Chapter 1 Fundamental Concept

1
Chapter 1 Fundamental Concept

• 1.1 What Is a Graph?
• 1.2 Paths, Cycles, and Trails
• 1.3 Vertex Degree and Counting
• 1.4 Directed Graphs

2
The KÖnigsberg Bridge Problem

• Königsber is a city on the Pregel river in
  Prussia
• The city occupied two islands plus areas on both
  banks
• Problem
• Whether they could leave home, cross every
  bridge exactly once, and return home.

3
A Model

• A vertex a region
• An edge a path(bridge) between two regions

4
What Is a Graph?

• A graph G is a triple consisting of
• A vertex set V(G)
• An edge set E(G)
• A relation between an edge and a pair of vertices

5
Loop, Multiple edges

• Loop An edge whose endpoints are equal
• Multiple edges Edges have the same pair of
  endpoints

Multiple edges
loop
6
Simple Graph

• Simple graph A graph has no loops or multiple
  edges

Multiple edges
loop
It is a simple graph.
It is not simple.
7
Adjacent, neighbors

• Two vertices are adjacent and are neighbors if
  they are the endpoints of an edge.
• Example
• A and B are adjacent.
• A and D are not adjacent.

B
A
C
D
8
Finite Graph, Null Graph

• Finite graph an graph whose vertex set and edge
  set are finite.
• Null graph the graph whose vertex set and edges
  are empty.

9
Complement

• Complement of G The complement G of a simple
  graph G
• A simple graph
• V(G) V(G)
• E(G) uv uv ?E(G)

G
10
Clique and Independent set

• A Clique in a graph a set of pairwise adjacent
  vertices (a complete subgraph)
• An independent set in a graph a set of pairwise
  nonadjacent vertices.
• Example
• x, y, u is a clique in G.
• u, w is an independent set.

u
G
v
y
w
x
11
Bipartite Graphs

• A graph G is bipartite if V(G) is the union of
  two disjoint independent sets called partite sets
  of G
• Also The vertices can be partitioned into two
  sets such that each set is independent
• Matching Problem
• Job Assignment Problem

Workers
Boys
Girls
Jobs
12
Chromatic Number

• The chromatic number of a graph G, written x(G),
  is the minimum number of colors needed to label
  the vertices so that adjacent vertices receive
  different colors

x(G) 3
13
Maps and coloring

• A map is a partition of the plane into connected
  regions
• Can we color the regions of every map using at
  most four colors so that neighboring regions have
  different colors?
• Map Coloring ? graph coloring
• A region ? A vertex
• Adjacency ? An edge

14
Scheduling and graph Coloring 1

• Two committees can not hold meetings at the same
  time if two committees have common member

Committee 2
Committee 1
common member
15
Scheduling and graph Coloring 1

• Model
• One committee being represented by a vertex
• An edge between two vertices if two corresponding
  committees have common member
• Two adjacent vertices can not receive the same
  color

Committee 2
Committee 1
common member
16
Scheduling and graph Coloring 2

• Scheduling problem is equivalent to graph
  coloring problem.

Committee 2
common member
Common Member Different Color
Committee 1
Committee 3
No Common Member Same Color OK Same time slot OK
17
Path and Cycle

• Path a sequence of distinct vertices such that
  two consecutive vertices are adjacent.
• Example (a, d, c, b, e) is a path
• (a, b, e, d, c, b, e, d) is not a path it is a
  walk.
• Cycle a closed Path
• Example (a, d, c, b, e, a) is a cycle

a
b
c
e
d
18
Subgraphs

• A subgraph of a graph G is a graph H such that
• V(H) ? V(G) and E(H) ? E(G) and
• The assignment of endpoints to edges in H is the
  same as in G.

19
Subgraphs

• Example H1, H2, and H3 are subgraphs of G

b
a
G
c
d
e
b
a
b
a
H3
c
H1
H2
e
d
d
e
20
Connected and Disconnected

• Connected There exists at least one path between
  two vertices.
• Disconnected Otherwise
• Example
• H1 and H2 are connected.
• H3 is disconnected.

H3
H1
H2
e
21
Adjacency, Incidence, and Degree

• Assume ei is an edge whose endpoints are (vj,vk)
• The vertices vj and vk are said to be adjacent.
• The edge ei is said to be incident upon vj
• Degree of a vertex vk is the number of edges
  incident upon vk . It is denoted as d(vk)

22
Adjacency matrix

• Let G (V, E), V n and Em
• The adjacency matrix of G written A(G), is the
  n-by-n matrix in which entry ai,j is the number
  of edges in G with endpoints vi, vj.

23
Incidence Matrix

• Let G (V, E), V n and Em
• The incidence matrix M(G) is the n-by-m matrix in
  which entry mi,j is 1 if vi is an endpoint of ei
  and otherwise is 0.

a b c d e 1 1 0 0 0 1
0 1 1 0 0 1 1 1 1 0 0
0 0 1
wxyz
24
Isomorphism

• An isomorphism from a simple graph G to a simple
  graph H is a bijection fV(G)?V(H) such that uv
  ?E(G) if and only if f(u)f(v) ? E(H)
• We say G is isomorphic to H, written G ? H

f1 w x y z c b d a
y
c
w
d
H
G
a
z
f2 w x y z a d b c
x
b
25
Complete Graph, Complete Bipartite Graph or
Biclique

• Complete Graph a simple graph whose vertices are
  pairwise adjacent.
• Complete bipartite graph (biclique) is a simple
  bipartite graph such that two vertices are
  adjacent if and only if they are in different
  partite sets.

Complete Bipartite Graph
Complete Graph
26
Petersen Graph 1.1.36

• The petersen graph is the simple graph whose
  vertices are the 2-element subsets of a 5-element
  set and whose edges are pairs of disjoint
  2-element subsets

27
Petersen Graph 1.1.37

• Assume the set of 5-element be (1, 2, 3, 4, 5)
• Then, 2-element subsets
• (1,2) (1,3) (1,4) (1,5) (2,3) (2,4) (2,5)
  (3,4) (3,5) (4,5)

45 (4, 5)
Disjoint, so connected
28
Petersen Graph 1.1.36

• Three drawings

29
Theorem If two vertices are non-adjacent in the
Petersen Graph, then they have exactly one common
neighbor. 1.1.38

• Proof

No connection, Joint, One common element.
3 elements in these vertices totally
x, z
x, y
Since 5 elements totally, 5-3 elements
left. Hence, exactly one of this kind.
30
Girth 1.1.39, 1.1.40

• Girth the length of its shortest cycle.
• If no cycles, girth is infinite

31
Girth and Petersen graph 1.1.39, 1.1.40

• Theorem The Petersen Graph has girth 5.
• Proof
• Simple ? no loop ? no 1-cycle (cycle of length 1)
• Simple ? no multiple ? no 2-cycle
• 5 elements ?no three pair-disjoint 2-sets ?no
  3-cycle
• By previous theorem, two nonadjacent vertices has
  exactly one common neighbor ?no 4-cycle
• 12-34-51-23-45-12 is a 5-cycle.

32
Walks, Trails1.2.2

• A walk a list of vertices and edges v0, e1, v1,
  ., ek, vk such that, for 1?i?k, the edge ei
  has endpoints vi-1 and vi.
• A trail a walk with no repeated edge.

33
Paths 1.2.2

• A u,v-walk or u,v-trail has first vertex u and
  last vertex v these are its endpoints.
• A u,v-path a u,v-trail with no repeated vertex.
• The length of a walk, trail, path, or cycle is
  its number of edges.
• A walk or trail is closed if its endpoints are
  the same.

34
Lemma Every u,v-walk contains a u,v-path 1.2.5

• Proof
• Use induction on the length l of a u, v-walk W.
• Basis step l 0.
• Having no edge, W consists of a single vertex
  (uv).
• This vertex is a u,v-path of length 0.
• 
  to be continued

35
Lemma Every u,v-walk contains a u,v-path 1.2.5

• Proof Continue
• Induction step l ? 1. (see the figure in the
  next page)
• Suppose that the claim holds for walks of length
  less than l.
• If W has no repeated vertex, then its vertices
  and edges form a u,v-path.
• If W has a repeated vertex w, then deleting the
  edges and vertices between appearances of w
  (leaving one copy of w) yields a shorter u,v-walk
  W contained in W. (see next page)
• By the induction hypothesis, W contains a
  u,v-path P,and this path P is contained in W.

36
Lemma Every u,v-walk contains a u,v-path 1.2.5

• An example

Delete
37
Components 1.2.8

• The components of a graph G are its maximal
  connected subgraphs.
• A component (or graph) is trivial if it has no
  edges otherwise it is nontrivial.
• An isolated vertex is a vertex of degree 0.

38
Theorem Every graph with n vertices and k edges
has at least n-k components 1.2.11

• Proof
• An n-vertex graph with no edges has n components
• Each edge added reduces this by at most 1
• If k edges are added, then the number of
  components is at least n-k

39
Theorem Every graph with n vertices and k edges
has at least n-k components 1.2.11

• Examples

n6, k3, 4 components
n6, k3, 3 components
n2, k1, 1 component
n3, k2, 1 component
40
Cut-edge, Cut-vertex 1.2.12

• A cut-edge or cut-vertex of a graph is an edge or
  vertex whose deletion increases the number of
  components.

41
Cut-edge, Cut-vertex 1.2.12

• G-e or G-M The subgraph obtained by deleting an
  edge e or set of edges M.
• G-v or G-S The subgraph obtained by deleting a
  vertex v or set of vertices S.

G-e
G
e
42
Induced subgraph 1.2.12

• An induced subgraph
• A subgraph obtained by deleting a set of
  vertices.
• We write GT for G- T, where T V(G)-T
• GT is the subgraph of G induced by T.
• Example
• Assume TA, B, C, D

GT
G
43
Induced subgraph 1.2.12

• More Examples
• G2 is the subgraph of G1 induced by (A, B, C, D)
• G3 is the subgraph of G1 induced by (B, C)
• G4 is not the subgraph induced by (A, B, C, D)

G3
G4
G1
G2
44
Induced subgraph 1.2.12

• A set S of vertices is an independent set if and
  only if the subgraph induced by it has no edges.
• G3 is an example.

G3
G1
45
Theorem An edge e is a cut-edge if and only if e
belongs to no cycles. 1.2.14

• Proof 1/2
• Let e (x, y) be an edge in a graph G and H be
  the component containing e.
• Since deletion of e effects no other component,
  it suffices to prove that H-e is connected if and
  only if e belongs to a cycle.
• First suppose that H-e is connected.
• This implies that H-e contains an x, y-path,
• This path completes a cycle with e.

46
Theorem An edge e is a cut-edge if and only if e
belongs to no cycles. 1.2.14

• Proof 2/2
• Now suppose that e lies in a cycle C.
• Choose u, v?V(H)
• Since H is connected, H has a u, v-path P.
• If P does not contain e
• Then P exists in H-e
• Otherwise
• Suppose by symmetry that x is between u and y on
  P
• Since H-e contains a u, x-path along P,
  the transitivity of the connection
  relation implies that H-e has a u, v-path.
• We did this for all u, v ? V(H), so H-e is
  connected.

47
Theorem An edge e is a cut-edge if and only if e
belongs to no cycles. 1.2.14

• An Example

48
Lemma Every closed odd walk contains an odd cycle

• Proof1/2
• Use induction on the length l of a closed odd
  walk W.
• l1. A closed walk of length 1 traverses a cycle
  of length 1.
• We need to prove the claim holds if it holds for
  closed odd walks shorter than W.

49
Lemma Every closed odd walk contains an odd cycle

• Proof 2/2
• Suppose that the claim holds for closed odd walks
  shorter than W.
• If W has no repeated vertex (other than first
  last), then W itself forms a cycle of odd length.
  
• Otherwise,
• Need to prove If repeated, W includes a shorter
  closed odd walk. By induction, the theorem hold
• If W has a repeated vertex v, then we view W as
  starting at v and break W into two v,v-walks.
• Since W has odd length, one of these is odd and
  the other is even. (see the next page)
• The odd one is shorter than W, by induction
  hypothesis, it contains an odd cycle, and this
  cycle appears in order in W.

50
Lemma Every closed odd walk contains an odd
cycle 1.2.15
Odd Odd Even
v
Odd
Even
51
Theorem A graph is bipartite if and only if it
has no odd cycle. 1.2.18

• Examples

A
B
A
B
F
C
D
E
D
C
A
B
A
B
C
D
C
D
E
F
52
Theorem A graph is bipartite if it has no odd
cycle. 1.2.18

• Proof (sufficiency1/3)
• Let G be a graph with no odd cycle.
• We prove that G is bipartite by constructing a
  bipartition of each nontrivial component H.
• For each v ? V(H), let f(v) be the minimum length
  of a u, v-path. Since H is connected, f(v) is
  defined for each v ? V(H) .

53
Theorem A graph is bipartite if it has no odd
cycle. 1.2.18

• Proof (sufficiency2/3)
• Let Xv ? V(H) f(v) is even and Yv ? V(H)
  f(v) is odd
• An edge v,v within X (or Y) would create a
  closed odd walk using a shortest u, v-path, the
  edge v, v within X (or Y) and the reverse of a
  shortest u, v-path.

• A closed odd walk using
• a shortest u, v-path,
• the edge v, v within X (or Y) , and
• the reverse of a shortest u, v-path.

54
Theorem A graph is bipartite if it has no odd
cycle. 1.2.18

• Proof (sufficiency3/3)
• By Lemma 1.2.15, such a walk must contain an odd
  cycle, which contradicts our hypothesis
• Hence X and Y are independent sets. Also X?Y
  V(H), so H is an X, Y-bipartite graph

Because even (or odd) even (or odd) even
even 1 odd Since no odd cycles, vv
doesnt exist. We have X and Y are independent
sets
Even (or Odd)
Odd Cycle
Even (or Odd)
55
Theorem A graph is bipartite only if it has no
odd cycle. 1.2.18

• Proof (necessity)
• Let G be a bipartite graph.
• Every walk alternates between the two sets of a
  bipartition
• So every return to the original partite set
  happens after an even number of steps
• Hence G has no odd cycle

56
Eulerian Circuits 1.2.24

• A graph is Eulerian if it has a closed trail
  containing all edges.
• We call a closed trail a circuit when we do not
  specify the first vertex but keep the list in
  cyclic order.
• An Eulerian circuit or Eulerian trail in a graph
  is a circuit or trail containing all the edges.
  

57
Even Graph, Even Vertex, and Maximal Path1.2.24

• An even graph is a graph with vertex degrees all
  even.
• A vertex is odd even when its degree is odd
  even.
• A maximal path in a graph G is a path P in G that
  is not contained in a longer path.
• When a graph is finite, no path can extend
  forever , so maximal (non-extendible) paths exist.

58
Lemma If every vertex of graph G has degree at
least 2, then G contains a cycle. 1.2.25

• Proof
• Let P be a maximal path in G, and let u be an
  endpoint of P
• Since P cannot be extended, every neighbor of u
  must already be a vertex of P
• Since u has degree at least 2, it has a neighbor
  v in V(P) via an edge not in P
• The edge uv completes a cycle with the portion of
  P from v to u

Must
u
P
v
u
P
Impossible
59
Theorem A graph G is Eulerian if and only if it
has at most one nontrivial component and its
vertices all have even degree. 1.2.26

• Proof (Necessity)
• Suppose that G has an Eulerian circuit C.
• Each passage of C through a vertex uses two
  incident edges, and the first edge is paired with
  the last at the first vertex.
• Hence every vertex has even degree. Also, two
  edges can be in the same trail only when they lie
  in the same component, so there is at most one
  nontrivial component.

60
Theorem A graph G is Eulerian if and only if it
has at most one nontrivial component and its
vertices all have even degree.1.2.26

• Proof (Sufficiency 1/3)
• Assuming that the condition holds, we obtain an
  Eulerian circuit using induction on the number of
  edges, m.
• Basis step m0. A closed trail consisting of one
  vertex suffices.
  ?

61
Theorem A graph G is Eulerian if and only if it
has at most one nontrivial component and its
vertices all have even degree. 1.2.26

• Proof (Sufficiency 2/3)
• Induction step mgt0.
• When even degrees, each vertex in the nontrivial
  component of G has degree at least 2.
• By Lemma 1.2.25, the nontrivial component has a
  cycle C.
• Let G be the graph obtained from G by deleting
  E(C).
• Since C has 0 or 2 edges at each vertex, each
  component of G is also an even graph.
• Since each component is also connected and has
  fewer than m edges, we can apply the induction
  hypothesis to conclude that each component of G
  has an Eulerian circuit. ?

62
Theorem A graph G is Eulerian if and only if it
has at most one nontrivial component and its
vertices all have even degree. 1.2.26

• Proof (Sufficiency 3/3)
• Induction step mgt0. (continued)
• To combine these into an Eulerian circuit of G,
  we traverse C, but when a component of G is
  entered for the first time we detour along an
  Eulerian circuit of that component.
• This circuit ends at the vertex where we began
  the detour. When we complete the traversal of C,
  we have completed an Eulerian circuit of G.

63
Proposition Every even graph decomposes into
cycles1.2.27

• Proof
• In the proof of Theorem 1.2.26
• It is noted that every even nontrivial graph has
  a cycle
• The deletion of a cycle leaves an even graph
• Thus this proposition follows by induction on the
  number of edges

64
Proposition If G is a simple graph in which
every vertex has degree at least k, then G
contains a path of length at least k. If k?2,
then G also contains a cycle of length at least
k1. 1.2.28

• Proof (1/2)
• Let u be an endpoint of a maximal path P in G.
• Since P does not extend, every neighbor of u is
  in V(P).
• Since u has at least k neighbors and G is simple,
  P therefore has at least k vertices other than u
  and has length at least k.

65
Proposition If G is a simple graph in which
every vertex has degree at least k, then G
contains a path of length at least k. If k?2,
then G also contains a cycle of length at least
k1. 1.2.28

• Proof (2/2)
• If k ? 2, then the edge from u to its farthest
  neighbor v along P completes a sufficiently long
  cycle with the portion of P from v to u.

d(u) ? k
v
u
At least k1 vertices Length ? k
66
Degree1.3.1

• The degree of vertex v in a graph G, written
  or d(v), is the number of edges incident
  to v, except that each loop at v counts twice
• The maximal degree is ?(G)
• The minimum degree is ? (G)

G
A
B
d(B) 3, d(C) 2 ?(G) 3, d(G) 2
F
C
D
E
67
Regular 1.3.1

• G is regular if ?(G) ? (G)
• G is k-regular if the common degree is k.
• The neighborhood of v, written Ng(v) or N(v)
  is the set of vertices adjacent to
  v.

3-regular
68
Order and size 1.3.2

• The order of a graph G, written n(G), is the
  number of vertices in G.
• An n-vertex graph is a graph of order n.
• The size of a graph G, written e(G), is the
  number of edges in G.
• For n?N, the notation n indicates the set
  1,, n.

69
Proposition (Degree-Sum Formula) If G is a
graph, then ?v?V(G)d(v) 2e(G) 1.3.3

• Proof
• Summing the degrees counts each edge twice,
• Because each edge has two ends and contributes
  to the degree at each endpoint.

70
Theorem If kgt0, then a k-regular bipartite graph
has the same number of vertices in each partite
set. 1.3.9

• Proof
• Let G be an X,Y-bigraph.
• Counting the edges according to their endpoints
  in X yields e(G)kX.

d (x) k
x
71
Theorem If kgt0, then a k-regular bipartite graph
has the same number of vertices in each partite
set. 1.3.9

• Proof
• Counting them by their endpoints in Y yields
  e(G)kY.
• Thus kXkY, which yields XY when kgt0.

y
d (y) k
d (x) k
x
72
A technique for counting a set 1/3 1.3.10

• Example The Petersen graph has ten 6-cycles
• Let G be the Petersen graph.
• Being 3-regular, G has ten copies of K1,3 (claw)
  . We establish a one-to-one correspondence
  between the 6-cycles and the claws.
• Since G has girth 5, every 6-cycle F is an
  induced subgraph.
• see below
• Each vertex of F has one neighbor outside F.
• d(v) 3, v ?V(G)

If Existing, Girth 3. But Girth5 so no such an
edge
73
A technique for counting a set 2/3 1.3.10

• Since nonadjacent vertices have exactly one
  common neighbor (Proposition 1.1.38), opposite
  vertices on F have a common neighbor outside F.
• Since G is 3-regular, the resulting three
  vertices outside F are distinct.
• Thus deleting V(F) leaves a subgraph with three
  vertices of degree 1 and one vertex of degree 3
  it is a claw.

Common neighbor of opposite vertices
If the neighbors are not distinct, d(v)gt3
74
A technique for counting a set 3/3 3.10

• It is shown that each claw H in G arises exactly
  once in this way.
• Let S be the set of vertices with degree 1 in H
  S is an independent set.
• The central vertex of H is already a common
  neighbor, so the six other edges from S reach
  distinct vertices.
• Thus G-V(H) is 2-regular. Since G has girth 5,
  G-V(H) must be a 6-cycle. This 6-cycle yields H
  when its vertices are deleted.

75
Proposition The minimum number of edges in a
connected graph with n vertices is n-1. 3.13

• Proof
• By proposition 1.2.11, every graph with n
  vertices and k edges has at least n-k components.
  
• Hence every n-vertex graph with fewer than n-1
  edges has at least two components and is
  disconnected.
• The contrapositive of this is that every
  connected n-vertex graph has at least n-1 edges.
  This lower bound is achieved by the path Pn.

76
Theorem If G is simple n-vertex graph with
?(G)?(n-1)/2, then G is connected. 1.3.15

• Proof 1/2
• Choose u,v ? V(G).
• It suffices to show that u,v have a common
  neighbor if they are not adjacent.
• Since G is simple, we have N(u) ? ? (G) ?
  (n-1)/2, and similarly for v.
• Recall ? (G) is the minimum degree, N(u)
  d(u) Hence N(u) ? ? (G)

77
Theorem If G is simple n-vertex graph with
?(G)?(n-1)/2, then G is connected. 1.3.15

• Proof 2/2
• When u and v are not connected, we have
  N(u)?N(v) ? n-2
• since u and v are not in the union
• Using Remark A.13 of Appendix A, we thus compute

78
Theorem Every loopless graph G has a bipartite
subgraph with at least e(G)/2 edges. 1.3.19

• Proof
• Partition V(G) into two sets X, Y.
• Using the edges having one endpoint in each set
  yields a bipartite subgraph H with bipartition X,
  Y.
• If H contains fewer than half the edges of G
  incident to a vertex v, then v has more edges to
  vertices in its own class than in the other
  class, as illustrated bellow.

79
Theorem Every loopless graph G has a bipartite
subgraph with at least e(G)/2 edges. 1.3.19

• Proof 2/2
• Moving v to the other class gains more edges of G
  than it loses.
• Using Iterative improvement approach
• When it terminates, we have dH(v) ? dG(v)/2 for
  every v?V(G) .
• Summing this and applying the degree-sum formula
  yields e(H) ? e(G)/2.

80
Example 1.3.20

• The algorithm in Theorem 1.3.19 need not produce
  a bipartite subgraph with the most edges, merely
  one with at least half the edges. - Local
  Maximum.
• Consider the graph in the next page.
• It is 5-regular with 8 vertices and hence has 20
  edges.
• The bipartition Xa,b,c,d and Ye,f,g,h
  yields a 3-regular bipartite subgraph with 12
  edges.
• The algorithm terminates here switching one
  vertex would pick up two edges but lose three .

81
Example(Cont.) 1.3.20
82
Example 1.3.20

• Nevertheless, the bipartition Xa,b,g,h and
  Yc,d,e,f yields a 4-regular bipartite subgraph
  with 16 edges.
• An algorithm seeking the maximal by local changes
  may get stuck in a local maximum.

Global Maximum
Local Maximum
83
Theorem The maximum number of edges in an
n-vertex triangle free simple graph is ? n2/4 ?
1.3.23

• Proof 1/6
• Let G be an n-vertex triangle-free simple graph.
• Let x be a vertex of maximum degree and d(x)k.
• Since G has no triangles, there are no edges
  among neighbors of x.

No edges between neighbors of x
84
Theorem The maximum number of edges in an
n-vertex triangle free simple graph is ? n2/4 ?
1.3.23

• Proof 2/6
• Hence summing the degrees of x and its
  nonneighbors counts at least one endpoint of
  every edge ?v?N(x)d(v) ? e(G).
• We sum over n-k vertices, each having degree at
  most k, so e(G) ? (n-k)k

85
Theorem The maximum number of edges in an
n-vertex triangle free simple graph is ? n2/4 ?
1.3.23
At least n-k vertices
Proof 3/6
Doesnt exist
No edges exist

• ?v?N(x) d(v) counts at least one endpoint of
  every edge

At most k vertices
86
Theorem The maximum number of edges in an
n-vertex triangle free simple graph is ? n2/4 ?
1.3.23

• Proof 4/6
• Since (n-k)k counts the edges in Kn-k, k, we have
  now proved that e(G) is bounded by the size of
  some biclique with n vertices.
• i.e. e(G) ? (n-k)k the edges in Kn-k, k

87
Theorem The maximum number of edges in an
n-vertex triangle free simple graph is ? n2/4 ?
1.3.23

• Proof 5/6
• Moving a vertex of Kn-k,k from the set of size k
  to the set of size n-k gains k-1 edges and loses
  n-k edges.
• The net gain is 2k-1-n, which is positive for
  2kgtn1 and negative for 2kltn1.
• Thus e(Kn-k, k) is maximized when k is ?n/2 ? or
  ?n/2? .

k
n-k
88
Theorem The maximum number of edges in an
n-vertex triangle free simple graph is ? n2/4 ?
1.3.23

• Proof 6/6
• The product is then n2/4 for even n and (n2-1)/4
  for odd n. Thus e(G) ? ?n2/4? .
• The bound is best possible.
• It is seen that a triangle-free graph with ?n2/4?
  edges is K?n/2?,?n/2 ?.

89
Degree sequence 1.3.27

• The Degree Sequence of a graph is the list of
  vertex degrees, usually written in non-increasing
  order, as d1? . ? d n .
• Example

x
v
Degree sequence d(w), d(x), d(y), d(z), d(v)
w
y
z
90
Proposition The nonnegative integers d1 ,, dn
are the vertex degrees of some graph if and only
if ?di is even. 1.3.28

• Proof ½ Necessity
• When some graph G has these numbers as its vertex
  degrees, the degree-sum formula implies that ?di
  2e (G), which is even.

91
Proposition The nonnegative integers d1 ,, dn
are the vertex degrees of some graph if and only
if ?di is even. 1.3.28

• Proof 2/2 Sufficiency
• Suppose that ? di is even.
• We construct a graph with vertex set v1,,vn and
  d(vi) di for all i.
• Since ? di is even, the number of odd values is
  even.
• First form an arbitrary pairing of the vertices
  in vi di is odd.
• For each resulting pair, form an edge having
  these two vertices as its endpoints
• The remaining degree needed at each vertex is
  even and nonnegative satisfy this for each i by
  placing di /2 loops at vi

92
Graphic Sequence 1.3.29

• A graphic sequence is a list of nonnegative
  numbers that is the degree sequence of some
  simple graph.
• A simple graph realizes d.
• means A simple graph with degree sequence d.

93
Recursive condition 1.3.30

• The lists (2, 2, 1, 1) and (1, 0, 1) are graphic.
  The graphic K2K1 realizes 1, 0, 1.
• Adding a new vertex adjacent to vertices of
  degrees 1 and 0 yields a graph with degree
  sequence 2, 2, 1, 1, as shown below.
• Conversely, if a graph realizing 2, 2, 1, 1 has a
  vertex w with neighbors of degrees 2 and 1, then
  deleting w yields a graph with degrees 1, 0, 1.

94
Recursive condition 1.3.30

• Similarly, to test 33333221, we seek a
  realization with a vertex w of degree 3 having
  three neighbors of degree 3.

3 3 3 3 3 2 2 1
Delete this Vertex
A new degree sequence
2 2 2 3 2 2 1
95
Recursive condition 1.3.30

• This exists if and only if 2223221 is graphic.
  (See next page)
• We reorder this and test 3222221.
• We continue deleting and reordering until we can
  tell whether the remaining list is realizable.
• If it is, then we insert vertices with the
  desired neighbors to walk back to a realization
  of the original list.
• The realization is not unique.
• The next theorem implies that this recursive test
  work.

96
Recursive condition 1.3.30

• 33333221 3222221 221111
  11100
• 2223221 111221
  10111

97
Theorem. For ngt1, an integer list d of size n is
graphic if and only if d is graphic, where d is
obtained from d by deleting its largest element ?
and subtracting 1 from its ? next largest
elements. The only 1-element graphic sequence is
d10. 1.3.31

• Proof 1/6
• For n1, the statement is trivial.
• For ngt1, we first prove that the condition is
  sufficient.
• Give d with d1?..?dn and a simple graph G with
  degree sequence d

For Example We have 1) d33333221
2) G with d 2223221 We show d
is graphic
G
98
Theorem. For ngt1, an integer list d of size n is
graphic if and only if d is graphic, where d is
obtained from d by deleting its largest element ?
and subtracting 1 from its ? next largest
elements. The only 1-element graphic sequence is
d10. 1.3.31

• Proof 2/6
• We add a new vertex adjacent to vertices in G
  with degrees d2-1,..,d?1-1.
• These di are the ? largest elements of d after
  (one copy of) ? itself,
• Note d2-1,..,d?1-1 need not be the ? largest
  numbers in d (see example in previous page)

d d1,d2, dn d d2-1,..,d?1-1, dn
G
May not be the ? largest numbers
New added vertex
99
Theorem 1.3.31 continue

• To prove necessity, 3/6
• Given a simple graph G realizing d , we produce
• A simple graph G realizing d.
• Let w be a vertex of degree ? in G, and let S be
  a set of ? vertices in G having the desired
  degrees d2,..,d?1.

w
d d1, d2, d?, d?1, dn
S ? vertices
d1?
100
Theorem 1.3.31

• Proof continue 4/6
• If N(w)S, then we delete w to obtain G.

w
d d1, d2, d?, d?1, dn
d1?

• Vertices, N(w)S
• i.e. They are connected to w

Delete w than we have d d2-1,..,d?1-1, dn
101
Theorem 1.3.31

• Proof continue 5/6
• Otherwise,
• Some vertex of S is missing from N(w).
• In this case, we modify G to increase N(w)?S
  without changing any vertex degree.
• Since N(w)?S can increase at most ? times,
  repeating this converts G into another graph G
  that realizes d and has S as the neighborhood of
  w.
• From G we then delete w to obtain the desired
  graph G realizing d.

w
d d1, d2, d?, d?1, dn
d1?

• Vertices, N(w)?S
• i.e. Some vertices are not connected to w.
• - We make them become connected to w without
  changing their degree.

102
Theorem 1.3.31

• Proof continue6/6
• To find the modification when N(w)?S , we choose
  x?S and z?S so that w z are connected and w x are
  not.
• We want to add wx and delete wz, but we must
  preserve vertex degrees. Since d(x)gtd(z) and
  already w is a neighbor of z but not x, there
  must be a vertex y adjacent to x but not to z.
  Now we delete wz,xy and add wx,yz to increase
  N(w)?S .

z
z
?
w
w
y
y
x
x
It becomes connected w
This y must exist.
103
2-switch 1.3.32

• A 2-switch is the replacement of a pair of edges
  xy and zw in a simple graph by the edges yz and
  wx, given that yz and wx did not appear in the
  graph originally.

104
Theorem If G and H are two simple graphs with
vertex set V, then dG(v)dH(v) for every v?V if
and only if there is a sequence of 2-switches
that transforms G into H. 1.3.33

• Proof
• Every 2-switch preserves vertex degrees, so the
  condition is sufficient.
• Conversely, when dG(v)dH(v) for all v?V , we
  obtain an appropriate sequence of 2-switches by
  induction on the number of vertices, n.
• If nlt3, then for each d1,..,dn there is at most
  one simple graph with d(vi)di.
• Hence we can use n3 as the basis step.

105
Theorem. 1.3.33 (Continue)

• Consider n?4 , and let w be a vertex of maximum
  degree,?.
• Let Sv1,..,v? be a fixed set of vertices with
  the ? highest degrees other than w.
• As in the proof of Theorem 1.3.31, some sequence
  of 2-switches transforms G to a graph G such
  that NG(w)S, and some such sequence transforms
  H to a graph H such that NH(w)S.
• Since NG(w)NH(w), deleting w leaves simple
  graphs GG-w and HH-w with dG(v)dH(v) for
  every vertex v.

106
Theorem. 1.3.33 Continue

• By the induction hypothesis, some sequence of
  2-switches transforms G to H. Since these do
  not involve w, and w has the same neighbors in G
  and H, applying this sequence transforms G to
  H.
• Hence we can transform G to H by transforming G
  to G, then G to H, then (in reverse order) the
  transformation of H to H.

107
Directed Graph and Its edges 1.4.2

• A directed graph or digraph G is a triple
• A vertex set V(G),
• An edge set E(G), and
• A function assigning each edge an ordered pair of
  vertices.
• The first vertex of the ordered pair is the tail
  of the edge
• The second is the head
• Together, they are the endpoints.
• An edge is said to be from its tail to its head.
• The terms head and tail come from the arrows
  used to draw digraphs.

108
Directed Graph and its edges 1.4.2

• As with graphs, we
• assign each vertex a point in the plane and
• each edge a curve joining its endpoints.
• When drawing a digraph, we give the curve a
  direction from the tail to the head.

109
Directed Graph and its edges 1.4.2

• When a digraph models a relation, each ordered
  pair is the (head, tail) pair for at most one
  edge.
• In this setting as with simple graphs, we ignore
  the technicality of a function assigning
  endpoints to edges and simply treat an edge as an
  ordered pair of vertices.

110
Loop and multiple edges in directed graph 1.4.3

• In a graph, a loop is an edge whose endpoints are
  equal.
• Multiple edges are edges having the same ordered
  pair of endpoints.
• A digraph is simple if each ordered pair is the
  head and tail of the most one edge one loop may
  be present at each vertex.

Multiple edges
Loop
111
Loop and multiple edges in directed graph 1.4.3

• In the simple digraph, we write uv for an edge
  with tail u and head v.
• If there is an edge form u to v, then v is a
  successor of u, and u is a predecessor of v.
• We write u?v for there is an edge from u to v.

Predecessor
Successor
112
Path and Cycle in Digraph 1.4.6

• A digraph is a path if it is a simple digraph
  whose vertices can be linearly ordered so that
  there is an edge with tail u and head v if and
  only if v immediately follows u in the vertex
  ordering.
• A cycle is defined similarly using an ordering of
  the vertices on the cycle.

113
Underlying graph 1.4.9

• The underlying graph of a digraph D
• the graph G obtained by treating the edges of D
  as unordered pairs
• the vertex set and edges set remain the same, and
  the endpoints of an edge are the same in G as in
  D,
• but in G they become an unordered pair.

The underlying Graph
A digraph
114
Underlying graph 1.4.9

• Most ideals and methods of graph theorem arise in
  the study of ordinary graphs.
• Digraphs can be a useful additional tool,
  especially in applications
• When comparing a digraph with a graph, we usually
  use G for the graph and D for the digraph. When
  discussing a single digraph, we often use G.

115
Adjacency Matrix and Incidence Matrix of a
Digraph 1.4.10

• In the adjacency matrix A(G) of a digraph G, the
  entry in position i, j is the number of edges
  from vi to vj.
• In the incidence matrix M(G) of a loopless
  digraph G, we set mi,j1 if vi is the tail of ej
  and mi,j -1 if vi is the head of ej.

116
Example of adjacency matrix 1.4.11

• The underlying graph of the digraph below is the
  graph of Example 1.1.19 note the similarities
  and differences in their matrices.

117
Connected Digraph 1.4.12

• To define connected digraphs, two options come to
  mind. We could require only that the underlying
  graph be connected.
• However, this does not capture the most useful
  sense of connection for digraphs.

118
Weakly and strongly connected digraphs 1.4.12

• A graph is weakly connected if its underlying
  graph is connected.
• A digraph is strongly connected or strong if for
  each ordered pair u,v of vertices, there is a
  path from u to v.

119
Eulerian Digraph 1.4.22

• An Eulerian trail in digraph (or graph) is a
  trail containing all edges.
• An Eulerian circuit is a closed trail containing
  all edges.
• A digraph is Eulerian if it has an Eulerian
  circuit.

120
Lemma. If G is a digraph with ?(G)?1, then G
contains a cycle. The same conclusion holds when
?-(G) ?1. 1.4.23

• Proof.
• Let P be a maximal path in G, and u be the last
  vertex of P.
• Since P cannot be extended, every successor of u
  must already be a vertex of P.
• Since ?(G)?1, u has a successor v on P.
• The edge uv completes a cycle with the portion of
  P from v to u.

121
Theorem A digraph is Eulerian if and only if
d(v)d-(v) for each vertex v and the underlying
graph has at most one nontrivial component. 1.4.24
122
De Bruijn cycles 1.4.25

• Application
• There are 2n binary strings of length n.
• Is there a cyclic arrangement of 2n binary digits
  such that the 2n strings of n consecutive
  digitals are all distinct?
• Example For n4, (0000111101100101) works.

0000 0001 0011 0111 1111 1110 1101 1011
123
De Bruijn cycles 1.4.25

• We can use such an arrangement to keep track of
  the position of a rotating drum.
• One drum has 2n rotational positions.
• A band around the circumference is split into 2n
  portions that can be coded 0 or 1.
• Sensors read n consecutive portions.
• If the coding has the property specified above,
  then the position of the drum is determined by
  the string read by the sensors.

124
De Bruijn cycles 1.4.25

• To obtain such a circular arrangement,
• define a digraph Dn whose vertices are the binary
  (n-1)-tuples.
• Put an edge from a to b if the last n-2 entries
  of a agree with the first n-2 entries of b.
• Label the edge with the last entry of b.
• Below we show D4. We next prove that Dn is
  Eulerian and show how an Eulerian circuit yields
  the desired circular arrangement.

125
De Bruijn cycles 1.4.25
Put an edge from a to b if the last n-2 entries
of a agree with the first n-2 entries of b.
Label the edge with the last entry of b.
126
Theorem. The digraph Dn of Application 1.4.25 is
Eulerian, and the edge labels on the edges in any
Eulerian circuit of Dn form a cyclic arrangement
in which the 2n consecutive segments of length n
are distinct. 1.4.26

• Proof
• We show
• first that Dn is Eulerian.
• Then the labels on the edges in any Eulerian
  circuit of Dn form a cyclic arrangement in which
  the 2n consecutive segments of length n are
  distinct.

127
Theorem. The digraph Dn is Eulerian 1.4.26

• Proof 1/2
• Every vertex has out-degree 2
• because we can append a 0 or a 1 to its name to
  obtain the name of a successor vertex.
• Similarly, every vertex has in-degree 2,
• because the same argument applies when moving in
  reverse and putting a 0 or a 1 on the front of
  the name.

011
001
101
111
110
128
Theorem. The digraph Dn is Eulerian 1.4.26

• Proof 2/2
• Also, Dn is strongly connected,
• because we can reach the vertex b(b1,..,bn-1)
  from any vertex by successively follows the edges
  labeled b1,..,bn-1.
• Thus Dn satisfies the hypotheses of Theorem
  1.4.24 and is Eulerian.

a
b
011
001
1
1
1
0
1
1
1
1
0
1
101
0
000
010
111
0
1
0
0
0
0
110
100
129
Theorem. The labels on the edges in any Eulerian
circuit of Dn form a cyclic arrangement in which
the 2n consecutive segments of length n are
distinct. 1.4.26

• Proof 1/4
• Let C be an Eulerian circuit of Dn. Arrival at
  vertex a(a1,..,an-1) must be along an edge with
  label an-1
• because the label on an edge entering a vertex
  agrees with the last entry of the name of the
  vertex.

a
b
011
001
1
1
1
0
1
1
1
0
1
101
0
000
010
111
0
1
0
0
0
0
110
100
130
Theorem. The labels on the edges in any Eulerian
circuit of Dn form a cyclic arrangement in which
the 2n consecutive segments of length n are
distinct. 1.4.26

• Proof 2/4
• The successive earlier labels (looking backward)
  must have been an-2,..,a1 in order.
• because we delete the front and shift the reset
  to obtain the reset of the name at the head

a
b
0 1 1
001
1
1
1
0
1
1
1
0
1
101
0
000
010
111
0
1
0
0
0
0
110
100
131
Theorem. The labels on the edges in any Eulerian
circuit of Dn form a cyclic arrangement in which
the 2n consecutive segments of length n are
distinct. 1.4.26

• Proof 2/4
• If C next uses an edge with label an, then the
  list consisting of the n most recent edge labels
  at that time is a1,..an.

a
b
011
001
1
1
1
0
1
1
1
0
1
1
101
1
0
000
010
111
0
0
1
0
0
0
110
100
132
Theorem. The labels on the edges in any Eulerian
circuit of Dn form a cyclic arrangement in which
the 2n consecutive segments of length n are
distinct. 1.4.26

• Proof 3/4
• Since
• the 2n-1 vertex labels are distinct, and
• the two out-going edges have distinct labels, and
  
• we traverse each edge exactly once

Distinct vertex label
011 0 011 1
Distinct labels on out-going edges
133
Theorem. The labels on the edges in any Eulerian
circuit of Dn form a cyclic arrangement in which
the 2n consecutive segments of length n are
distinct. 1.4.26

• Proof 4/4
• We have shown that the 2n strings of length n in
  the circular arrangement given by the edge labels
  along C are distinct.