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2023 Day 12 Complete!

This commit is contained in:
Brian Buller 2023-12-12 07:51:40 -06:00
parent 77d53011a1
commit ea9a19d034
4 changed files with 1337 additions and 0 deletions
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package main
import (
"fmt"
"strings"
h "git.bullercodeworks.com/brian/adventofcode/helpers"
)
func main() {
inp := h.StdinToStringSlice()
part1(inp)
fmt.Println()
part2(inp)
}
func part1(input []string) {
var records []string
var groups [][]int
for _, ln := range input {
pts := strings.Fields(ln)
records = append(records, pts[0])
var grp []int
for _, n := range strings.Split(pts[1], ",") {
grp = append(grp, h.Atoi(n))
}
groups = append(groups, grp)
}
var result int
for i := range records {
// Initialize cache for this row
cache := make(map[string]int)
// And solve the row
result += solveRow(0, 0, records[i], groups[i], cache)
}
fmt.Println("# Part 1")
fmt.Println(result)
}
func part2(input []string) {
var records []string
var groups [][]int
for _, ln := range input {
pts := strings.Fields(ln)
// Unfold the paper
pts[0] = fmt.Sprintf("%s?%s?%s?%s?%s", pts[0], pts[0], pts[0], pts[0], pts[0])
pts[1] = fmt.Sprintf("%s,%s,%s,%s,%s", pts[1], pts[1], pts[1], pts[1], pts[1])
records = append(records, pts[0])
var grp []int
for _, n := range strings.Split(pts[1], ",") {
grp = append(grp, h.Atoi(n))
}
groups = append(groups, grp)
}
var result int
for i := range records {
// Initialize cache for this row
cache := make(map[string]int)
// And solve the row
result += solveRow(0, 0, records[i], groups[i], cache)
}
fmt.Println("# Part 2")
fmt.Println(result)
}
func ck(i, j int) string { return fmt.Sprintf("%d_%d", i, j) }
func solveRow(rbIdx, groupIdx int, record string, group []int, cache map[string]int) int {
if rbIdx >= len(record) {
if groupIdx < len(group) {
return 0
}
return 1
}
if v, ok := cache[ck(rbIdx, groupIdx)]; ok {
return v
}
res := 0
if record[rbIdx] == '.' {
res = solveRow(rbIdx+1, groupIdx, record, group, cache)
} else {
if record[rbIdx] == '?' {
res += solveRow(rbIdx+1, groupIdx, record, group, cache)
}
if groupIdx < len(group) {
count := 0
for k := rbIdx; k < len(record); k++ {
if count > group[groupIdx] || record[k] == '.' || count == group[groupIdx] && record[k] == '?' {
break
}
count += 1
}
if count == group[groupIdx] {
if rbIdx+count < len(record) && record[rbIdx+count] != '#' {
res += solveRow(rbIdx+count+1, groupIdx+1, record, group, cache)
} else {
res += solveRow(rbIdx+count, groupIdx+1, record, group, cache)
}
}
}
}
cache[ck(rbIdx, groupIdx)] = res
return res
}

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[1]Advent of Code
• [2][About]
• [3][Events]
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• [5][Settings]
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br0xen [7](AoC++) 24*
   sub y{[8]2023}
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Our [14]sponsors help make Advent of Code possible:
[15]Tailscale - Secure remote access, powered by WireGuard. Replace your
VPN :)
--- 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.
References
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6. https://adventofcode.com/2023/auth/logout
7. Advent of Code Supporter
https://adventofcode.com/2023/support
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11. https://adventofcode.com/2023/sponsors
12. https://adventofcode.com/2023/leaderboard
13. https://adventofcode.com/2023/stats
14. https://adventofcode.com/2023/sponsors
15. https://tailscale.com/advent
16. https://adventofcode.com/2023/day/11
17. https://en.wikipedia.org/wiki/Onsen
18. https://adventofcode.com/2023
19. https://adventofcode.com/2023/day/12/input
20. https://twitter.com/intent/tweet?text=I%27ve+completed+%22Hot+Springs%22+%2D+Day+12+%2D+Advent+of+Code+2023&url=https%3A%2F%2Fadventofcode%2Ecom%2F2023%2Fday%2F12&related=ericwastl&hashtags=AdventOfCode
21. javascript:void(0);

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