'.
i::
:.~
ROTATABLE DESIG-NS OF SECOND AND THIRD ORDER
,
-j
_...
IN THREE OR MORE DIMENSIONS
..,..-.~
by
R. C. Bose and Norman R. Draper
;'
\0
University of North Carolina
This research was supported by the United
states Air Force through the Air Force
Office of Scientific Research of the Air
Research and Developmp,nt Command, under
Contract No. AF 18(600)-83. Reproduction
in whole or in part is permitted fer any
purpose of the United States Government.
\
Institute of Statistics
Mimeograph Series No. 197
May, 1958
11
ACKNOWLEDGMENTS
I am deeply grateful to Professor R. C. Bose who suggested
the research topic and provided many valuable comments and improvements during the course of the investigation.
To work under his
superb guidance has been both a privilege and a pleasure.
During the preparatory stages of this work I received financial assistance trom the Office of Naval Research and the United
states Air Force, to both of whom I acknowledge my appreciation.
In
the later stages, as a Bell Telephone Laboratories Graduate Fellow,
I received support from the Bell Telephone Laboratories.
I teel
greatly honored to have been selected tor an award and am very grate-
e,.
."
tUl tor the generous financial aid it prOVided.
My thanks are also due to two of my colleagues:
Dr. Jean
Engler, who typed and commented helpfUlly on the entire final draft;
and Mr. Dana Quade, who assisted With the computations of one section and also made helpfUl suggestions at the final stage.
In addition, I am gratetul to Miss Marianne Byrd tor her very
competent typing as well as her patience, consideration and help
which considerably eased the tediousness of the tinal preparation.
Finally I should like to acknowledge my indebtedness to Dr.
.
'
•
Walter L. Smith whose kindness several years ago made this whole
venture possible.
i11
TABLE OF CONTENTS
CHAPTER
PAGE
11
ACKNOWLEDGMENTS
I.
INTRODUCTION
v
SECOND ORDER ROTATABLE DESIGNS IN THREE DIMENSIONS
1
1. Definitions:
second order rotatable arrangements
and second order rotatable designs • • • • • • ••
1
2. A transformation group in three d!mensions and
its generated point sets • • • • • • • • • • • ••
8
3, The formation of rotatable arrangements and
rotatable designs by combination of the generated
point sets " • • • • • • • • • • • • • • • • • • •
•
17
4. The formation of further infinite classes of
second order rotatable designs by combination of
"
sets with positive and with negative excess. • ••
22
5. Classes of designs using sets with variable
excess • • • • • • • • • • • • • • • • • • • • • •
II.
27
FURTHER SECOND ORDER ROTATABLE DESIGNS IN THREE
DIMENSIONS
1. A second method of generating point sets suitable
for bUilding second order rotatable designs. • • •
•
2. Another construction method for second order
rotatable designs in three dimensions.
3. An extension of the method:
......
a sixteen point
design class • • • • • • • • • • • • • • • • • • •
62
PAGE
CHAFTER
•
III.
SECOND ORDER ROTATABLE DESIGNS IN POUR OR MORE
DIMENSIONS
1.
73
The generation of point sets in three or more
dimens ions • • • • • • • • • • • • • • •
.....
73
2 •. Infinite classes of second order designs in four
and more d~ens1ons • • • • • • • • • • • • • • • •
80
3. A method of constructing a second order design
in k
d~ensions
using a second order design in
(k-l) dimensions • • • • • • • • • • • • • • • • •
IV.
e•
THIRD ORDER ROTATABLE DESIGNS IN THREE DIMENSIONS
103
1.
Conditions for a third order rotatable design. • •
103
2.
The known third order rotatable designs in three
dimensions • • • • • • • • •
•
96
.. .. .. .. ...
108
3. The construction of infinite classes of third
order rotatable designs in three dimensions •• • •
109
4. Further infinite classes of third order rotata.ble
designs in three dtmens1ons.
BIBLIOGRAPHY
I)
•
...........
124
130
v
INTRODUCTION
The technique of fitting a response surface is one widely used
(especially in the chemical industry) to aid in the statistical analysis of experimental
in some unknown
wor~t
fa~hicn,
in which the "yield" of a product depends,
on one or more controllable variables.
Be-
fore the details of such an analysis can be carried out, experiments
must be performed at
prede~ermined
levels of the controllable factors,
1.e., an experiment.al deRign must be selected prior to experimentation.
Box and Hunter
[3 J
1 suggested Jesigns of a certain type,
which they called rotatable, as being suitable for such experimentation.
Very few of these designs were known a year ago.
time the work of R. L. Carter
•
•
Since that
[6 J has provided many new second order
rotatable designs in two factors.
However, additional methods were
needed which would provide both second and third order designs in
three and more factors.
meet this need.
The present work represents an attempt to
Several new construction methods for obtaining ro-
tatable designs of second and third order in three and higher dimensions are presented.
By use of these methods various infinite classes
of designs are obtained, and it is shown that all the rotatable designs previously known may be derived as special cases of these iufinite classes •
•
1 The numbers in square brackets refer to the bibliography listed at
the end.
vi
Most of the previous work on the construction of rotatable designs, by Box and Hunter ( [ ) J and
1:4 J ) and
by Gardiner, Grandage
and Hader ~9J, made use of the properties of regular figures and
•
solids.
In Chapter I of this dissertation certain sets of points are
introduced all of which have been obtained from a basic point set.·
The basic point set occurs as the result of applying a specified transformation group to a general point in three dimensions.
It is shown
that these generated point sets obey all the conditions tor a second
order rotatable design except one. There is then defined a function
called the excess of a point set, which describes the extent to which
the outstanding condition is not met,
e•
an~
it is shown that combina-
tions of point sets whose total excess is zero will provide second
order rotatable designs or infinite classes of rotatable designs;
several classes are derived in full.
•
In Chapter II is presented a second method of generating point
sets which may be used to build classes of second order designs in
three dimensions, and four more infinite classes are derived.
There
follows a particular construction method for single designs, which is
later shown to be extendable to the construction of second order designs in higher dimensions.
Another construction follows; two sets
are presented which Violate not one, but two, of the conditions tor
second order rotatability and it is shown that these sets may be advantageously combined to give an infinite class of second order designs
"
which contain only 16 points.
vii
Some of these methods of
or more dimensions.
construc~ion
can also be used in four
In Chapter III the necessary point sets are gen-
erated for a general dimension k and are combined in various ways to
give second order designs.
In addition the extension of a method used
in Chapter II to construct single second order designs is presented
in its general k-dimensional form.
Chapter IV deals with third order rotatable designs in three
dimensions.
A general theorem providing the exact conditions under
which a third order arrangement is non-singular is presented. The
previously known third order designs are tabulated.
It is then shown
that some of the second order design classes constructed earlier in
the dissertation may be combined in pairs to give infinite classes of
•
sequential third order rotatable designs in three dimensions. One
example of such a combination is worked out in full and it is established that two of the known designs are special cases of this class.
The chapter ends with a proof of the existence of two further infinite
classes of third order designs 1n three dtmensions.
ROTATABLE DESIGNS OF SECOND AND THIRD ORDER IN THREE OR MORE DIMENSIONS*
by
R. C. Bose and Norman R. Draper
CHAPl'ER I
'c;,
SECOND ORDER ROTATABLE DESIGNS IN THREE DIMENSIONS
1.1 Definitions:
second order rotatable arrangements and second order
rotatable designs.
It is frequently necessary in an experimental investigation to explore the relationship between a number of controllable factors coded as
x l ,x2 ' ••• ,xk ' say, (such as temperature, pressure and amount of material)
and some observable effect , usually called "yield", which depends in some way
on the factors involved.
Suppose , for example , that factor Xi were held
= 1/2 / ••• / k)j
.
at level x il (i
..
(x ll , x2l' ••• , ~l)
may be represented as a point in k-dimensional Euclidean space; this we
shall call the factor space.
then the vector
For each point
~u
= (xlU' ••• /~u)
in
the
factor space at which an experiment is performed I i.e. , for each experiment
at factor levels given by the coordinates of a point , there will be a
yield which we shall denote by y.
u
It is assumed that there exists a
functional relationship between y and
! denoted by
where Q is a vector of parameters and the form of the function is unknown.
..
The function is to be approximated over a restricted region of
interest of the controllable factors by a polynomial of degree d,
* 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. tJ l8( 600) -8;. Reproduction in whole
or in part is permitted for any purpose of the United States Government.
2
where d and the region of interest are determined prior to expertmentation from non-statistical considerations.
To carry out this approximation, we shall need to perform experiments at predetermined points of the factor space and to employ
the results to estimate the coefficients of a least squares polynomial
mode1 of degree d.
For example, if d = 2, we should fit the mode1
or
where
1.:
= (Yl'"
x ll
x21
2
x l1
2
x2l
x l1x21
x lu
2
x lu
2
x2u
x x
lu 2u
•
..•
x2u
1
x
x2N
x
2
2
x2N
xJJt'2N
1
•
"Yu '" ',YN)
.• .•
X
=
1
I
t
L
lN
lN
~'
= (t3 o '
t3 l , t3 2 , t3 11 , t3 22 , t3 l2 )
€'
= (€l'
•• " €u' •• " €N)
l
and .! ... N(O, 1;'),
where ;. is unknown.
When the model is of another order, the constituents of X and
~
will be different but the vector representation is still appropriate.
The normal equations for estimating the t3 coefficients will be
•
A
X'X
~
= X'!
where
A
so that
~
A
var(~)
while
= (X'X)·l
= (X'X)-l
A
~
•
is the estimate of
~1
X'i!
0
2
•
For all the coefficients to be estimable , X'X must be nonA
singular.
In order to esttnate the coefficient vector
~
by
~1
we
must first decide on an experimental design, namely, an aggregate of
points at which experiments are to be run.
Box and Hunter
~3~have
suggested that suitable designs for
the exploration of response surfaces are those for which
A
var y(~)
= f(~'~),
where f is an unspecified function which depends on the design used ,
•
i.e' l designs for which the variance of the
~
esti~ated
yield at a point
depends only on the distance in the factor space of that point from
the origin
they call
0
Box and Hunter [ ) ] have shown that such designs, which
rotatab~
designs, obey certain moment conditions to be
presented shortly and also give rise to an X'X matrix which depends on
very few parameters. These parameters we shall later define and call
~1
A4 1 A6 1 etc.
In general, a rotatable design which is used to in-
vestigate a polynomial model of degree d is called a rotatable design
•
of order d, and d A.-parameters define the X'X matrix completely. FUrthermore, the matrix is of a simplified and known form and the inverse
is easy to compute in terms of the A.jo
4
In any specific problem, the scale and location of the control!able factors with respect to the rotatable design are at our disposal. Thus, for example, we may choose the design origin of the
'L
temperature scale to be at, say, 700 and may let the unit of design
scale correspond to 100 • Then temperatures of 400 , 600 and 900 will
correspond to values of -3, -1 and +2 in the units of the experimental
design.
It follows that any particular design to be used can be made
to "fill" or "cover" any specified region of interest in the factor
space merely by an alteration in scale and location of the coded controllable variables. We now proceed to the mathematical definition of
a rotatable design. We shall define, in particular, a second order
rotatable design in k factors, that is, one used to estimate the coef-
,
ficients of a second order model in k variables.
Suppose, in an experimental investigation with k factors, N
(not necessarily all distinct) combinations of levels are employed.
Thus the group of N expertments which arises can be described by the
N points in k dimensions
(l.l.l)
(x ll , x2l '
... ,
~l)'
(x12 , x 22 '
•
... ,
~2)'
(xlu ' x2u '
•
•
•
e
•
... ,
•
...
xku ) ,
•
•
•
, ~) , say,
(xlN , x2N '
where, in the uth experiment, factor t is at level xtu •
5
The set of points (1.1.1) is said to form a rotatable arrangement of the second order in k factors if the following conditions,
which &ore equivalent to the condition var
9(.!)
= f(.!'.!),
are satis-
fied:
(1.1.2)
Ex
U
E
U
iu
x'
iu
= 0,
= 0,
2
E Xi u.,u
x..
U
EX
U
t
E x
= 0,
.. XII.U
i Ux "u
X-
4
u lu
= 0,
4
= uE x2u
=
1: x' x
u iu ju
• • I:
4
= Eu x..xu
=,
2
2
uE Xi ux j u
=, "'4N,
= 0,
where all summations are over u
= 1 to u = N and different numbers
1,j,Vand m (which take integer values from 1 to k) denote different
numbers.
The set (1.1.1) is said to form a
~~tab1e
design of second
order if the conditions (1.1.2) are satisfied and the matrix X'X used
6
in the least squares estimation is non-singular.
Box and Hunter
£31
show that the necessary and sufficient condition for this to be so is
(1.1.3)
For the determinant of X'X is proportional to (k+2);\4 - k{ and thus
if this latter quantity is zero, XIX is singular and some of the coefficients
~
are not estimablej in this case the set of points (1.1.1) is
said to torm a singular arrangement.
zero, since
var(~) ~
(X,x)-l
d2,
If (k+2);\4 -
2
k~
is very near to
some of the variances of the estimates
are very large. When this happens, the design 1s said to be almost
singular.
In fact
L3],
~
N
r
u=l
(1.1.4)
4
u.
N 2 2
( 1:: r )
k
.k+2
u=l u
..
where
r~ = xiu
+ '.0 +
~u' from which it follows that (1.1.3) 1s al-
ways satisfied unless all of the N points lie on a sphere in k dimensionsj otherwise equality is attained and the arrangement is singular.
Thus the possibility that the inequality (1.1.3) could be reversed is
excluded.
Even when a second order rotatable arrangement is spherical,
it can be made into a design merely by the addition of points at the
center (0,0,0) of the design.
This is seen from (1.1.4).
The addition
of points at (0,0,0) increases N but does not affect any of the other
parts of the right member.
The right member is thus increased by the
7
addition of center points so that it becomes possible to satisfy
(l.l.}) •
When presenting a rotatable design, it is customary to "scale ll
it. By this it is meant that the scale of the coded controllable
variables is chosen in such a way that
~
= 1.
The reason for this is as follows.
Given a second order de-
sign which satisfies the conditions (1.1.2) with a specified value of
A4' there are an infinite number of values possible for
~ ~
O.
Since
these design can be derived one from the other merely by change of
scale, we do not regard them as different.
As we have already pointed
out, any design can be made to fill the region of the experimenter's
interest. Thus the scaling condition
A
2
=1
fixes a particular design and enables better comparison between two
designs with different A4IS.
A great deal has been written about rotatable desj.gns,i more extensive infornl9.tion may be found in papers such as those by Box £1],
[2 J,
Box and Wilson
[5 J,
dissertation of R. L. Carter
Box and Hunter
[3 J, [4 J
and the Ph.D.
£6J.
We shall now turn to the main purpose of this chapter--the construction of second order rotatable designs for three factors.
We
shall add extensively to the designs now known and shall derive these
known designs as special cases.
8
1.2 A transformation group in three dimensions and its generated
point sets.
We shall define certain matrix transformations and consider
their effect upon a general point (x,y,z) in three dimensions.
Let
W=
r
0
1
0
I
0
0
1
1
0
0
0
0
1
0
0
1
0
I
t
~
;
'-
then
,-
I
w2
=
I 1
i
II
I
•
and
w3 = I,
Hence
0
the unit matrix.
,W
y
x
Y
I
-
=
Z
""t
w2
z
x
y
z
e
z
,
z
=
y
-.-
..,
X
X
w3
,
x
,
y
-.
I
r x ...
I-l:
-'
J
9
Thus I, W, w2 form a cyclical
g~oup
of linear transformations
in three dimensions.
Let
r
R_
J.
R2
R,
e"
=
=
=
I
-1
0
0
0
1
0
0
0
1
1
0
0
0
-1
0
0
0
1
1
0
0
0
1
0
0
0
-1
-.
,
Hence
x
..
,
Rl
-x
=
Y
z
Y
z
,
-;
X
R2
R,
Y
X
==
-y
z
z
x
x
y
z
==
,
y
-z
,
2
2
R1
= R
= R'2 = I .
2
10
~
The four transformations of coordinates represented by W, R ,
l
and
generate a group G of transformations of order 24 with ele-
R,
ments
(1.2.1)
J
J
j
j
J
J
j
j
W , W Rl , W R2 , W R" W R2R" W R,Rl' W R1R2 , W R1R2R,
(j = 1,2,').
It is easily seen that all the 24 elements in (1.2.1) are
distinct. While Rl' R and
2
For example,
R, commute, wJ and
x
R1R2
e..
R~l
Y
R
1
-y
=
-y
z
z
z
x
..x
-x
y
=
R2
Y
= 1,2,3).
-x
x
=
R do not (i,J
i
=
-y
z
z
z
x
-x
y
,
but·
WR1 Y = W
R1W
Y =
z
z
z
-x
X
Y
-y
Y =
z
Ii
z
x
=
I
z
X
1
A group table may be constructed, employing the identities
(1.2.2 )
11
and identities of the type WR l = R,W, to verify the statements above.
Because of the size of the group the table will not be reproduced here.
Given a general point (x,y,z) in three dimensions, we may apply
to it all the transformations of the group G.
In this way we obtain a
set of 24 points with coordinates
(!.x, !.y, !.z) ,
(:!:y, :!:z, :!:x) ,
(:!:z, :!:x, :.y)
4
We shall denote this set by
(1.2.4 )
G(x,y,z) •
Note that if <f,m,n) denotes any other points of the set,
G(x,y,z)
= G(f,m,n),
i.e., any point of the set, when operated on by G, will give rise to
the same set.
The set G(x,y,z) satisfies all the moment conditions (1.1.2)
except
N
4
t x
u=l iu
=3
N
2
2
t x x
u=l iu j u
(1'; j), (1,j
= l,2, ... ,k).
We now define a function K(x,y,z) of the point (x,y,z) as
(1.2.6) .
K(x,y,z)
144
= ;(x
+ y +
Z
4
-
2 2
2 2
2 2
3y z - 3z x - 3x y ).
This function is constant for all of the 24 points of G(x,y,z).
Furthermore, if it has the value zero, then G(x,y,z) is a rotatable
arrangement since the outstanding condition (1.2.5) becomes satisfied.
Write
x 2 = sz2,
y2 = tz 2 •
12
Then, if K(x,y,z) is
or
t
ze~o
and z
r 0,
22_
- 3t{S+1) + (8 -3s+1) - O.
This is the equation of a hyperbola.
If the point (s,t) lies
on the hyperbola and also in the first quadrant, G{x,y,z} is a rotatable arrangement. The following diagram shows points {s,t} for which
this is true. There is· complete symmetry about the line s
t
7
2
2
/ s +t +1-3s-3t-3st=O
6
e.
5
4
2
If we solve for t in terms of
(1.2.8)
5,
we obtain
= t.
13
This yields two non-negative solutions if
s2 _ 3s + 1 ~ 0"
which implies
s
> (3
o< 6 <
or
/5) /2
/5> /2
+
(3 -
= (cosec
=
4 81n
2
18 >;4
2
180
0
•
otherwise there is only one positive solution for each value of s
> O.
The reason for this is clear from the diagram.
The point set G(x"y,z) is clearly spherical" and thus equality
will be attained in the non-singularity condition unless additional
If no center points
points are added at the center to form the design.
are added,
N
= 24
+ no •
We find that
~N
= 8(x2+y2+z2 )
2
= 8(sf-t+l)z •
Thus if we apply the scaling condition A2
(1.2.10)
z2
= 1,
= Nj8(s+t+l)"
and we have an infinite class of second order designs which depends on
one parameter
6.
(1.2.11)
t
z
y
For if s
=~
~
0 is specified"
[3(s+1) :!:
j
5 (s2+ Gs+ l ) J,
(t
~
0 only),
= [NjB(s+t+1) j-/2
= t l /2z,
x = sl/2 z ,
and all design points are fixed.
Eac~
s gives rise to one or two
llt.
designs according as (1.2.9) 1s not satisfied or is satisfied.
For
this class,
2222222
A4/A = 8(x y +y Z +z X ),AI
2
= 8(st+s+t)z 4,AI.
Speeial cases:
(1)
= (3 :!:.15>/2 = (cosec 2 1.80)j4 or 4 sin2 180 ;
0, x = ( 15 :!:. l)zj2 = Qz or Q-lz, where 2Q = J5 +
= (3 + .j5)/2 = Q2.
s = 0, t
hence
y =
and
Qtl
Thus the set becomes
1
~(Q,l,O)
replicated once.
1,
As Coxeter
~8~ shows, this set of 12 points consists of the vertices of an
icosahedron, and these form a well known second order design.
(2)
s
= t = jiO
z
x
- 5, hence
= LN/(2 /fO - 5)~1/2 = ['(5
= Y = (N
- 8z
2
+ 2-.flO)N/120j-/2,
)/16.
This is the design referred to as the truncated cube by Gardiner,
Grandage and Hader
1:9J
and numbered (23) by them.
Let us now suppose that K(X,y,z)
r 0 for
the points of the
set G(x,y,z). We shall define
E K(x,y,z)
of that set and write it Ex(S).
(1.2.12)
r,
Ex~G(x,y,z)~
over a point set S to be the excess
Thus
= 8(x4+y4+z 4 -3y2z2-3z 2x2-3x2y2 ).
This can take both positive and negative values according to the
choice of x, y and z.
Clearly
15
where the notation
J6
j
means that points which belong to more than one
set 6.J contribute to the sum . each time they occur. The notation thus
does not denote the hunion" of sets in the usual sense.
Furthermore,
if a number of sets 6 l ,6 , ••• ,6m (say) satisfy, either separately or
2
together, the conditions for a second order rotatable arrangement except for the condition (1.2.5), then the condition
is a necessary and sufficient condition for the points of the whole
set 61 + 62 + ... + 6 to form a rotatable arrangement of second order.
m
We shall make use of this important fact in section 1.3.
For certain special choices of (x,y,z) in three dimensions,
the 24 points of G(x,y,z) will coincide in pairs or in triplets or in
quadruplets.
For example, G(p,q,O) consists of the twelve po1nts
(1.2.14)
each occurr1ng· twice" We may denote the 12 point set by
'21 G(p,q,O).
This set has excess
r
l
4
4
Ex _ ~(p ~ q, 0) J = 4 (p + q
22
- 3p q ),
a quantity which may be made positive or negative according to the
values of p and q.
The set ~(p,q,O) will itself form a rotatable arrangement if
16
p
422
4
-3pq +q
=0
p2 jq2 = (3!. J5)j2
or
=(cosec 2 18°)/4
2
or 4 sin 180
•
As was shown previously for the first special case of the design fonmed
by G(x,y,z), this gives rise to the icosahedron design.
Consider the set G(a,a,a); this consists of the eight points
(1.2.16)
each occurring three times. We may tberefore denote this set of 8
points by
'31
G(a,a,a).
1
Exf '3 G{a,a,a) J
(1.2.17)
==
4
-16a ,
which is always negative, hence this set alone cannot form a rotatable
arrangement.
Consider the set G(c,O,O); this consists of the six points
(1.2.18)
each occurring four times. The six points may be denoted by
1
'Ii: G(c, 0,0).
r '4
1
G(c,O,O) J
ExL,
==
4
2c ,
which is always positive, and so this set alone cannot form a rotatable arrangement.
For consistency of notation we may write the point (0,0,0) as
1
2lj: G(O,O,O).
17
In this case, no center points may be denoted by
no
2Ir G(O,O,O).
(1.2.20)
1.3 The formation of rotatable arrangements and rotatable designs
by comb:Jination of the generated point sets.
We have seen that some of the generated point sets, for example
(1.2.16) and (1.2.18), cannot form rotatable designs by themselves.
However, if we can combine sets with positive excess and sets with
negative excess in such a way that the total excess 1s zero, we have,
automatically, a rotatable arrangement which is easily converted to a
design by scaling and, i t necessary, by adding center points. To
demonstrate this we shall derive a well-known design.
Consider the combination of sets ~(a,a,a) and ia(c,O,O).
Then
Ex [
1
~(a,a,a)
=
1
]
+ ~(c,O,O)
= Ex!: ~(a,a,a)
= -168
This is zero if c 2
4+
= 2 J2
] +
EXL ia(c,O,O) J
4
2c •
a 2, in which case the 14 points form
a rotatable arrangement. For this arrangement,
~
~
A
2
= -~ •
N
~
2 4 = N!4(3 + 2
16(2+ fl) a
./2) > 5'3 '
since N > 14.
Thus the design is non-singular even without the addition of center
points.
However, in this case,
18
and the design is almost singular.
points with this design.
Hunter
Thus it is better to use some center
This point is discussed in a paper by Box and
C3J.
The actual design points are obtained by applying the scaling
condition
~
= 1.
This gives
2
2
8a + 2c = N
= 14 +
no ,
where no is the number of center points added.
4(2 + J2)a2
Thus
= N,
and both a and c are determined when N is given.
The method may now be extended.
We have seen that the combi-
nation of generated sets leads to a single design when only two
parameters are present, as in the example just given, since the two
conditions
Ex(set)
=
0,
~
=
1,
completely determine the design.
The first condition alone completely
determines the ratio of the two parameters and is sufficient to determine the design apart from scale. We now investigate combinations
of sets which contain three parameters.
We shall see that we obtain
a single infinity of designs which depend on a single parameter ratio.
Consider the 20 pointe
19
(:!.a , +a
(;tc l ,
(
(
o,
o,
(!,c 2 '
(
(
o,
o,
,
,
+a )
, }
,
,
0
)
:!.cl'
0
)
o,
o,
:!.c l
0
)
,
!,c 2 '
0
)
~
0
o,
'31 G(a,a,a),
1
'4 G(c1,O,O),
) ,
1
1j:
G(c 2,O,O).
:!:.c 2 ) ,
The excess of this whole set is
1
1
1
EX Lr 3G(a,a,a)
+ ~(cl'O,O)
+ ~(c2'O,O)
J
444
= -100 + 2c l + 2c 2 •
Note that since
1
Ex£ '3G(a,a,a)J
= -1004
is negative, we must combine
with it sets at least one of which has Fositive excess to compensate.
Thus the set has zero excess
4
cl +
2
2
Set
c = xa ,
l
2
Then
x +
if
4
c2
= &4.
2
c2
= ya2 •
y2 = 8.
Any positive values of x and y which satisfy this equation will give
rise to a rotatable arrangement of second order. Thus if (x,y) 1s a
point of the circle x2 + y2 = 8 and also lies in the first quadrant,
then we shall have a rotatable arrangement.
No additional center
points are required to make the arrangement into a design.
three radii of the three parts of the arrangement are
For the
20
or
If x := y, both have the value a j2 and if x = 0, y
= 2 3/4.
Thus the
radii are always different and the non-singularity condition will be
satisfied, even when degenerate sets are ignored.
N A2
Now
22
= 2c2
+ 2c 2 + 8a
l
= 2(x + Y + 4 )a2 •
Applying the scaling condition
~
= 1,
we obtain
a 2 = N/2(x+y+4).
We thus have a single infinity of second order rotatable designs dellendent on one :Parameter, say x.
0:$ x
~
For if' we choose x,
2, then
y = /a_x2 ,
a = eN /2 (x+y+4 ) j-/2,
c = xl /2a,
1
= yl/2a,
2
and the design becomes completely determined.
c
Since x and yare inter-
changeable, it is not necessary to consider the full range 0
~
x :$ 2 ..;2.
For this class,
We now derive, as special cases of the infinite class just obtained,
two designs which were previously known.
= 2 3/4a , c1 = O. We have ob2
tained the cube plus octahedron design, with 6 center points which are
(1)
x:=
o.
Then"
y = 2/2,
c
vertices ot the degenerate octahedron.
21
(2)
= y = 2.
x
Then c l
= c 2 = a~.
This gives rise to the
design numbered (27) by Gardiner" Grandage and Hader
t:9J"
which con-
sists of the vertices of a cube plus those of a doubled octahedron.
The derivation of the previous design suggests a combination
of the three sets
and
1
~(c,O,O).
This 22-point combination has zero excess if
221
The choice of any point on the circle x + y = ~ and also in the first
quadrant will give rise to a rotatable arrangement of second order.
The non-singularity condition is satisfied, since the three original
sets have radii
or
J5 a l , !3 a.2
and
c
$x c" .£ c
and
c,
and for no arrangement of the class are these equal.
Hence all arrange-
mente may be used as designs without center points, €"en
se:ha
have been
droppe('~.
A. N
2
NoW'
22
= 8(a2
+ a2 ) + 2c
l
=
2(1~x
2
+ 4y + l)c •
If we apply the scaling condition
~
= 1,
we find that
'W~mn
degenera.te
22
c
2
= Nj2(4x + 4y + 1).
We have now constructed a single infinity
or
signs dependent on one parameter, say x.
For.. suppose we shoose x,
o :s
x
:s 2"1
then
y =
A-
x
2
second order rotatable de-
,
c = ['N/2(4x+4y+1)
:j-/2,
al = xl /2c,
a = yl/2c ,
2
and the design 1s completely determined. Again, it is not necessary to
consider the full range of
Xj
if we do so we shall obtain all designs
twice, since x and y may be interchanged. For this class,
Special case:
x
= 0,
y
= 1/2/2.
Then c
= 23,14a2 ,
a l = O. We have the cube
plus octahedron design with eight center points which are the vertices
of the degenerate cube.
1.4 The formation of further infinite classes of second order rotatable desj.gns by combinations of sets with positive and with
We shall now consider other combinations of the generated
point sets.
Consider the 12 points (1.2.14) for the special case when
p=q=f
(say).
23
Thus we have 12 points
(1.4.1)
which we may denote by
21
G(O, 1', f).
Now
Ex! ~
G(O, 1', f)J=
_41'4,
a negative excess which must be neutralized by adding point sets with
positive excess, if a class of designs is to be formed.
We recall that the set of six points
(1.4.2)
denoted by
"Ii1 G(c i
2C~
has excess
, 0, 0),
•
Thus
1 1
EX[ 1
2 G(O, 1', f) + "Ii G(c!" 0, 0) + "Ii G(c 2 ,
0, 0)
J
444
= 2c l + 2c 2 - 41'
which is zero i f
(1.4.4)
When this condition is satisfied, the 24 points symbolized in
the left member of (1.4.3) form a rotatable arrangement.
2
2
2
2
Write
c l = xi' ,
c 2 = yf ;
then
2
2
x + Y
= 2,
and any point on the circle and lying in the first quadrant will proVide a rotatable arrangement.
24
Since x and yare interchangeable, we need consider only half
the points on the circle in the first quadrant; without loss of generality, we consider only x such that
o < x < 1.
The radii of the separate sets composing the arrangement are
and
or
f
j2
and
If x
= y,
•
each is unity; but if Y = 2, x
equal unless the degenerate set is retained.
= 0 and
the radii are
Thus, provided that when
1
the six degenerate points of ~
G(c l , 0, 0) are retained as center
points, no center points need be added to satisfy the non-singularity
x
=0
condition.
Now
Ni\.2
22
= 2{c 2
+ c 2 ) + 8f
l
2
= 2(x + Y + 4)f •
Applyi~g
the scaling condition
f2 = N j2(x
~
= 1,
we have
+ Y + 4) •
We thus have a single infinity of second order rotatable designs dependent on one parameter x, 0
calculate
< X < 1.
Given x in this range we can
25
and the design becomes completely determined.
2
"-4A2 = 4f
For this class,
4
/N.
Special case:
x
= 0,
y
= fl
= 2 1 ;4
so that c l = 0, c 2
t.
This is the design numbered (24) by Gariner, Grandage and
Hader
[9],
octahedron.
plus six center points from the vertices of the degenerate
Note that the value of b
= c2/f
quoted in their paper is
in error.
We may also make use of the negative excess set (1.4.1)
another way.
(1.4.11)
4
Ex[ ~ G(O, f, f)J= ;.4f ,
Ex[
! G(a, a, a) J = _168.4 ,
Ex[
i G(c,
0, 0)] =
2e
4
•
So
Ex [ ~1 G(O, f, f) + -;1 G(a, a, a) + ij:1 G(c, 0, 0)] =
if
(1.4.12)
e
444
=2f +8a.,
and if this condition is satisfied, the 26 points
(1.4.13)
(:ta , :ta, :!.a);
and
(:te, 0, 0) ,
form a rotatable arrangement.
(0,
:te,
0) ,
(0, 0,
:te) ,
°
26
Set
f
222
2
= xc, a = yc,
and we find that
2
2
2x + 8y = 1.
This is an ellipse; all points on the ellipse which lie in the first
quadrant will thus give rise to a second order rotatable arrangement.
The separate sets which constitute the design have radii
J3
.ft f,
or
c
/iX,
a
and
c
c.J!ii
and
c.
The first two are equal when 2x = 3 y, and then each has the
2
value Ji /5. When y = 0, 2x. = 1 and when x = 0, 8y2 = 1. It follows
that the three radii are never equal and so the addition of center
points is not required to satisfy the n?n-singularity condition, even
when degenerate sets are ignored.
222
~N = 8f + 8a + 2c
2
= 2(4x + 4y + 1}c •
If we apply the scaling condition
c
2
~
= 1,
therefore,
= N/2(4x + 4y + 1).
We now have a single infinity of second order rotatable designs
dependent on one parameter, say x. For we can choose x,
o~
x ~
1/ /2,
and obtain
2 )f8y/2,
y= £(1 _ 2x
c
= LN/2(4x +
t = x 1 /2c,
a = y1/2c.
4y + 1)_j/2,
27
ThuB the design is completely determined. For any design of the class,
= 4(f4 +
2
h4/A.2
4
2a ) fir
4
= 4(x2+ 2
2y )c IN.
Special case:
x = 1/ fl,
y
= 0,
so that
a
=0
and
Again we have the design numbered (27) by Gardiner, Grandage
and Hader
~9~,
this time with eight center points from the vertices
of the degenerate cube.
1.5 Classes of designs using sets with variable excess.
UP to this point all the sets we have used in combinations
have had a positive or negative excess. Let us now consider the set
of 12 points
(:!::.p, :!::.q, 0),
which may be denoted by
1
'2 G(p, q, 0).
Now
a quantity which may be made positive or negative according to the way
p and q are chosen. Thus ~ G(p, q, 0) may be combined with all of the
1 1 1
sets 3G(a, a, a), ~ G(c, 0, 0) and 2 G(f, f, 0) to obtain rotatable
arrangements and hence designs.
For example,
EXL ~ G(p, q, 0) + ~ G(a, a, a) J
if
=0
28
p4 + q4
Set
p
2
222
= xa,
q = ya.,
x
Any point
= 3p2q2 +
2
4a4.
and we have
2
- 3xy + Y = 4.
of this hyperbola which lies in the first quadrant will give
rise to a rotatable arrangement of second order.
y
7
6
5
4
3
2
1
0
1
2
4
5
6
7
8
x
If we solve for y in terms of x, we obtain
This yields two positive solutions if x
t1ve solution arises.
> 2;
This may easily be seen from the diagram.
radii of the separate point sets are
and
or
otherwise only one posi-
fi + ya
and
Aa
The
29
and these are equal when x + y :: .5.
Since the straight line x + y = 3
intersects the hyperbola in two points P and Q, equality occurs for two
arrangements of the class. For these two arrangements, the addition
of center points is necessary to satisfy the non-singularity condition.
Since points near P and Q on the hyperbola yield designs for which the
XIX matrix is almost singular, the addition of center points to de-
signs for which x + Y - 3 is very nearly zero is desirable.
points are also necessary when a
= 0,
Center
if the degenerate set is dropped.
As before, the number of design points N includes any center points
which may have been added.
Now
h'fil
= 4(p2 +
2
q2) + 8a
= 4(x + y + 2)a
2
Then the application of the scaling condition
a
2
= Nj4(x +
•
~
=1
yields
Y + 2).
Thus we obtain an infinite class of second order rota.ta.ble designs, each
design consisting of 20 points plus any center points which may have
been added.
The class depends on one parameter x, and if x
~
2, each
x gives rise to two separate designs; otherwise just one design exists.
Given x
~
0,
y =
[3x j./5x2
+
16]/2,
a = [Nj4(x + Y + 2):1/2
p
= x1/2a,
q
= y1/2a,
(if Y ~
°only),
30
and the design is completely determined. For this class l
2
4
2 2
A4ft.2 = (8a + 4p q ),m
= 4(2 + xy)a4,m •
This class has two well known special cases:
x
(1)
= co
1
Y = 00 j thus a = O.
Ignore the degenerate set.
We obtain p and q now from the equation
so that
where 2Q
or
= 1 + ./5.
As we pointed out previously (page 14)1 this is the icosahedron
designl which consists of the 12 points ot the set
1
~ G(Q 1 1 1 0).
( 2)
Choose a point on the hyperbola such that
x + y
= 3.
Two such points exist 1 as we have shown. Then
x
= )"-1 = (;:!: 13) /2 = Q2
or
Q-2 1
where 2Q = (1 +
and
g2 =
Q
J5)
+ 1.
Thus the 20 design points (center points omitted) consist of a constant
multiple of
As coxeter
['8 J shows 1 these are the vertices of a dodecahedron 1 which
form a well known second order rotatable design.
31
(Addition of center points is necessary, of course, when using
these designs, since they are spherical).
Consider now the set combination
21 G{p, q, 0)
(1.5.11)
+
41 G{c, 0, 0).
These 18 points have zero excess if
4(p4 + q4 _ 3p2q2) + 2c 4
(1.5.12)
= o.
222
2
= xc, q = yc ,
Set
P
221
x - 3xy + Y + -
then
2
= o.
Again we have a hyperbola. Any point of it which lies in the first
quadrant will give rise to a rotatable arrangement of second order.
y
7
6
~-3xy+y21
~o
X
5
4
-1---'-+---.-----,---,--.--..---.,.---r--~x
o
"3
5
6
7
8
32
Solving for y in terms of x, we find
2
There are two real positive solutions i f 5x
x
2
> 0.4.
otherwise there is no positive solution.
this is clear from the diagram.
~
2, i.e.,
The reason for
The radii of the two components
parts of the arrangement are
or
and
c
and
c.
These are equal only if x + Y = 1.
x real
=- y2
with x + Y
bola.
= 1;
~
But y real --:> x2
> 0.4
and
0.4. Clearly these cannot be satisfied together
geometrically, the line does not intersect the hyper-
Hence, prOVided c
f
0, it is not necessary to add center points
to the arrangement, since it is not spherical.
However, if c
=0
and the degenerate points are not retained, center points are necessary to make the XIX matrix nonsingular.
For this special case, see
below.
Since
222
4 (p + q) + 2c
= 2{2x + 2y + 1)c2 ,
~N =
the scaling condition ~ = 1 implies that
2
c = N/2 (2c + 2y + 1).
Thus we obtain an infinite class of second order rotatable designs which depends on one parameter x.
Given x
~
0.4, we obtain
33
y =
[3x :!: j5 x2
c
= LN/2(2x +
p
= xl /2c,
q
= yl/2C •
- 2]/2,
2y + 1) j-/2,
The 18 point design 1s now completely specified.
x ~ .632 gives rise to two designs.
222
A4ft.2 = 4p q IN
Each value of'
For this class,
::a
4xyc
4
IN.
Special case:
y=oo.
x=oo,
Again we obtain the icosahedron as on page 30.
The third of the "set combinations is
-1 G(p, q, 0)
2
1
+ '2
G(f, f, 0).
The excess of these 24 points will be zero if
p4 + q4 _ 3p2q2 _ f4 = O.
Write
Then
2
2
'2
2
p =xf,
q =yf.
x2 _ ;xy + y2 = 1.
This is a hyperbola similar to one previously obtained.
Any point of
it lying in the first quadrant will give rise to a second order rotatable arrangement.
y
o
1
x
4
2
In terms of x
These are two positive solutions if x
>1
and one if 0
< x < 1. The
diagram illustrates this clearly. The two sets comprising the arrangement have radii
or
j p2 + q2
and
j2 f
jx+yf
and
.fif.
Thus the arrangement is spherical if f
= 0 and
the degenerate set is
ignored (when we again have the icosahedron design as a special case)
or when x + y = 2, of which points there are two on the hyperbola.
As in a previous example, the arrangements arising from these points
require additional center points in order to form designs.
The
35
addition ot center points to designs arising from near points is
again desirable.
Now
A N = 4(p2 + q2) + 8t2
2
= 4(x + Y + 2 )t
When we put A2
= 1,
2
we obtain
t
2
= N,A (x +
Y + 2).
This gives us an infinite class ot second order designs with 24 points,
depending on x • Given x
y =
~
0,
[3x !../5x2 + 4 ]~,
t = ['N,A(x
P
= x l / 2t,
q
= yl/2t,
and the design is fixed.
(it Y
2:
0
only),
+ Y + 2) Jl/2,
For this class,
2
A4ft.2
= (4p2 q2
4
+ 4t ) IN
= 4(xy + l)f
2
/N.
Special case:
x =
(X),
y
= 00.
Then f
= 0 and we
obtain the icosahedron
design.
There is another usetul set which has variable excess and may
be combined with a set of positive or negative excess.
G(p, q, q), consisting ot points
tor which
ThiS is
= 8(p4
EX[G{p" q, q) J
4
- q
2 2
- 6p q ).
In view of the fact that this set contains 24 points, we shall combine
1
it only with the sets; G(a, a, a) and
41 G(c" 0, 0).
Now
EX[G(p" q, q) +
3
G(a" a, a)J = 0
if
4
p
2
Set
P
lI'hen
x
2
- q
4
22
4
- 2a
- 6p q
2
=xa"
q
2
- 6xy - y
2
= O.
2
=ya.
- 2
= O.
This is a hyperbola and all points of it which lie in the first quadrant will give rise to a second order rotatable arrangement.
y
5
2
x+2y=3
2
x -6xy-y -2=0
4
3
2
1
-I---.-...::::;.,....-~"r----Ti----,ir--...
I -~--.,.-~,---+)
o
I
2
3
4
5
6
7
8
9
x
37
In terms of x,
Since y
x ~
J2.
~
0, the negative root is inadmissible; also we require
This is clear from the diagram.
The component parts of the
arrangement have radii
and
or
and
Thus the arrangement is spherical only when a
=0
and the degenerate
set is dropped (see special case below) or when x + 2y
= 3.
This line
meets the hyperbola in two points, only one of which lies in the first
quadrant. For these two singular arrangements the addition of center
points is necessary; also it is desirable when using designs arising
from nearby points.
Now
~N
The scaling condition
=1
~
a
22
= 8{p2
+ 2q + a )
= S(x + 2y + l)a2 •
2
~plies,
= N/8(x +
therefore, that
2y + 1).
Thus we have an infinite class of second order rotatable designs depending on x, each design containing 32 points plUS any
center points which may have been added.
Given x, x ~
2
y = /lOx - 2 - 3x,
a
= LN/8(x +
p =
x1/2a,
q = yl/2a ,
2y + 1)
j-/2,
/2,
and all des-ign points are fixed.
There is just one design for each
value of x in the specified range.
For this class"
Special case:
x =
ex>..
y = ex> ,
a = O.
Ignore the degenerate set. We then
obtain p and q from the equation
p
= 6p22
q -
q2/p 2 =
Hence
and
4
p=
/fO -
4
= o.
3" (the other root is inadmissible),
LN/(2 jlO - 5)Jl/2 = L(5 + 2 jlo)N/120j-/2..
q= (N -
2
8p )/16.
Again we have obtained the
Grandage and Hader (
q
truncated cube design of Gardiner"
1.:9J,
design 23).
This design was previously
obtained as a special case of the design class G(p" q" r) when p
= q.
We now consider the combination
1
G(p" q.. q) + ~ G(c" 0" 0).
4
4{p
- q
4
2 2
4
- 6p q ) + c
221
x - 6xy - y + '4
the equation of a hyperbola.
= o.
= 0..
Any point (x,y) on this hyperbola and
\
39
also in the first quadrant will give rise to a second order rotatable
arrangement.
y
7
6
5
4
';)
I
1
~
x--6xy-y
x+2y=1
1
~=O
I
-------------
.......--~x
8
-+-~~--..---..-------r---r---r--
o
3
4
5
6
7
Solving for y in terms of x, we obtain
As is clear from the diagram, only one non-negative solution y exists
for each x
~
0, and so the negative root solution in the expression
for y above is not relevant.
The separate sets comprising the arrangement have radii
j p2
or
+ 2q2
jX + 2y
c
and
c
and
c•
It tollows that center points must be added to form a design when x
is chosen such that c
= 0 and
the degenerate points are dropped or
40
vhen x + 2y
= 1;
two of the latter points exist where the straight line
intersects the hyperbola.
The addition of center points to designs
arising from nearby points is desirable, as we have previously men·
tioned in similar cases.
Here,
2
"'2N = 8(p2 + 2q2) + 2c
2
= 2(4x + 8y + 1)c •
~
Therefore the scaling condition
c
2
= 1 implies that
= Nft (4x + 8y +
1).
Thus we have an infinite class of second order rotatable designs which depend on a single parameter x.
contains 30 points.
Given x
0, we have
~
2
y =)lOx
c
+
i - 3x,
= LN/2(4x +
p =
Each design of the class
8y + 1)
j-/2,
x l /2c,
q = yl/2c ,
and the design is fixed.
For each x
~
0, we obtain one design.
For
this class,
22 2
4
4
~P.2 = [8(2p q + q ) + 2c
= 2(8xy +4y2 +
l)c
J IN
4IN.
Special case:
x =
(l),
y=
(J),
c
= o.
We again obtain the special case G(p, q, q) previously mentioned as being design number (23) of Gardiner, Grandage and Hader
[9].
41
A summary table of the design classes presented in this chapter follows.
The table shows the generated sets used to form each
class together with the design coordinate values in terms of a s1ngls
parameter.
e
1
1
JiG(c,O,O) Number
of
points
(:!.p,!.q,O)
(!.C",O,O)
(!.a,!.a,!.a)
(!.p,!.'l.,!.r)
in
Points
design
etc.
etc.
etc.
class
No. of
6
24
12
8
points
Number of sets used for class
Set
e
·e
~(p,q,O)
G(p,q,r)
1
jG(a,a,a)
24
1
o ::;
t ::;
0-.15) /2
t ~ (3+./5)/2
1
2
2
1
2
1
20
22
24
Second parameter
ratio in terms
of first
Range of first
parameter ratio
on which class
depends
0::; x::;
s =
~D(t+1) !./5(t2+6t+1>.]
2./2
y =
O<x<-l..
-
y =
- 2./2
J8-i
J~ _x2
y = }2 _ x 2
O<x<l
p=q=f
1
p=q=f
1
1
1
1
20
1
1
26
18
o::;x::;~
x>O
x ~ 0.6,
I
Design point
coordinate values
in terms of N
and parameter
ratios
r =
c
c ,.
a
1
q" t 1 /2r
LN/2(X+y+-4):I/2
= x 1 /2a,
1
c
2
11
Sa!l- ftJ
19
= y1/2a
~ = y1/2c
f ,. LN/2(X+y+-4)jl/2
c 1 -- x1 /2f ,
c =
4
8( st+s+-t)r fii
LN/2 (!l-x+ !l-y+-1) :1/2
,. x 1/2c,
c
2
4
ftJ
21
4f!l- ftJ
23
4
2
4(x +2l)c ftJ
25
c
= l/2f
LN/2(4x+4Y+-1)y/2
y =
L(1-2x2 )fd:l/2
~[3x !.bX2+16 J
a"
y ,.
~Dx !.}5X2-2J
c =
y"
~Dx !.J,x2+4 J
f = [N!4(X+y+-2):I/2
y =
f = x 1 /2c,a ,. y1/2 c
LN!4(X+y+-2):I/2
P = x 1 /2a,
.'
!I-(2+xY)a
!I-
ftJ
27
q" l/2a
LN/2(2x+2y+-1):I/2
P ,. x 1 /2c,
Page
).41{
LNj8(s+-t+1):I/2
p = sl/2r,
a =
Value of
q"
l/2c
4xyc
4
ftJ
3l
I
i
I
1
2
(one
p=q=f
I
I
24
1
x>o
32
x ~fl
2
y" JlOx -2-3x
30
x>o
Y"
q=r
l
q=r
1
FoK1
lOX + t-3x
p = x 1 /2f,
a = LNj8(X+2Y+-l)y/2
p ,. x l /2a,
c ,.
q = l/2a
LN/2(4x+8y+-l):lt<
p = x1 /2c,
4(XY+-1)f4fii
33
~I(2xy+-y2+l)a 4 ftJ
36
q = y1/2f
q" l/2c
t
~(8xy+-4l+l)c!l- /If 38
CRAPI'ER II
FURTHER SECOND ORDER ROTATABLE DESIGNS IN THREE DIMENSIONS
2.1 A second method of generating point sets suitable for bUilding
second order rotatable designs.
Define
cos a
-sina
cos
Ct
o
a
T2 =
cos -2
0:
sin -2
0
0
0
,
1
0:
sin 2'
0
a
0
cos 2'
,
-1
0
where
Ct
= 21! /s •
Consider the effect of applying Tl and T2 to points of the form
(r, 0, b),
i.e., points on the plane y = 0, and to all other points obtained from
repeated applications of T 1 and T 2 • In this way we shall obtain 2s
points with coordinates
(2.1.1)
r sin ta,
(r cos ta,
1
(r cos(t + 2)a,
b),
1
r ain(t + 2)0:,
-b),
t = 0, 1, 2, ... , (s-l),
which we shall denote by
44
Although any point of the
se~
may be obtained in more than one way
from the others, we shall consider only the set of 2s points (2.1.1)
without duplications.
Provided s
~
5, the set Tz{r, 0, b) has the
following sums of powers and products:
,
2
!:z
Uu
= 2sb:2
4 /4,
tz4
= 2sb4
I;x2= r.y2 = sr2
u u u u
I;x4
uu
= r.y4 =
r:x.2 2
u uYu
u u
3sr
4
= ar /4
Uu
,
1:: 2 2
u
~uzu
,
2 2
= u!:zuxu = sr~2,
and all other sums of powers and products up to and including order
four are zero.
This is easily verified by using the fact that each
of the two a-gons in the set of 26 points is a second order rotatable
arrangement in two dimensions.
(It is well known [3] that a regu-
lar s-gon in two dimensions is a second order rotatable arrangement
if s
~
5). A rotation about the z axis of the complete point arrange-
ment will not affect the properties held by the sums of powers and
products.
We now recall the permutation group
2
(I, W, W )
generated by
W=
o
1
o
0
o
1
1
o
o
and apply this to T2 (r, 0, b) to give sets Tx(b, r, 0) and Ty(O, b, r).
In all we now have 68 points, which we denote by
T(r, 0" b)"
vith coordinates
,
,
b
),
(r cos(t + ~)a,
1
r sin(t + '2)a"
-b
(
r cos to
)"
),
(r cos to
b
"
,
(
-b
•
(r sin tct
(r sin(t +
where a
= 2n/s
r sin ta
,
1
r cos(t + '2)0',
r sin to
r sin(t + ~)a),
,
b
,
r cos to:
~)a"
-b
,
r cos(t + ~)a),
(s ~ 5) and t
= 0,
),
1, 2, ••• , (s-l).
The set T(r, 0, b) has sums of powers and products
Ez
222
= 2s(r + b ),
u u
rx2 y2
u u u
=
and all other sums of powers and products up to and including order
four are zero.
The formulae for the sums of powers and praducts will extend
to the case s
= 4,
provided we fix as the set
T (r, 0, b)
z
the points
, o
In the case s
= 4,
,
(0
+r
(-:.r/ ft.,
-:.r/./2,
b),
b),
-b).
rotation of the a-gons about the z axis will affect
46
the sums of' powers and products and thus cannot be permitted. This
point must be remembered whenever specific reference is made to the
case s
= 4.
From the properties of' sums of powers and products given
above, it follows that the excess of the set, defined in the same way
as before, is
Of course the excess of each single point varies in this case and it is
necessary to consider the total effect over all the points.
Since its
excess can be made positive or negative according to the choice of r
and b, it will be possible to combine the set T{r, 0, b) with sets of'
both positive and negative excess. Because of' the large number of'
points which would otherwise arise, we shall combine it only with
1
; G(a, a, a) and
41 G(c, 0, 0). We consider now the first of these
combinations.
Ex[T(r, 0, b) + ~ G(a, a, a) J = 0,
if
3r
Set
r
2
4
2.2
- 24ro + 8b
= xa2 ,
b
2
4
4
- 64a /s
= ya2 ,
= o.
and we have
2
3x - 24xy + 8y2 - 64/s ...
o.
This is the equation of afamil.y of hyperbolae.
Given s, any point
both on the resulting hyperbola and in the first quadrant will provide a rotatable arrangement of second order.
y
8
7
6
5
4
3
2
1
o
7
x
8
The diagram shows the hyperbola that arises for s
= 4.
As s - >
00,
the hyperbolae approach their common asymptotes which are given by
the equation
3x2 _ 24xy + 8y2 = 0
or
= 0,
(x - ay)(x - l3y)
where
13
0:,
= 4 + g /36
- 3
= 7.65 and 0.35
(approximately).
Solving for y in terms of x, we find
y =
f6x '!./30x2 + l28/s J /4.
There is always one positive root if x
roots if x>
Je./3s
(2.3 for s = 4).
~
0; there are two positive
This is clear from the diagram.
48
The separate sets which comprise the arrangement have radii
and
or
and
Thus singular designs are obtained when a
points are ignored, or when x + y
x
= 3,
= 0,
if the degenerate
i.e., when
= £60!
)40(27 + 56 /s) J/35,
L 45!
)40(27 + 56 / s ) J/35.
y =
These two :points are both in the first quadrant, as the graph shows.
Arrangements arising from these singular points require the addition
of center points; it is desirable to add center :points to designs
which arise from nearby points •. Now
AN
2
22
= 2s(r2
+ b ) + 8a
= !:2s(x + y) +
8J a2,
where N = 68 + 8 + no center points. Thus ~
2
a = N/2[s(X + y) + 41.
=1
implies
We thus have an infinite class of second order rotatable designs
dependi~g
on x, each design consisting of 63 + 8 points, omit-
ting center points. Given x
y=
2:
0,
[6x ! j;;ox2
+ l28/s]/4,
a = LN/2[s(x + y) + 4]
and all design points are fixed.
j-/2,
For this class,
(y
2:
0 only),
A.4/~
! sr2 (r2 + &2) + &4J /N
= [
4
= £sx{;x + By) + 32J a /4N.
Special case:
Let s
= 4"
x
= co.
= 00
Then y
and a
= 0.
We obtain rand
b from
3r
4
2. 2
- 24r-b + 8b
4
= 0"
whence
and from the application of the condition A.2 = 1"
2
r /2 = Nj4(lO
= c J2
This (with r
Grandage and Hader
L9] as
::!:.}30).
is the design obtained by Gardiner"
their design (26)" plus 8 extra center
points from the degenerate set
31 G(a,
a J a).
Consider now,
T(r, 0, b) +
41 G(c, 0, 0) •
ExLT(r, OJ b) +
if
Set
~ G(c J
0, O)J = 0
3r4 _ 24r~2 + & 4 + Be 4 /s
r
2
= xc
'2
b
J
2
2
= yc;
3x2 _ 24xy + 8y2 + 8/s
This is a family of hyperbolae.
= O.
then
'=
O.
Given sJ any point that lies both on
the resulting hyperbola and in the first quadrant gives rise to a
second order rotatable arrangement.
hyperbola which arises when s
= 4.
The following diagram shows the
50
y
ji~:=;::=~::4:::5::6::7==8:=~
e.
x
Again as s increases the family of hyperbolae approach their asymptotes,
which are given by the same lines as before.
This time, because of the
change in sign of the constant term, the hyperbolae occupy the other
portion of the plane, as is clear from the diagram.
Thus for real roots, x
2
~
In terms of x,
8/l5s and both roots are clearly positive,
so that for each value of x ~
18 /15s
(. 36 for s
= 4)
there are two
rotatable arrangements. This may be seen clearly in the diagram.
The separate sets comprising the arrangement have radii
51
or
Jx+yc
and
c
and
c.
Thus singular arrangements occur if e = 0 and we ignore the degenerate
set, of if x + Y = 1, i.e., when
x =
[20:, floc; •
= [15 -:;:
y
7/s )J/;5,
)40(; • 7/S ) J/;5.
Both these points lie in the first quadrant. As before, arrangements
corresponding to these singular points requ1rethe addition of center
points and a similar addition of center points for designs arising
from nearby points is desirable. We see that
~N = 2s(r2 + b2 ) + 2c 2
2
= 2[s(x + y) + l J e ,
where
The scaling condition
~
e
= 6s
+ 6 + n center points.
o
= 1 implies therefore that
N
2
= N/2[s(x +
y) + lJ.
Thus we have an infinite class of second order rotatable de.
signs depending on x, each design consisting of 6s + 6 points, omitt1ngcenter points.
Given
x ~ J81l5s,
we can find
y =
£6x :!:. /;Ox2
• 16/8
J/4,
C
= [N/2[s(X + y) + l J
r
= x1/2c,
j-/2,
52
and the desiGn is completely determined.
2
A4 fA; =
For the class,
22
2
['41sr
(r + 8b )
= LSx(x +
8y) + 8J
+ 2c
4
JIN
c4~N.
Special case:
s
= 4,
x
= CD,
Y
= CD.
again
We
(26) by Gardiner, Grandage and Hader
ob~ain
the desi,gn nt;!!lbered
[9].
In the same way that special choices of x, y, and z made it
possible to take fractions of G(x, y, z), a special choice of b will
enable us to use a smaller point set than T(r, 0, b).
e.'
Set b
= 0;
then
by employing only the transformation Tl and Wwe can produce a set of
3s points with suitable moment properties. We shall denote these 3s
points by the notation
TO(r, 0, 0).
The points will have coordinates
(r cos ta,
o
(r sin ta,
(
o
,
o
r sin ta,
r
,
cos ta,
t
= 0,
r
cos ta),
r sin t.et),
1, 2, ••• , (s-l) and s
The sums of powers and products of the set are
~••2 2
Y
U U
4A
U
),
= uuu
r.y2 z 2 = tz 2 x2 = st' 4/8 ,
uuu
~
5.
5'
and all other sums of powers and products up to and including order
four are zero.
Clearly any rotation of the , s-gons about their axes will
also give rise to the same moments, but we shall restrict attention
here to the set To(r, 0, 0).
From the sums of powers and products it
follows that
a positive excess.
Thus to form an infinite class of second order designs we must combine To(r, 0, 0) with sets at least one of which has negative excess.
Two examples of this follows.
Consider
To{r, 0, 0) +
31 G(al ,
a l , all +
31 G(a2 ,
a 2, a 2 )
This combination of sets has zero excess if
444
l28(a l + a 2 ) = 'sr = 0.
222
2
a l = xr, a 2 = yr , and we find that
Set
Given s
~
5, we can draw this circle centered at the origin.
Any point of it which lies in the first quadrant will give rise to a
rotatable second order arrangement.
Solving for y in terms of x, we obtain
Since x and yare interchangeable only half the points on each circle
in the first quadrant need be considered. Thus if we consider the
2
range ~ x ~
/256, we shall obtain all possible designs just once.
°
'S
The separate sets comprising the arrangement have radii
or
r,
a
15
and
r,
r.nx
and
l
The arrangement is spherical if 3x = 3y = 1 or i f x = 0 and 3y = 1
and the degenerate points are dropped.
sible I for if x
y = J3s/i28.
=y
each has the value /3s /256 and if x
A, N
2
= 8(a'2l
Now
'2
+ a 2 ) + sr
= C8(x + y) +
= 3s
= 0,
Thus none of the arrangements are spherical, even when
degenerate sets are ignored.
where N
Both these events are impos-
+ 16 + no center points.
S
2
J
2
r ,
The scaling condition A,2
=1
implies therefore that
r
2 = N/t:8 (x
al·
+ y) +
So we have obtained an infinite class of second order rotatable designs depending on x and s; each design has 3s + 16 points
apart from possible center points.
Given x,
°~ x ~
find
'2
- x ,
r = [N/(8x + 8y +
a
6)7/'2,
0
=
x1/-r,
l
a2
= y 1/2r,
which fixes all design points.
For the class,
)3;:'/256, we can
55
Consider now the set combination
'31 G(a, a, a)
To(r, 0, 0) +
+
4'1 G(c, 0, O}.
'I'his set of points has zero excess if
2
= xa2
Set
r
then
3sx2 + 16y2
c
I
2
= ya2
;
= 128.
This 1s an ellipse centered at the origin and whose axes are the coordinate axes. As s increases, the x axis of the ellipse shortens.
We find that
so that designs exist for
o~
x ~ jJ.28/3s.
The three component sets of the arrangement have radii
r
I
a..1'5
and
c
a
./X,
aJ3
and
a./Y.
Thus the arrangement is spherical if x
x
= 3,
or if x
= 0,
y
= 3 and
= y = 3,
and also if y
degenerate sets are removed.
=0
and
None of
these possibilities can occur. Thus the addition of center points is
not necessary even when degenerate sets are ignored.
~N
Therefore
~
=1
= sr2 +
= (ex +
tmplies that
2c
2
+ Sa
2
2
2y + 8}a
Now
'2
a· = N/( sx
+ 2y + 8).
We thus have an infinite class of second order rotatable designs depending on Xi each design has 3s + 14 points.
o < x < jI'28}3s,
Given x,
we can obtain
a
= LN/(sx
r
= x l / 2a,
c
= yl/2a,
+ 2y + 8)
J
1/2 ,
and the design is determined. For 'i;his class,
244
A4/~ = (64a
+ sr )/8N
2
4
= (64 + sx )a /8 N •
A summary table of the design classes in this section follows.
e
1
1
TO(r.O.O) y(a.a,a) jG(O.O.c) Number
of
points
(r l;OS to, (r cos to; (!a.!a,!a) (!c.O.O)
in
design
r sin to, r sin tA:i,
Points
(O.!c.O)
b)
0)
class
etc.
etc.
(O.O.!c)
No. of
~4
~
6
points
6s
8
3s
Set
e
·e
T(r.O.b)
Range of
first parameter
ratio on
which
class
depends
Design point
coordinate
values in
terms of N
and parameter
ratios
Second
parameter
ratio in
terms of
first
Page
Value
of
>"4~
NUIlIber of sets, used for elass
68+8
1
1
s>4
1
1
6s+6
x>O
-
x ?!8/15s
y'"
y
t["6x !l~(jx2+128/s
= t["6x !/3Ox2 -16/s
s>4
1
2
3s+16
°~ x ~J3s/256
y
°~ x ~/128/3s
Y '"
~ j3S /128-i·
s?5
I
1
1
1
3s+14
s?5
t
/128_3 s ,,2
= ["N /2(s(x+y)+4) 7/2
1
r'" x /2a. b = l/2a
a
c
= ["N/2(S(X+Y)+l)y/2
1
r '" x /2c.
b'" l/2c
r'" ["N/(8x+8Y+s).:r/2
1
a 1 '" x /2r,
a2
= l/2r
a '" ["N/(sx+2y+8)://2
1
r'" x /2a.
c '" l/2a
["sx(x+8Y)+32Ja
4N
["sx(x+8y)+8 Jc
4
46
4
4N
4
5sr /16N
2 4
(64+sx )a /BN
49
53
55
01
....
2.2 Another construction method for second order rotatable designs
in three dimensions.
It was mentioned, when forming T(r, 0, b) in the previous section, that an s-gon in two dimensions, defined by
(r cos ta,
r sin ta),
t = 0, 1, 2,
s
~
5,
0:
... , (s-l) ,
= 21{ /s
(or any rotation of it) is a second order rotatable design in two dimensions • This fact proved useful in constructing a three dimensional
point set sUitable for use in constructing second order designs.
Con-
sider now the 2s points
(2.2.1)
(r cos ta
r sin to
(r cos(t + ~)a ,
r sin(t +
,
b),
.t>o:,
2
For these points, as noted earlier,
(2.2.2)
2
IX
UU
r.x
4
u u
2
= UU
'£y =
sr2
= ur,y4u = Jsr4j4
24
r.x2
y = sr j4
uuu
,
Lz,2
Uu
= 2sb2
,
4
Ez
u u
= 2eb4
,
2 2
2 2
'£yz =I:zx = sr~2,
uuu uuu
,
and all other sums of powers and products up to and including order
four are zero. We could use this set as part of a second order design, if we could compensate for the peculiarity of moments which contain zu. Consider the four points
(0,
0,
:t,p),
(0,
0,
:tq) •
59
For these four points
(2.2.4)
J
4 = I:y4 = 0
r.x
u u u u
,
I:z
U
I:z
2
u
4
U U
= 2(p2 +
2
q )J
4
4
= 2(p + q )J
and all other sums of powers and products up to and including order
four are zero.
Thus if we combine the sets (2.2.1) and (2.2.3)Jfor
the combined set
sr2
,
44
r.x4
= i:;y = 3sr f4,
u u
I:z
,
224
= sr f4
Ex y
u u u
4 = 2(sb 4
44
+ P + q )J
uu
u u
22
22
Ey z = I:z x
uuu uuu
2-2
= sr-b
The conditions (2.2.5) will display symmetry provided that
222
2
sr = 2(sb + P + q ) J
(2.2.6)
3sr
sr
4J4 = 2 ( sb 4 + p4 + q4) ,
4
J4 2
= sr2
b •
From the third of these equations
whereupon
(2.2.8)
the other two equations give
222
+ q = sr f4,
P
p
4
+ q
and these have solutions
4
4
= 5sr /16 J
J
60
But now it is easily verified that the excess of the combined sets is
zero, that is to say, under the condition (2.2.9) the 2s +
(2.2.1) and (2.2.3) form a rota.table arrangement.
4 points
The radii of the
separate sets comprising the design are
for (2.2.1),
and
for the two parts of (2.2.3).
p and ..
When :p = q, s = 10 and thus their common value is r
q
= 0,
s
= 5 and p = r J5j.2.
./5/2;
also if
In both of these cases, therefore, the
arrangement is singular, if we drop degenerate sets, and additional
points at the origin are required to complete a design.
If n center
o
points are added,
N =
2s + 4 + no '
and then
2
ar •
Applying the scaling condition we have
ar 2
= N.
Hence
p2,
q2 = N(l !. )(10 _ s) 7s ) /8.
In order that both p2 and q2 be non-negative, it is necessary that
10
~
s
~
5.
Thus we have six separate second order designs, one for each
value of s in the range above.
(see below).
For these designs,
One of these 1s already well known
61
N/4s.
Special case:
s
As
= 5..
p2
= N/4 = 5r2 /4..
2
was shown above, p
= r '2 +
q2
= 0..
r
2
= N/5
and b
= N/20= r 2 /4.
2
2
b and so the arrangement is spherical
if we drop the degenerate points. We thus have two pentagons which
are parallel.. perpendicular to the z axis and equidistant from the
origin; and one is rotated
3SO
from the other ~e also have two points
on the z axis which lie on the sphere centered at the origin and defined 'by the two pentagons. The squared length of a side of either
pentagon is
e
•
The squared distance from the point (0, 0, p) to a vertex of the
pentagon in the plane z = b is
r
2 + (p _ b)2
= r 2{1 +
( )5 _ 1)'2/~= r 2 (5
- )5)/2.
Thus the vertex (0, 0, p) forms an eqUilateral triangle with any pair
of adjacent vertices of the pentagon in z
= 'b.
Hence the design must
be the icosahedron design.
The method we have just used, in which we employed a second
order design in two dimensions to obtain one in three dimensions, may
be extended.
In a later section we shall illustrate by an example how ..
in general.. a second order design in (k - 1) dimensions may be used
to obtain a second order design in k dimensions.
This set consists of all points of G(x, y, z) for which the product of
the coordinates is xyz.
.
It can be described as a
1
'2 replicate of
G(x, y, z) and we shall write it
(~)
G 2 (x, y, z).
The complementary set, where the product of the coordinates is -xyz,
we shall denote by
(_1:)
•
G 2 (x, y, z).
The set (2.3.2) satisfies all the conditions for a second order rotatable arrangement except two.
These are
(2.3.4)
(i,j ... 1, 2,
3),
(1:f j)
and
N
~
u=l
xl x2 Xx = O.
U
U.,IU
We recall that
(2.3.6)
N
4
N
2
2
ExL)oint set (x lu ,x2u ,xxu)' u=1,2, ••• ,N}: r x. - ~ xi X ji ,
.,I
u=l ~u u=l u
(i;' j).
Let us define a second excess function which rel..ates to the lett member of (2.3.5) as
63
Then if 8 is a
~oint
set or a combination of point sets which satis-
fies all of conditions (1.1.2) except (2.3.4) and (2.3.5), and if
(2.3.8)
Ex (8)
= 0,
Fx (a) = 0,
then 8 is a rotatable arrangement of the second order.
(+l)
4
4
Ex[G -2 (x, y, z) J = 4(x + y +
(2.3.10)
(+!)
FX['G -2 (x, y, z)J
Consider
G
=!.
Z
4
-
2 2
2 2
2 2
3Y z - 3z x - 3x y )
12 xyz.
(+!)
e
•
2 (a, a, a).
This consists of the four points
(2.3.11)
( a,
a,
a) ,
( a, -a, -a) ,
(-a,
a, -a),
(-a, -a, a),
each repeated three times. Thus we may denote the four points (2.3.11),
which form a half replicate of the 23 factorial design, by
1
'3
1
(2)
G
(a, a, a).
The second half replicate of the 2 3 factorial is denoted, similarly, by
(_!)
G 2 (a,
3
and consists of the four points
a, a),
64
(-a, -a, -a) ,
(-a, a, a),
a, -a, a),
(
( a,
a, -a) •
It is easily seen that
Ex[' .!
3
(+.!)
G -2 (a, a, a)J = _8a 4,
<:!:~) (a,
r 1
FX,- - G
3
a, a.) J =
:t.4a3 •
Consider the combination of sets of points
(+.!)
S
=G
2 (x, y, z) +
Ex(S)
= _&4 +
Fx(S)
= 12
3
(_1:.)
G 2 (a, a, a),
(16 points).
4(x 4 + y4 + z4 _ 3y2Z2 _ 3z2x2 _ 3x2 y2),
xyz - 4a 3 •
Thus S is a rotatable arrangement if, applying conditions (2.3.8),
444222222
+ Y +
x
Z
-
and
3(y z +
3 xyz
Z
x
= a3
•
+ x Y ) • 2a
Set
22222
= ua, y = va, z = wa2 ,
x
and the equations become
222
u + v + w -:; (uv + vw + wu) = 2,
9 uvw
= 1,
or
5(uv + vw + wu)
9 uvw
=
=
(u + v + w)2 - 2,
1.
4
Define
U+v+w=13·
uv + vw + wu
Then
= (13 2
- 2);?,
uvw = 1/9.
These three equations imply that u, v and ware the roots of the
cubic
t
If, given a 13
~
322
- (3t + (/3 - 2)t!5 - 1/9
= 0.
0, this cubic has three positive roots u, v and w, we
shall be able to use these values to obtain a rotatable arrangement
of the second order which contains only 16 points, using the relations
A sufficient condition for
(2.3.14)
Ax
3+
2
Bx
+ Cx + D =
°
to have three positive roots (provided all roots are real) is
A
Thus if 13
> 0,
B
< 0,
C
>
0,
D
< O.
> J2 and all three roots of (2.3.13) are real, they are all
positive.
The necessary and sufficient condition for (2.3.14) to have
three real roots is
6
(see Conkwright
= B2C2 +
~7~).
18ABCD - 4AC 3 - 27A2D2 - 4B 3D > 0
For the equation of (2.3.13) we find
It may be shown that
6(2.691376»)3645
= .0031,
6(2.691370)/3645
=-
.04,
66
so that a root of 6
= 0 lies
near
Furthermore
6(2.691376 + 8)/3645
i8 a sixth degree polynomial in
S
= .0031 + ~(8),
where ~(s)
with all coefficients positive.
4
6
2
A:L(S) = 96 + 145.3s5 + 1013.9s + 3846.78 3 + 7992.1s + 7089.88
Hence 8
>0
> ~ (s) > 0 => 6
(2.691376 + s)
> O.
In
addition,
6( )2) /3645
= -1789
and
>0
for
This means that the function 6 is convex
> )2 •
for f3 > fl
f3
and thus has only
one root in that range which must be at approximately
(nearly 2.7)
o
ft:
I
I
I
-1789 _. -- - - - -- - -- - - -'
67
Thus if
B
> 2.7,
the equation (2.3.13) gives rise to three real positive roots u, v and
wand the 16 points of S form a second order rotatable arrangement.
The radii of the two sets of points which comprise the arrangement are
jx2
+ y2 + z2
and
fja
or
Ju +v+wa
and
J3
a
or
~a
and
./3
a
Thus, when 13
= 3,
it will be necessary to add center points to the ar-
rangement in order to satisfy the non-singularity condition.
It is
desirable to add center points to arrangements which arise from values
of 13 near the singular value 3 in order that the variances of the esti-
..
mates of the model coefficients will not be large.
shall retain the degenerate points as center points.
A.2 N
When a = 0, we
Now
22
2
= 4(x2
+ Y + Z ) + 4a
= 4(u
2
+ v + w)a
+
4a2
.(
) 2
=4(3+la,
where N
= 16
+ no center points.
tion "'2
= 1,
we obtain
Hence if we apply the scaling condi-
Thus we have found an infinite class of second order rotatable designs
depending on a parameter 13; each design contains 16 points excluding
any center points which may have been added.
Given a value of
13 > 2.691376, we can find u, v and w, the positive roots of (2.3.13),
68
and then
a =~N/4(~ + 1)~1/2,
x
= u1/2a,
Y = v1/2a,
z
= w1/2a,
and the design is completely determined.
(2.3.16)
A4 / A2
For this class of designs,
22224/
= 4(x2
Y2
+ Y z + Z x ... a ) N
= 4(5(uv
= 4(~2
= (~2
+ vw + wu) +
5)a4/5N
+ 3)a4/ 5N
+
3)N/20(~ + 1)2.
Below is a table of some of the designs of this series.
was obtained by substituting for
solving the cubic equation.
sidered.
when no
~
The table
in (2.3.13) a specific value and
Only the range
~
> 2.691376 need be con-
The values given for x, y, z and a are those to be used
=0,
i.e., when
no center points are addedj for n0 center
points these values must be multiplied by the factor a
a short table of which is also given.
= (1
+ n /16)1/2,
o
The design points are obtained
from (2.3.1) and (2.3.11) with appropriate values for x, y, z and a
from the table.
The value of }..4/}..~ in the table is calculated from
(2.3.16) when N = 16.
For na center points these values must be mul-
A Selection of Designs from the 16 Point Series (when no
,
I
,
z
=0)
2'
A,4/A2
x
y
1.04096
.49090
.49090
1.56026
.60140
2.7
1.03975
.45968
.52238
1.56036
.60131
3
1.00000
.31645
.67348
1.56405
.60000
4
.894 43
.18375
.82366
1.57775
.60800
5
.81650
.12862
.88669
1.590 78
.62222
6
.75593
.09737
.92330
1.60206
.63673
7
.70711
.07722
.94697
1.61160
.65000
8
.66667
.06328
.96348
1.61965
.66173
9
.63246
.053 21
.97559
1.62647
.67200
11
.57735
.03951
.99212
1.63732
.68889
t3
a"
2.691376
,
II
I
I
14
.51640
.02767
1.00687
1.64887
.70756
19
.44721 I .01759
1.02001
1.66110
.72800
49
.28284
.00430
1.04018
1.68464
.76928
99
.20000
.00151
1.04601
1.69288
.78432
co
0
0
1.05146
1.7°130
.80000
I,
II
!
When no = 0, multiply a, x, y and z by ex and multiply
where
2
ex = 1 + no /16.
A.4/A~ by eP,
70
A Table of Factors for Adjusting the Coordinates of the
Design Points when there are n Center Points
o
no
f
ex
1
1.0625
1.°3°78
2
1.125
1.06066
3
1.1875
1.08972
4
1.25
1.11803
5
1·3125
1.14564
6
1.375
1.17260
7
1.4375
1.19896
8
1.5
1.22474
9
1.5625
1.25
10
1.625
1.27415
11
1.6815
1.299°4
12
1.15
1.32288
13
1.8175
1.34815
14
1.815
1.36931
15
1·9315
1.39194
16
2.
1.41421
The design given by
~
= 00
has a
= 0,
i.e., four of the design
points reduce to center points leaving a basic 12 point design plus
center points.
Unfortunately, this is not a new design; but it is
interesting to note that it is a rotation of the standard 12 point
71
icosahedron design, for, since
x
= 0,
y2
= 2(5
- /5)/5
and
z2
= 2(5
+.j5)/5,
Thus the 12 points consist of a constant multiple of
(0,
where 02
Coxeter (
=0
.:tl ,
~>.,
<20,
0,
.:tl ),
(.tl,
~,
0),
+ 1, and these correspond to the coordinates given by
C81,
equation 3.75).
The variation in the values of a, x, y and z is so well controlled that it is possible to use a graph to find their values for
e.
values of f3 other than those in the table.
The graph which follows,
in which «(3 + 1) is plotted on a log scale, may be used to extend
the calculated table of values if desired.
to the case n
o
=0
Once again this refers
and adjustments, as before, are necessary if
72
z
1.6
1.5
1-.4
1.3
1.2
1.11.0
e.
y
·9
.8
.7
.6
.491
.4
.3
.2
.1
e
->
x
4
5
6
8
10
20
(~1)
30
50
80 100
CHAPTER III
SECOND ORDER ROTATABLE DESIGNS IN FOUR OR MORE DIMENSIONS
3.1 The generation of point sets in three or more dimensions.
We shall now generate sets of points in k dimensions and combine them using the method we developed for the case of three dimensions.
Let (Xl' x2 ' ••• , Xg) be a point in k dimensions and let Pk be
the symmetric group of order k, that is, the group of all permutations
of k elements.
e
,
Thus we obtain kt points by operating upon
(Xl' x2 ' ••• , Xg) with the elements of Pk • Define
1
0 •••
o
o
1 • • •
o
•
• •
•
•
• 1
-1
1
o
o• • •
1
,
(i = 1,2, ...,k) ,
a diagonal matrix with -1 in the ith diagonal place, 1 in the other
diagonal places, and 0 elsewhere. From a single point (Xl'X2' ••• '~)'
by an application of
t~e k~
elements of Pk and/or the k transformations Rik , (i = 1,2, ••• ,k), we can obtain a set of 2kkl points all of
which are distinct, provided that Xl' x2 ' ••• ,
~
are all non-zero and
distinct.
The set, which we shall call
H(X l , x 2 ' ••• , ~)
and wh1ch consists of the points
(+X i ' +x i ' ... , +X. ),
- 1 - 2
- lk
where i l , i 2 , ••• , 1k run through every possible permutation of
1, 2, ••• , k, satisfies the following conditions:
EKiu = (k-l)t 2k(xi +
U
4
Ex
u iu
= (k-l)t
22
Ex x
u 1u j u
X~
2
+ ... + ~),
4
k 4
4
2 (xl + x 2 + ... + ~),
r
k
= (k-2). 2
k 22
E xix j '
1,.1=1
(i
f .1),
and all odd sums of squares and products up to and including order
four are zero, wherei,j = 1,2, ... ,k; and u is summed from 1 through
N, the total number of points.
Hence
The number of points in this set is too large for use in a design and
it will be necessary to reduce the size of the set by making several
of the Xi equal to one another and lor putting some of the Xi equal to
zero.
We note that we could have begun this discussion by considerk-L
ing a set of only 2
~t
points. The group of all permutations has,
as a subgroup, the group of all the
~
permutations.
The set ob-
tained from a point (xl,x2" •• '~)' when Xl' x 2 ' ••• , ~ are all distinct, by application of the even permutations only is such that its
75
moments are symmetrical in the way we desire.
However, nothing will
eventually be gained by this procedure, f'or, once we make two of' the
Xi'S equal in the more general set, we shall obtain double the set we
would have obtained from the set generated by use of' even permutations
Thus, except in the most general case when Xl' x2 ' ••• , ~
are all distinct, no additional reduction will be achieved by our
alone.
commencing With a half set.
> 3,
Note that when k
a cyclic permuta-
tion of coordinates does not achieve symmetry.
When there are k f'actors, the number of' constants to be estimated for a second order model is
1 + k + k + k(k-l) /2
or
(k 2 +3k+2)/2
For 4 ~ k ~ 7, we have the folloWing table:
k
4
5
6
7
15
21
28
36
•
To obtain a design consisting of' a number of' points equal to twice
the number of constants to be estimated will be regarded here as a
very desirable achievement.
Unfortunately, because of the large num-
ber of moments to be balanced when selecting design points, such an
achievement is rarely possible with the method of this section.
Thus
some of the designs to be presented are useful only when a f'airly
large number of design points is allowable.
In order to restrict the
number of points in a generated set, we shall consider only cases
where no more than three of' Xl' x2 ' ""
xk are distinct.
76
Consider the fraction of
H(p, ••• ,pj q, ••• ,qj r, ••• ,r)
which contains all possible points once and once only.
Let p occur x
times, q occur y times, and r occur z times, so that
x + Y+
= k.
Z
Let v be the number of zeros if any of p" q" and r are zero. For example" if p ., 0" q" 0" r
= 0,
then
'V
= z.
Hence the desired fraction
of the whole set" which may be denoted by H(pX, qY, r Z ), contains
points.
Therefore" the set may be written as
r, y.1 z.1
LX.
n
Co
V ~l
J
H( Px , q y , r z) ,
in notation consistent with earlier usage.
This set has sums of powers and products as follows:
4
(k-l)~ k-V[xp4 + yq4 + zr 4 7,
uIX.~u = x.1 y., z., 2
2
2
IX! X
u u ju
= X., (k-2)1
, ;
y. z.
k-v
2
4
4
4
+ y(y-l)q + z(z-l)r
2 2
2 2
2 2
+ 2xyp q + 2yzq r + 2zxr p J
~x(x-l)p
and all other sums of powers and products up to and including order
four are zero.
Hence
'
77
is the excess of this generated set of
2k-v points.
xt yt zt
By giving specific values to p, q, r, x, y and .z, we shall obtain the more useful sets of this type.
In particular, we shall re-
ject any set that contains more than 48 points in four dimensions.
4
p, q and r are distinct and all are non-zero, there are 4t 2 /2 =
points in four dimensions.
If prO, q
r 0, r = 0,
there are 4 t 22 /2 = 96 points in tour dimensions.
and z
If
192
= 1 = v,
Thus it there are
three distinct values for p, q and r, we must put r
=0
and allow p
and q to occur once only in order to maintain a reasonable number ot
points.
This leads us to consider the generated set
S(p, q, Ok-2)
= 0,
obtained by setting r
= ~4(k-2)t~
-1 R(p, q, 0, ••• , 0)
x = y = 1. The set has 4k(k-l) points and
its excess is
4 4
2 2
4(k-l) (p +q ) - 24p q •
A short table of the number of points in this set follows:
4
48
k
4k(k-l)
5
6
80
120
7
168
S ( p, q, 0k-2) by itself forms a rotatable arrangement if
or
4(k_l)(p4 + q4) _ 24p2q2
=0
= [3 ! /9 -
J /
p2/q2
(k_l)2
(k-l).
Since k 2: 4, this is possible only when k = 4 and p2/q2 =1. But if
2 2
p = q , the set can be reduced by half so that
S(p, q,
ok- 2 ) = ~8(k-2)~ -1 R(p,
p, 0, ••• , 0),
consisting of 2k(k-l) points, forms a rotatable arrangement. The
78
single design which arises from this calculation is already known and
1:9
is called the extension of (25) by Gardiner, Grandage and Hader
J.
2
If we consider only S{p, p, ak- ) to begin with, this result is
4
trivial, since the excess of S(p, PT ak- 2 ) is 4{k_4)p which can be
zero only if k
= 4,
since prO.
Although it would be llOssible to use the 4k(k-l) points of
S( p, q, 0k-2) in combination with other sets to form a rotatable design, we shall not do this because of the large number of points
which would be involved. This leads us to mention one other point set
that will not be used, given by
= 0,
z
e
x
= 1,
Y = k-l,
namely,
,
L:{k-l)l_7 -1 R(p, qk-l)o
This contains k2k points, too many for our purposes as the short table
which follows shows.
k
4
5
6
64
160
384
7
By the usual methods, it may be shown that when
and
p2
= C, + .j2k +
Ii )q2
q2
= N/2k (2 + k
+ J2k +1+ ),
a rotatable design is obtained, a design already quoted by Gardiner,
Grandage and Hader
[9 J
as an extension of their design (23).
Thus it becomes clear that the only
fraction of
po~.nt
sets which are a
79
x
z
y
R(p , q , r )
and which obey all the required moment conditions except that their
excess is not zero and which, in addition, contain what we shall consider a reasonable number of points are obtained by setting z
q
=0
= 0,
(i.e., letting the coordinates take two distinct values, one of
which is zero) or setting z
=y =0
value for the coordinate).
Proceeding in this way, we consider the
(i.e., allowing only one possible
following five sets as suitable for combination with one another for
the formation of rotatable designs. We shall not use the fractional
set notation here because of its unwieldiness.
e.
Rf NJ l~ \
k=4 k<.::5 k=6 k=7
~.!e
Set
8
1
Points of Set
(~a,~a,
••• ,~a)
1
"2 Sl one half
!h.of points N(k}
2k
16 32
2k - l
64 128
16 32 64
Excess
_2k+ l a 4
_2ka 4
(k>5 replicate of S1
only)
(~c,O,
S2
... ,O)
and permutations
S3
(O,:!:.f, ...,:!:.f)
and permutations
S4
(:!:.p,:!:.p,O, ... ,0)
and permutations
S5
(~P,:!:.P,!.P,0, ...,0)
1I
I
2k
k2k - 1
8 10 12 14
2c
I
j
I
4
k l 4
32 80 192 448 _(2k_5)2 - f i
I
2k(k-l)
24 40 60 84
4k(k-1) (k-2) /3 32
4(k-4)p
I
4
80 160 280 4(k-2}· (k-7}p
~
80
other possible sets are neglected on the grounds that they contain too
many points for our purposes.
~ediately
8everal features of the sets above are
noticeable:
81 and 8 have negative excess.
3
8 has negative excess if k < 7, positive excess if k
5
and zero excess if k=
7. (Thus if k
= 7,
>7
the points
of 8 form a rotatable arrangement; the design thus
5
formable will be derived later).
82 has positive excess.
84 has positive excess if k
> 4.
These facts determine the combinations of sets we shall choose to form
several infinite classes of rotatable designs analogous to those
formed in three dimensions.
3.2 Infinite classes of second order designs in four and more dimensions.
Consider the set combination
which has excess
e
1
0:
= 1 for
1
a full replicate of 81 and a = "2 if, when k ~ 5, a "2
replicate of 81 is used. Thus the set will have zero excess if
4
k
+ c~ = 2 aa •
where
ci
81
Set
It f'ollows that
2
2
x + Y
= 2k0:.
Thus the point (x,y) lies on a circle centered at the origin and of'
radiu
J2ka
j
to any point of' this circle that lies in the first quad-
rant corres1X)nds a rotatable arrangement of second order.
In terms of
Since x and yare interchangeable, we obtain all possible arrangements
by taking
The component sets of the arrangement have radii
or
The arrangement is spherical if x = Y = k or if x
= 0,
y
=k
and the
degenerate set S2(C )is dropped. Neither of these possibilities can
l
occur. Renee the addition of center points is not required when form.ing a design, even if degenerate sets are ignored.
AN
2
= 2(c 2l
= 2(x +
Applying the scaling condition
~
2
k 2
+ c 2 ) + 0:2 a
k 1 2
Y + 0:2 - )a •
= 1,
j: (
k-l) •
a2 =N2x+y+a2'
ThUS, we have obtained an int1nite class of' second order rotatable designs depending on x.
Given x,
y
=l2'ka _ x 2
,
k l
a = [N/2(x + y + a2 - )_j/2 ,
c l = x1/2a,
c 2= y 1/2a,
and the design is determined.
For this class,
2
k 4
~4A2 = 0.2 a ;N.
The number of design points, om1ttingcenter points, for 4 ~ k ~ 7 is
as follows:
k
4
5
6
32
52
88
7
92
(a
= 1)
(ex
=1)
2.
Spec ial cases:
(1)
k 2 l 2
x = 0, y = 2 / a / •
a = [N/(2(k+2)/2a l
J2
+ a2k ) J
1/2,
c = 0 ,
l
c
2
= yl/2a •
This gives the well-known measure polytope plus cross polytope
design with 2k center points from the vertices of the degenerate cross
polytope.
Ik-l
(2 ) x = y =.,2 a
•
a = [N/(cl/22(k+3)/2 + Q2k ) ] 1/2 ,
e
cl= c 2
= xl/2
a •
83
Then ci/a = (<X2k - l )114 and we obtain the design noted by
Gardiner l Grandage and Hader
[9 J
a.s an extension of their design
(27) I and formed by the vertices of measure polytope plus those of a
doubled cross polytope.
Consider now the set combination
which has excess
where 01 and <X are each 1 or
2
replicate of 8
1
!
according as to whether a full or half
2
is used.
This set will form a rotatable arrangement I
therefore I if
4 = xc 2 ,
Set
al
a 24 -- yc 2 ,.
we obtain
which is a circle or an ellipse centered at the origin according to
the values of <Xl and
°2 ,
Any point of the conic which lies in the
first quadrant will give a rotatable arrangement of second order •
•
Thus the range
will give all possible arrangements if <Xl
twice over if 01
changeable.
= 02'
r <X2 and all arrangements
since in this latter case x and y become inter-
The separate sets which form the arrangement have radii
84
or
cjXk,
c/Yk
and
c
Thus the arrangement is spherical if xk
or if Y = 0, xk
=1
•
=yk = 1 or
if x
and degenerate points are dropped.
possibilities can occur.
= 0,
yk
=1
None of these
Hence no additional center points need be
added, even when degenerate sets are ignored.
~ N
"'2
=~
c.C
2 + ~k(
2
2)
c. ala + a a
2 2
l
k
= [2 + 2 (a l x + Ct2 Y)
so that
~
=1
J
02,
implies
c
2
= N/2 £1 +
k l
2 - (a l x + Ct2 Y)
J.
We thus have an infinite class of second order rotatable designs depending on x.
Y
Given x, 0
= j(2~k
a
l
x
~ Rk,hl'
2
k l
2 - (a x +
1
a'2y )JJ
= x l / 2c,
1/2
a 2= Y c,
to determine the design points.
'2
"-4 A2
For this class,
4
= c IN·
The number of design points 1s as follows:
•
we obtain
- a 1x ) P'2 '
Ll +
c = [N/2
~
1/2,
85
k
k
N - n0 = (a 1 + a 2 )2 + 2k
4
5
6
7
°1
0:
40
74
140
270
1
58
108
206
58
108
206
1
1
2
1
42
76
142
1
2
2
1
1
2
1
2
Special cases:
(1)
If either x or y is zero, the design formed by the ver-
tices of a measure polytope plus
tho~e
of a cross polytope is obtained,
plus eight center points from the vertices of the degenerate measure
polytope.
The values of a and c for this design have already been
given.
(2)
If 01
= 0:2 = 0:
and x
= y,
we obtain a design consist-
ing of the vertices of a cross polytope plus those of a doubled measure polytope.
For this design,
x
J.v
= Y = /:2-k-l I~'
c
= [N/2(1 + 2(k-l)/2al / 2 ) ] 1/2
•
Consider the set combination
Its excess is
This is zero if
ci + c~ = (2k _ 5)2k - 2 4 ,
f
in which case the sets S2(c l ), S2(c 2 ) and S3(f) form a rotatable
86
arrangement when combined. Set
2
2
2
2
c l = xf,
c 2 = yf ;
then
Any point (x,y) on this circle and also in the first quadrant will
give a rotatable arrangement.
Now
•
Since x and yare interchangeable, only the range
need be considered.
The three sets which comprise the arrangement
have radii
fjX,
or
fJY
and
The arrangement will be spherical if x
the degenerate points are dropped.
fjk-l
= Y = k.l
or x
= 0,
y
= k-l
and
Neither possibility can occur.
Therefore the addition of center points is unnecessary, even when
degenerate sets are dropped.
A N = 2(ci + c~) + (k _ 1)2k - l f 2
2
= 2~(x +
Hence
~
=1
y) + (k _ 1)2k-2~ f2.
implies that
f2
= N/2
[x + y + (k - 1)2k -2_7
•
We thus have an infinite class of second order rotatable de·
k
signs depending on x. Given x, 0 ~ x s.j(2k - 5)2 - 3 ,we can find
87
_ 5)2k - 2 _ x2 ,
y
= /(2k
f
= [N/2(x +
k 2
y + (k - 1)2 - )
J 1/2,
c l = x1/2f,
c 2= y
1/2
f,
and so determine the design completely.
For this class"
and the number of design points is as below for 4 ~ k ~
k
N - no
=. k2k-l +
I
4k
7:
4
5
6
7
48
100
216
476
Special case:
k 2
x = 0" y = /(2k - 5)2 - ; hence
f2 = N/ f(k _ 1)2k - 1 + 2k / 2 j2k - 5] ,
cl =
Ignore the
o.
~egenerate
cross polytope.
Thus
c2/f =yl/2 = [2k - 2 (2k - 5) J
1/4,
and the design becomes the one called the extension of (24) by
Gardiner" Grandage and Hader
[9 1.
Note that on page 32 of this
reference" the value of b (which is our c 2 /f) given is incorrect
and should agree with the value above.
Consider the set combination
(k ~ 5 only)"
which has excess
88
and thus forms a rotatable arrangement if
Set
a
2
l
= xp2
9
ag
1
a x2 + Ogyg
l
then
= yP2
;
= 2 l - k (k
-
4),
which is a circle or ellipse according to the values of 01 and
a2 • Any
point of the conic which also lies in the first quadrant corresponds
to a rotatable arrangement of second order.
of
XI
Solving for y in terms
we obtain
so that we obtain all possible arrangements by choosing x in the range
if
a1
f:
If 01
02'
since then
X
=°2
1
the above range gives all arrangements tWice,
and yare interchangeable.
The sets which form the ar-
rangement have radii
aIJk,
ar/k
and
pfl
p/fi,
pjky
and
pfl
The arrangement is spherical if x
=Y=
or
x
= 2/k
and degenerate sets are dropped.
2/k or x = 0 1 Y
= 2/k
or y = 0,
Since none of these possi-
bilities can occur, no additional center points are needed l even when
degenerate sets are ignored.
2
~2N = 4(k - l)p
k
2
2
+ 2 (ct l a + 02a 2)
1
k
2
= [4(k - 1) + 2 (ctlx + ct2y)_7 p •
Applying the scaling condition, we find that
k
p2 = N/ [4(k - 1) + 2 (ct l x + ct2 Y)
J.
We thus have an infinite class of second order rotatable de1 k
signs depending on x. Given x" O:s x -S:/2 - (k - 4)Pl ' we can find
Y
=j [2l - k (k
P
= [N/ l"4(k
2
- 4) - ct l x
J / ct2
"
- 1) + 2k (ctl x + ct2Y)JJ 1/2,
1 2
a l = x / p"
1/2
a 2 = Y P"
and so determine the design completely.
2
4
~4/')I;2 = 2(k - 2)p
For this class,
IN·
A table of the number of design points for various situations follows:
N - n0
= 2k(k
5
6
7
ct
104
188
340
1
1
88
156
276
1
88
156
276
-21
-12
72
124
212
-21
4
k
- 1) + (exl + ct2 )2
k
Special cases:
(1)
1 k
x:: j2 - (k - 4}p
p2
1
= N/ [4(k
,
Y = O.
So
_ 1) + J(k _
4)2k+1o:~J "
l
ct
2
1
1
2
90
ai = x1 / 2
a
2
p,
= O.
This gives a design with C1.22k fewer points (omitting center
points) if we drop the vertices of the degenerate measure polytope.
The number of design points is then:
4
k
k
N - n0 - 2 0:2
(2)
k
= 4.
Then
8
1
7
5
6
72
124
212
(0:1
56
93
148
(0: = .:)
1
2
=8 2
= 1)
= 0 and we obtain the design formed
by S4(p) alone, which was derived earlier .. plus center points from the
degenerate sets.
Another combination of sets to be considered is
The total excess of this combination is
2c 4 _ C1.2k+l a 4 + 4(k _ 4)p4.
If the combination is to form a rotatable arrangement, its excess must
be zero .. that is ..
c
4
4
= 0:2k a 4 •
2
= ya2
+ 2(k - 4)p
Set
c
then
2
= xa2 ..
p
;
x2 + 2(k .. 4)y2 == a.2k ,
which is the equation of an ellipse. Any point which lies both on the
ellipse and also 1n the first quadrant corresponds to a rotatable arrangement of second order.
Solving for y in terms of x.. we obtain
91
,
and thus
os x sja2k
for y
~
O.
The three sets which form the arrangement have radii
or
ajk ,
c
and
a/k,
ajX
and
and the arrangement is spherical i f k
if Y = 0, x
=k
pfl
aJ2Y,
= x = 2y or
if x
and degenerate sets are dropped.
= 0,
2y
=k
or
Since none of these
situations is possible, no additional center points need be added,
even when degenerate sets are dropped.
= 2c 2 + a2k a2 +4(k
~N
k
= L2x +
Then
~
=1
2
- l)p
2
a •
4(k - l)y + 02 ]
implies that
2
k
a = N/ [2x + 4(k - l)y + a2
We thus have an infinite class of second
signs in k dimensions (k
~
5) depending on x.
OS
x
S !a2k
J
ord~r rot~table
Give~
x,
,
we can find
J
k
Y = (a2
2·
- x ) j2(k - 4)
a
= N/ £2x +
c
= x1'2
"a,
p
= y 1/2a,
and so determine the design.
,
4(k _ l)y + 02k
For this class,
J
1/2,
de-
92
4
""'4fA.22
k
= (4p + 02 a
4
k 4
= (4y + Q2 )a
),IN
ftl
The number of design points is as follows:
k
N -n
0
4
=2k2 +ex2k
5
6
7
82
136
226
(0 = 1)
66
104
162
1
(ex = -)
2
Special cases:
(1)
k
= 4"
c
= a =0
gives the design formed by the set
S4(P) alone with center points.
(2)
x =
la2k " so that y = 0 and hence p = O.
We obtain the
measure polytope plus cross polytope design plus center points from
the degenerate set.
Finally, consider the combination of sets
for k S 7.
S2(c l ) +S2(c 2 ) + S5(P)"
The excess of this set is
4
4
4
2(c l + c2 ) - 4(k - 2)(7 - k)p
and when this is zero" the points of the combined set form a rotatable
arrangement" which happens when
4
4
4
c l + c 2 = 2(k - 2)(7 - k)p •
Set
2
2
2
2
c l = xp ,
c 2 = yp ;
x2 + y2 = 2(k _ 2)(7 _ k).
then
Any point of this circle which also lies in the first quadrant corresponds to a rotatable arrangement of second order. Now
y
=J2(k
- 2)(7 - k) - x
2
Since x and y are interchangeable, we obtain all possible arrangements
by considering x such that
o~
x ~ j(k - 2) (7 - k) •
The separate sets which comprise the arrangement have radii
and" pJ3
pfi ,
or
and
pj3
We shall obtain a spherical arrangement if x =y
y
= 3 and
the degenerate set is dropped.
= 3 or
if x
= 0,
Neither of these possibili-
ties can occur and so no additional center points are needed, even
when degenerate sets are ignored.
~2N
Now
22
= 2(c2
1 + c2 ) + 4(k - l)(k - 2)p
= 2[x
Applying the scaling condition
p2 = N/2
+ Y + 2(k-l)(k-2)
~2
['x
= 1 we
+ y+
J
2
p •
find that
2(k - l)(k - 2)J.
We thus obtain our final infinite class of second order rotatable designs in k dimensions depending on the parameter x.
o ~x
~ j(k -
Given x,
2)(7 - k), we can find
y
=J2(k
- 2)(k - 7) _ x2
P = [N/2fx + Y + 2(k - l)(k - 2)JJ 1/2,
cl
c2
= x l /2PI
= y 1/2P,
and the design points are fully determined.
l
For this class I
2
A4/A2
=
8(k - 2)p
4
IN
and a table of design point numbers 1s as follows:
k
N - n
*~peC ial
o
= 4k(k2
- 3k + 5)
13
4
5
6
7
48
100
184
308*
case:
k = 7.
This implies that x = Y = c
the degenerate sets.
2
> k/(k
degenerate sets
= c
2
= 0 and we may drop
Hence we have a spherical arrangement formed by
the set S5(P) when p
that A4/A2
l
2
= N/120.
This set has 280 points.
+ 2) = 7/9, we must add center points.
S2(c ), (1
i
= 1,2),
are ~ removod, no
In order
If the two
= 28 and
no
additional center points are required.
This completes the list of designs to be obtained by this
method when k ~ 4.
Although it is possible to construct many other
examples, these would contain a large number of points.
table of the six designs presented follows.
A summary
e
e
·e
N-n
For additional information
0
on the sets re~er to the
Number ot points
table on page. 79
in each class
tor 4 < k < 7
as 82 8 3 84 80; (assuming full
set
replicate of 81'
CX .. 1)
Parameter a c f p p
Range
second parameter
ratio in terms
ot tirst
o~ t~st
parameter ratio
on which class
depends
Design point
coordinate
values in
terms of N
and the
parameter
ratios
I-
Value
of
~~
I
I
I
No .o~ sets used for c hss It=4 It=5 It=6 It=7
1
2
32
52
88 156
o~
~j2k-lcx
x
y .. }2kCX_i
a .. [N/2(x+y+02k-l):j-/2
c
I
l
.. x l /2a,
c
2
_ yl/2a
4
r::d-a IN
80
I
2
1
2
40 74 140 270
48 100 216 476
1
1
2
1
I
1
2
- 104 188 340
-
1
1
82 136 226
48 100 184 308
y .. j(2-k -C¥ x 2) 1x
2
l
o ~ x ~J2-k Ixl
o ~ x ~ J(2k-5):ft- 3
o~
x
~ /2 l - k (k_l) Ixl
o~
o~
x
x
~
~Ja2k
J(k-2)(7-k)
Y" ) (2k-5)2k-2 -x2
c .. [N/2(1+2k-l(CXlx+a2y»://2
a
~
c
y ..
[2l-k(k-4)~x2j-/2~
j-/2 /2(k-4)
2
Y .. j2(k-2)(7-k )-X
2
-l/2c
LN/2(X+~(k-l):ft-2://2
'. .
l
a
..
xl/2~, c ..
2
l/2f
k
p - LN/(4(k-l)+2 (CX1X+a2y»:j-/2
a
Y" [a2k _X2
l
.. x l /2c,
l
.. x l /2p , . a
a ..
c
l
l/2p
LN/(2x+4(k-l)~)_-;/2
c .. x l /2a,
P ..
2
..
4
!
IN
83
I
k l 4
(k_2)2 - f INI85
!
2(k-2)p
4
IN
87
(4y+02k)a4 IN
90
4
8(k-2)p N
'92
P" l/2a
[N/2(X+~(k-l)(k-2»://2
.. x 1/2p,
c
c
2
.. l/2p
~
~.3
A method of constructing a second order design in k dimensions
using a second order design in (k-l) dimensions.
We now deal with the extension to k dimensions of the method
used in section 2.2 to construct a design in three dimensions.
Select a second order rotatable arrangement of points in (k-l)
dimensions to which the scaling condition
~
=1
has not yet been ap-
plied. Then we shall have (say) Nt points
1
< u < Nt
,
,
for which
~~u
= A
~
4
uEx iu
3 2
= 3!'.x
u iux ju = 3C,
Nt,
say, (i ~ j),
i,j
= 1,2, ••• ,(k-l),
and all odd sums of powers and products up to and including order four
are zero.
ordinate
Consider all the points obtained by adding a further co~u
=~b
to the coordinates of the (k.l) dimensional points.
Thus we obtain a point set in k dimensions
(xlu ' x2u ' ••• , ~.l,u,-b),
consisting of 2Nt points.
Consider the point sets
(3.3.2)
(0, 0,
... ,
0, ~p)
(3.3.3)
(0, 0,
... ,
0, :!:.q).
Then the values of p and q may be so adjusted that the three sets
97
(3.3.1), (3.3.2) and (3.3.3), together with any center points which
may be added, form a second order rotatable design in k dimensions.
The number of points in the derived design is
N == 2N' + 4 + no •
This may be shown as follows.
ordinate
2
~
4
~u' ~u
u
The addition of the extra co-
to the (k-l) dimensional point set contributes only to
2 2
and Exiu~u' i == 1,2, ••• ,(k-l).
It is clear that moments
u
which were previously zero remain zero and that odd sums of powers and
products involving
~
are zero because
~
is constant
(~b).
Thus
these sums of powers and products will be zero either for each set of
N' points separately or else for the two sets combined.
For example,
if 8+ and 8 denote the separate sets, each of Nt points,
o
=
+
and
I:
u(8+ +8_)
2
xiu~u
b I: x2 + (-b) I: x2
iu
iu
8
82
1
=
==
b(A-A)
==
o.
Thus for all of the N == 2N' + 4+ no points,
(3.3.4)
2
Ex.
U J.U
4
Ex i
.
u
u
= 2A
,
1 ~ i ~ k-l,
= Ec ,
1 ~ i ~ k-l,
2
Ex x~
u iu JU
= 2C,
1 _< i
:I
j
_<
k-l,
o
98
1
~ i ~
k-l,
where u is summed from 1 through N, and all other sums of powers and
products up to and including order four are zero.
There will be symmetry in the moments up to fourth order provided that
p2 + q2 + N'b 2
P4
= A,
+ q4 + NIb 4 = 3C,
Ab
2
= C.
Thus, if these conditions can be satisfied by choice of p, q and b,
the N points will automatically form a second order rotatable arrangement, since the equations above imply that
N4
I: x .
usl
for i
:f
(3.;.6)
j
== 3
l.U
and i, j = 1,2, ... ,k.
b
2
N
2 2
1: Xi X
u Ju
u=l
= 6e
From (3.3.5)
= CiA,
q2 = (A2 _ N'C)/A,
q
4
= e (3A
2
- N Ie) /A
2
Solving the simultaneous equations of (3.3.6) we obtain
We now apply the scaling condition
or
~2
= 1,
2A
=
N
A
=
N/2,
which gives
99
and
this must be substituted into the expressions for p2 and q2 above.
Hence,
(3.3.8)
p2, q2
_ 4N'C) :. j&J(3N2_4NfC) _ (N2 .4N l C)2_7/ SA.
= l:(N2
2
2
In order that both p and q should be real, non-negative" i.e., in
order that a new design should be obtainable, the original design
must satisfy the condition
2 ~ 1; ~ 1,
¢=
where
(A
2 _ N'C)2/ (3A2 - N'C).
C
It is necessary to determine in the usual way for individual cases
whether or not the addition of center points is required.
Two illus-
trations of the use of this method follow.
Illustrations:
(1)
As a check on the calculations we reconsider the example
(from section 2.2) of two s-gons
(r cos
~
,
r
sin~
)"
(r cos(t+ ~)o:, r sin(t+ ~)o:),
which each form second order rotatable designs in two dimensions. For
these
A =
C
N'
=
=
2
sr /2,
4
sr /8,
s.
SUbstituting these values in (3.3.7), we obtain
which is the same result as before.
Since the scaling condition is
100
simply
N = 2A
= sr2 ,
it is easily applie<l to the values of p2 and q2 and, since
is restricted to 5
~
s
~
fJ
= s/5, s
10 in order that (3.}.12) should yield real
positive values for p2 and q2 •
(2)
We now derive a second order design in four dtmensions
from a second order design in three dtmensions.
Consider the well
known cube plus octahedron arrangement in three dimensions with no
center points, given by
(3.3.14)
::!:.a) ,
(::!:.a,
::!:.a,
(::!:.c,
0,
0),
0,
::!:.c,
0),
( 0,
0,
:!:.c) ,
(
In the notation of this section,
3C
44
= 8a4
+ 2c = 24a
= 3(c ) •
Hence
so that
c = &4,
A
=
&2 + 2c
2
= 4(2 + j2)a2 ,
N' = 14.
and
Thus ¢ .. 1.55, and use of the method is possible.
Cnnsider the
~oint
set in four dimensions given by
101
(:,a, !a, !a, !b)"
(:,c,
0,
0, !b),
( 0,
:!:c"
0, :!:b) ,
( 0"
0,
:!:c, :!:b) ,
( 0,
0"
0, :!:p) ,
( 0,
0,
0, :!:q) •
These points forma second order rotatable arrangement if the solutions for p2 and q2 which result from substitution of (3.3.15) into
(3.3.7) are real and non-negative, which they are, since 2
~ ~ ~
Performing the calculation, we find that
= 2[4/2 - 1 :!: 2j2ofl.
2
2
p = 4.196400 a ,
2
P , q2
or
q
2
2
- 22_7 a /(2 +
fl.)
2
= 1.259446 a "
so that
p = 2.049 a,
q = 1.122 a.
We recall that
c
= 1..682
b
= JClA = 0.765
a,
while
a.
Thus we have a second order rotatable arrangement in four
dimensions with 32 points given by
1.
102
,
+a
(::!:l.682 a,
0
(
(
(
(
+8
°
°
°
,
,
, ::!:l.682 a,
,
,
,
+a
, !.o.765 a),
0
, ::!:o.765 a),
0
, =.°.765 a),
°
, ::!:l.682 a, ::!:o.765 a),
°
,
,
, ::!:2.049 a),
°
, ::!:l.122 a),
°
°
where a is to be determined by application of the scaling condition
(
~
= 1.
°
The separate sets which comprise the arrangement have radii
p and q,
that is
1.189~ a
or
,
a.}3.414 ,
p and q,
,
1.848 a ,
2.049 a and 1.122 a.
Thus the arrangement is not spherical, and it can be used as a design
.
without addition of center points. However,
2
4
~4 /X
2 = 100 IN = .02144
. where N
= ~2
2
Hence ~4A2
+ no'
= .686,
than the singular value of .667 (for k
N,
when no
= 4),
= O.
This is greater
but not very much so; it
would therefore be preferable to use a few center points With this de2
sign. When, for example, no = 4, ~4/A.2 = .772. After deciding on the
number of center points to be used, we can determine the value of a
which specifies the design points from the scaling condition. This
gives
a
2
= (2 - j2}N/16 = .0~66l N.
CHAmR IV
THIRD ORDER ROTATABLE DESIGNS IN THREE DIMENSIONS
4.1 Conditions for a third order rotatable design.
Necessary and sufficient conditions for a third order rotatable design in k dimensions are as follows.
The N points
(4.1.1)
•
•
•
form a rotatable arrangement of the third order if
2
(4.1.2)
~lU = ...
2
=~
422
~iU = 3 ~iuXjU
= "l\2N, (say),
= 3"l\4N,
(say),
6
4 2
222
r.xiu
= 5Exi
u ux j u = l5L.x
u i UX.JUx/"XU = l5"l\6N"
U
where i, j, V= 1,2, ••• ,k, (i
1- j 1- VIs
(sa.y) ,
i) and U = 1,2, ... ,N, and all
other sums of powers and products up to and including order siX are
zero. Conditions (4.1.2) and the condition var
"y(~) = f(~t~)
are
equiva.lent. The N points form a design, if they giVe rise to a nonsingular XIX matrix and the scale is such that
~2
= 1.
The conditions
104
for non-singularity of the XIX matrix for the third order arrangement
are
A4/A~ > k/(k +
(4.1.3)
2)
(this, alone, is the condition that the matrix for a second order design should be non-singular) and
A.6~/A~ > (k
(4.1.4)
+ 2)/(k + 4).
These conditions are derived in the paper by Gardiner, Grandage and
Hader
[91.
For a third order design, the determinant of XIX is propor-
tional to L(k+2)A4 - kA.~J L(k+4)A6~ - (k+2)A.~J.
Thus, if either
of these factors is zero, XIX is singular and some of the coefficients
e.
~
are not estimable.
If either factor is very near to zero, some of
the variances of the estimates are large and the design is said to be
almost singular.
It is impossible for either fa.ctor to be less than
zero, i.e., for either of the inequalities (4.1.3) and (4.1.4) to be
reversed, as will be shown.
Since the left member of (4.1.3) is of order N, this first 1nequality may always be satisfied merely by an increase in N which
leaves the original points unaltered and which adds nothing to the
sums of powers and products, namely by the addition of center points.
However, the left member of (4.1.4) is of order zero in N and depends
only on the points xl'"
"XN'
Thus if a third order arrangement is
singular (i.e., gives rise to a singular matrix) because equality is
attained in (4.1.4), the situation cannot be altered by the addition
of center points.
The question now arises:
under what conditions
105
is a third order arrangement singular? An exact statement of the situation may be given as follows.
Theorem:
A third order rotatable design in k dimensions is singular
if and only i f all of its points lie on a k dimensional sphere centered at the origin.
Proof:
8 (i) =
Let
j
N
1:
u=l
J
xiu '
(i = 1,2,. u,k).
Then, if the points (4.1.1) satisfy the conditions (4.1.2),
(1)'
84
e.
N
4
N
'to
2
2
=u=l
~ xi = ~
u
xi x '
u=l u Ju
6
N 4 2
1: xi x
u=l u j u
~
~
~
(1,J = l,2, ••• ,k),
(i
r j)
and
8
( 1)
6
N
=u=l
1: x
=5
1u
(i,J,( = l,2, ••• ,k), (i';' J f(
r i),
= 15
N 2 2 2
t xi x x II..
u=l u j u ICy
and 8~i) may be denoted by Sj since
it is independent of i i f conditions (4.1.2) hold.
N
6
Hence
15 1:: r
or
15A6N = 86 = 15
u=l u
It follows that
= kS6L15 + 9(k-l) + (k-1)(k-2)
2
:=
kS 6 (k + 6k + 8)
~
u=l
r 6 /k(k+2)(k+4).
U
,
J
106
Similarly,
k
(i)
N
k
4
kS 4 = I: S4 = I: I: x.
i=l
u=l 1=1 ~u
=
4
N
I: (Xl
u=l
u
4
+
00.
ltU
N
2"
= u=lL
I: r{X +
lu
22
+x.
0 ..
=
+ x. )
xu
) -
22J
I: xi x.
irj
U
JU
N 4
N
2 2
I: r - k(k-l) I: xi x j
u=l U
u=l u u
N
4
= u=l
I: r
u
- k(k-l)S4/3.
Hence
3
4
~
r
u=l
U
e.
= kS 4 [3 + (k-l) J
= k{k+2)S4 '
and so
Clearly
N
Now
A
4
~
4
E ru
u"'l
NO{N
=
212
U:l ruJ
k
0i;2 0
(This expression is equal to k/{k+2) if and only if all points lie on
a sphere and it can be increased merely by an increase in N, ioeo l by
an addition of center points)
0
Also
107
N
6
t r
u=l u
N 2
t r
u
u=l
k (k+2) (k+4 )
k
k 2 (k+2)2
·(11 r~r
k+2
• k+4
Thus if' all the points of the arrangement l1e on a k dimensional
sphere, then r u = r (all u
duces to (k+2) /(k+4). Let
= 1,2, ••• ,N)
and the right hand side re-
Suppose that the N points are distributed in such a way that n. of
3.
them lie on a k dimensional sphere of radius r i. Then
6
6
6 . 2
2
2
(nlr l + n2 r 2 + • eo + nprp)(nlrl + n2 r 2 + ••• + nprp )
-=
42
')-...2
(nlrIil + n2 r Ii
4
2 + ... + npr p )
)..6~
k+2
·4
k+
2 2
2 2
t n.r i + t t nin .rir j
k+2
=i ~
itJ
J
4 2
• k+4
(I: n.r )
i
where i,j
3.
i
= 1,2, ••• ,p.
The first factor can reduce to unity only when
all of the r~~ are equal; otherwise the factor is greater than unity.
108
Thus the arrangement is singular if and only i f all the points lie on
a sphere in k dimensions.
Hence in order to get usable third order designs, we must combine several spherical sets of points with different radii.
4.2 The known third order rotatable designs in three dimensions.
We may divide third order designs into two groups, sequential
and non-sequential.
A sequential design can be performed in two
parts. One part is a second order rotatable design which may be run
first; then, if the second order polynomial approximation is found to
be inadequate, the trials of the second part may be run and a third
order surface fitted.
Such designs are more useful in practice than
the non-sequential type, of which all the trials must be run at one
time in order to make a rotatable least squares fitting possible.
Only four third order designs are known in three dimensions,
and these are to be found in the paper by Gardiner, Grandage and
Hader
[9]. The known designs contain points which are the vertices
of
(a) icosahedron plus dodecahedron
(32 points),
(b) cube plus two octahedra plus cuboctahedron
(32 points),
(c) (cube plus octahedron) plus (truncated cube plus
octahedron)
(44 points),
(d) (cube plus doubled octahedron) plus (truncated
cube plus octahedron)
(50 points).
109
We may translate these into our Chapter I notation as follows:
(4.2.2)
1 1 1
(a) ~(Pl,ql'O) + ~(P2'~'O) + ,G(a,a,a),
1 1 1 1 )
(b) ~(a,a,a) + ~(cl'O,O) + ~(c2'O,O) + ~(f,f,O ,
where in each case the values of the parameters are determined by the
rotatability conditions, and will not be quoted here.
Design (a) is such that the radii of the two spheres on
which all its points lie are very nearly equal.
"'2"'6
7=
4
In fact,
(1.00028) .2 ,
7
which means that the design is almost singular. Thus the variances
of' the linear and cubic coeffic ients are very large. This design is
of the sequential type. Design (b) is a combination of our basic
generated sets and is non-sequential. Designs (cl and (d) are 'both
sequential; (c) is almost singular, as was noted in the original
presentation •
.
'
4.,
The construction of infinite classes of third order rotatable
designs in three dimensions.
We Shall now proceed to obtain some infinite classes of third
order designs of the sequential type by making use of the previously
derived second order classes.
In order to do this, we shall find it
110
necessary to construct additional functions similar to the excess
functions previously introduced.
We recall that
EX(G(PI q, r»)
44222222
= 8(p4
+ q + r - 3p q - 3q r - 3r p ).
Additionally we define
(4.3.{)
r 2 )"
Ax{G(p, q, ~»)
= 8(p2 + q2 +
Gx{G(p" q, r»)
= 8(p4q2 + q 4r 2+ r 4p2 -
2 4 2 4 2 4
p q - q r - r p ),
Hx(G(p, q, r») = 8(p6 + q6 + r 6 _ 45p2q2),
Ix{G(p, ql r»)
= 4(p4q2+
q4r 2+ r 4p2+ p2q4+ q2r 4+ r 2p 4
_ l8p2q2r2).
e
Note tbat l for the point set G(p, q, r),
"
2
= Ax(G),
4
= 8(p4 + Q4 + r ~ ),
!'.xi
u u
!'.x.
u ~u
2 2
~iUX Ju
6
!'.xi
u u
= ;(p2q2 +
= 8(p6 +
q
6
2 2
q r
2 2
+ r p ),
6
+ r, ),
rx4 x2j
q4r 2 + r 4p2)"
r.x2 x~
= 8(p2q4 + q2r 4 + r 2p4),
u iu
1>.1
= 8(p4q2 +
u
u 1u JU
1>.1
22222 2
~1uxJt?fu = 24p q r ,
where i
rj
~ (~ i,
i,.1,(
= 1,2,; and u = 1,2, ••• ,N;
the notation
111
i>j here denotes that i is before j in cyclic order, i.e., 1 > 2,
2 > 3, 3 > 1. Consideration of (4.3.2) and (4.3.3) together with
(4.1.2) shows that if
(4.3.4)
Ex(G)
= GX(G) = Hx(G) = IX(G) = 0,
then the points of G(p, q, r) form a rotatable arrangement of the
third order. All of the excess functions we have defined operate
linearly on sets of points of the form G(p, q, r) and fractions of
G(p, q, r), that is to say
where q represents any of E, G, H or I. Thus if the four functions
of (4.3.4) are zero for any aggregate of points, then this aggregate
forms a third order rotatable arrangement. The addition of center
points and the satisfaction of the condition AX( ) =
N(~ =
1) makes
the arrangement into a design prOVided, of course, that the nonsingularity conditions are satisfied.
We shall now list the generated sets of the form G(p, q, r)
or fractions thereof which we used preViously, together with the
values of the excess functions for each set.
fe
e·
i
* G(p,q,r)
1 Point
1
*1
-~(p,q,O)
~(p,q,O)
Set
No. of
points
t
Ax
x
I
24
12
222
8(p +q+r )
2 2
4(p .+q )
8(p2+ q 2)
I
2 .2
8(p +2q )
8f
2 42 4 2 4
-p q -q r -r p )
t
666
8(p +q +r
H
x
Ix
1
24
4 2 4 2 4 2
8(p q +q r +r p
I
+ ~(q,p,O)
1
222
-45p q r )
!
1
~(f,f,O) "§G(a,a,a) ~(c,O,O)
12
22
22
22
-3p q -3q r -3r p )
Gx
G(p,q,q)
1
24
4 4 4
8(p +q +r
E
e
8
6
2
&2
2c
_4f 4
_16a 4
2c
2
i
I
4 4
2 2
442214422
4(p +q -3pq ) 8(p +q -3p q )! 8(p -q -6p q )
I
.
,i
4 2 2 4
4(p q -p q )
0
I
4
0
0
0
I
0
I
t
I
!
i
6 6
4(p+q )
424242
222
8(p q +q r +r p -9P q- r )
442
p q
or
-- 2 -4 2 4 2 4
2 2 2
8(p q +q r +r p -9P q r )
or
6 6
8(p +q )
4
81'6
-1l2a
4 2 6
2 4
8(p q +q -Bp q )
6
I
2c
6
,,
I
I
4 2 2 4
4(p q +p q )
24
pq
Ij8(p6+2q6 -45p2q4 )
4f
6
1
1- 6.
6
0
*
For these two sets a unique
for Ix does not exist since
the;t"e is a lack of symmetry. The two
possible values of the expression Ix
are shown; they are equal when p
q.
e~ression
=
l-'
l-'
I\)
113
It is, of course, possible to form non-sequential third order
designs and classes of designs by a skillful combination of these
sets. We have already shown that design (b) of the Gardiner,
Grandage and Hader paper is of this type. We shall leave aside this
possibility and instead form some infinite classes of designs that
may be performed sequentially.
Since, for sequential performance,
each of the two parts of the design must be itself a second order design, we shall employ some of the infinite classes of second order
designs already obtained. We now present a table of second order de.
sign classes which may be used, and the values of the various excess
functions for each class.
Since each class satisfies the second
order conditions, Ex{class)
= 0,
as is shown in the table. We shall
consider those classes, obtainable from the basic generated point
sets, which have Gx(class)
= O.
Each class contains three parameters
which give rise to two ratios connected by one equation (Ex
= 0).
If we combine two such classes and apply the other conditions of
(4.3.4), we shall have a set of points with six parameters giVing
five ratios connected by four equations. Thus we shall obtain a
single infinity of third order rotatable arrangements dependent on
one parameter ratio.
e·
I
D
1
Reference I
Set
composition
of
class
j No. of
I
i•
II
points
A
X
1
3G(a,a,a)
I
I
1
+ ~(cl'0,O)
I . 1
+ ~(c2'0,0)
I
D
3
D
2
1
1
3G(a1 ,a1 ,a1 )
~(f,f,O)
1
1
+ '3G(a2 ,a2 ,a2 ) + ~(cl'O,O)
1
1
+ ~(c,O,O)
+ ~(c2'0,0)
D4
D
~(f,f,O)
G(p,q,q)
1
+ 3G(a,a,al
Hx
Ix
1
+ 3G(a,a,a)
1
+ iG(c,O,O)
26
32
2
2 2
S(p +2q +a )
2+c2) 18(a
2+a
2)+2c
2 1Sf
2
22
2
2+2c
2
Sa2
+2(c
+2(c
+c )1Sf
+&
1 2
1
4
2(c 4+c 4 -Sa)
1 2
2
1
I
2
I
4
2(c 4 -S(a 4+a 4 » \. 2(c 4+c 4 -2f 4 ) 2(c 4 -2f4 -Sa)
1 2
1 2
I
2 2)
8(p4 ~q4 -6p.q
_169. 4
(zero)
Gx
5
1
24
22
20
I
Ex
e
-e
o
o
o
o
o
) 2.(c 6+c 6.
2(c 6+c 6 )-112a 61 2c 6 -112(a 6+a 61
)+8£ 61 2c 6+Sf 6 -1l2a 61 8(p 6+2q6 -45p2 q 4 )
1 2
1 2
1 2
_112a 6
_169. 6
6 6
-16(a +a )
1
2
4f
6
I
4r 6_16a 6
4262442 6
2 4
8(p q +q -8p q ) 8(p q +q -Bp q )
_169.
6
i
I-'
I-'
.;:-
115
We shall now illustrate the formation of infinite classes of
third order rotatable arrangements by the combination, in pairs, of
certain of D , D , ••• , D and the application of conditions (4.3.4).
2
l
5
Consider the combination Dl + D6' containing 50 points •
These points form a. sequential rotatable arrangement in three dimen-
."
sions if all the excess functions are zero, namely i f
(4.'.5)
GX(D l
Hx(D l
IX(D l
In
full~
= Ex(D6) = 0,
+ D6) = 0,
+ D6) = 0,
+ D6} = o.
Ex(D l )
these equations are
(4.3.6)
444
c l + c 2 - 8a
=0
4(p4 _ q4 _ 6p2q2) + c 4
c
e
= 0,
6 -I- c 6 - 56a6 + 4(p6 + 2q6 - 45p24
6
q ) + c = 0,
2
l
_ 2a.6+ p4q 2 + q6 _ 8p2q 4
<.
= o.
Make the substitution
c
p
q
'2
2
= ya ,
2
2
2
= uc ,
2
2
= vc ,
c
6 = ta 6 •
Since equations (4.3.6) are homogeneous, they may be put in the form
116
.
2
2
u - 6uv - v +
'41 = 0,
x 3 + y3 _ 56 + (4u 3 + 8v 3 _ l8uv2 + 1) t
2
2
-2 + (U v + v 3 - 8uv ) t
= 0,
= 0,
a system of four equations in five unknownsj thus if one variable is
specified, the values of the other variables are determined.
However,
we are interested only in solutions for which x, y, u, v and tare
all real and positive.
ment exist.
Only in such a case will a rotatable arrange-
Simple algebraic solution of the equations (4.3.8) is
not possible. We proceed by selecting one variable and obtaining the
others successively, applying the conditions for positive solutions
as we go.
Select v
> O.
Then from the second equation,
u -=
•
From the fourth equation,
2t-1
= u2
v +3
v - 8uv 2
2j
'2
= :!:. v40v
Now t
> O.
- 1 - 4v 3 - v j4 •
Thus the top root alternative is impossible, which means
that
and
(4.3.10)
...
'2t- l
= v2 j4Ov'2
u
> 0 => .025
t
>0
_ 1 _ 4v 3 _ v j4
~
v
2
~
.25,
2
>.143187 ~ v •
117
Thus we shall require
in order that all of t" u" and v shall be real and positive.
By sUb-
stituting for u and t in the third equation we find that
x3 + y3
(4.3.12)
=
f(v)"
where
(4.3.13)
f(v)
= 24 - 4~2(16v2_7v+l)
+ (1+8v_24v2 ) j4Ov 2 -1]Iv[)84v 4_48v 2-1].
But since
x
2
2
+ Y
= 8"
real" non-negative solutions exist for x and y only when
•
16
~
f(v)
~
16ft = 22.627424 •
The range of v for this to be true is more difficult to find and involves considerable computation. We find, considering only points in
the range
(4.;.11)" that
= 16"
f(.4663l6) = 16)2"
f(.4l9894)
and f(v) increases monotonically from its lower value
upper value (16ft) for v in the indicated range.
(16) to its
This may be observed
from the summary table of solutions to be presented later.
Thus we
see that whenever
.419894
i.e ."
then equations
Sv ~
.466316,
.176311 ~ v2~ .217451,
(4.3.8) have a solution that gives rise to a third
order rotatable arrangement. We have already obtained both t and u
118
in terms of v.
It remains only to express x and y in terms of v.
We
recall
(4.3.17)
x
2
2
+ Y
x 3 + y3
= 8,
= l' (v) •
Let
(4.3.18)
x + y
= 20,
xy =
/J ;
then
4(i -
and
20(8 - ¢)
2¢
=8
= f(v)
8Q(3 - Q2)
•
or
= f(V),
a cubic which, given v, may be solved for g
= g(v),
either intera-
tively or by the trigonometric method for solution of cubics.
From
(4.3.18)
(4.3.20)
x,y
=Q !
= 0
a function of v only.
J
g2 -
/J
!j4 _g2 ,
These calculations were carried out for 12
values of v in the range
(4.3.16) including the end points of the
range and the results were as follows:
119
t
x
v
u
.419894
.029596
61.248478
2
2
.420
.029553
61.069211
2.073576
1.923612
.425
.02750:;
5:;.553:;02
2.4:;8052
1.4:;:;84:;
.•430
.025484
47.517331
2.56:;986
1.19414:;
.435
.02:;497
42.568299
2.640568
1.013607
.440
.0215:;9
38.440873
2.695576
.856661
.445
.019610
34.949405
2.735256
.719982
.450
.017709
:;1.960134
2.765977
.590668
.455
.0158:;4
29.374247
2.790168
.46359:;
.460
.013984
27.117168
2.809441
.327169
.465
.012159
25.131560
2.824930
.140614
.466316
.011682
24.648:;:;1
2.828428
0
y
(
(All entries are correct to six decimal places)
The
diagr~
which follows indicates a typical solution for x
and y, the one corresponding to the value v
are interchangeable, the two solutions
= .435.
Since x and y
(intersections of curves)
shown are equivalent. The passage from one extreme solution,
x
=y
= 2 to the other, x
= 2./2,
y
= 0,
may easily be seen here.
120
y
2
2
~
x+Y=8~
2
•
•
1
1
2
x
121
The top line of the table corresponds to the case where x
= y,
that
1s to say the curves touch at the point x = y = 2. The bottom line
of the table corresponds to the case where the cubic curve is about to
pass outside the circle and the points of intersections are at
x
= 2/2,
Y
=0
and x
= 0,
y
= 2/2.
The case of the diagram lies be-
tween these extremes.
Any line of the table gives five ratios which may be employed
in (4.3.7) to give five of the parameters c l ' c2' a, p,q and c in
terms of the sixth. Thus we obtain a rotatable arrangement which becomes a design if the scaling condition is used to determine the free
parameter, prOVided that the non-singularity conditions are satisfied.
•
Since the first of these can be satisfied by the addition of center
points, it need not be considered further. We reqUire, then, that
.714286 •
By our theorem, this will be so unless all the points lie on one
sphere.
Now each design consists of five separate point sets of
s<2.uared radii
3a2 ,
or
It is easy to see from the table that the various radii are different.'
The actual values of the parameters are obtained as follows:
2222
2
A2N = 8a2
+ 2(c + c ) + 8(p + 2q ) + 2c
1
2
= a 2 (8 + 2x + 2y) + (8(u + 2v) + 2)c 2 ,
122
h4N
h6N
Since c
6
= ta6,
= Sa4 +
= 8a4 +
16p q + 8q
= 8a6 +
24p q
= 8a6 +
24uv2c6 •
2 2
4
L
(louv
+ 8v 2 )c 4,
2 4
these values may be found in terms of a, as the fol-
lowing table shows. This table may be considered an extension of the
preceding one, since their rows correspond.
¥a-2
•
h4Na
.4
-6
h6Na
h4/9
2
h6~/A4
51.29949'
".005920
15.670698
.012542
.7'7934
51.264057
32.963451
15.640735
.012543
.737912
49.742964
31.187123
14.384946
.012604
.735680
48.418222
29.705827
13.373640
.012671
.7'3796
47.249791
28.449,62
12.542421
.01274,
.732211
46.195292
27.368099
11.847274
.012825
.730679
45.242534
26.428609
11.257215
.012912
.729171
44.359086
25.599301
10.750681
.013010
.727717
43.531524
24.864635
10.310955
.013121
.726003
42.726269
24.208077
9.925752
.013261
.723665
41.867093
23.617604
9.585751
.013474
.719494
41.462397
23.471355
9.502710
.013653
.715197
(Recall that h6~/h~ShOU1d be greater than .714286.)
We now examine further the extreme casee of the table. The
•
bottom line gives a design consisting of
Lr ;1 G(a, a, a) + ~1 G(c l ,
1 G(O, 0, 0) J
0, 0) + 4
plus
~G(p, q, q) + ~ G(c, 0, O)J
with values of the parameter ratios, as derived above.
Reference to
(4.2.2) will show that this is known design (c) with six center points
(represented by
i G(O, 0, 0».
The top line gives a design consisting
of
~
1 1 1
3 G(a, a, a) + ~ G(e l , 0, 0) + i G(c l ,
0, O)~
plus
e·
'I
~G(p,
q, q) +
i G(c, 0, O)~
with the values of the parameter ratios as derived above.
(4.2.2) will show that this is known design (d).
Reference to
The details of the
verifications will not be reproduced here.
Thus the infinite class of third order rotatable designs obtained has as its two extreme cases two of the designs already known,
and the passage from one extreme to the other is by a continuous
infinite sequence of new third order designs for which the second nonsingUlarity condition becomes successively stronger.
The class of designs just obtained was chosen for a detailed
presentation precisely because of its link with the only sequential
type designs known up to this time (we are ignoring the claims of design (4.2.2) (a) which is almost singular as eVidenced by (4.2.3».
124
4.4 Further infinite classes ot third order rotatable designs in
three dimensions •
•
We shall now show how our method may be used to obtain other
new infinite classes of third order designs. The computations required to work out the details of a design class such as the one Just
presented are very great and would, in fact, provide work highly suitable for a small, easily programmed computer, for example the
Burroughs E101.
It is for this reason that we confine ourselves in
what follows to demonstrating the existence of the design classes we
shall derive.
•
Consider the combination D2 + D;. This contains 46 points
which form a sequential third order arrangement prOVided that all
of the excess functions are zero, i.e., if conditions (4.;.5) are
satisfied for D2 and D;. This implies that
(4.4.1)
c4
= 8(a4l +
a42 ),
444
c l + c 2 • 2f ,
c
6
6
6
6
6
6
-4(al + a 2 ) + f
Let
(4.4.2)
2
al
a
c
.
6
6
- 56 (a l + a 2 ) + (c l + c 2 ) + 4f
= xc 2 ,
2
2
= yc ,
2
2
l
2
= uf ,
2
c2
= vf2 ,
6
= tc 6;
f
= o.
6
= 0,
125
then equations (4.4.1) may be written
•
x
u
2
+ y2
2
+ v
2
= 1/8
= 2,
1 - 56(x 3 + y3) + (u 3 + v 3 + 4) t
_ 4(x3 + y3) +
= 0,
= o.
t
It tollows that
For a
real~
non-negative solution in x and y we thus require
1
•
16ft
>
1
4(10 _ u 3 _ v 3 )
>
-32
1
or
or
10 - 4ft ~ u 3 + v 3
But
>
2.
u2 + v 2 = 2;
hence tor a solution in u and v we require also
But
10 - 4.fi ~ 2.fi,
so the second range tor u3 + v3 is the smaller.
In addition, it the
above conditions hold, t is automatically positive.
Since u and v
are interchangeable, it tollows that it u is given, 0
~
u
~
tind a rotatable arrangement 'With parameter ratios given by
1, we can
126
u ,
•
- u
4t- l
x,y ==
==
2
,
10 _ u 3 _ (2 _ U)3/2 ,
Q
+}2:..
- (l ,
- 16
where Q is a real positive solution of the cubic
Q(3 - 32Q2)
==
2t.
This cubic is obtained in exactly the same way as was the cubic of
the first illustration.
It is now a matter of computation only to
obtain the ratios u, v, x, y and t, and then the actual parameters
•
cl' c2 , a, c, a l and a 2 by applying the scaling condition. We have
not yet verified that the second non-singularity condition is satisfied.
or
The squared radii of the separate sets are
2
; a 2 , ; a 2 , c,
l
2
nf2,
c;
2
;xc 2 , ;yc 2 , c,
nf2
c;
,
Clearly equality of radii cannot be attained.
value 114; and if u
=v
I
If x
= y,
both have the
they each ha"le the value 1. Thus we have
shown that a combination of the sets D2 and D; will provide an infinite class of third order rotatable designs.
Consider the combination D2 + D4 • This contains 48 points
which form a rotatable arrangement of the third order if
4 +4
c4
= 8(8.
8. ),
2
1
1
4
4
4
=C
2t
c
Be. ,
-
6 .. 56 (8. 6 +
l
1
.. 4 (8. 6 +
6
8. ) + c
2
6
+ f
8. )
2
1
6 + 4r 6 .. 56a6
6
4& 6
..
= 0,
= o.
Set
2
8.2
= YC 1 '
2
2
= uc ,
f
2
2
2
8. = vc ,
•
c
6
= tel6 •
The -equat.ions "then become
x
2
2u
= y2 = 1/8,
2
=1
2
.. 8v ,
~) + £1 + 4u' .. 56v'J
1 .. 56(x3 +
.. 4(x 3 + y3) +
Cu'
.. 4v' J t
t
= 0,
= 0.
Thus we find that
,-
Hence
'"
t
>0
v
~
0
> u' >
=>
u
~
0,
1
.1
=>
2
~ 2ti
v
>0
u ~ .464159,
::::=>
u
and
t ~ 0,
~
.707107.
provided that
128
.464159 :$ uS. 707107
2
i.e .,
Now
.215443 :$ u :$ .5 •
x3 + y3 = (u3 _ 4v 3 )j4(lou3 _ 1).
For a solution in x and y we thus require
or
u 3(10 - 4ft) + (1 - 2u2 )3/2
or
feu)
~
1
~
~
g(u),
1
~
2u' + ./2(1- 2u2 )3/2
say.
We need consider only the range of u given above, namely
•
.464159 SuS .707107. The following short table of values of feu)
and g(u) in this range is informative.
u2
feu)
g(u)
·50
1.536
.707
.45
1.343
.648
.40
1.188
.632 ~
.35
1.064
.647
.30
.967
.686
.25
.896
.750
.21
.860
.817
Note that g(u) attains a minimum of .632 for u2
Minimum
= .4
and increases
away from this point, but does not exceed unity in the range shown.
feu) has a minimum when u2 = .149 and, through the range shown above,
129
decreases to that value.
considered, but feu)
Thus g(u)
S1
for all u in the range being
~ 1 only when .5 ~ u2 ~ V, where .35 ~ f/ ~ .30 •
By an iteration procedure we find that
V=
when
f(f)
.318833,
=1.0000003
•
Hence, provided that
there exists a third order rotatable arrangement.
The parameter ratios
are obtained as follows, after a value of u has been specified:
•
tool == lOu,) _ 1 ,
x,y
where
Q
= Q "!aIR" rl
is any real positive root of the cubic
Once again this leads to a specified design after application of the
The various sets of the design have squared radii
scaling condition.•
as follows:
2
3a 1
or
.
,
2
3xc ,
l
2
a
3 2 '
2
3yc ,
l
,
2
cl '
2f2
2
cl '
2uc 2 ,
Clearly the radii cannot be all the same.
3a
2
3vc
2
and
c
and
c
2
2
•
Hence the second non-
singularity condition is satisfied and an infinite class of third
order rotatable designs of the type described has been shown to eXist.
Further designs of this type may be derived inan identical. manner.
130
BIBLIOGRAPHY
Box" G. E. P." "Multi-factor Designs of First Order,,"
Biometrika, 39 (19;2), 49-57.
Box" G. E. P., "The Exploration and Exploitation of Response
Surfaces:
Some General Considerations and Examples,,"
Biometrics" 10 (1954)" 16-60.
[3]
Box" G. E. P. and Hunter" J. S." "Multi-factor EXPerimental
Designs," Institute of Statistics Mimeo Series No. 92,
Raleigh, North Carolina" (1954); alsQ Annals of Mathematical Statistics"
[4J
•
28
(1957)" 195-241.
Box, G. E. P. and Hunter" J. S., "Experimental Designs for
the Exploration and Exploitation of Response Surfaces,"
Proceedings of Symposium on Design of Industrial Experiments" Institute of Statistics" Raleigh, North
Carolina, (1956).
[5]
Box, G. E. P. and Wilson, K. B." "On the Experimental Attainment of optimum Conditions," Journal of the Royal
statistical Society" (Series B), 13 (1951), 1-45.
['6]
carter, R. L., "New Designs for the Exploration of Response
Surfaces," Ph.D. Thesis, University of North Carolina,
Chapel Hill, North Carolina, (1957) •
•..
[7]
Conkwright, N. B." Introduction to the Theory of Equations,
Ginn" Boston, ~94l) •
131
L8J
.#
Coxeter, H. S. M., Regular Polytopes, Methuen and Co., Ltd.,
London (1948)
[9 J
•
Gardiner, D. A., Grandage, A. H. E. and Hader, R.
J.,
It
Some
Third Order Rotatable Designs,1t Institute of Statistics
Mimeo Series No.
149, Raleigh, North Carolina, (1956) •
•
. Mo-,
•
VP-
.h
YIJI";'(x'/; I ~ ~ tJ,,,,'Jf> I j, 7,(J, ""~
~ dil~t ~ t' wl"fwr ~
C>-
~~ ~ ~ ~ Ut~y&
rw~·
1 t&ejWL d· .
f1:4
A..