Advent of Code 2017 in R: Day 2

Day 2 of the Advent of Code provides us with a tab delimited input consisting of numbers 2-4 digits long and asks us to calculate its “checksum”. checksum is defined as the sum of the difference between each row’s largest and smallest values. Awesome! This is a problem that is well-suited for base R.

I started by reading the file in using read.delim, specifying header = F in order to ensure that numbers within the first row of the data are not treated as variable names.

When working with short problems like this where I know I won’t be rerunning my code or reloading my data often, I will use file.choose() in my read.whatever functions for speed. file.choose() opens Windows Explorer, allowing you to navigate to your file path.

input <- read.delim(file.choose(), header = F)

# Check the dimensions of input to ensure the data read in correctly.

After checking the dimensions of our input, everything looks good. As suspected, this is a perfect opportunity to use some vectorization via the apply function.

row_diff <- apply(input, 1, function(x) max(x) - min(x))
checksum <- sum(row_diff)

Et voilĂ , the answer is 45,972!

As was the case with Day 1, we are then prompted with a part two. In order to help out a worrisome computer, we now have to find the two evenly divisible numbers within each row, divide them, and add each row’s result.

This is a tad bit trickier but it’s clear we need to work with the modulo operator. We need to identify the two numbers a and b within each row such that a %% b == 0. If a < b, a %% b will just return a so my first thought is that we should sort the rows in ascending order.

# Sort rows of matrix
input <- t(apply(input, 1, sort))

You can avoid transposing the matrix if you use some helpful packages but per my previous post, I’m trying to stick to base R *sobs quietly*. I used loops to solve this because we need to iterate through each row, comparing each element to every other element. I did try using vectorization here via sapply,

# Compare all elements in first row of input matrix.
sapply(input[1,], function(x) x %% input[1,] == 0)

but this produces a 16 x 16 matrix for each row with a diagonal that needs to be ignored, on top of which we need to find the one TRUE element and map it back to the two matrix indices. I think I would pursue this method further if there was more than one pair of numbers we were searching for but since we areeeeen’t….

# Initialize vector to store each row value
row_val <- c(rep(NA, nrow(input)))

# For each row..
for(row in 1:nrow(input)){

  # Compare each element to its succeeding elements..
  for(col in 1:(ncol(input) - 1)){
    for(i in 1:(ncol(input) - col)){
      # If the modulo is equal to 0, 
      # set the vector element equal to the division result.
      if(input[row, col + i] %% input[row, col] == 0){ 
        row_val[row] <- input[row, col + i] / input[row, col]


Our sum is 326, the correct answer. I’d love to some alternative solutions to part 2, I feel like there is definitely a lot of optimization that could occur here!

Published by

Emma Davies Smith

Emma Davies Smith is currently a postdoctoral research fellow at the Harvard School of Public Health. Her current research interests include clinical trial methodology, nonparametric methods, missing data, data visualization, and communication. When she's not working on expanding her knowledge of statistics, she's busy petting cats and unsuccessfully convincing her husband to let her adopt them, hiking, and concocting indie and folk rock playlists.

Leave a Reply

Your email address will not be published. Required fields are marked *