CSE 202: Design and Analysis of Algorithms Lecture 8
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1 CSE 202: Deign and Analyi of Algorihm Lecure 8 Inrucor: Kamalika Chaudhuri
2 La Cla: Max Flow Prolem The Max Flow Prolem: Given direced graph G=(V,E), ource, ink, edge capaciie c(e), find an - flow of maximum ize An - flow i a funcion f: E R uch ha: - 0 <= f(e) <= c(e), for all edge e - flow ino node v = flow ou of node v, for all node v excep and, f(e) = e ino v e ou of v f(e) Size of flow f = Toal flow ou of = oal flow ino 2/2 u /3 /2 Size of f = 3 / v 2/5
3 La Cla: Flow and Cu The Max Flow Prolem: Given direced graph G=(V,E), ource, ink, edge capaciie c(e), find an - flow of maximum ize L* 2/2 / u v /2 /3 2/5 R* An - Cu pariion node ino group = (L, R).. in L, in R Capaciy of a cu (L, R) = c(u, v) Flow acro (L,R) = (u,v) E,u L,v R (u,v) E,u L,v R f(u, v) f(v, u) (v,u) E,u L,v R Size of f = 3 Propery: For any flow f, any - cu (L, R), ize(f) <= capaciy(l, R) Max-Flow <= Min-Cu Thu, a Min Cu i a cerificae of opimaliy for a flow Cu Flow
4 La Cla: Ford-Fulkeron Algorihm FF Algorihm: Sar wih zero flow Repea: Find a pah from o along which flow can e increaed Increae he flow along ha pah Example G: a In any ieraion, we have ome flow f and we are rying o improve i. How o do hi? : Conruc a reidual graph G f ( wha lef o ake? ) G f = (V, E f ) where E f E U E R For any (u,v) in E or E R, c f (u,v) = c(u,v) f(u,v) + f(v,u) [ignore edge wih zero c f : don pu hem in E f ] f: G f : a a 2: Find a pah from o in G f 3: Increae flow along hi pah, a much a poile
5 Analyi: Correcne FF algorihm give u a valid flow. Bu i i he maximum poile flow? Conider final reidual graph G f = (V, E f ) Le L = node reachale from in G f and le R = re of node = V L So L and R L z x w y R Edge from L o R mu e a full capaciy Edge from R o L mu e empy Therefore, flow acro cu (L,R) i c(u, v) (u,v) E,u L,v R Thu, ize(f) = capaciy(l,r) Recall: for any flow and any cu, ize(flow) <= capaciy(cu) Therefore f i he max flow and (L,R) i he min cu! Cu Thu, Max Flow = Min Cu Flow
6 An Oervaion: Inegraliy Inegral Flow: A flow f i inegral if f(e) i an ineger for all e Example: / a / 0.5/ a 0.5/ / v v / 0/ 0/ 0.5/ 0.5/ An Inegral Flow A Fracional Flow
7 An Oervaion: Inegraliy Inegral Flow: A flow f i inegral if f(e) i an ineger for all e Example: / a / 0.5/ a 0.5/ / v v / 0/ 0/ 0.5/ 0.5/ An Inegral Flow A Fracional Flow Propery: If all edge capaciie are ineger, hen, here i a max flow f which i inegral.
8 An Oervaion: Inegraliy Inegral Flow: A flow f i inegral if f(e) i an ineger for all e Example: / a / 0.5/ a 0.5/ / v v / 0/ 0/ 0.5/ 0.5/ An Inegral Flow A Fracional Flow Propery: If all edge capaciie are ineger, hen, here i a max flow f which i inegral. Proof: If he edge capaciie are ineger, hen, he FF algorihm alway find an inegral flow The FF algorihm alo alway find a max flow.
9 An Oervaion: Inegraliy Inegral Flow: A flow f i inegral if f(e) i an ineger for all e Example: / a / 0.5/ a 0.5/ / v v / 0/ 0/ 0.5/ 0.5/ An Inegral Flow A Fracional Flow Propery: If all edge capaciie are ineger, hen, here i a max flow f which i inegral. Proof: If he edge capaciie are ineger, hen, he FF algorihm alway find an inegral flow The FF algorihm alo alway find a max flow. Noe: All max flow are no necearily inegral flow!
10 Analyi: efficiency FF Algorihm: Sar wih zero flow Repea: Find a pah from o along which flow can e increaed Increae he flow along ha pah How many ieraion are needed o reach he maximum flow? Example: a A hillcliming procedure a Flow ize: max flow a Each ieraion i fa (O( E ) ime). 0 #ieraion can e Max Capaciy
11 Analyi: efficiency FF Algorihm: Sar wih zero flow Repea: Find a pah from o along which flow can e increaed Increae he flow along ha pah A hillcliming procedure Flow ize: max flow How many ieraion are needed o reach he maximum flow? Example: a Each ieraion i fa (O( E ) ime). 0 #ieraion can e Max Capaciy (wih ineger capaciie)
12 How o improve he efficiency? Ford-Fulkeron Syle Algorihm: Edmond Karp Capaciy Scaling Preflow-Puh
13 Edmond Karp Bad Example: FF Algorihm: Sar wih zero flow Repea: Find a pah from o along which flow can e increaed Increae he flow along ha pah a Bad Pah Sequence: (, a,, ), (,, a, ), (, a,, ),...
14 Edmond Karp Bad Example: FF Algorihm: Sar wih zero flow Repea: Find a pah from o along which flow can e increaed Increae he flow along ha pah a Bad Pah Sequence: (, a,, ), (,, a, ), (, a,, ),... EK Pah Selecion: Find he hore pah along which flow can e increaed (hore pah = hore in erm of #edge)
15 Edmond Karp EK Algorihm: Sar wih zero flow Repea: Find he hore pah from o along which flow can e increaed Increae he flow along ha pah Bad Example for FF: a f a Ieraion
16 Edmond Karp EK Algorihm: Sar wih zero flow Repea: Find he hore pah from o along which flow can e increaed Increae he flow along ha pah Bad Example for FF: a f a Gf a Ieraion
17 Edmond Karp EK Algorihm: Sar wih zero flow Repea: Find he hore pah from o along which flow can e increaed Increae he flow along ha pah Bad Example for FF: a f a Gf a Ieraion f a Ieraion 2
18 Edmond Karp EK Algorihm: Sar wih zero flow Repea: Find he hore pah from o along which flow can e increaed Increae he flow along ha pah Bad Example for FF: a f a Gf a Ieraion f a Gf a Ieraion 2
19 Edmond Karp FF Algorihm: Sar wih zero flow Repea: Find a pah from o along which flow can e increaed Increae he flow along ha pah Bad Example for FF: a Bad Pah Sequence: (, a,, ), (,, a, ), (, a,, ),... EK Pah Selecion: Find he hore pah along which flow can e increaed (hore pah = hore in erm of #edge) I can e hown ha hi require only O( V E ) ieraion (Proof no in hi cla)
20 Edmond Karp FF Algorihm: Sar wih zero flow Repea: Find a pah from o along which flow can e increaed Increae he flow along ha pah Bad Example for FF: a Bad Pah Sequence: (, a,, ), (,, a, ), (, a,, ),... EK Pah Selecion: Find he hore pah along which flow can e increaed (hore pah = hore in erm of #edge) I can e hown ha hi require only O( V E ) ieraion (Proof no in hi cla) Running Time: O( V E 2 )
21 How o improve he efficiency? Ford-Fulkeron Syle Algorihm: Edmond Karp Capaciy Scaling Preflow-Puh
22 Capaciy Scaling Bad Example: FF Algorihm: Sar wih zero flow Repea: Find a pah from o along which flow can e increaed Increae he flow along ha pah Bad Pah Sequence: (, a,, ), (,, a, ), (, a,, ),... a Capaciy Scaling: Find pah of high capaciy fir eween and
23 Capaciy Scaling Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Example: For f = 0 G a Gf(0) a
24 Capaciy Scaling: Correcne Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery: If all edge capaciie are ineger, algorihm oupu a max flow Proof: A D=, Gf(D) = Gf. So on erminaion, Gf(D) ha no more pah from o
25 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime
26 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E D E max flow curren
27 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Proof: Le L = node reachale from in G f (D) and le R = re of node = V L L R D E max flow curren
28 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Proof: Le L = node reachale from in G f (D) and le R = re of node = V L #edge in Gf(D) in he (L, R) cu = 0 D E L R max flow curren
29 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Proof: Le L = node reachale from in G f (D) and le R = re of node = V L #edge in Gf(D) in he (L, R) cu = 0 #edge in Gf in he (L,R) cu <= E D E L R max flow curren
30 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Proof: Le L = node reachale from in G f (D) and le R = re of node = V L #edge in Gf(D) in he (L, R) cu = 0 #edge in Gf in he (L,R) cu <= E D E Capaciy of each uch edge < D L R max flow curren
31 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Proof: Le L = node reachale from in G f (D) and le R = re of node = V L #edge in Gf(D) in he (L, R) cu = 0 #edge in Gf in he (L,R) cu <= E D E Capaciy of each uch edge < D L R Thu, ize(max flow) <= capaciy(l,r) <= ize(f) + D E max flow curren
32 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Propery 3: For any D, #ieraion of loop 2 in he D-caling phae <= 2 E
33 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Propery 3: For any D, #ieraion of loop 2 in he D-caling phae <= 2 E Proof: Afer previou (2D-caling) phae, ize(max flow) <= ize(curren flow) + 2D E
34 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Propery 3: For any D, #ieraion of loop 2 in he D-caling phae <= 2 E Proof: Afer previou (2D-caling) phae, ize(max flow) <= ize(curren flow) + 2D E Each ieraion of loop 2 increae flow ize y a lea D
35 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Propery 3: For any D, #ieraion of loop 2 in he D-caling phae <= 2 E Proof: Afer previou (2D-caling) phae, ize(max flow) <= ize(curren flow) + 2D E Each ieraion of loop 2 increae flow ize y a lea D Therefore, a mo 2 E ieraion in he D-caling phae
36 Capaciy Scaling: Running Time 2 Cmax = max capaciy edge. Sar wih D = Cmax Sar wih zero flow While D >=, repea: Gf(D) = D-reidual graph D caling phae While here i a pah from o in Gf(D) along which flow can e increaed Increae flow along ha pah Updae Gf(D) D = D/2 D-Reidual Graph: Sugraph of reidual graph wih only edge wih capaciy >= D Propery : While loop i execued + log2 Cmax ime Propery 2: A he end of a D-caling phae, ize(max flow) <= ize(curren flow) + D E Propery 3: For any D, #ieraion of loop 2 in he D-caling phae <= 2 E Toal Running Time: O( E 2 ( + log2 Cmax)) ( Recall: Time o find a flow pah in a reidual graph = O( E ) )
37 How o improve he efficiency? Ford-Fulkeron Syle Algorihm: Edmond Karp Capaciy Scaling Preflow-Puh
38 Preflow-Puh Main Idea: - Each node ha a lael, which i a poenial - Roue flow from high o low poenial Lael v w Idea: Roue flow along lue edge
39 Preflow Preflow: A funcion f: E R i a preflow if:. Capaciy Conrain: 0 <= f(e) <= c(e) 2. Inead of conervaion conrain: e ino v Exce(v) = f(e) f(e) e ou of v e ino v e ou of v f(e) 0 f(e) Example a exce = a 0 G f 0 0 exce =
40 Preflow-Puh: Two Operaion Preflow: A funcion f: E R i a preflow if:. Capaciy Conrain: 0 <= f(e) <= c(e) 2. Inead of conervaion conrain: e ino v Exce(v) = f(e) f(e) e ou of v e ino v e ou of v f(e) 0 f(e) l v w Laeling h aign a non-negaive ineger lael h(v) o all v in V Puh(v, w): Applie if exce(v) > 0, h(w) < h(v), (v, w) in Ef q = min(exce(v), cf(v, w)) Add q o f(v, w) Relael(v): Applie if exce(v) > 0, for all w. (v, w) in Ef, h(w) >= h(v) Increae l(v) y
41 Pre-Flow Puh: The Algorihm Sar wih laeling: h() = n, h() = 0, h(v) = 0, for all oher v Sar wih preflow f: f(e) = c(e) for e = (, v), f(e) = 0, for all oher edge e While here i a node (oher han ) wih poiive exce Pick a node v wih exce(v) > 0 If here i an edge (v, w) in Ef uch ha puh(v, w) can e applied Puh(v, w) Ele Relael(v) Puh(v, w): Applie if exce(v) > 0, h(w) < h(v), (v, w) in Ef q = min(exce(v), cf(v, w)) Add q o f(v, w) Relael(v): Applie if exce(v) > 0, for all w. (v, w) in Ef, h(w) >= h(v) Increae h(v) y
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