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

On 5/20/2024 4:01 AM, immibis wrote:

On 19/05/24 15:14, 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.

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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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>

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

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Nope. Not for this H
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>

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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.
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(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

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Your function H works like Richard's function H. You just called the variable "execution trace" instead of "flag".

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Since Richard did not respond to this post I am taking that
as he understands that he is incorrect.
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He has known this whole time that we having only been talking
about computable functions. Thus he knows his example using
static data is no good.
>

But the function H that you wrote also uses static data.

I am talking about a hypothetical function that does not
use static data. My current H does not use static data.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer