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 ‘leftright’ 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 twocandidate 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.
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], ShapleyOwen 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 .
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 axisparallel 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 topmost point on the circle, i.e. at .

Placing the remaining vertices around the circle so that the vertices are evenly spaced.
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 46, 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 comparisonbased decision algorithm that solves in time, provided the following two comparisonbased subroutines.
Subroutine 1.
A comparisonbased subroutine that, given a point and a regular sided polygon , decides if is outside in time.
Subroutine 2.
A comparisonbased 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 comparisonbased 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 .
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.
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 comparisonbased 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 comparisonbased decision algorithm that computes . Let be any comparison in the comparisonbased 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 comparisonbased 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 13.
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 13 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 comparisonbased 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 .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 comparisonbased algorithm and show that Property 2 holds.
Lemma 5.
Subroutine 1 has an time comparisonbased 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 .
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 comparisonbased 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 .
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.
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 comparisonbased 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 .
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 nearlinear 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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