MHB How many hours are needed for team meetings?

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The discussion revolves around calculating the minimum hours required for six teams to hold their meetings, given overlapping members among the teams. The user proposes that three hours are necessary, based on their analysis of a graph representing team memberships. Other participants confirm that the user's calculation appears correct, noting that they have effectively identified the chromatic number of the graph. The consensus is that three hours is indeed sufficient for all meetings to occur without conflicts. This conclusion highlights the importance of graph theory in solving scheduling problems.
evinda
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Hello! :)

I am looking at this exercise:

We have these teams : $E_{1},E_{2},...,E_{6}$ . Each team must have a meeting.
How many hours are needed (minimum) so that all the meetings take part, given that:

$$ E_{1}=\{A,B,C\} $$
$$ E_{2}=\{A,D,E\} $$
$$ E_{3}=\{B,C,Z\} $$
$$ E_{4}=\{Z,H,T\} $$
$$ E_{5}=\{E,H\} $$
$$ E_{6}=\{D,E,T\} $$

?

I tried to answer the question with this graph:

View attachment 2545

So, $3$ hours are needed,so that all the meetings take part...
Is it right or have I done something wrong? (Blush)
 

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evinda said:
Hello! :)

I am looking at this exercise:

We have these teams : $E_{1},E_{2},...,E_{6}$ . Each team must have a meeting.
How many hours are needed (minimum) so that all the meetings have been done, given that:

$$ E_{1}=\{A,B,C\} $$
$$ E_{2}=\{A,D,E\} $$
$$ E_{3}=\{B,C,Z\} $$
$$ E_{4}=\{Z,H,T\} $$
$$ E_{5}=\{E,H\} $$
$$ E_{6}=\{D,E,T\} $$

?

I tried to answer the question with this graph:

https://www.physicsforums.com/attachments/2545

So, $3$ hours are needed,so that all the meetings take part...
Is it right or have I done something wrong? (Blush)
Looks right to me. You found out the chromatic number of the graph. Nice!
 
caffeinemachine said:
Looks right to me. You found out the chromatic number of the graph. Nice!

Great! (Clapping) Thank you! :)
 
There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host...

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