Computing the Yolk in Spatial Voting Games without Computing Median Lines

02/13/2019 ∙ by Joachim Gudmundsson, et al. ∙ The University of Sydney 0

The yolk is an important concept in spatial voting games as it generalises the equilibrium and provides bounds on the uncovered set. We present near-linear time algorithms for computing the yolk in the spatial voting model in the plane. To the best of our knowledge our algorithm is the first algorithm that does not require precomputing the median lines and hence able to break the existing O(n^4/3) bound which equals the known upper bound on the number of median lines. We avoid this requirement by using Megiddo's parametric search, which is a powerful framework that could lead to faster algorithms for many other spatial voting problems.

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1 Introduction

Voting theory is concerned with preference aggregation and group decision making. A classic framework for aggregating voter’s preferences is the Downsian [7], or spatial model of voting.

In this model, voters are positioned on a ‘left-right’ continuum along multiple ideological dimensions, such as economic, social or religious. These dimensions together form the policy space. Each voter is required to choose a single candidate from a set of candidates, and a common voter preference function is a metric/distance function within the policy space. An intuitive reason behind using metric preferences is that voters tend to prefer candidates ideologically similar to themselves.

The spatial model of voting with metric preferences have been studied extensively, both theoretically [16, 17, 8, 20, 34] and empirically [23, 24, 25, 21, 29, 27, 28]. Recently, lower bounds were provided on the distortion of voting rules in the spatial model, and interestingly, metrics other than the Euclidean metric were considered [2, 30, 14].

We focus our attention on two-candidate spatial voting games, where the winner is the candidate preferred by a simple majority of voters. In a one dimension policy space, Black’s Median Voter Theorem [4] states that a voting equilibrium (alt. Condorcet winner, plurality point, pure Nash equilibrium) is guaranteed to exist and coincides with the median voter.

Naturally, social choice theorists searched for the equilibrium in the two dimensional policy space, but these attempts were shown to be fruitless [22]. The initial reaction was one of cynicism [16], but in response a multitude of generalisations were developed, with the yolk being one such concept [17, 20]. The yolk in the Euclidean  metric is defined as the minimum radius disk that intersects all median lines of the voters.

Figure 1: The  yolk intersects all median lines of voters.

The yolk is an important concept in spatial voting games due to its simplicity and its relationship to other concepts. The yolk radius provides approximate bounds on the uncovered set [10, 19, 20], limits on agenda control [12], Shapley-Owen power scores [9], the Finagle point [38] and the -core [37]. As such, studies on the size of the yolk [11, 15, 35] translate to these other concepts as well.

From the perspective of computational social choice, this raises the following problem: Are there efficient algorithms for computing the yolk? Fast algorithms would, for instance, facilitate empirical studies on large data sets. Tovey [33] provides the first polynomial time algorithm, which in two dimensions, computes the yolk in  time. De Berg et al. [3] provides an improved  time algorithm for the same.

The shortcoming of existing algorithms is that they require the computation of all limiting median lines, which are median lines that pass through at least two voters [31]. However, there are  [32] limiting median lines in the worst case. Moreover, the best known upper bound of  seems difficult to improve on [6]. It is an open problem as to whether there is a faster algorithm that computes the yolk without precomputing all limiting median lines .

Problem Statement

Given a set  of  points in the plane, a median line of  is any line that divides the plane into two closed halfplanes, each with at most  points. The yolk is a minimum radius disk in the  metric that intersects all median lines of .

Figure 2: Example of yolks in the  and  metrics.

We compute yolks in the  (Taxicab), the  (Euclidean), and the  (Uniform) metrics. As shown in Figure 2, the yolk in  is the smallest -rotated square and in  the smallest axis-parallel square, that intersects all median lines of .

Our Contributions and Results

Our contributions are, first, an algorithm that computes the yolk in the  and  metrics in  time, and second, an algorithm that computes a -approximation of the yolk in the  metric in  time.

We achieve the improved upper bounds by carefully applying Megiddo’s [18] parametric search technique, which is a powerful yet complex technique and that could be useful for other spatial voting problems.

The parametric search technique is a framework for converting decision algorithms into optimisation algorithms. For the yolk problem, a decision algorithm would decide whether a given disk intersects all median lines. If this decision algorithm satisifies the three properties as specified by the framework, then Megiddo’s result states that there is an efficient algorithm to compute the yolk.

For the purposes of designing a decision algorithm with the desired properties, we instead consider the more general problem of finding the smallest regular, -sided polygon that intersects all median lines of . The regular -sided polygon  is shown in Figure 3 and is defined as:

Definition 1.

Given an integer , construct the regular -sided polygon  by:

  • Constructing a circle with radius  and centered at .

  • Placing a vertex at the top-most point on the circle, i.e. at .

  • Placing the remaining  vertices around the circle so that the  vertices are evenly spaced.

Figure 3: The regular, -sided polygon .

In Section 2, we present the decision algorithm, which given a regular, -sided polygon , decides whether the polygon intersects all median lines of . Next, in Section 3

, we apply Megiddo’s technique to the decision algorithm and prove the convexity and parallelisability properties. This leaves one final property, the existence of critical hyperplanes, left to check. We prove this final property in Sections 4-6, thus completing the parametric search. Finally, in Section 7, we show that our general problem for the regular, 

-sided polygon  implies the claimed running times by setting  for  and , and  for .

2 Decision Algorithm

The aim of this section is to design an algorithm that solves the following decision problem:

Definition 2.

Given an integer  and a set  of  points in the plane, the decision problem  is to decide whether the polygon  intersects all median lines of .

We show that there is a comparison-based decision algorithm that solves  in  time, provided the following two comparison-based subroutines.

Subroutine 1.

A comparison-based subroutine that, given a point  and a regular -sided polygon , decides if  is outside  in  time.

Subroutine 2.

A comparison-based subroutine that, given points  outside a regular -sided polygon , computes the relative clockwise order of the four tangent lines drawn from  to  in  time.

Although the running time of these two subroutines are not too difficult to prove, we shall see in Section 3 that these subroutines must satisfy a stronger requirement for the parametric search technique to apply. We will formally define the stronger requirement in the next section. To avoid repetition, we simultaneously address the subroutine and the stronger requirement in Sections 5 and 6. But for now, we assume the subroutines exist and present the decision algorithm:

Theorem 1.

Given an integer  and a set  of  points in the plane, there is a comparison-based algorithm that solve the decision problem  in  time, provided that Subroutine 1 and Subroutine 2 exist.

Proof.

The proof comes in three parts. First, we transform the decision problem  into an equivalent form that does not have median lines in its statement. Then, we present a sweep line algorithm for the transformed version. Finally, we perform an analysis of the running time.

Consider for now a single median line  that has gradient . Construct two parallel lines  and  that also have gradient , but are tangent to  from above and below respectively. If the median line  intersects , as shown in Figure 4, then  must be in between  and .

Figure 4: The relative positions of  and  if  intersects the -sided regular polygon .

We will decide whether all median lines of gradient  are between  and , as this would immediately decide whether all median lines of gradient  intersects . We will solve this restricted decision problem by counting the number of points in  above  and the number of points in  below .

Let  be the number of points in  that are above , and similarly  for the points in  below . Suppose that  and . Then there cannot be a median line of gradient  above  or below , since one side of the median line, in particular the side that contains the polygon, will have more than points. Hence, if  and , then all median lines of gradient  must be between  and .

Conversely, suppose that all median lines of gradient  are between  and . Then if , we can move  continuously upwards until it becomes a median line, which is a contradiction. So in this case, we know  and .

In summary, we have transformed the decision problem into one that does not have median lines in its statement: All median lines intersect  if for all gradients , the pair of inequalities  and  hold.

We present a sweep line algorithm that computes whether the pair of inequalities hold for all gradients . Let  be an arbitrary line tangent to the polygon , and define  to be the open halfplane that has  as its boundary and does not include the polygon . Then all median lines intersect  if and only if for all positions of , the open halfplane  contains less than  points.

Figure 5: The rotating sweepline  and the open halfplane .

The tangent line  is a clockwise rotating sweep line and the invariant maintained by the sweep line algorithm is the number of points of  inside the region . Take any tangent line  to be the starting line, and calculate the number of points in . From here, define an event to be when the line  passes through a point. There are two events for each point outside ; there is one event for when the point enters the region , and one for when it exits. There are no events for points of  that lie inside . The unsorted set of event points can be computed by applying Subroutine 1 to each of point in .

We sort the set of event points in a clockwise fashion. If we consider only two voters, their associated events can be sorted using Subroutine 2. We can extend this to sort the associated events of all voters with any standard comparison-based sorting algorithm, for example Merge sort.

Once the sorted set of events is computed, we process the events in order. At each new event we maintain our invariant, the number of points inside the region . This value increases by one at “entry” events and decreases by one at “exit” events. Finally, we return whether our invariant remained less than at all events.

The running time analysis for the algorithm is as follows. Computing the points outside  takes  time per point by Subroutine 1, so in total this takes  time. Computing the sorted order of the event points takes  time per comparison by Subroutine 2, which adds up to  time. Processing the sorted event points takes  time. Adding these gives the stated bound. ∎

3 Parametric Search

Parametric search is a powerful yet complex technique for solving optimisation problems. The two steps involved in this technique are, first, to design a decision algorithm, and second, to convert the decision algorithm into an optimisation algorithm.

For example, our parameter space is , our decision algorithm is stated in Theorem 1, and our optimisation objective is to minimise .

Preliminaries

Megiddo’s (1983) states the requirements for converting the decision algorithm into an optimisation algorithm. First, let us introduce some notation. Let  be a parameter space, let  be a parameter and let  be a decision problem that either evaluates to true or false. Then the first requirement is for the decision problem .

Property 1.

The set of parameters that satisfies the decision problem is convex.

Convexity guarantees that the optimisation algorithm finds the global optimum.

The second property of the technique relates to the decision algorithm. Let  be a comparison-based decision algorithm that computes . Let be any comparison in the comparison-based decision algorithm . The comparison is said to have an associated critical hyperplane in if the result of the comparison is linearly separable with respect to . Formally, suppose that the comparison evaluates to either , or . Then we say that the -dimensional hyperplane is the associated critical hyperplane of if evaluates to , or if and only if is above, on, or below respectively. The comparisons of the decision algorithm must satisfy the following property.

Property 2.

Every comparison in the comparison-based decision algorithm either (i) does not depend on , or (ii) has an associated critical hyperplane in .

This requirement allows us to compute a large set of critical hyperplanes that determines the result of . Moreover, the optimum must lie on one of these critical hyperplanes, since the result of locally changes sign at the optimum. The new search space now has dimension instead of dimension , and we can recursively apply this procedure to reduce the dimension further. For details see [1].

The final property speeds up the parametric search.

Property 3.

The decision algorithm has an efficient parallel algorithm.

If the decision algorithm  runs in  time and runs on  processors in  parallel steps, then the parametric search over runs in  time [1].

Applying the technique

To apply the parametric search technique, we show that our decision problem  satisfies Properties 1-3.

Lemma 1.

Given an integer  and a set  of  points in the plane, the set of parameters that satisfies the decision problem is convex.

Proof.

Suppose we are given a convex combination  of the two parameters . Then the polygon  is a convex combination of the polygons  and . It is easy to check that if a line  intersects both  and , then the line  must also intersect the convex combination .

Now assume that both  and  are true. Then for any median line  both  and  intersect . By the observation above, the convex combination  must also intersects . Repeating this fact for all median lines implies that  intersects all median lines of . So  is true whenever  and  are true. Therefore, the set of parameters  is convex. ∎

Lemma 2.

Every comparison in the decision algorithm in Theorem 1 either (i) does not depend on , or (ii) has an associated critical hyperplane in .

Proof.

Theorem 1 consists of three steps, computing the points outside the polygon, computing the event order, and processing the events. For the first two steps, the comparisons do depend on and have associated critical hyperplanes. We defer the proof of this claim to Sections 5 and 6 respectively. For the third step, the comparisons do not depend on but rather the event order, so there is no requirement that comparisons have critical hyperplanes. ∎

Lemma 3.

The decision algorithm in Theorem 1 has an efficient parallel algorithm that runs on  processors and takes  parallel steps per processor.

Proof.

Given  processors, we decide which points are outside the polygon in parallel by assiging a processor to each point. By Subroutine 1, this takes  parallel steps per processor. We compute the event order in parallel using Preparata’s sorting scheme [26]. Each processor requires  calls to Subroutine 2, so it total, each processor requires parallel steps. Finally, processing the events generates no critical hyperplanes, so this step does not require parallelisation. ∎

Now we combine Properties 1-3 with Megiddo’s result to obtain an optimisation algorithm for the smallest, regular, -sided polygon  that intersects all median lines.

Theorem 2.

Given a set  of  points in the plane, there is an  time algorithm to compute the minimum  such that  is true for some regular, -sided polygon .

Proof.

Megiddo’s multidimensional parametric search implies that there is an efficient optimisation algorithm. It only remains to show the running time of the technique.

The parallel algorithm runs on processors in  parallel steps, whereas the decision algorithm runs in time. The dimension  of the parameter space is three. The running time of multidimensional parametric search is   [1]. Substituting our values into the above formula yields the required bound. ∎

4 Computing Critical Hyperplanes

The only requirement left to check is Property 2 for the comparisons in the comparison-based subroutines, that is, Subroutine 1 and Subroutine 2. Before launching into the analysis of the two subroutines, we first prove a tool. We will use the tool repeatedly in the next two sections to simplify checking Property 2.

Lemma 4.

Let gradient , point 

and vector 

be given, and let  be a variable parameter. Let  be the line of gradient  through the point . Then  is above, on, or below  if and only if the point  is above, on, or below its associated critical hyperplane .

Figure 6: Point  is above  if and only if parameter  is above .
Proof.

Let point  and vector . Now,  is above the line through  of gradient  if . Substituting the point , we get the inequality

This inequality can be rearranged into the form , where

In this form, we can see that the inequality is satisfied if and only if  lies above the hyperplane , where  are given above. Hence, the two conditions,  above a line and  above a hyperplane, can be decided with the same inequality, which completes the proof. ∎

Now we are ready to address the subroutines.

5 Subroutine 1

Subroutine 1 decides whether a given point  is outside the -sided, regular polygon . We present an  time comparison-based algorithm and show that Property 2 holds.

Lemma 5.

Subroutine 1 has an  time comparison-based algorithm, and comparisons in the algorithm that depend on the parameter  each have an associated critical hyperplane.

Proof.

We partition the polygon  into  triangles, and decide which partition the point  is in, if it indeed is in any of these partitions. For , the  partition of  is the triangle joining the  vertex, the  vertex and the center of . Figure 7 shows the  partition of .

Figure 7: The  partition of .

Assume for now that the point  is indeed in the polygon  and hence in one of the  partitions. We decide whether  is in the partition for some , or for some , and perform a binary search for the index . This can be done by deciding if the point  is above, on, or below the line joining the center of  and its  vertex. The comparison depends on , so we must compute its associated critical hyperplane using Lemma 4. Let be the -sided polygon of radius 1 and centered at the origin. Then set to be the gradient of the line joining the center to the  vertex of , and vector  in Lemma 4 to obtain the associated critical hyperplane.

We have searched for the partition that  is in if it is indeed in . Hence, it only remains to decide whether  is indeed in that partition. This requires a constant number of comparisons, each of which depend on . We have already computed associated critical hyperplanes for two of the sides. The last side joins two adjacent vertices of the polygon . Set  to be the gradient of the  side of polygon , and the vector  to be the  vertex of , to obtain the final associated critical hyperplane.

The running time is dominated by the binary search for the  partition, which takes  time. ∎

6 Subroutine 2

Subroutine 2 computes the relative clockwise order of four tangent lines drawn from two points to polygon .

Lemma 6.

Subroutine 2 has an -time comparison-based algorithm, and comparisons in the algorithm that depend on the parameter  each have an associated critical hyperplane.

Proof.

Draw two lines  tangent to  and parallel to , and let the points of tangency be vertex  and vertex . If there are multiple points of tangency then choose any such point. Then without loss of generality, set  to be horizontal, and assume further that  has a larger  coordinate than . Then the  and  partition the plane into the four regions, as shown in Figure 8. Region  is left of both tangents,  is right of both tangents,  is between the tangents and above , and  is between the tangents and below .

Figure 8: The lines  partition the plane into regions .

Then the relative clockwise order of the four lines drawn from  and  are determined by which of the four regions  or  the points  and  are located. See Figure 9.

Figure 9: The relative orders shown for when     and  .

Five cases follows. Let  and  points of tangency from  such that the points  are in clockwise order. If  are in the same region, then the containing region , and  correspond to the relative clockwise orders , and  respectively. If  are in different regions, then they must be in  and  respectively, and the relative order is . The proof for case analysis for the five cases is omitted, but the diagrams in Figure 9 may be useful for the reader.

The running time of the algorithm is as follows. Given the gradient of , there is an  time algorithm to binary search the gradients of the sides of  to compute the vertices  and . Then the remainder of the algorithm takes constant time: rotating the diagram so that  is horizontal, deciding whether  or  has a larger  coordinate, and computing the region  that points  are in.

The proof of existence of critical hyperplanes is as follows. Since the gradients of  and sides of  do not depend on , computing  and  generates no critical hyperplanes. Similarly, rotating the diagram so that  is horizontal and then deciding which of  or  have larger  coordinates also generates no critical hyperplanes. It only remains to decide which of the four regions  the point , and respectively , is in. Set  to the gradient of  and vector  to be the  vertex of  in Lemma 4 to decide if  is to the left of the tangent through . Do so similarly for  to decide if  is to the right of the tangent through . Finally, set  to the gradient of  and vector  to be either the  or  vertex of  to decide if  is above the chord . ∎

Checking that Property 2 holds for the comparison-based subroutines, Subroutine 1 and Subroutine 2, completes the proof to Theorem 2. In the final section we will prove that Theorem 2 implies that we have an efficient algorithm for computing the yolk in the and meetrics, and an efficient approximation algorithm for the metric.

7 Computing the Yolk in , and 

It remains to show that our general problem for  implies the results as claimed in the introduction.

Theorem 3.

Given a set  of  points in the plane, there is an  time algorithm to compute the yolk of  in the  and  metrics.

Proof.

Setting  in Theorem 2 gives an algorithm to compute the smallest  that intersects all median lines of  in  time. This rotated square coincides with yolk in the  metric, refer to Figure 2 and Definition 1.

Computing the yolk in the  metric requires one extra step. Rotate the points of  by  clockwise, compute the smallest , and then rotate the square  back  anticlockwise to obtain the yolk in the  metric. ∎

Theorem 4.

Given a set  of  points in the plane and an , there is an  time algorithm to compute a -approximation of the yolk in the  metric.

Proof.

Setting  in Theorem 2 gives an algorithm to compute the smallest  that intersects all median lines of  in the desired running time. It suffices to show that for this parameter set , the disk centered at  with radius  is a -approximation for the yolk in the  metric.

First, note that  intersects all median lines, and  encloses , so the disk must also intersect all median lines of . Hence, it suffices to show that the radius  of  satisfies , where  is the radius of the true yolk in the  metric.

Let the yolk in the  metric be the disk . Consider the regular, -sided polygon , so that by construction, all sides of this polygon are tangent to .

Figure 10: The polygon  is externally tangent to the disk .

Now since  is the  yolk, it intersects all median lines and so does its enclosing polygon . By the minimality of , we get . But for , we have . So,

which implies that , as required. ∎

8 Concluding Remarks

Cole’s [5] extension to parametric search states that the running time of the parametric search may be reduced if certain comparisons are delayed. This is a direction for further research that could potentially improve the running time of our algorithms.

An open problem is whether one can compute the yolk in higher dimensions without precomputing all median hyperplanes. Avoiding the computation of median hyperplanes yields even greater benefits as less is known about bounds on the number of median hyperplanes in higher dimensions.

Similarly, our approximation algorithm for the  yolk in the plane is optimal up to polylogarithmic factors, however, it is an open problem as to whether there is a near-linear time exact algorithm. Our attempts to apply Megiddo’s parametric search technique to the  yolk have been unsuccessful so far.

Finally, there are other solution concepts in computational spatial voting that currently lack efficient algorithms. The shortcomings of existing algorithms are: for the Shapley Owen power score there is only an approximate algorithm [13], for the Finagle point only regular polygons have been considered [38] and for the -core only a membership algorithm exists [36]. Since these problems have a close connection to either median lines or minimal radius, we suspect that Megiddo’s parametric search technique could also be useful for these problems.

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