Cumulative voting: Difference between revisions

=== Clone independence ===

Every form of <math>\ell_p</math>-norm cumulative voting except <math>p=\infty</math> (plain [[Range voting]]) is vulnerable to vote-splitting.

For continuous cumulative voting, suppose each candidate can be given any score between 0 and 1. Then let <math>x = \frac{\sqrt[p]{2} - 1}{2\big(\sqrt[p]{2} + 1\big)}</math> and consider the election

* <math>\frac{1}{2}+x-\epsilon</math>: A (1) B (0)

* <math>\frac{1}{2}-x+\epsilon</math>: A (0) B (1)

for some <math>\epsilon</math> where <math>x > \epsilon > 0</math>. A wins by majority rule no matter the value of p.

Clone A into A1 and A2 so that everybody rates the two clones equal. The <math>\ell_p</math> normalization then leads to the first faction giving each A candidate <math>y = \frac{1}{\sqrt[p]{2}}</math> points each. As a result, the two A candidates tie if <math>\epsilon=0</math>, and B wins for any <math>\epsilon > 0</math>, which demonstrates the clone failure.

=== Example ===

Suppose we want to find a clone failure for <math>p=2</math> (quadratic voting). Then <math>x = \frac{\sqrt{2}-1}{2+2\sqrt{2}} \approx 0.086</math> and <math>y = \frac{1}{\sqrt{2}} \approx 0.707</math>. Let <math>\epsilon=0.001</math> to compensate for the roundoff error. Before cloning, the ballots are

* 0.585: A (1) B (0)

* 0.415: A (0) B (1)

where A wins 0.585 points to 0.415.

After cloning:

* 0.585: A1 (0.707) A2 (0.707) B (0)

* 0.415: A1 (0) A2 (0) B (1)

A1 gets 0.413 points, A2 the same, and B gets 0.415 points, thus making B the winner.