5. Network flow. Network flow. Maximum flow problem. Ford-Fulkerson algorithm. Min-cost flow. Network flow 5-1

Transcription

1 Nework flow -. Nework flow Nework flow Mximum flow prolem Ford-Fulkeron lgorihm Min-co flow

2 Nework flow Nework N i e of direced grph G = (V ; E) ource 2 V which h only ougoing edge ink (or deinion) 2 V which h only incoming edge poiive inegrl cpciy c ij on ech direced edge (i ; j ) we cn define nework on n undireced grph y mking every edge i-direcionl he inegrl cpciy umpion cn e relxed ome ime, u i i necery oher ime flow f : E 7! R + i feile if edge cpciy limi: for ll (i ; j ) 2 E conervion of flow: for ll i 2 V n f; g 0 f ij c ij f ij = j k f ki

3 define. he vlue of he flow i he moun of flow leving he ource clim. equivlenly, he vlue of feile flow i he moun of flow enering he ink v(f ) = j f j = j f j proof. v(f ) = j = j = i2v = i2v f j h f j + f i i2v nf ;g f ij f ji j 2V i2v nfg j 2V j 2V f ij i2v j 2V f ji + j f j f i i = j he l line follow from he fc h, in he ummion, he flow coming from he ource re couned only once wih negive ign nd he flow going o he ink re couned once wih poiive ign. mx-flow prolem i o find feile flow wih mximum vlue f j

4 we denoe priion of node ino wo e S nd S c, - cu, y he noion (S; S c ), where 2 S nd 2 S c he vlue of he cu c(s; S c ) i he um of ll cpciie leving S nd enering S c c(s; S c ) = c 2S ;2S c clim. for ny cu (S; S c ) i cn e hown h proof. v(f ) = = = v(f ) = j f j = i 2S 2S 2S ;2S c f j f i + 2S ;2S c f f j + f i i 2S 2S c 2S ;2S c f i 2S nfg 2S ;2S c f h f i i 2S 2S f i f i i f i i 2S 2S c ince ll erm f ; wih oh nd in S pper wice, once wih poiive ign nd once wih he negive ign, cnceling ech oher, excep for hoe which cro over S nd S c

5 Nework flow - clim. for ny flow f nd ny cu (S; S c ), v(f ) c(s; S c ) proof. v(f ) = 2S ;2S c f f f c = c(s ; S c ) 2S ;2S c 2S ;2S c 2S ;2S c clim. mx-flow min-cu mx v(f ) min c(s; S c ) f feile S

6 mx-flow prolem for generl nework, here i polynomil ime lgorihm h find he mx-flow he mx-flow i lwy equl o he min-cu vlue mx v(f ) = min c(a; B) f feile (A;B) Greedy mx-flow lgorihm (u-opiml) Iniilize f ij = 0 for ll i ; j repe - find ph P eween nd uch h min (c ij f ij ) > 0 (i ;j )2P - cll uch ph unured - le df = min (i ;j )2P(c ij f ij ) - e f ij = f ij + df for ll (i ; j ) 2 P unil no more unured ph hi lgorihm i inefficien (numer of ph cn e exponenil in he ize of he grph) hi lgorihm doe no find he mx-flow (u-opiml)

7 Nework flow -7 reidul nework R(N; f ) = (V ; E r ) i verex e V wih weighed nd direced edge e E r conruced follow: if f ij < c ij plce n edge (i ; j ) wih cpciy cij 0 = c ij f ij if f ij > 0 plce n edge (j ; i) wih cpciy cji 0 = f ij (noe h hi i in he oppoie direcion) ide: ny ph P in R(N; f ) give ph which we cn incree he flow, including he one h revere previouly igned one Ford-Fulkeron lgorihm, 96 Iniilize f ij = 0 for ll i ; j while here i ph P from o in R(N ; f ) do - end flow of vlue df = min (i ;j )2P cij 0 in R(N ; f ) long P - upde f in (N ; f ): e f ij = f ij + df for ll (i ; j ) 2 P - reuild he reidul nework R(N ; f ) end while ph in direced grph i equence of node v v 2 v k uch h here i direced edge from v i o v i+

8 heorem. If Ford-Fulkeron lgorihm ermine, i oupu mximum flow proof. uppoe he lgorihm ermine ime T wih flow f, hen here i no ph from o in R(N ; f ) le S e he e of node rechle from clim. v(f ) = c(s ; S c ) hi prove h f i he mximum flow, ince we lredy know h he mximum flow i mo he minimum cu nd we found flow whoe vlue i he me cu (which i lo minimum cu) proof of clim. if exi n edge (i ; j ) uch h i 2 S, j 2 S c, nd f ij < c ij, hen here mu e n edge in R(N ; f ) from S o T if exi n edge (i ; j ) uch h i 2 S c, j 2 S, nd f ij > 0, hen here mu e n edge in R(N ; f ) from S o T herefore, v(f ) = f {z} ij i 2S ;j 2S c i 2S ;j 2S =c c ij = c(s ; S c ) f {z} ji =0

9 corollry. he minimum cu vlue in nework i he me he mximum flow vlue corollry. if ll edge cpciie in nework re non-negive ineger, hen here exi n inegrl mximum flow we re lef o how h he lgorihm ermine under mild umpion: clim. for nework where ll weigh re rionl numer, Ford-Fulkeron lgorihm ermine in finie ime i hould lo e noed h here re ce where he lgorihm doe no ermine: clim. here exi nework wih irrionl weigh uch h Ford-Fulkeron lgorihm never ermine

10 S M M M p r = e e3 e2 M M M reidul flow ep ugmening ph dded flow e e 2 e 3 0 r (; 2; 3; ) r 0 2 (; 4; 3; 2; ; ) r r 2 0 r 3 (; 2; 3; 4; ) r r 2 r 0 4 (; 4; 3; 2; ; ) r 2 0 r 3 r 2 (; ; 2; 3; ) r 2 r 2 r 3 0 r i choen uch h r 2 = r ince fer ep we hve he me reidul cpciy fer ep up o cling y r 2, he proce never op P he vlue of flow converge o v(f ) = + 2 i= r i = 3 + 2r, while he mximum flow i v(f ) = 2M + he lgorihm never ermine, nd furher he flow vlue doe no converge o he mximum Nework flow -0

11 Run-ime lemm. if ll edge cpciie of N re inegrl, hen Ford-Fulkeron lgorihm ermine in O(m v(f )) ime proof. he vlue of he flow incree le y one ech ierion, nd direced ph cn e found in O(m) ime how mny operion do we need o erch for ph from o in direced grph wih numer of edge mo m? how mny ierion re required in he wor-ce? Shore Augmening Ph (SAP) Chooe he ugmening ph in he reidul nework h h he fewe numer of edge lemm. If, cerin ierion, he lengh of hore ph from o in he reidul nework i `, hen every uequen ierion i i `. Furhermore, fer mo m ierion, he dince from o mu ecome ` +. lemm. he hore ugmening ph cn e found in O(m) ime uing redh-fir erch. heorem. SAP implemenion of Ford-Fulkeron lgorihm run in ime O(m 2 n), where m = jej nd n = jv j.

12 Applicion Projec elecion prolem i defined y e of projec ft ; : : : ; T ng, ech wih poiive profi P i e of equipmen fe ; : : : ; E mg, ech wih poiive co C j projec h ue of equipmen which re prerequiie find he opiml ue of projec h ify he prerequiie requiremen nd h mximum ol profi. conruc grph G where ll he projec re conneced o he ource, nd he weigh of h edge i P i connec ll he equipmen o he ink, nd he weigh of h edge i C j if E j i prerequiie for T i hen dd n edge (T i ; E j ) wih weigh clim. cu (A; B) wih finie vlue c(a; B) < define feile e of equipmen nd projec clim. find he minimum cu (A ; B ). B n i he e of projec nd equipmen h give mximum profi. c(a; B) = P + ( P i C j ) T i 2B E j 2B {z } ne profi P where P + = Pi i he ol profi chievle if one doe ll he i projec. Nework flow -2 ;

13 Tem eliminion prolem win: (NYY,93), (BOS,89), (BAL,86), (TOR,88) gme o ply: (NYY-BOS,), (NYY-TOR,6), (NYY-BAL,), (BOS-BAL,3), (TOR-BAL,) Queion: Cn Boon end up winning he diviion? Mx-flow formulion uppoe BOS h won ll he remining gme conider nework wih ource nd ink ech pir of em NYY-TOR i node, nd connec i o he ource wih weigh equl o gme o ply ech em BAL i node, nd connec mche o em wih infinie cpciy dd edge eween ech em nd he ink wih weigh W w i, where w i i how mny ime he em h won, nd W i he mo win BOS cn hve if i win ll he re of he gme. Thi enure h ech em doe no ccumule more hn W win. if mx-flow i le hn or equl o he ol numer of remining gme, hen BOS ill h chnce (ince here i wy o diriue he win of he remining gme uch h no em exceed BOS win) if mx-flow i ricly le hn he ol numer of remining gme, hen BOS hould e elimined (ince no mer how we diriue he win of he remining gme, here i no wy ll em cn hve le win hn BOS)

14 Nework flow -4 Minimum co flow moiving exmple: There re n worker nd n jo. Aigning worker i o jo j incur co ij. Mch worker o jo uch h he ol co i minimized (while mximizing he crdinliy of he mching). hi weighed mching prolem i pecil ce of minimum co flow prolem: minimum co flow prolem: given nework N, whoe edge ech hve co per uni flow ij nd n ineger cpciy c ij, find minimum-co flow of vlue v if we chnge hi prolem uch h he co i for uing he edge independen of he flow, hen he prolem ecome NP-hrd (i.e. no known lgorihm cn olve i in polynomil ime) given N, we cn lwy find he mximum flow vlue v, nd we lo know h for he vlue of v > v i i impoile o find flow of vlue v

15 Review of Ford-Fulkeron lgorihm for mx flow prolem 0 0 FLOW R(n,f) Nework flow -

16 Minimum co flow 2/0 4/ Ide: find he minimum co ugmening ph Definiion. The co of n ugmening ph i he um of he co of i edge. When conrucing reidul grph, pu c ij on he revere edge. Definiion. An ugmening cycle c i direced cycle, whoe edge ll hve poiive cpciy. The co of c i he um of he co of direced edge in c. If we hve negive co cycle, hen we cn ugmen he cycle o our flow o ge noher flow of me vlue u mller co. / / /0 co/cpciy FLOW R(n,f) 2/0 / / 4/3 -/2-4/2 /8 2/9 / 4 -/2-4/ /8 Nework flow -6

17 Nework flow -7 Minimum co flow Theorem. A flow f of vlue v h minimum co mong ll flow h hve vlue v if nd only if here i no negive co cycle in he reidul nework R(N ; f ). Proof. ()) Proof y rnpoiion. Suppoe here i negive co cycle, hen we cn dd h cycle o our flow o ge mller co flow of he me vlue. Then he originl flow i no minimum co flow. (() Proof y rnpoiion. Suppoe f i no minimum-co flow. Then, here i cheper flow g of vlue v. The flow g f cn e decompoed ino union of ugmening cycle. One of he cycle h negive co, ince g f h negive co. Thi give wy for checking wheher flow h minimum co or no. How do we find find he minimum co flow? co/cpciy 2/0 / / 4/ /0 flow f 2 2 min-co flow g 2 g-f 2/ / -4/

18 Nework flow -8 Minimum co flow 2/0 4/ Theorem. Le f e minimum co flow of vlue v. And le P e minimum-co ugmening ph in he reidul grph R(N ; f ). Then, ugmening f y dding P give minimum-co flow of vlue v + (le cll i g). Proof. Proof y rnpoiion. Suppoe g w no minimum co flow. Then, from he previou proof, we know h g dmi negive co cycle C in he reidul grph (which w no preen in he reidul grph R(N; f )). Since f dmi no negive-co cycle, here mu e ome edge (i; j ) 2 P wih (j; i) 2 C. Then, P i no he minimum co ugmening ph in R(N; f ), ince P + C i n ugmening ph wih mller co. / / /0 co/cpciy 2 FLOW R(n,f) -2/ / 2/9 -/ -/ 4/ /9 2/8 /4-2/2 -/ -/ -/2 4/ /9

19 Minimum co flow Succeive hore ph lgorihm. Sr wih f = 0. Repe finding minimum co ugmening ph in he reidul grph. Add h flow o curren flow unil he vlue i v.

20 Nework flow -20 Applicion: Communiy deecion in ocil nework Socil nework i nework of people conneced o heir friend Recommending friend i n imporn prcicl prolem oluion : recommend friend of friend oluion 2: deec communiie ide: ue mx-flow min-cu lgorihm o find minimum cu i fil when here re oulier wih mll degree ide2: find priion A nd B h minimize conducnce: min A;B c(a; B) jajjbj

21 Homework Prolem. A locl chriy wihe o orgnize erie of lind de. There re n mle nd n femle uden, nd ech mle uden i o e pired wih femle uden ech evening. Bed on form filled in y he pricipn, he we know which pir of uden re compile. Given hi informion, decrie n lgorihm o find he mximum numer of round of lind de h cn e orgnized, ujec o he following condiion: ech round coni of excly n de, no uden cn pricipe in more hn one de in he me round, no uden encouner noher uden more hn once, nd no wo incompile people re ever mched.

22 Homework Prolem 2. There re n uden in cl. We wn o chooe ue of k uden commiee. There h o e m numer of frehmen, m 2 numer of ophomore, m 3 numer of junior, nd m 4 numer of enior in he commiee. Ech uden i from one of k = m + m 2 + m 3 + m 4 deprmen. Excly one uden from ech deprmen h o e choen for he commiee. We re given li of uden, heir home deprmen, nd heir cl (frehmn, ophomore, junior, enior). Decrie n efficien lgorihm o elec who hould e on he commiee uch h he conrin re ified.

23 Homework Prolem 3. We hve n n n grid like elow. A ue of he cell re poin cell (mrked lck). A monoone ph in he grid r he op-lef cell nd ech cell cn only move o eiher righ or down y one ep (if here exi cell o he righ or elow). Evenully, he ph end he oom-righ cell. The gol i o cover mny poin cell poile, wih ingle monoone ph. Figure : Greedily covering he poin cell wih hree monoone ph. () Decrie n efficien lgorihm for finding monoone ph h cover he mximum numer of poin cell. () Now conider he prolem of covering ll poin cell uing muliple monoone ph. The gol i o cover ll poin cell uing mll numer of ph poile. A greedy heuriic i o ierively pply he ove lgorihm. A ech ep, find he monoone ph h cover mximum numer of poin cell h re no lredy covered, nd repe unil ll poin cell re covered. Prove y couner exmple h hi lgorihm doe no lwy give he opiml oluion. (c) Decrie n efficien lgorihm o compue he mlle e of monoone ph h cover every poin cell.

24 Homework Prolem 4. () Given nework G = (V ; E; ; ), give polynomil ime lgorihm o deermine wheher G h unique minimum cu.