Added Solution for 2019 day 14

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Chris Alge 2023-03-13 17:08:07 +01:00
parent 72914e6246
commit ddff0d564d
5 changed files with 325 additions and 0 deletions

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[package]
name = "day14_space_stoichiometry"
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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As you approach the rings of Saturn, your ship's *low fuel* indicator turns on. There isn't any fuel here, but the rings have plenty of raw material. Perhaps your ship's Inter-Stellar Refinery Union brand *nanofactory* can turn these raw materials into fuel.
You ask the nanofactory to produce a list of the *reactions* it can perform that are relevant to this process (your puzzle input). Every reaction turns some quantities of specific *input chemicals* into some quantity of an *output chemical*. Almost every *chemical* is produced by exactly one reaction; the only exception, `ORE`, is the raw material input to the entire process and is not produced by a reaction.
You just need to know how much `*ORE*` you'll need to collect before you can produce one unit of `*FUEL*`.
Each reaction gives specific quantities for its inputs and output; reactions cannot be partially run, so only whole integer multiples of these quantities can be used. (It's okay to have leftover chemicals when you're done, though.) For example, the reaction `1 A, 2 B, 3 C => 2 D` means that exactly 2 units of chemical `D` can be produced by consuming exactly 1 `A`, 2 `B` and 3 `C`. You can run the full reaction as many times as necessary; for example, you could produce 10 `D` by consuming 5 `A`, 10 `B`, and 15 `C`.
Suppose your nanofactory produces the following list of reactions:
```
10 ORE => 10 A
1 ORE => 1 B
7 A, 1 B => 1 C
7 A, 1 C => 1 D
7 A, 1 D => 1 E
7 A, 1 E => 1 FUEL
```
The first two reactions use only `ORE` as inputs; they indicate that you can produce as much of chemical `A` as you want (in increments of 10 units, each 10 costing 10 `ORE`) and as much of chemical `B` as you want (each costing 1 `ORE`). To produce 1 `FUEL`, a total of *31* `ORE` is required: 1 `ORE` to produce 1 `B`, then 30 more `ORE` to produce the 7 + 7 + 7 + 7 = 28 `A` (with 2 extra `A` wasted) required in the reactions to convert the `B` into `C`, `C` into `D`, `D` into `E`, and finally `E` into `FUEL`. (30 `A` is produced because its reaction requires that it is created in increments of 10.)
Or, suppose you have the following list of reactions:
```
9 ORE => 2 A
8 ORE => 3 B
7 ORE => 5 C
3 A, 4 B => 1 AB
5 B, 7 C => 1 BC
4 C, 1 A => 1 CA
2 AB, 3 BC, 4 CA => 1 FUEL
```
The above list of reactions requires *165* `ORE` to produce 1 `FUEL`:
* Consume 45 `ORE` to produce 10 `A`.
* Consume 64 `ORE` to produce 24 `B`.
* Consume 56 `ORE` to produce 40 `C`.
* Consume 6 `A`, 8 `B` to produce 2 `AB`.
* Consume 15 `B`, 21 `C` to produce 3 `BC`.
* Consume 16 `C`, 4 `A` to produce 4 `CA`.
* Consume 2 `AB`, 3 `BC`, 4 `CA` to produce 1 `FUEL`.
Here are some larger examples:
* *13312* `ORE` for 1 `FUEL`:
```
157 ORE => 5 NZVS
165 ORE => 6 DCFZ
44 XJWVT, 5 KHKGT, 1 QDVJ, 29 NZVS, 9 GPVTF, 48 HKGWZ => 1 FUEL
12 HKGWZ, 1 GPVTF, 8 PSHF => 9 QDVJ
179 ORE => 7 PSHF
177 ORE => 5 HKGWZ
7 DCFZ, 7 PSHF => 2 XJWVT
165 ORE => 2 GPVTF
3 DCFZ, 7 NZVS, 5 HKGWZ, 10 PSHF => 8 KHKGT
```
* *180697* `ORE` for 1 `FUEL`:
```
2 VPVL, 7 FWMGM, 2 CXFTF, 11 MNCFX => 1 STKFG
17 NVRVD, 3 JNWZP => 8 VPVL
53 STKFG, 6 MNCFX, 46 VJHF, 81 HVMC, 68 CXFTF, 25 GNMV => 1 FUEL
22 VJHF, 37 MNCFX => 5 FWMGM
139 ORE => 4 NVRVD
144 ORE => 7 JNWZP
5 MNCFX, 7 RFSQX, 2 FWMGM, 2 VPVL, 19 CXFTF => 3 HVMC
5 VJHF, 7 MNCFX, 9 VPVL, 37 CXFTF => 6 GNMV
145 ORE => 6 MNCFX
1 NVRVD => 8 CXFTF
1 VJHF, 6 MNCFX => 4 RFSQX
176 ORE => 6 VJHF
```
* *2210736* `ORE` for 1 `FUEL`:
```
171 ORE => 8 CNZTR
7 ZLQW, 3 BMBT, 9 XCVML, 26 XMNCP, 1 WPTQ, 2 MZWV, 1 RJRHP => 4 PLWSL
114 ORE => 4 BHXH
14 VRPVC => 6 BMBT
6 BHXH, 18 KTJDG, 12 WPTQ, 7 PLWSL, 31 FHTLT, 37 ZDVW => 1 FUEL
6 WPTQ, 2 BMBT, 8 ZLQW, 18 KTJDG, 1 XMNCP, 6 MZWV, 1 RJRHP => 6 FHTLT
15 XDBXC, 2 LTCX, 1 VRPVC => 6 ZLQW
13 WPTQ, 10 LTCX, 3 RJRHP, 14 XMNCP, 2 MZWV, 1 ZLQW => 1 ZDVW
5 BMBT => 4 WPTQ
189 ORE => 9 KTJDG
1 MZWV, 17 XDBXC, 3 XCVML => 2 XMNCP
12 VRPVC, 27 CNZTR => 2 XDBXC
15 KTJDG, 12 BHXH => 5 XCVML
3 BHXH, 2 VRPVC => 7 MZWV
121 ORE => 7 VRPVC
7 XCVML => 6 RJRHP
5 BHXH, 4 VRPVC => 5 LTCX
```
Given the list of reactions in your puzzle input, *what is the minimum amount of `ORE` required to produce exactly 1 `FUEL`?*
Your puzzle answer was `1582325`.
\--- Part Two ---
----------
After collecting `ORE` for a while, you check your cargo hold: *1 trillion* (*1000000000000*) units of `ORE`.
*With that much ore*, given the examples above:
* The 13312 `ORE`-per-`FUEL` example could produce *82892753* `FUEL`.
* The 180697 `ORE`-per-`FUEL` example could produce *5586022* `FUEL`.
* The 2210736 `ORE`-per-`FUEL` example could produce *460664* `FUEL`.
Given 1 trillion `ORE`, *what is the maximum amount of `FUEL` you can produce?*
Your puzzle answer was `2267486`.
Both parts of this puzzle are complete! They provide two gold stars: \*\*
At this point, you should [return to your Advent calendar](/2019) and try another puzzle.
If you still want to see it, you can [get your puzzle input](14/input).

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use std::collections::{HashMap, VecDeque};
type Chemical = usize;
struct Reaction {
input: Vec<(usize, Chemical)>,
output: (usize, Chemical),
}
impl Reaction {
fn from(line: &str, chemicals: &mut Vec<String>) -> Self {
let mut get_chemical = |name: &str| -> usize {
if let Some(idx) = chemicals.iter().position(|c| c == &name.to_string()) {
idx
} else {
chemicals.push(name.to_string());
chemicals.len()-1
}
};
let (in_str, out_str) = line.split_once(" => ").unwrap();
let in_components: Vec<_> = in_str.split(&[' ', ',']).chain([""].into_iter()).collect();
assert_eq!(in_components.len()%3, 0);
let out_components: Vec<_> = out_str.split(' ').collect();
assert_eq!(out_components.len(), 2);
let output = (out_components[0].parse().unwrap(), get_chemical(out_components[1]));
let input = in_components.chunks(3).map(|c| (c[0].parse::<usize>().unwrap(), get_chemical(c[1]))).collect();
Self {
input,
output,
}
}
}
pub fn run(input: &str) -> (usize, usize) {
let mut chemicals = Vec::new();
let reactions: Vec<_> = input.lines().map(|line| Reaction::from(line, &mut chemicals)).collect();
let fuel = chemicals.iter().position(|chem| chem == &String::from("FUEL")).unwrap();
let ore = chemicals.iter().position(|chem| chem == &String::from("ORE")).unwrap();
// dbg!(&chemicals);
let first = break_down(&reactions, fuel, ore, 1);
let second = bisection_find(1000000000000/first, 10000000000000/first, &reactions, fuel, ore, 1000000000000);
(first, second)
}
fn bisection_find(lower: usize, upper: usize, reactions: &[Reaction], target: usize, raw: usize, stock: usize) -> usize {
if upper-lower < 2 {
lower
} else {
let mid = (upper+lower)/2;
if break_down(reactions, target, raw, mid) > stock {
bisection_find(lower, mid, reactions, target, raw, stock)
} else {
bisection_find(mid, upper, reactions, target, raw, stock)
}
}
}
fn break_down(reactions: &[Reaction], target: Chemical, raw: Chemical, amount: usize) -> usize {
let mut current = VecDeque::from([(amount, target)]);
let mut leftovers = HashMap::new();
while !(current.len() == 1 && current[0].1 == raw) {
let (next_count, next_chem): (usize, Chemical) = current.pop_front().unwrap();
if next_chem == raw {
current.push_back((next_count, next_chem));
continue;
}
// dbg!(next_chem);
let reaction = reactions.iter().find(|r| r.output.1 == next_chem).unwrap();
let multiplier = (next_count + reaction.output.0 - 1)/reaction.output.0;
*leftovers.entry(next_chem).or_insert(0) += (reaction.output.0 * multiplier).saturating_sub(next_count);
// eprintln!("Breaking down {next_count} {next_chem} into");
for (input_count, input_chem) in &reaction.input {
let mut required = input_count * multiplier;
// eprintln!(" {required} {input_chem}");
if let Some(left) = leftovers.get_mut(input_chem) {
let consumed = required.min(*left);
required -= consumed;
*left -= consumed;
// eprintln!(" {required} after consuming leftovers. {left} left");
}
if required > 0 {
if let Some(idx) = current.iter().position(|c| c.1 == *input_chem) {
current[idx].0 += required;
} else {
current.push_back((required, *input_chem));
}
}
}
}
current[0].0
}
#[cfg(test)]
mod tests {
use super::*;
use std::fs::read_to_string;
fn read_file(name: &str) -> String {
read_to_string(name).expect(&format!("Unable to read file: {name}")[..]).trim().to_string()
}
#[test]
fn test_sample() {
let sample_input = read_file("tests/sample_input");
assert_eq!(run(&sample_input), (2210736, 460664));
}
#[test]
fn test_challenge() {
let challenge_input = read_file("tests/challenge_input");
assert_eq!(run(&challenge_input), (1582325, 0));
}
}

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1 ZQVND => 2 MBZM
2 KZCVX, 1 SZBQ => 7 HQFB
1 PFSQF => 9 RSVN
2 PJXQB => 4 FSNZ
20 JVDKQ, 2 LSQFK, 8 SDNCK, 1 MQJNV, 13 LBTV, 3 KPBRX => 5 QBPC
131 ORE => 8 WDQSL
19 BRGJH, 2 KNVN, 3 CRKW => 9 MQJNV
16 DNPM, 1 VTVBF, 11 JSGM => 1 BWVJ
3 KNVN, 1 JQRML => 7 HGQJ
1 MRQJ, 2 HQFB, 1 MQJNV => 5 VQLP
1 PLGH => 5 DMGF
12 DMGF, 3 DNPM, 1 CRKW => 1 CLML
1 JSGM, 1 RSVN => 5 TMNKH
1 RFJLG, 3 CFWC => 2 ZJMC
1 BRGJH => 5 KPBRX
1 SZBQ, 17 GBVJF => 4 ZHGL
2 PLGH => 5 CFWC
4 FCBZS, 2 XQWHB => 8 JSGM
2 PFSQF => 2 KNVN
12 CRKW, 9 GBVJF => 1 KRCB
1 ZHGL => 8 PJMFP
198 ORE => 2 XQWHB
2 BWVJ, 7 CFWC, 17 DPMWN => 3 KZCVX
4 WXBF => 6 JVDKQ
2 SWMTK, 1 JQRML => 7 QXGZ
1 JSGM, 1 LFSFJ => 4 LSQFK
73 KNVN, 65 VQLP, 12 QBPC, 4 XGTL, 10 SWMTK, 51 ZJMC, 4 JMCPR, 1 VNHT => 1 FUEL
1 BWVJ, 7 MBZM => 5 JXZT
10 CFWC => 2 DPMWN
13 LQDLN => 3 LBTV
1 PFZW, 3 LQDLN => 5 PJXQB
2 RSVN, 2 PFSQF => 5 CRKW
1 HGQJ, 3 SMNGJ, 36 JXZT, 10 FHKG, 3 KPBRX, 2 CLML => 3 JMCPR
126 ORE => 4 FCBZS
1 DNPM, 13 MBZM => 5 PLGH
2 XQWHB, 10 FCBZS => 9 LFSFJ
1 DPMWN => 9 PFZW
1 ZJMC, 3 TMNKH => 2 SWMTK
7 TZCK, 1 XQWHB => 5 ZQVND
4 CFWC, 1 ZLWN, 5 RSVN => 2 WXBF
1 BRGJH, 2 CLML => 6 LQDLN
26 BWVJ => 2 GBVJF
16 PJXQB, 20 SDNCK, 3 HQFB, 7 QXGZ, 2 KNVN, 9 KZCVX => 8 XGTL
8 PJMFP, 3 BRGJH, 19 MRQJ => 5 SMNGJ
7 DNPM => 2 SZBQ
2 JQRML, 14 SDNCK => 8 FHKG
1 FSNZ, 6 RFJLG, 2 CRKW => 8 SDNCK
2 CLML, 4 SWMTK, 16 KNVN => 4 JQRML
8 TZCK, 18 WDQSL => 2 PFSQF
1 LSQFK => 8 VTVBF
18 BRGJH, 8 ZHGL, 2 KRCB => 7 VNHT
3 TZCK => 4 DNPM
14 PFZW, 1 PFSQF => 7 BRGJH
21 PLGH, 6 VTVBF, 2 RSVN => 1 ZLWN
149 ORE => 2 TZCK
3 JSGM => 1 RFJLG
4 PFSQF, 4 DMGF => 3 MRQJ

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171 ORE => 8 CNZTR
7 ZLQW, 3 BMBT, 9 XCVML, 26 XMNCP, 1 WPTQ, 2 MZWV, 1 RJRHP => 4 PLWSL
114 ORE => 4 BHXH
14 VRPVC => 6 BMBT
6 BHXH, 18 KTJDG, 12 WPTQ, 7 PLWSL, 31 FHTLT, 37 ZDVW => 1 FUEL
6 WPTQ, 2 BMBT, 8 ZLQW, 18 KTJDG, 1 XMNCP, 6 MZWV, 1 RJRHP => 6 FHTLT
15 XDBXC, 2 LTCX, 1 VRPVC => 6 ZLQW
13 WPTQ, 10 LTCX, 3 RJRHP, 14 XMNCP, 2 MZWV, 1 ZLQW => 1 ZDVW
5 BMBT => 4 WPTQ
189 ORE => 9 KTJDG
1 MZWV, 17 XDBXC, 3 XCVML => 2 XMNCP
12 VRPVC, 27 CNZTR => 2 XDBXC
15 KTJDG, 12 BHXH => 5 XCVML
3 BHXH, 2 VRPVC => 7 MZWV
121 ORE => 7 VRPVC
7 XCVML => 6 RJRHP
5 BHXH, 4 VRPVC => 5 LTCX