[EM] Borda, Nanson, and "Crosscut" methods II

[Michael Rouse]

This is a second part to "Borda, Nanson, and Crosscut methods." (The reason 
I call the method "Crosscut" is from how I originally visualized it: a step 
function moving from left to right, removing one candidate at a time to a 
list, followed by a step function moving from right to left doing the same 
thing. The step function reminded me of a sawtooth moving back and forth.)

This got me to thinking: the step function might not be the best shape to 
cut the candidate list down to size. (ok, that's a truly awful pun. Sorry!) 
"Top" functions are the standard ones that take the highest ranked 
candidate at various points until only one candidate is missing from the 
list. "Bottom" functions are the more involved inverse functions that take 
a set of bottom candidates, drop the highest-ranked, and iterate over and 
over until only one bottom candidate remains. The loser from each method is 
compared; the overall loser of the pair is dropped and the process starts 
over again. To keep consistent with examples I used before, consider eight 
candidates:

Step function (square tooth)
"Top" candidates from left to right: 10000000, 11000000, 11100000, 
11110000, 11111000, 11111100, 11111110
"Bottom" candidates from right to left: 00000001, 00000011, 00000111, 
00001111, 00011111, 00111111, 01111111

Borda function (triangle tooth)
"Top" candidates from left to right: 10000000, 21000000, 32100000, 
43210000, 54321000, 65432100, 76543210
"Bottom" candidates from right to left:00000001, 00000012, 00000123, 
00001234, 00012345, 00123456, 01234567

Power of Two (pointy teeth with an inward curve)
"Top" candidates from left to right: 10000000, 21000000, 42100000, 
84210000, (16)8421000, (32)(16)842100, (64)(32)(16)84210
"Bottom" candidates from right to left: 00000001, 00000012, 00000124, 
00001248, 0001248(16), 001248(16)(32), 01248(16)(32)(64)

Here are three other graphs I've mentioned:
"Reflected" Power of Two: 1-2^(N-M), where N is the rank of the candidate 
and M is the number of candidates.
Power of Two + "Reflected" Power of Two: (1/2)+(2^-N)-(2^(N-M-1))  (I think 
I have that right)
Zipf Law: 1/N

The Borda function is superior (in my opinion) to the square function since 
it takes into account candidate rankings -- three firsts and a second place 
*should* be ranked higher than two firsts and two seconds. Any difference 
is probably extremely rare, but unless someone finds an improvement, I'll 
keep Borda in the final definition of the "Crosscut" method. The other 
graphs might change things a bit, but they may just be different rather 
than an improvement. It would be interesting to see where different 
functions gave different results with this method. About the only 
constraint I would put on which function to choose would be for symmetry -- 
I would require the top and bottom functions to be reflections of each other.

Having chosen the triangular "Borda" function, here's an example on how it 
would work as a Condorcet completion method. Say there were three 
candidates in the Smith set, ABC, where A>B>C>A. The "Top" function would 
first find the Plurality winner (100), then the Borda winner (210). 
Assuming they were different, the candidate remaining is the "Top" loser. 
The "Bottom" function would discover the Plurality loser (001) and then the 
Borda loser (012). It then compares them to come up with a "Bottom" loser. 
If the Top and Bottom loser are the same, it is dropped and the remaining 
two candidates are checked against each other. If they happen to be 
different they are compared and the overall loser dropped.

Michael Rouse
mrouse at cdsnet.net