Files
adventofcode/2021/src/day24.rs
2022-01-02 21:47:07 +01:00

160 lines
4.3 KiB
Rust

//! Very input-specific reverse-engineered solution
//!
//! # General implementation
//!
//! The code in the examples is a series of 14 times this:
//!
//! ```txt
//! inp w -> read digit
//! mul x 0
//! add x z
//! mod x 26 -> x = z % 26
//! div z $A -> pop Z (see below)
//! add x $B
//! eql x w -> x = ((z + $B) == w)
//! eql x 0 -> x = ((z + $B) != w)
//! mul y 0
//! add y 25
//! mul y x
//! add y 1 -> if x { 26 } else { 1 }
//! mul z y -> if x { z *= 26 } (push z, see below)
//! mul y 0
//! add y w
//! add y $C -> y = w + $C
//! mul y x
//! add z y -> if x { z += w + $C }
//! ```
//!
//! `$A` is either `1` or `26` which we can translate to a bool `$A == 26` for convenience. This
//! simplifies to the following rust.
//!
//! ```
//! fn validate<const A: bool, const B: i32, const C: i32>(mut z: i32, digit: i32) -> i32 {
//! let x = (z % 26 + B) != digit;
//! if A {
//! z /= 26;
//! }
//!
//! if x {
//! z = 26 * z + digit + C;
//! }
//!
//! z
//! }
//! ```
//!
//! In human terms, `z` is used to hold a base 26 number. When `$A` is `true`, we pop off the least
//! significant digit by dividing by 26. Then, depending on whether `(z + $B) % 26` is equal to our
//! digit, we push `digit + $C`. Ideally, we should pop as often as we push in order to arrive at `z
//! == 0` in the end. The input contains 7 pops, so we want each of those to not push.
//!
//! To solve this problem, we use a backtracking memoizing algorithm, where we cancel every sequence
//! that fails to pop at some point. A pop is failed whenever we execute a validation pop where we
//! can pop if `x` happened to be set to `0` at the time of the check. We can memoize this over the
//! running value of `z`.
//!
//! This implementation probably doesn't work on other people's input; to fix it, you'll want to
//! update the `parse_step` function. Good luck with that.
use std::collections::HashMap;
use std::io::Read;
use nom::branch::alt;
use nom::bytes::complete::tag;
use nom::character::complete::newline;
use nom::combinator::map;
use nom::multi::separated_list1;
use nom::sequence::delimited;
use nom::sequence::preceded;
use nom::sequence::tuple;
use nom::IResult;
use crate::common::read_input;
type Cache = HashMap<(usize, i32), Option<i64>>;
#[derive(Debug)]
struct Step(bool, i32, i32);
impl Step {
fn evaluate(&self, digit: i32, z: i32) -> Option<i32> {
if self.0 {
(z % 26 + self.1 == digit).then(|| z / 26)
} else {
Some(z * 26 + digit + self.2)
}
}
}
fn parse_step(input: &[u8]) -> IResult<&[u8], Step> {
use nom::character::complete::i32;
let parse_pop = preceded(
tag("inp w\nmul x 0\nadd x z\nmod x 26\ndiv z "),
alt((map(tag("1"), |_| false), map(tag("26"), |_| true))),
);
let parse_a = preceded(tag("\nadd x "), i32);
let parse_b = delimited(
tag("\neql x w\neql x 0\nmul y 0\nadd y 25\nmul y x\nadd y 1\nmul z y\nmul y 0\nadd y w\nadd y "),
i32,
tag("\nmul y x\nadd z y"),
);
map(tuple((parse_pop, parse_a, parse_b)), |(pop, a, b)| {
Step(pop, a, b)
})(input)
}
fn parse_input(input: &[u8]) -> IResult<&[u8], Vec<Step>> {
separated_list1(newline, parse_step)(input)
}
fn optimize(
current: usize,
steps: &[Step],
digits: &[i32],
z: i32,
cache: &mut Cache,
) -> Option<i64> {
if current >= steps.len() {
return (z == 0).then(|| 0);
}
if let Some(&memoized) = cache.get(&(current, z)) {
return memoized;
}
let result = digits.iter().find_map(|&digit| {
let z = steps[current].evaluate(digit, z)?;
let result = optimize(current + 1, steps, digits, z, cache)?;
Some(result + digit as i64 * 10i64.pow(13 - current as u32))
});
cache.insert((current, z), result);
result
}
fn parts_common(input: &mut dyn Read, digits: &[i32]) -> String {
let steps = read_input(input, parse_input);
let mut cache = Cache::new();
optimize(0, &steps, digits, 0, &mut cache)
.unwrap()
.to_string()
}
pub fn part1(input: &mut dyn Read) -> String {
let digits = [9, 8, 7, 6, 5, 4, 3, 2, 1];
parts_common(input, &digits)
}
pub fn part2(input: &mut dyn Read) -> String {
let digits = [1, 2, 3, 4, 5, 6, 7, 8, 9];
parts_common(input, &digits)
}