# Talk:Dominant mutual third set

## Possible multi-winner generalizations[edit | edit source]

User:Kristomun, I think this may have some kind of STV-PR generalization. In a 2-seat election, we know that anyone who has over 1/4th of the active votes at any point in STV is guaranteed to be one of the final 3 remaining candidates, since it's impossible for 3 other candidates to each have more votes than this candidate (since they'd each have to have over 1/4th of the active votes, resulting in more than 100% of votes total being allocated to different candidates), which is what would enable them to survive elimination longer. So, we can say that when over 1/3rds of the voters prefer someone from the "dominant mutual quarter" set (DMT but for 1/4th of the electorate) over anyone else who survives until the final round, then the dominant mutual quarter candidate must win. In general, someone who is preferred by a solid coalition of 1/(k+2)th of the voters (k being the number of seats) and preferred by 1/(k+1)th of all voters over any other given rival must win. I'm not sure if there's a way to extract more from this insight, though. BetterVotingAdvocacy (talk) 03:48, 27 March 2020 (UTC)

## Smith-efficiency[edit | edit source]

User:BetterVotingAdvocacy, I don't think what you said is true: that electing from the DMT set implies Smith when the Smith set is a subset of the DMT set is. Consider e.g. an election where more than a third of the voters vote ABCD in some order above everybody else, and that there are say, 26 candidates. Suppose furthermore that each voter in a majority votes (some random permutation of a random subset of candidates E..Z) > (some permutation of A, B, and C) > D > everybody else. Now {A,B,C,D} is the smallest DMT set, but D is beaten pairwise by A, B, and C, and thus D is not in the Smith set. So the Smith set is {A,B,C} which is a subset of the smallest DMT set {A,B,C,D}. Then our contrived DMT-passing method could elect D, which would be in the DMT set but not the Smith set.

E.g.

12: D>B>C>A>E>F>G>H>I 11: A>B>C>D>E>F>G>H>I 11: C>A>B>D>E>F>G>H>I 20: E>A>B>C>D 20: F>B>C>A>D 20: G>C>A>B>D

Or the obligatory anti-IRV example:

12:D>A>B>C>E>F>G 11:D>B>A>C>E>F>G 10:C>D>A>B>E>F>G 1:C>B>A>D>E>F>G 21:E>A>B>C>D>F>G 22:F>B>C>A>D>E>G 23:G>C>A>B>D>E>F

Here the smallest DMT set is {ABCD}. IRV elects D. The Smith set is {ABC}. Kristomun (talk) 22:10, 8 May 2020 (UTC)

## Speedup is not a speedup[edit | edit source]

I've removed the part where IRV is sped up by aborting early after finding the DMT winner. This because you need the pairwise matrix to determine if the candidate is a DMT winner, and compiling that pairwise matrix takes a longer time than just running IRV to completion. I've replaced it with a more general statement about potential speedups. Kristomun (talk) 19:28, 22 May 2020 (UTC)

- I did explicitly write "This never requires more rounds of counting than the regular IRV approach (ignoring the discovery of the pairwise comparison matrix)," Even if you don't want to call it a speedup, why not preserve the example in some form? Part of the reason I prefer to mention that DMT can be used to reduce rounds of counting is because it helps provide a regularity or predictability to these methods, which is important because their fickle order of elimination often makes it hard to understand them. In other words, if there's an IRV election that requires 10 rounds of counting under the regular approach but 5 with DMT, then it's less cognitively burdensome to look at it using the DMT approach. BetterVotingAdvocacy (talk) 21:48, 22 May 2020 (UTC)