Hi Doug: yes, this is a complex example intended to show the flavor of the algorithm. We have invested significant time in it that we probably don't want to toss it out. However, Boaz just added a much more intuitive bus simulation example from my talk at the beginning of the paper. >From dougo@ccs.neu.edu Fri Aug 29 16:47:24 1997 >From: Doug Orleans >To: lieber@ccs.neu.edu (Karl Lieberherr) >cc: dem@carbon.ccs.neu.edu >Subject: Re: strategy paper example > >Karl Lieberherr writes: > > Hi Doug: > > > > I think that for second example, we should visit three E-objects > > but only 2 are printed. > >Why do you think that? Here's the object: > > : A ( : B ( : D ( : E ( ) ) ) > : C ( : A ( : B ( : D ( : E ( ) ) ) >

: F ( : E ( ) ) ) ) >

: F ( : E ( ) ) ) > >There are four paths to E-objects: > > 1. A,l,B,o,D,e,E > 2. A,m,C,n,A,l,B,o,D,e,E > 3. A,m,C,n,A,p,F,q,E > 4. A,p,F,q,E > >Here's the strategy graph: > > { A -> C bypassing -> A,m,C, // edge 1 > C -> E, // edge 2 > A -> D, // edge 3 > D -> E bypassing A } // edge 4 > >Path 1 matches edge 3 (A,l,B,o,D) + edge 4 (D,e,E). >Path 2 matches edge 3 (A,m,C,n,A,l,B,o,D) + edge 4 (D,e,E). >Path 3 would match edge 1 (A,m,C) + edge 2 (C,n,A,p,F,q,E) > but for the bypassing clause, which rules out A,m,C. > It cannot match edge 3 because there is no D. When I look at the traversal graph on page 10 of the paper, it allows the path: A,m,C,n,A,p,F,q,E ^ in copy of graph corresponding to edge e2 Path 3 matches edge 1 (A) + transition (A,m,C) + edge 2 (C,n,A,p,F,q,E) So the bypassing has no effect! Another potential confusion which we will mention in the paper. >Path 4 cannot match edge 1 or edge 3 because there is no C or D. > >This is a pretty confusing example (as demonstrated by your >expectation); are you sure you want to use it in the paper? On the >other hand maybe it's more obvious when the object diagram is drawn >out graphically. > >--Doug > -- Karl