Re: Can D simulated by H terminate normally? --- Message_ID Provided

On 5/20/2024 6:24 AM, Richard Damon wrote:

On 5/20/24 1:17 AM, olcott wrote:

On 5/19/2024 11:37 PM, immibis wrote:

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.
>

>
*Quoted from page 4 of the paper linked below*
// Simplified Linz Ĥ (Linz:1990:319)
// Strachey(1965) CPL translated to C
void P(u32 x)
{
   if (H(x, x))HERE:
    goto HERE;
}
>
int main()
{
   Output("Input_Halts = ", H((u32)P, (u32)P));
}
>
That P is correctly simulated by H is proven by the fact that
every assembly language instruction of P is correctly simulated
by H in the order specified by the x86 assembly language of P
even when H correctly simulates itself simulating P.

 Which just means that you are admitting that you are using a non-standard definition of "Correct Simulation" from Computation theory, and thus can not link your simulations to non-halting.
 

*It is not at all unsound and has an*
*an equivalent in computation theory*
That P is correctly simulated by H is proven by the fact that
every state specified in the Turing Machine Description of P
is correctly simulated by H in the order specified by the Turing
Machine Description of P.
For Linz like proofs where P includes an embedded copy of H
this includes H simulating P having this embedded copy of H
as the above specifies and may involve recursive simulations
of copies of P.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer