Max Flow / Min Cut (undirected graph)

In summary, the equation provided states that the maximum flow over all possible s-t paths is equal to the minimum capacity over all possible s-t cuts. This is because for a given path, the largest flow that can be transmitted is equal to the smallest arc capacity along that path. The equation provides a way to understand and demonstrate this equality.
  • #1
goraemon
67
4

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.
 
Physics news on Phys.org
  • #2
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.
 

1. What is Max Flow / Min Cut in an undirected graph?

Max Flow / Min Cut is a graph algorithm used to find the maximum flow that can be sent from a source vertex to a sink vertex in an undirected graph. This algorithm also identifies the minimum cut, which is the minimum set of edges that, if removed, would disconnect the source from the sink.

2. How does Max Flow / Min Cut work?

The algorithm works by first finding a path from the source to the sink using a depth-first search or breadth-first search. Then, it calculates the maximum flow that can be sent along this path by finding the minimum capacity of the edges along the path. This process is repeated until there are no more paths from the source to the sink. The minimum cut is identified by finding all the edges that have a flow equal to their capacity.

3. What is the time complexity of Max Flow / Min Cut?

The time complexity of Max Flow / Min Cut is O(E * V^2), where E is the number of edges and V is the number of vertices in the graph. This is because the algorithm requires a loop through all the edges and a loop through all the vertices for each edge.

4. What are some applications of Max Flow / Min Cut?

Max Flow / Min Cut has many practical applications, such as optimizing network traffic flow, scheduling tasks, and resource allocation. It is also used in computer vision for image segmentation and in transportation and logistics for finding the most efficient routes.

5. What are some common variations of Max Flow / Min Cut?

Some common variations of Max Flow / Min Cut include the Ford-Fulkerson algorithm, the Edmonds-Karp algorithm, and the Dinic's algorithm. These algorithms use different approaches to find the maximum flow and minimum cut in an undirected graph, but they all have the same time complexity of O(E * V^2).

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
1
Views
979
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Classical Physics
Replies
6
Views
1K
Back
Top