1. Vertex Cut
Vertex Cut: A separating set or vertex cut of a graph G is a set SV(G) such that G-S has more than one component. d f b e a g c i h

2. 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

3. k-Connected Graph
k-Connected Graph: The graph whose connectivity is at least k.
(G)=2 a b c d e f g h i
G is a 2-connected graph
Is G a 1-connected graph ?

4. Connectivity of Kn
A clique has no separating set. And, Kn- S has only one vertex for S=Kn-1  (Kn)=n-1.

5. 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

6. Connectivity of Qk
K-dimensional Hypercube Qk :
K=1
K=2
K=3
K=4

7. Connectivity of Qk
For k>=2, the neighbors of one vertex in Qk form a separating set.  (Qk)<=k.
K=1
K=2
K=3
K=4

8. Connectivity of Qk
Every vertex cut has size at least k as proved by induction on k.  (Qk) =k.
Basic Step: For k<=1, Qk is a complete graph with k+1 vertices and has connectivity k.
Induction Hypothesis: (Qk-1)=k-1.
Induction Step: Consider as two copies Q and Q’ of Qk-1 plus a matching that joins corresponding vertices in Q and Q’. Let S be a vertex cut in Qk.

9. Connectivity of Qk Case 1: Q-S is connected and Q’-S is connected.
 S contains at least one endpoint of every match pair.
 |S|>= 2k-1.
 |S|>=k for k>=2. Q Q’

10. Connectivity of Qk Case 2: Q-S is disconnected .
 S contains at least k-1 vertices in Q.
(Induction Hypothesis)
 |S|>=k because S contains at least 1 vertices in Q’.
(If S contains no vertices of Q’, Qk-S is connected.) Q Q’

11. 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.
(Hk,n)=k.
|E(Hk,n)|= kn/2
H4,8

12. 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.
(Hk,n)=k.
|E(Hk,n)|= kn/2
H5,8

13. 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 ii+(n-1)/2 for 0<=i<=(n-1)/2.
In all cases, (Hk,n)=k. 2 1 3 4 5 6 7 8
|E(Hk,n)|= kn/2
H5,9
(Hk,n)=k.
|E(Hk,n)|= (kn+1)/2

14. Theorem 4.1.5
(Hk,n ) =k, and hence the minimum number of edges in a k-connected graph on n vertices is kn/2.
Proof. 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 SV(G) with |S|<k is not a vertex cut
since (Hk,n)=k.
H4,8

15. Theorem 4.1.5
3. Consider u,vV-S. The original circular has a clockwise u,v-path and a counterclockwise u,v-path along the circle.
4. Let A and B be the sets of internal vertices on these two paths.
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. .
H4,8 u v A B

16. Theorem 4.1.5
6. |S|<k.
 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. u v A B
H4,8

17. Theorem 4.1.5
7. Since Hk,n has kn/2 edges, we need to show a k-connected graph on n vertices has at least kn/2 edges.
8. Each vertex has k incident edge in k-connected graph.  k-connected graph on n vertices has at least kn/2 vertices.

18. Disconnecting Set
Disconnecting Set of Edges: A set of edges F such that G-F has more than one component.
Edge-Connectivity of G (’(G)): The minimum size of a disconnecting set.
k-Edge-Connected Graph: Every disconnecting set has at least k edges.

19. Edge Cut
Edge Cut: Given S,TV(G), [S,T] denotes the set of edges having one endpoint in S and the other in G. 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

20. Theorem 4.1.9 If G is a simple graph, then (G)<=’(G)<= (G).
Proof. 1. ’(G)<= (G)
since the edges incident to a vertex v of minimum degree form an edge cut.
2. We need to show (G)<=’(G).
3. Consider a smallest edge cut [S,V-S].
(’(G)= |[S,V-S]|)
4. Case 1: Every vertex of S is adjacent to every vertex of V-S.
 ’(G)=|[S,V-S]|=|S||V-S|>=n(G)-1.
 ’(G)>=k(G) since (G)<=n(G)-1.
5. Case 2: there exists xS and yV-S such that (x,y)E(G).

21. Theorem 4.1.9
5. Case 2: there exists xS and yV-S such that (x,y)E(G).
6. Let T consist of all neighbors of x in V-S and all vertices of S-{x} with neighbors in 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|. x T T
V-S S T T y T

22. Theorem 4.1.9
9. Pick the edges from x to TV-S and one edge from each vertex of TS to V-S yields |T| distinct edges of [S,V-S].
9. Pick the edges from x to TV-S and one edge from each vertex of TS to V-S yields |T| distinct edges of [S,V-S].
 ’(G)= |[S,V-S]|>=|T|. x T T
V-S S T T y T

23. Possibility of (G)<’(G)<(G)

24. Theorem 4.1.11 If G is a 3-regular graph, then (G) =’(G).
Proof. 1. Let S be a minimum vertex cut.
2. Let H1, H2 be two components of G-S. S H1 H2

25. Theorem 4.1.11 3. Each vS 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. v v v H1 H2 H1 H1 H1 H2 u
Case 1
Case 2
Case 3

26. Theorem (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. v v v H1 H2 H1 H1 H1 H2 u
Case 1
Case 2
Case 3
7. These (G) edges break all paths from H1 to H2 .