Maximum Weighted closed subgraph in Graph Theory

First, let's introduce it with one question. For details, refer to [linear planning and network flow 24 Question 2] space flight plan questions.

In this question, the experiment depends on the instrument, and both the experiment and the instrument have the right value, and the instrument is negative, and the experiment is positive.

Here, the concept of closed graphs is very good. In a graph, we select some vertices to form a set, which are marked as V, and the outbound edges in the Set (that is, the external arc of the points in the Set) point to the end point (ARC header) in V, we call V as a closed graph. The largest weight closure graph is the V with the largest sum of the weights in all the closed graphs. We call V the largest weight closure graph.

 

 

The closed graph has

{5}, {2, 5}, {4, 5}

{2, 4, 5}, {3, 4, 5}

{1, 2, 3, 4, 5}, {1, 2, 4, 5}

 

The maximum weight closure graph is {3, 4, 5 }.

 

 

For this question, we will link the experiment with the instrument with a directed edge. The experiment is the starting point (arc tail) and the instrument is the ending point (ARC head ). If we select a closed chart, the instrument required by the experiment contained in this closed chart is also the most closed chart. The largest weight closure diagram is the solution of the question.

After learning about the concept of the max weight closure graph, we need to know how to find the max weight closure graph.

First, we will convert it into a network (do not ask why now, and then we will prove that we can solve it using the network ). Construct a Source Vertex S, sink vertex T. We connect s to a point with a positive ownership value to an edge with a capacity of its weight value, and link a point with a negative ownership value to an edge with a capacity of the absolute value of its weight value, the original side sets its capacity to positive infinity.

 

Is converted to a network such as the left graph.

 

 

 

 

First, we introduce the conclusion that the Set (except S) of the Source Vertex s in the two sets generated by the minimal cut is the largest weight closed graph. Then we will explain some conclusions.

• Proof: The minimum cut is simple cut.

Introduce the concept of simple cut: each side of the cut set is associated with S or T. (The minimum cut and simple cut must be distinguished below, which is easy to confuse)

Why is the minimum cut simple? Because the edge capacity between points except S and T is positive infinity, the minimum cut capacity cannot be positive infinity. So, you have to prove it.

• It is proved that the simple cut in the network has a one-to-one correspondence with the closed graph in the source image. (That is, all closed graphs are simple cut, and simple cut must be a closed graph ).

It is proved that the closed graph is a simple cut: If the closed graph is not a simple cut (inverse method ). If one side is an infinite edge with a positive capacity, it means that the end point of an outbound edge in the closed graph is not in the closed graph. This is a contradiction.

It is proved that simple cut is a closed graph: Because simple cut does not contain positive and infinite edges, it does not contain vertices that are connected to another set (except T), so the outdegree ends of the edges are all in simple cut, meet the closed graph definition. .

• It is proved that the Set (except S) of the Source Vertex s in the two sets generated by the minimum cut is the largest weight closed graph.

First, we should note that the capacity of a simple cut is C, and the set of S is N, and the set of T is M.

Then, the ownership value in C = m is the right value of the point (that is, the size of the edge connecting S and M) + the absolute value of the Point Weight Value with a negative ownership value in N (that is, the capacity of the edge connecting the n-point and T-point ). Note (C = X1 + Y1); (this is easy to understand. If you do not understand it, you can understand it by drawing a graph or imagining it ).

Let's note that the weight of N is W.

Then W = n is the right of the point, and N is the absolute value of the right of the negative point. Note (W = x2-y2 );

Then W + c = X1 + Y1 + x2-y2.

Because of the obvious Y1 = Y2, W + c = X1 + x2;

X1 is the right of vertices whose ownership value is positive in m, and X2 is the right of vertices whose ownership value is positive in N.

Therefore, X1 + X2 = sum of the points with positive ownership values (marked as ToT ).

So we get W + c = Tot. Tidy up w = TOT-C.

Here we get the relationship between the weight of the closed graph and the capacity of the simple cut.

Because Tot is a fixed value, we want to minimize W, that is, C. In this case, the simple cut is the minimum cut. In this case, the closed graph is the set of S of Its Source Vertex (excluding S ). .

 

So far, we have transformed the largest weighted closed graph problem to the problem of least cut. The problem can be converted to the problem of maximizing the flow by finding the minimum cut capacity = the maximum flow.

 

PS: This article is transferred from the Internet

Maximum Weighted closed subgraph in Graph Theory