Advent of Code 2021: Day 8
The Data
Day 8 is really more a puzzle than a coding challenge. We’re given the inputs and outputs from a broken 7-segment display (like on a digital clock). We can imagine each segment of the clock as having a corresponding letter, from a (top) to g (bottom).
We can create a dictionary that will translate a string of letters into a corresponding number following the image above.
#Actual letter number combos
correct_codes <- 0:9
names(correct_codes) <- c("abcefg", "cf", "acdeg", "acdfg", "bcdf", "abdfg", "abdefg", "acf", "abcdefg", "abcdfg")The only problem is that the data we are given doesn’t have the correct letters! Each row of the input data we are given includes all 10 possible numbers (0-9) but the letters used are wrong. We have to work out a number of rules to decode this information! See the full explanation for today’s challenge here.
#Load data
day8_data <- readr::read_delim(file = "./data/Day8.txt", delim = " | ",
col_names = c("inputs", "outputs"), show_col_types = FALSE)
head(day8_data)## # A tibble: 6 × 2
## inputs outputs
## <chr> <chr>
## 1 fdceba bafdgc abeg afbdgec gbeacd abced bgc fcdge bg bedgc bafdec cgefd gcebd…
## 2 gbfac fegbda fcedagb bea ea abcdef dgbfe gfabe dgea gbdfec gdea bgefdc bea ef…
## 3 eg dagef gbcfeda ageb cegbfd gfe dbefa facdg abfged cedbaf befda daefb egf gc…
## 4 edgacfb gcfd dgb degfab bcega bdagc cgafbd fbacd gd fceabd fbdac gd gdbcaf dgb
## 5 eaf bedgaf dbafc bfceag fedcbg eafdcgb debfa ae adge gdebf abcdfeg febdg ae d…
## 6 afbdc aefg ea edbacfg dbefg eab gcbfde abecgd bgefad bfdae gfea bfdea gbdef f…Convert our data into nested lists so that it’s easier to work with.
day8_data <- day8_data %>%
mutate(inputs = stringr::str_split(inputs, pattern = " "),
outputs = stringr::str_split(outputs, pattern = " "))The Challenges
Challenge 1 & 2
In combination, the two challenges from Day 8 aim to decode the outputs so we’ll combine them together. The first rules we can use are simple. There are 4 numbers that can be easily identified just by the number of characters in the input.
- 2 characters = 1
- 3 characters = 7
- 4 characters = 4
- 7 characters = 8
So if we see an input ‘ab’ this must correspond to the number 1 or ‘cf’ when correctly translated to the 7-segment display. But we don’t know if ‘a’ is ‘c’ or ‘f’. We need some more rules to work this out. We’ll apply all these rules below:
numbers <- NULL
for (i in 1:nrow(day8_data)) {
#Split input into list with individual letters
split_input <- stringr::str_extract_all(day8_data$inputs[[i]], pattern = "[a-z]{1}")
split_output <- stringr::str_extract_all(day8_data$outputs[[i]], pattern = "[a-z]{1}")
#If there are two values, it must correspond to c and f
#Remove from the list afterwards
number1 <- split_input[unlist(lapply(split_input, \(letters) length(letters) == 2))][[1]]
split_input[unlist(lapply(split_input, \(letters) length(letters) == 2))] <- NULL
#If there are three values, it must correspond to a, c and f
#Remove from the list afterwards
number7 <- split_input[unlist(lapply(split_input, \(letters) length(letters) == 3))][[1]]
split_input[unlist(lapply(split_input, \(letters) length(letters) == 3))] <- NULL
#If there are four values, it must correspond to b, c, d and f
#Remove from the list afterwards
number4 <- split_input[unlist(lapply(split_input, \(letters) length(letters) == 4))][[1]]
split_input[unlist(lapply(split_input, \(letters) length(letters) == 4))] <- NULL
#If there are seven values, it must correspond to a-g
#Remove from the list afterwards
number8 <- split_input[unlist(lapply(split_input, \(letters) length(letters) == 7))][[1]]
split_input[unlist(lapply(split_input, \(letters) length(letters) == 7))] <- NULL
#Letter that's not shared between 1 and 7 must be a
a <- setdiff(number7, number1)
#Value of 6 characters that contains the letters in 4 must be 9.
#Remove from the list afterwards
number9 <- split_input[unlist(lapply(split_input, \(letters) length(letters) == 6 & all(number4 %in% letters)))][[1]]
split_input[unlist(lapply(split_input, \(letters) length(letters) == 6 & all(number4 %in% letters)))] <- NULL
#The missing letter in 9 is e
e <- setdiff(number8, number9)
#Remaining 6 letter character that includes the same letters as number1 must be 0.
#Remove from the list afterwards
number0 <- split_input[unlist(lapply(split_input, \(letters) length(letters) == 6 & all(number1 %in% letters)))][[1]]
split_input[unlist(lapply(split_input, \(letters) length(letters) == 6 & all(number1 %in% letters)))] <- NULL
#The missing value here is d
d <- setdiff(number8, number0)
#Last remaining length 6 is 6
number6 <- split_input[unlist(lapply(split_input, \(letters) length(letters) == 6))][[1]]
split_input[unlist(lapply(split_input, \(letters) length(letters) == 6))] <- NULL
#Missing value is c
c <- setdiff(number8, number6)
#The letter in 7 that wasn't just extracted is f
f <- setdiff(number1, c)
#Letter in 4 that hasn't yet been deciphered in b
b <- setdiff(number4, c(c, d, f))
#Remaining letter in 0 after removing all known ones is g
g <- setdiff(number0, c(a, b, c, e, f))
#Dictionary
dict <- letters[1:7]
names(dict) <- c(a, b, c, d, e, f, g)
#Now we can decode our output
output <- correct_codes[unlist(lapply(split_output, \(letters) paste(dict[letters][order(dict[letters])], collapse = "")))]
#Paste all the decoded numbers together
numbers <- append(numbers, as.integer(paste(output, collapse = "")))
}numbers## [1] 6233 4675 5572 5197 8513 4356 307 8507 4088 3756 9295 2569 5165 3061 2862
## [16] 1316 7879 2323 2593 5720 6695 7066 9013 9050 4501 2886 1326 734 9840 8003
## [31] 1752 4944 2038 9216 9250 4237 2145 7182 1101 5063 2136 8496 8523 3326 4598
## [46] 227 6537 3390 5065 9306 8460 6335 2664 6446 5550 4303 6955 3520 9376 2586
## [61] 6520 2121 6685 3109 2838 8257 3082 7739 7132 2770 5996 521 3470 940 2060
## [76] 6396 400 6050 2314 8464 262 9628 4719 706 8307 5355 4289 2422 2634 714
## [91] 2016 2778 6695 474 6970 3689 3305 12 5074 8060 5359 9331 8993 3307 9329
## [106] 458 9872 7089 6058 5030 3634 5702 3756 4506 2957 8995 5657 8375 415 2412
## [121] 3021 7241 8368 9298 6666 8567 2648 9621 3215 5533 4972 2236 8329 2582 2929
## [136] 2089 3436 3594 1383 5555 5461 2268 6377 1440 9494 8314 9483 3267 3037 5547
## [151] 9184 4900 6696 7304 1294 1349 8642 6126 7367 5525 3861 3393 2958 5312 2028
## [166] 5058 8513 4258 1780 3937 5568 6236 5159 518 8059 1014 1080 8538 40 6722
## [181] 3193 2519 5877 2460 3348 7329 8622 4380 3629 771 3734 6563 1900 4502 1472
## [196] 6293 7392 8285 9928 7535The answer of our final challenge is the sum of all these values.
sum(numbers)## [1] 982158See previous solutions here:
