[EM] Markus's '98 Cloneproof SSD wording
MIKE OSSIPOFF
nkklrp at hotmail.com
Fri Mar 16 22:38:16 PST 2001
Markus--
I checked message #2291, and I noticed that its way of dealing with
ties isn't what I'd believed. I must have misread it the 1st time.
I thought that you'd said:
If there are more than 1 winner, then eliminate the other candidates
and repeat the count, from the start, with just those winners.
Repeat till this doesn't reduce the set of winners. At that time
pick the final winner by Random Ballot.
That's what I thought it was, and that's what I've been telling people
all this time.
So what I've been advocating isn't the same as what you said. Would
the procedure that I describe above cause a problem, such as
nonmonotocity, or some other fault? If not, it's more decisive,
since it doesn't resort to Random Ballot till re-application of the
count rule among its ties no longer reduces the winner-set.
My main question here is whether that procedure that I describe above
has a fault.
I notice, in the message that I copy below, that you're saying to
use Random Ballot immediately if there are more than 1 winner, and then
, if Random Ballot doesn't reduce the winner set to 1, _then_ you'd
repeat the count, from the start, among only those winners. But
having just resorted to Random Ballot, then why not just stay with it
and if it doesn't reduce the winner set to 1, draw another random ballot,
till there's just 1 winner?
As you said, your procedure sounds the same as Cloneproof SSD, but
without mentioning cycles. But the price of not mentioning cycles seems
to be a requirement for more paragraphs. And cycles are easy to
check for while calculating the Schwartz set. But your wording might
be more accepted by audiences who don't like to hear about cycles.
I copy message # 2291 here:
Step 1:
Calculate the Schwartz Set of the remaining candidates and
eliminate all those candidates, who are not in the Schwartz
Set of the remaining candidates.
If there is only one candidate remaining, then go to Step 4.
Otherwise go to Step 2.
******
Step 2:
If there are still pairwise inequalities between remaining
candidates, then substitute the "weakest" pairwise inequality
between two remaing candidates with a pairwise equality and
go to Step 1. Otherwise go to Step 3.
[The "weakest" pairwise inequality is that pairwise inequality
with the smallest absolute number of votes for the winner of
this pairwise inequality. If there is more than one pairwise
inequality with the smallest absolute number of votes for the
winner of this pairwise inequality, then the "weakest" pairwise
inequality is that pairwise inequality (among those pairwise
inequalities with the smallest absolute number of votes for the
winner) with the largest absolute number of votes for the loser.
If there is more than one pairwise inequality with the smallest
absolute number of votes for the winner and the largest absolute
number of votes for the loser, then all these inequalities are
substituted with a pairwise equality simultaneously.]
******
[Here's the part that I'd misread]:
Step 3:
If there is a ballot, that hasn't yet been chosen randomly,
then -among those ballots that haven't yet been chosen randomly-
choose one ballot randomly. Restart the whole algorithm among
those remaing candidates, that are (among the remaining
candidates) top-ranked on this randomly chosen ballot and
eliminate the other remaining candidates. [Already eliminated
candidates stay eliminated even after the restart of the
algorithm.]
Otherwise, choose the winner randomly among the remaing
candidates and go tho Step 4.
******
Step 4:
The remaing candidate wins the election.
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