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Re: Smith and Schwartz
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Markus Schulze
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Oct 01, 2001 07:14 PDT
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Dear Pharamond,
suppose, that "X > Y" means that the number of voters who strictly
prefer candidate X to candidate Y is strictly larger than the
number of voters who strictly prefer candidate Y to candidate X.
Suppose, that "X >= Y" means that the number of voters who strictly
prefer candidate X to candidate Y is not strictly smaller than the
number of voters who strictly prefer candidate Y to candidate X.
Then "Local Independence from Irrelevant Alternatives" says that
when candidate A would have been elected when candidate B hadn't
run then --when candidate B does run-- at least one of the following
three statements must be valid:
1) Candidate A wins.
2) B >= A.
3) There is a set of candidates C[1],...,C[n] with
B >= C[1] >= ... >= C[n] >= A.
Every election method can be made compatible to Local Independence
from Irrelevant Alternatives simply by calculating the Smith set
and then applying this method only to the candidates of the Smith
set. However, one won't get a stronger version of Local Independence
from Irrelevant Alternatives by using the Schwartz set instead of
the Smith set. Therefore, there is no improvement when one uses the
Schwartz set instead of the Smith set.
******
Ranked Pairs cannot choose decisively a candidate who is not in
the Schwartz set. Whether Ranked Pairs can choose randomly a
candidate who is not a in the Schwartz set depends on the used
random tiebreaker. Actually, it is possible to define Ranked
Pairs in such a manner that it never chooses decisively or
randomly a candidate who is not in the Schwartz set.(Simply
use a TBRC where all Schwartz winners are ranked ahead of all
other candidates.)
And of course, without losing any of the important properties
of Ranked Pairs, it is also possible simply to calculate the
Schwartz set and then to apply Ranked Pairs only to the
candidates of the Schwartz set. But --as I said-- there is no
real reason why the winner should always be a candidate of the
Schwartz set.
Markus Schulze
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