Max Flow Min Cut Problem Directed Graph Applications

Max Flow – Min Cut Problem

Directed Graph Applications l l Shortest Path Problem (Shortest path from one point to another) Max Flow problems (Maximum material flow through conduit) l l l Liquid flow through pipes Parts through assembly line Current through electrical network, etc

Flow Networks l l l Directed graph G = (V, E) Only one Source(s) and one Sink(t) Weight on each edge = Capacity of the edge if (u, v) E then Capacity is non-negative, i. e. c(u, v) ≥ 0 if (u, v) E then Capacity is assumed zero, i. e. c(u, v) = 0

Flow Networks and Flows flow(f) in Flow Network(G) is a real-valued function f: V x V → R l f(u, v) is flow from vertex u to vertex v. l flow f(u, v) can be positive, negative or zero Constraints on flow : 1. Capacity constraint : f(u, v) ≤ c(u, v), for all u, v V 2. Skew symmetry constraint : f(u, v) = - f(v, u) l

Flow Networks and Flows (Contd. ) 3. Flow conservation constraint : Total net flow at vertex must equal 0. ∑j f(i, j) - ∑k f(k, i) = 0 for all i V – {s, t} k 1 j 1 k 2 j 2 k 3 i k 4 flow in equals flow out j 3 jn

Maximum Flow l l We refer a flow f as maximum if it is feasible and maximize ∑k f(s, k). Where f(s, k) is flow out of source s. (flow/capacity) Problem: 1 (8/10) (7/8) (1/1) s (5/6) 2 l t (6/10) Objective: To find a maximum flow

Multiple source – Multiple sink Network Factories 10 s 1 3 Warehouses 12 s 2 5 6 8 s 3 t 2 20 14 13 7 11 s 4 t 1 15 18 t 3 2 s 5 Convert this problem to single source - single sink.

Conversion to flow Network l l Add a super – source with infinite weighted edges emanting out to original sources Add a super – sink with edges of infinite weight from original sinks Factories 10 s 1 3 Warehouses Super-source s 12 s 2 5 13 7 11 s 4 2 s 5 t 2 20 14 18 6 8 s 3 t 1 15 t 3 Super-sink t

Another Example : Find a feasible match persons l tasks 1 5 2 6 3 7 4 8 Assign persons to tasks such that each task is assigned to a person and each person is assigned to a task in feasible manner.

Transformation to maximum flow problem persons tasks 1 5 2 6 1 1 s 1 1 1 3 7 1 4 8 t

Ford-Fulkerson Algorithm for Max Flow l Key ingredients: l l Residual Networks Augmenting Paths Cut Limitations : l l Flow should be integral or rational On each iteration residual capacity should be integral

The Residual Network Flow Network 1 (8/10) (7/8) (1/1) s (5/6) t u (flow / capacity) f (u, v) / c (u, v) v (6/10) 2 Residual Network 1 2 s 8 1 5 1 7 4 1 2 Residual capacity r (u , v) 6 c (u, f) – f (u, v) t u v f (u, v)

Augmenting Paths l l l An augmenting path is a path from s to t in the residual network. The Residual capacity of augmented path P is d(P) = min {r(i, j): (i, j) P} Augmentation along P l Add d(P) in each arc along P in flow network l Modify residual capacities in residual network r(u, v) = r(u, v) - d(P) and r(v, u) = r(v, u) + d(P) for u, v P 1 2 s 8 1 2 7 8 4 1 5 1 2 6 t s 1 1 8 4 6 2 6 t

Cut (9/10) 1 (8/8) (1/1) s (6/6) 2 t (7/10) An (S, T)-cut in a flow network G = (V, E) is a partition of vertices V into two disjoint subsets S and T such that s S, t T e. g. , S = { s, 1 } and T = { 2, t }. The capacity of a cut (S, T) is CAP(S, T) = u S v T c(u, v)

The Ford Fulkerson Maximum Flow Method Begin x : = 0; create the residual network G(x); while there is some directed path from s to t in G(x) do begin let P be a path from s to t in G(x); : = d(P); send units of flow along P; update the residual capacities; end {the flow x is now maximum}.

Proof of Correctness of the Algorithm Assume that all data are integral. Lemma: At each iteration all residual capacities are integral. Proof: By assumption it is true at beginning. Assume it is true after the first k-1 augmentations, and consider augmentation k along path P. The residual capacity of P is the smallest residual capacity on P, which is integral. After updating, we modify residual capacities by 0, or , and thus residual capacities stay integral.

Proof of finiteness of Algorithm Proof: The capacity of each augmenting path is at least 1. The augmentation reduces the residual capacity of some arc (s, j) and does not increase the residual capacity of (s, i) for any i. So, the sum of the residual capacities of arcs out of s keeps decreasing, and is bounded below by 0. Number of augmentations is O(n. U), where U is the largest capacity in the network.

Indication of Optimum Flow 1 (9/10) (8/8) (1/1) s (6/6) 2 t (7/10) There is no augmenting path in the residual network. 1 1 s 9 1 8 3 6 2 7 t

Indication of Optimum Flow (Contd. ) (9/10) 1 (8/8) (1/1) s (6/6) 2 t (7/10) Flow across the network is equal to the capacity of some cut (Max Flow Min Cut Theorem)

Weak Duality Theorem for the Max Flow Problem Theorem: If f is any feasible flow and if (S, T) is an (s, t)-cut, then the flow | from source to sink in the network is at most CAP(S, T). Proof: We define the flow across the cut (S, T) to be f(S, T) = i S j T f(i, j) - i S j T f(j, i) S T

Flows across different cuts (9/10) 1 (8/8) (1/1) s (6/6) S/T cut t 1 (6/6) 2 t (7/10) (9/10) 1 (8/8) (1/1) s (6/6) (8/8) (1/1) s T (7/10) 2 If S = {s, 1}, then the flow across (S, T) is 8 + 1 + 6 = 15 (9/10) S If S = {s}, then the flow across (S, T) is 9 + 6 = 15 If S = {s, 2}, then the flow across (S, T) is 9 + 7 – 1 = 15 2 t (7/10)

More on Flows Across Cuts Claim: Let (S, T) be any s-t cut. Then f(S, T) = | f | = flow into t. Proof: Add the conservation of flow constraints for each node i S - {s} to the constraint that the flow leaving s is |f|. The resulting equality is f(S, T) = |f|. Sj f(i, j) - Sk f(k, i) = 0 Sj f(s, j) = | for each i {S} - s S T

More on Flows Across Cuts Claim: The flow across (S, T) is at most the capacity of a cut. Proof: If i S, and j T, then f(i, j) c(i, j). If i T, and j S, then f(i, j) 0. f(S, T) = i S j T f(i, j) - i S j T f(j, i) CAP(S, T) = i S j T c(i, j) - i S j T 0

Max Flow Min Cut Theorem: (Optimality conditions for max flows). The following are equivalent. 1. A flow x is maximum. 2. There is no augmenting path in G(x). 3. There is an s-t cutset (S, T) whose capacity is the flow value of x. Corollary: (Max-flow Min-Cut). The maximum flow value is the minimum value of a cut. Proof of Theorem: 1 Þ 2. (not 2 Þ not 1) Suppose that there is an augmenting path in G(x). Then x is not maximum.

Max Flow Min Cut Theorem Contd. 3 Þ 1. Let v = Fx(S, T) be the flow from s to t. By assumption, v = CAP(S, T). By weak duality, the maximum flow is at most CAP(S, T). Thus the flow is maximum. 2 Þ 3. Suppose there is no augmenting path in G(x). Claim: Let S be the set of nodes reachable from s in G(x). Let T = NS. Then there is no arc in G(x) from S to T.

Max Flow Min Cut Theorem Contd. Saturated S Thus i S and j T i T and j S Þ Þ f(i, j) = c(i, j) f(i, j) = 0. T 0 . s Reachable from s There is no arc from S to T in G(x) . t Not Reachable from s It follows that Fx(S, T) = i S j T f(i, j) = i S j T c(i, j) - i S j T f(j, i) - i S j T 0 = CAP(S, T)

Limitation of Algorithm : example 1 • Termination even with integral flow and few nodes can take large no. of steps. • Depends on path selection and capacity of flow network. 1 M s 1 M M M t Flow M >> 1 2 Flow Network 27