Combinatorics - rotating people through different groups / tables
$begingroup$
I am trying to figure out a rotation schedule for 46 people. We have 6 tables. 5 tables have 4 groups of 2. 1 table has 3 groups of 2. We have 6 rounds and people are in groups of 2 for each round. The first round, 1 person in each pair presents their sales pitch and the other person listens. For the second round, they rotate to new tables in new groups and have new partners. The rep that presented is now a listener and the listener becomes a presenter. This goes on for all 6 rounds which means a person presents 3 times and listens 3 times, each time at a new table with a new partner.
Table 1 - 4 groups of 2
Table 2 - 4 groups of 2
Table 3 - 4 groups of 2
Table 4 - 4 groups of 2
Table 5 - 4 groups of 2
Table 6 - 3 groups of 2
I cannot, for the life of me,figure out how to do these rotations.
combinatorics
asked Dec 8 '15 at 5:32
Sandy McConnellSandy McConnell
11
$endgroup$
add a comment |
$begingroup$
I am trying to figure out a rotation schedule for 46 people. We have 6 tables. 5 tables have 4 groups of 2. 1 table has 3 groups of 2. We have 6 rounds and people are in groups of 2 for each round. The first round, 1 person in each pair presents their sales pitch and the other person listens. For the second round, they rotate to new tables in new groups and have new partners. The rep that presented is now a listener and the listener becomes a presenter. This goes on for all 6 rounds which means a person presents 3 times and listens 3 times, each time at a new table with a new partner.
Table 1 - 4 groups of 2
Table 2 - 4 groups of 2
Table 3 - 4 groups of 2
Table 4 - 4 groups of 2
Table 5 - 4 groups of 2
Table 6 - 3 groups of 2
I cannot, for the life of me,figure out how to do these rotations.
combinatorics
asked Dec 8 '15 at 5:32
Sandy McConnellSandy McConnell
11
$endgroup$
1
$begingroup$
If you want it exactly as you have said, then it is impossible. Reason (if I have understood the problem correctly): you have 46 people and they all have to be at every table once, in 6 rounds. But table 6 only holds 6 people, so in 6 rounds it will only get a maximum of 36 different people, so at least 10 people must miss out on table 6. So it seems that you will need to relax your requirements.
$endgroup$
– David
Dec 8 '15 at 5:37
$begingroup$
So I either have to bump it to 48 tables or someone will sit at a table 2x
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:20
$begingroup$
Even if I bump it to 48 tables or people have to sit at the same table twice, I would still need to figure out how to do it.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:30
$begingroup$
Well, if you change the conditions then it might be possible. But maybe it will still be impossible for different reasons. Sometimes this sort of scheduling task is impossible for very subtle and complicated reasons. It could be (emphasize: "could be" - I honestly don't have much of a clue) that you would have to make some pretty major changes.
$endgroup$
– David
Dec 8 '15 at 22:41
$begingroup$
David, it is not possible since there are 8 people at a table and only 6 tables. I haven't had to work through a problem like this in a while. Thank you for giving me some things to think about.
$endgroup$
– Sandy McConnell
Dec 9 '15 at 17:46
add a comment |
$begingroup$
I am trying to figure out a rotation schedule for 46 people. We have 6 tables. 5 tables have 4 groups of 2. 1 table has 3 groups of 2. We have 6 rounds and people are in groups of 2 for each round. The first round, 1 person in each pair presents their sales pitch and the other person listens. For the second round, they rotate to new tables in new groups and have new partners. The rep that presented is now a listener and the listener becomes a presenter. This goes on for all 6 rounds which means a person presents 3 times and listens 3 times, each time at a new table with a new partner.
Table 1 - 4 groups of 2
Table 2 - 4 groups of 2
Table 3 - 4 groups of 2
Table 4 - 4 groups of 2
Table 5 - 4 groups of 2
Table 6 - 3 groups of 2
I cannot, for the life of me,figure out how to do these rotations.
combinatorics
asked Dec 8 '15 at 5:32
Sandy McConnellSandy McConnell
11
$endgroup$
I am trying to figure out a rotation schedule for 46 people. We have 6 tables. 5 tables have 4 groups of 2. 1 table has 3 groups of 2. We have 6 rounds and people are in groups of 2 for each round. The first round, 1 person in each pair presents their sales pitch and the other person listens. For the second round, they rotate to new tables in new groups and have new partners. The rep that presented is now a listener and the listener becomes a presenter. This goes on for all 6 rounds which means a person presents 3 times and listens 3 times, each time at a new table with a new partner.
Table 1 - 4 groups of 2
Table 2 - 4 groups of 2
Table 3 - 4 groups of 2
Table 4 - 4 groups of 2
Table 5 - 4 groups of 2
Table 6 - 3 groups of 2
I cannot, for the life of me,figure out how to do these rotations.
combinatorics
combinatorics
asked Dec 8 '15 at 5:32
Sandy McConnellSandy McConnell
11
asked Dec 8 '15 at 5:32
Sandy McConnellSandy McConnell
11
asked Dec 8 '15 at 5:32
Sandy McConnellSandy McConnell
11
asked Dec 8 '15 at 5:32
Sandy McConnellSandy McConnell
11
asked Dec 8 '15 at 5:32
Sandy McConnellSandy McConnell
11
11
1
$begingroup$
If you want it exactly as you have said, then it is impossible. Reason (if I have understood the problem correctly): you have 46 people and they all have to be at every table once, in 6 rounds. But table 6 only holds 6 people, so in 6 rounds it will only get a maximum of 36 different people, so at least 10 people must miss out on table 6. So it seems that you will need to relax your requirements.
$endgroup$
– David
Dec 8 '15 at 5:37
$begingroup$
So I either have to bump it to 48 tables or someone will sit at a table 2x
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:20
$begingroup$
Even if I bump it to 48 tables or people have to sit at the same table twice, I would still need to figure out how to do it.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:30
$begingroup$
Well, if you change the conditions then it might be possible. But maybe it will still be impossible for different reasons. Sometimes this sort of scheduling task is impossible for very subtle and complicated reasons. It could be (emphasize: "could be" - I honestly don't have much of a clue) that you would have to make some pretty major changes.
$endgroup$
– David
Dec 8 '15 at 22:41
$begingroup$
David, it is not possible since there are 8 people at a table and only 6 tables. I haven't had to work through a problem like this in a while. Thank you for giving me some things to think about.
$endgroup$
– Sandy McConnell
Dec 9 '15 at 17:46
add a comment |
1
$begingroup$
If you want it exactly as you have said, then it is impossible. Reason (if I have understood the problem correctly): you have 46 people and they all have to be at every table once, in 6 rounds. But table 6 only holds 6 people, so in 6 rounds it will only get a maximum of 36 different people, so at least 10 people must miss out on table 6. So it seems that you will need to relax your requirements.
$endgroup$
– David
Dec 8 '15 at 5:37
$begingroup$
So I either have to bump it to 48 tables or someone will sit at a table 2x
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:20
$begingroup$
Even if I bump it to 48 tables or people have to sit at the same table twice, I would still need to figure out how to do it.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:30
$begingroup$
Well, if you change the conditions then it might be possible. But maybe it will still be impossible for different reasons. Sometimes this sort of scheduling task is impossible for very subtle and complicated reasons. It could be (emphasize: "could be" - I honestly don't have much of a clue) that you would have to make some pretty major changes.
$endgroup$
– David
Dec 8 '15 at 22:41
$begingroup$
David, it is not possible since there are 8 people at a table and only 6 tables. I haven't had to work through a problem like this in a while. Thank you for giving me some things to think about.
$endgroup$
– Sandy McConnell
Dec 9 '15 at 17:46
1
1
$begingroup$
If you want it exactly as you have said, then it is impossible. Reason (if I have understood the problem correctly): you have 46 people and they all have to be at every table once, in 6 rounds. But table 6 only holds 6 people, so in 6 rounds it will only get a maximum of 36 different people, so at least 10 people must miss out on table 6. So it seems that you will need to relax your requirements.
$endgroup$
– David
Dec 8 '15 at 5:37
$begingroup$
If you want it exactly as you have said, then it is impossible. Reason (if I have understood the problem correctly): you have 46 people and they all have to be at every table once, in 6 rounds. But table 6 only holds 6 people, so in 6 rounds it will only get a maximum of 36 different people, so at least 10 people must miss out on table 6. So it seems that you will need to relax your requirements.
$endgroup$
– David
Dec 8 '15 at 5:37
$begingroup$
So I either have to bump it to 48 tables or someone will sit at a table 2x
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:20
$begingroup$
So I either have to bump it to 48 tables or someone will sit at a table 2x
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:20
$begingroup$
Even if I bump it to 48 tables or people have to sit at the same table twice, I would still need to figure out how to do it.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:30
$begingroup$
Even if I bump it to 48 tables or people have to sit at the same table twice, I would still need to figure out how to do it.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:30
$begingroup$
Well, if you change the conditions then it might be possible. But maybe it will still be impossible for different reasons. Sometimes this sort of scheduling task is impossible for very subtle and complicated reasons. It could be (emphasize: "could be" - I honestly don't have much of a clue) that you would have to make some pretty major changes.
$endgroup$
– David
Dec 8 '15 at 22:41
$begingroup$
Well, if you change the conditions then it might be possible. But maybe it will still be impossible for different reasons. Sometimes this sort of scheduling task is impossible for very subtle and complicated reasons. It could be (emphasize: "could be" - I honestly don't have much of a clue) that you would have to make some pretty major changes.
$endgroup$
– David
Dec 8 '15 at 22:41
$begingroup$
David, it is not possible since there are 8 people at a table and only 6 tables. I haven't had to work through a problem like this in a while. Thank you for giving me some things to think about.
$endgroup$
– Sandy McConnell
Dec 9 '15 at 17:46
$begingroup$
David, it is not possible since there are 8 people at a table and only 6 tables. I haven't had to work through a problem like this in a while. Thank you for giving me some things to think about.
$endgroup$
– Sandy McConnell
Dec 9 '15 at 17:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I did a simple trial and error, looks to be correct to me, but someone should look it over again.
Assign a number to each person x=1-46.
After each rotation, new number(y) = mod(3x+8,46)+1.
Thus for Round 1.
Table 1: 1 2 3 4 5 6 7 8
Table 2: 9 10 11 12 13 14 15 16
Table 3: 17 18 19 20 21 22 23 24
Table 4: 25 26 27 28 29 30 31 32
Table 5: 33 34 35 36 37 38 39 40
Table 6: 41 42 43 44 45 46
Round 2
12 15 18 21 24 27 30 33
36 39 42 45 2 5 8 11
14 17 20 23 26 29 32 35
38 41 44 1 4 7 10 13
16 19 22 25 28 31 34 37
40 43 46 3 6 9
Round 3
45 8 17 26 35 44 7 16
25 34 43 6 15 24 33 42
5 14 23 32 41 4 13 22
31 40 3 12 21 30 39 2
11 20 29 38 1 10 19 28
37 46 9 18 27 36
Round 4
6 33 14 41 22 3 30 11
38 19 46 27 8 35 16 43
24 5 32 13 40 21 2 29
10 37 18 45 26 7 34 15
42 23 4 31 12 39 20 1
28 9 36 17 44 25
Round 5
27 16 5 40 29 18 7 42
31 20 9 44 33 22 11 46
35 24 13 2 37 26 15 4
39 28 17 6 41 30 19 8
43 32 21 10 45 34 23 12
1 36 25 14 3 38
Round 6
44 11 24 37 4 17 30 43
10 23 36 3 16 29 42 9
22 35 2 15 28 41 8 21
34 1 14 27 40 7 20 33
46 13 26 39 6 19 32 45
12 25 38 5 18 31
answered Dec 8 '15 at 5:55
Alvin ChangAlvin Chang
11
$endgroup$
$begingroup$
If I am following this correctly, person #1 ends up at table 4 twice and table 5 twice.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:38
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
I did a simple trial and error, looks to be correct to me, but someone should look it over again.
Assign a number to each person x=1-46.
After each rotation, new number(y) = mod(3x+8,46)+1.
Thus for Round 1.
Table 1: 1 2 3 4 5 6 7 8
Table 2: 9 10 11 12 13 14 15 16
Table 3: 17 18 19 20 21 22 23 24
Table 4: 25 26 27 28 29 30 31 32
Table 5: 33 34 35 36 37 38 39 40
Table 6: 41 42 43 44 45 46
Round 2
12 15 18 21 24 27 30 33
36 39 42 45 2 5 8 11
14 17 20 23 26 29 32 35
38 41 44 1 4 7 10 13
16 19 22 25 28 31 34 37
40 43 46 3 6 9
Round 3
45 8 17 26 35 44 7 16
25 34 43 6 15 24 33 42
5 14 23 32 41 4 13 22
31 40 3 12 21 30 39 2
11 20 29 38 1 10 19 28
37 46 9 18 27 36
Round 4
6 33 14 41 22 3 30 11
38 19 46 27 8 35 16 43
24 5 32 13 40 21 2 29
10 37 18 45 26 7 34 15
42 23 4 31 12 39 20 1
28 9 36 17 44 25
Round 5
27 16 5 40 29 18 7 42
31 20 9 44 33 22 11 46
35 24 13 2 37 26 15 4
39 28 17 6 41 30 19 8
43 32 21 10 45 34 23 12
1 36 25 14 3 38
Round 6
44 11 24 37 4 17 30 43
10 23 36 3 16 29 42 9
22 35 2 15 28 41 8 21
34 1 14 27 40 7 20 33
46 13 26 39 6 19 32 45
12 25 38 5 18 31
answered Dec 8 '15 at 5:55
Alvin ChangAlvin Chang
11
$endgroup$
$begingroup$
If I am following this correctly, person #1 ends up at table 4 twice and table 5 twice.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:38
add a comment |
$begingroup$
I did a simple trial and error, looks to be correct to me, but someone should look it over again.
Assign a number to each person x=1-46.
After each rotation, new number(y) = mod(3x+8,46)+1.
Thus for Round 1.
Table 1: 1 2 3 4 5 6 7 8
Table 2: 9 10 11 12 13 14 15 16
Table 3: 17 18 19 20 21 22 23 24
Table 4: 25 26 27 28 29 30 31 32
Table 5: 33 34 35 36 37 38 39 40
Table 6: 41 42 43 44 45 46
Round 2
12 15 18 21 24 27 30 33
36 39 42 45 2 5 8 11
14 17 20 23 26 29 32 35
38 41 44 1 4 7 10 13
16 19 22 25 28 31 34 37
40 43 46 3 6 9
Round 3
45 8 17 26 35 44 7 16
25 34 43 6 15 24 33 42
5 14 23 32 41 4 13 22
31 40 3 12 21 30 39 2
11 20 29 38 1 10 19 28
37 46 9 18 27 36
Round 4
6 33 14 41 22 3 30 11
38 19 46 27 8 35 16 43
24 5 32 13 40 21 2 29
10 37 18 45 26 7 34 15
42 23 4 31 12 39 20 1
28 9 36 17 44 25
Round 5
27 16 5 40 29 18 7 42
31 20 9 44 33 22 11 46
35 24 13 2 37 26 15 4
39 28 17 6 41 30 19 8
43 32 21 10 45 34 23 12
1 36 25 14 3 38
Round 6
44 11 24 37 4 17 30 43
10 23 36 3 16 29 42 9
22 35 2 15 28 41 8 21
34 1 14 27 40 7 20 33
46 13 26 39 6 19 32 45
12 25 38 5 18 31
answered Dec 8 '15 at 5:55
Alvin ChangAlvin Chang
11
$endgroup$
$begingroup$
If I am following this correctly, person #1 ends up at table 4 twice and table 5 twice.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:38
add a comment |
$begingroup$
I did a simple trial and error, looks to be correct to me, but someone should look it over again.
Assign a number to each person x=1-46.
After each rotation, new number(y) = mod(3x+8,46)+1.
Thus for Round 1.
Table 1: 1 2 3 4 5 6 7 8
Table 2: 9 10 11 12 13 14 15 16
Table 3: 17 18 19 20 21 22 23 24
Table 4: 25 26 27 28 29 30 31 32
Table 5: 33 34 35 36 37 38 39 40
Table 6: 41 42 43 44 45 46
Round 2
12 15 18 21 24 27 30 33
36 39 42 45 2 5 8 11
14 17 20 23 26 29 32 35
38 41 44 1 4 7 10 13
16 19 22 25 28 31 34 37
40 43 46 3 6 9
Round 3
45 8 17 26 35 44 7 16
25 34 43 6 15 24 33 42
5 14 23 32 41 4 13 22
31 40 3 12 21 30 39 2
11 20 29 38 1 10 19 28
37 46 9 18 27 36
Round 4
6 33 14 41 22 3 30 11
38 19 46 27 8 35 16 43
24 5 32 13 40 21 2 29
10 37 18 45 26 7 34 15
42 23 4 31 12 39 20 1
28 9 36 17 44 25
Round 5
27 16 5 40 29 18 7 42
31 20 9 44 33 22 11 46
35 24 13 2 37 26 15 4
39 28 17 6 41 30 19 8
43 32 21 10 45 34 23 12
1 36 25 14 3 38
Round 6
44 11 24 37 4 17 30 43
10 23 36 3 16 29 42 9
22 35 2 15 28 41 8 21
34 1 14 27 40 7 20 33
46 13 26 39 6 19 32 45
12 25 38 5 18 31
answered Dec 8 '15 at 5:55
Alvin ChangAlvin Chang
11
$endgroup$
I did a simple trial and error, looks to be correct to me, but someone should look it over again.
Assign a number to each person x=1-46.
After each rotation, new number(y) = mod(3x+8,46)+1.
Thus for Round 1.
Table 1: 1 2 3 4 5 6 7 8
Table 2: 9 10 11 12 13 14 15 16
Table 3: 17 18 19 20 21 22 23 24
Table 4: 25 26 27 28 29 30 31 32
Table 5: 33 34 35 36 37 38 39 40
Table 6: 41 42 43 44 45 46
Round 2
12 15 18 21 24 27 30 33
36 39 42 45 2 5 8 11
14 17 20 23 26 29 32 35
38 41 44 1 4 7 10 13
16 19 22 25 28 31 34 37
40 43 46 3 6 9
Round 3
45 8 17 26 35 44 7 16
25 34 43 6 15 24 33 42
5 14 23 32 41 4 13 22
31 40 3 12 21 30 39 2
11 20 29 38 1 10 19 28
37 46 9 18 27 36
Round 4
6 33 14 41 22 3 30 11
38 19 46 27 8 35 16 43
24 5 32 13 40 21 2 29
10 37 18 45 26 7 34 15
42 23 4 31 12 39 20 1
28 9 36 17 44 25
Round 5
27 16 5 40 29 18 7 42
31 20 9 44 33 22 11 46
35 24 13 2 37 26 15 4
39 28 17 6 41 30 19 8
43 32 21 10 45 34 23 12
1 36 25 14 3 38
Round 6
44 11 24 37 4 17 30 43
10 23 36 3 16 29 42 9
22 35 2 15 28 41 8 21
34 1 14 27 40 7 20 33
46 13 26 39 6 19 32 45
12 25 38 5 18 31
answered Dec 8 '15 at 5:55
Alvin ChangAlvin Chang
11
edited Dec 8 '15 at 6:20
answered Dec 8 '15 at 5:55
Alvin ChangAlvin Chang
11
answered Dec 8 '15 at 5:55
Alvin ChangAlvin Chang
11
answered Dec 8 '15 at 5:55
Alvin ChangAlvin Chang
11
11
$begingroup$
If I am following this correctly, person #1 ends up at table 4 twice and table 5 twice.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:38
add a comment |
$begingroup$
If I am following this correctly, person #1 ends up at table 4 twice and table 5 twice.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:38
$begingroup$
If I am following this correctly, person #1 ends up at table 4 twice and table 5 twice.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:38
$begingroup$
If I am following this correctly, person #1 ends up at table 4 twice and table 5 twice.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:38
add a comment |
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1
$begingroup$
If you want it exactly as you have said, then it is impossible. Reason (if I have understood the problem correctly): you have 46 people and they all have to be at every table once, in 6 rounds. But table 6 only holds 6 people, so in 6 rounds it will only get a maximum of 36 different people, so at least 10 people must miss out on table 6. So it seems that you will need to relax your requirements.
$endgroup$
– David
Dec 8 '15 at 5:37
$begingroup$
So I either have to bump it to 48 tables or someone will sit at a table 2x
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:20
$begingroup$
Even if I bump it to 48 tables or people have to sit at the same table twice, I would still need to figure out how to do it.
$endgroup$
– Sandy McConnell
Dec 8 '15 at 14:30
$begingroup$
Well, if you change the conditions then it might be possible. But maybe it will still be impossible for different reasons. Sometimes this sort of scheduling task is impossible for very subtle and complicated reasons. It could be (emphasize: "could be" - I honestly don't have much of a clue) that you would have to make some pretty major changes.
$endgroup$
– David
Dec 8 '15 at 22:41
$begingroup$
David, it is not possible since there are 8 people at a table and only 6 tables. I haven't had to work through a problem like this in a while. Thank you for giving me some things to think about.
$endgroup$
– Sandy McConnell
Dec 9 '15 at 17:46