Maybe None Clone-Independent

[Mike Ositoff]

Because Tideman said that his method could fail
Independence from Clones unless complicated modifications
are added, and since I now have an 8-candidate example
where Sequential Dropping fails that criterion, could
it be that none of these basic methods can always meet
that criterion without complicated modification?

Here's my example:

At first there are 6 alternatives. {D,E,F} is a clone-set
subcycle. 

{D,E,F}>A 1
A>G 3
G>{D,E,F} 2
{D,E,F}>H 4
H>G 5
A>H 6

{D,E,F}>A gets dropped first, and A wins.

D>E 10
E>F 20
F>D 3

***

Now, 2 other alternatives enter the race, B & C.

They form a clone-set subcycle with A:

A>B 100
B>C 200
C>A 300

Now, first {D,E,F}>{A,B,C} 1 is dropped.

The next greater defeat that conflicts with greater ones
is G>{D,E,F} 2. That defeat is dropped.

Next, D>E 10 is dropped, and E is unbeaten, and wins.

So the addition of the 2 clones defeated A.

***

With an 8-candidate example, I can't be sure that I didn't
make an error, of course, but it seems correct.

I haven't yet checked Schulze in that example, but even
if Schulze, unlike Tideman or Sequential Dropping, _never_
violates Independence From Clones, I still say that this
example is so improbable that, for all intents & purposes,
SD is clone-independent. And its greater simplicity, and
its decisiveness, and consequent freedom from need to further
add to its rules with a tiebreaker--those things outweigh
that highly improbable Clone Independence violation.

***

Mike