CSCI 4260 MATH 4150

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CSCI 4260MATH 4150

• GRAPH THEORY

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Last time

• We learned about BFS
• We will cover DFS after we learn more about
  connectivity and digraphs
• Today, connectivity properties

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Robustness of connectivity

• We can make a graph disconnected by deleting
  either edges or vertices
• An edge is a bridge of G if G-e is disconnected

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Theorem

• An edge e is a bridge of G iff e lies on no cycle
  on G
• e is a bridge ? e is on no cycle
• e is on a cycle ? e is not a bridge
• e is on no cycle ? e is a bridge
• e is not a bridge ? e is on a cycle

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bridges

• In a sense, if a graph contains a bridge, then
  its connectivity is not very robust
• Can you come up with a graph where every edge is
  a bridge?

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Cut vertices

• v is a cut vertex if G-v is disconnected
• Note that we delete edges incident to v as well

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Theorem

• Let v be a vertex incident with a bridge.Then, v
  is a cut vertex iff deg(v) ? 2
• v is a cut vertex ? deg(v) ? 2
• deg(v) lt 2 ? v is a not cut vertex
• deg(v) ? 2 ? v is a cut vertex

bridge
v
w
u
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Corollary

• Given connected G of order ? 3. If G contains a
  bridge, then G contains a cut vertex

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Theorem

• A vertex v of a connected graph G is a cut vertex
  of G iff there exists two vertices u and w (not
  equal to v) such that v lies on every u-w path.

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Non-cut vertices

• All edges of a tree are bridges
• Can you come up with a graph where every vertex
  is a cut vertex?

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Theorem

• Let G be a nontrivial, connected graph. Pick a
  vertex u. If v is a vertex that is farthest from
  u in G, then v is not a cut vertex.

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Corollary

• Every non trivial graph contains at least two
  non-cut vertices.

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Non-separable graphs

• A graph is non-separable if it contains no cut
  vertices
• Example?

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Theorem

• A graph of order at least 3 is non-separable if
  and only if every two vertices lie on a common
  cycle
• Common cycle ? non-separable
• Suppose not
• Non-separable ? common-cycle
• Suppose not. Pick u and v that do not lie on a
  common cycle. Among all possible pairs, pick two
  such that d(u,v) is as small as possible.

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Non-separable ? common-cycle

• Case 1 d(u,v) 1. Not possible. The edge (u,v)
  is a bridge

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Note w is not a cut vertex
v
w
u
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Blocks

• Maximal non-separable subgraphs
• If G is non-separable then it has only one block
• Analogy disconnected graphs are composed of
  components (connected subgraphs)

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Theorem

• Define a relation R on edges where
• e R f either ef or e and f lie on a common
  cycle.
• R is an equivalence relation

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Transitivity
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Corollary

• Every two distinct blocks B1 and B2 of a
  nontrivial connected graph have
• B1 and B2 are edge disjoint
• B1 and B2 have at most one vertex in common
• If v is the common vertex, then v is a cut vertex.

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B1 and B2 are edge disjoint

• Immediate because blocks correspond to the
  equivalence classes of R as we just saw

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B1 and B2 have at most one vertex in common

• Suppose not.

u
P
P
v
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If v is the common vertex, then v is a cut vertex.
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Vertex-cuts

• A set U of vertices is a vertex cut if G-U is
  disconnected

U
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Connectivity of G

• Minimum vertex cut
• Define the connectivity of G, ??(G)0 ? ?(G) ?
  n-1
• We say a graph is k-connected if ?(G) ? k

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Edge-cuts

• Are defined similarly
• Minimal vs. minimum?

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Edge-connectivity??

• ?(G) size of a min edge cut
• G is k-edge connected if ?(G) ? k

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Theorem

• For every graph G ?(G) ? ?(G) ? ?(G)

Minimum degree
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Proof

• First, special cases
• G is disconnected or trivial
• ?(G) ?(G) 0
• G is complete
• ?(G) ?(G) ?(G) n -1
• We may assume that ?(G) ? n-2

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Proof (cont)

• First ?(G) ? ?(G)
• Next ?(G) ? ?(G)
• Case 1 Let X be a min. edge cut
• X ?(G) ? n-2
• G X has two components, say G1 and G2
• Let k be the order of G1. So order(G2) n-k
• Note that every edge in X joins a vertex in G1 to
  a vertex in G2

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Order(G1) k, Order(G2) n-k, k?1 and n-k?1

• Case 1 Every vertex in G1 is adjacent to every
  vertex in G2
• X k(n-k)
• We have (k-1)(n-k-1) ? 0. Hence,
• (k-1)(n-k-1) k(n-k)-n1 ? 0
• Hence ?(G) k(n-k) ? n-1
• But we are studying the case ?(G) ? n-2
• So case 1 is not possible

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Case 2

• There exists u in G1 and v in G2 that are not
  adjacent in G
• Let us find a vertex cut U of size lt ?(G)
• For each edge of the edge cut, we add a vertex to
  U

u
v
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For this last case

• ?(G) ? U ? X ?(G).
• Done.

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Interesting cases

• For a cubic graph G, ?(G) ?(G).
• ?(G) ? ?(G) ?? ?(G) 3
• So there are 4 possible cases.
• ?(G) 0 Only if G is disconnected. ?(G) 0
• ?(G) 3 ?(G) 3
• Other cases Let U be a min vertex cut G1 and
  G2 be the two components in G-UU is either 1
  or 2. In either case, for every u?U, one of G1
  or G2 contains exactly one neighbor of u

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?(G) 1
G1
G2
U
An edge cut
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?(G) 2
G1
G1
G2
G2
U
U
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Other interesting cases

• Last time we proved that, ?(G) ? 2m/n
• It turns out that this bound is sharp.
• Define H(r,n) Harary graphs of order n and ?
  r

r/2
And r can be made almost as high as 2m/n Check
the book for details
r/2
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A deeper look into connectivity

• A set S of vertices separate u and v, if G-S is
  disconnected, and u and v are in different
  components
• Note If S separates u and v
• S is a vertex cut
• u and v are not adjacent

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Paths between u and v

• Let P u, u1, u2, , uk, v be a u-v path
• u1, u2, , uk are internal vertices
• A collection P1, P2, .., Pk of u-v paths are
  internally disjoint if every two paths have no
  internal vertices common.

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Mengers Theorem

• Let u and v be nonadjacent vertices in a graph G.
  The minimum number of vertices in a u-v
  separating set equals the maximum number of
  internally disjoint u-v paths in G

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Proof by induction on the size ( of edges)

• Basis m 0.
• Inductive step. Assume true for all graphs of
  size ? m
• Let U be a minimum u-v separating set.
• Clearly, the number of u-v disjoint paths is at
  most U k. So we need to show equality.

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Proof (cont)

• We look at all minimum u-v separating sets. There
  are three cases
• Case 1 There is a u-v separating set U that
  contains a vertex that is adjacent to both u and
  v
• Case 2 There is u-v separating set W with a
  vertex not adjacent to u and a vertex not
  adjacent to v
• What is the remaining case?
• Case 3 For each min. u-v separating set S,
  either (every vertex in S is adjacent to u but
  not to v) or (every vertex in S is adjacent to v
  but not to u)

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Case 1

• Consider G-x
• Its size is less than m
• U-x is a min. separating set for G-x. Why?
• Since U-x k -1, by the ind.
  hypothesis,there are k-1 internally disjoint u-v
  paths in G-x
• So in G, we have these paths plus u-x-v
• Done.

U
G1
G2
x
v
u
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Case 2

• Note x and y can be the same vertex

W
G1
G2
x
u
v
y
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Case 2

• W w1, , wk
• First lets construct G(u) which contains all
  u-wi paths for all wi ? W in G1 W (this must be
  added to the proof in the book)
• Make a new graph G(u) by adding a new vertex v
  to G(u) and connecting it to all wi
• Construct G(v) and G(v) similarly

W
G1
w1
u

wk
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Case 2

• size G(u) lt m
• W is a min u-v seperating set of size k.
• By the ind. hyp., there are k disjoint u-v
  paths.
• We take these paths and delete v from them. Call
  the resulting paths P1
• With similar reasoning, we conclude that G(v)
  has k disjoint v-u paths. Generate paths P2 in a
  similar fashion.
• Combine P1 and P2 using the vertices wi.
• We obtained k internally disjoint paths for G

W
w1
u
v

wk
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Case 3

• We have either the situation on the left or the
  symmetric case (where v is connected to all in S)

S
G1
G2
u
v
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Case 3

• Let P u,x,y, , v be a u-v geodesic in G
• Let e (x,y) and consider G-e
• Claim The size k of any minimum u-v separating
  set in G-e is also k.
• Clearly, k ? k-1. Why?
• Suppose, for contradiction, that k k-1 (i.e.
  the claim is false).
• Let Z be a min u-v separating set in G-e
• Z x is a min u-v separating set in G
• So all vertices in Z are adjacent to u (we are in
  case 3)
• Z y is a min u-v separating set in G
• So y is adjacent to v
• Can you see a contradiction?

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Case 3 (cont)

• G-e has a min. u-v separating set of size k.
• By ind. hyp. It has k internally disjoint u-v
  paths.
• So does G!

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We have just proved

• Let u and v be nonadjacent vertices in a graph G.
  The minimum number of vertices in a u-v
  separating set equals the maximum number of
  internally disjoint u-v paths in G

What does this say about connectivity?
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Theorem

• A non-trivial graph G (V,E) is k-connected (k
  ? 2) if and only if for each distinct u,v ?V
  there are at least k internally disjoint u-v
  paths in G

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k-connected ? k disjoint u-v paths ?u,v

• We need to prove for any u,v
• Case 1 u and v are not adjacent
• Follows from Mengers theorem
• Case 2 u and v are adjacent. Let e (u,v)
• Claim G-e is (k-1)-connected
• Then, there are (k-1) internally disjoint paths
  in G-e.
• In G, we also have the path u-v

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?u,v k disjoint u-v paths ? k-connected

• Take a min-vertex cut U. Pick u and v from
  different components
• There are at least k disjoint u-v paths
• Mengers theorem says that the min. size of a u-v
  separating set is at least k
• So U is at least k
• The graph is k-connected.

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Corollary

• Let G be a k-connected graph and S be any set of
  k vertices. If a graph H is obtained by adding a
  new vertex w to G and connecting w to all the
  vertices in S, then H is also k-connected.

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Corollary

• If G is a k connected graph and u, v1, , vk are
  k1 distinct vertics of G, then there exist
  internally disjoint u-vi paths for all i 1, , k

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Another way of looking at k-connected graphs

• Last lecture we proved
• A graph of order at least 3 is nonseparable iff
  every two vertices lie on a common cycle
• Note that non-separable means 2 connected.
• We now prove a generalization of this theorem

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Theorem

• If G is a k-connected graph, k ? 2, then every
  k-vertices of G lie on a common cycle.
• Proof suppose not.
• Suppose there exists S v1, , vk and there is
  no cycle containing S.
• Among all cycles in S, pick cycle C that contains
  most number of vertices of S.

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Proof (cont)

• Suppose C contains l lt k vertices from S.
• Once we choose C, we may assume that the vertices
  v1, , vl appear on C in this order (otherwise we
  rename the vertices).

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Proof (cont)
C
v1
P
v2
vl
Note that the graph is l-connected. Therefore by
the corollary, we have
u ? S But not in C