[EM] DMC Based Burial Resistance

[Forest Simmons]

Example 1.
49 C
26 A>B
25 B (Sincere B>A)

Respective IA's for A, B, C ... 26, 51, 49.
Respective MaxPS's: 26, 51, 49.

DMC winner X1=B. X2=A, defeating X1 26 to 25, X3=C, beating X2 49 to 26.
Note X1 beats X3 51 to 49.

As usual X1 is the manipulator favorite, X2 is the subverted sincere CW,
truncated/ side swipes by the bus X3.

Electing X3 punishes the B faction responsible for the false beat cycle.

A manual runoff between X1 and X2 results in a win for A.

The two stage manual runoff
(X1 vs X2) vs X3, that is (B vs A)vs C,  will (under rational well informed
voting) also be won by the Sincere CW, which is A.

How do we know this? By the theorem quoted in our previous message.

Example 2.

2 A>B
1 B>C
2 C>A

The pairwise win/ties are ...
A>B 4 to 1, B>C 3 to 2, and C>A 3 to 2.

The respective IA/MaxPS values are
4, 3, 3.

The DMC winner is X1= A, and we have X2=C, and X3=B.

We have noted that usually X3 is the "bus" ... and that would indeed be the
case if sincere votes in the A faction were 2 A>C, making C the buried CW
... buried under the bus B, that is.

So if our policy is to simply elect X3, then the burier's sincere anti-
favorite B gets elected ... they're not happy.

But what if the B supporters turned out to be the manipulators ... then
electing X3 would reward the manipulators ... which shows that the "more
often than not" assumption is not infallible.

Does the X1 vs X2 runoff rectify the situation?

Well, if the B faction is the faction guilty of creating the cycle by
unilateral burial, then A had to be the sincere CW that got buried under B.

So yes, the X1 vs X2 runoff does rectify the situation, since the sincere
CW is X1=A.

What if B were the sincere CW?

Then electing X3=B would neither punish nor reward the manipulating faction
... B would win regardless.

So our one stage runoff was good enough to rectify all unilateral burial
possibilities.

The two stage runoff shines when more than one faction manipulates, perhaps
in concert with other factions.

But as long as the sincere CW is among the three finalists, that two stage
runoff will elect it!

Is there any other manipulation rectification theorem like this?

We could re-cast it as ...
Any manipulation of an election that subverts a sincere CW ...  resulting
in a three member top cycle (that includes the Sincere CW) .... any such
mess can be rectified with a two stage runoff of the form X vs (Yvs Z).

If the Smith set resulting from subversion of a sincere CW has eight
members, then it can be rectified with a three stage runoff of the form ...

[(A vs B) vs (C vs D)] vs [(E vs F) vs (G vs H)].

If one of the eight is the sincere CW, then the order of the candidate
names doesn't matter ... all orders produce the unique sincere CW ... but
of course, it's better to have a well justified systematic way of ordering
them in general ... similar to our X1, X2, X3 method.

fws




On Sat, Jul 8, 2023, 1:30 PM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> The DMC candidate is the candidate with the least Implicit Approval that
> defeats every candidate with more IA.
>
> The original formulation of DMC was as Benham, IA: eliminate the lowest IA
> candidates one-by- one until there remains an undefeated candidate among
> the uneliminated.
>
> To base burial resistance on DMC, let X1 be the DMC candidate. Let X2 be
> the highest IA candidate to defeat X1, and let X3 be the highest IA
> candidate to defeat X2. [Assuming no undefeated candidate exists].
>
> More often than not, X1 will be the candidate of the buriers, X2 will be
> the buried candidate, and X3 will be the "bus" candidate under whom the X1
> faction buried X2.
>
> To punish the buriers, elect the bus X3.
>
> To fairly and squarely "unbury X2," i.e. restore the (likely) sincere CW
> to its rightful status, elect the winner of a sincere runoff between X1 and
> X2.
>
> To wean ourselves entirely from dependence on the "more often than not"
> assumption, do a sincere three candidate runoff of the form ... (X1 vs X2)
> vs X3.
>
> The first stage decides between two alternatives a1 and a2, where a1 is to
> elect the winner of a runoff between X1 and X2  (thereby rejecting
> alternative a2), and a2 is to reject a1 and elect X3.
>
> Optimal runoff strategy is to vote for a2 in the first stage of the runoff
> iff you prefer X3 to the projected winner of the a1 runoff.
>
> If a2 wins the first stage, then there is no second stage ... X3 is
> elected.
>
> If a1 wins the first stage, your optimum strategy in the second stage is
> to vote for your sincere preference between X1 and X2.
>
> If the voters are both rational and well informed about the preferences of
> the other voters, then they will vote according to the above strategy
> recommendations, and the Sincere CW will be elected, if one exists among
> the three candidates X1,X2, and X3.
>
> A slightly cleaner version DMC' of DMC can be formulated by replacing
> Implicit Approval with MaxPairwiseSupport: The DMC' candidate is lowest
> MaxPS candidate that defeats every candidate with greater MaxPS, etc.
>
> Examples ... coming soon!
>
> fws
>
>
>
>
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