Báo cáo toán học: "Splitting Numbers of Grids Dwight Duffus" pps

Splitting Numbers of Grids Dwight Duffus Mathematics and Computer Science Department Emory University Atlanta GA 30322 USA dwight@mathcs.emory.edu Bill Sands Mathematics and Statistics Department The University of Calgary Calgary AB T2N 1N4 CANADA sands@math.ucalgary.ca Submitted: Nov 10, 2003; Accepted: Apr 4, 2005; Published: Apr 13, 2005 MR Subject Classifications [2000]: 06A07, 06D99 Abstract For a subset S of a finite ordered set P ,let S ↑ = {x ∈ P : x ≥ s for some s ∈ S} and S ↓ = {x ∈ P : x ≤ s for some s ∈ S}. For a maximal antichain A of P ,let s(A)= max A=U∪D |U ↑|+ |D ↓| |P | , the maximum taken over all partitions U ∪ D of A,and s k (P )= min A∈A(P ),|A|=k s(A) where we assume P contains at least one maximal antichain of k elements. Finally, for a class C of finite ordered sets, we define s k (C)= inf P ∈C s k (P ). Thus s k (C) is the greatest proportion r satisfying: every k-element maximal an- tichain of a member P of C can be “split” into sets U and D so that U ↑∪D ↓ contains at least r|P| elements. In this paper we determine s k (G k ) for all k ≥ 1, where G k = {k × n : n ≥ k} is the family of all k by n “grids”. the electronic journal of combinatorics 12 (2005), #R17 1 1 Introduction Given a maximal antichain A of an ordered set P ,saythatA splits if there is a partition A = U ∪ D such that P = U ↑∪D ↓,where U ↑ = {x ∈ P : x ≥ u for some u ∈ U} and D↓ = {x ∈ P : x ≤ d for some d ∈ D}. Say that P has the splitting property if every maximal antichain of P splits. Ahlswede, Erd˝os and Graham introduced these notions in [1], and proved that every finite Boolean lattice has the splitting property. In [2] we used the splitting property to study maximal antichains in distributive lattices. More recently, in [3], we characterized the set of dis- tributive lattices with the splitting property, and also introduced the idea of a splitting number for any finite ordered set and any class of finite ordered sets. We restate the required definitions. For a maximal antichain A of a finite ordered set P ,let s(A)= max A=U∪D |U ↑|+ |D ↓| |P | , the maximum taken over all partitions U ∪ D of A. Define the splitting number of P to be s(P )= min A∈A(P ) s(A) , where A(P ) is the set of all maximal antichains of P . Furthermore, if C is a class of finite ordered sets, we define the splitting number of C to be s(C)= inf P ∈C s(P ). We also make analogous definitions when the antichains involved are restricted to a certain size: for a finite ordered set P ,oraclassC of finite ordered sets, let s k (P )= min A∈A(P ),|A|=k s(A)ands k (C)= inf P ∈C s k (P ), where k is a positive integer such that P contains at least one maximal antichain of k elements. Thus s k (C) is the greatest proportion r satisfying: every k-element maximal antichain of a member P of C canbe“split”intosetsU and D so that U ↑∪D ↓ contains at least r|P | elements. The same condition with the restriction on antichain size removed yields s(C). It is clear that s(P ) ≤ s k (P )ands(C) ≤ s k (C) for all k. Note that the Ahlswede–Erd˝os–Graham theorem [1] could be stated as: s(B)=1 where B is the class of all finite Boolean lattices. Also, it’s not difficult to see that if P is the class of all finite ordered sets, s(P)=s k (P)=1/2 for all k. The problem of determining s(C)ands k (C) for various classes is an interesting order-theoretic and combinatorial task. When C is the family of all finite distributive lattices, for instance, we only have bounds (and not very good ones) in [3]. But the more restricted family G k = {k × n : n ≥ k} of all k by n “grids”, where k is fixed and n ≥ k,appearedto the electronic journal of combinatorics 12 (2005), #R17 2 us to present a challenging but attainable goal, and in [3] we began to determine s k (G k ). It is not difficult to show that lim k→∞ s k (G k ) = 1, and at the time we had a guess for what s k (G k ) was, linked closely to the Pell numbers and a “Pascal-like” triangle. Here, we present verification of our guess. Theorem 1 For all positive integers k, s k (G k )=1− 1 k + 1 ky k , where y k is defined by: y 1 =2, y 2 =3, y 3 =6, and y k =  2y k−1 − y k−4 for k odd, 2y k−1 − y k−2 for k even. Thus the sequence y 1 ,y 2 , starts 2, 3, 6, 9, 16, 23, 40, 57, 98, 139, 238, and we get s 1 (G 1 )= 1 2 ,s 2 (G 2 )= 2 3 ,s 3 (G 3 )= 13 18 ,s 4 (G 4 )= 7 9 , and so on, as reported in [3]. The first two values are derived from general results for distributive lattices. The values for k = 3 through k = 6 were obtained with an early version of the strategy fully developed in this paper. Actually, the result we obtain is stronger: for each odd k there is, in a certain sense, a “unique” antichain which realizes the minimum splitting value. This is made precise in Theorem 2 in Section 5. For even k, there does not appear to be uniqueness, unless perhaps symmetry is imposed. Here is an outline of the paper. Section 2 contains definitions and notation. We employ matrix notation in the proof of Theorem 1 and the required material is provided in Sections 2 and 3. Section 3 also contains our proof that the value given in Theorem 1 is a lower bound for s k (G k ). The converse inequality is verified in Section 4. In Section 5 is the promised description and proof of uniqueness. Finally in Section 6 we show that for even k we cannot obtain the same uniqueness result and state some open problems. 2 Preliminaries We represent a k-element maximal antichains in the lattice L = k × n as a vector of nonnegative integers. Assume that the chains are labelled so that k = {1 < 2 < <k} and n = {1 < 2 < < n}.Givenak-element maximal antichain A = {a 1 ,a 2 , ,a k }, we can put a 1 =(k, n 1 ), a 2 =(k−1,n 1 +n 2 ), and in general a i =(k+1−i, n 1 +n 2 +···+n i ), where n i ≥ 1 for all 1 ≤ i ≤ k and  k i=1 n i ≤ n. Letting n k+1 = n −  k i=1 n i ≥ 0, we have a representation of A by the vector n =(n 1 ,n 2 , ,n k+1 ). It is clear that this provides a 1-1 correspondence between k-antichains of L and (k + 1)-vectors of integers n =(n 1 ,n 2 , ,n k+1 )where  k+1 i=1 n i = n, n i > 0(i =1, 2, ,k)andn k+1 ≥ 0. the electronic journal of combinatorics 12 (2005), #R17 3 Given a maximal antichain A of L, there is a corresponding natural partition {N(i, j):1≤ i ≤ k, 1 ≤ j ≤ k +1} of L into intervals N(i, j), where N(i, j)=  (i, v) ∈ L : j−1  t=1 n t <v≤ j  t=1 n t  for 1 ≤ i ≤ k and 1 ≤ j ≤ k +1. For all i, j, |N(i, j)| = n j . (See Figure 1.) An orientation o of a maximal antichain A is an ordered pair (U, D)whereA is partitioned by U and D.Wesaythato captures the elements in U ↑∪D ↓. We assign ↑’s to the elements of U and ↓’s to those of D. For instance, if o has U = {a 1 ,a 3 ,a 5 , } and D = {a 2 ,a 4 ,a 6 , },wedenoteo by a 1 ↑ a 2 ↓ a 3 ↑ a 4 ↓ . With the elements of A in their natural order, the a i ’s can be dropped and an orientation can be defined by a k-sequence of ↑’s and ↓’s — the “alternating” orientation above is just ↑↓↑↓↑↓ . a 1 a 2 a k+1−i a j a k k i k +1− j 2 1 n 1 n 1 + n 2 n 1 + ···+ n j n N(i, j) Figure 1: an interval N(i, j) the electronic journal of combinatorics 12 (2005), #R17 4 The reverse of an orientation o is the orientation o r obtained by both reversing the order of the arrows and replacing each ↑ by a ↓ and vice versa; so for o = ↑↑↓ we would get o r = ↑↓↓ for example. (See Figure 2.) An orientation is self-reversing if it is equal to its reverse. a 1 a 2 a 3 a 1 a 2 a 3 Figure 2: an orientation and its reverse Most of the time, if any element of N(i, j) is captured by an orientation o then all are. The only exception is that o can capture the greatest element (i, n 1 + ···+ n j )ofN(i, j) without capturing all of N(i, j). This happens for such an N(i, j) precisely if i+j ≥ k+1, a j ↑,anda r ↓ for all r such that k +1− i ≤ r<j; in this case all elements (r,  j l=1 n l ) for k +1− j ≤ r ≤ i are exceptional. In fact, o captures r 1 n 1 + r 2 n 2 + ···+ r k+1 n k+1 + r 0 elements of L,wherer j (j ≥ 1) is the number of indices i such that N(i, j) is captured by o,andr 0 is the number of exceptional elements (r, n 1 + ···+ n j ) captured by o but notinanintervalcapturedbyo.Notethat0≤ r 0 ≤ k, since there is at most one such exceptional element with a given first coordinate. Define the capture vector v o induced by o to be the (k + 1)-vector (r 1 ,r 2 , ,r k+1 ). Then the number of elements captured by o is v o · n + r 0 , the dot denoting the dot product of the vectors. Here is a simple result which we will need later. The reverse v r ofavectorv is obtained by writing the components of v in reverse order. Lemma 1 For any orientation o, v o r =(v o ) r . Finally, the methods we develop to prove Theorem 1 do not apply in case k =2. As noted above, for small values of k, the result in Theorem 1 was obtained in [3]. Where needed, we are free to assume k = 2 in Sections 3 and 4. 3 The lower bound Let s k =1− 1 k + 1 ky k the electronic journal of combinatorics 12 (2005), #R17 5 be the quantity given in Theorem 1. Our goal here is to prove that every maximal k- antichain of a lattice L = k × n ∈G k can be oriented so as to capture at least s k |L| elements of L. This will prove that s k (G k ) ≥ s k . To establish the lower bound, we show that for any antichain A with associated vector n as defined above, there is an orientation o such that v o · n ≥ s k |L| = s k kn.Ourmethod is to find a nonempty set O = {o 1 , ,o m } of orientations, and positive numbers λ i , 1 ≤ i ≤ m,sothat m  i=1 λ i (v o i · n)=s k kn m  i=1 λ i (1) for all n. It follows that at least one of the o i ’s in O satisfies v o i · n ≥ s k kn,sos k (G k ) ≥ s k as desired. It turns out we can select O and the λ i ’s independently of n. Arranging the capture vectors of the m orientations in the set O as rows of a matrix, we obtain the m by k +1 capture matrix M k of O. Now we can rewrite (1) in matrix form as nM t k λ = ns k kJλ, where n is a 1-by-(k + 1) row vector, λ =(λ 1 , ,λ m )isanm-by-1 column vector, and J is a matrix of 1’s of appropriate size, in this case (k +1)-by-m. This equation can be written as n(M t k − s k kJ)λ =0, so it certainly suffices to prove that (M t k − s k kJ)λ = 0, (2) where 0 is a zero column vector of length m. Now we will define the set O of m orientations and the associated m-vector λ. It turns out that we can let m = k +1. Define the orientations a ↑ :  ↑↓↑↓↑↓ ↑ for k odd, ↑↓↑↓↑↓ ↓ for k even, a ↓ :  ↓↑↓↑↓ ↓ for k odd, ↓↑↓↑↓ ↑ for k even, (3) and o i :                ↑↓↑↓ ↓ i ↑↑↓↑↓ ↑↓ for k odd, 1 ≤ i<k/2, i odd, ↓↑↓↑ ↓ i ↑↑↓↑↓ ↓↑ for k odd, 1 <i<k/2, i even, ↑↓↑↓ ↓ i ↑↑↓↑↓ ↓↑ for k even, 1 ≤ i<k/2, i odd, ↓↑↓↑ ↓ i ↑↑↓↑↓ ↑↓ for k even, 1 <i<k/2, i even. Also, for k even and at least 4, define the orientation oo : ↑↑↓↑↓↑ ↓↑↓↓ . The capture vectors of these orientations are given in the following tables, the first is for k odd, and the second is for k even. All vectors, including the constant vector k − 1=(k − 1,k− 1, ,k− 1), have length k +1. the electronic journal of combinatorics 12 (2005), #R17 6 orientation o capture vector v o a ↑ k − 1+(0, 1, −1, 1, − 1, ,−1, 1) a ↓ k − 1+(1, −1, 1, −1, ,−1, 1, 0) o 1 k − 1+(− 1, 0, 1, −1, 1, −1, 1, ,−1, 1, 0) o i ,1<i<k/2, i odd k − 1+(0, 1, −1, 1, −1, ,−1, 1, i −2, 0, 1, −1, 1, −1, ,−1, 1, 0) o i ,1<i<k/2, i even k − 1+(1, −1, 1, −1, ,−1, 1, i −2, 0, 1, −1, 1, −1, ,−1, 1), orientation o capture vector v o a ↑ k − 1+(0, 1, −1, 1, − 1, ,−1, 1, 0) a ↓ k − 1+(1, −1, 1, −1, ,−1, 1) o 1 k − 1+(−1, 0, 1, −1, 1, −1, 1, ,−1, 1) o i ,1<i<k/2, i odd k − 1+(0, 1, −1, 1, −1, ,−1, 1, i −2, 0, 1, −1, 1, −1, ,−1, 1) o i ,1<i<k/2, i even k − 1+(1, −1, 1, − 1, ,−1, 1, i −2, 0, 1, −1, 1, −1, ,−1, 1, 0), oo k − 1+(−1, 0, 1, −1, 1, −1, ,−1, 1, 0, −1) n 1 n 2 n 3 n k+1 Figure 3: finding the capture vector of a ↑ the electronic journal of combinatorics 12 (2005), #R17 7 Figure 3 shows the alternating orientation a ↑ applied to a maximal antichain of an arbitrary lattice in G k for odd k. The antichain elements are shown as small solid circles, and larger hollow circles show intervals which are not captured by the orientation. One sees that exactly k − 1intervals(outofk)oflengthn 1 are captured, all k of the length n 2 intervals are captured, k − 2 of the length n 3 intervals are captured, and so on, giving v a ↑ =(k − 1,k,k− 2,k,k− 2, ,k,k− 2,k)=k − 1+(0, 1, −1, 1, −1, ,−1, 1) as claimed. The other capture vectors can be similarly checked. Note that a ↓ is the reverse of a ↑ when k is odd, but not when k is even. When k is even, the orientations a ↑ , a ↓ and oo are all self-reversing. For k odd, we let O = {a ↑ ,o (k−1)/2 ,o (k−3)/2 , ,o 2 ,o 1 ,o r 1 ,o r 2 , ,o r (k−1)/2 ,a ↓ }. Note that |O| = k + 1. As noted in Lemma 1, the capture vector of o r is the reverse of the capture vector of o. Thus with the orientations of O ordered as listed, we obtain the (k +1)× (k + 1) capture matrix M k =(k − 1)J k+1 + C k , where J k+1 is the square all-ones matrix of order k +1, C k =                      01−11−1 −11−11−1 | 1 −11−1 −11−11 1 −11−11 1 −11−20| 1 −11−1 −11−11 01−11−1 −11−20 1|−11−11 1 −11 0 1 −11−11 1 −20 1−1 | 1 −11−1 −11−11 . . . . . . 01−20 1 1 −11−11|−11−11 1 −11 0 1 −20 1−1 −11−11−1 | 1 −11−1 −11−11 −10 1−11 1 −11−11|−11−11 1 −11 0 01−11−1 −11−11−1 | 1 −11−1 −11 0−1 1 −11−11 1 −11−11|−11−11 10− 21 01−11−1 −11−11−1 | 1 −11−1 0 −21 0 1 −11−11 1 −11−11|−11−11 −21−11 . . . . . . 01−11−1 −11−11−1 | 10−21 1 −11 0 1 −11−11 1 −11−11| 0 −21−1 −11−11 1 −11−11 1 −11−11|−11−11 1 −11 0                      when k ≡ 1 mod 4, and the electronic journal of combinatorics 12 (2005), #R17 8 C k =                      01−11−1 1 −11−11|−11−11 −11−11 01−11−1 1 −11−20| 1 −11−1 1 −11 0 1 −11−11 −11−20 1|−11−11 −11−11 01−11−1 1 −20 1−1 | 1 −11−1 1 −11 0 . . . . . . 01−20 1 −11−11−1 | 1 −11−1 1 −11 0 1 −20 1−1 1 −11−11|−11−11 −11−11 −10 1−11 −11−11−1 | 1 −11−1 1 −11 0 01−11−1 1 −11−11|−11−11 −11 0−1 1 −11−11 −11−11−1 | 1 −11−1 10−21 01−11−1 1 −11−11|−11−11 0 −21 0 1 −11−11 −11−11−1 | 1 −11−1 −21−11 . . . . . . 1 −11−11 −11−11−1 | 10−21 −11−11 01−11−1 1 −11−11| 0 −21−1 1 −11 0 1 −11−11 −11−11−1 | 1 −11−1 1 −11 0                      when k ≡ 3mod4. [Note: the horizontal and vertical lines divide C k into four square submatrices of order (k +1)/2.] For k even and at least 4, we similarly define O = {a ↑ ,o (k/2)−1 ,o (k/2)−2 , ,o 2 ,o 1 , oo, o r 1 ,o r 2 , ,o r (k/2)−1 ,a ↓ }. Again |O| = k + 1, and this time the (k +1)× (k + 1) capture matrix is M k =(k − 1)J k+1 + C k where C k =                          01−11−1 −11−11−1 | 1 |−11−11 1 −11 0 1 −11−11 1 −11−20| 1 |−11−11 1 −11 0 01−11−1 −11−20 1|−1 | 1 −11−1 −11−11 1 −11−11 1 −20 1−1 | 1 |−11−11 1 −11 0 . . . . . . 01−20 1 1 −11−11|−1 | 1 −11−1 −11−11 1 −20 1−1 −11−11−1 | 1 |−11−11 1 −11 0 −10 1−11 1 −11−11|−1 | 1 −11−1 −11−11 − 10 1−11 1 −11−11|−1 | 1 −11−1 −11 0−1 1 −11−11 1 −11−11|−1 | 1 −11−1 −11 0−1 01−11−1 −11−11−1 | 1 |−11−11 10−21 1 −11−11 1 −11−11|−1 | 1 −11−1 0 −21 0 01−11−1 −11−11−1 | 1 |−11−11 −21−11 . . . . . . 1 −11−11 1 −11−11|−1 | 10−21 1 −11 0 01−11−1 −11−11−1 | 1 | 0 −21−1 −11−11 1 −11−11 1 −11−11|−1 | 1 −11−1 −11−11                          when k ≡ 2 mod 4, and the electronic journal of combinatorics 12 (2005), #R17 9 C k =                          01−11−1 1 −11−11|−1 | 1 −11−1 1 −11 0 01−11−1 1 −11−20| 1 |−11−11 −11−11 1 −11−11 −11−20 1|−1 | 1 −11−1 1 −11 0 01−11−1 1 −20 1−1 | 1 |−11−11 −11−11 . . . . . . 01−20 1 −11−11−1 | 1 |−11−11 −11−11 1 −20 1−1 1 −11−11|−1 | 1 −11−1 1 −11 0 −10 1−11 −11−11−1 | 1 |−11−11 −11−11 − 10 1−11 −11−11−1 | 1 |−11−11 −11 0−1 1 −11−11 −11−11−1 | 1 |−11−11 −11 0−1 01−11−1 1 −11−11|−1 | 1 −11−1 10−21 1 −11−11 −11−11−1 | 1 |−11−11 0 −21 0 01−11−1 1 −11−11|−1 | 1 −11−1 −21−11 . . . . . . 01−11−1 1 −11−11|−1 | 10−21 −11−11 1 −11−11 −11−11−1 | 1 | 0 −21−1 1 −11 0 1 −11−11 −11−11−1 | 1 |−11−11 −11−11                          when k ≡ 0 mod 4. In each matrix the central row and column are flanked by four k/2 × k/2 submatrices. The next step is definition of the vector λ. For this we will use the sequence of integers (y k ) defined in Theorem 1, and also the well-known Pell numbers (u k ), defined by: u 1 =1, u 2 =2,andu i =2u i−1 + u i−2 for i ≥ 3, so that (u k )=(1, 2, 5, 12, 29, 70, ). We will also need the initial value u 0 = 0 in some circumstances. In case that k is even, we require yet another sequence of integers, closely related to the Pell numbers. Define (v k )by: v i = u i + u i−1 ,sothat (v 1 ,v 2 , )=(1, 3, 7, 17, 41, 99, ). See [5], for example, for information on these sequences. The following results are routine, and their proofs are left to the reader. Lemma 2 (a) u 2n =2 n  i=1 u 2i−1 ,u 2n+1 =2 n  i=1 u 2i +1. (b) u n+1 = n  i=1 v i +1. (c) v n =2v n−1 + v n−2 . (d) y n =            2 (n+1)/2  i=1 u i for n odd, 2 n/2  i=1 u i + u (n/2)−1 + u n/2 for n even. (e) y 2n =2 n  i=1 v i +1. the electronic journal of combinatorics 12 (2005), #R17 10 [...]... dot product of each column of Ck with λ equals 1 Let coli (Ck ) denote the ith column of Ck Our proof treats the four congruence classes of k modulo 4 separately Case (i): k ≡ 1 mod 4 Note that, for each i, the ith row of Ck is the reverse of the (k + 2 −i)th row by Lemma 1, since the corresponding orientations are reverses of each other Thus it is also true that coli (Ck ) is the reverse of colk+2−i... + 1 − u(k/2)−1 − uk/2 = (k − 1)yk + 1 This completes the proof of (7) and verification of the upper bound The proof of Theorem 1 is complete Note that, by (7) and the remark following the proof of (6), all the orientations o in the families O defined in §3 satisfy v o · nk = sk (kyk ) and so capture essentially the maximum number of elements of the lattice defined by the vector nk This helps to explain... electronic journal of combinatorics 12 (2005), #R17 29 family of vectors that are worth recording, particularly as they lead to some interesting open problems As was noted at the beginning of §2, any member o of Ok can be regarded as a ksequence of ↑’s and ↓’s Identify ↑ with 0 and ↓ with 1, so Ok is the set of all binary sequences of length k Let denote the usual lexicographic ordering of Ok Let’s also... an odd integer This completes the proof of Theorem 2 Incidentally, from the above proof we have (Mk − sk kJ)nk = Ck − 1 J nk = Ck nk − 1, yk so the property Mk nk = sk kJnk is equivalent to Ck nk = 1 t Compare this to the property Ck λ = 1 which we derived in §3 the electronic journal of combinatorics 12 (2005), #R17 27 6 Nonuniqueness of nk for k even; the set of capture vectors; open problems In... orientations o of A (Since x = 9 = y4 , this is (5) for k = 4 and with n4 replaced by x.) Therefore scalar multiples of each such vector will yield, in the corresponding grid, a maximal antichain which can capture at most 7/9 of the grid in the limit To show this we can mimic the methods of §4, although the lack of symmetry in x means we have more work to do To establish the counterpart of (7) when k... as row 5 of the triangular array at the end of §4 But other values of t give nonsymmetric solutions We did not consider the case k = 2 in this paper, as the matrix Mk is not defined for k = 2 But the value s2 = 2/3 was derived in [3], and the reader can easily check that, for any vector (t, 1, 2 − t) where 1/2 ≤ t ≤ 3/2, at most 2/3 of the 2 × n grid is captured by any of the four orientations of the... integer k For each of the 2k orientations o of an antichain in Gk = k × n, there is an associated (k + 1)-element capture vector v o Let Ok denote the set of all orientations, and let Vk denote the set of 2k (k + 1)-element capture vectors While it is easy to generate the 2k vectors in Vk recursively, we do not have a pleasing characterization of these integer vectors We do have a couple of observations... −2 1 −1 · · · 1 −1 1 0 The number of consecutive transpositions needed to move row 2 of Ck to the bottom of the matrix is k − 1, an even number, and two more transpositions are then needed the electronic journal of combinatorics 12 (2005), #R17 25 to move (original) row k to the bottom The number of transpositions needed to move column (k + 1)/2 to the right edge of Ck is the same as the number subsequently... verified (6) Note that, by the above proof, equality can only hold in (6) in Case (i), and then only for orientations o such that br−1 = ar−1 + 3, where ar−1 and br−1 are the (r − 1)th components of the capture vectors of o and o respectively The reader can check that this ˆ happens precisely if the two consecutive ↑’s in o are followed by a ↓ Let’s turn to the proof of (7), namely that v a · nk = sk (kyk... the last case, and the verification of the lower bound in Theorem 1 4 The upper bound Suppose that a k-antichain A in L = k×n is defined by the (k +1)-vector n (so n = n) We noted in Section 2 that the number of elements of L captured by an orientation o of A is v o · n + r0 , where r0 ≤ k Thus to prove sk (Gk ) ≤ sk for any fixed k, where sk is given at the beginning of §3, it is enough to verify that, . Splitting Numbers of Grids Dwight Duffus Mathematics and Computer Science Department Emory University Atlanta GA 30322 USA dwight@ mathcs.emory.edu Bill Sands Mathematics. by showing that the dot product of each column of C k with λ equals 1. Let col i (C k )denotetheith column of C k . Our proof treats the four congruence classes of k modulo 4 separately. Case. min A∈A(P ) s(A) , where A(P ) is the set of all maximal antichains of P . Furthermore, if C is a class of finite ordered sets, we define the splitting number of C to be s(C)= inf P ∈C s(P ). We also