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# Day 1: Sonar Sweep
[https://adventofcode.com/2021/day/1](https://adventofcode.com/2021/day/1)
## Description
### Part One
You're minding your own business on a ship at sea when the overboard alarm goes off! You rush to see if you can help. Apparently, one of the Elves tripped and accidentally sent the sleigh keys flying into the ocean!
Before you know it, you're inside a submarine the Elves keep ready for situations like this. It's covered in Christmas lights (because of course it is), and it even has an experimental antenna that should be able to track the keys if you can boost its signal strength high enough; there's a little meter that indicates the antenna's signal strength by displaying 0-50 _stars_.
Your instincts tell you that in order to save Christmas, you'll need to get all _fifty stars_ by December 25th.
Collect stars by solving puzzles. Two puzzles will be made available on each day in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants _one star_. Good luck!
As the submarine drops below the surface of the ocean, it automatically performs a sonar sweep of the nearby sea floor. On a small screen, the sonar sweep report (your puzzle input) appears: each line is a measurement of the sea floor depth as the sweep looks further and further away from the submarine.
For example, suppose you had the following report:
199
200
208
210
200
207
240
269
260
263
This report indicates that, scanning outward from the submarine, the sonar sweep found depths of `199`, `200`, `208`, `210`, and so on.
The first order of business is to figure out how quickly the depth increases, just so you know what you're dealing with - you never know if the keys will get <span title="Does this premise seem fishy to you?">carried into deeper water</span> by an ocean current or a fish or something.
To do this, count _the number of times a depth measurement increases_ from the previous measurement. (There is no measurement before the first measurement.) In the example above, the changes are as follows:
199 (N/A - no previous measurement)
200 (increased)
208 (increased)
210 (increased)
200 (decreased)
207 (increased)
240 (increased)
269 (increased)
260 (decreased)
263 (increased)
In this example, there are _`7`_ measurements that are larger than the previous measurement.
_How many measurements are larger than the previous measurement?_
### Part Two
Considering every single measurement isn't as useful as you expected: there's just too much noise in the data.
Instead, consider sums of a _three-measurement sliding window_. Again considering the above example:
199 A
200 A B
208 A B C
210 B C D
200 E C D
207 E F D
240 E F G
269 F G H
260 G H
263 H
Start by comparing the first and second three-measurement windows. The measurements in the first window are marked `A` (`199`, `200`, `208`); their sum is `199 + 200 + 208 = 607`. The second window is marked `B` (`200`, `208`, `210`); its sum is `618`. The sum of measurements in the second window is larger than the sum of the first, so this first comparison _increased_.
Your goal now is to count _the number of times the sum of measurements in this sliding window increases_ from the previous sum. So, compare `A` with `B`, then compare `B` with `C`, then `C` with `D`, and so on. Stop when there aren't enough measurements left to create a new three-measurement sum.
In the above example, the sum of each three-measurement window is as follows:
A: 607 (N/A - no previous sum)
B: 618 (increased)
C: 618 (no change)
D: 617 (decreased)
E: 647 (increased)
F: 716 (increased)
G: 769 (increased)
H: 792 (increased)
In this example, there are _`5`_ sums that are larger than the previous sum.
Consider sums of a three-measurement sliding window. _How many sums are larger than the previous sum?_

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# Day 2: Dive!
[https://adventofcode.com/2021/day/2](https://adventofcode.com/2021/day/2)
## Description
### Part One
Now, you need to figure out how to <span title="Tank, I need a pilot program for a B212 helicopter.">pilot this thing</span>.
It seems like the submarine can take a series of commands like `forward 1`, `down 2`, or `up 3`:
* `forward X` increases the horizontal position by `X` units.
* `down X` _increases_ the depth by `X` units.
* `up X` _decreases_ the depth by `X` units.
Note that since you're on a submarine, `down` and `up` affect your _depth_, and so they have the opposite result of what you might expect.
The submarine seems to already have a planned course (your puzzle input). You should probably figure out where it's going. For example:
forward 5
down 5
forward 8
up 3
down 8
forward 2
Your horizontal position and depth both start at `0`. The steps above would then modify them as follows:
* `forward 5` adds `5` to your horizontal position, a total of `5`.
* `down 5` adds `5` to your depth, resulting in a value of `5`.
* `forward 8` adds `8` to your horizontal position, a total of `13`.
* `up 3` decreases your depth by `3`, resulting in a value of `2`.
* `down 8` adds `8` to your depth, resulting in a value of `10`.
* `forward 2` adds `2` to your horizontal position, a total of `15`.
After following these instructions, you would have a horizontal position of `15` and a depth of `10`. (Multiplying these together produces _`150`_.)
Calculate the horizontal position and depth you would have after following the planned course. _What do you get if you multiply your final horizontal position by your final depth?_
### Part Two
Based on your calculations, the planned course doesn't seem to make any sense. You find the submarine manual and discover that the process is actually slightly more complicated.
In addition to horizontal position and depth, you'll also need to track a third value, _aim_, which also starts at `0`. The commands also mean something entirely different than you first thought:
* `down X` _increases_ your aim by `X` units.
* `up X` _decreases_ your aim by `X` units.
* `forward X` does two things:
* It increases your horizontal position by `X` units.
* It increases your depth by your aim _multiplied by_ `X`.
Again note that since you're on a submarine, `down` and `up` do the opposite of what you might expect: "down" means aiming in the positive direction.
Now, the above example does something different:
* `forward 5` adds `5` to your horizontal position, a total of `5`. Because your aim is `0`, your depth does not change.
* `down 5` adds `5` to your aim, resulting in a value of `5`.
* `forward 8` adds `8` to your horizontal position, a total of `13`. Because your aim is `5`, your depth increases by `8*5=40`.
* `up 3` decreases your aim by `3`, resulting in a value of `2`.
* `down 8` adds `8` to your aim, resulting in a value of `10`.
* `forward 2` adds `2` to your horizontal position, a total of `15`. Because your aim is `10`, your depth increases by `2*10=20` to a total of `60`.
After following these new instructions, you would have a horizontal position of `15` and a depth of `60`. (Multiplying these produces _`900`_.)
Using this new interpretation of the commands, calculate the horizontal position and depth you would have after following the planned course. _What do you get if you multiply your final horizontal position by your final depth?_

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# Day 3: Binary Diagnostic
[https://adventofcode.com/2021/day/3](https://adventofcode.com/2021/day/3)
## Description
### Part One
The submarine has been making some <span title="Turns out oceans are heavy.">odd creaking noises</span>, so you ask it to produce a diagnostic report just in case.
The diagnostic report (your puzzle input) consists of a list of binary numbers which, when decoded properly, can tell you many useful things about the conditions of the submarine. The first parameter to check is the _power consumption_.
You need to use the binary numbers in the diagnostic report to generate two new binary numbers (called the _gamma rate_ and the _epsilon rate_). The power consumption can then be found by multiplying the gamma rate by the epsilon rate.
Each bit in the gamma rate can be determined by finding the _most common bit in the corresponding position_ of all numbers in the diagnostic report. For example, given the following diagnostic report:
00100
11110
10110
10111
10101
01111
00111
11100
10000
11001
00010
01010
Considering only the first bit of each number, there are five `0` bits and seven `1` bits. Since the most common bit is `1`, the first bit of the gamma rate is `1`.
The most common second bit of the numbers in the diagnostic report is `0`, so the second bit of the gamma rate is `0`.
The most common value of the third, fourth, and fifth bits are `1`, `1`, and `0`, respectively, and so the final three bits of the gamma rate are `110`.
So, the gamma rate is the binary number `10110`, or _`22`_ in decimal.
The epsilon rate is calculated in a similar way; rather than use the most common bit, the least common bit from each position is used. So, the epsilon rate is `01001`, or _`9`_ in decimal. Multiplying the gamma rate (`22`) by the epsilon rate (`9`) produces the power consumption, _`198`_.
Use the binary numbers in your diagnostic report to calculate the gamma rate and epsilon rate, then multiply them together. _What is the power consumption of the submarine?_ (Be sure to represent your answer in decimal, not binary.)
### Part Two
Next, you should verify the _life support rating_, which can be determined by multiplying the _oxygen generator rating_ by the _CO2 scrubber rating_.
Both the oxygen generator rating and the CO2 scrubber rating are values that can be found in your diagnostic report - finding them is the tricky part. Both values are located using a similar process that involves filtering out values until only one remains. Before searching for either rating value, start with the full list of binary numbers from your diagnostic report and _consider just the first bit_ of those numbers. Then:
* Keep only numbers selected by the _bit criteria_ for the type of rating value for which you are searching. Discard numbers which do not match the bit criteria.
* If you only have one number left, stop; this is the rating value for which you are searching.
* Otherwise, repeat the process, considering the next bit to the right.
The _bit criteria_ depends on which type of rating value you want to find:
* To find _oxygen generator rating_, determine the _most common_ value (`0` or `1`) in the current bit position, and keep only numbers with that bit in that position. If `0` and `1` are equally common, keep values with a _`1`_ in the position being considered.
* To find _CO2 scrubber rating_, determine the _least common_ value (`0` or `1`) in the current bit position, and keep only numbers with that bit in that position. If `0` and `1` are equally common, keep values with a _`0`_ in the position being considered.
For example, to determine the _oxygen generator rating_ value using the same example diagnostic report from above:
* Start with all 12 numbers and consider only the first bit of each number. There are more `1` bits (7) than `0` bits (5), so keep only the 7 numbers with a `1` in the first position: `11110`, `10110`, `10111`, `10101`, `11100`, `10000`, and `11001`.
* Then, consider the second bit of the 7 remaining numbers: there are more `0` bits (4) than `1` bits (3), so keep only the 4 numbers with a `0` in the second position: `10110`, `10111`, `10101`, and `10000`.
* In the third position, three of the four numbers have a `1`, so keep those three: `10110`, `10111`, and `10101`.
* In the fourth position, two of the three numbers have a `1`, so keep those two: `10110` and `10111`.
* In the fifth position, there are an equal number of `0` bits and `1` bits (one each). So, to find the _oxygen generator rating_, keep the number with a `1` in that position: `10111`.
* As there is only one number left, stop; the _oxygen generator rating_ is `10111`, or _`23`_ in decimal.
Then, to determine the _CO2 scrubber rating_ value from the same example above:
* Start again with all 12 numbers and consider only the first bit of each number. There are fewer `0` bits (5) than `1` bits (7), so keep only the 5 numbers with a `0` in the first position: `00100`, `01111`, `00111`, `00010`, and `01010`.
* Then, consider the second bit of the 5 remaining numbers: there are fewer `1` bits (2) than `0` bits (3), so keep only the 2 numbers with a `1` in the second position: `01111` and `01010`.
* In the third position, there are an equal number of `0` bits and `1` bits (one each). So, to find the _CO2 scrubber rating_, keep the number with a `0` in that position: `01010`.
* As there is only one number left, stop; the _CO2 scrubber rating_ is `01010`, or _`10`_ in decimal.
Finally, to find the life support rating, multiply the oxygen generator rating (`23`) by the CO2 scrubber rating (`10`) to get _`230`_.
Use the binary numbers in your diagnostic report to calculate the oxygen generator rating and CO2 scrubber rating, then multiply them together. _What is the life support rating of the submarine?_ (Be sure to represent your answer in decimal, not binary.)

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# Day 4: Giant Squid
[https://adventofcode.com/2021/day/4](https://adventofcode.com/2021/day/4)
## Description
### Part One
You're already almost 1.5km (almost a mile) below the surface of the ocean, already so deep that you can't see any sunlight. What you _can_ see, however, is a giant squid that has attached itself to the outside of your submarine.
Maybe it wants to play [bingo](https://en.wikipedia.org/wiki/Bingo_(American_version))?
Bingo is played on a set of boards each consisting of a 5x5 grid of numbers. Numbers are chosen at random, and the chosen number is _marked_ on all boards on which it appears. (Numbers may not appear on all boards.) If all numbers in any row or any column of a board are marked, that board _wins_. (Diagonals don't count.)
The submarine has a _bingo subsystem_ to help passengers (currently, you and the giant squid) pass the time. It automatically generates a random order in which to draw numbers and a random set of boards (your puzzle input). For example:
7,4,9,5,11,17,23,2,0,14,21,24,10,16,13,6,15,25,12,22,18,20,8,19,3,26,1
22 13 17 11 0
8 2 23 4 24
21 9 14 16 7
6 10 3 18 5
1 12 20 15 19
3 15 0 2 22
9 18 13 17 5
19 8 7 25 23
20 11 10 24 4
14 21 16 12 6
14 21 17 24 4
10 16 15 9 19
18 8 23 26 20
22 11 13 6 5
2 0 12 3 7
After the first five numbers are drawn (`7`, `4`, `9`, `5`, and `11`), there are no winners, but the boards are marked as follows (shown here adjacent to each other to save space):
22 13 17 11 0 3 15 0 2 22 14 21 17 24 4
8 2 23 4 24 9 18 13 17 5 10 16 15 9 19
21 9 14 16 7 19 8 7 25 23 18 8 23 26 20
6 10 3 18 5 20 11 10 24 4 22 11 13 6 5
1 12 20 15 19 14 21 16 12 6 2 0 12 3 7
After the next six numbers are drawn (`17`, `23`, `2`, `0`, `14`, and `21`), there are still no winners:
22 13 17 11 0 3 15 0 2 22 14 21 17 24 4
8 2 23 4 24 9 18 13 17 5 10 16 15 9 19
21 9 14 16 7 19 8 7 25 23 18 8 23 26 20
6 10 3 18 5 20 11 10 24 4 22 11 13 6 5
1 12 20 15 19 14 21 16 12 6 2 0 12 3 7
Finally, `24` is drawn:
22 13 17 11 0 3 15 0 2 22 14 21 17 24 4
8 2 23 4 24 9 18 13 17 5 10 16 15 9 19
21 9 14 16 7 19 8 7 25 23 18 8 23 26 20
6 10 3 18 5 20 11 10 24 4 22 11 13 6 5
1 12 20 15 19 14 21 16 12 6 2 0 12 3 7
At this point, the third board _wins_ because it has at least one complete row or column of marked numbers (in this case, the entire top row is marked: _`14 21 17 24 4`_).
The _score_ of the winning board can now be calculated. Start by finding the _sum of all unmarked numbers_ on that board; in this case, the sum is `188`. Then, multiply that sum by _the number that was just called_ when the board won, `24`, to get the final score, `188 * 24 = 4512`.
To guarantee victory against the giant squid, figure out which board will win first. _What will your final score be if you choose that board?_
### Part Two
On the other hand, it might be wise to try a different strategy: <span title="That's 'cuz a submarine don't pull things' antennas out of their sockets when they lose. Giant squid are known to do that.">let the giant squid win</span>.
You aren't sure how many bingo boards a giant squid could play at once, so rather than waste time counting its arms, the safe thing to do is to _figure out which board will win last_ and choose that one. That way, no matter which boards it picks, it will win for sure.
In the above example, the second board is the last to win, which happens after `13` is eventually called and its middle column is completely marked. If you were to keep playing until this point, the second board would have a sum of unmarked numbers equal to `148` for a final score of `148 * 13 = 1924`.
Figure out which board will win last. _Once it wins, what would its final score be?_

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# Day 5: Hydrothermal Venture
[https://adventofcode.com/2021/day/5](https://adventofcode.com/2021/day/5)
## Description
### Part One
You come across a field of [hydrothermal vents](https://en.wikipedia.org/wiki/Hydrothermal_vent) on the ocean floor! These vents constantly produce large, opaque clouds, so it would be best to avoid them if possible.
They tend to form in _lines_; the submarine helpfully produces a list of nearby <span title="Maybe they're Bresenham vents.">lines of vents</span> (your puzzle input) for you to review. For example:
0,9 -> 5,9
8,0 -> 0,8
9,4 -> 3,4
2,2 -> 2,1
7,0 -> 7,4
6,4 -> 2,0
0,9 -> 2,9
3,4 -> 1,4
0,0 -> 8,8
5,5 -> 8,2
Each line of vents is given as a line segment in the format `x1,y1 -> x2,y2` where `x1`,`y1` are the coordinates of one end the line segment and `x2`,`y2` are the coordinates of the other end. These line segments include the points at both ends. In other words:
* An entry like `1,1 -> 1,3` covers points `1,1`, `1,2`, and `1,3`.
* An entry like `9,7 -> 7,7` covers points `9,7`, `8,7`, and `7,7`.
For now, _only consider horizontal and vertical lines_: lines where either `x1 = x2` or `y1 = y2`.
So, the horizontal and vertical lines from the above list would produce the following diagram:
.......1..
..1....1..
..1....1..
.......1..
.112111211
..........
..........
..........
..........
222111....
In this diagram, the top left corner is `0,0` and the bottom right corner is `9,9`. Each position is shown as _the number of lines which cover that point_ or `.` if no line covers that point. The top-left pair of `1`s, for example, comes from `2,2 -> 2,1`; the very bottom row is formed by the overlapping lines `0,9 -> 5,9` and `0,9 -> 2,9`.
To avoid the most dangerous areas, you need to determine _the number of points where at least two lines overlap_. In the above example, this is anywhere in the diagram with a `2` or larger - a total of _`5`_ points.
Consider only horizontal and vertical lines. _At how many points do at least two lines overlap?_
### Part Two
Unfortunately, considering only horizontal and vertical lines doesn't give you the full picture; you need to also consider _diagonal lines_.
Because of the limits of the hydrothermal vent mapping system, the lines in your list will only ever be horizontal, vertical, or a diagonal line at exactly 45 degrees. In other words:
* An entry like `1,1 -> 3,3` covers points `1,1`, `2,2`, and `3,3`.
* An entry like `9,7 -> 7,9` covers points `9,7`, `8,8`, and `7,9`.
Considering all lines from the above example would now produce the following diagram:
1.1....11.
.111...2..
..2.1.111.
...1.2.2..
.112313211
...1.2....
..1...1...
.1.....1..
1.......1.
222111....
You still need to determine _the number of points where at least two lines overlap_. In the above example, this is still anywhere in the diagram with a `2` or larger - now a total of _`12`_ points.
Consider all of the lines. _At how many points do at least two lines overlap?_

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# Day 6: Lanternfish
[https://adventofcode.com/2021/day/6](https://adventofcode.com/2021/day/6)
## Description
### Part One
The sea floor is getting steeper. Maybe the sleigh keys got carried this way?
A massive school of glowing [lanternfish](https://en.wikipedia.org/wiki/Lanternfish) swims past. They must spawn quickly to reach such large numbers - maybe _exponentially_ quickly? You should model their growth rate to be sure.
Although you know nothing about this specific species of lanternfish, you make some guesses about their attributes. Surely, <span title="I heard you like lanternfish.">each lanternfish creates a new lanternfish</span> once every _7_ days.
However, this process isn't necessarily synchronized between every lanternfish - one lanternfish might have 2 days left until it creates another lanternfish, while another might have 4. So, you can model each fish as a single number that represents _the number of days until it creates a new lanternfish_.
Furthermore, you reason, a _new_ lanternfish would surely need slightly longer before it's capable of producing more lanternfish: two more days for its first cycle.
So, suppose you have a lanternfish with an internal timer value of `3`:
* After one day, its internal timer would become `2`.
* After another day, its internal timer would become `1`.
* After another day, its internal timer would become `0`.
* After another day, its internal timer would reset to `6`, and it would create a _new_ lanternfish with an internal timer of `8`.
* After another day, the first lanternfish would have an internal timer of `5`, and the second lanternfish would have an internal timer of `7`.
A lanternfish that creates a new fish resets its timer to `6`, _not `7`_ (because `0` is included as a valid timer value). The new lanternfish starts with an internal timer of `8` and does not start counting down until the next day.
Realizing what you're trying to do, the submarine automatically produces a list of the ages of several hundred nearby lanternfish (your puzzle input). For example, suppose you were given the following list:
3,4,3,1,2
This list means that the first fish has an internal timer of `3`, the second fish has an internal timer of `4`, and so on until the fifth fish, which has an internal timer of `2`. Simulating these fish over several days would proceed as follows:
Initial state: 3,4,3,1,2
After 1 day: 2,3,2,0,1
After 2 days: 1,2,1,6,0,8
After 3 days: 0,1,0,5,6,7,8
After 4 days: 6,0,6,4,5,6,7,8,8
After 5 days: 5,6,5,3,4,5,6,7,7,8
After 6 days: 4,5,4,2,3,4,5,6,6,7
After 7 days: 3,4,3,1,2,3,4,5,5,6
After 8 days: 2,3,2,0,1,2,3,4,4,5
After 9 days: 1,2,1,6,0,1,2,3,3,4,8
After 10 days: 0,1,0,5,6,0,1,2,2,3,7,8
After 11 days: 6,0,6,4,5,6,0,1,1,2,6,7,8,8,8
After 12 days: 5,6,5,3,4,5,6,0,0,1,5,6,7,7,7,8,8
After 13 days: 4,5,4,2,3,4,5,6,6,0,4,5,6,6,6,7,7,8,8
After 14 days: 3,4,3,1,2,3,4,5,5,6,3,4,5,5,5,6,6,7,7,8
After 15 days: 2,3,2,0,1,2,3,4,4,5,2,3,4,4,4,5,5,6,6,7
After 16 days: 1,2,1,6,0,1,2,3,3,4,1,2,3,3,3,4,4,5,5,6,8
After 17 days: 0,1,0,5,6,0,1,2,2,3,0,1,2,2,2,3,3,4,4,5,7,8
After 18 days: 6,0,6,4,5,6,0,1,1,2,6,0,1,1,1,2,2,3,3,4,6,7,8,8,8,8
Each day, a `0` becomes a `6` and adds a new `8` to the end of the list, while each other number decreases by 1 if it was present at the start of the day.
In this example, after 18 days, there are a total of `26` fish. After 80 days, there would be a total of _`5934`_.
Find a way to simulate lanternfish. _How many lanternfish would there be after 80 days?_
### Part Two
Suppose the lanternfish live forever and have unlimited food and space. Would they take over the entire ocean?
After 256 days in the example above, there would be a total of _`26984457539`_ lanternfish!
_How many lanternfish would there be after 256 days?_

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# Day 7: The Treachery of Whales
[https://adventofcode.com/2021/day/7](https://adventofcode.com/2021/day/7)
## Description
### Part One
A giant [whale](https://en.wikipedia.org/wiki/Sperm_whale) has decided your submarine is its next meal, and it's much faster than you are. There's nowhere to run!
Suddenly, a swarm of crabs (each in its own tiny submarine - it's too deep for them otherwise) zooms in to rescue you! They seem to be preparing to blast a hole in the ocean floor; sensors indicate a _massive underground cave system_ just beyond where they're aiming!
The crab submarines all need to be aligned before they'll have enough power to blast a large enough hole for your submarine to get through. However, it doesn't look like they'll be aligned before the whale catches you! Maybe you can help?
There's one major catch - crab submarines can only move horizontally.
You quickly make a list of _the horizontal position of each crab_ (your puzzle input). Crab submarines have limited fuel, so you need to find a way to make all of their horizontal positions match while requiring them to spend as little fuel as possible.
For example, consider the following horizontal positions:
16,1,2,0,4,2,7,1,2,14
This means there's a crab with horizontal position `16`, a crab with horizontal position `1`, and so on.
Each change of 1 step in horizontal position of a single crab costs 1 fuel. You could choose any horizontal position to align them all on, but the one that costs the least fuel is horizontal position `2`:
* Move from `16` to `2`: `14` fuel
* Move from `1` to `2`: `1` fuel
* Move from `2` to `2`: `0` fuel
* Move from `0` to `2`: `2` fuel
* Move from `4` to `2`: `2` fuel
* Move from `2` to `2`: `0` fuel
* Move from `7` to `2`: `5` fuel
* Move from `1` to `2`: `1` fuel
* Move from `2` to `2`: `0` fuel
* Move from `14` to `2`: `12` fuel
This costs a total of _`37`_ fuel. This is the cheapest possible outcome; more expensive outcomes include aligning at position `1` (`41` fuel), position `3` (`39` fuel), or position `10` (`71` fuel).
Determine the horizontal position that the crabs can align to using the least fuel possible. _How much fuel must they spend to align to that position?_
### Part Two
The crabs don't seem interested in your proposed solution. Perhaps you misunderstand crab engineering?
As it turns out, crab submarine engines <span title="This appears to be due to the modial interaction of magneto-reluctance and capacitive duractance.">don't burn fuel at a constant rate</span>. Instead, each change of 1 step in horizontal position costs 1 more unit of fuel than the last: the first step costs `1`, the second step costs `2`, the third step costs `3`, and so on.
As each crab moves, moving further becomes more expensive. This changes the best horizontal position to align them all on; in the example above, this becomes `5`:
* Move from `16` to `5`: `66` fuel
* Move from `1` to `5`: `10` fuel
* Move from `2` to `5`: `6` fuel
* Move from `0` to `5`: `15` fuel
* Move from `4` to `5`: `1` fuel
* Move from `2` to `5`: `6` fuel
* Move from `7` to `5`: `3` fuel
* Move from `1` to `5`: `10` fuel
* Move from `2` to `5`: `6` fuel
* Move from `14` to `5`: `45` fuel
This costs a total of _`168`_ fuel. This is the new cheapest possible outcome; the old alignment position (`2`) now costs `206` fuel instead.
Determine the horizontal position that the crabs can align to using the least fuel possible so they can make you an escape route! _How much fuel must they spend to align to that position?_

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# Day 8: Seven Segment Search
[https://adventofcode.com/2021/day/8](https://adventofcode.com/2021/day/8)
## Description
### Part One
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](https://en.wikipedia.org/wiki/Seven-segment_display) in your submarine are malfunctioning; <span title="Yes, just the four-digit seven-segment ones. Whole batch must have been faulty.">they must have been damaged</span> 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?_
### Part Two
Through a little deduction, you should now be able to determine the remaining digits. Consider again the first example above:
acedgfb cdfbe gcdfa fbcad dab cefabd cdfgeb eafb cagedb ab |
cdfeb fcadb cdfeb cdbaf
After some careful analysis, the mapping between signal wires and segments only make sense in the following configuration:
dddd
e a
e a
ffff
g b
g b
cccc
So, the unique signal patterns would correspond to the following digits:
* `acedgfb`: `8`
* `cdfbe`: `5`
* `gcdfa`: `2`
* `fbcad`: `3`
* `dab`: `7`
* `cefabd`: `9`
* `cdfgeb`: `6`
* `eafb`: `4`
* `cagedb`: `0`
* `ab`: `1`
Then, the four digits of the output value can be decoded:
* `cdfeb`: _`5`_
* `fcadb`: _`3`_
* `cdfeb`: _`5`_
* `cdbaf`: _`3`_
Therefore, the output value for this entry is _`5353`_.
Following this same process for each entry in the second, larger example above, the output value of each entry can be determined:
* `fdgacbe cefdb cefbgd gcbe`: `8394`
* `fcgedb cgb dgebacf gc`: `9781`
* `cg cg fdcagb cbg`: `1197`
* `efabcd cedba gadfec cb`: `9361`
* `gecf egdcabf bgf bfgea`: `4873`
* `gebdcfa ecba ca fadegcb`: `8418`
* `cefg dcbef fcge gbcadfe`: `4548`
* `ed bcgafe cdgba cbgef`: `1625`
* `gbdfcae bgc cg cgb`: `8717`
* `fgae cfgab fg bagce`: `4315`
Adding all of the output values in this larger example produces _`61229`_.
For each entry, determine all of the wire/segment connections and decode the four-digit output values. _What do you get if you add up all of the output values?_

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# Day 9: Smoke Basin
[https://adventofcode.com/2021/day/9](https://adventofcode.com/2021/day/9)
## Description
### Part One
These caves seem to be [lava tubes](https://en.wikipedia.org/wiki/Lava_tube). Parts are even still volcanically active; small hydrothermal vents release smoke into the caves that slowly <span title="This was originally going to be a puzzle about watersheds, but we're already under water.">settles like rain</span>.
If you can model how the smoke flows through the caves, you might be able to avoid it and be that much safer. The submarine generates a heightmap of the floor of the nearby caves for you (your puzzle input).
Smoke flows to the lowest point of the area it's in. For example, consider the following heightmap:
2199943210
3987894921
9856789892
8767896789
9899965678
Each number corresponds to the height of a particular location, where `9` is the highest and `0` is the lowest a location can be.
Your first goal is to find the _low points_ - the locations that are lower than any of its adjacent locations. Most locations have four adjacent locations (up, down, left, and right); locations on the edge or corner of the map have three or two adjacent locations, respectively. (Diagonal locations do not count as adjacent.)
In the above example, there are _four_ low points, all highlighted: two are in the first row (a `1` and a `0`), one is in the third row (a `5`), and one is in the bottom row (also a `5`). All other locations on the heightmap have some lower adjacent location, and so are not low points.
The _risk level_ of a low point is _1 plus its height_. In the above example, the risk levels of the low points are `2`, `1`, `6`, and `6`. The sum of the risk levels of all low points in the heightmap is therefore _`15`_.
Find all of the low points on your heightmap. _What is the sum of the risk levels of all low points on your heightmap?_
### Part Two
Next, you need to find the largest basins so you know what areas are most important to avoid.
A _basin_ is all locations that eventually flow downward to a single low point. Therefore, every low point has a basin, although some basins are very small. Locations of height `9` do not count as being in any basin, and all other locations will always be part of exactly one basin.
The _size_ of a basin is the number of locations within the basin, including the low point. The example above has four basins.
The top-left basin, size `3`:
2199943210
3987894921
9856789892
8767896789
9899965678
The top-right basin, size `9`:
2199943210
3987894921
9856789892
8767896789
9899965678
The middle basin, size `14`:
2199943210
3987894921
9856789892
8767896789
9899965678
The bottom-right basin, size `9`:
2199943210
3987894921
9856789892
8767896789
9899965678
Find the three largest basins and multiply their sizes together. In the above example, this is `9 * 14 * 9 = 1134`.
_What do you get if you multiply together the sizes of the three largest basins?_

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# Day 10: Syntax Scoring
[https://adventofcode.com/2021/day/10](https://adventofcode.com/2021/day/10)
## Description
### Part One
You ask the submarine to determine the best route out of the deep-sea cave, but it only replies:
Syntax error in navigation subsystem on line: all of them
_All of them?!_ The damage is worse than you thought. You bring up a copy of the navigation subsystem (your puzzle input).
The navigation subsystem syntax is made of several lines containing _chunks_. There are one or more chunks on each line, and chunks contain zero or more other chunks. Adjacent chunks are not separated by any delimiter; if one chunk stops, the next chunk (if any) can immediately start. Every chunk must _open_ and _close_ with one of four legal pairs of matching characters:
* If a chunk opens with `(`, it must close with `)`.
* If a chunk opens with `[`, it must close with `]`.
* If a chunk opens with `{`, it must close with `}`.
* If a chunk opens with `<`, it must close with `>`.
So, `()` is a legal chunk that contains no other chunks, as is `[]`. More complex but valid chunks include `([])`, `{()()()}`, `<([{}])>`, `[<>({}){}[([])<>]]`, and even `(((((((((())))))))))`.
Some lines are _incomplete_, but others are _corrupted_. Find and discard the corrupted lines first.
A corrupted line is one where a chunk _closes with the wrong character_ - that is, where the characters it opens and closes with do not form one of the four legal pairs listed above.
Examples of corrupted chunks include `(]`, `{()()()>`, `(((()))}`, and `<([]){()}[{}])`. Such a chunk can appear anywhere within a line, and its presence causes the whole line to be considered corrupted.
For example, consider the following navigation subsystem:
[({(<(())[]>[[{[]{<()<>>
[(()[<>])]({[<{<<[]>>(
{([(<{}[<>[]}>{[]{[(<()>
(((({<>}<{<{<>}{[]{[]{}
[[<[([]))<([[{}[[()]]]
[{[{({}]{}}([{[{{{}}([]
{<[[]]>}<{[{[{[]{()[[[]
[<(<(<(<{}))><([]([]()
<{([([[(<>()){}]>(<<{{
<{([{{}}[<[[[<>{}]]]>[]]
Some of the lines aren't corrupted, just incomplete; you can ignore these lines for now. The remaining five lines are corrupted:
* `{([(<{}[<>[]}>{[]{[(<()>` - Expected `]`, but found `}` instead.
* `[[<[([]))<([[{}[[()]]]` - Expected `]`, but found `)` instead.
* `[{[{({}]{}}([{[{{{}}([]` - Expected `)`, but found `]` instead.
* `[<(<(<(<{}))><([]([]()` - Expected `>`, but found `)` instead.
* `<{([([[(<>()){}]>(<<{{` - Expected `]`, but found `>` instead.
Stop at the first incorrect closing character on each corrupted line.
Did you know that syntax checkers actually have contests to see who can get the high score for syntax errors in a file? It's true! To calculate the syntax error score for a line, take the _first illegal character_ on the line and look it up in the following table:
* `)`: `3` points.
* `]`: `57` points.
* `}`: `1197` points.
* `>`: `25137` points.
In the above example, an illegal `)` was found twice (`2*3 = 6` points), an illegal `]` was found once (_`57`_ points), an illegal `}` was found once (_`1197`_ points), and an illegal `>` was found once (_`25137`_ points). So, the total syntax error score for this file is `6+57+1197+25137 = 26397` points!
Find the first illegal character in each corrupted line of the navigation subsystem. _What is the total syntax error score for those errors?_
### Part Two
Now, discard the corrupted lines. The remaining lines are _incomplete_.
Incomplete lines don't have any incorrect characters - instead, they're missing some closing characters at the end of the line. To repair the navigation subsystem, you just need to figure out _the sequence of closing characters_ that complete all open chunks in the line.
You can only use closing characters (`)`, `]`, `}`, or `>`), and you must add them in the correct order so that only legal pairs are formed and all chunks end up closed.
In the example above, there are five incomplete lines:
* `[({(<(())[]>[[{[]{<()<>>` - Complete by adding `}}]])})]`.
* `[(()[<>])]({[<{<<[]>>(` - Complete by adding `)}>]})`.
* `(((({<>}<{<{<>}{[]{[]{}` - Complete by adding `}}>}>))))`.
* `{<[[]]>}<{[{[{[]{()[[[]` - Complete by adding `]]}}]}]}>`.
* `<{([{{}}[<[[[<>{}]]]>[]]` - Complete by adding `])}>`.
Did you know that autocomplete tools _also_ have contests? It's true! The score is determined by considering the completion string character-by-character. Start with a total score of `0`. Then, for each character, multiply the total score by 5 and then increase the total score by the point value given for the character in the following table:
* `)`: `1` point.
* `]`: `2` points.
* `}`: `3` points.
* `>`: `4` points.
So, the last completion string above - `])}>` - would be scored as follows:
* Start with a total score of `0`.
* Multiply the total score by 5 to get `0`, then add the value of `]` (2) to get a new total score of `2`.
* Multiply the total score by 5 to get `10`, then add the value of `)` (1) to get a new total score of `11`.
* Multiply the total score by 5 to get `55`, then add the value of `}` (3) to get a new total score of `58`.
* Multiply the total score by 5 to get `290`, then add the value of `>` (4) to get a new total score of `294`.
The five lines' completion strings have total scores as follows:
* `}}]])})]` - `288957` total points.
* `)}>]})` - `5566` total points.
* `}}>}>))))` - `1480781` total points.
* `]]}}]}]}>` - `995444` total points.
* `])}>` - `294` total points.
Autocomplete tools are an odd bunch: the winner is found by _sorting_ all of the scores and then taking the _middle_ score. (There will always be an odd number of scores to consider.) In this example, the middle score is _`288957`_ because there are the same number of scores smaller and larger than it.
Find the completion string for each incomplete line, score the completion strings, and sort the scores. _What is the middle score?_

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# Day 11: Dumbo Octopus
[https://adventofcode.com/2021/day/11](https://adventofcode.com/2021/day/11)
## Description
### Part One
You enter a large cavern full of rare bioluminescent [dumbo octopuses](https://www.youtube.com/watch?v=eih-VSaS2g0)! They seem to not like the Christmas lights on your submarine, so you turn them off for now.
There are 100 <span title="I know it's weird; I grew up saying 'octopi' too.">octopuses</span> arranged neatly in a 10 by 10 grid. Each octopus slowly gains _energy_ over time and _flashes_ brightly for a moment when its energy is full. Although your lights are off, maybe you could navigate through the cave without disturbing the octopuses if you could predict when the flashes of light will happen.
Each octopus has an _energy level_ - your submarine can remotely measure the energy level of each octopus (your puzzle input). For example:
5483143223
2745854711
5264556173
6141336146
6357385478
4167524645
2176841721
6882881134
4846848554
5283751526
The energy level of each octopus is a value between `0` and `9`. Here, the top-left octopus has an energy level of `5`, the bottom-right one has an energy level of `6`, and so on.
You can model the energy levels and flashes of light in _steps_. During a single step, the following occurs:
* First, the energy level of each octopus increases by `1`.
* Then, any octopus with an energy level greater than `9` _flashes_. This increases the energy level of all adjacent octopuses by `1`, including octopuses that are diagonally adjacent. If this causes an octopus to have an energy level greater than `9`, it _also flashes_. This process continues as long as new octopuses keep having their energy level increased beyond `9`. (An octopus can only flash _at most once per step_.)
* Finally, any octopus that flashed during this step has its energy level set to `0`, as it used all of its energy to flash.
Adjacent flashes can cause an octopus to flash on a step even if it begins that step with very little energy. Consider the middle octopus with `1` energy in this situation:
Before any steps:
11111
19991
19191
19991
11111
After step 1:
34543
40004
50005
40004
34543
After step 2:
45654
51115
61116
51115
45654
An octopus is _highlighted_ when it flashed during the given step.
Here is how the larger example above progresses:
Before any steps:
5483143223
2745854711
5264556173
6141336146
6357385478
4167524645
2176841721
6882881134
4846848554
5283751526
After step 1:
6594254334
3856965822
6375667284
7252447257
7468496589
5278635756
3287952832
7993992245
5957959665
6394862637
After step 2:
8807476555
5089087054
8597889608
8485769600
8700908800
6600088989
6800005943
0000007456
9000000876
8700006848
After step 3:
0050900866
8500800575
9900000039
9700000041
9935080063
7712300000
7911250009
2211130000
0421125000
0021119000
After step 4:
2263031977
0923031697
0032221150
0041111163
0076191174
0053411122
0042361120
5532241122
1532247211
1132230211
After step 5:
4484144000
2044144000
2253333493
1152333274
1187303285
1164633233
1153472231
6643352233
2643358322
2243341322
After step 6:
5595255111
3155255222
3364444605
2263444496
2298414396
2275744344
2264583342
7754463344
3754469433
3354452433
After step 7:
6707366222
4377366333
4475555827
3496655709
3500625609
3509955566
3486694453
8865585555
4865580644
4465574644
After step 8:
7818477333
5488477444
5697666949
4608766830
4734946730
4740097688
6900007564
0000009666
8000004755
6800007755
After step 9:
9060000644
7800000976
6900000080
5840000082
5858000093
6962400000
8021250009
2221130009
9111128097
7911119976
After step 10:
0481112976
0031112009
0041112504
0081111406
0099111306
0093511233
0442361130
5532252350
0532250600
0032240000
After step 10, there have been a total of `204` flashes. Fast forwarding, here is the same configuration every 10 steps:
After step 20:
3936556452
5686556806
4496555690
4448655580
4456865570
5680086577
7000009896
0000000344
6000000364
4600009543
After step 30:
0643334118
4253334611
3374333458
2225333337
2229333338
2276733333
2754574565
5544458511
9444447111
7944446119
After step 40:
6211111981
0421111119
0042111115
0003111115
0003111116
0065611111
0532351111
3322234597
2222222976
2222222762
After step 50:
9655556447
4865556805
4486555690
4458655580
4574865570
5700086566
6000009887
8000000533
6800000633
5680000538
After step 60:
2533334200
2743334640
2264333458
2225333337
2225333338
2287833333
3854573455
1854458611
1175447111
1115446111
After step 70:
8211111164
0421111166
0042111114
0004211115
0000211116
0065611111
0532351111
7322235117
5722223475
4572222754
After step 80:
1755555697
5965555609
4486555680
4458655580
4570865570
5700086566
7000008666
0000000990
0000000800
0000000000
After step 90:
7433333522
2643333522
2264333458
2226433337
2222433338
2287833333
2854573333
4854458333
3387779333
3333333333
After step 100:
0397666866
0749766918
0053976933
0004297822
0004229892
0053222877
0532222966
9322228966
7922286866
6789998766
After 100 steps, there have been a total of _`1656`_ flashes.
Given the starting energy levels of the dumbo octopuses in your cavern, simulate 100 steps. _How many total flashes are there after 100 steps?_
### Part Two
It seems like the individual flashes aren't bright enough to navigate. However, you might have a better option: the flashes seem to be _synchronizing_!
In the example above, the first time all octopuses flash simultaneously is step _`195`_:
After step 193:
5877777777
8877777777
7777777777
7777777777
7777777777
7777777777
7777777777
7777777777
7777777777
7777777777
After step 194:
6988888888
9988888888
8888888888
8888888888
8888888888
8888888888
8888888888
8888888888
8888888888
8888888888
After step 195:
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
0000000000
If you can calculate the exact moments when the octopuses will all flash simultaneously, you should be able to navigate through the cavern. _What is the first step during which all octopuses flash?_

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# Day 12: Passage Pathing
[https://adventofcode.com/2021/day/12](https://adventofcode.com/2021/day/12)
## Description
### Part One
With your <span title="Sublime.">submarine's subterranean subsystems subsisting suboptimally</span>, the only way you're getting out of this cave anytime soon is by finding a path yourself. Not just _a_ path - the only way to know if you've found the _best_ path is to find _all_ of them.
Fortunately, the sensors are still mostly working, and so you build a rough map of the remaining caves (your puzzle input). For example:
start-A
start-b
A-c
A-b
b-d
A-end
b-end
This is a list of how all of the caves are connected. You start in the cave named `start`, and your destination is the cave named `end`. An entry like `b-d` means that cave `b` is connected to cave `d` - that is, you can move between them.
So, the above cave system looks roughly like this:
start
/ \
c--A-----b--d
\ /
end
Your goal is to find the number of distinct _paths_ that start at `start`, end at `end`, and don't visit small caves more than once. There are two types of caves: _big_ caves (written in uppercase, like `A`) and _small_ caves (written in lowercase, like `b`). It would be a waste of time to visit any small cave more than once, but big caves are large enough that it might be worth visiting them multiple times. So, all paths you find should _visit small caves at most once_, and can _visit big caves any number of times_.
Given these rules, there are _`10`_ paths through this example cave system:
start,A,b,A,c,A,end
start,A,b,A,end
start,A,b,end
start,A,c,A,b,A,end
start,A,c,A,b,end
start,A,c,A,end
start,A,end
start,b,A,c,A,end
start,b,A,end
start,b,end
(Each line in the above list corresponds to a single path; the caves visited by that path are listed in the order they are visited and separated by commas.)
Note that in this cave system, cave `d` is never visited by any path: to do so, cave `b` would need to be visited twice (once on the way to cave `d` and a second time when returning from cave `d`), and since cave `b` is small, this is not allowed.
Here is a slightly larger example:
dc-end
HN-start
start-kj
dc-start
dc-HN
LN-dc
HN-end
kj-sa
kj-HN
kj-dc
The `19` paths through it are as follows:
start,HN,dc,HN,end
start,HN,dc,HN,kj,HN,end
start,HN,dc,end
start,HN,dc,kj,HN,end
start,HN,end
start,HN,kj,HN,dc,HN,end
start,HN,kj,HN,dc,end
start,HN,kj,HN,end
start,HN,kj,dc,HN,end
start,HN,kj,dc,end
start,dc,HN,end
start,dc,HN,kj,HN,end
start,dc,end
start,dc,kj,HN,end
start,kj,HN,dc,HN,end
start,kj,HN,dc,end
start,kj,HN,end
start,kj,dc,HN,end
start,kj,dc,end
Finally, this even larger example has `226` paths through it:
fs-end
he-DX
fs-he
start-DX
pj-DX
end-zg
zg-sl
zg-pj
pj-he
RW-he
fs-DX
pj-RW
zg-RW
start-pj
he-WI
zg-he
pj-fs
start-RW
_How many paths through this cave system are there that visit small caves at most once?_
### Part Two
After reviewing the available paths, you realize you might have time to visit a single small cave _twice_. Specifically, big caves can be visited any number of times, a single small cave can be visited at most twice, and the remaining small caves can be visited at most once. However, the caves named `start` and `end` can only be visited _exactly once each_: once you leave the `start` cave, you may not return to it, and once you reach the `end` cave, the path must end immediately.
Now, the `36` possible paths through the first example above are:
start,A,b,A,b,A,c,A,end
start,A,b,A,b,A,end
start,A,b,A,b,end
start,A,b,A,c,A,b,A,end
start,A,b,A,c,A,b,end
start,A,b,A,c,A,c,A,end
start,A,b,A,c,A,end
start,A,b,A,end
start,A,b,d,b,A,c,A,end
start,A,b,d,b,A,end
start,A,b,d,b,end
start,A,b,end
start,A,c,A,b,A,b,A,end
start,A,c,A,b,A,b,end
start,A,c,A,b,A,c,A,end
start,A,c,A,b,A,end
start,A,c,A,b,d,b,A,end
start,A,c,A,b,d,b,end
start,A,c,A,b,end
start,A,c,A,c,A,b,A,end
start,A,c,A,c,A,b,end
start,A,c,A,c,A,end
start,A,c,A,end
start,A,end
start,b,A,b,A,c,A,end
start,b,A,b,A,end
start,b,A,b,end
start,b,A,c,A,b,A,end
start,b,A,c,A,b,end
start,b,A,c,A,c,A,end
start,b,A,c,A,end
start,b,A,end
start,b,d,b,A,c,A,end
start,b,d,b,A,end
start,b,d,b,end
start,b,end
The slightly larger example above now has `103` paths through it, and the even larger example now has `3509` paths through it.
Given these new rules, _how many paths through this cave system are there?_

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# Day 13: Transparent Origami
[https://adventofcode.com/2021/day/13](https://adventofcode.com/2021/day/13)
## Description
### Part One
You reach another volcanically active part of the cave. It would be nice if you could do some kind of thermal imaging so you could tell ahead of time which caves are too hot to safely enter.
Fortunately, the submarine seems to be equipped with a thermal camera! When you activate it, you are greeted with:
Congratulations on your purchase! To activate this infrared thermal imaging
camera system, please enter the code found on page 1 of the manual.
Apparently, the Elves have never used this feature. To your surprise, you manage to find the manual; as you go to open it, page 1 falls out. It's a large sheet of [transparent paper](https://en.wikipedia.org/wiki/Transparency_(projection))! The transparent paper is marked with random dots and includes instructions on how to fold it up (your puzzle input). For example:
6,10
0,14
9,10
0,3
10,4
4,11
6,0
6,12
4,1
0,13
10,12
3,4
3,0
8,4
1,10
2,14
8,10
9,0
fold along y=7
fold along x=5
The first section is a list of dots on the transparent paper. `0,0` represents the top-left coordinate. The first value, `x`, increases to the right. The second value, `y`, increases downward. So, the coordinate `3,0` is to the right of `0,0`, and the coordinate `0,7` is below `0,0`. The coordinates in this example form the following pattern, where `#` is a dot on the paper and `.` is an empty, unmarked position:
...#..#..#.
....#......
...........
#..........
...#....#.#
...........
...........
...........
...........
...........
.#....#.##.
....#......
......#...#
#..........
#.#........
Then, there is a list of _fold instructions_. Each instruction indicates a line on the transparent paper and wants you to fold the paper _up_ (for horizontal `y=...` lines) or _left_ (for vertical `x=...` lines). In this example, the first fold instruction is `fold along y=7`, which designates the line formed by all of the positions where `y` is `7` (marked here with `-`):
...#..#..#.
....#......
...........
#..........
...#....#.#
...........
...........
-----------
...........
...........
.#....#.##.
....#......
......#...#
#..........
#.#........
Because this is a horizontal line, fold the bottom half _up_. Some of the dots might end up overlapping after the fold is complete, but dots will never appear exactly on a fold line. The result of doing this fold looks like this:
#.##..#..#.
#...#......
......#...#
#...#......
.#.#..#.###
...........
...........
Now, only `17` dots are visible.
Notice, for example, the two dots in the bottom left corner before the transparent paper is folded; after the fold is complete, those dots appear in the top left corner (at `0,0` and `0,1`). Because the paper is transparent, the dot just below them in the result (at `0,3`) remains visible, as it can be seen through the transparent paper.
Also notice that some dots can end up _overlapping_; in this case, the dots merge together and become a single dot.
The second fold instruction is `fold along x=5`, which indicates this line:
#.##.|#..#.
#...#|.....
.....|#...#
#...#|.....
.#.#.|#.###
.....|.....
.....|.....
Because this is a vertical line, fold _left_:
#####
#...#
#...#
#...#
#####
.....
.....
The instructions made a square!
The transparent paper is pretty big, so for now, focus on just completing the first fold. After the first fold in the example above, _`17`_ dots are visible - dots that end up overlapping after the fold is completed count as a single dot.
_How many dots are visible after completing just the first fold instruction on your transparent paper?_
### Part Two
<span title="How can you fold it that many times? You tell me, I'm not the one folding it.">Finish folding</span> the transparent paper according to the instructions. The manual says the code is always _eight capital letters_.
_What code do you use to activate the infrared thermal imaging camera system?_

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# Day 14: Extended Polymerization
[https://adventofcode.com/2021/day/14](https://adventofcode.com/2021/day/14)
## Description
### Part One
The incredible pressures at this depth are starting to put a strain on your submarine. The submarine has [polymerization](https://en.wikipedia.org/wiki/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 <span title="HO
HO -> OH">instructions</span> 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?_
### 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?_

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# Day 15: Chiton
[https://adventofcode.com/2021/day/15](https://adventofcode.com/2021/day/15)
## Description
### Part One
You've almost reached the exit of the cave, but the walls are getting closer together. Your submarine can barely still fit, though; the main problem is that the walls of the cave are covered in [chitons](https://en.wikipedia.org/wiki/Chiton), and it would be best not to bump any of them.
The cavern is large, but has a very low ceiling, restricting your motion to two dimensions. The shape of the cavern resembles a square; a quick scan of chiton density produces a map of _risk level_ throughout the cave (your puzzle input). For example:
1163751742
1381373672
2136511328
3694931569
7463417111
1319128137
1359912421
3125421639
1293138521
2311944581
You start in the top left position, your destination is the bottom right position, and you <span title="Can't go diagonal until we can repair the caterpillar unit. Could be the liquid helium or the superconductors.">cannot move diagonally</span>. The number at each position is its _risk level_; to determine the total risk of an entire path, add up the risk levels of each position you _enter_ (that is, don't count the risk level of your starting position unless you enter it; leaving it adds no risk to your total).
Your goal is to find a path with the _lowest total risk_. In this example, a path with the lowest total risk is highlighted here:
1163751742
1381373672
2136511328
3694931569
7463417111
1319128137
1359912421
3125421639
1293138521
2311944581
The total risk of this path is _`40`_ (the starting position is never entered, so its risk is not counted).
_What is the lowest total risk of any path from the top left to the bottom right?_
### Part Two
Now that you know how to find low-risk paths in the cave, you can try to find your way out.
The entire cave is actually _five times larger in both dimensions_ than you thought; the area you originally scanned is just one tile in a 5x5 tile area that forms the full map. Your original map tile repeats to the right and downward; each time the tile repeats to the right or downward, all of its risk levels _are 1 higher_ than the tile immediately up or left of it. However, risk levels above `9` wrap back around to `1`. So, if your original map had some position with a risk level of `8`, then that same position on each of the 25 total tiles would be as follows:
8 9 1 2 3
9 1 2 3 4
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
Each single digit above corresponds to the example position with a value of `8` on the top-left tile. Because the full map is actually five times larger in both dimensions, that position appears a total of 25 times, once in each duplicated tile, with the values shown above.
Here is the full five-times-as-large version of the first example above, with the original map in the top left corner highlighted:
11637517422274862853338597396444961841755517295286
13813736722492484783351359589446246169155735727126
21365113283247622439435873354154698446526571955763
36949315694715142671582625378269373648937148475914
74634171118574528222968563933317967414442817852555
13191281372421239248353234135946434524615754563572
13599124212461123532357223464346833457545794456865
31254216394236532741534764385264587549637569865174
12931385212314249632342535174345364628545647573965
23119445813422155692453326671356443778246755488935
22748628533385973964449618417555172952866628316397
24924847833513595894462461691557357271266846838237
32476224394358733541546984465265719557637682166874
47151426715826253782693736489371484759148259586125
85745282229685639333179674144428178525553928963666
24212392483532341359464345246157545635726865674683
24611235323572234643468334575457944568656815567976
42365327415347643852645875496375698651748671976285
23142496323425351743453646285456475739656758684176
34221556924533266713564437782467554889357866599146
33859739644496184175551729528666283163977739427418
35135958944624616915573572712668468382377957949348
43587335415469844652657195576376821668748793277985
58262537826937364893714847591482595861259361697236
96856393331796741444281785255539289636664139174777
35323413594643452461575456357268656746837976785794
35722346434683345754579445686568155679767926678187
53476438526458754963756986517486719762859782187396
34253517434536462854564757396567586841767869795287
45332667135644377824675548893578665991468977611257
44961841755517295286662831639777394274188841538529
46246169155735727126684683823779579493488168151459
54698446526571955763768216687487932779859814388196
69373648937148475914825958612593616972361472718347
17967414442817852555392896366641391747775241285888
46434524615754563572686567468379767857948187896815
46833457545794456865681556797679266781878137789298
64587549637569865174867197628597821873961893298417
45364628545647573965675868417678697952878971816398
56443778246755488935786659914689776112579188722368
55172952866628316397773942741888415385299952649631
57357271266846838237795794934881681514599279262561
65719557637682166874879327798598143881961925499217
71484759148259586125936169723614727183472583829458
28178525553928963666413917477752412858886352396999
57545635726865674683797678579481878968159298917926
57944568656815567976792667818781377892989248891319
75698651748671976285978218739618932984172914319528
56475739656758684176786979528789718163989182927419
67554889357866599146897761125791887223681299833479
Equipped with the full map, you can now find a path from the top left corner to the bottom right corner with the lowest total risk:
11637517422274862853338597396444961841755517295286
13813736722492484783351359589446246169155735727126
21365113283247622439435873354154698446526571955763
36949315694715142671582625378269373648937148475914
74634171118574528222968563933317967414442817852555
13191281372421239248353234135946434524615754563572
13599124212461123532357223464346833457545794456865
31254216394236532741534764385264587549637569865174
12931385212314249632342535174345364628545647573965
23119445813422155692453326671356443778246755488935
22748628533385973964449618417555172952866628316397
24924847833513595894462461691557357271266846838237
32476224394358733541546984465265719557637682166874
47151426715826253782693736489371484759148259586125
85745282229685639333179674144428178525553928963666
24212392483532341359464345246157545635726865674683
24611235323572234643468334575457944568656815567976
42365327415347643852645875496375698651748671976285
23142496323425351743453646285456475739656758684176
34221556924533266713564437782467554889357866599146
33859739644496184175551729528666283163977739427418
35135958944624616915573572712668468382377957949348
43587335415469844652657195576376821668748793277985
58262537826937364893714847591482595861259361697236
96856393331796741444281785255539289636664139174777
35323413594643452461575456357268656746837976785794
35722346434683345754579445686568155679767926678187
53476438526458754963756986517486719762859782187396
34253517434536462854564757396567586841767869795287
45332667135644377824675548893578665991468977611257
44961841755517295286662831639777394274188841538529
46246169155735727126684683823779579493488168151459
54698446526571955763768216687487932779859814388196
69373648937148475914825958612593616972361472718347
17967414442817852555392896366641391747775241285888
46434524615754563572686567468379767857948187896815
46833457545794456865681556797679266781878137789298
64587549637569865174867197628597821873961893298417
45364628545647573965675868417678697952878971816398
56443778246755488935786659914689776112579188722368
55172952866628316397773942741888415385299952649631
57357271266846838237795794934881681514599279262561
65719557637682166874879327798598143881961925499217
71484759148259586125936169723614727183472583829458
28178525553928963666413917477752412858886352396999
57545635726865674683797678579481878968159298917926
57944568656815567976792667818781377892989248891319
75698651748671976285978218739618932984172914319528
56475739656758684176786979528789718163989182927419
67554889357866599146897761125791887223681299833479
The total risk of this path is _`315`_ (the starting position is still never entered, so its risk is not counted).
Using the full map, _what is the lowest total risk of any path from the top left to the bottom right?_

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# Day 16: Packet Decoder
[https://adventofcode.com/2021/day/16](https://adventofcode.com/2021/day/16)
## Description
### Part One
As you leave the cave and reach open waters, you receive a transmission from the Elves back on the ship.
The transmission was sent using the Buoyancy Interchange Transmission System (<span title="Just be glad it wasn't sent using the BuoyancY Transmission Encoding System.">BITS</span>), a method of packing numeric expressions into a binary sequence. Your submarine's computer has saved the transmission in [hexadecimal](https://en.wikipedia.org/wiki/Hexadecimal) (your puzzle input).
The first step of decoding the message is to convert the hexadecimal representation into binary. Each character of hexadecimal corresponds to four bits of binary data:
0 = 0000
1 = 0001
2 = 0010
3 = 0011
4 = 0100
5 = 0101
6 = 0110
7 = 0111
8 = 1000
9 = 1001
A = 1010
B = 1011
C = 1100
D = 1101
E = 1110
F = 1111
The BITS transmission contains a single _packet_ at its outermost layer which itself contains many other packets. The hexadecimal representation of this packet might encode a few extra `0` bits at the end; these are not part of the transmission and should be ignored.
Every packet begins with a standard header: the first three bits encode the packet _version_, and the next three bits encode the packet _type ID_. These two values are numbers; all numbers encoded in any packet are represented as binary with the most significant bit first. For example, a version encoded as the binary sequence `100` represents the number `4`.
Packets with type ID `4` represent a _literal value_. Literal value packets encode a single binary number. To do this, the binary number is padded with leading zeroes until its length is a multiple of four bits, and then it is broken into groups of four bits. Each group is prefixed by a `1` bit except the last group, which is prefixed by a `0` bit. These groups of five bits immediately follow the packet header. For example, the hexadecimal string `D2FE28` becomes:
110100101111111000101000
VVVTTTAAAAABBBBBCCCCC
Below each bit is a label indicating its purpose:
* The three bits labeled `V` (`110`) are the packet version, `6`.
* The three bits labeled `T` (`100`) are the packet type ID, `4`, which means the packet is a literal value.
* The five bits labeled `A` (`10111`) start with a `1` (not the last group, keep reading) and contain the first four bits of the number, `0111`.
* The five bits labeled `B` (`11110`) start with a `1` (not the last group, keep reading) and contain four more bits of the number, `1110`.
* The five bits labeled `C` (`00101`) start with a `0` (last group, end of packet) and contain the last four bits of the number, `0101`.
* The three unlabeled `0` bits at the end are extra due to the hexadecimal representation and should be ignored.
So, this packet represents a literal value with binary representation `011111100101`, which is `2021` in decimal.
Every other type of packet (any packet with a type ID other than `4`) represent an _operator_ that performs some calculation on one or more sub-packets contained within. Right now, the specific operations aren't important; focus on parsing the hierarchy of sub-packets.
An operator packet contains one or more packets. To indicate which subsequent binary data represents its sub-packets, an operator packet can use one of two modes indicated by the bit immediately after the packet header; this is called the _length type ID_:
* If the length type ID is `0`, then the next _15_ bits are a number that represents the _total length in bits_ of the sub-packets contained by this packet.
* If the length type ID is `1`, then the next _11_ bits are a number that represents the _number of sub-packets immediately contained_ by this packet.
Finally, after the length type ID bit and the 15-bit or 11-bit field, the sub-packets appear.
For example, here is an operator packet (hexadecimal string `38006F45291200`) with length type ID `0` that contains two sub-packets:
00111000000000000110111101000101001010010001001000000000
VVVTTTILLLLLLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBBBBBB
* The three bits labeled `V` (`001`) are the packet version, `1`.
* The three bits labeled `T` (`110`) are the packet type ID, `6`, which means the packet is an operator.
* The bit labeled `I` (`0`) is the length type ID, which indicates that the length is a 15-bit number representing the number of bits in the sub-packets.
* The 15 bits labeled `L` (`000000000011011`) contain the length of the sub-packets in bits, `27`.
* The 11 bits labeled `A` contain the first sub-packet, a literal value representing the number `10`.
* The 16 bits labeled `B` contain the second sub-packet, a literal value representing the number `20`.
After reading 11 and 16 bits of sub-packet data, the total length indicated in `L` (27) is reached, and so parsing of this packet stops.
As another example, here is an operator packet (hexadecimal string `EE00D40C823060`) with length type ID `1` that contains three sub-packets:
11101110000000001101010000001100100000100011000001100000
VVVTTTILLLLLLLLLLLAAAAAAAAAAABBBBBBBBBBBCCCCCCCCCCC
* The three bits labeled `V` (`111`) are the packet version, `7`.
* The three bits labeled `T` (`011`) are the packet type ID, `3`, which means the packet is an operator.
* The bit labeled `I` (`1`) is the length type ID, which indicates that the length is a 11-bit number representing the number of sub-packets.
* The 11 bits labeled `L` (`00000000011`) contain the number of sub-packets, `3`.
* The 11 bits labeled `A` contain the first sub-packet, a literal value representing the number `1`.
* The 11 bits labeled `B` contain the second sub-packet, a literal value representing the number `2`.
* The 11 bits labeled `C` contain the third sub-packet, a literal value representing the number `3`.
After reading 3 complete sub-packets, the number of sub-packets indicated in `L` (3) is reached, and so parsing of this packet stops.
For now, parse the hierarchy of the packets throughout the transmission and _add up all of the version numbers_.
Here are a few more examples of hexadecimal-encoded transmissions:
* `8A004A801A8002F478` represents an operator packet (version 4) which contains an operator packet (version 1) which contains an operator packet (version 5) which contains a literal value (version 6); this packet has a version sum of _`16`_.
* `620080001611562C8802118E34` represents an operator packet (version 3) which contains two sub-packets; each sub-packet is an operator packet that contains two literal values. This packet has a version sum of _`12`_.
* `C0015000016115A2E0802F182340` has the same structure as the previous example, but the outermost packet uses a different length type ID. This packet has a version sum of _`23`_.
* `A0016C880162017C3686B18A3D4780` is an operator packet that contains an operator packet that contains an operator packet that contains five literal values; it has a version sum of _`31`_.
Decode the structure of your hexadecimal-encoded BITS transmission; _what do you get if you add up the version numbers in all packets?_
### Part Two
Now that you have the structure of your transmission decoded, you can calculate the value of the expression it represents.
Literal values (type ID `4`) represent a single number as described above. The remaining type IDs are more interesting:
* Packets with type ID `0` are _sum_ packets - their value is the sum of the values of their sub-packets. If they only have a single sub-packet, their value is the value of the sub-packet.
* Packets with type ID `1` are _product_ packets - their value is the result of multiplying together the values of their sub-packets. If they only have a single sub-packet, their value is the value of the sub-packet.
* Packets with type ID `2` are _minimum_ packets - their value is the minimum of the values of their sub-packets.
* Packets with type ID `3` are _maximum_ packets - their value is the maximum of the values of their sub-packets.
* Packets with type ID `5` are _greater than_ packets - their value is _1_ if the value of the first sub-packet is greater than the value of the second sub-packet; otherwise, their value is _0_. These packets always have exactly two sub-packets.
* Packets with type ID `6` are _less than_ packets - their value is _1_ if the value of the first sub-packet is less than the value of the second sub-packet; otherwise, their value is _0_. These packets always have exactly two sub-packets.
* Packets with type ID `7` are _equal to_ packets - their value is _1_ if the value of the first sub-packet is equal to the value of the second sub-packet; otherwise, their value is _0_. These packets always have exactly two sub-packets.
Using these rules, you can now work out the value of the outermost packet in your BITS transmission.
For example:
* `C200B40A82` finds the sum of `1` and `2`, resulting in the value _`3`_.
* `04005AC33890` finds the product of `6` and `9`, resulting in the value _`54`_.
* `880086C3E88112` finds the minimum of `7`, `8`, and `9`, resulting in the value _`7`_.
* `CE00C43D881120` finds the maximum of `7`, `8`, and `9`, resulting in the value _`9`_.
* `D8005AC2A8F0` produces `1`, because `5` is less than `15`.
* `F600BC2D8F` produces `0`, because `5` is not greater than `15`.
* `9C005AC2F8F0` produces `0`, because `5` is not equal to `15`.
* `9C0141080250320F1802104A08` produces `1`, because `1` + `3` = `2` \* `2`.
_What do you get if you evaluate the expression represented by your hexadecimal-encoded BITS transmission?_