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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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