[EM] Beatpath(Unc, wv), a version of Beatpath that always picks from the Uncovered set

[Jobst Heitzig]

Dear Forest,

you wrote:
> Let N be the number of ballots. In Beatpath(Unc, wv) the strength of
> a pairwise victory of X over Y is N+1 if X covers Y, else it is the
> number of ballots on which X is ranked above Y.

I fear a "Beatpath" or "River" method based on this definition of 
strength will not be monontonic. Let's assume the defeats 
X>B>Y>A>B>C>A>X>C>Y>X with winning votes of the following relative 
sizes:

          over...
wv        A  B  C  X  Y
of...
     A    -  1     4
     B       -  1     5
     C    1     -     2
     X       5  3  -   
     Y    5        1  -

Since every defeat is on a 3-cycle, the covering relation is empty and 
the beatpath strengths are of these relative sizes:

beatpath  over...
strength  A  B  C  X  Y
of...
     A    -  *  *  4  *
     B    *  -  *  4  *
     C    *  *  -  2  *
     X    5  5  3  -  5
     Y    *  *  *  4  -

Hence your Beatpath version will elect X.
(By the way, River would lock in the defeats X>B>Y>A and X>C, and Ranked 
Pairs would in addition lock in B>C>Y and C>A, both therefore also 
electing X.)

Now assume that weak defeat Y>X is reversed into a weak defeat X>Y, with 
winning votes like these:

          over...
wv        A  B  C  X  Y
of...
     A    -  1     4
     B       -  1     5
     C    1     -     2
     X       5  3  -  1! 
     Y    5           -

Then C covers Y and X covers B, hence the defeat strengths and beatpath 
strengths are now of these relative sizes:

defeat    over...
strength  A  B  C  X  Y
of...
     A    -  1     4
     B       -  1     5
     C    1     -     9!
     X       9! 3  -  1 
     Y    5           -

beatpath  over...
strength  A  B  C  X  Y
of...
     A    -  *  3  4  *
     B    *  -  3  4  *
     C    5  4  -  4! 9
     X    5  9  3  -  5
     Y    *  *  3  4  -

That is, X has a weaker beatpath against C than C has against X. The new 
winner is C, proving that the method is not monotonic!

Using your definition of defeat strength with River instead would lock 
in C>Y>A>X>B, also electing C. 

However, using it with Ranked Pairs would lock in C>Y and X>B, then 
B>Y>A, then X>C>A and B>C, giving the same social order as before in 
this example. 

So, could it be in this case the Ranked Pairs cycle-resolution method 
works better than the Beatpath and River cycle-resolution methods??

Yours, Jobst