[EM] Hello again -- and a new method for you!
Jobst Heitzig
heitzig-j at web.de
Sun Apr 11 03:16:01 PDT 2004
Dear Ernest,
you wrote:
> Hi Jobst,
>
> However, my impression is that the nasty nuts-and-bolts of most methods
> come from having to deal with same-size majorities, which is where (and
> why) they differ. I think several methods have suggested
> ordered-comparison rather than 'fixing'. Still, they all (including
> mine) get complicated when two incompatible decisions have the same
> size. Any thoughts on how to finesse that?
>
My first impulse would be to exclude all the weakest defeats (trying to
switch to the list's jargon...) at once and finally draw the winner at
random from the undefeated candidates. So the top-down version of the
River Method would then be this, in the list's jargon:
The river method
- top-down version fixed for the "non-general" case:
1. Allowing abstentions, determine all pairwise defeats.
Ties are not considered defeats.
2. Sort by descending strength (=size of supporting majority)
and process in groups of equal strength.
3. For each group of defeats of equal strength:
a) Add all of them as dashed arrows,
pointing from the defeated to the defeating option.
b) Remove those which originate at an option where already
a stronger arrow originates.
c) Then remove those which are on a cycle of equal or larger strength
(thus breaking some cycles at more than one point if necessary!)
4. The option uniquely undefeated at the end wins.
If there are more than one such option, draw at random.
I'm not absolutely sure that this is still monotone in the non-general
case - I will have to check that and will hopefully be able to post a proof.
Drawing at random in the end seems the easiest tiebreaker and perfectly
acceptable with larger groups. (By the way, did anyone ever estimate how
many voters are necessary to have the probability of equally strong
defeats less than, say, 1 percent?)
Alternatively, one could also start over in the end and apply the method
iteratively until stable (and then draw at random if there are still
more than one option left). However, I fear such iterative methods are
almost always non-monotone... (I think I proved this at least for the
iterative version of the beat path method some time).
PS: What is your method, Ernest? Could you tell me where I will find it
in the archives (they are somewhat too big to read everything :-)
Have a nice Easter!
Jobst
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