Chapter 9 Connectivity 连通度. 9.1 Connectivity. Consider the following graphs: G 1 : Deleting any edge makes it disconnected. G 2 : Cannot be disconnected by deletion of any edge; can be disconnected by deleting its cut vertex;
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Consider the following graphs:
A cut edge of G is an edge such that G-e has more components that G.
Theorem 9.1 Let G be a connected graph. The following are equivalent:
Let G be a nontrivial and loopless graph. A vertex v of G is a cut vertex if G-v has more components than G.
Theorem 9.2 Let G be a connected graph. The following propositions are equivalent:
1. A vertex v is a cut vertex of G
2. There are two distinct vertices u and w such that every path between u and w passes v;
3. The vertices of G can be partitioned into two disjoint vertex sets U and W such that every path between uU and wW passes v.
A vertex cut of G is a subset V’ of V such that G-V’ is disconnected. The connectivity , (G), is the smallest number of vertices in any vertex cut of G.
Let G be graph on n2 vertices. An edge cut is a subset E’ of E(G) such that G-E’ is disconnected.
The edge connectivity, (G), is the smallest number of edges in any edge cut.
Find (G), (G) and (G) for the following graphs
Theorem 9.3 For any connected graph G
where (G) is the smallest vertex degree of G.
Theorem 9.4 (Whitney) A graph G of order n(3) is 2-connected if and only if any two vertices of G are in a common cycle.
Theorem 9.5 (Whitney) A graph G of order n(3) is 2-edge-connected if and only if any two vertices of G are in a simple closed path.
Theorem 9.6’ A graph G is k-edge-connected if and only if any two distinct vertices of G are connected by at least k edge-disjoint paths.
Proof：If there are two vertices which are connected by less than k edge-disjoint paths, then G is not k-edge-connected. On the other hand, if G is not k-edge-connected, there are edge cut that contains less than k edges, hence there are two vertices which are connected by less than k edge-disjoint paths.
Theorem 9.7’ A graph of order n(k+1) is k-connected if and only if any two distinct vertices of G are connected by at least k vertex-disjoint paths.
Theorem 9.7 The maximum number of vertex-disjoint paths connecting two distinct non-adjacent vertices v and w of a connected graph G is equal to the minimum number of vertices whose removal disconnecting v and w.
Theorem 9.8 The maximum number of arc-disjoint paths from a vertex v to a vertex w in a digraph D is equal to the minimum number of arcs whose removal disconnecting v and w.
Proof: Assuming the capacity of every arc is an integer.
The network N can be seen as a digraph D in which
Lemma 9.9 Let N be a network with source s and sink t in which each arc has unit capacity. Then
Theorem 9.8 (Menger) Let s and t be two vertices of a digraph D. Then the maximum number of arc-disjoint directed (s,t)-paths in D is equal to the minimum number of arcs whose deletion destroys all directed (s,t)-paths in D.
Theorem 9.10 (Menger, undirected version) Let s and t be two vertices of a graph G. Then the maximum number of edge-disjoint (s,t)-paths in G is equal to the minimum number of edges whose deletion destroys all (s,t)-paths in G.
Proof: Apply the directed version of Menger’s theorem to the associated digraph D(G) of G (an edge becomes two directed edges). There is a one-one correspondence between paths in G and D(G). See Bondy and Murty.
Theorem 9.11 Let s and t be two vertices of a directed graph D such that s is not joined to t. Then the maximum number of vertex-disjoint (s,t)-paths in G is equal to the minimum number of vertices whose deletion destroys all directed (s,t)-paths in D.
Proof: by converting it to the arc-version of Menger’s theorem.
Proof : Let G=(V1,V2) be a bipartite graph. We have to prove that if |A||N(A)| for each subset A of V1, then there exists a complete matching form V1 to V2.
We add two extra vertices v and w (see the graph on previous page). Using Menger’s theorem of the vertex form, it is enough to prove that every vw-separating set (whose removal disconnect v and w) contains at least |V1|=k vertices.
Let S be a vw-separating set, consists of AV1 and B V2.
Since AB is a vw-separating set, there can be no edges joining a vertex of V1-A to a vertex of V2-B, that is, N(V1-A)B. It follows that (Hall’s condition)
so |S|=|AB|=|A|+|B|>=|V1|=k, as required.
证明: 设G=(V1,V2)是一个二分图. 证明如果对于V1的任意子集A, |A| <= |N(A)|, 则存在V1至V2的完全匹配.
在G上添加两个结点v,w, 使得v与V1的所有结点相邻, w与V2 的所有结点相邻. 显然, 存在V1至V2的完全匹配当且仅当v至w的结点不相交道路数目是|V1|=k.
因为V1构成分割v,w的一个集合，所以，根据Menger定理, 只需证明分割v,w的最少结点数是k. 设S是分割v,w的结点集, 且有V1的子集A和V2的子集B构成.
所以 N(V1-A)B, 再根据 Hall条件，