Graph Theory Reading Notes

Liu rujia's book on this chapter is also similar. When I was doing the question, I found that I had a lot of knowledge and thinking in graph theory. So I went to the library and borrowed two graph theory books, some notes are organized here.

I. Important ideas:
1. Makeup:
--
G is a graph of a V (G) point set, but the two points are connected in G only when they are not connected in G.
Eg
Proposition:In any group of six people, three people know each other, or three people do not know each other.
Proof:If A is one of them, three of the other five will know him. Otherwise, this phenomenon is true in the supplementary diagram. If none of the three knows each other, the proposition is true. Otherwise, when two people know each other, the two and A together constitute the three people who know each other. In this case, the proposition is true.

Drawing problems:Draw a simple graph with five vertices, three or seven edges.
Because a complete graph with five vertices has 5*4/2 = 10 edges, a complementary graph with three edges is a graph with seven edges, in this way, only one type can be drawn to generate another type.

Proposition:For any simple graph, this graph or its supplementary graph is connected.
Proof:Assume that the given graph is not connected. For any two points a and B in the graph, if the two points do not belong to the same block in the source image, in the supplementary graph, they are connected. If the two points belong to the same block in the source image, there is another center c in the source image, and the ac and bc are connected in the supplementary graph, therefore, a and B are connected in the supplementary graph, and the problem is proved.

2. Longest Path Method
Set L to one of the longest paths of graph G. Its length is m, and one of its endpoints is a. let's examine the edges associated with a in G, the other vertex of any edge must belong to L. Otherwise, adding the other edge to L will lead a longer path.
Eg
Proposition:A connected graph with 2 vertices is a ring.
Proof:Assume that the graph has a longest path L, and the endpoint is a. because the degree of a is 2, a and a C side are not in the L field, we can see from the longest path principle, the point at the other end of the edge must also belong to L, and the degree of only one vertex in L is 1, that is, point B at the other end, and the degree of the remaining vertex is 2, therefore, C can only be connected to B, so that the longest path L is bound to the first and end endpoints, that is, a ring

Proposition:At least one vertex of a directed acyclic graph has an outbound degree of 0.
Proof:Assume that the graph has a longest path L and an endpoint is a. If there are at least two edges connected to a, assume that the edges not in L are C. According to the longest path principle, the point at the other end of C can only be in L, and the question shows that the graph is a non-circled graph, so the side C connected to a does not exist, in this way, we can see that only one edge can be connected with a, so the outbound degree of a is 0.

3. Dual direction Principle
Inverse directed graph concept:Simply put, it is to reverse all edges of a directed graph.
Eg
Proposition:At least one vertex of a directed acyclic graph has an inbound degree of 0.
Proof:Through the inverse directed graph of the graph proved in the previous proposition, we can easily see that this proposition is also true (in fact, It is equivalent ).

4. Introduction of Matrices
A graph is determined by its joining or associativity. This information can be easily expressed using matrices.
Eg
Adjacent matrix
Theorem:If A is An adjacent matrix of G, the value of element A [I] [j] of An (n OF a) is the number of roads with the length from Vi to Vj equal to n.
Inference 1:In A2, a [I] [I] is the degree of Vi. The a [I] [I] of A3 is twice the number of triangles containing Vi.
Inference 2:If G is connected, for I! = J. The distance between Vi and Vj is the minimum integer n that makes a [I] [j] of An not equal to 0.

5. The graph degree is twice the number of edges.
Because graphs are composed of vertices and edges, and degrees are used to describe the quantitative relationship between these two quantities (because we usually know the degree of points), this equality relationship is too important, although it is very simple.
Eg
Proposition:If a graph is connected and the degree of each vertex is at least 2, graph G must contain a circle.
Proof:If the number of vertices in the graph is n, the total number of vertices in the graph is greater than or equal to 2 * n, so the number of edges is greater than or equal to n, because the generative tree of a graph has only n-1 edges, therefore, an extra edge will surely form a ring, and the proposition is proved.

Ii. Concept Review:
1. Cut Point:V is a cut point of graph G. There is a division that divides the point set V-{v} into subsets U and W, so that for any two points u <-U, and w <-W, point v is on every u-w Road.

2. Bridge:X is a bridge of G. X is not in any circle of G.

3. Block:G is a block of the graph. Any two points of g are in a public circle. For every three different points of g, there is a road connecting these two points, and there is a third point including (not included. (LA 3523)

4. Europe:
1). G is Euler's <=> G.
2). The degree of each vertex in an undirected graph G is an even number.
3). The inbound degree of directed graph G is equal to the outbound degree.
4). The edge set of the graph can be divided into circles.

5. Independent Set and coverage:
Point Coverage:Overwrite a "vertex" set of all "edges" of a graph.
Edge coverage:Overwrite an edge set of all vertices in a graph.
Minimum vertex coverage set:All vertices overwrite the set with the least number of elements, and the number of elements is Df.
Minimum edge cover set:The set with the least number of elements in all edges. The number of elements is Bf.
Point independence set:Is a point set of an image. No two points are adjacent to the source image.
Edge Independence set:Is the edge set of an image. No two edges are adjacent to the source image.
Maximum vertex independence set:The set with the largest number of elements in the independent set of all vertices. The number of elements is Dd.
Maximum Edge Independence set:The set with the largest number of elements in an independent set of all edges. The number of elements is Bd.

Theorem of importance for any connected graph G
Df + Dd = P = Bf + Bd (P is the number of vertices of G)

Important Expansion
If G is a bipartite graph, Bd = Df.
The Bd of the largest Edge Independence set is the maximum matching number.

Iii. Theorem
1. mingger's theorem:Separate the minimum number of s and t of two non-adjacent points to the maximum number of s-t PATHS where the points do not overlap.
Mingge theorem Deformation:For any two vertices in a graph, the maximum number of paths that link their edges is equal to the minimum number of edges that separate them.
(Point not intersection, that is, we generally know that the side is not intersection, that is, the path does not overlap .)

Max flow least cut theorem:In any network, if there is a path from s to t, the maximum stream from s to t is equal to the minimum cut-off.
To prove this theorem, we can replace the directed edge with n capacity from u to v into n non-capacity directed edges. In this way, we can use the deformation of the mingger theorem to prove this theorem.

2. A graph can be two-color only if it has no Odd Circle. (Used as the basis for determining the Bipartite Graph Algorithm)

3. The Edge Orientation of an undirected connected graph without bridges can be made into a strongly connected graph. (Ultraviolet A 10972)

A directed graph D is strongly connected. if and only if there is a loop in D, it passes through each vertex at least once. (La 4287)

An undirected graph G is edge-connected. if and only when G contains a loop, it passes through each vertex at least once. (Ultraviolet A 10972)

4. In an undirected graph, where the degrees of each vertex are even, some loops of the graph can be found so that each edge exactly belongs to one of these loops.
In a directed graph, when the inbound and outbound degrees of each vertex are equal, some loops of the graph can be found so that each edge exactly belongs to one of these loops.