3 Oct
2007
3 Oct
'07
9:33 p.m.
...and indeed it can't be done, except by the naive brute-force method of comparing every subtree, possibly optimized by cryptographically hashing a representation of every subtree, since sharing isn't an observable property.
i was thinking that instead of having a reference to a node, each node just holds an index from an array of nodes. Traversal would take an extra step, but it should fix the problem with encode/decode.