[EM] Cloneproof resistant set extensions: one's too weak, the other is too strong

[Kristofer Munsterhjelm]

The solid coalition cloneproof resistant set extension breaks the 
property that a voter who prefers some candidate outside the set to the 
winner has nothing to gain by burying the winner.

Suppose that B is the current winner and there's a coalition {A1, A2, 
A3} that is almost a clone set. Suppose that if {A1, A2, A3} were 
replaced with a single candidate so that each voter would replace his 
highest ranked set member with that single candidate, then that 
candidate (call him A) would disqualify B. But because a few voters vote 
something like

... > A1 > B > A2 > A3 > ...,

A1 does not obtain >1/k of the first preferences in every sub-election 
containing A1 and B (but not A2 or A3). Suppose furthermore that the 
method electing from the extended set elects B.

But then if these almost-voters bury B so that instead we have

... > A1 > A2 > A3 > B > ...

then the solid coalition and the "little more elegant" relation could 
both establish {A1, A2, A3} ~> B, thus kicking B off the set. The 
problem is that it's too weak, i.e. it doesn't consistently handle 
near-clone sets.

It is possible to be cloneproof and elect from the resistant set (e.g. 
IRV). But if IRV elects from the extended resistant set induced by the 
solid coalition/more elegant relations, then it has some additional 
property that makes it never elect B in the example above.

I don't know if cloneproof unburiability of the disqualification 
relation is incompatible with Condorcet, but I don't see a reason why it 
would be. Except that passing both LNHs is.


The first-preference one is too strong. As a result, it is cyclical. The 
example from the other post:

34: A>B>C
33: B>C>A
33: C>A>B

Now (A,B) ~> C since 34 + 33 > 50,
{B, C} ~> A since 33 + 33 > 50
and {C, A} ~> B since 33 + 34 > 50.

Now we could imagine altering the threshold to get around this problem. 
But for the cycle to be broken, then in the three-candidate case, it 
needs to be greater than v * floor(2/3) = 66, because that's the weakest 
link in this cycle. But if it's 2/3, then:

51: A>B
49: B>A

A ~> B, but clone A into A1 and A2:

26: A1>A2>B
25: A2>A1>B
49: B>A

the sum of A1 and A2's first preferences is less than 2/3 and so B 
wouldn't be disqualified.

There's a gap that's just too large for first preferences on its own to 
work: it can't simultaneously be acyclical in the Condorcet cycle 
example and work in the 26/25/49 example.

-km