<p dir="ltr"><br>
On Aug 3, 2016 12:25 PM, "Kristofer Munsterhjelm" <<a href="mailto:km_elmet@t-online.de">km_elmet@t-online.de</a>> wrote:<br>
><br>
> In wv, suppose A>B and B>A are equally strong. What happens? Does A>B<br>
> and B>A both go to 0 or do they both retain their pairwise strength?</p>
<p dir="ltr">(end quote)<br></p>
<p dir="ltr">That isn't a defeat. It's a pair-tie. The wv measure of defeat-strength only compares the strength of defeats.</p>
<p dir="ltr">That pair-tie merely does't give either candidate a defeat.</p>
<p dir="ltr">Of course in MMPO, it would count as pairwise opposition for both candidates.</p>
<p dir="ltr">Michael Ossipoff<br>
><br>
> =<br>
><br>
> I've been trying to work a bit more on the MAM-Hare multiwinner method,<br>
> but the inner function (the "magic" of the method) turned out to be very<br>
> hard to construct.<br>
><br>
> The idea of the MAM-Hare method is to sequentially find Hare-quota sized<br>
> groups of voters so that the group has particularly strong affirmed<br>
> majorities, then electing the candidate that this affirmed majority<br>
> suggests and excluding the voters who contributed to the majorities.<br>
><br>
> The problem lies in both finding a group of suitable voters and their<br>
> affirmed majorities at the same time.<br>
><br>
> For instance, if there are multiple candidate majorities supported by<br>
> more than a Hare quota, then choosing just which majorities to affirm to<br>
> get as many Hare-quota level majorities affirmed as possible can be<br>
> pretty difficult. If more than a Hare quota prefers A>B, then we have to<br>
> pick a Hare quota sized subset, but we want to choose a subset so that<br>
> we can affirm as many other Hare-quota sized majorities as possible<br>
> using only those voters. So the problem reduces to:<br>
><br>
> "Given a number of sets (voters who have some given pairwise<br>
> preference), determine which you can intersect (preferences you can<br>
> affirm) so that you intersect as many sets as possible (affirm as many<br>
> majorities as possible) while still keeping the intersection greater<br>
> than or equal to a Hare quota".<br>
><br>
> And that's apparently NP-complete (from clique).<br>
><br>
> The whole point of the method is to scale reasonably - otherwise, why<br>
> not just use Schulze STV? So if finding the right group of voters to<br>
> assign to a candidate is NP-complete, that's kind of a disappointment.<br>
><br>
> -<br>
><br>
> However, it seems the general idea can be salvaged. That is:<br>
><br>
> 1. Start with no winners<br>
> 2. Find a quota-sized group who most strongly prefer some candidate to win<br>
> 3. Elect that candidate and remove the voters from consideration<br>
> 4. If it's candidate-based (not party list) or the party who won has had<br>
> all its candidates elected (e.g. fielding 3 candidates and getting 3<br>
> seats), eliminate the winner from all the ballots.<br>
> 5. Repeat until all seats are full.<br>
><br>
> If I replace 2. by:<br>
> 2. For each remaining candidate:<br>
> 2.1. Measure how strong support we can get, from a Hare<br>
> quota, in favor of this candidate being elected.<br>
> Return the candidate that maximizes this measure.<br>
><br>
> then I basically get Monroe Algorithm A (if I've got it right), and it<br>
> forms a class of greedy multiwinner methods based on how step 2.1. works.<br>
><br>
> I don't know if it's possible to do this for more complex Condorcet<br>
> methods, but it should be possible for simpler ones. I made a linear<br>
> program for minmax(margins) for this purpose:<br>
> <a href="https://github.com/kristomu/voting-scripts/blob/master/linear_programming/sequential_minmax/multiwinner_minmax.mod">https://github.com/kristomu/voting-scripts/blob/master/linear_programming/sequential_minmax/multiwinner_minmax.mod</a><br>
><br>
> Getting the best minmax(margins) score of a candidate A (with only a<br>
> quota's worth of voters involved) can be phrased like this, in the<br>
> "minimize worst defeat" formulation:<br>
><br>
> Minimize max_defeat<br>
><br>
> subject to:<br>
> for every candidate Y != A:<br>
> max_defeat >= sum over all ballots b:<br>
> ballot_weight_chosen[b] * {voter voted Y>A}<br>
> - ballot_weight_chosen[b] * {voter voted A>Y}<br>
><br>
> for all ballots b:<br>
> 0 <= ballot_weight_chosen[b] <= ballot_weight[b]<br>
><br>
> (sum over all ballots b: ballot_weight_chosen[b]) = quota<br>
><br>
> where for the bth ballot, ballot_weight_chosen[b] is 0 if the ballot was<br>
> not made part of the group, otherwise it gives how much of the ballot<br>
> was chosen. E.g. if the ballot weight is 9, ballot_weight_chosen could<br>
> be anything between 0 and 9 inclusive.<br>
><br>
> Solving the linear program will then provide a suitable function for<br>
> step 2.1. above. The general method would, in each round, choose the<br>
> candidate with minimal max_defeat over all candidates. Ties can be<br>
> broken by random voter hierarchy based on the whole electorate (no<br>
> voters eliminated).<br>
><br>
> The "virtual constituencies" formed may contain fractional ballots, but<br>
> that is required for the LP to work. If we demanded the group be<br>
> assembled only from whole voters, the program would be an integer<br>
> program and (presumably) much more difficult to solve. Besides, not<br>
> permitting fractional vote assignment could lead to weird outcomes where<br>
> multiplying all voter weights by 100 could change the result.<br>
><br>
> Some other notes:<br>
><br>
> - It's rather easy to make a Droop MJ method based on this (and it ends<br>
> up looking a lot like STV).<br>
><br>
> - It's rather hard to make a Hare MJ method based on this, because 50%<br>
> of the voters in each Hare quota can vote for someone but your target<br>
> candidate, and how you choose them makes a difference as to whom will<br>
> get high scores later.<br>
><br>
> - If I recall correctly, way back when I made the proportionality/BR<br>
> simulator, I implemented something similar to this for Kemeny by brute<br>
> force. It ended up being very proportional but having very bad<br>
> (majoritarian) BR. So for party list, I'd suggest an ensemble-rule like<br>
> system with relatively more centrist tiebreakers/kingmakers.<br>
><br>
> - Minmax isn't that good a method as far as compliances go. It fails<br>
> Smith, independence from clones, and Condorcet Loser, for instance.<br>
><br>
> - Furthermore, any party list system based on quotas are subject to the<br>
> Alabama paradox. So the system isn't all that good, but it's a proof of<br>
> concept. Well, more of a sketch of a proof of concept :)<br>
> ----<br>
> Election-Methods mailing list - see <a href="http://electorama.com/em">http://electorama.com/em</a> for list info<br>
</p>