[EM] New method: Single-Member Virtual District Condorcet by Candidate Correlation (SMVDCCC) (was: BTR-STV)

[Dan Bishop]

Abd ul-Rahman Lomax wrote:
> Yes, each of four seats would, ideally, represent 25% of the 
> population, if the assembly is to be a peer assembly, with each 
> member equitably having the same voting power. However, this is the 
> very problem we are trying to solve. If might seem that if could 
> divide the electorate into N factions such that each faction had a 
> Condorcet winner, then we could create a perfect proportional 
> assembly with N members. However, never forget that when an election 
> method has winners, it has losers. If the factions were, for example, 
> randomly chosen, minorities might end up completely unrepresented.
> 
> So how would you define the factions? Mr. Oneala says, "each 25%," 
> but what we have is a collection of voters, all equal to each other. 
> How does one define which voter is in which faction?

Well, if we could arrange the ballots in a left-to-right order (or an 
authoritarian-to-libertarian order, or any other possible 
one-dimensional political spectrum), it's a simple matter of chopping up 
the spectrum into N equal intervals.  The hard part is constructing a 
political spectrum from the ballots.  But I think I've found a way:

EXAMPLE

3 seats, ballots are:

12.5%: A>B>C>D>E
12.5%: B>A>C>D>E
12.5%: B>C>A>D>E
12.5%: C>B>D>A>E
12.5%: C>D>B>E>A
12.5%: D>C>E>B>A
12.5%: D>E>C>B>A
12.5%: E>D>C>B>A

STEP 1: Arrange the candidates in order of their "Beatpath Copeland" 
scores.  If there is a tie, break it at random.

In the example,
* A has a beatpath tie (assigned ½ of a Copeland point) with E: ½ point
* B has beatpath victories against A and E and a beatpath tie against D: 
  2½ points
* C has beatpath victories against A, B, D, and E: 4 points
* D has beatpath victories against A and E and a beatpath tie against B: 
2½ points
* E has a beatpath tie against A: ½ point

The Beatpath ranking is C>B=D>A=E.  After two coin tosses to break the 
ties, this becomes C>B>D>E>A.

STEP 2: Compute the candidate correlations

Using third-order correlation (definition is on Electowiki):

corr(A, B) = 87.5%
corr(A, C) = 25%
corr(A, D) = 37.5%
corr(A, E) = 50%

corr(B, A) = 87.5%
corr(B, C) = 87.5%
corr(B, D) = 25%
corr(B, E) = 37.5%

corr(C, A) = 25%
corr(C, B) = 87.5%
corr(C, D) = 87.5%
corr(C, E) = 25%

corr(D, A) = 37.5%
corr(D, B) = 25%
corr(D, C) = 87.5%
corr(D, E) = 87.5%

corr(E, A) = 50%
corr(E, B) = 37.5%
corr(E, C) = 25%
corr(E, D) = 87.5%

STEP 3: Construct an ordering of the candidates.  Add candidates in the 
order computed in step 1.  Each candidate is added to its 
most-correlated end of the spectrum.

Recall that the beatpath ranking in the example is C>B>D>E>A.  Start 
with the winner C.  Next, put the second-place candidate (B) next to C, 
giving B-C.

D is 25% correlated to B but 87.5% correlated to C.  Clearly, C is the 
better match, so we put B on C's end, giving B-C-D.

Similarly, E is 37.5% correlated to B but 87.5% correlated to E.  E goes 
at D's end, giving B-C-D-E.

A is 87.5% correlated to B but only 50% correlated to E, so A goes at 
B's end.  This gives us a political spectrum of A-B-C-D-E, as you might 
have expected.

STEP 4: Assign ballots to positions on the political spectrum found in 
step 3.  Ballots can go between candidates.

One way to do this is to put each ballot at the position of its 
top-ranked candidate.

    A   B   C   D   E
---*---*---*---*---*---
    ^   ^   ^   ^   ^
    |   |   |   |   \______ E>D>C>B>A ballots
    |   |   |   \__________ D>C>E>B>A and D>E>C>B>A
    |   |   \______________ C>B>D>A>E and C>D>B>E>A
    |   \__________________ B>A>C>D>E and B>C>A>D>E
    \______________________ A>B>C>D>E


A simple refinement is to sort the ballots within each group by their 
second-choice candidate.  In left-to-right order,

A>B>C>D>E
B>A>C>D>E
B>C>A>D>E
C>B>D>A>E
C>D>B>E>A
D>C>E>B>A
D>E>C>B>A
E>D>C>B>A

STEP 5: Split the sorted list of ballots into N equal virtual districts.

If there are two seats to elect, the example ballots split into

Virtual district #1:

12.5: A>B>C>D>E
12.5: B>A>C>D>E
12.5: B>C>A>D>E
12.5: C>B>D>A>E
Winner: B

Virtual district #2:

12.5: C>D>B>E>A
12.5: D>C>E>B>A
12.5: D>E>C>B>A
12.5: E>D>C>B>A
Winner: D