P(w, v), then the search process can start again from v.
It appears that this fixed point voting procedure reduces the dependence upon budget changes. There can still be a dependence, but it is not as large as without the condition.
In Table 13, the Borda Fixed Point winner is A. With B the Borda winner, A is the alternative winner when B does not participate, and B loses from A in a pairwise match; starting the search from A, its alternative winner is B, and A wins from B.
More on this can be found in Colignatus (2001). That book has also been intended as a textbook and it developed Mathematica programs for the various voting schemes and data manipulations. Given the complexity of the matter, this working environment has appeared a great advantage.
Relation to Saari’s work
Donald Saari (2001ab) showed that Borda’s method is the only method that satisfies certain symmetries. His suggestion is that the Borda rule ‘therefor is best’. This argument does not convince by itself since ‘symmetry’ is not by itself a moral category. Dynamics is linked to morality, by the notion that morality presumes time, and thus seems a better angle.
Consider direct symmetry first. Suppose that your preference is A > B > C and that my preference is C > B > A. The direct symmetry consideration is that we might both abstain from a vote and stay home, since our preferences strictly oppose each other. Saari noted too that voting cycles can be catalogued under the mathematical concept of rotational symmetry. His subsequent suggestion is that cancellation should hold for all symmetries for all subsets of voters.
What happens when cancellation of ‘rotational symmetry’ is applied to subsets ? The following is an example by Saari that cancellation isn’t trivial then. In Table 14 there are 48 voters, and B is selected by both Borda and Condorcet. In Table 15, 27 voters have been added who have the mentioned rotational symmetry, with 9 for each subgroup. Now Borda still selects B, but Condorcet, and the Borda Fixed Point, select A. In Saari’s view, Borda satisfies symmetry, and ‘hence’ is the better method.
My reasoning is a bit different. First of all, note that I myself have used an argument similar to that of Saari. In my view, the typical Condorcet situation of three preferences A > B > C, B > C > A and C > A > B results into indifference rather than an inconsistency, and I use this against Arrow’s analysis. So I agree with Saari’s view that such votes cancel. I applaud Saari’s insight that if you apply cancellation for all cycles in all subsets, then the logic is to get rid of Condorcet’s method and to use Borda’s method.