Timeline for Maximize mixing in a 12 person dinner party [closed]
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| yesterday | history | closed |
gmvh Alexey Ustinov Daniele Tampieri Dave Benson Gerry Myerson |
Not suitable for this site | |
| yesterday | comment | added | user128807 | @DavidESpeyer Thanks so much for your answer. I'm not familiar with Hamiltonian cycles or with Mathematica. Can you give me the exact code? Is it FindHamiltonianCycle[12] or FindHamiltonianCycle[(0,1,2,3,4,5,6,7,8,9,10,11,12)]. I ask because I'm trying to find a Python implementation. | |
| yesterday | review | Close votes | |||
| yesterday | |||||
| yesterday | comment | added | David E Speyer | More generally, if $n=2r+1$, then you can have $r$ seatings such that everyone is next to all $2r$ guests, see en.wikipedia.org/wiki/Hamiltonian_decomposition#Complete_graphs for the solution. I haven't found a reference for the even case yet. | |
| yesterday | comment | added | David E Speyer | The question is clear enough, though likely to get closed as it is just a specific computation which a computer can do quickly. Here is one solution. The first seating is to seat them in order $(0,1,2,3,4,5,6,7,8,9,10,11)$. The second seating is $( 0,5,10,3,8,1,6,11,4,9,14,7 )$; i.e., every person sits next to the person who was 5 steps away from them in the previous seating. And the third seating is $(0,2,4,6,8,10,1,5,7,11,3,9,0)$. I found this by using the FindHamiltonianCycle[] command in Mathematica to find a cycle in the graph of permissible partners. | |
| S yesterday | review | First questions | |||
| yesterday | |||||
| S yesterday | history | asked | user128807 | CC BY-SA 4.0 |