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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

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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

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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?

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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.

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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

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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

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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

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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

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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.

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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.

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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.

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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.

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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.

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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

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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)].

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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]|)

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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).

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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].

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Possibility of (G)< ’(G)< (G) 1. (G) = 1. 2. ’(G) = 2. 3. (G) = 3.

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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.

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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

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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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