on the maximal number of factors and the enumeration of 3-symbol orthogonal arrays of strength 3 and...
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:• joumalof statistical planning
Journal of Statistical Planning and and inference Inference 58 (1997) 43-63 ELSEVIER
On the maximal number of factors and the enumeration of 3-symbol orthogonal arrays of strength 3
and index 2
A. H e d a y a t a, l, E s t h e r S e i d e n b, J o h n S t u f k e n c,, a University of Illinois at Chicago
b Hebrew University CDepartment of Statistics, Iowa State University, Ames, 1.4 50011, USA
Received 29 November 1993; revised 17 February 1995
In honor o f Paul Erdds
Abstract
We show that an orthogonal array for three symbols, of strength 3 and of index 2 can accommodate no more than five factors. We also show that there are exactly four nonisomorphic arrays with five factors.
A M S classifications: primary 05B15; secondary 62K15.
Keywords: Fractional factorials; Difference schemes; Nonisomorphic orthogonal arrays; Rao's bounds for orthogonal arrays
1. Introduction
An s-symbol orthogonal array o f strength t, index 2 and with k factors is a k x 2s t
array based on s symbols, say 0, 1 . . . . . s - 1, such that any t × 2s t subarray contains each
possible t-tuple based on 0, 1, . . . , s - 1 exactly 2 times as a column. We will denote such
an array by OA(2st , k , s , t ) . Besides being used for the construction o f various other
combinatorial configurations, orthogonal arrays have, since their introduction by Rao
(1947), gained prominence among statisticians for their properties as fractional factorial
designs. Especially, 2- and 3-symbol arrays of strength 2, 3 or 4 are o f interest in that
context, though more and more attention is being given to larger orthogonal arrays as well. Further information on orthogonal arrays may be found in Hedayat et al.
(1996).
* Corresponding author. i Research sponsored by National Science Foundation Grant # DMS-9304014.
0378-3758/97/$17.00 Q 1997 Elsevier Science B.V. All rights reserved PH S 0 3 7 8 - 3 7 5 8 ( 9 6 ) 0 0 0 5 9 - 6
44 A. Hedayat et al./Journal of Statistical Planning and Inference 58 (1997) 4343
Rao (1947) provides an upper bound, at least implicitly, for the possible num- ber of factors k in an orthogonal array for given values of s >~ 2, t >~ 2 and ). >t 1. The largest number of factors for which an orthogonal array with specified values for s, t and 2 exists has by some authors been denoted by f (2st ,s , t ) . Our in- terest will be in f (54 ,3 ,3) . By construction of an OA(54,5,3,3) it is known that f ( 5 4 , 3 , 3 ) / > 5. (See e.g. Fujii et al. (1987), who present three nonisomorphic ar- rays OA(54,5,3,3), found by a computer search.) Rao's upper bound conveys that
f (54 ,3 ,3 ) ~< 9, while the improvement by Bose and Bush (1952) on Rao's bound
shows that f (54 ,3 ,3 ) ~< 8. We will determine the exact value for f (54 ,3 , 3) and the number of nonisomorphic
arrays that attain this number of factors. The precise results are stated in Section 2. Since the proofs are rather long and technical, all proofs have been relegated to
appendices.
2. Results
Permuting rows or columns of a given orthogonal array results again in an orthogonal array. So does permuting or renaming symbols within a row. Any two orthogonal
arrays that can be obtained from each other through a combination of such operations are called isomorphic to each other.
The two main results of this paper are formulated below.
Theorem 1. The maximal number of factors in a 3-symbol orthogonal array of strength 3 and index 2 is equal to five.
Theorem 2. The number of nonisomorphic orthogonal arrays OA(54, 5,3,3) is equal to four.
In the appendices we will first establish the validity of Theorem 2, after which we will prove Theorem 1 by showing that none of the four arrays can be extended to an OA(54,6,3,3). A more direct proof of Theorem 1 can be found in Stufken (1991).
Since f (27 , 3, 3) = 4 and f (81 ,3 , 3) = 10, the practical implication of Theorem 1 is that a statistical experiment in which an orthogonal array of strength 3 for at least six 3-level factors is to be used requires at least 81 observations. With five such factors one need not use more than 54 observations.
The classification of the nonisomorphic orthogonal arrays OA(54,5,3,3) allows one to select one such array randomly from all available arrays with these parameters, or select one prudently based on extraneous considerations. For example, one might be able to avoid less desirable alias pattems or less desirable combinations of the levels for the five factors by a prudent selection.
We conclude this section with a presentation of the four nonisomorphic arrays OA(54, 5, 3, 3) and an argument to establish that they are indeed nonisomorphic.
A. Hedayat et al./Journal of Statistical Planning and Inference 58 (1997) 43~53 45
It is easy to see that the four arrays in Table 1 are nonisomorphic. Array I has two repeated columns (or rows in its presentation of Table 1 ), array II has only one repeated column, and arrays III and IV have none. This suffices, except that an additional argument is needed to establish that arrays III and IV are nonisomorphic. Both of these arrays contain nine pairs of columns for which the columns within a pair show the same symbol for four of the five factors. For example (0, 0, 0, 0, 1 )~ and (0, 0, 0, 0, 2) ~ form one such pair in array III, only differing for factor 5. (This also happens to be such a pair of columns for array IV.) By recording for each such pair the only factor for which the columns differ, we see that in array III this is in all nine cases for factor 5, while in array IV three different factors (1, 2 and 5) play this role, each three times. This implies that arrays III and IV are also nonisomorphic.
Some additional structure of the various OA(54, 5, 3, 3), in part due to the particular presentation we have given here, is presented in the appendices.
Table 1 Four nonisomorphic OA(54, 5,3, 3) (presented in transposed form)
1 ii lI1 IV
1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 0 1 1 2 2 0 1 2 1 2 0 1 2 2 1 0 2 1 2 0 2 1 2 1 0 2 2 1 1 0 1 2 2 1 0 2 2 1 0 2 2 1 2 0 1 1 2 2 0 1 2 1 2 0 2 1 1 1 1 0 2 2 1 2 0 2 1 2 0 2 1 2 1 0 2 2 1 0 2 1 2 2 0 1 1 1 2 0 2 1 2 1 0 2 1 2 2 0 1 2 1 1 0 2 2 1 2 0 1 2 2 1 0 1 1 1 2 2 0 1 2 1 2 0 1 2 2 1 0 2 1 1 2 0
0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 2 1 1 1 1 1 0 0 1 1 1 2 2 2 2 2 0 2 0 2 1 1 1 2 2 2 0 2 2 0 1 2 1 1 2 2 2 0 0 2 1 2 2 1 1 2 2 0 2 0 1 2 2 2 1 1 1 1 0 0 1 1 2 2 2 1 0 0 2 2 2 2 1 2 1 1 0 1 0 1 2 1 2 1 2 1 0 1 1 0 2 1 1 2 1 2 1 0 0 1 2 2 1 1 2 1 1 0 1 0 2 1 2 1 1 2 2 2 0 0 2 0 0 1 1 1 0 0 1 1 0 0 0 1 2 2 0 0 2 2 0 0 0 2 1 2 0 1 0 2 0 0 0 2 2 1 0 2 0 1 0 0 2 0 2 2 0 1 2 0 0 0 2 0 1 1 0 2 1 0 0 0 1 0 2 1 1 0 0 2 0 0 1 0 1 2 2 0 0 1 0 0 2 2 0 2 1 0 2 0 0 0 2 1 0 1 2 0 1 0 0 0 1 2 0 1 1 1 0 0 0 0 1 1 0 2 2 2 0 0 0 0 1 1 1 0 0 1 1 2 1 0 1 2 2 0 0 1 2 1 1 0 2 1 2 0 1 0 1 2 1 0 2 2 1 0 1 0 2 1 1 1 0 0 1 1 1 2 0 1 1 1 0 0 2 2 2 1 0 1 1
0 0 0 0 1 0 0 0 0 2 0 0 1 1 1 0 2 0 2 1 0 2 2 0 1 2 0 0 2 1 2 0 2 0 1 1 1 0 0 1 0 0 2 2 2 0 1 0 1 2 0 1 1 0 2 1 0 0 1 2 1 0 1 0 2 2 2 0 0 2 0 0 1 2 0 0 0 2 1 0 0 1 0 2 0 0 2 0 1 0 0 1 2 0 0 0 2 1 0 0 1 0 0 2 0 2 0 0 1 0 1 0 2 0 0 2 0 1 0 0 1 1 0 0 0 2 2 0 0 0 0 1 1 1 1 0 1 2 2 1 1 0 1 2 1 1 0 2 1 1 1 2 0 1 1 2 1 0 1 1
46 A. Hedayat et al./Journal of Statistical Plannin9 and Inference 58 (1997) 43~53
Table 1. Continued
I II Ill IV
2 1 2 1 0 2 0 0 1 2 1 2 1 0 1 1 2 1 0 1 2 2 1 1 0 2 0 0 2 1 2 1 1 0 1 2 1 1 0 1 0 0 0 1 1 2 0 2 0 2 0 2 2 1 2 0 2 2 1 2 0 0 1 0 1 2 0 1 0 1 0 2 1 2 2 0 2 1 2 2 0 1 0 0 1 1 0 2 0 1 2 0 2 1 2 2 0 2 2 2 1 0 0 0 1 1 0 1 0 2 2 0 1 2 2 2 0 1 1 2 0 0 1 1 0 2 0 2 2 0 2 1 0 2 2 2 1 0 2 2 0 1 0 1 0 2 0 1 1 0 1 2 0 2 2 1 2 0 2 2 1 0 0 1 0 1 0 2 1 0 2 1 2 0 2 2 1 2 0 2 0 1 1 0 0 1 0 1 2 0 1 2 2 0 2 1 2 2 0 2 1 0 1 0 0 1 1 0 0 1 1 1 2 2 1 1 1 2 2 1 1 1 0 0 0 1 2 0 0 2 2 2 2 2 1 2 2 2 1 1 0 0 0 2 2 2 1 0 0 2 2 2 1 1 1 2 2 1 2 1 0 0 2 0 2 2 2 0 0 1 2 2 1 1 2 2 2 1 1 2 0 2 0 0 2 2 2 0 2 0 1 1 1 1 2 1 1 1 2 2 2 0 0 0 2 2 1 0 1 0 1 1 2 2 2 1 1 2 1 2 0 0 2 2 0 1 2 0 1 0 1 1 1 1 0 1 1 1 1 0 0 2 0 2 0 1 1 0 2 0 1 2 1 2 0 1 2 1 1 0 2 0 0 2 0 1 1 1 0 0 1 2 2 1 0 1 2 2 2 0 0 2 2 0 0 1 2 2 0 0 2 2 2 2 0 2 1 2 1 0 2 0 2 0 0 2 1 2 0 0 2 1 2 1 0 2 2 2 2 0 2 2 0 0 0 2 2 1 0 0 2 1 1 2 0 2 1 1 2 0
Appendix A. The number of nonisomorphic orthogonal arrays OA(54, 5, 3, 3)
In this appendix we will prove Theorem 2. For clarity of presentation some claims
needed in the proof are shown in separate appendices, labeled A.1, A.2, and so on.
When we use the phrase "without loss of generality" when discussing properties or
structure of an orthogonal array, we will mean that any orthogonal array with said
properties or structure must be isomorphic to the one that we consider. As an example,
without loss of generality we may assume that every orthogonal array has a column
in which every symbol is 0, to be called an all-zero column.
Instead of symbols for a factor, we will use the terminology levels for a factor.
An ordered k-tuple based on the levels, or part of such a k-tuple when consider-
ing subarrays, will also be called a level combination for the factors, or just a level
combination. A column in an OA(Ls.t,k,s,t) may thus also be viewed as a level
combination.
By the number of coincidences of two level combinations we will mean the number
of entries that the two level combinations have in common. If c denotes a particular
column in an OA()],s t, k,s, t), we will use rti(¢ ) to denote the number of other columns
in the array that have exactly i coincidences with c. It is easily shown that
~ ( i ~ n i ( c ) = ( k . ~ ( 2 s t - J - 1 ) , j = 0 , 1 . . . . . t, (A.1) ~ J ] \ J . ] i=j
A. Hedayat et aL /Journal of Statistical Planning and Inference 58 (1997) 43-63 47
for any c (Bose and Bush, 1952). Further, if c is a level combination for k factors,
not necessarily appearing as a column in a given OA()tst,k,s,t), and if mi(c) denotes
the number o f columns in the array that have i coincidences with c, then we have that
k / ; \ / b \ ~ ( ~ ) m i ( c ) = (~)2s t-j, j = 0 , 1 . . . . . t. ,A.2,
\ J / \ J /
I f c does actually appear as a column in the given array, then mk(c) = nk(c) + 1 and
mi(c) = ni(c), 0 ~< i ~< k - 1. For brevity we will often write ni and mi instead of
hi(C) and me(C).
A column in an OA(2st, k,s,t) is said to be a repeated column if the level com-
bination in that column appears also in another column in the array. Since we will
only be considering arrays of index two, we will not have to deal with a case where
a level combination appears more than twice. When we use the phrase "two repeated columns", we will mean two different level combinations that both appear more than
once as a column. When presenting a column of an OA(2s~, k, s, t) in which not all entries are specified,
we will use two wildcard symbols: * will be used to denote that the entry belongs to
the set {1 . . . . . s - 1}, while • will be used for an entry from {0 . . . . . s - 1}.
To prove Theorem 2 each of the following claims will be established in the remainder
o f Appendix A:
I. There is a unique OA(54 ,5 ,3 ,3) with at least two repeated columns.
2. There is a unique OA(54 ,5 ,3 ,3 ) with exactly one repeated column.
3. There are two, and only two, nonisomorphic arrays OA(54, 5, 3, 3) without repeated
columns. To show claim 1, assume that we have an OA(54,5 ,3 ,3) with at least two repeated
columns. Without loss of generality the all-one column is one o f the repeated columns.
The choice o f the all-one column instead of the all-zero column is merely because it
will lead to an array that more clearly exhibits some additional structure. Taking c as
the all-one column in the equations in (A. 1 ) shows that n5 = 1, n2 = 40 and no = 12,
other ni 's being 0. A second repeated column must thus contain two or zero factors
at level 1. I f it were to contain two factors at level 1, without loss o f generality the
following columns ought to be present:
1 1 1 1 1 1 1 1
1 1 0 0
1 1 0 0
1 1 0 0
There should also be columns of the form ( . , •, 1,0, 0)'. Such a column has coinci- dences with both o f the repeated columns, and, again using the equations in (A. 1 ), must
therefore have two coincidences with both o f the repeated columns. Clearly this is not
possible. Therefore, we may conclude that two repeated columns in an OA(54, 5, 3, 3)
cannot have any coincidences. Thus, without loss of generality, let the all-one and
48 A. Hedayat et al./ Journal of Statistical Planning and Inference 58 (1997) 43~53
Table 2 Distributions for other columns in an OA(54, 5, 3, 3) with repeated all-one and all-two columns
level
0 1 2
1 2 2 3 2 0 3 0 2 5 0 0
al l - two columns be repeated columns. Every other column must now have zero or
two factors at level 1 and zero or two factors at level 2. This leaves four possible
distributions for each o f the remaining 50 columns in terms o f the number of factors
at level 0, 1 and 2, which are exhibited in Table 2.
It is now easily seen that the remaining 50 columns must consist o f all possible
combinations with two factors at level 1 and two at level 2 (30 columns), all possible
combinations with two factors at level 1 and zero at level 2 (10 columns) and all
possible combinations with zero factors at level 1 and two at level 2 (also 10 columns).
This results in the O A ( 5 4 , 5 , 3 , 3 ) labeled as I in Table 1.
Thus, there are at most two repeated columns in an OA(54 ,5 ,3 ,3 ) , and an array
with two repeated columns is unique up to isomorphism.
Note that the constructed array, say A, exhibits some additional structure. With the
levels considered as residues module 3, we see that the columns in A are, up to a
permutation, exactly the same as those i n - A . Moreover, the columns o f A may be
partit ioned as [X - X ] , where X is a 5 × 27 array. (But since f ( 2 7 , 3, 3) = 4, X is
not an orthogonal array o f strength 3. In fact, X is not even an orthogonal array of
strength 2).
To prove claim 2, consider an OA(54 ,5 ,3 ,3 ) with exactly one repeated column,
without loss o f generality the a l l -zero column. As in the proof of claim 1, we conclude
that n5 = 1, n2 = 40 and no = 12. First observe that any two columns with two factors
at level 0 can now have at most three coincidences. I f not, there would be a pair of
columns with four coincidences and we would, without loss o f generality, have the
following columns in our array:
0 0 1 1 * * * * * *
0 0 1 1 * * * • * •
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 * * * *
0 0 * * * * 0 0 0 0
Using that
g< * :¢ *
0 0 0 0
0 0 0 0
the index o f the array is 2, we see that columns 5 and 6 must become
( 2 , 2 , 0 , 0 , . ) ' . Considering factors 1, 2 and 3, we see now that columns 7 through 10
must twice show the pattern (1, 2, 0 , . , 0) ' and twice (2, 1,0, *, 0) ' . Similarly, considering
A. Hedayat et al./Journal of Statistical Planning and Inference 58 (1997) 4343 49
Table 3 The unique OA(12, 5, 2, 2) that can be embedded in an OA(54, 5, 3, 3 ) with the all-zero column as the only repeated column
1 2 1 2 2 2 1 2 2 1 1 1 I 2 1 1 2 2 2 I 1 2 2 1 l 2 2 1 1 2 2 2 1 1 1 2 l 2 2 2 1 1 2 1 2 2 1 I 1 2 2 2 2 1 1 1 1 1 2 2
factors 1, 2 and 4, the last four columns must twice admit the pattern (1 ,2 , , ,0 ,0 ) t
and twice (2, 1, , , 0, 0)'. But, among the last eight columns we see now, by considering factors 1, 2 and 5, a contradiction with the index of this array being 2.
Hence, the 12 columns in which two of the last three factors are at level 0 must use each of the level combinations (1,1), (1,2), (2,1) and (2,2) three times for factors 1 and 2. Since similar conclusions hold when the role of factors 1 and 2 in this
argument is played by any other pair of factors, it follows that the 12 columns with no factor at level 0 must form an OA(12,5,2,2). While an OA(12,5,2,2) is not unique, with the restriction that it does not contain a repeated column, which is a property
that the OA(12,5,2,2) that we are searching for must possess, it can be seen (see Appendix A. 1 ) that there is a unique choice for this OA(12, 5, 2, 2), which is exhibited in Table 3. It is also clear that the ones and twos in the forty columns with two
factors at level 0, which replicate each of the (52) = 10 possible combinations with two factors at level 0 exactly four times, are uniquely determined by this OA(12,5,2,2).
For example, the four columns with the pattern ( , , , , *, 0, 0) ' result in
1 1 2 2
1 2 1 2
1 2 2 1
0 0 0 0 0 0 0 0
where the choice for the first three factors consists of those level combinations that occur only once in the 3 × 12 subarray formed by the first three factors of the OA(12,5,2,2) in Table 3. Completing other columns with two factors at level 0 in a similar way results in the unique OA(54, 5, 3, 3) with one repeated column.
To prove claim 3 of this appendix, we first note that an OA(54,5,3,3) with-
out repeated columns must contain a pair of columns with four coincidences. (See Appendix A.2). Without loss of generality, the array must then contain the columns (0,0,0,0, 1) t and (0,0,0,0,2) ' . As in the proof of claim 1, the choice of these two particular columns is motivated by the fact that they will lead to an orthogonal array that clearly exhibits some additional structure. It is then easily seen that the structure displayed in Table 4 must exist. We have partitioned the other 52 columns in the array into eight sets, named A-H. Entries not given in Table 4 are either l or 2.
Fix one of the first four factors and let ~ denote the number of columns in part A where this factor is at level 1. Without loss of generality we have that ~ = 0 or ~ = 1.
50 A. Hedayat et al./ Journal of Statistical Plann•g and Inference 58 (1997) 4343
Table 4 A putative OA(54, 5,3,3) without repeated columns
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 O 0 0 O 0 O 0 0 0 0 0 O 0 0 0 0 0 0 0 O 0 O 0 0 O 0 0 O 0 0 O 0 O 0 0 0 0 l 2 ~ l 1 1 1 ~ 2 2 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0
0 0 O 0 0 0 O 0
0 0 0 0 0 0 0 0
1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 1 1 1 2 2 2 0 0 0 0 0 0
; [ F G fi
Moreover, as shown in Appendix A.3, the number of columns with this factor at level
1 in each o f parts B - H is now also determined: 3 - ~ columns in B, 3 columns in
C, 6 - 2 ~ columns in D, 2~ columns in E, ~ columns in F, 3 -c t columns in G, and
3 columns in H. Finally, again shown in Appendix A.3, it can be argued that ~ = 0
is not possible for an OA(54 ,5 ,3 ,3 ) without repeated columns, so that for each o f
the first four factors we have the following occurrences o f level 1 : 1 column in A,
2 columns in B, 3 columns in C, 4 columns in D, 2 columns in E, 1 column in F,
2 columns in G, and 3 columns in H.
Combined over the first four factors, the columns of A contain thus four times
level 1. It can be shown (see Appendix A.4) that these must occur in only two columns
of A. Therefore, without loss of generality, the columns in A are as follows:
0 0 0 2 2 1
0 2 2 0 0 1
1 0 2 0 2 0
1 2 0 2 0 0
1 1 1 1 1 1
These columns determine many other columns in Table 4, completely or in part. It
is shown in Appendix A.5 that Table 4 can now be completed to the structure shown
in Table 5. Entries not shown in Table 5 are either 1 or 2; in each rectangular box in
Table 5, level 1 should be used equally often as level 2.
For each of the two boxes in part C o f the array in Table 5 we have to choose
between entering (1 ,2) or (2,1). This results in four possibilities, discussed in the
following.
Case (i): Both boxes in part C o f the array in Table 5 are filled by (2, 1 ). We will
argue that for this choice the array cannot be completed to an OA(54, 5, 3 ,3) without
repeated columns. To see this, consider the completed part C and part H o f the array
in Table 5. Considering level combinations for any three factors that include factor 5
it follows that the level combination (1, 1)1 must occur twice in part H for any pair
A. Hedayat et al./ Journal of Statistical Planning and Inference 58 (1997) 43~53 51
of factors from the first four factors. It is easily seen, also keeping in mind that both
rectangular boxes require levels 1 and 2 to occur equally often, that H can only be
completed to meet this requirement if we allow repeated columns.
Case (ii): Both boxes in part C of the array in Table 5 are filled by (1,2). Consider the following eight columns that now appear in parts A, B and C of the array in Table 5:
0 0 2 2 0 0 1 l 0 0 2 2 0 0 l 1
2 2 0 0 1 1 0 0 2 2 0 0 1 1 0 0 0 2 0 2 0 1 0 1
Since the array is to be of index 2, any other column in the array can have at
most two coincidences among the first four factors with any of these eight columns. This immediately determines how the boxes in parts D and E are to be completed, and further simple considerations based on the index of the array being 2 result in a
unique way to complete the boxes in F, G and H. The resulting OA(54, 5, 3, 3) without repeated columns is that labeled as III in Table I.
This array exhibits a great deal of structure. We like to point out the following two features. Under addition modulo 3, the first two columns are each others additive
inverse. Similarly, columns in C and H come in pairs of additive inverses. Furthermore, A = - B , D = - E and F = - G . Obviously then, this OA(54, 5, 3, 3) can be represented as [X - X ] for an appropriate 5 × 27 array X.
Another feature of the array is that it may be represented as [Y Y + 1 Y + 2], where Y is an appropriately chosen 5 x 18 array and Y + i stands for the array obtained by adding i (modulo 3) to each element of Y. For example, Y may be taken to consist
of the columns of C and H.
Case (iii): In part C of the array in Table 5 we choose (1,2) in the box for factor 2 and (2, l ) in that for factor 4. Similarly as in Case (ii), this determines all other
Table 5
A putative OA(54, 5, 3, 3) without repeated columns, nearing completion
0 0 0 0 0 2 2 1 0 0 0 1 1 2 0 0
0 0 0 2 2 0 0 1 0 1 1 0 0 2 0 0 0 0 1 0 2 0 2 0 2 0 1 0 1 0 1 2
0 0 1 2 0 2 0 0 2 1 0 1 0 0 i J 1 2 1 1 1 1 1 1 2 2 2 2 2 2 0 0 , ,
2 0 0 I 1 1 2 1 2 0 0 2 2 2 1 2 1 1 I 0 0 i r i r 2 2 0 0 I J I J
1 2 1 2 0 0 1 1 2 1 2 1 0 0 2 2 t I i ~ 1 1 0 0 I J f J 2 2 0 0
1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2
0 0 0 0 1 2 1 2 1 2 1 2 1 2 0 0 0 0
0 0 2 1 0 0 2 1 0 0
2 1 0 0 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 2 2 2 1 1 1 1 1 2 2 2 ~ I 2 ~ i 1 [ I
2 2 1 1 1 2 1 1 2 2 2 1 2 ~ i 1 r - ~ [ I
I 1 1 2 2 2 0 0 0 0 0 0
F G
52 A. Hedayat et al./Journal of Statistical Plannino and Inference 58 (1997) 4343
entries for factor 2. (See also array IV in Table 1). It is important to observe that after completing factor 2 the structure obtained is invariant under permuting factors
1 and 2.
After completing factor 2, the following nine columns appear in parts C and H of the array in Table 5:
0 0 2 1 1 1 2 2 2
0 2 0 1 2 2 1 2 1
2 0 0 1 1 2 2 2 1
1 1 1
0 0 0 0 0 0 0 0 0
By considering triplets of factors that involve factors 4 and 5 it is clear that the
three additionally needed occurrences o f level 1 for factor 4 must be such that they coincide twice with level 1 for each of the first three factors. This means that level 1
must occur for factor 4 in the first column of part H and either in columns 2 and 4
o f H or in columns 3 and 6 of H. As a result of the earlier noted invariance under
permuting factors 1 and 2, we can thus without loss o f generality conclude that factor 4 is at level 1 in columns 1, 2 and 4 of part H.
After this all other entries for factor 4 are uniquely determined by the index of the
array being 2, and the resulting OA(54, 5,3, 3) is that labeled as IV in Table 1.
This array is not o f the form [X - X ] nor is it of the form [Y Y + I Y + 2 ] .
Appendix A. 1
In this appendix we will show that an OA(12 ,5 ,2 ,2) that does not have a repeated
column is unique. There are four solutions with n5 = 0 to the equations in (A.1), namely
(i) no = 0, nl = 1, n2 = 8, n3 = 0, n 4 = 2,
(ii) no = 0, n l = 2, n2 z 5 , n 3 = 3, n 4 = 1,
(iii) no = 0, nl = 3 , n2 = 2 , n3 = 6 , n4 -- 0 ,
(iv) no = 1, nl = 0, n2 = 5, n3 = 5 , n 4 = 0. We will first show that (i) and (iii) are not possible in an OA(12 ,5 ,2 ,2) without
repeated columns. We will then show that not every column in such an array can lead
to solution (ii), so that at least one column must lead to solution (iv). We will then argue that a column that leads to solution (iv) in such an array uniquely determines all other columns.
As in the proof of claim 2 o f Appendix A, we will use the levels 1 and 2 for the factors in the OA(12,5 ,2 ,2) . Suppose that a column in this array, say the all-one
column, leads to solution (i). Since n4 = 2, we must, without loss of generality, have
A. Hedayat et al./ Journal of Statistical Plannin9 and Inference 58 (1997) 4343 53
the following structure in the array:
1 1 1 1 l 1 2 2 2 2 2
1 1 1 2 2 2 1 1 1 2 2
1 1 1 2 2 2 2 2 2 1 1
1 1 2 • • • * • * • *
1 2 1 • , , * • * • •
By considering factors 1 and 4 it follows
in columns 4 - 6 . Similarly, by considering
2 2 1
:¢
that factor 4 is exactly once at level 1
factors 2 and 4 and factors 3 and 4 it
follows that factor 4 is exactly once at level 1 in columns 7-9 and in columns 10-12, respectively. Combining these observations, factor 4 will be at level 1 five times, and thus short o f the required six times. Hence solution (i) is not possible.
Next, suppose that the all-one column in an OA(12, 5, 2, 2) without repeated columns satisfies solution (iii). Since there are no repeated columns and since nl = 3, the only factor at level 1 in the three columns with one factor at level 1 must be different for all three columns. By counting pairs of l ' s that need to be formed it is easily seen
that these three factors must be four times at level 1 in the six columns with three factors at level 1. But that means that all three factors must be at level 2 in the two
columns with exactly two factors at level 1, implying that these two columns give rise to a repeated column. Hence, solution (iii) is not possible in an OA(12, 5, 2, 2) without
repeated columns. This leaves solutions (ii) and (iv) as the only possibilities. The columns of an
OA(12 ,5 ,2 ,2 ) can thus be grouped in pairs such that two columns in a pair have zero or four coincidences. Clearly, there are six such pairs. We claim that in at least one such pair the two columns that form the pair must have zero coincidences. Suppose this is not the case. Without loss of generality we must then have the following structure in the array:
1 1
1 1
1 1
1 1
1 2
1 1
2 2
2 2 2 2
1 2
To see this, i f for each of the six pairs the columns that form a pair have four co- incidences, there must be a factor, say factor 5, that is the only factor at which the columns in a pair differ for at least two of the pairs. Hence, without loss of generality, we have one pair as (1, 1, 1, 1, 1) ~ and (1, 1, I, 1,2) t and a second pair as ( . , . , * , . , 1)'
and ( . , , , . , . , 2 ) ~, where the first four factors are identical in the two columns in the last pair. In view of the two columns in the first pair, and using that the index is three, at most one of the first four factors can be at level 1 in the two columns in the second pair. On the other hand, i f all four factors are at level 2 in these two columns, both ( 1, 1, 1, 1, 1 )P and (2, 2, 2, 2, 2) t will be among the columns, contradicting that (1, 1, 1, 1, 1)' corresponds to solution (ii). Therefore, exactly one of the first four
54 A. Hedayat et al./ Journal of Statistical Planning and Inference 58 (1997) 4343
factors, say factor 1, must be at level 1 in the second pair of columns, leading to the structure as previously claimed.
In addition to these four columns, there must be two other columns in which factor
1 is at level 1. Each of the other four factors must be once at level 1 and once at
level 2 in those two columns, leading to the following structure:
1 1 1 1 1
1 1 2 2 *-- three times a 1 and 1 1 2 2 three times a 2
1 1 2 2
1 2 1 2 1 2
It is now easily seen that, avoiding repeated columns, one o f the last two columns will
have four coincidences with one o f the first two columns. But this means that one of
the first two columns corresponds to a solution with n4/> 2, which is not possible as
already shown.
Hence, in our six pairs of columns with zero or four coincidences for the columns
within a pair, there is at least one pair in which the two columns have zero coinci-
dences. Thus, the columns (1, 1, 1, 1, 1) r and ( 2 , 2 , 2 , 2 , 2 f are without loss o f generality
part of our OA(12, 5,2, 2). Since both of these columns must now correspond to solu-
tion (iv), there are five columns with exactly two factors at level 1 and three at level 2,
and five columns with exactly two factors at level 2 and three at level 1. By counting
l ' s that need to be formed, we see that in the five columns with two factors 1 each factor must be exactly twice at level 1. It is easily seen that, if no
are to be repeated, there is a unique way in which this can be accomplished:
2 2 2 1
1 2 2 2
1 1 2 2
2 1 1 2
2 2 1 1 remaining columns are now uniquely determined as well by considering pairs
pairs of at level
columns
1
1
2
2
2 The five
o f 2s:
2
1
2
1 1
2 1 1 1
1 2 2 1
1 1 1 2
2 2 1 1 1 1 2 2
Combining the previous 12 columns results in the unique OA(12,5 ,2 ,2) without
repeated columns as exhibited in Table 3.
Appendix A.2
The objective of this appendix is to show that an OA(54 ,5 ,3 ,3) without repeated columns must possess a pair of columns with four coincidences. I f not, we can find an
A. Hedayat et al./Journal of Statistical Planning and Inference 58 (1997) 43-63 55
OA(54,5 ,3 ,3 ) in which every pair o f columns has at most 3 coincidences. From the
equalities in (A.1), it follows then that n3 = 10, n2 = 20, n! = 15 and no = 8 for any
column in such an orthogonal array. Letting one column in the array be an all-zero
column, the remaining 53 columns can now be partitioned as follows: B3, a 5 × 10 array whose columns contain each pattern o f exactly three O's once;
B2, a 5 × 20 array whose columns contain each pattern of exactly two O's twice;
B~, a 5 × 15 array whose columns contain each pattern o f exactly one 0 thrice;
B0, a 5 × 8 array in which level 0 does not occur at all.
For every factor there are six columns in B3 where the factor is at level 0. We
claim that the factor must be twice at level 1 and twice at level 2 in the remaining
four columns. I f this were to be false, without loss of generality the following structure
should exist in B3:
0 0 0 0 0 0 * 1 1 1
0 0 0 * * * 0 0 0 *
0 * * 0 0 * 0 0 * 0
• 0 * 0 * 0 0 * 0 0
• * 0 * 0 0 * 0 0 0
Also consider the following eight columns that form part o f B2:
0 0 * * * * * *
• * 0 0 * * * *
• * * * 0 0 * *
• * * * * * 0 0
0 0 0 0 0 0 0 0
In the last six of
index of the entire
these eight columns, factor 1 must now be at level 2 since the
array is 2. Without loss o f generality, the level for factor 2 in the
last column of 83 is equal to 2. (see also Table 6). The two rectangular boxes for
factor 2 in Table 6 must each contain both levels 1 and 2 once since no two columns
can have more than three coincidences. Using that the index of the array is 2, we
arrive at the two occurrences of level 1 for factor 2 in the columns of B3 as shown
in Table 6. Again using that the index of the array is 2, these two l ' s imply the level 2 for
factor 2 in two of the columns in B2, as shown in Table 6. For factor 3 we must have
levels 1 and 2 once in each of the three rectangular boxes for factor 3 in Table 6, in
order to avoid a pair o f columns with more than three coincidences. Without loss of
generality, let the level for factor 3 in the next to last column in B3 be equal to 2.
Considering this column in B3 and the four columns in B2 corresponding to the last two rectangular boxes for factor 3 we find, using that the index of the array is 2, two
columns in B3 in which factor 3 must be at level 1. Thus, we have in B3 the columns (0, 0, 1, , , 0) t and (0, 1, 1,0, 0) ' , while B2 contains the column (0, 2, 1, *, 0) t. Considering
factors 1, 3 and 5, these columns violate that the index of the array is 2. Thus, we have shown that in B3 every factor must be twice at level 1 and twice at level 2.
56 A. Hedayat et al./ Journal of Statistical Planniny and Inference 58 (1997) 43~53
Table 6 Portion of a putative OA(54, 5, 3, 3 )
0 0 0 0 0 0 * 1 1 1 0 0 2 2 2 2 2 2 0 0 0 * 1 1 0 0 0 2 2 2 0 0 i* *t t,~ *~
0 * 1 0 0 1 0 0 2 0 t* *t i,~ *t 0 0 I* *1 • 0 * 0 * 0 0 * 0 0 * * * * * * 0 0
• * 0 * 0 0 * 0 0 0 0 0 0 0 0 0 0 0 Y
columns in ~3 eight of the columns in B 2
Table 7 Portion of a putative OA(54, 5, 3, 3) (Columns within each Bj are numbered for ease of reference)
]1 2 3 4 5 6 7 8 9 101 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
0 0 0 0 0 0 0 2 2 1 1 0 0 0 0 0 0 0 0 1 1 1 2 1 2 1 2 1 2 2 2
0 0 0 0 1 2 1 0 0 0 2 0 0 1 2 2 2 1 2 0 0 0 0 0 0 1 2 1 1 1 2 0 0 0 0 1 0 0 2 0 0 0 1 2 0 0 1 1 0 0 0 0 2 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 y
B3 B2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 151 1 2 3 4 5 6 7 8
0 0 0 1 2 2 1 2 2 1 1 2 1 1 2 1 1 1 1 2 2 2 2 1 1 2 0 0 0 2 1 2 1 2 1 1 2 2 1 1 2 2 1 1 2 2
0 0 0 2 2 1
0 0 0 0 0 0
Y
Bi B0
Without loss of generality, the pattern o f the O's and the occurrences of levels 1
and 2 for factor 1 must then be as displayed in Table 7. The pattern of the O's was
addressed earlier in this appendix; the two appearances o f levels 1 and 2 for factor 1
in the columns of B3 is a consequence o f the argument in the previous paragraphs,
and the positions for level 1 can be chosen arbitrarily without loss of generality; the
occurrences o f levels 1 and 2 for factor 1 in the other columns of the OA(54,5 ,3 ,3) follow now since the array is of index 2. Focusing on factor 2, it must also have level 1
exactly twice in the columns of B3. Without loss o f generality, we take the level o f
factor 2 in the last column of B3 (see Table 7) to be 2. Since no two columns can have more than three coincidences this implies then that columns 15 and 17 of B2 must have
factor 2 at level 1. In columns 19 and 20 of B2 factor 2 must be once at level 1 and
once at level 2, again to avoid a pair o f columns with more than three coincidences. Since the index is 2, this means now that the level for factor 2 in column 6 of B3 must be 1. In columns 4 and 5 o f B3, factor 2 must once be at level 1 and once at
level 2. Without loss o f generality, factor 2 is at level 1 in column 4 of B3 and at
A. Hedayat et al./Journal of Statistical Plannin 9 and Inference 58 (1997) 4343 57
level 2 in column 5. Further considering the index of the array, the other occurrences
of levels 1 and 2 for factor 2 as shown in Table 7 follow now easily.
In determining the l ' s and 2's for factors 1 and 2 in B1 and B0 we also use that
the array is of index 6 when considered as an array of strength 2. Next we turn to factor 3. Without loss of generality, let factor 3 be at level 2 in
column 9 of B3. This also determines the levels of factor 3 in columns 11 and 13 of B2
(to avoid four coincidences with column 9 in B3). Since factor 3 will be once at level 1
and once at level 2 in columns 19 and 20 of B2 (these have already three coincidences),
it follows that factor 3 must be at level 1 in column 6 of B3. Hence, the level in column
19 of B2 must be 2 and that in column 20 of B2 must be 1. By considering the number
of coincidences between columns 5 and 6 in B2, factor 3 must be at level 1 in one of
these and at level 2 in the other. To avoid four coincidences between columns 19 in
B2 and 12 in B1, factor 3 must be at level 1 in the latter column. Considering factors
2, 3 and 4 in columns 6 in B3, 10 in B1 and 12 in B1 it follows that factor 3 must
be at level 2 in column 10 of B1. Considering factors 2, 3 and 4 in columns 5 in B2,
20 in B2 and 11 in B1 we see that factor 3 must be at level 2 in the latter column.
But considering factors 1, 3 and 4 in columns 9 in B3, 10 in B1 and 11 in B1 we
find a contradiction with the index of the array being 2. Hence, we have shown that
an OA(54, 5, 3, 3) without repeated columns must contain a pair of columns with four
coincidences.
Appendix A.3
In this appendix it will be shown that we can assume without loss of generality that
for a putative OA(54,5,3,3) as in Table 5 we must for each of the first four factors
have the factor at level 1 in the following number of columns:
I column in A, 2 columns in B, 3 columns in C,
4 columns in D, 2 columns in E, 1 column in F,
2 columns in G, and 3 columns in H. (Letters A-H refer to the partition of the columns in Table 5).
Let one of the first four factors, without loss of generality factor 4, be c~ times at level 1 in part A. The a columns in A of the form (e,e, e, 1, 1) / contain 2a times
level 0. Hence, a column of the form (e, e, e, 1, 1)' must appear 6 - 2a times in part D
and a column (*,*,*, 1, 1) / must appear a times in part F.
Columns of the form (e, e,e, 1,0)' in part C must occur three times to capture all
level combinations (0, 1, 0) I for one of the first three factors, factor 4 and factor 5.
This also implies that factor 4 should have level 1 in three columns of part H. Finally,
columns of the form (e, e, e, 1, e) / in which two of the first three factors are at level 0 occur 3 + ~ times in A and C. Hence, such columns should occur 3 - c~ times
in B. This also implies the number of columns in parts E and G that have factor 4
at level 1. Thus, each of the first four factors should, for some e that may depend on the factor,
have level 1 appear in ~ columns in A, 3 - ~ columns in B, 3 columns in C,
58 A. Hedayat et al./ Journal of Statistical Planning and Inference 58 (1997) 4343
6 - 2~ columns in D, 2~ columns in E, :¢ columns in F, 3 - c~ columns in G, and 3 columns in H.
Clearly ~ E {0, 1,2, 3}, and by permuting levels for a factor we may assume without
loss of generality that ~ E {0, 1 }. We will now argue that if c~ = 0 for a factor among the first four, the structure in
Table 5 cannot be completed to an OA(54, 5, 3,3) without repeated columns.
Assume without loss of generality that ~ = 0 for factor 4. I f c~ = 0 for another factor as well, say for factor 3, then by the arguments in the previous paragraphs we will see three columns of the form ( . , . , 2 , 2 , 1 ) r in part F. Since the index of the array is 2,
this cannot happen. Thus, factors 1, 2 and 3 must now satisfy c~ -- 1.
For the one column in part A with factor 3 at level 1 there are now two different possibilities. Either factor 4 is at level 0 or it is at level 2 in that column. I f the first possibility holds, then we will see the column (0, 0, 2, 2, 1)t in part A and two columns
of the form ( . , *,2,2, 1) ~ in part F. Clearly this is not compatible with the index of the array being 2.
Hence, the only column in part A with factor 3 at level 1 must be (0,0, 1,2, 1) ~. This implies that there must be two columns of the form ( , , . , 1,0, 1) ~ in part D. But the same argument used for factor 3 applies also to factors 1 and 2. Hence, the column (1, 1, l, 0, 1)~ will be repeated in part D.
This establishes the claim of this appendix.
Appendix A. 4
For the putative OA(54 ,5 ,3 ,3 ) in Table 5 with each of the first four factors once at level 1 in part A, we will now show that the four l ' s in part A must occur in only two of the six columns.
We will first argue that the four l ' s in part A cannot be spread out over four columns. That will establish that there must be a column with two l ' s , and we will show that this implies that there must be a second column with two l ' s .
To begin, however, we will show some properties of the putative OA(54 ,5 ,3 ,3 ) that will also be useful in Appendix A.5. We will show the validity of the following
claims: (a) In part C, any pair of columns has at most three coincidences. (b) Under addition modulo 3, the columns of part B are obtained as the additive
inverse of those of part A. To see the first claim, it suffices to show that the two columns of the form
(*, , , 0 , 0 , 0 ) ~ have only three coincidences. I f they had four coincidences, we would have two columns of the form ( , , 1 ,0 ,0,0) r, say. But we must then also have two columns of each of the forms (e, 2, 0, *, 0) r and (e, 2 , . , 0, 0) ~. Since such columns can only occur in part C, the latter are actually of the form (0, 2, 0, *, 0) r and (0, 2 , . , 0, 0) ~. Hence, we would have four columns of the form (0,2, e ,e , 0) ~, which is obviously impossible.
A. Hedayat et al./ Journal of Statistical Planning and Inference 58 (1997) 4343 59
For claim (b) consider the columns
0 0 0 0 0 0 0 0 1 2 0 0
the first from part A, the second from part B, and the last two from part C. Clearly both factors 1 and 2 must occur exactly twice at level 1 in these columns. But, from
claim (a) we know that the level combinations for factors 1 and 2 have no coincidences in the last two columns. The same must thus hold for the first two columns.
Since the role of factors 1 and 2 in the above argument can be played by any pair
of factors from the first four, claim (b) follows also. Retuming to the essence of this appendix, assume first that all four l ' s in part A
appear in different columns. Then there are two columns in A with two 2's. Without loss of generality, one of the following possibilities must then occur:
(i) Part A contains the columns (2,2,0,0, 1) ~ and (2,0,2,0, 1)t;
(ii) Part A contains the columns (2,2,0,0, 1)' and (0,0,2,2, 1) t.
I f (i) occurs, A must without loss of generality consist of the following six columns:
2 2 1 0 0 0 2 0 0 2 1 0
0 2 0 1 0 2 0 0 2 0 2 l 1 1 1 1 t l
Since (1, o, o, 0, 1 )t and (e, 1, e, 0, 1 )~ do not appear in A, we must have two columns
of the form (1, 1, *, 0, 1) ~ in part D. However, by claim (b) we will also have the column (1, 1,0,0,2) ~ in part B. This means that we have three columns of the form
( 1, 1, e, 0, .)P, which is impossible.
I f (ii) occurs, A must without loss of generality consist of the following columns:
2 0 2 l 0 0 2 0 0 0 2 1 0 2 l 0 0 2 0 2 0 2 1 0 l 1 1 1 1 1
In part D we must have one column of the form (0, *, 1, 1, 1)t; in part B we must have the column (0,0, 1, 1,2) ~. Hence, also using claim (a), in part C we must have the columns (0,0, 1,2,0) ~ and (0,0,2, 1,0) t. Similarly, C will contain (1,2,0,0,0) ~ and (2, 1,0,0,0) ' .
In part D we must also have one column of the form (1, . ,2 ,0 , 1)'; in part B we must have the column (1,0,2,0,2) t. Hence, in part C we will have the columns (1,0, 1,0,0) t
and (2,0,2,0,0) ~. A similar conclusion holds if factors 1 and 2 or factors 3 and 4 are permuted.
60 A. Hedayat et al./ Journal of Statistical Planning and Inference 58 (1997) 4343
The ramifications for part H of the above arguments are as follows. For factors 1 and 2 and for factors 3 and 4 the level combinations (1, 1) ~ and (2,2) ' will occur
twice in H, while (1,2) ' and (2, 1) ' will come once. For any other pair o f factors from
the first four the level combinations (1, 1) I and (2,2) ~ will appear once, while (1,2) ~
and (2, l)t will each appear twice.
It is easily verified that H can only have these properties if we allow it to contain a
repeated column. Since we do not we can conclude that the four l ' s in part A cannot
be spread out over four columns.
Thus there should be at least one column in part A with two l 's . Assuming that
the statement to be shown in Appendix A.4 is false, suppose that there is exactly one
such column. Without loss of generality, the following structure must then exist in A:
1 2 2 0
1 0 0 2
0 1 0 2
0 0 0
1 1 1 1
0 0
2 0
0 2
1 1
Using that the index of the array is 2, it follows that D must contain two columns
of the form (0, 1, 1, . , 1)~. Also, using the relation between parts A and B, we see that
B should contain the column (0, 1, 1,0,2) ~. Three columns of the form (0, 1, 1 , o , . ) I is
of course not possible.
This establishes our claim that the four l ' s in part A must appear in just two of the
columns.
Appendix A.5
This appendix explains the steps that result in Table 5 starting from Table 4,
Based on Appendix A.4, we can take the columns of part A without loss o f gener-
ality as
0 0 0 2 2 1
0 2 2 0 0 1
1 0 2 0 2 0
1 2 0 2 0 0
1 1 1 1 1 1
As shown in Appendix
0 0 0 1 1 2
0 1 1 0 0 2
2 0 1 0 1 0
2 1 0 1 0 0
2 2 2 2 2 2
A.4, this means that the columns of part B are
A. Hedayat et al./Journal o f Statistical Plannin 0 and Inference 58 (1997) 43-63 61
Using the columns in part A and that the index of the array is 2 allows us to conclude that part D must contain two columns of each of the following forms: (0, 1 , , , . , 1) ~, (1 ,0 ,* , . , 1) ~, (*,*,0, 1, 1) ~ and (*,*, 1,0, 1) ~. Further, again considering the columns of part A and using that the index is 2, in each of these four pairs of columns the undetermined levels must once be 1 and once 2 for each factor. That explains part D of Table 5.
Similarly, part E in Table 5 is a consequence of part B. Part F in Table 5 is a consequence of part D. As seen in Appendix A.3, each of the
first four factors is once at level 1 in part F. Since columns of the form (1,e, 1,e, 1)' appear twice in D, these cannot appear in F. The same conclusion holds with the role of factors 1 and 3 played by factors 1 and 4, factors 2 and 3, and factors 2 and 4. This explains part F.
The arguments for part G are similar as those for part F, but now drawing conclusions by using part E instead of part D.
For part C, we know from Appendix A.4 that the two columns of the form (0,*,0,*,0) ~ should be completed to (0, 1,0, 1,0)' and (0,2,0,2,0) ' or to (0, 1,0,2,0) ~ and (0,2,0, 1,0) ~. But since (e, 1,0, 1,*)' appears already twice, once in B and once in D, the latter possibility is the only possibility. Similar arguments hold when consider- ing (0, *, *, 0, 0)', (*, 0, 0, *, 0)', and (*, 0, *, 0, 0) ~. But we cannot draw any conclusions regarding the two possibilities for (0, 0, *, *, 0) ~ and ( , , . , 0, 0, 0)'. This explains part C.
Part H is a simple consequence of part C. The validity of Table 5 has thus been shown.
Appendix B: The nonexistence of an OA(54, 6, 3, 3)
In this appendix we will prove Theorem 1 by showing that none of the OA(54, 5, 3, 3) can be extended to an OA(54,6,3,3).
By considering the equations in (A.1) for a possible OA(54,6, 3, 3), it is easily seen that n5 -- 0 for every possible solution in nonnegative integers. Hence an OA(54, 5, 3, 3) with one or more repeated columns cannot be extended to an OA(54, 6, 3, 3). It remains to be shown that the same is true for arrays III and IV in Table 1.
An OA(54, 5, 3, 3) can be extended to an OA(54, 6, 3, 3) if and only if its runs can be partitioned to form three arrays OA(18, 5, 3,2). Fix a level combination c that appears as a column in both array III and array IV in Table 1 and that is in both arrays part of a pair of columns with four coincidences. Say we take c -- (0, 0, 0, 0, 1)'. This column should go to one of our three arrays OA(18,5,3,2), requiring a solution with m5 = 1
for an OA(18,5,3,2) in the equations in (A.2). There are three such solutions, all of which are shown in Table 8.
All solutions in Table 8 have m4 = 0. But since arrays III and IV in Table 1 also contain the column (0,0,0,0,2) ~, a second solution for an OA(18,5,3,2) in our partition should have m4 = 1 and m5 --- 0. There are four solutions with this property, all shown in Table 9.
62 A. Hedayat et al./ Journal of Statistical Planning and Inference 58 (1997) 43-63
Table 8 Solutions with ms = 1 for an OA(18,5,3,2)
Solution m0 ml m2 m3 m4 m5
1 0 11 4 2 0 1 2 1 8 7 1 0 1 3 2 5 10 0 0 1
Table 9 Solutions with m 4 = 1 and rn5 = 0 for an OA(18,5,3,2)
Solution m0 ml m2 rn3 m4 m5
1 1 10 2 4 1 0 2 2 7 5 3 1 0 3 3 4 8 2 1 0 4 4 1 11 1 1 0
Table 10 Solutions with m4 = m5 = 0 for an OA(18,5,3,2)
Solution m0 ml m2 m3 m4 m5
1 2 8 2 6 0 0 2 3 5 5 5 0 0 3 4 2 8 4 0 0
Table 11 How an OA(54,5,3,3) may split into three arrays OA(18,5,3,2)
Solution in Solution in Solution in Solution Table 7 Table 8 Table 9
1 2 4 3 2 3 3 3 3 3 4 2
The th i rd O A ( 1 8 , 5 , 3 , 2 ) in the r equ i red pa r t i t ion o f the O A ( 5 4 , 5 , 3 , 3 ) shou ld n o w
p rov ide a so lu t ion w i t h m4 ~ m5 = 0. The three so lu t ions sa t i s fy ing th is r e q u i r e m e n t
are g i v e n in Tab le 10.
The ent i re O A ( 5 4 , 5 , 3 , 3 ) , b o t h for a r ray III and IV in Tab le 1, sat isf ies m0 = 9,
ml = 11, m2 = 26, m3 = 6, m4 = 1 a n d m5 = 1. Hence , we n e e d to select one so lu t ion
f r o m each o f Tab le s 8 - 1 0 so tha t these c o m b i n e to g ive the p r eced ing mi's. There are
th ree poss ib i l i t ies , g i v e n in Tab le 11.
W e wil l n o w argue tha t ne i the r so lu t ion in Tab le 11 is c o m p a t i b l e w i t h ar rays III
and IV in Tab le 1.
So lu t ion 3 in Tab le 11 requ i res five c o l u m n s w i th three co inc idences wi th
( 0 , 0 , 0 , 0 , 1 ) ~ for the O A ( 1 8 , 5 , 3 , 2 ) w i th m4 = m5 = 0. B o t h in array III and IV
in T a b l e 1 the on ly c o l u m n s w i t h th ree c o i n c i d e n c e s w i th ( 0 , 0 , 0 , 0 , 1) ~ h a v e fac tor
5 at leve l 1. Thus , there shou ld b e five runs w i t h fac tor 5 at leve l 1 and two o ther
factors at level 0 in th is O A ( 1 8 , 5, 3, 2) . For at leas t one o f the first four fac tors the leve l
A. Hedayat et al./Journal o f Statistical Planning and Inference 58 (1997) 4343 63
combination (0, 1)' for that factor and factor 5 will then appear at least three times,
contradicting that the index of an OA(18, 5, 3, 2) is 2. Thus solution 3 in Table 11 is
not possible.
For solutions 1 and 2 in Table 11, we concentrate again on the OA(18, 5,3,2) with rn 4 -~ m5 ---- 0. It must contain four columns that have three coincidences with (0, 0, 0, 0, 1 )', or equivalently four columns from part A of arrays III and IV in Table 1. (See Table 4 to identify part A and other parts.) It must then also contain two columns
from part F and no columns from part D. The level combination (1, 1)' for each of the first four factors and factor 5 appears
at most once in the four columns from part A, and should thus appear at least once in the two columns from part F. Clearly, this is not possible for array IV in Table 1. It is however possible for array 1II in Table 1, on which we concentrate our further efforts
now. It means that the columns (1, 1,2,2, 1)' and (2,2, 1, 1, 1)' should both appear in
our OA(18, 5, 3,2). Our OA(18,5,3,2) will also have a solution with m4 -- m5 = 0 when (0,0,0,0,2) '
is used for the level combination c. Hence, following the same arguments as above, it
must also contain the two columns (1, 1,2,2,2) ' and (2,2, 1, 1,2)' from part G. Our OA(18,5,3,2) will thus contain both (1, 1,2,2,1) ' and (1,1,2,2,2) ' ; but the
equations in (1), or the solutions in Table 8, tell us that an OA(18,5,3,2) cannot have a pair of columns with four coincidences. This concludes the proof, and establishes the
nonexistence of an OA(54, 6, 3, 3).
References
Bose, R.C. and K.A. Bush (1952). Orthogonal arrays of strength two and three. Ann. Math. Statist. 23, 508- 524.
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