Advent of Code 3 2017 part 1

3.rs

/*

the largest number in the square of (odd) width x is a(x) = x^2
the smallest number in the square of (odd) width x is b(x) = a(x-2)+1 = (x-2)^2+1 = x^2 - 4x5

the number x is in the square of width c(x) = ceil((x^0.5-1)/2)*2+1
the number of steps that x is from the smallest number in its square is d(x) = x - b(c(x))

assuming the number x is on the top row, its distance from the rightmost number is e(x) = d(x) - c(x) + 2
the distance from the centermost number out to the edge is then h(x) = floor(c(x)/2))
its distance from the centermost number is then f(x) = abs(e(x) - h(x))
the manhattan distance to the center is then g(x) = f(x) + h(x)

g(x) = abs(e(x) - floor(c(x)/2)) + floor(c(x)/2)
g(x) = abs(d(x) - c(x) + 2 - floor(c(x)/2)) + floor(c(x)/2)
g(x) = abs(x - b(c(x)) - c(x) + 2 - floor(c(x)/2)) + floor(c(x)/2)
g(x) = abs(x - b(ceil((x^0.5-1)/2)*2+1) - ceil((x^0.5-1)/2)*2+1 + 2 - floor(ceil((x^0.5-1)/2)*2+1/2)) + floor(ceil((x^0.5-1)/2)*2+1/2)
g(x) = abs(x - (ceil((x^0.5-1)/2)*2+1-2)^2+1 - ceil((x^0.5-1)/2)*2+1 + 2 - floor(ceil((x^0.5-1)/2)*2+1/2)) + floor(ceil((x^0.5-1)/2)*2+1/2)

g(x) = abs(x - (ceil((x^0.5-1)/2)*2+1-2)^2+1 - ceil((x^0.5-1)/2)*2+1 + 2 - floor(ceil((x^0.5-1)/2)*2+1/2)) + floor(ceil((x^0.5-1)/2)*2+1/2)



c(347991) = 591
b(c(347991)) = 591^2 - 4*591 + 5 = 346922
347991 - b(c(347991)) = 347991 - 346922 = 1069

1069 - 591 = 478
*/

/// The largest number in the square of (odd) width x
fn a(x: f64) -> f64 {
    x * x
}

/// The smallest number in the square of (odd) width x
fn b(x: f64) -> f64 {
    a(x - 2.) + 1.
}

/// The width of the square that x is in
fn c(x: f64) -> f64 {
    ((x.sqrt() - 1.) / 2.).ceil() * 2. + 1.
}

/// The number of steps that x is from the smallest number in its square
fn d(x: f64) -> f64 {
    x - b(c(x))
}

/// Assuming the number is on the top row, returns its distance from the
/// rightmost number.
fn e(x: f64) -> f64 {
    d(x) - c(x) + 2.
}

/// The distance from the centermost number to the edge
fn f(x: f64) -> f64 {
    (e(x) - h(x)).abs()
}

/// The Manhattan distance from the center to this number
fn g(x: f64) -> f64 {
    f(x) + h(x)
}

/// The distance from 1 out to the square that x is in
fn h(x: f64) -> f64 {
    (c(x) / 2.).floor()
}

fn main() {
    println!("{}", g(347991.));
}