23 Maximum Flows and Minimum Cuts

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1 A proce canno be underood by opping i. Underanding mu move wih he flow of he proce, mu join i and flow wih i. The Fir Law of Mena, in Frank Herber Dune (196) Conrary o expecaion, flow uually happen no during relaxing momen of leiure and enerainmen, bu raher when we are acively involved in a difficul enerprie, in a ak ha reche our menal and phyical abiliie.... Flow i hard o achieve wihou effor. Flow i no waing ime. Mihaly Cíkzenmihályi, Flow: The Pychology of Opimal Experience (1990) There a difference beween knowing he pah and walking he pah. Morpheu [Laurence Fihburne], The Marix (1999) 23 Maximum Flow and Minimum Cu In he mid-190, Air Force reearcher Theodore E. Harri and reired army general Frank S. Ro publihed a claified repor udying he rail nework ha linked he Sovie Union o i aellie counrie in Eaern Europe. The nework wa modeled a a graph wih 44 verice, repreening geographic region, and edge, repreening link beween hoe region in he rail nework. Each edge wa given a weigh, repreening he rae a which maerial could be hipped from one region o he nex. Eenially by rial and error, hey deermined boh he maximum amoun of uff ha could be moved from Ruia ino Europe, a well a he cheape way o dirup he nework by removing link (or in le abrac erm, blowing up rain rack), which hey called he boleneck. Their reul, including he drawing of he nework below, were only declaified in 1999.¹ Harri and Ro map of he Waraw Pac rail nework ¹Boh he map and he ory were aken fromfigure Alexander 2 Schrijver facinaing urvey On he hiory of combinaorial opimizaion From Harri(ill and Ro 1960). [19]: Schemaic diagram of he railway nework of he Weern Sovie Union and Eaern European counrie, wih a maximum flow of value 163,000 on from Ruia o Eaern Europe, and a cu of capaciy 163,000 on indicaed a The boleneck. Copyrigh 201 Jeff Erickon. Thi work i licened under a Creaive Common Licene (hp://creaivecommon.org/licene/by-nc-a/4.0/). Free diribuion i rongly encouraged; commercial diribuion i exprely forbidden. See hp:// for he mo recen reviion. The max-flow min-cu heorem 1 In he RAND Repor of 19 November 194, Ford and Fulkeron [194] gave (nex o defining he maximum flow problem and uggeing he implex mehod for i) he max-flow mincu heorem for undireced graph, aying ha he maximum flow value i equal o he

2 Thi one of he fir recorded applicaion of he maximum flow and minimum cu problem. For boh problem, he inpu i a direced graph G = (V, E), along wih pecial verice and called he ource and arge. A in he previou lecure, I will ue u v o denoe he direced edge from verex u o verex v. Inuiively, he maximum flow problem ak for he large amoun of maerial ha can be ranpored from o ; he minimum cu problem ak for he minimum damage needed o eparae from Flow An (, )-flow (or ju a flow if he ource and arge are clear from conex) i a funcion f : E 0 ha aifie he following conervaion conrain a every verex v excep poibly and : f (u v) = f (v w). u w In Englih, he oal flow ino v i equal o he oal flow ou of v. To keep he noaion imple, we define f (u v) = 0 if here i no edge u v in he graph. The value of he flow f, denoed f, i he oal ne flow ou of he ource verex : f := f ( w) f (u ). w I no hard o prove ha f i alo equal o he oal ne flow ino he arge verex, a follow. To implify noaion, le f (v) denoe he oal ne flow ou of any verex v: f (v) := f (u v) f (v w). u The conervaion conrain implie ha f (v) = 0 or every verex v excep and, o f (v) = f () + f (). v On he oher hand, any flow ha leave one verex mu ener anoher verex, o we mu have v f (v) = 0. I follow immediaely ha f = f () = f (). Now uppoe we have anoher funcion c : E 0 ha aign a non-negaive capaciy c(e) o each edge e. We ay ha a flow f i feaible (wih repec o c) if f (e) c(e) for every edge e. Mo of he ime we will conider only flow ha are feaible wih repec o ome fixed capaciy funcion c. We ay ha a flow f aurae edge e if f (e) = c(e), and avoid edge e if f (e) = 0. The maximum flow problem i o compue a feaible (, )-flow in a given direced graph, wih a given capaciy funcion, whoe value i a large a poible. u w 0/ /20 / 0/1 / /1 0/ /20 / An (, )-flow wih value. Each edge i labeled wih i flow/capaciy. 2

3 23.2 Cu An (, )-cu (or ju cu if he ource and arge are clear from conex) i a pariion of he verice ino dijoin ube S and T meaning S T = V and S T = where S and T. If we have a capaciy funcion c : E 0, he capaciy of a cu i he um of he capaciie of he edge ha ar in S and end in T: S, T := c(v w). v S w T (Again, if v w i no an edge in he graph, we aume c(v w) = 0.) Noice ha he definiion i aymmeric; edge ha ar in T and end in S are unimporan. The minimum cu problem i o compue an (, )-cu whoe capaciy i a large a poible An (, )-cu wih capaciy 1. Each edge i labeled wih i capaciy. Inuiively, he minimum cu i he cheape way o dirup all flow from o. Indeed, i i no hard o how he following relaionhip beween flow and cu: Lemma 1. Le f be any feaible (, )-flow, and le (S, T) be any (, )-cu. The value of f i a mo he capaciy of (S, T). Moreover, f = S, T if and only if f aurae every edge from S o T and avoid every edge from T o S. Proof: Chooe your favorie flow f and your favorie cu (S, T), and hen follow he bouncing inequaliie: f = f ( w) f (u ) by definiion w = f (v w) f (u v) v S w u = f (v w) f (u v) v S w T u u T by he conervaion conrain removing duplicae edge f (v w) becaue f (u v) 0 v S w T c(v w) v S w T becaue f (v w) c(v w) = S, T by definiion The wo inequaliie in hi derivaion are acually equaliie if and only if f (u v) = 0 and f (v w) = c(v w) for all v S and u, w T. Lemma 1 implie ha if f = S, T, hen f mu be a maximum flow, and (S, T) mu be a minimum cu. 3

4 23.3 The Maxflow Mincu Theorem Surpriingly, for any weighed direced graph, here i alway a flow f and a cu (S, T) ha aify he equaliy condiion. Thi i he famou max-flow min-cu heorem, fir proved by Leer Ford (of hore pah fame) and Delber Ferguon in 194 and independenly by Peer Elia, Amiel Feinein, and and Claude Shannon (of informaion heory fame) in 196. The Maxflow Mincu Theorem. In any flow nework wih ource and arge, he value of he maximum (, )-flow i equal o he capaciy of he minimum (, )-cu. Ford and Fulkeron proved hi heorem a follow. Fix a graph G, verice and, and a capaciy funcion c : E 0. The proof will be eaier if we aume ha he capaciy funcion i reduced: For any verice u and v, eiher c(u v) = 0 or c(v u) = 0, or equivalenly, if an edge appear in G, hen i reveral doe no. Thi aumpion i eay o enforce. Whenever an edge u v and i reveral v u are boh he graph, replace he edge u v wih a pah u x v of lengh wo, where x i a new verex and c(u x) = c(x v) = c(u v). The modified graph ha he ame maximum flow value and minimum cu capaciy a he original graph. Enforcing he one-direcion aumpion. Le f be a feaible flow. We define a new capaciy funcion c f : V V, called he reidual capaciy, a follow: c(u v) f (u v) if u v E c f (u v) = f (v u) if v u E. 0 oherwie Since f 0 and f c, he reidual capaciie are alway non-negaive. I i poible o have c f (u v) > 0 even if u v i no an edge in he original graph G. Thu, we define he reidual graph G f = (V, E f ), where E f i he e of edge whoe reidual capaciy i poiive. Noice ha he reidual capaciie are no necearily reduced; i i quie poible o have boh c f (u v) > 0 and c f (v u) > 0. 0/ /20 / 0/ 0/1 / /1 / / A flow f in a weighed graph G and he correponding reidual graph G f. Suppoe here i no pah from he ource o he arge in he reidual graph G f. Le S be he e of verice ha are reachable from in G f, and le T = V \ S. The pariion (S, T) i clearly an (, )-cu. For every verex u S and v T, we have c f (u v) = (c(u v) f (u v)) + f (v u) = 0, 4

5 which implie ha c(u v) f (u v) = 0 and f (v u) = 0. In oher word, our flow f aurae every edge from S o T and avoid every edge from T o S. Lemma 1 now implie ha f = S, T, which mean f i a maximum flow and (S, T) i a minimum cu. / 1 1 /20 / / 0/1 /1 0/ /20 / An augmening pah in G f wih value F = and he augmened flow f. On he oher hand, uppoe here i a pah = v 0 v 1 v r = in G f. We refer o v 0 v 1 v r a an augmening pah. Le F = min i c f (v i v i+1 ) denoe he maximum amoun of flow ha we can puh hrough he augmening pah in G f. We define a new flow funcion f : E a follow: f (u v) + F if u v i in he augmening pah f (u v) = f (u v) F if v u i in he augmening pah f (u v) oherwie To prove ha he flow f i feaible wih repec o he original capaciie c, we need o verify ha f 0 and f c. Conider an edge u v in G. If u v i in he augmening pah, hen f (u v) > f (u v) 0 and f (u v) = f (u v) + F by definiion of f f (u v) + c f (u v) by definiion of F = f (u v) + c(u v) f (u v) by definiion of c f = c(u v) Duh. On he oher hand, if he reveral v u i in he augmening pah, hen f (u v) < f (u v) c(u v), which implie ha f (u v) = f (u v) F by definiion of f f (u v) c f (v u) by definiion of F = f (u v) f (u v) by definiion of c f = 0 Duh. Finally, we oberve ha (wihou lo of generaliy) only he fir edge in he augmening pah leave, o f = f + F > 0. In oher word, f i no a maximum flow. Thi complee he proof! 23.4 Ford and Fulkeron augmening-pah algorihm Ford and Fulkeron proof of he Maxflow-Mincu Theorem ranlae immediaely o an algorihm o compue maximum flow: Saring wih he zero flow, repeaedly augmen he flow along any pah from o in he reidual graph, unil here i no uch pah. Thi algorihm ha an imporan bu raighforward corollary:

6 Inegraliy Theorem. If all capaciie in a flow nework are ineger, hen here i a maximum flow uch ha he flow hrough every edge i an ineger. Proof: We argue by inducion ha afer each ieraion of he augmening pah algorihm, all flow value and reidual capaciie are ineger. Before he fir ieraion, reidual capaciie are he original capaciie, which are inegral by definiion. In each laer ieraion, he inducion hypohei implie ha he capaciy of he augmening pah i an ineger, o augmening change he flow on each edge, and herefore he reidual capaciy of each edge, by an ineger. In paricular, he algorihm increae he overall value of he flow by a poiive ineger, which implie ha he augmening pah algorihm hal and reurn a maximum flow. If every edge capaciy i an ineger, he algorihm hal afer f ieraion, where f i he acual maximum flow. In each ieraion, we can build he reidual graph G f and perform a whaever-fir-earch o find an augmening pah in O(E) ime. Thu, for nework wih ineger capaciie, he Ford-Fulkeron algorihm run in O(E f ) ime in he wor cae. The following example how ha hi running ime analyi i eenially igh. Conider he 4-node nework illuraed below, where i ome large ineger. The maximum flow in hi nework i clearly 2. However, Ford-Fulkeron migh alernae beween puhing 1 uni of flow along he augmening pah u v and hen puhing 1 uni of flow along he augmening pah v u, leading o a running ime of Θ( ) = Ω( f ). v 1 u A bad example for he Ford-Fulkeron algorihm. Ford and Fulkeron algorihm work quie well in many pracical iuaion, or in eing where he maximum flow value f i mall, bu wihou furher conrain on he augmening pah, hi i no an efficien algorihm in general. The example nework above can be decribed uing only O(log ) bi; hu, he running ime of Ford-Fulkeron i acually exponenial in he inpu ize. 23. Irraional Capaciie If we muliply all he capaciie by he ame (poiive) conan, he maximum flow increae everywhere by he ame conan facor. I follow ha if all he edge capaciie are raional, hen he Ford-Fulkeron algorihm evenually hal, alhough ill in exponenial ime. However, if we allow irraional capaciie, he algorihm can acually loop forever, alway finding maller and maller augmening pah! Wore ye, hi infinie equence of augmenaion may no even converge o he maximum flow, or even o a ignifican fracion of he maximum flow! Perhap he imple example of hi effec wa dicovered by Uri Zwick. Conider he ix-node nework hown on he nex page. Six of he nine edge have ome large ineger capaciy, wo have capaciy 1, and one ha capaciy φ = ( 1)/ , choen o ha 1 φ = φ 2. To prove ha he Ford-Fulkeron algorihm can ge uck, we can wach he reidual capaciie of he hree horizonal edge a he algorihm progree. (The reidual capaciie of he oher ix edge will alway be a lea 3.) 6

7 Suppoe he Ford-Fulkeron algorihm ar by chooing he cenral augmening pah, hown in he large figure on he nex page. The hree horizonal edge, in order from lef o righ, now have reidual capaciie 1, 0, and φ. Suppoe inducively ha he horizonal reidual capaciie are φ k 1, 0, φ k for ome non-negaive ineger k. 1. Augmen along B, adding φ k o he flow; he reidual capaciie are now φ k+1, φ k, Augmen along C, adding φ k o he flow; he reidual capaciie are now φ k+1, 0, φ k. 3. Augmen along B, adding φ k+1 o he flow; he reidual capaciie are now 0, φ k+1, φ k Augmen along A, adding φ k+1 o he flow; he reidual capaciie are now φ k+1, 0, φ k+2. I follow by inducion ha afer 4n + 1 augmenaion ep, he horizonal edge have reidual capaciie φ 2n 2, 0, φ 2n 1. A he number of augmenaion grow o infiniy, he value of he flow converge o φ i = φ = 4 + < 7, even hough he maximum flow value i clearly i=1 1 1 ϕ A B C Uri Zwick non-erminaing flow example, and hree augmening pah. Pracically-minded reader migh wonder why we hould care abou irraional capaciie; afer all, compuer can repreen anyhing bu (mall) ineger or (dyadic) raional exacly. Good queion! One reaon i ha he ineger rericion i lierally arificial; i an arifac of acual compuaional hardware (or perhap he oherwie-irrelevan law of phyic), no an inheren feaure of he abrac compuaional problem. Bu a more pracical reaon i ha he behavior of he algorihm wih irraional inpu ell u omehing abou i wor-cae behavior in pracice given floaing-poin capaciie errible! Even wih very reaonable capaciie, a carele implemenaion of Ford-Fulkeron could ener an infinie loop imply becaue of round-off error Combining and Decompoing Flow Flow are normally defined a funcion on he edge of a graph aifying cerain conrain a he verice. However, flow have a econd repreenaion ha i more naural and ueful in cerain conex. 7

8 Conider an arbirary graph G wih ource verex and arge verex. Fix any wo (, )-flow f and g and any wo real number α and β, and conider he funcion h: E defined by eing h(u v) := α f (u v) + β g(u v) for every edge u v; we can wrie hi definiion more imply a h = αf + β g. Sraighforward definiion-chaing implie ha h i alo an (, )-flow wih value h = α f +β g. More generally, any weighed um of (, )-flow i alo an (, )-flow. I urn ou ha any (, )-flow can be wrien a a weighed um of flow wih a very pecial rucure. For any direced pah P from o, we define a correponding pah flow a follow; 1 if u v P, P(u v) = 1 if v u P, 0 oherwie. Sraighforward definiion-chaing implie ha he funcion P : E i indeed an (, )-flow wih value 1. I am deliberaely overloading he variable P o mean boh he pah (a equence of verice and direced edge) and he uni flow along ha pah. Similarly, for any cycle C, we define a correponding cycle flow 1 if u v C, C(u v) = 1 if v u C, 0 oherwie; again, i i eay o verify ha C : E i an (, )-flow wih value zero. Our earlier argumen implie ha any weighed um of hee pah and cycle flow give u anoher an (, )-flow; hi weighed um i called a flow decompoiion of he reuling flow. Moreover, every flow ha uch a decompoiion. Flow Decompoiion Theorem. Every feaible (, )-flow f can be wrien a a weighed um of direced (, )-pah and direced cycle. Moreover, a direced edge u v appear in a lea one of hee pah or cycle if and only if f (u v) > 0, and he oal number of pah and cycle i a mo he number of edge in he nework. Proof: We prove he heorem by inducion on he number of edge carrying non-zero flow. Fix an arbirary (, )-flow f in an arbirary flow nework G. There are hree cae o conider: If f (u v) = 0 for every edge u v, hen f i a weighed um of he empy e of pah and cycle. Suppoe f = 0, meaning flow i conerved a every verex, including and. Pick an arbirary edge u v wih f (u v) > 0. Conider an arbirary walk v 0 v 1 v 2 wih v 0 = u and v 1 = v, uch ha f (v i 1 v i ) > 0 for every index i. The conervaion conrain implie ha every verex ha ha incoming flow alo ha ougoing flow, o we can make hi walk arbirarily long; in paricular, he walk mu evenually vii ome verex more han once. Le j < k be he malle indice uch ha v j = v k. Then he ubwalk v j v j 1 v k i acually a imple direced cycle C. 8

9 Define f min (C) := min e C f (e), and conider he funcion f := f f min (C) C, or more verboely, f (u v) f min (C) if u v C, f (u v) := f (u v) + f min (C) if v u C, f (u v) oherwie. Sraighforward definiion chaing how ha f i indeed a feaible flow in G wih value 0. There i a lea one edge e C uch ha f (e) = f min (C) and herefore f (e) = 0. Thu, fewer edge carry flow in f han in f. The inducion hypohei implie ha f ha a valid decompoiion ino a mo E 1 pah and cycle. Adding C wih he appropriae weigh give u a flow decompoiion for f ; pecifically, f = f + f min (C) C. The final cae f > 0 i imilar o he previou cae. Conervaion implie ha here i a direced walk v 1 v 2 v l where every edge carrie poiive flow. By removing loop, we can aume wihou lo of generaliy ha hi walk i a imple pah P. Le f min (P) := min e P f (e); and define a new flow f := f f min (P) P. We eaily verify ha f i a feaible flow in G wih value f f min (P). The inducion hypohei implie ha f can be decompoed ino a mo E 1 pah and cycle. Adding P wih weigh give u a flow decompoiion for f. In all cae, we obain a valid flow decompoiion. The previou argumen implie ha we can renghen he decompoiion heorem in wo inereing pecial cae. Fir, we can decompoe any flow wih value zero ino a weighed um of cycle; no pah are neceary. Flow wih value zero are ofen called circulaion. On he oher hand, we can decompoe any acyclic (, )-flow ino a weighed um of (, )-pah; no cycle are neceary. The proof alo immediaely ranlae direcly ino an algorihm, imilar o he Ford-Fulkeron algorihm, o decompoe any (, )-flow ino pah and cycle in O(V E) ime. The algorihm repeaedly eek eiher a direced (, )-pah or a direced cycle in he remaining flow, and hen ubrac a much flow a poible along ha pah or cycle, unil he flow i empy. Each ieraion can be execued in O(V ) ime and remove a lea one edge from he graph, o he enire algorihm run in O(VE) ime. Flow decompoiion provide a naural lower bound on he running ime of any maximum-flow algorihm ha build he flow one pah or cycle a a ime. Every flow can be decompoed ino a mo E pah and cycle, each of which ue a mo V edge, o he overall complexiy of he flow decompoiion i O(V E). Moreover, i i eay o conruc flow for which every flow decompoiion ha complexiy Θ(V E). Thu, any maximum-flow algorihm ha (eiher explicily or implicily) conruc a flow a a um of pah or cycle in paricular, any implemenaion of Ford and Fulkeron augmening pah algorihm mu ake Ω(V E) ime in he wor cae Edmond and Karp Algorihm Ford and Fulkeron algorihm doe no pecify which pah in he reidual graph o augmen, and he poor behavior of he algorihm can be blamed on poor choice for he augmening pah. In he early 1970, Jack Edmond and Richard Karp analyzed wo naural rule for chooing augmening pah, boh of which led o more efficien algorihm. 9

10 Fa Pipe Edmond and Karp fir rule i eenially a greedy algorihm: Chooe he augmening pah wih large boleneck value. I a fairly eay o how ha he maximum-boleneck (, )-pah in a direced graph can be compued in O(E log V ) ime uing a varian of Jarník minimum-panning-ree algorihm, or of Dijkra hore pah algorihm. Simply grow a direced panning ree T, rooed a. Repeaedly find he highe-capaciy edge leaving T and add i o T, unil T conain a pah from o. Alernaely, one could emulae Krukal algorihm iner edge one a a ime in decreaing capaciy order unil here i a pah from o alhough hi i le efficien, a lea when he graph i direced. We can now analyze he algorihm in erm of he value of he maximum flow f. Le f be any flow in G, and le f be he maximum flow in he curren reidual graph G f. (A he beginning of he algorihm, G f = G and f = f.) We have already proved ha f can be decompoed ino a mo E pah and cycle. A imple averaging argumen implie ha a lea one of he pah in hi decompoiion mu carry a lea f /E uni of flow. I follow immediaely ha he fae (, )-pah in G f carrie a lea f /E uni of flow. Thu, augmening f along he maximum-boleneck pah in G f muliplie he value of he remaining maximum flow in G f by a facor of a mo 1 1/E. In oher word, he reidual maximum flow value decay exponenially wih he number of ieraion. Afer E ln f ieraion, he maximum flow value in G f i a mo f (1 1/E) E ln f < f e ln f = 1. (Tha Euler conan e, no he edge e. Sorry.) In paricular, if all he capaciie are ineger, hen afer E ln f ieraion, he maximum capaciy of he reidual graph i zero and f i a maximum flow. We conclude ha for graph wih ineger capaciie, he Edmond-Karp fa pipe algorihm run in O(E 2 log E log f ) ime, which i acually a polynomial funcion of he inpu ize Shor Pipe The econd Edmond-Karp rule wa acually propoed by Ford and Fulkeron in heir original max-flow paper; a varian of hi rule wa independenly conidered by he Ruian mahemaician Yefim Dini around he ame ime a Edmond and Karp. Chooe he augmening pah wih he malle number of edge. The hore augmening pah can be found in O(E) ime by running breadh-fir earch in he reidual graph. Surpriingly, he reuling algorihm hal afer a polynomial number of ieraion, independen of he acual edge capaciie! The proof of hi polynomial upper bound relie on wo obervaion abou he evoluion of he reidual graph. Le f i be he curren flow afer i augmenaion ep, le G i be he correponding reidual graph. In paricular, f 0 i zero everywhere and G 0 = G. For each verex v, le level i (v) denoe he unweighed hore pah diance from o v in G i, or equivalenly, he level of v in a breadh-fir earch ree of G i rooed a. Our fir obervaion i ha hee level can only increae over ime.

11 Lemma 2. level i+1 (v) level i (v) for all verice v and ineger i. Proof: The claim i rivial for v =, ince level i () = 0 for all i. Chooe an arbirary verex v, and le u v be a hore pah from o v in G i+1. (If here i no uch pah, hen level i+1 (v) =, and we re done.) Becaue hi i a hore pah, we have level i+1 (v) = level i+1 (u) + 1, and he inducive hypohei implie ha level i+1 (u) level i (u). We now have wo cae o conider. If u v i an edge in G i, hen level i (v) level i (u) + 1, becaue he level are defined by breadh-fir raveral. On he oher hand, if u v i no an edge in G i, hen v u mu be an edge in he ih augmening pah. Thu, v u mu lie on he hore pah from o in G i, which implie ha level i (v) = level i (u) 1 level i (u) + 1. In boh cae, we have level i+1 (v) = level i+1 (u) + 1 level i (u) + 1 level i (v). Whenever we augmen he flow, he boleneck edge in he augmening pah diappear from he reidual graph, and ome oher edge in he reveral of he augmening pah may (re-)appear. Our econd obervaion i ha an edge canno appear or diappear oo many ime. Lemma 3. During he execuion of he Edmond-Karp hor-pipe algorihm, any edge u v diappear from he reidual graph G f a mo V /2 ime. Proof: Suppoe u v i in wo reidual graph G i and G j+1, bu no in any of he inermediae reidual graph G i+1,..., G j, for ome i < j. Then u v mu be in he ih augmening pah, o level i (v) = level i (u) + 1, and v u mu be on he jh augmening pah, o level j (v) = level j (u) 1. By he previou lemma, we have level j (u) = level j (v) + 1 level i (v) + 1 = level i (u) + 2. In oher word, he diance from o u increaed by a lea 2 beween he diappearance and reappearance of u v. Since every level i eiher le han V or infinie, he number of diappearance i a mo V /2. Now we can derive an upper bound on he number of ieraion. Since each edge can diappear a mo V /2 ime, here are a mo EV /2 edge diappearance overall. Bu a lea one edge diappear on each ieraion, o he algorihm mu hal afer a mo EV /2 ieraion. Finally, ince each ieraion require O(E) ime, hi algorihm run in O(VE 2 ) ime overall Furher Progre Thi i nowhere near he end of he ory for maximum-flow algorihm. Decade of furher reearch have led o a number of even faer algorihm, ome of which are ummarized in he able below.² All of he algorihm lied below compue a maximum flow in everal ieraion. Each algorihm ha wo varian: a impler verion ha perform each ieraion by brue force, and a faer varian ha ue ophiicaed daa rucure o mainain a panning ree of he flow nework, o ha each ieraion can be performed (and he panning ree updaed) in logarihmic ime. There i no reaon o believe ha he be algorihm known o far are opimal; indeed, maximum flow are ill a very acive area of reearch. ²To keep he able hor, I have deliberaely omied algorihm whoe running ime depend on he maximum capaciy, he um of he capaciie, or he maximum flow value. Even wih hi rericion, he able i incomplee! 11

12 Technique Direc Wih dynamic ree Source Blocking flow O(V 2 E) O(V E log V ) [Dini; Sleaor and Tarjan] Nework implex O(V 2 E) O(V E log V ) [Danzig; Goldfarb and Hao; Goldberg, Grigoriadi, and Tarjan] Puh-relabel (generic) O(V 2 E) [Goldberg and Tarjan] Puh-relabel (FIFO) O(V 3 ) O(V 2 log(v 2 /E)) [Goldberg and Tarjan] Puh-relabel (highe label) O(V 2 E) [Cheriyan and Mahehwari; Tunçel] Peudoflow O(V 2 E) O(V E log V ) [Hochbaum] Peudoflow (highe label) O(V 3 ) O(V E log(v 2 /E)) [Hochbaum and Orlin] Compac abundance graph O(V E) [Orlin] Several purely combinaorial maximum-flow algorihm and heir running ime. The fae maximum flow algorihm known, announced by Jame Orlin in 2012, run in O(V E) ime, exacly maching he wor-cae complexiy of a flow decompoiion. The deail of Orlin algorihm are far beyond he cope of hi coure; in addiion o hi own new echnique, Orlin ue everal older algorihm and daa rucure a black boxe, mo of which are hemelve quie complicaed. (In paricular, orlin algorihm doe no conruc an explici flow decompoiion; in fac, for graph wih only O(V ) edge, an exenion of hi algorihm acually run in only O(V 2 / log V ) ime!) Neverhele, for purpoe of analyzing algorihm ha ue maximum flow, hi i he ime bound you hould cie. So wrie he following enence on your chea hee and cie i in your homework: Maximum flow can be compued in O(VE) ime. Exercie 1. Suppoe you are given a direced graph G = (V, E), wo verice and, a capaciy funcion c : E +, and a econd funcion f : E. Decribe an algorihm o deermine wheher f i a maximum (, )-flow in G. 2. Le (S, T) and (S, T ) be minimum (, )-cu in ome flow nework G. Prove ha (S S, T T ) and (S S, T T ) are alo minimum (, )-cu in G. 3. Decribe an efficien algorihm o deermine wheher a given flow nework conain a unique maximum flow. 4. Fix any flow nework G = (V, E). Our obervaion ha any weighed um of (, )-flow i ielf an (, )-flow implie ha he e of all (, )-flow in any graph acually define a vecor pace over he real. (a) Prove ha he dimenion of hi vecor pace i exacly E V + 2. (b) Le T be any panning ree of G. Prove ha he following collecion of pah and cycle define a bai for hi vecor pace: The unique pah in T from o ; The unique cycle in T {e}, for every edge e T. 12

13 (c) Le T be any panning ree of G, and le F be he fore obained by deleing any ingle edge in T. Prove ha he following collecion of pah and cycle define a bai for hi vecor pace: The unique pah in F {e} from o, for every edge e F ha ha one endpoin in each componen of F; The unique cycle in F {e}, for every edge e F wih boh endpoin in he ame componen of F.. Cu are omeime defined a ube of he edge of he graph, inead of a pariion of i verice. In hi problem, you will prove ha hee wo definiion are almo equivalen. We ay ha a ube of (direced) edge eparae and if every direced pah from o conain a lea one (direced) edge in. For any ube S of verice, le δs denoe he e of direced edge leaving S; ha i, δs := {u v u S, v S}. (a) Prove ha if (S, T) i an (, )-cu, hen δs eparae and. (b) Le be an arbirary ube of edge ha eparae and. Prove ha here i an (, )-cu (S, T) uch ha δs. (c) Le be a minimal ube of edge ha eparae and. (Such a e of edge i omeime called a bond.) Prove ha here i an (, )-cu (S, T) uch ha δs =. 6. Suppoe inead of capaciie, we conider nework where each edge u v ha a nonnegaive demand d(u v). Now an (, )-flow f i feaible if and only if f (u v) d(u v) for every edge u v. (Feaible flow value can now be arbirarily large.) A naural problem in hi eing i o find a feaible (, )-flow of minimum value. (a) Decribe an efficien algorihm o compue a feaible (, )-flow, given he graph, he demand funcion, and he verice and a inpu. [Hin: Find a flow ha i non-zero everywhere, and hen cale i up o make i feaible.] (b) Suppoe you have acce o a ubrouine MaxFlow ha compue maximum flow in nework wih edge capaciie. Decribe an efficien algorihm o compue a minimum flow in a given nework wih edge demand; your algorihm hould call MaxFlow exacly once. (c) Sae and prove an analogue of he max-flow min-cu heorem for hi eing. (Do minimum flow correpond o maximum cu?) 7. For any flow nework G and any verice u and v, le boleneck G (u, v) denoe he maximum, over all pah π in G from u o v, of he minimum-capaciy edge along π. (a) Decribe and analyze an algorihm o compue boleneck G (, ) in O(E log V ) ime. (b) Decribe an algorihm o conruc a panning ree T of G uch ha boleneck T (u, v) = boleneck G (u, v) for all verice u and v. (Edge in T inheri heir capaciie from G.) 8. Suppoe you have already compued a maximum flow f in a flow nework G wih ineger edge capaciie. 13

14 (a) Decribe and analyze an algorihm o updae he maximum flow afer he capaciy of a ingle edge i increaed by 1. (b) Decribe and analyze an algorihm o updae he maximum flow afer he capaciy of a ingle edge i decreaed by 1. Boh algorihm hould be ignificanly faer han recompuing he maximum flow from crach. 9. Le G be a nework wih ineger edge capaciie. An edge in G i upper-binding if increaing i capaciy by 1 alo increae he value of he maximum flow in G. Similarly, an edge i lower-binding if decreaing i capaciy by 1 alo decreae he value of he maximum flow in G. (a) Doe every nework G have a lea one upper-binding edge? Prove your anwer i correc. (b) Doe every nework G have a lea one lower-binding edge? Prove your anwer i correc. (c) Decribe an algorihm o find all upper-binding edge in G, given boh G and a maximum flow in G a inpu, in O(E) ime. (d) Decribe an algorihm o find all lower-binding edge in G, given boh G and a maximum flow in G a inpu, in O(EV ) ime.. A new aian profeor, eaching maximum flow for he fir ime, ugge he following greedy modificaion o he generic Ford-Fulkeron augmening pah algorihm. Inead of mainaining a reidual graph, ju reduce he capaciy of edge along he augmening pah! In paricular, whenever we aurae an edge, ju remove i from he graph. GreedyFlow(G, c,, ): for every edge e in G f (e) 0 while here i a pah from o π an arbirary pah from o F minimum capaciy of any edge in π for every edge e in π f (e) f (e) + F if c(e) = F remove e from G ele c(e) c(e) F reurn f (a) Show ha GreedyFlow doe no alway compue a maximum flow. (b) Show ha GreedyFlow i no even guaraneed o compue a good approximaion o he maximum flow. Tha i, for any conan α > 1, here i a flow nework G uch ha he value of he maximum flow i more han α ime he value of he flow compued by GreedyFlow. [Hin: Aume ha GreedyFlow chooe he wor poible pah π a each ieraion.] 14

15 (c) Prove ha for any flow nework, if he Greedy Pah Fairy ell you preciely which pah π o ue a each ieraion, hen GreedyFlow doe compue a maximum flow. (Sadly, he Greedy Pah Fairy doe no acually exi.) 11. A given flow nework G may have more han one minimum (, )-cu. Le define he be minimum (, )-cu o be any minimum cu (S, T) wih he malle number of edge croing from S o T. (a) Decribe an efficien algorihm o deermine wheher a given flow nework conain a unique minimum (, )-cu. (b) Decribe an efficien algorihm o find he be minimum (, )-cu when he capaciie are ineger. (c) Decribe an efficien algorihm o find he be minimum (, )-cu for arbirary edge capaciie. (d) Decribe an efficien algorihm o deermine wheher a given flow nework conain a unique be minimum (, )-cu. 12. We can peed up he Edmond-Karp fa pipe heuriic, a lea for ineger capaciie, by relaxing our requiremen for he nex augmening pah. Inead of finding he augmening pah wih maximum boleneck capaciy, we find a pah whoe boleneck capaciy i a lea half of maximum, uing he following capaciy caling algorihm. The algorihm mainain a boleneck hrehold ; iniially, i he maximum capaciy among all edge in he graph. In each phae, he algorihm augmen along pah from o in which every edge ha reidual capaciy a lea. When here i no uch pah, he phae end, we e /2, and he nex phae begin. (a) How many phae will he algorihm execue in he wor cae, if he edge capaciie are ineger? (b) Le f be he flow a he end of a phae for a paricular value of. Le S be he node ha are reachable from in he reidual graph G f uing only edge wih reidual capaciy a lea, and le T = V \ S. Prove ha he capaciy (wih repec o G original edge capaciie) of he cu (S, T) i a mo f + E. (c) Prove ha in each phae of he caling algorihm, here are a mo 2E augmenaion. (d) Wha i he overall running ime of he caling algorihm, auming all he edge capaciie are ineger? 13. An (, )-erie-parallel graph i an direced acyclic graph wih wo deignaed verice (he ource) and (he arge or ink) and wih one of he following rucure: Bae cae: A ingle direced edge from o. Serie: The union of an (, u)-erie-parallel graph and a (u, )-erie-parallel graph ha hare a common verex u bu no oher verice or edge. Parallel: The union of wo maller (, )-erie-parallel graph wih he ame ource and arge, bu wih no oher verice or edge in common. 1

16 Decribe an efficien algorihm o compue a maximum flow from o in an (, )-erieparallel graph wih arbirary edge capaciie. 14. In 1980 Maurice Queyranne publihed he following example of a flow nework where Edmond and Karp fa pipe heuriic doe no hal. Here, a in Zwick bad example for he original Ford-Fulkeron algorihm, φ denoe he invere golden raio ( 1)/2. The hree verical edge play eenially he ame role a he horizonal edge in Zwick example. (ϕ+1)/2 a 1/2 b ϕ/2 c (ϕ+1)/2 d ϕ/2 1/2 ϕ ϕ/2 ϕ (ϕ+1)/2 1 1/2 (ϕ+1)/2 e ϕ/2 f (ϕ+1)/2 g 1/2 h ϕ/2 Queyranne nework, and a equence of fa-pipe augmenaion. (a) Show ha he following infinie equence of pah augmenaion i a valid execuion of he Edmond-Karp fa pipe algorihm. (See he figure above.) QueyranneFaPipe: for i 1 o puh φ 3i 2 uni of flow along a f g b h c d puh φ 3i 1 uni of flow along f a b g h c puh φ 3i uni of flow along e f a g b c h forever (b) Decribe a equence of O(1) pah augmenaion ha yield a maximum flow in Queyranne nework. Copyrigh 201 Jeff Erickon. Thi work i licened under a Creaive Common Licene (hp://creaivecommon.org/licene/by-nc-a/4.0/). Free diribuion i rongly encouraged; commercial diribuion i exprely forbidden. See hp:// for he mo recen reviion. 16