[EM] Amalgamation details, hijacking, and free-riding

[Juho Laatu]

Tree building could be voluntary or mandatory. If voluntary, then parties and wings can stop free riding in their own area. If mandatory, then the most difficult part is to organize the parties as a tree (= party external tree). One should have rules on how to build a tree also in the case when there is no consensus on what the structure of the tree should be.

One simple approach would be to allow the already agreed (= voluntary) binary branches (= trees of a forest) to join themselves (or the bigger trees that they are already part of) into other trees of the forest in random order. Or maybe largest ones first into the largest tree, starting from the third largest, after joining the two largest ones together first. I assumed that the voluntary branches (that were agreed already before the forced phase) would be considered atomic (= no joining inside them).

Did you mean that there would be concrete strategic opportunities in the tree-bulding phase, or that one just needs to think a bit on how to form the tree or how to force the tree to be formed? Sincere strategy seems quite good to me. Or maybe one could nominate fake parties next to one's strongest competitors in the hope of making some of the voters of the competing party vote for the wrong party (that could get a seat if many enough voters make that mistake).

Juho


P.S. I might come back with a proposal of considering trees to be a good method that is simple and understandable to the voters, very strategy free, and even close to but better than plurality.



On 6.8.2011, at 10.46, Jameson Quinn wrote:

> Well, kinda; but in a sense, that pushes the strategy into the tree-building.
> 
> 2011/8/6 Juho Laatu <juho4880 at yahoo.co.uk>
> On 4.8.2011, at 2.09, Jameson Quinn wrote:
> 
> > Free riding in some form is inevitable in a good system. (That is, any system which avoids free riding entirely would be horribly warped by that necessity).
> 
> How about tree methods? If candidates are ordered as a binary tree (instead of an open list), then there are no choices between three or more branches, and related free riding becomes impossible.
> 
> Juho
> 
> 
> 
> 
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