# Support of a clamped B-spline

Let a B-spline of degree $p$ be defined by its parametric equation $$\mathbf{r}(t) = \sum_{i=0}^n N_i^p(t)\mathbf{P}_i$$

where the $n+1$ control points are denoted by $\mathbf{P}_i$. The basis functions $N_i^p$ are classically defined as

\begin{align} N_i^0(t) &= \begin{cases} 1 &\text{ if }t \in [t_i,t_{i+1}[ \\ 0 &\text{ otherwise } \end{cases} \end{align} \begin{align} N_i^p(t) = \frac{t - t_i}{t_{i + p} - t_i}N_i^{p-1} + \frac{t_{i + p + 1} - t}{t_{i + p + 1} - t_{i + 1}}N_{i+1}^{p-1} \end{align} where $\{t_i\}_{i = 0..m = n + p + 1}$ are the so-called spline knots.

It is my understanding that because of the compact support of those basis functions, one only needs to evaluate the parametric equation over $[t_p,t_{m-p}]$ to obtain the full spline.

My current implementation of these equations appears to work when the spline is left open (no knot has a multiplicity equal to p + 1):

For uniformly spaced knots $t = [ 0, 1/6, 1/3, 1/2, 2/3, 5/6, 1 ]$ (m = 6) and a control net comprised of $n + 1 = 4$ points, thus corresponding to a spline of degree $p =2$, I get the following curve (which as I said seems correct)

Opened B-spline

However, enforcing the clamping of the spline's ends by setting

$t= [ 0, 0, 0, 0.5, 1, 1, 1 ]$ only works on one end, as shown below

Clamped B-spline. Only one end is clamped properly

The end corresponding to $t_p = 0$ is clamped, but the other end corresponding to $t_{m-p} = 1$ goes to zero.

I believe this oddity is caused by the incorrect definition of the B-spline support in the clamped case. I found that changing $N_i^0(t) = 1$ if $t \in [t_i,t_{i+1}[$ to $N_i^0(t) = 1$ if $t \in [t_i,t_{i+1}]$ (i.e using a closed interval instead) was allowing the clamping to work.

This being said, I am not sure at all that changing the support of the basis functions is required in order to get a clamped B-spline. Am I missing something?

• Note, that you say "order" but you use "degree" $p$. I am almost sure that this is your problem. The order is the number of degrees of freedom between two knots the degree of the spline is the degree of the polynomial functions which represent the spline between its knots. E.g., a spline of degree 3 has order 4. The multiplicity of the knots at the ends must be equal to the order for clamped splines, i.e., for a spline of degree 3 you need 4 knots with equal value at both ends. May 27, 2016 at 16:49
• You should mention which value you used for $p$. May 27, 2016 at 16:55
• Thanks for your comment. You were right in correcting me for the confusion between 'order' and 'degree'. Yet, I can't find an error in my implementation based on the numbers I have (p = 2, n = 4 and m = 6). Keeping the definition of the 0-th degree basis function as in the first paragraph, I also noticed that evaluating the Bspline over the half-open interval $t = [u_p,u_{m-p})$ was circumventing the anomaly observed on the far end of the spline. Though I have not seen any explicit mention of this requirement (evaluating the clamped spline over a half-open interval)... May 27, 2016 at 17:00

If one starts from a definition of b(asis)-splines like that one in De Boor's book "A Practical Guide to Splines" basing on divided differences \begin{align*} N_i^p(t) = (t_{i+p+1}-t_i)[t_i,\ldots,t_{i+p+1}]_{\bar t} (\bar t - t)_+^p \end{align*} with \begin{align*} (x)_+^p = \begin{cases} 0 &@\ x\leq 0\\ x^p &@\ x > 0\end{cases} \end{align*} the convention $x^0:=0$ and a given knot sequence $(t_i)_{i=0}^m$ with $t_i\leq t_{i+1}$. Then this leads to right-continuous functions. Note, that your recursive formula works for this definition.

Case $p=0$ and $t_i<t_{i+1}$: \begin{align*} N_i^0(t) &= (t_{i+1}-t_i) [t_i,t_{i+1}]_{\bar t} (\bar t - t)_+^0\\ &=(t_{i+1}-t_i)\frac{(t_{i+1}-t)_+^0 - (t_i - t)_+^0}{t_{i+1}-t_i}\\ &=\begin{cases} 1 &@\ t\in [t_i,t_{i+1}[\\ 0 &@\text{ else} \end{cases} \end{align*}

Last base function for the case $p=1$: \begin{align*} N^1_{n}(t) &= (t_{n+1}-t_n) [t_n,t_{n+1},t_{n+1}]_{\bar t}(\bar t-t)_+\\ &= [t_{n+1},t_{n+1}](\bar t - t)_+ - [t_n,t_{n+1}](\bar t - t)_+\\ &= \left.\frac{d}{d \bar t}(\bar t - t)_+\right|_{\bar t=t_{n+1}} - \frac{(t_{n+1} - t)_+ - (t_n - t)_+}{t_{n+1}-t_n}\\ &= \begin{cases} 0 &@\ t < t_n\\ \displaystyle 1-\frac{(t_{n+1} - t)_+ }{t_{n+1}-t_n}&@\ t_n\leq t < t_{n+1}\\ 0 &@\ t_{n+1}\leq t \end{cases} \end{align*} This base function is discontinuous at $t=t_{n+1}$. The limit from the left is not equal to the value at $t=t_{n+1}$.

In praxis one does not really use the last base function directly but the continuous continuation of it from the half-open interval $[t_p,t_{m-p}[$ to the closed interval $[t_p,t_{m-p}]$.

This is simply done by limiting the search for the index $i$ with $t\in [t_i,t_{i+1}[$ to the largest of the indexes of the inner knots.

Beforehand one should check whether $t$ falls within the interval $[t_p,t_{m-p}]$. Places $t$ outside of this interval often get a special treatment such as linear extrapolation (this is only one of several alternatives to the nulling that the base functions cause there).