Re: [EM] Question about Schulze beatpath method

2012-04-05 Thread Markus Schulze

Hallo,

I rewrote section 5 ("Tie-Breaking") of my paper,
so that it is now more in accordance with the
other parts of my paper:

http://m-schulze.webhop.net/schulze1.pdf

Markus Schulze


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Re: [EM] Question about Schulze beatpath method

2012-02-20 Thread robert bristow-johnson

On 2/20/12 1:15 PM, Markus Schulze wrote:

Hallo,

in example 3 of my paper, the weakest link of the strongest
path from candidate A to candidate C is the same link as
the weakest link in the strongest path from candidate C
to candidate A:

http://m-schulze.webhop.net/schulze1.pdf

the thing that had been confusing me (until Kevin replied) is that the 
*direction* through that link is the same whether your defeat path is 
from A to C or the reverse.  that seems sorta counter-intuitive.


--

r b-j  r...@audioimagination.com

"Imagination is more important than knowledge."




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Re: [EM] Question about Schulze beatpath method

2012-02-20 Thread Markus Schulze
Hallo,

in example 3 of my paper, the weakest link of the strongest
path from candidate A to candidate C is the same link as
the weakest link in the strongest path from candidate C
to candidate A:

http://m-schulze.webhop.net/schulze1.pdf

Markus Schulze


Election-Methods mailing list - see http://electorama.com/em for list info


Re: [EM] Question about Schulze beatpath method

2012-02-17 Thread Kevin Venzke
Hi Robert,
 
Suppose there are four candidates ABCD. B beats A with strength of 10. C beats 
D with strength
of 20. With strength of 30, A beats C, B beats C, D beats A, and D beats B. 
Then every candidate
has a path to every other candidate, and the best path from A to B or from B to 
A involves traversing 
the C>D win (which is the weakest link in those paths).
 
Kevin
 

De : robert bristow-johnson 
À : election-methods@lists.electorama.com 
Envoyé le : Vendredi 17 février 2012 12h56
Objet : Re: [EM] Question about Schulze beatpath method

On 2/17/12 1:27 PM, Markus Schulze wrote:
> it can happen that the weakest link in the strongest path
> from candidate A to candidate B and the weakest link in the
> strongest path from candidate B to candidate A is the same link,
> say CD.

how can that be?  since a path is a *defeat* path.  you only traverse a 
beatpath from a candidate who beats the next candidate in the path.

is it that candidates C and D are exactly tied?  other than that, i cannot 
understand how the weakest link from A to B can be the same as *any* link from 
B to A.

-- 
r b-j                  r...@audioimagination.com

"Imagination is more important than knowledge."




Election-Methods mailing list - see http://electorama.com/em for list info
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Re: [EM] Question about Schulze beatpath method

2012-02-17 Thread robert bristow-johnson

On 2/17/12 1:27 PM, Markus Schulze wrote:

it can happen that the weakest link in the strongest path
from candidate A to candidate B and the weakest link in the
strongest path from candidate B to candidate A is the same link,
say CD.


how can that be?  since a path is a *defeat* path.  you only traverse a 
beatpath from a candidate who beats the next candidate in the path.


is it that candidates C and D are exactly tied?  other than that, i 
cannot understand how the weakest link from A to B can be the same as 
*any* link from B to A.


--

r b-j  r...@audioimagination.com

"Imagination is more important than knowledge."




Election-Methods mailing list - see http://electorama.com/em for list info


Re: [EM] Question about Schulze beatpath method

2012-02-17 Thread Markus Schulze
Hallo,

it can happen that the weakest link in the strongest path
from candidate A to candidate B and the weakest link in the
strongest path from candidate B to candidate A is the same link,
say CD.

I recommend that, in this case, the link CD should be declared
"forbidden" and the strongest path from candidate A to candidate B
and the strongest path from candidate B to candidate A, that does
not contain a "forbidden" link, should be calculated. If again the
weakest link in the strongest path from candidate A to candidate B
and the weakest link in the strongest path from candidate B to
candidate A is the same link (say EF), then also this link should
be declared "forbidden" and the paths from A to B and from B to A
should be calculated. This should be repeated until the weakest
link in the strongest path from A to B and the weakest link in the
strongest path from B to A are different links.

Markus Schulze


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