Re: HHH(DDD)==0 is correct

On 7/4/2025 8:22 AM, joes wrote:

Am Fri, 04 Jul 2025 07:50:23 -0500 schrieb olcott:

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:

 

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

If. But here it confuses that with not being able to simulate past the
recursive call.
 

It is the correct simulation of the input that
specifies the actual behavior of this input.
If this simulation cannot simulate past the recursive
call then this correctly simulated input cannot possibly
reach its own correctly simulated final halt state.
int main()
{
   DDD();  // Calls HHH(DDD) which is not accountable
}         // for the behavior of its caller

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?

DDD halts.
 

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.

*when executed directly, which is what should be simulated. HHH can't.
 

HHH1(DDD) has the same behavior as the directly executed DDD()

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.

Uh, the traces both show a call to HHH.
 

*This one page analysis by Clause.ai sum its up nicely*
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