# [EM] smith/schwartz/landau

Juho Laatu juho.laatu at gmail.com
Wed Mar 28 07:36:17 PDT 2018

> On 28 Mar 2018, at 14:05, Kristofer Munsterhjelm <km_elmet at t-online.de> wrote:
>
> On 03/28/2018 10:26 AM, Juho Laatu wrote:
>> Yes, works too. Includes "weak" votes that are not normalised to the 100...1 range.
>> My question is still "whether it is a natural extension of the majority rule to require that the winner should always be picked from the Smith Set, when observing the number of voters only”.
>
> I just thought of a way to see if the Smith criterion is a reasonable extension of majority; or rather, if it's a reasonable extension of Condorcet.
>
> You say that if ABC were one candidate, he should win. What if we make ABC originate as one candidate? One way of doing so would be by cloning an initial A into ABC. Then we know (if we value clone independence) that in any situation where A wins against d in a majority election (that's where majority comes in), then if we clone A into ABC, the winner should be one of ABC.

m: A=B=C > d
n: d > A=B=C

With these ballots, if n < m, we can say that majority extensions should not support electing d.

>
> And if we can turn any ABCd election (Smith set of size three, plus a Condorcet loser) into one where ABC are clones, then we get that any clone independent Condorcet method must be Smith in the four candidate case, just by extending majority, through Condorcet, into Smith.

Let's say we have matrix M that can be derived either from ballot set B1 or B2, where B1 contains no technical clones and in B2 A, B and C are technical clones. Being a technical clone means that those candidates are ranked together in all the ballots (the usual clone definition). It doesn't say that those candidates would be similar of that the voters would consider them to be closely related political alternatives (political clones) (they could be as well from three fiercely competing parties).

I note that clone criterion is a problematic criterion in the sense that it does not cover only political clones (does of course often miss also them), but sometimes also competing groups. In methods that are based on the matrix only it forces also ballots B1 to be treated as if A, B and C would be clones, although they can be very far from being political or technical clones. This means that especially in matrix based methods clone criterion can be an overkill.

m: mix(A, B, C) > d
n: d > mix(A, B, C)

Lets use margins as the default measure of preference strength (without losing generality). Expression mix(A, B, C) refers to a mixture of any preference orders between A, B and C. The strongest unavoidable defeat (sud) of that mix ((or any set of candidates)) is defined as the strongest defeat of the candidate whose strongest defeat is smallest (minmax style). The interesting ABCd cases (from the perspective on my question) could be the ones where sud(A, B, C) > m:n. I note also that if sud(A, B, C) > m:n, one can ask if A, B and C are really political clones, even in the case that they are technical clones in the ballots (B2).

>
> Are there any clone-independent methods that meet Condorcet but not Smith? The only one I can think of is the "obvious" method that collapses clones and then runs Minmax. Perhaps also BTR-IRV? I'm not sure. But both of these require access to the ballots themselves, which
> makes me suspect that you can't tell a clone scenario from a non-clone scenario if all you have is the pairwise matrix.

Yes, the pairwise matrix hides lots of ballot information, including having candidates next to each others in the ballots. (M vs. B1 and B2)

>
> If I'm right, then for any given (3-cycle + 1) Condorcet matrix, I could use my old tool, linear programming, to determine a clone ballot set that would produce that matrix. It would be something like:
>
> - Because ABC is a clone set, the only permitted ballots are xD or Dx where x is one of ABC ACB BAC BCA CAB CBA, i.e. ABCD ACBD BACD BCAD CABD CBAD DABC DACB DBAC DBCA DCAB DCBA.
>
> - We want the pairwise strengths to be as determined beforehand.
>
> - Usual constraints: no negative number of ballots, no infinities.
>
> In your ABCd case, we have:
>
> 17: A > B > d > C
> 16: A > d > B > C
> 17: B > C > d > A
> 16: B > d > C > A
> 17: C > A > d > B
> 16: C > d > A > B
>
> and the pairwise matrix is (for row, against column)
>
>   A  B  C  d
> A -- 66 33 50
> B 33 -- 66 50
> C 66 33 -- 50
> d 49 49 49 --
>
> Programming the LP gives the following clone solution:
>
> 33: A>B>C>d
> 17: B>C>A>d
> 16: d>B>C>A
> 33: d>C>A>B
>
> that produces the same pairwise matrix as above.
>
> If we collapse the clone set, we get
>
> 50: A>d
> 49: d>A
>
> so any clone-independent Condorcet method that only knows the pairwise matrix must elect from the ABC set here, and since it can't distinguish the LP solution (where ABC is a clone set) from the original ballot set, there's no way for it to elect d in your example.
>
> I haven't actually proven that clone independence + Condorcet + only look at the pairwise matrix implies 4-candidate Smith; that would require more linear algebra than I feel like using today.

Fine so far.

> But it's not that implausible; and if it's true, that means that whatever makes a Condorcet loser deserve to win, if anything, must come from information not provided by the pairwise matrix.

Note that the problems between matrix and ballot information mainly emerge from the clone criterion. One could say that pure clone independence / existence of clone candidates can not be measured from the matrix only (without doing the "overkill"). The non Smith Set arguments are more neutral (e.g. minmax style arguments) with respect to using the matrix only vs also the ballots (matrix is enough). The "overkill" is the problem that forces d not to be elected also when there are no technical clones. (Smith Set criterion is close to the clone independence criterion.)

The arguments in favour of d can be valid also when we have technical clones (B2). The sud(A, B, C) > m:n argument/criterion may work also with B2 (since the cycle among the technical clones / possible political clones would still cause lots of dissatisfaction if one of the technical clones would be elected).

BR; Juho

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