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Re: RFD: Reviving Constitutional amendment: Smith/Condorcet vote tallying



On Thu, Oct 17, 2002 at 11:48:17AM -0400, Raul Miller wrote:
> On Thu, Oct 17, 2002 at 01:47:35PM +1000, Anthony Towns wrote:
> > YM "Schwartz set" here? [0] There might be a "Schulze set" of some sort?

Sorry, there's a "Smith set", not a Schulze set. So presumably we mean the
Schwartz set.

> Remember that "innermost unbeaten set" is an ambiguous term if there
> are any pairwise ties in an innermost unbeaten set.

(Well, it can be; it isn't necessarily, but point taken)

> > It'd probably be more intuitive to say "A dominates B if A beats B,
> > or there is some other option C, where C dominates B and A beats C" or
> > something similar, so it's clear which direction the beat path goes in.
> > That rephrases the above as: "An option A is said to be in the Schultz
> > set if there is no option B where both B dominates A, but A does not
> > dominate B".
> "Dominates" invites non-technical comparisons between the proposed
> mechanism and the existing mechanism.  I'd like to avoid that term
> if possible.

*shrug* Then how about "An option A is said to master an option, B,
if A beats B, or if there is some other option, C, where A beats C and
C masters B." ? Or "transitively beats" ?

> > >     5. All options which do not beat the default option by their
> > >        supermajority ratio are discarded, and references to them
> > >        in ballot papers will be ignored. 
> > >     6. If a quorum is required, there must be at least that many votes
> > >        which prefer the winning option to the default option. If there
> > >        are not then the default option wins after all. For votes
> > >        requiring a supermajority, the actual number of Yes votes is used
> > >        when checking whether the quorum has been reached.
> > Shouldn't the quorom be counted at the same time the supermajority is?
> The quorum mechanism is structurally different from the supermajority
> requirement.

Sorry, I don't mean at the _exact_ same time, I just mean before the counting
method is applied. ie:

	1. Collate votes.
	2. Remove options that don't meet the supermajority requirement.
	3. Remove options that don't meet the quorum requirement.
	4. Count votes, according to cloneproof SSD.
	5. Break ties, if necessary.

ATM the quorum is done last ("check the winner meets quorum, if not,
the default option wins").

> > >     7. If no option beats the default option, the default option wins.
> > Why this special case? The Perl program I wrote for this uses the
> > following system:
> To deal with the case of no votes and on a ballot with no quorum
> requirement.

In which case every option will be listed as tied, and the tie-breaking
rules can come into play.

> > #     1. Calculate Schwartz set according to uneliminated defeats.
> > #     2. If there are no defeats amongst the Schwartz set:
> > #           2a. If there is only one member in the Schwartz set, it wins.
> > #           2b. Otherwise, there is a tie amongst the Schwatz set.
> > #           2c. End
> > #     3. If there are defeats amongst the Schwartz set:
> > #           3a. Eliminate the weakest defeat/s.
> > #           3b. Repeat, beginning at 1.
> > 
> > It might make sense to say:
> > 	2a. If there is only one member in the Schwartz set, it wins.
> > 	2b. If the default option is in the Schwartz set, it wins.
> > 	2c. Otherwise, the voter with a casting vote may choose a
> > 	    winner from the remaining options, or may choose to let the
> > 	    vote be retaken.
> In other words, don't bother dropping weakest defeats?

Huh? The weakest defeats are part of 3 -- 2(a,b,c) are just alternative
versions of 2(a,b,c) from above.

> > >     8. If only one option remains in the schultz set, that option is
> > >        the winner.
> > >     9. If all options in the schultz set are tied with each other,
> > >        the elector with the casting vote picks the winner from the
> > >        schultz set.
> > "tied with each other" doesn't seem particularly well defined, IMO.
> > Every single pairwise comparison has to be exactly balanced, or already
> > discarded.
> I'm not at all clear what you're objecting to, here.  Is there something
> ambiguous about that phrasing?

It's reasonable to consider all the options in the Scwartz set to be
"tied with each other" -- dropping the weakest defeats is an automatic
tie breaker, casting votes are a manual tie breaker. I'd be inclined
to swap 10 and 9, and say "If there are any pairwise defeats, drop the
weakest...", and "If there are no pairwise defeats, then either there
are no pairs and we have a winner, or we break the tie by...".

> > >           c. The schultz set is then refigured with the Beats of the
> > >              weakest defeats eliminated. 
> > "refigured" isn't well defined.
> I'm having trouble understanding this objection as well.

It's not well defined -- you have to use context and intuition to make
a guess at what it means, and what you're meant to do.

> Hmm.. can you propose some alternate interpretations of "refigured"?

Not really, but I wouldn't've been able to propose any of the alternate
interpretations of any of the voting stuff in the constitution when it
was proposed, either -- and that didn't stop it from decaying into an
ambiguous incomprehensible mess within a couple of years.

I'd suggest we be completely clear about what's going on. Something like:

	We define the "Current Schwartz Set" initially as the Schwartz
	Set calculated from the defeats of all options that have not been
	eliminated (This set is non-empty, since the default option is
	never eliminated by supermajority or quorum requirements)

	At this point we:
		1. Eliminate all options not in the Current Schwartz Set
		   from consideration.
		2. If there are any pairwise defeats amongst members of
		   the Current Schwartz Set, we:
			2a. Choose the weakest such defeat(s) [define]
			2b. Eliminate them from consideration [define]
			2c. Recalculate the Current Schwartz Set, ignoring
			    any options that have been eliminated, and
			    ignoring any pairwise comparisons that have
			    been eliminated, and return to step 1.
		3. If there were no pairwise defeats, then:
			3a. If there was only a single option in the Current
			    Schwartz Set, it is the winner.
			3b. If the default option is in the Current Schwartz
			    Set, it is the winner.
			3c. Otherwise, the vote is considered a tie,
			    and the elector with a casting vote may choose
			    to declare any of the tied options the winner,
			    or may declare the default option the winner.

Apart from the two lines marked [define], I think that's fairly
unambiguous. (3b) replaces "7" in the draft, (3c) might be being too
generous to whoever gets a casting vote?

Cheers,
aj

-- 
Anthony Towns <aj@humbug.org.au> <http://azure.humbug.org.au/~aj/>
I don't speak for anyone save myself. GPG signed mail preferred.

 ``If you don't do it now, you'll be one year older when you do.''

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