[EM] Quick and Clean Burial Resistant Smith

[Ted Stern]

Hi Forest,

Unfortunately, your new method does not handle Chicken-Dilemma types of
burial, which I am interested in. See for example Chris Benham's example:

46: A > B
44: B > C   (sincere B or B > A)
05: C > A
05: C > B

A is sincere CW. If B buries A, there is an A > B > C > A cycle. C or B are
both lowest basic score. If C is taken as lower, then C defeats A and B
wins, rewarding burial.

With approval cutoff at second rank, even Smith//Approval does better.

On Tue, Jan 4, 2022 at 7:25 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> 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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