RE: [EM] Condorcet vs Approval
Thanks to Martin Harper and Craig Layton for valuable critiques. In particular, Martin is right. The voters should be able to make distinctions among their unapproved candidates, too. Here's a more ideal version of a compromise between Condorcet and Approval, which could be considered a dyadic refinement of Approval. A voted ballot would look like this: ABC DE F GHIJ KLMNOP QRS TUVW XYZ Or expressed differently 111 ABC 110 DE 101 F 100 GHIJ 011 KLMNOP 010 QRS 001 TUVW 000 XYZ Candidates are compared head-to-head as in Condorcet. Candidates in the same group get scores of zero against each other. When candidates are in distinct groups, the more approved candidate (the one further to the left in the first representation or with the higher binary label in the second representation) gets a score corresponding to how big the biggest inequality symbol is that separates the two candidates being compared. In the second representation, the score would be the place of the most significant digit in which the two binary labels differ. For example, between C and F, the biggest inequality is a double inequality , so C beats F by two points. In the binary labels for C and F the most significant digit in which they differ is the second digit, so again, C beats F by two points. In this example, any letter before K in the alphabet beats any letter after J by three points because to get from one group to the other you have to cross the triple inequality and (from the other point of view) all such pairs differ in their most significant digit which is the third binary place (from the right). (Significance increases from right to left.) The marks the fundamental boundary between approved and unapproved. the marks the boundary between a refined approval within those two groups. Similarly, the gives further refinement. For strategy, figure the location first, as you would in normal approval. Then figure each as you would in normal approval if the only candidates running were the ones in the group to be divided by the . Similarly, put in each by pretending nothing exists outside the group being divided. That's it except for deciding how to break Condorcet cycles. I'm still not sure of the best method, but one possibility is to first ignore the single inequalities, and if that doesn't do it, then ignore the double inequalities. I have some other ideas for avoiding cycles which I will explain later. Obviously, this method can be extended to many levels of refinement. For practical purposes, I think three would be maximum. However, theoretical purposes abound, so we shouldn't neglect it for cumbersomeness alone. Forest
RE: [EM] Condorcet vs Approval
One other thing. In a zero information election, start by expressing your utilities in binary rounded to three binary digits, this takes you directly to the second representation of the dyadic refined approval ballot below, bypassing the strategic , , and boundary calculations. Forest On Fri, 2 Mar 2001, Forest Simmons wrote: Thanks to Martin Harper and Craig Layton for valuable critiques. In particular, Martin is right. The voters should be able to make distinctions among their unapproved candidates, too. Here's a more ideal version of a compromise between Condorcet and Approval, which could be considered a dyadic refinement of Approval. A voted ballot would look like this: ABC DE F GHIJ KLMNOP QRS TUVW XYZ Or expressed differently 111 ABC 110 DE 101 F 100 GHIJ 011 KLMNOP 010 QRS 001 TUVW 000 XYZ Candidates are compared head-to-head as in Condorcet. Candidates in the same group get scores of zero against each other. When candidates are in distinct groups, the more approved candidate (the one further to the left in the first representation or with the higher binary label in the second representation) gets a score corresponding to how big the biggest inequality symbol is that separates the two candidates being compared. In the second representation, the score would be the place of the most significant digit in which the two binary labels differ. For example, between C and F, the biggest inequality is a double inequality , so C beats F by two points. In the binary labels for C and F the most significant digit in which they differ is the second digit, so again, C beats F by two points. In this example, any letter before K in the alphabet beats any letter after J by three points because to get from one group to the other you have to cross the triple inequality and (from the other point of view) all such pairs differ in their most significant digit which is the third binary place (from the right). (Significance increases from right to left.) The marks the fundamental boundary between approved and unapproved. the marks the boundary between a refined approval within those two groups. Similarly, the gives further refinement. For strategy, figure the location first, as you would in normal approval. Then figure each as you would in normal approval if the only candidates running were the ones in the group to be divided by the . Similarly, put in each by pretending nothing exists outside the group being divided. That's it except for deciding how to break Condorcet cycles. I'm still not sure of the best method, but one possibility is to first ignore the single inequalities, and if that doesn't do it, then ignore the double inequalities. I have some other ideas for avoiding cycles which I will explain later. Obviously, this method can be extended to many levels of refinement. For practical purposes, I think three would be maximum. However, theoretical purposes abound, so we shouldn't neglect it for cumbersomeness alone. Forest
RE: [EM] Condorcet vs Approval
For this zero information direct utility conversion to work best, all utilities should be between zero and .99, and after the conversion to binary, truncate to three binary digits (instead of rounding). It's hard to say whether this method is more in the spirit of Approval or the spirit of Condorcet. The head-to-head feature is essential: otherwise strategy takes us right back to ordinary Approval. In other words, its the Condorcet aspect that makes this form of CR strategically different from ordinary Approval. On the other hand, the miniature approval strategies at the various levels makes this method seem more like Approval than Condorcet in spirit. In particular, it seems designed to handle those lopsided utility examples that Bart Ingles keeps bringing up in order to point out that even with a definite CW (no cycles to break) Condorcet can distort the combined will of the electorate. Bart Ingles and Joe Weinstein were the inspiration for this method: Bart because of his insightful critiques of Condorcet, and Joe for his great reasons to continue the hunt for a good Cardinal Ratings method, and his insight that summing and averaging are not always the best ways to aggregate individual utility into social utility. Forest On Fri, 2 Mar 2001, Forest Simmons wrote: One other thing. In a zero information election, start by expressing your utilities in binary rounded to three binary digits, this takes you directly to the second representation of the dyadic refined approval ballot below, bypassing the strategic , , and boundary calculations. Forest On Fri, 2 Mar 2001, Forest Simmons wrote: Thanks to Martin Harper and Craig Layton for valuable critiques. In particular, Martin is right. The voters should be able to make distinctions among their unapproved candidates, too. Here's a more ideal version of a compromise between Condorcet and Approval, which could be considered a dyadic refinement of Approval. A voted ballot would look like this: ABC DE F GHIJ KLMNOP QRS TUVW XYZ Or expressed differently 111 ABC 110 DE 101 F 100 GHIJ 011 KLMNOP 010 QRS 001 TUVW 000 XYZ Candidates are compared head-to-head as in Condorcet. Candidates in the same group get scores of zero against each other. When candidates are in distinct groups, the more approved candidate (the one further to the left in the first representation or with the higher binary label in the second representation) gets a score corresponding to how big the biggest inequality symbol is that separates the two candidates being compared. In the second representation, the score would be the place of the most significant digit in which the two binary labels differ. For example, between C and F, the biggest inequality is a double inequality , so C beats F by two points. In the binary labels for C and F the most significant digit in which they differ is the second digit, so again, C beats F by two points. In this example, any letter before K in the alphabet beats any letter after J by three points because to get from one group to the other you have to cross the triple inequality and (from the other point of view) all such pairs differ in their most significant digit which is the third binary place (from the right). (Significance increases from right to left.) The marks the fundamental boundary between approved and unapproved. the marks the boundary between a refined approval within those two groups. Similarly, the gives further refinement. For strategy, figure the location first, as you would in normal approval. Then figure each as you would in normal approval if the only candidates running were the ones in the group to be divided by the . Similarly, put in each by pretending nothing exists outside the group being divided. That's it except for deciding how to break Condorcet cycles. I'm still not sure of the best method, but one possibility is to first ignore the single inequalities, and if that doesn't do it, then ignore the double inequalities. I have some other ideas for avoiding cycles which I will explain later. Obviously, this method can be extended to many levels of refinement. For practical purposes, I think three would be maximum. However, theoretical purposes abound, so we shouldn't neglect it for cumbersomeness alone. Forest
Re: [EM] Multiple Winner Elections
I would like to make a suggestion for a multiple winner proportional method that is as good or better than any I have heard proposed so far, short of the Proportional Approval Voting (PAV) that Michael Welford and I proposed several weeks ago. (Full strength PAV would involve checking all of the two candidate subsets, one-by-one on each ballot, so it would probably be more elaborate than what you wanted.) The method I recommend is a simple sequential version of PAV: (1) Have the voters fill out approval ballots (identical to ordinary one vote per voter ballots with instructions to vote for as many of the candidates as they would like to support). (2) Completely eliminate all of the candidates that get fewer than 30% of the vote. (Do not even put them back in for step 4 below.) (3) Pick out the Approval Winner from among the remaining candidates (the candidate with the most votes). This is the first winner. (4) Recount the ballots with this twist: All of the ballots that did NOT approve of the first winner get counted twice. The winner of step 4 is the second winner. If one of the winners is to have special tie breaking privileges or be considered the chair, while the other is to be considered the co-chair, then no known method is superior to this one. It could well be worth designing the chair/co-chair position to fit this election method, since the method is superior in every way (including simplicity and ease of implementation) to all of the competing methods for that kind of arrangement. Even if the two co-chairs are to be considered equal in every way, this method is as good as any sequential method (like STV) for getting proportional (or better) representation, and is more likely to get winners with broad support than STV or any simple method (short of full PAV in simplicity). If you would like step by step examples, or have any questions, I would be glad to help. Forest On Mon, 19 Feb 2001, Moe St. EverGreen wrote: What are the best choice(s) of voting system(s) for a multiple winner election? I can believe there could be more than one depending on the type of election. For instance, we will have a 2 person co-chair in an organization, where the idea is to balance any opposing factions as they will likely occur. What would be the best form of election? I know Approval could be used (taking the top two), but it seems very easily for a faction to have a strategy of running multiple candidates. Cumulative seems a little like overkill, and I'm not sure how well it stands up against other systems. I don't see how we could possibly use any form of Proportional for this. And I worry that STV is just as bad as IRV. The other situation is that we will have a large county executive committee, which we want to elect using some form of PR. The idea is that each candidate must choose to run either unaffiliated, or as affiliated with one of our activist groups or neighborhood groups (affiliation being decided by a majority approval of that group). Whatever the method of election counting, the seats would be filled in the order of any top winning unaffiliated winners first, then the remaining seats would be proportioned out per the total support each group received (with affiliated being a group), with the seats for each group being filled in order of most support to least support. Since we are a county political party, we won't have parties within ourself, but we will very likely have factions and neighborhood groups which will need to have representation. Please let me know of the best alternative methods we may use, and if we did use the system above, should we rank the votes, or use something similar to approval voting, etc., for the actual ballot and ballot counting. - Moe.
[EM] Son of Condorcet vs Approval
111 ABC 110 DE 101 F 100 GHIJ 011 KLMNOP 010 QRS 001 TUVW 000 XYZ In the second representation, the score would be the place of the most significant digit in which the two binary labels differ. Oops, this means that there's a whopping difference between 100 and 011, merely because of an artifact of the representation. That is, if you dropped ABC from the election, 011 moves up to 100 and 100 moves up to 101, so the score becomes 1 instead of 3. How about just making the score the difference between the two values? Or if you want to give the score a logarithmic flavor, as your second proposal does, how about using the base 2 log of the difference? This gives roughly the same distribution of scores, but without the artifact.
Re: [EM] Condorcet vs Approval
If you solve circular ties by Approval, where candidates whom you've ranked get an Approval vote from you, then you have to worry about strategy, how far to extend you ranking, even if there's no danger of anyone using offensive order-reversal strategy. With Condorcet, in the form of PC, Smith//PC, SSD, Cloneproof SSD, or BeatpathWinner, a majority who prefer the sincere CW to B can be assured that B can't win, if that majority vote sincerely and if no one falsifies a preference. That can't be said for other methods. For SSD, Cloneproof SSD, BeatpathWinner, a majority who prefer a certain member of the sincere Smith set to B can be assured that B won't win, if that majority votes sincerely, and if B isn't a member of the sincere Smith set, and if no one falsifies a preference. For the purpose of these gurarantees, a voter votes sincerely if he doesn't falsify a preference or leave unvoted a preference that the balloting system in use would have allowed him to vote in addition to the preferences that he actually voted. Those gurantees that I mentioned are the criteria SFC GSFC. Complying methods are strategy-free under plausible conditions. Mike Ossipoff _ Get your FREE download of MSN Explorer at http://explorer.msn.com
