Day 19

Advent of Code: Worked Solutions

Setup

# Libraries
library(tidyverse)

# Read input from file
input <- read_lines("../input/day19.txt", skip_empty_rows = TRUE)

Part 1

Reformat the input as lists of robots with inputs and outputs for each blueprint:

ore      <- c("ore" = 1, "clay" = 0, "obsidian" = 0, "geode" = 0)
clay     <- c("ore" = 0, "clay" = 1, "obsidian" = 0, "geode" = 0)
obsidian <- c("ore" = 0, "clay" = 0, "obsidian" = 1, "geode" = 0)
geode    <- c("ore" = 0, "clay" = 0, "obsidian" = 0, "geode" = 1)
empty    <- c("ore" = 0, "clay" = 0, "obsidian" = 0, "geode" = 0)

unglue_pattern <- str_c(
  "Blueprint {blueprint}:",
  "Each ore robot costs {ore_ore} ore.",
  "Each clay robot costs {clay_ore} ore.",
  "Each obsidian robot costs {obsidian_ore} ore and {obsidian_clay} clay.",
  "Each geode robot costs {geode_ore} ore and {geode_obsidian} obsidian.",
  sep = " "
)

blueprints <- input |> 
  unglue::unglue_data(unglue_pattern, convert = TRUE) |> 
  pivot_longer(
    -blueprint, 
    names_to = c("robot", "cost_type"), 
    values_to = "cost_value",
    names_sep = "_"
  ) |> 
  pivot_wider(names_from = cost_type, values_from = cost_value, values_fill = 0) |> 
  mutate(geode = 0) |> 
  nest(cost = c(ore, clay, obsidian, geode)) |> 
  mutate(
    cost = map(cost, unlist),
    output = map(robot, get)
  ) |> 
  group_split(blueprint)

Define a set of functions to find the maximum number of geodes for each blueprint.

Optimizations to improve runtime were sourced from Reddit (1, 2).

max_geodes <- function(blueprint, inventory, production, time_left, max_cost, cur_best = 0) {
  
  # If the best possible set of geode robots in this branch can't outdo the current
  # best, then don't traverse this branch.
  theoretical_best <- inventory["geode"] + 
    production["geode"] * time_left +
    sum(1:time_left - 1)
  
  if (time_left == 0 | theoretical_best <= cur_best)
    return(inventory["geode"])
    
  # Determine the amount of time required to build each robot next
  time_to_build <- map_dbl(blueprint$cost, 
    ~ ceiling((.x - inventory) / production) |> 
      keep(.x > 0) |> 
      map_dbl(~ max(.x, 0) + 1) |> 
      max()
  )
  
  # Determine which robots are buildable within the remaining time
  idx <- which(
    is.finite(time_to_build) & 
      time_to_build < time_left & 
      production < max_cost
  )
  
  # Quit if there are no remaining options
  if (length(idx) == 0)
    return((inventory + time_left * production)["geode"])
  
  # Loop through each branch of options
  for (i in rev(idx)) {
    branch_best <- max_geodes(
        blueprint  = blueprint,
        inventory  = inventory + 
          production * time_to_build[[i]] - 
          blueprint$cost[[i]],
        production = production + blueprint$output[[i]],
        time_left  = time_left - time_to_build[[i]],
        max_cost   = max_cost,
        cur_best   = cur_best
      )
    if (branch_best > cur_best)
      cur_best <- branch_best
  }
  
  return(cur_best)
}

blueprint_geodes <- function(blueprints, total_time) {
  
  best <- c()
  
  for (blueprint in blueprints) {
    max_cost <- blueprint |> 
      pull(cost) |> 
      do.call(what = pmax) |> 
      modify_at(.at = "geode", ~ Inf)
    best <- c(
      best,
      max_geodes(
        blueprint  = blueprint,
        inventory  = empty,
        production = ore,
        time_left  = total_time, 
        max_cost   = max_cost,
        cur_best   = 0
      )
    )
  }
  
  unname(best)
}

Compute total quality score by multiplying the max geodes for each blueprint by its index:

blueprint_geodes(blueprints, total_time = 24) |> 
  imap_dbl(\(geodes, i) i * geodes) |> 
  sum()

[1] 2160

Part 2

Filter to the first 3 blueprints, increase the total time to 32 seconds, and take the product of the max geodes:

blueprints |> 
  keep_at(1:3) |> 
  blueprint_geodes(total_time = 32) |> 
  prod()

[1] 13340