Vertex Cut

1
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.

d
f
b
e
a
g
c
i
h
2
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
3
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 ?
4
Connectivity of Kn

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

5
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
6
Connectivity of Qk

• K-dimensional Hypercube Qk

K1
K2
K3
K4
7
Connectivity of Qk

• For kgt2, the neighbors of one vertex in Qk form
  a separating set. ? ?(Qk)ltk.

K1
K2
K3
K4
8
Connectivity of Qk

• Every vertex cut has size at least k as proved by
  induction on k. ? ?(Qk) k.

Basic Step For klt1, Qk is a complete graph with
k1 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.
9
Connectivity of Qk

• Case 1 Q-S is connected and Q-S is connected.

? S contains at least one endpoint of every match
pair.
? Sgt 2k-1.
? Sgtk for kgt2.
Q
Q
10
Connectivity of Qk

• Case 2 Q-S is disconnected .

? S contains at least k-1 vertices in Q.
(Induction Hypothesis)
? Sgtk because S contains at least 1 vertices
in Q.
(If S contains no vertices of Q, Qk-S is
connected.)
Q
Q
11
Harary Graph Hk,n

• Given 2ltkltn, 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
12
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
13
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 i?i(n-1)/2 for
  0ltilt(n-1)/2.

In all cases, ?(Hk,n)k.
E(Hk,n) ?kn/2?
H5,9
?(Hk,n)k.
E(Hk,n) (kn1)/2
14
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 k2r. (Leave the odd case as Exercise 12)

2. We need to show S?V(G) with Sltk is not a
vertex cut
since ?(Hk,n)k.
H4,8
15
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.
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 Sltk. .
16
Theorem 4.1.5

• 6. Sltk.

? 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
17
Theorem 4.1.5
7. Since Hk,n has ?kn/2? edges, we need to show a
k-connected graph on n vertices has at least
?kn/2? edges.
8. Each vertex has k incident edge in k-connected
graph. ? k-connected graph on n vertices has at
least ?kn/2? vertices.
18
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.
19
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 G. 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
20
Theorem 4.1.9

• If G is a simple graph, then ?(G)lt?(G)lt ?(G).
• Proof. 1. ?(G)lt ?(G)

since the
edges incident to a vertex v of minimum degree
form an edge cut.
2. We need to show ?(G)lt?(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-SSV-Sgtn(G)-1.
? ?(G)gtk(G) since ?(G)ltn(G)-1.
5. Case 2 there exists x?S and y?V-S such that
(x,y)?E(G).
21
Theorem 4.1.9
5. Case 2 there exists x?S and y?V-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)ltT.
8. It suffices to show S,V-SgtT.
x
T
T
V-S
S
T
T
y
T
22
Theorem 4.1.9
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.
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-SgtT.
x
T
T
V-S
S
T
T
y
T
23
Possibility of ?(G)lt?(G)lt?(G)
1. ?(G) 1. 2. ?(G) 2. 3. ?(G) 3.
24
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
25
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.
v
H1
H2
v
v
H1
H1
H1
H2
u
Case 1
Case 2
Case 3
26
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.
v
H1
H2
v
v
H1
H1
H1
H2
u
Case 1
Case 2
Case 3

• 7. These ?(G) edges break all paths from H1 to H2
  .