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 The Conversing Club 3 (Posted on 2004-03-28)
Many members of the club disliked the lack of variety and togetherness at the club. Although the club still had 12 members, some members were threatening to quit because each schedule was so short and there were so few people around each table.

To satisfy their request, the club decided to seat themselves around a big table and create a longer schedule. The twelve members of the club seated themselves in a schedule such that during each block of 55 days, no person was between the same pair of people. How was the schedule constructed?

(Based on The Round Table)

 No Solution Yet Submitted by Gamer Rating: 4.5714 (14 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Take heart Penny (with my 'research') | Comment 29 of 62 |
(In reply to Take heart Penny by brianjn)

Some thoughts here, but there was an earlier error, which some one would have found, but it still offers food for thought.

For groups of:

3 people there is 1 combination,

4 people there are 3 combinations,

5 people there are 6 combinations,

6 people there are 10 combinations,

7 people there are 15 combinations,

12 there are 55.

I note that the combinations are triangle numbers.  In the following arrangements I have 'split' before a 'sub-grouping' when a new element is added to the second column.  Note too, the last element of the group does not feature in this column.

Now for a group of  4

The arrangement is:

A     B     C      D

A     B     D      C

<o:p></o:p>

A    C      B      D

----------------

Arrangement for 5:

A   B   C   D  E

A   B   E   C   D

A   B   D   E   C

<o:p></o:p>

A   C   B   E   D

A   C   D   B   E

<o:p></o:p>

A   D   B   C   E <o:p></o:p>

I note that B has a diagonal 'property' in the second sub-group.

----------------

Arrangement for 6:

A   B   C   D   E   F

A   B   F   E   C   D

A   B   D   C   F   E

A   B   E   F   D   C

<o:p></o:p>

A   C   B   F   D   E

A   C   F   B   E   D

A   C   E   D   B   F

<o:p></o:p>

A   D   B   E   C   F

A   D   F   C   B   E

<o:p></o:p>

A   E   C   B   D   F

The B diagonal appears here also in the second sub-group (is it a feature when working with 7?).

----------------

There may be some relationship in the creation of pairs, after the first pair is chosen.

I note that where pairs are even, pair order reversal occurs; sometimes one pair, sometimes both.

<o:p></o:p>

Maybe there is something here for those who want to analyse how such schedules might be constructed.

Edited on July 29, 2004, 10:44 am

Corrected an error in the last line of the 5 element array.

Edited on July 29, 2004, 9:28 pm
 Posted by brianjn on 2004-07-29 02:01:48

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