3
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For a given set of data points

$$\{(x_i, y_i, z_i)\}$$

there exists some

$$f_{ABC}(x,y)=Ax^2+Bxy+Cy^2$$

that minimizes

$$\sum_i(f_{ABC}(x_i,y_i)-z_i)^2$$

$A$, $B$, and $C$ can be found quickly by directly solving/minimizing

$$(\begin{bmatrix}x_1^2&x_1y_1&y_2^2\\x_1^2&x_2y_2&y_2^2\\x_1^2&x_3y_3&y_3^2\\\vdots&\vdots&\vdots\end{bmatrix}\begin{bmatrix}A\\B\\C\end{bmatrix}-\begin{bmatrix}z_1\\z_2\\z_3\\\vdots\end{bmatrix})^2$$

However, suppose we want to reduce the model's degrees of freedom from 3 to 2 by constraining $f_{ABC}$ to be a parabolic sheet. I think this would mean that $\begin{bmatrix}A&B&C\end{bmatrix}$ has to be of the form $\begin{bmatrix}U^2&2UV&V^2\end{bmatrix}$ so that $f_{ABC}(x,y)=(Ux+Vy)^2$.

It seems this is no longer solvable directly as a linear system.

What would be a performant/fast way to solve this constrained version of the problem?

Background

The above question is a simpler version of my actual problem, which is to fit an explicit bicubic surface patch $g(x,y)=Ax^2+Bxy+Cy^2+D+Ex^3+Fy^3+Gx^2y+Hxy^2+Ix+Jy$ that is constrained to be quadratic (instead of cubic) along some (any) direction.

Thoughts

Thinking further, my best idea currently is to

  1. Directly fit an unconstrained quadratic surface i.e. find a solution in the span of $\begin{bmatrix}x_i^2&x_iy_i&y_i^2&1&x_i&y_i\end{bmatrix}$. Call this the "quadratic solution".
  2. Orthogonalize the remaining basis vectors $\begin{bmatrix}x_i^3&y_i^3&x_i^2y_i&x_iy_i^2\end{bmatrix}$ w.r.t. the quadratic solution.
  3. Somehow solve a constrained minimization within this reduced basis that is orthogonal to the quadratic solution, though I don't really know how to go about this.

I think the eventual cubic terms should be of the form $Ex^3+Fy^3+Gx^2y+Hxy^2=(Ux+Vy)^3$ so the constraint is $\begin{bmatrix}E&F&G&H\end{bmatrix}=\begin{bmatrix}U^3&3U^2V&3UV^2&V^3\end{bmatrix}$

How to do this constrained optimization?

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