Distance rationalization of voting rules

Abstract

The concept of distance rationalizability allows one to define new voting rules or rationalize existing ones via a consensus, i.e., a class of elections that have a unique, indisputable winner, and a distance over elections: A candidate is declared an election winner if she is the consensus candidate in one of the nearest consensus elections. Many classic voting rules are defined or can be represented in this way. In this paper, we focus on the power and the limitations of the distance rationalizability approach. Lerer and Nitzan (J Econ Theory 37(1):191–201, 1985) and Campbell and Nitzan (Soc Choice Welf 3(1):1–16, 1986) show that if we do not place any restrictions on the notions of distance and consensus then essentially all voting rules can be distance-rationalized. We identify a natural class of distances on elections—votewise distances—which depend on the submitted votes in a simple and transparent manner, and investigate which voting rules can be rationalized via distances of this type. We also study axiomatic properties of rules that can be defined via votewise distances.

Appendices

Appendix A: $\varvec{{\mathcal {M}}}$-Borda is not homogeneous

We show that Theorem 9 cannot be extended to $\ell _1$-votewise rules by showing that ${\mathcal {M}}$-Borda is not homogeneous. Consider election $E$:

$v_1$	$v_2$	$v_3$	$v_4$	$v_5$	$v_6$
$b$	$a$	$c$	$c$	$c$	$d$
$a$	$b$	$b$	$b$	$d$	$a$
$d$	$d$	$a$	$a$	$a$	$b$
$e$	$e$	$e$	$e$	$e$	$e$
$c$	$c$	$d$	$d$	$b$	$c$

A simple calculation shows that, to become a majority winner, $a$ needs $4$ swaps, $b$ needs $3$ swaps, $c$ needs $4$ swaps, and $d$ needs $5$ swaps. Thus, $b$ is the unique winner of this election under ${\mathcal {M}}$-Borda. However, in the election $2E$ candidate $b$ needs $5$ swaps to become a majority winner, while $c$ requires only $4$ swaps.

Appendix B: Proof of theorem 3

We will show that Plurality is $\ell _1$-votewise rationalizable with respect to strong unanimity. Given a set of candidates $C=\{c_1, \dots , c_m\}$ and two preference orders $u, v\in P(C)$, we say that $v$ can be obtained from $u$ by a cyclic shift if there exists an $i\in [m]$ and a permutation $\pi :C\rightarrow C$ such that $v=\pi (c_1)\succ \cdots \succ \pi (c_m)$, and $u=\pi (c_i)\succ \cdots \succ \pi (c_m)\succ \pi (c_1)\succ \cdots \succ \pi (c_{i-1})$.

We partition $P(C)$ into $m$ groups $L_1, \dots , L_m$ by setting $L_i=\{v\in P(C)\mid {\mathrm {pos}}(v, c_i)=1\}$. Set $s=(m-1)!$ and, for each $i\in [m]$, enumerate the preference orders in $L_i$ as $v_i^1, \dots , v_i^s$ so that for every $i, j\in [m]$ and every $t=1, \dots , s$ the vote $v_j^t$ can be obtained from the vote $v_i^t$ by a cyclic shift. To see that this is possible, note that for each $v_i^t$, $i\in [m]$, $t\in [s]$, and each $j\in [m]$, there is exactly one vote in $L_j$ that can be obtained from $v_i^t$ by a cyclic shift.

The distance $d$ is defined as follows. We set $d(v_i^t, v_j^r)=0$, if $i=j$ and $t=r$; $d(v_i^t, v_j^r)=1$, if $i=j$ or $t=r$ but $(i, t)\ne (j,r)$, and $d(v_i^t, v_j^r)=2$, if $i\ne j\text { and } t\ne r$. In other words, $d(u,u)=0$ for all $u\in P(C)$, $d(u,v)=1$ if and only if $u\ne v$ and $u$ and $v$ rank the same candidate first or $v$ is obtained from $u$ by a cyclic shift, and $d(u,v)=2$ in all other cases. Observe that since $d(u, v)\in \{1, 2\}$ for $u\ne v$, the mapping $d$ satisfies the triangle inequality; it is also obviously symmetric.

Consider an election $E=(C, V)$. For every $j\in [m]$ and $r\in [s]$ let $a_j^r$ be the number of voters in $V$ with preference order $v_j^r$. Let us calculate the distance from $E$ to the ${\mathcal {S}}$-consensus election $(C, X_i^t)$ with $X_i^t = |V|\times v_i^t$:

$$\begin{aligned} \widehat{d}(V,X_i^t)= \sum _{r\in [s]\setminus \{t\}} \!\!\!\!\!a_i^r \ \ + \sum _{j\in [m]\setminus \{i\}} \!\!\!\!\!\!a_j^t \ +\ 2\sum _{j\ne i}\sum _{r\ne t} a_j^r =\sum _{j\ne i}\sum _{r\in [s]} a_j^r \ \ + \sum _{j\in [m]}\sum _{r\ne t} a_j^r. \end{aligned}$$

As the first summand does not depend on $t$, the distance from $E$ to the nearest ${\mathcal {S}}$-consensus with winner $c_i$ is $\min _{t\in [s]} \widehat{d}(V,X_i^t)=\sum _{j\ne i} \sum _{r\in [s]} a_j^r + \min _{t\in [s]}\sum _{j\in [m]}\sum _{r\ne t} a_j^r.$ The second component of this expression does not depend on $i$, while its first component counts the number of voters who do not rank $c_i$ first. Thus, the nearest strong unanimity consensus to $E$ has $c_i$ as its winner if and only if $i$ minimizes the sum $\sum _{j\in [m]\setminus \{i\}}\sum _{r\in [s]} a_j^r$ over all $i\in [m]$, i.e., $c_i$ has the largest number of first-place votes. Thus, Plurality is $({\mathcal {S}}, \widehat{d})$-rationalizable.$\square$

Appendix C: Proof of theorem 5

As in the proof of Theorem 4, we set $C = \{a,b,c\}$ and use the same notation as throughout Sect. 4.2. We focus on STV, and define $V$ and $W$ as in the proof of Theorem 4. Also, we set $m_{sm}=\min \{S, M\}$, $m_{sb}=\min \{S, B\}$.

We will consider the majority consensus first. For the sake of contradiction, suppose that STV is $({\mathcal {M}}, \widehat{d}^N)$-rationalizable for some neutral distance $d$ over $P(C)$ and a symmetric norm $N$ that is monotonic in the positive orthant. Let $(C, M_a)$ be an ${\mathcal {M}}$-consensus with winner $a$ that is closest to $(C, V)$ and let $(C, M_b)$ be an ${\mathcal {M}}$-consensus with winner $b$ that is closest to $(C, V)$. We have $\widehat{d}^N(V,M_a) = N(0,\ldots ,0,\min \{m_{sb},m_{sm}\}) = \widehat{d}^N(V,M_b).$ However, this is a contradiction as $a$ is the unique winner of $(C, V)$. Thus, STV is not $({\mathcal {M}},\widehat{d}^N)$-rationalizable.

Similarly, suppose that STV is $({\mathcal {U}},\widehat{d}^N)$-rationalizable. Let $(C, U_a)$ be a ${\mathcal {U}}$-consensus with winner $a$ that is closest to $(C, V)$ and let $(C, U_b)$ be a ${\mathcal {U}}$-consensus with winner $b$ that is closest to $(C, V)$. We have

$$\begin{aligned} \widehat{d}^N(V, U_a)= & {} N(0, \dots , 0, m_{sm}, \dots , m_{sm}, m_{sb}). \\ \widehat{d}^N(V, U_b)= & {} N(m_{sb}, \dots , m_{sb}, 0, \dots , 0, m_{sm}). \end{aligned}$$

As $a$ is the unique winner of $(C, V)$, we have $\widehat{d}^N(V, U_a)<\widehat{d}^N(V, U_b)$, and therefore by symmetry and monotonicity $m_{sm}<m_{sb}$.

Now, let $(C, U'_a)$ be a ${\mathcal {U}}$-consensus with winner $a$ that is closest to $(C, W)$ and let $(C, U'_b)$ be a ${\mathcal {U}}$-consensus with winner $b$ that is closest to $(C, W)$. We have

$$\begin{aligned} \widehat{d}^N(W, U'_a)= & {} N(0, \dots , 0, m_{sm}, \dots , m_{sm}, m_{sm}), \\ \widehat{d}^N(W, U'_b)= & {} N(m_{sb}, \dots , m_{sb}, 0, \dots , 0, m_{sb}). \end{aligned}$$

As $b$ is the unique winner of $(C, W)$, we have $\widehat{d}^N(W, U'_a) > \widehat{d}^N(W, U'_b)$, and hence $m_{sm} > m_{sb}$, a contradiction. Thus, STV is not $({\mathcal {U}},\widehat{d}^N)$-rationalizable.

Appendix D: Condorcet consensus and homogeneity

The following example shows that the Condorcet consensus is not split-homogeneous. Consider the following election $E=(C, V)$ with $C=\{a, b, c, d, e\}$ and $V=(v_1, \dots , v_{12})$.

$v_1$	$v_2$	$v_3$	$v_4$	$v_5$	$v_6$	$v_7$	$v_8$	$v_9$	$v_{10}$	$v_{11}$	$v_{12}$
$a$	$b$	$c$	$d$	$e$	$c$	$e$	$a$	$b$	$c$	$d$	$c$
$b$	$c$	$d$	$e$	$a$	$a$	$d$	$e$	$a$	$b$	$c$	$a$
$c$	$d$	$e$	$a$	$b$	$b$	$c$	$d$	$e$	$a$	$b$	$b$
$d$	$e$	$a$	$b$	$c$	$d$	$b$	$c$	$d$	$e$	$a$	$d$
$e$	$a$	$b$	$c$	$d$	$e$	$a$	$b$	$c$	$d$	$e$	$e$

Set $V_1=(v_1, \dots , v_5)$, $V_2=(v_7, \dots , v_{11})$, $V'_1=(v_1, \dots , v_6)$, $V_2=(v_7, \dots , v_{12})$. The profile $V_1$ is a Condorcet cycle, and the profile $V_2$ is obtained from $V_1$ by reversing the preference order of each voter. Voters $v_6$ and $v_{12}$ are identical and rank candidate $c$ first. Since voters from the groups $V_1$ and $V_2$ ‘cancel’ each other, $c$ is the Condorcet winner in $E$. On the other hand, consider elections $E_1 = (C, V'_1)$ and $E_2= (C, V'_2)$. In $E_1$, $b$ is ranked above $c$ in $4$ votes, so $c$ is not a Condorcet winner in $E_1$. Similarly, in $E_2$, $d$ is ranked above $c$ in $4$ votes, so $c$ is not a Condorcet winner in $E_2$ either.

While Dodgson$^\infty$ is homogeneous, there are $\ell _\infty$-votewise distances that, paired with the Condorcet consensus, yield rules that fail homogeneity.

Proposition 10

There exists a set of candidates $C$ and a distance $d$ on $P(C)$ such that the voting rule rationalized by $({\mathcal {C}},\widehat{d}^\infty )$ is not homogeneous.

Proof

We will use two types of operations on preference profiles. Operations of the first type are the usual swaps of adjacent candidates. Let us now define an operation of the second type: a two-candidate shift. Given a preference order $v$ and two candidates $c$ and $d$ that are ranked consecutively in $v$, a two-candidate shift of $(c, d)$ pushes both $c$ and $d$ two positions upwards maintaining their relative ranking; if this is not possible, then the two-candidate shift is not defined. For example, given a preference order $u = a\succ b\succ c\succ d\succ e$ over the candidate set $C = \{a,b,c,d,e\}$, a two-candidate shift of $(c, d)$ transforms $u$ into $v = c\succ d\succ a\succ b\succ e$. Note that $v$ can be transformed back into $u$ by a two-candidate shift of $(a, b)$; generally, if a vote $u'$ can be transformed into a vote $v'$ by a two-candidate shift, then $v'$ can be transformed into $u'$ by a two-candidate shift as well.

We define the distance $d$ between two preference orders $u$ and $v$ as the minimum number of operations of both types that transform $u$ into $v$. It is easy to see that $d$ is indeed a distance, because it counts the number of reversible operations that are required to transform one preference order into the other. Let $\mathcal{{R}}$ be the voting rule that outputs all $({\mathcal {C}},\widehat{d}^\infty )$-winners.

We will now build an election $E = (C,V)$ such that $\mathcal{{R}}(E) \ne \mathcal{{R}}(2E)$. Set $C = \{a,b,c,d\} \cup S$, where $S=\cup _{j=1}^{18} S_j$ is disjoint from $\{a, b, c, d\}$ and $S_1, \dots , S_{18}$ are pairwise disjoint sets of candidates of size $4$ each. Let $V = (v_1, v_2, \ldots , v_6)$; the voters’ preference orders are given in the following table, where the candidates in each set $S_j$, $j=1, \dots , 18$, are listed in an arbitrary order and each symbol $\downarrow$ replaces all candidates that have not been listed explicitly in the vote (for instance, in the first vote $\downarrow$ replaces $S\setminus (S_1\cup S_6\cup S_{12})$).

$v_1$	$v_2$	$v_3$	$v_4$	$v_5$	$v_6$
$c$	$a$	$d$	$d$	$d$	$c$
$S_1$	$b$	$S_2$	$S_3$	$a$	$a$
$b$	$c$	$c$	$b$	$S_4$	$S_5$
$S_6$	$d$	$S_7$	$S_8$	$b$	$b$
$a$	$S_9$	$a$	$a$	$S_{10}$	$S_{11}$
$S_{12}$	$S_{13}$	$S_{14}$	$S_{15}$	$c$	$d$
$d$	$S_{16}$	$b$	$c$	$S_{17}$	$S_{18}$
$\downarrow$	$\downarrow$	$\downarrow$	$\downarrow$	$\downarrow$	$\downarrow$

Note that each candidate in $\{a,b,c,d\}$ is preferred to each candidate in $S$ by at least five voters. Further, four voters prefer $a$ to $b$ and only two voters prefer $b$ to $a$, while all other head-to-head contests among the candidates in $\{a,b,c,d\}$ are tied. Note also that one needs at least two operations (swaps or shifts) to change the relative order of two candidates separated by a set $S_j$, $j=1, \dots , 18$.

It is easy to verify that $\mathcal{{R}}(E) = \{a\}$. Indeed, $a$’s $({\mathcal {C}},\widehat{d}^\infty )$-score equals $1$ (to make $a$ the Condorcet winner, it suffices to swap it with $d$ in $v_5$ and with $c$ in $v_6$), whereas every other candidate requires at least two operations per vote to become the Condorcet winner.

On the other hand, we have $d\in \mathcal{{R}}(2E)$. Indeed, it is easy to see that each candidate’s $({\mathcal {C}},\widehat{d}^\infty )$-score in $2E$ is at least $1$. Further, the $({\mathcal {C}},\widehat{d}^\infty )$-score of $d$ in $2E$ is $1$: to make $d$ the Condorcet winner, it suffices to swap $d$ and $c$ in one copy of $v_2$, and perform a two-candidate shift of $(c, d)$ in the other copy of $v_2$.$\square$