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

On 4/28/2025 4:03 PM, olcott wrote:

On 4/28/2025 2:58 PM, dbush wrote:

On 4/28/2025 3:45 PM, olcott wrote:

On 4/28/2025 2:07 PM, dbush wrote:

On 4/28/2025 2:58 PM, olcott wrote:

On 4/28/2025 1:46 PM, dbush wrote:

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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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.

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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.
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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:
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Because the theory of computation was never previously
elaborated to make it clear that Turing computable
functions are required to derive their output by applying
finite string transformations to their input finite strings.
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And no such algorithm can derive this mapping:
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(<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
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*When we do this then a mapping suddenly appears*

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And that mapping is not the halting mapping, therefore the algorithm is not a halt decider.
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DD emulated by HHH according to the finite string
transformation rules of the x86 language DOES NOT HALT.

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In other words, HHH doesn't map the halting function.
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*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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And the halting function is not a computable function:
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Maybe you have ADD like Richard and can only
pay attention to a point when it is repeated many times
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I just proved that your halting function is incorrect.
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Category error.  The halting function below is fully defined, and this mapping is not computable *as you have explicitly admitted*.
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 Neither is the square root of an actual onion computable.

Category error.  There is no correct answer to "the square root of an onion", but there is a correct answer to whether any arbitrary algorithm X with input Y will halt when executed directly.  It's just that there's no algorithm that can tell us that, as Linz show and you have *explicitly* agreed is correct.

 Turing Computable Functions

i.e. mathematical mappings for which an algorithm exists which can compute them.

are required to apply finite
string transformations to their inputs.

Category error.  Mathematical mappings don't have behavior.  They are simply pairings from an input domain to an output domain.

The function defined
below ignores that requirement PROVING THAT IT IS INCORRECT.

Category error, as mathematical mappings don't have behavior.

 

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Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
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(<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
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