[EM] (Possibly) new method/request for voting paradoxes. :)

[Michael Rouse]

As usual with such posts, there is a good chance someone has come up 
with the same (or very similar) method, but I thought it had interesting 
properties, and was wondering what glaring voting paradoxes it had. In 
addition, the number of possible orders is overwhelming if there are a 
large number of candidates, and I'm not sure that can be simplified. 
Finally, Thunderbird sometimes seems to have weird formatting issues in 
email, which may screw up the following into unreadability.

With that in mind, here it is.

Step 1: For each ranked ballot, create a matrix for each pairwise vote, 
based on the distance and direction between each candidate. For example, 
on the ballot A>B>C, you would get:

        -2    -1    0    1    2
AB    0     0    0    1    0
BA    0     1    0    0    0
AC    0     0    0    0    1
CA    1     0    0    0    0
BC    0     0    0    1    0
CB    0     1    0    0    0

Taking the rows in order, this shows that A is one position higher than 
B on this ballot, which conversely makes B one position lower than A on 
the same ballot. Also, A is two positions above C (C is two positions 
below A), and B is one position above C (or C is one position below B). 
Such detail may be unnecessary -- simply looking at position 1 and above 
is sufficient, if you don't allow ties -- but I wanted to show the symmetry.

Step 2. Add all matrices together. As a simple example, let's consider 
the following 12 votes in a circular tie (to make it interesting):

5: A>B>C
3: B>C>A
4: C>A>B

 Taking the first line, A>B>C =

        -2    -1    0    1    2
AB    0     0    0    1    0
BA    0     1    0    0    0
AC    0     0    0    0    1
CA    1     0    0    0    0
BC    0     0    0    1    0
CB    0     1    0    0    0

Multiplied by 5 gives you:

        -2    -1    0    1    2
AB    0     0    0    5    0
BA    0     5    0    0    0
AC    0     0    0    0    5
CA    5     0    0    0    0
BC    0     0    0    5    0
CB    0     5    0    0    0


Taking the second line, 3: B>C>A =

        -2    -1    0    1    2
AB    3     0    0    0    0
BA    0     0    0    0    3
AC    0     3    0    0    0
CA    0     0    0    3    0
BC    0     0    0    3    0
CB    0     3    0    0    0


And finally, taking the third line, 4: C>A>B

        -2    -1    0    1    2
AB    0     0    0    4    0
BA    0     4    0    0    0
AC    0     4    0    0    0
CA    0     0    0    4    0
BC    4     0    0    0    0
CB    0     0    0    0    4

Adding these together gives you

(5: A>B>C, 3: B>C>A, 4: C>A>B) =

        -2    -1    0    1    2
AB    3     0    0    9    0
BA    0     9    0    0    3
AC    0     7    0    0    5
CA    5     0    0    7    0
BC    4     0    0    8    0
CB    0     8    0    0    4

Step 3: This is where a bit of a curve is thrown in. Looking at each 
position on the matrix, determine which is less -- the sum of the 
numbers above the current position, or the sum of the numbers below the 
current position -- and add that number to the value of the current 
position. In essence, this is adding the value of a position to the 
value of all other positions away from the median. Output that number to 
a new matrix in the appropriate spot. Using the numbers above, you end 
up with:

        -2    -1    0    1    2
AB    3     3    3    9    0
BA    0     9    3    3    3
AC    0     7    5    5    5
CA    5     5    5    7    0
BC    4     4    4    8    0
CB    0     8    4    4    4

Step 4: Looking at each possible preference order, add the values for 
the appropriate positions on the matrix, and choose the preference order 
with the highest score. Using the above values and excluding ties:

A>B>C = (A>B) [one position] + (B>C) [one position] + (A>>C) [two 
positions] = 9+8+5 = 22
B>C>A = 8+7+3 = 18
C>A>B = 7+9+4 = 20
A>C>B = 5+4+0 = 9
C>B>A = 4+3+0 = 7
B>A>C = 3+5+0 = 8

So the ranking of possible orders is

A>B>C = 22
C>A>B = 20
B>C>A =18
A>C>B = 9
B>A>C = 8
C>B>A = 7

A>B>C is the most preferred order. C>B>A is the least preferred order.

****************

As another example (ignoring a few pages of work), here is a set of 
ballots used in the Schulze Method entry in Wikipedia:

5:A>C>B>E>D
5:A>D>E>C>B
8:B>E>D>A>C
3:C>A>B>E>D
7:C>A>E>B>D
2:C>B>A>D>E
7:D>C>E>B>A
8:E>B>A>D>C

If I did the math correctly, the winning order is:
E>B>A>D>C = (27+25+30+28)+(23+26+13)+(8+16)+(8) = 204

For comparison, Schulze gives E>A>C>B>D, which this method would score 
(22+26+28+19)+(16+17+17)+(0+15)+(0) = 160

(Note, no claim is made as to which one is better, I just wanted to show 
the difference.)

**************

I've tried variants using distance multiplied by score (which seems to 
encourage strategic raising and lowering of ranks) and absolute ranking 
rather than relative ranking (which didn't seem to act as nicely in vote 
aggregation), but neither seemed to act as nicely on brief inspection. I 
haven't come up with a tiebreaker method yet, either -- like in the 
simple example of adding together 1: A>B>C and 1:B>C>A (both orders have 
the same score) -- but that can wait.

Any questions, comments, or criticisms (the latter most likely about my 
math!) are welcome. Especially welcome would be examples of paradoxes, 
and links to the same (or similar) method discussed in the past.

Michael Rouse