Flow
Chapter Maximum 26
Flow Graph
A common scenario is to use a graph to represent a flow network and use it to answer questions about material flows Flow is the rate that material moves through the network Each directed edge is a conduit for the material with some stated capacity Vertices are connection points but do not collect material
Flow into a vertex must equal the flow leaving the vertex, flow conservation
Sample Networks
Network communication circuits mechanical hydraulic financial transportation chemical Nodes Arcs Flow voice, video, packets current heat, energy fluid, oil money freight, vehicles, passengers energy telephone exchanges, cables, fiber optics, computers, satellites microwave relays gates, registers, processors joints reservoirs, pumping stations, lakes stocks, companies airports, rail yards, street intersections sites wires rods, beams, springs pipelines transactions highways, railbeds, airway routes bonds
Flow Concepts
Source vertex s
where material is produced
Sink vertex t
where material is consumed
For all other vertices what goes in must go out
Flow conservation
Goal: determine maximum rate of material flow from source to sink
Formal Max Flow Problem
Graph G=(V,E) a flow network
Directed, each edge has capacity c(u,v) 0 Two special vertices: source s, and sink t For any other vertex v, there is a path svt
Flow a function f : V V R
Capacity constraint: For all u, v V: f(u,v) c(u,v) Skew symmetry: For all u, v V: f(u,v) = f(v,u) Flow conservation: For all u V {s, t}:
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f (u, v) = f (u,V ) = 0, or f (v, u) = f (V , u) = 0
vV vV
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Cancellation of flows
We would like to avoid two positive flows in opposite directions between the same pair of vertices
Such flows cancel (maybe partially) each other due to skew symmetry
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Max Flow
We want to find a flow of maximum value from the source to the sink
Denoted by |f|
Lucky Puck Distribution Network
Max Flow, |f| = 19 Or is it? Best we can do?
Ford-Fulkerson method
Contains several algorithms:
Residue networks Augmenting paths
Find a path p from s to t (augmenting path), such that there is some value x > 0, and for each edge (u,v) in p we can add x units of flow
f(u,v) + x c(u,v)
Augmenting Path?
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Residual Network
To find augmenting path we can find any path in the residual network:
Residual capacities: cf(u,v) = c(u,v) f(u,v)
i.e. the actual capacity minus the net flow from u to v Net flow may be negative
Residual network: Gf =(V,Ef), where
Ef = {(u,v) V V : cf(u,v) > 0}
Observation edges in Ef are either edges in E or their reversals: |Ef| 2|E|
Sub-graph With c(u,v) and f(u,v)
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Residual Sub-Graph
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Residual Graph
Compute the residual graph of the graph with the following flow:
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Residual Capacity and Augmenting Path
Finding an Augmenting Path
Find a path from s to t in the residual graph The residual capacity of a path p in Gf: cf(p) = min{cf(u,v): (u,v) is in p}
i.e. find the minimum capacity along p
Doing augmentation: for all (u,v) in p, we just add this cf(p) to f(u,v) (and subtract it from f(v,u)) Resulting flow is a valid flow with a larger value.
Residual network and augmenting path
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 Gf network do 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.
Execution of Ford-Fulkerson (1)
Left Side = Residual Graph
Right Side = Augmented Flow
Execution of Ford-Fulkerson (2)
Left Side = Residual Graph
Right Side = Augmented Flow
Cuts
Does the method find the minimum flow?
Yes, if we get to the point where the residual graph has no path from s to t A cut is a partition of V into S and T = V S, such that s S and t T The net flow (f(S,T)) through the cut is the sum of flows f(u,v), where s S and t T
Includes negative flows back from T to S
The capacity (c(S,T)) of the cut is the sum of capacities c(u,v), where s
S and t T
The sum of positive capacities
Minimum cut a cut with the smallest capacity of all cuts. |f|= f(S,T) i.e. the value of a max flow is equal to the capacity of a min cut.
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Max Flow / Min Cut Theorem
Since |f| c(S,T) for all cuts of (S,T) then if |f| = c(S,T) then c(S,T) must be the min cut of G This implies that f is a maximum flow of G This implies that the residual network Gf contains no augmenting paths.
If there were augmenting paths this would contradict that we found the maximum flow of G
1231 and from 23 we have that the Ford Fulkerson method finds the maximum flow if the residual graph has no augmenting paths.
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
Edmonds Karp
Take shortest path (in terms of number of edges) as an augmenting path Edmonds-Karp algorithm
How do we find such a shortest path? Running time O(VE2), because the number of augmentations is O(VE) Skipping the proof her...