Arc connectivity and submodular flows in digraphs

,


Introduction
Graph Orientation is a rich area of Graph Theory.The basic problem consists in orienting the edges of an undirected graph in order to obtain a k-arc-connected digraph, and giving conditions under which such an orientation exists.Various constraints on the orientation can be imposed, leading to an extensive literature in the area; for examples, see [6], Chapter 9 of [8], and Chapter 61 of [21].One can set up the basic problem equivalently in terms of digraphs, which is more appropriate for this paper: start from a digraph and flip the orientation of some of the arcs in order to obtain desired connectivity properties.
Let D = (V, A) be a digraph, fixed throughout the rest of the introduction, unless stated otherwise.
For U ⊆ V denote by δ + D (U ) and δ − D (U ) the sets of arcs leaving and entering U , respectively.We shall drop the subscript D whenever it is clear from the context.For J ⊆ A and U ⊆ V , denote An important result is Nash-Williams' weak orientation theorem, stating that there exists a k-arcconnected flip if, and only if, the underlying undirected graph of D is 2k-edge-connected ( [18], also see [6,13]).Our main theorem strengthens (the nontrivial direction of) the weak orientation theorem in two ways.To state it we need to borrow a few notions from Submodular Optimization.
Let C be a family of subsets of V .Then C is a crossing family over ground set V if, for all Let C be a crossing family over ground set V , and let f : C → R be a crossing submodular function.
The linear system y(δ + (U )) − y(δ − (U )) ≤ f (U ) ∀U ∈ C is called a submodular flow system, and every feasible solution is called a submodular flow.Observe that for the crossing submodular function We shall prove Theorem 2 in §3.We shall discuss the complexity aspects following the theorem in §5.As we will see, the inequalities on the cuts of D above, can be verified in strongly polynomial time.Moreover, under some standard conditions on how C and f are provided, J can also be found in oracle strongly polynomial time.
Let us say a few more words about the inequalities on the cuts of D in Theorem 2. They are imposed for the simple reason that they are needed for our proof to work.That said, they possess some nice properties.
First, as a sanity check, we note that the inequalities readily imply that the underlying undirected graph is 2k-edge-connected: for every U ⊊ V, U ̸ = ∅, by adding the two inequalities corresponding to Secondly, we note that the inequalities are equivalent to asking that ȳ = k τ 1 satisfies y(δ As k increases, the inequalities become more strict.For k = 1 the inequalities ask precisely that every dicut (a cut with all arcs crossing in the same direction) has size at least τ (see Theorem 15), while for k = ⌊τ /2⌋ the inequalities almost ask that every cut has size at least τ (see Theorem 16).
In §4 we shall see applications of Theorem 2 to Graph Orientations and Combinatorial Optimization.For instance, we see that for τ = 2k this result strengthens the weak orientation theorem and its near-Eulerian sharpening, and for k = 1 it reduces to a recent result on decomposing A into a dijoin and a (τ − 1)-dijoin [1].Other applications of Theorem 2 include an extension of the weak orientation theorem in a different direction than above, a weaker version of Woodall's conjecture for digraphs with a τ -edge-connected underlying undirected graph [23], and a theorem on disjoint dijoins in 0, 1-weighted digraphs.
Main result II.Theorem 2 is proved by utilizing a result on submodular flows.To explain it, we need a few notions from Integer Programming.Let A ∈ Q m×n and b ∈ Q m .The linear system Ax ≤ b is totally dual integral (TDI) if for each w ∈ Z n , the dual of the linear program max{w ⊤ x : Ax ≤ b} has an integral optimal solution whenever the LP admits an optimum [5].The system Ax ≤ b is box-TDI if the system Ax ≤ b, ℓ ≤ x ≤ u is TDI, for all ℓ, u ∈ Z n such that ℓ ≤ u.An important result is that if Ax ≤ b is TDI and b ∈ Z m , then {x : Ax ≤ b} is an integral polyhedron, that is, every non-empty face of it contains an integral point [12,5].In particular, if Ax ≤ b is box-TDI and b ∈ Z m , then is an integral polyhedron for all ℓ, u ∈ Z n such that ℓ ≤ u.
A classic result of Edmonds and Giles states that a submodular flow system is box-TDI [5].This important theorem laid the basis for numerous min-max theorems and polynomial and strongly polynomial algorithms for submodular flows.For in-depth surveys see [20,9], and for a more recent treatment we recommend Chapter 60 of [21] and Chapter 16 of [8].
In contrast, given two crossing submodular functions f i : C i → Z, i = 1, 2 defined over (possibly different) crossing families C i , i = 1, 2 over the same ground set, the combined system y(δ , is not box-TDI and not even integral, as Example 4 at the end of this section shows.Furthermore, finding an integral solution to the system includes NP-complete problems; see §A in the appendix for details.
Against this backdrop, Theorem 2 is significant since it provides a 0, 1 solution to the intersection of two submodular flow systems.The theorem is a consequence of the following result, which provides a sufficient condition for the existence of capacitated integral solutions to the intersection of two submodular flow systems.
Theorem 3. Let D = (V, A) be a digraph.For i = 1, 2, let C i be a crossing family over ground set V , and let f i : C i → Z be a crossing submodular function, where min i=1,2 f i (U ) ≤ 0 for all Then the following statements hold: a.The system (1) defining P is TDI.In particular, P is an integral polyhedron.
We prove Theorem 3 in §2, and provide a delicate extension of it afterwards.We discuss the complexity aspects of the theorem in §5.We do not know whether condition (2) can be verified in polynomial time.However, as we shall see, under standard assumptions on how the crossing families C i and the submodular functions f i , i = 1, 2, are provided, there exists an oracle strongly polynomial time algorithm that returns either a vector y ⋆ as in the statement, or a subset U violating the cut condition (2).
It must be pointed out that, in contrast to the box-TDI-ness of submodular flow systems, the system (1) is not box-TDI.In fact, P ∩ [ℓ, u] is not necessarily integral even if ℓ, u satisfy the cut condition (2); for an example see §B of the appendix.
Finally, let us say a few words about the first set of inequalities imposed on min i=1,2 f i (U ).The condition that min i=1,2 f i (U ) ≤ 0 for all U ⊊ V, U ̸ = ∅ such that δ + (U ) = δ − (U ) = ∅, is necessary for P to be an integral polyhedron, as demonstrated by Example 4 below.Note that if P ̸ = ∅, then the inequalities are equivalent to min i=1,2 Note further that these inequalities are implied by the cut condition (2).
integer-valued crossing submodular functions over C ⋆ 1 , C ⋆ 2 , respectively.Then the corresponding system This system is not integral, and therefore not box-TDI, as (0.5, 0.5, 0.5) is a vertex of the polyhedron.
Outline of the paper.In §2 we prove Theorem 3, and provide an extension of it afterwards.In §3 we present three applications of Theorem 3: the original result of Edmonds and Giles, an application to digraphs with a connected underlying undirected graph, and of course Theorem 2. In §4 we present applications of Theorem 2. In §5 we discuss the complexity aspects of the two main results.Finally, in §6 we conclude with some open questions.

Intersection of two submodular flow systems
In this section we prove Theorem 3, for which we need two ingredients from Submodular Optimization and Network Flows.First we need the following result, essentially stating that the intersection of two base systems is box-TDI.
Theorem 5 (see Theorem 49.8 of [21], and §14.4 of [8]).For i = 1, 2, let C i be a crossing family over ground set V , let f i : C i → Z be a crossing submodular function, and let k be an integer.Then the We are now ready to prove Theorem 3.
Proof of Theorem 3. If P = ∅, then there is nothing to prove.Otherwise, P ̸ = ∅.Before we prove (a) and (b), let us set the scene.To this end, let c ∈ Z A be a cost vector such that max{c ⊤ y : y ∈ P } admits an optimal solution, let ω ⋆ be the optimal value, and let F be the optimal face.For i = 1, 2, let Define the polyhedron , and the face First we prove the second part as its proof is shorter and contains the crux of the argument.
for all U ⊊ V, U ̸ = ∅.Thus, by Theorem 6, there exists a b-transshipment y ⋆ ∈ Z A such that ℓ ≤ y ⋆ ≤ u.Since b ∈ F , it follows from Claim 1 that y ⋆ ∈ F .This is the desired point.
(a) To prove this part it suffices to show that the dual of max{c ⊤ y : y ∈ P } has an integral optimal solution.Let M ∈ {0, ±1} V ×A denote the node-arc incidence matrix of D. It is well-known that M is a totally unimodular matrix (see Theorem 13.9 of [21]).Observe that x = M ȳ.
Claim 2. There exists Claim 1, so c ⊤ y = ω ⋆ by definition.Thus, the linear system M y = x where y ∈ R A is a vector of variables, implies the equation c ⊤ y = ω ⋆ .Subsequently, there exists w ∈ R V such that w ⊤ M = c ⊤ and w ⊤ x = ω ⋆ .Since M is totally unimodular, and c ∈ Z A , we may choose w ∈ Z V such that By hypothesis, the right-hand side is at most 0 for all U ⊊ V, U ̸ = ∅ such that δ + (U ) = δ − (U ) = ∅, so for all such U , x ′ (U ) ≤ 0. Thus, by Theorem 6 with the choices of u = +∞ and ℓ = −∞, there exists an x ′ -transshipment y ′ ∈ R A , i.e.M y ′ = x ′ .Observe that y ′ ∈ P by Claim 1.Thus, , where the last inequality follows from the definition of ω ⋆ and the fact that y ′ ∈ P .♢ Now consider the dual of max{w ⊤ x : x ∈ P }: where µ ∈ R is the dual variable corresponding to 1 ⊤ x = 0, and χ U is the incidence vector of U as a subset of V .By Theorem 5, the system of constraints of P is TDI.Thus, since w is integral, it follows that (3) has an integral optimal solution (z, μ).By Claim 3 and LP Strong Duality, (3) has optimal value ω ⋆ .Thus, i=1,2 ,2 is an optimal solution to the dual of max{c ⊤ y : y ∈ P }: where χ δ + (U ) , χ δ − (U ) are the incidence vectors of δ + (U ), δ − (U ) as subsets of A.
Proof of Claim.By definition, ω ⋆ = max{c ⊤ y : y ∈ P }, so by LP Strong Duality, it suffices to prove that z is a feasible solution to (4) with objective value ω ⋆ .The latter is indeed the case as we argued above.Let us prove feasibility.Clearly, z ≥ 0.Moreover, where the first equality follows from M ⊤ 1 = 0, the second equality from the feasibility of (z, μ) for (3), and the third equality from the definition of w.Thus, z is feasible for (4), as required.♢ Claim 4 finishes the proof of the first part.
An extension.By a delicate analysis of the proof, we can show that Theorem 3 admits the following extension.Below, for a vector x ∈ R n , x + and x − denote the vectors in R n defined as Theorem 7. Let V be a finite set, and let M be a |V |-by-m totally unimodular matrix with rows indexed by the elements of V , such that M ⊤ 1 = 0.For i = 1, 2, let C i be a crossing family over ground set V , and let f i : C i → Z be a crossing submodular function, where Then the following statements hold: a.The system (5) defining P is TDI.In particular, P is an integral polyhedron.
Proof sketch.The proof is almost identical to that of Theorem 3 but with a few modifications which we highlight as follows.First, the optimal face F is now defined as Furthermore, P and its face F are defined as before, namely , and the face F := P ∩ {x : The statement of Claim 1 needs to be modified as follows: Proof of Claim.The claim follows from the definitions of P and F , and the fact that Furthermore, we need the following additional claim.
Proof of Claim.The "only if" direction of both statements is trivial, so we focus on the "if" statements.
By shifting z by some α1, α ∈ R, if necessary, we may assume that z ≥ 0. (Note that M ⊤ z = M ⊤ (z + α1) ) Furthermore, by scaling down z, if necessary, we may assume that z ≤ 1.Thus, we can only focus on the points z in the polyhedron In fact, by basic polyhedral theory, it suffices to focus on the extreme points z of Q.Since M is totally unimodular, z must be integral, so z = χ U for some U ⊆ V .
If U ∈ {∅, V }, then clearly b ⊤ z = 0, so we are done.Otherwise, since M ⊤ χ U = 0, it follows from the assumption that b(U ) ≤ 0. Since M ⊤ 1 = 0, we also have M ⊤ χ V \U = 0, so again by assumption For part ii), by Farkas' lemma, the system by shifting z by some α1, α ∈ R, if necessary, we may assume that z ≥ 0. Furthermore, by scaling down (z, v, w), if necessary, we may assume that z ≤ 1.As before, since M is totally unimodular, we can focus on z = χ U for some satisfies the condition (6), then for all x ∈ P , the system M y = x, ℓ ≤ y ≤ u is feasible.With these observations, the rest of the proof of the theorem follows exactly the proof of Theorem 3.
3 Applications of Theorem 3, and proof of Theorem 2 In this section we discuss three applications of Theorem 3, one of which is Theorem 2.
First application.Theorem 3 implies the classic theorem of Edmonds and Giles [5].One way to prove it is to use a recent characterization of box-TDI systems.Consider a polyhedron Q := {y : Proof.We need the following two claims.
Then C 2 is a crossing family and f 2 is a crossing submodular function Subsequently, it follows from Theorem 3 (a) that the combined system y(δ Observe that f 1 is a crossing submodular function, because it is the sum of a crossing submodular function f , and crossing modular functions ℓ(δ Observe also that f 2 (U ), as the sum of a crossing modular , is a crossing submodular function.
Consider the polyhedron and its face Hence, by Theorem 3 (b), F ′ contains an integer point y ⋆ ∈ R A ′ such that ℓ e ≤ y ⋆ e ≤ u e for all e ∈ A ′ .Extend the point y ⋆ to R A by defining y ⋆ e := ℓ e for all e ∈ A ℓ , and y ⋆ e := u e for all e ∈ A u .Then, by the definition of f 1 , we have y ⋆ ∈ F ∩ Z A , as desired.♢ It follows from Claim 2 that Q is principally box-integral.This, together with Claim 1 and the theorem of Chervet, Grappe, and Robert [3], implies that y(δ Second application.A digraph is weakly connected if its underlying undirected graph is connected.
The next application is the following result, which surprisingly seems to be new.Observe that the weak connectivity assumption cannot be dropped, as shown by Example 4.
Theorem 9. Let D = (V, A) be a weakly connected digraph and, for i = 1, 2, let C i be a crossing family over ground set V and f i : C i → Z be a crossing submodular function.Then the system in (1) is TDI, and in particular, the polyhedron P is integral.
Proof.Since D is weakly connected, the condition min i=1,2 f i (U ) ≤ 0 for all U ⊊ V, U ̸ = ∅ is vacuously true, because there is no such U .Thus, the result follows from Theorem 3 (a).
Third application.The final application is Theorem 2, which we restate for convenience.
Theorem 2. Let τ, k ≥ 1 be integers.Let D = (V, A) be a digraph where Let C be a crossing family over ground set V , and let f : C → Z be a crossing Observe that f 2 is a crossing submodular function.Consider the vector y ∈ R A that assigns k τ to every arc a ∈ A. Then y(δ + (U )) − y(δ − (U )) ≤ f 1 (U ) for all U ∈ C 1 , by one of our assumptions.
Moreover, for all U ⊊ V, U ̸ = ∅, our assumption implies that d which in turn can be written as y(δ implying in turn that J is a k-arc-connected flip.Thus, J is the desired set.

Applications of Theorem 2
We present several applications of Theorem 2 to Graph Orientations and Combinatorial Optimization.

An extension of the weak orientation theorem
For τ = 2k, Theorem 2 gives the following strengthening of the weak orientation theorem (that every digraph whose underlying undirected graph is 2k-edge-connected, has a k-arc-connected flip).
Theorem 10.Let D = (V, A) be a digraph whose underlying undirected graph is 2k-edge-connected.
Let C be a crossing family over ground set V , and let f : C → Z be a crossing submodular function A digraph is near-Eulerian if at every vertex the in-degree and out-degree differ by at most one.
Theorem 10 implies the following well-known extension of the weak orientation theorem.
Theorem 11 ([18]).Let D = (V, A) be a digraph whose underlying undirected graph is 2k-edgeconnected.Then there exists a k-arc-connected flip J such that after flipping its arcs the digraph becomes near-Eulerian.
true.Interestingly, however, every inclusionwise minimal dijoin is a 1-arc-connected flip (see [21], Theorem 55.1).For an integer k ≥ 1, a k-dijoin is an arc subset that intersects every dicut at least k times.Observe that the union of every k disjoint dijoins is a k-dijoin (the converse is not necessarily true, see Figure 1).Moreover, we have the following important observation.
Remark 12.Given a digraph and an integer k ≥ 1, every k-arc-connected flip is a k-dijoin.
(The converse of this remark is not necessarily true even if the k-dijoin is inclusionwise minimal, see Figure 1).
Figure 1: The solid arcs form an inclusionwise minimal 2-dijoin that cannot be decomposed into two dijoins [19], nor is it a 2-arc-connected flip.

Woodall's conjecture and a weaker variant
A seminal result of Lucchesi and Younger is that the minimum size of a dijoin is equal to the maximum number of pairwise disjoint dicuts [15].Woodall conjectured that the dual minimax relation also holds: the minimum size of a dicut is equal to the maximum number of pairwise disjoint dijoins [23]; this conjecture remains open.As a step towards the conjecture, it was recently shown that if the minimum size of a dicut is τ , then A may be decomposed into a dijoin and a (τ − 1)-dijoin [1].In fact, if Woodall's conjecture is true, then one should be able to decompose A into a k-dijoin and a (τ − k)dijoin, for every integer k ∈ {1, . . ., τ − 1}, but surprisingly even this remains open for k ̸ = 1, τ − 1.
This leads us to the following weaker conjecture.
Theorem 2 has the following consequence that relates to Conjecture 13.
Theorem 14.Let τ, k be integers such that τ − 1 ≥ k ≥ 1.Let D = (V, A) be a digraph where Then A can be decomposed into a k-arc-connected flip and a (τ − k)-dijoin.
Proof.Let C be the family of subsets U ⊆ V such that δ + (U ) is a dicut.Then C is a crossing family.
Let f : C → Z be the function defined as f (U Observe that the complement of a k-arc-connected flip is also a k-arc-connected flip, so for τ = 2k, Theorem 14 reduces simply to the weak orientation theorem, so our theorem extends this classic result in a different direction than Theorem 10.
Let us discuss two special cases of Theorem 14.In the special case k = 1, we obtain the following.
Proof.We claim that and in the latter case, the dicut must have size at least τ by assumption, so the claimed inequality holds.It therefore follows from Theorem 14 for k = 1 that A can be decomposed into a 1-arc-connected flip, which necessarily is a dijoin by Remark 12, and a (τ − 1)-dijoin, as required.This theorem was proved recently in an attempt to prove Woodall's conjecture by first reducing the problem to a special class of sink-regular (τ, τ + 1)-bipartite digraphs [1].The proof we have given here bypasses this reduction.
The reader may wonder why this theorem does not automatically prove Woodall's conjecture, as one may try to repeat the argument on the subdigraph D \ J.A key complication comes from the fact that deleting an arc from D may create a new dicut, whose size may unfavourably be smaller than τ − 1.Another comes from the fact that given the decomposition J ∪ J ′ , one may not necessarily be able to further decompose The second special case of Theorem 14 we consider is the case k = ⌊τ /2⌋.Theorem 16.Let τ ≥ 2 be an integer.Let D = (V, A) be a digraph where every dicut has size at least τ .Suppose further that every cut of D has size at least τ − 1, i.e., |δ and if equality holds, then the number of outgoing arcs is equal to the number of incoming arcs.Then A can be decomposed into a k-arc-connected flip and a (τ − k)-dijoin, for every k ∈ {1, . . ., ⌊τ /2⌋}.
If equality holds, then by assumption, A (U ) ≥ τ , proving the claimed inequality.The theorem now follows from Theorem 14.
In particular, this proves Conjecture 13 when the underlying undirected graph is τ -edge-connected: Proof.By symmetry we may assume that k ≤ τ − k, so k ∈ {1, . . ., ⌊τ /2⌋}.Thus, since every cut of D has size at least τ , we may apply Theorem 16 to decompose A into a k-arc-connected flip, which necessarily is a k-dijoin by Remark 12, and a (τ − k)-dijoin.
Theorem 17 suggests that it may be easier to prove Woodall's conjecture for τ -edge-connected instances.After all, if the underlying undirected graph has τ disjoint spanning trees, which is guaranteed by 2τ -edge-connectivity for instance [17,22], then the digraph has τ disjoint dijoins.It should be noted that Woodall's conjecture for τ = 3 has been proven for 4-edge-connected instances, that is, if the underlying undirected graph is 4-edge-connected, then the digraph contains 3 disjoint dijoins [16].

Packing dijoins in weighted digraphs
Finally, Theorem 2 leads to an intriguing extension of Theorem 14 to a setting where arcs are assigned nonnegative integer weights, viewed as capacities for packing dijoins.By replacing an arc of weight t ≥ 1 with t parallel arcs of weight 1, we may reduce to 0, 1 weights, so we can focus solely on them.
Given a digraph D = (V, A) and J ⊆ A, denote by D[J] the subdigraph with vertex set V and arc set J.
Proof.The proof is similar to that of Theorem 14.This time, however, we apply Theorem 2 to the digraph D[{a ∈ A : w a = 1}], denoted as D ′ = (V, A ′ ), the crossing family C := {U ⊊ V : δ − D (U ) = ∅, U ̸ = ∅}, and the crossing submodular function f : We include the proof for completeness.
The inequalities w(δ We may therefore apply Theorem 2 to get a k-arc-connected flip is the desired decomposition.
integer k, and w ∈ Z V , outputs an extreme optimal solution of max{w ⊤ x : Before we provide a proof sketch of this theorem, let us stress that it is well-known that max{w ⊤ x : is strongly polynomial time solvable; see for instance §47.4 of [21].Though the statement above is widely accepted, none of the algorithms we could find in the literature necessarily return an extreme optimal solution.However, this issue can be addressed by finding a lexicographically maximal optimal solution.
, and denote by F the optimal face of max{w ⊤ x : x ∈ P }.It follows from §49.3, §49.7, and §49.10 of [21] that one can construct, in strongly polynomial time, submodular functions f ′ i , i = 1, 2 defined over 2 V such that F is the optimal face of max{w ⊤ x : Theorem 47.4 of [21] shows how to construct, in strongly polynomial time, submodular functions f ′′ i , i = 1, 2 defined over 2 V such that This shows that the original problem can be reduced, in strongly polynomial time, to the problem of finding an extreme common base of two extended polymatroids.While there exist strongly polynomial time algorithms that can provide a common base [14] (see Theorem 47.1 in [21]), the common base returned by these algorithms may not be extreme.However, for any given v ∈ V , one can in strongly polynomial time find a point in F maximizing x v [7] (see Theorem 47.2 of [21]).This implies, as we discuss next, that we can pick an ordering 1, 2, . . ., n of the elements in V and compute a lexicographically maximal point in F , which will therefore be extreme.Indeed, we start by computing the value α := max{x n : x ∈ F }. To iterate, we need to compute the largest value of x n−1 for x ∈ F ∩ {x : x n = α}.This can be reduced again to the problem of finding a common base maximizing x n−1 for two extended polymatroids defined over {1, . . ., n − 1}, where the functions g i , i = 1, 2 are defined over 2 {1,...,n−1} by g i (U ) = min{f ′′ i (U ), f ′′ i (U ∪ {n}) − α} for all U ⊆ V \ {n}.It can be readily checked that g i , i = 1, 2 are indeed still submodular, and furthermore, a point z ∈ R {1,...,n−1} is a common base of g 1 and g 2 if and only if (z, α) is a common base of f ′′ 1 and f ′′ 2 .Hence we are able to apply Theorem 47.2 of [21] iteratively, repeating the process for 1, . . ., n − 2, and so on and so forth.
Complexity aspects of Theorem 3. To discuss the complexity aspects of part (b) of this theorem, we assume that for i = 1, 2, the crossing family C i is well-provided, and f i is provided via a value oracle.
Furthermore, the non-empty face of P is provided by some vector c ∈ Z A , namely, the face F is the optimal face of max{c ⊤ y : y ∈ P }.
The first question is whether the cut condition (2) can be verified in polynomial time.We do not know.The verification amounts to determining whether g(U However, g is not necessarily a crossing submodular function, making the verification problem challenging.That said, there is a way to circumvent this issue altogether. Following the proof of Theorem 3 (b), we can show that in strongly polynomial time, we can either find an integral point y ⋆ ∈ F such that ℓ ≤ y ⋆ ≤ u, or a subset U ⊆ V that violates the cut condition (2).
First we compute w ∈ Z V such that w ⊤ M = c ⊤ , which can be done in strongly polynomial time.
Then, if we let P and F be defined as in the proof of Theorem 3, we have that F is the optimal face of max{w ⊤ x : x ∈ P }.Next, we find a vertex b of F in strongly polynomial time, by Theorem 20.
Note that b ought to be integral by Theorem 5. Then we find either an integral b-transhipment y ⋆ satisfying ℓ ≤ y ⋆ ≤ u, or a certificate that such a b-transhipment does not exist, in the form of a set U , ; this can be done in strongly polynomial time (see Corollary 12.2d of [21]).In the first case, we get an integral point y ⋆ in F ∩{y ∈ R A : ℓ ≤ y ≤ u}, and in the second case, we have a certificate U that (2) is not satisfied.
Complexity aspects of Theorem 2. First, we claim that the assumed inequalities d This holds if, and only if, min{g(U ) : U ∈ C 2 } ≥ 0 for the crossing submodular function g over the crossing family C 2 defined as g(U ) := f 2 (U ) − ȳ(δ + (U )) + ȳ(δ − (U )).The claim now follows from the fact that the minimization problem can be solved in strongly polynomial time (see §49.10 of [21] for details).
Secondly, Theorem 2 also assumes the inequalities f ) for all U ∈ C. To be able to verify these inequalities in polynomial time, we would need some assumptions on how the data is provided.As such, suppose C is a well-provided crossing family, and f is provided via a value oracle.Then, just as above, the inequalities can be verified in strongly polynomial time (see §49.10 of [21]).
Finally, with the assumptions mentioned above on C = C 1 and f = f 1 , we can find the k-arcconnected flip J in strongly polynomial time, by an application of the algorithm we provided for Theorem 3 (b).To describe the algorithm, first we find b ∈ Z all U ∈ C i , then we find a b-transshipment y ⋆ ∈ {0, 1} A .The first step relies on the fact that every vertex of {x : integral by Theorem 5, and that a vertex b can be found in strongly polynomial time by Theorem 20.The second step can also be done in strongly polynomial time (see Corollary 12.2d of [21]).The subset J corresponds to the support of y ⋆ .

Open questions and concluding remarks
Theorem 19 connects intriguingly to a conjecture of Chudnovsky, Edwards, Kim, Scott, and Seymour [4] for 0, 1-weighted digraphs (D, w), that if every cut has nonzero weight and every dicut has weight at least τ , then {a ∈ A : w a = 1} can be decomposed into τ dijoins of D. In fact, another result of ours, namely Theorem 9, has a direct application to this conjecture for τ = 2.In this case, the conjecture can be equivalently formulated as below.

Conjecture 21 ([4]
).Let G = (V, E) be a tree, and let L be a lattice family over ground set V such that |δ(U )| ≥ 2 for all U ∈ L \ {∅, V }.Then there exists an orientation D of G such that To point out the connection to Theorem 9, let D = (V, A) be an arbitrary orientation of G. Observe that y = 1 2 • 1 is a feasible solution.Observe further that L \ {∅, V } is a crossing family.Since D is weakly connected, it follows from Theorem 9 that the system is TDI, and so it has an integral solution.Conjecture 21 states equivalently that the system has a 0, 1 solution.
The next open question comes from Theorem 9. We saw that the weak connectivity assumption could not be dropped due to Example 4.However, this example is not only disconnected but has three Clearly, to be able to do this, two conditions are necessary: (a) the underlying undirected graph of D must be 2k-edge-connected, and (b) every dicut of D must have size at least τ .Are these two conditions also sufficient?Finally, as we show in §B of the appendix, the intersection of two submodular flow systems with integral right-hand sides, is not necessarily box-integral.In March 2023, during a talk at the Combinatorics and Optimization workshop at ICERM, Brown University, the first author made the following "wild" conjecture: Let D = (V, A) be a weakly connected digraph and, for i = 1, 2, let C i be a crossing family over ground set V and f i : C i → Z be a crossing submodular function.Then the polyhedron P from (1) is box-half-integral.However, this conjecture has been refuted by Goemans and Pan [11].

B Theorem 3 and box constraints
Here we demonstrate that the system (1) from Theorem 3, together with box constraints, is not necessarily integral.To this end, consider the digraph D = (V, A) displayed in Figure 2 (left).To give some intuition, a 0, 1 vector ȳ belongs to Q if, and only if, {a ∈ A : ȳa = 0} is a 3-dijoin and {a ∈ A : ȳa = 1} is a 1-arc-connected flip of D. This gives a description of all the integral vertices of Q.However, Q may have fractional vertices.To see this, define y 1 ∈ 0, 1 2 , 1 A where for each a ∈ A, y 1 a = 0, 1 2 or 1 if a is dotted, dashed, or solid in Figure 2 (middle), respectively.Define y 2 ∈ 0, 1  2 , 1 A analogously with respect to Figure 2 (right).
Proof.It can be readily checked that y 1 , y 2 are feasible.To see that y i is a vertex, we need to exhibit 21 linearly independent tight inequalities at y i .We immediately get 15 from 0 ≤ y ≤ 1 as y i has as many coordinates set to 0 or 1.For y 1 the remaining 6 may be chosen as the follow-

Theorem 2 .
the incidence vector of a k-arc-connected flip of D is a submodular flow, by Definition 1. Main result I.The following theorem introduces a sufficient condition for the existence of a karc-connected flip whose incidence vector is also a submodular flow for another crossing submodular function.Let τ, k ≥ 1 be integers.Let D = (V, A) be a digraph where d + A (U ) + ( τ k − 1)d − A (U ) ≥ τ for all U ⊊ V, U ̸ = ∅.Let C be a crossing family over ground set V , and let f : C → Z be a crossing submodular function such that f (U ) ≥ k τ (d + A (U ) − d − A (U )) for all U ∈ C. Then D has a k-arcconnected flip J such that f (U ) ≥ d + J (U ) − d − J (U ) for all U ∈ C.

( b )
It suffices to find an integral point y ⋆ ∈ F satisfying ℓ ≤ y ⋆ ≤ u.By Theorem 5, P is an integral polyhedron, hence F contains an integral point b.Observe that b and since 1 ⊤ b = 0, we get that b ⊤ z = b(U ) = 0, as required.
that kQ is integral.This notion was coined recently by Chervet, Grappe, and Robert who proved that Ax ≤ b is box-TDI if, and only if, Ax ≤ b is TDI and Q is principally box-integral [3].Theorem 8 ([5]).Let D = (V, A) be a digraph, let C be a crossing family over ground set V , and let f : C → Z be a crossing submodular function.Then y(δ + (U )) − y(δ − (U )) ≤ f (U ) ∀U ∈ C is box-TDI.

Claim 2 .
For every integer k ≥ 1, kQ is box-integral.Proof of Claim.It suffices to prove this for k = 1, given that kf is a crossing submodular function for every integer k ≥ 1.Take ℓ, u ∈ Z A such that ℓ ≤ u and consider a nonempty face of Q ∩ [ℓ, u], say F = Q ∩ {y ∈ R A : y(δ + (U )) − y(δ − (U )) = f (U ) ∀U ∈ D, y e = ℓ e ∀e ∈ A ℓ , y e = u e ∀e ∈ A u }, where D ⊆ C; A ℓ , A u ⊆ A; A ℓ ∩ A u = ∅.We need to show that F contains an integer point.To this end, let C 1 := C and for every U ∈ C 1 define
connected components in its underlying undirected graph.An immediate open question is whether Theorem 9 extends beyond weakly connected digraphs to those digraphs with at most two connected components in their underlying undirected graph?Another open question comes from a closer look at Theorem 14.The theorem provides a sufficient condition for decomposing the arc set of a digraph D into a k-arc-connected flip and a (τ − k)-dijoin.
can be verified in strongly polynomial time.To elaborate, note that the inequalities are satisfied if, and only