[TASK] Solved day 8, but it wasn't easy
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164
src/day8.rs
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164
src/day8.rs
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use std::collections::HashMap;
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use num::{BigInt, Integer};
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use crate::day_solver::DaySolver;
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#[cfg(test)]
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use crate::util::read_file;
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pub struct Day8 {
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instructions: Vec<char>,
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nodes: HashMap<String, (String, String)>
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}
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impl Day8 {
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pub fn create(input: String) -> Self {
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let mut lines = input.lines();
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let instructions = lines.next().unwrap();
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// Skip blank line
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lines.next();
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let mut nodes = HashMap::new();
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for line in lines {
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let (node_id, options) = line.split_once(" = ").unwrap();
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let (left, right) = options[1..(options.len() - 1)].split_once(", ").unwrap();
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nodes.insert(node_id.to_string(), (left.to_string(), right.to_string()));
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}
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// Put the input into the day struct
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return Day8 {
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instructions: instructions.chars().collect(),
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nodes
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}
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}
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}
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impl Day8 {
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fn follow_instruction(&self, cur_loc: &str, instruction: &char) -> &String {
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let cur_options = self.nodes.get(cur_loc).unwrap();
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return if instruction == &'L' {
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&cur_options.0
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} else {
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&cur_options.1
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}
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}
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}
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#[derive(Debug, Clone)]
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struct Period {
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start: BigInt,
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repeat: BigInt
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}
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impl DaySolver for Day8 {
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fn solve_part1(&mut self) -> String {
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let mut i = 0usize;
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let mut cur_loc = "AAA";
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while cur_loc != "ZZZ" {
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let instruction = &self.instructions[i % self.instructions.len()];
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cur_loc = self.follow_instruction(cur_loc, instruction);
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i += 1;
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}
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return i.to_string();
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}
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fn solve_part2(&mut self) -> String {
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// Let's start with the naive approach
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let start_nodes: Vec<&String> = self.nodes.keys().filter(|n| n.ends_with('A')).collect();
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// let's assume periodicity, so we're gonna try and found out at which iterations we get a valid endpoint for each node
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let mut periods_per_start_node: Vec<Period> = Vec::new();
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for start_node in start_nodes {
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let mut cur_loc = start_node;
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let mut i = 0;
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let mut end_times: Vec<usize> = Vec::new();
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// Stop if i % instruction.len() == 0 && cur_loc == start_node, so continue if not that
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while i == 0 || !(i % self.instructions.len() == 0 && cur_loc == start_node) {
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let instruction = &self.instructions[i % self.instructions.len()];
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cur_loc = self.follow_instruction(cur_loc, instruction);
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i += 1;
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if cur_loc.ends_with('Z') {
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end_times.push(i);
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// Using this logging, we found out that each of the start_nodes has basically some time before it finds a valid end, and then it repeats that after a while:
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// println!("Found period at {} for {}", i, cur_loc);
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// if let Some(last_period) = last_period_option{
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// println!("iterations since last period: {}", i - last_period);
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// }
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// if periods.len() > 10 {
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// break
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// }
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if end_times.len() > 2 && end_times[2] - end_times[1] == end_times[1] - end_times[0] {
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break
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}
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}
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}
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// println!("Start node {}, after {}: {:?}", start_node, i, end_times);
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periods_per_start_node.push(Period {
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start: BigInt::from(end_times[0]),
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repeat: BigInt::from(end_times[1] - end_times[0])
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})
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}
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// Now we just need to find out when these periods fall at the same time
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// This appears to be the LCM of all periods
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let mut solution_period: Period = periods_per_start_node.first().unwrap().to_owned();
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for period in periods_per_start_node.iter().skip(1) {
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solution_period = Period {
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start: solution_period.start,
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repeat: solution_period.repeat.lcm(&period.repeat)
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};
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}
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// note: I don't understand why we can just ignore how long it takes to get to the end the first time...
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// let mut i = 0usize;
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//
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// while !cur_nodes.iter().all(|n| n.ends_with('Z')) {
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// let instruction = &self.instructions[i % self.instructions.len()];
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// for j in 0..cur_nodes.len() {
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// cur_nodes[j] = self.follow_instruction(&cur_nodes[j], instruction);
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// }
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// i += 1;
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// }
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return solution_period.repeat.to_string();
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}
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}
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#[test]
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fn test_part1() {
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let mut day = Day8::create(read_file("input/day08_example.txt"));
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assert_eq!("2", day.solve_part1());
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}
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#[test]
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fn test_part1_2() {
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let mut day = Day8::create(String::from("LLR
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AAA = (BBB, BBB)
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BBB = (AAA, ZZZ)
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ZZZ = (ZZZ, ZZZ)
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"));
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assert_eq!("6", day.solve_part1());
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}
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#[test]
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fn test_part2() {
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let mut day = Day8::create(String::from("LR
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11A = (11B, XXX)
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11B = (XXX, 11Z)
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11Z = (11B, XXX)
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22A = (22B, XXX)
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22B = (22C, 22C)
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22C = (22Z, 22Z)
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22Z = (22B, 22B)
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XXX = (XXX, XXX)"));
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assert_eq!("6", day.solve_part2());
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}
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@@ -8,10 +8,12 @@ use crate::day4::Day4;
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use crate::day5::Day5;
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use crate::day6::Day6;
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use crate::day7::Day7;
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use crate::day8::Day8;
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use crate::day_solver::DaySolver;
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use crate::util::read_file;
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mod util;
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mod day_solver;
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mod day1;
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mod day2;
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mod day3;
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@@ -19,9 +21,9 @@ mod day4;
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mod day5;
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mod day6;
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mod day7;
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mod day_solver;
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mod day8;
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const MAX_DAY: u8 = 7;
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const MAX_DAY: u8 = 8;
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const DEFAULT_BENCHMARK_AMOUNT: u32 = 100;
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fn main() {
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@@ -99,6 +101,7 @@ fn build_day_solver(day: u8, input: String) -> Option<Box<dyn DaySolver>> {
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5 => Some(Box::new(Day5::create(input))),
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6 => Some(Box::new(Day6::create(input))),
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7 => Some(Box::new(Day7::create(input))),
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8 => Some(Box::new(Day8::create(input))),
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_ => None
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}
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}
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