810MATHEMATICS TRAINING PROGRAM
On Tournaments Having
A Unique Hamiltonian Circuit*
by
Michael R. Shelor
Department of Statistics
University of North Carolina at Chapel Hill
Chapel Hill, North Carolina
*Preparation of this manuscript supported in part by the NSF grant .
MPS74-06975 AOI
May, 1977
ABSTRACT
A tournament is a directed complete graph.
A Hamiltonian circuit is a
circuit which passes through each vertex of the graph once and only once.
This
note examines the family T of all nonisomorphic tournaments on n vertices which
n
have a unique Hamiltonian circuit.
We let T
n
=
ITn I.
In [1] Douglas gives a
graphical characterization of the family T , and from this characterization
n
obtains
an involved formula for calculating the values T.
n
n =
Tn
=
In particular
3
4
s
6
7
8
9
10
1
1
3
8
21
ss
144
377
This note presents a graphical construction of the family Tn+ 1 from the family
Tn and, using this construction, obtains a proof of the recurrence
Tn+2 = 3 Tn+1 - Tn
en ~
4) .
1.
Preliminaries.
The outdegree of a vertex v is the number of edges incident with v and
•
directed away from it.
The indegree of v is the numher of incident edges
directed into it.
THEOREM 1:
(i)
If T
T , then T has at least one, and at most two, vertices with maximal
E
n
outdegree, i.e., with outdegree n-2.
(ii)
If T
Tn has 2 vertices vl,v Z of maximal olltdegree, then v l ,v 2 are suc-
E
cessive vertices of the unique Hamiltonian circuit in T.
PROOF:
Since T
E
T
n
has a Hamiltonian circuit, we first note that the maximal
outdegree of any vertex is n-2.
(i)
If T
E
T
then it follows from corollary 3 of Douglas [1] that at least one
n
vertex must have out degree n-2.
•
Furthermore, if v ,v are vertices of maximal
l 2
outdegree n-Z, and v any other vertex, then the indegree of v must be at least2 .
Consequently,
(ii)
the outdegree
of
v
is at most (n-l)-Z
=
n-3.
Let vl,v
be of maximal outdegree n-2. Without loss of generality, assume
Z
that the edge between v and v is v .-->--. v . Then this is the only incoming
Z
1
Z
1
edge for v .
2
Hence it must be an edge of the Hamiltonian circuit.
Thus, we can divide the family T into disjoint families according to whether a
n
tournament T
E
Tn has 1 or Z vertices of maximal outdegree.
A
= {T
E
T:
n
T has only 1 vertex of maximal outdegree}
B
n
= {T
E
T:
n
T has 2 vertices of maximal outdegree}
n
-e
We define
o
-2-
and
A
=
n
IAn I
B =
n
,
IBn I .
•
Then clearly
T = A u
n
n
Bn
= An
n
and
T
2.
n
+ B
Construction.
For n':: 3, if we are given the family T
n
ical construction to obtain the family T 1.
n+
then we may use the following graphFor each T c T we will adjoin an
n
_
•
•
(n+l)-st vertex of maximal outdegree (i.e., of outdegree n-l), while at the same
time maintaining all edge orientations of T, thus obtaining a tournament on
(n+l) vertices.
This vertex may be adjoined at either 2 or 3 different positions
(depending on whether T E A or T E
n
nonisomorphic.
Bn ) with the resulting tournaments being
The resulting tournaments will, of necessity, be in
Tn+ 1.
Conversely, we will show that every tournament in Tn+ 1 can be gotten by using
this construction.
tournaments
It will follow that T + is precisely the family of all
n l
obtained by applying this construction to T .
n
The graphical construction is as follows:
1.
Suppose TEA .
circuit
n
Let a be the vertex of maximal outdegree and the Hamiltonian
H be as shown.
e-
-3-
a
•
•
(i)
Adjoin to T an (n+l)-st vertex
only incoming edge is from boo
their same orientation.)
~
of maximal outdegree n-l and whose
(All other edges of T are retained with
We thus obtain a tournament T' on n+l vertices
containing
•
We wish to show that T'
E
Tn+ 1.
Obviously T' has a Hamiltonian circuit;
namely, the path (of H) from a to b O' along with the edges b ~ v
O
Call this Hami I tonian circuit H'.
Furthermore, H' will be the only Hamiltonian circuit of T'.
Hamiltonian circuit of T' must contain the edges b
only one incoming edge, and
-e
from b
O
~
O
.+
v
~
~
a.
Indeed any
a since v has
has only two incoming edges, one of which is
Thus any Hamiltonian circuit ofT' must look like
-4v
•
•
But then the path from
~
to b
O
along with the edge b
O
+~
would constitute
a Hamiltonian circuit of T, and by assumption there is only one such circuit,
namely H.
hence T'
Thus the circuit H' is the only Hami 1tonian circuit of T', and
E
Tn+ I .
Note that T'
namelr
E ~+l
since it has only one vertex of maximal outdegree,
~.
We will denote this particular constructive technique (applied only to
members of An) as C , and we write
I
.
(ii)
Similarly, if we adjoin to T an
(n+l)-st vertex v of maximal out-
degree and whose only incoming edge is from a
e·
-5-
then we will obtain another tournament T' in T
,
n+l
T'
E
Note that in this case
B
n+l since a and v both have maximal outdegree n-l.
•
We denote this constructive technique (applied only to members of
An ) as
C and write
2
II.
"
Suppose T
B . Let aD, a l be the vertices of maximal outdegree and the
n
Hamiltonian circuit as shown.
(i)
E
Adjoin an (n+l)-st vertex v of maximal outdegree whose only incoming
edge is from boo
We obtain a tournament T' on n+l vertices containing
Again the fact that T' will have one and only one Hamiltonian circuit
follows from the fact that the same property holds for T.
the only vertex of maximal degree, then TEA n+ I'
Since v will be
We call this procedure
-6-
(il1
Adjoin a vertex v of maximal outdegree n-l whose only incoming edge
is from a.
The resulting tournament T' on n+l vertices contains
o
v
t
As before, T' has one and only one Hamiltonian circuit.
T'
E
B + since v and a have maximal outdegree.
n l
O
Further,
This technique we call
C and we write
4
in this case.
(iii)
Adjoin a vertex
from a .
l
~
of maximal outdegree whose only incoming edge is
This tournament T' contains
v
Again, T'
outdegree.
E
T + and in fact, T'
n l
E
B + since v and a l both have maximal
n l
e-
-7-
This technique we denote as C and thus
5
•
T'
= C5 (T)
E Bn+ I for every T E Bn .
•
This completes the graphical construction procedure.
Notation:
In the following, we let
C.(T)
1
n
= {C.1 (T):
TEA }
for i
= 1,2
Bn }
for i
= 3,4,5
n
C.1 (Tn ) = {C.1 (T):
T
E
and
5
C = U C.1 (Tn ) .
(1)
i=l
In other words C is the family of all tournaments which can be constructed from
members of T .
n
3.
Distinctness of Elements of C.
We wish to show that all members of Care nonisomorphic.
Specifically, we
will show that each element of Tn+ 1 can be constructed from one and only one
member of T , and this must be done using a uniquely determined procedure C..
n
1
Since each member of C is in Tn+ l' and no two members of Tn+ 1 are isomorphic,
this will mean that no two different members of C(i.e., constructed by different
procedures C. or from different members of T ) can be isomorphic.
n
1
LEMMA 1:
Let T E Tn+ l '
Then there exists a unique SET n and a uniquely
determined procedure C.1 such that
-8-
= C.1 (S)
T
PROOF:
.
If T E A + , then T has only one vertex
n l
~
of maximal outdegree, so
must have been the vertex added in the constructive process.
~
Therefore, the
member of T from which T was constructed must have been the sub graph S of T
n
gotten by deleting the vertex v.
must have been C .
l
If SEA , then the constructive procedure
n
If S E B , it must have been C .
n
3
If T E B + then let v ,v be the vertices of maximal outdegree. By
l 2
n l
Theorem 1 (ii), vI and v must be adjacent so, without loss of generality,
2
assume that the edge v +v is in T.
l 2
From the description of the construction
(and specifically procedures C , C4 , C ) we see that v must have been the
2
s
2
vertex which was added.
Thus, again, the member of T from which T was conn
structed must have been the subgraph S of T gotten by deleting v
SEAn then the constructive procedure must have been C2 .
vertices of maximal outdegree vI' v
3
PROOF:
v3~vl
If
B and its
n
then the procedure had to
The members of Care nonisomorphic.
Follows from lemma 1, since each member of C is also a member of
0
Tn+ 1.
4.
E
.
are joined by the edge v +v then the
l 3
procedure had to be C4 ; if they are joined by
THEOREM 2:
If S
2
Equivalence of C and Tn+ 1.
Each object we construct from Tn is an element of Tn+ 1.
structed objects are nonisomorphic, then
C -c Tn+ 1.
Since the con-
•
,
-9-
We wish to show now that each member of Tn+ 1 may be gotten by applying the construction to some member of T .
n
•
THEOREM 3:
T
c C
n+l
PROOF:
Let T
E
Tn+ I .
Case I:
If T
E
B + with vertices v ,v of maximal outdegree and Hamiltonian
n l
l 2
circuit H, then T contains the following structure
a
b
~H
Consider the subgraph S gotten by deleting vertex v
2
and all its incident edges.
Then S is a tournament with at least one Hamiltonian circuit, namely
C
=
a,vl,b, ... ,a .
If we can show that C is the only Hamiltonian circuit of S, then it follows
immediately that SET
n
and that T can be constructed from S.
Suppose S has a second Hamiltonian circuit Co different from C.
b
-e
w
-10-
Then, for some
vertex~,
w
~
VI' in S we must have the edge VI
let Co be the circuit in T which is obtained by
Co by the two edges vI+vZ+w.
directed towards w.)
w in CO.
replacing the edge VI +
Now
W
in
(Recall that the edge between v 2 and w must be
Further, it is clear that Co differing from C (in S)
implies that Co will differ from H (in T).
unique.
+
But this is not possible since H is
Thus our supposition that S has two Hamiltonian circuits must be false
and the desired results follow.
Case II:
Let TEA
with V the vertex of maximum outdegree and H the unique
n+l
Hamiltonian circuit.
V
~H
Consider the subgraph S gotten by deleting
We must show that SET.
n
~
from T.
Again S is a tournament.
Observe that if we knew that the edge
~b
were in T
then we would obviously have a Hamiltonian circuit in S; namely, the circuit
gotten by replacing the edges a+V+b in H by the edge a+b.
In fact, we will prove that the edge a+b must be in T.
Suppose otherwise, i.e.,
suppose that b+a is an edge of T.
v
e-
-11-
Let a
be the first vertex preceding ~ (in the Hamiltonian circuit H) for which
O
the edge between a O and b is directed towards b. Thus if aO+alis an edge in H
then the edge between a
l
and
~
must be directed towards a
edge of H then the edge between
maximal outdegree.
~
and b
O
l
.
is directed toward b
Also if b+b
O
O
since v has
is an
Therefore the circuit
a o +b+a I+" . +a+v+b 0 + . . . +a 0
(where the dots indicate we follow the circuit H) will be another Hamiltonian
circuit in T, different from H.
This is not possible, so we conclude that the
edge b + a cannot be in T.
Thus, S has at least one Hamiltonian circuit:
C:
a + b + b + ... + a .
O
Suppose S has a second Hamiltonian circuit CO' different from C.
v
...-.....--f---IIl. b
Let a + b
l
be the edge of Co which is directed away from
~
(b l may equal
~).
If, in CO' we replace a+b l by the edges a + v + b l then we will have a
Hamiltonian circuit in T which is different from H.
-e
This is impossible, so we
conclude that we cannot have a second Hamiltonian circuit in S, Le.,
-12-
SET
Thus if TEA
n+
THEOREM 4:
C
=
n
1 then Tcan be gotten by applying the construction to SET . 0
n
Tn+l .
In words, by applying the construction of section 2 to Tn , we obtain Tn+ 1 and
each member of T
n+
PROOF:
5.
1
is constructed in a uniquely determined fashion.
o
Follows from (1), Lemma 1, and Theorem 3.
Recurrence.
From the description of the construction we see that
= CI (Tn)
A
n+I
B
n+l
=
C2 (Tn )
where the unions are disjoint, and by Theorem 2
n+I = IA n+ 11 =
A
=
IA n I
= A
n
and
+
+ B
n
IC I (Tn ) I
+
IC 3 (Tn ) I
IBn I
(2)
-13-
Bn+l
= 18n+ 11 = IC 2 (Tn ) I
+
IC 4 (Tn ) I
+
ICS(Tn ) I
+ 18n I + 18n I
= An + 2Bn
= An+l + Bn
= IAn I
j
(3)
(4)
where the least equality follows by applying (2).
THEOREM 5:
The sequences {A } and {B } both satisfy the recurrence
n
n
x
PROOF:
= 3 x
n+2
(n > 4)
n+l - xn'
From (2) and (4),
-e
An+2 - An+l
= Bn+l
= An+l
= An+l
=2
+ B
n
+ (A
An +l
n+ l
- A)
n
A
n
so that
A
=3
Bn+2
= An+ 1
n+2
An+ 1
A
n
Also from (3) and (4)
-e
=
(B
=3
n+l
+ 2 Bn+ 1
- B ) + 2 Bn+l
n
Bn+l - Bn
0
-14-
THEOREM 6:
The sequence {T } which counts the number of nonisomorphic tournan
ments on n vertices, which have a unique Hamiltonian circuit, satisfies
T
PROOF:
(n > 4)
n+2
Follows from Theorem 5 since T
n
= An
+
.
B for every n.
n
o
-15-
Bibliography
.)
1.
Douglas, Robert J., "Tournaments that admit exactly one Hamiltonian
circuit", Proceedings of the London Mathematical Society, 21 (1970),
pp. 716-730.
2.
Harary, Frank, Graph Theory, Addison-Wesley, Reading, Mass., 1969.
3.
Moon, John W., Topics on Tournaments, Holt, Rinehart and Winston, New
York, 1968.
For S