# How Many Freemasons Are There? The Consensus Voting Mechanism in Metric Spaces

We study the evolution of a social group when admission to the group is determined via consensus or unanimity voting. In each time period, two candidates apply for membership and a candidate is selected if and only if all the current group members agree. We apply the spatial theory of voting where group members and candidates are located in a metric space and each member votes for its closest (most similar) candidate. Our interest focuses on the expected cardinality of the group after T time periods. To evaluate this we study the geometry inherent in dynamic consensus voting over a metric space. This allows us to develop a set of techniques for lower bounding and upper bounding the expected cardinality of a group. We specialize these methods for two-dimensional metric spaces. For the unit ball the expected cardinality of the group after T time periods is Θ(T^1/8). In sharp contrast, for the unit square the expected cardinality is at least Ω(ln T) but at most O(ln T ·lnln T ).

There are no comments yet.

## Authors

• 4 publications
• 11 publications
• ### Favorite-Candidate Voting for Eliminating the Least Popular Candidate in Metric Spaces

We study single-candidate voting embedded in a metric space, where both ...
11/27/2019 ∙ by Xujin Chen, et al. ∙ 0

• ### Between Discord and Deadlock: Consensus Under a Deadline

Committees are an important scenario for reaching consensus. Beyond stan...
05/17/2019 ∙ by Marina Bannikova, et al. ∙ 0

• ### An Uncertainty- and Collusion-Proof Voting Consensus Mechanism in Blockchain

Though voting-based consensus algorithms in Blockchain outperform proof-...
12/25/2019 ∙ by Shengling Wang, et al. ∙ 0

• ### Representative Proxy Voting

We study a model of proxy voting where the candidates, voters, and proxi...
12/12/2020 ∙ by Elliot Anshelevich, et al. ∙ 0

• ### Justified Representation in Approval-Based Committee Voting

We consider approval-based committee voting, i.e. the setting where each...
07/31/2014 ∙ by Haris Aziz, et al. ∙ 0

• ### Of the People: Voting Is More Effective with Representative Candidates

In light of the classic impossibility results of Arrow and Gibbard and S...
05/04/2017 ∙ by Yu Cheng, et al. ∙ 0

• ### On Voting and Facility Location

We study mechanisms for candidate selection that seek to minimize the so...
12/18/2015 ∙ by Michal Feldman, et al. ∙ 0

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1. Introduction

This paper studies the evolution of social groups over time. In an exclusive social group, the existing group members vote to determine whether or not to admit a new member. Familiar examples include the freemasons, fraternities, membership-run sports and social clubs, acceptance to a condominium, as well as academia. To analyze the inherent dynamics we use the model of Alon, Feldman, Mansour, Oren and Tennenholtz (dyn, ). In each time period, two candidates apply for membership and the current members vote to decide if either or none of them is acceptable. The spatial model of voting is used: each person is located uniformly at random in a metric space and each group member votes for the candidate closest to them.

Alon et al. (dyn, ) analyze social group dynamics in a one-dimensional metric space, specifically, the unit interval . They examine how outcomes vary under different winner determination rules, in particular, majority voting and consensus voting. In consensus voting or unanimity voting a candidate is elected if and only if the group members agree unanimously. Equivalently, every member may veto a potential candidate.

Our interest lies in the evolution of the group size under consensus voting; that is, what is the expected cardinality of the social group after time periods? In the one-dimensional setting the answer is quite simple. There, Alon et al. (dyn, ) show that under consensus voting if a candidate is elected in round

then, with high probability, it is within a distance

of one endpoint of the interval. Because the winning candidate must be closer to all group members than the losing candidate, both candidates must therefore be near the endpoints. This occurs with probability . As a consequence, in a one-dimensional metric space, the expected size of the social group after time periods is . Here we use the notation if both and , where if for some constant .

Bounding the expected group size in higher-dimensional metric spaces is more complex and is the focus of this paper. To do this, we begin in Section 2 by examining the geometric aspects of consensus voting in higher-dimensional metric spaces. More concretely, we explain how winner determination relates to the convex hull of the group members and the Voronoi cells formed by the candidates. This geometric understanding enables us to construct, in Section 3, a set of techniques, based upon cap methods

in probability theory, that allow for the upper bounding and lower bounding of expected group size. In Sections

4 and 5, we specialize these techniques to two-dimensions for application on the fundamental special cases of the unit square and the unit ball. Specifically, for the unit square we show the following lower and upper bounds on expected group size.

###### Theorem 1.1 ().

Let the metric space be the unit square . Then the expected cardinality of the social group after periods is bounded by  .

Thus, expected group size for the two-dimensional unit square is comparable to that of the one-dimensional interval. Surprisingly, there is a dramatic difference in expected group size between the unit square and the unit ball. For the unit ball, the expected group size evolves not logarithmically but polynomially with time.

###### Theorem 1.2 ().

Let the metric space be the unit ball . Then the expected cardinality of the social group after periods is  .

### 1.1. Background and Related Work

Here we discuss some background on the spatial model and consensus voting. The spatial model of voting utilized in this paper dates back nearly a century to the celebrated work of Hotelling (harold1929stability, ). His objective was to study the division of a market in a duopoly when consumers are distributed over a one-dimension space, but he noted his work had intriguing implications for electoral systems. Specifically, in a two-party system there is an incentive for the political platforms of the two parties to converge. This was formalized in the median voter theorem of Black (Bla48, ): in a one-dimensional ideological space the location of the median voter induces a Condorcet winner111A candidate is a Condorcet winner if, in a pairwise majority vote, it beats every other candidate., given single-peaked voting preferences. The traditional voting assumption in a metric space is proximity voting where each voter supports its closest candidate; observe that proximity voting gives single-peaked preferences.

The spatial model of voting was formally developed by Downs (Dow57, ) in 1957, again in a one-dimensional metric space. Davis, Hinich, and Ordeshook (DHO70, ) expounded on practical necessity of moving beyond just one dimension. Interestingly, they observed that in two-dimensional metric spaces, a Condorcet winner is not guaranteed even with proximity voting. Of particular relevance here is their finding that, in dynamic elections, the order in which candidates are considered can fundamentally affect the final outcome (Bla48, ; DHO70, ).

There is now a vast literation on spatial voting, especially concerning the strategic aspects of simple majority voting; see, for example, the books (enelow1984spatial, ; arrow1990advances, ; merrill1999unified, ; Poo05, ; Sch07, ). There has also been a vigorous debate concerning whether voter utility functions in spatial models should be distance-based (such as the standard assumption of proximity voting used here), relational (e.g. directional voting (RS89, )), or combinations thereof (merrill1999unified, ). This debate has been philosophical, theoretical and experimental (Gro73, ; Mat79, ; LK00, ; TV08, ; Cla09, ; LP10, )

. Recently there has also been a large amount of interest in the spatial model by the artificial intelligence community

(ABJ15, ; FFG16, ; AJ17, ; SE17, ; BLS19, ). It is interesting to juxtapose these modern potential applications with the original motivations suggested by Black (Bla48, ), such as the administration of colonies!

Consensus is one of the oldest group decision-making procedures. In addition to exclusive social groups, it is familiar in a range of disparate settings, including judicial verdicts, Japanese corporate governance (vogel1975modern, ), and even decision making in religious groups, such as the Quakers (hare1973group, ). From a theoretic perspective, consensus voting in a metric spaces has also been studied by Colomer (colomer1999geometry, ) who highlights the importance the initial set of voters can have on outcomes in a dynamic setting.

## 2. The Geometry of Consensus Voting

In this section, we present a simple geometric interpretation of a single election using consensus voting in the spatial model. In the subsequent sections, we will apply this understanding, developed for the static case, to study the dynamic model. Specifically, we examine how a group grows over time when admission to the group is via a sequence of consensus elections.

Let denote the initial set of group members222We may take the cardinality of the initial group to be any constant . In particular, we may assume ., selected uniformly and independently from a metric space . In the consensus voting mechanism, for each round , a finite set of candidates applies for membership. Members of the group at the start of round , denoted , are eligible to vote. Assuming the spatial theory of voting, each group member will vote for the candidate who is closest to her in the metric space. That is, member votes for candidate if and only if for every candidate . Under the consensus (unanimity) voting rule, if every group member selects the candidate then is accepted to the group and ; otherwise, if the group does not vote unanimously then no candidate wins selection and .

As stated, to study how group size evolves over time, our first task is to develop a more precise understanding of when a candidate will be selected under consensus voting in a single election. Fortunately, there is a nice geometric characterization for this property in terms of the Voronoi cells (regions) formed in the metric space by the candidates (points) . Specifically, the Voronoi cell associated with point is . We will see that the convex hull of the group members , which we denote , plays an important role in winner determination. The characterization theorem for the property that a candidate is selected under the consensus voting mechanism is then:

###### Theorem 2.1 ().

Let be the candidates and let be the Voronoi cells on generated by . Then there is a winner under consensus voting if and only if for some candidate .

###### Proof.

Assume for some candidate . Then, for every voter , we have for any other candidate . Hence, every voter prefers candidate over all the other candidates. Thus candidate is selected. Conversely, assume that candidate is selected. Then, by definition of consensus voting, each voter voted for . Thus for all . Ergo, . ∎

Several useful facts can be derived from this characterization. These facts are stated in the subsequent corollary and lemma.

###### Corollary 2.2 ().

Let be set of candidates. If there is a candidate accepted with , then the same candidate is also accepted with for any convex set .

###### Proof.

By Theorem 2.1, a candidate is accepted if and only if the current convex hull is entirely contained within one of Voronoi regions, , generated by the candidates. Clearly, if then . Therefore, if there is an acceptance with then there would be an acceptance with . The result follows immediately. ∎

The next lemma requires the following definition: let denote the Euclidean ball centred at with radius . Furthermore, we denote by  the set of vertices (extreme points) of the convex hull  of the group members. Observe that and that .

###### Lemma 2.3 ().

Let be current set of group members and be its convex hull. Let be the current candidates. Under consensus, there is a winning candidate if and only if such that

 wi∈⋂k∈[n]∖i⋂vj∈∂SB(vj,wk)
###### Proof of Lemma 2.3.

If there is a consensus then there is a who is selected. Hence among all candidates, is closest to each group member. That is, , for each group member and each candidate . It follows that

 wi∈⋂k∈[n]∖i⋂vj∈GB(vj,wk)⊆⋂k∈[n]∖i⋂vj∈∂SB(vj,wk)

where the last inclusion holds since . Conversely, suppose there exists a such that . Thus, satisfies for each voter and . Hence we have , which implies as is convex. Therefore . Thus, by Theorem 2.1, candidate wins under consensus voting. ∎

Following Alon et al. (dyn, ), from now on we restrict attention to case of candidates in each round. The case is not conceptually harder and the ideas presented in this paper do extend to that setting, but mathematically the analyses would be even more involved than those that follow.

## 3. General Tools for Bounding Expected Group Size

In this section we introduce a general approach for obtaining both upper and lower bounds on the expected cardinality of the social group in round . These techniques apply for consensus voting in any convex compact domain . In the rest of the paper we will specialize these methods for the cases in which is either a unit ball or a unit square. In particular, lower bounds are provided for these two domains in Section 4 and upper bounds in Section 5.

Let be a convex compact set, and let be candidates appearing in round . We assume each candidate is distributed uniformly on . We may also assume that , as otherwise we can absorb the associated constant factor into our bounds. Note that the expected group size is , where denotes the event a new candidate wins in round .

Let’s first present the intuition behind our approach to upper bounding the probability of selecting a candidate in any round. Recall that, by Theorem 2.1, given two candidates in round , we accept candidate if and only if . Now in order for the convex hull to satisfy , it must be the case that in the previous round (i) , and (ii) a new candidate did not get accepted inside the complement , where  denotes set closure. Applying this argument recursively with respect to the worst case convex hulls for accepting candidates inside , we will obtain an upper bound on the probability of accepting a candidate. Such worst case convex hulls can be found by appropriately applying Corollary 2.2.

To formalize this intuition, we require some more notation. We denote by the event that a new candidate is selected inside . Let denote the probability of selecting a candidate inside given the convex hull of the group members is . The shorthand will be used when the context is clear. Note, by Lemma 2.3, the probability of acceptance depends only on the shape of the convex hull of the members, and not on the round. That is, if for two rounds then the probabilities of accepting a candidate inside a given region in the rounds and are exactly the same.

We say a set is a cap if there exists a half space such that . We remark that caps have been widely used for studying the convex hull of random points; see the survey article (baddeley2007random, ) and the references therein. Of particular relevance here is that, in the case of two candidates, the Voronoi regions for the candidates are caps. Furthermore, and vice versa.

###### Theorem 3.1 ().

Let be convex compact domain and let be any function which satisfies . Then

 Pr[Xt+1] ≲ ∫K∫Ke−tfK(w1,w2)dw1dw2
###### Proof.

Observe that, for any cap , we have the following inequality:

 Pr[St⊆A] (1)

Here the first inequality follows from Corollary 2.2. The second inequality is obtained by repeating the argument inductively for . Hence:

 Pr[Xt+1] = ∫∫Pr[Xt+1 | (w1,w2) are candidates]dw1dw2 = ∫∫(Pr[St⊆H1(w1,w2)]+Pr[St⊆H2(w1,w2)])dw1dw2 ≤ 2∑i=1∫∫(1−Pr[Z(Hi(w1,w2)) | ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯Hi(w1,w2) ])tdw1dw2 ≤ 2∫∫(1−fK(w1,w2))tdw1dw2 ≤ 2∫∫e−tfK(w1,w2)dw1dw2

The second equality holds by Theorem 2.1. The first inequality follows by combining inequality (1) and the facts and . Next, by assumption, for each ; the last two inequalities follow immediately. ∎

As stated, Theorem 3.1 allows us to upper bound the expected group size. However, the theorem is not easily applicable on its own. To rectify this, consider the following easier to apply corollary.

###### Corollary 3.2 ().

Let be compact space. If satisfies the conditions of Theorem 3.1 then

 Pr[Xt+1] ≲ 1t+∫ln(t)t0te−tλ⋅Φ(λ)dλ

where .

###### Proof of Corollary 3.2.

From Theorem 3.1, we have that,

 Pr[Xt+1] ≲ ∫∫I[fK(w1,w2)≤ln(t)t]⋅e−tfK(w1,w2)dw1dw2 + ∫∫I[fK(w1,w2)≥ln(t)t]⋅e−tfK(w1,w2)dw1dw2 (2) ≤ ∫∫I[fK(w1,w2)≤ln(t)t]⋅e−tfK(w1,w2)dw1dw2 + 1t

Next, consider the random variable

and denote its cumulative distribution function by . Then

 ∫∫I[fK(w1,w2)≤ln(t)t]⋅e−tfK(w1,w2)dw1dw2 = E[I[fK(w1,w2)≤ln(t)t]⋅e−tfK(w1,w2)] = EY[E[I[Y≤ln(t)t]⋅e−tY | Y]] (3) = ∫10I[λ≤ln(t)t]⋅e−tλ dFY(λ)

The second equality is due to the law of total expectation. Now, because is absolutely continuous it has a density function. Combining (2) and (3), we then have that

 Pr[Xt+1] ≲ 1t+∫ln(t)t0e−tλ⋅ddλΦ(λ)dλ = 1t+[e−tλΦ(λ)]ln(t)t0+∫ln(t)t0te−tλ⋅Φ(λ)dλ ≲ 1t+∫ln(t)t0te−tλ⋅Φ(λ)dλ

Here the last inequality was obtained by noting and . ∎

As alluded to earlier, when obtaining upper bounds for the unit ball and the unit square, we will apply Corollary 3.2. Of course, in order to do this, we must find an appropriate function which lower bounds the probability of acceptance inside a Voronoi region , given the current convex hull is . Finding such a function can require some ingenuity, but Lemma 3.3 below will be useful in assisting in this task. Moreover, as we will see in Section 4, this lemma can be used to obtain lower bounds as well as upper bounds on the expected cardinality of the group.

###### Lemma 3.3 ().

Given a two-dimensional convex compact domain  and a cap . If and are the endpoints of the line segment separating and then

 Pr[Z(A) | ¯¯¯A ] = 2⋅∫Avol(B(z1,ξ)∩B(z2,ξ)∩A) dξ
###### Proof.

Without loss of generality, let be the selected candidate. Thus, by Theorem 2.1, we have because, by assumption, the convex hull of the voters is . We claim . Otherwise, . But, by definition of the Voronoi regions, we have . Thus, . However, this cannot happen unless , a zero probability event. Therefore, we may assume that .

Next, consider the line separating and ; that is, is the convex hull . Given we claim if and only if . First assume . As is closed, and are in . Thus, by convexity, . On the other hand, assume and consider the Voronoi edge defined by . Then either or because separates and and, by assumption, . Furthermore, observe that the midpoint is in and in . So it must be the case that . But if then one of the Voronoi regions contains . In particular, because . Therefore,

 Pr[Z(A) | ¯¯¯A ] = 2 ∫A∫AI[l(z1,z2)⊆H2(w1,w2)] dw2dw1 = 2 ∫Avol(B(z1,w1)∩B(z2,w1)∩A)dw1

Here the first equality holds since two candidates are equally likely to win by symmetry and, by Theorem 2.1, the winning candidate satisfies . The second equality follows from the fact that if and only if when . Finally, the last equality holds because, by Lemma 2.3, we know that any candidate is selected with consensus voting. This is equivalent to , by Theorem 2.1. ∎

## 4. Lower bounds on Expected Group Size

In this section we provide lower bounds for the cases where is either a unit ball  or a unit square . Recall, we assumed that but for the unit ball. We remark that this is of no consequence as we may absorb the associated constant factor into our bounds.

### 4.1. Lower Bound for the Unit Ball

For the unit ball , a circular segment is the small piece of the circle formed by cutting along a chord. Evidently, this means that every cap of  is either a circular segment or the complement of a circular segment. Thus, to analyze the case of the unit ball we must study circular segments.

###### Lemma 4.1 ().

Let be a unit ball, and be a circular segment with height then

###### Proof.

Let and be the endpoints of the chord defining the line segment . By rotating the ball, we may assume the chord is horizontal. Furthermore, by translating the coordinates, we may assume that lies at the origin. This is illustrated in Figure 1.

To apply Lemma 3.3, let . Observe that the region contains a triangle of height and width . Again, this is shown in Figure 1. It follows that

 (4) vol(B(z1,ξ)∩B(z2,ξ)∩Jδ) ≥ 12y⋅(√x2+y2−x) ≥ y34(x+y)

To obtain inequality (4) first observe, by Bernoulli’s Inequality, that . This implies . Rearranging terms, we obtain . Equivalently, , as required. Combining Lemma 3.3 with inequality (4) then gives

 (5)

To lower bound this integral, rather than integrate over the entire circular segment , we will integrate over an inscribe rectangle. Specifically let be the rectangle of height inscribed inside . This rectangle, shown in Figure 1, has width and is centred at . Therefore,

 ∫Jδy34(x+y) dxdy ≥ ∫Ry34(x+y) dxdy = ∫δ20∫√δ(√2−δ+√1−δ4)√δ(√2−δ−√1−δ4)y34(x+y) dxdy ≥ 18⋅∫δ20∫√δ(√2−δ+√1−δ4)√δ(√2−δ−√1−δ4)y3x dxdy ≳ δ4⋅ln⎛⎜ ⎜⎝√2−δ+√1−δ4√2−δ−√1−δ4⎞⎟ ⎟⎠ (6) ≳ δ4

The second inequality holds because ; specifically, when the circular segment lies under the line . Thus, every point inside the circular segment satisfies . The last inequality follows by observing that the logarithmic term is lower bounded by . Combining the inequalities (5) and (4.1) completes the proof. ∎

Note that Lemma 4.1 can be used to prove lower bound on the expected cardinality of the group. It is also used in later sections to obtain appropriate function when using Corollary 3.2.

###### Theorem 4.2 ().

For the unit ball , the expected group size after rounds is .

###### Proof.

For each , we construct collection of disjoint circular segments on the unit ball. To do this, let the height of each circular segment in the collection be .Then we can fit of these segments into . To see this, observe that a circular segment of height has a chord of length . The central angle of the segment is then , implying the existence of at least disjoint circular segments of height . Now define to be the last round for which . Thus,

 (7) E[|Gτ|]≥N(τ)∑i=1Pr[Sτ∩Aτi≠∅]=N(τ)⋅Pr[Sτ∩Aτ1≠∅]≥12⌊πτ18⌋≳τ18

Here, the first inequality follows from the observation that if then there is a least one group member inside . The equality is due to symmetry; that is, for each pair . The second inequality follows because, by definition, . Next consider rounds . For these rounds, by definition of , we know which implies . Therefore,

 Pr[Xt+1] ≳ N(t)∑i=1Pr[Z(Ati) ∧ (St∩Ati=∅)] = N(t)⋅Pr[Z(Ati) | St∩Ati=∅]⋅Pr[ St∩Ati=∅ ] ≳ N(t)⋅Pr[Z(Ati) | ¯¯¯¯¯Ati]

Where the last inequality follows by Corollary 2.2. Finally by Lemma 4.1, we see , and thus for any . We may now lower bound the expected group size at the end of round . Specifically, for ,

 E[|GT|]=E[|Gτ|]+T∑t=τ+1Pr[Xt=1]≳τ18+T∑t=τ+1t−78≳T18

The last inequality was obtained using integral bounds. ∎

### 4.2. Lower bound for the Unit Square

For the unit square  caps are either right-angled triangles or right-angled trapezoids (trapezoids with two adjacent right angles). We can bound the probability of accepting a point inside a right-angled trapezoid by consideration of the largest inscribed triangle it contains. Thus, it suffices to consider only the case in which the cap forms a triangle.

###### Lemma 4.3 ().

Let be triangular cap on the unit square with perpendicular side lengths . Then

 Pr[Z(Ja,b) | ¯¯¯¯¯¯¯Ja,b ]≥1211a4⋅(1+ln(ba)).
###### Proof.

Take the unit square  and the cap . Without loss of generality, we may assume the hypotenuse is horizontal with one endpoint at the origin. This is shown in Figure 2.

Again, setting , we have that

 (8) vol(B(z1,ξ)∩B(z2,ξ)∩Jδ) ≥ 12y⋅(√x2+y2−x) ≥ y34(x+y)

Combining Lemma 3.3 with inequality (8) gives

 (9) Pr[Z(Ja,b) | ¯¯¯¯¯¯¯Ja,b ] ≥ ∫Ja,bvol(B(z1,ξ)∩B(z2,ξ)∩Ja,b) dξ ≥ ∫Ja,by34(x+y) dxdy

Recall the cap has perpendicular side lengths . Thus the height of is . Again, a lower bound can be obtained by integrating over an inscribed rectangle rather than over the entire cap . Specifically, let be the inscribed rectangle with half the height of ; see Figure 2. Then

 ∫Ja,by34(x+y) dxdy ≥ ∫Ry34(x+y) dxdy ≥ ∫ab2√a2+b20∫a2√a2+b2+b22√a2+b2a22√a2+b2y34(x+y) dxdy = 14⋅∫ab2√a2+b20y3⋅[ln(x+y)]a2√a2+b2a22√a2+b2 dy = 14⋅∫ab2√a2+b20y3⋅ln⎛⎜ ⎜⎝a2√a2+b2+b22√a2+b2+ya22√a2+b2+y⎞⎟ ⎟⎠dy = 14⋅∫ab2√a2+b20y3⋅ln(1+a2+b2a2+2y√a2+b2)dy (10) ≥ 14⋅∫ab2√a2+b20y3⋅ln(1+a2+b2a2+ab)dy

To simplify (4.2) recall that , by assumption. Thus . This implies that

 ∫Ja,by34(x+y)dxdy ≥ 14⋅∫ab2√a2+b20y3⋅ln(1+a2+b2a2+ab)dy (11) = 1211a4⋅(1+ln(ba))

Here the second inequality arises since

 (1+a2+b2a2+ab)2 = (ba+2ba+1)2 ≥ (ba+ab)2 ≥ 2+(ba)2 ≥ e⋅ba

Thus, taking the log on both sides gives the second inequality. Putting together inequalities (9) and (4.2) completes the proof. ∎

###### Theorem 4.4 ().

For the unit square , the expected group size after rounds is .

###### Proof.

For each , consider the triangle