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Subject: 
RE: Would-be hacker queries. (fwd)
Newsgroups: 
lugnet.robotics
Date: 
Tue, 11 May 1999 17:32:10 GMT
Original-From: 
Jim Choate <ravage@einstein.=IHateSpam=ssz.com>
Viewed: 
869 times
  
----- Forwarded message from Joel Shafer -----

Date: Tue, 11 May 1999 12:24:13 -0500
From: Joel Shafer <joel@connect.net>
Subject: RE: Would-be hacker queries. (fwd)

I'm afraid you are wrong.  That is exactly what it does.  The result of the
first replacement is dependent on the result of the second replacement
which is dependant on the result of the third replacement ...
It is exactly like the factorial example except the returned results are
concatenated instead of multiplied.

----- End of forwarded message from Joel Shafer ----

I guess we'll just have to agree to disagree.

The multiplication is irrelvant in what makes a factorial a recursive
function.

While you have certaily concatenated the individual replacements the actual
replacement function itself does not rely on the previous replacement. If it
did you wouldn't be able to replace the first 'the' until you'd defined some
axiom equivalent to 1!=1 and n!=n(n-1)!. You haven't done that so it isn't
recursive.

If anything all you've done is parallelize (again not recursive) the actual
replacement function so that they happen at one time. You haven't made
replacing 'the' with 'then' (or whatever) the iterative definition.

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