[EM] Condorcet How?

Kristofer Munsterhjelm km-elmet at broadpark.no
Sat Mar 20 11:54:14 PDT 2010

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:


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


(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


>> 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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