AN ABSTRACT OF THE THESIS OF
Stephen Kerron Prothero
for the
(Name)
Date thesis is presented
M.A.
(Degree)
May 14,
in Mathematics
(Major)
1963
Title ON THE ENUMERATION OF CERTAIN EQUIVALENCE CLASSES OF
EULER PATHS OF FULL GRAPHS
Abstract approved
This thesis treats the problem of enumerating equivalence classes of Euler paths of full graphs.
is a complete,
edges.
unordered,
A full graph
graph with no loops or repeated
Two Euler paths are equivalent if and only if one
can be transformed into the other by
a
finite sequence of
rotations and reflections of the path and permutations of
its vertices.
We obtain the number of equivalence classes
for full graphs on
3
and
5
vertices but obtain only
partial results for full graphs on
7
vertices.
We prove
theorems which enable us to obtain representatives of all
equivalence classes with relatively few repetitions for
any full graph.
Finally we prove
a
monotoneity theorem
for the number of equivalence classes.
ON THE ENUMERATION OF CERTAIN EQUIVALENCE CLASSES
OF EULER PATHS OF FULL GRAPHS
by
STEPHEN KERRON PROTHERO
A THESIS
submitted to
OREGON STATE UNIVERSITY
in partial fulfillment of
the requirements for the
degree of
MASTER OF ARTS
June 1963
APPROVED:
Associate Professor of Department of Mathematics
In Charge of
Major
chairman of Department of Mathematics
Dean of Graduate School
Date thesis is presented
91
Typed by Carolyn Joan Whittle
//
X73
Acknowledgement
The writer wishes.to thank Dr. Robert Stalley for
his help and guidance in the preparation of this thesis.
TABLE OF CONTENTS
Page
Chapter
I.
II.
III.
IV.
V.
Introduction
1
Basic Definitions and Theorems
3
Characteristic Sequences
8
Construction Theorems and Some Applications 15
A Monotoneity Theorem
30
Index of Notation
46
Bibliography
47
ON THE ENUMERATION OF CERTAIN EQUIVALENCE CLASSES
OF EULER PATHS OF FULL GRAPHS
Chapter
I
Introduction
Graph theory has its origin in the works of Leonhard
Euler in the early part of the eighteenth century [3].
The subject has experienced
decade.
a
strong revival in the past
Not only is graph theory being recognized as
a
powerful tool in applications but the theory is being
extended at an increasing rate.
The standard references
in graph theory are those by Berge,
in the bibliography [2],
[4],
König,
and Ore found
[5].
The statement of our main problem and its treatment
are graph theoretical.
state the problem as
as follows:
a
However, we may alternatively
problem in combinatorial analysis
"Enumerate the non -equivalent cyclic
arrangements of symbols taken from
a
set of
n
distinct
symbols where each unordered pair of distinct symbols
occurs in adjacent positions once and only once.
Two
arrangements are equivalent if and only if one may be
obtained from the other by
reflections,
a
finite sequence of rotations,
and permutations of the distinct symbols."
This problem has not received attention in the
literature.
The problem has its origin in the design of
computers [6], but no such application of the results of
2
this thesis are envisioned.
Let
n
be the number of distinct symbols from which
arrangements are formed.
We prove that there exists an
arrangement if and only if
n
is odd
and
n >
enumerate the non -equivalent arrangements for
3.
We
n =
3
and
but fail to obtain the enumeration for arbitrary
n =
5,
n.
We do obtain partial results for
We develop
n = 7.
several rules of construction which give us all arrangements with at most
a
small number of repetitions from
each equivalence class.
n =
7
However, even when we take
this method is prohibitive without
a
high speed
computer partly because of the necessity of eliminating
the repetitions mentioned.
Finally, we prove
a
monoto-
neity theorem for the number of pairwise non- equivalent
arrangements.
3
Chapter
II
Basic Definitions and Theorems
We are interested in full graphs in this thesis.
full graph
is defined by the equation,
G(V)
G(V) = 1(a, b}
where
G(V)
V =
j
<
1
j
<
n },
are unordered pairs.
finite,
A
unordered,
(
aeV,
n >
a/b },
beV,
2,
and the elements of
Full graphs are known also as
complete graphs with no loops and with
non- repeated edges.
The elements of
vertices and the elements of
An Euler path on
a
G(V)
V
are called
are called edges.
full graph is
cyclic sequence
a
of edges which contains each edge of the graph once and
for which the intersection of two consecutive edges is
single vertex.
Thus an Euler path is
(2.1)
.a1,a23, a2,a33,
ai-l' ail
where
aieV,
1
<
i
number of edges of
'
ai'ai+11,
< N,
...
and where
...
,
aN,al } },
N = 1,n(n -1)
is
G(V).
A simpler notation for the Euler path in (2.1)
(2.2)
the
al a2 a3 a4
...
ai-1 ai ai+1 ..*
is
aN a1.
This arrangement is also referred to as the Euler path.
It is now evident that the graph theoretical definition
a
4
of an Euler path is equivalent to the combinatorial defini-
tion given in the introduction.
Example
V = [1,2,33,
If
1.
{[1,2 },f2,3 }, {3,1 } },
(2.2),
2 3
1
an Euler path is
simpler notation of
or in the
1.
We will refer to an Euler path on
n
vertices as simply an Euler path on
first
symbols of
N
path on
where
<
1
i
vertices.
n
The
are the corners of the Euler
(2.2)
vertices.
n
full graph with
a
The symbol
is the
ai
corner
ith
Thus several corners may denote the
< N.
same vertex.
The next step in the investigation of Euler paths is
to determine necessary and sufficient conditions on the
order of the vertex set of the full graph for the existence
of an Euler path on that graph.
The following theorems
will present these conditions.
Theorem
graph
G(V),
If
1.
there exists an Euler path on
then the order of
V
is
a
full
odd and not less
than three.
Proof.
We first prove that the order,
less than three.
That
the definition of
a
n
is
full graph.
full graph contains one edge.
one edge with
(2.1),
al
at
n,
is not
least two follows from
If
n =
2,
then the
An Euler path would contain
as both the first and last vertex by
whereas by definition the vertices of an edge are
distinct.
5
That
is odd also follows from
n
the Euler path as defined in
that the vertex
Also,
(2.1).
a
consideration of
We have already noted
occurs in the first and last edges.
a1
if a vertex occurs in any other edge it must occur
either the preceding edge or the following edge.
since
st edge,
then
(2.1).
edges containing vertex
Thus
n -1.
N -1
must occur an even number of times in
al
the remaining edges of
is
Hence
does not occur in the second edge or the
a1
in
n -1
a1
of
However, the number of
full graph on
a
n
vertices
must be even which implies that
n
must be odd and the theorem is proved.
Theorem
graph
G(V)
2.
If the order of the vertex set
is odd and not
exists an Euler path on
of a full
less than three, then there
G(V).
The proof which will be given here is different from
known proofs which may be found in any standard reference
in graph theory and in some works of topology [1].
This
proof is useful in the construction of Euler paths for
study in later chapters.
The proof is by induction on the
order of the vertex set.
Proof.
Example
An Euler path on three vertices is given by
We assume that there is an Euler path on
1.
vertices denoted by (2.2) where
path on
n +
(2.3)
b
where
1
2
al = 1.
Then an Euler
vertices is given by
b2 b3
3 ...
2
bl = a1 = 1,
bi
...
bN bN
n
bN
bl'
6
(2<i<N),
bi = ai
k
bN+2k-1 =
(1<k<n),
n+l
bN+4k-2 =
-(
1<k<%( n+1)
)
,
n+2 (1<k<%(n-1)),
bN+4k =
bN+2n+1 = n+2.
Clearly consecutive edges contain
common vertex.
a
We need to show that each edge occurs once and only once.
We will begin by showing that each edge appears at least
The edges may be divided into four classes.
once.
Class
1
h
<
<
form
<
j
where
form
{b
2.
2 < h
n+l,h}
1
<
i
< n.
and
[n +1,h}
is odd they are
Class
3.
since
< N,
[h,n +l}
[n +2,h }.
and
and
Consider the edges
If
b
112.,=+
+2h- l'bN +2zfi }
h,n +2}
n +2,h}
and
[n +2,1}
The edge
is
h
is even
'
respectively, whereas
[h,n +l}
Edges of the forms
respectively.
and the edge
h,n +2}
and
These edges may also appear in the
{l,n +l}
bN+2n +l
,
b1}.
if
respectively,
{1,n+l}
These edges may also appear in the form
n+2,1}
b1 b2 ...
vertices.
n
Edges of the forms
N +2h -2' b N +2h -1
they are
h
where
forms an Euler path on
Class
where
All edges of this class are found in the
n.
bi,bi +l)
bN bN
h, j}
Edges of the form
1.
1,n+2 }.
and
{n +l,l}
is
and
{b N+1
,
b
N+ 2 }
7
Class
bN+2n'
n +l,n +21.
The edge
4.
This edge is
bN +2n +1 }'
Finally,
the number of edges of
(2.3)
N +2n +l.
is
Since
N +2n +l = %n(n -1)
=
%(n2
is the order of
+
G(V)
2n +
+
1
=
ß(n2
-
n + 4n +
2)
3n + 2) = %(n ±2)(n +1)
for
a
set
edge can occur more than once.
V
of order
n +2,
no
Thus each edge occurs
exactly once and the proof is complete.
Let
n
k
be the number of corners of an Euler path on
vertices which denote
N = nk
and
k =
1(n -1).
a
specific vertex.
We find that
8
Chapter
III
Characteristic Sequences
In this chapter we define equivalent Euler paths and
devise
a
method of determining whether or not two given
Euler paths are equivalent under this definition.
Two Euler paths
and
P
on the same full graph
P'
are defined to be equivalent if and only if
transformed into
by
P'
can be
P
finite sequence of rotations
a
and reflections of the path and permutations of the
vertices of
Here the terms rotation and reflection
V.
More precisely, two
are taken in the geometric sense.
Euler paths,
P =
bl b2 b3 b4
al
a2
bN bl,
and
aN al,
a3 a4
of the
- bi
a1,a2'a3'a4,
on the set
=
same full graph are defined
to be equivalent if and only if there exists
ai
P'
...,a
permutation
a
which is an
N
element of the permutation group generated by elements of
Vertex Permutation, that is
the following three types:
permutations for which corners which denote the same vertex
have images which are corners which denote the same vertex;
Rotations, that is permutations of the type,
1
<
i
< N;
b1 = a1
Reflections, that is permutations of the type,
and
bi = aN+2 -i
for an arbitrary integer
j
bi = ai +k for
r t (mod N)
and
1
<
for
t
j
<
by
N.
2 < i < N.
at =
aj
We define
where
at
9
The above definition defines an equivalence relation
on the set of all Euler paths on
Each
given full graph.
a
equivalence class of Euler paths contains paths which are
sequences of rotations, reflections,
of a given path.
Our problem is that of enumeration of
equivalence classes for each odd
three.
and vertex permutations
which is not less than
n
This statement of our problem agrees with the
statement given in the introduction.
We wish to devise
a
practical way of determining
whether or not two given Euler paths belong to the same
equivalence class.
For this purpose we define the right
representative sequence.
al a2 a3
For
...
a
ai ai
given Euler path,
+l
the right representative sequence,
...
aN al,
or
RRS,
r1 r2 r3 ... ri ri
where
ri
...
rN,
corresponds to the corner
(3.1)
ri = min
The element,
ri,
k
I
ai +k =
of this sequence
is
ai
ai,
is
and
k >
13.
called the right
characteristic number associated with the corner
Since each vertex is denoted by
if
n
> 3,
k =
1/2(n
-1)
a..
i
corners, then
each vertex is denoted by more than one corner
and will have more than one right characteristic number
corresponding to it.
A vertex permutation may be considered to be
relabelling of the elements of
V.
Thus, the
RRS
a
is
10
invariant under
vertex permutation since it is dependent
a
upon the order of the vertices and not upon the symbols
used to represent those vertices.
We always perform
a
vertex permutation upon each Euler path we are considering
in order to bring it to
a
denoting the vertex
precedes the first corner denoting
the vertex
if
j
If the
i
1
<
i
form such that the first corner
<
j
< n.
Euler path is subjected to
effect upon the
RRS
is
rotation then the
a
to operate upon it with the
permutation,
(3.2)
ri - ri +k.
We define
(mod N)
E t
j
rt
and
for any integer
1< j<
t
by
rt = rj
where
N.
For the purpose of characterizing Euler paths to
within
reflection, we also always perform
a
each Euler path to transform it to
a
form
a
al
rotation upon
aN al,
a2
such that the corresponding number,
N
r
NN-i
,
is maximal.
i=1
The right representative sequence of an Euler path
which has been transformed in the aforementioned way is
called the right characteristic sequence,
RCS
is
a
or
RCS.
The
characterization of an equivalence class
corresponding to an equivalence relation where two Euler
paths
P
and
P'
are equivalent if and only if
P
can
11
be transformed into
by
P'
and vertex permutations
finite sequence of rotations
a
as defined above.
If the Euler path is subjected
to
a
reflection, the
effect upon the right representative sequence is to operate
upon it with the permutation,
(3.3)
ri
s.
-.
.
i
1
Upon examination of this result we find that the sequence
of
may be defined in another way which is equivalent
s's
to the above.
For
given Euler path,
a
al a2 a3
...
ai
ai +l
...
aN al,
let the left representative sequence, or
si+l
-
corresponds to the corner
ai
s1
where
si
(3.4)
si = min
We define
j
E
a.
...
k
I
and
1< j<
upon comparing
= ai +r
si
then
,
1
ai =
=
ai-k
for any integer
st
(mod N)
t
Thus,
if
s2
t
LRS
be,
sN'
and where
a.,
k>13.
by
st =
sj
where
N.
with
(3.1)
a.
J
..
(3.4)
we see that
Moreover, we find that
-sj
the only difference between the right and left represen-
tative sequences is the sense of direction in which we
must count from
the same vertex.
a
corner to find the next corner denoting
Thus,
the effect of
a
reflection on an
Euler path is to interchange its right representative
sequence with its left representative sequence.
This result
12
will be used extensively in Chapter
The Euler path to which
a
V.
given left representative
sequence corresponds is sometimes transformed by
a
rotation
such that the number,
N
s.
Ni-1
i=1
is maximal.
The left representative sequence of an Euler path
which has been transformed in the aforementioned way is
called the left characteristic sequence,
or
When
LCS.
working with the left characteristic sequence but not
directly with the Euler path to which it corresponds,
will write the
LCS
in the following
...
sN
sN -1
sN -2
we
form:
...
sN -i
s2
2
By doing this we sacrifice the correspondence between
the left characteristic numbers and the corners of the
Euler path to which they correspond for the purpose of
sequences more easily.
being able to compare characteristic
Now to each Euler path correspond two sequences, the
right and left characteristic sequences,
and these
sequences
completely characterize an equivalence class of Euler paths
on
a
full graph.
Since the effect of
Euler path upon its
such
a
RRS
and
LRS
Thus the
RCS
of
rotation of an
is to interchange them
rotation also interchanges the
the path.
a
an Euler
RCS
and
LCS
of
path is identical to
13
the
LCS
Two Euler paths are equivalent if and only if the
once.
RCS
of that path after that path has been reflected
and
LCS
of
one are identical to the
LCS
Example
given by
of one equals the
RCS
are equivalent if and only if the
or
LCS
and
More simply, two Euler paths
of the other in either order.
RCS
RCS
of the other.
2.
(2.3)
three vertices.
If
n =
the following Euler path is
5,
using Example
as
1
an Euler path on
RRS
We also give the
LRS
and
for this
path.
14
Euler path:
1
RRS:
3 4 5 7 4 6
LRS:
7
2 3
6
5 3
2 5 3 4 5 1,
6 4
3 5 6 7,
7
5 4 3.
The right and left characteristic sequences of the Euler
path are,
RCS:
7 4 6 3
LCS:
7 4 6
5
6
3 5 6
3 4
5,
7 3 4
5.
7
The above procedure of writing the
for
a
RRS
and
LRS
path by writing their characteristic numbers directly
below the corners with which they are associated is used
throughout this thesis.
is identical to the
property of the
RCS
LCS.
and
Note that in Example
2
This is found later to be
LCS
RCS
the
a
of any Euler path on five
14
vertices.
LCS
It is not,
however,
of Euler paths in general.
a
property of the
RCS
and
15
Chapter
IV
Construction Theorems and Some Applications
The purpose of this chapter is first to developtheorems
which aid in the construction of the Euler paths which represent the equivalence classes described in Chapter
III.
These theorems are then to be used to construct representative Euler paths from each equivalence class for
to
three
We also prove
five.
and
equal
n
theorem which will
a
give some information about the number of equivalence
classes of Euler paths on seven vertices.
To construct the Euler paths we assign
the vertex set as
path,
value to each of the
a
vertex from
a
corners of the
N
We use
subject to the definition of an Euler path.
described in the preceding chapter,
the convention,
constructing Euler paths
such that the
aN al
al a2
of
N
number,
such that the first
is maximal and
ri NN -1,
i =1
corner denoting the vertex
i
denoting the vertex
1
j
if
precedes the first forner
<
i
<
j
<
n.
Let the segment associated with the corner
denotes the vertex
S(i,h) =
fai
1
<
which
be defined as the set
h
j
ai
<
s,
ai
/ ai
if
1
<
j
<
s,
as =
113.
The segment associated with the first corner of an Euler
path constructed according to the convention mentioned
16
above is called the principal segment of the Euler path.
segment associated with
corner is its
The length of
a
order.
the length of the segment associated with
a.
is
by
L.
Thus,
a
We denote the length of the principal segment
.
Clearly, the length of the principal segment is
greater than or equal to the lengths of the segments
associated with each of the other
N -1
the Euler path on three vertices,
1
2,3,1},
the principal segment,
Lemma
corners.
2 3
1,
Thus, in
the length of
is three.
The sum of the right characteristic numbers
1.
associated with the corners denoting
specific vertex is
a
N.
Proof.
i
be the empty set
S(i,h)
Let
does not denote the vertex
just,
U
S(i,h)
if the corner
Thus the Euler path is
h.
for any fixed
h.
Let
[A]
be the
1 <i <N
order of the set
associated with
A.
a
The sum of the characteristic numbers
specific vertex is just
N
[S(i'h)] =
I
U
This completes the proof of Lemma
Euler path on
Proof.
= N.
1<i<N
i=1
Theorem
S(i,h)
3.
n
If
n
>
5,
1.
the principal segment of an
vertices has length greater than
First we prove that
Consider the vertex
1.
L >
n.
Assume
This vertex is denoted by
n.
L < n.
1/2(n
-1)
17
corners in an Euler path on
Since
vertices.
n
the
is
L
length of the principal segment, the right characteristic
numbers of the other
vertex
W,
1/2(n
-1)
-
corners denoting the
1
must be less than or equal to
1
Thus the sum,
L.
of the characteristic numbers associated with corners
denoting the vertex
contrary to Lemma
1
Now we prove
W
<
%L(n -1)
which states that
W = N.
1
L >
satisfies
Assume
n.
L =
< 1/2n(n -1)
By Lemma
n.
1
=N
the
sum of the characteristic numbers associated with corners
denoting
a
specific vertex must equal
denoted by
corners throughout the Euler path.
1/2n(n -1)
All characteristic numbers of the
or equal to
less than
Each vertex is
N.
must be less than
RCS
If any of the characteristic numbers were
n.
the sum of the characteristic numbers of the
n,
corners denoting some vertex would be less than
all characteristic numbers must equal
consists only of
edge
an +l,an
since
n
Theorem
}.
If
n(n -1),
< N =
3
n's
n >
5
RCS
But if the
n.
a1,a2}
then the edge
Thus
N.
equals the
these edges are distinct
and we have
a
contradiction;
so
is proved.
Theorem
4.
Every vertex of an Euler path is denoted
by some corner in the principal segment.
Proof.
Assume the vertex
i
corner in the principal segment.
is not denoted by
Recall that the length
of the principal segment is denoted by
corner denoting the vertex
i
a
be corner
Let the first
L.
m
where
m > L.
18
Assume that the
is corner
Certainly
p.
corner denoting the vertex
1/2(n-1)st
p
is
less than or equal to
Consider the segment associated with corner
p.
i
N.
This
segment will contain all corners of the principal segment.
It will also contain corner one.
segment is at least
of the principal
Theorem
L+1
Thus the length of this
which contradicts the definition
segment and Theorem
4
is proved.
There is exactly one equivalence class
5.
of Euler paths on three vertices.
Proof.
Since each vertex is denoted by
corners in an Euler path on
denoted by
1/2(3-1)
vertices.
=
corner in an Euler path on three
1
The first corner denoting the vertex
precedes the corner denoting the vertex
1
<
i
<
denoting
each vertex is
vertices,
n
1/2(n-1)
<
j
a
3
in
j
i
if
Hence since every corner
this case.
vertex in an Euler path on three vertices is the
first corner denoting that vertex,
given by our construction is
1
2
the only Euler path
3
and the theorem
1,
is proved.
Let us examine the geometric significance of the results
of Theorem
5.
The full graph on three vertices may be
'
represented by,
5.
Z
The Euler path which we have found as
a
representative of
all Euler paths on three vertices may be represented by,
19
T
L
z.
In the above geometric interpretations we represent vertices
by points,
edges by line segments, and Euler paths by cyclic
sequences of directed line segments.
Theorem
where
n >
Suppose an Euler path on
6.
vertices,
n
is operated upon by a rotation permutation
5,
RRS
such that its
is its
Suppose further that
RCS.
a
vertex permutation is performed so that the first corner
denoting the vertex
the vertex
if
j
precedes the first corner denoting
i
1
<
i
<
j
corners denote the vertices
Proof.
By Theorem
3
<
Then the first four
n,
1,
2,
3,
and 4
that order.
ifi
the principal segment must be
of length greater than five.
Thus the vertex
which
1
occurs in the first corner cannot occur in corners two,
three,
or four.
The only
Consider the second corner.
possible vertex for the second corner to denote is
vertex,
the vertex
the vertex
vertex
2
2.
a
The third corner can denote either
or a new vertex,
namely
But if the
3.
is denoted by corner three we would have
2
occurring in the Euler path and this is not an edge.
the third corner must denote the vertex
corner may denote any of the vertices
vertex
because
4.
The vertex
3,33
is not
new
3
3.
2,
3,
[2,23
Thus,
The fourth
or the new
cannot be denoted by corner four
an edge.
The vertex
by corner four would cause the edge
2,3}
2
if denoted
to be repeated
20
Thus, the
which violates the definition of an Euler path.
only possible vertex which corner four can denote is the new
vertex
and the theorem is proved.
4,
A loop is
both of which denote the same vertex.
not an edge,
(2.2),
pair of consecutive corners of
a
easily seen that
it is
therefore an Euler path of
Because
loop is
full graph, and
a
cannot contain
full graph,
a
a
a
loop.
Theorem
The principal segment of an Euler path
7.
on five vertices cannot be of length six.
Proof.
Assume there is such an Euler path.
We
construct the representative Euler path of the equivalence
By Theorem
class to which it belongs.
corners must denote the vertices
1
by
a
Thus the vertex
5.
the first four
Corner five
2 3 4.
can possibly denote only two vertices,
the new vertex
6
the vertex
1
or
2
cannot be denoted
corner preceding the seventh corner whereas the
vertices
4
and
would give rise to
3
repeated edge, respectively.
The vertex
a
loop or
2
a
is impossible
also because the length of the segment associated with
corner five would then be seven,
a
number greater than
the length of the principal segment which is assumed to be
six.
This follows from Lemma
vertex is denoted by
on five vertices.
1
2 3 4 5 X X X X X
1
%(5 -1) = 2
and the fact that each
corners in an Euler path
The partial Euler path is now
1.
By
a
partial path we mean
a
path
21
partially determined.
undetermined value.
to have length six,
Each
X
denotes
corner with an
a
Since the principal segment is assumed
the seventh corner of the Euler path
must denote the vertex
1.
Leaving the sixth corner
undetermined, the partial path now becomes
1
2 3 4 5 X
X X X
1
1.
We now consider the possible
vertices which can be denoted by corner six.
2,
4,
and
Vertices
1,
are impossible because they would give rise
5
to loops or repeated edges.
The vertex
is impossible
3
also because the length of the segment associated with the
sixth corner would then be seven which would contradict
the assumption that the principal segment is of length six.
Thus there is no vertex which satisfies the conditions
required for
a
vertex to be denoted by corner six.
The
theorem is now proved.
Similar, though more complex, proofs can be devised
to show that the length of the principal segment of an
Euler path is greater than
n +
1
if
is seven or nine.
n
The lack of an inductive scheme keeps us from obtaining
this result for more general values of
Theorem
8.
n.
The number of equivalence classes of Euler
paths on five vertices is three.
Proof.
We use our construction procedure to construct
all representatives of equivalence classes of Euler paths
on five vertices.
We will then consider the
RCS
and
LCS
22
of each of these representative paths to
see if any are
Those Euler paths which remain will then be
equivalent.
the representatives of the various equivalence classes of
Euler paths on five vertices.
By Theorem
the first four corners of all repres-
6,
entative Euler paths must denote the vertices
The eighth corner denotes the vertex
and
7,
and we would have
1
the new vertex
edge or loop.
2 3 4
1
for otherwise we would have
5,
X
2 X X 1 X
denote the vertex
we now have
1
and
A=
A
1
is
2 3 4
because vertices
loop,
a
1
A.
Corner six of
2 3 4 5 X X 1 X X 1.
A
must
Corner seven
the vertex
and
2
5
so
3
or the
would give rise
respectively.
Thus partial
replaced by the two partial paths,
2 5 3
1
X X
1
and
Consider the partial path
determined
B=
2 3 4 2 5 X 1 X X 1.
1
repeated edge and
a
path
C =
4,
repeated
to avoid loops or repeated edges,
5
denotes either of two vertices,
vertex
a
or
2
Thus we have two partial paths,
Consider the partial path
to
repeated edge or
a
The fifth corner denotes either the vertex
loop.
3
for otherwise the ninth or tenth corner would
denote the vertex
A =
Theorems
by
1
2 3 4.
1
D=
1
2 3 4 2 5 4
RRS =
1
2 3 4 2 5 4
7 3
3 7
X X 1.
with its partially
D
RRS,
D=
1
1
7 3
X X
1
23
Note that for
RRS
(that is, for
RCS,
to be the
N
r.
the characteristic number for
to be maximal)
NN -1
This requires that the vertex
corner six must be three.
denoted by corner nine is the vertex
which in turn
5,
requires that corner ten denotes the vertex
Thus we have as an equivalence
loops and repeated edges.
class representative and corresponding
D=
1
2 3 4
RCS =
7
3
(4.1)
7 3
2 5 4
7 3
RCS,
1
5
3
1,
7 3
7
3
.
Now let us consider the partial path
its partial
to avoid
3
together with
C
RRS,
C=
1
2 3 4
2
5 3
RRS =
7
34
7
6
1
X X 1,
Note that the right characteristic number for corner six
is at least three
and the triple of right characteristic
numbers for corners five,
six,
and
seven is at least
7
The triple of characteristic numbers for corners one,
and
three is
7
3
4.
3 6.
two,
This fact contradicts the assumption
that the right representative sequence of the path we are
constructing is to be the
RCS
Thus
of the path.
C
is
not the representative of an equivalence class.
Next to be considered is the partial path
B =
1
2
3
4 5 X X
1
X X 1.
The vertices
4
and
5
are
not denoted by corner six otherwise we would have arepeated
edge or
a
loop.
Since the vertices 2 and
3
are acceptible,
24
partial path
E =
is replaced by two partial paths,
B
123452X1XX1
Path
F
F=
and
123453X1XX1.
is not the representative of an equivalence class
since if corner seven denotes any vertex, we have
seven of partial path
E=
must denote the vertex
E
or
3
Vertices
5.
and
1,
2,
So this
would give loops or repeated edges.
4
4
Corner nine may
2 3 4 5 2 4 1 X X 1.
1
denote either of the vertices
and
repeated
To avoid repeated edges and loops corner
edge or loop.
so we have
a
partial path is replaced by the two partial paths,
G =
1
2 3 4
5 2 4 1
3
X
1
H=
and
1234 5241
5 X
1.
Each of these partial paths may be continued by letting
corner ten denote the only vertex which will give no loops
and we obtain the partial paths
or repeated edges,
G =
1
2 3 4
5 2 4
1
and
3 5
H=
1
2 3 4 5 2 4 1
5 3.
Thus we have as equivalence class representatives and their
corresponding
RCS's,
G=
(4.2)
RCS =
H=
(4.3)
RCS =
Thus we have
classes.
1
2 3 4 5 2 4 1 3 5 1,
7 4
1
6
3 5
6
7 3 4
5
.
2 3 4 5 2 4 1 5 3 1,
7 4 7
(4.1),
3 4
6
(4.2),
7
3 6 3
and
We must now examine their
.
(4.3) as equivalence
RCS's
and
LCS's
to
see if some pair represents the same equivalence class.
The following is
a
list of the
the three Euler paths:
RCS
and
LCS
of each of
25
D
-
GH-
RCS =
3 7
3 7 3
6
3 5
6 7
7 4 7
3 4
6
7
3 7
RCS = 7 4
RCS =
7 3
3 4
7 3
6
and LCS = 7 3 7 3 7 3 7 3 7 3.
5
and LCS = 7 4
3 5
6
7 3 4
5.
3
and LCS = 7 4 7 3 4
6
7 3
3.
Thus we find that the three Euler paths
D,
6
G,
and
6
H
must represent different equivalence classes of Euler paths
on five vertices.
The three equivalence classes represented
by these three Euler paths are all that are possible on five
vertices and the theorem is proved.
Let us examine the geometric significance of this
result.
As for the case
n =
3,
points, edges by line segments,
we represent vertices by
and Euler paths by cyclic
sequences of directed line segments.
vertices may be represented by
The full graph on five
26
The three Euler paths which we have found as representatives
of each of the three equivalence classes of Euler paths on
five vertices,
path,
each followed by
a
representation of that
are as follows,
Equivalence Class One, path
D =
1
2 3 4
2 5 4 1
5
3 1:
1
2 3 4
5 2 4
3
5 1:
3
4
Equivalence Class Two,
path
1
G =
1
27
Equivalence Class Three, path
H =
1
2
3 4
ß-
2 4 1
5
5
3
1:
.Z
.3
4
This procedure of constructing all Euler paths which
are representatives
of
n
of equivalence classes for
becomes prohibitive
than or equal to seven.
n = 7
for values of
n
a
wlich
given value
are greater
The number of representatives for
which would be found using the theorems of this
chapter is estimated to be of the order of 10,000.
estimate was made on the basis of the results of
enumeration of the representatives.
a
This
partial
The following gives
a
lower bound on the number of equivalence classes of Euler
paths on seven vertices.
Theorem
9.
The number of equivalence classes of Euler
paths on seven vertices is greater than or equal to three.
Proof.
To prove this theorem we will display three
non -equivalent Euler paths on seven vertices.
The three
Euler paths we display are generated by using
(2.3)
and
the representative Euler paths of each of the three equiva-
lence classes of Euler paths on five vertices as found above.
The Euler path on seven vertices obtained by using
(2.3)
and path
for
G
D
as
and
H.
the Euler path on five vertices will be called
(RRS
and
Path
RRS:
LRS:
Path
G7:
RRS:
LRS:
Path
RRS:
LRS:
H7:
similarly
The three Euler paths on seven vertices, together with their
respective right and left representative sequence
D7:
and
D7
4
are,
LRS)
1
2
3
4
2
5
7
3
6
4
7
5
7
3
7
3
8
3(10) 3(10) 5(11) 4(10) 4
9
4
8
3
8(13)(14),
(11)(10)9
8
3
8
3
7
3
7
3(13)
8(14)
5
4(10) 4(10) 4
4
1
3
5
1
2
7
3
6
4
7
5
4
9
4
8
3
7(13)(14),
7(14)
6
4(10) 4
9
4
3.
2
6
7
1
5
3
1
6
2
1
2
3
4
5
2
7
4
6
3
5
7(10)
3
6
9(11) 4(10)
(11)(10)9
8
7
4
3
7
6
5
3(13)
4
1
5
3
1
6
6
6
7
1,
3.
7
1,
1
2
3
4
5
2
7
3
6
4
7
5
7
4
7
3
4
7(10) 3(10) 5(11) 4(10) 4
9
4
8
3
7(13)(14),
(11)(10)9
8
7
4
5
4(10) 4(10) 4
3
7
4
7
6
3(13) 7(14)
3.
1,
Now if we find that the
the
RCS
or
RCS
of each of
these three paths does not equal
of any other of these three paths,
LCS
say that these three
we can
paths are pairwise non -equivalent and therefore that there are at least three
equivalence classes of Euler paths on seven vertices.
The six characteristic sequences are
RCS of
D7:
(14)
7
LCS of
D7:
(14)
8(13)
RCS of
G7:
(14)
7
LCS of
GIs
(14)
7(13)
RCS of
H7:
(14)
7
LCS of
H7:
(14) 7(13)
3
4
4
7 3
8 3
3 7
3 7
6
3
5 7
3
5
6
7
7
3 4
7
3
7 4
7
(10)
3
(10)
3
(10)
3
3
(10)
5
(11)
8
3
8
3
6
9
4
7
8
9
5
(11)
3
4
(10)
7
8
9
(11)
9
4
(10)
4 9
(10)(11)
3 4
4 (10)
4 9
(10)(11)
4
(10)
(10)(11)
Since these six characteristic sequences are distinct,
4
8
3
8
4
(10)4
4
8
3 7
3 4
9
4
(10)4
4 9
4
8
3 7
4
(10)4
3
4
(10)
(10)
(13),
5.
(13),
6.
(13),
5.
there exist at least
three equivalence classes of Euler paths on seven vertices and the theorem is proved.
30
V
Chapter
A Monotoneity Theorem
The purpose of this chapter is to establish the
following general theorem.
Theorem
10.
The number of equivalence classes of
Euler paths on full graphs on
an arbitrary odd integer not
vertices, where
n
less than
ically increasing function of
Preliminary part of the proof.
is a monoton-
3,
where
k,
k =
1(n -1).
be the
r(n)
Let
number of equivalence classes of Euler paths on
vertices.
r(n +2)
n
n =
and
3
n
We must prove
(5.1)
where
is
n
is odd and
for
n = 5
Hence we assume that
n >
3.
>
r(n)
The validity of (5.1)
follows from Theorems
n >
7.
constructed from an Euler path on
n
and 9.
8,
5,
The proof of Theorem 2 is
constructive where an Euler path on
prove that Euler paths on
for
n +
n
2
vertices is
We will
vertices.
vertices which are from
different equivalence classes yield Euler paths on
n +
2
vertices which are from different equivalence
classes under this construction.
Our proof will be indirect.
We will assume that two Euler paths on
constructed from non -equivalent
n +
2
Euler paths on
vertices
n
vertices
are equivalent and will then show that this assumption
31
leads to
Chapter
a
We will use the notation of
contradiction.
2.
As the representative of
Euler paths on
given equivalence class of
a
vertices we pick the path whose right
n
representative sequence is identical to its
which the first corner denoting the vertex
first corner denoting the vertex
By Theorem
the path begins with
6
if
j
1
<
1
2 3 4
...}
(n -2)
5
(n +l)
(5.2)
is
or
(n +l)
4
(n -1)
(n+2)
n
(n +l)
(n +2)
each are denoted by
where
k =
%(n -1).
where
2 <
i
I
-part of
of the
n
(5.2).
I -part
of
the first corner.
twice.
(n +2)
1.
vertices from
n
This
The remaining
(5.2).
form what is called the induction part
Note that the vertices
<
vertices:
2
constructed are enclosed in braces.
(5.2)
I -part.
n +
3
is called the path part or P -part of
corners of
n.
Hence, our
(n +2)
All but the last corner of the path on
which
<
j
2
(n +l)
1
<
i
2 3 4.
construction yields the following path on
(5.2)
precedes the
i
1
and for
RCS
k +
1
n +
and
1
corners of the
n +
I -part
2
(5.2),
of
Note also that each of the vertices
i,
is denoted by exactly one corner of the
The vertex
(5.2),
1
is denoted by two corners
namely the
N +
1
st corner
and
The first corner of an Euler path appears
The second appearance here is needed to indicate
that the edge
(n +2),
1}
is contained in the Euler path
since these vertices are denoted by the
N + 2n +
corner and the first corner, respectively.
1
For the
st
32
remainder of this chapter we will assume that the I -part
contains only one corner denoting the vertex
st
corner of
a
N +
1
Hereafter we will enclose the path
(5.2).
part of such
the
1,
A path constructed in this
path in braces.
way is called an induction path and is denoted
Pn +2'
The right representative sequence of path
(5.2)
is
of the form
(5.3)
[X X X
(2n +1)
X]
X
(2n)
(2n -1) 4 (2n -2) 4
4
(N +4),
4 X 3 X (N +3)
X 4 X 4 X
where each
4
denotes
a
number which can be determined only
with additional information about the form of path
The part of
(5.2)
which corresponds to the path part of
(5.3)
This is called the path
is enclosed in brackets.
part or P -part and the rest of
tion part or
RRS
an induction RRS
(5.3)
is called
the induc-
Hereafter we will enclose the path
I -part.
part of such
(5.2).
and is denoted
Rn
is called
RRS
Such an
in brackets.
+2.
The left representative sequence of
(5.2)
is of the
form
[(2n +1) (2n)
(5.4)
(2n -1)
(2n -2) X X
(N+3) X (N +4) X 4 X 4 X
where each
X
denotes
a
X] X
4 X 4 X 4 3,
number which can be determined only
with additional information about the form of path
The part of
(5.2)
(5.4)
(5.2).
which corresponds to the path part of
is enclosed in brackets and is called the path part
or P -part.
The rest of
(5.4)
is
called the induction
33
part or
I -part.
We now prove four lemmas.
Preliminary lemmas.
Lemma
If
2.
n >
7,
N = %n (n -1) > 3n > 2n +
Proof.
Lemma
If
3.
N +
Lemma
n >
then N
>
2n + 1.
1.
7,
then
N + 3 > 4.
then by Lemma 2 we have
2n + 4 > 4.
3 >
(N +3,N +4),
7,
n >
If
4.
is odd,
n
Since
Proof.
pair
is odd,
n
is odd,
n
> 7,
n
then the unordered
occurs in adjacent positions of the path
part or in the last two positions of the induction part of
an induction
Proof.
Rn +2..
RRS
We examine all other pairs of consecutive
positions and find that always one of them is occupied by
an integer smaller than
integer
4
The integer
N + 3.
3
occurs in every consecutive pair of positions
Rn +2.
except the last pair of the induction part of
first integer of the path part of
part of
Rn
(N +3,N +4)
2n + 1.
is
is not greater
The integer preceding the pair
in the induction part of
%n(n -1) + 2 = N
than
Rn +2
The
The first integer of the induction
% n(n -1) = N.
than
or the
+
2.
Hence,
Rn
is not greater
by Lemmas
2
and
3
each of these other possible pairs of consecutive positions
is occupied by at
least one integer less than
so cannot be occupied by the pair
Lemma
i
1
1
<
i
5.
< n +
N + 3
(N +3,N+4).
There is no permutation on the set
23
so that a
sequence of consecutive
and
34
corners of the induction part of
denote the vertex
N +
is transformed into
a
Pn +2,
a
k +
of which
1
or which denote the vertex
1
n +
2,
sequence of corners with the same
order which are successive corners in the path part of
a
Pn +2'
By hypothesis at least one of the vertices
Proof.
n +
of
1,
n +
is denoted by
2,
k +
1
corners of the sequence
consecutive corners of the induction part of
Hence,
if
such
a
a
Pn +2.
permutation does exist, the common image
of the corners that denote that vertex is denoted by
corners of the path part of
of the path part
corners, which is
of
a
a
Pn +2.
a
Pn
1
However, each vertex
denoted by
is
k +
k =
contradiction, and Lemma
%(n -1)
is proved.
5
Our proof will be divided into two parts.
In Part A
we show that two induction paths which are constructed from
non -equivalent Euler paths on
n
vertices are not related
through rotations and vertex permutations.
In part B we
show that two induction paths which are constructed from
non -equivalent Euler paths on
through
a
reflection.
n
vertices are not related
The theorem will then follow since
we will have shown that two induction paths which are
constructed from non -equivalent Euler paths on
are not related through
reflections,
a
n
vertices
finite sequence of rotations,
or vertex permutations.
This shows that
representative Euler paths from different equivalence
classes on
n
vertices generate representatives of different
35
equivalence classes on
r(n +2)
>
n +
vertices and thus
2
r(n).
obtained
Pn +2'
The induction paths,
Proof, Part A.
by the aforementioned process must fall into three mutually
Each type is determined by the value of
exclusive types.
such that the right characteristic number denoted by
m
in the
r1
.
m = N + 2n +
A:
For
1.
a
type
A
r1 = N + 4.
Type
Pn
to the corner
The three mutually exclusive types are as follows.
Type
Pn +2'
of the Euler path corresponds
RCS
r1
For
N < m < N + 2n + 1.
B:
a
type
is dependent upon the Euler path on
B
vertices
n
which is used to construct the induction path.
Type
C:
<
1
m
For
N.
<
dependent upon the Euler path on
is
Pn +2, r1
type C
a
vertices which is used
n
to construct the induction path.
Thus after
a
rotation we obtain each type of induction
path with its corresponding
RCS
Type
path:
RCS:
(n +2)
(N+4)
path (cont):
RCS (cont):
1
4
(n +2)
(2n -2) 4
1
}
X]
X
'
3
2
(n +2)
4 (2n) 4
(2n+1)
(n +2)
5
(n +1)
n
(n +1)
(n+1)
3
(2n -1) 4
(n +2)
X (N +3).(N+4).
Euler path there is one vertex of the
-part occurring before the
vertices of the
A
3 4
[X X X X
Note that for Type A
I
2
follows:
as
P -part,
P -part,
followed by the
followed by the remaining
N
2n vertices
36
of the I -part.
Type
path:
u
RCS:
d
(n +2)
l
X (N +3)
(N +4)
[X X X X
n
3
path (cont):
RCS (cont):
Note that
(n +l)
(n +2)
...
(n +1)
1
(2n +1) 4
4
2 3 4
X]
4
(2n -2)
(2n -1) 4
of the I -part of
d >
.
a
We will
N + 4.
corner
jth
is denoted by the
u
that is, we will assume that
(5.2),
corners of the
and
u
u
(n +2)
(n+l) 4
3
}
is the right characteristic number of
d
assume that the vertex
-1
(n +2)
2
(2n)
corner denoting the vertex
j
B
follow the P -part of
I -part
a
Type
B
path.
Type
path:
v
RCS:
e
}
(n +l)
1
X](2n+1)
path (cont):
4
Note that
e
is
corner of the P -part of
I -part
4
(n +l)
3
(2n -1) 4
4
+2) 5
(n+2)
(2n -2)4
i -1
(n +2)
[1 2 3 4
v
X (N +3)
(N +4)
[X X X X
.
and that
v
v
(5.2),
e
> N +
is denoted by the
4.
a
We
ith
that is, we will assume
corners of the P -part which follow
of a type
We now consider
X
(n +l)
n
the right characteristic number of
will assume that the vertex
the
(2n)
3
corner denoting the vertex
that there are
(n +2)
2
(n +2)
RCS (cont):
C
C
a
path.
set of mutually exclusive cases
which will exhaust all induction paths which arise from the
37
aforementioned construction process.
In each case we
assume that the two induction paths, which have been
constructed from non- equivalent Euler paths on
tices,
are related through
and vertex permutations.
a
ver-
n
finite sequence of rotations
In other words,
we assume that
the RCS's of the two induction paths are identical.
obtain
contradiction by showing that the RCS's are not
a
identical,
and thus that if the induction paths are
equivalent they are related through
Case
a
reflection only.
The two induction paths are of the same type.
1.
We are assuming that the two Euler paths on
are non -equivalent.
where
We
<
1
<
a
n,
Therefore,
n
vertices
there is some vertex
a,
which is denoted by different sets of
corners for the two Euler paths on
vertex
a
P -parts
of the two induction
is denoted by different
n
vertices.
Thus,
the
sets of corners for the
paths Pn +2,
and so the sets
of right characteristic numbers which correspond to the
corners denoting the vertex
is because the vertex
corner of the
I -part
a
a
must be different.
This
must be denoted by the same
of the two induction paths.
Now assume
that the corner to which the right characteristic number
denoted by
Then the two
of
r1
corresponds is the same for both
RCS's
Pn +2's'
cannot be identical because the set
corners which correspond to the right characteristic
numbers of corners denoting the vertex
for the two
Pn +2's.
a
must be different
This means that the two Pn +2's cannot
38
since for type
A
be of type
to corner
N + 2n +
Assume the
A
paths
corresponds
r1
1.
are both of type
Pn +2's
B.
The corners
ri
to which the right characteristic numbers equal to
correspond cannot denote either the vertex
vertex
n +
n +
or the
1
This is because all right characteristic
2.
numbers corresponding to corners denoting these vertices
and if the greatest of them,
are known,
we would have
Thus
a
r1
were
path.
A
must correspond to different corners in each
r1
of the type
type
N + 4,
B
This is because each vertex
Pn +2's.
different from the vertex
n +
1
and the vertex
2
n +
is denoted by exactly one corner of the I -part of a type
B
These vertices are denoted by corners occupying
path.
actually consecutive odd, positions of the
different,
I -part
of the type
Thus the pair of right
path.
B
characteristic numbers
(N+3,N +4)
which corresponds to
the consecutive pair of corners denoting the pair of
vertices
(n +1,n +2)
of consecutive
RCS
of the
by Lemma
4,
of
must be identical with some other pair
right characteristic numbers from the
a
type
B
path.
But this is impossible
and so the two paths cannot be of type
Assume now that the
Pn +2's
I -part
are both of type
B.
C.
Let
the corner to which the right characteristic number denoted
by
ai +t
ri
corresponds be
ai
for the other path.
for one of the paths and
If
t
is greater than or equal
39
to
2n
n +
1
then the
,
k +
1
which occur in the
can be transformed by
corners which denote the vertex
Pn +2's
of one of the
I -part
2n
vertex permutation into
a
But this is
corners of the P -part of the other
Pn +2°
impossible by Lemma
must be less than
If
is less than
t
numbers
(N+3,N +4)
and so
5,
2n,
t
2n.
the pair of right characteristic
which corresponds to the consecutive
pair of corners which denote the vertex
n +
and the
1
-part of one of the
vertex
n +
Pn +2's
must be identical with some pair consisting of
which occurs in the
2
right characteristic numbers of
corners of the first
consecutive pair of
a
positions of the
I -part
But this is impossible by Lemma
Pn +2'
other
2n
I
both paths cannot be of type
Thus, Case
C.
4,
of the
and so
has been
1
proved to be impossible.
Case
and
B.
A
The two induction paths are of types
2.
The vertex
u
corner of the type
B
nor the vertex
2.
n +
which is denoted by the first
path is neither the vertex
n +
1
This is seen by considering the
right characteristic numbers of the corners which denote
these vertices.
All are less than
the last corner of the
n + 2.
have
a
I -part
N + 4
except that of
which denotes the vertex
But if this corner denoted the vertex
type
A
path instead of
is one of the vertices of the
a
set
type
B
u,
path.
[1,2,-,0.
we would
Thus,
u
This means
40
that
I -part
of the
RCS
the
4
or
A
Consider the last element of
Rn
of an
of the type
This number must be either
path.
B
This number is
path.
N + 3.
of the
RCS
Now consider the last element of the
3.
type
one of the unknown right characteristic numbers
is
d
RCS's
If the two
are to be identical these two right characteristic numbers
This is impossible by Lemma
must be equal.
3
and so this
case is impossible.
Case
C.
The two induction paths are of types
3.
Remembering that the vertex
corner of the P -part of
v
and
A
ith
is denoted by the
we consider three subcases
(5.2),
which exhause this case.
Subcase
rN
pair
Assume
3A.
i
<
2n.
of the
+2n +1- i'rN+2n +2 -i
This requires that the
RCS
of the type
RCS
of the
impossible by
of the type
I -part
Lemma
3B.
Assume
of the
RCS
of the type
of the
RCS
equation,
i
of the type
2n +
Subcase
Assume
3C.
part of the type
permutation into
A
a
=
2n.
A
path.
C
i
>
2n.
5.
path equals
rN +2n+2 -i
This leads to the
2.
This means that the I-
path may be transformed by
a
vertex
sequence which occupies consecutive
corners in the P -part of the type C
by Lemma
This requires that
which is impossible by Lemma
= N + 4,
1
This is
path.
4.
Subcase
rN
A
path
2n elements
must equal some consecutive pair form the first
of the
C
Thus Case
3
path, which is impossible
has been proved to be impossible.
41
Case
The two induction paths are of types
4.
Remembering that the vertex
C.
I -part
corner of the
jth
is denoted by the
v
we consider three subcases
(5.2),
of
and
B
which exhaust this case.
Subcase
N -i < 2n +
Assume
4A.
type
path must equal
B
r2n
the pair
of the
RCS
of the
RCS
of the type
B
path must equal some consecutive pair from the first
RCS
elements of the
of the
I -part
or
path,
C
of the type
rN +2 -i
+1- j,r2n +2 -j
We now have
j.
-
of the
r2n
two possibilities, either
1
of the type
j
-1
path.
C
The first of these possibilities leads to the equation,
N + 4 = 2n +
which is impossible by Lemma
1,
second possibility is impossible by Lemma
r2n
the pair
Subcase
that the
a
4B.
I -part
N
Assume
of the type
vertex permutation into
Subcase
that the
into
a
I
4C.
= 2n +
i
1
This means
j.
-
path can be transformed by
C
sequence which occupies
a
P -part of
successive corners in the
is impossible by Lemma
-
because
4,
(N +3,N +4).
pair
is the
+1- j,r2n +2 -j
The
2.
the type
path which
B
5.
Assume
N
-part of the type
-
>
i
B
2n +
1
We show
j.
-
path can be transformed
sequence which occupies successive corners of the
path, which is impossible by Lemma 5.
P -part of the type
C
Consider the first
2n + 2
-
j
corners of the type
These corners are from the
I -part
of the type
Since
these
2n +
N
-
i
>
2n +
1
-
j
1
-
j
B
B path.
path.
corners can be
42
transformed by
vertex permutation into
a
ing consecutive corners in the
Consider now the last
j
corners of the type
-1
rN+2n +2
-i
path also, because if they were not, one
C
Either
of the
RCS
of the type
C
path equals
the
RCS
of the type
B
path,
of
+2n +2 -j
rN+2n +l- i'rN +2n +2 -i
of the
RCS
of the
of the
j
path must
C
elements
-1
of the I -part of the type
RCS
B
The first possibility leads to the equation,
path.
= N + 4,
1
which is impossible by Lemma
second possibility is impossible by Lemma
rN
+2n +l- i'rN+2n +2 -i
Subcase
Part
or the pair
of the type
RCS
equal some consecutive pair from the first
pair
path.
B
vertex permutation
a
of the following would have to be true.
2n +
path.
C
sequence occupying consecutive corners in the P -part
a
of the type
rN
sequence occupythe type
P -part of
These corners can be transformed by
into
a
A
4C
is
The
2.
because the
4,
(N+3,N +4).
is the pair
Thus,
proved to be impossible which completes
of the proof.
Proof, Part
B.
In this part we show that two
induction paths formed from non -equivalent Euler paths on
n
vertices cannot be related through
this we will consider an induction path,
been reflected once.
path
Qn +2.
Qn
is
the
RCS
identical to the
Pn
which has
PN +2,
We will call this reflected induction
As was seen in Chapter
of the
To do
reflection.
a
LCS
with the
of
the
3,
Pn +2.
RCS
RCS
If we
of the
of
compare
Qn +2
and
43
show that they cannot be identical, we will have shown that
two
formed from non -equivalent Euler paths on
Pn +2's
vertices cannot be related through
show that the
RRS's
Pn +2
of
identical and since the
RCS
cannot be identical.
can never be
Qn
and
is just
specific element
a
RCS's
we will have shown that the
RRS's
of the set of all
We will
reflection.
a
n
RRS's
Consider the
and
Pn +2
of
Qn +2
RRS
The
of
is of the
form
X](2n +1) 4 (2n) 4 (2n -1) 4 (2n -2) 4 X
[X X X
4 X 4
X 4 X 3 X (N +3)
RRS
The
Pn +2
Qn +2
of
[X X X
X
4 X
(N +4).
is of the form
(2n -2)
(2n -1)
(2n)
(2n +1)]
3 4 X 4 X
X 4 X (N +4) X (N +3) X.
All the unknown right characteristic numbers of the
I -part
of
must be greater than or equal to
Pn +2
This follows from the fact that the vertex
occur before the
n +2.
cannot
i
position of the P -part because we
ith
are using only the representative Euler path of each
equivalence class on
n
vertices to form
Pn
+2.
Now
consider the known right characteristic number pair
of the I -part of
of the
I -part
of
All right characteristic numbers
Qn
Pn
+2
are greater than or equal to
except the known right characteristic number
3,
precedes the known right characteristic number
pair
(3,N +3)
(3,4)
cannot equal the pair
(3,4)
4
which
N +3.
by Lemma
The
3.
44
Since the first and last right characteristic numbers of
the I -part of
Pn
are
+2
which are greater than
following.
2n +
by Lemma
4
3,
we find the
No pair of consecutive right characteristic
numbers of the
RRS
which contains at least
Pn
of
one right characteristic number from the
RRS
of that path can equal the pair
RCS
of
Qn +2
the pair
both of
N + 4,
and
1
is
RRS
of the
RCS
I -part
of the
Thus if the
(3,4).
to be identical to the
of the
(3,4)
I -part
Pn +2
of
of
Qn
must equal some pair of right characteristic numbers from
the
RRS
of the P -part of
By Lemma
of
Qn +2
the first
2n -2
vertices of the
there will be
a
I -part
vertices of the
2n -4
all have their images in the P -part of
Qn
I -part
cannot all have their images in the P -part of
But unless the first
Pn
of
5,
Pn+2.
vertex which occurs in the
Pn +2'
P -part of an
induction path which occurs more than once in its
I -part.
This is impossible because of the way we constructed
these induction paths.
vertices of the
in the P -part of
I -part
Pn
of
Qn +2
must have their images
Qn
This means that the right
characteristic number of the
of
2n -2
nd
corner of the
I
-part
must equal the right characteristic number of
either the first or the second corner of the
Pn
2n -4
Thus we have that the first
I -part
of
This last statement leads to the equations,
N + 4 = 2n +
1,
and
N + 4 = 4,
respectively, which are
45
impossible by Lemmas
RRS
of
Pn+2
and
2
3,
respectively.
cannot be identical to the
and the proof of Part
the proof of Theorem
B
10.
is
complete.
RRS
Thus the
of
Qn +2
This also completes
46
INDEX
NOTATION
OF
n
-
The order of the vertex set of
N
-
The number of edges in
a
a
full graph.
full graph on
n
vertices.
Also the number of edges in an Euler path on
n
vertices.
k
-
The number of corners of an Euler path on
vertices which denote
a
specific vertex.
corner of an Euler path on
ith
n
vertices.
ai
-
The
ri
-
The right characteristic number associated with the
corner
si
-
ai
n
of an Euler path.
The left characteristic number associated with the
corner
ai
of
an Euler path.
RRS
-
Right representative sequence.
RCS
-
Right characteristic sequence.
LAS
-
Left representative sequence.
LCS
-
Left characteristic sequence,
L
-
The length of the principal segment of an Euler path.
47
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1.
Arnold,
Intuitive concepts in elementary topology.
H.
B.
Englewood Cliffs, N. J., Prentice -Hall, 1962.
2.
The theory of graphs and its applications.
Berge, Claude.
New York, Wiley, 1962.
3.
Euler,
8:128-140.
Commentarii Acadamiae Petropolitanae
1736.
Theorie der endlichen und unendlichen Graphen.
König, D.
New York, Chelsea, 1950.
5.
6.
I.,
R.
C.
Vol.
38.
Providence,
American Mathematical Society, 1962.
Rajchman, Jan A.
tube.
258p.
Theory of graphs.
Ore, Oystein.
R.
247p.
Solutio problematis ad geometriam situs
L.
pertinentis.
4.
182p.
A.
270p.
The selective electrostatic storage
Review
12 :53 -97.
1951.