Advent of Code 2023, Day 15

library(tidyverse)
library(here)
library(rlang)

Part 1

Another fairly straight forwad puzzle. We are given an algorithm that we need to follow. The algorithm is as follows:

• Determine the ASCII code for the current character of the string.
• Increase the current value by the ASCII code you just determined.
• Set the current value to itself multiplied by 17.
• Set the current value to the remainder of dividing itself by 256.

As always let’s read in the puzzle input and tidy the input into something we can work with.

data <- read_lines(
  here('posts', 'aoc-2023-d15', 'puzzle-input.txt')
)

data <- data |> 
  str_split(',') |> 
  unlist() |> 
  as_tibble()

data

# A tibble: 4,000 × 1
   value   
   <chr>   
 1 fzl-    
 2 tcjp=8  
 3 vkjr=9  
 4 xs-     
 5 jktcpk=3
 6 gzp-    
 7 kfsxsd- 
 8 zxkv=7  
 9 fxz-    
10 pj=7    
# ℹ 3,990 more rows

Now let’s write a helper function that can help us decode the hash based on the provided algorithm.

hash_decoder <- \(hash) {
  
  hash <- hash |> str_split('') |> unlist()
  
  current_value <- 0
  walk(hash, \(x) {
    ascii_code <- utf8ToInt(x)
    current_value <- (current_value + ascii_code) * 17
    current_value <<- current_value %% 256
    
  })
  
  return(current_value)
  
}

Finally let’s apply the function across all of the values.

data <- data |> 
  rowwise() |> 
  mutate(
    hash = hash_decoder(value)
  ) |> 
  ungroup()

data

# A tibble: 4,000 × 2
   value     hash
   <chr>    <dbl>
 1 fzl-        41
 2 tcjp=8      54
 3 vkjr=9     243
 4 xs-        200
 5 jktcpk=3    23
 6 gzp-       238
 7 kfsxsd-    160
 8 zxkv=7     119
 9 fxz-       149
10 pj=7        62
# ℹ 3,990 more rows

The solution of the puzzle is simply the sum of all the hash values.

sum(data$hash)