[EM] More formal DMTBR/stepping stone criteria?
Kristofer Munsterhjelm
km_elmet at t-online.de
Thu Jun 23 09:46:20 PDT 2022
In a previous post, I said that I my simulator might have been
interpreting the DMTBR criterion much more strictly than makes sense,
and thus I may have been setting a much too hard task for myself.
So let's think of some DMTBR-related criteria.
When we're burying DMT candidates under some candidates not in the DMT
set, it would be really nice to have the following property so that the
relative ordering of non-DMT candidates don't matter:
"Independence from DMT-dominated alternatives": Suppose W is the winner.
Then removing some subset of the candidates not in the innermost DMT set
can't change the winner from W. (I'll just all the innermost DMT set
"the" DMT set from this point on.)
But Benham and IRV both fail (example by James Green-Armytage):
6: D>A>B>C
5: B>C>A>D
4: C>A>B>D
The Smith set and DMT set is {A, B, C}. IRV eliminates in this order: A
then C then D, and B is elected. Benham stops after eliminating A
because B is then the CW.
But if D is removed, we have:
6: A>B>C
5: B>C>A
4: C>A>B
Benham eliminates C and then A beats B pairwise and wins.
...
So some easier and/or more precise DMTBR criteria:
Winner-loser DMTBR (as usual, I'm not very good at devising names):
Suppose W is the winner, and some voter who ranks X above W also ranks W
just above a candidate Y who's not in the DMT set. Then if this voter
lowers W by changing the W>Y on his ballot into Y>W, that shouldn't
change the winner to X.
(By repeating the procedure, this is the same as: lowering W below a
group of candidates, none of whom are in the DMT set, should not change
the outcome to X.)
"Double set" DMTBR:
Suppose W is the winner, and some voter who ranks X above W ranks a
subset C of n candidates containing W consecutively above some
candidates Y_1...Y_k not in the DMT set. Then if that voter changes the
relevant part of his ballot ballot from C_1>C_2>...>C_n>Y_1>...>Y_k to
Y_1>...>Y_k>C_1>C_2>...>C_n, that should not get X elected.
Burial-wise DMT independence (probably very hard):
Same as double set DMTBR, but C doesn't have to contain W.
None of these allow the winner to be buried below someone else who's in
the DMT set, so e.g. if B is in the DMT set and someone who votes
A>W>B>C changes his ballot to A>B>W>C and the winner turns into A, then
that's not covered. I think Smith-IRV hybrids are in some cases
resistant to those shenanigans, too, but I can't formalize it.
On the positive side, that the criteria don't allow for this kind of
change may make it possible to construct something that passes winner or
set DMTBR by considering solid coalitions, or some kind of path that
originates in first preferences, then goes to second, third, etc.
I think Benham passes both winner and double set, and I would guess it
also passes burial-wise independence, but I'm not sure.
A perhaps even easier variant is this: If a voter ranks W last out of
every DMT set member on his ballot, then if that voter lowers W to last
place, the winner shouldn't change to someone that voter prefers to W.
It's also possible to weaken the criteria above by not counting ties,
e.g. if changing A>B>W>C into A>B>C>W changes the winner from W to {B,
C}, then that doesn't count as a violation: the entire winner set has to
be of candidates originally ranked above W by the every voter
participating in the burial.
For instance, Copeland:
2: A>B>C>D
2: B>A>C>D
1: C>D>A>B
A is the CW and DMT candidate and thus wins with certainty. Then the B
faction buries A:
2: A>B>C>D
2: B>C>D>A <-- burial
1: C>D>A>B
and the Copeland winners are B and C. If the buriers' preferences are
actually B>A>>>C>D, then that is not to their benefit, but if they are
B>>>>A>C>D, it is. The strict (original) interpretation wants burial to
not pay no matter what the buriers' utilities are, while the
ties-allowed interpretation lets the burial pass if there's a
possibility that proper tiebreaking can lead to a method that passes the
strict interpretation.
-km
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