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Ma ksimum akış

Ma ksimum akış. Akış problemi. s kaynak köşesidir Akacak olanın (petrol, gaz veya veri) üretildiği yerdir t hedef köşesidir Akacak olanın ulaşması gereken yerdir Diğer tüm köşeler akacak olanın geçeceği ara köşelerdir Amaç : Kaynaktan hedefe olabildiğince maksimum akışı sağlamaktır.

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Ma ksimum akış

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  1. Maksimum akış

  2. Akış problemi • s kaynak köşesidir • Akacak olanın (petrol, gaz veya veri) üretildiği yerdir • t hedef köşesidir • Akacak olanın ulaşması gereken yerdir • Diğer tüm köşeler akacak olanın geçeceği ara köşelerdir • Amaç: Kaynaktan hedefe olabildiğince maksimum akışı sağlamaktır

  3. Örnek Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v3 v1 v1 S S t t kaynak v2 v4 v2 v4 hedef

  4. Kapasite 3 v3 v1 6 8 3 S t 3 6 8 v2 v4 6 c(u,v)=12 6 c(u,v)=6 u u v v Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v1 S t v2 v4 12 Büyük boru Küçük boru

  5. Introduction - capacity Representation Example: oil pipeline Flow network: directed graph G=(V,E) 3 v3 v3 v1 v1 6 8 3 S S t t 3 6 8 v2 v4 v2 v4 6 If (u,v)  E  c(u,v) = 0 6 v2 v4 0 0 v4 v3 0

  6. Akış 3 v3 v1 6 8 3 S t 3 6 8 v2 v4 6 f(u,v)=6 f(u,v)=6 u u v v Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v1 S t v2 v4 6/12 Kapasite altı akış 6/6 Tam kapasiteli akış

  7. Akış 3 v3 v1 6 8 3 S t 3 6 8 v2 v4 6 Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v1 S t v2 v4

  8. Introduction – flow 3 v3 v1 6 8 3 S t 3 6/6 6/8 v2 v4 6/6 Representation Example: oil pipeline Flow network: directed graph G=(V,E) v3 v1 S t v2 v4

  9. Akış 3 v3 v1 6 8 3 S t 3 6/6 6/8 v2 v4 6/6 Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v1 S t v2 v4

  10. Akış 3/3 v3 v1 3/6 3/8 3 S t 3 6/6 6/8 v2 v4 6/6 Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v1 S t v2 v4

  11. Akış 3/3 v3 v1 3/6 3/8 3 S t 3 6/6 6/8 v2 v4 6/6 Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v1 S t v2 v4

  12. Akış 3/3 v3 v1 3/6 5/8 2/3 S t 3 6/6 8/8 v2 v4 6/6 Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v1 S t v2 v4

  13. Akışı iptal etme 3/3 v3 v1 3/6 5/8 2/3 S t 3 6/6 8/8 v2 v4 6/6 Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v1 S t v2 v4

  14. Akışı iptal etme 3/3 v3 v1 4/6 6/8 3/3 S t 1/3 6/6 8/8 v2 v4 5/6 Model Örnek: petrol hattı Akış ağı: yönlü çizge G=(V,E) v3 v1 S t v2 v4 u u u u 8/10 3/4 5/10 4 10 4 8/10 4 v v v v

  15. Akış özellikleri

  16. The Ford-Fulkerson yöntemi 2 önemli kavram var:   1) kalan ağ     2) artırılan yol

  17. Ford Fulkerson – kalan ağ 5 11 5 8 cf(u,v) = c(u,v) – f(u,v) Akış ağı G = (V,E) Kalan ağ Gf = (V,Ef) 12/12 v3 v3 v1 v1 11/16 15/20 10 1/4 7/7 S t S t 4/9 4/4 8/13 v2 v4 v2 v4 11/14

  18. Ford Fulkerson – kalan ağ cf(u,v) = c(u,v) – f(u,v) Akış ağı G = (V,E) Kalan ağ Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 5 11/16 15/20 4 11 10 1/4 11 7/7 7 15 3 S t S t 4/9 5 5 4/4 4 3 8/13 8 v2 v4 v2 v4 11/14 11

  19. Ford Fulkersen-artırılan yol Definition: Kalan ağda s den t ye her hangi bir basit (döngüsüz) p yolu p yoluna göre kalan kapasite cf(p) = min{cf (u,v): (u,v) p üzerinde} Akış ağı G = (V,E) Kalan ağ Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 5 11/16 15/20 4 11 10 1/4 11 7/7 7 15 3 S t S t 4/9 5 5 4/4 4 3 8/13 8 v2 v4 v2 v4 11/14 11

  20. Ford Fulkerson – artırılan yol 12 12/12 v3 v3 v1 v1 5 5 11/16 15/20 4 11 10 1/4 11 7/7 7 15 3 S t S t 4/9 5 5 4/4 4 3 8/13 8 v2 v4 v2 v4 11/14 11 Artırılan yol

  21. Ford Fulkerson – artırılan yol Akış: fp: V x V  R: cf(p) eğer (u,v) p de ise fp(u,v) = - cf(p) eğer (v,u) p de ise 0 aksi durumda Akış ağı: G = (V,E) Kalan ağ Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 5 11/16 15/20 4 11 10 1/4 11 7/7 7 15 3 S t S t 4/9 5 5 4/4 4 3 8/13 8 v2 v4 v2 v4 11/14 11

  22. Ford Fulkerson – augmenting paths cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise We define a flow: fp: V x V  R such as: Flow network G = (V,E) residual network Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 4/5 11/16 15/20 4/4 11 10 1/4 11 7/7 7 -4/15 3 S t S t 4/9 -4/5 4/5 4/4 4 3 8/13 -4/8 v2 v4 v2 v4 11/14 11 Our virtual flow fp along the augmenting path p in Gf

  23. Ford Fulkerson – augmenting paths cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise We define a flow: fp: V x V  R such as: Flow network G = (V,E) residual network Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 4/5 11/16 15/20 4/4 11 10 1/4 11 7/7 7 -4/15 3 S t S t 4/9 -4/5 4/5 4/4 4 3 8/13 -4/8 v2 v4 v2 v4 11/14 11 Our virtual flow fp along the augmenting path p in Gf

  24. Ford Fulkerson – augmenting the flow cf(p) if (u,v) is on p fp(u,v) = - cf(p) if (v,u) is on p 0 otherwise We define a flow: fp: V x V  R such as: Flow network G = (V,E) residual network Gf = (V,Ef) 12 12/12 v3 v3 v1 v1 5 4/5 11/16 19/20 4/4 11 10 1/4 11 7/7 7 15 3 S t S t 0/9 5 4/5 4/4 4 3 12/13 8 v2 v4 v2 v4 11/14 11 New flow: f´: V x V  R : f´=f + fp Our virtual flow fp along the augmenting path p in Gf

  25. The Ford-Fulkerson method Ford-Fulkerson(G,s,t) 1 for each edge (u,v) in G.E do 2 f(u,v) ¬ f(v,u) ¬ 0 3 while there exists a path p from s to t in residual network Gfdo 4 cf = min{cf(u,v): (u,v) is in p} 5 for each edge (u,v) in p do 6 f(u,v) ¬ f(u,v) + cf 7 f(v,u)¬-f(u,v) 8 return f The algorithms based on this method differ in how they choose p in step 3. If chosen poorly the algorithm might not terminate.

  26. Örnek (1) Left Side = Residual Graph Right Side = Augmented Flow

  27. Örnek(2) Left Side = Residual Graph Right Side = Augmented Flow

  28. Worst Case Running Time • Assuming integer flow • Each augmentation increases the value of the flow by some positive amount. • Augmentation can be done in O(E). • Total worst-case running time O(E|f*|), where f* is the max-flow found by the algorithm. • Example of worst case: Augmenting path of 1 Resulting Residual Network Resulting Residual Network

  29. Application – Bipartite Matching • Example – given a community with n men and m women • Assume we have a way to determine which couples (man/woman) are compatible for marriage • E.g. (Joe, Susan) or (Fred, Susan) but not (Frank, Susan) • Problem: Maximize the number of marriages • No polygamy allowed • Can solve this problem by creating a flow network out of a bipartite graph

  30. Bipartite Graph • A bipartite graph is an undirected graph G=(V,E) in which V can be partitioned into two sets V1 and V2 such that (u,v)  E implies either u  V1 and v  V12 or vice versa. • That is, all edges go between the two sets V1 and V2 and not within V1 and V2.

  31. Model for Matching Problem • Men on leftmost set, women on rightmost set, edges if they are compatible A A A X X X B B B Y Y Y C C C Z Z Z D D D Men Women Optimal matching A matching

  32. Solution Using Max Flow • Add a supersouce, supersink, make each undirected edge directed with a flow of 1 A A X X B B t Y s Y C C Z Z D D Since the input is 1, flow conservation prevents multiple matchings

  33. Stable Marriage Problem

  34. Stable Marriage Problem • Given N men and N women, each person list in order of preference all the people of the opposite sex who would like to marry. • Problem: • Engage all the women to all the men in such a way as to respect all their preferences as much as possible.

  35. A B C D E 1 2 3 4 5 2 1 2 1 5 E D A C D 5 2 3 3 3 A E D B B 1 3 5 2 2 D B B D C 3 4 4 4 1 B A C A E 4 5 1 5 4 C C E E A Stable? • A set of marriages is unstable if • two people who are not married both prefer each other than their spouses • E.g. Suppose we have A1 B3 C2 D4 E5. This is unstable since • A prefer 2 more than 1 • 2 prefer A more than C

  36. A B C D E 1 2 3 4 5 2 1 2 1 5 E D A C D 5 2 3 3 3 A E D B B 1 3 5 2 2 D B B D C 3 4 4 4 1 B A C A E 4 5 1 5 4 C C E E A Naïve solution (2) • Does not work! • E.g. • A1 B3 C2 D4 E5 • A2B3 C1 D4 E5 • A3 B2 C1 D4 E5 • A3 B1 C2 D4 E5

  37. Solution • Let X be the first man. • X proposes to the best woman in the remaining on his list. (Initially, the first woman on his list!) • If α is not engaged • Pair up (X, α). Then, set X=next man and goto 1. • If α prefers X more than her fiancee Y, • Pair up (X, α). Then, set X=Y and goto 1. • Goto 1

  38. A B C D E 1 2 3 4 5 2 1 2 1 5 E D A C D 5 2 3 3 3 A E D B B 1 3 5 2 2 D B B D C 3 4 4 4 1 B A C A E 4 5 1 5 4 C C E E A Example A B C D E 2 1 2 1 5 5 2 3 3 3 1 3 5 2 4

  39. Time analysis • If there are N men and N women, • O(N2) time

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