[EM] Quick and Clean Burial Resistant Smith

[Forest Simmons]

Ted,

Thanks for providing a great example that illustrates how the burial
defense works.

I agree that it's probably better, especially when there are many
candidates, to eliminate the non-Smith candidates before counting the basic
scores.

In the case of public elections for political office we expect the Smith
set to be small ... usually a singleton, and occasionally a triplet ...
unless factions think they can get away with burial!

In general, I like ratings information (as in your version of Approval
Sorted Margins) better than rankings ... the Q&C and Q&D methods were
created to see what a minimal acceptable rankings based burial resistant
method would look like.

With ratings and lots of candidates I would go back to Range Based (or
Total Approval) Chain Climbing for a burial resistant method:

While there is no pairwise undefeated candidate among the remaining
candidates ... eliminate from the remaining candidates all of those that do
not pairwise defeat the remaining candidate X that has the lowest Range
score (or alternately .. lowest below midrange approval score ... with or
without renormalization as candidates are eliminated).

In the case of a three candidate Smith Set this method first eats away all
of the non-Smith candidates ... then the lowest score Smith candidate X
(the one that got buried) and finally Y, the one responsible for burying X,
leaving Z, the one not beaten by X, (i.e. the one that Y buried Z under) as
the sole survivor... someone not preferred over X by Y.

It would be interesting to see how this method works on Colin Champion's
five candidate example... especially to see if the renormalizations are
worth the trouble ... and if perhaps the below midrange approval scores (or
the ASM approval scores) work better than range scores.

That's a lot of work! Do you have any students that need a project?

My best,

Forest


El mar., 4 de ene. de 2022 3:23 p. m., Ted Stern <dodecatheon at gmail.com>
escribió:

> Your Q&CBRS works well in a situation in which Approval Sorted Margins
> does not: (due to Colin Champion):
>
> Sincere:
>
> 1:  B > D > A > E > C
> 1:  B > D > E > C > A
> 5:  C > A > D > B > E
> 1:  D > A > C > B > E
> 1:  D > C > B > A > E
> 2:  E > B > D > C > A
>
> D is the beats-all Condorcet voter.
>
> 5 C-first voters bury D:
>
> 1:  B > D > A > E > C
> 1:  B > D > E > C > A
> 5:  C > A > B > E > D    # Was C > A > D > B > E
> 1:  D > A > C > B > E
> 1:  D > C > B > A > E
> 2:  E > B > D > C > A
>
> Pairwise array:
> [-- 6. 2. 5. 8.]
> [5. -- 4. 9. 9.]
> [9. 7. -- 5. 7.]
> [6. 2. 6. -- 4.]
> [3. 2. 4. 7. --]
>
> Now, if I understand your method correctly, we first find the Smith Set.
> In the burial case, it is all 5 candidates, A, B, C, D, E.
>
> Then we find the basic scores for each Smith candidate:
>
> A: 8
> B: 11
> C: 10
> D: 6
> E: 9
>
> The Smith candidate with the smallest basic score is D (our previous CW).
>
> Smith candidates that defeat D are *B* and *E*.  B has highest basic
> score, therefore is the winner. C's strategy did not help C.
>
> With Approval Sorted Margins, C is able to win using Burial.
>
> The Basic Score needs to be tabulated separately from pairwise, and
> depends on the other rankings on the ballot: each candidate gets a point
> for any rating above minimum candidate rating, if using a ratings ballot.
> In the case of many candidates, this could lead to basic score ties, since
> each ballot would certainly have many candidates rated 0, so in those cases
> I would recommend recounting after eliminating all candidates outside the
> Smith Set.
>
> I've been thinking about the impracticality of computing pairwise arrays
> in "jungle" elections, those with, say, >9 candidates. If a ratings ballot
> were used, with rankings inferred, I would recommend a floating score
> threshold, starting at 1% of maximum approval, but rising until at most 9
> distinct candidate scores are above the threshold (allows score-tie
> clusters), then recounting to get the reduced pairwise array and basic
> scores. If the lowest basic score is tied, eliminate non-Smith candidates
> and recount basic scores.
>
> *For elections with 9 or fewer candidates and no lowest-basic score ties
> in the Smith set, this is summable, but requires recounts in the event of
> more candidates or lowest basic score ties.*
>
> Looking back at Colin's burial example, what happens if basic score is
> recalculated using ballots scoring candidate X above D?
>
> A: 5
> B: 9
> C: 5
> E: 7
>
> B still wins. So your overall basic score as a proxy for ballots scoring
> above the lowest Smith basic score candidate is a good proxy in this case.
>
> On Tue, Jan 4, 2022 at 1:15 PM Forest Simmons <forest.simmons21 at gmail.com>
> wrote:
>
>> Let's call it Q&CBRS.
>>
>> Pre-requisite background:
>>
>> Candidate X pairwise defeats candidate Y iff candidate X is ranked
>> above/before/ahead of candidate Y on more ballots than not.
>>
>> A defeat chain is a sequence of candidates in which each candidate
>> pairwise defeats the subsequent member of the chain.
>>
>> Using a "bubble sort" procedure to sort a list of candidates into
>> pairwise order produces a defeat chain of the listed candidates.
>>
>> In this way we can easily find a defeat chain that includes all of the
>> candidates. The first candidate in such a chain is an example of a Smith
>> candidate. More generally, any candidate who has a defeat chain to any
>> other candidate is a member of the Smith Set.
>>
>> Q&CBRS:
>>
>> First, find the "basic score" for each candidate defined as the number of
>> ballots on which it is ranked above one or more candidates.
>>
>> Then let X be the Smith candidate with the smallest basic score.
>>
>> Finally,  among the candidates not defeated by X, elect the one with the
>> greatest basic score.
>>
>> That's it! Quick & Clean!
>>
>> Note that if there is only one Smith candidate X, then X will be the only
>> candidate not defeated by X, and therefore the one elected .
>>
>> In general a candidate not defeated by X will defeat X, and thereby find
>> itself at the head of a defeat chain to every other candidate, i.e. it will
>> be a Smith candidate.
>>
>> When there is only one Smith candidate, that candidate will not be
>> pairwise defeated by any other candidate.
>>
>> I repeat the entire method procedure here:
>>
>> First, find the "basic score" for each candidate defined as the number of
>> ballots on which it is ranked above one or more candidates.
>>
>> Then let X be the Smith candidate with the smallest basic score.
>>
>> Finally,  among the candidates not defeated by X, elect the one with the
>> greatest basic score.
>>
>> Where else can you find such a simple, quick, and clean election method?
>>
>>
>>
>>
>>
>> El lun., 3 de ene. de 2022 4:19 p. m., Forest Simmons <
>> forest.simmons21 at gmail.com> escribió:
>>
>>> Good idea!
>>>
>>> Although it seems to me that the highest approval candidate would have
>>> to pairwise beat or tie the approval cutoff candidate X pairwise (which
>>> would be impossible for a non-Smith candidate to do) ...I could be wrong
>>> ... and in any case redundancy reinforces communication and understanding.
>>>
>>> El lun., 3 de ene. de 2022 3:04 p. m., Ted Stern <dodecatheon at gmail.com>
>>> escribió:
>>>
>>>> Hi Forest,
>>>>
>>>> This sounds like an interesting method to me!
>>>>
>>>> However, I would change the winning criteria to "Elect the most
>>>> approved member of the Smith Set".
>>>>
>>>> On Mon, Jan 3, 2022 at 11:06 AM Forest Simmons <
>>>> forest.simmons21 at gmail.com> wrote:
>>>>
>>>>> I apologize for the defiant tone at the end of the previous message
>>>>> ... I must have gotten carried away with the "Dirty Dozen"' theme.
>>>>>
>>>>> But isn't it frustrating to you when people use the 2nd law of
>>>>> thermodynamics (or Arrow and Gibbard-Satterthwaite in the EM context) to
>>>>> justify their stubborn resistance to any kind of engineering progress?
>>>>>
>>>>> In the previous message Q&D Burial Resistant Condorcet was formulate
>>>>> in the typical "stitched together" form ... "Elect the CW if there is one,
>>>>> Else ..."
>>>>>
>>>>> In this message I would like to formulate a seamless version:
>>>>>
>>>>> Let X be the Smith candidate who on the fewest ballots is ranked ahead
>>>>> of any other Smith candidate. On each ballot approve all candidates down to
>>>>> X, but include X only when no Smith candidate is ranked ahead of X.
>>>>>
>>>>> Elect the candidate approved on the most ballots.
>>>>>
>>>>> This method can be described as electing the approval winner when the
>>>>> approval cutoff is (at the rank of) the weakest of the Smith candidates,
>>>>> which itself is approved on (and only on) those ballots which do not
>>>>> approve any other Smith candidate.
>>>>>
>>>>> In other words, the approval cutoff is inclusive only when necessary
>>>>> to ensure approval of at least one member of Smith.
>>>>>
>>>>> Since a Smith member is approved on every ballot, the method satisfies
>>>>> the Condorcet Criterion, i.e. it elects the only Smith member when Smith is
>>>>> a singleton.
>>>>>
>>>>> How does that grab you?
>>>>>
>>>>> -FWS
>>>>>
>>>>> El dom., 2 de ene. de 2022 7:30 p. m., Forest Simmons <
>>>>> forest.simmons21 at gmail.com> escribió:
>>>>>
>>>>>> Is there any burial resistant Condorcet method simpler than this?
>>>>>>
>>>>>> The basic pre-requisite is to understand that whenever there is no
>>>>>> Condorcet Winner there will be a pairwise cycle, called a "top-cycle" of
>>>>>> candidates whose members are not defeated by any candidates outside of the
>>>>>> cycle, just as a Condorcet Winner is a candidate undefeated by any other
>>>>>> candidate.
>>>>>>
>>>>>> Here's the Q&D burial resistant method:
>>>>>>
>>>>>> Lacking a Condorcet Winner elect the candidate X having the greatest
>>>>>> pairwise victory over the top-cycle member Y that has the smallest ratio of
>>>>>> first to last place votes within the top cycle.
>>>>>>
>>>>>> Two examples illustrate the method:
>>>>>>
>>>>>> Example 1.
>>>>>>
>>>>>> 49 C
>>>>>> 26 A>B
>>>>>> 25 B (sincere B>A)
>>>>>>
>>>>>> The top cycle is ABCA
>>>>>>
>>>>>> Candidate A has the smallest ratio 26/74 of first to last place votes.
>>>>>>
>>>>>> Candidate C is the only candidate with a pairwise victory over it, so
>>>>>> C wins.
>>>>>>
>>>>>> Notice how our rule does not reward B for insincerely lowering A to
>>>>>> (equal) last?
>>>>>>
>>>>>> Example 2.
>>>>>>
>>>>>> 45 A>B (sincere A>C)
>>>>>> 35 B>C
>>>>>> 25 C>A
>>>>>>
>>>>>> Candidate C has the smallest ratio  25/45 of first to last.
>>>>>>
>>>>>> Candidate B wins as the only candidate with a pairwise victory over
>>>>>> C. So A's burial of C backfires.
>>>>>>
>>>>>> Typically, the faction A that buries or truncates a Condorcet Winner
>>>>>> C to create a top-cycle cannot by so doing become a pairwise victor over
>>>>>> the buried Condorcet Winner ... but must (in order to create a cycle) help
>>>>>> some other candidate B defeat C by insincerely voting B>C.
>>>>>>
>>>>>> Our Quick and Dirty method insures that if the sincere CW's rightful
>>>>>> victory is subverted, it goes to B, not to A.
>>>>>>
>>>>>> Is this method quick enough and dirty enough for the FairVote IRV
>>>>>> promoters?
>>>>>>
>>>>>> What objections/criticisms might they have?
>>>>>>
>>>>>> Do they have any counter proposal that rivals this one in any way?
>>>>>>
>>>>>> If so, let them educate us ... we'll gladly join them if they can
>>>>>> show us a better way!
>>>>>>
>>>>>> If not, then they should join us to educate the politicians, public,
>>>>>> and last but not least, the academics still stuck in the pre-EM era!
>>>>>>
>>>>> ----
>>>>> Election-Methods mailing list - see https://electorama.com/em for
>>>>> list info
>>>>>
>>>>
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