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

Connectivity of K m,n Every induced subgraph that has at least one vertex from X and from Y is connected.  Every separating set contains X or Y   (K m,n )= min(m,n) since X and Y themselves are separating sets (or leave only one vertex). K4,3

Harary Graph H k,n Given 2<=k<n, place n vertices around a circle, equally spaced. Case 1: k is even. Form H k,n by making each vertex adjacent to the nearest k/2 vertices in each direction around the circle. H 4,8  ( H k,n )=k. |E(H k,n )|= kn/2

Harary Graph H k,n Case 2: k is odd and n is even. Form H k,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. H 5,8  ( H k,n )=k. |E(H k,n )|= kn/2

Harary Graph H k,n (2/2) Case 3: k is odd and n is odd. Index the vertices by the integers modulo n. Form H k,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. 4 6 2 1 3 5 7 08 H 5,9  ( H k,n )=k. |E(H k,n )|= (kn+1)/2 In all cases,  ( H k,n )=k. |E(H k,n )|=  kn/2 

Theorem 4.1.5  (H k,n ) =k Proof. 1.  (H k,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 H 4,8 since  (H k,n )=k.

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. H 4,8 u v A B 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 4.1.5 6. |S|<k. u v A B H 4,8  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 H k,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,T  V(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). SV-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 F  E(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 4.1.9 If 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 x  S and y  V-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 T T 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 x  S and y  V-S such that (x,y)  E(G).

Theorem 4.1.9 x T T T T T y 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|. 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].

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. H1H2 S 2. Let H1, H2 be two components of G-S.

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. H1 H2 H1H2 Case 1Case 2Case 3 vv v 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 H2 H1H2 Case 1Case 2Case 3 vv v u 7. These  (G) edges break all paths from H1 to H2.

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