Topics in Combinatorial Optimization May 11, Lecture 22

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1 8.997 Topics in Combinaorial Opimizaion May, 004 Lecure Lecurer: Michel X. Goemans Scribe: Alanha Newman Muliflows an Disjoin Pahs Le G = (V,E) be a graph an le s,,s,,...s, V be erminals. Our goal is o fin isjoin pahs beween s i an i for each i, i. There are irece an unirece versions of his problem, i.e. G can be irece or unirece an we may wan o fin irece pahs from s i o i or unirece pahs beween hese erminal pairs. Aiionally, we specify if we wan o fin verex isjoin pahs or ege isjoin pahs (arc isjoin pahs for irece graphs). These isjoin pah problems can be viewe as specific cases of he muliflow problem.. Muliflows Suppose we are given he following inpus: a graph G = (V,E) (irece or unirece), erminals s,,s,,...s, V, emans i : i =,...,, + ineger (or raional) capaciies on he eges, c : E Z. For each i, fin an (s i, i )-flow f i of value i. Noe ha even for unirece graphs, flow is irece. Le f i (e) be he amoun of flow from s i o i ha uses ege e. A vali flow mus obey he capaciy consrain: for each ege e E, i= f i (e) c(e).. Ege Disjoin Pahs To fin ege isjoin pahs, we can se c(e) = for all e E an hen fin an ineger muliflow. The problem of fining verex isjoin pahs in a irece graph can be reuce o he problem of fining ege isjoin pahs in a irece graph; every verex v V unergoes he ransformaion shown in figure. Thus, a se of ege isjoin pahs in he moifie graph correspons o a se of pahs in he original graph in which each verex is use a mos once. Figure : Each verex unergoes he illusrae ransformaion. Toay, we focus on fining ege isjoin pahs in unirece graphs. Noe ha he problem of fining ege isjoin pahs is very ifferen in erms of complexiy for irece an unirece graphs. -

2 Ege Disjoin Pahs in Unirece Graphs: G is an unirece graph. Do here exis wo ege isjoin pahs beween s an? This problem can be solve easily by eermining if he minimum s- cu conains a leas wo eges. Arc Disjoin Pahs in Direce Graphs: G is a irece graph. Do here exis wo arc isjoin pahs, one from s o an one from o s? This problem is NP-har! The ege isjoin pahs problem in unirece graphs can be reuce o he arc isjoin pahs problem in irece graphs. Each ege in he original unirece graph is replace by he gage shown in figure. a b a b Figure : Each ege in he original unirece graph is replace by he above gage..3 Fracional Muliflow We focus on ege isjoin pahs (muliflows) in unirece graphs. When =, flow is easy. We can fin ineger flow using he max-flow min-cu heorem. In general, eciing if a muliflow exiss can be eermine by solving a linear program consising of flow an capaciy consrains. Le P i be se of all pahs beween s i an i. We have a variable x p for every such pah p P i. We have he following primal LP: max 0 x x p = i p Pi x p c(e) i:p P i,e p x p 0. Wha oes ual mean in his case? We use variables l(e) for each ege e E, an variables b i for i =,...,. min c(e)l(e) b i i () i= l(e) b i 0 p P i i =,... e p l(e) 0. To mae he erm ( i= b i i ) small, we shoul mae b i as large as possible. Fix he ege funcion l : E Q. Then b i is he (minimum) is l (s i, i ). The objecive funcion of he ual () can be rewrien: c(e)l(e) i is l (s i, i ). i= -

3 If he primal LP is feasible, hen here is no soluion for he ual LP wih a negaive objecive value. So here exiss a fracional muliflow if an only if l(e) 0,e E, he following hols: c(e)l(e) i is l (s i, i ). () i= Dualiy shows ha his is a necessary an sufficien for he exisence of a fracional muliflow. Ineger Muliflows In general, he problem of eermining when here is an ineger muliflow is NP-complee. However, here are special coniions ha imply he exisence of an ineger muliflow in cerain classes of graphs. Le R be a se of eges: R = {(s i, i ): i =,... }. (3) The se of eges in E ougoing from verex se U is enoe by δ E (U) an he se of eges in R ougoing from verex se U is enoe by δ R (U). A necessary coniion for he exisence of a muliflow (an hus of an ineger muliflow) is he cu coniion: c(δ E (U)) (δ R (U)), U V. In general, he cu coniion is no sufficien o guaranee he exisence of an ineger muliflow (or fracional muliflow) in a graph. However, in some cases of he muliflow problem, he cu coniion is sufficien for he exisence of a fracional muliflow. Furhermore, here are several cases nown where he cu coniion implies he exisence of an ineger muliflow when he Euler coniion is saisfie: c(δ E (v)) + (δ R (v)) is even, for each verex v. For example, when =, we have he following implicaions: (i) Cu coniion fracional muliflow. (ii) Cu coniion an ineger capaciies half-inegral muliflow. (iii) Cu coniion, ineger capaciies, an Euler coniion inegral muliflow. The firs proof of (i) an (ii) for he case when = was is ue o Hu. The proof of (iii) is ue o Rohschil an Winson. Noe ha (iii) implies (i) an (ii). For example, Consier he graph in Figure 3, le =, =. Le he capaciy of each ege be. Noe ha he cu coniion is saisfie bu he Euler coniion is no. However, suppose we ouble every capaciy an eman, hen he Euler coniion is saisfie. We can conver an ineger soluion for his laer problem o a half-inegral soluion for he original problem. Some goo cases in which coniions (i), (ii) an (iii) are saisfie are:. If here are wo commoiies, i.e. =, hen cu coniion an Euler coniion are sufficien for ineger muliflow. -3

4 s s Figure 3: When he capaciy of each ege in his graph is an, =, he Euler coniion is no saisfie. There exiss a half-inegral muliflow, bu no inegral muliflow.. G + D has no K 5 minor, e.g. G + D is planar, where D is he eman graph, D = (V,R) (see (3)). 3. {(s, ),...,(s, )} G is planar an all (s i, i ) are on bounary of ousie face. (Noe ha his oes no imply case.) 5. If here are faces an for each i, (s i, ) are boh on he insie face or boh on he ousie face. 3 Two-Commoiy Flows Theorem (Rohschil an Whinson) G = (V,E) is an unirece graph such ha c(e) Z + for e E. Terminals s,,s, are in V, an emans, are posiive inegers. Aiionally, he Euler coniion is saisfie for G. Then G has an ineger wo-commoiy flow if an only if he cu coniion is saisfie. Proof: Our goal is o fin flows from s o an from s o wih values an, respecively. We will show ha if he cu coniion is saisfie on G, hen we can fin such flows. G : s s s G : s s s Figure 4: The graphs G an G are consruce base on he given graph G. Firs, base on he graph G, consruc he graph G as shown in figure 4. Le he eges (s,s ) an (, )in G have capaciy an he eges (s,s )an(, )in G have capaciy.by he max-flow min-cu heorem, we can fin an ineger s - flow g wih value +, since he min-cu of G has value +. Noe ha his s - flow oes no necessarily give a wo-commoiy flow for he original problem (since some of he flow going hrough s may en up in ). Since he Euler coniion is saisfie, we will prove ha we can assume ha g(e) c(e) mo. To show his, firs noice ha he Euler coniion implies ha he oal capaciy incien o any verex of G is even. Furhermore, any inegral flow will use up an even amoun of capaciy incien o any verex. Now consier all he eges e E such ha g(e) c(e) mo. Since i is he case -4

5 ha e δ(v) (g(e) c(e)) = 0 mo, i follows ha an even number of eges ajacen o verex v have g(e) c(e) mo. Thus, he eges such ha g(e) c(e) mo mae up an Eulerian graph (an o no conain he arcs incien o s an ha we ae o G o mae up G ). We can ecompose his Eulerian graph ino cycles, an push push one uni of flow across all hese cycles (eiher increasing or ecreasing he flow by one uni along i epening on he orienaion), changing he pariy of g(e) for each such ege. Thus, for all eges e E, we have ha g(e) c(e) mo. For G, we have he same argumen. Thus, we fin an ineger flow h in G wih value + such ha h(e) = c(e), e E. Thus, for all eges e E, h(e) = g(e) mo. We arbirarily orien he eges of E o obain A. So for all a A, h(a) g(a) mo. Now we efine wo flows on he graph G: f (a) = [g(a)+ h(a)] f (a) = [g(a) h(a)]. The following properies are rue for he flows f an f :. f (a),f (a) are ineger flows (since f(a) an g(a) have he same pariy).. f (a) + f (a) = g(a)+ h(a) + g(a) h(a) max( g(a), h(a) ) c(a). 3. f is unis of flow from s o an f is unis of flow from s o. The las propery hols because we can show ha f (δ + (s )) f (δ (s )) = an f (δ ( )) f (δ + ( )) =. By conservaion of flow, if we consier he verex s in G, we have: Equaions (4) an (5) imply f (δ + (s )) f (δ (s )) =. g(δ + (s )) g(δ (s )) = (4) h(δ + (s )) h(δ (s )) = (5) g(δ ( )) g(δ + ( )) = (6) h(δ ( )) h(δ + ( )) =. (7) Equaions (6) an (7) imply f (δ ( )) f (δ + ( )) =. Similarly, we can show ha he las propery hols for flow f. If we consier verices s an in G, we have: g(δ + (s )) g(δ (s )) = (8) h(δ (s )) h(δ + (s )) = (9) g(δ ( )) g(δ + ( )) = (0) h(δ + ( )) h(δ ( )) =. () Equaions (8) an (9) imply f (δ + (s )) f (δ (s )) = an equaions (0) an () imply f (δ ( )) f (δ + (s )) =. As a final noe, consier he problem of maximizing he sum of he flow beween s an an beween s an. Thisishe max biflow problem. A bicu is a cu separaing s from an s from,husi iseiher a cu separaing s,s from, or a cu separaing s, from s,.one can show ha he following heorem follows from Theorem. Theorem The maximum biflow equals he minimum bicu. -5