# Advent of Code 2021: Days 6-10

My solutions to the #AdventOfCode2021 coding challenges, days 6 through 10.

Python
R
Author

Taylor Dunn

Published

December 6, 2021

R setup
library(tidyverse)
library(gt)
library(lubridate)
Python setup
library(reticulate)
use_virtualenv("r-reticulate")
import numpy as np
import pandas as pd

## Day 6: Lanternfish

day6 <- read_lines("day06-input.txt")
str_trunc(day6, 70)
[1] "5,1,4,1,5,1,1,5,4,4,4,4,5,1,2,2,1,3,4,1,1,5,1,5,2,2,2,2,1,4,2,4,3,3..."

### Part 1

A massive school of glowing 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, each lanternfish creates a new lanternfish 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.

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. Find a way to simulate lanternfish. How many lanternfish would there be after 80 days?

This is simple enough to do with a vector and a loop over 80 days:

fish_timers <- strsplit(day6, ",") %>% unlist() %>% as.numeric()

for (day in seq(1:80)) {
# Decrease each timer by one
fish_timers <- map_dbl(fish_timers, ~{.x - 1})

# Look for elapsed timers
elapsed_timers <- fish_timers < 0
# Reset them to 6
fish_timers[elapsed_timers] <- 6
# Add new fish with timers starting at 8
fish_timers <- c(fish_timers, rep(8, sum(elapsed_timers)))
}

# Count the number of fish after 80 days
length(fish_timers)
[1] 355386

For Python, I’ll use a list:

fish_timers = [int(x) for x in r.day6.split(',')]

for day in range(80):
fish_timers = [f - 1 for f in fish_timers]

n_new_fish = fish_timers.count(-1)
fish_timers = [6 if f == -1 else f for f in fish_timers]

if n_new_fish > 0:
fish_timers.extend([8] * n_new_fish)

len(fish_timers) 
355386

### Part 2

Suppose the lanternfish live forever and have unlimited food and space. Would they take over the entire ocean? How many lanternfish would there be after 256 days?

For this part, the list of fish would quickly become too large to store in memory after 256 days. Instead, find the number of fish at each timer value:

fish_timers_df <- tibble(
timer = strsplit(day6, ",") %>% unlist() %>% as.numeric()
) %>%
count(timer, name = "n_fish") %>%
# Need to fill in the missing days with 0 fish
bind_rows(tibble(timer = c(0, 6, 7, 8), n_fish = c(0, 0, 0, 0))) %>%
arrange(timer)
fish_timers_df
# A tibble: 9 × 2
timer n_fish
<dbl>  <dbl>
1     0      0
2     1     96
3     2     54
4     3     48
5     4     51
6     5     51
7     6      0
8     7      0
9     8      0

Now it is a matter of looping through the days and updating the counts:

for (day in 1:256) {
# Decrease the timer values by 1
fish_timers_df <- fish_timers_df %>% mutate(timer = timer - 1)
# Count new fish
new_fish <- fish_timers_df$n_fish[fish_timers_df$timer == -1]

# Add new fish with timers at 8
fish_timers_df <- fish_timers_df %>%
bind_rows(tibble(timer = 8, n_fish = new_fish))

# Reset elapsed timers to 6
fish_timers_df$n_fish[fish_timers_df$timer == 6] <-
fish_timers_df$n_fish[fish_timers_df$timer == 6] +
new_fish

# Remove the -1 values now that they've been accounted for
fish_timers_df <- fish_timers_df %>% filter(timer != -1)
}

fish_timers_df
# A tibble: 9 × 2
timer       n_fish
<dbl>        <dbl>
1     0 141979186737
2     1 163480438824
3     2 172999320549
4     3 189419502198
5     4 211569179655
6     5 220260612873
7     6 255636828738
8     7 118721109675
9     8 139349146560

That’s a lot of fish. To get the answer, I need to format the sum so that it is not printed in scientific notation:

format(sum(fish_timers_dfn_fish), scientific = FALSE) [1] "1613415325809" Python: fish_timers_df = pd.DataFrame({'timer': [int(x) for x in r.day6.split(',')]}) fish_timers_df = fish_timers_df.value_counts('timer') \ .rename_axis('n_fish') \ .reindex(range(9), fill_value = 0) for day in range(256): n_new_fish = fish_timers_df[0] for i in range(1, len(fish_timers_df)): fish_timers_df[i-1] = fish_timers_df[i] fish_timers_df[6] += n_new_fish fish_timers_df[8] = n_new_fish fish_timers_df.sum() 1613415325809 ## Day 7: The Treachery of Whales day7 <- read_lines("day07-input.txt") str_trunc(day7, 70) [1] "1101,1,29,67,1102,0,1,65,1008,65,35,66,1005,66,28,1,67,65,20,4,0,10..." ### Part 1 A giant 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. Each change of 1 step in horizontal position of a single crab costs 1 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? The amount of fuel spent by a crab $$i$$ at position $$x_i$$ moving to position $$x_0$$ can be represented as the absolute deviation $$|x_i - x_0|$$. We seek the position $$x_0$$ that minimizes the sum of absolute deviations for all $$N$$ crabs: $\sum_i^N |x_i - x_0|$ My mathematical proof skills are rusty (it has been about 8 years since I took Mathematical Reasoning at UPEI), but Google tells me that the median minimizes the sum of absolute deviations. The median position of the input is: crab_positions <- strsplit(day7, ",") %>% unlist() %>% as.numeric() median(crab_positions) [1] 331 And so the total fuel taken for each crab to move to this position is: sum(abs(crab_positions - median(crab_positions))) [1] 349769 Python: crab_positions = [int(x) for x in r.day7.split(',')] sum([np.abs(x - np.median(crab_positions)) for x in crab_positions]) 349769.0 ### Part 2 The crabs don’t seem interested in your proposed solution. Perhaps you misunderstand crab engineering? As it turns out, crab submarine engines don’t burn fuel at a constant rate. 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. 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? My first thought for calculating this fuel was the cumulative sum. For example, moving 7 steps would cost fuel equal to the last value of this cumsum(): cumsum(1:7) [1] 1 3 6 10 15 21 28 But then I figured there is probably a useful formula for this sum, and another lazy Google search tells me that these are called triangular numbers with the simple formula: $\sum_{i=1}^n k = \frac{n (n + 1)}{2}.$ And so we want to find $$x_0$$ that minimizes: $\sum_i^N \frac{|x_i - x_0|(|x_i - x_0| + 1)}{2}.$ I’m not aware of a simple solution (like the median in part 1) to this optimization problem. The brute force way is to loop over values of $$x_0$$, ranging from $$\text{min}(x_i)$$ to $$\text{max}(x_i)$$: fuel_spent <- tibble( x0 = seq(min(crab_positions), max(crab_positions)) ) %>% mutate( crab_dist = map(x0, ~abs(crab_positions - .x)), fuel_spent = map_dbl(crab_dist, ~sum(.x * (.x + 1) / 2)) ) %>% arrange(fuel_spent) head(fuel_spent) # A tibble: 6 × 3 x0 crab_dist fuel_spent <int> <list> <dbl> 1 479 <dbl [1,000]> 99540554 2 480 <dbl [1,000]> 99540639 3 478 <dbl [1,000]> 99541469 4 481 <dbl [1,000]> 99541724 5 477 <dbl [1,000]> 99543385 6 482 <dbl [1,000]> 99543809 We find that $$x_0$$ = 479 minimizes the fuel with 99540554 spent. Use a pandas.Series to do the same: fuel_spent = pd.Series(dtype = np.float64) for x0 in range(min(crab_positions), max(crab_positions)): crab_dist = [np.abs(x - x0) for x in crab_positions] fuel_spent.at[x0] = sum([d * (d + 1) / 2 for d in crab_dist]) fuel_spent.sort_values().head() 479 99540554.0 480 99540639.0 478 99541469.0 481 99541724.0 477 99543385.0 dtype: float64 ## Day 9: Smoke Basin day9 <- read_lines("day09-input.txt") head(day9) %>% str_trunc(70) [1] "4210129998765678999876598999987654567899891099876534567899989543458..." [2] "4321298987674567998965467898996543478998789987654323478998775412367..." [3] "5734987794543678987654356987987656567897678998943214589987654323456..." [4] "7649876543212347898765567896598767678976567899876323678998775454868..." [5] "8756989652103456999898678966459878989785456989965456789999976565699..." [6] "9897898773212367895939989656346989799654347778976587997896989876989..." ### Part 1 These caves seem to be lava tubes. Parts are even still volcanically active; small hydrothermal vents release smoke into the caves that slowly settles like rain. 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. 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.) The risk level of a low point is 1 plus its height. Find all of the low points on your heightmap. What is the sum of the risk levels of all low points on your heightmap? I’ll use a base R matrix to represent the height map of the cave floor: # Split the columns and convert to integers height_map <- day9 %>% map(~unlist(strsplit(.x, "")) %>% as.integer) # Convert it to a matrix height_map <- matrix(unlist(height_map), nrow = length(height_map), byrow = TRUE) height_map[1:10, 1:10]  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 4 2 1 0 1 2 9 9 9 8 [2,] 4 3 2 1 2 9 8 9 8 7 [3,] 5 7 3 4 9 8 7 7 9 4 [4,] 7 6 4 9 8 7 6 5 4 3 [5,] 8 7 5 6 9 8 9 6 5 2 [6,] 9 8 9 7 8 9 8 7 7 3 [7,] 5 9 9 8 9 9 9 7 6 5 [8,] 4 5 6 9 8 9 9 8 7 5 [9,] 3 4 5 6 7 8 9 9 7 6 [10,] 2 3 4 8 9 9 9 9 8 7 I’ll also define a helper function that finds neighboring points (while still within the bounds of the matrix): get_neighbor_coords <- function(row, col, max_row, max_col) { neighbor_coords <- cbind(row + c(-1, 1, 0, 0), col + c(0, 0, -1, 1)) in_bounds <- (neighbor_coords[,1] <= max_row) & (neighbor_coords[,1] > 0) & (neighbor_coords[,2] <= max_col) & (neighbor_coords[,2] > 0) neighbor_coords[in_bounds, , drop = FALSE] } Loop over all points and determine if it is a low point: low_point_heights <- c() for (row in 1:nrow(height_map)) { for (col in 1:ncol(height_map)) { height <- height_map[row, col] # Get adjacent points neighbor_coords <- get_neighbor_coords(row, col, nrow(height_map), ncol(height_map)) neighbor_heights <- height_map[neighbor_coords] # Is it the lowest point? if (all(height < neighbor_heights)) { low_point_heights <- c(low_point_heights, height) } } } low_point_heights  [1] 0 0 0 0 1 3 1 3 1 0 2 2 2 0 1 1 0 0 2 1 0 2 0 0 0 1 3 0 6 2 2 1 1 0 2 1 1 [38] 0 3 1 0 3 0 1 1 0 0 0 4 1 5 0 0 3 2 0 0 2 0 3 1 0 2 0 1 2 0 0 0 2 2 1 1 5 [75] 2 1 0 2 0 1 1 4 1 2 2 1 0 2 1 1 1 0 0 0 2 0 0 2 1 0 0 1 0 4 0 0 0 0 1 5 1 [112] 4 2 2 0 4 2 2 0 1 0 1 0 0 0 1 1 0 3 3 1 1 3 3 0 1 0 2 1 1 0 0 3 0 0 0 1 0 [149] 3 2 1 0 0 1 1 0 0 0 0 0 2 0 0 1 0 0 3 1 0 0 1 3 0 1 0 0 0 0 1 0 2 2 1 1 4 [186] 0 0 3 1 1 6 5 1 0 1 0 0 0 Then add 1 to get the risk level and sum it up: sum(low_point_heights + 1) [1] 417 In Python, represent the height map with a 2d numpy.array: height_map = np.array([[height for height in row] for row in r.day9], dtype = int) height_map[0:9, 0:9] array([[4, 2, 1, 0, 1, 2, 9, 9, 9], [4, 3, 2, 1, 2, 9, 8, 9, 8], [5, 7, 3, 4, 9, 8, 7, 7, 9], [7, 6, 4, 9, 8, 7, 6, 5, 4], [8, 7, 5, 6, 9, 8, 9, 6, 5], [9, 8, 9, 7, 8, 9, 8, 7, 7], [5, 9, 9, 8, 9, 9, 9, 7, 6], [4, 5, 6, 9, 8, 9, 9, 8, 7], [3, 4, 5, 6, 7, 8, 9, 9, 7]]) Then apply the same logic to get the sum of risk levels: def get_neighbor_coords(row, col, max_row, max_col): neighbor_coords = [[row + d for d in [-1, 1, 0, 0]], [col + d for d in [0, 0, -1, 1]]] neighbor_coords = np.array(neighbor_coords).T in_bounds = (neighbor_coords[:, 0] >= 0) & \ (neighbor_coords[:, 0] < max_row) & \ (neighbor_coords[:, 1] >= 0) & \ (neighbor_coords[:, 1] < max_col) return(neighbor_coords[in_bounds, :]) low_point_heights = [] for row in range(height_map.shape[0]): for col in range(height_map.shape[1]): height = height_map[row, col] neighbor_coords = get_neighbor_coords(row, col, height_map.shape[0], height_map.shape[1]) neighbor_heights = [height_map[nc[0], nc[1]] for nc in neighbor_coords] if all([height < nh for nh in neighbor_heights]): low_point_heights.append(height) sum([h + 1 for h in low_point_heights]) 417 ### Part 2 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. What do you get if you multiply together the sizes of the three largest basins? My strategy for part 2 is: • Loop over each point. • If that point has a height <9, start a new basin. • Rewrite that point with 9 (to prevent overcounting). • For each of that points neighbors: • If that point has a height <9, add it to the basin. • Re-write that point with height 9. • Add the points neighbors to a new list. • Repeat the above loop with new neighbors until no neighbors remain. • Compute the number of points in the basin. Basically, I’m finding points that are not 9, spreading out from those points, filling the basins with 9s, and counting those points along the way to get the basin size. basin_size <- c() for (row in 1:nrow(height_map)) { for (col in 1:ncol(height_map)) { height <- height_map[row, col] if (height < 9) { basin_points <- c(height) height_map[row, col] <- 9 neighbor_coords <- get_neighbor_coords(row, col, nrow(height_map), ncol(height_map)) while (nrow(neighbor_coords) > 0) { new_neighbor_coords <- matrix(nrow = 0, ncol = 2) for (neighbor in 1:nrow(neighbor_coords)) { nc <- neighbor_coords[neighbor, , drop = FALSE] height <- height_map[nc] if (height < 9) { basin_points <- c(basin_points, height) height_map[nc] <- 9 new_neighbor_coords <- rbind( new_neighbor_coords, get_neighbor_coords(nc[1], nc[2], nrow(height_map), ncol(height_map)) ) } } neighbor_coords <- new_neighbor_coords } basin_size <- c(basin_size, length(basin_points)) } } } basin_size  [1] 26 96 34 37 64 2 58 52 2 53 32 54 55 60 4 6 68 39 [19] 14 69 30 22 5 21 33 64 38 96 45 21 15 29 16 89 3 7 [37] 55 39 7 32 3 85 74 41 60 23 18 84 4 32 15 97 8 8 [55] 50 15 24 66 20 68 45 84 42 67 68 70 10 16 49 63 4 19 [73] 2 31 15 51 9 22 52 36 3 28 13 2 79 18 12 9 68 2 [91] 83 4 39 95 75 115 41 7 79 2 4 27 85 26 25 87 16 8 [109] 5 62 48 8 22 59 39 2 25 61 89 25 69 31 22 40 7 73 [127] 8 4 38 50 23 57 30 18 41 92 4 69 2 48 86 20 52 94 [145] 29 36 7 30 37 36 75 84 91 24 26 4 16 103 93 74 44 5 [163] 31 6 51 5 15 83 93 73 18 94 22 61 72 8 8 53 53 45 [181] 40 55 6 48 33 32 20 9 3 18 14 20 19 7 4 3 4 2 The product of the three largest basins: sort(basin_size, decreasing = TRUE)[1:3] %>% prod() [1] 1148965 And in Python: basin_size = [] for row in range(height_map.shape[0]): for col in range(height_map.shape[1]): height = height_map[row, col] if height < 9: basin_points = [height] height_map[row, col] = 9 neighbor_coords = get_neighbor_coords(row, col, height_map.shape[0], height_map.shape[1]) while len(neighbor_coords) > 0: new_neighbor_coords = np.empty((0, 2), dtype = int) for nc in neighbor_coords: height = height_map[nc[0], nc[1]] if height < 9: basin_points.append(height) height_map[nc[0], nc[1]] = 9 new_neighbor_coords = np.append( new_neighbor_coords, get_neighbor_coords(nc[0], nc[1], height_map.shape[0], height_map.shape[1]), axis = 0 ) neighbor_coords = new_neighbor_coords basin_size.append(len(basin_points)) basin_size.sort(reverse = True) np.prod(basin_size[0:3]) 1148965 ## Day 10: Syntax Scoring day10 <- read_lines("day10-input.txt") head(day10) %>% str_trunc(70) [1] "(<{([{([[[<{(<(){}>({}()))}[(<[]<>>{<>})[{{}()}<{}[]>]]><[[<{}{}>([..." [2] "<[<<{<[(([[[[<[][]>][(<>()){<>()}]]][<{<<>{}>(()<>)}><([<>{}]{()()}..." [3] "<(({(<{<({[<[[[]()][<>[]]]{{{}()}({}[])}>]{[{<<>[]>[()[]]}][[[{}[]]..." [4] "{{[{[[[{{{{{[([]())(<>())](<[]<>>{[]{}})}([[(){}]<<>()>]<<()()><{}{..." [5] "({([({([[({([{[]<>}]<<<>{}>[<>[]]>)})({{{[()[]]<()>}(([]<>)<[]()>>}..." [6] "<(((<<<{([[[<{()<>}{{}()}>[({}())([][])]]{{(()())(<>{})}(<{}><[]{}>..." ### Part 1 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. 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. Find the first illegal character in each corrupted line of the navigation subsystem. What is the total syntax error score for those errors? I’ll make heavy use of stringr and regular expressions to solve this problem. First, define a function that iterates over a line and determines if it is corrupt: # Regex representations of the character pairs (requires \\ to escape) pairs <- c("\$$\$$" = "", "\$\$" = "", "\\{\\}" = "", "\\<\\>" = "") remove_pairs <- function(line) { repeat { new_line <- line %>% str_replace_all(pairs) if (new_line == line) return(new_line) else line <- new_line } } line <- "([{<>}])[](<>)" line_corrupt <- "([{<>]])" remove_pairs(line) [1] "" remove_pairs(line_corrupt) [1] "([{]])" Then for corrupt lines, I need to find the location of the wrong closing characters: # Regex representations of mismatched pairs "[>", "(>", "{>", "[}", etc. mismatched_pairs <- c("[\$$\$\\{]\\>", "[\\(\\[\\<]\\}", "[\\(\\{\\<]\$", "[\\[\\{\\<]\$$") find_corrupt_loc <- function(line_reduced) { map_dfr( mismatched_pairs, ~remove_pairs(line_reduced) %>% str_locate(.x) %>% as.data.frame() ) %>% drop_na() } corrupt_loc <- find_corrupt_loc(remove_pairs(line_corrupt)) corrupt_loc  start end 1 3 4 Then I use the end location (the smallest value to get the first occurrence) to determine which character is incorrect: remove_pairs(line_corrupt) %>% str_sub(start = min(corrupt_locend), end = min(corrupt_loc$end)) [1] "]" Apply this to the full input: day10_df <- tibble(line = day10) %>% mutate( line_reduced = map_chr(line, remove_pairs), corrupt_loc = map(line_reduced, find_corrupt_loc) ) corrupt_chars <- day10_df %>% mutate(line_num = 1:n()) %>% # Filter down to just corrupt lines filter(map_int(corrupt_loc, nrow) > 0) %>% transmute( line_num, corrupt_char = map2_chr( line_reduced, corrupt_loc, ~str_sub(.x, start = min(.y$end), end = min(.y$end)) ) ) corrupt_chars # A tibble: 47 × 2 line_num corrupt_char <int> <chr> 1 1 } 2 5 > 3 9 } 4 12 } 5 14 } 6 16 ] 7 17 } 8 18 > 9 21 ] 10 22 ] # … with 37 more rows # ℹ Use print(n = ...) to see more rows Finally, add up the points of the characters: corrupt_chars %>% mutate(score = recode(corrupt_char, ) = 3, ] = 57, } = 1197, > = 25137)) %>% pull(score) %>% sum() [1] 469755 ### Part 2 Now, discard the corrupted lines. The remaining lines are incomplete. Incomplete lines don’t have any incorrect characters - instead, they’re missing some closing characters at the end of the line. To repair the navigation subsystem, you just need to figure out the sequence of closing characters that complete all open chunks in the line. You can only use closing characters ((, ], }, or >), and you must add them in the correct order so that only legal pairs are formed and all chunks end up closed. The score is determined by considering the completion string character-by-character. Start with a total score of 0. Then, for each character, multiply the total score by 5 and then increase the total score by the point value given for the character in the following table: ): 1 point. ]: 2 points. }: 3 points. >: 4 points. Autocomplete tools are an odd bunch: the winner is found by sorting all of the scores and then taking the middle score. (There will always be an odd number of scores to consider.) In this example, the middle score is 288957 because there are the same number of scores smaller and larger than it. Find the completion string for each incomplete line, score the completion strings, and sort the scores. What is the middle score? From part 1, we have these incomplete (and non-corrupt) lines: day10_incomplete <- day10_df %>% filter(map_int(corrupt_loc, nrow) == 0) %>% select(-corrupt_loc) glimpse(day10_incomplete) Rows: 47 Columns: 2$ line         <chr> "<[<<{<[(([[[[<[][]>][(<>()){<>()}]]][<{<<>{}>(()<>)}><([…
$line_reduced <chr> "<[<<{<[({{<{", "<(({(<{<((", "{{[{[[[{{((<<({", "<(((<<<… As expected, we have an odd number of lines so we can get a middle score. Because I’ve already reduced the lines down to line_reduced without matching pairs, I can simply reverse the order, and count the scores. For example: reversed_chars <- remove_pairs("[({(<(())[]>[[{[]{<()<>>") %>% strsplit("") %>% unlist() %>% rev() total_score <- 0 for (char in reversed_chars) { total_score <- total_score * 5 # No need to replace the opening with closing characters total_score <- total_score + switch(char, ( = 1, [ = 2, { = 3, < = 4) } total_score [1] 288957 day10_incomplete <- day10_incomplete %>% mutate( total_score = map_dbl( line_reduced, ~{ reversed_chars <- str_split(.x, "") %>% unlist() %>% rev() total_score <- 0 for (char in reversed_chars) { total_score <- total_score * 5 total_score <- total_score + switch(char, ( = 1, [ = 2, { = 3, < = 4) } total_score } ) ) sort(day10_incomplete$total_score)[(nrow(day10_incomplete) + 1) / 2]
[1] 2762335572

## Stats

My personal stats for this period:

tibble::tribble(
~Part, ~Day, ~Time, ~Rank,
1, 10, "11:11:48", 34029,
2, 10, "11:25:31", 32045,
1, 9, "14:58:48", 42395,
2, 9, "16:04:08", 33905,
1, 8, "16:37:13", 49102,
2, 8, "16:55:20", 34954,
1, 7, "11:23:20", 45816,
2, 7, "11:42:48", 43886,
1, 6, "10:38:42", 41671,
2, 6, "10:58:31", 34999
) %>%
pivot_wider(names_from = Part, values_from = c(Time, Rank),
names_glue = "Part {Part}_{.value}") %>%
mutate(
Time between parts = as.numeric(hms(Part 2_Time) - hms(Part 1_Time),
"minutes") %>% round(1)
) %>%
gt() %>%
tab_spanner_delim(delim = "_", split = "first") %>%
sub_missing(columns = "Time between parts", missing_text = "")
Day Time Rank Time between parts
Part 1 Part 2 Part 1 Part 2
10 11:11:48 11:25:31 34029 32045 13.7
9 14:58:48 16:04:08 42395 33905 65.3
8 16:37:13 16:55:20 49102 34954 18.1
7 11:23:20 11:42:48 45816 43886 19.5
6 10:38:42 10:58:31 41671 34999 19.8

## Reproducibility

Session info
 setting  value
version  R version 4.2.1 (2022-06-23 ucrt)
os       Windows 10 x64 (build 19044)
system   x86_64, mingw32
ui       RTerm
language (EN)
tz       America/Curacao
date     2022-10-27
pandoc   2.18 @ C:/Program Files/RStudio/bin/quarto/bin/tools/ (via rmarkdown)
python:         C:/Users/tdunn/Documents/.virtualenvs/r-reticulate/Scripts/python.exe
libpython:      C:/Users/tdunn/AppData/Local/r-reticulate/r-reticulate/pyenv/pyenv-win/versions/3.9.13/python39.dll
pythonhome:     C:/Users/tdunn/Documents/.virtualenvs/r-reticulate
version:        3.9.13 (tags/v3.9.13:6de2ca5, May 17 2022, 16:36:42) [MSC v.1929 64 bit (AMD64)]
Architecture:   64bit
numpy:          C:/Users/tdunn/Documents/.virtualenvs/r-reticulate/Lib/site-packages/numpy
numpy_version:  1.23.3

NOTE: Python version was forced by use_python function
Git repository
Local:    main C:/Users/tdunn/Documents/tdunn-quarto
Remote:   main @ origin (https://github.com/taylordunn/tdunn-quarto.git)
Head:     [4eb5bf2] 2022-10-26: Added font import to style sheet

## Citation

BibTeX citation:
@online{dunn2021,
author = {Taylor Dunn},
title = {Advent of {Code} 2021: {Days} 6-10},
date = {2021-12-06},