[EM] Fwd: agenda landau winner

[Susan Simmons]

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-------- Mensaje original --------
De: Susan Simmons <suzerainsimmons at outlook.com>
Fecha: 28/7/21 12:25 p. m. (GMT-08:00)
A: Kristofer Munsterhjelm <km_elmet at t-online.de>
Asunto: agenda landau winner









We work from an agenda of alternatives listed in order of “promise.” The agenda is “monotone” if increasing ballot support for an alternative moves it towards the promising end of the agenda without altering the relative order of the other candidates in the list.
We wish to show that the following method for choosing a winner from a monotone agenda satisfies the mono-raise property of election methods.
Agenda Landau Winner (ALW):
Let X be the least promising (remaining) alternative. Let Y be the winner of this ALW method applied recursively to the agenda restricted to the remaining alternatives with X (also) eliminated. If X covers Y, then elect X, else elect Y.
First suppose the winner W moves towards the more promising end of the agenda without changing the covering relation.Then, if anything, W attains an unassailable position at an earlier stage of the recursion.
Now suppose ballots are altered only by raising the ALW winner W relative to the other alternatives. We will show that this change cannot result in a new winner, by deriving a contradiction from the assumption that some W' other than W becomes the new winner.
First note that after the change W still covers every alternative it covered before the change, and W is still uncovered (its short beatpaths remain short).
Let A be the most promising agenda item. Could W' be A? No, because W still covers A, and the new winner W' has to be uncovered.
Does W' cover A? Yes, because an ALW winner (e.g. W') must cover the most promising agenda item (when not equal to it).
Let b1=A, b2, ... bN=W be the sequence of agenda items such that before the change (i.e. the raising of W) the agenda alternative b(k+1) was the most promising agenda item that covered b(k).
Let a1=A, a2, ... aN=W' be the sequence of agenda items such that a(k +1) is (after the change) the most promising item that covers a(k).
Let j be the smallest value of k, such that a(k) = b(k).
Then a(j+1) is a more promising alternative than b(j+1), so a(j+1) did not cover b(j) = a(j) before the change.
Let Z be the alternative that W beat pairwise after the change but not before. To simplify notation, let V = b(j) = a(j), and let U = a(j+1).
In summary, U covers V after the change but not before, and W beats Z after, but not before.
How can this be? U not covering V before entails V beating some alternative that U did not before, but now does. What alternative could that be?
The only pairwise beat change is W going from not beating Z to beating Z.
 Evidently W = U and Z = V.
 So W did not beat Z = V = b(j) before the change.
This contradicts the construction in which W covers every b(k) .... (by the transitivity of the covering relation).
This contradiction shows that the supposition W' not equal to W was untenable.
The ALW method satisfies mono-raise!



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