Complexity of color::is_dark(self: ARGB)

[ListeriaM]

for $c_1 \lt c_2$ the following holds:

$\sqrt{(x - c_1)^2} \lt \sqrt{(x - c_2)^2} \iff 2 \times x \lt c_2 + c_1$

proof:

$\sqrt{(x - c_1)^2} \lt \sqrt{(x - c_2)^2}$
$(x - c_1)^2 \lt (x - c_2)^2$
$x^2 - 2 \times c_1 \times x + c_1^2 \lt x^2 - 2 \times c_2 \times x + c_2^2$
$2 \times (c_2 - c_1) \times x \lt c_2^2 - c_1^2$
$2 \times x \lt c_2 + c_1$

using the fact that $W_r - B_r = W_g - B_g = W_b - B_b$ the same logic can
be applied to r, g, b, (B)lack and (W)hite, which gives us:

$\sqrt{(r - B_r)^2 + (g - B_g)^2 + (b - B_b)^2} \lt \sqrt{(r - W_r)^2 + (g - W_g)^2 + (b - W_b)^2} \iff$
$\iff 2 \times (r + g + b) \lt (W_r + B_r) + (W_g + B_g) + (W_b + B_b)$

with the following possible implementation for color::is_dark(self: ARGB):

pub fn is_dark(self) -> bool {
    // should cast to a larger integer type to avoid overflow
    2 * (self.r + self.g + self.b)
        < (Self::WHITE.r + Self::BLACK.r)
            + (Self::WHITE.g + Self::BLACK.g)
            + (Self::WHITE.b + Self::BLACK.b)
}

considering $W_r = W_g = W_b = 255 \land B_r = B_g = B_b = 0$ we can simplify:

pub fn is_dark(self) -> bool {
    2 * (self.r as u32 + self.g as u32 + self.b as u32) < 3 * 0xff
}