1 Introduction

The subject of human welfare and political institutions has been an ancient field of inquiry. However, we owe more to Kenneth Arrow than anybody else the bringing of formal reasoning and axiomatic rigor to the discipline (Arrow 1950, 1951). There had been other related works before that. There was the famous 1929 paper by Harold Hotelling, Arrow’s PhD adviser, who provided the tools for a purely positive exercise concerning the outcomes of democratic elections. There were the works of Bergson (1938) and Samuelson (1947) that created the concept of a social welfare function, which would lend rigor to early ideas pertaining to utilitarianism. This led to a substantial literature on the normative and positive aspects of using democratic processes to aggregate individual preferences (Sen 1970; Pattanaik 1971; d’Aspremont and Gevers 1977; Maskin 1978; Suzumura 1983). Our aim in this paper is to make a small but pointed contribution to this broad area of divergent preferences and democratic aggregation.

Starting with Hotelling’s (1929) classic paper, the economics of location, for instance, where and how firms position themselves on a line or on a surface, or the ideological space which political parties try to capture, has become an important tool for analyzing product quality differentiation and antitrust law, retail marketing and the theory of vendor positioning and, most prominently, political economy and electoral politics. In the area of political economy, it has led to the celebrated median voter theorem of Black (1948) for majority decisions, and of Downs (1957) for a representative democracy, which enabled us to take on some of the big questions of electoral democracy, and development (see, for instance, Stokes 1963; Osborne 1995; Congleton 2002; Acemoglu and Robinson 2005).

The present paper is an exercise in the pure theory of location choice over a circle or rotary. Given the wide use of location analysis in economics and political science it is hoped the paper will provide foundations for different kinds of practical studies on electoral democracy or retail trading. Each of these entail further complications. In the case of retail stores choosing their location, the analysis is complicated by the fact that they can, typically, choose the price as well. In electoral politics, personal attributes (for example, the appearance) of the candidate may matter as well and not just the ideological platform the candidate offers. However, the theorems reported in this paper should be of value for all such models. For the most part, we refer to the problem of electoral democracy and use its associated language, although our study is best viewed at a more abstract level.

The importance of the pivotal voter was recognized as early as 1785 by Condorcet. However, following the influential work of Hotelling, the bulk of the analysis got focused on voters spread over a line. At one level, our paper is a reminder that there can be other formations among voters and these can lead to related but distinct results. In our set up, for instance, it is not the case that candidates invariably converge on the same electoral platform but, instead, we get the result that no matter how many candidates contest an election, there will never be more than two candidates announcing the same platform.Footnote 1

As Stokes (1963, p. 368) had noted over a half a century ago, “The use of spatial ideas to interpret party competition is a universal phenomenon of modern politics. Such ideas are the common coin of political journalists...” Congleton (2002) points out in his survey of the median voter theorem, “Most analytical work in public choice is based upon relatively simple models of majority decision making. These models are widely used even though the researchers know that real political settings are more complex than the models seem to imply. The ... simple models provide us with engines of analysis that allow a variety of hypotheses about more complex phenomena to be developed, many of which would be impossible (or uninteresting) without the frame of reference provided by simple models.” It is in this spirit that we venture into investigating electoral democracy in this paper.

To understand the idea of using location analysis to investigate electoral democracy, suppose voters are interested in two aspects of candidates running for office: how anti-establishment they are, and their economic ideology, for instance, their position on economic conservatism. Clearly, each voter’s ideal can be represented as a point in a two-dimensional space. Suppose all voters ideals lie on a circle, of the kind shown in Fig. 1, and suppose each candidate has to announce his or her agenda which is basically a point on the circle. Thus we can think of candidates S and T announcing their respective agenda as points S and T on the circle as shown. From the vertical perspective, S and T look identical but they are far apart in terms of economic ideology, measured on the horizontal axis. (Any resemblance of S and T to Sanders and Trump is entirely spurious.) If we assume that voters opt for the candidate closest to his or her ideal, the closeness being measured in terms of distance along the circle, then we have a classic location-choice game.Footnote 2

Fig. 1
figure 1

Two dimensional attributes

Full size image

Before delving into the analysis, it is useful to make a few pertinent remarks about our choice of the circle as the space of voter ideals, and the related choice of the arc distance along the circle as the measure of the closeness of a candidate’s platform to a voter’s ideal.

The most general assumption would be to think of a finite number of individuals scattered arbitrarily in a specified subset of an M dimensional Euclidean space (where M is the number of distinct issues that a voter cares about); and then, to have a finite number of candidates choose their platforms as points in that subset. [For \(M=2,\) the specified subset could be a circle, a disc, or the entire two-dimensional Euclidean space]. In general, that would be very difficult to analyze and might be riddled with non-existence problems. Hotelling moved away from that framework by considering an infinite number of voters distributed on a finite length straight line (thereby focusing on a single over-riding issue that a voter cares about). That became such an innate part of our thinking that we seem to be somewhat rudderless in analyzing the rise of politics in which some candidate is conservative in terms of economics but politically a radical and vice versa for another candidate.

To analyze such cases, we need (for a start) to deal with at least two independent issues that voters care about. That is, it should not be possible to predict the voter’s view on the second issue by knowing her view on the first issue (otherwise, Hotelling’s formulation would suffice, as there would be only one over-riding issue). Our paper can be viewed as a first cut at formulating such a framework of analysis. If the space of voter ideals is a circle, it captures an essential aspect of this problem; as illustrated in Fig. 1, platforms S and T both reflect anti-establishment positions, but it is not possible to predict the economic ideology of the platforms from this, as that might vary widely.

We are, of course, aware that by treating the circle as the space of voter ideals we are using a strong assumption, which has some unintended features. To remedy the situation, one might explore in future research formulations in which (for instance) one moves from the unit circle as the space of voter ideals to the closed unit disc (which includes all points on and inside the unit circle). A pragmatic way to do this would be (as a generalization of our current framework), to consider the space of voter ideals to be a finite number of concentric circles, a framework which approaches the unit closed disc model as the number of concentric circles becomes large. The crucial issue one would have to face in such a formulation is how to define a reasonable notion of closeness between platforms which lie on distinct concentric circles.Footnote 3

It may also be pointed out that the set up we use has some important history in industrial organization theory, law and economics, as exemplified by the papers of Schmalensee (1978) and Salop (1979). Since the abstract results in our paper have as much bearing on political economy as on industrial organization, the paper may be viewed as a contribution to the latter as well.

The specific problem that we want to address in this paper is the following. If there are n candidates in an election how will they position themselves on the circle in equilibrium? And, once politicians have taken up positions, is there an easy way to check if this is an equilibrium? The entire exercise in this paper is done by defining equilibrium in the sense of Nash. What the paper shows is that there is a surprisingly easy answer to the question just posed (see Theorem 1).

Hotelling’s classic location game (for two vendors with exogenously fixed prices) has a unique Nash equilibrium. However, for many non-cooperative games, there is a multiplicity of Nash equilibria, and this is seen as a principal drawback of the equilibrium concept, as its predictive power can be limited as a result. In our setting too, there can be many (in fact, a continuum of) Nash equilibria. However, instead of focusing on equilibrium outcomes, if we shift our focus to equilibrium payoffs, a rather general result on the distribution of payoffs in Nash equilibria emerges in this model. We find that the entire set of Nash equilibria is subject to rather stringent bounds on the equilibrium payoffs. Specifically, if the sum of all players’ payoffs is equal to 1,  the Nash equilibrium payoff of each player must lie in the interval:

$$\begin{aligned} \Pi =\left[ \frac{1}{2(n-1)},\frac{2}{n+1}\right] \end{aligned}$$

for an arbitrary number n of players, and for an arbitrary Nash equilibrium (see Theorem 2). As an implication, in a four-way race, no candidate can get less than one-sixth of the vote in a Nash equilibrium. This makes one-sixth a useful benchmark to decide whether to change one’s strategy in a four-way race.

2 The model

2.1 Preliminaries

We consider a democratic election scenario in the following way. There are a number (say n) of political candidates. Each candidate has a political platform. When each candidate has a distinct platform, candidates can be identified by their political platforms, so that these words can be used interchangeably. Clearly, this is an abstraction, since there can be more to a candidate than her political platform, even from the point of view of an election. Further, the official platform can often be different from the actual platform, or the platform as perceived by the voters.

Platforms are typically multi-dimensional (as we have illustrated in Sect. 1), although in prevalent political discussions, one often discusses platforms in one (major) dimension (or in one dimension at a time), since this simplifies the argument one is trying to make. Actually, the distortion produced in such analysis is substantial, and this has long been recognized in the political science literature that has followed since the uni-dimensional median voter theorem.Footnote 4

We might describe a platform by a list of issues; when speaking of candidates, one might use the term attributes instead. Thus, for example, one might specify a platform by a list of seven major issues, by indicating a candidate’s stand on those seven issues. Unlike in the economics of location, where there would be fairly clear cut numerical measures to identify locations of vendors, and prices charged at those locations, the problem of quantitative representation of political issues can be a substantial one. On some issues (such as, what should be the magnitude of the capital gains tax), there might be clear-cut numerical measures to summarize a candidate’s stand on the issue. But, on others (such as, the extent of gun control), there might not be such summary measures. However, even in such cases, it is common to use not only ordinal measures (preference for more gun-control to less), but also measures which would be interpreted as cardinal (strongly opposed to gun control). While being aware of the abstraction involved, we choose to represent a platform by a point in M dimensional real space, if there are M clearly defined issues.Footnote 5

Not all points in M dimensional space may be considered to be platforms. It is more reasonable to suppose that there is a feasible set of platforms, \({\mathcal {F}},\) (a subset of \({\mathbb {R}}^{M})\) which is well understood by the candidates as well as the electorate.Footnote 6 Because we need to understand simpler structures first, in order to usefully investigate more complex and realistic structures later, we simplify by choosing \(M=2;\) whereas M can in reality take many values, there is a special significance to moving from 1 to any larger value, including 2. It immediately allows for political situations where candidates differ on some issues and agree on others. Further, again in the interest of simplification, we consider the feasible set of platforms \( {\mathcal {F}}\) to be a circle (with, say, origin at (0, 0) and circumference equal to 1),  just as represented in Figs. 2, 3, 4, 5, 6, 7 and 8 in Sects. 3 and 4 below. The circle is capable of including fine distinctions between similar platforms, as well as diametrically opposing platforms.Footnote 7

We turn now to the electorate. We suppose a voter has preferences over the set of feasible platforms, in the form of an “ideal platform”, which is best from the voter’s point of view, other platforms being inferior. A voter votes for the candidate whose platform is “closest” to the voter’s ideal platform. We define “closest” in terms of the length of the arc between the voter’s ideal and the candidate’s platform. [We elaborate on this measure in Sect. 3.1.3].

Finally, we need to consider the distribution of voters, or more precisely, the distribution of the ideal platforms of voters. As an extreme case, consider (in the context of Fig. 1) that even though there is a lot of variety of feasible platforms on a circle, voters don’t really care about whether a candidate is pro-establishment or anti-establishment. They would then be bunched up (in terms of their ideal platforms) at two points on the circle on the horizontal axis. Because we want to emphasize the importance of multi-dimensional platforms, we do not wish to investigate such extreme cases. A more reasonable assumption is that the voters ideal platforms are distributed with full support on the circle. Assuming a uniform distribution on the circle (which we do in our model below) is a useful place to start to keep the model tractable.

2.2 Description of the game

There are n players (political candidates). Each player has to set up a platform at some point on the circle (with origin at (0, 0) and circumference equal to 1). Each voter has an ideal platform on the circle and votes for the candidate whose platform is closest to his own. If a voter is equi-distant to several players’ locations, he chooses randomly from them by assigning equal probability to each of those players. There is a continuum of voters, and their ideal platforms are distributed uniformly on the circle.

Each player’s (political candidate’s) payoff is equal to the expected number of voters who vote for her and her aim is to maximize her expected payoff.

Players choose their platforms simultaneously. A choice of platform by each player will be called an n-tuple of strategies or simply a placement. The aim is to analyze the Nash equilibria of this game, that is a placement such that no player can do better (increase her payoff) through a unilateral deviation to another platform.

It is useful to coin a few more terms before we begin the analysis. A placement in which there is no platform that is chosen by more than one player will be referred to as a scattered placement. All other kinds of placements are called clustered placements. Given a placement, a stretch of the circle where no platform has been chosen by any of the players will be called an empty stretch. A Nash equilibrium in which the placement is scattered will be called a scatteredNash equilibrium. All other Nash equilibria are called clustered Nash equilibria.

3 Scattered Nash equilibria

This section provides a complete characterization of scattered Nash equilibria. If \(n=2\) it is easy to see that all placements are Nash equilibria. Henceforth, and without further mention, we will focus on cases where \(n\ge 3.\)

3.1 Notation and definitions of concepts

Recall that we want to present a criterion, such that given a placement, the criterion can be used to easily check whether the placement is (or is not) a Nash equilibrium. For this purpose, some notation is needed to formalize the concepts introduced above.

3.1.1 Circle and arcs

We consider a circle with circumference equal to C.Footnote 8 An arc of a circle is any connected set of points of the circle. Associated with any arc are its boundary points. The boundary points need not belong to the arc itself; for example, the arc can be open relative to the circle. If the closure of the arc is not the entire circle, it will have two boundary points.

Given any two distinct points A and B on the circle, two arcs can be associated with them. One arc is the set of points traversed as one moves from A to B clockwise; we denote this by arc[A, B]. [This set will be understood to include the points A and B].The other arc is the set of points traversed as one moves from A to B counter clockwise. This is, of course, the same as the set of points traversed as one moves from B to A clockwise. We denote this by arc[B, A]. The set \(arc[A,B]\backslash \{A,B\}\) is denoted by arc(A, B). Similar obvious notation applies to arcs containing B,  but not A,  and to arcs containing A,  but not B. Note that arc(A, B) is disjoint from arc[B, A],  and therefore from arc(B, A),  arc(B, A] and arc[B, A).

The arc length of arc[A, B] is defined to be equal to \((C\theta /360),\) if \(\theta \) is the angle (in degrees) which arc[A, B] subtends at the centre of the circle. This is also defined to be the arc length of arc(A, B). The arc length of arc[A, B] is denoted by a[A, B].Footnote 9When the two arc lengths a[A, B] and a[B, A] are not equal, the arc with the larger arc length is called the major arc, and the arc with the smaller arc length is called the minor arc. Thus, the arc length of the minor arc is \(\min \{a[A,B],\)\(a[B,A]\},\) and we denote it by ma[A, B];  it is also equal to ma[B, A]. In what follows, without loss of generality, we take the circumference C of the circle to be equal to 1,  and therefore the radius of the circle to be \(r=(1/2\pi ).\)

3.1.2 Candidates, platforms and neighborhoods

In a scattered placement, each candidate \(i\in I\equiv \{1,\ldots ,n\}\) can be identified with a distinct point (platform) on the circle. Thus, the placement itself can be described by an n-tuple of platforms \( \{s_{1},\ldots ,s_{n}\},\) where each \(s_{i}\) for \(i\in I\) is a distinct point on this circle.Footnote 10

Given a scattered placement \(s=\{s_{1},\ldots ,s_{n}\},\) there is a neighborhood\(Y_{i}(s)\) for each candidate \(i\in I\equiv \{1,\ldots ,n\},\) which is defined to be the maximal arc, containing \(s_{i},\) but not containing any \(s_{j}\ne s_{i};\) that is, it is an arc S,  containing \( s_{i},\) but not containing any \(s_{j}\ne s_{i},\) such that if \(S^{\prime }\) is any arc, containing \(s_{i},\) but not containing any \(s_{j}\ne s_{i},\) then \(S^{\prime }\subset S.\)Footnote 11 As we have defined it, a neighborhood \(Y_{i}(s)\) is a set which is open relative to the circle. The closure of \(Y_{i}(s)\) (relative to the circle) is denoted by \({\bar{Y}}_{i}(s).\) Thus, \({\bar{Y}}_{i}(s)\) contains \( Y_{i}(s)\) and the boundary points of \(Y_{i}(s).\)

Given a scattered placement \(s=\{s_{1},\ldots ,s_{n}\},\) for each candidate \( i\in I\equiv \{1,\ldots ,n\},\) we can define his neighbors to be:

$$\begin{aligned} s_{R(i)}=\arg \min _{s_{j}\in s\backslash \{s_{i}\}}a(s_{i},s_{j})\quad \text {and}\quad s_{L(i)}=\arg \min _{s_{j}\in s\backslash \{s_{i}\}}a(s_{j},s_{i}) \end{aligned}$$
(NBS)

These are the candidates with platforms closest to the platform of candidate i,  on the right and on the left respectively. Clearly, we have:

$$\begin{aligned} s_{R(i)}\ne s_{i}\quad \text {and}\quad s_{L(i)}\ne s_{i} \end{aligned}$$

and the neighbors are distinct:

$$\begin{aligned} s_{R(i)}\ne s_{L(i)}. \end{aligned}$$
(DNBS)

The statement in (DNBS) holds because if \(s_{R(i)}=s_{L(i)},\) then \({\bar{Y}} _{i}(s)\) is the circle itself, implying that there are just two distinct platforms on the circle, and this is impossible in a scattered placement with \(n\ge 3.\)

Using (NBS), it can be shown that:

$$\begin{aligned} Y_{i}(s)=arc(s_{L(i)},s_{R(i)})\quad \text {and}\quad {\bar{Y}} _{i}(s)=arc[s_{L(i)},s_{R(i)}]. \end{aligned}$$
(NBD)

Informally, the set \({\bar{Y}}_{i}(s)\) contains the platforms of \(i^{\prime }s\) “nearest” neighbors, while \(Y_{i}(s)\) does not. (NBD) is a useful characterization of a neighborhood; we make use of this characterization in our analysis below. We define \(y_{i}(s)\) to be the arc length of the arc \( Y_{i}(s).\)

3.1.3 Behavior of voters

Each voter has an ideal platform, v,  a point on the circle. The general principle that we wish to use with respect to voting behavior is that each voter votes for the candidate whose platform is “closest” to her ideal platform v. To implement this principle, we need to be explicit about what “closest” means.

However, without specifying a particular distance function, we can say that a necessary condition for the voter to vote for platform \(s_{i}\) (of candidate i) is that there be an arc joining v with \(s_{i}\) (either \( arc(v,s_{i})\) or \(arc(s_{i},v)\)) which contains no other platform \(s_{j}\) (for \(j\ne i).\) For if there is always some platform \(s_{j}\) in between v and \(s_{i}\) (that is, there is some \(s_{j}\) in \(arc(v,s_{i})\)and there is some \(s_{k}\) in \(arc(s_{i},v)\) ), then \(s_{i}\) cannot be closest to v by any reasonable notion of closeness.

A consequence of this is that a voter whose ideal platform v falls outside \({\bar{Y}}_{i}(s)\) will definitely not vote for candidate i,  since \( s_{L(i)}\) is in \(arc(v,s_{i})\) and \(s_{R(i)}\) is in \(arc(s_{i},v).\)

So, the only voters who can possibly vote for candidate i are those whose ideal platform v falls in \({\bar{Y}}_{i}(s).\) If \(v\in (s_{i},s_{R(i)}],\) we postulate that the voter votes for \(s_{i}\) when \(a(s_{i},v)<a(v,s_{R(i)}],\) votes for \(s_{R(i)}\) when \(a(s_{i},v)>a(v,s_{R(i)}],\) and votes for either platform with equal probability when \(a(s_{i},v)=a(v,s_{R(i)}].\) A similar rule is postulated if \(v\in [s_{L(i)},s_{i});\) that is, the voter votes for \(s_{i}\) when \(a(v,s_{i})<a[s_{L(i)},v),\) votes for \(s_{L(i)}\) when \(a(v,s_{i})>a[s_{L(i)},v),\) and votes for either platform with equal probability when \(a(v,s_{i})=a[s_{L(i)},v).\) Finally, if \(v=s_{i},\) then the voter votes for \(s_{i}.\)

Unlike the economics of location of vendors, and customers purchasing from the nearest vendor, where the above postulate would be entirely natural, it is a debatable postulate in electoral politics. The use of the arc length is not itself central to a possible objection to the postulate. It is the fact that a clockwise arc length is compared to an anti-clockwise arc length in deciding on voting behavior. If, for example, a voter always seeks the closest platform, moving in a clockwise direction from his ideal platform, then this postulate of voting behavior would not hold. Because closeness of a platform to another platform is primarily a mental concept, without a convenient physical counterpart, the analogy with spatial location theory need not carry over.

3.1.4 Payoffs of candidates

Given the postulate of voting behavior, and the assumption that the ideal platforms of voters are uniformly distributed on the circle, the payoff \( p_{i}(s)\) of candidate i is one-half the arc length of \(Y_{i}(s);\) that is, \(p_{i}(s)=y_{i}(s)/2.\) It measures the percentage of the votes that candidate i expects to get.

In studying scattered Nash equilibria, it is important to be able to deal with some clustered placements. This is because, for each player, one has to consider all unilateral deviations in platforms from her given platform (in the scattered placement). Consider a scattered placement \( \{s_{1},\ldots ,s_{n}\},\) and focus on player i. Player i can consider a unilateral deviation to a platform \(s_{i}^{\prime }.\) Now \(s_{i}^{\prime }\) can coincide with \(s_{k}\) for some \(k\ne i.\) Thus, after such a deviation, we have a clustered placement, and one must be able to compute the payoff to player i from such a deviation and compare that payoff to her payoff from the scattered placement \(\{s_{1},\ldots ,s_{n}\}\).Footnote 12 To see how this is done, notice that after such a unilateral deviation by player i,  we have a clustered placement \(s^{\prime }=\{s_{1},.,s_{i}^{\prime },..,s_{n}\}\) and we can define a neighborhood \(Y_{i}(s^{\prime })\) as above to be the maximal arc, containing \(s_{i}^{\prime },\) but not containing any \( s_{j}\ne s_{i}^{\prime }.\) Note that \(Y_{k}(s^{\prime })\) coincides with \( Y_{i}(s^{\prime }).\) The arc length of \(Y_{i}(s^{\prime })\) is denoted by \( y_{i}(s^{\prime }).\) The payoff to player i in the new placement \( s^{\prime }\) is defined to be \(p_{i}(s^{\prime })=y_{i}(s^{\prime })/4,\) since both i and k have to share the same neighborhood \(Y_{i}(s^{\prime })\) in the placement \(s^{\prime }.\)

3.2 Characterizing scattered Nash equilibria

Given a scattered placement \(s=\{s_{1},\ldots ,s_{n}\},\) an empty stretch is any arc that does not contain any \(s_{i}\) for \(i\in \{1,\ldots ,n\}.\) The arc length of the empty stretch of maximum arc length is denoted by x(s). We define:

$$\begin{aligned} y(s)=\min \{y_{1}(s),\ldots ,y_{n}(s)\}. \end{aligned}$$

A characterization of scattered Nash equilibria can be obtained by just comparing y(s) with x(s).

Theorem 1

  1. (A)

    Given a scattered placement \(s=\{s_{1},\ldots ,s_{n}\},\) if : 

    $$\begin{aligned} y(s)\ge x(s) \end{aligned}$$
    (1)

    then the placement is a Nash equilibrium.

  2. (B)

    Given a scattered placement \(s=\{s_{1},\ldots ,s_{n}\},\) if : 

    $$\begin{aligned} y(s)<x(s) \end{aligned}$$
    (2)

    then the placement is not a Nash equilibrium.

Proof

(A) Pick any \(j\in I,\) and fix it in what follows. By (1), and the definition of y(s),  we have:

$$\begin{aligned} p_{j}(s)=(y_{j}(s)/2)\ge (x(s)/2). \end{aligned}$$
(3)

Suppose j unilaterally changes his platform from \(s_{j}\) to some other point \(s_{j}^{\prime }\) on the circle. Denote the new placement created by \( s^{\prime }=\{s_{1},\ldots ,s_{j}^{\prime },\ldots ,s_{n}\}.\) We have either (a) \( s^{\prime }\) is a scattered placement, or (b) \(s^{\prime }\) is a clustered placement.

Case (a): If \(s_{j}^{\prime }\in Y_{j}(s),\) then the payoff of j does not change. If \(s_{j}^{\prime }\) is not in \(Y_{j}(s),\) then since \( s^{\prime }\) is a scattered placement, \(s_{j}^{\prime }\in Y_{k}(s),\) the neighborhood of some other player \(k\in I\) (with \(k\ne j).\) The neighborhood \(Y_{k}(s)\) consists of two empty stretches, and the new neighborhood of \(s_{j}^{\prime }\) will have to be one of these two empty stretches (no longer empty after it is occupied by \(s_{j}^{\prime }\)). Thus, we must have \(y_{j}(s^{\prime })\le x(s),\) and so:

$$\begin{aligned} p_{j}(s^{\prime })\le (x(s)/2). \end{aligned}$$
(4)

Clearly, (3) and (4) imply that there is no incentive for j to change his platform to \(s_{j}^{\prime }.\) This establishes that the placement s is a Nash equilibrium in case (a).

Case (b): In this case, there is some \(k\in I,\) with \(k\ne j,\) such that \(s_{k}=s_{j}^{\prime }.\) There are two subcases to consider: (I) \( Y_{j}(s)\) is disjoint from \({\bar{Y}}_{k}(s);\) (II) \(Y_{j}(s)\cap {\bar{Y}} _{k}(s)\) is non-empty. In subcase (I),

$$\begin{aligned} Y_{k}(s^{\prime })=Y_{k}(s) \end{aligned}$$
(5)

and so \(Y_{j}(s^{\prime })=Y_{k}(s).\)Footnote 13 The neighborhood \(Y_{k}(s)\) contains two empty stretches, and the point \(s_{k}.\) Thus, \(y_{k}(s)\le 2x(s),\) and we get:

$$\begin{aligned} p_{j}(s^{\prime })=y_{j}(s^{\prime })/4=y_{k}(s)/4\le 2x(s)/4=(x(s)/2). \end{aligned}$$
(6)

Clearly, (3) and (6) imply that there is no incentive for j to change his platform to \(s_{j}^{\prime }.\) This establishes that the placement s is a Nash equilibrium in case (b)(I).

In subcase (II), \(Y_{j}(s)\cap {\bar{Y}}_{k}(s)\) is non-empty. Let us write:

$$\begin{aligned} Y_{k}(s)=arc(s_{L(k)},s_{R(k)})\quad \text {and}\quad Y_{j}(s)=arc(s_{L(j)},s_{R(j)}). \end{aligned}$$
(7)

We now claimFootnote 14 that:

$$\begin{aligned} \text {Either }(\mathrm{i})\ s_{j}=s_{L(k)}\ \text {and}\ s_{R(j)}=s_{k};\quad \text {or}\quad (\mathrm{ii})\ s_{k}=s_{L(j)}\ \text {and}\ s_{R(k)}=s_{j}. \end{aligned}$$
(8)

Taking the claim for granted, we can proceed to complete the proof of case (b)(II) as follows. In case (i) of (8), after candidate j changes his platform from \(s_{j}\) to \(s_{j}^{\prime }=s_{k},\) the entire \( arc(s_{L(j)},s_{j}]\) becomes part of the neighborhood of player j ,  in addition to \(Y_{k}(s).\) That is,

$$\begin{aligned} Y_{j}(s^{\prime })= & {} arc(s_{L(j)},s_{j}]\cup Y_{k}(s) \nonumber \\= & {} arc(s_{L(j)},s_{L(k)}]\cup arc(s_{L(k)},s_{R(k)}) \nonumber \\= & {} arc(s_{L(j)},s_{L(k)}]\cup arc(s_{L(k)},s_{k})\cup arc[s_{k},s_{R(k)}) \nonumber \\= & {} arc(s_{L(j)},s_{L(k)}]\cup arc(s_{L(k)},s_{R(j)})\cup arc[s_{k},s_{R(k)}) \nonumber \\= & {} arc(s_{L(j)},s_{j}]\cup arc(s_{j},s_{R(j)})\cup arc[s_{k},s_{R(k)}) \nonumber \\= & {} Y_{j}(s)\cup arc[s_{k},s_{R(k)}) \end{aligned}$$
(9)

where we have used (8)(i) in lines 2,4 and 5 of (9). [This is also the new neighborhood \(Y_{k}(s^{\prime })\) of player k].

Since \(arc(s_{k},s_{R(k)})\) is one of the two empty stretches of \(Y_{k}(s)\), its arc length is at most x(s),  and so the arc length of \( arc[s_{k},s_{R(k)})\) is also at most x(s). The arc length of \(Y_{j}(s)\) is \(y_{j}(s).\) Thus, the arc length of \(Y_{j}(s^{\prime })\) is \(\le x(s)+y_{j}(s).\) Consequently,

$$\begin{aligned} p_{j}(s^{\prime })\le [x(s)+y_{j}(s)]/4\le 2y_{j}(s)/4=y_{j}(s)/2 \end{aligned}$$
(10)

where the second inequality in (10) follows from (1). Clearly, (3) and (10) imply that there is no incentive for j to change his platform to \( s_{j}^{\prime }.\) This establishes that the placement s is a Nash equilibrium, when case (i) of claim (8) holds.

If case (ii) of claim (8) holds, the proof is similar to the one just given for case (i) of claim (8). This establishes that the placement s is a Nash equilibrium in case (b)(II), and completes the proof of part (A) of the theorem.

(B) By (2), and the definition of y(s),  there is \(h\in I,\) such that \( y_{h}(s)=y(s)<x(s).\) Then:

$$\begin{aligned} p_{h}(s)<(x(s)/2). \end{aligned}$$
(11)

Now h can relocate from platform \(s_{h}\) to a platform \(s_{h}^{\prime }\) (say) in an empty stretch with arc length x(s). Then the new neighborhood of h is the empty stretch with arc length x(s) (no longer empty after it is occupied by \(s_{h}^{\prime }).\) Consequently,

$$\begin{aligned} p_{h}(s^{\prime })=(x(s)/2). \end{aligned}$$
(12)

Clearly, (11) and (12) imply that h can increase her payoff by a unilateral deviation from platform \(s_{h}\) to platform \(s_{h}^{\prime }.\) So, the (original) placement s cannot be a Nash equilibrium. This completes the proof of part (B) of the theorem. \(\square \)

A consequence of Theorem 1 is the following result, which provides a sufficient condition for a scattered placement s to be a Nash equilibrium. Note that, to use this criterion, one does not need to calculate y(s).

Corollary 1

Given a scattered placement \(s=\{s_{1},\ldots ,s_{n}\},\) if : 

$$\begin{aligned} x(s)\le \frac{1}{n-1} \end{aligned}$$
(13)

then the placement s is a Nash equilibrium.

Proof

We claim that:

$$\begin{aligned} y(s)\ge x(s). \end{aligned}$$
(14)

Suppose, on the contrary that \(y(s)<x(s).\) Then, there is \(j\in I\) such that:

$$\begin{aligned} y_{j}(s)=y(s)<x(s). \end{aligned}$$
(15)

Since x(s) is the arc length of the largest empty stretch, we can find a set of points X(s) which is an empty stretch, with arc length x(s). [There could be several such sets; we are picking one and calling it X(s)].

Using (13) and (15), we get:

$$\begin{aligned} x(s)+y_{j}(s)<2x(s)\le \frac{2}{n-1} \end{aligned}$$

and so:

$$\begin{aligned} 1-x(s)-y_{j}(s)>1-\frac{2}{n-1}=\frac{n-3}{n-1}\ge 0. \end{aligned}$$
(16)

If the neighborhood \(Y_{j}(s)\) overlaps with X(s) (that is \(Y_{j}(s)\cap X(s)\ne \phi ),\) then X(s) must be a subset of \(Y_{j}(s),\) and consequently \(y_{j}(s)\ge x(s),\) contradicting (9). Thus, \(Y_{j}(s)\) does not overlap with the empty stretch X(s),  with arc length x(s).

Note that \(Y_{j}(s)\) consists of two empty stretches (and the point \(s_{j})\) and neither one overlaps with X(s). If \(n=3,\) then there are no other empty stretches and we must have \(y_{j}(s)+x(s)=1,\) but this contradicts (16). Thus, we must have \(n>3\) and there are a total of \((n-3)\) empty stretches remaining (other than X(s) and the two empty stretches in \( Y_{j}(s)).\) Let z(s) be the maximum of the (arc) lengths of these remaining empty stretches. Consequently,

$$\begin{aligned} (n-3)z(s)\ge 1-x(s)-y_{j}(s). \end{aligned}$$
(17)

Using (16) in (17), we obtain:

$$\begin{aligned} z(s)>\frac{1}{n-1}. \end{aligned}$$
(18)

Clearly, (18) contradicts (13) and establishes our claim (14).

Now, the result follows directly from Theorem 1. \(\square \)

3.2.1 Geometric illustration for \(n=3\)

To hone our intuition for the analysis used in Theorem 1, it is useful to consider the special case of \(n=3.\)

Considered the scattered placement described in Fig. 2.

Fig. 2
figure 2

Illustration of Theorem 1(A)

Full size image

It is easy to see no player can do better by shifting platform on the same stretch (strictly) between the other two players. Note, for instance that player 1 will get exactly half the votes on the clockwise stretch traversed from her platform \(s_{1}\) to the platform \(s_{3}\) of player 3 and, likewise, half the votes on the anti-clockwise stretch traversed from her platform \(s_{1}\) to the platform \(s_{2}\) of player 2. In other words, player 1 gets half the votes on the stretch of rotary between \(s_{2}\) and \( s_{3}\) on the side where \(s_{1}\) is located. Since this makes no mention of where \(s_{1}\) is located on this stretch, any movement by player 1,  where she remains on the same side of the rotary leaves her payoff unchanged. [Analytically, we expressed this above by saying that if \(s_{1}\) is located anywhere on the \(arc(s_{2},s_{3}),\) then her payoff remains the same; this is the arc traversed as one moves from \(s_{2}\) to \(s_{3}\) in a clockwise direction].

Hence to check if a scattered placement is a Nash equilibrium all we have to check is if any player is better off by relocating her platform to another empty stretch. For instance, can player 1 do better by changing her platform to the southern stretch of rotary between \(s_{2}\) and \(s_{3}?\) [Analytically, we expressed this above by analyzing whether player 1 can do better by relocating from platform \(s_{1}\) to some platform \( s_{1}^{\prime }\) on the \(arc(s_{3},s_{2});\) this is the arc traversed as one moves from \(s_{3}\) to \(s_{2}\) in a clockwise direction]. Clearly, in Fig. 2, the southern stretch of rotary between \(s_{2}\) and \(s_{3}\) is less than (1 / 2),  and so player 1 cannot do better by such a relocation of her platform. Given the symmetry of the locations of the three players, the same argument applies to players 2 and 3 as well. So, the placement in Fig. 2 is a Nash equilibrium.

Now consider the case where the empty stretch \(arc(s_{3},s_{2})\) is more than (1 / 2). This is illustrated in Fig. 3.

Fig. 3
figure 3

Illustration of Theorem 1(B)

Full size image

It is obvious that such a placement cannot be Nash, since player 1 can shift her platform from \(s_{1}\) in the north connector between \(s_{2}\) and \( s_{3}\) to \(s_{1}^{\prime }\) in the south connector between \(s_{2}\) and \( s_{3}.\)

3.3 Symmetric and asymmetric scattered Nash equilibria

Corollary 1 has the following implication, which might be of interest. In the equally spaced case, the longest empty stretch is (1 / n),  and therefore such a scattered placement is a Nash equilibrium. So, for all placements sufficiently close to the equally spaced case, x is close to (1 / n),  and therefore less than \([1/(n-1)].\) Thus, all such scattered placements are also Nash equilibria by Corollary 1. That is, there is a continuum of scattered Nash equilibria, clustered around the equally spaced placement.

The fact that the players have no distinctive features ex ante (they can choose platforms anywhere on the circle, and once located the same payoff rule applies to all of them) does not, of course, mean that they receive the same or even similar payoffs in a Nash equilibrium. In fact, distinctly asymmetric Nash equilibria can arise in this model. We begin by describing an example with three players, which can be instructive.

3.3.1 The advantage of appearing different

Example 1

With Cartesian co-ordinates, consider the center of the circle to be at (0, 0). There are three players, and the scattered placement s is described as follows. The platform of player 1 is located at (r, 0),  where \(r=(1/2\pi ).\) The platforms of players 2 and 3 are located near \((-r,0),\) with the platform for 2 slightly above the axis on the circle, and the platform for 3 symmetrically slightly below the axis on the circle (see Fig. 4). To be precise, the platform of 2 is located at \((-(a^{2}-\varepsilon ^{2})^{1/2},\varepsilon )\) and the platform of 3 is located at \((-(a^{2}-\varepsilon ^{2})^{1/2},-\varepsilon ),\) where \( 0<\varepsilon <a,\) and we are to suppose that \(\varepsilon \) is close to zero. That is, candidates 2 and 3 are very alike in their platforms, while candidate 1 is very different from either one of them.

Fig. 4
figure 4

Scattered Nash equilibrium (Example 1)

Full size image

Note that \(y_{i}(s)\) exceeds (1 / 2) for all \(i\in I=\{1,2,3\},\) and x(s) is smaller than (1 / 2). Thus, s is a Nash equilibrium by Theorem 1. [In fact, Corollary 1 can also be applied directly in this case to reach this conclusion]. Now, notice that as \(\varepsilon \rightarrow 0,\) we have the payoff of player 1 approaching (1 / 2),  as her neighborhood approaches the entire circle, while the payoffs of players 2 and 3 approach (1 / 4). There is a big advantage from appearing different.

3.3.2 The distribution of payoffs in scattered Nash equilibria

It is a rather remarkable fact that the simple example just discussed illustrates a general result on the distribution of payoffs in Nash equilibria in this model. Non-uniqueness of Nash equilibria is the major drawback in its predictive power regarding the outcome of the game. In our next result, instead of focusing on equilibrium locations of platforms, we shift our focus to equilibrium payoffs, and we find that these are subject to rather stringent bounds. Specifically, the Nash equilibrium payoffs of all players must lie in the interval:

$$\begin{aligned} \Pi =\left[ \frac{1}{2(n-1)},\frac{2}{n+1}\right] \end{aligned}$$
(DOP)

for an arbitrary number n of players, and for an arbitrary Nash equilibrium.

As an implication, in a four-way race, no candidate can get less than one-sixth of the vote in a scattered Nash equilibrium. This makes (1 / 6) a useful benchmark to decide whether to change one’s strategy (platform). That is, it is all right to get less than a quarter of the vote (perform worse than average), but it is not all right to get less than a sixth of the vote.Footnote 15

We now proceed to establish the formula (DOP) using Theorem 1.

Proposition 1

If a scattered placement \(s=\{s_{1},\ldots ,s_{n}\}\) is a Nash equilibrium,  then : 

$$\begin{aligned} \frac{1}{2(n-1)}\le p_{i}(s)\le \frac{2}{n+1}\quad \text {for all }i\in I=\{1,\ldots ,n\}. \end{aligned}$$
(19)

Proof

Given the scattered placement \(s=\{s_{1},\ldots ,s_{n}\},\) recall that the arc length of the empty stretch of maximum arc length is denoted by x(s),  and we define:

$$\begin{aligned} y(s)=\min \{y_{1}(s),\ldots ,y_{n}(s)\};\;z(s)=\max \{y_{1}(s),\ldots ,y_{n}(s)\}. \end{aligned}$$

Note that, given the scattered placement \(s=\{s_{1},\ldots ,s_{n}\},\) there are n empty stretches. Each empty stretch is contained in the neighborhoods of two distinct players. Since the arc lengths of the n empty stretches must add up to 1 (the circumference of the circle), the arc lengths of the neighborhoods of the n players must add up to 2;  that is:

$$\begin{aligned} \sum _{i=1}^{n}y_{i}(s)=2. \end{aligned}$$
(20)

There is \(k\in I=\{1,\ldots ,n\}\) such that \(y_{k}(s)=z(s).\) Since \(Y_{k}(s)\) consists of two empty stretches (and the point \(s_{k}),\) we have:

$$\begin{aligned} z(s)=y_{k}(s)\le 2x(s). \end{aligned}$$
(21)

Using (20) and (21), and Theorem 1, we get:

$$\begin{aligned} 2= & {} z(s)+\sum _{i\ne k}y_{i}(s) \nonumber \\\ge & {} z(s)+(n-1)x(s) \nonumber \\\ge & {} z(s)+(n-1)\left[ z(s)/2\right] \nonumber \\= & {} z(s)\left[ 1+\frac{n-1}{2}\right] \nonumber \\= & {} z(s)\left[ \frac{n+1}{2}\right] \end{aligned}$$
(22)

where we used (20) in line 1, Theorem 1 in line 2 and (21) in line 3. This yields:

$$\begin{aligned} z(s)\le \left[ \frac{4}{n+1}\right] \end{aligned}$$

so that:

$$\begin{aligned} p_{i}(s)\le p_{k}(s)=\left[ z(s)/2\right] \le \left[ \frac{2}{n+1}\right] \quad \text {for all }i\in I. \end{aligned}$$
(23)

This establishes the right-hand inequality in (19).

We proceed now to establish the left-hand inequality in (19). Note that by using (21) and Theorem 1, we have:

$$\begin{aligned} z(s)=y_{k}(s)\le 2x(s)\le 2y(s). \end{aligned}$$
(24)

There is \(j\in I=\{1,\ldots ,n\}\) such that \(y_{j}(s)=y(s).\) The neighborhood \( Y_{j}(s)\) consists of two empty stretches (and the point \(s_{j}).\) Call these empty stretches \(Y_{j1}(s)\) and \(Y_{j2}(s).\) The corresponding arc lengths can be written as \([ y_{j}(s)/2] +\varepsilon \) and \( [ y_{j}(s)/2] -\varepsilon ,\) where \(-[ y_{j}(s)/2]<\varepsilon <[ y_{j}(s)/2].\) The empty stretch \(Y_{j1}(s)\) must be shared with another player \(r\ne j;\) that is, \(Y_{j1}(s)\) must be one of the two empty stretches of \(Y_{r}(s),\) the neighborhood of player r. Similarly, empty stretch \(Y_{j2}(s)\)must be shared with another player \( t\ne j,r;\) that is, \(Y_{j2}(s)\) must be one of the two empty stretches of \( Y_{t}(s),\) the neighborhood of player t. Using (20), we can now write:

$$\begin{aligned} 2= & {} y_{j}(s)+y_{r}(s)+y_{t}(s)+\sum _{i\ne j,r,t}y_{i}(s) \nonumber \\\le & {} y_{j}(s)+y_{r}(s)+y_{t}(s)+(n-3)z(s) \nonumber \\\le & {} y_{j}(s)+\{ [ y_{j}(s)/2] +\varepsilon +x(s)\} +\{ [y_{j}(s)/2] -\varepsilon +x(s)\} +(n-3)z(s) \nonumber \\= & {} 2y_{j}(s)+2x(s)+(n-3)z(s) \nonumber \\\le & {} 2y(s)+2y(s)+(n-3)z(s) \nonumber \\\le & {} 2y(s)+2y(s)+2(n-3)y(s) \end{aligned}$$
(25)

where we have used Theorem 1 in line 5 and (24) in line (6) of (25). Thus, we obtain:

$$\begin{aligned} y(s)\ge \frac{2}{2n-2}=\frac{1}{n-1}. \end{aligned}$$
(26)

Using (6), we have:

$$\begin{aligned} p_{i}(s)\ge p_{j}(s)=\left[ y(s)/2\right] \ge \left[ \frac{1}{2(n-1)} \right] \quad \text {for all }i\in I \end{aligned}$$

which establishes the left-hand inequality in (19). \(\square \)

Remark

In the context of Example 1 with \(n=3,\) we see that the lower bound of \(25\%\) in (19), and the upper bound of \(50\%\) in (19) are approached as we let \( \varepsilon \rightarrow 0.\)

4 Clustered Nash equilibria

To analyze clustered Nash equilibria, it is useful to have a more general set-up of the framework of Sect. 3, which will include all placements, including both scattered and clustered placements.

Consider as before that there are \(n\ge 3\) players. A placement will be described by (m, s, t) where \(m\le n\) is a positive integer, \( s=\{s_{1},\ldots ,s_{m}\}\) an m-tuple of distinct pointsFootnote 16 on the circle (with circumference equal to 1),  and \( t=\{t_{1},\ldots ,t_{m}\}\) an \(m-\)tuple of positive integers, with \(t_{1}+\cdots +t_{m}=n.\) In the placement \(c=(m,s,t),\) there are m locations on the circle which represent distinct platforms. For \(i\in J=\{1,\ldots ,m\},\) there are \(t_{i}\) candidates who share the platform \(s_{i}.\) A scattered placement is a special case, where \(m=n,\) and \(t_{i}=1\) for each \(i\in J.\) Concepts related to this framework can be defined as in Sect. 3.1, but with a few differences which justify our going over these concepts again.

Given a placement \(c=(m,s,t),\) with \(m\ge 2,\) there is a neighborhood\(Y_{i}(c)\) for each platform \(s_{i},\) where \(i\in J\equiv \{1,\ldots ,m\},\) which is defined to be the maximal arc, containing \(s_{i},\) but not containing any \(s_{j}\ne s_{i}.\)

Note that the subscript i in \(Y_{i}(c)\) keeps track of the platform \( s_{i}, \) and not a particular candidate who has announced platform \(s_{i}.\) All candidates sharing the same platform \(s_{i}\) have the same neighborhood \( Y_{i}(c).\) Note also the qualification \(m\ge 2\) in the above definition. If \(m=1,\) then all platforms are identical, and the neighborhood for each candidate announcing the only platform is the entire circle.

As in Sect. 3.1, a neighborhood is an arc, and we define \(y_{i}(c)\) to be the arc length of this arc. As we have defined it, when \(m\ge 2,\) a neighborhood \(Y_{i}(c)\) is a set which is open relative to the circle. The closure of \(Y_{i}(c)\) (relative to the circle) is denoted by \({\bar{Y}} _{i}(c). \) Thus, \({\bar{Y}}_{i}(c)\) contains \(Y_{i}(c)\) and the boundary points of \(Y_{i}(c).\)

Given a placement \(c=(m,s,t),\) with \(m\ge 2,\) for each platform \(s_{i}\) (with \(i\in J),\) we can define the neighboring platforms by:

$$\begin{aligned} s_{R(i)}=\arg \min _{s_{j}\in s\backslash \{s_{i}\}}a(s_{i},s_{j})\quad \text {and}\quad s_{L(i)}=\arg \min _{s_{j}\in s\backslash \{s_{i}\}}a(s_{j},s_{i}). \end{aligned}$$
(NBS)

Clearly, we have:

$$\begin{aligned} s_{R(i)}\ne s_{i}\quad \text {and}\quad s_{L(i)}\ne s_{i}. \end{aligned}$$

Unlike in Sect. 3.1, it is possible to have \(s_{R(i)}=s_{L(i)}.\) [In this case, there are exactly two distinct platforms].

Using (NBS), it can be shown that when \(m\ge 2,\)

$$\begin{aligned} Y_{i}(c)=arc(s_{L(i)},s_{R(i)})\quad \text {and}\quad {\bar{Y}} _{i}(c)=arc[s_{L(i)},s_{R(i)}]. \end{aligned}$$
(NBD)

When \(m\ge 2,\) the payoff \(p_{i}(c)\) of each candidate who has announced the platform \(s_{i}\) (with \(i\in J)\) is:

$$\begin{aligned} p_{i}(c)=\frac{y_{i}(c)}{2t_{i}}. \end{aligned}$$
(PO)

When \(m=1,\) there is exactly one platform and the payoff p(c) of each candidate is:

$$\begin{aligned} p(c)=\frac{1}{n}. \end{aligned}$$
(PO1)

4.1 An example of clustered Nash equilibrium

We start by providing an example of a clustered Nash equilibrium. This can be viewed as a limiting version of Example 1.

Example 2

With Cartesian co-ordinates, consider the center of the circle to be at (0, 0). There are three players, and the placement \(c=(m,s,t)\) is described as follows. The platform of player 1 is \(s_{1}=\) (r, 0). Players 2 and 3 have the same platform, specified by \(s_{2}=\)\((-r,0).\) Thus, we have \( m=2,\)\(s=\{s_{1},s_{2}\}\) and \(t=\{t_{1},t_{2}\}=\{1,2\}.\) This placement is illustrated in Fig. 5.

Fig. 5
figure 5

Clustered Nash equilibrium (Example 2)

Full size image

Here we have \(y_{1}(c)=1,\)\(y_{2}(c)=1,\) and so \(p_{1}(c)=(1/2),\) while the payoff of each player who have announced the same platform \(s_{2}\) is \( p_{2}(c)=(1/4).\) We can check that c is a Nash equilibrium by using the following elementary analysis.

A unilateral deviation by any player will create a new placement \(c^{\prime }=(m^{\prime },s^{\prime },t^{\prime }),\) where the platforms of the other two players remain the same as in the placement c. If player 1 changes his platform to \(s_{1}^{\prime }\ne s_{2},\) his neighborhood remains the same as in c,  and his payoff remains (1 / 2). If \(s_{1}^{\prime }=s_{2},\) then his neighborhood remains the same as in c,  but his payoff becomes (1 / 3). So, clearly, player 1 has no incentive to deviate from \(s_{1}.\)

Consider next a unilateral deviation by player 2 from her original platform of \(s_{2}\) to a new platform \(s_{2}^{\prime }\ne s_{2}.\) If \( s_{2}^{\prime }\ne s_{1},\) her new neighborhood is the semi-circle, and so her new payoff is (1 / 4). If, on the other hand, \(s_{2}^{\prime }=s_{1},\) then her neighborhood is the full circle, and her payoff is again (1 / 4). So, clearly, player 2 has no incentive to deviate from \(s_{2}.\) Similarly, player 3 has no incentive to deviate from \(s_{2}.\)

4.2 General properties of Nash equilibria

We now proceed to establish four general properties of Nash equilibria. Given that we already have a complete characterization of scattered Nash equilibria in Sect. 3, the focus of three of these properties is on clustered Nash equilibria.

First, in any clustered Nash equilibrium, there cannot be more than two candidates announcing the same platform. That is, for any Nash equilibrium placement \(c=(m,s,t),\) we must have \(t_{i}\in \{1,2\}\) for all \(i\in J.\) This implies that \(m\ge 2\) (since we have \(n\ge 3).\)

Second, if \(c=(m,s,t)\) is a clustered Nash equilibrium, and there are two candidates who have announced the same platform \(s_{j},\) then \(s_{j}\) must be precisely at the mid-point of its neighborhood \(Y_{j}(c).\)

Third, the analogue of part (B) of Theorem 1 is valid for all Nash equilibria. To describe it, let us define two key concepts. Given a placement \(c=(m,s,t),\) an empty stretch is any arc that does not contain any \(s_{i}\) for \(i\in J.\) The arc length of an empty stretch of maximum arc length (among all empty stretches) is denoted by x(c). We define for \(m\ge 2,\)

$$\begin{aligned} y(c)=\min \left\{ \frac{y_{1}(c)}{t_{1}},\ldots ,\frac{y_{m}(c)}{t_{m}}\right\} . \end{aligned}$$

If \(c=(m,s,t)\) is any Nash equilibrium, then it must be the case that \( y(c)\ge x(c).\)

Fourth, if \(c=(m,s,t)\) is any clustered Nash equilibrium, then it must be the case that \(y(c)=x(c).\)

Proposition 2

If a placement \(c=(m,s,t)\) is a Nash equilibrium,  then the following properties must hold : 

  1. (i)

    \(t_{i}\in \{1,2\}\) for all \(i\in J,\) and \(m\ge 2.\)

  2. (ii)

    For each location \(s_{i}\) with \(t_{i}=2,\)\(s_{i}\) must be at the mid point of its neighborhood \(Y_{i}(c).\)

  3. (iii)

    \(y(c)\ge x(c).\)

  4. (iv)

    If the placement c is clustered,  then \(y(c)=x(c);\) further,  for each location \(s_{i}\) with \(t_{i}=2,\) each of the two empty stretches of \( Y_{i}(c) \) has arc length x(c).

Proof

(i) Let \(c=(m,s,t)\) be a Nash equilibrium placement. Suppose there is some platform \(s_{j}\) which has been announced by \(k\ge 3\) candidates.

First consider the case where \(k=n;\) that is, all players have announced the same platform. Then the neighborhood of each player is the entire circle, and the payoff to each player is (1 / n). If one player deviates to another platform she will still have the entire circle as her neighborhood, but her payoff will be (1 / 2). Since \(n\ge 3,\) a deviation is worthwhile. Hence c is not a Nash equilibrium.

Now suppose \(3\le k<n,\) so that \(m\ge 2.\) The neighborhood of each player who has announced platform \(s_{j}\) is \(Y_{j}(c),\) and the payoff of each player who has announced platform \(s_{j}\) is \( p_{j}(c)=y_{j}(c)/2t_{j}=y_{j}(c)/2k.\) The neighborhood \(Y_{j}(c)\) consists of two empty stretches (and the point \(s_{j}).\) Call these empty stretches \( Y_{j1}(c)\) and \(Y_{j2}(c).\) Comparing the arc lengths of these empty stretches, pick an empty stretch whose arc length is not smaller than the arc length of the other empty stretch, and call it \(Y_{j1}(c)\) (without loss of generality).

If one of the k players who has announced platform \(s_{j}\) deviates to a platform \(s_{j}^{\prime }\in Y_{j1}(c),\) then this player’s neighborhood becomes \(Y_{j1}(c)\) and so this player’s payoff becomes half the arc length of \(Y_{j1}(c),\) which will therefore be greater than or equal to \( [y_{j}(c)/4].\) Since \(k\ge 3,\)

$$\begin{aligned} \frac{y_{j}(c)}{4}>\frac{y_{j}(c)}{2k}. \end{aligned}$$

Hence, the player benefits by deviating from \(s_{j}\) to \(s_{j}^{\prime }.\) So c cannot be a Nash equilibrium.

This establishes that \(t_{i}\in \{1,2\}\) for all \(i\in J.\) If \(m=1,\) then there would be at least three players who announce the same platform, since \( n\ge 3.\) But this is ruled out by the property just established. Thus, we must have \(m\ge 2.\)

(ii) Let \(c=(m,s,t)\) be a Nash equilibrium placement. Suppose there is some platform \(s_{j}\) which has been announced by \(t_{j}=2\) players. The neighborhood of each player who has announced platform \(s_{j}\) is \(Y_{j}(c),\) and the payoff of each player who has announced platform \(s_{j}\) is \( p_{j}(c)=y_{j}(c)/2t_{j}=y_{j}(c)/4.\) The neighborhood \(Y_{j}(c)\) consists of two empty stretches (and the point \(s_{j}).\) Call these empty stretches \( Y_{j1}(c)\) and \(Y_{j2}(c).\) Suppose the arc lengths of these two empty stretches are not the same. Pick the empty stretch whose arc length is the larger of the two, and call it \(Y_{j1}(c)\) (without loss of generality).

If one of the 2 players who has announced platform \(s_{j}\) deviates to a location \(s_{j}^{\prime }\in Y_{j1}(c),\) then this player’s neighborhood becomes \(Y_{j1}(c)\) and so this player’s payoff becomes half the arc length of \(Y_{j1}(c),\) which will therefore be greater than \([y_{j}(c)/4].\) Hence, the player benefits by deviating from \(s_{j}\) to \(s_{j}^{\prime }.\) So c cannot be a Nash equilibrium. This establishes that the arc lengths of \( Y_{j1}(c)\) and \(Y_{j2}(c)\) must be the same. That is, \(s_{j}\) must be at the mid point of its neighborhood \(Y_{j}(c).\)

(iii) Let \(c=(m,s,t)\) be a Nash equilibrium placement. Suppose \(y(c)<x(c).\) By the definition of y(c),  there is some platform \(s_{h}\) (with \(h\in J)\) such that \(y_{h}(c)/t_{h}=y(c)<x(c).\) Then the payoff of each player who has announced platform \(s_{h}\) is:

$$\begin{aligned} p_{h}(c)=y_{h}(c)/2t_{h}<(x(c)/2). \end{aligned}$$

Now a player (who has announced platform \(s_{h})\) can relocate from \(s_{h}\) to a platform \(s_{h}^{\prime }\) in an empty stretch with arc length x(c). Then the new neighborhood of this player is the empty stretch with arc length x(c) (no longer empty after it is occupied by this player). Consequently, the new payoff to this player is (x(c) / 2). Hence, this player benefits by deviating from \(s_{j}\) to \(s_{j}^{\prime }.\) So, the placement c cannot be a Nash equilibrium. This contradiction establishes that \(y(c)\ge x(c).\)

(iv) Let \(c=(m,s,t)\) be a Nash equilibrium placement, where the placement is clustered. Then, there is some platform \(s_{h}\) (with \(h\in J)\) which has been announced by more than one player. By (i), there are two players who have announced platform \(s_{j}.\) Thus, by definition of y(c),  we have:

$$\begin{aligned} \frac{y_{j}(c)}{2}\ge y(c)\ge x(c) \end{aligned}$$
(27)

where the second inequality in (27) follows from the result (iii).

The neighborhood of each player who has announced platform \(s_{j}\) is \( Y_{j}(c),\) and \(Y_{j}(c)\) consists of two empty stretches (and the point \( s_{j}).\)Thus,

$$\begin{aligned} y_{j}(c)\le 2x(c). \end{aligned}$$
(28)

Combining (27) and (28), we obtain:

$$\begin{aligned} \frac{y_{j}(c)}{2}\ge y(c)\ge x(c)\ge \frac{y_{j}(c)}{2} \end{aligned}$$
(29)

so that equality must hold in place of each inequality in (29). Thus, \(y(c)=x(c)\) must hold. Further, \(y_{j}(c)=2x(c),\) so by result (ii), each of the two (equal) empty stretches of \(Y_{j}(c)\) constitutes a largest empty stretch in the placement c. \(\square \)

Remark

Result (i) in Proposition 2 sheds interesting light on the median voter theorem which says that, when the available platforms are points on a straight line, two candidates will invariably choose the same platform (the median). What Example 2 and Proposition 2(i) show is that in an electoral democracy with voter preference on a circle of available platforms, two candidates may choose the same political agenda, but there will never be three candidates doing so.

4.3 A property of clustered Nash equilibria

Results (i), (ii) and (iv) of Proposition 2 place rather strong restrictions on clustered Nash equilibria. If \(n=3\) and \(c=(m,s,t)\) is a clustered Nash equilibrium, then there must be precisely two platforms. One candidate would announce one platform; call this \(s_{1}.\) The other two candidates would both announce a distinct platform, \(s_{2}.\) Given the platform \(s_{2},\) property (ii) of Proposition 2 demands that it be at the mid-point of its neighborhood. This implies that \(s_{1}\) must be diametrically opposite to \( s_{2}\) on the circle. In other words, Example 2 is essentially the only example of a clustered Nash equilibrium when \(n=3.\) In particular, the distribution of payoffs in any clustered Nash equilibrium must be:

$$\begin{aligned} \Omega =\left\{ \frac{1}{4},\frac{1}{4},\frac{1}{2}\right\} . \end{aligned}$$

More generally, for arbitrary \(n\ge 3,\) we note the following property for clustered Nash equilibria, which should be compared to Corollary 1 for scattered Nash equilibria. It is a useful property which is helpful in studying the distribution of payoffs in arbitrary Nash equilibria, which we take up in the next subsection.

Corollary 2

If a placement \(c=(m,s,t)\) is a clustered Nash equilibrium,  then : 

$$\begin{aligned} x(c)\ge \frac{1}{n-1}. \end{aligned}$$
(30)

Proof

Given the placement \(c=(m,s,t),\) there are m empty stretches. Since the arc lengths of the m empty stretches must add up to 1 (the circumference of the circle), we must have:

$$\begin{aligned} mx(c)\ge 1. \end{aligned}$$
(31)

Since the placement \(c=(m,s,t)\) is a clustered Nash equilibrium, there is at least one platform \(s_{k},\) which is announced by multiple candidates; by Proposition 2(i), we know in fact that there are precisely two candidates who announce the platform \(s_{k}.\) The remaining \((n-2)\) candidates announce the remaining \((m-1)\) platforms. In order that each of the remaining \((m-1)\) platforms is announced by at least one candidate each, we must have:

$$\begin{aligned} n-2\ge m-1 \end{aligned}$$

and consequently \(n-1\ge m.\) Using this in (31), we obtain (30). \(\square \)

4.4 Distribution of payoffs in Nash equilibria

We now present a generalization of Proposition 1, extending that result to all Nash equilibria. It will be noted that in our result, the bounds on the payoffs are determined in terms of n,  the number of players (candidates), and independent of m,  the number of platforms announced in a Nash equilibrium. Thus, one needs no information about the equilibrium placements to make use of the bounds.

Theorem 2

If a placement \(c=(m,s,t)\) is a Nash equilibrium,  then : 

$$\begin{aligned} \frac{1}{2(n-1)}\le p_{i}(c)\le \frac{2}{n+1}\quad \text {for all }i\in J=\{1,\ldots ,m\}. \end{aligned}$$
(DOP')

Proof

Given the placement \(c=(m,s,t),\) recall that the arc length of the empty stretch of maximum arc length is denoted by x(c),  and we define:

$$\begin{aligned} y(c)=\min \left\{ \frac{y_{1}(c)}{t_{1}},\ldots ,\frac{y_{m}(c)}{t_{m}}\right\} ;\quad z(c)=\max \left\{ \frac{y_{1}(c)}{t_{1}},\ldots ,\frac{y_{m}(c)}{t_{m}} \right\} . \end{aligned}$$

Note that, given the placement \(c=(m,s,t),\) there are m empty stretches. Each empty stretch is contained in the neighborhoods of two distinct platforms. Since the arc lengths of the m empty stretches must add up to 1 (the circumference of the circle), the arc lengths of the neighborhoods of the m platforms must add up to 2;  that is:

$$\begin{aligned} \sum _{i=1}^{m}y_{i}(c)=2. \end{aligned}$$
(32)

There is \(k\in J=\{1,\ldots ,m\}\) such that \((y_{k}(c)/t_{k})=z(c).\) By Proposition 2(i), there are two possibilities (a) \(t_{k}=2;\) (b) \(t_{k}=1.\) In case (a), we know by Proposition 2(iv) that \(y_{k}(c)=2x(c).\) Thus, we have:

$$\begin{aligned} z(c)=y_{k}(c)/t_{k}=2x(c)/t_{k}=x(c) \end{aligned}$$
(33)

and so by definition of z(c), 

$$\begin{aligned} y_{i}(c)/t_{i}\le z(c)=x(c)\quad \text {for all }i\in J. \end{aligned}$$
(34)

But, we also know from the definition of y(c) and Proposition 2(iii) that:

$$\begin{aligned} y_{i}(c)/t_{i}\ge y(c)=x(c)\quad \text {for all }i\in J. \end{aligned}$$
(35)

Thus, we have:

$$\begin{aligned} y_{i}(c)/t_{i}=y(c)=z(c)=x(c)\quad \text {for all }i\in J. \end{aligned}$$
(36)

Using (36) in (32), we obtain:

$$\begin{aligned} 2=\sum _{i=1}^{m}y_{i}(c)=\sum _{i=1}^{m}t_{i}\left[ \frac{y_{i}(c)}{t_{i}} \right] =\sum _{i=1}^{m}t_{i}x(c)=nx(c). \end{aligned}$$
(37)

Thus, (34) and (37) yield:

$$\begin{aligned} p_{i}(c)=y_{i}(c)/2t_{i}\le x(c)/2=(1/2)\left[ \frac{2}{n}\right] =\left[ \frac{1}{n}\right] \quad \text {for all }i\in J. \end{aligned}$$

Since \(n\ge 3,\) we have \((1/n)<[2/(n+1)],\) and so the right-hand inequality in (DOP’) holds in case (a).

In case (b), we have \(t_{k}=1,\) and so:

$$\begin{aligned} z(c)=y_{k}(c)/t_{k}=y_{k}(c)\le 2x(c). \end{aligned}$$
(38)

Using (32) and (38), and Proposition 2(iii), we get:

$$\begin{aligned} 2= & {} y_{k}(c)+\sum _{i\ne k}y_{i}(c) \nonumber \\\ge & {} z(c)+\sum _{i\ne k}t_{i}y(c) \nonumber \\\ge & {} z(c)+\sum _{i\ne k}t_{i}x(c) \nonumber \\\ge & {} z(c)+\sum _{i\ne k}t_{i}\left\{ z(c)/2\right\} \nonumber \\= & {} z(c)\left[ 1+\sum _{i\ne k}(t_{i}/2)\right] \nonumber \\= & {} z(c)\left[ \left\{ 1-(t_{k}/2)\right\} +\sum _{i=1}^{m}(t_{i}/2)\right] \nonumber \\= & {} z(c)\left[ (1/2)+(n/2)\right] \end{aligned}$$
(39)

where we used (32) in line 1, Proposition 2(iii) in line 3 and (38) in line 4. This yields:

$$\begin{aligned} z(c)\le \left[ \frac{4}{n+1}\right] \end{aligned}$$

so that:

$$\begin{aligned} p_{i}(c)=y_{i}(c)/2t_{i}\le z(c)/2\le \left[ \frac{2}{n+1}\right] \quad \text {for all }i\in J. \end{aligned}$$

This establishes the right-hand inequality in (DOP’) in case (b).

We proceed now to establish the left-hand inequality in (DOP’). We divide up our analysis into two cases (a) \(c=(m,s,t)\) is a scattered Nash equilibrium; and (b) \(c=(m,s,t)\) is a clustered Nash equilibrium. In case (a), the result follows directly from Proposition 1. We now analyze case (b). In case (b), we note that by Corollary 2,

$$\begin{aligned} x(c)\ge \frac{1}{n-1}. \end{aligned}$$
(40)

Thus, for all platforms \(s_{i}\) (with \(i\in J)\) we must have:

$$\begin{aligned} p_{i}(c)=y_{i}(c)/2t_{i}\ge y(c)/2\ge x(c)/2\ge \left[ \frac{1}{2(n-1)} \right] \end{aligned}$$

where the second inequality follows from Proposition 2(iii), and the last inequality from (40). This establishes the left-hand inequality in (DOP’) in case (b). \(\square \)

4.5 Best bounds in the distribution of payoffs

In the context of Example 2 with \(n=3,\) we see that the lower bound of \( [1/2(n-1)]=\)\(25\%\) in (DOP’), and the upper bound of \([2/(n+1)]=\)\(50\%\) in (DOP’) are attained. This shows that the bounds in (DOP’) are the best possible for \(n=3.\)

This raises the question of whether, in general, the bounds given in Theorem 2 (for the distribution of payoffs in an arbitrary Nash equilibrium) are the best possible. In this subsection, we provide answers to this question separately for the lower bound in (DOP’) and the upper bound in (DOP’).

4.5.1 Lower bound in the distribution of payoffs

We present an example to show that the lower bound of the interval, given in both Proposition 1 and Theorem 2, is the “best possible” in the sense that given any \(n\ge 3,\) it is possible to construct a scattered Nash equilibrium with n players, such that the payoff of some player in that equilibrium is exactly equal to \([1/2(n-1)]\).

Example 3

Let \(n\ge 3\) be given. Divide the circumference of the circle into \((n-1)\) equal parts. This will generate \((n-1)\) points on the circle, equally spaced.

Place the platforms \(\{s_{1},\ldots ,s_{n-1}\}\) at these \((n-1)\) points. To make the instruction more precise, place \(s_{1}\) at any one of the \((n-1)\) points. Place \(s_{2}\) at the next available point (of the remaining \(n-2\) points) as you go clockwise around the circle, starting from \(s_{1}.\) Continue similarly with \(s_{3},\ldots ,s_{n-1}.\) Finally, place the platform \( s_{n}\) at the mid point of the arc \(arc(s_{n-1},s_{1}).\) For an illustration of this procedure for \(n=5,\) see Fig. 6.

Fig. 6
figure 6

Lower bound of DOP (Example 3)

Full size image

The largest empty stretch is:

$$\begin{aligned} x(s)=\frac{1}{n-1}. \end{aligned}$$
(41)

The neighborhood with the smallest arc length is:

$$\begin{aligned} y(s)=y_{n}(s)=\frac{1}{n-1}. \end{aligned}$$
(42)

Thus, by Theorem 1, the placement \(s=\)\(\{s_{1},\ldots ,s_{n}\}\) described above is a Nash equilibrium.

Using (42), the payoff of player n in this Nash equilibrium is:

$$\begin{aligned} p_{n}(s)=\frac{y_{n}(s)}{2}=\frac{1}{2(n-1)} \end{aligned}$$
(43)

which is precisely the lower bound of \(\Pi \) in (DOP’).

Remark

The lower bound in (DOP’) is also attained when \(n=2,\) since in that case, the payoff of each player in any Nash equilibrium is equal to (1 / 2).

4.5.2 Upper bound in the distribution of payoffs

We now focus on the question whether the upper bound of the interval, given in both Proposition 1 and Theorem 2, is the “best possible”. When \(n\ge 3, \) and n is odd, we show that the upper bound in (DOP’) is the best possible in the following sense. Given \(n\ge 3\) and n odd, and any positive number \(\beta \) smaller than this upper bound:

$$\begin{aligned} \beta <\frac{2}{n+1} \end{aligned}$$
(44)

we can construct an example of a scattered Nash equilibrium, such that the payoff of some player exceeds \(\beta \) in that equilibrium. Thus, \([2/(n+1)]\) is not just an upper bound, it is the least upper bound of the payoff of the players in a Nash equilibrium, for \(n\ge 3\) and n odd.

Unlike the result for the lower bound, we do not claim that the bound \( [2/(n+1)]\) is actually attained in a Nash equilibrium for \(n\ge 3\) with n odd. This is principally because we work with scattered Nash equilibrium, for which there is a complete characterization provided by Theorem 1. The partial characterization result (in Proposition 2) of all Nash equilibria (scattered and clustered) does not provide us with a sufficient condition to check whether a proposed placement is a Nash equilibrium. It is certainly possible that the least upper bound \([2/(n+1)]\) is actually attained in a clustered Nash equilibrium, but we have not pursued this issue. [For \(n=3,\) this least upper bound is attained in a clustered Nash equilibrium, as we checked in Example 2].

Example 4

(The result for n odd) Let \(n\ge 3\) be given, with n odd. Then \((n+1)\) is even, and \(N=(n+1)/2\) is an integer, greater than or equal to 2. Let \(\beta \) be any positive number satisfying (44). Denote:

$$\begin{aligned} \varepsilon \equiv \frac{2}{n+1}-\beta \in \left( 0,\frac{2}{n+1}\right) . \end{aligned}$$
(45)

Divide the circumference of the circle into N equal parts. This will generate N points on the circle, equally spaced.

Place the platform \(s_{1}\) at any one of the N points. Place \(s_{3}\) at the next available point (of the remaining \(N-1\) points) as you go clockwise around the circle, starting from \(s_{1}.\) Continue similarly with \( s_{5},\ldots ,s_{n}.\) Notice that only the odd numbered platforms have been located at these N points. For example, with \(n=5,\) we will have \(N=3,\) and this procedure will place platforms \(s_{1},s_{3},\) and \(s_{5}\) at the \( N=3\) available points, moving clockwise from \(s_{1}\) to \(s_{3}\) and then clockwise from \(s_{3}\) to \(s_{5}.\) Observe that:

$$\begin{aligned} arc(s_{1},s_{3})=\cdots =arc(s_{n-2},s_{n})=arc(s_{n},s_{1})=\frac{1}{N}= \frac{2}{n+1}. \end{aligned}$$
(46)

For an illustration of this procedure for \(n=5,\) see Fig. 7.

We now proceed to place the even numbered platforms as follows. Place \(s_{2}\) in \(arc(s_{1},s_{3})\) so that \(arc(s_{2},s_{3})=\varepsilon .\) Since:

$$\begin{aligned} \varepsilon <\frac{2}{n+1}=\frac{1}{N}=arc(s_{1},s_{3}) \end{aligned}$$

this is certainly possible. Similarly, place \(s_{k}\) in \( arc(s_{k-1},s_{k+1}) \) with \(arc(s_{k},s_{k+1})=\varepsilon \) for all \(k\in \{2,4,\ldots ,n-1\}.\) See the figure again for an illustration of this procedure for \(n=5.\)

Fig. 7
figure 7

Upper bound of DOP (Example 4)

Full size image

It is now easy to check that:

$$\begin{aligned} y_{j}(s)=(1/N)\quad \text {for all }j\in \{2,3,\ldots ,n-1\}. \end{aligned}$$
(47)

Further, we have:

$$\begin{aligned} y_{n}(s)=(1/N)+\varepsilon ;\quad y_{1}(s)=(2/N)-\varepsilon . \end{aligned}$$
(48)

Using (45), (47) and (48), the neighborhood with the smallest arc length is:

$$\begin{aligned} y(s)=\frac{1}{N}. \end{aligned}$$
(49)

The largest empty stretch is:

$$\begin{aligned} x(s)=a(s_{n},s_{1})=\frac{1}{N}. \end{aligned}$$
(50)

Thus, by Theorem 1, the placement \(s=\)\(\{s_{1},\ldots ,s_{n}\}\) described above is a Nash equilibrium.

Using (48), the payoff of player 1 in this Nash equilibrium is:

$$\begin{aligned} p_{1}(s)=\frac{y_{1}(s)}{2}=\frac{1}{N}-\frac{\varepsilon }{2}=\frac{2}{n+1}- \frac{\varepsilon }{2}>\frac{2}{n+1}-\varepsilon =\beta . \end{aligned}$$
(51)

This shows that \([2/(n+1)]\) is the least upper bound of the payoff of the players in a Nash equilibrium, for \(n\ge 3\) and n odd.

Example 5

(The result for n even) For \(n\ge 3\) and n even, the upper bound in (DOP’) is not necessarily a least upper bound of the payoff of the players in a Nash equilibrium. For instance, with \(n=4,\) it can be shown that an upper bound of the payoff of the players in any scattered Nash equilibrium is equal to:

$$\begin{aligned} \frac{1}{3}=\frac{2}{n+2}<\frac{2}{n+1}=\frac{2}{5}. \end{aligned}$$
(52)

To see this, we proceed as follows. With \(n=4,\) let \(s= \{s_{1},s_{2},s_{3},s_{4}\}\) be an arbitrary scattered Nash equilibrium. Recall that the arc length of the empty stretch of maximum arc length is denoted by x(s),  and we define:

$$\begin{aligned} y(s)=\min \{y_{1}(s),\ldots ,y_{4}(s)\};\quad z(s)=\max \{y_{1}(s),\ldots ,y_{4}(s)\}. \end{aligned}$$

There is \(k\in I=\{1,\ldots ,4\}\) such that \(y_{k}(s)=z(s).\) Note that by Theorem 1, we have:

$$\begin{aligned} z(s)=y_{k}(s)\le 2x(s)\le 2y(s). \end{aligned}$$
(53)

Without loss of generality, we can take \(k=1.\) The neighbourhood \(Y_{1}(s)\) defines \(s_{j}\) and \(s_{q}\) with \(s_{j}\ne s_{q}\) such that:

$$\begin{aligned} Y_{1}(s)=arc(s_{j},s_{q}). \end{aligned}$$
(54)

Again without loss of generality, we can take \(j=2\) and \(q=3.\) It then follows that \(s_{4}\) belongs to the arc \(arc(s_{q},s_{j})=arc(s_{3},s_{2}).\) Thus,

$$\begin{aligned} Y_{4}(s)=arc(s_{3},s_{2}). \end{aligned}$$
(55)

See Fig. 8 for an illustration of the locations \(s_{1},s_{2},s_{3},\) and \( s_{4}.\)

Fig. 8
figure 8

Upper bound of DOP for n = 4 (Example 5)

Full size image

The neighborhood \(Y_{1}(s)\) consists of two empty stretches (and the point \( s_{1}),\)\(arc(s_{2},s_{1})\) and \(arc(s_{1},s_{3}).\)The corresponding arc lengths can be written as \(\left[ y_{1}(s)/2\right] +\varepsilon \) and \( \left[ y_{1}(s)/2\right] -\varepsilon \) respectively, where \(-\left[ y_{1}(s)/2\right]<\varepsilon <\left[ y_{1}(s)/2\right] .\) Similarly, the neighborhood \(Y_{4}(s)\) consists of two empty stretches (and the point \( s_{4}),\)\(arc(s_{3},s_{4})\) and \(arc(s_{4},s_{2}).\) The corresponding arc lengths can be written as \(\left[ y_{4}(s)/2\right] +\eta \) and \(\left[ y_{4}(s)/2\right] -\eta \) respectively, where \(-\left[ y_{4}(s)/2\right]<\eta <\left[ y_{4}(s)/2\right] .\) Note that \(\varepsilon \) and \(\eta \) are being allowed to be either negative or positive or zero. [In Fig. 8, \( \varepsilon =0\) while \(\eta <0].\)

With this notation we have:

$$\begin{aligned} y_{2}(s)=a(s_{4},s_{2})+a(s_{2},s_{1})=\left[ y_{4}(s)/2\right] -\eta +\left[ y_{1}(s)/2\right] +\varepsilon \end{aligned}$$
(56)

and:

$$\begin{aligned} y_{3}(s)=a(s_{1},s_{3})+a(s_{3},s_{4})=\left[ y_{1}(s)/2\right] -\varepsilon +\left[ y_{4}(s)/2\right] +\eta . \end{aligned}$$
(57)

Using (56) and (57), we can now write:

$$\begin{aligned} 2= & {} y_{1}(s)+y_{2}(s)+y_{3}(s)+y_{4}(s) \nonumber \\= & {} y_{1}(s)+\{\left[ y_{4}(s)/2\right] -\eta +\left[ y_{1}(s)/2\right] +\varepsilon \}+ \nonumber \\&\{\left[ y_{1}(s)/2\right] -\varepsilon +\left[ y_{4}(s)/2\right] +\eta \}+y_{4}(s) \nonumber \\= & {} 2y_{1}(s)+2y_{4}(s) \nonumber \\\ge & {} 2z(s)+2y(s) \nonumber \\\ge & {} 2z(s)+z(s) \nonumber \\= & {} 3z(s) \end{aligned}$$
(58)

where we have used (53) in the last but one line of (58). Thus, we obtain:

$$\begin{aligned} z(s)\le \frac{2}{3} \end{aligned}$$
(59)

and so:

$$\begin{aligned} p_{i}(s)=\frac{y_{i}(s)}{2}\le \frac{z(s)}{2}\le \frac{1}{3}\quad \text {for all }i\in I=\{1,2,3,4\}. \end{aligned}$$
(60)

5 Concluding remarks

This paper was an exercise in abstract geometric reasoning. When drawing on it for more applied models it will be important to add relevant complications. In applying it to retail trading and location choice or brand proliferation it will be important to introduce the option to vary prices. When applying to electoral politics we need to account for the fact that candidates typically come with a prior record of ideology and a sharp shift in this can cause voter dissonance. In all these models there may also be a case for considering sequential moves which would convert this into an extensive-form game. It is hoped that the simple model presented here can be of value for such extensions.