[EM] A Condorcet-like, STV-like, rank-based PR system

[Toby Pereira]

I was thinking about the problem Warren outlined here - 
http://rangevoting.org/PRcond.html - and how you might get around it using a 
ranked PR system. So here's my attempt:

Each voter ranks the candidates in order of preference. Ties are allowed. For n 
winners, each set of n candidates is compared against each other set. For two 
sets being compared, x and y, each member of x is compared against each member 
of y.

For example, in a two-seat election, candidates A and B are being compared 
against C and D. There are four pairwise comparisons - AC, AD, BC and BD. So 
under each result (A, B or C, D), a voter will have a score from 0 to 4. Based 
on these 0-4 scores, a representation score is calculated for each of the two 
results. The representation score is calculated in the same way as in my 
proportional range/approval voting (http://www.tobypereira.co.uk/voting.html) 
with each pairwise comparison treated as a "candidate" or qandidate. So in this 
example, four qandidates are elected.

The representation score for each of the two sets of candidates (the actual 
candidates, as opposed to qandidates) are then recorded next to each other. This 
is the pairwise result for the two sets of candidates.

Once every possible winning set of candidates has been compared against every 
other one, the winning set is calculated using an off- the-shelf pairwise 
Condorcet method such as Kemeny-Young or Schulze. If one set of candidates beats 
all others then it is the equivalent of a Condorcet winner.

When two sets of candidates are being compared against each other, the same 
candidate will often appear in both lists (e.g. AB versus AC). So when a 
candidate is compared against itself, you could give a score of, say, 0, 0.5 or 
1 to each voter for the tied result. The way the representation scores work, it 
won't make a difference to the numerical difference between the representation 
scores for each set of candidates.

If we use the above example, by giving 0 for the repeated candidate, the two 
sets of candidates may have a representation score of 0.5 and 0.4 respectively. 
By giving a score of 1, this will simply add 0.25 onto each score so the 
difference between them will still be 0.1. (0.25 because there are four 
qandidates and the overall representation score is out of 1.) So whatever score 
is given won't make any difference at least under Kemeny-Young (which I 
understand better than Schulze). It may actually be best to simply ignore these 
comparisons altogether, so when comparing AB against AC, there would be three 
qandidates (AC, BA and BC). This is what I will go with for now.

Where voters have ranked two different candidates at the same level, I think it 
may make more of a difference to things. These scores can't be ignored unless a 
voter has ranked every candidate in both sets at the same level (in which case 
we could ignore that voter entirely) or every voter has ranked the two 
candidates at the same level (not likely). To maintain equivalence with 
Kemeny-Young in the one-winner case, you can split the voter in two halves and 
award scores of 0 and 1 to each half respectively.

In the full version, every possible winning set of candidates would have to be 
compared against every other set, which would probably not be viable 
computationally in most cases. A sequential system could be used. The most basic 
version of this would be to find the best single winner, and then compare all 
two-winner sets that contain this winner and so on.

I'm not entirely sure what sort of results this would produce but in a few very 
simple cases, it seemed not to be too bad. One thing I've noticed - with loyal 
party voting and a comparison of (A1, A2 ... An) against (A1, A2, ... An-1, B), 
the B voters seem to be better off by ordering the A candidates in the same 
order as the A voters.

It could be tested against other proportional systems, to see if it produced 
similar results, including against my proportional range system.
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