A cubic Bézier curve is a polynomial
$$F(u) = \sum_{i=0}^{n} \mathbf{b}_i^n P_i \;\;\;\text{ with } u \in [0,1], P_i \in \mathbb{R}^2, n=3 \text{ and } \mathbf{b}_i^n = \begin{pmatrix}n\\i\end{pmatrix} u^i (1-u)^{n-i}$$
You get plots like this (source):
Having the set $\mathbb{R}^{4 \times 2}$ of all cubic Bézier curves defined by their control points and the set of all of their plots, I wondered: Are there any two Bézier curves which have different control points $P_i, P_i'$ but are the same function?
Obviously, to be the same function the point $P_0 = P_0'$ and $P_3'$ have to be the same. Also, $P_1, P_1'$ and $P_2, P_2'$ have to be on the same line, because $\overline{P_0 P_1}$ is a tangent on the curve. But besides that, I'm not too sure if there could be a combination where the points are different, but the curves are the same.
edit: I think one problem might be when all control points are on the same line. Is this actually a counter example? Are there others?
Best Answer
The equation for the cubic Bézier curve is $$ F(u) = (1-u)^3 P_0 + 3 (1-u)^2 u P_1 + 3 (1-u)u^2 P_2 + u^3 P_3 $$ and therefore $$F(0) = P_0, \quad F(1) = P_3$$ and, by differentiation, $$ F'(0) = 3(P_1 - P_0), \quad F'(1) = 3(P_3 - P_2) $$
Therefore all control points are uniquely determined by the function $F$.
But, as you already noted, there can be two different Bézier curves having the same image, e.g. if all control points are on a line. I don't have an answer for the question