Tiebreakers, Subcycle Rules

[Mike Ositoff]

Sorry--I did neglect to say the magnitude of D>E. It's 6, as
I was doing the example.

In case I miswrote it before, let me write the example in total:

{A,B,C} is a subcycle, but not a clone set.

A>B 2
B>C 3
C>A 1

{A,B,C}>D 7
E>A 8
E>B 5
E>C 4

D>E 6

***

B & C have Schulze wins over D & E, so D & E can't be Schulze
winners.

D & E have Schulze wins over A, so A can't be a Schulze wioner.

A has Schulze wins over B & C, so they can't be Schulze winner.

So there is no Schulze winner.

***

My fault, having not specified the magnitude of D>E. Though
my recent statements about which methods do what in 
that example seem correct (but I've missed errors lots of times),
I was mistaken previously when I said that none of those 4
methods can deal with that example. It's a Schulze indecisiveness
example. It also isn't terribly favorable for Sequential Dropping,
which declines to choose from the subcycle for the same reason
that Schulze doesn't choose. But, considering the relative rarity
& unpredictability of subcycle problems, that's forgivable. And,
in regards to Schulze's result there, it isn't that it's a problem
to sometimes use a tie-breaker--it's just that the need to have
one adds complication & words to the definition. There may well,
for all I know, be reasons why Schulze, with the right tie-breaker,
is better than Sequential Dropping. If that were determined for
sure, then maybe Schulze would be a contender for proposing in
meetings of an initiative committee, and it could be determined
how people would react. But explaining any but the simplest
rules to a political office-holder, or to the person-on-the-street,
is _daunting_.

In fact, when considering that task, that makes plain Condorcet(EM)
look pretty good. For the properties it affords, that method
is as simple as it can get.

Mike