Solve 2021/06
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# This file is automatically @generated by Cargo.
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# It is not intended for manual editing.
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version = 3
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[[package]]
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name = "day06"
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version = "0.1.0"
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[package]
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name = "day06"
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version = "0.1.0"
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edition = "2021"
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# See more keys and their definitions at https://doc.rust-lang.org/cargo/reference/manifest.html
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[dependencies]
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https://adventofcode.com/2021/day/6
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## \--- Day 6: Lanternfish ---
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The sea floor is getting steeper. Maybe the sleigh keys got carried this way?
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A massive school of glowing
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[lanternfish](https://en.wikipedia.org/wiki/Lanternfish) swims past. They must
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spawn quickly to reach such large numbers - maybe _exponentially_ quickly? You
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should model their growth rate to be sure.
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Although you know nothing about this specific species of lanternfish, you make
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some guesses about their attributes. Surely, each lanternfish creates a new
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lanternfish once every _7_ days.
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However, this process isn't necessarily synchronized between every lanternfish
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- one lanternfish might have 2 days left until it creates another lanternfish,
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while another might have 4. So, you can model each fish as a single number
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that represents _the number of days until it creates a new lanternfish_.
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Furthermore, you reason, a _new_ lanternfish would surely need slightly longer
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before it's capable of producing more lanternfish: two more days for its first
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cycle.
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So, suppose you have a lanternfish with an internal timer value of `3`:
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* After one day, its internal timer would become `2`.
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* After another day, its internal timer would become `1`.
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* After another day, its internal timer would become `0`.
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* After another day, its internal timer would reset to `6`, and it would create a _new_ lanternfish with an internal timer of `8`.
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* After another day, the first lanternfish would have an internal timer of `5`, and the second lanternfish would have an internal timer of `7`.
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A lanternfish that creates a new fish resets its timer to `6`, _not`7`_
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(because `0` is included as a valid timer value). The new lanternfish starts
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with an internal timer of `8` and does not start counting down until the next
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day.
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Realizing what you're trying to do, the submarine automatically produces a
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list of the ages of several hundred nearby lanternfish (your puzzle input).
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For example, suppose you were given the following list:
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[code]
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3,4,3,1,2
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[/code]
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This list means that the first fish has an internal timer of `3`, the second
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fish has an internal timer of `4`, and so on until the fifth fish, which has
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an internal timer of `2`. Simulating these fish over several days would
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proceed as follows:
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[code]
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Initial state: 3,4,3,1,2
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After 1 day: 2,3,2,0,1
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After 2 days: 1,2,1,6,0,8
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After 3 days: 0,1,0,5,6,7,8
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After 4 days: 6,0,6,4,5,6,7,8,8
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After 5 days: 5,6,5,3,4,5,6,7,7,8
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After 6 days: 4,5,4,2,3,4,5,6,6,7
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After 7 days: 3,4,3,1,2,3,4,5,5,6
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After 8 days: 2,3,2,0,1,2,3,4,4,5
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After 9 days: 1,2,1,6,0,1,2,3,3,4,8
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After 10 days: 0,1,0,5,6,0,1,2,2,3,7,8
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After 11 days: 6,0,6,4,5,6,0,1,1,2,6,7,8,8,8
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After 12 days: 5,6,5,3,4,5,6,0,0,1,5,6,7,7,7,8,8
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After 13 days: 4,5,4,2,3,4,5,6,6,0,4,5,6,6,6,7,7,8,8
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After 14 days: 3,4,3,1,2,3,4,5,5,6,3,4,5,5,5,6,6,7,7,8
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After 15 days: 2,3,2,0,1,2,3,4,4,5,2,3,4,4,4,5,5,6,6,7
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After 16 days: 1,2,1,6,0,1,2,3,3,4,1,2,3,3,3,4,4,5,5,6,8
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After 17 days: 0,1,0,5,6,0,1,2,2,3,0,1,2,2,2,3,3,4,4,5,7,8
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After 18 days: 6,0,6,4,5,6,0,1,1,2,6,0,1,1,1,2,2,3,3,4,6,7,8,8,8,8
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[/code]
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Each day, a `0` becomes a `6` and adds a new `8` to the end of the list, while
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each other number decreases by 1 if it was present at the start of the day.
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In this example, after 18 days, there are a total of `26` fish. After 80 days,
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there would be a total of `_5934_`.
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Find a way to simulate lanternfish. _How many lanternfish would there be after
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80 days?_
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3,4,3,1,2
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Initial state: 3,4,3,1,2
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After 1 day: 2,3,2,0,1
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After 2 days: 1,2,1,6,0,8
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After 3 days: 0,1,0,5,6,7,8
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After 4 days: 6,0,6,4,5,6,7,8,8
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After 5 days: 5,6,5,3,4,5,6,7,7,8
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After 6 days: 4,5,4,2,3,4,5,6,6,7
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After 7 days: 3,4,3,1,2,3,4,5,5,6
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After 8 days: 2,3,2,0,1,2,3,4,4,5
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After 9 days: 1,2,1,6,0,1,2,3,3,4,8
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After 10 days: 0,1,0,5,6,0,1,2,2,3,7,8
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After 11 days: 6,0,6,4,5,6,0,1,1,2,6,7,8,8,8
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After 12 days: 5,6,5,3,4,5,6,0,0,1,5,6,7,7,7,8,8
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After 13 days: 4,5,4,2,3,4,5,6,6,0,4,5,6,6,6,7,7,8,8
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After 14 days: 3,4,3,1,2,3,4,5,5,6,3,4,5,5,5,6,6,7,7,8
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After 15 days: 2,3,2,0,1,2,3,4,4,5,2,3,4,4,4,5,5,6,6,7
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After 16 days: 1,2,1,6,0,1,2,3,3,4,1,2,3,3,3,4,4,5,5,6,8
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After 17 days: 0,1,0,5,6,0,1,2,2,3,0,1,2,2,2,3,3,4,4,5,7,8
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After 18 days: 6,0,6,4,5,6,0,1,1,2,6,0,1,1,1,2,2,3,3,4,6,7,8,8,8,8
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5,1,1,4,1,1,4,1,1,1,1,1,1,1,1,1,1,1,4,2,1,1,1,3,5,1,1,1,5,4,1,1,1,2,2,1,1,1,2,1,1,1,2,5,2,1,2,2,3,1,1,1,1,1,1,1,1,5,1,1,4,1,1,1,5,4,1,1,3,3,2,1,1,1,5,1,1,4,1,1,5,1,1,5,1,2,3,1,5,1,3,2,1,3,1,1,4,1,1,1,1,2,1,2,1,1,2,1,1,1,4,4,1,5,1,1,3,5,1,1,5,1,4,1,1,1,1,1,1,1,1,1,2,2,3,1,1,1,1,1,2,1,1,1,1,1,1,2,1,1,1,5,1,1,1,1,4,1,1,1,1,4,1,1,1,1,3,1,2,1,2,1,3,1,3,4,1,1,1,1,1,1,1,5,1,1,1,1,1,1,1,1,4,1,1,2,2,1,2,4,1,1,3,1,1,1,5,1,3,1,1,1,5,5,1,1,1,1,2,3,4,1,1,1,1,1,1,1,1,1,1,1,1,5,1,4,3,1,1,1,2,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,3,3,1,2,2,1,4,1,5,1,5,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,5,1,1,1,4,3,1,1,4
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use std::collections::VecDeque;
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use std::io::stdin;
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fn main() {
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println!("Hello, world!");
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let args: Vec<String> = std::env::args().collect();
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if args.len() > 1 && args[1] == "part1" {
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part1();
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} else {
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part2();
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}
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}
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fn part1() {
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println!("Part 1:");
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let mut input = String::new();
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stdin().read_line(&mut input).ok();
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dbg!(&input);
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let count_fishes = count_fishes_after_days(&input, 80);
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dbg!(count_fishes);
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}
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fn part2() {
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println!("Part 2:");
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let mut input = String::new();
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stdin().read_line(&mut input).ok();
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dbg!(&input);
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let count_fishes = count_fishes_after_days(&input, 256);
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dbg!(count_fishes);
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}
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fn count_fishes_after_days(input: &str, after_days: i32) -> i64 {
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let fishes: Vec<usize> = input
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.split(',')
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.map(|s| s.trim())
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.map(|num| num.parse().unwrap())
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.collect();
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println!("Fishes: {:?}", fishes);
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let mut buf = VecDeque::from([0; 9]);
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for fish in fishes {
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buf[fish] += 1
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}
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println!("Initial state: {:?}", buf);
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for day in 1..after_days + 1 {
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buf.rotate_left(1);
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buf[6] += buf[8];
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let count_fishes: i64 = buf.iter().sum();
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println!(
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"After {:>2} days: {:?}, {:>3} fishes",
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day, buf, count_fishes
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);
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}
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let count_fishes: i64 = buf.iter().sum();
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return count_fishes;
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}
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