Here's a quick way to find the Condorcet Winner if there is one:
Use Rob LeGrand's ballot by ballot approval idea, but instead of ballot by ballot, use voter by voter.
For fairness, either randomize the order of polling the voters or else poll them twice, once from left to right, and once from right to left, so that each voter gets to vote twice.
Forest
From: "James Green-Armytage" <[EMAIL PROTECTED]> Subject: Re: [EM] James: counting time
No, pairwise counting takes longer than Approval, nearly always., whether done by ballot or show of hands.
Oh, I wasn't trying to argue that a rough pairwise count is quicker than approval. I was just saying that the extra time it takes is trivial unless there is less than a few minutes to make the decision, in which case it's likely to be an executive decision rather than a committee decision anyway.
Pairwise count time increases quadratically with the number of candidates. Approval counting time increases linearly with the number of candidates.
Here's what I'm suggesting for a quick pairwise count within a committee. First of all, you don't have to count all of the pairwise contests. Let's say that there are six options, A through F, and you want to make a quick majority decision between these options.
Step 1: Have everyone jot down their preference order on a sheet of paper in front of them. However, they keep the piece of paper, rather than handing it over to someone else.
Step 2: Start with two options that seem relatively likely to be Condorcet winners. Let's say that this is A and B. The chairperson (or whatever you call the person managing the process) asks for a show of hands for A>B, and then a show of hands for B>A. Let's say that B wins. The chairperson makes a note of the pairwise victory, preferably on a chalkboard or something everyone can see, drawing an arrow from the winner to the loser, with the arrow marked by the defeat strength.
Step 3: Since B is unbeaten, try comparing other options to B, starting with the options that seem most viable. If B is a Condorcet winner, then you can have the whole thing over with in 5 counts. There's no need to do the other 10. Or maybe candidate D beats candidate B. Then you want to do the comparisons with D to see if D is a Condorcet winner. It's possible that you'll uncover a cycle, in which case the process will take a bit longer, but there should still be pairwise comparisons that you can safely avoid without even compromising GeTChA-efficient methods like beatpath and ranked pairs. If you're using minimax (PC), you can get away with doing even fewer comparisons.
For example, let's say that the process goes like this, in a 15-member council (I don't know if these exact numbers are possible, but it doesn't matter): Count 1: A vs. B. B>A, 9-6 B is unbeaten Count 2: B vs. C. B>C, 11-4 B is unbeaten Count 3: B vs. D. D>B, 10-5 D is unbeaten Count 4: D vs. A. A>D, 8-7 D has the weakest defeat against it Count 5: D vs. C. D>C, 10-5 D has the weakest defeat against it Count 6: D vs. E. D>E, 12-3 D has the weakest defeat against it Count 7: D vs. F. D>F, 11-4 D has the weakest defeat against it. All pairwise comparisons with D have been explored. D is certainly the minimax winner; no more comparisons are necessary.
So, we've resolved a 3 candidate cycle in 7 counts, rather than doing the full 15 counts. Obviously more time-consuming than an approval election, but it can probably be done within 5 minutes, if the council members are on the ball. But anyway, I think that proving a CW in 5 or 6 counts is the most likely scenario, which makes things easier.
You're talking about when there are only very few candidates or alternatives. Maybe then handcounted Condorcert would work in a hurry.
Yeah, maybe I'm not thinking about a very large number of options. I don't think that semi-formal quick decision committee scenarios with more than 7 or 8 viable options are very common, but I'm not sure. Can you think of some good examples of this?
In a group voting on what movie to go to, I'd suggest Condorcet (PC) if there were only a few movies. If it were necessary to vote among many movies, and there weren't much time, or much pre-existing interest in voting systems, and no already-programmed computer, I'd use Approval instead of Condorcet.
Hmm, actually I'd rather be in a movie group that uses a proportional representation system of some kind. That is, even if we only get one movie per night, doing proportional representation for the movies chosen over a longer period of time. But if it was just a one-time thing, I don't know. I don't think the group should get a movie that isn't approved of by everyone; I'd like to think that it would be choosing between multiple movies that are acceptable to everyone. So, what if more than one movie gets a perfect approval score?
Sincerely, James fc.antioch.edu/~james_green-armytage/voting.htm
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Message: 2 Date: Fri, 14 Jan 2005 15:36:34 -0800 (PST) From: Forest Simmons <[EMAIL PROTECTED]> Subject: [EM] Why Three Candidate Methods Are So Important To: election-methods-electorama.com@electorama.com Message-ID: <[EMAIL PROTECTED]> Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed
Date: Tue, 11 Jan 2005 23:40:47 +0100 From: Jobst Heitzig <[EMAIL PROTECTED]> Subject: [EM] Re: Approval/Condorcet Hybrids
<snip>
Forest has made [a plausibility argument] that in public elections it will be paramountly probable that there is either a CW or a three-element covering set, that is, a cycle of three candidates of which each other beats at most one.
However, keep in mind that one or more of these three "candidates" could be a collapsed beat clone cycle of three real candidates.
In an election without primaries there might be a beat clone set of three Republican candidates forming a cycle, for example, and if there are other uncovered candidates, this "collapsed" Republican cycle could be one of the three "candidates" in the main cycle.
Hence I think that he is right in suggesting that we should not only reduce to the uncovered candidates but even to the members of the minimal covering set (=Dutta set). I conjecture that this set can be found in O(n^3) time. Anyway, once found, it can easily be shown to be the minimal covering set. Once we have reduced everything to at most three candidates, we can proceed as we like by either dropping the weakest defeat in the cycle or by drawing random ballots or by maximizing approval, etc.
If the resulting winner turns out to be a collapsed clone set, then we unfold that clone set and (recursively) apply the same method to it.
So we see that the study of three candidate methods is of greatest importance for single winner methods in general.
Now it seems that we could reduce the study of three candidate methods to a decision tree depending on the relative sizes of the twelve possible factions
A, B, C, A=B, A=C, B=C, A>B, A>C, B>C, B>A, C>A, C>B, C>A
There are twelve factorial possile orderings of these factions by size, but neutrality will cut down on the number that we need to consider.
Also requiring some form of monotonicity would cut down on the possibilities.
How about Reverse Cancellation? This means that wife can cancel husband's vote with hers no matter how he votes.
The ballot pairs that (should) cancel are of the form
X>Y>Z cancels Z>Y>X, and X=Y>Z cancels Z>X=Y.
This condition would reduce every three candidate election down to a six faction case.
It's tempting to also require that certain triples cancel:
{A,B,C} or {A>B, B>C, C>A} or {C>B, B>A, A>C} or {A=B, B=C, C=A} .
But there is a high price for this one. It leads irrevocably to Borda.
Still, Spruced Up Borda might not be so bad. It is just Black in the three candidate case.
How does the following three candidate proposal stack up?
First cancel reverse ballot pairs. Then between the most truncated (or worst rank in case of no truncations) candidate X and the candidate Y with the least top rank support, keep the one favored pairwise for the final pairwise comparison with the other candidate Z. If X=Y, then eliminate this candidate and take the pairwise winner from the other two.
Forest
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