[EM] Divisor-based individual PR method?

[Kristofer Munsterhjelm]

Here's a method I constructed today - it works by the concept of a 
divisor method, except that instead of using parties, it uses solid 
coalitions in the DSC/DAC sense.

First, define the score for a potential council. The score is the error 
measure between the desired fraction of a coalition and the actual 
fraction of a coalition, for all such coalitions that exist.

The desired fraction is simply the number of voters that support this 
coalition (or set), divided by a specified constant which we'll call the 
divisor. The actual fraction of a coalition is the number of candidates 
in the potential council that also exist in the coalition in question.

This definition gives some leeway, since the divisor is not actually 
specified. The score for a potential council is the error when the 
divisor is chosen so that the error is minimized, subject to that the 
desired fraction of the coalition consisting of all the candidates is 
equal to the council size.

The council that has the best score (lowest minimal error) wins. 
Different error measures may be used - most natural would be root mean 
square error or the Sainte-Laguë index. If the Sainte-Laguë index is 
used, one should not round the desired fractions.

-

Let's show an example. I'll use an election from the PSC-CLE post:

	33 A>D>B>C
	33 B>D>A>C
	32 C>D>A>B
	 2 D>A>B>C

which gives

	100: A B C D	
	 68: A B D	
	 32: A C D	
	 34: A D	
	 33: B D	
	 32: C D	
	 33: A		
	 33: B		
	 32: C		
	  2: D		

(2,4) election. Let's say we want to find out the score of the council 
{B, D} with a divisor of 41, and the error metric is root mean square 
error. The line marked with L is the limit (constraint): since the 
coalition consists of all the candidates, the desired fraction must be 
equal to the number of seats, which it is in this case.

r(x) is the ordinary rounding function.

We get
			desired		actual			diff
	100: A B C D	r(100/41) = 2	2 (B and D) << L	0
	 68: A B D	r(68/41) = 2	2 (B and D)		0
	 32: A C D	r(32/41) = 1	1 (D)			0
	 34: A D	r(34/41) = 1	1 (D)			0
	 33: B D	r(33/41) = 1	2 (B and D)		1
	 32: C D	r(32/41) = 1	1 (D)			0
	 33: A		1		0			1
	 33: B		1		0			1
	 32: C		1		0			1
	  2: D		0		1			1

The squared error is 5, and so the RMSE is sqrt(0.5) ~ 0.7. This is, 
incidentally, the best score for BD.

Another example, this time with the council {B, C} and divisor 47:

			desired		actual			diff
	100: A B C D	r(100/47) = 2	2 (B and C) << L	0
	 68: A B D	r(68/47) = 1	1 (B)			0
	 32: A C D	r(32/47) = 1	1 (C)			0
	 34: A D	r(34/47) = 1	0			1
	 33: B D	r(33/47) = 1	1 (B)	 		0
	 32: C D	r(32/47) = 1	1 (C)			0
	 33: A		1		0			1
	 33: B		1		1			0
	 32: C		1		1			0
	  2: D		0		0			0

The squared error is 2, so the RMSE is sqrt(0.2), ~ 0.45. This is also 
the best score for BC.

For this particular election, the method evaluates as this:

	AB	0.447214 at 41
	BC	0.447214 at 46
	AC	0.547723 at 46
	BD	0.632456 at 41
	AD	0.707107 at 41
	CD	0.707107 at 46

As one can see, it doesn't elect the compromise candidate D anywhere; 
instead it ties AB and BC. Using the Sainte-Laguë index gives

	AB	 1.99788 at 41
	BC	 2.26691 at 44
	AC	 4.11046 at 41
	BD	22.264   at 41
	AD	23.9981  at 41
	CD	24.4542  at 41

and thus picks AB.

-

Is this method any good? It clearly takes a very long time to calculate, 
especially for large councils, unless it shares Webster's monotonicity 
criteria, in which case it outputs a proportional ordering, meaning that 
one can first do a single-winner election, then lock the winner and only 
test councils including the weinner for a two-winner election, and so on 
down.

It does seem to do so at least in this particular election with the 
Sainte-Laguë index as error measure.
Single-winner returns
	B	 3.6688 at 68
	A	 3.7051 at 67
	C	 4.9064 at 67
	D	35.2989 at 67

thus electing B (showing the method is not Condorcet, by the way).
Two winners elect AB (retaining B), and three winners return:
	ABC	 0.9993 at 30
	ABD	15.9226 at 29
	BCD	15.9776 at 29
	ACD	18.4313 at 29

thus electing ABC (retaining AB).