My R take on Advent of Code – Day 1

Day 1 Puzzle

Here’s the original text of the riddle:

After feeling like you’ve been falling for a few minutes, you look at the device’s tiny screen. “Error: Device must be calibrated before first use. Frequency drift detected. Cannot maintain destination lock.” Below the message, the device shows a sequence of changes in frequency (your puzzle input). A value like +6 means the current frequency increases by 6; a value like -3 means the current frequency decreases by 3. For example, if the device displays frequency changes of +1, -2, +3, +1, then starting from a frequency of zero, the following changes would occur:

Current frequency 0, change of +1; resulting frequency 1.
Current frequency 1, change of -2; resulting frequency -1.
Current frequency -1, change of +3; resulting frequency 2.
Current frequency 2, change of +1; resulting frequency 3.

In this example, the resulting frequency is 3. Starting with a frequency of zero, what is the resulting frequency after all of the changes in frequency have been applied?

So what is it really saying? That basically we need to add all of the numbers from the sample together to get the solution. Sounds easy! But is it…?

Let’s load the packages we’ll need:

library(stringr)
library(dplyr)
library(tibble)

And let’s have a look at the puzzle input, what does it look like?

glimpse(changes)
##  chr "\n  +11\n  +14\n  +10\n  -8\n  -13\n  -2\n  +8\n  +14\n  -11\n  -4\n  +2\n  -17\n  -15\n  -12\n  -15\n  -16\n  "| __truncated__

OK, not too clean. It’s a character vector of positive and negative integers separated tabs and new lines. Something I learnt during The Advent of Code is: there’s nothing that strsplit() wouldn’t solve… 🙂 Just keep in mind that this function always returns a list, so we have to deal with it accordingly (i.e. use good old unlist()). In the following step I split the values by tabs and remove new lines, trim trolling spaces and bring the result to the vector format again.

# separate a character string into a vector of character numbers with + or -  
base <- strsplit(changes, '\t') %>% # splt by Tab
  str_replace_all('\n', '') %>% # remove all new lines
  stringr::str_trim() %>% # trim edges
  strsplit(' ') %>% # split by spaces
  unlist() # get to vector format

# check if it is a vector  
glimpse(base)
##  chr [1:2071] "+11" "" "+14" "" "+10" "" "-8" "" "-13" "" "-2" "" "+8" ...

We still see that there are some empty strings left, but once we get rid of them, we can turn the input to numeric and simply add it all up:

# generate a solution
base[nchar(base)>0] %>% # remove empty character strings 
  as.numeric() %>% # turn into numeric
  sum() # add it up
## [1] 402

And that’s it! First challenge done and you already can see how addictive it can become.. 🙂