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Hamilton's Method

Hamilton's Method

This is a Parliamentary Allocation method, also known as the Largest Remainder method with the Hare Quota.

The task is:

  • given
    • $p_i$ - the number of voters in favor of party $i$
    • $N$ - total number of parties, $i \in \{ 1, 2, ..., N\} \equiv P$
    • $n$ - total number of voters
    • the quota of $i$ is $q_i = S \cdot \cfrac{p_i}{n}$. Note that $q_i$ is a read number, not integer
  • allocate $S$ seats in parliament
    • $(s_1, ..., s_N)$ s.t. $\sum s_i = S$
    • $s_i$ must be an integer

So $s_i$ should be either $\lfloor q_i \rfloor$ or $\lfloor q_i \rfloor + 1$

  • it's always integer, so we either round down or up

Rule:

  • initially allocate $\lfloor q_i \rfloor$ seats
  • then order all parties by their decimal part of the quota
  • the party with the highest decimal part gets the seat

Formally,

  • $\forall i, j \in P: s_i = \lfloor q_i \rfloor + 1 \land s_j = \lfloor q_j \rfloor \iff q_i - \lfloor q_i \rfloor \geqslant q_j - \lfloor q_j \rfloor$
  • $i$ gets the additional seat if its decimal part is larger than $j$'s


Example

$S = 10$

$p_i$ $q_i$ $\lfloor q_i \rfloor$ $s_i$
$P_1$ 6373 6.373 6 6
$P_2$ 2505 2.505 2 2
$P_3$ 602 0.602 0 0
$P_4$ 520 0.520 0 0

There remain 2 places to allocate: $\sum_i s_i = 8, S = 10$

  • we order the parties by the decimal part of their quota:
  • ${\color{blue}{P_3: 0.602, P_4: 0.520}}, P_2: 0.505, P_1: 0.373$
  • so in this case $P_3$ and $P_4$ get the additional seats


$p_i$ $q_i$ $\lfloor q_i \rfloor$ $s_i$
$P_1$ 6373 6.373 6 6
$P_2$ 2505 2.505 2 2
$P_3$ 602 0.602 0 1
$P_4$ 520 0.520 0 1


It's not always possible to split the seats

$p_i$ $q_i$ $\lfloor q_i \rfloor$ $s_i$
$P_1$ 6373 6.373 6 6
$P_2$ 2512 2.512 2  ?
$P_3$ 603 0.603 0 1
$P_4$ 512 0.513 0  ?

We have a tie

  • ties are broken arbitrarily


Implementation

Python & Numpy

def hamilton_allocation(ratios, k):
    frac, results = np.modf(k * ratios)
    remainder = int(k - results.sum())
    
    indices = np.argsort(frac)[::-1]
    results[indices[0:remainder]] += 1
 
    return results.astype(int)

ratios here sum up to 1, and k is the total number of seats


Paradoxes

Alabama Paradox

At first we have only 7 positions to allocate: $S = 7$


$p_i$ $q_i$ $\lfloor q_i \rfloor$ $s_i$
$P_1$ 20 1.373 1 2
$P_2$ 34 2.333 2 2
$P_3$ 48 3.294 3 3
$\sum$ 102 6 7


But suddenly we have an additional seat: $S = 8$

$p_i$ $q_i$ $\lfloor q_i \rfloor$ $s_i$
$P_1$ 20 1.569 1 ${\color{red}{\fbox{1}}}$
$P_2$ 34 2.667 2 3
$P_3$ 48 3.765 3 4
$\sum$ 102 6 7


$P_1$ loses one seat!

  • no Monotonicity!
  • even with one additional vote $P_1$ cannot regain the seat!


$p_i$ $q_i$ $\lfloor q_i \rfloor$ $s_i$
$P_1$ 21 1.631 1 1
$P_2$ 34 2.641 2 3
$P_3$ 48 3.728 3 4
$\sum$ 103 6 7


This is called the Alabama paradox: increasing in the number of seats causes a party to lose a sit


Population Paradox

$S=20$

$p_i$ $q_i$ $\lfloor q_i \rfloor$ $s_i$
$P_1$ 1786 9.654 9 10
$P_2$ 1246 6.719 6 7
$P_3$ 671 3.627 3 3
$\sum$ 3700 18 20


But assume that we have forgotten about additional 65 votes:

  • 19 votes for $P_1$
  • 40 votes for $P_2$
  • and only 6 votes for $P_3$
$p_i$ $q_i$ $\lfloor q_i \rfloor$ $s_i$
$P_1$ 1805 9.588 9 9
$P_2$ 1283 6.815 6 7
$P_3$ 677 3.596 3 4
$\sum$ 3700 18 20

Even though $P_1$ was winning and received more additional votes than $P_3$, $P_3$ takes one seat from $P_1$

So the outcome is not robust and it is possible to manipulate the result.


Population Transfer Paradox

Shows that Independence to Third Alternatives is not satisfies

$S = 5$

$p_i$ $q_i$ $\lfloor q_i \rfloor$ $s_i$
$P_1$ 14 2.593 2 3
$P_2$ 10 1.667 1 2
$P_3$ 3 0.556 0 0
$\sum$ 27 3 5


Suppose $P_2$ lost one vote, and $P_3$ received this vote:

  • That has an impact on $P_1$!
$p_i$ $q_i$ $\lfloor q_i \rfloor$ s_i
$P_1$ 14 2.593 2 2
$P_2$ 9 1.667 1 2
$P_3$ 3 0.741 0 1
$\sum$ 27 3 5


$P_1$ loses one seat, even though its position remained unchanged


Links

See also

Sources

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