On Wed, Nov 14, 2012 at 7:35 PM, Alan Watson <[email protected]> wrote:
> There may well be better ways to say this. Alternatively, I think one > could say that the leafs are compared by eqv? (since structure has been > unfolded). > Almost. Not all leaf-less graphs are equal? to each other. I think that all leafless graphs consisting only of pairs are, but once you start interspersing vectors with pairs, this should affect equal?-ity. Consider "paths" which are sequences of operations like car, cdr, (vector-ref <node> 0), (vector-ref <node> 1), .... For equal? graphs, a path walked in one should be walkable in the other. If you arrive at a leaf, then the leaves are string-equal?, bytevector-equal?, or just eqv? This is my mental model, but someone please correct me if I'm mistaken.
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