Advent of Code 2023 - Day 3

Dec 3, 2023

Day 3 was no walk in the park; it had me revisiting my approach several times.

You can find the final code @ advent-of-code/2023/day3

We have the following input

467..114..
...*......
..35..633.
......#...
617*......
.....+.58.
..592.....
......755.
...$.*....
.664.598..

Part 1 of the problem wants us to return the sum of all the numbers that have a symbol as a neighbor (. is not a symbol).

The problem is not trivial, I would like to break it down as much as possible so that the solution is easier to read and understand.

At some point we will need to look at the neighbors of a char, so the first function that I would implement is one that returns neighboring coordinates of a coordinate.

type Coord = (usize, usize);

/// Returns neighboring coordinates that are inside of the matrix bounds
fn get_neighbors_coords(n: i32, m: i32, i: i32, j: i32) -> HashSet<Coord> {
    let coords = vec![
        (-1, -1),
        (-1, 0),
        (-1, 1),
        (0, -1),
        (0, 1),
        (1, 1),
        (1, 0),
        (1, -1),
    ];

    let mut nb = HashSet::new();
    for (x, y) in coords {
        nb.insert((i + x, j + y));
    }

    nb.into_iter()
        .filter(|(x, y)| !(*x < 0 || *y < 0 || *x >= n || *y >= m))
        .map(|(x, y)| (x as usize, y as usize))
        .collect()
}

I’ll also need a small utility function to check if a char is a valid symbol or not

fn is_symbol(c: &char) -> bool {
    c.is_ascii_punctuation() && *c != '.'
}

I would like to also have a function that returns a number given the coordinate of one of its digits. A number lies on a single row, so we can use a left and right pointer and move them respectively to the leftmost and rightmost digit value and parse the slice contained between the two pointers as a u32.

fn get_num_at_coord(matrix: &Vec<Vec<char>>, coord: &Coord) -> u32 {
    let row = coord.0;
    let (mut l, mut r) = (coord.1, coord.1);

    while l > 0 && matrix[row][l - 1].is_digit(10) {
        l -= 1;
    }

    while r < matrix[row].len() - 1 && matrix[row][r + 1].is_digit(10) {
        r += 1;
    }

    matrix[row][l..=r]
        .iter()
        .collect::<String>()
        .parse()
        .unwrap()
}

The last utility function that I would like to implement is the one that is going to actually return the neighboring numbers of a coordinate.

/// Returns all the neighboring numbers given a matrix and a coordinate.
fn neighboring_numbers(matrix: &Vec<Vec<char>>, i: usize, j: usize) -> Vec<u32> {
    let mut num = Vec::new();
    let (n, m) = (matrix.len() as i32, matrix[0].len() as i32);
    for (x, y) in get_neighbors_coords(n, m, i as i32, j as i32) {
        if matrix[x][y].is_digit(10) {
            num.push((x, y));
        }
    }

    // I'm using a HashSet because I have to consider a single number given
    // the coordinates of one of its digits, but num could contain multiple
    // coordinates that belong to the same number
    //
    // Consider for example `..23#111.`
    // I would end up with
    // num = (0,2), (0,3), (0,5), (0,6), (0,7)
    // but `get_num_at_coord((0,2)) == get_num_at_coord((0,3)) == 23` and
    // `get_num_at_coord((0,5)) == get_num_at_coord((0,6)) == get_num_at_coord((0,6)) == 111`
    let mut pairs: HashSet<u32> = HashSet::new();
    for coord in num {
        pairs.insert(get_num_at_coord(matrix, &coord));
    }

    pairs.into_iter().collect()
}

I've cheated a little bit in this case. This function will only return a correct answer if and only if the neighboring numbers all differ from each other. I come from the future and the input that is given to us seems to be okay with this assumption, bear with me.

The core logic of part 1 is now trivial, we have to iterate through all the matrix and find all the neighboring numbers of symbols

fn part1(input: &str) -> io::Result<u32> {
    let mut mat: Vec<Vec<char>> = Vec::new();
    for line in input.lines() {
        mat.push(line.chars().collect());
    }

    let mut nums = Vec::new();
    for i in 0..mat.len() {
        for j in 0..mat[0].len() {
            if is_symbol(&mat[i][j]) {
                nums.extend(neighboring_numbers(&mat, i, j))
            }
        }
    }

    Ok(nums.iter().sum())
}

This should be enough to get us through part 1, let’s run some tests to make sure everything is okay

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_get_neighbors_coors() {
        assert_eq!(
            HashSet::from_iter(vec![(1, 0), (1, 1), (0, 1)]),
            get_neighbors_coords(7, 7, 0, 0)
        );
        assert_eq!(
            HashSet::from_iter(vec![(5, 6), (5, 5), (6, 5)]),
            get_neighbors_coords(7, 7, 6, 6)
        );
        assert_eq!(
            HashSet::from_iter(vec![
                (2, 2),
                (2, 3),
                (2, 4),
                (4, 2),
                (4, 3),
                (4, 4),
                (3, 2),
                (3, 4)
            ]),
            get_neighbors_coords(7, 7, 3, 3)
        );
    }

    #[test]
    fn test_part1() {
        assert_eq!(
            4361,
            part1(
                r"467..114..
...*......
..35..633.
......#...
617*......
.....+.58.
..592.....
......755.
...$.*....
.664.598.."
            )
            .unwrap()
        );
    }
}

Bingo! Part 1 passes, let’s move on to Part 2.

Part 2 states that:

A gear is any * symbol that is adjacent to exactly two part numbers. Its gear ratio is the result of multiplying those two numbers together.

We have to return the sum of all the ratios in the input. We can reuse 100% of what we wrote before, I would add a single function to return only valid ratios neighboring numbers.

fn neighboring_number_pair(matrix: &Vec<Vec<char>>, i: usize, j: usize) -> Option<(u32, u32)> {
    let pairs = neighboring_numbers(matrix, i, j);

    match pairs.len() {
        2 => Some((pairs[0], pairs[1])),
        _ => None,
    }
}

With that, part 2 is very similar to part 1, we just have to iterate through the matrix and sum all the values of the ratios that we find.

fn part2(input: &str) -> io::Result<u32> {
    let mut mat: Vec<Vec<char>> = Vec::new();
    for line in input.lines() {
        mat.push(line.chars().collect());
    }

    let mut sum = 0;
    for i in 0..mat.len() {
        for j in 0..mat[0].len() {
            if mat[i][j] == '*' {
                if let Some((n1, n2)) = neighboring_number_pair(&mat, i, j) {
                    sum += n1 * n2;
                }
            }
        }
    }

    Ok(sum)
}

More tests

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_part2() {
        assert_eq!(
            467835,
            part2(
                "467..114..
...*......
..35..633.
......#...
617*......
.....+.58.
..592.....
......755.
...$.*....
.664.598.."
            )
            .unwrap()
        );
        assert_eq!(
            (467 * 35) + (617 * 2) + (755 * 598),
            part2(
                "467..114..
...*......
..35..633.
......#...
617*2.....
.....+.58.
..592.....
......755.
...$.*....
.664.598.."
            )
            .unwrap()
        );
    }
}

There we go, part 2 is done! As I noted above, this solution is not 100% correct as I’ve assumed that all the numbers that make up a ratio are different from each other. Indeed, my solution will fail with this input

467..114..
...*......
.467..633.
...$.*....
.664.598..

A quick test will show just that

#[test]
fn test_part2_alternative() {
    assert_eq!(
        (467 * 467) + (633 * 598),
        part2(
            "467..114..
...*......
.467..633.
...$.*....
.664.598.."
        )
        .unwrap()
    );
}

running 1 test
test tests::test_part2_alternative ... FAILED

---- tests::test_part2_alternative stdout ----
thread 'tests::test_part2_alternative' panicked at src/main.rs:215:9:
assertion `left == right` failed
  left: 596623
 right: 378534
note: run with `RUST_BACKTRACE=1` environment variable to display a backtrace

To solve this you just need to add more conditions to the neighboring_numbers function, but I’m lazy and I won’t do that today :)