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Vertex Cut Vertex Cut: A separating set or vertex cut of a graph G is a set SV(G) such that S has more than one component. Connectivity of G ((G)): The minimum size of a vertex set S such that S is disconnected or has only one vertex. k-Connected Graph: The graph whose connectivity is at least k.

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**Connectivity of Kn and Km,n**

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

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Connectivity of Qk K-dimensional Hypercube Qk :

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Connectivity of Qk (2/3) For k>=2, the neighbors of one vertex in Qk form a separating set. (Qk)<=k. 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.

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Connectivity of Qk (3/3) 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. Case 2: Q-S is disconnected. S contains at least k-1 vertices in Q by induction hypothesis. |S|>=k. Otherwise, S contains no vertices of Q’, implying Qk-S is connected.

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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. 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.

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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.

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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. Consider SV(G) with |S|<k. 3. Consider u,vS. The original circular has a clockwise u,v-path and a counterclockwise u,v-path along the circle. Let A and B be the sets of internal vertices on these two paths.

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Theorem (2/2) 4. |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 S via the set A. S is connected. S is not a vertex cut. 5. (Hk,n )>=k. (Hk,n) =k since (G)=k. 6. Each vertex has k incident edge in k-connected graph. k-connected graph on n vertices has at least kn/2 vertices. 7. Hk,n has kn/2 edges. The minimum number of edges in a k-connected graph on n vertices is kn/2.

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Edge Cut 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: Given S,TV(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, S], where S is a nonempty proper subset of V(G).

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Edge Cut (2/2)

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

Proof. 1. The edges incident to a vertex v of minimum degree form an edge cut. ’(G)<= (G). 2. Consider a smallest edge set [S, S]. 3. If every vertex of S is adjacent to every vertex of S, then ’(G)=|[S, S]|=|S||S|>=n(G)-1. ’(G)>=k(G) since (G)<=n(G)-1. 4. Otherwise, we choose xS and y S with (x,y)E(G).

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Theorem 4.1.9 5. Let T consist of all neighbors of x in S and all vertices of S-{x} with neighbors in S. 6. Every x,y-path pass through T. T is a separating set. (G)<=|T|. 7. Pick the edges from x to T S and one edge from each vertex of TS to S yields |T| distinct edges of [S, S]. ’(G)= |[S, S]|>=|T|. (G)<=’(G).

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**Possibility of (G)<’(G)<(G)**

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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. 2. Let H1, H2 be two components of G-S. 3. 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. For each vS with one neighbor in S, delete the edge from v to a member of {H1, H2} where v has only one neighbor.

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Theorem (2/2) 6. For each vS with no neighbor in S, delete the edge to H1. 7. These (G) edges break all paths from H1 to H2 .

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