Sandwich problems on orientations

Abstract

The graph sandwich problem for property Π is defined as follows: Given two graphs G₁=(V,E₁) and G₂=(V,E₂) such that E₁⊆E₂, is there a graph G=(V,E) such that E₁⊆E⊆E₂ which satisfies property Π? We propose to study sandwich problems for properties Π concerning orientations, such as Eulerian orientation of a mixed graph and orientation with given in-degrees of a graph. We present a characterization and a polynomial-time algorithm for solving the m-orientation sandwich problem.

Introduction

Given two graphs G₁=(V,E₁) and G₂=(V,E₂) with the same vertex set V and E₁⊆E₂, a graph G=(V,E) is called a sandwich graph for the pair G₁,G₂ if E₁⊆E⊆E₂. The graph sandwich problem for property Π is defined as follows [12]:

Graph Sandwich Problem for Property Π Instance: Given undirected graphs G₁=(V,E₁) and G₂=(V,E₂) with E₁⊆E₂. Question: Is there a graph G=(V,E) such that E₁⊆E⊆E₂ and G satisfies property Π?

We call E₁ the mandatory edge set, E₀=E₂∖E₁ the optional edge set and E₃ the forbidden edge set, where E₃ denotes the set of edges of the complementary graph $\overline{G}_{2}$ of G₂. Thus any sandwich graph G=(V,E) for the pair G₁,G₂ must contain all mandatory edges, no forbidden edges and may contain a subset of the optional edges. Graph sandwich problems have attracted much attention lately arising from many applications and as a natural generalization of recognition problems [1–3, 7, 22, 24]. The recognition problem for a class of graphs $\mathcal{C}$ is equivalent to the graph sandwich problem in which G₁=G₂=G, where G is the graph we want to recognize and property Π is “to belong to class $\mathcal{C}$”.

In this paper we propose to study sandwich problems for properties Π concerning orientations, such as Eulerian orientation of a mixed graph and orientation with given in-degrees of a graph, or more generally of a mixed graph.

The paper is organized as follows: Sect. 2 contains some basic definitions, notations and results. Section 3 contains some known results on degree constrained sandwich problems. We consider the undirected version and the directed version, the complexity, the characterization and the related optimization problems. We also define a simultaneous version and discuss its complexity. Section 4 focuses on Eulerian sandwich problems. We consider first undirected graphs and then directed graphs. These problems were already solved in [12], here we point out that the undirected case reduces to T-joins, while the directed case to circulations. We discuss the complexity of the problems and their characterizations and we also propose some mixed versions. In Sect. 5 we consider sandwich problems regarding an m-orientation, i.e., given undirected graphs G₁=(V,E₁) and G₂=(V,E₂) with E₁⊆E₂ and a non-negative integer vector m on V, we show that it is polynomial to decide whether there exists a sandwich graph G=(V,E) (E₁⊆E⊆E₂) that has an orientation $\vec{G}$ whose in-degree vector is m that is $d^{-}_{\vec{G}}(v)=m(v)$ for all v∈V. This result stands in contrast to the strongly connected m-orientation sandwich problem which we show is NP-complete. Section 6 is devoted to a new kind of sandwich problem where we may contract (and not delete) optional edges and property Π is being bipartite.

Definitions

Undirected graphs :

Let G=(V,E) be an undirected graph. For vertex sets X and Y, the cut induced by X is defined to be the set of edges of G having exactly one end-vertex in X and is denoted by . The degree (or d _E (X)) of X is the cardinality of the cut induced by X, that is, d _G (X)=|δ _G (X)|. The number of edges between X∖Y and Y∖X is denoted by . The number of edges of G having both (resp. at least one) end-vertices in X is denoted by or i _E (X) or simply i(X) (resp. ). It is well-known that (1) is satisfied for all X,Y⊆V,

(1)

We say that a vector m on V is the degree vector of G if d _G (v)=m(v) for all v∈V. For a vector m on V, we consider m as a modular function, that is, we use the notation . Let us recall that d _G (X) is the degree function of G. We define as the modular function defined by the degree vector d _G (v) of G. Note that $\hat{d}_{G}(X)=d_{G}(X)+2i_{G}(X)\forall X\subseteq V$.

We denote by the set of vertices of G of odd degree. For an edge set F of G, the subgraph induced by F, that is, (V,F), is denoted by . We say that G is Eulerian if the degree of each vertex is even, that is, if T _G =∅. Note that we do not suppose the graph to be connected.

Let T be a vertex set in G. An edge set F of G is called T-join if the set of odd degree vertices in the subgraph induced by F coincide with T, that is if T_G(F)=T. Given a cost vector on the edge set of G, a minimum cost T-join can be found in polynomial time by Edmonds and Johnson’s algorithm [5].

Let f be a non-negative integer vector on V. An edge set F of G is called an f-factor of G if f is the degree vector of G(F), that is, d _F (v)=f(v) for all v∈V. If f(v)=1 for all v∈V, then we say that F is a 1-factor or a perfect matching. An f-factor—if it exists—can be found in polynomial time, see [20]. The graph G is called 3-regular if each vertex is of degree 3. Note that for a 3-regular graph, the existence of two edge-disjoint perfect matchings is equivalent to the existence of three edge-disjoint perfect matchings which is equivalent to the 3-edge-colorability of the graph.

Directed graphs :

Let D=(V,A) be a directed graph. For a vertex set X, the set of arcs of D entering (resp. leaving) X is denoted by (resp. ). The in-degree (resp. out-degree) of X is the number of arcs of D entering (resp. leaving) X, that is $d^{-}_{D}(X)=|\varrho _{D}(X)|$ (resp. $d^{+}_{D}(X)=|\delta_{D}(X)|$). The set of arcs of G having both end-vertices in X is denoted by . The following equality will be used frequently without reference:

(2)

We say that a vector m on V is the in-degree vector of D if $d^{-}_{D}(v)=m(v)$ for all v∈V. Let us recall that $d^{-}_{D}(X)$ is the in-degree function of D. Let f be a non-negative integer vector on V. An arc set F of D is called a directedf-factor of D if f is the in-degree vector of D(F), that is, $d^{-}_{F}(v)=f(v)$ for all v∈V.

We say that D is Eulerian if the in-degree of v is equal to the out-degree of v for all v∈V, that is, $d^{-}_{D}(v)=d^{+}_{D}(v)$ for all v∈V. Note that we do not suppose the graph to be connected.

Let f and g be two vectors on the arcs of D such that f(e)≤g(e) for all e∈A. A vector x on the arcs of D is a circulation if (3) and (4) are satisfied.

(3)

(4)

Note that if f(e)=g(e)=1 for all e∈A, then D is Eulerian if and only if f is a circulation. We will use the following characterization when a circulation exists.

Theorem 1

(Hoffmann [15])

LetD=(V,A) be a directed graph andfandgtwo vectors onAsuch thatf(e)≤g(e)∀e∈A. There exists a circulation inDif and only if

(5)

We say that H=(V,E∪A) is a mixed graph if E is an edge set and A is an arc set on V. For an undirected graph G=(V,E), if we replace each edge uv by the arc uv or vu, then we get the directed graph . We say that $\vec{G}$ is an orientation of G.

Mixed graphs having Eulerian orientations are characterized as follows.

Theorem 2

(Ford, Fulkerson [8])

A mixed graphH=(V,E∪A) has an Eulerian orientation if and only if

(6)

(7)

The following theorem characterizes graphs having an orientation with a given in-degree vector.

Theorem 3

(Hakimi [13])

Given an undirected graphG=(V,E) and a non-negative integer vectormonV, there exists an orientation$\vec{G}$ofGwhose in-degree vector ismif and only if

(8)

(9)

Functions :

Let b be a set function on the subsets of V. We say that b is submodular if for all X,Y⊆V,

(10)

The function b is called supermodular if −b is submodular. A function is modular if it is supermodular and submodular. We will use frequently in this paper the following facts.

Claim 1

The degree functiond _G (Z) of an undirected graphGand the in-degree function$d^{-}_{D}(Z)$of a directed graphDare submodular and the functioni(Z) is supermodular.

Theorem 4

[17, 21]

The minimum value of a submodular function can be found in polynomial time.

Theorem 5

(Frank [9])

Letbandpbe a submodular and a supermodular set function onVsuch thatp(X)≤b(X) for allX⊆V. Then there exists a modular functionmonVsuch thatp(X)≤m(X)≤b(X) for allX⊆V. Ifbandpare integer valued thenmcan also be chosen integer valued.

A pair (p,b) of set functions on 2^V is a strong pair if p (resp. b) is supermodular (submodular) and they are compliant, that is, for every pairwise disjoint triple X,Y,Z,

$$b(X\cup Z) - p(Y\cup Z) \geq b(X) -p(Y).$$

Note that a pair (α,β) of modular functions is a strong pair if and only if α≤β. If (p,b) is a strong pair then the polyhedron

is called a generalized polymatroid (or a g-polymatroid). When α≤β are modular, we also call the g-polymatroid Q(α,β) a box.

Theorem 6

(Frank, Tardos [11])

The intersection of an integral g-polymatroidQ(p,b) and an integral boxQ(α,β) is an integral g-polymatroid. It is nonempty if and only ifα≤bandp≤β.

Matroids :

A set system $M=(V,{\mathcal{F}})$ is called a matroid if ${\mathcal{F}}$ satisfies the following three conditions:

• (I1) $\emptyset\in{\mathcal{F}}$,

• (I2) if $F\in{\mathcal{F}}$ and F′⊆F, then $F'\in{\mathcal{F}}$,

• (I3) if $F, F'\in{\mathcal{F}}$ and |F|>|F′|, then there exists f∈F∖F′ such that $F'\cup f\in{\mathcal{F}}$.

A subset X of V is called independent in M if $X\in{\mathcal{F}}$, otherwise it is called dependent. The maximal independent sets of V are the basis of M. Let ${\mathcal{B}}$ be the set of basis of M. Then ${\mathcal{B}}$ satisfies the following two conditions:

• (B1) ${\mathcal{B}}\neq\emptyset$,

• (B2) if $B, B'\in{\mathcal{B}}$ and b∈B∖B′, then there exists b′∈B′∖B such that $(B-b)\cup b'\in{\mathcal{B}}$.

Conversely, if a set system $(V,{\mathcal{B}})$ satisfies (B1) and (B2), then $M=(V,{\mathcal{F}})$ is a matroid, where ${\mathcal{F}}=\{F\subseteq V:\exists B\in{\mathcal{B}}, F\subseteq B\}$.

For S⊂V, the matroid M∖S obtained from M by deletingS is defined as $M\setminus S=(V\setminus S, {\mathcal{F}}|_{V\setminus S})$, where X⊆V∖S belongs to ${\mathcal{F}}|_{V\setminus S}$ if and only if $X\in{\mathcal{F}}$. For $S\in{\mathcal{F}}$, the matroid M/S obtained from M by contractingS is defined as $M/S=(V\setminus S, {\mathcal{F}}_{S})$, where X⊆V∖S belongs to ${\mathcal{F}}_{S}$ if and only if $X\cup S\in{\mathcal{F}}$. Let {V₁,…,V _l } be a partition of V and a₁,…,a _l a set of non-negative integers. Then $M=(V,{\mathcal{F}})$ is a matroid, where ${\mathcal{F}}=\{F\subseteq V: |F\cap V_{i}|\leq a_{i}\}$, we call it partition matroid. The dual matroid M^∗ of M is defined as follows: the basis of M^∗ are the complements of the basis of M.

Let $M=(V,{\mathcal{F}})$ be a matroid and c a cost vector on V={v₁,…,v _n }. We can find a minimum cost basis F _n of M in polynomial time by the greedy algorithm: take a non-decreasing order of the elements of V:c(v₁)≤…≤c(v _n ). Let F₀ be empty and for i=1,…,n, let F _i =F_i−1+v _i if $F_{i-1}+v_{i}\in {\mathcal{F}}$, otherwise let F _i =F_i−1.

If M₁ and M₂ are two matroids on the same ground set V, then we can find a common basis of M₁ and M₂ in polynomial time (if there exists one) by the matroid intersection algorithm of Edmonds [4].

Theorem 7

(Edmonds, Rota [18])

For an integer-valued, non-decreasing, submodular functionbdefined on a ground setS, the set {F⊆S;|F′|≤b(F′) for all ∅≠F′⊆F} forms the set of independent sets of a matroidM _b whose rank functionr _b is given by

$$r_b(Z) = \min\bigl\{b(X) + |Z-X|, X \subseteq Z\bigr\}.$$

Given an undirected graph G=(V,E) and a non-negative integer vector m on V, let $\bar{m}^{G}=\bar{m}$ be the set function defined on E by $\bar{m}(F) = m(V(F))$ where V(F) is the set of vertices covered by F. One can easily check that $\bar{m}$ is integer valued, non-decreasing and submodular. Thus, by Theorem 7, $\bar {m}$ defines a matroid $M_{\bar{m}}$. The following claim is straightforward.

Claim 2

The set {F⊆E:m(X)≥i _F (X),∀X⊆V} is the set of independent sets of the matroid$M_{\bar{m}}$.

Degree constrained sandwich problems

Before studying sandwich problems on orientations of given in-degrees, let us start as a warming up by considering sandwich problems for undirected and directed graphs of given degrees. These problems reduce to the undirected and directed f-factor problems. We mention that the directed case is much easier than the undirected case because the addition of an arc in a directed graph contributes only to the in-degree of the head and not of the tail, while the addition of an edge in an undirected graph contributes to the degree of both end-vertices. This section contains no new results, we added it for the sake of completeness.

Undirected graphs

Undirected Degree Constrained Sandwich ProblemInstance: Given undirected graphs G₁=(V,E₁) and G₂=(V,E₂) with E₁⊆E₂ and a non-negative integer vector f on V. Question: Does there exist a sandwich graph G=(V,E) (E₁⊆E⊆E₂) such that d _G (v)=f(v) for all v∈V?

Complexity: It is in P because the answer is Yes if and only if there exists an $(f(v)-d_{G_{1}}(v))$-factor in the optional graph G₀=(V,E₀).

Characterization: The general f-factor theorem due to Tutte [25] can be applied to get a characterization.

Optimization: The minimum cost f-factor problem in undirected graphs can be solved in polynomial time, see Schrijver [20].

Simultaneous Undirected Degree Constrained Sandwich ProblemInstance: Given two edge-disjoint graphs G₁=(V,E₁) and G₂=(V,E₂) in G₃=(V,E₃) and two non-negative integer vectors f₁ and f₂ on V. Question: Do there exist simultaneously sandwich graphs $\hat{G}_{1}=(V,\hat{E}_{1})$$(E_{1}\subseteq\hat{E}_{1}\subseteq E_{3})$ and $\hat{G}_{2}=(V,\hat{E}_{2})$$(E_{2}\subseteq\hat{E}_{2}\subseteq E_{3})$ such that $\hat{E}_{1}\cap\hat{E}_{2}=\emptyset$ and $d_{\hat{G}_{1}}(v)=f_{1}(v)$ and $d_{\hat{G}_{2}}(v)=f_{2}(v)$ for all v∈V?

Complexity: It is NP-complete because it contains as a special case whether there exist two edge-disjoint perfect matchings so 3-edge-colorability of 3-regular graphs. Indeed, let G=(V,E) be an arbitrary 3-regular graph. Let G₁ and G₂ be the edgeless graph on V, G₃=G and f₁(v)=f₂(v)=1 for all v∈V. Then the sandwich graphs $\hat{G}_{1}$ and $\hat{G}_{2}$ exist if and only if $\hat{E}_{1}$ and $\hat{E}_{2}$ are edge-disjoint perfect matchings of G or equivalently, if there exists a 3-edge-coloring of G. Since the problem of 3-edge-colorability of 3-regular graphs is NP-complete [16], so is our problem.

Directed graphs

Directed Degree Constrained Sandwich ProblemInstance: Given directed graphs D₁=(V,A₁) and D₂=(V,A₂) with A₁⊆A₂ and a non-negative integer vector f on V. Question: Does there exist a sandwich graph D=(V,A) (A₁⊆A⊆A₂) such that $d^{-}_{D}(v)=f(v)$ for all v∈V?

Complexity+Characterization: It is in P because the answer is Yes if and only if there exists a directed $(f(v)-d^{-}_{D_{1}}(v))$-factor in the optional directed graph D₀=(V,A₀), hence we have the following.

Theorem 8

TheDirected Degree ConstrainedSandwich Problemhas aYesanswer if and only if$d^{-}_{D_{2}}(v)\geq f(v)\geq d^{-}_{D_{1}}(v)$for allv∈V.

Optimization: The feasible arc sets form the basis of a partition matroid, so the greedy algorithm provides a minimum cost solution.

Simultaneous Directed Degree ConstrainedSandwich Problem 1Instance: Given two arc-disjoint directed graphs D₁=(V,A₁) and D₂=(V,A₂) in D₃=(V,A₃) and two non-negative integer vectors f₁ and f₂ on V. Question: Do there exist simultaneously sandwich graphs $\hat{D}_{1}=(V,\hat{A}_{1})$$(A_{1}\subseteq\hat{A}_{1}\subseteq A_{3})$ and $\hat{D}_{2}=(V,\hat{A}_{2})$$(A_{2}\subseteq\hat{A}_{2}\subseteq A_{3})$ such that $\hat{A}_{1}\cap\hat{A}_{2}=\emptyset$ and $d^{-}_{\hat{D}_{1}}(v)=f_{1}(v)$ and $d^{-}_{\hat{D}_{2}}(v)=f_{2}(v)$ for all v∈V?

Complexity: It is in P because the answer is Yes if and only if $d^{-}_{D_{3}}(v)\geq f_{1}(v)+f_{2}(v)$, $f_{1}(v)\geq d^{-}_{D_{1}}(v)$ and $f_{2}(v)\geq d^{-}_{D_{2}}(v)$ for all v∈V.

Simultaneous Directed Degree ConstrainedSandwich Problem 2Instance: Given directed graphs D₁=(V,A₁) and D₂=(V,A₂) with A₁⊆A₂ and two non-negative integer vectors f and g on V. Question: Does there exist a sandwich graph D=(V,A) (A₁⊆A⊆A₂) such that $d^{-}_{D}(v)=f(v)$ and $d^{+}_{D}(v)=g(v)$ for all v∈V.

Complexity: The feasible arc sets for the in-degree constraint form the basis of a partition matroid and the feasible arc sets for the out-degree constraint form the basis of a partition matroid. The answer is Yes if and only if there exists a common basis in these two matroids. Thus it is in P by the matroid intersection algorithm of Edmonds [4].

Eulerian sandwich problems

In this section we consider first two problems that were already solved in [12]: Eulerian sandwich problems for undirected and directed graphs. We point out that the undirected case reduces to T-joins, while the directed case to circulations. We show that in both cases the simultaneous versions are NP-complete.

Then we propose to study the problem in mixed graphs. We show two cases that can be solved. The first case will be solved by the Discrete Separation Theorem 5 of Frank [9], while the second case reduces to the Directed Eulerian Sandwich Problem. The general case, however, remains open.

Undirected graphs

Undirected Eulerian Sandwich ProblemInstance: Given undirected graphs G₁=(V,E₁) and G₂=(V,E₂) with E₁⊆E₂. Question: Does there exist a sandwich graph G=(V,E) (E₁⊆E⊆E₂) that is Eulerian?

Complexity: It is in P because the answer is Yes if and only if there exists a $T_{G_{1}}$-join in the optional graph G₀.

Characterization: The answer is Yes if and only if each connected component of G₀ contains an even number of vertices of $T_{G_{1}}$.

Optimization: The minimum cost T-join problem can be solved in polynomial time [5].

Simultaneous Undirected Eulerian Sandwich ProblemInstance: Given two edge-disjoint graphs G₁=(V,E₁) and G₂=(V,E₂) in G₃=(V,E₃). Question: Do there exist simultaneously Eulerian sandwich graphs $\hat{G}_{1}=(V,\hat{E}_{1})$$(E_{1}\subseteq\hat{E}_{1}\subseteq E_{3})$ and $\hat{G}_{2}=(V,\hat{E}_{2})$$(E_{2}\subseteq\hat{E}_{2}\subseteq E_{3})$ such that $\hat{E}_{1}\cap\hat{E}_{2}=\emptyset$?

Complexity: It is NP-complete because it contains as a special case whether there exist two edge-disjoint perfect matchings so 3-colorability of 3-regular graphs. Indeed, let G=(V,E) be an arbitrary 3-regular graph. Let G₃ be obtained from G by adding 2 edge-disjoint perfect matchings M₁ and M₂ to G, let G₁=(V,M₁) and G₂=(V,M₂). Then the Eulerian sandwich graphs $\hat{G}_{1}$ and $\hat{G}_{2}$ exist if and only if $\hat{E}_{1}\setminus M_{1}$ and $\hat{E}_{2}\setminus M_{2}$ are edge-disjoint perfect matchings of G or equivalently, if there exists a 3-edge-coloring of G. Since the problem of 3-edge-colorability of 3-regular graphs is NP-complete [16], so is our problem.

Directed graphs

Directed Eulerian Sandwich ProblemInstance: Given directed graphs D₁=(V,A₁) and D₂=(V,A₂) with A₁⊆A₂. Question: Does there exist a sandwich graph D=(V,A) (A₁⊆A⊆A₂) that is Eulerian?

Complexity: It is in P because it can be reformulated as a circulation problem: let f(e)=1,g(e)=1 if e∈A₁ and f(e)=0,g(e)=1 if e∈A₀. This way the arcs of A₁ are forced and the arcs of A₀ can be chosen if necessary.

Characterization: The answer is Yes if and only if $d^{-}_{D_{1}}(X)\leq d^{+}_{D_{2}}(X)$ for all X⊆V by Theorem 1.

Optimization: The minimum cost circulation problem can be solved in polynomial time, see Tardos [23].

Simultaneous Directed Eulerian SandwichProblemInstance: Given two arc-disjoint directed graphs D₁=(V,A₁) and D₂=(V,A₂) in D₃=(V,A₃). Question: Do there exist simultaneously Eulerian sandwich graphs $\hat{D}_{1}=(V,\hat{A}_{1})$$(A_{1}\subseteq\hat{A}_{1}\subseteq A_{3})$ and $\hat{D}_{2}=(V,\hat{A}_{2})$$(A_{2}\subseteq\hat{A}_{2}\subseteq A_{3})$ such that $\hat{A}_{1}\cap\hat{A}_{2}=\emptyset$?

Complexity: It is NP-complete, it contains as a special case (D₁=(V,t₁s₁), D₂=(V,t₂s₂) and D₃=D) the following directed 2-commodity integral flow problem that is NP-complete [6]: Given a directed graph D and two pairs of vertices, s₁,t₁ and s₂,t₂, decide whether there exist a path from s₁ to t₁ and a path from s₂ to t₂ that are arc-disjoint.

Mixed graphs

Mixed Eulerian Sandwich ProblemInstance: Given mixed graphs H₁=(V,E₁∪A₁) and H₂=(V,E₂∪A₂) with E₁⊆E₂,A₁⊆A₂. Question: Does there exist a sandwich mixed graph H=(V,E∪A) (E₁⊆E⊆E₂,A₁⊆A⊆A₂) that has an Eulerian orientation?

Complexity: We provide two special cases that can be treated, while the general problem remains open.

Special Case 1: E₁=E₂=E and $d^{+}_{A_{2}}(X)-d^{-}_{A_{1}}(X)+\hat{d}_{E}(X)$ is even for all X⊆V.

Characterization+Complexity: We show that the problem is in P and we provide a characterization.

Theorem 9

TheMixed Eulerian Sandwich ProblemwithE₁=E₂=Eand$d^{+}_{A_{2}}(X)-d^{-}_{A_{1}}(X)+\hat{d}_{E}(X)$is even for allX⊆Vhas aYesanswer if and only if

(11)

In particular, this problem is in P.

Proof

By the result of Sect. 4.2, the answer is Yes if and only if there exists an orientation $\vec{E}$ of E such that, ∀X⊆V, $d^{-}_{A_{1}\cup\vec{E}}(X)\leq d^{+}_{A_{2}\cup\vec{E}}(X)$ or equivalently

(12)

Let m be the in-degree vector of $\vec{E}$. Then $d^{-}_{\vec{E}}(X)-d^{+}_{\vec{E}}(X)=\sum_{v\in X}(d^{-}_{\vec{E}}(v)-d^{+}_{\vec{E}}(v))=\sum_{v\in X}(2d^{-}_{\vec{E}}(v)-d_{E}(v))= 2m(X)-\hat{d}_{E}(X)$, and (12) becomes

(13)

Let $b(X)=\frac{1}{2}(d^{+}_{A_{2}}(X)-d^{-}_{A_{1}}(X)+\hat{d}_{E}(X))$. Then b, being the sum of a modular function and a submodular function $(b(X)=\frac{1}{2}\sum_{v\in X}(d^{+}_{A_{1}}(v)-d^{-}_{A_{1}}(v)+d_{E}(v))+d^{+}_{A_{0}}(X))$, is a submodular function and, by the assumption, it is integer valued. By Theorem 3, an orientation $\vec{E}$ satisfying (12) exists if and only if there exists a vector m such that i _E (X)≤m(X)≤b(X), that is, by Claim 1 and Theorem 5, if and only if i _E (X)≤b(X). This is equivalent to (11) and can be decided in polynomial time by Theorem 4, namely the submodular function b′(X)=b(X)−i _E (X) must have minimum value 0. □

Special Case 2: E₁=∅.

Characterization+Complexity: It is in P because it can be reformulated as the following problem: We create two copies of each edge in E₂ and orient them in opposite directions. Denote this arc set by $\overrightarrow{E^{2}_{2}}$. It is not difficult to see that the graph (V,E₂∪A₂) has a subgraph containing (V,A₁) with an Eulerian orientation if and only if the graph $(V,\overrightarrow{E^{2}_{2}} \cup A_{2})$ has a directed Eulerian subgraph containing (V,A₁). Indeed, in such a graph, if every edge of E₂ is used at most once, we are done. If some edge of E₂ is used twice, as two arcs in opposite directions, we can just remove these two arcs, the obtained graph remaining Eulerian and containing (V,A₁). Now applying the result for Directed Eulerian Sandwich Problem we have

Theorem 10

TheMixed Eulerian Sandwich ProblemwithE₁=∅ has aYesanswer if and only if

(14)

In particular, this problem is in P.

Proof

Let D₁=(V,A₁) and $D_{2} = (V, A_{2} \cup\overrightarrow{E^{2}_{2}})$. By the arguments above, the Mixed Eulerian Sandwich Problem with E₁=∅ has a solution if and only if there is an Eulerian sandwich graph for D₁ and D₂ or equivalently, $d^{-}_{D_{1}}(X) \leq d^{+}_{D_{2}}(X)$ for all X⊆V. By $d^{+}_{D_{2}}(X) = d^{+}_{A_{2}}(X)+ d_{E_{2}}(X)$, we have $d^{-}_{A_{1}}(X) -d^{+}_{A_{2}}(X) \leq d_{E_{2}}(X)$ for all X⊆V. Note that $d_{E_{2}}(X)+d^{+}_{A_{2}}(X)-d^{-}_{A_{1}}(X)$ is a submodular function, and hence by Theorem 4, (14) can be verified in polynomial time. □

m-orientation sandwich problems

In this section we consider the sandwich problem where the property Π is to have an orientation of given in-degrees.

m-Orientation

m-Orientation Sandwich ProblemInstance: Given undirected graphs G₁=(V,E₁) and G₂=(V,E₂) with E₁⊆E₂ and a non-negative integer vector m on V. Question: Does there exist a sandwich graph G=(V,E) (E₁⊆E⊆E₂) that has an orientation $\vec{G}$ whose in-degree vector is m that is $d^{-}_{\vec{G}}(v)=m(v)$ for all v∈V?

Characterization: We prove the following theorem.

Theorem 11

The following assertions are equivalent.

• (a) Them-Orientation Sandwich Problemhas aYesanswer.

• (b) E₁is independent in$M_{\bar{m}}$and$M_{\bar{m}}$has an independent set of sizem(V).

• (c) $r_{\bar{m}}(E_{1}) = |E_{1}|$and$r_{\bar{m}}(E_{2})\geq m(V)$.

• (d) $i_{E_{1}}(X) \leq m(X) \leq e_{E_{2}}(X)$for allX⊆V.

Proof

(a) Implies (d). Let X⊆V. Since each edge of G₁ in X contributes 1 to m(X), we have $i_{E_{1}}(X) \leq m(X)$. On the other hand, the edges of G₂ that have no end-vertex in X cannot contribute 1 to m(X), so we have $m(X) \leq e_{E_{2}}(X)$.

(d) Implies (c). Let F be a subset of E₁ and X=V(F). The condition $i_{E_{1}}(X) \leq m(X)$ implies $|F| \leq m(V(F)) = \bar{m}(F)$, that is, $|E_{1}| \leq\bar{m}(F) + |E_{1}\setminus F|$. By Theorem 7, $r_{\bar{m}}(E_{1}) \geq|E_{1}|$, or equivalently $r_{\bar{m}}(E_{1}) = |E_{1}|$. Let now F be a subset of E₂ and X=V∖V(F). The condition $m(X) \leq e_{E_{2}}(X)$ implies that $m(V) \leq m(V(F)) +e_{E_{2}}(V-V(F))\leq\bar{m}(F) + |E_{2}\setminus F|$. By Theorem 7, $r_{\bar{m}}(E_{2}) \geq m(V)$.

(c) Implies (b). By definition.

(b) Implies (a). By (b), E₁ is independent in $M_{\bar{m}}$ and there exists an independent in $M_{\bar{m}}$ of size m(V). Therefore, by (I3), there exists an independent set E of size m(V) that contains E₁. By Theorem 3 and Claim 2, E is a solution of the m-orientation Sandwich Problem. □

We say that a subset F of E₀ is feasible if (V,F∪E₁) has an m-orientation. The next corollary of Theorem 11 characterizes the feasible sets.

Corollary 1

If them-Orientation Sandwich Problemhas aYesanswer, then a subsetFofE₀is feasible if and only ifFis a base of the matroid$M_{\bar{m}}/E_{1}$.

Complexity: The condition (d) of Theorem 11 can be verified in polynomial time by Theorem 4, so the m-Orientation Sandwich Problem is in P.

Optimization: The minimum cost version of the problem can be solved in polynomial time. First, we find an optimal feasible subset F by greedy algorithm. Then we can orient the edges of F∪E₁ using a known algorithm. (See [10] for example.)

Corollary 1 and the matroid intersection algorithm of Edmonds [4] imply that the two following simultaneous versions of the m-orientation Sandwich Problem are also in P.

Simultaneousm-Orientation Sandwich Problem 1Instance: Given two edge-disjoint undirected subgraphs G₁=(V,E₁) and G₂=(V,E₂) of an undirected graph G₃=(V,E₃) and two non-negative integer vectors m₁ and m₂ on V. Question: Do there exist simultaneously edge-disjoint sandwich graphs $\hat{G}_{1}=(V,\hat{E}_{1})$$(E_{1}\subseteq\hat{E}_{1}\subseteq E_{3})$ and $\hat{G}_{2}=(V,\hat{E}_{2})$$(E_{2}\subseteq\hat{E}_{2}\subseteq E_{3})$ such that $\hat{G}_{i}$ has an orientation whose in-degree vector is m _i for i∈{1,2}?

Note that the two input matroids for the matroid intersection algorithm must be taken as $(M^{G_{1}}_{\bar{m}_{1}}/E_{1})\setminus E_{2}$ and the dual matroid of $(M^{G_{2}}_{\bar{m}_{2}}/E_{2})\setminus E_{1}$.

Simultaneousm-Orientation Sandwich Problem 2Instance: Given two undirected subgraphs G₁=(V,E₁) and G₂=(V,E₂) of an undirected graph G₃=(V,E₃) and two non-negative integer vectors m₁ and m₂ on V. Question: Does there exist an edge set F in E₃∖(E₁∪E₂) such that the graph G _i =(V,E _i ∪F) admits an orientation whose in-degree vector is m _i for i∈{1,2}?

Strongly connected m-orientation

Strongly Connectedm-Orientation Sandwich ProblemInstance: Given undirected graphs G₁=(V,E₁) and G₂=(V,E₂) with E₁⊆E₂ and a non-negative integer vector m on V. Question: Does there exist a sandwich graph G=(V,E) (E₁⊆E⊆E₂) that has a strongly connected orientation $\vec{G}$ whose in-degree function is m?

Complexity: It is NP-complete because the special case E₁=∅,m(v)=1∀v∈V is equivalent to decide if G₂ has a Hamiltonian cycle.

(m₁,m₂)-orientation

(m₁,m₂)-Orientation Sandwich ProblemInstance: Given undirected graphs G₁=(V,E₁) and G₂=(V,E₂) with E₁⊆E₂ and non-negative integer vectors m₁ and m₂ on V. Question: Does there exist a sandwich graph G=(V,E) (E₁⊆E⊆E₂) that has an orientation $\vec{G}$ whose in-degree vector is m₁ and whose out-degree vector is m₂?

Complexity: The problem is NP-complete since it contains as a special case (E₁=∅) the NP-complete problem of [19].

Mixed m-orientation

Mixedm-Orientation Sandwich ProblemInstance: Given mixed graphs G₁=(V,E₁∪A₁) and G₂=(V,E₂∪A₂) with E₁⊆E₂, A₁⊆A₂ and an non-negative integer vector m on V. Question: Does there exist a sandwich mixed graph G=(V,E∪A) with E₁⊆E⊆E₂ and A₁⊆A⊆A₂ that has an orientation $\overrightarrow {G}=(V,\overrightarrow{E} \cup A)$ whose in-degree vector is m?

Characterization: Suppose that E₁⊆E⊆E₂ has been chosen and oriented, then the problem is reduced to the Directed Degree Constrained Sandwich Problem with $m_{1}(v)=m(v) - d^{-}_{\overrightarrow{E}}(v)$ which, by Theorem 8, has a solution if and only if $d^{-}_{A_{2}}(v) \geq m(v) -d^{-}_{\overrightarrow{E}}(v) \geq d^{-}_{A_{1}}(v)$ for all v∈V. Hence the Mixedm-orientation Sandwich Problem has a solution if and only if there exists E₁⊆E⊆E₂ which admits an orientation $\overrightarrow{E}$ with $m(v)-d^{-}_{A_{1}}(v) \geq d^{-}_{\overrightarrow{E}}(v) \geq m(v)-d^{-}_{A_{2}}(v)$ for all v∈V. Let m₂:V→ℤ satisfy $m(v)-d^{-}_{A_{2}}(v)\leq m_{2}(v)\leq m(v)-d^{-}_{A_{1}}(v)$. By Theorem 11, there exists E₁⊆E⊆E₂ which admits an orientation $\overrightarrow {E}$ with $d^{-}_{\overrightarrow{E}}(v) = m_{2}(v)$ if and only if $i_{E_{1}}(X) \leq m_{2}(X) \leq e_{E_{2}}(X)$ for all X⊆V. Therefore we have

Claim 3

TheMixedm-orientation Sandwich Problemhas aYesanswer if and only if there exists an integer-valued functionm₂:V→ℤ such that, ∀v∈Vand ∀X⊆V,

Claim 4

The pair$(i_{E_{1}}, e_{E_{2}})$is a strong pair.

Proof

Let X,Y,Z be three pairwise disjoint subset of V. We show that $e_{E_{2}}(X\cup Z)-i_{E_{1}}(Y\cup Z) \geq e_{E_{2}}(X) - i_{E_{1}}(Y)$. In fact, we have $i_{E_{1}}(Y\cup Z) - i_{E_{1}}(Y) = i_{E_{1}}(Z) +d_{E_{1}}(Y,Z) \leq i_{E_{2}}(Z) + d_{E_{2}}(Y,Z)$, and $e_{E_{2}}(X\cup Z) -e_{E_{2}}(X) = i_{E_{2}}(Z) + d_{E_{2}}(Z) - d_{E_{2}}(X,Z)$. As X,Y,Z are pairwise disjoint, $d_{E_{2}}(Y,Z) + d_{E_{2}}(X,Z) \leq d_{E_{2}}(Z)$. The claim follows by Claim 1. □

By Claim 3, 4 and Theorem 6 applied for $\alpha(v)= m(v)-d^{-}_{A_{2}}(v), \beta (v)=m(v)-d^{-}_{A_{1}}(v), p=i_{E_{1}}, b=e_{E_{2}}$, we have

Theorem 12

TheMixedm-orientation Sandwich Problemhas aYesanswer if and only if

(15)

for every subsetXofV.

Note that Theorem 12 implies Theorems 8 and 11.

Complexity: The condition (15) can be verified in polynomial time by Theorem 4. If it is satisfied, then a vector m₂ satisfying the conditions in Claim 3 can be found using a greedy algorithm for g-polymatroids. Then we find and orient an edge set E (E₁⊆E⊆E₂) with in-degree m₂ (m-orientation Sandwich Problem). Last, we choose an arc set A (A₁⊆A⊆A₂) such that $d^{-}_{A}(v)=m_{1}(v)=m(v)-m_{2}(v)$, for all v∈V (Directed Degree Constrained Sandwich Problem).

Contracting sandwich problems

In this section, we propose to consider a new type of sandwich problem. Instead of deleting edges from the optional graph, we are interested in contracting edges. We solve the problem for the property Π being a bipartite graph.

Contracting Sandwich ProblemInstance: Given an undirected graph G=(V,E) and E₀⊆E. Question: Does there exist F⊆E₀ such that contracting F results in a bipartite graph?

Complexity: Since a graph is bipartite if and only if all its cycles have an even length, the problem is equivalent to finding F⊆E₀ such that, for all cycles C, |C∩F|≡|C| mod 2.

Fix a spanning forest T of G. For e∈E∖T, denote C(T,e) the unique cycle contained in T∪e. By [18, Theorem 9.1.2], if C is a cycle of G then C=Δ_e∈CC(T,e), where Δ denotes the symmetric difference of sets. Therefore, |C∩F|≡∑_e∈C|C(T,e)∩F| mod 2. Let $\mathcal{C}_{T}$ denote the collection of cycles C(T,e) of G. The problem is reduced to finding F⊆E₀ such that, for all $C\in\mathcal{C}_{T}$, |C∩F|≡|C| mod 2, or equivalently, finding an F′(=E∖F)⊇E₁=E∖E₀ such that |F′∩C|≡0 mod 2, for all $C\in\mathcal{C}_{T}$.

Consider now the matrix M defined as the following. The rows of M correspond to $C \in\mathcal{C}_{T}$ and the columns correspond to the edges of G; the entry M _Ce is 1 if e∈C and is 0 otherwise. For X⊆E, let χ _X denote the characteristic vector of X. For a vector x∈{0,1}^E, let x_|X denote the projection of x on X. Let 1 be the all-one vector in {0,1}^E. A subset F′⊆E satisfies |F′∩C|≡0 mod 2, for all $C\in\mathcal{C}_{T}$, if and only if χ_F′∈KerM in $\mathbb{F}_{2}$. Such an F′ is the solution of the Contracting Sandwich Problem if and only if $\chi_{F'|E_{1}} = \mathbf {1}_{|E_{1}}$.

Let B be a basis of the kernel of M in $\mathbb{F}_{2}$. (This can be computed in polynomial time using the Gauss elimination.) Consider the projections B′ of B on E₁. Then the Contracting Sandwich Problem has a solution if and only if $\mathbf{1}_{|E_{1}}$ is in the subspace of $\{0,1\}^{E_{1}}$ spanned by B′, that is, $\mathrm{rank} B'= \mathrm{rank} B'\cup\mathbf{1}_{|E_{1}}$. This can be decided in polynomial time using the Gauss elimination. We conclude that the Contracting Sandwich Problem is in P.

We finish with a related problem. For a fixed integer k, solving the Contracting Sandwich Problem when E₀=E with extra requirement |F|≤k is known to be tractable in polynomial time [14]. However, the authors mention that finding a solution of minimum cardinality is NP-complete.