[EM] n-th order Copeland questions

[Kristofer Munsterhjelm]

Call the "iterated" Copeland variant I've mentioned, n-th order Copeland 
when it's iterated n times. First order Copeland is simply ordinary 
Copeland.

Each such nth order Copeland can be assigned a win and tie multiplier. 
Call the fully specified variant, (a, b, c)-Copeland, where a is the n 
in nth order, b is the win multiplier, and c is the tie multiplier. 
Thus, the sports type Copeland is (1, 3, 1)-Copeland.

On a given iteration, each candidate gets b points for every point of 
every candidate he beat in the last iteration, and c points for every 
point of every candidate he tied in the last iteration.

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For (2, x, 1)-Copeland, the method seems to be monotone if x >= 2. 
However, when x is even slightly below 2, it's not monotone. For 
instance, for x = 1.9999,

1: A > B > C > D
1: D > A > B > C
1: D > C > A > B
2: B > C > D > A
1: A > C > D > B

which is a four-way tie. Raise C:

1: A > B > C > D
1: D > A > B > C
1: C > D > A > B  (here)
2: B > C > D > A
1: C > A > D > B  (and here)

and now it's B > C > D > A, so by gaining support, C lost his chance of 
winning.

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Why is that the case, and what's the x for (3, x, 1)-Copeland, so that x 
is minimal and the method is monotone (if any such exists)?

The question doesn't have an immediate practical point, but it might be 
interesting to consider, even so.


It also seems that when x >= 1.5, the only sort of nonmonotonicity for 
(2,x,1)-Copeland involves an all-candidates tie. I say "seems" to both 
because I've used my simulator to find these things, and while it can 
show that a method is nonmonotonic, it can't show that a method is monotone.