Re: HHH(DD) --- COMPUTE ACTUAL MAPPING FROM INPUT TO OUTPUT --- Using Finite String Transformations

Op 24.apr.2025 om 21:45 schreef olcott:

On 4/24/2025 2:15 PM, Fred. Zwarts wrote:

Op 24.apr.2025 om 19:46 schreef olcott:

On 4/24/2025 3:11 AM, Fred. Zwarts wrote:

Op 24.apr.2025 om 05:34 schreef olcott:

On 4/23/2025 7:31 PM, Mike Terry wrote:

On 23/04/2025 16:38, olcott wrote:

On 4/23/2025 10:28 AM, Mike Terry wrote:

On 23/04/2025 10:02, Fred. Zwarts wrote:

Op 22.apr.2025 om 21:50 schreef olcott:

On 4/22/2025 2:30 PM, Fred. Zwarts wrote:

Op 22.apr.2025 om 21:14 schreef olcott:

On 4/22/2025 1:10 PM, Fred. Zwarts wrote:

Op 22.apr.2025 om 18:38 schreef olcott:

>
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
>
On Turing Machines inputs <are> finite strings, and
finite string transformation rules <are> applied to
these finite strings to derive corresponding outputs.
>

And it has been proven that no finite string transformations are possible that report the halting behaviour for all inputs that specify a correct program.

>
int sum(int x, int y) { return x + y; }
Only when people stupid assume the same thing as
sum(3,2) should return the sum of 5 + 3.
>

Therefore HHH should report on the actual input, the finite string that describes a halting program. Not on the hypothetical input that does not halt, because it is based on a hypothetical HHH that does not abort.
>
Why do you maintain that HHH should process the hypothetical input instead of the actual input.
Do you really believe that 3+2 equals 5+3?

>
I have proven that the directly executed DD and DD
emulated by HHH according to the semantics of the
x86 language have a different set of state changes
many hundreds of times for several years.

You never showed a proof. You only repeated a dream. You are dreaming many years without any logic. You failed to show the first state change where the direct execution is different from the simulation. You only showed an erroneous HHH that fails to reach the end of the simulation of a halting program.

>
Worse than this, on more than one occasion I've actually posted traces of computation DDD(DDD) executed directly and simulated by HHH side by side.  Both traces were of course /identical/, up to the point where HHH stops simulating.

>
*Factually incorrect* (You are usually very careful about these things)
The call to HHH(DD) from the directly executed DD returns.
The call to HHH(DD) from DD emulated by HHH cannot possibly return.
>

>
...because HHH stops simulating before reaching that step in the computation.  Note that I said
>
MT:  Both traces were of course /identical/,
      *up to the point where HHH stops simulating*
>
So I was factually correct.
>
>
Mike.
>

>
It *is not* up to the point where HHH stops simulating.
>
It is up to the point where the simulated versus directly
executed calls HHH(DD).
>

That is exactly the same point. If not, show the difference in the traces before that point.

>
As soon as the directly executed DD calls HHH(DD) this
call immediately returns.
>
When DD emulated by HHH calls HHH(DD) then HHH emulates
DD and also emulates itself emulating DD. This is one
whole recursive emulation than the directly executed
DD can possibly get to.

Again a lot of words, which hide that you cannot show where the traces differ up to that point.

 THEY DIFFER BY THE EMULATED DD REACHES RECURSIVE EMULATION
AND THE DIRECTLY EXECUTED DD NEVER DOES.
 

That is not visible in the trace up to that point. Both reach a finite recursion. The traces are identical up to that point. So, the difference is not in the trace of the the x86 emulation, but in the abort. The difference is that HHH aborts a finite recursion and the direct execution does not abort.