7. NETWORK FLOW II. Minimum cut application (RAND 1950s) Maximum flow application (Tolstoǐ 1930s) Max-flow and min-cut applications

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1 Minimum cu applicaion (RAND 90). NETWORK FLOW II Free world goal. Cu upplie (if Cold War urn ino real war). Lecure lide by Kevin Wayne Copyrigh 00 Pearon-Addion Weley biparie maching dijoin pah exenion o max flow urvey deign airline cheduling image egmenaion projec elecion baeball eliminaion hp:// La updaed on //8 : AM rail nework connecing Sovie Union wih Eaern European counrie (map declaified by Penagon in 999) Maximum flow applicaion (Toloǐ 90) Max-flow and min-cu applicaion Sovie Union goal. Maximize flow of upplie o Eaern Europe. flow capaciy rail nework connecing Sovie Union wih Eaern European counrie (map declaified by Penagon in 999) Max-flow and min-cu problem are widely applicable model. Daa mining. Open-pi mining. Biparie maching. Nework reliabiliy. Baeball eliminaion. Image egmenaion. Nework conneciviy. Markov random field. Diribued compuing. Securiy of aiical daa. Egaliarian able maching. Nework inruion deecion. Muli-camera cene reconrucion. Senor placemen for homeland ecuriy. Many, many, more. liver and hepaic vacularizaion egmenaion

2 Maching SECTION.. NETWORK FLOW II biparie maching dijoin pah exenion o max flow urvey deign airline cheduling image egmenaion projec elecion baeball eliminaion Def. Given an undireced graph G = (V, E), ube of edge M E i a maching if each node appear in a mo one edge in M. Max maching. Given a graph G, find a max-cardinaliy maching. Biparie maching Biparie maching: max-flow formulaion Def. A graph G i biparie if he node can be pariioned ino wo ube L and R uch ha every edge connec a node in L wih a node in R. Biparie maching. Given a biparie graph G = (L R, E), find a maxcardinaliy maching. Creae digraph Gʹ = (L R {, }, Eʹ ). Direc all edge from L o R, and aign infinie (or uni) capaciy. Add uni-capaciy edge from o each node in L. Add uni-capaciy edge from each node in R o. ' G ' ' ' ' ' ' ' L ' R ' maching: -', -', -', -' L R 8

3 Max-flow formulaion: proof of correcne Max-flow formulaion: proof of correcne Theorem. Max cardinaliy of a maching in G = value of max flow in Gʹ. Pf. Given a max maching M of cardinaliy k. Conider flow f ha end uni on each of he k correponding pah. f i a flow of value k. Theorem. Max cardinaliy of a maching in G = value of max flow in Gʹ. Pf. Le f be a max flow in Gʹ and le k denoe i value. 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 : apply flow-value lemma o cu (L {}, R {}) ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' ' G ' ' G G ' ' G 9 0 Nework flow II: quiz Perfec maching in biparie graph Wha i running ime of Ford Fulkeron algorihm o find a maximum maching in a biparie graph wih L = R = n? A. O(m + n) B. O(mn) C. O(mn ) D. O(m n) Def. Given a graph G = (V, E), a ube of edge M E i a perfec maching if each node appear in exacly one edge in M. Q. When doe a biparie graph have a perfec maching? Srucure of biparie graph wih perfec maching. Clearly, we mu have L = R. Which oher condiion are neceary? Which oher condiion are ufficien?

4 Perfec maching in biparie graph Hall marriage heorem Noaion. Le S be a ube of node, and le N(S) be he e of node adjacen o node in S. Obervaion. 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). Theorem. [Frobeniu 9, Hall 9] Le G = (L R, E) be a biparie graph wih L = R. Then, graph G ha a perfec maching iff N(S) S for all ube S L. Pf. Thi wa he previou obervaion. ' ' S = {,, } ' S = {,, } ' N(S) = { ', ' } N(S) = { ', ' } ' ' ' ' ' ' no perfec maching no perfec maching Hall marriage heorem Biparie maching Pf. Suppoe G doe no have a perfec maching. Formulae a a max-flow problem and le (A, B) be a min cu in Gʹ. By max-flow min-cu heorem, cap(a, B) < L. Define L A = L A, L B = L B, R A = R A. cap(a, B) = L B + R A R A < L A. Min cu can ue edge N(L A ) R A. N(L A ) R A < L A. Chooe S = L A. G ' LA = {,, } A ' LB = {, } RA = {', '} ' N(LA) = {', '} Problem. Given a biparie graph, find a max-cardinaliy maching. year wor cae echnique dicovered by 9 O(m n) augmening pah Ford Fulkeron 9 O(m n / ) blocking flow Hopcrof Karp, Karzanov 00 O(n.8 ) fa marix muliplicaion Mucha Sankowi 0 Õ(m 0/ ) elecrical flow Mądry 0xx running ime for finding a max-cardinaliy maching in a biparie graph wih n node and m edge ' '

5 Nework flow II: quiz Nonbiparie maching Which of he following are properie of he graph G = (V, E)? A. G ha a perfec maching. B. Hall condiion i aified: N(S) S for all ube S V. C. Boh A and B. Problem. Given an undireced graph, find a max-cardinaliy maching. Srucure of nonbiparie graph i more complicaed. Bu well underood. [Tue Berge formula, Edmond Galai] Bloom algorihm: O(n ). [Edmond 9] Be known: O(m n / ). [Micali Vazirani 980, Vazirani 99] D. Neiher A nor B. PATHS, TREES, AND FLOWERS JACK EDMONDS COMBINATORICA Akadmiai Kiad - Springer-Verlag COMBINATORICA (i) (99) -09. Inroducion. A graph G for purpoe here i a finie e of elemen called verice and a finie e of elemen called edge uch ha each edge mee exacly wo verice, called he end-poin of he edge. An edge i aid o join i end-poin. A maching in G i a ube of i edge uch ha no wo mee he ame verex. We decribe an efficien algorihm for finding in a given graph a maching of maximum cardinaliy. Thi problem wa poed and parly olved by C. Berge; ee Secion. and.8. A THEORY OF ALTERNATING PATHS AND BLOSSOMS FOR PROVING CORRECTNESS OF THE O(v/-VE) GENERAL GRAPH MAXIMUM MATCHING ALGORITHM VIJAY V. VAZIRANI Received December 0, 989 Revied June, 99 8 Hiorical ignificance (Jack Edmond 9) HACKATHON PROBLEM. Digreion. An explanaion i due on he ue of he word "efficien algorihm." Fir, wha I preen i a concepual decripion of an algorihm and no a paricular formalized algorihm or "code." For pracical purpoe compuaional deail are vial. However, my purpoe i only o how a aracively a I can ha here i an efficien algorihm. According o he dicionary, "efficien" mean "adequae in operaion or performance." Thi i roughly he meaning I wan in he ene ha i i conceivable for maximum maching o have no efficien algorihm. Perhap a beer word i "good." I am claiming, a a mahemaical reul, he exience of a good algorihm for finding a maximum cardinaliy maching in a graph. There i an obviou finie algorihm, bu ha algorihm increae in difficuly exponenially wih he ize of he graph. I i by no mean obviou wheher or no here exi an algorihm whoe difficuly increae only algebraically wih he ize of he graph. Hackahon problem. Hackahon aended by n Harvard uden and n Princeon uden. Each Harvard uden i friend wih exacly k > 0 Princeon uden; each Princeon uden i friend wih exacly k Harvard uden. I i poible o arrange he hackahon o ha each Princeon uden pair program wih a differen friend from Harvard? Mahemaical reformulaion. Doe every k-regular biparie graph have a perfec maching? Ex. Boolean hypercube. -regular biparie graph ʹ ʹ ʹ ʹ 9

6 Edge-dijoin pah SECTION.. NETWORK FLOW II biparie maching dijoin pah exenion o max flow urvey deign airline cheduling image egmenaion projec elecion baeball eliminaion Def. Two pah are edge-dijoin if hey have no edge in common. Edge-dijoin pah problem. Given a digraph G = (V, E) and wo node and, find he max number of edge-dijoin pah. Ex. Communicaion nework. digraph G Edge-dijoin pah Edge-dijoin pah Def. Two pah are edge-dijoin if hey have no edge in common. Edge-dijoin pah problem. Given a digraph G = (V, E) and wo node and, find he max number of edge-dijoin pah. Ex. Communicaion nework. Max-flow formulaion. Aign uni capaciy o every edge. Theorem. Max number of edge-dijoin pah = value of max flow. Pf. Suppoe here are k edge-dijoin pah P,, P k. Se f (e) = if e paricipae in ome pah P j ; ele e f (e) = 0. Since pah are edge-dijoin, f i a flow of value k. digraph G edge-dijoin pah

7 Edge-dijoin pah Nework conneciviy Max-flow formulaion. Aign uni capaciy o every edge. Theorem. Max number of edge-dijoin pah = value of max flow. Pf. Suppoe max flow value i k. Inegraliy heorem here exi 0 flow f of value k. Conider edge (, u) wih f(, u) =. - by flow conervaion, here exi an edge (u, v) wih f(u, v) = - coninue unil reach, alway chooing a new edge Def. A e of edge F E diconnec from if every pah ue a lea one edge in F. Nework conneciviy. Given a digraph G = (V, E) and wo node and, find min number of edge whoe removal diconnec from. Produce k (no necearily imple) edge-dijoin pah. can eliminae cycle o ge imple pah in O(mn) ime if deired (flow decompoiion) 8 Menger heorem Menger heorem Theorem. [Menger 9] The max number of edge-dijoin pah equal he min number of edge whoe removal diconnec from. Pf. 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 k. Theorem. [Menger 9] The max number of edge-dijoin pah equal he min number of edge whoe removal diconnec from. Pf. Suppoe max number of edge-dijoin pah i k. Then value of max flow = k. Max-flow min-cu heorem here exi a cu (A, B) of capaciy k. Le F be e of edge going from A o B. F = k and diconnec from. A 9 0

8 Nework flow II: quiz Edge-dijoin pah in undireced graph How o find he max number of edge-dijoin pah in an undireced graph? A. Solve he edge-dijoin pah problem in a digraph (by replacing each undireced edge wih wo aniparallel edge). Def. Two pah are edge-dijoin if hey have no edge in common. Edge-dijoin pah problem in undireced graph. Given a graph G = (V, E) and wo node and, find he max number of edge-dijoin pah. B. Solve a max flow problem in an undireced graph. C. Boh A and B. D. Neiher A nor B. digraph G Edge-dijoin pah in undireced graph Edge-dijoin pah in undireced graph Def. Two pah are edge-dijoin if hey have no edge in common. Edge-dijoin pah problem in undireced graph. Given a graph G = (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. Edge-dijoin pah problem in undireced graph. Given a graph G = (V, E) and wo node and, find he max number of edge-dijoin pah. digraph G ( edge-dijoin pah) digraph G ( edge-dijoin pah)

9 Edge-dijoin pah in undireced graph Edge-dijoin pah in undireced graph Max-flow formulaion. Replace each edge wih wo aniparallel edge and aign uni capaciy o every edge. Obervaion. Two pah P and P may be edge-dijoin in he digraph bu no edge-dijoin in he undireced graph. if P ue edge (u, v) and P ue i aniparallel edge (v, u) Max-flow formulaion. Replace each edge wih wo aniparallel edge and aign uni capaciy o every edge. Lemma. In any flow nework, here exi a maximum flow f in which for each pair of aniparallel edge e and eʹ : eiher f (e) = 0 or f (eʹ) = 0 or boh. Moreover, inegraliy heorem ill hold. Pf. [ by inducion on number of uch pair ] Suppoe f (e) > 0 and f (eʹ) > 0 for a pair of aniparallel edge e and eʹ. Se f (e) = f (e) δ and f (eʹ) = f (eʹ) δ, where δ = min { f (e), f (eʹ) }. f i ill a flow of he ame value bu ha one fewer uch pair. Edge-dijoin pah in undireced graph More Menger heorem Max-flow formulaion. Replace each edge wih wo aniparallel edge and aign uni capaciy o every edge. Lemma. In any flow nework, here exi a maximum flow f in which for each pair of aniparallel edge e and eʹ : eiher f (e) = 0 or f (eʹ) = 0 or boh. Moreover, inegraliy heorem ill hold. Theorem. Max number of edge-dijoin pah = value of max flow. Pf. Similar o proof in digraph; ue lemma. Theorem. Given an undireced graph and wo node and, he max number of edge-dijoin pah equal he min number of edge whoe removal diconnec and. Theorem. Given an undireced graph and wo nonadjacen node and, he max number of inernally node-dijoin pah equal he min number of inernal node whoe removal diconnec and. Theorem. Given a direced graph wih wo nonadjacen node and, he max number of inernally node-dijoin pah equal he min number of inernal node whoe removal diconnec from. 8

10 Nework flow II: quiz SECTION.. NETWORK FLOW II biparie maching dijoin pah exenion o max flow urvey deign airline cheduling image egmenaion projec elecion baeball eliminaion Which exenion o max flow can be eaily modeled? A. Muliple ource and muliple ink. B. Undireced graph. C. Lower bound on edge flow. D. All of he above. 0 Muliple ource and ink Muliple ource and ink: max-flow formulaion Def. Given a digraph G = (V, E) wih edge capaciie c(e) 0 and muliple ource node and muliple ink node, find max flow ha can be en from he ource node o he ink node. Add a new ource node and ink node. For each original ource node i add edge (, i) wih capaciy. For each original ink node j, add edge ( j, ) wih capaciy. Claim. correpondence beween flow in G and Gʹ. flow nework G 9 flow nework G

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Given a digraph G = (V, E) wih edge capaciie c(e) 0 and node demand d(v), a circulaion i a funcion f(e) ha aifie: For each e E: 0 f (e) c(e) (capaciy) For each v V: f(e) f(e) = d(v) (flow conervaion) e v e v 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. G ha circulaion iff Gʹ ha max flow of value D = d(v) = v : d(v)>0 v : d(v)<0 d(v) flow nework G 8 (upply node) 8 flow nework G 8 upply aurae all edge leaving and enering flow capaciy / / / 0 / / / / / 0 (demand node) 0 (ranhipmen node) demand Circulaion wih upplie and demand Circulaion wih upplie, 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-valued. Pf. Follow from max-flow formulaion + inegraliy heorem for max flow. Theorem. Given (V, E, c, d), here doe no exi a circulaion iff here exi a node pariion (A, B) uch ha Σ v B d(v) > cap(a, B). Pf kech. Look a min cu in Gʹ. demand by node in B exceed upply of node in B plu max capaciy of edge going from A o B Def. Given a digraph G = (V, E) wih edge capaciie c(e) 0, lower bound (e) 0, and node demand d(v), a circulaion f(e) i a funcion ha aifie: For each e E : (e) f (e) c(e) (capaciy) For each v V : f(e) f(e) = d(v) (flow conervaion) e v e v Circulaion problem wih lower bound. Given (V, E,, c, d), doe here exi a feaible circulaion?

12 Circulaion wih upplie, demand, and lower bound Max-flow formulaion. Model lower bound a circulaion wih demand. Send (e) uni of flow along edge e. Updae demand of boh endpoin. lower bound upper bound capaciy v [, 9] w v w d(v) d(w) d(v) + d(w) flow nework G flow nework G Theorem. There exi a circulaion in G iff here exi a circulaion in Gʹ. Moreover, if all demand, capaciie, and lower bound in G are ineger, hen here exi a circulaion in G ha i ineger-valued. Pf kech. f (e) i a circulaion in G iff f ʹ(e) = f (e) (e) i a circulaion in Gʹ. SECTION.8. NETWORK FLOW II biparie maching dijoin pah exenion o max flow urvey deign airline cheduling image egmenaion projec elecion baeball eliminaion Survey deign Deign urvey aking n conumer abou n produc. Can urvey conumer i abou produc j only if hey own i. Ak conumer i beween c i and c iʹ queion. Ak beween p j and p jʹ conumer abou produc j. Goal. Deign a urvey ha mee hee pec, if poible. Biparie perfec maching. Special cae when c i = c iʹ = p j = p jʹ =. one urvey queion per produc Survey deign Max-flow formulaion. Model a a circulaion problem wih lower bound. Add edge (i, j) if conumer j own produc i. Add edge from o conumer j. Add edge from produc i o. Add edge from o. All demand = 0. Ineger circulaion feaible urvey deign. [0, ] all upplie and demand are 0 [0, ] ʹ [c, c ʹ] ʹ [p, p ʹ] ʹ ʹ 9 conumer produc 0

13 Airline cheduling Airline cheduling.. N ETWORK F LOW II Complex compuaional problem faced by airline carrier. Mu produce chedule ha are efficien in erm of equipmen uage, biparie maching crew allocaion, and cuomer aifacion. dijoin pah One of large conumer of high-powered algorihmic echnique. exenion o max flow urvey deign Toy problem. Manage fligh crew by reuing hem over muliple fligh. airline cheduling Inpu: e of k fligh for a given day. image egmenaion Fligh i leave origin oi a ime i and arrive a deinaion di a ime fi. projec elecion SECTION.9 even in preence of unpredicable even, uch a weaher and breakdown Minimize number of fligh crew. baeball eliminaion Airline cheduling Airline cheduling: running ime Circulaion formulaion. [o ee if c crew uffice] Theorem. The airline cheduling problem can be olved in O(k log k) ime. For each fligh i, include wo node ui and vi. Pf. Add ource wih demand c, and edge (, ui) wih capaciy. k = number of fligh. Add ink wih demand c, and edge (vi, ) wih capaciy. c = number of crew (unknown). For each i, add edge (ui, vi) wih lower bound and capaciy. O(k) node, O(k) edge. if fligh j reachable from i, add edge (vi, uj) wih capaciy. A mo k crew needed. crew can begin day wih any fligh u a mo k augmenaion per circulaion problem. Overall ime = O(k log k). [0, ] u u ue c crew [, ] binary earch for opimal value c* Value of he flow i beween 0 and k. v [0, ] c olve lg k circulaion problem. crew can end day wih any fligh c v Remark. Can olve in O(k) ime by formulaing a minimum flow problem. v [0, ] u fligh i performed v ame crew can do fligh and

14 <laexi ha_bae="yqarqjdbdiekqhdxxfrk+jrru=">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</laexi> <laexi ha_bae="yqarqjdbdiekqhdxxfrk+jrru=">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</laexi> <laexi ha_bae="yqarqjdbdiekqhdxxfrk+jrru=">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</laexi> <laexi ha_bae="yqarqjdbdiekqhdxxfrk+jrru=">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</laexi> Airline cheduling: pomorem Remark. We olved a oy problem. Real-world problem model counle oher facor: Union regulaion: e.g., fligh crew can only fly cerain number of hour in given inerval. Need opimal chedule over planning horizon, no ju one day. Deadheading ha a co. Fligh don alway leave or arrive on chedule. Simulaneouly opimize boh fligh chedule and fare rucure. Meage. Our oluion i a generally ueful echnique for efficien reue of limied reource bu rivialize real airline cheduling problem. Flow echnique ueful for olving airline cheduling problem (and are widely ued in pracice). Running an airline efficienly i a very difficul problem. SECTION.0. NETWORK FLOW II biparie maching dijoin pah exenion o max flow urvey deign airline cheduling image egmenaion projec elecion baeball eliminaion Image egmenaion Image egmenaion Image egmenaion. Divide image ino coheren region. Cenral problem in image proceing. Ex. Separae human and robo from background cene. 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: foreground background i A a i + j B b j (i,j) E A {i,j} = p ij 8

15 Image egmenaion Image egmenaion Formulae a min-cu problem. Formulae a min-cu problem G ʹ = (Vʹ, Eʹ). Maximizaion. Include node for each pixel. No ource or ink. Ue wo aniparallel edge inead of Undireced graph. undireced edge. Turn ino minimizaion problem. ai + Maximizing bj i A pij wo aniparallel edge in G Add ource o correpond o foreground. pij Add ink o correpond o background. pij pij j B (i,j) E A {i,j} = <laexi ha_bae="yqarqjdbdiekqhdxxfrk+jrru=">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</laexi> i equivalen o minimizing ai + i V a conan edge in G aj bj ai j V bj + i A j B i pij (i,j) E pij j bi A {i,j} = <laexi ha_bae="auxnyvhvalgguwikq8vr+gri=">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</laexi> aj + or alernaively bi + j B i A pij G (i,j) E A {i,j} = <laexi ha_bae="lol9igpggvvnrymdybyzmo=">aaacqhicbvfdb9mwfhuypkbjbeelmiqhomvckeqqjaqmh8vgkugvueslbkwv/knr+ck/sbdlzj9e98m+zgopriui0gde/+gfbjqpnzx99ry/u89jwhpncuyiyxal0dhwkm8kawylumcznefgye0vufulgumilqmrecou+l/v/uliqfkcfhk8ntdepyao0bnsxa+ayglbudf9drfl9njf+xeldb0yz+auqnw8mpqbkpqpkwqg0anra8jhioac9uaf9qtsmocifbavnfkn0jdukvcbhkkjqhbjxywtxgldoxfw+hxeluq80uqrqjarhwdmmm+nx9pby/butflvjlplmxduyb8nzyfwqyqusoeadpmpqsxq+mzirbutsa8an8hcffum8/zdqunebfo0lupbc9numfkd+mmaymnzc8phe0jl+9hwer/zcji/jkjpynjetlxmijjwmfobfbs/c+eoniq/uqwwnxkbui9rjm0h</laexi> 9 0 Image egmenaion Grabcu image egmenaion Conider min cu (A, B) in G ʹ. Grabcu. [ Roher Kolmogorov Blake 00 ] A = foreground. cap(a, B) = aj + j B bi + i A pij if i and j on differen ide, pij couned exacly once (i,j) E i A, j B <laexi ha_bae="gcrfvkznab/z8lqbnechacxuvom=">aaacvhicbvhawfjw9ryanr9hizlwlhkqdlxgplo+0ygchsuijzhlmajjbjlgglgvxtbgybdbumhc+xk0lwy0ng+e/+dho8dppnj/8xkvnbvjsmywxotmqugbvcdwkvcw0upkkvegxxfxqqwp0yljcw9unzggbwosvq/gcgjymypokdqcz+0tmvimrwiruc0bprmgfq8ku9qsbpe8vu9vnfaixlirirzdfaxy/qrrnxwcfucbqhycgpyrukphcrpd8ixkfjcb+ehriqzshw8ynjwslswcwrmnawkgduw8eronfibugvthuvkkjqznbxztaazxlujldtyiksmowmnwekprmyp9p09loduokqk0qfhbafykdmngez8iwdpameauvdxvfnmbwqjscrrplus8vznuegk+yxqxxbzzvfb+cedmbhjw+d8nunlvknxllhiqkn8gx+uoyiqw8dhvxp/mxvmxdfa+leuxwzl/cybl0fu=</laexi> GrabCu Ineracive Foreground Exracion uing Ieraed Graph Cu Preciely he quaniy we wan o minimize. Caren Roher aj i A pij Vladimir Kolmogorov Microof Reearch Cambridge, UK Andrew Blake Figure : Three example of GrabCu. The uer drag a recangle looely around an objec. The objec i hen exraced auomaically. j Abrac bi G The problem of efficien, ineracive foreground/background egmenaion in ill image i of grea pracical imporance in image ediing. Claical image egmenaion ool ue eiher exure (colour) informaion, e.g. Magic Wand, or edge (conra) informaion, e.g. Inelligen Scior. Recenly, an approach baed on opimizaion by graph-cu ha been developed which uccefully combine boh ype of informaion. In hi paper we exend he graph-cu approach in hree repec. Fir, we have developed a more powerful, ieraive verion of he opimiaion. Secondly, he power of he ieraive algorihm i ued o implify ubanially he uer ineracion needed for a given qualiy of reul. Thirdly, a robu algorihm for border maing ha been developed o eimae imulaneouly he alpha-mae around an objec boundary and he colour of foreground pixel. We how ha for moderaely difficul free of colour bleeding from he ource background. In general, degree of ineracive effor range from ediing individual pixel, a he labour-inenive exreme, o merely ouching foreground and/or background in a few locaion.. Previou approache o ineracive maing In he following we decribe briefly and compare everal ae of he ar ineracive ool for egmenaion: Magic Wand, Inelligen Scior, Graph Cu and Level Se and for maing: Baye Maing and Knockou. Fig. how heir reul on a maing ak, ogeher wih degree of uer ineracion required o achieve hoe reul. Magic Wand ar wih a uer-pecified poin or region o compue a region of conneced pixel uch ha all he eleced pixel fall wihin ome adjuable olerance of he colour aiic of he pecified region. While he uer inerface i raighforward, finding

16 Projec elecion (maximum weigh cloure problem) SECTION.. NETWORK FLOW II biparie maching dijoin pah exenion o max flow urvey deign airline cheduling image egmenaion projec elecion baeball eliminaion Projec wih prerequiie. Se of poible projec P : projec v ha aociaed revenue pv. Se of prerequiie E : (v, w) E mean w i a prerequiie for v. A ube of projec A P i feaible if he prerequiie of every projec in A alo belong o A. Projec elecion problem. Given a e of projec P and prerequiie E, chooe a feaible ube of projec o maximize revenue. MANAGEMENT SCIENCE Vol., No., July, 9 Prined in U.SA. MAXIMAL CLOSURE OF A GRAPH AND APPLICATIONS TO COMBINATORIAL PROBLEMS* JEAN-CLAUDE PICARD Ecole Polyechnique, Monreal can be poiive or negaive Thi paper generalize he elecion problem dicued by J. M. Rhy [], J. D. Murchland [9], M. L. Balinki [] and P. Hanen []. Given a direced graph G, a cloure of G i defined a a ube of node uch ha if a node belong o he cloure all i ucceor alo belong o he e. If a real number i aociaed o each node of G a maximal cloure i defined a a cloure of maximal value. Projec elecion: prerequiie graph Projec elecion: min-cu formulaion Prerequiie graph. Add edge (v, w) if w i a prerequiie for v. Min-cu formulaion. Aign a capaciy of o each prerequiie edge. Add edge (, v) wih capaciy p v if p v > 0. Add edge (v, ) wih capaciy p v if p v < 0. For noaional convenience, define p = p = 0. w w u w p u p w p y y z p z v x v x p v p x { v, w, x } i feaible { v, x } i infeaible v x

17 Projec elecion: min-cu formulaion Open-pi mining Claim. (A, B) i min cu iff A { } i an opimal e of projec. Open-pi mining. [udied ince early 90] Infinie capaciy edge enure A { } i feaible. Max revenue becaue: cap(a, B) = Block of earh are exraced from urface o rerieve ore. pv + v B : pv >0 <laexi ha_bae="jovugo0bjapbuezlcgplwao=">aaacjicbvfntxxehuwl0kfhlioiskfo9qekh8knbluvgpaarpiedweinijarfyc/ozeixepccsgaky09vvjzzlr0ley/mkea+vxraen88/bd+w+9uaywrccbyldvllduuuoajcm8nbzliq8sg9+vplkvonc/bzgzhgb/sciifj08llrtcdi+/wmkoxpvxphzwwznrzauuilhhoikx8njpnceyrzqfe/wxl0uph0onkvqusos9ccqiflbfnsxxmy8zkwrosahuhpdkdq0krklkrtomhh0hbxw9wkhmgbprwecxh0+egcmkpgpp9pfhyzllzlnplxunarfyq8rnekdjqiuhsewjwytqoflemvloylrufqgexvvqgrjmlgvyxdkup9ucy9plwbtgfdyrbdk+dyngkm9hscf+fys/9btvkucgdbrojjycygalx/yf+w+aaewvhqfagjcxmrzuwel/usnefg==</laexi> = a conan pv Can remove block v unil boh block w and x are removed. pv v : pv >0 <laexi ha_bae="h0rdhkrdwfuvwoeplp+ycwevvu=">aaacbicbvdrshbfjwqqnepdhwpyasi0dzderreuyrqrvghwxyzicznbdpmgwzlfpffpffbugteyfyqnnhvvzikzjs/mpdw/iw+hfpeaxaxx8pry/psprkrbwpszc0qqbflkhrezwzeiu8im9/vfrvgivqfdkwzcb/rcigfj0dfrtmid8ndokqvzmjzjcrgefadnwtrgiblfgxeh/pqkhlmokyqfoxxe8b8eunm0zqonxmyseweocahulw9wm+oxbduigmfumepil9gz0hnet9ovvcd8dmbhazrbdxb0fbenseyccaerndcq8n9alfhqb+qojxhbnmwvuapvedcqbgwpiqncgonecmledrfkipzu/oumz8pljlrtpaodyrxdkeolsdoyzew8udzubwku9onb9m8l9lx9o9yahbzyfnznxsbypxgt+y8ej98by9elnjwnpfpp+crqimma==</laexi> Each block v ha ne value pv = value of ore proceing co. ( pv ) v A : pv <0 v A minimizing hi i equivalen o maximizing revenue w u pu w pw py A pv y z v pz x v px x 8 Baeball eliminaion. N ETWORK F LOW II biparie maching dijoin pah exenion o max flow urvey deign airline cheduling image egmenaion SECTION. projec elecion baeball eliminaion 0

18 Baeball eliminaion problem Baeball eliminaion problem Q. Which eam have a chance of finihing he eaon wih he mo win? i eam win loe o play ATL PHI NYM MON 0 Alana 8 8 Philly New York Monreal 8 0 Monreal i mahemaically eliminaed. Monreal finihe wih 80 win. Alana already ha 8 win. Remark. Thi i he only reaon por wrier appear o be aware of condiion are ufficien bu no neceary! Q. Which eam have a chance of finihing he eaon wih he mo win? i eam win loe o play ATL PHI NYM MON 0 Alana 8 8 Philly New York Monreal 8 0 Philadelphia i mahemaically eliminaed. Philadelphia finihe wih 8 win. Eiher New York or Alana will finih wih 8 win. Obervaion. Anwer depend no only on how many game already won and lef o play, bu on whom hey re again. Baeball eliminaion problem Baeball eliminaion problem: max-flow formulaion Curren anding. Se of eam S. Diinguihed eam z S. Team x ha won w x game already. Team x and y play each oher r xy addiional ime. Can eam finih wih mo win? Aume eam win all remaining game w + r win. Divvy remaining game o ha all eam have w + r win. Baeball eliminaion problem. Given he curren anding, i here any oucome of he remaining game in which eam z finihe wih he mo (or ied for he mo) win? game lef beween and eam can ill win hi many more game SIAM REVIEW Vol. 8, No., July, 9 0 w + r w g POSSIBLE WINNERS IN PARTIALLY COMPLETED TOURNAMENTS* BENJAMIN L. SCHWARTZ. Inroducion. In hi paper, we hall inveigae cerain queion in ournamen cheduling. For definiene, we hall ue he erminology of baeball. We hall be concerned wih he caegorizaion of eam ino hree clae during he cloing day of he eaon. A eam may be definiely eliminaed from pennan poibiliy; i may be in conenion, or i may have clinched he championhip. I will be our convenion ha a eam ha can poibly ie for he pennan i conidered ill in conenion. In hi paper neceary and ufficien condiion are developed o claify any eam properly ino he appropriae caegory. We game node (each pair of eam oher han ) eam node (each eam oher han )

19 Baeball eliminaion problem: max-flow formulaion Baeball eliminaion: explanaion for por wrier Theorem. Team no eliminaed iff max flow aurae all edge leaving. Pf. 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. Q. Which eam have a chance of finihing he eaon wih he mo win? i eam win loe o play NYY BAL BOS TOR DET 0 New York Balimore 8 0 Boon game lef beween and 0 0 eam can ill win hi many more game Torono 0 0 Deroi g w + r w game node (each pair of eam oher han ) eam node (each eam oher han ) AL Ea (Augu 0, 99) Deroi i mahemaically eliminaed. Deroi finihe wih win. Win for R = { NYY, BAL, BOS, TOR } = 8. Remaining game among { NYY, BAL, BOS, TOR } = =. Average eam in R win 0/ =. game. Baeball eliminaion: explanaion for por wrier Baeball eliminaion: explanaion for por wrier Cerificae of eliminaion. # remaining game!# " win #! $ " $ $ # T S, w(t ) := w i i T, g(t ) := g x y {x,y} T, Theorem. [Hoffman Rivlin 9] Team z i eliminaed iff here exi a ube T* uch ha w(t*)+ g(t*) w z + g z < T* Pf. w(t*)+ g(t*) Suppoe here exi T* S uch ha w z + g z <. T* Then, he eam in T* win a lea (w(t*) + g(t*)) / T* game on average. Thi exceed he maximum number ha eam z can win. Pf. Ue max-flow formulaion, and conider min cu (A, B). Le T* = eam node on ource ide A of min cu. Oberve ha game node x y A iff boh x T* and y T*. - infinie capaciy edge enure if x y A, hen boh x A and y A - if x A and y A bu x y A, hen adding x y o A decreae he capaciy of he cu by g xy y g xy x y x w z + r z w x 8

20 Baeball eliminaion: explanaion for por wrier Pf. Ue max-flow formulaion, and conider min cu (A, B). Le T* = eam node on ource ide A of min cu. Oberve ha game node x y A iff boh x T* and y T*. Since eam z i eliminaed, by max-flow min-cu heorem, g(s {z}) > cap(a, B) capaciy of game edge leaving capaciy of eam edge enering!## "### $ = 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 ) Rearranging erm: w z + g z < w(t*)+ g(t*) T* 9