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## Vertex Cut

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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 G-S has more than one component. d f b e a g c i h**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 minimumsize 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. (Hk,n)=k. |E(Hk,n)|= kn/2 H4,8**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**5**3 2 1 7 0 8 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. In all cases, (Hk,n)=k. 4 |E(Hk,n)|= kn/2 H5,9 6 (Hk,n)=k. |E(Hk,n)|= (kn+1)/2**Theorem 4.1.5**(Hk,n) =k 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 SV(G) with |S|<k is not a vertex cut since (Hk,n)=k. H4,8**B**u A H4,8 v Theorem 4.1.5 3. Consider u,vV-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.**B**u A v 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. H4,8**Theorem 4.1.5 (2)**• The minimum number of edges in a k-connected graph on n vertices is kn/2. • Since Hk,nhas 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. Edge-Connectivity of G (’(G)): The minimum size of a disconnecting set. k-Edge-Connected Graph: Every disconnecting set has at least k edges.**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.**Remark**• 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)]. • A minimum disconnected set is a minimum edge cut. • The edge connectivity of G is the minimum size of an edge cut.**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.**Theorem 4.1.9**5. Case 2: there exists xS and yV-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**Theorem 4.1.9**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]. 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|. x T T V-S S T T y T**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**Theorem 4.1.11**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. There are three cases for v. v v v H1 H2 H1 H1 H1 H2 u Case 1 Case 2 Case 3**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. 7. These (G) edges break all paths from H1 to H2 . 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