Advent of Code 2022 Day 19

Day 19: Not Enough Minerals

https://adventofcode.com/2022/day/19

You want to collect the geodes.

No manual labor here, let robots do that!

A geode-cracking robot collects geodes.
Those need obsidian to be made, collected by obsidian-collecting robots.
Those need clay to be made, collected by clay-collecting robots.
Those need ore to be made, collected by ore-collecting robots.
You have 1 ore-collecting robot in your backpack.
You have a portable robot factory in your backpack.

You can kickstart the robot-~~party~~collection!

Each robot collects 1 of its resource type per minute.
The factory can create 1 robot per minute, consuming the resources it needs to make that type of bot.

The amount of resources the factory needs for each bot depends on the used blueprint.

Todays input is a list of blueprints, a list of costs for the factory.

An example input looks like this (but each blueprint is on a single line in the real input):

Blueprint 1:
  Each ore robot costs 4 ore.
  Each clay robot costs 2 ore.
  Each obsidian robot costs 3 ore and 14 clay.
  Each geode robot costs 2 ore and 7 obsidian.

Blueprint 2:
  Each ore robot costs 2 ore.
  Each clay robot costs 3 ore.
  Each obsidian robot costs 3 ore and 8 clay.
  Each geode robot costs 3 ore and 12 obsidian.

Parsing

I want to represent one blueprint as a list of lists.

Each item in the outer list is a cost for a robot type. (4 types) Each cost is a list of ore amounts. (4 types)

An example for the first blueprint in the example:

Each ore robot costs 4 ore so the cost list is: [4, 0, 0, 0]
Each clay robot costs 2 ore so the cost list is: [2, 0, 0, 0]
Each obsidian robot costs 3 ore and 14 clay so the cost list is: [3, 14, 0, 0]
Each geode robot costs 2 ore and 7 obsidian so the cost list is: [2, 0, 7, 0]

Puttin all those cost lists together: [[4, 0, 0, 0], [2, 0, 0, 0], [3, 14, 0, 0], [2, 0, 7, 0]]

Used in parsing function:

An ore bot costs only ore
A clay bot costs only ore
An obsidian bot costs ore and clay
A geode bot costs ore and obsidian

// each cost is [ore_amount, clay_amount, obsidian_amount, geode_amount]
// [ore_bot_costs, clay_bot_costs, obsidian_bot_costs, geode_bot_costs]
fn parse() -> Vec<[[u16; 4]; 4]> {
    let input = std::fs::read_to_string("src/day19.txt").unwrap();
    let mut blueprints = Vec::new();

    for line in input.lines() {
        let mut iter = line.split_ascii_whitespace();

        // ore bots cost ore
        let ore_bot_costs = [iter.nth(6).unwrap().parse().unwrap(), 0, 0, 0];
        // clay bots cost ore
        let clay_bot_costs = [iter.nth(5).unwrap().parse().unwrap(), 0, 0, 0];
        // obsidian bots cost ore and clay
        let obsidian_bot_costs = [
            iter.nth(5).unwrap().parse().unwrap(),
            iter.nth(2).unwrap().parse().unwrap(),
            0,
            0,
        ];
        // geode bots cost ore and obsidian
        let geode_bot_costs = [
            iter.nth(5).unwrap().parse().unwrap(),
            0,
            iter.nth(2).unwrap().parse().unwrap(),
            0,
        ];

        let blueprint = [
            ore_bot_costs,
            clay_bot_costs,
            obsidian_bot_costs,
            geode_bot_costs,
        ];
        blueprints.push(blueprint);
    }

    blueprints
}

Part 1

The question asks for the sum of all blueprint quality levels.

A blueprint’s quality level can be calculated by multiplying the ID with the maximum amount of geodes you can open in 24 minutes using that blueprint.

In skeleton code:

let blueprints = parse();

blueprints
    .iter()
    .map(|blueprint| max_geodes(blueprint))
    .enumerate()
    .map(|(idx, geodes)| (idx + 1) * geodes)
    .sum()

That skeleton code is is alright. The complexity is in calculating the maximum amount of geodes a blueprint can produce.

Helpers

Today’s main event, a “search a bunch of states and take the maximum of something” algorithm!

It uses a queue filled with states so far.
We start by inserting a state with an empty inventory, 1 ore bot, and 0 elapsed time.

For each state that is popped off the queue:

we try to build every type of bot.
we gather resources
we add the possibly made bot to the bot-pool
we subtract the cost of the possibly made bot from the inventory

We keep adding new states to that queue and recording the maximum amount of geodes while we loop. If a state would exceed the time limit, we continue with the next item in the loop instead.

Eventually, the queue empties and the number in max_geodes at that time is the maximum possible amount of geodes for that blueprint.

use std::collections::VecDeque;

struct State {
    // [ore, clay, obsidian, geode]
    inventory: [u16; 4],
    // [ore_bots, clay_bots, obsidian_bots, geode_bots]
    bots: [u16; 4],
    // elapsed time in minutes
    elapsed: u16,
}

fn max_geodes(blueprint: &[[u16; 4]; 4]) -> u16 {
    let max_time = 24;
    let mut max_geodes = 0;

    let mut q = VecDeque::new();
    q.push_back(State {
        inventory: [0, 0, 0, 0],
        bots: [1, 0, 0, 0],
        elapsed: 0,
    });

    while let Some(State {
        inventory,
        bots,
        elapsed,
    }) = q.pop_front()
    {
        // for every bot cost, run simulation
        for i in 0..blueprint.len() {
            let costs = &blueprint[i];
            // Find the limiting resource type for the costs.
            let wait_time = (0..3)
                .map(|idx| {
                    match costs[idx] {
                        // state has enough of current resource in inventory to cover that part of the target bot cost. 0 wait time
                        cost if cost <= inventory[idx] => 0,
                        // no target bot type made yet
                        // we can't build it (it takes more than max_time to build it).
                        _ if bots[idx] == 0 => max_time + 1,
                        _ => (costs[idx] - inventory[idx] + bots[idx] - 1) / bots[idx],
                    }
                })
                .max()
                .unwrap();

            // if that choice would cause the time limit be to exceeded, skip
            // the + 1 is so the built bot has the chance to do something, it merely being built is not enough
            let new_elapsed = elapsed + wait_time + 1;
            if new_elapsed >= max_time {
                continue;
            }

            // gather ores with previously available bots
            let mut new_inventory = [0; 4];
            for idx in 0..bots.len() {
                new_inventory[idx] = inventory[idx] + bots[idx] * (wait_time + 1) - costs[idx];
            }

            // increase bot type for the bot we just built
            let mut new_bots = bots;
            new_bots[i] += 1;

            q.push_back(State {
                inventory: new_inventory,
                bots: new_bots,
                elapsed: new_elapsed,
            })
        }

        let geodes = inventory[3] + bots[3] * (max_time - elapsed);
        max_geodes = geodes.max(max_geodes);
    }

    max_geodes
}

Final code

pub fn part_1() -> usize {
    let blueprints = parse();

    blueprints
        .iter()
        .map(|blueprint| max_geodes(blueprint))
        .enumerate()
        .map(|(idx, geodes)| (idx + 1) * usize::from(geodes))
        .sum()
}

Part 2

An elephant ate most of the blueprints, only the first 3 are left.

The question asks for the product of all remaining blueprint’s maximum number of gathered geodes, in 32 minutes.

let blueprints = parse();

blueprints
    .iter()
    .take(3)
    .map(|blueprint| max_geodes(blueprint))
    .product()

Combined with changing the let max_time = 24; to let max_time = 32; aaaaaaaaaaand

That is not enough (unless you have a beefy computer, and lots of patience).

The key is not making bots you don’t need.

Bots you don’t need are bots that don’t help towards unbottlenecking production. If you need 5 ore to build a bot, it doesn’t matter if we have 5 in our inventory, or 5000, the bot gets built just as fast.

So, if we have the amount of bots that’s equal to the maximum cost of that type for any bot, we’re gathering enough of that resource every turn to never be bottlenecked. Producing more of that type would not be beneficial, they’d just do unnecessary work.

The changes:

fn max_geodes(blueprint: &[[u16; 4]; 4]) -> u16 {
    // calculate the maximum amount for every type of bot so that the creation of a new bot of any type is never bottlenecked
    // it doesn't make sense to build more bots than that maximum if the resources a bot type generates are
    // enough to cover that type (ore, clay, obsidian) cost for any possible bot (per question, you can only build 1 bot per turn)
    // for geode bots, there is no logical maximum amount
    // [ore, clay, obsidian, geode]
    let mut max_robots = [u16::MAX; 4];
    for i in 0..3 {
        max_robots[i] = blueprint.iter().map(|cost| cost[i]).max().unwrap();
    }

    let max_time = 32;
    let mut max_geodes = 0;

    let mut q = VecDeque::new();
    q.push_back(State {
        inventory: [0, 0, 0, 0],
        bots: [1, 0, 0, 0],
        elapsed: 0,
    });

    while let Some(State {
        inventory,
        bots,
        elapsed,
    }) = q.pop_front()
    {
        // for every bot cost, run simulation
        for i in 0..blueprint.len() {
            // if we already have enough of this bot type, skip
            if bots[i] == max_robots[i] {
                continue;
            }
            let costs = &blueprint[i];

            // Find the limiting resource type for the costs.
            let wait_time = (0..3)
                .map(|idx| {
                    match costs[idx] {
                        // state has enough of current resource in inventory to cover that part of the target bot cost. 0 wait time
                        cost if cost <= inventory[idx] => 0,
                        // no target bot type made yet
                        // we can't build it (it takes more than max_time to build it).
                        _ if bots[idx] == 0 => max_time + 1,
                        _ => (costs[idx] - inventory[idx] + bots[idx] - 1) / bots[idx],
                    }
                })
                .max()
                .unwrap();

            // if that choice would cause the time limit be to exceeded, skip
            // the + 1 is so the built bot has the chance to do something, it merely being built is not enough
            let new_elapsed = elapsed + wait_time + 1;
            if new_elapsed >= max_time {
                continue;
            }

            // gather ores with previously available bots
            let mut new_inventory = [0; 4];
            for idx in 0..bots.len() {
                new_inventory[idx] = inventory[idx] + bots[idx] * (wait_time + 1) - costs[idx];
            }

            // increase bot type for the bot we just built
            let mut new_bots = bots;
            new_bots[i] += 1;

            q.push_back(State {
                inventory: new_inventory,
                bots: new_bots,
                elapsed: new_elapsed,
            })
        }

        let geodes = inventory[3] + bots[3] * (max_time - elapsed);
        max_geodes = geodes.max(max_geodes);
    }

    max_geodes
}

There’s a further optimization you can apply right before pushing to the queue. If we theoretically only built geode bots every turn, and we still wouldn’t beat the current maximum. Don’t push that state the the queue, but skip to the next item.

let remaining_time = max_time - new_elapsed;
if ((remaining_time - 1) * remaining_time) / 2
    + new_inventory[3]
    + remaining_time * new_bots[3]
    < max_geodes
{
    continue;
}

Final code

pub fn part_2() -> usize {
    let blueprints = parse();

    blueprints
        .iter()
        .take(3)
        .map(|blueprint| usize::from(max_geodes(blueprint)))
        .product()
}

Final code

I turned the max_geode function into one that works for both part1 and part2 by passing in max_time as a parameter.

use std::collections::VecDeque;

struct State {
    // [ore, clay, obsidian, geode]
    inventory: [u16; 4],
    // [ore_bots, clay_bots, obsidian_bots, geode_bots]
    bots: [u16; 4],
    // elapsed time in minutes
    elapsed: u16,
}

// each cost is [ore_amount, clay_amount, obsidian_amount, geode_amount]
// [ore_bot_costs, clay_bot_costs, obsidian_bot_costs, geode_bot_costs]
fn parse() -> Vec<[[u16; 4]; 4]> {
    let input = std::fs::read_to_string("src/day19.txt").unwrap();
    let mut blueprints = Vec::new();

    for line in input.lines() {
        let mut iter = line.split_ascii_whitespace();

        // ore bots cost ore
        let ore_bot_costs = [iter.nth(6).unwrap().parse().unwrap(), 0, 0, 0];
        // clay bots cost ore
        let clay_bot_costs = [iter.nth(5).unwrap().parse().unwrap(), 0, 0, 0];
        // obsidian bots cost ore and clay
        let obsidian_bot_costs = [
            iter.nth(5).unwrap().parse().unwrap(),
            iter.nth(2).unwrap().parse().unwrap(),
            0,
            0,
        ];
        // geode bots cost ore and obsidian
        let geode_bot_costs = [
            iter.nth(5).unwrap().parse().unwrap(),
            0,
            iter.nth(2).unwrap().parse().unwrap(),
            0,
        ];

        let blueprint = [
            ore_bot_costs,
            clay_bot_costs,
            obsidian_bot_costs,
            geode_bot_costs,
        ];
        blueprints.push(blueprint);
    }

    blueprints
}

fn max_geodes(blueprint: &[[u16; 4]; 4], max_time: u16) -> u16 {
    // calculate the maximum amount for every type of bot so that the creation of a new bot of any type is never bottlenecked
    // it doesn't make sense to build more bots than that maximum if the resources a bot type generates are
    // enough to cover that type (ore, clay, obsidian) cost for any possible bot (per question, you can only build 1 bot per turn)
    // for geode bots, there is no logical maximum amount
    // [ore, clay, obsidian, geode]
    let mut max_robots = [u16::MAX; 4];
    for i in 0..3 {
        max_robots[i] = blueprint.iter().map(|cost| cost[i]).max().unwrap();
    }
    let mut max_geodes = 0;

    let mut q = VecDeque::new();
    q.push_back(State {
        inventory: [0, 0, 0, 0],
        bots: [1, 0, 0, 0],
        elapsed: 0,
    });

    while let Some(State {
        inventory,
        bots,
        elapsed,
    }) = q.pop_front()
    {
        // for every bot cost, run simulation
        for i in 0..blueprint.len() {
            // if we already have enough of this bot type, skip
            if bots[i] == max_robots[i] {
                continue;
            }

            let costs = &blueprint[i];

            // Find the limiting resource type for the costs.
            let wait_time = (0..3)
                .map(|idx| {
                    match costs[idx] {
                        // state has enough of current resource in inventory to cover that part of the target bot cost. 0 wait time
                        cost if cost <= inventory[idx] => 0,
                        // no target bot type made yet
                        // we can't build it (it takes more than max_time to build it).
                        _ if bots[idx] == 0 => max_time + 1,
                        _ => (costs[idx] - inventory[idx] + bots[idx] - 1) / bots[idx],
                    }
                })
                .max()
                .unwrap();

            // if that choice would cause the time limit be to exceeded, skip
            // the + 1 is so the built bot has the chance to do something, it merely being built is not enough
            let new_elapsed = elapsed + wait_time + 1;
            if new_elapsed >= max_time {
                continue;
            }

            // gather ores with previously available bots
            let mut new_inventory = [0; 4];
            for idx in 0..bots.len() {
                new_inventory[idx] = inventory[idx] + bots[idx] * (wait_time + 1) - costs[idx];
            }

            // increase bot type for the bot we just built
            let mut new_bots = bots.clone();
            new_bots[i] += 1;

            // extra optimization:
            // if we theoretically only built geode bots every turn, and we still don't beat the maximum, skip
            let remaining_time = max_time - new_elapsed;
            if ((remaining_time - 1) * remaining_time) / 2
                + new_inventory[3]
                + remaining_time * new_bots[3]
                < max_geodes
            {
                continue;
            }

            q.push_back(State {
                inventory: new_inventory,
                bots: new_bots,
                elapsed: new_elapsed,
            })
        }

        let geodes = inventory[3] + bots[3] * (max_time - elapsed);
        max_geodes = geodes.max(max_geodes);
    }

    max_geodes
}

pub fn part_1() -> usize {
    let blueprints = parse();

    blueprints
        .iter()
        .map(|blueprint| max_geodes(blueprint, 24))
        .enumerate()
        .map(|(idx, geodes)| (idx + 1) * usize::from(geodes))
        .sum()
}

pub fn part_2() -> usize {
    let blueprints = parse();

    blueprints
        .iter()
        .take(3)
        .map(|blueprint| usize::from(max_geodes(blueprint, 32)))
        .product()
}

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Advent of Code 2022 Day 19

Day 19: Not Enough Minerals

Parsing

Part 1

Helpers

Final code

Part 2

Final code

Final code