DEV Community

Nicky Meuleman
Nicky Meuleman

Posted on • Originally published at nickymeuleman.netlify.app

Advent of Code 2022 Day 12

#adventofcode #rust

Day 12: Hill Climbing Algorithm

https://adventofcode.com/2022/day/12

TL;DR: my solution in Rust

You want to get to a higher spot so your communication device gets better reception.

Today's input is a height map of the surrounding area.
Again, it's a perfect 2D grid of positions.

Each position has an elevation from "a" to "z".
"a" being the lowest and "z" the highest.

You start at the position marked "S".
It has an elevation of "a".

You want to go to the position marked "E".
It has an elevation of "z".

An example input looks like this:

Sabqponm
abcryxxl
accszExk
acctuvwj
abdefghi
Enter fullscreen mode Exit fullscreen mode

During each step, you can move exactly one square up, down, left, or right.

You can only go up one height per step.
So from "a" to "b" is possible.
But going from "a" to "c" is not.

You can go down as many heights as you want. (I guess you are wearing long fall boots from Portal.
So going from "y" to "a" is a perfectly valid move.

Parsing

I decided to extract a couple things from the input:

  • The height map, with numbers from 0-25 instead of letters.
  • The coordinates for the start (marked "S")
  • The coordinates for the end (marked "E")
  • The width of the map
  • The height of the map

I used a list of lists to store every height.

  • Each item in the outer list is a row.
  • Each item in a row is a height.

I used a Coord struct again to store coordinates.

struct Coord {
    x: usize,
    y: usize,
}
Enter fullscreen mode Exit fullscreen mode

I use the good 'ol double for loop to loop over each letter in the map.

When I encounter an "S".

  • I set its height to "a".
  • I set the start coordinate.

When I encounter an "E".

  • I set its height to "z".
  • I set the end coordinate.

I turn the height letter into a height number by using the ASCII values again. (as in multiple previous days)
That number gets stored in the height map at the corrent row and column index.

fn parse() -> (Coord, Coord, Vec<Vec<u8>>, usize, usize) {
    let input = std::fs::read_to_string("src/day12.txt").unwrap();
    let rows = input.lines().count();
    let cols = input.lines().next().unwrap().len();
    let mut map = vec![vec![0; cols]; rows];
    let mut start = Coord { x: 0, y: 0 };
    let mut end = Coord { x: 0, y: 0 };

    for (row, line) in input.lines().enumerate() {
        for (col, c) in line.chars().enumerate() {
            let letter = match c {
                'S' => {
                    start.x = col;
                    start.y = row;
                    'a'
                }
                'E' => {
                    end.x = col;
                    end.y = row;
                    'z'
                }
                'a'..='z' => c,
                _ => panic!("Invalid input"),
            };

            let val = letter as u8 - b'a';
            map[row][col] = val;
        }
    }

    (start, end, map, rows, cols)
}
Enter fullscreen mode Exit fullscreen mode

Part 1

The question asks for the fewest steps required to "E" if starting from "S".

Aha, a shortest path problem!
Time to bring out our good friend Dijkstra.

  • The start position is start.
  • The goal position is end.
  • Per step, the cost increases by 1.

It never makes sense to visit a location twice, you could have stepped in the correct direction the first time.
That's why I also keep track of coordinates the algoritm visited in a set, and only examine new coordinates.

in pseudocode:

let mut pq = priority_queue;
let mut visited = set;

while node = pq.pop() {
    // if goal is reached, return

    // get all next coordinates that fit the rules
    for candidate in candidates {
        if seen.includes(candidate) {
            // do not visit nodes twice
        } else {
            // insert new node into priority_queue
        }
    }
}
Enter fullscreen mode Exit fullscreen mode

Helpers

To make it a bit more readable I'll define a few helpers again.
To get all neighbours of a current position given the width and the height of the map:

impl Coord {
    fn neighbours(&self, rows: usize, cols: usize) -> Vec<Self> {
        let mut result = Vec::new();

        // up
        if self.y > 0 {
            result.push(Self {
                x: self.x,
                y: self.y - 1,
            });
        }
        // down
        if self.y < rows - 1 {
            result.push(Self {
                x: self.x,
                y: self.y + 1,
            });
        }
        // left
        if self.x > 0 {
            result.push(Self {
                x: self.x - 1,
                y: self.y,
            });
        }
        // right
        if self.x < cols - 1 {
            result.push(Self {
                x: self.x + 1,
                y: self.y,
            });
        }

        result
    }
}
Enter fullscreen mode Exit fullscreen mode

This doesn't care about the height at a coordinate, only that a coordinate exists on our map!

I also created a Node struct that will be the thing we insert into our priority queue.
It has a cost, and a coordinate.

struct Node {
    cost: u32,
    coord: Coord,
}
Enter fullscreen mode Exit fullscreen mode

The priority queue has to be ordered so the thing with the lowest cost gets popped first, so I implement Ord to get that behaviour.

impl Ord for Node {
    fn cmp(&self, other: &Self) -> Ordering {
        other.cost.cmp(&self.cost)
    }
}

impl PartialOrd for Node {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}
Enter fullscreen mode Exit fullscreen mode

Because this comes with a few other requirements (being able to compare Node, and because I plan on inserting a node into a set, I derive some traits on Coord and Node.

They're both small, so I derive Copy while I'm at it.

#[derive(PartialEq, Eq, Hash, Clone, Copy)]
struct Coord {
    x: usize,
    y: usize,
}

#[derive(PartialEq, Eq, Clone, Copy)]
struct Node {
    cost: u32,
    coord: Coord,
}
Enter fullscreen mode Exit fullscreen mode

Before looping over candidates in the algorithm, I calculate all neighbours.
Then, I filter those and only keep the ones that satisfy the rules.

let curr_height = map[coord.y][coord.x];
let neighbours = coord.neighbours(rows, cols);
let candidates: Vec<_> = neighbours
    .iter()
    .filter(|coord| {
        let height = map[coord.y][coord.x];
        height <= curr_height || height == curr_height + 1
    })
    .collect();
Enter fullscreen mode Exit fullscreen mode

Filling in the skeleton code gives the answer to part1.
A node at position end eventually gets popped off the priority queue.
At that point, we can be certain this node has the lowest cost to reach that coordinate.

Final code 

pub fn part_1() -> u32 {
    let (start, end, map, rows, cols) = parse();
    let mut pq = BinaryHeap::new();
    let mut visited = HashSet::new();

    pq.push(Node {
        cost: 0,
        coord: start,
    });
    visited.insert(start);

    while let Some(Node { coord, cost }) = pq.pop() {
        if coord == end {
            return cost;
        }

        let curr_height = map[coord.y][coord.x];
        let neighbours = coord.neighbours(rows, cols);
        let candidates: Vec<_> = neighbours
            .iter()
            .filter(|coord| {
                let height = map[coord.y][coord.x];
                height <= curr_height || height == curr_height + 1
            })
            .collect();

        for candidate in candidates {
            if visited.insert(*candidate) {
                pq.push(Node {
                    cost: cost + 1,
                    coord: *candidate,
                })
            }
        }
    }

    panic!("No path found");
}
Enter fullscreen mode Exit fullscreen mode

Part 2

The question asks for the fewest steps required to reach "E" if starting from any "a".

To do this, I flipped the logic from part 1, and adjusted the rules.

The parameters for the Dijkstra algorithm are now:

  • The start position is end.
  • The goal position is any a.
  • Per step, the cost increases by 1.

The rules from the initial problem also invert!

  • You can only go down one height per step.
  • You can go up as many heights as you want.

Translated into code, our return condition changes, and the spot we filter neighbours to apply the rules changes.

Making those changes, and boom, that's part2!

Final code 

pub fn part_2() -> u32 {
    let (start, end, map, rows, cols) = parse();
    let mut pq = BinaryHeap::new();
    let mut visited = HashSet::new();

    pq.push(Node {
        cost: 0,
        coord: end,
    });
    visited.insert(start);

    while let Some(Node { coord, cost }) = pq.pop() {
        let curr_height = map[coord.y][coord.x];

        if curr_height == 0 {
            return cost;
        }

        let neighbours = coord.neighbours(rows, cols);
        let candidates: Vec<_> = neighbours
            .iter()
            .filter(|coord| {
                let height = map[coord.y][coord.x];
                height >= curr_height || height == curr_height - 1
            })
            .collect();

        for candidate in candidates {
            if visited.insert(*candidate) {
                pq.push(Node {
                    cost: cost + 1,
                    coord: *candidate,
                })
            }
        }
    }

    panic!("No path found");
}
Enter fullscreen mode Exit fullscreen mode

Final code 

use std::{
    cmp::Ordering,
    collections::{BinaryHeap, HashSet},
};

#[derive(PartialEq, Eq, Hash, Clone, Copy)]
struct Coord {
    x: usize,
    y: usize,
}

#[derive(PartialEq, Eq, Clone, Copy)]
struct Node {
    cost: u32,
    coord: Coord,
}

impl Ord for Node {
    fn cmp(&self, other: &Self) -> Ordering {
        other.cost.cmp(&self.cost)
    }
}

impl PartialOrd for Node {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}

impl Coord {
    fn neighbours(&self, rows: usize, cols: usize) -> Vec<Self> {
        let mut result = Vec::new();

        // up
        if self.y > 0 {
            result.push(Self {
                x: self.x,
                y: self.y - 1,
            });
        }
        // down
        if self.y < rows - 1 {
            result.push(Self {
                x: self.x,
                y: self.y + 1,
            });
        }
        // left
        if self.x > 0 {
            result.push(Self {
                x: self.x - 1,
                y: self.y,
            });
        }
        // right
        if self.x < cols - 1 {
            result.push(Self {
                x: self.x + 1,
                y: self.y,
            });
        }

        result
    }
}

fn parse() -> (Coord, Coord, Vec<Vec<u8>>, usize, usize) {
    let input = std::fs::read_to_string("src/day12.txt").unwrap();
    let rows = input.lines().count();
    let cols = input.lines().next().unwrap().len();
    let mut map = vec![vec![0; cols]; rows];
    let mut start = Coord { x: 0, y: 0 };
    let mut end = Coord { x: 0, y: 0 };

    for (row, line) in input.lines().enumerate() {
        for (col, c) in line.chars().enumerate() {
            let letter = match c {
                'S' => {
                    start.x = col;
                    start.y = row;
                    'a'
                }
                'E' => {
                    end.x = col;
                    end.y = row;
                    'z'
                }
                'a'..='z' => c,
                _ => panic!("Invalid input"),
            };

            let val = letter as u8 - b'a';
            map[row][col] = val;
        }
    }

    (start, end, map, rows, cols)
}

pub fn part_1() -> u32 {
    let (start, end, map, rows, cols) = parse();
    let mut pq = BinaryHeap::new();
    let mut visited = HashSet::new();

    pq.push(Node {
        cost: 0,
        coord: start,
    });
    visited.insert(start);

    while let Some(Node { coord, cost }) = pq.pop() {
        if coord == end {
            return cost;
        }

        let curr_height = map[coord.y][coord.x];
        let neighbours = coord.neighbours(rows, cols);
        let candidates: Vec<_> = neighbours
            .iter()
            .filter(|coord| {
                let height = map[coord.y][coord.x];
                height <= curr_height || height == curr_height + 1
            })
            .collect();

        for candidate in candidates {
            if visited.insert(*candidate) {
                pq.push(Node {
                    cost: cost + 1,
                    coord: *candidate,
                })
            }
        }
    }

    u32::MAX
}

pub fn part_2() -> u32 {
    let (start, end, map, rows, cols) = parse();
    let mut pq = BinaryHeap::new();
    let mut visited = HashSet::new();

    pq.push(Node {
        cost: 0,
        coord: end,
    });
    visited.insert(start);

    while let Some(Node { coord, cost }) = pq.pop() {
        let curr_height = map[coord.y][coord.x];

        if curr_height == 0 {
            return cost;
        }

        let neighbours = coord.neighbours(rows, cols);
        let candidates: Vec<_> = neighbours
            .iter()
            .filter(|coord| {
                let height = map[coord.y][coord.x];
                height >= curr_height || height == curr_height - 1
            })
            .collect();

        for candidate in candidates {
            if visited.insert(*candidate) {
                pq.push(Node {
                    cost: cost + 1,
                    coord: *candidate,
                })
            }
        }
    }

    u32::MAX
}
Enter fullscreen mode Exit fullscreen mode

Top comments (2)

Collapse
 
aryuuu profile image
M Algah Fattah Illahi •

Nice solution!
I am curious, how long does it take to run the solution with the actual input?

Collapse
 
nickymeuleman profile image
Nicky Meuleman • • Edited

Just tested it, with Instant::now() so no rigorous benchmark.

Part 1: 370                 (took: 701.5µs)
Part 2: 363                 (took: 480.4µs)
Enter fullscreen mode Exit fullscreen mode

On a core i5 6600K with a bunch of programs open (VSCode, Steam, Chrome, Discord, ...).

Read next

bleuiot profile image

Seamless BLE Device Integration on Linux with BleuIO

Bleuio tech -

iammrgaurav profile image

Integrate Hashi Corp Vault, with GCS Backend Using Terraform and Helm in k8s cluster.

Gaurav Poudel -

basil-tech-star profile image

react export and import components

Basil -

meesmsik profile image

logical operator C++

Meesmsik -