`Dear all,`

`I have recently found a thread from 2002 from this mailing list dealing with "D'Hondt without lists".`

(see e.g. http://www.mail-archive.com/[EMAIL PROTECTED]/msg08214.html)

(see e.g. http://www.mail-archive.com/[EMAIL PROTECTED]/msg08214.html)

`It's about proportional election without party lists (or, you could say, the "lists" are determined by each voter)`

`I'm especially interested in Phragmén's method, as it seems "right" to me, which PAV doesn't.`

`Once nice thing about both Phragmén and PAV is that they degenerate into Approval Voting in the case of single-seat elections.`

One thing I don't like about PAV is that voters are "punished" for voting for candidates that are going to be elected anyway. This can tempt some voters into "letting the others ensure the victory of X" - not a very stable situation. Besides, PAV just feels wrong in my stomach ;-).

One thing I don't like about PAV is that voters are "punished" for voting for candidates that are going to be elected anyway. This can tempt some voters into "letting the others ensure the victory of X" - not a very stable situation. Besides, PAV just feels wrong in my stomach ;-).

`I have thought about a particular improvement of Phragmén, which seems even more "right" to me than plain Phragmén..`

`So, first let me describe Phragméns algorithm in terms that are closer to the modification I want to make, than the definition I have seen is.`

`--- Alternative description of Phragméns method ---`

`Each ballot contains an unordered list of candidates (like Approval).`

The result of the election is a set of N candidates, such that each winning candidate has a distribution of the total "candidate weight" of 1 across voters who voted for that candidate. That is, each voter who voted for the candidate, holds a specified part of his candidate weight, summing up to 1. The "voter weight" is defined as the sum of all the voter's candidate weight parts.

We want the weight of voters (the weight totalling N) to be shared as evenly as possible among the voters - corresponding to the elective power being almost evenly distributed.

Phragmén does this by minimizing the L-infinity norm of the vector of voter weights (this is my characterization, probably not his own).

This is done greedily, electing one candidate at a time, spreading his weight (of 1) over the voters with the smallest voter weight so far. Iteratively, the candidate that can keep the L-infinity norm lowest wins a round.

The result of the election is a set of N candidates, such that each winning candidate has a distribution of the total "candidate weight" of 1 across voters who voted for that candidate. That is, each voter who voted for the candidate, holds a specified part of his candidate weight, summing up to 1. The "voter weight" is defined as the sum of all the voter's candidate weight parts.

We want the weight of voters (the weight totalling N) to be shared as evenly as possible among the voters - corresponding to the elective power being almost evenly distributed.

Phragmén does this by minimizing the L-infinity norm of the vector of voter weights (this is my characterization, probably not his own).

This is done greedily, electing one candidate at a time, spreading his weight (of 1) over the voters with the smallest voter weight so far. Iteratively, the candidate that can keep the L-infinity norm lowest wins a round.

Example: voter#1: AB voter#2: A voter#3: C## Advertising

`In the first round, A is elected, because his weight of 1 can be shared among two voters, each getting a weight of 0.5. So the L-infinity norm after the first round is 0.5.`

In the second round, electing B would bring #1's weight up to 1.5, while electing C would bring #3's weight up to only 1. So C is elected.

In the second round, electing B would bring #1's weight up to 1.5, while electing C would bring #3's weight up to only 1. So C is elected.

I can see two ways to improve (in my opinion) Phragméns method:

(1) Why restrict ourselves to a sequential, greedy algorithm to find the optimal (lowest possible) L-infinity norm. Why not find the optimal winner-set from all possible sets of N candidates and all possible distributions of their weights among their voters. This is like replacing Sequential PAV with PAV. More computionally expensive, but more "right". It can serve as an ideal, with the sequential algorithm as a practical way to get close to that ideal.

(2) Why use the L-infinity norm? I would think L2 is "better", since it doesn't ignore everything of what goes on beneith the supremum value. Two distributions are not equally good, just because the have the same "maximum weight" at some voter.

I can see two ways to improve (in my opinion) Phragméns method:

(1) Why restrict ourselves to a sequential, greedy algorithm to find the optimal (lowest possible) L-infinity norm. Why not find the optimal winner-set from all possible sets of N candidates and all possible distributions of their weights among their voters. This is like replacing Sequential PAV with PAV. More computionally expensive, but more "right". It can serve as an ideal, with the sequential algorithm as a practical way to get close to that ideal.

(2) Why use the L-infinity norm? I would think L2 is "better", since it doesn't ignore everything of what goes on beneith the supremum value. Two distributions are not equally good, just because the have the same "maximum weight" at some voter.

When generalizing as in (1), the example above actually makes A+B, A+C and B+C equally good election results - all with an obtainable L-infinity norm of 1.

When changing as in (2) also, AC is best, with the norm of 1.22, whereas AB and BC both have an L2-norm of 1.41.

When generalizing as in (1), the example above actually makes A+B, A+C and B+C equally good election results - all with an obtainable L-infinity norm of 1.

When changing as in (2) also, AC is best, with the norm of 1.22, whereas AB and BC both have an L2-norm of 1.41.

Details: Electing AB: #1: A(0) B(1): total 1 #2: A(1): total 1 #3: C(0): total 0 norm: sqrt(1+1) = 1.41

Electing AC: #1: A(.5) B(0): total .5 #2: A(.5): total .5 #3: C(1): total 1 norm: sqrt(.5^2 + .5^2 + 1) = 1.22

Electing BC: #1: A(0) B(1): total 1 #2: A(0) #3: C(1): total 1 norm: 1.41

Note that Phragmén is a generalization of d'Hondt:

Imaging a two-party election, which in the Phragmén case is translated into each A-list voter voting for the (A1, A2, ...) list instead, and imagine that two seats should be filled.

In both d'Hondt and Phragmén, a party needs 2/3 of the votes to get both seats.

Unfortunately, with my modification (2), this is no longer so. Now a list needs 3/4 of the votes to win both seats.

Note that Phragmén is a generalization of d'Hondt:

Imaging a two-party election, which in the Phragmén case is translated into each A-list voter voting for the (A1, A2, ...) list instead, and imagine that two seats should be filled.

In both d'Hondt and Phragmén, a party needs 2/3 of the votes to get both seats.

Unfortunately, with my modification (2), this is no longer so. Now a list needs 3/4 of the votes to win both seats.

`Is this good or bad? :-)`

Do anyone know of ideas related to my (1) and (2) modifications?

Do anyone know of ideas related to my (1) and (2) modifications?

Kind regards, Bjarke

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