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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):

enter image description here

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?

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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

When are the control points uniquely determined by the image of a Bézier curve?

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  • $\begingroup$ An alternative to the polysemic word "image" is "trajectory" $\endgroup$ – Jean Marie Mar 6 '16 at 16:28
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Suppose $\mathbf{P}_0, \ldots \mathbf{P}_m$, and $\mathbf{Q}_0 \ldots \mathbf{Q}_m$ are two sets of control points that produce the same Bézier curve of degree $m$, in the sense that $$ \sum_{i=0}^m b^m_i(t)\mathbf{P}_i = \sum_{i=0}^m b^m_i(t)\mathbf{Q}_i \quad \text{for all $t \in[0,1]$} $$ Then we have $$ \sum_{i=0}^m b^m_i(t)(\mathbf{P}_i - \mathbf{Q}_i) = \mathbf{0} \quad \text{for all $t \in[0,1]$} $$ This implies that $\mathbf{P}_i = \mathbf{Q}_i$ for $i=0,1, \ldots, m$ since the Bernstein polynomials $b^m_0, \ldots b^m_m$ are linearly independent.

There are cases where two different sets of control points will produce the same trace/image/trajectory. Take a given curve, and compose it with two different polynomials that both map $[0,1]$ to itself. Then clearly you'll get the same image, but the parametric equations will be different, so the control points will also be different.

A specific example: consider the two degree 4 curves whose control points are

$(0,0)$, $(1,0)$, $(\tfrac53,\tfrac23)$, $(2,1)$, $(2,1)$

$(0,0)$, $(0,0)$, $(\tfrac13,0)$, $(1,0)$, $(2,1)$.

Routine calculations show that these two curves both represent the portion of the parabola $y = \tfrac14 x^2$ that lies between the points $(0,0)$ and $(2,1)$. The first one has equation $G(t) = (4t - 2t^2, 4t^2 - 4t^3 + t^4)$ and the second one has equation $H(t) = (2t^2, t^4)$.

These curves are formed by two different reparameterizations of the basic curve $F(t) = (2t, t^2)$: $$ F(t) = (2t, t^2) \;,\; t = 2u - u^2 \;\; \Rightarrow \;\; G(u) = (4u - 2u^2, 4u^2 - 4u^3 + u^4) $$ $$ F(t) = (2t, t^2) \;,\; t = v^2 \;\; \Rightarrow \;\; H(v) = (2v^2, v^4) $$ This is a form of degree elevation (though not the usual form). The degree of the final curve is the product of the degree of the original one and the degree of the reparametrization function. In the example above $2 \times 2 = 4$. So, this process can only produce a cubic curve in trivial cases where either the original curve or the reparameterization function is linear.

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  • $\begingroup$ You probably mean "... parabola $y = \frac14 t^2$ ..." $\endgroup$ – Martin R Mar 8 '16 at 10:23
  • $\begingroup$ You're right. Thanks. Fixed. $\endgroup$ – bubba Mar 8 '16 at 12:04
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Bézier curves with different control points could actually be the same function under one condition: one Bézier curve is degree elevated from the other.

A cubic Bézier curve is represented as (in Bernstein basis)

$C(u)=(1-u)^3P_0+3(1-u)^2uP_1+3(1-u)u^2P_2+u^3P_3$,

or as (in power basis)

$C(u)=P_0+3(P_1-P_0)u+3(P_2-2P_1+P_0)u^2+(P_3-3P_2+3P_1-P_0)u^3$.

From the power-basis representation, we can see that if $(P_3-3P_2+3P_1-P_0)=0$, the curve is actually of degree 2 only and therefore can be represented as a Bézier curve with 3 control points.

Basically, a Bézier curve with $K_1$ control points can always be exactly converted to a Bézier curve with $K_2$ ($K_2 > K_1$) control points. This process is called "degree elevation". However, it does not really change the degree of the underlying polynomial. It only makes the Bézier curve have more control points to be manipulated with.

In summary,
- Bézier curves with different number of control points might actually be the same function.
- Bézier curves of the same number of control points but different control point coordinates cannot be the same function but could be of the same image/trajectory.

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