Trying to Split Six People into Two Groups and Have Each Person Meet

In summary, the conversation discusses the challenge of creating a virtual meeting with six people in two groups of three, where everyone must see each other at least once. Despite attempting various solutions, it is determined that after three rounds, it is impossible for everyone to meet. The group discusses different possibilities and concludes that there is no easy formula or calculation to solve the problem, but a computer program can help.
  • #1
The Head
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Real World Application here. I'm creating a virtual meeting involving six people and will have the first round include two groups of three. Then we'll switch a few times. I tried hopelessly to do it such that after three rounds, everyone would see everyone else at least once. Was so close but always missed one or two pairs. It's not a straight 6C3 problem because if I have people A, B, C, D, E, F, I just care that A meets with D & F, but I don't necessarily need ADF to be a group. It could be ADE and then ABF.

Anyway, just trying to figure out how many rounds I'd need to make sure everyone interfaced with everyone else at least once, or if anyone has a suggestion that's better than guess and check.

Note, there may be a complication where only three of the six can "host" the meeting due to who has access to the full software, but honestly just want to figure this out first.
 
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  • #2
It's impossible after three rounds. Let's look at "A" first: They can meet two people per round and have to meet five people. Without loss of generality we can assume they meet B and C in the first round, D and E in the second, and F and someone else in the third round. This already fixes the first two rounds completely:
ABC DEF
ADE BCF
The third round has a group of "AF". So far B didn't see D and E, so we would need to make a group BDE, so our third round is ACF and BDE. But that means C doesn't see D and E.
 
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Likes The Head, PeroK and phinds
  • #3
mfb said:
It's impossible after three rounds. Let's look at "A" first: They can meet two people per round and have to meet five people. Without loss of generality we can assume they meet B and C in the first round, D and E in the second, and F and someone else in the third round. This already fixes the first two rounds completely:
ABC DEF
ADE BCF
The third round has a group of "AF". So far B didn't see D and E, so we would need to make a group BDE, so our third round is ACF and BDE. But that means C doesn't see D and E.

Thanks, that's helpful. I'll have to modify the format. I assume there's no formula or calculation I can do that's readily accessible that can help me check once I've adjusted?
 
  • #4
Not that I'm aware of. A computer program can do it, of course.
 
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