Subdivision Analysis via JSR

We already know the z-transform formulation of schemes:

To check if the scheme generates a continuous limit curve

( the scheme is ) , we have to check if the difference

scheme is contractive.

We hereby present the ‘joint spectral radius’ analysis (JSR) of schemes, and examine the relation between the two methods. Let us consider a univariate scheme with support [0,s].

)()()( 21 zFzazF mm

0C

mi

iij

mj faf

21 i

s

ii zaza

0

)(

)1/()()( zzazb

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2005 Subdivision Summer School

Since the support of the basic limit function is also [0,s],

and the limit function is , the values

of on [0,1] are fully determined by

f ts ffu ),...,( 0

00

10

)()( 0 ixfxf

ii

101-2-..1-

. 00

01

02

02

01

s

fffff ss

Since the support of the basic limit function is also [0,s],

and the limit function is , the values

of on [0,1] are fully determined by

The values on [0,1/2] are determined by

The values on [1/2,1] are determined by

f ts ffu ),...,( 0

00

10

)()( 0 ixfxf

ii

101-2-..1-

. 00

01

02

02

01

s

fffff ss

ts ffu ),...,( 1

01

10,0

21

21

21

10

11

12

12

11

0-1-..

.

s

ss fffff

ts ffu ),...,( 1

11

21,0

10-..

.

21

21

22

11

10

11

13

12

s

ss fffff

Let us recall the matrix representation of subdivision schemes, and use this representation to compute limit values of the subdivision process, and to analyze the convergence and the smoothness of the limit function.

We present everything for a mask with support [0,4], i.e.

the scheme coefficients are

and the symbol of the scheme is .

The scheme is:

The key model is the scheme generating the cubic B-spline:

43210 ,,,, aaaaai

i i zaza

4

0)(

.

,

131112

241201

2

mj

mj

mj

mj

mj

mj

mj

fafaf

fafafaf

.8

)1()(,

8

1,

2

1,

4

3,

2

1,

8

1 4

43210

zzaaaaaa

The subdivision matrix S: Infinite – 2 slanted

00000

000000

00000

000000

00000

000000

00000

000000

420

31

420

31

420

31

420

31

aaa

aa

aaa

aa

aaa

aa

aaa

aa

S

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m Sf

f

f

f

f

f

f

f

f

aa

aaa

aa

aaa

aa

aaa

aa

aaa

aa

f

f

f

f

f

f

f

f

f

f

.000000

00000

000000

00000

000000

00000

000000

00000

000000

2

1

0

1

2

3

4

5

13

024

13

024

13

024

13

024

13

13

12

11

10

11

12

13

14

15

1

The values of on [0,1] are determined by

The values on [0,1/2] are determined by

The values on [1/2,1] are determined by

f

101-2-..1-

. 00

01

02

02

01

s

fffff ss

ts ffu ),...,( 1

01

10,0

21

21

21

10

11

12

12

11

0-1-..

.

s

ss fffff

ts ffu ),...,( 1

11

21,0

10-..

.

21

21

22

11

10

11

13

12

s

ss fffff

ts ffu ),...,( 0

00

10

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m Sf

f

f

f

f

f

f

f

f

aa

aaa

aa

aaa

aa

aaa

aa

aaa

aa

f

f

f

f

f

f

f

f

f

f

.000000

00000

000000

00000

000000

00000

000000

00000

000000

2

1

0

1

2

3

4

5

31

420

31

420

31

420

31

420

31

13

12

11

10

11

12

13

14

15

1

The sub-matrix : 0At

st

s ffff ),...,(),...,( 10

11

00

01

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m

m Sf

f

f

f

f

f

f

f

f

aa

aaa

aa

aaa

aa

aaa

aa

aaa

aa

f

f

f

f

f

f

f

f

f

f

.000000

00000

000000

00000

000000

00000

000000

00000

000000

2

1

0

1

2

3

4

5

31

420

31

420

31

420

31

420

31

13

12

11

10

11

12

13

14

15

1

The sub-matrix : 1A ts

ts ffff ),...,(),...,( 1

11

20

00

1

The vector of initial values determines

the limit values at [0,1].

determines the values at [0,1/2] while determines the

values at [1/2,1] .

And so on, while

determines the limit function on the interval ,

determine the limit function on the left half and the right half

of respectively. It thus follows that at any point in [0,1],

the limit value of a convergent subdivision, starting with initial values,

is given by

tffffu ),,,( 00

01

02

03

0

00uA 0

1uA

tmj

mj

mj

mj

m ffffu ),,,( 123 ]2)2/1(,2[ mm jjI

mm uAuA 10 ,I

},1,0{,.....,]1,0[ 4321 kiiiiixx0f

....))(),(),(),(( 0

1234uAAAAxfxfxfxf iiii

t

First, a scheme must reproduce constants, therefore ,

Having that, we should make sure that the components in the subspace complementary to the constant eigenvector tend to 0.

Let us show how to check it, and let us also try to understand the relation to the difference scheme analysis.

The idea is to represent the operators and in another basis, the vectors comprising this basis are (e.g.) the columns of the following matrix:

0C

1A

1111

0111

0011

0001

V

.1eigenvaluewith1,1,1,1reigenvectohaveand 10tAA

0A

In the new basis, the two operators are:

Since is an eigenvector of and it follows that

Note that determine the behavior of the last 3 components in the new basis.

VAVBVAVB 11

101

0 ,

t)1,1,1,1( 10 AA

0

0

0

0

0***1

QB

1

1

0

0

0***1

QB

10 andonly QQ

Recalling the products , which in the

new basis appear as

The last three coordinates of this product will converge to

zero if the joint spectral radius is less than 1.

Each matrix has its own spectral radius. The

joint spectral radius is defined as :

If then the last three components of any vector of the form (#) will tend to zero.

0

1234... uAAAA iiii

#,... 0100

1234uVvvBBBB iiii

10 , QQ),( 10 QQ

1),( 10 QQ

mjiii

m

iQQQQQm

/110 }})1,0{:||...(max{||suplim),(

21

We already know that a scheme is convergent to a limit if the differences of the generated values tend to zero. In the new basis, namely, the columns of , it is clear that differences within the first column are zero. The differences within the other columns tend to zero simply because the corresponding coefficients tend to zero.

The number tells us how fast the differences are decaying to zero, and from it we can derive the Holder exponent of the limit function:

0C

1111

0111

0011

0001

V

V

),( 10 QQ

),(log 102 QQ

The choice of and the relation to the difference scheme.

As an exercise, show that with the

specific choice of as :

the two matrices

are just the two matrices generating

the limit values on [0,1] for the difference scheme related with

the generating polynomial

In fact, any other completion of the constant vector into a basis may be used here.

1111

0111

0011

0001

V

10 and QQ

V

V

.)1/()()( zzazb

The analysis of higher smoothness is very similar. First we have to show that if a scheme is , it must reproduce polynomials up to degree k. That is,

To show it we consider eigenvectors of the infinite matrix S

kC

}:)({ Zjjxspank

00000

000000

00000

000000

00000

000000

00000

000000

420

31

420

31

420

31

420

31

aaa

aa

aaa

aa

aaa

aa

aaa

aa

S

The non-zero eigenvalues of the infinite matrix S, due to its two-slanted structure, are just the non-zero eigenvalues of the finite matrix marked below.

00000

000000

00000

000000

00000

000000

00000

000000

420

31

420

31

420

31

420

31

aaa

aa

aaa

aa

aaa

aa

aaa

aa

S

tuuuuuu )...,,,,,,..(. 21012 Let be an eigen-vector of S,

and let denote the limit function of the subdivision process, starting with . It thus follows that

Exercise: Show that if then

Also, the entries of the eigen-vector are of the form

It can further be shown that if the scheme is non-degenerate, then all the monomials up to degree k are eigen-functions of the subdivision scheme.

,uSu uSfu

u

.)2()( xfxf uu ,0,0)0()( krf r

u

.)(,2 ru

r xxfand

01

1 ... ciciu rr

ri

In particular ,

To verify that a given scheme is , we first check that it satisfies the necessary conditions. I.e., its symbol has the factor , or, in the matrix language, it has the polynomial eigenvectors up to degree k.

Then, we have to show the decay to zero of all the

differences of the k-th order divided differences

One way is by showing that the scheme is contractive .

The other way

.0,2seigenvaluewith

rseigenvectopolynomialdegreehaveand 10

kr

rAAr

1)1( kz

1)1/()(2)( kk zzazb

, . . . why do we need the other way ?

kC

Here again we represent the operators and in another basis, where the first k+1 vectors in the basis are the polynomial

vectors (for a scheme with support [0,s]):

The differences of the k-th order divided differences within these vectors are all zero.

And, since these vectors span the polynomial eigenvectors of and , the operators now take the form:

.0,)1,2,...,)1(,()( kjssv tjjjj

0A

0A

1A

1A.2,1,

0000

0000

0000

0000****

****

****

****

000

*00

**0

***1

21

21

i

Q

B

i

ik

.2,1i,

0000

0000

0000

0000****

****

****

****

000

*00

**0

***1

i

21

21

i

Q

B k

To show that the scheme is it remains to verify thatkC

kQQ 2),( 10

kQQ ),(log 102 Holder exponent of the k-th derivative of the limit function:

An example : The 4-point scheme for w=1/16

0A 1A

V

m

mjiii

m

iQQQQQm

/110 }})1,0{:||...(max{||suplim),(

21

mjiii iQQQ

m

/1}})1,0{:||...(max{||:ofGraph21

It can be shown here that .25.0),( 10 QQ

Limit values at grid points

To represent the curve or surface at the m-th refinement level, we prefer to use the limit values and derivatives . Can we ?

We would like to use only values at the m-th level .

It is enough to show how to do it for the 0-th level : Using

)()( 0 ixfxfi

i

.0,,)()( )(0)( kZjijfjfi

i

Therefore, it is enough to know the values of the basic limit function and its derivatives at the integers!

Zjijfjfi

i ,)()( 0

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2005 Subdivision Summer School

Here we use the refinement equation:

.],0[)(supp,)2()( sjxaxj

j

.0,

)4(

)3(

)2(

)1(

)0(

2

)4(

)3(

)2(

)1(

)0(

0000

00

00

0000

)(

)(

)(

)(

)(

)(

)(

)(

)(

)(

4

234

01234

012

0

k

a

aaa

aaaaa

aaa

a

The values are left eigenvectors of the subdivision matrix S

.0,0,)2(2)( )()( ksijiaij

j

Example 1 : The limit values of the cubic B-spline scheme.4

81 )1()( zza

;

0

0

0

0

0000

00

00

0000

61

32

61

61

32

61

81

43

21

81

81

21

43

21

81

81

21

43

81

0

0

0

0

0

0

0000

00

00

0000

21

21

21

21

21

81

43

21

81

81

21

43

21

81

81

21

43

81

Normalization : Using the polynomial reproduction property

.0,!)()( kjjj

Example 2 : The limit derivatives of the 4-point scheme.

0

0

0

0

0

0

000000

00000

1000

010

0001

00000

000000

21

21

21

21

21

21

21

21

21

21

21

21

21

w

w

w

w

w

ww

www

wwww

www

ww

w

)()(2

1

41

2)2(' 2211

mi

mi

mi

mi

mm ffwff

wif

Approximation order and Quasi-interpolation

A local method has approximation order n+1 if it is exact on

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2005 Subdivision Summer School

:n

kn

k

k

n jhxk

jhfxThjjhx )(

!

)()( and,])1(,[Let :Proof

0

)(

bounded be let and,)R(,)(Let 1n0 Cfihffi

.,)()()( and,],[)supp(with i

n pxpixipsr

.)(|)(-)(|Then i

10 ni hOihxfxf

ii

0 |)()]()([)(-)(| |)(-)(| ihxihTihfxTxfihxfxf nni

.)(|)()(| termsfor these and

],1[for only 0)(])1(,[1

n

n hOihTihf

srijihxhjjhx

If the symbol of a convergent scheme has the factor then

1)1( nz)}({ jxspann

.,)( ,i.e.,Let n0

n0 pipff i

n000 )})({(,)())(( jfSZjijfjfS

ii

To see this let us start with an initial polynomial data

There exists a scheme transforming into

kn f1 11 kn f

n0

n1 .,0

fS

kfkf kkn

If the symbol of a convergent scheme has the factor then

1)1( nz)}({ jxspann

.)())})(({())((but,)})({( 0n

0 jpjipSjfSjfS

.)())})(({())((but, Also, 0n

0 xpxipSxfSfS

Interpolatory subdivision schemes reproduce polynomials. Non-interpolatory scheme usually do not.

.,)( ,i.e.,Let n0

n0 pipff i

n000 )})({(,)())(( jfSZjijfjfS

ii

.,,})())})(({({haveweYet 1 npxpxipS

Example: For the cubic B-spline scheme

We would like to find a local operator Q such that

;}{;}{;}{;1}1{ 333122 xxxSxxSxxSS

I.e., Q is the inverse of nS on

nppQpS ,

The ‘method’ will be local and exact on nQS

1:where,,On RRITn

)}{;}{exampletheRecall( 333122 xxxSxxS

.since

,onthen ,... Define1n

nn2

RITQ

ITQRRRIQ

.)( n ppQpS

))((;)())(( 000 jiToeplitzTfTjififSj

j

Constructing Q

Example : For the cubic B-spline scheme ( n=3 )

)(

))(]([))((016

10310

1610

0016

10320

1610

iiii

iii

ffff

ifRIfffiTf

.)( 3 ppQpS

016

10340

161

016

10310

161000 )())(]([))((

iii

iiii

fff

ffffifRIiQf

.)1(;)0(;))(()())(( 61

32000 ifTjififS

jj

016

10340

16100 ))(( iiii fffiQfg

Approximation order: Given values

This suggests ‘lifting’ the given data points, or control points, by the quasi-interpolation operator Q before applying subdivision.

That works if the control points are taken on a ‘uniform grid’ on a smooth curve (surface)

)R(,)( 40 Cfihffi

.|)()(| 1n0 chxfxQfS

.|)()(|,Then 40 chxfxgS

.)(|)()(|while 20 hOxfxfS

)R(,)(:casegeneraltheIn 1n0 Cfihffi

Subdivision for Surfaces

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2005 Subdivision Summer School

Refinement on regular rectangular meshes

Refinement on regular triangular meshes

Refinement on regular triangular meshes

Can we use a univariate scheme on each of the three directions?

Tensor Product Refinement

A univariate refinement along one direction

A univariate refinement along the second direction

MLK

cba

zzzzz

zzzzz

zzz

zzz

yxyx

zczbz

MLK )1()1()1(),a(z

)1()1()1(),a(z

:schemes splineBox

)1()1(),a(z

)1)(1(),a(z

:schemesproduct Tensor Some

)()(),(

)()(),a(z

:productTensor

212124

21

221

22

2116

121

22

214

121

2121

2121

The tensor product bi-cubic spline scheme with the symbol:

64

)1()1(),(

42

41

21

zzzza

41

41

41

41

161

161

161

161

161

161

161

161

169

83

83

83

83

323

323

323

323

641

641

641

641

The butterfly scheme on a grid:

12

1

21

21 w-

w-

w-

w-

w-

w-

w-

w-

w-

w- w-

w-

w2

w2

w2

w2w2

w2

21

21

21

The averaged butterfly scheme (w=1/16):

1

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

21

21

21

21

165

165

165

165

Analysis of bivariate schemes on regular grids

Let us consider a biivariate scheme

with the symbol

The generating functions

satisfy

),(),(),( 22

2121211 zzFzzazzF mm

21

221

21 21),(

,21 ),( ii

Zii

miim zzfzzF

21

21

21 2121ii

)σ(,ii,ii zza),za(z

a

mii

iiijij

mjj faf

212

21

221121 ,),(

2,21

,

We have four subdivision rules, according to the parity of

and the symbol should satisfy

Using the Laurent polynomial formulation

starting with we have

mii

iiijij

mjj faf

212

21

221121 ,),(

2,21

,

21

21

21 2121ii

)σ(,ii,ii zza),za(z

a

),( 21 jj

,1),( 210 zzF

,),(),(),( 22

2121211 zzFzzazzF mm

,),(),(),(),(),(11 2

221

42

41

22

212121

mm

zzazzazzazzazzFm

21

221

21 21),(

,21 ),( ii

Zii

miim zzfzzF

Plot coefficients’ array?

.0)1,1()1,1()1,1(;4)1,1( aaaa

An example . :

This scheme satisfies the necessary conditions for convergence. Does it converge to a continuous limit? How to check it?

Let us try to find a difference schemes :

If the symbol has the factor , we have a difference scheme transforming differences in one direction from one level

to the next.

)1( 1z

To verify convergence to a continuous limit we have to show that differences in two independent directions tend to zero.

In the example above, we do have a nice difference scheme for differences in the diagonal direction. Let us check it!

)1()1(),( 212

221

121

21

21 zzzzzza

),()1(1

),(),()1( 2

221

21

1

212111 zzFz

z

zzazzFz mm

.0)1,1()1,1()1,1(;4)1,1( aaaa

),()1(1

),(),()1( 2

221

22

21

21

2121121 zzFzz

zz

zzazzFzz mm

Hence, is the symbol of the

scheme taking differences into differences of the

same kind in the next level.

}{ 1,1,m

jimji ff

}{ 11,1

1,

mji

mji ff

222

112

121

21 )1(),( zzzzb

2141

221

1212

2812

181

21

21 ),( zzzzzzzzb

41

,2,22

1

,2,22

1

,2,124

3

,2,2 ||;||;||;||

jiji

jiji

jiji

jiji bbbb

Therefore, differences in the diagonal direction tend to zero .

)1()1(),( 212

221

121

21

21 zzzzzza

The symbol of the butterfly scheme:

12

1

21

21 w-

w-

w-

w-

w-

w-

w-

w-

w-

w- w-

w-

w2

w2

w2

w2w2

w2

21

21

21

2212

212121

1212

11

12

11

12

11

22

11

12

2121

12

112121212

121

2244412

2244422),(

)],(1)[1)(1)(1(),(

zzzzzzzz

zzzzzzzzzzzzzzc

zzzzcwzzzzzzaw

The symbol of the the averaged butterfly scheme:

1

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

321

21

21

21

21

165

165

165

165

)],()1()1(

)],(),([),(

212

22

121

21

12121

21

zzuzz

zzazzazza ww

Theorem 1.

Theorem 2.

.),()1)(1)(1(),( symbol thehave Let 21212121 zzuzzzzzzaSa

.convergent are

,3,2,1,2 symbols with theschemes theif functions

limit generatesThen above. as symbol thehave Let 1

ib

CSS

i

aa

e.contractiv are

schemes thoseof twoifcheck enough to isIt e.contractiv are

,1

),(),(;

1

),(),(;

1

),(),(

masks with theschemes difference theiff convergent is

21

21213

2

21212

1

21211 zz

zzazzb

z

zzazzb

z

zzazzb

Sa

.convergent are schemes thoseof twoifcheck enough to isIt

,),()1)(1)(1(),( symbol with For 21212121 zzuzzzzzzaSa

2212

212121

1212

11

12

11

12

11

22

11

12

2121

12

112121212

121

2244412

2244422),(

)],(1)[1)(1)(1(),(

zzzzzzzz

zzzzzzzzzzzzzzc

zzzzcwzzzzzzaw

For the butterfly scheme on a regular mesh

e.contractiv be tohave

,),()1(2;),()1(2;),()1(2

symbols following with theschemes the is ifcheck to

2121212211

1

zzuzzzzuzzzuz

CSa

)],(1)[1(),( 21121 zzcwzzzq

The scheme related to the following symbol should be contractive

21

21

21 2121Let ii

)σ(,ii,ii zzq),zq(z

q

))) consider Next we 22

212121 ,zq(z,zq(z,z(zq

.1|8||81||||| out that it turns

)||(max|||| norm theuse We,

2,21,0

wwwS

qS

q

Zjijlik

lkq

.08.001|||| out that it turns

)||max|||| norm theuse weNow,

4,43,0

wS

qS

q

Zjijlik

lkq

. cubics reproduces.for is particularIn

.is symbols other two for the holds same The

16116

11

1

aa

a

SwCS

CS

w

w

What if the symbol is not factorizable?

1 .We may use the JSR approach. Here one has to check the joint spectral radius of four local matrices.

2 .If the symbol satisfies the necessary conditions for convergence there still exists a difference scheme, but in general it is a matrix scheme. I.e., a scheme of the form where

For convergence, this matrix scheme has to be contractive .

,21

221

221121 ,),(

2,21

,mii

iiijij

mjj dBd

matrices.22 are}{and ,1,,

,1,,

jim

jimji

mji

mjim

ji Bff

ffd

.0)1,1()1,1()1,1(;4)1,1(satisfy symbol Let the aaaa

We want to show the existence of a difference scheme:

),()]1)(,()1)(,([),()1( 22

21

2221

21212111 zzFzzzczzzbzzFz mm

.),(),()1(),()1( know We 22

212112111 zzFzzazzzFz mm

.)1)(()1,(,)1)(()1,( 21121

21111 zzdzdzzdzd

.0)1,1( have we,),()1(),( Denote 21121 dzzazzzd

.1for 0)]()1()()1[(2

)1(),( 2122112

21

21

zzdzzdzz

zzd

. ),()cz-(1 )]()1()()1[(2

)1(),( 21

22122112

21

21 zzzdzzdzz

zzd

The existence of a matrix difference scheme

)1)(()1,( 21111 zzdzd

An example of a matrix difference scheme

)1()1(),( 212

221

121

21

21 zzzzzza

.)1()(;)3()( 218

112

218

111 zzdzzd

;),()1(),( 21121 zzazzzd

)1)((

)1)((),()1(

221616

3416

3165

2188448

381

211

2212121

22121

31

211

z

zzzazzzzzzz

zzzzzzz

Together with the difference scheme for the y-direction (using the symmetry) we have contraction for two independent directional differences, hence, the scheme is convergent.

Thank you!