Is there an equation to figure these groups out? Maybe a pattern?

[Beingmayaj]

Hello!
So I have no idea what type of equation this would be to figure out but I really need some help!

We have 5 people. These 5 people need to be split into groups of two over the course of 52 weeks to work in pairs on the Sunday of every week. Since the number is odd, the groups aren’t going to be the same every week of course and we also want everyone to work an equal amount of days without anyone having to work two weeks in a row. How do you split up everyone over 52 weeks to where each person is working the same amount as the other without having someone work two weeks in a row? Or can someone tell me if it only works out to where someone would have to work two weeks in a row? I really hope that makes sense.

Thank you to anyone that can help.

Here is an example of a pattern I started in my head where everyone is labeled as a number and grouped but I don’t think it works:

12
34
51
24
35
12
34
51
24
35
51
23

 

[Romsek]

well you have an immediate problem.

Assuming all the Sundays involve equal amounts of work.

You have 52 x 2 = 104 work slots to divide among 5 people.

5 does not divide 104 so 4 people are going to work 21 slots and 1 person is going work 20 slots.

How would you like to resolve that?

 

[Dr.Peterson]

Why can't you just write out 1234512345... and break it into pairs?

12​

34​

51​

23​

45​

...​

That way out of every 5 weeks, each person works twice, and never twice in a row. That's about as even as you can get; the last 2 weeks of the 52 would be

12​

34​

so #5 works one less day in the year.

Now, if you have an additional goal, such as that everyone works with everyone else more or less equally, you can just change things around after one cycle. There are ten possible pairs, and my five lines above use up half of them, so in the next 5 weeks you could just flip it around to 13524...:

13​

52​

41​

35​

24​

So in every 10 weeks, everyone has worked with everyone else once, and then you can repeat.

Oops -- #2 works twice in a row when you cycle back from 24 to 12 ...

So you can just move back one person in the next cycle, subtracting 1 from each (1 becoming 5) so the next 10 are:

51
23
45
12
34
52
41
35
24
13

and so on.

 

[pka]

We have 5 people. These 5 people need to be split into groups of two over the course of 52 weeks to work in pairs on the Sunday of every week.

There is no way that all five will work an equal number of times. In fact there will be one lucky person who will work ten weeks, while the rest will work eleven weeks. Look at this matrix of \(\displaystyle W's~\&~O's\); for Work or Off and people A, B, C, D, & E
\(\displaystyle \begin{array}{*{20}{c}}
A&B&C&D&E \\ \hline W&W&O&O&O \\ W&O&W&O&O \\ W&O&O&W&O \\ W&O&O&O&W \\ O&W&W&O&O \\ O&W&O&W&O \\
O&W&O&O&W \\ O&O&W&W&O \\ O&O&W&O&W \\ O&O&O&W&W \end{array}\) Note each of those ten rows contains 2W's & 3O's.
AND each of those five columns contains 4W's & 6O's.
Here you have an assignment scheme for five ten week of rotations,
Thus each for the five people works four times every ten weeks of the rotations.

*Now for the lottery: By what ever means pick at random one row of the matrix.
That is for week fifty-one.

Next you must eliminate any other row that shares an W with the one chosen row *above.
You should now have six remaining rows from which to randomly pick one for the fifty second week.

 

[pka]

We have 5 people. These 5 people need to be split into groups of two over the course of 52 weeks to work in pairs on the Sunday of every week.

This is a question that I just cannot let go. If you have read my other reply here is an amended tablr.
\(\displaystyle \begin{array}{*{20}{c}} {}&|&A&B&C&D&E \\ \hline i&|&W&W&O&O&O \\ {ii}&|&W&O&W&O&O \\ {iii}&|&W&O&O&W&O \\ {iv}&|&W&O&O&O&W \\ v&|&O&W&W&O&O \\ {vi}&|&O&W&O&W&O \\ {vii}&|&O&W&O&O&W \\ {viii}&|&O&O&W&W&O \\
{ix}&|&O&O&W&O&W \\ x&|&O&O&O&W&W \end{array}\)
I added a name of each row (bec. the LaTeX is poor at this site row viii in messedup).
The reason for this post script is that it occurred to me the using the table person A will work four weeks in a row in each rotation.
So here is a fix. LOOK AT THIS LINK. giving us a random permutation.
If it were (7 3 9 1 4 8 10 5 2 6) then the first week is row vii, second row iii,...,week ten will be row vi.

Here are four randomly generated permutations:
4 9 7 8 5 1 6 10 3 2
6 10 8 4 1 7 3 9 2 5
7 8 5 6 3 1 9 2 10 4
4 2 5 3 7 1 6 9 10 8
Now you have rotation sequences so that any each of the five will serve randomly.
You yourself can use that link to generate random sequences of ten.
Click on the the \(\displaystyle \bf\color{red}\boxed{=}\) at the end of the input window.
Scroll down to the "one-line notation". Copy plain-text.

 

[Dr.Peterson]

To make the pattern behind my suggestion more visible, here is a geometrical representation:



The five people are ABCDE, and the ten pairs are the segments in red. Taking them in numerical order, no one ever repeats twice in a row, and everyone works with everyone else every ten times. After each such cycle, we turn the whole figure one step to avoid a repetition.