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

Op 10.mei.2025 om 05:13 schreef olcott:

On 5/9/2025 9:40 PM, 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).
>

 Yes and then I moved on the next tiny incremental> step of my proof. Correctly simulated less than
an infinite number of instructions does not help
the simulated DDD to reach its "return statement"
final halt state.

The word 'tiny' is misleading. It is a fundamental change. Now HHH does no longer do a correct simulation, because it ignores the input part of the input that specifies that HHH simulates only a *finite* number of steps.

 

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.

 Right this takes a glimmering of understanding of
the x86 language. The x86 language it needed to
get an exactly precise understanding of the control
flow of DDD as directed graph state transition
diagram.

Yes and it takes into account that the input contains x86 instruction to abort and halt.

 

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".

 No much more than this you acknowledged that
when DDD is correctly simulated by HHH that
the simulated DDD cannot possibly reach its
own "return statement" (final halt state)

But when that HHH ignores an important part of its input, it is no longer correct.

 This is very important to computer science because
non-termination is entirely measured by the impossibility
of reaching a final halt state.
  From all of the we know that when HHH(DDD) reports
on the behavior of its correct simulation of its input
that it can correctly reject this input as not
specifying a sequence of configurations that halts.

It can only report that it aborted the simulation prematurely, without analysing the full input.