Vertex Cut

• Vertex Cut: A separating set or vertex cut of a graph G is a set SV(G) such that G-S has more than one component.

a

b

c

d

e

f

g

h i

Connectivity• Connectivity of G ((G)): The minimum size of a

vertex set S such that G-S is disconnected or has only one vertex. Thus, (G) is the minimum size of vertex cut. (X)

(G)=4(G)=2

k-Connected Graph• k-Connected Graph: The graph whose

connectivity is at least k.

(G)=2

G is a 2-connected graph

Is G a 1-connected graph?

Connectivity of Kn

• A clique has no separating set. And, Kn- S has only one vertex for S=Kn-1 (Kn)=n-1.

Connectivity of Km,n

Every induced subgraph that has at least one vertex from X and from Y is connected.

Every separating set contains X or Y (Km,n)= min(m,n) since X and Y themselves are

separating sets (or leave only one vertex).

K4,3

Harary Graph Hk,n

• Given 2<=k<n, place n vertices around a circle, equally spaced.

Case 1: k is even. Form Hk,n by making each vertex adjacent to the nearest k/2 vertices in each direction around the circle.

H4,8

(Hk,n)=k.

|E(Hk,n)|= kn/2

Harary Graph Hk,n

• Case 2: k is odd and n is even. Form Hk,n by making each vertex adjacent to the nearest (k-1)/2 vertices in each direction around the circle and to the diametrically opposite vertex.

H5,8

(Hk,n)=k.

|E(Hk,n)|= kn/2

Harary Graph Hk,n (2/2) Case 3: k is odd and n is odd. Index the vertices by

the integers modulo n. Form Hk,n by making each vertex adjacent to the nearest (k-1)/2 vertices in each direction around the circle and adding the edges ii+(n-1)/2 for 0<=i<=(n-1)/2.

4

62

1

3 5

70 8

H5,9

(Hk,n)=k.

|E(Hk,n)|= (kn+1)/2

In all cases, (Hk,n)=k.|E(Hk,n)|= kn/2

Theorem 4.1.5

(Hk,n ) =kProof. 1. (Hk,n ) =k is proved only for the even

case k=2r. (Leave the odd case as Exercise 12)2. We need to show SV(G) with |S|<k is not a vertex cut

H4,8

since (Hk,n)=k.

Theorem 4.1.53. Consider u,vV-S. The original circular has a clockwise u,v-path and a counterclockwise u,v-path along the circle.

H4,8

u

v

AB

5. It suffices to show there is a u,v-path in V-S via the set A or the set B if |S|<k.

4. Let A and B be the sets of internal vertices on these two paths.

Theorem 4.1.56. |S|<k.

u

v

A

B

H4,8

S has fewer than k/2 vertices in one of A and B, say A. Deleting fewer than k/2 consecutive vertices cannot block travel in the direction of A. There is a u,v-path in V-S via the set A.

Theorem 4.1.5 (2)The minimum number of edges in a k-connected graph on n

vertices is kn/2.

1. Since Hk,n has kn/2 edges, we need to show a k-connected graph on n vertices has at least kn/2 edges.

2. Each vertex has k incident edge in k-connected graph. k-connected graph on n vertices has at least kn/2

vertices.

Disconnecting Set• Disconnecting Set of Edges: A set of edges F such

that G-F has more than one component.

k-Edge-Connected Graph: Every disconnecting set has at least k edges.

Edge-Connectivity of G (’(G)): The minimum size of a disconnecting set.

Edge Cut• Edge Cut: Given S,TV(G), [S,T] denotes the set of

edges having one endpoint in S and the other in T. An edge cut is an edge set of the form [S,V-S], where S is a nonempty proper subset of V(G).

S V-S

Remark• Every edge cut is a disconnecting set, since G- [S,V-

S] has no path from S to V-S.• The converse is false, since a disconnecting set can

have extra edges.• Every minimal disconnecting set of edges is an edge

cut (when n(G)>1). If G-F has more than one component for some FE(G), then for some component H of G-F we have deleted all edges with exactly one endpoint in H. Hence F contains the edge cut [V(H),V-V(H)], and F is not minimal disconnecting set unless F=[V(H),V-V(H)].

Theorem 4.1.9If G is a simple graph, then (G)<=’(G)<= (G).Proof. 1. ’(G)<= (G)

3. Consider a smallest edge cut [S,V-S].

since the edges incident to a vertex v of minimum degree form an edge cut.

4. Case 1: Every vertex of S is adjacent to every vertex of V-S.

5. Case 2: there exists xS and yV-S such that (x,y)E(G).

’(G)>=k(G) since (G)<=n(G)-1. ’(G)=|[S,V-S]|=|S||V-S|>=n(G)-1.

2. We need to show (G)<=’(G).(’(G)= |[S,V-S]|)

Theorem 4.1.9

6. Let T consist of all neighbors of x in V-S and all vertices of S-{x} with neighbors in V-S.

x

T

TT

T

T

y

S V-S

7. Every x,y-path pass through T. T is a separating set. (G)<=|T|.

8. It suffices to show |[S,V-S]|>=|T|.

5. Case 2: there exists xS and yV-S such that (x,y)E(G).

Theorem 4.1.9

x

T

TT

T

T

y

S V-S

9. Pick the edges from x to TV-S and one edge from each vertex of TS to V-S yields |T| distinct edges of [S,V-S]. ’(G)= |[S,V-S]|>=|T|.

9. Pick the edges from x to TV-S and one edge from each vertex of TS to V-S yields |T| distinct edges of [S,V-S].

Possibility of (G)<’(G)<(G)

1. (G) = 1.2. ’(G) = 2.3. (G) = 3.

Theorem 4.1.11

If G is a 3-regular graph, then (G) =’(G).Proof. 1. Let S be a minimum vertex cut.

H1 H2

S

2. Let H1, H2 be two components of G-S.

Theorem 4.1.113. Each vS has a neighbor in H1 and a neighbor in H2.

Otherwise, S-{v} is a minimum vertex cut.4. G is 3-regular, v cannot have two neighbors in H1 and two in H2.5. There are three cases for v.

H1 H1 H1 H2 H1 H2

Case 1 Case 2 Case 3

v vv

u

Theorem 4.1.11 (2/2)5. For Cases 1 and 2, delete the edge from v to a member of {H1, H2} where v has only one neighbor.

6. For Case 3, delete the edge from v to H1 and the edge from v to H2.

H1 H1 H1 H2 H1 H2

Case 1 Case 2 Case 3

v vv

u

7. These (G) edges break all paths from H1 to H2 .