AN ABSTRACT OF THE THESIS OF
Santiago Sologuren P.
Science
Title:
Mathematics
in
Master of
for the degree of
presented on December 16, 1982
Systems of Incongruences in a Proof on Addition
mod m.
Abstract approved:
Redacted for Privacy
Robert D. Slitalley
Let
N = N(k,m),
that for any sequence of length
k
m
and
k
are positive
be the smallest positive integer so
k < m,
integers,
where
N
containing at least
integers which are distinct mod m
sequence with zero sum
there exists a sub-
mod m.
A result that we call the Fundamental Theorem
states:
If
m=11,13,16,17,
or
m > 19,
then
N(4,m)<m-6.
We obtain a set of 61 systems of incongruences any one of
which may lead to a variation of the proof of the Fundamental Theorem.
Also, such a new proof may supply us with
a method for obtaining corresponding results for N(k,m),
where
k > 5.
A relationship is explored between the above
systems of incongruences and a collection of families of
subsets of the set
{a,b,c,d}.
SYSTEMS OF INCONGRUENCES IN A PROOF ON ADDITION mod m
by
Santiago Sologuren P.
A THESIS
submitted to
Oregon State University
in partial fulfillment of
the requirements for the
degree of
Master of Science
Completed December 16, 1982
Commencement June, 1983
APPROVED:
Redacted for Privacy
Professor of
Mathematiin
charge of major
Redacted for Privacy
Chairman of Department of Mathematics
Redacted for Privacy
Dean of Grate School
Date thesis is presented
December 16, 1982
Typed by Jolan Eross for Santiago Sologuren P.
Acknowledgments
I wish to express my gratitude to my major
professor, Dr. Robert D. Stalley, a man of highly
noble human qualities, who has contributed a great
deal to my intellectual formation.
I also thank my friend, Jolin Eross who typed
this thesis in a very professional manner.
Table of Contents
Page
Introduction
1
Chapter I.
Fundamental Theorem and Preview
Chapter II.
An Analysis Yielding Representatives
.
of Eighteen S-10 Families
Chapter III.
10
Representatives of All Equivalence
Classes of S-10 Families
Chapter IV.
4
23
Representatives of All Equivalence
Classes of S-10 Systems
30
Epilogue
58
Bibliography
59
List of Tables
Page
Table I.
Some values of
N(k,m)
mainly for
k < 4
2
Table II.
S-10 Families
Table III.
S-10 Systems of Given
Table IV..
S-10 Families of Given S-10 Systems.
Table V.
Incongruences of Standard Repre-
24
S-10
Families
sentatives
Table VI.
33
36
40
S-10 Systems in Each Part of Rank Two
With More Than One S-10 System.
56
SYSTEMS OF INCONGRUENCES IN A PROOF ON ADDITION mod m
Introduction
In this thesis we are concerned with an aspect of
the problem of determining how long a sequence of integers
mod m must be in order that there will necessarily exist a
subsequence with zero sum
mod m
given a lower bound on
the number of integers in the sequence that are pairwise distinct
mod m.
Let us be more precise.
integers where
k < m.
Let
Let
k
N = N(k,m)
and
m
be positive
be the smallest
positive integer such that for any sequence of length
containing at least
k
N
integers which are distinct mod m
there exists a subsequence with zero sum
mod m.
Then, the
problem is that of evaluating the number theoretic function
N(k,m).
where
1 < k < m.
First, we list several known results, mainly for
k < 4,
in the following table.
(1)
2
Table I.
Some values of N(k,m)
mainly for k < 4
N (k,m)
(k,m)
2
k=1
1<k<m
3
m < 2k
4
k> 2; m>
1
5
6
7
8
9
10
11
12
in
k<N<m
k
1
2
k = 2; m = 2
k=2;m>3
k= 3 ; 3 >1. m > 5
k = 3; m = 6
k=3;m>7
k= 4; 4 <m<
<8
....
k = 4; m = 9, 10
k=4; m=11, 13, 16,
1
2
N> m-
k (k+1 )
2
m-1
3
MM.
4
m-3
.....
4
..._
5
17 or m>19
m- 6
k (k-1)
3
Entry 12 follows from entry 4 together with the
following theorem that will be called the Fundamental Theorem.
This theorem is the deepest of the known results for
N(k,m).
Fundamental Theorem.
m > 19,
then
N(4,m)
If
m = 11, 13, 16, 17, or
< m-6.
In this thesis we list a set of systems of incongruences each system of which may lead to a variation of the
proof of the Fundamental Theorem.
In Chapter I we give a part of the proof of the
Fundamental Theorem which serves to motivate the reminder
of this thesis.
We show that our proof of this theorem is
based on one of the above mentioned systems of incongruences
mod m.
Furthermore, we include a description of
the contents of Chapters II, III and IV.
In the epilogue
we discuss three related new problems to which one is lead
by this research.
4
Chapter I
Fundamental Theorem and Preview
We will start
by giving a partial proof of the
Fundamental Theorem to motivate the thesis, then we introduce in an informal but more detailed fashion than in the
introduction the problem of the thesis.
Throughout the
present work congruences are understood to have modulus m.
We restate the Theorem.
Fundamental Theorem.
m > 19,
then
N(4,m)
If
m = 11, 13, 16, 17, or
< m-6.
Partial proof (1)
.
Let
and suppose that the sequence
m=11, 13, 16, 17, or
m > 19,
Q = {al, a2, ... a m-6}
of
m-6 integers contains at least four integers which are
distinct
mod m.
with zero sum
We show that
Q
contains a subsequence
mod m.
We assume that no subsequence of
Let
distinct
m-9
has a zero sum
and obtain a contradiction.
mod m,
a
Q
a, b, c
mod m.
= a, a
m-8
and
d
be elements of
Q
which are
Suppose, without loss of generality, that
= b
a
'
m-7
Consider the sums
= c,
and
a
m-6
= d.
5
s
1
= a1
s2 =
al + a
= a
2
+ a
s
m-10
s
m-9 = a 1 + a 2
s
m-8
s
m-7
s
m-6
s
m-4
s
m-3
s
m-2
s
= a
= a
= a
m-5
s
s
= a
= a
= a
= a
= a
m-1
= a
m
1
+ a
1
1
1
1
1
+ a
+ a
+ a
+ a
+ a
1
1
1
+ a
+ a
1
+ a
2
2
2
2
+
+ a
+
+ a m-10 + a
+
+ a
+
+ a
+
+ a
+
+ a
+
+ a
+
+ a
+
+ a
+
+ a
+
+ a
2
2
2
2
2
2
m-10
m-10
m-10
m-10
m-10
m-10
+ b
+ c
+ d
+ a+b
+ c+d
m-10 + a+b+c
m-10
m-10
m-10
+
a+b+d
+ a+c+d
+ a+b+c+d
First, we establish conditions under which these
sums are distinct
m
mod m.
In this case, since there are
of these sums, one of them is zero
mod m,
which is a
contradiction.
If
1
< i
<
j
< m,
and
i < m-10.
then
s, - s
J
isthesumofasubsequenceofQ,andso,,
1
J
thatis,s,J ts.1 .fience, it suffices to consider the sums
where
m-9 < i <
j
< m.
6
Thus, it suffices to consider the sums:
a
b
d
a+b
c+d
a+b+c
a+b+d
a+c+d
a+b+c+d.
Since
and
a,b,c,
d
are distinct
the sum of the elements of any subset of
not zero
mod m,
mod m
{a,b,c,d}
and
is
by inspection, the conditions reduce to
a+b+c t d
a+b+d t c
a+c+d t b
a+b
c+d
a+b
c
a+b t d
c+d t a
c+d t b
We consider the
{a,b,c,d},
4: = 24
permutations of the set
and observe that we can replace the above sys-
tems of incongruences by any of the 24 transformed systems
of incongruences.
Notice that the above system of incongruences is invariant under the permutation
(c,d).
Thus, 24 transformed
systems can be paired off into 12 pairs of identical systems.
7
If any of the 12 systems of incongruences hold we
The other case to consider is where
have a contradiction.
That is, for each of the 12 systems
all 12 systems fail.
one incongruence is replaced by a congruence, and these
congruences are taken simultaneously.
Every such system
(1)
of congruences can be shown to lead to a contradiction.
Let S = {a,b,c,d }.
An S-10 family is a set of 10
nonempty subsets of
S
which includes the set
S.
It may happen that variations of the proof of the
Fundamental Theorem can be based on different choises of an
S-10 family than made in the above partial proof.
As in
the above partial proof of the Fundamental Theorem we impose the restriction that no two subsets in the family have
sums that are congruent
For this reason we are
mod m.
interested in identifying all
S-10
The reason we define an
set
S-10
families.
family to include the
is that otherwise the proof of the Fundamental Theo-
S
rem would be unnecessarily longer, more precisely, it would
contain all of the steps of the proof based on an
family, and also some other additional steps.
S-10
To show this,
consider any family that is defined in the same way as an
S-10
family, but where the set
element.
S
is not included as an
The collection of incongruences of the system
corresponding to such a family is a super-collection of the
collection of incongruences of the system corresponding to
8
S-10
some
family.
Thus we may take the
S-10
family
which is formed by replacing one of the proper subsets of
S
by
Hence, the collection of systems of congruences
S.
corresponding to the system of incongruences is a supercollection of the collection of systems of congruences
corresponding to the
S-10
system.
Next, we calculate the number of
Since an
S-10
family contains the set
S-10
families.
S, the remaining
9 elements must be chosen from the 14 proper non-empty subsets of
is
S.
The number of ways in which this can be done
95: = 2,002.
Two
S-10
families are equivalent if one can be
transformed into the other by a permutation on the set
S.
Therefore, variations of the proof of the Fundamental Theorem that are based on equivalent
tical except for the notation.
S-10
families are iden-
Hence, we restrict our at-
tention to a complete system of representatives from our
equivalence classes of
S-10
families.
In Chapters II and
III we develop a list of representatives of these equivalence classes.
There are 122 of them.
Two systems of incongruences which correspond to
S-10
families are equivalent if one can be transformed
into the other by a permutation on the set
S.
The variations to which we have been referring of
the proof of the Fundamental Theorem are based on the
9
several systems of congruences that are yielded from one
of the systems of incongruences corresponding to an
S-10
family as in our partial proof of the Fundamental Theorem.
It may happen that a system of incongruences does not lead
in this manner to a proof of the Fundamental Theorem.
(2)
On the other hand, equivalent systems of incongruences,
except for notation, lead to the same proof, while nonequivalent systems of incongruences may or may not lead to
the same proof.
For this reason we consider a complete
set of representatives from the equivalence classes of the
above systems of incongruences.
In Chapter IV we develop
a list of representatives of such equivalence classes.
is interesting that there are 61 of them.
It
10
Chapter II
An Analysis Yielding Representatives
of Eighteen
S-10
Families
Before stating the contents of Chapter II we give
three definitions.
We continue to let
Definition 1.
empty subsets of
S
An
S-10
.
family is a set of 10 non-
which contains the universe
Definition 2.
an
S-10
S = {a,b,c,d}
S.
A subfamily (of S) is a subset of
family.
Definition 3.
Two subfamilies (of S) are equi-
valent when one subfamily is transformed into the other
under a permutation on
S.
In this chapter we include a thorough analysis
yielding representatives of all equivalence classes of
S-10
families which contain all four singletons.
There
are 18 of these equivalence classes.
Before we go any further we give two more definitions.
Definition 4.
A singleton, a double, a triple and
a quadruple is a subset of
four elements of
S
Definition 5.
S
of one, two, three, and
respectively.
A (u-v) subfamily (of S) is a
subfamily that contains
u
triples and
v
doubles.
11
Since
u+v = 5
we have five cases.
need to find only the
In each case we
(u-v)
subfamily since then the
S-10 family is completely determined.
Case 1
(Four triples and one double.)
Consider the subfamilies:
= { {a,b,c },
-5
1
{a,b,d },
{a,c,d },
{b,c,d } },
= {{a,b} 1,
and
U .a
3
Theorem 1.
valent to family
Proof.
S.
Thus,
double of
Any (4-1) subfamily of
31.
Let 3 be an arbitrary (4-1) subfamily of
3 = 7 U .5,
S.
where
Then,
a, y
b.
is an arbitrary
.5 = {x,17}
is equivalent to
the following permutation on
x
is equi-
S
S transforms
(The other elements of
S
31
because
into
3
31:
are fixed under
the permutation.)
Case 2.
(Three triples and two doubles).
Consider the subfamilies:
3 = {{a,b,c},
13'
1
5
-13'
-&
2
3
4
{a,b,d },
= {{a,b},
{a,c } },
= { {b,c },
{b,d} },
= { {a,b },
{b,c } },
{ {a,b },
{c,d}},
=
{a,c,d}},
and
31 =
U
l'
3
2
= 7 U .6
2'
3
=3U.}
3
U
4
4'
12
Any (3-2) subfamily of
Part (a).
Theorem 2.
is equivalent to one of the subfamilies
31, 32, 33
S
and
3
Subfamilies
Part (b).
and
1
,
32'
are not equivalent to each other.
3
4
Proof of Part (a).
subfamily of
z, and
3
be an arbitrary (3-2)
as follows:
3
{{a,b,c},
{a,b,d },
{a,c,d}}
{{a,b,c},
{a,b,d},
{b,c,d}}
{{a,b,c},
{a,c,d},
{b,c,d}}
{{a,b,d},
{a,c,d},
{b,c,d}}
.
be the element of this singleton, and
x
w
3
This is clear by considering all possible
collections of triples in
Let
Let
The intersection of the triples in
S.
is a singleton.
y,
3
be the remaining elements of
S.
The
subfamilies are classified into four types.
doubles.
Type 1.
(The element
Let
and
y
be the other elements in these
Hence, the doubles are
Type 2.
Let
doubles.
z
is in each double.)
x
y
(The element
x
{x,z}.
is in neither double.)
be the element which is common to both
Hence, the doubles are
Type 3.
{x,y},
(The element
x
{y,z},
{y,w }.
is in just one double
and the intersection of the two doubles is nonempty.)
13
Let
be this intersection, and let
y
other element of the second double.
are
{x,y },
be the
z
Hence, the doubles
{y,z }.
Type 4.
(The element
is in just one double
x
and the intersection of the two doubles is empty.)
Let
y
be the other element in the first double.
{x,y },
Hence, the doubles are
{z,w }.
Suppose that a subfamily
is equivalent to subfamily 31,
3
permutation on
z
-0-
then
c,
3
3
into 31:
Similarly, if
0
is of type
2,
3
1'
a,
3
y+
or
b,
4,
or 34
This completes the proof of Part (a).
Proof of Part (b).
families
x -4-
is equivalent to subfamilies 32, 33,
respectively.
Then,
because the following
transforms
S
w -0- d.
is of type 1.
3
3
2'
3
and
3'
is a permutation on
3
4
are equivalent.
Subfamily
3j,
3i
i+j,
transformed into
Then there
transforming one of these sub-
S
families into the other.
into subfamily
Assume that two of the sub-
is transformed
only if
.&j.
is
Notice that the intersection of the
3
doubles in 51,.52,.83
{b},
and
.54
is the singleton
and the empty set respectively.
forming
&i.
into
.6j
{a}, {b},
A permutation trans-
1<i<4, 1<j<4, i+j,
transforms the
intersection of the doubles in
Si
into the intersection
of the doubles in
3
is not equivalent to
Hence,
3
4
14
either
or
31,
1
doubles in
.5
doubles in
5
3
is empty and the intersections of the
and
11 32
equivalent to either
transformed into
into
31
32
since the intersection of the
3
or
3
are not.
.5
3
or
2
3
Also,
3
is not
1
since otherwise
3
{a}
is
{b}
under the permutation transforming
33.
But, since the intersection of the
triples in the subfamily
is
j
{a },
u
is trans-
32
since the
then
formed into a different subfamily of triples.
Finally,
element
a
is not equivalent to
33
appears in one of the doubles of 33
neither of the doubles of
tion transforming
b, c
or
d.
.53
.52.
into
But then
q
That is, for any permuta-
32
the image of
a. must be
is transformed into a different
subfamily of triples.
Case 3.
(Two triples and three doubles.)
Consider the subfamilies,
= { {a,b,c },
{a,c },
{b,c }}
= { {a,b },
{arc},
{a,d } },
= { {a,b },
{a,c},
{b,d } },
= { {a,c },
{a,d },
{c,d}}
3 5 = {{a,c},
36 =
{b,c},
{c,d}}
{b,d},
{c,d }}
3
= {{a,b},
{a,c},
{c,d}}
= { {a,c },
{a,d },
{b,c } },
.5
.5
=
2
3
.5
4
7
38
and
{a,b,d} 1,
{ {a,b },
1
but in
15
2,
31
7U '&1'
32
35 =
51j.,5
36 = a U .56,
5,
Theorem 3.
U
Part (a).
S
33
37 = 7 U
36,
37
and
f
3
8
Any (2-3) subfamily of
3
5'
6'
3
7
and
3
8
S,
34,
Subfamilies
31, 32. 33' 34'
are not equivalent to each other.
Proof of Part (a).
subfamily of
33,
32,
31,
S
.
Part (b).
3
38 =
.19-71
is equivalent to one of the subfamilies
35 ,
7U
34
U
and let
Let
be an arbitrary (2-3)
3
and
{x,y},
be the inter-
{z,w}
section and its complement respectively of the two triples.
The subfamilies are classified into 8 types.
Type 1.
(fx,y1
is one of the doubles,
{z,w}
is
not, and the intersection of the other two doubles is a
singleton not in the first double.)
Let the intersection be
x.
Then
and
x
the other elements in these remaining doubles.
doubles are
{x,y }, {x,z },
Type 2.
(fx,y1
and
y
are
Hence, the
{y,z}.
is one of the doubles, and the
intersection of the other two doubles is a singleton in
this double.)
Note that
fz,w1
Let this intersection be
is not one of the doubles.)
x.
Then
z
and
the other elements in these remaining two doubles.
w
are
Hence,
16
the doubles are
Type 3.
{x,y },
({x,y}
{x,z},
and
fx,w1.
is one of the doubles, and the
Note
intersection of the other two doubles is empty.)
that
is not one of the doubles.
{z,w}
One of the elements of one of these other two
doubles is
z.
x.
Let the other element of this double in
Hence, the doubles are
Type 4.
({z,w}
{x,y}, {x,z},
and
{y,w}.
is one of the doubles,
{x,y}
is not, and the intersection of the other two doubles is
a singleton not in the first double.)
Let this intersection be
x.
Then
z
and w
the other elements in these remaining two doubles.
the doubles are
Type 5.
{x,z}, {x,w },
({z,w}
and
are
Hence,
{z,w }.
is one of the doubles, and the
intersection of the other two doubles is a singleton in
Note that
this double.)
{x,y}
is not one of these
doubles.
Let this intersection be
z.
Then
x
and
the other elements in these remaining two doubles.
the doubles are
Type 6.
{x,z}, {y,z},
( {z,w}
and
{x,y}
are
Hence,
{z,w }.
is one of the doubles, and the
intersection of the other two doubles is empty.)
that
y
Note
is not one of these doubles.
One of the elements of one of these other doubles
17
in
Let the other element of this double be
z.
Hence, the doubles are
Type 7.
{x,z}, {y,w},
and
({x,y}
and
x.
{z,w}.
are two of the
{z,w}
doubles.)
The remaining double must contain an element of
say
{x,y},
and an element of
x,
Hence, these doubles are
Type 8.
{x,y}, {x,z},
(Neither
nor
{x,y}
element of each of the doubles
and
permutation in
S
y
w
c,
-0-
3
3
or
y
is an
be the other
{x,z},
3
is of Type 1, then
because the following
into
3
3
1
x
31:
a,
d.
is of Type 2, 3, 4, 5, 6, 7, or
3
32, 33, 34, 35,
respectively.
8
Proof of Part (b).
families
z
is equivalent to subfamily
8, then
7'
3
31
transforms
Similarly, if
3
Then
Thus,
{y,z}.
3 is equivalent to subfamily
6'
{z,w}.
Hence, the doubles are
Suppose that a subfamily
z
and
{x,y}
are these doubles.
{x,w}
element of this double.
b,
are one of
{z,w}
element of the third double, and let
{x,w}
{z,w}.
is an element in two of the doubles.
x
and
{x,z}
and
z.
Each of the three doubles must contain an
the doubles.)
Suppose
say
{z,w},
2 ,
3
3'
3
4
,
Assume that two of the sub3
5'
3
6
and
3 7 ,
3
8
are
18
Let
equivalent.
be a permutation on
p
Subfamily 3i
one of these subfamilies into the other.
is transformed into subfamily
only if Pi
3
is transformed into
Notice that
and
{ a,b}
1<i<8,
,
transforming
S
1<j<8,
i+j,
p.j.
are the intersec-
{c,d}
tion and its complement respectively of the triples in
.
Thus, the doubles
under
P.
{a,b}
and 34. a5, .56
families
Hence, since 3 1' 3
3
1'
3
2'
3
of the subfamilies
since
3-7
subfamily
and
and
{a,b}
3
contains
37
33
and 38
and
3
2'
4'
3
3
5'
{c,d}
and
,5
contain
7
do not, each of the sub-
is not equivalent to each
7
3
are invariant
{ c,d}
and
6'
and
3
Similarly,
8.
,al, .52
and 33
do not,
is not equivalent to subfamilies
32'
33.
The double
and 33.
31, .52,
{a,b}
is an element of the subfamilies
The intersections of the remaining two
doubles in these subfamilies is the singleton
and the empty set respectively.
ing 3i
into 3j,
1<i<3,
{c}, {a},
A permutation transform-
1<j<3,
transforms the
i+j,
intersectionofthesecondandthirddoublesin3.into
the intersection of the second and third doubles in
Hence, since the image of element
or
then
d,
31
is not equivalent to
since the image of element
then
3
2
c
a
is not equivalent to
must be either
32
and
must be either
33.
33.
a
or
3..
c
Also,
b,
19
By an argument similar to the one in the last paragraph, since the double
families
and
J.54,
belongs to the sub-
{c,d}
9-6,
and the intersections of the
remaining doubles in these subfamilies are
and the empty set respectively,
35
and
and
36,
35
{c,d},
is not equivalent to
is not equivalent to
Since the subfamily P8
or
34
{a}, {c },
36.
does not contain {a,b}
and is the only subfamily of doubles with
this property,
is not equivalent to any of the re-
38
maining seven subfamilies, in particular subfamilies
34, 35,
and
36.
Case 4.
(One triple and four doubles.)
Consider the subfamilies:
=
{ {a,b,c } },
$1 = {{a,d},
{a,b},
{a,c},
{b,c}},
= {{a,d},
{b,d},
{c,d},
{a,b}1,
= {{a,d},
{b,d },
{a,b},
{a,c} },
= {{a,d},
{b,d},
{a,c },
{b,c}},
13'
.6
2
3
and
31 =JUP 1
Theorem 4.
3
2
=JU.i5 2
Part (a).
3
3
= 7U
34
=
4
3'
Any (1-4) subfamily of
is equivalent to one of the subfamilies
31, 32, 33
34.
Part (b).
and
3
4
Subfamilies
are not equivalent to each other.
,
32, 33
U,54.
4
S
and
20
Proof of Part (a).
subfamily of
S, and let
in the triple.
of
Let
Let
y,
x
z,
be an arbitrary (1-4)
3
be the element which is not
and w be the other elements
S.
The subfamilies are classified into 4 types.
Type 1.
Let
(The element
x
is in one double.)
be the other element of this double.
y
the remaining doubles are formed by
{y,x },
Hence, the doubles are
y, z,
{y,z},
and
{yew },
Then,
w.
and
fz,wl.
Type 2.
Then
(The element
y, z,
and
w
y
and
x
is in three doubles.)
are the other elements in
these doubles.
Let
fourth double.
Hence, the doubles are
be the elements in the
z
{y,x },
{z,x },
{y,z}.
{w,x},
Type 3.
(The element
x
is in two doubles and
the other two elements of these doubles form another
double.)
Let
and
y
z
be these other two elements, and
belong to the remaining double.
let
y
are
{y,x },
{z,x },
Type 4.
{17,z},
(The element
and
x
Hence, the doubles
{y,w }.
is in two doubles and the
other two elements of these doubles do not form another
double.)
21
Let
y
and
y
be these other two elements.
z
Thus,
belongs to one of the doubles that does not contain
and
belongs to the other double that does not contain
z
Hence, the doubles are
x.
{y,x},
{z,x },
and
{y,w },
{z,w}.
Suppose that a subfamily
3
is equivalent to subfamily 3
mutation on
w 4 c,
then
transforms
S
is equivalent to
3
then
because the following perinto
3
Similarly, if
x 4 d.
is of Type 1,
z 4 b,
is of Type 2, 3, or 4,
3
32, 3
y 4 a,
31:
3'
or
3
4
respectively.
Proof of Part (b).
families
31, 32, 33,
be a permutation on
and
then
7,
Subfamily
1<i<4,
1<j<4,
Thus, since
of
J5
3
3i
i+j
Since
d
leaves
P
d
is transformed into subfamily
only if
.9-i
not equivalent to either
32, 33,
not equivalent to either
33
which contain
doubles in
3A
d
are
p
invariant.
{a,d}
which contain
or
and
d.
3j.
is transformed into
and three doubles of
4'
Let
does not belong to the
appears in one double of
d
and
are equivalent.
34
transforming one of these sub-
S
families into the other.
triple of
Assume that two of the sub-
or
34.
two doubles
0-1,
2'
34,
then
and
3
32
The doubles of
{b,d },
is
1
is
3
the same
Thus, any permutation
3
22
However,
33
into
33
transforming
belongs to
{a,b}
is not equivalent to
{a,b}
leaves
34
.B.3,
but not to
invariant.
Hence,
.(94.
34.
(No triple and 5 doubles.)
Case 5.
Consider the subfamilies:
1
= { {a,b },
{a,d},
{a,c},
{b,d}},
{b,c},
and
Theorem 5.
Any (0-5) subfamily of
valent to subfamily
Proof.
be any subfamily of fine doubles.
3
There are altogether
3
is equi-
31.
Let
one double not in
S
Let
3.
is equivalent to
31
doubles, and so there is
= 6
(2)
be this double.
{x,y}
Then,
because the following permutation
x
y
transforms
3
into
other elements of
S
are fixed under the permutation.)
on
S
3
:
1
c,
d.
(The
23
Chapter III
Representatives of All Equivalence Class
of
S-10 Families.
In this chapter we give Table II which lists the
subsets of
class of
S
in a representative of each equivalence
S-10
The representatives of equi-
families.
valence classes of
families which contain fewer
S-10
than four singletons can be obtained by an analysis similar to that in Chapter II.
We first explain how to read Table II.
The main headings are the singletons, the first subheadings are the triples, and the second subheadings are
the doubles of an
family.
S-10
Since the set
is common to all
S
S-10 families
it has been omitted from the table.
To the left of the doubles of an
appears a symbol
.0 i ,
where
1<i<122
S-10
family
that denotes the
family.
For instance,
{c,d}
d3,5
consists of the doubles
in the second subheading, the triples
{a,b,d},
{a},
{b},
that
i05
{a,c,d}
{c},
{a,b},
{a b,c},
in the first subheading, and the singles
and
denotes the
{d}
S-10
in the main heading.
(Note
family that was used in
Chapter I in the partial proof of the Fundamental Theorem.)
24
Table II
S-10 Families
{c}, {d}
{a },
.
{a,b,c}, {a,b,d }, {a,c,d }, {b,c,d}
{a,b}
1
{a,b,c}, {a,b,d }, {a,c,d}
{a,b }, {a,c}
2
{b,c }, {b,d}
3
{a,b }, {b,c}
43,4
{a,b}, {c,d}
5
{a,b,c },
{a,b }, {a,c }, {b,c}
6
{a,b }, {a,c}, {a,d}
7
{a,b}, {a,c }, {b,d}
{a,c }, {a,d }, {c,d}
9
m10
{a,c }, {b,c }, {c,d}
11
{a,c }, {b,d }, {c,d}
$12
{a,b }, {a,c }, {c,d}
{a,c }, {a,d }, {b,c}
1 3
{a,b,c}
S14
{a,d }, {a,b }, {a,c }, {b,c}
.0
15
{a,d }, {b,d }, {c,d}, {a,b}
.4016
{a,d }, {b,d }, {a,b }, {a,c}
.0
17
{a,d }, {b,d }, {a,c }, {b,c}
18
{a,b }, {a,c }, {a,d }, {b,c },
{b },
{b,d}
{c}, {d}
{a,b c}, {a,b,d }, {a,c,d }, {b,c,d}
25
Table II (continued)
4
19
4 20
4
21
4 22
{a,b}, {a,c}
{a,b }, {b,c}
{a,b}, {c,d}
{b,c}, {c,d}
{a,c,d}
{a,b,c},
423
4 25
26
27
4 28
{a,c},
{a,d}
{a,b }, {a,c},
{b,c}
{a,b },
{a,b}, {a,c}, {b,d}
{a,b }, {b,c},
{a,b },
{b,c },
{c,d}
{b,d}
{b,d}, {c,d}
{b,c},
{a,b,d}, {a,c,d}, {b,c,d}
{a,b}, {a,c},
{b,c}
{b,d },
{b,c}
{a,b}, {b,c},
{c,d}
.;,32
{a,b }, {b,c},
{b,d}
433
{a,b}, {a,c},
{b,d}
{a,b },
{c,d}
435
{a,d}, {a,b},
{a,c}
436
437
{a,d}, {a,b},
{b,c}
{a,d},
{c,d}
438
{a,d}, {b,c},
{c,d}
.9
30
31
.134
{a,b,c },
{a,b,d}
439
440
{a,b}, {a,c}, {aid},
{c,d}
{a,b}, {b,c },
{a,c},
{a,d}
4 41
{a,b}, {b,c}, {a,c },
{c,d}
{a,b}, {b,c},
{c,d}
42
{a,d},
26
Table II (continued)
43
JJ
44
45
46
24
47
{ado}, {b,c}, {b,d},
{a,c}
{a,b}, {b,c}, {b,d},
{b,c}, {a,c}, {a,d},
{c,d}
{c,d}
{b,c}, {b,d},
{a,c},
{a,d}
{b,c }, {b,d},
{a,c},
{c,d}
{a,b}, {a,c},
{a,d},
{c,d}
{a,b}, {a,c},
{b,c},
{a,d}
{ado}, {a,c},
{b,c}, {c,d}
{a,c,d}, {b,c,d}
48
49
$50
{b,c},
5l
di,
4;,
do
53
54
56
{c,d}
{a,c}
{a,b }, {b,c },
52
55
.;v'
{a,d},
{ado}, {b,c},
{b,d},
{c,d}
{a,c}, {a,d},
{b,c},
{c,d}
{a,c}, {a,d },
{b,c},
{b,d }.
{b,c}, {b,d},
{a,c},
{c,d}
{a,b}, {a,c },
{a,d},
{b,c},
{a,c}, {a,d },
{b,c },
{a,b },
{a,c},
{a,d},
{b,d},
{c,d}
{a,c }, {a,d},
{b,c},
{b,d},
{c,d}
{a,b}, {a,c}, {a,d},
{b,c},
{b,d}
{b,c},
{b,d },
{c,d}
{a,b}, {a,c}, {a,d },
{b,c},
{b,d}, {c,d}
{b,c,d}
57
58
59
60
61
{a,b }, {a,c},
62
{b,d}
{c,d}
No triple.
63
{c}, {d}
{a,b,c},
{a,c,d }, {b,c,d}
27
Table II (continued)
64
65
66
67
S69
..!)
71
{a,b },
{a,c},
{b,c}
{a,b},
{a,c},
{a,d}
{a,b},
{a,c},
{b,d}
{a,b },
fa,c1,
{c,d}
{a,c},
{b,c},
{c,d}
{a,c},
{a,d},
{c,d}
fa,c1,
{b, d },
{c,d}
{a,c},
{a,d},
{b,c}
{a,b,c}, {a,b,d },
{a,c,d}
S72
{a,b},
{a,c},
{a,d },
S73
{a,b },
{b,c},
{a,c },
S74
{a,b },
{b,c},
{a,c}, {cad}
{a,b},
{b,c},
{a,d}, {c,d}
S76
{a,b},
{b,c},
{b,d },
S77
{a,b},
{b,c},
{b,d}, {c,d}
S78
{b,c },
{a,c},
{b,c},
{b,d},
{a,d}, {c,d}
{a,c}, {a,d}
{b,c},
{b,d},
{a,c },
75
79
80
{a,b,d}, {a,c,d },
{c,d}
{a,d}
{a,c}
{c,d}
{b,c,d}
{a,b},
{a,c },
{b,c}, {c,d}
S82
{a,b},
{a, c },
{aid}, {c,d}
S83
{a,b},
{a, d },
{b,c}, {c,d}
{a,b},
{a, d },
{b,d},
{c,d}
{a,b},
{a, c },
{a,d},
{b,c}
{a,b},
{a,c},
{a,d},
{b,d}
{a,c},
{a,d},
{b,c}, {c,d}
{a,c},
{a,d},
{b,d}, {c,d}
{a, c },
{a,d },
{b,c}, {b,d}
81
84
S 85
86
"87
S 88
89
{a,b,c}, {a,b,d}
90
{a,b}, {a,c },
{a,d},
{b,c},
{b,d}
{b,d},
{c,d}
:Y91
{a,c},
{a,d}, {b,c},
S92
{a,b},
{a,d },
{b,c},
{c,d}
28
Table II (continued)
{a,b,d }, {b,c,d}
.3(
S
93
94
95
{a,b }, {a,c }, {a,d }, {b,c },
tb,d}
{aic}, {a,d}, {b,c }, {b,d }, {c,d}
{a,b}, {a,d },
{b,c }, {b,d }, {c,d}
{a,b,c }, {a,c,d}
96
S 97
{a,b }, {a,c }, {a,d }, {b,c }, {b,d}
{a,b}, {a,c }, {a,d },
{a,b }, {a,c },
S 99
100
5101
{b,c}, {c,d}
{a,d }, {b,d },
{c,d}
{a,b }, {a,c }, {b,c }, {b,d }, {c,d}
{a,b}, {a,d }, {b,c}, {b,d }, {c,d}
{c,d}
.{a,c}, {a,d}, {b,c },
{b,c,d}
I./
102
{a,b },
{a,c }, {a,d }, {b,c }, {b,d }, {c,d}
{a,b,c}
S
103
{a,b },
{a,c }, {a,d }, {b,c }, {b,d }, {c,d}
{a,b,c }, {a,b,d }, {a,c,d }, {b,c,d}
104
S 105
S 106
S 107
{a,b },
{a,d },
{b,d },
{b,d },
{c,d}
{c,d}
fa,b1, {a,c },
{a,b }, {a,c }, {a,d }, {b,d}
{a,b },
{a,c }, {b,c },
{a,d}
{a,b,c }, {a,c,d }, {b,c,d}
5108
109
{a,c }, {a,d }, {b,c }, {b,d }, {c,d}
{a,b }, {a,c }, {bic}, {b,d }, {c,d}
{a,b,c }, {a,b,d }, {a,c,d}
S110
{a,c }, {a,d }, {b,c }, {b,d },
{c,d}
29
Table II (continued)
;
;
111
112
113
{c,d}
{ab}, {a,c }, {b,c },
{a,b}, {a,c}, {a,d},
{b,d },
{a,b}, {a,c}, {a,d},
{b,c}, {b,d}
{b,d}, {c,d}
{a,b,c}, {a,b,d}
;
114
{a,b }, {a,c},
{a,c,d },
;
115
{a,d },
{b,c },
{b,d},
{c,d}
{a,d },
{b,c},
{b,d },
{c,d}
{b,c,d}
{a,b }, {a,c},
No singleton
{a,b,c}, {a,b,d}, {a,c,d}, {b,c,d}
;116
{a,b}, {a,c}, {a,d}, {b,c}, {b,d}
.bl17
{a,b}, {a,c}, {a,d}, {b,c}, {c,d}
;118
{a,b}, {a,c}, {a,d}, {b,d}, {c,d}
{a,b}, {a,c}, {b,c}, {b,d}, {c,d}
119
;120
{a,b}, {a,d}, {b,c}, {b,d}, {c,d}
ii121
{a,c}, {a,d}, {b,c}, {b,d}, {c,d}
{a,b,c}, {a,b,d}, {a,c,d}
;122
{a,b}, {a,c}, {a,d}, {b,c}, {b,d }, {c,d}
30
Chapter IV
Representatives of All Equivalence Classes
Systems
S-10
of
Before stating the contents of Chapter IV we give
three definitions.
Definition 6.
An
system of an
S-10
S-10 family
is the set of all incogruences obtained by setting the
sums of the elements of every two members of the
S-10
family incongruent to each other, cancelling common terms
and discarding the resulting incongruences with zero on
one side.
Definition 7.
Two subsystems of
S-10
systems
are equivalent if one can be transformed into the other by
a permutation on
S.
If we transform the
tem by a permutation on
ing
S-10
family of an
S-10
S, the
S-10 system of the result-
family is identical to the system obtained from
a transformation of the original
permutation on
S.
Thus, an
by a permutation on
an
S-10
S-10 sys-
S
S-10
S-10
into another
system by the same
system is transformed
S-10
system.
Given
system, the equivalence class which contains
this system consists of all transforms of this system
under permutations on
of the 2,002
S-10
S.
Also, each of the
S-10 systems
families is a member of the equivalence
31
classes represented by
families
S-10
systems of one of the
S-10
Hence, there are at most 122 of these equi-
$i.
valence classes.
We will find out below that there are
just 61 of them.
Definition 8.
The standard representative
I.
,
3
of the equivalence class of
1<j<61,
systems which
S-10
containstheS-10systerriof2iiisthesarsternI.of
Table III, where the incongruences of
are given in
3
Table V.
In this chapter we show that the
systems,
I.
3
1<j<61,
form a complete set of representatives of the
equivalence classes of
S-10
systems.
First, we show that every
valent to an
I.
system.
S-10
system is equi-
This is done in Table III.
3
Then,weshowthateveryl.3
system is an
S-10
system.
This is done using Table IV.
Finallyfweshowthatnotwoofthel.3
are equivalent.
systems
This we do by listing them in Table V
in such an order that is apparent that no two are
equivalent.
Now we explain how to read Table III.
The
in the first column,
1<i<122,
represents
anS-10fartlilyasgiverlinTableII.TheI.
3
second column,
1<j<61,
denotes the standard
in the
32
representative from the equivalence class of
systems containing the
S-10
system ofi. The symbol
in the third column denotes the permutation on
transforms the
S-10
representative
I.3 .
S-10
system of
S
that
into the standard
Immediately following Table III is Table IV where
we list the
system
I.,
3
1<i<122,
1<j<61.
corresponding to each
33
Table III
S-10 Systems of Given
S-10
Families
Standard
r epresentative
1
I1
2
17
I
3
4
1
4
6
I
-3,
3
5
6
7
1
1
33
32
121
I
29
130
1
12
X13
14
15
16
S
17
18
1,
S
19
20
I
13
I
1
I
1
1
49
47
48
36
58
4
6
Families
Standard
represeFamilies tative
S -10
Permutation
e
e
.'.21
e
:Y22
17
(b,d)
,i,23
15
e
I
e
(b,d)
e
...
25
I
1
3
5
2
(c,d)
(b,d,c)
e
12
e
e
15
e
15
e
e
S(28
e
128
(b,d)
e
.8,30
134
(a,b,d)
e
.8,31
119
(a,b,d)
131
(b,d)
114
(a,b) (c,d)
18
120
1
Permutation
S-10
32
(c,d)
(a,c)
..*34
126
(b,c,d)
e
.8,35
128
(a,b,c)
7S
112
(b,d)
134
(b,d)
125
(b,c,d)
115
e
(a,b,c)
36
(a,c)
(b,d,c)
e
e
.*
.,.
38
39
.;:*40
117
(c,d)
34
Table III (continued
Standard
represenFamilies
tative
5 -I0
S 41
I
S 42
I
43
I
22
8
Pe rmu-
tation
e
(a,b)
Standard
S-10
r epresenFamilies
tative
Ji
61
S62
I
39
Permutation
(a,c)
242
e
155
e
24
e
S 44
227
e
.4,64
(45
216
(c,d)
S65
I29
(a,b)
S
I
e
$66
I13
e
e
.4,67
1
S 68
233
S 69
232
S70
I
S 71
I
S
46
47
48
S
49
.3,
50
ai51
S 52
53
.3,
54
S 55
.4,
56
I
I
I
I
9
23
45
44
50
136
246
252
251
I
1
36
54
(a,c)(b,d)
(b,d)
63
(a,c,b,d)
(a,b,d)
72
(a,b,d,c)
S73
(a,c). (b,d)
(a,b,d)
S 74
75
(b,d)
(a,b,d,c)
'776
,:*
57
:4,
58
S59
I
I
60
237
I
4°60
57
41
230
(a,d,b)
(a,c,d)
e
(a,b,c)
77
18
21
20
(c,d)
(a,b) (c,d)
(c,d)
(a,b)
e
(a,b)
227
e
223
(c,d)
222
(c,d)
1
1
I
8
16
15
S 78
I
S 79
1
s 80
217
24
9
(a,b)
e
e
(c,d)
e
e
35
Table III (continued
S-10
Familie
.b
81
82
83
84
85
86
87
ji
88
89
,L4
90
91
92
93
94
95
96
97
98
99
100
101
s
Standard
representative
I
I
45
50
1
I
I
36
52
44
I
1
1
I
I
I
I
I
I
I
1
I
I
46
51
53
36
Permutation
(c,d)
(c,d)
(c,d)
(c d)
I
102
.V
103
'104
.4$
(a,b,c,d)
105
106
(a,c,d,b)
107
(a,c,d)
.6,
(a,c,d)
109
110
10
56
e
(d,c)
(a,c)
'1112
6113
S
114
61
6115
59
'116
38
43
40
38
43
I
1
I
1
1
I
(a,b,c,)
117
(a,b)
(b,c)
118
.6,
119
120
(a,b)
(a,b,c)
121
.6(122
Permutation
(ab)
55
35
49
36
48
47
60
(b,c)
(a,c)
(a,d,c,b)
57
139
I
11
11
1
I
(b,c,d)
140
I
Standard
S-10
represenFamilies
tative
(a,c)
37
142
1
I
1
1
1
1
1
I
41
35
55
58
58
58
(a,c)
(a,b,c,d)
(a,b,d)
(a,b,d,c)
58
58
158
I
(a,b,c)
55
(c,d)
(b,d,c)
36
Table IV
S-10 Families of Given S-10 Systems
Standard
Representative
S-10
families
I
I
1
1
Standard
Representative
20
S13, 71
S
25, 26
2
I
S
3
S
5,
S
S
I4
221
19
I
S
23
224
I6
S
I
S2
17
18
1
9
I
I
1
1
1
S
4,
11
S46,S79
S
90, 91
S
12
S
11,
S
66
14
33
15
'Y39,*Y77
16
S
45,
217
18
19
S
1
I
I
1
36
I
27
28
S 12, S 6
s
38
S3
I
I
I
1
S 44, 572
S
S
29, 35
S
39
65
9,
S
10,
31
S
7,
S
S
34
35
S
64
32
32
6,
30,
569
S
S
68
37
103, 114
S17, S51,
36
S89,105
37
459,4111
S96, S100
7
31
4
29
s
40,80
S
43, 78
25
133
76
S
S
47, 73
226
I
592
S
I
20
S 2, S 22
S 42, S 75
10
113
1
24, 27, 28
70
22
SSS
3,
S
'p8,
S41,S74
21
25
I
S-10 families
38
S 83,
37
Table IV (continued)
Standard
Representative
I
1
I
1
39
S-10
Families
S
61 ,
110
40
S98 S99
41
S60 1113
112
42
62
143 3
97
I44
49
85
145
48
S 81
I
1
1
1
I
1
1
I
I
I
47
86
,115
107
48
:4,16 $106
49
-114 "v104
50
4'50,''82
51
5 4
S87
52
'°53,-184
53
,h,88
54
4v56
55
I56
1
Sl01
46
57
X63
63
Standard
Representative
102,S115, i/ 122
93
109
S-10 Families
'*1
1
58
'1116 ,S117 ,
S118,S120,121
159
I
1
60
61
95
S
S
58, 108
wh(
94
38
Note that the representative of the equivalence
class of
systems corresponding to S5
S-10
which is
used in the partial proof of the Fundamental Theorem is
the standard representative
13.
Note also that equivalent
equivalent
S-10 families have
systems while nonequivalent
S-10
families may or may not have equivalent
S-10
system.
S-10
See Table IV.
Next we introduce Table V.
First, we give several
definitions, next we explain how to read Table V, then
we give the table, and finally we describe the ordering
of the
I.
systems of Table V that makes it apparent
3
that no two of the systems are equivalent.
Definition 9.
An incongruence of type
sometimes simply called a
q > 1,
[p,q],
ruence whose left side is the sum of
whose right side is the sum of
for which no element of
S
q
p
[p,q], p > 1,
is an incongelements of
elements of
S,
S,
and
appears as a term in both
sides.
An
types
S-10
[3,1],
system consists of incongruences of
[2,2], and
[2,1],
as may be seen in
Table V.
Definition 10.
from an
S-10
same left side.
A pair of incongruences of type
family is two such in
[2,1]
with the
39
Definition 11.
from an
S-10
A single incongruence of type
[2,1]
family is an incongruence that does not
belong to a pair.
Definition 12.
The index of an
S-10
system is an
ordered quadruple whose first, second, third and fourth
entries are the number of incongruences of type
incongruences of type
type
[2,1],
[3,1],
[2,2], pairs of incongruences of
and singles of incongruences of type
[2,1],
respectively.
In Table V the heading for each system is the
symbol
I.,
1<j<61,
followed by the index.
The incong-
ruences of the system are grouped into incongruences of
the same type, with incongruences of type
[3,1]
appear-
ing first, incongruences of type
[2,2]
and incongruences of type
appearing at the end.
In turn
[2,1]
[2,1]
appearing second,
incongruences which form a pair are
grouped together.
40
Table V
Incongruences of Standard Representatives
I
1
(4,0,2,0)
12 (3,1,2,2)
a+b+cEd
a+b+dEc
a+c+dEb
b+c+dEa
a+btc
a+b d
c+dta
c+d b
1
3
(3,1,2,0)
a+b+cEd
a+b+dEc
a+c+dEb
a+b+cEd
a+b+dEc
a+c+dEb
a+btc+d
a+btc+d
a+btc
a+b c
a+bEc
a+bEd
a+dtb
a+d c
c+dta
c+d b
b+ctd
c +db
1
4
(3,0,4,0)
1
5
(3,0,3,3)
16 (3,0,3,1)
a+b+cEd
a+b+dEc
a+c+dEb
a+b+cEd
a+b+dEc
a+c+dEb
a+b+cEd
a+b+dEc
a+c+dEb
a+btc
a+b d
a+btc
a+b d
a+btc
a+b d
a+ctb
a+c d
a+4b
a+cd
a+dtb
b+dta
b+d c
a+dtb
a+d c
c+dta
c+d b
b+ctd
b+ctd
c+dta
c+d b
b+dtc
c+dtb
a+4c
41
Table V (continued)
1
(3,0,2,2)
7
a+b+cEd
a+b+dEc
a+c+dEb
a+btc
a+b d
a+cEb
a+cEd
b+dtc
c+dtb
10
(2,2,1,4)
(2,2,2,2)
1
8
1
9
(2,2,2,2)
a+b+cEd
a+b+dEc
a+b+ctd
a+b+d c
a+bEc+d
a+cEb+d
a+c b+d
a+d b+c
a+b c
a+bEd
a+c b
a+c d
a +cd
a+d b
a+d c
b+d ta
b+d c
c+dta
11
(2,2,1,4)
b+c*d
b +dc
1
12
(2,1,4,0)
a+b+ctd
a+b+dtc
a+b+cEd
a+b+dEc
a+b+cEd
a+b+dEc
a+btc+d
a+c b+d
a+c b+d
a+d b+c
a +bc +d
a+bEc
a+bEd
a+b c
a+b c
a +cd
a+ctd
a+dtc
a+dtc
b+ctd
b +dc
b+ctd
b+dtc
a+btc
a+b d
a+dtb
a+d c
b+c a
b+c d
c+d a
c+d b
42
Table V (continued)
1
13
(2,1,4,0)
1
14
(2,1,4,0)
1
15
(2,1,3,3)
a+b+cfd
a +b +cd
a+b+4c
a+b+dEc
a+b+ctd
a+b+dtc
a+ctb+d
a+ctb+d
a+btc+d
a+btc
a+b d
a+ctio
a+ctca
a+b*c
a+ctb
a+c d
a+dtb
a+4c
a+ctb
a+cd.
b+d
b+d
c+d
c+d
b+cta
b+ctd
a+dtb
a+dtc
b+dta
b+dte
b+ctd
a
c
a
b
a+W
b+dtc
c+dtb
1
16
1
(2
'
3)
3
'
'
1
17
(2,1,3,2)
1
18
(2,1,3,1)
a+b+ctd
a+b+d c
a+b+ctd
a+b+dtc
a+b+cr
a+ctb+d
a+ctb+d
a+btc+d
a+bte
a+1 4d
a+btc
a+bfd
a+btc
a+b d
a+ctb
a+c d.
a+ctb
a+ctd
a+ctb
a+c d
a+dtb
a+dtc
a+dr-b
b+dtc
b+ctd
b+dtc
b+ctd
c+dtb
a+d=c
b+dtc
a+b+d c
c+dta
c+d b
43
Table V (continued)
1
19
(2,1,3,1)
1
a +b +cd
1
22
20
1 3
(2
'
'
1)
1
'
21
(2,1,3,0)
a+b+dtc
a+b+c#d
a+b+dtc
a+b+c=d
a+b+d=c
a+ctb+d
a+ctb+d
a+ctb+d
a+btc
a+b*d
a+c#b
a+ctd
a+1:4c
a+c=b
a+c=d
a+d=b
a+d=c
a+ctb
a+ctd
b+dta
b+idta
b+ctcl
b+dtc
b+4a
c+dta
b+dtc
(2,1,2,4)
1
23
(2,1,2,4)
a+btd
b+4c
1
24
(2,1,2,3)
a+b+cid
a+b+d c
a+b+ctd
a+b+d c
a+b+ctd
a+b+d c
a+btc+d
a +cb +d
a+ctb+d
a+btc
a+b d
a+b c
a+b d
a+btc
a+b d
a+ctio
a+d
4
a+c b
a+c d
a+ctb
a+c d
a+dtc
a+dtc
a+dtc
b+ctsd
b+ctd
b+dtc
c+dtb
b+ctd
b +dc
c+d*b
b+dtc
44
Table V (continued)
I
25
(2,1,2,2)
I
(2,1,2,2)
I
27
"
(2 1 1 5)
a+b+ctd
a+b+d c
a+b+ctd
a+b+d c
a+b+ctd
a+b+d c
a+btc+d
a +cb +d
a+btc+d
a+ctb
a+c d
a+btc
a+b d
a+dtc
a+ctsd
b+ctd
a+dtc
b+dta
b+d c
b+cd
b+dc
c+db
a+bt c
a+bfc
a +cd
b +dc
c+dta
c+d b
I
26
28
(2
'0 5" 1)
I
29
(2
"
4
2)
'
I
30
(2,0,4,2)
a+b+ctd
a+b+d c
a+b+c d
a+b+d c
a+b+ctd
a+b+d c
a+btc
a+b d
a+btc
a+bfd
a+btc
a+b d
a+c b
a+c d
a+ctb
a+c d
a+clb
a+cEd
a+d-b
a+d c
a+d.tb
a+dtc
b+cla
b+c d
b+d a
b+d c
b+ctd
b+dtc
c+dta
c+d b
c+d4
a+dfc
b+cta
b+cfd
b +dc
c+dta
c+d b
45
Table V (continued
1
31
(2,0,3,3)
32
(2,0,3,2)
1
33
(2,0,3,2)
a+b+ctd
a+b+d c
a+b+ctd
a+b+d c
a+b+cEd
a+b+dEc
a+btc
a+b d
a+btc
a+b d
a+lotc
a+ctb
a+ctb
a+c d
a+ctb
a+c d
a+d tb
a+d c
a+dtc
a+d tb
a+d c
b+cta
b+c d
b+dtc
c+dtb
1
1
34
(2,0,1,5)
a+b d
b+ctd.
b+cta
b+c d
b +dc
b+dtc
I 35
(1I 3 1 14)
I
36
"
(1 2 4 0)
a+b+ctd
a+b+d c
a+b+ctd
a+b+ctd
a+btc
a+bfd
a+b-c+d
a+cEb+d
a+dEb+c
alc+d
a+ctd
a+c b+d
a+btc
a+b d
a+dtc
a+bEc
a+bEd
b+ctd
a +cd
b+dtc
a+dtc
c+dta
b+ctd
b+dta
b+d c
b +dc
c+dta
c+d b
a+ctb
a+cfd
46
Table V (continued)
1
37
(1,2,3,3)
1
(1,2,3,3)
1
39
(1,2,3,2)
a +b +cd
a+b+ctd
a+b+ctd
a+btc+d
a+c b+d
a+btc+d
a+ctb+d
a+btc+d
a+ctb+d
a+btc
a+b d
a+btc
a+b d
a+btd
a+ct b
a+c d
a+ctd
a+ctb
a+c d
b+c d
b+cta
b+c d
b +da
b +dEc
b+dta
b+d c
c+dta
c+d b
c+dtb
c+dta
a+d*c
a+sdtb
b+c ta.
a+d c
b +cd
1
38
40
(1,2,2,4)
1
41
(1,2,2,4)
1
42
(1,2,2,3)
a+b+ctd
a+b+ctd
a+b+ctd
a+btc+d
a +cb +d
a+blc+d
a+c b+d
a +bc
a+b:c
a+b d
a+btc
a+b d
a+ctb
a+c d
a+ctd
+d
a+c b+d
a +b c
a+btd
a+dtc
a+ctb
a+c d
b+ctd
b+ctd
c+dta
b+dtc
b+dtc
b+la
c+dtb
c+dtb
b+d c
a+dtb
b+ctd
47
Table V (continued)
1
43
(1,2,2,3)
1
44
(1,1,5,1)
I
45
"
(1 1 4 2)
a+b+ctd
a+b+ctd
a+b+ctd
a+4c+d
a+4b+d
a+btc+d
a+btc+d
a+btc
a+b c
a+bEc
a+bEc
a+c:b
a+c d
a+cEb
a+cEd
a+dtb
a+d c
a+dtc
a+btc
a+b d
a+ctd
b+ctd
b+dta
b+d c
c+dta
b+cta
b+c d
b+dta
c+dt a
c+d b
146
(1,1,4,2)
1
47
,,,
(1,1,4,2)
b+cEa
b+cEd
b+dtc
c+dta
c+d b
148
(1,1,4,1)
a +b +cd
alb+ctd
a+b+ctd
a+btc+d
a+btc+d
a+btc+d
a+btc
a+bfd
a+btc
a+b c
a+btc
a+ctb
a+c d
a+ctd
a+ctd
a+dtc
a+dtb
a+d c
b+ctcd
b+ctd
b+d a
b+d c
b+d a
b+d c
b+cta
b+c d
b+dta
b+d c
c+d a
c+d b
c+d a
c+d b
a+1:4d
c+dEa
c+dEb
48
Table V (continued)
I
49
"
(1 1 4 0)
150 (1,1,3,3)
a+b+ctd
a+b+ctd
a+b+ctd
a+btc+d
a+btc+d
a+btc+d
a+b c
a+b d
a+btc
a+b d
a+btc
a+b d
a+c b
a+c d
a+ctd
a+ctd
a+dtc
a+dtb
b+dta
b+cta
b +c d
b+c a
b+c d
b+c a
b+c=d
c+dEa
c+d-b
c+d a
c+d b
1
151 (1,1,3,2)
52
(1,1,2,4)
1
53
(1,1,2,3)
c+dta
c +d b
1
54
(1,1,2,2)
a+b+ctd
a+b+ctd
a+b+ctd
a+btc+d
a+btc+d
a+btc+d
a+btc
a+b d
a+iptc
a+1 *1
a +cd
a+ctd
a+btc
a+b d
a +cd
a+catb
b+ctd
a +dc
b+ctd
b+dtc
c+dta
b+dtsa
c+dta
c+4b
c+d t a
c+d b
c +d b
49
Table V (continued)
I
55
(0,3,3,3)
a+b c+d
a+c=b+d
a+dEb+c
a+btc
a+b d
a+ctb
a+c d
a+lb
a+d c
b+ctid
b+dtc
c+dtio
258 (0 2 5 0)
' "
I
56
(0,2,6,0)
1
a+bEc+d
57
(0,2,5,1)
a +c ±b +d
a+btc+d
a+c b+d
a+btc
a+b d
a+btc
a+b d
a+cEb
a+cEd
a+ctb
a+c d
a+dtb
a+d c
a+dtb
a+d c
b+cta
b+c d
b+dt a
b+d c
b+ctd
b+dta
b+d c
c+dta
c+d b
c+dt a
c+d b
I 59
(0,2,4,.2)
I
60
(0,2,4,1)
a+btc+d
a+ctb+d
a+btc+d
a+d b+c
a+btc+d
a+ctb+d
a+btc
a+b d
a+btc
a+b d
a+btc
a+b d
a+ctb
a+c d
a+ctb
a+dEb
a+dEc
b+dta
b+d c
c+dta
c+d b
a+dtb
a+d c
b+cta
b+c d
b+dtc
c+dta
c+d b
a+ctb
a+c d
a+dtb
b+dta
b+d c
c+dta
c+d b
50
Table V (continued)
1
61
(0,2,4,0)
a+ctb+d
a+d b+c
a+ctb
a+c d
a+dtb
a+catc
b+cta
b +c d
b+dta
b +d c
51
The way we obtain and order the
given in Table V is use the
S-10
systems
S-10 systems of the families
order the equivalence classes of
1<i<122,
I.
S-10
systems to which they belong, discard duplicates, and
then identify a certain representative from each of the
remaining equivalence classes which we have called the
standard representative
I.3 .
First, we partition the original 122
S-10
systems
by index and call the parts obtained parts of rank one.
We order the parts of rank one in a lexicographically decreasing manner on the index.
The
systems from
S-10
two different parts of rank one are not equivalent to each
In case a part of rank one consists of only one
other.
S-10 systems
this
S-10
standard representative
system is taken to be its
I.3 .
Next, we partition the parts of rank one with more
than one
S-10
system according to the structure of the
system of incongruences of types
[3,1]
and
parts obtained are called parts of rank two.
rank one with more than one
S-10
with only one incongruence of type
is performed on
formed into
S
[2,2].
The
If a part of
system has a system
[3,1],
a permutation
so that this incongruence is trans-
a+b+ctd.
If there are two
[3,1]'s,
a
52
permutation is performed on
a+b+ctd
formed into
and
so that they are trans-
S
If there are three
a+b-i-dtc.
[3,1] 's, a permutation is performed on
are transformed
into
In case there are no
a+b+0c,
a+b+ctd,
or four
[3,1] 's
S
not perform any permutation on
so that they
a+c+Ob.
and
[3,1] 's,
we do
S.
Once the above procedure is completed on each of
the parts of rank one that have more than one
5 -10
system, each of these parts consists of
systems
with identical
[3,1] 's.
Now, we consider the
analysis a permutation
system of
S-10
[3,1] 's
P
from an
In the following
[2,2] 's.
on
is allowed only if the
S
S-10
part of rank one is invariant under
system belonging to a
P.
If a part of rank one with more than one
S-10
system has a system with only one incongruence of type
this incongruence is transformed, if possible,
[2,2],
into
a+btc+d
by a permutation on
transformed, if possible, into
on
S.
Otherwise, it is
a+ctb+d
by a permutation
If this also is not possible the incongruence is
S.
a+dtb+c.
If a part of the above kind has an
where there are two incongruences of type
S-10
[2,2],
incongruences are transformed, if possible into
system
these
a+btc+d
53
and
a+cOp+d
by a permutation on
S.
are transformed, if possible, into
by a permutation on
are no
and
or three
[2,2]'s
permutation on
a+btc+d
and
a +d4b +c
If this also is not possible the
S.
a+ctb+d
incongruences are
Otherwise, they
a+dtb+c.
[2,2]'s
In case there
we do not perform any
S.
We have completed the second partitioning. A part
of rank two consists of
systems from a part of rank
S-10
one which have identical subsystems of
[3,1]
incongruences after the permutations on
and
[2,2]
mentioned in
S
the preceeding paragraph have been performed.
Next, we order the parts of rank two.
The parts of rank two, that contain one subsystem
of incongruences of type
[2,2]
are ordered in such a way
that a part that contains systems with incongruence
a+btc+d
ruence
preceds a part that contains systems with incong-
a+cb+d,
which in turn preceds a part that con-
tains systems with incongruence
a+dtb+c.
Similarly, the
parts of rank two that contain two subsystems of incongruences of type
[2,2]
are ordered in such a way that a
part that contains systems with incongruences
and
a+ctb+d
incongruences
a+btc+d
precedes a part that contains systems with
a+btc+d
and
a +db +c,
which in turn pre-
cedes a part that contains systems with incongruences
a+ctb+d
and
a+dtb+c.
54
In case a part of rank two consists of only one
S-10
system,
this
S-10
system is taken to be its
standard representative.
We now complete the process of ordering and identifying the
systems of each part of rank two, with
S-10
more than one system; these systems that have the same
index, the same
[3,11's
and the same
this purpose we label the 12 possible
[2,2]'s.
For
[2,1]'s of an
S-10
system of a part of rank two by increasing digits base 13
as follows:
Label
Incongruence
1
a+lotc
2
a+14d
3
a+ctho
4
a+ctid
5
a+dtlo
6
a +dEc
7
b+cta
8
b+ctcl
9
b+dta
a
b+dtc
c+dta
y
c+idta
55
Corresponding to each subsystem of incongruences
of type
there is a number that will be called a
[2,1]
This tag number is formed by increasing
tag number.
digits, where the digits are the labels of the
congruences.
system
a+btd,
c+dtb;
Next, we perform a permutation on
the
S-10
[3,1]'s
is
S
1278.
that trans-
system of each part of rank two, leaving
and the
number is minimized.
S-10
in-
For instance, the tag number of the sub-
a+btb+c,
forms the
[2,1]
invariant so that the tag
[2,2]'s
Once this is done, we order the
systems in each part of rank two in a lexicographi-
cally increasing manner on the tag numbers.
shows that no two of these
to each other.
S-10
Hence, these
This ordering
systems are equivalent
S-10
systems with smallest
tag number are the standard representatives,
of
Table V.
In Table VI that follows we list the
I.
S-10
systems
in each of the parts of rank two with more than one
3
S-10 system.
Included in the table are their indexes,
their incongruences of type
missible permutations on
and
5,
[2,2],
and the so called per-
that leave the
[3,1]'s,
[2,2]'s invariant, used to get the smallest tag
numbers.
56
Table VI
S-10 Systems in Each Part of Rank Two
With More Than One 5 -10 System
.
Incongruences
Permissible
Permutations
(2,1,4,0)
a+c t b+d
(a,b)(c,d),e
(2,1,3,1)
a+c t b+d
(a,b)(c,d),e
Index
I.
.1
I
13?
119'
I
29'
I
I
I
14
20
30
(2,0,4,2)
[2,2]
None
(a,b)(c,d),
(a,b) (c,d) ,e
1
321
1
33
(2,0,3,2)
None
(a,b),(c,d),
(a,b) (c,d) ,e
1
371
1401
1
I
45
42'
,I
1
I
1
38
41
43
1
46' 47
(1,2,3,3)
a+b*c+d,a+cb+d
(b,c), e
(1,2,2,4)
a+btc+d,a+ctb+d
(b,c), e
(1,2,2,3)
a+btc+d,a+ctb+d
(b,c) ,
(1,1,4,2)
a+b t c+d
(a,b), e
e
57
An alternating ordering of all the equivalence
classes of
of
the standard representatives
Table V
3
which makes clear that they are pairwise not equivalent
could have been obtained by a single lexicographical
ordering similar to the one used to order the standard
representatives
I.3
of Table VI.
However, the ordering
that we have chosen gives more insight into the structure
ofthesystemsI.,and it is obtained more easily.
58
Epilogue
We give several problems to which one is lead by
this thesis.
There are 61 entries in Table III, exactly half
the number of entries in Table II.
An interesting prob-
lem is to determine if this is a special case of a more
general theorem for families of subsets and the corresponding system of incongruences.
A further problem is
to obtain the relationship between the total number of
families and the total number of corresponding systems.
This latter problem has not even been solved for
families and
S-10
S-10
systems.
A different problem is to determine if any of the
61 systems of Table III lead to a variation in the proof
of the Fundamental Theorem, and if they do, if the proofs
are simpler.
Also, a problem is to find a new proof of the
Fundamental Theorem that will provide a method of evaluating the function
arbitrary
k > 5.
14(k,m)
for
k=5,
or perhaps even for
59
Bibliography
(1)
Not yet published proof of the Fundamental
Theorem, by Robert D. Stalley, Oregon State
University.
(2)
Unpublished notes by Robert D. Stalley.