[EM] Interactive Representation

[Juho Laatu]

Yes. It would be a quite natural approach to first count the number of seats that each party gets based on the number of votes that they got, and then use the rankings and more complex counting methods within each party separately. That would have brought the high numbers of 35 and 405 down to 11 and 35. Then, if needed, one could use tricks like combining several of the weakest candidates together as if they were one candidate that is ranked as high as the highest among those candidates in each vote. If that approach gives a result where none of the grouped candidates will be elected, we could safely assume that none of them would have won anyway. We would still have a small risk of not being able to count the results efficiently. There may be also other useful tricks. Maybe some special rules would be needed in that case (like STV like elimination of some number of weakest candidates).

Juho


On 6.11.2011, at 11.30, Kristofer Munsterhjelm wrote:

> Juho Laatu wrote:
>> Since you are building this on the single-seat district tradition,
>> three or four seats and 10 candidates is plenty. I'm used to numbers
>> like 6 seats with 108 candidates, and 35 seats with 405 candidates,
>> and at least eight parties in the parliament. (In that situation even
>> ranking all of the candidates, or even all of the candidates of one's
>> favourite party may be too tedious. One may however allow all votes
>> (also short ones) to be counted for the party.)
>> What would be a good (non-limiting) number of candidates? Maybe
>> something like (P * K1) * (S * K2), where P = current number of
>> parties with representatives, K1 = 1.5 or 2, S = number of seats, K2
>> = 1.
> 
> Maybe one can make shortcuts, too. In Kemeny (which is a hard single-winner method), you know that the method passes Smith. Therefore, only orderings where the winner is in the Smith set needs to be considered. (In practice, you can add such a constraint upon the mixed integer program; trying every possible ordering would take too much time).
> 
> So perhaps there are similar shortcuts one can make for the multiwinner method. That would help ward off candidate flooding attacks or simply unexpected popularity -- if most of the 405 candidates have no chance of winning, one might be able to prune them away and be left with a feasible problem even with extreme numbers of candidates.
> 
> That would, of course, not help if all 405 candidates are viable, but such a situation would be very unlikely to happen in practice, particularly if the voters have to rank the candidates manually. If the voters don't rank all the candidates manually but instead use some sort of party list with overrides hybrid, then that party list structure creates a pattern that might itself be used to reduce the scope of the problem.
> 
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