[EM] Borda, Nanson, and "Crosscut" methods II
Michael Rouse
mrouse at cdsnet.net
Tue Jun 19 15:30:43 PDT 2001
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
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