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

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.

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