# [EM] Dominant mutual third burial resistance

Kristofer Munsterhjelm km_elmet at t-online.de
Mon Jun 13 04:56:51 PDT 2022

```So I've been thinking a bit more about DMTBR. It seems that, in trying
to construct a method that passes actual DMTBR (and not just burial
resistance for the DMT *candidate*), I've been trying to achieve
something much more difficult than Benham was thinking of.

See e.g.
http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-April/081042.html
where he defines the weak burial resistance criterion, which seems to be
the same as what I've been calling DMTCBR, except only defined for three
candidates.

He then says:

> (You can probably generalize
> this from one candidate to a set of candidates, which
> I earlier defined as a "dominant mutual third" set.)

which would be DMTBR proper.

I also found something that I think is a snag to DMTBR. I'd like to get
feedback on this :-)

But it seems to not always be the case that if A wins and B>A voters
lower A below D, with A being in the innermost DMT set and D not, then
if B wins, that's a DMTBR failure. Consider this:

1: B>A>C>D  (ballot 1)

Suppose the innermost DMT set is {A, B, C} and A wins the election that
this ballot is a part of. Then burying like this:

1: B>C>D>A  (ballot 2)

doesn't just lower A below D, but also below C. So it's quite possible
that the weaker burial:

1: B>C>A>D  (ballot 3)

would make B win. Since DMTBR is about burying a candidate under someone
not in the innermost DMT set, it makes no guarantees about changing the
order of candidates *in* the innermost DMT set. So if going from ballot
1 to 2 makes B win, then it shouldn't be a DMTBR failure unless going
from ballot 1 to 3 *doesn't* make B win.

However, Smith-IRV seem to survive a lot of ballot 1 to ballot 2
transformations, so there may be a stronger burial resistance criterion
in there somewhere.

Now there's a stronger sense where going from ballot 1 to this ballot:

1: B>D>A>C  (ballot 4)

is a failure, because first we're lowering C:

1: B>A>D>C  (ballot 5)

then we're lowering D. If going from ballot 4 to ballot 5 changes the
winner from A to D, then it's a plain DMTBR failure. However, for it to
be okay to go from ballot 1 to ballot 4, we need a sort of "independence
of non-DMT candidate burial" criterion, where lowering someone (not
necessarily a winner) who's in the DMT set under someone who isn't
doesn't change the outcome.

I think Benham passes the independence criterion because all non-DMT
members get eliminated before the last DMT member is eliminated. So
perhaps we should just incorporate the independence criterion to keep
DMTBR as strong as possible. Any ideas?

On a final note, if the criteria are to be kept separate, then my
Smith,IRV DMTBR disproof needs to be modified. I said that going from

14: L>C>R>X>Y>Z
10: R>C>L>Y>Z>X
7: C>L>R>X>Z>Y
15: Y>Z>X>L>C>R
15: Z>C>R>L>Y>X
7: Z>X>C>R>L>Y
22: Y>X>R>C>L>Z

(where the DMT set is {L, C, R} and C wins being the CW), to

14: L>X>C>R>Y>Z   <-- burial
10: R>C>L>Y>Z>X
7: C>L>R>X>Z>Y
15: Y>Z>X>L>C>R
15: Z>C>R>L>Y>X
7: Z>X>C>R>L>Y
22: Y>X>R>C>L>Z

enlarges the Smith set to contain every candidate and so Smith,IRV
elects L. The problem is that the change buries C (the winner) under X,
but it also buries R under X, which is not a DMTBR thing, that's an
independence thing (since R isn't a winner). But this modification
should work:

2:L>C>X>R>Y>Z
14:L>C>R>X>Y>Z
10:R>C>L>Y>Z>X
7:C>L>R>X>Z>Y
14:Y>Z>X>L>C>R
14:Z>C>R>L>Y>X
9:Z>X>C>R>L>Y
20:Y>X>R>C>L>Z
2:Y>R>L>Z>X>C

the DMT set is {L, C, R} and C is the CW, then:

2:L>X>C>R>Y>Z		<-- burial
14:L>C>R>X>Y>Z
10:R>C>L>Y>Z>X
7:C>L>R>X>Z>Y
14:Y>Z>X>L>C>R
14:Z>C>R>L>Y>X
9:Z>X>C>R>L>Y
20:Y>X>R>C>L>Z
2:Y>R>L>Z>X>C

and the Smith set contains every candidate, so Smith,IRV elects L. (But
an elegant example, this ain't!)

-km
```