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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Can D simulated by H terminate normally? --- Message_ID Provided
Date: Sat, 18 May 2024 18:24:03 -0500
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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)

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 }

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)

Simulation invariant:
D correctly simulated by H cannot possibly reach past its own line 03.


-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer