Re: The execution trace of HHH1(DDD) shows the divergence

On 6/7/2025 10:58 AM, olcott wrote:

On 6/7/2025 9:56 AM, dbush wrote:

On 6/7/2025 10:54 AM, olcott wrote:

On 6/7/2025 9:51 AM, dbush wrote:

On 6/7/2025 10:32 AM, olcott wrote:

The execution trace of HHH1(DDD) shows the divergence
of DDD emulated by HHH from DDD emulated by HHH1.
>
int main()
{
   HHH1(DDD);
}
>
Shows that DDD emulated by HHH and DDD emulated by
HHH1 diverges as soon as HHH begins emulating itself
emulating DDD.
>
*From the execution trace of HHH1(DDD) shown below*
DDD emulated by HHH1              DDD emulated by HHH
[00002183] push ebp               [00002183] push ebp
[00002184] mov ebp,esp            [00002184] mov ebp,esp
[00002186] push 00002183 ; DDD    [00002186] push 00002183 ; DDD
[0000218b] call 000015c3 ; HHH    [0000218b] call 000015c3 ; HHH
*HHH1 emulates DDD once then HHH emulates DDD once, these match*
>
The next instruction of DDD that HHH emulates is at
the machine address of 00002183.
>
The next instruction of DDD that HHH1 emulates is at
the machine address of 00002190.

>
False.
>
The next instruction of DDD that both HHH and HHH1 emulates is at the machine address of 000015c3,

>
*That is not an instruction of DDD*
*That is not an instruction of DDD*
*That is not an instruction of DDD*
*That is not an instruction of DDD*

>
In other words, you're not operating on algorithms.

>
In other words you are not actually paying any attention.
>

 I'm very much paying to attention to the fact that you stated that the code of the function H is not part of the input and that you're therefore not working on the halting problem.