Idealness of k-wise intersecting families

A clutter is k-wise intersecting if every k members have a common element, yet no element belongs to all members. We conjecture that, for some integer \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$k\ge 4$$\end{document}k≥4, every k-wise intersecting clutter is non-ideal. As evidence for our conjecture, we prove it for \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$k=4$$\end{document}k=4 for the class of binary clutters. Two key ingredients for our proof are Jaeger’s 8-flow theorem for graphs, and Seymour’s characterization of the binary matroids with the sums of circuits property. As further evidence for our conjecture, we also note that it follows from an unpublished conjecture of Seymour from 1975. We also discuss connections to the chromatic number of a clutter, projective geometries over the two-element field, uniform cycle covers in graphs, and quarter-integral packings of value two in ideal clutters.


Introduction
Let V be a finite set of elements, and C be a family of subsets of V called members.The family C is a clutter over ground set V , if no member contains another one [11].A cover of C is a subset B ⊆ V such that B ∩ C ≠ ∅ for all C ∈ C. A cover is minimal if it does not contain another cover.The family of minimal covers forms another clutter over the ground set V , called the blocker of C and denoted b(C).It is well-known that b(b(C)) = C [15,11].Consider for w ∈ Z V + the dual pair of linear programs where the left and right LPs are denoted (P ) and (D), respectively.If the dual (D) has an integral optimal solution for every right-hand-side vector w ∈ Z V + , then C is said to have the max-flow min-cut (MFMC) property [7].By the theory of totally dual integral linear systems, for every MFMC clutter, the primal (P ) also admits an integral optimal solution for every cost vector w ∈ Z V + [12].This is why the class of MFMC clutters is a natural host to many beautiful min-max theorems in Combinatorial Optimization [8].Let us elaborate.
The packing number of C, denoted ν(C), is the maximum number of pairwise disjoint members.Note that ν(C) is equal to the maximum value of an integral feasible solution to (D) for w = 1.Furthermore, the covers correspond precisely to the 0 − 1 feasible solutions to (P ).The covering number of C, denoted τ (C), is the minimum cardinality of a cover.Notice that τ (C) is equal to the minimum value of an integral feasible solution to (P ) for w = 1.Also, by Weak LP Duality, τ (C) ≥ ν(C).The clutter C packs if τ (C) = ν(C) [9].Observe that if a clutter is MFMC, then it packs.
If the primal (P ) has an integral optimal solution for every cost vector w ∈ Z V + , then C is said to be ideal [10].Ideal clutters form a rich class of clutters, one that contains the class of MFMC clutters, as discussed above.This containment is strict, and in fact, the richest examples of ideal clutters are those that are not MFMC [14,19].Furthermore, unlike MFMC clutters, if a clutter is ideal, then so is its blocker [13,18,8].
A clutter is intersecting if every two members intersect yet no element belongs to every member [2].That is, a clutter C is intersecting if τ (C) ≥ 2 and ν(C) = 1.In particular, an intersecting clutter does not pack, and therefore is not MFMC.Intersecting clutters, however, may be ideal.For instance, the clutter whose elements are the edges and whose members are the triangles of K 4 , is an intersecting clutter that is ideal [23].In fact, Q 6 is the smallest intersecting clutter which is not ideal ([1], Proposition 1.2).It is worth pointing out that if a clutter and its blocker are both intersecting, then the clutter must be nonideal [3].
In this paper, we propose a sufficient condition for non-idealness that is purely combinatorial.We say that C is k-wise intersecting if every k members have a common element, yet no element belongs to all members.Note that for k = 2, this notion coincides with the notion of intersecting clutters.Furthermore, for k ≥ 3, a k-wise intersecting clutter is also (k − 1)-wise intersecting.The following is our main conjecture.Conjecture 1.There exists an integer k ≥ 4 such that every k-wise intersecting clutter is non-ideal.
A clutter is binary if the symmetric difference of any odd number of members contains a member [17].Equivalently, a clutter is binary if every member and every minimal cover have an odd number of elements in common [17].In particular, if a clutter is binary, then so is the blocker.Many rich classes of clutters are in fact binary [8].For example, given a graph G = (V, E) and distinct vertices s and t, the clutter of st-paths over ground set E is binary.The clutter Q 6 is also binary.As evidence for Conjecture 1, our main result is that it holds for all binary clutters.Theorem 2. Every 4-wise intersecting binary clutter is non-ideal.
We also show that 4 cannot be replaced by 3 in Conjecture 1, even for binary clutters.
The example from Proposition 3 comes from the Peterson graph, and also coincides with the clutter T 30 from [21], §79.3e.It has 30 elements and is the smallest such example that we are aware of.
Finally, as further evidence for Conjecture 1, we also show that it follows from an unpublished conjecture by Seymour from 1975 that was documented in [21], §79.3e.

Paper Outline
In Section 2, we show that a special class of clutters, called cuboids [5,1], sit at the heart of Conjecture 1. Cuboids allow us to reformulate Conjecture 1 in terms of set systems.
In Section 3, we prove Proposition 3 and Theorem 2. Two key ingredients of our proof of Theorem 2 are Jaeger's 8-flow Theorem [16] for graphs, and Seymour's characterization of the binary matroids with the sums of circuits property [25].
In Section 4, we propose a line of attack for tackling Conjecture 1, inspired by the recent work of [4].We also discuss two applications of Theorem 2 to ideal binary clutters.Each application goes hand-in-hand with two strengthenings of Conjecture 1.One strengthening is proposed by us, which we believe is the right strategy for tackling Conjecture 1.The other conjecture is the unpublished conjecture by Seymour.
We now describe what it means for cuboid(S) to be k-wise intersecting.We say that the points in S agree on a coordinate if S ⊆ {x ∶ x i = a} for some coordinate i ∈ [n] and some a ∈ {0, 1}.Remark 4. Let S ⊆ {0, 1} n .Then cuboid(S) is a k-wise intersecting clutter if, and only if, the points in S do not agree on a coordinate yet every k points do.
Next, we describe what it means for cuboid(S) to be ideal.Let conv(S) denote the convex hull of S.An inequality of the form ∑ i∈I x i + ∑ j∈J (1 − x j ) ≥ 1, for some disjoint I, J ⊆ [n], is called a generalized set covering inequality [8].The set S is cube-ideal if every facet of conv(S) is defined by x i ≥ 0, x i ≤ 1, or a generalized set covering inequality [1].
As a result, Conjecture 1 for cuboids reduces to the following conjecture: Conjecture 6.There is a constant k ≥ 4 such that for every cube-ideal set, either all the points agree on a coordinate, or there is a subset of at most k points that do not agree on a coordinate.Surprisingly, we now show that Conjecture 6 is equivalent to Conjecture 1! Let C be a clutter over ground set V .To duplicate an element u of C is to introduce a new element ū, and replace C by the clutter over ground set V ∪ {ū}, whose members are A duplication of C is a clutter obtained from C by repeatedly duplicating elements.It is easily checked that a clutter is ideal if and only if some duplication of it is ideal.Moreover, a clutter is k-wise intersecting if and only if some duplication of it is k-wise intersecting.
Let I and J be disjoint subsets of V .The minor C ∖ I J obtained after deleting I and contracting J is the clutter over ground set V − (I ∪ J) whose members are the minimal sets in If every k ≥ 2 members of a clutter have a common element, then so do every k members of a deletion minor.Furthermore, for every element and only if v belongs to some minimum cover of C. Motivated by these observations, we say that a clutter C is tangled if τ (C) = 2 and every element belongs to a minimum cover.
We require the following facts about tangled clutters.
Proposition 7. Let C be a binary tangled clutter.Then C is a duplication of a cuboid.
Proof.If {e, f } is a minimum cover, then for each C ∈ C, C ∩ {e, f } must be odd and therefore 1, since C is a binary clutter.Since C is tangled, this implies Moreover, if a clutter is ideal, then so is every minor of it [23].Thus, it suffices to prove Conjecture 1 for tangled clutters.Theorem 9 ([4], Theorem 5.5).Let C be an ideal tangled clutter.Then is also an ideal tangled clutter.Moreover, C ′ is a duplication of some cuboid.
We are now ready to prove that Conjectures 6 and 1 are equivalent.
Theorem 10.Conjecture 6 is true for k if, and only if, Conjecture 1 is true for k.
Proof.We already showed (⇐).It remains to prove (⇒).Suppose Conjecture 1 is false for some k ≥ 4. That is, there is an ideal k-wise intersecting clutter C. Let C ′ be a deletion minor of C that is minimal subject to τ (C ′ ) ≥ 2. By Remark 8, C ′ is an ideal tangled k-wise intersecting clutter.Let By Theorem 9, C ′′ is an ideal tangled clutter that is a duplication of some cuboid, say cuboid(S).As every k members of C ′ have a common element, so do every k members of C ′′ , so the latter is k-wise intersecting.As a result, cuboid(S) is an ideal k-wise intersecting clutter, so by Remark 4 and Theorem 5, S is a cube-ideal set whose points do not agree on a coordinate yet every k points do.Therefore, S refutes Conjecture 6 for k, as required.⊓ ⊔

Proof of Main Theorem
Let S ⊆ {0, 1} n .For x, y ∈ {0, 1} n , x △ y denotes the coordinate-wise sum of x, y modulo 2. We say that S is a vector space over GF (2), or simply a binary space, if a △ b ∈ S for all a, b ∈ S. Notice that a nonempty binary space necessarily contains 0.

The 8-flow Theorem
Let G = (V, E) be a graph where loops and parallel edges are allowed, where every loop is treated as an edge not incident to any vertex.A cycle is a subset C ⊆ E such that every vertex is incident with an even number of edges in C. A bridge of G is an edge e that does not belong to any cycle.The cycle space of G is the set where χ C denotes the incidence vector of C. As ∅ is a cycle, and the symmetric difference of any two cycles is also a cycle, it follows that cycle(G) is a binary space.We require the following two results on cycle spaces of graphs.
Remark 12. Let G = (V, E) be a graph.Then the points in cycle(G) agree on a coordinate if, and only if, G has a bridge.Moreover, for all k ∈ N, cycle(G) has a subset of at most k + 1 points that do not agree on a coordinate if, and only if, G has at most k cycles the union of which is E.
We need the following version of the celebrated 8-Flow Theorem of Jaeger [16].
Theorem 14 ([16]).Every bridgeless graph G = (V, E) contains at most 3 cycles the union of which is E. That is, given the set cycle(G) ⊆ {0, 1} E , either all the points agree on a coordinate, or there is a subset of at most 4 points that do not agree on a coordinate.
One may wonder whether the 3, 4 in Theorem 14 may be replaced by 2, 3? The answer is no, due to the Petersen graph (see Figure 1a): Remark 15 (see [26]).The edge set of the Petersen graph is not the union of 2 cycles.
As a consequence, an ideal 3-wise intersecting clutter does exist: Proof of Proposition 3. Let S be the cycle space of the Petersen graph, and let C ∶= cuboid(S).By Remarks 12 and 15, the points in S do not agree on a coordinate yet every 3 points do.Moreover, S is a cube-ideal set by Theorem 13.Therefore, by Remark 4 and Theorem 5, C is an ideal 3-wise intersecting clutter, as required.

⊓ ⊔
The cuboid of the cycle space of the Petersen graph has already shown up in the literature, and is denoted T 30 by Schrijver [21], §79.3e.Consider the graph obtained from Petersen by subdividing every edge once, and let T be any vertex subset of even cardinality containing all the new vertices.Then the clutter of minimal T -joins of this graft is precisely T 30 .This construction is due to Seymour ( [24], page 440).

Sums of Circuits Property
For basics on matroids, we refer the reader to Oxley [20].Let E be a finite set, S ⊆ {0, 1} E a binary space, and S ⊥ the orthogonal complement of S, that is, S ⊥ = {y ∈ {0, 1} E ∶ y ⊺ x ≡ 0 (mod 2) ∀x ∈ S}.Notice that S ⊥ is another binary space, and that (S ⊥ ) ⊥ = S. Therefore, there exists a 0−1 matrix A whose columns are labeled by E such that S = x ∈ {0, 1} E ∶ Ax ≡ 0 (mod 2) , and S ⊥ is the row space of A generated over GF (2).
Let S ∶= {C ⊆ E ∶ χ C ∈ S}.The pair M ∶= (E, S) is a binary matroid, and the matrix A is a representation of M .We call E the ground set of M .The sets in S are the cycles of M , and S is the cycle space of M , denoted by cycle(M ).The minimal nonempty sets in S are the circuits of M , and the circuits of cardinality one are loops.
Let S ⊥ ∶= {D ⊆ E ∶ χ D ∈ S ⊥ }.The binary matroid M ⋆ ∶= (E, S ⊥ ) is the dual of M .Notice that (M ⋆ ) ⋆ = M .The sets in S ⊥ are the cocycles of M , and S ⊥ is the cocycle space of M , denoted by cocycle(M ).The minimal nonempty sets in S ⊥ are the cocircuits of M , and the cocircuits of cardinality one are coloops of M .
Remark 16.Let M be a binary matroid.Then the points in cycle(M ) agree on a coordinate if, and only if, M has a coloop.Moreover, for every integer k ≥ 1, cycle(M ) has a subset of at most k + 1 points that do not agree on a coordinate if, and only if, M has at most k cycles the union of which is E.
Let G = (V, E) be a graph.The binary matroid whose cycle space is cycle(G) is a graphic matroid, and is denoted M (G).Notice the one-to-one correspondence between the cycles of M (G) and the cycles of G, between the loops of M (G) and the loops of G, between the cocycles of M (G) and the cuts of G, and between the coloops of M (G) and the bridges of G. Therefore, Remark 16 is an extension of Remark 12.
A binary matroid has the sums of circuits property if its cycle space is a cubeideal set.This notion is due to Seymour [24]. 5By Theorem 13, graphic matroids have the sums of circuits property [24].Seymour also proved a decomposition theorem [25] for binary matroids with the sums of circuits property.It turns out they can all be produced from graphic matroids and two other matroids, which we now describe.The Fano matroid F 7 is the binary matroid represented by the matrix in Figure 1b.The second matroid is M (V 8 ) ⋆ , where V 8 is graph in Figure 1c.Seymour [25] showed that F 7 and M (V 8 ) ⋆ both have the sums of circuits property.
To generate all binary matroids with the sums of circuits property, we require three composition rules.Let M 1 , M 2 be binary matroids over ground sets E 1 , E 2 , respectively.We denote by M 1 △M 2 the binary matroid over ground set E 1 △E 2 whose cycles are all subsets of , where e is neither a loop nor a coloop of M 1 or M 2 ; and Theorem 17 ([25], (6.4), (6.7), (6.10) and (16.4)).Let M be a binary matroid with the sums of circuits property.Then M is obtained recursively by means of 1-sums, 2-sums and Y -sums starting from copies of F 7 , M (V 8 ) ⋆ and graphic matroids.
We are ready to prove Conjecture 6 for cube-ideal binary spaces.
Theorem 18.Every binary matroid without a coloop and with the sums of circuits property has at most 3 cycles the union of which is the ground set.
Proof.A 3-cycle cover of a clutter is three (not necessarily distinct) cycles whose union is the ground set.
Subproof.Given the matrix representation of F 7 in Figure 1a, label the columns 1, . . ., 7 from left to right.Then ∅, {1, 2, 3, 7}, {4, 5, 6} is a 3-cycle cover of F 7 .Next, label the vertices of V 8 so that outer 8-cycle is labelled 1 . . .8. Then M (V 8 ) ⋆ has a 3-cycle cover given by the following cuts of V 8 : δ({1, 6, 7, 8}), δ({1, 7}), δ({2, 4}), where δ(X) is the set of edges with exactly one end in X.◇ has a 3-cycle cover.Then the following statements hold: Subproof.For i ∈ [2], let E i be the ground set of M i and C i 1 , C i 2 , C i 3 be a 3-cycle cover of M i .Clearly, (i) holds.For (ii), let E 1 ∩ E 2 = {e}.We may assume e ∈ C i 1 for all i ∈ [2].By replacing C i 2 by C i 1 △ C i 2 if necessary, we may assume e ∉ C i 2 for all i ∈ [2].Similarly, we may assume e ∉ C i 3 for i ∈ [2].But now for all i, j.Therefore, after possibly relabeling e, f, g simultaneously in M 1 and M 2 , and after possibly relabeling C i 1 , C i 2 , C i 3 for all i, we may assume that We leave the proof of the following claim as an easy exercise for the reader.
The proof is complete by combining the above claims with Theorems 14 and 17.

Proof of Main Theorem
Proof of Theorem 2. Let C be an ideal binary clutter such that τ (C) ≥ 2. Let C ′ be a deletion minor of C that is minimal subject to τ (C ′ ) ≥ 2. Since C is binary, so is C ′ [22].By Remark 4, C ′ is an ideal binary tangled clutter.Since a clutter is binary if and only if some duplication of it is binary, C ′ is a duplication of a cuboid containing 0, say cuboid(S), by Proposition 7. Note that cuboid(S) is an ideal binary cuboid with τ (cuboid(S)) ≥ 2. So, by Theorem 5 and Remark 11, S is a cube-ideal binary space whose points do not agree on a coordinate.By Theorem 18, S has ≤ 4 points that do not agree on a coordinate, thereby yielding ≤ 4 members of cuboid(S) without a common element.This gives ≤ 4 members of C ′ , and therefore of C, without a common element, as required.
4 Applications and two more conjectures

Embedding projective geometries
In this section, we propose a strengthening of Conjecture 1.We begin by motivating our strengthening.Conjecture 1 predicts that for some k ≥ 4, every ideal clutter with covering number at least two has k members without a common element.By moving to a deletion minor, if necessary, we may assume that our ideal clutter is tangled.Our stronger conjecture predicts that the clutter must actually have 2 k−1 members that "correspond to a projective geometry", and of these members, k many will not have a common element.Let A be the (k − 1) × (2 k−1 − 1) matrix whose columns are all the nonzero vectors in {0, 1} k−1 .The binary matroid represented by A is called a projective geometry over GF (2), and is denoted P G(k − 2, 2).Let r ∶= 2 k−1 − 1. Recall that cocycle(P G(k − 2, 2)) ⊆ {0, 1} r is the row space of A generated over GF (2).Note that the k − 1 points in cocycle(P G(k − 2, 2)) corresponding to the rows of A agree on precisely one coordinate, which is set to 1.These k − 1 points, together with the zero point 0, yield k points that do not agree on a coordinate.This yields the following remark.
Remark 19.There are k points of cocycle(P G(k − 2, 2)) that do not agree on a coordinate.In particular, cuboid(cocycle(P G(k − 2, 2))) has k members without a common element.
Let C be an ideal tangled clutter.We say that C embeds the projective geometry This notion is due to [4].We propose the following conjecture.Proof of Proposition 21.Assume Conjecture 20 holds for k.Let C be an ideal clutter with τ (C) ≥ 2. Let C ′ be a deletion minor of C that is minimal subject to τ (C ′ ) ≥ 2. Then C ′ is an ideal tangled clutter.Thus, C ′ embeds P G(n − 2, 2) for some n ∈ {2, . . ., k}.That is, a duplication of cuboid(P G(n − 2, 2)) is a subset of C ′ .By Remark 19, cuboid(P G(n − 2, 2)) has n members without a common element, so the duplication, and therefore C ′ , must have n members without a common element.Thus, C has n ≤ k members without a common element, as required.
As an application of our main theorem, we now prove Conjecture 20 for k = 3 for the class of binary clutters.
Proof.Let C be a binary tangled clutter.By Proposition 7, C is a duplication of a cuboid, say cuboid(S) for some S containing 0. It suffices to show that cuboid(S) embeds one of the three projective geometries.Note that cuboid(S) is an ideal binary cuboid with τ (cuboid(S)) ≥ 2. By Theorem 5 and Remark 11, S is a cube-ideal binary space whose points do not agree on a coordinate.By Theorem 18, S has a subset of at most 3 points that do not agree on a coordinate.Let S ′ be the binary space generated by these points.Note that S ′ ⊆ S, the points in S ′ do not agree on a coordinate, and S ′ has GF (2)-rank at most 3.By Proposition 22, cuboid(S ′ ), and therefore cuboid(S), embeds one of P G(0, 2), P G(1, 2), P G(2, 2), as desired.

⊓ ⊔
We now give an application of Theorem 23.Let G be a bridgeless graph.By applying Theorem 23 to cuboid(cycle(G)), G has 8 cycles where every edge is used in exactly 4 of the cycles.Since one of the 8 cycles may be assumed to be ∅, G has 7 cycles such that each edge is in exactly 4 of the cycles.This is Proposition 6 of [6].

Dyadic fractional packings
We finish by deriving another consequence of Theorem 2. Let C be a clutter over ground set V .A fractional packing of C is a vector y ∈ R C + such that ∑ (y C ∶ C ∈ C, v ∈ C) ≤ 1 for all v ∈ V .The value of y is 1 ⊺ y.For n ∈ N, the vector y is 1 n -integral if every entry is 1 n -integral.
This, combined with Theorem 23, implies the following: Theorem 25.Every ideal binary clutter C with τ (C) ≥ 2 has a 1 4 -integral packing of value 2.
Proof.Let C ′ be a deletion minor of C that is minimal subject to τ (C ′ ) ≥ 2. Then C ′ is an ideal binary tangled clutter, so by Theorem 23, C ′ embeds one of P G(0, 2), P G(1, 2), P G(2, 2).By Proposition 24, C ′ , and therefore C, has a

Conjecture 20 .
There exists an integer k ≥ 4 such that every ideal tangled clutter embeds one of P G(0, 2), . . ., P G(k − 2, 2).We now show that Conjecture 20 is indeed a strengthening of Conjecture 1. Proposition 21.If Conjecture 20 holds for k, then Conjecture 1 holds for k.

1 4 - 1 4 1 4, 2 4 , 3 4 , 1
integral packing of value 2, as required.⊓⊔In fact, Seymour conjectures a far-reaching generalization of this theorem:Conjecture 26(Seymour 1975, see[21], §79.3e).Every ideal clutter C has a -integral packing of value τ (C).This conjecture is open even for binary clutters, and in particular, for the clutter of minimal T -joins of a graft ([8], Conjecture 2.15).Proposition 27.If Conjecture 26 is true, then so is Conjecture 1 for k = 5.Proof.Assume Conjecture 26 is true.Let C be an ideal clutter with τ (C) ≥ 2.Let C ′ be a deletion minor of C with τ (C ′ ) = 2. Since C ′ is also ideal, it has a -integral packing y ∈ R C ′ + of value 2. Notice that y C ∈ 0, 1 4 for each C ∈ C ′ .In particular, {C ∶ y C > 0} ≤ 8. Pick a minimal subset C ′′ ⊆ {C ∶ y C > 0} such that ∑ C∈C ′′ y C > 1.Then C ′′ ≤ 5, and it is easily checked that the members of C ′′ cannot have a common element.As a result, C ′ , and therefore C, has a subset of at most 5 members without a common element, as required.⊓⊔