MOURAD BAÏOU AND FRANCISCO BARAHONA

Transcription

1 THE p-median POLYTOPE OF RESTRICTED Y-GRAPHS MOURAD BAÏOU AND FRANCISCO BARAHONA Abstract We further study the effect of odd cycle inequalities in the description of the polytopes associated with the p-median and uncapacitated facility location problems We show that the obvious integer linear programming formulation together with the odd cycle inequalities completely describe these polytopes for the class of restricted Y-graphs This extends our results for the class of Y-free graphs We also obtain a characterization of both polytopes for a bidirected path Introduction Let G = (V, A) be a directed graph, not necessarily connected, where each arc (u, v) A has an associated cost c(u, v) The p-median problem (pmp) consists of selecting p nodes, usually called centers, and then assign each nonselected node to a selected node The goal is to select p nodes that minimize the sum of the costs yield by the assignment of the nonselected nodes This problem has several applications such as location of bank accounts [6], placement of web proxies in a computer network [], semistructured data bases [, 0] When the number of centers is not specified and each opened center induces a given cost, this is called the uncapacitated facility location problem (UFLP) The facets of p-median polytope have been studied in [] and [8] The facets of the uncapacitated facility location polytope have been studied in [9], [7], [4], [5], [3] In [] we studied the effect of odd cycle inequalities in the description of the polytopes associated with the pmp and the UFLP for the class of Y -free graphs In this paper we further study these inequalities, namely we show that the obvious integer linear programming formulation together with the odd cycle inequalities completely describe these polytopes for the class of restricted Y-graphs Let G = (V, A) be a directed graph We are going to use variables y associated with the nodes in V, and variables x associated with the arcs in A For a directed cycle C = v, (v, v ), v, (v, v 3 ),, v k, (v k, v k ), v k, (v k, v ), v, we denote by A(C) the set of arcs in C We say that C is odd if k is odd We plan to study the following linear system: Date: February 8, 006 Key words and phrases p-median problem, uncapacitated facility location problem, odd cycle inequalities

2 M BAÏOU AND F BARAHONA () () (3) (4) (5) (6) y(v) = p, v V v:(u,v) A x(u, v) = y(u) u V, x(u, v) y(v) (u, v) A, 0 y(v) v V, x(u, v) 0 (u, v) A, x(a) A(C) a A(C) for each odd directed cycle C Inequalities ()-(5) give a linear programming relaxation of the pmp, by adding inequalities (6) we obtain a stronger relaxation Analogously ()-(5) give a linear programming relaxation of the UFLP and adding inequalities (6) yields a stronger relaxation Denote by P p (G) the polytope defined by ()-(5), let P C p (G) be the polytope defined by ()-(6), and let pmp (G) be the convex hull of P p (G) {0, } V + A In general we have pmp (G) P C p (G) P p (G) Also let P (G) be the polytope defined by ()-(5), let P C(G) be the polytope defined by ()-(6), and let UF LP (G) be the convex hull of P (G) {0, } V + A We have UF LP (G) P C(G) P (G) For a undirected graph G = (V, E) we denote by G = (V, A) the directed graph obtained from G by replacing each edge uv E by two arcs (u, v) and (v, u) A directed graph G = (V, A) is called -directed if (u, v) A (v, u) / A A directed graph G = (V, A), not necessarily connected, is called Y -free if it is -directed and it does not contain as induced subgraph the graph of Figure In [] we proved the following Theorem If G is a Y -free graph then pmp (G) = P C p (G) and UF LP (G) = P C(G) Figure The graph Y In Figure we show three graphs, that are not Y -free, and for each of them a fractional extreme point of P p (G) The numbers near the nodes correspond to the variables y and the numbers near the arcs correspond to the variables x In this paper we study graphs for which these three configurations are forbidden A directed graph G = (V, A), not necessarily connected, is called a Y -graph if it is -directed and it does not contain as induced subgraphs the graphs H, H and H 3 of Figure For a directed graph G = (V, A) and a set 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

3 THE p-median POLYTOPE OF RESTRICTED Y-GRAPHS 3 H H H 3 Figure 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 we use δ + G and δ G A node u with δ + (u) = is called a pendent node Remark Remark that for any arc (v, w) with w not a pendent node, δ (v) Thus in such graphs the only nodes v different from a pendent node that may have δ (v) = have the property: if (v, u) A then u is a pendent node Call such a node a Y-node and denote by Y G the set of Y -nodes in G A simple cycle C is an ordered sequence where v 0, a 0, v, a,, a p, v p, v i, 0 i p, are distinct nodes, a i, 0 i p, are distinct arcs, either v i is the tail of a i and v i+ is the head of a i, or v i is the head of a i and v i+ is the tail of a i, for 0 i p, and v 0 = v p By setting a p = a 0, we associate with C three more sets as below We denote by Ĉ the set of nodes v i, such that v i is the head of a i and also the head of a i, i p We denote by Ċ the set of nodes v i, such that v i is the tail of a i and also the tail of a i, i p We denote by C the set of nodes v i, such that either v i is the head of a i and also the tail of a i, or v i is the tail of a i and also the head of a i, i p Notice that Ĉ = Ċ A cycle will be called odd if C + Ĉ is odd, otherwise it will be called even A cycle C with C = C is a directed cycle A cycle C is called a Y -cycle if all the nodes in Ĉ are Y -nodes Remark that when Ĉ = then C is a directed cycle and also a Y -cycle A polyhedron is called integral if all its extreme points are integral A restricted Y - graph is a Y -graph that does not contain an odd Y -cycle C with Ĉ A restricted Y -graph may have directed odd cycles, and it may contain odd cycles such that some of the nodes in Ĉ are pendent nodes In this paper we prove that Theorem also holds for restricted Y -graphs We obtain as a corollary that if G is a -directed graph with no odd Y -cycle, then P p (G) is integral if and only if G does not contain any of the graphs H, H and H 3 as induced subgraphs Also we obtain as a corollary that for an undirected

4 4 M BAÏOU AND F BARAHONA graph G the polytope P p ( G ) is integral if and only if G is a path Other than the classes mentioned here, we do not know of any other classes of graphs for which the polytopes of the pmp or the UFLP have been characterized For simplicity, in what follows we use z to denote the vector (x, y), i e z(u) = y(u) and z(u, v) = x(u, v) This paper is organized as follows In Section we give some polyhedral preliminaries In Section 3 we prove our main result Section 4 is devoted to bidirected paths Consider a polyhedron P defined by Some basic polyhedral facts P = {x R n Ax b} Denote by A = x b = a maximal subsystem of Ax b such that A = x = b = for all x P Then the dimension of P is n rank(a = ) A face F of P is obtained by setting into equation some of the inequalities defining P Clearly F is a polyhedron An extreme point of P is a face of dimension 0 Lemma 3 Let P be a polyhedron defined by P = {x R n Ax b}, whose extreme points are all 0 vectors Let P be defined by P = {x P cx = d} If ˆx is an extreme point of P then all its components are in for some number α [0, ] {0,, α, α}, Proof Let A = x b = be a maximal subsystem of Ax b such that A =ˆx = b = If rank(a = ) = n then ˆx is an extreme point of P and it is a 0 vector If rank(a = ) = n then F = {x P A = x = b = } is a face of P of dimension Therefore ˆx is a convex combination of two extreme points of P 3 Characterization of pmp (G) and UF LP (G) when G is a restricted Y -graph In this section we show that if G is a restricted Y -graph then pmp (G) is defined by ()-(6), and UF LP (G) is defined by ()-(6) First we need several lemmas Lemma 4 Let G be a restricted Y -graph Let C be an even Y -cycle, then there is no intersection between the arc set of C and the arc set of any odd directed cycle

5 THE p-median POLYTOPE OF RESTRICTED Y-GRAPHS 5 Proof Let C = v 0, a 0,, a p, v p = v 0 and let C = v 0, a 0,, a k, v k = v 0 be an odd directed cycle If C intersects C and C is a directed cycle, then we would have the configuration H or H 3 So we should assume that C is not a directed even cycle Suppose that A(C) A(C ) We must have Ċ V (C ), otherwise the configuration H, H or H 3 is present If Ċ V (C ) =, then we can assume that Ċ V (C ) = {v 0 }, A(C) A(C ) = {a 0,, a r }, v 0 = v 0, and a 0 = a 0,, a r = a r Let C be the cycle whose arc set is A(C) A(C ) (the symmetric difference between A(C) and A(C )) Then the parity of A(C ) is different from the parity of A(C) Because G is a Y -graph we have that none of v,, v r+ is in Ĉ, thus Ĉ = Ĉ Thus C is an odd Y -cycle, we have a contradiction Now we use induction, so we assume that A(C) A(D) = when D is an odd directed cycle and Ċ V (D) m Consider now a directed odd cycle C with Ċ V (C ) = m + Consider a maximal directed path contained in A(C) A(C ) going from u to v So u Ċ V (C ) and v ( C V (C ) ), otherwise we have the configuration H or H 3 Since Ċ V (C ) there is a node w Ċ V (C ), w u, such that the directed path in C from v to w does not contain the node u Denote by C w v this path We can assume that V (C w v ) V (C) = {v, w} Consider the path in C from v to w that does not contain u, denote it by Cw v The junction of C v w and Cv w is a cycle C It contains at least one Y -node, because Cw v contains at least one Y -node Also notice that all the nodes in Ĉ are Y -nodes, since these nodes are in V (C) Thus C is a Y -cycle with Ĉ Since G is a restricted Y -graph C is even Notice that now w / Ċ, v Ċ and u / C Thus Ċ V (C ) m This implies A(C ) A(C ) =, which is impossible Now we study a vector z assuming that it is a fractional extreme point of P C(G) or P C p (G), we plan to arrive to a contradiction We denote by G z = (V z, A z ) the graph induced by the arcs (u, v) A such that 0 < z(u, v) < Below we state several properties of G z Lemma 5 We may assume that δ G z (v) = for every pendent node v in G z Proof If v is a pendent node in G z and δ G z (v) = {(u, v),, (u k, v)}, we can split v into k pendent nodes {v,, v k } and replace every arc (u i, v) with (u i, v i ) Then we define z such that z (u i, v i ) = z(u i, v), z (v i ) =, for all i, and z (u) = z(u), z (u, w) = z(u, w) for all other nodes and arcs Let G be this new graph We have that the constraints that are tight for z are also tight for z, so z is a fractional extreme point of P C p+k (G ) The lemma above implies that we can assume that every cycle is a Y -cycle Lemma 6 G z does not contain an even cycle Proof Let C = v 0, a 0, v, a,, a p, v p be an even cycle in G z, that is C + Ĉ is even If v Ĉ then v is not a pendent node in G z, since δ G z (v) > Hence v must be a Y -node in G z Also, for every node v C, δ G z (v) =, otherwise the configuration H is present Thus the unique arc directed into v belongs to C Assume v 0 Ċ Assign labels to the nodes and arcs of C as follows: l(v 0 ) 0; l(a 0 )

6 6 M BAÏOU AND F BARAHONA For i = to p do the following: If v i is the head of a i and is the tail of a i, then l(v i ) l(a i ), l(a i ) l(v i ) If v i is the head of a i and is the head of a i, then l(v i ) l(a i ), l(a i ) l(v i ) If v i is the tail of a i and is the head of a i, then l(v i ) l(a i ), l(a i ) l(v i ) If v i is the tail of a i and is the tail of a i, then l(v i ) 0, l(a i ) l(a i ) Now define z as follows For every arc a i of C, i = 0,, p, let z (a i ) = z(a i ) + l(a i )ɛ For every node v i, i = 0,, p, let z (v i ) = z(v i ) + l(v i )ɛ Also for every node u Ĉ, pick an arc (u, v) δ+ G z (u) and let z (u, v) = z(u, v) l(u)ɛ Finally let z (u) = z(u), z (u, v) = z(u, v) for all other nodes and arcs of G It should be clear that z satisfies constraints () for every node v v 0, and that every constraint (3) that is tight for z is also tight for z In order to show that constraint () with respect to v 0 is satisfied by z, and that equation () is also satisfied by z, we have to discuss some properties of the labeling Let v j(0), v j(),, v j(k) be the ordered sequence of nodes in Ċ, with v j(0) = v j(k) A path in C v j(i), a j(i),, a j(i+), v j(i+) from v j(i) to v j(i+) will be called a segment and denoted by S i A segment is odd (resp even) if it contains and odd (resp even) number of arcs Let l(s i ) = l(v) v S i V Let r be the number of even segments and t the number of odd segments We have that r + t = Ċ, and since the parity of C is equal to the parity of t, we have that t + Ċ is even Therefore r = Ċ t is also even The labeling has the following properties: a) If the segment is odd then l(a j(i) ) = l(a j(i+) ) b) If the segment is even then l(a j(i) ) = l(a j(i+) ) c) If S i is odd then l(s i ) = 0 d) If S i is even then l(s i ) = l(a j(i) ) e) Let S,, S r be the ordered sequence of even segments in C Then l(s i ) = l(s i+ ), for i =,, r Properties a) and b) imply that l(a p ) = l(a 0 ) It follows that constraints () are satisfied by z Properties c), d) and e) imply that equation () is satisfied by z It follows from the remarks above and Lemma 4 that any constraint among ()-(6) that was tight for z remains tight for z This contradicts the fact that z is an extreme point of P C p (G) or P C(G) Lemma 7 The graph G z must contain at least one Y -node Proof Suppose that z is an extreme point of P C p (G) Suppose that G z is Y -free Let G be the graph obtained by adding to G z all arcs (u, v) with z(u, v) =, and all nodes u with z(u) = It is easy to see that G is also Y -free Let z G be the restriction of z to G Theorem implies that P C p (G ) is an integral polytope Clearly z G P C p (G ) Since z G is fractional, we have that z G = /z +/z with z, z P C p (G ), z z Let z

7 THE p-median POLYTOPE OF RESTRICTED Y-GRAPHS 7 (resp z ) be the vector obtained by adding zeros to z (resp z ) for all (u, v) G\G We have to see that these two new vectors belong to P C p (G) For that we study constraints (6), it is easy to see that the other constraints are satisfied Suppose that we add a zero component associated with the arc (u, v) Consider the odd cycle C with A(C) = {(w i, w i+ ) i =,, l} {(u, v)}, where u = w l+ and v = w We have that z (w i, w i ) + z (w i, w i+ ), for i =,, l This implies a A(C) z (a) l The same is true for z Therefore z and z are in P C p (G) Since z = / z + / z we have a contradiction The same proof holds when z is an extreme point of P C(G) Lemma 8 There is a Y -node t in G such that: The arcs (u, t), (u, t), (t, w) are in A V can be partitioned into W and W so that {u, t, w} W and u W The unique arc in G z between W and W is (u, t) Any arc in G between W and W does not belong to an odd directed cycle In other words, all directed odd cycles have all their arcs either in A(W ) or in A(W ) See Figure 3 Proof Any Y -node in G z is also a Y -node in G Let t be a Y -node in G z, Lemma 7 shows that such a node exists The node t is incident to a pendent node w and there are exactly two arcs (u, t) and (u, t) directed into t Let G = (S, A ),, G r = (S r, A r ) be the connected components of G z Set S r+ = V \ V z Let G be the connected component that contains t It follows from lemmas 5 and 6 and from the definition of a restricted Y -graph that t does not belong to any cycle in G z Hence if we remove t from G z then we disconnect G into two connected components Let S and S be the node sets of these two components, containing u and u respectively Define S 0 = S {t, w} and redefine S = S The unique arc of G z that may have one endnode in S i and the other endnode in S j, i j, is (u, t) Define two sets W and W as follows Step 0 W S 0, W S Step If there is a set S i W such that there is an arc in G of an odd directed cycle having one endnode in W and the other endnode in S i, then set W W S i Step If there is a set S i W such that there is an arc in G of an odd directed cycle having one endnode in W and the other endnode in S i, then set W W S i Go to Step Step 3 For every set S i not included in W W include S i in W By definition we have W W = r+ i=0 S i We have to see that W W = For that suppose S j W W Notice that a directed cycle cannot go through w because we would have the configuration H or H 3 Then there is a cycle C in G containing a node in S j, the node u, the node u and t Because G is a restricted Y -graph this cycle should be even The cycle C contains arcs of odd directed cycles, this contradicts Lemma 4 Based on this last lemma, we define the graphs G and G as follows Let A(W ) and A(W ) be the set of arcs in G having their both endnodes in W and W, respectively

8 8 M BAÏOU AND F BARAHONA W W u t u w Figure 3 Let G = (W, A(W )) and G = (W {t, v, w }, A(W ) {(u, t ), (t, v ), (v, w )}), see Figure 4 Now we can prove the following W W u t w t v w u G G Figure 4 Theorem 9 If G is a restricted Y -graph then P C(G) is an integral polytope Proof The proof is by induction on the number of Y -nodes If the graph is Y -free the result follows from Theorem Then we assume that the result is true for any restricted Y -graph G with Y G < Y G Let z be the restriction of z to G Clearly z P C(G ) Define z P C(G ) as follows: z (u, t ) = z(u, t), z (t ) = z(t), z (t, v ) = z(t ), z (v ) = z(t ), z (v, w ) = z(t ), z (w ) = and z (u) = z(u), z (u, v) = z(u, v) for all other nodes and arcs of G We have that z P C(G ) Notice that G and G are both restricted Y -graphs Also Y G < Y G and Y G < Y G Since z and z are both fractional, by the induction hypothesis they are not extreme points of P C(G ) and P C(G ), respectively Thus there must exist a 0- vector z P C(G ) with z (t) = 0 and such that the same constraints that are tight for z are also tight for z Also there must exist a 0- vector z P C(G ) with z (t ) = 0 such that the same constraints that are tight for z are also tight for z Combine z and z to define a solution z P C(G) as follows z (u) = z (u), for every node of G, z (u, v) = z (u, v) for every arc of G, z (u, t) = 0, z (v) = z (v), for all node v W, z (u, v) = z (u, v), for all arc(u, v) A(W ), z (u, v) = z(u, v) = 0, for all arc having one endnode in W and the other in W Clearly that any constraint among ()-(5), that is tight for z is also tight for z By Lemma 8, any odd directed cycle is included either in G(W ) or in G(W ), thus any

9 THE p-median POLYTOPE OF RESTRICTED Y-GRAPHS 9 constraint (6) that is tight for z is also tight for z This contradicts the fact that z is an extreme point of P C(G) Now we plan to prove that if G is a restricted Y -graph then P C p (G) is integral The proof is by induction on the number of Y -nodes If the graph is Y -free the result follows from Theorem Then we assume that it is true for any restricted Y -graph G with Y G < Y G We keep working with the graphs G and G defined before Now let z be a fractional extreme point of P C p (G) We need the following lemmas Lemma 0 The values of z are in {0,, α, α}, for some number α [0, ] Proof Since P C(G) is an integral polytope and P C p (G) is obtained from P C(G) by adding exactly one equation, the result follows from Lemma 3 Lemma z(u, t) = z(u, t) = z(t) = Proof Suppose z(u, t) < z(t) Consider the graph G obtained from G by removing the arc (u, t) and adding the arc (u, w) Let z be defined as z (u, w) = z(u, t) and z (u) = z(u), z (u, v) = z(u, v) for all other nodes and arcs We have that z P C p (G ) Since G is a restricted Y -graph with Y G < Y G, then z is not an extreme point of P C p (G ) Hence, there exists a vector z P C p (G ), z z, such that all constraints that are tight for z are also tight for z Define z(u, t) = z (u, w) and z(u) = z (u), z(u, v) = z (u, v) for all other nodes and arcs Then z z and since the arc (u, t) does not belong to any odd directed cycle in G then all constraints that are tight for z are also tight for z This is impossible since z is an extreme point of P C p (G) (Notice that we do not need that z P C p (G)) The same may be done if z(u, t) < z(t) Thus we may assume that z(u, t) = z(u, t) = z(t) z(t) = Now we have to prove that Consider the graph G defined from G as follows Remove (u, t) and add (u, t ), (t, v ) and (v, w) Here t and v are new nodes, see Figure 5 u u t t w G v Figure 5 Define z to be z (u, t ) = z(u, t), z (t ) = z(u, t), z (t, v ) = z(u, t), z (v ) = z(u, t), z (v, w) = z(u, t) and z (u) = z(u), z (u, v) = z(u, v) for all other nodes and arcs We have that z P C p+ (G ) and G is a restricted Y -graph with Y G < Y G Hence z is not an extreme point of P C p+ (G ) Thus there must exist a 0- vectors z,, z r in P C p+(g ) such that z is a convex combination of z,, z r, and

10 0 M BAÏOU AND F BARAHONA all constraints that are tight for z are also tight for z,, z r Thus (7) (8) (9) r z = λ i z i, i= r λ i =, i= λ i 0, i =,, r If there exists a vector z k with z k (t) = z k (t ), then we can define from z k a 0- vector z P C p (G) such that the same constraints tight for z are also tight for z Thus we may suppose that for all z i, i =, r, we have z i (t) z i (t ) Let z i (t) =, z i (t ) = 0, for i =,, r, and z i (t) = 0, z i (t ) =, for i = r +,, r Then (0) () z (t) = z (t ) = r i= λ i, r i=r + λ i, and since by definition z (t) = z (t ) and r i= λ i =, the result is obtained Lemma If z is a fractional extreme point of P C p (G) then each component of z is in {0,, } Proof Immediate from Lemma 0 and Lemma Now we can state the final result of this section Theorem 3 The polytope P C p (G) is integral Proof Define p = v W z(v) and p = v W z(v), so p = p + p We distinguish two cases: Case The numbers p and p are integers Consider the graphs G and G of Figure 4, as defined above Let z be the restriction of z to G Clearly z P C p (G ) Define z as follows z (u, t ) = z(u, t) =, z (t ) =, z (t, v ) =, z (v ) =, z (v, w ) =, z (w ) = and z (u) = z(u), z (u, v) = z(u, v) for all other nodes and arcs of G We have that z P C p +(G ) G and G are both restricted Y -graphs and Y G < Y G, Y G < Y G Since z and z are both fractional, by the induction hypothesis they are not extreme points of P C p (G ) and P C p +(G ), respectively Thus there must exist a 0- vector z P C p (G ) with z (t) = 0 so that the same constraints that are tight for z are also tight for z Also there must exist a 0- vector z P C p +(G ) with z (t ) = 0 such that the same constraints that are tight for z are also tight for z Combine z and z to define a solution z P C p (G) as follows

11 THE p-median POLYTOPE OF RESTRICTED Y-GRAPHS z (u) = z (u), for every node u of G, z (u, v) = z (u, v), for every arc (u, v) of G, z (u, t) = 0, z (v) = z (v), for every node v W, z (u, v) = z (u, v), for every arc(u, v) A(W ), z (u, v) = z(u, v) = 0, for every arc (u, v) having one endnode in W and the other in W Remark that v V z (v) = p Also any constraint among ()-(5), that is tight for z is also tight for z By Lemma 8, any odd directed cycle is included either in A(W ) or in A(W ), thus any constraint (6) that is tight for z remains tight for z Then the same constraints of P C p (G) that are tight for z are also tight for z This contradicts the fact that z is an extreme point of P C p (G) Case The values of p and p are not integers Thus from Lemma, v W z(v) = p = α + and v W z(v) = p = β, where α and β are integers and α + β = p Define G and G from G as follows G = ( W {u }, (A(W )\{(u, t)}) {(u, u ), (u, t)}) and G = ( W {t, w }, A(W ) {(u, t ), (t, w )} ), see Figure 6 W u u u t t W w w G G Figure 6 Define z to be: z (u, u ) = z (u ) = z (u, t) =, Let z be defined by: z (u) = z(u) for all other nodes of G, z (u, v) = z(u, v) for all other arcs of G z (u, t ) = z (t ) = z (t, w ) =, z (w ) =, z (u) = z(u) for all other nodes of G, z (u, v) = z(u, v) for all other arcs of G Notice that z P C α+ (G ) and z P C β+ (G ) Notice also that G and G are restricted Y -graphs with Y G < Y G and Y G < Y G Thus there must exist a 0-

12 M BAÏOU AND F BARAHONA vector z P C α+ (G ) such that the same constraints that are tight for z are also tight for z, and such that z (u, u ) = 0 Also there must exist a 0- vector z P C β+ (G ) such that the same constraints that are tight for z are also tight for z and such that z (t ) = 0 Now from z and z define z P C p (G) as follows z(u, t) = 0, z(u) = z (u), z(u, v) = z (u, v), z(t) = 0, z(u, t) = 0, z(t, w) =, z(u) = z (u), for all u W, for all u W \ {t}, for all (u, v) A(W ) \ {(u, t), (t, w)}, z(u, v) = z (u, v), for all (u, v) A(W ), z(u, v) = z(u, v), for all other arcs It is easy to see that z P C p (G) and the same constraints that are tight for z are also tight for z We have a contradiction because z is an extreme point 4 Bidirected paths A bidirected path is a graph G = (V, A) where V = {u,, u n }, and A = {(u i, u i+ ), (u i+, u i ), i =,, n } In this section we show that pmp (G) = P p (G) and UF LP (G) = P (G) when G is a bidirected path For that we first consider a graph H = (V, A) where n V = {u,, u n } V i, with V i = {v i,, vi p(i) }, for i =,, n The set of arcs A is composed by two arc-subsets The arcs (u i, vj i ) for i =,, n and j =,, p(i) And the arc-subset between any two consecutive nodes u i and u i+, for i =,, n, that consists of one of the following possibilities: (u i, u i+ ) A and (u i+, u i ) / A, or (u i, u i+ ) / A and (u i+, u i ) A, or (u i, u i+ ) A and (u i+, u i ) A, or there is no arc between u i and u i+ Notice that H may not be connected Call such a graph an extended path For H an extended path, denote by P air(h) the set of pair of nodes {u i, u i+ } such that both arcs (u i, u i+ ) and (u i+, u i ) belong to the set of arcs of H Theorem 4 If H is an extended path then pmp (H) = P p (H) and UF LP (H) = P (H) i= Proof The proof is by induction on P air(h) If P air(h) = 0 then H is a restricted Y -graph with no odd directed cycle Hence from Theorem 3 we have that pmp (H) is defined by inequalities ()-(5) Suppose that the theorem is true for every extended path H with P air(h ) m Let H be an extended path with P air(h) = m + and assume that z is a fractional extreme point of P p (H) Define H z = (V z, A z ) to be the graph induced by the arcs (u, v) A, with 0 < z(u, v) < Remark that H z is also an extended path

13 Claim P air(h z ) THE p-median POLYTOPE OF RESTRICTED Y-GRAPHS 3 Proof Suppose P air(h z ) = 0 Then H z is a restricted Y -graph with no odd directed cycle Let H be the graph obtained by adding to H z all arcs (u, v) with z(u, v) =, and all nodes u with z(u) = It is easy to see that H is also a restricted Y -graph with no odd directed cycle Then P p (H ) is an integral polytope Let z H be the restriction of z to H Clearly z H P p (H ) Since z H is fractional, we have that z H = /z + /z with z, z P p (H ), z z Let z (resp z ) be the vector obtained by adding zeros to z (resp z ) for each (u, v) H \H It is easy to see that z and z belong to P p (H) Since z = / z +/ z, we have a contradiction From the claim above, we may assume that there are at least two arcs (u i, u i+ ) and (u i+, u i ) with 0 < z(u i, u i+ ) < and 0 < z(u i+, u i ) < Define H from H by removing the arc (u i, u i+ ) and adding the arc (u i, w) where w is new a pendent node Define z to be z (u i, w) = z(u i, u i+ ), z (w) = and z (u) = z(u), z (u, v) = z(u, v) for all other nodes and arcs of H Notice that H is an extended path with pair(h ) = m and that z P p+ (H ) Then by the induction hypothesis P p+ (H ) must be integral Hence there must exist a 0- solution z P p+ (H ) with z(u i+, u i ) =, so that the same constraints that are tight for z are also tight for z From z define z P p (H) as follows: z (u i, u i+ ) = z(u i, w) and z (u) = z(u), z (u, v) = z(u, v) for all other nodes and arcs All constraints that are tight for z are also tight for z To see this, it suffices to remark that z (u i+ ) = z(u i+ ) = 0 and z (u i, u i+ ) = z(u i, w) = 0 This contradicts the fact that z is an extreme point of P p (H) The proof for UF LP (H) is similar We also have the following two corollaries Corollary 5 Let G be a -directed graph with no odd Y -cycle Then P p (G) is integral for all p if and only if G does not contain any of the graphs H, H and H 3 as induced subgraph Proof Let G = (V, E) be a -directed graph with no odd Y -cycle If G does not contain none of the graphs H, H and H 3 as induced subgraph, then G is a restricted Y -graph with no odd directed cycle Thus P C p (G) = P p (G) and the integrality of P p (G) follows from Theorem 3 Now suppose that G contains H = (V, E ) as induced subgraph Call v the unique pendent node in H Define z as follows: z (v) = { if v V \ {v } otherwise, z (u, v) = { if (u, v) E 0 otherwise It is easy to verify that z is an extreme point of P V (G) By the same manner one can construct fractional extreme point of P p (G) when G contains H or H 3 as induced subgraph Corollary 6 Let G be a undirected graph Then P p ( G ) is integral for all p if and only if G is a path

14 4 M BAÏOU AND F BARAHONA Proof If G is a path the result follows from Theorem 4 Suppose G is not a path Thus G contains a node w of degree at least 3 Let w, w and w 3 three nodes adjacent to w The solution z defined below is an extreme fractional point of P V ( G ) { z (v) = if v {w, w, w, w 3 } otherwise, z (u, v) = if (u, v) {(w, w ), (w, w), (w, w), (w 3, w)} 0 otherwise References [] P Avella and A Sassano, On the p-median polytope, Math Program, 89 (00), pp [] M Baïou and F Barahona, On the p-median polytope of a special class of graphs, Technical Report RC3636, IBM Watson Research Center, 005 Available at [3] L Cánovas, M Landete, and A Marín, On the facets of the simple plant location packing polytope, Discrete Appl Math, 4 (00), pp 7 53 Workshop on Discrete Optimization (Piscataway, NJ, 999) [4] D C Cho, E L Johnson, M Padberg, and M R Rao, On the uncapacitated plant location problem I Valid inequalities and facets, Math Oper Res, 8 (983), pp [5] D C Cho, M W Padberg, and M R Rao, On the uncapacitated plant location problem II Facets and lifting theorems, Math Oper Res, 8 (983), pp [6] G Cornuejols, M L Fisher, and G L Nemhauser, Location of bank accounts to optimize float: an analytic study of exact and approximate algorithms, Management Sci, 3 (976/77), pp [7] G Cornuejols and J-M Thizy, Some facets of the simple plant location polytope, Math Programming, 3 (98), pp [8] I R de Farias, Jr, A family of facets for the uncapacitated p-median polytope, Oper Res Lett, 8 (00), pp 6 67 [9] M Guignard, Fractional vertices, cuts and facets of the simple plant location problem, Math Programming Stud, (980), pp 50 6 Combinatorial optimization [0] S Nestorov, S Abiteboul, and R Motwani, Extracting schema from semistructured data, in SIGMOD 98: Proceedings of the 998 ACM SIGMOD international conference on Management of data, New York, NY, USA, 998, ACM Press, pp [] F Toumani, Personal communication 00 [] A Vigneron, L Gao, M J Golin, G F Italiano, and B Li, An algorithm for finding a k-median in a directed tree, Inform Process Lett, 74 (000), pp 8 88 (M Baïou) CUST and Laboratoire LIMOS, Université Clermont II, Campus des cézeaux BP 5, 6373 Aubière cedex, France address, M Baïou: baiou@custsvuniv-bpclermontfr (F Barahona) IBM T J Watson research Center, Yorktown Heights, NY 0589, USA address, F Barahona: barahon@usibmcom