Condorcet tally shortcut?

[Mike Ossipoff]

Yes that seems like it would speed the count, but I'd add that
the alternatives should first be ordered according to their
1st choice vote totals, their Plurality counts, and the
winner-so-far should take on the highest alternative in
that list that it hasn't taken on yet.

This is especially important in hand-counts, or show-of-hands
pairwise voting in meetings, of course. For a computer program
I don't know if it's worth the added programming & explaining
complexity to reduce the count labor in that way, since the
computer can do a Condorcet count so fast anyway.

***

For the Smith set, my procedure included a shortcut, intended
mostly to save labor in hand-counts: Not only is something admitted
to the Smith set if it beats or ties something in the Smith set,
but also if it is beaten by as few as (or beats as many as)
something in the Smith set. (It seems to me that I found, at the
time, that one could use either how many it's beaten by or
how many it beats).

I'd arrange everyting according to how many it beats (or is
beaten by), as the beginning of the procedure.

The thing that beats most, or the set that's tied in that
respect, is the initial set. 

Then, everything is checked to determine if it beats a
Smith set member. If something does, then it & everything
that beats as many as it (or is beaten by as few as it)
is admitted to the Smith set.

Then, again, check everyting not in the Smith set to find
whether it beats something in the Smith set. Etc.

***

Yes, I'd considered that it might be best, when checking things
for whether they beat something in the Smith set, to start
with the things that beat the most (or are beaten by fewest).
That would of course speed finding that can do that. 

But, on the other hand, I also considered doing the opposite,
and starting from the bottom of the list, with the things beaten
by most, or which beat fewest, because then, when something is
admitted to the Smith set, and everything that beats more is
also admitted, that admits a lot more alternatives. I don't
know which would be better. That, as you said, is something
to maybe find out from spatial studies.

***

One reason why I no longer suggest the Schwartz set is because
I couldn't find an efficient procedure for finding the
Schwartz set by a systematic algorithm in a reasonable amount
of time when the Smith set is large. The Schwartz set search
can be confined to the Smith set, but if the Smith set is
large, the Schwartz set, it seems to me, could  quickly
become uncomputable for practical purposes. At least that
seems possible, because the only way I could find to 
determine the Schwartz set was by following every possible
beat-path, which reminded me of the "traveling salesman
problem" which becomes uncomputable when the number of cities
becomes large.

Because the Schwartz set isn't important for public elections,
I'm certainly not suggesting that you spend valuable time
working on an efficient Schwartz algorithm. I just though
I'd mention this though.

Of course I suppose it would be rare for the Smith set
to be large enolugh to cause problems in Schwartz determination.
Especially when EM needs to vote on a small set of alternatives.

I certainly wouldn't suggest the use of the Schartz set for
the vote Steve suggested, partly because it's more involvedto exsplain 
to the voters, and partly because conceivably there could be
a large Smith set, and a difficult-to-compute Schwartz set,
at least for hand-counts, and maybe (though maybe not likely)
even for computer counts.

Mike


 

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