# Re: Basis for bypassing the Halting Problem ?

On 2/8/2012 7:53 PM, Patricia Shanahan wrote:
On 2/8/2012 4:50 PM, Peter Olcott wrote:
On 2/8/2012 8:08 AM, Patricia Shanahan wrote:
Patricia Shanahan wrote:
On 2/8/2012 3:52 AM, Peter Olcott wrote:
On 2/7/2012 11:30 PM, Patricia Shanahan wrote:
On 2/7/2012 7:52 PM, Peter Olcott wrote:
On 2/7/2012 9:26 PM, Joshua Cranmer wrote:
On 2/7/2012 8:30 PM, Peter Olcott wrote:
On 2/7/2012 8:15 PM, Joshua Cranmer wrote:
On 2/7/2012 7:51 PM, Peter Olcott wrote:
The point is that no halting machine simulator can possibly
modify the
state a halt analyzer based on the output of this halt analyzer
if it
does not have access to the meaning of the output of this halt
analyzer.

And the point is that there must exist a simulator which does have

Yet since it has no way to access the meaning of this output it
has no
way to create the halting problem.

There exists a machine which could access the meaning. That's the
entire point.

It could not possibly have any way to determine the meaning of any
hidden key if this meaning is never provided, thus the Halting
Problem
can no longer be constructed.

Sure it can. All we need is for the meaning to be a computable
function
of the final state of your machine. We don't know *which* TM computes
it, but that TM does exist.

If there is at least one TM that computes a Yes/No answer to "does
this
TM computation halt" from the final state of your machine, there is
a TM
computation that halts if, and only if, your machine says it does not
halt.

We don't know which TM it is that your machine gets wrong, but that
does
not change the fact that it exists.

Patricia
The TM outputs a different integer number every time that it is
executed. The meaning of the integer number is only understood by this
TM and the human executing it. A TM simulator would contain the TM that
knows the meaning of this integer, yet it would not and could not know
the meaning itself, thus would have no way of knowing where to attach
the infinite loop graph cycle.

Please show the TM state machine lines you would use to implement that.

Ignoring the fact that such a machine would not represent an algorithm,
and is getting very far from being a decider for any language, let's
consider its practical use.

Suppose I were back to writing compilers. The program being compiled is
syntactically correct, using only libraries the compiler can find etc.
The user has checked the box for "report failure to halt", so the
compiler runs the great halting detection program. It gets back an
integer.

Now what?

Patricia

So when a software system returns a encrypted result this software
system is no longer considered to meet the definition of an algorithm?
Exactly what would be lacking?

I'm sorry, I misunderstood. I thought you were saying that the answer
was going to change from run to run, independently of any input.

If the result is encrypted, and could be decrypted by someone with a
key, then the actual result is a computable function of the result your
machine produces, there exists a TM that could compute the actual result
using your TM, and there exists a TM whose halting would be mispredicted.

Patricia

The answer will be encrypted differently from run to run such that any TM simulator would never know what the answer means. This prevents any TM simulator from creating the Halting Problem.
.

## Relevant Pages

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• Re: Basis for bypassing the Halting Problem ?
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(comp.theory)
• Re: Basis for bypassing the Halting Problem ?
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• Re: What is the Result from Invoking this Halt Function?
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• Re: What is the Result from Invoking this Halt Function?
... And, obviously, it's NOT an analytical impossibility, ... LoofIfHalts would always halt in each and every instance. ... it is natural to call this TM a halt analyzer. ... > TM itself but A STRING, THE SOURCE CODE FOR, or some other string DESCRIPTION ...
(sci.logic)