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On 10/30/2024 6:35 PM, Richard Damon wrote:But, you haven't removed yourself from the topic, so the definitions still apply.On 10/30/24 8:28 AM, olcott wrote:Do I have to repeat this a few hundred times in every postOn 10/30/2024 6:19 AM, Richard Damon wrote:>On 10/29/24 9:41 PM, olcott wrote:>On 10/29/2024 8:17 PM, Richard Damon wrote:>On 10/29/24 10:41 AM, olcott wrote:>On 10/29/2024 5:50 AM, Richard Damon wrote:>On 10/28/24 11:08 PM, olcott wrote:>On 10/28/2024 9:56 PM, Richard Damon wrote:>On 10/28/24 9:09 PM, olcott wrote:>On 10/28/2024 6:56 PM, Richard Damon wrote:>>>
It is IMPOSSIBLE to emulate DDD per the x86 semantics without the code for HHH, so it needs to be part of the input.
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*You seemed to be a totally Jackass here*
You are not that stupid
You are not that ignorant
and this is not your ADD
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_DDD()
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
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At machine address 0000217a HHH emulates itself emulating
DDD without knowing that it is emulating itself.
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Then how did it convert the call HHH into an emulation of DDD again?
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When HHH (unknowingly) emulates itself emulating DDD this
emulated HHH is going to freaking emulate DDD.
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Did you think it was going to play poker?
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Which is what it would do, get stuck and fail to be a decider. It might figure out that it is emulating an emulating decider, at which point it knows that the decider might choose to abort its conditional emulation to return, so it needs to emulate further.
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Only by recognizing itself, does it have grounds to say that if I don't abort, it never will, and thus I am stuck, so I need to abort.
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Counter-factual. This algorithm has no ability to its own code.
https://github.com/plolcott/x86utm/blob/master/Halt7.c // page 801
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*That people fail to agree with this and also fail to*
*correctly point out any error seems to indicate dishonestly*
*or lack of technical competence*
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DDD emulated by HHH according to the semantics of the x86
language cannot possibly reach its own "return" instruction
whether or not any HHH ever aborts its emulation of DDD.
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No, it knows its own code because it rule for "No conditional branches" excludes that code.
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Are you really so stupid that you think this will help
DDD reach its own return instruction?
DDD doesn't need any help to reach its own return instruction, as the HHH that it calls DOES abort and return to it.
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Are you really so stupid that you think you can keep getting
away with the strawman deception by changing the subject away
from DDD emulated by HHH?
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What strawman?
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I am just going to the defintions of the problem you claim to be solving.
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so that you can remember from one post to the next?
*AT THIS POINT HHH IS NOT A HALT DECIDER OR A TERMINATION ANALYZER*
*AT THIS POINT HHH IS NOT A HALT DECIDER OR A TERMINATION ANALYZER*
*AT THIS POINT HHH IS NOT A HALT DECIDER OR A TERMINATION ANALYZER*
HHH is each element of the set of x86 emulators that emulates zeroNo, it isn't, because your published HHH is not a "set of programs".
to infinity steps of DDD including zero to infinity emulations of
itself emulating DDD.
DDD emulated by HHH according to the semantics of the x86Which means HHH can't abort its emulaiton, or it fails to meet its requrements.
language cannot possibly reach its own "return" instruction
whether or not any HHH ever aborts its emulation of DDD.
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