Abstract
We study the districting problem from an axiomatic point of view in a framework with two parties, deterministic voter preferences and geographical constraints. The axioms are normatively motivated and reflect a notion of fairness to voters. Our main result is an “impossibility” theorem demonstrating that all anonymous districting rules are necessarily complex in the sense that they either use information beyond the mere number of districts won by the parties, or they violate an appealing consistency requirement according to which an acceptable districting rule should induce an acceptable districting of appropriate subregions.
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Notes
See, e.g., Tasnádi (2011) for an overview.
Since a districting forms a partition of the given region, it is evidently not possible to move from one districting to another districting by changing only one district.
Observe that overall determinacy, i.e. that \(\delta _A (F_\varPi )\) and \(\delta _B (F_\varPi )\) be singletons for every problem \(\varPi \), is a strictly stronger requirement than two-district determinacy; for instance, the least biased solution satisfies two-district determinacy but can easily be shown to violate overall determinacy.
To verify this, observe that if there exist admissible districtings \(D,D'\in \mathcal{D}_\varPi \) with \(\delta _A(D)=2\) and \(\delta _A(D')=1\), then one must have \(0.5<\mu _A(X)/\mu (X)<0.75\). Thus, \(D'\) must be chosen both by \(M\! E\) and \(L\! B\).
Clearly, this requirement has to be restricted to subregions that are unions of districts, since a given districting does in general not induce an admissible sub-districting on other subregions.
We would like to thank Dezső Bednay for suggestions that improved our original proof.
For a definition of overall determinacy see Footnote 3.
We call two subsets of the plane neighboring if they share a common boundary of positive length.
If \(\mu _A(e)\ne \mu (e)/2\), then \(\mu _A(e')\ne \mu (e')/2\) can be guaranteed by exchanging sets of sufficiently small measure \(\mu \) between \(d\) and \(e\). In addition, if \(\mu _A(e)=\mu (e)/2\) and \(\mu _A(e')=\mu (e')/2\), then we can repeat the exchange of territories between \(e'\) and \(d'\) to ensure that both sets satisfy (1).
Both pictures only show the two districts involved in a territorial exchange and not the entire districtings.
It might happen that \(d'\) or \(e'\) violate (1) since we only took care of the shapes and sizes of the two districts. However, Lemma 2 ensures that through an appropriate territorial exchange between \(d'\) and \(e'\) we can also ensure (1). In what follows we will carry out all territorial exchanges between districts so as to satisfy (1) without explicitly mentioning Lemma 2 each time.
In fact the number of required iterations is at most \(\lceil t\mu (Y)/\mu (X) \rceil +1\), where \(Y\) stands for the area “intertwined” by \(d\cup e\).
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We are most grateful to John Duggan, an associate editor and two anonymous referees for their thorough remarks and suggestions which helped to improve the first version of this paper.
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Appendix: Regular districting problems
Appendix: Regular districting problems
We have already seen examples of linked geographies in Figs. 1, 3 and 4. In this appendix we provide a natural and large class of further examples of districting problems with linked geographies.
A bounded subset \(A\) of \(\mathbb {R}^2\) will be called strictly connected if its boundary \(\partial A\) is a Jordan curve (i.e. a non self-intersecting continuous loop). A subset \(A\) of a strictly connected set \(B\subseteq \mathbb {R}^2\) separates \(B\) if \(B{\setminus } A\) is not strictly connected. We call a continuous function \(f:X\rightarrow \mathbb {R}\) nowhere constant if for any \(x\in X\) and any neighborhood \(N(x)\) of \(x\) there exists a \(y\in N(x)\) such that \(f(x)\ne f(y)\).
Definition 10
(Regular districting problems) A districting problem \(\varPi =(X,\mathcal{A},\mu ,\mu _A,\mu _B,t,G)\) is called regular if
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1.
\(X\) is a bounded and strictly connected subset of \(\mathbb {R}^2\),
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2.
\(\mathcal{A}\) equals the set of Borel sets on \(X\), i.e. following standard notation \(\mathcal{A}=\mathcal{B}(X)\),
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3.
\(\mu \) is a finite and absolutely continuous measure on \((X,\mathcal{B}(X))\) with respect to the Lebesgue measure,
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4.
\(G\) consists of all bounded, strictly connected and \(\mu (X)/t\) sized subsets lying in \(\mathcal{B}(X)\) and satisfying (1),
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5.
there exists a continuous nowhere constant function \(f:X\rightarrow \mathbb {R}\) such that \(\mu _A(C)=\int _C f(\omega )d\mu (\omega )\) for all \(C\in \mathcal{B}(X)\), and
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6.
\(\mu _B\) is given by \(\mu _B(C)=\mu (C)-\mu _A(C)\) for all \(C\in \mathcal{B}(X)\).
The fifth condition is a technical assumption to ensure that the districtings emerging in the proof of Lemma 3 below can be selected in a way that they satisfy (1). Specifically, we have the following lemma.
Lemma 2
If we have two neighboring,Footnote 8 bounded, strictly connected and \(\mu (X)/t\) sized sets \(d,e\in \mathcal{B}(X)\) such that \(\mu _A(d)=\mu (d)/2\) (i.e \(d\) violates (1)), then we can exchange territories between \(d\) and \(e\) in a way that the two resulting bounded, strictly connected and \(\mu (X)/t\) sized sets \(d',e'\in \mathcal{B}(X)\) satisfy (1).
Proof
Pick a point \(x\in \partial d\cap \partial e\) from the relative interior of the common boundary of \(d\) and \(e\). Since \(f\) is nowhere constant there exists a \(y\) arbitrarily close to \(x\) in the interior of \(d\) such that \(f(y)\ne f(x)\). Assume that \(f(y)>f(x)\). There exist a neighborhood \(N_{\varepsilon _y}(y)\) of \(y\) and a neighborhood \(N_{\varepsilon _x}(x)\) of \(x\) such that
by continuity of \(f\).
By establishing a sufficiently thin connection between \(N_{\varepsilon _y}(y)\) and \(N_{\varepsilon _x}(x)\), which shall be assigned to \(e'\), and exchanging a subset of \(N_{\varepsilon _y}(y)\) with a subset of \(N_{\varepsilon _x}(x)\cap e\) in a way such that \(\mu (d)=\mu (d')=\mu (e)=\mu (e')\), we can guarantee that \(\mu _A(d')\ne \mu (d')/2\).Footnote 9
Finally, the case of \(f(y)<f(x)\) can be handled in an analogous way. \(\square \)
In the following, we write \(D\sim D'\) if \(D,D'\in \mathcal{D}_\varPi \) and there exists a sequence \(D_1,\ldots ,D_k\) of districtings such that \(D=D_1, \left\{ D_2,\ldots ,D_{k-1}\right\} \subseteq \mathcal{D}_\varPi , D'=D_k\) and \(\# D_i\cap D_{i+1}=t-2\) for all \(i=1,\ldots ,k-1\). It is easily verified that \(\sim \) is an equivalence relation on the set of districtings.
Lemma 3
The geographies of regular districting problems are linked.
Proof
Linkedness is clearly satisfied if \(t=1\) or \(t=2\). We show that the linkedness of the geographies of all regular districting problems for \(t\le n\) implies the linkedness of the geographies of all regular districting problems for \(t=n+1\). From this, Lemma 3 follows by induction.
Take two arbitrary districtings \(D\) and \(E\) of a districting problem with \(t=n+1\). We can pick a district \(d\in D\) such that \(d\) and \(X\) have a non-degenerate curve as a common boundary, i.e. there exists a curve \(C\) of positive length such that \(C\subseteq \partial d\cap \partial X\). We divide our proof into three steps.
Step 1 We show that there exists a districting \(D'\sim D\) that contains a district \(d'\) which shares a common boundary of positive length with the boundary of \(X\) and which does not separate \(X\).
If \(d\) itself does not separate \(X\) we are done. Thus, assume that \(d\) separates \(X\). For simplicity, we start with the case in which \(d\) separates \(X\) into only two regions as shown in the picture on the left of Fig. 5.Footnote 10 By exchanging territories between the two districts \(d\) and \(e\), where \(e\) is a neighboring district of \(d\), as shown in the picture on the left of Fig. 5, we can arrive at districts \(d'\) and \(e'\) such that \(d'\) does not separate \(X\).Footnote 11
More generally, assume that \(d\) separates \(X\), where the number of strictly disconnected regions of \(X{\setminus }\{d\}\) equals \(k\le n\). We can find a district \(e\in D\) and a unique boundary element \(x\in \partial e\) such that \(x\in \partial d\cap \partial X\) and such that \(\partial d\) and \(\partial e\) have a common curve of positive length starting from \(x\). Hence, one can exchange territories between \(d\) and \(e\) so that for the resulting new districts \(d'\) and \(e'\) we have that \(d'\) separates \(X\) into at most \(k-1\) strictly disconnected regions. Clearly, \(D'=(D{\setminus }\{d,e\})\cup \{d',e'\}\sim D\). Repeating the described bilateral territorial exchange \(k-1\) times, we thus arrive at a districting \(D'\) that contains a district \(d'\) which shares a common boundary with \(X\) and which does not separate \(X\).
By Step 1, we may thus assume that \(d\in D\) shares a boundary of positive length with \(X\) and does not separate \(X\).
Step 2 We establish that there exists a districting \(E'\sim E\) containing a district \(e\in E'\) such that \(e, d\) and \(X\) have a nondegenerate common boundary, \(\mu (d\cap e)>0\) and \(d\cup e\) does not separate \(X\).
Clearly, there exist a district \(e\in E\) possessing a common boundary with \(d\) and \(X\), and satisfying \(\mu (d\cap e)>0\).
Assume that \(e\) separates \(X\), where the number of strictly disconnected regions of \(X{\setminus }\{e\}\) equals \(k\le n\) (see Fig. 6 to the left for a situation with \(k=3\)). Then \(d^c\cap \partial e\cap \partial X\ne \emptyset \). We can find a district \(e'\in E\) with a unique boundary element \(x\in \partial e'\) satisfying \(x\in d^c\cap \partial e\cap \partial X\) and that \(\partial e\cap \partial e'\) has a common curve of positive length starting from \(x\) (as illustrated in the left hand side of Fig. 6). Hence, one can exchange territories between \(e\) and \(e'\) so that for the resulting new districts \(h\) and \(h'\) we have that \(d\cap e\subset h, h\) separates \(X\) into at most \(k-1\) strictly disconnected regions (see the right hand side of Fig. 6). Clearly, \(E'=(E{\setminus }\{e,e'\})\cup \{h,h'\}\sim E\) and we can repeat the procedure to reduce the number of strictly disconnected regions by replacing \(E\) and \(e\) with \(E'\) and \(h\), respectively, until we arrive at a districting \(E'\sim E\) containing a district \(e'\) that does not separate \(X\) and has a common boundary with \(d\). Without loss of generality, we can thus replace \(e'\) and \(E'\) by \(e\) and \(E\), respectively.
We still have to ensure that \(d\cup e\) does not separate \(X\). A situation in which \(d\cup e\) separates \(X\) is shown in the picture on the left hand side of Fig. 7. In addition, the same picture contains (by the absolute continuity of \(\mu \)) a possible neighboring district \(e'\) to \(e\), which is drawn in a way such that \(e\cup e'\) does not separate \(X\), it covers an area from the separated regions and also an area within \(d\cup e\). A possible exchange of territories which reduces the separated area by \(d\cup e\) is illustrated in Fig. 7, where \(d\cup h\) separates a smaller area than \(d\cup e\).Footnote 12 Pick an arbitrary districting \(H\) of \(X{\setminus } (e\cup e')\) into \(n-1\) strictly connected districts and let \(E'=H\cup \{h,h'\}\). Observe that \(E{\setminus }\{e\}\sim H\cup \{e'\}\) by the induction hypothesis, \(h\cup h'= e\cup e'\) by construction, and therefore \(E\sim E'\). Replace \(e\) and \(E\) with \(h\) and \(E'\), respectively. After repeating the described territorial exchange finitely many timesFootnote 13 one arrives at a district \(e\) and a districting \(E\) such that \(d\cup e\) does not separate \(X\) and \(e\) still satisfies the other desired properties.
Step 3 Since \(d\cup e\) does not separate \(X\) and \(\mu \) is absolutely continuous, there exists a strictly connected set \(h\) such that \(\mu (h)=2\mu (X)/(n+1), d\cup e\subset h, d'=h{\setminus } d\in G\) and \(e'=h{\setminus } e\in G\) and \(h\) does not separate \(X\) (see Fig. 8). Let \(H\) be a districting of \(Y=X{\setminus } h\) into \(n-1\) strictly connected districts. Then \(\varPi \mid _{Y\cup d'}\) and \(\varPi \mid _{Y\cup e'}\) are regular districting problems, and therefore it follows by the induction hypothesis that \(D\sim H\cup \{d,d'\}\) and \(H\cup \{e,e'\}\sim E\). Clearly, \(\{d,d'\}\sim \{e,e'\}\), which gives \(H\cup \{d,d'\}\sim H\cup \{e,e'\}\). Finally, the statement of Lemma 3 follows from the transitivity of \(\sim \). \(\square \)
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Puppe, C., Tasnádi, A. Axiomatic districting. Soc Choice Welf 44, 31–50 (2015). https://doi.org/10.1007/s00355-014-0824-9
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DOI: https://doi.org/10.1007/s00355-014-0824-9