[EM] Pirate Party of Berlin adopts the Schulze method

robert bristow-johnson rbj at audioimagination.com
Tue Mar 1 13:43:25 PST 2011


Markus, here is a joke i've heard recently:

     Q: What do you call a person who knows two languages?
     A: Bilingual.

     Q: What do you call a person who knows three languages?
     A: Trilingual.

     Q: What do you call a person who knows four languages?
     A: Quadralingual.

     Q: What do you call a person who knows a single language?
     A: American.


that said, i can't figure out from the website who wins according to  
the Schulze algorithm.

and, if you know, could you tell us who would win if it was decided by  
Ranked Pairs?

and, lastly, do we know who would win if it were single-winner STV  
(commonly called IRV in the states)?

just curious.

thanx.

--

r b-j                  rbj at audioimagination.com

"Imagination is more important than knowledge."




On Mar 1, 2011, at 12:32 PM, Markus Schulze wrote:

> Hallo,
>
> magalski beats baum.
> magalski beats delius.
> magalski beats weiss.
> magalski beats höfinghoff.
> magalski ties wittich.
>
> baum beats mayer.
> baum beats delius.
> baum beats weiss.
> baum beats höfinghoff.
> baum beats wittich.
>
> mayer beats magalski.
> mayer beats delius.
> mayer beats weiss.
> mayer beats höfinghoff.
> mayer beats wittich.
>
> delius beats weiss.
> delius beats höfinghoff.
>
> weiss beats höfinghoff.
> weiss ties wittich.
>
> höfinghoff beats wittich.
>
> wittich beats delius.
>
> The Smith set consists of 7 candidates (magalski,
> baum, mayer, delius, weiss, höfinghoff, wittich).
> The Schwartz set consists of 3 candidates
> (magalski, baum, mayer).
>
> The Schulze method satisfies both criteria, the
> Smith criterion and the Schwartz criterion.
> Therefore, the Schulze method guarantees that
> all Smith winners are ranked ahead of all other
> candidates and that all Schwartz winners are
> ranked ahead of all other candidates.
>
> Markus Schulze
>
>
> ----
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>



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