Re: 195 page execution trace of DDD correctly simulated by HH0 ---Boilerplate Reply

On 6/22/24 2:53 PM, olcott wrote:

On 6/22/2024 1:50 PM, Fred. Zwarts wrote:

Op 22.jun.2024 om 15:11 schreef olcott:

On 6/22/2024 4:27 AM, Fred. Zwarts wrote:

Op 21.jun.2024 om 15:01 schreef olcott:

On 6/21/2024 2:44 AM, Fred. Zwarts wrote:

Op 20.jun.2024 om 16:12 schreef olcott:

On 6/20/2024 3:09 AM, Fred. Zwarts wrote:

Op 20.jun.2024 om 02:00 schreef olcott:

This shows all of the steps of HH0 simulating DDD
calling a simulated HH0 simulating DDD
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https://liarparadox.org/HH0_(DDD)_Full_Trace.pdf
*Some of the key instructions are color coded*
GREEN---DebugStep Address
RED-----HH Address
YELLOW--All of the DDD instructions
CYAN----Return from DebugStep to Decide_Halting_HH
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_DDD()
[000020a2] 55         push ebp      ; housekeeping
[000020a3] 8bec       mov ebp,esp   ; housekeeping
[000020a5] 68a2200000 push 000020a2 ; push DDD
[000020aa] e8f3f9ffff call 00001aa2 ; call H0
[000020af] 83c404     add esp,+04   ; housekeeping
[000020b2] 5d         pop ebp       ; housekeeping
[000020b3] c3         ret           ; never gets here
Size in bytes:(0018) [000020b3]
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Exactly which step of DDD emulated by H0 was emulated
incorrectly such that this emulation would be complete?
AKA DDD emulated by H0 reaches machine address [000020b3]
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If the simulation of a program with a loop of 5 iterations is aborted after 3 iterations, all instructions are correctly simulated. Nevertheless, it is an incorrect simulation, because it should simulate up to the final state of the program.
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It would be helpful if you answer the actual question being asked
right here and thus not answer some other question that was asked
somewhere else.

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If you do not understand that I answered the question why the simulation is incorrect, it is hopeless. The question which instruction is incorrect is not the right question.
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If you say that something is incorrect and can't be specific
then your rebuttal is pure bluster with no actual basis.
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If ..., but that condition is not present, so the 'then' does not apply.
This makes the sentence completely superfluous. I would expect better from someone who claims to be an experienced programmer.
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But since I pointed out in a very detailed way, why it is incorrect, your reply shows that you do not understand where you are talking about, which then becomes utterly nonsense.
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The question which instruction is incorrectly simulated already shows your error. The error is not that an instruction is simulated incorrectly, but that some instruction are not simulated at all.
Why is that already over your head?
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It is a verified fact that the behavior that finite string DDD presents
to HH0 is that when DDD correctly simulated by HH0 calls HH0(DDD) that
this call DOES NOT RETURN.
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It is a verified fact that the behavior that finite string DDD presents
to HH1 is that when DDD correctly simulated by HH0 calls HH1(DDD) that
this call DOES RETURN.
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I don't get why people here insist on lying about verified facts.
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We know that 'verified fact' for you means 'my wish'.

 Ignoramus?
 When we stipulate that the only measure of a correct emulation is the semantics of the x86 programming language then we see that when DDD is correctly emulated by H0 that its call to H0(DDD) cannot possibly return.
 _DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call H0(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
 When we define H1 as identical to H0 except that DDD does not call H1 then we see that when DDD is correctly emulated by H1 that its call to H0(DDD) does return. This is the same behavior as the directly executed DDD().
 

By a strict interpreation of your measure, this input has UNDEFIINED BEHAVIOR, so it is improper to ask about it.
The problem is the input DDD doesn't have all its x86 code defined, as it call a function not defined in its code.
When we include the contents of the all the memory, so we can correctly simulate the input, then the two "Correct Simulations" of H0 and H1 will be exactly IDENTICAL to each other up to the point that H0 stops simulatng, so all H0 can say is that it doesn't know the halting behavior of its input.
And you still need to properly define POOPing to know if it is correct for that, it might be, but we need the definitions,