<p>We have <span class="math-container">$a_i P_j a_k$</span> when <span class="math-container">$i-(j-1)\mod n&lt;k-(j-1)\mod n$</span>. Here, we represent <span class="math-container">$0$</span> by <span class="math-container">$n$</span>.</p>

CC BY-SA 4.0
2024-08-11T06:33:16.633Z
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?
EoDmnFOr3q
CC BY-SA 4.0
2024-08-11T08:23:20.293Z
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 $a_4 P_i a_1$
EoDmnFOr3q
CC BY-SA 4.0
2024-08-11T12:18:24.983Z
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 $N$.
Michael Greinecker
CC BY-SA 4.0
2024-08-11T12:24:22.223Z
Thank you very much for your comment. How can I ‘interpret’ $0$ as $N$? Is it standard? It seems a bit if a stretch to me
EoDmnFOr3q
CC BY-SA 4.0
2024-08-11T12:39:19.763Z
That is just the ordering used $1<2<3<...<0$.
Michael Greinecker
CC BY-SA 4.0
2024-08-11T13:01:03.963Z
Fair enough. However, isn’t there another way to defining what I need without having to re-interpret $0$ as $n$
EoDmnFOr3q
CC BY-SA 4.0
2024-08-11T15:48:47.357Z
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.
Michael Greinecker
CC BY-SA 4.0
2024-08-10T23:33:33.297Z
2024-08-14T12:51:42.270Z
2024-08-14T12:51:42.270Z
Michael Greinecker
<p>For <span class="math-container">$1\le i\le n$</span> let <span class="math-container">$a_{n+i}:= a_i$</span>. For <span class="math-container">$i\in N$</span> define <span class="math-container">$P_i$</span> by <span class="math-container">$a_iP_ia_{i+1}P_i\cdots P_ia_{i+n-1}$</span>.</p>

CC BY-SA 4.0
2024-11-02T07:42:04.033Z
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 $a_{n+i}=a_i$”?
EoDmnFOr3q
CC BY-SA 4.0
2024-11-03T08:35:07.263Z
@EoDmnFOr3q, you're right, I edited the answer.
VARulle
CC BY-SA 4.0
2024-08-14T12:52:02.680Z
2024-11-03T08:34:19.983Z
2024-11-03T08:34:19.983Z
VARulle
<p>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 &amp; four-alternative example.</p>
<p>Let <span class="math-container">$A=\{a_1,a_2,a_3,a_4\}$</span> be a set of four alternatives, let <span class="math-container">$N=\{1,2,3,4\}$</span> be a set of four players and let <span class="math-container">$P=(P_i)_{i\in N}$</span> be a preference profile (i.e., linear order profile) on alternatives such that
<span class="math-container">\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}</span></p>
<p>In the preference profile above, the top-ranked alternative of player <span class="math-container">$i$</span> becomes the bottom-ranked alternative of player <span class="math-container">$i+1$</span>, and so on.</p>
<p>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 <span class="math-container">$P=(P_i)_{i\in N}$</span> satisfy, for every player <span class="math-container">$i\in N$</span>,
<span class="math-container">\begin{gather}
a_iP_ia_{i+1}P_i\dots P_ia_{n+i-1}
\end{gather}</span>
Then, for player <span class="math-container">$i=1$</span>, we obtain <span class="math-container">$a_i=a_1$</span>, <span class="math-container">$a_{i+1}=a_2$</span>, <span class="math-container">$a_{i+2}=a_3$</span> and <span class="math-container">$a_{n+1-i}=a_{4+1-1}=a_4$</span>; and thus,
<span class="math-container">\begin{gather}
a_1P_1a_2P_1a_3P_1a_4
\end{gather}</span>
Hence, for player <span class="math-container">$i=1$</span>, this approach works.</p>
<p>However, for player <span class="math-container">$i=2$</span>, we obtain <span class="math-container">$a_i=a_2$</span>, <span class="math-container">$a_{i+1}=a_3$</span>, <span class="math-container">$a_{i+2}=a_4$</span> and <span class="math-container">$a_{n+1-i}=a_{4+1-2}=a_3$</span>; and thus,
<span class="math-container">\begin{gather}
a_2P_2a_3P_2a_4P_2a_3
\end{gather}</span>
Hence, for player <span class="math-container">$i=2$</span> (and by a similar token, for all players <span class="math-container">$i\neq 1$</span>), my approach does not work.</p>
<p>Hence, <strong>how to formally define the preference profile in which the top alternative of one player becomes the bottom alternative of the next player?</strong></p>
<p><strong>EDIT</strong> A friend of mine suggested using <a href="https://en.wikipedia.org/wiki/Modular_arithmetic" rel="nofollow noreferrer">modular arithmetic</a> to do what I need to do; and while I think he’s right, I don’t see how to proceed.</p>

<microeconomics><preferences><theory>
How to formally define the preference profile in which the top alternative of one player becomes the bottom alternative of the next player?
EoDmnFOr3q
# How to formally define the preference profile in which the top alternative of one player becomes the bottom alternative of the next player?

(Asked by: username_0 on 2024-08-10)

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](https://en.wikipedia.org/wiki/Modular_arithmetic) to do what I need to do; and while I think he’s right, I don’t see how to proceed.

**Tags:** microeconomics, preferences, theory

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## 2 answers

### Answer by username_1 on 2024-08-10. Score: 2 | Accepted answer

We have $a\_i P\_j a\_k$ when $i-(j-1)\mod n. Here, we represent $0$ by $n$.

> **Comments:**
> **username_0**: 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?  
> **username_0**: 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 $a_4 P_i a_1$  
> **username_1**: 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 $N$.  
> **username_0**: Thank you very much for your comment. How can I ‘interpret’ $0$ as $N$? Is it standard? It seems a bit if a stretch to me  
> **username_1**: That is just the ordering used $1<2<3<...<0$.  
> **username_0**: Fair enough. However, isn’t there another way to defining what I need without having to re-interpret $0$ as $n$  
> **username_1**: 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.  

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### Answer by username_2 on 2024-08-14. Score: 3

For $1\le i\le n$ let $a\_{n+i}:= a\_i$. For $i\in N$ define $P\_i$ by $a\_iP\_ia\_{i+1}P\_i\cdots P\_ia\_{i+n-1}$.

> **Comments:**
> **username_0**: 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 $a_{n+i}=a_i$”?  
> **username_2**: @EoDmnFOr3q, you're right, I edited the answer.  

---

