Problems and results on 1-cross intersecting set pair systems

The notion of cross intersecting set pair system of size $m$, $\Big(\{A_i\}_{i=1}^m, \{B_i\}_{i=1}^m\Big)$ with $A_i\cap B_i=\emptyset$ and $A_i\cap B_j\ne\emptyset$, was introduced by Bollob\'as and it became an important tool of extremal combinatorics. His classical result states that $m\le {a+b\choose a}$ if $|A_i|\le a$ and $|B_i|\le b$ for each $i$. Our central problem is to see how this bound changes with the additional condition $|A_i\cap B_j|=1$ for $i\ne j$. Such a system is called $1$-cross intersecting. We show that the maximum size of a $1$-cross intersecting set pair system is -- at least $5^{n/2}$ for $n$ even, $a=b=n$, -- equal to $\bigl(\lfloor\frac{n}{2}\rfloor+1\bigr)\bigl(\lceil\frac{n}{2}\rceil+1\bigr)$ if $a=2$ and $b=n\ge 4$, -- at most $|\cup_{i=1}^m A_i|$, -- asymptotically $n^2$ if $\{A_i\}$ is a linear hypergraph ($|A_i\cap A_j|\le 1$ for $i\ne j$), -- asymptotically ${1\over 2}n^2$ if $\{A_i\}$ and $\{B_i\}$ are both linear hypergraphs.

if |A i | ≤ a and |B i | ≤ b for each i.Our central problem is to see how this bound changes with the additional condition |A i ∩ B j | = 1 for i = j.Such a system is called 1-cross intersecting.We show that these systems are related to perfect graphs, clique partitions of graphs, and finite geometries.We prove that their maximum size is • asymptotically n 2 if {A i } is a linear hypergraph (|A i ∩ A j | ≤ 1 for i = j), • asymptotically 1 2 n 2 if {A i } and {B i } are both linear hypergraphs.
A cross intersecting set pair system of size m ≥ 2 consists of finite sets A 1 , . . ., A m and B 1 , . . ., B m such that We will consider further constrains but always keep these two basic properties.
Bollobás' theorem [4] states that must hold for any cross intersecting set pair system if we have |A i | ≤ a and |B i | ≤ b for each i.This size can be achieved by the standard example, taking all a-element sets of an (a+b)-element set for the A i -s and their complements as B i -s.Let A = {A i } m i=1 and B = {B i } m i=1 .The set pair system (SPS for short) is denoted by ( An SPS (A, B) is 1-cross intersecting if |A i ∩ B j | = 1 for each i = j.Our aim is to find good estimates for the size under this condition.This leads to interesting but seemingly difficult problems.
Our results are summarized in the next five subsections.In two warm-up sections we show that an 1-cross intersecting (n, n)-bounded SPS (A, B) can have exponential size and that its size is bounded by the sizes of the vertex sets of A (and B).We show how the latter provides an alternate ending of Gasparian's proof of Lovász's perfect graph theorem.The next two subsections present our main results: sharp bound of the size in the (2, n)-bounded case (Theorem 1.4) and asymptotically best bounds for the size in the (n, n)-bounded case when A, B are linear (Theorem 1.6) and when A, B are 1-intersecting (Theorem 1.7).Then we show the connection of 1-cross intersecting SPS-s with clique partition of graphs.
Although the main results of this article are about 1-intersecting families, we propose the problem in a very general setting in Section 2. The proof of the upper bounds are in Sections 3, 4. The constructions giving the lower bounds are in Section 5 and we conclude with some open problems.The proof of this, and most other proofs, are postponed to later sections.Starting from the standard example (with a = b = 1 and m = 2), Proposition 1.1 yields an (n, n)-bounded 1-cross intersecting SPS of size 2 n , exponential in n.Define the (2, 2)-bounded 1-cross intersecting SPS, called H(2, 2), using the edges of a fivecycle and its complement.The five pairs are ({i, i+1}, {i+2, i+4}) taken (mod 5).Then Proposition 1.1 gives the following.

1-cross intersecting SPS of exponential sizes
Corollary 1.2.There exists an (n, n)-bounded 1-cross intersecting SPS of size 5 n/2 if n is even and of size 2 • 5 (n−1)/2 if n is odd.This is the best lower bound we know.It remains a challenge to decrease essentially the upper bound 2n n in (1) for an (n, n)-bounded 1-cross intersecting SPS.Corollary 1.2 gives a (3, 3)-bounded 1-cross intersecting SPS of size 10, in fact two different ones, with 12 and with 15 vertices, depending on the order we apply Proposition 1.1.We have a third example, the pairs ({i, i+1, i+2}, {i+3, i+6, i+9}) taken (mod 10) has 10 vertices.Samuel Spiro (sspiro@ucsd.edu)informed us that his computer program successfully checked that 10 is indeed the largest size.

1-cross intersecting SPS and perfect graphs
One particular feature of a 1-cross intersecting SPS (A, B) is that its size is bounded by the sizes of the vertex sets of A (and B).This can be considered as a variant of Fischer's inequality, and does not hold for general SPS.
Proposition 1.3.Assume that (A, B) is 1-cross intersecting and V := ∪A.Then the characteristic vectors of the edges of A are linearly independent in R V .
A special case of Proposition 1.3 relates to perfect graphs and can be used in Gasparian's proof [11,6] of Lovász's characterization [18] of perfect graphs: a graph G is perfect if and only if holds for all induced subgraphs H of G.
To prove the nontrivial part, Gasparian showed that if a minimal imperfect graph G would satisfy (2) then there is a 1-cross intersecting SPS of size m = α(G)ω(G)+1 defined by independent sets and complete subgraphs of G.By Proposition 1.3,

(2, n)-bounded 1-cross intersecting SPS
Here we state the best bound for the size of (2, n)-bounded 1-cross intersecting SPS showing that the main term of the upper bound When A and B are both linear, and they form a 1-cross intersecting SPS then this bound can be approximately halved.
Theorem 1.6.Suppose that (A, B) is an (n, n)-bounded 1-cross intersecting SPS of size m such that both A and B are linear hypergraphs.Then m ≤ 1 2 n 2 + n + 1.A further small decrement comes if in addition A and B are both 1-intersecting hypergraphs.Then their union H = A ∪ B can be considered as a "geometry" where two lines intersect in at most one point, and every line has exactly one parallel line.
Theorem 1.7.Assume that (A, B) is an (n, n)-bounded 1-cross intersecting SPS of size m such that both A and B are 1-intersecting.Then m ≤ n 2 + 1 for n > 2. If n ≥ 4 and equality holds, then H is n-uniform and n-regular In Section 5 we give constructive lower bounds.Constructions 5.1, 5.2 and 5.3 show that the upper bounds in this subsection are asymptotically the best possible.

1-cross intersecting SPS and clique partitions of graphs
The notion of 1-cross intersecting SPS is closely related to the concept of clique and biclique partitions.A clique partition of a graph G is a partition of the edge set of G into complete graphs.Similarly, a biclique partition of a bipartite graph B is a partition of the edge set of B into complete bipartite graphs (bicliques).The minimum number of cliques (bicliques) needed for the clique (or biclique) partitions are well studied, see, for example [13].Our problem relates to another parameter of clique (biclique) partitions.The thickness of a clique (biclique) partition of a graph (bipartite graph) is the minimum s such that every vertex of the graph (bipartite graph) is in at most s cliques (bicliques).Let T 2m be the cocktail party graph, i.e., the complete graph K 2m from which a perfect matching is removed.Let B 2m be the bipartite graph obtained from the complete bipartite graph K m,m by removing a perfect matching.
Assume that (A, B) is an (n, n)-bounded 1-cross intersecting SPS of size m, and H = A ∪ B. The dual of this hypergraph, H * , has vertex set every pair x i , y j for i = j is covered exactly once by a hyperedge of H * .On the other hand, |A i ∩ B i | = 0 for every i so the pairs x i , y i are not covered by any hyperedge of H * .Thus the complete graphs induced by the hyperedges of H * form a biclique partition of thickness n of the bipartite graph B 2m .
If we have the additional assumption that A and B are both 1-intersecting then the pairs x i , x j and the pairs y i , y j are also covered exactly once by the hyperedges of H * .Thus in this case the complete graphs induced by the hyperedges of H * form a clique partition of thickness n of the cocktail party graph T 2m .
The above argument gives the following.
Theorem 1.8.The maximum m such that B 2m has a biclique partition of thickness n is equal to the maximum size of an (n, n)-bounded 1-cross intersecting SPS.The maximum m such that T 2m has a clique partition of thickness n is equal to the maximum size of an (n, n)-bounded 1-cross intersecting SPS in which A and B are also 1-intersecting.

Notation and general setting
Let a, b positive integers and I A , I B , I cross three sets of non-negative integers.We denote by m(a, b, I A , I B , I cross ) the maximum size m of a cross intersecting SPS (A, B) with the following conditions. i) To avoid trivialities we always suppose that 0 ∈ I cross , also that m ≥ 2. If a constraint in iv)-vi) is vacuous (i.e., either {0, 1, . . ., a} ⊆ I A or {0, 1, . . ., b} ⊆ I B or {1, . . ., min{a, b}} ⊆ I cross ) then we use the symbol * to indicate this.With this notation Bollobás' theorem [4]  Since in this paper the main results are about linear hypergraphs, we will have I A (and also I B ) is either {0, 1} (A is a linear hypergraph), or {1} (A is a 1-intersecting hypergraph), or * .Instead of writing I A = {1} we write '1-int', instead of I A = {0, 1} we write '01-int', and for I cross = {1} we use just '1' (as we did above).
Adding more restrictions can only decrease the maximum size, so we have In fact, we examined all 18 cases for m n (I A , I B , I cross ) where I A and I B are chosen from {1}, {0, 1}, or * and I cross is either {1} or * .By symmetry they define twelve functions.Summarizing our results, m n ( * , * , 1) and m n ( * , * , * ) are exponential as a function of n, the other cases are polynomial.Three of them, mentioned in (3), are asymptotically 1  2 n 2 while the other seven are asymptotically n 2 .Several problems under assumptions similar to 1-cross intersecting SPS have been studied before, see, e.g., [3,5,9,21] and more recently in [12,20].Proof.We have to show that Proposition 1.3.Assume that (A, B) is 1-cross intersecting and V := ∪A.Then the characteristic vectors of the edges of A are linearly independent in R V .
Proof.Let a i (resp.b i ) denote the characteristic vector of A i (resp.B i ), i.e. a i (v) = 1 for v ∈ V if and only if v ∈ A i .Otherwise the coordinates are 0. Suppose that Take the dot product of both sides of this equation with b j .Since Adding these for all j yields (m − 1) ( m i=1 λ i ) = 0. Consequently (using m > 1) m i=1 λ i = 0. Thus λ j = 0 for all j.Proof.The n-set B i must be an independent transversal for all edges other than A i (i.e., intersects all edges of A except A i but does not contain any edge of A) and disjoint from the edge A i .Suppose that the graph A contains an even cycle with edges A 1 = (x 1 , x 2 ), A 2 = (x 2 , x 3 ). . . .A 2k = (x 2k , x 1 ).Since B 1 is an independent transversal for all edges other than A 1 , we have x 3 ∈ B 1 which implies x 4 / ∈ B 1 , and so on, finally If there is an odd cycle C with k vertices, it cannot contain a diagonal, since any diagonal would create an even cycle, contradicting the previous paragraph.If there is an edge A i with exactly one vertex, say x 1 on C, then the argument of the previous paragraph implies x 2 ∈ B i , x 3 / ∈ B i , . . ., x 1 ∈ B i , contradiction.Also, if there is an edge A i with no vertex on C then B i must intersect all edges of C so it cannot be an independent transversal.Thus in this case m ≤ |C| ≤ 2n + 1.
Assume next that A is an acyclic graph.Lemma 3.2.Assume that T ⊆ A is a non-star tree component with t edges.Then Proof.Let P = x, y, z, z 2 , . . .be a maximal path of T , set A 1 = {x, y}, A 2 = {y, z}.
Let S ⊆ V (T ) the set of leaves connected to y.Note that t ≥ 3, |V (T )| = t + 1, N T (y) = S ∪ {z} and x ∈ S. Then B 1 ∩ V (T ) is the set X of vertices with odd distance from y in the tree T −x.On the other hand, B 2 ∩V (T ) is the set X ′ = S ∪D where D is the set of vertices with odd distance from z in the tree T − (S ∪ {y}).
Assume that there is a non-star tree component T in A with t edges, A 1 , . . ., A t , (t ≥ 3).We define another (2, n)-bounded 1-cross intersecting SPS (A ′ , B ′ ) of size m.Let A ′ be the graph defined by replacing T with S, where S is the union of two vertex disjoint stars S 1 and S 2 with centers s 1 , s 2 having t 2 and t 2 edges, respectively.We keep all edges of the other components of A, i.e., A ′ = (A \ E(T )) ∪ E(S).
For i = 1, . . ., t in case of A ′ i ∈ E(S α ) let C i be the complement of A ′ i in the star S α together with the center of the other star of S, i.e., 2 .Define B ′ as follows.
If i > t, we have where the inequality 2 ≤ |B i ∩ V (T )| holds because T is not a star.
Applying Claim 3.3 repeatedly, we may assume that all components of A are stars, S 1 , . . ., S k , where S i has t i ≥ 1 edges.For any edge A j ∈ S i , n ≥ |B j | = t i − 1 + k −1.Adding these inequalities for i = 1, . . ., k, we obtain that kn ≥ m−2k + k 2 which leads to k(n Taking together the bounds for odd cycles and acyclic graphs, we get that For n = 2, 3 the first term is larger, for n = 4 they are equal, and for n ≥ 5 the second term takes over.This proves the upper bound for m. The matching lower bound for n ≥ 4 comes from Proposition 1.1 applied to the standard construction with values (1, ⌈ n 2 ⌉) and (1, ⌊ n 2 ⌋).For n = 2 the hypergraph H(2, 2) works (defined in Subsection 1.1).For n = 3 we can define H(2, 3) as the pairs {i, i+1}, {i+2, i+4, i+6} taken (mod 7).

1-cross intersecting linear SPS -upper bounds
For v ∈ V we denote by d A (v), d B (v), d H (v) the degree of v in the hypergraphs A, B, H, respectively.Proposition 1.5.Suppose that (A, B) is an (n, n)-bounded cross intersecting SPS of size m such that A is a linear hypergraph.Then m ≤ n 2 + n + 1.
Proof.Our first observation here is the following.
Theorem 1.6.Suppose that (A, B) is an (n, n)-bounded 1-cross intersecting SPS of size m such that both A and B are linear hypergraphs.Then m ≤ 1 2 n 2 + n + 1.
Proof.Suppose that (A, B) is an (n, n)-bounded 1-cross intersecting SPS of size m such that both A and B are linear hypergraphs.We have m 2 (01-int, 01-int, 1) ≤ 5 by Theorem 1.4 so we may suppose that n ≥ 3.If m ≤ 2n + 2 then there is nothing to prove, so from now on, we may suppose that m ≥ 2n + 3. We claim that for every Let A i be the set of A j -s that intersect A i and different from A i .Our crucial observation is that if A i and A j do not intersect then (5) Indeed, the left hand side of ( 5) equals to ℓ:ℓ =i,j |A ℓ ∩ (A i ∪ A j )|.For two disjoint sets X, Y we say that a pair (x, y) joins X, Y if x ∈ X, y ∈ Y .For ℓ = i, j we have we select one pair (x, y) joining A i , A j , namely (x, y) = B ℓ ∩ (A i ∪ A j ).These pairs are distinct because Since there are n 2 pairs between A i and A j we obtain that ℓ:ℓ =i,j |A ℓ ∩ (A i ∪ A j )| ≤ n 2 , completing the proof of ( 5).If A i ∩ A j = {v} then we will prove that Indeed, as before, For every ℓ = i, j we select (at most) two pairs joining In this way we selected at least In the latter case we still have selected at least |A ℓ ∩ A i | + |A ℓ ∩ A j | − 1 pairs.So the left hand side of ( 6) is at most the number of pairs joining . This completes the proof of ( 6).
Next we prove that Add up inequalities ( 5) and ( 6) for all 1 ≤ i < j ≤ m Here the left hand side is The last two displayed formulas yield (7) and equality can hold only if (5) was not used.Note that similar upper bound must hold for v∈V d B (v) 2 , too.Apply (7) to A and to B and subtract the double of (4).We obtain As a last step we show that this inequality is strict completing the proof of the upper bound on m.Indeed, equality can hold only if (5) was never used to A neither to B. This implies that A and B are 1-intersecting and because of (6) there exists a v with d This contradicts the 1-intersection property.
Theorem 1.7.Assume that (A, B) is an (n, n)-bounded 1-cross intersecting SPS of size m such that both A and B are 1-intersecting.Then m ≤ n 2 + 1 for n > 2. If n ≥ 4 and equality holds, then H is n-uniform and n-regular Proof.Recall that H = A ∪ B. First, consider the case when there exists a vertex Then A n+2 cannot intersect all members of {A i , B i } 1≤i≤n+1 containing v, a contradiction.So in this case m = n + 1 and we are done.
From now on, we may suppose that m > n + 1, and and we conclude that m ≤ n 2 + 1.If n ≥ 4 and equality holds, then all vertices of A 1 (and of all other hyperedges) must have degree n.

Constructing cross intersecting linear hypergraphs
Here we give constructions of large cross intersecting SPS-s such that A is an intersecting linear hypergraph.Constructions 5.1 and 5.2 show that Since the right hand sides of these inequalities are bounded above by m n (01-int, * , * ) (which is at most n 2 + n + 1), Proposition 1.5 is asymptotically the best possible.Construction 5.3 shows that 1 2 Hence Theorems 1.6 and 1.7 are also asymptotically the best possible.
We use that the function m n (I A , I B , I cross ) is monotone increasing in n so we have to make constructions only for a dense set of special values of n.
Beyond Bertrand's postulate (for each real x > 1 there always exists a prime p with x < p < 2x) we need Hoheisel's theorem [14] about the density of primes: There are constants x 0 and 0.5 ≤ α < 1 such that for all x ≥ x 0 the interval The currently known best α is 0.525 by Baker, Harman and Pintz [2].

Building blocks: double stars and affine planes
The vertex set of a double star of size s consists of {v i,j | 1 ≤ i, j ≤ s, i = j} and two additional special vertices w a and w b .Define for i = 1, . . ., s sets A i = {w a } ∪ {v i,j | 1 ≤ j ≤ s, j = i} and B i = {w b } ∪ {v j,i | 1 ≤ j ≤ s, j = i}.Then (A, B) is a 1-cross intersecting SPS of size s containing s-element sets such that both A and B are 1-intersecting.The double star shows that m n (1-int, 1-int, 1) ≥ n for all n (consequently, m n (1-int, 1-int, * ) ≥ n and m n (1-int, * , 1) ≥ n).
The affine plane AG(2, q) = (P, L) is a q-uniform hypergraph with a q 2 element vertex set P , such that each edge L ∈ L (called line) has q vertices (points), and L can be split into q + 1 parts L = L 1 ∪ L 2 ∪ • • • ∪ L q+1 (directions or parallel classes of lines) such that each parallel class contains q lines, L δ = {L 1,δ , . . ., L q,δ }, the members of a parallel class are pairwise disjoint, but two lines from distinct classes always meet in a single point.It is known that an AG(2, q) exists if q is prime.
In the next subsection we give three different (but similar) constructions to prove the lower bounds ( 8)- (10).Each construction will use an associated Extension twice, where an Extension starts with a weaker construction of the same type and combine it with AG(2, q) for getting a stronger construction.In the following p and q will always denote odd primes.

Extensions of the affine plane
Extension I. Let (A ′ , B ′ ) be a cross-intersecting SPS of size at least q.For each 1 ≤ δ ≤ q + 1 take a new copy of (A ′ , B ′ ) so that the ground sets of the q + 1 copies are pairwise disjoint and also disjoint from AG(2, q).For i = 1, . . ., q let (A ′ i,δ , B ′ i,δ ) be the disjoint pairs in the δth copy.
Next we prove ( 9) and (10).The proofs are rather similar to the one presented above, so we leave out most of the details.
Note that C 2 (q, B ′ ) is not linear.

Conjectures, open problems
We conjectured [10] that there exists a positive ε such that m n ( * , * , 1) ≤ (1 − ε) 2n n for every n ≥ 2. This was proved by Holzman [15] in the following stronger form.If a, b ≥ 2, then m(a, b, 1) ≤ (29/30) a+b a .More recently Kostochka, McCourt, and Nahvi [17] showed that the factor 29/30 in this bound can be replaced by 5/6, which is the best possible since m(2, 2, 1) = 5. we strongly believe that the following is also true.We obtained some tight results for m(a, b, I A , I B , I cross ) in the case a = b and also in the case a = 2.There is plenty of room for further investigations.

1 .
states m(a, b, * , * , * ) = a + b a , and our Theorem 1.4 states (for n ≥ 4) m(2, n, * , * , In the rest of the results we deal with the case a = b = n and use the abbreviation of placing n as an index m n (I A , I B , I cross ) := m(n, n, I A , I B , I cross ).

Theorem 1 . 4 . 1 .Lemma 3 . 1 .
Let n ≥ 4, and let (A, B) be a (2, n)-bounded 1-cross intersecting SPS of size m.Then m ≤ This bound is the best possible.For n = 2, 3 the exact values are m = 5, 7. Proof.Let (A, B) be a (2, n)-bounded 1-cross intersecting SPS of size m.It is convenient to assume that A is two-uniform (a graph without multiple edges) and B is an n-uniform hypergraph.(For smaller sets dummy vertices can be added).Consider the simple graph A. If A contains a cycle then m ≤ 2n + 1.