3/3/2015. Chapter 7. Network Flow. Maximum Flow and Minimum Cut. Minimum Cut Problem

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1 // Chaper Nework Flow Maximum Flow and Minimum Cu Max flow and min cu. Two very rich algorihmic problem. Cornerone problem in combinaorial opimizaion. Beauiful mahemaical dualiy. Nonrivial applicaion / reducion. Daa mining. Open-pi mining. Projec elecion. Airline cheduling. Biparie maching. Baeball eliminaion. Image egmenaion. Nework conneciviy. Nework reliabiliy. Diribued compuing. Securiy of aiical daa. Nework inruion deecion. Muli-camera cene reconrucion. Many many more Minimum Cu Problem Flow nework. Abracion for maerial flowing hrough he edge. G = (V, E), direced graph, no parallel edge. Two diinguihed node: = ource, = ink. c(e) = capaciy of edge e. ource ink capaciy

2 // Cu Def. An - cu i a pariion (A, B) of V wih A and B. Def. The capaciy of a cu (A, B) i: cap( A, B) c(e) e ou of A A Capaciy = + + = Cu Def. An - cu i a pariion (A, B) of V wih A and B. Def. The capaciy of a cu (A, B) i: cap( A, B) c(e) e ou of A A Capaciy = = Minimum Cu Problem Min - cu problem. Find an - cu of minimum capaciy. A Capaciy = + + =

3 // Flow Def. An - flow i a funcion f 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 flow f i: v( f ) f (e). e ou of capaciy flow Value = Flow Def. An - flow i a funcion f 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 flow f i: v( f ) f (e). e ou of capaciy flow Value = Maximum Flow Problem Max flow problem. Find - flow of maximum value. capaciy flow Value =

4 // Flow and Cu Flow value lemma. Le f be any flow, and le (A, B) be any - cu. Then, he ne flow en acro he cu i equal o he amoun leaving. f (e) f (e) v( f ) e ou of A e in o A A Value = Flow and Cu Flow value lemma. Le f be any flow, and le (A, B) be any - cu. Then, he ne flow en acro he cu i equal o he amoun leaving. f (e) f (e) v( f ) e ou of A e in o A A Value = = Flow and Cu Flow value lemma. Le f be any flow, and le (A, B) be any - cu. Then, he ne flow en acro he cu i equal o he amoun leaving. f (e) f (e) v( f ) e ou of A e in o A A Value = =

5 // Flow and Cu Flow value lemma. Le f be any flow, and le (A, B) be any - cu. Then f (e) f (e) v( f ). e ou of A e in o A Pf. by flow conervaion, all erm excep v = are v( f ) va e ou of f ( e) e f ( e) ou of v e in o v f ( e) f ( e) e ou of A e in o A f ( e). Flow and Cu Weak dualiy. Le f be any flow, and le (A, B) be any - cu. Then he value of he flow i a mo he capaciy of he cu. Cu capaciy = Flow value A Capaciy = Flow and Cu Weak dualiy. Le f be any flow. Then, for any - cu (A, B) we have v(f) cap(a, B). Pf. v( f ) f (e) f (e) e ou of A e in o A f (e) e ou of A c(e) e ou of A cap(a, B) A B

6 // Cerificae of Opimaliy Corollary. Le f be any flow, and le (A, B) be any cu. If v(f) = cap(a, B), hen f i a max flow and (A, B) i a min cu. Value of flow = Cu capaciy = Flow value A Toward a Max Flow Algorihm Greedy algorihm. Sar wih f(e) = for all edge e E. Find an - pah P where each edge ha f(e) < c(e). Augmen flow along pah P. Repea unil you ge uck. Flow value = Toward a Max Flow Algorihm Greedy algorihm. Sar wih f(e) = for all edge e E. Find an - pah P where each edge ha f(e) < c(e). Augmen flow along pah P. Repea unil you ge uck. X X X Flow value =

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

8 // Ford-Fulkeron Algorihm G: flow capaciy Flow value = Ford-Fulkeron Algorihm G: flow capaciy X X X Flow value = G f : reidual capaciy Ford-Fulkeron Algorihm G: X X X X Flow value = G f :

9 // Ford-Fulkeron Algorihm G: X X X X Flow value = G f : Ford-Fulkeron Algorihm G: X X X X Flow value = G f : Ford-Fulkeron Algorihm X G: X X X X Flow value = G f :

10 // Ford-Fulkeron Algorihm G: Flow value = G f : Ford-Fulkeron Algorihm G: Cu capaciy = Flow value = G f : Augmening Pah Algorihm Augmen(f, c, P) { b boleneck(p) foreach e P { if (e E) f(e) f(e) + b ele f(e R ) f(e R ) - b } reurn f } forward edge revere edge Ford-Fulkeron(G,,, c) { foreach e E f(e) G f reidual graph while (here exi augmening pah P) { f Augmen(f, c, P) updae G f } reurn f }

11 // Max-Flow Min-Cu Theorem Augmening pah heorem. Flow f i a max flow iff here are no augmening pah. Max-flow min-cu heorem. [Elia-Feinein-Shannon, Ford-Fulkeron ] The value of he max flow i equal o he value of he min cu. Pf. We prove boh imulaneouly by howing he following aemen are logically equivalen: (i) There exi a cu (A, B) uch ha v(f) = cap(a, B). (ii) Flow f i a max flow. (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 flow. If here exi an augmening pah, hen we can improve f by ending flow along pah. Proof of Max-Flow Min-Cu Theorem (iii) (i) Le f be a flow wih no augmening pah in G f. Le A be e of verice reachable from in reidual graph. By definiion of A, A. By definiion of f, A. v( f ) e ou of A f ( e) e ou of A e in o A c( e) cap( A, B) f ( e) A B original nework Running Time Aumpion. All capaciie are ineger beween and C. Invarian. Every flow value f(e) and every reidual capaciy c f (e) remain an ineger hroughou he algorihm. Theorem. The algorihm erminae in a mo v(f*) nc ieraion. Pf. Each augmenaion increae value by a lea. Corollary. If C =, Ford-Fulkeron run in O(mn) ime, where n = V and m = E. Inegraliy heorem. If all capaciie are ineger, hen here exi a max flow f for which every flow value f(e) i an ineger. Pf. Since he algorihm erminae, he heorem follow from invarian.

12 //. Chooing Good Augmening Pah Ford-Fulkeron: Exponenial Number of Augmenaion Q. I generic Ford-Fulkeron algorihm polynomial in inpu ize? m, n, and log C A. No. If max capaciy i C, hen algorihm can ake C ieraion. X X X C C C C X X X C C X C C X X Chooing Good Augmening Pah Ue care when elecing augmening pah. Some choice lead o exponenial algorihm. Clever choice lead o polynomial algorihm. If capaciie are irraional, algorihm no guaraneed o erminae! Goal: chooe augmening pah o ha: Can find augmening pah efficienly. Few ieraion. Chooe augmening pah wih: [Edmond-Karp, Diniz ] Max boleneck capaciy. Sufficienly large boleneck capaciy. Fewe number of edge.

13 // Capaciy Scaling Inuiion. Chooing pah wih highe boleneck capaciy increae flow by max poible amoun. Don' worry abou finding exac highe boleneck pah. Mainain caling parameer. Le G f () be he ubgraph of he reidual graph coniing of only arc wih capaciy a lea. G f G f () Capaciy Scaling Scaling-Max-Flow(G,,, c) { foreach e E f(e) malle power of greaer han or equal o C G f reidual graph while ( ) { G f () -reidual graph while (here exi augmening pah P in G f ()) { f augmen(f, c, P) updae G f () } / } reurn f } Capaciy Scaling: Correcne Aumpion. All edge capaciie are ineger beween and C. Inegraliy invarian. All flow and reidual capaciy value are inegral. Correcne. If he algorihm erminae, hen f i a max flow. Pf. By inegraliy invarian, when = G f () = G f. Upon erminaion of = phae, here are no augmening pah.

14 // Capaciy Scaling: Running Time Lemma. The ouer while loop repea + log C ime. Pf. Iniially C <C. decreae by a facor of each ieraion. Lemma. Le f be he flow a he end of a -caling phae. Then he value of he maximum flow i a mo v(f) + m. proof on nex lide Lemma. There are a mo m augmenaion per caling phae. Pf. Le k be he number of augmenaion in he curren caling phae. Le f and f be he flow before and afer he curren caling phae, repecively. Lemma v(f*) v(f) + m (). Each augmenaion in a -phae increae v(f) by a lea. If k augmenion are done, v(f) + k v(f) v(f*) v(f) + m (). So k m Theorem. The caling max-flow algorihm find a max flow in O(m log C) augmenaion. I can be implemened o run in O(m log C) ime. Capaciy Scaling: Running Time Lemma. Le f be he flow a he end of a -caling phae. Then value of he maximum flow i a mo v(f) + m. Pf. (almo idenical o proof of max-flow min-cu heorem) We how ha a he end of a -phae, here exi a cu (A, B) uch ha cap(a, B) v(f) + m. Chooe A o be he e of node reachable from in G f (). By definiion of A, A. By definiion of f, A. v( f ) f (e) f (e) e ou of A e in o A (c(e) ) e ou of A e in o A c(e) e ou of A e ou of A e in o A cap(a, B) - m A B original nework. Biparie Maching

15 // 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 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 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 maching -', -', -' -' ' L ' R

16 // Biparie 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' ' ' ' ' L ' R Biparie Maching: Proof of Correcne Theorem. Max cardinaliy maching in G = value of max flow in G'. Pf. Given 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. ' ' ' ' ' ' ' ' G ' ' G' Biparie Maching: Proof of Correcne Theorem. Max cardinaliy maching in G = value of max flow in G'. Pf. Le f be a max flow in G' of value k. Inegraliy heorem k i inegral and can aume f i -. 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

17 // Perfec Maching Def. A maching M E i perfec 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. Wha oher condiion are neceary? Wha condiion are ufficien? Perfec Maching 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). ' ' ' No perfec maching: S = {,, } N(S) = { ', ' }. ' L ' R Marriage Theorem Marriage Theorem. [Frobeniu, Hall ] 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 previou obervaion. ' ' ' No perfec maching: S = {,, } N(S) = { ', ' }. ' L ' R

18 // Proof of Marriage Theorem Pf. Suppoe G doe no have 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. max maching G' -', -', -' -' ' A ' L A = {,, } L B = {, } ' R A = {', '} N(L A ) = {', '} ' ' Biparie Maching: Running Time Which max flow algorihm o ue for biparie maching? Generic augmening pah: O(m val(f*) ) = O(mn). Capaciy caling: O(m log C ) = O(m ). Shore augmening pah: O(m n / ). Non-biparie maching. Srucure of non-biparie graph i more complicaed, bu well-underood. [Tue-Berge, Edmond-Galai] Bloom algorihm: O(n ). [Edmond ] Be known: O(m n / ). [Micali-Vazirani ]. Dijoin Pah

19 // Edge Dijoin Pah Dijoin pah problem. Given 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 have no edge in common. Ex: communicaion nework. Edge Dijoin Pah Dijoin pah problem. Given 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 have no edge in common. Ex: communicaion nework. Edge Dijoin Pah Max flow formulaion: aign uni capaciy o every edge. Theorem. Max number edge-dijoin - pah equal max flow value. 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) =. Since pah are edge-dijoin, f i a flow of value k.

20 // Edge Dijoin Pah Max flow formulaion: aign uni capaciy o every edge. Theorem. Max number edge-dijoin - pah equal max flow value. Pf. Suppoe max flow value i k. Inegraliy heorem here exi - 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 (no necearily imple) edge-dijoin pah. can eliminae cycle o ge imple pah if deired Nework Conneciviy Nework conneciviy. Given a digraph G = (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. Edge Dijoin Pah and Nework Conneciviy Theorem. [Menger ] The max number of edge-dijoin - pah i equal o he min number of edge whoe removal diconnec from. Pf. Suppoe F = k i minimal and he removal of F E diconnec from. Every - pah ue a lea one edge in F. Hence, he number of edge-dijoin pah i a mo k.

21 // Dijoin Pah and Nework Conneciviy Theorem. [Menger ] The max number of edge-dijoin - pah i equal o he min number of edge whoe removal diconnec from. Pf. 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. Le F be e of edge going from A o B. F = k and diconnec from. A. Image Segmenaion Image Segmenaion Image egmenaion. Cenral problem in image proceing. Divide image ino coheren region. Ex: Three people anding in fron of complex background cene. Idenify each peron a a coheren objec.

22 // 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 i he credi if pixel i in foreground. b i i he credi if pixel i in background. p ij i eparaion penaly for labeling one of i and j a foreground, and he oher a background. Example. For a black-whie phoo, a whie pixel ha value and a black pixel ha value. We may define a i = v i and b i = v i, where v i i he value of pixel i. We alo le p ij = c/(+ a i -a j ), where c i a conan. 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. 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 i he credi if pixel i in foreground. b i i he credi if pixel i in background. p ij 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 b j p ij foreground background i A j B (i, j) E A{i, j} Image Segmenaion Formulae a min cu problem. Maximizaion. No ource or ink. Undireced graph. Turn ino minimizaion problem. Maximizing i equivalen o minimizing a i b j p ij i A j B (i, j) E A{i, j} i V a i j V b j a conan a i b j p ij i A j B (i,j) E A{i,j} or o alernaively minimizing a j b i p ij j B i A (i, j) E A{i, j}

23 // 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 G' a j i p ij j b i Image Segmenaion Conider min cu (A, B) in G'. A = foreground. cap(a, B) a j b i p ij j B i A (i, j) E i A, j B if i and j on differen ide, p ij couned exacly once Preciely he quaniy we wan o minimize. G' a j i p ij j A b i. Projec Selecion

24 // Projec Selecion can be poiive or negaive Projec wih prerequiie. Se P of poible projec. Projec v ha aociaed revenue p v. ome projec generae money: creae ineracive e-commerce inerface, redeign web page oher co money: upgrade compuer, ge ie licene Se of prerequiie E. If (v, w) E, can' do projec v, unle alo do projec w. A ube of projec A P i feaible if he prerequiie of every projec in A alo belong o A. Projec elecion. Chooe a feaible ube of projec o maximize revenue. Projec Selecion: Prerequiie Graph Prerequiie graph. Include an edge from v o w if can' do v wihou alo doing w. {v, w, x} i feaible ube of projec. {v, x} i infeaible ube of projec. w w v x v x feaible infeaible Projec Selecion: Min Cu Formulaion Min cu formulaion. Aign capaciy o all prerequiie edge. Add edge (, v) wih capaciy -p v if p v >. Add edge (v, ) wih capaciy -p v if p v <. For noaional convenience, define p = p =. p u u w -p w p y y z -p z p v v x -px

25 // Projec Selecion: Min Cu Formulaion Claim. (A, B) i min cu iff A { } i opimal e of projec. Infinie capaciy edge enure A { } i feaible. Max revenue becaue: cap(a, B) p v ( p v ) v B: p v v A: p v p v p v v: p v v A conan w u A p u -p w p v v p y y z x -p x Open Pi Mining Open-pi mining. (udied ince early ) Block of earh are exraced from urface o rerieve ore. Each block v ha ne value p v = value of ore - proceing co. Can' remove block v before w or x. w v x. Exenion o Max Flow

26 // Circulaion wih Demand Circulaion wih demand. Direced graph G = (V, E). Edge capaciie c(e), e E. Node upply and demand d(v), v V. demand if d(v) > ; upply if d(v) < ; ranhipmen if d(v) = Def. A circulaion i a funcion ha aifie: For each e E: f(e) c(e) (capaciy) For each v V: f (e) f (e) d(v) (conervaion) e in o v e ou of v Circulaion problem: given (V, E, c, d), doe here exi a circulaion? Circulaion wih Demand Neceary condiion: um of upplie = um of demand. d(v) d(v) : D v : d (v) v : d (v) Pf. Sum conervaion conrain for every demand node v. - - upply - flow capaciy demand Circulaion wih Demand Max flow formulaion. G: - - upply - demand

27 // Circulaion wih Demand Max flow formulaion. Add new ource and ink. For each v wih d(v) <, add edge (, v) wih capaciy -d(v). For each v wih d(v) >, add edge (v, ) wih capaciy d(v). Claim: G ha circulaion iff G' ha max flow of value D. aurae all edge leaving and enering G': upply demand Circulaion wih Demand 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 and inegraliy heorem for max flow. Characerizaion. Given (V, E, c, d), here doe no exi a circulaion iff here exi a node pariion (A, B) uch ha vb d v > cap(a, B) 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 wih Demand and Lower Bound Feaible circulaion. Direced graph G = (V, E). Edge capaciie c(e) and lower bound (e), e E. Node upply and demand d(v), v V. Def. A circulaion 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) (conervaion) e in o v e ou of v Circulaion problem wih lower bound. Given (V, E,, c, d), doe here exi a a circulaion?

28 // Circulaion wih Demand and Lower Bound Idea. Model lower bound wih demand. Send (e) uni of flow along edge e. Updae demand of boh endpoin. lower bound upper bound capaciy v [, ] w d(v) d(w) d(v) + d(w) - G G' v 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-valued. Pf kech. f(e) i a circulaion in G iff f'(e) = f(e) - (e) i a circulaion in G'. Lower bound on flow Edge wih minimum and maximum capaciy For all e: l(e) f(e) c(e) l(e) c(e) Flow wih Lower Bound Look for maximum flow wih for each e: l(e) f(e) c(e) Problem olved in wo phae Fir, find admiible flow Then, augmen i o a maximum flow Admiible flow: any flow f, wih Flow conervaion if v{,}, flow ino v equal flow ou of v Lower and upper capaciy conrain fulfilled: for each e: l(e) f(e) c(e)

29 // Finding admiible flow Fir, we ranform he queion o: find an admiible circulaion Finding admiible circulaion i ranformed o: finding maximum flow in nework wih new ource and new ink Tranlaed back Circulaion Given: digraph G, lower bound l, upper capaciy bound c A circulaion fulfill: For all v: flow ino v equal flow ou of v For all (u,v): l(u,v) f(u,v) c(u,v) Exience of circulaion: fir ep for finding admiible flow Circulaion v. Flow Model flow nework wih circulaion nework: add an arc (,) wih large capaciy (e.g., um over all c(,v) ), and ak for a circulaion wih f(,) a large a poible G G f (,) = value( f )

30 // Finding admiible flow Find admiible circulaion in nework wih arc (,) Conrucion: ee previou hee Remove he arc (,) and we have an admiible flow Finding admiible circulaion I ranformed o: finding a maximum flow in a new nework New ource New ink Each arc i replaced by hree arc Finding admiible circulaion a l(e) c(e) b Do hi for each edge l(e) T New ink New ource S a b Lower bound: c(e)-l(e) l(e)

31 // Finding admiible flow/circulaion Find maximum flow from S o T If all edge from S (and hence all edge o T ) ue full capaciy, we have admiible flow: f (u,v) = f(u,v) + l(u,v) for all (u,v) in G From admiible flow o maximum flow Take admiible flow f (in original G) Compue a maximum flow f from o in G f Here c f (u,v) = c(u,v) f(u,v) And c f (v,u) = f(u,v) l(u,v) If (u,v) and (v,u) boh exi in G: add (deail omied) f + f i a maximum flow from o ha fulfill upper and lower capaciy conrain Any flow algorihm can be ued. Survey Deign

32 // Survey Deign one urvey queion per produc Survey deign. Deign urvey aking n conumer abou n produc. Can only urvey conumer i abou 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. Goal. Deign a urvey ha mee hee pec, if poible. Biparie perfec maching. Special cae when c i = c i ' = p i = p i ' =. Survey Deign Algorihm. Formulae a a circulaion problem wih lower bound. Include an edge (i, j) if conumer j own produc i. Ineger circulaion feaible urvey deign. [, ] [, ] ' [c, c '] [p, p '] ' ' ' conumer ' produc. Baeball Eliminaion

33 // 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 have a chance of finihing he eaon wih mo win? Monreal eliminaed ince i can finih wih a mo win, bu Alana already ha. w i + r i < w j eam i eliminaed. Only reaon por wrier appear o be aware of. Sufficien, bu no neceary! 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 have a chance of finihing he eaon wih mo win? Philly can win, 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. Baeball Eliminaion 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?

34 // Baeball Eliminaion: Max Flow Formulaion 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. - game lef - - eam can ill win hi many more game r = - w + r -w - game node - eam node Baeball Eliminaion: Max Flow Formulaion Theorem. Team i no eliminaed iff max flow aurae all edge leaving 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 - - eam can ill win hi many more game r = - w + r -w - game node - eam node Baeball Eliminaion: Explanaion for Spor Wrier Team i Win w i Loe To play Again = r ij l i r i NY Bal Bo Tor NY - Balimore - Boon - Torono - Deroi De - - AL Ea: Augu, Which eam have a chance of finihing he eaon wih mo win? Deroi could finih eaon wih + = win.

35 // Baeball Eliminaion: Explanaion for Spor Wrier Team i Win w i Loe To play Again = r ij l i r i NY Bal Bo Tor NY - Balimore - Boon - Torono - Deroi De - - AL Ea: Augu, Which eam have a chance of finihing he eaon wih mo win? Deroi could finih eaon wih + = win. Cerificae of eliminaion. R = {NY, Bal, Bo, Tor} Have already won w(r) = game. Mu win a lea r(r) = more. Average eam in R win a lea / > game. Baeball Eliminaion: Explanaion for Spor Wrier Cerificae of eliminaion. # remaining game # win T S, w(t ): w i, g(t ): g xy, it {x, y} T If LB on avg # game won w(t ) g(t ) T w hen z i eliminaed (by ube T). z g z Theorem. [Hoffman-Rivlin ] 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 Pf of heorem. Ue max flow formulaion, and conider min cu (A, B). Define T* = eam node on ource ide of min cu. Oberve 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 game lef y eam x can ill win hi many more game r = x-y x w z + r z -w x

36 // Page : Problem - A popular web ie ha idenified k diinc demographic group, G, G,, Gk (hee group may overlap). The ie ha conrac wih m differen adverier, o how ad o he uer of he ie aifying he following conrain: (R) For he adverier i, i will how ad only o a ube Xi of {G, G,, Gk} and wan o how ad o a lea Yi uer each hour. The ie will how only one ad o each uer of he ie each hour. Suppoe a a given hour, here are n uer viiing he ie and we know each uer belong o which demographic group. I here a way o how a ingle ad o each uer aifying he conrain (R)? Page : Problem - A popular web ie ha idenified k diinc demographic group, G, G,, Gk (hee group may overlap). The ie ha conrac wih m differen adverier, o how ad o he uer of he ie aifying he following conrain: (R) For he adverier i, i will how ad only o a ube Xi of {G, G,, Gk} and wan o how ad o a lea Ri uer each hour. The ie will how only one ad o each uer of he ie each hour. Suppoe a a given hour, here are n uer viiing he ie and we know each uer belong o which demographic group. I here a way o how a ingle ad o each uer aifying he conrain (R)? Soluion: Nework G = (V, E,, ), where V = uer U adverier U {, } E = { (, x) x i a uer } U { (y, ) y i adverier } U { (x, y) x i a uer in a demographic group ha adverier y wan o how ad } U {(, ) } All edge have capaciy, excep c(y, ) = [Ri, infiniy]. Page : Problem - For he nex n day, a hopial need Pi docor for day i and ak each of i k docor o provide a li (Lj for docor j) of preferred working day for hee n day. Uing Pi and Lj, can you provide a chedule of docor uch ha each docor will work a lea number of non-preferred day?

37 // Page : Problem - For he nex n day, a hopial need Pi docor for day i and ak each of i k docor o provide a li (Lj for docor j) of preferred working day for hee n day. Uing Pi and Lj, can you provide a chedule of docor uch ha each docor will work a lea number of non-preferred day? Soluion : Nework G = (V, E,, ), where V = D U T U {, } Docor D = {d, d,, dk}, Day T = {,,, n} For each j in D, for each i in T, c(, j) = Lj, c(i, ) = Pi,c(j, i) = if i i in Lj. If G ha a flow of value P+P+ +Pn, hen no docor work on non-preferred day. Page : Problem - For he nex n day, a hopial need Pi docor for day i and ak each of i k docor o provide a li (Lj for docor j) of preferred working day for hee n day. Uing Pi and Lj, can you provide a chedule of docor uch ha each docor will work a lea number of non-preferred day? Soluion : Nework G = (V, E,, ), where V = D U T U {, } Docor D = {d, d,, dk}, Day T = {,,, n} For each j in D, for each i in T, c(, j) = Lj, c(i, ) = Pi,c(j, i) = if i i in Lj. If G ha a flow of value P+P+ +Pn, hen no docor work on non-preferred day. Soluion : Nework G = (V, E,, ), where V = D U C U T U {, } D, T a before and C = {c, c,, ck}. c(, cj) = b, where b i a conan. The value of b can be decided by binary earch. Page : Problem - Le M be a - n by n marix and Mij i he enry in row i and column j. Mii i called a diagonal enry. We ay M i rearrangeable if i i poible o wap ome of row or ome of column o ha afer all he wapping, all he diagonal enrie of M are equal o. Give a polynomial ime algorihm ha deermine wheher a - marix i rearrangeable.

38 // Page : Problem - Le M be a - n by n marix and Mij i he enry in row i and column j. M ii i called a diagonal enry. We ay M i rearrangeable if i i poible o wap ome of row o ha afer all he wapping, all he diagonal enrie of M are equal o. Give a polynomial ime algorihm ha deermine wheher a - marix i rearrangeable. Soluion: Le G = (X, Y, E) be a biparie graph, where X = {x, x,, xn} and Y = {y, y,, yn}. (xi, yi) in E iff M ij =. Claim: G ha a perfec maching iff M i rearrangeable. =>: If G ha a perfec maching, here i a permuaion f of {,,, n} uch ha x i mache o y f(i). We may wap row i o row f(i) in M. Afer wapping, M i,f(i) =. <=: Suppoe M i rearrangeable. If M i wapped o M and he correponding graph are G and G, repecively. Then G and G are iomorphic, and G ha a perfec maching iff G ha one. Since M ii i, G ha a perfec maching. Page : Problem In a group of n people, ome own money o oher people in he group. If A own money o B, hen B doen own o A. I ime o ele up hee deb by check. Can you deign a way o ha a mo a oal of n- check will be ued and A wrie a check o B only if A own B? Page : Problem In a group of n people, ome own money o oher people in he group. If A own money o B, hen B doen own o A. I ime o ele up hee deb by check. Can you deign a way o ha a mo a oal of n- check will be ued and A wrie a check o B only if A own B? Soluion: Creae a weighed digraph G = (V, E), where V = {,,,n} and w(x,y) = c if x own y $c, where c >. Repea he following: Search imple cycle C in G, reaing each edge a undireced. Decreae he flow along he cycle according o he minimum value. Delee he edge wih value. A he end, G ha no cycle, i.e., ha a mo n- edge.