Sujet : Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)
De : wyniijj5 (at) *nospam* gmail.com (wij)
Groupes : comp.theoryDate : 10. May 2025, 07:00:05
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <84b09a0d53d77e2a8fddf567226d05c0d65e60c0.camel@gmail.com>
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User-Agent : Evolution 3.54.3 (3.54.3-1.fc41)
On Sat, 2025-05-10 at 00:41 -0500, olcott wrote:
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 finite
string input to a simulating termination analyzer
POOH reads(takes) its input as a function, not 'finite string'.
Are you talking about POOH now? There is no POOH that takes
'finite string'.
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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