[EM] new working paper: (edit/second thought)

[Kristofer Munsterhjelm]

Kevin Venzke wrote:
> Hi Kristofer,
> 
> --- En date de : Sam 19.2.11, Kristofer Munsterhjelm <km-elmet at broadpark.no> a écrit :
>> However, if the method passes
>> LNHarm, then, to quote Woodall's definition, "adding a later
>> preference should not harm any candidate already listed". In
>> other words, because later ranks can't harm A, both
>> A>B>C>D (the honest ballot) and A>C>D>B
>> (the maximally burying ballot) has the same effect on
>> whether or not A wins, which is the effect that a bare-A
>> vote gives. Since you can't rig the field in favor of A by
>> rearranging later ranks, and burial tries to get A to win by
>> doing just that, LNHarm secures a method against this kind
>> of burial.
> 
> It sounds like you're attributing too much to LNHarm. It's quite possible
> under a LNHarm method for A>B>C>D to elect B and A>C>D>B to elect A. Here
> is a DSC example:
> 
> 49 B
> 17 A>C>D
> 3 A>B>C
> 17 C>D>A
> 14 D>A>C
> 
> B wins. But change the 3 votes to A>C>D and A wins. (If I made no errors.)

Let's see. We have

100 ABCD
49  B
48  ACD
20  A
17  CD
17  C
17  AC
14  D
14  AD
3   ABC

so B wins because of the second elimination. If the 3 voters vote ACD 
instead, we have

100 ABCD
51  ACD
49  B
20  A
...

so ACD gets affirmed (knocking out B), then B is skipped and A is elected.

But note that those voters can't make A win by truncating to a bullet 
vote. If they do so, A will have 23 votes which is not enough. That 
seems to support that there are two kinds of burial, the kind that makes 
use of LNHarm failure and the kind that makes use of LNHelp failure.

The LNHarm failure type burial exists if someone who votes A>B>C>D could 
make A win by merely truncating to a bullet vote. If the method passes 
LNHarm, then (I think) if it is the case that an A-only bullet vote 
makes A win, the voter can't make A lose by adding additional ranks.

The LNHelp failure type burial exists if A *doesn't* win when 
truncating, but burial makes A win. In this case, appending later ranks 
does help A win, which is a violation of LNHelp.


Let's check that with your and JGA's DSC examples. In your example, as 
I've shown, truncating A doesn't make A win. So if burial works (and we 
know it does), it should be of the LNHelp failure type -- which again 
seems reasonable since DSC fails LNHelp.

JGA's example is more tricky, since DAC and DSC is the same when there's 
no partial rank, yet the burial works both on the LNHarm and LNHelp 
kind. But let's look at it.

40: A>B>C
41: B>A>C
10: C>A>B

which is

91 ABC
81 AB
41 B
40 A
10 AC
10 C

so B wins at the third set since it's the only one remaining. Then the 
A>B>C voters bury by switching to A>C>B and we have:

40: A>C>B
41: B>A>C
10: C>A>B

91 ABC
50 AC
41 AB
41 B
40 A
...

and AC eliminates B, then AB excludes all but A (and B, which is already 
excluded), so A wins.

If the method is DSC, then truncating A would lead to

40: A
41: B>A>C
10: C>A>B

91 ABC
41 AB
41 B
40 A
10 AC
10 C

So truncation doesn't help, thus it's not the kind of burial nullified 
by LNHarm.

Similarly, if the method is DAC, truncating A would lead to

91 ABC
81 AB
50 AC
41 B
40 A
10 C

where the 81 AB would exclude C, and the 50 AC then excludes all but A, 
so A wins. So truncation here does help, so it's not the kind of burial 
nullified by LNHelp.

If I'm right about all that, it would show why a method like FPP and IRV 
(and all the variants of IRV one might make by below-mean elimination, 
below-median elimination, etc) would be impervious to burial. Because 
they meet LNHarm, there's no point in burying when truncation would 
work, and because they meet LNHelp, there's no point in burying when 
truncation won't work, either, and since truncation can either work or 
not work, that means there's no point in burying, period.

The degree to which LNHelp or LNHarm alone limits burial would depend on 
how often truncation works. But perhaps that is just a consequence, i.e. 
there is some sort of property that makes a method more or less burial 
resistant, and that property shows itself as LNHelp and LNHarm together 
if the resistance is 100%, and varying degrees of truncation resistance 
combined with the appropriate LNH if less than 100%.

>> Pushing B to
>> the bottom won't make A win, and filling stuff after a
>> truncated ballot won't make A win either. Yet that pair
>> comes at a great price: one can't have all of LNHelp,
>> LNHarm, mutual majority, and monotonicity (see http://www.mcdougall.org.uk/VM/ISSUE6/P4.HTM). Since
>> Smith implies mutual majority, any Smith-constrained version
>> of a method that meets both LNHs will fail to be monotone.
> 
> When you do Smith-constraint you already lose the LNHs though. If you
> want both LNHs you are basically stuck with FPP and IRV (and IRV variants
> like Craig Carey's IFPP).

True. One way of devising more strategy resistant methods would be to 
try and find sets that preserve more of the LNHs while limiting 
compromising incentive. Perhaps CDTT or CGTT would be such sets? I'm not 
sure. Of course, that trades away Smith-efficiency so it may not be 
worth it, and the methods still can't pass mutual majority and 
monotonicity while keeping the LNHs.

It's also possible to look at it from another direction. Limiting to 
Smith robs the methods of their strict LNH compliance. However, they're 
still more resistant to burial than most Smith-compliant methods 
considered. That's what JGA's paper is about. So there might be some 
"grand method" that is just as strategy-resistant yet monotone and 
Smith. Since Smith-compliance breaks LNH, Woodall's proof no longer 
holds and so the "grand method" wouldn't have to compete against 
perfection, only against methods like Smith-constrained IRV.

I have no idea what that grand method would be, or if it even exists, 
but there's the possibility.