B-spline/NURBS Curves: Knot Insertion

The meaning of knot insertion is adding a new knot into the existing knot vector without changing the shape of the curve. This new knot can be equal to an existing one and in this case the multiplicity of that knot is increased by one. Because of the fundamental equality m = n + p + 1, after adding a new knot, the value of m is increased by one, and, consequently, either the number of control points or the degree of the curve must also be increased by one. Changing the degree of the curve will change the shape of the curve globally and will not be considered. Therefore, inserting a new knot causes a new control point to be added. In fact, some existing control points are removed and replaced with new ones by corner cutting.

Although knot insertion looks not very interesting, it is one of the most important algorithms for B-spline and NURBS curves since many other useful algorithms will be based on knot insertion. Therefore, we shall start with knot insertion algorithms. In the following, we shall present an algorithm for inserting a single knot.

The left figure below shows a clamped B-spline curve of degree 4 with uniformly spaced knots. The right figure shows the result after a new knot u = 0.5 is inserted. As you can see, the shape of the curve does not change; however, the defining control polyline is changed. In fact, four new control points (in red) replace the original control points p₄, p₅ and p₆, and three line segments (in yellow) cut the corners at p₄, p₅ and p₆.

Inserting a Single Knot

Suppose the new knot t lies in knot span [u_k, u_k+1). From the strong convex hull property, p(t) lies in the convex hull defined by control points p_k, p_k-1, ..., p_k-p and all other basis functions are zero. Thus, the knot insertion computation can be restricted on control points p_k, p_k-1, ..., p_k-p. The way of inserting t is to find p new control points q_k on leg p_k-1p_k, q_k-1 on leg p_k-2p_k-1, ..., and q_k-p+1 on leg p_k-pp_k-p+1 such that the old polyline between p_k-p and p_k (in black below) is replaced by p_k-pq_k-p+1...q_kp_k (in orange below) by cutting the corners at p_k-p+1, ..., p_k-1. All other control points are unchanged. Note that p-1 control points of the original control polyline are removed and replaced with p new control points.

Fortunately, the positions of new control points q_i's are easy to compute. The formula for computing the new control point q_i on leg p_i-1p_i is the following:

where the ratio a_i is defined below:

In summary, to insert a new knot t, we first find the knot span [u_k, u_k+1) that contains this new knot. With k available, p new control points q_k-p+1, ..., q_k are computed with the above formula. Finally, the original control polyline between p_k-p and p_k is replaced with the new polyline defined by p_k-p, q_k-p+1, q_k-p+2, ..., q_k-1, q_k and p_k. Note that after inserting a new knot, the knot vector becomes u₀, u₁, ..., u_k, t, u_k+1, ..., and u_m. If the new knot t is equal to u_k, the multiplicity of u_k is increased by one.

The above computation scheme can be illustrated with the following diagram. First, the affected control points are listed on the left column. Then, the computed new control points are listed on the second column. Note that the number of new control points is one less than the affected control points. The computation of new control point q_i, where k-p+1 <= i <= k, requires two original control points p_i-1 and p_i with coefficients 1-a_i and a_i, respectively. After the computation, we shall use the points surrounded by the blue dotted line to replace those not in the region. All unaffected control points are preserved. Therefore, the original set of control points p_k-p, p_k-p+1, ..., p_k-1, p_k is replaced with p_k-p, q_k-p+1, ..., q_k-1, p_k.

Let us take a look at a geometric interpretation of a_i. From its definition, a_i is ratio of dividing interval [u_i, u_i+p) by the value of t as shown below:

There are k a_i's, each of which covers p knot spans (i.e., [u_i, u_i+p)). If we stack these intervals together and align at the value of t, we have the following diagram:

Therefore, the position of t divides knot spans [u_k, u_k+p), [u_k-1, u_k+p-1), ..., [u_k-p+1, u_k+1) into ratios a_k, a_k-1, ...,, a_k-p+1, which, in turn, provide the same ratios for dividing legs p_kp_k-1, p_k-1p_k-2, ... p_k-pp_k-p+1. ,

Example 1: Inserting a Knot in a Knot Span

Suppose we have a degree 3 B-spline curve with a knot vector as follows:

We want to insert a new knot t = 0.5. Since t = 0.5 lies in knot span [u₅,u₆), the affected control points are p₅, p₄, p₃ and p₂. To determine the three new control points q₅, q₄ and q₃, we need to compute a₅, a₄ and a₃ as follows:

a₅ = (t - u₅) / (u₈ - u₅) = (0.5 - 0.4) / ( 1 - 0.4) = 1/6
a₄ = (t - u₄) / (u₇ - u₄) = (0.5 - 0.2) / ( 0.8 - 0.2) = 1/2
a₃ = (t - u₃) / (u₆ - u₃) = (0.5 - 0) / ( 0.6 - 0) = 5/6

q₅ = (1 - 1/6)p₄ + (1/6)p₅
q₄ = (1 - 1/2)p₃ + (1/2)p₄
q₃ = (1 - 5/6)p₂ + (5/6)p₃

Since the original B-spline curve has p = 3 and m = 11, we have n = m - p - 1 = 11 - 3 - 1 = 7 and hence 8 control points. The following is a B-spline curve satisfying this condition and its basis functions:

The three blue rectangles are q₃, q₄ and q₅. After inserting t = 0.5, the corners at p₃ and p₄ are cut, yielding the following B-spline curve and its basis functions. Note that the shape of the curve does not change:

The following diagram shows the relationship between the old and new control points. Note that the new set of control points contains p₀, p₁, p₂, q₃, q₄, q₅, p₅, p₆ and p₇.

The relationship among a_j's, u_j's, and t are shown below:

Example 2: Inserting a Knot at an Existing Simple Knot

We want to insert a new knot t = 0.5, which is equal to an existing one (i.e., t = u₈ = 0.5). The following is a B-spline curve of degree 4 and its basis functions before the new knot t is inserted.

Since t is in [u₈,u₉), the affected control points are p₈, p₇, p₆, p₅ and p₄. The coefficients are computed as follows:

a₈ = (t - u₈) / (u₁₂ - u₈) = (0.5 - 0.5) / (1 - 0.875) = 0
a₇ = (t - u₇) / (u₁₁ - u₇) = (0.5 - 0.375) / (0.875 - 0.375) = 1/4
a₆ = (t - u₆) / (u₁₀ - u₆) = (0.5 - 0.25) / (0.75 - 0.25) = 1/2
a₅ = (t - u₅) / (u₉ - u₅) = (0.5 - 0.125) / (0.625 - 0.125) = 3/4

q₈ = (1 - 0)p₇ + 0p₈
q₇ = (1 - 1/4)p₆ + (1/4)p₇
q₆ = (1 - 1/2)p₅ + (1/2)p₆
q₅ = (1 - 3/4)p₄ + (3/4)p₅

a_k = (t - u_k) / (u_k+p - u_k) = 0

q_k = (1 - 0)p_k-1 + 0p_k = p_k-1

The figure shown at the beginning of this example shows the new control points and the new control polyline in orange, Please note that the new control points are p₀, p₁, p₂, p₃, p₄, q₅, q₆, q₇, q₈ = p₇, p₈, p₉, p₁₀ and p₁₁. The curve and its basis functions after new knot t = 0.5 inserted is shown below.

The relationship among a_j's, u_j's, and t are shown below:

Example 3: Inserting a Knot at an Existing Multiple Knot

What if the new knot t is inserted at a multiple knot? Suppose t is inserted at knot u_k of multiplicity s. Hence, we have s consecutive equal knots: u_k = u_k-1 = u_k-2 = .... = u_k-s+1 and u_k-s+1 being not equal to u_k-s. In the computation of coefficients a_k, ...., a_k-p+1, we have the following:

a_k = (t - u_k) / (u_k+p - u_k) = (u_k - u_k) / (u_k+p - u_k) = 0
a_k-1 = (t - u_k-1) / (u_k+p-1 - u_k-1) = (u_k - u_k-1) / (u_k+p-1 - u_k-1) = 0
..........
a_k-s+1 = (t - u_k-s+1) / (u_k-s+1+p - u_k-s+1) = (u_k - u_k-s+1) / = 0

q_k = (1 - a_k)p_k-1 + a_kp_k = p_k-1
q_k-1 = (1 - a_k-1)p_k-2 + a_k-1p_k-1 = p_k-2
..........
q_k-s = (1 - a_k-s+1) p_k-s + a_k-s+1p_k-s = p_k-s

Knot Insertion for NURBS Curves

Suppose we have n + 1 control points p₀, p₁, ..., p_n with associated weights w₀, w₁, ..., w_n, a knot vector U and a degree p. Let p_i = (x_i, y_i, z_i). Then, control points P_i = ( w_ix_i, w_iy_i, w_iz_i, w_i ), 0 <= i <= n, and knot vector U define a four dimensional B-spline of degree p. We can insert a knot t to this four dimensional B-spline yielding a new set of control points Q_i = ( X_i, Y_i, Z_i, W_i ), 0 <= i <= n. Projecting these control points back to the three-dimensional space by dividing the first three components with the fourth one yields the new set of control points of the given NURBS curve.

Let us take a look at an example. Suppose we have 9 knots

and a NURBS curve of degree 3 defined by the following 5 control points in the xy-plane:

The following shows the curve and its basis functions.

Let us insert a new knot t = 0.4. Since t is in knot span [u₃, u₄) and the degree of the NURBS curve is 3, the affected control points are p₃, p₂, p₁ and p₀. Since this is a NURBS curve, we shall use homogeneous coordinates by multiplying all control points with their corresponding weights. Call these new control points P_i:

Note that since P₄ is not affected, it is not computed in the above table. Then, we shall compute a₃, a₂ and a₁ as follows:

a₃ = (t - u₃)/ (u₆ - u₃) = (0.4 - 0)/(1 - 0) = 0.4
a₂ = (t - u₂)/ (u₅ - u₂) = (0.4 - 0)/(1 - 0) = 0.4
a₁ = (t - u₁)/ (u₄ - u₁) = (0.4 - 0)/(0.5 - 0) = 0.8

Q₃ = (1 - a₃)P₂ + a₃P₃ = ( 325.6, 26, 4.4 )
Q₂ = (1 - a₂)P₁ + a₂P₂ = ( 97.4, 142.5, 1.9 )
Q₁ = (1 - a₁)P₀ + a₁P₁ = ( -42, 14.8, 0.6 )

q₃ = ( 74, 5.9 ) with weight 4.4
q₂ = ( 51.3, 75 ) with weight 1.9
q₁ = ( -70, 24.6 ) with weight 0.6

The following is the resulting NURBS curve and its basis functions: