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

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Sujet : Re: 195 page execution trace of DDD correctly simulated by HH0 ---Boilerplate Reply
De : richard (at) *nospam* damon-family.org (Richard Damon)
Groupes : comp.theory sci.logic
Date : 22. Jun 2024, 15:38:25
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <v56k4h$onl4$4@i2pn2.org>
References : 1 2 3 4 5 6 7
User-Agent : Mozilla Thunderbird
On 6/22/24 9:11 AM, olcott wrote:
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.
 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.
 I don't get why people here insist on lying about verified facts.
 
The problem is that the "behavior" that the finite string DDD presents to HH0, is DEFINED by the problem. And if that problem is the Halting Problem, that behavior is the behavior of the machine the input represents. If HH0 treats the input as having a different behavior, then HH0 just isn't a Halting Decider, but something else.
If HH0 is supposed to be a Halting decider, but uses a method that makes it see something other than that behavior, then it is just an incorrect Halting Decider, and its algorithm just creates an incorrect recreation of the property of the input it is supposed to be working on.
A bit of a side note, the actual "Input" to HH0, is a pointer to memory, and as such it passes a reference to ALL of memory considering the starting point to be that address, so your "Input" isn't actually the few bytes of DDD, but ALL of memory and a starting point. If you actually mean that the input is just those few bytes pointed to by the address, then the input is improperly formed and is NOT a proper representation of the input machine, becuase it is incomplete.
The fact you don't understand this, seems to imply you are lacking the basic knowledge to be talking about this sort of thing.

Date Sujet#  Auteur
10 Nov 24 o 

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