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From: immibis <news@immibis.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Can D simulated by H terminate normally? --- Message_ID Provided
Date: Mon, 20 May 2024 06:37:17 +0200
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On 19/05/24 15:06, olcott wrote:
> On 5/19/2024 7:16 AM, immibis wrote:
>> On 19/05/24 05:59, olcott wrote:
>>> On 5/18/2024 6:38 PM, Richard Damon wrote:
>>>> On 5/18/24 7:24 PM, olcott wrote:
>>>>> On 5/18/2024 6:06 PM, Richard Damon wrote:
>>>>>> On 5/18/24 6:44 PM, olcott wrote:
>>>>>>> On 5/18/2024 3:02 PM, Richard Damon wrote:
>>>>>>>> On 5/18/24 3:57 PM, olcott wrote:
>>>>>>>>> On 5/1/2024 7:10 PM, Richard Damon wrote:
>>>>>>>>>> The second method uses the fact that you have not restricted 
>>>>>>>>>> what H is allowed to do, and thus H can remember that it is 
>>>>>>>>>> simulating, and if a call to H shows that it is currently 
>>>>>>>>>> doing a simulation, just immediately return 0. 
>>>>>>>>>
>>>>>>>>> Nice try but this has no effect on any D correctly simulated by H.
>>>>>>>>> When the directly executed H aborts its simulation it only returns
>>>>>>>>> to whatever directly executed it.
>>>>>>>>
>>>>>>>> Why? My H does correctly simulate the D it was given.
>>>>>>>>
>>>>>>>> You don't seem to understand how the C code actually works.
>>>>>>>>
>>>>>>>>>
>>>>>>>>> If the directly executed outermost H does not abort then none of
>>>>>>>>> the inner simulated ones abort because they are the exact same 
>>>>>>>>> code.
>>>>>>>>> When the directly executed outermost H does abort it can only 
>>>>>>>>> return
>>>>>>>>> to its own caller.
>>>>>>>>
>>>>>>>> WHAT inner simulatioin?
>>>>>>>>
>>>>>>>>
>>>>>>>> My H begins as:
>>>>>>>>
>>>>>>>> int H(ptr x, ptr y) {
>>>>>>>>    static int flag = 0;
>>>>>>>>    if(flag) return 0;
>>>>>>>>    flag = 1;
>>>>>>>>
>>>>>>>> followed by essentially your code for H, except that you need to 
>>>>>>>> disable the hack that doesn't simulate the call to H, but just 
>>>>>>>> let it continue into H where it will immediately return to D and 
>>>>>>>> D will then return.
>>>>>>>>
>>>>>>>>
>>>>>>>> Thus, your claim is shown to be wrong.
>>>>>>>>
>>>>>>>
>>>>>>> We are talking about every element of an infinite set where
>>>>>>> H correctly simulates 1 to ∞ steps of D thus including 0 to ∞
>>>>>>> recursive simulations of H simulating itself simulating D.
>>>>>>>
>>>>>>> *At whatever point the directly executed H(D,D) stops simulating*
>>>>>>> *its input it cannot possibly return to any simulated input*
>>>>>>
>>>>>> And my H never stops simulating, so that doesn't apply. It will 
>>>>>> reach the final state.
>>>>>
>>>>> *Show the error in my execution trace that I empirically*
>>>>> *proved has no error by H correctly simulating D to the*
>>>>> *point where H correctly simulates itself simulating D*
>>>>> (Fully operational empirically code proved this)
>>>>
>>>> See below:
>>>>
>>>>
>>>>>
>>>>> typedef int (*ptr)();  // ptr is pointer to int function
>>>>> 00 int H(ptr x, ptr y);
>>>>> 01 int D(ptr x)
>>>>> 02 {
>>>>> 03   int Halt_Status = H(x, x);
>>>>> 04   if (Halt_Status)
>>>>> 05     HERE: goto HERE;
>>>>> 06   return Halt_Status;
>>>>> 07 }
>>>>> 08
>>>>> 09 int main()
>>>>> 10 {
>>>>> 11   H(D,D);
>>>>> 12   return 0;
>>>>> 13 }
>>>>
>>>> For Reference
>>>>
>>>> 14 int H(ptr x, ptr y)
>>>> 15 {
>>>> 16   static int flag = 0
>>>> 17   if (flag)
>>>> 18      return 0
>>>> 19   ... continuation of H that simulates its input
>>>>
>>>>>
>>>>> In the above case a simulator is an x86 emulator that correctly
>>>>> emulates at least one of the x86 instructions of D in the order
>>>>> specified by the x86 instructions of D.
>>>>>
>>>>> This may include correctly emulating the x86 instructions of H
>>>>> in the order specified by the x86 instructions of H thus calling
>>>>> H(D,D) in recursive simulation.
>>>>>
>>>>> Execution Trace
>>>>> Line 11: main() invokes H(D,D);
>>>>>
>>>>> keeps repeating (unless aborted)
>>>>> Line 01
>>>>> Line 02
>>>>> Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)
>>>>
>>>> Line 03: Calls H (line 14)
>>>> Line 16: Static already inited, so not changed.
>>>> Line 17: Flag is 1, so
>>>> Line 18: Return 0
>>>> Line 03: Set Halt_Status to 0
>>>> Line 04: if (Halt_Status)      halts status is 0, so skip
>>>> Line 06: return Halt_Status
>>>>
>>>> Simulation completed, program halted.
>>>>
>>>>
>>>>>
>>>>> Simulation invariant:
>>>>> D correctly simulated by H cannot possibly reach past its own line 03.
>>>>>
>>>>>
>>>>
>>>> Nope. Not for this H
>>>>
>>>>
>>>
>>> (a) That idea might work yet you did not say it correctly.
>>> For example line 11 is the first one invoked.
>>> (b) Computable functions cannot alter their behavior this way.
>>>
>>> (1) the function return values are identical for identical arguments (no
>>> variation with local static variables, non-local variables, mutable
>>> reference arguments or input streams, i.e., referential 
>>> transparency), and
>>
>> Your function H works like Richard's function H. You just called the 
>> variable "execution trace" instead of "flag".
> 
> pages 4-5 (of a paper that I published 2021-09-26 09:39 AM)
> Show H simulating P and H simulating itself simulating P.
> 
> The 395 pages of the execution trace of the simulated H are
> screened out. No one here could ever understand the half page
> trace so embedding that in 395 more pages would not help.

The fact that you took 395 pages to get to "if(flag) return 0;" does not 
mean that you didn't use "if(flag) return 0;"

> 
> That P is simulated correctly is proven by the fact that the
> x86 assembly language instructions of P are correctly simulated
> and they are simulated in the order that the assembly language
> of P specifies.

This is a lie -> If every assembly instruction that was simulated is 
simulated correctly, the program was simulated correctly. <- This is a lie.