2020: Day 2

Setup

The original challenge

My data

Part 1

Part 2

Part 1: Number of letters

We need to find how many passwords are valid according to their policy. The policies and passwords are given as follows:

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.

library(tidyverse)

First load the libraries we’ll need. We then read in the data and use tidyr functions to separate out the parts of the policy and the password, making sure to convert the columns to numeric as appropriate:

passwords <- 
  read_tsv(here::here("2020", "day", "2", "input"),
           col_names = FALSE) %>%
  separate(X1, c("policy", "password"), sep = ":") %>%
  separate(policy, c("count", "letter"), sep = " ") %>%
  separate(count, c("min", "max")) %>%
  mutate(min = as.integer(min),
         max = as.integer(max))

Next, we use the stringr function str_count() to count how many times the given letter appears in the password, and conditional logic to check whether it is repeated within the specified number of times. Because TRUE has a numeric value of 1 and FALSE has a numeric value of 0, we can sum the resulting column to get a count of how many passwords are valid according to their policies.

passwords %>%
  mutate(count = str_count(password, letter)) %>%
  mutate(password_in_policy = if_else(
    count >= min & count <= max, TRUE, FALSE)) %>%
  summarise(correct = sum(password_in_policy)) %>%
  pull(correct)

[1] 625

Part 2

Now the policy is interpreted differently. Each policy actually describes two positions in the password, where 1 means the first character, 2 means the second character, and so on. Exactly one of these positions must contain the given letter. How many are valid now?

There were a couple of gotchas here. When I used separate() in the previous part, I had inadvertently left a leading whitespace in front of the password, something that was messing up my indexing with str_sub. Using str_trim() first cleared that up. Also, we need exactly one of the positions to match. | is an inclusive or. We need xor() for exclusive or instead.

passwords %>%
  mutate(password = str_trim(password)) %>%
  mutate(pos1_letter = str_sub(password, min, min),
         pos2_letter = str_sub(password, max, max)) %>%
  mutate(match_one = xor(pos1_letter == letter, 
                         pos2_letter == letter)) %>%
  summarise(correct = sum(match_one)) %>%
  pull(correct) 

[1] 391

Session info

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─ Session info ───────────────────────────────────────────────────────────────
 setting  value
 version  R version 4.3.1 (2023-06-16)
 os       macOS Sonoma 14.0
 system   aarch64, darwin20
 ui       X11
 language (EN)
 collate  en_US.UTF-8
 ctype    en_US.UTF-8
 tz       Europe/London
 date     2023-11-06
 pandoc   3.1.1 @ /Applications/RStudio.app/Contents/Resources/app/quarto/bin/tools/ (via rmarkdown)
 quarto   1.4.466 @ /usr/local/bin/quarto

─ Packages ───────────────────────────────────────────────────────────────────
 package     * version date (UTC) lib source
 dplyr       * 1.1.2   2023-04-20 [1] CRAN (R 4.3.0)
 forcats     * 1.0.0   2023-01-29 [1] CRAN (R 4.3.0)
 ggplot2     * 3.4.2   2023-04-03 [1] CRAN (R 4.3.0)
 lubridate   * 1.9.2   2023-02-10 [1] CRAN (R 4.3.0)
 purrr       * 1.0.1   2023-01-10 [1] CRAN (R 4.3.0)
 readr       * 2.1.4   2023-02-10 [1] CRAN (R 4.3.0)
 sessioninfo * 1.2.2   2021-12-06 [1] CRAN (R 4.3.0)
 stringr     * 1.5.0   2022-12-02 [1] CRAN (R 4.3.0)
 tibble      * 3.2.1   2023-03-20 [1] CRAN (R 4.3.0)
 tidyr       * 1.3.0   2023-01-24 [1] CRAN (R 4.3.0)
 tidyverse   * 2.0.0   2023-02-22 [1] CRAN (R 4.3.0)

 [1] /Users/ellakaye/Library/R/arm64/4.3/library
 [2] /Library/Frameworks/R.framework/Versions/4.3-arm64/Resources/library

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