Separable Preference Orders

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Separable Preference Orders Separable Preference Orders

Jonathan K. Hodge Western Michigan University

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SEPARABLE PREFERENCE ORDERS

byJonathan K. Hodge

A Dissertation Submitted to the

Faculty of The Graduate College in partial fulfillment of the

requirements for the Degree of Doctor of Philosophy

Department of Mathematics

Western Michigan University Kalam6izoo, Michigan

August 2002

Reproduced with permission of the copyright owner. Further reproduction prohibited without permission.

SEPARABLE PREFERENCE ORDERS

Jonathan K. Hedge, Ph.D.

Western Michigan University, 2002

Whenever a decision-maker must express simultaneously his or her preferences on

several possibly related issues, the existence of interdependence among these preferences

cem lezid to collective decisions that are unsatisfactory or even paradoxical. Intuitively,

an individual’s preferences me said to be separable on a subset of issues if they do not

depend on the choice of alternatives for issues outside the subset. Here we explore from a

mathematical standpoint the properties of separable and nonseparable preference orders.

We begin by formulating a general model of multidimensional preferences and we

formally introduce the notions of separability and noninfiuentiality. We study the structure

of interdependent preferences and explore connections to the previous notions of separability

studied in economics.

Next, we examine some of the properties and constructions related to separability. We

consider lexicographic and additive orders and use simple tools from set theory to study k-

majority aggregation of separable preferences. We show that, in contrast to famous results

such as Condorcet’s Voting Paradox, the property of separability is preserved by this natural

aggregation scheme.

We address the problem of enumerating separable preference orders and introduce the

notions of monoseparable, symmetric, preseparable and strongly preseparable preferences.

Counting formulas are developed for the latter three of these classes.

Finally, we consider the action of the symmetric group on the set of binary preference

orders. We characterize the group of symmetry-preserving permutations and the group of

separability-preserving permutations, providing useful insights into the extreme sensitivity

to small changes exhibited by separable preferences.


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© 2002 Jonathan K. Hodge


Acknowledgements

This dissertation is the culmination of an educational journey that would not have

been possible without the continual support and encouragement of family, friends, and

faculty. To all who have lent a helping hand or a listening ear, I am deeply appreciative.

I especially wish to acknowledge Dr. James Bradley, who introduced me to the joy

of mathematics; Dr. Arthur White, who persuaded me to keep pursuing my dreams, even

when it wasn’t pleasant; and Dr. Allen Schwenk, who was brave enough to venture into

unknown territory to help bring my goals to fruition. I could not have asked for a more

caring or supportive advisor and I am grateful for all that I have learned from you. Thanks

also to the other members of my committee, Dr. Clifton Ealy and Dr. Ping Zhang.

Many thanks to my family for their unfailing love and support. To my parents, for

believing in me and selflessly devoting their lives to me. To Jen and Brian, for many good

conversations and friendship over the years. To the extended Dauscher clan: it has been

wonderful sharing the excitement of the last few years with you. And last, but certainly

not least, to Melissa, who has made my life immeasurably rich and has never ceased to love

and encourage me. I could not have done any of this without you.

Finally, I would be remiss if I did not acknowledge the Creator of mathematics and

of all things for allowing me to take credit for some of His theorems.

m


Contents

Acknowledgements iii

1 Introduction 1

2 Preference Relations and Separability 6

2.1 Multidimensional Preferences............................................................................ 6

2.2 Separability.......................................................................................................... 7

2.3 Noninfiuentiality................................................................................................. 11

2.4 Set Theoretic P roperties..................................................................................... 12

2.5 An Application to Referendum Elections......................................................... 14

3 Properties Related To Separability 16

3.1 Basic Definitions and Properties ..................................................................... 16

3.2 Unions, Intersections, and Aggregation............................................................ 18

3.3 Lexicographic Sums ........................................................................................... 20

3.4 Additive O rders.................................................................................................... 24

4 Preseparable Orders 28

4.1 Preseparable Extensions.................................................................................... 28

4.2 Counting Binary Preseparable O r d e r s ............................................................ 31

4.3 Symmetric Extensions....................................................................................... 33

4.4 Counting Strongly Preseparable Orders ......................................................... 38

4.5 Other Combinatorial R e s u l ts ........................................................................... 42

4.6 Summary of Related P roperties........................................................................ 43

IV


Contents

5 Separability and The Symmetric Group 45

5.1 The Canonical A c t io n ........................................................................................ 45

5.2 Symmetry-Preserving Permutations................................................................... 46

5.3 Separability-Preserving Permutations................................................................ 50

5.4 The Structure of ........................................................................................... 53

6 Questions for Future Research 69

A Computer Code and Output 70

A.l The Separable Group C a lcu la to r...................................................................... 70

A.2 The Crosstab Q u e r y ........................................................................................... 78


Chapter 1

Introduction

When making decisions that depend on several possibly related criteria, individuals are of

ten required to express simultaneously their preferences on these criteria. A classic example

is the referendum election, in which a voter must choose between a yes or no vote for each

question or proposal. The predominant method of aggregating votes in such an election is

simultaneous voting, in which voters cast ballots for all issues at the same time and votes

are counted on an issue-by-issue basis. Thus, under simultaneous voting, a vote of YN in

an election on two issues would count as a yes vote for the first issue and a no vote for

the second issue. While this method of aggregation seems straightforward, it is not without

fiaws. To the contrary, whenever even a single voter’s preferences on one issue depend on

the outcome of another issue, election results can occur that are at best unsatisfactory and

at worst paradoxical (see [3], [4], [16] for examples).

At the heart of this paradoxical behavior lies the concept of separability, which has

been studied for many years by economists, politiceJ scientists, and mathematicieins (Kil-

gour 6md Bradley [14] provide a summary of this history). Intuitively, an individual’s pref

erences are said to be separable on a subset of issues if they do not depend on the choice

of alternatives for issues outside the subset. Thus, while preferences among several brands

of orange juice may be separable with respect to choices between competing teams in the

Superbowl, preferences on two closely related millage proposals in an upcoming election

may not be so clear-cut. For example, some voters may be in favor of both proposals but

prefer that only one pass, so as not to place an unreasonable financial burden on taxpay-

1


Chapter 1. Introduction

ers. Thus, if they knew the outcome of the election on one of the issues beforehand, they

would vote the opposite of that outcome on the remaining issue. Unfortunately, because

their preferences are nonseparable, they are doomed when voting day comes around. Since

simultaneous voting provides no adequate mechanism for expressing interdependent pref

erences, the votes they cast with the intention of producing a desirable election outcome

could just as easily contribute to an outcome which is far from desirable.

To illustrate this phenomenon, consider the following example: Suppose that, in a very

democratic family. Mom, Dad, and Ralphie are trying to finalize the details of an upcoming

automobile purchase. Due to Ralphie’s unfortunate performance in driving school, both of

their current vehicles are inoperable. Thus, they need to purchase two new cars, but are

having some trouble deciding which two cars to buy. After narrowing the competition

down to three choices, they agree to hold a referendum election to make the final decision.

Confident in their intent to promote democracy at the most basic levels of government,

they meet together on family night and vote on the following three questions:

• Question 1: Should we buy the BMW?

• Question 2: Should we buy the Ford?

• Question 3: Should we buy the Kia?

The ballots are cast without incident, and yet when the votes are tallied, something

seems terribly amiss. The outcome of the election is that all three questions pass, each by

a margin of 2 votes to 1! Dumbstruck by this strange outcome, but compelled by their

commitment to the democratic process, they agree to abide by the result of the election

and purchase all three cars. Unfortunately, democracy does not serve them well (at least

not in this manifestation). A few months pass, bills begin to accumulate, and Mom and

Dad finally realize that, even with Ralphie working overtime at the Burger Hut, they can

no longer afford to mzike the payments on their three new cars. Creditors come knocking

at their door, the cars get repossessed, and the happy democratic family finds themselves

in a situation that they would have never anticipated! Not only is Dad forced to endure

the humiliation of riding to work each day on his son’s tricycle, but, to top it all off, the

family’s credit has been badly damaged. They all agree that it would have been better if



they had bought no cars at all! In fact, anything would have been better than the situation

that transpired.

Determined to find out exactly what led to this most unfortunate situation, Mom,

Dad, and Ralphie begin to discuss how they voted. Dad is dismayed: he is certain that

his once perfect family has become corrupt and that someone must have greedily voted

for all three cars. He demands full disclosure of the ballots but is surprised to discover

that, contrary to his suspicions, each member of his family voted yes on exactly two of

the questions. He voted to buy the BMW and the Ford, his dear wife voted to buy the

BMW (for herself) and the Kia (for him), and Ralphie voted to buy the Ford and the Kia

(quite cleverly equating lower car payments with a higher allowance). Stupefied, Dad cries

out in frustration, “If I had known that you two were going to vote that way, I wouldn’t

have voted the way I did! I would have done anything to avoid the mess we’re in now!”

And in that very moment, he quite unintentionally stumbles upon a classic paradox of

nonseparable preferences and referendum elections. He realizes that the real trouble was

not with his family at all, but rather with a subtle, yet fundamental fiaw in the referendum

election itself. Because his family’s preferences on each of the questions in the election

were dependent on the outcomes of the other questions, and because they were required to

vote on all three questions simultaneously, they had been provided no adequate means for

expressing their true desires. As a consequence, the votes they cast with the intention of

producing a fair and democratic resolution to their dispute in fact led to the worst possible

outcome.

Now we admit that this example is a bit silly and contrived, but it is not too hard

to imagine how this same type of voting behavior, and its unfortunate consequences, could

arise in a much more serious context. The fact is that nonseparable preferences wreak

havoc on referendum elections and the decision-makers who rely on them to democraticzilly

solve important real-world problems. In recent years, researchers have begun to pay careful

attention to this troublesome issue and have made efforts to better understand nonseparable

preferences and their effects on the decision-making process.

Unfortunately, research on alternate voting methods, such as sequential voting (see

[14]), has been unable to shed much light on the separability problem. If any conclusion



has been made, it is that the best kind of preferences to have are those that are separable.

Furthermore, Kilgour and Bradley [14] suggest that “one way out of the difficulty [of non

separable preferences] may be to frame questions so as to avoid preference nonseparability.”

In order to do this, we must first understand the structure of separable and nonseparable

preferences both from a theoretical standpoint and as they occur in real-world situations.

Mathematics is a natural setting in which to cultivate understemding. By developing

reasonable preference models and studying the mathematical properties of these models,

we accomplish two goals. First, we build a theoretical foundation upon which future re

searchers can better understand the problem of nonseparablilty and work toward meaningful

and applicable solutions. Second, we uncover a surprisingly deep and interesting body of

mathematics that is worth studying in its own right.

With that in mind, the goal of this dissertation is to provide a mathematical view of

the interesting objects that we call separable preference orders. We do not claim to provide

a complete treatment of this emerging area of study. Indeed, there are many questions

that remain to be answered, and we have included several ideas for future research both

throughout the text in the concluding chapter. We hope that our explorations here will

provide a substantive starting point for future investigations.

We begin in Chapter 2 by formulating a general model of multidimensional preferences

and we formally introduce the notions of separability and noninfiuentiality. We study the

structure of interdependent preferences and explore connections to the previous notions of

separability studied in economics.

Next, we examine in Chapter 3 some of the properties and constructions related

to separability. We consider lexicographic and additive orders and use simple tools from

set theory to study A-majority aggregation of separable preferences. We show that, in

contrast to famous results such as Condorcet’s Voting Paradox, the property of separability

is preserved by this natural aggregation scheme.

In Chapter 4, we begin to focus our attention on discrete, and specifically binary, pref

erences. We consider the problem of enumerating separable preference orders, introducing

the notions of monoseparable, symmetric, preseparable and strongly preseparable preferences

and exhibit counting formulas for the latter three classes.



Finally, in Chapter 5, we consider the action of the symmetric group on the set of

binary preference orders. We characterize the group of symmetry-preserving permutations

and the group of separability-preserving permutations, providing useful insights into the

extreme sensitivity to small chzinges exhibited by separable preferences.


Chapter 2

Preference Relations and

Separability

2.1 Multidimensional Preferences

We consider the individual preferences associated with a general multistage or multiple-

criteria decision making process. We assume that there is a criteria set Q whose elements

will be called criteria and upon which we impose no particular structure. For each q G Q,

let Xg denote the alternative set for criteria q. Then an alternative is an element of the

Cartesian product set

X q = Y [ Xg.?6Q

For ease of notation, if x 6 X q and S C Q , then we denote by 15 the projection of x

onto 5, i.e.

Xs = (^g)qes 6 X s,

where X s is defined in a manner similar to X q above. Furthermore, if T’ = {5i, 52 , . . . , 5„}

is a partition of Q, then we often write x = (z s ,,z % ,...,x g ^ ) , reordering the criteria if

necessary.* Specifically, we use the notation - 5 to denote the complement of 5 in Q and

write X = (x s ,x -s ) .

* We allow the parts of P to be empty, in which case we take the notation x = (xs,, , • • •, xs„ )to mean x = (xs.^, xg.^, . . . , xj;^ ), where { :* :!< & < m} is the set of indices corresponding to

the nonempty parts of P.


Chapter 2. Preference Relations and Separability

To allow for the most general treatment of preference interdependence, we represent a

decision-maker’s preferences by an eirbitrary binary relation X on Xq.^ Historically, it has

been common to assume that X is a weak order or a strict partial order (see [8]), though

we presently impose no such restrictions.^ For a relation X on X q , we let >- be the relation

defined by x >- y <=>■ x > y and x ^^ y .

Example 2.1.1. Let Q be a finite set representing a number of offices to be filled in simul

taneous multi-candidate, single-winner elections. Then for each q e Q, Xg is a finite set

representing the candidates for office q. In this context, X represents the voter’s preference

between election outcomes differing on the choice of candidates for one or more offices.

Example 2 .1.2 . Let Q = {flour, sugar,butter} and let Xg represent the amounts of in

gredient q that a baker may put into his cookie recipe. Then each X g is an interval of

real numbers and y represents the taster’s preference between cookies made with differing

combinations of ingredients.

2.2 Separability

Intuitively, a decision-maker’s preferences on a subset S of the criteria set are said to be

separable if they do not depend on the choice of alternatives for criteria outside of S.

Another way of expressing this idea is to say that the decision-maker’s preferences on S

are invariant with respect to a chcuige of alternatives on the criteria in —S. Formally,

Definition 2.2.1. Let S C Q. Then S is said to be y-separable, or separable with respect

to y , if whenever two elements x s ,y s E X s have the property that (x s ,u -s ) b iy s ,u -s )

for some u - s 6 X - s , then X (j/s,u-s) for all u_s € A criterion g G Q is

said to be ^-separable if {q} is X-separable. The relation y is said to be separable if each

nonempty S C Q is ^-separable.

In addition to Definition 2.2.1, we adopt the convention that both Q and 0 are always

^We will occasionally (as in Example 2.2.11) have reason to restrict ^ and consider only a subset

of X q , which we will denote X q . In this context, we may replace Xs with X^ in all definitions

and theorems, where = {xs : x €*A relation y on X q is a weak order if it is transitive, reflexive and has the property that for

each X, y 6 X q , either x ^ y or y ^ x. It is a strict partial order if and only if it is transitive and

irreflexive (x ^ x for each x € X q ). See [15] for details.



^-separable. We also assume, unless otherwise indicated, that all subsets under considera

tion cire nonempty and proper.

Our definition of separability is based on the definition given by Kilgour and Bradley

in [14], which is originally due to Yu [21]. Kilgour and Bradley also note that the first

study of separability within the context of referendum elections seems to be a 1977 paper

by Schwartz [19]. According to economist W.M. Gorman, the earliest notions of separability

can be traced back to Leontief in 1948.

We should note that, until 1998, research on separable preferences in referendum

elections focused exclusively on the separability of individual criteria. Hodge and Bradley

[12] seem to have been the first to apply Yu’s more general definition to this particular

setting. To distinguish the single-criterion formulation of separability from the general

version above, we make the following definition:

Definition 2.2.2. A relation ^ on X q is said to be monoseparable if each q £ Q is

^-separable.

Yet another way of expressing the separability of a subset S with respect to X is to

say that the relation on Xg induced by X is well-defined; that is, it does not depend on

the choice of alternatives for criteria in - S . Formally, for each S Ç Q, let ^ 5 denote the

relation on X g defined by

Xg h s ys *=> {xg ,u-g) y {yg,u-g) for all u -g G X -g.^

Similarly, let denote the relation defined by

Xg yg (xg ,u -g) ^ (yg ,u-g) for all u -g € X -g .

It is important to note that does not denote the negation of X ,. Rather, it denotes

the relation induced on Xg by Rephrasing Definition 2.2.1, we see that if S is separable

and (xg ,u -g ) X {yg,u-g) for some xg, yg 6 Xg and some u -g 6 X -g , then it must be

the case that xg Vj yg. Thus,

Proposition 2.2.3. S C Q is separable if and only if, for each xg, yg € Xg, either xg ys

or Xg ys-

‘‘For ease of notation, we occasionally omit the subscript S on the induced relation ^s, relying on the context to clarify any ambiguity.



Proof. Suppose 5 is separable and let xs, ys 6 X s be given. If it is not the case that

x s ) t s ys, then there exists u - s € X - s such that (x s ,u -s ) b (î/SjW-s)- But then, by

our above comments, it follows that x s b s 2/s- The converse follows immediately from

Definition 2.2.1. □

Corollary 2.2.4. S Ç Q is ^-separable if and only if S is ^-separable.

Proof This follows immediately by noticing that ^ and ^ are interchzingeable in the state

ment of Proposition 2.2.3. □

Corollary 2.2.5. y is separable if and only if ^ is separable.

Proposition 2.2.6. Let y be a relation on X q and let S C Q be y-separable. I f T c S i s

y-separable, then T is yg-separable and (h s)r =

Proof Let T c 5 and suppose that ( x t , u s - t ) h s (2/r, u s - r ) for some x t , yr E X t

and some u s - t E X s - t - Let v s - t E X s - t be given. Then by the definition of Xg,

{ x T j U s - T f W - s ) h i y T , u s - T , w - s ) for all w - s E X - $ . But since T is X-separable, it

follows that { x t j V s - T j W - s ) X ( j/r ,u s - t , «^-s) for all w - s E X - s - Thus { x t , v s - t ) h s

{ y T , v s - T ) and, consequently, T is Xg-separable. To show (Xg)^ = X,., we observe that

XT (Xg)T y r {x t , u s - t ) Xg (yr,us_T) for all u s - t E X s - t

4=> (x tius-T ,w _s) X { y T , u s - T , v - s ) for all u s - t E X s - t , v - s E X - s

<=> {x t , w - t ) X ( y T , u j - T ) for all w - t E X - t

XT h r 2/r- □

Corollary 2.2.7. Let y be a separable relation on X q and let S C Q. Then Xg is a separable

relation on Xs-

Proof. This is 8ui immediate consequence of Proposition 2.2.6 □

Proposition 2.2.8. Equality is a separable relation on any alternative set.

Proof Let S C Q be given and suppose that {x s ,u -s ) = (ys,y—s) for some xs, ys E X s

and some u -s E X -s - Then xs = ys and so (x s ,v -s ) = (,ys,v-s) for all v - s E X -s .

Since our choice of 5 was arbitrary, it follows that = is separable. □

Corollary 2.2.9. Inequality is a separable relation on any alternative space.

Proof This follows directly from Corollary 2.2.5 and Proposition 2.2.8. □


Chapter 2. Preference Relations and Separability 10

Proposition 2.2.10. A separable relation X on X q is either reflexive or irreflexive.

Proof. Let x, y ^ X q and suppose that x > x. Now choose any subset S of Q. Since

( x s , x - s ) h i x s , x - s ) and S is ^-separable, it follows that y ( x s , y - s ) - But

since - S is also X-separable, it follows that { y s , y s ) h (j/S iÿ-s)- We have thus shown

that i i x y X for some x 6 X q , then j/ ^ y for all p € X q , which establishes our claim. □

Example 2.2.11. Let Q = R (the real line) and for each q e Q, let Xq = K. Then X q is

the set of all real-valued functions of a single real vziriable. Let X q be the subset of all

Lebesgue integrable functions and let y be the weak order on X q specified by

f t 9 [ f > f 9Jr Jr

Now let 5 Ç R be measurable and let F = (/s , h -s ) , G = {gs, h -s ) for some f s , gs € Xg

and some h - s 6 X ^s- Suppose also that F y G . Then

f f s + f h - s = f F > [ G - f 9s + f h -s Js J~s Jr Jr Js J - s

which implies that

93

from which it follows that

f f s + [ h - s > f g s + f k - s Js J - s Js J - s

for every k - s E Thus, S is separable. Since our choice of S was arbitrary, we see that

every measurable subset of the real numbers is separable with respect to the weak order

on X q induced by the Lebesgue integral.

Example 2.2.12. Let Q = {1,2,. . . ,n } cind let = {0,1} for each 1 < i < n. Then

a relation y on X q may be interpreted as an ordering of the 2" possible outcomes of

a referendum election on n questions, where a 1 in the i*'* component of an alternative

indicates passage of the i" ' issue and a 0 indicates failure. Specifically, let n = 3 and let >-

be the linear order on X q specified by the binary preference matrix

R —

A0

0

\

1

0

0

V

A1

1

1

0

0

0

0/



where i >- y if and only if the row of R corresponding to x appears higher than the row

corresponding to y. It is easy to see that {3} is separable with induced ordering 1 >- 0. In

contrast, {1} is nonseparable since 101 >- 001 but O il >- 111. Similarly, {2} is nonsepeirable

since 101 X 111 but O il X 001. Nevertheless, {1,2} is separable with induced ordering

10 X 01 X 00 X 11. Thus, we see that it is possible for a set to be separable even if none of

its proper subsets is.

2.3 Noninfluentîalîty

At this point, we hope that the reader has a decent intuitive understanding of the idea of

separability. Specifically, we hope that the reader sees that if a subset S of Q is nonseparable

with respect to some decision-maker’s preference, then the outcome of the decision-making

process on criteria outside of 5 exerts some sort of influence on the decision-maker’s prefer

ence on 5. In this situation, it seems logical to ask: On which criteria do the decision-maker’s

preferences on S depend?

We answer this question by identifying the subsets of Q on which the decision-maker’s

preferences on 5 are not dependent, i.e. those subsets for which a change of alternatives

would not influence the decision-maker’s preferences on 5. Formally,

Definition 2.3.1. Let S and T be disjoint subsets of Q. Then T is said to be y-noninfiuential

on S, or noninfluential on S with respect to X, if whenever two elements xs, ys e X s have

the property that (x s ,« t,îü -(su t)) h {ys,UT,w-^suT)) for some (ut,w_(suT)) E X -$ ,

then (xs,UT,io-(sur)) h iys,VT,‘u^-{SuT)) for all v t 6 X t- If T is X-noninfluential on

S for each nonempty S Ç Q with S C\T = 0, then T is said to be y -noninfluential, or

noninfluential with respect t o y . k criterion q € Q is said to be X-noninfluential if {g} is

X-noninfluential.

As in the definition of separability, we adopt the convention that both Q cind 0 are

always X-noninfluentied.

Example 2.3.2. Consider again the preference matrix from Exaunple 2.2.12. Notice that

{2} is influential on {1} since the induced preference on {1} depends on the choice of

alternatives on {2} (specifically, 1 x 0 given a choice of 0 on {2} and 0 x 1 given a choice

of 1 on {2}). Nevertheless, {3} is noninfluential on {1} since this induced preference is

independent of the choice of alternatives on {3}. Similarly, {3} is noninfluential on {2} and



on {1,2}, and so it follows that {3} is noninfluential. We can show in the same manner

that {1, 2} is noninfluential.

Lemma 2.3.3. Let S and T be disjoint subsets of Q.

(i) I f T is noninfluential on S , then T is noninfluential on each S ' Ç 5.

(ii) T is noninfluential if and only i fT is noninfluential on —T.

Proof, (i) Suppose that T is noninfluential on S and let 5 ' Ç S be given. Let xs ',y s ' E Xs>

and ( î u s - S ' , w t , u ' - ( s u T )) E X-s> be such that

( x s ‘ ,VJS-S' ,Ut ,W-^SUT)) b iyS’ ,WS-S',U 'T,‘W-{SuT))

and let vt E X t be given. Since T is noninfluential on 5 and (x s>,ujs- s '), {ys ',w s-s ') E X s,

it follows that

(xs> ,W S -S ’ ,Vt ,W -(S uT)) t . {yS ',‘Ws-S',VT,W-[SuT))-

Since our choice of vt was arbitrary, we have established that T is noninfluential on S'.

(ii) The forward implication follows immediately from Deflnition 2.3.1. For the con

verse, suppose that T is noninfluential on - T and let 5 Ç Q be disjoint from T. Then

S Ç - T , and so (i) implies that T is noninfluential on 5. Since our choice of S was arbi

trary, it follows that T is noninfluential. □

Proposition 2.3.4. Let S C Q . Then S is separable if and only if —S is noninfluential.

Proof. By Lemma 2.3.3 above, it suffices to show that S is separable if and only if —S is

noninfluential on S. This follows immediately from Definitions 2.2.1 and 2.3.1. □

2.4 Set Theoretic Properties

In the past, economists such as W.M. Gorman have studied the notion of separability from

within the context of utility theory. A question that was of interest to them, and will be

of interest to us, is that of whether the property of separability is preserved by certain set

operations. In order to consider previous results pertinent to this question, we must first

introduce some terminology.

Let S ,T Ç Q . Then we say that 5 and T overlap if all of 5 D T, 5 - T, and T - 5 are

nonempty. S is said to be essential with respect to V if there exists u -s € X - s and xs,



ys E X s such that (x s ,u -s ) X (yg,u_g), S is said to be strictly essential with respect to y

if, for all U -s € X-s> there exist x s , ys G X s such that (x s ,u -s ) h {vSjU-s )- We denote

by S A T the symmetric difference of S and T, defined precisely as S A T = (S —T )\J{T -S ).

Finally, we denote by 5^ the collection of all ^-separable subsets of Q and by the

collection of all ^-noninfluential subsets of Q.

The following theorem is due to Gorman [10].

Theorem 2.4.1. Suppose X is a continuous^ weak order on X q , and that for each q E Q,

Xq is topologically separable and arc-connected. Suppose further that S ,T C Q overlap and

that either S — T or T — S is strictly essential. I f S and T are y-separable, then 5 U T,

5 n r , 5 — r , T - 5 , and S A T are all y-separable and strictly essential.

Kilgour and Bradley [14] demonstrate that when the alternative sets are not arc-

connected (as is the case in a referendum election), Gorman’s Theorem may fail. Specifically,

they provide an example in which both S and T are separable and yet none of 5UT, 5 - T ,

T — S, or 5 A T is separable. It seems that the only portion of Gorman’s Theorem that

holds in general is the following:

Proposition 2.4.2. Sy is closed under finite intersections.

Proof. Suppose Si,5a 6 Sy, and let S = Si n Sg, Ti = Si - S, %2 = Sg - S, and

T = - (S i U S2). Suppose also that { x s ,u t i ,u t„ u t) h {vs,UTi,UTi,UT) for some u 6

X - s . Choose V e X -s - Since Si is separable and (u?^, u r) E X -s^ , it follows that

{ x s ,u t i ,v t„ u t ) y (2/s,wr,,UT2,nx). But since Sg is separable 2ind (ut’, ,u t ) E X -s , ,

we have that ( x s ,v t i ,v t 2 ,v t) h (ys,VTi,VT2 ,VT)- Because our choice of v was arbitrary,

it follows that S = Si n Sj is separable. Thus, Sy is closed under the binary operation of

set intersection. An easy induction argument then shows that Sy is closed under all finite

intersections. □

Corollary 2.4.3. Afy is closed under finite unions.

Proof. Let Si, S2, . . . ,S„ € N'y. Then by Proposition 2.3.4, - S i , - S 2, . . . , -S „ E Sy. But

by De’Morgan’s Laws and Proposition 2.4.2, it follows that

—(Si u S2 u . . . uS n) = —Si n —S2 n . . . n —S» g Sy.

Thus, by Proposition 2.3.4, Si U S2 U . . . U S« E N y . □

is con tin u o u s if t h e se ts { y € X q : r X y} a n d {y G X q : y t . x } a re c losed for each x G X q .



Given Proposition 2.4.2 and Corollary 2.4.3, Sy and Afy can be viewed as monoids®

under the binary operations of set intersection and set union, respectively. Consequently,

Proposition 2.4.4. The map tp :S y My defined by (p{S) = - S is a monoid isomorphism.

In particular, S y = M y .

Proof. Proposition 2.3.4 establishes that <p is well-defined. By De’Morgan’s Laws,

vp(5i n Sï) = - ( 5 i n S 2 ) = - S i U -5 2 = y)(5i) U tp{S2 ),

and so y is a homomorphism. Now (p is surjective since <p{—S) = 5 for all 5 Ç Q. Further

more, ip is injective since y (5 i) = ip{S2 ) <=# —S\ = - S 2 ■<=>• Si = 8 2 - Thus, p is an

isomorphism, as desired. □

We close this section by mentioning two open questions that are closely related to the

above results. The first concerns the extent to which Gorman’s Theorem may fail and the

second deals with a potential generalization of Proposition 2.4.2.

Open Question, (due to Allen Schwenk) Let X q be finite and let S be a subset ofV{Q)

that contains both Q and 0 and that is closed under finite intersections. Does there exist a

relation X on X q such that A Ç Q is separable if and only i f A Ç S?

Open Question. Is S y closed under arbitrary (i.e. infinite) intersections?

2.5 An Application to Referendum Elections

Proposition 2.5.1. Suppose Q is finite. Then X is separable if and only if each q S Q is

noninfluential.

Proof. Suppose that X is separable. Then for each g € Q, -{g} is separable. Thus, by

Proposition 2.3.4, g is noninfluential. For the converse, suppose that each q e Q is nonin-

fiuential and let S Ç Q be given. Then

- 5 = U {g}.q^S

Since each {g} is noninfluential and the union is taken over a finite number of sets, Corollary

2.4.3 implies that —5 is noninfluential. Thus, by Proposition 2.3.4, 5 is separable. Since

our choice of 5 was arbitrary, it follows that X is separable. □

monoid is a set 5 together with an associative binary operation * such that 5 contains an

element e, called an identity, for which n ♦ e = e ♦ a = a for all a € 5.



Proposition 2.5.1 provides a mechanism for identifying whether or not a voter’s prefer

ences are separable. This method could be implemented through a simple telephone survey

by asking questions such as: “If you knew beforehand the outcome of the election on pro

posal X, would that affect your preferences on the outcome of one or more of the other

proposals in the election?” If a respondent answers no to all such questions (one for each

proposal in the referendum), then we can state conclusively that his or her preferences are

separable. If a respondent answers yes to one of the questions, then follow-up questions

could be asked to help pinpoint the occurrence of nonsepmability and influentiality. We

note that, for a referendum election on n proposals, the method of Proposition 2.5.1 re

quires only n questions to be asked of the voter. The responses to these n questions can

provide information about all 2" subsets of Q.


Chapter 3

Properties Related To Separability

3.1 Basic Definitions and Properties

We pause now to formalize some of the notation introduced in the previous chapter.

Definition 3.1.1. Let X q be a countable alternative space and let >- be a linear order on

X q . The sequence (i*) of elements of X q defined by

Xi y Xj <=> i < j

is called the order sequence corresponding to X. Equivalently, >- is said to be the linear

order corresponding to (xjt). The element xi is said the be the leader of >-.

In certain settings, we are primarily interested in the case where preferences are binary

in nature, i.e. where Xg = {0,1} for each g 6 Q. We distinguish this context from other

more general settings by saying that the alternative space X q is binary. A specific case that

is of particular interest to us is when Q is finite (say |Q| = n), X q is binary, and preference

is represented by a linear order >- on X q (this is the model historically used in the context

of referendum elections, as in [2], [3], [4], [12], [14], [16]). In this case, we say that X is a

binary preference order on Q. The binary preference order >- is said to be normalized if its

leader is (1, 1, . . . , 1) and if (1, 0 , 0 , . . . , 0) >- (0 , 1, 0 , . . . , 0) >- . . . V (0, 0, . . . , 0 , 1).

We will often use the binary preference matrix model of Example 2.2.12, defined

formally below, as a convenient way of representing binary preference orders. It is clear

that the natural correspondence between binary preference orders and binary preference

matrices is bijective.

16


Chapter 3. Properties Related To Separability

Definition 3.1.2. Let Q = { 1 ,2 ,..., n}, let X q be binary, and let >- be a linear order on

X q with order sequence ( n ) . Then the 2" x n matrix (o ij) having the Oij entry equal to

the component of z,, i.e.

Oj.j =

is called the binary preference matrix corresponding to >-. Equivalently, >- is smd to be the

linear order corresponding to (ojj).

The binary preference matrix (oij) is said to be normalized if its corresponding lineztr

order is normalized. It is clear that every binary preference order can be obtained from

a normalized binary preference order by reordering the criteria and replacing a subset

of the components of each alternative with their bitwise complements. Thus, every binary

preference matrix can be obtained from a normalized binary preference matrix by permuting

and/or taking bitwise complements of the columns.

The following result is due to Bradley and Kilgour [2].

Proposition 3.1.3.

(i) I f IQI = 2, then there is exactly 1 separable normalized binary preference order on

Q.

(ii) I f \Q\ = 3, then there are exactly 2 separable normalized binary preference orders

on Q.

(iii) I f \Q\ = 4, then there are exactly 14 separable normalized binary preference orders

on Q.

The binary preference matrices corresponding to the orders in Proposition 3.1.3 are

listed below.

\Q\ = 2:

17

/ 1 1

1 0

0 1

0 0


Chapter 3. Properties Related To Separability 18

IQI = 3:

^ 1 1 f i 11 1 0 1 1 01 0 1 1 0 11 0 0 0 1 10 1 1 1 0 00 1 0 0 1 00 0 1 0 0 1

^0 0 oj ^0 0 Oy

\Q\ = 4:/ 1111\ I 1 1 1 0 \ 110 1

1100 10 11 10 10 0 111 0 110 1 0 0 1 1 0 0 0 0 10 1 0 10 0 0 0 11 0 0 10 0 0 0 1 0 0 0 0 '

/ I 1 1 Ivf I 1 1 0 \ 110 1 110 0 10 11 10 10 0 111 1 0 0 1 0 110 1 0 0 0 0 10 1 0 10 0 0 0 11 0 0 10 0 0 0 1 00 0 0'

/ I 1 1 Iv / 1 1 1 0 \

110 1 110 0 10 11 10 10 1 0 0 1 0 111 1 0 0 0 0 110 0 10 1 0 1 0 0 0 0 11 0 0 1 0 V 000 1

' 0 0 0 0 '

/ l l l l vI 1 1 1 0 \ 110 1

1 1 0 0 1 0 1 1 10 10 1 0 0 1 10 0 0 Dili 0 110 0 10 1 0 1 0 0 0 0 11 0 0 1 0 0 0 0 1 , loooo'

/ 1 111\ f 1 1 1 0 \

110 1 1100 10 11 0 111 10 10 0 110 10 0 1 0 1 0 1 1 0 0 0 0 10 0 0 0 11 0 0 10

V 0001 / ' 0000 '

' \ \V o \ 110 1 110 0 10 11 0 111 10 10 1 0 0 1 0 110 0 10 1 1 0 0 0 0 100 0 0 11 0 0 1 0 0 0 0 1 , loooo'

/ 1111\ / 1 I 1 0 \

110 1 1 0 11 1100 10 10 0 111 0 110 1001 1000 0 1 0 1 0 0 11 0 1 0 0 0 0 10 0 0 0 1 / 0 0 0 0 '

/ 1 1 1 1 \ f 1 1 1 0 \

1 1 0 1 1 0 11 1 1 0 0 1 0 1 0 0 111 1 0 0 1 0 1 1 0 1 0 0 0 0 1 0 1 0 0 11 0 1 0 0 0 0 10

\ 0 0 0 1 / ' 0000 '

/ I 1 1 Iv/ 1 1 I 0 \ 110 1

1 0 11 1100 10 10 1 0 0 1 0 111 1000 0 110 0 10 1 0 0 1 1 0 100 0 0 10 \ 0 0 0 1 ,

' 0 0 0 0 '

/ 1 111\ ( I 1 1 0 \ 110 1 10 11 110 0 10 10 10 0 1 1 0 0 0 0 111 0 110 0 1 0 1 0 0 11 0 1 0 0 0 0 10 0 0 0 1 / 0000 '

/ 1 111\ ( 1 I 1 0 \ 110 1

10 11 1 1 0 0 0 111 10 10 1 0 0 1 0 110 0 10 1 1 0 0 0 00 11 0 1 0 0 0 0 10

V 000 1 / ' 0000 '

' \ W l \ 110 1 10 11 1100 0 111 10 10 0 110 1 0 0 1 0 10 1 1 0 0 0 0 0 11 0 1 0 0 0 0 10 0001 0 0 0 o'

/ l l l l v f 1 1 I 0 \ 110 1 10 11 0 111 110 0 10 10 1 0 0 1 0 110 0 1 0 1 0 0 11 1000 0 1 0 0 00 10 0 0 0 1 0 0 0 0 '

/ 1111\ / I 1 I 0 \ 110 1

10 11 0 111 110 0 10 10 0 110 10 0 1 0 10 1 0 0 11 1 0 0 0 0 10 0 0 0 10 000 1 0 0 00 '

3.2 Unions, Intersections, and Aggregation

Definition 3.2.1. Let TZ= {ya | « € v4} be a collection of relations on X q for some index

set A.

(i) The union of the relations in R , denoted ha> is the relation y defined by

x y y x y a V for some A.

(ii) The intersection of the relations in denoted Haeyt relation y ' defined

by

I X' ÿ 4=#» X y a V for all a € .4.

Proposition 3.2.2. The union of any collection of separable relations on X q is separable.



Proof. Let % = | a G be a collection of separable relations on X q and let

y = Uq6^ ha- Now let S C Q be given and suppose that { x s , u - s ) b (ys,u_g) for

some x s , y s E X s and some u - s E X - s - Then it must be the case that, for some a £ A ,

{ x s , u - s ) bo, (j/s>«-s)- But then the separability of implies that { x s , v - s ) h a ( y s , v - s )

for all v - s E X - s , which certainly implies that ( x s , v - s ) b i v s , v - s ) for zdl v - s £ X - s -

Since our choice of S was arbitrary, it follows that X = b a is separable. □

Corollary 3.2.3. The intersection of any collection of separable relations on X q is separable.

Proof. This follows from Corollary 2.2.5 and Proposition 3.2.2 by observing that Dae-4

is the negation of Uae-4 O

Corollary 3.2.4. I f y is separable, then so is y .

Proof. This follows immediately from Corollary 3.2.3 and Corollary 2.2.9 since y is the

intersection of y and □

We conclude this section by applying the above results to the problem of aggregating

separable preferences. Suppose that n decision-makers are attempting to combine their

individual preferences into a collective preference relation that somehow represents their

overall views on the issues a t hand. A natural method of accomplishing this task would

be to choose some cutoff value, say k (where A: is an integer with 1 < k < n), and then

construct the collective preference relation so that the group prefers i to y if and only if at

least k of its members prefer x to y.^ Note that, i f &= [n/2j + 1, this method corresponds

to standard majority aggregation. If A: = n, then each member can singlehandedly block

a preference between any two alternatives. Similarly, if Ar = 1, then each member can

singlehandedly force a preference between any two alternatives. Of course, in this situation,

another member could also force the opposite of this preference, in which case the method

would be indecisive.

Condorcet demonstrated in 1785 that this seemingly natural method of aggregating

preferences can lead to some rather unexpected and paradoxical outcomes. Specifically, he

'In the recent literature, this aggregation scheme has been referred to as optimal majority

aggregation or k-majority aggregation. We prefer the latter, as it more specific2illy describes the

method by which the collective preference relation is obtained. Also, optimal majority aggregation

assumes that the veilue chosen for k is somehow optimal with regard to the decision being made,

a restriction that is not necessary for our investigations. See [11] or [17] for more details.



showed that even when all individual preferences are transitive, the aggregate preference

relation obtained by majority rule may be intransitive. This observation became known as

Condorcet’s paradox and is a standard example of the difficulties inherent in preference ag

gregation. Fortunately, and somewhat surprisingly, this undesirable behavior does not occur

when we consider the effect of aggregation on the property of separability. In fact. Corol

lary 3.2.5 below shows that, under the aggregation scheme described above, separability of

individual preferences is preserved in the resulting collective preference relation.

Corollary 3.2.5. Let Tl = {X], Xg, - -, hn} a collection of separable relations on X q

and, for each positive integer k < n , let be the relation defined by

r >[k) y <=> x ^ i y for at least k of the 6 TZ.

Then y^k) “ separable.

Proof. The result follows from Proposition 3.2.2 and Corollary 3.2.3 by observing that

= u f n °

3.3 Lexicographic Sums

Definition 3.3.1. Let Qi and be disjoint criteria sets and let ^ i, Xg be relations on

Xqi and Xq^, respectively. The lexicographic sum of and Xg is the relation Xi © X2

on Xq,uQ 2 defined by

(xQ,,XQ^) ( t l © b 2) (î/Quî/Qî) ^ XQ, y I yq, or {xq, = yq, and xq^ Xg g/qj.

Proposition 3.3.2. //X j and y 2 are separable relations on X q , andXq^, respectively, then

Xi © Xg is a separable relation on XqiuQj-

Proof. Let X = X © Xg and let S C Q1 UQ2 be given. Define Si = Q iCS and S2 = Q2 CS.

Then S = S 1 US2 and - 5 = (Qi — Si) U (Q2 - S2 ). Now suppose that

( X5 j , X 5 2 , n q , — , UQ2—52) ^ (2/S1 ) Î/S2 » —5 i ) ^ Q 2~ 5 2 )

for some x, y Ç X s and some u € X -s - Let v £ X -s - We must show that

{_xs,,xs2,vq,—s,,vq2~s2) ^ (2/Sijys2)^Qi— 5 2 )* (t)

By Definition 3.3.1 above, one of the following must occur:



(i) (xsj,UQj-Sj) >-i (j/Sii«Qi-Si), in which case the separability of >-i implies that

(xs,,VQ ,-Si) (ysi,VQi-s,), which then implies (t); or

(ii) (xs,,UQ,-S:) = (2 /s,,U Q ,-s,) and >2 (ySz.UQz-Sz), in which case

xsi = ysi, and so (a:si.«(Q i-s,)) = ( î /s , ,V(q,-Si))- The separability of ^2 then

implies that {xs2 ,v q ,- s 2 ) t .2 (!/% ,% -% ), which implies (f).

Since each case implies (f), we have shown that S is ^-separable. Since our choice of

S was arbitrary, it follows that X is separable. □

A partial converse follows:

Proposition 3.3.3. Let and ^2 be relations on X q , and Xq^, respectively. I f h i © h.2

is separable, then both >-i and h.2 are separable.

Proof. Let b = @ ba- By Corollaries 2.2.7 and 3.2.4, it suffices to show that >-i = >-q ,

and that h.2 = h.Q^- For the former, notice that

x ^ - i y <=> {x,u) x {y,u) for all u e Xq^ <=> x >-q, y.

Similarly, for the latter we have

x h .2 V <=> (u ,i) ^ (u,y) for all u 6 X q , <=> xh.Q 2 V- O

Proposition 3.3.4. The lexicographic sum is an associative binary operation. Specifically, let

Qi, Q2 , Qs be disjoint criteria sets and let X, be a relation on X q, for i = 1, 2 , 3. Then

(h i e ba) ® h 3 = h i © (ha © h a ) .

Proof. Let Q = Qi U Q2 U Q3 and let (11 , 12, 2:3), (j/i,ya,2/3) G X q , where Xi, yi € X q, for

i = 1, 2, 3. It is straightforward to verify that both (Xi © ^ 2) © hs and © (^2 © hs)

are equal to the relation h on X q defined by

( X i , 1 2 ,2 :3 ) h ( î / i , î / 2 , y 3 ) < = » 2:1 X i Î/1

or II = Î/1 and 12 >-2 Î/2

or I I = j/i, 12 = Î/2, and 13 X3 j/3. □

In light of Proposition 3.3.4, we will allow ourselves to omit parenthesis when tak

ing multiple lexicographic sums. For example, we will write h i © h 2 © hs rather than

(h i © ha) © hs- Note that the operation of taking lexicographic sums is not commutative.

Indeed, the order in which the component relations are summed plays an essential role in

Definition 3.3.1.



Proposition 3.3.5. LetQi and Q2 he disjoint finite criteria sets, let >-\ be a binary preference

order on Qi with order sequence ( n ) and heighP m, and let >-2 be a binary preference order

on Q2 with order sequence (y&) and height n. Let (zk) be the order sequence corresponding

to >~i ®>-2 - Then, for all integers i and j with 0 - = >-i © >-2- Observe that, by the definition of the lexicographic sum,

{Xk,yi) >■ (xi+uVj) k < i + l

or A: = i + 1 and I < j.

Since there are in elements of the first type and j - 1 elements of the second type, exactly

in + j - 1 elements precede {xi+i,yj) in the order sequence for X, as desired. □

We conclude this section by examining a special class of orders that will be of signifi

cant use to us in later chapters.

Definition 3.3.6. Let Q = {1 ,2 ,... ,n} and let >- be a linear order on X q . If, for each q e Q ,

there exists a linear order on Xq such that

>- = >-<r(l) ® >"<r(2) © • • • © >-<r(n)

for some cr ^ Sn, then >- is said to be a lexicographic order on Q. The permutation a is

called the importance permutation of the order.

Note that, in the case that X q is binary, a lexicographic order on Q is uniquely

determined by its importance permutation and its leader. Thus,

Proposition 3.3.7. Let |Ç| = n and let X q be binary. Then there are exactly 2" • n! distinct

lexicographic orders on Q.

Also notice that, since every linear order on a single criterion is vacuously separable.

Proposition 3.3.2 implies the following;

Corollary 3.3.8. Every lexicographic order is separable.

We should mention that Corollary 3.3.8 has been proved in the past (see [13], for

example), but follows here as a result of a more general theory of lexicographic sums than

*The height of a lineair order on a set S is simply the cardinality of S.



has been previously studied. We will see shortly that this corollary is also implied by the

upcoming Propositions 3.4.4 and 3.4.8.

Of special interest to us will be the normalized lexicographic order, i.e. the lexico

graphic order with leader (1, 1, . . . , 1) and importance permutation equal to the identity

permutation. We call this special order the standard lexicographic order on Q, denoted

>-iex- The standard lexicographic order has the useful property that the row of the

binary preference matrix corresponding to 'f-ux is, loosely speaking, the binary expansion

of 2" - fc (with leading zeros possibly added). More precisely.

Proposition 3.3.9. Let |Q| = n > 2 and let (ojj) he the binary preference matrix corre

sponding to y lex- Then, for each positive integer i < 2",

X ; a M - 2 " - ^ = 2 " - tj=i

Proof. We proceed by induction. For n = 2, the result can be easily verified by examining

the 4 rows of the standard lexicographic order on 2 criteria. Now suppose that it is true for

some integer n > 2. For each positive integer k, let >-* denote the standard lexicographic

order on k criteria, so that >-„+i = >-ti ® >-i- Let (oij), (bij) be the binary preference

matrices corresponding to >-„ and >-n+i, respectively. Proposition 3.3.5 implies that

b2i-ij = b2(i-i)+ij = <

Similarly,

Thus, by the induction hypothesis.

a>,j if 1 < i < n;

1 if J = n -t-1.

“ij if 1 < i < n;

0 if J = n + 1.

n + l n

5 3 b2 i - i j ■ 2"+^--' = 1 + 5 3 “‘-J • = 1 + 2(2" - i) = 2"+i - (2% - 1)j=i j=i

andn+l5 3 b2i,j ■ 2"+^-^ = 5 3 “ i.j • 2"+^-^ = 2(2" - i) = 2"+^ - 2i, j=i j=i

as desired. □



3.4 Additive Orders

For this section, we assume that Q is a finite criteria set.

Definition 3.4.1. Let X be a weak order on Xq. A value Junction for ^ is a function

V : X q -¥ M such that, for all x, y £ X q , x y y v(x) > v(y). The relation X is

said to be additive if, for each q € Q, there exists a value function u, : X , -> K such that

v{x) = is a value function for X.q€Q

In recent literature (see [2], [14]), exeunples of additive orders have typically involved

linear value Junctions, that is, value functions of the form v(x) = ^ where c, € R<î€Q

for each g E Q (of course, in this case it is necessary for each X , to be a subset of R). If

X is an additive order that admits such a linear value function, we say that y is linearly

additive. In light of this distinction, it seems reasonable to ask: Are there additive orders

that are not linearly additive? The answer to this question is yes, though we must look

beyond the binary preference orders to find them (see Proposition 3.4.3 below).

Take, for example, the additive order y on X{i,2> with value function v(x) = xj +x^,

where X% = Xg = {—1,2, -3}. Let ~ denote the indifference relation® corresponding to y .

Then (-3 , -3 ) ~ (2,2), since

u((—3, -3 )) = - 3 + 9 = 6 = 2 + 4 = u((2,2))

and ( - 1 , 2) ~ (2, - 1), since

u ( ( - l , 2)) = - l + 4 = 3 = 2 + l = v(2, —1).

By this token, any linezur value function v*{x) = ciXi +C2X2 for y would necessarily satisfy

-3ci - 3c2 = u*((-3, -3 )) = u*((2,2)) = 2ci + 2c2

and

-Cl + 2c2 = u " (( - l , 2)) = v*{{2 , - 1)) = 2ci - C2.

But some simple algebra shows that the only solution to this system of equations is

Cl = C2 = 0. If this were the case, then we would have x ~ y (i.e. x y y and y y x ) for every

*The indifference relation ~ corresponding to a weak order y is the relation defined by x ~

y < => x y y and y y x. Notice that, if ^ is an additive weak order with value function v, then

X ~ y < => v { x ) = v (y ) .



x ,y E X q . But this clearly does not occur, for, by our above calculations, (-3 ,3 ) X (-1 ,2)

but (-1 ,2 ) ^ (-3 ,3 ). Thus, there cannot exist a linear value function for X; it follows that

y is not linearly additive.

Proposition 3.4.3 below establishes our previous claim that we must look beyond the

binary preference orders to find examples of additive orders that are not linearly additive.

Lemma 3.4.2. Let Xj and Xg be additive orders on X q with respective value functions v

and w. I f V — w is a constant function, then X; = Xg.

Proof. Suppose that, for some c G R, v{x) - w{x) = c for all x G X q. Then

v {x i)> v {x 2 ) -*=»- + c > u)(a;2) + c <=> w (x i)> w {x 2 ),

and so xi Xa *=> xi ^2 Za. □

Proposition 3.4.3. Let Q = { 1 ,2 ,..., n}, let Xq = {o,, 5,} C K for each q G Q, and let X

be a weak order on X q . Then y is additive if and only if y is linearly additive.

Proof. The reverse implication is immediate. To prove the forward implication, assumen

that y is additive and let u(x) = ^ u ,(x ,) be a value function for y . Notice that, for each7=1

q GQ and for each x , G Xq (recall that Xq = {a„ 6,}), we have

Vq{Xq) = ^ ï^ ( X g - a ,) + Vq{üq).Oq Qq

Now let_ __ ~ ’ qi^q)

bq - Oq

Thenn n

= l ] c ,X , + '^{Vqiüq) - Cqùq).9=1 9=1

nSince the latter sum is a constant. Lemma 3.4.2 implies that w{x) = 53 C7X7 is a value

7=1function for X. □

Many authors have studied the connections between separability and additivity. We

state below some of their main findings. The first result has appeared in several locations;

its proof may be found in either [14] or [21]. The second is due to Gorman [10].

Proposition 3.4.4. Let y be a additive order on X q . Then y is separable.



Proposition 3.4.5. Suppose that, for each q £ Q ,q is strictly essential and Xq is topologically

separable and arc-connected. Suppose also that y is a continuous weak order on X q . Then

>~ is additive if and only if is separable.

It is known that Proposition 3.4.5 may fail when the alternative sets are not arc-

connected. The example we provide below to demonstrate this fact is due to Bradley zind

Kilgour [2].

Example 3.4.6. Let Q = {1,2,3,4,5} and let >- be any binary preference order on X q such

that

(O .l.l.O ,!);- (1,0,0,1,0)

( 1, 0 , 0 , 0 , 1) ; - ( 0 , 1, 1, 0 , 0)

( 0 ,0 ,1 ,1 ,0 ) X ( 0 ,1 ,0 ,0 ,1)

(0 , 1, 0 , 0, 0) ; - ( 0, 0, 1, 0, 1).

To show that >- is not additive, it suffices, by Proposition 3.4.3, to show that >- is not5

linearly additive. Now if there did exist a linear value function v{x) = CqXq for >-, we9=1

could conclude the following;

C2 + C3 -h Ci > Cl + C4

Cl 4- eg > C2 + C3

C3 + C4 > C2 -l- C5

C2 > C3 + C5

But adding the first three inequalities yields C3 + C5 > C2, which is a contradiction to the

fourth inequality.

It is worth mentioning that this example may be easily modified to demonstrate the

existence of a separable but non-additive order on n criteria for every integer n > 5.

Specifically, let Q = {1,2,3,. . . ,n } and let >- be any binary preference order on X q such

that

(0 , 1, 1, 0 , 1, 0 , . . . , 0) >- ( 1, 0 , 0 , 1, 0 , 0 , . . . , 0)

(1 ,0 ,0 ,0 ,1 ,0 , . . . ,0 ) V (0 ,1 ,1 ,0 ,0 ,0 , . . . ,0)

(0,0,1,1,0,0,...,0) V (0,1,0,0,1,0,...,0)

(0 ,1 ,0 ,0 ,0 ,0 ,...,0 ) >- (0 ,0 ,1 ,0 ,1 ,0 ,...,0 ).



In the same manner as above, it can be shown that >- is not additive. The following

proposition, which appears in [2], shows that Example 3.4.6 is the smallest counterexample

to Proposition 3.4.5.

Proposition 3.4.7. Let \Q\ < 4. Then every separable binary preference order on Q is

additive.

We conclude this chapter by considering the relationship between additive and lexi

cographic orders. The following proposition is due to Kilgour [13].

Proposition 3.4.8. Every leoncographic order is additive.

Kilgour also demonstrates that the converse of this proposition does not hold; that is,

there exist additive orders that are not lexicographic.


Chapter 4

Preseparable Orders

In this chapter, we consider some combinatorial problems related to the notion of sepa

rability. We introduce preseparable extensions as a mechanism for merging two separable

preference orders in a meinner that preserves the separability structure of the component

orders. We consider the effect of executing a finite sequence of these extensions and explore

counting problems related to the resulting orders.

4.1 Preseparable Extensions

Definition 4.1.1. Let Q\ and Qg be disjoint criteria sets and let >-i and >-2 be linear orders

on Xq, and Xq^, respectively. A preseparable extension of >-i by >-2 is a linear order >- on

Xq,uQ2 such that, for i = 1, 2,

XQi VQi = > (î/Q.,«-Qi) for all u_q, e X_q,.

If Q2 consists of a single element, then the extension is said to be simple. If Ç = {1,2 , . . . , n}

is a finite criteria set, then a linear order >- on X q is said to be preseparable if there exists

a sequence >-1, >-2, such that, for each positive integer i < n — 1, is a linear

order on {1, 2 , . . . , i} and >-,+i is a simple preseparable extension of >-(.

Note that if X is a preseparable order with the corresponding sequence >-i, >-2 , ■■■, >-n

of preseparable extensions, then each >-< is also preseparable.

Lemma 4.1.2. Let >- be a preseparable extension of >-i by >-2. If, for i = \ or i = 2 ,

(xQ i,u_Q i) X {vQ i,u -Q .) for some u _ q , € X _ q ,, then i q , > i yg...

28


Chapter 4. Preseparable Orders 29

Proof. We prove the contrapositive. Suppose that xq. yq .. Because >-i is a linear order,

it must be that yg,. >-i xq .. But then (y g ^ .u -g j >- (zg;,u_g.) for all u_g. € X _g ,, from

which it follows that ( ig ,,u _ g .) (yg,.,u_g() for all u_Çi 6 X -q .. □

Proposition 4.1.3. Preseparable extensions preserve separability. Specifically, let >~i and

>-2 be linear orders on X q , and X q j , respectively, and suppose that >- is a preseparable

extension of>-i by >-2. If, fo r i = 1 o ri = 2, S Ç Q, is >-i-separable, then S is y-separable

and >-s = (Vi)s-

Proof. Suppose that S' Ç Qj is >~j-separableandthat (x s ,u g ,-5 ,u _ g j >- (ys,ug,-g,u_g,.)

for some xs , ys £ X s and some u e X -s - Lemma 4.1.2 implies that (zg,ug._g) >-,•

iy s ,U Q ,-s ) . But then the >-i-separability of S implies that {x s , vq, s ) >-i (2/5, % - s )

for all v q .-s £ X q , s . Thus, by Definition 4.1.1, it follows that ( x s ,v q .- s ,v -q ,) y

i y s ,V Q ,- s ,v -Q ,) for all u e X - s and so S is separable. To show >-5 = we observe

that

Xs >-s ys {xs,UQi-s,u-Q ,) y (ys.ugj_g,u_g,) for all u e X - s

<=> {xs,UQi-s) y i iys,UQi-s) for all u q ,-s £ X q,s

Xs (Xi)s ys- □

Corollary 4.1.4. Let >-1, >-* be linear orders on X q ,, let >-2, y 2 be linear orders on Xq^,

and suppose that y is a preseparable extension o f y \ by >-2 and that >-* is a preseparable

extension of X* by y^- I f either Xi 7 >-î or >-2 i=- Xg, then y i^y*

Proof. We prove the contrapositive. Thus, suppose that >- = >-*. Then >-qj = >-* and

>-q2 = >-Qj. But Proposition 4.1.3 then implies that = >-î and >-i = >-g, as desired. □

Proposition 4.1.3 shows that preseparable extensions preserve the separability struc

ture of their component orders. Since preseparable orders are obtained through successive

simple presepzirable extensions and since any linear order on a single criterion is vacuously

separable, it follows that every preseparable order is monoseparable. The following example

demonstrates that, while monoseparability is a necessary condition for preseparability, it is

not sufficient.

Example 4.1.5. Let y be the linear order corresponding to the binary preference matrix



A 1 A1 1 00 1 11 0 10 0 11 0 00 1 0

V 0 °)

Upon inspection, we see that >- is separable on each criterion, with induced order

1 >- 0. Thus, >- is monoseparable. Now if >- were preseparable, then >- would be a simple

preseparable extension of some linear order on But since preseparable extensions

preserve separability and {91 , 92} is X'-separable with respect to any linear order >-' on

it follows that {91 , 92} would then be >--separable. Notice however that this is not

the case: since Oil >- 101 and yet 100 X 010, {91 , 92} is not separable. Consequently, y

cannot be preseparable.

Given our observations in Example 4.1.5, a natural question to ask is: W h a t co n d itio n s

m u s t be p la ced on m onoseparable o rders to o b ta in preseparable o rders? The proposition

below answers this question.

Proposition 4.1.6. L e t Q = {1,2,. . . ,n } a n d le t y be a lin ea r order o n X q . T h e n y is

preseparable i f a n d o n ly if:

(i) >- is m onoseparable , and

(ii) F o r each j e {1,2,. ..,n } , th e s e t Q j = {1,2,. . . , j } is y -sep a ra b le .

P roo f. ( ^ ) Suppose >- is preseparable and let >-i, X2 , - - , = >- be the corresponding

sequence of simple preseparable extensions. By our above remarks, >- is monoseparable.

Now since each y j is a linear order on X q . , it follows that each Q j is >-j-separable. But

since X is obtained from Xj through a finite sequence of simple preseparable extensions,

all of which preserve sepEirability, it follows that Q j is x-separable.

(<=) For the converse, we proceed by induction. The result is immediate for n = 1.

Now suppose that every lineeir order on Xq> having |Q'| = n - 1 and satisfying properties

(i) and (ii) is preseparable. Let 5 = Qn-i- By Proposition 2.2.6, X; is monoseparable and

Q j is Xs-separable for each j G {1 ,2 ,... ,n - 1}. Thus, by the inductive hypothesis, Xs is

preseparable. By Definition 4.1.1, it now suffices to show that X is a simple preseparable

extension of Xs by X{„}. This follows easily since - 5 = {n} and

Zs Xs y s = > {xs,Un) X (ys,Un) for all u„ G



and

Xn >-{.} Vn = > ( x „ ,u s ) X (j/n,us) for all U s e X s . □

Corollary 4.1.7. I f Q is finite and >- is a separable linear order on X q , then y is presepa

rable.

4.2 Counting Binary Preseparable Orders

Definition 4.2.1. A sequence (a&) of n (+ l)s and (—l)s is called a characteristic sequence if,k

for each positive integer k < n , the partial sum of (o*) is nonnegative, i.e. 5D o« > 0. If1=1

a characteristic sequence contains the same number of (+ l)s as ( - l ) s , then the sequence is

said to be complete. If a„_*+i = -a* for each k, then the sequence is Sciid to be symmetric.

Note that a characteristic sequence is one in which, at any point in the sequence,

the number of (+ l)s appearing up to that point is greater than or equal to the num

ber of (—l)s. Thus, the sequences (-Hi,4-1,-Hi, + 1 ,-1 ,—1), (-Hi,-Hi,—1 ,- 1 ,-Hi,—1) and

(-Hi, -1 , -Hi, -1 , -Hi, -1 ) are all characteristic sequences, the second being complete (but

not symmetric) and the third being symmetric (and complete - in fact, one can easily

verify that every symmetric chziracteristic sequence is complete). On the other hzmd, nei

ther (-Hi, -Hi, —1, —1, —1, —1) nor (—1, —1, -Hi, -Hi, —1, -Hi) nor (—1, -Hi, —1, -Hi, —1, -Hi) is

a characteristic sequence.

Proposition 4.2.2. Let Qi = {1 ,2 ,... ,n}, Q2 = {n-Hl}, |%Q,| = m < 0 0 , and |%n+i| = 2.

//>-! is a linear order on X q , and y 2 is a linear order on X n + i, then there is a bijection

between the set of preseparable extensions o f y \ by y 2 and the set of complete characteristic

sequences of length 2 m.

Proof Let Q = Q 1 U Q 2 . Let the elements of X„+i be labeled -Hi and - 1 so that -Hi >-2 -1 .

Let (a:*) be the order sequence corresponding to >-1. For each complete characteristic

sequence (o&) of length 2m, let (ÿ&) be the sequence of alternatives in X q defined by

j/fc = ( i j ,± l ) Ofc is they"* ±1 in (a/t).

Because (o*) is complete (and hence there are m (-Hl)s and m ( - l ) s in (a*)), each element

of X q appears in (j/t) exactly once. Thus, let >- be the linear order corresponding to (j/*)

and notice the following:



(i) (ofc) has all nonnegative partiîd sums (x j,+ 1) >-( i j , - 1) for all j .

(ii) For s = ±1, ( i j , s ) y ( x j , s ) <=> the s appears before the j''* s in (o*)

<=> i < j 4=* Xi >-i Xj.

These observations establish that the map (a*) •-> (y*), as defined above, is a well-

defined function from the set of all complete characteristic sequences of length 2n onto

the set of all preseparable extensions of >-i by >-2. We need only to show that this map

is injective. To do so, observe that if two characteristic sequences (a*) and (6&) differ, say

Oj 5 bi for some i, then their corresponding linear orders differ: specifically, the i'* most

preferred alternatives differ in the {n -f-1} component. This proves that our function is

bijective, as desired. □

To proceed, we require the following well-known result (paraphrased to suit our ter

minology and notation). Its proof can be found in Brualdi [6] (pages 252-254).

Proposition 4.2.3. The number of complete characteristic sequences of length 2n is the

Catalan number Cn, where

Example 4.2.4. Let >-i and >-2 be defined by 0 >-i 1 and 1 >-2 0 (so that both >-i and

>-2 are binary preference orders on a single criterion). Since there are exactly 2 complete

characteristic sequences of length 4, there are exactly 2 preseparable extensions of >-i by

>-2, as shown below:

( - f l , - 1 , + 1 , - 1 ) i— > (0,1) >-(0,0) >-(1,1) >-(1,0)

( + 1 , - b l , - 1 , - 1 ) i — > (0 ,1 ) >- (1 ,1) >- (0 ,0) >- (1,0)

We now have enough machinery to be able to count the number of binary presepmable

orders on a given finite criteria set.

Proposition 4.2.5. Let |Q| = n < 00 and suppose that X q is binary. Then the number of

distinct preseparable orders on X q is given by the formula

n —1

n(n) = 2" ]][ C2 >. i=0

Proof. Let Q = { 1 ,2 ,..., n} and let Q' = Q - {n}. Every preseparable order on X q is a

simple preseparable extension of a preseparable order on X q> . Thus, to form a preseparable

order on X q , we must



(i) choose a preseparable order Xi on X q',

(ii) choose a Unear order X2 on Xg„ (either 1 X 0 or 0 X 1), and

(iii) form a preseparable extension of Xi by Xz-

It is clear that (i) may be completed in 7r(n — 1) distinct ways, and (ii) in 2 distinct

ways. Since every linear order on X qi has height 2"“ , Proposition 4.2.3 implies that (iii)

may be completed in Cgn-i distinct ways. Furthermore, Corollary 4.1.4 implies that all

preseparable orders formed in this manner will be distinct. Thus,

7r(n) = 2 • Tr(n — I) • Cg—1.

Clearly ir(l) = 2 and so a simple induction argument finishes the proof. □

4.3 Symmetric Extensions

Let X q be a binary alternative set. For an alternative x e X q , let x denote the bitwise

complement of x (so that if, for example, x = (1,0,1), then x = (0,1,0)). Definition 4.3.1

and Propositions 4.3.2 and 4.3.3 are generalizations of results originally due to Bradley and

Kilgour [2].

Definition 4.3.1. Let X be a relation on a binary alternative set X q . Then X is said to

satisfy the mirror property if, for all x, y G X q , x X y = > ÿ X x.

Proposition 4.3.2. Let (t?( = n < 00 and let X q be binary. Then a linear order X on X q

satisfies the mirror property if and only if its order sequence (x&) is such that X2 "-i+i = x,

for each positive integer i < 2".

Proof. ( ^ ) Suppose that X satisfies the mirror property and let i < 2" be given. By

definition of the order sequence, xj X x, <=> j < i. But our assumption that X satisfies

the mirror property then implies that x j X Xj 4=> i < j . Since every Xk € (x&) can be

written as Xj for some j (namely, choose the unique j for which xj is the bitwise complement

of Xfc), it follows that x* X Xj for exactly 2" - i choices of k. Thus, x, = X2»_j+i, as desired.

( ^ ) For the converse, suppose that xg^-i+i = x l for each positive integer i < 2” .

Then

Xj X Xj <=> i < j

^ 2" - j + 1 < 2" - i 4-1

Xj “ Xg” —j+ i X Xgn_ j.^j — Xj. Q


Chapter 4. Preseparable Orders

Proposition 4.3.3. I f y is a separable relation on a binary alternative set X q , then y

satisfies the mirror property.

Proof. Let x ,y & X q and suppose that x y y . l i x = y, then Proposition 2.2.10 implies that

y is reflexive, from which it follows that ÿ y ÿ = x. Furthermore, i î x = ÿ, then x = y, and

soy = X y y = x. Now suppose that x ^ y and let 5 = {ç € Q 1 = %}. By assumption,

5 is a proper, nonempty subset of Q. Also notice that x s = ys and i_ s = ÿZg. Thus,

( î/s ,m ? ) = (xs ,x-s) = x y y = {ys ,y-s) ,

from which the separability of y implies that

V = (Vs, ÿZg) y {ÿs, y - s ) = (xs, xZs) = x. □

Example 4.3.4. Let >- be the linear order corresponding to the binary preference matrix

A 1 A 1 0 1 1 0 0

1 1 0

0 0 1

0 1 1

0 1 0

\ 0 0 0/

34

By Proposition 4.3.2, >- clearly satisfies the mirror property. Notice however that

{2} is nonseparable with respect to >- since 111 >- 101 but 100 >- 110. Thus, the mirror

property, though necessary for separability, is not sufficient. Indeed, we have seen that the

mirror property does not even guarantee monoseparability.

Example 4.3.5. (due to Bradley & Kilgour [2]) Let >- be the linear order corresponding to

the binary preference matrix



A 1 1 1

1 i \1 1

1 1 11 10 0

0 1

0 1 0 1 1 1

11 1

1 1

1 1

/

where the bottom half of the matrix is completed so that >- satisfies the mirror property.

It is straightforward, albeit quite tedious, to verify that:

• >- is monoseparable with induced order 1 >- 0 on each criterion.

• {1, 2} is >--separable with induced order 11 >- 10 >- 01 >- 00.

• {1,2,3} is separable with induced order

111 >- 110 >- 101 >- Oil X 100 >- 010 >- 001 000.

• {1, 2,3,4} is >--separable with induced order

1111 X 1110 >- 1101 1011 X 0111 >- 1100 >- 1010 >- 0110

>- 1001 >- 0101 X 0011 X 1000 >- 0100 X 0010 X 0001 >- 0000.

Thus, by Propositon 4.1.6, >- is preseparable. By construction, X satisfies the mirror

property. Nevertheless, {2,3,5} is not >--separable since 11010 >- 10111 but 10101 >-

11000. We see then that, while the mirror property and preseparability are both necessary

for separability, even both properties taken together are not sufficient.

Definition 4.3.6. Let Qi and Q2 be disjoint criteria sets, and suppose that |Xq, | = m and

= n. Let >-i and >-2 be linear orders on Xq, and Xq^, respectively and let (xk) and

(j/fc) be the order sequences corresponding to >-i and >-2. A symmetric extension of y 1 by



> -2 is a linear order >- on Xqûq^ such that, for all positive integers i, j , k and I with i,

k < m and j , I < n,

^ i^m—k+ltyn—l+l) ^ (^-m-i+l i î/n—j+l)-

Proposition 4.3.7. Let Xi and Xz be as in Definition 4-3.6. A linear order X on ^QiuQz

is a symmetric extension of Xi by Xz if ond only its order sequence [zk) = is

such that Zmn-p+i = (itn-tp+uPn-jp+i) for cach positive integer p< mn.

Proof. (=>•) Suppose that X is a symmetric extension of Xi by Xz and let p < m n be

given. For each q < mn, let z* = (xm -i,+ i,yn-j,+ i). We wish to show that Zmn-p+i = z*.

By definition of the order sequence, Zq X Zp -<=> q < p. But our assumption that X

is a symmetric extension of Xi by Xz then implies that z* X z* <=> q < p. Since

every z* e (z*) can be written as z* for some q (namely, choose the unique q for which

zj* = Zq), it follows that z* X z& for exactly p - 1 choices of k. Thus, z* x z* for exactly

m n - (p - 1) - 1 = m n - p choices of k, and so Zmn-p+i = z*.

(4=) For the converse, suppose that Zmn-p+i = Zp for each p < mn. We wish to show

that Zp X Zq z* X Zp. Observe that

Zp X Zq *=> p < q

<=> m n — g + 1 < mn — p + 1

>— ^ q ~ ^mn—q+ 1 X Z m n —p+1 — Zp,

as desired. □

Proposition 4.3.8. Symmetric extensions preserve the mirror property. Specifically, if Xi

and Xz are linear orders on finite binary alternative sets and X is a symmetric extension

of y 1 by y 2 , bhen X satisfies the mirror property if and only if both Xi and Xz satisfy the

mirror property.

Proof. Let Xi and Xz have heights m and n and order sequences (i*) and (%), respectively.

Let (zk) = be the order sequence corresponding to X.

(=>) Suppose that x satisfies the mirror property and let i < m and j < n h e given.

Since X is a linear order, there exists an integer k such that z& = (xi,yj). Thus

{xi,Vj) = ^k

= Zmn-k+i (by Proposition 4.3.2)

= (zm-:+i,!/n-j+i) (by Proposition 4.3.7),



from which it follows that r , = Xm-»+i and ÿj = yn-j+i- Since our choices of i and j were

arbitrary, it follows that both >-i and >-2 satisfy the mirror property.

(<=) Suppose that both Xi and y 2 satisfy the mirror property and let y be a sym

metric extension of >-i by >~2 - Then, for each k < mn,

Zk =

= {xm-i^+ï, Vn-j^+i ) (by Proposition 4.3.2)

= Zmn-k+i (by Proposition 4.3.7).

Thus y satisfies the mirror property. □

Proposition 4.3.9. Let Qi = {1,2,. . . ,n } , Q2 = {n + 1}, and let | = m < 00 and

|X„+il = 2. Let Xl ond y 2 be linear orders on Xq, and X„+i, respectively, and let y be a

preseparable extension of y i by Xa- Then X is a symmetric extension 0/ Xi by X2 if and

only its characteristic sequence (a*) is symmetric.

Proof. Let the elements of X„+i be labeled -f-1 and - 1 and let (x*), (y*)i and (z*) = (x< , )

be the order sequences corresponding to Xi, X2, and X, respectively. Note that, for ezich

k, = fl*. Also note that, since = ± 1, j/2-i*+i =

(=») Suppose X is a symmetric extension of Xi by X2. Then, by Proposition 4.3.7,

Z2m —k+ l ~ (®m—»*+liJ /2—jfc+l) — ( ^ m —«*+1 > “ J/j*)-

Thus, a2 m-k+\ = -Vju = as desired.

(•^) For the converse, assume that (a*) is symmetric. We wish to show that X is a

symmetric extension of Xi by X2. By Proposition 4.3.7, it suffices to show that

Z2m —k+ l — ( ^ fn —: i , + l i î / 2 - j i , + l ) — ( ^ m - « i ,+ l , ~î/jfc)

for each k < 2m. Choose such a k. We consider two cases, depending on whether i/j = +1

or j/ji, = -1 . In the first case, we have

Zk = (x,,,,-H) = > Ok is the 4* (-4-1) in (a&)

= > Ok is preceded by fy - 1 (+ l)s

= > a2 m-k+i = - 1 and a2 m-k+i is followed by ik - 1 ( - l ) s

= > a2m-k+i is preceded by m - (f& - 1) - 1 = m - ik ( - l ) s

=*- a2m-fc+i is the (m - û 4-1)'* ( - 1)

Z2m —k + l — (2^m-i),+l> ~ 1 ) — ( ^ m - ù + l i



as desired. A similar argument holds if jk = -1 . Thus, >- is a symmetric extension of >-i

by >-2. □

Proposition 4.3.10. Let Qi = {1 ,2 ,,.. ,n}, Q2 = { n + 1}, and let Xq^ and X„+i be binary.

Let >-i and >-2 be linear orders on Xq^ and Xn+i, respectively, and let >- be a preseparable

extension of>~i by y 2 - Then y satisfies the mirror property if and only if y is a symmetric

extension of y \ by >-2 and both >-i and >-2 satisfy the mirror property.

Proof. Suppose that X satisfies the mirror property and let (x*), (y*), and (z*) = (%,*, )

be the order sequences corresponding to >-j, >-2, and X, respectively. Note that >-], >-2

and >- have heights 2” , 2, and 2"+^. Without loss of generality, assume that 1 >-2 0 (if this

is not the case, then we can relabel the elements of A„+i to make it so). Let (a*) be the

characteristic sequence corresponding to >-. Then for each positive integer k < 2 "+ \

o.k = \+1 if = 1

- 1 if yj^ = 0 .

But since >- satisfies the mirror property, we have

and so = ÿJH- Thus, 020+1 _fe+i = -o&, and, by Proposition 4.3.9, it follows

that >- is a symmetric extension of >-i by >-2. Proposition 4.3.8 then implies that both

Xl and X2 satisfy the mirror property. The reverse implication follows immediately from

Proposition 4.3.8. □

4.4 Counting Strongly Preseparable Orders

Propositions 4.1.7 and 4.3.3 establish that every sepzirable linear order on a finite binary

alternative set must be presepaiable and satisfy the mirror property. Our goal in this section

is to count the orders satisfying these properties. For convenience, if X is a linear order on

a finite binary alternative set X q and X satisfies the mirror property, we say that X is a

symmetric order on Q. If X is both symmetric and preseparable, we say that X is strongly

preseparable.

Proposition 4.4.1. Let |Q| = n and let X q be binary. Then there are exactly 2^" * • 2"~M

symmetric orders on Q.



Proof. Let (a:*) be the order sequence corresponding to >-. Since X q is binary, >- has

height 2” . Thus, every symmetric order on Q can be uniquely determined by specifying

xi ,X2 , ■ ■. so that Xi ^ X j for all i, j < 2”~* (the bottom half of the order sequence

is determined by the mirror property). Furthermore, all such choices of xi,X2, . . .

correspond to symmetric orders. Now xi may be chosen in 2” ways, whereas X2 may be

chosen in 2” — 2 ways (x2 cannot be equal to xi or xf), xs may be chosen in 2" — 4 ways,

and so on. Thus, the number of symmetric orders on Q is equal to

2” - ( 2 ' * - 2 ) - ( 2 " - 4 ) - - 4 - 2 = 2^""'-2"~^l □

For each positive integer n, let P(n) denote the probability that a randomly selected

binary preference order on a finite criteria set of cardinality n is sepairable.

Corollary 4.4.2. P(n) 0 as n oo.

Proof. For each positive integer n, let s(n) denote the number of separable binary preference

orders on a finite criteria set of cardinality n. Note that the total number of binary prefer

ence orders on such a criteria set is 2"!. By Propositions 4.3.3 and 4.4.1, s[n) < 2^" ' •2""^!,

and so

^ 22""* -2"-*!- 2 n

_ 2 " - ( 2 " - 2 ) •(2'*-4) •••4-2 “ 2" (2" - 1) (2" - 2) . . . 2 1

1( 2 " - l ) . ( 2 " - 3 ) ( 2 " - 5 ) --3 1'

which approaches 0 (very quickly!) as n -> oo. □

Lemma 4.4.3. Let n be a positive integer. Then the Catalan number satisfies

Proof. Observe that



Proposition 4.4.4. The number of symmetric characteristic sequences of length 2n is the

middle binomial coefficient Wn, where

Proof. Every symmetric characteristic sequence of length 2n is uniquely determined by its

first n elements. Furthermore, every characteristic sequence of length n can be extended

uniquely to a symmetric characteristic sequence of length 2n. Thus, it suffices to show that

the number of characteristic sequences of length n is W„.

Let T){n, m) denote the number of characteristic sequences of length n whose sum Is

m and observe that 7j(n, m) = 0 whenever (i) n and m have different parity, (ii) m < 0, or

(iii) n < m . Also observe that, to form a characteristic sequence of length n whose sum is

m, we must either append a (+1) to a characteristic sequence of length n — 1 whose sum is

m — 1, or append a (-1 ) to a characteristic sequence of length n - 1 whose sum is m + 1 .

Thus,

7/(n, m) = j}{n - 1, m - 1) + t){n - 1, m + 1).

Finally, notice that Proposition 4.2.3 implies that r]{2n,0) = On-

We wish to count the number of characteristic sequences of length n. Let r]{n) denote

this quantity and observe thatn

?(n) = ^ 77(71, m).m =0

We now use induction on n to show that 77(71) = W„. For 7i = 1, our claim is immediate,

since the only characteristic sequence of length 1 is the sequence (+1). Now suppose that,

for some n > 1, we have 77(71 - 1) = W„_i. Then

’îW = H 7?Km)m—Q

n n= ^ 77(71 - 1 ,771-1) + 77(71 - 1,771 + 1)

771=0 m = 0

=77(77 - 1, - 1)+77(71 - 1, 0 ) + 2 1 77(77 - 1 ,7 7 7 ) I +77(77 - 1 ,7 7 ) + 7 7 ( 7 7 - 1 , 7 7 + 1).\T n = l



But we observed earlier that Tj(n — 1, -1 ) = T](n - 1, n) = jj(n - 1, n + 1) = 0. Thus,n —1

T}(n) = 2 7j(n - 1, m) + 77(71 - 1,0)171=1

n —1

= 2 ^ 77(71 - 1 , 777) - 77(77 - 1, 0 )m = 0

= 277(77- 1) - 77(77- 1,0)

= 2W n-i - 77(77 - 1,0).

If 77 is even, say n = 2k for some integer k, then 77 - 1 is odd and so 77(77 - 1 , 0 ) = 0. Thus,

”(") = = KT-i') = (ï-iVC 'O “ (?) “If 77 is odd, say 77 = 2A: + 1 for some integer k, then 77(77 - 1,0) = 77(2*, 0) = C*. In

this case. Lemma 4.4.3 implies that

77(77) = 2 K , _ i - C t = 2 ^ ^ ^ - C k = = W„.

In either case, 77(77) = W„, as desired. □

Proposition 4.4.5. Let |Q| = 77 < 00 077d let X q be binary. Then the number of strongly

preseparable orders on Q is given by the formula

k=l ' '

Proof. Let Q = {1,2, .. . , 77} and let Q' = ^ - { 77}. Proposition 4.3.10 establishes that every

strongly preseparable order >- on Ç is a symmetric preseparable extension of a strongly

preseparable order on Q'. Thus, to form a strongly preseparable order on Q, we must:

(i) choose a strongly preseparable order >-i on Q',

(ii) choose a linear order >-2 on (either 1 >- 0 or 0 >- 1), and

(iii) form a symmetric preseparable extension of by >-2.

It is clear that (i) may be completed in 07(77 - 1) distinct ways, and (ii) in 2 distinct

ways. Since every linear order on X q' has height 2"“ \ Propositions 4.3.9 and 4.4.4 imply

that (iii) may be completed in W^n-i = (jn-i) distinct ways (since we must choose a

symmetric characteristic sequence of length 2"). Furthermore, Corollary 4.1.4 implies that

all strongly preseparable orders formed in this manner will be distinct. Thus,

0,(77) = 2 - 07(7 7 - 1 ) . Q „ _ 2 ) -

It is clear that o,(l) = 2 and so a simple induction argument finishes the proof. 0



4.5 Other Combinatorial Results

The reader may wonder at this point why we have tackled the problem of counting sym

metric, preseparable and strongly preseparable orders, but have avoided the problem of

counting additive, separable, and monoseparable orders. Unfortunately, the answer to this

question is more pragmatic than it is enlightening. Simply put, the elements of the latter

three classes of orders seem to be rather difficult to count.

Consider, for example, the class of monoseparable orders. Kilgour [13] appecirs to have

been the first to observe that the problem of determining the number of monoseparable

binary preference orders on a criteria set Q of cardinality n is equivalent to the problem

of determining the number of linear extensions of the partial order on P(Q) induced by

inclusion (that is, the partial order y defined by Si y Sg 4=^ Si Ç Sz). In general,

the problem of counting linear extensions of a given partial order is very hard. In fact,

Brightwell and Winkler [5] have shown that this problem is #P-Complete, meaning that

it is among the hardest of the problems in the complexity class # P (the class of counting

problems associated with decision problems in NP). The # P problem corresponding to any

NP-Complete problem is NP-Hard, meaning intuitively that it is at least as hard as the

hardest NP problems.^

Sha and Kleitman [20] dealt specifically with the problem of counting linear extensions

of the partial order induced by subset inclusion, arriving at asymptotic upper and lower

bounds. Nevertheless, no progress has been made on a closed formula, and, given the

complexity of the general problem of counting linear extensions, it seems unlikely that such

a formula will be obtained.

We suspect that the problems of counting separable and additive orders are also quite

difficult, though this suspicion is presently nothing more than an educated guess. Thus, we

leave the following for future research:

Open Question. Is it possible to find a closed formula for the number of separable (or

additive) binary preference orders on a given finite criteria set?

'Formally, “a problem is in # P if there is a nondeterministic, polynomial-time Turing machine

that, for each instance I of the problem, has a number of accepting computations that is exactly

equal to the number of distinct solutions for instance I (quoted from [1]).”



The table below summarizes the known combinatorial results pertaining to binary

preference orders on smEill criteria sets.

\Q\-- 2 3 4 5 6

Monoseparable 8 384 26,886,144 ? ?

Preseparable 8 224 641,000 ~ 4.5 X 1013 ~ 5.0 X 103°

Symmetric 8 384 10,321,920 ~ 1.4 X lO'G ~ 1.1 X 1045

Strongly Preseparable 8 96 13,440 ~ 3.5 X 10» ~ 4.2 X 1017

Separable 8 96 5,376 ? ?

Additive 8 96 5,376 ? ?

Lexicographic 8 48 384 3,840 46,080

Table 4.1: Counts of certain classes of orders on small criteria sets.

4.6 Summary of Related Properties

We conclude this chapter by summarizing our results concerning the various properties

related to separability. Recall that we have shown the following inclusions between the

various classes of preference orders on a finite criteria set:

lexicographic Ç additive Ç separable

Ç strongly preseparable Ç preseparable Ç monoseparable.

For binary preferences on a criteria set having at least 5 elements, all of these inclusions

are proper. For smaller criteria sets, we note the following:

• When the criteria set has 4 elements, additivity and separability are equivalent and

all other inclusions are proper.

• When the criteria set has 3 elements, separability, strong preseparability and addi

tivity are equivalent and all other inclusions are proper.

• When the criteria set has 2 elements, we have the following result:

Proposition 4.6.1. Let |Q| = 2 and let y be a binary preference order on Q. The following

are equivalent:



(i) >- is lexicographic(ii) >- is additive

(iii) >- is separable(iv) >- is strongly preseparable(v) >- is preseparable

(vi) >- is monoseparable(vii) >- is symmetric

Proof. By our above remarks, (i) =>• (ii) => (iii) => (iv) =*- (v) => (vi). Since |Q| = 2,

the only proper subsets of Q are singletons. Thus, separability and monoseparability are

equivalent. But since the only normalized separable preference order on 2 criteria is the

standard lexicographic order, all separable preference orders on Q are lexicographic. Thus,

all monoseparable preference orders on Q are lexicographic, and so (vi) => (i). We now

show (vii) O’ (i). Notice that, by Proposition 4.4.1, there are exactly 8 symmetric orders

on Q. But this is exactly the number of lexicographic orders on Q, by Proposition 3.3.7.

Since every lexicographic order on Q is separable and thus symmetric, and it follows that

>- is symmetric if and only if >- is lexicographic, which completes the proof. □


Chapter 5

Separability and The Symmetric

Group

5.1 The Canonical Action

For this chapter, we assume that Q is a finite criteria set, say Q = {1,2, . . . , n}, and that

X q is binary. Unless otherwise noted, we assume also that n > 2. Let denote the set

of all linear orders on X q , i.e. the set of all binary preference orders on n criteria. We

wish to define an action of the symmetric group S2 " on the set On- Thus, let >- € C?„ be

given and let (i*) be its corresponding order sequence. For any a € 52», define a{>~) to be

the linear order whose corresponding order sequence (y&) is given by V a ( k ) = ik , or more

conveniently, ÿ* = x„-nk)- Observe that <7(y) is well-defined since a is a permutation and

is thus bijective. We clmm that the map {a, >-) i-> o-(>-) is indeed a group action of 52» on

On-

To verify this claim, first notice that, if we denote by 1 the identity permutation, then

!(>-) is the the linear order whose order sequence (t/*) is given by

Vk — ~ ^l(fc) — ^k-

Thus !(>-) = >-. We now must show that, for all o-,r e 52», (ctt)(>-) = a{T{y)). Recall

that (<rr)(>-) is the linear order whose order sequence is given by

V k = ^ ( < r r ) " ’ (fc) — 3 ^ (T -* ff-> )(A :) = ^ T - ' { < r - ^ { k ) ) -

45


Chapter 5. Separability and The Symmetric Group 46

Let x ' = T(x) and let (wk) be the order sequence corresponding to y' . Then, by definition,

Wk = Xr-i(k)- Now let (zk) be the order sequence corresponding to a{y') = ct(t(>-)). Then

Z k = W f f - n ^ k ) — 2 ^ t - ' ( o — ‘ ( * ) ) — V k t

which implies that (y*) = (2*). Thus (o’r)(>-) = c t ( t ( > - ) ) , as desired.

We call this action the canon ica l a c tio n of S 2 ” on C7„. It is straightforward to verify

that the canonical action is both faithful and transitive. Moreover, the stabilizer of any

order y £ On is trivial. '

5.2 Symmetry-Preserving Permutations

Recall that, if y is a symmetric order on X q with corresponding order sequence (z&), then

Xk = i 2"-*+i for each 1 < A: < 2". For convenience and to be consistent with this notation,

we define A: = 2" - A: 4-1, so that >- is symmetric if and only if Xk = a:* for each k.

Definition 5.2.1. A permutation a £ S2» is said to preserve s y m m e tr y if, whenever >- is a

symmetric order on X q , then <r(>-) is also symmetric.

We denote by 52» the set of symmetry-preserving permutations in 52». Similarly, we

denote by On the collection of all symmetric orders on X q .

Proposition 5.2.2. T h e s e t 52» o f sym m e try -p re se rv in g p e rm u ta tio n s is a subgroup o f 52».

P roo f. The identity permutation clearly preserves symmetry, and so 1 € 52». Since 52» is

a finite subset of 52», it now suffices to show that 52» is closed under function composition.

Thus, let a , T £ 52» and let >- be a symmetric order on X q . Since r preserves symmetry,

r{y) is symmetric. But since tr preserves symmetry, {crr){y) = cr(r(>-)) is then symmetric,

as desired. □

Proposition 5.2.3. A p e rm u ta t io n a £ 52» preserves s y m m e tr y i f a n d o n ly i f

a { r ) = s = y cx{f) = s

f o r a ll 1 < r , s < 2".

‘Recall that a group action (G,A) i-> A is said to be faithful if its kernel is the identity, i.e. if

{cr € G I o-(a) = a for all a 6 A} = {!}. The action is Scdd to be transitive if it has only one orbit,

i.e. if, for all a, 6 € A, there exists a £ G such that a{a) = b. The stabilizer Ga of an element

a e A is the set of all <r € G such that er(a) = a, i.e. Ga = {<r € G | cr(o) = a}.



Proof. Let >- be any symmetric order on X q , and let (a:*) and (j/*) be the order sequences

corresponding to >- and o-(>-), respectively.

( ^ ) Suppose that a preserves symmetry and that a(r) = s for some r and s. By the

definition of the canonical action, ys = ya{r) = ^r- But since a preserves symmetry, both >-

and <t(>-) are symmetric. Thus, y j = y i = = xr, which implies that a(f) = s, as desired.

(<=) For the converse, suppose that a{r) = s = > cr(f) = s for all r and s. Then for

each k, a~^{k) = a ^(k) and so

Vk = ^tr~Hk) = ~ â~^(k) ~

which implies that cr{y) is symmetric. Since our choice of >- was arbitrary, it follows that

(7 preserves symmetry. □

Proposition 5.2.4. Let >- be symmetric. Then <r{>~) is symmetric if and only if a preserves

symmetry.

Proof. The reverse implication is immediate. Thus, assume that a{>~) is symmetric and

assume, to the contrary, that a does not preserve symmetry. Let (x*) and (yk) be the

order sequences corresponding to >- and a(>-), respectively. By Proposition 5.2.3, there

exist r and s such that cr(r) = s and a{f) s. But then, by the definition of the canonical

action, y, = Xr and yj ^ x?. Since both >- and <r(>-) are symmetric, this then implies that

y l = y j ^ Xt = x^ = yl, a contradiction. □

Proposition 5.2.5. I f a preserves symmetry and <x(y) is symmetric, then >- is symmetric.

Proof. Suppose that a preserves symmetry and that a{>-) is symmetric. By Proposition

5.2.2, a~^ preserves symmetry. But then it must be the case that >-= is sym

metric, as desired. □

Note that the contrapositive of Proposition 5.2.5 says that symmetry-preserving per

mutations must also preserve asymmetry. Specifically, we know that if a preserves symmetry

and >- is asymmetric, then ct(>-) must also be asymmetric.

Proposition 5.2.6. There is a bijection between the set 52" of symmetry-preserving permu

tations and the set 0 „ of all symmetric orders on X q . In particular,

152-1 = |0 „ | = 22"”‘ .2”-M



P roo f. Define a map y : 5 2 " -> On by <p(cr) = o^(>-/ei)- Now - be any symmetric order on X q

and choose a 6 5 2 » so that <r(>-) = (the transitivity of the canoniccd action implies

that such a cr exists). Since both >-/ex and >- are symmetric, Proposition 5.2.4 implies that

a preserves symmetry. Thus, cr~^ preserves symmetry and

as desired. It follows that y; is a bijection and so Proposition 4.4.1 implies that

|52"| = 22”" ’ □

Proposition 5.2.7. L e t a G 5 2 » h a ve cyc le d eco m p o sitio n a = t i T 2 - ■ -Tq. T h e n a p reserves

s y m m e tr y i f a n d o n ly if, f o r each i < q , e ith e r

(i) Tj = (a:i,X2, .. . , X j , x i , X 2 , ■ ■ ■ , x j ) , w here x i ^ x* fo r a ll k , I < j ; o r

(ii) Tf = (x i, X2 , . . . , Xj ) , w here x, ^ x* fo r a ll k , I < j a n d f , = (xi,X2 , . . . , x j ) = Tp

f o r so m e p f ^ i -

P roof. (=») Suppose that a = tiT2 . . . r , preserves symmetry and let n = (xi , X2 , . . . , x„)

be given. Let j + 1 be the smallest index such that X j+ i = xj for some I < j + 1 . If there is

no such index, then (ii) occurs by Proposition 5.2.3 and the proof is finished. Thus, assume

that such an index exists. We claim that 1 = 1.

Suppose, to the contrary, that 1 ^ 1 . Since a { x j ) = x/. Proposition 5.2.3 implies

that cr(xj) = x/. But since 1 7 1, this implies that Xj — x/_i, from which it follows that

Xj = X/ - 1 with 1 - 1 < i , a contradiction to the minimedity of j + 1 .

Thus, 1 = 1 and so Proposition 5.2.3 now implies that Ti = (xi , X2 , . . . , Xj, x i , X2 , . . . , Xj )

with xj ^ Xk for all k , I < j , which is exactly case (i).

(*=) For the converse, suppose that, for each i < q, either (i) or (ii) holds. It is easy

to see, in this case, that a ( r ) = s = > <r(f) = s for all r and s . Thus, Proposition 5.2.3

implies that cr preserves symmetry, as desired. □

For a cycle r = (xi,X2 , . . . ,Xm) we define the dual of r , denoted f , by

T = (xi,X2 , . . . , Xm) , as in Proposition 5.2.7. If t = t (as in (i) above) then we say that

T is se lf-dua l. Using this terminology, we see that if cr preserves symmetry and r appears

in the cycle decomposition of a , then r is either self-dual or f also appears in the cycle

decomposition of cr. Since self-dual cycles necessarily have even length, we conclude the

following:



Corollary 5.2.8. I f the cycle decomposition of a contains an odd number of cycles of odd

length, then a does not preserve symmetry.

Another obvious corollary of Proposition 5.2.7 is this:

Corollary 5.2.9. The transposition {a, b) preserves symmetry if and only if it is self-dual.

There is a natural way to extend the notion of the dual of a cycle to an arbitrary

permutation a in 52». Specifically, if o- = t iT2 • • • where ti , T2, . . . , r„ are disjoint cycles,

we define the dual of a by ct = t i T2 - fg. Equivalently, a = 70-7 “ ^, where

7 — (^11 ^ 2 } ' ' ' (3:2» —l,X2»-l).

Using this latter definition, we observe that the map «7 o is a homomorphism, since

<TT i-> 7a r 7 "^ = (7 0 7 "^)(7 7 7 "^) = ô f for all <r, t € 52».

Proposition 5.2.10. The group 52» of symmetry-preserving permutations is isomorphic to

the group of symmetries of a 2"~*-dimensional hypercube.

Proof. Let 5 denote the group of symmetries of a 2"“ -dimensioned hypercube. Any such

symmetry can be obtained by permuting the 2"~* dimensions of the hypercube and then

reflecting over a subset of these dimensions. A straightforward argument thus shows that

5 S [%2] X 52»-i = Z2 I 52»-i ,

where Z2 denotes the cyclic group of 2 elements (see [9]). We claim that 52» is iso

morphic to this same group. Specifically, define 5^ = ( (a,5) | 1 < a < 2"“ ^) and

5^ = (oô I (7 6 52»-i ). By Proposition 5.2.7, both 5 ‘ and 5^ are subgroups of 52». We

will show that 5^ = [^2] " ’ , 5 s 52»-i, and that 5 = 5* x 5^.

To show that 5* = [^2] " *, we note that 5^ is an abelian group of order 2^" ' and

that each nonidentity element of 5^ has order 2. Up to isomorphism, there is only one such

group, namely [^2] " *, as desired.

Next, note that in order to define a homomorphism (f : S^ 52»-i, it suffices to

specify the action of on the generators of 5^ and then show that y> preserves multiplication

among these generators. Thus, we define y> by = a. For all C7, r G 52»-i, the cycle

decompositions of a and r are disjoint and so â and r commute. Thus,

ip{aâTT) = = (^[(c7t)(ctt)] = (7T = (p{aa)ip{Tf),



and therefore y is a homomorphism. It is clear that ip is bijective and so we have shown

that 5^ S 52»-!.

It remains to show that 5 = » 5^. First, we claim that S^ is a normal subgroup of

52". To verify this claim, let cr = {xi,xi){x2 ,X2 ) • ■ ■ {xm,Xm) € and let r e 52". Then

TFT-I = (r(ii),T(ïi))(T(a:2),r(Ï2)) • • • (T (a :„ ,),r(x„)).

But since r preserves symmetry, r(xi) = r(x,) for for all i,-. Thus

rcrr"* = (r(xi) ,r (xi))(r(x2) ,r(x2)) • • • (r(x,„),r(xm)) e S \

which proves that 5* is normal in 52». FVom this it follows that 5^5^ is a subgroup of 52».

But 5^ n 5^ = 1, and so 5^5^ = 5^ x 5^. Furthermore, Proposition 5.2.6 implies that

|5 '5 ^ | = |5 '||5 ^ | = 22"“‘ .2"-M = 152»|.

Thus 52» = 5 ^ X 5^, which concludes the proof.^ □

5.3 Separability-Preserving Permutations

We now turn our attention to the set of permutations that preserve the property of sepa

rability.

Definition 5.3.1. A permutation a € 52» is said to preserve separability if, whenever >- is a

separable order on X q , then a{>-) is also separable.

2W e co n fess th a t ou r n o ta t io n h ere is a b it a m b ig u o u s. B y 5 * x 5*, w e m ea n th e se m id irec t

p ro d u ct o f S i an d 5% w ith re sp e c t to th e h o m o m o rp h ism q : 5 A u t (5 *), w h ere a m a p s 7 6 5 *

t o th e a u to m o rp h ism o f co n ju g a tio n b y 7 o n 5 *. B y [^2]*” * x S g . - i w e m ean th e se m id irec t p rod

u c t o f Z2 x 2% x x Z2 a n d 5z n - i w ith resp ec t to th e h o m o m o rp h ism p : 5j n - i A u td ^ a )* ” ' )

d efin ed b y { p { a ) ) { z i , Z 2 , . . . , Z n ) = ( z , ( i ) , z , ( 2 ) , - B y d e fin it io n , th is sem id irec t p ro d u ct

is th e w rea th p ro d u ct o f Z 2 a n d S j n - i , w h ich w e d e n o te b y Z2 I 5 2 n - i . N o te th a t e le m e n ts o f

s ' h a v e th e form ( i i , t i ) ( t 2 , i 2 ) • ■ • ( i m , » m ) for 1 < »* < 2" " ' . T h u s , w e ca n e x p lic it ly sp ec ify th e

iso m o rp h ism ^ : S ' [Z2]*" * b y se n d in g th e e lem en t ( n , t i ) ( i2 ,« 2 ) • • • (* m .•»>) to th e v ecto r

h a v in g I s in th e i k c o m p o n e n ts a n d Os elsew h ere. U n d er th is iso m o rp h ism a n d th e iso m o rp h ism

y : S * S a n -I sp ec ified in ou r p ro o f, i t is stra igh tforw ard t o v er ify th a t th e h o m o m o rp h ism s a

a n d P in d u c e th e sa m e se m id ir e c t p r o d u c t. W e a lso n o te th a t w e d o n e e d to u se th is m ore co m p li

c a te d m ach in ery ( in s te a d o f th e s im p ler d irect p r o d u c t) to d esc r ib e th e grou p . In d eed , s in ce S * is

n o t n orm al in Sa» ( th is ca n b e e a s ily v erified ), it fo llow s th a t S2» ^ S ' x S* ^ S ' x S * .



We denote by SJ» the set of separability-preserving permutations. Similarly, we denote

by O* the collection of all separable orders on X q .

Proposition 5.3.2. The set S^n of separability-preserving permutations is a subgroup o/Sg».

Proof. Let a € SJn and choose some >- € O*. Since a preserves separability and >- is

separable, it follows that a{y) is separable. But then Proposition 4.3.3 implies that both y

and o-(y) are symmetric. It then follows from Proposition 5.2.4 that cr preserves symmetry.

Thus, a e 52" and so we have shown that 5 |" Ç S i" . The proof that S^n is a subgroup is

analogous to the proof of Proposition 5.2.2. □

Proposition 5.3.3. I f a preserves separability and cr(>-) is separable, then y is separable.

Proof. The result is immediate, since y = a~^{a{y)) and, by Proposition 5.3.2,

preserves separability. □

We note here (analogous to our remarks following Proposition 5.2.5), that Proposi

tion 5.3.2 implies that a separability-preserving permutation also preserves nonseparability.

Unfortunately, not all of the nice properties of S2" carry over to their analogs in 5jn. For

example, consider the action of the permutation a = (2,4,7,5) on >-iex-

A 1 A A 1 A A 1 A

1 1 0 0 1 1 0 0 1

1 0 1 1 0 1 1 0 1

1 0 0 a 1 1 0 0 0 1 1

0 1 1 0 0 1 1 0 0

0 1 0 0 1 0 0 1 0

0 0 1 1 0 0 1 1 0

0 0/ V 0 0/ v° 0 V

Notice that both y/ex and ar(>-iex) are separable, eind yet <r{(T{>~iex)) = cr^ly-Ux) is

nonseparable. It follows that a does not preserve separability.

We see then that, while the property of preserving symmetry is global in the sense

that if cr preserves the symmetry of one symmetric order, then cr preserves the symmetry of

all symmetric orders, the property of preserving separability is more localized. Specifically,

we see that it is possible for a permutation to preserve the separability of one separable

order while failing to preserve the separability of another. To formalize this local property,

we make the following definition:



Definition 5.3.4. Let >- be a separable order on X q . A permutation a € 52» is said to

(locally) preserve the separability of >- if cr{>-) is separable.

An obvious consequence of this definition is the following:

Proposition 5.3.5. A permutation a € S r, preserves separability if and only if a preserves

the separability of each >- G

For a separable order >-, we denote by 5^^ the set of permutations which preserve the

separability of >-. In the special case where y = >-iex, we write insteeid.

Notice that if cr preserves the separability of some separable preference order y , then

both >- and <r(>-) are symmetric (by Proposition 4.3.3), from which it follows by Proposition

5.2.4 that a preserves symmetry. Thus, 5 ^ Ç Sa» for each >- G O '.

Is S^„ a subgroup of Sa»? We know, by the previous example, that for >- = yux,

the answer to this question is no (since a preserves the separability of >-jex and does

not). On the other hand, it would not be entirely unreasonable to think that we might be

able to find some separable order >- for which things turn out dififerently and for which

is indeed a subgroup of 5a». The reality of the situation is that, for n > 3, no such

order exists: For each >- G O ', 5 ^ fails to be a group. We must, however, delay the proof

of this claim until we have built up some stronger machinery (the result and its proof are

presented near the end of this chapter as Proposition 5.4.37).

Proposition 5.3.6. Let y G 0„. Then |5 ^ | = |0* |.

Proof. We prove the result by exhibiting a bijection between and O '. Let tp : S^n -4 O '

be the map defined by ip{cr) = cr(y). Since >- is separable and a preserves the separability

of y is well-defined. To see that tp is injective, recall that

<7i(>-) = (Ta(>“) <=> stabilizes >- = 1 <=> a\ —a^.

To see that tp is surjective, note that, for any >-' G O ', there exists a G 5a" such that

(r{y) = y ' (by the transitivity of the canonical action). But since >- and y ' are both

separable, it follows that cr G 5 ^ . Thus tp is surjective, as desired. □

Corollary 5.3.7. For all Xi, >-a 6 O ', |5 ^ ‘| = |5 ^ “|.



5.4 The Structure of

We now turn our attention to determining the structure of the group ■ As we observed

earlier, this group does not behave as nicely as 52". As such, it will be considerably more

difficult to characterize. We will give the punchline first and then build up the theory

necessary to prove our claims.

Let V4 denote the Klein 4-group and let D4 denote the group of symmetries of the

square (the dihedral group of degree 4). Recall that V4 = Z2 x Z2 .

Theorem 5.4.1.

(i) Forn = 2, 5^. = 5 J S D 4 .

(ii) For n = 3, S^n = Sg — V4 x Sa-

(iii) For n > 4, S^n V4 .

The proof of Theorem 5.4.1 is, for the most part, combinatorial in nature. The cases

for n < 4 are verified by hand and/or computer using the machinery of Proposition 3.1.3.

An induction argument then completes the proof.

We should point out that we actually prove a stronger claim than that which is asserted

by Theorem 5.4.1. Specifically, we show that a e S^n if and only if its inversion number

inv(a) belongs to the set jo , 1, (^ ) - 1, (^ ) j . We then prove that this set of inversion

numbers corresponds uniquely to a set of permutations isomorphic to V4.

The glue that holds the induction together is the claim that if 5j„ contains a permu

tation (T for which 1 < inv(ff) < (^ ) - 1, then 5g._i contains a permutation a' for which

1 < inv(<r') < ( 2 ) " 1- The proof of this fact is lengthy and, at times, tedious. Never

theless, it is not overly complicated and the patient reader should have no trouble with the

details. Of course, one can skip many of the proofs of the intermediate results that follow

and still gain a good understanding of the machinery used to prove the main theorem.

We begin by developing some tools that will allow us to translate results about S^n

into results about

Let V € O* and let (z&) be the order sequence corresponding to >-. Then we define a

map Pi : O* -4 O^-x, called the i*'* projection map on C>*, by Pii>~) = >~Q-{qi]- This map

is well-defined by Corollary 2.2.7.



Lemma 5.4.2. Let >- e C7*_i and let >-i 6 O i (so either 1 >-i 0 or 0 >-i 1 / Then

(i) Pi(>-i @ X) = X.

(ii) P n { y © >"i) = X.

Proof. This follows directly from Definition 3.3.1. □

For a fixed >- € O*, the projection maps defined above induce maps Sj : S^n S^^-i

in the following manner: For cr 6 SJ™, let o' € 52—1 be the unique permutation for which

‘ '(Pi(>~)) = Pi(^(>~)}- Since >- is separable and a preserves separability, o' is well-defined.

Now define Si to be the map which sends o to o' ?

Definition 5.4.3. Let o e Sm and let a and b be integers such that 1 < a < 6 < m. We say

that o inverts the pair (a, 6 ) if o(a) > o{b). We denote by inv(a) the number of distinct

pairs inverted by o.*

Note that, for o- € 5m, 0 < inv(a) < (^2 )• Furthermore, inv(<z) = 0 if and only if

0 = 1 and inv(tr) = { \ ) if and only if cr = r^ , where Tr is the reflection permutation,

defined by Tr{a) = m — a -t- 1 = ô for all o.

Proposition 5.4.4. Let y E O* have order sequence (xk), let o E S \^ , and let i < n be

given. Let (y&) be the order sequence corresponding to pi{>~) and let a ,6 < 2” be such that

= (ÿoi) end Xb = i vd j ) for some c,d < 2"“ and some j € {0,1}. Then o inverts

(a,b} if and only if Si{o) inverts (c,d).

Proof. Let Xa = {Vcj) and Xb = (vd,j) be as above. Without loss of generality, assume

that Xa >■ Xb, so that a o(a) > o(b)

{Vdij) = Xb y Xa = {Vcij)

Vd >-i Vc-

®Note th a t Si d e p e n d s o n ou r ch o ice o f X. T h is n o ta t io n a l a m b ig u ity w ill ca u se n o rea l d iff icu ltie s ,

as th e choice o f X sh o u ld b e m ad e clecir by th e c o n te x t in w hich s< a p p ea rs .''In th is d e fin it io n , ou r p a irs are u n ord ered , in t h e se n se th a t w e d o n o t d is t in g u ish b e tw een

( a ,6 ) a n d { h , a ) . O u r co n v en tio n w ill b e to lis t th e sm a lle r o f o a n d h first w h en ev er th e ir re la tiv e

s iz e s are kn ow n .



But since Si{a) is the unique permutation for which

= S i (a ) ( p i { y ) ) = P i ( a ( y ) ) = P i ( y ' ) =

it follows that a inverts (a, 5) if and only if yd >~i He, if and only if Si(cr) inverts (c,d). □

Lemma 5.4.5. The permutation arinverts {a,b) if and only if one of the following condi

tions holds:

(i) T inverts {a,b) and a does not invert {T{b),T{a) ); or

(ii) T does not invert (a,b) and a inverts ( 7 (0 ), T(5) ).

Proof. If either (i) or (ii) holds, then n(r(a)) > n (%(&)), as desired. Now suppose conversely

that CTT inverts ( 0 , 6 ). Then cr(r(a)) > o-(r(6)). If t inverts ( 0 , 6 ), then t ( 6 ) < r(o), which

implies that a does not invert ( t ( 6 ) , t ( o ) ). On the other hand, if r does not invert (o ,5),

then r(a) < t ( 6 ) , which implies that a inverts ( r ( a ) ,r ( 6)). Since r must either invert or

not invert the pair ( a ,5), it follows that either (i) or (ii) must occur. □

Lemma 5.4.6. Let a G . Then inv(rr<T) = (^ ) — inv(n).

Proof. Since Tr inverts all possible pairs. Lemma 5.4.5 implies that XrO inverts ( a, 6 ) if and

only if a does not invert (a,b). The result then follows from the fact that there are (^ )

possible pairs. □

Lemma 5.4.7. If a inverts (a , 6 ), thena~^ inverts {<7 {b),(r{a)).

Proof. Suppose a inverts (o ,6 ). Then a{b) < <r{a) and a~^{a{b)) = b > a = cr~^{cr{a)),

which implies that inverts ( ff(b), a{a) ). □

The following proposition implies that a permutation is uniquely determined by the

set of pmrs that it inverts.®

Proposition 5.4.8. I f a ^ r, then there exists a pair ( a, 6 ) such that (a,b) is inverted by

exactly one of a or t .

Proof. Suppose a ^ t . Then ar~^ ^ 1. By our comments following Definition 5.4.3, this

implies that crr“ inverts some pair (a,b). Thus, by Lemma 5.4.5, it must be the case that

either

(i) T~ inverts ( 0 , 6 ) and cr does not invert (r~^(6),T“ ‘(a)); or

®This s e t o f p a irs is so m e t im e s referred to a s th e in v e r s io n tab le o f th e p erm u ta tio n .



(ii) does not invert ( a ,6 ) and a inverts (r"^(o),r~ ^(6}).

If (i) occurs, then Lemma 5.4.7 implies that r inverts (T~^(b),T~^(a) ) (and a does not). If

(ii) occurs, the same lemma implies that r does not invert (r"^(o),r~*(6)) (and cr does).

In either case, the pair ( r~* (a), r ” * (b) ) is inverted by exactly one of tr or r . □

Proposition 5.4.9. Let >-i E and let >-2 E Then, for any a £ SJ»,

(i) The permutation si(cr) induced by p i (>-2 ® >-1) does not depend on the choice of

Xi. Furthermore, si{a) E S^n-i-

(ii) The permutation s„(<t) induced by pn{>-\ ® >-2) does not depend on the choice of

>-1. Furthermore, Sn(o’) € S^n-i-

Proof. We will prove (i) and leave the analogous proof of (ii) to the reader. Assume, without

loss of generality, that 1 >-2 0 and let (a:*), (y*) be the order sequences corresponding to >-i

and >-2 ® >-1, respectively. Let a E SJn be given and let a' = 5i(<t) E 52»-1. Then a' is the

unique permutation for which cr'(pi(>-2 ® >-i)) = pi(cr(>-2 ® >-1)). But by Lemma 5.4.2,

Pi (>-2 ® >-1) = Xi and so a' is the unique permutation for which <r'(> i) = pi (<r(>-2 ® >-1)).

Thus, for emy pair ( a ,6 ), we have

cr'inverts ( 0 , 6 ) <=> X(, <r'(>-i)

-<=* Xb Pi(cr(>-2 ® >^l)) Xa

(l,Xft) cr(X2 ® ^ 1) (l,Za) and (0, x&) <r(>-2 ® >-1) (0 ,Xo)

Vb (r(X2 © Xi) Va and yb+2"-'- © Xi) î/o+2"->

(by Proposition 3.3.5)

or inverts both ( a ,6 ) and (o + 2"“ *,6 + 2'*“ ).

Thus, the inversion table for cr' is completely determined by the inversion table for

cr, which clearly does not depend on the choice of >-1. By Proposition 5.4.8, cr' is uniquely

determined by its inversion table. Consequently, a' = si(cr) is uniquely determined by a

alone and hence does not depend on the choice of >-1.

To see that cr' = si(cr) E observe that, for any >-i E 0*_i, cr(X2 © >-1) is

separable and thus, by Corollary 2.2.7, pi(cr(>-2 © Xi)) is separable. But then, by Lemma

5.4.2 and the definition of si(cr), we have

cr'(Vi) =cr'(p i()^2 © V i)) =pi(cr(^2 © Xi)),



which we just observed to be separable. Since >-i was chosen arbitrarily, it follows that

a ' = si(cr) e S^n-i- □

Lemma 5.4.10. I/cr inverts {a,b) and a < c < b , then a inverts either (a ,c) or {c,b).

Proof. Suppose, to the contrary, that a inverts neither (a ,c ) nor (c ,6 ). Then

a{a) < a(c) < a{b), a contradiction to the assumption that a inverts {a,b). □

Lemma 5.4.11. I f a inverts (a,b) and (b,c), then o inverts {a,c).

Proof If cr inverts (o, 6 ) and a inverts ( 6,c ), then a(a) > a{b) > a(c), as desired. □

Lemma 5.4.12. Let a 6 52». I f cr inverts {a,b), then a inverts (b,a).

Proof Suppose a inverts ( a ,6 ). Then a(a) > cr(b), which implies that

a{b) = a{b) > a{a) = cr(a),

as desired. □

Proposition 5.4.13. Let a € 52». J/inv(<7) = 1, then a = (2"“ , 2”“ + 1).

Proof. Suppose inv(o-) = 1 and let (o ,6 ) be the pair inverted by cr. By Lemma 5.4.10,

6 = o + 1, for otherwise there would exist c with a < c < b and hence another inversion.

By Lemma 5.4.12, a also inverts (S,o) = (a + l ,o ) , which implies that o = a + 1 =

2" — a. Thus, o = 2"“ and the pair inverted by a is exactly (2""^,2""^ + 1 ). Since the

permutation (2"~^, 2"~^ + 1) inverts this pair and no others, Proposition 5.4.8 implies that

<t = (2" - S 2" - i + 1). □

We call the permutation (2"“ S 2 "” + 1) the central transposition, denoted by Tc-

Lemma 5.4.14. //inv(a) = ( \ ) — 1, then a = t^Tc.

Proof. If inv(cr) = (^ ) - 1, then inv(rrcr) = 1 (by Lemma 5.4.6). Thus, TrO = Tc, which

implies that a = TrTc- □

Definition 5.4.15. Let >- Ç 0„ and let (x*) be the order sequence corresponding to >-. The

4-tuple (a,b,c,d), where a -) if for some S C Q, there exist y, z E X s and u, v £ X - s such that

Xa = iy, u) Xc = (y, v)

Xb = (Z, U) Xd = ( z , v )



Proposition 5.4.16. Let a 6 Sq" and let >- e O^. Then a preserves the separability of if

and only if, for each complete set (a,b,c,d), a inverts either both or neither of the pairs

(a,b), (c,d).

Proof. ( ^ ) We prove the contrapositive. Thus, suppose that there exists a complete set

(a,b,c,d) such that a inverts exactly one of ( a ,6 ) and (c ,d ). Without loss of generality,

assume that {a,b) is the inverted pair. Let S C Q and Xa, Xb, Xc, Xd be as in Definition

5.4.15 and let >-' = n(>-). Then Xb >' Xa (since Xa >■ and Xc >-' Xd- This, however,

implies that S is not X'-separable, since z y ' y given a choice of u on X - s but y y ' z

given a choice of v on X - s . Thus y ' = (x(y) is nonseparable and so o does not preserve

the separability of >-.

(4=) For the converse, suppose that a does not preserve the separability of >-; that

is, suppose that x ' = a{y) is not separable. Then for some S C Q, there exist y, z e X s

and u , v e X - s such that (z,u) y ' {y,u) and {y,v) y ' {z,v). But since y is separable,

either (y,u) y (z,u) and (y,v) y (z,v) or (z,u) >- (y,u) and (z,v) y (y,v). Without loss

of generality, assume the former. Let (x*) be the order sequence corresponding to y and

let a, b, c, d be the indices such that

Xa = {y,u) Xc = {y,v)

Xb = {z,u) Xd = {z,v).

Then (a,b,c,d) is a complete set. Furthermore a inverts (a,b) (since xj, y ' x„) but does

not invert (c ,d ) (since Xc >-' x^). □

Corollary 5.4.17. Let a E S2» and let >-i,>-2 6 O*. If y i and y 2 have the same complete

sets, then a preserves the separability of >-i if and only if a preserves the separability of

>~2-

Corollary 5.4.18. A permutation a belongs to S^n if and only if a preserves the separability

of each normalized binary preference order in O*.

Proof. The forward implication is immediate. For the converse, suppose that a preserves

the separability of each normalized binary preference order in O*. For any >- G C>* there

exists a normalized binary preference order y ' € such that >- can be obtained from y '

by reordering the criteria and replacing a subset of the components of each alternative with

their bitwise complements. Clearly this process preserves the complete sets of >-' and so

>- and y ' have the same complete sets. Since a preserves the separability of >-', Corollary

5.4.17 implies that <r preserves the separability of >-. Since >- was chosen arbitrarily, it

follows that cr € 5 |n . □



Corollary 5.4.19. The central transposition Tc belongs to S^^.

Proof. Let >- € O* and let (a:*) be the order sequence corresponding to >-. Then y e 0„

and so Ï 2" - ‘ = X2"->+i- Thus, there does not exist a complete set {a,b,c,d) for which

a = and b = 2""^ +1 or for which c = 2”“ * and d = 2"“ + 1. Since (2”“ S 2"“ + 1 )

is the only pair inverted by Tc, Proposition 5.4.16 implies that Tc preserves the separability

of >-. Since our choice of >- was arbitrary, it follows that Tc G . □

Corollary 5.4.20. The reflection permutation Tr belongs to S^n.

Proof. For any complete set (a,b,c,d), Tr inverts both {a,b) and (c ,d ). Thus, by Propo

sition 5.4.16, Tr preserves the separability of any y G C?*, from which it follows that

Tr G S^n . D

Corollary 5.4.21. Sg. contains a subgroup isomorphic to V4 .

Proof. Since |rd = \Tr\ = \TcTr\ = 2, the subgroup S = {l,Tc,Tr,Tc7 } contmns exactly

three nonidentity elements, all of which have order 2. Thus, 5 = V4. □

Let a G and let ( a ,6 ) be some pair. If both a and b are even, then we say that

the pair ( a ,6 ) is even. Similarly, if both a and b are odd, then we say that ( a ,6 ) is odd.

If a, 5 < 2"“ *, then we call {0 , 6 ) a top-half pair. Similarly, if o, 6 > 2”~ \ then we call

( 0 , 6 ) a bottom-half pair. The pair ( 2"“ *, 2”“ + 1 ) is called the central pair and all other

pairs are said to be non-central. We denote by inve(a), inVo(<r), invt(<r), and invfc(<7) the

respective numbers of even, odd, top-half, and bottom-heilf pairs inverted by a. Notice that

0 < inve(<r), invo(£T), invt(cr), inv6(<r) < 1 ^ 1.

We now continue our journey toward the proof of Theorem 5.4.1 with a long string of

technical lemmas. These lemmas help to establish Proposition 5.4.34, a critical ingredient

in our proof.

Lemma 5.4.22. Let a G S 2" • Then inve((r) = invo(<r) and invf (<r) = invi,(o-).

Proof. First observe that if ( a ,6 ) is an even pair, then ( 6, a ) is an odd pair (and vice

versa). Similarly, if ( a ,6 ) is a top-half pair, then ( 6, a ) is a bottom-half pair (and vice

versa). Now let a G S2" and suppose that cr inverts {a,b). Then, by Lemma 5.4.12, a also

inverts (b,a). Consequently, the map (a, 5) h-> ( 6,a ) is a bijection between the set of even

pairs inverted by <r and the set of odd pairs inverted by cr. It is also a bijection between

the set of top-half pairs inverted by a and the set of bottom-half pairs inverted by a. □



Lemma 5.4.23. Let a 6 52- and suppose that inv(a) > 2. Then either inve(<7) > 0 or

invt(o-) > 0 .

Proof. Suppose, to the contrary, that both inve(cr) = 0 and invt(a) = 0. Then for each

pair (a,b) inverted by tr, it must be that a < 2""^, b > 2" " \ and a and b have opposite

parity. Since inv(tr) > 2 , there exists a non-central pair (a,b) inverted by tr. Since the pair

is non-central and a and b have opposite parity, we have that b - a > 2 and s o a - f - l< 6 — 1.

Let c = t%4-l and d = b - l . Then a < c < d 2”" ‘

(ii) a , c < 2" - i and tf, 6 > 2" - i

(iii) a < 2" - i and c , d , b > 2”- i .

In case (i), we note that tr does not invert (ti,c) since it is a top-half pair and

invt(cr) = 0. Furthermore, tr does not invert (c,b) = (a + 1,6) since a + 1 and b have

the same parity and inv, (tr) = invg (tr) = 0. But tr does invert (o,i>) and so we have that

tr(c) < tr(6) < tr(a) < tr(c), a contradiction. Similar contradictions can be reached for cases

(ii) and (iii). □

Lemma 5.4.24. Let tr 6 52", where n > 5, and suppose that inv(tr) < l ( ^ ) . Then either

inve(tr) < (^"2 *) ~ 1 or invt(tr) < ') - 1-

Proof. We prove the contrapositive. Suppose that invg(tr),invt(tr) > g ) — 1- Let Se,

So, St, and Sb be the sets of even, odd, top-half, and bottom-half pairs inverted by tr.

Then |5e|, |5t| > ( 2 ) ~ and, by Lemma 5.4.22, |5o|, |5&| > ( 2 1- Notice that

SeC\ So = St n Sb = 0. Notice also that |5e n 5t| = |5o n 5&| < 2 ) that |5o n Sf | =

|S e n S 5 l< f 2 " ) .



Thus,

inv(a) > |5e U U 5t U 5 1

= lêl + |5o| + |Sfl + jSil — j5e n St| — |5o n Sftj — |5o n 5t| — j5e n 5b|

= 2 (2” " ^ )(2""^ - 1) - 2 (2""^)(2""2 _ 1) - 4

_ 22n-i _ 2" - 2^"“® + 2”“ - 4

_ 22n-l _ 2 "-l - - 4

= (2^"“ - 2"“^) + (2^"-2 _ 2”“ - 2^”“® - 4)

+ 2^"“ - (2”“ + 4)

> ^ ( 2" ) + - 2 "" ' (since n > 5 )

= 5 ( 2 " ) + 2 " “ ‘(2”- " - 1 )

>Kt)Lemma 5.4.25. Let a G S 2”, where n > 4.

(i) //inve(cr) > ( ') - 1, then invt(cr) > 5.

(ii) //invt(<7) > (^"2 ') - 1, then inve(<r) > 5.

Proof. We prove (i) and leave the analogous proof of (ii) to the reader. Suppose that

inve(a) > ( 2 ) - 1- Then there is at most one even pair that is not inverted by a. Since

(^2 ) of the top-half pairs are also even, a must invert at least - 1 top-half pairs.

But n > 4, and so it follows that invt(<T) > *) ~ 1 ^ (2) “ = 5, as desired. □

Lemma 5.4.26. For all odd integers a and b with o < 6 < 2” - 1, (a, 6, a + 1, 6 -f 1 ) is a

complete set with respect to >~ux ■

Proof. Let >-n-i be the standard lexicographic order on n - 1 criteria and let >-i be the

linear order on one criterion specified by 1 >-i 0. Then >-iex = >“ n - i ® >*i- Let (y&), (in)

be the order sequences corresponding to >iex and respectively. Then, by Proposition

3.3.5, y2k-i = (i&, 1) and p2k = (i&,0) for each k < 2"~*. Now let a = 2i - 1 and let



b = 2j - 1, where i < j . Then

Va = {Xi , l ) ya+ l = (xuO)

yb = { x j , i ) yb+i = ( xj , o)

and so ( o, 6, o + 1, 6 + 1 ) is a complete set. □

Lemma 5.4.27. Let X/g, be the standard lexicographic order on n criteria. Then for all

a < b< 2”"^, ( a, 6, a + 2"~*, b + 2”~* ) is a complete set with respect to >~ux ■

Proof. The proof is analogous to that of Lemma 5.4.26. □

Lemma 5.4.28. Let a 6 S^n. Ifinvgicr) = 1 and {a, b) is the unique odd pair inverted by cr,

then 0 + 6 = 2".

Proof. Suppose that invo(o) = 1 and let (o, 6 ) be the unique odd pair inverted by a. By

Lemma 5.4.26, (o, 6, a + 1, 6 + 1 ) is a complete set with respect to >-tex- But since a

preserves the separability of >-iex, Proposition 5.4.16 implies that a inverts (a + 1,6 + 1).

By Lemma 5.4.12, cr also inverts (6 + l ,o + l ) , which is an odd pair since both o + 1 and

6 + 1 are even. But ( o, 6 ) is the unique odd pair inverted by a and so it must be that

( 6 + 1,0 + 1) = (o ,6 ). Thus

fl — 6 + 1 — 2" — (6 + 1) + 1 = 2" — 6,

as desired. □

Lemma 5.4.29. Let a € S’Jr.- I f invt{a^) = 1 and (o ,6 ) is the unique top-half pair inverted

by a, then o + 6 = 2"“ * + 1.

Proof. Suppose that invf(cr) = 1 and let (o, 6 ) be the unique top-half pair inverted by a.

By Lemma 5.4.27, (o, 6, o + 2 " " \ 6 + 2"~* ) is a complete set with respect to >~iex. Thus cr

must also invert (o + 2" ^,6 + 2"“ ) and (6 + 2 " - \ o + 2"-i ). But (6 + 2""^,o + 2""^ )

is a top-half pair and so it must be that

0 = 6 + 2 " - i = 2" - (6 + 2" -^ ) + 1 = 2 " - i - 6 + 1. □

Lemma 5.4.30. Let >- 6 On and let >~i G C?i. 7/ (o, o + 1, 6, 6 + 1 ) is a complete set with

respect to y , then ( 2o, 2o + 1, 26, 26+ 1 ) is o complete set with respect to >- © >-i.

Proof. Without loss of generality, assume that 1 >-i 0. Let (x*), (y&) be the order sequences

corresponding to >- and >- © >-i, respectively. If (o, o + 1, 6, 6 + 1 ) is a complete set with



respect to >-, then there exists S C Q and y, z £ X s , u, v € X - s such that

Xa = {y,v) Xb = (y,v)

Xa+l = (z,u) Xb+l = {z,v).

Now by Proposition 3.3.5, we have

X2a = ( y , U, 0) X2b = {y , V, 0)

120+1 = (Z,U,1) l2b+l = (Z.V, 1).

But then ( 2a, 2o + 1 , 26, 26 + 1 ) is a complete set with respect to >- ® >-i, as desired. □

Lemma 5.4.31. Let a 6 where n > A . I f a inverts (2"“ , 2"“ + 1 ), then there is a

top-half pair (a,b) ^ ( 2"“ , 2”“ + 1 ) such that cr inverts (o ,6 ).

Proof. It suffices to show that, for each n > 4, there exists an order >- E O* such that

( 2"” ^, 2"“ +1, 2”“ * - 2"~^, 2"“ * - 2"“ +1 ) is a complete set with respect to >-. Indeed,

once we have established this fact, our result follows directly from Proposition 5.4.16, since

( 2" - i - 2"-^, 2" - i - 2"-< + 1 ) is a top-half pair.

We proceed by induction. For n = 4, consider the separable order >- corresponding to

the normalized bineiry preference matrix

A 1 1 1>

1 1 1 0

1 1 0 11 0 1 1

0 1 1 1

1 1 0 01 0 1 0

0 1 1 0

1 0 0 1

0 1 0 1

0 0 1 1

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

(o 0 0

Notice that (4 ,5 ,7 ,8 ) is a complete set with respect to as desired. Now suppose

that our claim is true for some n > 4 and let >-' 6 C>* be chosen so that

(2"~^, 2"“ -f 1, 2"~^ — 2"“ “*, 2"“ * — 2"~^ 4-1) is a complete set with respect to Let

>-i be the linear order on one criterion specified by 1 0. Then, by Lemma 5.4.30,

(2"~*, 2”“ 4- 1, 2" - 2"“ ^, 2” - 2"~^ 4-1) is a complete set with respect to >-' ® >-i.

Proposition 3.3.2 implies that >-' © >-i € O^+i, which completes the proof. □



Lemma 5.4.32. Let a € , where n > 4. Then invt(cr) ^ 1.

Proof. Suppose that inve(«r) = 1 and let (a,b) be the unique top-half pair inverted by cr.

Then, by Lemma 5.4.29, a + b = 2"“ + 1. We consider two cases:

Case 1. b = a + 1. Then o = 2”"^ and b = 2"“ +1. Consequently, by Lemma 5.4.31, there

exists a top-half pair (c ,d ) ^ ( a ,6 ) that is inverted by a. This, however is a contradiction

to the assumption that inv, (cr) = 1.

Case 2 . b > a + 1 . Then o < 6 — 1 < 6. Since a inverts ( a ,6 ), Lemma 5.4.10 implies that

a inverts either (a,b - 1) or (b - l ,b) , both of which are top-half pairs. This too is a

contradiction to the assumption that invt(cr) = 1. □

Lemma 5.4.33. Let a G SJn, where n > 4. //invo(a) = 1, then invt(£r) > 1.

Proof. Suppose that invo(o-) = 1 and let ( a ,6 ) be the unique odd pair inverted by a.

Then, by Lemma 5.4.28, o + 6 = 2". If 6 ^ o -t- 2, then there exists some odd integer c such

that a < c < b. But then Lemma 5.4.10 implies that a inverts either (o ,c) or (c ,6 ), a

contradiction to the assumption that invo(a) = 1. Thus, it must be the case that b = a + 2,

which implies that o = 2”“ — 1 and 6 = 2”“ 4-1. Let >~icx be the standard lexicographic

order on n criteria and let {xk), (Vk) be the order sequences corresponding to >~ux and

Tci>-iex), respectively. Then, by Proposition 3.3.9,

J/2"-* —1 — X2"” ' —1 — (1, 0 , 0,, . . . , 0 , 1)

J/2""*+l ~ X2"~* — (1; 0, 0, , . . . , 0, 0)

J/1 = X l = (1 ,1 ,1 ,..., 1,1)

j/2 =X 2 = (1, 1, 1, . . . , 1, 0).

Thus, (1 ,2 ,2”“ - 1,2"-^ 4-1 ) is a complete set with respect to Tc(>-iex)- Since a € SJ™

and a inverts ( a ,6 ) = (2"“ - 1,2"“ * -f- 1) , Proposition 5.4.16 implies that a inverts

(1 ,2 ), which implies that invf(tr) > 1. Lemma 5.4.32 then makes this inequality strict. □

Proposition 5.4.34. //cr € 5^» and 2 < invo < ^(^ ), then either 1 < invo(cr) < ( ') - 1

or 1 < inV((o) < *) - 1.

Proof. Suppose, to the contrary, that neither 1 < invo(<r) < ( 2 ) ~ nor 1 < inve(cr) <

(^2 ) - 1. Then one of the following must occur:

(i) inVo(<r), inv<(o-) < 1



(ii) inv<,(ff) < 1 and invt(a) > *) - 1

(iii) invo(cr) > (^”2 ~ and invt(<7) < 1

(iv) invo(o-), inve(tr) > ') - 1

Lemmas 5.4.24 and 5.4.25 rule out cases (ii) - (iv). Thus, it must be the case that

both invo(o-), invt(tr) < 1. Lemma 5.4.32 implies that invt(or) = 0. But then Lemma

5.4.23 implies that invo(cr) = 1. This, however, is a contradiction, since, by Lemma 5.4.33,

invo(o’) = 1 inv(((7) > 1. Thus, it must be the case that either 1 < invo(tr) < ( ') - 1

or 1 < invt(o-) < ( "2 ) — 1, as desired. □

We are now (finally!) ready to tie all of the pieces together and prove Theorem 5.4.1.

Proof of Theorem 5.4-1- For n = 2, separability and symmetry are equivalent by Proposi

tion 4.6.1. Thus, SI = S i, which is, by Proposition 5.2.10, isomorphic to D4 , the group of

symmetries of the square.

For n = 3 and n = 4, we use a brute-force method to calculate 5J». Specifically, we

note that, by Corollary 5.4.18, a permutation a belongs to S^n if and only if a preserves

the separability of each normalized binary preference order in O*, that is, if and only if

(T e 5 ^ for each normalized > -6 0*. Proposition 3.1.3 provides explicit descriptions of

all such orders. Furthermore, by taking permutations and/or bitwise complements of the

columns of these normalized binary preference orders, we can explicitly determine all of

the elements of O*. We then calculate S^n for each normalized >- 6 O^, using Proposition

5.3.6 to note that

= P i {it 6 52" : (t(>-) = >-'}.x'€0 ;

Finally, we obtain SJn by taking the intersection of these 5 ^ .

The case for n = 3 can be done by hand (albeit somewhat tediously), since only

2 sets of permutations must be intersected. The result is a group of 24 permutations,

generated by it = (18)(27)(36)(45), r = (18)(26)(37), and 7 = (12)(36)(45)(78). Notice

that CTT = (23)(45)(67), a j = (17)(28), and t j = (167)(283)(45). Thus,

(<T,r,7 I (T = = 7 = (irr)^ = (1T7 )- = (rj)^ = 1)

is a presentation for Sg. By Coxeter and Moser [7], it follows that Sg = Z2 x Dg = V4 x S3.

The situation is slightly more complicated for n = 4, since we are now intersecting 14

sets, each containing 5376 permutations. While this would be nearly impossible by hand.



a computer has no problem accomplishing the task. After a few minutes, we find that

% = {l,Tc,Tr,TcTr} = V4 (see Appendix A for more detml).

Indeed, our claim is that = {1,Tc, Tr,TcTr} = V4 for all n > 4. We use induction

to prove this claim.

We have already established the base case and so all that remains is the inductive

step. Thus, suppose that, for some n > 5, = {l,Tc,Tr,TcTr} and suppose also that

Sgn ^ {l,Tc,Tr,TcTr}. Corollaxy 5.4.21 implies that {l,Tc,Tr,rcTr} Ç S^n. Thus, it must be

the case that this containment is proper. By our comments following Definition 5.4.3 and by

Proposition 5.4.13 and Lemma 5.4.14, there exists a 6 S r, with 2 < inv <7 < - 2. Fur

thermore, we may assume, by Lemma 5.4.6, that 2 < inv tr < l ( ^ ) . But then Proposition

5.4.34 implies that either 1 < invo(cr) < ') - 1 or 1 < invt(tr) < ’) - 1.

Suppose that 1 < invp(tr) < - 1 and let >-i be the linear order on one crite

rion specified by 1 >-i 0. Then for any >>• 6 t h e permutation tr' = s „{(t ) induced

by p„(>- © >-i) is an element of (by Proposition 5.4.9). On the other hand. Propo

sitions 3.3.9 Eind 5.4.4 establish a bijection between the odd pairs inverted by tr and all

pairs inverted by tr', and so inv(tr') = invo(tr). This, however, is a contradiction, since

1 < invo(tr) < ( 2 ) ~ and tr' € 52„_,, which is equal to {1, Tc, r^, TcTt} by the induction

hypothesis.

A similar contradiction is obtained if we assume that 1 < inv,(tr) < ( ') — 1 (in

this case we consider the permutation tr' = si(tr) induced by pi(>-i © >-)). Since each case

leads to contradiction, our assumption that S^n {l,Tc,Tr,TcTr} must be false. □

The surprising conclusion of Theorem 5.4.1 (namely that 5^n contains only 4 elements

for n > 4) is the last in a trilogy of disturbing results about separable preferences. In

Chapter 1, we gave an example of the paradoxical behavior that can occur when preferences

are nonseparable. We also recalled previous research asserting that the best way out of the

problem of nonseparability may be to avoid the very possibility of nonseparable preferences.

Unfortunately, Corollary 4.4.2 indicates that this could be a formidable task, as the vast

majority of preference orderings are nonseparable. We now know also that even if we can

force all preferences to be separable, most changes to these preferences (even very small

modifications such as the interchanging of two non-central, adjacent alternatives) have the

potential to introduce nonseparability. Thus, an unfortunate practical implication of our

research is the following:



Separable preferences, which seem to be essential to effective multiple-criteria

decision-making, are rare and extremely sensitive to small changes.

Recall our earlier claim that, for n > 3, there does not exist a separable order >~

for which Sû is a subgroup of 52" • We now prove this result.

Lemma 5.4.35. For every n > 3, 5g" is not a subgroup of S2" .

Proof. Let n > 3 be given and let (j/*) be the order sequence corresponding to >~ux- Let

a = (2,2 + 2 "-i - 1){4,4 + 2 " - ' - 1) • • • (2""S 2”-^ + 2”"* - 1).

Proposition 3.3.9 implies that, for every x 6 pm = (1,1 , 0) if and only if m = 2 t

for some k < 2”~^. Now let y^k = (l,z ,0 ), as above. We clmm that 2/2*+2" - '- i = (0 ,i, 1).

To prove this, it suffices to show that

n —1 n —1

2" - (2& + 2 " - ' - 1) = 0 • 2 " - i + 5 ] Xj • 2 " - ' + 1 - 2° = 2 " ' ' + 1.

Now we know that

and so

j = 2 j = 2

n —1

2" - 2A = 2" - i + 5 ] x j - 2”- ^j=2

n - l

^ X j 2 " - j = 2" - 2 t - 2 " - ' .j = 2

Thusn —1

Y ^ X j ■ 2 " - j + 1 = 2” - 2ife - 2 " - i + 1 = 2” - (2fc + 2 " - ‘ - 1),j=2

as desired.

It follows from our above observations that a interchanges in (y&) all pairs of elements

of X q of the form ((l,x ,0 ), (0,x, 1)) for x e Clearly, the effect of this action is

to interchange the first and the criteria, while maintaining the original order specified

by >~iex- Thus, a(>~iex) is separable, and so ct € Sjn*.

Now let 7 = Tc(T. Then j{>~iex) = Tc{cr{>-iex)) is separable since Tc € S^n. Thus,

7 6 Sjn®. Notice also that

7 = (2”- \ 2 ”-^ + l)(2,2 + 2”-^ - l)(4,4 + 2”- ' - 1) • • • (2”- \ 2 ”"^ + 2”" ‘ - 1)

= (2,2”- \ 2 ” - l , 2 ”- ' + l)(4 ,4 + 2”- ' - 1 ) . . .(2”- ' - 2,2”"i - 2 + 1)



and so f + 1)(2,2” - 1) = Tc(2,2" - 1). We claim that f ÿ Since

Tc € Sgn, it suffices to show that (2,2" — 1) ^ Let (zk) be the order sequence corre

sponding to X = (2,2" - l)(>-/ei)- Then

■2i — Î/1 — (1) 1) • • • ) 1) 1)

Z2 = V ‘i " - i = (0,0, . . . ,0,1)

Z2" - ' —1 = 3/2'*“* — 1 ~ (1,0, . . . ,0,1)

22"->+1 = î/2““ '+l — (0 , 1, . . , 1, 1)'

Thus, (1 ,1 ,. . . , 1, 1) >- (0 , 1, . . . , 1, 1) but (0 ,0 ,...,0 ,1 ) y (1 ,0 ,...,0 ,1 ) , and so

y = (2,2" — l)(>-/ei) is not separable. It follows that 7 ^ 52»® and so is not a sub

group of 52". □

Lemma 5.4.36. Let >-i, >-2 E I f S^n is a subgroup of 8 2 ' , then so is S^n-

Proof. Suppose that 5 ^ ‘ is a subgroup of 52". We wish to show that is a also subgroup

of 52". Since is finite and 1 € S^n, it suffices to show that is closed under its

operation. To this end, choose ri,T2 € 5 ^ ' and let a E 52" be such that <t(>-i) = >-2. Since

(z(>-i) = >-2 € C?*, it follows that cr G S^n. Thus, E 5^*. Now

T i(r(X i) = T i(cr(> -i)) = T i(X 2) e d *

since t i E 5^» , from which it follows that ricr E 5^»*. Similarly, T2cr E 5 ^ . But then

TiT2<r = (ricr)((T“ ^)(r20-) E 5^„‘,

which implies that TiT2(>-2) = tiT20-(>-i) E O*. It follows that T1T2 E 5 ^ ', as desired. □

Proposition 5.4.37. Let n > 3 and let y E O*. Then S^n is not a subgroup of 8 2 ’' ■

Proof. If 5 ^ were a subgroup of 52", then, by Lemma 5.4.36,52»® would also be a subgroup

of 52». This, however, is a contradiction to Lemma 5.4.35. □

We remark here that, for n = 2, 5 ^ is always a subgroup of 8 2 " • In fact for every

y € O2 , 8 ^ = 8 4 . To see this, notice that 5^ Ç 8 for every >- E C?2 - Also notice that

|5 ^ | = IO2 I = 8 = |5 | | by Theorem 5.4.1 and by Propositions 3.3.7, 4.6.1, and 5.3.6. Thus,

S4 = 8 4 , as desired.


Chapter 6

Questions for Future Research

We have mentioned a few open questions within the preceding text; we now take a moment

to discuss in more detail some of the possible directions that future research in this area

might take.

In Chapter 2, we observed that some of Gorman's nice set-theoretic and structural

properties may fail to occur when the alternative sets are not arc-connected (as is the case

in a referendum election). In Chapter 3, we noted in this context the existence of preference

orders that are separable but not additive. This was also in contrast to a theorem of Gorman,

one which again required arc-connectedness of the alternative sets. Thus, it would seem

reasonable to explore the extent to which weaker versions of Gorman’s theorems might

hold in a discrete setting.

In Chapter 4, we mentioned that several combinatorial questions remain unanswered.

Counting sepsirable and additive preference orders seems to be a difficult and interesting

problem. We sense that further investigations into the action of the symmetric group on

the set of separable preference orders may provide some tools to help us attack these

and other related enumeration problems. We note here that a characterization of the set

of permutations that preserve the separability of the standeird lexicographic order could

quickly lead to a formula for the number of separable preference orders. Along these lines,

we imagine that it would be quite useful to study also the structure of the groups of

permutations that preserve other properties, such as monoseparability, preseparability, and

additivity.

69


Appendix A

Computer Code and Output

In this appendix, we discuss briefly the methods used to calculate Sj» for n = 4. We also

provide the Visual Basic code and a portion of the resulting Microsoft Access crosstab

query used for this calculation.

A .l The Separable Group Calculator

A Visual Basic program was used to calculate 5|» for n = 4. The interface for this program

is shown below in Figure A.I.

The Ccilculation requires three steps.

1. First, each of the 14 normalized separable binary preference matrices are input into the

program. The user enters each matrix into the provided text box and then clicks the

“Generate Matrices” button. This button calls a subroutine (cmdAddMatrix_Click)

that generates and stores in an array all of the preference matrices differing from the

input matrix by permutations and/or bitwise complements of columns. Thus, after

entering each of the 14 normalized separable binary preference matrices, the program

has generated and stored all 5,376 separable binary preference matrices.

2. Next, the user clicks the “Calculate Group!” button, calling a subroutine (cmdGroup.Click)

that writes to a Microsoft Access database the information necessary to calculate .

Specifically, for each pmr (>-, >-'), where >- is a normalized separable preference order

and >-' is any separable preference order, the program calculates the permutation

70


Appendix A. Computer Code and Output 71

» S e p a ia b l e Group Calculator

Figure A.l: The Separable Group Calculator interface

cr for which cr(>-) = >-' and stores the triple (cr, >-, >-') as a record in the database.

Note that a permutation cr preserves separability if and only if, for each normalized

>- G O'^, there exists >-'€ such that the triple (a, occurs as a record of the

generated database.

3. To complete the calculation, the user clicks the “Mark Separability Preserving Per

mutations” button, calling a subroutine(cmdMark_Click) that marks the records cor

responding to permutations for which the condition described in (ii) holds.

The detailed code for this process is listed below.

’D eclarations and constaints > ___________

Const nQuest = 4

Type PrefM atrixRowd To 16, 1 To 4) As Boolean

End Type

Dim SepM atricesO As PrefM atrix Dim OrigM atricesO As PrefM atrix Dim Perms() As S tring



Dim nMatrices As Integer

Dim oldMatrixO As Boolean Dim origMatrixO As Boolean Dim curMatrixO As Boolean Dim colPermsO As Integer Dim nPerms As Integer

Sub cmdAddMatrix_Click()

’This subroutine generates all separable preference ’matrices corresponding to a given normeilized matrix.

Dim cnt As IntegerIf Not InitializeMatrixO Then MsgBox "Error!": Exit Sub

ReDim Preserve OrigMatrices(l To nMatrices + 1) As PrefMatrix ReDim Preserve SepMatrices(l To (nMatrices + 1) * 384) As PrefMatrix nMatrices = nMatrices + 1

For 1 = 1 To 2 " nQuest For m = 1 To nQuest

OrigMatrices(nMatrices) .Row(l, m) = origMatrixd, m)Next m

Next 1

cnt = (nMatrices - 1) * 384 + 1 For k = 1 To nPerms

For j = 0 To 2 “ nQuest - 1 curRow = ""ResetCurMatrix colPerm k ColComp jFor 1 = 1 To 2 " nQuest

For m = 1 To nQuestSepMatrices(cnt) .Row(l, m) = curMatrixd, m)

Next m Next 1cnt = cnt + 1

Next j Next k

End Sub

Private Function InitializeMatrixO As Boolean }_____________________________________________________’This function converts the text entered in txtMatrix ’to the data type of a preference matrix (PrefMatrix)



Dim curPos As Long, oldPos As LongDim curRow As String, curRowNum As IntegerDim curCol As Integer

Dim permString As String

On Error GoTo errhandl

ReDim origMatrixd To 2 ~ nQuest, 1 To nQuest) As Boolean ReDim oldHatrix(l To 2 “ nQuest, 1 To nQuest) As Boolean ReDim curMatrixd To 2 ~ nQuest, 1 To nQuest) As Boolean

nPerms = Factorial(nQuest)ReDim colPermsd To nPerms, 1 To nQuest) As Integer

permString = ‘'4321,3421,3241,3214,4231,2431,2341,2314,4213,2413,2143,2134, 4312,3412,3142,3124,4132,1432,1342,1324,4123,1423,1243,1234"

For i = 1 To nPermsFor j = 1 To nQuest

colPermsd, j ) = Mid (permString, (nQuest + 1) * (i - 1) + j , 1)Next j

Next i

oldPos = 1 curPos = 1 curRowNum = 1 Do While curPos <> 0

curPos = InStr(oldPos, txtMatrix.Text, Chr(13))If curPos <> 0 Then

curRow = Mid(txtMatrix.Text, oldPos, curPos - oldPos) oldPos = curPos + 2

ElsecurRow = Right(txtMatrix.Text, Len(txtMatrix.Text) - oldPos + 1 )

End IfFor curCol = 1 To nQuest

origMatrix(curRowNum, curCol) = -Val(Mid(curRow, curCol, 1)) oldMatrix(curRowNum, curCol) = origMatrix(curRowNum, curCol)

Next curColcurRowNum = curRowNum + 1

Loop

InitializeMatrix = True

Exit Function errhandl;MsgBox "An error has occurred while attempting to initialize the matrix." Exit Function

End Function



Private Sub ResetCurMatrix()

’This routine clears the data in the current matrix.J __________________________________________________________ ______________________ __________________________________

For a = 1 To 2 “ nQuest For b = 1 To nQuest

curMatrix(a, b) = origMatrix(a, b)Next b

Next a

End Sub

Private Sub SetOldMatrixO

This routine writes the current matrix to the oldMatrix variable so that actions can be performed on it while retaining its original data.

For c = 1 To 2 “ nQuest For d = 1 To nQuest

oldMatrix(c, d) = curMatrix(c, d)Next d

Next c

End Sub

Private Sub colPerm(ByVal pNum As Long)) _________’This subroutine permutes the columns of the current’matrix according to the permutation number pNum.) ____________________________________________________

SetOldMatrixFor i = 1 To nQuest

For j = 1 To 2 ■ nQuestcurMatrixCj, i) = oldMatrix(j, colPermsCpNum, i))

Next j Next i

End Sub

Private Sub ColComp(ByVal colMask As Integer)) ________________________’This subroutine takes the bitwise complement of ’a subset of the criteria set determined by colMask.> ______________________________

For i = 1 To nQuest



If colMask And 2 ~ (nQuest - 1) Then For j = 1 To 2 “ nQuest

curMatrixCj, i) = Not curMatrixCj, i)Next j

End If Next i

End Sub

Private Sub cmdGroup.Click()

This subroutine populates the database used to calculate the group of separability-preserving permutations.

Dim db As Database Dim tabPerms As Recordset

Dim nMatrices As Integer Dim nOrig As Integer nMat = UBound(SepMatrices) nOrig = UBound(OrigMatrices)

Dim curMatrixPermd To 2 “ nQuest)Dim curCycleDecomp As StringDim match As Boolean, curSteurt As Integer, curPos As Integer

Dim curPerm As String, curPermCount As Integer

Set db = DBEngine.OpenDatabase(txtFile.Text)Set tabPerms = db.OpenRecordset("Perms", dbOpenTable)

For i = 1 To nMat ’For each separable matrixFor j = 1 To nOrig ’For each normalized separable matrix

For k = 1 To 2 “ nQuest ’Find out which permutation is requiredFor 1 = 1 To 2 “ nQuest ’to take normalized matrix j

match = True ’to separable matrix i.For m = 1 To nQuest

If SepMatrices(i).Row(k, m) <> OrigMatrices(j).Row(l, m) Then match = False Exit For

End If Next mIf match Then

curMatrixPerm(k) = 1 Exit For

End If Next 1

Next k



’Now find the cycle decomposition of the permutation.

curCycleDecomp = "" curStart = 1 curPos = 1

DoIf curCycleDecomp = "" Then

curStart = 1Else

curStcirt = 0For p = 1 To 2 ~ nQuest

If InStr(curCycleDecomp, "(" & p & = 0And InStr(curCycleDecomp, ", " & p & ")") = 0And InStr(curCycleDecomp, ", " & p 4 ",") = 0And InStr(curCycleDecomp, "(" 4 p 4 ")") = 0 Then

curStart = p Exit For

End If Next pIf curStart = 0 Then Exit Do

End If

For p = 1 To 2 ~ nQuestIf curMatrixPerm(p) = curStart Then curPos = p: Exit For

Next p

If curPos = curStart ThencurCycleDecomp = curCycleDecomp 4 "(" 4 curPos 4 ")"

ElsecurCycleDecomp = curCycleDecomp 4 "(" 4 curStart 4 ", " 4 curPos Do While curPos <> curStart

For p = 1 To 2 “ nQuestIf curMatrixPerm(p) = curPos Then curPos = p: Exit For

Next pIf curPos = curStart Then

curCycleDecomp = curCycleDecomp 4 ")"Else

curCycleDecomp = curCycleDecomp 4 ", " 4 curPos End If

Loop End If

Loop

’Write the information to the database as a new record.

tabPerms.AddNewtabPerms!Sepmatrix = i tabPerms!origMatrix = j tabPerms! Perm = curCycleDecomp



tabPerms.Update Next j

Next i

tabPerms.HoveFirst curPermCount = 0 curPerm = tabPerms!Perm

db.Close

End Sub

Private Sub cmdMark_Click()

This subroutine marks the separability-preserving permutations in the database created by cmdGroup.Click.

Set db = DBEngine.OpenDatabase(txtFile.Text)Set tabPerms = db.Execute("SELECT Sepmatrix, origMatrix, Perm, Preserves

FROM Perms ORDER BY Perm")

tabPerms.MoveFirst curPermCount = 0 curPerm = tabPermsIPerm

cnt = 0nOrig = UBound(OrigMatrices)

Do While Not tabPerms.EOFIf tabPermsI Perm = curPerm Then

curPermCount = curPermCount + 1Else

curPermCount = 1 curPerm = tabPerms! Perm

End IfIf curPermCount = nOrig Then

tabPerms.EdittabPermsI Preserves = True

tabPerms.Update End IftabPerms.HoveNext cnt = cnt + 1

Loop

db.Close

End Sub



A.2 The Crosstab Query

We should point out that the program used to calculate S^n forn = 4 generated a wealth

of data that could prove to be useful in future research. Specifically, the program calculated

exactly which permutations preserve the separability of eeich normalized separable prefer

ence order on 4 criteria. Thus, it seems that one could learn a great deal by e x a m in in g

these data and searching for patterns and correlations. We do not take up this task here,

but we do present two examples of the sort of information that can be garnered from this

type of investigation.

Table A.2 below lists the first 24 (of 59,488) rows of a query run by Microsoft Access to

cross-tabulate and count, for each a 6 , the number of normalized separable preference

orders whose separability is preserved by a. These rows correspond to all permutations in

San that preserve the separability of at least 8 of the 14 normeilized separable preference

orders. For ease of notation, we number the normalized separable preference matrices 1

through 14, where numbers 1 through 7 correspond to the first row of matrices for the

\Q\ = 4 case on page 18 and numbers 8 through 14 correspond to the second row.

Notice that, of the approximately 21 trillion (16!) permutations in Sie, only 59,488

preserve the separability of even one separable preference order. Also notice that only 24

of these preserve the separability of at least 8 of the 14 normalized separable preference

orders. A question that we can easily answer by looking at the results of our crosstab

query is: Given an integer n with 1 < n < 14, how many permutations in 5ie preserve the

separability of at least n of the 14 normalized separable orders on 4 criteria? The answer

to this question for different values of n is summarized in Table A.2.



or preserves the separability of... #1 1,2,3,4,5,6,7,8,9,10,11,12,13,14 14

(8,9) 1,2,3,4,5,6,7,8,9,10,11,12,13,14 14(1,16)(2,15)(3,14)(4,13)(5,12)(6,11)(7,10) 1,2,3,4,5,6,7,8,9,10,11,12,13,14 14

(1,16)(2,15)(3,14)(4,13)(5,12)(6,11)(7,10)(8,9) 1,2,3,4,5,6,7,8,9,10,11,12,13,14 14

(4,5)(8,9)(12,13) 1,2,3,4,5,6,7,8,9,10,11,12 12(4,5)(12,13) 1,2,3,4,5,6,7,8,9,10,11,12 12

(1,16)(2,15)(3,14)(4,12)(5,13)(6,11)(7,10)(8,9) 1,2,3,4,5,6,7,8,9,10,11,12 12(2,16)(2,15)(3,14)(4,12)(5,13)(6,11)(7,10) 1,2,3,4,5,6,7,8,9,10,11,12 12

(1,2)(3,6)(7,10)(8,9)(11,14)(15,16) 1,2,3,4,7,8,9,10 8(1,2)(3,6)(7,10)(11.14)(15,16) 1,2,3,4,7,8,9,10 8

(1,2)(3,6)(4,5)(7,10)(8,9)(11,14)(12,13)(15,16) 1,2,3,4,7,8,9,10 8(1,2)(3,6)(4,5)(7,10)(11,14)(12,13)(15,16) 1,2,3,4,7,8,9,10 8(1,15)(2,16)(3,11)(4,13)(5,12)(6,14)(8,9) 1,2,3,4,7,8,9,10 8

(1,15)(2,16)(3,11)(4,12)(5.13)(6,14) 1,2,3,4,7,8,9,10 8(1.15)(2,16)(3,11)(4,12)(5,13)(6,14)(8,9) 1,2,3,4,7,8,9,10 8

(1,15)(2,16)(3,11)(4,13)(5,12)(6,14) 1,2,3,4,7,8,9,10 8(4,5)(6,7)(8,9)(10,11)(12,13) 1,2,5,6,7,8,11,12 8

(4,5)(6,7)(10,11)(12,13) 1,2,5,6,7,8,11,12 8(6,7)(8,9)(1G,11) 1,2,5,6,7,8,11,12 8

(6,7)(10,11) 1,2,5,6,7,8,11,12 8(1,16)(2,15)(3,14)(4,13)(5,12)(6,10)(7,11)(8,9) 1,2,5,6,7,8,11,12 8(1,16)(2,15)(3,14)(4,12)(5,13)(6,10)(7,11) (8,9) 1,2,5,6,7,8,11,12 8

(1.16)(2,15)(3,14)(4,13)(5,12)(6,10)(7,11) 1,2,5,6,7,8,11,12 8(1,16)(2,15)(3,14)(4,12)(5,13)(6,10)(7,11) 1,2,5,6,7,8,11,12 8

Table A.l: Permutations that preserve the separability of at least 8 normalized separable

preference orders.

n 1 2 3 4 5 6 7,8 9-12 13, 14

Number of permutations 59488 13024 1436 1084 80 64 24 8 4

Table A.2: The number of permutations that preserve the separability of at least n nor

malized separable preference orders.


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Separable Preference Orders

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