[EM] Re: There is always a Condorcet Winner! (among all lotteries of candidates :-)

[Ted Stern]

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 A to every other candidate and from every other
> candidate to candidate A.
>
> Markus Schulze

But that still leaves you with another order of n to calculate the winner out
of n candidates, right?  So the answer would be no, by all appearances.

So 
-- 
Send real replies to
	ted stern at u dot washington dot edu

Frango ut patefaciam -- I break that I may reveal