Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
1199
Journal of Zhejiang University SCIENCE A
ISSN 1673-565X (Print); ISSN 1862-1775 (Online)
www.zju.edu.cn/jzus; www.springerlink.com
E-mail: jzus@zju.edu.cn
Convexity-preserving interpolation of trigonometric polynomial
curves with a shape parameter*
PAN Yong-juan1,2, WANG Guo-jin‡1
(1Department of Mathematics, Zhejiang University, Hangzhou 310027, China)
2
( Department of Applied Mathematics, Zhejiang University of Technology, Hangzhou 310032, China)
E-mail: panyongjuan@zjut.edu.cn; wanggj@zju.edu.cn
Received Dec. 1, 2006; revision accepted Feb. 28, 2007
Abstract: In computer aided geometric design (CAGD), it is often needed to produce a convexity-preserving interpolating curve
according to the given planar data points. However, most existing pertinent methods cannot generate convexity-preserving interpolating transcendental curves; even constructing convexity-preserving interpolating polynomial curves, it is required to solve a
system of equations or recur to a complicated iterative process. The method developed in this paper overcomes the above drawbacks. The basic idea is: first to construct a kind of trigonometric polynomial curves with a shape parameter, and interpolating
trigonometric polynomial parametric curves with C2 (or G1) continuity can be automatically generated without having to solve any
system of equations or do any iterative computation. Then, the convexity of the constructed curves can be guaranteed by the
appropriate value of the shape parameter. Performing the method is easy and fast, and the curvature distribution of the resulting
interpolating curves is always well-proportioned. Several numerical examples are shown to substantiate that our algorithm is not
only correct but also usable.
Key words: Computer aided geometric design (CAGD), α-trigonometric polynomial curves, Interpolation, Convexity-preserving,
Shape parameter
doi:10.1631/jzus.2007.A1199
Document code: A
CLC number: TP391
INTRODUCTION
In geometric shape design and reverse engineering, constructing a convexity-preserving interpolating curve according to the given planar data
points is a familiar and basal problem. So far, some
techniques have been invented to obtain this kind of
curves [see (Tai and Wang, 2004) and the references
therein], but almost all of them have to solve a system
of equations or recur to a complicated iterative process; although there is no mention of the fact that, in
spite of doing a lot of the above troublesome work,
‡
Corresponding author
Project supported by the National Basic Research Program (973) of
China (No. 2004CB719400), the National Natural Science Foundation of China (Nos. 60673031 and 60333010) and the National
Natural Science Foundation for Innovative Research Groups of China
(No. 60021201)
*
they could not generate convexity-preserving interpolating transcendental curves, such as cycloid, spiral
and some other common curves in shape design (Tai
and Wang, 2004). As for constructing convexitypreserving interpolating curve by using NURBS, a
user rarely adopts them because of the intangibility of
weight for curve shape and the complexity in computation of rational curve.
Just to overcome the above drawbacks, this paper constructs a kind of trigonometric polynomial
curve with a shape parameter and investigates its
convexity-preserving interpolation algorithm. Trigonometric spline was introduced by Schoenberg
(1964), who studied the interpolation but did not take
shape preservation into account. Subsequently, many
people explored further properties of trigonometric
splines (Lyche and Winther, 1979; Koch et al., 1995;
Walz, 1997a; 1997b; Peña, 1997). Later, Zhang (1996;
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Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
1997) developed C-curves using the basis {sin t, cos t,
t, 1}, and unified C-curves and H-curves that use
basis {sinh t, cosh t, t, 1} as a complete curve family,
named F-splines (Zhang et al., 2005; Zhang and
Krause, 2005). Some other properties of C-curves
were studied in (Wang and Chen, 2004; Wang and Li,
2006). However, they did not process interpolation.
This paper presents a novel trigonometric polynomial
curve with a shape parameter. The technique is based
on the Loe’s idea (Loe, 1996) using singular blending,
but Loe blended a cubic uniform B-spline and its
control polygon to get non-interpolatory α-B-spline.
Also our technique generalizes the work of (Tai and
Wang, 2004), which applied the singular blending
idea to actualize interpolation but cannot obtain interpolating transcendental curves.
In this paper we construct a new type of curve by
blending a parametrized polygon and trigonometric
polynomial splines with blending factor α which acts
as shape parameter; then a family of trigonometric
polynomial curves with C2 (or G1) continuity interpolating the given planar data points can be automatically generated without having to solve any system of equations or to go at any iterative computation;
next, we research the convexity-preserving property
of these curves, and find a range in which the shape
parameter α takes its value in order to make the corresponding interpolating curves convexity-preserving,
as well as to make its curvature distribution always
well-proportioned. Lastly, several numerical examples are shown to substantiate that our algorithm is
not only correct but also usable, and can generate
convexity-preserving interpolating transcendental
curves.
CONSTRUCTION OF α-TRIGONOMETRIC POLYNOMIAL INTERPOLATING CURVES
We will construct a kind of trigonometric polynomial curves with shape parameter to interpolate the
given planar data points { Pi }in=1 , n≥4. Firstly, to ensure the number of line segments formed by connecting the data points equals the number of curve
segments in the trigonometric polynomial curves to
be constructed, two auxiliary data points P0, Pn+1 are
introduced according to the method of (Tai and Wang,
2004); then for the C-B-splines in (Zhang, 1996), we
take π/2 as the parametric step, take ui=iπ/2
(i=1,2,…,n) as the parametric nodes, so the uniform
trigonometric polynomial curve can be constructed as
follows:
C (u ) = C j (u ) = B0,4 (t ) Pj −1 + B1,4 (t ) Pj +
B2,4 (t ) Pj +1 + B3,4 (t ) Pj + 2 ,
(1)
t = u − u j , u j ≤ u ≤ u j +1 , j = 1,2,...,n − 1,
where { Pi }in=+01 represent the control points, Bi,4(t) (i=0,
1,2,3) are the uniform trigonometric polynomial
splines basis:
 B0,4 (t ) = (π/2 − t − cost ) π ,

 B1,4 (t ) = (t − sint + 2cost ) π ,

 B2,4 (t ) = (π/2 − t + 2sint − cost ) π ,
 B (t ) = (t − sint ) π.
 3,4
(2)
Thus, the curve (1) has the following properties
(Zhang, 1996):
C (u j ) = (1/2 − 1/π)( Pj −1 + Pj +1 )+2 Pj /π,

C (u j +1 ) = (1/2 − 1/π)( Pj + Pj + 2 )+2 Pj +1 /π,
(3)

C ′ (u j ) = ( Pj +1 − Pj −1 )/π, C ′ (u j +1 ) = ( Pj + 2 − Pj )/π,

2
C (u ) ∈ C [u1 ,un ].
To avoid solving global equation systems and to
introduce a shape adjustment parameter without
modifying the order of continuity anywhere, we blend
a parametrized polygon and the above trigonometric
polynomial curve C(u) using a blending factor α. The
vertices of this polygon can be directly computed
from the interpolation conditions. In order to let the
curve to be constructed have the basis {sin t, cos t, t,
1}, like the curve C(u), we construct singular blending function (Loe, 1996) Sj(u) to parametrize the
polygon using the basis {sin t, cos t, t, 1}; so we take
Sj(u)∈C2[uj, uj+1] and satisfying the following conditions:
S j (u j ) = 0, S j (u j +1 ) = 1, S ′j (u j ) = S ′′j (u j ) = 0,
S ′j (u j +1 ) = S ′′j (u j +1 ) = 0, j =1,2,...,n − 1.
In this way, after some calculation, we get
(4)
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Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
∂
∂2
T (u j ,α ) = α C ′ (u j ), 2 T (u j ,α ) = α C ′′ (u j ). (8)
∂u
∂u
S j (u )= S (u − u j ), u j ≤ u ≤ u j +1 , j = 1, 2,…, n − 1,
where
6(2 + 3)(t − sint ) π ,
0 ≤ t ≤ π/6,

3(5 + 3 3)(sint − cost ) π − 6(3 + 2 3)t π
S (t ) = 
π/6 ≤ t ≤ π/3,
 + 5 + 3 3,

6(2 + 3)(t + cost ) π − 5 − 3 3, π/3 ≤ t ≤ π/2.
(5)
We call the reparametrized polygon a singular
polygon, whose yet-to-be-determined vertices Vj and
Vj+1 are dependent on α, and hence the singular
polygon is defined as follows:
L(u, α)=Lj(u, α)=[1− Sj(u)]Vj +Sj(u)Vj+1,
uj≤u≤uj+1, j=1,2,…,n−1.
By blending the singular polygon L(u, α) and the
uniform trigonometric polynomial curve C(u), both
defined on [u1, un], using α as the blending factor, we
obtain a uniform trigonometric polynomial curve with
an additional parameter α:
T(u, α)=αC(u)+(1−α)L(u, α), u1≤u≤un.
Each of its segments on [uj, uj+1] can be more explicitly represented as follows:
T (u ,α ) = T j (u ,α ) = α C j (u ) + (1 − α ) L j (u ,α )
= α C j (u ) + (1 − α ){[1 − S j (u )]V j + S j (u )V j +1}.
(6)
From the condition Eq.(4) of the singular blending
function Sj(u) and the interpolating condition T(uj,α)=
Pj (j=1,2,…,n), we obtain
(1 − α )V j = Pj − α C j (u j ),

(1 − α )Vn = Pn − α C n −1 (un ),
j = 1,2,...,n − 1,
j = n.
Finally, we get the following desired interpolating
uniform α-trigonometric polynomial curve
T (u ,α ) = T j (u ,α ) = α C j (u ) + [1 − S j (u )]
Thus, for a fixed shape parameter α, the curve
T(u, α) maintains the same order of continuity as that
of C(u) at every knot uj (j=1,2,…,n). Since, everywhere away from the knots, the components of the
uniform trigonometric polynomial curve C(u) are C∞
and the singular blending functions Sj(u) are C2, the
uniform α-trigonometric polynomial curve T(u, α)
∈C2.
For the purpose of practical shape design, we
only discuss the standard α-trigonometric polynomial
curve which has the restriction of α∈(0,1). When α=0,
the α-trigonometric polynomial curve is reduced to
the polygon P1P2…Pn connecting the data points
n ; when α=1, the α-trigonometric polynomial
{ Pi }i=
1
curve is reduced to the uniform trigonometric polynomial curve C(u), in which case the singular polygon
L(u, α) is non-existent.
CONVEXITY-PRESERVING
RELATIVE TO GLOBAL
POINTS
INTERPOLATION
CONVEX DATA
Definition 1 A set of planar data points denoted as
n is said to be convex if the close polygon
{ Pi }i=
1
P1P2…PnP1 is convex.
Definition 2 A planar curve is said to be convex if
there are no inflection points or singular points (including cusps and the double points) on it.
n
Let { Pi }i=
1 be convex data points in the plane
with any three consecutive points Pj−1, Pj, Pj+1 (j=2,
n
3,…,n−1) being not collinear and { Pi }i=
1 being arranged counterclockwise.
Convexity of jth curve segment Tj (u, α) on [uj, uj+1]
Let t=u−uj∈[0, π/2], u∈[uj, uj+1], and write Tj(u,
α) as T j (t , α ), Cj(u) as C j (t ) on [uj, uj+1], then
Eqs.(6) and (7) can be written as
⋅ [ Pj − α C j (u j )] + S j (u )[ Pj +1 − α C j (u j +1 )], (7)
T j (u,α ) = T j (t ,α )
u j ≤ u ≤ u j +1 , j = 1, 2,… , n − 1.
= α C j (t ) + (1 − α ){[1 − S (t )]V j + S (t )V j +1 }
The continuity of the curve T(u,α) follows directly
from Eq.(4) that
(9)
= α C j (t ) + [1 − S (t )][ Pj − α C j (0)] + S (t )[ Pj +1 − αC j (π/2)],
(10)
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Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
where 0 ≤ t ≤ π/2.
Denote the wedge product of two vectors ai and
aj as
[ ai , a j ] =
xi
xj
+α [7+3 3+h1 +(5+3 3 − 3h1 )sint
12
+sint +cost )w20
j +( − 1 − sint +cost )w j ], t ∈ [0,π/6];
M j (t ,α )
and let
F2(j ) (t ,α ) = {3(5+3 3)(1 − sint ) −
3(1+ 3)cost +α [1 − h2 − (10+6 3+h1 − 2h2 )sint +
ai=Pi−Pi−1, i=1,2,…,n+1,
cj=−(aj+aj+1+aj+2),
(10+6 3 − h1 )cost ]}w01
j +α [16+9 3 − h2 − (10+6 3
+h1 − h2 )(sint +cost )]w20
j +{3(5+3 3)(1 − cost )
then [cj, aj]≥0, [aj+2, cj]≥0 hold, so if we write
12
20
w01
j = [ a j , a j +1 ], w j = [ a j +1 , a j + 2 ], w j = [ a j + 2 , a j ],
and because {Pj−1PjPj+1Pj+2} are convex data points
arranged counterclockwise, we have
(11)
Moreover, it is easy to get
C j′ (t ) =[(1−sint)aj+(cost+sint)aj+1+(1−cost)aj+2]/π,
thus
∂
T j (t ,α )
∂t
= α C j′ (t )+S ′ (t ){( Pj +1 − Pj ) − α [C j (π/2) − C j (0)]}
= α [(1 − sint )/π − S ′ (t )(1/2 − 1/π)] a j
+[α (cost +sint )/π+S ′ (t ) − 2α S ′ (t )/π] a j +1
+α [(1 − cost )/π − S ′ (t )(1/2 − 1/π)] a j + 2 ,
∂2
T j (t ,α ) = α [ − cost/π − S ′′(t )(1/2 − 1/π)]a j +
∂t 2
[α (cost − sint )/π+S ′′(t ) − 2α S ′′(t )/π]a j +1 +
α [sint/π − S ′′(t )(1/2 − 1/π)]a j + 2 .
Let

π ∂
∂2
M j (t ,α ) =  T j (t ,α ), 2 T j (t ,α )  ,
α  ∂t
∂t

2
h1 = 6(2+ 3)/π, h2 = 6(5+3 3)/π,
after some algebraic calculation in subinterval [0, π/6],
[π/6,π/3] and [π/3,π/2], M j (t ,α ) can be expressed as
follows respectively:
F1(j ) (t ,α ) = {6(2+ 3)( − 1+sint +cost )
− (5+3 3+h1 )cost ]}w01
j +α (5+3 3 − h1 )[( − 1
yi
, ai=(xi, yi), aj=(xj, yj),
yj
20
12
20
01
12
w01
j ≥ w j , w j ≥ w j , w j > 0, w j > 0.
M j (t ,α )
− 3(1+ 3)sint +α [1 − h2 +(10+6 3 − h1 )sint
− (10+6 3+h1 − 2h2 )cost ]}w12j , t ∈ [π/6,π/3];
M j (t ,α )
F3(j ) (t ,α ) = α (5+3 3 − h1 )[( − 1+sint
20
− cost )w01
j +( − 1+sint +cost )w j ]+{6(2+ 3) ⋅
( − 1+sint +cost )+α [7+3 3+h1 − (5+3 3+h1 )sint
+(5+3 3 − 3h1 )cost ]}w12j , t ∈ [π/3,π/2].
Noting that the relative curvature of the curve
T j (t ,α ) (0 ≤ t ≤ π/2) is (Farin, 2005)
∂

∂2
k j (t ,α ) =  T j (t ,α ), 2 T j (t ,α ) 
∂t
 ∂t

3
∂
T j (t ,α ) ,
∂t
so, there are no inflection points on T j (t ,α ) (0≤t≤π/2)
if and only if M j (t ,α ) ≥ 0, ∀t∈[0,π/2]. In order to
obtain the range of the shape parameter α in which
M j (t ,α ) ≥ 0 holds ∀t∈[0,π/2] in the three subintervals of [0, π/2], the following three lemmas are needed.
Here we just prove Lemma 1, proofs of the other two
lemmas are similar.
Lemma 1
Let f(x)=f0+f1sinx+f2cosx, then f(x)≥0,
∀x∈[0,π/6] if and only if
R0(1) ≥ 0, R2(1) ≥ 0, R1(1) + 2 R0(1) R2(1) ≥ 0,
where
 R0(1) = f 0 + f 2 ,
 (1)
 R1 = f 0 + (2 − 3)f1 + f 2 ,
 (1)
 R2 = (4 − 2 3)f 0 + (2 − 3)f1 − (3 − 2 3)f 2 .
Proof Do trigonometric substitution to the functions
sin x and cos x, and let tan (x/2)=t, we can write f(x) as
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Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
f (x)=
f 0 + f 2 + 2 f1t + (f 0 − f 2 )t 2
, t ∈ [0, 2 − 3].
1+t 2
So
f (x) ≥ 0 , x∈[0, π/6] ⇔ g (t ) ≥ 0, t ∈ [0,2 − 3],
Next, we will use Lemmas 1~3 to obtain the
range of the shape parameter α in which M j (t ,α ) ≥ 0
holds ∀t∈[0,π/2].
Let
where
g (t )=f 0 + f 2 + 2 f1t + (f 0 − f 2 )t 2 , t ∈ [0,2 − 3].
Then let v = t/(2 − 3), t ∈ [0,2 − 3], and using
the conversion formula
1 0 0 
[1,v,v ] = [B (v),B (v),B (v)] 1 1/2 0 
1 1 1 
2
2
0
2
1
2
2
between the monomial basis and Bernstein basis, we
can convert g(t) into a Bernstein polynomial:
g (t ) = g (v) = [B02 (v),B12 (v),B22 (v)]
f 0 +f 2




⋅
f 0 +(2 − 3)f1 +f 2
.


 2[(4 − 2 3)f 0 +(2 − 3)f1 − (3 − 2 3)f 2 ] 
Hence, the positive conditions of quadratic Bernstein
polynomial (Chang, 1995) gives Lemma 1.
Lemma 2
Let f(x)=f0+f1sinx+f2cosx, then f(x)≥0,
∀x∈[π/6,π/3] if and only if
R0(2) ≥ 0, R2(2) ≥ 0, R1(2) + 2 R0(2) R2(2) 3 ≥ 0,
where
 R0(2) = (4 − 2 3)f 0 + (2 − 3)f1 − (3 − 2 3)f 2 ,

(2)
 R1 = 2(f1 + f 2 )+2 3 (f 0 − f1 − f 2 ) 3,
 (2)
 R2 =2 f 0 + 3 f1 +f 2 .
Lemma 3
Let f(x)=f0+f1sinx+f2cosx, then f(x)≥0,
∀x∈[π/3,π/2] if and only if
R0(3) ≥ 0, R2(3) ≥ 0, R1(3) + 2 R0(3) R2(3) /3 ≥ 0,
where
 R0(3) = 2 f 0 + 3 f1 +f 2 ,

(3)
 R1 = f 0 + f1 +f 2 + 3(f 0 + f1 − f 2 )/3,
 (3)
 R2 = f 0 + f1 .
20
 A1(j ) = [8 − 2 3 − 6(1+ 3)/π]w01
j + [2 − 6( 3 − 1)/π]w j


+[ − 2 3 + 6(3 − 3)/π]w12j ,

20
12
 A2(j ) = (3+ 3 − 18/π)w01
j +(1+ 3 − 6/π)(w j − w j ),

 A3(j ) = −[6+4 3 − 6(3+ 3)/π]w01
j +[4+2 3

12
− 6(1+ 3)/π]w20

j +[10+4 3 − 6(5+3 3)/π]w j ,
 (j )
01
12
 A4 = [ 3 − 6(3 − 3)/π](w j +w j )

+[3 − 6(3 − 3)/π]w20

j ,
 (j )
20
01
 A5 = [2+4 3/3 − 2(3 + 3)/π](w j − w j )

+[4+2 3 − 6(3 + 3)/π]w12j ,

k1(j ) = (A2( j ) )2 − 4 A1( j ) w01
j ,

( j)
01 2
k2(j ) = 12w01
j A2 − 24( 3 − 1)(w j ) ,
 (j )
01 2
( j)
( j) 2
( j) ( j)
k3 = 36(w j ) , k4 = (A4 ) − 3 A1 A3 ,
 (j )
01
12
( j)
12 ( j )
k5 = 6(3 − 3)(w j +w j )A4 − 18( 3 + 1)w j A1

( j)
− 18( 3 − 1)w01

j A3 ,
 (j )
01 12
2
01
12 2
k6 = (9 − 3 3) (w j +w j ) − 216w j w j ,
 (j )
( j) 2
( j ) 12
k7 = (A5 ) − 8 A3 w j 3,
 (j )
( j ) 12
12 2
k8 = 4(3+ 3)A5 w j − 16( 3 + 1)(w j ) ,
 (j )
2
12 2
k9 = 4(3+ 3) (w j ) ,
∆ 1(j ) = (k2(j ) ) 2 − 4k1(j ) k3(j ) , ∆ (2j ) = (k5(j ) )2 − 4k4(j ) k6(j ) ,
∆ 3(j ) = (k8(j ) ) 2 − 4k7(j ) k9(j ) ,
α1(j ) = 6(1 − 3)w01
A1(j ) , α 2(j ) = − 6w01
A2(j ) ,
j
j

α 3(j ) = −6(1+ 3)w12j A3(j ) ,

12
(j )
α 4(j ) = 3( 3 − 3)(w01
j +w j ) A4 ,

α 5(j ) = −2(3+ 3)w12j A5(j ) ,
(12)

α *(j ) ,α * (j ) = −k (j ) ± ∆ (j ) 2k (j ) , (α * (j ) ≤ α * (j ) ),
2
2
2
1
1
1
 1
 * (j ) * (j )
(j )
(j )
2k4(j ) , (α 3* (j ) ≤ α 4* (j ) ),
α 3 ,α 4 = −k5 ± ∆2

α * (j ) ,α * (j ) = −k (j ) ± ∆ (j ) 2k (j ) , (α * (j ) ≤ α * (j ) ).
6
8
3
7
5
6
 5
(
(
(
)
)
)
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Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
Then it is easy to see that Ai(j ) < 0 (i=1,2,3,4,5),
⇔ 0 < α ≤ min(α1(j ) ,α 3(j )α 4(j ) ,1).
k1(j ) > 0, k4(j ) < 0, k7(j ) > 0.
In the subinterval [0,π/6], F1(j ) (t ,α ) can be
F2(j ) (t ,α ) ≥ 0 , t ∈ [π/6, π/3]
max(α 4(j ) ,0) ≤ α ≤ min(α1(j ) ,α 3(j ) ,α 3∗(j ) ,1), or
F1(j ) (t ,α )=f 0(j ) + f1(j ) sin t + f 2(j ) cos t ,
max(α 4(j ) ,α 4∗(j ) ,0) ≤ α ≤ min(α1(j ) ,α 3(j ) ,1). (∆2(j ) ≥ 0)
where
01
f 0(j ) = −6(2 + 3)w01
j + α [(7+3 3 + h1 )w j
01
f1(j ) = 6(2 + 3)w01
j + α [(5+3 3 − 3h1 )w j
12
+ (5 + 3 3 − h1 )(w20
j − w j )],
= 6(2 + 3)w + α [ − (5+3 3 + h1 )w
01
j
01
j
Then we can see from Lemma 1 that F1(j ) (t ,α ) ≥ 0,
t∈[0,π/6] if and only if the following three formulae
hold synchronously:
(1) R0(1) = 2α w01
j ≥ 0;
(j )
(j )
(2) R2(1) = 6( 3 − 1)w01
j + α A1 ≥ 0 ⇔ α ≤ α1 ;
(3) R1(1) + 2R0(1) R2(1) ≥ 0
⇔ R1(1) ≥ 0 , or R1(1) < 0 and 2R0(1) R2(1) ≥ (R1(1) ) 2
(j )
01
(j )
⇔ 6 w01
j + α A2 ≥ 0 , or 6 w j + α A2 < 0 and
H1(j ) (α ) = k1(j )α 2 + k2(j )α + k3(j ) ≤ 0.
When the root discriminant ∆1(j ) ≥ 0 of the
quadratic polynomial H1(j ) (α ), H1(j ) (α ) has the zeroes α1* (j ) and α 2* (j ) . So combining (1) and (2), we get
F1(j ) (t ,α ) ≥ 0 , t∈[0,π/6]
(∆1(j ) < 0)
F1(j ) (t ,α ) ≥ 0 , t∈[0,π/6]
⇔ 0 < α ≤ min(α1(j ) ,α 2(j ) ,1), or
max(α 2(j ) ,α1∗(j ) ,0) ≤ α ≤ min(α1(j ) ,α 2∗(j ) ,1). (∆1(j ) ≥ 0)
Similarly, using Lemmas 2 and 3 on F2(j ) (t ,α )
and F3(j ) (t ,α ) respectively gives
(∆3(j ) < 0)
F3(j ) (t ,α ) ≥ 0 , t ∈ [π/3,π/2]
⇔ 0 < α ≤ min(α 3(j ) ,α 5(j ) ,1) , or
max(α 5(j ) ,α 5∗(j ) ,0) ≤ α ≤ min(α 3(j ) ,α 6∗(j ) ,1) . (∆3(j ) ≥ 0)
Summarizing the results on the above three
subintervals, if we denote
12
+ (5 + 3 3 − h1 )(w20
j + w j )].
⇔ 0 < α ≤ min(α1(j ) ,α 2(j ) ,1).
F3(j ) (t ,α ) ≥ 0 , t ∈ [π/3,π/2]
⇔ 0 < α ≤ min(α 3(j ) ,α 5(j ) ,1) .
12
− (5 + 3 3 − h1 )(w20
j + w j )],
f
(∆2(j ) < 0)
⇔ 0 < α ≤ min(α1(j ) ,α 3(j )α 4(j ) ,1), or
written as
(j )
2
F2(j ) (t ,α ) ≥ 0 , t ∈ [π/6, π/3]
α l(1j ) = 0, α l(2j ) = max(α 2(j ) ,α1* (j ) ,0),
 (j )
(j )
(j )
(j )
* (j )
α l 3 = max(α 4 ,0), α l 4 = max(α 4 ,α 4 ,0),
 (j )
(j )
* (j )
(j )
(j )
(j )
α l 5 = max(α 5 ,α 5 ,0), α l 6 = max(α l 2 ,α l 3 ),
α (j ) = max(α (j ) ,α (j ) ), α (j ) = max(α (j ) ,α (j ) ),
l2
l4
l8
l2
l5
 l7
α l(9j ) = max(α l(3j ) ,α l(5j ) ), α l(10j ) = max(α l(4j ) ,α l(5j ) ),
 (j )
(j )
(j )
(j )
(j )
(j )
(j )
(j )
α l11 = max(α l 2 ,α l 3 ,α l 5 ), α l12 = max(α l 2 ,α l 4 ,α l 5 ),
 (j )
(j )
(j )
(j )
(j )
(j )
α r1 = min(α1 ,α 2 ,α 3 ,α 4 ,α 5 ,1),
α (j ) = min(α (j ) ,α * (j ) ,α (j ) ,α (j ) ,α (j ) ,1),
1
2
3
4
5
 r2
(
)
(
)
(
)
(
)
(
)
j
j
j
j
*
j
α r 3 = min(α1 ,α 2 ,α 3 ,α 3 ,α 5(j ) ,1),
 (j )
(j )
(j )
(j )
(j )
(13)
α r 4 = min(α1 ,α 2 ,α 3 ,α 5 ,1),
 (j )
(j )
(j )
(j )
(j )
* (j )
α r 5 = min(α1 ,α 2 ,α 3 ,α 4 ,α 6 ,1),
α (j ) = min(α (j ) ,α ∗(j ) ,α (j ) ,α ∗(j ) ,α (j ) ,1),
1
2
3
3
5
 r6
(j )
(j )
(j )
(j )
∗(j )
α r 7 = min(α1 ,α 2 ,α 3 ,α 5 ,1),
 (j )
(j )
(j )
(j )
∗(j )
∗(j )
α r 8 = min(α1 ,α 2 ,α 3 ,α 4 ,α 6 ,1),
 (j )
∗(j )
( j)
( j)
( j)
* (j )
α r 9 = min(α1 ,α 2 ,α 3 ,α 3 ,α 6 ,1),
α (j ) = min(α (j ) ,α (j ) ,α (j ) ,α * (j ) ,1),
1
2
3
6
 r10
j)
α r(11
= min(α1(j ) ,α 2∗( j ) ,α 3(j ) ,α 3∗(j ) ,α 6∗(j ) ,1),
 (j )
(j )
(j )
∗(j )
∗(j )
α r12 = min(α1 ,α 2 ,α 3 ,α 6 ,1),
and set
∗(j )
(j )
 B1 = {α | 0 < α ≤ α r1 },
 ∗(j )
(j )
(j )
 Bi = {α | α li ≤ α ≤ α ri }, i =2,3,...,12,
we can state the following
(14)
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Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
Let { Pi }ij=+j2−1 be convex data points in
Theorem 1
the plane and any three consecutive points are not
collinear, let Tj (t ,α ) (0≤t≤π/2) be the uniform
α-trigonometric polynomial interpolating curve segment defined by Eq.(10), then there are no inflection
points on Tj (t ,α ) (0≤t≤π/2) if and only if α ∈ U ∗j ,
where
Then calculating the wedge product of both sides of
the above equation with aj+2 gives
12
−α w20
j (t 2 +cost2 − t1 − cost1 ) / π + α w j (sint2
−cost2 − sint1 + cost1 ) / π + [S (t2 ) − S (t1 )]
⋅[(1 − 2α / π)w12j + α (1/2 − 1/ π)w20
j ] = 0.
(15)
But, when w20
j ≥ 0, Eq.(15) can be written as
(1) When ∆1(j ) , ∆2(j ) , ∆3(j ) < 0, U ∗j = B1∗(j ) ;
(2) When ∆1(j ) ≥ 0 and ∆2(j ) , ∆3(j ) < 0,
∗
j
∗(j )
1
U =B
∗(j )
2
∪B
−2cost2 + 2cost1 +sint2 − sint1 )w20
j / π + [S (t2 ) − S (t1 )]
;
(3) When ∆ ≥ 0 and ∆ , ∆ < 0,
(j )
2
(j )
1
(j )
3
(4) When ∆3(j ) ≥ 0 and ∆1(j ) , ∆2(j ) < 0,
U ∗j = B1∗(j ) ∪ B5∗(j ) ;
(5) When ∆1(j ) , ∆2(j ) ≥ 0 and ∆3(j ) < 0,
(6) When ∆1( j ) , ∆3( j ) ≥ 0 and ∆2( j ) < 0,
U ∗j = B1∗(j ) ∪ B2∗(j ) ∪ B5∗(j ) ∪ B8∗(j ) ;
(7) When ∆ ,∆ ≥ 0 and ∆ < 0, ,
(j )
1
U ∗j = B1∗(j ) ∪ B3∗(j ) ∪ B4∗(j ) ∪ B5∗(j ) ∪ B9∗(j ) ∪ B10∗(j ) ;
12
(8) When ∆1(j ) , ∆2(j ) , ∆3(j ) ≥ 0, U ∗j = ∪ Bi∗(j ) .
i =1
∗(j )
i
Substituting “<” for “≤” in B
(16)
and, when w20
j < 0, if we choose α satisfying
U ∗j = B1∗(j ) ∪ B2∗(j ) ∪ B3∗(j ) ∪ B4∗(j ) ∪ B6∗(j ) ∪ B7∗(j ) ;
(j )
3
⋅[(1 − 2α / π)w12j + α (1/2 − 1/ π)w20
j ]=0.
The left side of Eq.(16) is larger than zero because of
the monotonicity of the function S (t ) on [0,π/2] and
Eq.(11). This contradiction shows that there are no
double points on the curve segment Tj (t ,α ) (0≤t≤π/2);
U ∗j = B1∗(j ) ∪ B3∗(j ) ∪ B4∗(j ) ;
(j )
2
α (w12j − w20j )(sint2 − cost2 − sint1 + cost1 ) / π + α (t1 − t2
(i=1,2,…,12),
and denoting the corresponding sets as Bi(j )
(i=1,2,…,12), U ∗j in Lemma 1 as Uj, then when α∈Uj,
M j (t ,α ) >0, that is ∂ T j (t ,α )/∂t ≠ 0, which indicates
that there are no cusp points on the curve segment
Tj (t ,α ) (0≤t≤π/2).
Next, let us consider the double points on this
curve segment. Suppose that there exist t1, t2, 0≤t1<t2≤
π/2 such that Tj (t1 ,α ) = Tj (t2 ,α ). From Eq.(10), after
a simple calculation, we have
α a j (t2 +cost2 − t1 − cost1 ) / π + α a j +1 (sint2 − cost2 − sint1
+cost1 ) / π + α a j + 2 (t2 − sint2 − t1 +sint1 ) / π + [S (t2 )
−S (t1 )][(1 − 2α / π)a j +1 − α (1/2 − 1/ π)(a j + a j + 2 )] = 0.
0 < α ≤ α 0(j ) = min {πw12j [2w12j + (1 − π / 2)w20
j ],1} ,
then the left side of Eq.(15) is larger than zero. So the
same reasoning shows the inexistence of double
points on the curve segment Tj (t ,α ) (0≤t≤π/2).
Combining the above discussion and Theorem 1, if
we let
V j = {α | 0 < α ≤ α 0(j ) } ,
(17)
we can state the following
Theorem 2 Let { Pi }ij=+j2−1 and Tj (t ,α ) (0≤t≤π/2) be
the same as in Theorem 1, if according to the different
cases in Theorem 1, we choose Dj as follows:
Dj = U j ,
Dj = U j ∩ Vj ,
(w20
j ≥ 0).
(w20
j < 0 ).
(18)
(19)
Then, when α∈Dj, the corresponding curve segment
T j (t ,α ) (0≤t≤π/2) is convex and C2-continuous.
Convexity of α-trigonometric polynomial interpolating curve T(u, α) (u1≤u≤un)
1. Taking uniform shape parameter for each
curve segment
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Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
From Section 3.1, we can get
n
Theorem 3 Let { Pi }i=
1 be convex data points in the
plane, with any three consecutive points being not
collinear; suppose that P0 and Pn+1 are two auxiliary
data points introduced according to the method of (Tai
and Wang, 2004) and T(u, α) (u1≤u≤un) is the uniform
α-trigonometric polynomial curve defined by Eq.(7).
If the shape parameter α satisfies
n −1
α ∈ ∩ Dj ,
Dj−1=Dj={α|α=0}. So, Tj−1(u,α) and Tj(u,α) are reduced to the straight line segments Pj−1Pj and PjPj+1
respectively. If we take the uniform shape parameter
for each curve segment according to Eq.(20), then α=0,
and T(u,0) (u1≤u≤un) is reduced to the polygon
P1P2…Pn connecting the data points {Pi }in=1 .
Pj+1
Pj
Pj−1
(20)
j =1
Pj+2
then the whole curve T(u, α) (u1≤u≤un) is convex and
C2-continuous.
Proof From Theorem 2 we can see each segment
Tj(u,α) (uj≤u≤uj+1), j=1,2,…,n−1 of the curve T(u,α)
n −1
(u1≤u≤un) with the shape parameter α ∈ ∩ D j is
j =1
convex. Also, from Eq.(8) and the fact that trigonometric polynomial curve is convexity-preserving
about its control polynomial (Zhang, 1997), we can
conclude there are no cusp points on the curve T(u,α)
(u1≤u≤un).
When the variable u∈[uj,uj+1], substituting t=π/4
into Eq.(10) gives
T j (π/4,α ) = ( Pj + Pj +1 )/2 + α ( 2 − 1)/π
⋅ [( Pj + Pj +1 )/2 − ( Pj −1 + Pj + 2 )/2].
Since 0<α<1, it is easy to prove that each data point Pj
(j=1,2,…,n) is not the inflection point of T(u,α)
(u1≤u≤un). Again, the definition of T(u,α) and the fact
of T(u,α)∈C2[u1,un] reveal that the joint point between
any two consecutive curve segments Tj(u,α) and
Tj+1(u,α) (j=1,2,…,n−2) do not comprise a double
point. So, curve T(u,α) (u1≤u≤un) is convex.
Remark
In Theorems 1~3, we suppose that any
n are not colthree consecutive points among { Pi }i=
1
Pj−2
Fig.1 Three collinear consecutive points Pj−1, Pj, Pj+1
2. Taking respective shape parameters for each
curve segment
In practical applications, when there is a large
number of data points, and some parameter value
ranges Dj calculated according to Eq.(18) or Eq.(19)
are very small, α obtained by Eq.(20) will be small
(close to zero). Then, the curve T(u,α) (u1≤u≤un)
cannot reveal the information of some curve segments,
and lacks flexibility. In this subsection, we will improve the algorithm in Section 3.2.1 and obtain a more
useful convexity-preserving algorithm. That is, for
each curve segment Tj(u,α) (uj≤u≤uj+1), j=1,2,…, n−1,
we may select the respective shape parameter α *j such
that α *j ∈ D j . Accordingly, write each curve segment
as T j (u,α *j ) (uj≤u≤uj+1), the whole curve as T(u)
(u1≤u≤un), and the singular polygon as L(u) (u1≤u≤un),
each of which is L j (u ,α *j ). It is worthwhile to note
that at this time the singular polygon here is discontinuous, however, T(u) is a convexity-preserving interpolating curve with G1 continuity and each curve
segment T j (u ,α *j ) ∈ C 2 [u j ,u j +1 ] [see Section 5 of
∩ D j = ∅ or ∩ D j = {α|α=0}. If there are three
(Tai and Wang, 2004)]. It is more suitable for application than the curve T(u,α) (u1≤u≤un) defined in
Section 3.2.1, with the uniform shape parameter α (see
the examples in Section 5).
points Pj−1, Pj and Pj+1 (j≥2) collinear (Fig.1), the case
is trivial: as the curve segments Tj−1(u,α) (uj−1≤u≤uj)
and Tj(u,α) (uj≤u≤uj+1), w12j −1 = w01
j = [a j , a j +1 ] = 0
CONVEXITY-PRESERVING INTERPOLATION
RELATIVE TO GENERAL DATA POINTS
linear. This supposition is necessary to avoid
n −1
n −1
j =1
j =1
hold, so α3(j −1) = α1(j ) = 0 from Eq.(12), and then
α
(j −1)
ri
= α = 0 in Eq.(13) (i=1,2,…,12). That is,
(j )
ri
In this section, we will discuss the convexitypreserving interpolation relative to general data points.
1207
Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
The detail is as follows. Search starting from P1, we
find the first convex subset I1 = { P1 P2 ... Pi1 } of the
data points according to Definition 1; as
I1 = { P1 P2 ... Pi1 } seems to be the same as { Pi }in=1 in
Example 1 The data points are shown in Table 1,
and the corresponding interpolating curves with different values of shape parameter α are plotted in Fig.2.
Section 2, we can obtain the auxiliary point P0; then
search starting from Pn reversely, we find the last
convex subset I m = { Pim−1 Pim−1 +1 ... Pn } of the data
points according to Definition 1, similarly, we can
obtain the auxiliary point Pn+1. Next, using { Pi }in=+01 as
the control vertices, we construct a uniform trigonometric polynomial curve C(u) by method similar to the
one mentioned in Section 2.
Finally, we construct uniform α-trigonometric
polynomial convexity-preserving interpolating curve
relative to the general data points.
Noting that construction of the jth curve segment,
interpolation of the data points Pj and Pj+1, of the
α-trigonometric polynomial interpolating curve, must
use the points Pj−1, Pj, Pj+1, Pj+2 (j=1,2,…,n−1). If the
point set {Pj−1PjPj+1Pj+2} (j=1,2,…,n−1) is convex, the
range of the shape parameter α: Dj, which makes the
jth curve segment convex, can be obtained according
to the algorithms in Section 3.1; else, we take
n −1
Dj={α|0<α<1}. Thus, when taking α ∈ ∩ D j , the
j =1
corresponding curve T(u,α) (u1≤u≤un) defined by
Eq.(7) is convexity-preserving and interpolates
n with C2 continuity.
{ Pi }i=
1
On the other hand, similar to Section 3.2.2, for
each curve segment Tj(u,α) (uj≤u≤uj+1), j=1,2,…,n−1,
we may select the respective shape parameter α *j such
that α *j ∈ D j . Then T(u) (u1≤u≤un) is a convexitypreserving interpolating curve with G1 continuity and
each curve segment T j (u ,α *j ) ∈ C 2 [u j ,u j +1 ] (j=1,2,…,
n−1).
NUMERICAL EXAMPLES
In this section, we will give some numerical
examples; in each example, the data points are
marked with small black points, we depict
α-trigonometric polynomial interpolating curve. In
Examples 3 and 4, we also give the corresponding
curvature plot.
(a)
(b)
(c)
(d)
Fig.2 α-trigonometric polynomial interpolating curve
with given α. (a) α=0.9; (b) α=0.74; (c) Each segment
having respective shape parameter αj* (j=1,2,3); (d)
α=0.44301
Table 1 The data points in Fig.2
xn
yn
12
28.7
23
13
25.7
12
43.5
19.4
Fig.2a shows the result with the uniform shape
parameter α=0.9. There is a double point on the
α-trigonometric polynomial interpolating curve.
Fig.2b shows the result with the uniform shape parameter α=0.74, there is a cusp point on the
α-trigonometric polynomial interpolating curve.
Fig.2c shows the result with respective shape parameter for each curve segment, where D1=D3=
{α|0<α<1}, D2={α|0<α<0.44302} according to
Theorems 1 and Theorem 2; and we take α 1* = 0.99,
α 2* = 0.44301, α3* = 0.99, which avoids cusp and double points, and the α-trigonometric polynomial interpolating curve is convexity-preserving. Fig.2d shows
the result with the uniform shape parameter α=
0.44302. The α-trigonometric polynomial interpolating curve is convexity-preserving. Apparently, Fig.2c
has better convexity-preserving effect than Fig.2d.
Example 2 The data points are shown in Table 2,
and the corresponding interpolating curves with different α are plotted in Fig.3.
Table 2 The data points in Fig.3
xn
yn
7.2
24.6
10.6
7.8
17.6
7.5
28.6
8.4
32.2
21.0
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Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
(a)
(b)
(a)
(b)
k(u)
0.168892
0
−0.115397
(c)
π/2
Fig.3 α-trigonometric polynomial interpolating curve
with given α. (a) α=0.8; (b) Each segment having respective shape parameter αj* ( j=1,2,3,4); (c) α=0.204647
α 4* = 0.725502; the α-trigonometric polynomial interpolating curve is convexity-preserving. Fig.3c shows
the result with the uniform shape parameter α=
0.204647, where the α-trigonometric polynomial
interpolating curve is convexity-preserving. Apparently, Fig.3b has better convexity-preserving effect
than Fig.3c.
Example 3 The data points are taken from (Fletcher
and McAllister, 1990) and shown in Table 3, and the
corresponding interpolating curves with different α
are plotted in Fig.4.
Table 3 The data points in Fig.4
xn
yn
0
0
6
13
15
−6
−6
−2
16
14
20
16
24
14
Fig.4 shows the convexity-preserving α-trigonometric polynomial interpolating curves and their
curvature plots. According to Theorems 1 and 2, we
can get Dj={α|0<α<1}, j=1,2,6; D3={α|0<α<
0.733471}, D5={α|0<α<0.603574}, and the 4th segment is an intergradation between two convex subsets
of the data points, and we take D4={α|0<α<1}. Fig.4a
shows the result with the respective shape parameter
for each curve segment, here, α *j = 0.99, j=1,2,6,
3π/2
2π 5π/2 3π 7π/2 4π
u
(c)
0.059104
k(u)
Fig.3a shows the result with the uniform shape
parameter α=0.8. There are inflection points on the
α-trigonometric polynomial interpolating curve.
Fig.3b shows the result with respective shape parameter for each curve segment, where D1={α|0<α<1},
D2={α|0<α<0.204648}, D3={α|0<α<0.244841}, D4=
{α|0<α<0.725503} according to Theorems 1 and 2;
and we take α1* = 0.99, α 2* = 0.204647, α 3* = 0.24484,
π
0
−0.067570
π/2
π
3π/2
2π 5π/2 3π
u
7π/2 4π
(d)
Fig.4 α-trigonometric polynomial interpolating curve
and their curvature plots k(u). (a) Each segment having respective shape parameter; (b) The whole curve
having the uniform shape parameter; (c) The curvature plots of curve in (a) with kmax=0.168892, kmin=
−0.115397; (d) The curvature plots of curve in (b) with
kmax=0.059104, kmin=−0.067570
α 3* = 0.733470, α 4* = 0.5, α 5* = 0.6. We must point
out that the value of α 4* does not affect the convexitypreserving property of α-trigonometric polynomial
interpolating curve, considering that α-trigonometric
polynomial curve is constructed by blending a parametrized polygon and the trigonometric polynomial
curve, we might as well take α 4* =0.5. Fig.4b shows
the result with the uniform shape parameter α=0.6.
Fig.4c shows the curvature plots of Fig.4a with
kmax=0.168892 and kmin=−0.115397. Fig.4d shows the
curvature plots of Fig.4b with kmax=0.059104 and
kmin=−0.067570.
Example 4 The data points are taken from the cycloid
 x = a (θ − sinθ ),
a = 100, 0 ≤ θ ≤ 2π,

 y = a (1 − cosθ ),
where θ = iπ/5, i = 0, 1, 2, … , 10.
Pan et al. / J Zhejiang Univ Sci A 2007 8(8):1199-1209
Fig.5 shows the convexity-preserving α-trigonometric polynomial interpolating curve with the respective shape parameter for each curve segment
(Fig.5a) and its curvature plot (Fig.5b). According to
Theorems 1 and 2, here, Dj={α|0<α<1} (j=1,3,4,5,6,7,
8,10), D2=D9={α|0<α<0.543712}, and we take
α *j = 0.99 (j=1,3,4,5, 6,7,8,10), α 2* = α 9* = 0.543711.
It is easy to see that for transcendental curves, the
effect of our method is very good.
y
200
x
0
100π
200π
(a)
k(u)
0
−0.000012
1209
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−0.135341
π/2 π 3π/2 2π 5π/2 3π 7π/2 4π 9π/2 5π 11π/2
u
(b)
Fig.5 (a) α-trigonometric polynomial interpolating curve
with the respective shape parameter for each curve segment; (b) The curvature plot k(u) of curve in (a) with
kmax=−0.000012, kmin=−0.135341
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