# Solution to Advent of Code "Day 3: Spiral Memory"

After an unsuccessful attempt at learning Rust earlier this year (I mainly read through the documentation without applying it in any project), I recently started to tackle the 2017 edition of Advent of Code, in order to practice Rust for real.

The 3rd challenge, Spiral Memory is interesting because you can bruteforce it, or solve it with math. I ended up doing the latter, even though math is really not my strong suit.

We're asked to calculate the Manhattan distance between a given point and the center, in a spiral reference. The problem amounts to finding the coordinates of any point $$P$$ in this spiral reference, as once we have the point coordinates, calculating the Manhattan distance is easy:

\begin{align*} D_P &= |X_P - X_0| + |Y_P - Y_0| \\ &= |X_P| + |Y_P| \end{align*}

## Nested shells

My approach was the following: a spiral has nested "shells", all centered around the center. In this image, the first shell is outlined in grey, and the second one in purple. Each of these spirals has a first value, called $$S_i$$, where $$i$$ is the index of the spiral. For any point $$(X_P, Y_P)$$ of value $$V$$, we know that it is located somewhere on the shell located right before the first shell with start value $$S$$ such as $$S > V$$. For example, if the input value was 23, we know that it's located on the second shell as $$S_2 ≤ 23 < S_3$$. We need to know the number of elements a shell of index $$i$$ is composed of, noted $$Δ_i$$ On this representation, the first shell is a square of side of length 3, the second shell is a square of side of length 5. We can generalize this to $$L = 2i + 1$$, where $$i$$ is the index of the shell. For any index $$i$$, the shell is composed of the following number of elements

\begin{align*} Δ_i &= (2i + 1)^2 - (2(i -1) + 1)^2 \\ &= 4i^2 + 4i + 1 - 4i^2 +4i - 1 \\ &= 8i \end{align*}

## Coordinates of the first element of a shell

Once we know on which shell a given point $$P$$ is located, we need to know the coordinates of the first point $$S_i$$ of this shell, so we can infer $$P$$'s coordinates. This first point will always be located after the center point, and all points composing the previous shells. We can thus infer

\begin{equation*} V_{S_{{}_i}} = 2 + \sum_{x=1}^{i-1}Δ_i \end{equation*}

We now need to get the coordinates of any given first shell point. By simply looking at the spiral itself, we can deduce that

\begin{equation*} (X_{S_{{}_i}}, Y_{S_{{}_i}}) = (n, -n + 1) \end{equation*}

## Navigating the spiral

The final piece of the puzzle is to infer the coordinates of the point $$P$$ given the coordinates of the start point $$S_i$$ of the shell it belongs to. To do that, we need to look at how the coordinates evolve along a shell. We can see that:

• on the first quarter of the shell, $$Y$$ coordinates increase by 1 for each increasing value
• on the second quarter of the shell, $$X$$ coordinates decrease by 1 for each increasing value
• on the third quarter of the shell, $$Y$$ coordinates decrease by 1 for each increasing value
• on the fourth quarter of the shell, $$X$$ coordinates increase by 1 for each increasing value

To calculate the coordinates of the point $$P$$, we just need to locate it on the shell, start from $$(X_{S_{{}_i}}, Y_{S_{{}_i}})$$ and increase/decrease the $$X$$ and $$Y$$ coordinates until we reach the target value.

## The implementation

The strategy is:

• calculate the values of the first shell points until we find a value greater than our target point
• backtrack to the previous shell
• compute the coordinates of the first point of the shell we backtracked to
• increase/decrease the $$X$$ and $$Y$$ coordinates until we reach the target value
• calculate the Manhattan distance using these coordinates
// advent_day03.rs

fn nb_elements_in_outer_level(level: i32) -> i32{
8 * level
}

fn start_element(level: i32) -> i32 {
if level == 0 {
1
} else {
let mut out = 0;
for i in 1..level {
out += nb_elements_in_outer_level(i);
}
out + 2
}
}

fn first_element_coordinates(level: i32) -> (i32, i32) {
(level, -level + 1)
}

fn number_coordinates(number: i32) -> (i32, i32) {
let mut level = 0;
let mut start: i32;

// Increase level until we found a starting value greater than
// input value. When such a value is found, backtrack a step.
loop {
start = start_element(level);
if start >= number {
level -= 1;
println!("{:?} is found on level {:?} of the spiral", number, level);
start = start_element(level);
break
} else {
level += 1;
}
}

// At this point, we've found the starting point of the spiral
// level we number belongs to.
let delta = number - start;
let (mut x, mut y) = first_element_coordinates(level);

if delta > 2 * level {
y += 2 * level;
} else {
y += delta;
return (x, y)
}

if delta > 4 * level {
x -= 2 * level;
} else {
x -= delta - (2 * level);
return (x, y)
}

if delta > 6 * level {
y -= 2 * level;
} else {
y -= delta - (4 * level);
return (x, y)
}

x += delta - (6 * level);
(x, y)
}

fn manhattan_distance(x: i32, y: i32) -> i32{
x.abs() + y.abs()
}

fn main() {
let number = 312051;
let (x, y) = number_coordinates(number);
println!("{:?} has coordinates {:?}", number, (x, y));
let distance = manhattan_distance(x, y);
println!("{:?} is at a distance of {:?} from the center", number, distance);
}


## The solution

312051 is found on level 279 of the spiral
312051 has coordinates (-152, -278)
312051 is at a distance of 430 from the center