[EM] Hopeland: a new Clone Free Copeland Variant

[Forest Simmons]

On Fri, Mar 17, 2023, 6:42 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> My "proof" leaks like a sieve, but here's the first step in a real proof
> by induction:
>
> Suppose that a hopelandwinner H needs zero friends to cover its enemies
> ... which means H is unbeaten. If H's ballot support is increased, then H
> still has zero enemies ... hence is still a Hopeland winner.
>
> Let's do another (less vacuous) step to get a feel how the general
> induction step might go:
>
> So this time the Hopeland winner H needs exactly one helper friend f to
> befriend it's only enemy e.
>
> When H's support is increased, does some other candidate X suddenly turn
> into a Hopeland winner needing less help than H? If not, then H remains a
> Hopeland winner.
>
> But if X suddenly finds itself with zero enemies then X, then evidently an
> erstwhile enemy e' of X suddenly became a friend of X because of H's
> increased ballot support.
>
> But how could increased support for H befriend to X the last enemy e' of X?
>
> For H to help X like this, e' and e would have to be the same candidate,
> and X advanced to unbeaten status, which means X beat H both before and
> after H's increased ballot support. So e'=e was already befriended by H
> before H's increased ballot support ... so H's increased ballot support
> could not turn X into a unique winner Hopeland winner.
>
> [needs some tidying up].
>
> For now, I want to give an instructive example of deducing a unique
> Hopeland winner on the assumption that if there is a unique candidate H in
> Hopeland that properly beats all of the other Hopeland candidates, then H
> must be elected.
>
> Start with the simple beat cycle ABCA.
>
> Then replace C with a clone set beat cycle C1C2C3C1.
>
> Then only the members of the clone cycle need two friends to befriend
> their enemies ... for example C1 needs its friends C2 and A to cover its
> three enemies C3, A, and B.
>
> On the other hand A's enemies (the C clones) are covered by (befriended
> by) on friend of A ... nam7ely B.
>
> And the enemy of B, namely A, is covered by any one of the C clones.
>
> So A and B form the Homeland set of tied winners. Of the two, A is the
> majority winner ... so A must be elected by the assumed principle alluded
> to above.
>

So in this example of one member C of a simple ABCA cycle replaced by a
cycle C1C2C3C1 of its clones, the Condorcet Criterion restricted to the
Hopeland tied set picks A.

Copeland, on the other hand, elects B, the only candidate with more than
two friends.

Hopeland discourages teaming, but seems to go too far in that direction
without taking defeat strengths into account ... eliminating out of hand
all of the clones instead of giving the strongest of them the same chance
that C had before it was cloned, which is the ideal standard of Clone
Freedom.

I'm starting to re-think this Hopeland idea.

-Forest


> -Forest
>
>
>
> On Fri, Mar 17, 2023, 11:05 AM Forest Simmons <forest.simmons21 at gmail.com>
> wrote:
>
>> In Kristofer's language of Friends ... the Copeland set is the set of
>> candidates tied for the most friends.
>>
>> To see Copeland's vulnerability to out of control clone nominations,
>> suppose that nobody but you has X for a friend ... meaning that X is an
>> enemy of all the other candidates ... meaning that X beats all of the other
>> candidates pairwise.
>>
>> If you had access to Hogan's Gonculator or Calvin's Transmogrifier, you
>> might be tempted to clone this friend of yours ...because each additional
>> clone increments your friend count ... so that eventually you would have
>> more friends than anybody else not also taking advantage of the wonderful
>> clone proliferation machinery.
>>
>> One way to remove this temptation is to reward candidates that need the
>> fewest friends in order to befriend all of their enemies.
>>
>> Let's call the candidates tied for the fewest friends needed to befriend
>> all of their enemies Hopeland Candidates.
>>
>> One of the nice things about Copeland Candidates is that all of their
>> enemies are friends of their friends ... that is they and their friends
>> together cover the entire field of candidates ... in other words, they are
>> Landau candidates ... they have a short beat-or-tie path to each enemy
>> candidate.
>>
>> This nice Landau property of Copeland extends to Hopeland ... in fact,
>> you could say that the Hopeland candidates are the ones that need the least
>> "little help from their friends" to befriend their enemies.
>>
>> Another nice thing about Copeland candidates is that when they get
>> increased support, they remain in Copeland. Their increase in ballot
>> support cannot decrease their number of friends relative the friend count
>> of anybody else.
>>
>> The same nice monotonicity property holds for Hopeland: if a Hopeland
>> candidate gets increased ballot support, it will remain in Hopeland.
>>
>> This a tad bit more subtle than the Copeland case:
>>
>> Suppose that H is an Hopeland candidate whose ballot support is
>> increased. If H is a friend of X, then H's increased ballot support could
>> very well result in H befriending some enemy E of X, thereby releasing
>> another friend of X from that responsibility ... thereby reducing the
>> number of friends needed by X to help befriend its enemies ... perhaps to
>> fewer than the number needed by H?
>>
>> Not to worry. Candidate X could not do this reduction without the help of
>> H.
>>
>> Now let F be the set of other friends besides H needed by X to cover its
>> enemies. In other words, X together with H and the friends in F must cover
>> the entire field of candidates.
>>
>> But that is the same as H together with X and the additional candidates
>> in F covering the field.
>>
>> The only problem is that not every member of F is necessarily a friend of
>> H.
>>
>> So let's partition F into F'+E, H's friends in F and H's enemies in F.
>>
>> But for each member e of E, H has a friend f(e) that covers it.
>>
>> In sum, H together with X and F'+ f(E) covers the entire field of
>> candidates, but this covering set is no larger than X together with H and
>> F'+E ...  because
>> f(E)={f(e)|e in E} has cardinality less than or equal to E ... with
>> equality only when f is one-to-one.
>>
>> So X does need as many or more friends to help cover the candidates as H
>> does. This means that H is still a member of the Hopeland set.
>>
>> Does that make sense? (i.e. enough sense to actually be correct?)
>>
>> Check me on this ... perhaps by induction on the number of friends needed
>> to cover H ... or else find a counter example!
>>
>> If this result is correct (as it probably is), then it can be the basis
>> for many new Landau methods ... one for each Hopeland tie breaker.
>>
>> But please make sure the tie breaker is both monotonic and clone free ...
>> after all our care to make the Hopeland definition both monotonic and clone
>> free ... don't let anybody just throw that away!
>>
>> -Forest
>>
>
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