[EM] Independence from clones, and Condorcet//FPP criterion failures

[Kristofer Munsterhjelm]

Since it's taking longer than I expected to write a long reply to Mike's 
post, I'll give some criterion failures for Condorcet//FPP while I work 
on the reply.

Independence from clones (as I understand it) is defined as follows 
(Tideman, 1987, "Independence of Clones as a Criterion for Voting Rules"):

     Let X denote the set of alternatives.
     Call Y (subset of) X a set of exact clones if, for all y,z (in) Y, 
every voter ranks y equal to z.
     Call Y (subset of) X a set of clones if, for all y,z (in) Y and all 
x (in) X\Y, every voter who ranks x
     over y also ranks x over z and every voter who ranks y over x also 
ranks z over x.
     For all Y (subset of) X which is a set of clones, all x (in) X\Y 
such that Y union {x} is not a set of
     exact clones, and all Y' which is a strict subset of Y, the 
probability that x is elected
     must not change if alternatives in Y' are deleted from all votes.

Or, less formally (but still somewhat, if I got it right):

Call Y a clone set if all candidates in Y have the property that:

- every voter who ranks a candidate in Y over some other candidate q, 
ranks all candidates in Y over q,

- and every voter who ranks some candidate q over a candidate in Y, 
ranks q over all candidates in Y.

Then eliminating a strict subset of Y should neither increase nor 
decrease the probability that r is elected the winner, where r is any 
candidate outside Y not equal-ranked with Y by every voter who ranks 
both Y and r.

While the formal proof does not explicitly say so, I think that "ranks A 
above B" means "ranks A strictly above B", since Tideman makes use of 
"ranks y equal to z" in his definition of a set of exact clones, and he 
does not mention equal-rank otherwise in the conditions for the 
independence from clones criterion.

-

On to some criterion failures:

If there is no truncation or no equal-rank, then (I've been told) ICT is 
equal to Condorcet//FPP. Hence, finding an example where Condorcet//FPP 
fails independence from clones with no equal-rank or truncation suffices:

	70: A > B > C
	68: B > C > A
	66: C > A > B

	There's no CW, so A wins.

	Now clone A into A1 and A2:

	35: A1 > A2 > B > C
	35: A2 > A1 > B > C
	68: B > C > A1 > A2
	66: C > A2 > A1 > B

	Now B wins. Cloning A made A lose (or B win).

And some other failures while I'm at it:

Reversal symmetry:
	1: A>B>C
	1: C>B>A
	1: B>A>C
	1: C>A>B

	There's no CW and reversing every ballot leads to the same ballot set. 
Hence, the outcome must be A=B=C (since the method can't distinguish 
between this set and its reverse), but C has a Plurality count of 2 and 
thus wins.

Condorcet loser, majority loser, mutual majority, Smith, Schwartz, 
Landau, others?:
	2: A>B>C>D
	2: B>C>A>D
	3: D>C>A>B

	There's no CW. The Smith set is {ABC}. 4 voters, constituting a 
majority, rank the candidates {ABC} above D. Yet D has the greatest FPP 
count and thus it wins.
I also think this example shows C//FPP failing Minimal Defense, but I'm 
not sure - I don't know MD that well. It might also show that C//FPP 
fails GSFC, but I'm even less sure of that.

Independence from Smith-dominated alternatives:
	Start with the Condorcet loser example above, but without D. Then C 
wins. Now add D, which is not in the Smith set. Then D wins, so the 
method fails ISDA.

Mono-add-top, Participation:
	51: A>B>C>D
	50: B>A>C>D
	50: C>B>A>D
	51: D>C>A>B
	 1: D>B>C>A

	There's no CW, so D wins by an FPP count of 52. Now two more D>B>C>A 
voters appear:

	51: A>B>C>D
	50: B>A>C>D
	50: C>B>A>D
	51: D>C>A>B
	 3: D>B>C>A

	B is now the CW and wins.