I've been perusing the Election Methods archive and noticed that more or less recently there was the need for simple legislative language to define Condorcet and/or Smith (Subjects: "Hay guys, look at this..." and "Another Simple Way to Find Smith"). Apologies if what I write following this paragraph has already been explored in the mailing list. Looking at the Wikipedia page for "Smith Set", it seems to contain a simple algorithm for finding the Smith Set... (1) Start with a set D (Dominating/Smith Set) with no candidates in it. (2) Add all candidate(s) who are in the Copeland (Winner) Set to D. (3) Add to D the candidate(s) who are NOT currently in D and who also pairwise beats any candidate currently in D. (4) Repeat step (3) until no candidate(s) are added by step (3). RESULT: Set D is the Smith Set. Copeland (Winner) Set are set of candidate(s) with the highest Copeland score. That's it! Read further if you want more details... The Wikipedia page for "Smith Set" ( https://en.wikipedia.org/wiki/Smith_set) uses propositional(?) logic to determine that any candidate in the Smith Set D will have at least a Copeland Score of θD (i.e. the threshold score for set D). If there is a θD, then there will be a maximum Copeland Score θW (i.e. the Copeland Winner(s) Score). θD and θW can be equal. I will sort of translate the propositional logic used in the Wikipedia page into something more layman like. > Choose d as an element of D with minimum Copeland score, and identify > this score with θD. Choose any candidate d in a pre-defined Smith Set D. This candidate d will have a Copeland Score of θD. > Now suppose that some candidate e ∉ D has a Copeland score not less > than θD. Now suppose there is some candidate e who is not in the Smith Set D and also has a Copeland Score equal to or more than θD. > Then since d belongs to D and e doesn't, it follows that d defeats e; Because candidate d is in the Smith Set D and candidate e is not, candidate d defeats candidate e. > and in order for e's Copeland score to be at least equal to d's, there > must be some third candidate f against whom e gets a better score than > does d. All the candidates in Smith Set D defeat candidate e. That is, candidate e's maximum possible Copeland Score is being eaten into. With candidate d being in Smith Set D, it will beat all candidates outside of Smith Set D. This sets the "high" threshold which candidate e needs to match. Therefore, for candidate e to at least equal candidate d's (threshold) Copeland Score, there will need to exist a candidate f who (1) candidate e defeats and (2) defeats candidate d. > If f ∈ D, then we have an element of D who does not defeat e, If candidate f was in the Smith Set D, then by definition, candidate f defeats candidate e. But we said in (1) that candidate e needs to defeat candidate f. > and if f ∉ D then we have a candidate outside of D whom d does not > defeat, leading to a contradiction either way. If candidate f was not in the Smith Set D, then by definition, candidate d defeats candidate f. But we said in (2) that there needs to be a candidate f who defeats candidate d. Thanks, Gervase.

On 4/8/23 03:18, Gervase wrote: > I've been perusing the Election Methods archive and noticed that more or > less recently there was the need for simple legislative language to > define Condorcet and/or Smith (Subjects: "Hay guys, look at this..." and > "Another Simple Way to Find Smith"). Apologies if what I write > following this paragraph has already been explored in the mailing list. > > Looking at the Wikipedia page for "Smith Set", it seems to contain a > simple algorithm for finding the Smith Set... > > (1) Start with a set D (Dominating/Smith Set) with no candidates in it. > (2) Add all candidate(s) who are in the Copeland (Winner) Set to D. > (3) Add to D the candidate(s) who are NOT currently in D and who also > pairwise beats any candidate currently in D. > (4) Repeat step (3) until no candidate(s) are added by step (3). > > RESULT: Set D is the Smith Set. > > Copeland (Winner) Set are set of candidate(s) with the highest Copeland > score. > > That's it! Read further if you want more details... That's a good idea. If there's a problem with it, I'd say it's the complexity: I think you worst case could end up doing n passes through the list of candidates, and each of these take n time. On the other hand, every Smith set candidate is in a three-cycle with some other Smith set candidate (absent ties), so perhaps I'm wrong about this. -km