980, 68 * 179, 64 * 456, and 64 * 379. Adding these together produces 126384.
Find the fewest number of button presses you'll need to perform in order to cause the robot in front of the door to type each code. What is the sum of the complexities of the five codes on your list?
Your puzzle answer was 211930.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
Just as the missing Historian is released, The Historians realize that a second member of their search party has also been missing this entire time!
A quick life-form scan reveals the Historian is also trapped in a locked area of the ship. Due to a variety of hazards, robots are once again dispatched, forming another chain of remote control keypads managing robotic-arm-wielding robots.
This time, many more robots are involved. In summary, there are the following keypads:
One directional keypad that you are using.
25 directional keypads that robots are using.
One numeric keypad (on a door) that a robot is using.
The keypads form a chain, just like before: your directional keypad controls a robot which is typing on a directional keypad which controls a robot which is typing on a directional keypad... and so on, ending with the robot which is typing on the numeric keypad.
The door codes are the same this time around; only the number of robots and directional keypads has changed.
Find the fewest number of button presses you'll need to perform in order to cause the robot in front of the door to type each code. What is the sum of the complexities of the five codes on your list?
| 10
|
--- Day 16: Flawed Frequency Transmission ---
You're 3/4ths of the way through the gas giants. Not only do roundtrip signals to Earth take five hours, but the signal quality is quite bad as well. You can clean up the signal with the Flawed Frequency Transmission algorithm, or FFT.
As input, FFT takes a list of numbers. In the signal you received (your puzzle input), each number is a single digit: data like 15243 represents the sequence 1, 5, 2, 4, 3.
FFT operates in repeated phases. In each phase, a new list is constructed with the same length as the input list. This new list is also used as the input for the next phase.
Each element in the new list is built by multiplying every value in the input list by a value in a repeating pattern and then adding up the results. So, if the input list were 9, 8, 7, 6, 5 and the pattern for a given element were 1, 2, 3, the result would be 9*1 + 8*2 + 7*3 + 6*1 + 5*2 (with each input element on the left and each value in the repeating pattern on the right of each multiplication). Then, only the ones digit is kept: 38 becomes 8, -17 becomes 7, and so on.
While each element in the output array uses all of the same input array elements, the actual repeating pattern to use depends on which output element is being calculated. The base pattern is 0, 1, 0, -1. Then, repeat each value in the pattern a number of times equal to the position in the output list being considered. Repeat once for the first element, twice for the second element, three times for the third element, and so on. So, if the third element of the output list is being calculated, repeating the values would produce: 0, 0, 0, 1, 1, 1, 0, 0, 0, -1, -1, -1.
When applying the pattern, skip the very first value exactly once. (In other words, offset the whole pattern left by one.) So, for the second element of the output list, the actual pattern used would be: 0, 1, 1, 0, 0, -1, -1, 0, 0, 1, 1, 0, 0, -1, -1, ....
After using this process to calculate each element of the output list, the phase is complete, and the output list of this phase is used as the new input list for the next phase, if any.
Given the input signal 12345678, below are four phases of FFT. Within each phase, each output digit is calculated on a single line with the result at the far right; each multiplication operation shows the input digit on the left and the pattern value on the right:
Input signal: 12345678
1*1 + 2*0 + 3*-1 + 4*0 + 5*1 + 6*0 + 7*-1 + 8*0 = 4
1*0 + 2*1 + 3*1 + 4*0 + 5*0 + 6*-1 + 7*-1 + 8*0 = 8
1*0 + 2*0 + 3*1 + 4*1 + 5*1 + 6*0 + 7*0 + 8*0 = 2
1*0 + 2*0 + 3*0 + 4*1 + 5*1 + 6*1 + 7*1 + 8*0 = 2
1*0 + 2*0 + 3*0 + 4*0 + 5*1 + 6*1 + 7*1 + 8*1 = 6
1*0 + 2*0 + 3*0 + 4*0 + 5*0 + 6*1 + 7*1 + 8*1 = 1
1*0 + 2*0 + 3*0 + 4*0 + 5*0 + 6*0 + 7*1 + 8*1 = 5
1*0 + 2*0 + 3*0 + 4*0 + 5*0 + 6*0 + 7*0 + 8*1 = 8
After 1 phase: 48226158
4*1 + 8*0 + 2*-1 + 2*0 + 6*1 + 1*0 + 5*-1 + 8*0 = 3
4*0 + 8*1 + 2*1 + 2*0 + 6*0 + 1*-1 + 5*-1 + 8*0 = 4
4*0 + 8*0 + 2*1 + 2*1 + 6*1 + 1*0 + 5*0 + 8*0 = 0
4*0 + 8*0 + 2*0 + 2*1 + 6*1 + 1*1 + 5*1 + 8*0 = 4
4*0 + 8*0 + 2*0 + 2*0 + 6*1 + 1*1 + 5*1 + 8*1 = 0
4*0 + 8*0 + 2*0 + 2*0 + 6*0 + 1*1 + 5*1 + 8*1 = 4
4*0 + 8*0 + 2*0 + 2*0 + 6*0 + 1*0 + 5*1 + 8*1 = 3
4*0 + 8*0 + 2*0 + 2*0 + 6*0 + 1*0 + 5*0 + 8*1 = 8
After 2 phases: 34040438
3*1 + 4*0 + 0*-1 + 4*0 + 0*1 + 4*0 + 3*-1 + 8*0 = 0
3*0 + 4*1 + 0*1 + 4*0 + 0*0 + 4*-1 + 3*-1 + 8*0 = 3
3*0 + 4*0 + 0*1 + 4*1 + 0*1 + 4*0 + 3*0 + 8*0 = 4
3*0 + 4*0 + 0*0 + 4*1 + 0*1 + 4*1 + 3*1 + 8*0 = 1
3*0 + 4*0 + 0*0 + 4*0 + 0*1 + 4*1 + 3*1 + 8*1 = 5
3*0 + 4*0 + 0*0 + 4*0 + 0*0 + 4*1 + 3*1 + 8*1 = 5
3*0 + 4*0 + 0*0 + 4*0 + 0*0 + 4*0 + 3*1 + 8*1 = 1
3*0 + 4*0 + 0*0 + 4*0 + 0*0 + 4*0 + 3*0 + 8*1 = 8
After 3 phases: 03415518
0*1 + 3*0 + 4*-1 + 1*0 + 5*1 + 5*0 + 1*-1 + 8*0 = 0
0*0 + 3*1 + 4*1 + 1*0 + 5*0 + 5*-1 + 1*-1 + 8*0 = 1
0*0 + 3*0 + 4*1 + 1*1 + 5*1 + 5*0 + 1*0 + 8*0 = 0
0*0 + 3*0 + 4*0 + 1*1 + 5*1 + 5*1 + 1*1 + 8*0 = 2
0*0 + 3*0 + 4*0 + 1*0 + 5*1 + 5*1 + 1*1 + 8*1 = 9
0*0 + 3*0 + 4*0 + 1*0 + 5*0 + 5*1 + 1*1 + 8*1 = 4
0*0 + 3*0 + 4*0 + 1*0 + 5*0 + 5*0 + 1*1 + 8*1 = 9
0*0 + 3*0 + 4*0 + 1*0 + 5*0 + 5*0 + 1*0 + 8*1 = 8
After 4 phases: 01029498
Here are the first eight digits of the final output list after 100 phases for some larger inputs:
80871224585914546619083218645595 becomes 24176176.
19617804207202209144916044189917 becomes 73745418.
69317163492948606335995924319873 becomes 52432133.
After 100 phases of FFT, what are the first eight digits in the final output list?
| 11
|
--- Day 10: Monitoring Station ---
You fly into the asteroid belt and reach the Ceres monitoring station. The Elves here have an emergency: they're having trouble tracking all of the asteroids and can't be sure they're safe.
The Elves would like to build a new monitoring station in a nearby area of space; they hand you a map of all of the asteroids in that region (your puzzle input).
The map indicates whether each position is empty (.) or contains an asteroid (#). The asteroids are much smaller than they appear on the map, and every asteroid is exactly in the center of its marked position. The asteroids can be described with X,Y coordinates where X is the distance from the left edge and Y is the distance from the top edge (so the top-left corner is 0,0 and the position immediately to its right is 1,0).
Your job is to figure out which asteroid would be the best place to build a new monitoring station. A monitoring station can detect any asteroid to which it has direct line of sight - that is, there cannot be another asteroid exactly between them. This line of sight can be at any angle, not just lines aligned to the grid or diagonally. The best location is the asteroid that can detect the largest number of other asteroids.
For example, consider the following map:
.#..#
.....
#####
....#
...##
The best location for a new monitoring station on this map is the highlighted asteroid at 3,4 because it can detect 8 asteroids, more than any other location. (The only asteroid it cannot detect is the one at 1,0; its view of this asteroid is blocked by the asteroid at 2,2.) All other asteroids are worse locations; they can detect 7 or fewer other asteroids. Here is the number of other asteroids a monitoring station on each asteroid could detect:
.7..7
.....
67775
....7
...87
Here is an asteroid (#) and some examples of the ways its line of sight might be blocked. If there were another asteroid at the location of a capital letter, the locations marked with the corresponding lowercase letter would be blocked and could not be detected:
#.........
...A......
...B..a...
.EDCG....a
..F.c.b...
.....c....
..efd.c.gb
.......c..
....f...c.
...e..d..c
Here are some larger examples:
Best is 5,8 with 33 other asteroids detected:
......#.#.
#..#.#....
..#######.
.#.#.###..
.#..#.....
..#....#.#
#..#....#.
.##.#..###
##...#..#.
.#....####
Best is 1,2 with 35 other asteroids detected:
#.#...#.#.
.###....#.
.#....#...
##.#.#.#.#
....#.#.#.
.##..###.#
..#...##..
..##....##
......#...
.####.###.
Best is 6,3 with 41 other asteroids detected:
.#..#..###
####.###.#
....###.#.
..###.##.#
##.##.#.#.
....###..#
..#.#..#.#
#..#.#.###
.##...##.#
.....#.#..
Best is 11,13 with 210 other asteroids detected:
.#..##.###...#######
##.############..##.
.#.######.########.#
.###.#######.####.#.
#####.##.#.##.###.##
..#####..#.#########
####################
#.####....###.#.#.##
##.#################
#####.##.###..####..
..######..##.#######
####.##.####...##..#
.#####..#.######.###
##...#.##########...
#.##########.#######
.####.#.###.###.#.##
....##.##.###..#####
.#.#.###########.###
#.#.#.#####.####.###
###.##.####.##.#..##
Find the best location for a new monitoring station. How many other asteroids can be detected from that location?
| 12
|
--- Day 8: Seven Segment Search ---
You barely reach the safety of the cave when the whale smashes into the cave mouth, collapsing it. Sensors indicate another exit to this cave at a much greater depth, so you have no choice but to press on.
As your submarine slowly makes its way through the cave system, you notice that the four-digit seven-segment displays in your submarine are malfunctioning; they must have been damaged during the escape. You'll be in a lot of trouble without them, so you'd better figure out what's wrong.
Each digit of a seven-segment display is rendered by turning on or off any of seven segments named a through g:
0: 1: 2: 3: 4:
aaaa .... aaaa aaaa ....
b c . c . c . c b c
b c . c . c . c b c
.... .... dddd dddd dddd
e f . f e . . f . f
e f . f e . . f . f
gggg .... gggg gggg ....
5: 6: 7: 8: 9:
aaaa aaaa aaaa aaaa aaaa
b . b . . c b c b c
b . b . . c b c b c
dddd dddd .... dddd dddd
. f e f . f e f . f
. f e f . f e f . f
gggg gggg .... gggg gggg
So, to render a 1, only segments c and f would be turned on; the rest would be off. To render a 7, only segments a, c, and f would be turned on.
The problem is that the signals which control the segments have been mixed up on each display. The submarine is still trying to display numbers by producing output on signal wires a through g, but those wires are connected to segments randomly. Worse, the wire/segment connections are mixed up separately for each four-digit display! (All of the digits within a display use the same connections, though.)
So, you might know that only signal wires b and g are turned on, but that doesn't mean segments b and g are turned on: the only digit that uses two segments is 1, so it must mean segments c and f are meant to be on. With just that information, you still can't tell which wire (b/g) goes to which segment (c/f). For that, you'll need to collect more information.
For each display, you watch the changing signals for a while, make a note of all ten unique signal patterns you see, and then write down a single four digit output value (your puzzle input). Using the signal patterns, you should be able to work out which pattern corresponds to which digit.
For example, here is what you might see in a single entry in your notes:
acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab |
cdfeb fcadb cdfeb cdbaf
(The entry is wrapped here to two lines so it fits; in your notes, it will all be on a single line.)
Each entry consists of ten unique signal patterns, a | delimiter, and finally the four digit output value. Within an entry, the same wire/segment connections are used (but you don't know what the connections actually are). The unique signal patterns correspond to the ten different ways the submarine tries to render a digit using the current wire/segment connections. Because 7 is the only digit that uses three segments, dab in the above example means that to render a 7, signal lines d, a, and b are on. Because 4 is the only digit that uses four segments, eafb means that to render a 4, signal lines e, a, f, and b are on.
Using this information, you should be able to work out which combination of signal wires corresponds to each of the ten digits. Then, you can decode the four digit output value. Unfortunately, in the above example, all of the digits in the output value (cdfeb fcadb cdfeb cdbaf) use five segments and are more difficult to deduce.
For now, focus on the easy digits. Consider this larger example:
be cfbegad cbdgef fgaecd cgeb fdcge agebfd fecdb fabcd edb |
fdgacbe cefdb cefbgd gcbe
edbfga begcd cbg gc gcadebf fbgde acbgfd abcde gfcbed gfec |
fcgedb cgb dgebacf gc
fgaebd cg bdaec gdafb agbcfd gdcbef bgcad gfac gcb cdgabef |
cg cg fdcagb cbg
fbegcd cbd adcefb dageb afcb bc aefdc ecdab fgdeca fcdbega |
efabcd cedba gadfec cb
aecbfdg fbg gf bafeg dbefa fcge gcbea fcaegb dgceab fcbdga |
gecf egdcabf bgf bfgea
fgeab ca afcebg bdacfeg cfaedg gcfdb baec bfadeg bafgc acf |
gebdcfa ecba ca fadegcb
dbcfg fgd bdegcaf fgec aegbdf ecdfab fbedc dacgb gdcebf gf |
cefg dcbef fcge gbcadfe
bdfegc cbegaf gecbf dfcage bdacg ed bedf ced adcbefg gebcd |
ed bcgafe cdgba cbgef
egadfb cdbfeg cegd fecab cgb gbdefca cg fgcdab egfdb bfceg |
gbdfcae bgc cg cgb
gcafb gcf dcaebfg ecagb gf abcdeg gaef cafbge fdbac fegbdc |
fgae cfgab fg bagce
Because the digits 1, 4, 7, and 8 each use a unique number of segments, you should be able to tell which combinations of signals correspond to those digits. Counting only digits in the output values (the part after | on each line), in the above example, there are 26 instances of digits that use a unique number of segments (highlighted above).
In the output values, how many times do digits 1, 4, 7, or 8 appear?
| 13
|
--- Day 18: RAM Run ---
You and The Historians look a lot more pixelated than you remember. You're inside a computer at the North Pole!
Just as you're about to check out your surroundings, a program runs up to you. "This region of memory isn't safe! The User misunderstood what a pushdown automaton is and their algorithm is pushing whole bytes down on top of us! Run!"
The algorithm is fast - it's going to cause a byte to fall into your memory space once every nanosecond! Fortunately, you're faster, and by quickly scanning the algorithm, you create a list of which bytes will fall (your puzzle input) in the order they'll land in your memory space.
Your memory space is a two-dimensional grid with coordinates that range from 0 to 70 both horizontally and vertically. However, for the sake of example, suppose you're on a smaller grid with coordinates that range from 0 to 6 and the following list of incoming byte positions:
5,4
4,2
4,5
3,0
2,1
6,3
2,4
1,5
0,6
3,3
2,6
5,1
1,2
5,5
2,5
6,5
1,4
0,4
6,4
1,1
6,1
1,0
0,5
1,6
2,0
Each byte position is given as an X,Y coordinate, where X is the distance from the left edge of your memory space and Y is the distance from the top edge of your memory space.
You and The Historians are currently in the top left corner of the memory space (at 0,0) and need to reach the exit in the bottom right corner (at 70,70 in your memory space, but at 6,6 in this example). You'll need to simulate the falling bytes to plan out where it will be safe to run; for now, simulate just the first few bytes falling into your memory space.
As bytes fall into your memory space, they make that coordinate corrupted. Corrupted memory coordinates cannot be entered by you or The Historians, so you'll need to plan your route carefully. You also cannot leave the boundaries of the memory space; your only hope is to reach the exit.
In the above example, if you were to draw the memory space after the first 12 bytes have fallen (using . for safe and # for corrupted), it would look like this:
...#...
..#..#.
....#..
...#..#
..#..#.
.#..#..
#.#....
You can take steps up, down, left, or right. After just 12 bytes have corrupted locations in your memory space, the shortest path from the top left corner to the exit would take 22 steps. Here (marked with O) is one such path:
OO.#OOO
.O#OO#O
.OOO#OO
...#OO#
..#OO#.
.#.O#..
#.#OOOO
Simulate the first kilobyte (1024 bytes) falling onto your memory space. Afterward, what is the minimum number of steps needed to reach the exit?
Your puzzle answer was 354.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
The Historians aren't as used to moving around in this pixelated universe as you are. You're afraid they're not going to be fast enough to make it to the exit before the path is completely blocked.
To determine how fast everyone needs to go, you need to determine the first byte that will cut off the path to the exit.
In the above example, after the byte at 1,1 falls, there is still a path to the exit:
O..#OOO
O##OO#O
O#OO#OO
OOO#OO#
###OO##
.##O###
#.#OOOO
However, after adding the very next byte (at 6,1), there is no longer a path to the exit:
...#...
.##..##
.#..#..
...#..#
###..##
.##.###
#.#....
So, in this example, the coordinates of the first byte that prevents the exit from being reachable are 6,1.
Simulate more of the bytes that are about to corrupt your memory space. What are the coordinates of the first byte that will prevent the exit from being reachable from your starting position? (Provide the answer as two integers separated by a comma with no other characters.)
| 14
|
--- Day 18: RAM Run ---
You and The Historians look a lot more pixelated than you remember. You're inside a computer at the North Pole!
Just as you're about to check out your surroundings, a program runs up to you. "This region of memory isn't safe! The User misunderstood what a pushdown automaton is and their algorithm is pushing whole bytes down on top of us! Run!"
The algorithm is fast - it's going to cause a byte to fall into your memory space once every nanosecond! Fortunately, you're faster, and by quickly scanning the algorithm, you create a list of which bytes will fall (your puzzle input) in the order they'll land in your memory space.
Your memory space is a two-dimensional grid with coordinates that range from 0 to 70 both horizontally and vertically. However, for the sake of example, suppose you're on a smaller grid with coordinates that range from 0 to 6 and the following list of incoming byte positions:
5,4
4,2
4,5
3,0
2,1
6,3
2,4
1,5
0,6
3,3
2,6
5,1
1,2
5,5
2,5
6,5
1,4
0,4
6,4
1,1
6,1
1,0
0,5
1,6
2,0
Each byte position is given as an X,Y coordinate, where X is the distance from the left edge of your memory space and Y is the distance from the top edge of your memory space.
You and The Historians are currently in the top left corner of the memory space (at 0,0) and need to reach the exit in the bottom right corner (at 70,70 in your memory space, but at 6,6 in this example). You'll need to simulate the falling bytes to plan out where it will be safe to run; for now, simulate just the first few bytes falling into your memory space.
As bytes fall into your memory space, they make that coordinate corrupted. Corrupted memory coordinates cannot be entered by you or The Historians, so you'll need to plan your route carefully. You also cannot leave the boundaries of the memory space; your only hope is to reach the exit.
In the above example, if you were to draw the memory space after the first 12 bytes have fallen (using . for safe and # for corrupted), it would look like this:
...#...
..#..#.
....#..
...#..#
..#..#.
.#..#..
#.#....
You can take steps up, down, left, or right. After just 12 bytes have corrupted locations in your memory space, the shortest path from the top left corner to the exit would take 22 steps. Here (marked with O) is one such path:
OO.#OOO
.O#OO#O
.OOO#OO
...#OO#
..#OO#.
.#.O#..
#.#OOOO
Simulate the first kilobyte (1024 bytes) falling onto your memory space. Afterward, what is the minimum number of steps needed to reach the exit?
| 15
|
--- Day 9: Encoding Error ---
With your neighbor happily enjoying their video game, you turn your attention to an open data port on the little screen in the seat in front of you.
Though the port is non-standard, you manage to connect it to your computer through the clever use of several paperclips. Upon connection, the port outputs a series of numbers (your puzzle input).
The data appears to be encrypted with the eXchange-Masking Addition System (XMAS) which, conveniently for you, is an old cypher with an important weakness.
XMAS starts by transmitting a preamble of 25 numbers. After that, each number you receive should be the sum of any two of the 25 immediately previous numbers. The two numbers will have different values, and there might be more than one such pair.
For example, suppose your preamble consists of the numbers 1 through 25 in a random order. To be valid, the next number must be the sum of two of those numbers:
26 would be a valid next number, as it could be 1 plus 25 (or many other pairs, like 2 and 24).
49 would be a valid next number, as it is the sum of 24 and 25.
100 would not be valid; no two of the previous 25 numbers sum to 100.
50 would also not be valid; although 25 appears in the previous 25 numbers, the two numbers in the pair must be different.
Suppose the 26th number is 45, and the first number (no longer an option, as it is more than 25 numbers ago) was 20. Now, for the next number to be valid, there needs to be some pair of numbers among 1-19, 21-25, or 45 that add up to it:
26 would still be a valid next number, as 1 and 25 are still within the previous 25 numbers.
65 would not be valid, as no two of the available numbers sum to it.
64 and 66 would both be valid, as they are the result of 19+45 and 21+45 respectively.
Here is a larger example which only considers the previous 5 numbers (and has a preamble of length 5):
35
20
15
25
47
40
62
55
65
95
102
117
150
182
127
219
299
277
309
576
In this example, after the 5-number preamble, almost every number is the sum of two of the previous 5 numbers; the only number that does not follow this rule is 127.
The first step of attacking the weakness in the XMAS data is to find the first number in the list (after the preamble) which is not the sum of two of the 25 numbers before it. What is the first number that does not have this property?
| 16
|
--- Day 1: No Time for a Taxicab ---
Santa's sleigh uses a very high-precision clock to guide its movements, and the clock's oscillator is regulated by stars. Unfortunately, the stars have been stolen... by the Easter Bunny. To save Christmas, Santa needs you to retrieve 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!
You're airdropped near Easter Bunny Headquarters in a city somewhere. "Near", unfortunately, is as close as you can get - the instructions on the Easter Bunny Recruiting Document the Elves intercepted start here, and nobody had time to work them out further.
The Document indicates that you should start at the given coordinates (where you just landed) and face North. Then, follow the provided sequence: either turn left (L) or right (R) 90 degrees, then walk forward the given number of blocks, ending at a new intersection.
There's no time to follow such ridiculous instructions on foot, though, so you take a moment and work out the destination. Given that you can only walk on the street grid of the city, how far is the shortest path to the destination?
For example:
Following R2, L3 leaves you 2 blocks East and 3 blocks North, or 5 blocks away.
R2, R2, R2 leaves you 2 blocks due South of your starting position, which is 2 blocks away.
R5, L5, R5, R3 leaves you 12 blocks away.
How many blocks away is Easter Bunny HQ?
| 17
|
--- Day 6: Chronal Coordinates ---
The device on your wrist beeps several times, and once again you feel like you're falling.
"Situation critical," the device announces. "Destination indeterminate. Chronal interference detected. Please specify new target coordinates."
The device then produces a list of coordinates (your puzzle input). Are they places it thinks are safe or dangerous? It recommends you check manual page 729. The Elves did not give you a manual.
If they're dangerous, maybe you can minimize the danger by finding the coordinate that gives the largest distance from the other points.
Using only the Manhattan distance, determine the area around each coordinate by counting the number of integer X,Y locations that are closest to that coordinate (and aren't tied in distance to any other coordinate).
Your goal is to find the size of the largest area that isn't infinite. For example, consider the following list of coordinates:
1, 1
1, 6
8, 3
3, 4
5, 5
8, 9
If we name these coordinates A through F, we can draw them on a grid, putting 0,0 at the top left:
..........
.A........
..........
........C.
...D......
.....E....
.B........
..........
..........
........F.
This view is partial - the actual grid extends infinitely in all directions. Using the Manhattan distance, each location's closest coordinate can be determined, shown here in lowercase:
aaaaa.cccc
aAaaa.cccc
aaaddecccc
aadddeccCc
..dDdeeccc
bb.deEeecc
bBb.eeee..
bbb.eeefff
bbb.eeffff
bbb.ffffFf
Locations shown as . are equally far from two or more coordinates, and so they don't count as being closest to any.
In this example, the areas of coordinates A, B, C, and F are infinite - while not shown here, their areas extend forever outside the visible grid. However, the areas of coordinates D and E are finite: D is closest to 9 locations, and E is closest to 17 (both including the coordinate's location itself). Therefore, in this example, the size of the largest area is 17.
What is the size of the largest area that isn't infinite?
| 18
|
--- Day 5: Alchemical Reduction ---
You've managed to sneak in to the prototype suit manufacturing lab. The Elves are making decent progress, but are still struggling with the suit's size reduction capabilities.
While the very latest in 1518 alchemical technology might have solved their problem eventually, you can do better. You scan the chemical composition of the suit's material and discover that it is formed by extremely long polymers (one of which is available as your puzzle input).
The polymer is formed by smaller units which, when triggered, react with each other such that two adjacent units of the same type and opposite polarity are destroyed. Units' types are represented by letters; units' polarity is represented by capitalization. For instance, r and R are units with the same type but opposite polarity, whereas r and s are entirely different types and do not react.
For example:
In aA, a and A react, leaving nothing behind.
In abBA, bB destroys itself, leaving aA. As above, this then destroys itself, leaving nothing.
In abAB, no two adjacent units are of the same type, and so nothing happens.
In aabAAB, even though aa and AA are of the same type, their polarities match, and so nothing happens.
Now, consider a larger example, dabAcCaCBAcCcaDA:
dabAcCaCBAcCcaDA The first 'cC' is removed.
dabAaCBAcCcaDA This creates 'Aa', which is removed.
dabCBAcCcaDA Either 'cC' or 'Cc' are removed (the result is the same).
dabCBAcaDA No further actions can be taken.
After all possible reactions, the resulting polymer contains 10 units.
How many units remain after fully reacting the polymer you scanned? (Note: in this puzzle and others, the input is large; if you copy/paste your input, make sure you get the whole thing.)
Your puzzle answer was 11636.
--- Part Two ---
Time to improve the polymer.
One of the unit types is causing problems; it's preventing the polymer from collapsing as much as it should. Your goal is to figure out which unit type is causing the most problems, remove all instances of it (regardless of polarity), fully react the remaining polymer, and measure its length.
For example, again using the polymer dabAcCaCBAcCcaDA from above:
Removing all A/a units produces dbcCCBcCcD. Fully reacting this polymer produces dbCBcD, which has length 6.
Removing all B/b units produces daAcCaCAcCcaDA. Fully reacting this polymer produces daCAcaDA, which has length 8.
Removing all C/c units produces dabAaBAaDA. Fully reacting this polymer produces daDA, which has length 4.
Removing all D/d units produces abAcCaCBAcCcaA. Fully reacting this polymer produces abCBAc, which has length 6.
In this example, removing all C/c units was best, producing the answer 4.
What is the length of the shortest polymer you can produce by removing all units of exactly one type and fully reacting the result?
| 19
|
--- Day 24: Crossed Wires ---
You and The Historians arrive at the edge of a large grove somewhere in the jungle. After the last incident, the Elves installed a small device that monitors the fruit. While The Historians search the grove, one of them asks if you can take a look at the monitoring device; apparently, it's been malfunctioning recently.
The device seems to be trying to produce a number through some boolean logic gates. Each gate has two inputs and one output. The gates all operate on values that are either true (1) or false (0).
AND gates output 1 if both inputs are 1; if either input is 0, these gates output 0.
OR gates output 1 if one or both inputs is 1; if both inputs are 0, these gates output 0.
XOR gates output 1 if the inputs are different; if the inputs are the same, these gates output 0.
Gates wait until both inputs are received before producing output; wires can carry 0, 1 or no value at all. There are no loops; once a gate has determined its output, the output will not change until the whole system is reset. Each wire is connected to at most one gate output, but can be connected to many gate inputs.
Rather than risk getting shocked while tinkering with the live system, you write down all of the gate connections and initial wire values (your puzzle input) so you can consider them in relative safety. For example:
x00: 1
x01: 1
x02: 1
y00: 0
y01: 1
y02: 0
x00 AND y00 -> z00
x01 XOR y01 -> z01
x02 OR y02 -> z02
Because gates wait for input, some wires need to start with a value (as inputs to the entire system). The first section specifies these values. For example, x00: 1 means that the wire named x00 starts with the value 1 (as if a gate is already outputting that value onto that wire).
The second section lists all of the gates and the wires connected to them. For example, x00 AND y00 -> z00 describes an instance of an AND gate which has wires x00 and y00 connected to its inputs and which will write its output to wire z00.
In this example, simulating these gates eventually causes 0 to appear on wire z00, 0 to appear on wire z01, and 1 to appear on wire z02.
Ultimately, the system is trying to produce a number by combining the bits on all wires starting with z. z00 is the least significant bit, then z01, then z02, and so on.
In this example, the three output bits form the binary number 100 which is equal to the decimal number 4.
Here's a larger example:
x00: 1
x01: 0
x02: 1
x03: 1
x04: 0
y00: 1
y01: 1
y02: 1
y03: 1
y04: 1
ntg XOR fgs -> mjb
y02 OR x01 -> tnw
kwq OR kpj -> z05
x00 OR x03 -> fst
tgd XOR rvg -> z01
vdt OR tnw -> bfw
bfw AND frj -> z10
ffh OR nrd -> bqk
y00 AND y03 -> djm
y03 OR y00 -> psh
bqk OR frj -> z08
tnw OR fst -> frj
gnj AND tgd -> z11
bfw XOR mjb -> z00
x03 OR x00 -> vdt
gnj AND wpb -> z02
x04 AND y00 -> kjc
djm OR pbm -> qhw
nrd AND vdt -> hwm
kjc AND fst -> rvg
y04 OR y02 -> fgs
y01 AND x02 -> pbm
ntg OR kjc -> kwq
psh XOR fgs -> tgd
qhw XOR tgd -> z09
pbm OR djm -> kpj
x03 XOR y03 -> ffh
x00 XOR y04 -> ntg
bfw OR bqk -> z06
nrd XOR fgs -> wpb
frj XOR qhw -> z04
bqk OR frj -> z07
y03 OR x01 -> nrd
hwm AND bqk -> z03
tgd XOR rvg -> z12
tnw OR pbm -> gnj
After waiting for values on all wires starting with z, the wires in this system have the following values:
bfw: 1
bqk: 1
djm: 1
ffh: 0
fgs: 1
frj: 1
fst: 1
gnj: 1
hwm: 1
kjc: 0
kpj: 1
kwq: 0
mjb: 1
nrd: 1
ntg: 0
pbm: 1
psh: 1
qhw: 1
rvg: 0
tgd: 0
tnw: 1
vdt: 1
wpb: 0
z00: 0
z01: 0
z02: 0
z03: 1
z04: 0
z05: 1
z06: 1
z07: 1
z08: 1
z09: 1
z10: 1
z11: 0
z12: 0
Combining the bits from all wires starting with z produces the binary number 0011111101000. Converting this number to decimal produces 2024.
Simulate the system of gates and wires. What decimal number does it output on the wires starting with z?
| 20
|
--- 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?
Your puzzle answer was 160040.
--- Part Two ---
Now, you just need to figure out how many orbital transfers you (YOU) need to take to get to Santa (SAN).
You start at the object YOU are orbiting; your destination is the object SAN is orbiting. An orbital transfer lets you move from any object to an object orbiting or orbited by that object.
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
K)YOU
I)SAN
Visually, the above map of orbits looks like this:
YOU
/
G - H J - K - L
/ /
COM - B - C - D - E - F
I - SAN
In this example, YOU are in orbit around K, and SAN is in orbit around I. To move from K to I, a minimum of 4 orbital transfers are required:
K to J
J to E
E to D
D to I
Afterward, the map of orbits looks like this:
G - H J - K - L
/ /
COM - B - C - D - E - F
I - SAN
YOU
What is the minimum number of orbital transfers required to move from the object YOU are orbiting to the object SAN is orbiting? (Between the objects they are orbiting - not between YOU and SAN.)
| 21
|
--- Day 10: Cathode-Ray Tube ---
You avoid the ropes, plunge into the river, and swim to shore.
The Elves yell something about meeting back up with them upriver, but the river is too loud to tell exactly what they're saying. They finish crossing the bridge and disappear from view.
Situations like this must be why the Elves prioritized getting the communication system on your handheld device working. You pull it out of your pack, but the amount of water slowly draining from a big crack in its screen tells you it probably won't be of much immediate use.
Unless, that is, you can design a replacement for the device's video system! It seems to be some kind of cathode-ray tube screen and simple CPU that are both driven by a precise clock circuit. The clock circuit ticks at a constant rate; each tick is called a cycle.
Start by figuring out the signal being sent by the CPU. The CPU has a single register, X, which starts with the value 1. It supports only two instructions:
addx V takes two cycles to complete. After two cycles, the X register is increased by the value V. (V can be negative.)
noop takes one cycle to complete. It has no other effect.
The CPU uses these instructions in a program (your puzzle input) to, somehow, tell the screen what to draw.
Consider the following small program:
noop
addx 3
addx -5
Execution of this program proceeds as follows:
At the start of the first cycle, the noop instruction begins execution. During the first cycle, X is 1. After the first cycle, the noop instruction finishes execution, doing nothing.
At the start of the second cycle, the addx 3 instruction begins execution. During the second cycle, X is still 1.
During the third cycle, X is still 1. After the third cycle, the addx 3 instruction finishes execution, setting X to 4.
At the start of the fourth cycle, the addx -5 instruction begins execution. During the fourth cycle, X is still 4.
During the fifth cycle, X is still 4. After the fifth cycle, the addx -5 instruction finishes execution, setting X to -1.
Maybe you can learn something by looking at the value of the X register throughout execution. For now, consider the signal strength (the cycle number multiplied by the value of the X register) during the 20th cycle and every 40 cycles after that (that is, during the 20th, 60th, 100th, 140th, 180th, and 220th cycles).
For example, consider this larger program:
addx 15
addx -11
addx 6
addx -3
addx 5
addx -1
addx -8
addx 13
addx 4
noop
addx -1
addx 5
addx -1
addx 5
addx -1
addx 5
addx -1
addx 5
addx -1
addx -35
addx 1
addx 24
addx -19
addx 1
addx 16
addx -11
noop
noop
addx 21
addx -15
noop
noop
addx -3
addx 9
addx 1
addx -3
addx 8
addx 1
addx 5
noop
noop
noop
noop
noop
addx -36
noop
addx 1
addx 7
noop
noop
noop
addx 2
addx 6
noop
noop
noop
noop
noop
addx 1
noop
noop
addx 7
addx 1
noop
addx -13
addx 13
addx 7
noop
addx 1
addx -33
noop
noop
noop
addx 2
noop
noop
noop
addx 8
noop
addx -1
addx 2
addx 1
noop
addx 17
addx -9
addx 1
addx 1
addx -3
addx 11
noop
noop
addx 1
noop
addx 1
noop
noop
addx -13
addx -19
addx 1
addx 3
addx 26
addx -30
addx 12
addx -1
addx 3
addx 1
noop
noop
noop
addx -9
addx 18
addx 1
addx 2
noop
noop
addx 9
noop
noop
noop
addx -1
addx 2
addx -37
addx 1
addx 3
noop
addx 15
addx -21
addx 22
addx -6
addx 1
noop
addx 2
addx 1
noop
addx -10
noop
noop
addx 20
addx 1
addx 2
addx 2
addx -6
addx -11
noop
noop
noop
The interesting signal strengths can be determined as follows:
During the 20th cycle, register X has the value 21, so the signal strength is 20 * 21 = 420. (The 20th cycle occurs in the middle of the second addx -1, so the value of register X is the starting value, 1, plus all of the other addx values up to that point: 1 + 15 - 11 + 6 - 3 + 5 - 1 - 8 + 13 + 4 = 21.)
During the 60th cycle, register X has the value 19, so the signal strength is 60 * 19 = 1140.
During the 100th cycle, register X has the value 18, so the signal strength is 100 * 18 = 1800.
During the 140th cycle, register X has the value 21, so the signal strength is 140 * 21 = 2940.
During the 180th cycle, register X has the value 16, so the signal strength is 180 * 16 = 2880.
During the 220th cycle, register X has the value 18, so the signal strength is 220 * 18 = 3960.
The sum of these signal strengths is 13140.
Find the signal strength during the 20th, 60th, 100th, 140th, 180th, and 220th cycles. What is the sum of these six signal strengths?
| 22
|
--- Day 17: Spinlock ---
Suddenly, whirling in the distance, you notice what looks like a massive, pixelated hurricane: a deadly spinlock. This spinlock isn't just consuming computing power, but memory, too; vast, digital mountains are being ripped from the ground and consumed by the vortex.
If you don't move quickly, fixing that printer will be the least of your problems.
This spinlock's algorithm is simple but efficient, quickly consuming everything in its path. It starts with a circular buffer containing only the value 0, which it marks as the current position. It then steps forward through the circular buffer some number of steps (your puzzle input) before inserting the first new value, 1, after the value it stopped on. The inserted value becomes the current position. Then, it steps forward from there the same number of steps, and wherever it stops, inserts after it the second new value, 2, and uses that as the new current position again.
It repeats this process of stepping forward, inserting a new value, and using the location of the inserted value as the new current position a total of 2017 times, inserting 2017 as its final operation, and ending with a total of 2018 values (including 0) in the circular buffer.
For example, if the spinlock were to step 3 times per insert, the circular buffer would begin to evolve like this (using parentheses to mark the current position after each iteration of the algorithm):
(0), the initial state before any insertions.
0 (1): the spinlock steps forward three times (0, 0, 0), and then inserts the first value, 1, after it. 1 becomes the current position.
0 (2) 1: the spinlock steps forward three times (0, 1, 0), and then inserts the second value, 2, after it. 2 becomes the current position.
0 2 (3) 1: the spinlock steps forward three times (1, 0, 2), and then inserts the third value, 3, after it. 3 becomes the current position.
And so on:
0 2 (4) 3 1
0 (5) 2 4 3 1
0 5 2 4 3 (6) 1
0 5 (7) 2 4 3 6 1
0 5 7 2 4 3 (8) 6 1
0 (9) 5 7 2 4 3 8 6 1
Eventually, after 2017 insertions, the section of the circular buffer near the last insertion looks like this:
1512 1134 151 (2017) 638 1513 851
Perhaps, if you can identify the value that will ultimately be after the last value written (2017), you can short-circuit the spinlock. In this example, that would be 638.
What is the value after 2017 in your completed circular buffer?
| 23
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--- 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?
Your puzzle answer was 12841.
--- Part Two ---
Out of curiosity, the debugger would also like to know the size of the loop: starting from a state that has already been seen, how many block redistribution cycles must be performed before that same state is seen again?
In the example above, 2 4 1 2 is seen again after four cycles, and so the answer in that example would be 4.
How many cycles are in the infinite loop that arises from the configuration in your puzzle input?
| 24
|
--- Day 24: Arithmetic Logic Unit ---
Magic smoke starts leaking from the submarine's arithmetic logic unit (ALU). Without the ability to perform basic arithmetic and logic functions, the submarine can't produce cool patterns with its Christmas lights!
It also can't navigate. Or run the oxygen system.
Don't worry, though - you probably have enough oxygen left to give you enough time to build a new ALU.
The ALU is a four-dimensional processing unit: it has integer variables w, x, y, and z. These variables all start with the value 0. The ALU also supports six instructions:
inp a - Read an input value and write it to variable a.
add a b - Add the value of a to the value of b, then store the result in variable a.
mul a b - Multiply the value of a by the value of b, then store the result in variable a.
div a b - Divide the value of a by the value of b, truncate the result to an integer, then store the result in variable a. (Here, "truncate" means to round the value toward zero.)
mod a b - Divide the value of a by the value of b, then store the remainder in variable a. (This is also called the modulo operation.)
eql a b - If the value of a and b are equal, then store the value 1 in variable a. Otherwise, store the value 0 in variable a.
In all of these instructions, a and b are placeholders; a will always be the variable where the result of the operation is stored (one of w, x, y, or z), while b can be either a variable or a number. Numbers can be positive or negative, but will always be integers.
The ALU has no jump instructions; in an ALU program, every instruction is run exactly once in order from top to bottom. The program halts after the last instruction has finished executing.
(Program authors should be especially cautious; attempting to execute div with b=0 or attempting to execute mod with a<0 or b<=0 will cause the program to crash and might even damage the ALU. These operations are never intended in any serious ALU program.)
For example, here is an ALU program which takes an input number, negates it, and stores it in x:
inp x
mul x -1
Here is an ALU program which takes two input numbers, then sets z to 1 if the second input number is three times larger than the first input number, or sets z to 0 otherwise:
inp z
inp x
mul z 3
eql z x
Here is an ALU program which takes a non-negative integer as input, converts it into binary, and stores the lowest (1's) bit in z, the second-lowest (2's) bit in y, the third-lowest (4's) bit in x, and the fourth-lowest (8's) bit in w:
inp w
add z w
mod z 2
div w 2
add y w
mod y 2
div w 2
add x w
mod x 2
div w 2
mod w 2
Once you have built a replacement ALU, you can install it in the submarine, which will immediately resume what it was doing when the ALU failed: validating the submarine's model number. To do this, the ALU will run the MOdel Number Automatic Detector program (MONAD, your puzzle input).
Submarine model numbers are always fourteen-digit numbers consisting only of digits 1 through 9. The digit 0 cannot appear in a model number.
When MONAD checks a hypothetical fourteen-digit model number, it uses fourteen separate inp instructions, each expecting a single digit of the model number in order of most to least significant. (So, to check the model number 13579246899999, you would give 1 to the first inp instruction, 3 to the second inp instruction, 5 to the third inp instruction, and so on.) This means that when operating MONAD, each input instruction should only ever be given an integer value of at least 1 and at most 9.
Then, after MONAD has finished running all of its instructions, it will indicate that the model number was valid by leaving a 0 in variable z. However, if the model number was invalid, it will leave some other non-zero value in z.
MONAD imposes additional, mysterious restrictions on model numbers, and legend says the last copy of the MONAD documentation was eaten by a tanuki. You'll need to figure out what MONAD does some other way.
To enable as many submarine features as possible, find the largest valid fourteen-digit model number that contains no 0 digits. What is the largest model number accepted by MONAD?
| 25
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--- Day 14: Restroom Redoubt ---
One of The Historians needs to use the bathroom; fortunately, you know there's a bathroom near an unvisited location on their list, and so you're all quickly teleported directly to the lobby of Easter Bunny Headquarters.
Unfortunately, EBHQ seems to have "improved" bathroom security again after your last visit. The area outside the bathroom is swarming with robots!
To get The Historian safely to the bathroom, you'll need a way to predict where the robots will be in the future. Fortunately, they all seem to be moving on the tile floor in predictable straight lines.
You make a list (your puzzle input) of all of the robots' current positions (p) and velocities (v), one robot per line. For example:
p=0,4 v=3,-3
p=6,3 v=-1,-3
p=10,3 v=-1,2
p=2,0 v=2,-1
p=0,0 v=1,3
p=3,0 v=-2,-2
p=7,6 v=-1,-3
p=3,0 v=-1,-2
p=9,3 v=2,3
p=7,3 v=-1,2
p=2,4 v=2,-3
p=9,5 v=-3,-3
Each robot's position is given as p=x,y where x represents the number of tiles the robot is from the left wall and y represents the number of tiles from the top wall (when viewed from above). So, a position of p=0,0 means the robot is all the way in the top-left corner.
Each robot's velocity is given as v=x,y where x and y are given in tiles per second. Positive x means the robot is moving to the right, and positive y means the robot is moving down. So, a velocity of v=1,-2 means that each second, the robot moves 1 tile to the right and 2 tiles up.
The robots outside the actual bathroom are in a space which is 101 tiles wide and 103 tiles tall (when viewed from above). However, in this example, the robots are in a space which is only 11 tiles wide and 7 tiles tall.
The robots are good at navigating over/under each other (due to a combination of springs, extendable legs, and quadcopters), so they can share the same tile and don't interact with each other. Visually, the number of robots on each tile in this example looks like this:
1.12.......
...........
...........
......11.11
1.1........
.........1.
.......1...
These robots have a unique feature for maximum bathroom security: they can teleport. When a robot would run into an edge of the space they're in, they instead teleport to the other side, effectively wrapping around the edges. Here is what robot p=2,4 v=2,-3 does for the first few seconds:
Initial state:
...........
...........
...........
...........
..1........
...........
...........
After 1 second:
...........
....1......
...........
...........
...........
...........
...........
After 2 seconds:
...........
...........
...........
...........
...........
......1....
...........
After 3 seconds:
...........
...........
........1..
...........
...........
...........
...........
After 4 seconds:
...........
...........
...........
...........
...........
...........
..........1
After 5 seconds:
...........
...........
...........
.1.........
...........
...........
...........
The Historian can't wait much longer, so you don't have to simulate the robots for very long. Where will the robots be after 100 seconds?
In the above example, the number of robots on each tile after 100 seconds has elapsed looks like this:
......2..1.
...........
1..........
.11........
.....1.....
...12......
.1....1....
To determine the safest area, count the number of robots in each quadrant after 100 seconds. Robots that are exactly in the middle (horizontally or vertically) don't count as being in any quadrant, so the only relevant robots are:
..... 2..1.
..... .....
1.... .....
..... .....
...12 .....
.1... 1....
In this example, the quadrants contain 1, 3, 4, and 1 robot. Multiplying these together gives a total safety factor of 12.
Predict the motion of the robots in your list within a space which is 101 tiles wide and 103 tiles tall. What will the safety factor be after exactly 100 seconds have elapsed?
Your puzzle answer was 221655456.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
During the bathroom break, someone notices that these robots seem awfully similar to ones built and used at the North Pole. If they're the same type of robots, they should have a hard-coded Easter egg: very rarely, most of the robots should arrange themselves into a picture of a Christmas tree.
What is the fewest number of seconds that must elapse for the robots to display the Easter egg?
| 26
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--- 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?
| 27
|
--- 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?
Your puzzle answer was 381699.
--- Part Two ---
The Elves in accounting are thankful for your help; one of them even offers you a starfish coin they had left over from a past vacation. They offer you a second one if you can find three numbers in your expense report that meet the same criteria.
Using the above example again, the three entries that sum to 2020 are 979, 366, and 675. Multiplying them together produces the answer, 241861950.
In your expense report, what is the product of the three entries that sum to 2020?
| 28
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--- Day 20: Particle Swarm ---
Suddenly, the GPU contacts you, asking for help. Someone has asked it to simulate too many particles, and it won't be able to finish them all in time to render the next frame at this rate.
It transmits to you a buffer (your puzzle input) listing each particle in order (starting with particle 0, then particle 1, particle 2, and so on). For each particle, it provides the X, Y, and Z coordinates for the particle's position (p), velocity (v), and acceleration (a), each in the format <X,Y,Z>.
Each tick, all particles are updated simultaneously. A particle's properties are updated in the following order:
Increase the X velocity by the X acceleration.
Increase the Y velocity by the Y acceleration.
Increase the Z velocity by the Z acceleration.
Increase the X position by the X velocity.
Increase the Y position by the Y velocity.
Increase the Z position by the Z velocity.
Because of seemingly tenuous rationale involving z-buffering, the GPU would like to know which particle will stay closest to position <0,0,0> in the long term. Measure this using the Manhattan distance, which in this situation is simply the sum of the absolute values of a particle's X, Y, and Z position.
For example, suppose you are only given two particles, both of which stay entirely on the X-axis (for simplicity). Drawing the current states of particles 0 and 1 (in that order) with an adjacent a number line and diagram of current X positions (marked in parentheses), the following would take place:
p=< 3,0,0>, v=< 2,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
p=< 4,0,0>, v=< 0,0,0>, a=<-2,0,0> (0)(1)
p=< 4,0,0>, v=< 1,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
p=< 2,0,0>, v=<-2,0,0>, a=<-2,0,0> (1) (0)
p=< 4,0,0>, v=< 0,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
p=<-2,0,0>, v=<-4,0,0>, a=<-2,0,0> (1) (0)
p=< 3,0,0>, v=<-1,0,0>, a=<-1,0,0> -4 -3 -2 -1 0 1 2 3 4
p=<-8,0,0>, v=<-6,0,0>, a=<-2,0,0> (0)
At this point, particle 1 will never be closer to <0,0,0> than particle 0, and so, in the long run, particle 0 will stay closest.
Which particle will stay closest to position <0,0,0> in the long term?
| 29
|
--- Day 7: Internet Protocol Version 7 ---
While snooping around the local network of EBHQ, you compile a list of IP addresses (they're IPv7, of course; IPv6 is much too limited). You'd like to figure out which IPs support TLS (transport-layer snooping).
An IP supports TLS if it has an Autonomous Bridge Bypass Annotation, or ABBA. An ABBA is any four-character sequence which consists of a pair of two different characters followed by the reverse of that pair, such as xyyx or abba. However, the IP also must not have an ABBA within any hypernet sequences, which are contained by square brackets.
For example:
abba[mnop]qrst supports TLS (abba outside square brackets).
abcd[bddb]xyyx does not support TLS (bddb is within square brackets, even though xyyx is outside square brackets).
aaaa[qwer]tyui does not support TLS (aaaa is invalid; the interior characters must be different).
ioxxoj[asdfgh]zxcvbn supports TLS (oxxo is outside square brackets, even though it's within a larger string).
How many IPs in your puzzle input support TLS?
| 30
|
--- Day 7: Camel Cards ---
Your all-expenses-paid trip turns out to be a one-way, five-minute ride in an airship. (At least it's a cool airship!) It drops you off at the edge of a vast desert and descends back to Island Island.
"Did you bring the parts?"
You turn around to see an Elf completely covered in white clothing, wearing goggles, and riding a large camel.
"Did you bring the parts?" she asks again, louder this time. You aren't sure what parts she's looking for; you're here to figure out why the sand stopped.
"The parts! For the sand, yes! Come with me; I will show you." She beckons you onto the camel.
After riding a bit across the sands of Desert Island, you can see what look like very large rocks covering half of the horizon. The Elf explains that the rocks are all along the part of Desert Island that is directly above Island Island, making it hard to even get there. Normally, they use big machines to move the rocks and filter the sand, but the machines have broken down because Desert Island recently stopped receiving the parts they need to fix the machines.
You've already assumed it'll be your job to figure out why the parts stopped when she asks if you can help. You agree automatically.
Because the journey will take a few days, she offers to teach you the game of Camel Cards. Camel Cards is sort of similar to poker except it's designed to be easier to play while riding a camel.
In Camel Cards, you get a list of hands, and your goal is to order them based on the strength of each hand. A hand consists of five cards labeled one of A, K, Q, J, T, 9, 8, 7, 6, 5, 4, 3, or 2. The relative strength of each card follows this order, where A is the highest and 2 is the lowest.
Every hand is exactly one type. From strongest to weakest, they are:
Five of a kind, where all five cards have the same label: AAAAA
Four of a kind, where four cards have the same label and one card has a different label: AA8AA
Full house, where three cards have the same label, and the remaining two cards share a different label: 23332
Three of a kind, where three cards have the same label, and the remaining two cards are each different from any other card in the hand: TTT98
Two pair, where two cards share one label, two other cards share a second label, and the remaining card has a third label: 23432
One pair, where two cards share one label, and the other three cards have a different label from the pair and each other: A23A4
High card, where all cards' labels are distinct: 23456
Hands are primarily ordered based on type; for example, every full house is stronger than any three of a kind.
If two hands have the same type, a second ordering rule takes effect. Start by comparing the first card in each hand. If these cards are different, the hand with the stronger first card is considered stronger. If the first card in each hand have the same label, however, then move on to considering the second card in each hand. If they differ, the hand with the higher second card wins; otherwise, continue with the third card in each hand, then the fourth, then the fifth.
So, 33332 and 2AAAA are both four of a kind hands, but 33332 is stronger because its first card is stronger. Similarly, 77888 and 77788 are both a full house, but 77888 is stronger because its third card is stronger (and both hands have the same first and second card).
To play Camel Cards, you are given a list of hands and their corresponding bid (your puzzle input). For example:
32T3K 765
T55J5 684
KK677 28
KTJJT 220
QQQJA 483
This example shows five hands; each hand is followed by its bid amount. Each hand wins an amount equal to its bid multiplied by its rank, where the weakest hand gets rank 1, the second-weakest hand gets rank 2, and so on up to the strongest hand. Because there are five hands in this example, the strongest hand will have rank 5 and its bid will be multiplied by 5.
So, the first step is to put the hands in order of strength:
32T3K is the only one pair and the other hands are all a stronger type, so it gets rank 1.
KK677 and KTJJT are both two pair. Their first cards both have the same label, but the second card of KK677 is stronger (K vs T), so KTJJT gets rank 2 and KK677 gets rank 3.
T55J5 and QQQJA are both three of a kind. QQQJA has a stronger first card, so it gets rank 5 and T55J5 gets rank 4.
Now, you can determine the total winnings of this set of hands by adding up the result of multiplying each hand's bid with its rank (765 * 1 + 220 * 2 + 28 * 3 + 684 * 4 + 483 * 5). So the total winnings in this example are 6440.
Find the rank of every hand in your set. What are the total winnings?
| 31
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--- Day 13: A Maze of Twisty Little Cubicles ---
You arrive at the first floor of this new building to discover a much less welcoming environment than the shiny atrium of the last one. Instead, you are in a maze of twisty little cubicles, all alike.
Every location in this area is addressed by a pair of non-negative integers (x,y). Each such coordinate is either a wall or an open space. You can't move diagonally. The cube maze starts at 0,0 and seems to extend infinitely toward positive x and y; negative values are invalid, as they represent a location outside the building. You are in a small waiting area at 1,1.
While it seems chaotic, a nearby morale-boosting poster explains, the layout is actually quite logical. You can determine whether a given x,y coordinate will be a wall or an open space using a simple system:
Find x*x + 3*x + 2*x*y + y + y*y.
Add the office designer's favorite number (your puzzle input).
Find the binary representation of that sum; count the number of bits that are 1.
If the number of bits that are 1 is even, it's an open space.
If the number of bits that are 1 is odd, it's a wall.
For example, if the office designer's favorite number were 10, drawing walls as # and open spaces as ., the corner of the building containing 0,0 would look like this:
0123456789
0 .#.####.##
1 ..#..#...#
2 #....##...
3 ###.#.###.
4 .##..#..#.
5 ..##....#.
6 #...##.###
Now, suppose you wanted to reach 7,4. The shortest route you could take is marked as O:
0123456789
0 .#.####.##
1 .O#..#...#
2 #OOO.##...
3 ###O#.###.
4 .##OO#OO#.
5 ..##OOO.#.
6 #...##.###
Thus, reaching 7,4 would take a minimum of 11 steps (starting from your current location, 1,1).
What is the fewest number of steps required for you to reach 31,39?
Your puzzle answer was 82.
--- Part Two ---
How many locations (distinct x,y coordinates, including your starting location) can you reach in at most 50 steps?
| 32
|
--- Day 11: Cosmic Expansion ---
You continue following signs for "Hot Springs" and eventually come across an observatory. The Elf within turns out to be a researcher studying cosmic expansion using the giant telescope here.
He doesn't know anything about the missing machine parts; he's only visiting for this research project. However, he confirms that the hot springs are the next-closest area likely to have people; he'll even take you straight there once he's done with today's observation analysis.
Maybe you can help him with the analysis to speed things up?
The researcher has collected a bunch of data and compiled the data into a single giant image (your puzzle input). The image includes empty space (.) and galaxies (#). For example:
...#......
.......#..
#.........
..........
......#...
.#........
.........#
..........
.......#..
#...#.....
The researcher is trying to figure out the sum of the lengths of the shortest path between every pair of galaxies. However, there's a catch: the universe expanded in the time it took the light from those galaxies to reach the observatory.
Due to something involving gravitational effects, only some space expands. In fact, the result is that any rows or columns that contain no galaxies should all actually be twice as big.
In the above example, three columns and two rows contain no galaxies:
v v v
...#......
.......#..
#.........
>..........<
......#...
.#........
.........#
>..........<
.......#..
#...#.....
^ ^ ^
These rows and columns need to be twice as big; the result of cosmic expansion therefore looks like this:
....#........
.........#...
#............
.............
.............
........#....
.#...........
............#
.............
.............
.........#...
#....#.......
Equipped with this expanded universe, the shortest path between every pair of galaxies can be found. It can help to assign every galaxy a unique number:
....1........
.........2...
3............
.............
.............
........4....
.5...........
............6
.............
.............
.........7...
8....9.......
In these 9 galaxies, there are 36 pairs. Only count each pair once; order within the pair doesn't matter. For each pair, find any shortest path between the two galaxies using only steps that move up, down, left, or right exactly one . or # at a time. (The shortest path between two galaxies is allowed to pass through another galaxy.)
For example, here is one of the shortest paths between galaxies 5 and 9:
....1........
.........2...
3............
.............
.............
........4....
.5...........
.##.........6
..##.........
...##........
....##...7...
8....9.......
This path has length 9 because it takes a minimum of nine steps to get from galaxy 5 to galaxy 9 (the eight locations marked # plus the step onto galaxy 9 itself). Here are some other example shortest path lengths:
Between galaxy 1 and galaxy 7: 15
Between galaxy 3 and galaxy 6: 17
Between galaxy 8 and galaxy 9: 5
In this example, after expanding the universe, the sum of the shortest path between all 36 pairs of galaxies is 374.
Expand the universe, then find the length of the shortest path between every pair of galaxies. What is the sum of these lengths?
| 33
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--- Day 6: Guard Gallivant ---
The Historians use their fancy device again, this time to whisk you all away to the North Pole prototype suit manufacturing lab... in the year 1518! It turns out that having direct access to history is very convenient for a group of historians.
You still have to be careful of time paradoxes, and so it will be important to avoid anyone from 1518 while The Historians search for the Chief. Unfortunately, a single guard is patrolling this part of the lab.
Maybe you can work out where the guard will go ahead of time so that The Historians can search safely?
You start by making a map (your puzzle input) of the situation. For example:
....#.....
.........#
..........
..#.......
.......#..
..........
.#..^.....
........#.
#.........
......#...
The map shows the current position of the guard with ^ (to indicate the guard is currently facing up from the perspective of the map). Any obstructions - crates, desks, alchemical reactors, etc. - are shown as #.
Lab guards in 1518 follow a very strict patrol protocol which involves repeatedly following these steps:
If there is something directly in front of you, turn right 90 degrees.
Otherwise, take a step forward.
Following the above protocol, the guard moves up several times until she reaches an obstacle (in this case, a pile of failed suit prototypes):
....#.....
....^....#
..........
..#.......
.......#..
..........
.#........
........#.
#.........
......#...
Because there is now an obstacle in front of the guard, she turns right before continuing straight in her new facing direction:
....#.....
........>#
..........
..#.......
.......#..
..........
.#........
........#.
#.........
......#...
Reaching another obstacle (a spool of several very long polymers), she turns right again and continues downward:
....#.....
.........#
..........
..#.......
.......#..
..........
.#......v.
........#.
#.........
......#...
This process continues for a while, but the guard eventually leaves the mapped area (after walking past a tank of universal solvent):
....#.....
.........#
..........
..#.......
.......#..
..........
.#........
........#.
#.........
......#v..
By predicting the guard's route, you can determine which specific positions in the lab will be in the patrol path. Including the guard's starting position, the positions visited by the guard before leaving the area are marked with an X:
....#.....
....XXXXX#
....X...X.
..#.X...X.
..XXXXX#X.
..X.X.X.X.
.#XXXXXXX.
.XXXXXXX#.
#XXXXXXX..
......#X..
In this example, the guard will visit 41 distinct positions on your map.
Predict the path of the guard. How many distinct positions will the guard visit before leaving the mapped area?
Your puzzle answer was 4988.
The first half of this puzzle is complete! It provides one gold star: *
--- Part Two ---
While The Historians begin working around the guard's patrol route, you borrow their fancy device and step outside the lab. From the safety of a supply closet, you time travel through the last few months and record the nightly status of the lab's guard post on the walls of the closet.
Returning after what seems like only a few seconds to The Historians, they explain that the guard's patrol area is simply too large for them to safely search the lab without getting caught.
Fortunately, they are pretty sure that adding a single new obstruction won't cause a time paradox. They'd like to place the new obstruction in such a way that the guard will get stuck in a loop, making the rest of the lab safe to search.
To have the lowest chance of creating a time paradox, The Historians would like to know all of the possible positions for such an obstruction. The new obstruction can't be placed at the guard's starting position - the guard is there right now and would notice.
In the above example, there are only 6 different positions where a new obstruction would cause the guard to get stuck in a loop. The diagrams of these six situations use O to mark the new obstruction, | to show a position where the guard moves up/down, - to show a position where the guard moves left/right, and + to show a position where the guard moves both up/down and left/right.
Option one, put a printing press next to the guard's starting position:
....#.....
....+---+#
....|...|.
..#.|...|.
....|..#|.
....|...|.
.#.O^---+.
........#.
#.........
......#...
Option two, put a stack of failed suit prototypes in the bottom right quadrant of the mapped area:
....#.....
....+---+#
....|...|.
..#.|...|.
..+-+-+#|.
..|.|.|.|.
.#+-^-+-+.
......O.#.
#.........
......#...
Option three, put a crate of chimney-squeeze prototype fabric next to the standing desk in the bottom right quadrant:
....#.....
....+---+#
....|...|.
..#.|...|.
..+-+-+#|.
..|.|.|.|.
.#+-^-+-+.
.+----+O#.
#+----+...
......#...
Option four, put an alchemical retroencabulator near the bottom left corner:
....#.....
....+---+#
....|...|.
..#.|...|.
..+-+-+#|.
..|.|.|.|.
.#+-^-+-+.
..|...|.#.
#O+---+...
......#...
Option five, put the alchemical retroencabulator a bit to the right instead:
....#.....
....+---+#
....|...|.
..#.|...|.
..+-+-+#|.
..|.|.|.|.
.#+-^-+-+.
....|.|.#.
#..O+-+...
......#...
Option six, put a tank of sovereign glue right next to the tank of universal solvent:
....#.....
....+---+#
....|...|.
..#.|...|.
..+-+-+#|.
..|.|.|.|.
.#+-^-+-+.
.+----++#.
#+----++..
......#O..
It doesn't really matter what you choose to use as an obstacle so long as you and The Historians can put it into position without the guard noticing. The important thing is having enough options that you can find one that minimizes time paradoxes, and in this example, there are 6 different positions you could choose.
You need to get the guard stuck in a loop by adding a single new obstruction. How many different positions could you choose for this obstruction?
| 34
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--- Day 2: Red-Nosed Reports ---
Fortunately, the first location The Historians want to search isn't a long walk from the Chief Historian's office.
While the Red-Nosed Reindeer nuclear fusion/fission plant appears to contain no sign of the Chief Historian, the engineers there run up to you as soon as they see you. Apparently, they still talk about the time Rudolph was saved through molecular synthesis from a single electron.
They're quick to add that - since you're already here - they'd really appreciate your help analyzing some unusual data from the Red-Nosed reactor. You turn to check if The Historians are waiting for you, but they seem to have already divided into groups that are currently searching every corner of the facility. You offer to help with the unusual data.
The unusual data (your puzzle input) consists of many reports, one report per line. Each report is a list of numbers called levels that are separated by spaces. For example:
7 6 4 2 1
1 2 7 8 9
9 7 6 2 1
1 3 2 4 5
8 6 4 4 1
1 3 6 7 9
This example data contains six reports each containing five levels.
The engineers are trying to figure out which reports are safe. The Red-Nosed reactor safety systems can only tolerate levels that are either gradually increasing or gradually decreasing. So, a report only counts as safe if both of the following are true:
The levels are either all increasing or all decreasing.
Any two adjacent levels differ by at least one and at most three.
In the example above, the reports can be found safe or unsafe by checking those rules:
7 6 4 2 1: Safe because the levels are all decreasing by 1 or 2.
1 2 7 8 9: Unsafe because 2 7 is an increase of 5.
9 7 6 2 1: Unsafe because 6 2 is a decrease of 4.
1 3 2 4 5: Unsafe because 1 3 is increasing but 3 2 is decreasing.
8 6 4 4 1: Unsafe because 4 4 is neither an increase or a decrease.
1 3 6 7 9: Safe because the levels are all increasing by 1, 2, or 3.
So, in this example, 2 reports are safe.
Analyze the unusual data from the engineers. How many reports are safe?
| 35
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--- Day 19: Go With The Flow ---
With the Elves well on their way constructing the North Pole base, you turn your attention back to understanding the inner workings of programming the device.
You can't help but notice that the device's opcodes don't contain any flow control like jump instructions. The device's manual goes on to explain:
"In programs where flow control is required, the instruction pointer can be bound to a register so that it can be manipulated directly. This way, setr/seti can function as absolute jumps, addr/addi can function as relative jumps, and other opcodes can cause truly fascinating effects."
This mechanism is achieved through a declaration like #ip 1, which would modify register 1 so that accesses to it let the program indirectly access the instruction pointer itself. To compensate for this kind of binding, there are now six registers (numbered 0 through 5); the five not bound to the instruction pointer behave as normal. Otherwise, the same rules apply as the last time you worked with this device.
When the instruction pointer is bound to a register, its value is written to that register just before each instruction is executed, and the value of that register is written back to the instruction pointer immediately after each instruction finishes execution. Afterward, move to the next instruction by adding one to the instruction pointer, even if the value in the instruction pointer was just updated by an instruction. (Because of this, instructions must effectively set the instruction pointer to the instruction before the one they want executed next.)
The instruction pointer is 0 during the first instruction, 1 during the second, and so on. If the instruction pointer ever causes the device to attempt to load an instruction outside the instructions defined in the program, the program instead immediately halts. The instruction pointer starts at 0.
It turns out that this new information is already proving useful: the CPU in the device is not very powerful, and a background process is occupying most of its time. You dump the background process' declarations and instructions to a file (your puzzle input), making sure to use the names of the opcodes rather than the numbers.
For example, suppose you have the following program:
#ip 0
seti 5 0 1
seti 6 0 2
addi 0 1 0
addr 1 2 3
setr 1 0 0
seti 8 0 4
seti 9 0 5
When executed, the following instructions are executed. Each line contains the value of the instruction pointer at the time the instruction started, the values of the six registers before executing the instructions (in square brackets), the instruction itself, and the values of the six registers after executing the instruction (also in square brackets).
ip=0 [0, 0, 0, 0, 0, 0] seti 5 0 1 [0, 5, 0, 0, 0, 0]
ip=1 [1, 5, 0, 0, 0, 0] seti 6 0 2 [1, 5, 6, 0, 0, 0]
ip=2 [2, 5, 6, 0, 0, 0] addi 0 1 0 [3, 5, 6, 0, 0, 0]
ip=4 [4, 5, 6, 0, 0, 0] setr 1 0 0 [5, 5, 6, 0, 0, 0]
ip=6 [6, 5, 6, 0, 0, 0] seti 9 0 5 [6, 5, 6, 0, 0, 9]
In detail, when running this program, the following events occur:
The first line (#ip 0) indicates that the instruction pointer should be bound to register 0 in this program. This is not an instruction, and so the value of the instruction pointer does not change during the processing of this line.
The instruction pointer contains 0, and so the first instruction is executed (seti 5 0 1). It updates register 0 to the current instruction pointer value (0), sets register 1 to 5, sets the instruction pointer to the value of register 0 (which has no effect, as the instruction did not modify register 0), and then adds one to the instruction pointer.
The instruction pointer contains 1, and so the second instruction, seti 6 0 2, is executed. This is very similar to the instruction before it: 6 is stored in register 2, and the instruction pointer is left with the value 2.
The instruction pointer is 2, which points at the instruction addi 0 1 0. This is like a relative jump: the value of the instruction pointer, 2, is loaded into register 0. Then, addi finds the result of adding the value in register 0 and the value 1, storing the result, 3, back in register 0. Register 0 is then copied back to the instruction pointer, which will cause it to end up 1 larger than it would have otherwise and skip the next instruction (addr 1 2 3) entirely. Finally, 1 is added to the instruction pointer.
The instruction pointer is 4, so the instruction setr 1 0 0 is run. This is like an absolute jump: it copies the value contained in register 1, 5, into register 0, which causes it to end up in the instruction pointer. The instruction pointer is then incremented, leaving it at 6.
The instruction pointer is 6, so the instruction seti 9 0 5 stores 9 into register 5. The instruction pointer is incremented, causing it to point outside the program, and so the program ends.
What value is left in register 0 when the background process halts?
Your puzzle answer was 1430.
--- Part Two ---
A new background process immediately spins up in its place. It appears identical, but on closer inspection, you notice that this time, register 0 started with the value 1.
What value is left in register 0 when this new background process halts?
| 36
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--- Day 21: RPG Simulator 20XX ---
Little Henry Case got a new video game for Christmas. It's an RPG, and he's stuck on a boss. He needs to know what equipment to buy at the shop. He hands you the controller.
In this game, the player (you) and the enemy (the boss) take turns attacking. The player always goes first. Each attack reduces the opponent's hit points by at least 1. The first character at or below 0 hit points loses.
Damage dealt by an attacker each turn is equal to the attacker's damage score minus the defender's armor score. An attacker always does at least 1 damage. So, if the attacker has a damage score of 8, and the defender has an armor score of 3, the defender loses 5 hit points. If the defender had an armor score of 300, the defender would still lose 1 hit point.
Your damage score and armor score both start at zero. They can be increased by buying items in exchange for gold. You start with no items and have as much gold as you need. Your total damage or armor is equal to the sum of those stats from all of your items. You have 100 hit points.
Here is what the item shop is selling:
Weapons: Cost Damage Armor
Dagger 8 4 0
Shortsword 10 5 0
Warhammer 25 6 0
Longsword 40 7 0
Greataxe 74 8 0
Armor: Cost Damage Armor
Leather 13 0 1
Chainmail 31 0 2
Splintmail 53 0 3
Bandedmail 75 0 4
Platemail 102 0 5
Rings: Cost Damage Armor
Damage +1 25 1 0
Damage +2 50 2 0
Damage +3 100 3 0
Defense +1 20 0 1
Defense +2 40 0 2
Defense +3 80 0 3
You must buy exactly one weapon; no dual-wielding. Armor is optional, but you can't use more than one. You can buy 0-2 rings (at most one for each hand). You must use any items you buy. The shop only has one of each item, so you can't buy, for example, two rings of Damage +3.
For example, suppose you have 8 hit points, 5 damage, and 5 armor, and that the boss has 12 hit points, 7 damage, and 2 armor:
The player deals 5-2 = 3 damage; the boss goes down to 9 hit points.
The boss deals 7-5 = 2 damage; the player goes down to 6 hit points.
The player deals 5-2 = 3 damage; the boss goes down to 6 hit points.
The boss deals 7-5 = 2 damage; the player goes down to 4 hit points.
The player deals 5-2 = 3 damage; the boss goes down to 3 hit points.
The boss deals 7-5 = 2 damage; the player goes down to 2 hit points.
The player deals 5-2 = 3 damage; the boss goes down to 0 hit points.
In this scenario, the player wins! (Barely.)
You have 100 hit points. The boss's actual stats are in your puzzle input. What is the least amount of gold you can spend and still win the fight?
| 37
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--- 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?
Your puzzle answer was 16106.
--- Part Two ---
Of course, in order to make this process even more secure, you've also implemented key stretching.
Key stretching forces attackers to spend more time generating hashes. Unfortunately, it forces everyone else to spend more time, too.
To implement key stretching, whenever you generate a hash, before you use it, you first find the MD5 hash of that hash, then the MD5 hash of that hash, and so on, a total of 2016 additional hashings. Always use lowercase hexadecimal representations of hashes.
For example, to find the stretched hash for index 0 and salt abc:
Find the MD5 hash of abc0: 577571be4de9dcce85a041ba0410f29f.
Then, find the MD5 hash of that hash: eec80a0c92dc8a0777c619d9bb51e910.
Then, find the MD5 hash of that hash: 16062ce768787384c81fe17a7a60c7e3.
...repeat many times...
Then, find the MD5 hash of that hash: a107ff634856bb300138cac6568c0f24.
So, the stretched hash for index 0 in this situation is a107ff.... In the end, you find the original hash (one use of MD5), then find the hash-of-the-previous-hash 2016 times, for a total of 2017 uses of MD5.
The rest of the process remains the same, but now the keys are entirely different. Again for salt abc:
The first triple (222, at index 5) has no matching 22222 in the next thousand hashes.
The second triple (eee, at index 10) hash a matching eeeee at index 89, and so it is the first key.
Eventually, index 22551 produces the 64th key (triple fff with matching fffff at index 22859.
Given the actual salt in your puzzle input and using 2016 extra MD5 calls of key stretching, what index now produces your 64th one-time pad key?
| 38
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--- Day 5: Sunny with a Chance of Asteroids ---
You're starting to sweat as the ship makes its way toward Mercury. The Elves suggest that you get the air conditioner working by upgrading your ship computer to support the Thermal Environment Supervision Terminal.
The Thermal Environment Supervision Terminal (TEST) starts by running a diagnostic program (your puzzle input). The TEST diagnostic program will run on your existing Intcode computer after a few modifications:
First, you'll need to add two new instructions:
Opcode 3 takes a single integer as input and saves it to the position given by its only parameter. For example, the instruction 3,50 would take an input value and store it at address 50.
Opcode 4 outputs the value of its only parameter. For example, the instruction 4,50 would output the value at address 50.
Programs that use these instructions will come with documentation that explains what should be connected to the input and output. The program 3,0,4,0,99 outputs whatever it gets as input, then halts.
Second, you'll need to add support for parameter modes:
Each parameter of an instruction is handled based on its parameter mode. Right now, your ship computer already understands parameter mode 0, position mode, which causes the parameter to be interpreted as a position - if the parameter is 50, its value is the value stored at address 50 in memory. Until now, all parameters have been in position mode.
Now, your ship computer will also need to handle parameters in mode 1, immediate mode. In immediate mode, a parameter is interpreted as a value - if the parameter is 50, its value is simply 50.
Parameter modes are stored in the same value as the instruction's opcode. The opcode is a two-digit number based only on the ones and tens digit of the value, that is, the opcode is the rightmost two digits of the first value in an instruction. Parameter modes are single digits, one per parameter, read right-to-left from the opcode: the first parameter's mode is in the hundreds digit, the second parameter's mode is in the thousands digit, the third parameter's mode is in the ten-thousands digit, and so on. Any missing modes are 0.
For example, consider the program 1002,4,3,4,33.
The first instruction, 1002,4,3,4, is a multiply instruction - the rightmost two digits of the first value, 02, indicate opcode 2, multiplication. Then, going right to left, the parameter modes are 0 (hundreds digit), 1 (thousands digit), and 0 (ten-thousands digit, not present and therefore zero):
ABCDE
1002
DE - two-digit opcode, 02 == opcode 2
C - mode of 1st parameter, 0 == position mode
B - mode of 2nd parameter, 1 == immediate mode
A - mode of 3rd parameter, 0 == position mode,
omitted due to being a leading zero
This instruction multiplies its first two parameters. The first parameter, 4 in position mode, works like it did before - its value is the value stored at address 4 (33). The second parameter, 3 in immediate mode, simply has value 3. The result of this operation, 33 * 3 = 99, is written according to the third parameter, 4 in position mode, which also works like it did before - 99 is written to address 4.
Parameters that an instruction writes to will never be in immediate mode.
Finally, some notes:
It is important to remember that the instruction pointer should increase by the number of values in the instruction after the instruction finishes. Because of the new instructions, this amount is no longer always 4.
Integers can be negative: 1101,100,-1,4,0 is a valid program (find 100 + -1, store the result in position 4).
The TEST diagnostic program will start by requesting from the user the ID of the system to test by running an input instruction - provide it 1, the ID for the ship's air conditioner unit.
It will then perform a series of diagnostic tests confirming that various parts of the Intcode computer, like parameter modes, function correctly. For each test, it will run an output instruction indicating how far the result of the test was from the expected value, where 0 means the test was successful. Non-zero outputs mean that a function is not working correctly; check the instructions that were run before the output instruction to see which one failed.
Finally, the program will output a diagnostic code and immediately halt. This final output isn't an error; an output followed immediately by a halt means the program finished. If all outputs were zero except the diagnostic code, the diagnostic program ran successfully.
After providing 1 to the only input instruction and passing all the tests, what diagnostic code does the program produce?
Your puzzle answer was 14522484.
--- Part Two ---
The air conditioner comes online! Its cold air feels good for a while, but then the TEST alarms start to go off. Since the air conditioner can't vent its heat anywhere but back into the spacecraft, it's actually making the air inside the ship warmer.
Instead, you'll need to use the TEST to extend the thermal radiators. Fortunately, the diagnostic program (your puzzle input) is already equipped for this. Unfortunately, your Intcode computer is not.
Your computer is only missing a few opcodes:
Opcode 5 is jump-if-true: if the first parameter is non-zero, it sets the instruction pointer to the value from the second parameter. Otherwise, it does nothing.
Opcode 6 is jump-if-false: if the first parameter is zero, it sets the instruction pointer to the value from the second parameter. Otherwise, it does nothing.
Opcode 7 is less than: if the first parameter is less than the second parameter, it stores 1 in the position given by the third parameter. Otherwise, it stores 0.
Opcode 8 is equals: if the first parameter is equal to the second parameter, it stores 1 in the position given by the third parameter. Otherwise, it stores 0.
Like all instructions, these instructions need to support parameter modes as described above.
Normally, after an instruction is finished, the instruction pointer increases by the number of values in that instruction. However, if the instruction modifies the instruction pointer, that value is used and the instruction pointer is not automatically increased.
For example, here are several programs that take one input, compare it to the value 8, and then produce one output:
3,9,8,9,10,9,4,9,99,-1,8 - Using position mode, consider whether the input is equal to 8; output 1 (if it is) or 0 (if it is not).
3,9,7,9,10,9,4,9,99,-1,8 - Using position mode, consider whether the input is less than 8; output 1 (if it is) or 0 (if it is not).
3,3,1108,-1,8,3,4,3,99 - Using immediate mode, consider whether the input is equal to 8; output 1 (if it is) or 0 (if it is not).
3,3,1107,-1,8,3,4,3,99 - Using immediate mode, consider whether the input is less than 8; output 1 (if it is) or 0 (if it is not).
Here are some jump tests that take an input, then output 0 if the input was zero or 1 if the input was non-zero:
3,12,6,12,15,1,13,14,13,4,13,99,-1,0,1,9 (using position mode)
3,3,1105,-1,9,1101,0,0,12,4,12,99,1 (using immediate mode)
Here's a larger example:
3,21,1008,21,8,20,1005,20,22,107,8,21,20,1006,20,31,
1106,0,36,98,0,0,1002,21,125,20,4,20,1105,1,46,104,
999,1105,1,46,1101,1000,1,20,4,20,1105,1,46,98,99
The above example program uses an input instruction to ask for a single number. The program will then output 999 if the input value is below 8, output 1000 if the input value is equal to 8, or output 1001 if the input value is greater than 8.
This time, when the TEST diagnostic program runs its input instruction to get the ID of the system to test, provide it 5, the ID for the ship's thermal radiator controller. This diagnostic test suite only outputs one number, the diagnostic code.
What is the diagnostic code for system ID 5?
| 39
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--- 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?
Your puzzle answer was 299.
--- Part Two ---
As the expedition reaches the far side of the valley, one of the Elves looks especially dismayed:
He forgot his snacks at the entrance to the valley!
Since you're so good at dodging blizzards, the Elves humbly request that you go back for his snacks. From the same initial conditions, how quickly can you make it from the start to the goal, then back to the start, then back to the goal?
In the above example, the first trip to the goal takes 18 minutes, the trip back to the start takes 23 minutes, and the trip back to the goal again takes 13 minutes, for a total time of 54 minutes.
What is the fewest number of minutes required to reach the goal, go back to the start, then reach the goal again?
| 40
|
--- Day 7: No Space Left On Device ---
You can hear birds chirping and raindrops hitting leaves as the expedition proceeds. Occasionally, you can even hear much louder sounds in the distance; how big do the animals get out here, anyway?
The device the Elves gave you has problems with more than just its communication system. You try to run a system update:
$ system-update --please --pretty-please-with-sugar-on-top
Error: No space left on device
Perhaps you can delete some files to make space for the update?
You browse around the filesystem to assess the situation and save the resulting terminal output (your puzzle input). For example:
$ cd /
$ ls
dir a
14848514 b.txt
8504156 c.dat
dir d
$ cd a
$ ls
dir e
29116 f
2557 g
62596 h.lst
$ cd e
$ ls
584 i
$ cd ..
$ cd ..
$ cd d
$ ls
4060174 j
8033020 d.log
5626152 d.ext
7214296 k
The filesystem consists of a tree of files (plain data) and directories (which can contain other directories or files). The outermost directory is called /. You can navigate around the filesystem, moving into or out of directories and listing the contents of the directory you're currently in.
Within the terminal output, lines that begin with $ are commands you executed, very much like some modern computers:
cd means change directory. This changes which directory is the current directory, but the specific result depends on the argument:
cd x moves in one level: it looks in the current directory for the directory named x and makes it the current directory.
cd .. moves out one level: it finds the directory that contains the current directory, then makes that directory the current directory.
cd / switches the current directory to the outermost directory, /.
ls means list. It prints out all of the files and directories immediately contained by the current directory:
123 abc means that the current directory contains a file named abc with size 123.
dir xyz means that the current directory contains a directory named xyz.
Given the commands and output in the example above, you can determine that the filesystem looks visually like this:
- / (dir)
- a (dir)
- e (dir)
- i (file, size=584)
- f (file, size=29116)
- g (file, size=2557)
- h.lst (file, size=62596)
- b.txt (file, size=14848514)
- c.dat (file, size=8504156)
- d (dir)
- j (file, size=4060174)
- d.log (file, size=8033020)
- d.ext (file, size=5626152)
- k (file, size=7214296)
Here, there are four directories: / (the outermost directory), a and d (which are in /), and e (which is in a). These directories also contain files of various sizes.
Since the disk is full, your first step should probably be to find directories that are good candidates for deletion. To do this, you need to determine the total size of each directory. The total size of a directory is the sum of the sizes of the files it contains, directly or indirectly. (Directories themselves do not count as having any intrinsic size.)
The total sizes of the directories above can be found as follows:
The total size of directory e is 584 because it contains a single file i of size 584 and no other directories.
The directory a has total size 94853 because it contains files f (size 29116), g (size 2557), and h.lst (size 62596), plus file i indirectly (a contains e which contains i).
Directory d has total size 24933642.
As the outermost directory, / contains every file. Its total size is 48381165, the sum of the size of every file.
To begin, find all of the directories with a total size of at most 100000, then calculate the sum of their total sizes. In the example above, these directories are a and e; the sum of their total sizes is 95437 (94853 + 584). (As in this example, this process can count files more than once!)
Find all of the directories with a total size of at most 100000. What is the sum of the total sizes of those directories?
| 41
|
--- Day 12: Garden Groups ---
Why not search for the Chief Historian near the gardener and his massive farm? There's plenty of food, so The Historians grab something to eat while they search.
You're about to settle near a complex arrangement of garden plots when some Elves ask if you can lend a hand. They'd like to set up fences around each region of garden plots, but they can't figure out how much fence they need to order or how much it will cost. They hand you a map (your puzzle input) of the garden plots.
Each garden plot grows only a single type of plant and is indicated by a single letter on your map. When multiple garden plots are growing the same type of plant and are touching (horizontally or vertically), they form a region. For example:
AAAA
BBCD
BBCC
EEEC
This 4x4 arrangement includes garden plots growing five different types of plants (labeled A, B, C, D, and E), each grouped into their own region.
In order to accurately calculate the cost of the fence around a single region, you need to know that region's area and perimeter.
The area of a region is simply the number of garden plots the region contains. The above map's type A, B, and C plants are each in a region of area 4. The type E plants are in a region of area 3; the type D plants are in a region of area 1.
Each garden plot is a square and so has four sides. The perimeter of a region is the number of sides of garden plots in the region that do not touch another garden plot in the same region. The type A and C plants are each in a region with perimeter 10. The type B and E plants are each in a region with perimeter 8. The lone D plot forms its own region with perimeter 4.
Visually indicating the sides of plots in each region that contribute to the perimeter using - and |, the above map's regions' perimeters are measured as follows:
+-+-+-+-+
|A A A A|
+-+-+-+-+ +-+
|D|
+-+-+ +-+ +-+
|B B| |C|
+ + + +-+
|B B| |C C|
+-+-+ +-+ +
|C|
+-+-+-+ +-+
|E E E|
+-+-+-+
Plants of the same type can appear in multiple separate regions, and regions can even appear within other regions. For example:
OOOOO
OXOXO
OOOOO
OXOXO
OOOOO
The above map contains five regions, one containing all of the O garden plots, and the other four each containing a single X plot.
The four X regions each have area 1 and perimeter 4. The region containing 21 type O plants is more complicated; in addition to its outer edge contributing a perimeter of 20, its boundary with each X region contributes an additional 4 to its perimeter, for a total perimeter of 36.
Due to "modern" business practices, the price of fence required for a region is found by multiplying that region's area by its perimeter. The total price of fencing all regions on a map is found by adding together the price of fence for every region on the map.
In the first example, region A has price 4 * 10 = 40, region B has price 4 * 8 = 32, region C has price 4 * 10 = 40, region D has price 1 * 4 = 4, and region E has price 3 * 8 = 24. So, the total price for the first example is 140.
In the second example, the region with all of the O plants has price 21 * 36 = 756, and each of the four smaller X regions has price 1 * 4 = 4, for a total price of 772 (756 + 4 + 4 + 4 + 4).
Here's a larger example:
RRRRIICCFF
RRRRIICCCF
VVRRRCCFFF
VVRCCCJFFF
VVVVCJJCFE
VVIVCCJJEE
VVIIICJJEE
MIIIIIJJEE
MIIISIJEEE
MMMISSJEEE
It contains:
A region of R plants with price 12 * 18 = 216.
A region of I plants with price 4 * 8 = 32.
A region of C plants with price 14 * 28 = 392.
A region of F plants with price 10 * 18 = 180.
A region of V plants with price 13 * 20 = 260.
A region of J plants with price 11 * 20 = 220.
A region of C plants with price 1 * 4 = 4.
A region of E plants with price 13 * 18 = 234.
A region of I plants with price 14 * 22 = 308.
A region of M plants with price 5 * 12 = 60.
A region of S plants with price 3 * 8 = 24.
So, it has a total price of 1930.
What is the total price of fencing all regions on your map?
| 42
|
--- Day 14: Extended Polymerization ---
The incredible pressures at this depth are starting to put a strain on your submarine. The submarine has polymerization equipment that would produce suitable materials to reinforce the submarine, and the nearby volcanically-active caves should even have the necessary input elements in sufficient quantities.
The submarine manual contains instructions for finding the optimal polymer formula; specifically, it offers a polymer template and a list of pair insertion rules (your puzzle input). You just need to work out what polymer would result after repeating the pair insertion process a few times.
For example:
NNCB
CH -> B
HH -> N
CB -> H
NH -> C
HB -> C
HC -> B
HN -> C
NN -> C
BH -> H
NC -> B
NB -> B
BN -> B
BB -> N
BC -> B
CC -> N
CN -> C
The first line is the polymer template - this is the starting point of the process.
The following section defines the pair insertion rules. A rule like AB -> C means that when elements A and B are immediately adjacent, element C should be inserted between them. These insertions all happen simultaneously.
So, starting with the polymer template NNCB, the first step simultaneously considers all three pairs:
The first pair (NN) matches the rule NN -> C, so element C is inserted between the first N and the second N.
The second pair (NC) matches the rule NC -> B, so element B is inserted between the N and the C.
The third pair (CB) matches the rule CB -> H, so element H is inserted between the C and the B.
Note that these pairs overlap: the second element of one pair is the first element of the next pair. Also, because all pairs are considered simultaneously, inserted elements are not considered to be part of a pair until the next step.
After the first step of this process, the polymer becomes NCNBCHB.
Here are the results of a few steps using the above rules:
Template: NNCB
After step 1: NCNBCHB
After step 2: NBCCNBBBCBHCB
After step 3: NBBBCNCCNBBNBNBBCHBHHBCHB
After step 4: NBBNBNBBCCNBCNCCNBBNBBNBBBNBBNBBCBHCBHHNHCBBCBHCB
This polymer grows quickly. After step 5, it has length 97; After step 10, it has length 3073. After step 10, B occurs 1749 times, C occurs 298 times, H occurs 161 times, and N occurs 865 times; taking the quantity of the most common element (B, 1749) and subtracting the quantity of the least common element (H, 161) produces 1749 - 161 = 1588.
Apply 10 steps of pair insertion to the polymer template and find the most and least common elements in the result. What do you get if you take the quantity of the most common element and subtract the quantity of the least common element?
Your puzzle answer was 2915.
--- Part Two ---
The resulting polymer isn't nearly strong enough to reinforce the submarine. You'll need to run more steps of the pair insertion process; a total of 40 steps should do it.
In the above example, the most common element is B (occurring 2192039569602 times) and the least common element is H (occurring 3849876073 times); subtracting these produces 2188189693529.
Apply 40 steps of pair insertion to the polymer template and find the most and least common elements in the result. What do you get if you take the quantity of the most common element and subtract the quantity of the least common element?
| 43
|
--- Day 17: Pyroclastic Flow ---
Your handheld device has located an alternative exit from the cave for you and the elephants. The ground is rumbling almost continuously now, but the strange valves bought you some time. It's definitely getting warmer in here, though.
The tunnels eventually open into a very tall, narrow chamber. Large, oddly-shaped rocks are falling into the chamber from above, presumably due to all the rumbling. If you can't work out where the rocks will fall next, you might be crushed!
The five types of rocks have the following peculiar shapes, where # is rock and . is empty space:
####
.#.
###
.#.
..#
..#
###
#
#
#
#
##
##
The rocks fall in the order shown above: first the - shape, then the + shape, and so on. Once the end of the list is reached, the same order repeats: the - shape falls first, sixth, 11th, 16th, etc.
The rocks don't spin, but they do get pushed around by jets of hot gas coming out of the walls themselves. A quick scan reveals the effect the jets of hot gas will have on the rocks as they fall (your puzzle input).
For example, suppose this was the jet pattern in your cave:
>>><<><>><<<>><>>><<<>>><<<><<<>><>><<>>
In jet patterns, < means a push to the left, while > means a push to the right. The pattern above means that the jets will push a falling rock right, then right, then right, then left, then left, then right, and so on. If the end of the list is reached, it repeats.
The tall, vertical chamber is exactly seven units wide. Each rock appears so that its left edge is two units away from the left wall and its bottom edge is three units above the highest rock in the room (or the floor, if there isn't one).
After a rock appears, it alternates between being pushed by a jet of hot gas one unit (in the direction indicated by the next symbol in the jet pattern) and then falling one unit down. If any movement would cause any part of the rock to move into the walls, floor, or a stopped rock, the movement instead does not occur. If a downward movement would have caused a falling rock to move into the floor or an already-fallen rock, the falling rock stops where it is (having landed on something) and a new rock immediately begins falling.
Drawing falling rocks with @ and stopped rocks with #, the jet pattern in the example above manifests as follows:
The first rock begins falling:
|..@@@@.|
|.......|
|.......|
|.......|
+-------+
Jet of gas pushes rock right:
|...@@@@|
|.......|
|.......|
|.......|
+-------+
Rock falls 1 unit:
|...@@@@|
|.......|
|.......|
+-------+
Jet of gas pushes rock right, but nothing happens:
|...@@@@|
|.......|
|.......|
+-------+
Rock falls 1 unit:
|...@@@@|
|.......|
+-------+
Jet of gas pushes rock right, but nothing happens:
|...@@@@|
|.......|
+-------+
Rock falls 1 unit:
|...@@@@|
+-------+
Jet of gas pushes rock left:
|..@@@@.|
+-------+
Rock falls 1 unit, causing it to come to rest:
|..####.|
+-------+
A new rock begins falling:
|...@...|
|..@@@..|
|...@...|
|.......|
|.......|
|.......|
|..####.|
+-------+
Jet of gas pushes rock left:
|..@....|
|.@@@...|
|..@....|
|.......|
|.......|
|.......|
|..####.|
+-------+
Rock falls 1 unit:
|..@....|
|.@@@...|
|..@....|
|.......|
|.......|
|..####.|
+-------+
Jet of gas pushes rock right:
|...@...|
|..@@@..|
|...@...|
|.......|
|.......|
|..####.|
+-------+
Rock falls 1 unit:
|...@...|
|..@@@..|
|...@...|
|.......|
|..####.|
+-------+
Jet of gas pushes rock left:
|..@....|
|.@@@...|
|..@....|
|.......|
|..####.|
+-------+
Rock falls 1 unit:
|..@....|
|.@@@...|
|..@....|
|..####.|
+-------+
Jet of gas pushes rock right:
|...@...|
|..@@@..|
|...@...|
|..####.|
+-------+
Rock falls 1 unit, causing it to come to rest:
|...#...|
|..###..|
|...#...|
|..####.|
+-------+
A new rock begins falling:
|....@..|
|....@..|
|..@@@..|
|.......|
|.......|
|.......|
|...#...|
|..###..|
|...#...|
|..####.|
+-------+
The moment each of the next few rocks begins falling, you would see this:
|..@....|
|..@....|
|..@....|
|..@....|
|.......|
|.......|
|.......|
|..#....|
|..#....|
|####...|
|..###..|
|...#...|
|..####.|
+-------+
|..@@...|
|..@@...|
|.......|
|.......|
|.......|
|....#..|
|..#.#..|
|..#.#..|
|#####..|
|..###..|
|...#...|
|..####.|
+-------+
|..@@@@.|
|.......|
|.......|
|.......|
|....##.|
|....##.|
|....#..|
|..#.#..|
|..#.#..|
|#####..|
|..###..|
|...#...|
|..####.|
+-------+
|...@...|
|..@@@..|
|...@...|
|.......|
|.......|
|.......|
|.####..|
|....##.|
|....##.|
|....#..|
|..#.#..|
|..#.#..|
|#####..|
|..###..|
|...#...|
|..####.|
+-------+
|....@..|
|....@..|
|..@@@..|
|.......|
|.......|
|.......|
|..#....|
|.###...|
|..#....|
|.####..|
|....##.|
|....##.|
|....#..|
|..#.#..|
|..#.#..|
|#####..|
|..###..|
|...#...|
|..####.|
+-------+
|..@....|
|..@....|
|..@....|
|..@....|
|.......|
|.......|
|.......|
|.....#.|
|.....#.|
|..####.|
|.###...|
|..#....|
|.####..|
|....##.|
|....##.|
|....#..|
|..#.#..|
|..#.#..|
|#####..|
|..###..|
|...#...|
|..####.|
+-------+
|..@@...|
|..@@...|
|.......|
|.......|
|.......|
|....#..|
|....#..|
|....##.|
|....##.|
|..####.|
|.###...|
|..#....|
|.####..|
|....##.|
|....##.|
|....#..|
|..#.#..|
|..#.#..|
|#####..|
|..###..|
|...#...|
|..####.|
+-------+
|..@@@@.|
|.......|
|.......|
|.......|
|....#..|
|....#..|
|....##.|
|##..##.|
|######.|
|.###...|
|..#....|
|.####..|
|....##.|
|....##.|
|....#..|
|..#.#..|
|..#.#..|
|#####..|
|..###..|
|...#...|
|..####.|
+-------+
To prove to the elephants your simulation is accurate, they want to know how tall the tower will get after 2022 rocks have stopped (but before the 2023rd rock begins falling). In this example, the tower of rocks will be 3068 units tall.
How many units tall will the tower of rocks be after 2022 rocks have stopped falling?
Your puzzle answer was 3098.
--- Part Two ---
The elephants are not impressed by your simulation. They demand to know how tall the tower will be after 1000000000000 rocks have stopped! Only then will they feel confident enough to proceed through the cave.
In the example above, the tower would be 1514285714288 units tall!
How tall will the tower be after 1000000000000 rocks have stopped?
| 44
|
--- Day 22: Monkey Map ---
The monkeys take you on a surprisingly easy trail through the jungle. They're even going in roughly the right direction according to your handheld device's Grove Positioning System.
As you walk, the monkeys explain that the grove is protected by a force field. To pass through the force field, you have to enter a password; doing so involves tracing a specific path on a strangely-shaped board.
At least, you're pretty sure that's what you have to do; the elephants aren't exactly fluent in monkey.
The monkeys give you notes that they took when they last saw the password entered (your puzzle input).
For example:
...#
.#..
#...
....
...#.......#
........#...
..#....#....
..........#.
...#....
.....#..
.#......
......#.
10R5L5R10L4R5L5
The first half of the monkeys' notes is a map of the board. It is comprised of a set of open tiles (on which you can move, drawn .) and solid walls (tiles which you cannot enter, drawn #).
The second half is a description of the path you must follow. It consists of alternating numbers and letters:
A number indicates the number of tiles to move in the direction you are facing. If you run into a wall, you stop moving forward and continue with the next instruction.
A letter indicates whether to turn 90 degrees clockwise (R) or counterclockwise (L). Turning happens in-place; it does not change your current tile.
So, a path like 10R5 means "go forward 10 tiles, then turn clockwise 90 degrees, then go forward 5 tiles".
You begin the path in the leftmost open tile of the top row of tiles. Initially, you are facing to the right (from the perspective of how the map is drawn).
If a movement instruction would take you off of the map, you wrap around to the other side of the board. In other words, if your next tile is off of the board, you should instead look in the direction opposite of your current facing as far as you can until you find the opposite edge of the board, then reappear there.
For example, if you are at A and facing to the right, the tile in front of you is marked B; if you are at C and facing down, the tile in front of you is marked D:
...#
.#..
#...
....
...#.D.....#
........#...
B.#....#...A
.....C....#.
...#....
.....#..
.#......
......#.
It is possible for the next tile (after wrapping around) to be a wall; this still counts as there being a wall in front of you, and so movement stops before you actually wrap to the other side of the board.
By drawing the last facing you had with an arrow on each tile you visit, the full path taken by the above example looks like this:
>>v#
.#v.
#.v.
..v.
...#...v..v#
>>>v...>#.>>
..#v...#....
...>>>>v..#.
...#....
.....#..
.#......
......#.
To finish providing the password to this strange input device, you need to determine numbers for your final row, column, and facing as your final position appears from the perspective of the original map. Rows start from 1 at the top and count downward; columns start from 1 at the left and count rightward. (In the above example, row 1, column 1 refers to the empty space with no tile on it in the top-left corner.) Facing is 0 for right (>), 1 for down (v), 2 for left (<), and 3 for up (^). The final password is the sum of 1000 times the row, 4 times the column, and the facing.
In the above example, the final row is 6, the final column is 8, and the final facing is 0. So, the final password is 1000 * 6 + 4 * 8 + 0: 6032.
Follow the path given in the monkeys' notes. What is the final password?
Your puzzle answer was 11464.
--- Part Two ---
As you reach the force field, you think you hear some Elves in the distance. Perhaps they've already arrived?
You approach the strange input device, but it isn't quite what the monkeys drew in their notes. Instead, you are met with a large cube; each of its six faces is a square of 50x50 tiles.
To be fair, the monkeys' map does have six 50x50 regions on it. If you were to carefully fold the map, you should be able to shape it into a cube!
In the example above, the six (smaller, 4x4) faces of the cube are:
1111
1111
1111
1111
222233334444
222233334444
222233334444
222233334444
55556666
55556666
55556666
55556666
You still start in the same position and with the same facing as before, but the wrapping rules are different. Now, if you would walk off the board, you instead proceed around the cube. From the perspective of the map, this can look a little strange. In the above example, if you are at A and move to the right, you would arrive at B facing down; if you are at C and move down, you would arrive at D facing up:
...#
.#..
#...
....
...#.......#
........#..A
..#....#....
.D........#.
...#..B.
.....#..
.#......
..C...#.
Walls still block your path, even if they are on a different face of the cube. If you are at E facing up, your movement is blocked by the wall marked by the arrow:
...#
.#..
-->#...
....
...#..E....#
........#...
..#....#....
..........#.
...#....
.....#..
.#......
......#.
Using the same method of drawing the last facing you had with an arrow on each tile you visit, the full path taken by the above example now looks like this:
>>v#
.#v.
#.v.
..v.
...#..^...v#
.>>>>>^.#.>>
.^#....#....
.^........#.
...#..v.
.....#v.
.#v<<<<.
..v...#.
The final password is still calculated from your final position and facing from the perspective of the map. In this example, the final row is 5, the final column is 7, and the final facing is 3, so the final password is 1000 * 5 + 4 * 7 + 3 = 5031.
Fold the map into a cube, then follow the path given in the monkeys' notes. What is the final password?
| 45
|
--- 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?
Your puzzle answer was 1466.
--- Part Two ---
While you were choosing the best blueprint, the elephants found some food on their own, so you're not in as much of a hurry; you figure you probably have 32 minutes before the wind changes direction again and you'll need to get out of range of the erupting volcano.
Unfortunately, one of the elephants ate most of your blueprint list! Now, only the first three blueprints in your list are intact.
In 32 minutes, the largest number of geodes blueprint 1 (from the example above) can open is 56. 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 ==
1 ore-collecting robot collects 1 ore; you now have 3 ore.
== Minute 4 ==
1 ore-collecting robot collects 1 ore; you now have 4 ore.
== Minute 5 ==
Spend 4 ore to start building an ore-collecting robot.
1 ore-collecting robot collects 1 ore; you now have 1 ore.
The new ore-collecting robot is ready; you now have 2 of them.
== Minute 6 ==
2 ore-collecting robots collect 2 ore; you now have 3 ore.
== Minute 7 ==
Spend 2 ore to start building a clay-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
The new clay-collecting robot is ready; you now have 1 of them.
== Minute 8 ==
Spend 2 ore to start building a clay-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
1 clay-collecting robot collects 1 clay; you now have 1 clay.
The new clay-collecting robot is ready; you now have 2 of them.
== Minute 9 ==
Spend 2 ore to start building a clay-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
2 clay-collecting robots collect 2 clay; you now have 3 clay.
The new clay-collecting robot is ready; you now have 3 of them.
== Minute 10 ==
Spend 2 ore to start building a clay-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
3 clay-collecting robots collect 3 clay; you now have 6 clay.
The new clay-collecting robot is ready; you now have 4 of them.
== Minute 11 ==
Spend 2 ore to start building a clay-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
4 clay-collecting robots collect 4 clay; you now have 10 clay.
The new clay-collecting robot is ready; you now have 5 of them.
== Minute 12 ==
Spend 2 ore to start building a clay-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
5 clay-collecting robots collect 5 clay; you now have 15 clay.
The new clay-collecting robot is ready; you now have 6 of them.
== Minute 13 ==
Spend 2 ore to start building a clay-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
6 clay-collecting robots collect 6 clay; you now have 21 clay.
The new clay-collecting robot is ready; you now have 7 of them.
== Minute 14 ==
Spend 3 ore and 14 clay to start building an obsidian-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 2 ore.
7 clay-collecting robots collect 7 clay; you now have 14 clay.
The new obsidian-collecting robot is ready; you now have 1 of them.
== Minute 15 ==
2 ore-collecting robots collect 2 ore; you now have 4 ore.
7 clay-collecting robots collect 7 clay; you now have 21 clay.
1 obsidian-collecting robot collects 1 obsidian; you now have 1 obsidian.
== Minute 16 ==
Spend 3 ore and 14 clay to start building an obsidian-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
7 clay-collecting robots collect 7 clay; you now have 14 clay.
1 obsidian-collecting robot collects 1 obsidian; you now have 2 obsidian.
The new obsidian-collecting robot is ready; you now have 2 of them.
== Minute 17 ==
Spend 3 ore and 14 clay to start building an obsidian-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 2 ore.
7 clay-collecting robots collect 7 clay; you now have 7 clay.
2 obsidian-collecting robots collect 2 obsidian; you now have 4 obsidian.
The new obsidian-collecting robot is ready; you now have 3 of them.
== Minute 18 ==
2 ore-collecting robots collect 2 ore; you now have 4 ore.
7 clay-collecting robots collect 7 clay; you now have 14 clay.
3 obsidian-collecting robots collect 3 obsidian; you now have 7 obsidian.
== Minute 19 ==
Spend 3 ore and 14 clay to start building an obsidian-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
7 clay-collecting robots collect 7 clay; you now have 7 clay.
3 obsidian-collecting robots collect 3 obsidian; you now have 10 obsidian.
The new obsidian-collecting robot is ready; you now have 4 of them.
== Minute 20 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
2 ore-collecting robots collect 2 ore; you now have 3 ore.
7 clay-collecting robots collect 7 clay; you now have 14 clay.
4 obsidian-collecting robots collect 4 obsidian; you now have 7 obsidian.
The new geode-cracking robot is ready; you now have 1 of them.
== Minute 21 ==
Spend 3 ore and 14 clay to start building an obsidian-collecting robot.
2 ore-collecting robots collect 2 ore; you now have 2 ore.
7 clay-collecting robots collect 7 clay; you now have 7 clay.
4 obsidian-collecting robots collect 4 obsidian; you now have 11 obsidian.
1 geode-cracking robot cracks 1 geode; you now have 1 open geode.
The new obsidian-collecting robot is ready; you now have 5 of them.
== Minute 22 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
2 ore-collecting robots collect 2 ore; you now have 2 ore.
7 clay-collecting robots collect 7 clay; you now have 14 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 9 obsidian.
1 geode-cracking robot cracks 1 geode; you now have 2 open geodes.
The new geode-cracking robot is ready; you now have 2 of them.
== Minute 23 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
2 ore-collecting robots collect 2 ore; you now have 2 ore.
7 clay-collecting robots collect 7 clay; you now have 21 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 7 obsidian.
2 geode-cracking robots crack 2 geodes; you now have 4 open geodes.
The new geode-cracking robot is ready; you now have 3 of them.
== Minute 24 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
2 ore-collecting robots collect 2 ore; you now have 2 ore.
7 clay-collecting robots collect 7 clay; you now have 28 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 5 obsidian.
3 geode-cracking robots crack 3 geodes; you now have 7 open geodes.
The new geode-cracking robot is ready; you now have 4 of them.
== Minute 25 ==
2 ore-collecting robots collect 2 ore; you now have 4 ore.
7 clay-collecting robots collect 7 clay; you now have 35 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 10 obsidian.
4 geode-cracking robots crack 4 geodes; you now have 11 open geodes.
== Minute 26 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
2 ore-collecting robots collect 2 ore; you now have 4 ore.
7 clay-collecting robots collect 7 clay; you now have 42 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 8 obsidian.
4 geode-cracking robots crack 4 geodes; you now have 15 open geodes.
The new geode-cracking robot is ready; you now have 5 of them.
== Minute 27 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
2 ore-collecting robots collect 2 ore; you now have 4 ore.
7 clay-collecting robots collect 7 clay; you now have 49 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 6 obsidian.
5 geode-cracking robots crack 5 geodes; you now have 20 open geodes.
The new geode-cracking robot is ready; you now have 6 of them.
== Minute 28 ==
2 ore-collecting robots collect 2 ore; you now have 6 ore.
7 clay-collecting robots collect 7 clay; you now have 56 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 11 obsidian.
6 geode-cracking robots crack 6 geodes; you now have 26 open geodes.
== Minute 29 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
2 ore-collecting robots collect 2 ore; you now have 6 ore.
7 clay-collecting robots collect 7 clay; you now have 63 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 9 obsidian.
6 geode-cracking robots crack 6 geodes; you now have 32 open geodes.
The new geode-cracking robot is ready; you now have 7 of them.
== Minute 30 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
2 ore-collecting robots collect 2 ore; you now have 6 ore.
7 clay-collecting robots collect 7 clay; you now have 70 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 7 obsidian.
7 geode-cracking robots crack 7 geodes; you now have 39 open geodes.
The new geode-cracking robot is ready; you now have 8 of them.
== Minute 31 ==
Spend 2 ore and 7 obsidian to start building a geode-cracking robot.
2 ore-collecting robots collect 2 ore; you now have 6 ore.
7 clay-collecting robots collect 7 clay; you now have 77 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 5 obsidian.
8 geode-cracking robots crack 8 geodes; you now have 47 open geodes.
The new geode-cracking robot is ready; you now have 9 of them.
== Minute 32 ==
2 ore-collecting robots collect 2 ore; you now have 8 ore.
7 clay-collecting robots collect 7 clay; you now have 84 clay.
5 obsidian-collecting robots collect 5 obsidian; you now have 10 obsidian.
9 geode-cracking robots crack 9 geodes; you now have 56 open geodes.
However, blueprint 2 from the example above is still better; using it, the largest number of geodes you could open in 32 minutes is 62.
You no longer have enough blueprints to worry about quality levels. Instead, for each of the first three blueprints, determine the largest number of geodes you could open; then, multiply these three values together.
Don't worry about quality levels; instead, just determine the largest number of geodes you could open using each of the first three blueprints. What do you get if you multiply these numbers together?
| 46
|
--- Day 19: A Series of Tubes ---
Somehow, a network packet got lost and ended up here. It's trying to follow a routing diagram (your puzzle input), but it's confused about where to go.
Its starting point is just off the top of the diagram. Lines (drawn with |, -, and +) show the path it needs to take, starting by going down onto the only line connected to the top of the diagram. It needs to follow this path until it reaches the end (located somewhere within the diagram) and stop there.
Sometimes, the lines cross over each other; in these cases, it needs to continue going the same direction, and only turn left or right when there's no other option. In addition, someone has left letters on the line; these also don't change its direction, but it can use them to keep track of where it's been. For example:
|
| +--+
A | C
F---|----E|--+
| | | D
+B-+ +--+
Given this diagram, the packet needs to take the following path:
Starting at the only line touching the top of the diagram, it must go down, pass through A, and continue onward to the first +.
Travel right, up, and right, passing through B in the process.
Continue down (collecting C), right, and up (collecting D).
Finally, go all the way left through E and stopping at F.
Following the path to the end, the letters it sees on its path are ABCDEF.
The little packet looks up at you, hoping you can help it find the way. What letters will it see (in the order it would see them) if it follows the path? (The routing diagram is very wide; make sure you view it without line wrapping.)
Your puzzle answer was VTWBPYAQFU.
--- Part Two ---
The packet is curious how many steps it needs to go.
For example, using the same routing diagram from the example above...
|
| +--+
A | C
F---|--|-E---+
| | | D
+B-+ +--+
...the packet would go:
6 steps down (including the first line at the top of the diagram).
3 steps right.
4 steps up.
3 steps right.
4 steps down.
3 steps right.
2 steps up.
13 steps left (including the F it stops on).
This would result in a total of 38 steps.
How many steps does the packet need to go?
| 47
|
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