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The Conversing Club 3 (Posted on 2004-03-28) Difficulty: 5 of 5
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)

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re(3): Partial seating arrangement generator V2.1 | Comment 55 of 62 |
(In reply to re(2): Partial seating arrangement generator by Bruce Brantley)

Bruce, thanks for insisting on the 5 times 11.

Before I got to the two concentric circles, I tried 3 and 4 circles.  The problem is that

you have to turn 2 circles at the time and that limits the future possibilities to much.
Your remark abouth 55 being 5 times 11 did me think again on how you can generate 11 seatings on a 11 numbers circle.  If you can, then you have 5 permutations on the 11 numbers.  This sounds like 5 times 11 and I think it is very promising.  

This is what I have at the moment:  draw a circle.  Set 1 at the top, +/- 33 degrees

further 2, and so on until you have placed 11 numbers.   Now you must find a way of going from one number to another.  If you continue clockwise with a fixed number of positions

(For example jumping from 1 to 2, then to 3,... or 1 to 3, to 5,...) You get in  trouble: Seating 1 would be 01, 02, 03, 04,... When starting the new seating at the 2 position, you would get 02, 03, 04 Bingo!  Or for the second example 01, 03, ...11, 02, 04, 06 and the second seating starts02, 04, 06 Bingo! 

So I was looking at a way to make always larger jumps, but only found it yesterday.  The trick is to jump one position clockwise, then 2 anti-clockwise, 3 clockwise and so on.
If you place the following numbers on the circle, according to the described method, you generate 11 seatings,
01 02 03 04 05 06 07 08 09 10 11
Take care: when following the circle normally (Without jumping for and backwards) the figures are on the circle as follows:
01 02 04 06 08 10 03 05 07 09 11

The seatings generated are:
02 04 01 06 03 08 05 10 07 11 09
04 06 02 08 01 10 03 11 05 09 07
06 08 04 10 02 11 01 09 03 07 05
08 10 06 11 04 09 02 07 01 05 03
10 11 08 09 06 07 04 05 02 03 01
11 09 10 07 08 05 06 03 04 01 02
09 07 11 05 10 03 08 01 06 02 04
07 05 09 03 11 01 10 02 08 04 06
05 03 07 01 09 02 11 04 10 06 08
03 01 05 02 07 04 09 06 11 08 10

The next 11 can be generated with
07 08 01 02 05 06 09 10 03 04 11

I haven't found a third seating yet.  My main problem is that I don't have a fast testing program.  I did wrote a macro in Excel, but have limited programming skils, so it is slow and not to trustworthy.  If somebody has something I can paste in VB5, I am very gratefull.

One more remark: since 11 persons can only give 45 seatings and I will have 55 solutions,  there will be at least 10 conflicting situations, where the figure 12 should be fitted in.
I am not completely sure, but I think taht in non conflicting solutions, the 12 can be written just before the number you started the seating with.

That's all for now, I feel that non-programmed solutions are only a couple days away!


  Posted by Hugo on 2004-09-23 14:14:49
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