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From: marcoj@ai.rl.af.mil (James D. Marco)
Subject: Re: P!=NP does NOT imply NPC cannot be solved
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In article <422oqi$1sk@tang.ccr-p.ida.org>, desj@ccr-p.ida.org (David
desJardins) wrote:

> Nick Maclaren <nmm1@cus.cam.ac.uk> writes:
> > For a generic example, consider the program that determines whether
> > an arbitrary K-parameter decision function (implemented as an oracle)
> > ever returns "yes" for values taken from within a set of size N.
> 
> I'm completely stumped trying to guess this could mean.  If a function
> is "implemented as an oracle" then how is a deterministic machine
> without access to the oracle supposed to compute it?  If a machine can't
> compute it then how are you going to define its complexity?
Infinite. Where the complexity of a problem exceedes the ability to compute
it, there is an effective infinite complexity, ie. you can continue trying 
to compute it forever.

eg: If each step taken towards a goal is of 0 length, then the goal cannot
be reached in less than an infinite number of steps. The computational 
answer to "Will the walker reach his goal?" is never. 
   Equivalently: a walker taking 0 steps of any length towards a goal will
never reach it.

The only solution, and the solution to both, otherwise intractable problems,
is the case that the walkers initial position is also his goal!

Hence, a single "position evaluation" algorithm solves the problem! This
evaluation works on facts known about the parameters of the problem, and
not 
the problem itself, effectivly changing the size of the computational 
universe, basically a form of complex math. The complexity of computing a
solution to a problem is relative to the computational viewpoint of the 
problem space within an n-dimensional universe. Note for the record, that
some problems still cannot be solved effeciently(NP) since they span an 
n-dimensional problem space where |n| is size aleph-nul (first order ifinite). 
I hypothesize: where the order of the problem space |n| is finite, and a
solution exists within the problem space,  then an n+1 problem space
exists, containing a log(n+1) order solution.    

> Obviously in a relativized model of computation it is easy to construct
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Exactly.

> decision problems with any particular lower bound, just by counting the
> number of necessary calls to the oracle.  This seems to have nothing to
> do with the stated problem though.
> 
> > This is trivially O(N^K) in all respects for a deterministic, serial
> > machine.  It is neither natural nor interesting.
> 
>                                         David desJardins
> -- 
> Copyright 1995 David desJardins.  Unlimited permission is granted to quote
> from this posting for non-commercial use as long as attribution is given.
