An Approval-Voting Polytope for Linear Orders

Journal of Mathematical Psychology  MP1155 journal of mathematical psychology 41, 171188 (1997) article no. MP971155

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 generates a collection of probability distributions on the family of all subsets of the set of alternatives. Focusing on the size-independent model proposed by Falmagne and Regenwetter, we recast the problem of characterizing these distributions as the search for a minimal system of linear equations and inequalities for a specific convex polytope. This approval-voting polytope, with n ! vertices in a space of dimension 2 n, is proved to be of dimension 2 n &n&1. Several families of facet-defining linear inequalities are exhibited, each of which has a probabilistic interpretation. Some proofs rely on special sequences of rankings of the alternatives. Although the equations and facet-defining inequalities found so far yield a complete minimal description when n4 (as indicated by the PORTA software), the problem remains open for larger values of n. ] 1997 Academic Press

1. INTRODUCTION

In the general procedure of approval voting, each voter selects some subset of a given set of alternatives, say a subset X of a fixed set C of n choice alternatives, or candidates. The literature on approval voting in political science and economics is substantial, including several books (e.g., Brams 6 Fishburn, 1983; Merrill, 1988; Nurmi, 1987) and countless 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 (set valued) random variable V (as in vote). A model proposed by Falmagne and Regenwetter (1996) for the probabilities P(V=X ) is based on the following, basic assumption: A We thank Jean-Claude Falmagne for his interest in this work, since he introduced us to the problem, and two anonymous referees for their careful reading. We acknowledge help from D. Bien, Computing Centre, U.L.B., in the implementation of the PORTA software. We are very grateful to the designers of PORTA for having made publicly available such a powerful tool and to A. A. J. Marley for very helpful feedback on an earlier draft. M. R. acknowledges postdoctoral fellowship support from A. A. J. Marley funded 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, 1050 Bruxelles, Belgium. E-mail: doignonulb.ac.be. 171

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

voter chooses subset X if she chooses as many alternatives as X contains and if the alternatives in X are hisher favorites. The reader who loves concrete examples is invited to first have a look at Sections 2 and 5, which start with the case C=[a, b, c], that is, n=3. To describe the model for general n, consider two more random variables which are jointly distributed with V. The first variable, S, takes its values in the set S=[0, 1, ..., n] of possible set-sizes, while the second variable, R, takes its values in the collection 6 of all rankings of C. Rankings of C are often identified with linear orders or permutations on C. For a given subset X of C, we use 6 X to denote the set of all rankings of C in which the elements of X are ranked before all the elements of C"X. When ? # 6 X , 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 the size-independent model of approval voting introduced by Falmagne and Regenwetter (1996) reads, for any subset X of C, as P(V=X )=P(S= |X| ) } P(R # 6 X ).

(1)

We will not discuss here the pros and cons of the size-independent model (see Falmagne 6 Regenwetter, 1996; Regenwetter 6 Grofman, in press, 1995; Regenwetter, Marley, 6 Joe, 1996), but rather address the problem of finding a characterization of this model. To be specific, how can we decide whether a given probability distribution on the family P(C) of all subsets of C results from the model? We investigate this question from a geometric perspective rather than a statistical testing approach. In geometric terms, the question is to find an analytical description of the collection Dn of all such distributions. The elements of Dn will be taken as vectors in the real vector space R P(C), with one vector component for each subset of C. The collection Dn , thus seen as a subset of the space R P(C), is not convex (see the next section). Note that for each subset X of C, the probability P(S= |X| ) is 0022-249697 25.00 Copyright  1997 by Academic Press All rights of reproduction in any form reserved.

172

DOIGNON AND REGENWETTER

uniquely determined by the values P(V=Y ). We can thus restrict our attention to the conditional probabilities P(V=X | S= |X|) with XC. The latter are equal to P(R # 6 X ). As we show in Section 2, the possible values of these conditional probabilities form the convex hull of the following subset V of R P(C): V=[ y # [0, 1] P(C) | _? # 6 with y X =1  ? # 6 X ]. The set V has n ! elements, namely one for each ? # 6, and can thus be identified with 6. In an abstract sense this solves the characterization problem for the size-independent model, since all possible ranking probabilities are specified by the convex hull of V. However, we seek to know more concretely what constraints this fact imposes on R, i.e. on the possible conditional probabilities. More precisely, we need a mechanism for determining whether a given realvalued function over P(C) does indeed lie in the convex hull of V. Furthermore, we desire the mechanism to be ``minimal'' (this will be defined formally below), in the sense of avoiding redundancies within a collection of necessary and sufficient conditions. The convex hull of finitely many points in a finite-dimensional real vector space is a convex polytope. We call the convex hull of V the approval-voting polytope for linear orders. This polytope results from the model for approval voting due to Falmagne and Regenwetter (1996); some other models of approval voting, e.g. based on weak orders or semiorders, are introduced in Regenwetter (1996). The AV-polytope, as we also call it, will be denoted by An where the subindex n is a reminder of the existence of one such polytope for each cardinality n of C. The next section will work out some technicalities about the passage from Dn to An , while the remainder of the paper then focuses on the convex polytope An and the interpretation of its properties in terms of probabilities over 6. The reader is referred to Grunbaum (1967) and Ziegler (1995) for more detailed terminology and results about affine geometry and convex polytopes. We limit ourselves to the basics here. A convex polytope in R d can also be characterized as a bounded intersection of finitely many halfspaces in R d, where each halfspace is defined by a linear inequality. Replacing such an inequality by the corresponding equality defines a hyperplane in R d, which is thus the intersection of two (opposite) halfspaces. A polytope may be contained in such a hyperplane. Thus, a convex polytope can be described by a system of linear equalities and inequalities. If such a system contains a maximum number of independent equalities and if removing one of its inequalities extends the polytope to a larger set, then the system is called a (minimal linear) description of the polytope. Indeed, our problem can now be rephrased as follows: find a description of An , that is, a minimal system of linear equations and linear inequalities on R P(C) whose set of solutions is An .

File: 480J 115502 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 6860 Signs: 5970 . Length: 56 pic 0 pts, 236 mm

Before we go into more detail, consider the following simple calculation: since the observables take the form of a probability distribution over P(C), they have 2 n &1 degrees of freedom. However, the probabilities P(S= |X| ) have n degrees of freedom. Thus, the conditional probabilities P(V=X | S= |X| ), which are in turn equal to the probabilities P(R # 6 X ), have altogether 2 n &n&1 degrees of freedom at most. The first step in our geometric analysis of the AV-polytope consists in determining the dimension of An . The dimension of a subset of R d is the maximum number of affinely independent points in that set, minus one. Hyperplanes of R d are of dimension d&1. The dimension of a polytope A in R d equals d minus the number of independent equations in a description of A. In Section 3 we prove dim An =2 n &n&1 and give a minimal system of n+1 linear equations for the affine subspace generated by An . This establishes from the geometric perspective that there are exactly 2 n &n&1 many degrees of freedom in the probabilities P(R # 6 X ). As the convex polytope An is not full dimensional in R P(C), a linear description of An is not unique. A linear inequality is called valid for a polytope A in R d if A is contained in the corresponding halfspace. A face of A consists of all the points of A that give equality in a linear inequality valid for A. A maximal proper face is called a facet; thus, a facet of An is a polytope of dimension 2 n &n&2. A linear inequality on R d satisfied by A is facet-defining if the corresponding equation (obtained by replacing the inequality sign with an equality sign) defines a hyperplane whose intersection with A is a facet. All linear inequalities in a (minimal linear) description of A are facet-defining, and there is one such inequality per facet. Thus the search for a minimal description of An involves determining the facet-defining inequalities for An . This is the desired ``minimal mechanism'' introduced above. We are able to determine which of the linear inequalities given in Falmagne and Regenwetter (1996) are facet-defining (Section 4). New facet-defining inequalities are offered in Sections 6 and 8, and more inequalities whose status (for arbitrary n) is still unknown are proposed in Sections 8 and 10. From outputs produced by the computer software PORTA (see Christof, 1996), we know that the facet-defining inequalities obtained so far entail a complete (minimal linear) description of An for n4. A vertex of a polytope A is a point which constitutes a face of A. The vertices of the approval-voting polytope An form the set V already encountered. They correspond to the n ! rankings of C. The latter property also holds for other polytopes in the literature, e.g., the ``permutohedron'' and the ``binary-choice polytope.'' The permutohedron has dimension n&1 (see, e.g., the references in Doignon 6 Falmagne, 1994); its facets correspond to weak orders of C having only two classes. The binary-choice polytope, which is intensively studied in many mathematical disciplines, has

173

AN APPROVAL-VOTING POLYTOPE FOR LINEAR ORDERS

dimension n(n&1)2. A complete characterization of its facets seems to be inaccessible at present, as witnessed by recent papers (e.g., Marley, 1990; Suck, 1992; Fishburn, 1992; and Koppen, 1995). (Related polytopes, obtained by replacing linear orders with, e.g., semiorders or interval orders, are studied in Suck, 1996). We cannot tell from our investigations whether the similar problem for the AVpolytope is of the same level of difficulty. Turning back to our original probabilistic setting, we now make an interesting observation. Each of the facetdefining linear inequalities obtained in this paper for the AV-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 short notation, P(E)0 for some event E in P(6). As an example, consider the (possibly facet-defining) inequality on R P(C)

generalize the conditions found so far into a new general class of conditions, some of which are proven to be facetdefining. We then move on to the case when n=5 for which the PORTA software provides a complete characterization. As an alternate perspective to the geometric approach taken throughout most of the paper, Section 10 provides probabilistic interpretations for all the previously discussed conditions. Furthermore, relying on probabilistic considerations, we derive a further class of conditions which generalize all previous conditions. This includes conditions that PORTA provided for n=5 and which had not been derived before. In this context, we also discuss open problems.

y [a, b] + y [a, c]  : y [a, x, y] ,

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

}}}

where a, b, and c are three alternatives, and the sum extends over all nonordered pairs [x, y] in C "[a] that contain at least one element of [b, c]. It is produced by the event consisting of all rankings in 6 that satisfy the conditions a and exactly one element of [b, c] are among the three most preferred alternatives, without being 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, for some event E in P(6). Related questions are raised in Section 10. This section provides the probabilistic interpretations of all of our facet-defining inequalities. It then infers a fairly general family of linear inequalities involving all of the previous ones. We leave it for further work to determine which of these new inequalities are facet-defining (for arbitrary n). Also left for future investigation is a more productive exploitation of the automorphism group of the AVpolytope. In Section 6, we use one particular automorphism to generate a new family of facet-defining inequalities. However, the full group of automorphisms is still to be determined, as well as its action on the family of all facets. The paper is organized as follows. The following section introduces further concepts and deals with some technicalities that are involved in getting from the data to the AVpolytope An . In Section 3 we compute the dimension of An . In Section 4 we review and classify the testable conditions of Falmagne and Regenwetter (1996) into those that are facetdefining and those that are not. We discuss the case of n=3 in detail in Section 5. The following section introduces further testable conditions. After providing the complete characterization of the case n=4 in Section 7 we then

File: 480J 115503 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 6178 Signs: 4439 . Length: 56 pic 0 pts, 236 mm

2. FROM THE SET Dn TO THE APPROVAL-VOTING POLYTOPE An

S=[0, 1, 2, 3] and 6=[abc, acb, bac, bca, cab, cba].

(2)

Note our shorthand for rankings of C, in which we assume that 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 in which we listed the elements of 6 in Eq. (2) and the elements of P(C) in Eq. (3). To each pair of probability distributions p: S  [0, 1]: s [ p s , q: 6  [0, 1]: ? [ q ? , we associate the probability distribution r: P(C)  [0, 1]: X [ p |X| } : q ? , ? # 6X

reminding the reader that 6 X is the set of all rankings having X as a beginning set. At the core of the size-independent model of Falmagne and Regenwetter (1996) is the mapping + sending the pair ( p, q) to r. For n=3, the mapping + is specified by (( p 0 , p 1 , p 2 , p 3 ), (q abc , q acb , q bac , q bca , q cab , q cba )) [ ( p 0 , p 1 } (q abc +q acb ), p 1 } (q bac +q bca ), p 1 } (q cab +q cba ), p 2 } (q abc +q bac ), p 2 } (q acb +q cab ), p 2 } (q bca +q cba ), p 3 ).

174

DOIGNON AND REGENWETTER

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

the image of T by + is a subset Dn of R P(C) formed by all probability distributions on P(C) that satisfy the size-independent model. The set Dn is not convex. Moreover, the affine subspace generated by Dn has dimension 2 n &1, as it is the hyperplane with equation

{

D3 = ( p 0 , p 1 } (q abc +q acb ), p 1 } (q bac +q bca ),

:

p 1 } (q cab +q cba ), p 2 } (q abc +q bac ), p 2 } (q acb +q cab ), p 2 } (q bca +q cba ), p 3 )

}

n

=

p0, : p i =1, q0, : q ? =1 , ?#6

i=0

where p0 means p i 0 for all i # S. From the definition, it should be clear that D3 is not convex. Indeed, take the point r 1 which is the image by + of the pair of distributions p on S and q on 6 characterized respectively by p 1 =1 and q abc =1. Then take also the point r 2 similarly obtained for p 2 =1 and q cba =1. We thus have r 1 =(0, 1, 0, 0, 0, 0, 0, 0) and

y X =1

(4)

X # P(C)

since any element in Dn clearly satisfies this equation: simply set y X =P(V=X ). Also, Dn contains the unit point of any axis. We now indicate how to get the ``relative interior'' of Dn . The relative interior E h of a set E of points is the (topological) interior of E taken in the affine subspace generated by that set. Theorem 1. The relative interior of Dn is formed by all images a=+( p, q), where p and q are strictly positive probability distributions on S and 6, respectively. In a formula, h Dh n =+(T ). Remark 2. To warn the reader that this statement is not as obvious as it might look, consider the following mapping: f : [(x, y) # R 2 | x0, y0, x+ y=1]  R:

r 2 =(0, 0, 0, 0, 0, 0, 1, 0).

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

These points r 1 and r 2 belong to D3 , while their midpoint r does not. Indeed, if

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

r=(0, 12 , 0, 0, 0, 0, 12 , 0)

[(x, y) # R 2 | x>0, y>0, x+ y=1]

were in D3 , then r=+( p, q) for some p, q. This would imply that p 1 >0 and also that q bca >0 or q cba >0, from which it would follow that the third or fourth component of +( p, q) is strictly positive, in contradiction with the actual value of the corresponding component of r. The nonconvexity of Dn holds for any size n of C with n3. Turning to the general case, we consider the mapping +: R S_R 6  R P(C): ( p, q) [ r, where r is the real-valued mapping on P(C) given by r(X )=r X =p |X| } : q ? , ? # 6X

for X # P(C). Setting

{

} q0, : q =1 , =

T= ( p, q) # R S_R 6 p0, : p i =1, i#S

?

?#6

File: 480J 115504 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 5609 Signs: 3105 . Length: 56 pic 0 pts, 236 mm

(more precisely, f ( 12 , 12 ) is not an interior point of the image of f ). The following notation will be useful in the rest of the paper: P(C, k)=[Y # P(C) | |Y| =k]. Proof (Theorem 1).

Let

U=[( p, q) # R S_R 6 | p0, q0], and N=+(U). Note that N is a positive cone, i.e., it contains *v whenever * # R + and v # N. Also, U is a positive cone. As moreover Dn =H & N, with H being the hyperplane given by Eq. (4), it suffices to prove N%=+(U%) (where this time we refer to topological interiors). Since + is obviously continuous, the inclusion N%+(U%) is clear. To prove the reverse inclusion, we remark first that + is continuously differentiable on the interior of U, which is U%=[( p, q) # R S_R 6 | p>0, q>0].

175

AN APPROVAL-VOTING POLYTOPE FOR LINEAR ORDERS

We will later show that the differential d+( p, q) of + at any point ( p, q) in U% has rank 2 n, which is the dimension of R P(C). Restricting + to an affine subspace L through ( p, q) complementary to the kernel of d+( p, q), we learn from the Inverse 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 R P(C). This completes the proof of the inclusion +(U%)N%, except for the computation of the rank of d+( p, q). The matrix of the linear mapping d+( p, q) in the canonical bases of R S_R 6 and of R P(C) (i.e., the Jacobian matrix) has rank at most 2 n, the number of its rows. To prove that the rank is exactly 2 n, we partition the columns of this matrix into the set C of columns indexed by sizes s # S, and the set D of columns indexed by rankings ? # 6. Let P and Q be respectively the subspaces of R P(C) generated by these two sets C and D of columns. To get the dimension of Q, we study the 2 n_n ! matrix N with rows indexed by subsets X of C and columns indexed by rankings ? of C, with N X, ? =

r X p |X| = q ? 0

{

if ? # 6 X , otherwise.

Since p>0, the matrix N has the same rank as the matrix M obtained by replacing (row by row) p |X| with 1. The value 2 n &n for the rank of M will be established in Section 3, where this matrix is intensively studied. From the linear equations that we obtain in Theorem 5 for the subspace generated by the columns of M, we see that the subspace Q (generated by the columns in D) is the set of vectors y # R P(C) which are solutions of the following system, having one linear equation for each size k{0: yX y< = . p p0 X # P(C, k) k :

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

C X, s =

r X = p s

{

: q?

if

|X| =s,

This vector v also belongs to the subspace Q iff it satisfies the equations for Q, thus iff for k=1, 2, ..., n vX v< = , p p0 X # P(C, k) k :

which implies *n * 0 *1 = = }}} = p0 p1 pn since  X # P(C, k)  ? # 6X q ? = ? # 6 q ? , a positive constant. Hence dim(P & Q)1. Finally, from the well-known equality dim(P+Q)+dim(P & Q)=dim P+dim Q, we get dim(P+Q)=2 n, in other words the rank of d+( p, q) equals 2 n at each point ( p, q) in U%. This completes the proof. K Instead of looking at the probabilities P(V=X ), for XC, 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 chooses subset X, under the condition that she selects s alternatives. Of course, P(V=X | S=s)=0 for s{ |X|, so we will always assume s= |X| in such a writing. Remember from the Introduction that P(V=X | S=s)=P(R # 6 X ). 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 p s =0. We are thus led to consider a mapping #: R P(C) "K  R P(C): r [

\

rX Y # P(C, |X| )

rY

+,

where

? # 6X

0

otherwise.

We now prove that the subspace P & Q has dimension at most 1. Any vector v in P is a linear combination of the columns of C, so there exists a family (* s ) s # S of real numbers such that for all X # P(C) v X =* |X| } : q ? . ? # 6X

File: 480J 115505 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 5689 Signs: 3568 . Length: 56 pic 0 pts, 236 mm

{

}

K= r # R P(C) _k # S:

: Y # P(C, k)

=

r Y =0 .

Although this mapping # is not defined everywhere on the collection Dn , it follows from Theorem 1 that it is defined (at least) on the relative interior of Dn . Moreover, a vector y from R P(C) belongs to #(D h n ) iff there exists a positive probability distribution q on 6 such that y X = ? # 6X q ? , for all X # P(C). We then say that q induces y (according to the

176

DOIGNON AND REGENWETTER

size-independent model). Thus #(D h n ) is the relative interior of the convex polytope

{

}

An = y # R P(C) _q # R 6: q0, : q ? =1, ?#6

=

and \X # P(C): y X = : q ? , ? # 6X

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 of a set V of n! points in R P(C). There is one such point v ? for each ranking ? of C. The X coordinate of v ? (with XC) 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 the sole point of An which satisfies the equality in the following inequality satisfied in turn by the whole polytope. To write this inequality, we use the notation B(?) for the collection of all beginning sets of ranking ?: :

P(C) arises from the size-independent model. The restriction to positive distributions is not harmful as regards practical usage of a possible characterization of the size-independent model. We now turn to the study of the polytope An , and first determine its dimension.

y X n+1.

X # B(?)

Since for each ranking % # 6, the point v % satisfies this inequality (remember y X # [0, 1] at such a point), so does the whole polytope An . Moreover, equality is obtained at point v % only for %=? (when all terms in the first sum equal 1). 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 of R P(C): F: : R P(C)  R P(C): r [ r$, with r$X =r :&1(X ) (thus F : just permutes the coordinate axes of R P(C) ). The image of the point v ? by F : is the point v :(?) , where :(?)=(:(a 1 ), :(a 2 ), ..., :(a n )) if ?=(a 1 , a 2 , ..., a n ); that is, F :(v ? )=v : b ? , if we view each ranking ? as a permutation of C (having picked an arbitrary reference enumeration of the alternatives in C). As each F : stabilizes the set [v ? | ? # 6] whose convex hull forms An , and moreover any given v ? can be mapped to any given v % by some F : , all the points of this set clearly are vertices of An . (In Section 6 we return to the group of affinities preserving An , 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 linear description of the AV-polytope An would deliver at once a way to check whether a positive probability distribution on

File: 480J 115506 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 5831 Signs: 3810 . Length: 56 pic 0 pts, 236 mm

3. THE DIMENSION OF THE APPROVAL-VOTING POLYTOPE

The following n+1 linear equations on R P(C) are clearly satisfied by the approval-voting polytope An , where n is as before the cardinality of the set C of alternatives: for s # S, y X =1.

: X # P(C, s)

As these n+1 equations are linearly independent, this yields dim An 2 n &n&1. We now proceed to show the reverse inequality, by considering again the matrix M met in the proof of Theorem 1, namely the binary matrix with rows indexed by subsets X of C and columns indexed by rankings ? of C, and M X, ? =

{

1 0

if ? # 6 X , otherwise

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

{

}

=

U= q # R 6 q0, : q ? =1 , ?#6

we see that An is the image of U by the linear mapping f which has matrix M in the canonical bases of R 6 and R P(C), respectively: f : R 6  R P(C): q [ r, where q induces r, that is r(X )= ? # 6X q ? , for X # P(C). To derive the dimension of An , we determine first the rank and image of f . Theorem 3. The rank of the linear mapping f , that is, the rank of the matrix M, equals 2 n &n. The image of f is formed by all vectors of R P(C) that are solutions of the following system of n linearly independent equations, with one equation for each k=1, 2, ..., n: :

yX= y< .

X # P(C, k)

Proof. Note first that each column of M satisfies the n equations in the statement, and as a consequence rank M2 n &n. To show the reverse inequality, and also the

177

AN APPROVAL-VOTING POLYTOPE FOR LINEAR ORDERS

second assertion in the statement, it remains to exhibit 2 n &n columns of M that are linearly independent. Let d=2 n &n. A collection C 1 , C 2 , ..., C d of columns of M is linearly independent as soon as C 1 is nonzero, and for each i=2, 3, ..., d, there exists a row index X such that C i has a 1 in row X, while all of C 1 , C 2 , ..., C i&1 have a 0 in the same row. The existence of a sequence of d columns satisfying the latter condition follows at once from the next lemma, since a column C of M has a 1 in the row indexed by X iff X is a beginning set of the ranking indexing the column. K Lemma 4. Let n be a positive natural number, and let d=2 n &n. There exists a sequence ? 1 , ? 2 , ..., ? d of rankings of 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, the result is trivial. For n=2, the following two rankings satisfy Condition (*) (with the subset formed by the underlined elements in the ranking): 1 2,  2 1.  To get the five desired rankings for n=3, we first add 3 at the end of the two rankings displayed above, and then add two rankings with 3 in the second position, and one ranking with 3 in the first position: 1  2  1  2  3 

2 3, 1 3, 3 2,  3 1,  1 2.

A similar construction can be used to build the required sequence of 2 n &n rankings over [1, 2, ..., n] from the list of 2 n&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, thus obtaining an initial sequence of 2 n&1 &(n&1) rankings, which all satisfy Condition (*) (with the same beginning sets as before). Then, we successively add subsequences, one for each value of k=n&1, n&2, ..., 1. In all rankings in the subsequence corresponding to the value k, we place n in position k. Moreover, one such ranking is selected for each subset X of [1, 2, ..., n&1] having cardinality k&1, with the only requirement that this ranking has X as a beginning set (the freedom left in positioning the elements will be exploited in Sections 4 and 6). Taking the beginning set X _ [n], it is easily seen that the newly added ranking

File: 480J 115507 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 5592 Signs: 3996 . Length: 56 pic 0 pts, 236 mm

satisfies Condition (*). After completion of the construction of all the successive subsequences, there results a sequence of f (n) rankings satisfying Condition (*), with moreover n&1

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

n&1

\k&1+

=2 n&1 &(n&1)+2 n&1 &1 =2 n &n. This sequence establishes the thesis for the value n.

K

Theorem 5. The dimension of the AV-polytope An equals 2 n &n&1. Proof. As indicated before the statement of Theorem 3, we use the equality An = f(U). As U is the simplex spanned by the unit points of the coordinate axes in R 6, we see that An is the convex hull of the images of these points. From the theorem, these images linearly span a subspace of R P(C) with dimension 2 n &n. Hence these images affinely span a subspace of dimension at least 2 n &n&1. This shows dim An 2 n &n&1. The reverse inequality was obtained at the beginning of this section. K 4. THE LINEAR INEQUALITIES FROM FALMAGNE AND REGENWETTER (1996)

From Section 2, we know that a description of the AVpolytope An will involve exactly n+1 linear equations (for instance, those listed in Theorem 3). Moreover, we already indicated in the Introduction that the description will have one linear inequality for each facet of An . We now inspect three families of linear inequalities satisfied by An that were given in Falmagne and Regenwetter (1996), and we determine which ones are facet-defining. While recalling these inequalities, we introduce names and symbols for them. Some notation has also been introduced in previous sections, e.g. P(X, k) denotes the collection of all subsets of X with cardinality k. Nonnegativeness Condition NNEG(X ). For any given subset X of C, y X 0.

(5)

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

(6)

178

DOIGNON AND REGENWETTER

Marginal Condition MARG(i, k). For any given alternative i in C and any given size k with 1
yY

:

yZ .

(7)

Z # P(C, k&1), i # Z

The latter inequality is also satisfied by An in case k=1 or k=n, but then coincides with a nonnegativeness condition NNEG(X) having respectively |X| =1 or |X| =n&1 (as a convention, a sum of zero terms equals 0). Note also that in the admitted case k=n&1, the marginal condition MARG(i, k) is equivalent to a general subset condition GSUB(X, l ) having l=n&2 and |X| =n&1. Falmagne and Regenwetter (1996) derive the above inequalities in the probabilistic setting. For instance, refering to notation from the Introduction, the marginal condition MARG(i, k) amounts to the nonnegativeness of the marginal probability that R ranks alternative i at position k. This interpretation motivates the name chosen for the condition. Another way of establishing all these inequalities consists in checking that they are satisfied by any vertex of An . To achieve this goal, remember that with each ranking ? of C is associated a vertex v ? having zero-valued coordinates everywhere except for the coordinates corresponding to the beginning 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-hand side equals 1, so does the left-hand side. For instance, take Formula (6) and assume y X =1 at some vertex v ? . Then the ranking ? begins with X. Therefore, it begins also with some subset Y of X of size l. Hence the left-hand side of the inequality equals 1. The proof is similar for the marginal condition MARG(i, k), and even simpler for the nonnegativeness condition NNEG(X ). Clearly, a linear inequality satisfied by An is facet-defining iff the corresponding equality is satisfied by 2 n &n&1 affinely independent vertices. This fact is exploited in the next three proofs. We will freely use the notation NNEG(X ) in order to specify not only the condition, but also the linear inequality and even the corresponding linear equation. A similar remark holds for GSUB(X, l ) and for MARG(i, k). Sequences of rankings similar to the one met in the proof of Lemma 4 will be most useful here. In this section, we set C=[1, 2, ..., n]. Let us repeat with additional comments the construction of 2 n &n rankings of C, proceeding by recurrence on n. The Generic Sequence Gn of Rankings of [1, 2, ..., n]. For n=2, the following two rankings constitute the sequence G2 : 1 2,  2 1. 

File: 480J 115508 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 6212 Signs: 4372 . Length: 56 pic 0 pts, 236 mm

To get the 2 n &n rankings that form the sequence Gn , we first copy the 2 n&1 &(n&1) rankings of Gn&1 , adding n at the end of each, thus obtaining what we call the recurrence part of Gn . Then, for each value of k=n&1, n&2, ..., 1, we form ( n&1 k&1 ) rankings with n in position k. Exactly one such ranking is selected for each subset X of [1, 2, ..., n&1] having cardinality k&1, with the only requirement that this ranking has X as a beginning set. Note that the selection is made among (k&1)! (n&k)! possible rankings whenever X is given. This is why we often speak of the ``generic'' sequence Gn . The freedom in the selection is illustrated below for n=4: alternatives within a pair of braces can be permuted freely, resulting in another choice of the ranking selected. The horizontal line separates the recurrence part from 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 2 n &n vertices v ? of An , for ? appearing in Gn , are affinely independent (see the proof of Theorem 3, which refers to beginning sets as those marked here by underlines). Theorem 6. The nonnegativeness inequality NNEG(X ), y X 0, for X # P(C), is facet-defining iff X differs from < and from C. Proof. If X=< or X=C, the inequality NNEG(X) is not facet-defining, since both coordinates y < and y C take the value 1 at each vertex. Conversely, it suffices to show that NNEG(X ) is facet-defining for X=[k, k+1, ..., n] and 1
179

AN APPROVAL-VOTING POLYTOPE FOR LINEAR ORDERS

rankings of Gn , all alternatives beyond n are taken in increasing order, only one of these rankings starts with X. The rankings ? in Gn distinct from this exceptional one give us 2 n &n&1 affinely independent vertices v ? satisfying equality NNEG(X ). K Theorem 7. GSUB(X, l ),

Among the general subset inequalities

:

yY yX ,

Y # P(X, l )

exactly those with 0
File: 480J 115509 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 6006 Signs: 4915 . Length: 56 pic 0 pts, 236 mm

n, then n has either rank s or a rank greater than s. In the first eventuality, the ranking begins with X and thus gives 1=1 in inequality GSUB(X, s&1). In the second eventuality, the alternative with rank s does not belong to X, and comes before alternative n. Because of the freedom in the construction of Gn , we may move this alternative to the first position, so that the ranking now gives 0=0. Now let us assume that ? starts with a subset of X of size s&1 that contains n. Since s
: Y # P(C, k), i # Y

yY

:

yZ ,

Z # P(C, k&1), i # Z

for i # C and k # S with 1
180

DOIGNON AND REGENWETTER

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 and AB+AC+BC=1 are satisfied by the AV-polytope A3 . Moreover, as y < = y C =1 holds, coordinates y < and y C 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+ACA,

(8)

AB+BCB,

(9)

AC+BCC.

(10)

where (11)(16) are the nonnegativeness conditions, and (17)(19) are the three marginal conditions, stated only in terms of the variables A, B, AB, AC. Moreover, for a given such solution (A, B, AB, AC), the proof builds the following explicit form for all probability distributions q on 6 that induce (A, B, AB, AC) according to the size-independent model: q abc q acb q bac q bca q cab q cba

\ +\ =

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

1+A&AC AC&1 AB+AC&A&1 1+A+B&AB&AC 1 0

+:

Theorem 9. If n=3, the six nonnegativeness inequalities and the three marginal inequalities, together with the four linear equations

&1 +1 +1 &1 &1 0

0 0 0 0 0 1

\ + \+ +;

+

,

(20)

where y < = y C =A+B+C=AB+AC+BC=1, : # [max(A+B, 1&AC, 1+A&AB&AC),

provide a linear description of A3 in R P([a, b, c]). The combinatorics of the polytope will be described at the end of this section (as well as another way of proving the theorems in this section). As regards analytical geometry, Theorem 9 states that the polytope A3 (now seen in the subspace with equations y < = y C =A+B+C=AB+AC+ BC=1) is the set of solutions (A, B, AB, AC) of the following system of nine linear inequalities: A0,

(11)

B0,

(12)

1A+B,

(13)

AB0,

(14)

AC0,

(15)

1AB+AC, AB+ACA, 1B+AC, A+BAB,

(16) (17) (18) (19)

File: 480J 115510 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 5073 Signs: 2489 . Length: 56 pic 0 pts, 236 mm

min(1+A&AC, 1+A+B&AB&AC, 1)] (21) and ;=:&A&B. Proof. The necessity of Conditions (11)(19) for a point to be in A3 was established in Section 4. As for the sufficiency, we consider the following inhomogeneous system of linear equations in the unknown probability distribution q, for a given quadruple (A, B, AB, AC):

\

1 0 1 0

1 0 0 1

0 1 1 0

0 1 0 0

0 0 0 1

0 0 0 0

q abc q acb q bac q bca q cab q cba

+\ + \ +

A B = . AB AC

(22)

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

181

AN APPROVAL-VOTING POLYTOPE FOR LINEAR ORDERS

TABLE 1

A=AB+AC



A=1&BC



BC=B+C, (23)

The FacetVertex Incidence Table Produced by Porta for A3

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:

(iii)

A=0,

B=0,

C=0,

BC=0,

AC=0,

AB=0;

the point r lies on the boundary of A3 ;

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

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

(21) amounts to the system (11)(19). This completes the proof of sufficiency. K Equations (20) and (21) together with ;=:&A&B also indicate when the probability distribution q is unique. Leaving details to the reader, we conclude: Theorem 10. Suppose n=3. Fix a probability distribution r on P(C) for which there exists a size-independent random utility representation q. Then

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

Remark 11. An analog of Sen's value restriction (1996) holds when the conditions of Theorem 10 are satisfied. In this case, an aggregate linear preference ordering, summarizing the overall preferences in the population, can be computed in a canonical way. For proofs and additional comments, see Regenwetter and Grofman (1995). Let us move to the combinatorial structure of the polytope A3 . By Theorem 9, A3 has nine facets, and of course six vertices. The PORTA-produced Table 1 displays the facetvertex incidence, with rows corresponding to facets and columns corresponding to vertices in the order corresponding to Eq. (2). Note that the data are formulated in terms of eight coordinates, respecting the ordering specified 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, we ran a Pascal implementation of an improvement of an algorithm due to Dowling (1993). It turns out that all pairs of vertices form an edge, i.e. A3 is a neighbourly polytope. The eighteen triangular, two-dimensional faces are all trios of vertices except two (namely, the trio consisting of vertices associated to even, resp. odd, permutations; here, assuming a fixed enumeration of the alternatives, we see a ranking as a permutation). The nine facets are three-dimensional simplices, thus A3 is simplicial. These nine facets are the complements 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 combinatorial automorphisms of A3 , that is, 72 permutations of the

182

DOIGNON AND REGENWETTER

vertices that map any face onto a face (see Grunbaum, 1967, for this notion of combinatorial automorphism). Taken with 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 remarkable that G acts transitively on the vertices, on the twodimensional faces, and on the facets, but not on the edges. All the assertions about a linear description and the combinatorial structure of A3 can also be routinely deduced from the following easy observations. We are back in the Euclidean space R P(C)"[<, C], and view again all rankings as permutations 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 vertices v ? , for ? odd, form the vertices of a triangle T &. These two triangles respectively lie in two (nonorthogonal) planes having only one point in common, which is the centroid of both triangles. As A3 is the convex hull of T + _ T &, all our assertions about A3 can thus also be obtained from these geometric considerations. Moreover, any combinatorial automorphism is induced by some affine automorphism of the affine subspace aff A3 generated by A3 . We should warn the reader that most of the nice properties of A3 fail to hold in general for the approval-voting polytope An .

Given a ranking ?=(a 1 , a 2 , ..., a n ) of C=[1, 2, ..., n], the reversed ranking of ? is r(?)=(a n , a n&1 , ..., a 1 ). We get the involutive map r: 6  6 (thus, r 2 is the identity on 6). The complement map P(C)  P(C): X [ C"X is also involutive. We deduce an involutive affine automorphism, which we name the reversing automorphism: F: R P(C)  R P(C): y [ z,

(26)

where by definition z X = y C"X , for X # P(C). The name given to F is motivated by the following property, which also shows that F preserves An : The vertex v ? of An is mapped by F to the vertex v r(?) . Facets defined by a nonnegativeness condition NNEG(X ) are easily seen to be transformed by F into facets of the same type, namely NNEG(C"X ). On the other hand, the facet defined by the subset condition SUBS(X) is transformed by F into a (generally) new facet-type. Using the covering relations & c and & C on P(C), defined by XC c X, & Y  (X/Y and |X| = |Y| &1)  Y &

(27)

the subset condition SUBS(X ) can be rewritten as :

yY  yX ,

(28)

Y # P(C), Y & CX

6. NEW FACET-DEFINING INEQUALITIES

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

where X is a subset of C with 1< |X|
v the nonnegativeness condition NNEG(X ) (with X # P(C)"[<, C]);

Superset Condition SUPS(X). with 1< |X|
v the subset condition SUBS(X ) (with X # P(C) and 1< |X|
File: 480J 115512 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 5934 Signs: 4277 . Length: 56 pic 0 pts, 236 mm

:

For a subset X of C

yY  yX .

(29)

Y # P(C), Y & cX

Note that for |X| =1, the latter inequality is equivalent to a marginal condition MARG(i, k) having k=2. Transforming in a similar manner a marginal inequality MARG(i, k) produces an inequality which is equivalent to some marginal inequality again, namely to MARG(i, n&k+1). 7. THE CASE OF FOUR ALTERNATIVES

Tables 2 and 3 show the output produced by the PORTA software for n=4, when the input is the lexicographic list of 4!=24 vertices in R 14. As in the case n=3, we leave out coordinates y < and y C . The other coordinates were relabelled from x1 to x14, and correspond to the nontrivial subsets of C=[a, b, c, d ] in the order

183

AN APPROVAL-VOTING POLYTOPE FOR LINEAR ORDERS

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

TABLE 3

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

The FacetVertex Incidence Table for A4 , as Produced by Porta

(30) A study of Table 2 teaches us that the 34 facet-defining inequalities all come from the nonnegativeness conditions (14), subset conditions (6), superset conditions (6), and marginal conditions (8). We skip a detailed description of the combinatorial structure of the polytope A4 , since only a few nice properties listed for A3 are shared by A4 . However, we display in Table 3 the facet-vertex incidence-matrix produced by PORTA. TABLE 2 Facet-Defining Inequalities for A4

8. MORE INEQUALITIES

We now introduce two fairly general families of inequalities satisfied by the approval-voting polytope An . Sandwich Condition SAND(U, V, k). For any two subsets U and V of C, and any size k with UV and |U| k |V|, : Note. Coordinates, with y < and y C left out, are listed as in Eq. (30).

File: 480J 115513 . By:XX . Date:01:07:01 . Time:07:51 LOP8M. V8.0. Page 01:01 Codes: 2412 Signs: 1013 . Length: 56 pic 0 pts, 236 mm

X # P(C, k), UXV

yX

: Y # P(C, k+1), UY, |V & Y| k

yY .

(31)

184

DOIGNON AND REGENWETTER

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

Similarly, the linear inequality RSAN(<, T, l ) in Formula (34) is facet-defining for 1
Reversed Sandwich Condition RSAN(W, T, l). For any two subsets W and T of C, and any size l with WT and |W| l |T|,

Proof. As the reversing automorphism transforms Inequality (33) into Inequality (34), we treat only the second assertion. It is easily checked that all vertices satisfy the inequality in Formula (34). Moreover, equality obtains at a vertex v ? in exactly two cases. Either both members equal 0, which occurs when at least one of the first l&1 alternatives of ? is not in T, or both members equal 1, when the first l alternatives of ? belong to T. Setting T=[n&t+1, n&t+2, ..., n], we turn again to the generic sequence (cf. Section 4). Of its rankings, those that do not give equality have both their first l&1 alternatives in T and their alternative ranked l not in T. We now use the freedom left in the construction of the generic sequence in order to modify rankings that do not give equalityexcept for one. If n is not in the last position of such a ranking, we may exchange the alternative in position l with an alternative in T (since 1
yX 

: X # P(C, l ), WXT

yY .

:

(32)

Y # P(C, l&1), YT, |W _ Y| l

It is not difficult to check that each vertex of An satisfies all of the (reversed) sandwich conditions, and thus that An does so also. At present, we are not able to decide which of these inequalities are facet-defining. As a matter of fact, all of the inequalities that were proved to be facet-defining, except for almost all of the nonnegativeness ones, are special cases of the (reversed) sandwich conditions. Indeed, the sandwich condition SAND(U, V, k), for v 1< |U| =k
reversed

sandwich

condition

v 1< |W| =l
: X # P(C, k), UX

yY ,

:

(33)

Y # P(C, k+1), UY

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

yX

:

yY .

(34)

Y # P(T, l&1)

At least for special cases, we achieve here new facetdefining inequalities. Theorem 12. The linear inequality SAND(U, C, k) in Formula (33) is facet-defining for 1< |U|
File: 480J 115514 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 5879 Signs: 4405 . Length: 56 pic 0 pts, 236 mm

9. THE CASE OF FIVE ALTERNATIVES

The Porta software output we get for n=5 indicates that A5 has 235 facets. From carefully scrutinizing the inequalities (and rewriting many of these), we came to the conclusion that 105 of these inequalities are among those we proved to be facet-defining, that 70 additional ones belong to the (reversed) sandwich inequalities, and finally that a new class of 60 inequalities is met. To describe the latter class, 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 result from applying to the above inequality all automorphisms F : induced by permutations : of the set C. Moreover, the

AN APPROVAL-VOTING POLYTOPE FOR LINEAR ORDERS

185

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 reversing automorphism. A general family of inequalities including this collection will be given in Section 10. Table 4 summarizes the Porta output for n=5 and provides additional information. We have grouped on one line all inequalities equivalent under the automorphisms F : of A5 . 10. PROBABILISTIC INTERPRETATION OF THE INEQUALITIES

Each of the (possibly facet-defining) linear inequalities found so far for the AV-polytope will be shown to mean that a certain event must have a nonnegative probability. We recall that the random variable R, which takes its values in the whole set 6 of rankings of C, models the latent ranking of the voter. Moreover, from the fundamental Eq. (1) of the size-independent model, we derived y X =P(V=X | S= |X| )=P(R # 6 X ).

File: 480J 115515 . By:XX . Date:01:07:01 . Time:07:52 LOP8M. V8.0. Page 01:01 Codes: 3130 Signs: 2078 . Length: 56 pic 0 pts, 236 mm

In the list below, the difference of the two terms in the mentioned inequality is the probability of the event that the value of R belongs to the set of rankings described. When we speak here of a beginning set, or of first alternatives, we refer to the ranking which is the outcome of R. v Nonnegativeness inequality NNEG(X): the set X is a beginning set. v Subset inequality SUBS(X ): the set X is not a beginning 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 position |X| is not. v Superset inequality SUPS(X ): all the elements of X are among the |X| +1 preferred alternatives, although they are not the |X| preferred ones. Equivalently, the first |X| +1 alternatives are in X, except for one alternative which is not the one in position |X| +1. v Marginal inequality MARG(i, k): alternative i is at position k; v Sandwich condition SAND(U, V, k): the first k+1 alternatives contain all those in U and at least k elements of

186

DOIGNON AND REGENWETTER

V, and either one of the first k is not a member of V or the alternative at rank k+1 is in U. In other words, all the elements from U, plus k& |U| or k& |U| +1 further elements from V, are among the first k+1 preferred alternatives, and either position k+1 contains an element of U or one of the first k elements is not in V. v Reversed sandwich condition RSAN(W, T, l): the first l&1 alternatives are in T, and they include all the elements of W except maybe one, although the first l alternatives do not contain W or are not all in T. In other words, all elements in W except maybe one are among the l&1 preferred alternatives which in turn are all in T, and at least one element of W is ranked higher than l or the alternative in position l is not in T. As stated in the Introduction, we do not know whether each facet of An can be described by an equality couched in the form P(E)0, for some E in P(6) (although its equation is for sure linear in the variables y X =P(R # 6 X ), for X # P(C), and can always be written with integer coefficients since the facet is affinely generated by points having integer coordinates). We now devise a new general class of necessary conditions for a point of R P(C) to be in An , thus obtaining for the first 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 l and k be two natural numbers with lk. Then the probability that some member of F appears within the first k positions, 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: FX], we define E=

\

6X

. X # PF(C, k)

+>\

+

6Y .

. Y # PF(C, l )

Then the necessary condition P(E)0 becomes the following linear inequality satisfied by the whole polytope An : yX 

: X # PF(C, k)

:

yY .

(35)

Y # PF(C, l )

The latter inequality is dubbed the generalized sandwich condition GSAN(F, l, k). It is easy to see that several of our conditions are special cases of this inequality, as shown below. Let us first assume that F contains exactly one set X. If moreover |X| =k>l, Inequality (35) becomes y X 0, which is the nonnegative condition NNEG(X ) when 0
yY  yX .

Y # P(X, k), XY

File: 480J 115516 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 6202 Signs: 4606 . Length: 56 pic 0 pts, 236 mm

As the reversing automorphism transforms this inequality into the Inequality (6) (as in the general subset condition, see Section 4), Theorems 7 and 8 imply that this inequality is facet-defining iff either 1
: Y # P(C, k), i # Y

yX .

: X # P(C, l ), i # X

This inequality is facet-defining iff either 1
y C"X 

:

y C "Y .

(36)

Y # PF(C, l )

Again, several of our inequalities are easily shown to be special cases of this inequality; we skip the details. It is important to note that all the facet-defining inequalities presented in this paper are particular cases of Conditions (35) and (36). However, for many of the other cases of these two inequalities, we still do not know whether they are facet-defining. The same question arises for other inequalities of the form P(E)0, for E a subset of 6. Another unanswered question is the following. Is there some family H of subsets of 6 such that for H # H the probability P(H) can be written in terms of values P(6 X ), where X # P(C), and such that moreover all inequalities P(H)0, for H # H, characterize the size-independent model? Remember that our main, original problem is to

AN APPROVAL-VOTING POLYTOPE FOR LINEAR ORDERS

characterize the mappings r on P(C) for which there exist probability distributions p and q on respectively S and 6, such that r X =p |X| } : q ? , ? # 6X

for all X in P(C). All these mappings r are probability distributions on P(C), and they altogether form the set Dn defined in the Introduction. Remember that the knowledge of r entails that of p, since for s # S ps=

rX .

: X # P(C, s)

The main problem was then recast as the following one: To characterize the mappings y on P(C) for which there exists a probability distribution q on 6, such that, for XC: y(X )= y X = : q ? . ? # 6X

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

187

11. SUMMARY AND CONCLUSIONS

The present paper investigates the approval-voting polytope An associated with the size-independent model of approval voting of Falmagne and Regenwetter (1996), where n is the number of candidates. We have shown that the dimension of An is 2 n &n&1, which implies that there are exactly 2 n &n&1 degrees of freedom in the marginal ranking probabilities P(R # 6 X ). Besides a classification of Falmagne and Regenwetter's (1996) necessary conditions for the model into those conditions that are facet-defining and those that are not, we obtained a complete (minimal linear) description of the model for n5 (relying heavily on the Porta software for n=4 or n=5). However, so far, we were able to prove only the conditions derived for n4 to be general facet-defining inequalities also for arbitrary n. The crucial method used for proving an inequality to be facet-defining lies in the recursive construction of the generic sequence Gn of rankings, and in demonstrating that it corresponds to 2 n &n&1 affinely independent vertices satisfying the corresponding equality. At the end of the paper, we discussed probabilistic interpretations of all inequalities encountered before, and derived a general set of inequalities from probabilistic considerations which encompasses all earlier cases. In this context some open problems were raised that future research needs to address.

h: [6 X | X # P(C)]  [0, 1]: 6 X [ y X , which is the restriction to the family E=[6 X | X # P(C)] of the probability measure associated to q. Conversely, y can be deduced from h since y X =h(6 X ). There follows still another reformulation of our main problem: To characterize the mappings h: [6 X | X # P(C)]  [0, 1] for which there exists a probability distribution q on 6 with h(6 X )= : q ? , ? # 6X

for all 6 X in E. Such a formulation shows that our specific problem is a particular case of the following general one: Given a family E of subsets of a set 0 together with a realvalued mapping h on E, under which conditions on E and h is there a probability measure P on the _-algebra of subsets of 0 generated by E, such that h(E)=P(E) for all E # E? In this connection, note that in our actual setting the _-algebra of events (subsets of 6) generated by the family (6 X ) X # P(C) equals P(6). Moreover, the geometric approach has shown in this paper that there is a minimal list of conditions; this fact is true for the general problem as well, at least when 0 is finite. However, we should mention that the general problem also involves the (presently intractable) problem of finding a description of the binary-choice polytope (cf. the Introduction).

File: 480J 115517 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 8048 Signs: 4879 . Length: 56 pic 0 pts, 236 mm

REFERENCES Brams, S. J., 6 Fishburn, P. C. (1983). Approval Voting. Boston: Birkhauser. Christof, T. (1996). PORTAA polyhedron representation transformation algorithm; available from ftp:ftp.zib-berlin.depubmathprog polythportaindex.html. Dieudonne, J. (1960). Foundations of Modern Analysis. New York: Academic Press. Doignon, J.-P., 6 Falmagne, J.-C. (1994). A polynomial time algorithm for unidimensional unfolding representations. Journal of Algorithms 16, 218233. Dowling, C. (1993). On the irredundant construction of knowledge spaces. Journal of Mathematical Psychology 37, 4962. Falmagne, J.-C., 6 Regenwetter, M. (1996). A random utility model for approval voting. Journal of Mathematical Psychology 40, 152159. Fishburn, P. C. (1992). Induced binary probabilities and the linear ordering polytope: A status report. Mathematical Social Sciences 23, 6780. Grunbaum, B. (1967). Convex Polytopes. New York: Wiley. Harary, F. (1969). Graph Theory. Reading, MA: AddisonWesley. Koppen, M. (1995). Random utility representations of binary choice probabilities: Critical graphs yielding critical necessary conditions. Journal of Mathematical Psychology 39, 2139. Marley, A. A. J. (1990). A historical and contemporary perspective on random scale representations of choice probabilities and reaction times in the context of Cohen and Falmagne's (1990, Journal of Mathematical Psychology 34) results. Journal of Mathematical Psychology 34, 8187. Merrill, S. I. (1988). Making Multicandidate Elections More Democratic. New Jersey: Princeton Univ. Press.

188

DOIGNON AND REGENWETTER

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). Mathematical Behavioral Sciences, University of California, Irvine. Regenwetter, M. (1996). Random utility representations of finite m-ary relations. Journal of Mathematical Psychology 40, 219234. Regenwetter, M., 6 Grofman, B. (1995). Choosing subsets: A size-independent probabilistic model and the quest for a social welfare ordering. Social Choice and Welfare (to appear); based on Technical Report MBS 94-20, IMBS, University of California, Irvine. Regenwetter, M., 6 Grofman, B. (in press). Approval voting, Borda winners and Condorcet winners: Evidence from seven elections. Management Science. Regenwetter, M., Marley, A., 6 Joe, H. (1996). Random utility threshold

File: 480J 115518 . By:DS . Date:03:07:01 . Time:11:19 LOP8M. V8.0. Page 01:01 Codes: 4086 Signs: 1606 . Length: 56 pic 0 pts, 236 mm

models of subset choice (submitted for publication); based on Technical Report MBS 94-26, IMBS, University of California, Irvine. Sen, A. K. (1966). A possibility theorem on majority decisions. Econometrica 34, 491499. Suck, R. (1992). Geometric and combinatorial properties of the polytope of binary choice probabilities. Mathematical Social Sciences 23, 81102. 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, 3949. Ziegler, G. M. (1995). Lectures on Polytopes. Berlin: Springer-Verlag. Received June 10, 1996