Combinatorics of Election Scores

Abstract

This paper presents a novel combinatorial approach for voting rule analysis. Applying reversal symmetry, we introduce a new class of preference profiles and a new representation (bracelet representation) of preference profiles. By applying an impartial, anonymous, and neutral culture model for the case of three alternatives, we obtain precise theoretical values for the number of election scores for the plurality rule, the Kemeny rule, the Borda rule, and the scoring rules in the extreme case.

5 Appendix

Proof of proposition 5

For $m=3$, a homogenous preference profile (n preference orders $P_0$) has vector of rank sums $\begin{pmatrix}2n\\ n\\ 0\end{pmatrix}$.Any preference profile has vector of ranks sums either $\begin{pmatrix}2n\\ n\\ 0\end{pmatrix}$ or vectors $\alpha$,$\beta$,$\gamma$,$\delta$,$\epsilon$, defined here:

$$\begin{aligned} \alpha = \begin{pmatrix}2n-x\\ n+x\\ 0\end{pmatrix}; \text { } \beta = \begin{pmatrix}2n\\ n-x\\ x\end{pmatrix}; \text { } \gamma = \begin{pmatrix}2n-y\\ n-(x-y)\\ x\end{pmatrix};\text { } \delta = \begin{pmatrix}2n-x\\ n+y\\ x-y\end{pmatrix}; \text { } \epsilon = \begin{pmatrix}2n-x\\ n\\ x\end{pmatrix}. \end{aligned}$$

For all vectors $\alpha$,$\beta$,$\gamma$,$\delta$,$\epsilon$, we have that the first component is not less than the second; the second component is not less than the third. Substituting x preference orders from $P_0$ to $P_5$, we obtain type $\alpha$ preference profile. There are $\lfloor \frac{n}{2}\rfloor$ preference profiles of type $\alpha$.Similarly, we have $\lfloor \frac{n}{2}\rfloor$ voting situations of type $\beta$.

Substituting $\frac{y}{2}$ preference orders from $P_0$ to $P_3$ and $x-y$ preference orders from $P_0$ to $P_1$, we obtain type $\gamma$ voting situations with even y.Substituting $\frac{(y-1)}{2}$ preference orders from $P_0$ to $P_3$, $x-y-1$ preference orders from $P_0$ to $P_1$, and one preference order from $P_0$ to $P_2$, we obtain type $\gamma$ voting situations with odd y.

We can construct all possible type $\gamma$ voting situations using these two design methods. There are two natural restrictions on x,y for saving the order of alternatives:

$$\begin{aligned} 2x-y \le n, \end{aligned}$$

$$\begin{aligned} 2y-x \le n. \end{aligned}$$

From this, we find the number of type $\gamma$ voting situations (if $n \ge 4$):

for $y \le \frac{x}{2}$, it is $\sum _{i=2}^{\lfloor \frac{n}{2}\rfloor }p_2(i)+\sum _{i=\lfloor \frac{n}{2}\rfloor +1}^{\lfloor \frac{2n}{3}\rfloor }p_{2,n-i}(i)$;

for $y \ge \frac{x}{2}$, it is $\sum _{i=2}^{\lfloor \frac{2n}{3}\rfloor }p_2(i) + \sum _{i=\lfloor \frac{2n}{3}\rfloor +1}^{n-1}\left[ p_2(i)-p_{2,2i-n-1}(i)\right]$;

for $y = \frac{x}{2}$, it is $\lfloor \frac{\lfloor \frac{2n}{3}\rfloor }{2}\rfloor$.

Thus, the number of type $\gamma$ voting situations is equal to

$$\begin{aligned} \sum _{i=2}^{\lfloor \frac{n}{2}\rfloor }p_2(i)+\sum _{i=\lfloor \frac{n}{2}\rfloor +1}^{\lfloor \frac{2n}{3}\rfloor }p_{2,n-i}(i) + \sum _{i=2}^{\lfloor \frac{2n}{3}\rfloor }p_2(i) + \sum _{i=\lfloor \frac{2n}{3}\rfloor +1}^{n-1}\left[ p_2(i)-p_{2,2i-n-1}(i)\right] - \lfloor \frac{\lfloor \frac{2n}{3}\rfloor }{2}\rfloor \end{aligned}$$

It is also the number of type $\delta$ voting situations.

Substituting $\frac{x}{2}$ preference orders from $P_0$ to $P_3$, we obtain type $\epsilon$ voting situations with even x. Substituting $\frac{(x-1)}{2}$ preference orders from $P_0$ to $P_3$, one preference order from $P_0$ to $P_1$, and one preference order from $P_0$ to $P_5$, we obtain type $\gamma$ voting situations with odd x.

We can construct all possible type $\epsilon$ voting situations using these two design methods. There is one restriction on x for saving the order of alternatives:

$$ x \le n. $$

If $n \ge 2$ then the number of type $\epsilon$ voting situations is equal to n.

Summing over all types, we have (if $n\ge 4$)

$$\begin{aligned}&1 + 2\left( {{ \lfloor }\frac{n}{2}{ \rfloor } + \sum \limits _{{i = 2}}^{{{ \lfloor }\frac{n}{2}{ \rfloor }}} {p_{2} } (i) + \sum \limits _{{i = { \lfloor }\frac{n}{2}{ \rfloor } + 1}}^{{{ \lfloor }\frac{{2n}}{3}{ \rfloor }}} {p_{{2,n - i}} } (i) + \sum \limits _{{i = 2}}^{{n - 1}} {p_{2} } (i)} \right. \\&\quad \quad \left. { - \sum \limits _{{i = { \lfloor }\frac{{2n}}{3}{ \rfloor } + 1}}^{{n - 1}} {p_{{2,2i - n - 1}} } (i) - { \lfloor }\frac{{{ \lfloor }\frac{{2n}}{3}{ \rfloor }}}{2}{ \rfloor }} \right) + n \\ \end{aligned}$$

Modifying this, we obtain

$$\begin{aligned} 3n-1&+ \lfloor \frac{2n}{3}\rfloor ^2-\lfloor \frac{2n}{3}\rfloor +\lfloor \frac{n}{2}\rfloor \left( \lfloor \frac{n}{2}\rfloor +3-2n\right) \\&+ 2\left( {\sum \limits _{{i = 2}}^{{{ \lfloor }\frac{n}{2}{ \rfloor }}} {{ \lfloor }\frac{i}{2}{ \rfloor }} + \sum \limits _{{i = 2}}^{{n - 1}} {{ \lfloor }\frac{i}{2}{ \rfloor }} - \sum \limits _{{i = { \lfloor }\frac{n}{2}{ \rfloor } + 1}}^{{{ \lfloor }\frac{{2n}}{3}{ \rfloor }}} {{ \lfloor }\frac{{i - 1}}{2}{ \rfloor }} } \right. \\&\quad \quad \left. { + \sum \limits _{{i = { \lfloor }\frac{{2n}}{3} + 1{ \rfloor } + 1}}^{{n - 1}} {\left[ {{ \lfloor }\frac{{i - 1}}{2}{ \rfloor }} \right] } - { \lfloor }\frac{{{ \lfloor }\frac{{2n}}{3}{ \rfloor }}}{2}{ \rfloor }} \right) . \\ \end{aligned}$$

Calculating the sums, we obtain the result, which is also correct for $n=2$, and $n=3$.

Proof of proposition 7

Let $f_0(k){:}\;\mathbb {N}_0 \rightarrow \mathbb {N}_0$ be the sum of distances between preference order $P_0$ and preference order $P_i$, where $i \equiv k$ (mod 6). Let $g_0 (k)=f_0 (k)+f_0 (k+2)$ and $h_0 (k)=(2g_0 (k)-g_0 (k+3))/6=(2f_0 (k)+2f_0 (k+2)-f_0 (k+3)-f_0 (k+5))/6$. This transformation is presented in Fig. 3a. The first circle is the circle of distances from order $P_0$ to all six preference orders. After transformation, each preference order is presented by three subsequent ones. This transformation has the inversion presented in Fig. 3b. We design one-to-one correspondence between $f_0 (k)$ and $h_0 (k)$.

Fig. 3

Correspondence between $f_{0}(k)$ and $h_{0}(k)$

In the same fashion, we define functions $f_j(k)$,$g_j (k)$,$h_j (k)$, $j=\bar{0,5}$. Instead of summing distances from preference orders $\sum _{j=0}^5f_j(k)n_j$, we sum transformed values $\sum _{j=0}^5h_j(k)n_j$. We have bisection between these sums. Any sum of transformed values $\sum _{j=0}^5h_j(k)n_j$ can be represented by the circle presented in Fig. 4.

Fig. 4

Circular partition of preference orders

Permutating one pair of alternatives (one or three swaps in a preference order) leads to Fig. 4 circle turnover. Two possibilities of permutating three alternatives (two swaps in a preference order) lead to the circle rotating. Reversing the preference profile leads to a rotating on 3 preference orders in a clockwise manner. By these operations we can always construct a circle, such that $x+y+z\le 3n-x-y-z$ and $x\ge y \ge z$. From this definition, we have the following additional restrictions on x,y,z

$$ x+y \ge n-z; $$

$$ n-y+n-z \ge x; $$

$$ z+n-y+n-z+y \ge 2x+2(n-x). $$

From these inequalities, we have $n \le x+y+z \le \lfloor \frac{3n}{2}\rfloor$. We will calculate the number of partitions of $i = \bar{n,\lfloor \frac{3n}{2}\rfloor }$ with 1,2, or 3 parts, such that each part does not exceed n (and the special case of $i = \frac{3n}{2}$).A complementary partition has sum $3n-i$.

If $x+y+z = \frac{3n}{2}$, then the partitions of $i = \frac{3n}{2}$ and $3n-i=\frac{3n}{2}$ can be the same. Partitions with $y= \frac{n}{2}$ are are symmetric ($n-x=z$, $n-z=x$, $n-y=y$). The number of such partitions is equal to

$$\begin{aligned} \frac{1}{2}&\left( p_{2,n}\left( \frac{3}{2}\right) +p_{3,n}\left( \frac{3}{2}\right) +\frac{n}{2}+1\right) =\nonumber \\&=\frac{1}{2}\left( 1+n-\lfloor \frac{n}{4}\rfloor \left( \frac{n}{2}-1\right) +\frac{n^2}{2}+\lfloor \frac{n}{4}\rfloor ^2-\lfloor \frac{3n-2}{4}\rfloor - \sum _{j=1}^{\frac{n}{2}}\lfloor \frac{3n-2j-2}{4}\rfloor \right) .\qquad {(*)} \end{aligned}$$

For all other i, we have

$$\begin{aligned} 1+ \sum _{i=n}^{\lfloor \frac{3n-1}{2}\rfloor }p_{2,n}(i) + \sum _{i=n}^{\lfloor \frac{3n-1}{2}\rfloor }p_{3,n}(i). \end{aligned}$$

For $n \ge 3$, we have

$$\begin{aligned} 1+ \lfloor \frac{n}{2}\rfloor + \sum _{i=n+1}^{\lfloor \frac{3n-1}{2}\rfloor }\left( n-\lfloor \frac{i-1}{2}\rfloor \right) +\sum _{i=n}^{\lfloor \frac{3n-1}{2}\rfloor }\sum _{j=1}^{\lfloor \frac{i}{3}\rfloor }\left( \min {i-2j,n}-\lfloor \frac{i-j-1}{2}\rfloor \right) .\qquad {(**)} \end{aligned}$$

Summing (*) for even n and (**) for all n, we obtain the result, which is also correct for $n=1$ and $n=2$.