1 The Network Flow Problem

Transcription

1 5-5/65: Deign & Anlyi of Algorihm Ooer 5, 05 Leure #0: Nework Flow I l hnged: Ooer 5, 05 In hee nex wo leure we re going o lk ou n imporn lgorihmi prolem lled he Nework Flow Prolem. Nework flow i imporn eue i n e ued o expre wide vriey of differen kind of prolem. So, y developing good lgorihm for olving nework flow, we immediely will ge lgorihm for olving mny oher prolem well. In Operion Reerh here re enire oure devoed o nework flow nd i vrin. Topi in ody leure inlude: The definiion of he nework flow prolem The i Ford-Fulkeron lgorihm The mxflow-minu heorem The iprie mhing prolem The Nework Flow Prolem We egin wih definiion of he prolem. We re given direed grph G, r node, nd ink node. Eh edge e in G h n oied non-negive piy (e), where for ll non-edge i i impliily umed h he piy i 0. For exmple, onider he grph in Figure elow. d Figure : A nework flow grph. Our gol i o puh muh flow poile from o in he grph. The rule re h no edge n hve flow exeeding i piy, nd for ny verex exep for nd, he flow in o he verex mu equl he flow ou from he verex. Th i, Cpiy onrin: On ny edge e we hve f(e) (e). Flow onervion: For ny verex v {, }, flow in equl flow ou: u f(u, v) = u f(v, u). Suje o hee onrin, we wn o mximize he ol flow ino. For inne, imgine we wn o roue mege rffi from he oure o he ink, nd he piie ell u how muh ndwidh we re llowed on eh edge. E.g., in he ove grph, wh i he mximum flow from o? Anwer: 5. Uing piy[flow] noion, he poiive flow look in Figure. Noe h he flow n pli nd rejoin ielf. How n you ee h he ove flow w relly mximum? Noie, hi flow ure he nd edge, nd, if you remove hee, you dionne from. In oher word, he grph h n - u of ize 5 ( e of edge of ol piy 5 uh h if you remove hem, hi dionne he ink from he oure). The poin i h ny uni of flow going from o mu ke up le uni of piy in hee pipe. So, we know we re opiml.

2 [] [] [] [] [] d [] [] Figure : A nework flow grph wih poiive flow hown uing piy[flow] noion. We ju rgued h in generl, he mximum - flow he piy of he minimum - u. An imporn propery of flow, h we will prove yprodu of nlyzing n lgorihm for finding hem, i h he mximum - flow i in f equl o he piy of he minimum - u. Thi i lled he Mxflow-Minu Theorem. In f, he lgorihm will find flow of ome vlue k nd u of piy k, whih will e proof h oh re opiml! To derie he lgorihm nd nlyi, i will help o e i more forml ou few of hee quniie. Definiion An - u i e of edge whoe removl dionne from. Or, formlly, u i priion of he verex e ino wo piee A nd B where A nd B. (The edge of he u re hen ll edge going from A o B). Definiion The piy of u (A, B) i he um of piie of edge in he u. Or, in he forml viewpoin, i i he um of piie of ll edge going from A o B. (Don inlude he edge from B o A.) Definiion I will lo e mhemilly onvenien for ny edge (u, v) o define f(v, u) = f(u, v). Thi i lled kew-ymmery. (We will hink of flowing uni on he edge from u o v equivlenly flowing uni on k-edge from v o u.) The kew-ymmery onvenion mke i epeilly ey o dd wo flow ogeher. For inne, if we hve one flow wih uni on he edge (, ) nd noher flow wih uni on he edge (, ), hen dding hem edge y edge doe he righ hing, reuling in ne flow of uni from o. In f, le now formlly define he um of wo flow. If f nd g re flow, hen h = f + g i defined h(u, v) = f(u, v) + g(u, v) for ll pir (u, v). Noie h if f nd g ify flow-onervion nd kew-ymmery, hen h doe oo. In f, uing kew-ymmery, if we wned we ould rewrie he flow onervion ondiion v {, }, u f(u, v) = 0, ine he ol flow ou of node will lwy e he negive of he ol flow ino node. How n we find mximum flow nd prove i i orre? Here very nurl regy: find ph from o nd puh muh flow on i poile. Then look he lefover piie (n imporn iue will e how exly we define hi, u we will ge o i in minue) nd repe. Coninue unil here i no longer ny ph wih piy lef o puh ny ddiionl flow on. Of oure, we need o prove h hi work: h we n omehow end up uopiml oluion y mking d hoie long he wy. Thi pproh, wih he orre definiion of lefover piy, i lled he Ford-Fulkeron lgorihm. Thi immediely implie he vlue of ny - flow mximum - flow minimum - u ny - u.

3 The Ford-Fulkeron lgorihm The Ford-Fulkeron lgorihm i imply he following: while here exi n ph P of poiive reidul piy (defined elow), puh he mximum poile flow long P. By he wy, hee ph P re lled ugmening ph, eue you ue hem o ugmen he exiing flow. Reidul piy i ju he piy lef over given he exiing flow, where we will ue kewymmery o pure he noion h if we puh f uni of flow on n edge (u, v), hi inree our iliy o puh flow on he k-edge (v, u) y f. Definiion Given flow f in grph G, he reidul piy f (u, v) i defined f (u, v) = (u, v) f(u, v), where rell h y kew-ymmery we hve f(v, u) = f(u, v). For exmple, given he flow in Figure, he edge (, ) h reidul piy. The k-edge (, ) h reidul piy, eue i originl piy w 0 nd we hve f(, ) =. Definiion 5 Given flow f in grph G, he reidul grph G f i he direed grph wih ll edge of poiive reidul piy, eh one leled y i reidul piy. Noe: hi my inlude k-edge of he originl grph G. Le do n exmple. Conider he grph in Figure nd uppoe we puh wo uni of flow on he ph. We hen end up wih he following reidul grph: d Figure : Reidul grph reuling from puhing uni of flow long he ph --- in he grph in Figure. The red edge denoe he hnge. If we oninue running Ford-Fulkeron, we ee h in hi grph he only ph we n ue o ugmen he exiing flow i he ph d. Puhing he mximum uni on hi ph we hen ge he nex reidul grph, hown in Figure. d Figure : Reidul grph reuling from puhing uni of flow long he ph ----d- in he grph in Figure. Agin, he red edge denoe he hnge. The lgorihm i due o Leer R. Ford Jr. nd Deler R. Fulkeron, hen he RAND orporion working on figuring ou he piy of he Sovie ril nework. They didn wn o find he mx-flow, hey wned he min-u. A we will oon ee, he wo prolem re inexrily enwined. They preened heir lgorihm in RAND repor of 955.

4 A hi poin here i no longer ph from o o we re done. We n hink of Ford-Fulkeron eh ep finding new flow (long he ugmening ph) nd dding i o he exiing flow, uing he definiion of dding flow we gve efore. The definiion of reidul piy enure h he flow found y Ford-Fulkeron i legl (doen exeed he piy onrin in he originl grph). We now need o prove h in f i i mximum. We ll worry ou he numer of ierion i ke nd how o improve h ler. Noe h one nie propery of he reidul grph i h i men h eh ep we re lef wih me ype of prolem we red wih. So, o implemen Ford-Fulkeron, we n ue ny lk-ox ph-finding mehod (e.g., DFS).. The Anlyi For now, le u ume h ll he piie re ineger. If he mximum flow vlue i F, hen he lgorihm mke mo F ierion, ine eh ierion puhe le one more uni of flow from o. We n implemen eh ierion in ime O(m + n) uing DFS. So we ge he following reul. Theorem 6 If he given grph G h ineger piie, Ford-Fulkeron ermine in ime O(F (m+ n)) where F i he vlue of he mximum - flow. Theorem 7 When i ermine, he Ford-Fulkeron lgorihm oupu mximum flow. Proof: Le look he finl reidul grph. Thi grph mu hve nd dionneed y definiion of he lgorihm. Le A e he omponen onining nd B e he re. Le e he piy of he (A, B) u in he originl grph o we know we n do eer hn. A d B The lim i h we in f did find flow of vlue (whih herefore implie i i mximum). Here why: le look wh hppen o he reidul piy of he (A, B) u fer eh ierion of he lgorihm. Sy in ome ierion we found ph wih k uni of flow. Then, even if he ph zig-zgged eween A nd B, every ime we wen from A o B we dded k o he flow from A o B nd ured k from he reidul piy of he (A, B) u, nd every ime we wen from B o A we ook wy k from hi flow nd dded k o he reidul piy of he u ; moreover, we mu hve gone from A o B exly one more ime hn we wen from B o A. So, he reidul piy of hi u wen down y exly k. So, he drop in piy i equl o he inree in flow. Sine he end he reidul piy i zero (rememer how we defined A nd B) hi men he ol flow i equl o. So, we ve found flow of vlue equl o he piy of hi u. We know we n do eer, o hi mu e mx flow, nd (A, B) mu e minimum u. Noie h in he ove rgumen we ully proved he nonoviou mxflow-minu heorem: Thi i where we ue he f h if we flow k uni on he edge (u, v), hen in ddiion o reduing he reidul piy of he (u, v) edge y k we lo dd k o he reidul piy of he k-edge (v, u).

5 Theorem 8 In ny grph G, for ny wo verie nd, he mximum flow from o equl he piy of he minimum (, )-u. We hve lo proven he inegrl-flow heorem: if ll piie re ineger, hen here i mximum flow in whih ll flow re ineger. Thi eem oviou, u i urn ou o hve ome nie nd non-oviou impliion. Tehnilly, we ju proved Theorem 8 only for ineger piie. Wh if he piie re no ineger? Firly, if he piie re rionl, hen hooe he mlle ineger N uh h N (u, v) i n ineger for ll edge (u, v). I i ey o ee h eh ep we end le /N moun of flow, nd hene he numer of ierion i mo NF, where F i he vlue of he mximum - flow. (One n rgue hi y oerving h ling up ll piie y N will mke ll piie ineger, whene we n pply our ove rgumen.) And hene we ge Theorem 8 for rionl piie well. Wh if he piie re irrionl? In hi e Ford-Fulkeron my no ermine. And he oluion i onverge o (in he limi) my no even e he mx-flow! (See here, here.) Bu he mxflow-minu Theorem 8 ill hold, even wih irrionl piie. There re everl wy o prove hi; here one. Suppoe no, nd uppoe here i ome flow nework wih he mxflow eing ɛ > 0 mller hn he minu. Chooe ineger N uh h ɛ, nd round ll piie down o he nere ineger N m muliple of /N. The minu wih hee new edge piie my hve fllen y m/n ɛ/, nd he mxflow ould e he me he originl flow nework, u ill here would e gp of ɛ/ eween mxflow nd minu in hi rionl-piy nework. Bu hi i no poile, eue ued Ford-Fulkeron o prove mxflow-minu for rionl piie in he previou prgrph. In he nex leure we will look mehod for reduing he numer of ierion he lgorihm n ke. For now, le ee how we n ue n lgorihm for he mx flow prolem o olve oher prolem well: h i, how we n redue oher prolem o he one we now know how o olve. Biprie Mhing Sy we wned o e more ophiied ou igning group o homework preenion lo. We ould k eh group o li he lo eple o hem, nd hen wrie hi iprie grph y drwing n edge eween group nd lo if h lo i eple o h group. For exmple: Group Slo Thi i n exmple of iprie grph: grph wih wo ide L nd R uh h ll edge go eween L nd R. A mhing i e of edge wih no endpoin in ommon. Wh we wn here in igning group o ime lo i perfe mhing: mhing h onne every poin in L wih poin in R. For exmple, wh i perfe mhing in he iprie grph ove? More generlly (y here i no perfe mhing) we wn mximum mhing: mhing wih he mximum poile numer of edge. We n olve hi follow: Biprie Mhing:. Se up fke r node onneed o ll verie in L. Conne ll verie in R o fke ink node. Orien ll edge lef-o-righ nd give eh piy of. 5

6 . Find mx flow from o uing Ford-Fulkeron.. Oupu he edge eween L nd R onining nonzero flow he deired mhing. Thi find legl mhing eue edge from R o hve piy, o he flow n ue wo edge ino he me node, nd imilrly he edge from o L hve piy, o you n hve flow on wo edge leving he me node in L. I mximum mhing eue ny mhing give you flow of he me vlue: ju onne o he hed of hoe edge nd onne he il of hoe edge o. (So if here w eer mhing, we wouldn e mximum flow). Wh ou he numer of ierion of ph-finding? Thi i mo he numer of edge in he mhing ine eh ugmening ph give u one new edge. Le run he lgorihm on he ove exmple. We fir uild hi flow nework. Then we ue Ford-Fulkeron. Sy we r y puhing flow on --- nd ---, herey mhing o nd o. Thee re d hoie, ine wih hee hoie nno e mhed. Bu he ugmening ph uomilly undoe hem i improve he flow! Mhing ome up in mny differen prolem like mhing up upplier o uomer, or ellphone o ell-ion when you hve overlpping ell. They re lo i pr of oher lgorihmi prolem. 6