adventofcode/2023/day12/problem

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--- Day 12: Hot Springs ---

   You finally reach the hot springs! You can see steam rising from secluded
   areas attached to the primary, ornate building.

   As you turn to enter, the [16]researcher stops you. "Wait - I thought you
   were looking for the hot springs, weren't you?" You indicate that this
   definitely looks like hot springs to you.

   "Oh, sorry, common mistake! This is actually the [17]onsen! The hot
   springs are next door."

   You look in the direction the researcher is pointing and suddenly notice
   the massive metal helixes towering overhead. "This way!"

   It only takes you a few more steps to reach the main gate of the massive
   fenced-off area containing the springs. You go through the gate and into a
   small administrative building.

   "Hello! What brings you to the hot springs today? Sorry they're not very
   hot right now; we're having a lava shortage at the moment." You ask about
   the missing machine parts for Desert Island.

   "Oh, all of Gear Island is currently offline! Nothing is being
   manufactured at the moment, not until we get more lava to heat our forges.
   And our springs. The springs aren't very springy unless they're hot!"

   "Say, could you go up and see why the lava stopped flowing? The springs
   are too cold for normal operation, but we should be able to find one
   springy enough to launch you up there!"

   There's just one problem - many of the springs have fallen into disrepair,
   so they're not actually sure which springs would even be safe to use!
   Worse yet, their condition records of which springs are damaged (your
   puzzle input) are also damaged! You'll need to help them repair the
   damaged records.

   In the giant field just outside, the springs are arranged into rows. For
   each row, the condition records show every spring and whether it is
   operational (.) or damaged (#). This is the part of the condition records
   that is itself damaged; for some springs, it is simply unknown (?) whether
   the spring is operational or damaged.

   However, the engineer that produced the condition records also duplicated
   some of this information in a different format! After the list of springs
   for a given row, the size of each contiguous group of damaged springs is
   listed in the order those groups appear in the row. This list always
   accounts for every damaged spring, and each number is the entire size of
   its contiguous group (that is, groups are always separated by at least one
   operational spring: #### would always be 4, never 2,2).

   So, condition records with no unknown spring conditions might look like
   this:

 #.#.### 1,1,3
 .#...#....###. 1,1,3
 .#.###.#.###### 1,3,1,6
 ####.#...#... 4,1,1
 #....######..#####. 1,6,5
 .###.##....# 3,2,1

   However, the condition records are partially damaged; some of the springs'
   conditions are actually unknown (?). For example:

 ???.### 1,1,3
 .??..??...?##. 1,1,3
 ?#?#?#?#?#?#?#? 1,3,1,6
 ????.#...#... 4,1,1
 ????.######..#####. 1,6,5
 ?###???????? 3,2,1

   Equipped with this information, it is your job to figure out how many
   different arrangements of operational and broken springs fit the given
   criteria in each row.

   In the first line (???.### 1,1,3), there is exactly one way separate
   groups of one, one, and three broken springs (in that order) can appear in
   that row: the first three unknown springs must be broken, then
   operational, then broken (#.#), making the whole row #.#.###.

   The second line is more interesting: .??..??...?##. 1,1,3 could be a total
   of four different arrangements. The last ? must always be broken (to
   satisfy the final contiguous group of three broken springs), and each ??
   must hide exactly one of the two broken springs. (Neither ?? could be both
   broken springs or they would form a single contiguous group of two; if
   that were true, the numbers afterward would have been 2,3 instead.) Since
   each ?? can either be #. or .#, there are four possible arrangements of
   springs.

   The last line is actually consistent with ten different arrangements!
   Because the first number is 3, the first and second ? must both be . (if
   either were #, the first number would have to be 4 or higher). However,
   the remaining run of unknown spring conditions have many different ways
   they could hold groups of two and one broken springs:

 ?###???????? 3,2,1
 .###.##.#...
 .###.##..#..
 .###.##...#.
 .###.##....#
 .###..##.#..
 .###..##..#.
 .###..##...#
 .###...##.#.
 .###...##..#
 .###....##.#

   In this example, the number of possible arrangements for each row is:

     • ???.### 1,1,3 - 1 arrangement
     • .??..??...?##. 1,1,3 - 4 arrangements
     • ?#?#?#?#?#?#?#? 1,3,1,6 - 1 arrangement
     • ????.#...#... 4,1,1 - 1 arrangement
     • ????.######..#####. 1,6,5 - 4 arrangements
     • ?###???????? 3,2,1 - 10 arrangements

   Adding all of the possible arrangement counts together produces a total of
   21 arrangements.

   For each row, count all of the different arrangements of operational and
   broken springs that meet the given criteria. What is the sum of those
   counts?

   Your puzzle answer was 7716.

--- Part Two ---

   As you look out at the field of springs, you feel like there are way more
   springs than the condition records list. When you examine the records, you
   discover that they were actually folded up this whole time!

   To unfold the records, on each row, replace the list of spring conditions
   with five copies of itself (separated by ?) and replace the list of
   contiguous groups of damaged springs with five copies of itself (separated
   by ,).

   So, this row:

 .# 1

   Would become:

 .#?.#?.#?.#?.# 1,1,1,1,1

   The first line of the above example would become:

 ???.###????.###????.###????.###????.### 1,1,3,1,1,3,1,1,3,1,1,3,1,1,3

   In the above example, after unfolding, the number of possible arrangements
   for some rows is now much larger:

     • ???.### 1,1,3 - 1 arrangement
     • .??..??...?##. 1,1,3 - 16384 arrangements
     • ?#?#?#?#?#?#?#? 1,3,1,6 - 1 arrangement
     • ????.#...#... 4,1,1 - 16 arrangements
     • ????.######..#####. 1,6,5 - 2500 arrangements
     • ?###???????? 3,2,1 - 506250 arrangements

   After unfolding, adding all of the possible arrangement counts together
   produces 525152.

   Unfold your condition records; what is the new sum of possible arrangement
   counts?

   Your puzzle answer was 18716325559999.

   Both parts of this puzzle are complete! They provide two gold stars: **

   At this point, you should [18]return to your Advent calendar and try
   another puzzle.

   If you still want to see it, you can [19]get your puzzle input.

   You can also [Shareon [20]Twitter [21]Mastodon] this puzzle.

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