COMP 382 Reasoning about Algorithms Fall 2015 Swarat

COMP 382: Reasoning about Algorithms Fall 2015 Swarat Chaudhuri

Soviet Rail Network, 1955 Reference: On the history of the transportation and maximum flow problems. Alexander Schrijver in Math Programming, 91: 3, 2002. 2

Maximum Flow and Minimum Cut Max flow and min cut. n n n Two very rich algorithmic problems. Cornerstone problems in combinatorial optimization. Beautiful mathematical duality. Nontrivial applications / reductions. n n n n Data mining. Open-pit mining. Project selection. Airline scheduling. Bipartite matching. Baseball elimination. Image segmentation. n n n n Network reliability. Distributed computing. Egalitarian stable matching. Security of statistical data. Network intrusion detection. Multi-camera scene reconstruction. Many more. . . Network connectivity. 3

Minimum Cut Problem Flow network. Abstraction for material flowing through the edges. G = (V, E) = directed graph, no parallel edges. Two distinguished nodes: s = source, t = sink. c(e) = capacity of edge e. n n 10 source s capacity 5 15 2 9 5 4 15 15 10 3 8 6 10 4 6 15 4 30 7 t sink 10 4

Cuts Def. An s-t cut is a partition (A, B) of V with s A and t B. Def. The capacity of a cut (A, B) is: 10 s 5 2 9 5 4 15 15 10 3 8 6 10 4 6 15 t A 15 4 30 7 10 Capacity = 10 + 5 + 15 = 30 5

Cuts Def. An s-t cut is a partition (A, B) of V with s A and t B. Def. The capacity of a cut (A, B) is: 10 5 s A 15 2 9 5 4 15 15 10 3 8 6 10 4 6 15 4 30 7 t 10 Capacity = 9 + 15 + 8 + 30 = 62 6

Minimum Cut Problem Min s-t cut problem. Find an s-t cut of minimum capacity. 10 5 s A 15 2 9 5 4 15 15 10 3 8 6 10 4 6 15 4 30 7 t 10 Capacity = 10 + 8 + 10 = 28 7

Flows Def. An s-t flow is a function that satisfies: For each e E: For each v V – {s, t}: (capacity) (conservation) n n Def. The value of a flow f is: 0 2 4 10 4 4 0 5 s 9 0 15 5 0 15 0 4 4 3 8 6 0 capacity 15 flow 0 4 0 6 15 0 0 4 30 10 7 10 t 0 10 Value = 4 8

Flows Def. An s-t flow is a function that satisfies: For each e E: For each v V – {s, t}: (capacity) (conservation) n n Def. The value of a flow f is: 6 2 10 10 4 4 3 5 s 9 0 15 5 6 15 0 8 8 3 8 6 1 capacity 15 flow 11 4 0 6 15 0 11 4 30 10 7 10 t 10 10 Value = 24 9

Maximum Flow Problem Max flow problem. Find s-t flow of maximum value. 9 2 10 10 4 5 s 9 1 15 5 9 15 0 9 8 3 8 6 4 capacity 15 flow 14 4 0 6 15 0 14 4 30 10 7 10 t 10 10 Value = 28 10

Flows and Cuts Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s. 6 2 10 10 4 4 3 s 5 9 0 15 5 6 15 0 8 8 3 A 8 6 1 15 4 0 11 6 15 0 11 4 30 10 7 10 t 10 10 Value = 24 11

Flows and Cuts Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s. 6 2 10 10 4 4 3 s 5 9 0 15 5 6 15 0 8 8 3 A 8 6 1 15 4 0 11 6 15 0 11 4 30 10 7 10 t 10 10 Value = 6 + 0 + 8 - 1 + 11 = 24 12

Flows and Cuts Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then, the net flow sent across the cut is equal to the amount leaving s. 6 2 10 10 4 4 3 s 5 9 0 15 5 6 15 0 8 8 3 A 8 6 1 15 4 0 11 6 15 0 11 4 30 10 7 10 t 10 10 Value = 10 - 4 + 8 - 0 + 10 = 24 13

Flows and Cuts Flow value lemma. Let f be any flow, and let (A, B) be any s-t cut. Then Pf. by flow conservation, all terms except v = s are 0 14

Flows and Cuts Weak duality. Let f be any flow, and let (A, B) be any s-t cut. Then the value of the flow is at most the capacity of the cut. Cut capacity = 30 10 s 5 Flow value 30 2 9 5 4 15 15 10 3 8 6 10 4 6 15 4 30 t A 15 7 10 Capacity = 30 15

Flows and Cuts Weak duality. Let f be any flow. Then, for any s-t cut (A, B) we have v(f) cap(A, B). Pf. A 4 8 s ▪ B t 7 6 16

Certificate of Optimality Corollary. Let f be any flow, and let (A, B) be any cut. If v(f) = cap(A, B), then f is a max flow and (A, B) is a min cut. Value of flow = 28 Cut capacity = 28 Flow value 28 9 2 9 4 1 15 10 10 0 4 5 s 5 9 15 0 9 8 3 8 6 4 A 15 4 0 14 6 10 15 0 10 t 10 10 14 4 30 7 17

Towards a Max Flow Algorithm Greedy algorithm. Start with f(e) = 0 for all edge e E. Find an s-t path P where each edge has f(e) < c(e). Augment flow along path P. Repeat until you get stuck. n n 1 0 0 20 10 30 0 s t 10 20 0 0 Flow value = 0 2 18

Towards a Max Flow Algorithm Greedy algorithm. Start with f(e) = 0 for all edge e E. Find an s-t path P where each edge has f(e) < c(e). Augment flow along path P. Repeat until you get stuck. n n 1 20 X 0 0 20 10 30 X 0 20 s t 10 20 0 X 0 20 Flow value = 20 2 19

Towards a Max Flow Algorithm Greedy algorithm. Start with f(e) = 0 for all edge e E. Find an s-t path P where each edge has f(e) < c(e). Augment flow along path P. Repeat until you get stuck. n n locally optimality global optimality 1 20 20 s 1 0 10 t 30 20 10 0 greedy = 20 20 20 s 20 20 2 10 10 opt = 30 t 30 10 20 20 2 20

Residual Graph Original edge: e = (u, v) E. Flow f(e), capacity c(e). n capacity u v 17 6 flow Residual edge. "Undo" flow sent. e = (u, v) and e. R = (v, u). Residual capacity: residual capacity n n n u 11 v 6 residual capacity Residual graph: Gf = (V, Ef ). Residual edges with positive residual capacity. Ef = {e : f(e) < c(e)} {e. R : c(e) > 0}. n n 21

Ford-Fulkerson Algorithm 2 4 4 10 2 8 6 10 10 3 9 5 10 capacity G: s t 22

Ford-Fulkerson Algorithm G: 0 10 s 0 10 2 0 4 2 0 0 8 3 0 9 4 60 0 10 5 0 10 flow capacity t Flow value = 0 23

Ford-Fulkerson Algorithm G: 8 X 0 10 s 0 10 2 0 4 2 0 0 8 X 8 3 0 9 4 60 5 0 10 8 X 0 10 flow capacity t Flow value = 0 2 4 4 2 8 6 10 3 9 5 10 Gf : 10 s 10 residual capacity t 24

Ford-Fulkerson Algorithm 2 G: 10 X 8 10 0 10 s 2 X 0 2 3 0 4 8 8 0 2 X 9 4 60 5 0 10 8 10 X 10 t Flow value = 8 Gf : 2 4 4 2 2 8 6 10 10 3 9 5 2 8 s t 8 25

Ford-Fulkerson Algorithm G: 10 10 s 0 6 X 10 2 0 4 2 2 8 8 3 2 8 X 9 4 0 6 X 6 5 0 6 X 10 10 10 t Flow value = 10 Gf : s 2 4 4 10 2 8 6 10 10 3 7 5 10 t 2 26

Ford-Fulkerson Algorithm 2 G: 10 10 s 6 8 X 10 2 X 2 0 3 0 2 X 4 8 8 8 9 4 66 6 8 X 10 5 10 10 t Flow value = 16 2 Gf : 4 4 6 s 10 2 8 6 4 4 3 1 5 10 6 t 8 27

Ford-Fulkerson Algorithm G: 10 10 s 8 9 X 10 2 2 3 X 4 2 0 8 7 X 8 3 8 9 X 9 4 66 8 9 X 10 5 10 10 t Flow value = 18 2 2 Gf : 2 4 8 s 10 2 8 6 2 2 3 1 5 10 8 t 8 28

Ford-Fulkerson Algorithm G: 10 10 s 9 10 2 3 4 2 0 7 8 3 9 9 4 66 9 10 5 10 10 t Flow value = 19 3 2 Gf : s 1 10 2 7 1 3 9 4 1 9 6 1 5 10 t 9 29

Ford-Fulkerson Algorithm G: 10 10 s 9 10 2 3 4 2 0 7 8 3 9 9 4 66 9 10 5 10 10 Cut capacity = 19 t Flow value = 19 3 2 Gf : s 1 10 2 7 1 3 9 4 1 9 6 1 5 10 t 9 30

Augmenting Path Algorithm Augment(f, c, P) { b bottleneck(P) foreach e P { if (e E) f(e) + b else f(e. R) f(e) - b } return f } forward edge reverse edge Ford-Fulkerson(G, s, t, c) { foreach e E f(e) 0 Gf residual graph while (there exists augmenting path P) { f Augment(f, c, P) update Gf } return f } 31

Max-Flow Min-Cut Theorem Augmenting path theorem. Flow f is a max flow iff there are no augmenting paths. Max-flow min-cut theorem. [Ford-Fulkerson 1956] The value of the max flow is equal to the value of the min cut. Proof strategy. We prove both simultaneously by showing that TFAE: (i) There exists a cut (A, B) such that v(f) = cap(A, B). (ii) Flow f is a max flow. (iii) There is no augmenting path relative to f. (i) (ii) This was the corollary to weak duality lemma. (ii) (iii) We show contrapositive. Let f be a flow. If there exists an augmenting path, then we can improve f by sending flow along path. n 32

Proof of Max-Flow Min-Cut Theorem (iii) (i) Let f be a flow with no augmenting paths. Let A be set of vertices reachable from s in residual graph. By definition of A, s A. By definition of f, t A. n n A B t s original network 33

Running Time Assumption. All capacities are integers between 1 and C. Invariant. Every flow value f(e) and every residual capacities c f (e) remains an integer throughout the algorithm. Theorem. The algorithm terminates in at most v(f*) n. C iterations. Pf. Each augmentation increase value by at least 1. ▪ Corollary. If C = 1, Ford-Fulkerson runs in O(mn) time. Integrality theorem. If all capacities are integers, then there exists a max flow f for which every flow value f(e) is an integer. Pf. Since algorithm terminates, theorem follows from invariant. ▪ 34

Choosing Good Augmenting Paths

Ford-Fulkerson: Exponential Number of Augmentations Q. Is generic Ford-Fulkerson algorithm polynomial in input size? m, n, and log C A. No. If max capacity is C, then algorithm can take C iterations. 1 1 1 0 X 0 C C 1 X 0 1 s t C C 0 0 1 X 2 1 X 0 0 1 X C C 1 0 1 X 0 X s t C C X 0 1 X 2 36

Choosing Good Augmenting Paths Use care when selecting augmenting paths. Some choices lead to exponential algorithms. Clever choices lead to polynomial algorithms. If capacities are irrational, algorithm not guaranteed to terminate! n n n Goal: choose augmenting paths so that: Can find augmenting paths efficiently. Few iterations. n n Choose augmenting paths with: [Edmonds-Karp 1972, Dinitz 1970] Max bottleneck capacity. Sufficiently large bottleneck capacity. Fewest number of edges. n n n 37

Capacity Scaling Intuition. Choosing path with highest bottleneck capacity increases flow by max possible amount. Don't worry about finding exact highest bottleneck path. Maintain scaling parameter . Let Gf ( ) be the subgraph of the residual graph consisting of only arcs with capacity at least . n n n 4 110 4 102 1 s 122 t 170 2 Gf 110 102 s t 122 170 2 Gf (100) 38

Capacity Scaling-Max-Flow(G, s, t, c) { foreach e E f(e) 0 smallest power of 2 greater than or equal to C Gf residual graph while ( 1) { Gf( ) -residual graph while (there exists augmenting path P in Gf( )) { f augment(f, c, P) update Gf( ) } / 2 } return f } 39

Capacity Scaling: Correctness Assumption. All edge capacities are integers between 1 and C. Integrality invariant. All flow and residual capacity values are integral. Correctness. If the algorithm terminates, then f is a max flow. Pf. By integrality invariant, when = 1 Gf( ) = Gf. Upon termination of = 1 phase, there are no augmenting paths. ▪ n n 40

Capacity Scaling: Running Time Lemma 1. The outer while loop repeats 1 + log 2 C times. Pf. Initially C < 2 C. decreases by a factor of 2 each iteration. ▪ Lemma 2. Let f be the flow at the end of a -scaling phase. Then the proof on next slide value of the maximum flow is at most v(f) + m . Lemma 3. There at most 2 m augmentations per scaling phase. Let f be the flow at the end of the previous scaling phase. L 2 v(f*) v(f) + m (2 ). Each augmentation in a -phase increases v(f) by at least . ▪ n n n Theorem. The scaling max-flow algorithm finds a max flow in O(m log C) augmentations. It can be implemented to run in O(m 2 log C) time. ▪ 41

Capacity Scaling: Running Time Lemma 2. Let f be the flow at the end of a -scaling phase. Then value of the maximum flow is at most v(f) + m . Pf. (almost identical to proof of max-flow min-cut theorem) We show that at the end of a -phase, there exists a cut (A, B) such that cap(A, B) v(f) + m . Choose A to be the set of nodes reachable from s in Gf( ). By definition of A, s A. By definition of f, t A. n n A B t s original network 42