Max Flow / Min Cut (undirected graph)

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SUMMARY

The discussion centers on the Max Flow / Min Cut theorem applied to undirected graphs, specifically the equality maxP∈P* mine∈P ce = minC∈C* maxe∈C ce. Participants clarify that the left-hand side (LHS) represents the maximum flow through the graph determined by the minimum capacity of edges along all s-t paths, while the right-hand side (RHS) denotes the minimum capacity of edges in all possible s-t cuts. The theorem asserts that the maximum flow from source s to sink t equals the minimum cut capacity separating them, a fundamental principle in network flow theory.

PREREQUISITES
  • Understanding of undirected graphs and their properties
  • Familiarity with flow networks and capacity concepts
  • Knowledge of the Max Flow / Min Cut theorem
  • Basic mathematical skills for parsing equations and inequalities
NEXT STEPS
  • Study the proof of the Max Flow / Min Cut theorem in detail
  • Explore algorithms for computing maximum flow, such as the Ford-Fulkerson method
  • Learn about network flow applications in real-world scenarios
  • Investigate variations of the Max Flow / Min Cut theorem in directed graphs
USEFUL FOR

Students and professionals in computer science, particularly those focusing on algorithms, network theory, and optimization problems involving flow networks.

goraemon
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Homework Statement


Here is an exercise I came across while studying MaxFlow / MinCut and I'm rather stumped:

Designate G = (V, E) as an undirected graph that has at least two vertices. Every edge e on this graph has capacity ce. Pick two arbitrary vertices from G and label them s and t respectively.

Now, we'll let P* be the set of s-t paths in the graph, and let C* be the set of s-t cuts -- meaning, C* refers to the subset of edges in G that constitute all possible s-t cuts.

Demonstrate that the following equality is true:

maxPP* mineP ce = minCC* maxe∈C ce

Homework Equations


Size of Max Flow = Min. Capacity of an s-t Cut

The Attempt at a Solution


My issue: I'm having trouble parsing the above equation. The LHS appears to be describing the longest(?) of all possible s-t paths, and the edge with the minimum capacity along that path. The RHS appears to be describing the minimum number of edges(?) possible in a subset of edges constituting an s-t cut, and the edge with the max. capacity within that subset? I'm frankly not even sure whether the above equation is meant to involve a multiplication.

Still, I figure there must be a way to parse and understand the equation correctly such that the equality would always hold, and that it can be demonstrated as such. I'd appreciate any help in this matter.
 
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goraemon said:

Homework Statement


Here is an exercise I came across while studying MaxFlow / MinCut and I'm rather stumped:

Designate G = (V, E) as an undirected graph that has at least two vertices. Every edge e on this graph has capacity ce. Pick two arbitrary vertices from G and label them s and t respectively.

Now, we'll let P* be the set of s-t paths in the graph, and let C* be the set of s-t cuts -- meaning, C* refers to the subset of edges in G that constitute all possible s-t cuts.

Demonstrate that the following equality is true:

maxPP* mineP ce = minCC* maxe∈C ce

Homework Equations


Size of Max Flow = Min. Capacity of an s-t Cut

The Attempt at a Solution


My issue: I'm having trouble parsing the above equation. The LHS appears to be describing the longest(?) of all possible s-t paths, and the edge with the minimum capacity along that path. The RHS appears to be describing the minimum number of edges(?) possible in a subset of edges constituting an s-t cut, and the edge with the max. capacity within that subset? I'm frankly not even sure whether the above equation is meant to involve a multiplication.

Still, I figure there must be a way to parse and understand the equation correctly such that the equality would always hold, and that it can be demonstrated as such. I'd appreciate any help in this matter.

For a given path ##P##, the largest flow that can be transmitted along that path equals the smallest arc capacity along the path. This is readily apparent: to send ##f## units of flow along the path, you must send ##f## units along each arc in the path, so ##f## cannot be bigger than the smallest arc-capacity in that path. You can actually achieve a flow equal to the smallest arc capacity, so you have a true minimum, not just a lower bound.
 

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