Algorithm or not

[Julmasara]

I have a variable amount of participants.
Each participant needs to talk to another participant in a MINIMAL AMOUNT OF TOTAL ROUNDS.

Currently my application generates:
R1____R2____R3____R4____R5___R6____R7
P2-P1,P3-P1,P4-P1,P5-P1,P6-P1,P5-P3,P6-P3
P4-P3,P4-P2,P3-P2,P6-P2,P5-P2,P6-P4,P5-P4
P6-P5

Though, it can be doen in 5 rounds (manually generated):
R1____R2____R3____R4____R5___R6____R7
P2-P1,P5-P1,P3-P1,P4-P1,P1-P6,-----,-----
P4-P3,P4-P2,P5-P2,P2-P6,P2-P3,-----,-----
P6-P5,P3-P6,P4-P6,P3-P5,P4-P5,-----,-----

(Rx = Round, Px = Participant)


It's quite easy to build a list ofunique combinations of participant-pairs and arrange them over each round.

But how do I continue?. Because of my lack of knowledge I can't seem to solve this puzzle.
Do I need some sort of algorithm for this? If so, can you point me in the right direction?

Bonus:
Of all the tables (2 seats per table) there's 1 with a digital camera, which records the conversation (educational purposes).
Every participant must be placed (at least once) at the camera-table.

[Zachm]

You can deal with each participant by itself, something like this:

1. For each round r, keep a list of pairs assigned to round r, lets call this list Pairs(r).

2. When algorithm starts, all the lists are empty, but you already know that if there are N participants, you only need N-1 rounds to cover all pairs, hence N-1 lists.

3. In addition keep a list of all unique pairs, AllPairs, and initialize it
with all unique pairs (order of participants inside a pair must not matter).

4. Algorithm pseudo code goes something like:

For each participant p do
{
   For each round r , do
   { 
      If p is not part of any pair in Pairs(r) Then
      For each participant q (q!=p), do
      { 
            If q is not part of any pair in Pairs(r) AND
              the pair (p,q) is in AllPairs Then
           {
                  1. Insert the pair (p,q) to Pairs(r).
                  2. Remove the pair (p,q) from AllPairs.
           }       
      }
   }
}

As for the bonus - make the "special table" the 1st one in the list of pairs for each round and you will get this requirement.

Regards,
Zachm

[Julmasara]

I've attached a screenshot of my application's datagrid, taken after the round-participant arrangement ran.

The result is based on the following/your logic

foreach participant p1
{
  for each round
  {
    foreach participant p2 != p1
    {
       if(
       p2 not in current round
       p1 not in current round
       p2 and p1 have not paired yet)
       then -> make unique pair and assign it to current round.
    }
  }
}

As you can see, I still end up with the situation that certain participants have to wait a round or two.
(round 6 and 7 are not in the grid because of the round limitation I've set in the code.)

Thanks alot for sharing your thoughts on this.
Very much appreciated!!

P.S. Deelnemer = Participant, RondeX = RoundX

[Zachm]

Looking at your table again, I just realized that the solution is much more trivial (and that my initial solution is wrong )

Participant       R1            R2          R3         R4       R5
  
    P1                P2            P3          P4         P5       P6

    P2                P3            P4          P5         P6       P1

    P3                P4            P5          P6         P1       P2

    P4                P5            P6          P1         P2       P3

    P5                P6            P1          P2         P3       P4

    P6                P1            P2          P3         P4       P5

Notice that if you take the sequence P1-P6 and rotate it one time each round, you'll get the matching you need for each of the participant.

In round one, the sequence P1 - P6 is rotated once to the left - the sequence goes:
P2-P3-P4-P5-P6-P1.

In round two, the sequence P1 - P6 is rotated twice to the left - the sequence goes:
P3-P4-P5-P6-P2-P1.

And so on.

I think it's quite easy to fill a "magic square" matrix like this.
Only seeing your output made realize how easy it really is ....

Regards,
Zachm

[Julmasara]

This won't work either.
According to your schema:
P1 meets P2 in round 1, but P2 is meeting P3 as well in that round... and so on.

I've asked my question on another forum and a member'sproposed solution seems to work. However, currently I'm walking into other issues.

If you want to contribute to, or follow the thread:
http://www.codeproject.com/script/Fo...59&msg=2757023

Thanks for your effort to help me out!