When I started looking at Condorcet winners, Smith sets, circular ties and the like, I never saw any examples higher than a three-way tie. Just in case someone wanted a few possibilities to try their new election method on, here are four circular ties involving five candidates (a standard version, an "interlaced" version, and inverse methods of both). Of course, there are many other possibilities involving five candidates, but I don't think they are as neatly symmetric. I also found circular ties involving seven candidates but not six -- but that's probably because of the geometric way I figure out the relationships. To generate the ties, just make sure the sum of the three smallest values are larger than the sum of the two largest -- the values 25-30-35-40-45 work well in any order, since 25+30+35>40+45. Some other interesting possibilities are 1-2-2-2-2, 2-2-3-3-3, and 5-6-7-8-9. With a little effort you can create circular ties where the sum of the two largest values is greater than the sum of the three smallest, but this way you don't need to look at every possibility to see if the tie unravels. Tie 1 (standard) A B C D E B C D E A C D E A B D E A B C E A B C D Tie 2 (interlaced) A B C D E C D E A B E A B C D B C D E A D E A B C Tie 3 (inverse of tie 2) A B C D E D E A B C B C D E A E A B C D C D E A B Tie 4 (inverse of tie 1) A B C D E E A B C D D E A B C C D E A B B C D E A Here are the ten relations for each tie (I hope they show up correctly): Tie 1 Tie 2 Tie 3 Tie 4 A>B A>C A>B A>D A>C A>E A>D A>E B>C B>A B>C B>A B>D B>D B>E B>E C>D C>B C>A C>A C>E C>E C>D C>B D>E D>A D>B D>B D>A D>C D>E D>C E>A E>D E>A E>C E>B E>B E>C E>D I thought I might as well post them because it might save someone some time. It's a fun way of generating some voting paradoxes. If a voting method handles a five-way circular tie without blowing up, it'll probably handle anything (grin). Mike Rouse [EMAIL PROTECTED]