Topic: Applications of Network Flow Date: 9/14/2007
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1 CS787: Advanced Algorihm Scribe: Daniel Wong and Priyananda Shenoy Lecurer: Shuchi Chawla Topic: Applicaion of Nework Flow Dae: 9/4/ Inroducion and Recap In he la lecure, we analyzed he problem of finding he maximum flow in a graph, and how i can be efficienly olved uing he Ford-Fulkeron algorihm. We alo came acro he Min Cu-Max Flow Theorem which relae he ize of he maximum flow o he ize of he minimal cu in he graph. In hi lecure, we will be eeing how variou differen problem can be olved by reducing he problem o an inance of he nework flow problem. The fir applicaion we will be eeing i he Biparie Maching problem. 5.2 Biparie Maching A Biparie Graph G(V, E) i a graph in which he verex e V can be divided ino wo dijoin ube X and Y uch ha every edge e ɛ E ha one end in X and he oher end in Y. A maching M i a ube of edge uch ha each node in V appear in a mo one edge in M. A Maximum Maching i a maching of he large ize. X Y Fig : Biparie Graph A clarificaion abou he erminology: Definiion 5.2. (Maximal Maching) A maximal maching i a maching o which no more edge can be added wihou increaing he degree of one of he node o wo; i i a local maximum. Definiion (Maximum Maching) A maximum maching i one which i he large poible; i i globally opimal.
2 I can be hown ha for any maximal maching M i, M i 2 M where M i he maximum maching. Given hi, i i raighforward o obain a 2-approximaion o he maximum maching. The problem of finding he maximum maching can be reduced o maximum flow in he following manner. Le G(V, E) be he biparie graph where V i divided ino X, Y. We will conruc a direced graph G (V, E ), in which V which conain all he node of V along wih a ource node and a ink node. For every edge in E, we add a direced edge in E from X o Y. Finally we add a direced edge from o all node in X and from all node of Y o. Each edge i given uni weigh. X Y Fig 2: The Biparie Graph convered ino a flow graph Le f be he maximum inegral flow of G, of value k. Then we can make he following obervaion:. There i no node in X which ha more han one ougoing edge where here i a flow. 2. There i no node in Y which ha more han one incoming edge where here i a flow. 3. The number of edge beween X and Y which carry flow i k. By hee obervaion, i i raighforward o conclude ha he e of edge carrying flow in f form he maximum maching for he graph G. The running ime for he Ford-Fulkeron algorihm i O(mF ) where m i he number of edge in E and F = e δ() (c e). In cae of biparie maching problem, F V ince here can be only V poible edge coming ou from ource node. So he running ime i O(mn) where n i he number of node. An inereing hing o noe i ha any pah in he reidual graph of he problem will have alernaing mached and unmached edge. Such pah are called alernaing pah. Thi propery can be ued o find maching even in general graph Perfec Maching A perfec maching i a maching in which each node ha exacly one edge inciden on i. One poible way of finding ou if a given biparie graph ha a perfec maching i o ue he above 2
3 algorihm o find he maximum maching and checking if he ize of he maching equal he number of node in each pariion. There i anoher way of deermining hi, uing Hall Theorem. Theorem A Biparie graph G(V,E) ha a Perfec Maching iff for every ube S XorS Y, he ize of he neighbor of S i a lea a large a S, i.e Γ(S) S We will no dicu he proof of hi heorem in he cla. 5.3 Scheduling Problem We will now ee how he Max Flow algorihm can be ued o olve cerain kind of cheduling problem. The fir example we will ake will be of cheduling job on a machine. Le J = {J, J 2,..., J n } be he e of job, and T = {T, T 2,.., T k } be lo available on a machine where hee job can be performed. Each job J ha a e of valid lo S j T when i can be cheduled. The conrain i ha no wo job can be cheduled a he ame ime. The problem i o find he large e of job which can be cheduled. Thi problem can be olved by reducing i o a biparie maching problem. For every job, creae a node in X, and for every imelo creae a node in Y. For every imelo T in S j, creae an edge beween J and T. The maximum maching of hi biparie graph i he maximum e of job ha can be cheduled. We can alo olve cheduling problem wih more conrain by having inermediae node in he graph. Le u conider a more complex problem: There are k vacaion period each panning muliple coniguou day. Le D j be he e of day included in he j h vacaion period. There are n docor in he hopial, each wih a e of vacaion day when he or he i available. We need o maximize he aignmen of docor o day under he following conrain:. Each docor ha a capaciy c i which i he maximum oal number of day he or he can be cheduled. 2. For every vacaion period, any given docor i cheduled a mo once. We will olve hi problem by nework flow. A wa done earlier, for every docor i we creae a node u i and for every vacaion day j we creae a node v j. We add a direced edge wih uni capaciy from ar node o u i and from v j o ink. To include he conrain, he way he graph i conruced in he following way: The docor capaciie are modeled a capaciie of he edge from he ource o he verex correponding o he docor. To preven he docor o be cheduled more han once in a vacaion period, we inroduce inermediae node. For any docor i and a vacaion period j, we creae an inermediae node w ij. We creae an edge wih uni capaciy from u i o w ij. For every day in he vacaion period ha he docor i available, we creae an edge from w ij o he node correponding o ha day wih uni capaciy. 3
4 docor inermediae node vacaion period ci Fig 3: Scheduling Problem wih conrain Le u ee if he an inegral flow hrough he graph produce a valid chedule. Since he edge connecing o he node correponding o he docor ha he capaciy equal o he oal availabiliy of he docor, he flow hrough he docor node canno exceed i. So he fir conrain i aified. From any docor o any vacaion period he flow i amo one, ince he inermediae node ha only one incoming edge of uni capaciy. Thi make ure ha he econd crieria i me. So he flow produced aifie all he conrain. If k i he ize of an inegral flow hrough he graph, hen he oal number of vacaion day which have been aigned i alo k, ince he edge which connec he node correponding o he vacaion day o he ink node have uni capaciy each. So he ize of he cheduling i equal o he ize of he flow. From hi we can conclude ha he large valid cheduling i produced by he maximum flow. 5.4 Pariioning Problem Now we will ee how cerain kind of pariioning problem can be olved uing he nework flow algorihm. The general idea i o reduce i o min-cu raher han max-flow problem. The fir example we will ee i he Image egmenaion problem Image Segmenaion Conider he following implificaion of he image egmenaion problem. Aume every pixel in an image belong o eiher he foreground of an image or he background. Uing image analyi echnique baed on he value of he pixel i poible o aign probabilie ha an individual pixel belong o he foreground or he background. Thee probabilie can hen be ranlaed o co for aigning a pixel o eiher foreground or background. In addiion, here are co for egmening pixel uch ha pixel from differan region are adjacen. The goal hen i o find an aignmen of all pixel uch ha he co are minimized. Le f i be he co of aigning pixel i o he foreground. Le b i be he co of aigning pixel i o he background. 4
5 Le ij be he co of eparaing neighboring pixel i and j ino differen region. Problem: Find a pariion of pixel ino, p S, he e of foreground pixel and S, he e of background pixel,uch ha he global co of hi aignmen i minimized. Le (S, S) be he pariion of pixel ino foreground and background repecively. Then he co of hi pariion i defined a: co(s) = i S b i + i/ S f i + i S,j / S i and j are neighbour ij (5.4.) We need o find an S which minimize co. Thi problem can be reduced o a min-cu max-flow problem. The naural reducion i o a min-cu problem. Le each pixel be a node, wih neighboring pixel conneced o each oher by undireced edge wih capaciy ij. Creae addiional ource and ink node, and, repecively which have edge o every pixel. The edge from o each pixel have capaciy b i. The edge from each pixel o have capaciy f i. background vi foreground fi bi ij Fig 3: Image egmenaion flow graph Any cu in he graph will hen naurally eparae he pixel ino wo group. Wih he arrangemen he capaciy of he cu i he co aociaed wih ha pariion. The min-cu max-flow algorihm will find uch a pariion of minimum co Image Segmenaion con. Conider a modificaion of he original problem. Replace f i and b i wih value repreening he benefi/value for aigning he pixel i o eiher he foreground or background. Problem: Find an aigmen uch ha for all pixel, p S, he e of foreground pixel, or S, he e of background pixel,uch ha he global value i maximized. Where value i: 5
6 val(s) = i S f i + i/ S b i i S,j / S i and j are neighbour ij (5.4.2) Since he min-cu approach i a minimizaion problem, we need o conver i ino a minimizaion problem in order o reduce i o min-co. Le (S, S) be he pariion of pixel ino foreground and background repecively. We can relae co (defined a per 5.4.) and val by auming co o mean lo benefi : co(s) = i (b i + f i ) val(s) (5.4.3) Since i (b i + f i ) i a conan for any given inance, he problem of maximiing val reduce o minimizing he co. We can reformulae co a co(s) = i S b i + i/ S f i + i S,j / S i and j are neighbour ij (5.4.4) Comparing hi o he previou problem, we ee ha hi i he ame a inerchanging b i and f i in he graph we ued o olve he fir problem; i.e. he weigh of he node from o v i i e a f i and weigh of v i o i e a b i. Solving hi uing min-cu will give u he pariion which maximize he benefi. background vi foreground bi fi ij Fig 4: Flow graph for maximizing benefi 5.5 Max-Flow wih Co Le u conider an exenion of he perfec maching problem where differen edge have differen co. The naurally lead o he min-co max-flow problem. In hi cae, no only do we have o find a max flow hrough he graph, bu alo have o find one wih he lea co. 6
7 Likewie, in he min-co perfec maching problem, he co of a perfec maching M i eɛm c(e)f(e) where f(e) i he co of including he edge e in he maching. I can be hown ha by direcing he flow hrough he minimum co augemening pah fir,we will find he maximum flow wih he lea co. Reference [] Eva Tardo, Jon Kleinberg Algorihm Deign. Pearon Educaion,
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