Hello,
On Fri, Nov 15, 2002 at 06:51:47PM -0500, Raul Miller wrote:
Definition: An option F transitively defeats an option G if G
defeats F or if there is some other option H where H defeats
G AND F transitively defeats H.
There is a mistake: ... if G defeats F
Hello
On Fri, Nov 15, 2002 at 06:51:47PM -0500, Raul Miller wrote:
Definition: A proposition is a defeat, or a pair of options
where both have received votes explicitly comparing the two
options but neither option is able to defeat the other.
Sorry, but I do
Hello,
even more confusion on my side :-(
On Fri, Nov 15, 2002 at 06:51:47PM -0500, Raul Miller wrote:
1. Each ballot orders the options being voted on in the order
specified by the voter. If the voter does not rank some options,
this means that the voter prefers all ranked
Heres an attempt to describe the winner of a CSSD election logically,
using the beatpath winner implimentation. I've tryed to structure it so
that removing the parenthesis gives a plain english definition. Any
comments would be appresiated.
---
In the following definition all occurances of A
On Fri, Nov 15, 2002 at 09:48:55PM -0500, Branden Robinson wrote:
I am sorry if this tries your patience, but I do not understand this
statement. Can you give an example with numbers? Specifically, I don't
understand what would distinguish a weak pairwise tie from a non-weak
one.
Well.. I'm
On Fri, Nov 15, 2002 at 06:51:47PM -0500, Raul Miller wrote:
Definition: An option F transitively defeats an option G if G
defeats F or if there is some other option H where H defeats
G AND F transitively defeats H.
On Sat, Nov 16, 2002 at 09:04:42AM +0100,
On Sun, Nov 17, 2002 at 03:47:01AM +1100, Clinton wrote:
Heres an attempt to describe the winner of a CSSD election logically,
using the beatpath winner implimentation. I've tryed to structure it so
that removing the parenthesis gives a plain english definition. Any
comments would be
Hi,
Raul Miller:
On Sat, Nov 16, 2002 at 05:23:13AM +0100, Matthias Urlichs wrote:
Note that ALL propositions are considered here, not just those
participating in the Schwartz set.
Hmm.. that's not what I was trying to say. Thanks.
The reason for this to be a problem is that the
More silliness fixed: The word tie is now only used in one context.
The definition of transitive defeats is fixed. The language defining
propositions has been cleaned up.
If anyone feels that draft is too hard to understand, please write me
a letter indicating the first part that you have
On Sat, Nov 16, 2002 at 06:56:37PM +0100, Matthias Urlichs wrote:
The reason for this to be a problem is that the dropping step only considers
the _innermost_ Schwartz set. There may be more than one, though off-hand
I couldn't invent a suitable example -- I'm sure one exists in the
literature
Hello,
On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
1. [...] Any options unranked by the voter are treated
as being equal to all other unranked options.
I still do not understand the other unranked options.
Could we simply write Any options unranked by the voter are
Hello,
On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
1. [...] Any options unranked by the voter are treated
as being equal to all other unranked options.
On Sat, Nov 16, 2002 at 07:47:39PM +0100, Jochen Voss wrote:
I still do not understand the other unranked
Hello again!
Another one:
On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
3. If an option has a quorum requirement, that option must defeat
the default option by the number of votes specified in the quorum
requirement, or the option is eliminated.
We did not
Hi,
Raul Miller:
The Schwartz set is the set of innermost unbeatened sets. There cannot
be more than one Schwartz set at any one time for any one set of votes.
See http://www.barnsdle.demon.co.uk/vote/condor2.html for a more extensive
discussion of this subject.
Thanks for the link.
Hi,
Raul Miller:
Definition: The dominant strength of a proposition is the
count of votes in a proposition which is not smaller than
the other vote count in that proposition.
Please replace not smaller with larger or equal to, and vice versa,
throughout the
On Sat, Nov 16, 2002 at 09:31:08PM +0100, Jochen Voss wrote:
Is the default option supposed to be in the list of options for which
we do Cloneproof Schwartz Sequential Dropping below? Maybe we
should replace Options by Non-default options in this paragraph?
For an option to be in the Schwartz
On Sat, Nov 16, 2002 at 10:47:43PM +0100, Matthias Urlichs wrote:
Please replace not smaller with larger or equal to, and vice versa,
throughout the text. Those negatives make thinking about whatever it is
that the text actually means ;-) more difficult.
Actually, based on a suggestion by
Hello Raul,
trying to actually implement the algorithm from the draft turns out to
be a good test :-)
On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
ii. Unless this would eliminate all options in the Schwartz set,
the weakest propositions are eliminated.
Hello Raul,
On Sat, Nov 16, 2002 at 05:44:40PM -0500, Raul Miller wrote:
On Sat, Nov 16, 2002 at 09:31:08PM +0100, Jochen Voss wrote:
Is the default option supposed to be in the list of options for which
we do Cloneproof Schwartz Sequential Dropping below? Maybe we
should replace Options
Hello,
I spent some time implementing the voting mechanism as described
in Rauls draft. Maybe this is useful for somebody who wants to
play with the new voting system.
I append a preliminary version of the program here. The second
attachment is an example input file. The program is modelled
On Sun, Nov 17, 2002 at 12:23:11AM +0100, Jochen Voss wrote:
Do we intend that the default option actually can win the vote?
Am I correct that this only could happen via step 5?
You're right, I should it so that the default option doesn't need to
defeat the default option.
The interpretation
On Sat, Nov 16, 2002 at 05:54:59AM +0100, Matthias Urlichs wrote:
Another alternative might have been to have the default option win if
it's _ever_ a member of the Scwartz set, rather than if it's a member
of the Schwartz set after the sequential dropping phases are complete.
That rule
On Sat, Nov 16, 2002 at 05:23:13AM +0100, Matthias Urlichs wrote:
v. If this new Schwartz set contains only one option, that
option wins.
As per the definition, there is no such thing as a one-option Schwartz
set, so actually you'll have to restart with step 5. You might
Hi,
Anthony Towns:
I'd rather run the algorithm with the full set of votes first, and _then_,
if the default option wins, have a separate rule on what to do next.
Nope: see the first vote listed above. You _don't_ want to declare a
result when the majority of developers would prefer
Hello,
On Fri, Nov 15, 2002 at 06:51:47PM -0500, Raul Miller wrote:
Definition: An option F transitively defeats an option G if G
defeats F or if there is some other option H where H defeats
G AND F transitively defeats H.
There is a mistake: ... if G defeats F
Hello
On Fri, Nov 15, 2002 at 06:51:47PM -0500, Raul Miller wrote:
Definition: A proposition is a defeat, or a pair of options
where both have received votes explicitly comparing the two
options but neither option is able to defeat the other.
Sorry, but I do
Hello,
even more confusion on my side :-(
On Fri, Nov 15, 2002 at 06:51:47PM -0500, Raul Miller wrote:
1. Each ballot orders the options being voted on in the order
specified by the voter. If the voter does not rank some options,
this means that the voter prefers all ranked
Heres an attempt to describe the winner of a CSSD election logically,
using the beatpath winner implimentation. I've tryed to structure it so
that removing the parenthesis gives a plain english definition. Any
comments would be appresiated.
---
In the following definition all occurances of A
On Fri, Nov 15, 2002 at 09:48:55PM -0500, Branden Robinson wrote:
I am sorry if this tries your patience, but I do not understand this
statement. Can you give an example with numbers? Specifically, I don't
understand what would distinguish a weak pairwise tie from a non-weak
one.
Well.. I'm
Raul Miller:
iv. If eliminating the weakest propositions would not eliminate
all votes, a new Schwartz set is found based on the newly
revised set of propositions.
On Sat, Nov 16, 2002 at 05:23:13AM +0100, Matthias Urlichs wrote:
Note that ALL propositions are
On Fri, Nov 15, 2002 at 06:51:47PM -0500, Raul Miller wrote:
Definition: An option F transitively defeats an option G if G
defeats F or if there is some other option H where H defeats
G AND F transitively defeats H.
On Sat, Nov 16, 2002 at 09:04:42AM +0100,
On Sun, Nov 17, 2002 at 03:47:01AM +1100, Clinton wrote:
Heres an attempt to describe the winner of a CSSD election logically,
using the beatpath winner implimentation. I've tryed to structure it so
that removing the parenthesis gives a plain english definition. Any
comments would be
Hi,
Raul Miller:
On Sat, Nov 16, 2002 at 05:23:13AM +0100, Matthias Urlichs wrote:
Note that ALL propositions are considered here, not just those
participating in the Schwartz set.
Hmm.. that's not what I was trying to say. Thanks.
The reason for this to be a problem is that the
More silliness fixed: The word tie is now only used in one context.
The definition of transitive defeats is fixed. The language defining
propositions has been cleaned up.
If anyone feels that draft is too hard to understand, please write me
a letter indicating the first part that you have
Hi,
Anthony Towns:
v. If this new Schwartz set contains only one option, that
option wins.
As per the definition, there is no such thing as a one-option Schwartz
set
Uh, if C is the Condorcet winner, or becomes the Condorcet winner after
ignoring some defeats of
On Sat, Nov 16, 2002 at 06:56:37PM +0100, Matthias Urlichs wrote:
The reason for this to be a problem is that the dropping step only considers
the _innermost_ Schwartz set. There may be more than one, though off-hand
I couldn't invent a suitable example -- I'm sure one exists in the
literature
Hello,
On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
1. [...] Any options unranked by the voter are treated
as being equal to all other unranked options.
I still do not understand the other unranked options.
Could we simply write Any options unranked by the voter are
Hello,
On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
1. [...] Any options unranked by the voter are treated
as being equal to all other unranked options.
On Sat, Nov 16, 2002 at 07:47:39PM +0100, Jochen Voss wrote:
I still do not understand the other unranked
Hello
On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
2. Options which do not defeat the default option are eliminated.
Definition: Option A defeats option B if more voters prefer A
over B than prefer B over A.
Is the default option supposed to be in the list
Hi,
Raul Miller:
The Schwartz set is the set of innermost unbeatened sets. There cannot
be more than one Schwartz set at any one time for any one set of votes.
See http://www.barnsdle.demon.co.uk/vote/condor2.html for a more extensive
discussion of this subject.
Thanks for the link.
Hi,
Raul Miller:
Definition: The dominant strength of a proposition is the
count of votes in a proposition which is not smaller than
the other vote count in that proposition.
Please replace not smaller with larger or equal to, and vice versa,
throughout the
On Sat, Nov 16, 2002 at 09:31:08PM +0100, Jochen Voss wrote:
Is the default option supposed to be in the list of options for which
we do Cloneproof Schwartz Sequential Dropping below? Maybe we
should replace Options by Non-default options in this paragraph?
For an option to be in the Schwartz
On Sat, Nov 16, 2002 at 10:47:43PM +0100, Matthias Urlichs wrote:
Please replace not smaller with larger or equal to, and vice versa,
throughout the text. Those negatives make thinking about whatever it is
that the text actually means ;-) more difficult.
Actually, based on a suggestion by
Hello Raul,
trying to actually implement the algorithm from the draft turns out to
be a good test :-)
On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
ii. Unless this would eliminate all options in the Schwartz set,
the weakest propositions are eliminated.
Hello Raul,
On Sat, Nov 16, 2002 at 05:44:40PM -0500, Raul Miller wrote:
On Sat, Nov 16, 2002 at 09:31:08PM +0100, Jochen Voss wrote:
Is the default option supposed to be in the list of options for which
we do Cloneproof Schwartz Sequential Dropping below? Maybe we
should replace Options
Hello,
I spent some time implementing the voting mechanism as described
in Rauls draft. Maybe this is useful for somebody who wants to
play with the new voting system.
I append a preliminary version of the program here. The second
attachment is an example input file. The program is modelled
On Sun, Nov 17, 2002 at 12:12:53AM +0100, Jochen Voss wrote:
trying to actually implement the algorithm from the draft turns out to
be a good test :-)
Ok. Note that I've not sat down and read your implementation, yet.
On Sat, Nov 16, 2002 at 12:48:07PM -0500, Raul Miller wrote:
ii.
On Sun, Nov 17, 2002 at 12:23:11AM +0100, Jochen Voss wrote:
Do we intend that the default option actually can win the vote?
Am I correct that this only could happen via step 5?
You're right, I should it so that the default option doesn't need to
defeat the default option.
The interpretation
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