robert bristow-johnson wrote:
On Mar 20, 2010, at 12:08 PM, Dave Ketchum wrote:
Counting: Besides the N*N matrix,
i dunno why the common layout of the NxN matrix is popularly used. it
should be like a triangle, e.g. for the 2009 Burlington election:
[snip]
It's used to handle equal ranking. If nobody equal ranks, then the
preference for B above A is equal to (number of voters) - (number
preferring A to B). However, if some voters equal-rank, then this may
not be true. For instance:
1: A > B > C
1: B > A > C
1: A = B > C
gives
(row beats column)
A B C
A - 1 3
B 1 - 3
C 0 0 -
since the third voter expresses no preference between A and B. If you
were to just take a triangle:
A B C
A -
B 1 -
C 0 0 -
you would come to the conclusion that B>A is 2, which is not the case.
it seems to me that this is far more easy to see the result of each
pairwize race. it can be easily modified to provide <defeatStrength> (is
that what you mean by margins?).
The margins for A>B is the number of voters who preferred A>B minus the
number of voters who preferred B>A, or zero, whichever is greater.
Winning votes, the common alternative, counts the number of votes on the
strongest side: A>B is either the number of voters who preferred A>B, or
zero if more preferred B>A.
at a glance, this is much better for my eyes than the NxN matrix that
seems common for Condorcet results. if you wanted to sort by beat
strength, the data is right there.
You could show it explicitly. For instance:
Score for A:
Voters preferring A to B: 10
Voters preferring A to C: 8
Voters preferring A to D: 3
Score for B:
Voters preferring B to A: 10
Voters preferring B to C: 7
Voters preferring B to D: 18
etc.
I would add an N array to optimize this.
not sure exactly what that is.
Count each ranked candidate in the array. Later the array will be
added into the matrix as if the ranked candidates won in every one of
their pairs. This is correct for pairs with no ranking, and for pairs
with one ranked. For pairs w/winner and loser, give loser a negative
count to adjust; for ties can leave both winning; or mark both losing
via negative count.
can you be a little more explicit about this? i can't tell what this
"negative count" is about. and why is it needed? i think that the RP
procedure is pretty well cut-and-dried. if it were me, i would not use
*any* cycle-breaking procedure unless a cycle exists and then use
whatever resolution (whether it be Tideman or Schulze or whoever).
I'm not sure what this is, either.
I'll note in response to you, though, that for Tideman (and Schulze, for
that matter), the method *is* the tiebreaker. Tideman/RP/MAM elects the
CW if there is one, and if there isn't, it just proceeds to elect
another winner. The voters don't have to care about cycles unless they
(the voters) are curious.
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