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

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Discussion Overview

The discussion revolves around the challenge of organizing a virtual meeting for six people, aiming to ensure that each participant meets every other participant at least once over a series of rounds. The focus is on the combinatorial aspects of group formation and the constraints involved in the meeting setup.

Discussion Character

  • Exploratory, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant describes the goal of creating groups of three from six people and ensuring that everyone meets everyone else at least once, noting the complexity of the problem.
  • Another participant argues that it is impossible to achieve this goal in three rounds, providing a specific example of how the meetings would need to be structured and highlighting the limitations of the groupings.
  • There is a reiteration of the impossibility claim with a detailed breakdown of the meeting structure, emphasizing that certain participants would not meet others within the proposed rounds.
  • A participant expresses a need to modify the meeting format and inquires about the existence of a formula or calculation to assist with adjustments.
  • Another participant suggests that while there may not be a straightforward formula, a computer program could potentially solve the problem.

Areas of Agreement / Disagreement

Participants generally agree on the complexity of the problem and the challenges involved in ensuring all meetings occur. However, there is a clear disagreement regarding the feasibility of achieving the goal within three rounds, with some asserting it is impossible.

Contextual Notes

The discussion does not resolve the mathematical steps involved in determining the necessary rounds or the implications of the hosting constraints mentioned by the original poster.

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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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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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?
 
Not that I'm aware of. A computer program can do it, of course.
 

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