Cumulative voting: Difference between revisions
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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. |
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. |
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=== Example === |
==== Example ==== |
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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 |
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 |
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