Re: DDD correctly emulated by H0 -- Ben agrees that Sipser approved criteria is met

On 6/27/2024 6:34 AM, Richard Damon wrote:

On 6/26/24 11:34 PM, olcott wrote:

On 6/26/2024 10:16 PM, Richard Damon wrote:

On 6/26/24 10:51 PM, olcott wrote:

>
Is disabled. It is commented out.
It was only ever used so that humans could see the depth.

>
But, if it can measure the fact that this is the top level decider, that means that it sees something that it can't know.
>

>
The top level decider simply reaches its infinite
recursion behavior pattern first. It need not know
that it is first.
>

 But it if abortts, then the pattern ISN'T infinite recursion,

void Infinite_Recursion()
{
   Infinite_Recursion();
}
It is the exact same code that recognizes this as non-halting
SO IT IS THE INFINITE RECURSION BEHAVIOR PATTERN AS A MATTER OF FACT.

 as a correct emulation of the code it was emulating will have finite behaivior.
 

void Infinite_Loop()
{
   HERE: goto HERE;
}
Only in the same way and for the same reason that Infinite_Loop()
and Infinite_Recursion() have finite behavior. H0 stops simulating
them because it correctly determines that they would not otherwise
stop.

You can't have an infinite level of recursion in a finite number of steps.
 Your problem is your emulator doesn't look at the program it is actually given, but thinks of it as something different.
 

Liar.

Remember, either it can't look past the call instruction as there is nothing there to look at, or the code after the call instruction is part of the input, and thus you can't think about changing it and still having the same input.
 

Unless the outermost H0(DDD) aborts none of them do. The
outermost one reaches its abort criteria one recursive
emulation before the next one.

 

EVERY level of dicider should think that it is, or at least could be, the top level, as it can't know any differently.
>

That Is how they work.

 Then why isn't that what your traces show?
 

The traces do show that, they are simply above your degree
of technical competence.

A why does the comment ask about at the global top, since any one decider doesn't know where it
 

SO THAT HUMANS CAN SEE THIS AS I HAVE ALREADY TOLD YOU MANY TIMES
SO THAT HUMANS CAN SEE THIS AS I HAVE ALREADY TOLD YOU MANY TIMES
SO THAT HUMANS CAN SEE THIS AS I HAVE ALREADY TOLD YOU MANY TIMES
SO THAT HUMANS CAN SEE THIS AS I HAVE ALREADY TOLD YOU MANY TIMES
IT IS FREAKING DISABLED AS I HAVE TOLD YOU MANY TIMES
IT IS FREAKING DISABLED AS I HAVE TOLD YOU MANY TIMES
IT IS FREAKING DISABLED AS I HAVE TOLD YOU MANY TIMES
IT IS FREAKING DISABLED AS I HAVE TOLD YOU MANY TIMES

>

>

a decider shouldn't be

able to know that it isn't the top level decider.

>
This doesn't have any effect on its computation thus irrelevant.
>

>
It does if it knows that it isn't being simulated, which is knowledge that no simulated machine is allowed to have, as that means the simulation isn't correct. BY DEFINITION.

>
It only knows this to decide whether to call Allocate
or not. It never uses this for anything else.
>

 Why does that make a difference? Each level needs to allocate the buffer in its own memory space for it to use.
 If they share a buffer, that is improper state sharing.

That is the way that ALL UTMs work knucklehead.
UTMs cannot possibly operate in any other way.
Message-ID: <rLmcnQQ3-N_tvH_4nZ2dnZfqnPGdnZ2d@brightview.co.uk>
On 3/1/2024 12:41 PM, Mike Terry wrote:
 >
 > Obviously a simulator has access to the
 > internal state (tape contents etc.) of the
 > simulated machine. No problem there.
 >
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer