Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)

On Fri, 2025-05-09 at 23:18 -0500, olcott wrote:

On 5/9/2025 10:43 PM, wij wrote:

On Fri, 2025-05-09 at 22:24 -0500, olcott wrote:

On 5/9/2025 10:13 PM, wij wrote:

On Fri, 2025-05-09 at 19:40 -0700, Keith Thompson wrote:

olcott <polcott333@gmail.com> writes:

On 5/9/2025 4:40 PM, Richard Heathfield wrote:

On 09/05/2025 21:15, olcott wrote:

On 5/9/2025 3:07 PM, Richard Heathfield wrote:

On 09/05/2025 20:46, olcott wrote:

We have not begun to get into any of those points.
We are only asking can DDD correctly simulated
by any HHH that can exist ever reach its own
"return" instruction.

 
DDD can't be correctly simulated by itself (which is effectively
what you're trying to do when you fire up the simulation from
inside DDD).

 
How the Hell did you twist my words to say that?

I haven't touched your words. What I have done is to observe that
DDD's /only/ action is to call a simulator. Since DDD isn't itself a
simulator, there is nothing to simulate except a call to a
simulator.
It's recursion without a base case - a rookie error.
HHH cannot successfully complete its task, because it never regains
control after the first recursion. To return, it must abort the
simulation, which means the simulation fails.
 

void DDD()
{
     HHH(DDD);
     return;
}
 
When 1 or more statements of DDD are correctly
simulated by HHH then this correctly simulated
DDD cannot possibly reach its own “return statement”.

On what grounds can you persuade an extraordinarily sceptical
readership that HHH 'correctly simulated' DDD?

 
Any competent C programmer can see that
the call from DDD to HHH(DDD) (its own simulator)
is equivalent to infinite recursion.
 
On 5/8/2025 8:30 PM, Keith Thompson wrote:

Assuming that HHH(DDD) "correctly simulates" DDD, and assuming it
does nothing else, your code would be equivalent to this:
 
       void DDD(void) {
           DDD();
           return;
       }
 
Then the return statement (which is unnecessary anyway) will never be
reached.  In practice, the program will likely crash due to a stack
overflow, unless the compiler implements tail-call optimization, in
which case the program might just run forever -- which also means the
unnecessary return statement will never be reached.

 
I had not intended to post again, but I feel the need to make
a clarification.
 
I acknowledged that the return statement would never be reached
*given the assumption* that HHH correctly simulates DDD.  Given
that assumption, a call to DDD() should be equivalent to a call
to HHH(DDD).
 
I did not address whether the assumption is valid.  I merely
temporarily accepted it for the sake of discussion, just as I would
accept that if I were ten feet tall I would bump my head against
the ceiling in my house.
 
The discussion I had with olcott did not reach the point of
discussing *how* HHH could correctly simulate DDD, or whether it
would even be logically possible for it to do so.  I also did not
address any issues of partial simulation, where olcott claims that
HHH can "accurately simulate" only a few x86 instructions rather
than simulating its entire execution.  I did not participate in
any discussion that would require knowledge of x86 machine or
assembly code.  (I have no doubt that I could learn x86 machine
and assembly code reasonably well if motivated to do so, but I am
not so motivated.)
 
What I acknowledged was barely more than "if HHH correctly simulates
DDD, then HHH correctly simulates DDD".  (My understanding from
posts by others, whom I presume to be sufficiently knowledgeable,
is that HHH logically cannot accurately simulate DDD.)  I would
prefer that olcott refrain from using my words to support any of
his arguments beyond the scope of what he and I directly discussed.

 
Don't know why you people stick on the 'simulation' stuff so long.
The HP simply asks for such an H (in function form. POOH does not
resemble TM):
   H(D)=1 if D() halt.
   H(D)=0 if D() not halt.

 
My invention of a simulating termination
analyzer shows exactly how to compute the
mapping that the input that HHH(DD) specifies
into a correct answer for the halting problem's
otherwise impossible input.
 
All rebuttals are based on failing to compute
this mapping correctly.
 

 
What is the correct mapping?
 

 
_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]
 
Computing the mapping of DDD emulated by HHH
according the the rules of the x86 language
to its behavior by HHH actually emulating DDD.

The above says you have no idea what the mapping is.

If POOH are not talking about the mapping:
 
   H(D)=1 if D() halt.
   H(D)=0 if D() not halt.
 

 
The way that simulating termination analyzers process
their input by showing all of the steps of how the mapping
must be computed refutes the above simplistic view.

No (real) problem with that. But the HP asks:

H(D)=1 if D() halt.
H(D)=0 if D() not halt.

You still evade the question: Is POO H anything to do with the HP?

POOH is likely nothing to do with HP