On Mon, 3 Mar 2003, Steve Barney wrote:
Forest:
In message # 10970, why did you say wisely, as follows?:
Kemeny (wisely) doesn't believe in cyclic symmetry removal
[...]
Do you mean to imply that the KR tie between ACB and BAC is more
reasonable than the ABC outcome yielded by both
To focus on the order is to miss the boat. Sometimes one order is most
efficient and sometimes another.
In the example you gave me the most efficient order is removal of five
copies of the cycle.
The other order that you suggested did not reduce the ballot set to the
minimum of three ballots.
Forest:
In message # 10970, why did you say wisely, as follows?:
Kemeny (wisely) doesn't believe in cyclic symmetry removal
[...]
Do you mean to imply that the KR tie between ACB and BAC is more
reasonable than the ABC outcome yielded by both Saari's and your
decomposition of my example?:
Forest:
What do you mean by the order of removal isn't important as along as you
recognize that whenever you have two non-adjacent factions left, more symmetry
reduction is possible.
I showed you with my example:
3:ABC
5:ACB
0:CAB
5:CBA
0:BCA
5:BAC
that the order of those operations matters
=3
SB
- Forwarded Message -
From: Forest Simmons [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: RE: [EM] Saari's Basic Argument
On Fri, 21 Feb 2003, Steve Barney wrote:
Forest:
How do you decompose my example (from my last email, #10873), and what do
you
get?:
3:ABC
5:ACB
0
On Thu, 27 Feb 2003, Steve Barney wrote:
Forest:
Apparently, as I thought, your method of decomposition is to simply to remove
cycles first, and then reversals. My point remains, then, that your
decomposition method does NOT NECESSARILY yield the same outcome as Saari's
matrix decomposition
[I am resending this, because nobody replied yet.]
Forest:
How do you decompose my example (from my last email, #10873), and what do you
get?:
3:ABC
5:ACB
0:CAB
5:CBA
0:BCA
5:BAC
SB
From: Forest Simmons [EMAIL PROTECTED]
To: EM-list [EMAIL PROTECTED]
Subject: RE: [EM] Saari's Basic Argument
On Fri, 21 Feb 2003, Steve Barney wrote:
Forest:
How do you decompose my example (from my last email, #10873), and what do you
get?:
3:ABC
5:ACB
0:CAB
5:CBA
0:BCA
5:BAC
Subtract out five copies of the cycle ACB+CBA+BAC.
That leaves 3*ABC.
Forest
For more information about this
On Thu, 20 Feb 2003, Alex Small wrote:
snip
However, people electing politicians are clearly not machines. We have
our idiosyncracies and legitimate differences of opinion, and we debate
matters that don't have obvious, objectively correct answers. Because we
don't behave or think like
Forest:
How do you decompose my example (from my last email, #10873), and what do you
get?:
3:ABC
5:ACB
0:CAB
5:CBA
0:BCA
5:BAC
SB
From: Forest Simmons [EMAIL PROTECTED]
To: EM-list [EMAIL PROTECTED]
Subject: RE: [EM] Saari's Basic Argument
On Tue, 18 Feb 2003, Steve Barney wrote:
Here
On Tue, 18 Feb 2003, Steve Barney wrote:
Here is a simpler example to illustrate the difference that the order in which
cyclic and reversal terms are canceled does not matter when using the strictly
correct method - as opposed to the method used by Forest Simmons and Alex
Small, and in some
One note:
Below I state that these symmetries preserve the Borda count. That's
because I use the numbers 1,0, and -1 for the three rank positions, so
that the symmetrical distributions all give a Borda count of zero to all
three candidates.
So when you add or subtract symmetrical sets of
Forest Simmons said:
But worrying about the details of symmetry cancellations is to bark up
the wrong tree.
Amen.
This result may make sense in the context of dispassionate decision
making such as in robotics when a robot is trying to decide what
movement to make or whether a visual image
Steve Barney said:
IF YOU DO THE DECOMPOSITION WITH THESE
MATRIXES, IT DOES NOT MATTER IF YOU FIND THE REVERSAL TERMS OR THE
CONDORCET (CYCLIC) TERMS FIRST.
That approach gives us profiles where -12 voters or whatever have a given
preference order. It is undoubtedly a mathematical fact that
I want to add something else to my argument:
For normative reasons I consider the Borda Count to be an unsatisfactory
election method. I have certain behavioral, strategic, and political
criteria that Borda fails miserably (no method is perfect, but Borda gets
a flat F while some methods gets
Here is a simpler example to illustrate the difference that the order in which
cyclic and reversal terms are canceled does not matter when using the strictly
correct method - as opposed to the method used by Forest Simmons and Alex
Small, and in some of Saari's popular expositions where he is
On Sat, 18 Jan 2003, Steve Barney wrote:
If you don't like Condorcet's example, how about this one, which I have looked
at before:
5 ABC
3 BCA
Can you give me the decomposition profile, T(p), for this example?
This example is essentially the same as the
66% ABC
34% BCA
example.
Alex:
If you go to the online copy of my message (message # 10757), and hit Unwrap
Lines at the top, the link to Saari's article should work. It works for me.
Go to:
http://groups.yahoo.com/group/election-methods-list/message/10757?unwrap=1
You might be able to find it yourself by searching
Steve Barney said:
this is why you get two different decompositions when you do it in
different orders. Try using Saari's decomposition matrix with your
examples, and see if you get the same decomposition profile as you get
with your method.
Decomposing a vector into its projections onto
Alex:
See my comments between your lines, below.
SB
--- In [EMAIL PROTECTED], Alex Small [EMAIL PROTECTED]
wrote:
Steve Barney said:
[...]
If you don't like Condorcet's example, how about this one, which I have
looked at before:
5 ABC
3 BCA
Can you give me the decomposition
First, I tried to get to the paper that you referenced but the link was
bad. Rather than the full link, maybe it's best to send me instructions
on how to search for the paper.
Steve Barney said:
p=profile=
[[5]
[0]
[0]
[0]
[3]
[0]]
T(p)=(1/6)(7,8,3,-2,8,8)
I'll have to look again
]
Subject: RE: [EM] Saari's Basic Argument
I hate to beat a dead horse, but in order to see the fallacy of Saari's
symmetry arguments let's take this example a little further:
66 ABC
34 BCD
The 12 o'clock and 8 o'clock positions representing these two factions are
non adjacent on the clock face. Between
On Thu, 16 Jan 2003, Steve Barney wrote:
Forest:
In your example,
66 ABC
34 BCA
If you give second preferences any more than 16/33 of the weight which you
give to the first prefs, the winner is B; since:
No need of giving weights to see all the mischief that could come from
giving the
Forest Simmons said:
On Thu, 16 Jan 2003, Steve Barney wrote:
Forest:
In your example,
66 ABC
34 BCA
No need of giving weights to see all the mischief that could come from
giving the win to B.
Moreover, if candidate C weren't there then we'd all agree that A trounced
B conclusively.
I hate to beat a dead horse, but in order to see the fallacy of Saari's
symmetry arguments let's take this example a little further:
66 ABC
34 BCD
The 12 o'clock and 8 o'clock positions representing these two factions are
non adjacent on the clock face. Between them is the fully ranked order
On Fri, 17 Jan 2003, Alex Small wrote:
Forest Simmons said:
66 ABC
34 BCA
No need of giving weights to see all the mischief that could come from
giving the win to B.
Moreover, if candidate C weren't there then we'd all agree that A trounced
B conclusively. Then we throw in C, and
Forest:
Isn't that just another way of saying Kemeny's Rule does not respect cyclic
symmetry?
SB
From: Forest Simmons [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: RE: [EM] Saari's Basic Argument
On Wed, 15 Jan 2003, Alex Small wrote:
...
I'm not convinced
that symmetry
On Thu, 16 Jan 2003, Steve Barney wrote:
Forest:
Isn't that just another way of saying Kemeny's Rule does not respect cyclic
symmetry?
Or we could say that cyclic symmetry doesn't respect the minimal
distance criterion, since that is what Kemeny's rule is.
A more neutral statement is that
On Thu, 16 Jan 2003, Forest Simmons wrote:
On Thu, 16 Jan 2003, Steve Barney wrote:
Forest:
Isn't that just another way of saying Kemeny's Rule does not respect cyclic
symmetry?
Or we could say that cyclic symmetry doesn't respect the minimal
distance criterion, since that is what
message in which you did what you describe below.
SB
From: Forest Simmons [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: RE: [EM] Saari's Basic Argument
I've already given an example in which Borda gives the wrong answer after
the symmetry is removed. Now you have given an example
Steve Barney said:
In this case, a second pref must be given less than 1/1,000,001 of the
weight of a first pref, if A should win. That's a stretch.
So, what is the appropriate weight for a second choice? Half the weight
of first? Equal to first? Maybe a third? How about 1/Pi?
Possible
I've already given an example in which Borda gives the wrong answer after
the symmetry is removed. Now you have given an example in which symmetry
removal shows the CW to be wrong. So that evens the score :-)
In other words, neither Borda nor Condorcet can claim to be superior on
the basis of
Forest Simmons said:
So here's a new method (for three candidate races only): first remove
all of the symmetry, and then the candidate with a majority of first
place votes (on the remaining ballots) is the winner.
The order in which one removes symmetry matters. Canceling out reversal
On Wed, 15 Jan 2003, Alex Small wrote:
I'm not convinced
that symmetry is a particularly compelling reason to pick an election
method,
especially not the symmetry of {ABC,BCA,CAB}, which has a rotational bias.
True, it favors no candidate, but it does favor its three orders over
In EM archives message #8999 Alex recounts in his own words Saari's idea
of subtracting out the symmetrical part of a ballot pattern and then
deciding the winner on the basis of the residual ballots.
Let's do an example:
7 ABC, 5 ACB , 9 CAB, 3 CBA, 7 BCA, 8 BAC.
If we subtract three of each
Forest Simmons said:
In the case of full rankings of three candidates, this residual method
seems to always gives the same result as the Kemeny order, MinMax,
Ranked Pairs, SSD, etc. for the original problem.
That's what I expected. It's quite reasonable to assume that a _ranked_
method will
Alex:
You may be thinking of Condorcet's example, the profile which Condorcet used
to decredit the Borda Count by pointing out that the BC-winner was not the
Condorcet-winner in that case. Saari argues that rather than supporting the
Condorcet winner, these examples expose a flaw, and shows
On Tue, 26 Mar 2002, Steve Barney wrote:
Forest:
From: Forest Simmons [EMAIL PROTECTED]
Date: Fri Mar 15, 2002 10:59 am
Subject: Re: [EM] Saari's Basic Argument
At best Saari proves that Borda is the best choice method based on
rankings in situations where there can
Forest:
From: Forest Simmons [EMAIL PROTECTED]
Date: Fri Mar 15, 2002 10:59 am
Subject: Re: [EM] Saari's Basic Argument
At best Saari proves that Borda is the best choice method based on
rankings in situations where there can be no stacking of the deck (clones)
or insincere rankings
At best Saari proves that Borda is the best choice method based on
rankings in situations where there can be no stacking of the deck (clones)
or insincere rankings.
Unfortunately these two conditions are usually important considerations in
elections. So Borda's method should be restricted to
Alex:
Welcome to the Donald Saari reading club! More comments below.
--- In [EMAIL PROTECTED], Alex Small [EMAIL PROTECTED] wrote:
[...]
So, if you subtract out the ballots that form a perfectly symmetric
cycle, Saari has proven that applying the Borda count to what remains will
satisfy the
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