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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: DDD correctly emulated by H0 -- Ben agrees that Sipser approved
 criteria is met
Date: Thu, 27 Jun 2024 07:34:14 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <v5jinm$19368$2@i2pn2.org>
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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, as a 
correct emulation of the code it was emulating will have finite behaivior.

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.

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.


>> 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?

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

> 
>>>
>>>>> 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.