15.082J & 6.855J & ESD.78J October 7, Introduction to Maximum Flows

Transcription

1 5.0J &.55J & ESD.7J Ocober 7, 00 Inroducion o Maximum Flow

2 The Max Flow Problem G = (N,A) x ij = flow on arc (i,j) u ij = capaciy of flow in arc (i,j) = ource node = ink node Maximize v Subjec o j x ij - k x ki = 0 for each i, j x j = v 0 x ij u ij for all (i,j) A.

3 Maximum Flow We refer o a flow x a maximum if i i feaible and maximize v. Our objecive in he max flow problem i o find a maximum flow A max flow problem. Capaciie and a nonopimum flow. 3

4 The feaibiliy problem: find a feaible flow warehoue reailer I here a way of hipping from he warehoue o he reailer o aify demand? 4

5 5 Tranformaion o a max flow problem warehoue warehoue reailer There i a - correpondence wih flow from o wih 4 uni (why 4?) and feaible flow for he ranporaion problem

6 ending flow along - pah One can find a larger flow from o by ending uni of flow along he pah

7 A differen kind of pah One could alo find a larger flow from o by ending uni of flow along he pah ---. (Backward arc have heir flow decreaed.) Decreaing flow in (, ) i mahemaically equivalen o ending flow in (, ) w.r.. node balance conrain. 7

8 The Reidual Nework i u ij x ij j u - x ij ij i x ij j The Reidual Nework G(x) We le r ij denoe he reidual capaciy of arc (i,j)

9 A Ueful Idea: Augmening Pah An augmening pah i a pah from o in he reidual nework. The reidual capaciy of he augmening pah P i (P) = min{r ij : (i,j) P}. To augmen along P i o end (P) uni of flow along each arc of he pah. We modify x and he reidual capaciie appropriaely. r ij := r ij - (P) and r ji := r ji + (P) for (i,j) P

10 The Ford Fulkeron Maximum Flow Algorihm x := 0; creae he reidual nework G(x); while here i ome direced pah from o in G(x) do le P be a pah from o in G(x); := (P); end uni of flow along P; updae he r'; Ford- Fulkeron Algorihm Animaion 0

11 To prove correcne of he algorihm Invarian: a each ieraion, here i a feaible flow from o. Finiene (auming capaciie are inegral and finie): The reidual capaciie are alway ineger valued The reidual capaciie ou of node decreae by a lea one afer each updae. Correcne If here i no augmening pah, hen he flow mu be maximum. max-flow min-cu heorem.

12 Inegraliy Aume ha all daa are inegral. Lemma: A each ieraion all reidual capaciie are inegral. Proof. I i rue a he beginning. Aume i i rue afer he fir k- augmenaion, and conider augmenaion k along pah P. The reidual capaciy of P i he malle reidual capaciy on P, which i inegral. Afer updaing, we modify reidual capaciie by 0, or, and hu reidual capaciie ay inegral.

13 Theorem. The Ford-Fulkeron Algorihm i finie Proof. The capaciy of each augmening pah i a lea r j decreae for ome j. So, he um of he reidual capaciie of arc ou of decreae a lea per ieraion. Number of augmenaion i O(nU), where U i he large capaciy in he nework. 3

14 Menal Break Wha are agle? The plaic hing on he end of hoelace. How fa doe he quarz cryal in a wach vibrae? Abou 3,000 ime per econd. If Barbie (he doll) were life ize and 5 9 all, how big would her wai be? inche. Incidenally, Barbie full name i Barbara Millicen Rober

15 Menal Break True or fale. In Alaka i i illegal o hoo a mooe from a helicoper or any oher flying vehicle. True. True or fale. In Ahen, Georgia, a driver licene can be aken away by law if he driver i deemed eiher unbahed or poorly dreed. Fale. However, i i rue for Ahen, Greece. In Helinki, Finland ha don give parking icke o illegally parked car. Wha do hey do inead? They deflae he ire of he car.

16 To be proved: If here i no augmening pah, hen he flow i maximum 0, 9,,, 0,7 G(x) = reidual nework for flow x. x* = final flow If here i a direced pah from i o j in G, we wrie i j. S* = { j : j in G(x*)} T* = N\S*

17 Lemma: here i no arc in G(x*) from S* o T* S* = { j : j in G(x*)} T* = N\S* i Proof. If here were uch an arc (i, j), hen j would be in S*. j We will ue hi Lemma in lide. 7

18 Cu Dualiy Theory 0, 9,,, 0,7 An (,)-cu in a nework G = (N,A) i a pariion of N ino wo dijoin ube S and T uch ha S and T, e.g., S = {, } and T = {, }. The capaciy of a cu (S,T) i CAP(S,T) = i S j T u ij

19 The flow acro a cu We define he flow acro he cu (S,T) o be F x (S,T) = i S j T x ij - i S j T x ji 0, 9, 0, 9,,,, 0,7, 0,7 If S = {, }, hen F x (S,T) = + + = 5 If S = {, }, hen F x (S,T) = = 5 9

20 Max Flow Min Cu Theorem. (Max-flow Min-Cu). The maximum flow value i he minimum value of a cu. Proof. The proof will rely on he following hree lemma: Lemma. For any flow x, and for any - cu (S, T), he flow ou of equal F x (S, T). Lemma. For any flow x, and for any - cu (S, T), F x (S, T) CAP(S, T). Lemma 3. Suppoe ha x* i a feaible - flow wih no augmening pah. Le S* = {j : j in G(x*)} and le T* = N\S. Then F x* (S*, T*) = CAP(S*, T*). 0

21 Proof of Theorem (uing he 3 lemma) Le x be a maximum flow Le v be he maximum flow value Le x* be he final flow. Le v* be he flow ou of node (for x*) Le S* be node reachable in G(x*) from. Le T* = N\S*.. v* v by definiion of v. v = F x (S*, T*) by Lemma. 3. F x (S*, T*) CAP(S*, T*) by Lemma. 4. v* = F x* (S*, T*) = CAP(S*, T*) by Lemma,3. Thu all inequaliie are equaliie and v* = v.

22 Proof of Lemma Proof. Add he conervaion of flow conrain for each node i S - {} o he conrain ha he flow leaving i v. The reuling equaliy i F x (S,T) = v. 0, 9, 0, 9,,,, 0,7, 0,7 x + x = v x + x x = 0 x + x + x = v x + x = v x x x = 0 x -x + x = v

23 Proof of Lemma Proof. If i S, and j T, hen x ij u ij. If i T, and j S, hen x ij 0. F x (S,T) = i S j T x ij - i S j T x ji CAP(S,T) = i S j T u ij - i S j T 0 0, 9, 0, 9,,,, 0,7, 0,7 CAP(S, T) = 5 CAP(S, T) = 3

24 Proof of Lemma 3. We have already een ha here i no arc from S* o T* in G(x*). i S* and j T* x* ij = u ij and x* ji = 0 i x* ij = u ij x* ji = 0 j Oherwie, here i an arc (i, j) in G(x*) Therefore F x* (S*, T*) = CAP(S*, T*) 4

25 Review Corollary. If he capaciie are finie ineger, hen he Ford-Fulkeron Augmening Pah Algorihm erminae in finie ime wih a maximum flow from o. Corollary. If he capaciie are finie raional number, hen he Ford-Fulkeron Augmening Pah Algorihm erminae in finie ime wih a maximum flow from o. (why?) Corollary. To obain a minimum cu from a maximum flow x*, le S* denoe all node reachable from in G(x), and T* = N\S* Remark. Thi doe no eablih finiene if u ij = or if capaciie may be irraional. 5

26 A imple and very bad example M M M M

27 Afer augmenaion M- M M M- 7

28 Afer wo augmenaion M- M- M- M-

29 Afer 3 augmenaion M- M- M- M- 9

30 And o on 30

31 Afer M augmenaion M M M M 3

32 An even wore example In Exercie.4, here i an example ha ake an infinie number of augmenaion on irraional daa, and doe no converge o he correc flow. Bu we hall oon ee how o olve max flow in a polynomial number of operaion, even if daa can be irraional. 3

33 Summary and Exenion. Augmening pah heorem. Ford-Fulkeron Algorihm 3. Dualiy Theory. 4. Nex Lecure: Polynomial ime varian of FF algorihm Applicaion of Max-Flow Min-Cu 33

34 MIT OpenCoureWare hp://ocw.mi.edu 5.0J /.55J / ESD.7J Nework Opimizaion Fall 00 For informaion abou ciing hee maerial or our Term of Ue, vii: hp://ocw.mi.edu/erm.