6/3/2009. CS 244 Algorithm Design Instructor: t Artur Czumaj. Lecture 8 Network flows. Maximum Flow and Minimum Cut. Minimum Cut Problem.

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1 Maximum Flow and Minimum Cu CS lgorihm Deign Inrucor: rur Czumaj Lecure Nework Max and min cu. Two very rich algorihmic problem. Cornerone problem in combinaorial opimizaion. Beauiful mahemaical dualiy. Nonrivial applicaion / reducion. Daa mining. Nework reliabiliy. Open-pi mining. Diribued compuing. Projec elecion. Egaliarian able maching. irline cheduling. Securiy of aiical daa. Biparie maching. Nework inruion deecion. Baeball eliminaion. Muli-camera cene reconrucion. Image egmenaion. Many many more... Nework conneciviy. Minimum Cu Problem Cu Flow nework. bracion for maerial ing hrough he edge. G = (V, E) = direced graph, no parallel edge. Two diinguihed node: = ource, = ink. c(e) = capaciy of edge e. ource = where maerial originae ink = where maerial goe Def. n - cu i a pariion (, B) of V wih and B. Def. The capaciy of a cu (, B) i: cap(, B) c(e) edge i no couned ource ink capaciy Capaciy = =

2 Minimum Cu Problem Flow Min - cu problem. Find an - cu of minimum capaciy. Def. n - i a funcion ha aifie: For each e E: f (e) c(e) (capaciy) For each v V {, : f (e) f (e) (conervaion) e in o v e ou of v Def. The value of a f i: Equalize in and ou a every inermediae node. v( f ) f (e). e ou of Capaciy = + + = capaciy Value = Maximum Flow Problem Flow and Cu Max problem. Find - of maximum value. Flow value lemma. Le f be any, and le (, B) be any - cu. Then, he ne en acro he cu i equal o he amoun leaving. f (e) f (e) v( f ) e in o capaciy Value = Value = =

3 Flow and Cu Flow and Cu Weak dualiy. Le f be any, and le (, B) be any - cu. Then he value of he i a mo he capaciy of he cu. Weak dualiy. Le f be any. Then, for any - cu (, B) we have v(f) cap(, B). Proof: Cu capaciy = Flow value v( f ) f (e) f (e) e in o f (e) c(e) cap(, B) B Capaciy = Cerificae of Opimaliy Toward a Max Flow lgorihm Corollary. Le f be any, and le (, B) be any cu. If v(f) = cap(, B), hen f i a max and (, B) i a min cu. Value of = Cu capaciy = Flow value Greedy algorihm. Sar wih f(e) = for all edge e E. Find an - pah P where each edge ha f(e) < c(e). ugmen along pah P. Repea unil you ge uck. Flow value =

4 Toward a Max Flow lgorihm Toward a Max Flow lgorihm Greedy algorihm. Sar wih f(e) = for all edge e E. Find an - pah P where each edge ha f(e) < c(e). ugmen along pah P. Repea unil you ge uck. Greedy algorihm. Sar wih f(e) = for all edge e E. Find an - pah P where each edge ha f(e) < c(e). ugmen along pah P. Repea unil you ge uck. locally opimaliy global opimaliy X X X Flow value = greedy = op = Reidual Graph Ford-Fulkeron lgorihm Original edge: e = (u, v) E. Flow f(e), capaciy c(e). u capaciy v capaciy Reidual edge. "Undo" en. e = (u, v) and e R reidual capaciy = (v, u). Reidual capaciy: u v c(e) f (e) c f (e) if e E reidual capaciy f R f (e) if e E Reidual graph: G f = (V, E f ). Reidual edge wih poiive reidual capaciy. E f = {e : f(e) < c(e) {e R : c(e) >.

5 Ford-Fulkeron lgorihm Ford-Fulkeron lgorihm capaciy X X capaciy X Flow value = Flow value = reidual capaciy Ford-Fulkeron lgorihm Ford-Fulkeron lgorihm X X X X X X X X Flow value = Flow value =

6 Ford-Fulkeron lgorihm Ford-Fulkeron lgorihm X X X X X X X X X Flow value = Flow value = Ford-Fulkeron lgorihm Ford-Fulkeron lgorihm Flow value = Cu capaciy = Flow value =

7 ugmening Pah lgorihm Max-Flow Min-Cu Theorem ugmen(f, c, P) { b boleneck(p) foreach e P { if (e E) f(e) f(e) + b ele f(e R ) f(e) - b reurn f Ford-Fulkeron(G,,, c) { foreach e E f(e) G reidual graph G f while (here exi augmening pah P) { f ugmen(f, c, P) updae G f reurn f forward edge revere edge ugmening pah heorem. Flow f i a max iff here are no augmening pah. Max- min-cu heorem. [Ford-Fulkeron ] The value of he max i equal o he value of he min cu. Proof raegy. We prove boh imulaneouly by howing he TFE: (i) There exi a cu (, B) uch ha v(f) = cap(, B). (ii) Flow f i a max. (iii) There i no augmening pah relaive o f. (i) (ii) Thi wa he corollary o weak dualiy lemma. (ii) (iii) We how conrapoiive. Le f be a. If here exi an augmening pah, hen we can improve f by ending along pah. Proof of Max-Flow Min-Cu Theorem Running Time (iii) (i) Le f be a wih no augmening pah. Le be e of verice reachable from in reidual graph. By definiion of,. By definiion of f,. v( f ) f (e) f (e) e in o c(e) cap(, B) B umpion. ll capaciie are ineger beween and C. Invarian. Every value f(e) and every reidual capaciie c f (e) remain an ineger hroughou he algorihm. Theorem. The algorihm erminae in a mo v(f*) nc ieraion. Proof: Each augmenaion increae value by a lea. Corollary. If C =, Ford-Fulkeron run in O(mn) ime. Inegraliy heorem. If all capaciie are ineger, hen here exi a max f for which every value f(e) i an ineger. Proof: Since algorihm erminae, heorem follow from invarian. original nework