[EM] [Fwd: No response was requested from you]

[Richard Moore]

Dave Ketchum wrote:

> Aha!  Levels seems clear from Condorcet for 10 candidates having 10 levels.
> 
> But then you say you can have something in 2 levels that you also call Condorcet.


Now I understand where the confusion comes from. I really 
didn't say something with only two levels could be called
Condorcet. If I were to paraphrase what I said, it would be
that if you start with Condorcet, and add a constraint to
the ballots such that the number of levels of expression is
two, then you get Approval. That does not mean that the
resulting thing is still Condorcet, anymore than a globe
forced through a trash compacter can still be called a
globe.

> There are enough words here to see that is what you are saying.  Looking
> back to what Craig was quoting on Thu, 27 Sep 2001 08:29:41 +1200, there
> were less words, what looks like a contradiction, and I side with Craig
> on need for clarification.


By expressing my idea more tersely as "Condorcet forced
into two levels", I may have made it easy to miss my
intended meaning. It was more of an analogy than an 
equation, and in that light I don't believe it is a 
contradiction.


> Said another way, Condorcet requires the N levels, and you do nothing
> but confuse by labeling something else a variation of Condorcet.
> 
> Modification:  Some have wanted to allow giving multiple candidates
> identical ranks, as in A>B=C where A is preferred, B and C are equally
> liked second best, and D and E, assuming such exist, share being liked
> less than B and C.  Thus this ballot fits in 3 levels while another
> voter might have used all 5 levels in this election.  I see this
> particular stretch as both compatible with Condorcet philosophy and
> explainable to voters.


I don't recall calling it a "variation of Condorcet" either, 
although in the sense of meeting the weaker Condorcet 
criterion suggested by Forest, it might be considered
that. But the usefulness of that weaker criterion is 
dubious. It is better just to say that there is a parallel 
between the two methods.

However, I should also add that we could generalize 
Condorcet to ballots with M levels, where M is not
necessarily equal to N.

If M > N, then there is no effective difference from the M = 
N case. If M < N, then voters have more constraints placed 
on their ballot choices, but the same counting algorithm
can be employed. Perhaps this needs a name to distinguish it 
from standard Condorcet methods; "constrained ballot 
Condorcet" is properly descriptive (but awkward). As long as 
M > 2, I suspect many of the properties of Condorcet
voting are intact. If we look at the "special case" where M 
is 2, although the counting algorithm remains unchanged, the 
response becomes linear, and some fundamental changes in 
properties occur. Continuing to call this "constrained 
ballot Condorcet" would be a stretch.

Oops, now I'll have to define what I meant by "linear 
response". Actually, I'm referring to the scoring that has a 
linear response; the final selection is necessarily a 
non-linear function. Anyway, here then is my definition of 
an election method with linear response:

A method is represented by two functions. The first 
function, S(B), where B is a set of ballots, produces the 
vector [s1(B), s2(B), ..., si(B), ..., sN(B)], where each 
scalar function si(B) is single-valued for all B. (S is the 
scoring function.)

The second function, M(S), produces a single-valued integer 
result for each N-element vector S, such that 0 < M(S) <= N.
Also, if 0 < i <= N and si(B) > sj(B) for all j, 0 < j <= N, 
then M(S(B)) = i. Furthermore, if 0 < k <= N and sk(B) < 
sj(B) for some j, 0 < j <= N, then M(S(B)) != k. This 
definition does not specify a tiebreaker but implies that 
one exists. (M is the selecting function.)

For each i, 0 < i <= N, there is one candidate, Ci. For a 
set of ballots, B, the result of an election is Cj, where
j = M(S(B)).

Now, iff, for any two disjoint sets of ballots B1 and B2 in 
an N-candidate election, it is true that

	S(B3) = S(B1) + S(B2),

where B3 is the union of B1 and B2, then the method has 
linear scoring response.

The best-known linear methods are Plurality, Borda, CR, and 
Approval.

One last remark: When I noted that Approval is CR in two 
levels, this is more than an analogy. The only property 
distinguishing Approval and higher-resolution CR is, well,
resolution.

Richard