Sudoku-based space-filling designs

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1 Biometrika (2011), 98,3,pp C 2011 Biometrika Trust Printed in Great Britain doi: /biomet/asr024 Sudoku-based space-filling designs BY XU XU Department of Statistics, University of Wisconsin Madison, Wisconsin 53706, U.S.A. xuxu@stat.wisc.edu BEN HAALAND Office of Clinical Sciences, Duke-National University of Singapore Graduate Medical School, Singapore benjamin.haaland@duke-nus.edu.sg AND PETER Z. G. QIAN Department of Statistics, University of Wisconsin Madison, Wisconsin 53706, U.S.A. peterq@stat.wisc.edu SUMMARY Sudoku is played by millions of people across the globe. It has simple rules and is very addictive. The game board is a nine-by-nine grid of numbers from one to nine. Several entries within the grid are provided and the remaining entries must be filled in subject to no row, column, or three-by-three subsquare containing duplicate numbers. By exploiting these three types of uniformity, we propose an approach to constructing a new type of design, called a Sudoku-based space-filling design. Such a design can be divided into groups of subdesigns so that the complete design and each subdesign achieve maximum uniformity in univariate and bivariate margins. Examples are given illustrating the proposed construction method. Applications of such designs include computer experiments with qualitative and quantitative factors, linking parameters in engineering and crossvalidation. Some key words: Computer experiment; Design of experiment; Latin hypercube design; Latin square; Orthogonal array; Popular puzzle. 1. INTRODUCTION The now internationally popular game Sudoku was invented by Howard Garns and first appeared in Dell Magazines in the 1970s under the name Number Place. It became widespread in Japan during the 1980s when introduced under its current name, Sudoku. The game became popular almost 20 years later in Europe and the United States when several major newspapers began offering daily Sudoku puzzles. Figure 1 presents a completed Sudoku grid, which forms a Latin square, but with subsquare uniformity in addition to row and column uniformity. Based on these three types of uniformity, we propose an approach to constructing a new type of design, called a Sudoku-based space-filling design. Our approach has two steps. The first generates a set of doubly orthogonal Sudoku Latin squares in which the whole squares are orthogonal and the subsquares are orthogonal after suitable levelcollapsing. The second uses this set of squares to generate a Sudoku-based space-filling

2 712 XU XU, BEN HAALAND AND PETER Z. G. QIAN Fig. 1. A completed Sudoku grid. design. Such a design has two desirable properties: (i) the complete design achieves maximum uniformity in both univariate and bivariate margins, and (ii) it can be divided into groups of subdesigns with maximum stratification in any one- or two-dimensional projection. These properties make this type of design differ from existing Latin hypercube types of designs, such as Latin hypercube designs (McKay et al., 1979), U designs (Tang, 1993), orthogonal Latin hypercube designs (Ye, 1998) and optimal Latin hypercube designs (Butler, 2001). Sudoku-based space-filling designs differ from the sliced designs constructed in Qian & Wu (2009), for which each slice has attractive stratification in two-dimensional projections but cannot achieve maximum one-dimensional uniformity. The proposed designs are useful for computer experiments with qualitative and quantitative factors (Qian et al., 2008; Han et al., 2009), linking parameters in engineering and crossvalidation. 2. NOTATION AND DEFINITIONS A Latin square of order s is an s s square with entries from a set of s symbols such that, in each row and column, every symbol appears precisely once. The Sudoku grid in Fig. 1 is a Latin square of order 9 but with additional subsquare uniformity. That is, there is no duplicate number in each subsquare. As in Pedersen & Vis (2009), for s 1 = s2 2, define a Sudoku Latin square L of order s 1 to be a special Latin square consisting of s 2 s 2 subsquares in which each symbol appears precisely once in each subsquare, where L (ij) denotes the (i, j)th subsquare of L. For example, the Sudoku Latin square in Fig. 1 contains nine 3 3 subsquares. Note that Sudoku Latin squares are closely related to gerechte designs (Behrens, 1956). As a stepping stone for constructing doubly orthogonal Sudoku Latin squares in 3, we build a connection between the subsquares and the whole grid of a Sudoku Latin square via the subfield projection in Qian & Wu (2009), originally proposed for obtaining sliced orthogonal arrays using the Rao Hamming method and other methods. For every prime p and every integer u 1, there exists a Galois field GF(p u ) of order p u.lets 1 = p 2u and s 2 = p u be powers of the same prime p for an integer u 1, with s 1 corresponding to the number of rows of a Sudoku Latin square of order s 1 and s 2 corresponding to the number of rows of its subsquares. Let F = GF(s 1 ) with a primitive polynomial p 1 (x), whereα is a primitive element. The subfield projection φ is constructed as follows. Letting β = α (s 1 1)/(s 2 1), G ={0,β,β 2,...,β s2 1 } is a subfield of F, where β is a primitive element of G and β s2 1 = 1. Let β 0 = 0andβ i = β i,fori = 1,...,s 2 1. For u = 1, G becomes the residue field {0, 1,...,p 1} mod p. Then, as F is a vector space

3 Sudoku-based space-filling designs 713 over G with respect to the polynomial basis {1,α} (Lidl & Niederreiter, 1997, Ch. 3), represent any a F as a = b 0 + b 1 α (b 0, b 1 G) and define Lemma 1 captures some important properties of φ. φ(a) = b 0 + b 1 β (b 0, b 1 G). (1) LEMMA 1(Qian & Wu, 2009). For the subfield projection φ, we have that (i) φ(b) = b, for any b G; (ii) φ(a 1 + a 2 ) = φ(a 1 ) + φ(a 2 ), for any a 1, a 2 F; (iii) φ(ba) = bφ(a), for any b G and a F. For a F, define a new projection δ: F G as { φ(a 1 ) (a = 0), δ(a) = 0 (a = 0). For a matrix A, A T denotes its transpose, A(:, j) denotes the jth column, A(i, :) denotes the ith row, A(i, j) denotes the (i, j)th entry and a + A denotes the elementwise sum of A and a scalar a. ForamatrixD with entries in F, δ(d) denotes the matrix obtained from D after the levels of its entries are collapsed according to δ. For any set B F, letδ(b) ={δ(a) : a B}. For any b G, define its preimage to be δ 1 (b) ={a F : δ(a) = b}. ForD G, letδ 1 (D) = {a F : δ(a) D}. Define the kernel matrix Ɣ of δ to be an s 2 s 2 matrix whose ith row is (2) Ɣ(i, ) = δ 1 (β i 1 ), (3) where the elements in each row are arranged in lexicographical order. This matrix has the following properties: (a) each element of F appears precisely once in Ɣ;(b)δ collapses the entries in the same row of Ɣ into a common element in G, i.e.fori = 1,...,s 2, δ{ɣ(i, j)}=β i 1 ( j = 1,...,s 2 );and(c)for j = 1,...,s 2, δ{ɣ(:, j)}=g. Example 1. Let p = 3andu = 1. Use p 1 (x) = x 2 + x + 2forF = GF(9) with α as a primitive element. Take G to be the subfield {0, 1, 2} of F with β = 2. Here φ is: {0,α+ 1, 2α + 2} 0, {1,α+ 2, 2α} 1, {2,α,2α + 1} 2, δ is: {0,α,2α} 0,{1, 2α + 1, 2α + 2} 1, {2,α+ 1,α+ 2} 2 and the kernel matrix Ɣ of δ is Ɣ = 0 α 2α 2 α + 1 α α + 1 2α CONSTRUCTION OF DOUBLY ORTHOGONAL SUDOKU LATIN SQUARES We propose a method for constructing doubly orthogonal Sudoku Latin squares, serving as a basis for the generation of Sudoku-based space-filling designs in 4. Lets 1, s 2, F and G be as defined in 2. Fora F, a + G ={a + b : b G} is an additive coset of G for which a is a representative. Since G is a subfield of F, F can be partitioned into s 2 additive cosets of G, P 1,...,P s2,wherec i is a representative of P i and c 1 is the zero element of G (Lidl & Niederreiter, 1997, Ch. 1). Denote by A 0 the s 2 s 2 addition table of G. Asin

4 714 XU XU, BEN HAALAND AND PETER Z. G. QIAN Pedersen & Vis (2009), let A ij = c i + c j + A 0,fori, j = 1,...,s 2. Partition the s 1 s 1 addition table A of F as A 11 A 1s2 A =.... (4) A s2 1 A s2 s 2 Since a Sudoku Latin square is a Latin square with some special structure, mutual orthogonality can be defined for such squares as well. Two Sudoku Latin squares of the same order are called orthogonal if, when one is superimposed onto the other, all level combinations appear precisely once. Example 2. Let p = 3andu = 1 with s 1 = 9ands 2 = 3. Use p(x) = x 2 + x + 2forGF(9) = {0, 1, 2,α,α+ 1,α+ 2, 2α, 2α + 1, 2α + 2} with α as a primitive element. The following are two orthogonal Sudoku Latin squares of order 9 based on GF(9): α α + 1 α + 2 2α 2α + 1 2α + 2 2α 2α + 1 2α α α + 1 α + 2, α + 1 α + 2 α 2α + 1 2α + 2 2α α + 2 α α + 1 2α + 2 2α 2α α + 2 2α 2α α + 2 α α α + 2 α α + 1 2α + 2 2α 2α α α + 1 α + 2 2α 2α + 1 2α + 2 α + 2 α α + 1 2α + 2 2α 2α α + 2 2α 2α α + 2 α α + 1. α + 1 α + 2 α 2α + 1 2α + 2 2α α + 2 α α + 1 2α + 2 2α 2α + 1 2α 2α + 1 2α α α + 1 α + 2 Construction of Sudoku-based space-filling designs requires Sudoku Latin squares with a more elaborate structure than mutual orthogonality. As described in Definition 1, thekeyhere is to construct doubly orthogonal Sudoku Latin squares so that the whole squares are mutually orthogonal and the subsquares in the same location are mutually orthogonal after suitable levelcollapsing. DEFINITION 1. For s 1 = s 2 2, a set of Sudoku Latin squares of order s 1 is doubly orthogonal if the whole squares are mutually orthogonal and for any given location, all s 2 s 2 subsquares are mutually orthogonal after level-collapsing according to the projection φ in (1).

5 Sudoku-based space-filling designs 715 Section 4 will show that double orthogonality guarantees that for a Sudoku-based space-filling design, both the whole design and each slice achieve attractive space-filling properties. The proposed method proceeds in two steps. Step 1 uses the procedure from Pedersen & Vis (2009) to obtain a set of orthogonal Sudoku Latin squares. For a F and the matrix A in (4), denote by r a its row whose first entry is a. Denote by ξ i the ith entry of the first column of A. For any η F\G, i.e., the subset of F lying outside G, obtain a Sudoku Latin square L(η) whose ith row is r ξi η 1. (5) Let L (ij) (η) denote the (i, j)th subsquare of L(η). By Theorem 1 in Pedersen & Vis (2009), L ={L(η) : η F\G} (6) is a set of orthogonal Sudoku Latin squares. Orthogonal Sudoku Latin squares are not automatically doubly orthogonal. For example, after the levels are collapsed according to the projection φ in Example 1, the3 3 subsquares in the upper-left corners of the two orthogonal Sudoku Latin squares in Example 2 are not orthogonal. Guided by the projection δ in (2), Step 2 generates a set of doubly orthogonal Sudoku Latin squares from (6). Using the kernel matrix Ɣ in (3), pick γ i to be an arbitrary element of Ɣ(i, )\G (i = 1,...,s 2 ). (7) By properties (b) and (c) of Ɣ discussed in 2, δ(γ 1 ),...,δ(γ s2 ) constitute the subfield G. Put where L(γ i ) is constructed by (5), for i = 1,...,s 2. THEOREM 1. For S defined above, we have that S ={L(γ 1 ),...,L(γ s2 )}, (8) (i) the squares L(γ 1 ),...,L(γ s2 ) are orthogonal Sudoku Latin squares of order s 1 ; (ii) for two different elements a 1, a 2 {γ 1,...,γ s2 }, φ{l (ij) (a 1 )} and φ{l (ij) (a 2 )} are orthogonal, for i, j = 1,...,s 2. Proof. The result in (i) follows readily from Theorem 1 in Pedersen & Vis (2009). For two different elements a 1, a 2 {γ 1,...,γ s2 }, we now show that the result in (ii) holds for φ{l (11) (a 1 )} and φ{l (11) (a 2 )}.Fork, m = 0,...,s 2 1, let s km,l be the (k + 1, m + 1)th entry of L (11) (a l )(l = 1, 2).By(5), s km,l = β k a 1 l + β m (l = 1, 2), (9) where β 0,...,β s2 1 are the elements of the subfield G and appear as the first s 2 entries in the first column of A in (4).

6 716 XU XU, BEN HAALAND AND PETER Z. G. QIAN If φ{l (11) (a 1 )} and φ{l (11) (a 2 )} are not orthogonal, then there exist two different pairs (k 1, m 1 ) and (k 2, m 2 ) such that φ(s k1 m 1,l) = φ(s k2 m 2,l) (l = 1, 2). (10) Applying (9)to(10) yields φ(β k1 a 1 l + β m1 ) = φ(β k2 a 1 l + β m2 ) (l = 1, 2), (11) which together with Lemma 1 (ii) imply φ{β k1 (a1 1 a2 1 )}=φ{β k 2 (a1 1 a2 1 )}. By Lemma 1 (i) and (iii), φ{(β k1 β k2 )(a1 1 a2 1 )}=(β k 1 β k2 ){δ(a 1 ) δ(a 2 )}=0. Thus, β k1 = β k2 since δ(a 1 ) = δ(a 2 ) by property (c) of Ɣ discussed in 2. Applying this to (11) yields φ(β k1 a β m1 ) = φ(β k1 a β m2 ), which implies β m1 = β m2, thus contradicting the fact that (k 1, m 1 ) and (k 2, m 2 ) are different. This proves that (ii) holds for φ{l (11) (a 1 )} and φ{l (11) (a 2 )}. Next, we show φ{l (11) (a 1 )} and φ{l (11) (a 2 )} are orthogonal if and only if φ{l (ij) (a 1 )} and φ{l (ij) (a 2 )} are orthogonal, for i, j = 1, 2,...,s 2. By the construction of L(a l )(l = 1, 2), L (ij) (a l ) = c j + L (i1) (a l ) and L (i1) (a l ) = c i a1 1 + L (11) (a l ),fori, j = 1,...,s 2. Hence, for l = 1, 2, φ{l (ij) (a l )}=φ{c i a 1 l + c j }+φ{l (11) (a l )} (i, j = 1,...,s 2 ), which implies that φ{l (11) (a 1 )} and φ{l (11) (a 2 )} are orthogonal if and only if φ{l (ij) (a 1 )} and φ{l (ij) (a 2 )} are orthogonal, for i, j = 1,...,s 2. This proves the result in (ii). As suggested by one referee, φ in Theorem 1 can be replaced with a projection ϕ that changes (1)toϕ(a) = b 0. Lemma 1 holds for ϕ.in(7), now choose γ i ϕ 1 (β i 1 )\G. Then, the theorem still holds with φ replaced by ϕ for L(γ1 1 ),...,L(γs 1 2 ). Partition F into s 2 additive cosets of G, P 1,...,P s2. Theorem 1 (ii) with this modification can be verified by showing that for all b 1, b 2 G, the following system has a unique solution (x, y) with x P i and y P j, { ϕ(a 1 1 x + y) = b 1, ϕ(a 1 2 x + y) = b 2. Example 3. Let p = 3andu = 1. Use p 1 (x) = x 2 + x + 2forF = GF(9) with α asaprimitive element. Take G to be the subfield {0, 1, 2} of F with β = 2. The projection φ from F to G

7 was given in Example 1. The addition table of F is Sudoku-based space-filling designs α α + 1 α + 2 2α 2α + 1 2α α + 2 α α + 1 2α + 2 2α 2α + 1 A = α + 1 α + 2 α 2α + 1 2α + 2 2α 1 2 0, α + 2 α α + 1 2α + 2 2α 2α α 2α + 1 2α α α + 1 α + 2 2α + 2 2α 2α α + 2 α α + 1 where the first row is denoted as r 0 and the fourth row is denoted as r α and so on. With γ 1 = α, γ 2 = α + 1andγ 3 = 2α + 1, Theorem 1 gives three doubly orthogonal Sudoku Latin squares: α α + 1 α + 2 2α 2α + 1 2α + 2 α + 1 α + 2 α 2α + 1 2α + 2 2α α + 2 2α 2α α + 2 α α + 1 L(γ 1 ) = α + 2 α α + 1 2α + 2 2α 2α , 2α 2α + 1 2α α α + 1 α α + 2 α α + 1 2α + 2 2α 2α α α + 1 α + 2 2α 2α + 1 2α + 2 2α 2α + 1 2α α α + 1 α + 2 L(γ 2 ) =, α + 1 α + 2 α 2α + 1 2α + 2 2α α + 2 α α + 1 2α + 2 2α 2α α + 2 2α 2α α + 2 α α α + 2 α α + 1 2α + 2 2α 2α α α + 1 α + 2 2α 2α + 1 2α + 2 α + 2 α α + 1 2α + 2 2α 2α α + 1 α + 2 α 2α + 1 2α + 2 2α L(γ 3 ) = 2α 2α + 1 2α α α + 1 α + 2, α + 2 α α + 1 2α + 2 2α 2α + 1 2α + 2 2α 2α α + 2 α α + 1 where φ{l (ij) (γ 1 )}, φ{l (ij) (γ 2 )} and φ{l (ij) (γ 3 )} are mutually orthogonal, for i, j = 1,...,3.

8 718 XU XU, BEN HAALAND AND PETER Z. G. QIAN 4. OBTAINING SUDOKU-BASED SPACE-FILLING DESIGNS We present a two-step procedure for using the set of doubly orthogonal Sudoku Latin squares S in (8) to generate a Sudoku-based space-filling design. The first step obtains a special sliced orthogonal array B from S. An orthogonal array OA(n, d, s, t) is an n d matrix with entries from a set of s symbols such that for every n t submatrix, all level combinations occur equally often. A sliced orthogonal array is an orthogonal array that can be divided into several subarrays, each of which becomes a smaller orthogonal array after some level-collapsing (Qian & Wu, 2009). The second step uses B to generate a Sudoku-based space-filling design via some elaborate level-relabelling scheme. Recall that s 1 = s 2 2. The first step constructs B as follows: (i) for k = 1,...,s 2,convert L (ij) (γ k ) to a column vector ζ k of length s 1 by stacking the s 2 columns of the subsquare, for i, j = 1,...,s 2 ; (ii) combine ζ 1,...,ζ s2 column by column to form an s 1 s 2 array B ij ;and (iii) combine all B ij together row by row to form an s 2 1 s 2 array B. The process for converting orthogonal Latin squares into orthogonal arrays is well known (Hedayat et al., 1999, Ch 8). Here, we convert the subsquares of doubly orthogonal Sudoku Latin squares sequentially according to their locations and obtain a special sliced orthogonal array. The subsquares of L(γ 1 ),...,L(γ s2 ) in the same location correspond to one slice of B. The double orthogonality of these Sudoku Latin squares guarantees that B is a sliced orthogonal array. THEOREM 2. For B and B ij constructed above, we have that (i) the matrix B is an OA(s 2 1, s 2, s 1, 2); (ii) the submatrices B ij form a partition of B and each φ{b ij } is an OA(s 1, s 2, s 2, 2); (iii) each B ij is a Latin hypercube of s 1 levels. The theorem can be verified as follows: (i) holds because L(γ 1 ),...,L(γ s2 ) are mutually orthogonal; (ii) holds because each B ij is from L (ij) (γ 1 ),...,L (ij) (γ s2 ) and φ{l (ij) (γ 1 )},...,φ{l (ij) (γ s2 )} are orthogonal; and (iii) follows because of the subsquare uniformity of L(γ 1 ),...,L(γ s2 ). The arrays obtained in Theorem 2 differ from the families of sliced orthogonal arrays constructed in Qian & Wu (2009). Unlike our construction, their method cannot guarantee the obtained array has property (iii) of Theorem 2, which is critical to constructing a sliced spacefilling design with additional one-dimensional uniformity for each slice. Example 4. Consider the doubly orthogonal Sudoku Latin squares, L(γ 1 ), L(γ 2 ), L(γ 3 ),in Example 3. For k = 1,...,3, convert L (ij) (γ k ) to ζ k of length 9 by stacking the three columns of the subsquare, for i, j = 1,...,3. Put ζ 1,...,ζ 3 together to form a 9 3arrayB ij. Combining nine B ij s together by rows yields a sliced orthogonal array B given in Table 1. The second step permutes the levels of B to produce a Sudoku-based space-filling design D. First, relabel the s 1 levels of B as 1,...,s 1, such that the group of levels of GF(s 1 ) that are mapped to the same level according to φ form a consecutive subset of {1,...,s 1 }. Label the s 2 groups as groups 1,...,s 2 and the s 2 levels within the ith group as (i 1)s 2 + 1,...,(i 1)s 2 + s 2, for i = 1,...,s 2.AsinTang (1993), in each column of B, replace the s 1 positions with entry k by a random permutation of (k 1)s 1 + 1,...,(k 1)s 1 + s 1,fork = 1,...,s 1, to generate a Latin hypercube C = (c ij ).Finally,D = (d ij ) is generated through d ij = s1 2 (c ij u ij ),for i = 1,...,s1 2, j = 1,...,s 2, where the u ij are independent U(0, 1] random variables. Let D ij be the subdesign of D corresponding to B ij,fori, j = 1,...,s 2. As described in Proposition 1,

9 Sudoku-based space-filling designs 719 Table 1. The orthogonal array B in Example 4 Run# Run# Run# α α α 55 2α 2α 2α 2 α + 1 α α α + 1 2α 2α α + 2 2α 2α α + 2 α α α + 1 α + 1 α α + 1 2α + 1 2α α + 2 α + 1 α 32 2α + 2 2α + 1 2α α 2α + 1 2α α α + 1 α α + 2 α + 2 α α + 2 2α + 2 2α α α + 2 α α 2α + 2 2α α + 1 2α + 2 2α α + 1 α + 2 α α + 1 α α α α + 1 α α α 38 2α + 2 α α + 1 α 12 2α α α + 1 α α 1 2α α + 2 α α α α + 2 α α 2 2α α α α + 2 α α + 1 α α + 2 α 69 α α α α 43 α 0 2α 70 2α α 0 17 α α α + 1 α α α α + 2 α α α α α α + 2 2α α + 2 2α α α α 2α α 2 α α + 2 2α α α 48 1 α + 2 2α 75 α + 1 2α α 2α 49 α 2α α 0 α 23 α + 1 2α α α α 2α α α α 2α α + 2 2α α + 1 2α α + 1 2α α α α + 2 2α α α 80 2 α + 1 2α 27 2α 1 α α + 1 2α α 2α D is a Sudoku-based space-filling design in which both the whole design and each slice achieve maximum uniformity in any one- or two-dimensional projection. PROPOSITION 1. Consider D and D ij constructed above. Then (i) the design D achieves maximum uniformity on s 1 s 1 grids in two dimensions, in addition to achieving maximum uniformity in one dimension; (ii) the subdesigns D ij form a partition of D and achieve maximum uniformity on s 2 s 2 grids in two dimensions, in addition to achieving maximum uniformity in one dimension. Theorem 2 (i) gives Proposition 1 (i) and Theorem 2 (ii) and (iii) imply Proposition 1 (ii). Example 5. Consider the array B in Example 4. Divide the nine levels into three groups: {0,α+ 1, 2α + 2}, {1,α+ 2, 2α} and {2,α,2α + 1}. Label the levels of these three groups as: {0,α+ 1, 2α + 2} {4, 6, 5}, {1,α+ 2, 2α} {7, 9, 8}, {2,α,2α + 1} {3, 1, 2}. Then, B is used to produce a Sudoku-based space-filling design D of 81 points, which can be divided into nine subdesigns D ij, each corresponding to one slice B ij of B. In any one dimension of the bivariate projections of D, each of the 81 equally spaced intervals of (0,1] contains exactly one point and in any two dimensions, each of the 9 9 square bins contains precisely one point. Figure 2 presents the bivariate projections of D 11, where in any one dimension, each of the nine equally spaced intervals of (0,1] contains exactly one point and in any two dimensions, each of the 3 3 square bins contains precisely one point.

10 720 XU XU, BEN HAALAND AND PETER Z. G. QIAN x x Fig. 2. Bivariate projections of a subdesign of nine points of a Sudoku-based space-filling design of 81 points, where the displayed numbers are the values of the nine points in the other dimension. ACKNOWLEDGEMENT The authors thank the editor and referees for comments that have improved the article. This work is supported by the National Science Foundation of the United States and an IBM Faculty Award. Ben Haaland is supported by a Singapore National Medical Research Council grant. REFERENCES BEHRENS,W.U.(1956). Feldversuchsanordnungen mit verbessertem ausgleich der bodenunterschiede. Zeitschrift für Landwirtschaftliches Versuchsund Untersuchungswesen 2, BUTLER, N.A.(2001). Optimal and orthogonal Latin hypercube designs for computer experiments. Biometrika 88, HAN, G., SANTNER, T.J., NOTZ, W.I.& L., B. D. (2009). Prediction for computer experiments having quantitative and qualitative input variables. Technometrics 51, HEDAYAT, A.S., SLOANE, N.J.A.& STUFKEN, J.(1999). Orthogonal Arrays: Theory and Applications. NewYork: Springer. LIDL, R.& NIEDERREITER, H.(1997). Finite Fields (Encyclopedia of Mathematics and Its Application), 2nd ed. Cambridge: Cambridge University Press. MCKAY, M.D., BECKMAN, R.J.& CONOVER, W.J.(1979). A comparison of three methods for selecting values of input variables in the analysis of output from a computer code. Technometrics 21, PEDERSEN,R.M.& VIS,T.L.(2009). Sets of mutually orthogonal Sudoku Latin squares. College Math. J. 40, QIAN,P.Z.G.& WU, C.F.J.(2009). Sliced space-filling designs. Biometrika 96, QIAN, P.Z.G., WU, H.& WU, C.F.J.(2008). Gaussian process models for computer experiments with qualitative and quantitative factors. Technometrics 50, TANG,B.(1993). Orthogonal array-based Latin hypercubes. J. Am. Statist. Assoc. 88, YE,K.Q.(1998). Orthogonal column Latin hypercubes and their application in computer experiments. J. Am. Statist. Assoc. 93, x [Received November Revised March 2011]