Re: Defining a correct simulating halt decider

On 9/9/24 4:43 PM, olcott wrote:

On 9/9/2024 3:24 PM, olcott wrote:

On 9/9/2024 3:02 PM, joes wrote:

Am Mon, 09 Sep 2024 13:19:26 -0500 schrieb olcott:

On 9/8/2024 9:53 AM, Mikko wrote:

On 2024-09-07 13:57:00 +0000, olcott said:

On 9/7/2024 3:29 AM, Mikko wrote:

On 2024-09-07 05:12:19 +0000, joes said:

Am Fri, 06 Sep 2024 06:42:48 -0500 schrieb olcott:

On 9/6/2024 6:19 AM, Mikko wrote:

On 2024-09-05 13:24:20 +0000, olcott said:

On 9/5/2024 2:34 AM, Mikko wrote:

On 2024-09-03 13:00:50 +0000, olcott said:

On 9/3/2024 5:25 AM, Mikko wrote:

On 2024-09-02 16:38:03 +0000, olcott said:

>

A halt decider needn't compute the full behaviour, only
whether that behaviour is finite or infinite.
>

Local Halt Decider: Infinite Recursion Detected Simulation
Stopped
Hence  HHH(DDD)==0 is correct

>
Nice to see that you don't disagree with what said.
Unvortunately I can't agree with what you say.
HHH terminates, so DDD obviously terminates, too.

>
DDD emulated by HHH never reaches it final halt state.

>
If that iis true it means that HHH called by DDD does not return
and therefore is not a ceicder.

The directly executed HHH is a decider.

What does simulating it change about that?

If the simulation is incorrect it may change anything.
>

PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR

However, a correct simultation faithfully imitates the original
behaviour.
>

A correct emulation obeys the x86 machine code even if this machine code
catches the machine on fire.

I don’t see an HCF instruction above.
>

It is impossible for an emulation of DDD by HHH to reach machine address
00002183 AND YOU KNOW IT!!!

I know that HHH1 does it.
>

>
Right [and] Bill is guilty of robbing the liquor store because you
saw his identical twin brother Harry rob the store and you knew
that it was Harry that you saw rob the store and not Bill.
>

 HHH1 and HHH are identical code yet have different
behavior in input DDD BECAUSE IT IS MORONICALLY
STUPID TO JUST FREAKING PRETEND THAT THE FACT THAT
DDD CALLS HHH AND DOES NOT CALL HHH1 DOES NOT MAKE
ANY FREAKING DIFFERENCE.
 

No, it is moronically stupid to beleive that truely identical programs given the same input would behave differently.
This just PROVES that you are the LYING MORON.
Since the input to both HHH(DDD) and HHH1(DDD) is the same, its behavior is the same, because it is the same program.
Only a MORON thinks a program can change its behavior based on what emulator correctly emulates it.
If the results differ, then one of the emulation must have been incorrect.
In fact, if we look at an actually correct partial emulation done by HHH, it will be EXACTLY equal to the same number of steps emulated by HHH1, and thus your claim that the will bedifferent is proven to be a LIE.
Your failure, and refusal, to show where that is wrong by showing the first instruction that is ACTUALLY EMULATED that gives a different result, just proves you are wrong.
Note, saying that HHH has "correctly emulated" the call HHH by starting a new emulation is just incorrect, as that isn't the definition of the call HHH instruction, but that is the root of your claim, which is just the lie you are based on, because you don't understand what CORRECT means.