[EM] Proportional Approval Method

[raphfrk at netscape.net]

Proportional Approval Method

I think this system is a way to do approval voting so that votes are 
split in a Droop Quota like way.

1) The weighting given to a vote is the product of the weightings for 
each elected candidate approved

2) All unelected candidates have a weighting of one

3) The votes are counted and the candidate with the most votes is 
considered elected.

4) A weighting is assigned to each elected candidate such that each if 
those weighting were used for the count, all elected candidates would 
have the same total number of votes as the most approved unelected 
candidate

5) Each candidate is assigned a weighting of 1 minus the above 
weighting the count is re-run until enough candidates have been elected

The question is if step 4 will give a unique weighting for each 
candidate.  Also, I have a feeling that that the calculation will get 
pretty complex.

Step 4 doesn't distinguish between candidates based on the order of 
election which is good (kinda like Meek's).

The theory behind it is that you work out what percentage of the vote 
is required to keep the candidate elected and then consider that part 
of the vote exhausted.  The remainder can be used in subsequent rounds.

So a simple example electing 2 candidates and 2 parties (A and B).  
Assume that all voters 100% support one party or the other.

Votes are

5: A1
50: A1 A2
40: B1 B2
5: B1

Round 1:

All candidates weighted at 1

Result:
A1: 55
A2: 50
B1: 45
B2: 40

Winner:  A1

Setting the weighting for A1 to 50/55 will reduce A1 to 50, but will 
also reduce A2 to 45.45, so A1's vote still doesn't match the 2nd 
highest vote.

Setting the weighting for A1 to 45/55 will reduce A1 to 45 which is 
equal the highest unelected candidate (B1):

Weight:  45/55 = 0.82
A1: 45
A2: 36.8
B1: 45
B2: 40

0.82 is the weight for A1 that if used will make all elected candidates 
equal to the highest unelected candidate.

A1 is therefore assigned a weight of 0.18 ( 1 - the above )

Round 2:

A1:  10
A2:  9.1
B1:  45
B2:  40

B1 wins

Thus A1, B1 are the winners and this is valid based on the Droop quota 
as both parties got more than 1/3.

Another example is with party A have more than 2/3 of the vote

Votes are

3: A1
67: A1 A2
25: B1 B2
5: B1

Round 1:

A1: 70
A2: 67
B1: 30
B2: 25

A1 wins as before.  However, because the win is stronger the weighting 
applied is not as much of a penalty.

The weight is 30/70 to match A1 with the highest non-elected candidate 
(B1).

Weight:  30/70
A1: 30
A2: 28.71
B1: 30
B2: 25

Thus the weighting applied to votes with A1 is 40/70 = 0.5714

Round 2

A1:  40
A2:  38.3
B1:  30
B2:  25

A2 wins (as A1 already elected)

Thus party A gets both seats.

If party A has 2/3 and party B had 1/3, then the weighting for step 4 
would be 0.5.  This would reduce party A to 1/3 after reweighting which 
is equal to party B.  Thus, 2/3 is the required support necessary to 
get the 2 seats.

The question is if there is a unique set of weighting when there 2 or 
more elected candidates.

For 2 elected candidates, there are 3 types of votes.

Vab ( votes which vote for both of them)
Va (votes for candidate a)
Vb (votes for candidate b)

The equation would be

ab(Vab) + aVa = Target
ab(Vab) + bVb = Target

This should work as its 2 unknowns with 2 equations.  For higher 
numbers of candidates, there is still one equation for each unknown.  
However, it gets more complex.

Also, there is the problem that there is a moving target.  As was seen 
in the first example, the initial 2nd place candidate's votes were 
reduced so that in the end it was the 3rd candidate (B1) who the winner 
was matched against.

Finally, if there are candidates each polling station would have to 
announce more counts.  For one candidate, it just needs to announce a 
total for each candidate.  For 2 candidates, it also needs to announce 
totals for each pair.  It would probably be easier just to do the 
counting for each round after the results of the previous round.