More Network Flow Applications

Transcription

1 Chaper 19 More Nework Flow Applicaions CS 473: Fundamenal Algorihms, Spring 2011 April 5, Edge disjoin pahs Edge-disjoin pahs in a direced graphs Quesion Given a graph G (eiher direced or undireced), wo verices s and, and a parameer k, he ask is o compue k pahs from s o in G, such ha hey are edge disjoin; namely, hese pahs do no share an edge. To solve his problem, we will conver G (assume G is a direced graph for he ime being) ino a nework flow graph H, such ha every edge has capaciy 1. Find he maximum flow in G (beween s and ). We claim ha he value of he maximum flow in he nework H, is equal o he number of edge disjoin pahs in G. Lemma If here are k edge disjoin pahs in G beween s and, hen he maximum flow in H is a leas k. Proof : Given k such edge disjoin pahs, push one uni of flow along each such pah. The resuling flow is legal in h and i has value k. Definiion (0/1-flow.) A flow f is 0/1-flow if every edge has eiher no flow on i, or one uni of flow. Lemma Le f be a 0/1 flow in a nework H wih flow value µ. Then here are µ edge disjoin pahs beween s and in H. Proof : By inducion on he number of edges in H ha has one uni of flow assigned o hem by f. If µ = 0 hen here is nohing o prove. 1

2 Oherwise, sar raversing he graph H from s raveling only along edges wih flow 1 assigned o hem by f. We mark such an edge as used, and do no allow one o ravel on such an edge again. There are wo possibiliies: (i) We reached he arge verex. In his case, we ake his pah, add i o he se of oupu pahs, and reduce he flow along he edges of he generaed pah π o 0. Le H be he resuling flow nework and f he resuling flow. We have f = µ 1, H has less edges, and by inducion, i has µ 1 edge disjoin pahs in H beween s and. Togeher wih π his forms µ such pahs. (ii) We visi a verex v for he second ime. In his case, our raversal conains a cycle C, of edges in H ha have flow 1 on hem. We se he flow along he edges of C o 0 and use inducion on he remaining graph (since i has less edges wih flow 1 on hem). The value of he flow f did no change by removing C, and as such i follows by inducion ha here are µ edge disjoin pahs beween s and in H. Since he graph G is simple, here are a mos n = V (H) edges ha leave s. As such, he maximum flow in H is n. Thus, applying he Ford-Fulkerson algorihm, akes O(mn) ime. The exracion of he pahs can also be done in linear ime by applying he algorihm in he proof of Lemma As such, we ge: Theorem Given a direced graph G wih n verices and m edges, and wo verices s and, one can compue he maximum number of edge disjoin pahs beween s and in H, in O(mn) ime. As a consequence we ge he following cue resul: Lemma In a direced graph G wih nodes s and he maximum number of edge disjoin s pahs is equal o he minimum number of edges whose removal separaes s from. Proof : Le U be a collecion of edge-disjoin pahs from s o in G. If we remove a se F of edges from G and separae s from, hen i mus be ha every pah in U uses a leas one edge of F. Thus, he number of edge-disjoin pahs is bounded by he number of edges needed o be removed o separae s and. Namely, U F. As for he oher direcion, le F be a se of edges has is removal separaes s and. We claim ha he se F form a cu in G beween s and. Indeed, le S be he se of all verices in G ha are reachable from s wihou using an edge of F. Clearly, if F is minimal hen i mus be all he edges of he cu (S, T ) (in paricular, if F conains some edge which is no in (S, T ) we can remove i and ge a smaller separaing se of edges). In paricular, he smalles se F wih his separaing propery has he same size as he minimum cu beween s and in G, which is by he max-flow mincu heorem, also he maximum flow in he graph G (where every edge has capaciy 1). Bu hen, by Theorem , here are F edge disjoin pahs in G (since F is he amoun of he maximum flow). 2

3 Edge-disjoin pahs in undireced graphs We would like o solve he s- disjoin pah problem for an undireced graph. Problem Given undireced graph G, s and, find he maximum number of edgedisjoin pahs in G beween s and. The naural approach is o duplicae every edge in he undireced graph G, and ge a (new) direced graph H. Nex, apply he algorihm of Secion o H. So compue for H he maximum flow f (where every edge has capaciy 1). The problem is he flow f migh use simulaneously he wo edges (u v) and (v u). Observe, however, ha in such case we can remove boh edges from he flow f. In he resuling flow is legal and has he same value. As such, if we repeaedly remove hose double edges from he flow f, he resuling flow f has he same value. Nex, we exrac he edge disjoin pahs from he graph, and he resuling pahs are now edge disjoin in he original graph. Lemma There are k edge-disjoin pahs in an undireced graph G from s o if and only if he maximum value of an s flow in he direced version H of G is a leas k. Furhermore, he Ford-Fulkerson algorihm can be used o find he maximum se of disjoin s- pahs in G in O(mn) ime Applicaions Survey design We would like o design a survey of producs used by consumers (i.e., Consumer i: wha did you hink of produc j? ). The ih consumer agreed in advance o answer a cerain number of quesions in he range [c i, c i]. Similarly, for each produc j we would like o have a leas p j opinions abou i, bu no more han p j. Each consumer can be asked abou a subse of he producs which hey consumed. In paricular, we assume ha we know in advance all he producs each consumer used, and he above consrains. The quesion is how o assign quesions o consumers, so ha we ge all he informaion we wan o ge, and every consumer is being asked a valid number of quesions. The idea of our soluion is o reduce he design of he survey o he problem of compuing a circulaion in graph. Firs, we build a biparie graph having consumers on one side, and producs on he oher side. Nex, we inser he edge beween consumer i and produc j if he produc was used by his consumer. The capaciy of his edge is going o be 1. Inuiively, we are going o compue a flow in his nework which is going o be an ineger number. As such, every edge would be assigned eiher 0 or 1, where 1 is inerpreed as asking he consumer abou his produc. 3

4 1: Boson (depar 6 A.M.) - Washingon DC (arrive 7 A.M,). 2: Urbana (depar 7 A.M.) - Champaign (arrive 8 A.M.) 3: Washingon (depar 8 A.M.) - Los Angeles (arrive 11 A.M.) 4: Urbana (depar 11 A.M.) - San Francisco (arrive 2 P.M.) 5: San Francisco (depar 2:15 P.M.) - Seale (arrive 3:15 P.M.) 6: Las Vegas (depar 5 P.M.) - Seale (arrive 6 P.M.) (i) (ii) 5 Figure 19.1: (i) a se F of flighs ha have o be served, and (ii) he corresponding graph G represening hese flighs. The nex sep, is o connec a source o all he consumers, where he edge (s i) has lower bound c i and c i, c i p j, p j 0, 1 upper bound c i. Similarly, we connec all he producs o he desinaion, where (j ) has lower bound p j and upper bound p s j. We would like o compue a flow from s o in his nework ha comply wih he consrains. However, we only know how o compue a circulaion on such a nework. To overcome his, we creae an edge wih infinie capaciy beween and s. Now, we are only looking for a valid circulaion in he resuling graph G which complies wih he aforemenioned consrains. See figure on he righ for an example of G. Given a circulaion f in G i is sraighforward o inerpre i as a survey design (i.e., all middle edges wih flow 1 are quesions o be asked in he survey). Similarly, one can verify ha given a valid survey, i can be inerpreed as a valid circulaion in G. Thus, compuing circulaion in G indeed solves our problem. We summarize: Lemma Given n consumers and u producs wih heir consrains c 1, c 1, c 2, c 2,..., c n, c n, p 1, p 1,..., p u, p u and a lis of lengh m of which producs where used by which consumers. An algorihm can compue a valid survey under hese consrains, if such a survey exiss, in ime O((n + u)m 2 ) Airline Scheduling Problem Given informaion abou flighs ha an airline needs o provide, generae a profiable schedule. The inpu is a deailed informaion abou legs of fligh ha he airline need o serve. We denoe his se of flighs by F. We would like o find he minimum number of airplanes needed o carry ou his schedule. For an example of possible inpu, see Figure 19.1 (i). 4

5 We can use he same airplane for wo segmens i and j if he desinaion of i is he origin of he segmen j and here is enough ime in beween he wo flighs for required mainenance. Alernaively, he airplane can fly from des(i) o origin(j) (assuming ha he ime consrains are saisfied). Example As a concree example, consider he flighs: 1. Boson (depar 6 A.M.) - Washingon D.C. (arrive 7 A.M,). 2. Washingon (depar 8 A.M.) - Los Angeles (arrive 11 A.M.) 3. Las Vegas (depar 5 P.M.) - Seale (arrive 6 P.M.) This schedule can be served by a single airplane by adding he leg Los-Angeles (depar 12 noon)- Las Vegas (1 P,M.) o his schedule Modeling he problem The idea is o model he feasibiliy consrains by a graph. Specifically, G is going o be a direced graph over he fligh legs. For i and j, wo given fligh legs, he edge (i j) will be presen in he graph G if he same airplane can serve boh i and j; namely, he same airplane can perform leg i and aferwards serves he leg j. Thus, he graph G is acyclic. Indeed, since we can have an edge (i j) only if he fligh j comes afer he fligh i (in ime), i follows ha we can no have cycles. We need o decide if all he required legs can be served using only k airplanes? Soluion k s k 1, 1 u 1 v 1 1, 1 u 2 v 2 1, 1 u 3 v 3 u 1, 1 4 v 4 1, 1 u 5 v 5 1, 1 u 6 v 6 Figure 19.2: The resuling graph H for he insance of airline scheduling from Figure The idea is o perform a reducion of his problem o he compuaion of circulaion. Specifically, we consruc a graph H, as follows: For every leg i, we inroduce wo verices u i, v i V (H). We also add a source verex s and a sink verex o H. We se he demand a o be k, and he demand a s o be k (i.e., k unis of flow are leaving s and need o arrive o T ). Each fligh on he lis mus be served. This is forced by inroducing an edge e i = (u i v i ), for each leg i. We also se he lower bound on e i o be 1, and he capaciy on e i o be 1 (i.e., l(e i ) = 1 and c(e i ) = 1). 5 k

6 If he same plane can perform fligh i and j (i.e., (i j) E(G)) hen add an edge (v i u j ) wih capaciy 1 o H (wih no lower bound consrain). Since any airplane can sar he day wih fligh i, we add an edge (s u i ) wih capaciy 1 o H, for all flighs i. Similarly, any airplane can end he day by serving he fligh j. Thus, we add edge (v j ) wih capaciy 1 o G, for all flighs j. If we have exra planes, we do no have o use hem. As such, we inroduce a overflow edge (s ) wih capaciy k, ha can carry over all he unneeded airplanes from s direcly o. Le H denoe he resuling graph. See Figure 19.2 for an example. Lemma There is a way o perform all flighs of F using a mos k planes if and only if here is a feasible circulaion in he nework H. Proof : Assume here is a way o perform he flighs using k k flighs. Consider such a feasible schedule. The schedule of an airplane in his schedule defines a pah π in he nework H ha sars a s and ends a, and we send one uni of flow on each such pah. We also send k k unis of flow on he edge (s ). Noe, ha since he schedule is feasible, all legs are being served by some airplane. As such, all he middle edges wih lower-bound 1 are being saisfied. Thus, his resuls is a valid circulaion in H ha saisfies all he given consrains. As for he oher direcion, consider a feasible circulaion in H. This is an ineger valued circulaion by he Inegraliy heorem. Suppose ha k unis of flow are sen beween s and (ignoring he flow on he edge (s )). All he edges of H (excep (s )) have capaciy 1, and as such he circulaion on all oher edges is eiher zero or one (by he Inegraliy heorem). We conver his ino k pahs by repeaedly raversing from he verex s o he desinaion, removing he edges we are using in each such pah afer exracing i (as we did for he k disjoin pahs problem). Since we never use an edge wice, and H is acyclic, i follows ha we would exrac k pahs. Each of hose pahs correspond o one airplane, and he overall schedule for he airplanes is valid, since all required legs are being served (by he lower-bound consrain). Exensions and limiaions. There are a lo of oher consideraions ha we ignored in he above problem: (i) airplanes have o undergo long erm mainenance reamens every once in awhile, (ii) one needs o allocae crew o hese flighs, (iii) schedule differ beween days, and (iv) ulimaely we ineresed in maximizing revenue (a much more fluffy concep and much harder o explicily describe). In paricular, while nework flow is used in pracice, real world problems are complicaed, and nework flow can capure only a few aspecs. More han undermining he usefulness of nework flow, his emphasize he complexiy of real-world problems. 6

7 (i) (ii) Figure 19.3: The (i) inpu image, and (ii) a possible segmenaion of he image Image Segmenaion In he image segmenaion problem, he inpu is an image, and we would like o pariion i ino background and foreground. For an example, see Figure The inpu is a bimap on a grid where every grid node represens a pixel. We cover his grid ino a direced graph G, by inerpreing every edge of he grid as wo direced edges. See he figure on he righ o see how he resuling graph looks like. Specifically, he inpu for ou problem is as follows: A bimap of size N N, wih an associaed direced graph G = (V, E). For every pixel i, we have a value f i 0, which is an esimae of he likelihood of his pixel o be in foreground (i.e., he larger f i is he more probable ha i is in he foreground) For every pixel i, we have (similarly) an esimae b i of he likelihood of pixel i o be in background. For every wo adjacen pixels i and j we have a separaion penaly p ij, which is he price of separaing i from j. This quaniy is defined only for adjacen pixels in he bimap. (For he sake of simpliciy of exposiion we assume ha p ij = p ji. Noe, however, ha his assumpion is no necessary for our discussion.) Problem Given inpu as above, pariion V (he se of pixels) ino wo disjoin subses F and B, such ha q(f, B) = f i + b i p ij. i F i B 7 (i,j) E, F {i,j} =1

8 is maximized. We can rewrie q(f, B) as: q(f, B) = i F f i + j B b j (i,j) E, F {i,j} =1 p ij = i v (f i + b i ) i B f i j F b j (i,j) E, F {i,j} =1 Since he erm i v (f i + b i ) is a consan, maximizing q(f, B) is equivalen o minimizing u(f, B), where u(f, B) = f i + b j + p ij. (19.1) i B j F (i,j) E, F {i,j} =1 How do we compue his pariion. Well, he basic idea is o compue a minimum cu in a graph such has is price would correspond o u(f, B). Before dwelling ino he exac deails, i is useful o play around wih some oy examples o ge some inuiion. Noe, ha we are using he max-flow algorihm as an algorihm for compuing minimum direced cu. To begin wih, consider a graph having a source s, a verex i, and f s i b i i a sink. We se he price of (s i) o be f i and he price of he edge (i ) o be b i. Clearly, here are wo possible cus in he graph, eiher ({s, i}, {}) (wih a price b i ) or ({s}, {i, }) (wih a price f i ). In paricular, every pah of lengh 2 in he graph beween s and forces he algorihm compuing he minimum-cu (via nework flow) o choose one of he edges, o he cu, where he algorihm prefers he edge wih lower price. Nex, consider a bimap wih wo verices i an j ha are adjacen. f i i b i Clearly, minimizing he firs wo erms in Eq. (19.1) is easy, by generaing s f j j lengh wo parallel pahs beween s and hrough i and j. See figure b j on he righ. Clearly, he price of a cu in his graph is exacly he price of he pariion of {i, j} ino background and foreground ses. However, his ignores he separaion penaly p ij. To his end, we inroduce wo new edges (i j) and (j i) ino he graph and se heir price o be p ij. Clearly, a price of a cu in he graph can be inerpreed as he value of u(f, B) of he s f i p ij f j corresponding ses F and B, since all he edges in he segmenaion from nodes of F o nodes of B are conribuing heir separaion price o he cu price. Thus, if we exend his idea o he direced graph G, he minimum-cu in he resuling graph would corresponds o he required segmenaion. Le us recap: Given he direced grid graph G = (V, E) we add wo special source and sink verices, denoed by s and respecively. Nex, for all he pixels i V, we add an edge e i = (s i) o he graph, seing is capaciy o be c(e i ) = f i. Similarly, we add he edge e i = (j ) wih capaciy c(e i) = b i. Similarly, for every pair of verices i.j in ha grid ha are adjacen, we assign he cos p ij o he edges (i j) and (j i). Le H denoe he resuling graph. The following lemma, follows by he above discussion. 8 i j b i p ij b j p ij.

9 Lemma A minimum cu (F, B) in H minimizes u(f, B). Using he minimum-cu max-flow heorem, we have: Theorem One can solve he segmenaion problem, in polynomial ime, by compuing he max flow in he graph H Projec Selecion You have a small company which can carry ou some projecs ou of a se of projecs P. Associaed wih each projec i P is a revenue p i, where p i > 0 is a profiable projec and p i < 0 is a losing projec. To make hings ineresing, here is dependency beween projecs. Namely, one has o complee some infrasrucure projecs before one is able o do oher projecs. Namely, you are provided wih a graph G = (P, E) such ha (i j) E if and only j is a prerequisie for i. Definiion A se A P is feasible if for all i A, all he prerequisies of i are also in A. Formally, for all i A, wih an edge (i j) E, we have j A. The profi associaed wih a se of projecs A P is profi(a) = i A p i. Problem (Projec Selecion Problem.) Selec a feasible se of projecs maximizing he overall profi. The idea of he soluion is o reduce he problem o a minimum-cu in a graph, in a similar fashion o wha we did in he image segmenaion problem The reducion The reducion works by adding wo verices s and o he graph G, we also perform he following modificaions: For all projecs i P wih posiive revenue (i.e., p i > 0) add he e i = (s i) o G and se he capaciy of he edge o be c(e i ) = p i. Similarly, for all projecs j P, wih negaive revenue (k.e., p j e j = (j ) o G and se he edge capaciy o c(e j) = p j. < 0) add he edge Compue a bound on he max flow (and hus also profi) in his nework: C = i P,p i >0 p i. Se capaciy of all oher edges in G o 4C (hese are he dependency edges in he projec, and inuiively hey are oo expensive o be broken by a cu). Le H denoe he resuling nework. Le A P Be a se of feasible projecs, and le A = A {s} and B = (P \ A) {}. Consider he s- cu (A, B ) in H. Noe, ha no edge in E(G) is of (A, B ) since A is a feasible se. 9

10 Lemma The capaciy of he cu (A, B ), as defined by a feasible projec se A, is c(a, B ) = C i A p i = C profi(a). Proof : The edges of H are eiher: (i) Edges of G, (ii) edges emanaing from s, and (iii) edges enering. Since A is feasible, i follows ha no edges of ype (i) conribue o he cu. The edges enering conribue o he cu he value X = The edges leaving he source s conribue Y = i/ A and p i >0 p i = i P,p i >0 by he definiion of C. The capaciy of he cu (A, B ) is as claimed. X + Y = i A and p i <0 ( p i ) + C i A and p i <0 p i p i. i A and p i >0 p i = C i A and p i >0 p i. = C i A i A and p i >0 p i, p i = C profi(a), Lemma If (A, B ) is a cu wih capaciy a mos C in G, hen he se A = A \ {s} is a feasible se of projecs. Namely, cus (A, B ) of capaciy C in H corresponds one-o-one o feasible ses which are profiable. Proof : Since c(a, B ) C i mus no cu any of he edges of G, since he price of such an edge is 4C. As such, A mus be a feasible se. Puing everyhing ogeher, we are looking for a feasible se ha maximizes i A p i. This corresponds o a se A = A {S} of verices in H ha minimizes C i A p i, which is also he cu capaciy (A, B ). Thus, compuing a minimum-cu in H corresponds o compuing he mos profiable feasible se of projecs. Theorem If (A, B ) is a minimum cu in H hen A = A \ {s} is an opimum soluion o he projec selecion problem. In paricular, using nework flow he opimal soluion can be compued in polynomial ime. 10

11 Baseball eliminaion There is a baseball league aking place and i is nearing he end of he season. One would like o know which eams are sill candidaes o winning he season. Example There 4 eams ha have he following number of wins: New York: 92, Balimore: 91, Torono: 91, Boson: 90, and here are 5 games remaining (all pairs excep New York and Boson). We would like o decide if Boson can sill win he season? Namely, can Boson finish he season wih as many poin as anybody else? (We are assuming here ha a every game he winning eam ges one poin and he losing eam ges nada. 1 ) Firs analysis. Observe, ha Boson can ge a mos 92 wins. In paricular, if New York wins any game hen i is over since New-York would have 93 poins. Thus, o Boson o have any hope i mus be ha boh Balimore wins agains New York and Torono wins agains New York. A his poin in ime, boh eams have 92 poins. Bu now, hey play agains each oher, and one of hem would ge 93 wins. So Boson is eliminaed! Second analysis. As before, Boson can ge a mos 92 wins. All hree oher eams ges X = (5 2) poins ogeher by he end of he league. As such, one of hese hree eams will ge X/3 = 93 poins, and as such Boson is eliminaed. While he analysis of he above example is very cue, i is oo edious o be done each ime we wan o solve his problem. No o menion ha i is unclear how o exend hese analyses o oher cases Problem definiion Problem The inpu is a se S of eams, where for every eam x S, he eam has w x poins accumulaed so far. For every pair of eams x, y S we know ha here are g xy games remaining beween x and y. Given a specific eam z, we would like o decide if z is eliminaed? Alernaively, is here away such z would ge as many wins as anybody else by he end of he season? Soluion Firs, we can assume ha z wins all is remaining games, and le m be he number of poins z has in his case. Our purpose now is o build a nework flow so we can verify ha no oher eam mus ge more han m poins. 1 nada = nohing. 11

12 To his end, le s be he source (i.e., he source of wins). For every remaining game, a flow of one uni would go from s o one of he eams playing i. Every eam can have a mos m w x flow from i o he arge. If he max flow in his nework has value α = x,y z,x<y (which is he maximum flow possible) hen here is a scenario such ha all oher eams ges a mos m poins and z can win he season. Negaing his saemen, we have ha if he maximum flow is smaller han α hen z is eliminaed, since here mus be a eam ha ges more han m poins. Consrucion. Le S = S \ {z} be he se of eams, and le α = g xy. We creae a {x,y} S nework flow G. For every eam x S we add a verex v x o he nework G. We also add he source and sink verices, s and, respecively, o G. For every pair of eams x, y S, such ha g xy > 0 we creae a node u xy, and add an edge (s u xy ) wih capaciy g xy o G. We also add he edge (u xy v x ) and (u xy v y ) wih infinie capaciy o G. Finally, for each eam x we add he edge (v x ) wih capaciy m w x o G. How he relevan edges look like for a pair of eams x and y is depiced on he righ. g xy s g xy u xy v x v y m w x m w y Analysis. If here is a flow of value α in G hen here is a way ha all eams ge a mos m wins. Similarly, if here exiss a scenario such ha z ies or ges firs place hen we can ranslae his ino a flow in G of value α. This implies he following resul. Theorem Team z has been eliminaed if and only if he maximum flow in G has value sricly smaller han α. Thus, we can es in polynomial ime if z has been eliminaed A compac proof of a eam being eliminaed Ineresingly, once z is eliminaed, we can generae a compac proof of his fac. Theorem Suppose ha eam z has been eliminaed. Then here exiss a proof of his fac of he following form: 1. The eam z can finish wih a mos m wins. 2. There is a se of eams Ŝ S so ha w x + g xy > m Ŝ. s Ŝ {x,y} Ŝ (And hence one of he eams in Ŝ mus end wih sricly more han m wins.) 12

13 Proof : If z is eliminaed hen he max flow in G has value γ, which is smaller han α. By he minimum-cu max-flow } heorem, here exiss a minimum cu (S, T ) of capaciy γ in G, and le {x Ŝ = v x S Claim For any wo eams x and y for which he verex u xy exiss, we have ha u xy S if and only if boh x and y are in Ŝ. Proof : ( ) x / Ŝ or y / Ŝ = u xy / S : If x is no in Ŝ hen v x is in T. Bu hen, if u xy is in S he edge (u xy v x ) is in he cu. However, his edge has infinie capaciy, which implies his cu is no a minimum cu (in paricular, (S, T ) is a cu wih capaciy smaller han α). As such, in such a case u xy mus be in T. This implies ha if eiher x or y are no in Ŝ hen i mus be ha u xy T. (And as such u xy / S.) x Ŝ and y Ŝ = u xy S : Assume ha boh x and y are in Ŝ, hen v x and v y are in S. We need o prove ha u xy S. If u xy T hen consider he new cu formed by moving u xy o S. For he new cu (S, T ) we have c(s, T ) = c(s, T ) c ( (s u xy ) ). s g xy u xy v x v y m w x m w y Namely, he cu (S, T ) has a lower capaciy han he minimum cu (S, T ), which is a conradicion. See figure on he righ for his impossible cu. We conclude ha u xy S. The above argumenaion implies ha edges of he ype (u xy v x ) can no be in he cu (S, T ). As such, here are wo ype of edges in he cu (S, T ): (i) (v x ), for x Ŝ, and (ii) (s u xy ) where a leas one of x or y is no in Ŝ. As such, he capaciy of he cu (S, T ) is c(s, T ) = x Ŝ (m w x ) + {x,y} Ŝ However, c(s, T ) = γ < α, and i follows ha m Ŝ w x x Ŝ Namely, x Ŝ w x + {x,y} Ŝ g xy = m Ŝ w x + α x Ŝ {x,y} Ŝ g xy > m Ŝ, as claimed. 13 g xy < α α = 0. {x,y} Ŝ g xy.