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

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.
 

My original H(P,P) does simulate itself simulating P.
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.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer