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On Sat, 2025-05-10 at 00:41 -0500, olcott wrote:It <is> a finite string of x86 bytes.On 5/10/2025 12:27 AM, wij wrote:On Sat, 2025-05-10 at 00:19 -0500, olcott wrote:>On 5/10/2025 12:13 AM, wij wrote:>On Sat, 2025-05-10 at 00:06 -0500, olcott wrote:>>>When mathematical mapping is properly understood>
it will be known that functions computed by models
of computation must transform their input into
outputs according to the specific steps of an
algorithm.
>
_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]
>
For example HHH(DDD) only correctly map to the
behavior that its input actually specifies by correctly
emulating DDD according to the rules of the x86 language.
>
This causes the first four instructions of DDD
to be emulated followed by HHH emulating itself
emulating the first three instructions of DDD.
>
It is right at this recursive simulation just
before HHH(DDD) is called again that HHH recognizes
the repeating pattern and rejects DDD.
Yes, but you still did not answer the question: Is POOH exactly about HP?
>
>>>>> H(D)=1 if D() halt.
>>>>> H(D)=0 if D() not halt.
>
Right now it is mostly about proving the
above requirements are is mistaken.
>
Why is the requirement invalid?
>
H(D)=1 if D() halt.
H(D)=0 if D() not halt.
>The notion that the behavior specified by the finitePOOH reads(takes) its input as a function, not 'finite string'.
string input to a simulating termination analyzer
Are you talking about POOH now? There is no POOH that takes
'finite string'.
This is a verified fact.does sometimes differ from the behavior of its direct
execution. It is a provably different sequence of steps.
So, you are talking about the behavior of the 'simulating termination analyzer'--
i.e. POOH? (not really about the HP)
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