On Jun 6, 2009, at 11:59 AM, Paul Kislanko wrote:
Besides the obvious problem with the notion of a fraction of a bit,
you're
still confusing the number of possible ballots with the amount of
information conveyed by a single ballot.
There's no problem, really, with fractional bits. It's useful to be
able to say that a ballot requires (say) 2.5 bits (rather than 3),
because it tells us that we can represent 10 ballots in 25 bits
(rather than 30).
We're just generalizing the number of bits to log2(n). When n is a
power of two, we get an integer; else not.
If there are 3 candidates, in approval a ballot only needs 3 bits.
Ranked
ballots need to carry the order the voter selected, and that
requires 2 bits
per alternative. I.e for ballots with ABC, you need 11, 01, 10 to
indicate
B>C>A. You cannot do that with fewer than 6 bits, even though it
only takes
3 bits to count the 3! = 6 possible ballots.
As has already been pointed out, all we need is a lookup table with
six entries for the six possible ballots.
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