# A New 5-Point Ternary Interpolating Subdivision Scheme and Its Differentiability

Authors: Ghulam Mustafa; Jayyada Irum; Mehwish Bari
Journal: ISRN Computational Mathematics
 (2012)

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## Abstract

A new 5-point ternary interpolating scheme with a shape parameter is
introduced. The resulting curve isC2 for a certain range of
parameters. The differentiable properties of the proposed scheme to extend its application in the generation of smooth curves are explored. Application of the proposed scheme is given to show its visual smoothness. The scheme is also extended to a 5-point tensor product ternary interpolating scheme, and its numerical examples are also included.


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## Body

## 1. Introduction

Geometric modeling plays a significant role to cover up the gap between computer and industry. It has a pivotal importance in the fields of aircraft manufacturing, automobile industry, and general product design.One of the most important tools of computer aided geometric design is “Subdivision.” Subdivision is a well flourished field. It is a process of taking unrefined shape and to polish it up to produce another shape that is more visually tempting. Due to the comprehensibility and simplicity of this method, it is used in the fields of 3D geometrical measurement, computer graphics, computer animation, and computer aided geometric design.In 1986, Dubuc [1] presented a interpolation through an iterative scheme. Dyn et al. [2] introduced a 4-point interpolating subdivision scheme for curve design. Later on, Deslauriers  and Dubuc [3] introduced a symmetric iterative interpolation process. Weissman [4] also offered a 6-point interpolating scheme in 1990. In 2002, Hassan et al. [5, 6] gave ternary three-point and 4-point interpolatory schemes. Further analysis of ternary three-point univariate scheme was given in technical report by Hassan and Dodgson [7] in 2004. Dyn [8] has given the analysis of the convergence and smoothness of interpolating and approximating schemes by Laurent’s polynomial method.In 2007, Beccari et al. [9] presented an interpolating 4-point ternary nonstationary scheme with tension control. They also offered a nonstationary uniform tension controlled interpolating 4-point scheme reproducing conics [10] in 2007. Ko [11] in his Ph.D. thesis presented a detailed study on subdivision scheme. Zheng et al. [12] presented the method to find the differentiability of a four-point ternary scheme. Lian [13] extended 3-point and 5-point interpolating schemes into a-ary subdivision scheme for curve design. Conti et al. [14] derived symmetric subdivision masks of the Hurwitz type to the interpolating scheme masks. In this paper, we present a new 5-point ternary interpolating subdivision scheme with one parameter.


## 2. Preliminaries

Letfik∈ℝN, i∈ℤ, denote a sequence of points in ℝN, N⩾2, where k is a nonnegative integer, then a univariate ternary subdivision scheme S which maps coarse points fik to refined points fik+1 is defined by
(1)f3ik+1=∑j=0m‍a3jfi+jk,f3i+1k+1=∑j=0m‍a3j+1fi+jk,f3i+2k+1=∑j=0m‍a3j+2fi+jk,
where m>0 and the set a={ai∣i∈ℤ} of coefficients is called mask of the scheme. The z-transform of the mask a of subdivision scheme S can be given as
(2)a(z)=∑i∈ℤ‍aizi,
which is called the symbol or the Laurent polynomial of the scheme. The Laurent polynomial of a convergent subdivision scheme satisfies
(3)a(e2iπ/3)=a(e4iπ/3)=0,a(1)=3.
The existence of associated subdivision scheme Sn to S for the divided differences of the original control polygon and of related Laurent polynomial an(z) is assured by the following condition:
(4)an(z)=(3z2z2+z+1)an-1,(z)=(3z2z2+z+1)na(z),n⩾1,∥(13Sn)L∥∞=max{∑j∈ℤ‍|bi+3Lj[n,L]|:i=0,1,…,3L-1},
where
(5)b[n,L](z)=13L∏j=0L-1‍an(z3j).
By Hassan and Dodgson [6], Cn continuity of the scheme S requires that an(z) satisfy (4) and ∥((1/3)Sn+1)L∥∞<1, for the first integer value of L>0.


## 3. A 5-Point Ternary Interpolating Scheme

In this section, we construct a 5-point ternary interpolating subdivision scheme and give its analysis.


### 3.1. Construction of the Scheme

Consider the following three recursive relations which refine given thekth level polygon fk={fik}i∈ℤ to (k+1)th level polygon fk+1={fik+1}i∈ℤ(6)f3i-1k+1=a0fi-2k+a1fi-1k+a2fik+a3fi+1k+a4fi+2k,f3ik+1=fik,f3i+1k+1=a4fi-2k+a3fi-1k+a2fik+a1fi+1k+a0fi+2k.
We have the following sequence from the above recurrence relation:
(7)a={…,0,a4,0,a0,a3,0,a1,a2,1,a2,a1,0,a3,a0,0,a4,0,…}.
The Laurent polynomial of this sequence is
(8)a(z)=a4z7+a0z5+a3z4+a1z2+a2z+1+a2z-1+a1z-2+a3z-4+a0z-5+a4z-7.
Let {Si:i=0,1,2,…} be the divided difference subdivision schemes of S corresponding to the Laurent polynomial ai(z), then by Hassan and Dodgson [6], the necessary condition for Ci continuity of scheme S is that  ai(z) satisfy (3), which gives for  i=0,1,2,3 and by (4)
(9)a0+a1+a2+a3+a4=1,6a0+3a1-3a3-6a4=1,36a0+9a1+9a3+36a4=1,-9a0-9a3-27a4=1,-54a0+27a3+162a4=1,162a0-162a4=-8,(10)a1(z)=3a4z7-3a4z6+3a0z5+3(a3+a4-a0)z4-3(a3+a4)z3+3(a0+a1)z2+3(1-2a0-2a1)z+3(a3+a4-a0)z-2+3(a0+a1)-3(a3+a4)z-1+3a0z-3-3a4z-4+3a4z-5,(11)a2(z)=9a4z7-18a4z6+9(a0+a4)z5+9(a3+2a4-2a0)z4+9(a0-2a3-4a4)z3+9(2a0+a1+a3+2a4)z2+9(a0-2a3-4a4)z+9(a3+2a4-2a0)+9(a0+a4)z-1-18a4z-2+9a4z-3,(12)a3(z)=27a4z7-81a4z6+(27a0+81a4)z5+(-81a0+27a3+54a4)z4+(81a0-81a3-243a4)z3+(-81a0+27a3+54a4)z2+(27a0+81a4)z-81a4+27a4z-1,(13)a4(z)=27a4z7-108a4z6+(27a0+162a4)z5+(-108a0+27a3)z4+(27a0+162a4)z3-108a4z2+27a4z.
Solving the system of liner equation (9) by taking a4=ω, we get  a0=ω-(4/81), a1=-4ω+(10/27), a2=6ω+(20/27), and a3=-4ω-(5/81).Now we propose the following 5-point ternary interpolating subdivision scheme:(14)f3i-1k+1=a0fi-2k+a1fi-1k+a2fik+a3fi+1k+a4fi+2k,f3ik+1=fik,f3i+1k+1=a4fi-2k+a3fi-1k+a2fik+a1fi+1k+a0fi+2k,
where a0=ω-(4/81), a1=-4ω+(10/27), a2=6ω+(20/27), a3=-4ω-(5/81), and a4=ω.


### 3.2. Smoothness Analysis Necessary Conditions

By using the matrix formalism of Hassan and Dodgson [6], we derive necessary conditions for a scheme to be C0, C1, and C2 by using the eigenvalues of the subdivision matrices for, mid-point rule and the vertex rule. By (14), we have the following subdivision matrices for mid-point and the vertex rules:
(15)(a0a1a2a3a40001000a4a3a2a1a000a0a1a2a3a40001000a4a3a2a1a0)(Mid-point  rule)(0010000a4a3a2a1a0000a0a1a2a3a4000010000a4a3a2a1a0000a0a1a2a3a40000100)(Vertex rule)
where a0=ω-(4/81), a1=-4ω+(10/27), a2=6ω+(20/27), a3=-4ω-(5/81), and a4=ω. The above subdivision matrices have the following eigenvalues, respectively:
(16)1,13,19,127,-481+11ω,-481+3ω,1,13,19,127,-4ω,-6ω+11162±1162472392ω2-9720ω+121.
Now we find the bounds on the parameter for continuity of the scheme. From [15], the necessary condition for 5-point scheme to be C0 is that all eigenvalues must be less than 1. This implies that the proposed scheme is C0 continuous for -7/81<ω<85/891. The necessary condition for C1 continuity is that all eigenvalues except 1 must be less than 1/3. This is true for (-8+210)/81<ω<31/891. The necessary condition for the scheme to be C2 is that all eigenvalues except 1 and 1/3 must be less than 1/9. This gives the combine range (-2+5)/81<ω<13/891 for C2 continuity of the proposed scheme.


### 3.3. Smoothness Analysis Sufficient Conditions

In this section, we calculate sufficient conditions forC0, C1, and C2 continuity of our proposed scheme by the Laurent polynomial method.By substituting values ofai, i=0,1,2,3,4, in ai(z) and then by (5), we get
(17)∥13S1∥∞=max{|ω-481|+|3ω-2681|+|3ω+581|+|ω|,|ω-481|+|3ω-2681|+|3ω+581|+|ω|,2|ω|+2|4ω+181|+|6ω+2981|},∥13S2∥∞=max{2|6ω-427|+|12ω-1727|,|9ω|+|15ω+29|+|-19+12ω|,|9ω|+|15ω+29|+|12ω-19|},∥13S3∥∞=max{9|ω|+|45ω-79|+|36ω-49|,2|27ω|+|54ω+13|,9|ω|+|45ω-79|+|36ω-49|},∥13S4∥∞=max{|108ω|+|189ω-43|,|108ω|+|189ω-43|,2|27ω|+|-216ω+113|}.
Since for -7/81<ω<53/324, ∥(1/3)S1∥∞<1, then scheme is C0. Similarly for -1/324<ω<2/81, ∥(1/3)S2∥∞<1, then scheme is C1, and for 1/324<ω<1/162,∥(1/3)S3∥∞<1, then scheme is C2.C
3-continuous scheme must satisfy the condition ∥(1/3)S4∥∞<1; this is true for 1/243<ω<7/891 and 4/243<ω<7/405, but there is no combine range of ω for which ∥(1/3)S4∥∞<1. Similarly we can prove that for L=2,3,…, we have ∥((1/3)S4)L∥∞≮1; therefore the scheme is not C3-continuous. By summarizing the above discussion, we have the following theorem.Theorem 1.
Let{fik}i∈ℤ be defined by (14) for k≥0, where {fi0}i∈ℤ are given initial control points. The values correspond to i/3k, and let f(t) be the limit function of this process, then f(t) is C0, C1, and C2-continuous for the improved ranges -7/81<ω<53/324, -1/324<ω<2/81,  and 1/324<ω<1/162, respectively.Remark 2.
From the Laurent polynomial method, we have the improved ranges for continuity as compared to the ranges derived by using eigenanalysis method.


### 3.4. Basic Properties of the Scheme

The proposed 5-point scheme has the following properties.(i)
It is exact for cubic interpolation.(ii)
Its approximation order is four.(iii)
Its highest smoothness is achieved atω=2/405 and its Hölder exponent is RH=R(2/405)=2.130256.(iv)
The basic limit function of proposed scheme has support width 7, which implies that it vanishes outside the interval [−7/2, 7/2].


### 3.5. A Modified 5-Point Ternary Interpolating Scheme

In this section, we present a modified form of the 5-point ternary interpolating scheme which interpolates end-points without using any auxiliary control points in the case of open polygon.Theorem 3.
Let{fjk} be the values corresponding to j/3k for 0≤j≤3kn at level k≥0 and {fj0:j=0,…,n} the set of initial control points of initial polygon f0, then new values at level k+1 can be defined recursively as
(18)f3j-1k+1=a0fj-2k+a1fj-1k+a2fjk+a3fj+1k+a4fj+2k,2≤j≤3kn-2,(19)f3jk+1=fjk,0≤j≤3kn,(20)f3j+1k+1=a4fj-2k+a3fj-1k+a2fjk+a1fj+1k+a0fj+2k,2≤j≤3kn-2,(21)f1k+1=(821-3a0-57a4)f0k+(53-3a1-57a3)f1k+(83-267a2)f2k+(-57a1-3a3)f3k+(-57a0-3a4)f4k,(22)f2k+1=(221-2a0-37a4)f0k+(53-2a1-37a3)f1k+(53-177a2)f2k+(-37a1-2a3)f3k+(-37a0-2a4)f4k,(23)f4k+1=(-1105+65a0-17a4)f0k+(13+65a1+17a3)f1k+(-23+4735a2)f2k+(17a1+65a3)f3k+(17a0+65a4)f4k,(24)f3k+1n-4k+1=(17a0+65a4)f3k+1n-4k+(17a1+65a3)f3k+1n-3k+(-23+4735a2)×f3k+1n-2k+(13+65a1+17a3)f3k+1n-1k+(-1105+65a0+17a4)f3k+1nk,(25)f3k+1n-2k+1=(-37a0-2a4)f3k+1n-4k+(-37a1-2a3)f3k+1n-3k+(53-177a2)×f3k+1n-2k+(53-2a1-37a3)f3k+1n-1k+(221-2a0-37a4)f3k+1nk,(26)f3k+1n-1k+1=(-57a0-3a4)f3k+1n-4k+(-57a1-3a3)f3k+1n-3k+(83-267a2)×f3k+1n-2k+(53-3a1-57a3)f3k+1n-1k+(821-3a0-57a4)f3k+1nk.


## 3.1. Construction of the Scheme

Consider the following three recursive relations which refine given thekth level polygon fk={fik}i∈ℤ to (k+1)th level polygon fk+1={fik+1}i∈ℤ(6)f3i-1k+1=a0fi-2k+a1fi-1k+a2fik+a3fi+1k+a4fi+2k,f3ik+1=fik,f3i+1k+1=a4fi-2k+a3fi-1k+a2fik+a1fi+1k+a0fi+2k.
We have the following sequence from the above recurrence relation:
(7)a={…,0,a4,0,a0,a3,0,a1,a2,1,a2,a1,0,a3,a0,0,a4,0,…}.
The Laurent polynomial of this sequence is
(8)a(z)=a4z7+a0z5+a3z4+a1z2+a2z+1+a2z-1+a1z-2+a3z-4+a0z-5+a4z-7.
Let {Si:i=0,1,2,…} be the divided difference subdivision schemes of S corresponding to the Laurent polynomial ai(z), then by Hassan and Dodgson [6], the necessary condition for Ci continuity of scheme S is that  ai(z) satisfy (3), which gives for  i=0,1,2,3 and by (4)
(9)a0+a1+a2+a3+a4=1,6a0+3a1-3a3-6a4=1,36a0+9a1+9a3+36a4=1,-9a0-9a3-27a4=1,-54a0+27a3+162a4=1,162a0-162a4=-8,(10)a1(z)=3a4z7-3a4z6+3a0z5+3(a3+a4-a0)z4-3(a3+a4)z3+3(a0+a1)z2+3(1-2a0-2a1)z+3(a3+a4-a0)z-2+3(a0+a1)-3(a3+a4)z-1+3a0z-3-3a4z-4+3a4z-5,(11)a2(z)=9a4z7-18a4z6+9(a0+a4)z5+9(a3+2a4-2a0)z4+9(a0-2a3-4a4)z3+9(2a0+a1+a3+2a4)z2+9(a0-2a3-4a4)z+9(a3+2a4-2a0)+9(a0+a4)z-1-18a4z-2+9a4z-3,(12)a3(z)=27a4z7-81a4z6+(27a0+81a4)z5+(-81a0+27a3+54a4)z4+(81a0-81a3-243a4)z3+(-81a0+27a3+54a4)z2+(27a0+81a4)z-81a4+27a4z-1,(13)a4(z)=27a4z7-108a4z6+(27a0+162a4)z5+(-108a0+27a3)z4+(27a0+162a4)z3-108a4z2+27a4z.
Solving the system of liner equation (9) by taking a4=ω, we get  a0=ω-(4/81), a1=-4ω+(10/27), a2=6ω+(20/27), and a3=-4ω-(5/81).Now we propose the following 5-point ternary interpolating subdivision scheme:(14)f3i-1k+1=a0fi-2k+a1fi-1k+a2fik+a3fi+1k+a4fi+2k,f3ik+1=fik,f3i+1k+1=a4fi-2k+a3fi-1k+a2fik+a1fi+1k+a0fi+2k,
where a0=ω-(4/81), a1=-4ω+(10/27), a2=6ω+(20/27), a3=-4ω-(5/81), and a4=ω.


## 3.2. Smoothness Analysis Necessary Conditions

By using the matrix formalism of Hassan and Dodgson [6], we derive necessary conditions for a scheme to be C0, C1, and C2 by using the eigenvalues of the subdivision matrices for, mid-point rule and the vertex rule. By (14), we have the following subdivision matrices for mid-point and the vertex rules:
(15)(a0a1a2a3a40001000a4a3a2a1a000a0a1a2a3a40001000a4a3a2a1a0)(Mid-point  rule)(0010000a4a3a2a1a0000a0a1a2a3a4000010000a4a3a2a1a0000a0a1a2a3a40000100)(Vertex rule)
where a0=ω-(4/81), a1=-4ω+(10/27), a2=6ω+(20/27), a3=-4ω-(5/81), and a4=ω. The above subdivision matrices have the following eigenvalues, respectively:
(16)1,13,19,127,-481+11ω,-481+3ω,1,13,19,127,-4ω,-6ω+11162±1162472392ω2-9720ω+121.
Now we find the bounds on the parameter for continuity of the scheme. From [15], the necessary condition for 5-point scheme to be C0 is that all eigenvalues must be less than 1. This implies that the proposed scheme is C0 continuous for -7/81<ω<85/891. The necessary condition for C1 continuity is that all eigenvalues except 1 must be less than 1/3. This is true for (-8+210)/81<ω<31/891. The necessary condition for the scheme to be C2 is that all eigenvalues except 1 and 1/3 must be less than 1/9. This gives the combine range (-2+5)/81<ω<13/891 for C2 continuity of the proposed scheme.


## 3.3. Smoothness Analysis Sufficient Conditions

In this section, we calculate sufficient conditions forC0, C1, and C2 continuity of our proposed scheme by the Laurent polynomial method.By substituting values ofai, i=0,1,2,3,4, in ai(z) and then by (5), we get
(17)∥13S1∥∞=max{|ω-481|+|3ω-2681|+|3ω+581|+|ω|,|ω-481|+|3ω-2681|+|3ω+581|+|ω|,2|ω|+2|4ω+181|+|6ω+2981|},∥13S2∥∞=max{2|6ω-427|+|12ω-1727|,|9ω|+|15ω+29|+|-19+12ω|,|9ω|+|15ω+29|+|12ω-19|},∥13S3∥∞=max{9|ω|+|45ω-79|+|36ω-49|,2|27ω|+|54ω+13|,9|ω|+|45ω-79|+|36ω-49|},∥13S4∥∞=max{|108ω|+|189ω-43|,|108ω|+|189ω-43|,2|27ω|+|-216ω+113|}.
Since for -7/81<ω<53/324, ∥(1/3)S1∥∞<1, then scheme is C0. Similarly for -1/324<ω<2/81, ∥(1/3)S2∥∞<1, then scheme is C1, and for 1/324<ω<1/162,∥(1/3)S3∥∞<1, then scheme is C2.C
3-continuous scheme must satisfy the condition ∥(1/3)S4∥∞<1; this is true for 1/243<ω<7/891 and 4/243<ω<7/405, but there is no combine range of ω for which ∥(1/3)S4∥∞<1. Similarly we can prove that for L=2,3,…, we have ∥((1/3)S4)L∥∞≮1; therefore the scheme is not C3-continuous. By summarizing the above discussion, we have the following theorem.Theorem 1.
Let{fik}i∈ℤ be defined by (14) for k≥0, where {fi0}i∈ℤ are given initial control points. The values correspond to i/3k, and let f(t) be the limit function of this process, then f(t) is C0, C1, and C2-continuous for the improved ranges -7/81<ω<53/324, -1/324<ω<2/81,  and 1/324<ω<1/162, respectively.Remark 2.
From the Laurent polynomial method, we have the improved ranges for continuity as compared to the ranges derived by using eigenanalysis method.


## 3.4. Basic Properties of the Scheme

The proposed 5-point scheme has the following properties.(i)
It is exact for cubic interpolation.(ii)
Its approximation order is four.(iii)
Its highest smoothness is achieved atω=2/405 and its Hölder exponent is RH=R(2/405)=2.130256.(iv)
The basic limit function of proposed scheme has support width 7, which implies that it vanishes outside the interval [−7/2, 7/2].


## 3.5. A Modified 5-Point Ternary Interpolating Scheme

In this section, we present a modified form of the 5-point ternary interpolating scheme which interpolates end-points without using any auxiliary control points in the case of open polygon.Theorem 3.
Let{fjk} be the values corresponding to j/3k for 0≤j≤3kn at level k≥0 and {fj0:j=0,…,n} the set of initial control points of initial polygon f0, then new values at level k+1 can be defined recursively as
(18)f3j-1k+1=a0fj-2k+a1fj-1k+a2fjk+a3fj+1k+a4fj+2k,2≤j≤3kn-2,(19)f3jk+1=fjk,0≤j≤3kn,(20)f3j+1k+1=a4fj-2k+a3fj-1k+a2fjk+a1fj+1k+a0fj+2k,2≤j≤3kn-2,(21)f1k+1=(821-3a0-57a4)f0k+(53-3a1-57a3)f1k+(83-267a2)f2k+(-57a1-3a3)f3k+(-57a0-3a4)f4k,(22)f2k+1=(221-2a0-37a4)f0k+(53-2a1-37a3)f1k+(53-177a2)f2k+(-37a1-2a3)f3k+(-37a0-2a4)f4k,(23)f4k+1=(-1105+65a0-17a4)f0k+(13+65a1+17a3)f1k+(-23+4735a2)f2k+(17a1+65a3)f3k+(17a0+65a4)f4k,(24)f3k+1n-4k+1=(17a0+65a4)f3k+1n-4k+(17a1+65a3)f3k+1n-3k+(-23+4735a2)×f3k+1n-2k+(13+65a1+17a3)f3k+1n-1k+(-1105+65a0+17a4)f3k+1nk,(25)f3k+1n-2k+1=(-37a0-2a4)f3k+1n-4k+(-37a1-2a3)f3k+1n-3k+(53-177a2)×f3k+1n-2k+(53-2a1-37a3)f3k+1n-1k+(221-2a0-37a4)f3k+1nk,(26)f3k+1n-1k+1=(-57a0-3a4)f3k+1n-4k+(-57a1-3a3)f3k+1n-3k+(83-267a2)×f3k+1n-2k+(53-3a1-57a3)f3k+1n-1k+(821-3a0-57a4)f3k+1nk.


## 4. Differentiability of  the  5-Point Ternary Scheme

In this section, we find exact expressions for the first and the second derivatives of limit function of the proposed scheme.Theorem 4.
For given initial control points{fi0}, let {fik} be the values defined by the ternary 5-point interpolating subdivision scheme (14) corresponding to i/3k(i,k∈ℤ,k≥0) and f∈C2 with 1/324<ω<1/162; be the corresponding limit function, then for arbitrarily fixed m,n0∈ℤ, the limit function f has the derivatives
(27)f′(n03m)=3m+k144ω+12×{3ω(f3kn0-3m+k-f3kn0+3m+k)+(f3kn0-2m+k-f3kn0+2m+k)-(81ω+8)(f3kn0-1m+k-f3kn0+1m+k)},(28)f′′(n03m)=32(m+k)6561ω2+324ω-1×{(729ω2-18ω)(f3kn0-3m+k+f3kn0+3m+k)+(81ω-2)(f3kn0-2m+k+f3kn0+2m+k)+(162ω+7)(f3kn0-1m+k+f3kn0+1m+k)-(729ω2+225ω+5)f3kn0m+k}.Proof.
Let us denote(29)Fk=(f3kn0-3m+k,f3kn0-2m+k,f3kn0-1m+k,f3kn0m+k,f3kn0+1m+k,f3kn0+2m+k,f3kn0+3m+k)T.As the scheme is interpolatory, we havef3kn0m+k=fn0m. At each subdivision step, new control points are obtained by using the rule
(30)Fk+1=AFk,
where
(31)A=(0010000a4a3a2a1a0000a0a1a2a3a4000010000a4a3a2a1a0000a0a1a2a3a40000100),
is vertex subdivision matrix with a0=ω-(4/81), a1=-4ω+(10/27), and a2=6ω+(20/27), a3=-4ω-(5/81), a4=ω. Matrix A has seven eigenvalues  λ1=1, λ2=1/3, λ3=1/9, λ4=1/27, λ5=-6ω+(11/162)+(1/162)472392ω2-9720ω+121, λ6=-4ω, and λ7=-6ω+(11/162)-1/162472392ω2-9720ω+121 and has seven orthogonal eigenvectors. Let r^i and l^i be the right and left eigenvectors of matrix A corresponding to the eigenvalues λi, i=1,2,3,…,7, then direct computation leads to
(32)r^1=(1111111),r^2=(-3-2-10123),r^3=(9410149),r^4=(-27-8-101827),(33)r^5=(1-729ω2-60ω+162λ5ω+(77/162)-(7/162)472392ω2-9720ω+12181ω-2λ50λ5-729ω2-60ω+162λ5ω+(77/162)-(7/162)472392ω2-9720ω+12181ω-21),r^6=(-1324ω2+35ω4ω0-4ω-(324ω2+35ω)1),(34)r^7=(1-729ω2-60ω+162λ7ω+(77/162)+(7/162)472392ω2-9720ω+12181ω-2λ70λ7-729ω2-60ω+162λ7ω+(77/162)+(7/162)472392ω2-9720ω+12181ω-21),and left eigenvectors are(35)l^2=(3ω144ω+12,1144ω+12,-(81ω+8)144ω+12,0,81ω+8144ω+12,-1144ω+12,-3ω144ω+12),(36)l^3=(9ω(81ω-2)2(6561ω2+324ω-1),81ω-22(6561ω2+324ω-1),162ω+72(6561ω2+324ω-1),-(729ω2+225ω+5)6561ω2+324ω-1,162ω+72(6561ω2+324ω-1),81ω-22(6561ω2+324ω-1),9ω(81ω-2)2(6561ω2+324ω-1)).
If the limit curve generated by 5-point ternary interpolating subdivision scheme defines a C2-continuous function for 1/324<ω<1/162, then necessarily(37)limk→∞Fk=fn0mr^1=f(n03m)r^1,(38)limk→∞f3kn0+jm+k-fn0mj/3m+k=limk→∞f((3kn0+j)/3m+k)-f(n0/3m)j/3m+k=f′(n03m),j=±1,±2,±3,(39)limk→∞f3kn0+jm+k+f3kn0-jm+k-2fn0m(j/3m+k)2=limk→∞f((3kn0+j)/3m+k)+f((3kn0-j)/3m+k)-2f(n0/3m)(j/3m+k)2=f′′(n03m),j=±1,±2,±3.So we have(40)limk→∞3m+kD(Fk-fn0mr^1)=f'(n03m)e^,
where e^=(1,1,1,0,1,1,1)T, D=diag(-1/3,-1/2,-1,1,1,1/2,1/3). This implies that
(41)limk→∞32(m+k)D2(Fk+JFk-2fn0mr^1)=f′′(n03m)e^,
where D2=diag(1/9,1/4,1,1,1,1/4,1/9). Since matrix A has linearly independent eigenvectors, there exist scalars α1,α2,…,α7, such that F0 can be written as
(42)F0=∑i=17‍αir^i.
This implies that  α1=fn0m; therefore Fk can be written as
(43)Fk=AkF0=λkF0=∑i=17‍λikαir^i=fn0mr^1+α2(13)kr^2+α3(19)kr^3+α4(127)kr^4+∑i=57‍αi(λi)kr^i.
By using (43) in (40), we have
(44)f'(n03m)e^=limk→∞3m+kD(Fk-fn0mr^1)=limk→∞3m+kD{fn0mr^1+α2(13)kr^2+α3(19)kr^3+α4(127)kr^4+∑i=57‍αi(λi)kr^i-fn0mr^1}=limk→∞3m{α2Dr^2+α3(13)kDr^3+α4(19)kDr^4+3k∑i=57‍αi(λi)kDr^i}.
This implies that
(45)f'(n03m)e^=3mα2Dr^2.
In view of
(46)Jr^i={r^i,i=1,3,5,7,-r^i,i=2,4,6,
and using (43) in (41) we have
(47)f′′(n03m)e^=limk→∞32(m+k)D2(Fk+JFk-2fn0mr^1)=limk→∞32(m+k)D2×{fn0mr^1+α2(13)kr^2+α3(19)kr^3+α4(127)kr^4+∑i=57‍αi(λi)kr^i+fn0mr^1-α2(13)kr^2+α3(19)kr^3-α4(127)kr^4+α5(λ5)kr^5-α6(λ6)kr^6+α7(λ7)kr^7-2fn0mr^1(13)k}.
By simplification, we get
(48)f′′(n03m)e^=limk→∞32m(2α3D2r^3+2α5(9λ5)kD2r^5+2α7(9λ7)kD2r^7).
So, we have
(49)f′′(n03m)e^=32m2α3D2r^3.
Since we have
(50)Dr^2=(-130000000-120000000-10000000100000001000000012000000013)(-3-2-10123)=(1110111)=e^,(51)D2r^3=(19000000014000000010000000100000001000000014000000019)(9410149)=(1110111)=e^.
By using the above facts in (45) and (49), we get
(52)f'(n03m)=3mα2,(53)f′′(n03m)=2(3)2mα3.
Now by multiplying (43) with the left eigenvector l^2, corresponding to eigenvalue λ2=1/3, and in view of
(54)l^jr^i={1,i=j,0,i≠j,
we have
(55)l^2Fk=l^2(fn0mr^1+α2(13)kr^2+α3(19)kr^3+α4(127)kr^4+∑i=57‍αi(λi)kr^i).
This implies that
(56)l^2Fk=α2(13)k.
So, we have
(57)α2=3kl^2Fk.
Using the above result in (52), we have
(58)f'(n03m)=3mα2=3m3kl^2Fk.
Now by (29) and using the value of l^2, we have
(59)f'(n03m)=3m+k144w+12(3ω1-(81ω+8)0(81ω+8)-1-3ω)T(f3kn0-3m+kf3kn0-2m+kf3kn0-1m+kf3kn0m+kf3kn0+1m+kf3kn0+2m+kf3kn0+3m+k).
By simplification, we get
(60)f'(n03m)=3m+k144ω+12×{3ω(f3kn0-3m+k-f3kn0+3m+k)+(f3kn0-2m+k-f3kn0+2m+k)-(81ω+8)×(f3kn0-1m+k-f3kn0+1m+k)}.
Similarly, by multiplying (43) with the left eigenvector l^3, corresponding to λ3=1/9, we get
(61)l^3Fk=α3(19)k.
This implies that
(62)α3=9kl^3Fk=32kl^3Fk.
Using the above equation in (53), we have
(63)f′′(n03m)=32(m+k)2l^3Fk.
So by (29) and using the value of,  l^3, we have
(64)f′′(n03m)=32(m+k)6561ω2+324ω-1×(9ω(81ω-2)81ω-2162ω+7-729ω2-225ω-5162ω+781ω-29ω(81ω-2))T(f3kn0-3m+kf3kn0-2m+kf3kn0-1m+kf3kn0m+kf3kn0+1m+kf3kn0+2m+kf3kn0+3m+k).
After simplification, we get
(65)f′′(n03m)=32(m+k)6561ω2+324ω-1×{f3kn0m+k(729ω2-18ω)×(f3kn0-3m+k+f3kn0+3m+k)+(81ω-2)(f3kn0-2m+k+f3kn0+2m+k)+(162ω+7)(f3kn0-1m+k+f3kn0+1m+k)-(729ω2+225ω+5)f3kn0m+k}.


## 5. Comparison and Application of the Scheme

We give the comparison of some properties of the proposed 5-point ternary scheme with other existing binary and ternary schemes in Table1.Table 1
Comparison of proposed 5-point ternary scheme with some of existing binary and ternary schemes.
Subdivision scheme
Type
Approximation order
Support
C
n
4-point binary [10]
Interpolating
4
6
1
4-point binary [2]
Interpolating
4
6
1
DD 4-point binary [3]
Interpolating
3
3
1
4-point ternary [9]
Interpolating
3
5
2
4-point ternary [5]
Interpolating
3
5
2
5-point ternary [13]
Interpolating
6
7
1
Proposed 5-point ternary
Interpolating
4
7
2In this section, we also display the performance of the proposed schemes (14) and the scheme introduced in Theorem 3, by applying these schemes on open and closed polygons. Figure 1(a) shows initial open polygon and its limit curve. In this figure we also show the initial closed polygon and the results after first and 2nd subdivision level with parametric value ω=1/200.(a) The dashed lines show initial polygon while the solid curve is constructed by the proposed scheme with parameterω=1/200, (b) initial polygon, (c) result after first subdivision step, and (d) result after 2nd subdivision step.
(a)
(b)
(c)
(d)


## 6. A 5-Point Tensor Product Ternary Scheme

In this section, we present a 5-point tensor product ternary interpolating subdivision scheme. Some numerical examples of this tensor product scheme are also included. By taking the tensor product of scheme (14), we have the following nine rules representing a 5-point tensor product ternary interpolating scheme:
(66)f3i,3jk+1=∑s=04‍∑t=04‍asatfi+s-2,j+t-2k,f3i,3j+1k+1=∑s=04‍asfi+s-2,jk,f3i,3j+2k+1=∑s=04‍∑t=04‍asa4-tfi+s-2,j+t-2k,f3i+1,3jk+1=∑s=04‍asfi,j+s-2k,f3i+1,3j+1k+1=fi,jk,f3i+1,3j+2k+1=∑s=04‍a4-sfi,j+s-2k,f3i+2,3jk+1=∑s=04‍∑t=04‍a4-satfi+s-2,j+t-2k,f3i+2,3j+1k+1=∑s=04‍a4-sfi+s-2,jk,f3i+2,3j+2k+1=∑s=04‍∑t=04‍a4-sa4-tfi+s-2,j+t-2k.Figure2 shows how does the proposed 5-point tensor product ternary interpolating scheme work on control polygon with control points fi,jk, i,j∈ℝN, N≥2? In this figure, solid lines show the faces of initial polygon while dotted lines show the faces of refined polygon.Figure 2
Solid lines show initial polygon while dotted lines are refined polygon.Figure3(a) shows the outline of the cup like shape which is considered to be an initial control polygon of proposed scheme. Figures 3(b) and 3(c) show the results of the proposed scheme after first and second subdivision levels. This figure shows that the results of a 5-point tensor product ternary interpolating scheme are visually smooth.(a) Initial polygon, (b) result after first level, (c) result after the second level.
(a)
(b)
(c)


## 7. Conclusion

In this paper, we have presented a new 5-point ternary interpolating subdivision scheme with one parameter. The differentiability of the proposed scheme is also calculated. Some important properties of the scheme such as support, approximation order, conditions ofC0-, C1-, C2-continuity, and Hölders exponent are derived. A modified 5-point ternary scheme to deal with open polygons is also proposed to enhance the modeling ability of the scheme. Comparison of the proposed scheme with the other existing scheme is given. Some numerical examples are presented to show the visual performance of our scheme. In this paper, we have also introduced a 5-point tensor product interpolating scheme.



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*Source: 924839-2012-11-14.xml*