Vertex Cut
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Transcript of Vertex Cut
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.
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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
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.
H4,8
(Hk,n)=k.
|E(Hk,n)|= kn/2
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.
H5,8
(Hk,n)=k.
|E(Hk,n)|= kn/2
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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(Hk,n)=k.
|E(Hk,n)|= (kn+1)/2
In all cases, (Hk,n)=k.|E(Hk,n)|= kn/2
Theorem 4.1.5
(Hk,n ) =kProof. 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
H4,8
since (Hk,n)=k.
Theorem 4.1.53. Consider u,vV-S. The original circular has a clockwise u,v-path and a counterclockwise u,v-path along the circle.
H4,8
u
v
AB
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 4.1.56. |S|<k.
u
v
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B
H4,8
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 Hk,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,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 4.1.9If 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.
’(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.
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T
TT
T
T
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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 xS and yV-S such that (x,y)E(G).
Theorem 4.1.9
x
T
TT
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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|.
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].
Theorem 4.1.11
If G is a 3-regular graph, then (G) =’(G).Proof. 1. Let S be a minimum vertex cut.
H1 H2
S
2. Let H1, H2 be two components of G-S.
Theorem 4.1.113. 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.
H1 H1 H1 H2 H1 H2
Case 1 Case 2 Case 3
v vv
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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 H1 H1 H2 H1 H2
Case 1 Case 2 Case 3
v vv
u
7. These (G) edges break all paths from H1 to H2 .