Re: DDD emulated by HHH diverges from DDD emulated by HHH1

Am Wed, 04 Jun 2025 11:38:10 -0500 schrieb olcott:

On 6/4/2025 4:20 AM, Fred. Zwarts wrote:

Op 04.jun.2025 om 04:37 schreef olcott:

On 6/3/2025 9:13 PM, Richard Damon wrote:

On 6/3/25 3:47 PM, olcott wrote:

On 6/3/2025 3:28 AM, Fred. Zwarts wrote:

Op 02.jun.2025 om 17:52 schreef olcott:

DDD correctly emulated by HHH diverges from DDD correctly emulated
by HHH1 as soon as HHH begins emulating itself emulating DDD,
marked below.
>
*HHH1 never emulates itself emulating DDD*

*This is the beginning of the divergence of the behavior*
*of DDD emulated by HHH versus DDD emulated by HHH1*

Misleading words when you change the meaning of diverging.
Mike showed the traces side by side. Even after many requests, you
still cannot show the first instruction that is interpreted
differently by HHH and HHH1. The only difference is that HHH gives
up the simulation too early.

>
As soon as HHH begins emulating itself and HHH1 NEVER begins
emulating itself THIS IS THE DIVERGENCE.
>

So, WHAT INSTRUCTION, correctly simulated did that.
>

HHH emulates DDD two times and HHH1 emulates DDD one time the whole
second time is the divergence.
>

That did not answer the question: WHAT INSTRUCTION, correctly simulated
did that?

 
When HHH1(DDD) simulates DDD it never simulates itself. When HHH(DDD)
simulates DDD then simulates itself simulating DDD the first instruction
that this simulated HHH simulates diverges from the simulation that HHH1
did.

What? No. There is simply no instruction to compare to, because HHH
doesn't simulate that far.

You cannot point to any instruction interpreted differently by the two
simulators.

There are no instructions interpreted differently. It is the fact that
HHH(DDD) emulates DDD twice and HHH1(DDD) only emulates DDD once that is
the key difference.

Yes, there is no divergence. You couldn't tell the simulator just from
the trace. HHH just aborts the simulation where HHH1 continues it. You
could have said this way earlier. Disambiguation is not your strong suit.

It seems you understand that the trace does not show a difference, but
that the divergence is only in your interpretation of the trace.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.