16 Max-Flow Algorithms

Transcription

1 A process canno be undersood by sopping i. Undersanding mus move wih he flow of he process, mus join i and flow wih i. The Firs Law of Mena, in Frank Herber s Dune (196) There s a difference beween knowing he pah and walking he pah. Morpheus [Laurence Fishburne], The Marix (1999) 16 Max-Flow Algorihms 16.1 Recap Fix a direced graph G = (V, E) ha does no conain boh an edge u v and is reversal v u, and fix a capaciy funcion c : E IR 0. For any flow funcion f : E IR 0, he residual capaciy is defined as c(u v) f (u v) if u v E c f (u v) = f (v u) if v u E. 0 oherwise The residual graph G f = (V, E f ), where E f is he se of edges whose non-zero residual capaciy is posiive. 0/ /20 s / 0/1 / /1 s 1 0/ /20 1 / A flow f in a weighed graph G and is residual graph G f. In he las lecure, we proved he Max-flow Min-cu Theorem: In any weighed direced graph nework, he value of he maximum (s, )-flow is equal o he capaciy of he minimum (s, )-cu. The proof of he heorem is consrucive. If he residual graph conains a pah from s o, hen we can increase he flow by he minimum capaciy of he edges on his pah, so we mus no have he maximum flow. Oherwise, we can define a cu (S, T) whose capaciy is he same as he flow f, such ha every edge from S o T is sauraed and every edge from T o S is empy, which implies ha f is a maximum flow and (S, T) is a minimum cu. / s 1 /20 s / 0/1 /1 0/ 1 / /20 / An augmening pah in G f and he resuling (maximum) flow f. 1

2 16.2 Ford-Fulkerson I s no hard o realize ha his proof ranslaes almos immediaely o an algorihm, firs developed by Ford and Fulkerson in he 190s: Saring wih he zero flow, repeaedly augmen he flow along any pah s in he residual graph, unil here is no such pah. If every edge capaciy is an ineger, hen every augmenaion sep increases he value of he flow by a posiive ineger. Thus, he algorihm hals afer f ieraions, where f is he acual maximum flow. Each ieraion requires O(E) ime, o creae he residual graph G f and perform a whaever-firs-search o find an augmening pah. Thus, in he words case, he Ford-Fulkerson algorihm runs in O(E f ) ime. If we muliply all he capaciies by he same (posiive) consan, he maximum flow increases everywhere by he same consan facor. I follows ha if all he edge capaciies are raional, hen he Ford-Fulkerson algorihm evenually hals. However, if we allow irraional capaciies, he algorihm can loop forever, always finding smaller and smaller augmening pahs. Worse ye, his infinie sequence of augmenaions may no even converge o he maximum flow! Perhaps he simples example of his effec was discovered by Uri Zwick. Consider he graph shown below, wih six verices and nine edges. Six of he edges have some large ineger capaciy, wo have capaciy 1, and one has capaciy φ = ( 1)/ , chosen so ha 1 φ = φ 2. To prove ha he Ford-Fulkerson algorihm can ge suck, we can wach he residual capaciies of he hree horizonal edges as he algorihm progresses. (The residual capaciies of he oher six edges will always be a leas 3.) s 1 1 ϕ A B C Uri Zwick s non-erminaing flow example, and hree augmening pahs. The Ford-Fulkerson algorihm sars by choosing he cenral augmening pah, shown in he large figure above. The hree horizonal edges in order from lef o righ, now have residual capaciies 1, 0, φ. Suppose inducively ha he horizonal residual capaciies are φ k 1, 0, φ k for some non-negaive ineger k. 1. Augmen along B, adding φ k o he flow; he residual capaciies are now φ k+1, φ k, Augmen along C, adding φ k o he flow; he residual capaciies are now φ k+1, 0, φ k. 3. Augmen along B, adding φ k+1 o he flow; he residual capaciies are now 0, φ k+1, φ k Augmen along A, adding φ k+1 o he flow; he residual capaciies are now φ k+1, 0, φ k+2. 2

3 Thus, afer 4n + 1 augmenaion seps, he residual capaciies are φ 2n 2, 0, φ 2n 1. As he number of augmenaion seps grows o infiniy, he value of he flow converges o i=1 φ i = φ = 4 + < 7, even hough he maximum flow value is clearly Picky sudens migh wonder a his poin why we care abou irraional capaciies; afer all, compuers can represen anyhing bu (small) inegers or (dyadic) raionals exacly. Good quesion! One reason is ha he ineger resricion is lierally arificial; i s an arifac of acual compuaional hardware 1, no an inheren feaure of he absrac mahemaical problem. Anoher reason, which is probably more convincing o mos pracical compuer scieniss, is ha he behavior of he algorihm wih irraional inpus ells us somehing abou is wors-case behavior in pracice given floaing-poin capaciies errible! Even wih very reasonable capaciies, a careless implemenaion of Ford-Fulkerson could ener an infinie loop simply because of round-off error! 16.3 Edmonds-Karp: Fa Pipes The Ford-Fulkerson algorihm does no specify which alernaing pah o use if here is more han one. In 1972, Jack Edmonds and Richard Karp analyzed wo naural heurisics for choosing he pah. The firs is essenially a greedy algorihm: Choose he augmening pah wih larges boleneck value. I s a fairly easy o show ha he maximum-boleneck (s, )-pah in a direced graph can be compued in O(E log V ) ime using a varian of Jarník s minimum-spanning-ree algorihm, or of Dijksra s shores pah algorihm. Simply grow a direced spanning ree T, rooed a s. Repeaedly find he highes-capaciy edge leaving T and add i o T, unil T conains a pah from s o. Alernaely, once could emulae Kruskal s algorihm inser edges one a a ime in decreasing capaciy order unil here is a pah from s o alhough his is less efficien. We can now analyze he algorihm in erms 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 residual graph G f. (A he beginning of he algorihm, G f = G and f = f.) Le e be he boleneck edge in he nex augmening pah. Le S be he se of verices reachable from s hrough edges in G wih capaciy greaer han c(e) and le T = V \ S. By consrucion, T is non-empy, and every edge from S o T has capaciy a mos c(e). Thus, he capaciy of he cu (S, T) is a mos c(e) E. On he oher hand, he maxflow-mincu heorem implies ha S, T f. We conclude ha c(e) f /E. The preceding argumen implies ha augmening f along he maximum-boleneck pah in G f muliplies he maximum flow value in G f by a facor of a mos 1 1/E. In oher words, he residual flow decays exponenially wih he number of ieraions. Afer E ln f ieraions, he maximum flow value in G f is a mos f (1 1/E) E ln f < f e ln f = 1. (Tha s Euler s consan e, no he edge e. Sorry.) In paricular, if all he capaciies are inegers, hen afer E ln f ieraions, he maximum capaciy of he residual graph is zero and f is a maximum flow. We conclude ha for graphs wih ineger capaciies, he Edmonds-Karp fa pipe algorihm runs in O(E 2 log E log f ) ime. 1...or perhaps he laws of physics. Yeah, righ. Whaever. Like realiy acually maers in his class. 3

4 16.4 Dinis/Edmonds-Karp: Shor Pipes The second Edmonds-Karp heurisic was acually proposed by Ford and Fulkerson in heir original max-flow paper, and firs analyzed by he Russian mahemaician Dinis (someimes ranslieraed Dinic) in Edmonds and Karp published heir independen and slighly weaker analysis in So naurally, almos everyone refers o his algorihm as Edmonds-Karp. 2 Choose he augmening pah wih fewes edges. The correc pah can be found in O(E) ime by running breadh-firs search in he residual graph. More surprisingly, he algorihm hals afer a polynomial number of ieraions, independen of he acual edge capaciies! The proof of his upper bound relies on wo observaions abou he evoluion of he residual graph. Le f i be he curren flow afer i augmenaion seps, le G i be he corresponding residual graph. In paricular, f 0 is zero everywhere and G 0 = G. For each verex v, le level i (v) denoe he unweighed shores pah disance from s o v in G i, or equivalenly, he level of v in a breadh-firs search ree of G i rooed a s. Our firs observaion is ha hese levels can only increase over ime. Lemma 1. level i+1 (v) level i (v) for all verices v and inegers i. Proof: The claim is rivial for v = s, since level i (s) = 0 for all i. Choose an arbirary verex v s, and le s u v be a shores pah from s o v in G i+1. (If here is no such pah, hen level i+1 (v) =, and we re done.) Because his is a shores pah, we have level i+1 (v) = level i+1 (u) + 1, and he inducive hypohesis implies ha level i+1 (u) level i (u). We now have wo cases o consider. If u v is an edge in G i, hen level i (v) level i (u) + 1, because he levels are defined by breadh-firs raversal. On he oher hand, if u v is no an edge in G i, hen v u mus be an edge in he ih augmening pah. Thus, v u mus lie on he shores pah from s o in G i, which implies ha level i (v) = level i (u) 1 level i (u) + 1. In boh cases, 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 disappears from he residual graph, and some oher edge in he reversal of he augmening pah may (re-)appear. Our second observaion is ha an edge canno appear or disappear oo many imes. Lemma 2. During he execuion of he Dinis/Edmonds-Karp algorihm, any edge u v disappears from he residual graph G f a mos V /2 imes. Proof: Suppose u v is in wo residual graphs G i and G j+1, bu no in any of he inermediae residual graphs G i+1,..., G j, for some i < j. Then u v mus be in he ih augmening pah, so level i (v) = level i (u) + 1, and v u mus be on he jh augmening pah, so level j (v) = level j (u) 1. By he previous lemma, we have level j (u) = level j (v) + 1 level i (v) + 1 = level i (u) To be fair, Edmonds and Karp discovered heir algorihm a few years before publicaion geing ideas ino prin akes ime, especially in he early 1970s which is why some auhors believe hey deserve prioriy. I don buy i; Dinis also presumably discovered his algorihm a few years before is publicaion. (In Sovie Union, resul publish you.) On he gripping hand, Dinis s paper also described an improvemen o he algorihm presened here ha runs in O(V 2 E) ime insead of O(V E 2 ), so maybe ha ough o be called Dinis s algorihm. 4

5 In oher words, he disance from s o u increased by a leas 2 beween he disappearance and reappearance of u v. Since every level is eiher less han V or infinie, he number of disappearances is a mos V /2. Now we can derive an upper bound on he number of ieraions. Since each edge can disappear a mos V /2 imes, here are a mos EV /2 edge disappearances overall. Bu a leas one edge disappears on each ieraion, so he algorihm mus hal afer a mos EV /2 ieraions. Finally, since each ieraion requires O(E) ime, Dinis algorihm runs in O(VE 2 ) ime overall. Exercises 1. A new assisan professor, eaching maximum flows for he firs ime, suggess he following greedy modificaion o he generic Ford-Fulkerson augmening pah algorihm. Insead of mainaining a residual graph, jus reduce he capaciy of edges along he augmening pah! In paricular, whenever we saurae an edge, jus remove i from he graph. GREEDYFLOW(G, c, s, ): for every edge e in G f (e) 0 while here is a pah from s o π an arbirary pah from s 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 else c(e) c(e) F reurn f (a) Show ha his algorihm does no always compue a maximum flow. (b) Prove ha for any flow nework, if he Greedy Pah Fairy ells you precisely which pah π o use a each ieraion, hen GREEDYFLOW does compue a maximum flow. (Sadly, he Greedy Pah Fairy does no acually exis.) 2. Describe and analyze an algorihm o find he maximum-boleneck pah from s o in a flow nework G in O(E log V ) ime. 3. Describe a direced graph wih irraional edge capaciies, such ha he Edmonds-Karp fa pipe heurisic does no hal. 4. Describe an efficien algorihm o check wheher a given flow nework conains a unique maximum flow.. For any flow nework G and any verices u and v, le boleneck G (u, v) denoe he maximum, over all pahs π in G from u o v, of he minimum-capaciy edge along π. Describe an algorihm

6 o consruc a spanning ree T of G such ha boleneck T (u, v) = boleneck G (u, v). (Edges in T inheri heir capaciies form G.) One way o hink abou his problem is o imagine he verices of he graph as islands, and he edges as bridges. Each bridge has a maximum weigh i can suppor. If a ruck is carrying suff from u o v, how much can he ruck carry? We don care wha roue he ruck akes; he poin is ha he smalles-weigh edge on he roue will deermine he load. 6. We can speed up he Edmonds-Karp fa pipe heurisic, a leas for ineger capaciies, by relaxing our requiremens for he nex augmening pah. Insead of finding he augmening pah wih maximum boleneck capaciy, we find a pah whose boleneck capaciy is a leas half of maximum, using he following capaciy scaling algorihm. The algorihm mainains a boleneck hreshold ; iniially, is he maximum capaciy among all edges in he graph. In each phase, he algorihm augmens along pahs from s o in which every edge has residual capaciy a leas. When here is no such pah, he phase ends, we se /2, and he nex phase begins. (a) How many phases will he algorihm execue in he wors case, if he edge capaciies are inegers? (b) Le f be he flow a he end of a phase for a paricular value of. Le S be he nodes ha are reachable from s in he residual graph G f using only edges wih residual capaciy a leas, and le T = V \ S. Prove ha he capaciy (wih respec o G s original edge capaciies) of he cu (S, T) is a mos f + E. (c) Prove ha in each phase of he scaling algorihm, here are a mos 2E augmenaions. (d) Wha is he overall running ime of he scaling algorihm, assuming all he edge capaciies are inegers? c Copyrigh 2009 Jeff Erickson. Released under a Creaive Commons Aribuion-NonCommercial-ShareAlike 3.0 License (hp://creaivecommons.org/licenses/by-nc-sa/3.0/). Free disribuion is srongly encouraged; commercial disribuion is expressly forbidden. See hp:// for he mos recen revision. 6