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344
Solutions to Exercises
od:
for j
to n do
indeg[i]
.
indeg[i] +
A[j,i]:
>
t
:= 1:
> found := 0:
#found = 1 when we have found
#the
next
vertex
>
while(t <= n) do
i
:= 0:
found := 0:
while((found =0) and
(i <n)) do
i
:= i+1:
if (member
(i,V)=true)
then
fi:
if (indeg[i]=O)
v[t] :=
i;
t
:=
t+1:
found
. 1:
V := V minus
{i};
in V do
#remove
vertex i
#find
indegrees
for
if
(A[i,j]=l)
indeg[j]
indeg [ j ]

1:
S :=
seq(v[i],i=l
..
n):
lprint(/A
canonical
labeling
is ,
S);
new V
5. (a) [BB]
If
g
has a cycle, then there is no canonical labeling so (with notation as in Theorem 12.4.3)
some
S
=
V
"
{vo,
VI,
V2,
.•• ,
Vk}
has the property that
Sv
#
0
for all
V
E
S;
that is, for every
v
E
S,
there exists an arc of the form
xv
with
xES.
Let
Uo
E
S
be arbitrary and choose
UI E
S
such that
UI
is an arc. Choose
U2
E
S
such that
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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