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

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.

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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.

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Why? My H does correctly simulate the D it was given.
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You don't seem to understand how the C code actually works.
>

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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.

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WHAT inner simulatioin?
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My H begins as:
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int H(ptr x, ptr y) {
   static int flag = 0;
   if(flag) return 0;
   flag = 1;
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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.
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Thus, your claim is shown to be wrong.
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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.
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*At whatever point the directly executed H(D,D) stops simulating*
*its input it cannot possibly return to any simulated input*

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And my H never stops simulating, so that doesn't apply. It will reach the final state.

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*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)

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See below:
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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 }

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For Reference
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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
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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.
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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.
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Execution Trace
Line 11: main() invokes H(D,D);
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keeps repeating (unless aborted)
Line 01
Line 02
Line 03: simulated D(D) invokes simulated H(D,D) that simulates D(D)

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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
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Simulation completed, program halted.
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Simulation invariant:
D correctly simulated by H cannot possibly reach past its own line 03.
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Nope. Not for this H
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 (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".