Re: HHH(DDD)==0 is correct

On 7/4/2025 2:35 AM, Mikko wrote:

On 2025-07-03 12:56:42 +0000, olcott said:
 

On 7/3/2025 3:57 AM, Mikko wrote:

On 2025-07-03 02:50:40 +0000, olcott said:
>

On 7/1/2025 11:37 AM, Mr Flibble wrote:

On Mon, 30 Jun 2025 21:12:48 -0400, Richard Damon wrote:
>

On 6/30/25 2:30 PM, Mr Flibble wrote:
>
PO just works off the lie that a correct simulation of the input is
different than the direct execution, even though he can't show the
instruction actually correctly simulated where they differ, and thus
proves he is lying.
>
The closest he comes is claiming that the simulation of the "Call HHH"
must be different when simulated then when executed, as for "some
reason" it must be just because otherwise HHH can't do the simulation.
>
Sorry, not being able to do something doesn't mean you get to redefine
it,
>
You ar4e just showing you are as stupid as he is.

>
No. A simulator does not have to run a simulation to completion if it can
determine that the input, A PROGRAM, never halts.
>
/Flibble

>
The most direct way to analyze this is that
HHH(DDD)==0 and HHH1(DDD)==1 are both correct
because DDD calls HHH(DDD) in recursive simulation and
DDD does not call HHH1(DDD) in recursive simulation.

>
Either "no" (encoded as 0) or "yes" (encoded as any other number) is the
wrong asnwer to the quesstion "does DDD specify a halting computation?".

>
That is *not* the actual question.

 THe actual question is whatever someone asks.

What is the area of a square circle with a radius of 2?

However, if the question is
not "does DDD specify a halting computation?" or the same about some
other computation then it is not in the scope of the halting problem
or the termination problem.
 

The halting problem has always been flatly incorrect
by making that the question. So I am reframing the
question the same way that ZFC reframed Russell's Paradox.
HHH computes the mapping from its actual inputs to the
behavior specifies by these inputs by correctly simulating
them.

HHH(DDD) is asked: Does your input specify a computation that halts?
DDD correctly simulated by HHH cannot possibly reach its own "return"
statement final halt state, so NO.

 

int sum(int x, int y) { return x + y; }
sum(3,4) derives 7 only from its actual inputs.
When we make a rule that sum(3,4) must derive
the sum of 5 + 6, then this requirement is wrong.
int main()
{
   DD(); // calls HHH(DD)
}
Likewise for HHH(DD) to report on the behavior of its caller.

THat is the same question if the input specifies the computation as
DDD. If it does not then HHH(DDD) is irrelevant and either the user's
manual of HHH species another input for the purpose or HHH is not
relevant to the halting problem.
 

HHH1(DDD) is asked: Does your input specify a computation that halts?
DDD correctly simulated by HHH1 reaches its own "return" statement final halt state, so YES.

 The user's manual of HHH1 apparently dpecifies different encoding rules.
 

The full execution trace of the input to HHH1(DDD) is
different than The full execution trace of the input to HHH(DDD)
because DDD calls HHH in recursive simulation and does not
call HHH1 in recursive simulation.
Claude.ai figured this out on its own
proving that it is smarter than ChatGPT
https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer