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Subject: Re: Guarded normal form (part 1)
I think I made it, although it has to be cooked better. First, let me change the notation from Pos{foo}<...> to foo{Pos}<...> <choice> and <intersection>(concur) are symmetric operations, so I choose to use them (and <not>) for the algorithm. As I wrote, <diff> A B </diff> => <intersection> A <not> B </not> </intersection>. 1. <choice> NC1 NC2 </choice> ------------------------------------------------ case NC1=PosSet<...> & NC2=PosSet<...> return PosSet< union of Namespaces >; if two Namespaces foo<S>{X} and foo<T>{Y} collide, switch(S,T){ case Pos,Pos => foo{Pos}<X \cup Y> case Pos,Neg => foo{Neg}<Y-X> case Neg,Neg => foo{Neg}<X \cap Y> case NC1=PosSet<A> & NC2=NegSet<B> return NegSet< B > if A and B has a common Namespace, do as above. case NC1=NegSet<...> & NC2=NegSet<...> return NegSet< intersection of Namespaces > if two Namespaces collide, do the same as above. 2. <intersection> NC1 NC2 </intersection> ------------------------------------------------ case NC1=PosSet<...> & NC2=PosSet<...> return PosSet< intersection of Namespaces >; for every collided Namespace foo{S}<X> and foo{T}<Y>, switch(S,T){ case Pos,Pos => foo{Pos}<X \cap Y> case Pos,Neg => foo{Pos}<X-Y> case Neg,Neg => foo{Neg}<X \cup Y> case NC1=PosSet<A> & NC2=NegSet<B> return PosSet< A > if A and B has a common Namespace, do as above. case NC1=NegSet<...> & NC2=NegSet<...> return NegSet< union of Namespaces > if two Namespaces collide, do the same as above. 3. <not> NC </not> ------------------------------------------------ case NC=PosSet<...> return NegSet< negated Namespaces > case NC=NegSet<...> return PosSet< negated Namespaces > To negate Namespace: foo{Pos}<a,b,...> => foo{Neg}<a,b,...> foo{Neg}<a,b,...> => foo{Pos}<a,b,...> -- Kohsuke KAWAGUCHI +1 650 786 0721 Sun Microsystems kohsuke.kawaguchi@eng.sun.com
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