A Note on Subdivision

Kwan Pyo Ko

Dongseo University

kpko@dongseo.ac.kr

What is Subdivision?• Subdivision produces a smooth curve or surface as the limit of

a sequence of successive refinements• We can repeat a simple operation and obtain a smooth result

after doing it an infinite number of times

pj +1 = Spj

(Sp)a =X

b®a¡ 2bpb

Subdivision Surfaces

historyClassical schemes:

• De Rahm(1947)• Chaikin(1974)- An algorithm for high speed curve generation• Riesenfeld(1975)- On Chaikin’s algorithm• Catmull-Clark (1978)- Recursively generated B-spline surfaces on arbitrary topological meshes• Doo-Sabin (1978)- Behavior of recursive division surfaces near extraordinary points• Ball-Storry(1986)- A matrix approach to the analysis of recursively generated B-spline surfaces• Loop (1987)- Smooth subdivision surfaces based on triangles (new domain, eigen analysis)• Dyn,Levin,Gregory(1987)- A 4-point interpolatory subdivision scheme for curve design• (1990)- A butterfly subdivision scheme for surface interpolation with tension control• Reif(1995)- A unified approach to subdivision algorithms near extraordinary vertices.

history

New Schemes: Kobbelt(1996)- Interpolatory subdivision on open quadrilateral nets with arbitrary topology (2000)- subdivision Velho(2000)- Quasi 4-8 subdivision Velho-Zorin(2000)- 4-8 subdivision Dogson,Ivrissimtzis,Sabin(2003)- Characteristic of dual triangular subdivision (2004)- subdivision Non-uniform schemes: Warren(1995)- Binary subdivision schemes for functions of irregular knot sequences Gregory,Qu(1996)- Non-uniform corner cutting Sederberg,Sewell,Sabin(1998)- Non-uniform recursive subdivision surfaces Dyn(1999)- Using Laurent polynomial representation for the analysis of non-uniform binary SS Non-stationary schemes:Dyn,Levin(1995)- Analysis of asymptotically equivalent binary subdivision schemesMorin,Warren,Weimer(2001)- A subdivision scheme for surfaces of revolutionDyn,Levin,Luzzatto(2003)- Non-stationary interpolatory SS reproducing spaces of exponential polys.

p3

p3p

5

Schemes• Catmull-Clark (1978)• Doo-Sabin (1978)• Loop (1987)• Butterfly (1990)• Kobbelt (1996)• Mid-edge (1996 / 1997)• Subdivision (2000)p

3

Why Subdivision?• Arbitrary topology

• Multi-resolution

• Simple/Efficient code

• Construct Wavelet

• Analysis

Chaikin’s Algorithm

-converges to the quadratic B-Spline.

pk+12i+1 = 1

4pki + 3

4pki+1

pk+12i = 3

4pki + 1

4pki+1

Cubic Spline Algorithm

-converges to the cubic B-Spline.-converges to the cubic B-Spline.

pk+12i+1 = 1

8pki + 3

4pki+1 + 1

8pki+2

pk+12i = 1

2pki + 1

2pki+1

4-point Interpolatory Scheme (N. Dyn, D.Levin and J.Gregory)

-continuous forcontinuous for-CC11 for for

pk+12i = pk

i

jwj < 14

0< w < 18

pk+12i+1 =

µ 12 + w

¶(pk

i +pki+1) ¡ w(pk

i ¡ 1 + pki+2)

Deslauriers-Dubuc Scheme

• Finding a minimally supported mask such that (2N+1)th degree polynomial filling property (de Villiers, Goosen and Herbst)

• DD mask has the explicit formulation

X

kaj ¡ 2kp(k) = p( j

2); j 2 Z;p2 ¼2N +1

aj = 0; jj j ¸ 2N + 2

a2j +1 = N + 124N +1

µ2N + 1N

¶ (¡ 1)j

2j +1µ 2N +1

N + j +1¶

;j = ¡ N ¡ 1;::: ;N

a2j = ±j ;0; j = ¡ N;::: ;N

Polynomial Reproducing• DD scheme obtained the mask by using

polynomial reproducing property

• 4-point DD scheme:f k+1

2i = f ki ;

f k+12i+1 = 9

16(f ki + f k

i+1) ¡ 116(f k

i¡ 1 +f ki+2):

Ternary Interpolatory Subdivision Scheme

pj +13j = pi

j

pj +13j +1 = api

j ¡ 1 + bpij +cpi

j +1 +dpij +2

pj +13j +2 = dpi

j ¡ 1 +cpij + bpi

j +1 + apij +2

a = ¡ 118 ¡ 1

6w;b= 1318 + 1

2w;c= 718 ¡ 1

2w;d= ¡ 118 + 1

6w:

C2 for 115 < w < 1

9

where the weights are given by

Mask of binary 2n-point SS• Y.Tang, K.P.Ko and B.G.Lee obtained the mask of inte

rpolatory symmetric SS by using symmetry and the necessary condition for smoothness.

[w+ 12;¡ w]

[2w+ 916;¡ 3w¡ 1

16;w]

[5w+ 75128;¡ 9w¡ 25

256;5w¡ 3256;¡ w]

[14w+ 12252048;¡ 28w¡ 245

2048;20w¡ 492048;¡ 7w¡ 5

2048;w]

General form for mask of 2n-point ISSS

• The coefficients of w can be expressed in a matrix form

A =

2666666666664

1 0 0 0 0 0 ¢¢¢ 01 ¡ 1 0 0 0 0 ¢¢¢ 02 ¡ 3 1 0 0 0 ¢¢¢ 05 ¡ 9 5 ¡ 1 0 0 ¢¢¢ 014 ¡ 28 20 ¡ 7 1 0 ¢¢¢ 042 ¡ 90 75 ¡ 35 9 ¡ 1 ¢¢¢ 0... ... ... ... ... ... ... 0

¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ ¢¢¢ (¡ 1)n¡ 1

3777777777775

• We found out that matrix A satisfied the particular rule.

A =

26664

A11 0 ¢¢¢ 0

A21 A2

2 ¢¢¢ 0... ... ... ...

An1 An

2 ¢¢¢ Ann

37775 :

Ani =

8>><>>:

An¡ 1i ¡ An¡ 1

i+1 i = 1;n = 2;3:::(¡ 1)n¡ 1 i = n2An¡ 1

i ¡ An¡ 1i ¡ 1 ¡ An¡ 1

i+1 i = 2;3:::;n;n = 3;4;:: :0 i > n:

Relation between 2n-point ISSS and DD scheme

• We have the mask of 2n-point ISSS:

aI SSS;2n2i¡ 1 = aI SSS;2n

¡ 2i+1

aI SSS;2n2i¡ 1 = aD D ;2n¡ 2

2i¡ 1 +Ani w; i = 1;2;3:::;n;n 2 N +;

aD D ;21 = 1

2; aD D ;23 = ¢¢¢= aD D ;2n¡ 2

2n¡ 1 = 0

Example (for n=3)

we get 6-point ISSS:

aI SSS;61 = aD D ;4

1 +A31w = 9

16 +2w;

aI SSS;63 = aD D ;4

3 +A32w = ¡ 1

16 ¡ 3w;aI SSS;6

5 = aD D ;45 +A3

3w = w:

[w;¡ 116 ¡ 3w; 9

16 +2w; 916 +2w;¡ 1

16 ¡ 3w;w]:

Matrix formula for the mask of 2n-point

DD scheme• The formula can be written in the form:

a =

26664

aD D ;2n1

aD D ;2n3 ...

aD D ;2n2n¡ 1

37775 =

26664

A11 A2

1 ¢¢¢ An1

0 A22 ¢¢¢ An

2... ... ... ...0 0 0 An

n

37775

26664

w1w2...

wn

37775 = AT ¤w:

wi = i24i ¡ 3

µ2i ¡ 1i ¡ 1

¶ 12i ¡ 1

Example (for n=3)

we obtain 6-point DD scheme:

A =

24

A11 0 0

A21 A2

2 0A3

1 A32 A3

3

35 =

24

1 0 01 ¡ 1 02 ¡ 3 1

35 ;

wn = [w1;w2;w3] = [12; 116; 3

256]

[ 3256;¡ 25

256; 150256; 150

256;¡ 25256; 3

256]:

Mask of D-D scheme

2 (n=1) 12 0 0 0 0

4 (n=2) 916 ¡ 1

16 0 0 06 (n=3) 150

256 ¡ 25256

3256 0 0

8 (n=4) 12252048 ¡ 245

204849

2048 ¡ 52048 0

10 (n=5) 1984532768 ¡ 2250

16384567

16384 ¡ 40516384

3565536

Generalization of Mask (G.J.Yoon)

• Consider the problem of finding mask reproducing polynomial of degree (2N+1)

We let

a = faj g2N +3j =¡ 2N ¡ 3

v = a2N +2; w = a2N +3

X

kaj ¡ 2kp(k) = p( j

2); j 2 Z; p2 ¼2N +1:

For we get the maskj=-N-1,¢¢¢;N;

a2j = ±j ;0 +(¡ 1)j +N +1µ 2N +2

N + j +1¶

v:

a2j +1 = N +124N +1

µ2N +1N

¶ (¡ 1)j

2j + 1µ 2N + 1

N +j + 1¶

+ (¡ 1)j +N +1wµ 2N + 1

N +j + 1¶ (2N +2)(2j +1)

(N + j + 2)(N ¡ j +1)

Remark• We obtain symmetric SS which reproduce all polynomi

al degree (2N+1) and which is not interpolatory.

• In case v=0, it becomes (2N+4) interpolatory SS.

• In case v=w=0, it becomes (2N+2) DD scheme.

Example • For N=1, we get

• In case v=0, we have

• In case v=w=0, it becomes 4-point DD scheme

[w; v; ¡ 116¡ 3w; ¡ 4v; 9

16+2w; 1+6v; 916+2w; ¡ 4v; ¡ 1

16¡ 3w; v; w]:

[w; 0; ¡ 116 ¡ 3w; 0; 9

16 + 2w; 1; 916 +2w; 0; ¡ 1

16 ¡ 3w; 0; w]:

[¡ 116; 0; 9

16; 1; 916; 0; ¡ 1

16]:

Approximation OrderDefinition A subdivision scheme S has

approximation order n if for initial data with smooth F enough

Theorem The approximation order of a convergent SS S, which is exact for , is n+1.

Q) Determine the approximation order of the 4-point DD scheme.

f 0i = F (ih)

j(F ¡ S1 f 0)(x)j · Chn

¼n

Subdivision ZooClassification :

• stationary or non-stationary

• binary or ternary

• type of mesh (triangle or quadrilateral)

• approximating or interpolating

• linear or non-linear

Classification• Vertex insertion (primal)

– Insert a vertex on the interior of each edge and one on the interior of each face for quads

– Loop, Kobbelt, Catmull-Clark, Modified Butterfly

• Corner cutting (dual)– Insert a face in the middle of each old face and con

nect faces in adjacent old faces– Doo-Sabin

Classification• Interpolating

– Can control the limit surface in a more intuitive manner

– Can simplify algorithms (efficient)

• Approximating

– Higher quality surfaces– Faster convergence

Face split (primal type)

Vertex split (dual type)

Triangular meshes Quad.meshes

approximating Loop Catmull-Clark

interpolating Butterfly Kobbelt

Quad. meshes

approximating

Doo-Sabin , Midedge

(C2)(C2)

(C1) (C1)

(C1)(C1)

Binary Subdivision of B-Spline

• Univariate B-Spline :

where : normalized B-Spline.

(¤) Sr (u) =X

i2Zdr

i N r (u ¡ i);

N r

Properties • Partition of unity :• Positivity :• Local support :• Continuity :• Recursion :

Pi N r (u ¡ i) = 1

N r (u) ¸ 0N r (u ¡ i) = 0 if u =2 [i; i + r +1]

N r (u) 2 C (r ¡ 1)

N r (u ¡ i) = u¡ ir N r ¡ 1(u ¡ i) + i+r+1¡ u

r N r ¡ 1(u ¡ i ¡ 1)

Idea behind Subdivision• Rewrite the curve (*) as a curve over a refined knot se

quence• (*) becomes

• Determine A single B-Spline can be decomposed into similar B-

Spline of half the support.

Z=2:

Sr (u) =X

j 2Z=2dr

j N r (2(u ¡ j ))

drj

• This results in

where

• For example

Chaikin’s algorithm

N r (u) =X

j 2Z=2crj N r (2(u ¡ j ));

drj =

X

i2Zcrj ¡ i dr

i ; j 2 Z=2

r = 2d2

0 =X

i2Zc2¡ i d2

i = ¢¢¢+ (3=4)d2¡ 1 + (1=4)d2

0 +¢¢¢

d21=2 = ¢¢¢+(1=4)d2

¡ 1 +(3=4)d20 + ¢¢¢

crj = 2¡ r ¡r+1

2j¢

Tensor Product B-Spline Surface

Sr;s(u) =X

i2Z 2

dr;si N r;s(u ¡ i)

A tensor product B-Spline is the product of two independently univariate B-Splines, i.e

N r;s(u ¡ i) = N r (u ¡ i)N s(v ¡ j )

• Example 1

• The mask set :

dr;sj =

X

i2Z 2

cr;sj ¡ i dr;s

i ; j 2 Z2=2

r = s = 2d2;2

0;0 = ¢¢¢+( 916)d2;2

¡ 1;¡ 1 + ( 316) +d2;2

0;¡ 1 + ¢¢¢

¢¢¢+( 316)d2;2

¡ 1;0 +( 116) + d2;2

0;0 +¢¢¢

cr;sj = cr

i csj

= 2¡ (r+s)µr +1

2i¶µs + 1

2j¶

• Example 2

The mask for bi-cubic tensor product B-Spline

r = s = 3

Triangular Spline

• A triangular spline surface :

where

: a normalized triangular spline of degree (r+s+t-2)

(¤) Sr;s;t(u) =X

idr;s;t

i N r;s;t(u ¡ i); u 2 R2; i 2 Z2

N r;s;t(u)

Subdivision of Triangular Spline

• The surface (*) can be rewritten over the refine grid.

• A single triangular spline is decomposed into splines of identical degree over the refined grid.

Sr;s;t(u) =X

jdr;s;t

j N r;s;t2(u ¡ j)

N r;s;t(u) =X

j2Z=2cr;s;tj N r;s;t(2(u ¡ j));

Where

The subdivision masks for the triangular spline

dr;s;tj =

X

i2Zcr;s;tj ¡ i dr;s;t

i

N 2;2;2(u)

cr;s;tj = 2¡ (r+s+t)

tX

k=0

µ r2i ¡ k

¶µ s2j ¡ k

¶µ tk¶

Doo-Sabin Scheme

• Approximating Scheme

• Tensor Product Quadratic B-Spline

• Continuity– C1 regular regions and extraordinary vertices

• Dual Quadrilateral

Doo/Sabin algorithm

ProblemSubdivision for tensor product quadratic B-spline surface has rigid restrictions on the topology.

Each vertex of the net must order 4.This restriction makes the design of many surfaces

difficult

Doo/Sabin presented an algorithm that eliminated this restriction by generalizing the bi-quadratic B-Spline subdivision rules to include arbitrary topologies.

P0

P1

P3

P2

P’’2

P’3

P’0 P’2

P’1

P’b

P’’1

P’’3P’’0

Doo-Sabin Scheme

SD S4 = 1

16

2664

9 3 1 33 9 3 11 3 9 33 1 3 9

3775

(P0\ ;P1\ ;P2\ ;P3\ )T

= SD S4 (P0;P1;P2;P3)T

• The subdivision masks for bi-quadratic B-Spline :

• Geometric view of bi-quadratic B-Spline subdivision : the new points are centroids of the sub-face formed by the face centroid, a corner vertex and the two mid-edge points next to the corner.

arbitrary topology

• For an n-sided face Doo/Sabin used subdivision matrix

• As subdivision proceeds, the refined control point mesh becomes locally rectangular everywhere except at a fixed number of points.

• Since bi-quadratic B-splines are , the surfaces generated by the Doo/Sabin algorithm are locally

SD Sn = (®i j )n£ n;

C1

C1

®i j = 5+n4n ; i = j

®i j =3+2cos

³2¼(i¡ j )

n

´

4n ; i 6= j

Maple Code:• http://

www.mrl.nyu.edu/projects/subdivision/maple/doosabin.pdf

Catmull-Clark Scheme

• Approximating Scheme

• Tensor Product Bicubic B-spline

• Continuity

- C2 regular regions– C1 extraordinary vertices

• Primal Quadrilateral

q11 : New face pointsq12 : New edge pointsq22 : New vertex points

P41P42

x

x

x

x

x

x

x

x

xx

x xx

xx

xx x x

xxx

x x x

P21

P31

P11 P12P13

P14

P22 P23

P32

P43

P44

P34P33

P24

q11 q12

q22

Catmull/Clark Algorithm

• New face points :

• New edge points :

• New vertex points :q12 =

C +D2 + P12+P22

22 ; C = q11;D = q13

q22 = Q4 + R

2 + P224

q11 = P11 + P12 + P21 +P224

Q = q11 + q13 +q31 + q334

R = 14

· µ p22 +p122

¶+

µ p22 +p212

¶+

µ p22 + p322

¶+

µ p22 + p232

¶¸

• The subdivision masks for bi-cubic B-Spline :Approach : generalization of bi-cubic B-Spline

arbitrary topology

• New face points : the average of all of the old points defining the face.

• New edge points : the average of the mid points of the old edge with average of the two new face.

• New vertex points : Q : the average of the new face points of all faces sharing an old vertex point.R : the average of the midpoints of all old edges incident on the old vertex point.S : the old vertex point.

14Q + 1

2R + 14S;

Note : tangent plane continuity was not maintained at extraordinary points.

N : order of the vertex

tangent plane continuity at extraordinary points.

S = 1N Q + 2

N R + N ¡ 3N S

Maple Code:• http://www.mrl.nyu.edu/projects/subdivision/maple/

catmullclark.pdf

Butterfly Scheme• Interpolating Scheme

• General Scheme

• Continuity– C1 regular regions

• Primal Triangular Quadrisection

Butterfly scheme

qke = 1

2(pke;1 +pk

e;2) +2w(pke;3 +pk

e;4) ¡ w8X

j =5pk

e;j

pk+1i = pk

i [ qke

The mask of the butterfly scheme:

The mask of the butterfly scheme:

Well-known Fact On the three directional grid, if the values given are co

nstant in one of the three grid directions (1,0),(0,1),(1,1), then the butterfly scheme inserts new values which are constant in the same direction and could be computed by univariate 4-point scheme along the other two grid directions.

For a new vertex in the (1,0) direction, the rule is

f k+12i+1;j = 1

2(f ki ;j + f k

i+1;j ) +2w(f ki+1;j +1 + f k

i ;j ¡ 1)¡ w(f k

i+2;j +1 + f ki ;j +1 +f k

i¡ 1;j ¡ 1 + f ki+1;j ¡ 1)

• For a new vertex in the direction (0,1)

• For (1,1)

• Under the assumption

f k+1i ;2j +1 = 1

2(f ki ;j + f k

i ;j +1) +2w(f ki+1;j +1 + f k

i ¡ 1;j )¡ w(f k

i+1;j +2 + f ki+1;j +f k

i¡ 1;j ¡ 1 + f ki ¡ 1;j +1)

f k+12i+1;2j +1 = 1

2(f ki ;j + f k

i+1;j +1) +2w(f ki+1;j + f k

i ;j +1)¡ w(f k

i+2;j +1 + f ki+1;j +2 + f k

i¡ 1;j + f ki ;j ¡ 1)

f ki ;j = f k

0;j

• We get from the first insertion rule

• For a new vertex (0,1) direction we get

• The third insertion rule gives the same rule as above.

f k+12i+1;j = f k

0;j

f k+1i ;2j +1 = (1

2 +w)(f k0;j +f k

0;j +1) ¡ w(f k0;j +2 +f k

0;j ¡ 1)

Maple Code:• http://

www.mrl.nyu.edu/projects/subdivision/maple/butterfly.pdf

Loop Scheme• Approximating Scheme

• Three Directional Box-Spline

• Continuity

– C2 regular regions– C1 extraordinary vertices

• Primal Triangular Quadrisection

Loop Scheme• A generalized triangular subdivision surface.• The subdivision masks for

• mask A generates new control points for each vertex• mask B generates new control points for edge of the

original regular triangular mesh.

N 2;2;2(u)

• Mask B : generalization is to leave this subdivision rule intact (why?)

• Mask A : The new vertex point can be computed as a convex combination of the old vertex and all old vertices that share an edge with it.

V : the old vertex point.Q : the average of the old points

that share an edge with V.

This same idea may be applied to an arbitrary triangular mesh.

V = 58V + 3

8Q;

• Note : tangent plane continuity is lost at the extraordinary points.

• Loop scheme

where

- curvature continuity at regular point.

- tangent plane continuity at extaordinary point.

V = ®N V + (1¡ ®N )Q;

®N =µ 3

8 + 14cos 2¼

N¶2

+ 38

Maple Code• http://

www.mrl.nyu.edu/projects/subdivision/maple/loop.pdf

Mid-Edge SchemeP0

P1

P3

P2

P’3

P’0 P’2

P’1

P’’0

P’’1

P’’3

P’’2

SM E4 = 1

4

2664

2 1 0 11 2 1 00 1 2 11 0 1 2

3775

(P0\ ;P1\ ;P2\ ;P3\ )T

= SM E4 (P0;P1;P2;P3)T

• For an n-sided face, subdivision matrix :

SM En = (¯ i j )n£ n;

¯ i j =1+ cos 2¼(i¡ 1)

nn

Circulant matrix

eigenvalues of can be calculated by evaluating the polynomial.

SGn =

26664

a0 a1 ¢¢¢ an¡ 1an¡ 1 a0 ¢¢¢ an¡ 2

... ... ... ...a1 a2 ¢¢¢ a0

37775

SGn

z = wj ; w = ei 2¼n ; j = 0;1;¢¢¢;n ¡ 1

Pn(z) = a0 +a1z +¢¢¢+an¡ 1zn¡ 1;

Further Subdivision Schemes

• Non-uniform scheme.• Shape preserving scheme.• Hermite-type scheme.• Variational scheme.• Quasi-linear scheme.• Poly-scale scheme.• Non-stationary scheme.• Reverse subdivision scheme.

Convexity Preserving ISS• A constructive approach is used to derive convexity pr

eserving subdivision scheme.

1. interpolatory.2. local : four points scheme.

define the first and second differences.df i = f i+1 ¡ f idi = d2f i = f i+1 ¡ 2f i + f i ¡ 1

f k+12i = f k

i

f k+12i+1 = 1

2¡f k

i +f ki+1

¢¡ F1¡f k

i¡ 1; f ki ; f k

i+1; f ki+2

¢

3. invariant under addition of affine functions.

(★)f k+1

2i = f ki

f k+12i+1 = 1

2¡f k

i +f ki+1

¢¡ F (dki ;dk

i+1): subdivision functionF

4. continuous5. homogeneous, i.e. , 6. symmetric

F (¸x;¸y) = ¸F (x;y)

f k+12i = f k

i

f k+12i+1 = 1

2¡f k

i +f ki+1

¢¡ F2

µ 12(f k

i +f ki+1);df k

i ;dki ;dk

i+1

¶

• Theorem 1 (Convexity)A subdivision scheme of type (*) satisfying conditions 1 to 6 is convex preserving for all convex data F satisfies

Note : if F=0 linear SS. only

(Question)under what conditions, SS(*) with conditions 1 to 6 and convexity condition generate continuously differentiable limit functions.Answer

0· F (x;y) · 14minfx;yg;8x;y ¸ 0

C0

• Theorem 2 (Smoothness)Under the same conditions hold as in Theorem 1.The scheme given by

continuously differentiable function which is convex and interpolate the data

• Theorem 3 (Approximation order)The convexity preserving subdivision scheme has approximation order 4.

f k+12i = f k

i

f k+12i+1 = 1

2¡f k

i +f ki+1

¢¡ 14

11

dki

+ 1dk

i + 1

Analysis of convergence and smoothness of binary SS by the

formalism of Laurent polynomials

A binary univariate SS:

f k+1i =

X

j 2Zai ¡ 2j f k

j

The general form of an interpolatory SS:

f k+12i = f k

if k+1

2i+1 =X

j 2Za1+2j f k

i¡ j

For each scheme , we define the symbol

Theorem 1 Let be a convergent SS, then

Theorem 2 Let denote a SS with symbol satisfying (*). Then there exists a SS with the property

where and

S

a(z) =X

i2Zaizi

S

a(¡ 1) = 0; a(1) = 2

a(z)

df k = S1df k¡ 1

f k = Skf 0

S

(¤)X

j 2Za2j =

X

j 2Za2j +1 = 1

S1

df k = f (df k)i = 2k(f ki+1 ¡ f k

i )g

Theorem 3 is a uniformly convergent SS, if only if

converges uniformly to the zero function

Theorem 4 Let . If is convergent, then for any initial data .

S12S1

limk! 1

(12S1)kf 0 = 0

a(z) = (1+z)m

2m b(z) SbS1

a f 0 2 Cm(R) f 0

DraftObjective: obtain the mask of interpolatory subdivision schemes

(binary 2n-point scheme, ternary 4-point scheme, butterfly scheme) using symmetry and necessary condition for smoothness.

• A binary univariate SS:

f k+1i =

X

j 2Zai ¡ 2j f k

j

The general form of an interpolatory SS:f k+1

2i = f ki

f k+12i+1 =

X

j 2Za1+2j f k

i¡ j

4-point interpolatory SS• The 4 –point interpolatory SS:

• The Laurent-polynomial of this scheme is

• By symmetry and necessary condition, we get mask

a(z) = a¡ 3z¡ 3 + a¡ 1z¡ 1 +1+a1z +a3z3

a1 +a3 = 12 [¡ w; 1

2 +w; 12 +w;¡ w]

f k+12i = f k

if k+1

2i+1 = a3f ki¡ 1 +a1f k

i + a¡ 1f ki+1 +a¡ 3f k

i+2

6-point interpolatory SS• The 6 –point interpolatory SS:

• The Laurent-polynomial of this scheme is

• By symmetry and necessary condition, we get mask

a(z) = a5z¡ 5 +a¡ 3z¡ 3 + a¡ 1z¡ 1 +1+ a1z +a3z3 + a¡ 5f ki+3

a1 +a3 + a5 = 12

24a5 +8a3 + 12 = 0

[w;¡ 116 ¡ 3w; 9

16 +2w; 916 + 2w;¡ 1

16 ¡ 3w;w]

f k+12i = f k

if k+1

2i+1 = a5f ki¡ 2 +a3f k

i ¡ 1 +a1f ki + a¡ 1f k

i+1 +a¡ 3f ki+2 +a¡ 5f k

i+3

Ternary interpolatory SS

• A ternary SS

• The general form of an interpolatory SS:

f k+1i =

X

j 2Zai ¡ 3j f k

j

f k+13i = f k

if k+1

3i+1 =X

j 2Za1+3j f k

i¡ k

f k+13i+2 =

X

j 2Za2+3j f k

i¡ k

For each scheme , we define the symbol

Theorem 5 Let be a convergent SS, then

(**) Let us consider a n-ary convergent scheme. The associated polynomial satisfies the properties.

S

a(z) =X

i2Zaizi

S

(¤)X

j 2Za3j =

X

j 2Za3j +1 =

X

j 2za3j +2 = 1

a(e2i¼=3) = a(e4i¼=3) = 0; a(1) = 3

a(1) = n; a(wpn) = 0;wp

n = e2i p¼n ;p2 f1;2;:::;n ¡ 1g

Theorem 6 Let denote a SS with symbol satisfying (*). Then there exists a SS with the property

where and

Theorem 7 is a uniformly convergent SS, if only if

converges uniformly to the zero function

a(z)

df k = S1df k¡ 1

f k = Skf 0

SS1

df k = f (df k)i = 3k(f ki+1 ¡ f k

i )g

S13S1

limk! 1

(13S1)kf 0 = 0

Ternary 4-point interpolatory SS

• The ternary 4 –point interpolatory SS:

• The Laurent-polynomial of this scheme is

a(z) = a¡ 5z¡ 5 + a¡ 4z¡ 4 +a¡ 2z¡ 2 + a¡ 1z¡ 1 +1+ a1z + a2z2 + a4z4 +a5z5

f k+13i = f k

if k+1

3i+1 = a4f ki¡ 1 +a1f k

i + a¡ 2f ki+1 +a¡ 5f k

i+2f k+1

3i+2 = a5f ki¡ 1 +a2f k

i + a¡ 1f ki+1 +a¡ 4f k

i+2

• By symmetry and necessary condition, we get mask

• Let

a1 +a2 +a4 + a5 = 13a2 ¡ 3a4 +6a5 = 1

a4 + a5 = ¡ 19

a5 = ¡ 118 + 1

6¹

f k+13i+1 : [¡ 1

18 ¡ 16¹ ; 13

18 + 12¹ ; 7

18 ¡ 12¹ ;¡ 1

18 + 16¹ ]

f k+13i+2 : [¡ 1

18 + 16¹ ; 7

18 ¡ 12¹ ; 13

18 + 12¹ ;¡ 1

18 ¡ 16¹ ]

Butterfly subdivision scheme

A =

2666666664

¢ ¢ ¢ ¢ ¡ w ¡ w ¢¢ ¢ ¡ w ¢ 2w ¢ ¡ w¢ ¡ w 2w 1

212 2w ¡ w

¢ ¢ 12 1 1

2 ¢ ¢¡ w 2w 1

212 2w ¡ w ¢

¡ w ¢ 2w ¢ ¡ w ¢ ¢¢ ¡ w ¡ w ¢ ¢ ¢ ¢

3777777775

• The bivariate Laurent polynomial of this scheme:

• The symbol of the butterfly scheme:

a(z1;z2) =3X

i=¡ 3

3X

j =¡ 3ai ;j zi

1zj2

a(z1;z2) = 12(1+ z1)(1+ z2)(1+ z1z2)(1¡ wc(z1;z2))(z1z2)¡ 1;

c(z1;z2) = 2z¡ 21 z¡ 1

2 + 2z¡ 11 z¡ 2

2 ¡ 4z¡ 11 z¡ 1

2 ¡ 4z¡ 11 ¡ 4z¡ 1

2 +2z¡ 11 z2

+ 2z1z¡ 12 +12¡ 4z1 ¡ 4z2 ¡ 4z1z2 +2z2

1z2 +2z1z22

• The verification that the scheme generates requires checking the contractivity of three schemes:

• Butterfly scheme satisfies:

2(1+z1)b(z1;z2);2(1+ z2)b(z1;z2);2(1+z1z2)b(z1;z2)

w¡ w = 0;w¡ 2w+w = 0;¡ w¡ 2w+ 12 ¡ 1

2 +2w+w = 0;

¡ 12 + 1¡ 1

2 = 0

Symmetric 8-point Butterfly SS

(1;2= ®);(3;4) = ¯);(5;6;7;8= °)

A =

2666666664

¢ ¢ ¢ ¢ ° ° ¢¢ ¢ ° ¢ ¯ ¢ °¢ ° ¯ ® ® ¯ °¢ ¢ ® 1 ® ¢ ¢° ¯ ® ® ¯ ° ¢° ¢ ¯ ¢ ° ¢ ¢¢ ° ° ¢ ¢ ¢ ¢

3777777775

• To converges to a surface with , we have

• This is the same mask of Butterfly scheme:

C0

1;2= 12 3;4= 2w 5;6;7;8= ¡ w

2° +¯ = 0;¡ 2®+1= 0

Symmetric 10-point Butterfly SS

(1;2= ®);(3;4= ¯);(5;6;7;8= °);(9;10= w)

A =

2666666664

¢ ¢ ¢ w ° ° w¢ ¢ ° ¢ ¯ ¢ °¢ ° ¯ ® ® ¯ °w ¢ ® 1 ® ¢ w° ¯ ® ® ¯ ° ¢° ¢ ¯ ¢ ° ¢ ¢w ° ° w ¢ ¢ ¢

3777777775

• To converges to a surface with , we have

• We find the mask of 10-point Butterfly scheme:

• This mask is exact with a modified Butterfly scheme

C0

° = 116 + w

1;2= 12 ¡ w 3;4= 2° 5;6;7;8= ¡ ° 9;10= w

2° +¯ = 0;¡ 2®¡ 2w+ 1= 0

Non-stationary Subdivision Scheme

• Construction an insertion rule by interpolation with the span of the functions

• Consider the k-th level insertion rule at the point based on the values at the points the solution of system leads to the

insertion rulef k+1

2i+1 = ¡ 116cos2(µ2¡ k¡ 2) cos(µ2¡ k¡ 1)(f k

i ¡ 1 +f ki+2)

+ (1+2cos(µ2¡ k¡ 1))2

16cos2(µ2¡ k¡ 2) cos(µ2¡ k¡ 1)(f ki + f k

i+1)

f1;t;cost;sintg

µ2¡ k¡ 1

¡ µ2¡ k;0;µ2¡ k;2µ2¡ k

Non-stationary Butterfly Scheme

• Objective: given a well-known non-stationary 4-point scheme as the following rule:

• Our objective is to extend this non-stationary 4-point scheme to 2D Butterfly scheme, we denote

f k+12i+1 = ¡ 1

16cos2(µ2¡ k¡ 2) cos(µ2¡ k¡ 1)(f ki ¡ 1 +f k

i+2)

+ (1+2cos(µ2¡ k¡ 1))2

16cos2(µ2¡ k¡ 2) cos(µ2¡ k¡ 1)(f ki + f k

i+1)

wk = 116cos(µ2¡ k¡ 2) cos(µ2¡ k¡ 1)

Definition Two SSs and are asymptotically equivalent if

where

Theorem 1 Let and be two asymptotically equivalent SSs having finite masks of the same support. Suppose is non-stationary and

is stationary. If is and

Then the non-stationary scheme is .

$S$

fSak g fSbk g1X

k=0kSak ¡ Sbk k < 1

kSak k1 = maxfX

ja(k)2®j;

Xja(k)

2®+1jg

fSak g fSag

fSak g fSagfSag

fSak g

Cm

Cm

1X

k=02mkkSak ¡ Sak < 1

Lemma For

The mask of :

$S$

k ¸ 0; 0 · µ · ¼=2

116 · wk · 1

8 jwk ¡ 116j · C

22k

2666666664

¢ ¢ ¢ ¢ ¡ wk ¡ wk ¢¢ ¢ ¡ wk ¢ 2wk ¢ ¡ wk¢ ¡ wk 2wk 1

212 2wk ¡ wk

¢ ¢ 12 1 1

2 ¢ ¢¡ wk 2wk 1

212 2wk ¡ wk ¢

¡ wk ¢ 2wk ¢ ¡ wk ¢ ¢¢ ¡ wk ¡ wk ¢ ¢ ¢ ¢

3777777775

fSkg

Subdivision operator associated with butterfly scheme has mask:

Theorem The non-stationary butterfly scheme is asymptotically equivalent to the stationary butterfly scheme . Moreover, the limit function belong to .

S

2666666664

¢ ¢ ¢ ¢ ¡ 1=16 ¡ 1=16 ¢¢ ¢ ¡ 1=16 ¢ 1=8 ¢ ¡ 1=16¢ ¡ 1=16 1=8 1

212 1=8 ¡ 1=16

¢ ¢ 12 1 1

2 ¢ ¢¡ 1=16 1=8 1

212 1=8 ¡ 1=16 ¢

¡ 1=16 ¢ 1=8 ¢ ¡ 1=16 ¢ ¢¢ ¡ 1=16 ¡ 1=16 ¢ ¢ ¢ ¢

3777777775

fSkg

fSgC1

Subdivision (Kobbelt)

• Approximating Scheme

• Continuity

– C2 regular regions– C1 extraordinary vertices

• Triangular Trisection

• Adaptive Refinement

p3

Subdivision• Subdivision schemes on triangle meshes are usually

based on the 1-to-4 split operation which inserts a new vertex for every edge of the given mesh and then connects the new vertices.

p3

Subdivision• This scheme is based on a split operation which first

inserts a new vertex for every face of the given mesh. Flipping the original edges. Applying this scheme twice leads to a 1-to-9 refinement of the original mesh.

p3

Smoothing Rules• Two smoothing rules, one for the placement of the

newly inserted vertices and one for the relaxation of the old ones.

• Reverse Engineering Process: instead of analyzing a given scheme, we derive one which by construction satisfies the known necessary criteria.

q:= 1

3(pi +pj + pk)

S(p) := (1¡ ®n)p+ ®n1n

n¡ 1X

i=0pi

Subdivision Matrix• When analyzing the eigen-structure of this matrix, it is not suitable

all eigenvalues are complex

.

S = 13

26666666664

u v v v ¢¢¢ v1 1 1 0 ¢¢¢ 01 0 ... ... ... ...... ... ... ... ... 01 0 ... ... 11 1 0 ¢¢¢ 0 1

37777777775

u = 3(1¡ ®n); v = 3®=n

Subdivision MatrixThe matrix has its eigen-values:

The necessary condition for C1:. The choice for the eigen-values:

R =

26666664

1 0 ¢¢¢ 0 00 0 ¢¢¢ 0 10 1 ... ... 0... ... ... ... ...0 0 1 0

37777775

~S = RS2

19[9;(2¡ 3®n)2;2+2cos(2¼1

n);¢¢¢;2+ 2cos(2¼n ¡ 1n )]

¸1 = 1> ¸2 = ¸3 > ¸ i ; i = 4;:: : ;n +1

¸4 = ¸22 ) ®n = 4¡ 2cos(2¼=n)9

Boundaries

• Edge flipping at the boundaries is not possible.

• We have to apply a univariate tri-section rule to the boundary polygon and connect the new vertices to the corresponding interior ones such that a uniform 1-to-9 split is established for each boundary triangle.

• We can obtain tri-section mask for cubic splines by convolution

13[1;2;3;2;1]¤(1

3[1;1;1])2 = 127[1;4;10;16;19;16;10;4;1]

Interpolatory -Subdivision

Stencil for a new vertex and scheme:

p3

qk+1 = 3281(pk

1 +pk2 +pk

3) ¡ 181(pk

4 +pk5 +pk

6) ¡ 281(pk

7 +pk8 +pk

9 +pk10 +pk

11 +pk12)

Interpolatory -Subdivision

Modification for extraordinary vertices:

For

For

For

p3

qk+1 = ®pk +n¡ 1X

i=0®i pk

i

®= 89; ®n

i =19 + 2

3 cos(2¼i=n) + 29 cos(4¼i=n)

n

n ¸ 5

n = 3 ®= 89;®0 = 7

27;®1 = ®2 = ¡ 227

®= 89;®0 = 7

36;®1 = ®3 = 127;®2 = ¡ 5

36n = 4

• Stencil for boundary subdivision

pk+13i¡ 1 = ¡ 4

81pki ¡ 2 + 10

27pki ¡ 1 + 20

27pki ¡ 5

81pki+1

pk+13i+1 = ¡ 5

81pki ¡ 1 + 20

27pki + 10

27pki+1 ¡ 4

81pki+2

pk3i = pk

i

• Reflection vertices across the boundary of the mesh

Eigen Structure of SS

• Successive applying of a matrix relates to eigenvectors and values.

• Applying of a matrix to a vector produces a transformation on the vector

• Why are eigenvalues and vectors so important?

same limit¸v;¸2v;¸3v;:: :

Sv = ¸v

Sv;S2v;S3v;:::

Affine Invariance Sum of row elements must be unit, so

• What is the interpretation of above relation? is an eigenvector of S and is the corresponding eigenvalue

S

26664

11...1

37775 =

26664

11...1

37775

1=

26664

11...1

37775

¸ = 1

Subdivision Limit Position• If then S is .

1= ¸0 > ¸ i ;f i = 1;ng¸ 0 C0

S = V¤V ¡ 1

=£ v0 v1 ¢¢¢vn

¤26664

¸0 0 ¢¢¢ 00 ¸1 ¢¢¢ 0... ... ... ...0 0 ¢¢¢ ¸n

37775

26664

v¡ 10

v¡ 11...

v¡ 1n

37775

• To find the limit neighborhood , we must compute

All of the vertices of the limit local neighborhood converges to the same position

P 1

S1

S1 = V¤1 V ¡ 1

=£ 1 v1 ¢¢¢vn

¤26664

1 0 ¢¢¢ 00 0 ¢¢¢ 0... ... ... ...0 0 ¢¢¢ 0

37775

26664

v¡ 10

v¡ 11...

v¡ 1n

37775 =

2664

v¡ 10

v¡ 10¢¢¢v¡ 1

0

3775

P 1 = S1 P 0 =

2664

v¡ 10

v¡ 10¢¢¢v¡ 1

0

3775 P 0

Subdivision Derivatives The 1st derivatives of the subdivision manifold

for a parameter direction u defines:

We define the 2nd and 3rd derivatives

@S@u = lim

g! 11¸g

1

pgi+1 ¡ pg

ikp0

i+1 ¡ p0i k

@2S@u2 = lim

g! 11¸g

2

(pgi+1 ¡ pg

i ) ¡ (pgi ¡ pg

i¡ 1)kp0

i+1 ¡ p0i k2

@3S@u3 = lim

g! 11¸g

3

(pgi+2 ¡ 2pg

i+1 +pgi ) ¡ (pg

i+1 ¡ 2pgi +pg

i ¡ 1)kp0

i+1 ¡ p0i k3

Example: Uniform Cubic Spline

• Eigen Decomposition of S:

S = 18

266664

1 6 1 0 00 4 4 0 00 1 6 1 00 0 4 4 00 0 1 6 1

377775

= V¤V ¡ 1

266664

1 ¡ 1 1 1 01 ¡ 1=2 2=11 0 01 0 ¡ 1=11 0 01 1=2 2=11 0 01 1 1 0 1

377775

266664

1 0 0 0 00 1=2 0 0 00 0 1=4 0 00 0 0 1=8 00 0 0 0 1=8

377775

266664

0 1=6 4=6 1=6 00 ¡ 1 0 1 00 11=6 ¡ 22=6 11=6 01 ¡ 3 3 ¡ 1 00 ¡ 1 3 ¡ 3 1

377775

Sg = V¤gV ¡ 1

= V

266664

1 0 0 0 00 ¸g

1 0 0 00 0 ¸g

2 0 00 0 0 ¸g

3 00 0 0 0 ¸g

3

377775

V ¡ 1

Sg =

26666664

6¸ g3

61+6¸ g

1 +11¸ g2 ¡ 18¸ g

36

4¡ 22¸ g2 +18¸ g

36

1¡ 6¸ g1+11¸ g

2 ¡ 6¸ g3

6 00 1+3¸ g

1+2¸ g2

64¡ 4¸ g

26

1¡ 3¸ g1 +2¸ g

26 0

0 1¡ ¸ g2

64+2¸ g

26

1¡ ¸ g2

6 00 1¡ 3¸ g

1 +2¸ g2

64¡ 4¸ g

26

1+3¸ g1 +2¸ g

26 0

0 1¡ 6¸ g1 +11¸ g

2 ¡ 6¸ g3

64¡ 22¸ g

2 +18¸ g3

61+6¸ g

1 +11¸ g2 ¡ 18¸ g

36

6¸ g3

6

37777775

Limit Position• We derive the formula for the limit position

S1 =

266664

0 1=6 4=6 1=6 00 1=6 4=6 1=6 00 1=6 4=6 1=6 00 1=6 4=6 1=6 00 1=6 4=6 1=6 0

377775

P 1 = S1 P 0

V0;0;0 = 16V¡ 2;¡ 1;0 + 4

6V¡ 1;0;1 + 16V0;1;2

1st DerivativesDifferencing and is the same as

differencing the corresponding row and of

pgi+1 pg

ii +1 i Sg

@Sg

@u = DuSg Du =

2664

¡ 1 1 0 0 00 ¡ 1 1 0 00 0 ¡ 1 1 00 0 0 ¡ 1 1

3775

@Sg

@u =

26664

¡ 2¸ g3

2¡ ¸ g

1 ¡ 3¸ g2+6¸ g

32

6¸ g2 ¡ 6¸ g

32

¸ g1 ¡ 3¸ g

2 +2¸ g3

2 00 ¡ ¸ g

1 ¡ ¸ g2

22¸ g

22

¸ g1 ¡ ¸ g

22 0

0 ¡ ¸ g1 +¸ g

22

¡ 2¸ g2

2¸ g

1 +¸ g2

2 00 ¡ ¸ g

1 +3¸ g2 ¡ 2¸ g

32

¡ 6¸ g2+6¸ g

32

¸ g1 +3¸ g

2 ¡ 6¸ g3

22¸ g

32

37775

• We derive the subdivision process is C1:

@P 1

@u = @S1

@u P 0

@V0;0;0u = ¡ 1

2V¡ 2;¡ 1;0 + 12V0;1;2

@S1

@u = limg! 1

1¸g

1

@Sg

@u =

2664

0 ¡ 12 0 1

2 00 ¡ 1

2 0 12 0

0 ¡ 12 0 1

2 00 ¡ 1

2 0 12 0

3775

2nd DerivativesThe same finite differencing can be done to the

1st derivatives matrix to compute 2nd derivatives:

@2Sg

@u2 = Duu@Sg

@uDuu =

24

¡ 1 1 0 00 ¡ 1 1 00 0 ¡ 1 1

35

@2Sg

@u2 =

24

¸g3 ¸g

2 ¡ 3 g3 ¡ 2 g

2 + 3 g3 ¸g

2 ¡ ¸g3 0

0 ¸g2 ¡ 2 g

2 ¸g2 0

0 ¸g2 ¡ ¸g

3 ¡ 2 g2 + 3 g

3 ¸g2 ¡ 3 g

3 ¸g3

35

• We derive the subdivision process is C2:

@2S1

@u2 = limg! 1

1¸g

2

@2Sg

@u2 =

24

0 1 ¡ 2 1 00 1 ¡ 2 1 00 1 ¡ 2 1 0

35

@2P 1

@u2 = @2S1

@u2 P 0

@2V0;0;0u2 = V¡ 2;¡ 1;0 ¡ 2V¡ 1;0;1 +V0;1;2

3rd DerivativesWe can iterate again to check if the subdivision

process is C3:

The rows are not the same, so the process is not C3

@3Sg

@u3 = Duuu@2Sg

@u2Duuu =

· ¡ 1 1 00 ¡ 1 1

¸

@3Sg

@u3 =· ¡ ¸g

3 3 g3 ¡ 3 g

3 ¸g3 0

0 ¡ ¸g3 3 g

3 ¡ 3 g3 ¸g

3

¸

Catmull-Clark Subdivision

Loop Subdivision

Subdivisionp

3