CSE 521: Design & Analysis of Algorithms I

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1 CSE 52: Deign & Analyi of Algorihm I Nework Flow Paul Beame

2 Biparie Maching Given: A biparie graph G=(V,E) M E i a maching in G iff no wo edge in M hare a verex Goal: Find a maching M in G of maximum poible ize 2

3 Biparie Maching 3

4 Biparie Maching 4

5 The Nework Flow Problem 5 a 4 3 x b 4 y c 5 z How much uff can flow from o? 5

6 Biparie maching a a pecial cae of flow a x b y c z 6

7 Ne Flow: Formal Definiion Given: A digraph G = (V,E) Two verice, in V (ource & ink) A capaciy c(u,v) 0 for each (u,v) E (and c(u,v) = 0 for all non-edge (u,v)) Find: A flow funcion f: E R.., for all u,v: 0 f(u,v) c(u,v) [Capaciy Conrain] if u,, i.e. f ou (u)=f in (u) [Flow Conervaion] Maximizing oal flow ν(f) = f ou () Noaion: in f (v) = e = f(u, v) (u,v) E ou f (v) = e = f(v, w) (v,w ) E 7

8 Example: A Flow Funcion flow/capaciy, no /2 2/3 u f in (u)=f(,u)=2=f(u,)=f ou (u) 8

9 Example: A Flow Funcion 4/5 a 3/4 /3 x 3/3 7 6 b 4 y /4 7 /6 c 5 z No hown: f(u,v) if = 0 Noe: max flow 4 ince f i a flow funcion, wih ν(f) = 4 9

10 Max Flow via a Greedy Alg? While here i an pah in G Pick uch a pah, p Find c, he min capaciy of any edge in p Subrac c from all capaciie on p Delee edge of capaciy 0 Thi doe NOT alway find a max flow: a 2 b 3 2 If pick b a fir, flow uck a 2. Bu flow 3 poible. 0

11 A Brief Hiory of Flow bound n = # of verice m= # of edge U = Max capaciy Source: Goldberg & Rao, FOCS 97

12 Greed Reviied: Reidual Graph & Augmening Pah a 2/2 / a +/2 2/3 /3 2/2 b 2/2 b / a 2 a b Reidual Graph 2 b 2

13 Greed Reviied: An Augmening Pah / a +/2 /3 2/2 b / a b New Reidual Graph 3

14 Reidual Capaciy The reidual capaciy (w.r.. f) of (u,v) i c f (u,v) = c(u,v) - f(u,v) if f(u,v) c(u,v) and c f (u,v)=f(v,u) if f(v,u)>0 4/5 a 3/4 /3 x 3/3 7 6 b 4 y /4 7 /6 c 5 z e.g. c f (,b)=7; c f (a,x) = ; c f (x,a) = 3 4

15 Reidual Graph & Augmening Pah The reidual graph (w.r.. f) i he graph G f = (V,E f ), where E f = { (u,v) c f (u,v) > 0 } Two kind of edge Forward edge f(u,v)<c(u,v) o c f (u,v)=c(u,v)-f(u,v)>0 Backward edge f(u,v)>0 o c f (v,u) -f(v,u)=f(u,v)>0 An augmening pah (w.r.. f) i a imple pah in G f. 5

16 A Reidual Nework 4/5 a 3/4 /3 x 3/3 7 6 b 4 y /4 7 /6 c 5 z a b c x y 3 z

17 An Augmening Pah 4/5 a 3/4 /3 x 3/3 7 6 b 4 y /4 7 /6 c 5 z a b c x y 3 z

18 Augmening A Flow augmen(f,p) c P min (u,v) P c f (u,v) boleneck(p) for each e P if e i a forward edge hen increae f(e) by c P ele (e i a backward edge) decreae f(e) by c P endif endfor reurn(f) 8

19 9 Augmening A Flow a b c x y z 4/5 /6 7 3/4 /3 4 /5 3/3 /7 /6 4 a b c x y z 4/ /4 / /3 7 /6 /4 a b c x y z

20 Claim 7. If G f ha an augmening pah P, hen he funcion f =augmen(f,p) i a legal flow. Proof: f and f differ only on he edge of P o only need o conider uch edge (u,v) 20

21 Proof of Claim 7. If (u,v) i a forward edge hen f (u,v)=f(u,v)+c P f(u,v)+c f (u,v) = f(u,v)+c(u,v)-f(u,v) =c(u,v) If (u,v) i a backward edge hen f and f differ on flow along (v,u) inead of (u,v) f (v,u)=f(v,u)-c P f(v,u)-c f (u,v) = f(v,u)-f(v,u)=0 Oher condiion like flow conervaion ill me 2

22 Ford-Fulkeron Mehod Sar wih f=0 for every edge While G f ha an augmening pah, augmen Queion: Doe i hal? Doe i find a maximum flow? How fa? 22

23 Obervaion abou Ford-Fulkeron Algorihm A every age he capaciie and flow value are alway ineger (if hey ar ha way) The flow value ν(f )=ν(f)+c P >ν(f) for f =augmen(f,p) Since edge of reidual capaciy 0 do no appear in he reidual graph Le C=Σ (,u) E c(,u) ν(f) C F-F doe a mo C round of augmenaion ince flow are ineger and increae by a lea per ep 23

24 Running Time of Ford-Fulkeron For f=0, G f =G Finding an augmening pah in G f i graph earch O(n+m)=O(m) ime Augmening and updaing G f i O(n) ime Toal O(mC) ime Doe i find a maximum flow? Need o how ha for every flow f ha in maximum G f conain an --pah 24

25 Cu A pariion (A,B) of V i an --cu if A, B Capaciy of cu (A,B) i {} c= a b c {,b,c} c=5 4 x y z V-{} c=6 c(a,b) {,x} c=2 = a b c u A v B c(u,v) x 3 y z 25

26 Convenien Definiion f ou (A)=Σ v A, w A f (v,w) f in (A)=Σ v A, u A f (u,v) 26

27 Claim 7.6 and 7.8 For any flow f and any cu (A,B), he ne flow acro he cu equal he oal flow, i.e., ν(f) = f ou (A)-f in (A), and he ne flow acro he cu canno exceed he capaciy of he cu, i.e. f ou (A)-f in (A) c(a,b) Corollary : Max flow Min cu Cu Cap = 3 Ne Flow = Cu Cap = 2 Ne Flow = 27

28 Proof of Claim 7.6 Conider a e A wih A, A f ou (A)-f in (A) =Σ v A, w A f (v,w)-σ v A, u A f (u,v) We can add flow value for edge wih boh endpoin in A o boh um and hey would cancel ou o f ou (A)-f in (A)= Σ v A, w V f (v,w)-σ v A, u V f (u,v) = Σ v A (Σ w V f (v,w) - Σ u V f (u,v)) =Σ v A fou (v) - f in (v) =f ou ()-f in () ince all oher verice have f ou (v)=f in (v) ν(f) = f ou () and f in ()=0 28

29 Proof of Claim 7.8 ν(f)=f ou (A)-f in (A) f ou (A) = Σ v A, w A f (v,w) Σ v A, w A c(v,w) Σ v A, w B c(v,w) =c(a,b) 29

30 Max Flow / Min Cu Theorem Claim 7.9 For any flow f, if G f ha no augmening pah hen here i ome --cu (A,B) uch ha ν(f)=c(a,b) (proof on nex lide) We know by Claim 7.6 & 7.8 ha any flow f aifie ν(f ) c(a,b) and we know ha F-F run for finie ime unil i find a flow f aifying condiion of Claim 7.9 Therefore by 7.9 for any flow f, ν(f ) ν(f) Corollary () F-F compue a maximum flow in G (2) For any graph G, he value ν(f) of a maximum flow = minimum capaciy c(a,b) of any --cu in G 30

31 Claim 7.9 Le A = { u an pah in G f from o u } B = V - A; A, B A B auraed f(u,v)=c(u,v) u x v w no flow f(w,u)=0 Thi i rue for every edge croing he cu, i.e. ou f ( A) = f( u,v) = c( u,v) = c( A,B) and f in (A)=0 o u A v B u A v B ν(f)=f ou (A)-f in (A)=c(A,B) 3

32 Flow Inegraliy Theorem If all capaciie are ineger The max flow ha an ineger value Ford-Fulkeron mehod find a max flow in which f(u,v) i an ineger for all edge (u,v) 0.5/ 0.5/ / 0.5/ 0.5/ 32

33 Corollarie & Fac If Ford-Fulkeron erminae, hen i found a max flow. I will erminae if c(e) ineger or raional (bu may no if hey re irraional). However, may ake exponenial ime, even wih ineger capaciie: c a c c = 0 9, ay c b c 33

34 Biparie maching a a pecial cae of flow a x b y c z Ineger flow implie each flow i ju a ube of he edge Therefore flow correpond o a maching O(mC)=O(nm) running ime 34

35 Conequence of Ford-Fulkeron: Hall Theorem Def: For a graph G=(V,E) and A V le he neighborhood of A be A} Γ(A)={w V (v,w) E for ome v Hall Theorem: A graph G=(V,E) wih bipariion V=X Y where X = Y ha a perfec maching if and only if for every e A X we have Γ(A) A. Proof: : If Γ(A) < A for ome e A 35

36 Hall Theorem Proof: Par a b c A w x Γ(A) Γ(A) X - A d y z Min cu X - A + Γ(A) < X o maching i no perfec 36

37 Hall Theorem Proof: Par 2 A a b w c x d y z If no perfec maching here i ome -cu (A,B) wih cu value < X Le A=par of A in X 37

38 Hall Theorem Proof: Par 2 A a b w c x d y Modify A o include all verice inγ(a) Value of mincu canno go up o ill < X # of edge from cu i X - A # of edge o cu i Γ(A) Toal i X - A + Γ(A) bu < X o Γ(A) < A z 38

39 Conequence of Ford-Fulkeron: Edge-dijoin pah Given a direced graph G=(V,E) and verice and find a maximum e of edge-dijoin (imple) pah from o. Lemma: Any 0- flow f in he nework flow graph on G=(V,E),, wih all capaciie= conain a e of ν(f) edgedijoin imple pah from o. Proof: Follow he edge of poiive flow hrough he graph. Flow conervaion 39

40 Edge-dijoin pah in direced graph Since a collecion of k edge-dijoin pah i alway a legal 0- flow of value k, by Maxflow=Mincu we have: Direced Menger Theorem: The maximum number of edge-dijoin pah from o i equal o he minimum number of edge ha need o be cu o eparae from. We can ue Ford-Fulkeron o find he flow in O(mn) ime (ince capaciie are all ) and hen prune he flow a on previou lide in O(m+n) exra ime. 40

41 Edge-dijoin pah in undireced graph Given an undireced graph G=(V,E) and node and, we can modify G o creae a direced graph G =(V,E ) where we include boh (u,v) and (v,u) in E whenever {u,v} i in E. Obervaion: If in a flow f we have f(v,u)=f(u,v) >0 hen we can e f(v,u)=f(u,v)=0 and keep he ame flow value Algorihm: Run F-F on G in O(mn) ime and hen remove flow cycle in O(m+n) ime. Undireced Menger Theorem: The maximum number of edge-dijoin pah from o i equal o he minimum number of edge ha need o be cu o eparae from. 4

42 Polynomial running ime? F-F ue O(mC) ime where C i he oal capaciy leaving. If all capaciie are a mo ome upper bound U hen C < nu bu each inpu capaciy ake only log 2 U bi o repreen o he running ime i no polynomial in he inpu ize We will give wo modificaion ha yield polynomial-ime algorihm for general nework flow problem 42

43 Capaciy-caling algorihm General idea: Chooe augmening pah P wih large capaciy c P Can augmen flow along a pah P by any amoun b c P Ford-Fulkeron ill work Ge a flow ha i maximum for he highorder bi fir and hen add more bi laer 43

44 Capaciy Scaling 5 a 4 3 x b 4 y c 5 z 44

45 Capaciy Scaling 0 a 00 0 x 0 0 b y 0 c 0 z 45

46 Capaciy Scaling Bi 0 a 00 0 x 0 0 b y 0 c 0 z Capaciy on each edge i a mo (eiher 0 or ime =4) 46

47 Capaciy Scaling Bi 0 a 00 0 x 0 / /0 b / y / /0 c /0 z O(nm) ime 47

48 Capaciy Scaling Bi 2 0 a 00 0 x 0 0/ b 0/0 0/ y 0/ 0/0 c 0/0 z Reidual capaciy acro min cu i a mo m (eiher 0 or ime =2) 48

49 Capaciy Scaling Bi 2 0/0 a 0/00 0/0 x 0/0 0/ b 0/0 0/ y / 0/0 c 0/0 z Reidual capaciy acro min cu i a mo m m augmenaion 49

50 Capaciy Scaling Bi 3 00/0 a 00/00 00/0 x 00/0 00/ b 00/00 y 0/ 00/ /0 c 00/0 z Reidual capaciy acro min cu i a mo m (eiher 0 or ime =) 50

51 Capaciy Scaling Bi 3 a 0/0 0/ b 0/0 c 00/00 00/00 00/00 0/0 0/0 00 x / y z 0/0 00/0 Afer m augmenaion 5

52 Capaciy Scaling Final 5/5 a 2/4 3/3 x 2/3 5/7 b 4/4 y 7/7 5/6 / 4 6/6 c 5/5 z 52

53 Capaciy Scaling Min Cu 5/5 a 2/4 3/3 x 2/3 5/7 b 4/4 y 7/7 5/6 / 4 6/6 c 5/5 z 53

54 Toal ime for capaciy caling log 2 U round where U i large capaciy A mo m augmenaion per round Le c i be he capaciie ued in he i h round and f i be he maxflow found in he i h round For any edge (u,v), c i+ (u,v) 2c i (u,v)+ i+ round ar wih flow f = 2 f i Le (A,B) be a min cu from he i h round ν(f i )=c i (A,B) o ν(f)=2c i (A,B) ν(f i+ ) c i+ (A,B) 2c i (A,B)+m =ν(f)+m O(m) ime per augmenaion Toal ime O(m 2 log U) 54

55 Edmond-Karp Algorihm Ue a hore augmening pah (via Breadh Fir Search in reidual graph) Time: O(n m 2 ) 55

56 BFS/Shore Pah Lemma Diance from in G f i never reduced by: Deleing an edge Proof: no new (hence no horer) pah creaed Adding an edge (u,v), provided v i nearer han u Proof: BFS i unchanged, ince v viied before (u,v) examined v a back edge u 56

57 Key Lemma Le f be a flow, G f he reidual graph, and P a hore augmening pah. Then no verex i cloer o afer augmenaion along P. Proof: Augmenaion along P only delee forward edge, or add back edge ha go o previou verice along P 57

58 Augmenaion v BFS G: G f G f 5/9 x 3/0 u 0/5 x u x u 3/3 2/5 v v v 58

59 Theorem The Edmond-Karp Algorihm perform O(mn) flow augmenaion Proof: Call (u,v) criical for augmening pah P if i cloe o having min reidual capaciy I will diappear from G f afer augmening along P In order for (u,v) o be criical again he (u,v) edge mu re-appear in G f bu ha will only happen when he diance o u ha increaed by 2 (nex lide) I won be criical again unil farher from o each edge criical a mo n/2 ime 59

60 Criical Edge in G f >c P >c P Shore - pah P in G f c P u v x w Afer augmening along P >0 >0 c P criical edge d f (,v)=d f (,u)+ ince hi i a hore pah u v x w For (u,v) o be criical laer for ome flow f i mu be in G f o mu have augmened along a hore pah conaining (v,u) u v x w Then we mu have d f (,u)=d f (,v)+ d f (,v)+=d f (,u)+2 60

61 Corollary Edmond-Karp run in O(nm 2 ) ime 6

62 Projec Selecion a.k.a. The Srip Mining Problem Given a direced acyclic graph G=(V,E) repreening precedence conrain on ak (a ak poin o i predeceor) a profi value p(v) aociaed wih each ak v V (may be poiive or negaive) Find a e A V of ak ha i cloed under predeceor, i.e. if (u,v) E and u A hen v A, ha maximize Profi(A)=Σ v A p(v) 62

63 Projec Selecion Graph Each ak poin o i predeceor ak 63

64 Exended Graph

65 Exended Graph G For each verex v If p(v) 0 add (,v) edge wih capaciy p(v) If p(v)<0 add (v,) edge wih capaciy p(v)

66 Exended Graph G Wan o arrange capaciie on edge of G o ha for minimum --cu (S,T) in G, he e A=S-{} aifie precedence conrain ha maximum poible profi in G Cu capaciy wih S={} i ju C=Σ v: p(v) 0 p(v) Profi(A) C for any e A To aify precedence conrain don wan any original edge of G going forward acro he minimum cu Tha would correpond o a ak in A=S-{} ha had a predeceor no in A=S-{} Se capaciy of each of he edge of G o C+ The minimum cu ha ize a mo C 66

67 Exended Graph G Capaciy C

68 Exended Graph G Cu value = =3+3 +C

69 Projec Selecion Claim Any --cu (S,T) in G uch ha A=S-{} aifie precedence conrain ha capaciy c(s,t)=c - Σ v A p(v) = C - Profi(A) Corollary A minimum cu (S,T) in G yield an opimal oluion A=S-{} o he profi elecion problem Algorihm Compue maximum flow f in G, find he e S of node reachable from in G f and reurn S-{} 69

70 Proof of Claim A=S-{} aifie precedence conrain No edge of G croe forward ou of A ince hoe edge have capaciy C+ Only forward edge cu are of he form (v,) for v A or (,v) for v A The (v,) edge for v A conribue Σ v A:p(v)<0 -p(v) = - Σ v A:p(v)<0 p(v) The (,v) edge for v A conribue Σ v A: p(v) 0 p(v)=c-σ v A: p(v) 0 p(v) Therefore he oal capaciy of he cu i c(s,t) = C - Σ v A p(v) =C-Profi(A) 70

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