Mth60)
2
assignment
1) Eulerian trail PG K5 K2
:
circuit
Eulerian
Ke not
Ks,
trial
w
circuit
or
are
the
I
odd degree
that
vertises
them
between
.
handshaking
the
From
5
,
and
path
no
:
,
eulerian
a
2) Assume U
have
,
& degin)
Koma,
=
IE)
vEU
we
know that
be
even
the
.
If u and
Component B
A
Component
I
containing
the same
must be in
be the
the walk :
p = (v = Vo , , ... ()
even
then
j
path
odd
.
Since
have
-i + ]
maximal
vertex of P
the endpoint
is
I
.
let
so
V
,
path
Since
Vi Uj
,
,
component
connected
3) Let f
i
add degree.
an
be
should
all
of
sum
a
vartex
have
both
apply
-
degree
of
6
in G the
length of P is given
P must
stop
.
·
up
must
Kil ;
be
then the cycle
(Vo Y U2 -- Vi
Vit vo)
,
i
is
even
this
,
-
,
implies that
j-i + 2 edges
because
is
means
even
which
G
other
have
the
at
3
has
i
is
also
Up
its
even
contains
even
is
.
edges
the
,
an
So
edges connecting back
of v. If
which
cycle
(Vo Vi With ...
j
,
.
as
which
neighbors
,
even
odd
.
vertices
to
v
vertex
degrees in the components
contradiction Therefore, a graph with
its a
the
can
separately
I
let
must
components
containing
and
degrees
vertex
disconnected then we
are
v
assumption ,
our
But the
of all
sum
to both of their
Koma
From
the
,
would
Vol.
length
is
cycle
Mu
4)
for
removing
components for removing a
components
2
5a) Let T have
cycles
,
6) Let
T haven
and
is
vertices
forming
and thus
6) Proving
Base
case
a)
creating
a
bl =
connected
Cu
free
the
aligns
induction
Proving
m
.
vertices , this
a
vertex
/
E , So
Therefore the
.
adding
edge
Only
.
G
G is
has
is
have
I
an
one
is
edge
free
a
edge
edge
results
also
was
added
proof
and use
in
unique path
a
must
connect
so
they
connected
cut
a
alread
↓
edges
o
have
vertices
m
Cm-1 + 1
is
expected
formula
with a
vertices
must
or
has
G
more
.
is
edges so
1-1 > K = 1
(j -1) + (jc-1)
,
-
a
and
.
to
only
between
2
existing
I
cycle
induction .
by
tree
·
.
stated
has j
=
a
edges
.
vertex
:
is
added, there
# of edges = (m+ )
.
.
vertex
new
-
to
not
are
m+
1
m
=
m
has In
1- k
I
connectivity while also
new
edges Assume
n-1
G
that
.
beginning
at the
vertices and
(jp-1)
edges Add
maintain
Since
n-1
have
means
Ma
m
n-1
m-l
components. If G
+
.....
This
=
cycles and
no
disconnected
acyclic If a tree
connected and
follows
which
edge must be added to
new
all K components
it would be
was
·
the
2
meaning it has
n-1 =
free there
So
# of edges
with
free
I
a) Assume
because
a
.
of
number
,
E
edges Adding
and n-1
free
vertices,
cycle So
a
remove
cycle
a
every
the tree
.
.
(n = 1) :
with
is
you
b) Assume
=
Inductive : Let
By
E
longer
formed ,
was
removing
as
2 vertices in the free
any
be
after
vertices
no
T
in
components Since trees can't
created between the I endpoints so
.
edges
edge
any
.
can't
increases
components
he
I
more
2
in
tree
the
connected
longer
no
are
path
another
edges Let
2 or
.
graph .
connected
a
.
disconnects
Removing
E
graph
connected
a
from
vertex
and n-1
vertices
n
edge from
cut
a
cut
G
components, then
is
each
not
of those
j-l edges . The
total
# of edges for
From
know
b has n-1
only has
this
we
I component Therefore ,
.