Algorithm Design and Analysis

LECTURES18-19 • Network Flows • Flows, cuts • Ford-Fulkerson • Min-cut/max-flow theorem Algorithm Design and Analysis CSE 565 Adam Smith A. Smith; based on slides by E. Demaine, C. Leiserson, S. Raskhodnikova, K. Wayne

Detecting Negative Cycles • Theorem. Can find negative cost cycle in O(mn) time. • Add new node t and connect all nodes to t with 0-cost edge. • Check if OPT(n, v) = OPT(n-1, v) for all nodes v. • if yes, then no negative cycles • if no, then extract cycle from shortest path from v to t • Last lecture we proved: if OPT(n, v) < OPT(n-1, v) then we can find a neg.-cost cycle on a path from v to t • Questions to think about: • Prove that if there is a negative-cost cycle,then OPT(n, v) < OPT(n-1, v) for some v • (Hint: what would happen ifwe kept running Bellman-Ford for many iterations?) • Why do we bother with the extra node t? t 0 0 0 0 0 18 2 6 -23 5 -11 v -15

Network Flow and Linear Programming A. Smith; based on slides by S. Raskhodnikova and K. Wayne

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

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

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. 2 5 9 10 15 15 10 4 source sink 5 s 3 6 t 8 10 15 4 6 10 15 capacity 4 7 30

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: we don’t count edges into A 2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 A 15 4 6 10 15 Capacity = 10 + 5 + 15 = 30 4 7 30

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: 2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 A 15 4 6 10 15 Capacity = 9 + 15 + 8 + 30 = 62 4 7 30

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

2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 4 7 30 Flows • Def. An s-t flow is a function f from E to real numbers that satisfies: • For each e E: [capacity] • For each v V – {s, t}: [conservation] water flowing from source to sink 0 4 0 0 0 4 0 4 4 0 0 0 0 capacity flow 0 0

2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 4 7 30 Flows • Def. An s-t flow is a function f from E to real numbers that satisfies: • For each e  E: [capacity] • For each v  V – {s, t}: [conservation] • Def. The value of a flow f is: water flowing from source to sink 0 4 0 0 0 4 0 4 4 0 0 0 0 capacity flow 0 0 Value = 4

2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 4 7 30 Flows • Def. An s-t flow is a function f from E to real numbers that satisfies: • For each e  E: [capacity] • For each v  V – {s, t}: [conservation] • Def. The value of a flow f is: water flowing from source to sink 6 10 6 0 0 4 3 8 8 1 10 0 0 capacity flow 11 11 Value = 24

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

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

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

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

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

Question • Two problems • min cut • max flow • How do they relate?

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. Big Optimization Idea #1: Look for structural constraints, e.g. max flow min-cut Cut capacity = 30  Flow value  30 2 5 9 10 15 15 10 4 5 s 3 6 t 8 10 A 15 4 6 10 15 Capacity = 30 4 7 30

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

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

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. 1 0 0 20 10 30 0 t s 10 20 Flow value = 0 0 0 2

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. 1 20 0 0 X 20 10 20 30 0 t X s 10 20 Flow value = 20 20 0 0 X 2

How can we get from greedy to opt here? What if we push water back across middle edge? 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. locally optimality  global optimality 1 1 20 0 20 10 20 10 20 10 30 20 30 10 t t s s 10 20 10 20 0 20 10 20 2 2 greedy = 20 opt = 30

Residual Graph • Original edge: e = (u, v)  E. • Flowf(e), capacity c(e). • Residual edge. • e= (u, v) and eR = (v, u). • Residual capacity: • If e in E: • unused capacity • If eR in E: • ability of sent less water, or “undo“ flow • Residual graph: Gf = (V, Ef). • Residual edges with positive residual capacity. • Ef = {e : f(e) < c(e)} {eR : c(e) > 0}. capacity u v 17 6 flow residual capacity u v 11 6 residual capacity

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

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

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

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

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

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

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

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

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

Augmenting Path Algorithm Ford-Fulkerson(G, s, t, c) { foreach e  E f(e)  0 Gf residual graph while (there is an s-t path P in Gf) { f  Augment(f, c, P) update Gf } return f } Augment(f, c, P) { b  bottleneck(P) foreach e  P { if (e  E) f(e)  f(e) + b else f(e)  f(e) - b } return f } Min residual capacity of an edge in P forward edge reverse edge

Ford-Fulkerson: Summary so far • Ford-Fulkerson: • While you can, • Greedily push flow • Update residual graph • Lemma 1: This outputs a valid flow. • Proof: (Check conservation conditions… see book.) • Still to do: • Running time (in particular, termination!) • How good a flow?

Running Time • Assumption. All capacities are integers between 1 and C. • Invariant. Every flow value f(e) and every residual capacity cf(e) remains an integer throughout the algorithm. • Proof: exercise. • Theorem. The algorithm terminates in at most v(f*) nC iterations. • Pf. Each augmentation increase value by at least 1. ▪ • Running time of Ford-Fulkerson: O(mnC). Space: O(m+n). • Corollary. If C = 1, Ford-Fulkerson runs in O(mn) time.

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. [Elias-Feinstein-Shannon 1956, Ford-Fulkerson 1956]The value of the max flow is equal to the value of the min s-t cut. • Pf. We prove both simultaneously by showing (i) -- (iii) are equivalent. (i) There exists an s-t cut (A, B) such that v(f) = cap(A, B). (ii) Flow f is a max flow, and (A,B) is min s-t cut. (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.

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. • Observation: No edges of the residual graph go from A to B. • Claim 1: If e goes from A to B, then f(e) =c(e).Proof: Otherwise there would be residual capacity, and the residual graph would have an edge A to B. • Claim 2: If e goes from B to A, then f(e)=0.Proof: Otherwise residual edge would go from A to B. original network A B t s

Summary • Assumption. All capacities are integers between 1 and C. • Running time: The FF algorithm terminates in at most v(f*) nC iterations. Running time = O(mnC). Space: O(m+n). • Important special case: 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.

Algorithm Design and Analysis