Geometrical Constructions for Ordered Orthogonal
Arrays and (T, M, S)-Nets
Ryoh Fuji-Hara and Ying Miao
Institute of Policy and Planning Sciences
University of Tsukuba
Tsukuba 305-8573, Japan
[email protected] and [email protected]
Abstract
The concept of a linear ordered orthogonal array is introduced, and its equivalent
geometrical configuration is determined when its strength is 3 and 4. Existence of
such geometrical configurations is investigated. They are also useful in the study
of (T, M, S)-nets.
1
Introduction
In 1987, Niederreiter [13] introduced the concept of a (T, M, S)-net in base b. This new
concept significantly generalized that of a family of low discrepancy point sets in the
S-dimensional unit cube [0, 1)S due to Sobol’ [15], which are useful for quasi-Monte Carlo
methods such as numerical integration.
Let S ≥ 1 and b ≥ 2 be integers. An elementary interval in base b is an interval of
the form
E=
S
Y
[ai b−di , (ai + 1)b−di ),
i=1
where ai and di are non-negative integers such that 0 ≤ ai < bdi for 1 ≤ i ≤ S. The
volume of E is
S
Y
b−di = b−
PS
i=1
di
.
i=1
For integers 0 ≤ T ≤ M , a (T, M, S)-net in base b is a set N of bM points in the
S-dimensional unit cube [0, 1)S such that every elementary interval E in base b having
volume bT −M contains exactly bT points of N .
The use of (T, M, S)-nets in the computation of definite integrals has had particular
impact in the area of finance [2]. The principal current application is in the finalcial arena,
1
where they are used to price exotic options. They also have found applications in pseudorandom number generation, which in turn are employed in cryptographic protocols [13].
(T, M, S)-nets have received considerable attention in recent literature. For a good
survey of known results, see [3]. (T, M, S)-nets have been characterized by combinatorial
objects called orthogonal arrays.
Let A be an m × s array over the set V of b symbols, and R be a subset of columns of
A. We say R is orthogonal if R contains every |R|-tuple over V exactly m/b|R| times as
rows. If A is orthogonal for any t-subset R of columns of A, then A is called an orthogonal
array (OA) of strength t, denoted by OAλ (t, s, b), where λ = m/bt .
Schmid [14, 11] proved that (T, M, S)-nets are equivalent to a type of orthogonal
arrays called orthogonal orthogonal arrays. Lawrence [7, 8] independently showed an
equivalent result in terms of generalized orthogonal arrays. In this paper, we will present
our results in terms of ordered orthogonal arrays. The definition we adopt here is from
Edel and Bierbrauer [5], which is equivalent to Schmid’s, and has been used by Martine
and Stinson [9, 10]. Note that such arrays were also called cubical orthogonal arrays by
Colbourn, Dinitz and Stinson in [4].
Let V be a set of b symbols. An ordered orthogonal array (OOA) A of strength t is
an m × sl array over V which satisfies the following properties:
1. The columns of A are partitioned into s groups of l columns, denoted by G1 , G2 ,
· · ·, Gs ;
P
2. Let (t1 , t2 , ..., ts ) be an s-tuple of non-negative integers such that si=1 ti = t, where
0 ≤ ti ≤ l for 1 ≤ i ≤ s. If R is the subset of the columns of A obtained by taking
the first ti columns within each group Gi , 1 ≤ i ≤ s, then A is orthogonal for such
an R.
We use the notation OOAλ (t, s, l, b), where λ = m/bt .
Note that we can assume without loss of generality that l ≤ t ≤ sl when we study
OOAλ (t, s, l, b). When l = 1, the above definition reduces to that of an orthogonal array,
i.e., an OOAλ (t, s, 1, b) is equivalent to an OAλ (t, s, b). When l = t, we have the following
important result originally due to Lawrence [7] and Schimid [14].
Theorem 1.1 ([7, 14]) There exists a (T, M, S)-net in base b if and only if there exists
an OOAbT (M − T, S, M − T, b).
As an immediate consequence of Theorem 1.1, we have the following corollary.
Corollary 1.2 There exists an OOAbT (2, S, 2, b) if and only if there exists an OAbT (2, S, b).
Proof This is clear since Niederreiter [13] and Mullen and Whittle [12] showed that the
existence of a (T, T + 2, S)-net in base b is equivalent to the existence of an OAbT (2, S, b).
2
2
Also observe that if we take t1 = t2 = · · · = ts = 1 in the definition of an ordered
orthogonal array, then we get an orthogonal array OAλ (t, s, b).
Theorem 1.3 If there exists an OOAλ (t, s, l, b), then there exists an OAλ (t, s, b).
In this paper, we will introduce the concepts of a linear orthogonal array and a linear
ordered orthogonal array, and then try to determine their equivalent configurations in
finite projective geometries. These geometrical configurations will be used to investigate
the existence problems of such linear arrays. These in turn can yield the existence of the
corresponding (T, M, S)-nets in base b. Some computational results will be also reported
with discussion.
2
Linear Orthogonal Arrays and Finite Projective
Geometries
Let C be the set of rows of an m × s orthogonal array (or ordered orthogonal array,
respectively) A over the finite field V = GF(b), where b is a prime power. If C satisfies
the following conditions, then C is called a linear space over GF(b):
1. x + y ∈ C for any x , y ∈ C ;
2. αx ∈ C for any x ∈ C and α ∈ GF(b).
If C is a linear space, then A is said to be a linear orthogonal array (or linear ordered
orthogonal array, respectively). Note that, since C is a set of rows, no repeated rows
are permitted to occur in any linear orthogonal array (or linear ordered orthogonal array,
respectively). It can be seen that C is a linear space over GF(b) if and only if there exists
an n×s non-degenerate matrix G, called generator matrix, with entries from GF(b), such
that
C = {x G : x ∈ GF(b)n }.
Therefore |C | = m = bn holds for some non-negative integer n.
There is an important result due to Bose [1] for the construction of orthogonal arrays
and codes.
Theorem 2.1 ([1]) Let R be a t-subset {i1 , . . . , it } of the coordinats in a linear space C.
Then the set of t columns obtained by taking coordinats of R from C is orthogonal if and
only if the column vectors gi1 , . . . , git of G are linearly independent.
From Theorem 2.1, we can say that any t columns of G are linearly independent if
and only if C is a linear orthogonal array of strength t.
3
The points of a finite projective geometry PG(n − 1, b), b being a prime power, can
be represented by n-dimensional vectors over GF(b). Two n-dimensional vectors x , y
represent the same point of PG(n − 1, b) if and only if there is an element λ ∈ GF(b) \
{0} such that x = λy . Therefore, if column vectors of G are from distinct points of
PG(n − 1, b), then the linear space generated by G is an orthogonal array of strength 2.
Generally, there exists an n × s non-degenerate matrix G such that any of its t columns
are linearly independent if and only if there exists a set of s points in PG(n − 1, b) no t
of which are in a (t − 2)-flat. This imples the following result:
Theorem 2.2 ([1]) There exists a linear orthogonal array OAbn−t (t, s, b) if and only if
there exists a set of s points in PG(n − 1, b) no t of which are in a (t − 2)-flat.
3
Linear OOA of Strength 3
Let S be a set of points in PG(n − 1, b) and P be a point of S. A tangent line to S is a
line meeting S in exactly one point. When the intersection point is P , it is said to be a
tangent line to S at P . A line meeting S in exactly two points is called a secant of S.
Theorem 3.1 There exists a linear OOAbn−3 (3, s, 3, b) if and only if there exists a tangent
line to S at each point P ∈ S, where S is a set of s points in PG(n − 1, b) no three of
which are collinear.
Proof Suppose that S is a set of s points in PG(n−1, b) no three of which are collinear,
and there is a tangent line to S at each point P ∈ S. Then we can construct a sequence of
points Q = (a1 , b1 , c1 , . . . , as , bs , cs ) in PG(n−1, b) which satisfies the following properties:
1. S = {a1 , a2 , . . . , as } is the set of points in PG(n−1, b) no three of which are collinear;
2. li is the tangent line to S at the point ai , and bi is a point in li \ {ai } for each
1 ≤ i ≤ s;
3. ci is a point of PG(n − 1, b) not on li for each 1 ≤ i ≤ s.
Let Gi = (ai , bi , ci ) for 1 ≤ i ≤ s. From Theorems 2.1 and 2.2, the array A obtained
from the points of Q is a linear OOAbn−3 (3, s, 3, b) with groups Gi , 1 ≤ i ≤ s, if and only
if each triple of (i) {ai , aj , ak }, (ii) {ai , aj , bj } and (iii) {ai , bi , ci } , 1 ≤ i 6= j 6= k ≤ s,
are not collinear. The non-collinearity of the three points in the triple (i) is from the
assumption. For the three points in the triple (ii), since lj is a tangent line to S at aj , lj ,
which passes through aj and bj , never meets any other point of S except aj . Therefore
ai , aj and bj , 1 ≤ i 6= j ≤ s, are not collinear. The non-collinearity of the three points in
the triple (iii) is also clear since ci is not on the tangent line li = ai bi for each 1 ≤ i ≤ s.
Conversely, suppose that A is a linear OOAbn−3 (3, s, 3, b). By Theorems 1.3 and 2.2,
the subset of the s columns of A obtained by taking the first column within each group
4
Gi , 1 ≤ i ≤ s, has the property that any of its three columns are linearly independent, or
equivalently, there exists a set S of s points in PG(n−1, b) no three of which are collinear.
Let ai , 1 ≤ i ≤ s, be the point of S which corresponds to the first column of the group
Gi , and bi , 1 ≤ i ≤ s, be the point in PG(n − 1, b) corresponding to the second column of
Gi . Then for any 1 ≤ i 6= j ≤ s, the three points in {ai , aj , bj } are not collinear. For each
1 ≤ i ≤ s, let li be a line passing through ai and bi . Then the line li is either a tangent
or a secant to S. If it is a secant, then there is a point d in S such that d, aj and bj are
collinear, which is obviously impossible. So lj must be a tangent line to S at aj .
2
Let m(N, b) denote the maximum number of points in PG(N, b) such that no three
are collinear. Then we have the following known results, which can be found in [1] and
Section 3.3 of [6].
Lemma 3.2 For N ≥ 2,
m(N, b) = bN ,
for b = 2,
N −1
m(N, b) ≤ b
+ 1, for b 6= 2.
Theorem 3.3 When b 6= 2, there exists a linear OOAλ (3, s, 3, b) if and only if there
exists a linear OAλ (3, s, b), where λ = bn−3 , n ≥ 3.
Proof The construction of a linear OAλ (3, s, b) from a linear OOAλ (3, s, 3, b) is trivial
as the linear case of Theorem 1.3. We will prove the converse. Suppse that there exists
a set S of s points in PG(n − 1, b) no three of which are collinear. This is, by Theorem
2.2, equivalent to the existince of a linear OAbn−3 (3, s, b). Let P be an arbitrarily fixed
point of S. The number of lines passing through the point P is (bn−1 − 1)/(b − 1) in
PG(n − 1, b). The number of secants of S passing through the point P is s − 1. From
Lemma 3.2, we have the inequality s − 1 ≤ bn−2 . Since
bn−2 <
bn−1 − 1
,
b−1
there is a tangent line at the point P of S. Applying Theorem 3.1, we obtain a linear
OOAbn−3 (3, s, 3, b).
2
Theorem 3.4 When b = 2, there exists a linear OOAλ (3, s, 3, b) if and only if 1 ≤ s ≤
2n−1 − 1, where λ = 2n−3 , n ≥ 3.
Proof From Lemma 3.2, there is a set S of 2n−1 points in PG(n − 1, 2) no three of
which are collinear. Since 2n−1 − 1 = (bn−1 − 1)/(b − 1) if b = 2, there is no tangent line
at any point of S in this case. Let P be a point of S and S 0 = S \ {P }. The line P ai for
ai ∈ S 0 is a tangent line to S 0 at ai . So by Theorem 3.1, Smax = 2n−1 − 1 is the maximum
of the value of s for the existence of a linear OOAλ (3, s, 3, 2) in PG(n − 1, 2).
2
5
4
Linear OOA of Strength 4
Suppose, just as in Section 3, that S is a set of s points in PG(n − 1, b), and P a point of
S. A tangent plane to S at P is a plane π such that π ∩ S = {P }. For points P1 , P2 ,. . .,Pk
of PG(n − 1, b), if there is a plane containing these points, then P1 , P2 , . . . , Pk are said to
be coplanar. Clearly the following assertion holds.
Lemma 4.1 Let P1 , P2 , P3 , P4 be four distinct points in PG(n − 1, b). These points are
not coplanar if and only if lines P1 P2 and P3 P4 are disjoint.
Let F be a set of points in PG(n − 1, b). If any four points in F are not coplanar,
then F is said to be a 4-independent set. We consider the following configuration in
PG(n − 1, b):
(C1) S = {a1 , a2 , . . . , as } is a 4-independent set of points in PG(n − 1, b);
(C2) S̄ is the point union of all secants of S, and li is a tangent line to S̄ at each point
ai ∈ S for 1 ≤ i ≤ s;
(C3) l1 , l2 , . . . , ls are mutually disjoint;
(C4) πi is a tangent plane to S at each point ai ∈ S containing the tangent line li for
1 ≤ i ≤ s.
Such a set {S; l1 , l2 , . . . , ls ; π1 , π2 , . . . , πs } of a 4-independent set of points, tangent
lines and tangent planes is called a theta configuration for S, and is denoted by Θ(S).
Theorem 4.2 There exists a linear OOAbn−4 (4, s, 4, b) if and only if there exists a theta
configuration Θ(S) for a 4-independent set S of s points in PG(n − 1, b).
Proof Suppose that Θ(S) = {S; l1 , l2 , . . . , ls ; π1 , π2 , . . . , πs } is a theta configuration for
a 4-independent set S of s points in PG(n − 1, b). Then we can construct a sequence
of points Q0 = (a1 , b1 , c1 , d1 , . . . , as , bs , cs , ds ) in PG(n − 1, b) which satisfies the following
properties:
1. S = {a1 , a2 , . . . , as } is the 4-independent set of s points in PG(n − 1, b);
2. bi is a point in li \ {ai } for each 1 ≤ i ≤ s;
3. ci is a point in πi \ li for each 1 ≤ i ≤ s;
4. di is a point of PG(n − 1, b) not on πi for each 1 ≤ i ≤ s.
6
Let G0i = (ai , bi , ci , di ) for 1 ≤ i ≤ s. From Theorems 2.1 and 2.2, the array A0 obtained
from the points of Q0 is a linear OOAbn−4 (4, s, 4, b) with groups G0i , 1 ≤ i ≤ s, if and
only if each of the following quadruples is not coplanar: (i) {ai , aj , ak , al }, 1 ≤ i 6= j 6=
k 6= l ≤ s; (ii) {ai , aj , ak , bk }, 1 ≤ i 6= j 6= k ≤ s; (iii) {ai , bi , aj , bj }, 1 ≤ i 6= j ≤ s; (iv)
{ai , aj , bj , cj }, 1 ≤ i 6= j ≤ s; and (v) {ai , bi , ci , di }, 1 ≤ i ≤ s.
The non-coplanarity of the four points in the quadruple (i) is from the assumption.
For the four points in the quadruple (ii), since lk is a tangent line to S̄ at ak , lk , which
passes through ak and bk , never intersects the line ai aj , which, by Lemma 4.1, implies
that ai , aj , ak and bk , 1 ≤ i 6= j 6= k ≤ s, are not coplanar. For the four points in the
quadruple (iii), since {ai , bi } and {aj , bj }, 1 ≤ i 6= j ≤ s, are two disjoint tangent lines,
by Lemma 4.1, ai , bi , aj , bj , 1 ≤ i 6= j ≤ s, are not coplanar. For the case (iv), the plane
generated by aj , bj and cj is the tangent plane π to S at aj . From the definition of a
tagent plane, ai cannot be in π. Finally, consider the quadruple (v). Since ai , bi and ci
are not collinear, and di is not in the plane π, it is clear that the four points ai , bi , ci , di
are not coplanar.
Conversely, suppose that A is a linear OOAbn−4 (4, s, 4, b). By Theorems 1.3 and 2.2,
the subset of the s columns of A obtained by taking the first column within each group
Gi , 1 ≤ i ≤ s, has the property that any of its four columns are linearly independent, or
equivalently, there exists a 4-independent set S of s points in PG(n − 1, b). For 1 ≤ i ≤ s,
let ai , bi , and ci be the point in PG(n − 1, b) corresponding to the first, second, and third
column of the group Gi respectively. Then for any 1 ≤ i 6= j 6= k ≤ s, the four points
in {ai , aj , ak , bk }, {ai , bi , aj , bj } and {ai , aj , bj , cj } are not coplanar, respectively. For each
1 ≤ i ≤ s, let li be the line passing through ai and bi . Then the line li must be disjoint to
any secant of S, that is, li must be a tangent line to S̄, the point union of all secants of
S. Also, by Lemma 4.1, the tangent lines l1 , . . . , ls must be mutually disjoint. For each
1 ≤ i ≤ s, we can construct a plane πi from the tagent line li = ai bi and the point ci .
This plane πi can be easily seen to be a tangent plane to S at ai . Therefore, stating from
a linear OOAbn−4 (4, s, 4, b), we can obtain a theta configuration Θ(S) for a 4-independent
set S of s points in PG(n − 1, b).
2
Theorem 4.3 If there exists a linear OOAbn−4 (4, s, 4, b), then the following inequality
holds:
à !
s
bn − 1
(b + 1)s + (b − 1)
≤
.
2
b−1
Proof Let S = {a1 , a2 , . . . , as } be the 4-independent set of s ≥ 4 points in PG(n − 1, b)
constructed from the first column within each group of the linear OOAbn−4 (4, s, 4, b),
and S̄ be the point union for all secants of S. Since every two points in PG(n − 1, b)
determine a unique line
³ ´ in PG(n − 1, b), and every line contains exactly b + 1 points, we
have |S̄| = s+(b−1) 2s . By Theorem 4.2, there exists a tangent line li to S̄ at each point
ai ∈ S, such that l1 , l2 , . . . , ls are mutually disjoint. There are (b + 1)s distinct points
contained in these
³ ´tangent lines, s of which being from the set S. Therefore, totally
(b + 1)s + (b − 1) 2s distinct points are needed in PG(n − 1, b) for the existence of such
7
a linear OOAbn−4 (4, s, 4, b). This implies that
à !
bn − 1
s
(b + 1)s + (b − 1)
≤
.
2
b−1
2
Theorem 4.4 Let S be a 4-independent set of s points in PG(n − 1, b), and Θ0 (S) be a
configuration in PG(n − 1, b) satisfying the conditions (C1) to (C3). Θ0 (S) is extendable
to Θ(S) (i.e. Θ0 (S) also satisfies the condition (C4)) if and only if the following inequality
holds:
bn−2 − 1
s≤
.
b−1
Proof Assume that S = {a1 , a2 , . . . , as } is a 4-independent set of s points in PG(n −
1, b), n ≥ 4, and S̄ is the point union for all secants of S. By our hypothesis, there
is a tangent line li to S̄ at ai ∈ S for each 1 ≤ i ≤ s. Let Πi be the set of planes
containing li in PG(n − 1, b), then clearly we have |Πi | = (bn−2 − 1)/(b − 1). If there
exists a plane of Πi which meets S \ {ai } in more than one point, say, {aj , ak , . . .}, where
1 ≤ j 6= k 6= . . . 6= i ≤ s, then {aj , ak , ai , bi }, bi ∈ li \ {ai }, would be coplanar, which is
impossible. So any plane of Πi can meet S \ {ai } in at most one point. Meanwhile, the
plane generated by li and any point ak ∈ S \ {ai } is obviously contained in Πi . Therefore,
in this case, |Πi | is greater than s − 1 if and only if there is a tangent plane to S at the
point ai in Πi , that is, Θ0 (S) is extendable to Θ(S) if and only if the desired inequality
holds.
2
It is clear from the proof of Theorem 4.4 that the following is true.
Corollary 4.5 If there exists a linear OOAbn−4 (4, s, 4, b), then the following inequality
holds:
bn−2 − 1
.
s≤
b−1
Proof
Apply Theorem 4.2.
2
It should be noted here that the bound for s obtained from Theorem 4.5 is larger than
that obtained from Theorem 4.3 except for the case when n = 4. This imples that Θ0 (S)
is extendable to Θ(S) if the dimention n − 1 of the projective geometry is greater than
or equal to 4. As a result, we can say that usually we do not have to pay attention to
the condition (C4) of the theta configuration.
5
Some Computational Results and Remarks
We have perfomed a small number of computational experiments for b = 2 and 3. The
following tables are their corresponding results:
8
n, b : the parameters of PG(n − 1, b);
B1 : the bound for s in Theorem 4.3;
B2 : the bound for s in Theorem 4.5;
s1 : the largest size |S| = s1 of the found 4-independent sets S;
s2 : the largest size |S| = s2 of the found constructable Θ(S).
n b
4 2
5 2
6 2
7 2
8 2
9 2
4 3
5 3
6 3
B1
3
5
9
13
20
29
5
9
17
B2
3
7
15
31
63
127
4
13
40
s1
5
6
8
11
17
23
5
11
14
s2
3
5
8
11
17
23
4
8
14
It can be observed that when the dimension of the projective geometry increases, s1
(B2 too) increases more slowly than the number of points in PG(n − 1, b) does. This gives
us more flexibility for finding disjoint tangent lines. From this observation, we can state
the following conjecture.
Conjecture 5.1 For each prime power b, there is a finite positive integer db such that for
any positive integer n ≥ db , there exists a theta configuration Θ(S) for any 4-independent
set S of points in PG(n − 1, b).
For linear OOAs with strength higher than 4, it is also possible to define an equivalent
geometrical configuration. However, the high strength makes the conditions which the
equivalent geometrical configuration should satisfy more complicated. It is not a good idea
to define such an equivalent configuration for each strength. It is strongly recommended
to find a simple general geometrical configuration which is sufficient (but not necessarily
necessary) to guarantee the existence of linear OOA.
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10