I would like to know if there are any serious flaws, and if so, how the method can be improved.
AN EXAMPLE
Using some random ballots, wherein Party X={A,B,C} and Party Y={D,E,F} are clone sets.
A>B>C>E>F>D A>B>C>E>F>D A>C>B>E>F>D A>C>B>F>E>D B>A>C>D>E>F B>A>C>E>F>D C>A>B>F>D>E C>B>A>F>E>D D>E>F>A>B>C D>F>E>A>C>B F>D>E>B>A>C F>D>E>B>A>C
First, count the first-choice votes, and elect candidates that meet quota (in this case, both the Hare and Droop quotas are 4 votes) as in STV.
A: 4 -> elected with a surplus of 0 B: 2 C: 2 D: 2 E: 0 F: 2
Next, construct a pairwise matrix for all non-elected candidates, weighting the ballots as in the Newland-Britton system. In this example, A exactly met the quota, so A's ballots are not transferred at all, so there are effectively only 8 ballots left:
B>C>D>E>F B>C>E>F>D C>B>F>D>E C>B>F>E>D D>E>F>B>C D>F>E>C>B F>D>E>B>C F>D>E>B>C
The corresponding pairwise matrix is:
B C D E F B 0 5 4 4 4 C 3 0 4 4 4 D 4 4 0 6 3 E 4 4 2 0 3 F 4 4 5 5 0
Next, elect the Condorcet winner (or the winner of a Condorcet completion method) from this matrix. In this case, there is no Condorcet winner; there are 2 weak Condorcet winners, B and F. Suppose that F is elected.
Weight F's ballots so that the number of total votes is reduced by one quota. Note that this gives the ballots a negative weight.
B>C>D>E B>C>E>D C>B>D>E C>B>E>D D>E>B>C D>E>C>B D>E>B>C weight = -1 D>E>B>C weight = -1
(Note: The last 2 ballots are still considered F ballots, not D ballots, so at this point D is considered to have 2 votes, not zero, which would complicate things a great deal if D were elected.)
The new pairwise matrix is:
B C D E B 0 1 4 4 C 3 0 4 4 D 0 0 0 2 E 0 0 2 0
C is the new Condorcet winner, and gets elected. The winning committee is ACF, which contains 2 members of Party X and 1 member of Party Y, as expected.
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