# A New Paradigm to Design a Class of Combined Ternary Subdivision Schemes

Authors: Humaira Mustanira Tariq; Rabia Hameed; Ghulam Mustafa
Journal: Journal of Mathematics
 (2021)

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## Abstract

Subdivision schemes play a vital role in curve modeling. The curves produced by the class of2n+2-point ternary scheme (Deslauriers and Dubuc (1989)) interpolate the given data while the curves produced by a class of 2n+2-point ternary B-spline schemes approximate the given data. In this research, we merge these two classes to introduce a consolidated and unified class of combined subdivision schemes with two shape control parameters in order to grow versatility for overseeing valuable necessities. However, the proposed class of subdivision schemes gives optimal smoothness in the final shapes, yet we can increase its smoothness by using a proposed general formula in form of its Laurent polynomial. The theoretical analysis of the class of subdivision schemes is done by using various mathematical tools and using their coding in the Maple environment. The graphical analysis of the class of schemes is done in the Maple environment by writing the codes based on the recursive mathematical expressions of the class of subdivision schemes.


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## Body

## 1. Introduction

Nowadays, subdivision schemes have got great importance in the field of geometric modeling. It is an innovation which creates smooth shapes. This technique produces a sequence of refined polygons which converge to a limiting shape. New points are added at each refinement level in order to get a smooth final shape. The number of points which are inserted at each refinement level is known as the arity of the curve subdivision scheme. Binary schemes insert two new points at each refinement level between every old consecutive pair of points of the previous refinement level [1]. Similarly, ternary subdivision schemes insert three new points at each refinement level between every old consecutive pair of points of the previous refinement level. Hence, ternary schemes smooth the given sketch in fewer subdivision steps as compared to the binary subdivision schemes. If we move the single point of the control polygon, then the shape of the curve changes over the specific region. This region is known as the support of the scheme. Subdivision schemes give us local control on the shapes. Schemes with small support give better local control on shapes as compared to the schemes with large support size. Ternary schemes give us small support as compared to the binary subdivision scheme and hence give better local control on the shapes. Because of these two main characteristics of ternary schemes, we consider them superior to the binary subdivision schemes.Subdivision schemes can also produce the shapes which pass through the initial data. These types of subdivision schemes are known as interpolatory subdivision schemes. These schemes have been introduced by [2–6]. The other types of subdivision schemes produce shapes which do not pass through the initial data. This type of subdivision schemes is known as approximating subdivision schemes and was introduced by [7–10]. The analysis of both types of schemes is done by [11, 12]. In [13–16], the subdivision schemes are introduced with parameters which can produce both the interpolatory and approximating shapes. The combined behaviour of these combined subdivision schemes got more attention in curve modeling.In this paper, we present a class of ternary combined schemes which is obtained from the well-known classes of interpolatory and approximating subdivision schemes. We unify interpolatory and approximating classes of subdivision schemes into a single class of combined schemes by applying certain mathematical operations on their refinement rules. Consequently, we get a class of2n+2-point ternary combined subdivision schemes. Two shape parameters are also inserted in the refinement rules to control the appearance of the limiting shapes. A sketch of our construction procedure is shown in Figure 1.Figure 1
Flowchart for the construction of combined subdivision schemes.The remaining layout of the paper is as follows: in Section2, basic definitions, notations, and established tools are introduced. In Section 3, we construct a parametric class of 2n+2-point ternary relaxed subdivision schemes. This class depends on two parameters. In Section 4, we discuss several features of this class. In Section 5, we present the geometrical influence of the schemes on the shapes. The conclusion of this research is presented in Section 6.


## 2. Preliminaries

In this section, we review some fundamental definitions and known realities about subdivision schemes, which structure the basis of the remainder of this paper. Ifqk=qiki∈ℤ and qk+1=qik+1i∈ℤ are two polygons at kth and k+1 the level, then the ternary subdivision scheme SSa which produces qk+1 from qk is defined as(1)SSaqki=qik+1=∑j∈ℤai−3jqjk,i∈ℤ,where the set a=ai:i∈ℤ of coefficients is called the mask of the subdivision scheme.For the ternary subdivision schemeSSa, a point at k+1-th level is calculated by(2)SSaqk3i+l=q3i+lk+1=∑γ∈ℤa3γ+lqi−γk,i∈ℤ,0≤l≤2.A necessary condition of the subdivision scheme for uniform convergence is that(3)∑j∈ℤa3j=∑j∈ℤa3j+1=∑j∈ℤa3j+2=1.In order to analyze the characteristics of the subdivision scheme, thez-transform of the mask is(4)az=∑i∈ℤaizi,z∈ℂ\0,which is also known as the Laurent polynomial of the scheme.Definition 1.
A subdivision scheme is uniformly convergent if, for any initial dataq0=qi0:i∈ℤ, there exists a continuous function f, such that for any closed interval I⊂ℝ that satisfies(5)limk⟶∞supj∈3kIfj3k−qjk=0,where f=S∞q0.Definition 2.
If we rewrite the Laurent polynomial ofSSa as az=1+z+z2/3kδz, where δz has no factor of the form 1+z+z2 and m¯0,…,m¯l are nonzero coefficients of δz. Let F0,F1,…,Fl be l×l matrices with the elements given by Fℏi,j=m¯ℓ+i−3j+ℏ, where ℏ=0,1,…,l while i,j=1,…,l. Hölders regularity/continuity is defined as r=k−lognμ, whereas the parameter μ is known as a joint spectral radius of F0,F1,…,Fl. It is bounded below by spectral radii and from above by the norm of matrices F0,F1,…,Fl. If lower and upper bounds coincide, an explicit formula for the joint spectral radius is obtained:(6)maxρF0,ρF1,…,ρFl,≤μ≤maxF0∞,F1∞,…,Fl∞.Theorem 1.
(see [17]). If Sa is the subdivision scheme with Laurent polynomial az, then Sa is said to be Cj-continuous if the subdivision scheme 1/3Sn is contractive. That is,(7)13SnL∞=13Lmax∑j∈ℤai+3jLn,L;i=0,1,2,…,3L−1<1,where(8)an,L=∏j=1Laz3j−1,anz=3z21+z+z2an−1z.


## 3. Formulation of the Class of Combined Schemes

Here, we construct and unify the generalized form of the refinement rules of the class of interpolating and approximating schemes. The construction of the class of combined subdivision schemes is based on two classes of subdivision schemes. One of which is the class of2n+2-point ternary interpolatory subdivision schemes (CETISS), which is constructed by the Lagrange interpolating polynomial, and the other one is the class of 2n+2-point relaxed ternary approximating subdivision schemes (CETASS), which is constructed by the B-spline basis function. We merge these classes of schemes in a specific manner such that we get a single class of 2n+2-point combined schemes from them.


### 3.1. Mathematical Representation of the Construction Procedure

In this subsection, we present the stepwise procedure to construct the class of combined ternary subdivision schemes. The construction procedure of the class of combined subdivision schemes is given below.Step 1.
We define the generalized form of the refinement rules of the CETISS; that is, if we have a shape which is obtained by joining the 2D pointsqi,2n+2k:i∈ℤ, then, to refine this shape by CETISS, we use the following refinement rules:(9)L3i,2n+2k+1=qi,2n+2k,L3i+1,2n+2k+1=133n+1∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+jqi+j,2n+2k,L3i+2,2n+2k+2=133n+1∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1qi+j,2n+2k,and, hence, we get a refined shape by joining the 2D refined points Li,2n+2k+1:i∈ℤ, where nr=n!/r!n−r!.Step 2.
By Pitolli [18], ternary B-spline schemes of degree-M can be written as(10)Bδ,Mk+1=13M∑η=δ+1/2δM+1ηηδ−η,0≤δ≤2M+1.
By using (10), we can generalize the refinement rules of the CETASS; that is, if we have a shape which is obtained by joining the 2D points qi,2n+2k:i∈ℤ, then to refine this shape by CETASS, we use the following refinement rules:(11)S3i,2n+2k+1=133m+1∑η=02m∑γ=3η+3/23η+23m+2γγ3η+2−γqi+η−m,2n+2k,S3i+1,2n+2k+1=133m+1∑η=02m+1∑γ=3η+2/23η+23m+2γγ3η+1−γqi+η−m,2n+2k,S3i+2,2n+2k+1=133m+1∑η=02m+1∑γ=3η+1/23η+23m+2γγ3η−γqi+η−m,2n+2k,and thus we get a refined shape by joining the 2D refined points Si,2n+2k+1:i∈ℤ.Step 3.
Now we calculate the displacement vectors from the refined points of CETISS, that is,L3i+α,2n+2k+1:α=0,1,2 to the refined points of CETASS, that is, S3i+α,2n+2k+1:α=0,1,2. We denote these displacement vectors by Δ3i+α,2n+2k+1 where α=0,1,2, respectively. Hence the displacement vectors are(12)Δ3i+α,2n+2k+1=L3i+α,2n+2k+1−S3i+α,2n+2k+1,where α=0,1,2.Step 4.
Now we find the new refinement pointq3i+α,2n+2k+1:α=0,1,2 by using the displacement vectors Δ3i+α,2n+2k+1:α=0,1,2 and the refined points L3i+α,2n+2k+1:α=0,1,2, respectively. We control the position of the refinement points q3i+α,2n+2k+1:α=0,1,2 by controlling the direction and size of these displacement vectors. The shape parameters μ and ν control the direction and size of the displacements vectors. The proposed refined points are given below:(13)q3i,2n+2k+1=L3i,2n+2k+1−μ+νΔ3i,2n+2k+1,q3i+1,2n+2k+1=L3i+1,2n+2k+1−μΔ3i+1,2n+2k+1−νΔ3i+2,2n+2k+1,q3i+2,2n+2k+1=L3i+2,2n+2k+1−νΔ3i+1,2n+2k+1−μΔ3i+2,2n+2k+1.Step 5.
Hence we get the refinement rulesq3i+α,2n+2k+1:α=0,1,2 by using (9), (12), and (13). That is,(14)q3i,2n+2k+1=1−μ−νqi,2n+2k+μ+ν×133m+1∑η=02m∑γ=3η+3/23η+23m+2γγ3η+2−γqi+η−m,2n+2k,q3i+1,2n+2k+1=133n+1∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+jqi+j,2n+2k−μ133n+1∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+jqi+j,2n+2k−133m+1∑η=02m+1∑γ=3η+2/23η+23m+2γγ3η+1−γqi+η−m,2n+2k−ν133n+1∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1qi+j,2n+2k−133m+1∑η=02m+1∑γ=3η+1/23η+23m+2γγ3η−γqi+η−m,2n+2k,q3i+2,2n+2k+1=133n+1∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1qi+j,2n+2k−ν133n+1∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+jqi+j,2n+2k−133m+1∑η=02m+1∑γ=3η+2/23η+23m+2γγ3η+1−γqi+η−m,2n+2k−μ133n+1∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1qi+j,2n+2k−133m+1∑η=02m+1∑γ=3η+1/23η+23m+2γγ3η−γqi+η−m,2n+2k.where ⌊⋅⌋ denotes the floor function. In this article, we denote the class of combined subdivision schemes (14) by CSSa2n+2.


### 3.2. The Graphical Clarification of the Parameters and the Refinement Rules

In this subsection, we graphically explain the construction of the class of subdivision schemes. For this, we restrict the value ofn to be 1. Let qi+α,4k=qi+αk:α=−1,0,1,2 be the control points of the polygon qkqk; then, for n=1, the class of subdivision schemes (9) gives the following 4-point ternary interpolatory subdivision scheme:(15)L3ik+1=L3i,4k+1=qik,L3i+1k+1=L3i+1,4k+1=−581qi−1k+6081qik+3081qi+1k−481qi+2k,L3i+2k+1=L3i+2,4k+1=−481qi−1k+3081qik+6081qi+1k−581qi+2k.Similarly, forn=1, the class of scheme (11) gives the following 4-point relaxed approximating subdivision scheme:(16)S3ik+1=S3i,4k+1=527qi−1k+1727qik+527qi+1k,S3i+1k+1=S3i+1,4k+1=581qi−1k+59qik+1027qi+1k+181qi+2k,S3i+2k+1=S3i+2,4k+1=181qi−1k+1027qik+59qi+1k+581qi+2k.Now, we find the displacement vectors from the refined points defined in (15) to the refined points defined in (16); thus, we have(17)Δ3ik+1=Δ3i,4k+1=L3ik+1−S3ik+1=1027qik−527qi−1k−527qi+1k,Δ3i+1k+1=Δ3i+1,4k+1=L3i+1k+1−S3i+1k+1=−1081qi−1k+527qik−581qi+2k,Δ3i+2k+1=Δ3i+2,4k+1=L3i+2k+1−S3i+2k+1=−581qi−1k+527qi+1k−1081qi+2k.Now, we find the refined points by applying operations on vectors−μ+νΔ3ik+1, −μΔ3i+1k+1, −νΔ3i+2k+1 and refined points L3i+αk+1:α=0,1,2, where μ and ν are real numbers. Hence, we get(18)q3ik+1=q3i,4k+1=L3ik+1−μ+νΔ3ik+1,q3i+1k+1=q3i+1,4k+1=L3i+1k+1−μΔ3i+1k+1−νΔ3i+2k+1,q3i+2k+1=q3i+2,4k+1=L3i+2k+1−νΔ3i+1k+1−μΔ3i+2k+1.By simplifying (18), we get(19)q3ik+1=5μ27+5ν27qi−1k+1−10μ27−10ν27qik+5μ27+5ν27qi+1k,q3i+1k+1=−581+5ν81+10μ81qi−1k+2027−5μ27qik+1027−5ν27qi+1k+−481+5μ81+10ν81qi+2k,q3i+2k+1=−481+5μ81+10ν81qi−1k+1027−5ν27qik+2027−5μ27qi+1k+−581+5ν81+10μ81qi+2k.Scheme (19) is the first member of the class of combined subdivision schemes, that is, 4-point combined subdivision scheme. The graphical sketch of the refined points of (19) is shown in Figure 2. In this figure, green bullets show the points qik:i∈ℤ, green lines show the polygon qk, red squares show the refined points L3i+αk+1:α∈ℝ, red lines show the refined polygon Lk+1, blue squares show the refined points S3i+αk+1:α∈ℝ, blue lines show the refined polygon Sk+1, black arrows show the displacement vectors, and black solid circles show the refined points qi,2n+2k+1=qik+1:i∈ℤ.Figure 2
Graphical sketch of the construction of the refinement points of scheme (19).Remark 1.
Whenμ=ν=0, the first member of the class of combined subdivision schemes which is defined in (19) reduces to the first member of the CETISS. And when μ=1 and ν=0, the first member of the class of combined subdivision schemes which is defined in (19) reduces to the first member of the CETASS.


## 3.1. Mathematical Representation of the Construction Procedure

In this subsection, we present the stepwise procedure to construct the class of combined ternary subdivision schemes. The construction procedure of the class of combined subdivision schemes is given below.Step 1.
We define the generalized form of the refinement rules of the CETISS; that is, if we have a shape which is obtained by joining the 2D pointsqi,2n+2k:i∈ℤ, then, to refine this shape by CETISS, we use the following refinement rules:(9)L3i,2n+2k+1=qi,2n+2k,L3i+1,2n+2k+1=133n+1∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+jqi+j,2n+2k,L3i+2,2n+2k+2=133n+1∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1qi+j,2n+2k,and, hence, we get a refined shape by joining the 2D refined points Li,2n+2k+1:i∈ℤ, where nr=n!/r!n−r!.Step 2.
By Pitolli [18], ternary B-spline schemes of degree-M can be written as(10)Bδ,Mk+1=13M∑η=δ+1/2δM+1ηηδ−η,0≤δ≤2M+1.
By using (10), we can generalize the refinement rules of the CETASS; that is, if we have a shape which is obtained by joining the 2D points qi,2n+2k:i∈ℤ, then to refine this shape by CETASS, we use the following refinement rules:(11)S3i,2n+2k+1=133m+1∑η=02m∑γ=3η+3/23η+23m+2γγ3η+2−γqi+η−m,2n+2k,S3i+1,2n+2k+1=133m+1∑η=02m+1∑γ=3η+2/23η+23m+2γγ3η+1−γqi+η−m,2n+2k,S3i+2,2n+2k+1=133m+1∑η=02m+1∑γ=3η+1/23η+23m+2γγ3η−γqi+η−m,2n+2k,and thus we get a refined shape by joining the 2D refined points Si,2n+2k+1:i∈ℤ.Step 3.
Now we calculate the displacement vectors from the refined points of CETISS, that is,L3i+α,2n+2k+1:α=0,1,2 to the refined points of CETASS, that is, S3i+α,2n+2k+1:α=0,1,2. We denote these displacement vectors by Δ3i+α,2n+2k+1 where α=0,1,2, respectively. Hence the displacement vectors are(12)Δ3i+α,2n+2k+1=L3i+α,2n+2k+1−S3i+α,2n+2k+1,where α=0,1,2.Step 4.
Now we find the new refinement pointq3i+α,2n+2k+1:α=0,1,2 by using the displacement vectors Δ3i+α,2n+2k+1:α=0,1,2 and the refined points L3i+α,2n+2k+1:α=0,1,2, respectively. We control the position of the refinement points q3i+α,2n+2k+1:α=0,1,2 by controlling the direction and size of these displacement vectors. The shape parameters μ and ν control the direction and size of the displacements vectors. The proposed refined points are given below:(13)q3i,2n+2k+1=L3i,2n+2k+1−μ+νΔ3i,2n+2k+1,q3i+1,2n+2k+1=L3i+1,2n+2k+1−μΔ3i+1,2n+2k+1−νΔ3i+2,2n+2k+1,q3i+2,2n+2k+1=L3i+2,2n+2k+1−νΔ3i+1,2n+2k+1−μΔ3i+2,2n+2k+1.Step 5.
Hence we get the refinement rulesq3i+α,2n+2k+1:α=0,1,2 by using (9), (12), and (13). That is,(14)q3i,2n+2k+1=1−μ−νqi,2n+2k+μ+ν×133m+1∑η=02m∑γ=3η+3/23η+23m+2γγ3η+2−γqi+η−m,2n+2k,q3i+1,2n+2k+1=133n+1∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+jqi+j,2n+2k−μ133n+1∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+jqi+j,2n+2k−133m+1∑η=02m+1∑γ=3η+2/23η+23m+2γγ3η+1−γqi+η−m,2n+2k−ν133n+1∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1qi+j,2n+2k−133m+1∑η=02m+1∑γ=3η+1/23η+23m+2γγ3η−γqi+η−m,2n+2k,q3i+2,2n+2k+1=133n+1∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1qi+j,2n+2k−ν133n+1∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+jqi+j,2n+2k−133m+1∑η=02m+1∑γ=3η+2/23η+23m+2γγ3η+1−γqi+η−m,2n+2k−μ133n+1∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1qi+j,2n+2k−133m+1∑η=02m+1∑γ=3η+1/23η+23m+2γγ3η−γqi+η−m,2n+2k.where ⌊⋅⌋ denotes the floor function. In this article, we denote the class of combined subdivision schemes (14) by CSSa2n+2.


## 3.2. The Graphical Clarification of the Parameters and the Refinement Rules

In this subsection, we graphically explain the construction of the class of subdivision schemes. For this, we restrict the value ofn to be 1. Let qi+α,4k=qi+αk:α=−1,0,1,2 be the control points of the polygon qkqk; then, for n=1, the class of subdivision schemes (9) gives the following 4-point ternary interpolatory subdivision scheme:(15)L3ik+1=L3i,4k+1=qik,L3i+1k+1=L3i+1,4k+1=−581qi−1k+6081qik+3081qi+1k−481qi+2k,L3i+2k+1=L3i+2,4k+1=−481qi−1k+3081qik+6081qi+1k−581qi+2k.Similarly, forn=1, the class of scheme (11) gives the following 4-point relaxed approximating subdivision scheme:(16)S3ik+1=S3i,4k+1=527qi−1k+1727qik+527qi+1k,S3i+1k+1=S3i+1,4k+1=581qi−1k+59qik+1027qi+1k+181qi+2k,S3i+2k+1=S3i+2,4k+1=181qi−1k+1027qik+59qi+1k+581qi+2k.Now, we find the displacement vectors from the refined points defined in (15) to the refined points defined in (16); thus, we have(17)Δ3ik+1=Δ3i,4k+1=L3ik+1−S3ik+1=1027qik−527qi−1k−527qi+1k,Δ3i+1k+1=Δ3i+1,4k+1=L3i+1k+1−S3i+1k+1=−1081qi−1k+527qik−581qi+2k,Δ3i+2k+1=Δ3i+2,4k+1=L3i+2k+1−S3i+2k+1=−581qi−1k+527qi+1k−1081qi+2k.Now, we find the refined points by applying operations on vectors−μ+νΔ3ik+1, −μΔ3i+1k+1, −νΔ3i+2k+1 and refined points L3i+αk+1:α=0,1,2, where μ and ν are real numbers. Hence, we get(18)q3ik+1=q3i,4k+1=L3ik+1−μ+νΔ3ik+1,q3i+1k+1=q3i+1,4k+1=L3i+1k+1−μΔ3i+1k+1−νΔ3i+2k+1,q3i+2k+1=q3i+2,4k+1=L3i+2k+1−νΔ3i+1k+1−μΔ3i+2k+1.By simplifying (18), we get(19)q3ik+1=5μ27+5ν27qi−1k+1−10μ27−10ν27qik+5μ27+5ν27qi+1k,q3i+1k+1=−581+5ν81+10μ81qi−1k+2027−5μ27qik+1027−5ν27qi+1k+−481+5μ81+10ν81qi+2k,q3i+2k+1=−481+5μ81+10ν81qi−1k+1027−5ν27qik+2027−5μ27qi+1k+−581+5ν81+10μ81qi+2k.Scheme (19) is the first member of the class of combined subdivision schemes, that is, 4-point combined subdivision scheme. The graphical sketch of the refined points of (19) is shown in Figure 2. In this figure, green bullets show the points qik:i∈ℤ, green lines show the polygon qk, red squares show the refined points L3i+αk+1:α∈ℝ, red lines show the refined polygon Lk+1, blue squares show the refined points S3i+αk+1:α∈ℝ, blue lines show the refined polygon Sk+1, black arrows show the displacement vectors, and black solid circles show the refined points qi,2n+2k+1=qik+1:i∈ℤ.Figure 2
Graphical sketch of the construction of the refinement points of scheme (19).Remark 1.
Whenμ=ν=0, the first member of the class of combined subdivision schemes which is defined in (19) reduces to the first member of the CETISS. And when μ=1 and ν=0, the first member of the class of combined subdivision schemes which is defined in (19) reduces to the first member of the CETASS.


## 4. Analysis of the Class of Combined Subdivision Schemes by Using Mathematical Tools

In this section, we present the characteristics of the class of combined subdivision schemes. We use the mathematical tools to analyze the class of schemes.


### 4.1. Affine Combination of Control Points

A linear combination in which the sum of coefficients is equal to one is called the affine combination. The necessary condition for the convergence of a subdivision scheme is that each refinement rule of it must be the affine combination of the control points in the previous subdivision step. To check this property forCSSa2n+2, we prove the following theorem.Theorem 2.
Each refined point of theCSSa2n+2 at k+1-th subdivision step which is defined in (14) is the affine combination of control points of k-th subdivision step.Proof.
To prove this theorem, we have to prove that the maskai,2n+2:i∈ℤ of CSSa2n+2 satisfies the relation ∑i∈ℤa3i+α,2n+2=1, where α=0,1,2.
Let us simplify the expressions:(20)∑η=02m∑γ=3η+3/23η+23m+2γγ3η+2−γ=33m+1,∑η=02m+1∑γ=3η+2/23η+23m+2γγ3η+1−γ=33m+1,∑η=02m+1∑γ=3η+1/23η+23m+2γγ3η−γ=33m+1.
We also calculate(21)∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+j=33n+1,∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1=33n+1.
Hence, to prove that each of the refined pointsq3i,2n+2k+1, q3i+1,2n+2k+1, and q3i+2,2n+2k+1 is the affine combinations of qik:i∈ℤ, we use (20) and (21). Thus,(22)∑i∈ℤa3i,2n+2=1−μ−ν+μ+ν133m+133m+1=1,∑i∈ℤa3i+1,2n+2=133n+133n+1−μ133n+133n+1−133m+133m+1−ν133n+133n+1−133m+133m+1=1,∑i∈ℤa3i+2,2n+2=133n+133n+1−ν133n+133n+1−133m+133m+1−μ133n+133n+1−133m+133m+1=1,which completes the proof.


### 4.2. Support of the Class of Combined Schemes

In this subsection, we present the support size of the class of combined schemes. The support of a subdivision scheme represents how far one vertex is affected by its neighboring points. Its size represents the local support property of the subdivision curve.Theorem 3.
The support of the class of subdivision schemesCSSa2n+2 is −3n+2/2,3n+2/2.Proof.
Letqi,2n+20=qi0∈ℝi∈ℤ be the initial data such that q00=1 and qα0≠0 for α∈ℤ\0; that is,(23)qα,2n+20=1,forα=0,0,otherwise.
In order to calculate the support size ofCSSa2n+2, we have to calculate the distance between two corresponding subscripts belonging to the leftmost vertex and rightmost vertex after k-steps of subdivision.
If we operate on initial dataqi,2n+20=qi0∈ℝi∈ℤ by CSSa2n+2 one time, then we get(24)qα,2n+21=≠0,forα=−303n+2,−303n+2+1,…,303n+2−1,303n+2,0,otherwise.
If we operate on the dataqi,2n+21∈ℝi∈ℤ which is defined in (24) by CSSa2n+2 one time, that is, we operate on initial data qi,2n+20=qi0∈ℝi∈ℤ by CSSa2n+2 two times, then we get(25)qα,2n+22=≠0,forα=−∑j=013j3n+2,−∑j=013j3n+2+1,…,∑j=013j3n+2−1,∑j=013j3n+2,0,otherwise.
Similarly, if we operate on initial dataqi,2n+20=qi0∈ℝi∈ℤ which is defined in (23) by CSSa2n+2k-times, then we get(26)qα,2n+2k=≠0,forα=−∑j=0k−13j3n+2,−∑j=0k−13j3n+2+1,…,∑j=0k−13j3n+2−1,∑j=0k−13j3n+2,0,otherwise.
Hence, afterk-steps of subdivision, the leftmost nonzero vertex is q−3n+2∑j=0k−13j,2n+2k and the rightmost nonzero vertex is q3n+2∑j=0k−13j,2n+2k. Since k-times, the subdivision process is applied and, at each step, we have relabeled the subscripts of the vertices; thus, qi/3k,2n+2k=qi/3k−1,2n+2k−1=qi/3k−2,2n+2k−2=⋯=qi/3,2n+21=qi,2n+20. Hence, after k-steps, the distance between leftmost and rightmost nonzero vertices is equal to the support of CSSa2n+2; that is,(27)support=3n+21+3+32+⋯+3k−13k−−3n+21+3+32+⋯+3k−13k=23n+213+132+⋯+13k=23n+21/31−1/3=3n+2.
Since the class of subdivision schemes is symmetric, then obviously the support region is−3n+2/2,3n+2/2.


### 4.3. Generation Degree

One of the necessary conditions for generatingCn- continuous limit curves is the order of polynomial generation of the subdivision scheme. A ternary subdivision scheme can generate polynomials up to degree d if and only if its Laurent polynomial can be written in the following form:(28)az=1+z+z23d+1φz,z∈ℂ\0.Here, φz is the Laurent polynomial with φ1=3 and it does not have the factor 1+z+z2.Theorem 4.
The class of combined schemesCSSa2n+2 generates polynomials of degree up to 2 for all μ and ν.Proof.
The Laurent polynomial ofCSSa2n+2 is(29)a2n+2z=1+z+z233b2n+2z,where b2n+2z is a Laurent polynomial obtained by b2n+2z=3a2n+2z/1+z+z2. Hence, the degree of polynomial generation of CSSa2n+2 is 2.


### 4.4. Properties of the 4-Point Subdivision Scheme (19)

From Theorem2, the subdivision scheme which is defined in (14) and denoted by CSSa4 satisfies the necessary conditions for convergence. Moreover, the Laurent polynomial of CSSa4 is(30)a4z=181z51+z+z2310ν−4+5μz4+−25ν+7−5μz3+30ν+3z2+−25ν+7−5μz+10ν−4+5μ.By using Laurent polynomial ofCSSa4, we prove the following theorem.Theorem 5.
Letqi0i∈ℤ be the initial data; then, the refinement rules defined in (19) of CSSa4 produce the C2- continuous shapes if ν∈−2/5,1/5 and its corresponding μ∈1/5−7ν/2,2−7ν/2.Proof.
To prove thatCSSa4 produces the C2-continuous shapes, first, we have to prove that this scheme produces C0-continuous and C1-continuous shapes in the given intervals of μ and ν.
Thus, by using (30), we calculate(31)a41z=3z21+z+z2a40z.
Hence, from the above equation, we have(32)a41z=127z2−4+5μ+10νz5+−1+5μ−5νz4+10ν+5μ+5z3+26−10μ−20νz2+29+10ν−10μz1+26−10μ−20νz0=∑j=−23aj,4zj,where the subdivision scheme corresponding to the Laurent polynomial a41z is denoted by CSSa41; then, by Theorem 1, we have(33)13CSSa411∞=13max∑j∈ℤai+3j,41,1;i=0,1,2.
This implies that(34)13CSSa411∞=13max∑j∈ℤa3j,41,1,∑j∈ℤa1+3j,41,1,∑j∈ℤa2+3j,41,1.
By substituting values, we get(35)13CSSa411∞=max181−4+5μ+10ν+26−10μ−20ν+10ν+5μ+5,−1+5μ−5ν+29+10ν−10μ+−1+5μ−5ν.
Hence,1/3CSSa411∞<1 for μ+2ν∈−14/5,53/10 and μ−ν∈5/2,28/5.
This implies thatCSSa41 is contractive and the scheme CSSa4 produces C0-continuous shapes.
Now, we again calculate(36)a42z=3z21+z+z2a41z,which implies(37)a41z=19z210ν−4+5μz5+3−15νz4+6+15νz3+−10μ+17−20νz2+6+15νz1+3−15νz0+10ν−4+5μz−1,where the subdivision scheme corresponding to the Laurent polynomial a42z is denoted by CSSa42; then, by Theorem 1, we have(38)13CSSa421∞=max127.−4+5μ+10ν+17−10μ−20ν+10ν+5μ−4,3−15ν+6+15ν.
Thus,1/3CSSa421∞<1 for ν∈−1,4/5 and μ∈−1/10−2ν,−2ν+13/5.
Therefore,CSSa42 is contractive and hence CSSa4 produces C1-continuous shapes.
Now, we find the following Laurent polynomial by usinga43z=3z2/1+z+z2a42z. Thus, we have(39)a43z=1310ν−4+5μz5+−25ν+7−5μz4+30ν+3z3+−25ν+7−5μz2+10ν−4+5μz.
LetCSSa43 be the subdivision scheme corresponding to the Laurent polynomial a43z; then, by Theorem 1, we get(40)13CSSa431∞=max19−4+5μ+10ν+7−5μ−25ν,3+30ν.
Thus,1/3CSSa431∞<1 for ν∈−2/5,1/5 and μ∈1/5−7ν/2,2−7ν/2.
This implies thatCSSa43 is contractive and hence CSSa4 is C2-continuous. This completes the proof.If we putν = 0 in the Laurent polynomial (30), then it gives one more factor of 1+z+z2. Hence, the following corollary is proved.Corollary 1.
Ifν=0 and μ∈4/5,6/5, then CSSa4 produces C3-continuous shapes.Corollary 2.
The continuity ofCSSa4 can be increased if we use the refinement rules corresponding to the following Laurent polynomial which is derived from the Laurent polynomial of (30):(41)αz=13n+2C0nC0n10ν−4+5μ+C0nC0n−15ν+3+C1n10ν−4+5μz+C0nC0n15ν+6+C1n−15ν+3+C2n10ν−4+5μ+C1nC0n−110ν−4+5μz2+C0nC0n−20ν−10μ+17+C1n15ν+6+C2n−15ν+3+C3n10ν−4+5μ+C1nC0n−1−15ν+3+C1n−110ν−4+5μz3+C0nC0n15ν+6+C1n−20ν−10μ+17+C2n15ν+6+C3n−15ν+3+C1nC0n−115ν+6+C1n−1−15ν+3+C2n−110ν−4+5μ+C2nC0n−210ν−4+5μz4+⋯++Cnn−2C2210ν−4+5μz2n+4+Cn−1n10ν−4+5μ+−15ν+3+Cnn−15ν+3z2n+5+Cnn10ν−4+5μz2n+6,where μ and ν are shape parameters and n=1,2,….Remark 2.
If we putn=2, n=3, and n=4 in (41), we get the Laurent polynomials of 3-point, 4-point, and 5-point combined subdivision schemes which produce continuous shapes up to C2, C3, and C4 smoothness.Theorem3 gives the following result.Corollary 3.
The support ofCSSa4 is −5/2,5/2.Now, here we use the method proposed by Rioul [19] to find out the Ho¨ölders regularity of CSSa4.Theorem 6.
If we chooseμ=1 and ν=−1/5, then the lower and upper bound of Ho¨lder regularity of CSSa4 is 2.1079 and 2.369, respectively.Proof.
From (30), we have(42)a4z=1+z+z233z−5∑j=04mjzj,where m0=10ν−4+5μ/3, m1=−25ν+7−5μ/3, m2=30ν+3/3, m3=−25ν+7−5μ/3, m4=10ν−4+5μ/3, k=3, ℓ=4, n=3, and ℏ=0,1,2,3,4. F0, F1, F2, F3, and F4 are matrices given by(43)Fℏi,j=mℓ+i−3j+ℏ.
SinceF0i,j=m4+i−3j,F1i,j=m5+i−3j,F2i,j=m6+i−3j,F3i,j=m7+i−3j,F4i,j=m8+i−3j, where i,j=1,2,3,4, thus, we have(44)F0=m2000m3m000m4m1000m200,F1=m3m000m4m1000m2000m3m00,F2=m4m1000m2000m3m000m4m10,F3=0m2000m3m000m4m1000m20,F4=0m3m000m4m1000m2000m3m0.
LetρFj:j=0,1,2,3,4 be the joint spectral radii of matrices F0, F1, F2, F3, and F4, respectively; then, we have(45)maxρF0,ρF1,ρF2,ρF3,ρF4≤ρ≤maxF0∞,F1∞,F2∞,F3∞,F4∞.
The eigenvalues ofF0, F1, F2, F3, and F4 are required to calculate their joint spectral radii; therefore, ρF0=ρF2=ρF3=ρF4=,maxmo,m2, ρF1=,maxm3+m0,m1+m4.
The infinity norm ofF0, F1, F2, F3, and F4 is(46)Fj∞=maxm2,m3+m0,m4+m1,where j=0,1,2,3,4.(47)maxm0,m2,m3+m0,m1−m4≤ρ≤maxm2,m3+m0,m4+m1.
If we useμ=1 and ν=−1/5 in (47), we get(48)max1,0.333,2,2.6667≤ρ≤max1,2.
Hence, the lower bound of Ho¨ölders exponent is(49)r≥3−log32.6667=2.10719.
The upper bound of Ho¨lders exponent is(50)r≤3−log32=2.369.
This completes the proof. □Theorem 7.
The limit stencil ofCSSa4 which is defined in (18) is(51)5μ+5ν−348,27−5μ−5ν48,27−5μ−5ν48,5μ+5ν−348.Proof.
To find out the limit stencil ofCSSa4, we substitute i=−1 and i=0 in the refinement rules (19) of CSSa4. Hence, we obtain the following refined points:(52)q−3k+1=5μ27+5ν27q−2k+1−10μ27−10ν27q−1k+5μ27+5ν27q0k,q−2k+1=−581+5ν81+10μ81q−2k+2027−5μ27q−1k+1027−5ν27q0k+−481+5μ81+10ν81q1k,q−1k+1=−481+5μ81+10ν81q−2k+1027−5ν27q−1k+2027−5μ27q0k+−581+5ν81+10μ81q1k,q0k+1=5μ27+5ν27q−1k+1−10μ27−10ν27q0k+5μ27+5ν27q1k.
The matrix form of the above system of four linear equations is(53)q−3k+1q−2k+1q−1k+1q0k+15μ27+5ν271−10μ27−10ν275μ27+5ν270−581+5ν81+10μ812027−5μ271027−5ν27−481+5μ81+10ν81−481+5μ81+10ν811027−5ν272027−5μ27−581+5ν81+10μ8105μ27+5ν271−10μ27−10ν275μ27+5ν27q−3kq−2kq−1kq0k.
Symbolically, it can be written as(54)qk+1=R1qj.
Hence, the local subdivision matrix of the 4-point schemeCSSa4 is(55)R1=5μ27+5ν271−10μ27−10ν275μ27+5ν270−581+5ν81+10μ812027−5μ271027−5ν27−481+5μ81+10ν81−481+5μ81+10ν811027−5ν272027−5μ27−581+5ν81+10μ8105μ27+5ν271−10μ27−10ν275μ27+5ν27.
Now we calculate the eigenvalues of the subdivision matrixR1 and denote them by λi:i=0,1,2,3. Thus, we have(56)λi=1,13,19,10ν+127.
The matrix of the eigenvector corresponding to the eigenvalues is(57)W=1−11−15μ+5ν−35μ+5ν−27−1315ν−5μ+115μ+15ν−275μ+5ν−35μ+5ν−27131−5ν−5μ+115μ+15ν−271111.
By taking the inverse of the above matrix, we get(58)W−1=27−5μ−5ν485μ+5ν−27485μ+5ν−274827−5μ−5ν48−5ν−5μ+120ν−16−35ν+5μ−920ν−1635ν+5μ−920ν−165ν−5ν+120ν−165μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−348−5ν+5μ−920ν−1635ν+5μ−920ν−16−35ν+5μ−920ν−165ν+5μ−920ν−16.
Letδ be the diagonal matrix of eigenvalues; then,(59)δ=1000013000019000010ν+127.
SinceR1=WδW−1, therefore R1k=WδkW−1. Also, we have(60)qk+1=R1qk=R12qk−1=R13qk−2=⋯=R1kq0,which implies that(61)qk+1=WδkW−1q0.
Hence,(62)limk⟶∞qk+1=Wlimk⟶∞δkW−1q0.
Thus, we have(63)q−3∞q−2∞q−1∞q0∞=5μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−3485μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−3485μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−3485μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−348q−30q−20q−10q00.
Hence, we get the limit stencil ofCSSa4, that is,(64)5μ+5ν−348,27−5μ−5ν48,27−5μ−5ν48,5μ+5ν−348.
Hence, it is proved.


### 4.5. Properties of the 6-Point Subdivision Scheme

If we putn=2 in (14), then we get the following 6-point combined subdivision scheme. We denote this scheme by CSSa6:(65)q3ik+1=4μ243+4ν243qi−2k+56μ243+56ν243qi−1k+1−40μ81−40ν81qik+56μ243+56ν243qi+1k+4μ243+4ν243qi+2k,q3i+1k+1=−16μ2187−20ν2187+8729qi−2k+−70729+280ν2187+476μ2187qi−1k+560729−664μ2187−56ν2187qik+280729−56μ2187−664ν2187qi+1k+−56729+280μ2187+476ν2187qi+2k+7729−20μ2187−16ν2187qi+3k,q3i+2k+1=7729−20μ2187−16ν2187qi−2k+−56729+280μ2187+476ν2187qi−1k+280729−56μ2187−664ν2187qik+560729−664μ2187−56ν2187qi+1k+−70729+280ν2187+476μ2187qi+2k+−16μ2187−20ν2187+8729qi+3k.The Laurent polynomial ofCSSa6 denoted by a6z=a60z is defined:(66)a6z=12187z81+z+z2321−20μ−16νz10+−39+44μ+28νz9+24μ+48ν−9z8+−54+276ν+84μz7+−108μ−936ν+153z6+99+1200ν−48μ×z5+−108μ−936ν+153z4+−54+276ν+84μz3+24μ+48ν−9z2+−39+44μ+28νz+21−20μ−16ν.Theorem 8.
Letqi0i∈ℤ be the initial data; then, the refinement rules defined in (65) of CSSa6 produce the C2- continuous shapes if ν∈−1/4,1/8 and its corresponding μ∈3/32−157ν/32,255/128−157ν/32. It gives C3-continuous shapes for ν=0 and μ∈93/128,87/64. It produces C4-smooth shapes for ν=0 and μ∈3/5,51/40. And it gives C5-smooth shapes for ν=0 and μ∈57/62,33/31.Proof.
By Theorem1, we have(67)a6nz=3z21+z+z2a6n−1z,i=1,2,3,4,5,6.
Hence,(68)a61z=172921−20μ−16νz8+3−4ν+4μz7+−24+56ν+52μz6+224μ+424ν−147z5+−39−200ν+200μz4+280ν+186+80μz3+−336μ+693−744νz2+801+408ν−408μz1+−336μ+693−744νz0+280ν+186+80μz−1+−39−200ν+200μz−2+224μ+424ν−147z−3+−24+56ν+52μz−4+3−4ν+4μz5+21−20μ−16νz−6.a62z=124321−20μ−16νz8+12ν+24μ−18z7+60ν+48μ−27z6+152μ−102+352νz5+90−612νz4+198−72μ+540νz3+405−264μ−672νz2+198−72μ+540νz1+90−612νz0+152μ−102+352νz−1+60ν+48μ−27z−2+12ν+24μ−18z−3+21−20μ−16νz−4.a63z=181z221−20μ−16νz10+−39+44μ+28νz9+24μ+48ν−9z8+−54+276ν+84μz7+−108μ−936ν+153z6+99+1200ν−48μz+−108μ−936ν+153z4+−54+276ν+84μz3+24μ+48ν−9z25+−39+44μ+28νz+21−20μ−16ν.
And whenν=0, we get the following Laurent polynomials from (67):(69)a64z=12721−20μz8+−60+64μz7+−20μ+30z6+40μ−24z5+147−128μz4+40μ−24z3+−20μ+30z2+−60+64μz+21−20μ,a65z=1921−20μz6+−81+84μz5+−84μ+90z4+40μ−33z3+−84μ+90z2+−81+84μz1+21−20μ,α6z=3z21+z+z21921−20μz6+−81+84μz5+−84μ+90z4+40μ−33z3+−84μ+90z2+−81+84μz1+21−20μ.
Then, by using Theorem1 and denoting the subdivision schemes corresponding to the Laurent polynomials a6jz:j=0,1,2,3,4,5,6 by CSSa6j:j=0,1,2,3,4,5,6, respectively, we get(70)13CSSa6j1∞=13max∑j∈ℤαi+3jj,1;i=0,1,2,j=0,1,2,3,4,5,6.
By solving1/3CSSa6j1∞<1 for j=0,1,2,3,4,5,6, we get the required result.Corollary 4.
The order of smoothness ofCSSa6 can be increased if we use the refinement rules corresponding to the following Laurent polynomial which is derived from the Laurent polynomial of (66):(71)αz=13n+5C0nC0n21−16ν−20μ+C0nC0n−18+12ν+24μ+C1n21−16ν−20μz+C0nC0n48μ+60ν−27+C1n−18+12ν+24μ+C2n21−16ν−20μ+C1nC0n−121−16ν−20μz2+C0nC0n152μ+352ν−102+C1n×48μ+60ν−27+C2n−18+12ν+24μ+C3n21−16ν−20μ+C1n×C0n−1−18+12ν+24μ+C1n−121−16ν−20μz3+C0nC0n−612ν+90+C1n152μ+352ν−102+C2n48μ+60ν−27+C3n−18+12ν+24μ+C1n×C0n−148μ+60ν−2+C1n−1−18+12ν+24μ+C2n−121−16ν−20μ+C2nC0n−221−16ν−20μz4⋯+Cn−2nC2221−16ν−20μ+Cn−1nC1121−16ν−20μ+C01−18+12ν+24μ+Cnn48μ+60ν−27z2n+10+Cn−1n×21−16ν−20μ+Cnn−18+12ν+24μz2n+11Cnn21−16ν−20μz2n+12,where μ and ν are shape parameters and m∈ℕ.Theorem3 gives the following result.Corollary 5.
The support ofCSSa6 is −4,4.Theorem 9.
The limit stencil ofCSSa6 is ξ1,ξ2,ξ3,ξ4,ξ3,ξ2,ξ1, where(72)ξ1=64μ3+μ2ν−μν2−ν3−580μ2−930μν+555μ+350ν2+555ν75620μ+145ν−6μν−6ν2+240,ξ2=−12μ3+40μ2ν−317μ2+44μν2+45μ+16ν3−644ν2+45ν−961μν4220μ+145ν−6μν−6ν2+240,ξ3=32μ3+88μ2ν−1396μ2+80μν2−2138μν+6525μ+24ν3−742ν2+6525ν8420μ+145ν−6μν−6ν2+240,ξ4=−68μ3+μ2ν−μν2−ν3+3719μ2+303μν−25455μ−3416ν2−1830ν+4536018920μ+145ν−6μν−6ν2+240.Theorem 10.
If we chooseμ=1 and ν=−1/5, then the lower and upper bound of Ho¨ölder regularity of CSSa6 are 2.1302 and 2.455, respectively.


## 4.1. Affine Combination of Control Points

A linear combination in which the sum of coefficients is equal to one is called the affine combination. The necessary condition for the convergence of a subdivision scheme is that each refinement rule of it must be the affine combination of the control points in the previous subdivision step. To check this property forCSSa2n+2, we prove the following theorem.Theorem 2.
Each refined point of theCSSa2n+2 at k+1-th subdivision step which is defined in (14) is the affine combination of control points of k-th subdivision step.Proof.
To prove this theorem, we have to prove that the maskai,2n+2:i∈ℤ of CSSa2n+2 satisfies the relation ∑i∈ℤa3i+α,2n+2=1, where α=0,1,2.
Let us simplify the expressions:(20)∑η=02m∑γ=3η+3/23η+23m+2γγ3η+2−γ=33m+1,∑η=02m+1∑γ=3η+2/23η+23m+2γγ3η+1−γ=33m+1,∑η=02m+1∑γ=3η+1/23η+23m+2γγ3η−γ=33m+1.
We also calculate(21)∑j=−nn+1−1jn−j+21−3j3n+2n2n+2n+j=33n+1,∑j=−n−1n+1−1j+1n+j+13j−23n+2n2n+2n+j+1=33n+1.
Hence, to prove that each of the refined pointsq3i,2n+2k+1, q3i+1,2n+2k+1, and q3i+2,2n+2k+1 is the affine combinations of qik:i∈ℤ, we use (20) and (21). Thus,(22)∑i∈ℤa3i,2n+2=1−μ−ν+μ+ν133m+133m+1=1,∑i∈ℤa3i+1,2n+2=133n+133n+1−μ133n+133n+1−133m+133m+1−ν133n+133n+1−133m+133m+1=1,∑i∈ℤa3i+2,2n+2=133n+133n+1−ν133n+133n+1−133m+133m+1−μ133n+133n+1−133m+133m+1=1,which completes the proof.


## 4.2. Support of the Class of Combined Schemes

In this subsection, we present the support size of the class of combined schemes. The support of a subdivision scheme represents how far one vertex is affected by its neighboring points. Its size represents the local support property of the subdivision curve.Theorem 3.
The support of the class of subdivision schemesCSSa2n+2 is −3n+2/2,3n+2/2.Proof.
Letqi,2n+20=qi0∈ℝi∈ℤ be the initial data such that q00=1 and qα0≠0 for α∈ℤ\0; that is,(23)qα,2n+20=1,forα=0,0,otherwise.
In order to calculate the support size ofCSSa2n+2, we have to calculate the distance between two corresponding subscripts belonging to the leftmost vertex and rightmost vertex after k-steps of subdivision.
If we operate on initial dataqi,2n+20=qi0∈ℝi∈ℤ by CSSa2n+2 one time, then we get(24)qα,2n+21=≠0,forα=−303n+2,−303n+2+1,…,303n+2−1,303n+2,0,otherwise.
If we operate on the dataqi,2n+21∈ℝi∈ℤ which is defined in (24) by CSSa2n+2 one time, that is, we operate on initial data qi,2n+20=qi0∈ℝi∈ℤ by CSSa2n+2 two times, then we get(25)qα,2n+22=≠0,forα=−∑j=013j3n+2,−∑j=013j3n+2+1,…,∑j=013j3n+2−1,∑j=013j3n+2,0,otherwise.
Similarly, if we operate on initial dataqi,2n+20=qi0∈ℝi∈ℤ which is defined in (23) by CSSa2n+2k-times, then we get(26)qα,2n+2k=≠0,forα=−∑j=0k−13j3n+2,−∑j=0k−13j3n+2+1,…,∑j=0k−13j3n+2−1,∑j=0k−13j3n+2,0,otherwise.
Hence, afterk-steps of subdivision, the leftmost nonzero vertex is q−3n+2∑j=0k−13j,2n+2k and the rightmost nonzero vertex is q3n+2∑j=0k−13j,2n+2k. Since k-times, the subdivision process is applied and, at each step, we have relabeled the subscripts of the vertices; thus, qi/3k,2n+2k=qi/3k−1,2n+2k−1=qi/3k−2,2n+2k−2=⋯=qi/3,2n+21=qi,2n+20. Hence, after k-steps, the distance between leftmost and rightmost nonzero vertices is equal to the support of CSSa2n+2; that is,(27)support=3n+21+3+32+⋯+3k−13k−−3n+21+3+32+⋯+3k−13k=23n+213+132+⋯+13k=23n+21/31−1/3=3n+2.
Since the class of subdivision schemes is symmetric, then obviously the support region is−3n+2/2,3n+2/2.


## 4.3. Generation Degree

One of the necessary conditions for generatingCn- continuous limit curves is the order of polynomial generation of the subdivision scheme. A ternary subdivision scheme can generate polynomials up to degree d if and only if its Laurent polynomial can be written in the following form:(28)az=1+z+z23d+1φz,z∈ℂ\0.Here, φz is the Laurent polynomial with φ1=3 and it does not have the factor 1+z+z2.Theorem 4.
The class of combined schemesCSSa2n+2 generates polynomials of degree up to 2 for all μ and ν.Proof.
The Laurent polynomial ofCSSa2n+2 is(29)a2n+2z=1+z+z233b2n+2z,where b2n+2z is a Laurent polynomial obtained by b2n+2z=3a2n+2z/1+z+z2. Hence, the degree of polynomial generation of CSSa2n+2 is 2.


## 4.4. Properties of the 4-Point Subdivision Scheme (19)

From Theorem2, the subdivision scheme which is defined in (14) and denoted by CSSa4 satisfies the necessary conditions for convergence. Moreover, the Laurent polynomial of CSSa4 is(30)a4z=181z51+z+z2310ν−4+5μz4+−25ν+7−5μz3+30ν+3z2+−25ν+7−5μz+10ν−4+5μ.By using Laurent polynomial ofCSSa4, we prove the following theorem.Theorem 5.
Letqi0i∈ℤ be the initial data; then, the refinement rules defined in (19) of CSSa4 produce the C2- continuous shapes if ν∈−2/5,1/5 and its corresponding μ∈1/5−7ν/2,2−7ν/2.Proof.
To prove thatCSSa4 produces the C2-continuous shapes, first, we have to prove that this scheme produces C0-continuous and C1-continuous shapes in the given intervals of μ and ν.
Thus, by using (30), we calculate(31)a41z=3z21+z+z2a40z.
Hence, from the above equation, we have(32)a41z=127z2−4+5μ+10νz5+−1+5μ−5νz4+10ν+5μ+5z3+26−10μ−20νz2+29+10ν−10μz1+26−10μ−20νz0=∑j=−23aj,4zj,where the subdivision scheme corresponding to the Laurent polynomial a41z is denoted by CSSa41; then, by Theorem 1, we have(33)13CSSa411∞=13max∑j∈ℤai+3j,41,1;i=0,1,2.
This implies that(34)13CSSa411∞=13max∑j∈ℤa3j,41,1,∑j∈ℤa1+3j,41,1,∑j∈ℤa2+3j,41,1.
By substituting values, we get(35)13CSSa411∞=max181−4+5μ+10ν+26−10μ−20ν+10ν+5μ+5,−1+5μ−5ν+29+10ν−10μ+−1+5μ−5ν.
Hence,1/3CSSa411∞<1 for μ+2ν∈−14/5,53/10 and μ−ν∈5/2,28/5.
This implies thatCSSa41 is contractive and the scheme CSSa4 produces C0-continuous shapes.
Now, we again calculate(36)a42z=3z21+z+z2a41z,which implies(37)a41z=19z210ν−4+5μz5+3−15νz4+6+15νz3+−10μ+17−20νz2+6+15νz1+3−15νz0+10ν−4+5μz−1,where the subdivision scheme corresponding to the Laurent polynomial a42z is denoted by CSSa42; then, by Theorem 1, we have(38)13CSSa421∞=max127.−4+5μ+10ν+17−10μ−20ν+10ν+5μ−4,3−15ν+6+15ν.
Thus,1/3CSSa421∞<1 for ν∈−1,4/5 and μ∈−1/10−2ν,−2ν+13/5.
Therefore,CSSa42 is contractive and hence CSSa4 produces C1-continuous shapes.
Now, we find the following Laurent polynomial by usinga43z=3z2/1+z+z2a42z. Thus, we have(39)a43z=1310ν−4+5μz5+−25ν+7−5μz4+30ν+3z3+−25ν+7−5μz2+10ν−4+5μz.
LetCSSa43 be the subdivision scheme corresponding to the Laurent polynomial a43z; then, by Theorem 1, we get(40)13CSSa431∞=max19−4+5μ+10ν+7−5μ−25ν,3+30ν.
Thus,1/3CSSa431∞<1 for ν∈−2/5,1/5 and μ∈1/5−7ν/2,2−7ν/2.
This implies thatCSSa43 is contractive and hence CSSa4 is C2-continuous. This completes the proof.If we putν = 0 in the Laurent polynomial (30), then it gives one more factor of 1+z+z2. Hence, the following corollary is proved.Corollary 1.
Ifν=0 and μ∈4/5,6/5, then CSSa4 produces C3-continuous shapes.Corollary 2.
The continuity ofCSSa4 can be increased if we use the refinement rules corresponding to the following Laurent polynomial which is derived from the Laurent polynomial of (30):(41)αz=13n+2C0nC0n10ν−4+5μ+C0nC0n−15ν+3+C1n10ν−4+5μz+C0nC0n15ν+6+C1n−15ν+3+C2n10ν−4+5μ+C1nC0n−110ν−4+5μz2+C0nC0n−20ν−10μ+17+C1n15ν+6+C2n−15ν+3+C3n10ν−4+5μ+C1nC0n−1−15ν+3+C1n−110ν−4+5μz3+C0nC0n15ν+6+C1n−20ν−10μ+17+C2n15ν+6+C3n−15ν+3+C1nC0n−115ν+6+C1n−1−15ν+3+C2n−110ν−4+5μ+C2nC0n−210ν−4+5μz4+⋯++Cnn−2C2210ν−4+5μz2n+4+Cn−1n10ν−4+5μ+−15ν+3+Cnn−15ν+3z2n+5+Cnn10ν−4+5μz2n+6,where μ and ν are shape parameters and n=1,2,….Remark 2.
If we putn=2, n=3, and n=4 in (41), we get the Laurent polynomials of 3-point, 4-point, and 5-point combined subdivision schemes which produce continuous shapes up to C2, C3, and C4 smoothness.Theorem3 gives the following result.Corollary 3.
The support ofCSSa4 is −5/2,5/2.Now, here we use the method proposed by Rioul [19] to find out the Ho¨ölders regularity of CSSa4.Theorem 6.
If we chooseμ=1 and ν=−1/5, then the lower and upper bound of Ho¨lder regularity of CSSa4 is 2.1079 and 2.369, respectively.Proof.
From (30), we have(42)a4z=1+z+z233z−5∑j=04mjzj,where m0=10ν−4+5μ/3, m1=−25ν+7−5μ/3, m2=30ν+3/3, m3=−25ν+7−5μ/3, m4=10ν−4+5μ/3, k=3, ℓ=4, n=3, and ℏ=0,1,2,3,4. F0, F1, F2, F3, and F4 are matrices given by(43)Fℏi,j=mℓ+i−3j+ℏ.
SinceF0i,j=m4+i−3j,F1i,j=m5+i−3j,F2i,j=m6+i−3j,F3i,j=m7+i−3j,F4i,j=m8+i−3j, where i,j=1,2,3,4, thus, we have(44)F0=m2000m3m000m4m1000m200,F1=m3m000m4m1000m2000m3m00,F2=m4m1000m2000m3m000m4m10,F3=0m2000m3m000m4m1000m20,F4=0m3m000m4m1000m2000m3m0.
LetρFj:j=0,1,2,3,4 be the joint spectral radii of matrices F0, F1, F2, F3, and F4, respectively; then, we have(45)maxρF0,ρF1,ρF2,ρF3,ρF4≤ρ≤maxF0∞,F1∞,F2∞,F3∞,F4∞.
The eigenvalues ofF0, F1, F2, F3, and F4 are required to calculate their joint spectral radii; therefore, ρF0=ρF2=ρF3=ρF4=,maxmo,m2, ρF1=,maxm3+m0,m1+m4.
The infinity norm ofF0, F1, F2, F3, and F4 is(46)Fj∞=maxm2,m3+m0,m4+m1,where j=0,1,2,3,4.(47)maxm0,m2,m3+m0,m1−m4≤ρ≤maxm2,m3+m0,m4+m1.
If we useμ=1 and ν=−1/5 in (47), we get(48)max1,0.333,2,2.6667≤ρ≤max1,2.
Hence, the lower bound of Ho¨ölders exponent is(49)r≥3−log32.6667=2.10719.
The upper bound of Ho¨lders exponent is(50)r≤3−log32=2.369.
This completes the proof. □Theorem 7.
The limit stencil ofCSSa4 which is defined in (18) is(51)5μ+5ν−348,27−5μ−5ν48,27−5μ−5ν48,5μ+5ν−348.Proof.
To find out the limit stencil ofCSSa4, we substitute i=−1 and i=0 in the refinement rules (19) of CSSa4. Hence, we obtain the following refined points:(52)q−3k+1=5μ27+5ν27q−2k+1−10μ27−10ν27q−1k+5μ27+5ν27q0k,q−2k+1=−581+5ν81+10μ81q−2k+2027−5μ27q−1k+1027−5ν27q0k+−481+5μ81+10ν81q1k,q−1k+1=−481+5μ81+10ν81q−2k+1027−5ν27q−1k+2027−5μ27q0k+−581+5ν81+10μ81q1k,q0k+1=5μ27+5ν27q−1k+1−10μ27−10ν27q0k+5μ27+5ν27q1k.
The matrix form of the above system of four linear equations is(53)q−3k+1q−2k+1q−1k+1q0k+15μ27+5ν271−10μ27−10ν275μ27+5ν270−581+5ν81+10μ812027−5μ271027−5ν27−481+5μ81+10ν81−481+5μ81+10ν811027−5ν272027−5μ27−581+5ν81+10μ8105μ27+5ν271−10μ27−10ν275μ27+5ν27q−3kq−2kq−1kq0k.
Symbolically, it can be written as(54)qk+1=R1qj.
Hence, the local subdivision matrix of the 4-point schemeCSSa4 is(55)R1=5μ27+5ν271−10μ27−10ν275μ27+5ν270−581+5ν81+10μ812027−5μ271027−5ν27−481+5μ81+10ν81−481+5μ81+10ν811027−5ν272027−5μ27−581+5ν81+10μ8105μ27+5ν271−10μ27−10ν275μ27+5ν27.
Now we calculate the eigenvalues of the subdivision matrixR1 and denote them by λi:i=0,1,2,3. Thus, we have(56)λi=1,13,19,10ν+127.
The matrix of the eigenvector corresponding to the eigenvalues is(57)W=1−11−15μ+5ν−35μ+5ν−27−1315ν−5μ+115μ+15ν−275μ+5ν−35μ+5ν−27131−5ν−5μ+115μ+15ν−271111.
By taking the inverse of the above matrix, we get(58)W−1=27−5μ−5ν485μ+5ν−27485μ+5ν−274827−5μ−5ν48−5ν−5μ+120ν−16−35ν+5μ−920ν−1635ν+5μ−920ν−165ν−5ν+120ν−165μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−348−5ν+5μ−920ν−1635ν+5μ−920ν−16−35ν+5μ−920ν−165ν+5μ−920ν−16.
Letδ be the diagonal matrix of eigenvalues; then,(59)δ=1000013000019000010ν+127.
SinceR1=WδW−1, therefore R1k=WδkW−1. Also, we have(60)qk+1=R1qk=R12qk−1=R13qk−2=⋯=R1kq0,which implies that(61)qk+1=WδkW−1q0.
Hence,(62)limk⟶∞qk+1=Wlimk⟶∞δkW−1q0.
Thus, we have(63)q−3∞q−2∞q−1∞q0∞=5μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−3485μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−3485μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−3485μ+5ν−34827−5μ−5ν4827−5μ−5ν485μ+5ν−348q−30q−20q−10q00.
Hence, we get the limit stencil ofCSSa4, that is,(64)5μ+5ν−348,27−5μ−5ν48,27−5μ−5ν48,5μ+5ν−348.
Hence, it is proved.


## 4.5. Properties of the 6-Point Subdivision Scheme

If we putn=2 in (14), then we get the following 6-point combined subdivision scheme. We denote this scheme by CSSa6:(65)q3ik+1=4μ243+4ν243qi−2k+56μ243+56ν243qi−1k+1−40μ81−40ν81qik+56μ243+56ν243qi+1k+4μ243+4ν243qi+2k,q3i+1k+1=−16μ2187−20ν2187+8729qi−2k+−70729+280ν2187+476μ2187qi−1k+560729−664μ2187−56ν2187qik+280729−56μ2187−664ν2187qi+1k+−56729+280μ2187+476ν2187qi+2k+7729−20μ2187−16ν2187qi+3k,q3i+2k+1=7729−20μ2187−16ν2187qi−2k+−56729+280μ2187+476ν2187qi−1k+280729−56μ2187−664ν2187qik+560729−664μ2187−56ν2187qi+1k+−70729+280ν2187+476μ2187qi+2k+−16μ2187−20ν2187+8729qi+3k.The Laurent polynomial ofCSSa6 denoted by a6z=a60z is defined:(66)a6z=12187z81+z+z2321−20μ−16νz10+−39+44μ+28νz9+24μ+48ν−9z8+−54+276ν+84μz7+−108μ−936ν+153z6+99+1200ν−48μ×z5+−108μ−936ν+153z4+−54+276ν+84μz3+24μ+48ν−9z2+−39+44μ+28νz+21−20μ−16ν.Theorem 8.
Letqi0i∈ℤ be the initial data; then, the refinement rules defined in (65) of CSSa6 produce the C2- continuous shapes if ν∈−1/4,1/8 and its corresponding μ∈3/32−157ν/32,255/128−157ν/32. It gives C3-continuous shapes for ν=0 and μ∈93/128,87/64. It produces C4-smooth shapes for ν=0 and μ∈3/5,51/40. And it gives C5-smooth shapes for ν=0 and μ∈57/62,33/31.Proof.
By Theorem1, we have(67)a6nz=3z21+z+z2a6n−1z,i=1,2,3,4,5,6.
Hence,(68)a61z=172921−20μ−16νz8+3−4ν+4μz7+−24+56ν+52μz6+224μ+424ν−147z5+−39−200ν+200μz4+280ν+186+80μz3+−336μ+693−744νz2+801+408ν−408μz1+−336μ+693−744νz0+280ν+186+80μz−1+−39−200ν+200μz−2+224μ+424ν−147z−3+−24+56ν+52μz−4+3−4ν+4μz5+21−20μ−16νz−6.a62z=124321−20μ−16νz8+12ν+24μ−18z7+60ν+48μ−27z6+152μ−102+352νz5+90−612νz4+198−72μ+540νz3+405−264μ−672νz2+198−72μ+540νz1+90−612νz0+152μ−102+352νz−1+60ν+48μ−27z−2+12ν+24μ−18z−3+21−20μ−16νz−4.a63z=181z221−20μ−16νz10+−39+44μ+28νz9+24μ+48ν−9z8+−54+276ν+84μz7+−108μ−936ν+153z6+99+1200ν−48μz+−108μ−936ν+153z4+−54+276ν+84μz3+24μ+48ν−9z25+−39+44μ+28νz+21−20μ−16ν.
And whenν=0, we get the following Laurent polynomials from (67):(69)a64z=12721−20μz8+−60+64μz7+−20μ+30z6+40μ−24z5+147−128μz4+40μ−24z3+−20μ+30z2+−60+64μz+21−20μ,a65z=1921−20μz6+−81+84μz5+−84μ+90z4+40μ−33z3+−84μ+90z2+−81+84μz1+21−20μ,α6z=3z21+z+z21921−20μz6+−81+84μz5+−84μ+90z4+40μ−33z3+−84μ+90z2+−81+84μz1+21−20μ.
Then, by using Theorem1 and denoting the subdivision schemes corresponding to the Laurent polynomials a6jz:j=0,1,2,3,4,5,6 by CSSa6j:j=0,1,2,3,4,5,6, respectively, we get(70)13CSSa6j1∞=13max∑j∈ℤαi+3jj,1;i=0,1,2,j=0,1,2,3,4,5,6.
By solving1/3CSSa6j1∞<1 for j=0,1,2,3,4,5,6, we get the required result.Corollary 4.
The order of smoothness ofCSSa6 can be increased if we use the refinement rules corresponding to the following Laurent polynomial which is derived from the Laurent polynomial of (66):(71)αz=13n+5C0nC0n21−16ν−20μ+C0nC0n−18+12ν+24μ+C1n21−16ν−20μz+C0nC0n48μ+60ν−27+C1n−18+12ν+24μ+C2n21−16ν−20μ+C1nC0n−121−16ν−20μz2+C0nC0n152μ+352ν−102+C1n×48μ+60ν−27+C2n−18+12ν+24μ+C3n21−16ν−20μ+C1n×C0n−1−18+12ν+24μ+C1n−121−16ν−20μz3+C0nC0n−612ν+90+C1n152μ+352ν−102+C2n48μ+60ν−27+C3n−18+12ν+24μ+C1n×C0n−148μ+60ν−2+C1n−1−18+12ν+24μ+C2n−121−16ν−20μ+C2nC0n−221−16ν−20μz4⋯+Cn−2nC2221−16ν−20μ+Cn−1nC1121−16ν−20μ+C01−18+12ν+24μ+Cnn48μ+60ν−27z2n+10+Cn−1n×21−16ν−20μ+Cnn−18+12ν+24μz2n+11Cnn21−16ν−20μz2n+12,where μ and ν are shape parameters and m∈ℕ.Theorem3 gives the following result.Corollary 5.
The support ofCSSa6 is −4,4.Theorem 9.
The limit stencil ofCSSa6 is ξ1,ξ2,ξ3,ξ4,ξ3,ξ2,ξ1, where(72)ξ1=64μ3+μ2ν−μν2−ν3−580μ2−930μν+555μ+350ν2+555ν75620μ+145ν−6μν−6ν2+240,ξ2=−12μ3+40μ2ν−317μ2+44μν2+45μ+16ν3−644ν2+45ν−961μν4220μ+145ν−6μν−6ν2+240,ξ3=32μ3+88μ2ν−1396μ2+80μν2−2138μν+6525μ+24ν3−742ν2+6525ν8420μ+145ν−6μν−6ν2+240,ξ4=−68μ3+μ2ν−μν2−ν3+3719μ2+303μν−25455μ−3416ν2−1830ν+4536018920μ+145ν−6μν−6ν2+240.Theorem 10.
If we chooseμ=1 and ν=−1/5, then the lower and upper bound of Ho¨ölder regularity of CSSa6 are 2.1302 and 2.455, respectively.


## 5. Geometrical Influence ofCSSa2n+2 in Curve Modeling

In this section, we present models produced by the new combined 4-point and 6-point subdivision schemes. For modeling of 2D shapes by the combined schemes, we use the procedure which is explained in Figure3 through a flowchart. This procedure is used in the following numerical examples for curve modeling.Figure 3
Flowchart to explain the geometric examples.Example 1.
In this example, we draw the initial sketchq0 by using the initial data qi0:i∈ℤ. The initial sketch and the initial data are shown by black polygons and red bullets in Figure 4. Curves fitted by the subdivision schemes CSSa4 and CSSa6 after four refinements steps for various values of μ and ν are shown in this figure. Figure 4(a) shows the curves fitted by a 4-point ternary combined scheme for ν=−0.2 and μ=1,1.5,2. Figure 4(b) shows the curves fitted by a 4-point ternary combined scheme for ν=0.1 and μ=−0.5,−0.8,1. Figure 4(c) shows the curves fitted by a 6-point ternary combined scheme for ν=−0.1 and μ=0.8,1,1.5. Figure 4(d) shows the curves fitted by a 4-point ternary combined scheme for ν=0.1 and μ=−0.2,−0.5,1.Figure 4
Coloured lines show the curves fitted by the combined subdivision schemes.
(a)(b)(c)(d)Example 2.
In this experiment, we input the initial data taken from the discontinuous function:(73)hx=+cosπx,ifx∈0.5,0,−cosπx,ifx∈0,0.5.
Curves fitted by the first two members of our class of schemes are shown in Figures5(b) and 5(d), respectively, and the curves fitted by the first two members of the CETISS are shown in Figures 5(a) and 5(c), respectively. We see that the CETISS produce Gibbs oscillations in the limit curves close to the discontinuity. It is also noted that our combined schemes do not produce Gibbs oscillations in the limit curves for certain values of the shape parameters.Figure 5
Red bullets are initial control points. Black solid lines are initial control polygons. Coloured lines show the fitted curves by our schemes in (b) and (d).
(a)(b)(c)(d)Example 3.
In this example, we take the initial data such that if we join this data by straight lines, we get a close polygonal shape. We smooth this shape by the first two members of our class of schemes and results are shown in Figure6. This figure shows the flexible behaviour of our combined subdivision scheme. It also shows that shape parameters in the combined subdivision schemes provide grip on the limit curves.Figure 6
Red bullets show the initial control points. Black solid lines are the initial control polygon. Coloured lines show curves fitted by our subdivision scheme after two refinement steps. (a) 4-point combined schemeν=−0.1,μ=0.1,0.5,1.5; (b) 4-point combined scheme ν=0.1,μ=0.1,0.5,1.5; (c) 6-point combined scheme ν=−0.1,μ=1,1.5,1.9; and (d) 6-point combined scheme ν=−0.3,μ=1,1.3,1.5.
(a)(b)(c)(d)


## 6. Conclusion

In this research, we have constructed and analyzed a class of combined ternary subdivision schemes. The construction procedure is very simple and consists of only a few simple mathematical operations on the refinement rules of the two well-known classes of the schemes. But it gives us amazing results that are discussed theoretically and geometrically in detail. The class of schemes gives good smoothness and local control on the limit shapes. It also provides us a flexible environment to fit curves. The unnecessary oscillations in the limit curves of our combined schemes can be removed by changing the values of the shape parameters.



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*Source: 6679201-2021-02-16.xml*