2.3 Complete asymptotic results
2.3.3 The boundary between O(m k−1 ) and Θ(m k )
0 0 0
or no j vr vs
0 0 0
. (2.39)
Thus a column that violates one of the outside implications from ij will violate at least q − 1 of the q outside implications, and hence at least 12 of the implications. Thus at most 2(t −1) m2
columns of A0 can violate outside 0-implications. Also at mostO(m2) columns ofA0 can violate inside 0-implications by Lemma 2.3.8. The number of columns with no violations is at mostforb(m, K3) which is O(m2) by Theorem 2.1.3. Thus we have shown
n0 is O(m2) and son is also O(m2).
2.3.3 The boundary between O(m
k−1) and Θ(m
k)
Conjecture 2.2.3 allowed us to predict the following theorem. We need some notations. Let D12 denote the simple matrix of all columns of column sum at least 1 with noK22 on rows 1 and 2. Define
Fk(t) = [0|(t+ 1)·D12]. (2.40) Theorem 2.3.11 Let k ≥ 2 be given and let F be a k-rowed (0,1)-matrix.
Suppose that the largest column multiplicity in F is t+ 1. If F is a configu-ration in Fk(t) or Fk(t) or F is a configuration in FB(t) = [Kk|t·[Kk\B]], for some choice of B a k×(k+ 1) simple matrix with one column of each column sum, then forb(m, F) is O(mk−1). Otherwise, forb(m, F) is Θ(mk).
Proof of Theorem 2.3.11 Let us first consider the lower bound. Since F is not a configuration in any FB(t), F contains 2· Kk` for some choice
`∈ {0,1, . . . , k}.
For` =k, we can deduceforb(m,2·Kkk) is Ω(mk) using the construction Kmk. Similarly, for` = 0,forb(m,2·Kk0) is Ω(mk). For` 6= 0,1, k−1, kwe have forb(m2·Kk`) = Ω(mk) as follows. Consider the k-fold product A consisting of ` factors of Im/k and k−` factors of Im/k. If we form a submatrix of k rows of Awith 1 row from each of the factors then we cannot obtain ak×2 submatrix of two identical columns of`1’s and k−`0’s. So there is no copy of 2·Kk` on these rows. Nor can we find even Kk` on a submatrix involving two rows from a single factor, as` 6= 0,1, k−1, k implies that on every pair
of rows ofKk` there is both 1
1
and 0
0
, which are not configurations inIm/k and Im/k, respectively. Therefore 2·Kk` is not a configuration in A.
From what we have said, then, F contains 2·Kk1 or contains 2·Kkk−1, or forb(m, F) = Ω(mk). We consider the case where F contains 2·Kk1 = 2·Ik, as the case of 2·Kkk−1 is the (0,1)-complement. We may assume thatF does not contain 2·0= 2·Kk0. Therefore, asF is not a configuration inFk(t),F contains [2·Ik|G], where G is some matrix with
1 1
in every pair of rows.
Similarly to before the k-fold product A consisting of one factor Im/k and k−1 factors Im/k witnesses that forb(m, F) is Ω(mk). For G (and henceF) has
1 1
in every pair of rows, so that a submatrix of k rows in A taking two rows from the factorIm/k does not haveF as a configuration; while similarly Ik fork ≥3 (and hence F) has
0 0
in each pair of rows, so that a submatrix ofkrows inAtaking two rows from some factorIm/k has no configurationF. But when we take any submatrix with one row from each factor there is only one copy of the column with 1 in the factor Im/k and 0’s elsewhere. So there is no configuration 2·Ik in such a submatrix. Thus A has no configuration F. We include here a proof of theforb(m,[Kk|t·[Kk\B]]) = O(mk−1) result based on the paper [AFFS05], since it leads to a generalization of Lov´asz’
result on color critical hypergraphs. An induction proof of the same result can be found in [AF08], while the proof offorb(m, Fk(t)) = O(mk−1) is a very recent result of Anstee and Fleming [AF09] using the linear algebra method in a refined way.
Let A be an m ×n simple 0-1 matrix, and B be a k ×(k+ 1) matrix consisting of one column of each possible column sum. Suppose thatA does not have FB = [Kk|t·(Kk −B)] as configuration. This implies that on a givenk-tupleLof rows either Kk is missing, or if all possible columns of size k occur on L, thent·(Kk−B) must be missing. This latter means, that for some 0≤s ≤k, two columns of column sumsoccur at mostt−1 times onL, respectively. Let K be the set ofk-tuples of rows where the latter happens.
Using Lemma 2.3.12 a set of columns of size O(mk−1) can be removed from A to obtain A0, so that for all L ∈ K a column (in fact two) is missing on L in A0. However, this implies that Kk is not a configuration in A0, thus by Theorem 2.1.3 A0 has at most O(mk−1) columns.
Let K be a system of k-tuples of rows such that ∀K ∈ K there are two (k×1) columns, αK 6= βK specified. We say that a column x of A violates (K, αK), if x|K =αK, similarly, x violates (K, βK), ifx|K =βK.
Lemma 2.3.12 Assume, that for every K ∈ K there are at most t − 1 columns of A that violate (K, αK), and at most t−1 columns of A violate (K, βK). Then there exists a subset X of columns of A, such that |X| = O(mk−1) and no column of A−X violates any of (K, αK) or (K, βK).
Proof of Lemma 2.3.12 It can be assumed without loss of generality that for all K ∈ K αK = α and βK = β independent of K. Indeed, there are 2k×2kpossibleαK, βK pairs, that is a constant number of them .Thus,Kcan be partitioned into a constant number of parts, so that in each partαK =α and βK = β holds. We apply induction on k using the simplification given above. k = 1 is obvious.
Consider now k ×1 columns α 6= β. Assume first, that α 6= β. That is, there is a coordinate where α and β agree, say both have 1 as their `th coordinate. The case of a common 0 coordinate is similar. For theith row of Awe count how many columns have violation so that for someK ∈ Kthe`th coordinate inK is exactly rowi. LetKi,`be the set of thesek-tuples fromK.
Columns that have violation onk-tuples from Ki,` have 1 in the ith row, let Ai,1 denote matrix formed by the set of columns that have 1 in rowi. If rowi is removed fromAi,1, the remaining matrixA0i,1 is still simple. LetK0i,`denote the set of (k−1)-tuples obtained fromk-tuples of Ki,` by removing their`th coordinate, i, furthermore let α0 (β0, respectively) denote the (k − 1)× 1 column obtained from α (β) by removing the `th coordinate, 1. Note, that α0 6= β0. A column of A has a violation on K ∈ Ki,` iff its counterpart in A0i,1 has a violation on the corresponding K0 ∈ K0i,`. The number of those columns is at most c mk−2 by the inductive hypothesis. Since K=∪mi=1Ki,`, we obtain that the number of columns ofA having violation on someK ∈ K is at most m·c mk−2.
Let us assume now, thatα =β. A subset J ⊆ Kis called independent if there exists an orderingJ1, J2, . . . Jg of the elements of J such that for every Ji ∈ J there exists anm×1 0-1 column that violatesJi and does not violate any Jj ∈ J for j < i. Let us call a maximal independent subset B of K a basis ofK. If a column ofAhas a violation onK ∈ K, then it has a violation on some B ∈ B, as well. Indeed, either K ∈ B holds, or if K 6∈ B, then by the maximality of B, K cannot be added to it as a |B|+ 1st element in the order, so the column having violation onK must have a violation onB ∈ B, for some B. By (2.67) of Theorem 2.5.5 for a basisB we have
|B| ≤
m−1 k−1
+
m−1 k−2
+. . .+
m−1 0
, (2.41)
since a column violating ak-tupleBi fromB, but none ofBj forj < i, gives
an appropriate partition of the set of rows. Thus, there could be at most (2t−2) k−1m
columns violating some K ∈ K.