Users browsing this forum: Bing [Bot]MCNP and 4 guests. I tried to work this out and couldn't vating up with an answer. The *xkcd speed dating* question is, given N people, how do you rotate them between tables such that each one meets each other person exactly once? Last edited by Random on Tue Jul 29, 9: I'm not about to write out the solutions again, but it's definitely possible for 4 and 8.

I suspect you can do it for any even number. This is basically a specific case of BIBD s. Spoilers I xkcd speed dating, since this is logic puzzles now? Suppose you have a way to do it for n people. To do it for 2n people, break everyone off into n pairs, say x 1 ,y 1 through x n xkcd speed dating n. Use the scheme for those n pairs, but xkcd speed dating the xkcd speed dating twice as long.

Whenever you would have i dating j, instead have 2 dates: After all that's over, speex each x i date y i. Since you knew how to do it for 2k for any odd number k, we can now do it for any even number. Last edited by Buttons on Tue Jul 29, 8: So it turns out if a topic gets moved while you're writing a reply, the reply gets lost. How did your solution for those numbers work?

Of your 2k people, seat 2k-1 in a circle, and put the 2kth person in the middle. We number the people in the circle from 0 to 2k-2, and all arithmetic will be done modulo 2k There will be 2k-1 rounds of good username ideas for dating sites, numbered 0 to 2k Person i dates the person in the middle of the circle.

This also lets you determine how to minimize the number of rounds needed for an odd number of people: Dammit - I actually had a solution will write up later, once Sperd sure it'll actually WORK unlike the speeed I thought I had before for odd numbers that this would work for yours doesn't, by the way; e. Person 1 dates all of the other people, 2 through N, which is a total of N-1 dates. So every person X has N-X xkcd speed dating. Every combination of person X with person N-X will equal N I thought it looked really simple I'm actually supposed to find table arrangements.

How many people to a table? Can't we all just sit at one big round table and everybody meets everybody? What about this one which works for all odd numbers: This only woks for odd numbers. Assign every person a number. Draw a perfect circle with each number on the circle xkcd speed dating order an equal distance from either number next to it. Then draw xkcd speed dating line from 1 to N, from 2 to N-1, etc.

After all of the people with lines drawn between them meet, rotate the lines clockwise. Continue to do this until everybody has met. Last edited by crzftx on Wed Jul 30, 4: All arithmetic is all taken mod 2k Diagram of Cauchy's solution - which actually is basically the same as the one I was considering for odd numbers, if D is omitted: Daging see Cauchy had a solution. There would be 7 dates per round, so that would be 91 dates in 13 rounds. I may have read his solution **xkcd speed dating,** or he incorrectly figured 2k-1 k Cauchy doesn't make errors.

Now, the question remains - is there a way to do it while having everyone move sleed same way? Who is online *Xkcd speed dating* browsing this forum: Board index All times are UTC Delete all board cookies The team Contact us.

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