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Chapter 9 Connectivity 连通度

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Chapter 9 Connectivity连通度

Consider the following graphs:

- G1: Deleting any edge makes it disconnected.
- G2: Cannot be disconnected by deletion of any edge; can be disconnected by deleting its cut vertex;
- Intuitively, G2 is more connected than G1, G3 is more connected thant G2, and G4 is the most connected one.

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:

- An edge e of G is a cut edge
- e is not contained in any cycle of G.
- There are two vertices u and w such that e is on every path connecting u and w.

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.

- A complete graph has no vertex cut. Define (Kn)=n-1;
- For disconnected graph G, define (G) = 0;
- G is said to be k-connected if (G)k;
- It is easy to see that all nontrivial connected graphs are 1-connected.
- (G)=1 if and only if G=K2 or G has a cut vertex.

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.

- For trivial and disconnected graph G, define (G)=0;
- G is said to be k-edge-connected if (G)k ;
- All nontrivial connected graphs are 1-edge-connected.

Find (G), (G) and (G) for the following graphs

Theorem 9.3 For any connected graph G

(G) (G)(G)

where (G) is the smallest vertex degree of G.

容易说明(G)(G)，只需将最小度数结点的关联边删除可得非连通图。如左图所示。

如何说明(G) (G)？

- 设E={e1,e2,…,ek}是割边集，删除每个边ei的一个端点便删除了E，从而得到非连通图？
- E={(u,w),(v,x)}是割边集，删除u,v后图仍然是连通的。
- 假定删除E后有连通分支G1, G2, 必有v1V(G1),v2V(G2), v1,v2不相邻(见课本)，应该保留v1,v2，删除E的k个结点使得E被删除。如右图，保留u,x, 删除w,v.

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.

- The theorem can be proved by induction on the length of the paths. See the textbook for the proof.
- State it in another way: A graph G of order n(3) is 2-connected if and only if any two vertices of G connected by at least 2 vertex-disjoint paths.

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.

- State it in another way: A graph G of order n(3) is 2-edge-connected if and only if any two vertices of G are connected by at least 2 edge-disjoint paths.

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Let v and w are two vertices in a graph. A collection of paths from v to w are called edge-disjoint paths if no two paths in it share an edge.

Count the number of edge-disjoint paths from v to w in the graph above. Find the edge connectivity of the graph.

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Similarly, we can define vertex-disjoint paths.

Find the connectivity of the graph and the number of vertex-disjoint paths from v to w.

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Theorem 9.6 The maximum number of edge-disjoint paths connecting two distinct vertices v and w in connected graph G is equal to the minimum number of edges whose removal disconnecting v and w.

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

- the capacities represent the number of arcs connecting the various vertices.
- The maximum flow corresponds to the total number of arc-disjoint path form s to t in D;
- The capacity of a minimum cut refers to the minimum number of arcs in a st-disconnecting set of D.

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Lemma 9.9 Let N be a network with source s and sink t in which each arc has unit capacity. Then

- The value of a maximum flow in N is equal to the maximum number m of arc-disjoint directed (s,t)-paths in N; and
- The capacity of a minimum cut in N is equal to the minimum number n of arcs whose deletion destroys all directed (s,t)-paths in N.

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.

- Construct a new digraph D’ from D by splitting each vertex vV-{s,t} such that v becomes an arc v’->v’’, arcs leading to v now leading to v’ and arcs leaving v now leaving from v’’;
- To each edge-disjoint (s,t)-path in D’ there corresponds a vertex-disjoint directed (s,t)-path in D;and vice verse; and See Bondy and Murty.

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Theorem 9.12 Menger’s theorem implies Hall’s theorem.

There is a complete matching from V1 to V2 if and only if the number of vertex-disjoint paths from v to w is equal to |V1|=k.

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)

|V1-A|<=|N(V1-A)|<=|B|.

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.

因为AB是一个vw分离集, 不存在V1-A与V2-B之间的边,

所以 N(V1-A)B, 再根据 Hall条件，

|V1-A|<=|N(V1-A)|<=|B|.

所以, |S|=|AB|=|A|+|B|>=|V1|=k.

- A graph representing a communication network, the connectivity (or edge-connectivity) becomes the smallest number of stations (or links) whose breakdown would jeopardise the system.
- The higher the connectivity and edge connectivity, the more reliable the network.
- Let k be a given positive integer and let G be a weighted graph. Determine a minimum-weight k-connected spanning subgraph of G.
- For k=1, this is solved by Kruskal’s algorithm, for example. For k>1, the problem is unsolved.

- Vertex cut and edge cut;
- Connectivity and edge-connectivity;
- Menger Theorem;
- Equivalence of Menger Theorem and the Max-flow Min-cut Theorem;
- Menger Theorem implies Hall Theorem.