[EM] Quick and Clean Burial Resistant Smith

[Forest Simmons]

Ted,

We may need a different example ... let's go through this one slowly and
carefully ...
46: A > B
44: B > C   (sincere B or B > A)
05: C > A
05: C > B

A gets 46+5=51 basic points from the first and third factions, resp.

B gets 46+44+5=95 points from the first, second, and fourth factions, resp.

C gets 44+5+5=54 points from the last three factions.

If this is right, A has the lowest score, and C wins as the only candidate
pairwise undefeated by A.

I don't doubt that this method could fail Chicken Defense, but it would
have to be a marginal failure that the attacked faction could easily
counter with a partial counter defection.

For a range of examples let's look at

p: C
q: A>B
r: B (sincere B>A)

Where r+q>p>q>r

A beats B because q>r
B beats C because r+q>p
C beats A because p>q

The respective basic scores for A, B, & C are q, p+q, & p.

The lowest of these is q which is A's score. So C wins, as the only
candidate pairwise undefeated by A.

So the method robustly punishes Chicken defection over a range of standard
examples.

Not bad for Quick and Dirty/Clean!




El mié., 5 de ene. de 2022 1:52 p. m., Ted Stern <dodecatheon at gmail.com>
escribió:

> 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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