Network Flows: Introduction & Maximum Flow

Transcription

1 CSC lgorihm Deign, nalyi, and Complexiy Summer 2016 Lalla Mouaadid Nework Flow: Inroducion & Maximum Flow We now urn our aenion o anoher powerful algorihmic echnique: Local Search. In a local earch algorihm, we ar wih an arbirary oluion o our problem, hen keep refining hi oluion by making mall repeaed local change ha will increae he qualiy of our oluion. Once a oluion converge o a local opimum, hen no number of mall change will increae i qualiy and we are done. u hi doe no alway mean ha hi local opimum i he opimal oluion o our problem. In hi lecure, we focu on a well rucured inance of local earch where he local opimum i indeed he global opimum. Definiion flow nework i a direced graph G(V, E) wih he following properie: 1. There i a poiive weigh funcion on E, called capaciy, c : E R There are wo deignaed verice and in V, uch ha. i he ource of he nework and he erminal. 3. There are no incoming edge oward and no ougoing edge from. We ue (G,,, c) o denoe a flow nework, wih ource, erminal, and edge capaciie c. Given a flow nework (G,,, c), a flow in G i a funcion f : E R + ha aifie he following conrain: 1. Capaciy conrain : For all edge e E, 0 f(e) c(e). Meaning no edge ge a flow ha exceed i capaciy. 2. Flow conervaion : For every verex v V \{, }, he flow going ino v equal he flow coming ou of v: u:(u,v) E f(u, v) = w:(v,w) E f(v, w) The value of he flow, denoed val(f) i he oal value of he flow leaving he ource : val(f) = f(, u) u:(,u) E Example : Conider he nework below 1. val(f) = 4, a we can end 2 uni of flow down he pah and. 1 noe ha he label on an edge f/c denoe he flow and he capaciy of ha edge 1

2 2 Given a flow nework, our goal i o end a much flow a poible from o. We ll ee how a local earch algorihm olve hi problem. Formally, we have he following problem: The Maximum Flow Problem: Inpu: flow nework (G,,, c) Oupu: flow f on G where val(f) i maximized. Since we aid hi i an inance of local earch, we hould be able o ar wih any oluion and ry o improve i over a number of ieraion. For inance, we could ar by end wo uni of flow down he pah in he example above, hen improve on hi oluion by ending anoher 2 uni of flow down. Since he capaciie of he ource are now auraed, we know we can improve on hi oluion; and hu val(f) = 4 i he local opimum. I urn ou for hi example, 4 i alo an opimal oluion for hi nework Now uppoe our iniial aring oluion wa he pah, hi pah give u a flow of value 1. We can improve i by ending anoher uni of flow down and anoher uni of flow down. The flow nework would hen look like: 1/1 The oal value of hi flow i val(f) = 3, bu we know here i a beer oluion. Which pah can we ue o increae he flow? If we claim ha local earch on he inance of max flow i opimal, hen we hould be able o end one more uni of flow he nework. efore we aack hi problem, we need a few more definiion. Given a flow nework (G,,, c), an augmening pah P i an, pah in G where we can puh more flow down he nework. Tha i, P = v 1, v 2,..., v k where c(v i, v i+1 ) f(v i, v i+1 ) > 0 for all 1 i < k and v 1 =, v k =. In he example above, here are no augmening pah o puh ha 4 h uni of flow, bu we ge around hi

3 3 problem by redirecing he flow in he nework. There are wo ype of augmening pah in a nework: Cae 1. n, pah where every edge ha ome unued capaciy: c(e) f(e) > 0 for every edge e P. We ju puh a lea one more uni of flow, hu increaing val(f). Cae 2. n, pah where ome edge have unued capaciy and ome are auraed, bu we can redirec he flow on he auraed edge. To illurae hi 2 nd cae, conider he example above where we val(f) = 3. Since here i no augmening pah ha fall in cae 1, we will ry o redirec ome flow in G. In paricular, we will augmen he flow along he pah and in doing o, redirec 1 uni of flow down he edge (, ). Noice ha when we redireced he flow down (, ), we decreaed he flow on he edge (, ) ince i i he exac flow we redireced. 1/1 To formalize hi concep, we inroduce a daa rucure ha allow u o keep rack of augmening pah in a nework. Definiion: Le (G,,, c) be a flow nework and f a flow on G. reidual graph G f of G i an edge-weighed direced graph on he ame verex e V a G and he edge e E where E i conruced a follow: If here i an edge (u, v) G, we add wo edge (u, v) and (v, u) o E wih capaciie: c f (u, v) = c(u, v) f(u, v) : he amoun of available capaciy lef on he edge (u, v) c f (v, u) = f(u, v) : The amoun of flow we are allowed o redirec c f denoe he capaciy of an edge in he reidual graph G f. If an edge e E ha capaciy c f (e ) = 0, we ju remove e from G f. Now we can redefine an augmening pah wih repec o he reidual graph: n augmening pah P on G f i an, pah on G f coniing of edge wih poiive capaciy: c f (e) > 0 for all e P. For furher clariy, le conider he example above and produce wha G f look like a every ieraion:

4 4 Send 1 uni of flow uing P = : 1/1 Send 1 uni of flow uing P = 1/1 Send 1 uni of flow uing P =

5 5 1/1 Send 1 uni of flow uing P = 1/1 There are no ougoing edge from wih lefover capaciy, o we can ar an augmening pah, and we are done. I i no alway he cae however ha we aurae he ougoing edge from before we are done, here could be lefover capaciy on an (, v) bu no augmening pah in G f. So in general, we keep performing augmenaion along pah P by increaing he flow on edge wih available capaciy, and decreaing he flow when redirecing. How much flow can we end a every augmenaion ep? Le P = v 1, v 2,..., v k be an augmening pah wih v 1 =, v k =. Since P i an augmening pah, we know ha c f (v i, v i+1 ) > 0 for all edge (v i, v i+1 ) in P wih 1 i < k. Therefore he mo flow we can end down P i ju min 1 i<k c f (v i, v i+1 ). Why? So far, we haven aid how o olve our Maximum Flow problem! Ford and Fulkeron propoed hi local earch algorihm o olve i: Keep augmening f uing augmening pah unil here are no more augmening pah! The claim i when he algorihm hal, he flow we end up wih i maximized. lgorihm 1 Ford-Fulkeron Inpu: flow nework (G,,, c) Oupu: flow f on G where val(f) i maximized 1: Se f(e) = 0 for every edge in G. 2: while There i an augmening pah P in G f do 3: ugmen f uing P 4: end while 5: Reurn f We ll prove he correcne of hi algorihm in he nex lecure. The proof i an immediae corollary of a well know heorem: The Max-Flow/Min-Cu Theorem.