constructions of nested orthogonal arrays

Constructions of Nested OrthogonalArrays

Kun Wang and Yang LiDepartment of Mathematics, Soochow University, Suzhou 215006, P. R. China,E-mail: [email protected]

Received June 13, 2012; revised December 19, 2012

Published online 6 February 2013 in Wiley Online Library (wileyonlinelibrary.com).DOI 10.1002/jcd.21343

Abstract: A symmetric nested orthogonal array, denoted by NOA((N, M), k, (s, r), t), is anOA(N, k, s, t) which contains an OA(M, k, r, t) as a subarray, where N > M, s > r . Nestedorthogonal arrays are useful in designing an experimental setup consisting of two experiments,the expensive one of higher accuracy being nested in a relatively less expensive one of loweraccuracy. In this paper, some combinatorial constructions of nested orthogonal arrays are pro-vided. By employing these constructions, the existence spectrum of NOA((2s2, r2), 4, (s, r), 2)is completely determined. C© 2013 Wiley Periodicals, Inc. J. Combin. Designs 21: 464–477, 2013

Keywords: orthogonal array; nested; constructions; existence spectrum

1. INTRODUCTION

Let S be a set of s symbols. A symmetric orthogonal array OA(N, k, s, t) over S isan N × k array with entries from S, in which all t-tuples of S appear equally often asrows in every N × t subarray, where 2 ≤ t ≤ k. Here, we call the integer λ = N/st theindex of the array. In fact, every t-tuple of S appears exactly λ times in every N × t

subarray of the orthogonal array. Orthogonal arrays belong to an important and high-profile area of statistics and combinatorics and have been studied extensively, see [3,6] forexamples.

Let R be a subset of symbol set S where |R| = r and |S| = s. A symmetric nestedorthogonal array over (S, R), denoted by NOA((N, M), k, (s, r), t), where N > M, s >

r , is an OA(N, k, s, t) over S which contains an OA(M, k, r, t) over R as a subarray. The

Contract grant sponsor: Tian Yuan Mathematical Foundation of China; Contract grant number: 11126283; Contractgrant sponsor: research foundation of Soochow University; Contract grant number: SDY2011A01.

Journal of Combinatorial Designs464 C© 2013 Wiley Periodicals, Inc.

CONSTRUCTIONS OF NESTED ORTHOGONAL ARRAYS 465

studies of nested orthogonal array are motivated by its application in the constructionof nested space-filling designs, i.e., designing an experimental setup consisting of twoexperiments, the expensive one of higher accuracy being nested in a relatively lessexpensive one of lower accuracy, due to its low-dimensional orthogonality and nestedstructure. For the detail descriptions, the interested reader may refer to [9, 10], andso on.

The question of the existence of nested orthogonal arrays has been investigated byMukerjee et al. [8], Aloke Dey [4, 5], and Wang et al. [11] and some other authors, anda series of constructions and existence results were given. In this paper, we continuestudying the construction and existence problem of symmetric nested orthogonal arrays.More combinatorial constructions are provided in Section 2 and the existence spectrumof NOA((2s2, r2), 4, (s, r), 2) is completely determined in Section 3.

2. COMBINATORIAL CONSTRUCTIONS OF NOA

In this section, we will establish some combinatorial constructions of nested orthogonalarray. For this, we require the following definition and notations.

Let R1, . . . , Rn be n pairwise disjoint subsets of symbol set S, where |Ri | = ri , i =1, . . . , n and |S| = s. An incomplete orthogonal array (IOA) over (S, R1, . . . , Rn) of in-dex (λ,μ1, . . . , μn), strength t , degree k, denoted by IOA(λ,μ1,...,μn)(t, k, (s, r1, . . . , rn)),is an (λst − ∑n

i=1 μirti ) × k array with entries from S, in every (λvt − ∑n

i=1 μirti ) × t

subarray of which each t-tuple of St \ (⋃n

i=1 Rti ) appears exactly λ times as row

and each t-tuple of Rti appears exactly λ − μi times for any i, 1 ≤ i ≤ n. When

r1 = · · · = rn = 0, it is just an OA(λst , k, s, t). When μ1 = · · · = μn = λ, it is oftendenoted by IOAλ(t, k, (s, r1, . . . , rn)) ( or IOA(t, k, (s, r1, . . . , rn)) if λ = 1). IOA canbe used as input ingredient in the construction of NOA. It is obvious that by startingwith an IOA(λ,μ)(t, k, (s, r)), one can obtain an NOA((λst , μrt ), k, (s, r), t) by filling anOA(μrt , k, r, t).

Like Theorem 1.1 of [2], the basic idea of the following construction of NOAcomes from Wilson’s recursive construction of mutually orthogonal Latin square(see [12]).

Theorem 2.1. Suppose that A is an IOA(λ,μ)(2, k + l, (m + n, n)) over (G,T ), whereG = {g1, . . . , gm, gm+1, . . . , gm+n}, T = {gm+1, . . . , gm+n}. Let u, uij be nonnegativeintegers where i ∈ I = {1, . . . , l}, j ∈ J = {1, . . . m + n}. If

(1) for any row (a1, . . . , ak, ak+1, . . . , ak+l) of A, where (ak+1, . . . , ak+l) =(gj1, . . . , gjl

), an IOA(2, k, (u + ∑li=1 uiji

, u1j1, . . . , uljl)) exists; and

(2) for any i ∈ I , an IOA(λ,μ)(2, k, (pi, qi)) with pi = ∑m+nj=1 uij , qi = ∑m+n

j=m+1 uij

exists,

then so does an IOA(λ,μ)(2, k, (s, r)), where s = (m + n)u + ∑li=1 pi, r = nu +∑l

i=1 qi . Further, if an OA(μr2, k, r, 2) also exists, then so does anNOA((λs2, μr2), k, (s, r), 2).

Proof. Let B = {b1, . . . , bu} ∩ G = ∅, Dij = {dijt |t = 1, . . . , uij } ∩ G = ∅, i ∈I, j ∈ J are pairwise disjoint sets. Write Yi = ⋃m+n

j=1 Dij , Zi = ⋃m+nj=m+1 Dij ,

Y = ⋃li=1 Yi , Z = ⋃l

i=1 Zi , S = (G × B) ∪ Y , R = (T × B) ∪ Z. For the given

Journal of Combinatorial Designs DOI 10.1002/jcd

466 WANG AND LI

IOA(λ,μ)(2, k + l, (m + n, n)) A over (G,T ), we denote by Egj1 ,...,gjlthe subarray of A

consisting of all those rows with the form (a1, . . . , ak, ak+1 = gj1, . . . , ak+l = gjl). From

the assumption, for any row (a1, . . . , ak, gj1, . . . , gjl) ∈ Egj1 ,...,gjl

, an IOA(2, k, (u +∑li=1 uiji

, u1j1, . . . , uljl)) over (B ∪ (

⋃li=1 Diji

),D1j1, . . . , Dljl) exists. In this IOA, re-

place any b ∈ B with (ai, b) if b is the ith element of a row. Denote the array obtained byjuxtaposing all those new arrays according to Egj1 ,...,gjl

by Fj1,...,jl. As for each i ∈ I , an

IOA(λ,μ)(2, k, (pi + qi, qi)) over (Yi, Zi), say Ci exists, then by juxtaposing all Fj1,...,jl’s,

all Ci’s, we obtain an array denoted by H . Further, an OA(μr2, k, r, 2) over R, say L

exists. We claim that H is an IOA(λ,μ)(2, k, (s, r)) over (S,R) and V = (H ′|L′)′ is thedesired NOA((λs2, μr2), k, (s, r), 2).

To prove the above conclusions, we only need to check that V is an OA(λs2, k, s, 2),i.e., for any k1 = k2 in {1, . . . , k}, the subarray formed by the k1th and k2th columnsof V contains each ordered pair (x, y) ∈ S × S exactly λ times. This procedure willbe done under some cases. For example, if (x, y) = ((a, b), (a′, b′)) ∈ ((G \ T ) × B) ×((G \ T ) × B), as there are exactly λ rows in A having a in the k1th column and a′

in the k2th column, and in every IOA(2, k, (u + ∑li=1 uiji

, u1j1, . . . , uljl)) according to

these λ rows, there is exactly one row having b in the k1th column and b′ in the k2thcolumn, thus (x, y) appears exactly λ times in the subarray formed by the k1th andk2th columns of all those Fj1,...,jl

. Considering the structures of Ci and L, then (x, y)appears exactly λ times in the subarray formed by the k1th and k2th columns of V . Andfor example, if (x, y) = (di1j1t1, di2j2t2 ) ∈ (Zi1 × Zi2 ), as there are exactly λ − μ rows inA having gj1 in the (k + i1)th column and gj2 in the (k + i2)th column, and in everyIOA(2, k, (u + ∑l

i=1 uiji, u1j1, . . . , uljl

)) according to these λ − μ rows, there is exactlyone row having di1j1t1 in the k1th column and di2j2t2 in the k2th column, thus (x, y)appears exactly λ − μ times in the subarray formed by the k1th and k2th columns of H ,and exactly λ times in the subarray formed by the k1th and k2th columns of V . Similarly,after a routine work, the conclusion holds for other choices of the pair (x, y). Then theproof is completed. �

Example 2.2. An NOA((72, 9), 4, (6, 3), 2) constructed by using Theorem 2.1 withm = 1, n = 1, λ = 2, μ = 1, k = 4, l = 1, u = 2, u11 = 1, u12 = 1.

Start with an IOA(2,1)(2, 5, (2, 1)) over (G = {g1 = a, g2 = b}, T = {b}):

A =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

a b b a a

a b a b a

b a a b a

b a b a a

b b a a b

a a a a b

a a b b b

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

=(

Ea

Eb

)

An IOA(2, 4, (u + u11, u11)) over (B ∪ D11), an IOA(2, 4, (u + u12, u12)) over (B ∪D12), where B = {c, d}, D11 = {e}, and D12 = {f }, an IOA(2,1)(2, 4, (2, 1)) C1 over(Y1, Z1), where Y1 = D11 ∪ D12, Z1 = D12, and an OA(9, 4, 3, 2) L over (T × B) ∪ Z1



are given, respectively, as follows:

⎛⎜⎜⎜⎝

e e c c c d d d

c d e c d e c d

c d c d e d e c

d c c e d d c e

⎞⎟⎟⎟⎠

′ ⎛⎜⎜⎜⎝

f f c c c d d d

c d f c d f c d

c d c d f d f c

d c c f d d c f

⎞⎟⎟⎟⎠

′

⎛⎜⎜⎜⎝

f e e e e f f

f e f e f e e

e e f f e e f

e e e f f f e

⎞⎟⎟⎟⎠

′ ⎛⎜⎜⎜⎝

f f bc bc bc bd bd bd f

bc bd f bc bd f bc bd f

bc bd bc bd f bd f bc f

bd bc bc f bd bd bc f f

⎞⎟⎟⎟⎠

′

C1 L

Here, (x, y) is denoted by xy for convenience. Then, F1 and F2 in the proof of Theorem2.1 are as follows:⎛⎜⎜⎜⎝

e e ac ac ac ad ad ad e e ac ac ac ad ad ad e e bc bc bc bd bd bd e e bc bc bc bd bd bd

bc bd be bc bd be bc bd bc bd be bc bd be bc bd ac ad e ac ad e ac ad ac ad e ac ad e ac ad

bc bd bc bd e bd e bc ac ad ac ad e ad e ac ac ad ac ad e ad e ac bc bd bc bd e bd e bc

ad ac ac e ad ad ac e bd bc bc e bd bd bc e bd bc bc e bd bd bc e ad ac ac e ad ad ac e

⎞⎟⎟⎟⎠

′

⎛⎜⎜⎜⎝

f f bc bc bc bd bd bd f f ac ac ac ad ad ad f f ac ac ac ad ad ad

bc bd f bc bd f bc bd ac ad f ac ad f ac ad ac ad f ac ad f ac ad

ac ad ac ad f ad f ac ac ad ac ad f ad f ac bc bd bc bd f bd f bc

ad ac ac f ad ad ac f ad ac ac f ad ad ac f bd bc bc f bd bd bc f

⎞⎟⎟⎟⎠

′

From Theorem 2.1, H = (F1|′F ′2|C ′

1)′ is an IOA(2,1)(2, 4, (6, 3)) over ({ac, ad,

e, bc, bd, f }, {bc, bd, f }) and V = (H ′|L′)′ is the desired NOA((72, 9), 4, (6, 3), 2).Our next construction embodies the identification technique in design theory.

Theorem 2.3. Let λ, λ′, and μ be positive integers satisfying that λ = λ′(λ − μ).Suppose that an IOA(λ−μ)(2, k, (λ′d + r, λ′d)) and an OA(λd2, k, d, 2) both exist, then sodoes an IOA(λ,μ)(2, k, (s, r)), where s = d + r . Further, if an OA(μr2, k, r, 2) also exists,then so does an NOA((λs2, μr2), k, (s, r), 2).

Proof. Let symbol sets V = {a1, . . . , aλ′d}, R = {g1, . . . , gr}, where V ∩ R = ∅. LetT = {a1, . . . , ad}, G = V ∪ R, S = T ∪ R. Suppose that A is an IOA(λ−μ)(2, k, (λ′d +r, λ′d)) over (G,V ), B be an OA(λd2, k, d, 2) over T , C be an OA(μr2, k, r, 2) over R.Consider a mapping f from G to S, where

f (x) ={

aj ifx = ai, i ≡ j (mod d) and 1 ≤ j ≤ d;

x ifx ∈ R.

Denote by D the array obtained by replacing each entry a of A with f (a). Then it is easyto see that in every subarray formed by two columns of D, any ordered pair of S appears


468 WANG AND LI

exactly (λ − μ) times if the two symbols both from R, λ′(λ − μ) = λ times if one symbolfrom T and the other from R, and does not appear if the two symbols both from T . TakeE = (B ′|D′)′ and F = (B ′|D′|C ′)′. Then any ordered pair of S appears exactly λ timesin every subarray formed by two columns of E. As C is an OA(μr2, k, r, 2), then E isan NOA((λs2, μr2), k, (s, r), 2) over (S, R). �

Example 2.4. An NOA((50, 16), 4, (5, 4), 2) constructed by using Theorem 2.3 withs = 5, r = 4, d = 1, λ = 2, μ = 1, k = 4.

An IOA(2, 4, (6, 2), 2) A over (G = V ∪ R, V ), where V = {a1 = 4, a2 = 5},R = {g1 = 0, g2 = 1, g3 = 2, g4 = 3}, an OA(2, 4, 1, 2) B over T = {4}, and anOA(16, 4, 4, 2) C over R are given as follows:

⎛⎜⎜⎜⎜⎝

4 1 1 2 2 1 3 2 0 3 2 5 2 4 4 1 5 2 0 5 1 0 0 5 0 1 3 0 3 3 3 4

0 2 1 3 0 0 4 5 4 5 4 2 1 1 2 3 1 2 5 3 5 1 3 0 2 4 0 0 1 2 3 3

0 5 4 5 4 1 3 2 0 0 1 2 3 1 3 0 0 0 1 1 3 5 3 3 4 2 5 2 2 1 4 2

0 2 0 0 3 5 0 2 2 3 1 0 4 2 3 4 1 5 0 3 1 3 5 2 1 3 1 4 5 4 2 1

⎞⎟⎟⎟⎟⎠

′

A

⎛⎜⎜⎜⎜⎝

4 4

4 4

4 4

4 4

⎞⎟⎟⎟⎟⎠

′ ⎛⎜⎜⎜⎜⎝

0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3

0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3

0 1 2 3 1 0 3 2 2 3 0 1 3 2 1 0

0 1 2 3 2 3 0 1 3 2 1 0 1 0 3 2

⎞⎟⎟⎟⎟⎠

′

B C

Using the mapping f , we have

D =

⎛⎜⎜⎜⎝

4 1 1 2 2 1 3 2 0 3 2 4 2 4 4 1 4 2 0 4 1 0 0 4 0 1 3 0 3 3 3 4

0 2 1 3 0 0 4 4 4 4 4 2 1 1 2 3 1 2 4 3 4 1 3 0 2 4 0 0 1 2 3 3

0 4 4 4 4 1 3 2 0 0 1 2 3 1 3 0 0 0 1 1 3 4 3 3 4 2 4 2 2 1 4 2

0 2 0 0 3 4 0 2 2 3 1 0 4 2 3 4 1 4 0 3 1 3 4 2 1 3 1 4 4 4 2 1

⎞⎟⎟⎟⎠

′

Then , E = (B ′|D′)′ and F = (B ′|D′|C ′)′ are the desired IOA(2,1)(2, 4, (5, 4)) andNOA((50, 16), 4, (5, 4), 2) over (T ∪ R, R), respectively.

Finally, we also present the other two easy, but very useful constructions.

Theorem 2.5. Suppose that an IOAμ(2, k, (s, r)), an OA((λ − μ)s2, k, s, 2), and anOA(μr2, k, r, 2) all exist, then so does an NOA((λs2, μr2), k, (s, r), 2).

Proof. Suppose that A is an IOAμ(2, k, (s, r)) over (S, R), B is an OA((λ − μ)s2,

k, s, 2) over S, and C is an OA(μr2, k, r, 2) over R, where the symbol set R ⊂ S

and |S| = s, |R| = r . Let D = (A′|B ′|C ′)′. Then one can easily check that all orderedpair of V appear exactly λ times as rows in every subarray formed by two columns



of D. As C is an OA(μr2, k, r, 2) over R, then D is an NOA((λs2, μr2), k, (s, r), 2)over (S, R). �

Theorem 2.6. Suppose that an IOA(λ,μ)(2, k, (m + n, n)), an OA(u2, k, u, 2), andan OA(μ(nu)2, k, nu, 2) all exist, then so does an NOA((λs2, μr2), k, (s, r), 2), wheres = (m + n)u, r = nu.

Proof. Suppose that A is an IOA(λ,μ)(2, k, (m + n, n)) over (G ∪ T , T ), B is anOA(u2, k, u, 2) over U, and C is an OA(μ(nu)2, k, nu, 2) over T × U , where G andT are two disjoint symbol set with |G| = m, |T | = n and U is a u-symbol set. Let D

be an array consisting of all rows in the set {((a1, b1), . . . , (ak, bk))|(a1, . . . , ak) is a rowof A, (b1, . . . , bk) is a row of B}. Then one can easily check that for any ordered pair of(G ∪ T ) × U , it appears exactly (λ − μ) times if the two symbols both from T × U inevery subarray formed by two columns of D, and λ times otherwise. Thus, E = (D′|C ′)′

is an NOA((λs2, μr2), k, (s, r), 2) over ((G ∪ T ) × U, T × U ). �

3. THE EXISTENCE SPECTRUM OF NOA((2s2, r2), 4, (s, r), 2)

In this section, we will employ our constructive methods to determine the existencespectrum of NOA((2s2, r2), 4, (s, r), 2).

In order to use the above constructions, we need some existence results on OA andIOA.

Lemma 3.1 [1, 3]. An OA(λs2, 4, s, 2) exists except when (λ, s) = (1, 2) or (1, 6).

Lemma 3.2 [7]. An IOA(2, 4, (s, r)) exists if and only if s ≥ 3r > 0, (s, r) = (6, 1).

Lemma 3.3. An IOA(2,1)(2, k, (s, r)) exists for each triple (k, d, r) = (k, s − r, r) listedin Table I.

Proof. The corresponding IOA is constructed directly, see the Appendix. �

Lemma 3.4. An IOA(2,1)(2, 4, (s, r)) exists when s − r = 1 or 2.

Proof. Let r0 = 3 if s − r = 1, and r0 = 7 if s − r = 2. It follows from Lemma 3.3 thatan IOA(2,1)(2, 4, (s, r)) exists when r ≤ r0. When r > r0, we can apply Theorem 2.3 withλ = 2, μ = 1, d = s − r, k = 4 as from Lemma 3.2 and Lemma 3.1, an IOA(2, 4, (2d +r, 2d)), an OA(2d2, 4, d, 2) both exist. Thus an IOA(2,1)(2, 4, (s, r)) exists. �

Now we try to determine the existence spectrum of NOA((2s2, r2), 4, (s, r), 2).

TABLE I. Triple (k,d,r) in Lemma 3.3.

k 4 4 4 5 5 5 5 6d 2 3 5 1 2 3 4 1r 6,7 5,8,11 3,7 1-3,5 1-5 2,4 2 3


470 WANG AND LI

First, from [8], we know that an NOA((λs2, μr2), k, (s, r), t) can exist only ifλst ≥ μrt

∑uj=0

(kj

)(r−1s − 1)j if t = 2u ≥ 2 and λst ≥ μrt (

∑uj=0

(kj

)(r−1s − 1)j +(

k−1u

)(r−1s − 1)u+1) if t = 2u + 1 ≥ 3. Notice the conclusion in Lemma 3.1, we then

have:

Lemma 3.5. An NOA((2s2, r2), 4, (s, r), 2) exists only if s > r and r = 2, 6.

Next, we try to construct NOA((2s2, r2), 4, (s, r), 2) step by step.

Lemma 3.6. An NOA((2s2, r2), 4, (s, r), 2) exists when s ≥ 3r and r = 2, 6.

Proof. When (s, r) = (6, 1), an OA(72, 4, 6, 2) exists from Lemma 3.1. When (s, r) =(6, 1), we can apply Theorem 2.5 with λ = 2, μ = 1 as an IOA(2, 4, (s, r)) exists fromLemma 3.2 , an OA(s2, 4, s, 2) and an OA(r2, 4, r, 2) exist from Lemma 3.1. Then anNOA((2s2, r2), 4, (s, r), 2) exists. �

Lemma 3.7. An NOA((2s2, r2), 4, (s, r), 2) exists when 5r/4 < s < 3r, r = 2, 6 ex-cept for each pair (d, r) = (s − r, r) listed in Table II.

Proof. We apply Theorem 2.1 with λ = 2, μ = 1, k = 4, l = 1 in six cases, respec-tively, as shown in Table III where δ = 1+(−1)d

2 , γ = 1+(−1)r+1

2 , μi = 0 or 1 accordingto r − id ∈ {2, 6} or not for each i ∈ {1, 2, 3}, and 0 ≤ η ≤ 2, −1 ≤ ρ ≤ 1 satisfy-ing that η ≡ ρ ≡ −r (mod 3). What remains is to show that the required ingredientsexist. It is obvious from Lemma 3.3 that an IOA(2,1)(2, 5, (m + n, n)) exists for eachpair (m, n) ∈ {(1, 1), (1, 2), (1, 3), (2, 1)} and so does an OA(2, 4, r) with r = 2, 6 fromLemma 3.1. Therefore, it is suffice for us to show that an IOA(2, 4, (u + u1j , u1j ))for each j ∈ {1, . . . , m + n} and an IOA(2,1)(2, 4, (

∑m+nj=1 u1j ,

∑m+nj=m+1 u1j )) exist for

each case. And for Case 1, an IOA(2, 4, ( d+1−δ2 , 1)), an IOA(2, 4, ( d−1+δ

2 , δ)), anIOA(2, 4, (r, r − d−1−δ

2 )), and an IOA(2,1)(2, 4, (r − d−3−3δ2 , r − d−1−δ

2 )) all exist from

TABLE II. Pair (d, r) in Lemma 3.7 and Lemma 3.8.

d r d r d r d r d r d r

1 3 4 5,7-15 7 4,8-20 10 11-15 13 7,8,14-19 16 9,102 3,4,5,7 5 3,7-19 8 5,9-23 11 7,12-16 14 8,9,15-21 17 9-113 4,5,7-11 6 5,7-23 9 5,10-13 12 7,13-18 15 8,9

TABLE III. Six cases in the proof of Lemma 3.7.

Case (d, r) (m, n, u, n1,1, . . . , n1,m+n)

1 d2 < r < 2d

3 , d ≥ 18 (2, 1, d−1−δ2 , 1, δ, r − d−1−δ

2 )2 2d

3 ≤ r ≤ d, (d, r) = (6, 5) (1, 1, r, d − r, 0)3 d < r ≤ 3d

2 , d ≥ 15 (1, 1, d − μ1, μ1, r − d + μ1)4 3d

2 < r ≤ 2d, d ≥ 9 (1, 2, r−γ

2 , d − r−γ

2 , γ, 0)5 2d < r < 3d, d ≥ 9 (1, 2, d − μ2, μ2,

r−2(d−μ2)+γ

2 , r−2(d−μ2)−γ

2 )6 3d ≤ r < 4d, d ≥ 7 (1, 3, d − μ3, μ3,

r−3(d−μ3)+η3 , r−3(d−μ3)+ρ

3 , r−3(d−μ3)−η−ρ3 )



Lemmas 3.1, 3.2, and 3.4. Thus, an NOA((2s2, r2), 4, (s, r), 2) exists for this case. Forother cases, by using Lemmas 3.1–3.4, one can also easily check that the required ingre-dients in Theorem 2.1 all exist, thus the conclusion holds. �

Lemma 3.8. An NOA((2s2, r2), 4, (s, r), 2) exists for each pair (d, r) = (s − r, r)listed in Table II.

Proof. We divide those (d, r) in Table II into three parts: those (d, r) in Table IV, inTable V, and (d, r) = (4, 14). When (d, r) = (4, 14), we apply Theorem 2.1 with k =4, l = 2, λ = 2, μ = 1, and (m, n, u, u11, u12, u13, u14, u21, u22, u23, u24) = (1, 3, 3,

1, 1, 1, 0, 0, 1, 1, 1). Then the input designs need to contain those IOA(2, 6, (4, 3));IOA(2, 4, (4, 1)), IOA(2, 4, (5, 1, 1)); IOA(2,1)(2, 4, (3, 2)), IOA(2,1)(2, 4, (3, 3)), andOA(169, 4, 14, 2). They all exist from Lemmas 3.1–3.3 except that IOA(2, 4, (5, 1, 1))can be obtained by adding three OA(1, 4, 1, 2) to an OA(2, 4, (5, 1, 1, 1, 1, 1)) whichexists from [3]. Thus there is an NOA((648, 196), 4, (18, 14), 2).

TABLE IV. Five-tuple (d,r,m,n,u) in the proof of Lemma 3.8.

(d, r,m, n, u) (d, r,m, n, u) (d, r,m, n, u) (d, r,m, n, u) (d, r,m, n, u) (d, r,m, n, u)

(1,3,1,3,1) (2,3,2,3,1) (2,4,2,4,1) (2,5,2,5,1) (2,7,2,7,1) (3,5,3,5,1)(3,8,3,8,1) (3,11,3,11,1) (5,3,5,3,1) (5,7,5,7,1) (3,9,1,3,3) (4,8,1,2,4)(4,12,1,3,4) (5,10,1,2,5) (5,15,1,3,5) (6,9,2,3,3) (6,12,2,4,3) (6,15,2,5,3)(6,18,2,6,3) (6,21,2,7,3) (7,14,1,2,7) (8,12,2,3,4) (8,16,2,4,4) (8,20,2,5,4)(9,12,3,4,3) (10,15,2,3,5) (12,16,3,4,4) (14,21,2,3,7) (15,9,5,3,3)

TABLE V. Five-tuple (d,r,m,n,u) in the proof of Lemma 3.8.

(d, r,m, n, u) (d, r,m, n, u) (d, r,m, n, u) (d, r,m, n, u) (d, r,m, n, u) (d, r,m, n, u)

(3,4,1,1,3) (3,7,1,2,3) (3,10,1,3,3) (4,5,1,1,4) (4,7,1,2,3) (4,9,1,2,4)(4,10,1,3,3) (4,11,1,2,4) (4,13,1,3,4) (4,15,1,3,4) (5,8,1,2,4) (5,9,1,2,4)(5,11,1,2,4) (5,12,1,2,4) (5,13,1,3,4) (5,14,1,2,5) (5,16,1,3,4) (5,17,1,3,4)(5,18,1,3,4) (5,19,1,3,5) (6,5,1,1,4) (6,7,2,2,3) (6,8,1,2,4) (6,10,1,2,4)(6,11,1,2,4) (6,13,1,3,4) (6,14,1,3,4) (6,16,1,3,4) (6,17,1,3,4) (6,19,2,5,3)(6,20,1,5,4) (6,22,1,5,4) (6,23,1,5,4) (7,4,2,1,3) (7,8,1,1,6) (7,9,1,1,6)(7,10,1,1,7) (7,11,2,3,3) (7,12,2,3,3) (7,13,2,4,3) (7,15,1,2,6) (7,16,1,2,6)(7,17,1,2,6) (7,18,1,2,6) (7,19,1,2,7) (7,20,2,5,3) (8,5,2,1,4) (8,9,1,1,6)(8,10,1,1,7) (8,11,1,1,8) (8,13,2,3,4) (8,14,1,2,6) (8,15,1,2,6) (8,17,1,2,6)(8,18,1,2,6) (8,19,1,2,7) (8,21,1,2,8) (8,22,1,3,6) (8,23,1,2,8) (9,5,2,1,4)(9,10,1,1,7) (9,11,1,1,8) (9,13,1,1,9) (10,11,1,1,8) (10,12,1,1,8) (10,13,1,1,9)(10,14,1,1,10) (11,7,3,2,3) (11,12,1,1,8) (11,13,1,1,9) (11,14,1,1,10) (11,15,1,1,10)(11,16,1,1,11) (12,7,3,2,3) (12,13,1,1,9) (12,14,1,1,10) (12,15,1,1,10) (12,17,1,1,12)(12,18,1,2,8) (13,7,4,2,3) (13,8,3,2,4) (13,14,1,1,10) (13,15,1,1,10) (13,16,1,1,11)(13,17,1,1,12) (13,18,1,1,12) (13,19,1,2,9) (14,8,2,1,6) (14,9,2,1,6) (14,15,1,1,10)(14,16,1,1,11) (14,17,1,1,12) (14,18,1,1,12) (14,19,1,1,13) (14,20,1,2,10) (15,8,2,1,6)(16,9,2,1,6) (16,10,2,1,7) (17,9,2,1,6) (17,10,2,1,7) (17,11,2,1,8)


472 WANG AND LI

For each pair (s, r) listed in Table IV, we apply Theorem 2.6 with k = 4, λ = 2, μ = 1,and the given (m, n, u) in that table. The corresponding input designs all exist fromLemmas 3.1 and 3.3, so there is an OA((2s2, r2), 4, (s, r), 2).

For each pair (s, r) listed in Table V, we apply Theorem 2.1 with k = 4, l = 1, λ =2, μ = 1, and the given (m, n, u) and u11 = · · · = u1α = s−r−mu−α

m+ 1, u1(α+1) = · · · =

u1m = s−r−mu−αm

, u1(m+1) = · · · = u1(m+β) = r−nu−βn

+ 1, u1(m+β+1) = · · · = u1(m+n) =r−nu−β

n, where α ≡ s − r − mu (mod m), β ≡ r − nu (mod n), and α < m, β < n. The

corresponding input designs all exist from Lemma 3.1, 3.3, and 3.4, thus there is anOA((2s2, r2), 4, (s, r), 2). �

Lemma 3.9. An NOA((2s2, r2), 4, (s, r), 2) exists when s ≤ 5r/4 and r = 2, 6.

Proof. We can apply Theorem 2.3 with λ = 2, μ = 1 as an IOA(2, 4, (2d + r, 2d))with d = s − r exists from Lemma 3.2, an OA(2s2, 4, s, 2) and an OA(r2, 4, r, 2) bothexist Lemma 3.1. Thus an NOA((2s2, r2), 4, (s, r), 2) exists. �

Then, from Lemmas 3.6–3.9, we have:

Lemma 3.10. If s > r and r = 2, 6, then an NOA((2s2, r2), 4, (s, r), 2) exists.

Finally, by combining Lemma 3.5 and Lemma 3.10, we then determine the existencespectrum of NOA((2s2, r2), 4, (s, r), 2) and have the following main result.

Theorem 3.11. An NOA((2s2, r2), 4, (s, r), 2) exists if and only if s > r and r = 2, 6.

ACKNOWLEDGMENT

We would like to thank the referees very much for their useful comments and suggestions.

REFERENCES

[1] R. C. Bose, S. S. Shrikhande, and E. T. Parker, Further results on the construction of mutuallyorthogonal Latin squares and the falsity of Euler’s conjecture, Canad J Math 12 (1960), 189–203.

[2] A. E. Brouwer and G. H. J. van Rees, More mutually orthogonal Latin squares, Discrete Math39 (1982), 263–281.

[3] C. J. Colbourn and J. H. Dinitz, The CRC Handbook of Combinatorial Designs, Boca Raton,FL, 2007.

[4] A. Dey, Construction of nested orthogonal arrays, Discrete Math 310 (2010), 2831–2834.

[5] A. Dey, On the construction of nested orthogonal arrays, Australas J Combin 54 (2012), 37–48.

[6] A. S. Hedayat, N. J. A. Sloane, and J. Stufken, Orthogonal Arrays: Theory and Applications,Springer, New York, 1999.

[7] K. Heinrich and L. Zhu, Existence of orthogonal Latin squares with aligned subsquares, DiscreteMath 59 (1986), 69–78.

[8] R. Mukerjee, P. Z. G. Qian, and C. F. J. Wu, On the existence of nested orthogonal arrays,Discrete Math 308 (2008), 4635–4642.

[9] P. Z. G. Qian, M. Ai, and C. F. J. Wu, Construction of nested space-filling designs, Ann Statist37(6A) (2009), 3616–3643.



[10] P. Z. G. Qian, B. Tang, and C. F. J. Wu, Nested space-filling designs for computer experimentswith two levels of accuracy, Statist Sinica 19(6A) (2009), 287–300.

[11] K. Wang and J. Yin, Further results on the existence of nested orthogonal arrays, Des. CodesCryptogr., published online 2012 in http://www.springerlink.com, DOI: 10.1007/s10623-011-9603-0.

[12] R. M. Wilson, Concerning the number of mutually orthogonal Latin squares, Discrete Math 9(1974), 181–198.

APPENDIX

IOA(2,1)(2, k, (s, r)) CONSTRUCTED IN THE PROOF OF LEMMA 3.3

For each triple (k, s, r), let S = {0, 1, . . . , s − 1}, R = {s − r, . . . , s − 1}. The desiredIOA(2,1)(2, k, (s, r)) A is given directly or can be obtained from the given array byreplacing every row whose entries are not constant with its k cyclic shifts if the triple(k, s, r) is marked with an asterisk.

*(k,s,r) = (4,8,6)

⎛⎜⎜⎜⎝

6 4 4 1 0 0 2 3 4 5 6 7 0 0 0 5 1 3 1 0 0 6 4 5 1 4 2 0 5

0 1 2 2 2 3 2 3 4 5 6 7 0 0 1 1 7 6 4 2 1 7 5 2 0 3 6 7 7

7 1 1 5 3 7 2 3 4 5 6 7 0 0 6 4 1 5 0 1 5 3 3 7 5 2 3 2 6

4 3 6 1 5 1 2 3 4 5 6 7 0 0 2 7 3 0 6 7 6 0 0 0 4 0 1 4 1

⎞⎟⎟⎟⎠

′

*(k,s,r) = (4,9,7)

⎛⎜⎜⎜⎝

2 1 7 1 7 1 2 0 4 8 8 1 4 0 5 7 8 8 2 1 6 3 5 0 0 0 2 3 4 5 6 7 8 0 0

5 6 6 0 1 3 1 3 0 2 6 1 5 8 7 5 5 0 0 8 0 1 6 1 5 6 2 3 4 5 6 7 8 0 0

4 1 2 2 4 0 3 2 3 1 3 4 0 7 0 0 1 7 6 3 4 0 8 6 3 5 2 3 4 5 6 7 8 0 0

1 7 3 8 6 5 5 7 8 1 0 3 2 2 1 4 7 4 1 4 2 7 1 4 6 8 2 3 4 5 6 7 8 0 0

⎞⎟⎟⎟⎠

′

*(k,s,r) = (4,8,5)

⎛⎜⎜⎜⎝

0 3 5 1 4 1 2 2 0 3 0 1 0 4 5 6 2 3 0 1 3 0 0 2 3 4 5 6 7 0 0

1 4 2 5 2 2 7 5 6 1 7 1 3 6 7 0 7 2 5 3 0 1 2 2 3 4 5 6 7 0 0

7 0 6 1 1 3 1 3 7 7 6 3 1 2 2 4 2 0 4 7 5 6 4 4 3 4 5 6 7 0 0

5 7 1 4 0 6 1 0 1 0 5 5 6 1 2 5 6 6 2 4 2 4 7 3 3 4 5 6 7 0 0

⎞⎟⎟⎟⎠

′

*(k,s,r) = (4,11,8)

⎛⎜⎜⎜⎝

6 10 3 1 4 4 0 10 7 10 7 4 1 3 7 4 6 6 1 1 2 9 7 1 9 2 1 0 0 4 6 7 10 5 3 0 9 0 8 1 1 2 3 4 5 6 7 8 9 10 0 0

5 4 10 7 0 7 6 7 4 0 10 9 1 2 6 10 1 9 2 8 1 5 9 0 1 3 5 10 2 2 0 5 5 0 6 2 8 5 7 8 3 8 3 4 5 6 7 8 9 10 0 0

8 8 8 8 3 0 4 1 5 4 6 2 6 6 3 1 10 1 5 2 9 10 4 9 5 0 3 3 10 5 5 1 2 3 1 2 6 6 2 9 2 3 3 4 5 6 7 8 9 10 0 0

0 2 1 0 5 8 2 9 1 1 2 3 8 7 0 0 1 4 7 4 7 2 0 3 0 10 1 7 9 2 9 2 8 8 4 6 2 10 0 0 2 9 3 4 5 6 7 8 9 10 0 0

⎞⎟⎟⎟⎠

′


474 WANG AND LI

(k,s,r) = (4,14,11) A = (B|C)′

B =

⎛⎜⎜⎜⎝

8 1 2 5 6 2 1 4 0 0 0 4 3 4 0 11 7 0 0 3 12 11 9 1 13 6 6 3 10 10 0 3

13 10 0 7 0 11 3 7 0 6 12 5 8 0 7 2 2 8 5 2 0 5 8 1 0 11 4 2 2 5 6 12

9 7 10 0 13 4 4 13 2 0 6 9 11 11 1 9 4 2 8 6 11 1 1 5 2 12 2 13 7 0 12 0

11 6 9 12 11 10 12 8 11 1 5 1 1 13 9 3 9 3 10 2 7 4 12 13 10 8 10 5 11 13 11 4

⎞⎟⎟⎟⎠

C =

⎛⎜⎜⎜⎝

0 12 5 2 1 1 12 10 2 2 0 7 8 8 3 12 10 1 1 4 11 3 5 2 6 13 6 1 8 6 1 2

3 2 2 4 2 0 7 11 5 12 0 0 5 7 13 10 3 2 8 1 1 0 4 12 10 2 13 9 3 8 11 9

7 1 0 13 13 10 10 3 11 4 9 3 2 2 1 1 9 6 9 8 7 9 8 1 8 8 7 4 10 2 2 10

4 5 7 6 3 4 1 1 9 2 13 6 7 2 7 13 5 9 0 0 1 12 0 13 1 12 9 6 0 1 5 12

⎞⎟⎟⎟⎠

*(k,s,r) = (4,8,3)

⎛⎜⎜⎜⎝

0 1 2 5 1 5 0 1 3 4 4 1 2 1 4 2 6 4 3 0 4 0 2 7 1 5 5 0 1 5 6 7

1 1 6 3 4 2 1 0 0 6 5 3 7 3 5 7 4 2 2 3 6 7 5 3 2 2 0 4 5 5 6 7

6 2 7 2 7 2 6 5 6 3 4 6 3 0 0 4 4 5 4 1 2 2 6 5 3 1 6 3 7 5 6 7

4 6 2 1 1 0 1 4 0 3 3 5 6 7 0 0 7 1 2 7 4 0 0 3 3 7 3 7 4 5 6 7

⎞⎟⎟⎟⎠

′

(k,s,r) = (4,12,7)A = (B|C)′

B =

⎛⎜⎜⎜⎝

2 4 1 1 0 8 3 0 11 6 9 3 1 4 3 5 6 0 3 6 8 3 1 8 2 1 9 0 7

5 6 10 1 6 7 9 1 4 3 11 9 4 4 5 3 5 8 0 2 0 8 1 3 8 3 2 0 3

10 10 0 11 1 0 4 7 8 8 4 10 7 0 2 11 9 2 7 7 2 4 4 3 4 2 7 5 5

0 2 11 10 5 4 2 1 6 1 2 1 9 5 10 1 3 2 6 11 6 7 6 0 10 9 1 8 3

⎞⎟⎟⎟⎠

C =

⎛⎜⎜⎜⎝

4 0 3 5 0 10 7 2 3 4 11 9 3 0 3 4 3 10 10 2 2 4 2 2 11 2 10 1 1

9 4 6 1 10 6 8 4 1 11 5 2 1 11 4 11 11 8 7 0 11 1 1 11 3 2 5 2 2

8 9 7 8 9 0 1 5 8 7 0 5 6 0 5 8 9 2 0 9 2 0 7 0 3 3 7 8 10

10 6 10 9 4 3 9 11 11 4 1 6 4 9 4 3 0 1 0 3 6 6 5 7 10 7 4 5 4

⎞⎟⎟⎟⎠

(k,s,r) = (5,2,1)

⎛⎜⎜⎜⎜⎜⎝

1 0 0 0 0 1 1

1 0 1 0 1 0 0

0 0 1 1 0 0 1

0 0 0 1 1 1 0

1 1 0 1 0 0 0

⎞⎟⎟⎟⎟⎟⎠

′



(k,s,r) = (5,3,2)

⎛⎜⎜⎜⎜⎜⎝

0 1 1 0 2 0 0 2 1 1 0 2 0 2

2 2 0 1 1 1 0 0 0 1 0 2 2 0

2 1 2 0 2 1 1 0 0 0 2 0 0 1

2 0 0 0 0 1 2 2 1 2 1 1 0 0

0 1 0 1 2 0 2 1 2 0 1 0 2 0

⎞⎟⎟⎟⎟⎟⎠

′

*(k,s,r) = (5,4,3)

⎛⎜⎜⎜⎜⎜⎝

0 0 0 0 1 2 3

2 0 0 1 1 2 3

3 2 3 0 1 2 3

0 1 1 3 1 2 3

1 3 2 2 1 2 3

⎞⎟⎟⎟⎟⎟⎠

′

*(k,s,r) = (5,6,5)

⎛⎜⎜⎜⎜⎜⎝

3 3 4 0 0 4 1 0 0 4 5

1 2 3 0 3 2 1 1 2 4 5

3 0 5 1 3 3 2 4 5 4 5

0 5 2 5 4 0 0 5 0 4 5

5 0 2 4 1 0 2 1 4 4 5

⎞⎟⎟⎟⎟⎟⎠

′

*(k,s,r) = (5,3,1)

⎛⎜⎜⎜⎜⎜⎝

0 0 0 0 1

2 0 2 0 1

2 1 1 0 1

0 2 1 0 1

1 1 2 0 1

⎞⎟⎟⎟⎟⎟⎠

′

(k,s,r) = (5,4,2)

⎛⎜⎜⎜⎜⎜⎝

1 2 0 1 1 1 3 1 3 0 2 2 2 0 0 0 1 3 0 1 0 1 3 3 2 0 2 3

2 0 3 0 1 3 0 0 2 3 1 1 2 2 1 2 1 1 1 2 0 3 3 1 3 0 0 0

0 0 0 1 3 1 2 3 1 3 1 2 3 2 1 1 2 3 0 0 3 2 0 0 1 2 0 1

3 1 0 2 2 1 0 0 1 1 0 1 0 2 2 0 3 1 3 1 3 0 2 0 3 1 2 3

0 2 3 2 3 0 0 1 1 2 3 0 0 1 0 2 2 1 1 3 0 1 0 2 1 3 1 3

⎞⎟⎟⎟⎟⎟⎠

′


476 WANG AND LI

*(k,s,r) = (5,5,3)

⎛⎜⎜⎜⎜⎜⎝

2 2 1 2 0 0 0 1 4

1 2 4 0 3 4 2 2 4

1 3 0 0 4 3 0 1 4

3 0 1 1 0 1 4 2 4

3 0 3 4 3 1 1 4 4

⎞⎟⎟⎟⎟⎟⎠

′

*(k,s,r) = (5,6,4)

⎛⎜⎜⎜⎜⎜⎝

3 5 4 5 3 1 0 1 4 1 0 5

4 1 4 0 0 2 1 4 1 0 1 5

0 3 1 4 0 2 4 3 5 2 1 5

0 3 3 2 3 0 5 2 1 5 0 5

2 1 0 0 5 3 4 1 2 2 5 5

⎞⎟⎟⎟⎟⎟⎠

′

*(k,s,r) = (5,7,5)

⎛⎜⎜⎜⎜⎜⎝

0 2 1 5 2 1 1 0 2 6 0 0 0 0 4 5 6

5 4 3 0 2 2 5 3 1 5 2 0 4 6 4 5 6

6 0 1 1 1 3 4 5 5 0 0 3 6 4 4 5 6

0 0 4 3 4 3 2 1 3 1 6 4 3 1 4 5 6

4 5 5 2 1 6 6 6 0 1 2 3 1 1 4 5 6

⎞⎟⎟⎟⎟⎟⎠

′

*(k,s,r) = (5,5,2)

⎛⎜⎜⎜⎜⎜⎝

4 1 2 2 4 0 0 0 0 1 2 0 1 2

3 2 3 4 2 1 2 3 0 1 2 0 1 2

0 0 1 1 3 3 1 3 0 1 2 0 1 2

2 4 0 0 0 2 3 1 0 1 2 0 1 2

1 4 3 4 1 4 4 2 0 1 2 0 1 2

⎞⎟⎟⎟⎟⎟⎠

′

*(k,s,r) = (5,7,4)

⎛⎜⎜⎜⎜⎜⎝

2 1 1 4 3 0 1 4 3 5 2 6 0 6 0 0 0 6

2 2 1 1 1 2 5 2 5 2 4 3 4 0 3 5 0 6

5 6 2 4 1 1 1 6 5 1 1 2 4 5 3 1 0 6

3 4 5 0 3 6 6 2 2 5 0 0 2 0 6 3 0 6

0 3 6 1 2 5 0 2 0 4 6 0 3 4 1 4 0 6

⎞⎟⎟⎟⎟⎟⎠

′



*(k,s,r) = (6,4,3)

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

2 0 3 0 2 2 3 3 2 0 1 0 0 0 0 1 3 1 0 1 1 2 3

0 0 0 1 3 1 3 0 2 3 0 2 1 3 0 1 1 3 2 0 2 0 2

1 0 1 3 3 0 2 0 0 0 3 2 2 1 3 1 0 0 1 2 0 2 3

0 2 1 0 0 1 0 3 3 2 3 1 3 3 1 2 0 1 0 0 0 2 2

3 1 1 2 1 0 0 2 0 3 3 3 1 0 0 0 3 2 2 0 1 2 0

0 2 1 1 0 2 2 0 1 1 2 0 0 3 3 0 3 0 2 1 3 3 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

′


constructions of nested orthogonal arrays

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