a
l
6
9 2 4
8 9 3
7 8 9
8
7 8 9
7 8
5 a 2 7
4 6 1 3
a
7
8
9
1
3
5
2
4
6
6 5 4
1 0 6
7 2 1
8 7 3
9 8 7
2 9 8
4 3 9
3 4 5
5 6 0
0 1 2
9 8 7 1 2 3
5 9 8 2 3 4
0 6 9 3 4 5
2
0 4 5 6
4 3 2 5 6 0
7 5 4 6 0 1
2
8 7 6 0
5 0 1 7 8 9
1 2 3 8 9 7
3 4 5 8 7 8
3 5 2 4 6
3 5 a
4 6 1
5 a 2
1
6 1 3
7 0 2 4
6 1 3 5
0 7 8 9
1 9 7 8
2 8 9 7
COM BIN A TOR I A L
MAT HEM A TIC S
YEA R
1
I
1
February 1969 - 1970
1
TOURNAMENTS THAT ADMIT EXACTLY ONE HAMILTONIAN CIRCUIT
by
ROBERT J. DOUGLAS
Department of Statistics
University of North Carolina at Chapel Hill
Institute of Statistics Mimeo Series No. 600.9
JUNE 1969
Research partially sponsored by the Air Force Office of Scientific
Research, Office of Aerospace Research, United States Air Force under
Grant AFOSR-68-l406,
TOURNAMENTS THAT ADMIT EXACTLY ONE HAMILTONIAN CIRCUIT
by
Robert J. Douglas
ABSTRACT
In this paper is given a characterization of those tournaments
that admit exactly one Hamiltonian circuit.
In addition, we determine
the number of non-isomorphic tournaments, with
n
vertices, that admit
a unique Hamiltonian circuit.
I.
INTRODUCTION
A tournament is a directed graph where, between each two vertices
v
and
w,
not both.
either the ordered pair
(v, w)
or
(w, v)
is an edge, but
Tournaments have been studied by a number of authors, and
John Moon's book [3] contains a good review and an extensive bibliography on the subject.
The present paper gives a solution to a problem
raised by Branko GrUnbaum (oral communication), namely finding a characterization of those tournaments that contain a unique Hamiltonian circuit.
(All paths and circuits are simple and directed, and a path or
circuit is Hamiltonian if it passes through all the vertices of the
graph.)
Frequently in combinatorial mathematics, once a characterization
of a class of objects is found, an enumeration theorem follows.
Such
is the case here where we also give the exact number of non-isomorphic
tournaments, with
n
vertices, which admit a unique Hamiltonian circuit.
Research partially sponsored by the Air Force Office of Scientific
Research, Office of Aerospace Research, United States Air Force under
Grant AFOSR-68-1406.
2
The analogous problem for Hamiltonian paths is easy, and is given
by the following theorem due to Redei [4].
THEOREM.
tonian path
The tournament
n
vI' v 2 ' ••• , v
n
following way:
i
T
with
n
vertices has a unique Hamil-
if and only if
the directed edge
is oriented in the
n
is in
(v., v.)
1.
J
T
if and only if
n
< j.
II.
If
S
IS I
is any set,
will denote a graph
G
the number of edges with
... ,
P: v I'
P(v. , v j) ,
1.
v
k
for
v
::;
j
S.
G:
and edge set
E.
T : (V, E)
n
Ivi
n.
=
V
For
G: (V, E)
v I'
... , v k '
(V, E)
p(v) [a(v)]
E V,
is
If
vi' v i + I '
... ,
V'
-=-
V,
then
V', i.e., the graph with
set and whose edges are all those edges in
and
v. ,
J
w2 ' ... , wm provided
is a graph, and
section graph determined by
in
v
as initial [terminal] vertex.
denotes the path
P + Q denotes the path
Finally, i f
will be the cardinality of
Q: wI' ... , w are paths in a graph, then
m
and
i
SOME NOTATION
with vertex set
will denote a tournament with
v
S (V ')
V'
.
= wI
k
is the
as vertex
E which connect two vertices
V'.
III.
THEOREM 1.
If
STATEMENT OF THE RESULTS
C: vI' .•• , v ' v + 1
k
k
in a strongly connected tournament
X
T
E
V
~
C,
which has
= vI)
T : (V, E),
n
there exists a (simple) path
x
(
is a (simple) circuit
then for any vertex
P: wI' ... , w
in
m
V
~
C
as starting vertex or terminal vertex, and such that
can be inserted between some two adjacent vertices of
C
to obtain
P
3
a longer circuit, Le., such that there exists an
and
(w ' v + I ) E E
m
i
is a circuit.
(P
COROLLARY 1.
(simple) circuit
C
where
may consist of only the vertex
E
x.)
In any strongly connected tournament
C
in
(vi' wI)
and hence
T ,
n
can be extended to a Hamiltonian circuit
there exists a Hamiltonian circuit
vertices of
i
H
H
every
H,
i.e.,
such that the order of the
is the same as their order in
C.
We thus obtain the following result which abounds in the literature
(cf. P. Camion [1], J.D. Foulkes [2], J.W. Moon [3]):
COROLLARY 2.
A tournament admits a Hamiltonian circuit if and only
if it is strongly connected.
PROPOSITION.
For
~
n
5,
unique Hamiltonian circuit
a tournament
H
T : (V, E)
n
admits a
if and only if the following three con-
ditions hold:
C=
There exist vertices
and
V
= {d 1 '
... ,
d }
m
{ cl'
•
" e ,
ck1 =I rP
I
iEossibly empty)
I
and a circuit
H:
x, c 1 '
•
$
a ,
c k ' y, d,l ,
• c •
,
d
m'
x
I
I
I
~
where
n
=
mtk+2
and
p
(x)
1
a(y)
I
I
and, for
Ii - jl
~
2,
I
(c. , c.)
~
J
E
E
if
i > j
(d. , d )
~
j
E
E
if
i < j
I
(1)
E
4
and
(2)
and
If
i
<
j,
r S s,
and
(3)
then
(c., d )
J
E
s
E.
c
C.
~~''--'-........
/
C.
~
I
J
/
!
I
x
H
/
J
Y
/
\
',I'
"
d
s
\
/
•
~
H
/
/
/
J
,/
'---._-
u
-------'
d
r
V
Figure 1.
Figure 2.
Condition 3
Notice that (1) completely determines the section graphs
S({x, y, cl' ... , c })
k
and
S({x, y, d I
,
".,
COROLLARY 3.
v,
or
o(v)
circuits.
~
2
If
T
n
for all
j
then
T
n
Also notice
= i + 1.
is strongly connected, and
v,
while (2) and (3)
C and V.
are conditions on the edges joining the sets
that (3) holds if and only if (3) holds for
d }),
m
p(v)
~
2
for all
admits at least two Hamiltonian
5
We would like to replace the "forbidden subgraph" condition(3) in
the Proposition by an explicit description of the orientation on the
edges with one end point in
C
and the other end point in
accomplish this,we now define (see figure 3) a class
T : (V, E)
on
n
~
n
(possibly empty),
V = r/J,
then
To
Tn of tournaments
5 vertices:
T.
n
DEFINITION OF
V.
Let
C
= {x, y}
V
T : (V, E)
u
V,
C u
and let (1) and (2) hold.
is completely described by (1).
n
If
If
V i= r/>,
fix
o
~
p
~
min(m-1, k-1)
< i
P
~
k
(4)
1
For
~
j
i
define
{c.+' ... , c
J
1.
1
lc., c.], [d., d.],
1.
J
to mean
J
1.
lc., c.[
1.
[c., c.l
1.
J
(sl' s2)
E
j -1
etc.
E
If
51
for all
the edges between
C and V.
1.
0
J
1.
similar definitions are used for
"(52' 51)
[c l , c. l
{c., ... , c.},
};
inition holds for
in
=
sl
E".)
E
C and
c
E
51
and
52':" V, define
s2
E
52'
(A similar def-
We will inductively describe all
We first give the edges with one end point
and the other in
V.
6
Let
(c.
~o
,
d. )
Jp
E
E
I
([ d 1 ' d. [ ,
Jp
C. )
~o
E
(0, [C 1 ' C.
[)
E
~o
1
Also let
So
=>
E
E
I
1
I
I
m of j .
p
Ir
]d. , d ]
J
m
be any subset of
(5)
,
p
I
subject only to the condition that
and let
This starts the inductive definition.
Now, for
edges with one end point in
]
are given.
sets
]c
i
1
i
1
i
~
q
~
p,
assume the
and the other end point in
0
q-l
We will extend the definition to give the edges joining the
[c ' c
,c
q-l
[c ' c
1
]
q
i
]
and
q
and
O.
O.
To do this, we need to give the edges between
7
For
1
~
q
p,
~
(] C.
1
define
, c. ], [d.
, d ])
l
J p _ q+ 1
m
q
q-I
lc.
([d , d.
[,
l
J p - q+ 1
(c.
l
q
,d.
)
J _
p q
l
q_ 1
E
E
( [d l' d.
[, c )
i q
J p_q
Also let
S
E
,c. [)
E
E
E
q
E.
(6)
be any subset of
q
l
E
]d.
J _q
p
,d.
J p _ q+ 1
[
D
r
q
i
and write
1
(c.
1
(D
q
, S ) E
q
"" S
q
E
c. )
1
q
E
c ], V)
k
E
q'
E.
)
I
Finally we let
(] c
REMARK.
With
n
i
'
P
For
~
5,
q
(7)
E.
a,l, ... ,p,
each choice of the parameters
k, p, i o '
"', i
p, 1
= jo' j l '
... , j
p , SO' ... , S p
(8)
defines, by means of (1), (2), (4), (5), (6), and (7), a tournament
T : (V, E).
n
Also it will be shown (see the proof of Theorem 2 in
Section IV) that any two different choices of parameters (8) result in
isomorphically different tournaments.
naments is denoted by
T •
n
The collection of these tour-
8
S([C p
k
'1<])
=5
c.
~
q-l
c.
~
p(x)
p
a (y)
=1
d.
Jo
Figure 3.
1
9
THEOREM 2.
n
T
~
n
5
(MAIN THEOREM) :
An arbitrary tournament
vertices, admits a unique Hamiltonian circuit
Tn .
is isomorphic to an element in
H
T ,
on
n
if and only if
Furthermore, every two different
choices for the parameters (8) define isomorphica1ly different tournaments in
T.
n
THEOREM 3.
For
n
~
5,
there exist exactly
n-3
1
+
l:
k=l
T ,
many non-isomorphic tournaments
unique Hamiltonian circuit
n
H.
on
n
vertices, that admit a
(Here
if
b > a.)
IV. THE PROOFS OF THEOREM 1, THE PROPOSITION, THE REMARK, AND
THEOREMS 2 AND 3.
PROOF OF THEOREM 1:
Let
C:
be any circuit
C.
For each
Case
1:
--
i
For each
=
E
1,
There exists a simple path
the last vertex in
is a circuit in
w
and suppose
T : (V, E) ,
n
i, (vi' x)
T
n
is the vertex in
vI' ... , vk ' vi
E
... ,
or
k,
Q in
is a vertex of
x
(x, v.)
E
1.
T
n
from
Q which is also in
Q that follows
v. .)
1.
E.
to
vi
x.
Let
v.
1.
be
Then
C.
which is longer than
not in
E.
E
1.
(x, v. )
T
n
C.
(Here
P
Q(w, x)
where
10
Case 2:
Say
(v., x)
~
E
E,
=
i
1, ..• , k •
Reverse the orientation on all the edges of
T ,
apply Case 1, and then
n
again reverse the orientation on all the edges of
T
to obtain the
n
required circuit.
Assume there exist
Case
3:
-(x, v )
q
E
p, q
and
(v
p'
x)
E
E
and
E.
Without loss of generality
that
such that
p
=
1-
Let
(VI' x), (v 2 ' x), •.. , (Vi' x)
(x, v.+ )
~
E.
E
1
i
be the largest integer such
are each in
E.
Hence
1 ~ i < q
Consequently
is a circuit longer than
C.
(Here
P
consists of just the vertex
x.)
This completes the proof of Theorem 1.
If
DEFINITION.
H:
a tournament
T : (V, E)
n
c-vertex) if
(v
(Vi_I' v + )
i 1
[ II
c"
i+I'
E.
E
n
1
we define
v) E E
i-I'
v.
~
to be a
~
c
vertex (or a
and a tvne a vertex (or an a-vertex) if
~
These definitions depend on the given circuit
is for the circuit
" anticircuit".]
is a Hamiltonian circuit in
v, ... , v,
II
a"
is for
See Figure 4.
PROOF OF THE PROPOSITION:
Assume
T : (V, E)
n
has exactly one Hamiltonian circuit
H:
H.
11
c)
v.J- I
H
V
2
':'j+1
V
2
VI
V2l
\ ~Vi+1
.-/
/
V _
i 1
V 2 £'_2
v. (type a)
H
H
Figure 4.
LEMMA 1.
PROOF:
k:./
V2£,_2
1
Figure 5.
/
Figure 6.
H has at least one c-vertex and one a-vertex.
If each
v.
is of type
1
c~
and
n
2£.-1
is
odd~
then
(see Figure 5)
is a Hamiltonian circuit different from
v.
1
is of type
can be
extended~
ferent from
Figure 7.
H.
c
and
n
= 2£. is
H,
even~
a contradictiono
If each
then (see Figure 6) the circuit
by Corollaries 1 and 2, to a Hamiltonian circuit difNow assume each
v.
1
is of type
If
(V
I
,
V
0
2-{..-2
)
E
E~
a~
n = 2£.,
and see
12
then
is another Hamiltonian circuit; while if
(v
0
2,(,.-2
,v)
1
E
E,
then
is a Hamiltonian circuit different from
type
a,
and
n
=
VI'
2£-1,
H.
Finally, i f each
v.~
is of
then
v 3' ..., v 2£- l ' v 2' v 4' ..., v 2£-2' v 1
is another Hamiltonian circuit.
(See Figure 8.)
This final contradiction
proves Lemma 1.
H
H
)
)
v 2£-3
V 2£_1
v 2£-2
Figure 7.
Figure 8.
13
Let
t (v,)
be the type (with respect to
].
If
LEMMA 2.
t(v,]. )
=
c
and
t(v i + 1 )
=
H) of the vertex
a,
then
v, •
].
L
a(v,+
)
]. 1
(See Figure 9.)
Proof:
there is a
Without loss of generality
such that
j
n
~.
j
>
4
and
i = 2.
Say
(v, , v )
3
J
E
a(v )
3
E.
>
l',
so
But then (see
Figure 10) the circuit
can be extended, by Corollaries 1 and 2, to a Hamiltonian circuit different from
H,
and the contradiction proves Lemma 2.
v,
J
H
Figure 9.
Figure 10.
As a consequence of Lemmas 1 and 2, we have that there exists
exactly one
t(v,+ )
]. 2
= a.
i
such that
t(v,)
= c
].
and
t(v,+
)
]. I
= a;
furthermore,
Without loss of generality, we can, therefore, say
that the following situation holds:
14
For some
1
k,
:0;;
k
= 1,
t(v.)
=
c
for
j
t(v.)
=
a
for
j
J
J
Since
n-2,
:0;;
•
"l
k+1,
"
,
•
c
1
k
•
,
n.
)
t(v )
n
=
a
and
c,
t(v] )
(9)
r
I
it follows from Lemma 2 that
(10)
1.
p (v )
n
To see this, reverse the orientation on all the edges of
Lemma 2,
T ,
apply
n
and then again reverse all orientations.
Therefore, using (9), (10) and Lemma 2, we can write
H in the
following way:
where
=
p(x)
C = {c 1 '
and
V
{x, y}
u
... ,
c }
k
+ ,p
1
a(y)
is the set of all the c-vertices, and
is the set of all the a-ve rti ces , where
(which may be empty) .
So
k
~
1
and
m + k + 2 = n.
V = {d , ,
y
y
Figure 11.
"
"
,
d }
m
(See Figure 2. )
H~+
J I
x
e
Figure 12.
15
li-j I
Hence (2) holds, and (1) will hold if for
(c.,c.)EE
for
i
>
(d., d.)
1
J
for
i
< j
J
1
Assume there exist
(c ' c )
j
i
E;
E
i
E
E
and
see Figure 11.
j
~
3
we can prove that
j
such that
.
i - j
~
3
and
Then
is a circuit which can be extended to a Hamiltonian circuit different
from
H,
replace
a contradiction.
by
C.
J- 1
such that
j
-
i
~
(Here we assumed
x
in the above.)
3
and
J
1
1
E
1
y, d. , d. + H(d. ,d.
J-
1)
>
1;
if
j
Now say there exist
(d. , d. )
J
j
E·,
see Figure 12.
1,
and
i
j
Then
+ d.J- l ' d j + 1 + H(d j + 1 ,y)
is a circuit which can be extended to a Hamiltonian circuit different
from
H.
(If
j
m,
replace
d j +1
by
in the above. )
x
This
contradiction proves (1) •
We now will show that (3) holds.
there exist
i < j
r s s
and
where
Say this is not the case, i. e. ,
(c. , d )
r
1
E
E,
yet
(d , c. )
s
J
H
~
ci
c1
x
x
Y
d
k
Y
s
d
k1
Figure 13.
Figure 14.
E
E.
16
Then (see Figure 13) the circuit
d
+ H(d ,d ) + d , c. + H(c.,y) + y, x + H(x, c,)
r
rs
s
J
l.
J
can be extended to another Hamiltonian circuit; hence (3) is proved.
To complete the proof of the Proposition, we must show that if a
tournament
T,
on
n
n
vertices, satisfies (1), (2), and (3), then
is the only Hamiltonian circuit in
circuit different from
i
~
1
H.
Edge
Assume
T .
n
(x, c
1
)
H'
E
H'
H
is a Hamiltonian
as
= 1.
p(x)
Let
be the largest integer such that the path
P:
H'
is a subpath of
x, c
1'
... ,
c
, yet
E
H'
is a path in
k
because there do not exist
(d , d )
s
r
(c, , c + ) ~ H'
i 1
l.
Let
E.)
does not exist a
r
(ci ' d r )
j
>
i
,
then
an
E
with
s
E
must be a path in
> i.
as
H'
s
i
E
u
H'.
r
>
As
and
By (3) , there
a(y)
and
1
v
>
u
<
...
<
k
E
E which con-
such that
v
and
and
a(y) = 1
and see Figure 14.
(d , c,)
r
J
exists as i f
and
E,
P
j
H'
H
such that
because, by (1) , there do not exist
(d , d )
v
u
(Such an
H'
+ cl."
d , ~ , (t
r
for some
But then we have
••• , d
,
1
1k 2
s
where
(c. , d )
l.
r
E
E
and
r
<
ks
k
' c.
J
(~
s
k
<
1
,
2
c, )
J
This proves the Proposition.
tradicts (3) •
PROOF OF THE REMARK:
A
q
Let
and
B
q
[d,
J _
p q
,d].
m
s
and
17
Now
(A, B )
P
P
E
E
by (7), so it suffices to show that
(Aq-l' B
) E E
q-l'
implies
(A, B
whence
q
q-l
)
E
E,
Aq -c Aq-l
Note that
(A
and we are done as
(A, B )
q
E
Bq-l -c Bq'
and
- A ,B
q-l
E
q
q
q-l
)
E
E
by (6).
PROOF OF THEOREM 2:
We notice that by the Proposition we are to show that a tournament
Tn
belongs to
assume (1), (2), and (3) hold for a tournament
that
T
n
E
T.
n
V
We are done if
smallest index such that
~,
(c. , d.)
1
E
E
J
0
so say
T ,
(ck ' d 1 )
index
t
q
>
i
E
q-j
E.
Given
such that
i
V +~.
for some
(c. ,d. )
1
q J
as the smallest index
t
<
t
for some
E
E
, let
q-l
i
q
j
= 0, 1, ... , p,
Figure 3.)
jo
=
1
as
(ck,d 1 )
E
q
(c i ' d. )
Jp
0
E
exists
<
t
q-l'
and define
Continue
i
P
and
t.
P
E.
,
Clearly (4) holds.
Now
i
as
io
to
define
t
and notice that
be the
be the smallest
such that
q-l
io
and define
j,
this procedure until we are forced to stop, say with
q
Let
Certainly
and
q-l
and we will show
n
as the smallest index such that
as
First we
if and only if (1), (2), and (3) are true.
E
and
o
=
1
(d , c )
1
m
=>
E
E.
jp
+m
Define
(See
For
18
{d. :
and for
q
1, ••. , p,
sq
{d. :
J
j
and
m
$
J
, d.)
(c.
1
E} ,
E
J
0
we define
p-q
< •
< j
and
J p - q+ l
(c
i q
' d.)
E},
E
J
and note that
as
(d , c l )
m
E
E by (2).
From the above definitions, we immediately
obtain (5), and, except for
(11)
we also obtain (6).
(3) holds for
But so far we have not used the assumption that
Using (3), we conclude that (7) and (11) hold, and
T .
n
it therefore follows, using the above parameters, that
We next show that if
(3).
T
n
E
T,
then
n
The only thing to prove is that if
For some
By our definition of
d
s
E
[d.
J p- q+ l
the Remark,
, d ],
m
s
q
$
Tn ,
and
(c., d)
J
1
q,
E
d
i
E.
p,
$
r
E
< j
c.
V
+¢,
E
[d.
J p- q+ l
implies
[c
i
q-l
, d ).
m
c.
J
E
T •
n
then (3) holds.
E
r
1
1
n
satisfies (1), (2), and
(c., d )
and
Case 1:
--
T
n
T
E
Let
E.
c. [
1
q
Now
)c
i
q-1
r
$
s
ck )·
implies
Hence, by
19
Case
2:
-Thus
c.
E
[ C.
1.
1.
c.
J
E
] C.
1.
,
P
c ].
k
, c ] , whence, by
k
P
(c ., d )
s
J
(7) ,
E.
E
Case 3:
Tn ,
By definition of
(D, [c 1 ' c.1. [)
0
E
whence
E,
(d , c.)
r
E
1.
E,
which shows that the hypothesis of (3) cannot hold in Case 3.
Since these three cases exhaust all the possibilities, we have
proved the first part of Theorem
1.
Turning to the second part of Theorem 2, we first note that, by
the definition of
Tn ,
( 1) and ( 2) hold for any
(1) and (2) we see that the collection
c-vertices in
T
n
(with respect to
a-vertices (with respect to
tournaments in
T.
n
For
H).
i
Hi:
be the Hamiltonian circuit in
under the isomorphism must be
circuit.
i
T
n
H2
j.
Then, for all
Hl) if and only if
Tn'
and from
{x, y} u D is the set of
H) and
TI
n
T2
n
and
be isomorphic
let
... ,
i
v n'
i
VI
given by (1) •
because
T2
n
Without loss of generality, we assume
spond, for all
to
i
vI'
E
in (1) is the set of all
C
Now let
1, 2,
T
n
J' , v j1
has only
v.I
J
~
and
v~
J
n
is a c-vertex (with respect
is a c-vertex (with respect to
are equal, which proves Theorem 2.
Hamiltonian
corre-
H2 ).
this, it easily follows that the corresponding parameters for
T2
HI
The image of
Using
and
20
PROOF OF THEOREM 3:
Fix
n
5.
~
T' [Til]
n n
as the set of all those tournaments
and
>
iO
IT
By Theorem 2, we are to determine
T
n
T
n
E
Define
I.
n'
V t r/J
such that
l[i o = 1].
IT'n I:
We first determine
We have
k
~
V
1,
~~,
and
Section III, the section graphs
S({x, y, d l
,
... ,
dm})
~
m= n - k - 2
1.
~})
S({x, y, c 1 ' •.. ,
are uniquely determined by
1 < i
1
i
<
0
jo < j
<
1
1
...
...
<
<
<
i
jp
and
nand
k.
We have
k
~
P
:::;
As noted in
n-k-2
where
o : :;
Also, for
q
p
~
min(k-l, n-k-3).
1, ... , p,
S
-c ] d.J , dm]
0
and
[
S c- ] d.
, d.
q
J _
J p _ q+ 1
p q
p
Hence, for fixed
k
sequence
... , i p '
m-j
2
p
jo' jl' ... , jp,
2jp-q+l-jp_q-l
and
(k-1)
p+l
there exist
p,
(
n-k-3
P
choices for the sequence
)
choices for
choices for each
S
choices for the
q
So'
and
(q
1, ... , p).
Therefore
we
have the following:
IT'I
n
n-3
min(k-l,n-k-3)
L
L
k=l
p=O
<p(n,k,p)
L:
l=j
1
o
< i
<
O
j
< •• <
1
< •• < i
p
j
p
:::; k
~
n-k-2
(12)
21
where the inner sum, for fixed
i
O
'
••• , i
P
k
and
p,
is taken over all sequences
such that the indicated equality and
,
inequalities hold, and where
2
cjJ(n, k, p)
m-j
P
P
IT
q=l
is the number of choices for the sequence
. .. ,
j
p
But
.
m-jp
and hence is independent of the choice of
i
p
min(k-1,n-k-3)
n-3
L
L
k=1
p=o
0
then
= 1,
<
i
1
jo <
o
< i
p
jo' .. ', jp
2
(and of course
< jp
2n- k - p- 3 • (n-k-3)(k-1)
P
P+1
and
dm <t SO'
Hence
k,
<
n-k-2.
and
(
n-k-4
P )
So c- ] d.J , dm[
p
choices for
choices for
:s;
P
+m
jp
m-j -1
2
and there exist
1
n-k-p-3
2jp-jo-P
ITiln I:
We now determine
i
•
Thus (12) reduces to
).
IT'I
n
If
for a given
reduces to
2
.. . ,
SO' Sl' .'0' Sp
choices for
With these changes, we proceed as in the
first case to obtain:
n-3
IT"I
n
L
k=1
n-3
=
L
k=1
min(k-1, n-k-3)
L
L
m-j -1
2
P
<jp < n-k-2
l=i < •• <i :s; k
0
P
min(k-1,n-k-3)
n-k-p-4 n-k-4 (k-l)
l:
2
( p )
P
p=o
p=o
l=j
0
< ••
.
j -jo-P
2 P
,
22
Finally, as there exists exactly one
T
n
E:
Tn with
v
have
which proves Theorem 3.
V
REMARKS
From Theorem 3, we obtain the following table:
T(n),
with
the number of non-isomorphic tournaments,
n
vertices, that admit a unique Hamiltonian
circuit
~--1_3--:--:--:--2-:--5-:-l-4-:-3-~-:
The three non-isomorphic tournaments in
Figure
15.
Ts :
¢,
we
23
The eight non-isomorphic tournaments in
Figure 16.
T5 :
24
REFERENCES
1.
P. Carnion, Chemins et Circuits Hami1toniens des
G~aphes
Comp1ets,
Comptes Rendus, 249 (1959), 2151-2152.
2.
J.D. Foulkes, Directed Graphs and Assembly Schedules, Proc.
Symp. Applied Math., Combinatorial Analysis, 10 (1960),
281-289.
3.
J.W. Moon, Topics on Tournaments, Holt, Rinehart and Winston,
1968.
4.
L. Redei, Ein kombinatorischer Satz. Acta Sci. Math. (Szeged) ,
7 (1934), 39-43.