on the p-median polytope and the directed odd cycle

Discrete Optimization 22 (2016) 206–224

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Discrete Optimization

www.elsevier.com/locate/disopt

On the p-median polytope and the directed odd cycle inequalities:Triangle-free oriented graphs

Mourad Baıoua,∗, Francisco Barahonab

a CNRS, and Universite Clermont II, Campus Universitaire des Cezeaux, 1 rue de la Chebarde TSA 60125CS 60026, 63178 AUBIERE CEDEX, Franceb IBM T. J. Watson research Center, Yorktown Heights, NY 10589, USA

a r t i c l e i n f o

Article history:Available online 21 December 2015

Keywords:p-median problemUncapacitated facility locationproblemOdd cycle inequalities

a b s t r a c t

We study the effect of the odd directed cycle inequalities in the description of thepolytope associated with the p-median problem. We treat oriented graphs, i.e., if(u, v) is in the arc-set, then (v, u) is not in the arc-set. We characterize the orientedgraphs for which the obvious linear relaxation together with the directed odd cycleinequalities describe the p-median polytope. This study has two parts: in this paperwe treat triangle-free graphs, then in a second paper we use induction on the numberof triangles to treat general oriented graphs.

© 2015 Elsevier B.V. All rights reserved.

1. Introduction

Let G = (V,A) be a directed graph, not necessarily connected, where each arc (u, v) ∈ A has an associatedcost c(u, v). The p-median problem (pMP) consists of selecting p nodes, usually called centers, and then assigneach nonselected node to a selected node. The goal is to select p nodes that minimize the sum of the costsyielded by the assignment of the nonselected nodes. This problem has several applications such as locationof bank accounts [1], placement of web proxies in a computer network [2], semistructured data bases [3,4].When the number of centers is not specified and each opened center induces a given cost, this is called theuncapacitated facility location problem (UFLP).

The facets of p-median polytope have been studied in [5,6]. The facets of the uncapacitated facilitylocation polytope have been studied in [7–11]. For more references on the p-median problem see [12]. In [13]we studied the effect of the odd directed cycle inequalities in the description of the polytopes associatedwith the pMP and the UFLP for the class of Y -free graphs. In this paper we treat oriented graphs, i.e., if(u, v) ∈ A =⇒ (v, u) ∈ A, for each arc (u, v) ∈ A. Our goal is to characterize the oriented graphs for

∗ Corresponding author.E-mail addresses: [email protected] (M. Baıou), [email protected] (F. Barahona).

http://dx.doi.org/10.1016/j.disopt.2015.07.0061572-5286/© 2015 Elsevier B.V. All rights reserved.

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M. Baıou, F. Barahona / Discrete Optimization 22 (2016) 206–224 207

which the obvious integer linear programming formulation together with the odd directed cycle inequalitiescompletely describe the polytope associated with the pMP. This is done in two parts: in this paper we treattriangle-free graphs, then in a second paper we use induction on the number of triangles to treat generaloriented graphs.

If we associate the variables y to the nodes, and the variables x to the arcs, the following is a linearrelaxation of the pMP.

v∈Vy(v) = p, (1)

y(u) +

v:(u,v)∈A

x(u, v) = 1 ∀u ∈ V, (2)

x(u, v) ≤ y(v) ∀(u, v) ∈ A, (3)0 ≤ y(v) ∀v ∈ V, (4)x(u, v) ≥ 0 ∀(u, v) ∈ A. (5)

For a directed cycle

C = v1, (v1, v2), v2, (v2, v3), . . . , vk−1, (vk−1, vk), vk, (vk, v1), v1,

we denote by A(C) the set of arcs in C. We say that C is odd if k is odd. We consider the following additionalset of inequalities

a∈A(C)

x(a) ≤ |A(C)| − 12 for each odd directed cycle C. (6)

Inequalities (6) are 0− 1/2 Chvatal–Gomory cuts, in the terminology of [14]. They can be obtained from(1)–(5) using Chvatal–Gomory procedure with multipliers 0 or 1/2. They are among the simplest sets ofinequalities that one can add to improve the relaxation (1)–(5).

This type of inequalities are classical and they appear as valid inequalities for several combinatorialoptimization problems, for example for the stable set [15] and the max-cut [16] problems. They have beenintroduced for the uncapacitated facility location problem in [9,10,8] together with their lifting to obtainstronger inequalities. Their separation problem may be reduced to the separation problem of the odd holeinequalities that may be solved in polynomial time [17]. More general inequalities may be obtained from theg-odd Y -cycles that are introduced later, for which their separation may be solved in polynomial time [14].The separation problem for the lifted odd hole inequalities as defined in [9,10,8] has been studied in [18] andit has been shown that they may be separated in polynomial time. Here we study the oriented graphs forwhich adding inequalities (6) is sufficient to obtain the p-median polytope.

Denote by Pp(G) the polytope defined by (1)–(5), let PCp(G) be the polytope defined by (1)–(6), andlet pMP (G) be the convex hull of Pp(G) ∩ {0, 1}|V |+|A|. The p-median polytope of a graph G is pMP (G).In general we have pMP (G) ⊆ PCp(G) ⊆ Pp(G).

Our goal is to characterize all the graphs such that

pMP (G) = PCp(G).

In [19] we characterized the graphs for which pMP (G) = Pp(G). The optimization problem over Pp(G) ispolynomially solvable. Moreover Pp(G) is used in most methods for solving the p-median problem, as a toolfor providing tight lower bounds. Since the odd directed cycle inequalities may be separated in polynomialtime, the optimization over PCp(G) is also polynomial and provide tighter lower bounds. Therefore it isinteresting to characterize those graphs for which optimizing over PCp(G) suffices to solve the problem.Unfortunately, the techniques developed in [19] are not enough for this case. Of course characterizing

208 M. Baıou, F. Barahona / Discrete Optimization 22 (2016) 206–224

Fig. 1. Forbidden configurations for PCp(G).

the graphs for which pMP (G) = PCp(G) is harder than the characterization of the graphs for whichpMP (G) = Pp(G). As we will see, in this paper we will develop a new way to tackle the problem thatpermits the use of the result in [19]; this considerably helps to simplify the proof. We will focus on orientedgraphs. The main result in this paper is the characterization of triangle-free oriented graphs, which is notan easy question even when restricted to this class of graphs, since the p-median problem still NP-hard intriangle-free graphs [20].

We need the following definitions. Let P (G) be the polytope defined by (2)–(5), let PC(G) be the polytopedefined by (2)–(6), and let UFLP (G) be the convex hull of P (G) ∩ {0, 1}|V |+|A|. We have

UFLP (G) ⊆ PC(G) ⊆ P (G).

In Fig. 1 we show two graphs and for each of them a fractional extreme point of PCp(G). The numbersnear the nodes correspond to the variables y and the x variables take value 1/2 in the graphs H1 and H2.These two graphs play a crucial role in our characterization.

We have to describe one more configuration where PCp(G) is not integral. First we need some definitionsand notations.

A simple cycle C is an ordered sequence v0, a0, v1, a1, . . . , at−1, vt, where

• vi, 0 ≤ i ≤ t− 1, are distinct nodes,• either vi is the tail of ai and vi+1 is the head of ai, or vi is the head of ai and vi+1 is the tail of ai, for

0 ≤ i ≤ t− 1, and• v0 = vt.

Let V (C) and A(C) denote the nodes and the arcs of C, respectively. By setting at = a0, we associatewith C three more sets as below.

• We denote by C the set of nodes vi, such that vi is the head of ai−1 and also the head of ai, 1 ≤ i ≤ t.• We denote by C the set of nodes vi, such that vi is the tail of ai−1 and also the tail of ai, 1 ≤ i ≤ t.• We denote by C the set of nodes vi, such that either vi is the head of ai−1 and also the tail of ai, or vi

is the tail of ai−1 and also the head of ai, 1 ≤ i ≤ t.

Notice that |C| = |C|. A cycle will be called g-odd if |C|+ |C| is odd, otherwise it will be called g-even.A cycle C with V (C) = C is a directed cycle, otherwise it is called a non-directed cycle. The notion of g-odd(g-even) cycles generalizes the notion of odd (even) directed cycles. The size of a cycle C is |V (C)|.


Fig. 2. A non-directed g-odd Y -cycle with an arc outside the cycle.

A path is defined in a similar way to a cycle, but without asking the condition v0 = vt. In a similar waywe define P , P and P , excluding v0 and vt. For a path P from v0 to vt, the nodes v1, . . . , vt−1 are calledinternal and the arc of P incident to v0 (resp. vt) is called the first (resp. last) arc of P . We say that P isg-odd (resp. g-even) if |P |+ |P | is odd (resp. even).

Definition 1. A simple cycle is called a Y -cycle if for every v ∈ C there is an arc (v, v) in A, where v is inV \ C.

Notice that when C = ∅, then C is a directed cycle and also a Y -cycle.In Fig. 2 we show a fractional extreme point of PCp(G) different from those given in Fig. 1. We have a

non-directed g-odd Y -cycle with an arc having its both endnodes not in the cycle. The values reported neareach node correspond to the y variables. The x variables take all the value 1/2. Here p = 4. This examplemay be extended to any configuration consisting of a non-directed g-odd Y -cycle and one extra arc.

Definition 2. A polytope is called integral if all its extreme points are integral.

Now we are ready to present the main result of this paper.

Theorem 3. Let G = (V,A) be a triangle-free oriented graph. Then PCp(G) is integral for any integer p ifand only if

(C1) it does not contain as a subgraph neither of the graphs H1 nor H2 of Fig. 1, and(C2) it does not contain a non-directed g-odd Y -cycle C and an arc (u, v) with neither u nor v in V (C).

We need the theorem below as a first step in the proof of Theorem 3.

Theorem 4. Let G = (V,A) be an oriented graph without a non-directed g-odd Y -cycle. Then PCp(G) isintegral for any integer p if and only if

(C1) it does not contain as a subgraph neither of the graphs H1 nor H2 of Fig. 1.

This paper is organized as follows. In Section 2 we give some definitions and preliminary results. InSection 3 we prove the necessity for both theorems. Then the proof of sufficiency is considered. First weprove it for Theorem 4. This is done in Section 4. Then we treat graphs with a non-directed g-odd Y -cycleand no triangles, that is the sufficiency for Theorem 3. This is presented in Section 5.

2. Preliminaries

In this section we present several definitions and technical lemmas.


2.1. Definitions

Let G = (V,A) be a simple directed graph. For W ⊂ V , we denote by δ+(W ) the set of arcs (u, v) ∈ A,with u ∈ W and v ∈ V \W . Also we denote by δ−(W ) the set of arcs (u, v), with v ∈ W and u ∈ V \W .We write δ+(v) and δ−(v) instead of δ+({v}) and δ−({v}), respectively. If there is a risk of confusion weuse δ+G and δ−G . A node u with δ+(u) = ∅ is called a pendent node. A node v is called an articulation nodeif its removal increases the number of connected components of G.

Let l : V ∪A→ Z be a labeling function that associates integer values to each node and arc of G. A vector(x, y) ∈ Pp(G) will be denoted by z, i.e., z(u) = y(u) for all u ∈ V and z(u, v) = x(u, v) for all (u, v) ∈ A.Given a vector z and a labeling function l, we define a new vector zl from z as follows:

zl(u) = z(u) + l(u)ϵ, for all u ∈ V, andzl(u, v) = z(u, v) + l(u, v)ϵ, for all (u, v) ∈ A,

where ϵ is a sufficiently small positive scalar.

Remark 5. When we assign labels to some nodes and arcs without specifying the labels of the remainingnodes and arcs, it means that they are assigned the label 0.

Definition 6. Let G(s, t) denote the graph obtained from G by removing the arc (s, t) and adding the arc(s, t′) with t′ a new pendent node.

Definition 7. When dealing with a vector z ∈ PCp(G), we say that the arc (u, v) is tight if

z(u, v) = z(v).

2.2. The labeling procedure

Let C = v0, a0, v1, a1, . . . , at−1, vt be a g-even cycle, not necessarily a Y -cycle. The following labelingprocedure will assign labels to the nodes and arcs in C.

• If C is a directed cycle then, set l(v0) ← 1; l(a0) ← −1. Otherwise, assume v0 ∈ C and setl(v0)← 0; l(a0)← 1.• For i = 1 to t− 1 do the following:

– If vi is the head of ai−1 and is the tail of ai, then l(vi)← l(ai−1), l(ai)← −l(vi).– If vi is the head of ai−1 and is the head of ai, then l(vi)← l(ai−1), l(ai)← l(vi).– If vi is the tail of ai−1 and is the head of ai, then l(vi)← −l(ai−1), l(ai)← l(vi).– If vi is the tail of ai−1 and is the tail of ai, then l(vi)← 0, l(ai)← −l(ai−1).

Next we give two useful properties of the labeling procedure. The first is given in the following remarkand is easy to see.

Remark 8. If C is a directed even cycle then l(at−1) = l(v0) = 1 andl(vi) = 0.

The second property is given in the following lemma and it concerns cycles that are not necessarilydirected.

Lemma 9 ([19]). If C is a g-even cycle of size t, then l(at−1) = −l(a0) = −1 andl(vi) = 0.


Fig. 3. Forbidden configurations for Pp(G).

2.3. Y -free graphs

Here we discuss a special class of graphs studied in [13]. A node t is called a Y -node, if there exists threedifferent nodes u1, u2, w in V such that (u1, t), (u2, t) and (t, w) belong to A. The graph G is called Y -freeif it does not contain a Y -node.

The following theorem is proved in [13]. The same result is given in [21] where the proof is based on thematching polytope.

Theorem 10. If G is a Y -free graph, then PCp(G), for any integer p, and PC(G) are integral.

2.4. On the linear relaxation of the pMP

In [19] we characterized the oriented graphs for which Pp(G) is integral, as follows.

Theorem 11. Let G = (V,A) be an oriented graph, then Pp(G) is integral if and only if

(i) it does not contain as a subgraph any of the graphs F1, F2 or F3 of Fig. 3, and(ii) it does not contain a g-odd Y -cycle C and an arc (u, v) with neither u nor v in V (C).

2.5. The intersection property for UFLP (G)

Let P be an integral polytope in Rn. Let q be an integer valued row vector in Rn such that the g.c.d. ofits components is one. For an integer p let Hp = {x ∈ Rn : qx = p}.

We say that P has the intersection property with respect to q, if for every integer p the polytope P ∩Hpis integral.

A stable set of an undirected graph H is a set of pairwise non adjacent nodes in H. The stable set polytopeof H denoted by SSP (H), is the convex hull of incidence vectors of stable sets of H. A claw-free graph is agraph that does not contain the bipartite graph K1,3 as an induced subgraph. The following result has beenproved in [22].

Theorem 12. The stable set polytope of an undirected graph H = (V,E) has the intersection property withrespect to

v∈V x(v) = p if and only if H is a claw-free graph.

Given a directed graph G = (V,A), we can build an undirected graph L(G) where for each arc of G thereis a node in L(G). Let a1 and a2 be two arcs in G, we put an edge between the two corresponding nodes ofL(G) in the following two cases:


(i) a1 and a2 have the same tail, or(ii) the tail of one coincides with the head of the other.

It is easy to see that there is a one to one correspondence between solutions of the UFLP in G and stablesets of L(G). It is easy to check the following lemma.

Lemma 13. An oriented graph G does not contain any of the graphs H1 nor H2 as a subgraph if and only ifL(G) is claw-free.

Combining Theorem 12 and Lemma 13, we obtain the following:

Theorem 14. For an oriented graph G, UFLP(G) has the intersection property with respect tov∈V y(v) =

p, if and only if G does not contain H1 nor H2 as a subgraph.

Before introducing some useful lemmas for the next section, the relationship of the uncapacitated facilitylocation polytope with some known results on the stable set polytope will be discussed. One can easily checkthat given an oriented triangle-free graph G satisfying conditions (C1) and (C2), its intersection graph L(G)is a quasi-line graph, that is the neighbors of any vertex may be divided into two cliques. From [23], onecan obtain a complete characterization of the stable set polytope for quasi-line graphs, then combining withTheorem 12 a complete characterization for pMP (G) may also be obtained. However such a characterizationinvolves other inequalities different from (6), among them the odd hole inequalities that come from g-oddY -cycles. Our result shows that only the odd hole inequalities that come from directed odd cycles (of coursewith the clique inequalities) are needed even when the underlying graph contains g-odd Y -cycles. In otherwords, any other inequality different form (6) is redundant in this case. This shows that L(G) is a h-prefectgraph, a graph is h-perfect if the clique and the odd hole inequalities suffice to describe its associated stableset polytope.

2.6. Some useful lemmas

Here we assume that the oriented graph G = (V,A) satisfies conditions (C1) and (C2) of Theorem 3 andz is a fractional extreme point of PCp(G).

Lemma 15. We may assume that z(u, v) > 0 for all (u, v) ∈ A.

Proof. Let (u, v) ∈ A with z(u, v) = 0. Let G′ = (V,A \ (u, v)), and z′ the restriction of z on G′. It is clearthat z′ ∈ PCp(G′). Suppose that z′ = 1/2z1 + 1/2z2 where z1, z2 ∈ PCp(G′), z1 = z2. Let z1 (resp. z2)be the vector obtained by adding a zero component to z1 (resp. z2). We have that z = 1/2z1 + 1/2z2 andz1 = z2.

Now let us see that z1 and z2 are in PCp(G). Clearly they satisfy (1)–(5), so we just have to see thatconstraints (6) are satisfied. Consider the directed odd cycle

C = {(wi, wi+1) | i = 1, . . . , 2l} ∪ {(u, v)},

where u = w2l+1 and v = w1. We have that

z1(w2i−1, w2i) + z1(w2i, w2i+1) ≤ 1, for i = 1, . . . , l.

This implies z1(C) ≤ l. The same is true for z2. Therefore z1 and z2 are in PCp(G).

We have then a contradiction because z is an extreme point. So z′ must be an extreme point of PCp(G′).Thus one may consider the pair z′ with G′ instead of z with G, since G′ has the same required propertiesas G. �


Lemma 16. We may assume that z(v) > 0 for all v ∈ V with |δ−(v)| ≥ 1.

Proof. It is straightforward from Lemma 15 and constraints (4). �

Lemma 17. We may assume that |δ−(v)| = 1 for every pendent node v.

Proof. If v is a pendent node in G and δ−(v) = {(u1, v), . . . , (uk, v)}, we can split v into k pendent nodes{v1, . . . , vk} and replace every arc (ui, v) with (ui, vi). Then we define z′ such that z′(ui, vi) = z(ui, v),z′(vi) = 1, for all i, and z′(u) = z(u), z′(u,w) = z(u,w) for all other nodes and arcs. Let G′ be this newgraph. Notice that none of the arcs (ui, v) belongs to an odd directed cycle nor to a Y -cycle. It is easy tocheck that z′ is a fractional extreme point of PCp+k−1(G′) and that G′ satisfies conditions (C1) and (C2)of Theorem 3. �

We also need the following for PC(G) as a relaxation of UFLP (G).

Remark 18. The analogs of these three lemmas hold for PC(G).

Remark 19. It follows from the results in [24] that when dealing with UFLP (G) and PC(G), we can assumethat G is two-connected.

3. Necessity proof of Theorem 3

To prove that conditions (C1) and (C2) are necessary for Theorems 3 and 4, we just have to show afractional extreme point of PCp(G) whenever G violates (C1) or (C2).

In the following we give a fractional vector (x∗, y∗) of PCp(G) in each case. The proof that (x∗, y∗) is infact an extreme point is easy and it is left to the reader.

• Suppose that G = (V,A) contains H1 or H2 as a subgraph. Extend the fractional vector defined in eachcase of Fig. 1 by setting y∗(u) = 1 for all other nodes, and x∗(u, v) = 0 for all other arcs. Then (x∗, y∗)is a fractional extreme point of PC|V |−2(G).• Let C be a non-directed g-odd Y -cycle in G = (V,A) and let (t, w) ∈ A such that neither t nor w is inV (C). Define x∗(u, v) = 1/2 for each arc in A(C). Define y∗(v) = 1/2 if v ∈ C ∪ C and y∗(v) = 0 if v ∈ C.By the definition of a Y -cycle, if v ∈ C then there must exist an arc (v, v) ∈ A with v ∈ V (C). Thus foreach v ∈ C, set x∗(v, v) = 1/2. Set y∗(t) = 1/2 and x∗(t, w) = 1/2. Set to zero all remaining componentsof x∗ and to one the remaining components of y∗. Thus (x∗, y∗) is a fractional extreme point of Pp(G)with p = |V | − |C| − (|C|+ |C|+ 1)/2.

Now it remains to prove that conditions (C1) and (C2) are sufficient.

4. Graphs without a non-directed g-odd Y -cycle

We first prove the theorem below for the uncapacitated facility location polytope, then we combine thiswith the intersection property to obtain Theorem 4.

Theorem 20. Let G be an oriented graph that satisfies condition (C1) and does not contain a non-directedg-odd Y -cycle. Then PC(G) is integral.

To prove this we assume that z is a fractional extreme point of PC(G). It is assumed that G is two-connected. Also notice that the analogs of Lemmas 15–17 hold for PC(G). We need several lemmas below.


Fig. 4. A g-odd cycle of size four.

Lemma 21. The graph G does not contain a non-directed g-odd cycle. Notice that a g-odd cycle is differentfrom a g-odd Y -cycle.

Proof. Notice first that any non-directed g-odd cycle must be of size at least four. Recall that G is two-connected and does not contain a non-directed g-odd Y -cycle. If G contains a non-directed g-odd cycle C(which is not a Y -cycle) of size at least five, then the configuration H1 is present. To see this notice thatLemma 17 implies |δ+(v)| ≥ 1 for any node v ∈ C. Then at least one node v ∈ C adjacent to a node v ∈ C.

Now assume that C is a non-directed g-odd cycle (that is not a Y -cycle) of size four. We have onlyone possible configuration for that as shown in Fig. 4. Call a1 = (v1, v2); a2 = (v2, v3); a3 = (v4, v3); a4 =(v1, v4); a5 = (v3, v1) and let ∆1 = {a1, a2, a5} and ∆2 = {a4, a3, a5} be the two triangles in Fig. 4.

Here we have

δ+(v3) = {a5}, (7)

otherwise C would be a Y -cycle, which is impossible. With the feasibility of z we have that z(a5) = 1−z(v3).If z(a2) = z(v3) or z(a3) = z(v3) then, since z(∆1) ≤ 1 and z(∆2) ≤ 1, we obtain z(a1) = 0 or z(a4) = 0,which is not possible from Lemma 15. Thus we may assume that

z(a2) < z(v3), (8)z(a3) < z(v3). (9)

Notice also that

δ+(v1) = {a1, a4}, (10)

otherwise we have the graph H1.

Now we need the following three claims.

(i) δ−(v2) = {a1} and δ−(v4) = {a4}.

Proof. Let us show the first equality, the second hold by symmetry. Let (u, v2) ∈ A. Since G is two-connected we must have a path P from v2 to a node v in V (C) that uses the arc (u, v2). Notice thatu ∈ V (C), if u = v4 then the cycle formed by the arcs a1, (u, v2), a3, a5, is a non-directed g-odd Y -cycle,which is impossible. Thus P must have at least one internal node. We have three cases to consider.– v = v1. From (10) the last arc of P incident to v1 is directed into v1. In this case if P is g-even (resp.g-odd) then, the cycle defined by P ∪ {a5, a2} (resp. P ∪ {a4, a3, a2}) is a g-odd cycle of size ≥ 5,which is not possible from above.


– v = v3. Notice that, from (7), the last arc of P must be directed into v3, which is not possible sinceG does not contain H2.

– v = v4. First assume that the last arc in P is directed into v4. Remark that in this case P has at leasttwo internal nodes, otherwise there is a g-odd Y -cycle. If P is g-even (resp. g-odd) then, the cycledefined by P ∪ {a2, a3} (resp. P ∪ {a4, a1}) is a g-odd cycle of size ≥ 5, which is not possible. Nowassume that the last arc in P is leaving v4. Here we must also have at least two internal nodes in P .If P is g-odd (resp. g-even) then, the cycle defined by P ∪ {a3, a5, a1} (resp. P ∪ {a4, a1}) is a g-oddcycle of size ≥ 5, which is again not possible.

(ii) An odd directed cycle contains a1 if an only if it contains a2.

Proof. The arc a2 belongs to only one odd directed cycle that is ∆1, this comes from (7) and (i). LetC ′ be a directed odd cycle containing a1 but not a2. It is different from ∆1. Let (u, v1) be the arc ofC entering v1. Assume u different from v3. Let (v2, v) the arc in C ′ leaving v2. From (7), v3 is not inV (C ′). If v = u then, the cycle formed by (u, v1), a5, a2 and (v2, u) is a g-odd Y -cycle. If v = u, then thearcs (u, v1), a5, a1 and (v2, v) induce H1. In both cases we have a contradiction. Thus we may assumethat the arc in C ′ entering v1 is a5. Since G cannot contain H2, the arc in C ′ entering v3 is a3. Call Pthe directed path in C ′ that joins v2 to v4. So C ′ is composed by P plus the arcs a3,a5 and a1. Now it iseasy to check that the cycle formed by P plus the arc a3 and a2 is a g-odd Y -cycle, which is not possible.

(iii) An odd directed cycle contains a3 if an only if it contains a4. This follows by symmetry.

To complete the proof consider the following labeling function l that assigns +1 to the arcs a1, a3 and tothe node v2; it assigns −1 to the arcs a2, a4 and to the node v4 and assigns 0 to all other arcs and nodes.Using the claims above together with (8) and (9) it is straightforward to see that zl satisfies as equation allthe constraint that are satisfied as equation by z. This contradicts the fact that z is an extreme point. �

Lemma 22. Any g-even cycle in G is a Y -cycle.

Proof. Let C be non-directed cycle. If C is not a Y -cycle and the graph H1 is not present, then C is a g-oddcycle of size four. �

Lemma 23. G does not contain a directed odd cycle of size greater than or equal to five.

Proof. Let C = v0, a0, . . . , at−1, vt = v0 be a directed odd cycle with t ≥ 5. If G consists of only this cyclethen G is a Y -free graph and from Theorem 10, PC(G) is integral. Assume G is different from C. Since G istwo-connected we must have a path P = vk, b1, . . . , bs, vl, from a node vk ∈ V (C) to another node vl ∈ V (C)where b1, . . . , bs are arcs not in A(C). The nodes vk and vl are both the tails of b1 and bs, respectively.Otherwise, H1 is present, or there is a non-directed g-odd Y -cycle. Let C1 and C2 be the two paths in Cgoing from vk to vl. Then one of the cycles C1∪P or C2∪P is non-directed g-odd cycle, which is impossibleby Lemma 21. �

Lemma 24. Let C be a g-even cycle in G with V (C) minimum. Let ∆ = {a1, a2, a3} be a triangle in G.Then

(i) |A(C) ∩∆| ≤ 1,(ii) let A(C) ∩∆ = {a1} with a1 = (u, v), then u ∈ C and v ∈ C. Moreover, v does not belong to another

triangle, and(iii) z(a2) < z(w) where w is the head of a2.


Fig. 5. Intersection of a g-even cycle with a triangle.

Fig. 6. Intersection of a g-even cycle with a triangle.

Proof. (i) Suppose that A(C) ∩ ∆ = {a1, a2}, a1 = (u, v), a2 = (v, w). The possible four cases are:u ∈ C, w ∈ C;u ∈ C, w ∈ C;u ∈ C, w ∈ C;u ∈ C, w ∈ C. See Fig. 5. Replace the arcs a1 anda2 of C by a3 and call the resulting cycle C ′. In cases (1) and (2), C ′ is a non-directed g-odd cyclewhich contradicts Lemma 21. In cases (3) and (4), C ′ is a g-even cycle with |V (C ′)| < |V (C)|, whichcontradicts the lemma hypothesis.

(ii) Let a2 = (v, w), then w ∈ V (C). Otherwise we obtain a g-odd cycle or a g-even cycle C ′ with|V (C ′)| < V (C). In both cases we have a contradiction.

There are four possible cases as follows: u ∈ C, v ∈ C;u ∈ C, v ∈ C;u ∈ C, v ∈ C;u ∈ C, v ∈ C. SeeFig. 6. Call C ′′ the cycle obtained by removing a1 and by adding a2 and a3. In Cases (1) and (2) C ′′ isa non-directed g-odd cycle, which not possible.

Consider Case (3) and let b = (s, u) be the arc of C entering u and c = (v, t) the arc of C leaving v.If t = s, then the arcs b, a3, a1 and c induce H1, and if t = s, the cycle C is a triangle. Both cases areimpossible.

We conclude that the only possibility is Case (4) shown in Fig. 6. Suppose now that v belongs toanother triangle different from ∆. We need the following observations.• There is no other arc entering v. Because H2 is not present.• There is no other arc from v to a non-pendent node. To see this, suppose that the arc (v, t) exists,

with t = w, a2 = (v, w). We assume that t is not pendent, and H1 is not present, then either the arc(t, u) exists, or the arc (t, r) exists, where (r, v) is the other arc in C entering v. In both cases weobtain a g-odd cycle, that is impossible.• We have that (v, w) is the only arc leaving v that could be in a triangle containing v. Consider another

arc leaving w, to avoid H1 it should be the arc (w, r). Recall that (r, v) is the other arc in C enteringv. The arcs (r, v), (u, v), (w, u) and (w, r) form a non-directed g-odd cycle, which is impossible. Thusδ+(w) = {a3}, and the only triangle containing (v, w) is ∆.

(iii) We have just seen that δ+(w) = {a3}. So we must have z(a3) = 1 − z(w) and if z(a2) = z(w), fromz(∆) ≤ 1 we obtain z(a1) = 0. This is impossible because of Lemma 15. �

Lemma 25. G does not contain a g-even cycle.


Proof. Let C = v0, a0, v1, a1, . . . , at−1, vt be a g-even cycle in G. By Lemma 22 we may assume that C is aY -cycle. Among such cycles let C be one with V (C) minimum. Let l be the labeling function that assignslabels to the nodes and arcs of C following the labeling procedure of Section 2. Next we will extend thislabeling to some arcs not in A(C), but before let us note the following remark.

• If (v, v) ∈ A and v ∈ C, then v ∈ V (C).

Proof. If v ∈ V (C), then we must have a triangle (v, v), (v, w), (w, v) with both (v, w), (w, v) are in A(C),otherwise H1 or H2 is present. But in this case we have a contradiction with Lemma 24.

For each v ∈ C: if v belongs to a (unique) triangle ∆ let (v, v) ∈ ∆; otherwise there is an arc (v, v) withv a pendent node. In either case, assign the label −l(v) to (v, v).

Now we have to see that each constraint that is tight for z it remains tight for zl. By definition this truefor constraints (2) with respect to the nodes in V (C) \ v0. Remark 8 and Lemma 9, imply that this alsoholds for v0.

Consider now an inequality (3) with respect to (u, v). We have the following cases.

• If both (u, v) and v receive a label then (u, v) ∈ A(C). By definition of the labeling procedure we havezl(u, v) = z(u, v) + l(u, v) and zl(v) = z(v) + l(u, v).• If neither (u, v) nor v receives a label, then zl(u, v) = z(u, v) and zl(v) = z(v).• Assume that (u, v) receives a label but no v. This is true only when u ∈ C and v is pendent in which casez(u, v) < z(v) or (u, v) is an arc of a triangle and in this case Lemma 24 (iii) implies that z(u, v) < z(v).• Finally suppose that v receives a label but not (u, v). We cannot have v ∈ C, otherwise H2 is present. So

we must have v ∈ C. In this case we must have H1 or an intersection of C with a triangle that contradictsLemma 24(i) and (ii).

Let us consider an odd directed cycle inequality (6). From Lemma 23, inequality (6) is associated with atriangle. Call this triangle ∆ = {a1, a2, a3} and a1 = (u, v), a2 = (v, w) and a3 = (w, u). We have two cases:

• If ∆ ∩A(C) = ∅, then zl(∆) = z(∆).• When ∆ ∩ A(C) = ∅, Lemma 24(ii) implies that A(C) ∩ ∆ = {a1} with u ∈ C and v ∈ C, and v

does not belong to any other triangle. Then by the extension of the labeling above the arc a2 receivesthe label −l(v) and by the labeling procedure a1 receives the label l(v) (a3 receives the label 0). Hencezl(∆) = z(∆). �

Now we can put all pieces together. Lemma 25 shows that there is no g-even cycle. Lemma 21 shows thatthere is no non-directed g-odd cycle. Thus the only possibility for a cycle is to be a directed odd cycle, andLemma 23 implies that it should be a triangle. Since the graph is two-connected, the only possibility is thatthe entire graph G is a triangle. In this case PC(G) is integral, to see this notice that L(G) is a completegraph with three nodes, then consider the Stable set polytope of L(G). This proves Theorem 20, and fromTheorem 14 we obtain Theorem 4.

5. Graphs containing no triangle

Here we assume that G = (V,A) satisfies (C1) and (C2) and contains no triangle. We assume that z is afractional extreme point of PCp(G), based on Lemma 15, we assume that z(u, v) > 0 for each arc (u, v) ∈ A.Also we make the assumption in the lemma below.


Lemma 26. Let z be a fractional extreme point of PCp(G). We may assume that there is no arc (u, v) ∈ A, vnot pendent, where the following statements hold:

• z(u, v) < z(v),• (u, v) does not belong to any odd directed cycle, and• G(u, v) satisfies conditions (C1) and (C2).

Proof. If such an arc (u, v) exists, we keep working with G(u, v). The new vector is also a fractional extremepoint of PCp(G(u, v)). �

Let C be a non-directed g-odd Y -cycle. Condition (C2) implies that any arc not in A(C) should haveat least one endnode in V (C). So if there is an odd directed cycle, there is a directed path P between twonodes in V (C) consisting of one or two arcs not in A(C).

If there is no odd directed cycle, then G satisfies the conditions of Theorem 11 and so Pp(G) is integral.Now we assume that there is a directed path P from vl to vk, with vl, vk ∈ V (C), so that the arcs in P arenot in A(C). This path consists of one or two arcs. In what follows we treat these two cases.

5.1. The path P contains one arc

We start with the cases that are easy to dismiss.

• If vk ∈ C we would have the graph H2.• If vk ∈ C.

– If vl ∈ C, in order to avoid H1 we obtain two triangles.– If vl ∈ C, we obtain a smaller g-odd Y -cycle C ′. Condition (C2) implies V (C) \ V (C ′) = {vr}, andvr ∈ C. There is an arc (vr, w), and (C2) implies w ∈ V (C ′). Since w ∈ C, we obtain H1.

– If vl ∈ C, this creates a smaller g-odd Y -cycle C ′. If C ′ is non-directed, condition (C2) leads to theexistence of a triangle or the presence of H1. In both cases we have a contradiction. The case when C ′is a directed odd cycle will be treated in Section 5.3.

• If vk ∈ C.– If vl ∈ C, we obtain H1 or a triangle, so we have a contradiction.– If vl ∈ C, we obtain H1 or a triangle.– If vl ∈ C, again we obtain H1 or a triangle.

5.2. The path P consists of (vl, v) and (v, vk)

Again here, there are several cases that are easy to eliminate. See below.

• If vk ∈ C, then H2 is present.• If vk ∈ C.

– If vl ∈ C, then H1 is present.– If vl ∈ C, then this creates a smaller g-odd Y -cycle C ′. Condition (C2) implies that |V (C)\V (C ′)| = 1,

but then C cannot be g-odd.– If vl ∈ C, this will be treated in Section 5.3.• If vk ∈ C.

– If vl ∈ C, then we obtain H1.– If vl ∈ C, here we obtain H1 or a triangle.– If vl ∈ C, again we obtain H1 or a triangle.


Fig. 7. The cycle C is in bold. The cycles C1 and C2 are non-directed g-even. The cycle C3 is non-directed g-odd. The cycles C4and C5 are odd directed.

5.3. The three remaining cases

One case left is when P contains one arc, vk ∈ C, vl ∈ C, and adding P to C creates a directed odd cycle.This will be treated below and called Case (1). See Fig. 7 (1). The directed cycle obtained by adding P isdenoted by C4, the non-directed cycle obtained by adding P is denoted by C1.

Consider now the case when P consists of two arcs, vk ∈ C and vl ∈ C. Let (vk, p) and (vk, q) be the twoarcs in C incident to vk. Let (r, vl) and (vl, s) be the two arcs in C incident to vl. Assume that one of thepaths in C from vl to vk contains s and q, denote this by P1. Assume that the other path in C from vl tovk contains r and p, denote this by P2. We have the following three cases.

• If P2 and P form a non-directed g-odd Y -cycle, then condition (C2) implies q = s. This contradicts theparity of C.• If P2 and P form a directed odd cycle, this will be treated below, and called Case (2). See Fig. 7 (2).

The directed cycle formed by P2 and P is denoted by C5. The non-directed cycle formed by P1 and P isdenoted by C2.• If P1 and P form a non-directed g-odd Y -cycle, then condition (C2) implies p = r. This will be treated

below, and called Case (3). We will use u to denote p = r. See Fig. 7 (3). The non-directed cycle formedby P1 and P is denoted by C3.

Next we will show that in fact none of these remaining three cases can occur.

Remark 27. Because of condition (C1), and since there is no triangle, the following holds.

• Any arc entering a node v ∈ C ∪ C should be in A(C).• There is no arc other than (vl, v) entering v.• There is at most one arc entering a node in C.

Consider Case (3). Notice that from the remark above, none of the arcs (vl, v), (v, vk), (vk, u) and (u, vl)can belong to a directed odd cycle. Then if we apply the labeling procedure of Section 2.2, we obtain asolution zl that satisfies with equation each constraint of PCp(G) that is satisfied with equation by z.Therefore Case (3) cannot occur. Now we concentrate on Cases (1) and (2).

Lemma 28. The only directed path crossing C is P .


Fig. 8. The cycle D is shown with a thick line. The cycle D′ is shown with a dashed line.

Proof. Let P ′ be another directed path crossing C. Then P ′ should have at most two arcs. From all casesthat have been eliminated, we have that P ′ should go from a node in r ∈ C to a node in s ∈ C, and whenadded to C it should create an odd directed cycle as in Case (1) or (2).

If s = vk, to avoid H1 we obtain a triangle. Otherwise, let D (resp. D′) be the directed cycle obtainedafter adding P (resp. P ′) to C. Let C ′ be the cycle obtained by removing form C the arcs in D and D′ andadding P and P ′. The cycle C ′ is a non-directed g-odd Y -cycle. Then condition (C2) leads to the existenceof a triangle. The several possible cases are depicted in Fig. 8, when P and P ′ consist of one arc. The pathP ′ and the cycle D′ are drawn with dashed lines. �

As a consequence of Lemma 28, we have the following.

Remark 29. The unique directed odd cycles in Cases (1) and (2) are C4 and C5.

Consider Case (2). Let l be the labels obtained by the labeling procedure of Section 2.2 when applied tothe cycle C2 of Case (2). Notice that if the odd directed cycle inequality associated with C5 is tight by z,then it remains tight. This is because the only arcs of C5 that receive a label are (vl, v) and (v, vk) and theyare opposite labels. In fact, with Remarks 27 and 29, we can see that is true for every constraint of PCp(G).This contradicts the fact that z is an extreme point.

It is left to show that Case (1) cannot appear. If the odd directed cycle inequality associated with C4is not tight, then one can use the labeling procedure as in the previous case, but applied to C1. Hence weshould assume that

z(C4) = |C4| − 12 . (11)

Lemma 30. The following statements hold.

(i) There is at most one arc directed from a node in V (C4) to a pendent node.(ii) There is at most one non-tight arc in A(C4).

Proof. (i) Notice that any arc leaving a node in C4 must be directed to a pendent node, this follows fromLemma 28. Now let t and t′ be two pendent nodes, so that (vi, t) and (vj , t′) are two arcs in A, with vi andvj in V (C4), vi = vj . One of the directed paths in C4 between vi and vj contains an even number of arcs,call it P ′. Add an artificial arc (t, t′). Call C(P ′) the cycle obtained by the junction of P ′, (vi, t), (vj , t′)and (t, t′). This cycle is g-even, so we may apply the labeling procedure to C(P ′). Let l the resultinglabeling and let l′ the restriction of l to G, with l′(t) = l′(t′) = 0. It is a simple matter to see thata∈A(C4) l

′(a) = 0 and thatv∈V l

′(v) = 0. Therefore, the inequality associated with C4 is satisfiedwith equation by zl′ , the same is true for every other constraint that is satisfied with equation by z. Wehave a contradiction.


(ii) The proof uses the same argument as in (i). Let (vi, vj) and (v′i, v′j) two non-tight arcs in A(C4), thenone of the paths directed from vj to v′i or the one directed from v′j to vi has an odd number of arcs.Assume that this path is directed from vj to v′i. Then assign the label +1 to vj and extend this labelingthrough this directed path such that each arc receive the opposite label of his tail and the same as hishead, we stop this labeling once (v′i, v′j) get a label. Call this labeling l. It has the same property as thelabeling l′ in (i) and yields to the same contradiction. �

Lemma 31. Exactly one of the following holds.

(i) There is exactly one non-tight arc in C4.(ii) There is exactly one node w ∈ V (C4) so that (w, t) ∈ A, where t is a pendent node.

Proof. Suppose that neither (i) nor (ii) holds. Then from Lemma 30, each arc in C4 is tight, and each nodein V (C4) \ {vl, vk} is only incident to its neighbors in C4. Consequently, z(C4) = |C4|−1

2 + z(vl, vk), which isimpossible.

Assume now that both (i) and (ii) hold. Suppose that (u, v) ∈ A(C4) and z(u, v) < z(v). Let w ∈ V (C4)with (w, t) ∈ A and t a pendent node.

Let P1 be the directed path in C4 from v to w, and P2 the directed path in C4 from w to u. Thesetwo paths have the same parity. We first apply the labeling procedure to C1. Since C1 is g-even, the newvector zl satisfies with equation the same constraints as z, with the exception of the directed cycle inequalityassociated with C4. To deal with this we apply a second labeling starting from zl as follows.

• If P1 and P2 have an odd number of arcs. We add the artificial arcs (t, s) and (s, v), where s is also artificial.Let C ′ the even directed cycle formed by these two arcs, (w, t) and P1. We give the label l(vl, vk) to vand extend the labels around C ′, starting from the vector zl. Let l′ be the new labels. We remove theartificial arcs and we reset l′(t) = 0. The new vector zl′ satisfies with equation the same constraints as z.• If P1 and P2 have an even number of arcs. We add the artificial arc (t, v), and create a g-even cycle C ′

formed by this artificial arc, (w, t), (u, v) and P2. We give the label l(vl, vk) to t and extend the labelsthrough C ′. Let l′ be the new labels. We remove the artificial arc, and reset l′(t) = l′(v) = 0. The newvector zl′ satisfies with equation the same constraints as z.

In the two cases above we have a contradiction because z is an extreme point. �

Lemma 32. We have z(vl, vk) = z(vk).

Proof. Assume the contrary. Then from Lemma 31, it follows that each arc in A(C4) \ {(vl, vk)} is tight,and that each node in V (C4) \ {vl, vk} is adjacent to only its neighbors in C4. But this immediately impliesthat z(C4) = |C4|−1

2 + z(vl, vk), which is not possible. �

Let (u, v), (v, w) two arcs in A(C4) where u and w are the unique neighbors of v and δ+(u) = {(u, v)}.Construct the graph Guw by shrinking the nodes u and w, that is we remove the arcs (u, v) and (v, w) andthe node v, and we create a new node uw where each arc incident to u or to w now is incident to uw. Fromz define zuw its restriction on Guw, where zuw(uw) = z(u). We have the following lemma.

Lemma 33. If z is an extreme point of PCp(G) and both arcs (u, v) and (v, w) are tight, then zuw is anextreme point of PCp−1(Guw).


Fig. 9. The dashed arc is the non-tight arc of C4.

Proof. We have z(u, v) = z(v); z(v, w) = z(w) and z(v) = 1 − z(v, w) and since δ+(u) = {(u, v)}, thevalidity of z implies that z(u) = 1 − z(u, v). So z(v) + z(w) = 1, which implies that

r zuw(r) = p − 1.

We have zuw(uw) = z(u), but by construction the arcs that leave uw are those that leave w in G, and soz(δ+(uw)) = 1− z(w). By combining the equations above, one can see that z(w) = z(u). Therefore

z(δ+(uw)) + z(uw) = z(δ+(uw)) + z(u)= z(δ+(uw)) + z(w)= 1.

It is also easy to see that Guw contains one odd directed cycle call it C ′4, obtained from C4 and ofsize |C4| − 2. We have A(C ′4) = A(C4) \ {(u, v), (v, w)}, and since z(u, v) + z(v, w) = 1, we conclude thatz(A(C ′4)) = |C′4|−1

2 . Hence zuw ∈ PCp−1(Guw). If zuw is not an extreme point, then there is z′ ∈ PCp−1(Guw)where each constraint that is tight by zuw is also tight by z′. From z′ one can define z′′ as follows:z′′(u) = z′(uw); z′′(u, v) = 1 − z′(uw); z′′(v) = 1 − z′(uw); z′′(v, w) = z′(uw) and z′′(w) = z′(uw) andfor each other arc and node we have z′′(·) = z′(·). It is straightforward, that each constraint that is tight byz is also tight by z′′. �

To conclude this section we need the following two lemmas.

Lemma 34. Case (i) of Lemma 31 cannot occur.

Proof. Assume now that we are Case (i) of Lemma 31. Let (u, v) be the non-tight arc in C4. If the paths of C4directed from v to vl and from vk to v are both even, then z(A(C4) = |C4|−1

2 + z(vk, vl), which is impossible.Assume they are both odd. Then if we apply Lemma 33 recursively to an appropriate choice of the nodes inC4 that satisfies the hypothesis we may assume that C4 is (a) a triangle composed by the arcs (vl, vk), (vk, v)and (v, vl), where u = vk or (b) C4 is of size five composed by the arcs (vl, vk), (vk, v′k), (v′k, u), (u, v) and(v, vl), see Fig. 9. Notice that the arc (vl, vk) is always tight, this comes from Lemma 32.

Consider the case (a). Let G(vk, v) the graph obtained by removing the arc (vk, v) and adding the arc(vk, t) with t a new pendent node. Let z′ be the restriction of z on G(vk, v), with z′(vk, t) = z(vk, v) andz′(t) = 1. Notice that the graph G(vk, v) satisfies the conditions of Theorem 11, so Pp+1(G(vk, v)) is integral.Hence there is a zero–one solution z1 in Pp+1(G(vk, v)), with z1(vl, vk) = 1, that satisfies with equality eachconstraint that is satisfied with equality by z′. Since z′(v, vl) = z′(vl), we have that z1(v, vl) = z1(vl) = 0.Therefore, the solution z obtained from z1 by setting the variable z(vk, v) to zero and z(·) = z1(·) for allother nodes an arcs, is in PCp(G). Moreover, it satisfies with equality each tight constraint for z. The samemay be done in case (b), when considering the graph G(u, v) instead of G(vk, v). �

Lemma 35. Case (ii) of Lemma 31 cannot occur either.


Proof. Assume that we are in Case (ii) of Lemma 31. Let u be the node in V (C4) so that (u, t) ∈ A, wheret is a pendent node. The paths in C4 that are directed from vk to u and from u to vl must contain anodd number of arcs, otherwise z(A(C4)) = |C4|−1

2 + z(vk, vl). If we apply Lemma 33 recursively with anappropriate choice of the nodes in C4 that satisfies the lemma’s hypothesis, we may assume that C4 is atriangle composed by the arcs (vl, vk), (vk, u) and (u, vl), where u is directed to a pendent node call it t.Moreover, each arc in C4 is tight and that z(A(C4)) = 1, if we combine this with the validity of z, we obtainthat z(vk) + z(vl) + z(u) = 1 and z(δ+({vl, vk, u})) = 1.

Shrink the nodes vk, vl and u and let u∗ the resulting node and G∗ the resulting graph. Let z∗ be Therestriction of z on G∗, with z∗(u∗) = 0. We have z∗ ∈ Pp−1(G∗). Notice that from Remark 29, G∗ does notcontain any directed odd cycle, so conditions of Theorem 11 hold and it follows that Pp−1(G∗) is integral.Then, there is a zero–one solution in Pp−1(G∗), call it z1 with z1(u∗, t) = 1, that satisfies with equationeach constraint, among the constraints of Pp−1(G∗), that is satisfied with equation by z∗.

Define z from z1 as follows, z(u, t) = 1; z(u) = z(vk, u) = z(u, vl) = z(vl) = 0; z(vl, vk) = z(vk) = 1; andfor all other nodes and arcs z(·) = z1(·). It is a simple matter to check that z satisfies with equality eachconstraint of PCp(G) that is satisfied with equality by z. Thus we have a contradiction. �

Now the proof of Theorem 3 is complete.

Acknowledgments

We are grateful to the referees for their comments that helped us to improve the presentation.The work of Mourad Baıou was partially supported by Agence Nationale de Recherche for the project

ATHENA with reference ANR-13-BS02-0006-01.

References

[1] G. Cornuejols, M.L. Fisher, G.L. Nemhauser, Location of bank accounts to optimize float: an analytic study of exact andapproximate algorithms, Manag. Sci. 23 (1976/77) 789–810.

[2] A. Vigneron, L. Gao, M.J. Golin, G.F. Italiano, B. Li, An algorithm for finding a k-median in a directed tree, Inform.Process. Lett. 74 (2000) 81–88.

[3] F. Toumani, personal communication, 2002.[4] S. Nestorov, S. Abiteboul, R. Motwani, Extracting schema from semistructured data, in: Proceedings of the 1998 ACM

SIGMOD International Conference on Management of Data, SIGMOD ’98, ACM Press, New York, NY, USA, 1998,pp. 295–306.

[5] P. Avella, A. Sassano, On the p-median polytope, Math. Program. 89 (2001) 395–411.[6] I.R. de Farias Jr., A family of facets for the uncapacitated p-median polytope, Oper. Res. Lett. 28 (2001) 161–167.[7] M. Guignard, Fractional vertices, cuts and facets of the simple plant location problem, Math. Program. Stud. (1980)

150–162. Combinatorial optimization.[8] G. Cornuejols, J.-M. Thizy, Some facets of the simple plant location polytope, Math. Program. 23 (1982) 50–74.[9] D.C. Cho, E.L. Johnson, M. Padberg, M.R. Rao, On the uncapacitated plant location problem. I. Valid inequalities and

facets, Math. Oper. Res. 8 (1983) 579–589.[10] D.C. Cho, M.W. Padberg, M.R. Rao, On the uncapacitated plant location problem. II. Facets and lifting theorems, Math.

Oper. Res. 8 (1983) 590–612.[11] L. Canovas, M. Landete, A. Marın, On the facets of the simple plant location packing polytope, in: Workshop on Discrete

Optimization, Piscataway, NJ, 1999, Discrete Appl. Math. 124 (2002) 27–53.[12] J. Reese, Solution methods for the p-median problem: An annotated bibliography, Networks 48 (2006) 125–142.[13] M. Baıou, F. Barahona, On the p-median polytope of Y -free graphs, Discrete Optim. 5 (2008) 205–219.[14] A. Caprara, M. Fischetti, {0, 1/2}-chvatal-gomory cuts, Math. Program. 74 (1996) 221–235.[15] M. Padberg, On the facial structure of set packing polyhedra, Math. Program. 5 (1973) 199–215.[16] F. Barahona, A.R. Mahjoub, On the cut polytope, Math. Program. 36 (1986) 157–173.[17] M. Grotschel, L. Lovasz, A. Schrijver, Geometric Algorithms and Combinatorial Optimization, Springer, 1988.[18] A. Caprara, J.J.S. Gonzalez, Separating lifted odd-hole inequalities to solve the index selection problem, Discrete Appl.

Math. 92 (1999) 111–134.[19] M. Baıou, F. Barahona, On the linear relaxation of the p-median problem, Discrete Optim. 8 (2011) 344–375.

http://refhub.elsevier.com/S1572-5286(15)00046-8/sbref1



















[20] M. Baıou, L. Beaudou, Z. Li, V. Limouzy, Hardness and algorithms for variants of line graphs of directed graphs, in: L. Cai,S.-W. Cheng, T.-W. Lam (Eds.), Algorithms and Computation, in: Lecture Notes in Computer Science, vol. 8283, Springer,Berlin Heidelberg, 2013, pp. 196–206.

[21] G. Stauffer, The p-median polytope of Y -free graphs: an application of the matching theory, Oper. Res. Lett. 36 (2008)351–354.

[22] G. Calvillo, The concavity and intersection properties for integral polyhedra, Ann. Discrete Math. 8 (1980) 221–228.[23] F. Eisenbrand, G. Oriolo, G. Stauffer, P. Ventura, The stable set polytope of quasi-line graphs, Combinatorica 28 (2008)

45–67.[24] M. Baıou, F. Barahona, On the integrality of some facility location polytopes, SIAM J. Discrete Math. 23 (2009) 665–679.






on the p-median polytope and the directed odd cycle

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