On Tue, May 4, 2010 at 8:12 AM, Peter Zbornik <[email protected]> wrote: > I am affraid that this is not possible. First we have mostly odd-numbered > council sizes, and secondly the gender rule does not require that half of > the men should be men and the other half women. > Our current gender rule goes as following: "for every three members of the > body, there has to be one person of each sex". A five member council thus > has to have one woman and one man. For seven members it is two men and two > women.
For elimination based PR-STV, I think my suggestion would be the most reasonable. - Set the threshold at one larger than the required number - protect from elimination members of a gender if elimination would reduce their number below the threshold - prohibit from election members of a gender if that election would leave less than a threshold for the other gender - on the last round, remove remove the restrictions In a 5 person council, that means that there must be at least 1 man and 1 woman. If the candidates were Men: M1 M2 M3 M4 M5 and Women W1 W2 W3 then an election might go something like Round 1 M1: 20 M2: 15 M3: 15 M4: 10 M5: 10 W1: 10 W2: 8 W3: 12 Total: 100 Quota: 17 M1 gets elected + 3 are distributed Round 2 M1: 17* M2: 16 (+1) M3: 15 M4: 10 M5: 11 (+1) W1: 11 (+1) W2: 8 W3: 12 W2 is lowest, so is eliminated, +8 are distributed Round 3 M1: 17* M2: 16 M3: 15 M4: 13 (+3) M5: 14 (+3) W1: 13 (+2) W2: 0 W3: 12 W3 is lowest. However, eliminating W3 would reduce the number of women below 2, so the lowest man is eliminated. M4 is eliminated + 13 are distributed Round 4 M1: 17* M2: 17 (+1) M3: 17 (+2) M4: 0 M5: 16 (+2) W1: 17 (+4) W2: 0 W3: 16 (+4) M2, M3 and W1 all meet the quota, so all are elected, but no surplus is distributed. Round 5 M1: 17* M2: 17* M3: 17* M4: 0 M5: 16 W1: 17* W2: 0 W3: 16 If this had been a previous round, W3 would be protected from elimination, as there are only 2 women left. However, since this is the last round, (only 1 seat left to fill and 2 candidates for the seat), the restriction is lifted. Both W3 and M5 have 16 votes, so a tie break rule (say coin toss), would decide which one is eliminated. If M5 is eliminated, then the results are: M1+M2+M3+W1+W3 if W3 is eliminated, then the results are M1+M2+M3+M5+W1 In both cases, the requirement for at least 1 man and 1 woman is met. >> Methods like Schulze STV work by comparing possible councils to determine >> which are best. Thus, it may be possible to limit them to only consider >> "balanced" councils. I'm not sure how to do this in ordinary STV, however, >> since it doesn't work that way, and in any case, this would be untested. Yes, you can. The software would just need to be updated. A council with 5 men and 0 women would be considered to lose to a council of 4 men and 1 woman. ---- Election-Methods mailing list - see http://electorama.com/em for list info
