Sandwich problems on orientations
 Olivier Durand de Gevigney^{1}Email author,
 Sulamita Klein^{2},
 VietHang Nguyen^{1} and
 Zoltán Szigeti^{1}
https://doi.org/10.1007/s1317301200657
© The Brazilian Computer Society 2012
Received: 1 February 2012
Accepted: 3 February 2012
Published: 28 February 2012
Abstract
The graph sandwich problem for property Π is defined as follows: Given two graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) such that E_{1}⊆E_{2}, is there a graph G=(V,E) such that E_{1}⊆E⊆E_{2} 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 indegrees of a graph. We present a characterization and a polynomialtime algorithm for solving the morientation sandwich problem.
Keywords
1 Introduction
Given two graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) with the same vertex set V and E_{1}⊆E_{2}, a graph G=(V,E) is called a sandwich graph for the pair G_{1},G_{2} if E_{1}⊆E⊆E_{2}. The graph sandwich problem for property Π is defined as follows [12]:
Graph Sandwich Problem for Property Π Instance: Given undirected graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) with E_{1}⊆E_{2}. Question: Is there a graph G=(V,E) such that E_{1}⊆E⊆E_{2} and G satisfies property Π?
We call E_{1} the mandatory edge set, E_{0}=E_{2}∖E_{1} the optional edge set and E_{3} the forbidden edge set, where E_{3} denotes the set of edges of the complementary graph \(\overline{G}_{2}\) of G_{2}. Thus any sandwich graph G=(V,E) for the pair G_{1},G_{2} 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 G_{1}=G_{2}=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 indegrees 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 Tjoins, 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 morientation, i.e., given undirected graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) with E_{1}⊆E_{2} and a nonnegative integer vector m on V, we show that it is polynomial to decide whether there exists a sandwich graph G=(V,E) (E_{1}⊆E⊆E_{2}) that has an orientation \(\vec{G}\) whose indegree vector is m that is \(d^{}_{\vec{G}}(v)=m(v)\) for all v∈V. This result stands in contrast to the strongly connected morientation sandwich problem which we show is NPcomplete. 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.
2 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 endvertex 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) endvertices in X is denoted by or i_{ E }(X) or simply i(X) (resp. ). It is wellknown that (1) is satisfied for all X,Y⊆V,
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 Tjoin 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 Tjoin can be found in polynomial time by Edmonds and Johnson’s algorithm [5].
 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 indegree (resp. outdegree ) 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 endvertices in X is denoted by . The following equality will be used frequently without reference:
We say that a vector m on V is the indegree vector of D if \(d^{}_{D}(v)=m(v)\) for all v∈V. Let us recall that \(d^{}_{D}(X)\) is the indegree function of D. Let f be a nonnegative integer vector on V. An arc set F of D is called a directedffactor of D if f is the indegree vector of D(F), that is, \(d^{}_{F}(v)=f(v)\) for all v∈V.
We say that D is Eulerian if the indegree of v is equal to the outdegree 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.
Theorem 1
(Hoffmann [15])
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])
The following theorem characterizes graphs having an orientation with a given indegree vector.
Theorem 3
(Hakimi [13])
Claim 1
The degree functiond_{ G }(Z) of an undirected graphGand the indegree function\(d^{}_{D}(Z)\)of a directed graphDare submodular and the functioni(Z) is supermodular.
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.
Theorem 6
(Frank, Tardos [11])
The intersection of an integral gpolymatroidQ(p,b) and an integral boxQ(α,β) is an integral gpolymatroid. 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 \(F'\cup f\in{\mathcal{F}}\).
 (I1)
 (B1)
\({\mathcal{B}}\neq\emptyset\),
 (B2)
if \(B, B'\in{\mathcal{B}}\) and b∈B∖B′, then there exists b′∈B′∖B such that \((Bb)\cup b'\in{\mathcal{B}}\).
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 {V_{1},…,V_{ l }} be a partition of V and a_{1},…,a_{ l } a set of nonnegative 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={v_{1},…,v_{ n }}. We can find a minimum cost basis F_{ n } of M in polynomial time by the greedy algorithm: take a nondecreasing order of the elements of V:c(v_{1})≤…≤c(v_{ n }). Let F_{0} be empty and for i=1,…,n, let F_{ i }=F_{i−1}+v_{ i } if \(F_{i1}+v_{i}\in {\mathcal{F}}\), otherwise let F_{ i }=F_{i−1}.
If M_{1} and M_{2} are two matroids on the same ground set V, then we can find a common basis of M_{1} and M_{2} in polynomial time (if there exists one) by the matroid intersection algorithm of Edmonds [4].
Theorem 7
(Edmonds, Rota [18])
Given an undirected graph G=(V,E) and a nonnegative 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, nondecreasing 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\(M_{\bar{m}}\).
3 Degree constrained sandwich problems
Before studying sandwich problems on orientations of given indegrees, 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 ffactor 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 indegree of the head and not of the tail, while the addition of an edge in an undirected graph contributes to the degree of both endvertices. This section contains no new results, we added it for the sake of completeness.
3.1 Undirected graphs
Undirected Degree Constrained Sandwich ProblemInstance: Given undirected graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) with E_{1}⊆E_{2} and a nonnegative integer vector f on V. Question: Does there exist a sandwich graph G=(V,E) (E_{1}⊆E⊆E_{2}) 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 G_{0}=(V,E_{0}).
Characterization: The general ffactor theorem due to Tutte [25] can be applied to get a characterization.
Optimization: The minimum cost ffactor problem in undirected graphs can be solved in polynomial time, see Schrijver [20].
Simultaneous Undirected Degree Constrained Sandwich ProblemInstance: Given two edgedisjoint graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) in G_{3}=(V,E_{3}) and two nonnegative integer vectors f_{1} and f_{2} 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 NPcomplete because it contains as a special case whether there exist two edgedisjoint perfect matchings so 3edgecolorability of 3regular graphs. Indeed, let G=(V,E) be an arbitrary 3regular graph. Let G_{1} and G_{2} be the edgeless graph on V, G_{3}=G and f_{1}(v)=f_{2}(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 edgedisjoint perfect matchings of G or equivalently, if there exists a 3edgecoloring of G. Since the problem of 3edgecolorability of 3regular graphs is NPcomplete [16], so is our problem.
3.2 Directed graphs
Directed Degree Constrained Sandwich ProblemInstance: Given directed graphs D_{1}=(V,A_{1}) and D_{2}=(V,A_{2}) with A_{1}⊆A_{2} and a nonnegative integer vector f on V. Question: Does there exist a sandwich graph D=(V,A) (A_{1}⊆A⊆A_{2}) 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 D_{0}=(V,A_{0}), 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 arcdisjoint directed graphs D_{1}=(V,A_{1}) and D_{2}=(V,A_{2}) in D_{3}=(V,A_{3}) and two nonnegative integer vectors f_{1} and f_{2} 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 D_{1}=(V,A_{1}) and D_{2}=(V,A_{2}) with A_{1}⊆A_{2} and two nonnegative integer vectors f and g on V. Question: Does there exist a sandwich graph D=(V,A) (A_{1}⊆A⊆A_{2}) such that \(d^{}_{D}(v)=f(v)\) and \(d^{+}_{D}(v)=g(v)\) for all v∈V.
Complexity: The feasible arc sets for the indegree constraint form the basis of a partition matroid and the feasible arc sets for the outdegree 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].
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 Tjoins, while the directed case to circulations. We show that in both cases the simultaneous versions are NPcomplete.
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.
4.1 Undirected graphs
Undirected Eulerian Sandwich ProblemInstance: Given undirected graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) with E_{1}⊆E_{2}. Question: Does there exist a sandwich graph G=(V,E) (E_{1}⊆E⊆E_{2}) 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 G_{0}.
Characterization: The answer is Yes if and only if each connected component of G_{0} contains an even number of vertices of \(T_{G_{1}}\).
Optimization: The minimum cost Tjoin problem can be solved in polynomial time [5].
Simultaneous Undirected Eulerian Sandwich ProblemInstance: Given two edgedisjoint graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) in G_{3}=(V,E_{3}). 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 NPcomplete because it contains as a special case whether there exist two edgedisjoint perfect matchings so 3colorability of 3regular graphs. Indeed, let G=(V,E) be an arbitrary 3regular graph. Let G_{3} be obtained from G by adding 2 edgedisjoint perfect matchings M_{1} and M_{2} to G, let G_{1}=(V,M_{1}) and G_{2}=(V,M_{2}). 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 edgedisjoint perfect matchings of G or equivalently, if there exists a 3edgecoloring of G. Since the problem of 3edgecolorability of 3regular graphs is NPcomplete [16], so is our problem.
4.2 Directed graphs
Directed Eulerian Sandwich ProblemInstance: Given directed graphs D_{1}=(V,A_{1}) and D_{2}=(V,A_{2}) with A_{1}⊆A_{2}. Question: Does there exist a sandwich graph D=(V,A) (A_{1}⊆A⊆A_{2}) 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∈A_{1} and f(e)=0,g(e)=1 if e∈A_{0}. This way the arcs of A_{1} are forced and the arcs of A_{0} 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 arcdisjoint directed graphs D_{1}=(V,A_{1}) and D_{2}=(V,A_{2}) in D_{3}=(V,A_{3}). 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 NPcomplete, it contains as a special case (D_{1}=(V,t_{1}s_{1}), D_{2}=(V,t_{2}s_{2}) and D_{3}=D) the following directed 2commodity integral flow problem that is NPcomplete [6]: Given a directed graph D and two pairs of vertices, s_{1},t_{1} and s_{2},t_{2}, decide whether there exist a path from s_{1} to t_{1} and a path from s_{2} to t_{2} that are arcdisjoint.
4.3 Mixed graphs
Mixed Eulerian Sandwich ProblemInstance: Given mixed graphs H_{1}=(V,E_{1}∪A_{1}) and H_{2}=(V,E_{2}∪A_{2}) with E_{1}⊆E_{2},A_{1}⊆A_{2}. Question: Does there exist a sandwich mixed graph H=(V,E∪A) (E_{1}⊆E⊆E_{2},A_{1}⊆A⊆A_{2}) that has an Eulerian orientation?
Complexity: We provide two special cases that can be treated, while the general problem remains open.
Special Case 1: E_{1}=E_{2}=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
Proof
Special Case 2: E_{1}=∅.
Characterization+Complexity: It is in P because it can be reformulated as the following problem: We create two copies of each edge in E_{2} 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,E_{2}∪A_{2}) has a subgraph containing (V,A_{1}) 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,A_{1}). Indeed, in such a graph, if every edge of E_{2} is used at most once, we are done. If some edge of E_{2} is used twice, as two arcs in opposite directions, we can just remove these two arcs, the obtained graph remaining Eulerian and containing (V,A_{1}). Now applying the result for Directed Eulerian Sandwich Problem we have
Theorem 10
Proof
Let D_{1}=(V,A_{1}) and \(D_{2} = (V, A_{2} \cup\overrightarrow{E^{2}_{2}})\). By the arguments above, the Mixed Eulerian Sandwich Problem with E_{1}=∅ has a solution if and only if there is an Eulerian sandwich graph for D_{1} and D_{2} 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. □
5 morientation sandwich problems
In this section we consider the sandwich problem where the property Π is to have an orientation of given indegrees.
5.1 mOrientation
mOrientation Sandwich ProblemInstance: Given undirected graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) with E_{1}⊆E_{2} and a nonnegative integer vector m on V. Question: Does there exist a sandwich graph G=(V,E) (E_{1}⊆E⊆E_{2}) that has an orientation \(\vec{G}\) whose indegree vector is m that is \(d^{}_{\vec{G}}(v)=m(v)\) for all v∈V?
Characterization: We prove the following theorem.
Theorem 11
 (a)
ThemOrientation Sandwich Problemhas aYesanswer.
 (b)
E_{1}is 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 G_{1} in X contributes 1 to m(X), we have \(i_{E_{1}}(X) \leq m(X)\). On the other hand, the edges of G_{2} that have no endvertex 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 E_{1} 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, \(r_{\bar{m}}(E_{1}) \geqE_{1}\), or equivalently \(r_{\bar{m}}(E_{1}) = E_{1}\). Let now F be a subset of E_{2} and X=V∖V(F). The condition \(m(X) \leq e_{E_{2}}(X)\) implies that \(m(V) \leq m(V(F)) +e_{E_{2}}(VV(F))\leq\bar{m}(F) + E_{2}\setminus F\). By Theorem 7, \(r_{\bar{m}}(E_{2}) \geq m(V)\).
(c) Implies (b). By definition.
(b) Implies (a). By (b), E_{1} 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 E_{1}. By Theorem 3 and Claim 2, E is a solution of the morientation Sandwich Problem. □
We say that a subset F of E_{0} is feasible if (V,F∪E_{1}) has an morientation. The next corollary of Theorem 11 characterizes the feasible sets.
Corollary 1
If themOrientation Sandwich Problemhas aYesanswer, then a subsetFofE_{0}is 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 mOrientation 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∪E_{1} 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 morientation Sandwich Problem are also in P.
SimultaneousmOrientation Sandwich Problem 1Instance: Given two edgedisjoint undirected subgraphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) of an undirected graph G_{3}=(V,E_{3}) and two nonnegative integer vectors m_{1} and m_{2} on V. Question: Do there exist simultaneously edgedisjoint 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 indegree 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 \((M^{G_{2}}_{\bar{m}_{2}}/E_{2})\setminus E_{1}\).
SimultaneousmOrientation Sandwich Problem 2Instance: Given two undirected subgraphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) of an undirected graph G_{3}=(V,E_{3}) and two nonnegative integer vectors m_{1} and m_{2} on V. Question: Does there exist an edge set F in E_{3}∖(E_{1}∪E_{2}) such that the graph G_{ i }=(V,E_{ i }∪F) admits an orientation whose indegree vector is m_{ i } for i∈{1,2}?
5.2 Strongly connected morientation
Strongly ConnectedmOrientation Sandwich ProblemInstance: Given undirected graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) with E_{1}⊆E_{2} and a nonnegative integer vector m on V. Question: Does there exist a sandwich graph G=(V,E) (E_{1}⊆E⊆E_{2}) that has a strongly connected orientation \(\vec{G}\) whose indegree function is m?
Complexity: It is NPcomplete because the special case E_{1}=∅,m(v)=1∀v∈V is equivalent to decide if G_{2} has a Hamiltonian cycle.
5.3 (m_{1},m_{2})orientation
(m_{1},m_{2})Orientation Sandwich ProblemInstance: Given undirected graphs G_{1}=(V,E_{1}) and G_{2}=(V,E_{2}) with E_{1}⊆E_{2} and nonnegative integer vectors m_{1} and m_{2} on V. Question: Does there exist a sandwich graph G=(V,E) (E_{1}⊆E⊆E_{2}) that has an orientation \(\vec{G}\) whose indegree vector is m_{1} and whose outdegree vector is m_{2}?
Complexity: The problem is NPcomplete since it contains as a special case (E_{1}=∅) the NPcomplete problem of [19].
5.4 Mixed morientation
MixedmOrientation Sandwich ProblemInstance: Given mixed graphs G_{1}=(V,E_{1}∪A_{1}) and G_{2}=(V,E_{2}∪A_{2}) with E_{1}⊆E_{2}, A_{1}⊆A_{2} and an nonnegative integer vector m on V. Question: Does there exist a sandwich mixed graph G=(V,E∪A) with E_{1}⊆E⊆E_{2} and A_{1}⊆A⊆A_{2} that has an orientation \(\overrightarrow {G}=(V,\overrightarrow{E} \cup A)\) whose indegree vector is m?
Characterization: Suppose that E_{1}⊆E⊆E_{2} 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 Mixedmorientation Sandwich Problem has a solution if and only if there exists E_{1}⊆E⊆E_{2} 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 m_{2}: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 E_{1}⊆E⊆E_{2} 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
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
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 m_{2} satisfying the conditions in Claim 3 can be found using a greedy algorithm for gpolymatroids. Then we find and orient an edge set E (E_{1}⊆E⊆E_{2}) with indegree m_{2} (morientation Sandwich Problem). Last, we choose an arc set A (A_{1}⊆A⊆A_{2}) such that \(d^{}_{A}(v)=m_{1}(v)=m(v)m_{2}(v)\), for all v∈V (Directed Degree Constrained Sandwich Problem).
6 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 E_{0}⊆E. Question: Does there exist F⊆E_{0} 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⊆E_{0} 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∈C}C(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⊆E_{0} such that, for all \(C\in\mathcal{C}_{T}\), C∩F≡C mod 2, or equivalently, finding an F′(=E∖F)⊇E_{1}=E∖E_{0} 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 allone 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 E_{1}. 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 E_{0}=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 NPcomplete.
Declarations
Acknowledgements
This research was partially supported by CNPq, CAPES (Brazil)/COFECUB (France), FAPERJ.
Authors’ Affiliations
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