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On 4/28/2025 3:13 PM, Richard Heathfield wrote:Nothing /has/ to be measured. P's behaviour (halts, doesn't halt) when given D as input must be /established/. If you can do that by measuring the tape, great! But if you can do it by parsing the program's symbols and analysing the parse tree, that's great too. /How/ you do it doesn't matter as long as you express it as a Turing Machine that can handle any P and any D and produce a result of either 'halts' or 'doesn't halt'.On 28/04/2025 19:30, olcott wrote:Yet it is H(P,D) and NOT P(D) that must be measured.On 4/28/2025 11:38 AM, Richard Heathfield wrote:>On 28/04/2025 16:01, olcott wrote:>On 4/28/2025 2:33 AM, Richard Heathfield wrote:>On 28/04/2025 07:46, Fred. Zwarts wrote:>
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>So we agree that no algorithm exists that can determine for all possible inputs whether the input specifies a program that (according to the semantics of the machine language) halts when directly executed.>
Correct?
Correct. We can, however, construct such an algorithm just as long as we can ignore any input we don't like the look of.
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The behavior of the direct execution of DD cannot be derived
by applying the finite string transformation rules specified
by the x86 language to the input to HHH(DD). This proves that
this is the wrong behavior to measure.
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It is the behavior THAT IS derived by applying the finite
string transformation rules specified by the x86 language
to the input to HHH(DD) proves that THE EMULATED DD NEVER HALTS.
The x86 language is neither here nor there.
Computable functions are the formalized analogue
of the intuitive notion of algorithms, in the sense
that a function is computable if there exists an
algorithm that can do the job of the function, i.e.
*given an input of the function domain it*
*can return the corresponding output*
https://en.wikipedia.org/wiki/Computable_function
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*Outputs must correspond to inputs*
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*This stipulates how outputs must be derived*
Every Turing Machine computable function is
only allowed to derive outputs by applying
finite string transformation rules to its inputs.
In your reply to my article, you forgot to address what I actually wrote. I'm not sure you understand what 'reply' means.
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Still, I'm prepared to give you another crack at it. Here's what I wrote before:
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What matters is whether a TM can be constructed that can accept an arbitrary TM tape P and an arbitrary input tape D and correctly calculate whether, given D as input, P would halt. Turing proved that such a TM cannot be constructed.
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This is what we call the Halting Problem.
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Computer science has been wrong about this all ofComputer science is way ahead of you, as the proof long pre-dates the x86 chip family.
these years. When I provide the 100% concrete example
of the x86 language there is zero vagueness to slip
through the cracks of understanding.
Even calling the Turing Machine language the TuringI've never called it that.
Machine description language make this confusing.
*This is a verified fact*Your claim is that DD never halts? Fine. You could have proved that by rewriting DD to have a while(1); at the end of main()..
When DD is emulated by HHH according to the finite
string transformation rules of the x86 language
DD cannot possibly reach its own final state no
matter what HHH does.
How outputs are made to conform to inputs is neither here nor there. What matters is whether incomputable functions exist.Whatever you think you've proved, you haven't solved the Halting Problem. There are *no* solutions. We know this because there is a simple well-known proof. So the only way to devise a solution is to re- define the problem.It ultimately is only a confused view because key
details about how outputs are made to conform to
inputs:
You aren't paying enough attention because youWord salad is not a proof. You can talk till you're blue in the face, but you can't (correctly) prove that 2 + 2 = 5.
are too sure that I am wrong. I proved my point
above.
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