🚀 go-pugleaf

Pardon my impertinence, Dr. O'Keefe, but how does goal caching make
an implementation no longer ``a Prolog implementation''?  It seems to
me that whatever implementation observes the externally observable
language semantics is ``a Prolog implementation''.  If the approach you
seem to suggest is taken, then any FORTRAN compiler that does loop 
invariant floating is not a FORTRAN implementation.  That surely is not
the approach taken by the FORTRAN community.

Since the Tak benchmark is side effect free with only one solution, to all 
external appearances the execution is the same with or without goal caching.
Is there some subtle semantic point I've missed somewhere? I know that 
tracing/debugging would be different, but that doesn't seem to be at issue
here.

Now the fact that there's not enough redundancy in Tak to account for 
that much of a speedup is more proof that there's no goal caching going
on in the example at hand.  Of course, the fibonacci example cited in
a subsequent posting is much more suspect for this, although I'll take
Dr. van Roy at his word on both examples.

David