We investigate a packing problem in M-dimensional grids, where bounds are given for the number of allowed entries in different axis-parallel directions. The concept is motivated from error correcting codes and from more-part Sperner theory. It is also closely related to orthogonal arrays. We prove that some packing always reaches the natural upper bound for its size, and even more, one can partition the grid into such packings, if a necessary divisibility condition holds. We pose some extremal problems on maximum size of packings, such that packings of that size always can be extended to meet the natural upper bound. 1 The concept of full transversals Let us be given positive integers n1,n2,...,nM and L1,L2,...,LM, such tha

Aydinian, H.

Czabarka, É.

Engel, K.

Erdős, Péter

Székely, L.A.

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AUSTRALASIAN JOURNAL OF COMBINATORICSVolume 48 (2010), Pages 133–141A note on full transversals and mixedorthogonal arrays∗Harout Aydinian†University of BielefeldPOB 100131, D-33501 BielefeldGermanyayd@mathematik.uni-bielefeld.deE´va CzabarkaUniversity of South CarolinaColumbia, SC 29208U.S.A.czabarka@math.sc.eduKonrad EngelUniversita¨t RostockInstitut fu¨r Mathematik, 18051 RostockGermanykonrad.engel@uni-rostock.dePe´ter L. Erdo˝s‡Alfre´d Re´nyi Institute13-15 Rea´ltanoda u., 1053 BudapestHungaryelp@renyi.huLa´szlo´ A. Sze´kely§University of South CarolinaColumbia, SC 29208U.S.A.szekely@math.sc.edu134 HAROUT AYDINIAN ET AL.AbstractWe investigate a packing problem in M-dimensional grids, where boundsare given for the number of allowed entries in diﬀerent axis-parallel di-rections. The concept is motivated from error correcting codes and frommore-part Sperner theory. It is also closely related to orthogonal arrays.We prove that some packing always reaches the natural upper bound forits size, and even more, one can partition the grid into such packings, if anecessary divisibility condition holds. We pose some extremal problemson maximum size of packings, such that packings of that size always canbe extended to meet the natural upper bound.1 The concept of full transversalsLet us be given positive integers n1, n2, . . . , nM and L1, L2, . . . , LM , such thatL1n1 + 1≤ L2n2 + 1≤ · · · ≤ LMnM + 1≤ 1. (1)Let Π = [0, n1] × · · · × [0, nM ], and let I denote a subset of Π. We call an I ⊆ Πan (L1, L2, . . . , LM)-transversal, if there are no Li + 1 elements of I, any two ofthem diﬀering from each other only in the ith coordinate, for any i = 1, 2, . . . ,M .Identifying I with a 0-1 valued function deﬁned on Π, namely the indicator functionof I, an alternative deﬁnition of the (L1, L2, . . . , LM )-transversal is that for any k,ﬁxing i1, . . . , îk, . . . , iM in an arbitrary fashion (where ̂ denotes a missing entry),nk∑ik=0I(i1, . . . , ik, . . . , iM) ≤ Lk (2)holds. We talk about L-transversals, if L1 = L2 = · · · = LM = L.Lemma 1.1. If I is an (L1, . . . , LM)-transversal, then for every i = 1, . . . ,M|I| ≤ Lini + 1M∏j=1(nj + 1). (3)Proof. The statement is almost trivial: deleting the ith coordinate from the elementsof I leaves at most (n1 + 1) · · · (nM + 1)/(ni + 1) distinct elements. No one of these∗ E´va Czabarka, Pe´ter L. Erdo˝s and La´szlo´ A. Sze´kely were supported in part by the HUBIMTKD-CT-2006-042794.† Harout Aydinian was supported in part by the EU Project: Fist (Finite Structures) MarieCurie Host Fellowship for the Transfer of Knowledge, while visiting the Alfre´d Re´nyi Institute ofthe Hungarian Academy of Sciences, Budapest, Summer 2007; and by the DFG Project AH46/8-1.‡ Pe´ter L. Erdo˝s was supported in part by the Hungarian NSF contracts Nos. T37846, T34702,T37758.§ La´szlo´ A. Sze´kely was supported in part by the NSF DMS contracts Nos. 0701111 and 1000475.A NOTE ON FULL TRANSVERSALS 135elements may have more than Li diﬀerent pre-images, since otherwise I would failbeing an (L1, . . . , LM)-transversal.We call an (L1, . . . , LM)-transversal I full, if I shows equality in (3) for i = 1.One motivation for this research is that (L1, L2, . . . , LM )-transversals deﬁne ho-mogeneous M-part (n1, n2, . . . , nM ;L1, L2, . . . , LM)-Sperner families, see [1], [2], [6]or [7]. Although full transversals were important for the study of 2-part Spernerfamilies [1], [6], [7], they turned less useful for M-part Sperner families [2]. Howeversome positive results for the existence problem of full transversals might lead to newresults in that theory. In the last section we will discuss the connection between fulltransversals and (generalized) orthogonal arrays.For L = 1, a more general notion of a d-fold transversal is a subset I ⊂ Π with theproperty that for every two distinct elements of the d-fold transversal, there exist atleast d + 1 coordinates in which they diﬀer. In other words the minimum Hammingdistance of I is d+1, which means that I is a⌊d2⌋- error correcting code. The d-foldtransversal is related to d-fold M-part Sperner families in [2] like transversals arerelated to M-part Sperner families.The main concern of this note is the existence of full transversals. If we relax thedeﬁnition of an (L1, L2, . . . , LM)-transversal to a non-negative real function deﬁnedon Π, and modify the deﬁnition of |I| to∑i1∑i2· · ·∑iMI(i1, i2, . . . , iM),then the analogue of (3) still holds. Deﬁning the relaxation of full transversals byequality in the analogue of (3) for i = 1, and setting I(i1, i2, . . . , iM) =L1n1+1, oneﬁnds that the relaxed transversals always exist. Therefore our 0-1 full transversalproblem is about the feasibility of a particular integer program, which is a packingproblem. We give a very explicit solution to the 0-1 full transversal problem, unlike[8], which invoked matching theory to settle the special case M = 2, L1 = L2.Furthermore, if L1 divides n1 + 1, which is a necessary condition for partitioning Πinto full transversals, we construct such a partition.We raise an extremal problem: what is the smallest size of a transversal in Πthat cannot be extended into a full transversal of Π?2 Existence of full transversalsFor a real number α, let 〈α〉 be its fractional part, i.e.〈α〉 = α− α	.We will prove the following lemma:Lemma 2.1. Let α, β, μ ∈ R and 0 ≤ μ < 1, 0 ≤ β ≤ 1− μ. Then∣∣∣∣∣{i ∈ Z : 0 ≤ i ≤ n,〈α +in + 1〉∈ [β, β + μ)} ∣∣∣∣∣≤ (n + 1)μ (4)136 HAROUT AYDINIAN ET AL.and if (n + 1)μ is integral, then equality holds.Proof. α, α + 1n+1, α + 2n+1, . . . , α + nn+1are n + 1 evenly spaced points at distance1n+1in the interval [α, α + 1).Divide the interval [β, β + μ) into k + 1 := (n + 1)μ subintervals, k of whichare of length 1n+1, as follows:[β, β +1n + 1), . . . ,[β +k − 1n + 1, β +kn + 1),[β +kn + 1, β + μ).Since any half-open interval that is contained in [0, 1) and has length at most 1n+1contains at most one of the points of the form〈α + jn+1〉, and intervals of length 1n+1do contain one such point, the lemma follows.Theorem 2.2. Let n1, . . . , nM and L1, . . . , LM be positive integers satisfying inequal-ity (1) and set μ = L1n1+1. Then, for any β with 0 ≤ β ≤ 1− μ,I ={(i1, . . . , iM) ∈ Π :〈M∑j=1ijnj + 1〉∈ [β, β + μ)}is a full transversal.Proof. Let j ∈ {1, 2, . . . ,M} and ﬁx i1, . . . , ij−1, ij+1, . . . , iM such that ik ∈{0, 1, . . . , nk} for all k ∈ {1, 2, . . . ,M} \ {j}. Setαj =∑k∈{1,2,...,M}\{j}iknk + 1.Then (nj + 1)μ ≤ (nj + 1) · Ljnj+1 = Lj , so by Lemma 2.1, we have that∣∣∣∣∣{ij ∈ Z : 0 ≤ ij ≤ nj,〈αj +ijnj + 1〉∈ [β, β + μ)} ∣∣∣∣∣≤ Lj,with other words I is a transversal.Now assume that j = 1. Then (n1+1)μ = L1, which is integral, and by Lemma 2.1we have that∣∣∣∣∣{i1 ∈ Z : 0 ≤ i1 ≤ n1,〈α1 +i1n + 1〉∈ [β, β + μ)} ∣∣∣∣∣= L1,with other words I is a full transversal.Corollary 2.3. Let n1, . . . , nM and L1, . . . , LM be positive integers satisfying in-equality (1) and set A =⌊n1+1L1⌋. Then Π can be decomposed into A full transversalsand one additional transversal, where this additional transversal is empty if n1+1L1isintegral.A NOTE ON FULL TRANSVERSALS 137Proof. Let with μ = L1n1+1, for k = 0, 1, . . . , A− 1,Ik ={(i1, . . . , iM) ∈ Π :〈M∑j=1ijnj + 1〉∈ [kμ, (k + 1)μ)}andIk+1 ={(i1, . . . , iM) ∈ Π :〈M∑j=1ijnj + 1〉∈ [Aμ, 1)}.Clearly, Π =⋃k+1j=1 Ij. By Theorem 2.2, I1, . . . , Ik are full transversals and, since1− Aμ ≤ μ, Ik+1 is a transversal.3 Extendability of transversals into full transversalsThere are many transversals that cannot be extended into full transversals. SetM = 2, n1 = n2 = n ≥ 2 and let L be an integer with 2 ≤ L ≤ n. Set m =max(n−L+1, L−1), so 1 ≤ n−m ≤ L−1. Consider a full L-transversal on [0,m]2(since L ≤ m+1, such a transversal exists). This is an L-transversal of [0, n]× [0, n],which cannot be extended into a full L-transversal of [0, n]2: as we can only addelements from [m + 1, n]2, no extension can have more elements thanL(m + 1) + (n −m)2 = L(n + 1) + (n −m)(n −m− L) < L(n + 1).We give below a suﬃcient condition for the case M = 2, which guarantees thatevery transversal can be extended into a full transversal. Note that for M = 2, (1)boils down to L1(n2 + 1) ≤ L2(n1 + 1). We will use a stronger condition than this,namelyL1(n2 + 1) + (L1 − 1)(L2 − 1) ≤ L2(n1 + 1). (5)Theorem 3.1. If Condition (5) holds, then any (L1, L2)-transversal can be extendedinto a full transversal.Proof. Assume that we have an (L1, L2)-transversal I, which is not full, so |I| <L1(n2 + 1). There is a column, say column j, such that there are only x ≤ L1 − 1elements in this column from I. We claim that there is a row i with at most L2 − 1elements in I and with (i, j) ∈ I — now transversal I can be extended by adding(i, j). Were the claim false, I had x entries in column j and n1+1−x rows containingL2 additional entries each, so |I| ≥ x + (n1 + 1 − x)L2 = (n1 + 1)L2 − x(L2 − 1) ≥(n1 + 1)L2 − (L1 − 1)(L2 − 1) ≥ L1(n2 + 1), where the last inequality follows from(5). We have the contradiction.Observe that the opening example of this section fails inequality (5) just by 1when L = 2. We are left with an intriguing open problem: if not all transversals ofΠ can be extended into a full transversal of Π, then what is the smallest size of atransversal of Π that cannot be extended into a full transversal of Π? More precisely,138 HAROUT AYDINIAN ET AL.we deﬁne P := P(n1, . . . , nM ;L1, . . . , LM) as the set of (L1, . . . , LM)-transversals inΠ that can not be extended into a full transversal, andt(n1, . . . , nM ;L1, . . . , LM) ={minI∈P |I|, if P = ∅undeﬁned otherwise.Obviously, t(n1, . . . , nM ;n1 + 1, . . . , nM + 1) is undeﬁned. The beginning para-graph of this section shows that t(n, n;L,L) ≤ Lmax(n − L + 2, L) for 2 ≤ L ≤ n .Theorem 3.1 implies that t(n, n; 1, 1) is undeﬁned, which does not extend to M = 3:Proposition 3.2. For 1 ≤ L ≤ n, 2 ≤ n we have thatt(n, n, n;L,L, L) ≤{L(⌈n2⌉+ 1)2, if L ≤ ⌈n2⌉+ 1L3, otherwise.Proof. Letm ={⌈n2⌉, if L ≤ ⌈n2⌉+ 1L− 1, otherwise.Since L ≤ m + 1, by Theorem 2.2 there is a full L-transversal of [0,m]3. Let I besuch a full L-transversal, then |I| = L(m+1)2. Also, I is an L-transversal in [0, n]3.To prove our theorem, it is enough to show that I can not be extended into a fullL-transversal of [0, n]3.Assume, contrary to our statement, that I ′ is a full L-transversal in [0, n]3 withI ⊆ I ′. Then |I ′| = L(n + 1)2. For k = 1, 2, 3, letΠ(1)k = A1 × A2 × A3, where Ak = [0, n] and for j = k,Aj = [m + 1, n],Jk = I′ ∩ Π(1)k = {(i1, i2, i3) ∈ I ′ : if j = k then ij > m} ,Ik = {(i1, i2, i3) ∈ I ′ : |{j : ij > m}| = k} ,Π(2)k = B1 ×B2 × B3, where Bk = [0, n] and for j = k,Bj = [0,m].Clearly, |I ′| = |I|+ |I1|+ |I2|+ |I3|, and, since Jk is an L-transversal on Π(1)k , |Jk| ≤L(n−m)2. Since I ′ ∩Π(2)k is an L-transversal on Π(2)k , it has at most L(m+1)2 = |I|elements, and, since I ⊆ Π(2)k , I ′ ∩Π(2)k = I. But then I = I ′ ∩⋃k Π(2)k = I ∪ I1, fromwhich |I1| = 0. Also,⋃k Jk = I2 ∪ I3 and for k = k′, Jk ∩ Jk′ = I3, implying3L(n−m)2 ≥∑k|Jk| =∣∣∣⋃kJk∣∣∣+∑k =k′|Jk ∩ Jk′ | − |J1 ∩ J2 ∩ J3|= |I2 ∪ I3|+ 3|I3| − |I3| = |I2|+ 3|I3| ≥ |I2|+ |I3|+ |I1|= |I ′| − |I| = L(n + 1)2 − L(m + 1)2.Thus, we get(n + 1)2 ≤ (m + 1)2 + 3(n −m)2. (6)A NOTE ON FULL TRANSVERSALS 139Assume ﬁrst that L ≤ ⌈n2⌉+ 1. Equation (6) gives that(n + 1)2 ≤(⌈n2⌉+1)2+ 3⌊n2⌋2.For even n, this means (n + 1)2 < n2 + n + 1, a contradiction. For odd n, we get(n + 1)2 ≤(n + 32)2+3(n − 1)24=4n2 + 124= n2 + 2 + 1≤ n2 + n + 1,which is also a contradiction.Now, if⌈n2⌉+ 1 < L ≤ n, using k = n−m equation (6) gives(n + 1)2 ≤ (n + 1− k)2 + 3k2 = (n + 1)2 + 2k(2k − (n + 1)),which is a contradiction, since 2k < n.Therefore I cannot be extended to a full transversal.4 Connections with orthogonal arraysIn this section we give evidence of the close connection between full transversalsand orthogonal arrays. Consider r sets Si (i = 1, ..., r) of si symbols and take amatrix T of size N × r where the ith column draws its elements from the set Si.This matrix is a mixed (or asymmetrical) orthogonal array (the notion of orthogonalarray with variable numbers of symbols is also used), of strength d, constraints r andindex set L if for any choice of d diﬀerent columns j1, . . . , jd each possible sequence(aj1 , . . . , ajd) ∈ Sj1 × · · · × Sjd appears exactly λ(j1, . . . , jd) ∈ L times. In case ofequal symbol set sizes and ﬁxed λ we have the classical deﬁnition of orthogonalarrays, introduced by C.R. Rao (see [10, 11]). The name was coined by K.A. Bush[3, 4]. C.-S. Cheng [5] seems to be the ﬁrst to consider variable sizes. The standardreference work for orthogonal arrays is the book of A.S. Hedayat, N.J.A. Sloane andJ. Stufken [9]. These arrays are widely used in planning experiments or fractionalfactorial designs.It turns out that certain full transversals are suitable to design mixed orthogonalarrays:Proposition 4.1. Assume that parameters M ;L1, ...LM ;n1, ..., nM, in Theorem 2.2satisfy n1+1L1= ... = nM+1LM. Then the construction in Theorem 2.2 is a mixed orthogo-nal array with the symbol sets Si = {0, ..., ni} of strength M − 1, constraints M andindex set L = {L1, ..., LM}.If L1 divides n1 + 1, using Corollary 2.3, we can partition the set of row vectorsrepresenting the elements of Π into (n1 + 1)/L1 disjoint orthogonal arrays with thesymbol sets Si = {0, . . . , ni} of strength M − 1, constraints M and index set L ={L1, . . . , LM}.140 HAROUT AYDINIAN ET AL.Proof. In view of Theorem 2.2 we can construct full transversals with these param-eters. Such a full transversal can be described as an N × M matrix X = (xij),whereN = L1(n2 + 1) · · · (nM + 1) = Li(n1 + 1) · · · ̂(ni + 1) · · · (nM + 1),and the ith row vector (xi1, . . . , xiM ) is the ith element of our full transversal. Clearly,the jth column of this matrix draws its elements from Si. Taking M−1 columns of thematrix is the same as leaving out one column. If we leave out the jth column from thematrix, then — since we started with a full transversal — each (a1, . . . , âj, . . . , aM) ∈S1×· · ·× Ŝj×· · ·×SM appears exactly Lj times in the remaining matrix. Thus, thismatrix actually is a mixed orthogonal array with the symbol sets Si = {0, . . . , ni} ofstrength M−1, constraints M and index set L = {L1, . . . , LM}. The rest follows.Even those transversals that are not full have a connection to the theory of arrays,namely to packing arrays (see B. Stevens and E. Mendelsohn [12]).References[1] H. Aydinian and P.L. Erdo˝s, All maximum size 2-part Sperner systems — inshort, Comb. Prob. Comp. 16 (4) (2007), 553–555.[2] H. Aydinian, E´. Czabarka, P.L. Erdo˝s and L. A. Sze´kely, A tour of M-partL-Sperner families, submitted (2009), pp. 30.[3] K.A. Bush, Orthogonal Arrays, Ph.D.Thesis, North Carolina State University,(1950).[4] K.A. Bush, A generalization of a theorem due to MacNeish, Ann. Math. Stat.23 (2) (1952), 293–295.[5] C.-S. Cheng, Orthogonal arrays with variable numbers of symbols, Ann. Stats.8 (2) (1980), 447–453.[6] K. Engel, Sperner Theory, Cambridge University Press, Cambridge, New York;Encyclopedia of Mathematics and its Applications 65, (1997) pp. x+417.[7] P.L. Erdo˝s and G.O.H. Katona, Convex hulls of more-part Sperner families,Graphs and Combinatorics 2 (1986), 123-134.[8] Z. Fu¨redi, J.R. Griggs, A.M. Odlyzko and J.M. Shearer, Ramsey-Sperner theory,Discrete Math. 63 (1987), 143–152.[9] A.S. Hedayat, N.J.A. Sloane and J. Stufken, Orthogonal Arrays: Theory andApplications, Springer Series in Statistics. Springer-Verlag, New York, (1999)pp. xxiv+416.[10] C.R. Rao, M.A. Thesis, Calcutta University (1943).A NOTE ON FULL TRANSVERSALS 141[11] C.R. Rao, Factorial experiments derivable from combinatorial arrangements ofarrays, Suppl. J. Royal Stat. Soc. 9 (1) (1947), 128–139.[12] B. Stevens and E. Mendelsohn, Packing arrays and packing designs, Des. CodesCrypt. 27 (2002), 165–176.(Received 2 Dec 2009; revised 12 Apr 2010)

A note on full transversals and mixed orthogonal arrays

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A note on full transversals and mixed orthogonal arrays

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