A non-stationary uniform tension controlled

interpolating 4-point scheme

reproducing conics

C. Beccari a, G. Casciola b, L. Romani b,∗aDepartment of Pure and Applied Mathematics, University of Padova,

Via G. Belzoni 7, 35131 Padova, ItalybDepartment of Mathematics, University of Bologna,P.zza di Porta San Donato 5, 40127 Bologna, Italy

Abstract

In this paper we propose a non-stationary uniform tension controlled interpolating4-point scheme which provides users with a single tension parameter that, for anyvalue in the range of definition, generates a C1-continuous limit curve. This curvescheme unifies non-stationary interpolating 4-point rules that are capable of re-presenting elements of the linear space spanned by trigonometric, polynomial andhyperbolic functions. As a consequence, for special values of the tension parameter,such a scheme will be capable of reproducing all conic sections exactly. Additio-nally, progressively increasing the value of the tension parameter, we will be able totighten the limit curve to the piecewise linear curve between the data points.

Key words: Subdivision; Interpolation; Curves; Locality; Tension control; Conicreproduction

1 Introduction

The subdivision schemes proposed in the literature up to now cannot satisfyat the same time all the geometric properties we have decided to include inour scheme. To be more precise, there exist schemes that possess a tensionparameter (Dyn et al., 1987) but which can not reproduce conic sections, andthere exist schemes for reproducing circles (Zhang, 1996; Ivrissimtzis et al.,2002; Jena et al., 2003) and generating arcs of ellipses (Zhang, 1996) which

∗ Corresponding author.Email addresses: beccari@math.unipd.it (C. Beccari),

casciola@dm.unibo.it (G. Casciola), romani@dm.unibo.it (L. Romani).

Preprint submitted to Elsevier Science 13 July 2005

do not possess any tension parameter. The only one which possesses a tensionparameter and can reproduce circles as well, is the non-stationary schemepresented in Morin et al. (2001). However this is not interpolatory. Aim ofthis work is therefore to generate a novel interpolatory subdivision schemewith a tension parameter, that is capable of reproducing circles and all otherconic sections exactly.

The paper is structured as follows. In Section 2 and 3 we respectively defineour novel subdivision scheme and we analyze its smoothness. In Section 4 weprove that the same regularity is achieved whenever we generate open curveswith a special endpoint rule. Next, in Section 5, we show some geometricproperties of the basic limit function and in Section 6 we demonstrate the roleof the tension parameter by a few examples.

2 Definition of the scheme

In this section we are going to define an interpolating 4-point scheme thatunifies three different curve schemes which are capable of representing trigono-metric, polynomial and hyperbolic functions respectively. We begin by intro-ducing the three schemes separately and then we derive a common insertionrule which unifies all the proposals.

2.1 The interpolating 4-point scheme exact for trigonometric functions

We start constructing a binary 4-point insertion rule which interpolates cosineand sine functions exactly. This means that, if we sample points from one ofthese trigonometric functions and then apply such a rule, we get in the limitthat function from which the initial control points were sampled. Such aninsertion rule can be obtained by interpolation with a function from the linearspace spanned by {1, x, cos(x), sin(x)}, working in the following way. Since thefunction interpolating the points pk

i+h at 2−kth + t0, h = −1, 0, 1, 2, t0 < t,coincides with a t0-translation of the function interpolating the same pointsat 2−kth, without any loss of generality we can define the points {pk

i | i ∈Z}k∈Z+ at refinement level k, on a grid 2−ktZ. Thus, interpolating the data(2−kth, pk

i+h), h = −1, 0, 1, 2 by a function of the form

f(x) = a0 + a1x + a2 cos(x) + a3 sin(x), (1)

2

we get the following system of equations

f(− t2k ) = pk

i−1

f(0) = pki

f( t2k ) = pk

i+1

f( t2k−1 ) = pk

i+2

from which it follows

pki−1 = a0 − a1

t2k + a2 cos( t

2k )− a3 sin( t2k )

pki = a0 + a2

pki+1 = a0 + a1

t2k + a2 cos( t

2k ) + a3 sin( t2k )

pki+2 = a0 + 2a1

t2k + a2(cos2( t

2k )− sin2( t2k )) + 2a3 sin( t

2k ) cos( t2k ).

(2)

Solving (2) with respect to a0, a1, a2, a3, we get the coefficients in (1):

a0 =−pk

i−1 + 2 cos( t2k )pk

i − pki+1

2(cos( t2k )− 1)

a1 =pk

i−1 − (1 + 2 cos( t2k ))pk

i + (1 + 2 cos( t2k ))pk

i+1 − pki+2

t2k−1 (cos( t

2k )− 1)

(3)a2 =pk

i−1 − 2pki + pk

i+1

2(cos( t2k )− 1)

a3 =− cos( t

2k )pki−1 + (1 + 2 cos( t

2k ))pki − (2 + cos( t

2k ))pki+1 + pk

i+2

2 sin( t2k )(cos( t

2k )− 1).

To get the insertion rule now, we only need to compute the value of theinterpolating function f(x) at the grid point t

2k+1 , defining the new point pki+ 1

2

as a linear combination of the four consecutive points pki−1, pk

i , pki+1, pk

i+2 inthe last set:

f(

t

2k+1

)=

2 cos( t2k ) cos( t

2k+1 )− cos( t2k+1 )− 1

4 cos( t2k+1 )(cos( t

2k )− 1)(pk

i + pki+1) +

− 1

8 cos( t2k+1 )

(1 + cos( t

2k+1 )) (pk

i−1 + pki+2) ≡ pk

i+ 12. (4)

3

In this way we will define the set of points at the (k+1)-th level of refinement,as:

pk+12i = pk

i (5)

pk+12i+1 = pk

i+ 12

=(

1

2+ wk

)(pk

i + pki+1)− wk (pk

i−1 + pki+2)

where

wk =1

8 cos(

t2k+1

) (1 + cos

(t

2k+1

)) . (6)

Remark 1 The subdivision scheme (5) with weights in (6) coincides withthe interpolating 4-point scheme on the circle presented in Ivrissimtzis et al.(2002).

2.2 The interpolating 4-point scheme exact for polynomial functions

This time we derive the insertion rule by interpolation with a function fromthe span of the four functions {1, x, x2, x3}, proceeding in the same way asabove. Given the points {pk

i | i ∈ Z}k∈Z+ , which are defined on a grid 2−ktZ,and interpolating the data (2−kth, pk

i+h), h = −1, 0, 1, 2 through a function ofthe form

f(x) = a0 + a1x + a2x2 + a3x

3, (7)

we get the following system of equations

f(− t2k ) = pk

i−1

f(0) = pki

f( t2k ) = pk

i+1

f( t2k−1 ) = pk

i+2

which, once solved, gives the coefficients

a0 = pki

a1 =(−2pk

i−1 − 3pki + 6pk

i+1 − pki+2)2

k

6t

4

(8)

a2 =(pk

i−1 − 2pki + pk

i+1)22k

2t2

a3 =(−pk

i−1 + 3pki − 3pk

i+1 + pki+2)2

3k

6t3.

To get the insertion rule now, we only need to compute the value of theinterpolating function f(x) at the grid point t

2k+1 , defining the new point pki+ 1

2

as a linear combination of the four consecutive points pki−1, pk

i , pki+1, pk

i+2 inthe last set:

f(

t

2k+1

)=

9

16(pk

i + pki+1) −

1

16(pk

i−1 + pki+2) ≡ pk

i+ 12. (9)

In this way we will define the set of points at the (k+1)-th level of refinement,as:

pk+12i = pk

i (10)

pk+12i+1 =

(1

2+ wk

)(pk

i + pki+1)− wk (pk

i−1 + pki+2)

where

wk =1

16. (11)

Remark 2 Note that in this case the scheme reduces to the special case ofDyn and Levin’s stationary interpolating 4-point scheme (Dyn et al., 1987)which coincides with Dubuc-Deslauriers’s proposal (Dubuc, 1986; Deslauriersand Dubuc, 1989).

2.3 The interpolating 4-point scheme exact for hyperbolic functions

Finally we derive the insertion rule by interpolation with a function from thespan of the four functions {1, x, cosh(x), sinh(x)}, proceeding in the same wayas before. Given the points {pk

i | i ∈ Z}k∈Z+ , which are defined on a grid 2−ktZ,and interpolating the data (2−kth, pk

i+h), h = −1, 0, 1, 2 through a function ofthe form

f(x) = a0 + a1x + a2 cosh(x) + a3 sinh(x), (12)

5

we get the following system of equations

f(− t2k ) = pk

i−1

f(0) = pki

f( t2k ) = pk

i+1

f( t2k−1 ) = pk

i+2

which, once solved, gives the coefficients

a0 =−pk

i−1 + 2 cosh( t2k )pk

i − pki+1

2(cosh( t2k )− 1)

a1 =pk

i−1 − (1 + 2 cosh( t2k ))pk

i + (1 + 2 cosh( t2k ))pk

i+1 − pki+2

t2k−1 (cosh( t

2k )− 1)

(13)a2 =pk

i−1 − 2pki + pk

i+1

2(cosh( t2k )− 1)

a3 =− cosh( t

2k )pki−1 + (1 + 2 cosh( t

2k ))pki − (2 + cosh( t

2k ))pki+1 + pk

i+2

2 sinh( t2k )(cosh( t

2k )− 1).

To get the insertion rule now, we only need to compute the value of theinterpolating function f(x) at the grid point t

2k+1 , defining the new point pki+ 1

2

as a linear combination of the four consecutive points pki−1, pk

i , pki+1, pk

i+2 inthe last set:

f(

t

2k+1

)=

2 cosh( t2k ) cosh( t

2k+1 )− cosh( t2k+1 )− 1

4 cosh( t2k+1 )(cosh( t

2k )− 1)(pk

i + pki+1) +

− 1

8 cosh( t2k+1 )

(1 + cosh( t

2k+1 )) (pk

i−1 + pki+2) ≡ pk

i+ 12. (14)

In this way we will define the set of points at the (k+1)-th level of refinement,as:

pk+12i = pk

i (15)

pk+12i+1 =

(1

2+ wk

)(pk

i + pki+1)− wk (pk

i−1 + pki+2)

6

where

wk =1

8 cosh(

t2k+1

) (1 + cosh

(t

2k+1

)) . (16)

2.4 The unified interpolating 4-point scheme reproducing conics

The common thread linking all three of these subdivision schemes is that theycan be expressed by a single insertion rule of the form

pk+12i = pk

i (17)

pk+12i+1 =

(1

2+ wk

)(pk

i + pki+1)− wk (pk

i−1 + pki+2)

where

wk =1

8vk+1(1 + vk+1)(18)

with vk+1 equal to cos(

t2k+1

), 1, cosh

(t

2k+1

)depending on the scheme we

choose. However, the following result makes these three cases unnecessaryonce the initial parameter v0 has been chosen.

Proposition 3 If v0 ∈]− 1, +∞[ , for all three of the previously stated casesthe parameters vk and vk+1 satisfy the recurrence

vk+1 =

√1 + vk

2∀k ∈ Z+. (19)

PROOF. In the three cases above, equation (19) follows from the fact that

√1 + cos( t

2k )

2= cos

(t

2k+1

)∈ ]− 1, 1[ ∀k ∈ Z+

√1 + 1

2= 1

√1 + cosh( t

2k )

2= cosh

(t

2k+1

)∈ ]1, +∞[ ∀k ∈ Z+

respectively. ¤

In this way, we can formulate a unified non-stationary subdivision schemewhich combines the three previous schemes in a very elegant manner.

7

Definition 4 Given a set of control points P 0 = {p0i | i ∈ Z} at refinement

level 0, we define a unified subdivision scheme that generates a new set ofcontrol points P k+1 = {pk+1

i | i ∈ Z}k∈Z+ at the (k + 1)-th level of refinement,by the subdivision rules

pk+12i = pk

i (20)

pk+12i+1 =

(2vk+1 + 1)2

8vk+1(1 + vk+1)(pk

i + pki+1)−

1

8vk+1(1 + vk+1)(pk

i−1 + pki+2)

where the weight vk+1 ∈ ]0, +∞[ is easily updated at each subdivision stepthrough equation (19).

Thus, given the parameter vk, the subdivision rules (20) are derived by firstcomputing vk+1 using equation (19) and by then substituting vk+1 into equa-tion (20). As a consequence, depending on the choice of the initial parameterv0, we can generate the interpolatory 4-point scheme exact for trigonometric,polynomial and hyperbolic functions whenever −1 < v0 < 1, v0 = 1, v0 > 1respectively (note that for every choice of the initial value v0, we automaticallyhave vk+1 ∈]0, +∞[ for any k ∈ Z+).

Remark 5 Whenever v0 = 1, the subdivision scheme defined in (20) coincideswith the special case of Dyn and Levin’s subdivision scheme (Dyn et al., 1987).Since it has been proved to have cubic precision, by choosing four initial pointson the parabola (t, at2), the limit curve we obtain is exactly the parabola itself.

Conversely, for the other conic sections the following result holds.

Proposition 6 Choosing the initial tension parameter v0 = cos(t), t ∈ ]− π,0 [, the subdivision scheme defined in (20) reproduces exactly the trigonomet-ric functions f(t) = cos(t) and f(t) = sin(t) whenever the given data points(it, p0

i ), i = −1, 0, 1, 2 lie on such functions.Analogously choosing the initial tension parameter v0 = cosh(t), t 6= 0, thesubdivision scheme defined in (20) reproduces exactly the hyperbolic functionsf(t) = cosh(t) and f(t) = sinh(t) whenever the given data points (it, p0

i ), i =−1, 0, 1, 2 lie on such functions.

PROOF. The result above is trivially verified by applying the definition ofthe scheme.We show here how to work out the computation for the cosine function. Inall the other cases it will be sufficient to proceed analogously. Taken pk

i =

cos(i t2k ), i ∈ Z, k ∈ Z+, we have to show that pk+1

2i = cos(2i t

2k+1

)and

pk+12i+1 = cos

((2i + 1) t

2k+1

). The first relation follows directly from

pk+12i = pk

i = cos(i

t

2k

)= cos

(2i

t

2k+1

).

8

For the second relation we have

pk+12i+1 =

(2vk+1 + 1)2

8vk+1(1 + vk+1)(pk

i + pki+1)−

1

8vk+1(1 + vk+1)(pk

i−1 + pki+2)

with

pki + pk

i+1 = cos(

it

2k

)+ cos

((i + 1)t

2k

)= 2 cos

((2i + 1)t

2k+1

)cos

(t

2k+1

),

pki−1 + pk

i+2 = cos

((i− 1)t

2k

)+ cos

((i + 2)t

2k

)= 2 cos

((2i + 1)t

2k+1

)cos

(3t

2k+1

).

Thus

pk+12i+1 =

2(2vk+1 + 1)2 cos

(t

2k+1

)

8vk+1(1 + vk+1)−

2 cos(

3t2k+1

)

8vk+1(1 + vk+1)

cos

((2i + 1)t

2k+1

).

Now since vk+1 = cos(

t2k+1

), then

2(2vk+1+1)2 cos( t

2k+1 )8vk+1(1+vk+1)

− 2 cos( 3t

2k+1 )8vk+1(1+vk+1)

= 1 and

pk+12i+1 = cos

((2i + 1)

t

2k+1

).

This proves our thesis. ¤

Fig. 1. Reproduction of trigonometric and hyperbolic functions: f(t) = cos(t),f(t) = sin(t), f(t) = cosh(t), f(t) = sinh(t).

Fig. 2. Reproduction of conic sections: (t, at2), (a cos t, b sin t), (a cosh t, b sinh t).

9

Corollary 7 From the parametric representation of the circle with center 0and radius 1, if we take as initial data the four points p0

i = (cos(it), sin(it)),i = −1, 0, 1, 2 which, for t = π

2, are equidistant points on the unit circle, we

get pki =

(cos

(i t2k

), sin

(i t2k

)). Thus choosing the initial tension parameter

v0 = cos(π2) = 0, the resulting limit curve is the unit circle itself.

Analogously, if we take as initial data four points p0i = (a cos(iπ

2), b sin(iπ

2)),

i = −1, 0, 1, 2 from the parametric representation of the ellipse with center 0and radii a, b, by choosing the initial tension parameter v0 = cos(π

2) = 0, the

resulting limit curve is exactly the ellipse itself (see Fig.2).In the same way, by taking as initial data four points p0

i = (a cosh(it), b sinh(it)),i = −1, 0, 1, 2 on the parametric representation of the hyperbola (x(t), y(t)) =(a cosh(t), b sinh(t)), by choosing the initial tension parameter v0 = cosh(t),the resulting limit curve is the hyperbola itself (see Fig.2).

Remark 8 Figs. 1-2 have been obtained by extending uniformly on each sidethe open control polygon P 0 = {p0

i | i = 0, ..., n} by two extra segments whoseendpoints lie on the curve. Whenever two auxiliary points have been definedfor each endpoint, we can deal with open polygons leaving the subdivision rules(20) unmodified. In this way the open curve generated in the limit through (20)with i = −1, ..., n, trivially turns out to have p0

0 as first endpoint and p0n as the

last one (see Section 4).

3 Convergence analysis

Aim of this section is to show that, given an initial polygon P 0, the subdivi-sion rules in (20) allow us to define an increasingly dense collection of polygonsP k that converges to a C1 limit curve. In order to prove this, we analyze theconvergence properties of our scheme by exploiting the well-known theorem in(Dyn, N., Levin, D., 1995), which relates the convergence of a non-stationaryscheme to its asymptotically equivalent stationary scheme. To this aim, weobserve that the parameters vk converge to 1 as k → ∞, i.e., in practicethe non-stationary subdivision scheme (20) converges to Dyn and Levin’s sta-tionary scheme with w = 1

16, and there exists a useful bound on the rate of

convergence 1−vk

1−vk−1.

Lemma 9 Due to the definition of the parameter vk in the three separatedcases, it holds

limk→+∞

vk = 1. (21)

10

PROOF. Since

vk =

cos( t2k ) if −1 < v0 < 1

1 if v0 = 1cosh( t

2k ) if v0 > 1, (22)

equation (21) trivially follows. ¤

Lemma 10 Due to the recurrence (19), the parameters vk−1 and vk satisfythe relation

1− vk

1− vk−1

<1

2(23)

for any vk−1 ∈]− 1, +∞[ \{0}.

PROOF. Exploiting recurrence (19) we can write the ratio 1−vk

1−vk−1in the

following way:

1− vk

1− vk−1

=1−

√1+vk−1

2

1− vk−1

=

√2−√1 + vk−1√2(1− vk−1)

=1

2 +√

2√

1 + vk−1

and observe that since√

2√

1 + vk−1 > 0 for any k ∈ Z+\{0}, thus

1− vk

1− vk−1

<1

2. ¤

Proposition 11 The non-stationary scheme (20) is asymptotically equiva-lent to the stationary scheme (10). Moreover, it generates C1-continuous limitcurves.

PROOF. Let’s start compactly writing the subdivision step (20) in a singleequation of the form

pki =

∑

j∈Zmk−1

2j−i pk−1j ∀k ∈ Z+\{0} (24)

where

mk−1 =

[− 1

8vk(1 + vk), 0,

(2vk + 1)2

8vk(1 + vk), 1,

(2vk + 1)2

8vk(1 + vk), 0, − 1

8vk(1 + vk)

]

(25)

11

is the so-called mask of the unified scheme satisfying

∑

i∈Zmk−1

2i = 1,∑

i∈Zmk−1

2i+1 = 1.

Now, in order to prove that the unified subdivision scheme converges to a C1

limit curve, we compute the first divided difference mask and we show that thelimit curves associated with the divided difference scheme are C0-continuous.In fact, since the mask of the first divided difference scheme related to (25) isgiven by

dk−1 =

[− 1

8vk(1 + vk),

1

8vk(1 + vk),

1

2,

1

2,

1

8vk(1 + vk), − 1

8vk(1 + vk)

],

by applying Lemma 9 it follows that

d∞ ≡ limk→+∞

dk−1 =[− 1

16,

1

16,

1

2,

1

2,

1

16, − 1

16

],

that is in the limit our scheme tends to Dubuc-Deslauriers’s interpolating4-point scheme (Dubuc, 1986; Deslauriers and Dubuc, 1989) whose first dif-ference is given by d∞. Now, by computing dk−1 − d∞ we have that

‖ dk−1 − d∞ ‖∞=|1− vk|(2 + vk)

8vk(1 + vk)(26)

where vk ∈ ]0, +∞[ , ∀k ∈ Z+\{0}. Next if

+∞∑

k=1

‖ dk−1 − d∞ ‖∞< ∞, (27)

the two difference schemes are asymptotically equivalent, and then since d∞

belongs to C0, we can conclude that the scheme associated with dk−1 is C0

too (see Dyn, N., Levin, D., 1995).To show this, we study the convergence of (27) as the parameter vk varies inthe interval ]0, +∞[.

• When 0 < vk < 1,

‖ dk−1 − d∞ ‖∞=1

8

(1− vk)(2 + vk)

vk(1 + vk)

and

+∞∑

k=1

‖ dk−1 − d∞ ‖∞ =1

8

+∞∑

k=1

(1− vk)(2 + vk)

vk(1 + vk).

12

Thus, by applying Lemma 10,

+∞∑

k=1

‖ dk−1 − d∞ ‖∞ <(1− v0)

8

∞∑

k=1

1

2k

2 + vk

vk(1 + vk)

<3

8(1− v0)

∞∑

k=1

1

2k

1

vk(1 + vk)

<3

8(1− v0)

∞∑

k=1

1

2k

1

vk

=3

8(v0 − 1)

∞∑

k=1

1

2k

(− 1

vk

)

<3

8(v0 − 1)

∞∑

k=1

1

2k< ∞.

• When vk = 1,

‖ dk−1 − d∞ ‖∞= 0

and+∞∑

k=1

‖ dk−1 − d∞ ‖∞= 0 < ∞.

• When vk > 1,

‖ dk−1 − d∞ ‖∞=1

8

(vk − 1)(2 + vk)

vk(1 + vk)

and

+∞∑

k=1

‖ dk−1 − d∞ ‖∞ =1

8

+∞∑

k=1

−(1− vk)(2 + vk)

vk(1 + vk).

Thus, by applying Lemma 10,

+∞∑

k=1

‖ dk−1 − d∞ ‖∞ <−(1− v0)

8

∞∑

k=1

1

2k

2 + vk

vk(1 + vk)

<−1

8(1− v0)

∞∑

k=1

1

2k(2 + vk)

=1

8(1− v0)

∞∑

k=1

1

2k(−2− vk)

<−3

8(1− v0)

∞∑

k=1

1

2k< ∞.

So the schemes dk−1 and d∞ are asymptotically equivalent. In this way, sincedk−1 belongs to C0, we can conclude that mk−1 belongs to C1, that is theoriginal scheme produces C1 limit curves. ¤

13

4 A subdivision rule for curve endpoints

Equation (20) gives subdivision rules for C1-continuous closed curves. In caseof open curves, such rules can be applied only on the interior of the curve,while for the endpoints we should include an alternative rule. In fact, whendealing with open polygons, it is not possible to refine the first or the lastedge by the original 4-point scheme, since, by the rule in the second line ofequation (20), it turns out that to insert a new point pk+1

2i+1 in the refinedpolygon, it is necessary to possess a well-defined 2-neighborhood, given bytwo points on its left and on its right side in the coarsest polygon. To this aimwe observe that if we define just two auxiliary points p0

−2, p0−1 in the coarsest

polygon P 0 = {p0i | i = 0, ..., n}, each new point pk+1

2i+1 has a well-defined 2-neighborhood and the open curve generated in the limit through (20) withi = −1, ..., n trivially turns out to have p0

0 as first endpoint and to be C1-continuous.However, the extension of this strategy to surface subdivision, implies thedefinition of two rings of points around the boundary control net. Thus, toavoid so many computations when subdividing the first edge pk

0, pk1, we propose

here a special rule for computing the point pk+11 independently of the two

auxiliary points p0−2, p0

−1; to work out the endpoint rule for the last edge, itwill be sufficient to proceed analogously.Let pk

0, pk1 be the first edge of the non-refined polygon P k = {pk

i | i = 0, ..., 2kn}.Once defined an auxiliary point pk

−1, we can compute the point pk+11 through

(20) with i = 0, applying subdivision to the subpolygon pk−1, pk

0, pk1, pk

2.We choose here

pk−1 = 2pk

0 − pk1. (28)

In this way the additional rule for the endpoint can be expressed as the follo-wing stationary linear combination of points from the non-extrapolated openpolygon pk

0, pk1, pk

2:

pk+11 =

4v2k+1 + 4vk+1 − 1

8vk+1(1 + vk+1)pk

0 +4v2

k+1 + 4vk+1 + 2

8vk+1(1 + vk+1)pk

1 −1

8vk+1(1 + vk+1)pk

2. (29)

Proposition 12 Rule (29) does not affect the convergence of the originalscheme to a continuously differentiable limit.

PROOF. Our goal is proving that there exist two auxiliary points p0−2, p0

−1

such that, refining the polygon P 0 through (20), after k rounds of subdivisionthe expression of the point pk

−1 turns out to coincide with (28).Let

p0−2 = 2p0

0 − p02 (30)

14

and

p0−1 = 2p0

0 − p01. (31)

Now applying (20) with i = −1 to the subpolygon p0−2, p0

−1, p00, p0

1, we get apoint p1

−1 which coincides with the definition in (28) for k = 1.Next, by induction on the subdivision level k, it is easy to verify that, refiningk times the subpolygon p0

−2, p0−1, p0

0, p01 through (20) with i = −1, the point

pk−1 computed turns out to be the same given by (28). Thus, through the

convergence properties of (20), we can conclude that, starting from an opencontrol polygon and refining the boundary by (29), we get a C1-continuouslimit curve. ¤

5 Basic limit function

The basic limit function defined by scheme (20) is the limit function of thescheme for the data

p0i =

{1, if i = 0,0, if i 6= 0.

(32)

By Proposition 11 it follows that the basic limit function belongs to C1(R).We show now that it is symmetric about the Y -axis and it possesses a compactsupport over the interval [-3,3] (see Fig. 3).

Fig. 3. Basic limit functions for: v0 = −0.75, v0 = 0, v0 = 1, v0 = 100.

Proposition 13 The basic limit function F defined by the scheme (20) issymmetric about the Y -axis.

PROOF. Let us define the set Dn := { i2n | i ∈ Z} such that the restriction

of the basic limit function F to Dn satisfies F(

i2n

)= pn

i for all i ∈ Z andprove the thesis by induction on n.

First of all we observe that since F (i) = p0i = p0

−i = F (−i) ∀i ∈ Z, we have

F(

i2n

)= F

(− i

2n

)∀i ∈ Z, n = 0.

15

Now, assuming that F(

i2n

)= F

(− i

2n

)∀i ∈ Z, it follows that

F(

2i2n+1

)= pn+1

2i = pni = F

(i

2n

)= F

(− i

2n

)= pn

−i = pn+1−2i = F

(− 2i

2n+1

)

and

F(

2i+12n+1

)= pn+1

2i+1 =

= (2vn+1+1)2

8vn+1(1+vn+1)

(pn

i + pni+1

)− 1

8vn+1(1+vn+1)

(pn

i−1 + pni+2

)=

= (2vn+1+1)2

8vn+1(1+vn+1)

[F

(i

2n

)+ F

(i+12n

)]− 1

8vn+1(1+vn+1)

[F

(i−12n

)+ F

(i+22n

)]=

= (2vn+1+1)2

8vn+1(1+vn+1)

[F

(−i2n

)+ F

(−i−12n

)]− 1

8vn+1(1+vn+1)

[F

(−i+12n

)+ F

(−i−22n

)]=

= (2vn+1+1)2

8vn+1(1+vn+1)

(pn−i + pn

−i−1

)− 1

8vn+1(1+vn+1)

(pn−i+1 + pn

−i−2

)=

= pn+1−2i−1 = F

(− 2i+1

2n+1

).

Hence F(

i2n

)= F

(− i

2n

)∀i ∈ Z and n ∈ Z+.

As a consequence, from the continuity of F we have that F (x) = F (−x) forall x ∈ R, which completes the proof. ¤

Proposition 14 The basic limit function F defined by the scheme (20) van-ishes outside the interval [-3,3].

PROOF. Let us define the set Dn := { i2n | i ∈ Z} such that the restriction

of the basic limit function F to Dn satisfies F(

i2n

)= pn

i for all i ∈ Z.

Take t0 = 0 and define tn recursively by tn = tn−1 + 32n . Observe that tn ∈ Dn,

but tn /∈ Dn−1. We claim that the restriction of F to Dn vanishes outside[−tn, tn]. Since F is symmetric, it is enough to prove that F vanishes outside[0, tn]. We prove this by induction on n.

Observe that F (t0) = F (0) = p00 = 1 and F (t1) = F

(t0 + 3

2

)= F

(32

)= p1

3.

Thus F (t1) = p13 = (2v1+1)2

8v1(1+v1)(p0

1 + p02)− 1

8v1(1+v1)(p0

0 + p03) = − 1

8v1(1+v1)6= 0.

Analogously, for all k > 0 we have that

F(t1 + 2k−1

2

)= F

(2(t1+k)−1

2

)= p1

2(t1+k)−1 =

= (2v1+1)2

8v1(1+v1)(p0

t1+k−1 + p0t1+k)− 1

8v1(1+v1)(p0

t1+k−2 + p0t1+k+1) = 0

and

16

F(t1 + 2k

2

)= F

(2(t1+k)

2

)= p1

2(t1+k) = p0t1+k = 0.

This proves our claim for n = 1.

We now assume that

F (tn) = − 1

8vn(1 + vn)F (tn−1) 6= 0

and for all k > 0

F

(tn +

2k − 1

2n

)= 0 and F

(tn +

2k

2n

)= 0.

Therefore, since

F(tn+1 + 2k

2n+1

)= F

(tn + 2k+3

2n+1

)= F

(tn + 2(k+1)+1

2n+1

)=

(2vn+1+1)2

8vn+1(1+vn+1)

[F

(tn + k+1

2n

)+ F

(tn + k+2

2n

)]−

18vn+1(1+vn+1)

[F

(tn + k

2n

)+ F

(tn + k+3

2n

)]

and

F(tn+1 + 2k+1

2n+1

)= F

(tn + k+2

2n

),

we get that

F (tn+1) = (2vn+1+1)2

8vn+1(1+vn+1)

[F

(tn + 1

2n

)+ F

(tn + 2

2n

)]

− 18vn+1(1+vn+1)

[F (tn) + F

(tn + 3

2n

)]

= − 18vn+1(1+vn+1)

F (tn) 6= 0

and for all k > 0

F(tn+1 + 2k−1

2n+1

)= F

(tn+1 + 2(k−1)+1

2n+1

)= F

(tn + (k−1)+2

2n

)= F

(tn + k+1

2n

)= 0

and

F(tn+1 + 2k

2n+1

)= (2vn+1+1)2

8vn+1(1+vn+1)

[F

(tn + k+1

2n

)+ F

(tn + k+2

2n

)]−

18vn+1(1+vn+1)

[F

(tn + k

2n

)+ F

(tn + k+3

2n

)]= 0.

Since [−tn, tn] is contained in [−3, 3] for all n and limn→∞ [−tn, tn] = [−3, 3]

(as tn =∑n

k=1 3(

12

)k=⇒ limn→∞ tn =

∑∞k=1 3

(12

)k= 3

12

1− 12

= 3 ), we

have the required result. ¤

17

Remark 15 The translations of the basic limit function F form a partitionof unity.

PROOF. If f(x) is the limit function interpolating the initial sequence ofpoints {p0

i }i∈Z ≡ 1, then f(x) =∑

i∈Z p0i F (x − i) ∀x ∈ R, but since p0

i =1 ∀i ∈ Z and f(x) = 1 (as the scheme defined in (20) reproduces exactly theunit function whenever the given data lie on such a function), it follows that1 =

∑i∈Z F (x− i) ∀x ∈ R. ¤

6 Applications and examples

The following examples show open and closed curves which pass through a setof given points. The control polygons (corresponding to the piecewise linearcurve between the given points) are drawn by a dashed line, and the smoothcurves obtained by our algorithm by a full line.Figure 4, depicting the open limit curves obtained with v0 = −0.4, 1, 1000,demonstrates the increase in the tightness (tension) of the curve with theincrease in v0.

Fig. 4. Increasing the tightness (tension) of an open curve.

Analogously, figure 5, depicting the closed limit curves obtained with v0 =−0.5, 0, 1, 5, 50, 500, demonstrates the increase in the tightness (tension)of the curve with the increase in v0.The value of v0 which is relevant to be mentioned is v0 = 1, since in this casethe limit curve obtained exactly defines the Dubuc-Deslauriers interpolatory4-point curve (Dubuc, 1986; Deslauriers-Dubuc, 1989).

18

Fig. 5. Increasing the tightness (tension) of a closed curve.

The following figures show the effect of the tension parameter v0 when ouralgorithm is applied on a regular N -sided control polygon inscribed in theunit circle. While choosing v0 < 1 the parameter acts as a looseness and,smaller it is, looser the limit curve is, for high values of the tension parameterv0, the limit curve tends to shrink to the initial control polygon. In addition,whenever we choose v0 = cos(2π

N), in the limit we obtain exactly the unit circle

(see Figs. 6, 7).

Fig. 6. Interpolation of the vertices of a square with the uniform tension controlledinterpolating 4-point scheme defined by the following values of the parameter v0:-0.95, -0.75, -0.5, 0, 1, 5, 25, 500.

19

Fig. 7. Interpolation of the vertices of a regular pentagon with the uniform ten-sion controlled interpolating 4-point scheme defined by the following values of theparameter v0: -0.5, -0.25, 0, cos(2π

5 ), 1, 5, 25, 500.

7 Conclusions and Future Work

This paper describes a simple and efficient non-stationary subdivision schemefor curve interpolation depending on a single tension parameter, that is ca-pable of reproducing conic sections exactly. The subdivision algorithm (20) isactually an insertion algorithm since all the points at stage k of the algorithmare carried over to stage k + 1 and new points are inserted in between theold ones. Evidently, the resulting limit curve interpolates the initial points.The local nature of the scheme, the possibility of reproducing a certain classof trigonometric, polynomial and hyperbolic functions, and the control of thetension by the associated parameter, are important features for curve design.The algorithm proposed here is unique in combining these five ingredients: sub-division, locality, interpolation, global tension control, reproduction of conicsections.An interesting generalization of this proposal could include the possibility ofworking with a different tension parameter v0 for each segment of the initialpolygon P 0. In this way, since during each subdivision step each segment issplit into two new segments, these two will inherit a new tension via equation(19). The resulting subdivision scheme will allow therefore different tensionson distinct curve segments.The curve scheme proposed can also be naturally extended to tensor-productsurfaces. Next step will be therefore generalizing the univariate scheme to asurface scheme over arbitrary quadrilateral meshes.

20

Acknowledgements

This research was supported by MIUR-PRIN 2004 and by University of Bologna”Funds for selected research topics”.

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