[EM] Alternate view of Phragmén´s method

Bjarke Dahl Ebert bjarke2003 at trebe.dk
Thu Oct 2 14:30:02 PDT 2003

Dear all,

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

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 

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.

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

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.

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.

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 
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?

Kind regards,

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