Cancellation Conditions for Finite Two-Dimensional Additive Measurement

Journal of Mathematical Psychology 45, 226 (2001) doi:10.1006jmps.1999.1285, available online at http:www.idealibrary.com on

Cancellation Conditions for Finite Two-Dimensional Additive Measurement Peter C. Fishburn AT 6 T Labs-Research

It is well known that a weak order  on a finite set X=X 1 _X 2 has an additive real-valued order-preserving representation if and only if  on X satisfies a denumerable scheme of cancellation conditions C(2), C(3), ... . Condition C(K) is based on K distinct ordered pairs in X_X. Given fixed cardinalities m and n for X 1 and X 2 , there is a largest K, denoted by f (m, n), such that some  on X satisfies C(2) through C(K&1) but violates C(K). It has been known for some time that f (2, n)=2 for all n2, and f (3, 3)=3. It was proved recently that f (3, n)n for all even n4 and that f (m, n) m+n&1 for all m, n2. The present paper shows that f (3, 4)= f (4, 4)=4, f (5, n)n+1 for all odd n5, and f (m, n)m+n&10 for all odd m and n greater than or equal to 11. The last result in conjunction with the upper bound of m+n&1 shows that f (m, n) for most (m, n) is approximately m+n.  2001 Academic Press

1. INTRODUCTION

This paper considers the extent to which finite truncations of a denumerable scheme of cancellation conditions that are necessary for additive conjoint measurement are sufficient for additivity when the set X=X 1 _X 2 _ } } } _X N of N-tuples for comparison is finite. Our new results apply to the two-dimensional case of N=2 with |X 1 | =m and |X 2 | =n, so that |X| =mn. The main result is that for most values of m and n, cancellation conditions of orders 2 through approximately m+n are sufficient as well as necessary for additive representability. Old and new results on cancellation conditions for finite X and N2 are outlined in the next section. The rest of the present section offers general background on the additive conjoint model and defines terms used in the paper. We assume throughout that N2 and that each X i has at least two members. Cardinalities of the X i are otherwise unrestricted except as noted in context. Address correspondence and reprint requests to Peter C. Fishburn, Information Science Research, AT 6 T Research, 180 Park Avenue, Room C227, Florham Park, NJ 07932. 0022-249601 35.00 Copyright  2001 by Academic Press All rights of reproduction in any form reserved.

2

CANCELLATION CONDITIONS

3

A binary relation  on a product set X=X 1 _X 2 _ } _X N is said to be additively representable if for each i in [1, 2, ..., N] there is a real-valued function u i on X i such that, for all x=(x 1 , x 2 , ..., x N ) and y=( y 1 , y 2 , ..., y N ) in X, N

N

x  y  : u i (x i ) : u i ( y i ). i=1

(1.1)

i=1

It is common and useful to think of X as a set of decision outcomes or alternatives, each of which is characterized by a level or value of each of N attributes or criteria. The attribute sets are X 1 through X N , and x i or y i denotes a level or value of X i . Within the multiattribute setting we will interpret  as a nonstrict preference relation: x  y signifies that x is not preferred to y. Relations of strict preference O and indifference t are defined from  in the usual way by xOy

if x  y and not ( y  x)

xty

if x  y and y  x.

We say that x is less preferred than y, or that y is preferred to x, when x O y, and that x is indifferent to y when xty. The relationship of incomparability between x and y, defined by not (x  y) and not ( y  x), will be excluded by assuming that is complete, i.e., that for all x, y # X either x  y or y  x. Completeness implies also that  is reflexive (x  x). Axioms for  on X that imply additive representability include necessary ordering, cancellation, and Archimedean conditions, and can also involve nonnecessary solvability and connectedness assumptions that endow (X,  ) with structure which facilitates derivation of the u i for (1.1). Because Archimedean conditions are irrelevant when X is finite, they have no role later. Solvability assumptions are also excluded from our analysis, which focuses solely on the necessary ordering and cancellation conditions. Important contributions for (1.1) that involve Archimedean and solvability or connectedness conditions include Debreu (1960), Luce 6 Tukey (1964), Krantz (1964), Fishburn (1970, 1992), Krantz, Luce, Suppes, 6 Tversky (1971), Wakker (1989, 1991, 1993), Gonzales (1996, 1999), and Karni 6 Safra (1998). A key implication of these references is that the special conditions in conjunction with the first few cancellation conditions lead to (1.1). This can encourage the mistaken impression that higher-order cancellation conditions are not generally needed for additive representability. The purpose of the present paper is see how many of those higher-order conditions must be assumed to ensure (1.1) for finite structures when nonnecessary conditions like solvability are not presumed. The natural ordering condition for additive representability and many other models is that  on X is a weak order. This means that  is transitive (x  y and y  z O x  z) and complete. These properties are implied by (1.1), and we assume henceforth that  on X is a weak order. We also consider the weak-order specialization in which xty  x= y, and we refer to  or O in this case as a linear order. Weak order implies that t on X is an equivalence relation (reflexive, transitive, and symmetric: xty O ytx) and that

4

PETER C. FISHBURN

the equivalence classes in X determined by t are linearly ordered by the natural extension of O from individuals in X to equivalence classes. When  itself is a linear order, each equivalence class is a singleton subset of X. Cancellation conditions arose in the theory of webs as early as the 1920s (Thomsen, 1927; Blaschke, 1928), but they were not integrated into additive measurement as in (1.1) until the late 1950s and early 1960s. This was first done in Kraft, Pratt 6 Seidenberg (1959), which specified a denumerable scheme of cancellation conditions for additive subjective probabilities on finite state spaces. This work, which is continued by Fishburn (1996, 1997a), applies to our present setting when |X i | =2 for all i. General cancellation conditions which apply to attribute sets of all cardinalities and are sufficient for (1.1) when X is finite were first presented in Scott (1964), Tversky (1964), and Adams (1965). The ensuing formulation is based on their work. We define a cancellation condition C(K) for each integer K2. Condition C(K) considers precisely K distinct pairs (x 1, y 1 ), (x 2, y 2 ), ..., (x K, y K ) of N-tuples in X. If K> |X| 2, then C(K) holds trivially for that X. Given X, K2, distinct pairs (x 1, y 1 ), (x 2, y 2 ), ..., (x K, y K ) in X_X, and positive integers a 1 , a 2 , ..., a K , let 7 K a k(x k, y k ) # CK mean that for every i # [1, 2, ..., N], the sequence a 1 x 1i [x 1i repeated a 1 times], a 2 x 2i , ..., a K x Ki of 7a k terms from X i is a permutation of the sequence a 1 y 1i , a 2 y 2i , ..., a K y Ki . Here, 7 K a k(x k, y k ) # CK indicates that the a k -weighted x ki are in perfect balance with the a k -weighted y ki for each attribute. An immediate consequence of this for every set [u i ] of utility functions is that N

7 K a k(x k, y k ) # CK O :

K

N

: a k u i (x ki )= :

i=1 k=1

K

: a k u i ( y ki ).

i=1 k=1

Because (1.1) requires  on its right side when x k  y k, and < when x k Oy k, the following Kth-order cancellation condition must hold when  on X is additively representable. C(K). For all 7 K a k(x k, y k ) # CK , it is false that x k  y k for k=1, 2, ..., K and x k O y k for some k # [1, 2, ..., K]. The main thrust of the theorems of Scott (1964), Tversky (1964), and Adams (1965) is that the denumerable sequence C(2), C(3), ... of cancellation conditions is sufficient for additive representability when X is finite. Their sufficiency proofs are based on standard theory for the existence of solutions to finite systems of linear inequalities. It is easily seen that in the first few C(K), for example C(2) and C(3), it suffices to take a k =1 for each k. However, larger K may require different values for the a k to demonstrate a failure of cancellation. In particular, Fishburn (1996) shows that there are situations in which C(2), C(3), and C(4) hold, C(5) holds when a 1 = } } } =a 5 =1, but a failure of C(5) occurs when a 1 = } } } =a 4 =1 and a 5 =2.

5

CANCELLATION CONDITIONS

Conditions C(2) and C(3) are used extensively in prior references and are known by various names, including independence, first-order independence, preferential independence, strong separability, and coordinate independence for C(2), and double cancellation and triple cancellation for C(3). An equivalent version of C(2) is the following: Coordinate independence. For all x, y, z, w # X, x  y  z  w if for each i # [1, 2, ..., N] either (x i = y i and z i =w i ) or (x i =z i and y i =w i ). This condition has been discussed in consumer demand theory for more than a century with explicit formulations in Samuelson (1940), Sono (1943), and Debreu (1960), and the last of these shows that it is sufficient for additive representability when N3 in a topological formulation with appropriate Archimedean and connectedness assumptions. A similar formulation for N=2, or the algebraically oriented alternative for N=2 in Luce 6 Tukey (1964), also requires C(3) to derive (1.1) in the presence of solvability or connectedness conditions. Recent extensions along this line are in Gonzales (1996, 1999) and Karni 6 Safra (1998). Relationships between coordinate independence and independence axioms in expected utility theory are described in Fishburn 6 Wakker (1995). An important implication of coordinate independence or C(2) that is used later is that it induces an unambiguous weak order  i on X i defined by xi i yi

if x  y

whenever x j = y j

for all j{i.

We also write x i Oi y i if x i  i y i and not ( y i  i x i ), and x i ti y i if x i  i y i and y i  i x i . Given C(2), ti is an equivalence relation on X i , and the natural extension of Oi from individuals in X i to equivalence classes of X i determined by ti is a linear order on those classes. The theme of the present paper and Fishburn (1996, 1997a, 1997b) is to determine the maximum number of cancellation conditions that need to be checked to ascertain whether a weak order on finite X is additively representable. This number depends on the cardinalities of the attribute sets. Accordingly, we define the size of finite X=X 1 _X 2 _ } } } _X N as (n 1 , n 2 , ..., n N ), where n i = |X i | 2 for every i. The key quantity we wish to determine for each size (n 1 , n 2 , ..., n N ) is the unique K2 such that (i) every weak ordered set of size (n 1 , n 2 , ..., n N ) that satisfies C(2), C(3), ..., C(K) is additively representable; (ii) some weak ordered set of size (n 1 , n 2 , ..., n N ) that satisfies all cancellation conditions (if any) prior to C(K) is not additively representable. We denote the unique K that satisfies (i) and (ii) by f (n 1 , n 2 , ..., n N ). A specific instance of f whose proof is outlined later is f (4, 4)=4. Hence, with X=X 1 _X 2 and |X 1 | = |X 2 | =4, the definition of f says that (i) every  on X that satisfies C(2), C(3), and C(4) is additively representable, and (ii) some  on X that satisfies coordinate independence and C(3) is not additively representable.

6

PETER C. FISHBURN

It seems extremely hard to determine f (n 1 , n 2 , ..., n N ) precisely except in a few cases. The main reason is that the problem is highly combinatoric, as suggested perhaps by the computations in Arbuckle 6 Larimer (1976), McCelland (1977), and Ullrich 6 Wilson (1993) for the number of ways of ordering cells in finite arrays for N=2 and N=3 that satisfy coordinate independence and more complex cancellation conditions. I have found it a very delicate task, using a combination of linear algebra and combinatorial notions, to construct instances of weak orders or linear orders that satisfy C(2), C(3), ..., C(K&1) but violate C(K) when K appears to be in the neighborhood of f(n 1 , n 2 , ..., n N ). As a consequence, many of the results on f obtained thus far specify lower and upper bounds rather than exact values for f. Fortunately, the bounds are often close together and provide good asymptotic information as one or more of the arguments becomes large. As a final point of introduction, it should be mentioned that the present work is devoted to the mathematical structure of cancellation conditions rather than to the construction of additive representations when they exist. Sherman (1977) presents an elegant algorithm for constructing additive representations that merits close examination by those who are interested in the algorithmic side of finite conjoint measurement. 2. RESULTS FOR FINITE STRUCTURES

We begin with an upper bound on f. Theorem 2.1. f (n 1 , n 2 , ..., n N ) N i=1 n i &(N&1). This is proved in Fishburn (1997b) from elementary facts about linear inequalities and linear dependence. The proof makes no use of finer combinatorial structures of the problem, and for this reason the bound seems generally larger than f. However, it is very close to known lower bounds in several cases that are noted shortly. The first significant results for f were obtained in the context of additive representations of subjective probability by Kraft, Pratt, 6 Seidenberg (1959). They include: Theorem 2.2. f (2, 2, 2)= f (2, 2, 2, 2)=2 and f (2, 2, 2, 2, 2)4. This says that conjoint independence implies additivity when N4 and all attribute sets are binary. Moreover, when X is composed of five binary attributes, there are weak orders on X that satisfy C(2) and C(3) but violate C(4). Fishburn (1996, 1997a) extends Theorem 2.2 by determining the exact value of f(2, 2, 2, 2, 2) and bounding all other binary cases. Let 2 N denote the N-tuple of N 2s. Theorem 2.3. f (2 5 )=4. N&1 f (2 N )N+1 for all N6. The upper bound comes from Theorem 2.1. The lower bound for each N is established by a linear order on 2 N that satisfies C(2) through C(N&2) but violates C(N&1). I conjecture that f (2 N )=N&1 for N6 but have had no success in reducing the upper bound of N+1. The only other result for N3 that I am aware of besides Theorem 2.1 and implications by monotonicity from other results is f (2, 3, 3)4, which is implied by the computations in Ullrich 6 Wilson (1993).

CANCELLATION CONDITIONS

7

We assume henceforth that N=2 and let (n 1 , n 2 )=(m, n). An initial result for N=2 was established in Krantz et al. (1971). Theorem 2.4. f (2, n)=2 for all n2. The upper bound of Theorem 2.1 is not close to f in this case since it says that f(2, n)n+1. However, when we increase 2 to 3, the picture changes dramatically. We denote by wxx the largest integer that does not exceed x. Theorem 2.5. 2 wn2x f (3, n)n+2 for all n3. The upper bound is from Theorem 2.1. The lower bound is proved in Fishburn (1997b). Theorem 2.5 implies that f (3, n)n Ä 1 as n gets large. Our first new result replaces 3 by 5 in the preceding theorem and gives the first verified instances for which f (m, n)>max[m, n]. Theorem 2.6. 2 w(n+1)2x f (5, n)n+4 for all n5. The lower bound of n+1 for odd n is proved in the next section. The lower bound for even n is the same as the lower bound for f (3, n) in Theorem 2.5. We note shortly that f (3, 3)=3 and f (4, 4)=4 so that n=5 is the smallest n for which f(n, n)>n. The main result of the paper has a larger difference between the bounds on f(m, n) than the two preceding theorems, but that difference remains fixed as m and n increase. Theorem 2.7. m+n&10 f (m, n)m+n&1 for all odd m and n greater than or equal to 11. The upper bound comes again from Theorem 2.1. The lower bound is proved in Section 4. As m and n get large, Theorem 2.7 shows that f (m, n) is approximately equal to m+n. Asymptotically, f (n, n)(2n) Ä 1. Our final theorem reverts back to small values of m and n for some exact values of f. Theorem 2.8. f (3, 3)=3 and f (3, 4)= f (4, 4)=4. The result for f (3, 3) is not new since it was noted in Table 2c in Arbuckle 6 Larimer (1976) and is verified in Table 1 in Ullrich 6 Wilson (1993). However, I include a proof of f (3, 3)=3 in Section 5, followed by an outline of a proof of f(4, 4)=4. Because my full proof of f (4, 4)=4 is embarrassingly long and is not given here, some readers may prefer to think of f (4, 4)=4 as a claim. The paper concludes in Section 6 with notes on open problems and suggestions for further research.

3. LOWER BOUNDS

This section develops a theorem, patterned after Theorem 6.1 in Fishburn (1996), that helps to verify the lower bounds for N=2 in the preceding theorems. We then use it to show that f (3, 3)3, f (3, 4)4, and f (5, n)n+1 for each odd n5.

8

PETER C. FISHBURN

We say that a sequence (x 1, y 1 ), (x 2, y 2 ), ..., (x K, y K ) of pairs of pairs in X= X 1 _X 2 is a unitary balance sequence if 7 K (x k, y k ) # CK and every x k and y k pair is distinct, i.e., |[x 1, ..., x K, y 1, ..., y K ]| =2K. A unitary balance sequence (x 1, y 1 ), ..., (x K, y K ) is irreducible if no member of CJ for 2J
for

k=2, ..., K.

This implies v 1( y 11 )+v 2( y 12 )
i=1, 2,

where =>0 is sufficiently small so that =[ max |v 1(a)&v 1(b)| + max |v 2(a)&v 2(b)| ] a, b # X1

a, b # X2


[x, y] Â [[x 1, y 1 ], [x 2, y 2 ], ..., [x K, y K ]]].

9

CANCELLATION CONDITIONS

Also define O* on X by x O* y  w 1(x 1 )+w 2(x 2 )
and no x k1

for k{i

or y k1 for k{ j

equals x i1 ;

or x i2 = y 2j

and no x k2

for k{i

or y k2 for k{ j

equals x i2 ;

or x 1j = y i1

and no x k1

for k{ j

or y k1 for k{i equals x 1j ;

or x 2j = y i2

and no x k2

for k{ j

or y k2 for k{i equals x 2j .

In other words, [(x i, y i ), (x j, y j )] # E if one of the four connections denoted by lines in the array

defines an equality and the member of X 1 (or X 2 ) in that equality is not used as the member of X 1 (or X 2 ) in any other pair in X in the unitary balance sequence. We follow the usual definition by saying that a graph G=(V, E) is connected if for any distinct vertices u, v # V there is a sequence u=v 1 , v 2 , v 3 , ..., v r =v for r2 such that [v i , v i+1 ] # E for i=1, ..., r&1. Lemma 3.2. connected.

A unitary balance sequence is irreducible if its connection graph is

Proof. Let G=(V, E) be the connection graph of a unitary balance sequence (x 1, y 1 ), ..., (x K, y K ). Suppose the orders pairs (x, y) in a member 7 of CJ come entirely from [(x 1, y 1 ), ..., (x K, y K )]. If (x i, y i ) is used in 7 and [(x i, y i ), (x j, y j )] # E, then (x j, y j ) must also be used in 7, or else the balance needed for membership in CJ

10

PETER C. FISHBURN

would be contradicted. Hence, if G is connected, then every (x k, y k ) must be used in 7, so in fact CJ =CK and the unitary balance sequence is irreducible. K It should be noted that an irreducible unitary balance sequence of length K does not imply that f (m, n)K if the condition on u 1 and u 2 of Theorem 3.1 fails. We illustrate with an example. Example 3.3. Let X 1 =X 2 =[1, 2, 3, 4, 5] with m=n=5, and define (x 1, y 1 ), ..., (x , y 6 ) as follows: 6

k

xk

yk

1

(1, 4)

(2, 1)

2

(2, 2)

(1, 5)

3

(3, 1)

(2, 3)

4

(2, 4)

(3, 2)

5

(5, 3)

(4, 4)

6

(4, 5)

(5, 4) .

Because 123254 is a permutation of 212345 for component 1, 421435 is a permutation of 153244 for component 2, and all x k and y k are different, (x 1, y 1 ), ..., (x 6, y 6 ) is a unitary balance sequence. Moreover, when we use k as a surrogate of (x k, y k ), the connection graph of the sequence includes edges [1, 2], [2, 4], [3, 4], [3, 5], and [5, 6], so the graph is connected and, by Lemma 3.2, (x 1, y 1 ), ..., (x 6, y 6 ) is irreducible. However, if u 1 and u 2 are defined on [1, 2, 3, 4, 5] so that u 1(1)+u 2(4)=u 1(2)+ u 2(1) for k=1, u 1(2)+u 2(2)=u 1(1)+u 2(5) for k=2, and so on through k=6, it follows that u 1(4)+u 2(2)=u 1(5)+u 2(1). Since [(4, 2), (5, 1)] is not one of the [x k, y k ], the condition in Theorem 3.1 for u 1 and u 2 does not hold. Suppose in fact that the six-term irreducible unitary balance sequence gives a violation of C(6) with x k  y k for all k and, for definiteness, x 1 Oy 1. Let x 0 =(4, 2) and y 0 =(5, 1). Regardless of whether x 0 O y 0 or y 0 O x 0, we get a violation of C(4) as follows: x 0 O y 0:

(4, 2) O (5, 1)

y 0  x 0:

(5, 1)  (4, 2)

k=3

(3, 1)  (2, 3)

k=1

(1, 4) O (2, 1)

k=4

(2, 4)  (3, 2)

k=2

(2, 2)  (1, 5)

k=5

(5, 3)  (4, 4)

k=6

(4, 5)  (5, 4).

Hence, with the failure of the u i condition of Theorem 3.1, the presumed C(6) violation implies that there is also a C(4) violation. We now use Theorem 3.1 and Lemma 3.2 to prove that f (3, 3)3, f (3, 4)4, and f (5, n)n+1 for each odd n5. The bulk of the proof is given in Fig. 1. For

11

CANCELLATION CONDITIONS

convenience we take X 1 =[1, 2, ..., m], X 2 =[1, 2, ..., n], and define u i as strictly increasing over X i . The matrix entry in row i and column j is u 1(i)+u 2( j). The general structure is as follows. columns (X 2 )

1

u 1(1)

rows

2

u 1(2)

(X 1 )

b

b

m

u 1(m)

1

2

3

}}}

n

u 2(1)

u 2(2)

u 2(3)

}}}

u 2(n)

u 1 +u 2

A unitary balance sequence (UBS) is shown next to each matrix in Fig. 1. The (x, y) pairs in the UBS are the cell coordinates at the ends of the sloping lines. The orders of the two cells are chosen to ensure complete balance. If we designate the left cell x in a UBS pair (x, y) by & and the right cell by +, balance obtains when the number of minus signs equals the number of plus signs in every row and column.

FIGURE 1

12

PETER C. FISHBURN

It follows by inspection that each claimed UBS list is indeed a unitary balance sequence and that the u i condition of Theorem 3.1 holds. For u 2 in the m=5 matrix, u 2(2)&u 2(1)=u 2(2k+1)&u 2(2k)=1, and u 2( j+1)&u 2( j)=2 for the other adjacent u 2 differences. It remains only to note that each unitary balance sequence is irreducible, and this is where we use Lemma 3.2. The connection graphs for the top two UBS's are easily seen to be connected. The edges in the connection graph of the UBS for (5, n) include [((1, 2k), (2, 1), ((2, 2), (1, 2k+1))] [((2, 2), (1, 2k+1)), ((2, 3), (3, 2))] [((2, 3), (3, 2)), ((2, 4), (4, 3))] b [((2, k), (3, k&1)), ((2, k+1), (3, k))] [((1, 2k), (2, 1)), ((3, 2k&1), (2, 2k))] [((3, 2k&1), (2, 2k)), ((3, 2k&2), (2, 2k&1))] b [((3, k+2), (2, k+3)), ((3, k+1), (2, k+2))] [((1, 2k), (2, 1)), ((5, 1), (4, k+1))] [((5, 1), (4, k+1)), ((4, 2k+1), (5, k+1))], and these show that the graph is connected.

4. A GENERAL LOWER BOUND

This section shows that Theorem 3.1 and Lemma 3.2 imply f (2j+5, 2k+5)2( j+k)

for 3 jk,

which proves the lower bound of Theorem 2.7. We explain the matrix setup which defines the 2( j+k)-term UBS and observe that the UBS's connection graph is connected. The proof is completed by demonstrating assignments for u 1 and u 2 that verify the u i condition of Theorem 3.1. Figure 2 shows the cell pairs of the UBS. There are 2k&2 pairs in the first three rows, 2j&2 pairs in the first three columns, and 4 in the lower right corner, for 2( j+k) overall. Each pair has a & and a + for (&, +) in the UBS. For example, by cell coordinates, the two UBS members that involve row 1 are ((1, 2k+2), (2, 4)) and ((2, 5), (1, 2k+3)), and the four in the lower right corner are ((2j+4, 2k), (2j, 2k+5)), ((2j+1, 2k+5), (2j+5, 2k+1)), ((2j+2, 2k+4), (2j+4, 2k+2)), and ((2j+5, 2k+3), (2j+3, 2k+4)).

CANCELLATION CONDITIONS

FIG. 2.

13

UBS definition.

Because the number of minus signs equals the number of plus signs in each row and column, the alleged UBS is in fact a unitary balance sequence. It is irreducible because its connection graph is connected. This is implied by the fact that if we begin with a + in cell (2, 4), then the sign of every cell in the UBS is uniquely determined. For example, cell (2, 4)=+ implies (1, 2k+2)=& and (3, 4)=&, cell (3, 4)=& implies (2, 6)=+, which in turn implies (3, 6)=& and so forth, (1, 2k+2)=& implies (1, 2k+3)=(2j+4, 2k+2)=+, (2j+4, 2k+2)=+ implies (2j+2, 2k+4)=&, which implies (2j+3, 2k+4)=(2j+2, 1)=+, ... . The lower right corner is used to connect the first three rows and the first three columns. To specify u 1 and u 2 , we first define 2=2k+3 a=(4j+4) 2 b=a+4k+4 c=b 2.

14

PETER C. FISHBURN

The following assignments are clearly increasing: u 1(1)=0

u 2(1)=0

u 1(2)=2k

u 2(2)=2 j2

u 1(3)=2k+2

u 2(3)=(2 j+2) 2

u 1(4)=b u 1(5)=b+2 u 1(6)=b+22

u 2(4)=a u 2(5)=a+1 u 2(b)=a+2

u 1(7)=b+32

u 2(7)=a+3

b

b

u 1(2 j&1)=b+(2 j&5) 2 u 1(2 j)=b+(2 j&4) 2 u 1(2 j+1)=b+(2 j&3) 2

u 2(2k&1)=a+2k&5 u 2(2k)=a+2k&4 u 2(2k+1)=a+2k&3

u 1(2 j+2)=b+2 j2

u 2(2k+2)=a+2k

u 1(2 j+3)=b+(2 j+1) 2

u 2(2k+3)=a+2k+1

u 1(2 j+4)=c+2 j2+2k+4 u 1(2 j+5)=c+(2 j+1) 2+2k+3

u 2(2k+4)=c u 2(2k+5)=c+42&4.

These imply that u 1(x 1 )+u 2(x 2 )=u 1( y 1 )+u 2( y 2 ) for each (x, y) in the UBS: cells for x 6 y

value of u 1 +u 2 for x 6 y

(1, 2k+2), (2, 4)

a+2k

(1, 2k+3), (2, 5)

a+2k+1

(2, 6), (3, 4)

a+2k+2

b (2, 2k+1), (3, 2k&1)

a+4k&3

(2 j+2, 1), (4, 2)

b+2 j2

(2 j+3, 1), (5, 2)

b+(2 j+1) 2

(6, 2), (4, 3)

b+(2 j+2) 2

b (2 j+1, 2), (2 j&1, 3)

b+(4 j&3) 2

(2 j+4, 2k), (2 j, 2k+5)

a+c+2 j2+4k( =b+c+2 j2&4)

(2 j+1, 2k+5), (2 j+5, 2k+1)

b+c+(2 j+1) 2&4( =a+c+(2 j+1) 2+4k)

(2 j+2, 2k+4), (2 j+4, 2k+2)

b+c+2 j2( =a+c+2 j2+4k+4)

(2 j+5, 2k+3), (2 j+3, 2k+4)

a+c+(2 j+1) 2+4k+4( =b+c+(2 j+1) 2) .

15

CANCELLATION CONDITIONS

The u 1 +u 2 values in the preceding display are all different, so it remains only to check that every other cell in the matrix has a u 1 +u 2 value that is not duplicated by another cell. We begin with the (2j+3)_(2k+3) submatrix that excludes the last two rows and last two columns. The smallest nine u 1 +u 2 values are in the upper left 3_3 corner: they go from 0 to (2j+2) 2+2k+2 in cell (3, 3), and all are different. The next larger value is a=(4j+4) 2 in cell (1, 4), then the values increase by increments of 1 across the top row to a+2k&3 in cell (1, 2k+1). The other six u 1 +u 2 values that are not involved in the UBS in rows 2 and 3 toward the right end of the (2j+3)_(2k+3) submatrix are a+4k&2, a+4k&1, a+4k, a+4k+1, a+4k+2, and a+4k+3, all of which exceed the value of a+4k&3 in the UBS cells (2, 2k+1) and (3, 2k&1). Because b in cell (4, 1) exceeds a+4k+3 in (3, 2k+3), every u 1 +u 2 value in rows 4 through 2j+3 exceeds the largest value in row 3 prior to the excluded last two columns. The situation in the first three columns and rows 4 through 2j+3 is similar to that in the first three rows for columns 4 through 2k+3. Next, the value in (4, 4) of a+b exceeds b+(4j+3) 2 in (2j+3, 3), so every entry in (rows 4 through 2j+3)_(columns 4 through 2k+3) exceeds all entries to the left of (and above) this 2j_2k submatrix. Within that submatrix, the values increase lexicographically with rows dominant since the entry in ( j, 2k+3) is less than the entry in ( j+1, 4). This concludes the analysis of the main (2j+3)_(2k+3) submatrix. The smallest entry in the final two rows and final two columns is c=b 2 in (1, 2k+4). This exceeds the largest entry considered previously, a+b+(2j+1) 2+2k+1 in (2j+3, 2k+3), so everything in the last two rows and columns is larger than everything in the (2j+3)_(2k+3) matrix. Because c is large compared to b, the largest four u 1 +u 2 values in the entire matrix are in the lower right 2_2 corner, and they are all different: 2c+2j 2+2k+4, 2c+(2j+1) 2+2k+3, 2c+(2j+4) 2+2k, and 2c+(2j+5) 2+2k&1. This leaves the last two rows and columns, excluding the lower right corner, for final consideration. Each entry there has one c, so we omit c for convenience. The rescaled values are: row 2 j+4

row 2 j+5

column 2k+4

column 2k+5

(2 j+1) 2+1 (4 j+1) 2+1 (4 j+3) 2+1 (6 j+5) 2+1 (6 j+5) 2+2 b (6 j+5) 2+2k&4 (6 j+5) 2+2k &3 (6 j+5) 2+2k&2 (6 j+5) 2+2k +1 (6 j+5) 2+2k+2

(2 j+2) 2 (4 j+2) 2 (4 j+4) 2 (6 j+6) 2 (6 j+6) 2+1 b (6 j+6) 2+2k&5 (6 j+6) 2+2k&4 (6 j+6) 2+2k &3 (6 j+6) 2+2k (6 j+6) 2+2k +1

0 2&3 2&1 (4 j+5) 2+2k+1 (4 j+6) 2+2k+1 b 6j2+2k+1 (6 j+1) 2+2k+1 (6 j+2) 2+2k+1 (6 j+5) 2+2k +1 (6 j+6) 2+2k +1

42&4 52&7 52&5 (4 j+9) 2+2k&3 (4 j+10) 2+2k&3 b (6 j+4) 2+2k&3 (6 j+5) 2+2k &3 (6 j+6) 2+2k &3 (6 j+9) 2+2k&3 (6j+10) 2+2k&3.

The UBS values are shown in boldface. It is clear that the smallest 12 values in this tableau are in the first three rows, and all are different. Since (6j+5) 2+2k+2< (6j+6) 2, there is no duplication in the first two columns (for rows 2j+4 and

16

PETER C. FISHBURN

2j+5), and no duplication in the last two columns because (2k+1)&(2k&3)=4 and no multiple of 2 equals 4. There is also no duplication in the last four rows apart from the boldfaced UBS pairs, and no value there is duplicated in a higher row. When we exclude the first three and last four rows, all remaining entries in columns three and four (for columns 2k+4 and 2k+5) are less than all remaining entries in columns 1 and 2. It follows that all non-UBS values differ and do not duplicate a UBS value.

5. EXACT VALUES

We conclude our proofs by arguing that f (3, 3)3 and f (4, 4)4. Because f(3, 3)3 and f (4, 4) f (3, 4)4, it follows that f (3, 3)=3 and f (3, 4)= f (4, 4) =4, which completes the proof of Theorem 2.8. The proofs in this section use an exhaustive search procedure to bound the largest T for which a sequence (x 1, y 1 ), (x 2, y 2 ), ..., (x T, y T ) of not necessarily distinct (x, y) pairs violates cancellation for some  and can not be reduced to a shorter sequence that also violates cancellation for the same  . The procedure shows that max T3 for (m, n)=(3, 3) and max T4 for (m, n)=(4, 4). We precede its description by some general definitions and observations. We take X 1 =[1, 2, ..., m] and X 2 =[1, 2, ..., n]. It is assumed without loss of generality that  on X satisfies coordinate independence and that O1 and O2 are the natural linear orders defined by 1O1 2 O1 } } } O1 m

and

1 O2 2 O2 } } } O2 n.

(5.1)

We use linear coordinate orders because nontrivial instances of t1 and t2 reduce (m, n) to a smaller domain. For example, if (5.1) holds except that 3t1 4 instead of 3 O1 4 for (m, n)=(4, 4), then every instance of 4 for an x 1 or y 1 in a sequence that violates cancellation can be replaced by 3, thus reducing the 4_4 violation to a 3_4 violation. Given (5.1), let x< y mean that (x{ y, x 1  y 1 , x 2  y 2 ), and let x & y mean that (x 1 < y 1 , y 2
CANCELLATION CONDITIONS

17

for CV lists, discarding reducible candidates in a branch-and-bound fashion as it proceeds. It will be guided by the following definition and lemma. For each (m, n)(3, 3), let L(m, n) denote the set of all sequences S=(x 1, y 1 ), 2 (x , y 2 ), ..., (x K, y K ) that satisfy the following properties: P1. K3. P2. S is balanced. P3. x k & y k for k=1, ..., K. P4. y j {x k for all j, k # [1, ..., K]. P5. No distinct j, k # [1, ..., K] have y j
18

PETER C. FISHBURN

Corollary 5.2.

f (m, n)l(m, n).

There is no assurance that f (m, n)=l(m, n), for f (m, n)
y

12

21

V

13

.

This leaves three possibilities for V that satisfy P3P5: I

or

II

or

III

x

y

x

y

x

y

12 22

21 13

12 31

21 13

12 32

21 13

.

CANCELLATION CONDITIONS

19

For I we take y 32 =2 since there are 2s for x 2 but none thus far for y 2 . Then, because 12 and 22 are xs, y 3 must be 32. To have x 3 & y 3, we need x 32 =3, and since 13 is a y and 33 is out by P3, x 3 =23. Hence, with no loss of generality, we take (x 3, y 3 )=(23, 32) in I. Then I has (12, 21), (23, 32) as an almost balanced subsequence with completion pair (31, 13). This yields the three-term member (12, 21), (23, 32), (31, 13) of L(3, 3). We move on to II, which is itself almost balanced and yields the same member of L(3, 3). For III we can take x 32 =3 because y 22 =3 and no x 2 is yet 3. Then, because 13 is already a y, P3 and P4 require x 3 =23: P3 requires y 31 =3 and, since 32 is an x, y 3 must be 31. But then [(x 2, y 2 ), (x 3, y 3 )]= [(32, 13), (23, 31)], which violates P5 and completes our consideration of III. We conclude that, up to permutation of terms, a unique member of L(3, 3) contains (12, 21). It has exactly three terms. We discard (12, 21) from further use as a member of sequences in L(3, 3). At this point we pause to consider symmetry implications of the Hasse diagram of the partial order < on X=[1, 2, 3]_[1, 2, 3] shown in Fig. 3. If x< y, a connected series of upward sloping lines goes from x to y. If there is no such series for distinct x and y then x& y. By P3, only pairs (x, y) with x & y are eligible for sequences in L(3, 3). The diagram has two axes of rotational symmetry, H and V. Symmetry around V implies that (21, 12) can be discarded once (12, 21) has been fully analyzed, so we ignore (21, 12) henceforth as a term in sequences in L(3, 3). Operationally, V symmetry corresponds to interchanging the two coordinates. Let x$ denote the image of x around H. The H action maps (x, y) into ( y$, x$). We denote this by h(x, y)=( y$, x$). The reversal of x$ and y$ in ( y$, x$) reflects the sense of the preference order under (5.1). For example, 12<13 maps into y$=13< 23=x$, and x=31 O 12= y maps into y$=23<31=x$. Note also that the H action preserves P5: the preceding failure of P5 is preserved by h since [h(32, 13), h(23, 31)]=[(13, 21), (31, 12)] with 21<31 and 12<13.

FIG. 3.

Partial order < on [1, 2, 3] 2 with axes of rotational symmetry.

20

PETER C. FISHBURN

Because of H symmetry, our prior completion of (12, 21), along with (21, 12) by V symmetry, says that we can also disregard h(21, 12)=(23, 32) and h(12, 21)= (32, 23) henceforth as terms in sequences in L(3, 3). We now begin our second round with (x 1, y 1 )=(13, 21). Take y 21 =1 without loss of generality, so y 2 =12 because 13 is x 1. Then x 22 must be 1 to satisfy P3, and x 2 =31 because (21, 12) has been discarded. However [(13, 21), (31, 12)] violates P5, so we are done with (x 1, y 1 )=(13, 21). By V and H symmetry, (31, 12), (23, 31), and (32, 13) are also discarded from further consideration as terms in L(3, 3) sequences. Let (x 1, y 1 )=(12, 31) to begin the third round. Take x 21 =3 without loss of generality, so x 2 =32 because 31= y 1. Then P3 requires y 22 =3, so y 2 is 13 or 23. However, both (32, 13) and (32, 23) have been discarded, so we are done with (x 1, y 1 )=(12, 31). By V and H symmetry, (21, 13), (31, 23), and (13, 32) are also discarded henceforth. At this point, all (x, y) that satisfy P3 and have x or y in [12, 21, 23, 32] have been discarded. Only 13, 31, and 22 remain, so the only (x, y) we consider further for L(3, 3) construction are (13, 31), (31, 13), (13, 22), (22, 13), (31, 22), and (22, 31). Regardless of which is taken as (x 1, y 1 ) in round four, it is impossible to satisfy P1P4, and the proof of l(3, 3)=3 is complete; L(3, 3) has exactly two members up to permutations of terms, (12, 21), (23, 32), (31, 13), and, by either V or H symmetry, (21, 12), (32, 23), (13, 31). K

FIG. 4.

Partial order < on [1, 2, 3, 4] 2 with axes of rotational symmetry.

CANCELLATION CONDITIONS

21

Outline of the proof of l(4, 4)=4. Figure 4 shows the Hasse diagram of the partial order < on X=[1, 2, 3, 4]_[1, 2, 3, 4]. By P3, only its incomparable pairs with x & y can be used to construct sequences in L(4, 4). For example, if x=21, y # [12, 13, 14]; if y=14, x # [23, 22, 33, 21, 32, 43, 31, 42, 41]; and so forth. The full proof has 16 rounds, R1 through R16. We show (x 1, y 1 ) for each round, followed by other (x, y) that can be discarded due to V symmetry and H symmetry once (x 1, y 1 ) has been processed. The order of the (x 1, y ) was chosen in an attempt to simplify their processing as much as possible, but even then, several rounds, including R4, R5, and R7, required extensive computations. R1.

(14, 21) : (41, 12), (43, 14), (34, 41).

R2.

(21, 14) : (12, 41), (14, 43), (41, 34).

R3.

(14, 31) : (41, 13), (42, 14), (24, 41).

R4.

(31, 14) : (13, 41), (14, 42), (41, 24).

After R4, 14 as x or y was never paired with 21, 31, 42, or 43 as y or x, and 41 was never paired with 12, 13, 24, or 34. R5.

(24, 31) : (42, 13). (The V and H actions are identical.)

R6.

(13, 21) : (31, 12), (43, 24), (34, 42).

R7.

(31, 24) : (13, 42).

R8.

(22, 14) : (22, 41), (14, 33), (41, 33).

R9.

(14, 22) : (41, 22), (33, 14), (33, 41).

R10.

(23, 31) : (32, 13), (42, 23), (24, 32).

R11.

(21, 13) : (12, 31), (24, 43), (42, 34).

R12.

(31, 23) : (13, 32), (23, 42), (32, 24).

After R12, x and y in every pair used in a member of L(4, 4) are from the same level of the lattice in Fig. 4: a level is a set of ij with constant i+ j. R13.

(12, 21) : (21, 12), (43, 34), (34, 43).

R14.

(13, 31) : (31, 13), (42, 24), (24, 42).

R15.

(13, 22) : (31, 22), (33, 24), (33, 42).

R16.

(22, 13) : (22, 31), (24, 33), (42, 33).

After R16, the only members of X available as x or y are the midlevel points 41, 32, 23, and 41, which cannot be used to form a member of L(4, 4). The analysis of each round shows that no member of L(4, 4) has more than four terms, so l(4, 4)=4. Figure 5 illustrates four four-term CV lists that use all four points of each coordinate.

22

PETER C. FISHBURN

FIG. 5.

Four violations of C(4).

I conclude this outline by summarizing R1, R2, and R10. P3 and P4 are used without special mention and, as in the proof for l(3, 3)=3, branches will be followed in ways needed for balance (P2). The symbol h signifies that the analysis for a branch or subbranch is complete. R1. (x 1, y 1 )=(14, 21). Take y 21 =1. Then y 2 # [12, 13]. Suppose y 2 =12. Then x 2 # [31, 41], but both [(14, 21), (31, 12)] and [(14, 21), (41, 12)] contradict P5. h Suppose y 2 =13. Then x 2 # [22, 31, 32, 41, 42], but all possible choices for x 2 give a contradiction of P5. h R2. (x 1, y 1 )=(21, 14). Take x 21 =1. Then x 2 # [12, 13]. Suppose x 2 =12. Then y 2 # [31, 41]. Each choice of y 2 violates P6 because 21<(31 or 41) and 12<14. h Suppose x 2 =13. Then y 2 # [22, 31, 32, 41, 42], but each choice of y 2 violates P6. h R10. (x 1, y 1 )=(23, 31). According to the preceding rounds, 14 never appears with 21, 22, 31, 33, 42, or 43 in an (x, y), 41 never appears with 12, 22, 13, 33, 24, or 34 in an (x, y), 13 and 42 are never paired, 31 and 24 are never paired, and (13, 21), (31, 12), (43, 24), or (34, 42) are never (x, y) in a sequence in L(4, 4) henceforth. These, exclusions are accounted for in what follows.

23

CANCELLATION CONDITIONS

Given (x 1, y 1 )=(23, 31), take x 22 =1, so x 2 # [21, 41]. If x 2 =21 then y 2 # [12, 13], and if x 2 =41 then y 2 # [32, 14]. This yields four main branches: I

II

III

IV

x

y

x

y

x

y

x

y

23 21

32 12

23 41

32 32

23 41

32 14

23 21

32 12

.

I. Take x 31 =1, so x 3 # [12, 14]. If x 3 =12 then y 3 =31, and (21, 13) and (12, 31) violate P6. h If x 3 =14 then y 3 # [32, 41]. If y 3 =41 then (21, 13) and (14, 41) are almost balanced. h This leaves (x 3, y 3 )=(14, 32). Take x 41 =3, so x 4 # [33, 34]. If x 4 =34 then y 4 =43 and (23, 31) and (34, 43) are almost balanced. h If x 4 =33 then y 4 # [24, 42]. If y 4 =24 then (14, 32) and (33, 24) are almost balanced. h This leaves 23 21 14 33

31 13 32 42 .

Take x 52 =2, so x 5 # [12, 22]. If x 5 =12 then y 5 =31 and (33, 42) and (12, 31) are almost balanced. h If x 5 =22 then y 5 # [31, 13]. If y 5 =31 then (33, 42) and (22, 31) are almost balanced. h If y 5 =13 then (14, 32), (33, 42), (22, 13) is almost balanced. h II. Take x 31 =3, so x 3 # [33, 34]. If x 3 =34 then y 3 =43 and (23, 31) and (34, 43) are almost balanced. h If x 3 =33 then y 3 # [24, 42]: 23 31 41 32 33 24

or

23

31

41

32

33

42.

Suppose y 3 =24. Take y 42 =3, so y 4 # [13, 43]. If y 4 =13 then x 4 =21 and (33, 24) and (21, 13) are almost balanced. h If y 4 =43 then x 4 =34, and (23, 31) and (34, 43) are almost balanced. h Suppose y 3 =42. Take y 42 =3, so y 4 # [13, 43]. If y 4 =13 then x 4 =21, so (x 4, y 4 ) =(21, 13). Take y 51 =2, so y 5 # [22, 24]. There is no feasible x 5 for y 5 =22. h If y 5 =24 then x 5 =33 and (21, 13) and (33, 24) are almost balanced. h If y 4 =43 then x 4 # [34, 24]. If x 4 =34 then (23, 31) and (34, 43) are almost balanced. h This leaves (x 4, y 4 )=(24, 43). Take y 51 =2, so y 5 # [21, 22]. If y 5 =21 then x 5 =12 and (41, 32), (33, 42), (12, 21) is almost balanced. h If y 5 =22 then x 5 =13 and (24, 43) and (13, 22) are almost balanced. h III. Take x 31 =1, so x 3 # [12, 13]. If x 3 =12 then y 3 # [21, 31], and both cases of (x 3, y 3 ) give an almost balanced sequence in conjunction with (41, 14). h If x 3 =13 then y 3 # [22, 32]. For both y 3 take y 42 =3, so y 4 # [33, 43].

24

PETER C. FISHBURN

Suppose (x 3, y 3 )=(13, 22). If y 4 =33 then x 4 # [24, 42]. If x 4 =24 then (24, 33) and (13, 22) are almost balanced. h If x 4 =42, so (x 4, y 4 )=(42, 33), take x 51 =3. Then x 5 # [32, 34]. If x 5 =34 then y 5 =43, and (34, 43) and (23, 31) are almost balanced. h If x 5 =32 then y 5 # [14, 24]. If y 5 =14 then (32, 14) and (13, 22) are almost balanced. h If y 5 =24, so (x 5, y 5 )=(32, 24), take x 62 =4. Then x 6 =34, y 6 =43, and (34, 43) and (23, 31) are almost balanced. h Suppose (x 3, y 3 )=(13, 22) and y 4 =43. Then x 4 # [24, 34]. If x 4 =24 then (24, 43) and (13, 22) are almost balanced. h If x 4 =34 then (34, 43) and (23, 31) are almost balanced. h Suppose henceforth that (x 3, y 3 )=(13, 32). If y 4 =43 then x 4 # [24, 34]: (24 or 34, 43) and (41, 14) are almost balanced. h If y 4 =33 then x 4 # [24, 42]. In each x 4 case take x 51 =3. Then x 5 =34, y 5 =43, and (34, 43) and (23, 31) are almost balanced. h IV. Take x 31 =1, so x 3 # [13, 14]. If x 3 =13 then y 3 # [31, 22, 32]. If y 3 = 31 then (13, 31) and (21, 12) are almost balanced. h If y 3 =22, so (x 3, y 3 )= (13, 22), take y 42 =3. Then y 4 # [33, 43]. If y 4 =33 then x 4 # [24, 42] : (24, 33) and (23, 31) are almost balanced, and (42, 33), (21, 12), (13, 22) is almost balanced. h If y 4 =43 then x 4 # [24, 34] : (24, 43) and (13, 22) are almost balanced, and (34, 43) and (23, 31) are almost balanced. h This leaves y 3 =32 for x 3 =13: 23 31 21 12 13 32. Take y 42 =3, so y 4 # [33, 43]. If y 4 =33 then x 4 # [24, 42], and if y 4 =43 then x 4 # [24, 34]. If (x 4, y 4 )=(34, 43) then (34, 43) and (23, 31) are almost balanced. h The other three cases for (x 4, y 4 ) are 23 21 13 24

31 12 32 33

23 21 13 42

31 12 32 33

23 21 13 24

31 12 32 43

In each take x 51 =3. Then x 5 =34, y 5 =43, and (34, 43) and (23, 31) are almost balanced in the first two cases. h For the third, x 5 # [33, 34]. If x 5 =34 then y 5 =43, and (23, 31) gives almost balance. h If x 5 =33 then y 5 =42, so (x 5, y 5 )= (33, 42). Take x 62 =2. Then x 6 =22, y 6 =31, and (22, 31) and (33, 42) are almost balanced. h Suppose henceforth that x 3 =14. Then y 3 # [32, 41]. If y 3 =41 then (14, 41) and (21, 12) are almost balanced. h Assume (x 3, y 3 )=(14, 32) henceforth. Take x 42 =2. Then x 4 # [22, 42]. If x 4 =22 then y 4 # [13, 31], and if x 4 =42 then y 4 # [24, 33, 34].

25

CANCELLATION CONDITIONS

If (x 4, y 4 )=(22, 13), this pair is almost balanced with (14, 32), and if (x 4, y 4 )= (42, 24), this pair is almost balanced with (21, 12). h This leaves three cases for (x 4, y 4 ): 23 21 14 22

31 12 32 31

23 21 14 42

31 12 32 33

23 21 14 42

31 12 32 34

Take x 51 =3 in each case. The first case allows x 5 # [33, 34]. If x 5 =33, then y 5 # [24, 42] : (33, 24) and (14, 32) are almost balanced, and (33, 42) and (22, 31) are almost balanced. h If x 5 =34 for the first case, then y 5 =43, and (34, 43) and (23, 31) are almost balanced. h The second case requires x 5 =34, then y 5 =43 with (34, 43) and (23, 31) almost balanced. h The third case requires x 5 =33, then y 5 =24 with (33, 24) and (14, 32) almost balanced. h This completes the processing of R10.

6. DISCUSSION

The main purpose of this paper has been to assess the maximum possible K for which a weak order  on finite X=X 1 _X 2 can satisfy cancellation conditions C(2), ..., C(K&1) but violate C(K). For m= |X 1 | and n= |X 2 |, the maximum K, denoted by f (m, n), is approximately m+n so long as min[m, n]3. Exact values of f (3, 3)=3 and f (4, 4)=4 were noted along with other results for two dimensions. For the more general case of N2 dimensions with X=X 1 _X 2 _ } } } _X N and n i = |X i | for each i, we recalled the upper bound of 7n i &(N&1) on f (n 1 , n 2 , ..., n N ) along with f (2, 2, ..., 2)N&1 for N5 binary factors. I am aware of no other general results for N3, so this is a prime topic for further work. Several interesting open questions also remain for N=2, including exact values of f (m, n) for small (m, n) such as (3, 5), (4, 5), and (5, 5), a better lower bound on f(m, n) for fixed m # [4, 6, 7, ...] than is given in Theorem 2.7, and a slightly smaller upper bound than m+n&1 from Theorem 2.1. The difficulties of proving exact values and of reducing the upper bound suggest that new mathematical insights are needed to make headway on these problems.

REFERENCES Adams, E. W. (1965). Elements of a theory of inexact measurement. Philosophy of Science, 32, 205228. Arbuckle, J., 6 Larimer, J. (1976). The number of two-way tables satisfying certain additivity axioms. Journal of Mathematical Psychology, 13, 89100. Blaschke, W. (1928). Topologische Fragen der Differential-Geometrie. I. Thomsens Sechseckgewebe. Zueinander diagonale Netze. Mathematische Zeitschrift, 28, 150157. Debreu, G. (1960). Topological methods in cardinal utility theory. In K. J. Arrow, S. Karlin, 6 P. Suppes (Eds.), Mathematical methods in the social sciences, 1959, (pp. 1626). Stanford, CA: Stanford Univ. Press.

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