<div dir="auto"><div><br><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Thu, Nov 10, 2022, 9:40 PM Richard, the VoteFair guy <<a href="mailto:electionmethods@votefair.org">electionmethods@votefair.org</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">On 11/9/2022 12:05 AM, Forest Simmons wrote:<br>
<br>
> A candidate is uncovered iff it has a beatpath of<br>
> only two steps to each candidate (if any) that beats it.<br>
<br>
Should the "only two steps" be interpreted as "two or fewer steps?" or <br>
"two or more steps?" or "exactly two steps?"<br></blockquote></div></div><div dir="auto">In this context, exactly two, because when a candidate beats it, it cannot beat that candidate in one step, since it is not possible for X to outrank Y on more ballots than Y outranks X, and at the same time have Y out rank X on more ballots than X outranks Y.</div><div dir="auto"><br></div><div dir="auto">A candidate X has a beatpath to every other candidate if and only if X is in the Smith set.</div><div dir="auto"><br></div><div dir="auto">So the Landau is a subset of the Smith set.</div><div dir="auto"><br></div><div dir="auto">You see Landau is more special because it requires shorter beatpaths. Any length beatpath will do for Smith.</div><div dir="auto"><br></div><div dir="auto">Suppose the cycle ABCA is augmented with a candidate D that beats only A, and is beaten by B and C.</div><div dir="auto"><br></div><div dir="auto">Then all four candidates are in Smith because the beat cycle DABCD includes a beatpath from each candidate to each of the other candidates.</div><div dir="auto"><br></div><div dir="auto">Is D a Landau candidate? Well C beats it. So does it have a two-step path back to C? No. So it is covered by C, and covered means not in Landau.</div><div dir="auto"><br></div><div dir="auto">Is A a Landau candidate? Yes. Only D and C beat it, and ABD and ABC are the respective required two-step beatpaths that get back to D and C, respectively.</div><div dir="auto"><br></div><div dir="auto">Does that help?</div><div dir="auto"><br></div><div dir="auto"><br></div><div dir="auto"><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
<br>
I'm still trying to understand the "uncovered" (and "covered") concept. <br>
Especially, what is its relationship to the Smith set?<br>
<br>
Thanks,<br>
<br>
Richard Fobes<br>
<br>
<br>
On 11/9/2022 12:05 AM, Forest Simmons wrote:<br>
> In this context the most relevant question is what do we mean by <br>
> "uncovered", since that's the word used in the method definition ...<br>
> <br>
> Repeatedly eliminate the (remaining) candidate with fewest votes until <br>
> there remains only one uncovered candidate to elect.<br>
> <br>
> No need to know what covering means, although you can figure it out <br>
> indirectly from the definition of "uncovered:"<br>
> <br>
> A candidate is uncovered iff it has a beatpath of only two steps to each <br>
> candidate (if any) that beats it.<br>
> <br>
> Any candidate X who complains that they should have won because they <br>
> beat the winner W pairwise will get this truthful and obviously relevant <br>
> rejoinder:<br>
> <br>
> When you were eliminated, you had fewer transferred votes than I.<br>
> <br>
> I fact, I beat every candidate pairwise that was not already eliminated <br>
> (like you) on the basis of two few (transferred) votes.<br>
> <br>
> It is very easy to discern if some candidate X is uncovered:<br>
> <br>
> Just check each candidate Y that beats it (X) to see if it has a two <br>
> step beatpath via some Z, back to Y:<br>
> <br>
> X beats Z beats Y<br>
> <br>
> Only Smith candidates can be uncovered because only Smith candidates <br>
> have beatpaths back to the candidates that beat them. So the candidates <br>
> you have to check are the Smith candidates ... at most three, and rarely <br>
> more than one, in a public election.<br>
> <br>
> If you want, you can run IRV all the way through ... then if the IRV <br>
> winner is uncovered, you are done. If not, back up until you cone to an <br>
> uncovered candidate ... that's your winner!<br>
> <br>
> It's just a matter of doing regular IRV, and backing up (if necessary) <br>
> until you get to an uncovered candidate.<br>
> <br>
> Forest<br>
> <br>
> <br>
> On Tue, Nov 8, 2022, 11:18 AM Kristofer Munsterhjelm <br>
> <<a href="mailto:km_elmet@t-online.de" target="_blank" rel="noreferrer">km_elmet@t-online.de</a> <mailto:<a href="mailto:km_elmet@t-online.de" target="_blank" rel="noreferrer">km_elmet@t-online.de</a>>> wrote:<br>
> <br>
> On 08.11.2022 18:02, Richard, the VoteFair guy wrote:<br>
> > Forest, what do you mean by "covered"? Is there a Wikipedia or<br>
> > Electowiki article (or section of an article) that explains it? <br>
> Or is<br>
> > there a dictionary reference you can point to?<br>
> ><br>
> > Yes, you've used the words "covered" and "uncovered" many times<br>
> but I<br>
> > don't recall ever seeing a clear explanation of what you mean. I<br>
> > presume it involves pairwise counts, but that's as far as I can<br>
> guess.<br>
> <br>
> The short answer is: A covers B if A pairwise beats everybody B<br>
> pairwise<br>
> beats and then some.<br>
> <br>
> An uncovered candidate is someone who is not covered by anyone else.<br>
> <br>
> This definition works when there are no pairwise ties. Things get<br>
> trickier with pairwise ties, as I found out when generalizing Friendly<br>
> Cover.<br>
> <br>
> -km<br>
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</blockquote></div></div></div>