From: "Jobst Heitzig" <[EMAIL PROTECTED]>
Subject: [EM] Sprucing up vs. Condorcet Lottery vs. immunity: The
"twisted prism" example
Jobst wrote:
Dear Forest!
Your sprucing up technique is a very nice idea since it can simplify the
tallying of those methods which fulfil beat-clone-proofnes
On 6 Jan 2005 at 15:08 PST, Markus Schulze wrote:
> Dear Ted,
>
> Jobst Heitzig wrote (5 Jan 2005):
>> Although this means that it may in bad cases be complicated
>> to find p, it is always easy to prove that the result is
>> correct once it is found by just showing that it beats each
>> of the n p
On 6 Jan 2005 at 15:15 PST, Paul Kislanko wrote:
>> On 6 Jan 2005 at 14:31 PST, Markus Schulze wrote:
>> > Well, with the Dijkstra algorithm you can calculate
>> > in O(n^2) time the strengths of the strongest paths from
>> > candidate A to every other candidate and from every other
>> > candi
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED]
> ] On Behalf Of Ted Stern
> Sent: Thursday, January 06, 2005 4:52 PM
> To: [EMAIL PROTECTED]
> Subject: [EM] Re: There is always a Condorcet Winner! (among
> all lotteriesof candidates :-)
>
> On 6 Jan 2005 at
Dear Ted,
Jobst Heitzig wrote (5 Jan 2005):
> Although this means that it may in bad cases be complicated
> to find p, it is always easy to prove that the result is
> correct once it is found by just showing that it beats each
> of the n pure strategies. This can be done in O(n^2) time
> just as i
On 6 Jan 2005 at 14:31 PST, Markus Schulze wrote:
> Dear Jobst,
>
> you wrote (5 Jan 2005):
>> By the way, is it possible to prove a Beatpath winner
>> in O(n^2) time also?
>
> Well, with the Dijkstra algorithm you can calculate
> in O(n^2) time the strengths of the strongest paths from
> candidate
Dear Jobst,
you wrote (5 Jan 2005):
> By the way, is it possible to prove a Beatpath winner
> in O(n^2) time also?
Well, with the Dijkstra algorithm you can calculate
in O(n^2) time the strengths of the strongest paths from
candidate A to every other candidate and from every other
candidate to ca
Sorry, the diagram didn't come out right, so here it is again:
A1
10/| \ clockwise cycle
/ | \
A3--A2
| | | upward beats
|9 | | (diagonal downward beats omitted)
| B1 |
| / \ | counter-clockwise cycle
| /10 \|
B3--B2
__
Dear Forest!
Your sprucing up technique is a very nice idea since it can simplify the
tallying of those methods which fulfil beat-clone-proofness and
uncoveredness. However, some of which you wrote has confused me
completely: Did I understand you right in that you claim that the
technique shou
