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

Am Tue, 03 Jun 2025 14:47:23 -0500 schrieb olcott:

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.

Yes, that is exactly the point where HHH aborts. There is no divergence
because there are no further simulated instructions that could be
compared. If there were, they were the same.

Would you say that HHH simulating DDD and HHH1 simulating
DDD1() {HHH1(DDD1)} are the same thing?

You start to call that as diverging, but that is not a divergence in
simulation, but an error in the decision to give up. There is no
divergence in the simulation up to the point when that erroneous
decision is made, proving that the simulation up to that point gives no
reason for that decision.

--
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.