advent

func AbsInt ¶

func AbsInt(input int) int

func Day1 ¶

func Day1()

--- 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?

--- 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?

func Day10 ¶

func Day10()

--- Day 10: Adapter Array --- Patched into the aircraft's data port, you discover weather forecasts of a massive tropical storm. Before you can figure out whether it will impact your vacation plans, however, your device suddenly turns off!

Its battery is dead.

You'll need to plug it in. There's only one problem: the charging outlet near your seat produces the wrong number of jolts. Always prepared, you make a list of all of the joltage adapters in your bag.

Each of your joltage adapters is rated for a specific output joltage (your puzzle input). Any given adapter can take an input 1, 2, or 3 jolts lower than its rating and still produce its rated output joltage.

In addition, your device has a built-in joltage adapter rated for 3 jolts higher than the highest-rated adapter in your bag. (If your adapter list were 3, 9, and 6, your device's built-in adapter would be rated for 12 jolts.)

Treat the charging outlet near your seat as having an effective joltage rating of 0.

Since you have some time to kill, you might as well test all of your adapters. Wouldn't want to get to your resort and realize you can't even charge your device!

If you use every adapter in your bag at once, what is the distribution of joltage differences between the charging outlet, the adapters, and your device?

For example, suppose that in your bag, you have adapters with the following joltage ratings:

16 10 15 5 1 11 7 19 6 12 4 With these adapters, your device's built-in joltage adapter would be rated for 19 + 3 = 22 jolts, 3 higher than the highest-rated adapter.

Because adapters can only connect to a source 1-3 jolts lower than its rating, in order to use every adapter, you'd need to choose them like this:

The charging outlet has an effective rating of 0 jolts, so the only adapters that could connect to it directly would need to have a joltage rating of 1, 2, or 3 jolts. Of these, only one you have is an adapter rated 1 jolt (difference of 1). From your 1-jolt rated adapter, the only choice is your 4-jolt rated adapter (difference of 3). From the 4-jolt rated adapter, the adapters rated 5, 6, or 7 are valid choices. However, in order to not skip any adapters, you have to pick the adapter rated 5 jolts (difference of 1). Similarly, the next choices would need to be the adapter rated 6 and then the adapter rated 7 (with difference of 1 and 1). The only adapter that works with the 7-jolt rated adapter is the one rated 10 jolts (difference of 3). From 10, the choices are 11 or 12; choose 11 (difference of 1) and then 12 (difference of 1). After 12, only valid adapter has a rating of 15 (difference of 3), then 16 (difference of 1), then 19 (difference of 3). Finally, your device's built-in adapter is always 3 higher than the highest adapter, so its rating is 22 jolts (always a difference of 3). In this example, when using every adapter, there are 7 differences of 1 jolt and 5 differences of 3 jolts.

Here is a larger example:

28 33 18 42 31 14 46 20 48 47 24 23 49 45 19 38 39 11 1 32 25 35 8 17 7 9 4 2 34 10 3 In this larger example, in a chain that uses all of the adapters, there are 22 differences of 1 jolt and 10 differences of 3 jolts.

Find a chain that uses all of your adapters to connect the charging outlet to your device's built-in adapter and count the joltage differences between the charging outlet, the adapters, and your device. What is the number of 1-jolt differences multiplied by the number of 3-jolt differences?

--- Part Two --- To completely determine whether you have enough adapters, you'll need to figure out how many different ways they can be arranged. Every arrangement needs to connect the charging outlet to your device. The previous rules about when adapters can successfully connect still apply.

The first example above (the one that starts with 16, 10, 15) supports the following arrangements:

(0), 1, 4, 5, 6, 7, 10, 11, 12, 15, 16, 19, (22) (0), 1, 4, 5, 6, 7, 10, 12, 15, 16, 19, (22) (0), 1, 4, 5, 7, 10, 11, 12, 15, 16, 19, (22) (0), 1, 4, 5, 7, 10, 12, 15, 16, 19, (22) (0), 1, 4, 6, 7, 10, 11, 12, 15, 16, 19, (22) (0), 1, 4, 6, 7, 10, 12, 15, 16, 19, (22) (0), 1, 4, 7, 10, 11, 12, 15, 16, 19, (22) (0), 1, 4, 7, 10, 12, 15, 16, 19, (22) (The charging outlet and your device's built-in adapter are shown in parentheses.) Given the adapters from the first example, the total number of arrangements that connect the charging outlet to your device is 8.

The second example above (the one that starts with 28, 33, 18) has many arrangements. Here are a few:

(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31, 32, 33, 34, 35, 38, 39, 42, 45, 46, 47, 48, 49, (52)

(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31, 32, 33, 34, 35, 38, 39, 42, 45, 46, 47, 49, (52)

(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31, 32, 33, 34, 35, 38, 39, 42, 45, 46, 48, 49, (52)

(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31, 32, 33, 34, 35, 38, 39, 42, 45, 46, 49, (52)

(0), 1, 2, 3, 4, 7, 8, 9, 10, 11, 14, 17, 18, 19, 20, 23, 24, 25, 28, 31, 32, 33, 34, 35, 38, 39, 42, 45, 47, 48, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45, 46, 48, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45, 46, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45, 47, 48, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45, 47, 49, (52)

(0), 3, 4, 7, 10, 11, 14, 17, 20, 23, 25, 28, 31, 34, 35, 38, 39, 42, 45, 48, 49, (52) In total, this set of adapters can connect the charging outlet to your device in 19208 distinct arrangements.

You glance back down at your bag and try to remember why you brought so many adapters; there must be more than a trillion valid ways to arrange them! Surely, there must be an efficient way to count the arrangements.

What is the total number of distinct ways you can arrange the adapters to connect the charging outlet to your device?

func Day11 ¶

func Day11()

func Day12 ¶

func Day12()

func Day13 ¶

func Day13()

func Day14 ¶

func Day14()

func Day15 ¶

func Day15()

func Day16 ¶

func Day16()

func Day17 ¶

func Day17()

func Day18 ¶

func Day18()

func Day19 ¶

func Day19()

for each instruction list explicit dependencies (variables) if instruction has no dependencies assign level 0 if all dependencies are level 0, assign level 1 if all dependencies are level n or below, assign level n + 1 in example: - level 0: 4, 5 - level 1: 2, 3 - level 2: 1 - level 3: 0

compile level n instructions starting with zero descend down levels looking for a match for each part

does each instruction output have a unique length? each subinstruction does have unique length use this to determine spacing for three part instructions

Part 2: add extra rules 8: can be any combination of rules satisfying 8 11: 42 11 31 becomes 42 42 ... 42 31 ... 31 31 Total pattern: 42 * (m + n) + 31 * n

func Day2 ¶

func Day2()

--- Day 2: Password Philosophy --- Your flight departs in a few days from the coastal airport; the easiest way down to the coast from here is via toboggan.

The shopkeeper at the North Pole Toboggan Rental Shop is having a bad day. "Something's wrong with our computers; we can't log in!" You ask if you can take a look.

Their password database seems to be a little corrupted: some of the passwords wouldn't have been allowed by the Official Toboggan Corporate Policy that was in effect when they were chosen.

To try to debug the problem, they have created a list (your puzzle input) of passwords (according to the corrupted database) and the corporate policy when that password was set.

For example, suppose you have the following list:

1-3 a: abcde 1-3 b: cdefg 2-9 c: ccccccccc Each line gives the password policy and then the password. The password policy indicates the lowest and highest number of times a given letter must appear for the password to be valid. For example, 1-3 a means that the password must contain a at least 1 time and at most 3 times.

In the above example, 2 passwords are valid. The middle password, cdefg, is not; it contains no instances of b, but needs at least 1. The first and third passwords are valid: they contain one a or nine c, both within the limits of their respective policies.

How many passwords are valid according to their policies?

--- Part Two --- While it appears you validated the passwords correctly, they don't seem to be what the Official Toboggan Corporate Authentication System is expecting.

The shopkeeper suddenly realizes that he just accidentally explained the password policy rules from his old job at the sled rental place down the street! The Official Toboggan Corporate Policy actually works a little differently.

Each policy actually describes two positions in the password, where 1 means the first character, 2 means the second character, and so on. (Be careful; Toboggan Corporate Policies have no concept of "index zero"!) Exactly one of these positions must contain the given letter. Other occurrences of the letter are irrelevant for the purposes of policy enforcement.

Given the same example list from above:

1-3 a: abcde is valid: position 1 contains a and position 3 does not. 1-3 b: cdefg is invalid: neither position 1 nor position 3 contains b. 2-9 c: ccccccccc is invalid: both position 2 and position 9 contain c. How many passwords are valid according to the new interpretation of the policies?

func Day21 ¶

func Day21()

func Day22 ¶

func Day22()

func Day23 ¶

func Day23()

func Day24 ¶

func Day24()

func Day3 ¶

func Day3()

--- Day 3: Toboggan Trajectory --- With the toboggan login problems resolved, you set off toward the airport. While travel by toboggan might be easy, it's certainly not safe: there's very minimal steering and the area is covered in trees. You'll need to see which angles will take you near the fewest trees.

Due to the local geology, trees in this area only grow on exact integer coordinates in a grid. You make a map (your puzzle input) of the open squares (.) and trees (#) you can see. For example:

..##....... #...#...#.. .#....#..#. ..#.#...#.# .#...##..#. ..#.##..... .#.#.#....# .#........# #.##...#... #...##....# .#..#...#.# These aren't the only trees, though; due to something you read about once involving arboreal genetics and biome stability, the same pattern repeats to the right many times:

..##.........##.........##.........##.........##.........##....... ---> #...#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#.. .#....#..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#. ..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.# .#...##..#..#...##..#..#...##..#..#...##..#..#...##..#..#...##..#. ..#.##.......#.##.......#.##.......#.##.......#.##.......#.##..... ---> .#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....# .#........#.#........#.#........#.#........#.#........#.#........# #.##...#...#.##...#...#.##...#...#.##...#...#.##...#...#.##...#... #...##....##...##....##...##....##...##....##...##....##...##....# .#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.#.#..#...#.# ---> You start on the open square (.) in the top-left corner and need to reach the bottom (below the bottom-most row on your map).

The toboggan can only follow a few specific slopes (you opted for a cheaper model that prefers rational numbers); start by counting all the trees you would encounter for the slope right 3, down 1:

From your starting position at the top-left, check the position that is right 3 and down 1. Then, check the position that is right 3 and down 1 from there, and so on until you go past the bottom of the map.

The locations you'd check in the above example are marked here with O where there was an open square and X where there was a tree:

..##.........##.........##.........##.........##.........##....... ---> #..O#...#..#...#...#..#...#...#..#...#...#..#...#...#..#...#...#.. .#....X..#..#....#..#..#....#..#..#....#..#..#....#..#..#....#..#. ..#.#...#O#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.#..#.#...#.# .#...##..#..X...##..#..#...##..#..#...##..#..#...##..#..#...##..#. ..#.##.......#.X#.......#.##.......#.##.......#.##.......#.##..... ---> .#.#.#....#.#.#.#.O..#.#.#.#....#.#.#.#....#.#.#.#....#.#.#.#....# .#........#.#........X.#........#.#........#.#........#.#........# #.##...#...#.##...#...#.X#...#...#.##...#...#.##...#...#.##...#... #...##....##...##....##...#X....##...##....##...##....##...##....# .#..#...#.#.#..#...#.#.#..#...X.#.#..#...#.#.#..#...#.#.#..#...#.# ---> In this example, traversing the map using this slope would cause you to encounter 7 trees.

Starting at the top-left corner of your map and following a slope of right 3 and down 1, how many trees would you encounter?

--- Part Two --- Time to check the rest of the slopes - you need to minimize the probability of a sudden arboreal stop, after all.

Determine the number of trees you would encounter if, for each of the following slopes, you start at the top-left corner and traverse the map all the way to the bottom:

Right 1, down 1. Right 3, down 1. (This is the slope you already checked.) Right 5, down 1. Right 7, down 1. Right 1, down 2. In the above example, these slopes would find 2, 7, 3, 4, and 2 tree(s) respectively; multiplied together, these produce the answer 336.

What do you get if you multiply together the number of trees encountered on each of the listed slopes?

func Day4 ¶

func Day4()

--- Day 4: Passport Processing --- You arrive at the airport only to realize that you grabbed your North Pole Credentials instead of your passport. While these documents are extremely similar, North Pole Credentials aren't issued by a country and therefore aren't actually valid documentation for travel in most of the world.

It seems like you're not the only one having problems, though; a very long line has formed for the automatic passport scanners, and the delay could upset your travel itinerary.

Due to some questionable network security, you realize you might be able to solve both of these problems at the same time.

The automatic passport scanners are slow because they're having trouble detecting which passports have all required fields. The expected fields are as follows:

byr (Birth Year) iyr (Issue Year) eyr (Expiration Year) hgt (Height) hcl (Hair Color) ecl (Eye Color) pid (Passport ID) cid (Country ID) Passport data is validated in batch files (your puzzle input). Each passport is represented as a sequence of key:value pairs separated by spaces or newlines. Passports are separated by blank lines.

Here is an example batch file containing four passports:

ecl:gry pid:860033327 eyr:2020 hcl:#fffffd byr:1937 iyr:2017 cid:147 hgt:183cm

iyr:2013 ecl:amb cid:350 eyr:2023 pid:028048884 hcl:#cfa07d byr:1929

hcl:#ae17e1 iyr:2013 eyr:2024 ecl:brn pid:760753108 byr:1931 hgt:179cm

hcl:#cfa07d eyr:2025 pid:166559648 iyr:2011 ecl:brn hgt:59in The first passport is valid - all eight fields are present. The second passport is invalid - it is missing hgt (the Height field).

The third passport is interesting; the only missing field is cid, so it looks like data from North Pole Credentials, not a passport at all! Surely, nobody would mind if you made the system temporarily ignore missing cid fields. Treat this "passport" as valid.

The fourth passport is missing two fields, cid and byr. Missing cid is fine, but missing any other field is not, so this passport is invalid.

According to the above rules, your improved system would report 2 valid passports.

Count the number of valid passports - those that have all required fields. Treat cid as optional. In your batch file, how many passports are valid?

--- Part Two --- The line is moving more quickly now, but you overhear airport security talking about how passports with invalid data are getting through. Better add some data validation, quick!

You can continue to ignore the cid field, but each other field has strict rules about what values are valid for automatic validation:

byr (Birth Year) - four digits; at least 1920 and at most 2002. iyr (Issue Year) - four digits; at least 2010 and at most 2020. eyr (Expiration Year) - four digits; at least 2020 and at most 2030. hgt (Height) - a number followed by either cm or in: If cm, the number must be at least 150 and at most 193. If in, the number must be at least 59 and at most 76. hcl (Hair Color) - a # followed by exactly six characters 0-9 or a-f. ecl (Eye Color) - exactly one of: amb blu brn gry grn hzl oth. pid (Passport ID) - a nine-digit number, including leading zeroes. cid (Country ID) - ignored, missing or not. Your job is to count the passports where all required fields are both present and valid according to the above rules. Here are some example values:

byr valid: 2002 byr invalid: 2003

hgt valid: 60in hgt valid: 190cm hgt invalid: 190in hgt invalid: 190

hcl valid: #123abc hcl invalid: #123abz hcl invalid: 123abc

ecl valid: brn ecl invalid: wat

pid valid: 000000001 pid invalid: 0123456789 Here are some invalid passports:

eyr:1972 cid:100 hcl:#18171d ecl:amb hgt:170 pid:186cm iyr:2018 byr:1926

iyr:2019 hcl:#602927 eyr:1967 hgt:170cm ecl:grn pid:012533040 byr:1946

hcl:dab227 iyr:2012 ecl:brn hgt:182cm pid:021572410 eyr:2020 byr:1992 cid:277

hgt:59cm ecl:zzz eyr:2038 hcl:74454a iyr:2023 pid:3556412378 byr:2007 Here are some valid passports:

pid:087499704 hgt:74in ecl:grn iyr:2012 eyr:2030 byr:1980 hcl:#623a2f

eyr:2029 ecl:blu cid:129 byr:1989 iyr:2014 pid:896056539 hcl:#a97842 hgt:165cm

hcl:#888785 hgt:164cm byr:2001 iyr:2015 cid:88 pid:545766238 ecl:hzl eyr:2022

iyr:2010 hgt:158cm hcl:#b6652a ecl:blu byr:1944 eyr:2021 pid:093154719 Count the number of valid passports - those that have all required fields and valid values. Continue to treat cid as optional. In your batch file, how many passports are valid?

func Day5 ¶

func Day5()

--- Day 5: Binary Boarding --- You board your plane only to discover a new problem: you dropped your boarding pass! You aren't sure which seat is yours, and all of the flight attendants are busy with the flood of people that suddenly made it through passport control.

You write a quick program to use your phone's camera to scan all of the nearby boarding passes (your puzzle input); perhaps you can find your seat through process of elimination.

Instead of zones or groups, this airline uses binary space partitioning to seat people. A seat might be specified like FBFBBFFRLR, where F means "front", B means "back", L means "left", and R means "right".

The first 7 characters will either be F or B; these specify exactly one of the 128 rows on the plane (numbered 0 through 127). Each letter tells you which half of a region the given seat is in. Start with the whole list of rows; the first letter indicates whether the seat is in the front (0 through 63) or the back (64 through 127). The next letter indicates which half of that region the seat is in, and so on until you're left with exactly one row.

For example, consider just the first seven characters of FBFBBFFRLR:

Start by considering the whole range, rows 0 through 127. F means to take the lower half, keeping rows 0 through 63. B means to take the upper half, keeping rows 32 through 63. F means to take the lower half, keeping rows 32 through 47. B means to take the upper half, keeping rows 40 through 47. B keeps rows 44 through 47. F keeps rows 44 through 45. The final F keeps the lower of the two, row 44. The last three characters will be either L or R; these specify exactly one of the 8 columns of seats on the plane (numbered 0 through 7). The same process as above proceeds again, this time with only three steps. L means to keep the lower half, while R means to keep the upper half.

For example, consider just the last 3 characters of FBFBBFFRLR:

Start by considering the whole range, columns 0 through 7. R means to take the upper half, keeping columns 4 through 7. L means to take the lower half, keeping columns 4 through 5. The final R keeps the upper of the two, column 5. So, decoding FBFBBFFRLR reveals that it is the seat at row 44, column 5.

Every seat also has a unique seat ID: multiply the row by 8, then add the column. In this example, the seat has ID 44 * 8 + 5 = 357.

Here are some other boarding passes:

BFFFBBFRRR: row 70, column 7, seat ID 567. FFFBBBFRRR: row 14, column 7, seat ID 119. BBFFBBFRLL: row 102, column 4, seat ID 820. As a sanity check, look through your list of boarding passes. What is the highest seat ID on a boarding pass?

--- Part Two --- Ding! The "fasten seat belt" signs have turned on. Time to find your seat.

It's a completely full flight, so your seat should be the only missing boarding pass in your list. However, there's a catch: some of the seats at the very front and back of the plane don't exist on this aircraft, so they'll be missing from your list as well.

Your seat wasn't at the very front or back, though; the seats with IDs +1 and -1 from yours will be in your list.

What is the ID of your seat?

func Day6 ¶

func Day6()

--- Day 6: Custom Customs --- As your flight approaches the regional airport where you'll switch to a much larger plane, customs declaration forms are distributed to the passengers.

The form asks a series of 26 yes-or-no questions marked a through z. All you need to do is identify the questions for which anyone in your group answers "yes". Since your group is just you, this doesn't take very long.

However, the person sitting next to you seems to be experiencing a language barrier and asks if you can help. For each of the people in their group, you write down the questions for which they answer "yes", one per line. For example:

abcx abcy abcz In this group, there are 6 questions to which anyone answered "yes": a, b, c, x, y, and z. (Duplicate answers to the same question don't count extra; each question counts at most once.)

Another group asks for your help, then another, and eventually you've collected answers from every group on the plane (your puzzle input). Each group's answers are separated by a blank line, and within each group, each person's answers are on a single line. For example:

abc

a b c

ab ac

a a a a

b This list represents answers from five groups:

The first group contains one person who answered "yes" to 3 questions: a, b, and c. The second group contains three people; combined, they answered "yes" to 3 questions: a, b, and c. The third group contains two people; combined, they answered "yes" to 3 questions: a, b, and c. The fourth group contains four people; combined, they answered "yes" to only 1 question, a. The last group contains one person who answered "yes" to only 1 question, b. In this example, the sum of these counts is 3 + 3 + 3 + 1 + 1 = 11.

For each group, count the number of questions to which anyone answered "yes". What is the sum of those counts?

--- Part Two --- As you finish the last group's customs declaration, you notice that you misread one word in the instructions:

You don't need to identify the questions to which anyone answered "yes"; you need to identify the questions to which everyone answered "yes"!

Using the same example as above:

abc

a b c

ab ac

a a a a

b This list represents answers from five groups:

In the first group, everyone (all 1 person) answered "yes" to 3 questions: a, b, and c. In the second group, there is no question to which everyone answered "yes". In the third group, everyone answered yes to only 1 question, a. Since some people did not answer "yes" to b or c, they don't count. In the fourth group, everyone answered yes to only 1 question, a. In the fifth group, everyone (all 1 person) answered "yes" to 1 question, b. In this example, the sum of these counts is 3 + 0 + 1 + 1 + 1 = 6.

For each group, count the number of questions to which everyone answered "yes". What is the sum of those counts?

func Day7 ¶

func Day7()

func Day8 ¶

func Day8()

func Day9 ¶

func Day9()

--- 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?

--- Part Two --- The final step in breaking the XMAS encryption relies on the invalid number you just found: you must find a contiguous set of at least two numbers in your list which sum to the invalid number from step 1.

Again consider the above example:

35 20 15 25 47 40 62 55 65 95 102 117 150 182 127 219 299 277 309 576 In this list, adding up all of the numbers from 15 through 40 produces the invalid number from step 1, 127. (Of course, the contiguous set of numbers in your actual list might be much longer.)

To find the encryption weakness, add together the smallest and largest number in this contiguous range; in this example, these are 15 and 47, producing 62.

What is the encryption weakness in your XMAS-encrypted list of numbers?