An Approval-Voting Polytope for Linear Orders

File: 480J 115501 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01Codes: 6793 Signs: 4769 . Length: 60 pic 11 pts, 257 mm

Journal of Mathematical Psychology � MP1155

journal of mathematical psychology 41, 171�188 (1997)

An Approval-Voting Polytope for Linear Orders

Jean-Paul Doignon

Universite� Libre de Bruxelles, Brussels, Belgium

and

Michel Regenwetter

McGill University, Montreal, Canada

A probabilistic model of approval voting on n alternatives generatesa collection of probability distributions on the family of all subsets ofthe set of alternatives. Focusing on the size-independent modelproposed by Falmagne and Regenwetter, we recast the problem ofcharacterizing these distributions as the search for a minimal system oflinear equations and inequalities for a specific convex polytope. Thisapproval-voting polytope, with n ! vertices in a space of dimension 2n,is proved to be of dimension 2n&n&1. Several families of facet-defin-ing linear inequalities are exhibited, each of which has a probabilisticinterpretation. Some proofs rely on special sequences of rankings of thealternatives. Although the equations and facet-defining inequalitiesfound so far yield a complete minimal description when n�4 (asindicated by the PORTA software), the problem remains open for largervalues of n. ] 1997 Academic Press

1. INTRODUCTION

In the general procedure of approval voting, each voterselects some subset of a given set of alternatives, say a subsetX of a fixed set C of n choice alternatives, or candidates. Theliterature on approval voting in political science andeconomics is substantial, including several books (e.g.,Brams 6 Fishburn, 1983; Merrill, 1988; Nurmi, 1987) andcountless articles (see Weber, 1995, for a recent paper).

We regard the subset X selected by a given individual,which can be empty or equal to C, as the value of a (setvalued) random variable V (as in vote). A model proposedby Falmagne and Regenwetter (1996) for the probabilitiesP(V=X ) is based on the following, basic assumption: A

voter chooses subset X if s�he chooses as many alternativesas X contains and if the alternatives in X are his�herfavorites. The reader who loves concrete examples is invitedto first have a look at Sections 2 and 5, which start with thecase C=[a, b, c], that is, n=3. To describe the model forgeneral n, consider two more random variables which arejointly distributed with V. The first variable, S, takes itsvalues in the set S=[0, 1, ..., n] of possible set-sizes, whilethe second variable, R, takes its values in the collection 6 ofall rankings of C. Rankings of C are often identified withlinear orders or permutations on C. For a given subset X ofC, we use 6X to denote the set of all rankings of C in whichthe elements of X are ranked before all the elements of C"X.When ? # 6X , we say that X is a beginning set of ?, or that? begins with X. Throughout the paper we set 6<=6.

With this notation, the fundamental equation in thesize-independent model of approval voting introduced byFalmagne and Regenwetter (1996) reads, for any subset Xof C, as

P(V=X )=P(S=|X| ) } P(R # 6X). (1)

We will not discuss here the pros and cons of the size-inde-pendent model (see Falmagne 6 Regenwetter, 1996;Regenwetter 6 Grofman, in press, 1995; Regenwetter,Marley, 6 Joe, 1996), but rather address the problem offinding a characterization of this model. To be specific, howcan we decide whether a given probability distribution onthe family P(C) of all subsets of C results from the model?We investigate this question from a geometric perspectiverather than a statistical testing approach. In geometricterms, the question is to find an analytical description of thecollection Dn of all such distributions.

The elements of Dn will be taken as vectors in the realvector space RP(C), with one vector component for eachsubset of C. The collection Dn , thus seen as a subset of thespace RP(C), is not convex (see the next section). Note thatfor each subset X of C, the probability P(S=|X| ) is

article no. MP971155

171 0022-2496�97 �25.00

Copyright � 1997 by Academic PressAll rights of reproduction in any form reserved.

We thank Jean-Claude Falmagne for his interest in this work, since heintroduced us to the problem, and two anonymous referees for their carefulreading. We acknowledge help from D. Bien, Computing Centre, U.L.B., inthe implementation of the PORTA software. We are very grateful to thedesigners of PORTA for having made publicly available such a powerfultool and to A. A. J. Marley for very helpful feedback on an earlier draft.M. R. acknowledges postdoctoral fellowship support from A. A. J. Marleyfunded through NSERC Collaborative Research Grant CGP0164211.

Correspondence and reprint requests should be addressed to J.-P.Doignon, Universite� Libre de Bruxelles, c.p. 216, Bd. du Triomphe, 1050Bruxelles, Belgium. E-mail: doignon�ulb.ac.be.


uniquely determined by the values P(V=Y ). We can thusrestrict our attention to the conditional probabilitiesP(V=X | S=|X| ) with X�C. The latter are equal toP(R # 6X). As we show in Section 2, the possible values ofthese conditional probabilities form the convex hull of thefollowing subset V of RP(C):

V=[ y # [0, 1]P(C) | _? # 6 with yX=1 � ? # 6X].

The set V has n ! elements, namely one for each ? # 6, andcan thus be identified with 6. In an abstract sense this solvesthe characterization problem for the size-independentmodel, since all possible ranking probabilities are specifiedby the convex hull of V. However, we seek to know moreconcretely what constraints this fact imposes on R, i.e. onthe possible conditional probabilities. More precisely, weneed a mechanism for determining whether a given real-valued function over P(C) does indeed lie in the convex hullof V. Furthermore, we desire the mechanism to be ``minimal''(this will be defined formally below), in the sense of avoid-ing redundancies within a collection of necessary andsufficient conditions.

The convex hull of finitely many points in a finite-dimen-sional real vector space is a convex polytope. We call theconvex hull of V the approval-voting polytope for linearorders. This polytope results from the model for approvalvoting due to Falmagne and Regenwetter (1996); someother models of approval voting, e.g. based on weak ordersor semiorders, are introduced in Regenwetter (1996). TheAV-polytope, as we also call it, will be denoted by An wherethe subindex n is a reminder of the existence of one suchpolytope for each cardinality n of C. The next section willwork out some technicalities about the passage from Dn toAn , while the remainder of the paper then focuses on theconvex polytope An and the interpretation of its propertiesin terms of probabilities over 6.

The reader is referred to Gru� nbaum (1967) and Ziegler(1995) for more detailed terminology and results aboutaffine geometry and convex polytopes. We limit ourselves tothe basics here. A convex polytope in Rd can also be charac-terized as a bounded intersection of finitely many halfspacesin Rd, where each halfspace is defined by a linear inequality.Replacing such an inequality by the corresponding equalitydefines a hyperplane in Rd, which is thus the intersection oftwo (opposite) halfspaces. A polytope may be contained insuch a hyperplane. Thus, a convex polytope can bedescribed by a system of linear equalities and inequalities. Ifsuch a system contains a maximum number of independentequalities and if removing one of its inequalities extends thepolytope to a larger set, then the system is called a (minimallinear) description of the polytope. Indeed, our problem cannow be rephrased as follows: find a description of An , thatis, a minimal system of linear equations and linearinequalities on RP(C) whose set of solutions is An .

Before we go into more detail, consider the following sim-ple calculation: since the observables take the form of aprobability distribution over P(C), they have 2n&1 degreesof freedom. However, the probabilities P(S=|X| ) have ndegrees of freedom. Thus, the conditional probabilitiesP(V=X | S=|X| ), which are in turn equal to theprobabilities P(R # 6X), have altogether 2n&n&1 degreesof freedom at most. The first step in our geometric analysisof the AV-polytope consists in determining the dimension ofAn . The dimension of a subset of Rd is the maximum numberof affinely independent points in that set, minus one. Hyper-planes of Rd are of dimension d&1. The dimension of apolytope A in Rd equals d minus the number of independentequations in a description of A. In Section 3 we provedim An=2n&n&1 and give a minimal system of n+1linear equations for the affine subspace generated by An .This establishes from the geometric perspective that thereare exactly 2n&n&1 many degrees of freedom in theprobabilities P(R # 6X).

As the convex polytope An is not full dimensional inRP(C), a linear description of An is not unique. A linearinequality is called valid for a polytope A in Rd if A is con-tained in the corresponding halfspace. A face of A consistsof all the points of A that give equality in a linear inequalityvalid for A. A maximal proper face is called a facet; thus, afacet of An is a polytope of dimension 2n&n&2. A linearinequality on Rd satisfied by A is facet-defining if the corre-sponding equation (obtained by replacing the inequality signwith an equality sign) defines a hyperplane whose intersec-tion with A is a facet. All linear inequalities in a (minimallinear) description of A are facet-defining, and there is onesuch inequality per facet. Thus the search for a minimaldescription of An involves determining the facet-defininginequalities for An . This is the desired ``minimal mecha-nism'' introduced above.

We are able to determine which of the linear inequalitiesgiven in Falmagne and Regenwetter (1996) are facet-defin-ing (Section 4). New facet-defining inequalities are offered inSections 6 and 8, and more inequalities whose status (forarbitrary n) is still unknown are proposed in Sections 8 and10. From outputs produced by the computer softwarePORTA (see Christof, 1996), we know that the facet-defininginequalities obtained so far entail a complete (minimallinear) description of An for n�4.

A vertex of a polytope A is a point which constitutes aface of A. The vertices of the approval-voting polytope An

form the set V already encountered. They correspond to then ! rankings of C. The latter property also holds for otherpolytopes in the literature, e.g., the ``permutohedron'' andthe ``binary-choice polytope.'' The permutohedron hasdimension n&1 (see, e.g., the references in Doignon 6Falmagne, 1994); its facets correspond to weak orders of C

having only two classes. The binary-choice polytope, whichis intensively studied in many mathematical disciplines, has

172 DOIGNON AND REGENWETTER


dimension n(n&1)�2. A complete characterization of itsfacets seems to be inaccessible at present, as witnessed byrecent papers (e.g., Marley, 1990; Suck, 1992; Fishburn,1992; and Koppen, 1995). (Related polytopes, obtained byreplacing linear orders with, e.g., semiorders or intervalorders, are studied in Suck, 1996). We cannot tell from ourinvestigations whether the similar problem for the AV-polytope is of the same level of difficulty.

Turning back to our original probabilistic setting, wenow make an interesting observation. Each of the facet-defining linear inequalities obtained in this paper for theAV-polytope An can be rewritten in the form P(R # E)�0,for some subset E of the set 6 of all rankings of C; in shortnotation, P(E)�0 for some event E in P(6). As an exam-ple, consider the (possibly facet-defining) inequality onRP(C)

y[a, b]+ y[a, c]� :} } }

y[a, x, y] ,

where a, b, and c are three alternatives, and the sum extendsover all nonordered pairs [x, y] in C"[a] that contain atleast one element of [b, c]. It is produced by the event con-sisting of all rankings in 6 that satisfy the conditions

a and exactly one element of [b, c] are amongthe three most preferred alternatives, withoutbeing the two most preferred alternatives.

It is an open problem to decide whether every facet of An

can be specified by an inequality of the form P(E)�0, forsome event E in P(6). Related questions are raised in Sec-tion 10. This section provides the probabilistic interpreta-tions of all of our facet-defining inequalities. It then infers afairly general family of linear inequalities involving all of theprevious ones. We leave it for further work to determinewhich of these new inequalities are facet-defining (forarbitrary n).

Also left for future investigation is a more productiveexploitation of the automorphism group of the AV-polytope. In Section 6, we use one particular automorphismto generate a new family of facet-defining inequalities.However, the full group of automorphisms is still to bedetermined, as well as its action on the family of all facets.

The paper is organized as follows. The following sectionintroduces further concepts and deals with some techni-calities that are involved in getting from the data to the AV-polytope An . In Section 3 we compute the dimension of An .In Section 4 we review and classify the testable conditions ofFalmagne and Regenwetter (1996) into those that are facet-defining and those that are not. We discuss the case of n=3in detail in Section 5. The following section introducesfurther testable conditions. After providing the completecharacterization of the case n=4 in Section 7 we then

generalize the conditions found so far into a new generalclass of conditions, some of which are proven to be facet-defining. We then move on to the case when n=5 for whichthe PORTA software provides a complete characterization.As an alternate perspective to the geometric approach takenthroughout most of the paper, Section 10 providesprobabilistic interpretations for all the previously discussedconditions. Furthermore, relying on probabilistic con-siderations, we derive a further class of conditions whichgeneralize all previous conditions. This includes conditionsthat PORTA provided for n=5 and which had not beenderived before. In this context, we also discuss openproblems.

2. FROM THE SET Dn TO THE APPROVAL-VOTINGPOLYTOPE An

To start with a concrete example, let C=[a, b, c] be a setof three alternatives. We then have n=3,

S=[0, 1, 2, 3]and

6=[abc, acb, bac, bca, cab, cba]. (2)

Note our shorthand for rankings of C, in which we assumethat the most preferred alternative is written first. Also,

P(C)=[<, [a], [b], [c], [a, b], [a, c], [b, c], C]. (3)

We will in the following refer to the (lexicographic) order inwhich we listed the elements of 6 in Eq. (2) and theelements of P(C) in Eq. (3). To each pair of probability dis-tributions

p: S � [0, 1]: s [ ps ,

q: 6 � [0, 1]: ? [ q? ,

we associate the probability distribution

r: P(C) � [0, 1]: X [ p |X| } :? # 6X

q? ,

reminding the reader that 6X is the set of all rankings hav-ing X as a beginning set. At the core of the size-independentmodel of Falmagne and Regenwetter (1996) is the mapping+ sending the pair ( p, q) to r. For n=3, the mapping + isspecified by

(( p0 , p1 , p2 , p3), (qabc , qacb , qbac , qbca , qcab , qcba))

[ ( p0 , p1 } (qabc+qacb), p1 } (qbac+qbca),

p1 } (qcab+qcba), p2 } (qabc+qbac),

p2 } (qacb+qcab), p2 } (qbca+qcba), p3).

173AN APPROVAL-VOTING POLYTOPE FOR LINEAR ORDERS


All the probability distributions r on P([a, b, c]) obtainedin this way, and in no other, satisfy the size-independentmodel. It is their set D3 that we seek to characterize in a sim-ple way, avoiding the reference to p and q made in

D3={( p0 , p1 } (qabc+qacb), p1 } (qbac+qbca),

p1 } (qcab+qcba), p2 } (qabc+qbac), p2 } (qacb+qcab),

p2 } (qbca+qcba), p3) }p�0, :

n

i=0

pi=1, q�0, :? # 6

q?=1= ,

where p�0 means pi�0 for all i # S. From the definition, itshould be clear that D3 is not convex. Indeed, take the pointr1 which is the image by + of the pair of distributions p onS and q on 6 characterized respectively by p1=1 andqabc=1. Then take also the point r2 similarly obtained forp2=1 and qcba=1. We thus have

r1=(0, 1, 0, 0, 0, 0, 0, 0)

and

r2=(0, 0, 0, 0, 0, 0, 1, 0).

These points r1 and r2 belong to D3 , while their midpoint rdoes not. Indeed, if

r=(0, 12 , 0, 0, 0, 0, 1

2 , 0)

were in D3 , then r=+( p, q) for some p, q. This would implythat p1>0 and also that qbca>0 or qcba>0, from which itwould follow that the third or fourth component of +( p, q)is strictly positive, in contradiction with the actual value ofthe corresponding component of r.

The nonconvexity of Dn holds for any size n of C withn�3. Turning to the general case, we consider the mapping

+: RS_R6 � RP(C): ( p, q) [ r,

where r is the real-valued mapping on P(C) given by

r(X )=rX=p |X| } :? # 6X

q? ,

for X # P(C). Setting

T={( p, q) # RS_R6 } p�0, :i # S

pi=1,

q�0, :? # 6

q?=1= ,

the image of T by + is a subset Dn of RP(C) formed by allprobability distributions on P(C) that satisfy the size-inde-pendent model. The set Dn is not convex. Moreover, theaffine subspace generated by Dn has dimension 2n&1, as itis the hyperplane with equation

:X # P(C)

yX=1 (4)

since any element in Dn clearly satisfies this equation: simplyset yX=P(V=X ). Also, Dn contains the unit point of anyaxis.

We now indicate how to get the ``relative interior'' of Dn .The relative interior E h of a set E of points is the (topologi-cal) interior of E taken in the affine subspace generated bythat set.

Theorem 1. The relative interior of Dn is formed by allimages a=+( p, q), where p and q are strictly positive prob-ability distributions on S and 6, respectively. In a formula,Dh

n =+(Th).

Remark 2. To warn the reader that this statement is notas obvious as it might look, consider the following mapping:

f : [(x, y) # R2 | x�0, y�0, x+ y=1] � R:

(x, y) [ (x& 12) } ( y& 1

2).

The relative interior of the image of f is not the image by fof

[(x, y) # R2 | x>0, y>0, x+ y=1]

(more precisely, f ( 12 , 1

2) is not an interior point of the imageof f ).

The following notation will be useful in the rest of thepaper:

P(C, k)=[Y # P(C) | |Y|=k].

Proof (Theorem 1). Let

U=[( p, q) # RS_R6 | p�0, q�0],

and N=+(U). Note that N is a positive cone, i.e., it con-tains *v whenever * # R+ and v # N. Also, U is a positivecone. As moreover Dn=H & N, with H being the hyper-plane given by Eq. (4), it suffices to prove N%=+(U%)(where this time we refer to topological interiors). Since + isobviously continuous, the inclusion N%�+(U%) is clear. Toprove the reverse inclusion, we remark first that + is con-tinuously differentiable on the interior of U, which is

U%=[( p, q) # RS_R6 | p>0, q>0].



We will later show that the differential d+( p, q) of + at anypoint ( p, q) in U% has rank 2n, which is the dimension ofRP(C). Restricting + to an affine subspace L through ( p, q)complementary to the kernel of d+( p, q), we learn from theInverse Function Theorem (see e.g. 10.2.5 in Dieudonne� ,1960) that + injectively transforms some neighborhood of( p, q) in L into some neighborhood of +( p, q) in RP(C). Thiscompletes the proof of the inclusion +(U%)�N%, except forthe computation of the rank of d+( p, q).

The matrix of the linear mapping d+( p, q) in the canoni-cal bases of RS_R6 and of RP(C) (i.e., the Jacobian matrix)has rank at most 2n, the number of its rows. To prove thatthe rank is exactly 2n, we partition the columns of thismatrix into the set C of columns indexed by sizes s # S, andthe set D of columns indexed by rankings ? # 6. Let P andQ be respectively the subspaces of RP(C) generated by thesetwo sets C and D of columns.

To get the dimension of Q, we study the 2n_n ! matrix Nwith rows indexed by subsets X of C and columns indexedby rankings ? of C, with

NX, ?=�rX

�q?={p |X|

0if ? # 6X ,otherwise.

Since p>0, the matrix N has the same rank as the matrix Mobtained by replacing (row by row) p |X| with 1. The value2n&n for the rank of M will be established in Section 3,where this matrix is intensively studied. From the linearequations that we obtain in Theorem 5 for the subspacegenerated by the columns of M, we see that the subspace Q(generated by the columns in D) is the set of vectorsy # RP(C) which are solutions of the following system, hav-ing one linear equation for each size k{0:

:X # P(C, k)

yX

pk=

y<

p0

.

On the other hand, the dimension of the space P generatedby the columns in C is easily seen to be n+1: remember thata column from the set C corresponds to a size s from S, andhas entries indexed by subsets X with

CX, s=�rX

�ps={ :

? # 6X

q? if |X|=s,

0 otherwise.

We now prove that the subspace P & Q has dimension atmost 1. Any vector v in P is a linear combination of thecolumns of C, so there exists a family (*s)s # S of real num-bers such that for all X # P(C)

vX=* |X| } :? # 6X

q? .

This vector v also belongs to the subspace Q iff it satisfies theequations for Q, thus iff for k=1, 2, ..., n

:X # P(C, k)

vX

pk=

v<

p0

,

which implies

*0

p0

=*1

p1

= } } } =*n

pn

since �X # P(C, k) �? # 6Xq?=�? # 6 q? , a positive constant.

Hence dim(P & Q)�1. Finally, from the well-knownequality

dim(P+Q)+dim(P & Q)=dim P+dim Q,

we get dim(P+Q)=2n, in other words the rank of d+( p, q)equals 2n at each point ( p, q) in U%. This completes theproof. K

Instead of looking at the probabilities P(V=X ), forX�C, we now consider as in Falmagne and Regenwetter(1996) the conditional probabilities P(V=X | S=s). Thus,P(V=X | S=s) is the probability that the voter choosessubset X, under the condition that s�he selects s alternatives.Of course, P(V=X | S=s)=0 for s{|X|, so we will alwaysassume s=|X| in such a writing. Remember from the Intro-duction that P(V=X | S=s)=P(R # 6X). We clearly have

P(V=X | S=s)=P(V=X )

�Y # P(C, s) P(V=Y ).

Note that the denominator in the last formula can vanish,namely when ps=0. We are thus led to consider a mapping

#: RP(C)"K � RP(C): r [ \ rX

�Y # P(C, |X| ) rY+ ,

where

K={r # RP(C) } _k # S: :Y # P(C, k)

rY=0= .

Although this mapping # is not defined everywhere on thecollection Dn , it follows from Theorem 1 that it is defined (atleast) on the relative interior of Dn . Moreover, a vector yfrom RP(C) belongs to #(Dh

n ) iff there exists a positive prob-ability distribution q on 6 such that yX=�? # 6X

q? , for allX # P(C). We then say that q induces y (according to the



size-independent model). Thus #(Dhn ) is the relative interior

of the convex polytope

An={y # RP(C) } _q # R6: q�0, :? # 6

q?=1,

and \X # P(C): yX= :? # 6X

q?= ,

which we call the approval-voting polytope for linear orders,or in short, the AV-polytope.

By its definition, the AV-polytope An is the convex hull ofa set V of n! points in RP(C). There is one such point v? foreach ranking ? of C. The X coordinate of v? (with X�C)takes value 1 if X is a beginning set of ? and 0 otherwise.

The vertices of An are among the n ! points v? . In fact,each point v? is a vertex of the polytope An , since it is thesole point of An which satisfies the equality in the followinginequality satisfied in turn by the whole polytope. To writethis inequality, we use the notation B(?) for the collectionof all beginning sets of ranking ?:

:X # B(?)

yX�n+1.

Since for each ranking % # 6, the point v% satisfies thisinequality (remember yX # [0, 1] at such a point), so doesthe whole polytope An . Moreover, equality is obtained atpoint v% only for %=? (when all terms in the first sum equal1). As a conclusion, any point v? is a vertex of An .

Another way to show this relies on affine automorphisms.Each permutation : of C induces a linear permutation ofRP(C):

F: : RP(C) � RP(C): r [ r$,

with r$X=r:&1(X ) (thus F: just permutes the coordinate axesof RP(C)). The image of the point v? by F: is the point v:(?) ,where :(?)=(:(a1), :(a2), ..., :(an)) if ?=(a1 , a2 , ..., an);that is, F:(v?)=v: b ? , if we view each ranking ? as a per-mutation of C (having picked an arbitrary referenceenumeration of the alternatives in C). As each F: stabilizesthe set [v? | ? # 6] whose convex hull forms An , andmoreover any given v? can be mapped to any given v% bysome F: , all the points of this set clearly are vertices of An .(In Section 6 we return to the group of affinities preservingAn , which contains other elements than the F:'s.)

We will often refer in the following to ``the vertex v?'',where ? is a ranking of C.

As a conclusion for this section, we may state that a lineardescription of the AV-polytope An would deliver at once away to check whether a positive probability distribution on

P(C) arises from the size-independent model. The restric-tion to positive distributions is not harmful as regards prac-tical usage of a possible characterization of the size-inde-pendent model. We now turn to the study of the polytopeAn , and first determine its dimension.

3. THE DIMENSION OF THE APPROVAL-VOTINGPOLYTOPE

The following n+1 linear equations on RP(C) are clearlysatisfied by the approval-voting polytope An , where n is asbefore the cardinality of the set C of alternatives: for s # S,

:X # P(C, s)

yX=1.

As these n+1 equations are linearly independent, this yieldsdim An�2n&n&1. We now proceed to show the reverseinequality, by considering again the matrix M met in theproof of Theorem 1, namely the binary matrix with rowsindexed by subsets X of C and columns indexed by rankings? of C, and

MX, ?={10

if ? # 6X ,otherwise

(in other words, the column indexed by ? contains the coor-dinates of the vertex v? of An). For

U={q # R6 } q�0, :? # 6

q?=1= ,

we see that An is the image of U by the linear mapping fwhich has matrix M in the canonical bases of R6 and RP(C),respectively:

f : R6 � RP(C): q [ r,

where q induces r, that is r(X )=�? # 6Xq? , for X # P(C). To

derive the dimension of An , we determine first the rank andimage of f .

Theorem 3. The rank of the linear mapping f , that is, therank of the matrix M, equals 2n&n. The image of f is formedby all vectors of RP(C) that are solutions of the followingsystem of n linearly independent equations, with one equationfor each k=1, 2, ..., n:

:X # P(C, k)

yX= y< .

Proof. Note first that each column of M satisfies the nequations in the statement, and as a consequence rankM�2n&n. To show the reverse inequality, and also the



second assertion in the statement, it remains to exhibit2n&n columns of M that are linearly independent. Letd=2n&n. A collection C1 , C2 , ..., Cd of columns of M islinearly independent as soon as C1 is nonzero, and for eachi=2, 3, ..., d, there exists a row index X such that Ci has a 1in row X, while all of C1 , C2 , ..., Ci&1 have a 0 in the samerow. The existence of a sequence of d columns satisfying thelatter condition follows at once from the next lemma, sincea column C of M has a 1 in the row indexed by X iff X is abeginning set of the ranking indexing the column. K

Lemma 4. Let n be a positive natural number, and letd=2n&n. There exists a sequence ?1 , ?2 , ..., ?d of rankingsof the n-element set [1, 2, ..., n], such that, for i=1, 2, ..., d,the ranking ?i satisfies the following condition:

(*) ?i has a beginning set X, with X�[1, 2, ..., n],which is not a beginning set of ?1 , ?2 , ..., ?i&1.

Proof. The proof is by recurrence on n. For n=1, theresult is trivial. For n=2, the following two rankings satisfyCondition (*) (with the subset formed by the underlinedelements in the ranking):

1�

2,

2�

1.

To get the five desired rankings for n=3, we first add 3 atthe end of the two rankings displayed above, and then addtwo rankings with 3 in the second position, and one rankingwith 3 in the first position:

1�

2 3,

2�

1 3,

1�

3�

2,

2�

3�

1,

3�

1 2.

A similar construction can be used to build the requiredsequence of 2n&n rankings over [1, 2, ..., n] from the list of2n&1&(n&1) rankings obtained so far for [1, 2, ..., n&1].First, we add n at the end of each of the given rankings, thusobtaining an initial sequence of 2n&1&(n&1) rankings,which all satisfy Condition (*) (with the same beginningsets as before). Then, we successively add subsequences, onefor each value of k=n&1, n&2, ..., 1. In all rankings in thesubsequence corresponding to the value k, we place n inposition k. Moreover, one such ranking is selected for eachsubset X of [1, 2, ..., n&1] having cardinality k&1, withthe only requirement that this ranking has X as a beginningset (the freedom left in positioning the elements will beexploited in Sections 4 and 6). Taking the beginning setX _ [n], it is easily seen that the newly added ranking

satisfies Condition (*). After completion of the constructionof all the successive subsequences, there results a sequenceof f (n) rankings satisfying Condition (*), with moreover

f (n)=2n&1&(n&1)+ :n&1

k=1\n&1

k&1+=2n&1&(n&1)+2n&1&1

=2n&n.

This sequence establishes the thesis for the value n. K

Theorem 5. The dimension of the AV-polytope An

equals 2n&n&1.

Proof. As indicated before the statement of Theorem 3,we use the equality An= f (U). As U is the simplex spannedby the unit points of the coordinate axes in R6, we see thatAn is the convex hull of the images of these points. From thetheorem, these images linearly span a subspace of RP(C)

with dimension 2n&n. Hence these images affinely span asubspace of dimension at least 2n&n&1. This showsdim An�2n&n&1. The reverse inequality was obtained atthe beginning of this section. K

4. THE LINEAR INEQUALITIES FROM FALMAGNEAND REGENWETTER (1996)

From Section 2, we know that a description of the AV-polytope An will involve exactly n+1 linear equations (forinstance, those listed in Theorem 3). Moreover, we alreadyindicated in the Introduction that the description will haveone linear inequality for each facet of An . We now inspectthree families of linear inequalities satisfied by An that weregiven in Falmagne and Regenwetter (1996), and we deter-mine which ones are facet-defining. While recalling theseinequalities, we introduce names and symbols for them.Some notation has also been introduced in previous sec-tions, e.g. P(X, k) denotes the collection of all subsets of Xwith cardinality k.

Nonnegativeness Condition NNEG(X ). For anygiven subset X of C,

yX�0. (5)

General Subset Condition GSUB(X, l ). For anygiven subset X of C, and any given size l with 0<l<|X|,

:Y # P(X, l )

yY� yX . (6)



Marginal Condition MARG(i, k). For any givenalternative i in C and any given size k with 1<k<n,

:Y # P(C, k), i # Y

yY� :Z # P(C, k&1), i # Z

yZ . (7)

The latter inequality is also satisfied by An in case k=1 ork=n, but then coincides with a nonnegativeness conditionNNEG(X) having respectively |X|=1 or |X|=n&1 (as aconvention, a sum of zero terms equals 0). Note also that inthe admitted case k=n&1, the marginal conditionMARG(i, k) is equivalent to a general subset conditionGSUB(X, l ) having l=n&2 and |X|=n&1. Falmagne andRegenwetter (1996) derive the above inequalities in theprobabilistic setting. For instance, refering to notation fromthe Introduction, the marginal condition MARG(i, k)amounts to the nonnegativeness of the marginal probabilitythat R ranks alternative i at position k. This interpretationmotivates the name chosen for the condition.

Another way of establishing all these inequalities consistsin checking that they are satisfied by any vertex of An . Toachieve this goal, remember that with each ranking ? of C

is associated a vertex v? having zero-valued coordinateseverywhere except for the coordinates corresponding to thebeginning sets of ? (one per size). The latter are equal to 1.Thus each side of any of the inequalities in Formulae(5)�(7) takes the value 0 or 1 exactly at the vertices.Moreover, the reader can easily check that if the right-handside equals 1, so does the left-hand side. For instance, takeFormula (6) and assume yX=1 at some vertex v? . Then theranking ? begins with X. Therefore, it begins also with somesubset Y of X of size l. Hence the left-hand side of theinequality equals 1. The proof is similar for the marginalcondition MARG(i, k), and even simpler for the non-negativeness condition NNEG(X ).

Clearly, a linear inequality satisfied by An is facet-definingiff the corresponding equality is satisfied by 2n&n&1affinely independent vertices. This fact is exploited in thenext three proofs. We will freely use the notation NNEG(X )in order to specify not only the condition, but also the linearinequality and even the corresponding linear equation. Asimilar remark holds for GSUB(X, l ) and for MARG(i, k).

Sequences of rankings similar to the one met in the proofof Lemma 4 will be most useful here. In this section, we setC=[1, 2, ..., n]. Let us repeat with additional commentsthe construction of 2n&n rankings of C, proceeding byrecurrence on n.

The Generic Sequence Gn of Rankings of [1, 2, ..., n]. Forn=2, the following two rankings constitute the sequenceG2 :

1�

2,

2�

1.

To get the 2n&n rankings that form the sequence Gn , wefirst copy the 2n&1&(n&1) rankings of Gn&1 , adding n atthe end of each, thus obtaining what we call the recurrencepart of Gn . Then, for each value of k=n&1, n&2, ..., 1, weform ( n&1

k&1) rankings with n in position k. Exactly one suchranking is selected for each subset X of [1, 2, ..., n&1] hav-ing cardinality k&1, with the only requirement that thisranking has X as a beginning set. Note that the selection ismade among (k&1)! (n&k)! possible rankings whenever Xis given. This is why we often speak of the ``generic''sequence Gn . The freedom in the selection is illustratedbelow for n=4: alternatives within a pair of braces can bepermuted freely, resulting in another choice of the rankingselected. The horizontal line separates the recurrence partfrom the rest of the sequence G4 .

1�

2 3 4,

2�

1 3 4,

1�

3�

2 4,

2�

3�

1 4,

3�

[1 2] 4,

[1�

2�] 4

�3,

[1�

3�] 4

�2,

[2�

3�] 4

�1,

1�

4�

[2 3],

2�

4�

[1 3],

3�

4�

[1 2],

4�

[1 2 3].

The generic sequence Gn is important because the 2n&nvertices v? of An , for ? appearing in Gn , are affinelyindependent (see the proof of Theorem 3, which refers tobeginning sets as those marked here by underlines).

Theorem 6. The nonnegativeness inequality NNEG(X ),

yX�0,

for X # P(C), is facet-defining iff X differs from < andfrom C.

Proof. If X=< or X=C, the inequality NNEG(X) isnot facet-defining, since both coordinates y< and yC takethe value 1 at each vertex. Conversely, it suffices to showthat NNEG(X ) is facet-defining for X=[k, k+1, ..., n] and1<k<n since we may always relabel the alternatives to getthis description of X. In the generic sequence Gn of rankings,this subset X cannot occur as a beginning set in any rankingfrom the recurrence part (because such a ranking assignsrank n to n). Moreover, if we make sure that in the other



rankings of Gn , all alternatives beyond n are taken inincreasing order, only one of these rankings starts with X.The rankings ? in Gn distinct from this exceptional one giveus 2n&n&1 affinely independent vertices v? satisfyingequality NNEG(X ). K

Theorem 7. Among the general subset inequalitiesGSUB(X, l ),

:Y # P(X, l )

yY� yX ,

exactly those with 0<l=|X|&1<n&1 are facet-defining.

When 1<|X|<n&1 and l=|X|&1, the general subsetinequality GSUB(X, l ) will be called the subset inequalitySUBS(X ). (We discard the value |X|=n&1 since theinequality is then equivalent to MARG(i, n&1).)

Proof. It is not restrictive to assume X=[1, 2, ..., s],with 0<l<s�n, in Condition GSUB(X, l ). We first showthat the inequality GSUB(X, l ) is not facet-defining whenl<s&1. To this end, we compare the vertices satisfying theequality GSUB(X, l ) with those satisfying the equalityGSUB(X, s&1). Any vertex satisfying the first equalitysatisfies also the second. Indeed, if inequality GSUB(X, l )becomes equality 0=0 at vertex v? , then the ranking ? doesnot begin with any subset of X of size l. As ? cannot thenbegin with a subset of X of size s&1, we see that equalityGSUB(X, s&1) is satisfied also at vertex v? . Again, ifinequality GSUB(X, l ) becomes equality 1=1, then ?begins with X, and inequality GSUB(X, s&1) becomes alsoequality 1=1 at v? . Moreover, any vertex associated to aranking starting with 1, 2, ..., l, s+1 in that order does notsatisfy equality GSUB(X, l ) (the left-hand side equals 1, theright-hand side 0), although it gives 0=0 inGSUB(X, s&1). As a consequence, inequality GSUB(X, l )cannot be facet-defining for l<s=|X| because the set ofvertices satisfying equality GSUB(X, l ) is strictly includedin the set of vertices satisfying equality GSUB(X, s&1), andthe latter is not the full set of vertices of An . The same con-clusion holds for |X|=n, that is X=C (since all verticessatisfy equality then).

There remains to show that inequality GSUB(X, s&1) isfacet-defining, whenever 1<s=|X|<n. We may takeX=[n&s+1, n&s+2, ..., n] this time. In order to exhibit2n&n&1 affinely independent vertices satisfying equalityGSUB(X, s&1), we turn again to the generic sequence Gn ofrankings. The vertex v? gives 0=0 in inequalityGSUB(X, s&1) iff the ranking ? does not begin with anysubset Y of X of size s&1.

We first inspect rankings ? in the nonrecurrence part. If ?starts with a subset of X of size s&1 that does not contain

n, then n has either rank s or a rank greater than s. In thefirst eventuality, the ranking begins with X and thus gives1=1 in inequality GSUB(X, s&1). In the second even-tuality, the alternative with rank s does not belong to X, andcomes before alternative n. Because of the freedom in theconstruction of Gn , we may move this alternative to the firstposition, so that the ranking now gives 0=0. Now let usassume that ? starts with a subset of X of size s&1 that con-tains n. Since s<n, there are at least two alternativesbeyond alternative n in ranking ?. We may permute alter-natives beyond n in such a way that (i) if n has rank s&1,then ? starts with X, so that we get 1=1; (ii) if n has rankless than s&1, then ? does not start anymore with any sub-set of X of size s&1, so that we get 0=0.

Now we prove that only one ranking ? in the recurrencepart will not give equality at vertex v? , assuming that whena selection had to be made along the construction of therecurrence part of Gn , the alternatives were rankedincreasingly. So let ? begin with a subset of size s&1 ofX=[n&s+1, n&s+2, ..., n]. Since n is at the end of ?,this subset must be [n&s+1, n&s+2, ..., n&1]. Ass&1<n&1, ranking ? must be in the nonrecurrence part(in Gn&1) of the recurrence part of Gn . Finally, note that onlyone ranking of the nonrecurrence part of Gn&1 begins with[n&s+1, n&s+2, ..., n&1] (here we use the increasing-ness assumption). K

Theorem 8. All marginal inequalities MARG(i, k),

:Y # P(C, k), i # Y

yY� :Z # P(C, k&1), i # Z

yZ ,

for i # C and k # S with 1<k<n, are facet-defining.

Proof. Inequality MARG(i, k) becomes the equality0=0 (resp. 1=1) at any vertex v? for which i is not amongthe first k alternatives in the ranking ? (resp., i is among thefirst k&1 alternatives). In other words, we have equalityexcept if i has rank k in ?.

We take i=1. According to the construction of thegeneric sequence Gn , the rank of 1 is constrained in only2(n&1) rankings, to values which are 1 (this happens n&1times), 2, ..., n. Indeed, for n�4, this is apparent in the dis-play before Theorem 6. Moreover, when we build Gn fromGn&1 , the position of 1 is constrained in only two additionalrankings, and at respective ranks 1 and n. As a consequence,we may assume that the sequence Gn is such that only one ofits 2n&n rankings places 1 at rank k. This completes theproof. K

Generalizations of the subset conditions and of themarginal conditions will be given in Section 8.



5. THE CASE OF THREE ALTERNATIVES

When |C|=3, say C=[a, b, c], we will write A for y[a] ,AB for y[a, b] , etc. Thus, equalities A+B+C=1 andAB+AC+BC=1 are satisfied by the AV-polytope A3 .Moreover, as y<= yC=1 holds, coordinates y< and yC

will be ignored in all the computations of this section.Since 1<|X|<3&1 is impossible, no subset condition

remains for n=3. The marginal conditions boil down to

AB+AC�A, (8)

AB+BC�B, (9)

AC+BC�C. (10)

It is easy to establish that Inequalities (8)�(10) are inde-pendent by providing appropriate 6-tuples (A, B, C, AB,AC, BC) (see Regenwetter, 1995), or by listing the verticesof the corresponding facets (see Table 1 below).

Theorem 9. If n=3, the six nonnegativeness inequalitiesand the three marginal inequalities, together with the fourlinear equations

y<= yC=A+B+C=AB+AC+BC=1,

provide a linear description of A3 in RP([a, b, c]).

The combinatorics of the polytope will be described at theend of this section (as well as another way of proving thetheorems in this section). As regards analytical geometry,Theorem 9 states that the polytope A3 (now seen in the sub-space with equations y<= yC=A+B+C=AB+AC+BC=1) is the set of solutions (A, B, AB, AC) of the follow-ing system of nine linear inequalities:

A�0, (11)

B�0, (12)

1�A+B, (13)

AB�0, (14)

AC�0, (15)

1�AB+AC, (16)

AB+AC�A, (17)

1�B+AC, (18)

A+B�AB, (19)

where (11)�(16) are the nonnegativeness conditions, and(17)�(19) are the three marginal conditions, stated only interms of the variables A, B, AB, AC. Moreover, for a givensuch solution (A, B, AB, AC), the proof builds the followingexplicit form for all probability distributions q on 6 thatinduce (A, B, AB, AC) according to the size-independentmodel:

\qabc

+=\1+A&AC

+qacb AC&1qbac AB+AC&A&1qbca 1+A+B&AB&ACqcab 1qcba 0

+: \&1

++;\0

+ , (20)

+1 0+1 0&1 0&1 00 1

where

: # [max(A+B, 1&AC, 1+A&AB&AC),

min(1+A&AC, 1+A+B&AB&AC, 1)] (21)

and ;=:&A&B.

Proof. The necessity of Conditions (11)�(19) for a pointto be in A3 was established in Section 4. As for the suf-ficiency, we consider the following inhomogeneous systemof linear equations in the unknown probability distributionq, for a given quadruple (A, B, AB, AC):

\1010

1001

0110

0100

0001

0000+ \

qabc

+=\AB

ABAC+ . (22)

qacb

qbac

qbca

qcab

qcba

Its set of solutions in R6 is given by Eq. (20), with :, ; # R.Now, a solution is a probability distribution on 6 iff all itscoordinates are nonnegative and add up to 1. This latterrequirement translates as ;=:&A&B, together with For-mula (21). The nonemptiness of the interval in Formula


File: 480J 115511 . By:XX . Date:01:07:01 . Time:07:51 LOP8M. V8.0. Page 01:01Codes: 4209 Signs: 2760 . Length: 56 pic 0 pts, 236 mm

TABLE 1

The Facet�Vertex Incidence Table Produced by Porta for A3

Note. The 24 vertices and 8 coordinates are in the orderings correspond-ing to Eqs. (2) and (3), resp.

(21) amounts to the system (11)�(19). This completes theproof of sufficiency. K

Equations (20) and (21) together with ;=:&A&Balso indicate when the probability distribution q is unique.Leaving details to the reader, we conclude:

Theorem 10. Suppose n=3. Fix a probability distri-bution r on P(C) for which there exists a size-independentrandom utility representation q. Then

A=AB+AC � A=1&BC � BC=B+C, (23)

B=AB+BC � B=1&AC � AC=A+C, (24)

C=AC+BC � C=1&AB � AB=A+B, (25)

and the following statements are equivalent:

(i) the probability distribution q is unique;

(ii) equality holds in at least one of the Eqs. (23)�(25),or in at least one of the following equations:

A=0, B=0, C=0,

BC=0, AC=0, AB=0;

(iii) the point r lies on the boundary of A3 ;

(iv) picking an arbitrary reference ranking and viewingall others as permutations of the reference, at least one evenand one odd permutation have probability zero.

Remark 11. An analog of Sen's value restriction (1996)holds when the conditions of Theorem 10 are satisfied. Inthis case, an aggregate linear preference ordering, sum-marizing the overall preferences in the population, can becomputed in a canonical way. For proofs and additionalcomments, see Regenwetter and Grofman (1995).

Let us move to the combinatorial structure of thepolytope A3 . By Theorem 9, A3 has nine facets, and ofcourse six vertices. The PORTA-produced Table 1 displaysthe facet�vertex incidence, with rows corresponding tofacets and columns corresponding to vertices in the ordercorresponding to Eq. (2). Note that the data are formulatedin terms of eight coordinates, respecting the orderingspecified by Eq. (3).

We now identify a face with the set of vertices it contains.To obtain all the faces of A3 as all intersections of facets, weran a Pascal implementation of an improvement of analgorithm due to Dowling (1993). It turns out that all pairsof vertices form an edge, i.e. A3 is a neighbourly polytope.The eighteen triangular, two-dimensional faces are all triosof vertices except two (namely, the trio consisting of verticesassociated to even, resp. odd, permutations; here, assuminga fixed enumeration of the alternatives, we see a ranking asa permutation). The nine facets are three-dimensional sim-plices, thus A3 is simplicial. These nine facets are the com-plements of the following pairs of vertices (where we write? instead of v?):

[abc, acb], [bac, cab], [bca, cba],

[bac, abc], [bca, acb], [cba, cab],

[bac, bca], [abc, cba], [acb, cab].

As a consequence, there are 72 combinatorialautomorphisms of A3 , that is, 72 permutations of the



vertices that map any face onto a face (see Gru� nbaum, 1967,for this notion of combinatorial automorphism). Takenwith composition, they form a group G isomorphic to the``Gruppenkranz'' S2[S3] (see Harary, 1969, p. 164, with Sn

denoting the symmetric group on n letters). It is remarkablethat G acts transitively on the vertices, on the two-dimensional faces, and on the facets, but not on the edges.

All the assertions about a linear description and the com-binatorial structure of A3 can also be routinely deducedfrom the following easy observations. We are back in theEuclidean space RP(C)"[<, C], and view again all rankings aspermutations of a reference ranking. The three vertices v? ,for ? an even permutation, form the vertices of a triangle T+

(which happens to be equilateral). Similarly, the three ver-tices v? , for ? odd, form the vertices of a triangle T&. Thesetwo triangles respectively lie in two (nonorthogonal) planeshaving only one point in common, which is the centroid ofboth triangles. As A3 is the convex hull of T+ _ T&, all ourassertions about A3 can thus also be obtained from thesegeometric considerations.

Moreover, any combinatorial automorphism is inducedby some affine automorphism of the affine subspace aff A3

generated by A3 .We should warn the reader that most of the nice proper-

ties of A3 fail to hold in general for the approval-votingpolytope An .

6. NEW FACET-DEFINING INEQUALITIES

Up to here, we have exhibited three families of facet-defin-ing linear inequalities, named and designated respectively as

v the nonnegativeness condition NNEG(X ) (withX # P(C)"[<, C]);

v the subset condition SUBS(X ) (with X # P(C) and1<|X|<n&1); and

v the marginal condition MARG(i, k) (with 1<k<nand i # C)

(cf. Section 4).By definition, any combinatorial automorphism of the

approval-voting polytope An transforms a facet of An into afacet. We exploit this idea to get one new family of facets. Aswe make use of an affine automorphism of the affine spaceaff An , we easily derive the corresponding (new) linearinequality from any known one.

At the end of Section 2, we explained that each permuta-tion : of the set C of alternatives produces an affineautomorphism F: of aff An that (globally) preserves An .However, each of the three known families of facet-defininginequalities obtained in Section 4 is transformed into itselfby any such automorphism. Fortunately, however, here isanother automorphism that will prove fruitful.

Given a ranking ?=(a1 , a2 , ..., an) of C=[1, 2, ..., n],the reversed ranking of ? is r(?)=(an , an&1, ..., a1). We getthe involutive map r: 6 � 6 (thus, r2 is the identity on 6).The complement map P(C) � P(C): X [ C"X is alsoinvolutive. We deduce an involutive affine automorphism,which we name the reversing automorphism:

F: RP(C) � RP(C): y [ z, (26)

where by definition zX= yC"X , for X # P(C). The namegiven to F is motivated by the following property, whichalso shows that F preserves An : The vertex v? of An ismapped by F to the vertex vr(?) . Facets defined by a non-negativeness condition NNEG(X ) are easily seen to betransformed by F into facets of the same type, namelyNNEG(C"X ). On the other hand, the facet defined by thesubset condition SUBS(X) is transformed by F into a(generally) new facet-type.

Using the covering relations c& and C& on P(C), definedby

XC&Y � (X/Y and |X|=|Y|&1) � Yc&X, (27)

the subset condition SUBS(X ) can be rewritten as

:Y # P(C), YC&X

yY� yX , (28)

where X is a subset of C with 1<|X|<n&1. It is trans-formed by F into the following condition, a new family offacet-defining inequalities.

Superset Condition SUPS(X). For a subset X of C

with 1<|X|<n&1,

:Y # P(C), Yc&X

yY� yX . (29)

Note that for |X|=1, the latter inequality is equivalent to amarginal condition MARG(i, k) having k=2.

Transforming in a similar manner a marginal inequalityMARG(i, k) produces an inequality which is equivalentto some marginal inequality again, namely toMARG(i, n&k+1).

7. THE CASE OF FOUR ALTERNATIVES

Tables 2 and 3 show the output produced by the PORTAsoftware for n=4, when the input is the lexicographic list of4!=24 vertices in R14. As in the case n=3, we leave outcoordinates y< and yC . The other coordinates wererelabelled from x1 to x14, and correspond to the nontrivialsubsets of C=[a, b, c, d] in the order



[a], [b], [c], [d], [a, b], [a, c], [a, d], [b, c],

[b, d], [c, d], [a, b, c], [a, b, d], [a, c, d], [b, c, d].

(30)

A study of Table 2 teaches us that the 34 facet-defininginequalities all come from the nonnegativeness conditions(14), subset conditions (6), superset conditions (6), andmarginal conditions (8).

We skip a detailed description of the combinatorial struc-ture of the polytope A4 , since only a few nice propertieslisted for A3 are shared by A4 . However, we display in Table3 the facet-vertex incidence-matrix produced by PORTA.

TABLE 2

Facet-Defining Inequalities for A4

Note. Coordinates, with y< and yC left out, are listed as in Eq. (30).

TABLE 3

The Facet�Vertex Incidence Table for A4 , as Produced by Porta

8. MORE INEQUALITIES

We now introduce two fairly general families ofinequalities satisfied by the approval-voting polytope An .

Sandwich Condition SAND(U, V, k). For any twosubsets U and V of C, and any size k with U�V and|U|�k�|V|,

:X # P(C, k), U�X�V

yX� :Y # P(C, k+1), U�Y, |V & Y|�k

yY . (31)



By applying the reversing automorphism F of Formula(26), we derive:

Reversed Sandwich Condition RSAN(W, T, l ). Forany two subsets W and T of C, and any size l with W�T and|W|�l�|T|,

:X # P(C, l ), W�X�T

yX� :Y # P(C, l&1), Y�T, |W _ Y|�l

yY . (32)

It is not difficult to check that each vertex of An satisfiesall of the (reversed) sandwich conditions, and thus that An

does so also. At present, we are not able to decide which ofthese inequalities are facet-defining. As a matter of fact, allof the inequalities that were proved to be facet-defining,except for almost all of the nonnegativeness ones, are specialcases of the (reversed) sandwich conditions. Indeed, thesandwich condition SAND(U, V, k), for

v 1<|U|=k<n&1, gives the superset conditionSUPS(U( ) (see Inequality (29));

v 1<|V|=k<n&1, gives the superset conditionSUPS(V );

v U=[i], 0<k<n&1, and V=C, gives the marginalcondition MARG(i, k&1) (see Theorem 8).

Similarly, the reversed sandwich conditionRSAN(W, T, l ), for

v 1<|W|=l<n&1, gives the subset conditionSUBS(W ) (see Theorem 7);

v 1<|T|=l<n&1, gives the subset conditionSUBS(T );

v W=<, 1<l<n, and T=C"[i], gives the marginalcondition MARG(i, l ) (see Theorem 8).

From the above two lists of special cases, it is clear thatthe sandwich and reversed sandwich conditions can coin-cide for certain values of their parameters.

We mention in passing two other interesting subcases,namely SAND(U, C, k), which obviously generalizesMARG(i, k+1),

:X # P(C, k), U�X

yX� :Y # P(C, k+1), U�Y

yY , (33)

and RSAN(<, T, l ), a generalization of MARG(i, l ):

:X # P(T, l )

yX� :Y # P(T, l&1)

yY . (34)

At least for special cases, we achieve here new facet-defining inequalities.

Theorem 12. The linear inequality SAND(U, C, k) inFormula (33) is facet-defining for 1<|U|<k<n&1.

Similarly, the linear inequality RSAN(<, T, l ) in Formula(34) is facet-defining for 1<l<|T|<n&1.

Proof. As the reversing automorphism transformsInequality (33) into Inequality (34), we treat only thesecond assertion. It is easily checked that all vertices satisfythe inequality in Formula (34). Moreover, equality obtainsat a vertex v? in exactly two cases. Either both membersequal 0, which occurs when at least one of the first l&1alternatives of ? is not in T, or both members equal 1, whenthe first l alternatives of ? belong to T. SettingT=[n&t+1, n&t+2, ..., n], we turn again to the genericsequence (cf. Section 4). Of its rankings, those that do notgive equality have both their first l&1 alternatives in T andtheir alternative ranked l not in T. We now use the freedomleft in the construction of the generic sequence in order tomodify rankings that do not give equality��except for one.

If n is not in the last position of such a ranking, we mayexchange the alternative in position l with an alternative inT (since 1<l<|T| ), and in this way get equality.

If n is in the last position, we are in the recurrence part ofthe generic sequence. If n&1 is not in the before-last posi-tion, and l�|T|&1, a similar argument applies.

We may repeat such an argument until we haven, n&1, ..., n&t+l all fixed in the position equal to theirrespective value (and still l&1 other elements in T ). Clearly,among the remaining rankings to examine, only one will notgive equality (since in the generic sequence on m elements,with 0<l&1<m, if alternatives for which a choice ispossible are put in increasing order, there is exactly oneranking that begins with [m&l+2, m&l+3, ..., m]).

Consequently, we can list 2n&n&1 rankings givingequality in Formula (34). Since the corresponding verticesare affinely independent, the thesis follows. K

9. THE CASE OF FIVE ALTERNATIVES

The Porta software output we get for n=5 indicates thatA5 has 235 facets. From carefully scrutinizing theinequalities (and rewriting many of these), we came to theconclusion that 105 of these inequalities are among those weproved to be facet-defining, that 70 additional ones belongto the (reversed) sandwich inequalities, and finally that anew class of 60 inequalities is met. To describe the latterclass, we display only one example. Assuming C=[a, b, c, d, e], and writing e.g. AB instead of y[a, b] , we have

ADE+BCE+CDE

�AD+AE+BC+BE+CD+CE+DE

(another equivalent form is given in the last row of Table 4).It is easily seen that 59 distinct but similar inequalities resultfrom applying to the above inequality all automorphisms F:

induced by permutations : of the set C. Moreover, the



TABLE 4

The List of All Facet-Defining Inequalities for A5

Note. The last two columns respectively give the number of vertices satisfying the inequality example and the number of such inequalities.

resulting collection of 60 facets is stable under the reversingautomorphism. A general family of inequalities includingthis collection will be given in Section 10.

Table 4 summarizes the Porta output for n=5 andprovides additional information. We have grouped on oneline all inequalities equivalent under the automorphisms F:

of A5 .

10. PROBABILISTIC INTERPRETATION OFTHE INEQUALITIES

Each of the (possibly facet-defining) linear inequalitiesfound so far for the AV-polytope will be shown to mean thata certain event must have a nonnegative probability. Werecall that the random variable R, which takes its values inthe whole set 6 of rankings of C, models the latent rankingof the voter. Moreover, from the fundamental Eq. (1) of thesize-independent model, we derived

yX=P(V=X | S=|X| )=P(R # 6X).

In the list below, the difference of the two terms in the men-tioned inequality is the probability of the event that thevalue of R belongs to the set of rankings described. When wespeak here of a beginning set, or of first alternatives, we referto the ranking which is the outcome of R.

v Nonnegativeness inequality NNEG(X): the set X is abeginning set.

v Subset inequality SUBS(X ): the set X is not a begin-ning set, although all of its elements but one come in the first|X|&1 positions (in any order). Equivalently, the first|X|&1 alternatives are all in X, but the alternative in posi-tion |X| is not.

v Superset inequality SUPS(X ): all the elements of Xare among the |X|+1 preferred alternatives, although theyare not the |X| preferred ones. Equivalently, the first |X|+1alternatives are in X, except for one alternative which is notthe one in position |X|+1.

v Marginal inequality MARG(i, k): alternative i is atposition k;

v Sandwich condition SAND(U, V, k): the first k+1alternatives contain all those in U and at least k elements of



V, and either one of the first k is not a member of V or thealternative at rank k+1 is in U. In other words, all theelements from U, plus k&|U| or k&|U|+1 furtherelements from V, are among the first k+1 preferred alter-natives, and either position k+1 contains an element of Uor one of the first k elements is not in V.

v Reversed sandwich condition RSAN(W, T, l): thefirst l&1 alternatives are in T, and they include all theelements of W except maybe one, although the first l alter-natives do not contain W or are not all in T. In other words,all elements in W except maybe one are among the l&1preferred alternatives which in turn are all in T, and at leastone element of W is ranked higher than l or the alternativein position l is not in T.

As stated in the Introduction, we do not know whethereach facet of An can be described by an equality couched inthe form P(E)�0, for some E in P(6) (although its equa-tion is for sure linear in the variables yX=P(R # 6X), forX # P(C), and can always be written with integer coef-ficients since the facet is affinely generated by points havinginteger coordinates).

We now devise a new general class of necessary condi-tions for a point of RP(C) to be in An , thus obtaining for thefirst time a class that includes the last inequality in Table 4(for n=5). Let F be a collection of subsets of C, and let land k be two natural numbers with l�k. Then the probabil-ity that some member of F appears within the first k posi-tions, but no member of F is within the first l positions,has to be nonnegative. Formally, writing PF(C, m)=[X # P(C, m) | _F # F: F�X], we define

E=\ .X # PF(C, k)

6X+>\ .Y # PF(C, l )

6Y+ .

Then the necessary condition P(E)�0 becomes the follow-ing linear inequality satisfied by the whole polytope An :

:X # PF(C, k)

yX� :Y # PF(C, l )

yY . (35)

The latter inequality is dubbed the generalized sandwichcondition GSAN(F, l, k). It is easy to see that several of ourconditions are special cases of this inequality, as shownbelow.

Let us first assume that F contains exactly one set X. Ifmoreover |X|=k>l, Inequality (35) becomes yX�0,which is the nonnegative condition NNEG(X ) when0<k<n also. If, on the other hand, |X|=l, then Inequality(35) becomes

:Y # P(X, k), X�Y

yY� yX .

As the reversing automorphism transforms this inequalityinto the Inequality (6) (as in the general subset condition,see Section 4), Theorems 7 and 8 imply that this inequalityis facet-defining iff either 1<k&1=|X|<n&1 (yieldingthe superset condition SUPS(X )), or k&1=|X|=1 (yield-ing the marginal condition MARG(i, k) with k=2).

If F contains exactly one set which is now assumed to bethe singleton [i], we derive

:Y # P(C, k), i # Y

yY� :X # P(C, l), i # X

yX .

This inequality is facet-defining iff either 1<k=l+1<n, inwhich case it is the marginal condition MARG(i, k), ork # [1, n], in which case the inequality is a nonnegative con-dition NNEG(X ) with |X|=k.

Taking F equal to the collection of all subsets of size l ofC that contain a given subset U, we obtain a generalizationof Inequality (33) (which results for l=k&1).

Moreover, given subsets U, V of C and a natural numbers such that |U|�s�|V|, we may form the collection F ofall subsets F with |F|=s and U�F�V. Then condition(35) provides a generalization of the sandwich conditionSAND(U, V, k) (which is obtained for s=l=k&1).

Finally, for n=5, with

F=[[1, 2], [1, 3], [2, 4]],

l=2, and k=3, Formula (35) becomes the inequality in thelast row of Table 4.

Not surprizingly, the reversing automorphism generatesa reversed generalized sandwich condition RGSA(F, l, k).Let F again be a collection of subsets of C and l�k. Thenthe probability that some member of F appears within thelast k positions, but no member of F is within the last l posi-tions, has to be nonnegative. The resulting inequality is

:X # PF(C, k)

yC"X� :Y # PF(C, l )

yC "Y . (36)

Again, several of our inequalities are easily shown to be spe-cial cases of this inequality; we skip the details.

It is important to note that all the facet-defininginequalities presented in this paper are particular cases ofConditions (35) and (36). However, for many of the othercases of these two inequalities, we still do not know whetherthey are facet-defining. The same question arises for otherinequalities of the form P(E)�0, for E a subset of 6.

Another unanswered question is the following. Is theresome family H of subsets of 6 such that for H # H theprobability P(H) can be written in terms of values P(6X),where X # P(C), and such that moreover all inequalitiesP(H)�0, for H # H, characterize the size-independentmodel? Remember that our main, original problem is to



characterize the mappings r on P(C) for which there existprobability distributions p and q on respectively S and 6,such that

rX=p |X| } :? # 6X

q? ,

for all X in P(C). All these mappings r are probability dis-tributions on P(C), and they altogether form the set Dn

defined in the Introduction. Remember that the knowledgeof r entails that of p, since for s # S

ps= :X # P(C, s)

rX .

The main problem was then recast as the following one: Tocharacterize the mappings y on P(C) for which there existsa probability distribution q on 6, such that, for X�C:

y(X )= yX= :? # 6X

q? .

All those y 's constitute the AV-polytope An .For a given y, we may at once deduce the following map-

ping:

h: [6X | X # P(C)] � [0, 1]: 6X [ yX ,

which is the restriction to the family E=[6X | X # P(C)]of the probability measure associated to q. Conversely, ycan be deduced from h since yX=h(6X). There follows stillanother reformulation of our main problem: To charac-terize the mappings h: [6X | X # P(C)] � [0, 1] for whichthere exists a probability distribution q on 6 with

h(6X)= :? # 6X

q? ,

for all 6X in E. Such a formulation shows that our specificproblem is a particular case of the following general one:Given a family E of subsets of a set 0 together with a real-valued mapping h on E, under which conditions on E andh is there a probability measure P on the _-algebra of sub-sets of 0 generated by E, such that h(E)=P(E) for allE # E? In this connection, note that in our actual setting the_-algebra of events (subsets of 6) generated by the family(6X)X # P(C) equals P(6). Moreover, the geometricapproach has shown in this paper that there is a minimal listof conditions; this fact is true for the general problem aswell, at least when 0 is finite. However, we should mentionthat the general problem also involves the (presently intrac-table) problem of finding a description of the binary-choicepolytope (cf. the Introduction).

11. SUMMARY AND CONCLUSIONS

The present paper investigates the approval-votingpolytope An associated with the size-independent model ofapproval voting of Falmagne and Regenwetter (1996),where n is the number of candidates. We have shown thatthe dimension of An is 2n&n&1, which implies that thereare exactly 2n&n&1 degrees of freedom in the marginalranking probabilities P(R # 6X). Besides a classification ofFalmagne and Regenwetter's (1996) necessary conditionsfor the model into those conditions that are facet-definingand those that are not, we obtained a complete (minimallinear) description of the model for n�5 (relying heavily onthe Porta software for n=4 or n=5). However, so far, wewere able to prove only the conditions derived for n�4 tobe general facet-defining inequalities also for arbitrary n.The crucial method used for proving an inequality to befacet-defining lies in the recursive construction of the genericsequence Gn of rankings, and in demonstrating that itcorresponds to 2n&n&1 affinely independent verticessatisfying the corresponding equality. At the end of thepaper, we discussed probabilistic interpretations of allinequalities encountered before, and derived a general set ofinequalities from probabilistic considerations which encom-passes all earlier cases. In this context some open problemswere raised that future research needs to address.

REFERENCES

Brams, S. J., 6 Fishburn, P. C. (1983). Approval Voting. Boston:Birkha� user.

Christof, T. (1996). PORTA��A polyhedron representation transforma-tion algorithm; available from ftp:��ftp.zib-berlin.de�pub�mathprog�polyth�porta�index.html.

Dieudonne� , J. (1960). Foundations of Modern Analysis. New York:Academic Press.

Doignon, J.-P., 6 Falmagne, J.-C. (1994). A polynomial time algorithm forunidimensional unfolding representations. Journal of Algorithms 16,218�233.

Dowling, C. (1993). On the irredundant construction of knowledge spaces.Journal of Mathematical Psychology 37, 49�62.

Falmagne, J.-C., 6 Regenwetter, M. (1996). A random utility model forapproval voting. Journal of Mathematical Psychology 40, 152�159.

Fishburn, P. C. (1992). Induced binary probabilities and the linear order-ing polytope: A status report. Mathematical Social Sciences 23, 67�80.

Gru� nbaum, B. (1967). Convex Polytopes. New York: Wiley.Harary, F. (1969). Graph Theory. Reading, MA: Addison�Wesley.Koppen, M. (1995). Random utility representations of binary choice

probabilities: Critical graphs yielding critical necessary conditions.Journal of Mathematical Psychology 39, 21�39.

Marley, A. A. J. (1990). A historical and contemporary perspective on ran-dom scale representations of choice probabilities and reaction times inthe context of Cohen and Falmagne's (1990, Journal of MathematicalPsychology 34) results. Journal of Mathematical Psychology 34, 81�87.

Merrill, S. I. (1988). Making Multicandidate Elections More Democratic.New Jersey: Princeton Univ. Press.



Nurmi, H. (1987). Comparing Voting Systems. The Netherlands: Kluwer.Regenwetter, M. (1995). Random utility representations and probabilistic

models of subset choice and ranking data (Ph.D. dissertation). Mathe-matical Behavioral Sciences, University of California, Irvine.

Regenwetter, M. (1996). Random utility representations of finite m-aryrelations. Journal of Mathematical Psychology 40, 219�234.

Regenwetter, M., 6 Grofman, B. (1995). Choosing subsets: A size-inde-pendent probabilistic model and the quest for a social welfare ordering.Social Choice and Welfare (to appear); based on Technical Report MBS94-20, IMBS, University of California, Irvine.

Regenwetter, M., 6 Grofman, B. (in press). Approval voting, Borda win-ners and Condorcet winners: Evidence from seven elections. Manage-ment Science.

Regenwetter, M., Marley, A., 6 Joe, H. (1996). Random utility threshold

models of subset choice (submitted for publication); based on TechnicalReport MBS 94-26, IMBS, University of California, Irvine.

Sen, A. K. (1966). A possibility theorem on majority decisions.Econometrica 34, 491�499.

Suck, R. (1992). Geometric and combinatorial properties of the poly-tope of binary choice probabilities. Mathematical Social Sciences 23,81�102.

Suck, R. (1996). Random utility representations based on semiorders,interval orders and partial orders (submitted for publication).

Weber, R. J. (1995). Approval voting. Journal of Economic Perspectives 9,39�49.

Ziegler, G. M. (1995). Lectures on Polytopes. Berlin: Springer-Verlag.

Received June 10, 1996


Date post:	20-Jan-2023
Category:	Documents
Upload:	ulb
View:	0 times
Download:	0 times

An Approval-Voting Polytope for Linear Orders

Documents