Manipulating Districts to Win Elections: Fine-Grained Complexity

Manipulating Districts to Win Elections: Fine-Grained Complexity

Eduard Eiben,

¹

Fedor V. Fomin,

²

Fahad Panolan,

³

Kirill Simonov

²

1

Department of Computer Science, Royal Holloway, University of London, UK

2

Department of Informatics, University of Bergen, Norway

3

Department of Computer Science and Engineering, IIT Hyderabad, India

eduard.eiben@rhul.ac.uk, {fedor.fomin, kirill.simonov}@uib.no, fahad.panolan@gmail.com 19th February 2020

Abstract

Gerrymandering is a practice of manipulating district boundaries and locations in order to achieve a political advantage for a particular party. Lewenberg, Lev, and Rosenschein [AAMAS 2017] initiated the algorithmic study of a geographically-based manipulation problem, where voters must vote at the ballot box closest to them.

In this variant of gerrymandering, for a given set of possible locations of ballot boxes and known political pref- erences ofnvoters, the task is to identify locations forkboxes out ofmpossible locations to guarantee victory of a certain party in at least`districts. Here integerskand`are some selected parameter.

It is known that the problem isNP-complete already for4political parties and prior to our work only heuristic algorithms for this problem were developed. We initiate the rigorous study of the gerrymandering problem from the perspectives of parameterized and fine-grained complexity and provide asymptotically matching lower and upper bounds on its computational complexity. We prove that the problem isW[1]-hard parameterized byk+n and that it does not admit anf(n, k)·m^o(

√

k)algorithm for any functionfofkandnonly, unless the Exponential Time Hypothesis (ETH) fails. Our lower bounds hold already for 2parties. On the other hand, we give an algorithm that solves the problem for a constant number of parties in time(m+n)^O(^√k).

1 Introduction

In 1812, Massachusetts governor Elbridge Gerry immortalized his name by signing a bill that created a mytho- logical salamander shaped district in the Boston area to the benefit of his political party. Since then the term gerrymandering has been used for the practice of establishing a political advantage by manipulating voting district boundaries. Wikipedia article Gerrymandering contains a lot of notable gerrymandering examples in electoral systems of many countries. One of possible responses of such manipulations is a more “rational” definition of voting districts, for example the system where voters should always go to a ballot box (or central area) that is closest to them [19].

However, even in the “rational” geographical settings manipulations are still possible. [15] considered the variant of the gerrymandering problem in which the agent in charge of the design of voting districts has to follow a clear rule:all voters vote in the ballot box nearest to them.As it was shown by [15], even with clear and rational rules, manipulation is possible. This brought Lewenberg et al. to the following natural question. Is there an efficient algorithmic procedure that would allow an agent, who is obliged to follow the rules, to optimize division for his personal preferred outcome? The good news here is that the existence of such a procedure is highly unlikely, because as it was shown in [15], the problem isNP-complete already for4political parties.

On the other hand, when the number of districtskis small, the problem is trivially solvable in timem^kn^O(1), wheremis the number of all possible locations of ballot boxes andnis the number of voters, by a brute-force enumerating all possible locations forkboxes. Thus in the situation when the number of districts is small, there is an efficient procedure for finding an optimal manipulation. This immediately brings us to the following question which serves as the departure point of our work.

Question 1. Is it possible to solve the gerrymandering problem faster than the brute-force?

To answer this question, we study GERRYMANDERINGthrough the lens of Parameterized Complexity, which is a multivariate paradigm of algorithm analysis introduced by [7]. In Parameterized Complexity, one measures the

arXiv:2002.07607v1 [cs.DS] 18 Feb 2020

performance of algorithms not merely in terms of the size of the input, but also with respect to certain properties of the input or the output (captured by one or several numericalparameters,k). This gives rise to two notions of tractability, both of which correspond to polynomial-time tractability in the classical setting, when the parameter is constant. First is the classFPTthat contains all problems that can be solved in timef(k)·n^O(1) (for some computable functionf), while the (asymptotically less efficient) classXPcontains all problems that can be solved in timen^f(k). Aside from these complexity classes, we will make use of the complexity classW[1], the class of parameterized problems that can be in timef(k)·n^O(1) reduced to INDEPENDENTSETparameterized by the solution size. FPT6=W[1] is the working hypothesis of Parameterized Complexity and it is widely believed that noW[1]-hard problem is inFPT. Finally, our more fine-grained lower bound result depends on the Exponential Time Hypothesis (ETH) [13, 16], which states that the satisfiability ofk-CNFformulas (fork≥3) is not solvable in subexponential-time2^o(n), wherenis the number of variables in the formula. We refer to the recent books [4]

and [8] for more exposition on these complexity notions.

Contribution [15] proved that GERRYMANDERINGplurality(the Gerrymandering problem when the winner at each district is decided bypluralityvoting rule) isNP-complete, even when the number of candidates|C|is4.

We enhance this intractability result by providing a fine-grained complexity of the problem. Also note that our reduction also implies that the problem is NP-complete for any number|C| ≥2of candidates.

Theorem 1. For any number|C| ≥ 2of candidates,GERRYMANDERINGplurality isW[1]-hardparameterized byk+n. Moreover, there is no algorithm solvingGERRYMANDERINGpluralityin timef(k, n)·m^o(

√k)for any computable functionf, unlessETHfails.

The tools we used to obtain lower bounds are based on the grid-tiling technique of [17]. We refer to the book of [4] for further discussions of subexponential algorithms.

Interestingly, the ETH lower bound from Theorem 1 is asymptotically tight. We give an algorithm whose running time matches our lower bound.

Theorem 2. For any constant number of candidates and for any voting ruleg,GERRYMANDERINGgis solvable in time(m+n)^O(^√k).

To obtain our algorithm, we use the machinery developed by [18]. The main idea is to guess roughly√ kmany ballot boxes from an optimal solution that separate the plane into two parts such that each part contains between k/3and2k/3ballot boxes of the solution.

Related work The GERRYMANDERINGproblem was introduced by [15]. It is an example of general control problem, where a central agent may influence the outcome using its power over the voting process [1, 2, 9, 12, 14].

The problem also belongs to the large family of voting manipulation problems, whose computational aspects are widely studied in the computational social choice community, see [3, 10, 11, 20] for further references. In particular, parameterized complexity of manipulation was studied in [5, 6].

2 Preliminaries

For an integeri, we let[i] = {1,2, . . . , i}. We denote byNthe set of natural numbers and byRthe set of real numbers. Given two pointsp₁ = (x₁, y₁),p₂ = (x₂, y₂)in the planeR², thedistancebetweenp₁ andp₂ is dist(p₁, p₂) =p

(x₂−x₁)²+ (y₂−y₁)².

2.1 Voting and G

ERRYMANDERING

Given a set ofvotersV = {v1, . . . , vn}and a set ofcandidatesC, thepreference rankingof voterviis a total order (i.e., transitive, complete, reflexive, and antisymmetric relation)ioverC. The interpretation ofc1ic2is that the votervistrictly prefers candidatec1over candidatec2. We denote byπ(C)the set of preference rankings for the set of candidate C. A voting ruleis then a functionf : π(C)ⁿ → C. Anelection Ef = (C, V)is then comprised of a set of votersV, a set of candidatesC, a preference rankingioverCfor each voter and a voting rulef. In this paper, we will only consider voting rules that do not distinguish between voters. That is the outcome of the voting rule only depends on the number of voters with each preference ranking and not their order.

An example of such rule isplurality, in which for each voter only the topmost candidate counts and the winner is the candidate that got the most votes.

In adistrict-basedelectionEf,gvoters are divided into disjoint setsV1, . . . , Vssuch thatS

i∈[s]Vi=V. These sets define electionsE_fⁱ = (C, Vi,). The ultimate winner from amongst the winners ofE_fⁱ is determined by voting ruleg, which for us will be athreshold function.

Definition 1(GERRYMANDERING[15]). The input of theGERRYMANDERINGf problem is:

• A set of candidatesC.

• A set ofnvotersV ={v1, . . . , vn} ⊂R², where every votervi ∈V is identified by their location on the plane, and a preferred rankingi∈π(C).¹

• A set ofmpossible ballot boxesB={b1, . . . , bm} ⊂R².

• Parameters`, k∈N, such that`≤k≤m.

• A target candidatep∈C.

The task is to decide whether there is a subset ofkballot boxesB⁰ ⊆ B, such that they define a district-based election, in which every voter votes at their closest ballot box inB⁰, the winner at every ballot box is determined by voting rulef, andpwins in at least`ballot boxes.

We assume that there are no two possible ballot boxes and a voter such that the voter is equidistant from the two boxes, meaning that no voter could ever be exactly on the boundary between two districts and so a fixed set of ballot boxes unambiguously defines the partition of voters into districts.

In Figure 1, we give an example of an input of GERRYMANDERINGgand two possible solutions.

Figure 1: The example of [15]. The green circle and red cross are voters voting for a different candidates. The squares are ballot boxes. In the bottom are two examples with 3 ballot boxes opened giving a different outcome.

2.2 Voronoi diagrams

To describe our algorithm, we require the notion of a Voronoi diagram. For a finite set of pointsP on the plane, the Voronoi regionof a pointp∈Pis the set of those points of the plane that are closer topthan to any other point inP. We may also usethe Voronoi cellas a synonym. A Voronoi region is convex and its boundary is a polyline consisting of segments and possibly infinite rays, since the Voronoi region ofp∈Pis essentially the intersection of|P| −1half-planes.The Voronoi diagramofP is the tuple of the Voronoi regions of all the elements ofP.

The Voronoi diagram could be considered a planar graph. The faces of the graph correspond to the Voronoi regions, the edges correspond to the segments which form the boundaries of the individual regions, and the vertices are the endpoints of these segments. The technical details of this correspondence can be found in [18].

Voronoi diagrams appear naturally in GERRYMANDERING problem, since when the set S of ballot boxes chosen is fixed, the Voronoi diagram ofS defines exactly which regions of the plane are assigned to vote in a particular ballot box.

3 Hardness

To obtain our ETH lower bound and proveW[1]-hardness of GERRYMANDERING, we reduce from the following W[1]-hard problem.

Definition 2(GRIDTILLING[17]). The input of theGRIDTILINGproblem is:

1Lewenberg et al. [15] also define the weighted problem, where every votervhas weightwv. We focus on the version without weights.

• Integersk, n∈N, and a collectionSofk²nonempty setsSi,j⊆[n]×[n](1≤i, j≤k).

The task is to decide whether there is a set of pairss_i,j ∈S_i,j, for each1≤i, j≤k, such that:

• Ifs_i,j= (p, q)ands_i+1,j = (p⁰, q⁰), thenp=p⁰.

• Ifs_i,j= (p, q)ands_i,j+1= (p⁰, q⁰), thenq=q⁰.

Lemma 1([17], see also Theorem 14.28 in [4]). GRIDTILINGisW[1]-hardparameterized bykand, unlessETH fails, it has nof(k)n^o(k)-time algorithm for any computable functionf.

1 2 3 3 2 1

3 2 1 1 2 3 1 2 3

1 2 3

1 23

3 2 1 1

25

25 125 1 5

125 5

5 1

1 5

25 25

125

125

Figure 2: Example of a hardness reduction from Grid Tiling for n = 3 and k = 2. The sets are S_1,1 = {(1,1),(2,2),(3,3)}, S_1,2 = {(1,3),(2,1),(2,3),(3,2)}, S_2,1 = {(1,2),(2,1),(3,2)}, and S_2,2 = {(1,2),(3,1)}. The voters are denoted by×, where our candidate is red and the other candidate is blue. The multiplicity of red voters in each cell is1more than the sum of the blue voters in the same cell. The ballot boxes are depicted as•. A solution to an instance is the set of red ballot boxes.

Overview of the Proof. Before we give a formal proof of Theorem 1, we first give an informal description of our reduction from GRIDTILINGproblem that excludes some tedious computations that are necessary for the proof to work, but are not important for understanding where the difficulty of the problem lies. The main idea is to put all voters and ballot boxes inside largek×kgrid such that the cell(i, j)of this grid correspond to the setSi,j. In the center of each cell is one much smallern×ngrid corresponding to the setSi,j(see Figure 2). The ballot boxes are placed on appropriate positions inside small grid, with a small catch – for two neighboring cells of large grid, the small grids are “mirrored”. This way, if we select precisely the same pair(p, q)for each setSi,j, then the voters inside each large cell vote in the ballot box inside the same cell.

We will place a number of the “other candidate” voters, determined by a calculation later in the proof, very close to the center of each of 4 borders of the cell (see Figure 2). By selecting the right distances, for the size of the large grid, size of the small grid, and distance of the “other candidate” voters to the border of the cell, we can force that if we select ballot boxes corresponding tos_i,j = (p, q) ∈ S_i,j ands_i+1,j = (p⁰, q⁰) ∈ S_i+1,j (resp.

s_i,j+1 = (p⁰, q⁰)∈S_i,j+1), then the voters near the border between cells corresponding toS_i,jandS_i+1,j (resp.

S_i,j+1) vote inside their corresponding cells if and only ifp=p⁰(resp.q=q⁰).

Finally, we place in the center of each large cell the number of the “our candidate” voters that is just 1 larger than the number of the “other candidate” voters inside the cell. As we placed “our candidate” voters on k²

different positions, to wink²ballot boxes, in each box in a solution has to vote all voters from precisely one of these positions. Furthermore, by selecting the number of voters in each cell to grow exponentially, we get that the only way to win all ballot boxes is if all voters in the same cell vote in the same ballot box. This is only possible if we choose in each cell precisely one ballot box and if the corresponding selection of pairs inS_i,j’s is actually a solution to the original GRIDTILINGinstance.

In the remaining of the section we formalize above intuition by assigning the voters and ballot boxes specific positions in the plane and computing distances between positions of ballot boxes and “other candidate” voters.

Theorem 1. For any number|C| ≥ 2of candidates,GERRYMANDERINGplurality isW[1]-hardparameterized byk+n. Moreover, there is no algorithm solvingGERRYMANDERINGpluralityin timef(k, n)·m^o(

√

k)for any computable functionf, unlessETHfails.

Proof. We will prove the theorem by a reduction from GRIDTILING. Letk, n∈Nbe integers andSbe a collec- tion ofk²nonempty setsS_i,j⊆[n]×[n](1≤i, j≤k). We will construct an instanceI= (C, V, B, k², k²,“red”) of GERRYMANDERINGplurality with two candidates “red” and “blue”,2^O(^k2)voters andO k²n²

ballot boxes and all voters and boxes positioned at the integral points inside[O k·n²

]×[O k·n² ]grid.

Ballot boxes. Let (p, q) ∈ Si,j, we letx^i,j_(p,q) = (10n²+ 4n)(i−1) + 5n² + 4pifi is even and x^i,j_(p,q) = (10n²+ 4n)(i−1) + 5n²+ 4(n−p)ifiis odd. Similarly, we lety^i,j_(p,q)= (10n²+ 4n)(j−1) + 5n²+ 4qifjis even andy^i,j_(p,q)= (10n²+ 4n)(j−1) + 5n²+ 4(n−q)ifjis odd. We then identify the pair(p, q)∈Si,jwith a ballot boxb^i,j_(p,q)= (x^i,j_(p,q), y_(p,q)^i,j ). Finally, we will denote byBi,j the set of ballot boxes associated with pairs in Si,j.

Voters. For each pair of1 ≤i, j ≤ kwe will place5(i−1)·k+(j−1)“blue” voters at each of the following four positions:((10n²+4n)(i−1)+1,(10n²+4n)(j−1)+5n²+2n),((10n²+4n)(i−1)+5n²+2n,(10n²+4n)(j−

1)+1),((10n²+4n)i−1,(10n²+4n)(j−1)+5n²+2n), and((10n²+4n)(i−1)+5n²+2n,(10n²+4n)j−1).

We denote the sets of voters at the above four positions byV_←^i,j,V_↑^i,j,V_→^i,j, andV_↓^i,j, respectively. And we will place4·5(i−1)·k+(j−1)+ 1“red” voters at the position((10n²+ 4n) + 5n²+ 2n,(10n²+ 4n)(j−1) + 5n²+ 2n), we denote the set of these votersV_R^i,j.

The goal is to assignk²ballot boxes such that in each ballot box “red” has majority.

Correctness. We will show that(k,n, S)is a YES-instance of GRID TILLINGif and only if(C,V,B,k², k², “red”)is a YES-instance of GERRYMANDERINGplurality. Letsi,j be a pair inSi,j for alli, j ∈ [k]and let B⁰ ={b^i,j_si,j |i, j ∈[k]}. The following two claims show that ifS

i,j∈[k]{si,j}is a solution to GRIDTILLING, thenB⁰is a solution to GERRYMANDERINGplurality. Moreover, ifB⁰ is a solution such that for alli, j∈[k]the voters inS

∈{R,←,→,↑,↓}vote inb^i,j_s

i,j, thenS

i,j∈[k]{s_i,j}is a solution to GRIDTILLING.

5n²−1 4n−4p

4p 2n−4q

4n−4p⁰ 5n²−1

4q⁰

2 b^i,j_(p,q)

b^i+1,j_(p0,q⁰)

4q⁰−2n V_→^i,j

V_←^i+1,j

Figure 3: The situation in Claim 1. The distance dist(b^i,j_(p,q), V_→^i,j) = p

(5n²+ 4n−4p−1)²+ (2n−4q)². Which is clearly between5n²+ 4n−4p−1and5n²+ 4n−4p. Similarly,dist(b^i+1,j_(p0,q⁰), V_→^i+1,j)is between 5n²+ 4n−4p⁰−1and5n²+ 4n−4p⁰.

Claim 1. Letb^i,j_(p,q) andb^i+1,j_(p0,q⁰) be two ballot boxes. Then for every voterv ∈ V_→^i,j it holdsdist(v, b^i,j_(p,q)) <

dist(v, b^i+1,j_(p0,q⁰))if and only ifp≤p⁰. Similarly, for everyv⁰∈V_←^i+1,jit holdsdist(v⁰, b^i+1,j_(p0,q⁰))<dist(v⁰, b^i,j_(p,q))if and only ifp≥p⁰.

Proof. Let us assume that bothi andj are even. The remaining3 cases are analogous. Thenb^i,j_(p,q) is at po- sition((10n²+ 4n)(i−1) + 5n²+ 4p,(10n²+ 4n)(j−1) + 5n²+ 4q)andb^i+1,j_(p0,q⁰)is at position((10n²+ 4n)i+ 5n²+ 4(n−p⁰),(10n²+ 4n)(j−1) + 5n²+ 4q⁰). Moreover, voters inV_→^i,jandV_←^i+1,jare at positions

(10n²+ 4n)i−1,(10n²+ 4n)(j−1) + 5n²+ 2n

and (10n²+ 4n)i+ 1,(10n²+ 4n)(j−1) + 5n²+ 2n , respectively.

Since both q andq⁰ are numbers between 0 andn, it follows from Pythagorean theorem that the distance betweenb^i,j_(p,q)andV_→^i,jis betweend¹_min= 5n²+ 4n−4p−1andd¹_max=p

(d¹_min)²+ (2n)²< d¹_min+ 1(see Figure 3). For the same reasoning, the distance betweenb^i,j_(p,q)andV_←^i+1,jis betweend¹_min+ 2andd¹_min+ 3, the distance betweenb^i+1,j_(p0,q⁰)andV_←^i+1,jis betweend²_min= 5n²+ 4n−4p⁰−1andd²_min+ 1, and finally the distance betweenb^i+1,j_(p0,q⁰)andV_→^i,jis betweend²_min+ 2andd²_min+ 3. It follows that wheneverp > p⁰, thend²_min+ 3< d¹_min and the voters inV_→^i,jare closer tob^i+1,j_(p0,q⁰)than tob^i,j_(p,q). Similarly, ifp < p⁰thend¹_min+ 3< d²_minand the voters inV_←^i+1,jare closer tob^i,j_(p,q)than tob^i+1,j_(p0,q⁰).

Repeating the same argument for ballot boxesb^i,j_(p,q)andb^i,j+1_(p0,q⁰), we obtain also the following claim.

Claim 2. Letb^i,j_(p,q) andb^i,j+1_(p0,q⁰) be two ballot boxes. Then for every voterv ∈ V_↓^i,j it holdsdist(v, b^i,j_(p,q)) <

dist(v, b^i,j+1_(p0,q⁰))if and only ifq≤q⁰. Similarly, for everyv⁰ ∈V_↑^i,j+1it holdsdist(v⁰, b^i,j+1_(p0,q⁰))<dist(v⁰, b^i,j_(p,q))if and only ifq≥q⁰.

From the above claims it directly follows that if we start with a YES-instance of GRID TILING, we end up with a YES-instance of GERRYMANDERING.

For the reverse direction, it remains to show that any solutionB⁰ to the reduced instanceIof GERRYMAN-

DERINGhas to select exactly one ballot boxb^i,j_(p,q)in eachBi,jand that for each such box, the set of voters voting there is precisely the set of voters associated with(i, j).

First note that there are precisely k² positions of “red” voters. Hence each ballot box in a solution has to be closest to precisely one of these positions. Moreover, there are onlyk²more “red” voters than “blue” voters.

Hence, if “red” wins in all boxes, then by pigeon-hole principle “red” has to win in each ballot box by precisely 1. Now letb ∈ B⁰ be a ballot box closest toV_R^i,j. SinceV_R^i,j is precisely the set of “red” voters that vote inb, none of voters inS

∈{←,→,↑,↓}V^i0,j0

such that(i⁰−1)k+j⁰−1 >(i−1)k+j−1can vote inb. Now the number of the remaining “blue” voters is precisely5·5(i−1)k+(j−1)−1 = 4·P(i−1)k+(j−1)

`=0 5<sup>`</sup>. Therefore, if one group of voters amongV_←^i,j,V_↑^i,j,V_→^i,j, andV_↓^i,jdoes not vote atb, then “red” winsbby more than1vote and by pigeon-hole principle, there will be a ballot box where “red” does not win. Finally, it is easy to see that if there is more than one ballot box selected fromBi,j, then the box whereV_R^i,j vote cannot take all four groups of voters V_←^i,j,V_↑^i,j,V_→^i,j, andV_↓^i,j. Therefore, it follows that in eachBi,jexactly one ballot box is selected inB⁰and the set of voters in this ballot box is precisely the set of voters associated with(i, j). It follows from Claims 1 and 2 that the set of pairs associated with any such solution is a solution to GRIDTILING.

4 Algorithm

Our algorithm uses the machinery developed by Marx and Pilipczuk [18] for guessing a balanced separator in the Voronoi diagram of a potential solution. The main idea of our algorithm is as follows. The Voronoi diagram of Scould be viewed as a 3-regular planar graph withkfaces andO(k)vertices. From [18] we know that for such a graph there exists a polygonΓwhich goes throughO√

k

faces and vertices and there are at most ²₃kfaces strictly insideΓand at most ²₃kfaces strictly outsideΓ. Moreover,Γcould be defined by the sequence ofO√

k elements ofB. This allows us to guess all possible variants ofΓwithout actually knowing the solutionS, and for eachΓwe recurse into two smaller subproblems defined inside and outside ofΓ. Since the separatorΓis balanced and the recurrenceT(k) = n^O(^√k)T(²₃k)solves toT(k) = n^O(^√k), we obtain the desired time bound. See Figure 4 for the illustration.

We assume that no four elements ofB lie on the same circle. From this assumption, it follows that no vertex of any Voronoi diagram corresponding to a possible solution has degree more than 3. We also assume that for any four distinct points inB, one is never exactly the center of the unique circle passing through the three others. It

Figure 4: The Voronoi diagram of a potential solution. The solution is represented by large squares, small squares are the other possible ballot boxes. A separating polygonΓis in red and dashed, the boxes onΓare also in red.

Our algorithm guessesΓ, takes the red boxes in the solution and then solves the inside (blue and red squares) and the outside (black and red squares) separately.

means that for any subsetSofB, the vertices of the Voronoi diagram ofSare disjoint fromB. Both assumptions are for the ease of presentation and could be achieved by a small perturbation of the coordinates.

Next we state the result of Marx and Pilipczuk about finding small balanced separators in Voronoi diagrams.

Lemma 2([18]). LetS be a set ofkpoints on the plane. Consider the Voronoi diagram ofS, and the planar graphGassociated with it. There is a polygonΓwhich has lengthO√

k

and its vertices alternate between elements ofSand vertices ofGand each segment ofΓlies inside a face ofG. At most ²₃kfaces lie strictly inside Γ, and at most ²₃kstrictly outside.

Note that sinceΓpasses throughO√ k

faces, the number of faces which lie non-strictly inside (outside)Γ is bounded by³₄kwhenkis greater thanγ, whereγis some constant.

Now we are ready to prove Theorem 2 which we restate for convenience.

Theorem 2. For any constant number of candidates and for any voting ruleg,GERRYMANDERINGgis solvable in time(m+n)^O(^√k).

Proof. Our algorithm is a recursive brute-force search. An input to the recursion step is a particularstateR= (V, B,`,k,F,v,∆). A state consists of a set of votersV, a set of possible ballot boxesB, parameters`,ksuch that

`≤k ≤ |B|, a subset of boxesF ⊂B, for anyf ∈F andσ∈π(C)there is an integerv(f, σ)from0to|V|, and a set of segments∆, the segments form the boundary of the region containingV andB. Each segmentδ∈∆ has exactly one endpoint inF, and we denote this endpoint byδF. The algorithm returns whether the given state isvalid.

Definition 3(Valid state). A state is valid if there exists a subsetS ⊂ B\F such that|S|+|F| = k, and in the election with votersV and boxesS∪F the target partypwins in at least`boxes ofS, and for eachf ∈F andσ∈π(C)there are exactlyv(f, σ)voters with preference listσvoting in boxf. Additionally, in the Voronoi diagram ofS∪F, each segmentδof∆lies completely inside the cell (touching the border) corresponding to the boxδF.

If the state is valid, the algorithm also returns a particular subsetS⊂B\F satisfying Definition 3.

Intuitively,V andBcorrespond to a set of voters and possible boxes in the current subtask,Fcorresponds to a set of boxes which we have already decided to take into the solution on the previous steps of the recursion, and

for any box inFthe number of voters with a particular preference list is also fixed throughv. Among the boxes ofB\Fwe are supposed to select at mostk− |F|to go into the solution, and in at least`of them the target party should win. The condition that segments of∆lie inside the corresponding cells enforces that these cells actually separate the whole state from the outside of the region defined by∆.

Initially, to run our algorithm having the input(C, V, B, `, k, p)to GERRYMANDERINGf, we proceed as fol- lows. First, find an equilateral triangleT such that V andBare completely insideT. Then mirror each vertex of the triangle against the opposing side, and denote the resulting set of three points asF. Consider the Voronoi diagram ofB∪F, by construction the outer faces are exactly the cells ofF, and these faces are disjoint fromB andV. Construct a polygon∆in the following way. Start from a point ofF and go to a next clockwise point ofF by a sequence of two segments having as the common point a vertex of the diagram on the border between the corresponding outer faces; repeat until the polygon reaches the starting point again.The initial state is defined as R0 = (V,B ∪F, `,k+ 3,F,v, ∆)wherev(f, σ) = 0for any f ∈ F andσ ∈ π(C). Since for any v ∈V anyb ∈Bis closer tovthan anyf ∈F,R0is valid if and only if(C, V, B, `, k, p)is a YES-instance of GERRYMANDERINGf.

The recursive step.On the given stateR= (V,B,`,k,F,v,∆)the algorithm proceeds as follows. Ifk− |F|is at mostγ, whereγis the constant mentioned below the statement of Lemma 2, we try all possibleS⊂Bof size k− |F|and for each check the conditions of validity in polynomial time, sinceγ =O(1)the whole procedure works in polynomial time as well. Ifk− |F|> γ, try all possible polygonsΓof the form as in Lemma 2, that is, alternating between elements ofBand potential vertices of the Voronoi diagram of the solution, and having length at mostα√

kwhereαis the constant underOin the lemma. Since a vertex of a Voronoi diagram constructed over any subset ofBis equidistant from three elements from this subset and is uniquely determined by these three elements, there are at most|B|³ potential locations for a Voronoi diagram vertex. Therefore there are at most

|B|^4α

√kvariants forΓ. We consider only theseΓwhich do not go out of the region defined by∆.

For each possibleΓ, we split the instance in two parts. Denote byQthe set of boxes onΓ. LetV1(V2) be the set of voters inside (outside) ofΓ, andB1(B2) be the set of possible boxes inside (outside) ofΓ,V1∪V2=V, V1∩V2 =∅,B1∪B2 =B. Fori∈ {1,2}, letFi =Q∪(F∩Bi)and let∆ibe those segmentsδin∆such thatδ_F ∈F_i. The fixed preference countsv₁andv₂forF₁andF₂respectively are defined as follows. For boxes inF \Qthe value ofv is transferred directly tov₁orv₂depending on to which part the box went. For boxes inF∩Qand for any preference listσ∈ π(C)we guess how the value ofv(f, σ)is split betweenv₁(f, σ)and v₂(f, σ). For boxes inQ\Fwe guess separately the value ofv₁(f, σ)andv₂(f, σ).

Finally, guess howk+|Q\F|is split between the two partsk₁andk₂,k₁andk₂must be each at most ³₄k.

Also guess how`−wis split between`₁and`₂, wherewis the number of boxes inQ\Fwhere the target party pwins if voters fromV1andV2vote according tov1andv2.

Next, we run the recursion on the stateR1 = (V1,B1,`1,k1,F1,v1,∆1∪Γ)and on the stateR2 = (V2, B2,`2,k2,F2,v2,∆2∪Γ). If bothR1 andR2are reported as valid states, we return thatRis valid and take S=S1∪S2∪(Q\F). Otherwise, we continue to the next choice ofΓ. If for all choices ofΓwe do not succeed, we return thatRis not valid.

The correctness. Now we prove the correctness of the algorithm above. The proof is by induction onk. In the base casek− |F| ≤γthe algorithm tries all possible ways to select the ballot boxes and then checks the definition directly. In the casek− |F|> γ, assume first that the algorithm reports a given stateR= (V,B,`,k,F,v,∆)as valid. So for someΓthe corresponding splitting statesR1= (V1,B1,`1,k1,F1,v1,∆1∪Γ)andR2= (V2,B2,

`₂,k₂,F₂,v₂,∆₂∪Γ)are reported as valid by the algorithm, and by induction that means that they are actually valid states since bothk₁andk₂are at most ³₄k. Consider setsS₁ ⊂B₁\F₁andS₂⊂B₂\F₂returned by the recursive calls, by inductionS₁andS₂also satisfy Definition 3 on the validity ofR₁andR₂, respectively. As before, denoteS₁∪S₂∪(Q\F)byS. We claim thatSsatisfies Definition 3 forR, and thereforeRis valid.

By construction ofR₁andR₂,k=k₁+k₂− |Q\F|, and sinceS₁,S₂, andQ\F are pairwise disjoint, the size ofSis indeed equal tok− |F|. Next, we show that in the election with boxesS∪Fthe target partypwins in at least`districts ofSand that the votes in districts ofFare distributed according tov. The next claim shows that for voters in each part the box where they vote is the same in the election with boxesSi∪Fi and in the election with boxesS∪F.

Claim 3. Fori∈ {1,2}and any voterx∈Vi, the closest box toxinS∪Fbelongs toSi∪Fias well.

Proof. Leti= 1, the other case is analogous. Assume the contrary, there is a voterx∈V1such that there is a box b∈S2∪F2\Qwhich is strictly closer toxthan any box inS1∪F1, since two boxes cannot be at the same distance tox. Consider a straight line segment fromxtob, sincexandblie on the different sides ofΓ, the segment crosses Γat least once, denote any intersection point byy. SinceR2is valid, the segments ofΓlie completely inside the

cells ofQin the Voronoi diagram ofS2∪F2. That means that there isq∈Qsuch thatyis at least as close toqas to any other box inS2∪F2, includingb. But thendist(x, q)≤dist(x, y) + dist(y, q)≤dist(x, y) + dist(y, b) = dist(x, b), where the first inequality is the triangle inequality, and this contradicts the initial assumption.

By Claim 3, for eachi∈ {1,2}and for each box inS_iexactly the same set of voters votes there in the election with boxesS∪Fcompared to the election with boxesS_i∪F_i. It means that among the boxes inS₁andS₂the target partypwins in exactly`<sub>1</sub>+`₂ of them. For eachi ∈ {1,2}and each boxf inF_i\Qit also holds that exactly the same set of voters votes inf in the election with boxesS∪Fas votes inf in the election with boxes Si∪Fi. Sincev(f, σ) = vi(f, σ)for anyσ∈π(C), the vote distribution onFi\Qis as desired. For each box q∈Qand eachσ∈π(C)by constructionv(q, σ) =v1(q, σ) +v2(q, σ). Since voters fromV1andV2who vote inqare exactly preserved,v(q, σ)is indeed equal to the number of voters with preference listσwho vote inq.

This finishes the proof thatRis valid andSsatisfies Definition 3.

Finally, we show that for eachδ∈∆,δlies inside the cell ofδF in the Voronoi diagram ofS∪F. IfδF ∈Q, sinceR1andR2are valid,δlies inside the cell ofδF in the Voronoi diagram ofSi∪Fi fori ∈ {1,2}, then no other point inS∪F is closer to each point onδthanδF. IfδF ∈/ Q,δis completely inside or outside ofΓ, and the same argument as in Claim 3 shows that for any point onδthe closest box is preserved, then it has to beδF, sinceδis in∆ifor somei∈ {1,2}, andRiis valid.

Towards the other direction, assume thatRis valid, we show that the algorithm correctly reports the validity ofR. Consider a particularS from Definition 3, by Lemma 2 for the Voronoi diagram ofS∪F, there exists a polygonΓof lengthO(√

k)which is a balanced separator forS∪F. Since the algorithm tries all polygons of this form, it will tryΓas well, or report thatRis valid earlier. When considering the polygonΓ, the algorithm will guess the right values ofk₁,k₂,`<sub>1</sub>,`₂, andv₁,v₂on the boxes ofQ\F, according to the elections onS∪F.The new statesR1andR2are valid since the elections onS∩F exactly induce conditions onRi, and there exists a valid selection of boxesSi= (S∪F)∩Bi\Q, for eachi∈ {1,2}. Therefore the recursive call will findRivalid by induction. Thus the proof of correctness is finished.

Running time analysis.Denote byT(k)the worst-case running time of our algorithm wherekis the respective parameter of the input state. Ifk≤γ, the algorithm does a brute force in polynomial time soT(k) = (n+m)^O(1). Ifk > γ, we try at mostm^4α

√kpolygonsΓ, for each of them we in time at mostk`n^2α

√k guess the parameters of the new two instancesk₁,`<sub>1</sub>,v<sub>1</sub>andk<sub>2</sub>,`₂,v₂. We run our algorithm on the two instances recursively, and in both instances the value of the parameter is bounded by ³₄k, so we get the following recurrence relation,

(T(k)≤(n+m)^β

√

kT(³₄k), k > γ T(k)≤(n+m)^β, k≤γ,

whereβis some constant. Now, by applying the first upper bound until the expression underT is at mostγwe get T(k)≤(n+m)^β

√

k(1+q+q²+...+q^t)

≤(n+m)^β

√k/(1−q)= (n+m)^O(^√k),

whereq=p

3/4andt=dlog_4/3(k/γ)e.

5 Concluding Remarks

We initiated study of a gerrymandering problem from the perspective of fine-grained and parameterized complex- ity. Our main result is asymptotically tight algorithm with respect to parameterk, the number of opened ballot boxes, together with a matching lower bound.

For future work, it would be interesting to investigate the complexity of finding an approximate solution for the problem. That is the question whether there is an efficient (polynomial time) algorithm that either finds a set ofkballot boxes such that pwins <sup>`</sup><sub>c</sub> districts, for some constantc, or correctly decides that in no elections withkdistricts the candidatepwins`of them. Another interesting question is what happens, when we introduce obstacles, such as lakes or hills, in the instance so that the distances are not anymore Euclidean distances in the plane. Can we still get better than trivialm^kalgorithm in this case? Finally, it is also natural to impose restrictions on sizes of districts, so they are not too disproportional.

6 Acknowledgments

This work is supported by the Research Council of Norway via the project “MULTIVAL”.

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