Sujet : Re: The execution trace of HHH1(DDD) shows the divergence
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory comp.ai.philosophy sci.logicDate : 07. Jun 2025, 15:58:42
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <1021k32$3327l$9@dont-email.me>
References : 1 2 3 4
User-Agent : Mozilla Thunderbird
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.
-- Copyright 2025 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
The execution trace of HHH1(DDD) shows the divergence
Re: The execution trace of HHH1(DDD) shows the divergence