Maximum Flow Applications

Transcription

1 Maimum Flow Applicaion Conen Maimum Flow Applicaion Ma flow eenion and applicaion. Dijoin pah and nework conneciiy. Biparie maching. Circulaion wih upper and lower bound. Cenu abulaion (mari rounding). Airline cheduling. Image egmenaion. Projec elecion (ma weigh cloure). Baeball eliminaion. Princeon Unieriy COS Theory of Algorihm Spring 00 Kein Wayne Dijoin Pah Dijoin pah nework: G = (V, E,, ). Direced graph (V, E), ource, ink. Two pah are edge-dijoin if hey hae no arc in common. Dijoin Pah Dijoin pah nework: G = (V, E,, ). Direced graph (V, E), ource, ink. Two pah are edge-dijoin if hey hae no arc in common. Dijoin pah problem: find ma number of edge-dijoin - pah. Applicaion: communicaion nework. Dijoin pah problem: find ma number of edge-dijoin - pah.

2 Dijoin Pah Dijoin Pah Ma flow formulaion: aign uni capaciy o eery edge. Ma flow formulaion: aign uni capaciy o eery edge. Theorem. There are k edge-dijoin pah from o if and only if he ma flow alue i k. Proof. Suppoe here are k edge-dijoin pah P,..., P k. Se f(e) = if e paricipae in ome pah P i ; oherwie, e f(e) = 0. Since pah are edge-dijoin, f i a flow of alue k. Theorem. There are k edge-dijoin pah from o if and only if he ma flow alue i k. Proof. Suppoe ma flow alue i k. By inegraliy heorem, here ei {0, } flow f of alue k. Conider edge (,) wih f(,) =. by coneraion, here ei an arc (,w) wih f(,w) = coninue unil reach, alway chooing a new edge Produce k (no necearily imple) edge-dijoin pah. Nework Conneciiy Nework Conneciiy Nework conneciiy nework: G = (V, E,, ). Direced graph (V, E), ource, ink. A e of edge F E diconnec from if all - pah ue a lea on edge in F. Nework conneciiy nework: G = (V, E,, ). Direced graph (V, E), ource, ink. A e of edge F E diconnec from if all - pah ue a lea on edge in F. Nework conneciiy: find min number of edge whoe remoal diconnec from. Nework conneciiy: find min number of edge whoe remoal diconnec from. 8

3 Dijoin Pah and Nework Conneciiy Dijoin Pah and Nework Conneciiy Menger Theorem (9). The ma number of edge-dijoin - pah i equal o he min number of arc whoe remoal diconnec from. Menger Theorem (9). The ma number of edge-dijoin - pah i equal o he min number of arc whoe remoal diconnec from. Proof. 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 Dijoin Pah and Nework Conneciiy Maching Menger Theorem (9). The ma number of edge-dijoin - pah i equal o he min number of arc whoe remoal diconnec from. Proof. Suppoe ma number of edge-dijoin pah i k. Then ma flow alue i k. Ma-flow min-cu cu (S, T) of capaciy k. Le F be e of edge going from S o T. F = k, and definiion of cu implie F diconnec from. Maching. Inpu: undireced graph G = (V, E). M E i a maching if each node appear in a mo edge in M. Ma maching: find a ma cardinaliy maching.

4 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. Ma maching: find a ma 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. Ma maching: find a ma cardinaliy maching. Maching Maching -, -, - -, -, -, - L R L R Biparie Maching Ma flow formulaion. Creae direced graph G = (L R {, }, E ). Direc all arc from L o R, and gie infinie (or uni) capaciy. Add ource, and uni capaciy arc from o each node in L. Add ink, and uni capaciy arc from each node in R o. Biparie Maching: Proof of Correcne Claim. Maching in G of cardinaliy k induce flow in G of alue k. Gien maching M = { -, -, - } of cardinaliy. Conider flow ha end uni along each of pah: -- -, -- -, f i a flow, and ha cardinaliy. L R

5 Biparie Maching: Proof of Correcne Claim. Flow f of alue k in G induce maching of cardinaliy k in G. By inegraliy heorem, here ei {0, }-alued flow f of alue k. 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 ) Perfec maching. Perfec Maching Inpu: undireced graph G = (V, E). A maching M E i perfec if each node appear in eacly one edge in M. Perfec biparie maching. Inpu: undireced, biparie graph G = (L R, E), L = R = n. Can deermine if biparie graph ha perfec maching by running maching algorihm. I here an eay way o conince omeone ha a biparie graph doe no hae a perfec maching? Need good characerizaion of uch graph. 8 Perfec Maching Le X be a ube of node, and le N(X) be he e of node adjacen o node in. Oberaion. If a biparie graph G = (L R, E), ha a perfec maching, hen N(X) X for eery X L. Each node in X ha o be mached o a differen node in N(X). Perfec Maching Hall Theorem. Le G = (L R, E) be a biparie graph wih L = R. Then, eiher (i) G eiher ha a perfec maching, or (ii) There ei a ube X L uch ha N(X) < X. No perfec maching: X = {,, }, N(X) = {, }. L R L R 9 0

6 Perfec Maching Proof. Suppoe G doe no hae perfec maching. Then, here ei a ube X L uch ha N(X) < X. Le (S, T) be min cu. By ma-flow min-cu, cap(s, T) < n. Perfec Maching Proof. Suppoe G doe no hae perfec maching. Then, here ei a ube X L uch ha N(X) < X. Le (S, T) be min cu. By ma-flow min-cu, cap(s, T) < n. Define X = L S = L S, L T = L S, R S = R S. cap(s, T) = L T + R S R S < L S. For all arc (, w) E: S w S. (min cu can ue arc) N(L S ) R S N(L S ) R S. L S = {,, } L T = {, } R S = {, } S Dancing Problem (k-regular Biparie Graph) Dancing problem. Ecluie Iy league pary aended by n men and n women. Each man know eacly k women. Each woman know eacly k men. Acquainance are muual. I i poible o arrange a dance o ha each man dance wih a differen woman ha he know? Mahemaical reformulaion: doe eery k-regular biparie graph hae a perfec maching? Dancing Problem (k-regular Biparie Graph) Mahemaical reformulaion: doe eery k-regular biparie graph hae a perfec maching? Slick oluion: Size of ma maching i equal o ma flow in nework G. Conider following flow: flow f /k / k if (,w) E if = f (, w) = if w = 0 oherwie f i a flow and f = n. Inegraliy heorem inegral flow of alue n perfec maching. G

7 Vere Coer Vere Coer Gien an undireced graph G = (V, E), a ere coer i a ube of erice C V uch ha: Eery arc (, w) E ha eiher C or w Cor boh. Gien an undireced graph G = (V, E), a ere coer i a ube of erice C V uch ha: Eery arc (, w) E ha eiher C or w Cor boh. C = {,,,, } C = C = {,,, } C = Vere Coer Vere Coer: König-Egeráry Theorem Gien an undireced graph G = (V, E), a ere coer i a ube of erice C V uch ha: Eery arc (, w) E ha eiher C or w Cor boh. Gien an undireced graph G = (V, E), a ere coer i a ube of erice C V uch ha: Eery arc (, w) E ha eiher C or w Cor boh. Oberaion. Le M be a maching, and le C be a ere coer. Then, M C. Each ere can coer a mo one edge in any maching. König-Egeráry Theorem: In a biparie, undireced graph he ma cardinaliy of a maching i equal o he min cardinaliy of a ere coer. M = { -, -, - } M = M* = { -', -', -', -' } M* = C* = {, ', ', ' } C* = L R 8

8 Vere Coer: Proof of König-Egeráry Theorem Vere Coer: Proof of König-Egeráry Theorem König-Egeráry Theorem: In a biparie, undireced graph, he ize of ma maching and min ere coer are equal. Suffice o find maching M* and coer C* uch ha M* = C*. Ue ma flow formulaion, and le (S, T) be min cu. König-Egeráry Theorem: In a biparie, undireced graph, he ize of ma maching and min ere coer are equal. Suffice o find maching M* and coer C* uch ha M* = C*. Ue ma flow formulaion, and le (S, T) be min cu. Define L S = L S, L T = L T, R S = R S, R T = R T, and C* = L T R S. Define L S = L S, L T = L T, R S = R S, R T = R T, and C* = L T R S. Claim. C* i a ere coer. Claim. C* i a ere coer. conider (, w) E Claim. C* = M*. L S, w R T impoible ince infinie capaciy ma-flow min-cu heorem M* = cap(s, T) hu, L T or w R S or boh only arc of form (, ) or (w, ) conribue o cap(s, T) M* = u(s, T) = L T + R S = C*. 9 0 Biparie Maching and Vere Coer Uni Capaciy Simple Nework Which ma flow algorihm o ue for biparie maching / ere coer? Uni capaciy imple nework. Generic augmening pah: O( m f * ) = O(mn). Eery arc capaciy i one. Capaciy caling: O(m log U ) = O(m ). Shore augmening pah: O(m n ). Eery node ha eiher: (i) a mo one incoming arc, or (ii) a mo one ougoing arc. Seem o indicae "more cleer" algorihm are no a good a we fir hough. If G i imple uni capaciy, hen o i G f, auming f i {0, } flow. No - ju need more cleer analyi! For biparie maching, hore augmening pah algorihm run in O(m n / ) ime. Shore augmening pah algorihm. Normal augmenaion: lengh of hore pah doen change. Special augmenaion: lengh of hore pah ricly increae. Theorem. Shore augmening pah algorihm run in O(m n / ) ime. L. Each phae of normal augmenaion ake O(m) ime. L. Afer a mo n / phae, f f * - n /. L. Afer a mo n / addiional augmenaion, flow i opimal.

9 Uni Capaciy Simple Nework Uni Capaciy Simple Nework Lemma. Phae of normal augmenaion ake O(m) ime. Sar a, adance along an arc in L G unil reach or ge uck. if reach, augmen and delee ALL arc on pah if ge uck, delee node and go o preiou node Lemma. Phae of normal augmenaion ake O(m) ime. Sar a, adance along an arc in L G unil reach or ge uck. if reach, augmen and delee ALL arc on pah if ge uck, delee node and go o preiou node Augmen Delee node and rerea Leel graph Leel graph Uni Capaciy Simple Nework Uni Capaciy Simple Nework Lemma. Phae of normal augmenaion ake O(m) ime. Sar a, adance along an arc in L G unil reach or ge uck. if reach, augmen and delee ALL arc on pah if ge uck, delee node and go o preiou node Lemma. Phae of normal augmenaion ake O(m) ime. Sar a, adance along an arc in L G unil reach or ge uck. if reach, augmen and delee ALL arc on pah if ge uck, delee node and go o preiou node Augmen STOP Lengh of hore pah ha increaed. Leel graph Leel graph

10 Uni Capaciy Simple Nework Uni Capaciy Simple Nework Lemma. Phae of normal augmenaion ake O(m) ime. Sar a, adance along an arc in L G unil reach or ge uck. if reach, augmen and delee ALL arc on pah if ge uck, delee node and go o preiou node O(m) running ime. O(m) o creae leel graph O() per arc, ince each arc raered a mo once O() per node deleion ARRAY pred[ V] L G leel graph of G f, pred[] nil REPEAT WHILE (here ei (,w) L G ) pred[w], w IF ( = ) P pah defined by pred[] f augmen(f, P) updae L G, pred[] nil delee from L G UNTIL ( = ) RETURN f AdanceRerea(V, E, f,, ) adance augmen rerea 8 Uni Capaciy Simple Nework Lemma. Afer a mo n / phae, f f * - n /. Afer n / phae, lengh of hore augmening pah i > n /. Leel graph ha more han n / leel. Le h n / be layer wih min number of node: V h n /. Uni Capaciy Simple Nework Lemma. Afer a mo n / phae, f f * - n /. Afer n / phae, lengh of hore augmening pah i > n /. Leel graph ha more han n / leel. Le h n / be layer wih min number of node: V h n /. S := { : l() < h} { : l() = h and ha ougoing reidual arc}. cap f (S, T) V h n / f f * - n /. Leel graph Reidual arc Leel graph V V h V 0 V V h Vn / 9 0 V 0 Vn /

11 Baeball Eliminaion Oer on he radio ide he producer aying, "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." 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 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? 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! 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. Anwer depend no ju on how many game already won and lef o play, bu alo on whom hey re again.

12 Baeball eliminaion problem. Se of eam X. Diinguihed eam X. Baeball Eliminaion Team i ha won w i game already. Team i and j play each oher r ij addiional ime. I here any oucome of he remaining game in which eam finihe wih he mo (or ied for he mo) win? Baeball Eliminaion: Ma Flow Formulaion 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 - - Baeball Eliminaion: Ma Flow Formulaion Baeball Eliminaion: Eplanaion for Spor Wrier Theorem. Team i no eliminaed if and only if ma flow aurae all arc leaing ource. Inegraliy heorem each remaining game beween i and j added o number of win for eam i or eam j. Capaciy on (, ) arc enure no eam win oo many game. game lef r = eam can ill win hi many more game w + r -w 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

13 Baeball Eliminaion: Eplanaion for Spor Wrier Baeball Eliminaion: Eplanaion for Spor Wrier 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 Cerificae of eliminaion. R X, LB on ag # game won 8 win w( R) : = w, # # remaining game i R i 8 r( R) : = r, ij i, j R 8 w( R) + r( R) If > w + r hen i eliminaed (by R). R Theorem (Hoffman-Rilin, 9). Team i eliminaed if and only if here ei a ube R ha eliminae. Proof idea. Le R = eam node on ource ide of min cu. Conider ube R = {NY, Bal, Bo, Tor} Hae already won w(r) = 8 game. Mu win a lea r(r) = more.! Aerage eam in R win a lea 0/ > game. 9 0 Baeball Eliminaion: Eplanaion for Spor Wrier Proof of Theorem. Ue ma flow formulaion, and conider min cu (S, T). Define R = eam node on ource ide of min cu = T S. Claim. i-j S if and only if i R and j R. infinie capaciy arc enure if i-j S hen i S and j S if i S and j S bu i-j T, hen adding i-j o S decreae capaciy of cu Baeball Eliminaion: Eplanaion for Spor Wrier Proof of Theorem. Ue ma flow formulaion, and conider min cu (S, T). Define R = eam node on ource ide of min cu = T S. Claim. i-j S if and only if i R and j R. r( X { }) > = cap( S, T ) r( X { }) r( R) + ( w i R + r w ) i - eam can ill win hi many more game - eam can ill win hi many more game game lef - game lef - r = - w + r -w r = - w + r -w

14 Baeball Eliminaion: Eplanaion for Spor Wrier Proof of Theorem. Ue ma flow formulaion, and conider min cu (S, T). Define R = eam node on ource ide of min cu = T S. Claim. i-j S if and only if i R and j R. r( X { }) > cap( S, T ) = = Rearranging erm: r( X { }) r( R) + r( X { }) r( R) w + r ( w i R w( R) w( R) + r( R) <. R + r w ) + R ( w i + r ) Circulaion wih Demand Circulaion wih demand. Direced graph G = (V, E). Arc capaciie u(e), e E. Node upply and demand d(), V. demand if d() > 0; upply if d() < 0; ranhipmen if d() = 0 A circulaion i a funcion f: E R ha aifie: For each e E: 0 f(e) u(e) (capaciy) For each V: f ( e) f ( e) = d( ) (coneraion) e in o e ou of Circulaion problem: gien (V, E, u, d), doe here ei a circulaion? Circulaion wih Demand A circulaion i a funcion f: E R ha aifie: For each e E: 0 f(e) u(e) (capaciy) For each V: f ( e) f ( e) = d( ) (coneraion) e in o e ou of Circulaion wih Demand A circulaion i a funcion f: E R ha aifie: For each e E: 0 f(e) u(e) (capaciy) For each V: f ( e) f ( e) = d( ) (coneraion) e in o e ou of Neceary condiion: um of upplie = um of demand. d( ) : d ( ) > 0 = d( ) : d ( ) < 0 = : D Proof: um coneraion conrain for eery demand node. Capaciy Demand Flow

15 Circulaion wih Demand Ma flow formulaion. Add new ource and ink. For each wih d() < 0, add arc (, ) wih capaciy -d(). For each wih d() > 0, add arc (, ) wih capaciy d(). Circulaion wih Demand Ma flow formulaion. Add new ource and ink. For each wih d() < 0, add arc (, ) wih capaciy -d(). For each wih d() > 0, add arc (, ) wih capaciy d(). G: -8 - Demand G : 8 Demand Ma flow formulaion. Circulaion wih Demand Graph G ha circulaion if and only if G ha ma flow of alue D (aurae all arc leaing and enering ). Ma flow formulaion. Circulaion wih Demand Graph G ha circulaion if and only if G ha ma flow of alue D (aurae all arc leaing and enering ). Moreoer, if all capaciie and demand are ineger, and here ei a circulaion, hen here ei one ha i ineger-alued. G : Demand Characerizaion. Gien (V, E, u, d), here doe no ei a circulaion if and only if here ei a node pariion (A, B) uch ha: d( ) > u( A, B). B demand by node in B eceed upply of node in B plu ma capaciy of arc going from A o B Proof idea: look a min cu in G. 9 0

16 Circulaion wih Demand and Lower Bound Feaible circulaion. Direced graph G = (V, E). Arc capaciie u(e) and lower bound l(e), e E. force flow o make ue of cerain edge Node upply and demand d(), V. Circulaion wih Demand and Lower Bound A circulaion i a funcion f: E R ha aifie: For each e E: l(e) f(e) u(e) (capaciy) For each V: ( ) ( ) ( ) (coneraion) f e f e = d e ou of e in o A circulaion i a funcion f: E R ha aifie: For each e E: l(e) f(e) u(e) (capaciy) For each V: f ( e) f ( e) = d( ) (coneraion) e in o e ou of Gien (V, E, l, u, d), i here a circulaion? Idea: model lower bound wih upply / demand. Creae nework G. u ( e) = u( e) l( e) d ( ) = d( ) + l( e) e ou of Lower bound l( e in o d() e), 9 Capaciy w d(w) Flow Capaciy d() + w d(w) - Circulaion wih Demand and Lower Bound A circulaion i a funcion f: E R ha aifie: For each e E: l(e) f(e) u(e) (capaciy) For each V: (coneraion) f ( e) f ( e) = d( ) e ou of e in o Creae nework G. u ( e) = d ( ) = u( e) l ( e) d( ) + l( e) l( e ou of e in o e) Feaible mari rounding. Mari Rounding Gien a p q mari D = {d ij } of real number. Row i um = a i, column j um b j. Round each d ij, a i, b j up or down o ineger o ha um of rounded elemen in each row (column) equal row (column) um. Original applicaion: publihing US Cenu daa. Theorem: for any mari, here ei a feaible rounding. Theorem. There ei a circulaion in G if and only if here ei 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. Proof idea: f(e) i a circulaion in G if and only if f (e) = f(e) - l(e) i a circulaion in G Original Daa Poible Rounding

17 Mari Rounding Mari Rounding Feaible mari rounding. Gien a p q mari D = {d ij } of real number. Row i um = a i, column j um b j. Ma flow formulaion. Original daa proide circulaion (all demand 0). Inegraliy heorem here alway ei feaible rounding! Round each d ij, a i, b j up or down o ineger o ha um of rounded elemen in each row (column) equal row (column) um. Original applicaion: publihing US Cenu daa. Theorem: for any mari, here ei a feaible rounding. Noe: "hrehold rounding" doen work Row, Column, 8, , 0, ,, Original Daa Poible Rounding Lower bound Upper bound Airline Scheduling Airline Scheduling Airline cheduling. Comple compuaional problem faced by naion airline carrier. Produce chedule for houand of roue each day ha are efficien in erm of: equipmen uage, crew allocaion, cuomer aifacion in preence of unpredicable iue like weaher, breakdown One of large conumer of high-powered algorihmic echnique. "Toy problem." Manage fligh crew by reuing hem oer muliple fligh. Inpu: li of ciie V. Trael ime (, w) from ciy o w. Ma flow formulaion. For each fligh i, include wo node u i and i. Add ource wih demand -c, and arc (, u i ) wih capaciy. Add ink wih demand c, and arc ( j, ) wih capaciy. For each i, add arc (u i, i ) wih lower bound and capaciy. if fligh j reachable from i, add arc ( i, u i ) wih capaciy. crew can begin day wih any fligh -c 0, u u 0, crew can end day wih any fligh c Fligh i: (o i, d i, i ) coni of origin and deinaion ciie, and deparure ime. ue c crew u, 0, u fligh i performed ame crew can do fligh and 8

18 Running ime. Airline Scheduling: Running Time k = number of fligh. O(k) node, O(k ) edge. A mo k crew needed ole k ma flow problem. Arc capaciie beween 0 and k a mo k augmenaion per ma flow compuaion. Oerall ime = O(k ). Remark. Can ole in O(k ) ime by formulaing a "minimum flow problem." Airline cheduling. Airline Scheduling: Pomorem We oled oy problem. Real problem addree counle oher facor: union regulaion: e.g., fligh crew can only fly cerain number of hour in gien ineral need opimal chedule oer planning horizon, no ju day deadheading ha a co imulaneouly rying o re-work fligh chedule and re-opimize fare rucure Meage. Our oluion i a generally ueful echnique for efficien re-ue of limied reource bu riialize real airline cheduling problem. Flow echnique ueful for oling airline cheduling problem, and are genuinely ued in pracice. Running an airline efficienly i a ery difficul problem. 9 0 Image egmenaion. Image Segmenaion Cenral problem in image proceing. Diide image ino coheren region. hree people anding in fron of comple background cene idenify each of hree people a coheren objec Image Segmenaion Foreground / background egmenaion. Label each piel in picure a belonging o foreground or background. V = e of piel, E = pair of neighboring piel. a = likelihood piel in foreground. b = likelihood piel in background. p w = eparaion penaly for labeling one of and w a foreground, and he oher a background. Goal. Accuracy: if a > b, in iolaion we prefer o label piel in foreground. Smoohne: if many neighbor of are labeled foreground, we hould be inclined o label a foreground. Find pariion (S, T) ha maimize: a + b pw S T {, w } E S I {, w } =

19 Formulae a min cu problem. Maimizaion. No ource or ink. Undireced graph. Turn ino minimizaion problem. Image Segmenaion Since a + b i a conan, V V Image Segmenaion Formulae a min cu problem: G = (V, E ). Maimizaion. No ource or ink. add ource o correpond o foreground add ink o correpond o background Undireced graph. add wo ani-parallel arc p w p w p w maimizing a S + b T p w {, w } E S I {, w} = i equialen o minimizing a T + b S + p w {, w } E S I {, w } = a p w b w Image Segmenaion Conider min cu (S, T) in reuling graph. cap( S, T ) = a T b S p (, w w E S, ) w T Preciely he quaniy we wan o minimize. noe if and w on differen ide, p w couned eacly once a + Projec wih prerequiie. Projec Selecion Se P of poible projec. Projec ha aociaed reenue p. ome projec generae money: creae ineracie e-commerce inerface, redeign c web page oher co money: upgrade compuer, ge ie licene for encrypion ofware Se of prerequiie E. If (, w) E, can do projec and unle alo do projec w. can ar on e-commerce opporuniy unle you e go encrypion ofware A ube of projec A P i feaible if he prerequiie of eery projec in A alo belong o A. for each P, and (, w) E, we hae w P p w b Projec elecion (ma weigh cloure) problem: chooe a feaible ube of projec o maimize reenue. G w

20 Projec Selecion Prerequiie graph. Include an arc from o w if can do wihou alo doing w. {, w, } i feaible ube of projec. {, } i infeaible ube of projec. Projec Selecion Projec elecion formulaion. Aign infinie capaciy o all prerequiie arc. Add arc (, ) wih capaciy p if p > 0. Add arc (, ) wih capaciy -p if p < 0. For noaional conenience, define p = p = 0. w w u w -p w y z Feaible Infeaible p -p 8 Projec Selecion Claim. (S, T) i min cu if and only if S \ {} i opimal e of projec. Infinie capaciy arc enure S \ {} i feaible. Ma reenue becaue: cap( S, T ) = = p + ( p ) T : p > 0 S: p < 0 p p : p > 0 S conan Projec Selecion Open-pi mining (udied ince early 90 ). Block of earh are eraced from urface o reriee ore. Each block ha ne alue p = alue of ore - proceing co. Can remoe block before w or. w u p u p y y z -p w w p -p 9 80