Normal Reconstruct Z: Could make normalization optional

[OwenTheProgrammer]

Because of the way this function is structured, we can guarantee the length of the resulting vector will be 1 unit in length, thus normalized. This will not be recognized by the compiler, meaning the normalization will not be optimized away automatically.

Math Proof

\begin{aligned}
\sqrt{x^2+y^2+z^2} &= 1\quad \text{Solve for } z \\
\left(\sqrt{x^2+y^2+z^2}\right)^{\textcolor{cyan}{2}} &= 1^{\textcolor{cyan}{2}} \quad \left(\sqrt{v}\right)^2 = |v| \\
|x^2+y^2+z^2| &= 1  \quad \{ \forall v \in \mathbb{R} \;|\; 0 \le v^2 \} \implies |v^2| = v^2\\
x^2 + y^2 + z^2 &= 1 \\
x^2+y^2+z^2 \textcolor{cyan}{- \left(x^2+y^2\right)} &= 1 \textcolor{cyan}{- \left(x^2+y^2\right)} \\
z^2 &= 1 - x^2 + y^2 \\
\textcolor{cyan}{\sqrt{z^2}} &= \sqrt{1 - x^2 + y^2} \quad \sqrt{z^2} = z \\
z &= \sqrt{1 - x^2 + y^2} = \textcolor{green}{\sqrt{1 - \left(x, y\right) \cdot \left(x, y\right)}}
\end{aligned}

Shader Assembly Difference

Current Node	Proposed New Node

This removes 3 shader instructions for free

• 5: dp3 or dot product of vector3's
• 6: rsq or reverse square root as in 1/sqrt(x)
• 7: mul which is just multiply

This will also put this node ahead of SG in terms of performance, as they also implement it the same way

[OwenTheProgrammer]

Those nodes by request

[OwenTheProgrammer]

Ah, for safety of the xy components not being within the ranges and out of bounds like NaN values, I'll change the topic to request normalization as an option

[Dawie3565]

http://paste.amplify.pt/view/raw/ad521618 --- add function switch to make option --- set default as included for backward compatible --- added sticky notes