Re: What is the Result from Invoking this Halt Function?
From: Will Twentyman (wtwentyman_at_read.my.sig)
Date: 08/14/04
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Date: Sat, 14 Aug 2004 14:24:55 -0400
Peter Olcott wrote:
> "Will Twentyman" <wtwentyman@read.my.sig> wrote in message news:411a779e$1_4@newsfeed.slurp.net...
>>Peter Olcott wrote:
>>>"Will Twentyman" <wtwentyman@read.my.sig> wrote in message news:4118e82c$1_5@newsfeed.slurp.net...
>>Here is where your analysis begins to fail. First, HALT may or may not
>>be running on a UTM. If it is not running on a UTM, it cannot query the
>>UTM.
>
> I am only required to provide a single case where my method is not
> prohibited from working. I assume as an integral part of this single
> case a UTM.
And you must also show that your case is not equivalent to a case where
it does not work. You have failed to do this.
>>Second, HALT does its processing bases only on its input. Context
>>is meaningless because there is only one context, the machine itself.
>
> This is merely a False and Unsupported Assumption (FUA).
You would do well to drop these things. A TM is *defined* to operate
only on its input. It is assumed to be a machine. The fact that it can
be emulated on a UTM is coincidental.
> This a purely arbitrary restriction that has nothing at all to do with providing
> the design for a machine that can correctly determine whether or not each
> and every element of the universal set of Turing Machines halts on a specific
> input.
The assumed context is that it is not being called, but is a stand-alone
machine. The fact that it can be emulated is where the power of TMs arises.
>>If it is being emulated on a UTM, that fact cannot affect its results or
>>you are not emulating HALT.
>
> I am not emulating your pure arbitrary (and thus meaningless) restriction.
> Emulating arbitrary restrictions is not required.
Then you need to start clarifying what you are doing. Preferably on
your website with technical details of *how* things are being done. You
would also need to demonstrate how that work applies to the standard
situation.
>>HALT does not have three possible outputs.
>> It has two possible outputs <TM(input) halts> and <TM(input) does not
>>halt>. The method of encoding those outputs is irrelevant.
>
> That is the way that it now works. There is no longer a third output.
> I will update my website to provide this update.
I look forward to seeing the updated version.
>>If you make the modifications that you propose, several things result:
>>First, you are no longer discussing the Halting Problem.
>
> We are discussing any and all proof that a TM that can determine
> whether or not any TM halts on a specific input is impossible.
> Every single published source that I have encountered calls this
> the Halting Problem, including Turing himself.
>
> Halting Problem according to Alan Turing:
> The problem of finding out whether a given number is the D.N of a
> circle-free machine, and we have no general process for doing this
> in a finite number of steps. In fact, by applying the diagonal process
> argument correctly, we can show that there cannot be any such general
> solution.
>
>
>>Second, you are no longer discussing UTMs.
>
> I augmented the basic definition of a UTM slightly. This is not anywhere
> near an impossible task. For anyone that knows finite automatons, this
> is trivial.
It's not impossible, just a different system.
>>Third, any conclusions you draw from the new situation are unrelated to
>>the Halting Problem.
>
> I have completely refuted each point that you made, point-by-point.
> I have already completely refuted this latter point in my refutations above.
> I have refuted the proof that solving the Halting Problem is impossible
> according to Alan Turing's own definition of the Halting Problem.
You would do well to stop making such claims until you have a
significant number of people (more than 2) agreeing with you. Until
then, you come across as arrogant and/or ignorant.
-- Will Twentyman email: wtwentyman at copper dot net
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