P.S. One can use trees / exact clones also with the vote set that was discussed.
> 48 A > 27 C>B > 25 B>C The tree structure could be just a forest here. One tree is A alone. The other tree is (B, C). The idea is that support to B/C means also support to C/B. If I try to vote B>A>C, that would be corrected to B>C>A. If I try to vote B, that would be corrected to B>C (or maybe I did that intentionally, knowing that all the clones will get my vote). The clones will thus always inherit whatever the other clones get. If there is a third clone D, then vote D>A will be corrected / complemented to D>B=C>A. This approach thus sets some explicit limits to how voters are allowed to vote. Or in other words, how the votes to the explicit clones will be transferred / shared between the clones. If the voter forgets that these candidates are explicit clones, their vote can be corrected (rather than rejected). The intention of this mail is just to point out that although the most straight forward approach with trees is to use bullet votes only, one can use the tree structure (and the explicit clone approach) also with more complex votes like ranked votes. Juho On 7.8.2011, at 10.37, Juho Laatu wrote: > On 7.8.2011, at 2.04, Jameson Quinn wrote: > >> >> >> 2011/8/6 <fsimm...@pcc.edu> >> Jan, >> >> IRV elects C like all of the other methods if the B faction doesn't >> truncate. But IRV elects A when the B >> faction truncates. Of course, with this knowledge, the B faction isn't >> likely to truncate, and as you say C >> will be elected. >> >> The trouble with IRV is that in the other scenario when the B faction >> truncates sincerely because of >> detesting both A and C, IRV still elects A instead of B. >> >> Also, if the A faction votes A>B, then B clearly should win, but does not >> under IRV. So yes, IRV solves the chicken dilemma, but in so doing causes >> other problems. (This same argument, as it happens, works against tree-based >> methods.) >> >> I still claim that SODA is the only system I know of that can solve the >> chicken dilemma without over-solving it and making other problems. > > I wouldn't say that trees "over-solve" the problem. The tree approach to the > chicken problem could be called "explicit clones". That's quite natural. Some > candidates just announce that they are clones and that they will support each > others. That sounds like a pretty exact solution, not an over-solution. > > Do trees "cause other problems" then? They do not allow the voter to support > one of the clones without supporting the other. But this is exactly what the > intention of the explicit clone approach is. Also the need to declare a > branch in the tree could be considered to be a practical problem / increased > complexity. And the need to identify the clones is an extra task / problem. > But maybe not really. What other (more serious) problems would the trees > cause? > > Juho > > > > > ---- > Election-Methods mailing list - see http://electorama.com/em for list info
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