]> Hello, On Sun, 28 Jul 2002, John Max Skaller wrote: > Given a recursive type > > Fix 'a . T (where 'a occurs in T) > > we can unfold the type to T' = T['a -> Fix 'a.T], > we define unfold t = t, if t doesn't start with a fixpoint operator. > > Any ideas how to best implement fold, the inverse isomorphism? > > Brute force method: examine every subterm, and compare with > the main term using equi-recusive comparison .. this seems quadratic > in the number of nodes .. smarter method: only compare arguments > of fixpoint binders .. can we do any better? You can keep in each node of the term a hash value for the corresponding subterm; this hash value should be invariant by folding/unfolding (you can for instance look at a fixed depth to compute this hash value and unfold when necessary). These hash values avoid most equi-recursive comparisons (and elegate most of the remaining ones). My Recursive module uses the same technique; it may do what you want: http://www.eleves.ens.fr:8080/home/frisch/soft#recursive It helps manipulating recursive structures (and recursive types was indeed the main motivation) with maximal sharing and unique representation (that is: two terms that have the same infinite unfolding will be represented by the same value). You can also have a look at the following papers, which describe another solution: [1] Improving the Representation of Infinite Trees to Deal with Sets of Trees, Laurent Mauborgne; http://www.di.ens.fr/~mauborgn/publi/esop00.ps.gz [2] Efficient Hash-Consing of Recursive Types, Jeffrey Considine; http://www.cs.bu.edu/techreports/2000-006-hashconsing-recursive-types.ps.Z Hope this helps. Alain ------------------- To unsubscribe, mail caml-list-request@inria.fr Archives: http://caml.inria.fr Bug reports: http://caml.inria.fr/bin/caml-bugs FAQ: http://caml.inria.fr/FAQ/ Beginner's list: http://groups.yahoo.com/group/ocaml_beginners