Mixed-integer vertex covers on bipartite graphs

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1 Mxed-nteger vertex covers on bpartte graphs Mchele Confort, Bert Gerards, Gacomo Zambell November, 2006 Abstract Let A be the edge-node ncdence matrx of a bpartte graph G = (U, V ; E), I be a subset the nodes of G, and b be a vector such that 2b s ntegral. We consder the followng mxed-nteger set: X(G, b, I) = {x : Ax b, x 0, x nteger for all I}. We characterze conv(x(g, b, I)) n ts orgnal space. That s, we descrbe a matrx (A, b ) such that conv(x(g, b, I)) = {x : A x b }. Ths s accomplshed by computng the projecton onto the space of the x-varables of an extended formulaton, gven n [1], for conv(x(g, b, I)). We then gve a polynomal algorthm for the separaton problem for conv(x(g, b, I)), thus showng that the problem of optmzng a lnear functon over the set X(G, b, I) s polynomally solvable. 1 Introducton 1.1 The problem Gven a bpartte graph G = (U, V ; E), a vector b = (b e ) e E, wth the property that b s half-ntegral,.e. 2b e Z, e E, and a set I (U V ), we consder the problem of characterzng the convex hull of all nonnegatve x R V such that x + x j b for every E, x Z for every I. That s, gven the edge-node ncdence matrx A of a bpartte graph G, a partton (I, L) of ts column-set, and an half-ntegral vector b, we consder the followng mxed-nteger set: X(G, b, I) = {x : Ax b, x 0, x nteger for all I}. (1) Dpartmento d Matematca Pura e Applcata, Unverstá d Padova, Va Treste 63, Padova, Italy. Centrum voor Wskunde en Informatca, Kruslaan 413, 1098 SJ Amsterdam, The Netherlands & Technsche Unverstet Endhoven, Den Dolech 2, Endhoven 1

2 In ths paper we provde a formulaton for the polyhedron conv(x(g, b, I)). A formulaton for a polyhedron P (n ts orgnal space) s a descrpton of P as the ntersecton of a fnte number of half-spaces. So t conssts of a fnte set of nequaltes Cx d such that P = {x : Cx d}. In [1] a general technque was ntroduced to descrbe an extended formulaton for the set of solutons of a system Ax b, when A s a network matrx and some of the varables are restrcted to be nteger. A formulaton of P s extended whenever t defnes a polyhedron P n a hgher dmensonal space the ncludes the orgnal space, so that P s the projecton of ths polyhedral descrpton onto the orgnal space. In Secton 2 we derve the extended formulaton for conv(x(g, b, I)), whle n Secton 3 we descrbe a formulaton n the orgnal space by explctly computng the projecton of the polyhedron defned by the extended formulaton. Fnally, n Secton 4, we gve a polynomal tme algorthm to solve the separaton problem for conv(x(g, b, I)). 1.2 The man result Gven a bpartte graph G = (U, V ; E), a partton (I, L) of U V and an half-ntegral vector b, we say that a path P of G s an I-path f at least one endnode of P s n I, and no ntermedate node of P s n I. We say that P s odd f P has an odd number of edges e such that b e = 1 2 mod 1. In ths paper we show the followng: Theorem 1 The polyhedron conv(x(g, b, I)) s defned by the followng nequaltes: x + x j b E (2) 2x(V (P ) L) + x(v (P ) I) b(p ) P odd I-path (3) x 0 V (4) Esenbrand [2] conjectured that the nequaltes n (2)-(4) are suffcent to characterze conv(x(g, b, I)) when G s a path. So Theorem 1 shows that ths conjecture holds n a qute more general settng (and t certanly cannot be extended beyond that). Prelmnary results for the path case were obtaned by Skutella [9] and Esenbrand [2]. 1.3 Frst Chvátal closure The followng observaton allows us to descrbe X(G, b, I) n terms of a pure nteger set. Observaton 2 Let x be a vertex of conv(x(g, b, I)). Then 2 x s ntegral. 2

3 Proof: If not, let U and V be the sets of nodes n U and V, respectvely, such that 2 x s not nteger. Then, for ɛ small enough, the vectors x+ɛχ U ɛχ V and x ɛχ U + ɛχ V are both n conv(x(g, b, I)), where we denote by χ S the ncdence vector of S for any S U V. Let b = 2b, A be obtaned form A by multplyng by 2 the columns correspondng to nodes n I. By Observaton 2, the lnear transformaton x = x, I, x = 2x, L, maps X(G, b, I) nto {x : A x b, x 0, x ntegral} whch s a pure nteger set. Let P = v 1,... v n be an I-path. Notce that b(p ) = 1 2 mod 1 s equvalent to b (P ) odd. Then the nequalty b x (P ) (5) 2 V (P ) s a Gomory-Chvátal nequalty of {x : A x b, 0}. Indeed, assume v 1 I. If v n I, then (5) s obtaned from n (2x v 1 +x v 2 b 1 v 1 v 2 )+ 2 (x v +x v +1 b v v +1 )+ 1 2 (x v n 1 +2x v n b v n 1 v n ) =2 by roundng up the rght-hand-sde. If x n / I, then (5) s obtaned from n (2x v 1 + x v 2 b 1 v 1 v 2 ) + 2 (x v + x v +1 b v v +1 ) (x v n 0) =2 by roundng up the rght-hand-sde. Furthermore the nequaltes n (5) are equvalent to the nequaltes n (3). 1.4 The motvaton A (general) mxed-nteger set s a set of the form s {x Ax b, x nteger I} (6) where I s a subset of the columns of A and b s a vector that may contan fractonal components. In [1], t s shown that the problem of decdng f the above set s nonempty s NP-complete, even f b s an half-ntegral vector and A s a network matrx. (We refer the reader to [5] or [8] for defntons and results related to network matrces and, more generally, totally unmodular matrces.) However, t may be possble that, when A s the transpose of a network matrx, the assocated mxed-nteger programmng problem s polynomally solvable. Indeed, let MIX 2T U be a mxed-nteger set of the form (6) when 3

4 A s a network matrx. An extended formulaton of the polyhedron conv(mix 2T U ) was descrbed n [1]. The extended formulaton nvolves an addtonal varable for each possble fractonal parts taken by the varables at any vertex of conv(mix 2T U ). If ths number s polynomal n the sze of (A, b), t s shown n [1] that the formulaton s compact,.e. of polynomal sze n the sze of (A, b). Therefore the problem of optmzng a lnear functon over MIX 2T U can be effcently solved n ths case. However, t seems to be rather dffcult to compute the projecton n the orgnal x-space. It follows from Observaton 2 that f x s a vertex of conv(x(g, b, I)), then x x {0, 1 2 }. Therefore the extended formulaton for conv(x(g, b, I)) (whch wll be ntroduced n Secton 2) s compact. The man contrbuton of ths paper s the explct descrpton of the projecton of the polyhedron defned by ths extended formulaton n the orgnal x-space. The mxed-nteger set X(G, b, I) s related to some mxed-nteger sets that arse n the context of producton plannng (see [7]). The case when G s a star wth center node n L and leaves n I has been studed by Pochet and Wolsey n [6], where they gave an extended formulaton for the convex hull of feasble solutons whch s compact. Günlük and Pochet [3] projected ths formulaton onto the orgnal space, thus showng that the famly of mxng nequaltes gves the formulaton n the x-space. Mller and Wolsey [4] extended the results n [6] to general bpartte graphs, wth the restrcton that the partton (I, L) corresponds to the bpartton (U, V ) of the graph. Ther result shows that the mxng nequaltes assocated wth every sngle star of G havng center a node n L and leaf nodes all nodes n I gve a formulaton for ths case. 2 The extended formulaton We use here a modelng technque ntroduced by Pochet and Wolsey [6] and extensvely nvestgated n [1]. Observaton 2 allows to express each varable n L as x = µ δ, µ 0 nteger, 0 δ 1 nteger. (7) For now, we assume I =, that s, L = (U V ). Lemma 3 Let E, and suppose x, x j satsfy (7). If b = 1 2 mod 1, x, x j satsfy x + x j b f and only f µ + µ j b µ + δ + µ j + δ j b. If b = 0 mod 1, x, x j satsfy x + x j b f and only f µ + δ + µ j b µ + µ j + δ j b. (8) (9) 4

5 Proof. Assume x, x j satsfy (7). Then, f b = 1 2 mod 1, constrant x + x j b s satsfed f and only f µ + µ j b and δ + δ j = 1 whenever µ + µ j = b. If b = 0 mod 1, the constrant s satsfed f and only f µ + µ j b 1 and δ = δ j = 1 whenever µ + µ j = b 1. It s easy to see that these two condtons are modeled by the above constrants. Observaton 4 Gven E, the constrants (8) and (9) belong to the frst Chvátal closure of the polyhedron defned by µ δ + µ j δ j b µ, µ j 0 δ, δ j 1 δ, δ j 0 whenever b = 1 2 mod 1 and b = 0 mod 1, respectvely. By applyng the unmodular transformaton µ 0 = µ, µ 1 = µ + δ, the constrants x = µ δ, µ 0, 0 δ 1 become x 1 2 (µ0 + µ 1 ) = 0 (10) and constrants (8) and (9) become: µ µ 1 µ0 1 µ 0 + µ0 j b µ 1 + µ1 j b (11) (12) µ 1 + µ0 j b µ 0 + µ1 j b (13) Theorem 5 The projecton onto the space of the x varables of the polyhedron Q defned on the space of the varables (x, µ 0, µ 1 ) by the nequaltes (10), (11) for every U V, (12) for every E s.t. b = 1 2 mod 1 (13) for every E s.t. b = 0 mod 1 s the polyhedron conv(x(g, b, )). Proof: Snce the varable x s determned by (10) for all U V, we only need to show that the polyhedron defned by nequaltes (11) for every U V, (12) for every E s.t. b mod 1, and (13) for every 5 = 1 2

6 E s.t. b = 0 mod 1, s ntegral. Let A µ be the constrant matrx of the above system. Snce G s a bpartte graph, then the matrx Ā, obtaned by multplyng by 1 the columns of A µ relatve to the varables µ 0, µ1, V, has at most a 1 and at most a 1 n each row. Therefore Ā s the transpose of a network matrx, so A µ s totally unmodular (see [8]). Snce the left-hand-sdes of (11)-(13) are all nteger, the statement follows from the theorem of Hoffman and Kruskal. Observaton 6 Varable x s nteger valued f and only f δ = 0, U V. Therefore, for a gven I (U V ), the polyhedron conv(x(g, b, I)) s the projecton on the space of the x varables of the face Q I of Q defned by the equatons µ 1 1 µ0 = 0, I (whch correspond to δ = 0, I). 3 The formulaton n the orgnal space In ths secton we prove Theorem 1 by projectng onto the x-space the polyhedron Q I. Let p = µ0 µ1 2. The µ 0 = x + p and µ 1 = x p. The nequaltes defnng Q become: p + p j b x x j, b = 1 2 mod 1 p p j b x x j, b = 1 2 mod 1 p p j b x x j, b = 0 mod 1 p + p j b x x j, b = 0 mod 1 p 1 2 U V p 0 U V p x U V By Observaton 6, conv(x(g, B, I)) s the projecton onto the x-space of the polyhedron defned by the above nequaltes and by p = 0 for every I. Assocate multplers to the above constrants as follows: u ++ p + p j b x x j u p p j b x x j u + p p j b x x j u + p + p j b x x j u 1 2 p 1 2 u 0 p 0 u x p x (14) Any vald nequalty for conv(x(g, b, I)) has the form α u x β u, where α u x = + u )(x + x j ) + (u ++ b = 1 2 mod 1 6

7 b =0 mod 1 (u + + u + )(x + x j ) + u x x (15) β u = (u b + u ++ b ) + b = 1 2 mod 1 (u + + u + )b 1 2 u 1 2 (16) L b =0 mod 1 for some nonnegatve vector u = (u ++, u, u +, u +, u 1 2, u 0, ux ) such that up = 0, where P s the column-submatrx of the above system nvolvng columns correspondng to varables p, L (see e.g. Theorem 4.10 n [5]). For nstance the nequalty x + x j b where b = 1 2 mod 1 s obtaned by settng u ++ = u = 1 2, and all other entres of u to be 0. We are nterested n characterzng the nonnegatve vectors u such that up = 0 and α u x β u s facet defnng for conv(x(g, b, I)), and such that the nequalty α u x β u s not of the form x + x j b, for some E, or x 0, for some U V. From now on we wll assume, w.l.o.g., that the entres of u are nteger and relatvely prme. We defne an auxlary graph Γ u = (L {d}, F ), where d s a dummy node not n U V, and F u s defned as follows. For every edge E such that, j L, there are u ++ + u + u + parallel edges between and j n F, each edge correspondng to u + a varable among u ++, u, u +, u +. + For each node L, there are u u 0 + ux + j I : E (u++ + u + u + + u + ) parallel edges between d and n F, each edge correspondng to a varable among u 1 2, u 0, ux u +, for some j I., or u++, u, u +, We mpose a b-orentaton ω on Γ, that s, to each edge e F, and each endnode of e that belongs to L, we assocate the value ω(e, ) = tal f e corresponds to an nequalty of (14) where p has coeffcent 1, whle we assocate the value ω(e, ) = head f e corresponds to an nequalty of (14) where p has coeffcent +1. The dummy node d s nether a tal nor a head of any edge. Thus, each edge of Γ u can have one head and one tal, two heads, two tals, or, f d s one of the two endnodes, only one head and no tal or only one tal and no head. For each L, we denote wth δω n () the number of edges n F u of whch s a head, and wth δ out w () the number of edges n F of whch s a tal. ω () for every L. We say that Γ u s ω-euleran f δ n ω () = δ out 7

8 Observaton 7 Γ u s ω-euleran f and only f up = 0. We defne a closed ω-euleran walk n Γ u as a closed-walk n Γ u, v 0, e 0, v 1, e 1,..., v k, e k, v k+1, where v 0 = v k+1, wth the property that ω(e h 1, v h ) ω(e h, v h ) for every h such that v h s n L, h = 0,..., k, k + 1, where the ndces are taken modulo k. That s, f v h L, then v h s a head of e h 1 f and only f v h s a tal of e h. Observaton 8 Γ u s ω-euleran f and only f Γ u s the dsjont unon of closed ω-euleran walks. In partcular, every node n L {d} has even degree n Γ u. Observe that, f v 0, e 0,..., e k, v k+1 s a closed ω-euleran walk n Γ u, then both graphs Γ, Γ on L {d} wth edge-sets F = {e 1,..., e k } and F = F \ F, respectvely, are ω-euleran. Suppose F. Then there are nonnegatve nteger vectors u and u, both dfferent from zero, such that u P = 0, u P = 0, Γ = Γ u and Γ = Γ u, and u = u + u. By the fact that Γ and Γ are ω-euleran, and by the structure of the nequaltes n (14), the vectors (α u, β u ) and (α u, β u ) are both non-zero. Furthermore α u = α u + α u and β u = β u + β u, contradctng the fact that α u x β u s facet defnng and the entres of u are relatvely prme. Hence we have shown the followng. Observaton 9 Every closed ω-euleran walk of Γ u traverses all the edges n F. In partcular, there exsts a closed ω-euleran walk v 0, e 0,..., e k, v k+1 of Γ u such that F = {e h h = 1,..., k}. Suppose d has postve degree n Γ. Then we may assume, w.l.o.g., that v 0 = v k+1 = d. Suppose d = v h for some h = 1,..., k. Then v 0, e 0, v 1,..., e h 1 v h s a closed ω-euleran walk, contradctng the prevous observaton. Hence we have the followng. Observaton 10 Node d has degree 0 or 2 n Γ u. Next we show the followng. Lemma 11 Every node n L {d} has degree 0 or 2 n Γ u. Proof: We have already shown d has degree 0 or 2 n Γ u. If d has degree 2, we assume d = v 0 = v k+1, else v 0 s arbtrarly chosen. If there s a node n L wth degree at least 4, then there exsts dstnct ndces s, t {1,..., k} such that v s = v t. We choose s and t such that t s s postve and as small 8

9 as possble. Therefore C = v s, e s,..., e t 1, v t s a cycle of Γ u contanng only nodes n L. Snce G s a bpartte graph, C has even length, hence the edges n C can be parttoned nto two matchngs M 0, M 1 of cardnalty C /2. We wll denote wth HH, T T, HT the sets of edges of F wth, respectvely, two heads, two tals, one head and one tal. If v s s the head of exactly one among e s and e t 1, then C s a closed ω-euleran walk, contradctng Observaton 9. Hence v s s ether a head of both e s and e t 1 or a tal of both e s and e t 1. Ths shows that C T T = C HH ± 1. Therefore there s an odd number of edges e n C such that b e = 1 2 mod 1. By symmetry, we may assume e M 0 b e e M 1 b e Then the nequalty 2 x b e + 1 (17) 2 V (C) e C s vald for conv(x(g, b, I)), snce t s mpled by the vald nequaltes x + x j b, M 0, because 2 V (C) x = 2 (x +x j ) 2 b b e + b e = b e M 0 M 0 e M 0 e M 1 e C Case 1: Node v s s a tal of both e s and e t 1. Then C T T = C HH + 1, hence b e + b e + b e = b e (18) e C T T e C HH e C HT e C Let u be the vector obtaned from u as follows { u = u 1 u 0 v s = u 0 v s + 2 for every C all other components of u and u beng dentcal, where u among u ++, u, u +, u + correspondng to edge of C. s the varable Then one can easly see that Γ u s the graph obtaned from Γ u by removng the edges e s,..., e t, and addng two parallel edges v s d both wth tal n v s, hence Γ u s ω-euleran and u P = 0. By (18) β u = β u e C b e 1 2, whle by constructon α u x = α u x + 2 x. V (C) 9

10 Thus α u x β u can be obtaned by takng the sum of α u x β u and (17), contradctng the assumpton that α u x β u s facet defnng. Case 2: Node v s s a head of both e s and e t 1. Then C T T = C HH 1, hence e C T T b e + e C HH b e + e C HT Let u be the vector obtaned from u as follows b e = e C b e 1 2. (19) { u = u 1 u 1 2 vs = u 0 v s + 2 for every C all other components of u and u beng dentcal. Then one can easly see that Γ u s the graph obtaned from Γ u by removng the edges e s,..., e t, and addng two parallel edges v s d both wth head n v s, hence u P = 0. By (19) β u = β u e C b e , whle by constructon α u x = α u x + 2 x. V (C) Thus α u x β u can be obtaned by takng the sum of α u x β u and (17), contradctng the assumpton that α u x β u s facet defnng. We are now ready to gve the proof of the man theorem. Proof of Theorem 1. We show that all facet defnng nequaltes α u x β u, where u s nonnegatve, ntegral, and wth entres that are relatvely prme, that are not nequaltes n (2) or (4), are of the form (3). Frst we show the followng. E u > E u ++ In fact, we can wrte the nequalty α u x (u + u ++ )b + b = 1 2 mod V b =0 mod 1 u 1 2 (20) (u + + u + )b

11 as nonnegatve combnaton of nequaltes of the form (2) or (4), therefore we must have β u > (u + u ++ )b + (u + + u + )b. b = 1 2 mod 1 b =0 mod 1 Thus 0 < β u (u + u ++ )b + b = 1 2 mod 1 = 1 2 ( E whch proves (20). u E u ++ V u 1 2 ) b =0 mod 1 (u + + u + By Lemma (11) and Observaton (9), Γ u conssts of an nduced cycle C and solated nodes, where every node n V (C) L s a head of exactly one edge and a tal of exactly one edge. If d s an solated node, then each edge of C corresponds to a varable of the form u, and snce the total number of heads n C equals the number of tals, then E u = E u++ and V u 1 2 = 0, contradctng (20). Thus we may assume that C = v 0, e 0,..., e k, v k+1 where d = v 0 = v k+1. Clam: The followng are the only possble cases. 1. Edges dv 1, dv k of Γ u correspond to varables u x v 1 and u x v k, respectvely; 2. dv 1 corresponds to varable u or u + for some w I, and dv k corresponds to u x v k ; 3. dv 1 corresponds to varables u or u + for some w I, and dv k corresponds to varable u w v k or u + w v k for some w I. Proof of clam If v 1 s a head of e 0 and v k s a head of e k, then the number of edges among e 1,..., e k 1 wth two tals s one plus the number of edges wth two heads. Snce the former correspond to varables of type u for some E, and the latter correspond to to varables of type u ++ for some E, then by (20) dv 1 does not correspond to varable u 1 2 v1 )b or to a varable u ++ for any w I, and dv k does not correspond to varable u 1 2 vk or to a varable u ++ wv k for any w I, thus one of the above three cases holds. If v 1 s a tal of e 0 and v k s a head of e k, then the number of edges among e 1,..., e k 1 wth two tals s equal the number of edges wth two heads. By (20), dv 1 corresponds to varable u for some w I, and dv k corresponds to ether u x v k or to a varable u + w v k for some w I, thus case 2 or 3 holds. If v 1 s a tal of e 0 and v k s a tal of e k, then the number of edges among e 1,..., e k 1 wth two tals s equal one mnus the number of edges wth two 11

12 heads. By (20), dv 1 corresponds to varable u for some w I, and dv k corresponds to a varable u w v k for some w I, thus case 3 holds. Ths completes the proof of the clam. Case 1: Edges dv 1, dv k of Γ u correspond to varables u x v 1 and u x v k, respectvely. In ths case the path P = v 1, e 1,..., e k 1, v k of Γ u s also a path of G contanng only nodes n L, and P contans an odd number of edges e such that b e = 1 2 mod 1. The nequalty α ux β u s then 2x(V (P )) b(p ) The edges of P can be parttoned nto two matchngs M 0 and M 1, thus we may assume, w.l.o.g., e M 0 b e e M 1 b e Thus 2x(V (P )) 2 M 0 (x + x j ) 2 M 0 b e M 0 b e + e M 1 b e = b(p ) + 1 2, hence α u x β u s not facet defnng. Case 2: dv 1 corresponds to varable u or u + corresponds to u x v k. for some w I, and dv k In ths case, P = w, v 1, e 1,..., e k 1, v k s an odd I-path of G between w I and v k L. The nequalty α u x β u s 2x(V (P ) L)+x w b(p )+ 1 2, whch s one of the nequaltes n (3). Case 3: dv 1 corresponds to varables u or u + for some w I, and dv k corresponds to varable u w v k or u + w v k for some w I. If w w, then the path P = w, v 1, e 1,..., e k 1, v k, w s an odd I-path of G between w I and w I. The nequalty α u x β u s 2x(V (P ) L) + x w + x w b(p ) + 1 2, whch s one of the nequaltes n (3). If w = w, then we must have v 1 v k, snce otherwse v 1 would be ether the head or the tal of both edges of Γ u ncdent to v 1. Thus C = w, v 1,..., v k, w s a cycle of G. Snce G s a bpartte graph, C has even length, hence the edges n C can be parttoned nto two matchngs M 0, M 1 of cardnalty C /2. Snce C contans an odd number of edges e such that b w = 1 2 mod 1, then we may assume, w.l.o.g., e M 0 b e e M 1 b e The nequalty α u x β u s 2x(V (C )) b(c ) But 2x(V (C )) = 2 M 0 (x + x j ) 2 M 0 b e M 0 b e + e M 1 b e = b(c ) + 1 2, hence α u x β u s not facet defnng. 4 Separaton Theorem 5 and Observaton 6 mply that the problem of mnmzng a lnear functon over the set X(G, b, I) s solvable n polynomal tme, snce t reduces to solvng a lnear programmng problem over the set of feasble ponts for (10)-(13). In ths secton we gve a combnatoral polynomal tme algorthm for the separaton problem for the set conv(x(g, b, I)), thus gvng an alternatve 12

13 proof that the problem of optmzng a lnear functon over such polyhedron, and thus over X(G, b, I), s polynomal. Clearly, gven a nonnegatve vector x, we can check n polynomal-tme whether x satsfes (2) for every edge. Thus, by Theorem 1, we only need to descrbe a polynomal tme algorthm that, gven a nonnegatve vector x satsfyng (2), ether returns an nequalty of type (3) volated by x, or proves that none exsts. For every E, let s = x + x j b. Snce x satsfes (2), then s e s nonnegatve for every e E. Let P = v 1,... v n be an odd I-path. Clam 12 The vector x satsfes 2x (V (P ) L) + x (V (P ) I) b(p ) f and only f s (P ) + x ({v 1, v n } L) 1 2. Indeed, assume v 1 I. If v n I then n 1 n 1 s v v +1 = (x v + x v +1 b v v +1 ) =1 =1 gves the equalty s (P ) = 2x (V (P ) L) + x (V (P ) I) b(p ), hence 2x (V (P ) L) + x (V (P ) I) b(p ) f and only f s (P ) 1 2. If v n / I, then n 1 n 1 s v v +1 + x v n = (x v + x v +1 b v v +1 ) + x v n =1 =1 gves the equalty s (P )+x v n = 2x (V (P ) L)+x (V (P ) I) b(p ), hence 2x (V (P ) L) + x (V (P ) I) b(p ) f and only f s (P ) + x v n 1 2. Ths completes the proof of the Clam. Therefore, f we assgn length s e to every e E, we need to gve an algorthm that, for any two nodes r, t such that r I, ether determnes that the shortest odd I-path between r and t (f any) has length at least 1 2 x ({t} L), or returns an odd I-path P for whch 2x (V (P ) L) + x (V (P ) I) < b(p ) Observe that any walk W between r and t that contans an odd number of edges e such that b e = 1 2 mod 1 ether contans a sub-path P that s an odd I-path or t contans a cycle C that contans an odd number of edges e such that b e = 1 2 mod 1. In the former case, ether both endnodes of P are n I, or t s the only endnode of P n L. Hence, f s (W ) < 1 2 x ({t} L), then also s (P ) < 1 2 x ({t} L), hence 2x (V (P ) L)+x (V (P ) I) < b(p ) In the second case, snce G s bpartte, the edges of C can be parttoned nto two matchngs M 0 and M 1 such that b(m 0 ) b(m 1 ) Thus s (C) = C (x + x j b ) = 2x (V (C)) b(c) 2(x (V (C)) b(m 0 )) = 2 M 0 (x + x j b ) , hence s (W )

14 Thus we only need to fnd, for every par r, t V wth r I, the shortest walk W between r and t, w.r.t. the dstance s, among all such walks contanng an odd number of edges e such that b e = 1 2 mod 1. If, for a gven choce of r, t, s(w ) < 1 2 x ({t} L), then by the above argument we can fnd n polynomal tme a sub-path P of W such that P s an odd I-path and 2x (V (P ) L) + x (V (P ) I) < b(p ) + 1 2, otherwse we can conclude that x conv(x(g, b, I)). To conclude, we only need to show a polynomal tme algorthm that, gven an undrected graph Γ wth nonnegatve lengths on the edges l e, e E(Γ), a subset F E(Γ), and a par of nodes r, t V (Γ), determnes the walk W of mnmum length between r and t such that E(W ) F s odd, or determnes that no such walk exsts. Notce that ths problem easly reduces to the case where F = E(Γ), snce we can construct a graph Γ by subdvdng each edge uv E(Γ) \ F nto the path u, w, v, where w s a new node, and assgn lengths to uw and wv n such a way that the sum of such lengths equals l uv. Clearly, a walk W between two nodes r and t n Γ contans an odd number of edges n F f and only f the correspondng walk W n Γ has an odd number of edges. Furthermore, W and W have the same length. By the prevous argument, we are nterested n the problem of fndng a shortest walk wth an odd number of edges between a gven par of nodes. Ths problem can be solved n polynomal tme. Snce, as far as we know, ths fact s folklore, we brefly descrbe an algorthm. We construct a new graph Γ as follows. For every node v V (Γ), we have a par of nodes v, v n V (Γ ). For every edge uv E(Γ), we have two edges uv and u v n Γ, both wth length l uv. One can verfy that a walk W wth an odd number of edges between r and t exsts n Γ f and only f there exsts a walk of the same length between r and t n Γ. Hence we only need to fnd a shortest path between r and t n Γ, f any exsts, and output the correspondng walk n Γ. References [1] M. Confort, M. D Summa, F. Esenbrand, L.A. Wolsey, Network formulatons of mxed-nteger programs, In preparaton, [2] F. Esenbrand, Mxed Integer Programmng over TU systems, Manuscrpt, [3] O. Günlük and Y. Pochet, Mxng mxed nteger nequaltes, Mathematcal Programmng 90 (2001), [4] A. Mller and L.A. Wolsey, Tght formulatons for some smple MIPs and convex objectve IPs, Mathematcal Programmng B 98 (2003),

15 [5] G.L. Nemhauser, L.A. Wolsey, Integer and Combnatoral Optmzaton, Wley Interscence, New York, [6] Y. Pochet and L.A. Wolsey, Polyhedra for lot-szng wth Wagner-Whtn costs, Mathematcal Programmng 67 (1994), [7] Y. Pochet and L.A. Wolsey, Producton Plannng by Mxed Integer Programmng, Sprnger Seres n Operatons Research and Fnancal Engneerng, New York, [8] A. Schrver, Theory of Lnear and Integer Programmng, Wley, New York, [9] M. Skutella, Mxed Integer vertex cover on paths, Manuscrpt,