[EM] Participation criterion and Condorcet rules

Kristofer Munsterhjelm km_elmet at t-online.de
Sat Sep 1 15:41:34 PDT 2018

On 2018-08-16 06:23, VoteFair wrote:
> On 8/14/2018 7:28 AM, Kristofer Munsterhjelm wrote:
> > As I said, it's always NP-hard, because NP-hard is a *worst-case 
> > property*.
> If you mean that the Condorcet-Kemeny METHOD is NP-hard, yes, I agree.

Method X being NP-hard means "if you're given an algorithm that can
solve any instance of X in polynomial time, you can solve any decision
problem that is in NP in polynomial time, as well". Since these
reductions are on a method level rather than on an instance level,
NP-hardness is only defined for problems as a whole.

So you can't say that a problem instance is or isn't NP-hard. Hence "the
METHOD is NP-hard" is the only way one can speak of NP-hardness.

> Yet critics claim (or imply) that if the Condorcet-Kemeny method were 
> used in real-life elections, then the computation time would be too long.
> As I've pointed out, long computation times for calculating a single 
> winner when there are more than about 50 (or maybe 200) ballots would be 
> very rare (regardless of how many candidates there are).

The Conitzer et al. paper I linked to in my previous post makes a
similar point. Because the mixed integer programming approach can
quickly handle instances up to 40 candidates (using a good solver), even
with voting patterns close to a tie, runtime is not the problem in
itself. The only ways this could count against Kemeny is if we can find
another method that is not NP-hard and is otherwise as good as Kemeny,
or if the context is so that we deal with enormous numbers of candidates
(e.g. web page ranking).

My argument against Kemeny would probably be a combination of the former
above (there are simpler methods that do just as well) and that Kemeny
itself isn't cloneproof, so it's more vulnerable to tactical nomination
than methods that are cloneproof. Thus I'd prefer MAM/Ranked Pairs,
which is cloneproof, simpler to implement, and always polytime if you
accept tiebreaking by random voter hierarchy.

Of course, implicit in this is my value judgement that consistency is
not all that important a criterion. Since every Condorcet method fails
reinforcement, it seems unlikely that someone would say "I like methods
where, if you have two districts and the same winner comes up in both,
the election with the two ballot sets combined should also have the same
winner. No Condorcet method does that, but at least Kemeny does it if
the complete social order is the same in both districts, so I'm going to
choose that one". Others might disagree with me.

>  > ....
>  > 1: A>B>D>E>C
>  > 1: A>C>B>D>E
>  > 1: A>D>C>B>E
>  > 1: A>E>B>D>C
>  > 1: B>E>C>A>D
>  > 1: B>E>C>A>D
>  > 1: C>A>D>B>E
>  > 1: C>B>E>A>D
>  > 1: C>B>E>A>D
>  > 1: D>C>A>B>E
>  > 1: D>C>B>E>A
>  > 1: D>C>E>B>A
>  > 1: D>E>A>B>C
>  > 1: D>E>A>B>C
>  > 1: E>C>A>D>B
>  > 1: E>C>D>B>A
>  > 1: E>D>B>C>A
>  >
>  > According to Kemeny score, there's a tie for first between D and E
>  > (score 93 for orderings D>C>B>E>A and E>C>A>D>B), but VoteFair says that
>  > C (score 92 for C>E>A>D>B) is the unique winner.
> The method created by John Kemeny does not specify how to deal with 
> ties.  (If someone knows otherwise, I'd love to read what Kemeny wrote 
> about handling ties.) >
> The VoteFair ranking software DOES deal with ties.

I was under the impression that VoteFair's standard was to rank
candidates according to their Kemeny score, e.g. if there's an order
starting with A with Kemeny score 91 and another starting with B with
Kemeny score 92, you get B>A. This would make sense given how you said
that VoteFair was primarily concerned with winners rather than full
orderings, because ranking by Kemeny score would show how good a claim
each candidate has on being the winner.

But from what you're saying, it seems that I'm wrong, and that you're
instead aiming to preserve the actual Kemeny ordering, i.e. if A>B>C>D
is the unique ordering with the highest Kemeny score, VoteFair should
return A>B>C>D in that order, even if, say, the greatest Kemeny score
ordering starting in C has a greater Kemeny score than the greatest
ordering starting in B.

As for dealing with ties, I can think of two ways of doing so. The first
is to use random voter hierarchy to construct a tiebreaker ordering and
adding an epsilon (less than 1/numvoters^2) to every pairwise contest
that agrees with the tiebreaker ordering. The second is to find every
ordering that has maximum Kemeny score and recursively apply Kemeny to
the orderings. Both tiebreaking methods preserve (their own) variant of

The former would pass that if the tiebreak is the same for both, or if
one district has a certain ordering be the unambiguous winner (no ties)
and the other has that ordering tied with some others, and has that
ordering as the RVH tiebreak, then the result for the combined district
will be the same as for those two districts.

The latter would pass it if both districts are ties with the same set of
orderings tied for maximum Kemeny score, then the combined result would
also have that set of orderings tied for maximum Kemeny score, and so
the tiebreaker would provide the same result.

It seems your approach is more in the spirit of the latter (as combining
A>B>C>D and A>B>D>C to A>B>C=D looks like some kind of aggregation),
except that the Kemeny method only returns strict orderings, and so
would still need some final tiebreak or adjustment to deal with that

> Actually measurements of HOW OFTEN each vote-counting method fails each 
> fairness criterion is not well-known.

The problem with such a calculation is that it depends on the
distribution of ballots. Formally speaking, the chance of getting a
wrong scenario with k voters and n candidates is

integral over all k-voter, n-candidate ballot sets x: p(x) * L(x) dx

where L(x) is 1 if the method gets it wrong and 0 otherwise. We may
replace the integral with an averaging operation if voter weights are
always integer-valued (as in a real election).

However, p(x), the probability of encountering ballot x, depends on how
voters behave, and how the method induces the voters to behave, and
opens up a considerable element of ambiguity. That's the problem with
determining how often something happens. The criterion compliance
approach to getting around this problem is to ensure that L(x) is zero
for all possible x, as then the integral (or mean) will be zero no
matter what.

If we just want a raw count, we can let p be some predefined probability
distribution, e.g. the Dirichlet distribution corresponding to choosing
n! random numbers subject to that they sum up to k, or the discrete
impartical culture which consists of picking a random ordering of the
candidates and letting that be a ballot, k times. But such choices are
still vulnerable to the claim that this amplifies/attenuates realistic
regions and attenuates/amplifies unrealistic ones.

For VoteFair, I think the value of the integral will be small for any
reasonable probability distribution, at least unless there are too many
candidates. But if we, instead of using a method that always runs in
polynomial time but sometimes gets it wrong, use a method that always
gets it right but sometimes takes a long time, then we would sidestep
the whole issue entirely. So why not do so?

> The tricky part of such an analysis is for the criteria that involve two 
> related scenarios, such as with and without a clone candidate.

There are two ways of doing that, I think. To use monotonicity as an

The first is to consider ballot set X to exhibit a monotonicity failure
if there exists some ballot set Y so that:
	- the method elects some candidate A when given ballot set X
	- Y is equal to X, except that on some ballots, A has been raised
(moved up in rank)
	- the method elects someone else than A when given ballot set Y

(and similarly for X not electing A and then lowering A makes A win)

The second is to consider all pairs X, Y so that
	- the method elects A when given ballot set X
	- Y is equal to X, except that on some ballots, A has been raised
(moved up in rank)

and then consider how many of these (X,Y) pairs have the method elect
someone else than A when Y is given to the method, relative to the
number of (X,Y) pairs possible.

I prefer the former, because the latter doesn't mirror what really
happens. The voters vote, and then the result is called; they don't
vote, get a result, and then modify their ballots to (e.g.) raise the

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