Can it be done?

[Guest]

I have been trying many different methods hat do not get me what I'm looking for. We have a group of 12 representatives
coming to our two day seminar. They don't know each other and I'm trying to set up a "meet & greet" exercises using 3 tables with 4 individuals at each table to meet & greet each other. At the end of 20 minutes I want everyone to go to their next assigned table and meet 3 more (different) individuals. The same thing happens 2 more times so at the end all 12 individuals will have met the other 11 individuals exactly once. It would look something like the following:

Table 1 Table 2 Table 3 Table 4
Session 1 ABCD EFGH IJKL MNOP
Session 2 ?
Session 3 ?
Session 4 ?

I can get it close, but now realize it's not as easy as I thought.

Any suggestions?

bobr7@live.com

[Guest]

If you have 12 people with 4 people per table then it is not possible for every person to meet every other person exactly once.
In each session person A will meet 3 new people so the number of people person A meets at the end must be a multiple of 3, however you want person A to meet 11 people in total which is clearly not a multiple of 3.

If instead you had 16 people with 4 people per table (as your diagram suggests) then it is possible. (As before person A must meet a multiple of 3 people however now we want person A to meet 15 people in total so our previous argument does not cause us any problems. Note that this does not prove a schedule exists, to do that we still have to write one down.)

If you want I can post the schedule for 16 people with 4 per table.

Hope that helped,

R. Baber.

[Guest]

I assume there are greeters and meeters..?
Will A, E, I, M be classified as greeters (or leaders) to meet and greet the 3 others at each table.
Then only 12 need to circulate and meet each other. Let A, E, I, M stay at their original tables.
Every time 3 new meeters appear at a new table they must meet 2 other new meeters and of course the greeter and they can only meet each greeter once as well.......Is that correct..?

[Guest]

If there are only 3 tables, then there must only be 3 greeters, one at each table, but your layout indicates 4 tables with 4 at each table.

[syed20]

In each session person A will meet 3 new people so the number of people person A meets at the end must be a multiple of 3, however you want person A to meet 11 people in total which is clearly not a multiple of 3.


___________
sultan

[Guest]

Table1 Table2 Table3 Table4

ABCD EFGH IJKL MNOP

AEIM BFJN CGKO DHIP

AFKP EJOD INCH MBOI

AJNG FOCI KDHM PEIB

[Guest]

Just from a very quick look at the first column of your sessions................

A has not met either H or L

There are probably gaps in the others as well.................................

[Guest]

For those still interested the solution for 4 tables with 4 people per table is the following:

Session 1: ABCD EFGH IJKL MNOP
Session 2: AEIM BFJN CGKO DHLP
Session 3: AFKP BELO CHIN DGJM
Session 4: AGLN BHKM CEJP DFIO
Session 5: AHJO BGIP CFLM DEKN

The solution is unique (up to reordering of tables, sessions, labelling of people).

As I stated previously there is no solution for 3 tables with 4 people per table.

Hope this helped,

R. Baber.

[Guest]

Figure 3 on the site:
http://www.mathpuzzle.com/MAA/54-Golf%2 ... 14_07.html
gives a nice visual representation.

R. Baber.