[EM] Fragmented Condorcet doesn't imply DPC

[Kristofer Munsterhjelm]

 From off-list conversation, I discovered an example of that my 
tentative multiwinner criterion, Fragmented Condorcet, doesn't imply DPC.

Consider this bullet-voting situation:

400: A
400: B
400: C
300: D

Three to be elected.

The Droop quota is 375. So, according to DPC, A, B, and C should be elected.

There's at least one way of splitting these votes into three bundles so 
that "the right" candidates (A, B, and C) get elected, and so that each 
contains 500 ballots (1500/3). For instance,

first bundle:  400 A, 100 D
second bundle: 400 B, 100 D
third bundle:  400 C, 100 D

but there's also a way that isn't proportional:

first bundle:  300 A, 100 B,  100 C     A beats B and C, A wins
second bundle: 300 C, 100 B,  100 A     C beats A and B, C wins
third bundle:  300 D, 200 B             D beats B, D wins

The parallels to packing and cracking are obvious. I suppose I shouldn't 
be surprised, since Condorcet doesn't imply mutual majority, either, but 
this allows for the possibility that Fragmented Condorcet contradicts 
the DPC.

If it does, the construction would probably be something like: arrange a 
setup so that there's only one way of arranging Condorcet winners in 
each bundle, all others causing cycles in at least one bundle. Then 
modify this arrangement so that the only CW-permitting partitioning 
contradicts the DPC.

I don't know if that's possible, though, and it would have to use full 
preference votes (e.g not just bullet votes).

-

In general, I think my surprise at this confirms what I've suggested 
before: that it's not enough to technically satisfy criteria, one must 
also gracefully fail towards them. Clone independence isn't worth much 
if the system is "remove clones then run Borda". Similarly, if it's 
possible to pass both FC and the DPC, then the method, for a ballot set 
where DPC and FC provide no constraints, must elect results that are in 
some fashion "close" to a ballot set where they would provide constraints.