If x and are vertices of a

groh t, we say X is adjacent try it there is

y
and y denote such edge by xy
on edge between X . We an

incident with on edge e it X is an endpoint of e.

me say that ce vertex X is
incident with
number of edges
v.
v is the
wetex
The degree of a

isokalet ⑧

degree degree
O ⑧
2

4 16
of degree 4 3 3

·
2
+ +

sum
+
= + =

↳ Sumofdegree in a
graph
:
2 x number of edges

⑧
2

Im
let di , da, ---dp be the degrees of
Let V,, Ve ,
....
Up be the vertice of graph 5 and
,

Then
the vertical respectuely Let .
a be the number of edges of 5 .

29
di +det ---
+dp=

integes called graphic it there exists a graph

of non-negative
is
· A sequence
Precisely that sequence
.

is
Sequence
whole degree 2 , 2, 2
3 3
6 , 5, 5, 4 ,

(3)
, ,

2 2, 1
(1) 5 , 413 , ,

4 3, 3 , 0
(4)6 ,
6,6 , 6 ,
,

5, 5 , 4 , 4, 0
(2)

=>

3 2 1 17 is not graphic .

1) 2 =

7
+
+ +
+

5 +

18 is not a graph .

5 5 +h 3=
+ 0 =

2) 6 refine
+

at leat
degree 5 verfex ->

Existence of a
 32
3) degree tem
:

3
desresum
=

4)
Hakimi)
-Have ,

two sequences and assure sequence (1) is a descending sequence .

consider the following

(I) is graphic
its dis du The Sequence
(I) Sit,, tz ,
---
,
,

graphic
---

ift sequence (2) is

----its-1 dis-- d ,

(2) til Ez-1

,
,
,

Ro (2)
with degree sequence
.

Then there
is graph z
graphic
a

Assume (2) is .

it to vertices
vortex and connecting
from be by adding
a
5.
constract
t , -1 gr
with degrees
.

Suppose
,

(1)
---
,

graph H with degree

Sequence
·

There is
graphic
a

suppose() is .

Di di
des
=

degS=s ,
desTi= ti ,

adjacent to
Tis removes .

I S is
some i.
adjat to Tifr
S is not
Suppose
Dj
adjacent to Some
exchange Ti and Bi
.

S
If tirdi ,
is
ti di
=>

is descending , .

Since the sequence

Ti has more neighbors the Di

then
If tibds not
we and Dj
,
is
Ti adjacent to
There is a
retex losit .

=>

Tiw
SD ; and
remove edges
....... ....... Sti and Diw
add edges
Dj
·

so ↑

necessary
removes and repeat this process if
.

Obtain graph He with the sare sequence now

#3) from above ,
6 5 , 5, 413 / 3 , 2 , 2, 2
,

2 1 2, 2
4 4 ,3, ,
2
, , ,

2, 1
↳ , 493 , 2, 2 , 2,
1 2 211
32 1 , ,
,

1 1 1
2, ,
3 , 2 , ,

1 , ,
1
, 1, 1
,

e
an
an
·

* ul from abore ,
6 ,
6 6 , 6 , 4, 3, 3
5 5,
,

5 , 2 ,
2 ,
2
,

,
0

1
,

1 , , 0
4 4, 2 ,
,
1 01 0 ,
0 net graphic
,
3 ,

is
refices of add degree even

the number of
·In a griph
all degree differnt .

·
No graph has
graph
number < 5 there exists
a
E
that
,

for every
,

prove h
which have degree
.

with a refices all of,

* X 11 21 1 , 5,

ec
with 0 &X 15 the Sequence ,

x
intege
that for every

is not graphic .
 and Isomerphi Erohs
-
siph 5 is a graph I
A subgraph of a

of and
It is vertex
Sit every latex of
,
a

5
ever edge of It is an edge of .

· · La
LI I
·

1i
⑳
⑧
-

He
a ·

13
② 0

Hi
- nation
said to be pomorphe if the
5 and 52 with particl as
Two graphs ,
I to
number from
52 can be labeled with the ;

adjacent to
G the ratex
5, and is
of wortex ; in
latex ; in 51
,

is adjacent to -

Wheneve vetexi adjacency

sit that preserves
.

V152)
.

between ~(5)
and

-1 correspondence
isomorphic
L
↳
two graphs are

it
SY:?
⑧

degree

I
⑧

have the some

then they

-
⑳
-

Sequence

↳
.

.
40
degree sequences
:

EX 43 sane

in
-

..
I -
!
6 ⑧

!
&

-7
⑳

⑳ "

I 3
⑯ ⑧
4
· ·
/-I
/-
Vertex
each degree 4
vertex two other vertical
each degree is adjacent to
offer
adjoet to of degree far
one .

is
four.
vertical of degree
There is no isonerphom .
· I
-
k4

Kn : Complete graph on nvertices

-

↳ valex is adjacent to
every
every
other vertex .

kn has (2) =
A) edges

isnapmeto
In has a subgraphy

bipartite gran

:
sets
m+ 1 vertices divided into two

say
Mi red verte blue nerfox
.

evera
adjacent to

red vertex is
every
the color
and no retics of sane

are
adjacent .
 # ⑥

k4 ,
4

a
/
A
. ... ⑤

·↑↑
⑧

↑
&-
S >
.
-

-
- ⑧
⑧
⑦

labeled Xn and edges XoX , Xix2 XmXn

The graph with nil vertices Xo , x , .... ,
---
,

denoted by On
is called a path of length n ,

Yo and An are called and verfiles

Y
"
⑧
·

of Pr , they are connected by the path Pn .

Du
.. = y
and
en the graph with vertices xo Y .... n

of length n ,
is a .

cycte
,

The

XoX n-YO
the edges X, X2 1. --
, ,

15
⑳ &

⑧
⑧
 REES
and
b, there is path from
two reticen
a

t connected it for a
A graph is
any

a to b.

contains cycle removing edge from the Cycle

If connected graph a on
,
a

will not disconnect the graph .

rem
t connected graph with vertices and a edges the P-7911
If is a ,

froInduction on the number of edges

q <1-1
q =
n

(i) 5 cycle
n(n + 1
P ((d 1) 1 =
- +

subgraph isomorphic to cycle

a .

contains
connected graph that no

A tree is a

·
. 8
⑧

cor

that contains no agale .

A forest is a sraph
is tree
forest
·

Every connected subgraph of a

edges the p 9+1

=

and
tree with prefines a
If E is a

Proc
- the number of
exeges
Induction on

free with one edge ,

the thm is tree .

If 5 is a

Assume thi is true for all vertical with emer than n edges .
 Let 5 be a free with n edges Select ·

a longest path in 5 with a and b be the

the path could be made
degree 1. Since otherwise
ends of the path Vertex a most have .

in E
there would cycle
.

or be a
longe with obtain tree It with
with the edge incident a
We a

G together
.

Dekk vertex a
from
and n-ledges .

P-lvetical
1 1
hyp
n
Ind p-
- +
=
=

P =

n +
1

5 odd then the number

Exerc tree, and all degrees of vertice
in
are ,

-show that if 5 is a

of edges is odd .

P =
q
+
1

Ihm
If 5 connected and P 9 +1 then 5 is a tree .

is =

least two cyctel

has
.

get Prove that

at
2 5
degree of graph 5 be greater then
.

a
the average

di dz + +
-
- - -
+ dr =
29
are
-

P P
tree .

5 is not a

P<q =

leat one cycle .

=> E contains of

Im
tree iff there exists exactly one path between
A sraph 5 is a

two vertices
any
Prod
~ 5 is tree and ViUEVIE)
Assume
a

V, to Un
are connected , so
there is a path from
Trees
P, UnV2
suppose there are two paths .

= VIU . Ue - -- ,

Wm Ve
Pz 4 w , wz --
=
.
,
 Hi M , then follow until send that
If y
me a vertex in ,
is in P2 .

(this vertex can be V2)

now, we
have a calle .
I
then look at Un For some i ,
di Fwi
If Hi W,
.
=

There are
two paths by assumption .

Det It
It of Est
-

subgraph
-

A spanning subgraph of a graph 5 is a

schsuch
of 5 is a spanning
5. A spanning free
contains all netics of .

5 that is a free .

of

m
spanning tree
.

5 contains
a

Every connected graph

5 contains Let esbeaned

Prod tree ,
then a
cycle
-If
.

not a
is
5

and Let Hir Ere,

of the cycle ,

not tree then Hi contains a

cycle ...

If It, is a

Haw curches
Exerci
- connected graph with a vertices and edges .
many
be a
Let 5

I have ?
doo

=> xactly one

Exercise prove or disprove

-
then 5 is cycle
grih has degree
.

2 ,
a

vetex of
a

If every
&

Falst
0

:
&
&
⑧
O