Re: Can D simulated by H terminate normally?

On 4/29/2024 7:22 AM, Mikko wrote:

On 2024-04-28 12:56:36 +0000, olcott said:
 

On 4/28/2024 3:23 AM, Mikko wrote:

On 2024-04-28 00:17:48 +0000, olcott said:
>

Can D simulated by H terminate normally?

>
One should not that "D simulated by H" is not the same as
"simulation of D by H". The message below seems to be more
about the latter than the former. In any case, it is more
about the properties of H than about the properties of D.

>
D specifies what is essentially infinite recursion to H.

 The recursion is infinite only if H calls D in infinite recursion.
If H actually calls D in infinite recursion it fails to complete
in finite time and therefore is not a halt decider.
 

A simulating halt decider is defined to abort its simulation and report
non-halting when-so-ever D simulated by H would otherwise never stop
running. H(D,D)==0 matches this requirement.

Several people agreed that D simulated by H cannot possibly
reach past its own line 03 no matter what H does.

 They probably  mean that H cannot continue the simulation of D
past line 03 of D. Whether that as any significance is another
question.
 

They mean that D specifies non-halting behavior to H.
00 int H(ptr x, ptr x)  // ptr is pointer to int function
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 void main()
10 {
11   H(D,D);
12 }
Execution Trace
Line 11: main() invokes H(D,D);
keeps repeating (unless aborted)
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