Isotropic Boxsplines on the A* lattice
.In i Kim and Jorg Peters
February 27, 2008
Abstract
\\* show that the isotropic nvariate boxspline M, on the A* lat
tice naturally generalizes bivariate boxsplines on the hexagonal lattice
and trivariate boxsplines on the BCC (BodyCentered Cubic) lattice and
shares the desirable properties of the boxspline i. ,I!! Mr. Reconstruc
tion from samples on the A* lattice with M, is informationtheoretically
more t i. i for bounded frequency data than reconstruction with tensor
product Bsplines on the Cartesian lattice.
1 Introduction
As a generalization of uniform Bsplines to several variables, boxsplines have
been used in i ii~i areas. In most applications, shifts of one boxspline on
the Cartesian lattice are used. In particular, a family Mr of nvariate box
splines is popular due to their linear independence and approximation proper
ties (Section 3.2). Members of Mr are defined by rfold convolution, in the n
directions of the Cartesian grid plus a diagonal, so that the footprint of these
boxsplines is anisotropically distorted in the diagonal direction. To improve re
construction of vector fields by isotropic convolution, convolution and shifts on
2 and 3dimensional nonCartesian lattices have only recently been considered
[ .1, 53, 52, 29, 28, 30, 271.
In this paper, we propose a generalization of isotropic bivariate boxsplines on
the hexagonal lattice and trivariate isotropic boxsplines on the BCC lattice
 to isotropic nvariate boxsplines M, (Section 6.2) by leveraging a natural
connection to the A*, lattice (Section 4). This lattice is wellknown in crystal
1,i '1,, and discrete geometry. We apply a newly derived square generator
matrix for the A* lattice in R", rather than nonsquare generator matrix for
embedding in ]R"+1 favored in i I 11 _ i 1.1 Then geometric construction of
the shifts of the isotropic linear boxspline J i1 on the A* lattice simplifies to the
classical construction of boxsplines by projection: The shifts of the isotropic
linear boxspline on A, are the orthogonal projection of a slab of thickness 1
decomposed into unit cubes along the diagonal direction of the cubes (1 i,.i 1).
(b) n 2
1 ,Ii. 1: Orthogonal projection of a slab along the diagonal direction for (a)
n 1 and (b) n = 2.
By comparison, M1 has the same preimage, but its support is distorted by its
anisotripic direction matrix. Therefore M, on the A* lattice shares the desir
able properties of the family Mr on the Cartesian grid, and we can characterize
support, partition and quasiinterpolants for i,!! number of variables n.
2 Previous Work
Piecewise linear hat functions, the shifts of the 3directional linear boxspline
in two variables and of the 4directional linear boxsplines in three variables are
popular basis functions for the 2D and 3D FEM (I !,,I. 1.1. I ii Method), re
spectively. They apply to arbitrary triangular, respectively tetrahedral meshes
in an intuitive way. Higherdegree boxsplines are obtained by repeated convo
lution along the mesh directions, hence require structured meshes. For a small
sample of the literature on the bivariate 3directional boxspline see [20, 21,
22, 23, 39, 12, 13, 18, 19, 37, 15, 38, 25,;., 7, 48, 3, 2, 16]. ('lLI et al. [14]
and Lai [42] gave ti !. i I evaluation of convolutions of hat functions via the
BB(BernsteinBezier)form. The analogous trivariate 4directional boxspline
evaluation has, for example, been discussed by He et al. [';it], and Casciola et
al. [8] extended the approach of ('! Ii et al. [14] to three variables. (' I! et al.
[9, 10, 11] proposed a volumetric subdivision scheme based on the 8directional
trivariate boxspline, M2.
Interpolation on the Cartesian grid by Mr was discussed by Arge et al. [1] and
!, et al. [49] discussed the associated spline space. Neuman [47] proposed a
closed but intricate formula for evaluation of Mr. The literature refers to the
space decomposition corresponding to the polynomial pieces of M, as (n + 1)
directional mesh.
The lattices A., and A* are wellknown in I I 11.,i i'1~ , discrete geometry
and related areas. A standard treatise of the I.l. I is [171. Here the lattices
(a) n 1
are derived embedded in IR"~ Hamitouche et al. [34] recognized the need for
square generator matrices that embed the A, and A* lattices in R". Their
definition, in iterative bottomup fashion, is more complex and the resulting
matrices are more complicated than the one to be presented below.
Frederickson [33, 31, 32] first discussed the isotropicc) bivariate splines on the
hexagonal lattice. The hexagonal lattice is known to be the optimal sampling
lattice in 2 dimensions and is equivalent to the A* lattice. Van de Ville et al.
[1, 53, 52] proposed hexsplines on the hexagonal lattice which share 11 ii!
properties with the boxsplines on the hexagonal lattice. Similarly, the BCC
lattice is the optimal 3D sampling lattice [.11, 28, 27] and is equivalent to the
A* lattice [17]. Entezari et al. [29, 28, 30, 27] were the first to investigate the
isotropicc) 4 and 8directional trivariate boxsplines on the BCC lattice.
3 Notation and Background
The dimension of vectors and matrices will be determined by context. Some of
the specific vectors and matrices are:
ie the kth unit vector,
I the 'l, iI li matrix,
j :=(1, ,1) the diagonal vector,
J : jjT the matrix composed of Is only,
0 : (0, 0) the zero vector,
and the dot product is defined as x y := xTy E R.
Hj is the 1l i d i 1i !defined as j x = 0.
Lattices are I I. . in calligraphic upper case; e.g., n and An.
Following the convention in [24], an n x m matrix will be interpreted as
a set of column vectors allowing iii.LIi idl i or
a linear transformation ]RI ]R'.
Column vectors are used as either vectors or points depending on the
context.
Lattice generator matrices (Section 3.1), such as G, A* and A, are i , l
in bold upper case.
Boxspline direction matrices (Section 3.2), such as E and T,, are I" .
in upper case unless they are derived from generator matrices; e.g., P.
Linear ,, ... ,,., ....., ... are I I.. . I in upper case, e.g., B but are I" . l
in bold upper case when related to generator matrices; e.g., P.
C'(P) is the convex hull of the points in P.
A matrix B e Zmx" is unimodular [24, II .7] if
det Z = 1, VZ C B : Z is square and rankZ = n.
If m= n then B has an integer inverse B 1 E Znn.
3.1 Lattice packing and optimal sampling
Given an m x n matrix G with m > n and rank(G) n, all integer linear com
binations of its columns, GZ", define (the points of) an ndimensional lattice,
, embedded in Rm:
{:= Gj e :j Z}.
G is called a generator matrix of n, and we call the columns of G a basis of
,. The choice of a generator matrix for a lattice is not unique.
Lemma 1. If B E Znxn is unimodular then G and GB generate the same
lattice points: GZ" GBZ".
...... Since BZ" C Z" and every integer linear combination of the basis is a
lattice point, G(BZ") C GZ". Conversely, (GB)(B 1Z) GZ C GBZ"
since B 1Z" C Z". O
If one lattice can be obtained from another by a rotation, reflection and change
of scale we , 11. are equivalent, or similar, written [17]. Any ndimensional
lattice , has a dual (or reciprocal, or polar) lattice given by
{ xC ." : x u E Z,Vue ,}. (1)
If G is a square generator matrix of n,, then GT is a square generator matrix
of C [171.
A Dirac comb or ~!!... function on the lattice GZ", GE Rnx", is defined as
fIIG(x) := 6(x Gk) (2)
kCZ
where l is the Dirac delta function. Its Fourier transform is (see [26])
27x
IG() {UIG}() detGI L 6(uw 2 Tk) (3)
where F{} is the Fourier transform operator. Sampling a function f on the
lattice GZ" is the same as iilli .1ll1 i; f with IIIG,
(flllG)(x) = f (Gk)(x Gk).
kCZn
sampling. Although both lattices (a) and (c) can reconstruct the original signal,
(c) corresponds to a denser sampling (d).
Therefore
S{f lIG}(w) f()* IIG(w)
2
rdet G f(w 2G k) (4)
kCZ
where f(w) : {f} (w). In other words, in the Fourier domain the (scaled)
Fourier transform of f, 27rf(c)/I det G is replicated on the (scaled) dual lattice
of GZ, 27G T [26].
The sphere packing problem, "! ... ,I ..... I., we can pack identical spheres in R"?",
is one of the oldest problems in geometry [17]. The lattice packing problem is
to find the lattice that induces the densest sphere packing when the spheres are
located at the lattice points.
The lattice packing problem is closely related to the optimal sampling lattice for
multidimensional signal processing. Assuming the input signal has bounded
f. 1i1. i, we can reconstruct the original signal using a sphereshaped till I in
the frequency domain. Since the lattice in the frequency domain is the dual of
the sampling lattice, the more densely we can pack the spheres (reconstruction
t!il i i in the frequency domain, the sparser a sampling lattice we can choose
in the space domain to reconstruct the original signal (1 !,.i! 2). Therefore,
for input signals with bounded f!r n! the optimal ndimensional sampling
lattice is the dual of the optimal sphere packing lattice [28, 41, 27].
A 1.. '!,, of a lattice packing is the proportion of the space occupied by the
spheres when packed. A center .1 ,. "i,, of a lattice is the number of the lattice
points per unit volume, which can be obtained by dividing its I iiil by the
volume of the unit sphere [17]. Therefore, larger 1. !iil (or center .1. i i I'
implies that its dual is a more IT, '. " sampling lattice. Table 1 and I ,,.i! 3
respectively show the center ,l. I.i and the l1. ii of several important lat
tices, i '1,! ii1 poorer sampling !tn! i i, of the traditional Cartesian lattice Z"
compared to other lattices.
2 2 /2(n+ 1) 1/2
nn/2
2"(n + 1)("1)/2
231.525
2(n+2)/2 31 2
2(n1)
Table 1: Center l. iiil of several lattices. D :
[171. See Section 4 for AS, and A*k.
{j E Z : EC j(k) is even}
i i, i.. 3: D. !!i I of the lattice packing for several lattices up to dimension 10.
3.2 The Boxsplines Mr
A boxspline ME is defined by a direction matrix E. Unless mentioned specifi
cally, we assume S E ZnX (m > n) and ranE = R". Geometrically, the value
at x E ranE of the boxspline ME is defined as the (normalized) shadow 1. i i I
of the (m n)dimensional volume of the intersection between the preimage of
x and the mdimensional halfopen unit cube L := [0..1)": [24, I3] (see e.g.,
1i ,,u. 1)
Ms(x) := vol n ( l{x} n )/I det E (5)
where E is viewed as a linear transformation E : RI R" and the preimage of
x is defined as [24, I7]
 l{x} =T(TEE) l{x}+kerE.
Let H (E) be the collection of all the hyperplanes spanned by the columns of E.
[24, page 8] We call the shifts of all the hyperplanes in H (E) knot planes: [24,
page 16]
r(): U H+z". (7)
ffH(3)
(n 3)
(n > 3)
A, P,
ME with cE Znxm is a piecewise polynomial function on ranE, delineated
by the knot planes, of degree less than or equal to [24, page 9]
k () := m dimranS. (8)
Specifically, k (E) = m n if ranE = R".
The centered boxspline Mc of M7 is [24, I211
ME : M3(. + E /2). (9)
Given an invertible linear map L on R", [24, I23]
Ms= I det LIML o L. (10)
The Fourier transform of M7 is [24, I17]
1 exp(i C())
ME (w): F ME } (w) [ i ix := i V). (11)
Specifically, if M7 is centered, [24, page 11]
AMs() since( c w). (12)
By [24, page 9], the (closed) support of M7 consists of the set
suppME E {I t : 0 < t < 1}
where 0 := [0..1]" is the closed unit cube and t is the element of t associated
with by Et. Assuming ranE = R", let [24, page 81
B(E) : {Z C : #Z = rankZ n} (13)
In other words, B(E) is the set of all bases of E. The support of M7 is composed
of the parallelepipeds spanned by Z e B(E):
Theorem 1 (Support decomposition; [241 I.';). Let ranE = R". I!. .. exists
points az E E{0, 1}m, Z E B(E), so that E is the essentially 1ii ..il union of
the sets
ZO+az, Z e B(E). (14)
The cardinal spline space [24, II1]
S := span (Ms( j))jc (15)
is the spline space spanned by the shifts of Ms on Z".
The shifts of Ms are linearly independent if and only if E is unimodular.
The spline space S, contains some polynomials. In other words, Ms can re
produce the polynomials in [24, page 52]
HMI := H n S,
where H is the set of all the polynomials on R". Specifically,
Hm,() c LHMs
where
Ha is the set of polynomials of (total) degree up to a,
m() : min{#Z : Z C A()} 1 and
A(E) : {Z C : E\Z does not span}.
In other words, ME can reproduce all the polynomials up to (total) degree m(E).
The next section gives a concrete construction.
A quasiinterpolant Q7 for a boxspline ME provides a fast way of approx
imating a function f with a spline QIf E S7 [24]. Here we focus on the
quasiinterpolant which provides the optimal approximation order m(E) + 1:
[24, page 721
(Qf) (x) := M(x j)A (f( + j)) (16)
jCZ
where A. is a linear functional [24, III22]
A7 (f( + j)) := (O) (Dof) (j) (17)
and a E Z" is a multiindex with a := cEa(). The Appell sequence
{ga} in (17) can be computed either recursively as
{go a n ([24, III19])
.9111 II
where
S(f) : M (j) f(j), (18)
or from the Fourier transform M!: [24, III34]
ga(O) ([ iD (1/M )) (0). (19)
Note that I is the normalized apower function
a'vi )
The BoxSpline M,. Boxsplines defined by possibly repeated (n+1) distinct
convolution directions are also called boxsplines on the (n + 1)directional mesh
[1]. Given the n x (n + 1) direction matrix
T := [I I ]
[ ii " in
j ] E nx(n+1),
the boxsplines with ...l! '"l'. !', r in each direction is defined by the n x r(n+1)
direction matrix [24, page 80]
T : T1.
j 1
As pointed out in Section 2, this family of boxsplines has been widely used.
Note that TI [1 1] in the univariate case and M. can be viewed as a
generalization of the uniform Bsplines of odd degree to arbitrary dimensions.
Following the notation established in [24, page 80], we define
Mr*: MT,.
4 The Lattices An and Ai* as subsets of _' 1
The (ndimensional) lattice A,, embedded in ]R"+1 has points {x E Zn+1 :j x =
0} and can be generated by the (n + 1) x n matrix [17, page 109]
1
1
Ac : .. Z(n 1)xn. (21)
1
1
We note that j Ac 0,
easily check (see also [17,
(n + 1) x n matrix
i.e., all the basis elements lie in Hj C ]R"+1. We can
page 115]) that its dual A* can be generated by the
1 n/(n + 1)
1/(n + 1)
1 1/(n + 1)
1/(n + 1)
n/(n + 1)
j/(n + 1)
1/(n + 1)
I R(n +1)xn
Some examples of A,n and A* are:
A2 and A are equivalent to the hexagonal lattice.
A3s D is equivalent to the FCC (FaceCentered Cubic) lattice.
A DV is equivalent to the BCC (BodyCentered Cubic) lattice.
The BCC lattice and hence A*, is the optimal sampling lattice in 2 and in 3
dimensions [46, .In, 28, 29, 41, 30, 1, 27]. In dimensions higher than 3, 1 ,!.. 3
shows that An, packs spheres better than the Cartesian lattice, making A* a
better sampling lattice than Z".
The basis of A, can be taken from an ndimensional equilateral simplex [34].
Lemma 2 (Geometric construction of An, in R"+1). Let oa be an equilateral
ndimensional simplex one of whose vertices is located at the origin. IP. , the
n edges of ~Tn emanating from the origin form a basis of a lattice equivalent to
4 ....f By Lemma 1,
AcB [I Z(n+ )xn (22)
AcB j
with
1
1 1
B := and B1
1
1
also generates An. Since
I. I _, = Vv E AcB
t V2 Vvj, Vk e AcB, vj Yk,
the simplex C'(0 U UveAcB v) is equilateral hence equivalent to ;!u an. O
Lemma 3 (Geometric construction of An in IR"+1). An can be generated by
P: I 1 JER(n+l)x(n+l), (23)
n+l
the orthogonal projection of the (n+1) dimensional Cartesian lattice Z"+1 along
the diagonal direction j.
J ...... For
B := 0 1 1 Z(n+1)x(n+1) where B1 = ] j
we verify that
A*B= A n (n+l)I J R(n+1)x", (24)
C P ; n + 1 j I
where A* is the matrix of the first n columns of P. The last column of P
is a linear combination of the first n columns, A*. By Lemma 1, the claim
holds. E
5 Boxsplines on nonCartesian lattices
By (10), given a square generator matrix G, a spline generated by the shifts of
the (scaled) boxspline
M := det GMGa (25)
on the (possibly nonCartesian) lattice GZ" can be expressed as the one gener
ated by the shifts of Ms on the Cartesian lattice Z" with change of variable:
SMH(. j)a(j) = E (G k)a(Gk) (26)
jcGZ" kcZ"
where a : GZ" R is the mesh function (spline i.... h i I , on GZ". De Boor
[21, page .' 1] already pointed to this relationship in the bivariate case.
5.1 Spline space
Let
SG : span Ms( j))
\ /j'eGZ"
be the spline space spanned by the shifts of Ms on GZ". This notation becomes
consistent with (15) by omitting G I and defining
5.2 Quasiinterpolant
Lemma 4. Let D' := IoeG D" be the composition of directional derivatives
D, := v g v(j)Dj along the columns of G and {ga} the Appell sequence of
A7 (17). 1!b. quasiinterpolant QG for SG ., ....I by the functional
A (f (. +j)) : A ((fo G) (. + G j)) (27)
= g ,(O) (Dof) (j), j GZn, (28)
a
provides the same optimal approximation power as does Q7 .. 1; .... by A7 for
S7.
i .f'... If we define
(Qf) (x): (Q, (f o G)) (Gx)
then, since f =f o Go G1,
(f Qf)() (= (( o G) Q (fo G)) (G x) = (f QSf) (),
for f := foG and x := G 1 ox, i.e., QG has the the same approximation power
as Q,. Since
(Q (f o G)) (G x)
SM(G x k)A, ((fo G) ( + k))
keZ'
I detGMG(Z j)A ((f oG)(. + Gj)),
jcGZ"
f (G +hGik)
Dk (fo G)= lim
htO h
the corresponding functional AG is for j E GZ"
A (f ( + j))
A ((foG) (+ Glj))
Sg(0() (Do (f o G)) (G 1j)
E g,(O) (DGf)(j).
a~ ,z
6 The isotropic boxspline family M1, on A*
The boxspline family Mr and the A*, lattice have a close relationship, that is
apparent when considering the linear boxspline M1 and A*. The lattice A*
can be constructed by projecting the Cartesian lattice Z"+1 along the diagonal
direction j. By (20) and (6)
TiT(TiTiT) 1
where A* are the first n columns of P := I
Lemma 3. Therefore
1 J E R(n+1 )x(n+1) defined in
Ti 1{x} Ax + span(j), x R,j E n+1
i.e. the preimage of suppMi Ti C R" is PO C Hj C ]R+ 1 and PD is
isotropic since the directions, i.e. the columns of P := I  J, are
isometric: they have the same lengths and
isotropic: the inner product (hence the angle) between ;,i! two directions
is the same.
f(G)
(DGikf) o G,
(by (17))
1 [ (n +l )IJ ] Ap* e R(n+l)xn
n + 1 jJ p
LL
(a) Z2 (b) AZ2 A2
1 L ,1i" 4: Geometric construction of A, in R".
However, suppMi is anisotropic, (1 ,.i. 6d), since A, : R'" Hj C ]R"+1 is
not an orthogonal transformation:
A TA~ I J.
n+1
We can use P directly as the direction matrix of a linear boxspline
MI := Mp : i C R"'+l  R.
Evidently M+ (and M? defined by U31 P) has an isotropic support (1 iii. 6b
and 6e). But neither Mr nor M+ are useful since
1. M, is anisotropic and
2. the domain of M, is not R" but Hj embedded in ]R"+1.
The remedy (1 ','i. 6c and 6f) is a square generator matrix of A, to be presented
next.
6.1 The new square generator matrices A and A
To construct square generator matrices for A,, and A* in RI, we consider a
linear map that simply scales along the diagonal j. The map transforms a point
x E R" according to
x + (j )j
n
where c is the scaling factor.
Lemma 5 (Geometric construction of AS, in R"). AS, can be generated by
A:= I + Jc with c, : 1 / 1.
n
(b)XZ A'
1 ,!,.i 5: Geometric construction of A* in R".
JF,..,, Any vector ij ik for i / k is parallel to Hj and hence its length remains
/2, unchanged by A and regardless of the dimension n. To show that the n
dimensional simplex Ct({Aij : 1 < j < n} U 0) is equilateral, we verify that the
vectors ij satisfy
+ 12
Cn+1) +(n 1)
(n /n
42. (30)
The claim follows by Lemma 2.
Note that two 11 I, !I choices of c, produce the equivalent result with respect
to Hj because I J/n projects ij on Hj.
Lemma 6 (Geometric construction of A* in ]R"). A* can be generated by
: I + c E .' with c*
n
1
v1n+
F ..... Since
1 ( 1) ( =1) (Cn + 1) (c; + 1)
cc* + c, + c* = 0 and hence
ATrA I + (cc*n + cn + c)J
cc + c, + c + 1,
D
Under the diagonal scaling A, the length of j becomes the same as those of the
unit vectors (1 [,Li!, 5):
IA*j IAij, V j < n.
As with A, two roots of c* result in equivalent transformations with respect to
Hj.
Aij 2= 2 Aij Aij
(a) Z2
(b) M+ on Hj ; R2
(e) A41 on Hj R3
(c) M* on XZ C R
(f) M* on AZ2 C R2
1i io. 6: (top) l,11it of linear univariate boxsplines and (bottom) shifts of (the
support) of linear bivariate boxsplines.
(a) Mi on Z
(d) M1 on Z2
For example, for n = 2,
1 1 l//3 1 l/v/3
2 1 1//3 1 1//3
and for n = 3, the BCC lattice, the two choices are
1 5 1 1 1 
A : 1 5 1 or 1 1
1 1 5 1 1
6.2 A as the domain lattice of M*
We now interpret the columns of the matrix T,* := A*T, as direction vectors in
Rn.
Lemma 7. T* is isometric and isotropic.
F ......f Since (31) implies isometry, we need only verify isotropy,
1 1
(Az (j)) (Ai,) 1 Vi and (A*i) (A*i) Vii 7 i .
n+' n+'
Therefore MT* is isotropic and AZ" can serve as a domain lattice for the box
spline family (1 Iii, 6c and 6f)
l1, := IdetA* MT. Mr oA*1 (32)
as in (25). Alternatively, the isotropy is verified by observing that the transfor
mation A)A* 1 is orthogonal
(AA* )(A A* 1) I
and hence, in contrast to (29) for M1, isotropy of PO is preserved when com
puting the preimage,
T* {x} A* {x} + ker T*. (33)
By ('i) a spline generated by the shifts of M* on AZ" An* can be expressed
as
S1, ( j)a(j) = M,( 1 j)a(Aj). (34)
jeA*Z jcz"
Therefore M* inherits most of the properties of Mr.
Lemma 8. M* is centered.
,,,,[.f By (9),
JI *+ E ^ I, (+u")
since EET,* = 0. E
Lemma 9. Mr* M T*
.,,.,,,f By (12),
F {MT } (w) J since (Tw)
CT,*
and
F {M *} (w) sin( ) since T) ( Tw) = since (Tw)
^T* CT* CT*
because sine is an even function. The claim holds since the Fourier transform
is invertible. E
Lemma 10 (even function). M, = M,*().
f,,.,' By (10) and Lemma 9,
MT, = det (I) M T, o (I) = MT,(). (36)
Lemma 11 (basis functions). I ., shifts of M,* on A*Z" are linearly indepen
dent.
Pf....f Due to (34),
S := span ( l, (.j5))j
and ST, share the same linear independence property. Note that
B(T) = B(T) U T \( (37)
since ; n directions in T1 span R". The claim holds by Section 3.2 since
detZ = l, VZ B(T,).
By (8), Mr, hence M*, is piecewise polynomial of (total) degree less than or
equal to (n + 1)r n.
Lemma 12 (polynomial reproduction). M,* can reproduce all the ;'.1',......'..,
of (total) degree up to 2r 1:
m(T,) 2r 1.
jF...... Due to (34), m(T*) = m(T,). For M1, we have to remove at least 2
directions so that the remaining directions in Ti no longer span R', hence
m(Ti) ((n + 1) (n 1)) 1 2 1 1.
In the same way, at most r(n 1) directions in T, span a hyperplane, therefore
m(T,) = (r(n + 1) r(n 1)) 1 = 2r 1.
Note that m(T,*) does not depend on the dimension n.
Also, since the knot planes generated by T* are linear transformed ones of F (T,),
the mesh inherits the I. 1. .1. ._ of the (n + 1)directional mesh.
Lemma 13. 1!, .'* are n(n + 1)/2 nonparallel planes in H (T,).
'! ..... There are n planes generated by the n unit vectors in I and (, 2) ad
ditional nonparallel planes are spanned by the diagonal direction j and n 2
additional unit vectors yielding a total of
n + n2 n + 2n( 1)= n(n +1)
nonparallel planes in H (T,). D
Next, we characterize the partition of R" into simplices by the knot planes in
F (TT).
Lemma 14 (Partition of (n + 1)directional mesh = Kuhn triangulation). 1.
knot planes in H (T,) partition the unit cube 0 into n! simplices
n i
a := C(H(V,), V : U J i(J), 7E n (38)
i= j=1
where n" is the set of all the permutations of {1, n ,n}.
The partition {af }Jtrn is called Kuhn triangulation [40, Lemma 11, [43, page 140]
and [1, 5, 6].
jF.4.,.f Recall that T1 [I, j]. All planes with normal direction ij ik, j / k,
intersect the interior of [ and are generated by Ti\ {ij, ik} i.e., as knot planes
of M1 generated by n 1 vectors including j. Unless two vertices vj, vk E V,
for some permutation 7, there exist indices a and f so that
vj(a) l, v (a) 0 and vj(,3) 0,vk(,3) =1.
and hence the knot plane with normal i, i3 separates them,
(ia ip) vj = 1 > 0 and (ia i3) vk = 1 < 0. (39)
Conversely, since knot planes excluding j are axisaligned, neither they nor their
shifts on Z" intersect the interior of the unit cube 1. It remains to show that
no shifts of the knot planes with normal i, i3 separate vertices in of a simplex
Ca for the same fixed permutation 7. Since shifts by j E Z" within the knot
plane, i.e. j (ia i3) j(a) j(B) 0 result in (ia i3) (x j) = 0, we can
assume j (i, i3) j(a) j(f) > 0. Then, for all v E {0, 1}",
j (a) +j(/ )+ < 0 v(a) = 1, v() 0
(i i) (v ) j(a) +j() 1 < 1 v(a) =0, v() 1
j(a) +j(() < 0 v(a) =v(B)
<0.
The case j (i, i3) < 0 corresponds to a flipped normal and yields (ia i3)
(v j) > O
We can investigate the structure of ,q 11, by first decomposing it into paral
lelepipeds. I ii we consider suppM*. There are two decompositions.
Lemma 15. 1! (closed) support of M* is the i .. ... '.,ll, disjoint union of the
(n + 1) parallelepipeds
{ZD: Z e B(T*)} (40)
or, .I .... i ', 1,.
{Z[+Cz : Z e B(Tf*)} (41)
where (z := T*\Z. In either decomposition, all the parallelepipeds are congruent.
F'...." See Section 7.1. O
This result is easily extended to T* since
T,* { :0 < t:0
SE T'* ETJ*
For Z E B(T*), the pair (Z, (z) is a linear transformation of the pair (I, j).
Therefore Z is decomposed in the same way as the unit cube 0 is by Kuhn
triangulation and suppM* consists of (n + 1)! simplices. This count also agrees
with the number of modular cells in the first neighbor i, '. I of A* [34].
The two i of the decomposition of suppMr in Lemma 15 can be viewed
as cubical meshes such that one is the flip of the other [4] since each cubical
mesh can be viewed as the projection of the (n + 1)dimensional cube along two
opposite directions of its diagonal.
Here we investigate M2 that can reproduce all the cubic polynomials. (Lemma 12)
Lemma 16 (Quasiinterpolant for M*). 1., optimal quasiinterpolant of M2*
is 1. 7; ... by the functional
A (f ( + j)) : A (f(.+ )): f Df (j), j AZ". (42)
J',,,,f See Section 7.2. E
For discrete input f : A*Z ' R, we approximate the directional derivative
along C E R" by finite differences, e.g.,:
D f f( + ) + f( ) 2f. (43)
Therefore
A (f( + j)) f (j) (f( + ) + f(j 2f(j))
21
(+nf^l)f)_ f(j) (44)
where f(j) is the average of 2(n + 1) ;i1i ,1. 0 neighbor data. When specialized
to two variables, this agrees with Levin's formula [44].
7 Appendix
7.1 Proof of Lemma 15
J..,,f Due to the relation (34), we need only consider M1. Let Zj E B(T1) be
a basis of Ti and
:= Ti\Z = .
For aj := azj in Theorem 1, there are only two choices, aj E {0, j } since no
C E Zj doubled fits into the support
Zj3+(C 9C(+( TIo, V(cZj.
Now assume
a = 0 and ak = C for Zj, Zk E B(TI),Zj / Zk.
This leads to a contradiction as we prove that the two parallelepipeds Zj 0 +aj
and Zk e +ae are not essentially 1i ii., I but share a point p interior to both.
Let
CezTnZh
Then p E Z (0, 1)" + a, since
p1
z\ z
Zj\z ( 4 Czj + a
where aj = 0, Zj = (Zj\Z) + (Zj n Zk) and + denotes .1iii..i union. Also
p E Z2(0, 1)" + a~ since
1 1 1
2 Z 4 2
Cezjnz, ceT1
12 1 1 1
2 CZZ 4Z 2 2c
(( znZk CZk
(ECTI, C 0)
(3
4c
E + j z +akz
znZk (CZc\zj )
This establishes that there are only two alternatives. Next, we prove that all
parallelepipeds are congruent. Let's consider the decomposition
{Z*D: Z* e (T,)}.
The following lemma makes the proof easier.
Lemma 17. Let the matrices A,, Ba E Rnxn 1, 6', 1 as follows:
A(,k) k a
0 otherwise
and Ba(j,k)
Sj k =a
0 otherwise.
I! .. for Zj : I Aj Bj,
Zj (I+ J)Z,
I + J and Z2
i
S....f Both can be verified using the following relations.
AjJ = J, BjJ = A, j AT = J, AjB = A, BjA,
By, A2 Aj.
I I note that A2 = I + J. For Z: := AZj and Z: : A*Zk,
Zy = (AZjZ A 1)Zk
where we can verify that
(A*ZjZ A 1)(A*ZjZkXA 1)T = ZjZA2ZTZTXA
A (I + J)A*
(A = A)
(Lemma 17)
Therefore AZiZA* 1 is an orthonormal (rigid) transformation hence all the
parallelepipeds Z* ,Z* e B(T,) are congruent. It's quite straightforward when
one of the basis is AI. Note that ;, permutation of the columns of Z* e
B(T,) can be done with a permutation matrix, which is also an orthonormal
transformation.
The other decomposition can be verified in the same way. E
7.2 Proof of Lemma 16
We first derive the quasiinterpolant QT2 for ST2 defined by AT, (17). Then QA
for Si defined by A* can be derived by (27).
j.,.,,f Following lemma will make the computation easier.
Lemma 18. For an odd function f, T2f = 0 (18).
j.,,,,,f
T2 f = M,(j) f (j)
(by (10))
M2(j)f(j)
M,(j) f(j)
j
(f f( ))
(change of index)
therefore pT2 f = 0.
We compute ga(0) for each degree a.
1. 0a =0
ga(0) = go(0) = 1 By [24, page 68].
2. a = 1
By Lemma 18, p7.  
g 1 a1 E P
3'a
III (III1111 )90 111
therefore ga(0) 0.
3. lal 2
By [24, page 11],
fT(w) : {MT2}(w)
Yi sinc(TwC) = sinc2(~w).
CT2 sCT1
Therefore, By (19), for j / k,
1 ) (1
MT2 e\j,ij,ik sinc2 (Tw)
since2 (jTw)sinc2 (~j )sinc2 (wk)
Since sinc(0) = 1, with the help of MAPLE, we can compute
DjDk (0)
( 1 M)
(DDkyD sc2 ( )sc2 (l)sc2( ) (0)
since (ci(j)sin (ci(j)sinc (; + CI))
Also,
(D 1 ) 1
Sit ) h since2 (Tw)
~C T1e \j,ij }
Again, with the help of MAPLE, we can compute
( 2 (0)
1 M )
D s1c4 (0)
jsinC4 ,))
By (19), for j / k,
gij+i, (o) iDl'+i' 1 (0)
1T2 M
g2ij (0) (I
1
DjDk (0)
MT2
1D ) (0)
2 M)
0 and
(DjDk
iD i 1 (0)
MT2
4. la 3
By [24, III19],
g.= I E (T,I
3a
III ( ( 11 II
)go + (?1 II
/31 1
) 3+ 1 (Ii=2
1/312
hence g(O) = 0 because
PT II II 0 by Lemma 18,
gg = 0 for l 1 and
p  IIT
Summing up,
AT2 (f( +j))
0 for 131 = 2 hence Ia f3 = 1 by Lemma 18.
E g) (OD) (j)
a
1
=2
a2
Df) (j)
D 2ff D U()
(SDJ)21))J^
J
Now, by (28),
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