[EM] Hopeland: a new Clone Free Copeland Variant

[Forest Simmons]

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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