Re: HHH(DDD)==0 is correct

On 7/5/2025 2:36 AM, Fred. Zwarts wrote:

Op 05.jul.2025 om 00:17 schreef olcott:

On 7/4/2025 3:33 PM, Richard Damon wrote:

On 7/4/25 2:24 PM, olcott wrote:

On 7/4/2025 12:48 PM, Richard Damon wrote:

On 7/4/25 9:32 AM, olcott wrote:

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.

>
Right, and that means correctly simulating EVERY instruction that makes up the PROGRAM, which must include the code of *THE* HHH, or you can't "correctly simulate" the call instruction.
>
SIn
>

>
If this simulation cannot simulate past the recursive
call then this correctly simulated input cannot possibly
reach its own correctly simulated final halt state.
>

>
But it can, as the call goes into HHH, and you then just simulate the code of HHH.
>

>
Do you mean like this? (as I have been telling you for three years)
https://liarparadox.org/HHH(DDD)_Full_Trace.pdf

>
Yes, so where in that trace does HHH's "correct simulation" differ from the exact same simulation generated by HHH1 before HHH aborts?
>
>

>
When we compare DDD emulated by HHH and DDD emulated
by HHH1 SIDE-BY-SIDE. (Mike didn't do it this way).
>
*The difference is when*
HHH begins to simulate itself simulating DDD and
HHH1 NEVER begins to simulate itself simulating DDD.

 No its does the same thing as HHH does: it simulates HHH.

*SIDE-BY-SIDE not top down*
*SIDE-BY-SIDE not top down*
*SIDE-BY-SIDE not top down*
HHH(DDD) Simulates itself simulating DDD.
HHH1(DDD) NEVER simulates itself simulating DDD.
As soon as HHH begins simulating itself simulating
DDD the execution trace of HHH1(DDD) and HHH(DDD)
diverge. HHH aborts its simulation many hundreds
of steps later.

This means that there is no difference in the trace. The are simulating the same input.
So, since the are simulating the same input, the result must be the same if the simulations are correct. If the simulations differ, at least one of the must be incorrect.
 

--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
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