Re: Turing Machine computable functions apply finite string transformations to inputs

On 4/28/2025 2:30 PM, olcott wrote:

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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<snip>
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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?

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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.

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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
 *Outputs must correspond to inputs*
 *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.
 

And no turing machine exists that can derive the following mapping (i.e. the mapping is not a computable function), as proven by Linz and others:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly