4/12/12. Applications of the Maxflow Problem 7.5 Bipartite Matching. Bipartite Matching. Bipartite Matching. Bipartite matching: the flow network
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1 // Applicaion of he Maxflow Problem. Biparie Maching Biparie Maching Biparie maching. Inpu: undireced, biparie graph = (, E). M E i a maching if each node appear in a mo one edge in M. Max maching: find a max cardinaliy maching. Biparie Maching Biparie maching. Inpu: undireced, biparie graph = (, E). M E i a maching if each node appear in a mo one edge in M. Max maching: find a max cardinaliy maching. ' ' ' ' ' ' maching -', -', -' ' max maching -', -', -' -' ' ' ' ' ' ' ' ' Biparie Maching Biparie maching: he flow nework Biparie maching: he flow nework How o olve uing maxflow? Need a ource and ink raph need o be direced Capaciie? How o inerpre he flow? Source + ink: ' ' Capaciie: ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' '
2 // Biparie Maching Max flow oluion. Creae direced graph = ( {, }, E' ). Direc all edge from o, and aign infinie (or uni) capaciy. Add ource, and uni capaciy edge from o each node in. Add ink, and uni capaciy edge from each node in o. ' ' Biparie Maching: Proof of Correcne Theorem. Max cardinaliy maching in = value of max flow in. Proof. iven max maching M of cardinaliy k. Conider flow f ha end uni along each of k pah. f i a flow, and ha value k. ' ' ' ' Biparie Maching: Proof of Correcne Theorem. Max cardinaliy maching in = value of max flow in. Proof. e f be a max flow in of value k. Inegraliy heorem k i inegral and can aume f i 0-. Conider M = e of edge from o wih f(e) =. each node in and paricipae in a mo one edge in M M = k: conider cu (, ) ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' 8 9 Biparie Maching: unning Time unning ime? O(m val(f*) ) = O(mn). educion We ook problem A (biparie maching) and howed ha i can be olved uing an inance of problem B (maxflow). Thi i an example of reducing problem A o problem B educion: mapping from an inance of A o an inance of B, uch ha from he oluion o he inance of B we can conruc he oluion o he inance of A. Dijoin Pah If A reduce o B, which one i harder? ' ' ' ' ' ' ' ' ' ' 0 biparie maching maxflow
3 // Edge Dijoin Pah Dijoin pah problem. iven a digraph = (V, E) and wo node and, find he max number of edge-dijoin - pah. Def. Two pah are edge-dijoin if hey have no edge in common. Example: communicaion nework. Edge Dijoin Pah Dijoin pah problem. iven a digraph = (V, E) and wo node and, find he max number of edge-dijoin - pah. Def. Two pah are edge-dijoin if hey have no edge in common. Example: communicaion nework. Edge Dijoin Pah Max flow formulaion: aign uni capaciy o every edge. Theorem. Max number edge-dijoin - pah equal max flow value. Proof. Suppoe here are k edge-dijoin pah P,..., P k. Se f(e) = if e paricipae in ome pah P i ; ele e f(e) = 0. Since pah are edge-dijoin, f i a flow of value k. Edge Dijoin Pah Nework Conneciviy Edge Dijoin Pah and Nework Conneciviy Max flow formulaion: aign uni capaciy o every edge. Theorem. Max number edge-dijoin - pah equal max flow value. Proof. Suppoe max flow value i k. Inegraliy heorem here exi 0- flow f of value k. Conider edge (, u) wih f(, u) =. by conervaion, here exi an edge (u, v) wih f(u, v) = coninue unil reach, alway chooing a new edge Produce k edge-dijoin pah. Nework conneciviy. iven a digraph = (V, E) and wo node and, find min number of edge whoe removal diconnec from. Def. A e of edge F E diconnec from if every - pah ue a lea one edge in F. Theorem. [Menger 9] The max number of edge-dijoin - pah i equal o he min number of edge whoe removal diconnec from. Proof. Suppoe he removal of F E diconnec from, and F = k. Every - pah ue a lea one edge in F. Hence, he number of edge-dijoin pah i a mo k. 8
4 // Edge Dijoin Pah and Nework Conneciviy Theorem. [Menger 9] The max number of edge-dijoin - pah i equal o he min number of edge whoe removal diconnec from. Proof. Suppoe max number of edge-dijoin pah i k. Then max flow value i k. Max-flow min-cu cu (A, B) of capaciy k. e F be e of edge going from A o B. F = k and diconnec from.. Exenion o Max Flow Circulaion wih Demand Circulaion wih demand. Direced graph = (V, E). Edge capaciie c(e), e E. Node upply and demand d(v), v V. demand if d(v) > 0; upply if d(v) < 0; conervaion if d(v) = 0 Def. A circulaion i a funcion ha aifie: For each e E: 0 f(e) c(e) (capaciy) For each v V: f (e) f (e) = d(v) (conervaion) e in o v e ou of v A Circulaion problem: given (V, E, c, d), doe here exi a circulaion? 9 Circulaion wih Demand Neceary condiion: um of upplie = um of demand. d(v) v : d (v) > 0 = d(v) =: D v : d (v) < 0 Proof. Sum conervaion conrain for every demand node v. Max flow formulaion? Circulaion wih Demand Circulaion wih Demand Max flow formulaion. Add new ource and ink. For each v wih d(v) < 0, add edge (, v) wih capaciy -d(v). For each v wih d(v) > 0, add edge (v, ) wih capaciy d(v). Claim: ha circulaion iff ha max flow of value D. aurae all edge leaving and enering upply 9 capaciy demand : upply 9 demand : 8 upply demand flow
5 //.0 Image egmenaion Image Segmenaion Image egmenaion. Cenral problem in image proceing. Divide image ino coheren region. Example: Image Segmenaion Foreground / background egmenaion. abel each pixel in picure a belonging o foreground or background. V = e of pixel, E = pair of neighboring pixel. a i 0 i likelihood pixel i in foreground. b i 0 i likelihood pixel i in background. 0 i eparaion penaly for labeling one of i and j a foreground, and he oher a background. oal. Accuracy: if a i > b i in iolaion, prefer o label i in foreground. Smoohne: if many neighbor of i are labeled foreground, we hould be inclined o label i a foreground. Find pariion (A, B) ha maximize: a i + b j foreground background i A j B (i, j) E A{i, j} = Image Segmenaion Image Segmenaion Image Segmenaion Formulae a min cu problem. Maximizaion. No ource or ink. Undireced graph. Turn ino minimizaion problem. Formulae a min cu problem. = (V', E'). Add ource o correpond o foreground; add ink o correpond o background Ue wo direced edge inead of undireced edge. Conider min cu (A, B) in. A = foreground. cap(a, B) = a j + b i + j B i A (i, j) E i A, j B Preciely he quaniy we wan o minimize. if i and j on differen ide, couned exacly once Maximizing a i + b j i A j B (i, j) E A{i, j} = i equivalen o maximizing ( a i V i + b j V j ) a i b j i B j A or alernaively minimizing a j + + j B a conan b i i A (i, j) E A{i, j} = (i, j) E A{i, j} = i a j b i j A i a j b i j 8 9 0
6 // educion We have een he following problem ha can be reduced o maxflow/mincu: v v v v Biparie maching Number of dijoin pah Circulaion wih demand Image egmenaion There are many more! educion a a problem olving ool: If you have an algorihm for a very general problem, ha give you a ool for olving lo of oher problem.
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