How about this for independence of clone criteria that are compatible with
Droop proportionality?
Cloning a winning candidate should not turn a loosing candidate into a winner.
Cloning a loosing candidate should not turn a winning candidate into a looser.
--- On Mon, 4/8/13, Ross Hyman
I think that the clone properties to look for in proportional representation
methods are:
Increasing the clones in a clone set should not decrease the number of winners
from the clone set and should not make a looser that is outside the clone set
into a winner.
and
Decreasing the clones in a
More general variant:
Candidate sets of N candidates are notated by Greek letters.
Define V(Alpha,Beta) by performing your favorite version of STV on a modified
ballot set where all candidates have been removed from ballots except those in
set Alpha that are ranked higher than all candidates in
I think this will be the final version.
For sequential STV, as defined by Hill and Gazeley in Voting Matters Issue 20,
a candidate set of N candidates should win if it is the only candidate set
where “any set of N+1 candidates, consisting of those N and 1 more, will result
in the election of
Here is a tentative proposal for an STV method based on an extension to
multiple seats of the clone-proof Condorcet method I introduced earlier. It is
based on several ideas from Schulze STV.
Start with a variant way of defining the previous single winner method that is
exactly equivalent
the number of
candidates is just one more than the number of winners. What about the general
case of N+K candidates and N winners?
--- On Thu, 4/4/13, Ross Hyman rahy...@sbcglobal.net wrote:
From: Ross Hyman rahy...@sbcglobal.net
Subject: STV method based on previously introduced clone-proof
only one candidate set
remains.
--- On Thu, 4/4/13, Ross Hyman rahy...@sbcglobal.net wrote:
From: Ross Hyman rahy...@sbcglobal.net
Subject: Re: STV method based on previously introduced clone-proof Condorcet
single winner method
To: election-meth...@electorama.com
Date: Thursday, April 4
V_AB is the number of ballots that rank A above B.
V_A is the number of ballots that rank A at the top.
S_A = sum_B M_AB V_B is the score for
candidate A.
M_AB can be any antisymmetric function of V_AB and V_BA that is positive if
V_AB V_BA.
examples:
M_AB = V_AB - V_BA
M_AB = (V_AB -
Here is a clone independent modification of Baldwin.
Has this been discussed before?
V_AB is the number of ballots that rank A above B.
V_A is the number of ballots that rank A at the top.
S_A = sum_B (V_AB - V_BA)V_A V_B is the score for candidate A. The V_AV_B
factor makes it a
Just realized silly mistake. S_A1+SA2+SA3 etc = S_A does not mean that some
S_Aj have to be above S_A and some below S_A. Does not effect clone
independence.
--- On Sun, 2/3/13, Ross Hyman rahy...@sbcglobal.net wrote:
From: Ross Hyman rahy...@sbcglobal.net
Subject: clone independent
Oops. Need to show that if S_A is negative, one of the S_Aj will be more
negative. Haven't done so. Without that, clone independence isn't proved.
Back to the drawing board.
--- On Sun, 2/3/13, Ross Hyman rahy...@sbcglobal.net wrote:
From: Ross Hyman rahy...@sbcglobal.net
Subject: Re
http://www.knesset.gov.il/elections19/eng/list/results_eng.aspxThe
official Israeli election results show that of the parties receiving
more than the 2% threshold needed to get into the Knesset, the
center-left parties actually got a higher percentage of the vote,
46.67%, than the right
for the effective number of parties
To: Ross Hyman
Cc: election-meth...@electorama.com
Date: Friday, December 14, 2012, 9:59 AM
How would that work using the other formula?
(I know I'm kinda just being lazy here, but I think other people would be
interested.)
Jameson
2012/12/14 Ross Hyman
example using
Consider that there are a number of parties, with the ith party having
vote fraction P_i. Now consider that you can divide the parties into
two, left parties and right parties. Call the vote fraction for the left
parties P_L and the vote fraction for the right parties P_R. Use the
effective
of parties n_c= 4
n_b/n_a = n_c/n_b = sqrt(2)
Splitting one party has the same effect on the ratio, regardless if the other
party has split or not.
--- On Fri, 12/14/12, Ross Hyman wrote:
Consider that there are a number of parties, with the ith party having
vote fraction P_i. Now consider
Here is a physics alternative to the effective number of parties formulas
mentioned on the Wikipedia page:
http://en.wikipedia.org/wiki/Effective_number_of_parties
Based on the concept of entropy, a sensible formula for the effective number of
parties = exp(-sum_i P_i log(P_i))
where P_i is
If one removed all pairwise defeats that contradict the Schulze beathpath order
and then constructed a new beatpath order from the reduced set of defeats,
would the new beatpath order always be consistent with (although not
necessarily the same as) the previous beatpath order? Could this
there is no candidate, C, with a candidate in
its set that does not contain C in its set. The count ends since all sets are
complete so no changes can occur.
--- On Fri, 12/9/11, Ross Hyman rahy...@sbcglobal.net wrote:
From: Ross Hyman rahy...@sbcglobal.net
Subject: Re: finding the beat path
)
B is the second place winner.
--- On Fri, 12/9/11, Ross Hyman rahy...@sbcglobal.net wrote:
From: Ross Hyman rahy...@sbcglobal.net
Subject: finding the beat path winner with just one pass through the ranked
pairs
To: election-methods@lists.electorama.com
Date: Friday, December 9, 2011, 4
Refinement: Don't determine winner until the end.
C_i is the ith candidate. Initially M is the Identity matrix of size equal to
the number of candidates.
The pairs are ranked in order. Affirm each group of equally ranked pairs in
order, from highest to lowest. To Affirm a group of equally
Markus is right.
One way of retaining monotonicity, I think, is to replace the Sets with objects
that record the number of times that a A has beaten B.
Then for the pair ordering
AC, BC
CD
DA, DB, AB
Affirming AC and B C
A(W):A(W)
B(W):B(W)
C(L):A(W),B(W),C(L)
D(W):D(W)
Affirming CD
How so?
--
Dear Ross Hyman,
you wrote (28 Nov 2011):
One way of retaining monotonicity, I think, is to replace
the Sets with objects that record the number of times that
a A has beaten B.
I guess that this tie-breaking strategy will violate
independence of clones.
Markus Schulze
But is that the only monotonic clone independent method? The method I describe
elects D instead of A in accordance with DA. But I don't see why it would
violate clone independence.
Consider the matrix in which the elements are the number of times the column
candidate has defeated the row
Candidates are classed in two categories: Winners and Losers. Initially, all
candidates are Winners. C_i is the ith candidate. A matrix M contains how
many times candidates have defeated each other. The element M_ij equals the
number of times that C_i defeats C_j. Initially M is the
When beat path produces a tie, this method can produce a single winner unless
the tie is genuine. It is the same method I presented earlier except for the
addition of the Removing step, which resolves the ties.
Candidates are classed in two categories: Winners and
Losers. Initially, all
for each ballot is w_i = sum_a w_i,a.
5) The new v_i,a = w_i,a / w_i. Proceed to step 2.
The candidate set with the highest score wins the election.
--- On Sat, 10/1/11, Toby Pereira tdp2...@yahoo.co.uk wrote:
From: Toby Pereira tdp2...@yahoo.co.uk
Subject: Re: [EM] PR approval voting
To: Ross Hyman
priority relations have
determined that AD CD. According to this procedure, candidates A and
D are elected.
-Ross Hyman
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