[EM] Hopeland: a new Clone Free Copeland Variant

[Forest Simmons]

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.

-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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