[EM] Condorcet Flavored PR Methods

[Forest Simmons]

In this installment I would like to summarize (by example) the "how to" of
the current Condorcet Flavored Proportional Representation (CFPR) method
that takes into account the constructive criticism of Adam Tarr:


Here's information from one typical ballot:

C1>C2=C3>C4>C5=C6=C7>C8=C9>C10>C11>C12=C13=C14=C15

This information might have come from Grade Ballot information:

Grade Candidates
  A   C1
  B   C2,C3
  C   C4
  D   C5,C6,C7
  E   C8,C9
  F   C11
  G   C12
  H   C13,C14,C15

Now suppose that the two subsets being compared head-to-head are

S1={C2,C5,C6,C7,C14,C15} and S2={C1,C3,C8,C9,C12,C13}.

To score these subsets relative to this ballot, first we make a copy of
the ballot restricted to the members of the union of S1 and S2:

C1>C2=C3>C5=C6=C7>C8=C9>C12>C13=C14=C15

We see that these candidates are divided into six equivalence classes.

The respective subsets S1 and S2 have the following respective numbers in
each of the six classes:

X1=(0,1,3,0,0,2) and X2=(1,1,0,2,1,1) .

Now we take the difference of these two ordered six-tuples and let Y1 and
Y2 be the respective positive and negative parts (with negative signs
removed):

X1-X2=(-1,0,3,-2,-1,1) ,  Y1=(0,0,3,0,0,1), and  Y2=(1,0,0,2,1,0).

Next we find the median positions in Y1 and Y2 respectively.


The higher and lower ranking medians M and m are in positions 3 and 4,
respectively.

Now we calculate k1 and k2, the number of candidates in S1 and S2,
respectively, ranked higher than M (while counting 1/2 for each candidate
with rank M):

            k1=2.5,  and k2=2 .

Next, we calculate j1 and j2, the number of candidates in S1 and S2,
respectively, ranked lower than m (counting 1/2 for each candidate on the
boundary):

           j1=2,  and  j2=3 .


Now the sums  k1+j2=5.5  and k2+j1=4


Since k1+j2 is greater than k2+j1, this ballot supports S1 over S2.

Since k1+j2 is an odd multiple of .5, the margin is calculated as
follows:

-ln(4)+1/.5+1/1/5+1/2.5+1/3/5+1/4.5+1/5.5 - (1+1/+1/3+1/4)


which is approximately equal to  .286793663 .


This last value is awarded to S1 in the head-to-head comparison of S1 to
S2, while S2 gets zero in this head-to-head comparison.

That's it.

Forest


----
For more information about this list (subscribe, unsubscribe, FAQ, etc), 
please see http://www.eskimo.com/~robla/em