Phragmén's method for unordered (approval) ballots can be presented in several equivalent ways.
Load balancing Each elected candidate creates a "load" of 1 unit. The load of a candidate must be born by voters who support him. The goal is to find a committee for which the load can be divided among the voters in the most "balanced" way. Depending on the exact definition of "balanced" several rules are possible: •
Leximax-Phragmen: Minimizing the maximum load, and subject to that the second-maximum load, etc. (using
lexicographic max-min optimization). •
Leximin-Phragmen: Maximizing the minimum load, and subject to that the second-minimum load, etc.. •
var-Phragmen or '''Ebert's method''': Minimizing the
variance of the load. Each of these variants has two sub-variants: • A
global optimization variant, which is usually NP-hard to compute; • A
sequential variant, in which candidates are selected sequentially, and in each turn, the next elected candidate is the one who attains the optimal measure among all candidates (i.e., a
greedy algorithm). Phragmen's original method is the sequential method that minimizes the maximum load, which is currently known as
Seq-Phragmen. In practice, the rules that have the best axiomatic guarantees in the global-optimization category are leximax-Phragmen and var-Phragmen. Among the sequential variants, the best guarantees are given by Seq-Phragmen. Phragmen illustrated his method by representing each voter as a vessel. The already-elected candidates are represented by water in the vessels. To elect another candidate, 1 liter of water has to be poured into the vessels corresponding to voters who voter for that candidate. The water should be distributed such that the maximum height of the water is as small as possible.
Virtual money Seq-Phragmen can alternatively be described as the following continuous process: • Each voter starts with 0 virtual money, and receives money in a constant rate of 1 per day. • At each time
t, we define a not-yet-elected candidate
x as
affordable if the total money held by voters who approve
x is at least 1. • At the first time in which some candidate is affordable, we choose one affordable candidate
y arbitrarily. We add
y to the committee, and reset the virtual money of voters who approve
y (as they have now "used" their virtual money to fund
y). • Voters keep earning virtual money and funding candidates until all
k committee members are elected.
Examples Party-list The following simple example resembles party-list voting. There are k=6 seats and 9 candidates, denoted a,b,c,d,e,f,g,h,i. There are 63 voters with the following preferences: 31 voters approve a,b,c; 21 voters approve d,e,f; and 11 voters approve g,h,i.
Round 1 Let candidates be given a score equal to the reciprocal of the number of approvers. We get the following: We select the candidate with the lowest score. In this case, it's tied between a, b, and c. Let's suppose that a was selected.
Round 2 Now, candidate scores are increased by the product of 1/31 (a's score) and the proportion of approvers who also approve of a. We select the candidate with the lowest score. In this case, it's tied between d, e, and f. Let's suppose that d was selected.
Round 3 Now, candidate scores are increased by the sum of the product of 1/31 (a's score) and the proportion of approvers whose 'most recent' approval was a, and the product of 1/21 (d's score) and the proportion of approvers whose 'most recent' approval was d. We select the candidate with the lowest score. In this case, it's tied between b and c. Let's suppose that b was selected.
Round 4 Now, candidate scores are increased by the sum of the product of 1/31 (a's score) and the proportion of approvers whose 'most recent' approval was a, and the sum of the product of 1/21 (d's score) and the proportion of approvers whose 'most recent' approval was d, and the product of 2/31 (b's score) and the proportion of approvers whose 'most recent' approval was b. We select the candidate with the lowest score. In this case, it's tied between g, h, and i. Let's suppose that g was selected.
Round 5 We select the candidate with the lowest score. In this case, it's tied between e and f. Let's suppose that e was selected.
Round 6 We select the candidate with the lowest score, which is c. • Voters start earning money at a fixed rate of 1 per day. After 1/31 or ~0.0323 days, the 31
abc voters have 0.0323 each, so together they can fund one of their approved candidates. One of a,b,c is chosen arbitrarily; suppose it is a. • After 1/21 or ~0.0476 days, the 31
abc voters have only ~0.015 each, but the 21
def voters have 0.0476 each, so together they can fund one of their approved candidates. One of d,e,f is chosen arbitrarily; suppose it is d. • After ~0.0645 days, the
abc voters again have 0.0323 each, so they buy another one of their approved candidates, say b. • After 1/11 or ~0.0909 days, the
ghi voters have 0.0909 each, so together they can fund one of their approved candidates, say g (at this point, the
abc voters have only 0.0264 each and the
def voters have 0.0434 each, so none of them can by another candidate). • After 0.0952 days, the
def voters again have 0.0476 each, so they can buy another candidate, say e. • After 0.0968 days, the abc voters again have 0.0323 each, so they can buy another candidate, say c. The final committee is a,b,c; d,e; g. Note that each "party" is represented approximately in proportion to its size: 3 candidates for 31 voters, 2 candidates for 21 voters, and 1 candidate for 11 voters.
Small, non-{party-list} As an example without a party structure, consider the following instance with 4 candidates, denoted by a,b,c,d, and 5 voters with approval sets 1: a; 2: b; 3: b and c; 4: a,b, and c; 5: d.
Round 1 Again, let candidates be given a score equal to the reciprocal of the number of approvers. We get the following: We select the candidate with the lowest score, which is b.
Round 2 Now, candidate scores are increased by the product of 1/3 (b's score) and the proportion of approvers who also approve of b. We select the candidate with the lowest score, which is a.
Round 3 Now, candidate scores are increased by the sum of the product of 1/3 (b's score) and the proportion of approvers whose 'most recent' approval was b, and the product of 2/3 (a's score) and the proportion of approvers whose 'most recent' approval was a. We select the candidate with the lowest score. In this case, it's tied between c and d. • Voters again start earning money at a fixed rate of 1 per day. After 1/3 days, the approvers of b have enough to buy b, resetting their money to 0, leading to a money distribution of (1/3, 0, 0, 0, 1/3). • After 2/3 days, the money distribution is (2/3, 1/3, 1/3, 1/3, 2/3). Thus, the approvers of a can buy the candidate, with their money afterward being reset to 0 leading to a distribution of (0, 1/3, 1/3, 0, 2/3). • Finally, after 1 day, the money distribution is (1/3, 2/3, 2/3, 1/3, 1). Thus, either c or d can be bought according to the tie-breaking used. Hence, for the committee size k = 3 both {a,b,c} and {a,b,d} are valid seq-Phragmén committees.
Realistic Here is a more 'realistic' example. There are
k=3 seats and 6 candidates, denoted by A, B, C, P, Q, R. The ballots are: 1034 vote for ABC, 519 vote for PQR, 90 vote for ABQ, 47 vote for APQ. The winners are elected sequentially as follows:
Round 1 Again, let candidates be given a score equal to the reciprocal of the number of approvers. We get the following: We select the candidate with the lowest score, which is A.
Round 2 Now, candidate scores are increased by the product of 1/1171 (A's score) and the proportion of approvers who also approve of A. We select the candidate with the lowest score, which is Q.
Round 3 Now, candidate scores are increased by the sum of the product of 1/1171 (A's score) and the proportion of approvers whose 'most recent' approval was A, and the product of 327/192044 (Q's score) and the proportion of approvers whose 'most recent' approval was Q. We select the candidate with the lowest score, which is B. == Computation ==