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Combined solution for parts 1+2 of 2022/day19, both finishing in <2 sec

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Tobias Marschner
2024-04-10 10:51:19 +02:00
parent aadb672532
commit 19d5613bd9
2 changed files with 133 additions and 25 deletions
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2022/day19/Cargo.toml Normal file
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[package]
name = "day19"
version = "0.1.0"
edition = "2021"
# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
[dependencies]

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2022/day19/src/main.rs Normal file
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#[derive(Debug)]
struct Blueprint {
// The id and costs parsed from the input.
id: u16,
ore_robot_ore_cost: u16,
clay_robot_ore_cost: u16,
obsidian_robot_ore_cost: u16,
obsidian_robot_clay_cost: u16,
geode_robot_ore_cost: u16,
geode_robot_obsidian_cost: u16,
// The maximal number of geodes that can be collected by this blueprint.
// Initialized to 0 and overwritten by the solver, once it concludes.
optimal_geode_count: u16,
}
impl Blueprint {
// Solve the given blueprint using BFS.
fn solve_bfs(&mut self, total_runtime: u16) {
// For performance reasons we will search the solution space using breadth-first search.
// vec_a has the RecursionStates for the current timeslot, while vec_b has the next slot's states.
let mut vec_a: Vec<RecursionState> = Vec::with_capacity(2u64.pow(20) as usize);
let mut vec_b: Vec<RecursionState> = Vec::with_capacity(2u64.pow(20) as usize);
// Initialize the simulation with the starting state.
vec_a.push(RecursionState {
ore_robots: 1,
clay_robots: 0,
obsidian_robots: 0,
geode_robots: 0,
ore: 0,
clay: 0,
obsidian: 0,
geode: 0,
});
// Iterate over all timeslots.
// Building a robot at t=1 cannot influence the final geode-count,
// so it's omitted from the simulation here.
let mut early_exit = false;
for ts in (2u16..=total_runtime).rev() {
// Have we reached >= 2^20 elements on the input? Time to go for DFS instead.
// Additionally, the queue-overhead shouldn't be worth it for the last few timesteps.
if vec_a.len() >= 2u64.pow(20) as usize || ts <= 3 {
// println!("Switching to recursive solving ...");
// Iterate over all possibilities and run recursively.
for rs in &vec_a {
self.solve_recursive(*rs, ts);
}
// And now we're done proper, no need to run the remaining loop iterations.
early_exit = true;
break;
}
// Process every RS of the past timeslot
// to find all the states for the current timeslot.
for rs in &vec_a {
// Go through all five options and branch down them, if possible.
// Specifically, we can either:
// -> Build one of the four robot types, if resources permit.
// -> Don't build anything at all.
// It's important to note that the optimal solution may include waiting in the middle,
// i.e. letting resources accumulate so we can build one of the more expensive robots
// down the line instead of immediately spending the resources on a cheaper robot type.
// Copy over the current state and let time for it pass.
// This is the same no matter what type of robot we build since the robot will
// go live at the end of the timeslot, not at its beginning.
let mut next_rs = *rs;
next_rs.ore += next_rs.ore_robots as u16;
next_rs.clay += next_rs.clay_robots as u16;
next_rs.obsidian += next_rs.obsidian_robots as u16;
next_rs.geode += next_rs.geode_robots as u16;
// Check whether we can build the different robots, using `rs` and not `next_rs`
// since the resources have to be allocated at the beginning of the turn.
// (1) Ore Robot
if rs.ore >= self.ore_robot_ore_cost {
let mut nrs = next_rs;
nrs.ore -= self.ore_robot_ore_cost;
nrs.ore_robots += 1;
vec_b.push(nrs);
}
// (2) Clay Robot
if rs.ore >= self.clay_robot_ore_cost {
let mut nrs = next_rs;
nrs.ore -= self.clay_robot_ore_cost;
nrs.clay_robots += 1;
vec_b.push(nrs);
}
// (3) Obsidian Robot
if rs.ore >= self.obsidian_robot_ore_cost
&& rs.clay >= self.obsidian_robot_clay_cost
{
let mut nrs = next_rs;
nrs.ore -= self.obsidian_robot_ore_cost;
nrs.clay -= self.obsidian_robot_clay_cost;
nrs.obsidian_robots += 1;
vec_b.push(nrs);
}
// (4) Geode Robot
if rs.ore >= self.geode_robot_ore_cost
&& rs.obsidian >= self.geode_robot_obsidian_cost
{
let mut nrs = next_rs;
nrs.ore -= self.geode_robot_ore_cost;
nrs.obsidian -= self.geode_robot_obsidian_cost;
nrs.geode_robots += 1;
vec_b.push(nrs);
}
// (5) Build nothing and let time pass.
vec_b.push(next_rs);
}
// Done!
// println!("Finished simulation round for t = {}", ts);
// println!(" inserted elements: {}", vec_b.len());
// Prune elements.
prune_states(&mut vec_b, &mut vec_a);
// println!(" elements after prune: {}", vec_a.len());
// Clear vec_b since all the relevant states have been copied over to vec_a.
vec_b.clear();
}
// Collect and print the final geode count.
// Remember that we still have to simulate the geode-collection for t=1,
// hence `e.geode + e.geode_robots as u16`.
if !early_exit {
self.optimal_geode_count = vec_a
.iter()
.map(|e| e.geode + e.geode_robots as u16)
.max()
.unwrap();
}
// println!("Found optimal geode count: {}", self.optimal_geode_count);
}
// Solve the task recursively, providing the current state and remaining time.
// Essentially, and in contrast to `solve_bfs`, this recursive solver performs
// depth-first-search (DFS) on the solution space instead of BFS.
// This removes our ability to prune redundant elements, but doesn't require
// keeping a queue of elements, making for a *much* lighter memory footprint.
// Recommended for the final few timesteps.
fn solve_recursive(&mut self, rs: RecursionState, t: u16) {
// print!("t = {}, ", t);
// rs.print();
// println!();
// Exit condition. If t == 1, we're basically done.
// No need to build the final robot, it can't influence the final geode result.
// Simply add one more round of harvesting (rs.geode_robots) and check for improvements.
if t == 1 {
let next_geode_count = rs.geode + rs.geode_robots as u16;
if next_geode_count > self.optimal_geode_count {
// Update the optimal result, if improved.
self.optimal_geode_count = next_geode_count;
}
return;
}
// Check a cutoff-condition, in case this branch is not worth it.
let upper_bound = rs.geode // The resources we already have.
// The resource the already existing robots would produce.
+ rs.geode_robots as u16 * t
// The resources we would get if we produced one robot every timeslot.
// This is the triangular number for (t - 1).
+ (t - 1) * t / 2;
// Now check if this would be an improvement.
if upper_bound <= self.optimal_geode_count {
// No point continuing.
return;
}
// The following section is basically the same as in `solve_bfs`.
// Copy over the current state and let time for it pass.
// This is the same no matter what type of robot we build since the robot will
// go live at the end of the timeslot, not at its beginning.
let mut next_rs = rs;
next_rs.ore += next_rs.ore_robots as u16;
next_rs.clay += next_rs.clay_robots as u16;
next_rs.obsidian += next_rs.obsidian_robots as u16;
next_rs.geode += next_rs.geode_robots as u16;
// Check whether we can build the different robots, using `rs` and not `next_rs`
// since the resources have to be allocated at the beginning of the turn.
// (1) Ore Robot
if rs.ore >= self.ore_robot_ore_cost {
let mut nrs = next_rs;
nrs.ore -= self.ore_robot_ore_cost;
nrs.ore_robots += 1;
self.solve_recursive(nrs, t - 1);
}
// (2) Clay Robot
if rs.ore >= self.clay_robot_ore_cost {
let mut nrs = next_rs;
nrs.ore -= self.clay_robot_ore_cost;
nrs.clay_robots += 1;
self.solve_recursive(nrs, t - 1);
}
// (3) Obsidian Robot
if rs.ore >= self.obsidian_robot_ore_cost && rs.clay >= self.obsidian_robot_clay_cost {
let mut nrs = next_rs;
nrs.ore -= self.obsidian_robot_ore_cost;
nrs.clay -= self.obsidian_robot_clay_cost;
nrs.obsidian_robots += 1;
self.solve_recursive(nrs, t - 1);
}
// (4) Geode Robot
if rs.ore >= self.geode_robot_ore_cost && rs.obsidian >= self.geode_robot_obsidian_cost {
let mut nrs = next_rs;
nrs.ore -= self.geode_robot_ore_cost;
nrs.obsidian -= self.geode_robot_obsidian_cost;
nrs.geode_robots += 1;
self.solve_recursive(nrs, t - 1);
}
// (5) Build nothing and let time pass.
self.solve_recursive(next_rs, t - 1);
}
}
// Store all of the state that's passed up and down the recursion in one struct.
// Derive PartialOrd + Ord for lexicographic sorting, something we'll use during pruning.
#[derive(Debug, Copy, Clone, PartialEq, Eq, PartialOrd, Ord)]
struct RecursionState {
// The currently active fleet of robots.
ore_robots: u8,
clay_robots: u8,
obsidian_robots: u8,
geode_robots: u8,
// Our resources.
ore: u16,
clay: u16,
obsidian: u16,
geode: u16,
}
impl RecursionState {
#[allow(dead_code)]
fn print(&self) {
print!("{:>3} OR, ", self.ore_robots);
print!("{:>3} CR, ", self.clay_robots);
print!("{:>3} BR, ", self.obsidian_robots);
print!("{:>3} GR, ", self.geode_robots);
print!("{:>3} O, ", self.ore);
print!("{:>3} C, ", self.clay);
print!("{:>3} B, ", self.obsidian);
print!("{:>3} G, ", self.geode);
}
}
// Copy over states for the next simulation round, pruning a lot of (but not all)
// RecursionStates that are "strictly inferior" in terms of Pareto optimality.
// Will clear any previously present elements in `dest`.
//
// A few words on the general idea here:
// The goal here is to check for Pareto improvements. An example:
// RS1: 2 ore, 2 clay, 1 ore robot, 1 clay robot, 20 minutes left
// RS2: 2 ore, 1 clay, 1 ore robot, 1 clay robot, 20 minutes left
// RS1 is just as "good" in terms of ore, robot counts and time left
// but is "strictly better" in terms of clay. It makes no sense to continue
// running the simulation for RS2 b/c it cannot possibly produce a better
// outcome than RS1.
// Having more resources, robots or time can only ever lead to better outcomes.
// If, however, RS1 had, say, one more ore but less clay than RS2 we cannot say
// that RS1 is "strictly better". It is different, having made a different tradeoff in
// resource collection, which may or may not lead to a better outcome overall.
// This comparison is used to cut off redundant simulation paths in the solver.
fn prune_states(source: &mut [RecursionState], dest: &mut Vec<RecursionState>) {
// Begin by sorting the source lexicrgraphically and clearing the destination.
source.sort_unstable();
dest.clear();
// Iterate through it from smallest to largest element and look at every pair of states.
for (a, b) in source.iter().zip(source.iter().skip(1)) {
// Don't copy a over if it is strictly inferior or equal to b.
// Compare from bottom to top.
if a.geode > b.geode
|| a.obsidian > b.obsidian
|| a.clay > b.clay
|| a.ore > b.ore
|| a.geode_robots > b.geode_robots
|| a.obsidian_robots > b.obsidian_robots
|| a.clay_robots > b.clay_robots
|| a.ore_robots > b.ore_robots
{
dest.push(*a);
}
}
// Copy over the very last element, too.
dest.push(*source.last().unwrap());
}
fn main() {
// Use command line arguments to specify the input filename.
let args: Vec<String> = std::env::args().collect();
if args.len() < 2 {
panic!("Usage: ./main <input-file>\nNo input file provided. Exiting.");
}
// Next, read the contents of the input file into a string for easier processing.
let input = std::fs::read_to_string(&args[1]).expect("Error opening file");
// First, parse all the blueprints.
let mut blueprints = input
.lines()
.map(|l| l.split_whitespace().collect::<Vec<_>>())
.collect::<Vec<_>>()
.iter()
.map(|l| Blueprint {
id: l[1].trim_end_matches(':').parse::<u16>().unwrap(),
ore_robot_ore_cost: l[6].parse::<u16>().unwrap(),
clay_robot_ore_cost: l[12].parse::<u16>().unwrap(),
obsidian_robot_ore_cost: l[18].parse::<u16>().unwrap(),
obsidian_robot_clay_cost: l[21].parse::<u16>().unwrap(),
geode_robot_ore_cost: l[27].parse::<u16>().unwrap(),
geode_robot_obsidian_cost: l[30].parse::<u16>().unwrap(),
optimal_geode_count: 0,
})
.collect::<Vec<_>>();
// PART ONE
// Solve for every blueprint with time 24.
for bp in &mut blueprints {
// Solve every blueprint with TOTAL_RUNTIME minutes of time.
// println!("Solving Blueprint {}", bp.id);
bp.solve_bfs(24u16);
}
println!(
"Total Quality Level for Part 1: {}",
blueprints
.iter()
.map(|b| b.id * b.optimal_geode_count)
.sum::<u16>()
);
// PART TWO
// Now solve the first three blueprints again, but for 32 minutes.
for bp in blueprints.iter_mut().take(3) {
bp.solve_bfs(32u16);
}
println!(
"Multiplied Geode Counts for Part 2: {}",
blueprints
.iter()
.take(3)
.map(|b| b.optimal_geode_count as u64)
.product::<u64>()
);
}