Hi,
Buddha Buck:
> I think we need to come up with better, understandable, language.
>
Right.
> Is it accurate to say that if x is in the Set, and y>>x, then y is in
> the set?
>
Yes.
> Is it accurate to say that if there is no y such that x>>y (i.e., x
> defeats nothing), then x is NOT in the set?
>
Yes.
> If we have options A, B, C, D, an A>>B,C,D, B>>C>>D>>B, then {B,C,D} is
> not the Schwartz Set because A>>B? A is in the Schwartz Set because
> there is no x such that x>>A?
>
Yes.
No -- either there's an x>>A => A is in the Schwartz set,
or there is no such x => A is th winner.
> What is the Schwartz set for:
>
> A==B; A>>C,D,E; B>>C,D,E; C>>D>>E>>C?
A and B are the winners. Since these are not reducible, you have a tie.
No rule (with the possible exception of "toss a coin") is going to get
you out of that one.
NB, you can define x>>y as to include the x==y case. It's not particularly
helpful in this case because you can't apply the usual rules to a
two-member Schwartz set, but you could end up with an A==B, B>>C and C>>A
situation.
--
Matthias Urlichs | noris network AG | http://smurf.noris.de/
