Geometrical Constructions for Ordered Orthogonal Arrays and (T, M, S)-Nets

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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 fujihara@sk.tsukuba.ac.jp and miao@sk.tsukuba.ac.jp 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 S E = [a i b d i, (a i + 1)b d i ), i=1 where a i and d i are non-negative integers such that 0 a i < b d i for 1 i S. The volume of E is S b d i = b S i=1 d i. i=1 For integers 0 T M, a (T, M, S)-net in base b is a set N of b M points in the S-dimensional unit cube [0, 1) S such that every elementary interval E in base b having volume b T M contains exactly b T 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 OA λ (t, s, b), where λ = m/b t. 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 G 1, G 2,, G s ; 2. Let (t 1, t 2,..., t s ) be an s-tuple of non-negative integers such that s i=1 t i = t, where 0 t i l for 1 i s. If R is the subset of the columns of A obtained by taking the first t i columns within each group G i, 1 i s, then A is orthogonal for such an R. We use the notation OOA λ (t, s, l, b), where λ = m/b t. Note that we can assume without loss of generality that l t sl when we study OOA λ (t, s, l, b). When l = 1, the above definition reduces to that of an orthogonal array, i.e., an OOA λ (t, s, 1, b) is equivalent to an OA λ (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 OOA b T (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 OOA b T (2, S, 2, b) if and only if there exists an OA b T (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 OA b T (2, S, b). 2

Also observe that if we take t 1 = t 2 = = t s = 1 in the definition of an ordered orthogonal array, then we get an orthogonal array OA λ (t, s, b). Theorem 1.3 If there exists an OOA λ (t, s, l, b), then there exists an OA λ (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 = {xg : x GF(b) n }. Therefore C = m = b n 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 {i 1,..., i t } 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 g i1,..., g it 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 OA b n 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 OOA b n 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 = (a 1, b 1, c 1,..., a s, b s, c s ) in PG(n 1, b) which satisfies the following properties: 1. S = {a 1, a 2,..., a s } is the set of points in PG(n 1, b) no three of which are collinear; 2. l i is the tangent line to S at the point a i, and b i is a point in l i \ {a i } for each 1 i s; 3. c i is a point of PG(n 1, b) not on l i for each 1 i s. Let G i = (a i, b i, c i ) for 1 i s. From Theorems 2.1 and 2.2, the array A obtained from the points of Q is a linear OOA b n 3(3, s, 3, b) with groups G i, 1 i s, if and only if each triple of (i) {a i, a j, a k }, (ii) {a i, a j, b j } and (iii) {a i, b i, c i }, 1 i j 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 l j is a tangent line to S at a j, l j, which passes through a j and b j, never meets any other point of S except a j. Therefore a i, a j and b j, 1 i j s, are not collinear. The non-collinearity of the three points in the triple (iii) is also clear since c i is not on the tangent line l i = a i b i for each 1 i s. Conversely, suppose that A is a linear OOA b n 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

G i, 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 a i, 1 i s, be the point of S which corresponds to the first column of the group G i, and b i, 1 i s, be the point in PG(n 1, b) corresponding to the second column of G i. Then for any 1 i j s, the three points in {a i, a j, b j } are not collinear. For each 1 i s, let l i be a line passing through a i and b i. Then the line l i 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, a j and b j are collinear, which is obviously impossible. So l j must be a tangent line to S at a j. 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) = b N, for b = 2, m(n, b) b N 1 + 1, for b 2. Theorem 3.3 When b 2, there exists a linear OOA λ (3, s, 3, b) if and only if there exists a linear OA λ (3, s, b), where λ = b n 3, n 3. Proof The construction of a linear OA λ (3, s, b) from a linear OOA λ (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 OA b n 3(3, s, b). Let P be an arbitrarily fixed point of S. The number of lines passing through the point P is (b n 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 b n 2. Since b n 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 OOA b n 3(3, s, 3, b). Theorem 3.4 When b = 2, there exists a linear OOA λ (3, s, 3, b) if and only if 1 s 2 n 1 1, where λ = 2 n 3, n 3. Proof From Lemma 3.2, there is a set S of 2 n 1 points in PG(n 1, 2) no three of which are collinear. Since 2 n 1 1 = (b n 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 = S \ {P }. The line P a i for a i S is a tangent line to S at a i. So by Theorem 3.1, S max = 2 n 1 1 is the maximum of the value of s for the existence of a linear OOA λ (3, s, 3, 2) in PG(n 1, 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 P 1, P 2,...,P k of PG(n 1, b), if there is a plane containing these points, then P 1, P 2,..., P k are said to be coplanar. Clearly the following assertion holds. Lemma 4.1 Let P 1, P 2, P 3, P 4 be four distinct points in PG(n 1, b). These points are not coplanar if and only if lines P 1 P 2 and P 3 P 4 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 = {a 1, a 2,..., a s } is a 4-independent set of points in PG(n 1, b); (C2) S is the point union of all secants of S, and l i is a tangent line to S at each point a i S for 1 i s; (C3) l 1, l 2,..., l s are mutually disjoint; (C4) π i is a tangent plane to S at each point a i S containing the tangent line l i for 1 i s. Such a set {S; l 1, l 2,..., l s ; π 1, π 2,..., π s } 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 OOA b n 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; l 1, l 2,..., l s ; π 1, π 2,..., π s } 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 Q = (a 1, b 1, c 1, d 1,..., a s, b s, c s, d s ) in PG(n 1, b) which satisfies the following properties: 1. S = {a 1, a 2,..., a s } is the 4-independent set of s points in PG(n 1, b); 2. b i is a point in l i \ {a i } for each 1 i s; 3. c i is a point in π i \ l i for each 1 i s; 4. d i is a point of PG(n 1, b) not on π i for each 1 i s. 6

Let G i = (a i, b i, c i, d i ) for 1 i s. From Theorems 2.1 and 2.2, the array A obtained from the points of Q is a linear OOA b n 4(4, s, 4, b) with groups G i, 1 i s, if and only if each of the following quadruples is not coplanar: (i) {a i, a j, a k, a l }, 1 i j k l s; (ii) {a i, a j, a k, b k }, 1 i j k s; (iii) {a i, b i, a j, b j }, 1 i j s; (iv) {a i, a j, b j, c j }, 1 i j s; and (v) {a i, b i, c i, d i }, 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 l k is a tangent line to S at a k, l k, which passes through a k and b k, never intersects the line a i a j, which, by Lemma 4.1, implies that a i, a j, a k and b k, 1 i j k s, are not coplanar. For the four points in the quadruple (iii), since {a i, b i } and {a j, b j }, 1 i j s, are two disjoint tangent lines, by Lemma 4.1, a i, b i, a j, b j, 1 i j s, are not coplanar. For the case (iv), the plane generated by a j, b j and c j is the tangent plane π to S at a j. From the definition of a tagent plane, a i cannot be in π. Finally, consider the quadruple (v). Since a i, b i and c i are not collinear, and d i is not in the plane π, it is clear that the four points a i, b i, c i, d i are not coplanar. Conversely, suppose that A is a linear OOA b n 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 G i, 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 a i, b i, and c i be the point in PG(n 1, b) corresponding to the first, second, and third column of the group G i respectively. Then for any 1 i j k s, the four points in {a i, a j, a k, b k }, {a i, b i, a j, b j } and {a i, a j, b j, c j } are not coplanar, respectively. For each 1 i s, let l i be the line passing through a i and b i. Then the line l i must be disjoint to any secant of S, that is, l i must be a tangent line to S, the point union of all secants of S. Also, by Lemma 4.1, the tangent lines l 1,..., l s must be mutually disjoint. For each 1 i s, we can construct a plane π i from the tagent line l i = a i b i and the point c i. This plane π i can be easily seen to be a tangent plane to S at a i. Therefore, stating from a linear OOA b n 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). Theorem 4.3 If there exists a linear OOA b n 4(4, s, 4, b), then the following inequality holds: ( ) s (b + 1)s + (b 1) bn 1 2 b 1. Proof Let S = {a 1, a 2,..., a s } 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 OOA b n 4(4, s, 4, b), and S be the point union for all secants of S. Since every two points in PG(n 1, b) determine a unique line in PG(n 1, b), and every line contains exactly b + 1 points, we have S = s+(b 1) ( ) s 2. By Theorem 4.2, there exists a tangent line li to S at each point a i S, such that l 1, l 2,..., l s are mutually disjoint. There are (b + 1)s distinct points contained in these tangent lines, s of which being from the set S. Therefore, totally (b + 1)s + (b 1) ( ) s 2 distinct points are needed in PG(n 1, b) for the existence of such 7

a linear OOA b n 4(4, s, 4, b). This implies that ( ) s (b + 1)s + (b 1) 2 bn 1 b 1. Theorem 4.4 Let S be a 4-independent set of s points in PG(n 1, b), and Θ (S) be a configuration in PG(n 1, b) satisfying the conditions (C1) to (C3). Θ (S) is extendable to Θ(S) (i.e. Θ (S) also satisfies the condition (C4)) if and only if the following inequality holds: s bn 2 1 b 1. Proof Assume that S = {a 1, a 2,..., a s } is a 4-independent set of s points in PG(n 1, b), n 4, and S is the point union for all secants of S. By our hypothesis, there is a tangent line l i to S at a i S for each 1 i s. Let Π i be the set of planes containing l i in PG(n 1, b), then clearly we have Π i = (b n 2 1)/(b 1). If there exists a plane of Π i which meets S \ {a i } in more than one point, say, {a j, a k,...}, where 1 j k... i s, then {a j, a k, a i, b i }, b i l i \ {a i }, would be coplanar, which is impossible. So any plane of Π i can meet S \ {a i } in at most one point. Meanwhile, the plane generated by l i and any point a k S \{a i } is obviously contained in Π i. Therefore, in this case, Π i is greater than s 1 if and only if there is a tangent plane to S at the point a i in Π i, that is, Θ (S) is extendable to Θ(S) if and only if the desired inequality holds. It is clear from the proof of Theorem 4.4 that the following is true. Corollary 4.5 If there exists a linear OOA b n 4(4, s, 4, b), then the following inequality holds: s bn 2 1 b 1. Proof Apply Theorem 4.2. 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 Θ (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); B 1 : the bound for s in Theorem 4.3; B 2 : the bound for s in Theorem 4.5; s 1 : the largest size S = s 1 of the found 4-independent sets S; s 2 : the largest size S = s 2 of the found constructable Θ(S). n b B 1 B 2 s 1 s 2 4 2 3 3 5 3 5 2 5 7 6 5 6 2 9 15 8 8 7 2 13 31 11 11 8 2 20 63 17 17 9 2 29 127 23 23 4 3 5 4 5 4 5 3 9 13 11 8 6 3 17 40 14 14 It can be observed that when the dimension of the projective geometry increases, s 1 (B 2 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 d b such that for any positive integer n d b, 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. References [1] R.C. Bose, Mathematical theory of the symmetrical factorial design, Sankhyā 8 (1947), 107 166. [2] P.P. Boyle, M. Broadie and P. Glasserman, Monte Carlo methods for security pricing, J. Economic Dynamics and Control 21 (1997), 1267 1321. 9

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