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A group of $n$ people organizes recurring parties, $n-1$ events in total. At each event, each person offers a present to one other person, and each person receives a present ($n$ presents exchanged in total per event). Is it possible to organize the parties so that

  1. each person has given a present to each other person after $n-1$ parties,
  2. each person has received a present from each other person after $n-1$ parties,
  3. at each party, the "gives a present" relation forms a single cycle (i.e., the group cannot be subdivided in subgroups so that no presents are exchanged between persons in different subgroups)?

To me, it looks as if this is possible if and only if $n$ is prime, but a proof remains elusive to me. Is this a well-known problem? What would be a good formulation of the problem that allows for an elegant proof?

Pieter21 mentioned Graeco-Latin squares: for given $n$ we search for a Graeco-Latin square with w.l.o.g. the first column equal to $A\alpha, B\beta, C\gamma, \ldots$ and the other columns such that each forms a cycle, e.g., $A\beta \rightarrow B\gamma \rightarrow C\delta \rightarrow D\alpha \rightarrow A\beta$ is a cycle for $n=4$.

This is from a real-life setting where $n$ happens to be $6$ (a solution can be ruled out due to the nonexistence of such a Graeco-Latin square). Please feel free to re-tag as appropriate.

krlmlr
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  • Look at 'Mutual Orthogonal Latin Squares', or Graeco-Latin squares. I recall from my study (25 years ago) that prime powers are also OK, and in composite numbers, the lowest prime power - 1 can be proven, but is not necessarily the lower bound. (E.g. with 6 you can't get past 1 party, but with 10 you can) – Pieter21 Nov 29 '15 at 21:27
  • Isn't this equivalent to asking if the group $<Z_n,+>$ is generated by every non-zero element? Because if so, that only happens when $n$ is prime. – Ben Nov 29 '15 at 21:31
  • @Pieter21: Thanks, I'll take a look. Still, it seems that I can have at least two such parties for any $n > 2$: $1 \leftarrow 2 \leftarrow 3 \leftarrow \ldots n \leftarrow 1$ and $1 \leftarrow n \leftarrow n-1 \leftarrow \ldots 1$. – krlmlr Nov 29 '15 at 23:02
  • @Ben: Currently I'm only convinced about one direction: Group generation $\rightarrow$ party formulation. The other direction might hold if we could prove that there is at most one possible arrangement (except for permutation). – krlmlr Nov 29 '15 at 23:10
  • @Pieter21: Close, but not quite. The Graeco-Latin squares would need the additional property that all columns (except the first) form a cycle; the order 4 example on Wikipedia doesn't have this property, e.g., $B\gamma \rightarrow C\beta \rightarrow B\gamma$ in the second column. But of course the non-existence of a solution for $n=6$ shows that there's no solution for my particular instance :-) – krlmlr Nov 29 '15 at 23:21

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Let's rephrase this question into a graph theory question.

Basically you are asking if we have the directed complete graph, $K_n^*$, then there a decomposition into directed cycles all having length $n$. If you are not used to graph theory, the directed complete graph is a graph on $n$ vertices where each vertex has a directed arc to all other vertices. This question was solved in a paper "Cycle decompositions IV: complete directed graphs and fixed length directed cycles". A special case of the main theorem in the paper is

Given $K_n^*$, there exists a cycle decomposition where all cycles have length $n$ if and only if $n$ divides the number of edges, and $n \not = 4$ or $n \not =6$.

The number of edges in $K_n^*$ is $(n-1)(n)$, therefore $n$ always divides the number of edges. Thus we can conclude that for all $n$ not $4$ or $6$ such a decomposition exists. I haven't read the paper entirely as it is $44$ pages long so I'm not sure how they actually construct the decomposition (if they did so constructively).

Ben
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