295
Perfect's Theorem.
(a,bi)'
that
G
Let
i = l,2, ... ,k-l
be a
k-connected
graph.
are independent paths from
Suppose
a
to
B ~ VG\{a} with
IBI <'= k.
Then there exist k independent paths from
to B,
k-1 of whose endvertices are {b.: i=1,2, ... ,k-1}.
A COROLLARY TO PERFECT'S THEOREM
a
l
Proof.
D.A. HOLTON, B,D, MC KAY AND M,D, PLUMMER
In a 3-connected graph Perfect's Theorem shows how two independent
Here we show, again for 3-connected graphs, how two
independent paths from a path with given endvertices may be extended.
of
G
The need for the theorem in this paper arose in a first proof [2]
Let C be a set of paths in G and define <C> to be subgraph
containing just those vertices and edges of G which are used
by elements of
C.
Define
to be the set of those vertices of
V(C)
which have degree greater than two, or are endvertices of an
<C>
This can be done in one of two ways.
D
We need some notation before stating our main result.
paths from a given vertex to a given subgraph may be extended to three
independent paths.
See [4].
element of C.
C is a configuration if
<C> whose endvertices are in V(C).
contains every path in
C
of the fact that there is a cycle through any nine vertices in a 3connected cubic graph.
result has been found which is independent of the
theorem~
However, it
is used in [l] in showing that there is a cycle through any
in a
k-connected
,Ck
vertices.
<'=
3)
where the path
and
vertices
(a 1 ,a 2
)
k-connected
has more than two specified
and
a 1 ,a 2
b,
where
b
(a,b)
path which passes through
a.
If there is any doubt which
possible exception of
c,
(a,c,b)
to denote the
We also use
a
or we will mean the
and
a
and
(a,b)
a,b
b.
u
{a}
b.
For
•
b 1 ,b 2
E
V(S)
i = 1,2,
let
G
(ii)
It will be clear in the context which use is
),
A path from a vertex
=
(a,b)
Further, we use
(a,b 2
a
)
[a,b)
(a,b]
and
to denote the vertex
containing both
Ci
and
s.
We say that
<P 1
:
C1
+
C2
(iii)
B
is a path
maps the endvertices of each
<P 1 CP),
for each v E VG,
v E V(C 2 ) n V<S>.
v
E
V(C 1
)
P
n V<S>
C1
E
onto the endvertices
if and only if
For notational convenience, whenever we are dealing with a configuration C 2 + S which is equivalent to
of <jl 2 (z) for all z E V(C 1 ).
C1
+ S,
we will write
z
instead
Finally we define the configurations W, X and Y as in Figure
1, where b 3 i {b 1 ,b 2 } and where, in Y, u 1 = b 1 and u 1 = b 2 are
allowed.
[a,b].
are said to be independent if
to a set
<P 2
of
When a new path is introduced it will
and we similarly define
Two paths (a,b 1
[a,b 1 J n [a,b 2 ] = {a}.
B
and that
C1 + S and C2 + S are equivalent if there are bijections
and <P 2 : V(C 1 ) + V(C 2 ) such that
path which
to represent the vertices of the path
being made of the notation.
n
b 1 f: b 2
and
smallest configuration in
a,b
its endvertices.
(a,b]
G,
a 1 f: a 2
G such that {a 1 ,a 2 ,b 1 ,b 2 } ~ V(Ci)'
E<C.>
n E<S> = \1.
Define C.l +S to be the
l
a,b
be understood to contain no previously named vertices except possibly
(a,b)
is a configuration in
where
Ci be a configuration in
V<C.>
n V<S> c V(C.)
and
l
l
avoids all previously named vertices of the graph in question with the
set
S
VG\V<S>,
to denote a path between the vertices
may equal
path we mean, then we will either write
excluding
E
At this stage we have been unable to prove such a result.
Throughout we use
a
k+4
k-regular graph.
We believe that the theorem can be generalised to
graphs
Suppose that
Subsequently an alternative proof [3] of this
(a,b)
such that
in
G
Theorem. Let G be a 3-connected graph and
such that
IV<S>I <'= 3.
If G contains the configuration
uration equivalent to X', or Y.
W,
then
S
G
a configuration
contains a config-
{b}.
We can now state Perfect's Theorem for
k-connected
graphs.
Proof. Let Q = V<S>. By Perfect's Theorem, there exists a path
(a 15 y 1 ) from a 1 to y 1 E (Q\{b 1 } ) u (a 20 b 2 ) .
If y 1 E Q\{bpb 2 } ,
then we have a configuration equivalent to X.
297
296
For 6 2 e Ca 1 ,b 1 J and a 2 e Ca 1 ,y 2 J we again contradict the
minimality of
1Ca 1 ,y 2 JI.
If a 2 e [a 1 ,y 2 ) u [a 2 ,b 2 ) u (al'b 2 ) u (a 2 ,y 2 ),
If a 2 e (a 1 ,a 2 ), we have a
we have a configuration equivalent to X.
configuration equivalent to
1.3:
Y 2 e (a 1 ,b 2 ).
Y.
The proof follows a similar pattern to that
of Case 1.2.
Case 2:
y 1 e (a 2 ,b 2 ). Amongst all equivalent configurations, we
choose one which minimises
1Cy 1 ,b 2 JI. We note that
1Cy 1 ,b 2 JI # 1
since otherwise we apply Case 1.
')
By Perfect's Theorem, there exists a path
we argue as in Case 1.2. Hence we may assume that
y1
= b2.
Using Perfect's Theorem again we see that there
(a 2 ,y 2 ) from a 2 to Y2 e Q\{b 2 } u Ca 1 ,b1) u Cal,b2).
y 2 e Q\{b 1 ,b 2 }, then we have a configuration equivalent to X.
exists a path
If
= b1.
y2
1.1:
Since
{b 1 ,b 2 }
is not a cut set, there must exist
a path (a 1 ,6 1 )
in G such that
a 1 e [a 1 ,b 1 ) u (a 1 ,a 2 ) u [a 2 ,b 2 }u (a 1 ,b 2 ) u (a 2 ,b 1 )
and 6 1 e Q\{b 1 ,b 2 }.
For a 1 e [a 1 ,b 1 ) u [a2'b 2 ) u (al'b 2 ) u (a2'b 1 ) we have a configuration
equivalent to
X and for
equivalent to
Y.
1. 2:
a
1
e
(a 1 ,a 2 )
E
1 ,
b1)
•
·
Agaln
{b 1 , b 2 }
is not a cutset so we have
and
(a 2 ,b 1 )
If
a
1
i
1.2.2:
Suppose
a 1 e (b 1 ,y 2 ].
(a 2 ,y 2 ,b 1
that
{a 1 ,b 2 }
a
e [a
2
62
1
61 )
Y.
We thus have
in
that
a
,y 2 ) u (a 15 a 2 ) u [a 2 ,b 2
e CQ\{b 1 }) u Capb 1 ) u (a 1
)
u Capb 2
)
(a 1 ,y 2 ) u [a 1 ,a 2 J u (a 2 ,y 1
{b 1 ,b 2 }
) u (a 1 ,y 1 ) u (a 2 ,y 2 ).
If
is not a cutset, there exists a path
a 1 eAu (b 1 ,y 2 J u (b 2 ,y 1 ]
then we have either configuration
a 1 e (b 1 ,y 2 J u (b 2 ,y 1 J.
e (bl'y 2 J.
1
and
61
e Q\{b 1 ,b 2 }.
X or configuration
By equivalence we may assume
Amongst all configurations which are equivalent to the surviving
1Ca 1 ,y 2 JI.
{a 1 ,b 2 }
is not a cutset and so there exists a path
(a 2 ,S 2 )
a 2 e A u (b 2 ,y 1 J u Ca 1 ,y 2 J to
6 2 e CQ\{bpb 2 }) u CapS 1 ) u [b 15 a 1 ) . For a 2 eAu (al'y 2 ] we argue as
in Case 1.2.2 with minor modifications. Hence we have a configuration
in
joining
G
C,
61 = 62 ,
or
D
E
of Figure 2 where in
D we may have
and in
C
and
E
we may
62 = b 1 .
Amongst all equivalent configurations which now arise, we choose
Amongst all configurations equiv-
is not a cutset, there is a path
=
Then
G•
a 1 e A,
If
).
alent to this one, we choose a configuration which minimises
Since
(a 1 ,
have
(b 1 ,y 2 J, then we argue as in Case 1.1 with
adjusted where necessary to
A
then since
),
equivalent to
Q\{bpb 2 }.
1.2.1:
(apb 1
e
y 2 e (a 1 ,b 1 ) u (a 1 ,y 1 ).
configuration of this case choose one which minimises
we have a configuration
a path (a 1 ,6 1 )
in G,
such that
a 1 e [al'b 1 ) u Capa 2) u [a 2 ,b 2 ) u (al'b 2 ) u (a 25 y 2 ) u (bl'y 2 J
61
Let
2.1:
y2
Now
y 2 e (a
with
y 2 e (yl'b 2 J u CQ\{b 1 ,b 2 }) u (al'b 1 J u (a 15 y 1 ) .
If Y2 e (y 1 ,b 2 J,
then
Ca 1 ,y 1 ,y 2 ) intersects
we contradict the minimali ty of
I [y 1 ,b 2 J I since
If y 2 e Q\{b 2 }, then
(a 2 ,y 2 ,b 2 ) at y 2 with
1Cy 2 ,b 2 JI < 1Cy 1 ,b 2 JI.
Figure 1
Case 1:
(a 2 ,y 2 )
(a 2 ,6 2 )
u (a 25 Y2 ) u CapY2 J
G such
and
Ca 35 6 3
)
in
{a 1 ,a 2 }
=
joining
G
2.1.1:
C.
ent to
Y
If
or
+ 1Ca 2 ,y 1 JI.
a
)
3
Hence there exists a path
eAu (al'y 2 ] u (a 2 ,y 1 ]
to
u (a 2 ,6 2 ) u (al'b 1 J u (a2'b 2 ].
Suppose the configuration giving rise to minimum
equivalent to
X
1Ca 1 ,y 2 JI
is not a cutset.
6 3 e CQ\{bl'b 2 }) u CapS 1
,6 1 ) .
If 6 2 e CQ\{b 1 ,b 2}) u (a 1 ,6 1 ) , then we proceed as in Case 1.2.1
or we contradict the minimality of
1Ca 1 ,Y 2 JI ·
G,
In
1Ca 1 ,y2JI ·
in
t
one which minimises
If
a 3 e A,
t
is
then we achieve-configurations equival-
with arguments similar to those of Case 1.2.2.
6 3 e (a 1 ,S 1 J and a 3 e (a 1 ~y 2 J, then we produce a configuraC which contradicts the minimality of t.
tion equivalent to
298
299
through some configuration which is equivalent to
only difficult cases here arise when
Or
f3 3 E (a 2 ,b 2 ].
a 3 e: (a 1 ,y 2 ]
C,.
D
and
or
E.
The
S 3 e: (a 2 ,S 2 )
In both cases we get a configuration equivalent to D which
contradicts the minimality of !/,. In the first case it is necessary to
Cb 1 ,_B 2 ,(3 3 ,a 3 ,y 2 )
map
b2
---------- -s
-~--3----
c
(a 3 ,a 1 ,13 1
~
)
to
(a 1 ,B 1
to
)
(bl'y 2 ),
(a 20 (3 3 ) to (a 2 ,S 2 ) and
and in the second case the homeomorphism takes
(bl,S2,a2,yl) to (bl,y2),
(bl,(33,a3,y2)
(a 2 ,(3 2 ) and Capap(3 1 ) to (al'B 1 ) .
I
to
(b2,yl),
(a3,(32)
to
D
to
X
2.2: If Y2 E (a 1 ,y 1 ) , then we achieve configurations equivalent
or Y by arguments similar to those in Case 2.1.
D
REFERENCES [1]
D.A. Holton, Cycles through specified vertices in k-regular
graphs,
[2]
\
s
--------_..,.,....
E
a3
equivalent to
E
(al,bl]
D
and
a3
ity of
!/,
a3
[3]
D.A. Holton, B.D. McKay, M.D. Plummer and C. Thomassen, A nine
point theorem for 3-connected cubic graphs, Combinato~iaa,
to appear.
(a2,yl],
E
then we have a configuration
[4]
JL
H. Perfect, Application of Menger's graph theorem, J. Math. AnnaZ.
AppZ. 22 (1968) 96-111.
(a2,b2].
a3 E Q\{bl,b2} and a3 E (al'y2J u (a2,ylJ, then the minimalis contradicted by a configuration equivalent to c.
Department of Mathematics
University of Melbourne
Parkville
a3 E (al'S 1 ) u (a 2 ,S 2 ) , then the minimality of !/, is contradieted via a configuration equivalent to C or by one equivalent to E.
If
2.1.2: Suppose the configuration giving rise to minimum
equivalent to E. Then we argue as in Case 2.1.1.
2.1.3:
equivalent to
Suppose the configuration giving rise to minimum
D.
If
a3
E
A,
(a 2 ,S 2 )
tion occurs when
S3
having origin in
(a 1 ,b 1 ] .
E
For all other positions of
!/,
!/,
X
or
Y.
Victoria
is
Department of Computer Science
is
Vanderbilt University
Nashville
then by the standard arguments we achieve
a configuration equivalent to either
U.S.A.
The only minor varia-
when we treat the path
((3 2 ,S 3 ,a 3 )
Mathematics Department
as
Vanderbilt University
Nashville
a3
to appear.
specified vertices in 3-connected cubic graphs, Univ. of
Melbourne, Maths Research Report:, 38, 1979.
I
;-..I
which contradicts the minimality of
By symmetry we deal with
If
E
Combinato~ia,
D.A. Holton, B.D. McKay and M.D. Plummer, Cycles through
Figure 2
If
A~s
we contradict the minimality of
!/,
U.S.A.