1 Motivation and Basic Definitions

Transcription

1 CSCE : Deign and Analyi of Algorihm Noe on Max Flow Fall 20 (Baed on he preenaion in Chaper 26 of Inroducion o Algorihm, 3rd Ed. by Cormen, Leieron, Rive and Sein.) Moivaion and Baic Definiion Conider applicaion ha need o model maerial flowing hrough a nework, e.g.: liquid in pipe par in an aembly line curren in an elecrical nework informaion in a communicaion nework In all cae, we wan o find he greae rae of hipmen. Define a flow nework like hi: direced graph G = (V, E) if (u, v) i in E, hen (v, u) i no in E diinguihed verice (ource) and (ink) in V each verex i on ome pah each edge (u, v) E ha a capaciy c(u, v) 0 Example: v A flow in G i a funcion f : V V R aifying hee wo condiion:. capaciy conrain: for all u, v V, 0 f(u, v) c(u, v) (o if (u, v) i no in E, hen f(u, v) mu be 0)

2 2. flow conervaion: For all u V oher han and, f(v, u) = f(u, v) v V v V. The value of flow f, denoed f, i he value of all he flow going ou of he ource minu he value of any flow going ino he ource. (So far, we have no flow going ino he ource, bu laer hi will change.) In mahemaical erm: f = v V f(, v) v V f(v, ). The maximum flow problem i, given a flow nework, o find a flow wih maximum value. Example of a flow (noaion i flow/capaciy on each edge) i given nex; noe ha he value of he flow i + 8 = 9. /6 v 2/2 5/20 / /9 7/7 8/3 / / 2 Ford-Fulkeron Mehod Ford-Fulkeron mehod o find maximum flow ieraively increae he value of he flow, aring wih nohing. iniialize flow f o all 0 while here i an "augmening pah" p in he "reidual nework" G_f augmen flow f along p reurn f Sep : Conruc reidual nework. The reidual capaciy of any pair of verice u and v, denoed c f (u, v), i: c(u, v) f(u, v) if (u, v) E c f (u, v) = f(v, u) if (v, u) E (allow decreaing flow on (u, v)) 0 oherwie The reidual nework G f of G w.r.. flow f i he graph (V, E f ) wih he ame verice a G bu poenially differen edge: E f = {(u, v) V V : c f (u, v) > 0}. Noe ha E f 2 E. 2

3 The reidual nework i very imilar o a flow nework excep ha i can have ani-parallel edge (i.e., boh (u, v) and (v, u)). Sep 2: Find an augmening pah. An augmening pah p i a imple pah (no cycle) from o in he reidual nework G f. Such a pah can be found by, ay, breadh-fir-earch or deph-fir earch. Define a flow in G f ha goe along he augmening pah p and whoe value i he malle capaciy of any edge in he pah (he boleneck). I.e., le and define flow f p like hi: c f (p) = min{c f (u, v) : (u, v) i on p}. f p (u, v) = { cf (p) if (u, v) i on p 0 oherwie Claim (Lemma 26.2): f p i a flow in G f wih value c f (p) > 0. Sep 3: Augmen original flow. Now augmen he original flow f in G uing he new flow f p in G f. Thi combinaion of flow i denoed (f f p ), and defined like hi: { f(u, v) + fp (u, v) f (f f p )(u, v) = p (v, u) if (u, v) E 0 if (u, v) i no in E. Claim (Lemma 26. and Corollary 26.3): (f f p ) i a flow in G wih value f + f p > f. I.e., we have increaed he flow in G! Example: Reidual nework for flow given in previou figure. v An augmening pah in he picure above i ; boleneck capaciy i, from he edge (, ). Afer augmening he original flow wih hi flow, we ge hi new flow (noe ha he new flow ha value + 2 = 23): /6 v 2/2 9/20 / 0/9 7/7 2/3 / / 3

4 3 The Max-Flow Theorem How do we know ha when Ford-Fulkeron op we have found he maximum flow? We need ome mahemaical definiion and reul o how hi. A cu (S, T ) of a flow nework G = (V, E) i a pariion of V ino S and T = V S uch ha S and T. The ne flow acro cu (S, T ) i denoed f(s, T ) and i he um of he flow for all pair from S o T, minu he um of he flow for all pair from T o S, i.e., f(s, T ) = f(u, v) f(v, u). The capaciy of cu(s, T ) i denoed c(s, T ) and i he um of he capaciie for all pair from S o T, i.e., c(s, T ) = c(u, v). Example of a cu; S = {, v, } and T = {,, }: /6 v 2/2 5/20 / /9 7/7 8/3 / / The ne flow acro hi cu i 2 + = 9. The capaciy of hi cu i 2 + = 26. Claim (Lemma 26.): If G ha flow f and cu (S, T ), hen he ne flow acro (S, T ) i f. The proof i baed on ome ideniie for manipulaing flow and cu. The inuiion i ha everyhing in he flow f ha o cro from S o T omewhere. Claim (Corollary 26.5): The value of he maximum flow i a mo he minimum capaciy of any cu. So we can never ge a flow ha i beer han he capaciy of he minimum cu. Ex: The cu S = {, v,, }, T = {, } ha capaciy 23. So we have found he large poible flow. You can check ha here i no cu wih maller capaciy. The main reul, hown nex, i ha hi upper bound can alway be achieved. Max-flow min-cu Theorem (26.6): If f i a flow in G, hen he following are eiher all rue or all fale:. f i a maximum flow in G. 2. G f ha no augmening pah. 3. f = c(s, T ) for ome cu (S, T ) of G.

5 Proof: Show implie 2: If f i a max flow and G f ha an augmening pah p, hen (f f p ) i a larger flow for G, conradicion. Show 2 implie 3: Suppoe G f ha no augmening pah. We ll define a cu wih he deired propery ha i capaciy i he value of he flow f. Le S = {v V : here exi an v pah in G f }. Le T = V S. Why i hi a cu? Obviouly S. Alo T ince here i no pah in G f becaue G f ha no augmening pah. So i i a cu. We now how ha he capaciy of hi cu i f. f = f(s, T ) by Lemma 26. = f(u, v) f(v, u) by definiion of flow acro a cu = (f(u, v) f(v, u)) by algebra = c(u, v) explained nex = c(s, T ) by definiion of capaciy of a cu For he fourh ep above, conider any u S and any v T. Cae : Suppoe (u, v) E. Then f(u, v) = c(u, v), oherwie here would be reidual capaciy on ha edge and u and v would no be on oppoie ide of he cu. Since (v, u) / E, f(v, u) = 0. Thu f(u, v) f(v, u) = c(u, v) 0 = c(u, v). Cae 2: Suppoe (v, u) E. Since u and v are on oppoie ide of he cu, (u, v) / G f, and hu he reidual capaciy on (u, v) i 0. By definiion of G f, he reidual capaciy on (u, v) i f(v, u). Thu f(v, u) = 0. On he oher hand, ince (u, v) / E (by aumpion on original flow nework), boh f(u, v) and c(u, v) are 0. Cae 3: Neiher (u, v) nor (v, u) i in E. Then f(u, v), f(v, u), and c(u, v) are all 0. Show 3 implie : Suppoe f = c(s, T ). By Claim above (Corollary 26.5), f c(s, T ) for every cu (S, T ). So (S, T ) mu be a malle capaciy cu, and hu f i a maximum flow. Inaniaing he Ford-Fulkeron Mehod: Edmond-Karp Now le look a he Ford-Fulkeron mehod in more deail. (I a mehod and no an algorihm ince no all he deail are pelled ou.) Aume f(u, v) i alway 0 if (u, v) i no in E. f(u,v) := 0 for each (u,v) in E while here i an -> pah p in G_f do exra := min {c_f(u,v) : (u,v) i in p} for each (u,v) in p do if (u,v) i in E hen f(u,v) := f(u,v) + exra ele f(v,u) := f(v,u) - exra endif endfor endwhile See Figure 26.6 on pp for example execuion. 5 // deail miing here

6 Baic mehod o find augmening pah: Keep a daa rucure repreening G = (V, E ), where E = {(u, v) : (u, v) E or (v, u) E}. Mainain capaciie and flow on he edge of G. To produce G f from hi daa rucure, include all edge (u, v) in E uch ha c f (u, v) > 0. Find any pah in G f by running DFS or BFS aring a. If we allow any equence of choice for he augmening pah, he algorihm can fail o converge if he capaciie are irraional number. However, if all capaciie are ineger (or raional), hen any equence of choice for he augmening pah will work, alhough he running ime migh be bad. Running Time: Iniializaion ake O(E) ime. Each ieraion of he while loop ake O(E ) = O(E) ime o find augmening pah, and O(E) ime o find minimum reidual capaciy and adju he flow. Number of ieraion of he while loop: The flow alway increae by a lea ince capaciie are ineger. So he number of ieraion i a mo f, where f i he flow found by he algorihm (he maximum flow). Thu he ime i O(E f ). Thi running ime can be errible. Suppoe V and E are mall bu he capaciie are very large. Example of bad running ime for Ford-Fulkeron: Keep alernaing beween uing augmening pah u v and augmening pah v u. Thi will caue he flow o be incremened by a every ieraion, requiring wo million ieraion.,000,000 u,000,000,000,000 v,000,000 To improve, we have he Edmond-Karp algorihm: Alway chooe he augmening pah o be he hore pah, in erm of he number of edge. Do hi uing BFS. Check ha he previou bad example canno occur now. Running ime of Edmond-Karp: Claim (Theorem 26.8): The oal number of augmenaion i O(V E). The idea for proving he claim i ha each augmenaion ha a criical edge, he edge on he augmening pah wih minimum reidual capaciy. I can be hown ha each edge in he graph can only be criical in O(V ) differen augmenaion. So he oal number of augmenaion i O(V E). Uing he claim, ince each augmenaion ake O(E) ime, we ge a oal ime for Edmond- Karp of O(V E 2 ). 6

7 5 Applicaion: Maximum Maching in a Biparie Graph An undireced graph G = (V, E) i biparie if V can be pariioned ino wo e, L and R, uch ha all edge in E go beween L and R. Example of a biparie graph (L coni of verice on he lef, R of hoe on he righ): v 0 v v 5 v 6 A maching of an undireced graph G = (V, E) i a ube of edge uch ha each verex i an endpoin of a mo one edge in he ube. Example: The hick dark edge in he picure below form a maching. v 0 v v 5 v 6 A maximum maching i a maching wih he large poible number of edge. 7

8 Example: Add he edge {v, } in he picure above. A pracical applicaion of finding a maximum maching in a biparie graph i a follow. Le L be a e of machine, each of which can perform one ak. Le R be a e of ak. A paricular machine i only capable of performing cerain ak. Thi can be capured by a biparie graph G = (V, E) in which V = L R, and (u, v) i in E if and only if machine u i capable of performing ak v. We wan o do a many ak a poible a one ime. Such an opimal aignmen of ak o machine would correpond o a maximum maching in G. We can ranform he problem of finding a maximum maching in a biparie graph ino he max-flow problem! Take he maching problem, conver i ino an inance of he max-flow problem, olve he max-flow inance, and hen conver he max-flow oluion o a oluion o he maching problem. In more deail: Given an undireced biparie graph G = (V, E), wih V G = (V, E ) like hi: V = V {, } (new verice for ource and ink) E coni of edge: from o every verex in L from every verex in R o edge in E direced from L o R Aign each edge capaciy. Example: Here i he flow nework for he biparie graph given above. = L R, define a flow nework v 0 v v 5 v 6 And here i a maximum flow for he example. The maximum maching for he original graph coni of he hick dark edge in he middle column (hoe wih poiive flow). 8

9 v 0 / / 0/ 0/ / 0/ v v 5 / / / / / 0/ 0/ v 6 / Lemma (26.9): () If M i a maching in G, hen here i an ineger-valued flow f in G wih f = M. (Inegervalued mean he flow on every edge i an ineger.) (2) If f i an ineger-valued flow in G, hen here i a maching M in G wih M = f. Proof: If a verex v i an endpoin of an edge in a maching M, we ay ha v i in M. () Suppoe M i a maching in G. Define flow f in G o be for all edge from o any verex in M, from any verex in M o, and correponding o an edge in M. The remaining flow are 0. Thi i an ineger-valued flow and he value of he flow i he number of edge in he maching. (2) Suppoe f i an ineger-valued flow in G. Define M o be {(u, v) : u L, v R, f(u, v) > 0}. To how M i a maching, we mu how no verex in L ha wo ougoing edge wih poiive flow, and no verex in R ha wo incoming edge wih poiive flow. For L: verex u in L ha only one incoming edge (from ) and i ha capaciy. Since f i ineger-valued, he flow on (, u) i eiher 0 or. If he flow on (, u) i 0, hen no flow leave u. If he flow on (, u) i, hen exacly one edge ougoing from u ha poiive flow, and i ha flow, ince f i ineger-valued. The argumen for R i imilar. Finally, we mu how M = f. Conider he cu in which S = {} L and T = {} R. By Lemma 26., he ne flow acro hi cu i f (he value of he enire flow). Since he flow i ineger-valued, every edge croing hi cu i aigned flow eiher 0 or. Thu he number of edge croing hi cu wih flow i equal o boh f and o M. Thu if we can find a max flow in G, we will have a maximum maching in G, a long a he maximum flow i ineger-valued. Theorem 26.0 ae ha Ford-Fulkeron produce an inegervalued flow; he proof i an exercie. Finally, Corollary 26. combine hee wo reul (Lemma 26.9 and Theorem 26.0) o conclude ha he cardinaliy of a maximum maching in a biparie graph equal he value of a maximum flow in he flow nework correponding o he graph. 9