7.5 Bipartite Matching. Chapter 7. Network Flow. Matching. Bipartite Matching

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1 Chaper. Biparie Maching Nework Flow Slide by Kein Wayne. Copyrigh 00 Pearon-Addion Weley. All righ reered. Maching Biparie Maching Maching. Inpu: undireced graph G = (V, E). M E i a maching if each node appear in a mo edge in M. Max maching: find a max cardinaliy maching. Biparie maching. Inpu: undireced, biparie graph G = (L R, E). M E i a maching if each node appear in a mo edge in M. Max maching: find a max cardinaliy maching. ' ' ' maching -', -', -' ' L ' R

2 Biparie Maching Biparie Maching Biparie maching. Inpu: undireced, biparie graph G = (L R, E). M E i a maching if each node appear in a mo edge in M. Max maching: find a max cardinaliy maching. Max flow formulaion. Creae digraph G' = (L R {, }, E' ). Direc all edge from L o R, and aign infinie (or uni) capaciy. Add ource, and uni capaciy edge from o each node in L. Add ink, and uni capaciy edge from each node in R o. ' G' ' ' max maching -', -', -' -' ' ' ' ' ' L ' R L ' R Biparie Maching: Proof of Correcne Biparie Maching: Proof of Correcne Theorem. Max cardinaliy maching in G = alue of max flow in G'. Pf. Gien max maching M of cardinaliy k. Conider flow f ha end uni along each of k pah. f i a flow, and ha cardinaliy k. Theorem. Max cardinaliy maching in G = alue of max flow in G'. Pf. Le f be a max flow in G' of alue k. Inegraliy heorem k i inegral and can aume f i 0-. Conider M = e of edge from L o R wih f(e) =. each node in L and R paricipae in a mo one edge in M M = k: conider cu (L, R ) ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' G ' ' G' G' ' ' G 8

3 9 Perfec Maching Perfec Maching Def. A maching M E i perfec if each node appear in exacly one edge in M. Noaion. Le S be a ube of node, and le N(S) be he e of node adjacen o node in S. Q. When doe a biparie graph hae a perfec maching? Srucure of biparie graph wih perfec maching. Clearly we mu hae L = R. Wha oher condiion are neceary? Wha condiion are ufficien? Oberaion. If a biparie graph G = (L R, E), ha a perfec maching, hen N(S) S for all ube S L. Pf. Each node in S ha o be mached o a differen node in N(S). ' ' ' No perfec maching: S = {,, } N(S) = { ', ' }. ' L ' R 0 Marriage Theorem Proof of Marriage Theorem Marriage Theorem. [Frobeniu 9, Hall 9] Le G = (L R, E) be a biparie graph wih L = R. Then, G ha a perfec maching iff N(S) S for all ube S L. Pf. Thi wa he preiou oberaion. L ' ' ' ' ' R No perfec maching: S = {,, } N(S) = { ', ' }. Pf. Suppoe G doe no hae a perfec maching. Formulae a a max flow problem and le (A, B) be min cu in G'. By max-flow min-cu, cap(a, B) < L. Define L A = L A, L B = L B, R A = R A. cap(a, B) = L B + R A. Since min cu can' ue edge: N(L A ) R A. N(L A ) R A = cap(a, B) - L B < L - L B = L A. Chooe S = L A. G' A ' ' ' L A = {,, } L B = {, } ' R A = {', '} N(L A ) = {', '} '

4 Biparie Maching: Running Time Which max flow algorihm o ue for biparie maching? Generic augmening pah: O(m al(f*) ) = O(mn). Capaciy caling: O(m log C ) = O(m ). Shore augmening pah: O(m n / ).. Dijoin Pah Non-biparie maching. Srucure of non-biparie graph i more complicaed, bu well-underood. [Tue-Berge, Edmond-Galai] Bloom algorihm: O(n ). [Edmond 9] Be known: O(m n / ). [Micali-Vazirani 980] Edge Dijoin Pah Edge Dijoin Pah Dijoin pah problem. Gien a digraph G = (V, E) and wo node and, find he max number of edge-dijoin - pah. Dijoin pah problem. Gien a digraph G = (V, E) and wo node and, find he max number of edge-dijoin - pah. Def. Two pah are edge-dijoin if hey hae no edge in common. Def. Two pah are edge-dijoin if hey hae no edge in common. Ex: communicaion nework. Ex: communicaion nework.

5 Edge Dijoin Pah Edge Dijoin Pah Max flow formulaion: aign uni capaciy o eery edge. Max flow formulaion: aign uni capaciy o eery edge. Theorem. Max number edge-dijoin - pah equal max flow alue. Pf. 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 alue k. Theorem. Max number edge-dijoin - pah equal max flow alue. Pf. Suppoe max flow alue i k. Inegraliy heorem here exi 0- flow f of alue k. Conider edge (, u) wih f(, u) =. by coneraion, here exi an edge (u, ) wih f(u, ) = coninue unil reach, alway chooing a new edge Produce k (no necearily imple) edge-dijoin pah. can eliminae cycle o ge imple pah if deired 8 Nework Conneciiy Edge Dijoin Pah and Nework Conneciiy Nework conneciiy. Gien a digraph G = (V, E) and wo node and, find min number of edge whoe remoal diconnec from. Theorem. [Menger 9] The max number of edge-dijoin - pah i equal o he min number of edge whoe remoal diconnec from. Def. A e of edge F E diconnec from if all - pah ue a lea on edge in F. Pf. Suppoe he remoal of F E diconnec from, and F = k. All - pah ue a lea one edge of F. Hence, he number of edgedijoin pah i a mo k. 9 0

6 Dijoin Pah and Nework Conneciiy Theorem. [Menger 9] The max number of edge-dijoin - pah i equal o he min number of edge whoe remoal diconnec from.. Exenion o Max Flow Pf. Suppoe max number of edge-dijoin pah i k. Then max flow alue i k. Max-flow min-cu cu (A, B) of capaciy k. Le F be e of edge going from A o B. F = k and diconnec from. A Circulaion wih Demand Circulaion wih Demand Circulaion wih demand. Direced graph G = (V, E). Edge capaciie c(e), e E. Node upply and demand d(), V. Neceary condiion: um of upplie = um of demand. d() : d () > 0 = d() =: D : d () < 0 Pf. Sum coneraion conrain for eery demand node. demand if d() > 0; upply if d() < 0; ranhipmen if d() = 0 Def. A circulaion i a funcion ha aifie: For each e E: 0 f(e) c(e) (capaciy) For each V: f (e) f (e) = d() (coneraion) e in o e ou of Circulaion problem: gien (V, E, c, d), doe here exi a circulaion? upply flow 9 capaciy demand

7 Circulaion wih Demand Circulaion wih Demand Max flow formulaion. Max flow formulaion. Add new ource and ink. For each wih d() < 0, add edge (, ) wih capaciy -d(). For each wih d() > 0, add edge (, ) wih capaciy d(). Claim: G ha circulaion iff G' ha max flow of alue D. aurae all edge leaing and enering G: -8 - upply G': 8 upply demand 0 0 demand Circulaion wih Demand Circulaion wih Demand and Lower Bound Inegraliy heorem. If all capaciie and demand are ineger, and here exi a circulaion, hen here exi one ha i ineger-alued. Pf. Follow from max flow formulaion and inegraliy heorem for max flow. Characerizaion. Gien (V, E, c, d), here doe no exi a circulaion iff here exi a node pariion (A, B) uch ha Σ B d > cap(a, B) Feaible circulaion. Direced graph G = (V, E). Edge capaciie c(e) and lower bound l (e), e E. Node upply and demand d(), V. Def. A circulaion i a funcion ha aifie: For each e E: l (e) f(e) c(e) (capaciy) For each V: f (e) f (e) = d() (coneraion) e in o e ou of Pf idea. Look a min cu in G'. demand by node in B exceed upply of node in B plu max capaciy of edge going from A o B Circulaion problem wih lower bound. Gien (V, E, l, c, d), doe here exi a a circulaion? 8

8 9 Circulaion wih Demand and Lower Bound Idea. Model lower bound wih demand. Send l(e) uni of flow along edge e. Updae demand of boh endpoin..8 Surey Deign lower bound upper bound capaciy [, 9] w d() d(w) d() + d(w) - G G' w Theorem. There exi a circulaion in G iff here exi a circulaion in G'. If all demand, capaciie, and lower bound in G are ineger, hen here i a circulaion in G ha i ineger-alued. Pf kech. f(e) i a circulaion in G iff f'(e) = f(e) - l(e) i a circulaion in G'. Surey Deign Surey Deign Surey deign. Deign urey aking n conumer abou n produc. Can only urey conumer i abou a produc j if hey own i. Ak conumer i beween c i and c i ' queion. Ak beween p j and p j ' conumer abou produc j. Algorihm. Formulae a a circulaion problem wih lower bound. Include an edge (i, j) if cuomer own produc i. Ineger circulaion feaible urey deign. [0, ] Goal. Deign a urey ha mee hee pec, if poible. [0, ] ' Biparie perfec maching. Special cae when c i = c i ' = p i = p i ' =. [c, c '] ' [p, p '] ' ' conumer ' produc

9 Image Segmenaion.0 Image Segmenaion Image egmenaion. Cenral problem in image proceing. Diide image ino coheren region. Ex: Three people anding in fron of complex background cene. Idenify each peron a a coheren objec. Image Segmenaion Image Segmenaion Foreground / background egmenaion. Label 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. p ij 0 i eparaion penaly for labeling one of i and j a foreground, and he oher a background. Goal. 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 + foreground background i A b j j B p ij (i, j) E AI{i, j} = Formulae a min cu problem. Maximizaion. No ource or ink. Undireced graph. Turn ino minimizaion problem. Maximizing a i + i A b j j B i equialen o minimizing p ij (i, j) E AI{i, j} = ( i V a + b i j V ) j or alernaiely a j + + b j + j B a conan b i i A p ij (i, j) E AI{i, j} = a i i A j B p ij (i,j) E AI{i, j} =

10 Image Segmenaion Image Segmenaion Formulae a min cu problem. G' = (V', E'). Add ource o correpond o foreground; add ink o correpond o background Ue wo ani-parallel edge inead of undireced edge. p ij p ij p ij Conider min cu (A, B) in G'. A = foreground. cap(a, B) = a j + b i + j B i A p ij (i, j) E i A, j B Preciely he quaniy we wan o minimize. if i and j on differen ide, p ij couned exacly once a j a j i p ij j i p ij j b i A b i G' G' 8 Projec Selecion. Projec Selecion Projec wih prerequiie. can be poiie or negaie Se P of poible projec. Projec ha aociaed reenue p. ome projec generae money: creae ineracie e-commerce inerface, redeign web page oher co money: upgrade compuer, ge ie licene Se of prerequiie E. If (, w) E, can' do projec and unle alo do projec w. A ube of projec A P i feaible if he prerequiie of eery projec in A alo belong o A. Projec elecion. Chooe a feaible ube of projec o maximize reenue. 0

11 Projec Selecion: Prerequiie Graph Projec Selecion: Min Cu Formulaion Prerequiie graph. Include an edge from o w if can' do wihou alo doing w. {, w, x} i feaible ube of projec. {, x} i infeaible ube of projec. Min cu formulaion. Aign capaciy o all prerequiie edge. Add edge (, ) wih capaciy -p if p > 0. Add edge (, ) wih capaciy -p if p < 0. For noaional conenience, define p = p = 0. w w p u u w -p w p y y z -p z feaible x infeaible x p x -p x Projec Selecion: Min Cu Formulaion Open Pi Mining Claim. (A, B) i min cu iff A { } i opimal e of projec. Infinie capaciy edge enure A { } i feaible. Max reenue becaue: cap( A, B) = B: p p + > 0 A: p< 0 ( p ) Open-pi mining. (udied ince early 90) Block of earh are exraced from urface o reriee ore. Each block ha ne alue p = alue of ore - proceing co. Can' remoe block before w or x. = p : p > 0 conan A p w u A p p u p y y z x -p w -p x w x

12 Baeball Eliminaion. Baeball Eliminaion "See ha hing in he paper la week abou Einein?... Some reporer aked him o figure ou he mahemaic of he pennan race. You know, one eam win o many of heir remaining game, he oher eam win hi number or ha number. Wha are he myriad poibiliie? Who' go he edge?" "The hell doe he know?" "Apparenly no much. He picked he Dodger o eliminae he Gian la Friday." - Don DeLillo, Underworld Team i Win w i Which eam hae a chance of finihing he eaon wih mo win? Monreal eliminaed ince i can finih wih a mo 80 win, bu Alana already ha 8. w i + r i < w j eam i eliminaed. Only reaon por wrier appear o be aware of. Sufficien, bu no neceary! Loe To play Again = r ij l i r i Al Phi NY Mon Alana Philly New York Monreal Baeball Eliminaion Baeball Eliminaion Team i Win w i Loe To play Again = r ij l i r i Al Phi NY Mon Alana Philly New York Monreal Which eam hae a chance of finihing he eaon wih mo win? Philly can win 8, bu ill eliminaed... If Alana loe a game, hen ome oher eam win one. Remark. Anwer depend no ju on how many game already won and lef o play, bu alo on whom hey're again. 8

13 9 Baeball Eliminaion Baeball Eliminaion: Max Flow Formulaion Baeball eliminaion problem. Se of eam S. Diinguihed eam S. Team x ha won w x game already. Team x and y play each oher r xy addiional ime. I here any oucome of he remaining game in which eam finihe wih he mo (or ied for he mo) win? Can eam finih wih mo win? Aume eam win all remaining game w + r win. Diy remaining game o ha all eam hae w + r win. - game lef - - eam can ill win hi many more game r = - w + r - w - game node - eam node 0 Baeball Eliminaion: Max Flow Formulaion Baeball Eliminaion: Explanaion for Spor Wrier Theorem. Team i no eliminaed iff max flow aurae all edge leaing ource. Inegraliy heorem each remaining game beween x and y added o number of win for eam x or eam y. Capaciy on (x, ) edge enure no eam win oo many game. game lef r = - w + r - w eam can ill win hi many more game Team i Win w i Loe To play Again = r ij l i r i NY Bal Bo Tor NY Balimore 8 - Boon Torono 0 - Deroi AL Ea: Augu 0, 99 Which eam hae a chance of finihing he eaon wih mo win? Deroi could finih eaon wih 9 + = win. De game node - eam node

14 Baeball Eliminaion: Explanaion for Spor Wrier Baeball Eliminaion: Explanaion for Spor Wrier Team i Win w i Which eam hae a chance of finihing he eaon wih mo win? Deroi could finih eaon wih 9 + = win. Cerificae of eliminaion. R = {NY, Bal, Bo, Tor} Hae already won w(r) = 8 game. Mu win a lea r(r) = more. Loe To play Again = r ij l i r i NY Bal Bo Tor NY Balimore 8 - Boon Torono 0 - Deroi AL Ea: Augu 0, 99 Aerage eam in R win a lea 0/ > game. De Cerificae of eliminaion. # remaining game # win 8 8 T S, w(t ) := w i, g(t ) := g x y, i T {x,y} T LB on ag # game won 8 w(t)+ g(t) If > w z + g z hen z i eliminaed (by ube T). T Theorem. [Hoffman-Rilin 9] Team z i eliminaed iff here exi a ube T* ha eliminae z. Proof idea. Le T* = eam node on ource ide of min cu. Baeball Eliminaion: Explanaion for Spor Wrier Baeball Eliminaion: Explanaion for Spor Wrier Pf of heorem. Ue max flow formulaion, and conider min cu (A, B). Define T* = eam node on ource ide of min cu. Obere x-y A iff boh x T* and y T*. infinie capaciy edge enure if x-y A hen x A and y A if x A and y A bu x-y T, hen adding x-y o A decreae capaciy of cu Pf of heorem. Ue max flow formulaion, and conider min cu (A, B). Define T* = eam node on ource ide of min cu. Obere x-y A iff boh x T* and y T*. g(s {z}) > cap(a, B) capaciy of game edge leaing capaciy of eam edge leaing 8 8 = g(s {z}) g(t*) + (w z + g z w x ) x T* = g(s {z}) g(t*) w(t*) + T* (w z + g z ) game lef y eam x can ill win hi many more game w(t*)+ g(t*) Rearranging erm: w z + g z < T* r = x-y x w z + r z - w x