Max Flow, Min Cut. Maximum Flow and Minimum Cut. Soviet Rail Network, Minimum Cut Problem


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1 Maximum Flow and Minimum u Max Flow, Min u Max flow and min cu. Two very rich algorihmic problem. ornerone problem in combinaorial opimizaion. eauiful mahemaical dualiy. Minimum cu Maximum flow Maxflow mincu heorem FordFulkeron augmening pah algorihm dmondkarp heuriic iparie maching Nonrivial applicaion / reducion. Nework conneciviy. iparie maching. aa mining. Openpi mining. irline cheduling. Image proceing. Projec elecion. aeball eliminaion. Nework reliabiliy. ecuriy of aiical daa. iribued compuing. galiarian able maching. iribued compuing. Many many more... Princeon Univeriy O lgorihm and aa rucure pring Kevin Wayne hp://www.princeon.u/~co ovie Rail Nework, Minimum u Problem Nework: abracion for maerial FLOWING hrough he edge. ireced graph. apaciie on edge. ource node, ink node. Min cu problem. elee "be" e of edge o diconnec from. ource ink capaciy ource: On he hiory of he ranporaion and maximum flow problem. lexander chrijver in Mah Programming, :,.
2 u u cu i a node pariion (, T) uch ha i in and i in T. capaciy(, T) = um of weigh of edge leaving. cu i a node pariion (, T) uch ha i in and i in T. capaciy(, T) = um of weigh of edge leaving. apaciy = apaciy = Minimum u Problem Maximum Flow Problem cu i a node pariion (, T) uch ha i in and i in T. capaciy(, T) = um of weigh of edge leaving. Min cu problem. Find an  cu of minimum capaciy. Nework: abracion for maerial FLOWING hrough he edge. ireced graph. ame inpu a min cu problem apaciie on edge. ource node, ink node. Max flow problem. ign flow o edge o a o: qualize inflow and ouflow a every inermediae verex. Maximize flow en from o. ource capaciy ink apaciy =
3 Flow Flow flow f i an aignmen of weigh o edge o ha: apaciy: f(e) u(e). Flow conervaion: flow leaving v = flow enering v. excep a or flow f i an aignmen of weigh o edge o ha: apaciy: f(e) u(e). Flow conervaion: flow leaving v = flow enering v. excep a or capaciy flow Value = capaciy flow Value = Maximum Flow Problem Flow and u Max flow problem: find flow ha maximize ne flow ino ink. Obervaion. Le f be a flow, and le (, T) be any  cu. Then, he ne flow en acro he cu i equal o he amoun reaching. capaciy flow Value = Value =
4 Flow and u Flow and u Obervaion. Le f be a flow, and le (, T) be any  cu. Then, he ne flow en acro he cu i equal o he amoun reaching. Obervaion. Le f be a flow, and le (, T) be any  cu. Then, he ne flow en acro he cu i equal o he amoun reaching. Value = Value = Flow and u Max Flow and Min u Obervaion. Le f be a flow, and le (, T) be any  cu. Then he value of he flow i a mo he capaciy of he cu. Obervaion. Le f be a flow, and le (, T) be an  cu whoe capaciy equal he value of f. Then f i a max flow and (, T) i a min cu. u capaciy = Flow value u capaciy = Flow value Flow value =
5 MaxFlow Minu Theorem Toward an lgorihm Maxflow mincu heorem. (FordFulkeron, ): In any nework, he value of max flow equal capaciy of min cu. Proof IOU: we find flow and cu uch ha Obervaion applie. Find  pah where each arc ha f(e) < u(e) and "augmen" flow along i. Min cu capaciy = Max flow value = flow Flow value = capaciy Toward an lgorihm Toward an lgorihm Find  pah where each arc ha f(e) < u(e) and "augmen" flow along i. Greedy algorihm: repea unil you ge uck. Find  pah where each arc ha f(e) < u(e) and "augmen" flow along i. Greedy algorihm: repea unil you ge uck. Fail: need o be able o "backrack." flow Flow value = flow Flow value = capaciy X X X capaciy X X X oleneck capaciy of pah = Flow value =
6 Reidual Graph ugmening Pah Original graph. Flow f(e). dge e = vw v flow = f(e) capaciy = u(e) w ugmening pah = pah in reidual graph. Increae flow along forward edge. ecreae flow along backward edge. Reidual edge. dge e = vw or wv. "Undo" flow en. Reidual graph. ll he edge ha have ricly poiive reidual capaciy. v reidual capaciy = u(e) f(e) w reidual capaciy = f(e) reidual original X X X X X ugmening Pah FordFulkeron ugmening Pah lgorihm Obervaion. If augmening pah, hen no ye a max flow. Q. If no augmening pah, i i a max flow? FordFulkeron algorihm. Generic mehod for olving max flow. reidual while (here exi an augmening pah) { Find augmening pah P ompue boleneck capaciy of P ugmen flow along P original Flow value = X X X X X Queion. oe hi lead o a maximum flow? ye How do we find an augmening pah?  pah in reidual graph How many augmening pah doe i ake? How much effor do we pending finding a pah?
7 MaxFlow Minu Theorem Proof of MaxFlow Minu Theorem ugmening pah heorem. flow f i a max flow if and only if here are no augmening pah. Maxflow mincu heorem. The value of he max flow i equal o he capaciy of he min cu. We prove boh imulaneouly by howing he following are equivalen: (i) f i a max flow. (ii) There i no augmening pah relaive o f. (iii) There exi a cu whoe capaciy equal he value of f. (i) (ii) equivalen o no (ii) no (i), which wa Obervaion (ii) (iii) nex lide (iii) (i) hi wa Obervaion (ii) (iii). If here i no augmening pah relaive o f, hen here exi a cu whoe capaciy equal he value of f. Proof. Le f be a flow wih no augmening pah. Le be e of verice reachable from in reidual graph. conain ; ince no augmening pah, doe no conain all edge e leaving in original nework have f(e) = u(e) all edge e enering in original nework have f(e) = f e ou of e ou of f ( e) u( e) capaciy (, T) e in o f ( e) T reidual nework Max Flow Nework Implemenaion FordFulkeron lgorihm: Implemenaion dge in original graph may correpond o or reidual edge. May need o ravere edge e = vw in forward or revere direcion. Flow = f(e), capaciy = u(e). Iner wo copie of each edge, one in adjacency li of v and one in w. FordFulkeron main loop. // while here exi an augmening pah, ue i while (augpah()) { public cla dge { privae in v, w; privae in cap; privae in flow; // from, o // capaciy from v o w // flow from v o w public dge(in v, in w, in cap) {... public in cap() { reurn cap; public in flow() { reurn flow; public boolean from(in v) { reurn hi.v == v; public in oher(in v) { reurn from(v)? hi.w : hi.v; public in capro(in v) { reurn from(v)? flow : cap  flow; public void addflowro(in v, in d) { flow += from(v)? d : d; // compue boleneck capaciy in bole = INFINITY; for (in v = ; v!= ; v = T(v)) bole = Mah.min(bole, pred[v].capro(v)); // augmen flow for (in v = ; v!= ; v = T(v)) pred[v].addflowro(v, bole); // keep rack of oal flow en from o value += bole;
8 FordFulkeron lgorihm: nalyi hooing Good ugmening Pah umpion: all capaciie are ineger beween and U. Ue care when elecing augmening pah. Invarian: every flow value and every reidual capaciie remain an ineger hroughou he algorihm. Theorem: he algorihm erminae in a mo f * V U ieraion. orollary: if U =, hen algorihm run in V ieraion. no polynomial in inpu ize! Inegraliy heorem: if all arc capaciie are ineger, hen here exi a max flow f for which every flow value i an ineger. Original Nework hooing Good ugmening Pah hooing Good ugmening Pah Ue care when elecing augmening pah. Ue care when elecing augmening pah. X X X Original Nework Original Nework
9 hooing Good ugmening Pah hooing Good ugmening Pah Ue care when elecing augmening pah. Ue care when elecing augmening pah. X X X Original Nework Original Nework ieraion poible! hooing Good ugmening Pah hore ugmening Pah Ue care when elecing augmening pah. ome choice lead o exponenial algorihm. lever choice lead o polynomial algorihm. Opimal choice for real world problem??? eign goal i o chooe augmening pah o ha: an find augmening pah efficienly. Few ieraion. hooe augmening pah wih: dmondkarp () Fewe number of arc. (hore pah) Max boleneck capaciy. (fae pah) hore augmening pah. ay o implemen wih F. Find augmening pah wih fewe number of arc. while (!q.impy()) { in v = q.dequeue(); InIeraor i = G.neighbor(v); while(i.hanex()) { dge e = i.nex(); in w = e.oher(v); if (e.capro(w) > ) { // i vw a reidual edge? if (w[w] > w[v] + ) { w[w] = w[v] + ; pred[w] = e; // keep rack of hore pah q.enqueue(w); reurn (w[] < INFINITY); // i here an augmening pah?
10 hore ugmening Pah nalyi Fae ugmening Pah Lengh of hore augmening pah increae monoonically. ricly increae afer a mo augmenaion. mo V oal augmening pah. O( V) running ime. Fae augmening pah. Find augmening pah whoe boleneck capaciy i maximum. eliver mo amoun of flow o ink. olve uing ijkrayle (PF) algorihm. X v reidual capaciy w if (w[w] < Mah.min(w[v], e.capro(w)) { w[w] = Mah.min(w[v], e.capro(w)); pred[w] = v; Finding a fae pah. O( log V) per augmenaion wih binary heap. Fac. O( log U) augmenaion if capaciie are beween and U. hooing an ugmening Pah Hiory of Worae Running Time hooing an augmening pah. ny pah will do wide laiude in implemening FordFulkeron. Generic prioriy fir earch. ome choice lead o good worcae performance. hore augmening pah fae augmening pah variaion on a heme: PF verage cae no well underood. Reearch challenge. Pracice: olve max flow problem on real nework in linear ime. Theory: prove i for worcae nework. Year... icoverer Mehod ympoic Time anzig implex V U Ford, Fulkeron ugmening pah V U dmondkarp hore pah V dmondkarp Max capaciy log U ( + V log V) iniz Improved hore pah V dmondkarp, iniz apaciy caling log U inizgabow Improved capaciy caling V log U Karzanov Preflowpuh V leaortarjan ynamic ree V log V GoldbergTarjan FIFO preflowpuh V log (V / ) GoldbergRao Lengh funcion / log (V / ) log U V / log (V / ) log U rc capaciie are beween and U.
11 n pplicaion iparie Maching Jon placemen. ompanie make job offer. uden have job choice. an we fill every job? iparie maching. Inpu: undireced and biparie graph G. e of edge M i a maching if each verex appear a mo once. Max maching: find a max cardinaliy maching. an we employ every uden? licedobe obyahoo arolhp avepple lizaim Frankun Maching M , ,  L R iparie Maching iparie Maching iparie maching. Inpu: undireced and biparie graph G. e of edge M i a maching if each verex appear a mo once. Max maching: find a max cardinaliy maching. Reduce o max flow. reae a direced graph G'. irec all arc from L o R, and give infinie (or uni) capaciy. dd ource, and uni capaciy arc from o each node in L. dd ink, and uni capaciy arc from each node in R o. L R Maching M , , ,  L R G G'
12 iparie Maching: Proof of orrecne iparie Maching: Proof of orrecne laim. Maching in G of cardinaliy k induce flow in G' of value k. Given maching M = { , ,  of cardinaliy. onider flow f ha end uni along each of pah: f i a flow, and ha cardinaliy. laim. Flow f of value k in G' induce maching of cardinaliy k in G. y inegraliy heorem, here exi / valued flow f of value k. onider M = e of edge from L o R wih f(e) =. each node in L and R inciden o a mo one edge in M M = k L R L R G G' G G' Reducion Reducion. Given an inance of biparie maching. Tranform i o a max flow problem. olve max flow problem. Tranform max flow oluion o biparie maching oluion. Iue. How expenive i ranformaion? O( + V) I i beer o olve problem direcly? O( V / ) biparie maching oom line: max flow i an exremely rich problemolving model. Many imporan pracical problem reduce o max flow. We know good algorihm for olving max flow problem.
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