[EM] A few more Bucklin variants, because why not?

[Etjon Basha]

Thanks Kristofer,



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.


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.


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.


To do this, we must see that within the confines of this scenario, there
are only 15 possible votes that can be cast:


Vote 1, ABC

Vote 2, ACB

Vote 3, AB

Vote 4, AC

Vote 5, A

Vote 6, BAC

Vote 7, BCA

Vote 8, BA

Vote 9, BC

Vote 10, B

Vote 11, CAB

Vote 12, CBA

Vote 13, CA

Vote 14, CB

And Vote 15, C


We do not know how many of each Vote type were cast, but we know certain
things.


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.


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.


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.


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.


Now rewrite the third inequality to Votes 3 + 5 + 8 – 7 -9 -11 -12 -13 must
exceed Votes 14 +15.


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.


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.


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).


Still, I simulated very many random scenarios, and couldn’t find an
election where the Set was empty.


Regards,


Etjon Basha



On Wed, Oct 2, 2024 at 5:25 AM Kristofer Munsterhjelm <
km-elmet at munsterhjelm.no> wrote:

> On 2024-09-28 13:06, Etjon Basha wrote:
> > Thanks Kristofer,
> >
> > On why the CW must be in the Serious Candidate Set: once explained in
> > those terms, it really seems quite intuitive indeed, and I should have
> > thought a bit harder before posting.
> >
> > I also think that explanation is sufficient to model the observed
> > relationship between the Smith and SC Sets: using your argument, setting
> > the approval cut-off at a Smith Set member will elect some Smith Set
> > member. Setting to the cutoff at some non-member may elect any
> > candidate, inside or outside the Smith Set.
>
> It almost is. You'd also have to show that no cycle can appear, so that
> the Smith set members don't exclude each other from the SC set by having
> another Smith set member win when the approval cutoff is placed just
> below their rank.
>
> I'm pretty sure that the relation is acyclical, so that that can't
> happen: my thoughts are something like that one of the Smith set members
> has more votes closer to the top than the other Smith set members have,
> and therefore will win his own contest. But I haven't proven it in any
> formally rigorous way :-)
>
> -km
>
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