progressive dinner dilema
 

I have 24 couples taking part in a progressive dinner! 6 couples (houses)
serve aperitif, 6 starter, 6 main and 6 desert. with 4 couples eating at each
house including the host. The dilema is that no 2 couples can meet twice is
there a way of getting excel to work this out? I have numbered the couples
1-24 then gone in blocks of 6 by 4, but each time I get some duplication????
Please help

progressive dinner dilema
 

This is the best I could come up with:

App St Main Des
House 1
01 01 01 01
02 22 18 14
03 19 11 23
04 16 24 04
House 2
05 05 05 05
06 02 22 18
07 23 15 03
08 20 04 08
House 3
09 09 09 09
10 06 02 22
11 03 19 07
12 24 08 12
House 4
13 13 13 13
14 10 06 02
15 07 23 11
16 04 12 16
House 5
17 17 17 17
18 14 10 06
19 11 03 15
20 08 16 20
House 6
21 21 21 21
22 18 14 10
23 15 07 19
24 12 20 24

There are six pairs who have two courses together, so I have arranged
for this to happen for the first and last courses - they are person
numbers 01/04, 05/08, 09/12, 13/16, 17/20 and 21/24.

Hope this helps.

Pete

On Feb 12, 8:44*am, Debbie wrote:
I have 24 couples taking part in a progressive dinner! 6 couples (houses)
serve aperitif, 6 starter, 6 main and 6 desert. with 4 couples eating at each
house including the host. The dilema is that no 2 couples can meet twice is
there a way of getting excel to work this out? I have numbered the *couples
1-24 then gone in blocks of 6 by 4, but each time I get some duplication????
Please help

progressive dinner dilema
 

Thanks Pete, perhaps I didnt explain very well, once everyone has had their
aperitif they then move onto the next location for starter as do the people
who hosted, therefore, every couple will host a course. so you have 24 houses
and 24 couples, each course has 4 couples including the host and not one
couple can be duplicated. It's definately one of those puzzles that drives
you insane, I sort of hoped excel might have a solution???????

"Pete_UK" wrote:

This is the best I could come up with:

App St Main Des
House 1
01 01 01 01
02 22 18 14
03 19 11 23
04 16 24 04
House 2
05 05 05 05
06 02 22 18
07 23 15 03
08 20 04 08
House 3
09 09 09 09
10 06 02 22
11 03 19 07
12 24 08 12
House 4
13 13 13 13
14 10 06 02
15 07 23 11
16 04 12 16
House 5
17 17 17 17
18 14 10 06
19 11 03 15
20 08 16 20
House 6
21 21 21 21
22 18 14 10
23 15 07 19
24 12 20 24

There are six pairs who have two courses together, so I have arranged
for this to happen for the first and last courses - they are person
numbers 01/04, 05/08, 09/12, 13/16, 17/20 and 21/24.

Hope this helps.

Pete

On Feb 12, 8:44 am, Debbie wrote:
I have 24 couples taking part in a progressive dinner! 6 couples (houses)
serve aperitif, 6 starter, 6 main and 6 desert. with 4 couples eating at each
house including the host. The dilema is that no 2 couples can meet twice is
there a way of getting excel to work this out? I have numbered the couples
1-24 then gone in blocks of 6 by 4, but each time I get some duplication????
Please help

progressive dinner dilema
 

Houses 1 to 6 are not the same for different courses - I suppose they
could be labelled House A, B, C etc as in the following:

App St Main Des

A G M S
01 01 01 01
02 22 18 14
03 19 11 23
04 16 24 04

B H N T
05 05 05 05
06 02 22 18
07 23 15 03
08 20 04 08

C I O U
09 09 09 09
10 06 02 22
11 03 19 07
12 24 08 12

D J P V
13 13 13 13
14 10 06 02
15 07 23 11
16 04 12 16

E K Q W
17 17 17 17
18 14 10 06
19 11 03 15
20 08 16 20

F L R X
21 21 21 21
22 18 14 10
23 15 07 19
24 12 20 24

where A to F are the 6 houses used for aperitif, G to L are the houses
for the starter course etc.

My numbers 01 to 24 represent the couples, four of whom get together
for one "course" in one house (i.e. each vertical group of 4), and
then split up into another house for another course.

Hope this makes it a bit clearer.

Pete

On Feb 12, 11:03*am, Debbie wrote:
Thanks Pete, perhaps I didnt explain very well, once everyone has had their
aperitif they then move onto the next location for starter as do the people
who hosted, therefore, every couple will host a course. so you have 24 houses
and 24 couples, each course has 4 couples including the host and not one
couple can be duplicated. It's definately one of those puzzles that drives
you insane, I sort of hoped excel might have a solution???????



"Pete_UK" wrote:
This is the best I could come up with:

App * St * Main * Des
House 1
01 * * 01 * * 01 * * 01
02 * * 22 * * 18 * * 14
03 * * 19 * * 11 * * 23
04 * * 16 * * 24 * * 04
House 2
05 * * 05 * * 05 * * 05
06 * * 02 * * 22 * * 18
07 * * 23 * * 15 * * 03
08 * * 20 * * 04 * * 08
House 3
09 * * 09 * * 09 * * 09
10 * * 06 * * 02 * * 22
11 * * 03 * * 19 * * 07
12 * * 24 * * 08 * * 12
House 4
13 * * 13 * * 13 * * 13
14 * * 10 * * 06 * * 02
15 * * 07 * * 23 * * 11
16 * * 04 * * 12 * * 16
House 5
17 * * 17 * * 17 * * 17
18 * * 14 * * 10 * * 06
19 * * 11 * * 03 * * 15
20 * * 08 * * 16 * * 20
House 6
21 * * 21 * * 21 * * 21
22 * * 18 * * 14 * * 10
23 * * 15 * * 07 * * 19
24 * * 12 * * 20 * * 24

There are six pairs who have two courses together, so I have arranged
for this to happen for the first and last courses - they are person
numbers 01/04, 05/08, 09/12, 13/16, 17/20 and 21/24.

Hope this helps.

Pete

On Feb 12, 8:44 am, Debbie wrote:
I have 24 couples taking part in a progressive dinner! 6 couples (houses)
serve aperitif, 6 starter, 6 main and 6 desert. with 4 couples eating at each
house including the host. The dilema is that no 2 couples can meet twice is
there a way of getting excel to work this out? I have numbered the *couples
1-24 then gone in blocks of 6 by 4, but each time I get some duplication????
Please help- Hide quoted text -

- Show quoted text -

progressive dinner dilema
 

Thanks for your help, unfortunately it doesnt work, you have 4 blocks of
numbers 1-24, the top row of each block 1 is 1-6, then block 2 is 7-12, then
block 3 is 13-18 and finally block 4 is 19-24, these represent the couples
that are cooking then underneath each one there are 3 couples

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
22 23 24


"Pete_UK" wrote:

Houses 1 to 6 are not the same for different courses - I suppose they
could be labelled House A, B, C etc as in the following:

App St Main Des

A G M S
01 01 01 01
02 22 18 14
03 19 11 23
04 16 24 04

B H N T
05 05 05 05
06 02 22 18
07 23 15 03
08 20 04 08

C I O U
09 09 09 09
10 06 02 22
11 03 19 07
12 24 08 12

D J P V
13 13 13 13
14 10 06 02
15 07 23 11
16 04 12 16

E K Q W
17 17 17 17
18 14 10 06
19 11 03 15
20 08 16 20

F L R X
21 21 21 21
22 18 14 10
23 15 07 19
24 12 20 24

where A to F are the 6 houses used for aperitif, G to L are the houses
for the starter course etc.

My numbers 01 to 24 represent the couples, four of whom get together
for one "course" in one house (i.e. each vertical group of 4), and
then split up into another house for another course.

Hope this makes it a bit clearer.

Pete

On Feb 12, 11:03 am, Debbie wrote:
Thanks Pete, perhaps I didnt explain very well, once everyone has had their
aperitif they then move onto the next location for starter as do the people
who hosted, therefore, every couple will host a course. so you have 24 houses
and 24 couples, each course has 4 couples including the host and not one
couple can be duplicated. It's definately one of those puzzles that drives
you insane, I sort of hoped excel might have a solution???????



"Pete_UK" wrote:
This is the best I could come up with:

App St Main Des
House 1
01 01 01 01
02 22 18 14
03 19 11 23
04 16 24 04
House 2
05 05 05 05
06 02 22 18
07 23 15 03
08 20 04 08
House 3
09 09 09 09
10 06 02 22
11 03 19 07
12 24 08 12
House 4
13 13 13 13
14 10 06 02
15 07 23 11
16 04 12 16
House 5
17 17 17 17
18 14 10 06
19 11 03 15
20 08 16 20
House 6
21 21 21 21
22 18 14 10
23 15 07 19
24 12 20 24

There are six pairs who have two courses together, so I have arranged
for this to happen for the first and last courses - they are person
numbers 01/04, 05/08, 09/12, 13/16, 17/20 and 21/24.

Hope this helps.

Pete

On Feb 12, 8:44 am, Debbie wrote:
I have 24 couples taking part in a progressive dinner! 6 couples (houses)
serve aperitif, 6 starter, 6 main and 6 desert. with 4 couples eating at each
house including the host. The dilema is that no 2 couples can meet twice is
there a way of getting excel to work this out? I have numbered the couples
1-24 then gone in blocks of 6 by 4, but each time I get some duplication????
Please help- Hide quoted text -

- Show quoted text -

progressive dinner dilema
 

I think we are looking at the data differently. If you take my first
block, i.e.:

App St Main Des

A G M S
01 01 01 01
02 22 18 14
03 19 11 23
04 16 24 04

this means that in House A (couple 01) will serve Apperitifs to the
group made up of couples 01, 02, 03 and 04. Then in house G (couple
22) will serve the starter course to couples 01, 22, 19 and 16. Then
in House M (couple 11) will serve the main course to couples 01, 18,
11 and 24, and finally dessert will be served in House S, by couple 04
to the group 01, 14, 23 and 04.

The house owner is derived along the diagonal of the 4 x 4 grid of
numbers, so that in the second grid, the houses B, H, N and T are
owned by couples 05, 02, 15 and 08 respectively. If you want the house
owner to be shown on the top row, then just move them to the top.

As I said before, it doesn't work out completely as there will be one
pair of couples who are present for both apperitifs and for dessert,
but this is very close to what you require.

Hope this helps.

Pete


On Feb 12, 4:01*pm, Debbie wrote:
Thanks for your help, unfortunately it doesnt work, you have 4 blocks of
numbers 1-24, the top row of each block 1 is 1-6, then block 2 is 7-12, then
block 3 is 13-18 and finally block 4 is 19-24, these represent the couples
that are cooking then underneath each one there are 3 couples

1 *2 *3 *4 *5 *6 *7 *8 *9 *10 *11 *12 *13 *14 *15 *16 *17 *18 *19 *20 *21 *
22 *23 * 24



"Pete_UK" wrote:
Houses 1 to 6 are not the same for different courses - I suppose they
could be labelled House A, B, C etc as in the following:

App * * St * * * Main * * Des

A * * * * *G * * * * M * * * * S
01 * * * *01 * * * *01 * * * *01
02 * * * *22 * * * *18 * * * *14
03 * * * *19 * * * *11 * * * *23
04 * * * *16 * * * *24 * * * *04

B * * * * *H * * * * N * * * * T
05 * * * *05 * * * *05 * * * *05
06 * * * *02 * * * *22 * * * *18
07 * * * *23 * * * *15 * * * *03
08 * * * *20 * * * *04 * * * *08

C * * * * *I * * * * *O * * * * U
09 * * * *09 * * * *09 * * * *09
10 * * * *06 * * * *02 * * * *22
11 * * * *03 * * * *19 * * * *07
12 * * * *24 * * * *08 * * * *12

D * * * * *J * * * * *P * * * * V
13 * * * *13 * * * *13 * * * *13
14 * * * *10 * * * *06 * * * *02
15 * * * *07 * * * *23 * * * *11
16 * * * *04 * * * *12 * * * *16

E * * * * *K * * * * Q * * * * W
17 * * * *17 * * * *17 * * * *17
18 * * * *14 * * * *10 * * * *06
19 * * * *11 * * * *03 * * * *15
20 * * * *08 * * * *16 * * * *20

F * * * * *L * * * * R * * * * *X
21 * * * *21 * * * *21 * * * *21
22 * * * *18 * * * *14 * * * *10
23 * * * *15 * * * *07 * * * *19
24 * * * *12 * * * *20 * * * *24

where A to F are the 6 houses used for aperitif, G to L are the houses
for the starter course etc.

My numbers 01 to 24 represent the couples, four of whom get together
for one "course" in one house (i.e. each vertical group of 4), and
then split up into another house for another course.

Hope this makes it a bit clearer.

Pete

On Feb 12, 11:03 am, Debbie wrote:
Thanks Pete, perhaps I didnt explain very well, once everyone has had their
aperitif they then move onto the next location for starter as do the people
who hosted, therefore, every couple will host a course. so you have 24 houses
and 24 couples, each course has 4 couples including the host and not one
couple can be duplicated. It's definately one of those puzzles that drives
you insane, I sort of hoped excel might have a solution???????

"Pete_UK" wrote:
This is the best I could come up with:

App * St * Main * Des
House 1
01 * * 01 * * 01 * * 01
02 * * 22 * * 18 * * 14
03 * * 19 * * 11 * * 23
04 * * 16 * * 24 * * 04
House 2
05 * * 05 * * 05 * * 05
06 * * 02 * * 22 * * 18
07 * * 23 * * 15 * * 03
08 * * 20 * * 04 * * 08
House 3
09 * * 09 * * 09 * * 09
10 * * 06 * * 02 * * 22
11 * * 03 * * 19 * * 07
12 * * 24 * * 08 * * 12
House 4
13 * * 13 * * 13 * * 13
14 * * 10 * * 06 * * 02
15 * * 07 * * 23 * * 11
16 * * 04 * * 12 * * 16
House 5
17 * * 17 * * 17 * * 17
18 * * 14 * * 10 * * 06
19 * * 11 * * 03 * * 15
20 * * 08 * * 16 * * 20
House 6
21 * * 21 * * 21 * * 21
22 * * 18 * * 14 * * 10
23 * * 15 * * 07 * * 19
24 * * 12 * * 20 * * 24

There are six pairs who have two courses together, so I have arranged
for this to happen for the first and last courses - they are person
numbers 01/04, 05/08, 09/12, 13/16, 17/20 and 21/24.

Hope this helps.

Pete

On Feb 12, 8:44 am, Debbie wrote:
I have 24 couples taking part in a progressive dinner! 6 couples (houses)
serve aperitif, 6 starter, 6 main and 6 desert. with 4 couples eating at each
house including the host. The dilema is that no 2 couples can meet twice is
there a way of getting excel to work this out? I have numbered the *couples
1-24 then gone in blocks of 6 by 4, but each time I get some duplication????
Please help- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -