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From: immibis <news@immibis.com>
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Subject: Re: Can D simulated by H terminate normally? --- Message_ID Provided
Date: Sun, 19 May 2024 14:16:17 +0200
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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".