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//! # Day 13: Packet Scanners
//!
//! You need to cross a vast **firewall**. The firewall consists of several layers, each with a
//! **security scanner** that moves back and forth across the layer. To succeed, you must not be
//! detected by a scanner.
//!
//! By studying the firewall briefly, you are able to record (in your puzzle input) the **depth** of
//! each layer and the **range** of the scanning area for the scanner within it, written as
//! `depth: range`. Each layer has a thickness of exactly `1`. A layer at depth `0` begins
//! immediately inside the firewall; a layer at depth `1` would start immediately after that.
//!
//! For example, suppose you've recorded the following:
//!
//! ```txt
//! 0: 3
//! 1: 2
//! 4: 4
//! 6: 4
//! ```
//!
//! This means that there is a layer immediately inside the firewall (with range `3`), a second
//! layer immediately after that (with range `2`), a third layer which begins at depth `4` (with
//! range `4`), and a fourth layer which begins at depth `6` (also with range `4`). Visually, it
//! might look like this:
//!
//! ```txt
//! 0 1 2 3 4 5 6
//! [ ] [ ] ... ... [ ] ... [ ]
//! [ ] [ ] [ ] [ ]
//! [ ] [ ] [ ]
//! [ ] [ ]
//! ```
//!
//! Within each layer, a security scanner moves back and forth within its range. Each security
//! scanner starts at the top and moves down until it reaches the bottom, then moves up until it
//! reaches the top, and repeats. A security scanner takes **one picosecond** to move one step.
//! Drawing scanners as `S`, the first few picoseconds look like this:
//!
//! ```txt
//!
//! Picosecond 0:
//! 0 1 2 3 4 5 6
//! [S] [S] ... ... [S] ... [S]
//! [ ] [ ] [ ] [ ]
//! [ ] [ ] [ ]
//! [ ] [ ]
//!
//! Picosecond 1:
//! 0 1 2 3 4 5 6
//! [ ] [ ] ... ... [ ] ... [ ]
//! [S] [S] [S] [S]
//! [ ] [ ] [ ]
//! [ ] [ ]
//!
//! Picosecond 2:
//! 0 1 2 3 4 5 6
//! [ ] [S] ... ... [ ] ... [ ]
//! [ ] [ ] [ ] [ ]
//! [S] [S] [S]
//! [ ] [ ]
//!
//! Picosecond 3:
//! 0 1 2 3 4 5 6
//! [ ] [ ] ... ... [ ] ... [ ]
//! [S] [S] [ ] [ ]
//! [ ] [ ] [ ]
//! [S] [S]
//! ```
//!
//! Your plan is to hitch a ride on a packet about to move through the firewall. The packet will
//! travel along the top of each layer, and it moves at **one layer per picosecond**. Each
//! picosecond, the packet moves one layer forward (its first move takes it into layer 0), and then
//! the scanners move one step. If there is a scanner at the top of the layer **as your packet
//! enters it**, you are **caught**. (If a scanner moves into the top of its layer while you are
//! there, you are **not** caught: it doesn't have time to notice you before you leave.) If you were
//! to do this in the configuration above, marking your current position with parentheses, your
//! passage through the firewall would look like this:
//!
//! ```txt
//! Initial state:
//! 0 1 2 3 4 5 6
//! [S] [S] ... ... [S] ... [S]
//! [ ] [ ] [ ] [ ]
//! [ ] [ ] [ ]
//! [ ] [ ]
//!
//! Picosecond 0:
//! 0 1 2 3 4 5 6
//! (S) [S] ... ... [S] ... [S]
//! [ ] [ ] [ ] [ ]
//! [ ] [ ] [ ]
//! [ ] [ ]
//!
//! 0 1 2 3 4 5 6
//! ( ) [ ] ... ... [ ] ... [ ]
//! [S] [S] [S] [S]
//! [ ] [ ] [ ]
//! [ ] [ ]
//!
//!
//! Picosecond 1:
//! 0 1 2 3 4 5 6
//! [ ] ( ) ... ... [ ] ... [ ]
//! [S] [S] [S] [S]
//! [ ] [ ] [ ]
//! [ ] [ ]
//!
//! 0 1 2 3 4 5 6
//! [ ] (S) ... ... [ ] ... [ ]
//! [ ] [ ] [ ] [ ]
//! [S] [S] [S]
//! [ ] [ ]
//!
//!
//! Picosecond 2:
//! 0 1 2 3 4 5 6
//! [ ] [S] (.) ... [ ] ... [ ]
//! [ ] [ ] [ ] [ ]
//! [S] [S] [S]
//! [ ] [ ]
//!
//! 0 1 2 3 4 5 6
//! [ ] [ ] (.) ... [ ] ... [ ]
//! [S] [S] [ ] [ ]
//! [ ] [ ] [ ]
//! [S] [S]
//!
//!
//! Picosecond 3:
//! 0 1 2 3 4 5 6
//! [ ] [ ] ... (.) [ ] ... [ ]
//! [S] [S] [ ] [ ]
//! [ ] [ ] [ ]
//! [S] [S]
//!
//! 0 1 2 3 4 5 6
//! [S] [S] ... (.) [ ] ... [ ]
//! [ ] [ ] [ ] [ ]
//! [ ] [S] [S]
//! [ ] [ ]
//!
//!
//! Picosecond 4:
//! 0 1 2 3 4 5 6
//! [S] [S] ... ... ( ) ... [ ]
//! [ ] [ ] [ ] [ ]
//! [ ] [S] [S]
//! [ ] [ ]
//!
//! 0 1 2 3 4 5 6
//! [ ] [ ] ... ... ( ) ... [ ]
//! [S] [S] [S] [S]
//! [ ] [ ] [ ]
//! [ ] [ ]
//!
//!
//! Picosecond 5:
//! 0 1 2 3 4 5 6
//! [ ] [ ] ... ... [ ] (.) [ ]
//! [S] [S] [S] [S]
//! [ ] [ ] [ ]
//! [ ] [ ]
//!
//! 0 1 2 3 4 5 6
//! [ ] [S] ... ... [S] (.) [S]
//! [ ] [ ] [ ] [ ]
//! [S] [ ] [ ]
//! [ ] [ ]
//!
//!
//! Picosecond 6:
//! 0 1 2 3 4 5 6
//! [ ] [S] ... ... [S] ... (S)
//! [ ] [ ] [ ] [ ]
//! [S] [ ] [ ]
//! [ ] [ ]
//!
//! 0 1 2 3 4 5 6
//! [ ] [ ] ... ... [ ] ... ( )
//! [S] [S] [S] [S]
//! [ ] [ ] [ ]
//! [ ] [ ]
//! ```
//!
//! In this situation, you are **caught** in layers `0` and `6`, because your packet entered the
//! layer when its scanner was at the top when you entered it. You are **not** caught in layer `1`,
//! since the scanner moved into the top of the layer once you were already there.
//!
//! The **severity** of getting caught on a layer is equal to its **depth** multiplied by its
//! **range**. (Ignore layers in which you do not get caught.) The severity of the whole trip is the
//! sum of these values. In the example above, the trip severity is `0*3 + 6*4 = 24`.
//!
//! Given the details of the firewall you've recorded, if you leave immediately, **what is the
//! severity of your whole trip**?
use anyhow::Result;
pub const INPUT: &str = include_str!("d13.txt");
pub fn solve_part_one(input: &str) -> Result<i64> {
Ok(0)
}
pub fn solve_part_two(input: &str) -> Result<i64> {
Ok(0)
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn part_one() {}
#[test]
fn part_two() {}
}