Decision Support
On the complexity of Slater’s problems

https://doi.org/10.1016/j.ejor.2009.07.034Get rights and content

Abstract

Given a tournament T, Slater’s problem consists in determining a linear order (i.e. a complete directed graph without directed cycles) at minimum distance from T, the distance between T and a linear order O being the number of directed edges with different orientations in T and in O. This paper studies the complexity of this problem and of several variants of it: computing a Slater order, computing a Slater winner, checking that a given vertex is a Slater winner and so on.

Introduction

A tournament T is an asymmetric and complete directed simple graph: between two distinct vertices x and y, there is one and only one of the two arcs (i.e. directed edges) (x,y) and (y,x). Such a structure arises for instance in voting theory, to model the result of a pairwise comparison method, as the one suggested by Condorcet [11]. Indeed, assume that we are given a set X of n candidates and a collection Π=(O1,O2,,Om), called a profile, of the opinions Oi of m voters (1im) who want to rank the n candidates; we assume moreover that the individual preferences Oi(1im) of the m voters are linear orders over X. An order O defined on X will be represented under the form x1>x2>>xn for an appropriate numbering of the n elements xj(1jn) of X, with the agreement that x>y means that x is preferred to y; with this notation, we say that x1 is the first element of O. More generally, if O and O denote two linear orders defined on different sets X and X,O>O will denote the linear order defined on XX as the concatenation of O followed by O.

In order to aggregate these m linear orders into a linear order which can be considered as the collective preference, Condorcet suggested to compute, for each pair of candidates {x,y} (with xy), the number mxy of voters who prefer x to y and the number myx of voters who prefer y to x; x is then collectively preferred to y if we have mxy>myx. We may model this collective preference by a graph T, with the set of candidates as its set of vertices, and with an arc from x to y when x is collectively preferred to y, i.e. when we have mxy>myx. If there is no tie (it is in particular the case when m is odd, since we have mxy+myx=m), as it will be assumed in the sequel, the obtained graph T=(X,A) is a tournament, called the majority tournament of the election, where X is the set of candidates and where A is thus defined by: (x,y)Amxy>myx (see [9] and the references therein for more details).

Sometimes T is a linear order O, i.e. a complete directed graph without directed cycles, providing the desired ranking of the candidates of the election. It is well-known (see [6], [20], [21], [22], [23] for the definitions and the basic results about graphs and tournaments) that a linear order is a transitive tournament and conversely (remember that a tournament T is transitive if the existence in T of the arcs (x,y) and (y,z) implies the one of (x,z)), and that a tournament is transitive if and only if it is without any circuit (i.e. any directed cycle). But, as discovered by Condorcet himself, the majority tournament T=(X,A) of an election may not be transitive, even if the preferences of the voters are all assumed to be linear orders. This is the so-called “voting paradox” or also “Condorcet effect” [14]. The following example illustrates this situation.

Example 1

Assume that m=9 voters must rank n=4 candidates a,b,c, and d. The profile Π0 of the preferences of the voters is assumed to be given by the following linear orders:

  • the preferences of three voters are: a>b>c>d;

  • the preferences of two voters are: b>d>c>a;

  • the preference of one voter is: c>d>a>b;

  • the preference of one voter is: d>a>c>b;

  • the preference of one voter is: d>b>a>c;

  • the preference of one voter is: c>d>b>a.

Thus we have Π0=(a>b>c>d,a>b>c>d,a>b>c>d,b>d>c>a,b>d>c>a,c>d>a>b,d>a>c>b,d>b>a>c,c>d>b>a). The quantities mxy implied in Condorcet’s procedure are the following, where the bold values show the ones greater than or equal to the strict majority (m+1)/2, here equal to 5:

  • mab=5;mba=4;

  • mac=5;mca=4;

  • mad=3;mda=6;

  • mbc=6;mcb=3;

  • mbd=5;mdb=4;

  • mcd=5;mdc=4.

Here, the majority tournament is not a linear order, but the tournament T0 of Fig. 1.

Even if T is not transitive, it may happen that there exists a Condorcet winner: a Condorcet winner is a candidate ζ collectively preferred to any other candidate: xX with xζ,mζx>mxζ. From the graph theoretic point of view, it means that all the arcs involving ζ go from ζ to the other vertices: the out-degree of ζ is equal to n  1 and its in-degree is equal to 0. Notice that, when there exists a Condorcet winner, there is only one.

When there exists a Condorcet winner of the election, we may consider him as the winner of the election (though it is not the case for many voting procedures). When there is no Condorcet winner (as it is the case for the example above), the question arises to know who can be considered as the winner of the election. Several methods, known under the name of tournament solutions, have been designed to answer this question (see for instance [20] for a survey on these solutions and [17] for their complexities). Among them, we find for example the solution proposed by Banks [3]. It consists in considering as winners (called the Banks winners of T) the first elements of the maximal (with respect to inclusion) transitive subtournaments of T (for instance, the Banks winners of the tournament of Fig. 1 are a because of the maximal transitive subtournament a>b>c,b because of the maximal transitive subtournament b>c>d, and d because of the maximal transitive subtournament d>a).

Another solution, that we are going to consider in this paper, is the one studied by P. Slater, called Slater’s solution ([24]; for a survey and references on this problem, see [9], [10]). Slater’s solution consists in transforming Tinto a linear order by reversing a minimum number of arcs of T. For instance, it is easy to see that it is necessary and sufficient to reverse the arc (d,a) in the tournament of Fig. 1 to transform it into the linear order a>b>c>d.

More formally, in the sequel, T=(X,A) will denote a tournament with n vertices. The set of linear orders defined on X will be ω(X), and O will represent a linear order defined on X (in other words, O is an element of ω(X)). We define the symmetric difference distance δ (in fact, half this distance) between T and O=(X,B) by:δ(T,O)=12|AΔB|,where Δ denotes the usual symmetric difference between sets. As T and O are complete and asymmetric, we have also:δ(T,O)=(x,y)A-B=(x,y)B-A(which shows that δ(T,O) is an integer). This distance, which owns good axiomatic properties (see [4]), measures the number of disagreements between T and the ranking defined by O. From the graph theoretic point of view, δ(T,O) can also be interpreted as the number of arcs which do not have the same orientation in T and in O, or, equivalently, as the number of arcs that must be reversed in T to obtain the linear order O.

So, Slater’s problem consists in computing a linear order O which minimizes the distance δ from T over the set ω(X) :δ(T,O)=minOω(X)δ(T,O),or, equivalently, a linear order which minimizes the number of arcs which must be reversed in T to obtain O. This minimum number will be called the Slater index of T and is noted i(T); an order O(T) which minimizes the distance δ to T will be called a Slater order of T; a Slater winner of T is the first element of any Slater order of T; notice that there always exists at least one Slater winner.

The aim of this paper is to study the algorithmic complexity of Slater’s problem and some of its variants. Section 2 defines these variants. The complexity results are proved in Section 3, and summarized in the conclusion, in Section 4.

Section snippets

Slater’s problems

From the previous considerations, we may define several problems.

  • PROBLEM (P1). Given a tournament T, compute the value of the Slater index i(T) of T.

  • PROBLEM (P2). Given a tournament T, compute a Slater order O(T) of T.

  • PROBLEM (P3). Given a tournament T, compute all the Slater orders O(T) of T.

  • PROBLEM (P4). Given a tournament T, compute a Slater winner of T.

  • PROBLEM (P5). Given a tournament T, compute all the Slater winners of T.

  • PROBLEM (P6). Given a tournament T and a vertex v of T, determine

Complexity results for Slater’s problems

The complexity results of this section all come from the NP-hardness of (P8). This problem, when extended to any directed graph (i.e., T is not necessarily a tournament), has been known to be NP-hard for a long time (see [19]). Recently, Alon [2], Charbit et al. [7], and Conitzer [12] proved independently that (P8) is NP-hard. More precisely, they proved that the decision problem, called FAST below, associated with problem (P8), is NP-complete:

  • Name: Feedback Arc Set for Tournaments (FAST)

Summary

As a conclusion, let us summarize the results obtained above.

  • The computation of the Slater index of a tournament (Problem (P1)) is an NP-hard (Corollary 2) and NP-easy (Theorem 7) problem, so it is NP-equivalent. This problem belongs to FLNP (Theorem 7). The decision problem that is associated with it (problem SI) is NP-complete (Theorem 1).

  • The determination of a Slater order of a tournament (Problem (P2)), of a Slater winner of a tournament (Problem (P4)), or of all the Slater winners of a

Acknowledgement

I would like to thank Lucile Belgacem-Denœud for her help in writing this text. I thank also the anonymous referees for their valuable comments.

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