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I need to generically define a strict preference profile with Condorcet cicles when the number of players and alternatives coincide. To illustrate my problem, consider the following four-player & four-alternative example.

Let $A=\{a_1,a_2,a_3,a_4\}$ be a set of four alternatives, let $N=\{1,2,3,4\}$ be a set of four players and let $P=(P_i)_{i\in N}$ be a preference profile (i.e., linear order profile) on alternatives such that \begin{gather} a_1P_1a_2P_1a_3P_1a_4\\ a_2P_2a_3P_2a_4P_2a_1\\ a_3P_3a_4P_3a_1P_3a_2\\ a_4P_4a_1P_4a_2P_4a_3 \end{gather}

In the preference profile above, the top-ranked alternative of player $i$ becomes the bottom-ranked alternative of player $i+1$, and so on.

In order to generalise the strict preference profile above to an arbitrary (but equal) number of players and alternatives, I had thought of writing the following: let $P=(P_i)_{i\in N}$ satisfy, for every player $i\in N$, \begin{gather} a_iP_ia_{i+1}P_i\dots P_ia_{n+i-1} \end{gather} Then, for player $i=1$, we obtain $a_i=a_1$, $a_{i+1}=a_2$, $a_{i+2}=a_3$ and $a_{n+1-i}=a_{4+1-1}=a_4$; and thus, \begin{gather} a_1P_1a_2P_1a_3P_1a_4 \end{gather} Hence, for player $i=1$, this approach works.

However, for player $i=2$, we obtain $a_i=a_2$, $a_{i+1}=a_3$, $a_{i+2}=a_4$ and $a_{n+1-i}=a_{4+1-2}=a_3$; and thus, \begin{gather} a_2P_2a_3P_2a_4P_2a_3 \end{gather} Hence, for player $i=2$ (and by a similar token, for all players $i\neq 1$), my approach does not work.

Hence, how to formally define the preference profile in which the top alternative of one player becomes the bottom alternative of the next player?

EDIT A friend of mine suggested using modular arithmetic to do what I need to do; and while I think he’s right, I don’t see how to proceed.

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2 Answers 2

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We have $a_i P_j a_k$ when $i-(j-1)\mod n<k-(j-1)\mod n$. Here, we represent $0$ by $n$.

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  • $\begingroup$ Thank you very much for your helpful answer. However, could you please add some context that helps me understand it? Also, is the modulus $N$ or $N+1$? I personally came with $a_iP_ia_{(i+1)\bmod n}P_ia_{(i+2)\bmod n}P_i\cdots P_ia_{(i+n-1)\bmod n}$, but my and your expressions seem to be quite different. Could you please point out the error in mine? $\endgroup$
    – EoDmnFOr3q
    Commented Aug 11 at 6:33
  • $\begingroup$ Also, for $n=4$ and $i=2$, I obtain $(1-(i-1)) \bmod (n+1)=0<3=(4-(i-1)) \bmod (n+1)$; and thus, $a_1P_ia_4$. But instead, we should have $a_4 P_i a_1$ $\endgroup$
    – EoDmnFOr3q
    Commented Aug 11 at 8:23
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    $\begingroup$ You are right, it should be mod $N$. Then $i-(j-1)$ mod $N$ is the rank of alternative $j$ for $i$ when there are $N$ alternatives, and a lower rank is better. And you need to interpret $0$ as $N$. $\endgroup$
    – Michael Greinecker
    Commented Aug 11 at 12:18
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    $\begingroup$ That is just the ordering used $1<2<3<...<0$. $\endgroup$
    – Michael Greinecker
    Commented Aug 11 at 12:39
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    $\begingroup$ Given that the labels are $1,2,3,\ldots,n$ instead of $0,1,2,\ldots,n-1$, that is the natural thing. Most people I know with an analog watch have 12 at the top, not 0. $\endgroup$
    – Michael Greinecker
    Commented Aug 11 at 15:48
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For $1\le i\le n$ let $a_{n+i}:= a_i$. For $i\in N$ define $P_i$ by $a_iP_ia_{i+1}P_i\cdots P_ia_{i+n-1}$.

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    $\begingroup$ Thank you very for your answer. However, shouldn’t it be “For $1<i \leq n$, let $a_{n+i}=a_i$” instead of “For $1\leq i \leq n-1$, let $a_{n+i}=a_i$”? $\endgroup$
    – EoDmnFOr3q
    Commented Nov 2 at 7:42
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    $\begingroup$ @EoDmnFOr3q, you're right, I edited the answer. $\endgroup$
    – VARulle
    Commented Nov 3 at 8:35

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