On 4/23/2025 9:41 PM, Richard Damon wrote:Yes, the input specifies FINITE recusive PARTIAL emulation, as the HHH that DD calls will emulate only a few instructions of DD and then return,On 4/23/25 11:32 AM, olcott wrote:The input to HHH(DD) does specify the recursive emulationOn 4/23/2025 6:25 AM, joes wrote:>Am Tue, 22 Apr 2025 13:51:48 -0500 schrieb olcott:>On 4/22/2025 1:07 PM, Fred. Zwarts wrote:>Op 22.apr.2025 om 18:28 schreef olcott:On 4/22/2025 7:57 AM, joes wrote:Am Tue, 15 Apr 2025 15:44:06 -0500 schrieb olcott:No, DD halts (when executed directly). HHH is not a halt decider, not even>>libx86emu <is> a correct x86 processor and does emulate its inputsYou continue to stupidly insist that int sum(int x, int y) {return xWhat else is it missing that the processor uses to execute it?
+ y; }
returns 7 for sum(3,2) because you incorrectly understand how these
things fundamentally work.
>
It is stupidly wrong to expect HHH(DD) report on the direct
execution of DD when you are not telling it one damn thing about
this direct execution.
>
correctly.
The key thing here is that Olcott consistently does not understand that
HHH is given a finite string input that according to the semantics of
the x86 language specifies a halting program,
That is stupidly incorrect.
for DD only.
>People here stupidly assume that the outputs are not required toBut the direct execution of DD is computable from its description.
correspond to the inputs.
>
Not as an input to HHH.
But neither the "direct execution" or the "simulation by HHH" are "inputs" to HHH. What is the input is the representation of the program to be decided on.
>When HHH computes halting for DD is is only allowed>
to apply the finite string transformations specified
by the x86 language to the machine code of DD.
It is only ABLE to apply them.
>
of DD including HHH emulating itself emulating DD when
one applies the finite string transformation rules of the
x86 language to THE INPUT to HHH(DD).
Of course, part of the problem is that in your input you exclude the code for HHH, and try to make it only implicit, which isn't allowed. HHH can not assume behavior for the HHH it sees except by what is defined in the input, or by the rules of the environment.
You can't pop any other execution trace from the inputYou can get *ANY* trace of the input beyond the call to HHH, as HHH isn't defined in the input.
to HHH(DD) than that.
If you include HHH in the input, then the actual execution trace of the input, as exactly shown by an actual correct emulation of it, will see that DD calls HHH(DD), which will emulated its input for a while then stop and return 0, and then DD will halt.
Now, HHH's problem is it stops before it get there, as that is what its code says to do, as seen in what is now part of the input you have defined, and thus it needs to "guess" as to the answer, and DD is programmed so what ever HHH guesses will be wrong.
Your problem is you just don't undetstand the fundamental of the problem, logic, or truth,
| Date | Sujet | # | Auteur | |
| 21 Apr 26 |  … |
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