# [EM] Saari's Basic Argument

Steve Barney barnes99 at vaxa.cis.uwosh.edu
Thu Feb 27 14:53:06 PST 2003

```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 method.

There is a lot of disagreemnt over the correct outcome, in this example. Even
Kemeny's Rule (which is also Condorcet's method, according to Hannu Nurmi, as
I quoted previously in message # 10496) disagrees with both your decomp and
Saari's. According to my calculations, Kemeny's Rule goes like this:

Voting Vector:

p=[ [3] [5] [0] [5] [0] [5] ]
[This is a vertical column matrix.]

Matrix (M) for finding the tally for Kemeny's Rule:

M=
[[ 0 1 2 3 2 1 ]
[ 1 0 1 2 3 2 ]
[ 2 1 0 1 2 3 ]
[ 3 2 1 0 1 2 ]
[ 2 3 2 1 0 1 ]
[ 1 2 3 2 1 0 ]]

The KR tally is:

M(p)=[ [25] [23] [31] [29] [31] [23] ], or

25:A>B>C [0(3) + 1(5) + 2(0) + 3(5) + 2(0) + 1(5) = 25]
23:A>C>B [1(3) + 0(5) + 1(0) + 2(5) + 3(0) + 2(5) = 23]
31:C>A>B [2(3) + 1(5) + 0(0) + 1(5) + 2(0) + 3(5) = 31]
29:C>B>A [3(3) + 2(5) + 1(0) + 0(5) + 1(0) + 2(5) = 29]
31:B>C>A [2(3) + 3(5) + 2(0) + 1(5) + 0(0) + 1(5) = 31]
23:B>A>C [1(3) + 2(5) + 3(0) + 2(5) + 1(0) + 0(5) = 23]

With the KR, the lower tally is the better, so we have a KR tie between A>C>B
and B>A>C. Interesting enough, that is similar to the end result that you get
when you first remove the reversals, and then the cycles:

0:A>B>C: 3-3-0=0
3:A>C>B: 5-2-0=3
0:C>A>B: 0-0-0=0
0:C>B>A: 5-3-2=0
0:B>C>A: 0-0-0=0
3:B>A>C: 5-0-2=3

SB

>----- Forwarded Message -----
>From: Forest Simmons <fsimmons at pcc.edu>
>To:  <election-methods-list at eskimo.com>
>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:A>B>C
>> 5:A>C>B
>> 0:C>A>B
>> 5:C>B>A
>> 0:B>C>A
>> 5:B>A>C
>
>Subtract out five copies of the cycle ACB+CBA+BAC.
>
>That leaves 3*ABC.
>
>Forest

Steve Barney

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