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Re: [EM] Name of this Criterion

MS
Markus Schulze
Sat, Apr 23, 2016 8:12 PM

Hallo,

I have now added a proof that the Schulze method
satisfies this criterion. See section 4.12 of
my paper:

http://m-schulze.9mail.de/schulze1.pdf

Markus Schulze

Hallo,

I remember that we discussed the following
criterion at this mailing list. Unfortunately,
I forgot the name of this criterion. Could
someone please tell me the name of this criterion?

Suppose M is the number of candidates.

Suppose there is a k with 2 <= k <= (M-1) such
that candidate A wins every sub-election between
candidate A and (k-1) other candidates. Then
candidate A should also be the overall winner.

Markus Schulze

Hallo, I have now added a proof that the Schulze method satisfies this criterion. See section 4.12 of my paper: http://m-schulze.9mail.de/schulze1.pdf Markus Schulze > Hallo, > > I remember that we discussed the following > criterion at this mailing list. Unfortunately, > I forgot the name of this criterion. Could > someone please tell me the name of this criterion? > > Suppose M is the number of candidates. > > Suppose there is a k with 2 <= k <= (M-1) such > that candidate A wins every sub-election between > candidate A and (k-1) other candidates. Then > candidate A should also be the overall winner. > > Markus Schulze
RB
robert bristow-johnson
Sat, Apr 23, 2016 9:12 PM

---------------------------- Original Message ----------------------------

Subject: Re: [EM] Name of this Criterion

From: "Markus Schulze" markus.schulze@alumni.tu-berlin.de

Date: Sat, April 23, 2016 4:12 pm

To: election-methods@lists.electorama.com


Hallo,

I have now added a proof that the Schulze method

satisfies this criterion. See section 4.12 of

my paper:

Markus Schulze

Hallo,

I remember that we discussed the following

criterion at this mailing list. Unfortunately,

I forgot the name of this criterion. Could

someone please tell me the name of this criterion?

Suppose M is the number of candidates.

Suppose there is a k with 2 <= k <= (M-1) such

that candidate A wins every sub-election between

candidate A and (k-1) other candidates. Then

candidate A should also be the overall winner.


sorry, but just on the surface this doesn't seem right.
for instance, a run-of-the-mill no-cycle Condorcet case (actually it was IRV and the Condorcet winner did not win) would be the Burlington 2009 election most of us are familiar with. �in that case M=5. �now if you
picked k=3 and chose candidate A to be Kurt Wright (who was the plurality winner and neither the IRV winner nor the CW), there is a set of (k-1) candidates (those would be Smith and Simpson) that Wright beat consistently. �yet he is not the overall winner.
you can go to the Warren Smith
page or i can scarf up the numbers from the defeat matrix again.
how am i reading this wrong?

r b-j � � � � � � � � �rbj@audioimagination.com

"Imagination is more important than knowledge."

---------------------------- Original Message ---------------------------- Subject: Re: [EM] Name of this Criterion From: "Markus Schulze" <markus.schulze@alumni.tu-berlin.de> Date: Sat, April 23, 2016 4:12 pm To: election-methods@lists.electorama.com -------------------------------------------------------------------------- > Hallo, > > I have now added a proof that the Schulze method > satisfies this criterion. See section 4.12 of > my paper: > > http://m-schulze.9mail.de/schulze1.pdf > > Markus Schulze > > > Hallo, > > > > I remember that we discussed the following > > criterion at this mailing list. Unfortunately, > > I forgot the name of this criterion. Could > > someone please tell me the name of this criterion? > > > > Suppose M is the number of candidates. > > > > Suppose there is a k with 2 <= k <= (M-1) such > > that candidate A wins every sub-election between > > candidate A and (k-1) other candidates. Then > > candidate A should also be the overall winner. � sorry, but just on the surface this doesn't seem right. for instance, a run-of-the-mill no-cycle Condorcet case (actually it was IRV and the Condorcet winner did not win) would be the Burlington 2009 election most of us are familiar with. �in that case M=5. �now if you picked k=3 and chose candidate A to be Kurt Wright (who was the plurality winner and neither the IRV winner nor the CW), there is a set of (k-1) candidates (those would be Smith and Simpson) that Wright beat consistently. �yet he is not the overall winner. you can go to the Warren Smith page or i can scarf up the numbers from the defeat matrix again. how am i reading this wrong? -- r b-j � � � � � � � � �rbj@audioimagination.com � "Imagination is more important than knowledge." �