Bivariate polynomial to Bezier surface

Question

Please, does anyone know how to obtain the coordinates of control points of cubic Bezier surface for the following cubic bivariate polynomial:

$$p(x, y) = ax^3 + by^3 + cx^2 + dy^2 + ex + fy + g$$

Answer 1

The Bézier patch can be written as $$ \left[\begin{matrix} (1-x)^3 & 3x(1-x)^2 & 3x^2(1-x) & x^3 \end{matrix}\right] \left[\begin{matrix} z_{00} & z_{01} & z_{02} & z_{03} \\ z_{10} & z_{11} & z_{12} & z_{13} \\ z_{20} & z_{21} & z_{22} & z_{23} \\ z_{30} & z_{31} & z_{32} & z_{33} \end{matrix}\right] \left[\begin{matrix} (1-y)^3 \\ 3y(1-y)^2 \\ 3y^2(1-y) \\ y^3 \end{matrix}\right] $$ We need to choose the $z_{ij}$ so that this will be identically equal to $$ ax^3 + by^3 + cx^2 + dy^2 + ex + fy + g $$ In effect, this is just a change-of-basis problem in the vector space of polynomials of degree $3 \times 3$: we have the coefficients in the power basis (aka Taylor basis or monomial basis), and we want to find the coefficients in the Bernstein basis. There are a couple of possible approaches (at least).

Firstly, you could just work out the coefficients $x^3$, $y^3$, $x^2$, $y^2$, etc. in the Bézier patch equation, and equate these to the corresponding coefficients $a$, $b$, $c$, $d$, etc. This will give a linear system of equations that you can solve to get the $z_{ij}$.

Another approach is to just substitute 16 different values for $(x,y)$, which will give you 16 linear equations that you can solve to get the $z_{ij}$. For example, if you substitute $(x,y) = (0,0)$, you immediately get $z_{00} =g$. Similarly, substituting $(x,y) = (1,1)$, gives $z_{33} =a+b+c+d+e+f+g$.