[EM] More easily hand-counting three-slot Condorcet

Kevin Venzke stepjak at yahoo.fr
Thu Nov 4 20:48:50 PST 2004

Hello all,

I accidentally came up with an easier way to hand-count 
three-slot Condorcet.  By "easier" I mean fewer tally marks 
(potentially by far) per ballot.

The intuitive way to count Condorcet methods is to take a
ballot, and mark a vote in a matrix at X,Y for every candidate
X ranked above another candidate Y.

If there are three ranks, and R1 is the number of candidates
placed in the first rank, R2 for second, and R3 the number of
unmarked candidates, then using the above method the manual
counter will have to mark

(R1*R2) + (R1*R3) + (R2*R3) tallies.

E.g.: For the ballot AB>CD>EF, mark
A,C A,D A,E A,F;
B,C B,D B,E B,F;
C,E C,F;
D,E D,F.  12 tallies.

Consider this method instead: We still have the matrix, and
also use another column (or perhaps the diagonal, like X,X)
to store "approval," which will be the number of non-last
votes received by a candidate.

For a given ballot, mark a vote at X,X for every candidate X
placed in the first or second slot.  Mark a vote at X,Y for
every candidate X placed in the first slot above a candidate
Y placed in the middle slot.

This means that for each ballot, we will only make:

R1 + R2 + (R1*R2) tallies.

E.g.: For the ballot AB>CD>EF, mark
A,A B,B C,C D,D;
A,C A,D;
B,C B,D.  8 tallies.

Once we have counted all the ballots, we have enough info
to determine whether there is a Condorcet winner.

A beats B pairwise if A,A - B,A > B,B - A,B.

In my example ballot, the reduction is only 33%.  But if
we suppose that there may be many candidates unmarked by
almost all of the voters, the savings could be huge.  Look
at a bullet vote with many candidates:


By the usual counting method this is 25 tallies!  But using
the method I suggest here, there is just 1 tally, at A,A.

I looked long and hard, but I don't believe it is possible
to recover the magnitudes of victory.  I think, though, that
the approval measure is almost as good.  I also believe that
when the approval of the pairwise winner is used as the
defeat strength, then all methods I know of will elect the
same candidate (RP, Schulze, Raynaud, MinMax, probably River).

I believe all rules are then equivalent to "Elect the least-
approved candidate who pairwise beats every candidate with
greater approval."


9 A>B>C
8 B>C>A
7 C>A>B

Hand-counted matrix will be:
    A    B    C
A  16    9    0
B   0   17    8
C   7    0   15

16-0 > 17-9, so A>B pairwise
17-0 > 15-8, so B>C pairwise
15-0 > 16-7, so C>A pairwise

Approval order is B>A>C.  Cycle is B>C>A>B.
Starting at the bottom, C can't win because B beats C.
A beats B pairwise and so A is the winner.

49 B>>GN
24 G>>BN
27 N>G>B

Hand-counted matrix will be:
    B    G    N
B  49    0    0
G   0   51    0
N   0   27   27

49-0 < 51-0, so B<G pairwise
51-27 < 27-0, so G<N pairwise
27-0 < 49-0, so N<B pairwise

Approval order is G>B>N.  Cycle is G>B>N>G.
N can't win because B beats N.
B can't win because G beats B.
So G is the winner.

I hope someone finds this interesting.

Kevin Venzke
stepjak at yahoo.fr


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