Number of spline basis functions

[robertdj]

Hi

Thank you for this package. I consider using it in another package, but there is a thing I don't understand -- from simple examples I expect to have fewer functions.

Consider the knot vector

using BSplines
knots = [0; 0; 0; 1; 1; 1]
basis = BSplineBasis(1, knots)

Here, length(basis) returns 5 as expected and plotting the function look right. But when increasing the order I run into trouble

basis = BSplineBasis(2, knots)

Here, length(basis) returns 6, but I expect 4. Consider the output of plot(basis)

I would index the functions from 0 through 3. I expect the 0th and 3rd functions to be constantly 0. The 1st and 2nd should be the two straight lines.
What are the rest?

This becomes a problem when constructing a Spline because it insists on having a larger coeffs vector than the number of basis functions I anticipate.

[hyrodium]

Here, length(basis) returns 6, but I expect 4.

It seems the constructor BSplineBasis automatically inserts additional knots on the endpoints. (i.e. the knot vector will be converted from [0,0,0,1,1,1] to [0,0,0,0,1,1,1,1] for order 2 (polynomial degree 1). ) With these converted knot vector, number of B-spline functions is 6.

If you would like to have four B-spline functions with knots (0,1) and polynomial degree 1, the following script seems working fine here.

julia> basis = BSplineBasis(2, [0,0,1,1])
4-element BSplineBasis{Vector{Int64}}:
 order: 2
 breakpoints: [0, 0, 1, 1]

julia> length(basis)
4

Btw, I'm also working on another B-spline package, named BasicBSpline.jl. With this package, you can solve this issue like this:

julia> using BasicBSpline

julia> k = KnotVector(0,1)*3
KnotVector([0.0, 0.0, 0.0, 1.0, 1.0, 1.0])

julia> P = BSplineSpace{1}(k) # polynomial degree 1
BSplineSpace{1, Float64}(KnotVector([0.0, 0.0, 0.0, 1.0, 1.0, 1.0]))

julia> dim(P)
4

[robertdj]

Thanks @hyrodium ! It makes sense that the constructor extends the knots based on the order.