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Single new-knot insertion preserves the B-spline curve

Statement

Let t0≤⋯≤tn+p be a clamped knot vector of degree p as in B-spline spaces and knot conventions, with a=t0 and b=tn+p, and let ξ(v)j=vj−1 be the one-based version of any zero-based knot vector v. Write Bi,q,v=Bi+1,q,ξ(v) for the basis in B-spline spaces and knot conventions. Let a B-spline curve be C(x)=∑i=0n−1PiBi,p,t(x) with d∈N and Pi∈Rd. Insert one new knot u strictly inside a nonempty span by Single new-knot insertion for a B-spline curve, obtaining knots t′ and controls Qi. Write Bi,q′=Bi,q,t′. Then, for every x∈[a,b], C(x)=∑i=0nQiBi,p,t′(x).

Facts & Assumptions

Given: A clamped degree-p knot vector, controls Pi, and a new knot u strictly inside a nonempty span [tk,tk+1].

[F1]

B-spline spaces and knot conventions defines degree-zero B-splines as half-open interval indicators and defines higher degrees by the Cox–de Boor recurrence with zero-denominator terms set to zero.

[F2]

Single new-knot insertion for a B-spline curve uses Qi=Pi for i≤k−p, Qi=λiPi+(1−λi)Pi−1 for k−p+1≤i≤k, and Qi=Pi−1 for i≥k+1, where λi=(u−ti)/(ti+p−ti) in the middle range.

[F3]

A single new knot refines the B-spline basis gives Bi,p,t=λi(p)Bi,p,t′+(1−λi+1(p))Bi+1,p,t′ for every old degree-p basis index.

[A1]

u is strictly inside the nonempty span, so it is not already a knot and every denominator in the middle blend range is strictly positive.

Proof

1.1A1F1F2F3algebraconstruct

Since the knot vector is clamped and the insertion span is nonempty, p≤k≤n−1. For bookkeeping only, set P−1=Pn=0 and write λj=λj(p). The copy, blend, and shift ranges in [F2] give Qj=λjPj+(1−λj)Pj−1 for every 0≤j≤n: in the copy range λj=1, in the blend range the algorithm's αj is λj, and in the shift range λj=0. This also covers p=0, when the blend range is empty. Therefore the finite refined sum is ∑j=0nQjBj,p′=∑i=0n−1Pi[λiBi,p′+(1−λi+1)Bi+1,p′]: the first control sum uses Pn=0, and in the second the change of index i=j−1 drops the P−1=0 term.

2.1A1F1F3step 1.1algebra∎

By [F3], each bracketed basis combination in step 1.1 equals Bi,p,t. Substitution in that finite sum gives ∑j=0nQjBj,p,t′=∑i=0n−1PiBi,p,t=C(x). The recurrence and control algebra hold at interior knots under the right-cell convention, and at b they hold by the prescribed left limits. Hence the refined expansion equals the original curve for every x∈[a,b].

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