[EM] hey Markus, can you confirm something?...

[Juho]

On Feb 6, 2010, at 3:28 PM, Juho wrote:

> On Feb 6, 2010, at 5:34 AM, robert bristow-johnson wrote:
>
>>
>> On Feb 5, 2010, at 10:24 PM, Warren Smith wrote:
>>
>>> robert bristow-johnson:
>>>> In a Condorcet election in which a cycle occurs, if there are  
>>>> only three
>>>> candidates in the Smith set, or even if there are more but the  
>>>> defeat
>>>> path is a simple single loop, is the outcome of the election any
>>>> different if decided by Schulze rules than if decided by Tideman  
>>>> ruules
>>>> (Ranked Pairs)?
>>>>
>>>> Of course, this question is open for anyone to answer.
>>>
>>> Kristofer Munsterhjelm:
>>> AFAIK, if there are three candidates in a cycle, RP and Schulze  
>>> returns
>>> the same: they break the cycle by its weakest defeat. This is  
>>> obvious
>>> for RP, because the weakest defeat (least victory) gets sorted  
>>> last. For
>>> Schulze, consider the Schwartz set implementation (CSSD) as  
>>> described on
>>> http://en.wikipedia.org/wiki/Schulze_method#The_Schwartz_set_heuristic 
>>>  .
>>> Since there's a cycle and all three candidates are involved in it,  
>>> steps
>>> one and two can't be done, so 3 is done, which also discards the  
>>> weakest
>>> defeat.
>>>
>>> Warren D Smith:
>>> To complete KM's answer, the case where
>>> "there are more [than three in the Smith set]
>>> but the defeat  path is a simple single loop"
>>> does not exist.
>>
>> that, i didn't expect.
>>
>>> That is because if there is a directed N-cycle
>>> for some N>3, you can always "draw a chord" in the "circle"
>>
>> <smacking forehead>
>>
>>> and no matter which way the arrow on that chord points, the result  
>>> is always
>>> at least one shorter directed cycle, i.e. with smaller N.    
>>> Consequently, no
>>> matter what the value of N>3, there is always a 3-cycle.
>>
>> so everyone in the smith set is related to everyone else via a 3- 
>> cycle?  or is that also not general?
>
> In a 4 member Smith set we can without loss of generality assume  
> that there is a 4-cycle A>B>C>D>A and that the remaining "chords"  
> are A>C and B>D. This means that B and C are not part of any 3- 
> cycle. All 4 member Smith sets are thus isomorphic, and a 4 member  
> Smith set always has exactly two members that are not connected to  
> each others with a 3-cycle.
>
> Here are some more general results. A Smith set (n>=3) may have two  
> members that are not both part of any of its 3-cycles. A n member  
> Smith set may have all pairs of its members connected to to each  
> others via some of its 3-cycles if n>=5. For all m, 3 <= m <= n, all  
> members of a n member Smith set are part of a m-cycle.

> For all m, 4 <= m <= n, any two members of a n member Smith set are  
> both part of one of the m-cycles.

This last sentence is unfortunately not true. Circular connection  
between any two Smith set members can be guaranteed only for the full  
n-cycle.

Juho

>
> (All this is based on the assumption that there are no pairwise  
> ties. I hope I got the equations otherwise right.)
>
> Juho
>
>
>>
>>> So... do Schulze & Ranked Pairs always return the same winner? No.
>>> If the Smith set has 3 or fewer members? Yes.
>>
>>
>> i suspected these two answers but don't mind getting misconceptions  
>> enlightened.
>>
>> thanks, Warren.
>>
>> --
>>
>> r b-j                  rbj at audioimagination.com
>>
>> "Imagination is more important than knowledge."
>>
>>
>>
>>
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