I apologize for taking so long to respond to Forest's post of 15 May: > On Fri, 3 May 2002, Rob LeGrand wrote in part: > > > Speaking of "the chaos of the cyclic patterns", I've written a program > > to draw Craig-Carey-style triangle graphs. Each of the three corners > > corresponds to a certain ranked ballot; each point inside the triangle > > corresponds to an election with those ballots in the same proportions > > as the distances from the point to the corners. Each point is colored > > according to the winner of its election given the election method. > > I've never had the patience to try and decipher any of Craig Carey's > writings enough to figure out where he was coming from. Thanks for giving > a simple explanation.
Actually, my explanation isn't precisely correct. Strictly speaking, each point inside the triangle corresponds to an election with those ballots in the same proportions as the distances from the point to the sides of the triangle. For example, say the vertices of the equilateral triangle in question, X, Y and Z, correspond to the ballots A>B>C, B>A>C and C>A>B. If a point is, say, 4 cm from the XY side, 3 cm from the XZ side and 2 cm from the YZ side, then that point corresponds to the election 2:A>B>C 3:B>A>C 4:C>A>B My program would then calculate the winner of that election by whatever method is being used and color the point the color of the winner. So if it were drawing a Schulze triangle and A's color were blue, then the point would be blue in this case. > How does he handle six voting factions in three candidate elections? Does > he use five dimensional simplices? I guess he'd have to. I haven't tried to go beyond the two-dimensional computer screen, but I have used more than three voting factions. It's just that only three of them vary in strength. For example, I might try a:A>C>B>D b:B>A>D>C c:C>D>B>A 350:D>A>C>B where a is the distance from the current point to the YZ side, b is the distance to the XZ side, c to the XY side and a + b + c = 1200. In this example, the number of D>A>C>B voters never changes. Candidate A wins decisively in the region near vertex X, B wins near vertex Y and C near Z, but the interior of the triangle varies widely depending on the method used. Besides the usual ranked-ballot methods, I use a CRAB-like method. It results in beautifully semichaotic triangles, especially in the regions without a Condorcet winner. If anyone's interested in seeing some of the triangles I've generated, let me know and I'll try to convert a few into graphics files and post them on my website. ===== Rob LeGrand [EMAIL PROTECTED] http://www.onr.com/user/honky98/campaign.html __________________________________________________ Do You Yahoo!? Yahoo! - Official partner of 2002 FIFA World Cup http://fifaworldcup.yahoo.com ---- For more information about this list (subscribe, unsubscribe, FAQ, etc), please see http://www.eskimo.com/~robla/em
