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Advent of Code 2020-03 with R

colinfay profile image Colin Fay Originally published at colinfay.me on ・2 min read

Solving Advent of Code 2020-03 with R

[Disclaimer] Obviously, this post contains a big spoiler about Advent of Code.

Instructions

Step 1

  • The input is a map (x, y), where . is a square, # is a tree.

  • We can move this way: 3 right (x + 3), 1 down (y - 1).

  • The y repeats itself indefinitely.

  • We start from the top left corner (coordinate x = nrow(map), and y = 1).

  • We move until we’ve reached x == 1, i.e by taking nrow(map) steps.

  • How many trees # have we met along the way?

Step 2

Repeat this but modify the step schema.

Find the complete instructions at:https://adventofcode.com/2020/day/3.

R solution

Part one

library(purrr)

n_trees <- function(
  by_x = 1, 
  by_y = 3
){
  ipt <- read.delim( "2020-03-aoc.txt", header = FALSE, stringsAsFactors = FALSE)
  # Building the path
  # x need to go from 1 to nrow(ipt), 1 by 1
  x <- seq(1, nrow(ipt), by = by_x)
  # y needs to start at 1, and go by 3 steps until we have length(x) steps
  y <- seq(1, by = by_y, length.out = length(x))
  list(
    x = x, 
    y = y
  ) %>%
    pmap_dbl(~ {
      # Which x step are we in?
      row <- ipt[..1,] 
      # If the row isn't wide enough, expand it until it is
      # There is probably a better way to do that but I'm not 
      # sure I want to spend 5 minutes trying stuff that will
      # save me half a micro second
      while(nchar(row) < ..2){
        row <- paste0(row, row)
      }
      # Split the damn thing
      row <- strsplit(row, "")[[1]]
      # look for the y, is it a tree?
      if (row[..2] == ".") return(0)
      if (row[..2] == "#") return(1)
    }
    ) %>% sum()
}
n_trees()

## [1] 278
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Part two

data.frame(
  x = c(1, 1, 1, 1, 2),
  y = c(1, 3, 5, 7, 1)
) %>%
  pmap_dbl(~ n_trees(..1, ..2)) %>%
  reduce(`*`)

## [1] 9709761600
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