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