Download D.K. Ray Chaudhuri; (1959)On the application of the geometry of quadrics to the construction of partially balanced incomplete block designs and error correcting binary codes."

ON THE APPLICATION OF THE GEOMETRY OF Q,UADRICS
TO THE CONSTRUCTION OF PARTIALLY BALANCED
INCOMPLETE BLOCK DESIGNS AND ERROR
CORRECTING BINARY CODES
by
D. K.Ray-Chaudhuri
Un~ersity
of
North~arolina
This research was supported by the United
States Air Force through the Air Force
Office of Scientific Research of the Air
Research and Development Command, under
Contract No. AF 49(638)-213. Reproduction
in whole or in part is permitted for any
purpose of the United States Government.
f
~,
'.,.
Institute of Statistics
Mimeograph Series No. 230
June, 1959
.
•
ACKNOWLEDGMENTS
It is a great pleasure to express my deep gratitude to
Professor R. C. Bose, for suggesting the problem herein considered
and providing guidance and encouragement during the course of my
investigation.
My thanks are due to the Un!ted States Air Force for its
financial assistance during my stay in Chapel Hill.
To Miss Marianne Byrd" Miss Jane Rogers and Mrs. Ouida
Taylor I am extremely grateful for their skillful and quick
.
typing of the manuscript.
.e
for a variety of forms of aid •
To Miss Martha Jordan I extend thanks
Finally, I wish to express my deep indebtedness to
Professor S. N. Roy whose fascinating lectures in the Indian
Statistical Institute at Calcutta in the year 1956 inspired
me to come to Chapel Hill for higher studies and research in
Statistics •
•
e·
ii
..
TABLE OF CONTENTS
.
·.·........ ......
· ........ ·....
• • • · . ·. . .. • • · . · . . • •
· . ..
Acknowledgments • •
Introduction. •
Notation• • •
,
•
ii
v
ix
CHAPTER
I.
SOME PRELIMINARY RESULTS ON THE GEOMETRY OF QUADRICS IN
FINITE PROJECTIVE SPACE • • • • • • • • • • • • • • • •
1. Summ.ary. • • • • • . • . . . • • . • • .
·.
·.·... • •
·· ·• • •
• ·
4. Stereograpbic projection and its use • • • • • • • • •
·
5. Linear spaces contained in a nondegenerate quadric ~
in PG(n, s) • • • • • • • • • • • • • • • • • • • • •
·•
6. Canonical forms of quadrics. . • • • •
• • • · • ·
·
·
·
m
7. Nucleus of polarity of a quadric in PG(2k, 2 ) • • • ·
·
2. Quadrics in finite projective geometry • •
•
3. Conjugate points, polar space and tangent space.
.
II.
1
1
1
7
21
26
29
34
SOME CLASSES OF PBIB DESIGNS WITH TWO ASSOCIATE CLASSES
OBTAINED FROM THE CONFIGURATION OF LINEAR SPACES CONTAINED
IN A QUADRIC. • • • • • • • • • • • • • • • • • • • • • • •
1. Summ.ary• • • • • • • • • •
2. Introduction
..·.•
·.·..
•
• •
• • • • • • • • • • • • • • • • •
38
38
39
3. PBIB designs from the configuration of generators for
blocks and points for treatments of a quadric. • • • • •
4. PBIB designs from the configuration of points of a
quadric for blocks and generators of a quadric for
treatments • • • • • • • • • • • • • • • • • • • •
5. PBIB designs from the configuration of generators
on generators. • • • • • • • • • • • • • • • • • •
·..
·. .
60
6. PBIB designs from the configuration of lines and points
•
of PG( 3, s) truncated by a quadric.
••
64
7. Concluding remarks • • • • • • • • • • • • • • • • • ••
68
iii
• • • • •
iv
•
CHAPTER
Page
III. SOME CLASSES OF PBIB DESIGNS WITH THREE ASSOCIATE CLASSES 69
1. S'l1II11D.ary. • • • • • • • • • • • • • • • • • • • • •
69
2. A theorem on three associate PBIB designs. • • • • ••
70
3. Some PBIB designs with three associate classes obtained from the configuration of' generators and
points of a cone • • • • • • • • • • • • • • • • •
74
4. Some PBIB designs with three associate classes obtained from the configuration of secants and external
points of a quadric. • • • • • • • • • • • • • • • ••
84
IV.
•
.e
..
A CLASS OF THO ERROR CORRECTING CODES WITH RATE OF TRANSMISSION ARBITRARILY CLOSE TO UNITY AND FRACTIONAL REPLICATIONS PRESERVING MAIN EFFECTS AND TWO FACTOR INTERACTIONS 94
1. Summary. • • • • • • • • • • • • • • •
94
2. General problem of information theory.
• • • • ••
95
3. Binary channel • • • • • •
• • • •
98
4. Statement of the problem
• • • • • • • • • • • • • 100
5. Some prelimary results on group codes. • • • • • • • • 103
6. Relationship between error-correcting binary group
codes and fractional replications of factorial experiments at two levels. • • • • • • • • • • • • • • • 106
7. A correspondence between the points of PG(n,2) and
the elements of GF ( 2n+l) • • • • • • • • • • • • • • • 109
8. An R4-set in PG(2m-l, 2) containing (2m - 1) points
and a sequence of two error correcting codes with
asymptotic rate of transmission equal to unity • •
111
9. An R4-set in PG(2m, 2) containing ~ + N4(m - 1)
points and a sequence of two error correcting codes
with asymptotic rate of transmission equal to un!ty. • 115
10. Examples illustrating the method of constructing
R4-sets. . . . . . . . . . . . . . . . . . . . . . . .
BIBLIOGRAPHY • • • •
.......
..
..
121
124
INTRODUCTION
The theory of linear spaces in finite projective geometry has
been used very profitably by several authors in solving combinatorial
problems of statistical interest.
The properties of linear spaces
have been used for the construction of
designs (Bose
IJ.J),
(ii) partially balanced incomplete block (PBIB)
[JJ),
designs (Bose and Nair
(Bose and Nair
(i) balanced incomplete block
L-a.J),
(iii) set of orthogonal latin squares
(iv) designs with block confounding or fractional
replication which preserve all main effects and interactions up to a
certain order (Bose
•
[3.J) and
(v) orthogonal arrays.
Bose I)J.ofirs-t
used the properties of quadric surfaces in finite projective geometry
of two and three dimensions for constructing e:x;perimental designs.
Primrose
iJ.g* studied some properties
of a quadric in a general finite
projective space and used them to construct series of balanced incomplete
block designs Which, however, do not include any design with parameters
in the practical range.
The works of Bose and Primrose suggested that
a detailed study of quadrics in a general finite projective space will
be very useful for solving combinatorial problems of statistical origin.
In this thesis an attempt is made to study the quadrics in finite
projective space systematically and the results so obtained in the theory
of quadrics are applied for the construction of PBIB designs.
A detailed
summary of the work done in each chapter is given in the beginning of
*The numbers in square brackets refer to the bibliography listed
at the end.
v
vi
that chapter.
Below we give only a very brief summary of the work done
in various chapters of the thesis.
In chapter I the theory of quadrics is dealt with.
results have been obtained.
Several new
The general formula for the number of
p-flats contained in a nondegenerate quadric PG(n,s), the finite projective geometry of n dimensions based on a Galois field GF(s) , is
obtained for the elliptic as well as the hyperbolic quadric.
The canon-
ical forms for elliptic and hyperbolic nondegenerate quadrics are obtained.
The polar of a k-flat with respect to a nondegenerate quadric
is defined and various properties of the polar spaces are derived,
The
properties of the nucleus of polarity of a nondegenerate quadric in
m
PG(2k,2 ) are studied.
In chapter II several series of PBIB designs with two associate
classes are given.
PBIB designs were introduced by Bose and Nair
C7J
to fulfill the need of incomplete block designs which do not require too
many replications.
PBIB designs have been used very profitably in ex-
perimental situations when the complete blocks cannot be used and also
no balanced incomplete block design exists for the particular parameters
C7J, Bose [4J, Bose
and Clatworthy L-5J, Bose and Shimamato C9J, Bose, Shrikhande and
Bhattacharya fJ.91, Bose Clatworthy and Shrikhande C6J J. Roy and
of interest.
Various authors (Bose and Nair
1
R. Laha
1
[2'§, and many others) have made an almost exhaustive treatment
of the problem of construction of PBIB designs with two associate
classes with parameters in the practical range.
In chapter II we have
vii
given a very general method of constructing PBIB designs using classes
of sets in PG(n , s) • A series of two associate PBIB designs is obtained from the configuration of generators and points of a nondegenerate quadric taking generators for blocks and points for treatments.
Three other series of two associate PBIB designs are obtained.
series contain many designs with r
These
and k not greater than 10.
Of
these designs several are new and the rest are already obtained by other
authors by different methods.
In chapter III two series of PBIB designs with 3 associate classes
are given,
•
These two series are obtained from the configuration of
generators and points of a cone and the configuration of secants and
external points of a quadric in the finite projective plane.
These
two series contain 8 new designs with rand k not greater than 10.
In
chapter IV we have considered the problem of construction of
error correcting binary codes which is also a combinatorial problem of
statistical interest.
L"2'f1 which
16,
217
From the general existence theorems of Shannon
are completed and extended by various authors ["1, 18, 20,
it is known that under certain conditions it is possible to find
a method of encoding which transmits information through a noisy channel
with probability of correct transmission of the message arbitrarily close
to unity and also with a rate of transmission arbitrarily close to the
capacity of the channel.
However 1 no method is known for actual construc-
tion of codes with the reqUired characteristic.
Slepian
~-V
considers
the problem of construction of (n,k) binary group codes with n places
viii
and k information places which maximizes the probability of correct
transmission of the message.
values of n and k.
Slepian's problem for
Slepian solved his problem only for small
Knebler
k
LI<jJ obtained a general solution of
= 3 and 4.
However, the codes obtained by
Slepian and Knebler have poor rate of transmission.
In this chapter we
have obtained a class of two error correcting binary group codes with
rate of transmission arbitrarily close to unity.
Denoting by Nt(m)
the maximum number of points that can be packed in PG(m,2)
no t
such that
of the points lie in a (t - 2)-flat, it is shown that at-error
correcting (n,k) group code exists if and only if Nt(n - k - 1) ~ n •
..
Two error correcting codes are constructed,for (i) n
m
m
k = 2 - 1 - 2m , (ii) n = 2 + N4 (m - 1), k = ~ +
A theorem is proved showing that a ~ fraction of a
= ~ - 1,
N4 (m- 1) - 1 - 2m •
2n e:x;periment
2
preserving main effects and interactions up to (t - l)th order exists
if and only if a t-error correcting (n,k) binary group code exists.
ix
NOTATION
standard Symbol
u
Meaning of the Symbol
Union (of sets)
n
Intersection (of sets)
c
'is a subset of'
e
'belongs to'
'does not belong to'
'implies'
'is not a subset of'
the set containing the single point P
EG(n"s)
Finite Euclidean geometry based on
a Galois field GF(s)
PG(n"s)
Finite projective geometry based on
a Galois field GF(s)
CHAPTER I
SOME PRELIMINARY RESULTS ON THE GEOMETRY OF
QUADRICS IN FINITE PROJECTIVE SPACE
1, Summary.
The theory of quadrics in finite projective geometry is found
to be very useful in the study of combinatorial problems of statistical
interest.
In this chapter a brief systematic treatment of the theory
of quadrics in finite projective geometry is presented and some new
results in the theory of quadrics are derived.
the polar spaces are proved.
Several properties of
The explicit formulae for the number of
p-flats contained in nondegenerate quadrics in pG(n,s) , the finite
projective geometry of n dimensions based on a Galois field GF{s),
are obtained.
The canonical forms for the elliptic and hyperbolic
nondegenerate quadrics and several properties of the nucleus of polarity
of a nondegenerate quadric in PG(2k, 2m) are given. The results obtained in this chapter are used in the later chapters to construct
PBIB designs and error-correcting codes.
2. Quadrics in finite projective geometry.
Quadric
A quadric
Q in PG(n,s), the
dimensions based on a Galois field
is the set of all points
equation
x'
= (x',
o
ti~te
projective geometry of n
GF(s) ,where
xl', ••• ,
x')
n
s
is a prime power,
which satisfies the
2
n
(1)
where
E a, .xix,
j:::i=O ~J
J
aij's are elements of GF(s)
=
0
and the operations of addition
n
and multiplication are in GF(s)
The expression
E aijx,xj
j:::i=O
~
to be the quadratic form. or, in short, the form. of' Q.
teristic of the field
GF(s)
equation of any quadric
If the charac-
is not 2, then it is possible to write the
Q as
n
E
(2)
is said
i,j=O
ai,x,xj
J ~
where
i,j
=0
= 0,1, •.• ,n
In this case in matrix notation the equation of Q can be written as
x'
.e
A
x
=0
(1 x n+l) (n+l x n+I) (n+l x 1)
where
A is a symmetric matrix.
However, if the characteristic of the
field is (2), the equation of Q cannot always be written in the form
(3). Since we are especially interested in the case of a field with
characteristic 2, we shall use the form
(1) of the equation of Q.
Example
Consider
The elements of
Q in
PG(2, 22)
with the equation
2
GF(22 )
can be represented as O,l,t and t ,the
minimum function being 1 + t + t 2 • Then it can easily be checked that
Q consists of the following
5 points
3
P
1
= (0
0 1)
P
= (0
1 0)
P
3
= (1
1 1)
P4
=
2
(1 t t 2 )
P = (1 t 2 t)
5
Degenerate Quadric
A quadric Q in pG(n,s)
is said to be degenerate if there
exists a nonsingu1ar transformation
x
(ii+I x 1)
.e
=
y
B
(n+l x n+l) (n+l x 1)
r
which transforms the form of Q to
E c .. Y'Y , r < n
.>.
J.J
J. j
J_J.=o
Nondegenerate Quadric
A
quadric
Q is said to be nondegenerate if it is not degenerate.
The form of a quadric is said to be nondegenerate in PG(n,s)
if the
corresponding quadric is nondegenerate in PG(n,s) •
Rank
of a Quadri c
A quadric
Q in PG(n,s)
is said to have rank r
if there
exists a nonsingu1ar transformation
x
(n+1 x 1)
=
which transforms the form of Q to
l.
(ii+I x ii+I) (ii+I x 1)
B
r-1
E
j~i=O
a .. y.y.
J.J J. J
which is a nondegen-
erate form in PG(r-1, s).
It is easily seen that a quadric
~
if and only if the rank of Q is (n + 1).
in pG(n,s)
is nondegenerate
If the characteristic of the
4
field is not 2, the rank of a quadric
symmetric matrix of its form.
acteristic is not 2.
Q is equal to the rank of the
However, this is not true if the char-
The following example will make it clear.
Example
Consider
Q
2
in PG( 2,2)
with the equation
222
xo+~+x2=O
~
The matrix of the form of
.e
'Which has rank 3.
is
However, the rank of
Q
2
is
1
since the nonsingular
transformation
Yl =
+ x2 + x
~
3
= x2
Y2
Y = x
3
reduces the form of Q
2
to
3
2
Y •
l
Cone
A quadric
Q in
PG(n,s)
of rank r
is said to be a cone of
order (n + 1 - r) .
vertex and base of a cone
Consider a cone
Q in
PG(n,s)
of order
(n + 1 - r).
there exists a nonsingular transformation which transforms
~-l with the equation
r-l
r:
j:::i=O
= 0
Then
Q to
5
~-l
where
E
n-r
is a nondegenerate quadric in
PG(r-l, s).
The (n-r)-flat
determined by the equations
y
o
=y = ••• =Y
=0
1
r-l
Any (r-l)-flat E 1 which does
ris called the base of the cone. In
is said to be the vertex of the cone.
not intersect the vertex E
n-r
particular the (r-l)-flat E 1
r-
determined by the equation
If P is a point of the cone lying on. the base,
is a base of the cone.
then the (n-r+l)-flat determined by the vertex E
n-r
and the point
P
is completely contained in the cone .
.e
Example
Q in
Consider the quadric
Obviously
Q
plane E
determined by the equation
2
has rank 3.
PG(3,3)
So
Q
with the equation
is a cone of order 1 in
PG(3,3).
The
x = 0
o
is a base of the cone.
The base contains the following four points of
the cone.
Pl
= (0 1 1 1)
P2
= (0 1 2 1)
P
=
3
P
5
(0 1 2 2)
= (0
1 1 2)
6
The vertex is the point
P
= (1
0 0 0)
The cone consists of the 13 points lying on the lines
and PP4 .
PP1 , PP2 , PP
3
The cone can be represented by the following diagram.
.e
HYPerbolic and elliptic nondegenerate quadrics
It has been shown by Primrose
quadric in PG(2k, s)
IJ.!J that
every nondegenerate
contains linear spaces of dimensionality
(k - 1)
and does not contain any linear space of higher dimensionality.
So with
respect to the maximum dimensionality of a linear space contained in the
quadric, the nondegenerate quadrics in PG(2k, s)
type.
belong to only one
However, the nondegenerate quadrics in PG(2k-l, s)
two different types" hyperbolic or elliptic.
quadric in PG(2k-l, s)
belong to
If a nondegenerate
contains (k-l)-flats and does not contain any
linear space of higher dimensionality, then the quadric is said to be a
7
hyperbolic nondegenerate quadric.
PG(2k-l, s)
If a nondegenerate quadric in
contains (k-2)-flats and does not contain any linear
space of higher dimensionality, then the quadric is said to be elliptic.
Primrose
1217 uses
the words unruled and ruled quadric, for elliptic
and hyperbolic quadrics.
Tallini L2§J uses the names elliptic
.
and hyperbolic quadrics.
3. Conjugate points, polar space and tangent space.
Conjugate points
n
Consider a quadric
Q
in PG(n, s)
with the form E
j~i=O
A point
••• I
-e
a ij xixj •
ex }
n
is said to be conjugate to a point
13 = (13 0 , 131' ••• , I3 n )
with respect to
Q
if
Obviously the relationship of conjugacy is symmetrical, i.e., if ex
conjugate to
~,
then 13
is
is conjugate to ex .
Polar space of a point
The polar space of a point ex with respect to
all points which are conjugate to ex with respect to
Q is the set of
Q.
If
it is easily seen that the polar space of ex is an (n-l)-flat determined
by the equation
8
n
I:
(1)
j~i=O
a .. (a.x. + xia.) = 0
1J
~ J
J
If the characteristic of the field is 2, every point is self conjugate
and the equation of the polar of a: reduces to
n
I:
a . (a:ix. + a:jxi ) = 0
j>i=O i J
J
If the characteristic of the field GF(s) is not 2, a point is self
conjugate if and only if it is a point of the quadric.
the equation of the polar of a
In this case
can be written as
where
-e
i
F j,
i,j
= O,l, ••• ,n
.
In matrix notation the equation of the polar can be written as
(4)
a:
(1 x n+I)(n+l x n+I)(n+I x 1)
Xl
A
where A is a symmetric matrix.
=
0
Since our treatment will cover both
the cases of characteristic of the field equal to 2 and the characteristic
not equal to 2, we shall use the equation (1) for the polar of a: • The
polar of a point a: will be denoted by T(a:) •
Tan§ent space
•
If a: is a point of the quadric
respect to
Q
Q,
the polar of a: with
is said to be the tangent space of
Q
at the point a: •
9
Conjugacy of two linear spaces
Two linear spaces t
with respect to a quadric
and t
p
are said to be mutually conjugate
q
Q if every point of t p
every point of E with respect to
q
is conjugate to
Q.
Polar of a p-flat
The polar space of a p-flat t
p
with respect to a quadric
the set of all points which are conjugate to every point of t p
respect to
is
Q
with
Q.
Lemma
3.1.1
,
If a point
P
respect to a quadric
.e
is conjugate to the points
Q, then
1\0'~.••• ,
Ap ' with
P is conjugate to the linear space
determined by the points Ao ' ~, ••• , Ap •
Proof. We have to show that P is conjugate to every point of
the linear space determined by the points
is conjugate to the points
A, A_, ••• , A •
o -"1.
P
Since
P
A,
A..., ••• , Ap ,the polar hyperplane of
o --1.
contains all the points of A,
A..., ••• , A
o -"1.
p
Then by a flmdamental
property of linear space, the polar hyperplane of ex
space determined by the points A,
••• , A •
o A...,
-"1.
P
contains the linear
Hence the lemma follows.
All the considerations in the following are with respect to a.
fixed quadric
Q
in PG(n, s).
Theorem 3.1
The polar of a p-flat t
Ao ' ~, ••• , Ap
in t
p
where
p
is the intersection of the polars of
A ' ~, •.• , A
o
p
are
P
(p+l) independent points
10
Proof.
Let
Let
To prove the theorem we need to show that
T(E )
P
= T' (EP )
Let a be a point of T(Ep ) ' Then a is conjugate to the
points Ao I -~
A_I " ' 1 A
p since all these points belong to Ep • So we
have
(5)
.e
From (5) it follows that
a
(6)
T' (I: )
€
P
So we have
T(E ) C or' (E )
P
P
Conversely let
Then a
a
is conjugate to the points AOI
is conjugate to the p-flat I:p • Hence
a
€
T(I: )
P
So we have
(8)
~I
or' (Ep ) C
or(I:p )
The theorem follows from (1) and (8).
••• I
Ap '
So by lemma 3.1.1
-e
11
Theorem 3.2
Let
PG(n,s) .
A
A , ••• , A
'
be independent points of a quadric
p
l
The p-flat
O
Q
in
r:p determined by these points is completely con-
tained in the quadric if and only if the (p+l) points are pairwise
conjugate.
Proof.
SUfficiency.
We shall prove sufficiency by induction.
First we shall prove the result for
= (<:xo '
Ao
<:Xl'
p
= 1.
Let
••• I
... ,
Since
A
o
and A
l
are points of the quadric and are mutually conjugate,
we have the following.
n
r:
j~i=O
a .. ex.a:. = 0
J.J
J. J
n
r:
j~i=O
a .. f'it'.
J.J
J
=0
n
r:
j~i=O
Any point on the line
a. j(a.t'. + a:jt'.)
J.
J.
AoA
l
J
=0
J.
other than A
o
and
1\
can be represented
as
where
A
is a non-zero element of GF(s) •
Using the equations (9) it can be easily seen that for any
non-zero element
Ao~
A,
A +
o
is contained in Q.
~
is a point of
Q.
Hence the line
12
A:t.'
Now suppose the result is true for (p - 1), i.e., if Ao '
••• , A _
p l
are points of the quadric and pairwise conjugate, the
(p-l )-flat E.. 1
determined by these points is contained in
p-
point of E
p
not lying on E 1
p-
A is a point of Eland ).
GF( s).
and A are points of the quadric and by assumption
to
Ao '
A:t.' ••• ,
So from the first part of the proof the line
So
Ap
+).A
\>
Hence by lemma. 3.1.1
Ap _l '
is a point of
A
Both A
P
is conjugate
p
is conjugate to
A A is contained in
p
Hence the p-flat
Q.
A + A.A where
p
can be represented as
is any element of
p-
Any
Q.
E
p
A.
Q
is contained in
Q.
This completes the proof of sufficiency.
Necessity.
••. , Ap
Aj
Assume that the p-flat Ep
is completely contained in
If possible, suppose
Q.
are not mutually conjugate, i ~ j, i, j
loss of generality let us assume
n
E
(10)
j~i=O
i
determined by Ao '
= 0,1, ••. ,p
= 0 and
j
= 1.
•
A:t.,
Ai
and
Without any
Then we have
a'j(a.~. + a.~J.') ~ 0
J.
J.
Using (10) and the fact that
A
o
J
J
and ~
are points of the quadric,
A +).A:t. is not a point of Q for some
o
non-zero X -which contradicts the assumption that·, E' is contained in Q.
it is easy to see that the point
p
Corollary 3.2.1
Let Ep
quadric
and E
q
be two linear spaces which are contained in the
Q and are mutually conjugate with respect to
linear space determined by E
p
and E
q
Q.
is contained in
Q.
Then the
•
13
Theorem 3.3
Let
Q
n
be a nondegenerate quadric in pG(n,s)
~.
point of the quadric
Let
T(p)
n be an (n-l)-flat not passing through
1)
~
n
T(P)
2)
Q
n
T(P)
n
be a
be the tangent space at
P.
P
and
Then
is a cone of order 1 on the (n-l)-flat
n
and P
T(P)
n is a nondegenerate quadric in PG(n-2, s)
which is elliptic or hyperbolic according as
~
is elliptic or
hyperbolic.
Proof. (1)
Let
R be any point of
T(p)
n ~.
Then
are mutually conjugate and both are points of the quadric.
theorem 3.2, the line
PR
is contained in the quadric.
n
Q
""n
be
E
a ..
j2:i =O J.J
P
and the equation of
T( P )
n
n
E
i=O
J.
= (ao'
n
E clix.
i=O
J.
Let the equation of
X'X
'
j
Let
a l , ... , an)
be
=0
be
=0
c oi xi
The following two relations follow easily.
n
E
c
n
E
c .
. 0
J.=
(11)
. 0
J.=
li a.J.
oJ.
=0
a.J. t
0
and R
So by
Using this
geometrical fact, we shall prove the theorem algebraically.
form of
P
Let the
14
Let
be
C. :;: (c, O' c. l ' .•. ,
J
J
J
C
j
n
),
j
= 1,2, ••• ,n,
n independent points on the (n-l)-flat I: 1 with the equation
nn
I:
i=O
=0
a,x
J. i
.
Consider the transformation
y.
J
= c.JOx0
+ c'lxl + ••. + C. x ,
J
In n
= 0,l,2, ••• ,n
j
It can easily be seen that the transformation is nonsingular.
.
Under
this transformation
goes to
P
0 0 ... 0)
T(p) goes to the (n-l)-flat
T'
with the equation Yl
goes to the (n-l)-flat
1[1
with the equation yo
'Jf
-e
= (1
pI
=0
=0
,
n
and the form of
~
goes to
b .. Y.y.
j:::i=O J.J J. J
n
Qn
T(p) is transformed to a quadric
and has the equation
(12)
2
b
y
00 0
(say)
I:
Q~-l
on the (n-l)-flat
Tf
n
+ Y (b02Y2 + b3Y3 + ••• + b y ) + I:
b.. y. y, = 0
0
on n
j:::i=2 J.J J. J
Using the fact that the (n-l)-flat with the equation Y
l
tangent space at the point P' , we can easily get that
b
00
=b
02
=b
03
=0
is the
= ••• =b,=0
on
So the equation of ~-l reduces to
n
I:
j:::i=2
b .. yiy,
J.J
J
=0
To prove part (1) of the theorem it is sufficient to show that the form
15
n
E
b . YiYj
.>. 2 i J
J_J.=
that if
It can easily be shown
is nondegenerate in PG(n-2, s).
n
is degenerate in PG(n-2, s), then
E b .. y.y .
.""",. 2 J.J J. J
J;::::J.=
degenerate in PG(n, s)
n
is
This completes the
'Which is a contradiction.
(1).
proof of part
(2) Under the transformation used in part (1), Q
n
~-2
the quadric
Q
in the (n-2)-flat, Yo
n
= Yl = 0
T(P)
n
1t
goes to
, with the equation
n
E
b .. y.y. = 0
.>. 2 J.J J. J
J_J.=
I
~-2
We have already shown that
.e
Q
n
n
n
PG(n-2, s).
Hence
PG(n-2, s).
It remains to show that
hyperbolic according as
Assume
~
~
is elliptic.
hyperbolic.
point
T(p)
is a nondegenerate quadric in
P and E _
k 2
is a nondegenerate quadric in
~
n
T(p)
n
1t
is elliptic or
is elliptic or hyperbolic for
If possible, suppose
n
~n T(p)
Then
1t
1t
Q
n
n T(p) n
contains a (k-2)-flat
Hence
Q
n
n T(p) n
~
1t
- 1 .
is
E _ .
k 2
The
•
~
and P
is
Q is elliptic.
n
Similarly we can show that if ~
This contradicts the fact that
~.
is hyperbolic,
= 2k
are mutually conjugate and are contained in
Hence by corollary 3.2 the (k-l)-flat determined by E _
k 2
contained in
n
1t
n
is elliptic.
T(p)
n
1t
is hyperbolic in
PG(n-2, s).
Lemma 3.3.a
Let
~
be a nondegenerate quadric in PG(n,s).
any (k-l )-flat contained in
secting E _ .
k l
~
Let E _
k l
be
and E _ be an (n-k)-flat not intern k
Suppose the following hold.
•
16
(1)
T(E _ ) , the polar of E _
k l
k l
(2)
T(Ek_l)n Ln _
k
(3)
~
n
T(Ek _l )
is an (n-2k)-flat
n En _k
Then for any k-flat
(n-k-l)-flat
L _ _
n k l
T(E )
k
(2)
T(E )
k
Ek contained in
not intersecting E
~
in
and an
k
is an (n-k-l)-flat
n En _k _l
Proof.
point of E
~-2k
is a nondegenerate quadric
PG(n-2k, s).
(1)
is an (n-k)-flat
is an (n-2k-2)-flat.
Let E _
k l
be any (k-l)-flat in E
k
and P
k
is a
not contained in L _ • Then the k-flat E
is deterk l
k
mined by E _
and P • We shall write this fact symbolically as
k l
k
.e
k
By theorem 3.1
T(P )
k
is an (n-l)-flat and
T(E _ )
k l
is an (n-k)-flat.
Hence
T(E )
k
is an (n-k-l)-flat unless
(14)
If possible, suppose (14) is true.
Let
(1)
E _
n k
n
(2)
P
k
E _
n k
pen)
in ~.
€
E _
k l
=
CD
Let E
n-
k
be an (n-k)-flat such that
,the null set
denote the maximum dimensionality of a linear space contained
Since
~
n
T(E _ )
k l
PG(n-2k, s) , there exists a
n
En _
k
is a nondegenerate quadric in
p(n-2k)-flat
Ep (n_2k)
such that
17
and
Pk ~ Ep (n_2k)
since from (14)
T(E _ )<:: T(P )
k
k l
and
Ep (n_2k)<:: T(Ek _l )
(15)
Ep (n_2k)
C
1
T(Pk )
From (15), we have
Hence E (n_2k)
p
and E
k
spaces are contained in
-e
are mutually conjugate.
~.
Also both the linear
Hence by corollary 3.2.1
It is easy to see that
Ep (n_2k)
Hence
Ep (n_2k)
n
(!)
Ek
=
Ek
CD, the null set.
p(n-2k) + k + 1 -flat
is a
1
so we have
(16)
p(n)
= pen
- 2k) + k + 1
However 1 it is known that
pen)
= pen
So (16) contradicts (17).
- 2k) + k
S.o (14) cannot be true.
the first part of the lemma, Le., that
It remains to prove that
T(E )
k
n
T(E )
k
E _ _
n k l
This establishes
is an (n-k-l)-flat.
is an (n-2k-2)-flat.
Eke T(Ek )
and
EnE
k
n-l'l:-l
= CD
, from hypothesis,
Since
18
we have
(18)
Part two of the lemma follows immediately from (18) and the fact that
T(~k)
is an (n-k-l)-flat.
Theorem 3.3.a
Let
in PG(n, s).
T(~k)
(1)
~
n
T(~k)
be a k-flat contained in a nondegenerate quadric
~k
Let ~n-k-l be an (n-k-l)-flat not intersecting ~k • Then
T(~k)
is an (n-k-l)-flat,
n
~
~n-k-i
n
~n-k-l
is an (n-2k-2)-flat and
is a nondegenerate quadric
~-2k-2 in
PG(n - 2k - 2, s) which is elliptic or hyperbolic according as
~
is
elliptic or hyperbolic
-e
~
(2)
with
n
~k
T(~k)
is a cone of order (k+l) on the (n-k-l)-flat
T(~k)
as the vertex.
Proof.
We shall prove the theorem by finite induction on k.
We have proved the theorem for
k
=0
in theorem 3.3.
Hence it will be
sufficient to show that if the theorem is assumed to be true for (k-l),
it is true for
k.
The proof given here holds for the case when the
characteristic of the field is 2 as well as for the case when the
characteristic is not 2.
However, it is possible to give a shorter
proof for the case when the characteristic is not 2.
Assume that the theorem is true for
T(~k)
Let
n
k - 1.
Then by lemma 3.3.a
is an (n-k-l)-flat and T(E k )
En _k _l is an (n-2k-2)-flat.
T(Ek ) be determined by the following (k + 1) independent equations.
19
n
E c" x '
j=O J.J J
=°
,
= 0,1,2, ... ,k
i
Let E _ _ be determined by the eQ.uations
n k 1
n
(20)
E ci ' x ' =
i = n-k, n-k+1, ••• , n
j=O J J
°,
Consider the {n-k-1)-f1at determined by the eQ.uations
n
(21)
E
0:"
j=O J.J
X.
J
=
°
,
i=O,l, ... ,k,
,
where
are (k + 1) independent points iIi E
k
belong to the {n-k-1)-f1at (21).
= 0,1, ••• ,k
• The points
,
-e
i
Let
1
= 0,1, I I.,k
co' c1 ' .•• , Ck' c k+1 ' ""
cn _k _1
be a set of (n-k) independent points lying on the (n-k-1)-f1at (21).
Consider the transformation
yi
=
,
c.J.
x
(1 x n+1)(n+1 x 1)
It can be easily seen that this transformation is a nons1ngular one ,
Under this transformation T(E k ) goes to the (n-k-1)-flat T'
the eQ.uations y, = 0, i = O,l" •• ,k ,
J.
r: n-k-l goes to E'n-k-1 determined by the eQ.uations
with
° ,
with the eQ.uation r: b,. YiY = ° ,
'>. ° J.J
J_J.=
Yn-k = Yn-k+1
= ••.
= Yn
=
n
Q
goes to
Q'
--n""n
...
the point Ai
goes to the point
'
J
Bi = (O,o" •• ,O'~i~-k' ""~in) ,
1=0,1, ... ,k,
Qn
...
20
n
n En _k_l
(E k )
goes to
~-2k-2 with the equation
n-k-l
(22)
E
b .. Y Y .
i
J
= 0
j=::i=k+l J.J
and ~
n
goes to ~-2k-l with the equation
T(E k )
n-k-l
E
b .. YiYj + terms involving Y -k'Yn - k+l "" 'Yn
j=::i~k+l J.J
n
= 0
(23)
I
It can be shown that if the form of the quadric
~-2k-2
in PG(n-2k-2,s), then ~ is also degenerate.
Hence
nondegenerate.
~
So
n
T{E )
k
n En_k_l
is degenerate
~-2k-2
is
is nondegenerate in
PG{n-2k-2, s).
..
-e
Also by a corresponding method used in the proof of theorem 3.3,
it can be shown that
~
n
T(E )
n
En _k is elliptic or hyperbolic
k
is elliptic or hyperbolic. This completes the proof
according as
~
of part (1).
To prove part (2), it is sufficient to show that the
coefficients of Yn- k' Yn- k+l' ••• , Yn
in (23) are all zero.
be the k-flat determined by Bo ' Bl , .•• , ~.
~-2k-l' Also if
is contained in
pI
E~ is contained in
is any point of Q~-2k-l' the (k+l)-flat
I
""'n-2k-l • Using this fact we can show that all the
Q"
terms involving Y k' Y k l' ••• , Y
nn- +
n
..
l
Let Ek
in (23) are zero.
Let Ek be a k-flat contained in a nondegenerate quadric .~
in PG{n,s). Let Ep be any linear space which is contained in ~
and contains E • Then Ep is contalned in T(E ) , the polar of E •
k
k
k
21
..
Proof.
in L
k
.
Ao ' ~, .•• , ~ be (k + 1) independent points
Let
Since L
k
is contained in
conjugate by theorem 3.2.
~,
is contained in
J these points must be pairwise
Let the p-flat Lp
be determined by the
Ao ' ~, •.• , i\J ~+l' •.• , Ap •
(1' + 1) independent points
Lp
~
Since
these points are also pairwise conjugate.
By theorem 3.1
k
n
T(L ) =
k
Since
Ai
T(A )
i
i=O
Aj , i ~ jJ iJj
is conjugate to
= 0,1, •.• ,1'
, we have
Hence
(24)
A
j
€
T(L )
k
,
j
= 0,1, •.• ,1'
From (24) it follows that
EpC T(E k )
Corollar,r 3.4
The tangent hyperplane through a point
P
of a nondegenerate
Q contains all the linear spaces that contain
Q.uadric
P and are con-
tained in Q.
4.
Stereographic projection and its use
Let
0
be a point in PG(n,s)
passing through
intersects
of
P
PG( n Js )
on
on
Let
through
1C
through
1C
be an (n-l)-flat not
P be a point other than
at a point
1C
1C
o.
and
pl.
0.
°
pI
0.
The line
OP
is called the stereograph projection
The stereographic projection of a set
A in
is defined to be the set of all points whi ch
are stereographic projections of the points of
A on
1C
through
o.
22
The stereographic projection of A on 1t through
by So ,1t (A) •
be written as
If 0
S(A).
and A are assumed to be fixed,
If
S(C)
SO,1t(A)
will
C is a set of points containing 0, then
the stereographic projection of the set
written as
0 will be denoted
C-
{o}
through 0 will be
for convenience.
Lemma 4.1.1
Let P be a point on a nondegenerate quadric
and
T
be the tangent space at
P
and
~
in ro(n,s)
1t be an (n-l)-flat not passing
through P.
In the following any stereographic projection is on
through P.
Then
(a)
S(~
n
T)
is
1t
~-2' a nondegenerate quadric on the (n-2)-flat
Tn
(b)
If Ep
is a p-flat containing P and contained in ~, then
S(Ep )
and
( c)
If Ep
then
and
(d)
If Ep
and
1t •
=Ep-1
' a (p-l)-flat
S(Ep ) C ~-2
is a p-flat not containing P and contained in
S(E) = E'
p
p
, a p-flat
S(Ep ) C ~-2
•
is a p-flat contained in
S(Ep )
= E;
SeEp)
et:.
~
~-2
~
but not in
T,
then
nT,
23
(a) We have shown in theorem 3.3 that
Proof.
a nondegenerate quadric ~-2 in PG(n-2, s).
~
n
T ('\ n is
Hence it Will be suf-
ficient to show that
S(~
(:1)
n T) = ~ n T n n
(1) follows immediately from the definition of sterographic projection.
(b) Since
and
EpC ~
by corollary 3.4
EpC T(P)
So we have
EpC
It follows that
n
T
S(Ep ) C S( ~
n
~
T) = ~-2
Now to prove (b) it is sufficient to show that
(2)
It is easy to check that
SeEp )
= Ep +n rc
I
Hence (2) follows from the fact that Ep
(c)
is not contained in n.
Let
Ep +l
and
I
Ep
=pe
Ep
= Ep +l n
n
It is easy to check that
t
seEp)
= Ep
Since
EpC T(P)
by corollary 3.4
Ep+1C
So
~
n
T{P)
E;C ~ n T(p)n
n
= ~-2
•
24
(d)
Let
and
It is easy to check that
s(z)
p
To show that
S(Ep )
= z'P
cr- ~-2
it is sufficient to show that
E; ct Tn
0)
7t
(3) follows immediately from the fact that
Ep
C/-.
T
Stereographic projection of a class of sets
Let
point and
d(
be a class of sets in PG(n, s).
Let P be a given
be an (n-l)-flat not passing through P.
7t
graphic projection of the class
(JL
on
7t
The stereo-
through 0 is defined to be
the class consisting of the sets which are the stereographic projections
on
1f
through P of the sets of
(Q
and 1 s denoted by S( (J2.)
•
Lemma 4.1.2
Let
02
in PG(n,s)
be a class of distinct p-flats passing through a point P
and
7t
be an (n-l)-flat not passing through P.
there exists a one to one correspondence between the two classes
Then
(Q
and S( (Q.> •
Proof.
Let
Q,
be any set in the class
Let
Q
be made to
correspond to S( C) • We shall show that this correspondence is unique.
25
It will be sufficient to show that for a:ny two different sets
:.:.>1'
Q
and 0 I
the class
(4)
If possible, suppose
(4)
is not true.
=
o f
S(o)
Since
Then
S(OI)
01
there exists a point
R
Let
RI = S(R) , the sterographic projection of R .
Then
R'
so the line
the line
PH l i s
PR' •
belonging to 0
€
S( g I)
,
P
€
contained in 0 t
Hence
R
but not belonging to
ot
,
Obviously R
is a point of
ot
QI.
is a point on
which is a contradiction.
Theorem 4.1
P be a point of a nondegenerate quadric
Let
-r{ P )
be the tangent space at
through
P.
Let
-4
and passing through
n,p
P
and
in
PG(n,s) ,
be an (n-l) -flat not passing
1t
denote the class of p-flats contained in
e
and ,1:J n,p
P
'~
be the class of all p-flats of
~
Q
il
Then there exists a one to one correspondence between the classes
'~n,p
and
(V ~-2,p-l
and hence the number of elements in each class
is the same.
Proof.
lemma 4.1.2
Since each p-flat of
n,
p
passes through
P, by
it Will be sufficient to show that
S(~~n,p) = (~n-2,p-l
(5)
We shall show that (5) is true it for
quadric
~
"'l~
~
n -rep) n
1t
in
Q
il-2
PG(n-2, s) •
we take the nondegenerate
•
26
Let
Ep E:
seEp) =
and
By part
~~ n,p
0,
(b) of lemma 4.1.1
o = Ep _l C
Hence
o
E:
~-2 .
tt n-2,p-l
It follows that
(6)
Ep _l E: CQ n-2,p-l
Then it can be easily seen that
Conversely let
Ep = P (f) Ep _l
and
C
~
SeEP ) = Ep- 1
Hence
Ep _l
S({n,p)
E:
and
a
n_2,p_1 C SC(n,p) •
(5) follows from (6) and (7).
5.
Linear spaces contained in a nondegenerate quadric
~
in PG(n, s)
Theorem 5.1
Let N(p,n)
denote the number of different p-flats contained in
a nondegenerate quadric
~
in PG(n,s)·.
~(p,k)
N(p,n)
=
~2(p,k)
Then
,for n
= 2k,
for n
= 2k
,for n
= 2k
p ~ k - 1 ,
elliptic
and p~k-2 ,
- 1 ,
~
- 1 , ~ h.y:perbolic
and p ~ k - 1 ,
27
~(p,k) =
where
P
IT
(S2(k-p+r) _ 1.)
r=O (sP+l-r
~l(P,k) =
p
IT
_ 1)
(sn-2p+2r + sk-p+r-l. sk-p+r _ 1)
~-----------_ .... , p ~ k - 2
(sP+l-r _ 1)
r=O
=
The e.x;pressions for
Proof.
(sn-2p+2r _ sk-p+r-l + sk-p+r _ 1)
IT
~----..;;...--------~,
r=O
(sP+l-r _
1)
N(o,n) were o'btained by Primrose
p ~ k - 1 ,
ffrJ-J .
First we shall establish the following difference equation.
_ N(p-l, n-2)N(o,n)(s - 1)
N( p,n ) 1
(sP+ - 1)
(1)
Let
p
P
be a point of
~.
of p-flats contained in
~
From theorem 4.1 it follows that the number
and passing through
P
is
Let us count the points in the p-flats contained in
p+l
contributes
~ is
s
N(p,n).
s-
-11
Through every point of
and the number of points of
~
Primrose
Every p-flat
p+l 1
Hence this collection of p-flats contain N(p,n)Ss --1
In this collection every point 'Will
be repeated as many times as there are p-flats of
p-flats of
~.
points and the number of p-flats contained in
points which are not all different.
a point.
N(p-l, n-2) •
contains
L21J
~
~
is
~
there passes
N(o,n).
N(o,n)N(p-l, n-2)
passing through
N(p-l, n-2)
p-flats
Hence the collection of
points.
Hence (1) follows.
has obtained the following formulae.
28
<fl(o"k)
(2)
2k
= s s -- 11
<fll(o"k) =
(2k
k-l - sk + 1)
s
+ 6
(6 - 1)
<fl 2 (o"k) =
( s 2k - s k-l + s k - 1)
(s - 1)
Applying the difference equation (1) repeatedly and using the formulae
(2)" we get the required e~ressions for
<fl(p"k)" <fll(p"k) and O2 (P,,k) .
Theorem 5.2
~
The number of p-flats contained in a nondegenerate quadric
in PG(n"s)
.e
which pass through a given k-flat
N(p-k-l" n-2k-2) "where
N(p"n)
E
k
contained in
~
is
denotes the number of p-flats con-
tained in a nondegenerate quadric of the type (elliptic or hyperbolic)
of
~.
Proof.
Let
denote the polar of E
T(E )
k
(n-k-l)-flat which does not intersect Ek •
class of p-flats contained in
JJ k,p
E _ _
n k l
denote the class of
•
~
and E _ _
n k l
fl k"p
(p-k-l)-flats contained in ~
~-2k-2
~
n
T(E )
k
be an
denote the
and passing through E •
k
By theorem 3.3a it is known that
a nondegenerate quadric
Let
k
n
in PG(n - 2k - 2" s).
Let
n T(Ek ) n
E _ _
n k l
is
Hence to
prove the theorem it will be sufficient to show that there is a one to
one correspondence between the classes ---f k
.
~ "P
Let
Ep e {; k,p
Then
(3)
and
9:t k
~- "p
of p-flats.
29
(4)
I:
k
n I:n _k_l
=
~,the
null set
From (3) and (4) it follows that
So !:
p
n I:n- k- 1
I:
p
p
n
I:
k 1
I: _ _
n k l
has dimensionality at least equal to
the dimensionality of I:
p
I:
¢
n
I: k 1
n- -
is a (p-k-l)-flat I:
n- -
(p-k-l) •
cannot exceed (p-k-l).
k l'
Since
Hence
Since
p- -
I:ke I:p C.~
I: C T(I: )
p
k
by theorem (3.4)
So
.e
I:p
n I:n _k _l C
So
~
e)J- k,p
I:p _k _l
Let us make !:p
of
-& k,p
n T(I:k ) n
correspond to
I:n _k _l
= ~-2k-2
•
!:p-k-l
of
/:t k,p'
It can
easily be seen that the correspondence is unique.
6. Canonical forms of quadrics
Consider a quadric
n
I:
a. xix
j
j~i=O 1. j
Q in
be the form of
PG(n,s)
Q.
with rank
r.
Let
It is known that if the character-
istic of the field is not 2, then by a nonsingular transformation
x
(ii+I x 1)
the form of
Q
=
B
y
(ii+I x ii+I}(n+l x 1)
can be reduced to the canonical form
are non-zero elements of the field.
r-l
2
I: A.iY. where
t=O
It is shown by Dickson
A.' s
1.
L'f.21 that
a
30
nondegenerate quadric
~ in PG(n,
tD')
can be reduced to one of the
folloWing canonical forms:
(1)
~ x2 + x3x4 + ••• + x~k_l x 2k = 0,
0.
= 2k - 1
2) + ~ x + ~x4 + ••• + x _ x
2k 1 2k
2
2
=0
~(~ + x2
where
,
~(xi + x~) + ~x2 is irreducible in GF(zn).
n=2k-l
In the folloWing
theorem we have obtained certain canonical forms for the elliptic and
h;yperbolic nondegenerate quadrics in PG(2k-l, s)
for any prime power s •
Theorem 6.1
Let
Let
(3
(1)
be a Galois field With characteristic not equal to 2.
a be a non-zero element of GF( s) such that -a is a square and
be a non-zero element such that -(3
element of
GF(
GF(s)
zn).
m
GF(2)
:2
A(~
such that
is a square.
Let
~
be an
2
+ x2 ) + ~~ is irreducible in
Then
The qUadric
2
~-l in PG(2k-l, s), s ~
2
2
2
zn , With the
2
equation
2
X;t + ax2 + x + Cl:X4 + ... + x2k_l + ax2k = 0
3
is a hyperbolic nondegenerate quadric.
(2)
The quadric
2
~
+
m
~k-l in PG(2k-1,s) , s ~ 2
2
(3~
, With the equation
:2:2
2
:2
+ x + Cl:X4 + ••• + x2k_l + aX2k
3
=0
is an elliptic nondegenerate quadric.
(3)
The quadric
Q2k-l
in PG(2k-l,
is a h;yperbolic nondegenerate quadric.
zn)
With the equation
31
(4)
m
Q2k-l in PG(2k-l, 2 ) with the equation
The quadric
2
A(~ +
x22) + XJ.x2 + x3x4 +
+ x _ x
2k l 2k
=0
is an elliptic nondegenerate quadric.
(1)
Proof.
It is obvious that
~-l
shall show that
We
is hyperbolic by finite induction on k .
we prove the result for
-a
First
Since -a is a square element of
k = 1.
A of GF(s)
GF(s) , there exists an element
The equation of Q
l
Q2k-l is nondegenerate.
such that
2
A
=
is
2
2
~+ax2=O
.e
It can be easily seen that
~
(-A, 1). Since
i.e. points,
~
~
conta.ins the two points
(A, 1) and
conta.ins linear space of dimensionality
is hyperbolic.
Assume that
~k-3
0
(= k-l)
is hyperbolic.
Consider the nonsingular transformation
It is easy to see that under this transformation
~k-l
transforms to
c.:.•
with the equation
'""2k-l
Since the incidence properties in a projective geometry remain invariant
over nonsingular transformations, it is sufficient to show that
is hyperbolic.
Consider the point
~k-l
32
P
of Q;k-l.
=
(:0 0 ••• 0 1 0)
The equation of
T(P)
=0
Y2k
Let
is obviously
n be the (n-l)-flat with the equation
Y2k-l
Then
Q;k-l
n Tn
=0
if h~.6 the equation
2
2
••• + Y2k-3 + ay2k-2
By assumption
Q~k-l
(k-2)-flat E _ •
k 2
n Tn n
.e
is hyperbolic and hence contains a
The point
P and the (k-2)-flat E _
k 2
,
contained in the quadric
Q2k-l
by corollary 3.2 the (k-l)-flat
Hence
and are mutually conjugate.
peEk_2
is contained in
result for
k = 1.
The quadric
2
2
Xl + I3x2
will be elliptic if
contains a point
Q
l
Q
l
Hence
Q~k-l •
First we prove the
in PG(l, s)
2
With the equation
=0
(~, x;).
2J.
xr
Then
x~ ~
o.
If possible, suppose
We can easily get
2
=-13
which contradicts the assumption that
~
k.
does not contain any point.
I
Hence
are both
~k-l is hyperbolic.
(2) We shall prove the result by induction on
~
=0
-13 is a non-square element.
is elliptic.
Assume that the result is true for
(k - 1)
so that
~k-3
is
33
a non1.egenerate elliptic quadric.
Applying the nonsingular transfor-
mation used in part 1, we can transform
•
~k-l
to
t
Q2k-l
with the
equation
222
2
Yl + /3Y2 + Y3 + C¥Y4 +
~k-l
As before it will be sufficient to show that
possible, suppose
(k-l )-flats •
Q;k-l
The equation of
1t
Then
~k-l
= (0
contains
0 ••• 010)
is
T(P)
be the (n-l) -fle,t with the equation
.e
Y2k- l
~k-l
Then the equation of
=0
n n
T
1t
is
2
2
••• + Y2k- 3 + Qy2k-2
which is elliptic by induction assumption.
number of p-flats passing through a point
P
the
of a nondegenerate quadric
P of the quadric.
Q~k-l contains (k-l)-flats, it follows that there exists a
(k-l)-flat Ek _1
contained in
~k-l n
-rep)
n
contains' a' (k-2~~flat
~k-ln T(P)
n
Q~k-l
and passing through P.
is contained in Q~k-l n
theorem 3.4 E _
k l
sects
,
=0
By theorem 3.5.2
and contained in the quadric is equal for every point
Since
If
Consider the point
P
Let
is hyperbolic.
is elliptic.
1t
1t
E _
k 2
in a (k-2)-flat
-r(p).
By
Hence E _ interk l
So ~k_l'n -rep)
Ek _2 •
.. This contradicts' ,~he assumption :that .
is elliptic.
n
This completes the proof of part (2).
1t
34
Parts (3) and (4) of the theorem can be proved by arguments exactly
similar to arguments used in parts (1) and (2) respectively•
•
7. Nucleus of polarity of a quadric in PG( 2k,
tJl).
Theorem 7.1
For every nondegenerate quadric
exists a point
S not lying on the quadric such that every line through
S intersects the quadric
Q
in a single point.
2k
called the nucleus of polarity of
was proved by Bose
Proof.
.e
m
~k in PG( 2k, 2 ) there
Let
IJ,
Q2k'
For
The point
PG(2,
tfl)
S
is
this result
p. l5q.
m
Q~k be a nondegenerate quadric in PG(2k" 2 ) •
Then according to Dickson
LI'[f there exists a nonsingular transformation
which transforms ~k to
~k with the equation
2
Xo
+ ~x2 + x3x4 + ••• + x2k_lx2k
=0
Since the incidence properties in a projective geometry are invariant
over nonsingular transformations, it will be sufficient to prove the
theorem for
Q2k.
Let
S
We shall show that
to
Q2k.
= (1
0 0 ••• 0)
S possesses the required properties With respect
other point not in Q2k.
mutually conjugate.
It is easily seen that
quadric.
Let R be any ..
S and Rare
Then by theorem 7.2, which follows, the line
intersects the quadric in a single point.
•
~k.
ObViously S is not a point of
It is easy to see that
Let R'
S and R'
SR
be a point of the
are mutually conjugate.
35
If possible, suppose the line
point R"
point
of
Q2k'
Since
SRI
intersects the quadric in another
S and R'
are mutually conjugate" the
S occurs in T(R')" the tangent space at
contains R' •
So the line
occurs in T(R')
SR'
is contained in
and RII
and R'
R' .
T(R' ).
RIIR'
is contained in
T(R')
Hence
are mutually conjugate.
are points of the quadric and are mutually conjugate.
the line
Also
~.
So
R'
R"
and RII
So by theorem 3.2
S is a point of
Q
2k
which is a contradiction.
Theorem 7.2
Let
P and R be two points of
Q
in PG( 2k" ~) .
2k
nondegenerate quadric
.e
m
PG(2k" 2 )
not lying on a
The line
the quadric -in a single point if and only if the points
PR
intersects
P and Rare
mutually conjugate.
Proof.
Sufficiency.
P
Let
= Cao " al" ••• " an)
Let the equation of Q
be
2k
n
1:
j:;:i=O
Since
a .. x,x
J.J
J.
j
=0
P and R are not points of the quadric and are mutually conjugate"
we have the following.
36
n
E
j~i=O
a .. a.a. ~ 0
J. J
~
J
J.
n
= 2k
(l)
Any point on the line
P + AR
where
>..
intersect
.e
PR
= (ao
other than P and R can be expressed as
+ A~ 0 ~ a l + ~l~ ..• ~ a n + A~n )
is a non-zero element of
GF(2m}.
So the line
PR will
Q
in a single point if the equation
2k
n
E
aij(a. + A~. }(aj + ~j) = 0
'>.
J_J.=
0
J.
J.
has a single solution in A .
Using (l)~ (2) can be simplified as
n
2 2k
E a ij aiaj + A
E a' j ~'~j.
j~i=O
j~i=O J.
J.
m
From (l) and the fact that in GF(2 )
=0
every non-zero element has a
unique square root ~ it follows that (3) has a unique solution in A
Necessity.
Assume that the line
single point Rt
•
PR
intersects the quadric
If possible ~ suppose
~
in a
P and R are not mutually
conjugate.
Noting that every point is self conjugate ~ it follows that
Rt
cannot be mutually conjugate.
and P
Let
R' = (ao~ al' .•• ~ an)
P
= (~o~
~l~ ... ~ ~n)
37
Then we have
(4)
Let
R"
=P
+
~I
where
A = - ----------
.e
It can easily be seen that R"
is a point of
~.
This contradicts
the assumption that the line PR intersects Q2k in a single point.
This completes the proof of necessity.
Theorem 7.3
If P is a point not lying on a qUadric
~
in PG(n"s) and
R is a point of the quadric such that P and R are not mutually
conjugate" then the line PR intersects the quadric in two points.
Proof is very simple and hence is omitted.
CHAPrER II
SOME CLASSES OF PBIB DESIGNS WITH TWO ASSOCiATE
CLASSES OBTAINED FROM THE CONFIGURATION OF
LINEAR SPACES CONTAINED IN A QUADRIC
1. Summary
In the present chapter a general method of constructing
Partially Balanced Incomplete Block Designs is debeloped. Let Band
D be two classes of linear spaces in PG(n/s) such that every member
of each class intersects a nondegenerate quadric Qn in FG(n,s) in a
quadric of the same nature. Then the configuration of the two classes
Band D prOVides a P.BIB design. The configuration of the generators
and points
of a nondegenerate quadric Qn in PG(n,s) gives a class of
PBIB design with two associate classes. Many other series with two
associate classes are obtained. These series contain several new PBIB
designs with rand k not greater than 10.
2.
Introduction.
PBIB designs with two associate classes were introduced by Bose
and Nair
L7 J.
Bose and Shimamato
L'9J have
rephrased the definition
so as to stress the distinction between the association scheme and the
design. Bose and Shimamato defintion for the P.BIB design with m associate
classes is substantially as follows.
A PBIB design with m associate classes is an arrangement of v
treatments in b blocks such that
(1) Each of the v treatments is replicated r times in b blocks
each of size k and no treatment occurs more than once in every block.
(2) There exists a relationship of association between every
pair of the v treatments satisfying the following conditions:
(a) Any two treatments are either first associates
or second associates ••• or m-th associates.
(b) Each treatment has n first associates, n2 second
l
associates,
nm m-th associates.
0 ••
(c) Given any two treatments which are i-th associates,
the number P~k of treatments which are j-th associates of the first
and k-th associates of the second is independent of the pair of
treatments with which we start.
i,j,k
i
Furthermore Pjk
i
= Pkj'
for
= l,2,.•••m.
(3) Any pair of treatments which are i-th associates occur
togetherm exactly
~i
blocks for i
= 1,2, •••m.
Bose and Clatworthy L-5~have given a less demanding definition for
the PBIB design with two associate classes which is as follows:
A PBIB design with two associate classes is an arrangement of
v treatments in b blocks such that
(1) Each of the v treatments is replicated r times in b blocks
each of size k and no treatment occurs more than once in any block.
40
(2)
There exists a relationship of association between the v
treatments satisfying the following conditions:
(a) Any two treatments are either first associates or
second associates.
(b) Each treatment has nl first associates and n2 second
associates.
(c) For any pair of the v treatments which are i-th associate the number pil of treatments common to the first associates of
the first and first associates of the second is independent of the
pair.of treatments with which we start, i = 1,2.
(3)
in exactly
Any pair of treatments which are i-th associates occurs
~i
blocks, i=1,2.
Since it is easier to check the conditions of Bose-Clatworthy definition, we shall use this definition.
The following relations hold be-
tween the parameters of a PBIB design and are useful for computing
some parameters when others are known.
vr
=bk
,
for 1
, i,j,k
:I J
1,J=1,2, ••• m,
= 1,2, •••m.
41
A general method of constructing PBIB designs.
Let B denote the class consisting of the sets Bl , B2 , ••• Bb
whereB j , j=l,2, •••b, is a set of points in PG(n,s). Let V denote
another class consisting of the sets Vl , V2 , ••• Vv where Vi' i=1,2, •• uv
is a set of points in PG(n;s). These two classes generate a design
D(B,V) with the following incidence matrix
where
nij
= 1,
if the sets Vi and Bj intersect each other,
· 0, if the sets Vi and Bj do not intersect.
Then we have the following.
ri
=
b
~ n
= Number of sets of the class B which intersect Vi'
j=l ij
v
k
j =
~ii'
~ n
ij
i=l
= Number
b
= j~~ijni~
of sets of the class V which intersect Bj •
= Number of sets of the class B which intersect
both Vi and Vi'
i ~
i'
= 1,2,~ ••v.
Let Cl , C2 , ••• Om be m classes of sets in Fa (n,s) 'such that
The sets Vi and
Vl,~re
said to be J-th associates if
42
Let P~V(Vi'V~) denote the number of sets of the class V which are u-th
associates of Vi and y-th associates ·of Vi"
t,u,v
il= i', i,i'=1,2,. ••v,
= 1,2, •••m.
Theorem 2.1
The design D(B,V) is a PBIB design with two associate classes
if the following are true.
(1) Any two sets are either first associates or second associates
• • • or m-th associates •
(2) Each set Vi (i=1,2, •••v) has nl first associates, n2 second
associates, ••• nm m-th associates.
t
(3) puv(Vi,V ,) is independent of the t-th associate pair of sets
i
t
t
(Vi,V i , ) and Puv = Pvu' tiv,u = 1,2, •••m.
(4) r i
= r,
i = 1,2, •••v and k j
= k,
j=1,2, •••b.
(5) For any pair of t-th associate sets
t
(vi,V i ,) ~ii'
= ~t'
= 1,2, •• om.
Proof is obvious and hence omitted. As a consequence of Bose -Clatworthy
defin1tion of PBIB designs with two associate classes we have the
following result which for the sake of reference is presented as corollary
Corollary 2.1
The design D(B,V) is a PBIB design with two associate classes
if
the conditions (1), (2), (4) and (5) are satisfied for m=2 and the
following condition (3)' is satisfied instead of (3).
(3)'. The number pil (Vi,V i ,) is equal for every pair of t-th associate sets (vi,Vi ,), t=1,2.
3. PBIB designs from the configuration of generators for blocks and
points for treatments of the quadric.
Definition. Generator. A line which is contained in a quadric
is called a generator of the quadric.
Lemma 3.1.1
Let Pl and P2 be two points of a nondegenerate quadric ~ in
PG(nls) such that the line P1P2 is not a generator. The number of
points P such that both the lines PPl and PP2 are generators of the
quadric is N(Oln-2) where N{Pln) denotes the number of p-flats contained in a
non-degene~ate
quadric of the type of
~(elliptiC
or hyper-
bolic) in PG (n l s).
Proof.
Since the line P1P2 is not a generator, by theorem 3.2
of Chapter I the points Pl and P2 are not mutually conjugate. Let
Tl and T2 denote the tangent spaces at Pl and P2 respectively. Let
P be a point of ~ such that both PPl and PP2 are generators of ~.
Since PPl is a generator l by theorem 3.2 of Chapter I P must be
conjugate to Pl " Hence P must be a point of Tl " Similarly, P must
be a point of T 2 • Hence P is a point of ~ n T l
T2 " Conversely
if P is a point of Qnf'\ T l f1 "2 1 Pis cooJugate to both Pl and P2 and
n
hence both PPl and PP2 are generators of Qn " It follows from the
above argument that the required number is equal to the number of
points in Qnn Tln. T2 •
Since Pl and P2 are mutually not conjugate T2 does not pass
through Tl " So Tl is a tangent space at Pl and T2 is an (n-l)-flat
•
44
not passing through Pl'
By theorem 3.3 of Chapter I Qn
n
TIn T2
is a nondegenerate quadric Q 2in PG(n-2~s} which is elliptic or
n-
byperbolic according Qn is elliptic or hyperbolic.
Hence the lemina
follows.
Lemma ,.1.2
Let P and P be two points of Q in PG(n,s) such that P1P2
2
l
n
is a generator of Qn' Then the number of points P other than Pl and
P2 such that both PP and PP are generators of Q is
l
2
n
2
(s-l) + s N(O,n-4)
where N(p,n) has the same meaning as in lemma 3.1.1.
.e
Proof.
Let T and T be the tangent spaces at P and P respectively •
2
l
l
2
Let ~n-2 be an (n-2)-flat not intersecting the line ~ determined by
Pl and P2 •
of lemma
Let P be a point of Q,n other than P and P • As in the pX'tiof
2
l
we can show that both PP and PP are generators of Q
n
l
,.,.1,
if and only if P is a point of Qn II T1
r\ T2'
of points is equal to the number of points of
2
Hence the required number
~
n T1 n T2 other than
Pl and P2 • By theorem '.1 of Chapter I
Then by th~orem '.3.a ~nT(I1)n~n_2 is Qn-4' a nondegenerate quadric:
in PG(n-4,a) and
~
Qn
T(1).)
~
T(l)
is a cone of order 2 With Zl as the vestex and
1: _ as the base.
n 2
"(1).) is
It follows easily that the number of points
2
(s+l) + s N(O,n-4).
Then the lemma follows from the fact that both Pl and P2 are points
of Qnn T(I:l ).
Theorem 3.1
Let B be the class of generators of Qn' a nondegenerate quadric
in PG(n,s) and V be the class of points of
as a point set.
~"
each point being regarded
Then D(B,V) is a PBIB design with two associate classes
with the following parameters.
b
= N(O,n),
= N(l,n)"
k
= s+l,
v
r = N(0,n-2),
.e
>-:t. = 1,
~ = 0,
nl = sN(0"n-2),
pi
11
= (s-l)
+ s2N(O,n-4)"
2
Pll = N( 0,n-2) •
The other parameters are easily obtained from the relations (2.1) between
~he
parameters.
In the above ex.pressions N(p,n) denotes the
number of p-flats contained in a nondegenerate quadric of the type of
~(elliptic
or hybolic).
Proof. We sball apply corollary 2.1 to prove the theorem.
is easy to see the following:
It
46
b
= Number
of generators of Qn'
= N(l"n).
k
= Number
of points on a generator"
= (s+l).
v
= Number
of points of
~"
= N(O"n).
r
= Number
of generators passing through a point,
= N(O,n-2)
by theorem 5.2 of Chapter I.
The association scheme is defined as follows.
Two points Pl and P2
of Q are first associates of each other if the line P P2 is a generator
1
n
.e
and are second associates of each other i f the line P1 P2 is not a
generator.
Since there can be at most one generator passing through
two points of
~I
we have
Let Pl be a particular point of
~.
The number of points P which are
first associates to P is equal to the number of points P such that
l
PP
l
is a generator and hence equal to the number of points lying on
the generators passing through Pl'
From the above argument, we have
~ = sN(O,n-2).
Let Pl and P be two first associate points.
2
of
~.
Then P P2 is a generator
1
The number of points P which are first associates of both
Pl and P2 is equal to the number of points P other than Pl and P2
such that both PPl and PP2 are generators. By lemma 3.3.2 this number
is equal to
(s-l) + s
2
N(O,n-4)~
From the above argument it follows that
~l
= (8-1)
+ s2 N(O,n- 4 ).
Let Pl and P2 be two second associate points. Then the line P1P2 is
not a generator. The number of points P which are first associates
.e
to both Pl and P2 1s equal to the number of points P which are such
that both PPl and PP2 are generators. By lemma 3.3.1 this number does
not depend on the particular pair of second associate points and is
equal to N(O,n-2). From the above argument it follows that
Corollary 3.1.1
Taking n=2t , we get the series of PBIB designs with two associate
classes with the following parameters
48
s2t_1
v = s- l '
,
k
= s+l,
b
I:
~
-1,
~
= 0,
n1
.e
(s2t_ 1 )(s2t-2_1 )
,
(e_1)2 (8+1)
I:
e(e 2t - 2 _1)
s-l
'
e2 (s2t-4_ 1 )
(e-1)'
1
P11
= (e-1)
2
P11
s2t-2_1
= s-l •
+
Putting t=2, we get the following symmetric seriee.
= e+1,
r =k
h..r
= 1,
A = 0,
2
n
2
= s +
1
S,
1
P11 = (a-1),
2
11 = s+1.
P
Thia series was obtained by C1atworthy [1;] by a different method.
Corollary 3.1.2
Taking n=2t-l., t >
- 31 and Qn elliptic we get the series of
PBIB designs with two associate classes with the following parameters.
V
=
S
2t-l-s t +8 t-l - 1
8-1
1
8
2t-3 -8 t-l+ S t-2 - 1
8-1
1
r
=
k
= 6+1 1
~= 1, ~ = 01
Putting t=3, we get the following series.
v = (s3+l )(s+1)
r = s2+l
k = 8+1
b =
(s3+ 1 )(62+l)
~= 1,
A2 = 0,
nl = s(s 2+1)1
50
1
= (s-l),
P11
2
2
Pll = (s +1).
This series contains the following two designs with r .and knot
greater than 10.
Design
Number
Dl
D
2
-
.
v
r
k
b
27 .
5
3
45
10
4
280
112
2
P1l
~
n
l
1
Pll
1
0
10
1
5
1
0
30
2
10
\
The design Dl is included in BeS (Bose, Clatworthyand Shrikhande)
catalogue Lb J. The design D2 is new •
Corollary 3.1.3
Taking n
= 2t-l,t~
2, and
~
hyperbolic, we get the series of
PBIB designs with two associate classes with the following parameters.
V
a
s 2t-l+8 t -s t-l - 1
s-l
'
k= 8+1,
b
~
"'2
=
(s2t-l+ s t_ s t-l_ l ) (s2t-3+st-l.8t-2.l)
= 1,
= 0,
(s-l) 2 (s+l)
'
51
s(s2t-3+ s t -1_ s t-2_ 1 )
n1 =
(s-l)
1
(
)
s2(s2t- 5+s t-2_ s t-3_ 1 )
(s-l)
P11
=
2
P11
s2t-3+ s t -1_ s t-2_ 1
=
(s-l)
s-l +
Putting t = 2, we get the following one parameter family with the parameters.
v = (S+1)2 ,
k = (s+l),
r
= 2,
b
= 2(s+1),
~ = 1
X = 0
2
n1
1
P11
= 2s,
= (6-1),
2
P11
1:1
2.
This family is same as the simple lattice family.
Putting t = 3, we get the following series.
v
I:
(s2+1 ) (8 2+s+l),
r
I:
(6+1)2
k=(s+l),
b
= (6+1)(6 2+1)(s2+s+1 ),
~ = 1,
52
~= 0,
s(s+l) 2 ,
(s-1)+2s 2 ,
(s+l) 2 •
This family contains only one design with rand k not greater
than 10 with the following parameters.
v
= 35,
r
1
nl = 18, Pll
= 9,
= 9,
k ==
2
Pll
==
3,
b
= 105,
A:J.
==
1,
h2 ==
°
9.
This design is new.
Example.
The actual method of writing down the blocks is illustrated in
the following example. Consider the design Dl of the series given in
corollary 3.1.1. The parameters are
This design is obtained by taking generators of Q4' a nondegenerate
quadric in FG(4,2), as blocks and points ofQ4 as treatments.
equation of Q4 can be taken as
The 15 points of Q4 are
The
53
Pl
P
2
P3
P
4
P
5
P6
= (00001)"
IS
P
9
(00010) ,
P10 = (00110)"
= (10011)"
II:
Pll
= (01000),
==
= (01010)"
P12 == (11110)"
P = (10111),
13
P14 • (11011) ,
(00100)"
= (11100) ,
P7 = (00101)"
Pa
= (11101)"
P15 == (01111) 0
(01001)"
To write down the blocks systematically we can proceed as follows.
Consider the treatment 1. The blocks which contain treatmentl correspond
to the generators containing the point Pl.
.e
So find out the generators
n 'T(Pl ) where
n T{Pl )" PPl
passing through Pl' we find out Q4
space at Pl. For any point P of Qq.
T(Pl ) is the tangent
is a generator.
In
this way we can exhaust all the blocks containing treatment 1. Next
we proceed to treatment 2 and by a similar procedure can find out the
blocks containing treatment 2 which are not already included. We continue
in this manner until all the blocks are obtained.
In our example the 15 generators are P1P4 " P1P , P1P6, P2PV
5
P2P5 , P2P6, P3P4" P3P5 " P3P6" P7Pll , P7P12 " PaP10' Pa P12 , P9P10" and
P Pll " So the 15 blocks of the designs are
9
(1,4,7),
(1,5,8),
(1,6,9),
(2,4,10), (2,5,11), (2,6,12),
C~,4,13),
(3,5,14), (3,6,15),
(,,11,15), (7,12,14), (8,10,15),
(8,12,13), (9,10,14) and (9,11,13).
The quadric Q4 can be represented by the following diagram where
points of Q4 are represented by small circles and three points on a
generator are joined together by a straight line or a continuous curve.
'l.
.e
4. PBIB designs from the configuration of 'Points of a quadric for blocks
and generators of a quadric for treatments.
Let Qn be a nondegenerate quadric in PG(nl's) which contains lines
but does not contain planes. Thus Qncan be a nonde~nerate quadric in
PG(4,s) or an elliptic quadric in PG(5,s) or a hyperbolic nondegenerate
..
quadric in PG(3,s) •
55
Lemma 4.1.1
Let
secting
f/ be
a genera.tor of ~. The number of generators inter-
V (other
than
V)
is
N(0,n-2)-1
(s+l).
The line 1/ contains (s+1) points.
Proof.
By theorem 5.2 of
Chapter I, the number of generators through every point of '1/ is
N( 0,n-2) of which
f/
is one.
Hence the lemma follows.
Lemma 4.1.2
Let
f/l
and "2 be two intersecting generators of ~.
Then the
number of generators other than "1 and "2 which intersect 'both "1 and "2 is
-e
N(O,n-2)-2
Proof.
Suppose the two' generators in "1, and "2 intersect
at the point P.
~""O----nv--~~'l.-
L
,(..'}..
Then there cannot be a.ny generator which intersects both "1 and 1/2 at
points other than p.
If possible, suppose there exists a generator
which intersects "1 at Pl and "2 at P2 • Then the three points P, Pl
and P2 are mutually conjugate and are points of ~. So by theorem '.2
of Chapter I" the plane determined by them is contained in Qn"
contradicts the assumption that
~
This
does not contain any planes.
So it
follows that the required generators are those which intersect both
"'1 and "'2 at the point P.
By theorem 5.2 of Chapter I the number of
generators passing through P is N( 0"n-2) of which "'1 ~nd "'2 are two
generators.
Lemma
Hence the lemma follows
4.1.,
Let "1 and "2 be two nonintersecting generators of ~.
Then
the number of generators which intersect both "1 and (2 is (8+1).
Proof •
.e
Let P be a point of the generator "2.
Consider the generators which
intersect "2 at P and also intersect "1'
Any such generator will lie
in the plane 1:2 determined by "1 and the point P.
The plane 1:2 is not
contained in ~ and contains a generator "1 of ~ and a point P of Qn
not lying on "1'
~
From these facts it follows easily that 1:2 intersects
in a pair of lines, Hence there exists one and only one line pass-
ing through P and intersecting "1'
This is true for every point of "2
and the number of points of "2 is (s+l).
Hence the lemma follow6.
57
Theorem 4.1
Let B be the class of points of Qn a nondegenerate quadric in
PG(n,s) which does not contain planes.
Let V be the clan of generators of Qn. Then D{B,V} is a PBm design
with two associate classes with the following parameters.
v = N(l,n),
r = s+l,
k
111
N(0,n-2),
b = N(O,n),
~ =
.e
~
1,
= 0,
nl = ['N(O,n-2)-lJ(s+1),
1
P1l = N(0,n-2)-2,
2
Pll = (s+l).
The other parameters can be obtained from the relations (2.1) between
the parameters of PBm design.
Proof. We shall apply corollary 2.1 to prove this theorem.
The following results can be obtaned easily.
v • Number of generators of
~
• N(l,n).
r
= Number
of points of
~
intersecting a generator,
• (s+l).
k = Number of generators of Qn intersecting a point,
= N(O,n-2).
b • Number of points of
~
=N(O,n).
Two generators "'1 and V2
are first associates if they intersect and are second associates if
The assoc iation scheme is defined as follows.
of
.e
~
they do not
inters~cti
most one point, we have
Let
Vl
Since two generators of Qn can intersect at
~=l, ~=O.
be a given generator. The number of generators which
are first associates to "'1 is equal to the number of generators which
intersect "'1. Hence we have from lemma 4.1.1
1
2
The constancy of the parameters Pll and Pll and their expressions
follow from lemma 4.1.2 and 4.1.3 by using an argument exactly
similar to that used in the proof of theorem 2.1.
Corollary 4.1.1
Taking n
= 5 and Q5 an
elliptic nondegenerate quadric in PG(5,.)
we get the follOWing series of PBIB designs with two associate classes.
59
v • (8 3+1 )(8 2+1)"
= (6+1)"
r
k
b
= (8 2+1)"
= (8 3+1) (8+1)"
~
= 1"
~
= 0"
nl
1
Pll
1:11
8
2
(8+1)"
= 8 2 -1"
2
Pll = s+1.
This series contains the following two designs with rand k not greater
~
than 4•
.e
Design
Number
v
r
k
b
~
~
nl
1
Pll
2
Pll
D
l
45
3
5
27
1
0
12
3
3
D2
280
4
10
112
1
0
36
8
4
The design Dl is included in the BCe catalogue.
Corollary 4.1.2
Taking n
= 3 and
The design D2 is new.
Q an hyperbolic nondegenrate quadric in
3
FG(3"s) we get the following series of PBIB designs with two associate
classes.
60
v = 2(6+1)"
r
k
b
= 6+1"
= 2"
= (6+1) 2 ,
\
= 1"
~
= 0"
= 6+1,
n1
1
Pll = 0"
2
P11 = 6+1.
This series is given by C1atworthy ~14~.
,.
PBIB designs from the configuration of generators on generators.
Theorem 5.1.
Let B be the class of generators of a nondegenerate quadric
Q which contains lines but does not contain planes.
If two coincident
n
generators are regarded as nonintersecting" D(B"B) is a PBIB design with
two associate classes with the following parameters.
v
r
= b = N(l"n)"
= k = f!1(O"n-2)-lJ
\
= N(Ol n - 2 )-2"
~
= (s+1)"
nl
1
P11
2
Pl1
=
N(0"n-2)-1
= N(0"n-2)-2,
= (s+l).
(S+l)1
(s+l)"
61
Proof. We shall apply corollary 2.1 to prove this theorem.
The following results can be easily obtained.
v
= b = Number
of generators of
~,
= N(l,n).
r =k
= Number
of generators intersecting a given
generator excluding the given generator
=
["N(O,n-2)-J] (s+l) by lemma 4.1.1.
The association scheme is defined as follows.
sect, they are first associates.
If two generators inter-
If two generators do not intersect,
they are second associates.
Let "1 and
"2 be two generators which intersect each other.
By
lemma 4.1.2 the number of generators other than "1 and "2 which intersect both (11 and "2 is N(O"n-2)-2.
h:L
Hence we have
= Number of generators otherthan "1 and
"2
which intersect both "1 and ((2"
= N(O,n-2)-2.
Similarly by lema 4.1.3
~ =
(s+l).
1
2
The constancy of Pll and Pll and their expressions also follow from
lemma 4.1.2 and 4.1.3.
62
Corollary 5.1.1.
Taking Q a nondegenerate quadric in PG(4,s) we get the following
n
series of PBIB designs with two associate classes.
r
=k
::
s(s+l),
A:J.
= (s-l),
~
= (s+l),
nl = s(s+l),
1
Pll = (s-l),
2
Pll
= (s+l)
0
This series contains only one design with rand k not greater than 10.
This design is new. The parameters are
Theorem 5.2.
Let B be the class of generators of a nondegenrate quadric Qn in
PG(n,s) which contains lines but does not contain planes. It two coincident generators are considered as intersecting,
~e
D(B,B) is a PBIB
design with two associate classes with the following paraneters.
v == b = N(l,n),
r =k
~ =
= (s+1)N(O,n-2)-s,
N(O,n-2),
~ = (8+1),
nl = [N(O,n-2)-1_7 (8+1),
1
= N(O,n-2)-2,
Pll
2
Pll = (s+l).
~.
The association scheme is defined as in theorem '.1.
If two generators of
~
intersect, they are first
do not intersect, they are second associates.
.e
associat~
If they
The paramters of the
association scheme are obtained by the same argument as used in the
proof of theorem '.1. The expressions for v and b are obvious. Also,
we have
r =k
= Number
of generators which intersect a given
generator including the given generator,
==
~
[N(O,n-2)-!.7 (s+l)+l by lemma 4.1.1.
= Number
of generators which intersect two given
mutually intersecting generators including the
two given generators,
= N(O,n-2)
by lemma 4.1.2.
2 == Number of generators which intersect two mutually
non-intersecting generators,
A
==
This
co~p1etes
(8+1) by lemma 4.1.3.
the proof of the theorem.
64
Corollary 5.2.1.
Taking Qn as a hyperbolic nondegenerate quadric in PG(3,s)
we get the following series.
v
= b = 2(s+1),
nl
r
= k = (s+2),
= b = 2(s+1),
pile 0,
This series is given by Clatworthy
~
= 2,
~
= (s+l),
2
Pll • (s+l).
[i2J.
6. PBrB deSignS obtained from the configuration of lines and points
of PG(3 J s) truncated by a quadric.
.e
Let Ql' Q2' ••• , Qs+l be (s+l) points on a nondegenerate quadric
on a plane 1:2 in PG(3,s), s=2m• Let Qo be the nucleus of polarity of
the quadric.
Lemma 6.1.1
Let Pl be a point of PG(3,s) not lying on 1:2 • The number of
points P other than Pl not lying on ~ such that PPl is a line passing
through one of the points Qo' ~, ••• , QS+ l is
(s+2)(s-1)
Proof is simple and hence omitted.
Lemma 6.1.2
Let Pl and P2 be two points which do not lie on ~ and which are
such that the line P1P2 passes through one of the points Qo' Ql' ••• , Qs+
The number of points P other than Pl and P2 not lying on 1: such that
2
t
both the lines PPl and PP2 pass through one of the points QOI Q11
is (s-2).
'''1
QS+l
Proof.
The points on the line P1P2 other than Pl/ P2 and Q11 the point of
intersection of P1 P2 and I:21 possess the required property. So the
line P1P2 contributes (s-l) points. If possible suppose there exists
a point P not lying on the line P1P2 which possesses the required
property. Let Q2 and Q, be respectively the intersections of the
lines PPl and PP2 with I:2 " ObViously the three points Ql' Q2 and Q,
are collinear. Bose
L-'] has
shown that the (6+2) points QO'~/'",Q"'l
are such that no three of these points are collinear.
Hence our assump-
tion that there exists a point P outside the line P1P2 satisfying the
required conditions leads to a contrdiction. This completes the proof
of the lemma."
Lemma 6.1.'
Let Pl and P2 be two points not lying on the plane I:2 such that
the line P1P2 does not pass through one of the points QOI ~" . . . , Qs+l"
Then the number of points P other than Pl and P2 not lying on
~
such
66
...
that both the lines PPl and PP2
pass through one of'the points
Qo' ~, ••• , Qs+l is (s+2).
Proof'.
Let S be the point of' intersection of' P1P2 and 1:2.
(J..~
f7¥!=~---H---~O
~
Since Qo is the nucleus of polarity of' the quadric containing the
points Ql'
~I .... ,
Qs+l in
~,
by theorem 7.1 of Chapter I, the
line S~o will intersect the quadric in a single point ~(say). Consider
the line SQ2.
Since the line SQ' intersects the quadric in a single point,
o
by theorem 7.2 of Chapter I Sand Qo are mutually conjugate and SQ0
is the polar of' S with respect to the quadric. Since Q2 is not on the
line SQo' Sand Q2 are not mutually conjugate and the line S~ will be
a secant. So the line S~ contains another point Q, say Q,. Similarly,
any line through S containing one of the points of
Qo'~'" .,Qs+l
will
s+2
contain a second point Q. Suppose the (s+2 ) points Q lie on the T
lines, SQo~, SQ2Q"
.0.,
SQsQs+lo
Any point P satisfying the required property must lie on one of
67
From the diagram given below we can easily see that any such
plane contributes two points P.
s
~()
Hence the total number of points P is (s+2).
Theorem 6.1
-e
Let B be the clas8 of lines in
roC~,8)
not lying in
~
and
passing through one of the points Q0
I ~I ••• Q +1"
Let V be the
8 ,
class of points of PG(3, s) not lying on L2 • Then D(B,V) is a PBIB
design with two associate classes with the following parameters.
s='f1AI
r
= s+2 ,
k = s,
2
b = 8 (8+2),
A:t = 1,
~
= 0,
nl = (s+2)(s-1),
1
Pil = (s ..2),
2
Pll = (s+2).
68
Proof.
The association scheme is defined as follows.
Two points
Pl and P2 are first associates if the line P1P2 passes through one of
the points Qo'~' ••• ' Qs+l and second associates otherwise. Then the
proof of the theorem follows easily from lemmas 6.1.1, 6.1.2, and 6.1.3.
This series contains the following 3 designs with rand knot
greater than 10.
-e
2
Pll
"'2
n
l
1
Pll
1
0
4
0
4
96
1
0
18
2
6
640
1
0
70
6
10
Design
Number
v
r
k
b
Dl
8
4
2
16
D2
64
6
4
D
3
512
10
8
The design Dl is given in Clatworthy
new.
\
L1.4].
The designs D2 and D are
3
7. Concluding remarks.
The method of constructing PBIB designs developed in this chapter
is very general.
It is possible to obtain many series of PBIB designs
with two associate classes from the configuration of linear spaces of
higher dimensiona.lity contained in the quadric.
These series are not
included since they do not contain new designs with r and k not greater
than 10. However, these designs may be useful in experiments other than
varietal trial, for instance in asymmetrical factorial experiments.
CHAPTER III
SOME CLASSES OF PBIB DESIGNS WITH
THREE ASSOCIATE CLASSES
1.
Summarl.
In this chapter a theorem is proved about PBIB designs with
three associate classes which in effect gives a much less demanding
definition of PBIB design with three associate classes.
This theorem
is applied to construct PBIB designs with three associate classes.
A
series of PBIB designs with three associate classes is Obtained from
the configuration of generator and points of cone with a point vertex.
.e
Another series of PBIB designs with three associate classes is obtained
from the configuration of secants and external points of a nondegenerate
quadric in finite projective plane.
These two series contain the fol-
lowing designs with rand k not greater than 10.
The following table
1
1
i
gives the parameters v, r, k, b, AI' A2 , A , nl , n2 , I'll' 1'12' 1'22 for
3
i = 1, 2, 3. The other parameters can be obtained from the well known
relations between the parameters of a PBIB design.
All these designs are new.
70
D
1
15 2 3 10 1 0
D
2
18 4 3 24
D
3
0
4 2
1
0
0
0
0
1
1
1
0
0 0
8 1
2
1
0
8
0
0
4
0
0
30 6 3 60 1 0
0 12 1
2
1
0 12
0
0
6
0
0
D4
48 6 4 72 1
0 18 2
6
2
0 18
0
1
6
0
0
D
5
54 10 3 180 1 0
0
20 1
2
1
0 20
0
0 10
0
0
D6
63
1 0 0
24 6
9
2
0
8
0
5
9
3
0
2.
0
D
100 8 5 160 1 0
0 32 3
12
3
0 32
0
2
8
0
0
D8
180 10 6 300 1
0 50 4 20
4
0 50
0
3 10
0
0
7
.e
4 7 36
1
0
A theorem on three associate PBIB designs.
In this section a theorem on three associate
PBIB designs is
proved which actually is an extension of a theorem of Bose and
C1atworthy ~5_7 for PBIB designs with two associate classes.
Before
proving this theorem we shall prove a lemma.
Lemma 2.1.1.
Let P~k(Q,¢) denote the number of treatments which are j-th associates of Q and k-th associates of
associate treatments, i,j,k
¢ where
(Q,¢) is a pair of i-th
= 1,2,3.
Let there exist a relationship of association between every pair
among v treatments satisfying the following:
71
(a)
Any two treatments are either first associates or second
associates or third associates.
(b)
Each treatment has n first associates" n2 second associates
l
and n third associates.
3
(c) The numbers pil{Q"¢),, p~(Q,,¢) and P~2(Q,,¢) are independent
i
.
of the particular pair of i-th associates (Q,,¢) and P12{Q,,¢)
i
i
.
= P2l{Q,,¢),
= 1,2,3.
Then the numbers pi (Q,,¢), P~l{Q,¢)" P~3{Q,¢)" P~2(Q,¢) and P~3{Q,,¢)
3
are independent of the particular pair of i-th associates (Q,¢) and
pi3 (Q,,¢) = P~l{Q,¢), P~3{Q,¢) = P~2(Q,¢), i = 1,2,3.
Proof. Consider the pair of treatments (Q.lI¢) which are first
.e
associates.
The n first associates of Q are made up ¢, pil{Q,¢) treatl
ments which are first associates of both Q and ¢, pi2(Q,¢) treatments
which are first associates of Q and second associates of ¢ and pi {Q,¢)
3
treatments which are first associates of Q and third associates of ¢.
So
we have the identity
(2.1)
Similarly considering the first associates of ¢, we get
•••••
ApplYing similar considerations to the second associates of Q and the
second associates of ¢ and remembering that Q and ¢ are first associates
we get
(2.3)
0 ••••
P~l(Q,¢)
+
P~2(Q,¢)
+
P~3(Q,¢) = n2 '
72
Similarly considering the third associates we get
•••••
P~l(Q,¢) + P~2(Q,¢) + p~,{Q,¢) = n, ,
•••••
pi,(Q,¢) +
p~,{Q,¢) + p;,{Q,¢) = n, •
Solving these equalities we get
pi,{Q,¢)
(2.7)
•••••
= P;l{Q,¢) = nl
- pil(Q,¢) -
p~(Q,¢)
- 1
p~,{Q,¢) = P;2(Q,¢) = n2 - pi2(Q,¢) - p~(Q,¢)
p;, (Q,¢) = n, - nl - n2 + pil (Q,¢) +2pi2(Q,¢) + P~2(Q,¢) + 1 •
By assumption pil(Q,¢), pi2(Q,¢) and P~2(Q,¢) are independent of the
.e
particular pair of first associates (Q,¢).
Hence from (2.7) it fol-
lows that pi,(Q,¢), p~,(Q,¢) and P;3(Q,¢) are independent of the particular pair of first associates (Q,¢).
the lemma for i
= 1.
This completes the proof of
Similarly the lemma can be proved for i
=2
and ,.
Theorem 2.1.
I:f an arrangement of v treatments in b blocks is such that the
following conditions are satisfied:
(i)
Each of the v treatments is replicated r times in b blocks
each of size k and no treatment occurs more than once in any block.
(ii)
There exists a relationship of association between every
pair of the v treatments satisfying the following conditions:
(a)
Any two treatments are either first or second or
third associates.
73
(b)
Each treatment has nl first associates, n2 second
associates and n third associates e
3
(c) Given any two treatments which are i-th associates,
the numbers pil' P~ and P~2 are independent of the pair of i-th
= 1 , 2,3
associates with which we start , i
(iii)
and Pi2
= P~l'
= 1,2,3.
i
Any pair of treatments which are i-th associates occur
together in Ai blocks for i
= 1,2,3,
then this arrangement is a PBIB design with three associate classes with
the following parameters.
V,
b,
r,
k,
111
P13 = n1 - Pll - P12 - 1,
1
P2l
111
P22' P23
= P12 ,
1
1
= n2 - P22 - P12 ,
1
1
111
1
1
1
P31 = P13 , P32 = P23 , P = n - nl - n2 + 2p12 + P22 + P11 + 1,
3
33
22222
222
P11, P12' P13 = n1 - P1l - P12 , P21 = P12 ' P22 ,
222
222
2
P23 = n2 - P22 - P12 - 1, P31 = PlY P32 = P23 ,
222
3
3
2
P33 = n3 - n1 - n2 + 2P12 + Pl1 + P22 + 1, Pi l' P12 ,
333
3
33_
3
3
3 _
P13 - n1 - P11 - P12 , P21 = P12 , P22 , P23 - n2 - P12 - P22 ,
3_33_33
P31 - PlY P32 - P23 , P33
= n,
- n1
3
3
3
- n2 - 1 + 2P12 + P11 + P22 •
Proof follows easily from lemma 201.1.
Theorem 2.1 provides a very useful way of checking whether a
given arrangement is a PBIB design With 3 associate classes or not.
The usual definition of PBIB designs with three associate classes requires the constancy of 27 parameters of the second kind P~k'
74
i,j,k
= 1,2,3
whereas due to theorem (2.1) we need to check the coniii
stancy of 9 parameters of the second kind namely Pll' P12 and P13'
i
= 1,2,3.
Also the usual definition requires 9 s;ymmetry conditions
i
of the type Pjk
i
= Pkj'
i,j,k
= 1,2,3.
In theorem 2.1 we need to check
only 3 symmetry conditiona.
Corollary 2.1.1.
Let B and V be two classes of sets in PG(n,s) and D(B,V) be the
design as defined in section 2 of Chapter II. Let r i' k.1 and \1' have
the same meaning as in section 2 of Chapter It. Then D(B, V) is a PBIB
design with three associate classes if the following are true.
(i)
(ii)
r
and k j are constants for i = 1,2, ••• ,v and j = 1,2, ••• ,b.
i
Any pair of sets are either first associates or second as-
sociates or third associates.
(iii)
and
~
Each set Vi has n first associates, n2 second associates
l
third associates, i = 1,2, ••• ,v.
(iv)
Given a pair of sets (Vi' Vi' ) which are t-th associates
the numbers pil (Vi' Vi')'
P~(Vi'Vi I) and P~2(Vi' Vi 1) are independent of
t
the particular pair of t-th associates ( Vi,Vi ,) and P12
(v)
t
3.
t
= P2l
,
For any pair of t-th associates (Vi,Vi ,), ~iil
t =1,2,3.
= ~t'
= 1,2,3.
SOme PBIB designs with three associate classes obtained from the
configuration of generators and points of a cone.
Let ~-l be a nondegenerate quadric on an (n-l)-flat Ln _l in
PG(n,s) and
~
be the cone with
~~l
as the base and a point 0 outside
75
En- 1 as the vertex.
As in Chapter II we shall use N(p, n-1-2t) to de-
note the number of p-tlats contained in a nondegenerate quadric at the
type of Qn- 1 (hyperbolic or elliptic) in PG{n-1-2t).
prove some lemmas which will be used later.
Let p be any point at
~
other than O.
Now we shall
Then the number of genera-
tors which pass through P but do not pass through 0 is sN(O, n-3).
Proot.
Let PR be a generator at
~
not passing through O.
o
~--~--------e)l<-,
Then the three points 0, P and R are points of
~
and mutually conjugate.
So by theorem 302 of Chapter I, the plane OPR is contained
PtR t be the intersection of OPR and E 10
n-
Qn-l.
in~.
Let
Then PtR t is a generator of
Hence any such generator PR 1s contained in a plane OptR' where
pi is the intersection of OP and En _l and pt Rt 1s a generator ot 'b -1"
The number of such planes OP'R t is equal to the number of generators of
~-l
passing through P' and hence by theorem 5.2 is equal to N{O, n-3).
Every plane OPtR t contains s generators passing through p but not through
O.
Hence the lemma follows.
Let Pl and P2 be two points of Qn other than 0 such thatP1P is
2
a generator not passing through O. Then the number of points P such
76
that both PP and PP2 are generators not passing through 0 is
l
s2 _ s + s3 N{O, n-5) •
Proof.
~(
L-------t---7 "
By theorem 302 of Chapter I the plane OP1P2 is contained in
.e
~.
Hence the points of the plane other than those lying on the line OPl and
OP possess the required property. So the plane OP1P2 contributes
2
(s2+s +1) - (2s+1)
points Po Let P be a point not lying on the plane
OP1P2 which possesses the required property. By theorem '.2 of Chapter
I it Will follow that the '-flat OPP1P2 is contained in~. Let the
plane PiP2P' be the intersection of the '-flat OPP1P2 and En _l where
Pi and P2 are respectively the intersections of OPl and OP2 with En _r
Hence we have shown that every such point P lies in a '-flat
OPiP2P ' where PiP2P' is a plane of ~-l passing through PiP2.
ber of planes contained in
of Chapter I equal to
~-l
passing through PiP2 is by theorem 5.2
N(O, n-5). Hence the number of ,-spaces of the
type OPiP2P' is also N{O, n-5).
•
The num-
Any point P of the '-flat OPiP2P' not
lying on the plane OP1P contributes
2
6' points.
Hence it follows that
77
I
the total number of points P is
. (s2+s +l ) - (2s+l) + s3N(O" n-5) = s2_ s +s3N(o"n-5) ..
Let Pl and P2 be two points of ~ such that P1P2 is a generator
passing through O. Then the number of points P such that both PP l and
PP2 are generators not passing through 0 is s2N(O, n-3).
Proof.
Let Pi be the intersection of the line OP1P2 and En_l •
Let P be a point such that both PP and PP are generators not passing
l
2
through O. By theorem 3.2 of Chapter I it can be easily seen that the
plane PP1P2 is contained in~. Let PiP' be the intersection of the
plane PP1P2 and En_l • ObViously PiP' is contained in ~_p the base of
the cone ~. Hence we have obtained that every such point Plies 1n a
plane OPiP' 'Where PiP' is a generator of
~-l
passing through Pi.
The
number of planes OPj?' is equal to the number of generators PiP' of
~-l
passing through Pi and hence is equal to N(O" n-3) by theorem 5.2
of Chapter I.
Every plane OPiP' obViously contributes s2 points P
2
such that both PP and PP are generators not passing through O.
l
the lemma follows.
tor.
Let Pl and P2 be two points of ~ such that P1P is not a genera2
Then the number of points P such that both PPl and PP2 are genera-
tors not passing through 0 1s sN(O, n-3) •
..
Hence
78
Proof.
Let Pi and P be respectively the intersection of OPl and OP2
2
with I:
n-
1"
Let P be a poLLt such that both
Let P' be the intersection of OP and L
that pIPJ_and p t P
2 is
n-
1.
pp~
~
and PP,., are generators.
~
Then it can be easily seen
a pair of intersecting generators of'
~-l
and also
PiP2 is not a generator of Qn-l" Hence every point P such that both
......
.~
PP
l
and PP
where pi
2
are generators not passing through 0 lies on a line opt
is a point of
tors of ~-l.
~ .. l
2 are
such that both plPi and PiP
genera-
The points Pi and P are points of ~J;,-1 and are not
mutually conjugate.
2
Hence by lemma 3.101 the number of points pi such
that both P'Pi and p i P
2 are
generators of ~-l is N(O, n-3).
opt contributes s points P possessing the required property.
Every line
Bence
the lemma follows.
Let Pl and P2 be two points of
not passing through O.
~-l
such that P1P2 is a generator
Then the number of points P other than P and
l
P2 such that PPl is a generator not passing through 0 and PP2 is a
79
generator passing through 0 is (a-I).
Proof.
Since Pl P is a generator not passing through 0" by
2
theorem 3.2 of Chapter I we can easily see that the plane OP l P2 is
contained
in~.
Let P be a point possessing the required property.
is a generator passing through O. So P must be a point of OP •
2
2
Since the plane OPl P is contained in ~" every point P of OP2 is such
2
that PPl is a generator Of~. Hence the required points P are the
Then PP
points of the line OP other than 0 and P '
2
2
Lemma 3.1.6.
.e
Hence the lemma follows.
Let P and P be two points of ~ such that P1P is a generator
l
2
2
passing through O. Then the number of points P such that PP is a
l
generator not passing through 0 and PP is a generator passing through
2
o is O.
Proof is obvious'and hence omitted.
Let Pl and P2 be two points of ~ such that P P is not a genera1 2
tor. Then the number of points P such that PP is a generator not
l
passing through 0 and PP2 is a generator passing through 0 is O.
Proof.
Since PP2 is a generator passing through 0" P must be a
point of OP2 • If possible" suppose there is a point P on the line OP
2
such that PPl is a generator. Then the three points 0, P and P are
l
mutually conjugate and are points of~. Hence by theorem 3.2 of Chapter I the plane OPPl is contained
the plane OPP1'
in~.
ObViously P P is a line of
1 2
80
~:::'-----'l!If7.-
1\.
We have shown that the plane OP1P is contained in~. So it follows
that P P2 is a generator of ~ which is a contradiction. Hence the
1
lemma follows.
Lemma 3.1.8.
_e
Let P1 and P be two points of Qn such that P1P2 is a generator
2
not passing O. Then the number of points P other than 0 such that both
PP1 and PP are generators passing through 0 is O.
2
Proof is simple and hence omitted.
Let P1 and P2 be two points of ~ such that P1P2 is a generator
passing through O. Then the number of points P other than P and P
1
2
and 0 such that both PP and PP are generators passing through 0 is
1
2
(~-2).
Proof is simple and hencli) omitted.
Lemma
3.1.1,0~
Let P1 and P2 be two points of ~ such that P1P2 is not a generator. Then the number of points P other than 0, P and P such that
1
2
both PP1 and PP are generators passing through 0 is O.
2
81
Proof is simple and hence omitted..
Theorem 3.1..
Let B be the class of generators of'
~
V be the class of points of'
~
other than 0.
not passing through
° and
Then D(B,V) is a PBIB
design with three associate classes with the f'ollowing parameters.
= sm(o,
k =s + 1
A. = 1
l
v
n-l),
,
,
b
r
= s 2N(l, n-l)"
= sm(o, n-3) ,
A.2
= A.3 =
°
,
nl = s2N(0, n-3)"
n2
123
Pll = s -s+s N(O" n-5),
1
.e
°
P22
=
2
P12
=0
1
P12
- 1,
,
= (s-l)
,
2
Pll = s2N(0, n-3)"
,
2
P22
3 _ (
Pll - sN 0" n-3),
=s
= (s-2)
3 _ p3
P12 - 22
=0
•
The other parameters can be obtained from the relations between the
parameters of a PBIB design.
Proof.
We shall apply corollary 2,,1.1 to prove the theorem.
f'ollowing results can be seen easily.
v
= Number
of' points of' the cone
~
other than 0"
= sN(O, n-l).
b
= Number
= s 2N(l,
of' generators of' the cone
n-l).
~
not passing through 0,
The
82
k = Number of points on a generator,
;:: (s+l).
r
= Number
of generators passing through a point other than 0,
= sN(O, n-3) by lemma 3.1.1.
The association scheme is defined as follows.
Two points Pl and P2 are
first associates if P P2 is a generator not passing through 0, second
1
associates if P P is a generator passing through 0 and third associ1 2
ates if P1P is not a generator. Obviously A. = 1 and 7\.2 = A. = O.
2
l
3
The e~ressions for the parameters of the association scheme follow
from lemmas 3.1.2 to 3.1.10.
Corollary 3.1.1.
= 4 and
Taking n
'~
a nondegenerate elliptic quadric in PG(3,s)
we get the following aeries:
6(S+1)2,
v
=
r
= 26
7\.1
=1
pl = s2 • s
11
2
2
pU= 26
,
,
,
,
,
p3 = 2s
11
,
nl = 2s
2
20 2 (S+1),
b
-=:
k
-= a + 1
,
n2 = (a-l)
p~= (6-1)
,
,
,
2
P22= (s-2)
,
A.
2
= 7\.3 = 0
2.
1
3
Pl2= P22 = P12 -- p322 -- 0 •
This series contains the following four designs with r and k not greater
than 10.
Some of the nonzero parameters are given below.
83
v
r
k
b
A
1
n1
n2
1
P11
1
P12
2
P11
2
P22
~1
D1
D2
18
4
3
24
1
8
1
2
1
8
0
4
48
6
4
72
1
18
2
6
2
18
1
6
D
100
8
5
160
1
32
3
12
3
32
2
8
D4
180
10
6
300
1
50
4
20
4
50
3
10
Design
Number
3
Corollary 3.1.2.
Taking n=5 and Q4 .a nondegenerate quadric we get the following
series of PBIB designs with three associate classes.
v
= s(s3+s 2+s +1 ),
k
= (s+l)
A
=1
= s2(s+1)
1
~e
n
,
,
,
1
1
2
P1l= s - s
p3 = s(s+l)
11
= s2(s3+62 +s +1 ),
r
= 6(s+1)
A
= A3
n2
= (6-1)
2
= 0
1
P12 = (s-l)
2
P22 = (s-2)
,
,
P2ll = s 2( s+l)
b
,
,
,
,
,
p1 _ p2 _ p3 _ p3 _ 0
22- 12 - 12 - 22 - •
This series contains the following design with r and k not greater than
10.
v = 30,
b = 60,
k = 3,
r = 6,
nl = 12,
De = 1,
112
312
2
3
3
Pll = 2, P12 = 1, Pll = 12, P11 = 6, P22 = P12 = P22 = P12 = 0, P22 =
.
Corollary 3.1.3 •
Taking n
=6
and
~
a nondegenerate elliptic quadric in PG(5,s)
we get the following series:
o.
84
= S(S3+1 )(S+1), b = S2(S3+1)(s2+1), k = (8+1),
r = s(s 2 +1),
Al = 1,
A2 = A = 0,
n1 = s 2( s 2+1)"
3
2
2 2
n2 = (a-l),
pi1 = 8 _S"
pt2 = (a-1),
piI = 8 (6 +1),
v
•
p~ = (s-2),
pi1
= S(6 2+1),
P~2 = Pi2
=
P~
=
P~2
= 0.
This series gives the following design with r and k not greater than
10.
v
= 54,
b
= 180,
= 3,
n1 = 20 , n2
112
P11 = 2"
P12 = 1,
P11 = 20,
k
r
= 10,
= 1,
Al
= 1,
I _p2 _p2 _p3 _p3 -0
P22
- 12 - 22 - 12 - 22 - c
4. Some PBIB deSignS with three associate classes obtained from the
configuration of secants and an external Foint of a guadric.
m
Let ~ be a nondegenerate quadric in PG(2"S), s = 2 • Any line
of PG(2,s) which intersects the quadric in two points is called a
secant.
We shall now prove a few lemmas which will be useful later.
Lemma 4.1.1.
Let P be a point other than the nucleus of polarity of
PG(2,,8), s =
.;n,
which do not lie on Q2.
~
in
The number of secants passing
through P is ~ •
Proof.
P with respect
...
We shall assume that m > 1.
to~.
T(p) is a line.
Let T(p) denote the polar of
It is known that a line can be
either completely contained in a quadric or can intersect the quadric
in two points or can intersect the quadric in one point or can intersect
85
the quadric in no points.
Since P is an external point and T(p) con-
tains P, T(p) is not contained in~.
tains another external point P
external points of
~
l
(say).
Hence it follows that T(p) conThen the points P and Pl are
and are mutually conjugate.
So by theorem 7.2
of Chapter I the line PPl' i.e. T(P}, contains only one point of the
quadric~.
Let R be any point of the quadric not lying on T(p).
l
P and R are mutually not conjugate.
l
Then
Hence by theorem 7.3 of Chapter I
the line PR contains another point of the quadric. So PR is a secant
l
l
of the quadric. Hence for every point R of the quadric not lying on
T(p) PR is a secant.
The number of points of the quadric not lying on
T(p) is s and every secant contains two points of the quadric.
.e
Hence
the lemma follows •
Let S denote the nucleus of polarity
of~.
Let P be any point
of PG(2 , s) not lying on ~.. Then T(p) contains S.
By lemma 4.1.1 T(p) contains a single point RO of ~.
Proof.
Then the points P and R are mutually conjugate.
O
contains P.
So the polar of
Hence it follows that the polar of RO or the tangent line
at R is the line PRO.
O
same as T(P) •
Hence T(RO) I the tangent line at ROI is the
Then the lemma follows from theorem 7.1 of Chapter I.
Let P be a point other than 8, the nucleus of polarity of
l
..
which do not lie on
P
l
Ra
such that PP
l
~.
~I
Then the number of external points P other than
is a secant is
86
s(s-2)
2
Proof.
•
Obviously the required points are those which lie on a
secant passing through P but do not lie
l
on~.
The number of secants
passing through P is ~ by lemma 4.1.1 and every secant contains two
l
points of
~e
Hence the lemma follows.
Lemma 4.1.4.
Let P be an external point of Q other than 5. Then the number
2
l
of external points P other than 5 and P such that PPl intersects the
l
quadric in a single point is (s-2).
Proof.
Obviously the required points P lie on T(Pl ).
4.1.2 T.(P;L) contains 5 and one point Of~.
.e
By lemma
Hence the lemma follows •
Lemma 4.1.5.
Let P1 and P2 be two external points of
PlP2 is a secant.
~
other than 5 such that
Then the number of external points P other than 5,
P and P such that both PP and PP2 are secants is
2
l
l
(!2 Proof.
2)2 + (s-3) •
By theorem 7.2 of Chapter I, P1 and P2 are mutually non-
conjugate.
Let R and R be respectively the points at which T(Pl )
l
2
and T(P ) intersects the quadric. Then P R must be a secant. If pos1 2
2
sible, suppose P R is a tangent line intersecting ~ in a single point
1 2
R •
2
Then P an~ R are mutually conjugate.
l
2
the polar of Pl'
Then ~ occurs in T(Pl ),
So T(P ) intersects the quadric at two points R
l
l and
R which contradicts theorem 7.2 of Chapter I since T(p1) is a line.
2
87
So it follows that P R is a secant.
t =~.
and
Similarly P2Rl is a secant. Let
1 2
Let the t secants passing through Pl and P2 respectively be
Pl~l'
P1P2~'''·'
P1Mlt
P2~l'
P1P2~21.'"
Pht
where
Obviously the required points P must lie on one of the secants P1Mll,
P1P2~'
... , P1Mlt • Let us count the required number of points P
lying on each of the secants. Consider the common secant P1P2~2 which
contains two points of
~,
the points P1 and P2.
Hence the common
secant contains (s-3) points P satisfying the required conditions.
.e
Next we consider P1Mll • The external points lying on Pl~l which are
points of intersection of P1Mll and a secant passing through P2 will
possess the required property.
P1Mll contains two points
~l
and Mil
of Q2.
1'1-
The line P2R2 is a tangent line and P2Mh is a secant. Also P1P2M22
intersects Pl~l at Pl. Hence it follows that (t-2) of the secants
passing through P intersect P1Mll at external points other than PI' P2
and S. So the secant P1Mll contributes (t-2) points P possessing the
..
required property.
Next we consider the secant P M which contains
1 13
two points M13 and Mi.3 of Q2.
88
').
Both the line Pt1.3 and P~, are secants.
,
Pl.
P1P2M22 intersects P1M at
13
SO (t-3) of the secants passing through P intersects P1M at ex13
ternal points other than Pl' P2 and S. So Pl~3 contributes (t-3)
points P possessing the required property. Similarly each of the remaining secants contributes (t-3) points P.
Counting together the
points, we get the total number of points P as given in the lemma.
Lemma 4.1.6.
.-
Let P and P be two external points other than S such that P1P2
l
2
is a tangent line intersecting the quadric in a single point.
Then the
number of external points P other than Pl' P and S such that both PP l
2
and PP
2
are secants is
!(!
- 2) •
2 2
Proof.
Let R be the point at which P P
1 2
O
is any other, point of
of Chapter I.
~I
intersects~.
If R
both P1R and P2R are secants by theorem 7.3
Let the ~ secants passing through Pl and P respectively
2
be
and
An~
P2R2l ,
P2R21'
.. 01
P2R2t
where t
= 2's •
required point P must lie on one of the secants P1Rl !' P1R12 ,
... , P1Rlt •
Let us consider the points P which lie on P1Rll and possess
the required property.
The secant P1Rll contains two points Rll and Ril
89
Of~.
Both the lines P2Rll and P2Ril are secants.
So
(~
- 2) of the
secants passing through P intersect P1Rll at external points other
2
than P1" P2 and S (note that S, the nucleus of polarityI cannot lie on
any secant).
So the line P R
1 1l
contributes (~ - 2) required points P.
This is true for all the secants passing through Pl.
Hence the lemma
follows.
Let P1 and P be two external points of
2
~
other than S such that
P1P2 is an external line intersecting the quadric in no point.
Then the
number of external points P other than Pl" P and S such that both PPl
2
and PP are secants is
2
(~ _ 1)2 •
Proof.
Let R and R respectively be the points at which T(P )
l
l
2
and T(P ) intersect Q2. Since PI and P are mutually nonconjugate, by
2
2
theorem 7.2 of Chapter I, it follows that P R and P R are secants.
1 2
2 I
Let the secants passing through PI and P be respectively
2
and
where
P1M1I"
PI M12 ,
"
Pl~t
P2M2l"
P2~2'
"
P2M2t
MIl = R2
and
M21 = R1 •
By an argument exactly similar to that used in lemma 4.1.5 and 4.1.6,
we can easily see that PI M11 will contribute (~ - 1) points P possessing the required property and the remaining secants passing through PI
•
will contribute (~ - 2) points each.
Hence the lemma follows.
90
Lemma 4.1.8.
Let P and P2 be two external points of
l
P P is a secant.
l 2
~
other than S such that
Then the number of points P other than PI' P2 and S
such that pp1 is a tangent line and PP2 is a secant is
(~ - 2) •
~.
Let P1R be the polar of P
r
l
Then obviously the required
points P must lie on the line PlR "
l
Since Pl and P are nonconjugate,
2
it can be easily seen that P R will be a secant where R is the point
l
2 l
of intersection of the polar of P
l
and~.
So it the ~ secants passing
through P , (~- 2) secants, namely the secants other t~an P Pl and
2
2
P2Rl intersects P1Hl in An· !S2rt.ernal point other than Pl.
Hence the
lemma follows.
two external points of Q other than S such that
2
is a tangent line. Then the number of external points P other than
Let P and P
l
2
P1P
2
b~
P l , P and S such that PP is a tangent line and PP is a secant is 0.
l
2
2
Proof is simple and hence omitted.
Lemma. 4.1.10.
Let Pl and P be
2
~wo
Pl P is an external line.
2
external points of Q other than S such that
2
Then the number of external points Pother
than 8, Pl and P such that PP is a tangent line and PP is a secant is
l
2
2
(~ - 1) •
•
Proof.
Let P1Rl be
th~
the polar of P intersects Q2.
l
polar of PI' R1 being
~h~
point at which
Arguing as in lemma. 4.1.8, we can see
91
that P R U. a secant. Hence out of the ~ secants passing through P ,
2
2 l
."
(2s - 1) se:ants,
namely the secants other t~an P2Rl intersects P1Rl in
an externs,lpoint.
Hence the lemma follows.
Lemma. 4.1.11.
Let
P1 and P:2 be two external points of ~ other than S such that
P P2 is a secant.
1
P
2
Then the number of external points P other than Pl ,
and S such that both PP
Proof.
l
and PP are tangent lines is O.
2
Let P1Rl be the polar of P and P R2 be the polar of P2 •
l
2
It can be easily seen that every point P satisfying the required conditiona lie on both the lines P1Rl and P:2R2.
the
f~ct
P1R
l
Hence the lemma follows from
and P R intersects at the points, the nucleus of polarity
2 2
of ~.
Lemma 4.1.12.
Let Pl and P be two external points of
2
P P is a tangent line.
1 2
~
other than S such that
Then the number of external points P other than
P 'p2 and S such that both PP and PF are tangent lines is (s-3).
1
2
1
Proof follows easily from the fact that every point P satisfying
the required condition must lie on the line P P and the
1 2
tains all the three points P , P and S and a point of
1
2
~ine
P1P con2
~.
Lemma 4.1.13.
Let P , and P be two external points of
2
1
external line.
~
such that P P is an
1 2
Then the number of external points P
ot~er
•
and S such that both PP and PP are tangent lines is O.
1
2
Proof is exactly similar to that of lemma 4.1.11.
than P1 , P
2
Theorem 4.1.
Let B be the class of lines of PG(2,s) which are secants of Q2
and V be the class of external points of ~ other than 8, the nucleus
of polarity
Of~.
Then D(B,V) is a PBIB design with three associate
classes with the following parameters:
= s2
- 1,
r
= '2
b
= (S+l)~,
k
= (a-l),
=1
,
~2
"S
2
"'1
1
s
v
Pll =(s...~) +(2- 2 ) ,
'
= A3 = 0,
P~ = (~ -
nl
Pll
s(s
= 22
-
2
P12
P~
=
3
Pll
= (82
pie = (~ - 1),
=
-1
)2
¥
2
(s-3)
~ = (s-2),
2),
° ,
=°
,
1
P22
= ~(a-2),
)
2 ,
,
P~2 = 0
0
The other parameters can be obtained from the relation between
the parameters of a PBIB design.
Proof.
We shall apply corollary 2.1 to prove the theorem.
The
following results can be obtained easily.
v
= Number
of points of PG(2,s) other than S which do not lie
on ~'.
= s2
r
- 1.
= Number
of secants passing through a given external point
other than 8 ,
= ~ by
lemma 4.1.1.
93
k
= Number of external points lying on a secant,
= (6-1).
The association scheme is defined as follows.
Two external points PI
and P are first associates if the line P P is a secant, second asI 2
2
sociates i f the line PI P is a tangent line and third associates if the
2
line PI P is an external line. Since there can be at most one secant
2
passing through two external points, we have
A2 = A = o.
3
iii
The constancy of the parameters Pll' P12 p~ (i = 1,2,3) and their ex'
pressions follow from lemmas 4.1.5 to 4.1.13. The expressions for nl
Al
.e
=1
and
and n2 follow from lemmas 4.1.3 and 4.1.4.
Corollary 4.1.1•
The series given in theorem 4.1.1 contains the following two designs with r and k not greater than 10.
The values of some of the non-
zero parameters of these designs are given below.
Design
Number
Dl
D
2
v r
k
b
A
l
15
2 3 10 1
63
4
7 36 1
2
1
1
2
3
n n Pll
P12 Pll P22 Pll ;12
1 2
4 2
1
0
0
1
1
1
6
9
2
8
5
9
3
24
CHAPTER IV
A ClASS OF TWO ERROR CORRECTmG CODES WITH RATE OF TRANSMISSION
ARBITRARILY CLOSE TO UNITY AND FRACTIONAL REPLICATIONS
PRESEBVmG MAIN, EFFECTS .Am> IW PAEJTOB DTERACTIONS
1.
summary.
A set of points in PG(m,2) is defined to be a Rt-set if no t
points of the set lie in a(t-2)-flat.
points in a ma.x1ma.l Rt-set in PG(m,2).
Let Nt(m) denote the number of
It is proved that there eXists
a t-error correcting binary (n,k)-group codes with n places and k in-
.e
formation places if and only if N2t (n-k-l) ~ n. It is shown there
exists e. ~ fraction of' a 2n experiment, an experiment with n factors
2
each at two levels, Which preserves all main effects and c-factor interactions for cst if and only if there eXists a t-error correcting
(n,k) binary group code.
We have proved that
m
~ 2 -1
m
N (2m) ~ 2 + N (m-l)
4
4
N (2m-l)
4
•
Actual method of construction of R4-sets containing 2m_l points in
m
PG(2m-l, 2) and 2 +Nl,.(m-l) points in PG(2m, 2) are given Where Nl,.(m-l)
denote the number of points in the R4-set in PO(m-l, 2) constructed by
our method. These R4-sets give two error correcting codes, With
95
where c is any non-negative interger or zero.
So two-error correcting codes can be obtained with rate of trans-
mission kin arbitrarily close to unity. Also these R4-sets give a ~k
fraction of a 2n e.x;periment for preserving all main effects and two
factor interactions for values of n and k given in (1.1).
The follow-
ing table gives the values of n and k for some of the two error correcting codes that have been obtained.
For the same values of n and k we
can also have a ~ fraction of a 2n experiment preserving main effects
and two factor interactions.
n
,e
2.
Rate of transmission of the group
code ~ bits per symbol
k
11
4
.}6
15
7
.47
21
12
.57
}l
21
.68
}7
26
.70
6}
51
.81
74
61
.82
127
ll}
.89
148
1}3
.90
General problem of information theory.
Information theory deals with the problem of sending information
from one place to another through a channel subject to random disturbances in such a way that the probability of correct transmission of the
96
information is very close to unity and yet economic utilization of the
channel is made.
There is a source which is generating pieces of in-
formations or messages one after another in a sequence.
The messages
of the source are encoded into symbols which are transmitted through
the channel and the received symbols at the other end are decoded into
the original
Ire Bsage 0
The channel is subject to disturbances or noise
as a result of which it is not always possible to obtain the transmitted
information correctly from the received signal.
The entire system can
be described by the follOWing diagram.
(
.e
Informtion
Source
TransmUted
C
\!J
Encoding
Dest1nation
Receiver
I
el
T
\V
Decoding
Source of
Noise
For a mathematical treatment of the subject it is assumed that the output of the source can be described by a stochastic process.
Also for
any given sequence of informations transmitted through the channel, the
output at the receiving end can be described by a stochastic process.
Mathematically encoding is a function which assigns a unique sequence
of symbols of the transmitter to any sequence of messages of the source
and decoding is a function which assigns a unique sequence of messages
of the output to every sequence of symbols of the receiver.
97
Shannon in his fundamental paper
basic concepts ot information theory.
L23_7 tirst
introduced the
Entropy H of a source is a
measure of uncertainty about the information generated by a source,
rate of transmissionR is a measure of information transmitted per
symbol of the transmitter and capacity C ot a channel is the maximum
rate of transmission tor the given channel.
Shannon
["23J first
proved
that whenever the entropy ot the source is less than the capacity of
the given channel, it is possible to find a method of encoding and decoding such that the probability of correct transmission ot information
is arbitrarily close to unity and yet the rate of transmission is arbitrarily close to C provided certain conditions about the stochastic
.e
processes describing the source and the channel are satisfied.
Macmillan
L20J gave rigorous proofs ot some ot the Shannon results. Feinstein
L16J, Khinchin L18J, Woltowitz 1:27_7 and Blackwell, Breiman and
Thomasia.n L1J extended Shannon's results in various directions. However the 'WOrks of all these authors only prove the existence ot codes
With the required optimum propsrties but no method is available tor
actual construction of such codes.
Khinchin
L18J in page
116 com-
ments, "From the purely practical point of view, we must note again
that in both the Feinstein and Shannon methods the construction of codes
with the required properties is not given,; the existence of such a code
is proved but no indication is given of how to actually find it."
In
the present chapter we shall be concerned with the construction of codes
with some optimum property for a binary channel.
98
3.
Binary channel.
A binary channel is one which can transmit two symbols denoted by
o and
1.
Each piece of information is encoded into a sequence of n bi-
nary digits which are transmitted through the channel one after another.
A binary channel is called symmetrical if the probability of receiving
o when
1 is transmitted is equal to the probability of receiving 1
when 0 is transmitted.
For a symmetric binary channel in which the
noise operates independently on every symbol transmitted" the capacity
C of the channel is given by
L'23J
C = 1 + P log2 P + (l-p) log2 (l-p) bits/symbol"
bits meaning binary digits" where p denotes the probability of receiving
1 when 0 is transmitted.
SUppose the source can generate one of v possible messages or
letters at a time.
1he collection of the v letters is called the output
alphabet of the source. If v = '2 k" it is possible to transmit the messages by k-place binary sequences.
In this case the probability of re-
ceiving a transmitted letter correctly is (l_p)k which may not be close
enough to unity to be acceptable in actual practice.
of redundancy is introduced.
So the principal
Instead of sending k-place binary se-
quences" n-place binary sequences are used to transmit the information.
Roughly speaking" out of the n-pla.ces of the binary sequence" k-places
a.re used as information positions and (n-k)-places are used as check
positions as a result of which the probability of correct transmission
of a letter is increased. The 2k messages or letters of the alphabet
99
ot the source are identified With 2k n-place binary sequences chosen
out of 2n n-place binary sequenceso
This identification is called en-
Also a rule ot decoding is given which assigns a unique letter
coding.
or message to everyone ot the 2n possible received binary sequences.
The basic problem is that of finding a method of encoding and decoding
such tha.t the probability of correct transmission ot a letter is close
to unity and yet the price paid by way of redundancy is not very high.
Since we are using n digits to transmit information that could
be basically sent by k digits, it is reasonable to detine the rate of
information transmitted for such a binary code as
Cl
= kin
bits per symbol.
The basic existence theorem of Shannon
L23J for
the case of a binary
symmetric channel reduces to the following statement
fixed Cl
=C -
8 (8 > 0) and any fixed
E
£25J.
Given any
> 0, there eXists a code which
will transmit information at the rate C bits per symbol and Will decode
l
it With an error probability per block of n symbols Q (n,k,p)
l
Cl
<
E.
It
> C, no such N eXists.
However as already noted in section 1, the actual construction
ot such binary codes is also not known.
The problem ot construction of
error correcting binary codes was first considered by Hamming £17J.
Hamming solved the problem ot construction ot (n,k) binary codes which
can transmit the information correctly it there is at most one error in
the n binary digits transmitted.
.
Slepian
1:25_7 considers
(n,k) binary
group codes and gives a maximum likelihood decoder which maximizes the
100
probability of correct transmission of a letter for a given method of
encoding.
Slepian poses the problem of finding the (n,k) binary group
code which maXimizes the probability of correct transmission for given
nand k.
He solves this problem for small values of n and k.
['"19_7 solves Slepian's problem for k
= 3,4
and general n.
Kuebler
However the
codes obtained by Slepian and Kuebler have poor rate of information
transmittal and hence is not of much practical use.
For instance the
highest rate of information transmittal among the two error correcting
codes given by Slepian is .36 bits per symbol which is attained by the
code with n
= 11
and k
= 4.
In this chapter we have considered the
problem of finding two error correcting group codes with rate of information transmittal arbitrarily close to unity.
In the following sec-
tion we shall give a formal mathematical statement of the problem under
investigation.
4.
Statement of the problem.
Before stating the problem under investigation in formal mathe-
matical terms, we shall need a few definitions.
An n-place binary
sequence is denoted by a = (a ,a , ... ,a ) where each a is 0 or 1.
l 2
n
i
Weight of a sequence. The weight of an n place binary sequence a is
defined to be the number of places in which the binary sequence has the
number 1 and is denoted by w(a).
Hamming distance.
The Hamming distance d(a"l3) between two sequences a
and 13 is defined to be the number of places in which the sequence a has
1 and 13 has 0 or Vice versa.
101
Let B denote the set of 2n n-place b1na.ry sequences.
n
It is
known that B is a group with respect to vector addition modulo 2 of
n
the sequences.
Binary encoder.
A v letter n-place binary encoder E is a subset of Bn
cons:1s ting of v n-pla.ce sequences a p a , ••• , av •
2
Binary decoder.
between
th~
A v-letter n-place binary decoder D is a correspondence
v sequences a , a , ••• , a and v mutually disjoint sets
l
2
v
5~V
52' ••• , 5v of n-place binary sequences such that 51 U
•••
U.
5v
52
V
= Bn •
Binary code.
A v-letter n-place binary code C is the combination (E,D)
of a v-letter n-place binary encoder E and a v-letter n-place decoder D.
Suppose the source alphabet consists of the v letters (or messages) Al , A , ... , A •
2
v
When the binary code C is used to transmit
through a binary channel, the sequence a
will be transmitted if Ai is
i
the letter to be sent, i = l,2,o .. ,v, and at the receiver the letter
transmitted will be taken as Aj if the received n-place binary sequence
is a member of the set 5j , j
Minimum distance decoder e
= l,2, •• o,v.
A binary decoder D is said to be a minimum
distance decoder if for any sequence 13 of the set 5
i
d(ai , 13) ~ d(a j ,l3) for i,j
It is known that 5lepian's
= l,2, ••• ,v.
L25J maximum likelihood decoder
is a mini-
mum distance decoder.
A t-error correcting binary code.
A V-letter n-place binary code C is
said to be a t-error correcting code if the decoder is a minimum distance
102
decoder and for any n-place sequence
~,
8i ,
i = 1,2, .. • ,v.
SUppose the sequence 0i is transm1tted through a binary channel and there
d(Oi,f3) ~ t~ ~
€
is a disturbance in t or less of the places of 0i.
Then the received
sequence will be
0.1 + 'Y
=~
where 'Y is an n-place
sequence of weight not greater than t and addition is vector addition of
the two sequences modulo 2.
most t places.
80
the sequence
~
Will differ from 0i in at
Hence it follows that d(Oi'~) ~ t and consequently f3 is .
an element of 81 " Therefore the received signal will be read as
the transmission will be correct.
Rate of information transmittal.
01 and
For an n-place, v-letter, binary code
C the rate of information transmittal is defined as'
R
=
10g2 v
n
bits per symbol.
R will also be referred to as rate of transmission.
Rt-set and Nt(m).
A set of points in PG(m,2) is defined to be aRt-set
i f no t points of the set lie on a (t-2)-flat.
The number of points in
a maximal Rt-set in PG(m,2) is denoted by Nt(m).
Sequence of t-error correcting codes with asymptotic rate of transmission
equal to unity.
Consider a sequence {Cm} of binary codes such that the
m-th code of the sequence is a t ..error correcting code with n places
m
and km information places. The sequence of binary codes
Cm is said
to have asymptotic rate of transmission equal to 1 if
lim R
m->CD m
=1
103
where R denotes the rate of transmission for the code em. In this
m
chapter we have constructed such sequences of codes for t = 2 and in
the course of this study certain bounds on N4 (m) are obtained.
5.
Some preliminary results on group codes.
A v-letter n-place binary code is said to be a group code if the
sequences
°1 , °2,
••• , 0v of the encoder form a group with respect to
vector addition modulo 2.
For a group code, v is of the form 2k •
k
A 2 -letter n-place group
code is also referred to as an (n,k) group code.
Theorem 5.1.
There exists a t-error correcting (n"k) group code if and only if
there exists a R -set in PG(r-l, 2) containing n points and hence if and
2t
only if N2t (r-l) ~ n where r = n-k. This theorem is proved by Bose
in a different form. The proof given here is new and simpler.
Proof.
Necessity.
the encoder and let al"
encoder.
GF(2).
()J
Let. 00' 01' ••• , a2k_l be the elements of
°2"
• 00"
Ok be
k independent generators of the
Consider every n-place sequence as a vector with elements in
Let Vk den~te the vector space determined by the k independent
vectors 01'
°2"
'1t.
Obviously the dimensionality of Vk is k.
Hence the dimensionallty of the orthogonal vector space Vr is r where
r
= n-k.
Let
... ,
~l' ~2"
••• ,
~r
be r independent vectors in Vr •
i
-
.
= 1"2, ••• ,,r•
Let
104
Let
j
= l,2, ••• ,r.
Then Cj can be regarded as a point in PG(r-l, 2).
consisting of the points C1' C , ••• , Cn •
2
2t -set.
R
Consider the set
We shall show that
1-2.
11 is
a
Kuebler J:19Jh8S ehawa that an (n,k) group co4e with a minimum
distance decoder is a t-error correcting code if and only if
w(aj ) ~ 2t + 1
for any nonnull element a
j
= l,2, ••• ,v,
set.
v
j
= 2k_l.
of the encoder.
If possible, suppose the set
Then there exists pointe C1'
a (2t-2)-flat.
Hence we have w(aj ) ~ 2t + 1,
C ,
2
not a R2t -
.. ~, C2t of 1~2. which lie on
Hence at least one of the points C ' ... ,C2t will lie on
l
the linear space determined by the remaining points.
Cl lies on the
1-2. is
~inear
space determined by C ' ••• , C2t •
2
Let us assume that
Then there exists
constants ~2' ••• '~2t where ~ts are elements of GF(2) and all ~'s are not
o such
that
Then we have
for
Let ~
= (~1'~2' ••• '~2t
i
= 1,2, ••• ,r,
' 0,0, ••• ,0).
where
~l
= 1.
It can be easily seen that for any
vector 13 of Vr" we have
~
13' = 0
(lin) (nil)
105
So the vector ~ is orthogonal to the vector space Vr'
vector of V and hence
k
is a sequence of the encoder. This is a con-
S 2t
w(~)
tradiction since
~
Hence A is a
whereas the code is a t-error correcting
code and hence we!::) ~ 2t+l.
Sufficiency.
Assume that there exists a R2t-set
1-1 in PG(r-l,
SUppose ~1 consists of the points Cj , j
taining n points.
2) con-
= 1,2, ••• ,
n
I3lj
132j
=
where Cj
.
n-vectorsl3l, 132 1
"'1
Let (Xl' (X2' ••• ,
~
where k
= n-r.
Let V be the vector space determined by the r
r
•
•..
I3rj
I3r where l3i
= (l3il,l3i21 ••• ,l3in)'
i
= 1,2, ••• ,r.
be k n-vectors orthogonal to the vector space Vr
Let (Xo
= (0,0,. .. ,0)
be t~e null vector.
the vector space determined by al' (X2' • u,
'1t.
Let Vk denote
Let (XOj(Xl'" o,'1t'
k
~+l,.o.'(Xv_l
be the 2 n-vectors lying on the vector space Vk • Let C
be the (n, k) group code whose encoder consists of the v sequences
aO'
a l , ••• , (Xv_l and whose decoder is a minimum distance decoder.
code C will be said to be the code induced by the set
by
c(1-1)
also.
The
1-1 and denoted
We shall show that C is a t-error correcting group code.
Obviously the sequences (XO' (Xl' .... , (Xv_l form a group With vector addition modulo 2 as the group operation.
a t-error correcting code.
If possible, suppose C is not
Then by Kuebler's result there exists an
element of the encoder whose weight is less than (2t+l).
an n-vector !::
= (0
0 ••• Ai 0 Ai
1
0 ••• Ai
2
So there exists
0 ••• 0) With at most 2t
2t
non-zero elements which is a. sequence of the encoder.
So!:: belongs to
106
the vector space V • Without any loss of generality let us take
k
~
= (Al'~2""'~2t'0,O, ••• ,O).
Since ~ is an element of Vk, ~ must be
orthogonal to all the vectors of Vr • Hence we have
~l~il + ~2~i2 + ••• + ~2t~i2t
=0
for i
= l,2, ••• ,r.
Now it follows easily that the points Cl , C2 , Q'" C2t of 1-1 lie on a
(2t-2)-flat which contradicts the fact that 1-1 is a R2t-set.
Corollary 5.1.
If there exists an (n,k) t-error correcting group code, then
there eXists an (n-l, k-l) t-error correcting group code.
6. Relationship between error-correcting binary group codes and
fractional replications of factorial experiments at two levels.
The problem of finding ·.!-k fraction of a. 2n experiment, an experi2
.
ment with n factors each at two levels, which preserves all main effects and interactions up to (t-l)-th orderuualiassed with main effects
and interactions up to (t-l)-th order is closely related to the problem
of finding error correcting (n,k)-group code.
combinations x
= (xl,~, ••• ,xn)
Suppose the treatment
of the n factors are represented by
points in EG(n,2), the finite Euclidean geometry of n dimensions based
on GF(2).
let
l
=
i l xl
+ 12 x2 + • u
denote a pencil of (n-l)-flats in EG(n,2).
trasts corresponding to the pencil
I
+ lnxn = constant
It is known that the con-
belong to a main effect or a
t-factor or (t-l)-th order interaction according as the number of l's
among 11 ,
t2 ,
••• ,
tn
is 1 or t.
107
Let 01" 02' ••• ,
0i
= Oil~
~
be k independent linear forms where
+ 0i2 x2 + ••• + 0inxn' i
= 1,2, ... ,k.
Consider the (n-k)-
flat determined by the equations
(6.1)
01
it
= 02 = ••• = Ok = 0
•
Consider the
fraction of the 2n experiment which contains only the
2
2n - k treatments corresponding to the 2n- k points lying on the (n-k)flat (6.1). If' a fractional experiment is conducted containing only
these 2n- k treatments, it is known that a contrast belonging to the
pcncil
l
will be aliassed (or confounded) with the corresponding con-
trasts of all the pencils
l
+ "'101 + ••• + "'k~
are elements in GF(2) and ("'1,"'2, ••• ,"'k)
= constant where ""s
1 (0,0, ••• ,0).
Theorem 6.1.
There exists a -\ fraction of a 2n experiment which preserves all
2
main effects and interactions up to (t-l)-th order unaliassed with main
effects and interactions up to (t-l)-th order if' and only if there
exists a t-error correcting (n,k) group code.
Proof.
Suff1ciencl_
Assume there exists an (n,k) t-error cor~I .... ,
Ok
recting group code.
Let 01'
the encoder where 0i
= (Oi1'0i21. c. 'Oin)'
a
be k independent generators of
i
= 1,2, ••• ,k.
Let i denote the linear form 0il~ + ai2~ + ... + ain~.
I: _ denote the (n-k)-flat in EG(n,2) determined by the equations
n k
al = a2
= ••• ~ =
Let
o.
Consider the fractional replication containing the 2n- k treatments corresponding to the 2n- k points of I:n- k. We shall show that this
108
fraction preserves all main effects and interactions up to (t-l)-th
order unaliassed with main effects and interactions up to (t-l)-th
order.
In this fraction the contrasts belonging to a pencil
I = f l Xl
+ 12 x2 + ... + lnxn
= constant
will be aliassed with the corresponding contrasts of all the pencils
1 + ~lcil
+ ~ii2 + ... + ~k~ = constant
where ~IS are elements of GF(2) and (~P~2'''.'~k)
I:
(0,0"00.,0).
HenCE!
to prove that the fraction corresponding to En- k possesses the required
property it will be sufficient to show that none of the linear forms
t
+ ~lal + • eo + ~k~ has less than. (t+l) non-zero coefficients when-
ever
l
has at most t non-zero coefficientso
Since C is an (n"k) t-error correcting group code" by KUebler's
result any of the 2k_l non-zero sequences ~lQl + .0 .. + ~kC1t has weight
at least equal to (2t+l).
Hence anyone of the linear forms
~lcil + ••• + ~~! (~l'· •• '~k)
zero coefficients.
Also
1 has
I: (0".0.,0),
has at least (2t+l) non-
at most t non-zero coefficients.
it follows that the linear form
1 + ~lal
+
000
Hence
+ ~~ has at least
(t+l) non-zero coefficients.
Necessity.
Assume there exists a ~ fraction of a 2n experiment
2
which preserves all main effects and interactions up to (t-l)-th order
unaliassed with main effects and interactions up to {t-l)-th order.
Let
this fraction be determined by an (n-k)-flat En _k of EG(n,2) whose equations are
'".
...
...
• ••
109
Let 0i denote the n-place sequence (Oil'0i2'."'Oin)' i = 1,2,. .. ,k.
Let C denote the (n,k) group code whose encoder consists of the sequences ~lal + ... + ~'1t where ~'s are elements in GF(2) and addition
is vector addition modulo 2 and whose decoder is a minimum distance
decoder.
Since the fraction
corre~nding
to E k preserves main effects
n-
and interactions up to (t-l)-th order, it can be easily seen that any
of the linear forms ~l~ + ••• + ~k~ for (Al'."''k) ~ (0, ... ,0), has
at least (2t+l) non-zero coefficients.
~101 +
~ (0,
.e
Hence every sequence
+ ~'1t has weight at least equal to (2t+l) for (Al ,. .. ,Ak )
,0).
Now it follows from Kuebler's result that C is at-error
correcting code •
7. A correspondence between the points of PG(n,2) and the elements
of PG(2n +l ).
We shall set up a correspondence between the points of PG(n,2)
and the elements of GF(2n +l ).
a
We shall make
= (aO,al,~.o,an)
e.
corrEispond
a
Let
be a point of PG(n,2).
to the element
a of GF(2n +l ) where
= a O + alx + ••• + an>f
and vice versa and x is an undetermined element of GF(2).
Obviously
the correspondence is unique.
Gt-set.
.
A set of elements of GF(2n +l ) is said to be a Gt-set if the sum
of no c elements of the set is 0 for c
~t •
110
Theorem 7.1.
Let ~1 be a set of points in PG(n,2) and A be the corresponding
set of elements of GF(2n +l ).
The set
the set A is a Gt-set prov~ded
!!22f•
Sufficiency.
n l
GF(2 + ) is a Gt-set.
1-1 is
1-1 contains
an Rt-set if and only if
at least t independent points.
Assume that the set A of the elements of
poss.ib~e" suppose
1-1 is not an Rt-set, Then
of 1-1 which lie on a (t-2)-flat.
If
there exists t points a , a 2 , ••• , at
l
Then at least one of the points will lie on the linear space determined
by the other points.
Without any loss of generality we can assume that
a l lies in the linear space determined by a , a"
2
.
(7.1)
al
= 'l\.2a2
... , at.
+ 'l\.,a, + ••• + 'l\.tat
where 'l\.' s are elements in GF(2) and not all 'l\.' s are O.
of the 'l\.' s are non-zero.
Then we have
Suppose (c-l)
Without any loss of generality, we can assume
that
'l\.2 = ~
= .'0 = 'l\.c = 1,
c < t.
Then it follows easily from (7.1) that the sum of the elements al , a2 "
... " a of A is 0 where Q. is the element of GF(2n +l ) corresponding to
c
ai" i
1
~
1,,2"'G,to
But this contradicts the assumption that A is a
Gt-set.
Assume
1-1 is
an Rt-set.
Then there
ex1~ts
elements a l ,
Necessity.
not a Gt-set.
If possible, suppose A is
c
a2 ,
:s
•• e"
Q
c of A such that
t,
and the sum of the d's of these elements is non-zero for d < c.
denote the point of
1-1 corresponding
to the element Qi of A"
Let a i
111
i
= 1,2, ... ,t.
Then it follows easily from (7.2) that
= a2
+ ~.. + ac.
3
From (7.3) we can see that the points al' a 2 , ... , a c determine a (C-2)(7.3)
al
+a
Iet ac+l' a c +2 ' ... , at be (t-c) mutually independent points
lying in 1-1 which are independent of the points a , a , ••• , a c ' Then
l
2
it is easy to see that the points a , a , ••• , at of 1-1 lie on a (t-2)l
2
flat which contradicts the assumption that 1-1 is an Rt-set.
flat.
8.
m
An RJ.,.-set in PG(2m-l, 2) containig (2 _l) points and a sequence of
two error correcting codes with asymptotic rate of transmission
equal to unityo
..
In this section we shall obtain a sequence of two error correct-
ing codes with asymptotic rate of transmission equal to unity.
we shall prove a few lemmas
First
0
Lemma 8.. 1.1.
Let
ing
Ct
be a primitive element of GF(s2), the Galois field contain-
s2 elements where s is a prime power.
o
,
/'Vs+l
~
2(s+l)
,or
..
".~~"
Consider the set of elements
/'V(s+l) (s-l) .
~
•
This set of elements constitute a subfield of GF(s2) with the operations
of addition and multiplication as in GF(s2).
This lemma is a standard result in the theory of Galois field
and hence the proof 1s omitted.
Lemma 8.1.2.
Let x" y" z be three non-zero elements of GF(s) such that
112
x +y + Z
= 0,
s
= pm,
p
13
•
Then
This lemma is a result in elementary algebra which holds for any field
and hence the proof is omitted.
Lemma 8.1.3.
Let x, y, z and w be four non-zero distinct elements of GF(s),
s = 2Q, such that
x + y + z +W
~ + .; + z3 +
Then
Proof.
=0.
w3 1 0.
Since the characteristic of the field is 2, we have from
the assumption of the lemma
x +
y+
Z
= o.
Combining both sides and ex,panding by binomial theorem for positive index which obviously holds for a Galois field, we get
~ + .; + z3 + 3(X+Y)(Y+z)(z+x)
(8.1)
= w3.
The lemma follows from (8.1) using the fact that the characteristic of
the field is 2 and the addition inverse of any element in such a field
is the element itself.
Theorem 8.1.
Let
.
Q
be a primitive element of GF(s2), s
consisting of the elements
Q+
,"
= 2m..
where
Q
~, Q2 +
Q
Q6,
Q
0.0'
Q(s-l) + a ~(S-l)
= a(s+l)
0
Let A be the set
113
Then A is a G4-set consisting of (2m_l) elements.
Proof. We have to show that the sum of no c elements of A is 0
for c < 4.
Let us consider two elements Qi + a Q3 i and Q3 + a ~j, i ~ j,
i,j
= l,2, ••• ,(s-1).
The sum of these two elements is
The sum is zero if and only if either (8.2) or (8.3) given below is true.
Qi + Qj
(8.2)
{ ~i + Q3j
Qi + Qj
.
Since i
=0
~
,
=0
•
= a(Q3 i
+ ~j) •
j and in a field of characteristic 2, the addition inverse of
any element is the element itself, (8.2) cannot be true.
By lemma (8.1.1) the set of elements 0, Q, Q2, ••• , Q(s-l) constitute a subfield of GF(s2) with respect to the operations of addition
and multiplication as in GF(s2)" So both (Qi+Qj) and Q3 i +Q3j are elements of the subfield. Since a is not an element of the subf1eld, it
follows that a(Q3 i +Q3j) is not an element of the subfield. Since
Qi + Qj is an element of the subfield and a(Q3i~j) is not an element
of the subfield, (8.3) cannot be true..
This proves that the sum. of no
two elements of A can be zero.
Now let us consider three elements Qi + a Q3 i , Qj + a ~j and
..
Qk + a ~k, i ~
j
~ k, i,j,k
= 1,2, ••• ,s-1.
The sum of these three ele-
ments can be zero if and only if either (8.4) or (8.5) given below is
true.
114
Q1 + Qj + Qk
(8.4)
= 0,
{. ~i + g3j + ~k
g1 + Qj + gk
(8.5)
= 0,
= a(Q3i~j~k).
By lemma. 8.102, (8.4) cannot be true.
Using the fact that the set of
elements 0, g, Q2, ... , Qs-l constitute a subfield of GF(s2) and a is
not an element of this subfield, we can show that (8.5) is not true.
This proves that the sum of no three elements is 0.
Using lemma. 8.1.3
and an exactly similar argument, we can show that the sum of no four'
elements is 0.
Corollary 8.1.1.
N4 (2m-l) ~ 2m_1The existence of a G -set A in GF(22m ) containing (2m_l) elements im-
4
plies the existence of an R4-set in PG(2n-l, 2) containing (2m_l) points.
Hence the corollary follows from the fact N ,(2m-l) denotes the number
4
of points in a ~x1mal R4-set in PG(2m-l, 2)0
Corollary 8.102.
Let Am denote the G4 -set in GF(22m ) containing (2m.l) elements.
Let
denote the corresponding R -set in PG(2m-l, 2). Let C denote
4
m
the (n,k) group code induced by the set
with n = 2~-1.1 and
m
k = 2 _1_2m as used in the sufficiency part of theorem 5.1. Then {Cm}
i-1m
i-2m
is a sequence of two error correcting grO\4P codes with asymptotic
•
rate of transmission equal to unity•
The corollary follows from theorem 5.1 and the fact that the rate
of transmission of the code C is
m
115
m
Rm = ..m
~ -1
2 -1·2m
which tends to unity as m goes to infinity.
9. An R4-set in PG(2m,,2) containing tJ1+N4(m-l) points and a sequence
of two error correcting codes with asymptotic rate of transmission
equal to unity.
Defin1tions.
sum
of points.
Let Pl " P " ..
2
Pi
.
e,
P be k points in PG(n,,2).
k
= (aiO,ail, ••• "ain)'
i
Let
= 1"2" ••• ,,k.
The sum of these k points is the point
where addition is in GF(2) and th~. ~um is denoted by P + P + ... + Pke
l
2
Null point. The vector
= (0,0" 88.,,0) containing all zero elements is
called the null point.
Pl " P2 "
eo."
°
° is not a point of PG(n,,2).
For k points
Pk " if' we ha~ ..
P2
= Pl
+ ••• + P i - l + Pi +l +
we shall say that the sum of the k points is 0.
distinct collinear points in PG(n,,2) is 0.
points of' PG(n,2) which are coplanar is
+ Pk "
o.e
i
= 1"2, ••• "k,,
Thus the sum of three
The sum of' four distinct
° if no three of the points are
collinear•
.
"
Image of a set.
Let
outside the set
1-1.
1-1 be
a set of points in PG(n,,2) and P be a point
Then the image of the set
n
With respect to P is
116
defined to be the set of all points P + X where X is a point of 1-1 and
is denoted by
P + 1-1
•
Linear envelope of a set.
The linear envelope of a set 1-1 of points in
PG(n,2) is defined to be the set of all points in PG(n,,2) which lie on
a line determined by two points of 1-1 and is denoted by L(1-1).
It is
easily seen that L(1-1) consists of the points of 1-1 and points which
are sums of two points of 1-1.
Lemma 9.1.1.
Let 1-1 be an R -set in PG(n,,2) lying on an (n-l)-flat En _r
4
Then the set
1-1*
=
{p}
U
P + 1-1
is an R4 -set where P is a point not lying on E _ •
n l
Proof. It will be sufficient to show that the sum of no three
points of 1-1* is 0 and the sum of no three points occurs in 1-1*.
Every point of 1-1* can be represented as P + X where X is the null
point or a point of
.
P.
11.
When X is the null point" P + X is the point
.
Consider three points P+Xl.' P+~" p+x, of 1-1*.
three points is P + Xl +
~
+ x,.
Since each Xi" i
the null point or a point of En _l " it follows that
point of En:- l •
cannot be O.
o or
o or
"
The sum of the
= 1,2,,3 ,
JS.
+
~
+
is either
X,
is a
Also P is a. point outside En _l • Hence P + ~ + ~ + X,
The sum can occur in 1-1* if and only if ~ + ~ + X, is
Since 1-1 is an R4 -set" ~ + ~ + X, cannot be
This completes the proof of the lemma.
is a point of 1-1.
a point of 1-1.
117
I.et
..
1-11 be
an R -set in PG(n,2) lying on an (n-l)-flat E _ and
n l
4
1-~ be another R -set in PG(n,,2) such that no point of 1-~ lies in
4
E _l -
n
Then
I.et
1-1 is
an R4-set if
L(1-1l ) n L(1-2e) = ¢,
the null set_
Proof _ It will be sufficient to show that the sum of no , points
of
1-1 is
0 or occurs in
for the points of
points of
1-11
1-1 can be
(i)
1-1-
and
We shall use X and Y as generic notations
1-2e respectively_
The triplets of the
of the following four kinds:
(~"
X2" X,) "
(ii)
(Yl , Y2 " Y,) ,
(iii)
(Xl" Y2 " Y,) ,
(iv)
(Xl' X2" Y,) "
Consider a triplet of the kind (i).
Since all the three points
1-11'
which is an R4-set, the sum ~ + ~ + X, cannot be
o and the sum does not belong to
1 - Also since 111 is contained in
En _l , the sum of three points of 1-11 is a point of En _1 does
not contain any point of En _l • So Xl + ~ + X, cannot be a point of
are points of
.r1
1-4
12e-
Therefore ~ + ~ +
X,
is not a point of
Consider a triplet of the kind (ii)_
Since all the three points are points
sum is not 0 and does not belong to
•
1-1-
The sum :Is
of.1le which is
1-12 -
Yl + Y2 + Y,_
an R4-set, the
Also as all the three points
are points lying outside E _l' it follows easily that Yl + Y2 + Y, is a
n
118
1-~ is completely contained in E _ •
point outside E _
n r
n l
1-11 ,
l + Y2 + Y3 cannot be a point of
a point of 1-1,
Y
Hence Yl + Y2
-I-
So
Y cannot be
3
The sum is Xl + Y2 + Y ,
3
Consider a triplet of the kind (iii).
If possible, suppose
~ + Y2 + Y
3
(9.1)
Xl is a, point of
L(1-1l )
tion (9.1) implies that
=0
,
and Y2 + Y is a point of
3
L(1-1l )
L(!.-12).
The equa-
and L(1~) have a common point which
contradicts the assumption that
L(1-1l ) n
~(1-12)
So
JS.
JS.
+ Y2 + Y3 cann~t be a point of
+ Y2 + Y cannot be O.
3
is a point of
1-1.
= ¢,
the null set.
By similar arguments we will show that
1-1·
If possible, s~ppose
JS. + Y2
+Y
3
Then either (902) or (9.3) is true.
(9.2)
Xl + Y2 + Y3
= Y4 "
(9.3)
~
+ Y2 + Y
= X2
~
= Y2
3
0
From (9.2), we have
(9.4)
+
Y + Y4,
3
1-)e
and hence points outside E _1"
n
Y2 + Y + Y4 is a ~oint outside E _ , Also 'S. is a point of E _ • So
n l
n l
3
it follows that (9.4) is a contradictiono
Since Y2 , Y and Y4 are points of
3
From (9.3) we have
(9,5 )
~ + X
2
Si~ce Xl + X2 is a point of
(9.5) is a contradiction.
= Y2
+ Y
3
L(1-1l )
Q
and (Y2 + Y ) is a point of L(1-~),
3
So ~ + Y + Y cannot be a point of 1-1.
2
3
119
Similarly we can show that the sum of three points o:f the kind
(iV) cannot be 0 or a point of
.
-C1.
This completes the proof of the lemma•
Let I: _ be a (2m-l)-flat in PG(2m,2). Let the points of I: _l
2m l
2m
2m
be made to correspond to the elements o:f GF(2 ) as in section 7. Let
a be a primitive element of GF(22m ).
a~,
Q+
Q2 + a Q6, '. . ~,
Let A denote the set of elements
l
Q(s-l) + a ~(s-l) where Q=a(S+l), s=2m•
Let F1 denote the set of elements
Q, g2 , ~, ••• , g(s-l)
which form the non-zero elements of GF(s).
Let A2 be a G4-set in GF(22m ) which is a subset of Fl containing
N (m-l) points. Let 1-11 and
respectively denote the set of points
i-le
4
of ZZlll-l corresponding to the sets Al and A2 of elements of GF(2
where P
i~ ~
1-~ =
i p} U
.C1 =
1~11 U 1-1~
P +
1-12
2m
). let
'
point outside E2m _1 o
Theorem 9.1.
m
The set 1-1 in PG(2m,2) is an R4-set contai~ing 2 +N (m-1) points
4
where N4 (m-l,) denotes the n~ber of points in 1-~.
Proof.
.
By theorem 8.1 ~, is a G4-set in GF(22m ).
By assumption
A2 is, a G4-set. ~ by theorem 7.1, both ~-ll and 1-12 are R4-sets in
E _ • By lemma 9.1.1, 1-1~ is an R -set. So the theorem will :form
2m 1
4
lemma 9.1.2 i:f we can show that
L(1-1 )
1
fi
L<,{-~)
='/;,
the null set.
120
Let X and Y be generic notations for points of
respectively.
sinc~
are points of E _ •
2m 1
1-11
and
1-,k
1-11
is a subset of E2m_l " the points of L(1-1l )
The points of L(1-~) w~ch lie in E2m_l are
either of the form (i) Y or (ii) Yl + Y2 • Renee if L{1-1l ) and L{1~~)
l
have a common point one of the folloWing four equations must be true.
JS.
(9.6)
= Yl •
.
(9.7)
Xl • Yl + Y2 •
(9.8)
Xl + ~
(9.9)
JS. + x2 = Yl
= Yl
•
+ Y2 "
SUppose (9.6) is true. Let the element of A corresponding to Xl be
l
i
(gi + Ct g3 ) and the element of A corresponding to Y be gj"
l
2
i"j = 1,,2, ••• ,s-1. Then (9.6) implies
(9.l0)
Which is a contradiction by lemma 80101.
SUppose (9.7) is true.
Let the elements of A corresponding to
2
and 1 be respectively gak and. gk Let the element of A correspondl
2
1
ing to Xl be gi + a g3i. Then (9.7) implies
1
0
(9.11)
gi + gj + gk
= Ct g3i
which is a contradiction by lemma 8.1.1.
Similarly we can show that each
of (9.8) and (9.9) leads to contradiction.
Corollary 9.1.1.
m
Nt,. (2m.) ~ 2
•
Proof.
+ N~ (m-l).
By theorem 7.1 there exists a Gt,. -set in GF{2m) containing
Nt,. (m-l) points Where Nt,. (m-l) denotes the number of points in a maximal
121
R4-set in PG(m-l,2).
Also by lemma 8.1.1 the set Fl of elemente
2 ...
(e-l)
0, Q, Q , ••• , Q
constitute a subfield Which actually is
isomorphic with GF(2m), s~.
Hence there ie a subset ~f Fl which is a G4-set and cont~inS N4 (m-l)
pointe. SO in theorem 9.1 we can take N4(m-l) = N4 (m-l). Hence the
corollary follows from the fact
th~t
theorem 9.1 provides an R4-set
1-1 contain.ing
~ + N4(m-l) points.
Corollary 9.1.2.
let
i-lm be
the R4-eet in PG(2m,2) containing 2m + N4(m-l) points.
i-lm
let Cm be the (n,k) group code induced by the R4-set
as introduced
in the sufficiency part. of theorem 5.1 with n = 2m + N4(m-l) and
k =
'if!
+ N4(m-l) - 2m-l.
Then {Cm} is a sequence of two error cor-
recting codes with asymptotic rate of transmission equal to un! ty.
Proof follows from theorem 5.1 and the fact that the rate of
transmission of the code Cm is I\n where
2m + N4(m-l) - 2m-l
R =
m
+ N4(m-l)
zn
10.
•
Examples illustrating the method of constructing R4-aets.
Example I.
R4-set in PG(3,2) by the method of theorem 8.1.
The minimum function for GF(24 ) according to Carmichel
is
rex) = x4
•
Let x denote a primitive root.
+ x + 1
Then we have
J:llJ
122
6+1
I')
Q=x
=X"'"
s = 22 ..
Using the minimum function we obtain the following po1ynomia.1 representations for the powers of Q
Q
= x2
+ x"
= x2
Q2
~ = 1.
+ x + 1,
The G4-set A in GF(24 ) consists of the following three elements.
-~
= x2 ,
()1 = Q + x ~
6
2
()2 = Q + x, Q
a, = Q3 + x Q9
=1
= x2
+ 1,
+ x.
SO the corresponding R -set consists of the following three points.
4
(0 0
Example II.
1
0),
(1
0 1 0),
(1 1
0 0).
R -sets in PG(5,2) by the method of theorem 8.1.
4
The minimum function in GF(2 6) a.s given by Carmichel £11
J
= x6 + x
f(x)
Let x denote a primitive root.
+1
is
~
Then we have
Using the minimum function, we obtain the following polynomial expressions for the powers of Q.
Q = x4 +
Q4
= x4 + x 2
7 _
Q
~,
-
Q2 =
+ x,
.;>
Q5
~
= x4
+
+ x + 1,
+
.;>
Q3 =
+ 1, Q6
.;>
+ x'2 + x,
= x4 + x'2
+ x + 1
1.
Hence the G4-set A in GF(26) consists of the following 7 elements.
()1 = Q + x Q3
= x2 ,
Q'2 = Q2 + x Q6 =
:p
+ 1,
394
4121')
'2
a,=Q +xQ =x,~=Q +xQ
=X"'+X,
a, = ~
+ xQ15 =
~
+
~
+ 1"
a:., = Q7
06 = Q6 + xQ18
+ xQ'21
=1
=~
+ x4. +
~
+ x + 1"
+ x.
The corresponding R4.-set in PG(5,,'2) consists of the following 7 points.
(001000)"
(100001)"
(100101),
(000010)"
(110111)"
(001001)"
(11000).
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