Kathy Dopp wrote:
On Sat, Jan 22, 2011 at 5:44 PM, Kristofer Munsterhjelm
<[email protected]> wrote:
This leaves the first step. At first glance, that seems to be prohibitive.
If we have n candidates, there are n! possible ways to rank them,
There are only n! possible ways to rank them if you prohibit partial
rankings. There are many more than n! when partial rankings are
permitted, as in US elections. I include the formula for calculating
the number of rankings in my "Realities Mar Instant ..." linked below.
However, perhaps you are simplifying for the purposes of your
simplification, so it doesn't matter?
Yes, I am simplifying. Because this method finds the proportion with a
near infinity of voters (depends on the precision of the implementation
of erf), the line between the regions where voters will equal rank will
shrink to zero. Thus, equality is out.
As for truncation, I can't really see how to implement it. I could have
voters randomly deciding to truncate, but that seems a bit arbitrary.
But let's say I decide to implement it somehow. Then, assuming
truncations are monotone, that would multiply the number of ballots with
(at most) n, the number of candidates.
By "assuming truncations are monotone", I mean that if someone's full
ranking was, say, A > B > C > D, he would partial-rank A > B > C or A >
B, but never A > D.
Such a multiplication of n would make things even worse if the
combinatorial explosion was in force. It's good, then, that it isn't.
Leon Smith said there would be n^4 distinct ballots in the worst case,
so with truncation, that would make n^5. Completely impractical for a
large number of candidates, but Yee diagrams usually have few candidates.
But to sum up after that digression: unless you add in voter ignorance,
equal-rank would vanish under the model, and truncation seems too
arbitrary to implement. Thus I'm staying with full rank. I may consider
the other things later, but I haven't even written the program to do
full rank yet.
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