<html><head><style type="text/css"><!-- DIV {margin:0px;} --></style></head><body><div style="font-family:times new roman, new york, times, serif;font-size:12pt"><DIV>I was thinking about the problem Warren outlined here - <A href="http://rangevoting.org/PRcond.html" target=_blank><FONT color=#000080>http://rangevoting.org/PRcond.html</FONT></A> - and how you might get around it using a ranked PR system. So here's my attempt:<BR><BR>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.<BR><BR>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 (<A href="http://www.tobypereira.co.uk/voting.html" target=_blank><FONT color=#000080>http://www.tobypereira.co.uk/voting.html</FONT></A>) with each pairwise comparison treated as a "candidate" or qandidate. So in this example, four qandidates are elected.<BR><BR>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.<BR><BR>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.<BR><BR>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.<BR><BR>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.<BR><BR>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.<BR><BR>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.<BR><BR>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.<BR><BR>It could be tested against other proportional systems, to see if it produced similar results, including against my proportional range system.</DIV></div></body></html>