<div dir="ltr"><div dir="ltr"><p class="MsoNormal"><span lang="EN-US">Thanks Kristofer,<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US"> </span></p>
<p class="MsoNormal"><span lang="EN-US">Proving the
impossibility of a cycle - not just within Smith Set members either - within
the SC Set is rather important, as failing to do so would leave open the opportunity
for the SC Set to be empty. I wouldn’t like that at all.<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">I’ll try
below to prove that a three-way cycle between three candidates only (ignoring
all others) is indeed impossible, at least if equal ranking is disallowed. <span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">The scenario
is this: three candidates only (A, B, and C) are ranked, with truncation
allowed but equal-ranking disallowed, and with the count being approval: if we
assume that setting the cutoff at A elects B, and that setting the cutoff at B
elects C, I shall try to prove that Setting the cutoff at C can never elect A.<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">To do this,
we must see that within the confines of this scenario, there are only 15
possible votes that can be cast: <span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 1, ABC<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 2, ACB<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 3, AB<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 4, AC<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 5, A<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 6, BAC<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 7, BCA<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 8, BA<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 9, BC<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 10, B<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 11,
CAB<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 12,
CBA<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 13, CA<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">Vote 14, CB<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US">And Vote
15, C<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">We do not know
how many of each Vote type were cast, but we know certain things. <span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">If setting the
cutoff at A elects B, then it must follow that Votes 9 +10 + 14 must exceed
Votes 1 + 2 + 3 + 4 +5 + 11 +13, with Votes 6, 7, 8 and 15 being irrelevant.<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">If setting the
cutoff at B elects C, then it must follow that Votes 4 +13 + 15 must exceed
Votes 1 + 3 + 6 + 7 +8 +9 +10, with Votes 2, 5, 11, 12 and 14 being irrelevant.<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">If we
assume that setting the cutoff at C would elect A, then it would follow than Votes
3 + 5 +8 must have exceeded Votes 7 + 9 + 11 +12 +13 +14 +15, with Votes 1, 2,
4, 6 and 10 being irrelevant. But this cannot be. <span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">To see why,
add the first two true expressions and remove common terms from both ends of
the inequality, leaving the fact that Votes 14 +15 must exceed Votes 1 + 1 + 2
+ 3 + 3 + 5 + 6+ 7 +8 + 11.<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">Now rewrite
the third inequality to Votes 3 + 5 + 8 – 7 -9 -11 -12 -13 must exceed Votes 14
+15.<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">If we
combine these last two, we can write that Votes 3 + 5 + 8 – 7 -9 -11 -12 -13 must
exceed Votes 14 +15, which in turn must exceed Votes 1 + 1 + 2 + 3 + 3 + 5 + 6+
7 +8 + 11.<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">Removing the
middle bit and common terms, leaves the “fact” that the negative sum of Votes 7,
9, 11, 12 and 13 must somehow exceed the very positive sum of Votes 1,1,2,3,6,7
and 11. This cannot be, so the third term cannot be possible, hence there
cannot be a three-way cycle under these conditions.<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">There’s almost
certainly some error in there, and even if there isn't, you may still have a cycle between more than 3 candidates, or if equal rankings are allowed (too lazy to think that through).</span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p><p class="MsoNormal"><span lang="EN-US">Still, I simulated very many random scenarios, and
couldn’t find an election where the Set was empty. <span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">Regards,<span></span></span></p><p class="MsoNormal"><span lang="EN-US"><br></span></p>
<p class="MsoNormal"><span lang="EN-US">Etjon Basha<span></span></span></p>
<p class="MsoNormal"><span lang="EN-US"> </span></p></div></div><br><div class="gmail_quote"><div dir="ltr" class="gmail_attr">On Wed, Oct 2, 2024 at 5:25 AM Kristofer Munsterhjelm <<a href="mailto:km-elmet@munsterhjelm.no">km-elmet@munsterhjelm.no</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0px 0px 0px 0.8ex;border-left:1px solid rgb(204,204,204);padding-left:1ex">On 2024-09-28 13:06, Etjon Basha wrote:<br>
> Thanks Kristofer,<br>
> <br>
> On why the CW must be in the Serious Candidate Set: once explained in <br>
> those terms, it really seems quite intuitive indeed, and I should have <br>
> thought a bit harder before posting.<br>
> <br>
> I also think that explanation is sufficient to model the observed <br>
> relationship between the Smith and SC Sets: using your argument, setting <br>
> the approval cut-off at a Smith Set member will elect some Smith Set <br>
> member. Setting to the cutoff at some non-member may elect any <br>
> candidate, inside or outside the Smith Set.<br>
<br>
It almost is. You'd also have to show that no cycle can appear, so that <br>
the Smith set members don't exclude each other from the SC set by having <br>
another Smith set member win when the approval cutoff is placed just <br>
below their rank.<br>
<br>
I'm pretty sure that the relation is acyclical, so that that can't <br>
happen: my thoughts are something like that one of the Smith set members <br>
has more votes closer to the top than the other Smith set members have, <br>
and therefore will win his own contest. But I haven't proven it in any <br>
formally rigorous way :-)<br>
<br>
-km<br>
</blockquote></div>