# Assigning tables for Speed Networking session

I'm planning a structured Speed Networking activity for an event.

Here are the details:

• 100 attendees participating
• Split into groups of 4
• 12 rotations
• 25 tables

One person at each table never moves. How do I rotate each other person and ensure no one meets each other twice?

• I am absolutely opposed to closing this question. What's required is perfectly clear and a good answer might well involve discrete mathematics in some way, and will definitely involve some level of mathematical reasoning. The fact that "the method should be practical" isn't formally definable shouldn't be an issue. – Jack M Jan 20 '15 at 21:31
• @Jack: Despite it being removed from the title, I'm in favor of putting anything that comes with "Need answer ASAP" on hold for at least 48 hours. – Asaf Karagila Jan 20 '15 at 21:33
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did Jan 25 '15 at 10:29
• @Jyrki Lahtonen: the major remaining problem is that the question is still a PSQ, and the OP has not improved it (or even been active since Jan 20). Unlike delete/undelete wars, close/reopen wars don't need moderator attention, because each person only can vote one time. – Carl Mummert Jan 26 '15 at 23:45
• @JyrkiLahtonen I do not follow at all. To me that solution makes it likely this is not a real "real world" problem. It is perhaps not a homework problem in a strict sense, might also be some problem in or to prepare some aptitude test/hiring process or so. The numbers seem set up for exactly this to work/to be found. It works for more rounds but it is simpler for $12$ rounds, in the equivalent variant where one would let the 4th not move 3 clockwise but 1 counterclockwise. To be clear, I have no problem with this; actually as hinted at I'd prefer it. Now, moving on. :-) – quid Jan 27 '15 at 9:28

Since each participants will meet only 36 of the 99 other participants, there's plenty of room to set up a schedule.

One simple way is to divide the participants into four sets of 25, and arrange the tables (at least virtually) in a circle. In each round every table contains one person from each set.

Everyone in set 0 stay where they are for the entire event. Everyone in set 1 will move 1 table clockwise after each round. Everyone in set 2 will move 2 tables clockwise after each round. And everyone in set 3 moves 3 tables clockwise after each round.

• This is very clever. Of course it works because $1$, $2$ and $3$ are all coprime to $25$. – Jack M Jan 20 '15 at 21:42
• I suggest putting the oldest 25 participants in set 0, the youngest 25 in set 3, and so forth, since the latter will have to travel the farthest! – hardmath Jan 20 '15 at 23:10

Look up duplicate bridge movements: What you are asking for is a 25-table individual rainbow Mitchell movement. The full movement is impossible with a composite number of tables (25) but with only 12 rotations, you will find that no individual meets another individual more than once.

I believe a working solution would be at table 1 to instruct N to remain stationary, E to move to table 2 E, South to move to table 3 S, and W to move to table 4 W.

• Actually, we can continue up to 25 rotations with no trouble. The key fact is that the difference between any two partitipants' speed is either 0 (and then they never meet, obviously) or coprime to 25. – hmakholm left over Monica Jan 20 '15 at 21:51