[EM] Manual Construction of Smith Set

[Forest Simmons]

See a slight correction in the first method, below ...

Also, once you have Smith, the TopTwoRunoff is no longer necessary; just
elect the  Smith candidate ranked on the most ballots.

If you wanted to go the extra mile so as to have a burial resistant method,
Elect the candidate ranked on the most ballots that is not defeated
pairwise by the Smith candidate ranked on the fewest ballots.

-Forest

El sáb., 8 de ene. de 2022 7:01 p. m., Forest Simmons <
forest.simmons21 at gmail.com> escribió:

> Richard,
>
> Here are two hand count methods for constructing the Smith set (off the
> top of my head):
>
> Method 1:
>
> List the candidates in any convenient order. Then sort the list pairwise.
> The Smith set consists of all of the candidates in the sorted list above
> some cutoff level.
>
> Usually that cutoff level can be quickly found by "inspection." But here's
> a systematic procedure.
>
> Start with the tentative cutoff just below the candidate at the top of the
> sorted list. Then ..
> While there remains any candidate with questionable Smith status,  let C
> be the highest such candidate. If C pairwise defeats a candidate above the
> current cutoff, then lower the cutoff to the position immediately below C.
> Else mark C as "non-Smith".
>

Change this designation to "not a Smith set cutoff candidate"

EndWhile
>
> Method 2:
>
> Borrow a handheld calculator having matrix multiplication capability.
> Initialize an n×n matrix (where n is the number of candidates) with all
> ones, including the main diagonal.
>
> For each (i, j) combination where i  and j are not equal, if candidate i
> does not defeat candidate j pairwise, then zero out the j_th element of the
> i_th row.
>
> Now square this matrix repeatedly until the number of zero entries
> stabilizes.
>
> [This will take fewer than ceiling(log(n)/log(2)) squarings. So for 1000
> candidates, fewer than ten squarings.]
>
> The rows that end up with all positive entries are the Smith candidate
> rows. Rows with one or more zero entries correspond to non-Smith candidates.
>
> If you are doing this by hand, you can speed this up drastically by
> replacing the sums of products that define matrix multiplication with "sups
> of infs":
>
> For squaring the matrix M, replace the (i,k) entry of M^2  given by the
> sum of products
>
> Sum (over j) of m(i,j)*m(j,k)
>
> with the max of mins
>
> Max(over j) of Min(m(i,j), m(j,k))
>
> This limits all entries to zeros and ones, making the computations trivial
> (though still tedious in the case of hundreds of candidates). It gets
> faster with practice... the limit to your speed is how fast you can write
> zeros and ones into n×n arrays.
>
> It's kind of fun once you get the hang of it ... you haven't really lived
> without having mastered this technique!
>
>
> El vie., 7 de ene. de 2022 5:42 p. m., Richard, the VoteFair guy <
> electionmethods at votefair.org> escribió:
>
>> On 1/7/2022 4:27 PM, Forest Simmons wrote:
>>  > Actually, neither RP nor Schulze has better Condorcet efficiency than
>>  > Smith//TopTwoRunOff, which is the simple method you should be aiming
>> for.
>>
>> To anyone, I have some questions (that might also be in the minds of a
>> few lurkers).
>>
>> How are the "top two" candidates determined according to the
>> Smith//TopTwoRunoff method?
>>
>> Or, where is Smith//TopTwoRunoff described? (I didn't find it in
>> Electowiki.)
>>
>> Since the topic is simplicity, how can ballots be hand counted to
>> determine the Smith set?
>>
>> Yes I know that pairwise counting can be done by having each person at a
>> table keep track of only one pair of candidates -- as each ballot is
>> passed from person to person.  But how can those pairwise vote counts be
>> simply(!) converted into the Smith set -- for any set of ballots?
>>
>> As a related question, how does "//" differ from "/"?  In other words,
>> is Smith/TopTwoRunoff different from Smith//TopTwoRunoff, and if so, how?
>>
>> Thanks!
>>
>> Richard Fobes
>>
>>
>> On 1/7/2022 4:27 PM, Forest Simmons wrote:
>> > Robert,
>> >
>> > You opined ...
>> >
>> > "Probably Schulze or RP is the best thing to do for those cases when
>> > there is no Condorcet winner.  But getting that into legislative
>> > language is difficult, which is why I have advocated for BTR-STV."
>> >
>> > Actually, neither RP nor Schulze has better Condorcet efficiency than
>> > Smith//TopTwoRunOff, which is the simple method you should be aiming
>> for.
>> >
>> > Here's the typical example that I used earlier today:
>> >
>> > 40 A>C
>> > 35 B>C
>> > 25 C>A
>> >
>> > The sincere CW is C, which any Condorcet method will elect in an ideal
>> > neighborhood, but only a burial resistant method like Q&D/C will
>> > reliably elect in a saavy, scrappy neighborhood.
>> >
>> > All of the standard (head-in-the-sand) Universal Domain methods like RP,
>> > Schulze, MinMax, etc. are more or less likely to elect A, depending on
>> > how politically saavy/street smart the A faction is.
>> >
>> > So those highly vaunted methods are no better than Condorcet completed
>> > byTopTwoRunoff, which produces the exact same result as they do 100
>> > percent of the time in this context, but much more simply.
>> >
>> > So TopTwoRunoff works just as well as Schulze for cycle resolution in
>> > this context, yet it is by far the most adoptable proposal for Condorcet
>> > completion.... because of its simplicity and familiarity.
>> >
>> > Don't worry about the legislative language .. just copy the language
>> > from existing jurisdictions where it is already in use ... with the
>> > simple tweak of replacing the phrase, "In the event there is no absolute
>> > majority candidate ..."  ... with the phrase, "In the event there is no
>> > head-to-head majority candidate ..."
>> >
>> > There is no valid excuse for proposing Plurality as a Condorcet finisher
>> > when this more adoptable proposal is there for the taking.
>> >
>> > Plurality as a finisher would be a huge liability/embarrassment even if
>> > you could get it adopted, which is doubtful.
>> >
>> > Fair Vote people would rightly mock a Condorcet method that in principle
>> > allows a Condorcet Loser to win.
>> >
>> > How would we prevent that happening?
>> >
>> > Of course we could say ... "In the event there is no majority
>> > head-to-head winner, elect the FPTP winner, unless all Plurality counts
>> > fall short of 50 percent, in which case complete the Plurality finisher
>> > with a TopTwoRunOff finisher, finisher."
>> >
>> > That wouldn't even make sense, because there could never be a 50% plus
>> > Plutality winner without already having a Condorcet Winner... since a 50
>> > percent plus Plurality winner is automatically a Condorcet Winner.
>> >
>> > I hate to be pedantic, but let's not squander our opportunity on an
>> > atrocious proposal. If there were not a simpler, more adoptable proposal
>> > readily available, I would say, "Go ahead, leave your keys in the car,
>> > cross your fingers, and take your chances!"
>> >
>> > In my humble, but expert opinion, there are only two tenable proposals
>> > for Burlington, Vt. at this time ...
>> >
>> > 1. Condorcet completed by TopTwoRunoff when necessary... of course under
>> > a more attractive name.
>> >
>> > 2. TopTwoRunoff restricted to the "top cycle" or "Smith Set" ... even
>> > more important to get a better name.
>> >
>> > This second method is just as good as Schulze for public elections, so
>> > don't pine for the day when the world is safe for Kemeny-Young, for
>> > Pete's sake!
>> >
>> > You should put the choice to the people who control the decision
>> > orocess. After making clear the prestige of an ISDA upgrade at
>> > practically no extra cost, let them decide between the two options.
>> >
>> > Either choice will result in a respectable method that cannot be easily
>> > gainsayed.
>> >
>> > Good Luck!
>> >
>> > -Forest
>> ----
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>> info
>>
>
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